TITLE: beyond differentially algebraic power series
QUESTION [5 upvotes]: In a recent question, we learned about the existence of functions that do not satisfy any algebraic differential equation.  
One nice property of such equations is that there is a good way to enumerate a basis: we can produce the stream of "monomials" $\left(\prod_i D^{\lambda_i-1}f(x)\right)_\lambda$, where $D$ is the differentiation operator and $\lambda=(\lambda_1,\lambda_2,\dots)$ runs over the integer partitions in lexicographic order:
$$
1, f(x), f(x)^2, f'(x), f(x)^3, f(x)f'(x), f''(x), f(x)^4, f(x)^2f'(x),\dots
$$
I'm wondering: is there a "natural" class of equations, more general than ADEs, that has a similar basis.  (Natural meaning: equations that specify many functions occurring in "nature")?  Or, alternatively, just another class of equations.
I realise this is vague, please bear with me...
edit: I should add that I'm aware of "algebraic recurrences"  (i.e., shift instead of differentiation) and "Mahler-type functional equations" (i.e.,  $f(x^{k+1})$ instead of $f(x)^{(k)}$).
Martin Klazar mentions that a few interesting sequences (eg. the ordinary generating function for Bell numbers) satisfy functional equations of the form
$$p_1(x)f(x)=p_2(x)+p_3(x)f(\frac{x}{1-x}),$$
with polynomials p1, p2, p3 (and concludes that they are not differentially algebraic), but I'm not sure how common such equations are.
edit: the motivation for this question comes from the desire of being able to guess a formula (or recurrence, differential or functional equation) for a given sequence (of numbers or polynomials, etc.), as pioneered be GFUN, see also Section 7 in my preprint with Waldek Hebisch.
For example, given the first few (say 100) terms of the sequence, we compute it's (truncated) generating function $f_1 := f(x)$, and also $f_2 := f(x)^2, f_3 := f'(x), f_4 := f(x)^3, f_5 := f(x)f'(x), f_6 := f''(x), f_7 := f(x)^4, f_8 := f(x)^2f'(x), \dots$.  We fix the maximal degree, say $N$ of the coefficient polynomials $p_1, p_2, \dots, p_m$, and then try to solve the linear system of equations obtained by equating coefficients in
$$
ord(p_1 f_1 + \dots + p_m f_m)\geq\sigma
$$
for large sigma. If we get a solution, and the given sequence is somehow naturally defined, chances are good that the equation holds for all terms of the sequence.

REPLY [5 votes]: I am not quite sure whether the question is about a "natural" graded algebra which is infinitely generated, or finitely generated algebras are fine as well. Because there are nice examples of algebras of multiple zeta values, but also of multiple polylogarithms and of finite multiple harmonic sums, as well as the algebra of classical modular forms. The latter gives rise to a certain structure which is presumably richer than the algebra of differential monomials, and I could probably try to explain this point.
The Eisenstein series $E_2=1-24\sum_{n=1}\sigma_1(n)q^n$, $E_4=1+240\sum_{n=1}\sigma_3(n)q^n$ and $E_6=1-504\sum_{n=1}\sigma_5(n)q^n$, where $\sigma_k(n)=\sum_{d\mid n}d^k$, generate a differentially stable ring over $\mathbb Q$ with respect to the differentiation $D=q\dfrac{d}{dq}$ (a result usually attributed to Ramanujan). The weights $2,4,6$ are assigned to $E_2,E_4,E_6$ respectively, and $D$ increases the weight by $2$. The graded ring $\mathbb Q[E_2,E_4,E_6]$ possesses an additional structure coming from the functional equations for replacing $q$ by $q^k$ where $k$ is a positive integer, although it's very hard to write down the structure explicitly. Let me call the corresponding scale operators (substitutions $q^k$ for $q$) $D_k$. They do not change weights.
The counterpart consists of the infinite family $F_{2m+1}(q)=\sum_{n=1}^\infty\sigma_{2m}(n)q^n$, $m=0,1,2,\dots$, which are known to be linearly independent over $\mathbb Q$ and even over the field of meromorphic functions on $\mathbb C$. We can formally assign the weight $2m+1$ to each $F_{2m+1}(q)$, although there could be reasons to normalize them in a way used for the Eisenstein series. Again, the differential operator $D$ increases weights by 2, and the open problem here is to show that the $F_{2m+1}(q)$ are all algebraically differentially independent over $\mathbb C$ (or $\mathbb Q$). An expanded version of the problem is to show that the ring of all $D$- and $D_k$-monomials have no nontrivial relations at all. In a sense this includes both the algebraic differential structure from the problem, as well as all kind of Mahler-type equations. If one restricts to considering $D$- and $D_2$-monomials (or $D_k$ monomials for a finite set of $k$'s), the corresponding set of monomials of finite weight will be finite.<|endoftext|>
TITLE: Maximum number of perfect matchings in a graph 
QUESTION [11 upvotes]: What is the maximal number of perfect matchings a graph $G(V,E)$ can have if $|V|$ and $|E|$ are fixed? I am particularly interested in a case when $|E| = c|V|^2$.

REPLY [13 votes]: I think this is exactly the main result of this recent paper we just published in Discrete Mathematics.  Just in case the link doesn't work: this is "Graphs with the maximum or minimum number of 1-factors" by D. Grossa, N. Kahl and J.T. Saccoman.   I have read only the abstract.  Let me know if this is what you were looking for.<|endoftext|>
TITLE: What are the maximal subgroups of GSp(2g,l)?
QUESTION [7 upvotes]: Is there a nice description of the maximal subgroups of $GSp_{2g}(\mathbb{F}_l)$? 
When $g = 1$ this is $GL_2(\mathbb{F}_l)$, and Serre (in his 72 Inventiones paper) classifies its maximal subgroups (Borel, Normalizer of Cartans, and a few exceptional subgroups). I expect a classification for general g to be longer, but maybe its managable when g = 2?

REPLY [5 votes]: "I expect a classification for general g to be longer, but maybe its managable when g = 2?"
The Experimental Mathematics paper of Dieulefait for g=2 uses this. He quotes Mitchell from 1914.
http://www.expmath.org/expmath/volumes/11/11.4/pp503_512.pdf
Kleidman and Liebeck is on Google Books. There is also a brief survey of it.
http://www.springerlink.com/index/X631112H21U82040.pdf<|endoftext|>
TITLE: How to prove that $w_1(E)=w_1(\det E)$?
QUESTION [5 upvotes]: How to prove that the first Stiefel-Whitney class $w_1 (E)$ of a real rank $n$ vector bundle over a manifold M is equal to $w_1(\det E)$, where $\det E$ is the $n$-th wedge power of $E$?
(I want to assume the "axiomatic" definition of Stiefel-Whitney classes, as given e.g. in the book by Milnor and Stasheff).
I have just been asked an analogous question by a younger guy, but I think I could only find a proof starting from a different definition of the $w_i$'s. Perhaps I'm just missing something? Of course, feel free to close it if you find it's to homework-ish for MO standards.

REPLY [10 votes]: This is really a long comment regarding some of the above discussion.
Hatcher (in his Vector Bundles notes) certainly proves that the characteristic classes defined using Leray-Hirsch satisfy the axioms from Milnor-Stasheff.  But $w_1$ is a more basic object: the axioms specify its value on the tautological bundle over ${\Bbb R}P^1$ (= $S^1$) and this immediately determines its values on all line bundles (see p. 81 of Hatcher's notes).  One can then see that as an element of 
$H^1(X; {\Bbb Z}/2) = Hom(H_1 X, {\Bbb Z}/2) = Hom(\pi_1 X, {\Bbb Z}/2)$, 
$w_1(L)$ simply answers the question: "Along which loops is L trivial?"  (Actually, this is true for all bundles, not just lines.)  From this point of view, multiplicativity ($w_1 (L\otimes K) = w_1(L) + w_1(K)$) is a quick exercise (hmmm... what should a homomorphism from a multiplicative group to an additive group be called?  Anyway, I just mean it's a homomorphism from the Picard group of line bundles to $H^1$.).  Alternatively, it follows from the H-space structure on ${\Bbb R}P^\infty$ defined via the map ${\Bbb R}P^\infty\times {\Bbb R}P^\infty \to {\Bbb R}P^\infty$ classifying $\gamma^1_\infty \otimes \gamma^1_\infty$.  This is spelled out in  my notes on vector bundles  (it's written in the complex case but works the same way in the real case).  I couldn't quickly see where Hatcher discusses this point.  
Incidentally, my notes also discuss the relationship between orientability and $w_1$.  I've never been crazy about discussions of this point in the literature (e.g. Hatcher states this relationship only for spaces of the homotopy type of a CW complex, although he doesn't seem to use that assumption in his proof).<|endoftext|>
TITLE: Existence of certain Hall-subgroups based on knowledge of the degrees of irreducible characters
QUESTION [5 upvotes]: I am trying to find out what results are already out there in this direction.
For example, from the Ito-Michler theorem, if $\rho(G)$ is the set of prime divisors of irreducible characters of $G$ and either $\rho(G)\subseteq \pi$ or $|\rho(G)\cap\pi|\leq 1$ then we are guaranteed the existence of a $\pi$-Hall subgroup.
Also, there is a result by J. Thompson that if some prime $p$ divides the degrees of all non-linear irreducible characters of a finite groups, then that group has a normal $p$-complement.
Are there other results like these, which give a relations between some set connected to the degrees of irreducible characters of a group and a set $\pi$ of prime numbers, such that we are guaranteed the existence of a $\pi$-Hall subgroup?

REPLY [7 votes]: Background:

(Ito–Michler) p does not divide the order of the degree of any (absolutely) irreducible (ordinary) character of G iff G has a normal, abelian, Sylow p-subgroup.

For any subset π of ρ(G)′, it follows that G has a normal, abelian, Hall π-subgroup, and hence also a Hall π′-subgroup.  Flipping the roles of π and π′ gives your first existence claim.
Chapter 14 of Isaacs's Character Theory textbook contains this and further results, which I assume are already known.  For instance, the classical:

(Blichfeldt) if G has a faithful (possibly reducible, ordinary) character of degree n, then for any π consisting of primes p > n, G has an abelian, Hall π-subgroup.

Blichfeldt's paper below contains the result that G has an abelian, normal, Hall π-subgroup for π a collection of primes p > (n-1)(2n+1).  This was refined to:

(Feit–Thompson) if G has a faithful (possibly reducible, ordinary) character of degree n, then for any π consisting of primes p > 2n+1, G has an abelian, normal, Hall π-subgroup.

The more detailed information alluded to in Isaacs's textbook is likely Winter's 1964 paper below.  It gives a complete list of groups with non-normal Sylow p-subgroups for p > (2n+1)/3.  A non-abelian but normal Sylow π-subgroup is implied by the following:

(Feit) if G has a a faithful, irreducible (ordinary) character of degree p-2 ≥3, then either the Sylow p-subgroup of G is normal, G is a direct product of SL(2,p) and an abelian group, or G is a direct product of 3.Alt(6) and abelian group.

Now each of these results required faithful characters.  If you are OK with that, then Berkovich (MR1189113) has a Thompson like hypothesis and an Ito like conclusion (normal Hall π-subgroup containing the nilpotent residual), but it looks at π(G/Z(χ)) too.
The Ito–Michler result is also valid in some sense over finite fields:

(Manz) Every irreducible, p-modular character degree is coprime to p iff G has a normal Sylow p-subgroup.
(Manz–Wolf) If G is p-solvable and every irreducible, p-modular character degree is coprime to q, then Oq′(G), the subgroup generated by the Sylow q-subgroups, is solvable of q-length at most 2 (π-length roughly measures how far away from having a normal Hall π-subgroup you are).

(Navarro, et al.) has generalized both the Ito and Thompson results to blocks in MR1810119, MR1956546, and MR2159761.
A more recent article by (Matterei) gives a reasonable survey of results.  The way he rephrases results may be quite interesting to you.  He also gives the Carter–Hawkes (MR1199667) results that continue the ideas of Thompson.  A recent paper of Dolfi et. al (MR2469367) derives Ito-like results using the orders of a non-vanishing elements g in G, such that χ(g) ≠ 0 for any any ordinary, irreducible character χ and derives Ito's and Thompson's results as corollaries.

Blichfeldt, H. F. "On the order of linear homogeneous groups."
Trans. AMS. 4 (1903), 387–397. JFM 34.0176.02 JSTOR 1986408
Feit, Walter; Thompson, John G. "Groups which have a faithful representation of degree less than (p-1)/2". Pacific J. Math. 11 (1961), 1257–1262. MR133373 euclid.pjm/1103036911
Winter, David L. "Finite groups having a faithful representation of degree less than (2p+1)/3." Amer. J. Math. 86 (1964), 608–618. MR183788 DOI: 10.2307/2373026
Feit, Walter. "On finite linear groups. II." J. Algebra 30 (1974), 496–506. MR357634 DOI: 10.1016/0021-8693(74)90220-8
Manz, Olaf. "On the modular version of Ito's theorem on character degrees for groups of odd order." Nagoya Math. J. 105 (1987), 121–128. MR881011 euclid.nmj/1118780642
Manz, Olaf; Wolf, Thomas R. "Brauer characters of q′-degree in p-solvable groups."
J. Algebra 115 (1988), no. 1, 75–91. MR937602 DOI: 10.1016/0021-8693(88)90283-9
Mattarei, Sandro. "Retrieving information about a group from its character degrees or from its class sizes." Proc. Amer. Math. Soc. 134 (2006), no. 8, 2189–2195. MR2213690 DOI: 10.1090/S0002-9939-06-08274-8<|endoftext|>
TITLE: The geometry behind the ICM 2010 Logo
QUESTION [5 upvotes]: The logo for this year's ICM show the inequality $ |\tau(n)| \leq n^{11/2} d(n)$ where $\sum \tau(n)q^n = q \prod_{n \neq 1} (1 - q^n)^{24}$ is the tau function.  Wikipedia says this bound was conjectured by Ramanujan (appropriate for a conference in Hyderabad) and proven by Deligne in '74 in the process of proving as a corollary of the Weil Conjectures (which I also don't get).  The background of the ICM logo looks like Ford circles (or sun rays).  What is the hyperbolic geometry behind the Tau Conjecture and its proof?  
EDIT: It would also be nice to see the proof of this bound, but the Weil conjectures and l-adic cohomology are a topic in themselves.

REPLY [9 votes]: One way to answer this question is as follows: Ramanujan's conjecture is a special case of
a much more general conjecture that any cuspidal automorphic representation of $GL_n$ over
a number field is tempered.  This is a technical but fundamental notion, which in the
special case of the automorphic representation of $GL_2$ attached to the $\Delta$ function,
reduces to Ramanujan's original conjecture. In fact, many people working in the theory of automorphic representations refer to this very general conjecture simply as the Ramanujan conjecture.
When applied to other cuspforms on $GL_2$ (namely Maass forms) it includes Selberg's conjecture that on congruence quotients of the upper half-plane, the spectrum of the hyperbolic Laplacian is bounded below by $1/4$.
The appearance of hyperbolic geometry can be understood in the following way:
the quotient of $SL_n(\mathbb R)$ by $SO(n)$ is a non-compact symmetric space,
which in the particular case of $SL_2$ is the hyperbolic plane.  So, while the particular appearance of hyperbolic geometry may be a bit of a red herring, the appearence of highly
symmetric geometry is a reflection of the group representation theory that is underlying the picture.
As of the current moment, no purely representation-theoretic approach to the (general form of) Ramanujan's conjecture is known.  (Or rather, a proof strategy involving what is called symmetric power functoriality is known, but the requisite results on symmetric power functoriality seem very much out of reach at the moment).  The only cases that are proved at the moment are cases when one can relate the group theoretic picture of automorphic forms to algebraic geometry (first over $\mathbb C$, then over a number field, and then ultimately over finite fields, so that the Weil conjecture apply).  This is how Deligne's proof proceeds.  This connection between the geometry of symmetric spaces and arithmetic and geometry over finite fields is one of the profound points of investigation of modern number theory, but despite many positive results related to it, it remains essentially mysterious, even to experts.<|endoftext|>
TITLE: Anything going on for a mathematician stuck at New York?
QUESTION [14 upvotes]: First of all, apologies for the really non-standard question/announcement. I know this is not what MO was intended for, but in this situation it is the easiest way to reach (perhaps) the right person.
On my way back to London from some workshop in Ohio, I got stuck in NYC because of that infamous volcano. No definite news except there is no chance I can get back to Europe before next Friday. I thought I could make something productive with all this time, but have no contact with any mathematicians in any of the Universities nearby.
So, if anybody in NYC or surroundings is interested on hearing about my work (lately mostly about the field with one element, plus some quantum groups and noncommutative geometry stuff), you have the perfect occasion to bring me to your department. For free! I am open to discussions, seminars or in general anything that is going around, if anybody is up, just send an email!

REPLY [18 votes]: Just walk into any seminar you find interesting:
http://math.columbia.edu/calendar/main/one/next_week.html
http://www.cims.nyu.edu/events/
http://www.math.poly.edu/news/seminars.phtml
http://math.gc.cuny.edu/seminars/bulletin04_13.pdf
No need to contact anyone in advance. If you need directions on how to get to any of these places, just ask and I'll post them.
ADDED: It's also very easy to do day trips by train to neighboring universities:
http://www.math.sunysb.edu/html/seminars.shtml
http://www.math.rutgers.edu/seminars/calendar.php
http://www.math.princeton.edu/seminars/
https://www.math.upenn.edu/cgi-bin/calendar/view.pl?item=NextWeek
MORE: If you want to give a talk instead of listening to one, I suggest you find a seminar where the speaker isn't going to make it from Europe and where the organizers know either you or your work. You could then contact them and offer to be a substitute.<|endoftext|>
TITLE: Are there any "homotopical spaces"?
QUESTION [40 upvotes]: This is a somewhat vague question; I don't know how "soft" it is, and even if it makes sense. 
[Edit: in the light of the comments, we can state my question in a formally precise way, that is: "Is the homotopy category of topological spaces a concrete category (in the sense, say, of Kurosh and Freyd)?". You may still want to read what follows, for a bit of motivation]
Historically, I'd bet people started to look at concrete metric spaces $(X,d)$ before exploring the utility of the abstraction given by general topological spaces $(X,\tau)$ ($\tau$ is a topology). The heuristic idea captured by the concept of a topological space is endowing a set $X$ with a "geometry" that forgets the rigidity of a hypothetical metric structure $(X,d)$, though retaining the "qualitative" aspects of the geometry given by the metric. 
Of course there are non-metrizable topological spaces, but let stick to metrizable ones for the moment. I think it should be possible to see a topological space $(X,\tau)$ as an equivalence class of metric spaces: $(X,[d])$, where $[d]$ is the class of all metrics on the set $X$ that give rise to the same topology. So, an $(X,\tau)$ just has several "rigid" models $(X,d)$, and a morphism of topological spaces $f:(X,\tau) \rightarrow (Y,\tau')$ is given by taking any map $(X,d) \rightarrow (Y,d^{'})$ of "representatives" which is continuous according to the "metric ball" definition.
[Please correct the above picture if it is imprecise or even just wrong!!]
The (perhaps naive) way I have always thought about homotopy is that it is an even "further step" in making the geometry more "qualitative" and less rigid: you can "collapse positive dimensional appendices" of a space as far as they are contractible, and so forth. When trying to make this formal, you consider "homotopy types", which are equivalence classes $[(X,\tau)]$ of topological spaces, where $(X,\tau) \sim (X',\tau')$ if there is a homotopy equivalence $\varphi:(X,\tau) \rightarrow (X',\tau')$. What are morphisms in the homotopy category? Just morphisms $f$ between "representatives", but now you have also to consider them up to homotopy, i.e. you take $[f]$ where $f \sim f'$ if there's a homotopy $\alpha: f \rightarrow f'$.
It's ugly to think of topological spaces as $[(X,d)]$ (or, rather, $(X,[d])$): it's better to use the simpler and more expressive abstraction $(X,\tau)$.
So, the question is:

Is there some kind of "homotopology" $h$ (whatever it is) that one can put on sets $S$ so that each homotopy type $[(X,\tau)]$ is fully described by a "homotopical space" (whatever it is) $(S,h)$ and homotopy classes of morphisms $[f]$ correspond to "$h$-compatible" (whatever it means) set-theoretic maps $F:S \rightarrow S'$?

(I don't even dare asking about the existence of "non-topologizable homotopical spaces" because the above question is already by far too vague!)

REPLY [59 votes]: No.<|endoftext|>
TITLE: Classification of 1-dimensional manifolds (not second-countable)
QUESTION [26 upvotes]: It is easy to see that every connected $1$-dimensional second-countable manifold (that is, what is often called just a manifold) is either homeomorphic to $\mathbb{R}$ or to $S^1$. Now let's drop the secound-countable-condition.

How do you prove that every connected $1$-dimensional manifold homeomorphic to $\mathbb{R}, S^1$, the long line or the long ray? And why are the long line and the long ray not homeomorphic?

A good survey about the latter spaces can be found in the wikipedia entry. Basically, a long ray is built up of $\omega_1$-many intervals pasted together, and the the long line consists of two long rays in both directions.

REPLY [17 votes]: Here is a response to the first boxed question (the second was already answered by Robin Chapman). (Much belated of course, but I only just saw this question.) 
Suppose that $Y$ is a connected (nonempty) topological 1-manifold without boundary; let $y$ be a point. Unless $Y$ is a circle, the complement $Y - \{y\}$ has two open connected components $U$ and $V$, and $Y$ can be reconstructed by gluing together $U \cup \{y\}$ and $V \cup \{y\}$, which are 1-manifolds with one boundary point each. 
I found it technically easier to analyze the possibilities for such connected 1-manifolds with (at least) one boundary point. Recall that a 1-manifold with boundary is a topological space where every point has a neighborhood homeomorphic to an open subset of the interval $[0, 1]$. In conjunction with the gluing above, it suffices to establish the following result. 
Theorem: Suppose $X$ is a connected 1-manifold with at least one boundary point. Then $X$ is homeomorphic to one of the following types of spaces: 

A closed interval $[0, 1]$. 
A half-open interval (homeomorphic to $\mathbb{R}_{\geq 0}$). 
A long half-open ray. 

(I should say right away that a fully rigorous proof, with all i's dotted and t's crossed, would be somewhat lengthy. So I will content myself with a proof outline. See also reference [1], which should help fill most if not all the gaps.) 
Proof: Observe that $X$ is path-connected, since it is connected and locally path-connected. 
Let $0$ denote a boundary point, and order $X$ as follows: say $x \lt y$ if $x$ and $0$ belong to the same path component of $X - \{y\}$. It is not hard to show that $X$ is linearly ordered under $\lt$, with bottom element $0$. Every interval $[0, x]$ is a compact connected manifold with two endpoints (compact because there is a path from $0$ to $x$), and thus homeomorphic to the standard interval. 
Suppose a closed subset $D \subset X$ is well-ordered under the order it inherits from $X$. The order type of such $D$ must be $\omega_1$ (the first uncountable ordinal) or less. For otherwise, there would be an initial segment $S$ of $D$ of order type $\omega_1 + 1$. In that case, if $s$ is the top element of $S$, the interval $[0, s)$, which is homeomorphic to $\mathbb{R}_{\geq 0}$, would contain $\omega_1$ as a suborder -- but this is absurd since $\mathbb{R}_{\geq 0}$ has a countable cofinal set. 
We can now classify the possibilities for $X$, according to the smallest ordinal $\xi$ which does not occur as a well-ordered closed subset of $X$. This dictates what well-ordered closed subsets $D$ that are cofinal in $X$ look like. 

If $\xi = \omega_1 + 1$, then any closed well-ordered cofinal $D$ must be of type $\omega_1$, and $X$ is a topological union (a directed colimit) of open sets $[0, d)$ where $d$ ranges over $D$. This union is homeomorphic to a long half-open ray. 
If $\xi = \omega_1$, then any closed well-ordered cofinal $D$ is countable. This forces $X$ to be homeomorphic to $\mathbb{R}_{\geq 0}$. 

(For an easy induction argument shows that for any countable ordinal $\alpha$, the lexicographically ordered set $\alpha \times [0, 1)$ with the order topology is homeomorphic to $\mathbb{R}_{\geq 0}$). 

If $\xi = \omega_0$, then $X$ is homeomorphic to $[0, 1]$. 

(End of proof) 
[1] David Gale, The Classification of 1-Manifolds: A Take-Home Exam, Amer. Math. Monthly, Vol. 94 No. 2 (February 1987), 170-175.<|endoftext|>
TITLE: Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?
QUESTION [14 upvotes]: I was told that a polynomial group law on (all of) $\mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group. 
Is it true? Where can I find a proof?
A counterexample for open subsets of $\mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law.

REPLY [15 votes]: This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"<|endoftext|>
TITLE: What is the "correct" category of multisets
QUESTION [18 upvotes]: During seminar the other day, when speaking about subobject classifiers, I asked if the subobject classifier for the category of multisets would have integer truth values, corresponding to the number of times and element is in the set. We attempted to show this, but quickly realized that we were not even sure of the "correct" category for multisets. 
To clarify, when I say correct I want my category to 

Have objects identified by multisets
Have maps between the multisets be on the level of elements in the multiset, and forget the order of those elements, e.g. there is only one map {111223}->{55}
The subobject classifier will behave as I had hoped, with {1} having truth value 3 in {111}

My question is;

Can you construct a category satisfying these properties?

Thanks in advance!
EDIT: First, sorry about not checking nLab, I forget about that site far too often. Second, I should say that I have a little bit of motivation for my property two. So let me clarify what I meant in property two. Given a multiset, it can be thought of as a pair $S\times\mathbb{N}$ for a set $S$. Now, when considering morphisms between multisets I want the maps $f,g:\lbrace 1122\rbrace\rightarrow\lbrace34\rbrace$ such that $f$ sends
$\begin{eqnarray*}
1&\mapsto& 3,\\
1&\mapsto& 4,\\
2&\mapsto& 3,\\
2&\mapsto& 4\\
\end{eqnarray*}$
and $g$ sends
$\begin{eqnarray*}
1&\mapsto& 4,\\
1&\mapsto& 3,\\
2&\mapsto& 4,\\
2&\mapsto& 3\\
\end{eqnarray*}$
to be the same morphism. But if $h$ sends 
$\begin{eqnarray*}
1&\mapsto& 4,\\
1&\mapsto& 4,\\
2&\mapsto& 4,\\
2&\mapsto& 3\\
\end{eqnarray*}$
then $h$ is not the same as $g$ or $f$. Further I would like it such that $\lbrace 112\rbrace$ is not a subobject of $\lbrace 12\rbrace$ but it is a subobject of $\lbrace 11122\rbrace$.
Hopefully this will clear it up.

REPLY [2 votes]: I suspect that the notion of "multiset" is actually ambiguous. For instance, different notions will be appropriate when talking about bosons vs. fermions vs. distinguishable (but intrinsically identical) particles. Ultimately, I suspect that if you really want there to be a reasonable category of multisets, you should assume that the multiple copies of a given element are distinguishable (though identical). So I'd be led to one of the following two candidates for the category of multisets:

The category $Set^\to$ of maps of sets. An object consists of two sets $A,B$ and a function $f: A \to B$; a morphism is a commuting square of maps.

The category $Set^{epi}$ of surjective maps of sets. This is the full subcategory of $Set^\to$ where $f: A \to B$ is required to be surjective.


To be clear, if $A \to B$ is an object of one of these categories, we are thinking of it as a multiset whose elements are the elements $b \in B$, with multiplicity given by the size of the fiber $f^{-1}(b) \subseteq A$. The passage from (1) to (2) is motivated by the idea that if $b \in B$ occurs with multiplicity 0, then we shouldn't think of $b$ as occurring in $B$ at all. The category $Set^{epi}$ of case (2) is equivalent to the category of equivalence relations suggested by Owen Biesel.
In case (1), we are working with the category of presheaves on the arrow category (a Grothendieck topos) and we can calculate the subobject classifier $\Omega$ in the standard way. It turns out that $\Omega$ is the multiset $\{110\}$, with the universal subobject $1 \to \Omega$ picking out one of the copies of $1$.
In case (2), we are dealing with a full subcategory $Set^{epi}$ of case the category $Set^\to$ from case (1), but I'm pretty sure this category of multisets doesn't have pullbacks. Nevertheless, any pullback square in $Set^\to$ whose objects happen to lie in $Set^{epi}$ will still be a pullback in $Set^{epi}$, so there is a non-negligible supply of pullbacks, and it's conceivable there might be a subobject classifier. But $\Omega$ from case (1) is not a subobject classifier -- it classifies subobjects in $Set^\to$, and there are almost always more of these than subobjects in $Set^{epi}$. So I don't think this category has a subobject classifier, though I wouldn't claim to have a proof.<|endoftext|>
TITLE: Krull dimension of a completion
QUESTION [6 upvotes]: How does one study Krull dimension of some I-adic completion of a ring or, more generally, a module? I know that in case of Noetherian local ring Krull dimension of its completion equals Krull dimension of the ring, but what can we say in general case?

REPLY [11 votes]: For a Noetherian ring R, the Krull dimension of its $I$-adic completion, $\hat{R}$ is given by $\sup h(J)$, where $J$ ranges over all maximal ideals of $R$ containing $I$ and $h(J)$ is the height of $J$. Therefore $\dim \hat R\le \dim R$ with equality only when $I\subset \operatorname{rad} R$. A reference is "Topics in $\mathfrak m$-adic topologies" by S.Greco, P.Salmon<|endoftext|>
TITLE: What would be the ramifications of homotopy theory being as easy as homology theory?
QUESTION [10 upvotes]: Greg Muller, in a post called Rational Homotopy Theory on the blog "The Everything Seminar" wrote 
"I tend to think of homotopy theory a little bit like ‘The One That Got Away’ from mathematics as a whole.  Its full of wistful fantasies about how awesome it would have been if things could only have worked out.  Imagine if homotopy groups of spaces and homotopy classes of maps were as easy to compute as homology groups…"
What are the wonderful consequences that he is referring to?

REPLY [12 votes]: Homology groups and homotopy groups are two sides of the same story.  Homotopy groups tell us all the ways we can have a map Sn → X, and in particular describe all ways we can attach a new cell to our space.  On the other side, the homology groups of a space change in a very understandable way each time we attach a new cell, and so they tell us all the ways that we could build a homotopy-equivalent CW-complex.  In cases where we can understand both of them, we can get things like complete theorems about classification of spaces.
Here's an example where we can compute: classification of the homotopy types of compact, orientable, simply-connected 4-manifolds.  (I originally saw this is Neil Strickland's bestiary of topological spaces.)
Poincare duality tells us that the homology groups are finitely generated free in degree 2, ℤ in degrees 0 and 4, and zero elsewhere.  We can cut out a closed ball, and get an expression of the manifold as obtained from a manifold-with-boundary N by attaching a 4-cell.  The Hurewicz theorem tells us that we can construct a map from a wedge of copies of S2 to N which induces an isomorphism on H*, and by the (homology) Whitehead theorem this is a homotopy equivalence.  So our original manifold is obtained, up to homotopy equivalence, by attaching a 4-cell to $\bigvee S^2$.
How many ways are there to do this?  It is governed by $\pi_3 (\bigvee S^2)$, which we can compute because it's low down enough.  This homotopy group is naturally identified with the set of symmetric bilinear pairings $H^2(\bigvee S^2) \to \mathbb{Z}$, and this identification is given by seeing how the cup product acts after you attach a cell.  So these 4-manifolds are classified up to homotopy equivalence by the nondegenerate symmetric bilinear pairing in their middle-dimensional cohomology.
Some of what we used here is general and well-understood machinery about homology, homotopy, and their relationships.  Wouldn't it be nice if the standard tools were always so effective?  But the real meat is that we have a complete understanding of homology and homotopy in the relevant ranges.  It turns our questions about classification into questions about pure algebra.  For questions that require specific knowledge about higher homotopy groups of spheres (or even lower homotopy groups of complicated spaces), it is much harder to get answers.  There aren't a lot of spaces where we have complete understanding of both the homology groups and the homotopy.  We have tools for reconstructing the former from the latter but their effectiveness wears down the farther out you try to go.
There are categories that are somewhat like the homotopy category of spaces where we can get an immediate and specific understanding of both sides of the coin.
One such example is the category of chain complexes over a ring R.  There, our fundamental building block is R itself.  The homology of any chain complex tells us both how R can be mapped in modulo chain homotopy, and how complicated any construction of the underlying chain complex must be.  A more complicated example would be the category of differential graded modules over a DGA, where the divide between how things can be constructed and how things can have new cells attached is, at the very least, governed by the complexity of H* A as a ring, and then by the secondary algebraic operations if A is far from being anything like formal.
Another such example is the rational homotopy theory of simply-connected spaces you mentioned.  There, homology and homotopy are roughly something like the difference between a ring's underlying abelian group structure and how you build it using generators and relations.
So you might think of the complexity of homotopy groups as telling us how much more complicated spaces are than chain complexes.<|endoftext|>
TITLE: Haar Measure on a Quotient
QUESTION [8 upvotes]: Suppose you have a locally compact group G with a discrete subgroup H.  Of course G has a unique (up to scalar) Haar measure, but it seems that G/H has and induced Haar measure as well.  
How does one induce a measure on G/H from the Haar measure on G? Any sources that describe this?

REPLY [7 votes]: I couldn't figure out which other answer to make this be a comment for... so: 
Already mentioned in other answers, but really working better than might have been acknowledged, is that the averaging map $\alpha:C^o_c(G)\rightarrow C^o_c(H\backslash G)$ by $\alpha f(x)=\int_H f(hx) dh$ is (readily provably) a surjection, for a closed subgroup $H$ of $G$. Thus, to attempt to define a right $G$-invariant integral/measure on $C^o_c(H\backslash G)$, define $\int_{H\backslash G} \alpha f dx=\int_G f$. The immediate issue, of course, is well-definedness, which holds only when the modular function of $G$ restricted to $H$ is equal to the modular function of $H$. 
This is one of those quasi-standard riffs that gets submerged... perhaps I saw it in Weil's book on integration on topological groups.<|endoftext|>
TITLE: The etale fundamental group of a field
QUESTION [7 upvotes]: Background and motivation:
I am teaching the "covering space" section in an introductory algebraic topology course.  I thought that, in the last five minutes of my last lecture, I might briefly sketch how to compute the "fundamental group of a field," primarily as a way of illustrating the analogy between Galois theory and covering space theory, but also because this is the sort of thing that makes me eager to learn more about a subject when I am on the receiving end of a lecture.  
Unfortunately, as yet, I myself know little more about the etale fundamental group than the definition and a bit of motivation (although I am certainly planning to learn more).  In particular, I realized that there is an obvious question I don't know how to answer.  When defining the fundamental group in an algebraic setting, we define it as an inverse limit because there is, in general, no natural analogue of a universal cover.  However, if we are looking at Spec k, there is an obvious candidate, namely the Spec of the algebraic closure of k.
Question:
Let $k$ be a field, with algebraic closure $K$.  How does the etale fundamental group of Spec $k$ compare to the automorphism group of $K$ over $k$?  Assuming they are different, what (in very general terms) are the reasons for working with one rather than the other?

REPLY [9 votes]: Also for the étale fundamental group there is in fact always some universal
cover. However, in the abstract way that Grothendieck formulated the theory of
coverings a universal cover would only exist as a pro-object. It then depends on
the precise situation (and the precise definitions) whether this pro-object can
be realised as an actual covering. Consider the case of a field $k$, the
question is whether 
$$\mathrm{Spec}(\overline{k}) \rightarrow \mathrm{Spec}(k)$$
where $k\subseteq\overline k$ is a separable closure, is considered an étale map
or not (depending on your preferences only finite separable $k$-algebras may
give you étale covers).
Here is, I think, an even more instructive example. Consider $\mathbb C^\ast$ as
the algebraic variety $\mathrm{Spec}(\mathbb C[\mathbb Z])$, where the group
algebra $\mathbb C[\mathbb Z]$ is nothing but the Laurent polynomial ring
$\mathbb C[t,t^{-1}]$. A topological universal cover is given by
$\exp\colon\mathbb C \rightarrow \mathbb C^\ast $ whereas an étale universal cover
is given by the pro-covering 
$$\{\mathrm{Spec}(\mathbb C[1/n\mathbb Z])\rightarrow
\mathrm{Spec}(\mathbb C[\mathbb Z])\}$$ 
(with transition maps induced by inclusions $1/n\mathbb Z\subseteq 1/mn\mathbb Z$). If one accepts such maps as étale covers one may take the direct limit of the rings involved and get an étale universal cover 
$$\mathrm{Spec}(\mathbb C[\mathbb Q]) \rightarrow\mathrm{Spec}(\mathbb{C}[\mathbb Z]).$$ 
This algebraic universal cover is not (analytically) isomorphic to $\exp\colon\mathbb C \rightarrow \mathbb C^\ast $; the fibres of the algebraic
cover can be identified with the profinite completion of $\mathbb Z$ while the
fibres of the analytic/topological universal have fibres $\mathbb Z$
itself. There is a map of covers 
$$\mathbb C^\ast \rightarrow \mathrm{Spec}(\mathbb C[\mathbb Q])$$ 
which on fibres correspond to the inclusion of $\mathbb Z$ into its profinite completion.<|endoftext|>
TITLE: Wanted: Quadratic Space in Characteristic 2 as a Counterexample to a Theorem of Arf
QUESTION [23 upvotes]: Hi. Peter Roquette sent me an email asking for an example of a quadratic space in characteristic 2 having certain features.  I have no idea on this, but maybe someone reading this does. 
He would like an example of a field $K$ of characteristic 2 with the following two properties:
(1) The quaternion algebras over $K$ form a group (within the Brauer group of $K$).
(2) There exists an anisotropic quadratic form of
dimension > 4 over $K$ that is "completely regular in the sense of Arf". That means: there is a quadratic space $(V,Q)$ over $K$ such that 
(a) $\dim(V) > 4$
(b) if $Q(v) = 0$ then $v = 0$ 
(c) for the associated bilinear form $B(v,w) = Q(v+w) - Q(v) - Q(w)$, if
$B(v,w) = 0$ for all $w$ in $V$ then $v = 0$.
Here's the background. Roquette has written a paper with Falko Lorenz on the historical development of the Arf invariant, and they include in the paper a counterexample to the method of proof of a theorem of Arf. (The paper is at Roquette's homepage at http://www.rzuser.uni-heidelberg.de/~ci3/arf.pdf and it has also appeared in "Mathematische 
Semesterberichte" vol. 57 (2010) pp. 73--102.) Their counterexample to Arf's method of proof is not actually a counterexample to his main theorem.  In order to find a counterexample to the main theorem itself they want a quadratic space with the properties above (in characteristic 2). Roquette has asked around but nobody has yet been able to give him an example. 
EDIT (Aug. 16): After Roquette learned about the answer posted below, he has posted on his website http://www.rzuser.uni-heidelberg.de/~ci3/manu.html a paper (number 46) with Lorenz which discusses the solution to his question in the context of Arf's paper.  Anyone interested in this topic is encouraged to look at the paper. He wrote to me "Your idea to put my question on the website Math Overflow has worked wonderfully."

REPLY [17 votes]: I forwarded this question to Detlev Hoffmann, who says that such examples exist.  Specifically, you can produce such an example where there is, say, an anisotropic form of dimension 8 using characteristic 2 analogues of the techniques in Merkurjev's 1992 article "Simple algebras and quadratic forms".  He says details can be found in the PhD thesis of his student Frederic Faivre at Université Franche-Comté.
Detlev further explained the source of Arf's confusion, which I will now recap.  Over a field $F$ of any characteristic, the tensor product of two quaternion algebras is not a division algebra if and only if the two quaternion algebras contain a common quadratic extension.  In characteristic different from 2, this is an essentially complete criterion.  But in characteristic 2, one has to wonder if this quadratic extension is separable or inseparable.
Draxl showed that if two quaternion algebras contain a common quadratic extension, you can always find one that is separable.  That is, the property of containing a common inseparable extension is much stronger (because it implies that they contain a common separable extension).  A nice exposition of this can be found in T.Y. Lam's 2002 article "On the linkage of quaternion algebras".
Detlev asserts: If you demand that every pair of quaternion division algebras over $F$ (of characteristic 2) share an inseparable quadratic extension, then there are no anisotropic regular quadratic forms of dimension > 4.  "This is essentially due to Baeza.  In fact, in some sense even to Arf, except that he didn't realize there's a difference between [sharing separable and inseparable subfields]."  So presumably this is what Arf was claiming to have proved.
In contrast to this, the requirement that the quaternion algebras form a subgroup of the Brauer group is just that every pair of quaternion division algebras share a separable quadratic extension.  This is a much weaker hypothesis, and allows for examples of fields like in Faivre's thesis.
Here are some precise references for the theorem "every pair of quaternion algebras over $F$ share an inseparable quadratic extension implies every regular quadratic form of dimension > 4 is isotropic":

Arf (with confusion mentioned above): Satz 11 in "Untersuchungen über quadratische Formen in Körpern der Charakteristik 2. I." J. reine angew. Math. 183, 148-167 (1941)"
Baeza: Theorem 3.1 in "Comparing $u$-invariants of fields of characteristic $2$." Bol. Soc. Brasil. Mat. 13 (1982), no. 1, 105--114.
Faivre's thesis: Proposition 3.3.5 (with complete proof)<|endoftext|>
TITLE: Knot complement diffeomorphism groups and embedding spaces
QUESTION [7 upvotes]: I'm interested in the following collection of questions: Let $S^n_k = \sqcup_k S^n$ be a disjoint union of $k$ distinct $n$-dimensional spheres.  Write $Emb(S_k^n, S^{n+2})$ for the space of embeddings of these spheres into $S^{n+2}$.  Pick your favorite embedding $e: S_k^n \to S^{n+2}$, and let $X_e = S^{n+2} \setminus im(e)$ be the complement of the image of the embedding.

What is $\pi_1(Emb(S_k^n, S^{n+2}), e)$?  Since this is probably unknown, what is known?
How is this related to the mapping class group $\pi_0(Diff(X_e))$ of $X_e$?

I ask #2 because in dimension $n=0$, they are the same: the space of embeddings is the configuration space of points in the sphere.  Its fundamental group is the (spherical) braid group, which is the same as the mapping class group of the punctured sphere.  My guess is that life is not so simple in higher dimensions.  Lastly, does any of this simplify out when you get into the range of dimensions where surgery theory starts working well?

REPLY [8 votes]: There is a locally-trivial fibre bundle
$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$
here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$.  $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$. 
First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic.  It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$.  Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere.  What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things. 
So now you have a fibration:
$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$
where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$.  This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$. 
$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there.  In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(X_e)$ allows on the spherical normal bundle to $L$.  There's probably a decent answer to that which doesn't take too much work but the above is a start. 
edit: by Cerf's pseudoisotopy theorem, the kernel of the map $\pi_0 Diff(S^{n+2}, L) \to \pi_0 Diff(X_e)$ contains the exotic sphere "part" of $\pi_0 Diff(S^{n+2}, L)$.<|endoftext|>
TITLE: Additive commutators and trace over a PID
QUESTION [11 upvotes]: I would like to find an example of principal ideal domain $R$, such that there exists a square matrix $A\in \mathfrak{M}_n(R)$ with zero trace that is not a commutator (i.e. for all $B,C \in \mathfrak{M}_n(R)$, $A\neq BC-CB$).
I know that such a PID (if it can be found) cannot be a field, or $\mathbb{Z}$.

REPLY [3 votes]: I'm a bit (?) late on that one, but the theorem for general PIDs has been proven after this question was asked.
The reference is (see here for the ArXiv version):

Stasinski, Alexander, Similarity and commutators of matrices over principal ideal rings, Trans. Amer. Math. Soc. 368 (2016), no. 4, 2333–2354.<|endoftext|>
TITLE: What is an antiequivalence of two categories?
QUESTION [7 upvotes]: Hello everyone, I am a newbie to the category theory. While reading some paper about infinite Galois theory, the 'anti-equivalence' of two categories showed up. Could anyone give me a 'good' explanation what this means? Thanks a lot.

REPLY [15 votes]: All it means that one of the categories is equivalent to the opposite of the other.
Wikipedia has informative pages on opposites of categories and equivalences
of categories:
http://en.wikipedia.org/wiki/Opposite_%28category_theory%29 ,
http://en.wikipedia.org/wiki/Equivalence_of_categories .<|endoftext|>
TITLE: The difference between the Recursive and the Effective topos.
QUESTION [11 upvotes]: I would like to know which is the real difference between the Recursive topos (in the sense of Mulry) and the Effective topos (in the sense of Hyland). Especially what is related to recursive functions. Do they have the same semantic power?
I will be gratefull with some hints about texts related to this.
Thanks in advance.
PS: I don't give definitions because of their big extension, but I could give them if anybody wants.

REPLY [8 votes]: I am aware that I am answering an old question, but for completeness of MathOverflow, I'd like to point out that G. Rosolini's 1986 thesis "Continuity and effectiveness in topoi" contains (chapter 6) a description of both the effective and the recursive topoi, and a comparison of them by constructing a functor from the effective to the recursive (preserving limits, finite coproducts, images and the natural number object, faithful on the full subcategory of countable objects and full and faithful on the full subcategory of countable separated objects).  According to the author, "the major difference between the effective topos and the recursive topos is that the [natural number object] in [the effective topos] does not generate the topos", and the functor takes an object of the effective topos to the set of maps from the natural numbers object to it, which can be viewed as an object of the recursive topos.<|endoftext|>
TITLE: Relation of Lie Groups and Cohmology Theories via Formal Group Laws
QUESTION [5 upvotes]: There is a standard process (for example explained here) to obtain a formal group law form a complex oriented cohomology theory.
For a Lie group G one can choose coordinates at the unit and expand the multiplication map to a power series which also gives a (possibly higher) formal group law.
What is know about the two realization problems and their relation? Which group laws come from complex oriented cohomology theories? Which from Lie groups? Is it sensible to associate Lie groups to cohomology theories or vice versa because of their group laws?

REPLY [12 votes]: Completions of Lie groups at the unit element will give you formal groups defined over the reals. If you want to compare those with the formal groups obtained from cohomology theories, you'll probably want to look for one-dimensional factors, but that's not going to lead to anything interesting since all one-dimensional formal groups over R are isomorphic to the additive group by the logarithm map.
On the other hand, looking at group schemes is much more promising. Elliptic curves give you elliptic cohomology, and one-dimensional summands of completions of higher abelian varieties have been studied by e.g. Ravenel.
As for your realization problem: I don't know about Lie groups, but for cohomology theories, the Landweber exact functor theorem gives you a partial, but still quite satisfying result.<|endoftext|>
TITLE: Is there a schemetical construction for modular curves over the rationals?
QUESTION [12 upvotes]: One can get modular curves by the following procedures: first take the uper half plane and the rationan numbers on the x-axis, then we consider the quotient by a congruence subgroup. Now we get a compact Riemann surface, and by Chow's results, it is algebraic--an algebraic variety. Here we start with an analytic object and finally we get an algebraic one. But can we use algebraic methords only (e.g. by quotients of group schemes) to get modular curves? Or, can we find a meanful moduli problem solved by a modular curve?

REPLY [3 votes]: Passing comment: Mumford's GIT constructs modular curves as quotients---not of the upper half plane, but of some parameter space of subspaces of projective space, by an algebraic group. As for the last part of your question: sure there is a moduli problem solved by a modular curve! Isomorphism classes of elliptic curves plus level structure (e.g. point of order N) is represented by a modular curve (Y_1(N) in this case, as long as N>=4).<|endoftext|>
TITLE: Spectrum of a generic integral matrix.
QUESTION [5 upvotes]: My collaborators and I are studying certain rigidity properties of hyperbolic toral automorphisms.
These are given by integral matrices A with determinant 1 and without eigenvalues on the unit circle.
We obtain a result under two additional assumptions
1) Characteristic polynomial of the matrix A is irreducible
2) Every circle contains no more than two eigenvalues of A (i.e. no more than two eigenvalues have the same absolute values)
We feel that the second assumption holds for a "generic" matrix. Is it true?
To be more precise, consider the set X of integral hyperbolic matrices which have determinant 1 and irreducible characteristic polynomial.
What are the possible ways to speak of a generic matrix from X? Does assumption 2) hold for generic matrices?
Comments:

Assumption 1) doesn't bother us as it is a necessary assumption.
Probably it is easier to answer the question when X is the set
off all integral matrices. In this case we need to know that
hyperbolicity is generic, 2) is generic and how generic is
irreducibility.

REPLY [6 votes]: Yes, a generic integer matrix has no more than two eigenvalues of the same norm. More precisely, I will show that matrices with more than two eigenvalues of the same norm lie on a algebraic hypersurface in $\mathrm{Mat}_{n \times n}(\mathbb{R})$. Hence, the number of such matrices with integer entries of size $\leq N$ is $O(N^{n^2-1})$. 
Let $P$ be the vector space of monic, degree $n$ real polynomials. Since the map "characteristic polynomial", from $\mathrm{Mat}_{n \times n}(\mathbb{R})$ to $P$ is a surjective polynomial map, the preimage of any algebraic hypersurface is algebraic.
Thus, it is enough to show that, in $P$, the polynomials with more than two roots of the same norm lie on a hypersurface. Here are two proofs, one conceptual and one constructive.
Conceptual: Map $\mathbb{R}^3 \times \mathbb{R}^{n-4} \to P$ by 
$$\phi: (a,b,r) \times (c_1, c_2, \ldots, c_{n-4}) \mapsto (t^2 + at +r)(t^2 + bt +r) (t^{n-4} + c_1 t^{n-5} + \cdots + c_{n-4}).$$
The polynomials of interest lie in the image of $\phi$. Since the domain of $\phi$ has dimension $n-1$, the Zariski closure of this image must have dimension $\leq n-1$, and thus must lie in a hyperplane.
Constructive: Let $r_1$, $r_2$, ..., $r_n$ be the roots of $f$. Let 
$$F := \prod_{i,j,k,l \ \mbox{distinct}} (r_i r_j - r_k r_l).$$
Note that $F$ is zero for any polynomial in $\mathbb{R}[t]$ with three roots of the same norm. Since $F$ is symmetric, it can be written as a polynomial in the coefficients of $f$. This gives a nontrivial polynomial condition which is obeyed by those $f$ which have roots of the sort which interest you.<|endoftext|>
TITLE: Oriention-Reversing Diffeomorphisms of a Manifold
QUESTION [12 upvotes]: I am trying to figure out when a closed, oriented manifold admits an orientation reversing diffeomorphism. My naive argument that the orientation cover should allow you to switch orientations is apparently wrong, since not every manifold admits such a diffeomorphism.
Can anyone give me some criteria for when such a morphism should exist, or why some of the standard counterexamples (such as $\mathbb{P}^{2n}$) fail to admit one?
Thanks

REPLY [16 votes]: A large number of manifolds of dimension $4k$ can't admit an orientation-reversing diffeomorphism just because of their cobordism type. That is, if $f: M\rightarrow \overline{M}$ is an orientation preserving diffeomorphism, then the cobordism class $[M^n]$ is a 2-torsion element of the cobordism group of oriented $n$-manifolds: Since $M\sqcup M \cong M\sqcup\overline{M}$ bounds the cylinder $M\times[0,1]$, thus $2[M] = [M\sqcup \overline{M}] = 0 \in \Omega^{\rm SO}_n$. By the Thom-Pontryagin theorem, if $M$ has a nonzero Pontryagin number (which requires that the dimension of $M$ to be a multiple of 4), then $[M]$ is generates a free abelian subgroup of $\Omega^{\rm SO}_n$ and is not a 2-torsion element. Thus, $M$ will not admit an orientation-reversing diffeomorphism.
In particular, this applies if the signature of $M$ is nonzero, since by Hirzebruch's signature theorem the signature is computable in terms of Pontryagin numbers. The previously mentioned examples of $\mathbb{CP}^{2k}$ and $\mathbb{HP}^k$ are special cases of this statement, since both have nonzero signature and hence are do not represent 2-torsion elements of the oriented cobordism group.<|endoftext|>
TITLE: Generalized Chinese Remainder Theorem
QUESTION [5 upvotes]: Let $U,V$ be submodules of a $R$-module $M$. Then the diagonal induces an isomorphism
$M/(U \cap V) \to M/U \times_{M/(U+V)} M/V.$
This is a (useful!) generalization of the Chinese Remainder Theorem and the proof is very easy. But I'm interested what happens when we take finitely many submodules $U_1,...,U_n$. How can we relate $M/(U_1 \cap ... \cap U_n)$ with the $M/U_i$? I think the case $n=2$ can not be used for an induction, there are more compatiblities to check for an element in $\prod_i M/U_i$ to come from $M$. I wonder if there is a nice description.
For $M=R$, this question asks for a sort of sheaf condition for sections on closed subschemes.

REPLY [4 votes]: So this is what's in Kleinert's paper "Some remarks on the Chinese Remainder Theorem" that I mentioned in the comments.
If $\mathcal F=\{U_1,U_2,\dots,U_n\}$ is a family of submodules of the $R$-module M, then there is an embedding $\phi(\mathcal F)$ of $M/U_1\cap \cdots \cap U_n$ into 
$$M(\mathcal F):= \{(u_i)\in \prod M/U_i \quad \rvert u_i\equiv u_j \mod (U_i+U_j),\forall i,j\}.$$
Let the cokernel of $\phi$ be
$$O(\mathcal F)=M(\mathcal F)/\phi(M/U_1\cap \cdots \cap U_n).$$
$O(F)$ is thought of as the obstruction against the ability to solve simultaneous congruences, and so we say that the generalized Chinese Remainder Theorem holds if $O(\mathcal F)=0$. He proceeds to the following sheaf-theoretical interpretation of the problem:
Let $X$ be the discrete topological space $\{1,2,\dots,n\}$, and define the presheaf $\mathcal P(\mathcal F)$ on $X$ by $\mathcal P(V)=M/\sum_{i\notin V}U_i$, for $V\subset X$. If $V\subset W$ the restriction map is given by the residue map
$$\mathcal P(W)=M/\sum_{i\notin W}U_i\to M/\sum_{i\notin V}U_i=\mathcal P(V).$$
Now let $\mathcal U$ be the covering $\{X/\{i\}\}$. It follows that $M(\mathcal F)$ is the set of cocycles $C^0(\mathcal U,\mathcal P)$ and that $O(\mathcal F)=0$ iff $\mathcal P$ satisfies the second sheaf axiom with respect to $\mathcal U$. He also makes the remark that when $n=2$ , which you described in the question, this is always the case and so the generalized Chinese Remainder Theorem always holds, even though it doesn't always in the general case.<|endoftext|>
TITLE: `Topos' with alternate subobject lattice?
QUESTION [7 upvotes]: We know that for any topos E, and for any object A in E, the subobjects of A, Sub(A), form a Heyting lattice.
Does anybody know of any sort of modification of the definition of a topos that makes Sub(A) a different type of lattice? Could we get an incomplete lattice, or maybe a quantum lattice?
I'm curious because I know a lot(all?) of logical systems can be realized as a lattice, and I think this may be an interesting way to look at some alternative logics.

REPLY [2 votes]: The fact that there is always a Heyting algebra structure on Sub(A) doesn’t exclude the possibility that another lattice structure can coexist with the natural one. As an example take C to be the three element poset with a least element, and take as the topos the functor category from C to Set.  You can visualize this as a category of bipartite multigraphs. The subobject classifier has five arrows  and two vertices on each of the two sides.  The Heyting algebra structure at the arrow level has the form $2^2+1$, and the other  lattice structure compatible with tail and head maps has the form $N_5$, a minimal five element non modular lattice. Each Sub(A)
inherits this structure, with the joins and meets here formed as prescribed by the subobject classifier.<|endoftext|>
TITLE: General Equilibrium for Mathematicians
QUESTION [20 upvotes]: I've been reading up a lot on the recent financial crisis, and central to the story is the existence of general equilibrium models in economics, say, as proven by Arrow and Debreu (and MacKenzie?). Regardless of the real-world validity of these models (they're not looking so hot these days), I'm interested in them as a purely mathematical exercise. So, that said, does anyone know of a good (preferably available online) exposition of the existence of a general equilibrium aimed at mathematicians?

REPLY [2 votes]: Debreu is certainly a superb exposition, but I think a better choice is McKenzie, for two reasons:
1)  Debreu's model assumes a fixed number of firms, which means that it's implicitly a model only of short-run equilibrium.  McKenzie's model (which was contemporaneous with Debreu's) allows the number of firms to vary (at the cost of assuming constant returns to scale) and is therefore a better introduction to the kind of models you seem to be interested in.
2)  Debreu's book, well written as it is, dates from the 1950s.  Because McKenzie's book is much more recent, it's able to offer some historical perspective that I think will be helpful to you, as well as incorporating a lot of material (e.g. uncertainty) that wasn't well worked out until more recently.
Like Debreu, McKenzie was a good and careful expositor.<|endoftext|>
TITLE: Is there a notion of integration over the algebraic numbers?
QUESTION [11 upvotes]: For reasons which are hard to articulate (due to they not being very clear in my mind), but having to do with the eprint From Matrix Models and quantum fields to Hurwitz space and the absolute Galois group by  Robert de Mello Koch and Sanjaye Ramgoolam, I have been wondering whether there is a notion of integration over the algebraic numbers $\overline{\mathbb{Q}}$, as there is over the p-adic completions of the rationals.

REPLY [2 votes]: Pick an algebraic number at random.  What is its expected degree?  As noted, since the set is countable, we cannot expect to do conventional integrals on it.  But we still have things like this:  pick a natural number at random---then it is square-free with probability $6/\pi^2$.
So, maybe the question is:  What are "natural" Folner sets for the algebraic numbers?  The usual Folner sets for $\{1,2,3,\dots\}$  are the sets $\{1,2,\dots,n\}$, which define the "density" (not an actual measure)
$$
\delta(A) = \lim_{n\to\infty}\frac{\\#(A \cap \{1,\dots,n\})}{n}
$$
for sets $A \subseteq \{1,2,\cdots\}$.  See, for example, http://en.wikipedia.org/wiki/F%F8lner%5Fsequence for information on Folner sequences.<|endoftext|>
TITLE: Proving existence of non-special divisors of a given degree d on compact Riemann surfaces
QUESTION [5 upvotes]: I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.
I think it should be true, but I want to prove the following concretely: 
"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."
(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)
Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $\deg D = g-1$.
Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.
More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?

REPLY [10 votes]: Take $g+1$ general points $p_1, \dots, p_{g+1}$ on your curve. The divisor $p_1+ \cdots +p_g - p_{g+1}$ is non-special. The proof is easy from the following lemma: if $D$ is a divisor such that $\mathrm{h}^0(D) > 0$, then $\mathrm{h}^0(D - p) = \mathrm{h}^0(D) - 1$ for all but finitely many points $p$. First you use Riemann-Roch to deduce that $\mathrm{h}^0(p_1+ \cdots +p_g) = 1$, then you apply the lemma once again to $p_1+ \cdots +p_g$. This also works for $g = 0$ and $g = 1$.<|endoftext|>
TITLE: Can a non-algebraic complex manifold be embedded meromorphically into projective space?
QUESTION [5 upvotes]: Background: By Chow's theorem, if a complex manifold can be embedded holomorphically into complex projective space, then this complex manifold must be algebraic.
Question: Suppose X is a compact complex manifold (not necessarily algebraic). Let $f:X--> {\mathbb{CP}}^n$ be a meromorphic map that is injective on its domain. Does this imply that X is algebraic?

REPLY [11 votes]: If $X$ is not compact there are loads of problems so I follow you in adding compactness as a condition.
Under the assumption of the question  $X$ is bimeromorphic to a projective variety and hence by a result of Artin it is an algebraic space. Hence if you accept algebraic spaces as being algebraic the answer is yes. As there are smooth proper algebraic spaces that are not schemes if you don't the answer is no.<|endoftext|>
TITLE: Fixed points sets of pushouts
QUESTION [5 upvotes]: Let $G$ be a group and $X \to Y, X \to Z$ morphisms of $G$-sets with pushout $P=Y \cup_X Z$. Is then $P^G$ the pushout of $X^G \to Y^G, X^G \to Z^G$? This is not clear from general category theory, because pushouts do not have to commute with equalizers. I hope it's true when $X \to Y$ is injective (thus also $Z \to P$). Then $Y^G \cup_{X^G} Z^G \to P^G$ is injective, but I don't know why an element of $P^G$ which comes from $Y$ lies in the image...
Now let's work with $G$-spaces instead of $G$-sets. Then it is not clear at all to me if the canonical map from the pushout to $P^G$ is open. Is it true when $G$ is finite? Or if $X,Y,Z$ are $G$-CW-complexes? By the way, this was used in a lecture about how to compute the homology of the classifying space $BG$ with the help of $\underline{E} G$.
EDIT: I was told that it is true if $X,Y;Z$ are $G$-CW-complexes and $X \to Y$ is a $G$-cofibration. How is this proven?

REPLY [5 votes]: No. Consider the case where your sets happen to be abelian groups with an action of G. Pushing out the zero morphism over any map gives the cokernel, so a special case of your question is whether taking G-invariants is right exact, and it isn't. This should give an idea of where counterexamples might come from.
Concretely, lets say your group is $\mathbb{Z}/2$ and you push out $\mathbb{Z}/2 \to \ast$ over itself. The ordinary pushout is a point (with the trivial action). But if you take invariants first you find you are pushing two points out over the empty set, which results in two points.<|endoftext|>
TITLE: Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$
QUESTION [19 upvotes]: I have a question about vector bundles on the algebraic surface $\mathbb{P}^1\times\mathbb{P}^1$.  My motivation is the splitting theorem of Grothendieck, which says that every algebraic vector bundle $F$ on the projective line $\mathbb{P}^1$ is a direct sum of $r$ line bundles, where $r$ is the rank of $F$.
My question is: to what extent is this true for $\mathbb{P}^1\times\mathbb{P}^1$, if at all?  At first glance I imagine this is classical, but I haven't had luck working it out, nor do I have a good reference for where this might be done.
Irrespective of the content of the answer, would it follow that the answer would be the same if the question were asked for $(\mathbb{P}^1)^k$?
Thanks!

REPLY [23 votes]: There is a precise sense in which the theory of vector bundles on $P^1\times P^1$ is exactly as complicated as that for $P^2$ (and this is true for any hypersurface in $P^3$, and conjecturally for all surfaces):
For any bundle $E$ on $P^1\times P^1$, the (singly graded) cohomology table of $E$ is the collection of numbers
$$
h^i(E(p)); \quad p,q\in Z\times Z, \quad i=0,1,2.
$$
Let $C(P^1\times P^1)$ be the positive rational convex cone generated by the Betti tables of bundles on $P^1\times P^1$. By a Theorem of mine with Schreyer (soon to be on the arXiv), the cone $C(P^1\times P^1)$ is identical to the corresponding cone for $P^2$. 
The idea of the proof is simple: there exist Ulrich sheaves on $P^1\times P^1$---these are sheaves (bundles in this case) with the property that under a finite map to $P^2$ they push forward to a direct sum of copies of the structure sheaf. Pulling back a vector bundle from $P^2$ and tensoring with an Ulrich sheaf only multiplies the cohomology table by the rank of the Ulrich sheaf. This gives one inclusion. For the other, observe that pushing a bundle forward by a finite linear projection $P^1\times P^1 \to P^2$ preserves the cohomology table of the bundle.<|endoftext|>
TITLE: Why is the prime spectrum not useful in non-archimedean analytic geometry?
QUESTION [11 upvotes]: This semester I am attending a reading seminar on non-archimedean analytic geometry (a subject I know nothing about), roughly following the notes of Conrad. 
Reading Conrad's notes (and e.g. those of Bosch) it struck me that the prime spectrum of affinoid algebras never seems to appear, only the maximal spectrum. Can somebody explain the reason for this?

REPLY [9 votes]: Another point to bear in mind, in addition to those raised by Brian and Kevin, is that generic points (in the sense of non-maximal prime ideals) don't make sense in analytic geomtery.
For example, the Tate algebra $\mathbb Q_p\langle\langle x\rangle \rangle$ contains 
one non-maximal prime ideal, the zero ideal.  Geometrically it corresponds to the closed 
disk $|x| \leq 1$.  Where in this disk would the generic point corresponding to the zero
ideal live?  The point is that, unlike in algebraic geometry, in rigid analytic geometry one can find disjoint open subsets of irreducible spaces such as the closed disk.
In Berkovich's theory, one does have generic points, but they consist of more data than just a prime ideal; one must also choose a norm on the residue field.  (This relates to Brian's comment.)  Geometrically, this choice of norm pins down where on the rigid space the generic point lives.<|endoftext|>
TITLE: weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?
QUESTION [19 upvotes]: A weight 2 modular form which happens to be a normalised cuspidal eigenform with rational coefficients has a natural geometric avatar---namely an elliptic curve over the rationals. It seems to be a subtle question as to how this notion generalises---this question was raised by Tony Scholl in conversation with me the other day. For example I guess I wouldn't expect a weight 3 normalised cuspidal eigenform with rational coefficients to be the $H^2$ of a smooth projective surface, because any such surface worth its salt would have (1,1)-forms coming from a hyperplane section, whereas the Hodge numbers of the motive attached to a weight 3 form are 0 and 2.
But in weight 4 one can again dream. A rigid Calabi-Yau 3-fold defined over the rationals has 2-dimensional $H^3$ and the Hodge numbers match up. Indeed there are many explicit examples of pairs $(X,f)$ with $X$ a rigid Calabi-Yau 3-fold over $\mathbf{Q}$ and $f$ a weight 4 cuspidal modular eigenform, such that the $\ell$-adic Galois representation attached to $f$ is isomorphic to $H^3(X,\mathbf{Q}_\ell)$ for all $\ell$.

The question: Is it reasonable to expect that (the motive attached to) every weight 4 normalised cuspidal eigenform with rational coefficients is associated to the cohomology of a rigid Calabi-Yau 3-fold over $\mathbf{Q}$?

REPLY [3 votes]: To follow up the question of the intermediate Jacobian, there is indeed a later survey by Noriko Yui (Arithmetic of Calabi-Yau varieties.  Mathematisches Institut, Georg-August-Universität Göttingen: Seminars Summer Term 2004,  9--29) where she does conjecture it to be defined over $\mathbb Q$ for any (modular) rigid CY3 over $\mathbb Q$. 
Moreover, she refers to a joint paper with X. Xarles in preparation (still unpublished) which claims to settle this in the CM case:
Let $X$ be a rigid CY3 of CM type (i.e. with commutative Hodge group) over some number field $F$. Then the intermediate jacobian $J^2(X)$ is an elliptic curve with CM by an order in an imaginary quadratic field (understood: the same field, since there will be a relation of Hecke characters over some extension), and it has a model over $F$.
Let's apply this to rigid CY3's over $\mathbb Q$ and assume that there is one for each newform of weight 4 with rational coefficients. Pick one of the weight 4 newforms with rational coefficients and CM of class number 3, i.e. induced by a Hecke character for an imaginary quadratic field $K$ of class number 3 such as $\mathbb Q(\sqrt{-23})$ (or generally of class group exponent 3). By assumption there is an associated CY3 $X$ over $\mathbb Q$ (which ought to have CM). But then, by the above result, its intermediate Jacobian is an elliptic curve over $\mathbb Q$ with CM in $K$, contradiction.
Of course, one can still ask whether all non-CM newforms can be realized in some CY3's over $\mathbb Q$, but I would find it surprising if this would only fail at certain CM forms. And after all, I would be glad to allow non-rigid CY3's over $\mathbb Q$ admitting the right submotive, but this might still not be sufficient.<|endoftext|>
TITLE: When is every "solid" perfect complex faithful?
QUESTION [16 upvotes]: Let $R$ be a noetherian commutative ring. Consider $D^{perf}(R)=K^b(R-proj)$ the category of bounded complexes of finitely generated projective $R$-modules, with maps of complexes up to homotopy.
QUESTION: For which $R$ is it true that every complex $E \in D^{perf}(R)$ whose support is the whole spectrum, $Supph(E)=Spec(R)$, is also $\otimes$-faithful?
Explanations: 
(1) For simplicity, let us call an object $E$ "solid" if it is supported everywhere: $Supph(E)=Spec(R)$. This means that for every $P\in Spec(R)$, the object $E_P\in K^b(R_P-proj)$ is non-zero (i.e. not null-homotopic). Equivalently, the image of $E$ in $D(\kappa(P))$ is non-zero for every residue field $\kappa(P)$. Equivalently again, the ordinary support of the total homology of $E$ is the whole $Spec(R)$, as a finitely generated $R$-module. ($Supph$ is for support of the $h$omology and is called the "homological support".)
(2) An object $E$ in a tensor-category can be called "$\otimes$-faithful" (simply "faithful") if $-\otimes E$ is a faithful functor, i.e., for every map $f:E'\to E''$, the map $f\otimes E: E'\otimes E\to E''\otimes E$ is zero only if $f$ is zero.
(3) It is an exercise on rigid tensor-triangulated categories to show that $E\in D^{perf}(R)$ is faithful if and only if the unit $\eta:R\to  E\otimes E^\vee$ is split injective, where $E^\vee$ is the dual of $E$ and of course $R=R[0]=\cdots 0\to 0\to R\to 0\to 0\cdots$ is the unit of the tensor on $D^{perf}(R)$. (Use that $\eta$ is split injective after applying $-\otimes E$ by the unit-counit relation.) Split injectivity of $\eta:R\to E\otimes E^\vee$ forces $E$ to be solid. In short, we always have faithful$\implies$ solid and the question is: For which $R$ is the converse true?
(4) It is well-known (and non-trivial) that a map $f:E'\to E''$ in $D^{perf}(R)$ which goes to zero in every residue field must be $\otimes$-nilpotent: $\exists\ n>0$ such that $f^{\otimes n}=0$. So, if $f\otimes E=0$ with $E$ solid, one can conclude $f^{\otimes n}=0$ for some $n>0$ but not necessarily $f=0$ in general.
(5) Note that it is necessary for "solid$\implies$ faithful" that $R$ be reduced. Indeed, let $a\in R$ with $a^2=0$ and let $E=cone(a)$ be the complex $E=\cdots0\to 0\to R\overset{a}\to R\to 0\to 0\cdots$. Then $f=a:R\to R$ satisfies $f\otimes E=0$ and $E$ is solid.
(6) Reduced is not enough though. I've an example of a non-faithful solid complex over $R=k[T^2,T^3]\subset k[T]$ ($k$ a field).
(7) When $R$ is Dedekind, every solid complex has a non-zero projective summand, hence is faithful. So, that is a class of examples where solid does imply faithful.
(8) In particular, I would be very interested in a regular ring $R$ with a non-faithful solid object. (Call that a conjecture in disguise, if you want.)
References: 
M. Hopkins, "Global methods in homotopy theory", Homotopy theory (Durham, 1985),  73--96.
A. Neeman, "The chromatic tower for $D(R)$", Topology 31 (1992), no. 3, 519--532. Article (requires access).
R. Thomason, "The classification of triangulated subcategories",
Compositio Math. 105 (1997), no. 1, 1--27. Article (requires access).

REPLY [3 votes]: OK, I think you are right that there are non-faithful solid guys even over a two dimensional regular local ring. For example, say $R = k[[x, y]]$ and we consider the complex $E$ given by
$$
A = \left(
\begin{matrix}
t_1 & t_2 \\
t_3 & t_4
\end{matrix}
\right)
: R^{\oplus 2} \longrightarrow R^{\oplus 2}
$$
where $t_1, t_2, t_3, t_4$ are pairwise distinct monomials in $x$ and $y$ all of the same degree (for example $17$) such that $t_1 t_4 - t_3 t_2 = 0$. Such a quadruple of monomials exists and then $E$ is solid as follows from the discussion in the question. Then we consider $E \otimes E^\wedge$ and we get the complex
$$
M_2(R) \longrightarrow M_2(R) \oplus M_2(R) \longrightarrow M_2(R)
$$
where the first arrow maps $X$ to $(XA, AX)$ and the second arrow maps $(Y, Z)$ to $AY - ZA$. There is the map $\eta$ from $R$ placed in degree $0$ to the displayed complex, sending $1$ to $(\mathbf{1}_2, \mathbf{1}_2)$. But, instead of using the characterization with $\eta$ you give, we can use the dual characterization with a trace map which sends $(Y, Z)$ to $\text{Tr}(Y) - \text{Tr}(Z)$. The characterization is that the dual complex $E^\wedge$ is faithful if and only if there exists an element $(Y, Z)$ of the kernel of the boundary with $\text{Tr}(Y) - \text{Tr}(Z) = 1$. If you are confused about the signs, dear reader, then I suggest you let $k$ have characteristic $2$. OK, but now let's look at the leading terms... say
$$
Y = \left(
\begin{matrix}
y_1 & y_2 \\
y_3 & y_4
\end{matrix}
\right)
+ h.o.t
\quad\text{and}\quad
Z = \left(
\begin{matrix}
z_1 & z_2 \\
z_3 & z_4
\end{matrix}
\right)
+ h.o.t
$$
with $y_i, z_i \in k$. Multiplying out we get
$$
AY - ZA =
\left(
\begin{matrix}
t_1(y_1 - z_1) + t_2 y_3 - t_3 z_2 & ... \\
... & ...
\end{matrix}
\right)
+ \text{terms of degree} > 17
$$
OK, so for this to be zero we see that we have $y_3 = z_2 = 0$ and $y_1 = z_1$ because the monomials $t_1, t_2, t_3, t_4$ are $k$-linearly independent. My millennia long experience with these types of things now suggests that working out the other dots we find that similarly $y_4 = z_4$. However, this contradicts the assumption that $\text{Tr}(Y) - \text{Tr}(Z) = 1$.<|endoftext|>
TITLE: The norm of a non-Galois extension of local fields
QUESTION [6 upvotes]: Background and Motivation
Local Class Field Theory says that abelian extensions of a finite extension $K/\mathbb{Q}_p$ are parametrized by the open subgroups of finite index in $K^\times$.  The correspondence takes an abelian extension $L/K$ and sends it to $N_{L/K}(L^\times)$, and this correspondence is bijective.
If one starts instead with a galois extension $L/K$ that isn't abelian, one can then ask "What abelian extension does $N_{L/K}(L^\times)$ correspond to?"  The answer is the maximal abelian extension of $K$ contained in $L$.

The hypothesis of being galois isn't necessary in the statement of the non-abelian theorem: both the question and the answer still make sense.  I am thus asking

Assume that $L/K$ as above.  Is the abelain extension of $K$ corresponding to $N_{L/K}(L^\times)$ the maximal abelian extension of $K$ contained in $L$?

A couple examples to illustrate this problem (including the example that I was told would sink this):
If $p > 2$, then consider $L = \mathbb{Q}_p(\sqrt[p]{p})$.  The norm subgroup that I am anticipating is all of $\mathbb{Q}_p^\times$.  Moving into the galois closure and using the theorem, one gets that $N_{L/\mathbb{Q}_p}(L^\times)$ contains $p^\mathbb{Z} \times (1 + p\mathbb{Z}_p)$.  Moreover, the norm of a number $a \in \mathbb{Z}_p$ will just be $a^p$, which is congruent to $a$ mod $p$, so the norm will also hit something congruent to any given root of unity in $\mathbb{Q}_p$, and that was all that I was missing from the earlier note.
One can also, using the same idea (moving into the galois closure and getting a lot of information from that, and then using the explicit structure of the field that one started with) show that this works for $K(\sqrt[n]p)$ with $(n, p^{f(K/\mathbb{Q}_p)}) = (n,p) = 1$ as well.

REPLY [6 votes]: As an addition to the answers given above: these theorems were first formulated (and proved) by Arnold Scholz in

 Die Abgrenzungssätze für Kreiskörper und Klassenkörper
(Limitation theorems for cyclotomic fields and class fields), 
Sitzungsberichte Akad. Berlin 1931, 417 - 426<|endoftext|>
TITLE: Definition of an algebra over a noncommutative ring
QUESTION [26 upvotes]: I've tried in vain to find a definition of an algebra over a noncommutative ring. Does this algebraic structure not exist? In particular, does the following definition from http://en.wikipedia.org/wiki/Algebra_(ring_theory) make sense for noncommutative $R$?

Let $R$ be a commutative ring.   An algebra is an $R$-module $A$ together with a binary operation
  $$ [\cdot,\cdot]: A\times A\to A $$
  called $A$-multiplication, which satisfies the following axiom:
  $$ [a x + b y, z] = a [x, z] + b [y, z], \quad  [z, a x + b y] = a[z, x] + b [z, y] $$
  for all scalars $a$, $b$ in $R$ and all elements $x$, $y$, $z$ in $A$.

So, is there a common notion of an algebra over a noncommutative ring?

REPLY [3 votes]: This is a follow-up to the answer by Zoran Skoda, applicable to the case where one requires "algebras" to be associative and have a unit.
I'd like to point out that a monoid $(A,\mu,\eta)$ internal to the monoidal category of $R$-bimodules is exactly the same thing as a ring $A$ along with a ring homomorphism $\eta:R\rightarrow A$. And a morphism $(A,\mu,\eta)\rightarrow (A',\mu',\eta')$ is exactly the same thing as a ring morphism $A\rightarrow A'$ that makes the triangle with $R$ commute.
In other words, this viewpoint can be greatly simplified if we just view an "algebra over $R$" as an object $\eta:R\rightarrow A$ in the category of rings under $R$. One can require that $\eta:R\rightarrow A$ have central image, or not, depending on one's needs.<|endoftext|>
TITLE: Jerome Franel's Chair at ETH
QUESTION [5 upvotes]: In a review of a book by Ferdinand Gonseth in the Spring-Summer 2006 (Volume XI, Issue 1) of the HOPOS Newsletter it is said that Gonseth was the 
"successor on Jerome's (sic) Franel's chair for Mathematics in French language at the ETH"
1) Was this a chair that Franel held or was it a chair Franel endowed?  If the former, does the chair have a name?
2) What is the nature of a chair "for Mathematics in the French language"?  Is it a history of French mathematics chair or is it a chair for a mathematician that writes in French?
Thanks for any insight.
Cheers, Scott

REPLY [6 votes]: Franel was a number theorist who gave introductory calculus lectures in French
at the ETH. Maybe the ETH offered courses in different languages, because of its location in a multilingual country. Does anybody know? Anyway, for a little more information, see this excerpt from Jerry Alexanderson's The Random Walks of George Polya.
[Added later] At this site there is a list of all professors in the history of the ETH, between 1855 and 2005.  Franel is not the only professor for "Mathematik in französischer Sprache" -- there were some before Franel, and others in more recent times were Beno Eckmann and Armand Borel. There were also professors of geometry, mechanics and statistics "in französischer Sprache". So, Franel was not at all special in this respect, which strengthens my suspicion that the ETH made a point of offering courses in French.<|endoftext|>
TITLE: Origin of the term "localization" for the localization of a ring
QUESTION [23 upvotes]: I'm curious if the term localization in ring theory comes from algebraic geometry or not.  The connection between localization and "looking locally about a point" seems like it should be the source for the notion of localization.  It seems plausible, but it seems like we would have had to wait until Zariski defined the Zariski topology for the connection to become apparent.  That seems hard to believe given the amount of work done in commutative algebra before 20th century, especially given the importance of localization in commutative algebra.  
Then this raises the question: Where and when was the term 'localization' first used to describe the adjunction of inverses, and does it originate from algebraic geometry or from somewhere else?  Was the notion of localization used regularly with a different name before it was given this name?

REPLY [8 votes]: I don't remember the history too well, but the answers above perfectly fit one of my favorite quotes from V.I. Arnold on this very question that illustrates the gulf between (1) axiomatic training and (2) hands-on approach.

 there's the obvious stupid answer: on an affine scheme, restriction to distinguished open sets corresponds to localization of the ring. It seems rather clear that localization is a good name for this, especially since you can look at smaller and smaller open sets around a point.  (Ilya Grigoriev)
 If $M$  is a manifold and $x\in M$ , then the ring of smooth germs in $x$ is canonically isomorphic to the localization  […]  Geometric localization is expressed algebraically in introducing inverses for functions for which it makes sense, thus you can call it localization.  (Martin Brandenburg)


Студенты высшей нормальной школы в Париже спросили меня: "Почему вы называете кольцо формальных степенных рядов локальным? Разве оно удовлетворяет аксиомам локального кольца?" Для неспециалистов поясню, что заданный вопрос аналогичен вопросу "Почему вы называете окружность коническим сечением?" Это были лучшие студенты-математики Франции. По-видимому, какой-то преступный алгебраист обучил их аксиомам колец (и даже локальных колец), не приводя ни одного примера (и, в частности, не объяснив происхождение термина "локальное").

(В.И. Арнольд, Топологические проблемы теории распространения волн, УМН, т.51, вып.1 (307), 1996, с.5)
Here is my rather literal translation:

Students from École Normale Supérieure, Paris asked me: "Why are you referring to the formal power ring as local? Does it really satisfy axioms of a local ring?" Let me remark for the non-experts that their question is analogous to the question: "Why do you call the cirlce a conic section?" Those were the best mathematics students in France. Apparently, some criminal algebraist taught them ring axioms (and even local ring axioms) without giving a single example (and, in particular, without explaining the origin of the term "local").

(V.I. Arnold,  Topological problems in the theory of wave propagation, Russian Math Surveys, 51:1, 1996)<|endoftext|>
TITLE: When can a freely moving sphere escape from a 'cage' defined by a set of impassible coordinates?
QUESTION [10 upvotes]: To ask this question in a (hopefully) more direct way:
Please imagine that I take a freely moving ball in 3-space and create a 'cage' around it by defining a set of impassible coordinates, $S_c$ (i.e. points in 3-space that no part of the diffusing ball is allowed to overlap).  These points reside within the volume, $V_{cage}$, of some larger sphere, where $V_{cage}$ >> $V_{ball}$.  Provided the set of impassible coordinates, $S_c$, is there a computationally efficient and/or nice way to determine if the ball can ever escape the cage?   

Earlier version of question:
In Pachinko one shoots a small metal ball into a forest of pins, then gravity then pulls it downwards so that it will either fall into a pocket (where you win a prize) or the sink at the bottom of the machine.  The spacing and distribution of the pins will help to insure that one only wins certain prizes with low probability.
Now imagine that we have a more general game where:
(1) - The ball is simply diffusing in 3-space (like a molecule undergoing Brownian motion).  I.e. there is no fixed downward trajectory due to gravity.  
(2) - You win a prize if the ball diffuses over a particular coordinate, just like one of the pockets in regular pachinko.
(3) - We generalize he pins as a set of impassible coordinates.
(4) - We define a 'sink' as an always accessible coordinate.
(5) - We define a starting coordinate for the sphere.
Given access the 3-space coordinates for (2), (3), (4), & (5), what's the most efficient way to find whether the game is 'winnable', or if the ball will fall into the 'sink' with a probability of unity?  How can we find the minimum set from (3) that prevents the ball from reaching the pocket?

REPLY [22 votes]: Replace the pins by balls of radius $R_{ball}$ and the ball by a point. This is a logically equivalent formulation. The question, then, is: given a finite set of balls, $B_1$, $B_2$, ...., $B_k$ in $\mathbb{R}^n$, and a point $x$, how to determine where $x$ is in the unbounded component of $\mathbb{R}^n \setminus \bigcup B_i$.
I don't know the answer to this, but here is an easy way to compute the number of connected components of $\mathbb{R}^n \setminus \bigcup B_i$. In other words, I can determine whether there is some place from which a ball cannot escape. 
By Alexander duality, the number of bounded components of $\mathbb{R}^n \setminus \bigcup B_i$ this is the dimension of $H_n(\bigcup B_i)$. 
Cover $\bigcup B_i$ by the $B_i$. Every intersection of finitely many $B_i$ is convex, hence contractible. So $\bigcup B_i$ is homotopic to the nerve of this cover. That is a simplicial complex, so it is easy to compute its homology.
One final practical idea: I have used painting software where I could click on a point and it would color every point which was connected to that one. Maybe the algorithms used to make that software could solve this problem as well?

REPLY [5 votes]: I recently heard a beautiful talk by Yuliy Baryshnikov on the general question of when an object can be pinned by some set of fixed points.  They consider arbitrary objects in 2D and prove the following theorem:

Let D be a planar domain.  Either one can pull a configuration C of two points $\{p_1,p_2\}$ around D, or there exists a full rotation of C entirely within D, that is a loop π′: $S^1$ → E (E being the Euclidean group of transformations) such that the vector $π′\circ p_1 − π′ \circ p_2$ turns around the origin (perhaps, several times). 

They use a topological approach which uses Mayer-Vietoris sequences in homology; apparently to generalize to 3D one must use Mayer-Vietoris spectral sequences, though this is "future work".
The slides are here and do include some discussion of computing the possibility of caging / linking effectively, but again, they focus on the 2D problem.<|endoftext|>
TITLE: What is the actual meaning of a fractional derivative?
QUESTION [43 upvotes]: We're all use to seeing differential operators of the form $\frac{d}{dx}^n$ where $n\in\mathbb{Z}$. But it has come to my attention that this generalises to all complex numbers, forming a field called fractional calculus which apparently even has applications in physics!
These derivatives are defined as fractional iterates. For example, $(\frac{d}{dx}^\frac{1}{2})^2 = \frac{d}{dx}$ or $(\frac{d}{dx}^i)^i = \frac{d}{dx}^{-1}$
But I can't seem to find a more meaningful definition or description. The derivative means something to me; these just have very abstract definitions. Any help?

REPLY [4 votes]: Probabilistically, you can give a perfectly clear meaning to many fractional derivatives.
I will look at definitions of fractional\nonlocal derivatives that are Markovian generators of stochastic processes with jumps. I hope to convince the reader that

Different definitions arise naturally,
there is a clear interpretation of many properties (like nonlocality or killing/not-killing constants), and
generalizations are natural and meaningful for applications.

It is useful to look at the most simple stochastic  jump process and its corresponding generator.  Take a Markov chain $P=\{p_{i,j}\}_{i,j\in \text{State space}}$ (which is intrinsically jumpy)  and write out its generator
$$
\mathcal G f(x):=(P-I)f(x)=\sum_{y\in\text{ State space}}(f(y)-f(x))p_{x,y},\quad x\in\text{ State space}.
$$
Here the intuition is clear: the infinitesimal jump (working with unit time in this case) from $x$ to $y$ is assigned intensity/probability $p_{x,y}$. The operator $\mathcal G$ is non-local. If we modify the process (impose boundary conditions), say by forcing the process to be absorbed at $a\in\text{ State space}$ once it tries to jump to a state $y\notin \Omega\subset \text{State space},$ we obtain a new generator
$$
\mathcal G^{\text{abs}} f(x):=(P^{\text{abs}}-I)f(x)=\sum_{y\in\Omega}(f(y)-f(x))p_{x,y}+(f(a)-f(x))\sum_{y\notin\Omega}p_{x,y},\quad x\in\Omega.
$$
If we instead decide to kill it (by testing against functions with $f(a)=0$, for example), the new generator will be
$$
\mathcal G^{\text{kill}} f(x):=(P^{\text{kill}}-I)f(x)=\sum_{y\in\Omega}(f(y)-f(x))p_{x,y}-f(x)\sum_{y\notin\Omega}p_{x,y},\quad x\in\Omega.
$$
So from one single process we can obtain many different generators/fractional derivative (as mentioned in a comment above, the boundary conditions are reflected in the representation of the operator away from the boundary due to the non-locality of $\mathcal G$).
Let us now move to the Riemann-Liouville and Caputo derivatives of order $\beta\in(0,1)$. Consider the  three fractional derivatives for $x<a$
\begin{align}
D^{\beta}_{\infty}f(x)&:= \int_0^{\infty}(f(x+y)-f(x))\nu(y)dy, \\
^{C}D^{\beta}_a f(x):&= \int_0^{a-x}(f(x+y)-f(x))\nu(y)dy &+(f(a)-f(x))\int_{a-x}^\infty\nu(y)dy,\\
^{RL}D^{\beta}_af(x)&:= \int_0^{a-x}(f(x+y)-f(x))\nu(y)dy &-f(x)\int_{a-x}^\infty\nu(y)dy,
\end{align}
where $\nu(y):=\frac{-\Gamma(-\beta)^{-1}}{y^{1+\beta}}$. Similarly as for the  Markov chain above: the operator $D^{\beta}_{\infty}$ is the generator of a $\beta$-stable subordinator $X^\beta(s)$, the operator $^{C}D^{\beta}_a$ is the generator of a $\beta$-stable subordinator $X^\beta(s)$ absorbed at $\{a\}$ on the first attempt to jump outside $\Omega:=(-\infty,a)$, and the operator $^{RL}D^{\beta}_a$ is the generator of a $\beta$-stable subordinator $X^\beta(s)$ killed on the first attempt to jump outside $\Omega:=(-\infty,a)$.
Integrating by parts we can rewrite the three operators above in their Riemann-Liouville integral representation, namely
\begin{align}
D^{\beta}_{\infty}f(x)&= \int_x^{\infty}f'(y)\frac{(y-x)^{-\beta}}{\Gamma(1-\beta)}dy \\
^{C}D^{\beta}_a f(x)&= \int_x^{a}f'(y)\frac{(y-x)^{-\beta}}{\Gamma(1-\beta)}dy,\\
^{RL}D^{\beta}_af(x)&= \frac{d}{dx}\int_x^{a}f(y)\frac{(y-x)^{-\beta}}{\Gamma(1-\beta)}dy,
\end{align}
where the last two operators are your standard definitions of Caputo and Riemann-Liouvile derivatives (right and left versions will correspond to the processes $X^\beta(s)$ and $-X^{\beta}(s)$ respectively).  We can now say that the Caputo derivative $^{C}D^{\beta}_a$ (Riemann-Liouville derivative  $^{RL}D^{\beta}_a$) kills (does not kill) constants  as it is the generator of a process (killed process). Again you can see that (naturally) $^{C}D^{\beta}_a$ and $^{RL}D^{\beta}_a$ contain boundary information in their representation away from the boundary (in sharp difference with local differential operators).
Some references: Caputo, Riemann-Liouville, and Grünwald-Leitnikov derivatives from a stochastic point of view in this book. Reflecting boundary conditions and other options for Caputo derivatives of order $\beta\in(1,2)$ here and here.
By substituting a general Lévy measure $\nu(x,dy)$ in the formulas above (generalizing fractional derivatives), many meaningful stochastic processes and their versions on a bounded domain can be studied through their generators (see book, article ). Similar arguments can be carried over for some fractional Laplacians (see this book for example).<|endoftext|>
TITLE: abelian sylow-p-subgroups
QUESTION [5 upvotes]: Can anyone help me with this problem?
if $G$ has abelian Sylow-p-subgroups, prove that $p$ does not divide the order of $G'\cap Z(G)$, 
where $G'$ and $Z(G)$ are as usual, the subgrup generated by the set of all commutators and the 
center, respectively.
thanks a lot. :D

REPLY [7 votes]: There is also a character-theoretic argument.  Suppose $G' \cap Z(G)$ has a subgroup $U$ of order $p$. We want a contradiction. Let $\lambda$ be a nonprinciipal linear character of $U$. Since $U \subseteq P$ and $P$ is abelian, $\lambda$ has an extension to $\mu$, a linear character of $P$. The induced character $\mu^G$ has degree $|G:U|$, which is prime to $p$, so some irreducible constituent $\chi$ of $\mu^G$ has degree not divisible by $p$. Then $\mu$ is a constituent of the restriction $\chi_P$ by Frobenius reciprocity, and thus $\lambda$ is a constituent of $\chi_U$. But $U$ is central, so $\chi_U = \chi(1)\lambda$. Now let $\sigma$ be the linear character det$(\chi)$. Then $\sigma_U = \lambda^{\chi(1)}$, which is nontrivial since $p$ does not divide $\chi(1)$. This is a contradiction, however, since $U \subseteq G' \subseteq {\rm ker}(\sigma)$.  [Note that transfer proofs can often be replaced  by arguments using the determinant of a character.]<|endoftext|>
TITLE: How to formally -- and cleanly -- express relationships of limits of diagrams?
QUESTION [8 upvotes]: Requesting: a good reference for formal manipulation of limits of diagrams, with respect to maps of index diagrams.
As an example, consider the following result, for some "nice enough" category C (say, Top), there is a natural isomorphism
$$(A \times_B C) \times_C D \cong A \times_B D.$$
This is a nice, intuitively true result that if we stick two pullback squares next to each other, we get another pullback square. I can easily convince myself of this on paper, chasing around the requisite number of arrows -- and it's a simple enough result that no wizardry is required; all maps could be named and the necessary commutativity relations and existence/uniqueness conditions can be checked.
However, it's a bit unsatisfactory. There's a blow-up of notation required to prove a relatively simple result. What happens when the diagrams get more complicated? It becomes less clear how proving basic results about limits can be written up with an acceptable amount of rigor, while remaining concise.
A general question of this type: let $C$ be a complete category, and let $D,E$ be small categories. A functor $f : D \to E$ induces a functor $f^* : C^E \to C^D$. For which $f$ is it the case that $\lim_E = \lim_D \circ f^*$ as functors $C^E \to C$?
It is at least intuitively clear that if $f(D)$ sufficiently "sits above" $E$, then this will hold. I am interested in formalizing this intuition.
I'm sure this question must have a standard answer in category theory, but a sufficient reference escapes me at the moment.

REPLY [3 votes]: As was mentioned in the comments, such general isomorphisms can be reduced by Yoneda to the case of sets, and in your example $((a,c),d) \mapsto (a,d)$ is an isomorphism, simply because $c$ is already determined by $d$. As for me, I always manipulate arrows without caring about single-use names and use Yoneda. This worked very good in the past years. For example, for every test object $T \in C$, we have canonical bijections
$\{T \to (A \times_B C) \times_C D\} = \{T \to A \times_B C \to C = T \to D \to C\}$
$= \{T \to A \to B = T \to C \to B, T \to C = T \to D \to C\}$
$=\{T \to A \to B = T \to D \to C \to B\} = \{T \to A \times_B D\}$
You may argue that it's not clear at all what is given etc., but there is only one plausible interpretation: Everything not given before belongs to the data of the morphism sets. Every equation is a condition on this data. It's even nicer to draw everything in commutative diagrams (I don't know how to draw them here). This method also works when you want to simplify a universal object without knowing the result. You just reduce the diagrams as above. Every step is almost forced. Of course, this is just another way of writing down the proof mentioned first.
Regarding your second question: Your intuition is absolutely correct. A sufficient and useful condition is that $f$ is a final functor. You can read about them in Mac Lane, Categories for the working mathematician, IX.3.<|endoftext|>
TITLE: What can you do with a compact moduli space?
QUESTION [30 upvotes]: So sometime ago in my math education I discovered that many mathematicians were interested in moduli problems.  Not long after I got the sense that when mathematicians ran across a non compact moduli they would really like to compactify it.  
My question is, why are people so eager to compactify things?  I know compactness is a great property of a space to have because it often makes other results much easier to prove.  So I think my question is better stated as: what are some examples of nice/good/cool results related to a moduli spaces that were (only) able to be proved once there was a compactification of the space?

REPLY [18 votes]: The answers here are all excellent examples of things that can only be proved once a moduli space is compactified. I would like to add a perhaps more basic reason for compactifying moduli spaces, involving something simpler than theoretical applications such as defining enumerative invariants. The moral is the following: 

If you study families of geometric objects then either you are almost certain to encounter the boundary of the moduli space, or you must have some very good reason to rule it out. 

For example, to find a non-trivial compact family of smooth complex curves is actually quite awkward and such families are very rare. (The first examples were due to Atiyah and Kodaira.) From this point of view the "ubiquity of the compactification" amounts to the fact that the boundary divisor of singular curves in the compactified moduli space is positive in a certain sense, so it intersects almost all curves in the moduli space. It is this positivity of the boundary which forces us to study it!
Some more examples explain - I hope! - the way compactification enters when considering pseudoholomorphic curves as in Gromov-Witten theory, without ever coming close to trying to define an enumerative invariant. Just by looking at a conic in $\mathbb{CP}^2$, which degenerates into two lines, one sees that when moving a pseudoholomorphic curve around, one is almost certain to encounter bubbling, unless one has a very good reason to know otherwise. Understanding how to compactify the moduli space, we see that this bubbling phenomenon is the main thing which can go wrong. What is interesting here is that often one tries to prove this compactification is not actually necessary, by ruling out bubbling somehow. Two examples follow - taken from Gromov's original use of pseudoholomorphic curves in his Inventiones paper - which exploit this idea. 
Firstly, Gromov's proof of his non-squeezing theorem. Here the key point in the argument is that one can find a certain pseudoholomorphic disc for a standard almost complex structure on $\mathbb{C}^n$. One would like to know that as one deforms the almost complex structure the disc persists so that we have such a disc for a special non-standard almost complex structure. It is standard in this kind of "continuity method" that you can always deform the disc for a little while because the problem is elliptic. But to push the deformation indefinitely you need to show compactness - why doesn't the disc break up? Thanks to our knowledge of the compactification of the moduli space, we understand that the only thing that can go wrong is bubbling and in this case bubbles cannot form because the symplectic structure is exact.
The second example is of the following type: suppose one knows the existence of one pseudoholomorphic curve in a symplectic manifold; then one can try and use it to investigate the ambient space, moving it around and trying to sweep out as much of the space as possible. In this way you can prove, for example, that any symplectic structure on $\mathbb{CP}^2$ which admits a symplectic sphere with self-intersection 1 must be the standard symplectic structure. The reason is you can find an almost complex structure which makes this sphere a pseudoholomorphic curve. Then you move the curve around until is sweeps out the whole space, doing it carefully enough to give a symplectomorphism with the standard $\mathbb{CP}^2$. Here you can push the curve wherever you want because it wont break. Bubbles can't form because the curve has symplectic area 1 and so there is no "spare area" to make bubbles with.

REPLY [8 votes]: Thurston's compactification of Teichmueller space gives an elegant proof of the Nielsen--Thurston classification of surface automorphisms.  The compactified Teichmueller space of the surface of genus g is homeomorphic to a ((6g-6)-dimensional) ball, on which an automorphism $\phi$ acts.  By Brouwer's Fixed Point Theorem, $\phi$ fixes a point $x$.

If $x$ is an interior point then $\phi$ is an isometry of some hyperbolic structure, and it's easy to deduce that $\phi$ is periodic.
If $x$ is a point on the boundary then it corresponds to some measured lamination.  If the lamination has a closed leaf then $x$ is reducible.
Finally, if the lamination has no closed leaves then $\phi$ is pseudo-Anosov.

Admittedly, it is possible to give a proof of the classification that doesn't use the compactification.  But why would you want to?<|endoftext|>
TITLE: The closure of a generic ultrapower
QUESTION [5 upvotes]: Let $I$ be a normal ideal on $P_{\kappa} (\lambda)$. Let $V$ denote our ground model. Now we force with the $I$-positive sets, and if $G$ is the resulting generic filter, it can be shown that $G$ is a $V$-ultrafilter on $P_{\kappa} (\lambda)$ extending the dual filter of $I$, which is normal if $I$ is normal.
We can build now the so called generic ultrapower, i.e. we construct inside $V[G]$,  $Ult_{G} (V)$ the class of all functions in $V$ with domain $\kappa$ and the usual binary relation $\in^{\ast}$. We assume that the generic ultrapower is well-founded, so we identify it with its transitive collaps $M \cong Ult_{G} (V)$
My question now is: Is this $M$ closed under sequences from $V[G]$ of length $\lambda$, i.e. $M^{\lambda} \cap V[G] = M^{\lambda} \cap M$?
$M$ is closed under sequences from $V$ of length $\lambda$, this is clear to me but I don't have a good argument for the sequences of $V[G]$.

REPLY [5 votes]: If your ideal is normal, fine, precipitous and has the disjointing property (a consequence of saturation), then the answer is yes. As you likely know, you need more assumptions than you had stated, just in order to know that the ultrapower is well-founded. The difference in closure for the ultrapower that you mentioned appears to be related to the difference between having the disjointing property or not.
For a reference, I recommend Matt Foreman's chapter for the Handbook of Set Theory, which  states the following theorem (it is Theorem 2.25 in the preliminary version I have here, but the published number may differ). 
Theorem. Suppose I is a normal, fine, precipitous ideal on $Z\subset P(X)$, where $|X|=\lambda$. Let $G\subset P(Z)/I$ be generic, and $M$ the generic ultrapower of $V$ by $G$. Then $P(\lambda)\cap V\subset M$. Further, if $I$ has the disjointing property, then $M^\lambda\cap V[G]\subset M$. 
Note that this theorem covers your case of $Z=P_\kappa(\lambda)$. 
To prove the first part, you simply observe that $[id]$ represents $j " \lambda$, and then for any $A\subset\lambda$ you can get $j"A$ using the function $g(z)=z\cap A$. Now, from $j"\lambda$ and $j"A$ you can easily build $A$ in $M$. 
For the second part, the part you were interested in, you use the disjointing property in order to know that a term for a $\lambda$-sequence of elements of $M$ can be transformed into a $\lambda$-sequence of terms in $M$. That is, if $\langle\dot a_\alpha :\alpha<\lambda\rangle$ is a $\lambda$-sequence of terms for objects in $M$, then disjointing allows us to find in $V$ a sequence of functions $\vec g = \langle g_\alpha: \alpha<\lambda\rangle$ such that $[g_\alpha]^G = \dot a_\alpha^G$. From this, it follows that the function $g(z) = \langle g_\alpha(z) | \alpha\in z\rangle$ represents $j(\vec g)(j"\lambda)$, which is $\langle j(g_\alpha)_\beta(j"\lambda) | \beta\in j"\lambda\rangle$, from which we can construct $\langle j(g_\alpha)(j"\lambda) | \alpha <\lambda\rangle$, which is the desired $\lambda$-sequence.<|endoftext|>
TITLE: Why is Kleene's notion of computability better than Banach-Mazur's?
QUESTION [15 upvotes]: In this post about the difference between the recursive and effective topos, Andrej Bauer said:

If you are looking for a deeper explanation, then perhaps it is fair to say that the Recursive Topos models computability a la Banach-Mazur (a map is computable if it takes computable sequences to computable sequences) and the Effective topos models computability a la Kleene (a map is computable if it is realized by a Turing machine). In many respects Kleene's notion of computability is better, but you'll have to ask another question to find out why :-)

So I'm asking: 
1) What is "computability a la Banach-Mazur"? I would guess it has something to do with Baire spaces and computable analysis, but I don't really know.
2) Why is Kleene's notion of computability better?

REPLY [3 votes]: Jacques asks about a "kernel" idea for the three counter-examples. I am not sure I can answer the question well. The most general result, namely the third one, is really a consequence of Friedberg's result. Peter Hertling's construction is an adaptation of Friedberg's result (although not an obvious one by any means), too. So the question is, what did Friedberg do? I must admit I cannot explain this succinctly, perhaps it's best if you have a look at Section 9 of

P. Hertling, "A Banach–Mazur computable but not Markov computable function on the computable real numbers", Annals of Pure and Applied Logic 132 (2005) 227–246.

Suffice it to say that the construction involves showing that a certain set of indices of B-M computable functionals is a $\Pi^0_4$-set, but the corresponding set of indices of Markov computable functionals is $\Sigma^0_4$-complete and contained in the first set, therefore the two sets differ. Now if there is anyone around who can explain in simple terms what $\Sigma^0_4$-complete sets are like, I would love to hear it.<|endoftext|>
TITLE: A-infinity structure on the ribbon graph complex and more general graph complexes
QUESTION [12 upvotes]: Moduli spaces of curves (with nonempty boundary or at least one marked point) admit cell decompositions in which the cells are labelled by ribbon graphs.  In fact, the moduli space of normalised metric ribbon graphs with $n$ boundary components is homeomorphic to the product of the moduli space of curves and an $(n-1)$-simplex.  
From this geometric statement one can show that the ribbon graph complex (a chain complex spanned by ribbon graphs, with the differential given by summing over all edge contractions) computes the cohomology of moduli spaces of curves.
My question is: does anyone know how to describe the cochain level cup product structure on the ribbon graph complex?  By general machinery, if a complex computes the cohomology of a space then it carries an $A_\infty$ structure for which it is equivalent to the cochains on the space with their usual $A_\infty$ structure.
I would like to know if there is a way to write down a combinatorial formula for this $A_\infty$ structure on the ribbon graph complex.
Similarly, the Lie graph complex (in which vertices are labelled by words in a generic Lie algebra) computes the cohomology of the spaces $BOut(F_n)$ (the classifying spaces of outer automorphism groups of free groups).  Is there a way to describe the resulting $A_\infty$ structure on the Lie graph complex?

REPLY [3 votes]: This is basically an expansion on Ben Cooper's answer.
As far as I know (I'm very happy for anyone to correct me...) this is what's known so far:

In Graph cohomology and Kontsevich cycles, Igusa introduces a certain category $Fat$. The objects of this category are ribbon graphs and the morphisms are edge contraction. He then proves (Theorem 1.22) that there is a rational equivalence
$$ \phi : C_*(Fat) \to \mathcal{G}_* $$
between the chain complex of the nerve of $Fat$ and the ribbon graph complex. Now, like any simplicial complex, there is a canonical diagonal on $C_*(Fat)$, namely the Alexander-Whitney diagonal which gives a description of the cup product.
It is well-known that $\mathcal{G}_*$ with rational coefficients computes $H_*(\mathcal{M};\mathbb{Q})$ where
$$\mathcal{M} := \coprod \mathcal{M}_{g,n}$$
the union of moduli spaces of smooth Riemann surfaces of genus $g$ and $n$ unlabeled boundary components.
Given a cyclic $A_\infty$-algebra, with finite dimensional cohomology, inner product is even and symmetric, Kontsevich shows in Feynman diagrams and low-dimensional topology how to associate to it a ''characteristic cocycle" $c_A$ in the graph complex (hence a characteristic class in $H_*(\mathcal{M};\mathbb{Q})$ as well). A good review of this construction is
Charcteristic classes of $A_\infty$-algebras by Lazarev and Hamilton.
Since there is such a strong connection between the moduli of curves and the $A_\infty$-operad (see e.g. Are there graph models for other moduli spaces?), we might try and look into diagonals in the latter. In the context of $A_\infty$-algebras there are (at least?) 3 different constructions of diagonals: Saneblidze and Umble, Markl and Shnider, and Loday. Maybe there are homotopic but (as far as I know?) nobody proved that. They are also quite painful to work with directly.
However, in the context of cyclic $A_\infty$-algebras, the situation is different. Lino Amorim and Junwu Tu proved in Tensor products of cyclic $A_\infty$-algebras and their Kontsevich cycles that there is a unique diagonal up to cyclic homotopy. Their diagonal and the Alexander-Whitney diagonal from 1. are homotopic (thus providing an explicit formula as you requested). This diagonal has an explicit combinatorical description, which allows you to write it (at least the initial terms), but it is a very complicated formula.<|endoftext|>
TITLE: Locally complete space is topologically equivalent to a complete space
QUESTION [8 upvotes]: Can someone please tell me where I can find a citeable reference for the following result:
Call the metric space $(X, d)$ "locally complete" if for every $x \in X$ there a neighbourhood of $x$ which is complete under $d$.
If $(X,d)$ is locally complete and separable then there exists a metric $d'$ on $X$ such that $(X,d) \to (X,d')$ is a homeomorphism and $(X, d')$ is complete.
This result follows immediately from Alexandrov's theorem that a $G_\delta$ subset of a Polish space is Polish, but I'd rather find a statement in the literature of the above more straightforward result if there is one.

REPLY [3 votes]: For a reference: this paper has a reference [30] that has a proof. The author cites your result and refers to it. I don't have access to these papers, so I cannot verify exactly.<|endoftext|>
TITLE: Can the circle be characterized by the following property?
QUESTION [13 upvotes]: In the Euclidean plane, is the circle the only simple closed curve that has an axis of symmetry in every 
direction?

REPLY [18 votes]: A slightly different argument is as follows. Choose two symmetries $\sigma,\tau$
with axes
intersecting at a point $P$ and forming an angle of $2\pi \lambda$ with $\lambda$ irrational.
The composition $\rho=\sigma\circ \tau$ is then a rotation of infinite order generating
a dense subgroup of the group of all rotations centered at $P$.
Any closed subset left invariant under $\rho$ is thus a union of concentric circles centered at $P$. A simple closed curve invariant under $\rho$ is thus such a circle.<|endoftext|>
TITLE: Stacks vs. Groupoids
QUESTION [10 upvotes]: I'm not an algebraic geometer, but I've been doing a little armchair reading on stacks just to see what all the fuss is about.  It appears that stacks are used in algebra in a similar way that groupoids are used in analysis, e.g. to handle pathological group actions and moduli spaces geometrically.  I'm wondering if the language of groupoids is also available in algebraic geometry (it seems like it should be) and, if so, what some of the technical advantages and disadvantages it has in comparison to stacks.
Thanks!

REPLY [11 votes]: I will try to answer this question in a way relevant to more than one field, however, to be honest, I'm rather unconventional in the sense that my experience in this area stems from topological and differentiable stacks rather than algebriaic ones. However, from a formal view point, everything is the same.
So let us work in a "background Grothendieck site", which can be topological spaces, differentiable manifolds, or schemes over a fixed base (the first two with the "open cover topology"). Let's call an object in category a "space".
If G is a group object, and X is a space with an action of G, we can take the corase qoutient. However, this is generally not a "nice space" in the senes that the quotient loses a lot of information about the action. In the context of topological spaces, a "nice quotient" would be one that makes the map $X \to X/G$ into a principal G-bundle. However, you need some really nice conditions on the action for this to work in general. E.g., the action needs to be free.
Note, if we consider the projection $X \to X/G$ "coming from the left and the top" and take the pullback, the action is free if and only if the pullback is $G \times X$.
Now, from G acting on X, we can construct the so-called "action groupoid", which has objects X, and arrows $G \times X$, where $(g,x):x \to gx$. This is a groupoid object in spaces, denote it by Act_G(X). Given a space T, we can pretend it's a groupoid object, with all idenity arrows.  We can consider Hom(T,Act_G(X)), where the Hom is taken in the 2-category of groupoid objecs, hence, this Hom gives a groupoid, not just a set (the 2-cells are internal natural transormations). The assignment $T \mapsto Hom(T,Act_G(X))$ defines a presheaf of groupoid on spaces. Moreover, there is a canonical morphism $X \to Hom(Blank,Act_G(X))$ of presheaves of groupoids (where X is identified with its representable presheaf). If form a weak 2-pullback by having this morphism "coming from the left and the top", the pullback becomes $G \times X$, one projetion becomes the "source map" and another the "target map". If we say that $Hom(Blank,Act_G(X))$ is our new qoutient, then "the action becomes weakly free".
So far, everything I did was using groupoids. So where to stacks enter the game? Well, $Hom(Blank,Act_G(X))$ is not a very good quotient because if Y is another space, maps from Y to it don't see $Hom(Blank,Act_G(X))$ as "being like a space". E.g. if we are in topological spaces, we can't define maps from Y into it by defining them on the opens of Y in a way that agrees. (For more explanation see my answer to Stacks in the Zariski topology?). What we have to do is "stackify" the presheaf of groupoids $Hom(Blank,Act_G(X))$, (call its stackification X//G). This makes  X//G behaves like a spacein the sense that, e.g. in topological spaces, we can defined maps into it by mapping out of opens in a way that agrees. Since stackification preserves finite weak 2-limits, if we form the same pullback diagram but insetad with respect to $X \to X//G$, we still recover the action grouoid and the action is still "weakly free". Morevoer, the projection $X \to X//G$ becomes a G-torosor (principal G-bundle).
So, just using the groupoid, allowed us to keep track of the isotropy data, but not in a way that we get something like a space. For that we need stacks.
If instead of using the groupoid $Act_G(X)$, we used any groupoid object, we can still stackify its associated presheaf of groupoids. The stacks we get in this way are "geometrical", and give rise to topological, differentiable, and Artin stacks respectively.
A final remark. In the comments, it was said that in some sense groupoids are "atlases for stacks". To see this, let's go to manifolds. Given a manifold M described in terms of an atlas, we can construct a Lie groupoid whose objects are the disjoint union of the elements of the atlas, and where we have an arrows from (x,U_a) to (x,U_b) whenever x is in the intersection of these two. This Lie groupoid's associated stack is the same as the manifold M. More generally, given an orbifold described in terms of charts, we can also construct a Lie groupoid with respect to these charts, and its associated stack "represents the orbifold". In general, you can think of Lie groupoids as "generalized atlases" which describe the geometric object which is their associated stack. Of course, just as a manifold can be described by more than one atlas, a differenitbale stack can be described by more than one Lie groupoid.<|endoftext|>
TITLE: A question about the proof of Mostow rigidity
QUESTION [18 upvotes]: I have recently been studying a proof of Mostow rigidity (along the lines of Mostow's original argument), and I'm left a little confused about something.  We start with an isomorphism $\alpha: \Gamma \to \Gamma'$ between cocompact lattices in $\mathbb{H}^n$, $n \geq 3$, and observe that such a map lifts to a quasi-isometry $\phi: \mathbb{H}^n \to \mathbb{H}^n$.  We declare that two quasi-isometries are close if the pointwise distance between them is uniformly bounded, and we prove that the group $QI$ of quasi-isometries on $\mathbb{H}^n$ modulo closeness is isomorphic to the group $QC$ of quasi-conformal homoemorphisms of the boundary sphere.  With the aid of some detailed analysis of quasi-conformal maps and an ergodic theorem, we complete the proof by showing that the map on the boundary sphere induced by $\phi$ is actually conformal.
My question is about the part of the argument where we prove that $QI$ is isomorphic to $QC$.  To prove that every quasi-isometry on $\mathbb{H}^n$ extends to a homeomorphism of the boundary sphere, that this extension map is injective, and that its image lies in $QC$ involves only standard ideas in large scale geometry and hyperbolic geometry (mainly the fact that hyperbolic space is Gromov hyperbolic).  Surjectivity, on the other hand, is much harder; it basically involves a tricky compactness theorem in analysis.  Most of the references that I have used make a point to discuss this argument, but it is not clear to me why we even need surjectivity since in the proof of Mostow rigidity we show that an a-priori quasi-conformal map is secretly conformal and conformal maps are much more easily seen to be induced by honest isometries.  Could someone familiar with this argument explain why we need surjectivity (assuming we do)?
Thanks!

REPLY [10 votes]: As you say, $QI\cong QC$ is not necessary for the proof of Mostow rigidity. I'm not sure which reference you are referring to, but  Gromov's proof (which is popular among topologists) does not lead to this result. 
I suspect it is mentioned because there is a general program to try to classify quasi-isometry groups of $\delta$-hyperbolic spaces, and this is the motivating example. There is a nice discussion of how to extend quasiconformal isotopies of the sphere to quasi-isometries of hyperbolic space by "visual extension" in Appendix B of Curt McMullen's book "Renormalization and 3-manifolds which fiber over the circle".<|endoftext|>
TITLE: Does the exact pair phenomenon for partial orders occur in your area of mathematics? 
QUESTION [31 upvotes]: Suppose that I have a partial order P and an increasing sequence $a_0< a_1<a_2<\cdots$ of elements of $P$. A pair of elements (b,c) from P is said to be an exact pair for this sequence, if 

Both $b$ and $c$ are upper bounds for the sequence, so that $a_n<b$ and $a_n<c$ for every $n$, and  
Whenever $d\leq b$ and $d\leq c$ then $d\leq a_n$ for some $n$.  


           b     c
              :
              : 
              :
             a_2
             a_1
             a_0


More generally, an ideal $I$ in $P$ admits an exact pair $(b,c)$, if $I=\{\ a \mid a< b\text{ and } a<c\ \}$. 
Note that if a strictly increasing sequence has an exact pair, then the sequence cannot have a least upper bound, since such a bound would be below both b and c and therefore have to be below some $a_n$, and consequently not an upper bound of the sequence after all. Note also that if $b$ and $c$ form an exact pair for a strictly increasing sequence, then there can be no greatest lower bound for $\{b,c\}$, and so orders with such exact pairs are not lattices. 
The exact pair property arises in computability theory, because in the hierarchy of Turing degrees, every increasing sequence admits an exact pair.  This is one way of seeing that the set of Turing degrees is not a lattice. More generally, every countable ideal in the Turing degrees has an exact pair, and in the case of principal ideals, this implies that every Turing degree is the greatest lower bound of a pair of incomparable degrees. It also arises for certain hierarchies of complexity theory. 
The exact pair property is so beautifully structural, serving as an alternative to completeness, and for this reason I have always wondered whether it could have applications in other contexts, but I have only ever heard of it in connection with the computability degrees. Therefore,
My question is:  Does the exact pair phenomenon arise elsewhere in mathematics? Are there other natural orders studied outside logic that have the exact pair property?
Perhaps other natural hierarchies in mathematics exhibit the exact pair property? Or perhaps they do but this remains undiscovered...

REPLY [3 votes]: In trying to understand better a certain weak choice axiom, I once concocted an example in sheaf theory which involves an exact pair in the sense of this question, without knowing it was a "thing" elsewhere. 
In constructive mathematics, one choice principle which has been found useful is called the Presentation Axiom, Pax for short, which says roughly that any object $A$ (of one's given category) can be covered by an object $P$ which is projective: any further covering $Q \twoheadrightarrow P$ must split. You can imagine this applying for example to certain categories of sheaves of modules where one might want to consider projective resolutions. Or you can imagine it applying to certain categories of Heyting-valued sets or other toposes where the full axiom of choice is out of the question. 
In constructive mathematics, it is also known as "CoSHEP" (Category of Sets Has Enough Projectives); a more philosophical justification (quoting the nLab) is 

every set $A$ should have a ‘completely presented’ set of ‘canonical’ elements, that is elements given directly as they are constructed without regard for the equality relation imposed upon them. For canonical elements, equality is identity, so the BHK interpretation of logic justifies the axiom of choice for a completely presented set. This set is $P$, and $A$ is obtained from it as a quotient by the relation of equality on $A$. This argument can be made precise in many forms of type theory (including those of Martin-Löf and Thierry Coquand), which thus justify CoSHEP, much as they are widely known to justify dependent choice. 

Now, Pax as stated above (for let's say a topos $E$) was actually the external version: an object $P$ is externally projective if the external hom-functor $\hom(P, -): E \to Set$ takes epimorphisms to epimorphisms. There's also an internal version of Pax which is often more appropriate: if $(-)^P: E \to E$ denotes the internal hom-functor, then we say $P$ is internally projective if $(-)^P$ preserves epimorphisms, and internal Pax would say that every object $A$ is covered by an internally projective $P$. What I had wanted was an example of a topos which validates external Pax but not internal Pax. 
Presheaf toposes $E = Set^{C^{op}}$ (the category of functors $C^{op} \to Set$, where $C$ is a small category and $Set$ is just a vanilla category of sets in a ZFC background) always validate external Pax. It's easy to show that representable functors $\hom_C(-, a)$ are externally projective, as are coproducts of representable functors, and any object $F: C^{op} \to Set$ is covered by such a coproduct via a canonical map $\sum_{A \in Ob(C)} \sum_{x \in F A} hom_C(-, A) \twoheadrightarrow F$. So I wanted a $C$ which would force internal Pax to fail for $Set^{C^{op}}$. 
The example I came up with is where $C$ is a generic exact pair, i.e., the poset $\mathbb{N} \cup \{a, b\}$ where the natural numbers $\mathbb{N}$ is ordered as usual and $a, b$ are incomparable elements that dominate every element of $\mathbb{N}$. We consider $C$ as a 'thin' category as is customary for posets. I claim that $A = \hom_C(-, a)$, or indeed any object $P$ covering $A$, cannot be internally projective. Specifically, that such $(-)^P: E \to E$ cannot preserve an evident epimorphism $F \twoheadrightarrow G$ where $F(n) = \{m \in \mathbb{N}: m \geq n\}$ and $F(a), F(b)$ are empty, where the restriction maps are given by the inclusions $F(n+1) \hookrightarrow F(n)$, and where $G$ is the presheaf with $G(n) = \ast$ (a one-point set) and $G(a), G(b)$ are empty. 
To show the induced natural transformation $F^P \to G^P$ isn't an epimorphism, it suffices to show that the component $F^P(b) \to G^P(b)$ isn't a surjection between sets; in fact one may calculate that $F^P(b)$ is empty and $G^P(b)$ has exactly one element. Such calculations aren't exactly a spectator sport, so I'll wrap it up by making a few observations: 

For general $F$ one may identify $F^P(b)$ with $Nat(\hom_C(-, b) \times P, F)$, where the right side indicates a set of natural transformations. 
In the case where $P = \hom_C(-, a)$, we have in fact $\hom_C(-, b) \times \hom_C(-, a) \cong G$, and a natural transformation $G \to F$ boils down to having an element that belongs to the intersection of all the $F(n)$. Of course no such element exists since the intersection is empty. 
So $F^{\hom_C(-, a)}(b)$ is empty, and $G^{\hom_C(-, a)}(b)$ has an element (in fact just one). 
For general $P$ covering $\hom_C(-, a)$, there is a section $s: \hom_C(-, a) \to P$ (Yoneda lemma), and this section induces a map $F^P \to F^{\hom_C(-, a)}$, and thence a function $F^P(b) \to F^{\hom_C(-, a)}(b)$. The domain $F^P(b)$ has to be empty since the codomain is. On the other hand, $G^P(b)$ is nonempty because there is a map $\hom_C(-, b) \times P \to \hom_C(-, b) \times \hom_C(-, a) \cong G$.<|endoftext|>
TITLE: References for modular polynomials
QUESTION [6 upvotes]: I am teaching a graduate "classical" course on modular forms. I try to achieve the most elementary level for presenting modular polynomials. Serge Lang's "Elliptic functions" cover the topic quite well, except for some inconsistence (from my point of view) in using left/right cosets for actions of the modular group on the set of matrices of determinant $n$. I overcome this trouble by using one more simple arithmetic lemma which relates left and right cosets. I wonder whether there exist another version of the Kronecker--Weber approach for modular polynomials, or maybe even another elementary proof.

REPLY [2 votes]: Hi,
I find Silverman's exposition in his "Advanced Topics in the Arithmetic of Elliptic Curves" very clear, and concise. Look at the section on the integrality of the j-invariant, in particular Theorem 6.1 of Chapter II, Section 6, pages 140-151. You can find three proofs of the integrality of the j-invariant and, more concretely, you should have a look at the "Analytic proof of Theorem 6.1", starting in page 143, which is the "classical" one I believe you are interested in (it is also a proof using matrices of determinant n, but the exposition is great).<|endoftext|>
TITLE: The word problem for fundamental groups of smooth projective varieties
QUESTION [20 upvotes]: While attending a very nice talk on the geometric group theory of fundamental groups of Kahler manifolds by Pierre Py last weekend, I realized that I don't know the answer to the following question.  Let $X$ be a smooth projective variety over $\mathbb{C}$.  Is the word problem for $\pi_1(X)$ solvable?
Here are a couple of relevant facts.  Taubes proved that every finitely presentable group is the fundamental group of a compact complex manifold of complex dimension 3.  Earlier, Gompf proved that every finitely presentable group is the fundamental group of a compact symplectic manifold of real dimension 4.  Thus the word problem is not solvable for fundamental groups of compact complex manifolds.  Also, Toledo has an example of a smooth compact projective variety whose fundamental group is not residually finite.  This rules out using maps to finite groups to solve the word problem, and also shows that $\pi_1(X)$ need not be linear.
EDIT : Another relevant remark is that the answers to the question here show that presentations for $\pi_1(X)$ are computable, so there are no issues there.

REPLY [7 votes]: I was going through my old questions and realized that this one did not have a good answer (as far as I know, the reference that Ben Wieland gave in his answer does not work).  I've since learned that it is a well-known open question.  However, I thought I'd point out the recent paper
Kapovich, Michael,
Dirichlet fundamental domains and topology of projective varieties. 
Invent. Math. 194 (2013), no. 3, 631–672. 
which proves that every finitely presentable group is the fundamental group of an irreducible projective variety with very mild singularities (only normal crossings and Whitney umbrellas), and thus the word problem is unsolvable for the fundamental groups of such varieties.  If you don't require irreducibility, the paper
Kapovich, Michael and Kollár, János, 
Fundamental groups of links of isolated singularities. 
J. Amer. Math. Soc. 27 (2014), no. 4, 929–952. 
manages to do this with only normal crossing singularities.<|endoftext|>
TITLE: Random Walk anecdote.
QUESTION [9 upvotes]: I'm looking for an anecdote about a mathematician who studied random walks. I'm attempting to write an article and hope to include the story (but only if I can get the details correct). I'll try to do my best describing it in hopes someone else has heard it and knows a name or the full story.
A mathematician was walking through the park and entertaining mathematical whims. He noticed that he kept running into this same couple as he wandered around aimlessly. He wasn't sure if this behaviour was expected by chance or whether perhaps the female of the couple thought that he was cute. He rushed home to analyse the situation in terms of random walks in two dimensions.

REPLY [4 votes]: My favorite is this one attributed to Kakutani:
"A drunk man will find his way home, but a drunk bird may get lost forever."
referring to the fact that the simple random walk in $Z^2$ is recurrent while it is transient in $Z^d$ for $d>2$.
Here you can find a reference to the anecdote: http://m759.xanga.com/122558830/item/<|endoftext|>
TITLE: K-Theory and the Stack of Vector Bundles
QUESTION [15 upvotes]: I have some understanding that vector bundles provide a basic, familiar example of what I should call a stack.  Namely, consider the functor $Vect$ that assigns to a space $X$ the set of isomorphism classes of vector bundles on $X$.  This functor isn't local, in the sense that the isoclass of a vector bundle isn't determined by its restriction to an open cover, but rather by gluing data on overlapping sets in a cover.  Since for any space $Y$ a map $X \to Y$ is determined by what it does when restricted to a cover of $X$, this tells us there is no space $Y$ that represents the functor $Vect$ in this fashion.  However, I can also consider $Vect$ as a stack, which assigns to $X$ the groupoid of vector bundles on $X$.  This gadget is fancy enough to understand how vector bundles glue together, and so recovers the locality missing from our earlier functor.
In K-theory, we attach to a space $X$ a ring $K(X)$ whose underlying group is the the free abelian group on the set of isoclasses of vector bundles on $X$, mod short exact sequences.  It turns out that one can describe $K(X)$ as the set of homotopy classes of maps from $X$ to $\mathbb{Z} \times BU(\infty)$.
At this point my meager understanding of K-theory seems to be contradict what I said in the first paragraph.  The fact that $K(X)$ has a classifying space seems at odds with the observation that vector bundles aren't determined by their restrictions to open covers, whereas maps to another space are.  Is something wrong with what I've said so far? If not, perhaps there isn't a contradiction because either 1) $K(X)$ isn't quite the set of isoclasses of vector bundles, but rather a group completion thereof, or 2) we're looking at homotopy classes of maps to $\mathbb{Z} \times BU(\infty)$, so what I said in the first paragraph doesn't apply?

REPLY [10 votes]: Johannes Ebert and I wrote about this in our paper http://arxiv.org/abs/0712.0702.  
The groupoid of vector bundles doesn't quite give you topological K-theory directly - you have to first pass through the group completion of this with respect to the monoidal product given by direct sum of vector bundles.  Alternatively, you can instead work with virtual bundles $E-F$, which can be made to form a groupoid in a reasonable way, and on compact bases this is enough, though for noncompact bases topological K-theory classes are non quite represented by virtual bundles.
Getting on to stacks here, a stack in the category of manifolds which admits an atlas (i.e. a representable surjective map from a topological space) has a homotopy type.  An easy way to build the homotopy type is the take a Lie groupoid that presents the stack and then form the classifying space of this groupoid.  Passing to homotopy types gives a well-defined functor to the homotopy category of topological spaces.
You can make a classifying stack for vector bundles, and its homotopy type will be $\coprod_n BU(n)$.  Direct sum makes this stack into a monoid, and its homotopy type then has the usual monoidal structure given by $\oplus$.  If you group complete the homotopy type then you get the representing space $Z\times BU$ for topological K-theory.
You can also avoid having to apply group completion by making a classifying stack for virtual bundles.  Johannes and I did exactly this in our paper.  (Technically, if you allow noncompact bases then it is only a pre-stack and needs to be stackified, which adds in the objects slightly more general than virtual bundles that I alluded to above.)  It turns out that the homotopy type of the classifying stack of virtual bundles is exactly $Z\times BU$.
Thus, if you have a virtual bundle on $X$, this represents a class in $K(X)$, but it also is classified by a map to the classifying stack of virtual bundles, and taking homotopy types gives the corresponding map $X \to Z\times BU$.<|endoftext|>
TITLE: Special divisors on hyperelliptic curves
QUESTION [5 upvotes]: I was reading a proof that used the following result
Let $C$ be a hyperelliptic of genus $\ge 3$ and $\tau \colon C \to C$ the hyperelliptic involution.  If $D$ is an effective divisor of degree $g-1$ such that $h^0(D)>1$ then $D = x + \tau(x) + D'$ where $D'$ is an effective divisor.
My question is, how is this result proved?  It seems equivalent to showing that $|D|$ contains the unique $g^1_2$ and this made me think of Clifford's theorem but this didn't lead to much.  For $g = 3$ the result holds because then $|D| = g^1_2$.  But already for $g = 4$ I'm stuck.  I tried playing around with the Riemann-Roch theorem but didn't get far.

REPLY [12 votes]: Suppose that $D=x_1+x_2+\cdots+x_{g-1}$. We may assume that $\tau(x_i)\neq x_j$
for all $i\neq j$. Now, assume that $D'=y_1+\cdots+y_{g-1}$ is an element of
$|K-D|$. Then $x_1+x_2+\cdots+x_{g-1}+y_1+\cdots+y_{g-1}$ is an element of $|K|$
but we know that any such element is of the form
$z_1+\tau(z_1)+\cdots+z_{g-1}+\tau(z_{g-1})$. After possibly renumbering the $z_i$ (as
well as possibly replacing $z_i$ by $\tau(z_i)$) we may assume $x_1=z_1$ and
then $x_2=z_2$ and so on. This gives $D'=\tau(D)$ which implies
$h^0(K-D)\leq 1$. However, R-R and gives $h^0(D)=h^0(K-D)\leq1$ which
contradicts assumptions.

REPLY [6 votes]: It is easy to see that $D + \tau^*D$ is a canonical divisor. Suppose that $p$ is a ramification point of the $g^{1}_{2}$; since there are $2g-2$ such points, we may assume that $p$ is not a base point of $|D|$. Then we have effective divisors $D'$ and $D''$ in $|D|$, such that $p$ lies on $D'$ but not on $D''$. So $D' + \tau^*D''$ is an effective canonical divisor containing $p$; by Riemann-Roch, it has to contain $2p$. Since $p$ does not lie on $\tau^*D''$, the divisor $D'$ has to contain $2p$, and we are done.<|endoftext|>
TITLE: Example of a smooth morphism where you can't lift a map from a nilpotent thickening?
QUESTION [19 upvotes]: Definition. A locally finitely presented morphism of schemes $f\colon X\to Y$ is smooth (resp. unramified, resp. étale) if for any affine scheme $T$, any closed subscheme $T_0$ defined by a square zero ideal $I$, and any morphisms $T_0\to X$ and $T\to Y$ making the following diagram commute

    g
T0 --> X
|      |
|      |f
v      v
T ---> Y

there exists (resp. exists at most one, resp. exists exactly one) morphism $T\to X$ which fills the diagram in so that it still commutes.

For checking that $f$ is unramified or étale, it doesn't matter that I required $T$ to be affine. The reason is that for an arbitrary $T$, I can cover $T$ by affines, check if there exists (a unique) morphism on each affine, and then "glue the result". If there's at most one morphism locally, then there's at most one globally. If there's a unique morphism locally, then there's a unique morphism globally (uniqueness allows you to glue on overlaps).
But for checking that $f$ is smooth, it's really important to require $T$ to be affine in the definition, because it could be that there exist morphisms $T\to X$ locally on $T$, but it's impossible to find these local morphisms in such a way that they glue to give a global morphism.

Question: What is an example of a smooth morphism $f\colon X\to Y$, a square zero nilpotent thickening $T_0\subseteq T$ and a commutative square as above so that there does not exist a morphism $T\to X$ filling in the diagram?

I'm sure I worked out such an example with somebody years ago, but I can't seem to reproduce it now (and maybe it was wrong). One thing that may be worth noting is that the set of such filling morphisms $T\to X$, if it is non-empty, is a torsor under $Hom_{\mathcal O_{T_0}}(g^*\Omega_{X/Y},I)=\Gamma(T_0,g^*\mathcal T_{X/Y}\otimes I)$, where $\mathcal T_{X/Y}$ is the relative tangent bundle. So the obstruction to finding such a lift will represent an element of $H^1(T_0,g^*\mathcal T_{X/Y}\otimes I)$ (you can see this with Čech cocycles if you want). So in any example, this group will have to be non-zero.

REPLY [24 votes]: Using some of BCnrd's ideas together with a different construction, I'll give a positive answer to Kevin Buzzard's stronger question; i.e., there is a counterexample for any non-etale smooth morphism.
Call a morphism $X \to Y$ wicked smooth if it is locally of finite presentation and for every (square-zero) nilpotent thickening $T_0 \subseteq T$ of $Y$-schemes, every $Y$-morphism $T_0 \to X$ lifts to a $Y$-morphism $T \to X$.

Theorem: A morphism is wicked smooth if and only if it is etale.

Proof:
Anton already explained why etale implies wicked smooth.  
Now suppose that $X \to Y$ is wicked smooth.  In particular, $X \to Y$ is smooth, so it remains to show that the geometric fibers are $0$-dimensional.  Wicked smooth morphisms are preserved by base change, so by base extending by each $y \colon \operatorname{Spec} k \to Y$ with $k$ an algebraically closed field, we reduce to the case $Y=\operatorname{Spec} k$.  Moreover, we may replace $X$ by an open subscheme to assume that $X$ is etale over $\mathbb{A}^n_k$ for some $n \ge 0$.
Fix a projective variety $P$ and a surjection $\mathcal{F} \to \mathcal{G}$ of coherent sheaves on $P$ such that some $g \in \Gamma(P,\mathcal{G})$ is not in the image of $\Gamma(P,\mathcal{F})$.  (For instance, take $P = \mathbb{P}^1$, let $\mathcal{F} = \mathcal{O}_P$, and let $\mathcal{G}$ be the quotient corresponding to a subscheme consisting of two $k$-points.)  Make $\mathcal{O}_P \oplus \mathcal{F}$ an $\mathcal{O}_P$-algebra by declaring that $\mathcal{F} \cdot \mathcal{F} = 0$, and let $T = \operatorname{\bf Spec}(\mathcal{O}_P \oplus \mathcal{F})$.  Similarly, define $T_0 = \operatorname{\bf Spec}(\mathcal{O}_P \oplus \mathcal{G})$, which is a closed subscheme of $T$ defined by a nilpotent ideal sheaf.  We then may view $g = 0+g \in \Gamma(P,\mathcal{O}_P \oplus \mathcal{G}) = \Gamma(T_0,\mathcal{O}_{T_0})$.
Choose $x \in X(k)$; without loss of generality its image in $\mathbb{A}^n(k)$ is the origin.  Using the infinitesimal lifting property for the etale morphism $X \to \mathbb{A}^n$ and the nilpotent thickening $P \subseteq T_0$, we lift the point $(g,g,\ldots,g) \in \mathcal{A}^n(T_0)$ mapping to $(0,0,\ldots,0) \in \mathbb{A}^n(P)$ to some $x_0 \in X(T_0)$ mapping to $x \in X(k) \subseteq X(P)$.  By wicked smoothness, $x_0$ lifts to some $x_T \in X(T)$.  The image of $x_T$ in $\mathbb{A}^n(T)$ lifts $(g,g,\ldots,g)$, so each coordinate of $x_T$ is a global section of $\mathcal{F}$ mapping to $g$, which is a contradiction unless $n=0$.  Thus $X \to Y$ is etale.<|endoftext|>
TITLE: When Are Quotients Complete Intersections?
QUESTION [11 upvotes]: Let $S_{n}$ denote the permutation group on $n$ letters and $G\subset S_{n}$ a transitive subgroup.  The inclusion of $G$ in $S_{n}$ defines an action of $G$ on $\mathbb{C}^{n}$.  By finding a generating set of invariant polynomials and relations among these polynomials we may realize the quotient space $\mathbb{C}^{n}/G$ as an algebraic variety.
I have noticed that for $n=2,3,4$ such quotients are always complete intersections.  That is, the difference between the order of a minimal generating set for the invariant polynomials, and the order of a minimal generating set for the relations, is equal to $n=\mathrm{dim}(\mathbb{C}^{n}/G)$.
I would like to know if this fact persists for all $n$.  If so, what is the property of the group action which makes the quotient a complete intersection? If not what is a counterexample?

REPLY [14 votes]: This paper of Kac and Watanabe may be of interest:
Kac, Victor; Watanabe, Keiichi, Finite linear groups whose ring of invariants is a complete intersection
http://www.ams.org/mathscinet-getitem?mr=640951
From the review:
"The following theorems are proved: Let $k$ be a field and $G$ be a finite subgroup of $\text{GL}(n,k)$. The group $G$ acts naturally on the polynomial ring $S=k[x_1,\cdots,x_n]$. Let $R=S^G$. 
Theorem A: If $R$ is a complete intersection then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq 2\rbrace$. 
Theorem B: Let $k={\bf C}$. If $R$ has $m$ generators such that their ideal of relations is generated by $m-n+s$ elements, then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq s+2\rbrace$. 
Theorem C: If $R$ is a complete intersection then each stabilizer $G_x$, $x\in k^n$, is generated by $\lbrace g\in G_x\colon\text{rank}(g-I)\leq 2\rbrace$. "
In this case, the cyclic group of $S_4$ generated by the long cycle acting on ${\bf C}^4$ should be a counterexample. I should probably calculate the ring of invariants, but I'll leave it at that for now.<|endoftext|>
TITLE: What is descent theory?
QUESTION [58 upvotes]: I read the article in wikipedia, but I didn't find it totally illuminating. As far as I've understood, essentially you have a morphism (in some probably geometrical category) $Y \rightarrow X$, where you interpret $Y$ as being the "disjoint union" of some "covering" (possibly in the Grothendieck topology sense) of $X$, and you want some object $\mathcal{F'}$ defined on $Y$ to descend to an object $\mathcal{F}$ defined on $X$ that will be isomorphic to $\mathcal{F}'$ when pullbacked to $Y$ (i.e. "restricted" to the patches of the covering). To do this you have problems with  $Y\times_{X}Y$, which is interpreted as the "disjoint union" of all the double intersections of elements of the cover.
I'm aware of the existence of books and notes on -say- Grothendieck topologies and related topics (that I will consult if I'll need a detailed exposition), but I would like to get some ideas in a nutshell, with some simple and maybe illuminating examples from different fields of mathematics. 
I also know that there are other MO questions related to descent theory, but I think it's good that there's a (community wiki) place in which to gather instances, examples and general picture. 
So,

What is descent theory in general? And what are it's unifying abstract patterns?
In which fields of mathematics does it appear or is relevant, and how does it look like in each of those fields? (I'm mostly interested in instances within algebraic geometry, but having some picture in other field would be nice).
Could you make some examples of theorems which are "typical" of descent theory? And also mention the most important and well known theorems?

REPLY [4 votes]: If you're interested in seeing this in the context of algebraic geometry, you should check "Fundamental Algebraic Geometry: Grothendieck's FGA explained", by Fantechi et al. The first part of this book is a self contained but still quick intro to these topics (Gorthendieck topologies, fibered categories, descent theory, and stacks.)<|endoftext|>
TITLE: Are any two K3 surfaces over C diffeomorphic?
QUESTION [5 upvotes]: Let $S$ be a K3 surface over $\mathbb{C}$, that is, $S$ is a simply connected compact smooth complex surface whose canonical bundle is trivial. I recall reading somewhere that any two such surfaces are diffeomorphic, however I can't for the life of me remember where, or how the proof goes. 
Does anybody know a good reference to a proof, or can provide a proof?
Thanks,
Dan

REPLY [6 votes]: I find this reference quite readable:
Le Potier: "Simple connexité des surfaces K3", in Asterisque 126, 1985.
I haven't read Kodaira's paper (in Joel's answer), so I don't know whether it is the same argument, but Le Potier also deforms to quartic surfaces.<|endoftext|>
TITLE: Is the complement of an affine variety always a divisor?
QUESTION [16 upvotes]: Let $X$ be a connected affine variety over an algebraically closed field $k$, and let $X \subset Y$ be a compactification, by which I mean $Y$ is a proper variety (or projective if you prefer), and $X$ is embedded as an open dense subset. 
I am guessing that it is not always the case that $Y\setminus X$ is a divisor, one could imagine it being a single point with a horrible singularity. But if $Y$ is smooth or even normal, is it the case that $Y\setminus X$ is always a divisor? Does anybody know a proof of such a result?
Thanks,
Dan

REPLY [2 votes]: Goodman wrote a paper on a related subject entitled: "Affine open subsets of algebraic varieties and ample divisors". You might find something in there that's relevant.<|endoftext|>
TITLE: To what extent is convexity a local property?
QUESTION [12 upvotes]: A polyhedron is the intersection of a finite collection of halfspaces.  These halfspaces are not assumed to be linear, i.e. their bounding hyperplanes are not assumed to contain the origin.  The support Supp(M) of a collection M of polyhedra is the union of the polyhedra in M.
I can prove the following theorem:
Theorem.  Let M be a finite set of n-dimensional polyhedra in Rn.  Suppose:
(i)  The interior of Supp(M) is path-connected; and
(ii) For every x in the boundary of Supp(M), there exists a closed
halfspace H+ bounded by a hyperplane H such that x is in H, and such that H+ contains every P in M such that x is in P.
Then Supp(M) is convex.
(Acknowledgment:  I proved a characterization of coarsenings of a given polyhedral complex and Ezra Miller remarked that part of my argument amounted to some sort of local criterion for convexity.  The theorem above is that criterion.)
The point here is that you only need to check, at each point x of the boundary, that Supp(M) looks sufficiently like a convex set near x, and (ii) says exactly what "sufficiently like a convex set" means in this case.
The question is:  

Is this a special case of some general theorem that says that convexity is
  somehow a local condition?

I suspect that I'm asking for a reference to something known.  One convexity person that I asked about felt that it is "highly likely..., that this result is a special case
of a result in functional analysis, once properly understood."  The same person suggested that there might be a connection to the theory of tight manifolds in topology.  For that reason I have added the tags fa.functional-analysis and gt.geometric-topology.  My apologies if these tags turn out not to be appropriate.

REPLY [3 votes]: "Locally convexity implies convexity" holds more generally for CAT(0) spaces. See e.g. Convex rank 1 subsets of Euclidean Buildings Proposition 4.1<|endoftext|>
TITLE: Mixed Hodge structure on the rational homotopy type
QUESTION [27 upvotes]: A mixed Hodge structure (mHs) on a commutative differential graded algebra (cgda) over $\mathbf{Q}$ is a mixed Hodge structure on the underlying vector space such that the product and the differential are maps of mixed Hodge structures. If a cdga is equipped with an mHs, then so is its cohomology. Now let $A$ be, say, the algebra of piecewise polynomial cochains of a complex algebraic variety $X$ and let $M$ be the minimal model of $A$. Is there a mixed Hodge structure on $M$ which would be functorial in some sense and such that the comparison quasi-isomorphism $M\to A$ induces a map of mixed Hodge structures? Here is a guess as to what the "suitable sense" should be, inspired by the case when we have the weight filtration alone. Given a morphism $X\to Y$ of two varieties and given a minimal model $M_X\to A_{PL}(X), M_Y\to A_{PL}(Y)$ for each variety, there is a map of the minimal models, unique up to a homotopy preserving both filtrations and such that the obvious diagram commutes up to homotopy. (This would imply the uniqueness of the mHs on the minimal model.)
Here is a related (and maybe equivalent?) question: due to Kadeishvili, the cohomology $H^\ast$ of any cdga carries an $A_{\infty}$ structure, and in fact, a $C_{\infty}$ structure. See e.g. Keller, Introduction to $A_\infty$ algebras and modules, theorem on p. 7. If now $H^\ast$ is the cohomology of a complex algebraic variety, can one choose the structure maps $(H^\ast)^{\otimes n}\to H^*$ to be morphisms of mixed Hodge structures? Once again, this should be functorial in an appropriate sense.
Here are some remarks:

If one considers weight filtrations alone, then two things happen. First, one has to slightly modify the above definition: a weight filtration on a cdga is a filtration such that the differential and the product are strictly compatible with it (this is automatic in the mixed Hodge case). Second, the answer to the similar question is yes: namely, there is a weight filtration on the minimal model of an algebraic variety, it induces the ``right'' weight filtration in cohomology and the functoriality property can be stated as follows: a map of algebraic varieties gives a map of filtered minimal models, unique up to filtration-preserving homotopy. This is essentially due to J. Morgan, The algebraic topology of smooth algebraic varieties, Publ. IHES 48, 1978 and Navarro-Aznar, Sur la th\'eorie de Hodge-Deligne, Inv. Math. 90, 1987.
In the above-mentioned paper Navarro-Aznar claims (7.7, p.38) that the answer to the question of this posting is positive as well. He does not give a proof however, instead referring the reader to Morgan (ibid) and `la deuxi\eme partie de cet article''. It is not clear to me how to deduce the statement (if it is true) from either one of these sources.
Let $X$ be a smooth complete curve and let $K$ be a finite subset with $\geq 2$ elements. Then according to Morgan, a filtered model of $X-K$ is the $E_2$-term of the Leray spectral sequence of the embedding $X-K\subset X$, equipped with the Leray filtration. This indeed determines the cohomology of $X-K$ as a filtered algebra. But the $E_2$-term viewed as a mixed Hodge structure does not determine $H^\ast(X-K,\mathbf{Q})$ as a mixed Hodge structure since it does not depend on $K$ and $H^\ast(X-K,\mathbf{Q})$ does.

REPLY [17 votes]: I think these questions are very satisfactorially answered in a paper of joana cirici on arxiv 2013.
this paper was published in 2015
dennis sullivan<|endoftext|>
TITLE: Algebraic properties of the algebra of continuous functions on a manifold.
QUESTION [11 upvotes]: Does the algebra of continuous
  functions from a compact manifold to
  $\mathbb{C}$ satisfy any specific
  algebraic property?

I'm not sure what kind of algebraic property I expect, but I feel that because of the Gel'fand transform, it may not be unreasonable to expect something. We can drop the compactness condition if we switch to continuous functions to $\mathbb{C}$ that vanish at infinity.
I'm really hoping for some necessary and sufficient condition, but if anybody knows of any sort of condition, that would be appreciated.

REPLY [4 votes]: I found a reference for a necessary property that might be called algebraic.
Browder proved a theorem relating the number of generators of a complex commutative Banach algebra to the Čech cohomology with complex coefficients of the maximal ideal space, and as a corollary concluded that if $M$ is a compact orientable $n$-dimensional manifold, then $C(M)$ cannot be generated as a Banach algebra by fewer than $n+1$ elements. The paper is very short, but for an even shorter summary here's the MR review.

Just some comments, added later:
One obtains the compact Hausdorff space $X$ (up to homeomorphism) from $C(X)$ by considering the maximal ideal space of $C(X)$ with Gelfand topology, but clearly you want something less tautological than "the maximal ideal space is a manifold."  A small step in this direction would be to try to formulate the topological properties of $X$ in terms of the closed ideals of $C(X)$.  As alluded to in Qiaochu's comment, there is an analogue of Nullstellensatz: each closed ideal in $C(X)$ consists of all functions vanishing on a (uniquely determined) closed subset of $X$.  So for example, the locally Euclidean property could be reformulated for a commutative C*-algebra $A$ as follows: There is an $n$ such that for every maximal ideal $M$ of $A$ there is a closed ideal $I$ of $A$ such that $I$ is not contained in $M$ and $I$ is $*$-isomorphic to $C_0(\mathbb{R}^n)$.  Second countability of the maximal ideal space is equivalent to $A$ being separable in the norm topology; that's not algebraic, but might be considered more intrinsic to the C*-algebra.
But this only leads to another, more specific question: Is there a useful or interesting (C*-)algebraic characterization of $C_0(\mathbb{R}^n)$?`<|endoftext|>
TITLE: A Simple Generalization of the Littlewood Conjecture
QUESTION [9 upvotes]: The Littlewood Conjecture asserts that for all real numbers $r$ and $s$, and for every
$\epsilon > 0$, the inequality $|x(rx-y)(sx-z)| < \epsilon$ is solvable in integers $x, y, z$ with $x > 0$.
The Littlewood conjecture is clearly a consequence of the following:
For all real numbers $r$ and $s$, and for every $\epsilon > 0$ the inequalities 
$|x(rx-y)| < 1, \,\,|sx-z| < \epsilon$ are solvable in integers $x, y, z$ with $x > 0$.
Does anyone know a counter-example to the latter statement? Does anyone know of any references to it in the literature?  
Note that the inequality $|x(rx-y)|  <  1$ always has infinitely many solutions $(x,y)$ with $x > 0$. This is a consequence of Dirichlet's Approximation Theorem. So it is natural to ask:  "How does $sx$ behave mod 1 as $(x,y)$ runs through the solutions of  $|x(rx-y)|  <  1$?"  For example, can  the closure of the  $sx$ mod 1 contain some non-empty open set and be disjoint from another?
Experiments with Sage seem to "suggest" that  the numbers $sx$ are either dense mod 1 or have just finitely many limit points mod 1,  depending on whether the numbers $1,r,s$ are linearly independent over the rationals.  Again, any counter-examples or references relevant to this statement would be  appreciated.

REPLY [3 votes]: This answer pieces together the various comments made by Roland Bacher, SJR and 
gowers previously. The proposed generalization of Littlewood's conjecture is false.
As suggested by Roland and SJR, take $r$ to be the golden ratio $(1+\sqrt{5})/2$.
Then $x|rx-y| \le 1$ only when $x$ runs over the Fibonacci numbers $F_k$.  Now 
we want to show that there is an irrational number $s$ such that $F_k s$ is 
bounded away from integers.  
As suggested by gowers there do exist such $s$ for any lacunary sequence $n_k$ 
(that is a sequence with $n_{k+1}/n_k \ge 1+ c>1$ for all large $k$), and so in 
particular for the Fibonacci numbers.  This is related to a conjecture of Erdos, 
that for any lacunary sequence $n_k$ there exist irrational numbers $\alpha$ with 
$n_k \alpha$ not being dense $\mod 1$.  Erdos's problem was settled independently 
by de Mathan and Pollington in a stronger form, showing that there exist many such $\alpha$.  We need in fact that the values $\mod 1$ are bounded away from $0$.  This 
is worked out in detail using Pollington's argument in a recent nice preprint of 
Haynes and Munday (see Lemma 1 of http://arxiv.org/abs/1308.0208 ).<|endoftext|>
TITLE: Connectedness of random distance graph on integers
QUESTION [12 upvotes]: This is not my field, a friend needs the answer for the following question. Suppose we have a decreasing probability function, $p: N \rightarrow [0,1]$ such that $sum_n p(n) = \infty$. Take the graph where we connect two integers at distance d with probability $p(d)$. Will this graph be connected with probability one?
I see that if the sum is convergent, then we almost surely have an isolated vertex (unless $p(1)=1$), so this would be "sharp". One possible approach would be to take the path that starts from the origin and if it is at n after some steps, then next goes to the smallest number that is bigger than $n$ and is connected to $n$ and to show that this path has a positive density with probability one. Is this second statement true?
I am sure that these are easy questions for anyone who knows about this.

REPLY [9 votes]: All right. Here goes, as promised. We shall work with a big circle containing a huge number $N$ of points and a sequence of probabilities $p_1,\dots,p_L$ such that $\sum_j p_j=P$ is large (so we never connect points at the distance greater than $L$ but connect points at the distance $d\le L$ with probability $p_d$). If $N\gg L$ and $p_j<1$ for all $j$, the probability of a connected path going around the entire circle is extremely small, so the problem is essentially equivalent to the one on the line. I chose the circle just to make averaging tricks technically simple (otherwise one would have to justify some exchanges of limits, etc.). Fix $\delta>0$.
Our aim will be to show that with probability at least $1-2\delta$, we have $\sum_{j\in E_0} p_{|j|}\ge P$ where $E_0$ is the connected component of $0$ and integers are understood modulo $N$, provided that $P>P(\delta)$. This, clearly, implies the problem (just consider the connected component of $0$ in the subgraph with even vertices only; whatever it is, the edges going from odd vertices to even vertices are independent of it, so we get $0$ joined to $1$ with probability $1$ in the limiting line case with infinite sum of probabilities).
We shall call a point $x$ good if $\sum_{y\in E_x}p_{|y-x|}\ge P$. We will call a connected component $E$ with $m$ points good if at least $(1-\delta)m$ its points are good. 
Fix $m$. Let's estimate the average number of points lying in the bad components. To this end, we need to sum over all bad $m$-point subsets $E$ the probabilities of the events that the subgraph with the set of vertices $E$ is connected and there are no edges going from $E$ elsewhere and then multiply this sum by $m$. For each fixed $E$ these two events are independent and, since $E$ is bad, there are at least $\delta m$ vertices in $E$ for which the probability to not be connected with a vertex outside $E$ is at most $e^{-P}$ (the total sum of probabilities of edges emanating from a vertex is $2P$ and only the sum $P$ can be killed by $E$). Thus, the second event has the probability at most $e^{-\delta P m}$ for every bad $E$ and it remains to estimate the sum of probabilities to be connected.
We shall expand this sum to all $m$-point subsets $E$. Now, the probability that subgraph with $m$ vertices is connected does not exceed the sum over all trees with the set of vertices $E$ of the probabilities of such trees to be present in the graph. Thus, we can sum the probabilities of all $m$-vertex trees instead. 
We need an efficient way to parametrize all $m$-trees. To this end, recall that each tree admits a route that goes over each edge exactly twice. Moreover, when constructing a tree, in this route one needs to specify only new edges, the returns are defined uniquely as the last edge traversed only once by the moment. Thus, each $m$ tree can be encoded as a starting vertex and a sequence of $m-1$ integer numbers (steps to the new vertex) interlaced with $m-1$ return commands. For instance, (7;3,2,return,-4,return,return) encodes the tree with vertices 7,10,12,6 and the edges 7--10, 10--12, 10--6. Well I feel a bit stupid explaining this all to a combinatorist like you...
Now when we sum over all such encodings, we effectively get $N$ (possibilities for the starting vertex) times the sum the products of probabilities over all sequences of $m-1$ integers multiplied by the number of possible encoding schemes telling us the positions of the return commands. (actually a bit less because not all sequences of integers result in a tree). Since there are fewer than $4^{m-1}$ encoding schemes, we get $4^{m-1}(2P)^{m-1}$ as a result. Thus the expected number of bad $m$-components is at most $N\cdot 4^{m-1}(2P)^{m-1}e^{-\delta Pm}$. Even if we multiply by $m$ (which is not really necessary because each tree is counted at least $m$ times according to the choice of the root) and add up over all $m\ge 1$, we still get less than $\delta N$ if $P$ is large. 
Now we see that the expected number of bad points is at most $2\delta N$ (on average at most $\delta N$ points lie in the bad components and the good components cannot contain more than $\delta N$ points by their definition). Due to rotational symmetry, we conclude that the probability of each particular point to be bad is at most $2\delta$.
The end.<|endoftext|>
TITLE: Line bundles vs. Cartier divisors on a non-integral scheme
QUESTION [17 upvotes]: It is well-known that if $X$ is an integral scheme, then there is an isomorphism $CaCl(X)\to Pic(X)$ taking $[D]$ to $[\mathcal{O}_X(D)]$. Does anyone know any simple examples where the above map fails to be surjective, i.e., a line bundle on a scheme $X$, not isomorphic to $\mathcal{O}_X(D)$ for any Cartier divisor D?

REPLY [12 votes]: An example is given in this note (it was credited to Kleiman).<|endoftext|>
TITLE: Constant curvature manifolds
QUESTION [8 upvotes]: In two different books I found these two related statements.

The book by Jost defines a ``locally symmetric space" as one for which the curvature tensor is constant and which is geodesically complete. 
Andreas' book attributes to Cartan a theorem that a space is locally symmetric if and only if its curvature tensor is constant. Where he defines a space to be locally symmetric if about every point there is a geodesic reflecting isometry. 

I could not trace this theorem of Cartan anywhere else nor does Andreas' book give a reference. I would like to know some reference which would prove this stuff and resolve the apparent definition conflict between the above two statements.
Further I came across these 3 statements about classifying constant curvature spaces,

If the isometry group is transitive on points on the manifold, then the scalar
curvature  is constant.
If the isometry group is transitive on all one-dimensional subspaces
of tangent spaces, then the Ricci curvature tensor is a scalar multiple of the metric.
tensor
If the isometry group is transitive on all two-dimensional subspaces
of tangent spaces, then the sectional curvature is constant on all
two-dimensional subspaces of tangent spaces.

I have an understanding of why the first statement is true but I would like to know some reference which explains the other two statements.
Further call a space to be homogeneous if it is quotient of some Lie Group mod a closed subgroup. Then in physics literature one finds a statement of the kind that an "isotropic and homogeneous space-time is of constant curvature". I would like to know what is the precise mathematical meaning of this statement.
Further I would also like to know a reference for the fact that a maximally symmetric space always has constant curvature and how being maximally symmetric fits in with being "isotropic and homogeneous".
Many sources refer me to the volumes by Kobayashi and Nomizu or the book by Besse or Spivak for discussion on constant curvature spaces but unfortunately I don't have access to these books. It would be great if some online reference like lecture notes/expository review paper on constant curvature spaces could be linked which clarifies the above questions.
In the books on Riemannian Geometry that I have access to like the ones by Gallot et al,Andreas or Jost, I can't see much of any discussion on these topics. 
{In all the above statements I suppose the Riemann Christoffel connection is being assumed. It would be interesting to know how much of the things hold for a general affine connection.}

REPLY [6 votes]: First, a clarification.  When you say "the isometry group is transitive on all one-dimensional subspaces of tangent spaces", what this really means is that for any $v\in T_p M$ and $w\in T_q M$ with $|v| = |w|$, there is an isometry $f:M\rightarrow M$ with $f(p) = q$ and $d_p f v = w$.
This immediately implies that $Ric_p(v) = Ric_q(w)$ for any vectors $v$ and $w$ of the same length.  This implies that the polarization of this to $Ric(\cdot, \cdot)$ are also the same (meaning $Ric_p(v,v') = Ric_q(d_p f v, d_p f v')$ for any isometry $f$ mapping $p$ to $q$).
Now, fix a $p\in M$ and let $G_p\subseteq $Iso$(M)$ be the subgroup of all isometries which fix $p$.  By the assumption of the second bullet point, $G_p$ acts transitively on the sphere in $T_p M$ (i.e., $G_p$ takes any unit vector to any other unit vector).  It follows that the representation of $G_p$ on $T_p M$ is irreducible.
Now, let $g$ denote the inner product at $T_p M$ and let $P:T_pM\rightarrow T_p M$ be defined by $g(P v, w) = Ric_p(v,w)$.  It's easy to see that $P$ will be $G_p$ invariant, and hence, by Schur's lemma, a multiple of the identity.  That is, $Ric_p$ is a multiple of the metric.
The statement about sectional curvatures is easier:  Curvature is an invariant of the metric, and hence so is sectional curvature. By the assumption of bullet point 3, there is an isometry of $G$ which takes any point to any other point and moves any 2-plane to any other 2-plane.  But then the sectional curvature of any 2-plane at any point is equal to the sectional curvature of any 2-plane at any other point, that is, $M$ has constant sectional curvature.<|endoftext|>
TITLE: Infinite product experimental mathematics question.
QUESTION [7 upvotes]: A while ago I threw the following at a numerical evaluator (in the present case I'm using wolfram alpha)
$\prod_{v=2}^{\infty} \sqrt[v(v-1)]{v} \approx 3.5174872559023696493997936\ldots$
Recently, for exploratory reasons only, I threw the following product at wolfram alpha
$\prod_{n=1}^{\infty} \sqrt[n]{1+\frac{1}{n}} \approx 3.5174872559023696493997936\ldots$
(I have cut the numbers listed above off where the value calculated by wolfram alpha begins to differ)
Are these products identical or is there some high precision fraud going on here?

REPLY [5 votes]: Aha, I get Gjerji's insight, and I should have seen it sooner, but I was stuck on dealing with series representations by logarithms. 
The second product looks like this:
$\sqrt[1]{\frac{2}{1}}\sqrt[2]{\frac{3}{2}}\sqrt[3]{\frac{4}{3}}\sqrt[4]{\frac{5}{4}}\ldots$, and it can be rewritten like this: $2^{1-\frac{1}{2}}3^{\frac{1}{3}-\frac{1}{4}}4^{\frac{1}{4}-\frac{1}{5}}5^{\frac{1}{5}-\frac{1}{6}}\ldots$, and then one can employ Gjerji's observation to convert the second into the first.<|endoftext|>
TITLE: A question about axes of symmetry in the plane.
QUESTION [5 upvotes]: Suppose J is a Jordan curve in the Euclidean plane E and that there are two perpendicular straight lines 
in E, each of which is an axis of symmetry of J. Does the intersection point of these two lines always
necessarily lie in the interior of J?

REPLY [7 votes]: Here is a more general fact: if $J$ is central-symmetric w.r.t. a point $O$ (in our case, this is the intersection of the axes), then $O$ is inside.
Indeed, let $D$ be the domain bounded by $J$. We know it's a topological disc. And it's central symmetric w.r.t. $O$. Now if $O$ is outside, then the symmetry is a map from $D$ to itself without fixed points, contrary to Brouwer fixed point theorem.
If $O$ is on the boundary (i.e. on $J$), the the central symmetry is an involution of $J$ with exactly one fixed point. But there are no such involutions of the circle.<|endoftext|>
TITLE: Details of Perelman's example about soul of Alexandrov space
QUESTION [7 upvotes]: Reading Perelman's preprint(1991) Alexandrov space II now. Got confused about the last section 6.4, which contains an example which indicate that the statement ".... manifold is diffeomorphic to the normal bundle over soul" in Cheeger-Gromoll's Soul Theory won't hold for Alexandrov spaces. I also read BBI's book(A course in Metric Geometry) (page 400-401) which also contains the description of this example. I am using the notation in BBI's book, the example goes as follows:
Let $\pi: K_0(\mathbb{CP}^2)\to K_0(\mathbb{CP}^1)$ be the projection. $\bar{B}_0(1)$ be the unit ball in $\mathbb{CP}^1$ (Note: here should be $K_0(\mathbb{CP}^1$), right?). Let $X^5=\pi^{-1}(\bar{B}_0(1))$. Take double of $X^5$ and it will be the example
The picture in my mind is $K_0(\mathbb{CP}^1)$ is sub-cone of $K_0(\mathbb{CP}^2)$, so the projection is the projection on the second factor if we write the coordinate in cone as $(t, x)$ for $t\in \mathbb R$ and $x\in \mathbb{CP}^2$.
My question is what is the topology of $X^5$? 
1) Is $\bar{B}_0(1)$ a close ball? If so then $\bar{B}_0(1)$ will have boundary $\mathbb{CP}^1$, right? and $X^5$ will be a closed cone over $\mathbb{CP}^2$, right?
2) Is $X^5$ compact?

REPLY [9 votes]: No $X^5$ is not a cone over $CP^2$ and is not compact. The projection has nothing to do with the cone structure. In fact, it's better to forget about the cone structure altogether (until you ask what is the topology of the thing).
The spaces are just quotients of $\mathbb C^3$ and $\mathbb C^2$ by the standard circle action, and the projection is induced by the coordinate projection $\mathbb C^3\to \mathbb C^2$. For example, there is a whole half-line in the pre-image of the origin.
I suggest you visualize a similar construction with $\mathbb R$ in place of $\mathbb C$ or lower the dimensions by 1 (or both) to see what is going on.<|endoftext|>
TITLE: Are groups in (Var/k, rational maps) necessarily algebraic groups?
QUESTION [8 upvotes]: Let $RVar$ be the category which has complex algebraic verieties as objects and rational maps as morphisms [Edit: for $RVar$ to be a category, the rational maps have to be dominant]. 
Let's consider a group object $G$ in this category, i.e. a tuple $(G,\mu,e,\iota)$ where $\mu:G\times G \rightarrow G$ , $e:* \rightarrow G$, and $\iota: G \rightarrow G$ are rational maps satisfying the usual commutative diagrams defining a group structure.
[Edit: in the light of the comments, e.g. the observation that rational maps have better be dominant, the identity $e:*\rightarrow G$ doesn't seem to make much sense; same for the diagram for the inverse, then; what people, among whom Weil, actually considered were "rational group chunks" in which there's only an associative rational $\mu$, and you can ask the same question(s)].
Just out of curiosity, two natural questions:

Is such a $G$ necessarily an algebraic group?
That is: is it the case that for any $(G,\mu,e,\iota) \in Grp(RVar)$, there exists an algebraic group $(G',\mu',e',\iota')$ and a birational map $\varphi: G\rightarrow G'$ such that "$\varphi$ intertwines the operations of $G$ and $G'$"?
Analogous question in the holomorphic/meromorphic setting.

REPLY [18 votes]: Yes.  This result usually goes under the name of Weil's theorem on group chunks.
The original reference is:
A. Weil, On algebraic groups of transformations, Am. J. Math, 77, 1955, 355 - 381.
In model theory this theorem has taken on a life of its own and is the basis of Hrushovski's theorem on generically presented groups.  A model theoretic treatment of Weil's theorem, including a weakening where one is allowed to use negative powers of the Frobenius, may be found in 
E. Bouscaren, Model theoretic versions of Weil's theorem on pregroups, in The Model Theory of Groups, Notre Dame Mathematical Lectures, Number 11
The analogous results are known at least for compactifiable complex analytic groups and real algebraic groups.  The following paper gives a good general treatment of these problems.
L. van den Dries, Weil's group chunk theorem: a topological setting, Ill. J. Math, 34 no. 1, 1990, 127 - 139.<|endoftext|>
TITLE: Extending vector bundles on a given open subscheme
QUESTION [24 upvotes]: Let $U$ be a dense open subscheme of an integral noetherian scheme $X$ and let $E$ be a vector bundle on $U$. Suppose that the complement $Y$ of $U$ has codimension $\textrm{codim}(Y,X) \geq 2$. Let $F$ be a vector bundle on $X$ extending $E$, i.e., $F|_{U} = E$. 
Is any   extension of $E$ to $X$   isomorphic to $F$?

REPLY [16 votes]: Let $i:U \to X$ be the embedding. Assume that $i^*F = E$. Then by the adjunction we have a map $F \to i_*E$ which is an embedding, since the kernel is zero on $U$, so it is a torsion sheaf, and a vector bundle doesn't have torsion subsheaves (if $X$ doesn't have embedded components). So, we have an exact triple $0 \to F \to i_*E \to G \to 0$ for some $G$, which is supported on $X \setminus U$. If $X$ satisfies and $codim_X (X\setminus U) \ge 2$, then (provided $S_2$ condition) we have $G^* = {\mathcal Ext}^1(G,O_X) = 0$, so dualizing the sequence we see that $F^* = (i_* E)^\ast$, hence $F = (i_* E)^{\ast\ast}$, so $F$ has to be the reflexive envelope of $i_* E$. This proves the uniquencess. This also allows to construct an example of a vector bundle on $U$ which does not extend to a vector bundle on $X$.<|endoftext|>
TITLE: Are there "reasonable" criteria for existence/non-existence of Levi factors or their conjugacy in prime characteristic?
QUESTION [10 upvotes]: Classical theorems attributed to Levi, Mal'cev, Harish-Chandra for a finite
dimensional Lie algebra over a field of characteristic 0 state that it has a Levi decomposition (semisimple subalgebra plus solvable radical) and that all such semisimple subalgebras (Levi factors) are conjugate in a strong sense: see Jacobson, Lie Algebras, III.9, for example.    This carries over to connected linear algebraic
groups, but in prime characteristic there are counterexamples going back perhaps
to Chevalley that involve familiar group schemes like $SL_2$ over rings of Witt
vectors.    Recent posts here have somewhat ignored that difficulty, having just characteristic 0 in mind.   Borel and Tits redefined "Levi factor" to be a reductive complement to the unipotent radical, which is makes no real difference in characteristic 0 but allows them to concentrate on positive answers for parabolic subgroups of reductive groups in general.    Other familiar subgroups of reductive groups like the identity component of the centralizer of a unipotent element require much more subtle treatment, as in work of George McNinch.
Whether or not the characteristic $p$ question is important, it has remained
open for many decades (say over an algebraically closed field).   I gave up after one forgettable paper (Pacific J. Math. 23, 1967).   The problem is still
easy to state: 

Are there effective necessary or sufficient conditions for existence or uniqueness of Levi factors in a connected linear algebraic group over an algebraically closed field of prime characteristic?

It's clear that a scheme-theoretic viewpoint may be needed.   Possibly the known
counterexamples using Witt vectors suggest in some way all possible counterexamples?    (Or is the question hopeless to resolve completely?)
EDIT: For online access to my 1967 paper, via Project Euclid, see
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102991730.   Here
Chevalley's counterexample is mentioned only in the abstract, but in remarks
later on it is noted that Borel-Tits (III.15) gave an example involving two
Levi subgroups which fail to be conjugate; see NUMDAM link to PDF version of
Publ. Math. IHES 27 (1965) at http://www.numdam.org:80/?lang=en
In April
1967 Tits responded to my inquiry with a letter outlining the behavior of the
group scheme $SL_2$ over the ring of Witt vectors of length 2, which gives a
6-dimensional algebraic group over the underlying field with unipotent radical of dimension 3 but no Levi factor.  He remarked that he got this counterexample from P. Roquette but had also been told about Chevalley's counterexample.
ADDED: The question as formulated probably doesn't have a neat answer, but meanwhile George McNinch has delved much deeper (over more general fields) in his new arXiv preprint
1007.2777.  Some technical steps rely on the forthcoming book  Pseudo-reductive groups (Cambridge, 2010) by Conrad-Gabber-Prasad.

REPLY [3 votes]: [See Edit below.]
This isn't really an answer, but I believe it is relevant.
Work geometrically, so $k$ is alg. closed. Let $G$ reductive over $k$, and let
$V$ be a $G$-module (linear representation of $G$ as alg. gp.).
If $\sigma$ is a non-zero class in $H^2(G,V)$, there is a non-split extension
$E_\sigma$ of $G$ by the vector group $V$ -- a choice of 2-cocyle representing
$\sigma$ may be used to define a structure of alg. group on the variety
$G \times V$. Here "non-split" means "$E_\sigma$ has no Levi factor".
And if $H^2(G,V) = 0$, then any $E$ with reductive quotient $G$ and
unipotent radical that is $G$-isomorphic to $V$ has a Levi factor.
You can look at the $H=\operatorname{SL}_2(W_2(k))$ example from this viewpoint;
$H$ is an extension of $\operatorname{SL}_2$ by the first Frobenius twist
$A = (\mathfrak{sl}_2)^{[1]}$ of its adjoint representation. Of course, this point of view doesn't really help to see that $H$ has no Levi factor; the fact that $H^2(\operatorname{SL}_2,A)$ is non-zero only tells that it might be interesting (or rather: that there is an interesting extension).
The extension $H$ determines a class in that cohomology group, and the argument
in the pseudo-reductive book of Conrad Gabber and Prasad -- or a somewhat clunkier representation theoretic argument I gave some time back -- shows this class to be non-zero, i.e. that $H$ has no Levi factor.
So stuff you know about low degree cohomology of linear representations comes up. And this point of view can be used to give examples that don't seem to be related to Witt vectors.
A complicating issue in general is that there are actions of reductive $G$ on a product of copies of $\mathbf{G}_a$ that are not linearizable, so one's knowledge of the cohomology of linear representations of $G$ doesn't help...
Edit: It isn't clear I was correct last April about that "complicating issue". See this question.
Also: the manuscript arXiv:1007.2777 includes a "cohomological" construction
of an extension $E$ of SL$_3$ by a vector group of dim $(3/2)(p-1)(p-2)$ having no Levi factor in char. $p$, and an example of a group having Levi factors which aren't geometrically conjugate.<|endoftext|>
TITLE: What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes?
QUESTION [11 upvotes]: More specifically, I was wondering if there are well-known conditions to put on $X$ in order to make $K_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?

REPLY [4 votes]: Asking $K^0(X)$ to be isomorphic to $K_0(X)$ is not always "good enough". Of course, it will allow you to carry over constructions for $K_0(X)$ to $K^0(X)$, but not canonically. And it can happen that $K^0(X)\cong K_0(X)$ without $X$ being regular. For example, take $X= \textrm{Spec} A$, $A=k[x]/(x^n)$ with $n\geq 2$. Then you have an infinite resolution as given in David's answer for $k$. Computing $Tor^A_i(k,k)$ shows that $k$ has no finite resolution. (In fact, $Tor_i^A(k,k) = k^2$ for all $i>0$.) Now, although the above "existence of finite resolution" fails, it is not hard to see that $K^0(X)\cong \mathbf{Z}\cong K_0(X)$ in this case. (Use that $A$ is a local ring and the length map on $A$.) Of course, the natural map $K^0(X) \longrightarrow K_0(X)$ is not an isomorphism. (It is given by $1\mapsto n$.) 
[Edit: I added another example]
[Edit 2: There was something wrong with the example below as noted by Michael. I fixed the problem]
Let me also add to my answer the following "snake in the grass". If you work with general schemes, even if regular, one requires the extra assumption of "finite-dimensionality". For example, take the scheme $X=\textrm{Spec} (k \times k[t_1]\times k[t_1,t_2] \times \ldots)$. Now, even though $A = k\times k[t_1]\times\ldots$ is regular, there is an infinite resolution for $k$ of the form $$\ldots \longrightarrow A\longrightarrow A\longrightarrow A \longrightarrow k \longrightarrow 0$$ which corresponds geometrically to taking a point, then adding a line, then adding a plane, etc.  Again, take the Tor's to see that $k$ has no finite resolution. Do note that $X$ is not noetherian.
[Edit 3: I added the following for completeness]
Let $X$ be a regular finite-dimensional scheme. Assume that $X$ has enough locally frees. (This notion also arose in  Are schemes that "have enough locally frees" necessarily separated ). Then the canonical morphism $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. In the second example, $X=\textrm{Spec} \ A$ is regular, but not finite-dimensional. Does $X$ have enough locally frees?<|endoftext|>
TITLE: On Alexandrov embedding theorem
QUESTION [9 upvotes]: Consider a complete $C^\infty$ Riemannian metric on $\mathbb R^2$ of positive sectional curvature. 

Is the metric embeddable as the boundary of a convex subset of $\mathbb R^3$?
Is the embedding unique? 
Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?
What are good references for these matters?

UPDATE: 
$\bullet\ $ after doing some reading on the subject I found that the assertion 1 is true in the sense that the surface is isometric, as a metric space, to the boundary of a convex body in $\mathbb R^3$ (as proved by Alexandrov back in 1942). The matter of uniqueness is well-understood.
$\bullet\ $ However, one should not expect the boundary to be smooth, e.g. there are examples of $C^\infty$ metrics of nonnegative curvature on $S^2$ which cannot be isometrically $C^3$-embedded into $\mathbb R^3$. 
$\bullet\ $ If the curvature is positive, then smoothness can be achieved as proved by Pogorelov and Nirenberg (independently in the 1950s). 
$\bullet\ $ Local smooth isometric embedding for nonnegatively curved surfaces was established by Lin in 1985. 
$\bullet\ $ A more recent reference for these matters is the book by Burago and Zalgaller, Geometry III, Encyclopedia of Mathematical Sciences.

REPLY [10 votes]: Is the metric embeddable as the boundary of a convex subset of 3?
YES, it is a limit case of standard Alexandrov's theorem. Moreover one can choose any embedding of cone at infinity and construct the embedding. This is a theorem of Olovyanishnikov --- one of three students of Alexandrov who died in the war.
Is the embedding unique?
NO, but I suspect it is unique once you fixed the convex embedding of the cone at infinity. It might follow from the proof of Pogorelov's theorem but I was not able to check his proof.
Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?
I'm not sure what you mean --- if it has strictly positive curvature at one point then it is automatically $\mathbb R^2$. If it is $\mathbb R^2$ then it is all the same.<|endoftext|>
TITLE: exotic differentiable structures on manifolds in dimensions 5 and 6
QUESTION [13 upvotes]: It's a result of low-dimensional topology that in dimensions 3 and lower, two manifolds are homeomorphic if and only if they are diffeomorphic. Milnor's 7-spheres give nice counterexamples to this result in dimension 7, and exotic $\mathbb{R}^4$'s give nice counterexamples in dimension 4. But I don't know about dimensions 5 and 6. Is the result true or false in dimensions 5 and 6? And, if false, what are some classic counterexamples, and do stronger constraints -- say compactness or closedness -- happen to make it true?

REPLY [17 votes]: It is false in dimension 5 and 6. Spheres happen to be standard, but some other (compact and closed) manifolds happen to admit different smooth (and PL) structures.
Simple example are tori. For example, $\mathbb T^5$ admits 3 different PL structures that give rise to 3 different differentiable structures. See, e.g., Hsiang, Shaneson "Fake tori" or Wall's book on surgery.

REPLY [15 votes]: Any PL-manifold of dimension $\le 7$ is smoothable, and the smooth structure is unique in dimensions $5,6$. See e.g. remark 6.7 in   Rudyak's paper for details.
EDIT: To explain the above, the smooth structures on a PL manifold $M$ of dimension $\ge 5$ are in 1-1 correspondence with $[M, PL/O]$, homotopy classes of maps from $M$ to the space $PL/O$, which is $6$-connected. This implies the claim in the previous paragraph. Similarly, PL structures on a topological  manifold $M$ of dimension $\ge 5$ are in 1-1 correspondnece with $[M,TOP/PL]$, and $TOP/PL$ is $K(\mathbb Z_2,3)$. Thus $[M,TOP/PL]$ is simply $H^3(M;\mathbb Z_2)$, the third cohomology group with $\mathbb Z_2$ coefficients, and if $H^3(M;\mathbb Z_2)$ is nonzero, then $M$ admits more than one PL structure. See Madsen-Milgram "Classifying spaces for surgery and cobordism of manifolds".<|endoftext|>
TITLE: Can curves induced by analytic maps wiggle infinitely across a line?
QUESTION [6 upvotes]: Let $f$ be a function analytic on an open subset $D\subset \mathbb{C}$, and let $\gamma:[0,1] \to D$ be a line segment. $g = f\circ\gamma$ is another curve in the complex plane; is it possible to for $g$ to cross a straight line infinitely often, where the crossing points accumulate towards a point? That is, does there exist a point $\alpha$ and a ray $R$ emanating from the point $\alpha$ such that for all $\epsilon > 0$, $g$ crosses the ray $R$ infinitely many times in the $\epsilon$-ball around $\alpha$?
We're trying to show something about analytic continuation, but we cannot rule out pathological beasts like these.
Thanks!

REPLY [8 votes]: The image of $[0,1]$ is compact and so must contain the purported
accumulation point. It makes no loss to assume that $\gamma(t)=t$,
$f(0)=0$ is the accumulation point, and the line in question
is the real axis. Then $f(z)=a_n z^n+a_{n+1}z^{n+1}+\cdots$ where
$a_n$ is nonzero and $n$ is a positive integer.
At this stage I'll assume there is a sequence of reals $t_1>t_2>\cdots$
tending to zero with each $f(t_j)$ real. we want to show that all the $a_k$
are real. Then considering $f(t_j)/t_j^n$ we get $a_n$ real. Now consider
$f(z)-a_n z^n$ in place of $f(z)$. We get $a_{n+1}$ real etc. So $f$
takes reals to reals so there's no "crossing" of the real axis.
In general, there must be a sequence of distinct points $(t_j)$ in $[0,1]$
where the curve crosses the line and which tends (by Bolzano-Weierstrass)
to a point $t\in[0,1]$. Replace $[0,1]$ by $[0,t]$ or $[t,1]$
(one of these has infinitely many $t_j$) and rescale the interval
to $[0,1]$.

REPLY [4 votes]: I am little worried by Gowers' comment as this seems fairly straightforward to me.
Let $L:{\mathbb C} \to {\mathbb R}$ be a nonconstant affine map vanishing on your ray.  I assume that by "line segment" you really mean that $\gamma$ is linear (or affine).  In any case, as long as $\gamma$ is real analytic (by which I mean that it extends to a real analytic function on some open neighborhood of $[0,1]$), so is $h := L \circ f \circ \gamma$.  A real analytic function on a compact interval is either always zero or has only finitely many zeros.  If $h$ is identically zero, then your curve will stay on the line containing your ray and can enter or leave the ray only finitely many times.  If $h$ is not identically zero, then your curve meets the ray only finitely many times.<|endoftext|>
TITLE: Classification of mapping tori
QUESTION [5 upvotes]: Assume $M$ is a topological space and $f\in \operatorname{Homeo}(M)$, then $f$ obviously plays a significant role in the torus bundle 
$$M_f = M\times I/\{(x,0)\sim (f(x),1)\mid x\in M\}.$$
Hence there should be some general result of the following type: $M_f$ and $M_g$ are bundle isomorphic (resp. diffeomorphic) if and only if "W", where W is a relation between $f$ and $g$.
Can someone help give W and explain? Thank you!

REPLY [5 votes]: Mapping tori come equipped with a projection to the circle: $p\colon M_f\to S^1$. Bundle isomorphism is isomorphism of the total space that commutes with the projection. A weaker notion of isomorphism for mapping tori is an isomorphism that commutes with the projection only up to homotopy. Pseudoisotopy is the name for the relation on isomorphisms of $M$ that replaces isotopy in this coarser notion of equivalence of mapping tori. 
If $M$ is a simply connected manifold of high dimension, then pseudoisotopy implies isotopy and preserving the homotopy class of the map to the circle isn't a big deal, so isomorphism of mapping tori is pretty much isotopy of mapping classes. But if $M$ is not simply connected, there are two ways that mapping tori may be isomorphic without being bundle isomorphic: there may be more pseudoisotopies than isotopies and there may be isomorphisms that do not preserve the homotopy class of the map to the circle.<|endoftext|>
TITLE: What is the German translation of "catenary ring"?
QUESTION [7 upvotes]: I am looking for the correct technical term in German for the notion of catenary ring in commutative algebra.
Does anyone know?

For those who don't know what a catenary ring is but would like to: A Noetherian commutative ring A is called catenary if the following codimension formula holds for irreducible closed subsets T ⊆ Y ⊆ Z of Spec A:

codim(T, Z) = codim(T, Y) + codim (Y, Z).

REPLY [13 votes]: It should be "Kettenring", see for example p. 148 in Brodmann, Algebraische Geometrie.<|endoftext|>
TITLE: Regular borel measures on metric spaces
QUESTION [28 upvotes]: When teaching Measure Theory last year, I convinced myself that a finite measure defined on the Borel subsets of a (compact; separable complete?) metric space was automatically regular.  I used the Borel Hierarchy and some transfinite induction.  But, typically, I've lost the details.
So: is this true?  Are related questions true?  What are some good sources for this sort of questions?  As motivation, a student pointed me to http://en.wikipedia.org/wiki/Lp_space#Dense_subspaces where it's claimed (without reference) that (up to a slight change of definition) the result is true for finite Borel measures on any metric space.
(I'm normally only interested in Locally Compact Hausdorff spaces, for which, e.g. Rudin's "Real and Complex Analysis" answers such questions to my satisfaction.  But here I'm asking more about metric spaces).
To clarify, some definitions (thanks Bill!):

I guess by "Borel" I mean: the sigma-algebra generated by the open sets.
A measure $\mu$ is "outer regular" if $\mu(B) = \inf\{\mu(U) : B\subseteq U \text{ is open}\}$ for any Borel B.
A measure $\mu$ is "inner regular" if $\mu(B) = \sup\{\mu(K) : B\supseteq K \text{ is compact}\}$ for any Borel B.
A measure $\mu$ is "Radon" if it's inner regular and locally finite (that is, all points have a neighbourhood of finite measure).

So I don't think I'm quite interested in Radon measures (well, I am, but that doesn't completely answer my question): in particular, the original link to Wikipedia (about L^p spaces) seems to claim that any finite Borel measure on a metric space is automatically outer regular, and inner regular in the weaker sense with K being only closed.

REPLY [6 votes]: Here's a reason why it appears hard to come up with an example of a non-tight probability measure on a complete metric space:
Theorem: Let X be a complete metric space. Denote by w(X) the smallest cardinality of a basis for the topology on X. Then there is a non-tight probability measure on the class of borel subsets of X iff w(X) is a measurable cardinal (i.e. there is a non-atomic measure on the power set of w(X)).
A proof can be found in Fremlin's Measure theory, volume 4, page 244.<|endoftext|>
TITLE: Largest rank submatrix of a skew symmetric matrix
QUESTION [8 upvotes]: Is the following statement true?
Given a skew symmetric matrix M, among all of its largest rank sub-matrix, there must be one that is the principal submatrix of M.

REPLY [2 votes]: Another proof: For any $I$ and $J$ two subsets of $\{1,2,\ldots,n\}$ of the same cardinality, let $D(I,J)$ be the minor in rows $I$ and columns $J$. Let $Pf(I)$ be the Pfaffian $\sqrt{D(I,I)}$. We set $D(\emptyset, \emptyset) = Pf(\emptyset) =1$.
Lemma: Every $D(I,J)$ can be written as a quadratic polynomial in the $Pf(K)$'s.
Proof of Lemma: I claim that
$$D(I,J) = \sum_{S \subset J \ |J| \ \mbox{even}} (-1)^{|S|/2} \ Pf(I \cup (J \setminus S)) \ Pf(S).$$
(There may be some sign errors.)
Every term on the right can be visualized as corresponding to a perfect matching on the indices $J \sqcup I$. Focus on one particular term. Let $(j_1, j_1')$, $(j_2, j_2')$, ..., $(j_r, j_r')$ be the matchings which lie within $J$. This term will appear only when $S$ is a union of some set of these matchings. The total contribution of this term is thus $\sum_{k=0}^r (-1)^k \binom{r}{k}$. This is $1$ for $r=0$ and $0$ for $r>0$.
Application of lemma:  Suppose that $Pf(K)=0$ for all $K$ with $|K| > r$. Let $|I|=|J|>r$. Write $D(I,J)$ as a quadratic polynomial in the $Pf(K)$'s. Then, by degree considerations, for every monomial $Pf(K) Pf(K')$ in that quadratic polynomial, one of $Pf(K)$ and $Pf(K')$ is $0$. So imposing that all the principal minors of size $>r$ be $0$ implies that the same is true for non-principal minors.<|endoftext|>
TITLE: Complete tree invariants?
QUESTION [7 upvotes]: If we take a graph invariant to be "a property that depends only on the abstract structure, not on graph representations such as particular labellings or drawings of the graph" (from Wikipedia), I have the feeling that Harrsion's question for complete graph invariants remained basically unanswered, since Greg's answer is mainly about (canonical) labellings. 
Thinking - as Harrison did - of "the usual ones (the Tutte polynomial, the spectrum, whatever)", I'd like to repeat Harrison's question, but restrict it to trees:

Are there any known complete tree invariants?

REPLY [7 votes]: @Hans: The path sequence of $G$ is $(p_1, p_2, ...)$ where $p_i$ is the number of $i$-edge path subgraphs of $G$ isomorphic to an $i$-edge path.  For a tree, this is the same as the number of pairs of vertices at mutual distance $i$.  Morin, Wagner and I proved [arXiv:math/0609339] that the path and degree sequences of a tree (actually, something a but more general) can be determined from its chromatic symmetric function, but OTOH the path and degree sequences do not together determine the tree up to isomorphism; there is an 11-vertex counterexample in our article (due to Eisenstat and Gordon).<|endoftext|>
TITLE: Jacobson-Morozov on the algebraic group level
QUESTION [7 upvotes]: Let $k$ be a field of characteristic 0 and $G$ a connected reductive algebraic group defined over $k$. 
Let $g \in G(k)$ be a unipotent element. Is it true that there is a homomorphism $\varphi: SL_2(k) \to G(k)$ mapping some unipotent element $u \in SL_2(k)$ to $g$?
Background: There is some nilpotent element $n \in Lie\, G$ which corresponds to $u$. If $Lie\, G$ is semisimple, the Jacobson-Morozov lemma states that $n$ is part of a $sl_2$-triple, so ''integrating'' would yield the desired homomorphism.
(Apparently, Jacobson-Morozov holds more generally for completely reducible Lie subalgebras of $gl(V)$, though I don't have a precise reference.)

REPLY [2 votes]: $\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\SL{SL}$I am just writing some details on the comment by BCnrd. We use various statements from chapter 7 page 105ff in Borel's 'Linear Algebraic Groups, second enlarged edition'.
We start with a unipotent element $g$ in a linear algebraic group $G$. After conjugation we may assume that $g$ is upper triangular (with $1$'s on diagonal). We can apply the matrix logarithm
$$
X\mathrel{:=} \log (g) \mathrel{:=} \sum_{n=1}^{\infty} (-1)^{k+1} \frac{(g-I)^n}{n},
$$
which is well defined in characteristic 0 since there are only finitely many powers of $(g-I)$ (so it is a finite sum). The matrix $X$ is nilpotent and we can use it to construct an exponential function from the additive group to $G$ (as in 7.4(1) Borel)
\begin{align}
\operatorname{G}_{a} & \to G \\
t &\mapsto \exp (t\cdot X) = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!}
\end{align}
which is an algebraic morphism in characteristic 0 since there are only finitely many powers of $X$. Since we are in characteristic 0, it is also injective. Since $g=\exp(1\cdot X)$, and we are in char. 0, the closed one-dimensional algebraic subgroup of $G$ generated by $g$ is $U:=\exp (\operatorname{G}_a)$. The Lie algebra $\Lie(U)$ consists of nilpotent elements and we can apply Jacobson–Morozow to get an $\mathfrak{sl}_2$-triple (let's call its span $M$) that contains $\Lie(U)$. Following Borel, we define the closed algebraic subgroup
$$
\mathcal{A}(M) := \bigcap_{\text{$H$ closed subgroup of $G$} \\ M \subset \Lie(H) \subset \Lie(G)} H
$$
and take its commutator group
$$
\SL_g \mathrel{:=} [\mathcal{A}(M),\mathcal{A}(M)] \mathrel{:=} \left< ghg^{-1}h^{-1} : g,h \in \mathcal{A}(M) \right>.
$$
Using Prop 7.8 and Cor 7.9 in Borel, we get
$$
\Lie(\SL_g) = \Lie([\mathcal{A}(M),\mathcal{A}(M)]) = [\Lie(\mathcal{A}(M)),\Lie(\mathcal{A}(M))] = [M, M] = M,
$$
where the last equality follows since $M$ is the span of a Lie-triple system. The Lie-algebra of $\SL_g$ is thus isomorphic to $\mathfrak{sl}_2$ and since we are in char. 0, this means that $\SL_g$ has to be isomorphic to $\SL_2$ (or contain a subgroup isomorphic to $\SL_2$).
In other words, an algebraic exponential map $ \operatorname{G}_a \to G$, $t \mapsto \exp(t\cdot \tilde{X})$ factors through $\SL_2$ via
$$
t \mapsto \left( \begin{matrix} 1 & t \\ 0 & 1\end{matrix}\right),
$$
so indeed, the original element $g$ comes from a unipotent upper triangular element in $\SL_2$.<|endoftext|>
TITLE: Finiteness property of automorphism scheme
QUESTION [34 upvotes]: Some time ago I mentioned a certain open question in an MO answer, and Pete Clark suggesting posting the question on its own.  OK, so here it is:
First, the setup. Let $X$ be a projective scheme over a field $k$.  By Grothendieck, there is a locally finite type $k$-scheme $A = {\rm{Aut}}_ {X/k}$ representing the functor assigning to any $k$-scheme $T$ the group of $T$-automorphisms of $X_T$.  (Artin proved a related result with projectivity relaxed to properness, even allowing $X$ to be an algebraic space.)  The construction uses Hilbert schemes, so at most countably many geometric connected components can occur. 
In some cases the automorphism scheme is connected (such as for projective space, when the automorphism scheme is ${\rm{PGL}}_n$), and in other cases the geometric component group $\pi_0(A) = (A/A^0)(\overline{k})$ can be infinite.  For the latter, a nice example is $X = E \times E$ for an elliptic curve $E$ without complex multiplication over $\overline{k}$; in this case $A$ is an extension of ${\rm{GL}}_ 2(\mathbf{Z})$ by $E \times E$, so $\pi_0(A) = {\rm{GL}}_ 2(\mathbf{Z})$.  This latter group is finitely presented. 
Question: is the geometric component group $\pi_0(A)$ of the automorphism scheme $A$ of a projective $k$-scheme $X$ always finitely generated?  Finitely presented? And with projectivity relaxed to properness, and "scheme" relaxed to "algebraic space"?
Feel free to assume $X$ is smooth and $k = \mathbf{C}$, since I believe that even this case is completely wide open.
Remark: Let me mention one reason one might care (apart from the innate appeal, say out of analogy with finite generation of Neron-Severi groups in the general proper case).  If trying to study finiteness questions for $k$-forms of $X$ (say for fppf topology, which amounts to projective $k$-schemes $X'$ so that $X'_K = X_K$ for a finite extension $K/k$), then the language of ${\rm{H}}^1(k, {\rm{Aut}}_{X/k})$ is convenient.  To get hands on that, the Galois cohomology of the geometric component group intervenes.  So it is useful to know if that group is finitely generated, or even finitely presented.

REPLY [17 votes]: The answer is negative even for smooth projective varieties over $\mathbb C$: a counterexample is constructed in http://arxiv.org/abs/1609.06391.  The example is a smooth, uniruled sixfold of Picard rank somewhere in the high 30s.
Here's the basic idea.  One should try to find a variety $X$ whose automorphism group is something like a free group. It's possible to rig things so that there's a point $x$ on $X$ whose stabilizer in $\operatorname{Aut}(X)$ is a non-finitely generated subgroup; free groups have lots of these.  If you blow up $x$, the automorphisms of $X$ that lift to the blow-up are precisely the ones that fix $x$.  One also needs to check that the blow-up doesn't have any automorphisms other than the ones lifted from $X$, but this isn't too hard.  Things don't work out quite this cleanly in the linked example, but this is the strategy.
There's also the issue of how to actually prove the automorphism group isn't finitely generated.  We've arranged that the "obvious" automorphisms we can write down are a non-finitely generated group, but we might worry that there are other automorphisms we don't know about.  Maybe we actually just found a non-finitely generated subgroup of some larger finitely generated group.  The solution is to arrange that there's a rational curve $C$ on $X$ and prove that every automorphism of $X$ must fix $C$, and in fact restrict to $C$ as a map of the form $z \mapsto z+c$.
This means we have a restriction map $\rho : \operatorname{Aut}(X) \to \operatorname{Aut}(C)$, and the image lies in the abelian subgroup of translations fixing $\infty$.  We construct explicit automorphisms $\mu_n$ (for every $n$) whose restrictions to $C$ are the maps $z \mapsto z + 1/3^n$.  The means the image of $\rho$ is an abelian group containing $\mathbb Z[1/3]$, so it's not finitely generated.  This means that $\operatorname{Aut}(X)$ isn't finitely generated either, though I don't know exactly what it is.  If I had to guess, it's probably free on countably many generators.
(I apologize that this is very vague, but the paper is not so long, and kind of fun!)<|endoftext|>
TITLE: What is Serre's condition (S_n) for sheaves? 
QUESTION [37 upvotes]: The Serre's condition $(S_n)$, especially $(S_2)$, has been mentioned in a few MO answers: see  here and here for example. I am pretty sure I have seen it in other questions as well, but could not remember exactly. 
I have always been confused by this condition, especially for a sheaf $F$ on a locally Noetherian scheme $X$, mainly because as far as I know, there are at least three different definitions in the literature. Let us look at them, $F$ is said to satisfy condition $(S_n)$ if:
$$ depth_x (F_x) \ge \min (\text{dim} \mathcal O_{X,x},n) \  \forall x\in X  \ (1) $$ 
$$ depth_x (F_x) \ge \min (\dim F_x,n) \ \forall x\in X  \ (2)  $$ 
$$ depth_x (F_x) \ge \min (\dim F_x,n) \ \forall x\in \text{Supp}(F) \ (3) $$ 
Definition (1) can be found  in Evans-Grifffith book "Syzygies", definition (2) is given in EGA IV (definition 5.7.2) or Bruns-Herzog book "Cohen-Macaulay modules". Definition (3) is  what VA used in his answer to the second question quoted above (and I certainly have seen it in papers or books, but can't locate one right now, so references would be greatly appreciated). 
When $F$ is the structure sheaf or a vector bundle (of constant positive rank), then they all agree. However, they can differ when $E$ is a sheaf. For example, (1) allows us to say that if $X$ is normal, then $E$ is reflexive if and only if it is $(S_2)$. But according to (2) or (3), if $X=Spec(R)$ for $(R,m)$ local, then $k=R/m$ would satisfy $(S_n)$ for all $n$. (2) and (3) are equivalent if we assume that the depth of the $0$ module is infinity, but I have seen papers which do not use that convention, adding to the confusion. 
Since a result surely depends on which definition we use (I have certainly made mistakes because of this confusion, and I think I am not alone), I would like to ask:
Question: Is there an agreement on what exactly is condition $(S_n)$ for sheaves? If not, what are the advantages and disadvantage for each of the different definition? 
Some precise references: 
Bruns-Herzog "Cohen-Macaulay modules" : after Theorem 2.1.15, version (2)
Evans-Griffith "Syzygies": part B of Chapter 0, version (1)
Kollar-Mori "Birational geometry of algebraic varieties": definition 5.2, version (1)
EGA, Chapter 4, definition 5.7.2, version (2).

REPLY [3 votes]: The following is too long for a comment, but I hope it is useful. I would appreciate any additional information anyone might have concerning these notions and their history.
As far as I can tell, definition (1) for $(S_n)$ first appeared in Definition 1.1 of the following paper of Vasconcelos:

Wolmer V. Vasconcelos, Reflexive modules over Gorenstein rings, Proc. Amer. Math. Soc. 19 (1968), 1349–1355. DOI: 10.2307/2036210. MR: 237480.

In this paper, Vasconcelos shows that reflexive modules behave particularly well on rings satisfying $(G_1)$ and $(S_2)$ (in the notation of [Hartshorne 1994, p. 291]), which Hartshorne utilized later in a geometric context. Here, we note that rings satisfying $(G_1)$ and $(S_2)$ are also known as $(G_2)$ rings (following [Ischebeck 1969, Definition 3.16]) or 2-Gorenstein rings (following Auslander; see [Reiten and Fossum 1972, p. 35]).
One reason to use definition (1) for $(S_n)$ is the following:
Theorem (Samuel; see [Malliavin 1968, Proposition 2]). Let $A$ be a noetherian local ring and let $M$ be a finitely generated $A$-module. For every integer $n \ge 0$, the following are equivalent:

$(\mathrm{i}_n)$ For every prime ideal $\mathfrak{p}$ in $A$, we have $\operatorname{depth}(M_\mathfrak{p}) \ge \min\{n,\operatorname{depth}(A_\mathfrak{p})\}$.
$(\mathrm{ii}_n)$ Every regular sequence on $A$ of length $\le n$ is a regular sequence on $M$.

Note that if $A$ satisfies $(S_n)$, then $M$ satisfies $(\mathrm{i}_n)$ if and only if $M$ satisfies $(S_n)$ in the sense of definition (1). In fact, definition (1) for $(S_n)$ is exactly how Samuel originally stated the condition $(\mathrm{i})_n$ in [Samuel 1964, Proposition 6].
In [Ischebeck 1969, Satz 4.4 and Satz 4.6], Ischebeck proved that Samuel's conditions $(\mathrm{i})_n$ and $(\mathrm{ii})_n$ are also equivalent to $M$ being an $n$-th syzygy module under the assumption that $A$ satisfies $(G_n)$ in the sense of [Ischebeck 1969, Definition 3.16].
Ischebeck denotes Samuel's conditions $(\mathrm{i}_n)$ and $(\mathrm{ii}_n)$ by $(\mathrm{b}_n)$ and $(\mathrm{a}_n)$, respectively. Because of the Theorem above, Malliavin says that a module satisfies Samuel's condition $(\mathrm{a}_n)$ if it satisfies either of the equivalent conditions $(\mathrm{i}_n)$ or $(\mathrm{ii}_n)$.<|endoftext|>
TITLE: Which properties of finite simplicial sets can be computed?
QUESTION [11 upvotes]: A simplicial set $X$ is a a combinatorial model for a topological space $|X|$, its realization, and conversely every topological space is weakly equivalent to such a realization of a simplicial set. I am wondering, which properties of finite simplicial sets can be effectively (or even theoretically) computed using a computer program. Maybe there are some implementations and I am not aware of their existence, so I would be very pleased about any information.
The first problem is how to input the simplicial set (maybe that's not really a problem).
Does one know terminating algorithms for the following problems then? Are there other problems for which anything is known in this direction?

Is a given finite simplicial set a Kan complex?
Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homeomorphic?  
Given two finite simplicial sets $X$ and $Y$, are $|X|$ and $|Y|$ homotopy equivalent?
What are the homotopy/homology groups of a given Kan complex? 
Given a finite simplicial set $X$, what are the homotopy/homology groups of $|X|$?

etc.

REPLY [5 votes]: In the simply connected case, essentially everything is in principle computable, by some very early work of E.H. Brown:
\bib{MR0083733}{article}{
   author={Brown, Edgar H., Jr.},
   title={Finite computability of Postnikov complexes},
   journal={Ann. of Math. (2)},
   volume={65},
   date={1957},
   pages={1--20},
   issn={0003-486X},
   review={\MR{0083733 (18,753a)}},
}

In particular, if $X$ and $Y$ are finite simplicial complexes then $[\Sigma^n X,\Sigma^n Y]$ is computable for $n\geq 2$, and for large $n$ this gives the group of stable homotopy classes of maps.  However, I do not think that there are practical algorithms for many such questions, although I am not up to date on this.  Probably the simplest case that I do not know is as follows: is there a practical algorithm to compute the complex $K$-theory $K^0(X)$ for a finite simplicial complex $X$?<|endoftext|>
TITLE: Is there any analogs of Vassiliev invariants in higher dimensions?
QUESTION [8 upvotes]: http://www.artofproblemsolving.com/Forum/blog.php?u=55354&b=32193
Today I asked my TA for the knot theory class whether there exists any analogs for Vassilev invariants, like at least in $ (2,4) $ case (to me the question is how to define singularities properly?). Also I asked him whether there exists any basic classifications (I guess we still use something like triangular moves). He told me to ask the professor, but the professor is not accessible, so I ask in here. I know the question seemed primitive (definitely not a research type question), I just don't know where I can find the needed reference. 
Another question I want to ask is whether there exists any good integral type invariants. I feel it is highly unlikely to find one, but I don't know why. I think one criterion such an integral has to satisfy is it can distinguish small knotted parts in comparison with unknot parts, and it should be able to distinguish images with its mirror images.

REPLY [5 votes]: First, let me recall briefly Vassiliev's theory (the way I understand it). I will skip many important details on the way.
Vassiliev's invariants first appeared as certain cohomology classes of the space of all knots in the 3-space. The strategy is as follows: instead of the space of all knots, which is infinite-dimensional, consider the space $V_d$ of maps from $\mathbf{R}$ to $\mathbf{R}^3$ given by polynomials of some fixed degree $d$ behaving in a prescribed way at infinity. This is a finite dimensional space and non-knots form a hypersurface $\Sigma_d$ in it. Using the Alexander duality one can reduce the computation of the cohomology of the complement of $\Sigma_d$ in $V_d$ to computing the Borel-Moore homology of $\Sigma_d$. [Note: there is a price to pay for replacing an open set with its complement -- we loose the information of how $\Sigma_d$ lies in $V_d$.] The Borel-Moore homology of $\Sigma_d$ can be computed by constructing a suitable semi-simplical resolution of $\Sigma_d$ (i.e. a semi-simplicial space with the geometric realization properly homotopy equivalent to $\Sigma_d$).
The geometric realization of a semi-simplicial space admits a natural filtration, which gives a spectral sequence converging to the Borel-Moore homology of $\Sigma_d$. There is a way to embed $V_d$ into some $V_{d'},d'>d$ so that knots are mapped to knots and unknots to unknots. Moreover, spectral sequences for $d'$  and $d$ are mapped to one another. This corresponds to the restriction map in cohomology, but constructing the maps on the level of spectral sequences is non-trivial. As $d$ goes to $\infty$ a part of the $E^1$ term of the spectral sequence stabilizes. It is a difficult theorem (Kontsevich-Vassiliev) that (at least over the rationals) the second and higher differentials vanish on the stable part and that if an element in the stable part is killed, it is killed in the first sheet by some stable element for all sufficiently large $d$.
So letting $d$ go to $\infty$ we get a vector subspace (in fact, a subring) in the cohomology of the space of all knots (equipped with the Whitney topology), together with a filtration on it. Vassiliev's conjecture says that this subspace is in fact the whole of the cohomology. By taking the $H^0$ part we get Vassiliev invariants and the Vassiliev filtration on them.
The most non-trivial part of the above is the actual construction of the semi-simplicial resolution of $\Sigma_d$ (since this gives a spectral sequence that collapses very fast). 
The same strategy can be applied in many other cases, such as spaces of smooth functions without complicated singularities, spaces of mappings from $m$-dimensional CW-complexes to $m-1$ connected ones, spaces of smooth projective hypersurfaces, classical Lie groups etc. It can also be applied to spaces of knots in 3-manifolds other than $\mathbf{R}^3$ and in higher-dimensional manifolds. For $\mathbf{R}^n,n>3$ this gives a complete description of the cohomology: it is isomorphic to the cohomology of the Hochschild complex of the Poisson operad for $n$ odd and of the Gerstenhaber operad for $n$ even, as shown by V. Turchin arXiv:math/0010017. The case $n=3$ is much trickier. Some of these applications can be considered higher dimensional analogs of Vassiliev knot invariants.
Most of the above is described in detail in "Complements of the Discriminants of Smooth Maps" by Vassiliev. A brief synopsis can be found in Vassiliev's ICM 1994 talk.<|endoftext|>
TITLE: Usefulness of Frechet versus Gateaux differentiability or something in between.
QUESTION [13 upvotes]: If you have a function $V: L \rightarrow \mathbb{R}$, where $L$ is an infinite dimensional topological vector space, there are multiple notions of differentiability.  For $x,u \in L$, $V$ is Gateaux differentiable at $x$ in the direction $u$ if the limit $\underset{t \rightarrow 0}{\lim} \frac{V(x + tu) - V(x)}{t}$ exists.  Supposing that you are in a Banach space, $V$ is Frechet differentiable if the above limit exists for all $u$ in a ball around $x$, and importantly, with the convergence being uniform over this neighborhood.  
The question is, what difference does it make for a function to be Frechet differentiable versus Gateaux differentiable, maybe with respect to proving theorems that generalize the finite-dimensional setting, where the two notions of differentiability more or less agree.  What kind of pathological behavior can functions exhibit that are merely Gateaux differentiable in every direction?  There are also intermediate forms of differentiability between Frechet and Gateaux, defined in terms of uniform convergence of the difference quotients over some preferred family of sets (a bornology).  Are there any intermediate kinds of differentiability that are important?

REPLY [3 votes]: I know this question has been inactive for such a long time, but I will try to add something.
Dieudonne in Treatise on analysis defines an "intermediate" concept of differentiability between Fréchet and Gâteaux ones called quasi-differentiability. Using your notation, $V$ is quasi-differentiable at $x$ if $V \circ g$ is differentiable at 0, for all $g : [0,1] \to L$ which are continuous, right-differentiable in $0$, and $g(0) = x$.
(I am not sure about it, but i think you could caracterize quasi-differentiability in terms of uniform convergence on compact sets, i.e. the limit $ \lim_{t\to 0} \frac{V(x+tu) - V(x)}{t}$ ` is uniform on compact sets.)
If $L$ is finite dimensional, then quasi-differentiability implies Fréchet differentiability. If $V$ is Lipschitz, then Gâteaux differentiability implies quasi-differentiability.
Nice thing about quasi-differentiability is that the theorem of differentiation of composite functions holds (wich is not the case with Gâteaux differentiability). Moreover, if $F$ is quasi-differentiable and $G$ is Gâteaux differentiable, then $F \circ G$ is G-differentiable.
Unfortunately, it's possible to construct a function $F : L^2 \to L^2$ which is everywhere quasi-differentiable, its differential is a right inverse (thus injective), but is nowhere locally injective. This is not possible if $F$ is Fréchet differentiable.<|endoftext|>
TITLE: How to write down the determinant of a quasi-isomorphism?
QUESTION [9 upvotes]: This   question about the determinant of a perfect complex reminded me of an old question that I had.
The construction of the determinant (as in  MR1914072  or  MR0437541) is a difficult piece of mathematics.  However, the definition of the determinant of a single complex is easy.  Indeed,  Jonathan Wise  wrote down the definition in his (2-sentence)  question.  I have read that the difficult part of the construction lies in defining the determinant of a quasi-isomorphism in a functorial way.  Reading the papers, the construction certainly looks difficult, but it would be nice to have a concrete example to convince skeptics.  
Recall how functoriality works.  To every quasi-isomorphism $f \colon E_\bullet \to F_\bullet$, one wants to assign an isomorphism $\operatorname{det}(f) \colon \operatorname{det}(E_\bullet) \to \operatorname{det}(F_{\bullet})$ in such a way that $\operatorname{det}$ preserves composition.  I would like a concrete computation of $\operatorname{det}(f)$ for a non-trivial example.
To make life easy, let's just work with complexes of vector spaces over $\mathbb{C}$.
Here are cases that I can work out:
1) Both $E_{\bullet}$ and $F_{\bullet}$ are both concentrated in a fixed degree.  
Say $E_{\bullet} = F_{\bullet}$ are both complexes concentrated in degree $0$.  Then $f$ is a genuine automorphism, and the map  $\operatorname{def}(f)$ is just the usual determinant that I learned in linear algebra.  In other words, if we fix bases and represent $f$ as a matrix, then, with respect to the natural bases, $\operatorname{det}(f)$ is given by multiplication by the determinant of that matrix.
If we instead consider the case where both complexes are concentrated in degree $d$, then I think $\operatorname{det}(f)$ is $(-1)^{d}$ times the map just described.
2) The complex $E_{\bullet}$ is the zero complex and $F_{\bullet}$ is a 2-terms complex.  
Then $\operatorname{det}(E_{\bullet}) = \mathbb{C}^1$, $\operatorname{det}(F_{\bullet})= \operatorname{Hom}(\bigwedge F_{1}, \bigwedge F_{0})$, and $\operatorname{def}(f)$ is the map that sends $1$ to the top exterior power of the differential $\partial$ of $F_{\bullet}$.  In other words, $\operatorname{det}(f)$ is basically the determinant of the differential for $F_{\bullet}$
Question:  Say that $f \colon E_{\bullet} \to F_{\bullet}$ is a quasi-isomorphism of 2-term complexes.  Fix bases for all of the relevant vector spaces and represent $f$ and the two differentials by 4 different matrices.  How do you explicitly write down the determinant map $\operatorname{det}(f) \colon \operatorname{det}(E_{\bullet}) \to  \operatorname{det}(F_{\bullet})$ in terms of these matrices? 
Note: Tracing through the construction in  MR1914072  or  MR0437541, I think that one gets a description of $\operatorname{def}(f)$ in terms of standard triangulated category constructions.  I am not looking for an explanation of that, I really want a formula that I could show to, say, an undergraduate linear algebra student.
Edit @darij grinberg, thanks for the response.  You are correct that Wise only defined the determinant of a complex.  The answers to that question discussed the fact that there is a natural functor from the category of vector bundles (with morphisms taken to be quasi-isomorphisms) to the category of line bundles with the property that the map on objects is given by Wise's formula.  I am asking for a description of the map on morphisms.  
The relevant functor is defined in the two articles that I referred to as well as the  slides that "YLB" linked to.  I edited the post to include a link to the mathsci review of the articles (and fixed a typo).

REPLY [2 votes]: Denote the basis of $E_0$ (resp. $E_1$, $S_0$, $S_1$) (sorry that I changed the notation because I do not want to have too many f's) by $e_{0,1},\cdots,e_{0,n_0}$ (resp. $e_{1,1},\cdots,e_{1,n_1}$, $s_{0,1},\cdots,s_{0,m_0}$, $s_{1,1},\cdots,s_{1,m_1}$, and the matrix corresponds to $f_0$ (resp. $f_1$) by $F_0$ (resp. $F_1$). It is always possible to choose bases of $E_1$ and $S_1$ such that the matrices corresponding to differentials are of the form 
$$\begin{pmatrix} 
I & 0
\end{pmatrix}$$ 
or its transpose.
Case 1: if $f$ is split injective, then there is an exact sequence
$$0\rightarrow E_.\rightarrow S_. \rightarrow H_. \rightarrow 0$$
with $H_.$ acyclic. Then the determinant of $f$ can be obtained by the combine the isomorphisms $\det E_. \otimes\det H_. \xrightarrow{\sim} \det S_.$ and $\det S_. \xrightarrow{\sim} \mathbb{1}$. 
In this case, you just have invertible matrices $F_0$ and $F_1$, and the basis of $S_.$ can be chosen suitable such that the first half is mapped from $E_.$ and the lower half of basis of $S_.$ contributes to the basis of $H_.$. Also, the map from $S_.$ to $H_.$ is projection. By above, $\det f$ is just a scaler multiple whose scaler is given by $\det F_0 (\det F_1)^{-1}$.
Case 2: otherwise, we can construct an intermediate complex, namely $C_.$, which is defined as $$C_i:=E_i\oplus S_i \oplus E_{i-1}$$ with differentials 
$$d_{C,i}:=
\left( \begin{array}{ccc} 
d_{E,i} & 0 & -1\\ 
0 & d_{S,i} & f_{i+1} \\ 
0& 0& -d_{E,i-1}
\end{array} \right)$$ 
and the maps
$$
\begin{aligned}
&\alpha: E_. \rightarrow C_., e \mapsto (e,0,0)\\
&\beta: S_.\rightarrow C_., s \mapsto (0,s,0)\\
&\gamma: C_.\rightarrow S_., (e,s,e') \mapsto f(e)+s.
\end{aligned}
$$
Then it is easy to check that $\gamma\beta=\operatorname{id}.$ and $\gamma\alpha=f$. So we can apply the method in case 1 to construct $\det f$, which is $\det^{-1}\beta \det\alpha$. 
In your case, by choosing the bases wisely, the only non-trivial part during computing $\det f$ is the determinant of $\alpha$, which is also not too complicated when written as matrix. And the final conclusion is that $$\det f=\det F_0(\det F_1)^{-1}.$$<|endoftext|>
TITLE: Can I relate the L1 norm of a function to its Fourier expansion?
QUESTION [5 upvotes]: I would like to express the integral of the absolute value of a real-valued function $f$ (over a finite interval) in terms of the Fourier coefficients of $f$. Failing that, I would like to know of any constraints or statistical correlations (in a sense explained in the motivation) relating these quantities.
Motivation: This comes from a biophysics application, but is perhaps best explained as follows.  If a rubber band of tension $t$ is stretched along the $x$ axis from $0$ to $L$, then it is easy to calculate the thermal fluctuations of its arc-length by letting $z(x)$ be the (small) deviation from the $x$ axis, and then writing the energy (arc-length times tension) in terms of the Fourier coefficients of $z(x)$.  The Boltzmann weight turns out to be a Gaussian since in the limit of small deviations the arc-length becomes a sum of squares of the Fourier coefficients.  My problem is more complicated: We have two rubber bands stretched over the same interval, with deviations $z_{1}(x)$ and $z_{2}(x)$.  The energy includes not only the stretching of the rubber bands, but also a term proportional to the (positive) area enclosed between them, which is
$\int_{0}^{L}|z_{1}(x) - z_{2}(x)|dx$
Hence my question.  So it would be nice to know how this area can be related to the Fourier coefficients of $z_{1}$ and $z_{2}$ or perhaps just to the arc-lengths of the rubber bands.  By "statistical correlations" I am referring to the Boltzmann probability distribution with energy equal to the stretching energy plus the area-energy.
Edit: Specifics on the Boltzmann probability distribution, more motivation.
The state of the system is the pair of functions $z_{1}(x)$ and $z_{2}(x)$ describing the deviation of the two rubber bands from the x axis.  Let's say it's the set of pairs of functions defined on [0,L] and that these functions are identified with a finite number of Fourier coefficients - I am a physicist and would like to avoid nasty functions or mathematically honest discussions of path integrals.
The probability of occurrence of a state (z_{1}, z_{2}) is (before normalization)
$\exp(-\beta E\left[z_{1},z_{2}\right])$
where $E\left[z_{1},z_{2}\right]$ is the energy of the system, which in this case is the functional
$E\left[z_{1},z_{2}\right] = \frac{t}{2}\int_{0}^{L}\left[(\frac{dz_{1}}{dx})^{2}+(\frac{dz_{2}}{dx})^{2}\right]dx + \kappa \int_{0}^{L}|z_{1}(x)-z_{2}(x)|dx$
Where the tension t and the "surface tension" $\kappa$ are just numbers; set them equal to 1 if you wish.  The first integral is the energy cost of stretching the rubber bands (in a linearized regime) and the second is the strange term proportional to the area enclosed between them.  Without the second term, it is easy to diagonalize this functional in terms of the Fourier series of the two functions. That is why I was interested in writing the second term in terms of Fourier coefficients.  That may be too much to ask, but perhaps it is still possible to calculate some quantities such as the statistical average of $z_1(x)^{2}$ - that is the kind of thing I ultimately want to know.
I realize this is an unnatural-looking problem, so I will just mention that it's not really about rubber bands, but rather about fluctuating interfaces which occur in lipid bilayers with coexisting phases.  There are two phase boundaries (one for each monolayer) with their respective "tensions" but there is also a term proportional to the area between them.

REPLY [5 votes]: This answer really has to do with the physics of it: are you sure about the area energy term?  
Let me simplify a little bit: Consider a case where $z_1(x)=0$ and $z_2(x)=w(x)$. According to your formula I should get a term proportional to $w'(x)^2$ for the kinetic  term  which is typical and no one will object to you for that. You potential energy would then be proportional to $\int_0 ^L |w(x)| dx$. That does raise an alarm: nonlinear problem.
Unless you really have nonlinear physics going on (which is likely, see below) you should have had $w(x)^2$. That would bring back the $L_2$ norm and everything will be very simple. Just transform $z_2(x)=z_1(x)+w(x)$, get rid of $z_2$ and you will get two uncoupled (continuum of) modes, one (corresponding to $z_1$) is a free particle and the other a Harmonic oscillator. Your Boltzmann statistic will determine how these modes are filled up as you know.
If you think of your rubber band as a collection of springs and masses, $w(x)^2$ is the actual term but collections of springs and masses hardly exist outside textbooks and problem sets. As a physicist you know: molecules interact. The actual expansion (if you are still interested in writing an effective field theory) will involve (interacting) terms of the type $w(x)^2+O[w(x)^4]$. Quantitative results in this model might involve renormalization. See the wikipedia page on quartic interaction that even describes how you should quantize it (Fourier transform). My field theory is rusty so you might already know more than I do.
What if $|w(x)|$ is what you have: If $|w(x)|$ is not too small or too large, just approximate it with $[w(x)]^2$ to linearize the problem. People might point out that this is not a mathematically good approximation but it will physically make sense: very small stretchings are not physically possible and very large stretchings would be ruled out by the Boltzmann statistics as they would correspond to exponentially rare high energy modes. So I would just introduce a factor so that $w(x)$ and $w(x)^2$ coincide where $\kappa w(x)^2 L\approx kT$.
There are a famous nonlinear equations that come cheerfully close to your problem but miss it. Example: Sine-Gordon Equation where instead of $|w(x)|$ you would have $1-\cos [w(x)]$. Actually  more up to the point would be the Sinh-Gordon... which reminds me of the Toda field theory which describes a Toda lattice. A Toda lattice is a nonlinear set of coupled equations that describe a set of nonlinearly coupled particles. If you are working in the liquid state, I doubt that they would be relevant. $w(x)^2$ should be good enough but the dissipative terms will be more troublesome.
Edit: Couldn't resist the pun.  Seems that the relevant equation is a "SIGN-Gordon equation":
$$\varphi_{tt}- \varphi_{xx} + \text{sgn}(\varphi) = 0.$$ 
Not sure if it is really simple, messy, or plain difficult to solve in your case. An option is to try solve it like a wave equation with a sign changing external force term and then parametrize the solutions based on their energy and apply the Boltzmann statistics. Forgo the Hamiltonian altogether. I think it will be messy.
Another Edit: as the commenters mentioned before, the solution to the absolute value potential Schrodinger eqn. involves Airy functions. So you could in principle use the Airy function eigenstates to find the time-independent modes and fix their energy. This could quantize your problem. The solution in terms of the Airy functions can be found for example here, [PRL, 94, 176805 (2005)].
Yet Another Update: Apparently you have the "Signum-Gordon" equation. Thanks to this answer.<|endoftext|>
TITLE: Intuition behind Riemann's bilinear relations 
QUESTION [43 upvotes]: I'm trying to understand Riemann's bilinear relations on the normalized period matrix of a Riemann surface.  Recall that they say the following.  Let $X$ be a compact Riemann surface of genus $g>0$.  Fix a standard basis $\{a_1,b_1,\ldots,a_g,b_g\}$ for $H_1(X,\mathbb{Z})$.  We can then choose a basis $\omega_1,\ldots,\omega_g$ for the space of holomorphic $1$-forms on $X$ with the following property.  Define $A_{i,j}=\int_{a_j} \omega_i$.  Then $A_{i,j} = \delta_{i,j}$.  We call $\omega_1,\ldots,\omega_g$ a normalized basis for the set of holomorphic 1-forms on $X$.
Riemann's bilinear relations say that if $\omega_1,\ldots,\omega_g$ is a normalized basis for the holomorphic 1-forms on $X$ and if we define $B_{i,j}=\int_{b_j} \omega_i$, then the matrix $B=(B_{i,j})$ has the following two properties.  First, it is symmetric.  Second, its imaginary part is positive definite.
I understand the proof of this result, but I feel like I have very little geometric intuition as to why it is true.  This leads to the following three questions.

What is the geometric meaning behind the fact that we can choose a normalized basis?
What is the geometric meaning behind the bilinear relations?
One important consequence of the bilinear relations is that the Jacobian of a Riemann surface is an abelian variety.  What is the geometric intuition behind the relationship between the bilinear relations and the fact that we can make the Jacobian into a variety?

Thank you very much for any help.

REPLY [8 votes]: I don't know if this will make matters better or worse for you, but this can be expressed in
basis free language as follows: A  Hodge structure of type $\{(1,0), (0,1)\}$ consists
of a lattice $H$ together with decomposition $H\otimes \mathbb{C}= H^{10}\oplus H^{01}$
such that $\overline{H^{10}}= H^{01}$.When $X$ is a compact Riemann surface $H=H^1(X,\mathbb{Z})$ carries such a Hodge structure where $H^{10}$ is the space of holomorphic $1$-forms. 
A polarization is a skew symmertric integer valued pairing  $\langle,\rangle$ on $H$ satisfying the Riemann bilinear relations, that is it vanishes on $H^{10}$ (i.e. the space is isotropic) and $-i\langle \alpha,\overline{\alpha}\rangle>0$ for $\alpha\in H^{10}$ nonzero. For a Riemann surface, the cup product pairing
$\langle \alpha,\beta\rangle= \int_X \alpha\wedge \beta$ does the job as explained in David Speyer's answer. 
Note that for any Hodge structure as above, there is a
torus $J=H^{01}/H$ which coincides with the
 Jacobian $H^{01}/H= H^1(X,\mathcal{O}_X)/H^1(X,\mathbb{Z})$ when this comes from  a Riemann surface. In general, there is not much more one can do. But when we have a polarization, we can construct sufficiently many quasi-periodic holomorphic functions (called theta functions)
to embed $J$ into projective space. In other words, $J$ is an abelian variety,
This can be done by explicitly writing down certain Fourier series which
can be shown to converge using the Riemann relations. Alternatively, under the identification
$\wedge^2H\cong H^2(J,\mathbb{Z})$, a polarization defines an integral cohomology class.
The Riemann bilinear relations are exactly the conditions for this to be represented
by a positive $(1,1)$ form, so $J$ can be embedded into projective space by Kodaira's
embedding theorem as Mathwonk indicated.<|endoftext|>
TITLE: Why does this sum depend on the Axiom of Choice?
QUESTION [8 upvotes]: On page 168 of Mathematical Fallacies and Paradoxes, it states that the fact that the series 
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $
has a sum depends on the Axiom of Choice. Where does the AC come in to play? I know that if the terms are permuted, we can get any sum we want, and I can see how the AC might be involved there, but just the fact that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $ converges?

REPLY [21 votes]: It doesn't seem to me that you need any choice principle at all to prove that this series converges. The Alternating Series Test that appears in any elementary calculus books seems to do the job, and doesn't seem to require any amount of AC. If $\Sigma_n (-1)^n a_n$ is an alternating series, with $a_n$ descending to $0$, then the finite partial sums up to positive term are descending and the partial sums up to a negative term are increasing, and that these two sequences converge to the same limit. 
So I'm not sure which theorem had been "called upon to show" that the series converges, but unless I am mistaken, it must have been something other than what our students would call the Alternating Series Test.

Edit. More generally, I claim that the convergence of a series can never depend on the Axiom of Choice. Suppose that $r=\langle a_0,a_1,\ldots\rangle$ is a sequence of real numbers, and we are consider the series $\Sigma a_n$. The assertion "$\Sigma a_n$ converges" is a statement about $r$ having complexity $\Sigma^1_1(r)$, that is, an analytic fact about $r$, and therefore has the same truth value in the set theoretic universe $V$ as it has in the relativized constructible universe $L[r]$, where AC holds. In particular, the series converges in $V$ if and only if it converges in $L[r]$. 
Conclusion. If you have a series $\Sigma a_n$ defined in a sufficiently concrete manner and you can prove in ZFC that it converges, then you can prove in ZF that it converges.
(The technical requirement here about sufficiently concrete is that the description of $r=\langle a_0,a_1,\ldots\rangle$ should be absolute from $V$ to $L[r]$. This would be true of any arithmetically definable series as in the question, defined by an arithmetic formula, or even a Borel definition or a $\Sigma^1_2$ definition.)<|endoftext|>
TITLE: Stable graphs: Feynman diagrams and Deligne-Mumford space
QUESTION [9 upvotes]: I do not know very much about quantum field theory, but I have seen, in my reading, that stable graphs can appear in QFT in the form of, I think, Feynman diagrams. By stable graph I mean a "graph with tails", whose vertices are labelled by nonnegative integers, and such that each vertex with labeling 0 has valence at least 3, and each vertex with labeling 1 has valence at least 1.
Algebraic geometers of course know that stable graphs also give a stratification of the Deligne-Mumford spaces $\overline{M}_{g,n}$: Vertices with label $g$ correspond to genus $g$ curves; edges correspond to nodes; tails correspond to marked points. Valency conditions correspond to finitude of automorphism group of the nodal curve.
Is there an explanation for this coincidence?
I guess there is probably some kind of explanation via Gromov-Witten theory. But I get the impression that stable graphs show up in QFTs more generally, and beyond Gromov-Witten theory. Do they? If so, how? And where?

REPLY [6 votes]: I'm not sure exactly what the question is, but let me comment that lots of Feynman graphs with lots of different rules for labelling the vertices come up in QFT, as explained by Theo.  From the QFT point of view, the labelling you're describing is not particularly common; you have infinitely many terms in your Lagrangian, for instance.  So I would say that the answer is no, stable graphs do not come up much beyond the cases that are clearly related to Gromov-Witten theory or other string theories.<|endoftext|>
TITLE: Topological Content of the Kakutani Fixed Point Theorem
QUESTION [14 upvotes]: Reading about general equilibria, the Kakutani fixed point theorem seems to be a central tool. It states (following Wikipedia)
For $S \subset \mathbb{R}^n$, non-empty, compact and convex, and $\phi : S \to 2^S$, if $\phi$ has a closed graph and, for all $s \in S$, $\phi(s)$ is nonempty and convex, then there exists an $x \in S$ such that $x \in \phi(x)$.
It is often called a generalization of the Brouwer fixed point theorem, but I'm not sure I exactly see that as Brouwer holds for any subset homeomorphic to the closed ball, and plenty of those are nonconvex. Regardless, the proofs that I've found on the internet usually proceed by reducing to a simplex. Then, for every subdivision of that simplex into smaller simplices of side length $1/n$, you can generate a function to $S$ (not $2^S$) that agrees with $\phi$ on the vertices. Brouwer that thing, and you get a series of fixed points for each sized subdivision. Pass to a convergent subsequence, take the limit and you prove that you get a fixed point in the sense of the theorem by appealing to convexity of the target sets. Or something like that; I might have missed a detail or three.
Yuck. Maybe I'm betraying my anti-analysis prejudices, but compared to the proof of the Brouwer fixed point theorem using homology, I'm left a little unsatisfied. So, the question is, is there a way to think about the Kakutani fixed point theorem topologically? Perhaps not, given the role convexity seems to play in the proof, but then you could ask, like Brouwer for convex sets, is Kakutani a special case of a more topological theorem?

REPLY [2 votes]: I'm still a student so a bit hesitant to try to make a contribution here, but as an economist quite enamored by topology, I've wondered similar things.  The below is based on an old attempt at a solution to an exercise in a rather challenging book (reference at end).  
Consider first a result on coincidences of certain mappings.  

Theorem Let $X$ be a finite dimensional compact Hausdorff space, $S$ a compact, convex subset of Euclidean space.  Let $f,g: X\to S$ be continuous maps, with $f$ a closed, surjective map with $\tilde{H}^q(f^{-1}(s))=0$ for all $q\ge 0$ and $s \in S$ (where $\tilde{H}^*$ denotes reduced Cech cohomology with coefficients in some module $G$).  Then there exists a coincidence point $f(x) = g(x)$.  

Proof: Define $\partial S$ as the boundary of $S$ and $\partial X = f^{-1}(\partial S)$.  Consider the two long exact sequences in relative cohomology for the pairs $(X, \partial X)$ and $(S, \partial S)$.  By the Vietoris-Begle Theorem and the Five Lemma, $f^*$ acting on the relative groups is an isomorphism, and in particular, is non-zero, given $\tilde{H}^{n-1}(S,\partial S; G) \neq 0$.  
Now, if there were no $x \in X$ for which $f(x) = g(x)$ then using the ray from $g(x)$ through $f(x)$ we can define a function $\bar{f}: X \to \partial S$ that is homotopic to $f$, and which agrees with $f$ restricted to $\partial X$.  But $\bar{f}^*$ is zero on the relative groups because $\bar{f}$ takes X onto $\partial S$, hence $f^*$ is zero, contradicting the above.  QED
Now, consider an upper-hemicontinuous correspondence $\phi: S\to S$.  Let the graph of this correspondence play the role of $X$ in the above, the projection to the domain $f$, and the projection to the codomain $g$. Call $\phi$ "acyclic valued" if the projection to the domain obeys the conditions on $f$ above.  Then coincidence points of $f$ and $g$ are precisely those fixed points $s^* \in \phi(s^*)$.
If you are interested, the above proof is set as an exercise in Repeated Games by Mertens, Sorin, and Zamir (p. 54-55), as well as a number of extensions.<|endoftext|>
TITLE: p-adic L-functions
QUESTION [14 upvotes]: For modular forms, it is known that you can construct p-adic L-functions by interpolating (p-power conductor) twists of their associated L-functions at special values. Similarly, Kubota-Leopoldt's p-adic L-function can be constructed by interpolating Dirichlet L-functions at negative integers. My question is: how general is this method? For example, for which (L-functions of) automorphic representations have p-adic L-functions been constructed?

REPLY [7 votes]: The following is more a long comment than an answer per se.
One thing to keep in mind when discussing $p$-adic $L$-functions is that to a given algebraic automorphic representation $\pi$ or Galois representation $\rho$ is potentially attached several objects which could reasonably called the $p$-adic $L$-function of $\rho/\pi$. Largely for historical reasons, when one speaks of the $p$-adic $L$-function of $\rho$ without further comment, one generally speaks of the $p$-adic $L$-function coming from the cyclotomic $\mathbb Z_{p}$-extension, as I assume you do in your question. The most natural object from a strictly mathematical point of view seems to me to be the $p$-adic $L$-function attached to the universal deformation ring of $\bar{\rho}$ (at least when this universal deformation ring exists). 
Even restricting yourself to the simplest case of the cyclotomic $p$-adic $L$-function, the case of $GL_{n}$ over $\mathbb Q$ has not been done (that I know of) and I doubt (euphemism) that it will follow from the work of Eischen, Emerton, Harris, Li and Skinner (Emerton claims nothing of the sort). Unless I am very much mistaken, the cyclotomic case for $GL_{n}$ over $\mathbb Q$ would be an extremely impressive progress. Somehow, the case of the anticyclotomic $\mathbb Z_{p}$-extension of a CM field is sometimes easier because one can use the Rankin-Selberg method to prove that special values are algebraic and the Rankin-Selberg method is quite amenable to $p$-adic methods. I imagine that this is an ingredient in the work of EEHLS (but I know nothing about it, so please M.Emerton correct me if I'm wrong). 
Leaving the real world for a second: conjecturally, cyclotomic $p$-adic $L$-functions are now constructed for any motive over $\mathbb Q$ (though you will have a really hard time finding this in the literature, as one has to combine an impressive series of very involved papers). Of course, the conjectural construction would not tell you much in way of an actual construction (the conjectural construction gives you an element in some local cohomology group and you will have somehow to identify it as a global element), even though I admit I have been more than mildly impressed by an answer of Idoneal to a question here on MO about $p$-adic $L$-functions here which seems to indicate that analytic argument allows you to do just that in the case of modular forms.
Kevin Buzzard, sure the anticyclotomic $p$-adic $L$-function of an elliptic curve is part (technically, a specialization) of a two-variable $p$-adic $L$-function. In this setting and at least in the ordinary case, this has been known for more than 25 years (it was done in his thesis by S.Haran and later widely expanded by H.Hida in his Invent. Math. 79 paper). And further, this two-variable $p$-adic $L$-function is a specialization of a three-variable $p$-adic $L$-function taking into account variation of the weight in the Hida family passing through this elliptic curve. Even in the finite slope non-ordinary case, I think this three-variable $p$-adic $L$-function is known to exist by the work of A.Panciskin.<|endoftext|>
TITLE: Algorithm for the class field tower problem?
QUESTION [8 upvotes]: This is a spur of the moment algebraic number theory question prompted by a side remark I made in a course I'm teaching:
Let $K$ be a number field.  The (Hilbert) class field tower of $K$ is the sequence defined by $K^0 = K$ and for all $n \geq 0$, $K^{n+1}$ is the Hilbert class field of $K^n$.  Put $K^{\infty} = \bigcup_n K^n$.  We say that the class field tower is infinite if $[K^{\infty}:K] = \infty$ (equivalently $K^{n+1} \supsetneq K^n$ for all $n$).  Golod and Shafarevich gave examples of number fields with infinite class field tower, and thus which admit everywhere unramified extensions of infinite degree.  It is now known that a number field with "sufficiently many ramified primes" has infinite class field tower.
My question is this: is there a known algorithm which, upon being given a number field, decides whether the Hilbert class field tower of $K$ is infinite?

REPLY [5 votes]: There's little if nothing to add to Cam's answer, except that I want to point out that there is a big technical difference between class field towers and $p$-class field towers.
I have never seen any conjecture in the direction of the statement "if $K$ has infinite class field tower, then some subfield of the class field tower has infinite $p$-class field tower for some prime $p$". All known infinite class field towers in fact come from some $p$-class field tower, for which Golod-Shafarevich applies.
Thus general class field towers are a very difficult topic. For $p$-class field towers, on the other hand, I would guess that most specialists indeed think that if such a tower is infinite, then some subfield satisfies the Golod-Shafarevich bound. In this connection, see

F. Hajir, On the growth of $p$-class groups in $p$-class field towers,
J. Algebra 188, No.1, 256-271 (1997)

But even if this were known, there would not be a terminating algorithm for deciding the finiteness of the $p$-class field tower. There are nontrivial cases in which the $2$-tower was shown to be finite; for some recent calculations see e.g. 

H. Nover, Computation of Galois groups associated to the 2-class towers of some imaginary quadratic fields with 2-class group $C_2 \times C_2\times C_2$,
J. Number Theory 129, No. 1, 231-245 (2009)

This approach shows that certain types of class groups in small subfields prevent the $p$-class field tower from becoming infinite for group theoretic reasons. But there's a large gap between these results and Golod-Shafarevich, where no one really knows what is happening.<|endoftext|>
TITLE: Fermat numbers and the infinitude of primes
QUESTION [16 upvotes]: Wonder whether any of you guys know why it is that the proof of the infinitude of primes that is based on the coprimality of any pair of (distinct) Fermat numbers is commonly attributed to Pólya.
In the first paragraph of this letter from Golbach to Euler, there already appears the approach along those lines but since documents crediting it to Pólya altogether are not rare out there, it seems like it's passed unnoticed by a nonzero number of persons.
So, what do you think about this? It's not like Fermat numbers are essential to the proof or that there are no other demonstrations of the result... It's just that I'd really like to know about the origins of this discrepancy between the sources.
UPDATE: Robert Haas implies in 1 that it was Adolf Hurwitz the first mathematician that stated explicitly the fact that the coprimality of any two (distinct) Fermat numbers implies the infinitude of the prime numbers. According to Mr. Haas, Adolf Hurwitz began, in the summer of 1891, a
compilation of number-theoretic problems which he would eventually entitle
"Übungen zur Zahlentheorie" (follow the link if you wish to download a PDF copy of it): the coprimality of any pair of (distinct) Fermat numbers and its relation to the infinitude of the primes is the subject matter of the second entry of this compilation.
In Mr. Haas's paper one can even find a potential explanation as to
why it is that the proof of the infinitude of the primes via the coprimality
of any two (distinct) Fermat numbers is usually attributed to Pólya (or Pólya & Szegö, while we are at it):
"Through most of the twentieth century, until Hurwitz's book [Die Übungen zur Zahlentheorie] was printed in 1993, the primes proof was attributed to Pólya and Szegö, who presented it (without references or claim of originality) as a problem and solution in their famous 1925 'Aufgaben und Lehrsätze aus der Analysis'. But considering that Pólya was Hurwitz's colleague and posthumous editor, the idea may well have come directly from Hurwitz's 'Übungen zur Zahlentheorie'. At any rate, Hurwitz had at least 7 year's priority [the last entry of the 'Übungen zur Zahlentheorie' was added sometime in 1918]."
Nevertheless, in the light of Mr. Lemmermeyer's answer below, I consider that the real priority dispute in this matter is not between Hurwitz and Pólya (or Pólya & Szegö) but between Hermann Scheffler and Adolf Hurwitz. Oddly enough, Scheffler's "Beiträge zur Zahlentheorie, insbesondere zur Kreis und Kugeltheilung, mit einem Nachtrage zur Theorie der Gleichungen" was published in the same year in which Hurwitz began putting together his "Übungen zur Zahlentheorie"!
Do you think that it is possible to determine at this stage of the game whether Scheffler's book had something to do with Hurwitz's impulse to recognize in print what Goldbach apparently never did, i.e., that the pairwise relatively prime sequence of Fermat numbers guarantees the infinitude of the prime numbers? What is more: did Hurwitz have in his possession a copy of Scheffler's book once?
Let me close this update by quoting the paragraph of Mr. Haas's paper wherein he tells us why it is that Goldbach has never received full credit for this approach to the infinitude of the prime numbers:
"Goldbach, having showed that the Fermat numbers are pairwise relatively
prime, clearly had a proof of the infinitude of primes in his hands. But being
absorbed in whether the Fermat numbers are absolutely prime, he overlooked
that consequence of his work. Holding a mathematical proof to be a DELIBERATE
act of reasoned argument, one must therefore award shared credit to his
"collaborator" 160 years later who did notice it, Hurwitz. Goldbach dug out
the ore, and Hurwitz spotted the diamond and showed it off."  
References
[1] R. Haas, Goldbach, Hurwitz, and the Infinitude of Primes: Weaving a Proof
across the Centuries. Math Intelligencer, Vol. 36, 1, 2014.

REPLY [9 votes]: On p. 167 of   Beiträge zur Zahlentheorie, insbesondere zur Kreis- und 
     Kugeltheilung, mit einem Nachtrage zur Theorie der Gleichungen (1891), 
Scheffler deduces the infinitude of primes from the fact that Fermat numbers are pairwise coprime. I don't think that Scheffler's book was widely read, however.<|endoftext|>
TITLE: Partition of R into midpoint convex sets
QUESTION [9 upvotes]: We say that a subset $X$ of $\mathbb{R}$ is midpoint convex if for any two points $a,b\in X$ the midpoint $\frac{a+b}{2}$ also lies in $X$. 
My question is: is it possible to partition $\mathbb{R}$ into two midpoint convex sets in a non-trivial way? 
(trivial way is $\mathbb{R}=(-\infty,a]\cup(a,+\infty)$ or $\mathbb{R}=(-\infty,a)\cup[a,+\infty)$)

REPLY [13 votes]: well, if we assume that $A$ and $B$ are measurable then at least one of them (say A) should have positive measure, and since A+A for a set of positive measure contains an interval, A contains and interval, say $(a,b)$. Then I think it is easy to show that for the maximal interval of this type $a$ or $b$ must be infinity, since otherwise, by taking two sequences in $B$ converging from different sides, you get a contradiction.
Solovey has constructed models of ZF in which every set is measurable, so I think this implies you need something like AC.<|endoftext|>
TITLE: Baer's criterion for functors
QUESTION [7 upvotes]: Baer's criterion can be generalized as follows: Let $A$ be an abelian category satisfying (AB3-5) with a generator $R$ and let $T : A^{o} \to Set$ be a continuous functor such that $T(R) \to T(I)$ is surjective for all subobjects $I \subseteq R$. Then for every monomorphism $M \to N$ the map $T(N) \to T(M)$ is surjective. If $T$ is representable and $A=R-Mod$, this becomes the usual Baer's criterion. The proof is simple.

Does anyone has come across this theorem?
Are there applications to non-representable functors?
Can we somehow put the proof into a general pattern of the form: A statement about monomorphisms can be proven on a "generating system"?
There are striking similarities with the proof that the cohomology of flabby sheaves vanishes. Is there a common generalization?

REPLY [2 votes]: Regarding the original question (with $A=R$-$\mathbf{Mod}$), I think that
by SAFT any continuous functor $A^{\mathrm{op}}\to \mathbf{Set}$ is representable,
and hence the assertion in the original question does not generalize
Baer's theorem.
In detail (with $A=R$-$\mathbf{Mod}$):
(*)  $R$ is a generator in $A$, and hence a cogenerator in
$A^{\mathrm{op}}$.
(*)  $A$ is co-well-powered, because there is a bijection between the
quotient objects of $M\in A$ and the set of submodules of $M$, and the
latter set is small (since by assumption $M$ is small).  It follows
that $A^{\mathrm{op}}$ is well-powered.
(*) $A$ is small cocomplete (as is any $\tau$-algebra, for
$\tau=$(operations, identities)), and hence $A^{\mathrm{op}}$ is small
complete.
(*) Both $A^{\mathrm{op}}$ and $\mathbf{Set}$ have small hom-sets.
So, all the conditions of SAFT hold for a functor
$A^{\mathrm{op}}\to\mathbf{Set}$, 
and hence any such continuous functor has a left adjoint. Now, if a
functor $G\colon A^{\mathrm{op}}\to\mathbf{Set}$ has a left adjoint
then it is surely representable: Saying that a functor $G$ is
representable is like saying that there is a universal arrow from a
one-object set $1$ to $G$ (Prop. 3.2.2, p. 60 in Mac Lane), and for
this we can take the unit $\eta_1\colon 1\to G(F1)$ (with $F$ the left
adjoint of $G$). 
(See also the discussion on Watt's theorem on p. 131 of Mac Lane).
I am not sure about the general case of the edited question
(where $A$ is an arbitrary abelian category with a generator +
AB3--AB5).  Cocompleteness holds by AB3 (as I have seen in
Wikipedia
), but I do
not know enough to say anything about the question of being co-well-powered.<|endoftext|>
TITLE: Kodaira-Spencer map in a concrete instance
QUESTION [10 upvotes]: Let $\pi:X_{\epsilon} \rightarrow \Delta$ be a family of (say smooth) projective plane curves parametrized by $\Delta:=\operatorname{Spec}(k[\epsilon])$, and let $X=X_0$ be the closed fiber.
Suppose that $X_\epsilon$ is given by a polynomial $f(x,y,z;\epsilon)$ homogeneus in $x,y,z$.
Let $\phi=\operatorname{ks}(\pi)=\operatorname{ks}(\partial/\partial\epsilon) \in H^1(X,T_X)=H^0(X,K_X^2)^{*}$ be the Kodaira-Spencer image of the above family.

Is it possible to characterize $\phi$ concretely in terms of the polynomial $f$?

If you want, feel free to restrict to the hyperelliptic case: 
$f(x,y,1;\epsilon):=y^2-\Pi_{i=1}^{2g+2}(x-\lambda_i(\epsilon))$,
(which I think has to be desingularized though)
in which case a basis of $H^0(X,K_X)$ is given by $\frac{x^{k}}{y}dx$ for $k=0 \cdots g-1$.

REPLY [10 votes]: Here is an attempt. Based on your comment to Kevin Lin's post, I think that you know the first part of what I have written, but I included this for the sake of completeness.

Some Generalities on $\phi$:   Any deformation of an affine hyperelliptic curve such as

$$
  y^2 = \prod (x - \lambda_i(\epsilon))
$$
is trivial and hence corresponds to the zero cohomology class.  Indeed, any deformation of a smooth, affine scheme (separated and of finite type over a field?) is trivial.  Given a deformation $X_{\epsilon} \to \Delta$ as you describe, the Kodaria-Spencer map is computed by fixing an open affine cover $U_i$ of $X_0$ and isomorphisms $\phi_i \colon X_{\epsilon}|_{U_i} \to U_{i} \otimes k[\epsilon]$ of the restriction of $X_{\epsilon}$ to $U_i$ with the trivial deformation of $U_{i}$.  The automorphism $\phi_{i} \circ \phi_{j}^{-1}$ of determines an explicit Cech cocyle that represents a class in $H^{1}(X_0, TX_0)$, and one checks that this class is independent of the choices made.  The main point: the Kodaira-Spencer class comes from deforming the gluing data NOT from deforming the equations. 

Computation of $\phi$: As you wrote, it is not clear from that description how everything works in a concrete cases.  Here is how it works out in the case of a general genus $2$ hyperelliptic curve.  Working over the field $k$, this curve
can be described as the curve obtained by gluing the two affine schemes

$$
 U_1 := \operatorname{Spec}(k[x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)),
$$
$$
 U_2 := \operatorname{Spec}(k[x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)),
$$
over the usual opens via the isomorphism $g$ defined by the rules
$$
 x_1 \mapsto x_2^{-1},
$$
$$
 y_1 \mapsto y_2 x_2^{-3}.
$$
Here $r_1, \dots, r_6$ are general scalars.
Associated to the affine open cover $\{U_1, U_2\}$
is the usual Cech complex, and we can use this complex
to compute $H^{1}(X, TX)$.  Some elements of this cohomology
group are given by the Cech cocycles
$$
 y_1/x_1 \frac{\partial}{\partial x_1},  y_1/x_1^{2} \frac{\partial}{\partial x_1},  y_1/x_1^{3} \frac{\partial}{\partial x_1} \in H^{0}(U_{12}, TX).
$$
Here $U_{12}$ denotes the intersection of $U_1$ and $U_2$.  Note: one needs to check that these vector fields are regular on $U_{12}$.
The vector field $\frac{\partial}{\partial x_1}$ has simple poles at ramification points of the degree $2$ to $\mathbb{P}^1$, and the $y_1$ terms
are needed to cancel these poles.  I think these elements form a basis, but 
you just asked for an example so I guess we don't care about this.
Let's compute the 1st order deformation of $X$ associated to $D:= y_1/x_1 \frac{\partial}{\partial x_1}$. To construct the deformation, we take 
the trivial deformations of $U_1$ and $U_2$ and deform the gluing automorphism.  The trivial deformations are
$$
 \operatorname{Spec}(k[\epsilon, x_1, y_1]/(y_1^2 = \prod_{i=1}^{6} (x_1-r_i)), 
$$
$$
 \operatorname{Spec}(k[\epsilon, x_2, y_2]/(y_2^2 = \prod_{i=1}^{6} (1-r_i x_2)).
$$
The general rule is that the deformed gluing map $\tilde{g}$ is given by $\tilde{g}(a) = g(a) + \epsilon \cdot g(D(a))$. For our particular choice of
$D$, I think this yields:
$$
 x_1 \mapsto x_2^{-1} + y_2 x_2^{-2} \epsilon,
$$
$$
 y_1 \mapsto y_2 x_2^{-3} + y_2 x_2^{-2} \frac{-x_2^{-1} q'(x_2) + 6 x_2^{-2} q(x_2)}{2 y_2} \epsilon.
$$
Here $q(x_2) = \prod_{i=1}^{6} (1-r_i x_2)$.
The expression for the image of $y_1$ is quite complicated, but it hopefully is just
$g(y_1/x_1 \frac{\partial y_1}{\partial x_1})$.
One can work our a similar description for the deformations coming from the other cohomology classes that I wrote down.  Assuming these form a basis, this completely describes the map $\phi$.
It is easy to reverse this construct as well.  Every deformation arises by deforming the map $g$ to a map $\tilde{g}$ as we have done.  The associated cohomology class can be described by writing $\tilde{g} = g + \epsilon \cdot D$ for some function $D$.  One can show that $D$ defines a regular vector field on $U_{12}$ and hence represents an element of $H^{1}(X, TX)$.<|endoftext|>
TITLE: why haven't certain well-researched classes of mathematical object been framed by category theory?
QUESTION [22 upvotes]: Category theory is doing/has done a stellar job on Set, FinSet, Grp, Cob, Vect, cartesian closed categories provide a setting for $\lambda$-calculus, and Baez wrote a paper (Physics, Topology, Logic and Computation: A Rosetta Stone) with Mike Stay about many of the interconnections between them. 
But there are mathematical objects that aren't thought of in a category-theoretic fashion, at least the extant literature doesn't tend to treat them as such. For instance nobody talks about Series, Products, IndefInt as being categories in their own right. (infinite series, infinite products, and indefinite integrals, respectively). (google searches for the phrase "the category of infinite series" in both the web and book databases have no hits whatsoever).  I suppose my question is: why not?

REPLY [3 votes]: I really like to think of $\mathbb{N}$-graded $R$-modules as power series $\bigoplus_{n \in \mathbb{N}} M_n \otimes X^{\otimes n}$, where each "coefficient" $M_n$ is an $R$-module and $X$ is the graded module concentrated degree $1$ and which is $R$ there. Therefore, we have a category of power series, where a morphism is just a family of morphisms between the coefficients. Actually this is a symmetric monoidal category - the tensor product is given by some convolution. And the same works if we replace $\mathsf{Mod}(R)$ by any cocomplete symmetric monoidal category. This is spelled out for example in Section 5.4 of my thesis.
In order to get a connection to power series in analysis, we might endow the unit interval $[0,1]$ with the structure of a cocomplete symmetric monoidal category (cf. Example 3.1.6 in loc.cit.): We use the usual ordering to make it a (thin) category and the usual multiplication to make it a symmetric monoidal category. Colimits are given by suprema. Therefore, we get a cocomplete symmetric monoidal category of sequences valued in $[0,1]$. This is again just an order with a multiplication, where we have $(a_n) \leq (b_n)$ iff $a_n \leq b_n$ for all $n$, and $((a_p) \cdot (b_q))_n = \sup_{p+q=n} a_p \cdot b_q$.
Also notice that coends in category theory capture some ideas of (definite) integration. See MO/78471 for some intuition.<|endoftext|>
TITLE: Closed subschemes and pulling back the structure sheaf via the inclusion map
QUESTION [6 upvotes]: I would just like a clarification related to closed subschemes.
If $(X,{\cal O}_X)$ is a locally ringed space and $A\subset X$ is any subset with the subspace topology then $i^{-1}{\cal O}_X$ will be a sheaf of rings on $A$ where $i:A\rightarrow X$ is the inclusion map. (Recall that the inverse image $i^{-1}{\cal O}_X$ is the sheafification of the presheaf $U \mapsto \lim_{V\supset i(U)} {\cal O}_X(V)$  for $U\subseteq A$ open, where the inductive limit is over all open subsets $V$ of $X$ containing $U$.)
Is the reason why we don't do this (and instead start talking about closed subschemes, etc. etc.) just that $(A,i^{-1}{\cal O}_X)$ need not be a scheme even when $X$ is?
Put differently: given any closed subset of a scheme there will be many ways to make it a closed subscheme. What is the relation between the locally ringed spaces on a closed subset making it a closed subscheme and the locally ringed space I have described above which we obtain by pulling back the structure sheaf via the inclusion map.

REPLY [24 votes]: It might help to consider the extreme case when $x$ is a closed point of $X$,
and $i$ is the inclusion $\{x\} \hookrightarrow X$.  The pullback $i^{-1}\mathcal O_X$
is then the stalk of $\mathcal O_X$ at $x$, i.e. the local ring $A_{\mathfrak m}$,
if Spec $A$ is an affine n.h. of $x$ in $X$, and $\mathfrak m$ is the maximal ideal in $A$
corresponding to the closed point $x$.
Now a single point, with a local ring $A_{\mathfrak m}$ as structure sheaf, is not a scheme
(unless $A_{\mathfrak m}$ happens to be zero-dimensional).    
Moreover, the restriction map from sections of $\mathcal O_X$ over $X$ to
section of $i^{-1}\mathcal O_X$ over $x$ is not evaluation of functions at $x$
(which corresponds to reducing elements of $A$ modulo $\mathfrak m$), but is rather
just passage to the germs of functions at $x$.
The idea in scheme theory is that sections of $\mathcal O_X$ should be functions,
and restriction to a closed subscheme should be restriction of functions.  In particular,
restriction to a closed point should be evaluation of the function (if you like, the constant term of the Taylor series of the function), not passage to the germ (which is like
remembering the whole Taylor series).   
If you bear this intuition in mind, and think about the case of a closed point, you will
soon convince yourself that the general notion of closed subscheme is the correct one:
If we restrict functions to the locus cut out by an ideal sheaf $\mathcal I$, or
(in the affine setting) by an ideal $I$ in $A$, then two sections will give the
same function on this locus if they coincide mod $\mathcal I$ (or mod $I$ in the
affine setting), and so it is natural to define the structure sheaf to then be
$\mathcal O_X/\mathcal I$ (or to take its global sections to be $A/I$ in the affine
settin), rather than $i^{-1}\mathcal O_X$.<|endoftext|>
TITLE: Is formal smoothness a local property?
QUESTION [9 upvotes]: Is the following statement true?


Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps $R\to S_{\mathfrak{p}}$ are formally smooth for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally smooth.  


I've looked all over, and I have not been able to find a proof or counterexample for this claim.  It might even be an open problem.  This statement is true for formally unramified and formally étale morphisms, but the proof for formally étale morphisms uses the fact that the module of Kähler differentials is $0$, so it doesn't seem like the same approach will work for the formally smooth case.

REPLY [16 votes]: I think this works. Suppose we have a ring $R$ and an $R$-module $M$ all of
whose localisations are projective and consider $S=S^\ast_RM$, the symmetric algebra
on $M$. Then $R \rightarrow S$ is formally smooth precisely when $M$ is
projective. Let $\mathfrak p$ be a prime ideal of $S$ and $\mathfrak q$ the
prime ideal of $R$ lying below $\mathfrak p$. Then $S_{\mathfrak p}$ is a
localisation of $S_{\mathfrak q}=S^\ast_{R_{\mathfrak q}}M_{\mathfrak q}$ which is a
formally smooth $R_{\mathfrak q}$-algebra and hence $S_{\mathfrak p}$ is a
formally smooth $R$-algebra (localisations being formally étale). To show that
the statement is false it thus is enough to find a non-projective $R$-module all
of localisation are projective.
If $R$ is a boolean ring (i.e., $r^2=r$ for all $r\in R$) then all its
localisations are isomorphic to $\mathbb Z/2$ all of whose modules are
projective. Hence it suffices to find a boolean ring $R$ and a non-projective
$R$-module. This is easy: Let $S$ be a totally disconnected compact topological
space and $s\in S$ a non-isolated point and put $R$ equal to the boolean ring of
continuous maps $S \rightarrow \mathbb Z/2$. Evaluation at $s$ gives a ring
homomorphism $R \rightarrow \mathbb Z/2$ making $\mathbb Z/2$ an $R$-module. If
it were projective there would be a module section $\mathbb Z/2 \rightarrow
R$. Let $e$ be the image of $1$. Then for any $f\in R$ we have $fe=f(s)e$. This
means that $e$ must be the characteristic function of $s$. Indeed, as $e \mapsto
1$ under evaluation at $s$ we have $e(s)=1$. On the other hand, if $t\neq s$
there is an $f\in R$ with $f(t)=1$ and $f(s)=0$ giving $e(t)=f(t)e(t)=f(s)e(t)=0$.
Hence $\{s\}=\{t\in S:e(t)=1\}$ is open so that $s$ is isolated.
Note that we may take for $S$ any infinite compact totally disconnected set as
some point of it must be non-isolated. Nice examples are the Cantor set (all of whose points are non-isolated) and the
spectrum of $\prod_T\mathbb Z/2$ for any infinite set $T$ (which is the
ultrafilter compactification of $T$, any non-principal ultrafilter is
non-isolated I think).<|endoftext|>
TITLE: Global description of the Levi-Civita connection
QUESTION [10 upvotes]: I'm interested in finding a global (coordinate-free) description of the Levi-Civita connection on a (possibly infinite-dimensional) Riemannian manifold X.
I'm not looking for a description of this object as a differential operator.  
Instead, I'm looking for a splitting of the natural map
$\alpha = (\pi_{T(TX)}, D\pi_{TX}): T(TX) \to TX \oplus TX$, where $\pi_{T(TX)}$ is the structure map of the double tangent bundle and $D\pi_{TX}$ is the map on tangent bundles induced by the structure map $\pi_{TX} : TX \to X$. 
Noting that $TX \oplus TX= (\pi_{TX})^* (TX)$, a splitting of $\alpha$ is the analogue, for the vector bundle $TX$, of the standard notion of a connection on a principle bundle: it's a way of lifting tangent vectors on the manifold $X$ up to tangent vectors on the bundle $TX$.
Lang (in GTM 160, Differential and Riemannian Manifolds) explains how to obtain this splitting using the metric spray, which is a map $F: TX \to T(TX)$ that splits both of the above maps $T(TX) \to TX$ (and satisfies another "quadratic" condition).  Lang gives a global description of $F$ as the vector field on $TX$ corresponding, under the metric, to the 1-form -dK, where $K(v) = (1/2)\langle v,v\rangle$ is the kinetic energy functional on $TX$.  However, he doesn't really give a coordinate-free extension of F to the desired splitting.  From studying the discussion in Lang, it seems to me that there is a unique splitting $H: T(TX) \to TX \oplus TX$ satisfying $F(v+w) = (F(v) + H(w,v)) + (F(w) + H(v,w))$ and such that in any local chart U on X, H has the form $H(x, v, w) = (x, v, w, B(x, v, w))$ (as a map $U\times E \times E \to (U \times E) \times (E\times E)$) with $B(x, -, -)$ a symmetric bilinear mapping.  Here E is the Hilbert space on which X is modeled.
The parentheses in the expression $(F(v) + H(w,v)) + (F(w) + H(v,w))$ are important: inside the parentheses, + means addition in the fibers of the map $D\pi_{TX}$, whereas outside the parentheses, + means addition in a fiber of $\pi_{T(TX)}$.  Note that H itself is definitely not symmetric, so I don't think it's clear from the global formula that H exists.
Establishing existence of the map H seems to depend on the rather ugly change-of-coordinate formulas for the "quadratic part" of the spray F, given by Lang.
Lang mentions that the book Symmetric Spaces (Loos, 1969) gives some discussion of this material in terms of second-order jet bundles, and I suspect that may be what I'm looking for.  However, this book is hard to come by.  I can't find any previews on-line, and it's not in our library.  Lang also mentions Pohl's paper "Differential geometry of higher order" (Topology 1 1962 169--211) but I couldn't see anything about the Levi-Civita connection in there.
Does anyone know if Loos has what I'm looking for?  Are there other discussions of these ideas in the literature? Does anyone have other suggestions for how to think about the splitting H?
I'll point out, as motivation, that the splitting $H$ gives a decomposition of $T(TX)$ as a direct sum $\pi^* (TX) \oplus \pi^* (TX)$ of bundles over $TX$ (because the kernel of $\alpha$ is isomorphic to $\pi^* (TX)$, and so this is one way to think about the standard fact that $TX$ is an orientable manifold, with a Riemannian metric inherited from the one on $X$.

REPLY [7 votes]: See my response (number 4) to the MO question:
Exponential map and covariant derivative
There is a Math Review article by Kuranishi there of the paper "Sprays" by Ambrose, Singer, and Palais (in which sprays were first defined). I think the approach taken there (and described in Kuranishi's review) is pretty much what Dan Ramras is asking for.<|endoftext|>
TITLE: What is a "block" in an abelian category?
QUESTION [11 upvotes]: In the literature and in some posts here, there has been variation in the undefined use of the term "block" for a category of modules over a ring, or more abstractly an abelian category (all of which are categories of modules by Freyd-Mitchell).     This raises the natural question:

What is meant by a "block" in an abelian category?

The concept originates to some extent in the modular representation theory of finite groups or their group algebras pioneered by Richard Brauer.
Here a block is just an indecomposable two-sided ideal of the group algebra,
corresponding to a primitive central idempotent.   But in later developments the
language of homological algebra plays a greater role than the group algebra or its center: the category of modules decomposes into a direct sum of subcategories,
which are as small as possible relative to permitting no nontrivial extensions among their simple objects (irreducible representations).    This approach  generalizes well to other situations, where a center or central characters may be elusive and where the decomposition may be infinite, etc.   By now "blocks" occur in many areas of representation theory influenced by classical Lie theory: algebraic groups, restricted enveloping algebras, quantum analogues,
finite $W$-algebras, Cherednik algebras, Kac-Moody algebras and groups, Lie superalgebras.
There is some inconsistency in the literature about allowing "blocks" which might be further decomposed into direct sums.    In classical or Kac-Moody Lie theory
this usually reflects the special influence of infinitesimal/central characters.    But full centers in Lie theory and related quantum groups may be unknown or unneeded, e.g., the approach Kac took to his analogue of the Weyl character formula for integrable modules of an affine Lie algebra relied just on a single Casimir-type operator (which works equally well in the classical finite-dimensional case).  In Jantzen's book Representations of Algebraic Groups (AMS, 2003), the discussion of blocks  for algebraic group schemes in II.7.1 is careful but not completely general.
In practice looser definitions of "block" than the homological one work well enough in many settings, but it creates some confusion when the word is used with no definition at all.  Is there a single convention which reduces in familiar cases to older usage?   In a recent comment to another post I paraphrased  Humpty Dumpty ("my remote ancestor"), who actually said: "When I use a word it means just what I choose it to mean --- neither more nor less." 
But communication is better when the short and convenient word "block" starts out with a common meaning.

REPLY [7 votes]: The classic situation in which blocks are defined is for an artinian ring $A$.  In this case, writing $1=e_1+\cdots+e_n$ as a sum of central primitive idempotents, the blocks are the (two-sided) ideals $A_i=Ae_i$.  For each non-zero indecomposable module $M$, there is a unique $1 \leq i \leq n$ for which $e_iM \neq 0$, and in this case $M$ belongs to the block $A_i=eA_i$.  In particular this partitions the set of irreducible $A$-modules into blocks (more categorically, decomposes $A$-mod as a direct sum of $A_i$-mod's).
There are (at least) two ways in which one might try to generalize this setup:
(1) Weaken the Artinian hypothesis.
(2) Weaken the hypothesis that we are looking at the category of modules for a ring. 
It looks to me like (2) is treated in Torsten's answer, but it looks like no one has mentioned the (by now standard, at least for Noetherian ring theorists) answer for (1): here the replacement for blocks is "cliques". 
Suppose $R$ is a Noetherian ring.  "Ideal" will mean two-sided unless otherwise specified.  Prime ideals $P$ and $Q$ of $R$ are linked if there is a an ideal $I$ with $QP \subseteq I \subset P \cap Q$, and $P \cap Q/I$ non-zero, torsion free as a right $R/P$-module and as a left $R/Q$-module.  The cliques of $R$ are the connected components of the graph with vertex set $\mathrm{Spec}(R)$ and with edges given by links.  (The graph is really directed b/c of the assymetry in the definition: really it should be an arrow from $Q$ to $P$ but for defining cliques this makes no difference). 
To make the definition a little more intuitive, assume $R$ is an algebra over a field $F$, that $M$ and $N$ are irreducible $R$-modules which are $F$-finite dimensional, and that $P$ and $Q$ are the annihilators of $M$ and $N$.  Then it should be a not-too-hard exercise to check that $P$ and $Q$ are linked iff there is a non-trivial extension between $M$ and $N$.
In an Artinian ring, the cliques are the same thing as the blocks (see e.g. pages 142 to 144 of Jategaonkar's book "Localization in Noetherian Rings").  For Noetherian ring theory, the original purpose of introducing cliques was to figure out which multiplicative subsets were "good" for localizing (K. Brown's survey "Ore Sets in Noetherian Rings" is readable and discusses this motivation).<|endoftext|>
TITLE: Compactification theorem for differentiable manifolds ?
QUESTION [28 upvotes]: Just parallelling this question, that seemed not to admit an easy answer at all, let's "soft down" the category and ask the same thing in the case of $\mathcal{C}^{\infty}$-differentiable manifolds [Edit: we consider only manifolds without boundary].
Well, so:

Is every differentiable manifold diffeomorphic to an open submanifold of a compact one?

Edit: As some comments have pointed out, there are manifolds for which the compactification theorem fails, so someone has suggested to change the question to the more meaningful:

Which differentiable manifolds are diffeomorphic to an open submanifold of a compact one?

REPLY [5 votes]: I'm trying to answer the second question. Most of the results can be found in the joint paper with Guilbault https://arxiv.org/abs/1712.05995.
Here I just wanted to make a quick summary.
An $m$-manifold $M^m$ with (possibly empty) boundary is completable if there exists a compact manifold $\widehat{M^m}$ and a compactum $C \subseteq \partial \widehat{M^m}$ such that $\widehat{M^m}-C$ is homeomorphic to $M^m$. In this case $\widehat{M^m}$ is called a (manifold) completion of $M^m$. One can change the homeomorphism to diffeomorphism or PL homeo. for other categories.
Dim = 0,1 are obvious.
Dim = 2, we can't find a complete classification in the literature, so we provided a theorem in that paper. That is, a connected 2-manifold is completable iff it has finitely generated first homology. This is mainly based on classical work of Kerekjarto and Richards.
Dim = 3, it's mainly due to Tucker, where he showed that a
3-manifold can be completed if and only if each component of each clean neighborhood
of infinity has finitely generated fundamental group.
Let me talk about dimensions $\geq 6$ first. Then I'll go back to dimensions 4 and 5. The first breakthrough regarding this problem was due to Siebenmann in 1965. In his PhD thesis, he proved that an open n-manifold $M^n$ is completable (it was called collarable by that time) iff
(1) M is inward tame, (2) the end is pro-$\pi_1$ stable and (3) the Wall finiteness obstruction of the end vanishes.
In 1983, O'Brien generalized the theorem to one-ended manifold with possibly non-empty boundary.
In our paper, we dropped the O'Brien's assumption on that manifolds are one-ended. By properly generalizing Siebenmann's conditions, we proved that manifolds of dimension at least 6 are completable iff they are inward tame, peripherally $\pi_1$-stable at infinity, of zero Wall and Whitehead torsion. Our proof is based on PL manifolds, but one can employ standard techniques such as "rounding off corners" to handle the other catergories.
Our theorem is still true in dimension = 5 provided that the fundamental groups are good in the sense of Freedman and Quinn.
The theorem fails in dimension = 4. Kwasik-Schultz and (independently) Weinberger discovered that there are open 4-manifold satisfying Siebenmann's condition but fail to be collarable.
Just a quick comment on Ian's reference to a contractible open manifold $M'$ constructed by Kister-McMillan which doesn't embed in $S^3$. The example was first proposed by R. H. Bing. Haken further proved that $M'$ doesn't embed in any 3-manifold using his finiteness theorem. Recently, I showed that $M'$ can't be embedded in any compact, locally connected and locally 1-connected metric space. https://arxiv.org/abs/1809.02628<|endoftext|>
TITLE: Does anyone have access to a copy of Yury G. Teterin's 1984 (Russian) preprint "Representation of numbers by spinor genera"
QUESTION [8 upvotes]: Encouraged by
Does anyone have an electronic copy of Waldspurger's "Sur les coefficients de Fourier des formes modulaires de poids demi-entier"?
I realized I could ask for this rare item here.
Again, it is  Yury G. Teterin's 1984 (Russian) preprint "Representation of numbers by spinor genera."
For whatever reason (possibly length?) it never made it into the usual
Zapiski Nauchnykh Seminarov Leningradskogo Otdeleniya Matematicheskogo Instituta im. V. A. Steklova AN SSSR.
Let me give some detail. I do not know the number of pages. The preprint probably never appeared elsewhere, at least not under the same title. It was probably never translated into English. No American mathematics library admits to having it. The easiest place to find mention of the preprint is in Math Reviews MR0732548 (86d:11042) which is a review, by Oleg M. Fomenko, of Schulze-Pillot's 1984 "Thetareihen positiv definiter quadratischer Formen." (I do not know how to make the MR reference a link).
The preprint is mentioned in a later related item by the same author that has been translated, see
http://www.springerlink.com/content/t701481j73531761/
as well as one by Elena P. Golubeva
http://www.springerlink.com/content/w740g82624753417/
Finally, I have emailed Teterin with no result so far. If nothing happens for a long time I could write to O. M. Fomenko and see what happens. The three people in St. Petersburg that I have mentioned are
http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=25046
http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=22736
http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=33524
Well, thanks for any assistance. Individual replies are always welcome, one of my email addresses can be found through   http://www.ams.org/cml
EDIT, May 2021:  the pdf is on one of my websites
KAP
direct link
TETERIN
Let me point out that there are the strongly related Golubeva_Fomenko_1984.pdf in English translation, as well as Schulze_Pillot_1984_Darstellungsmasse.pdf and
Schulze_Pillot_1984_Thetareihen.pdf in these cases still in German. Also Math Reviews of most of the papers for which I lack a translation into English.

REPLY [2 votes]: I think you can try to call to the library of PDMI.
Most libraries copy materials and mail it to you for a reward.<|endoftext|>
TITLE: Connected subset of matrices ?
QUESTION [9 upvotes]: Let $m,n$ be positive integers with $m \leqslant n$, and denote by $\mu_M$ the minimal polynomial of a matrix.
Do we know for which $m$ the set $E_m$ of $M \in \mathfrak{M}_n(\mathbb{R})$ such that $\deg(\mu_M) = m$ is connected?

REPLY [5 votes]: The answer is always yes. Indeed the set is path-connected.
Let $C(f)$ denote the companion matrix associated to the monic
polynomial $f$. Every matrix $A$ is similar to a matrix in rational
canonical form:
$$B=C(f_1)\oplus C(f_1 f_2)\oplus\cdots\oplus C(f_1 f_2,\cdots f_k)$$
where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of
$f_1 f_2\cdots f_k$. Starting with $B$ deform each $f_i$ into a power
of $x$. We get a path from $B$ to
$$B'=C(x^{a_1})\oplus C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$
inside $E_m$. There's a path from $B'$ in $E_m$ given by
$$(1-t)C(x^{a_1})\oplus (1-t)C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$
ending at
$$B_m=O\oplus C(x^m).$$
Thus there is a path in $E_m$ from $A$ to $UB_mU^{-1}$
where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there is
a path in $GL_n(\mathbf{R})$ from $U$ to $I$ and so a path in $E_m$
from $A$ to $B_m$. If $m< n$ then there is a matrix $V$ of negative determinant
with $VB_m V^{-1}=B_m$ so that we may take $U$ to have positive determinant.
The only case that remains is when $m=n$. In this case $E_m$ contains
diagonal matrices with distinct entries, and each of these commutes
with a matrix of negative determinant.<|endoftext|>
TITLE: What are some examples of interesting uses of the theory of combinatorial species?
QUESTION [79 upvotes]: This is a question I've asked myself a couple of times before, but its appearance on MO is somewhat motivated by this thread, and sigfpe's comment to Pete Clark's answer.
I've often heard it claimed that combinatorial species are wonderful and prove that category theory is also useful for combinatorics. I'd like to be talked out of my skepticism!
I haven't read Joyal's original 82-page paper on the subject, but browsing a couple of books hasn't helped me see what I'm missing. The Wikipedia page, which is surely an unfair gauge of the theory's depth and uses, reinforces my skepticism more than anything.
As a first step in my increasing appreciation of categorical ideas in fields familiar to me (logic may be next), I'd like to hear about some uses of combinatorial species to prove things in combinatorics.
I'm looking for examples where there is a clear advantage to their use. To someone whose mother tongue is not category theory, it is not helpful to just say that "combinatorial structures are functors, because permuting the elements of a set A gives a permutation of the partial orders on A". This is like expecting baseball analogies to increase a brazilian guy's understanding of soccer. In fact, if randomly asked on the street, I would sooner use combinatorial reasoning to understand finite categories than use categories of finite sets to understand combinatorics.
Added for clarification: In my (limited) reading of combinatorial species, there is quite a lot going on there that is combinatorial. The point of my question is to understand how the categorical part is helping.

REPLY [6 votes]: To see a specific example of species applied to a fairly difficult enumeration problem (counting bipartite blocks), see my paper with Andrew Gainer-Dewar, Enumeration of Bipartite Graphs and Bipartite Blocks, Electronic Journal of Combinatorics, Volume 21, Issue 2 (2014) Paper #P2.40. If you're really serious you might compare our approach to counting bipartite graphs with Hanlon's.<|endoftext|>
TITLE: Free division rings?
QUESTION [13 upvotes]: Does it make sense to talk about, say, the free division ring on 2 generators? If so, does the free division ring on countably many generators embed into the free division ring on two generators?

REPLY [17 votes]: There is a notion of free division ring, due to Paul Cohn.  I don't know a good online reference, but this AMS bulletin article talks about the construction and gives references.  They go by the name of "free skew field" rather than "free division ring".
In the usual category of division rings and ring homomorphisms, there is no free division ring in the sense of category theory, but Cohn's free division rings are pretty close.  If you consider the category of division rings with specializations as morphisms, then Cohn's construction gives you precisely the free objects.
Let $D, D'$ be division rings.  Then a specialization is a homomorphism $R \to D'$, where $R$ is a subring which generates $D$ as a division ring.  (In other words, the smallest division ring within $D$ containing $R$ is $D$ itself.)  Specializations are a reasonably natural idea.  For example, let $C$ be the complex numbers, and $C(x)$ be the field of rational functions over $C$.  Then all specializations between $C(x)$ and $C$ are given by sending $p(x)$ in $C(x)$ to the value $p(a)$ for some $a$ in $C$.  This is not defined for all rational functions, but only for the rational functions that don't have a pole at $a$.
If you restrict to the category of fields over a fixed base field $k$ and specializations, the free fields are just of the form $k(x_1, \ldots, x_n)$, where $x_1, \ldots, x_n$ are indeterminates.  In the noncommutative case, you get something much more complicated, but free division rings can be realized as subrings of the noncommutative analogue of Laurent series.<|endoftext|>
TITLE: Request:  intermediate-level proof: every 2-homology class of a 4-manifold is generated by a surface.
QUESTION [11 upvotes]: Hi, everyone:
For the sake of context, I am a graduate student, and I have taken classes in
algebraic topology and differential geometry. Still, the 2 proofs I have found
are a little too terse for me; they are both around 10 lines long, and each line
seems to pack around 10 pages of results. Of course, I am considering cases for
"reasonable" spaces, being the beginner I am at this point.
It would also be great if someone knew of similar results for H_1 (equiv. H_3).
Thanks in Advance.

REPLY [14 votes]: Torsten's answer was good, but there are also more elementary answers.  Here's one, which is essentially a big transversality argument, followed by a mild de-singularization.
Let me consider $M$ to be a triangulated 4-manifold, and represent a class in $H_2(M)$ as a sum of 2-faces of the triangulation.  For each 2-face appearing with multiplicity $n$ in the sum, take $n$ copies of the face, pushed off of each other slightly and meeting only at the edges.
Now, along each edge (1-face) of the triangulation, since we started with a 2-cycle the total number of triangles meeting there is $0$.  (This is a signed count if we are working in $H_2(M; \mathbb{Z})$, and means that there are an even number if we are in $H_2(M; \mathbb{Z}/2)$.)  Along each edge, pair up the incident triangles in an arbitrary way that's compatible with the orientations, and resolve the intersections along the interior of the edge according to that pairing.  (A neighborhood of the edge looks like $D^3 \times I$; the incident triangles are coming in from fixed directions, i.e., at fixed points in $S^2 \times I$; so given any pairing of the points on $S^2$, we can just join them up and avoid the edge altogether.  It's easier to think about what happens in a 3-manifold, where it's very similar, you just have to be more picky about how you pair the incident triangles.)
After the last step, we have a surface $S$ with codimension-2 singularities at points in $M$.  For each such singularity, consider a small ball $B$ in $M$ and consider $S \cap \partial B$.  This is a link $L$ in a 3-manifold (oriented or not, depending on which homology we look at).  Replace $S \cap B$ with a Seifert surface for $L$, and we're done.  (You could also use any surface with boundary $L$ inside the 4-ball $B$ instead of the Seifert surface, of course.  Frequently you can get lower genus that way.)<|endoftext|>
TITLE: Sigma algebra without atoms ?
QUESTION [11 upvotes]: I'm looking for an example of a set S, and a sigma algebra on it, which has no atoms.
Motivation: It seems to me that a lot of definitions in probability and stochastic processes - conditional probability, filtrations, adapted processes - become a lot simpler if phrased in terms of a sample space partitioned into atoms.
In the book I'm reading, this is done for finite sample spaces.
But the only problem I find with extending the definitions to general spaces seems to be that, in general, you might have a sigma algebra without any atoms.
(Note: All this definitions can be made without introducing a measure, so objections on grounds of uncountability don't apply.)
Does anyone have an example ?
2) Related question: What if you replace sigma algebra by algebra - closed under finite, rather than countable unions ?
I suspect that the algebra of subsets of R generated by open intervals has no atoms, but can't prove it rigorously.

REPLY [8 votes]: A good source for results on atomless $\sigma$-algberas is chapter 3 of Borel Spaces (1981) by Rao and Rao (Diss. Math.). Here are two remarks, taken from this text: 

There is no atomless countably generated $\sigma$-algebra.
Let $\mathcal{C}$ be a countable set of generators for a $\sigma$-algebra on $X$. W.l.o.g., we can assume that $\mathcal{C}$ is closed under complements. Then for every $x\in X$, the set  $A(x)=\bigcap \{C:x\in C,C\in\mathcal{C}\}$ is measurable in $\sigma(\mathcal{C})$ as a countable intersection of measurable sets. Since one cannot separate points by $\sigma(\mathcal{C})$ that one cannot separate by $\mathcal{C}$, for each $x$, $A(x)$ is an atom of $\sigma(\mathcal{C})$
On every uncountable set, there exists an atomless $\sigma$-algebra that separates points.
Let $X$ be an uncountable set. Clearly, it suffices to show that an atomless $\sigma$-algebra that separates points exists on a set with the same cardinality as $X$. The set $Y$ of all finite subsets of $X$ has the same cardinality as $X$, so we work with $Y$. We verify that the $\sigma$-algebra on $Y$ generated by elements of the form $G_x=\{F:x\in F\}$ for some $x\in X$ does the job. 
It is obvious that this $\sigma$-algebra separates the elements of $Y$. So if there were any atom, it would be a singleton $\{F\}$ and it would be generated by countably many of the $G_x$. So let $C$ be a countable subset of $X$ such that $\{F\}\in\sigma\{G_c:c\in C\}$. For the reason pointed out in 1., $\{F\}$ would be the intersection of all elements in $\{G_c:c\in C\}\cup\{G_c^C:c\in C\}$ that contain $F$. Now $F\in G_c$ only if $c\in F$. So let $x\notin F\cup C$. Then $F\cup\{x\}\in G_c$ for all $c\in F$ and $F\cup\{x\}\in G_c^C$ for $c\notin F$. So the intersection doesn't contain only $F$, which is a contradiction.<|endoftext|>
TITLE: Can analysis detect torsion in cohomology?
QUESTION [72 upvotes]: Take, for example, the Klein bottle K.  Its de Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}_2$ in dimension 1.  It is in general true that de Rham cohomology ignores the torsion part of singular cohomology.  This is not a big surprise since de Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion.  I feel like I should know this, but what can I say...
Thanks!

REPLY [43 votes]: You can compute the integer (co)homology groups of a compact manifold from a Morse function $f$ together with a generic Riemannian metric $g$; the metric enters through the (downward) gradient flow equation
$$ \frac{d}{dt}x(t)+ \mathrm{grad}_g(f) (x(t)) = 0 $$ 
for paths $x(t)$ in the manifold.
After choosing further Morse functions and metrics, in a generic way, you can recover the ring structure, Massey products, cohomology operations, Reidemeister torsion, functoriality.
The best-known way to compute the cohomology from a Morse function is to form the Morse cochain complex, generated by the critical points (see e.g. Hutchings's Lecture notes on Morse homology). Poincaré duality is manifest.
Another way, due to Harvey and Lawson, is to observe that the de Rham complex $\Omega^{\ast}(M)$ sits inside the complex of currents $D^\ast(M)$, i.e., distribution-valued forms. The closure $\bar{S}_c$ of the the stable manifold $S_c$ of a critical point $c$ of $f$ defines a Dirac-delta current $[\bar{S}_c]$. As $c$ varies, these span a $\mathbb{Z}$-subcomplex $S_f^\ast$ of $D^*(M)$ whose cohomology is naturally the singular cohomology of $M$. 
The second approach could be seen as a "de Rham theorem over the integers", because over the reals, the inclusions of $S_f\otimes_{\mathbb{Z}} \mathbb{R}$ and  $\Omega^{\ast}_M$ into $D^\ast(M)$ are quasi-isomorphisms, and the resulting isomorphism of $H^{\ast}_{dR}(M)$ with $H^\ast(S_f\otimes_{\mathbb{Z}}\mathbb{R})=H^\ast_{sing}(X;\mathbb{R})$ is  the de Rham isomorphism.<|endoftext|>
TITLE: Pointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials?
QUESTION [13 upvotes]: The following describes an extension of the Descartes Rule of Signs to polynomials with complex coefficients.
First, I need to define the notion of a "sweep"... Given a complex polynomial p(z) := c0 zm0 + c1 zm1 + ... + cn zmn with non-zero ci and strictly increasing mi, let {αi} be the coefficients' "non-decreasing argument sequence": αi = arg ci (mod 2π), and α0 ≤ α1 ≤ ... ≤ αn, with each αi+1-αi taken as small as possible; then αn-α0 computes the "angular sweep" --which I'll denote sweep(p)-- of a needle with one end anchored at the origin of the Complex Plane that begins pointing at c0 and then spins counter-clockwise to point at each ci in turn ("stalling in place" when consecutive coefficients have equal arguments).[*]
With this, we have:
The Descartes Rule of Sweeps. The number of positive real roots of p is at most $\lfloor \frac{1}{\pi} sweep(p)\rfloor$.
(The Descartes Rule of Signs represents a special case: each sign change in a polynomial's real coefficient sequence contributes π to the sweep, so that $\frac{1}{\pi}sweep(p)$ exactly counts those sign changes.)
Now, I can prove the Rule of Sweeps using the Descartes Rule of Signs itself, but that approach sheds no light on why the Rule of Signs works (which is what I'm really after). The result seems to be one clever contour away from falling directly out of the Cauchy Argument Principle --and it even seems curiously appropriate that the formula provides a kind of half-winding number from the coefficient sequence-- but apparently I'm not sufficiently clever. :(
Suggestions?
[*] Note: One can also compute a sweep of the coefficients taken in reverse order (or, equivalently, spinning the needle clockwise). Differently-directed sweeps are usually not equal, so we can optimize the bound in the Rule of Sweeps by taking "the" sweep as the minimum of the two directed sweeps.

REPLY [3 votes]: This paper is related to generalizing Sturm's theorem rather than Descartes' rule of signs, but I think it is relevant:
http://arxiv.org/abs/0808.0097<|endoftext|>
TITLE: Sign-Gordon Equation
QUESTION [5 upvotes]: What can be said and done about the "SIGN-Gordon equation"?
$$\varphi_{tt}- \varphi_{xx} + \text{sgn}(\varphi) = 0.$$ 
It came up here.

REPLY [10 votes]: Turns out it appears in literature as the "signum-Gordon equation". For example the paper 
Signum-Gordon wave equation and its self-similar solutions<|endoftext|>
TITLE: For a representation V of a finite group G, when is Hom(W, W⊗V) trivial for all irreps W?
QUESTION [6 upvotes]: This is probably really easy, but I just need someone to help me get mentally unstuck.  As part of a description of the McKay correspondence, I want to show that if $G$ is a finite subgroup of $SU(2)$ and $V$ the corresponding 2-dimensional representation, then $\dim \text{Hom}(W, W \otimes V) = 0$ for any irreducible representation $W$ of $G$.  I suspect the result is true in slightly greater generality, but it clearly can't always be true.  Since $\dim \text{Hom}(W, W \otimes V) = \dim \text{Hom}(W \otimes W^{\ast}, V)$, the result is false if, for example, $V \simeq W \otimes W^{\ast}$ or is a direct summand thereof for some $W$.
So I am wondering when, for a given $G$ and $V$, it is always true that $\dim \text{Hom}(W, W \otimes V) = 0$ for all irreducible representations $W$.  One can easily reduce to the case that $V$ is irreducible.

REPLY [2 votes]: For a compact group $G$ one can define the following equivalence: given  two irreps $X$ and $Y$,  $X \sim Y$  if they both appear as summands in  a finite string of tensor products of irreps $X_1 \otimes X_2 \otimes \dots X_n$. The equivalence classes have the structure of an abelian group which turns out to be the dual of the centre of $G$. This was conjectured in  http://arXiv.org/abs/math/0311170 and proven in http://arxiv.org/abs/math/0312257.
Thus $Hom(W \otimes W^*,V) \neq 0$ iff V is in the identity class (i.e. the centre acts trivially on $V$).<|endoftext|>
TITLE: Potential semi-stability of etale cohomology of etale covers.
QUESTION [13 upvotes]: Let $\mathbf{Q}_p$ denote the field of $p$-adic numbers.
Suppose that $K/\mathbf{Q}_p$ is a finite extension, and let $O_K$ denote the ring of integers of $K$. Suppose that $X$ is proper over $O_K$, with smooth generic fibre. Consider the following three statements:
A. There exists a finite extension $L/K$ such that $X/L$ has a semi-stable model over $O_L$.
B. There exists a finite extension $L/K$ such that, for every proper etale map $Y \rightarrow X$, the etale cohomology $H^i(Y/\overline{K},\mathbf{Q}_p)$ is semi-stable when considered as a representation of $\mathrm{Gal}(\overline{K}/L)$.
C. There exists a finite extension $L/K$ such that $H^i(X/\overline{K},\mathbf{Q}_p)$ is semi-stable when considered as a representation of $\mathrm{Gal}(\overline{K}/L)$.
In light of Tsuji's proof of the semi-stable conjecture of Fontaine and Jannsen, we know that  $A \Rightarrow B \Rightarrow C$ (the point being that a semi-stable model for $X$ pulls back to one for $Y$). On the other hand, Tsuji also proved $C$ without proving $A$ by using de Jong's theory of alterations.
My question: Can one also use Tsuji's arguments to prove $B$? One could imagine some argument with alterations being compatible with taking etale covers, but this is not my field so I would rather ask an expert.

REPLY [4 votes]: Answered to my satisfaction. Well, except for the bit about typesetting $\mathbf{Q}_p$, that still confuses me.<|endoftext|>
TITLE: Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$
QUESTION [8 upvotes]: For a finite dimensional $k$-algebra $A$, each $A^{\otimes n}, n \geq 0$ is a $k$-algebra ($A^0 = k $).  Let $T= \prod_{n \geq 0} A^{\otimes n}$.  This is a $k$-alegbra with unit $(1,1,\dots)$ and multiplication is component-wise.  Let $\Delta^{(n)} : A^{\otimes n} \to T \otimes T$ be the deconcatenation map
$$ \Delta^{(n)}(a_1 \otimes \dots \otimes a_n) = \sum_{i=0}^n (a_1 \otimes \dots \otimes a_i) \otimes (a_{i+1} \otimes \dots \otimes a_n ). $$
I want to extend these $\Delta^{(n)}$ to a comultiplication $\Delta : T \to T \otimes T $.  This does not seem to work in a straightforward way because if $t = ( t_0, t_1, \dots ) \in T, $ then $\sum_n \Delta^{(n)}(t_n)$ may not be a finite sum of pure tensors in $T \otimes T$ (I have not shown this sum can be infinite, but suspect it can be).

Is there a way to make $T$ into a Hopf algebra so that $\Delta(t) = \Delta^{(n)}(t)$ when $t_i=0$ for $i \ne n$?  If not, is there an algebra similar to $T$ where this does work?
Is there a standard way to complete the tensor product and instead get a map $\Delta$ from $T$ to the completion?  Does this give rise to a genuine Hopf algebra or some generalization of Hopf alegbras?

REPLY [9 votes]: You can indeed complete the tensor product and get a good comultiplication, but it's not strictly speaking a Hopf algebra. An algebraic geometer would call it an affine formal group. If you think of the infinite product $T=\prod_n A^{\otimes n}$ as a pro-object indexed by $\mathbb{N}$ with $T(n) = \prod_{i=0}^n A^{\otimes i}$, then you can tensor $T$ with itself and get a pro-object indexed by $\mathbb{N} \times \mathbb{N}$. Take the limit and that's the completed tensor product you want. Alternatively, you can do it with topologized rings and topologically-complete the tensor product.
For the uncompleted tensor product, there is no comultiplication extending your rule. There is, however, a cofree Hopf algebra (as Anton points out, consult Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint?). That's a sub-Hopf-algebra of the completed Hopf algebra $T$.<|endoftext|>
TITLE: How often are irrational numbers well-approximated by rationals?
QUESTION [27 upvotes]: Suppose $x\in \mathbb{R}$ is irrational, with irrationality measure $\mu=\mu(x)$; this means that the inequality $|x-\frac{p}{q}|< q^{-\lambda}$ has infinitely many solutions in integers $p,q$ if and only if $\lambda < \mu$. A beautiful theorem of Roth asserts that algebraic numbers have irrationality measure $2$. For $\lambda<\mu$, let $\mathcal{Q}(x,\lambda) \subset \mathbb{N}$ be the (infinite) set of all $q$ occuring in solutions to the aforementioned inequality.
Question: For which pairs $(x,\lambda)$ does $\mathcal{Q}(x,\lambda)$ have positive relative density in the positive integers? For which pairs $(x,\lambda)$ does the cardinality of $\mathcal{Q}(x,\lambda) \cap [1,N]$ grow like a positive power of $N$?

REPLY [8 votes]: An answer to your second question may be found in an old paper of Erdos: Some results on diophantine approximation. Acta Arith. 5 1959 359–369 (1959). (MathSciNet)
He proves that for $1 < \lambda < 2$, for almost every $x$, the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$.  More precisely, Erdos proves that a one-sided version of your set $$\mathcal{Q}'(x,\lambda) = \big\{ q \geq 1 \ \big| \ \text{there exists } p, \ (q,p) = 1, \text{ such that } 0 < x-\frac pq < q^{-\lambda} \big\}$$ satisfies $$\lim_{n \to \infty} \frac{\big |\mathcal{Q}'(x,\lambda) \cap [1,n] \big| }{\sum_{q=1}^n q^{1-\lambda}} = \frac{12}{\pi^2}.$$ That the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$ is consistent with the fact that $\mathcal{Q}(x,1)$ is all of $\mathbb{N}$ while the numbers in $\mathcal{Q}(x,2)$ coming from continued fraction convergent denominators grow exponentially.<|endoftext|>
TITLE: Numbers characterized by extremal properties
QUESTION [6 upvotes]: The golden ratio $\phi=\frac{1+\sqrt5}2$ is sometimes said to be one of the most difficult numbers to approximate with rational numbers, because its continued fraction development $$\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}}$$has all entries equal to $1$, so that its convergents are as far from $\phi$ as possible. This can be quantified and made precise (I guess: I'd love to know the precisely how!) It shares this property with all the numbers $\frac{a+b\phi}{c+d\phi}$ with $\left(\begin{smallmatrix}a&b\\\c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb Z)$ (and no others?)
Are there other numbers which are characterized by such extremal properties? Is there a second worst aproximable number?

REPLY [13 votes]: Certainly there is one type of study of this question, coming under the heading of the Markoff Spectrum, going back to the German tranliteration of Markov. There is a very nice book called "The Markoff and Lagrange Spectra" by Thomas W. Cusick and Mary E. Flahive. It is this topic that lead to the Markoff Numbers, which are any of the triple $x,y,z$ of a positive integer solution to $$ x^2 + y^2 + z^2 = 3 x y z , $$ see 
http://en.wikipedia.org/wiki/Markov_number
If you are curious, there is my paper with Kaplansky in the Illinois Journal, pdf at:
http://zakuski.math.utsa.edu/~jagy/bib.html 
I am especially proud of section 8.3, called "Other Families," as I was able to construct a mirror image of our "Markov Ratios" 
$$ 9 - \frac{4}{m^2} $$
in forms that gave
$$ 9 + \frac{4}{m^2} $$
To clarify, any indefinite binary quadratic form with integer coefficients and non-square discriminant has a nonzero "minimum" which is the smallest nonzero absolute value of any value obtained with integral values for the arguments. Our "Markov Ratio" was simply the discriminant divided by the square of this minimum. 
Well, your Golden Ratio has the smallest possible Markov Ratio, with 5, the form being equivalent to $x^2 + x y - y^2.$ The second smallest is 8, form $2 x^2 + 4 x y - 2 y^2$ which is visibly imprimitive. 
By the way, the "reduced" forms with the rather inflated middle coefficient (see chapter 3, Duncan A. Buell, Binary Quadratic Forms) correspond with purely periodic continued fractions, which I gradually came to realize are the best way to compute things: the cycle of a reduced form eventually arrives back at its precise starting point, so there is no need to compare with a long list of previously computed forms. If you simply compute the continued fraction for a number such as $\sqrt{5}$ you get some non-repeated coefficients or "quotients" $a_0 ; a_1 $ at the start.
If you look at the Cusick and Flahive book, you will see how the Markoff and Lagrange spectra coincide with "Markov Ratio" below 9 but are rather different above, where the description using binary forms is no longer adequate.
For the hasty:
http://en.wikipedia.org/wiki/Markov_spectrum<|endoftext|>
TITLE: Are group schemes in Char 0 reduced? (YES)
QUESTION [15 upvotes]: A Theorem of Cartier (e.g.  Mumford, Lecture 25) states that every separated, finite type group scheme $G/k$ over a field $k$  of characteristic $0$ is reduced.  Does this result remain valid if we drop the assumption that $G/k$ is separated and of finite type?
Frans Oort (MR0206005) observed that one can use limit formalism to argue that every affine group scheme over $k$ is reduced.
Edit: BCnrd pointed out that group schemes over a field are automatically separated.  Furthermore, the proof of Cartier's Theorem in Mumford's book remains valid for a locally Noetherian $k$-group scheme.

REPLY [12 votes]: The answer is yes - every group scheme over a field of characterstic zero is reduced: see Schémas en groupes quasi-compacts sur un corps et groupes henséliens (especially Thm. 2.4 in part II and Thm. 1.1 and Cor. 3.9 in part V of the 1st part), and for a summary of the relevant results see 4.2 (in particular 4.2.8) of Approximation des schémas en groupes, quasi compacts sur un corps.<|endoftext|>
TITLE: Diameter of a circle in an embedded Riemannian manifold
QUESTION [5 upvotes]: This question was inspired by an answer to the "Magic trick based on deep mathematics" question. I wanted to post it as a comment, but I ran out of characters! I'm sure there must be a collection of standard results related to this question, but I don't know where to start looking.
First, a quick definition. The diameter of a set $S \in \mathbb{R}^n$ is $\sup\{d(x, y) \mid x, y \in S\}$.
A sheet of paper is a good physical example of a Riemannian 2-manifold with boundary, and a table is a good physical model of (a subset of) $\mathbb{R}^2$. Embed the paper isometrically in $\mathbb{R}^2$ by laying flat on a table.
Draw the outline of a circular cup on the paper. It seems obvious that no matter how you embed the paper in $\mathbb{R}^2$, the outline of the cup will always be a metric circle, and it will always have the same diameter $D$.
Now, lift the paper into the air, embedding it isometrically in $\mathbb{R}^3$. If you let the paper flop around, the outline of the cup might not be a metric circle anymore... but will it still have diameter $D$?
Finally, cut along the outline of the cup, removing an open disk from the sheet of paper. The paper now has a second boundary component, and it's no longer simply connected. The paper has also gained a surprising property: you can bend it around in midair (that is, embed it isometrically in $\mathbb{R}^3$) so that the outline of the cup has diameter greater than $D$! What's the important property of the paper that we changed to make this possible?

Comments
I don't think you need to cut along the outline of the cup to make this work... you could probably just cut out any disk contained within the outline of the cup. So maybe simply-connectedness is the important property?
My gut tells me that if you draw two dots on the sheet of paper, the distance between the dots is maximized when the paper is flat on the table. When you bend the paper around in midair, the dots can get closer together, but they can never get farther apart. I think this is equivalent to the statement that if $\delta$ is the natural distance function on the paper, $d$ is the distance function in $\mathbb{R}^3$, and $F$ is an isometric embedding of the paper in $\mathbb{R}^3$, $d(Fx, Fy) \le \delta(x, y)$ for all points $x$ and $y$ on the paper.

REPLY [6 votes]: For the second question: 
As commented above, if you remove an open disc from the inside of the region bounded by a circle, you can bend the paper so as to increase the diameter of the circle in $\mathbb{R}^3$.  This doesn't work if you remove a point rather than an open disc - but with a point removed you can instead bend the paper so as to decrease the diameter of the circle by making the removed point a cone point of a wavy cone.  This gives us a good way of thinking about what's going on here: embedding $\mathbb{R}^2$ into $\mathbb{R}^3$ preserves diameter of circles, but when you ignore part of $\mathbb{R}^2$, (such as when your circle is on a small piece of paper, or when you've removed a point inside the circle) you can introduce singularities there.  Perhaps what's important is probably not the fundamental group, but the type and location of singularities that would arise if the embedding was extended to all of $\mathbb{R}^2$.   Any thoughts?<|endoftext|>
TITLE: Fiber functor of category of D-module on affine Grassmannian.
QUESTION [5 upvotes]: Geometric Satake correspondence allows us to construct Langlands dual group in a canonical way. In Mirkovic-Vilonen's paper, they prove that category of spherical perverse sheaves is an commutative tensor category, and hypercohomology functor gives a fiber functor. By Tannakian formalism, we can construct an algebraic group, which is exactly the Langlands dual group.
We can also realize the category to be the category of spherical D-module on affine Grassmannian. Now my quesion is, is there any nice construction of fiber functor on this category, without using Riemann-Hilbert correspondence?
Edit:
On a smooth finite dimensional variety, given a D-module, one can associate a deRham complex, and then take hypercohomology. The problem is that on smooth variety, we have sheaf of differential forms, which is canonical. However, on affine Grassmannian, D-module is actually not concret, so we can't associate a deRham complex canonically, I mean it depends on the realization of D-module. 
Can someone answer this question?

REPLY [6 votes]: The very short answer is that you just replace hypercohomology with global de Rham cohomology of a D-module.
The short answer is to read Theorem 3.5 in Mirkovic-Vilonen (I am referring to the arXiv paper here) and use their definition of the functor $F$ via the weight functors $F_\nu$.
The longer answer is that you can phrase this construction in a slightly more geometric way.  Suppose you choose in $G$ a Borel subgroup $B$ and denote by $T$ the quotient of $B$ by its unipotent part.  Then the maps $B \to G$ and $B \to T$ induce maps on the affine grassmannians, giving a diagram $\operatorname{Gr}\_G \xleftarrow{b} \operatorname{Gr}\_B \xrightarrow{t} \operatorname{Gr}\_T$.  Recall the structure of $\operatorname{Gr}_T \cong X_*(T)$ (at least, topologically and as a group), and let's call (as in the paper) $2\rho$ the sum of the positive roots of $G$ with respect to $B$.  Then that Theorem 3.5 can be understood as saying that $t_* b^! \mathcal{F}[2\rho(\lambda)]$ is a vector space (inside the derived category of vector spaces) whenever $\mathcal{F}$ is a spherical perverse sheaf (or D-module) on $\operatorname{Gr}_G$.
That is, we can define $F(\mathcal{F})$ to be the vector space $t_* b^! \mathcal{F}[2\rho(\lambda)]$ on the component $\{\lambda\}$ of $\operatorname{Gr}_T$; it is a vector space graded by $X_*(T)$ and this gives a faithful exact tensor functor from spherical D-modules to $\mathbf{Rep}({}^L T)$ (thus, a little more specific than just a fiber functor).
You might want to read the notes (written by me) from February 16th and 23rd at this seminar page.<|endoftext|>
TITLE: Why torsion is important in (co)homology ? 
QUESTION [77 upvotes]: I've once been told that "torsion in homology and cohomology is regarded by topologists as a very deep and important phenomenon". I presume an analogous statement could be said in the context of algebraic geometry.
In this community wiki question I would like to gather examples, in geometrical fields such as algebraic topology and algebraic geometry, of phenomena that manifest themselves by the presence of torsion in (co)homology groups and whose trace is consistently lost if we simply disregard the torsion part of those groups. 
As guidelines for the answers:

Which kind of information is lost disregarding torsion in (co)homology? (provide examples)
What does the torsion part of (co)homology tell us about the geometric object involved? 
  (provide examples)

Here "(co)homology" should be understood in any relevant sense, from singular cohomology of cw complexes to étale cohomology of algebraic varieties and so on and so forth. 
It may well be true that the algebro geometric examples have nothing to do, conceptually, with the topological ones: I'm not interested in a unifying pattern per se, but if such a unifying pattern does appear in some answers, well, it's just good.

REPLY [5 votes]: In his seminal work on the topology of Lie groups, Armand Borel discovered a connection between odd torsion in the integral homology of simply connected compact Lie groups and the failure of the homology Hopf algebra to be commutative.
Borel's observation led to work of William Browder, and Alex Zabrodsky tracing the issue back to homotopy associativity of the multiplication.  One nice result,due to Richard Kane, is that if p is an odd prime and there is p-torsion in the integral homology of a simply connected finite H-space admitting a homotopy associative multiplication,then the rank is at least 2p-2.  The same conclusion can be stated for p-localizations or p-completions.<|endoftext|>
TITLE: Algebras over the little disks operad
QUESTION [31 upvotes]: Hello,
The so-called "recognition principle" of Boardman-Vogt and May leaves me unsatisfied.
My problem is the following: 
The "recognition principle" says that every "group-like" algebra over the little $k$-disks operad is equivalent (as an algebra over the  little $k$-disks operad) to a $k$-fold loop space. However, if I start with a space homotopy equivalent to a $k$-fold loop space, then it is not a priori equipped with an action of the little disks. So:
1) First, can someone give me an explicit example of a space homotopy equivalent, say, to a double loop space and such that it admits no (compatible) actions of the little $2$-disks operad?
2) Can you characterize among the spaces homotopy equivalent to double loop spaces those which are algebras over the little disks operad?
3)I heard that the problem was related to the fact that the little disks operads are not cofibrant (in the homotopy category of operads), and that the cofibrant replacements would be the so-called Fulton-McPherson operads. These are "compactifications" of configuration spaces of points in $\mathbf{R}^k$ (modulo the action of the affine group maybe) defined using a variant of a construction due to Fulton and McPherson.
Is it obvious from the definition that this operad indeed acts on iterated loop spaces?
Many thanks,
K.

REPLY [15 votes]: I'll give an answer from the old days.  The $W$ construction of Boardman and Vogt 
corresponds to the modern notion of cofibrant replacement of operads.  If an
operad $\mathcal{C}$ acts on $X$ and $Y$ is homotopy equivalent to $X$, then 
$W\mathcal{C}$ acts on $Y$.  By the way, there is an inherent flaw in the little 
discs operads $\mathcal{D}_n$, namely there is no map of operads 
$\mathcal{D}_n\longrightarrow \mathcal{D}_{n+1}$ that is compatible with suspension
(in the obvious sense: consider  $\Omega^n \longrightarrow \Omega^{n+1}\Sigma$).  The little
$n$-cubes operads do not have this problem, but have others not shared by the 
$\mathcal{D}_n$.  The Steiner operads have all the good properties of both the
$\mathcal{C}_n$ and the $\mathcal{D}_n$.  In practice, that is in actual 
applications, such geometric differences are far more important than the
questions of cofibrancy and homotopy invariance.<|endoftext|>
TITLE: P-adic local Langlands for non-unitary representations?
QUESTION [10 upvotes]: In Colmez's work on the p-adic local Langlands correspondence for ${\rm GL}_2(\mathbb{Q}_p)$, he works with ${\rm GL}_2(\mathbb{Q}_p)$-representations on $p$-adic Banach spaces which admit an invariant norm, so the reduction modulo $p$ makes sense. To each irreducible admissible representation of this kind (let's call these "unitary" Banach representations), he attaches a rank 2 $(\varphi, \Gamma)$-module, and hence a 2-dimensional p-adic representation of ${\rm Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$.
The unitary condition is quite strict -- it rules out all nontrivial finite-dimensional algebraic representations of ${\rm GL}_2$, for instance. Is there any natural way to extend the correspondence to non-unitary admissible Banach space representations of ${\rm GL}_2(\mathbb{Q}_p)$, and what sort of Galois-theoretic objects would these match up with?

REPLY [7 votes]: This is a natural question.  For example, using Colmez's results, as completed by Paskunas (who shows that Colmez's p-adic local Langlands describes all topologically irreducible unitary admisisble Banach space representations of $GL_2(\mathbb Q_p)$) one can start to prove purely representation-theoretic facts about unitary admissible Banach space reps.
of $GL_2(\mathbb Q_p)$, using Colmez's description in terms of $(\phi,\Gamma)$-modules.  Now while some of these might naturally be related to unitarity,
there are certainly results that now seem accessible in the unitary case, which I suspect
don't actually require unitarity in order to hold.  However, if one is going to use Colmez's and Paskunas's results, one needs unitarity as a hypothesis.
One could imagine (and here I am talking at the vaguest level) working with some kind of Weil group representations rather than Galois representations in order to include the non-unitary representations.  I think that Schneider and Teitelbaum may have pondered this at some point, but I don't know what came of it.  And I don't know how reasonable it is to hope for such a correspondence.  I am just making the most absolutely naive guess, which you've probably also made yourself!
(One thing that makes me nervous is that when one works with unitary reps., there is a natural way to go from locally analytic reps. to Banach ones, by passing to universal unitary completions, and this is sometimes sensibly behaved, e.g. in the case of locally analytic inductions attached to crystabelline reps., by Berger--Breuil.  But if one starts 
to imagine completions that are not unitary, then I could imagine that they are much more wild; but again, this is just speculation.)<|endoftext|>
TITLE: Perverse Sheaves - MacPherson Lecture Notes
QUESTION [12 upvotes]: I keep running across papers that refer to a set of lecture notes by Robert MacPherson at MIT during the fall of 1993 on Perverse Sheaves. There might also be a set of notes from lectures in Utrecht in 1994 taken by Goresky. For references to these elusive, unpublished notes see the work of Maxim Vybornov and A. Polishchuk.
Does anyone have a copy or know how I could obtain a copy?
Thanks!

REPLY [16 votes]: The notes on Friedman's page are great -- they were very helpful when I was learning about perverse sheaves as a graduate student, since they explain how to think about middle perversity perverse sheaves on complex analytic spaces with complex stratifications (the case of most interest for most people) without having to first define the derived category.  Unfortunately the main result there, although it is a theorem, is not fully proved in the notes.  One of Bob's students, Mikhail Grinberg, had a project to fill in the details and flesh out the notes into a book, but he left mathematics for a job at Renaissance Technologies.  He taught a course at MIT around 2000 where he went through the steps needed to fill in the proof.  Perhaps someone has usable notes from that class?
The 1993 course was quite different.  Again the goal was to see perverse sheaves and their abelian category structure without going through the derived category, but he was working with arbitrary perversities on a regular cell complex.  By putting weird "phantom dimensions" on the cells and putting arrows between cells of adjacent phantom dimensions you get a quiver whose representations are perverse sheaves constructible for the cell complex structure.  But it's hard to use this to compute beyond simple examples, because the cell complexes would have to be very large, and it's hard to define the condition that gives constructibility for a coarser stratification like a complex analytic one.<|endoftext|>
TITLE: Are There Primes of Every Hamming Weight?
QUESTION [44 upvotes]: That is, for every integer $n \in \mathbb{Z}_{>0}$ does there exist a prime which is the sum of $n$ distinct powers of $2$?
In this case, the Hamming weight of a number is the number of $1$s in its binary expansion.
Many problems of this sort have been considered, but perhaps not in such language. For instance, the question "Are there infinitely many Fermat primes?" corresponds to asking, "Are there infinitely many distinct primes with Hamming weight exactly $2$?" Also related is the question of whether there are infinitely many Mersenne primes.
These examples suggest a class of such problems, "Do there exist infinitely many primes which are the sum of exactly $n$ distinct powers of two?"
Since this question is open even for the $n=2$ case, I pose a much weaker question here.
What is known is that for every $n \leq 1024$ there is such a prime.
The smallest such prime is listed in the Online Encyclopedia of Integer Sequences A061712.
The number of zeros in the smallest such primes are listed in A110700. The number of zeros in a number with a given Hamming weight is a reasonable measure of how large that number is. The conjecture at OEIS is quite a bit stronger than the question I pose.
Is there a theorem ensuring such primes for every $n \in \mathbb{Z}_{>0}$?

REPLY [3 votes]: For this context, though not so highbrow: Wagstaff, Prime Numbers with a Fixed Number of One Bits or Zero Bits in Their Binary Representation, Experiment Math 10 (2001), 267-273.
http://www.expmath.org/restricted/10/10.2/wagstaff.ps<|endoftext|>
TITLE: Can we prove set theory is consistent?
QUESTION [34 upvotes]: Disclaimer
Of course not, I'm aware of Gödel's second incompleteness theorem. Still there is something which does not persuade me, maybe it's just that I've taken my logic class too long ago. On the other hand, it may turn out I'm just confused. :-)
Background
I will be talking about models of set theory; these are sets on their own, so a confusion can arise, since the symbol $\in$, viewed as "set belonging" in the usual sense, may have a different meaning from the symbol $\in$ of the theory. So, to avoid confusion, I will speak about levels.
On the first level is the set theory mathematicians use all day. This has axioms, but is not a theory in the usual sense of logic. Indeed, to speak about logic we already need sets (to define alphabets and so on). In this naif set theory we develop logic, in particular the notions of theory and model. We call this theory Set1.
On the second level is the formalized set theory; this is a theory in the sense of logic. We just copy the axioms of the naif set theory and take the (formal) theory which has these strings of symbols as axioms. We call this theory Set2.
Now Gödel's result tells us that if Set2 is consistent, it cannot prove its own consistence. Well, here we need to be a bit more precise. The claim as stated is obvious, since Set2 cannot prove anything about the sets in the first level. It does not even know that they exist.
So we repeat the process that carried from Set1 to Set2: we define in Set2 the usual notions of logic (alphabets, theories, models...) and use these to define another theory Set3.
A correct statement of Gödel's result is, I think, that

if Set2 is consistent, then it cannot prove the consistence of Set3.

The problem
Ok, so we have a clear statement which seems to be completely provable in Set1, and indeed it is. This doesn't tell us, however that

if Set1 is consistent, then it cannot prove the consistence of Set2.

So I'm left with the doubt that what one can do "from the outside" may be a bit more than what one can do in the formalized theory. Compare this with Gödel's first incompleteness theorem, where one has a statement which is true for natural numbers (and we can prove it from the outside) but which is not provable in PA.
So the question is:

is there any reason to believe that Set1 cannot prove the consistence of Set2? Or I'm just confused and what I said does not make sense?

Of course one could just argue that Set1, not being formalized, is not amenable to mathematical investigation; the best model we have is Set2, so we should trust that we can always "shift our theorems one level". But this argument does not convince me: indeed Gödel's first incompleteness theorem shows that we have situations where the theorem in the formalized theory are strictly less then what we can see from the outside.
Final comment
In a certain sense, it is far from intuitive that set theory should have a model. Because models are required to be sets, and sets are so small...
Of course I know about universes, and how one can use them to "embed" the theory of classes inside set theory, so sets may be bigger than I think. But then again, existence of universes is not provable from the usual axioms of set theory.

REPLY [51 votes]: Your question certainly makes sense and it is a point that I feel is too often glossed over in textbooks.
Let me rephrase your question.  Goedel's second theorem says that, assuming that a certain formal system (ZFC, say) has a certain property that we call "consistency," then there is no formal proof in ZFC of a certain string, commonly denoted by "Con(ZFC)."  Fine.  But why on earth should this theorem say anything about whether the consistency of ZFC can be proved mathematically?  The theorem is just a theorem about abstract strings of symbols, not about what human beings can and cannot do.  The string denoted "Con(ZFC)" is commonly taken to "say" that "ZFC is consistent," but what is the justification for doing so?  A string is just a string, and doesn't "say" anything.  If we choose to think of the string as "meaning" something then that's our business, but surely that kind of human social activity is not something we can prove mathematical theorems about?
The answer is that, underlying the usual discussions of Goedel's second theorem, there is the following Key Assumption: If someone were to come up with a mathematical proof of the consistency of ZFC, then by mimicking that proof, we could produce a formal proof of Con(ZFC) from the axioms of ZFC.  The Key Assumption is crucial.  Without it, we cannot make the leap from Goedel's second theorem to a meta-mathematical statement about the (im)possibility of proving the consistency of mathematics.  And note that the Key Assumption is not a purely mathematical one; it cannot be, because it is a statement linking something that is not purely mathematical (namely, mathematical proof, which is a product of human activity) and something that is purely mathematical (namely, ZFC and theorems of ZFC).  Therefore the Key Assumption is not susceptible to mathematical proof, and the reasons we have for accepting it must be in part philosophical.
So what reasons do we have for accepting the Key Assumption?  The chief reason is that long experience has taught us that all mathematical proofs that mathematicians come up with can indeed be mimicked by formal proofs in ZFC.  This may seem obvious to us today, but it is not at all a trivial statement.  Prior to the set-theoretic revolution, it was by no means obvious that all the diverse areas of mathematics could be formulated in a single common language (i.e., set theory) and deduced from a short list of axioms.  It is only through the hard work of those working in the foundations of mathematics that we now take for granted that for any precise mathematical statement we want to make, there exists a formal sentence $S$ in the first-order language of set theory with the property that any mathematically acceptable proof of the original mathematical statement can be mimicked to produce a formal proof of $S$ from the axioms of ZFC.  And if you had any lingering doubts about whether this formal mimicry existed only in theory and not in practice, then in recent years, the advent of formal theorem-proving software such as Mizar, HOL Light, Coq, Isabelle, etc., should have swept away such doubts by demonstrating concretely that large areas of mathematics can be mimicked formally in practice, and not just in theory.
Finally, let me mention that although I believe it is very reasonable to accept the Key Assumption, it is possible to reject it.  Perhaps most notably, the philosopher Michael Detlefsen has challenged the standard claim that the string Con(ZFC) properly mimics the statement "ZFC is consistent" in the sense of the Key Assumption, and has suggested that Hilbert's program to prove the consistency of mathematics is not yet dead.  I believe that Detlefsen is simply mistaken and that there is nothing unsatisfactory about the standard string Con(ZFC), but he is at least correct that there is something to be checked here, and it is not a purely mathematical point but a partially philosophical one.<|endoftext|>
TITLE: Picard groups of reductive group schemes
QUESTION [7 upvotes]: There exists information on the Picard (and Brauer) group of a reductive algebraic group over a number field k. For example, Sansuc shows (in his big Crelle paper of 1980) that if G is  connected and semisimple over a number field k, then Pic G is the group of k-rational points of the character group of the fundamental group of G. In particular, if G is semisimple and simply connected, then Pic G=0. My question is: do there exist results of this type over more general bases? For example, a natural generalization of the equality "Pic G=0 if G semisimple and simply connected over a number field" would be "Pic G=Pic U if G is a semisimple and simply connected group scheme over a Dedekind scheme U". Is the latter true?

REPLY [3 votes]: Assuming you know that Pic of the generic fibre is trivial, this seems to follow immediately from the localisation sequence: since G is a group there is a section, so the map Pic U to Pic G is an injection. On the other hand (since G is smooth so Pic = Cl) there is an exact sequence:
$\ \ \ \oplus_x \mathbb{Z}_x \to Pic \ G \to Pic \ G_K \to 0 $
where X runs over the closed points of U and $G_K$ denotes the generic fibre. Since $Pic\  G_K$ is trivial it follows from this that the map Pic U to Pic G is surjective.
(Note that the only property of the closed fibres that is used is that they are reduced and irreducible.)<|endoftext|>
TITLE: Is there a combinatorial way to factor a map of simplicial sets as a weak equivalence followed by a fibration?
QUESTION [20 upvotes]: Background on why I want this:
I'd like to check that suspension in a simplicial model category is the same thing as suspension in the quasicategory obtained by composing Rezk's assignment of a complete Segal space to a simplicial model category with Joyal and Tierney's "first row functor" from complete Segal spaces to quasicategories. Rezk's functor first builds a bisimplicial set (I will say simplicial space) with a very nice description in terms of the model category, then takes a Reedy fibrant replacement in the category of simplicial spaces. Taking this fibrant replacement is the only part of the process which doesn't feel extremely concrete to me.
To take a Reedy fibrant replacement $X_*'$ for a simplicial space $X_*$ you can factor a sequence of maps of simplicial sets as weak equivalences followed by fibrations. The first such map is $X_0 \rightarrow *$.  If I factor that as $X_0 \rightarrow X_0' \rightarrow *$ the next map to factor is $X_1 \rightarrow X_0' \times X_0'$ and factors as $X_1 \rightarrow X_1' \rightarrow X_0' \times X_0'$. (The maps are all from $X_n$ into the $n$th matching space of the replacement you're building, but these first two matching spaces are so easy to describe I wanted to write them down explicitly.)

Background on the question:
The small object argument gives us a way to factor any map $X \rightarrow Y$ in a cofibrantly generated model category as a weak equivalence $X \rightarrow Z$ followed by a fibration $Z \rightarrow Y$. (In fact, as an acyclic cofibration followed by a fibration.) However, the object $Z$ this produces is in general hard to understand.
If the model category we're interested in is the usual one on simplicial sets (weak equivalences are weak homotopy equivalences on the geometric realizations, fibrations are Kan fibrations, and cofibrations are inclusions) we have some special ways of factoring maps $X \rightarrow *$ as a weak equivalence followed by a fibration. Probably the most familiar is using the singular chains on the geometric realization of $X$ to make $X \rightarrow S(\mid X\mid ) \rightarrow *$. This is again big and rather hard to understand.
Another method is Kan's Ex$^\infty$ functor, which I learned about in Goerss and Jardine's book. Ex is the right adjoint to the subdivision functor, which is defined first for simplices in terms of partially ordered sets, so the subdivision functor and Ex are both very combinatorial. There is a natural map $X \rightarrow$ Ex $X$, and Ex$^\infty X$ is the colimit of the sequence $X \rightarrow$ Ex $X \rightarrow$ Ex$^2 X \rightarrow$... . It turns out that $X \rightarrow $Ex$^\infty X \rightarrow *$ is a weak equivalence followed by a fibration.
I would like a way of factoring a map $X \rightarrow Y$ of simplicial sets as a weak equivalence followed by a fibration that is similar in flavor to the use of Kan's Ex$^\infty$ functor to find a fibrant replacement for a simplicial set $X$.

Question:

In the standard model category structure on simplicial sets, is there a combinatorial way of factoring a map as a weak equivalence followed by a fibration?

REPLY [15 votes]: I don't know if this does what you want, but it's one thing I know how to do.
If $X\to Y$ is a map between Kan complexes, then you can build a factorization using the path space construction. Thus, $P=Y^I\times_Y X$, and $P\to Y$ is given by evaluation, while $X\to P$ is produced using the constant path.  Here $I=\Delta[1]$, and $Y^I$ is the mapping object in simplicial sets.  Since $Y$ is a Kan complex, $P\to Y$ is a Kan fibration (using the fact that $Y^I\to Y^{\partial I}$ is a Kan fibration).
For a general map $f: X\to Y$, consider $Ex^\infty(f): Ex^\infty X\to Ex^\infty Y$.  This is a map between Kan complexes, and therefore the path space construction on $Ex^\infty(f)$ gives a Kan fibration $P\to Ex^\infty Y$.  Now pull back along $Y\to Ex^\infty Y$ to get a Kan fibration $Q\to Y$.  Because pullbacks of fibrations in simplicial sets always are homotopy pullbacks, it should be possible to see that $X\to Q$ is a weak equivalence. 
(I think this construction goes back to Quillen's book on model categories.)<|endoftext|>
TITLE: Area of cross-section (at midpoint perpendicular to longest diagonal) in the unit cube of dimension N 
QUESTION [8 upvotes]: Take a unit cube (of side 1) in N dimensions. Construct the cross-section at the midpoint of the longest diagonal. What is the area of this N-1 dimensional region? I can compute this, but it would be nice to have a reference and a formula to cross-check.

REPLY [7 votes]: This is a very old problem and there is a classical analytic approach to it. You can express the volume of sections of a convex body in terms of the Fourier transform of powers of the Minkowski functional. 
Let $Q_n=[-1/2,1/2]^n$ be the unit cube in $\mathbb R^n$ and let $[\xi^\perp]$ denote the hyperplane orthogonal to the vector $\xi\in S^{n-1}$. Then 
$$Vol_{n-1}(Q_n\cap [\xi^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty}
\prod\limits_{k=1}^n\frac{\sin(r \xi_k)}{r\xi_k}dr.$$
In our case $\xi=\xi_*=n^{-1/2}(1,1,\dots,1)$ so the integral becomes
$$Vol_{n-1}(Q_n\cap [\xi_*^\perp])=\frac{1}{\pi}\int_{-\infty}^{\infty}
\left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr.$$
In fact, the latter formula was already known to Laplace! The integral tends to $\sqrt {6/\pi}$, as $n\to\infty$ (e.g. by Laplace's method).  This can be also justified via the probabilistic interpretation suggested by Michael Lugo.
The Fourier analytic approach to sections of convex bodies is nicely presented in The Interface between Convex Geometry and Harmonic Analysis by A. Koldobsky and V. Yaskin. The derivation of the formula for volumes is available here.
EDIT (15.01.2011). In fact both integrals can be calculated explicitly. The sinc integrals were studied by Borwein & Borwein (see also Multi-Variable sinc Integrals and the Volumes of Polyhedra).
For any $n\in \mathbb N$, $n>1$, we have
$$\frac{1}{\pi}\int_{-\infty}^{\infty}
\left(\frac{\sin(r/\sqrt n)}{r/\sqrt n}\right)^ndr =\frac{\sqrt n}{2^{n-1}(n-1)!}\sum_{k=0}^{n/2}(-1)^k{n \choose k}(n-2k)^{n-1}$$<|endoftext|>
TITLE: Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint?
QUESTION [24 upvotes]: (Alternate title: Find the Adjoint: Hopf Algebra edition)
I was chatting with Jonah about his question Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$. It's very closely related to the following question:

For a $k$-algebra $A$, is there a Hopf Algebra $H(A)$ such that for any Hopf algebra $B$, we have
  $\mathop{Hom}_{k\text{-alg}}(B,A)\cong \mathop{Hom}_{k\text{-Hopf-alg}}(B,H(A))$?

In other words, does the forgetful functor from $k$-Hopf-algebras to $k$-algebras have a right adjoint? There are a few related questions, any of which I'd be interested in knowing the answer to:

Does the forgetful functor from Hopf algebras to augmented algebras (sending the counit to the augmentation) have a right adjoint?
Does the forgetful functor from Hopf algebras to algebras with distinguished anti-automorphism (sending the antipode to the anti-automorphism) have a right adjoint?
Does the forgetful functor from Hopf algebras to algebras with augmentation and distinguished anti-automorphism have a right adjoint?

Unfortunately, I don't feel like I can motivate this question very well. My motivation is that the better I know which forgetful functors have adjoints, the better I sleep at night.

REPLY [13 votes]: The problem is famous and you will find in the book Sweedler. Ana Agore (one of my student in master) prove the result in june 2009 -- and her paper will apper in Comm. Algebra. Is this one http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.2613v3.pdf
However, related to the problem is also another paper of her more recently: http://front.math.ucdavis.edu/1003.0318 
On the other hand, two months latter (july 2009) a colleque of her (Alex Chirvasitu) give the explicit construction of it in this paper http://arxiv.org/PS_cache/arxiv/pdf/0907/0907.2881v2.pdf that allready appered in J. Algebra at the begining of 2010. 
Cheers! and a nice site.
Gigel Militaru<|endoftext|>
TITLE: Does every hyperbolic curve over a finite field have an etale cover with a real Frobenius eigenvalue?
QUESTION [18 upvotes]: More precisely:  let X/F_q be a smooth projective algebraic curve of genus at least 2.  Does there always exist a curve Y/F_{q^d} with a finite etale projection Y -> X, such that one of the Frobenius eigenvalues on H^1(Y,Z_ell) is q^{d/2}?
I mentioned this on my blog about three weeks BMO, but this seems a much better venue.  The question is motivated by some (analogous, I think?) conjectures in topology; see the blog post for more about the motivation, and what (little) I know about the question.

REPLY [7 votes]: Hi Jordan,
I have a guess on how to go looking for it,  although the process touches on matters I know little about. 
First, let's fix one power $d$, so that  the Frobenius action will have eigenvalues of a fixed size. This enables us to compute mod $l$ for large $l$. Then, we should be able to look for an $l$-adic local system $V$ instead of a curve, say of rank $n$. This is because we can reduce $V$ mod $l$ (as usual, by way of compactness of the fundamental group $\pi_1$) to get a sheaf $V_l$. But then, the kernel of the corresponding mod $l$ $\pi_1$-representation will give a curve $f:Y \rightarrow X$. Since $V_l$ trivializes over $Y$, there is an exact sequence
$0 \rightarrow V_l \rightarrow f_*((Z/l)^n) \rightarrow M \rightarrow 0,$
with $M$ just defined as the quotient, so the $q^{d/2}$ part of $H^1(\bar{X}, V_l)$ should inject into $H^1(\bar{Y}, (Z/l)^n)$, giving you what you want.
I admit that several parts of the argument are sloppily written. It seems to me plausible to fix it all up to be rigorous.
Now, where does one get $V$ having $q^{d/2}$ inside $H^1(\bar{X}, V)$? It suffices to have $q^{-d/2}$ as a zero of the $L$-function $L(X,V, t)$.  So we go searching for a rank $n$ unramified automorphic form for $K(X)$ whose $L$-function has the right property. This is where my shaky expertise starts to fail, but I've always had the impression that constructing automorphic forms over function fields was a rather accessible combinatorial matter. Perhaps real experts can now comment on whether this is at all plausible.<|endoftext|>
TITLE: Natural setting for characteristic classes?
QUESTION [22 upvotes]: In my mind, algebraic topology is comprised of two components:
Chain complex information, which is intrinsic information concerning how your object may be built up out of simple "lego blocks".
Characteristic classes (bundle information) which give information on how your object might stably embed in some sufficiently big standard object.
Chain complexes make sense over any abelian category.
I have no corresponding intuitive understanding of what the "natural setting" for characteristic classes should be. The classical theory looks to me like a concession to the sad fact that, at its very basis, manifolds are locally modeled on Euclidean space and are not intrinsically defined objects. This is reflected in the central role played by specific concrete spaces such as the Thom spaces MSO(n), the real and complex Grassmanians used to define Wu classes, and the classifying spaces BU and BO concerning which we have Bott periodicity.
I realize that I have no understanding of any of this. Part of this feeling is because I really don't understand what forces us to consider these specific concrete spaces, to the exclusion of all others. If constants appearing in physics ought to be conceptually explained, I'd like to understand these "constants" in mathematics. Can one work with characteristic classes in a more general setting, to parallel abelian categories? What about over number fields, over arbitrary rings, or in finite characteristic? Can I replace Lie groups such as SO(n), U(n), and O(n) by groups of Lie type for instance, and still have a "useful" theory?
My question is then:
What is the most general categorical setting for a "useful" theory of characteristic classes? In particular, are all of those special concrete spaces really necessary, and if so, why?

REPLY [5 votes]: I hope that "A note on characteristic classes: Euler, Stiefel-Whitney, Chern and Pontrjagin" is a useful addition to the discussion.<|endoftext|>
TITLE: Quotient of a category by a group action
QUESTION [7 upvotes]: Let a group $G$ act on a small category $C$. If $G$ acts freely on objects, there is a sensible construction of the quotient $C/G$ (this is briefly spelt out here)
What about the non-free case?

REPLY [3 votes]: Here is a particular special case that seems to be useful to me:
Let $F: D \to G$-Sets be a diagram of $G$-sets, indexed by some small category D.  Then ignoring the G-actions, one can form the Grothendieck wreath product $D\wr F$.  Objects are pairs (d, x) with $x\in F(d)$, and morphisms $(d,x)\to (d', x')$ are arrows $a:d\to d'$ such that $F(a)(x) = x'$.  This category inherits an action of G from the actions on the sets in the diagram; $g\cdot (d,x) = (d, gx)$ (on morphisms, $g$ looks like the identity: $g\cdot (a:(d,x)\to(d',x')) = a:(d,gx)\to(d',gx')$, which makes since $F(a)$ is $G$-equivariant.  
There is another diagram $F/G : D\to$ Sets, which takes $d\in D$ to $F(D)/G$, and there's a natural functor $D\wr F \to D\wr (F/G)$.  I claim that this functor satisfies the universal property of the colimit, i.e. $(D\wr F)/G = D\wr (F/G)$.  
This special case has the interesting property that the nerve of $D\wr (F/G)$ is precisely $N_\cdot (D\wr F)/G$.  I don't think that will hold for arbitrary group actions on categories.<|endoftext|>
TITLE: "Why the heck are the homotopy groups of the sphere so damn complicated?" 
QUESTION [50 upvotes]: This is a quote from a dear friend asking the rest of us on Facebook. I gave him some half-baked response, but the truth is I don't really know enough about this to give him a good response. 
So why ARE they so complicated? The topologists here want to give a few responses so I can give him some feedback to his desperate query?

REPLY [11 votes]: I might be misinterpreting your friend's question, but if your friend knows very little about the subject and is expressing surprise that something so seemingly simple turns out to be so complicated, then maybe you should concentrate on explaining the Hopf fibration to him or her.  There is an interesting visualization in Thurston and Levy's book Three-Dimensional Geometry and Topology.  I think that anyone who successfully goes through the exercise of visualizing the Hopf fibration will be disabused of any preconception that the homotopy groups "should" be simple.<|endoftext|>
TITLE: If a tensor product of modules is semi-simple, are the tensor factors semi-simple ?
QUESTION [5 upvotes]: Suppose $M$ and $M'$ are two $R$-moules (I am most interested in the case of $R$ a DVR).   If $M\otimes M'$ is a semi-simple module (i.e., every submodule is a direct summand) then is it true that the tensor factors $M$ and $M'$ are semi-simple ?
I.e., if this is not true then it would be nice to see a counter-example.

REPLY [7 votes]: I can't resist providing the following comment and references, even though it is likely not really relevant to the (already old by now) original post.
Quick background: Let $k$ be a field of char. $p>0$ and let $G$ be a(ny) group. 
Serre proved [Invent. Math. 116 (1994), no. 1-3, 513--530] that if $V$ and $W$ are finite dimensional, semisimple $kG$-modules and $$(\dim V - 1) + (\dim W -1) < p$$ then $V \otimes_k W$ is a semisimple $kG$-module. (The argument is quite nice - one reduces to alg. closed $k$, and replaces $G$ by a certain algebraic group
over $k$ whose identity component is reductive. One then has to argue that $[G:G^0]$ has order prime to $p$, so one is reduced to consideration of connected reductive $G$. And that case is handled by some "weight combinatorics" and the linkage principle.)
In a subsequent paper [Semisimplicity and tensor products of group representations: converse theorems. With an appendix by Walter Feit. J. Algebra 194 (1997), no. 2, 496--520] Serre proved some "converse theorems". For example, he shows that 
$$V \otimes_k W \quad \text{semisimple} \implies V \ \text{semisimple if $\dim W \not \equiv 0 \pmod{p}$}$$ 
Examples (due to Feit, and included in the paper) show that one can't get rid of the assumption on $\dim W$.<|endoftext|>
TITLE: What are the implications of torsion in H^2 for geometric quantization?
QUESTION [11 upvotes]: Given a real manifold $M$ with symplectic $2$-form $\omega$,
one can ask whether the cohomology class $[\omega] \in H^2(M;{\mathbb R})$ lies in the image of 
$H^2(M;{\mathbb Z})$. If so, one can ask for a line bundle ${\mathcal L}$
with $c_1({\mathcal L}) = [\omega]$ (or even better, 
a connection $\alpha$ on $\mathcal L$ whose curvature $curv(\alpha)$ is $\omega$).
In the weakest definition of "geometric quantization", one puts a compatible
almost complex structure on $M$ and uses it to define the pushforward
of $\mathcal L$ to a point in $K$-theory. Call this $Q(M)$.
Are there examples worked out somewhere in which $H^2(M;{\mathbb Z})$
has torsion, so that $\mathcal L$ is not uniquely determined by $\omega$?
Can $Q(M)$ depend on the choice of $\mathcal L$?
I don't have any very good reason for asking this, other than I've felt
it to be a hole in my understanding of geometric quantization. The spaces I care about quantizing never seem to have torsion in $H^2$.

REPLY [4 votes]: It is the group of periods of a closed 2-form $\omega$ which plays a role on the different quantizations. Every closed 2-form $\omega$ on a manifold $M$ (more generally on a diffeological space) is the curvature of a connexion on an integration bundle, a principal bundle with group the torus of periods $T_\omega = {\bf R}/P_\omega$, where $P_\omega$ is the group of periods
$$
P_\omega = \{ \int_\sigma \omega \mid \sigma \in H_2(M,{\bf Z})\} \subset {\bf R}.
$$
The different integration structures (bundle + connexion) are classified by $H^1(M,T_\omega)$, the different integration bundles only are classified by ${\rm Ext}(H_1(M,{\bf Z}),P_\omega)$. The torsion plays a role at the level of $H_1(M,{\bf Z})$. For example, if this group has no torsion and $M$ is compact then the integration bundle is unique, up to an equivalence.
The torus of periods $T_\omega$ is a priori equipped with the quotient diffeology of $\bf R$. Of course $T_\omega$ is a manifold only if the group of periods is generated by one number (we say that $\omega$ is integer), that is, $P_\omega = a{\bf Z}$ one says then that $\omega$ is quantizable if $a$ is a multiple of $\hbar$, $a = k\hbar$, $k \in {\bf Z}$ (it's just a normalization). But note that the construction of the integration structure doesn't need $\omega$ to be integer.
Actually this proposition is true in general if the group of periods is a strict subgroup of $\bf R$, which is always the case for a second countable manifold, but it may be not happen if $M$ is just a general diffeological space, or some very special manifold.

References

La trilogie du moment, Annales de l'Institut Fourier, t. 45, n°3 (1995)
Diffeology, in Mathematical Surveys and Monographs, 185, AMS Providence RI, (2013).<|endoftext|>
TITLE: Philosophy : seeking examples illuminating deeper geometric ideas behind base change of schemes 
QUESTION [6 upvotes]: So for certain base changes, it's clear that 'base change' really means base change from high school : for example, if a curve is defined over a field, it will be of course defined over any extension of that field, and 'base change' as defined for schemes agrees with that, essentially because of properties of tensor product.  Similarly, if we define projective space over $\mathbb{Z}$, or the general linear group scheme over $\mathbb{Z}$, then changing base by the unique $Spec(A)\to Spec(Z)$ (for any ring $A$) gives projective space over $A$ and the general linear group scheme over $A$ as well.
It would be nice to see some examples of base change of $X/S$ through some stranger change of base maps $S'\to S$, and to see how $X/S$ is deformed by $S'\to S$ in ways that are 'fundamentally' not like the above ; i.e., the examples I listed above don't really capture the 'spirit' of base change in general.  I'm looking for a list of other central geometric ideas behind the notion of base change that go beyond the above ideas.
For example, if $X/S$ is finite etale, $s$ in the image, then changing base by the strict localization at $s$ gives a trivial finite étale covering (i.e. the base change is a finite union of copies of the strict localization at $s$), thus showing that (surjective?), finite étale maps are locally trivial for the étale topology (and therefore the name 'étale covering' is meaningful in this sense).  
Can someone give a motivated discussion of the geometry of base change ?  E.g., examples-based observation and pondering.

REPLY [3 votes]: I think that thinking about nontrivial pullbacks in the category of topological spaces might help you out.  The answer might be below you, but hopefully it will be of some use to people thinking about this stuff for the first time.
In topological spaces, the pullback of $X \overset{f}{\rightarrow} S$ and $Y \overset{g}{\rightarrow} S$ is just that subspace of $X \times Y$ where $f(x) = g(y)$.  A similar description in all other categories is possible:  The pull back always has a natural monic arrow to the product induced by the projection maps.  
Examples:
1) The pullback of $X \rightarrow 1$ and $Y \to 1$ is just the ordinary product.  In general the pullback of two maps to the terminal object of a category is just the categorical product of the domains of those maps.  This is why we define products of schemes to be their fiber product as $\mathbb{Z}-schemes$.
2) If $1 \overset{s}{\to} S$ just picks out an element $s$ of $S$ and $X \overset{f}{\to} S$ is any morphism, then the pullback of $s$ with $f$ is just the fiber $f^{-1}(s)$ considered as a subspace of $X$.
Really nontrivial pullbacks are kind of hard to actually visualize.  Here is one example where it is possible, and might be useful to think about: 
3) It is difficult to draw in this format, but consider the map $\pi: S^1 \to [-1,1]$ from the unit circle centered at the origin to the interval given by just projecting down to the x-axis,  and the map $\rho: D^1 \to [-1,1]$ from the unit disk to the interval also given by projection.
The product space $D \times S$ is just a solid torus.  The pullback in this case will be a two-dimensional surface sitting inside this torus with two cusp points corresponding to the endpoints of the interval.  You will have to just draw this out and see what subset of $D \times S$ I am talking about.
So it is a bit weird that "extending the base" from the line segment to the disk results in this singular surface contained in the solid torus, but it is really  the data which the projection maps are carrying around that matters.  The geometry seems to get a bit lost here compared with the first two examples.
Pullbacks in the category of schemes are even crazier because you also have morphisms of sheaves to worry about, and so all of this algebraic data enters the mix, and actually kills the geometric interpretation we developed for topological spaces (The product of two schemes is not defined on the product of the topological spaces!).  But hopefully getting a handle on the topological space first will help you think about schemes.<|endoftext|>
TITLE: Why are finitely generated modules over principal artin local rings direct sums of cyclic modules?
QUESTION [19 upvotes]: I am looking for a proof of the following fact:
If $R$ is a principal artin local ring and $M$ a finitely generated $R$-module, then $M$ is a direct sum of cyclic $R$-modules.
(Apparently such rings $R$ are called, e.g. in Zariski-Samuel, special principal ideal rings.)
I almost didn't ask this question for fear that I might just be missing something obvious, but I've been unable to come up with a proof myself and I can't find one anywhere (if I am just being stupid I certainly don't mind being told). Zariski-Samuel proves a structure theorem for principal ideal rings (they are products of PID's and special PIR's), as well as the fact that a submodule of a principal ideal ring generated by n elements is generated by $\leq n$ elements. I thought perhaps the statement above could be deduced from this last fact, in a similar manner to the way one proves the corresponding result for finite modules over PID's, but the obstruction to this (as I see it) is that submodules of free $R$-modules will not in general be free (for instance, the maximal ideal of $R$ is not free, assuming $R$ is not a field, i.e., has length $\geq 2$). Essentially the naive induction on the number of generators doesn't seem to work because I can't be sure that the relevant exact sequence splits...again, maybe I'm just being foolish. I think a proof might be found in chapter VIII of Bourbaki's Algebra text, but this chapter isn't in my copy (I think maybe it's only available in French).
Incidentally, it's straightforward to show that, if such a decomposition exists, the number of times a factor of the form $R/(\pi^i)$, where $(\pi)$ is the maximal ideal of $R$ and $1\leq i\leq k$ ($k$ being the length of $R$, i.e., the index of nilpotency of $(\pi)$) is uniquely determined as the length of a quotient of $M$, for instance.
The reason I'm interested in this is because I want to know that the isomorphism type of $M$ is completely determined by the function sending $i$, $1\leq i\leq k$, to the length of $M[\pi^i]$ (the kernel of multiplication by $\pi^i$), which, assuming such a decomposition exists, is definitely the case.
Edit: I realized that my module M can (being artinian) be written as a finite direct sum of indecomposable submodules, so I guess this reduces my question to: must an indecomposable submodule of $M$ be a cyclic?

REPLY [9 votes]: The "local Artinian" assumption in OP's question can be removed as the following more general result holds (rings are supposed unital and commutative):

If $M$ is a finitely generated module over a principal ideal ring $R$ then $M$ is the direct sum of finitely many cyclic $R$-modules $R/(a_i)$ with $a_i \in R$, such that $R \neq (a_1) \supset \cdots \supseteq (a_n)$. In addition, the principal ideals $(a_i)$ are unique with respect to this property.

A principal ideal ring (PIR) is a ring whose ideals are principal, equivalently a PIR is a Noetherian Bézout ring.
The aforementioned general result follows from the existence and uniqueness of the invariant factor decomposition of Noetherian elementary divisor rings see [Theorems 9.1 and 9.3, 1]. The fact that a PIR is an elementary divisor ring in the sense of I. Kaplansky is an immediate consequence of [Theorem 12.3 and subsequent remark, 1] and can also be inferred from [Theorem 1, 2]. This fact is stated without proof in [Notes on Chapter I, 3].
By the way, it is rather unfair to allude to Zariski-Samuel's structure theorem (as T. W. Hungerford in [2] and subsequently in the wikipedia article on PIRs) when quoting the result about PIR decomposition in direct sum of domains and Artinian local rings. Indeed, this theorem was originally proved by W. Krull in 1924 and reproved by I. Kaplansky in 1949, cf. [Theorem 12.3, 1] and related comments therein.

[1] "Elementary divisors and modules", I. Kaplansky, 1949 (MR0031470).
[2] "On the structure of principal ideal rings", T. W. Hungerford, 1968 (MR0227159).
[3] "Serre's problem on projective modules", T. Y. Lam, 2006 (MR2235330).<|endoftext|>
TITLE: How to know if somebody else is also working on your problem?
QUESTION [50 upvotes]: Once you have spotted a mathematical problem that (presumably) fits your degree of expertise, whether you are a phd student or an established professor, you have to deal with the following non mathematical problems:

How to know if somebody else in the world is already working (or has already been working) on the same problem? 

If the other guy has already completed a certain amount of (say, not yet published) work on that specific topic, knowing this would help you to avoid waisting time to try to re-do something that has already been done (at least with the same methods).
On the other hand, if the problem is broad enough, knowing of some other's interest in the same context would also be fruitful because you'd may have somebody with whom to talk and to whom to ask questions, without overlapping the specific research goals. Or you may even find a collaborator.

In some cases the very choice of an interesting specific problem can be a nontrivial task by itself. So, in case you want to ask around if some previous/present effort has/is been made in that specific direction or related ones, should you worry about the possibility that somebody with a higher degree of expertise would just "take your problem" and solve it faster than you would do? 

I'd expect the obvious answers, such as "have a look to mathscinet/arxiv" or "search the literature" or "talk to people (your advisor if you're phd)", to be enriched -if possible- by some more elaborate viewpoint or more specific suggestion.

REPLY [21 votes]: Recently I'd started working on a project with a leading expert in the relevant field where the result we'd worked on turned out to be already in the literature unbeknownst to either of us.  So this can happen even to leading experts.  (Our method of proof was slightly different, but that part of the project instead of being a new result just becomes the pushout of a known result with a known technique proving a similar result, and so not really worth publishing.  (Though it was still a project where I learned valuable things!)  Fortunately we hadn't actually started writing.)
So how did we figure out that this result was in the literature?  One day I decided to read the mathscinet reviews of every paper with primary subject classification 46L37.  The reason I did this was actually to try to learn where subfactor papers were published, but a nice side affect was that I learned about this prior excellent work so we didn't put in a bunch of effort only to discover that someone else had done it several years ago!
I highly highly recommend at least skimming through the mathscinet listings of subject classifications that you tend to publish in.  It really gives a nice birds eye view of the field (albeit a couple years out of date).  (Of course, this is easier to do in fields that have only existed for 30 years.)

REPLY [11 votes]: I'm surprised that no one has posted the following response yet:
Post a question on MO asking what is known about your problem.
I'm curious to hear people's opinions about this.  Is this within the intended purpose of MO?  In a sense, this is certainly a "research-level question".  Some people may be more cautious about broadcasting what they're working on, so as to avoid attracting others to the problem.  But we certainly have a large community of active mathematicians here, and one might gain a lot of insight from the answers to such a question.<|endoftext|>
TITLE: normalizer of algebras and groups
QUESTION [8 upvotes]: Hi,
I am looking at inclusion of discrete groups $H\subset G$ such that $H$ is abelian and $(hgh^{-1},h\in H)$ is infinite if $g\in G-H$. If you have this, $LH\subset LG$ is a maximal abelian subalgebra of a finite von Neumann algebra.
Suppose that $LH\subset LG$ is a Cartan subalgebra, i.e. the group of unitary of $LG$ that normalize the algebra $LH$ generates $LG$. Do we have necessarily that $H$ is a normal subgroup of $G$?
Thanks for your help.

REPLY [6 votes]: This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith.  One can also give the following alternative argument based on ideas of Popa:
If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$.  (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)
Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$.  Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.
Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$.
Lemma:
For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.
Proof.  This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).  
Let's show this by induction on $n$.  For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$.  Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ (see Popa, op. cit.).  In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$.  The same inequality holds for the other minimal projection $1 - p_k$.
Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$.  This proves the induction step.  QED
Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810.  Let's give the proof here.
Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$.  From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$.  Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$.
Denote by $B_0$ the commutant of $A_0$ in $LG$.
Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$.  Therefore
$| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= 
\| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v,  u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 )  + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.
Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal.  Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.<|endoftext|>
TITLE: How can we realize different combinatorial objects as the dimension of a construction on vector spaces? Are the resulting algebras useful?
QUESTION [14 upvotes]: Fix a vector space $V$ of dimension $n$ over some field $F$. Here are three commonly seen constructions:

its $k$th tensor power, $T^kV$, which has dimension $n^k$
its $k$th exterior power, $\Lambda^k(V)$, which has dimension $\binom{n}{k}$
its $k$th symmetric power, $S^k(V)$, which has dimension $\binom{n+k-1}{k}$

Let $X$ denote one of these constructions. Then the "$X$ algebra", namely $\bigoplus_{k=0}^\infty X^k(V)$ with an algebra structure on it, tends to be rather important (for example, the tensor, exterior, and symmetric algebras all satisfy universal properties, and are used all over mathematics).
I recently noticed that the dimensions of the above three constructions were all answers to one of the questions of the "twelvefold way" of combinatorics (specifically, see here). The 12 questions ask for the number of functions from a $k$-element set $K$ to an $n$-element set $N$, under 3 different restrictions (no restriction, injective, surjective) and up to 4 different equivalence relations (equality, up to permutation of $K$, up to a permutation of $N$, up to a permutation of both). 
The tensor power has dimension $n^k$, and so would correspond to (no restriction, equality).
The exterior power has dimension $\binom{n}{k}$, and so would correspond to (injective, up to permutation of $K$).
The symmetric power has dimension $\binom{n+k-1}{k}$, and so would correspond to (no restriction, up to permutation of $K$)
So my questions are:

Can we provide a unified combinatorial explanation of why the tensor, exterior, and symmetric powers have the dimensions that they do, using the framework of the twelvefold way? I am guessing that we want $N$ to be a basis for $V$ (hence having $n$ elements) and $K$ to be any $k$-element set; but in what way does the tensor power correspond to all functions from $K$ to $N$? How does the construction of the exterior power from the tensor power correspond to the restricting attention to the injective functions from $K$ to $N$, and only up to permutation of $K$?
Given this (hypothetical) unified explanation, what are the constructions on $V$ (presumably, they will be some quotients of the tensor power) that correspond to the other 9 combinatorial questions of the twelvefold way?
Where do the resulting algebras of these new constructions, formed as before by direct summing over all $k$, show up? For each, can we work out what universal property it satisfies? Or (perhaps too optimistically), can we try to reverse engineer the correct universal property that will produce a construction with the desired dimension / combinatorial interpretation?


EDIT: Spurred by Gowers's recent question, I decided to bump this question to the front page to get some fresh eyes on it. Igor's answer is very helpful, in that it says not to necessarily expect an explicit algebra / "power" construction, and Richard's answer also sounds great (though I'm afraid the reference he pointed to went too fast for me to follow), but ultimately I'm still wondering if any piece of the twelvefold way other than the three I listed does have such a construction.
Richard - I don't know anything about Young Tableaux beyond the Wikipedia page, but it seems like there aren't any other "natural" arrangements of $n$ squares other than the ones you mentioned. Might this explain why there doesn't appear to be any other "power" constructions of the kind I'm looking for? Could you explain where the "$k$" is in this approach?
Igor - I will certainly take your word that the "right" approach is through spaces of invariants, but while you explain how $S_k$ acting on $\mathbb{C}[S_k]$ by conjugation can get us a space of dimension $p(k)$, this is still not quite the $p_n(k)$ or $p_n(n+k)$ that occur in the twelvefold way. Could you explain for example the group algebra and action on it that corresponds to the exterior or symmetric power - and if possible in a way that highlights why that choice of algebras and actions is related to considering functions $f:K\rightarrow N$ under (injective, up to permutation of $K$) or (no restriction, up to permutation of $K$)? 
If we were to switch from algebra / "power" constructions to spaces of invariants, I suppose I would restate my goal as: is there a single group $G$ such that there are twelve group actions on $\mathbb{C}[G]$, the spaces of invariants of which had the dimensions appearing in the twelvefold way, and with the definition of each action clearly showing its relationship with the corresponding (restriction, equivalence). Perhaps $G=(\mathbb{Z}/n\mathbb{Z})^k$ ?
Of course, I will continue to welcome any ideas on the problem as previously stated.

REPLY [8 votes]: Tensor power, exterior power, and symmetric power are example of polynomial functors. The general theory of polynomial functors unifies them combinatorially in terms of Young tableaux, RSK, etc. Tensor power corresponds to the skew shape consisting of $n$ disjoint squares; exterior power to a column of length $n$; and symmetric power to a row of length $n$. A standard reference (though not so user-friendly) is the appendix to Chapter 1 of Macdonald's Symmetric Functions and Hall Polynomials.<|endoftext|>
TITLE: Classification of l-adic representations
QUESTION [6 upvotes]: Either the following is a really stupid question or it is a really really stupid question, but here goes:
Does there exist a classification of $\ell$-adic 2-dimensional representations of $\mathrm{Gal}(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$, where $\ell\neq p$? 
I did a quick search of the internet that came up rather empty. 
What about the subtler case of $\ell=p$?
References?

REPLY [2 votes]: A small post script to Emerton's post (that would not fit in the comment box): as you suggest, there is a nicer (easier to understand) classification of potentially semi-stable representations. Basically the idea is that via B_st semi-stable representations are easy to understand, and a potentially semi-stable representation can be given in terms of a semi-stable representation of some field extension and a descent datum, to get you back to where you started.
A nice exposition of the potentially crystalline case (with a nice application) can be found in Volkov's paper, "A class of p-adic Galois representations arising from abelian varieties over Q_p".<|endoftext|>
TITLE: Is the area of a polygonal linkage maximized by having all vertices on a circle?
QUESTION [11 upvotes]: Consider a (non-stellated) polygon in the plane.  Imagine that the edges are rigid, but that the vertices consist of flexible joints.  That is, one is allowed to move the polygon around in such a way that the vertices stay a fixed distance from their adjacent neighbors.  Such a system is called a polygonal linkage.  
As the linkage varies in its embedding in the plane, the area of the interior varies.  The question is, When is the area maximized?
I have a specific answer I suspect is correct, but I am having trouble showing.  I believe it is true that every polygonal linkage has as embedding where all the vertices lie on one circle (this isn't hard to show in the case when the linkage starts non-stellated).  My claim is that the area is maximized exactly when all the vertices lie on a circle.  
I can show this for a 4-sided polygon, but with techniques that do not generalize.
Also, my requirement that the polygon be non-stellated was only so that it was clear that there was a way to flex it to have all vertices on a circle.  This question extends to the stellated case, but the question there is whether every stellated linkage can be flexed to one which is non-stellated.

REPLY [3 votes]: This is proved in I. M. Yaglom and V. G. Boltyanskii's book Convex Figures, on page 208. The essence of the argument is suppose you have two polygons $M$ and $N$ whose corresponding sides have the same lengths, but one of them, $M$, is inscribed in a circle. Then you can glue the segments of the circle outside $M$ to the corresponding sides of $N$. The resulting figure won't be a circle, but it is known that amongst all compact convex figures of given perimeter, the circle is the one with largest area (the solution to the isoperimetric problem), So $M$ plus bits has more area than $N$ plus the same bits, so $M$ has more area than $N$.
A lot of this goes back to the work of Steiner.<|endoftext|>
TITLE: How to find a closest integer point to the intersection of two lines?
QUESTION [29 upvotes]: Here's a question that originates from StackOverflow.
Given are two lines on a plane, specified by equations ($a x + b y = c$) with integer coefficients. The lines aren't parallel and they don't necessarily pass through any integer points.  Given also is an integer point $(x,y)$ (x and y are both in $\mathbb{Z}$) that lies on neither line.
The problem is to find the integer point $(x',y')$ closest to the intersection of the lines that lines in the same quadrant of the plane as $(x,y)$.  It may be $(x,y)$ itself, but we're interested in the nontrivial case when an even closer point exists.
Other than being a way to specify the quarter of interest, the (x,y) point is seemingly needed to ensure the problem is NP or easier.  Without this point, it seems, the answer might not be polynomial wrt the line parameters $a_1$, $a_2$, $b_1$ etc.
The problem is claimed to have a polynomial solution, but I doubt that it really exists.  Is there a solution or a proof of its NP-completeness?

REPLY [2 votes]: Let $n$ be the input size, i.e. the number of input digits in some base, or the sum or max of the logs of the magnitudes of the input numbers-- these are all equivalent for the purposes of big-O.
Strategy: apply at most $O(n)$ shears (area preserving linear transformations leaving one coord axis fixed) to
get an easy problem, solve the easy problem, and then apply the inverse sequence of shears to get the solution back in the original
coordinate space.
EASY PROBLEM #1: when at least one of the two bounding rays is axis-aligned.
EASY PROBLEM #2: when the two bounding ray directions point into the interiors of different quadrants of the plane.
Details of the easy problems are omitted, since they are straightforward and not the interesting or challenging part of this problem.
So assume we have a non-easy case of the problem: that is, both bounding ray directions point into the interior of the same quadrant;
w.l.o.g. both point into the first quadrant.
Choose one of the two bounding rays and call it "primary" and the other "secondary".
If the primary ray has slope $< 1$, swap the coordinate axes, so that the primary ray has slope $\ge 1$.
Let $a/b \ge 1$ be the slope of the primary bounding ray, with $a \ge b \gt 0$ taken directly from the input equations.
Let $q,r$ be the quotient and remainder of dividing $a$ by $b$.  That is, $q=\mathrm{floor}(a/b), r=a-q b, a \ge b \gt r \ge 0$.
Apply to the problem geometry the shear that leaves the $y$ axis fixed and takes $(1,q)$ to $(1,0)$;
in other words, the linear transformation that leaves $(0,1)$ fixed and takes $(1,0)$ to $(1,-q)$.
Maintain the designation "primary" on the image of the primary bounding ray.  This shear transformation decreases the primary ray's slope to r/b, either keeping its direction into the first quadrant (if $r>0$), or parallel to the $x$ axis (if $r=0$).
Note that the solution before this shear transformation can be described by any of the following
equivalent characterizations (equivalent because both bounding ray dirs point into the first quadrant):

(a) "the integer point in the quarter closest to the intersection point" (by definition)
(b) "the integer point in the quarter with minimal $x+y$",
(c) "the integer point in the quarter with minimal $x$ among all points with minimal $y$".
(d) "the integer point in the quarter with minimal $y$ among all points with minimal $x$"

There are three cases.
Case 1: After the shear, both rays still point into the interior of the first quadrant.
In this case, the solution to the sheared problem is, again, the following equivalent characterizations:

(a') "the integer point in the transformed quarter closest to the transformed intersection point"
(b') "the integer point in the transformed quarter with minimal $x+y$"
(c') "the integer point in the transformed quarter with minimal $x$ among all points with minimal $y$".
(d') "the integer point in the transformed quarter with minimal $y$ among all points with minimal $x$"

By focusing on (d) and (d') it's clear that the solution to the transformed problem
is the transformed solution to the original problem.
In the transformed problem, the primary bounding ray has slope $r/b$ with $b > r > 0$ (still in Case 1 here) so its slope is $< 1$;
therefore as we start analysis of the transformed problem as before,
we'll swap the coords so that its slope $b/r > 1$.
Notice that what we've done so far is perform one iteration of the euclidean algorithm;
that is, $a$ has been replaced by $b$, and $b$ has been replaced by the remainder of $a$ divided by $b$.
Repeat as often as we find ourselves in Case 1 (keeping the "primary" designation on the same one of the two rays even as they get transformed).  Since the $a,b$ values are following the euclidean
algorithm starting with two of the original input numbers, well-known analysis of the euclidean algorithm
tells us Case 1 can happen at most $O(n)$ times: that is, if we get all the way to the end of the euclidean algorithm,
$b$ will become 0 which will take us out of Case 1 (if we haven't already left it before that).
Case 2: The shear makes the primary bounding ray horizontal, and the sheared secondary ray still points into the first quadrant.
Exactly as in Case 1, (a')=(b')=(c')=(d') and so the solution to the transformed problem
is the transform of the solution to the original problem.  Furthermore the transformed problem
is now a case of EASY PROBLEM #1; solve that, inverse-transform the solution back to the original space; done.
(When this is encountered recursively from Case 1, it means we've made it all the way to the end of the euclidean algorithm.)
Case 3: The shear takes the secondary bounding ray direction out of the interior of the first quadrant, to either the $x$ axis or the fourth quadrant. (This is regardless of whether the primary bounding ray direction became horizontal or stayed in the first quadrant).
This case is relatively easy to solve, but we have to be careful--
in this case the transform of the point inside the quarter closest to the intersection (i.e. satisfying (a))
is not necessarily the point in the transformed quarter closest to the transformed intersection (i.e. satisfying (a')).
However, it is true that the transform of the point satisfying (d) is the point satisfying (d').
Therefore the transformed problem, though easy, is not in the form of the original problem.  Instead, it's a case of:
EASY PROBLEM #3: The bounding ray directions both point into the +x half plane,
but do not both point into the interior of the same quadrant,
and we are asked to find the point satisfying (d) rather than (a) (which are not necessarily the same in this case).
So I've described an algorithm for transforming the original problem
into one of three easy problems, in $O(n)$ iterations.
Complexity analysis
The number of iterations is $O(n)$, so the overall runtime is $O(n)$ times the cost per iteration.
So how expensive is an iteration?
Each iteration consists of a small constant number of arithmetic operations, each of which is at most $O(M(\mathrm{num\ digits\ in\ operands}))$ where $M$ stands for the cost of one multiplication (see the wikipedia article "Computational cost of mathematical operations").  But how many digits is that?
Well, at first glance it appears that intermediate and final values
could have up to $n^2$ digits which would make the overall runtime $O(n M(n^2)))$.
Furthermore one might suspect the $n$-digit witness point could help decrease this bound.
But the following closer analysis reveals
it's not as bad as that, and the $n$-digit witness actually makes no difference at all.
Each transform is a shear followed by a coord axis swap:
$$
\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
- q_i & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
- q_i & 1 \\
1 & 0 \\
\end{pmatrix}
$$
where $q_0, q_1, ...$ is the sequence of quotients $q$ occurring
in the euclidean algorithm (possibly cut short by Case 3).
So the matrix norm of the cumulative matrix is $\leq$ the product of the $q_i$'s,
which is at most the original value of $a$, which has $n$ digits.
And the same can be said about the cumulative inverse matrix.
That tells us the maximum number of digits occuring in the answer or any intermediate value is $O(n)$.
Therefore the overall runtime is $O(n M(n))$.
Incidentally, the sequence of 2x2 integer matrix computations being followed is actually well known
as the EEA (Extended Euclidean Algorithm) for CRT (Chinese Remainder Theorem).
If taken all the way to completion,
the cumulative matrix represents an invertible linear transformation with integer coeffs
that takes the original primary ray direction to a coordinate axis,
which can be a useful building block for many applications (notably Project Euler problems :-) ).
So one might think we could simply use a prepackaged EEA implementation as a building block;
but I don't see how to make that work for this problem,
due to the possibility of needing to stop early due to Case 3.<|endoftext|>
TITLE: Given a smooth algebraic curve in CP^n, why is the variety of trisecants at most two-dimensional?
QUESTION [8 upvotes]: I'm trying to understand the following statement which I read somewhere without proof.  Let $C$ be a smooth algebraic curve in $\mathbb{CP}^n$.  Define $X_k$ to be the subvariety of $\mathbb{CP}^n$ consisting of points that lie on lines that intersect $C$ at least $k$ times.  Then $X_3$ is at most $2$ dimensional.
Here's how I've been trying to understand this result.  It is easy to see that $X_2$ is at most $3$-dimensional.  Indeed, let $Y \subset C \times C \times \mathbb{CP}^n$ be the closure of the set of points $(v,v',w)$, where $v \neq v'$ and $w$ lies on the line through $v$ and $v'$.  Then $Y$ is clearly at most $3$ dimensional and $X_2$ is the projection of $Y$ onto its third factor.
We have $X_3 \subset X_2$, and the desired statement would follow if we could prove that $X_3 \neq X_2$.  However, this need not hold -- for instance, $C$ could be a high degree curve in some $\mathbb{CP}^2$ in $\mathbb{CP}^n$ (in that case, of course, the desired result is trivial!).  How do I complete this proof?

REPLY [7 votes]: Dear Undergraduate Student, first of all, congratulations on the beautiful geometry you chose to study so early in your studies.
To prove $dim X_3 \leq 2$ it is indeed enough to prove that there exists a secant to $C$ which is not a trisecant, since the space of secants is irreducible (Harris, Algebraic Geometry, p.144). 
This existence is proved on page 110 of the book Geometry of Algebraic Curves by Arbarello, Cornalba, Griffiths, Harris (Springer, Grundlehren 267).They assume the curve is nondegenerate i.e. not contained in a hyperplane of $\mathbb P^n, n \geq 3$, and you can reduce to this for a non plane curve.
For a plane curve the result $dim X_3 \leq 2$  is evident.<|endoftext|>
TITLE: Is the group of integer points on a finite-type group scheme over Z finitely presented?
QUESTION [33 upvotes]: Let $G$ be a group scheme of finite type over $\mathbf{Z}$.  Must $G(\mathbf{Z})$ be finitely presented?

(The question is inspired by a not yet successful attempt to answer a question of Brian Conrad.)
A few special cases:

If $G$ is the Néron model of an abelian variety over $\mathbf{Q}$, then a positive answer amounts to the Mordell-Weil theorem (combine with restriction of scalars to get the full Mordell-Weil theorem).
If $G$ is the restriction of scalars of $\mathbf{G}_m$ from the ring of integers of a number field down to $\mathbf{Z}$, then a positive answer follows from Dirichlet's unit theorem.

REPLY [14 votes]: It follows from Theorem 6.12 of Borel and Harsh-Chandra, "Arithmetic subgroups of algebraic groups", that $G(\mathbb{Z})$ is finitely generated if $G$ is affine. Perhaps one can combine this with Chevalley's theorem to deduce finite generation in the general (not necessarily affine) case.
EDIT  (Added idea for proof of general case; see also Torsten Ekedahl's comment below)
EDIT (Proof completed (assuming $G$ is separated) and simplfied using comments of BCnrd below)
As discussed in the comments below, we may assume $G$ is flat and we may also assume it is connected. By Chevalley's theorem, there is an affine subgroup scheme $H_{\mathbb{Q}}$ of the generic fibre $G_\mathbb{Q}$ of $G$
such that the quotient $G_{\mathbb{Q}}/H_\mathbb{Q}$ is an abelian variety. Let $H$ be the Zariski closure of $H_{\mathbb{Q}}$ in $G$ with the reduced induced structure. Then $H$ is a closed subgroup scheme of $G$. By a theorem of Raynaud (see comment of BCnrd below for the reference) $H$ is also affine.
We have an incusion of groups 
$G(\mathbb{Z})/H(\mathbb{Z}) \subset G(\mathbb{Q})/H(\mathbb{Q}) \subset (G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$.
Since $G_{\mathbb{Q}}/H_{\mathbb{Q}}$ is an abelian variety, by the Mordell-Weil theorem $(G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$ is a finitely generated abelian group, hence so is $G(\mathbb{Z})/H(\mathbb{Z})$. Since $H(\mathbb{Z})$ is finitely generated by the Borel-Harish-Chandra theorem, it follows that $G(\mathbb{Z})$ is finitely generated.<|endoftext|>
TITLE: Must a locally compact group be Hausdorff in order to possess a Haar measure?
QUESTION [5 upvotes]: Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?

REPLY [6 votes]: No. Simon Rubinstein-Salzedo's "On the existence and uniqueness of invariant measures on locally-compact groups" presents a proof of existence (and uniqueness up to a multiplicative strictly positive constant) of a left Haar measure given a locally-compact, not necessarily Hausdorff, group.<|endoftext|>
TITLE: Self-homomorphisms of surface groups
QUESTION [10 upvotes]: Let $X$ be a closed, orientable surface of genus at least 2, and let $\phi: \pi_1(X) \to \pi_1(X)$ be a surjective homomorphism.  Is $\phi$ necessarily injective?

REPLY [10 votes]: Yes.  Surface groups are Hopfian.  More generally, all residually finite groups are Hopfian -- see Theorem IV.4.10 in Lyndon and Schupp's book "Combinatorial Group Theory".<|endoftext|>
TITLE: Upper bound of period length of continued fraction representation of very composite number square root
QUESTION [14 upvotes]: Given natural numbers of special very composite form, like primorials or factorials, how to give some useful upper bound limit of continued fraction period length of their square roots?
I'm not a professional, any advices are welcome.

REPLY [11 votes]: The continued fraction length is usually a small constant factor away from the regulator. A more precise version can also be achieved, but I don't remember a reference, so if anyone does...
Then, we know the regulator times the class number is usually a small constant factor away from the discriminant (of the order, not necessarily the field).
In addition, if a discriminant has $n$ prime factors in its squarefree part, the class number will be divisible by $2^{n-1}$.
Finally, most positive numbers of some size don't have many prime factors, and we suspect the real quadratic fields composed from these to have relatively small class number (look up Cohen-Lenstra).
Combining these facts and heuristics we get that primorials, and even factorials (still have large squarefree part), will have larger class number, hence small regulator, and therefore smaller continued fraction period length.
That said, we can dig even deeper. For factorials, there is a hefty squareful part. When we go from the maximal order of the field $\mathbb{Q}(\sqrt{n!})$ to the order $\mathbb{Z}(\sqrt{n!})$, the discriminant is enlarged hugely. Each prime (even) power $p^{2m}\ ||\ n!$ adds $p^{m-1}(p-(squarefree(n!)/p))$ to the original $h_K\times R_K$. So for each such factor, something goes into the class number of the smaller order, and something goes into the regulator of the smaller order.
Here comes the interesting bit. The factors that make up the new regulator tell us how far the unit group in the small order is from the unit group of the maximal order. Since we are in a real quadratic field the unit groups are of rank 1, so this distance is just the exponent to which the fundamental unit is powered by in order to enter the smaller order. Say $p_1-1$ and $p_2-1$ (say Legendre symbol is 1) have a large gcd. Once we power the fundamental unit to, say, $p_1-1$ to get (locally) into the $p$-part of the smaller order, we can slack off when getting into the $p_2$-part of the smaller order, because we've already done some of the work.
So in the factorial case, or in any number with many square factors (not dividing the squarefree part), since many of the above mentioned factors will have a large gcd, most of the factors will have to go into the class number of the small order - hence the very small regulator and continued fraction expansion.

We can write the above explicitly, but we need another notion. When a prime divides the discriminant to an even power greater than 2, the power of $p$ from the above mentioned factor that goes into the regulator measures how the fundamental unit is far p-adically from being 1. For most $p$, the fundamental unit will not be $1\ (mod\ p^2)$, so pretty much all of the $p^{m-1}$ goes into the regulator.
Hence, we expect:
$$ \frac{period-length(n!)}{\sqrt{n!}} \sim \frac{lcm(\{\ p-(p/squarefree(n!))\ |\ p^{2m}\ ||\ n!\ \}}{\prod_{p^{2m}||n!} (p-(p/squarefree(n!))} $$
And how that factor behaves - I have no idea. Sounds like a combinatorical answer could exist.<|endoftext|>
TITLE: Are extensions of linear groups linear?
QUESTION [18 upvotes]: A group $G$ is said to be linear if there exists a field $k$, an integer $n$ and an injective homomorphism $\varphi: G \to \text{GL}_n(k).$ 
Given a short exact sequence 
$1 \to K \to G \to Q \to 1$ of groups where $K$ and $Q$ are linear (over the same field), is it true that $G$ is linear too?
Background: Arithmetic groups are by definition commensurable with a certain linear group, so they are finite extensions of a linear group, and finite groups clearly are linear (over any field).

REPLY [11 votes]: Erschler has shown that there exists a central extension $G$ of 
$\mathbb{Z} \mathbin{wr} \mathbb{Z}$ by a finite group $F$ which
is not residually finite. Thus the short exact sequence $1 \to F \to G \to \mathbb{Z} \mathbin{wr} \mathbb{Z} \to 1$
provides an example of a non-linear group which is an extension of two linear groups over $\mathbb{C}$.
A. Erschler, Not residually finite groups of intermediate growth,
commensurability and non-geometricity, J. Alg. 272
(2004), 154--172.<|endoftext|>
TITLE: Ideals of statements?
QUESTION [8 upvotes]: The following is a somewhat vague question concerning logic, but with ideas from algebraic geometry (see in particular the example at the end).  The vagueness is in the notion of "language".
Let $A$ be a set and fix some language in which to write propositions which assign a value of true or false to each element of A.  Let $R$ be the set of such propositions about elements of $A$, and suppose that $R$ contains "True" and "False" and is closed under "and" ($\wedge$) and "or" ($\vee$).
Now let $V\subseteq A$ be a subset and consider the set $I(V)\subseteq R$ such that $r\in I(V)$ iff $r$ holds of every element $v\in V$.  This set $I(V)$ has the following properties:

"True" is in $I(V)$,
If $i$ and $j$ are in $I(V)$ then $i\wedge j$ is in $I(V)$, and
If $i$ is in $I(V)$ and $r\in R$ then $r\vee i$ is in $I(V)$.

Thus $I(V)$ satisfies the axioms of an "ideal" in the "ring" $R$ if we take the following "strange correspondence" between statements and algebraic operations: $T\mapsto 0$, $F\mapsto 1$, $\wedge\mapsto +$ and $\vee\mapsto \times$. Note that in fact $R$ is a commutative "Rig" (ring without negatives) under this "strange correspondence."  The correspondence is "strange" because we usually think of "or" as $+$, "and" as $\times$, "true" as $1$, and "false" as $0$.  But we can find intuition for this correspondence via Example 2 below.
Suppose we define an ideal in $R$ to be a subset $I\subseteq R$ satisfying conditions 1,2,3.  Given an ideal $I$ we can consider the set $Z(I)\subseteq A$ of all elements $a\in A$ such that $i$ holds of $a$ for each $i\in I$.  Call such subsets "closed".  
Observation 1: There is an (order-reversing) correspondence between the ideals of $R$ and the closed subsets of $A$.
Example 2:  Suppose we take the language of commutative rings.  Say $A$ is the set  ${\mathbb R}^n$ and $R$ is the set of statements of the form $f(x_1,\ldots,x_n)=0$ where $f$ is a polynomial with real coefficients.  The set of such statements forms a ring, by operating on the polynomials; that is $R\cong {\mathbb R}[x_1,\ldots,x_n]$.
Given a subset $V\subseteq A$ we can consider those equations that are true of all points in $V$.  Such a subset will satisfy conditions 1,2,3 above (where "True" is 0=0).  We see that the "strange correspondence" discussed above does make sense: if $f=0$ and $g=0$ hold of every point in $V$ then so does $f+g=0$; for any equation $r=0$ in $R$, if $f=0$ holds of every point in $V$ then so does $rf=0$. The empty subset of $A$ is satisfied by $1=0$ (or "false").  In fact, Observation 1 is the basic observation of algebraic geometry in this context.  
Question 3: Has anyone looked at this correspondence?  Can it be made more rigorous?  Can algebro-geometric notions (like schemes) be applied to other "languages" in an interesting way?

REPLY [9 votes]: Yes, of course, the algebraic aspects of logic have been very well studied. There is a lot to say about this, but since I am supposed to be on a voluntary MO hiatus until the end of the semester, I will only mention a few things.
You might want to ask for your "ideals" to be closed under logical equivalence too. Otherwise, statements of length 17 or more form a rather silly ideal. With this change, your "ideals" are called deductively closed theories. 
The space you're describing is basically the Stone space of the Lindenbaum algebra of your language. The Lindenbaum algebra is the Boolean algebra which consists of all sentences of the language modulo logical equivalence. This construction also makes sense over a nontrivial base theory T, where logical equivalence is replaced by T-provable equivalence.
Another name for this Stone space is the space of (complete) 0-types over the theory T. The more general space of (complete) n-types is obtained in the same way by using formulas in n fixed variable symbols x1,...,xn (instead of sentences, which have no free variables). These spaces of types and their topology are a central concept in model theory. One usually takes T to be a the complete theory of a structure M. Then the space of 0-types has only one point since T is complete and n-types correspond to coherent things that one could say about a n-tuple of elements of M.
Several theorems of model theory have interesting meanings in this context. For example, the Compactness Theorem corresponds to the fact that Stone spaces are compact, and the Omitting Types Theorem corresponds to the Baire Category Theorem (for compact Hausdorff spaces). Some leading model theorists have supported the view that model theory should become more and more algebraic/geometric, which is indeed a current trend.<|endoftext|>
TITLE: Percolation Model and Complex Probabilities 
QUESTION [5 upvotes]: Let $d>0$ be an integer and consider the first neighbors independent bond percolation model in $\mathbb Z^d$, where each edge is open with probability $p\in[0,1]$. 
I would like to know, if can we generalize the construction of the probability measure of this model for a complex parameter $p\in \mathbb C$ in some neighborhood of zero. 
In other words, 
Let be  $E=\{\{x,y\}\subset\mathbb Z^d: \|x-y\|_1=1\}$, where $\|x-y\|_1=\sum_{j=1}^d|x_j-y_j|$. 
If $\{0,1\}$ a complex measure space, such that $\mu(\{1\})=p\in\mathbb C$ and $\mu(\{0\})=1-p$, is there any domain $D$ on the complex plane, for which it is possible to construct a product measure, formally given by  
$$\mathbb P_p=\prod_{e\in E} \mu $$ 
defined on the sigma algebra generated by the cylinder sets of $\Omega=\{0,1\}^{E}$ with $\mathbb P_p(\{0,1\}^{E})=1$ 
for any $p\in D$ ? 
If the answer is no. What is the best set-function fitting as much as possible the properties of a measure we could construct for a $p$ in some domain of the complex plane ? 
This question it was motivated by the possibility, in case of a positive answer ( or a good construction near a probability), to use complex analysis results, to study the behavior of the probability of certain events as a function of $p$.

REPLY [3 votes]: Not an answer to your original question, but more a reaction to the previous answer. It does make a lot of sense to analyse the behaviour of various quantities as function of complex-valued physical parameters (here p). For example, analysis of a statistical mechanical system (say, an Ising model) as a function of a complex magnetic field or complex temperature provides many important information about the system. To cite some: the Lee-Yang theorem (about possible locations of singularities, and thus possible locations of phase transitions) or Isakov's theorem (existence of an essential singularity at 0 of the free energy of the Ising model as a function of a complex magnetic field, thus showing that stable phases cannot be analytically continued into the metastable phases, contrarily to what mean-field theory suggests). Actually, there is a old version of Isakov theorem's for percolation (by Kunz and Souillard).<|endoftext|>
TITLE: Complex analytic vs algebraic families of manifolds
QUESTION [7 upvotes]: I'm studying the deformation theory of compact complex manifolds as developed by Kodaira and Spencer. On the side I'm reading as much about deformation theory in general as I can get my hands on (and understand), and I've been wondering about the relationship between the basic definitions in the analytic and algebraic categories. To summarize:
Analytic side: A complex analytic family of smooth compact manifolds is a holomorphic map $\pi : \mathcal X \to S$ of smooth complex manifolds $\mathcal X$ and $S$ such that $\pi$ is a proper submersion and each fiber $X_t = \pi^{-1}(t)$ is a compact complex manifold. This implies some other conditions, like that $\mathcal X$ is locally trivial over $S$.
Algebraic side: A family of schemes is a proper flat morphism $\pi : X \to Y$ of schemes.
I've been asking myself what the relationship between these definitions is. To get something like the algebraic definition in the analytic category we just replace "scheme" by "complex space".
Now, a complex manifold is a smooth complex space, and local triviality of $\mathcal X$ along with compactness of the fibers implies that $\pi : \mathcal X \to S$ is proper (edit: unnecessary). I'm also fairly certain that $\pi$ is flat (my algebraic side is weak), so $\pi : \mathcal X \to S$ will be a family of complex spaces in the algebraic sense.
My question is: what conditions do we need on $\pi : X \to Y$ to pass in the other direction? Is it enough that the complex spaces $X$ and $Y$ be smooth? I've been thinking about this and I've got this vague idea that flatness of $\pi$ and coherence of the structure sheaves will lead to local triviality, but I haven't been able figure out how.

REPLY [4 votes]: The standard situation in Kodaira-Spencer's work is the following:
If you're on the algebraic side and you have a smooth ("smooth" in the sense of algebraic geometry) and proper (proper in the sense of algebraic geometry) map $\pi: X \to Y$ , then when you translate this to the analytic side, "smooth" turns into "submersion" (in the sense of: pushforward of vector fields is surjective), and "proper" turns into "proper" (in the sense of: inverse image of compact set is compact). And "map" turns into "holomorphic map". Then you can use, for instance, the "preimage theorem" (be careful to not get confused by the usage of "smooth" in that article --- there smooth means $C^\infty$) to deduce that the fibers are holomorphic complex manifolds. Strictly speaking we must use the holomorphic version of the "preimage theorem". But the holomorphic version does hold, as do holomorphic versions of other standard theorems like implicit function theorem and inverse function theorem. Perhaps this is in Chapter 0 of Griffiths-Harris, or Chapter 1 of Huybrechts.
The fibers are compact because a point is compact. :)<|endoftext|>
TITLE: Big picture concerning Ito integral, Stratonovich integral and standard results in probability theory
QUESTION [8 upvotes]: I am confused and don't get the big picture concerning the connection between

Ito integral
Stratonovich integral
Standard results in probability theory concerning skewed distributions.

Example: Take e.g. the mean of the log-normal distribution $e^{\mu+{\sigma^2\over 2}}$. The extra term $\sigma^2\over 2$ is a result of Jensen's inequality. The same result can be produced by Ito's lemma, it is the result of the famous extra term. Therefore this term is sometimes also called the Ito correction term.
Question: What I don't understand is why we need a strange integral which can't be used in the classical way (e.g. the standard chain rule doesn't hold any more) to come to a result which could as well be derived by standard techniques. On the other hand if we use an integral where the classical rules still apply (Stratonovich integral) we miss this additional term.
Why can't we just transform the stochastic process into an equivalent probability distribution and use the classical integrals to arrive at the right results (and, too, stick to the classical calculus rules). In a way we then would integrate not with respect to a stochastic process but with respect to the resulting probability distribution. It all seems like outsmarting ourselves and overcomplicating matters?!?
Addendum: Don't get me wrong: I think I understand (most) of these results piece by piece, what I miss is the bigger picture how everything fits together - perhaps somebody can enlighten me - thank you!

REPLY [9 votes]: It's not quite clear where exactly you have a difficulty. Of course, using stochastic calculus and Ito's integral (which is central to the modern theory of stochastic processes) to derive properties of the log-normal distribution is an overkill, but it might be a nice exercise.
Some random quick points on why Ito's integral is important:

Ito introduced his integrals to describe continuous Markov processes. This was successfully carried out so that now we have a nice view of Markov processes as solutions of stochastic differential equations.
Ito's integral seen as a process is a martingale which is very convenient since martingales "have a lot of structure", and there are lots of martingale tools. As for Stratonovich's integral, the change of variables formula for it is the same as in classical calculus, so it might appear more natural, but it is harder to work with due to the lack of martingale structure.<|endoftext|>
TITLE: Improvements of the Baire Category Theorem under (not CH)?
QUESTION [5 upvotes]: The Baire category theorem implies that a nonempty complete metric space without isolated points must be uncountable.  In many situations I have encountered, the "natural examples" of complete metric spaces without isolated points (of a certain type, or possibly with some additional structure) in fact have at least continuum cardinality.  This is not so surprising, since if Cantor's continuum hypothesis holds, then uncountable is equivalent to at least continuum cardinality.
However, if we do not wish to assume CH -- and, ever since Godel and Cohen proved that (G)CH is independent of ZFC set theory, this seems to be the prevalent attitude -- what can be said about the existence of such spaces of uncountable cardinality less than the continuum?  
I asked this question to someone before, and I seem to remember that it is known that one cannot unconditionally improve the conclusion of this application of Baire category to say "continuum cardinality".  But could someone say a little bit about how this goes?  Preferably in words that are comprehensible to a non-set theorist like myself?
Addendum: Thanks to Sergei Ivanov for a quick and convincing answer: evidently I was making things much more complicated than I needed to.  Just to get myself reoriented properly, I would like to try to remember where the set-theoretic subtleties come in.  Suppose I ask about the conclusion of BCT itself, rather than this particular corollary: not assuming CH, what can we say about the minimal cardinality of a covering family of nowhere dense subsets in a complete metric space?  
Second Addendum: I was even more turned around than I had realized: I was (i) worrying needlessly about uncountable cardinals smaller than the continuum and (ii) not worrying enough about cardinals greater than the continuum! In particular, I was under the misimpression that for any cardinal $\kappa \geq 2^{\aleph_0}$, $\kappa^{\aleph_0} = \kappa$.  This led me to incorrectly guess the strong form a classical theorem of F.K. Schmidt.  I think I have it right now: if you are interested, see pp. 13-16 of 
http://math.uga.edu/~pete/8410Chapter3.pdf

REPLY [6 votes]: The number of meager sets needed to cover the real line is a "cardinal invariant of the continuum". It is one of the invariants in Cichoń's diagram. In particular, it is Cov(K) in Cichoń's diagram on Wikipedia. 
Looking at nowhere-dense sets instead of meager sets would not change this invariant, because of basic cardinal arithmetic. 
I am not certain, off the top of my head, if the invariants are the same for every uncountable complete separable metric space.<|endoftext|>
TITLE: What are the uses of the homotopy groups of spheres?
QUESTION [78 upvotes]: Pete Clark threw down the challenge in his comment to my answer on Why the heck are the homotopy groups of the sphere so damn complicated?:

Have the homotopy groups of spheres ever been applied to anything, including in algebraic topology itself?

It started to get some answers in those comments, but comments are a lousy place to record answers to a question like this so I'm reposting it as a question.
In order to add some more value to the question (and justify my reposting it), let me say that I can foresee answers coming in several different flavours and I'd like the answers to explicitly say which flavour they use.
Firstly, there is the distinction between stable and unstable homotopy groups.  Briefly, there is a natural map $\pi_k(S^n) \to \pi_{k+1}(S^{n+1})$ and eventually (you will see the phrase, "in the stable range") this becomes an isomorphism.  Once it is an isomorphism, we refer to them as the stable homotopy groups.  So there are more unstable homotopy groups than stable ones, but to balance that, the stable ones are better behaved.
Secondly, there is the point that I was trying to make in the aforementioned question: the fact that the homotopy groups are so complicated is correlated with their usefulness.  So there may be some uses of the homotopy groups of spheres that explicitly rely on their complexity: if they weren't so complicated, they wouldn't be able to detect X.
Thirdly, and partly in converse to the above, we do know some of the homotopy groups of spheres.  So a use might be: because we know $\pi_7(S^{16})$ then we know X.
So in your answer, please indicate which of the above best fits (or if none do, try to classify it in some way).  Also, please note that this is a question about the homotopy groups of spheres, not homotopy theory in general, and that although I'm an algebraic topologist (some of the time), answers outside algebraic topology will be more useful in "selling" our subject!

This question is a fairly obvious one for community wiki: it wasn't originally my question (though I hope that I've expanded it a little to add extra value) and I appear to be asking for a "big list".  However, I suspect that the really good answers will involve some work to explain to a non-expert the key idea of why the homotopy groups of spheres are so important - merely linking to a paper will not be very satisfactory because it is likely that that paper is written for algebraic topologists rather than a general audience, and I would like to reward such efforts with the only coinage MO has.  If the only answers I get are "see this paper" then I will gladly hit the "community wiki" button (indeed, if that was all I got, I'd consider closing the question).

REPLY [4 votes]: I am just reporting something I am not familiar with, but apparently the existence of the Breen-Deligne resolution for Abelian groups requires finite generation of stable homotopy groups of spheres.
The result is the following
Theorem (Breen–Deligne). For an abelian group $A$, there exists a resolution of the form
$$ \dots → \bigoplus_{j = 0}^{n_i} ℤ[A^{r_{i,j}}] → \dots →  ℤ[A^2] → ℤ[A] → A → 0 $$
that is functorial in $A$.
The proof appears for instance in Appendix to Lecture IV in Scholze's lectures on condensed mathematics and I learned about this example from this answer.
In particular, the fact that the proof (that I have not read) requires finite generation of stable homotopy groups of spheres is a comment by @ReidBarton<|endoftext|>
TITLE: How do you select an interesting and reasonable problem for a student?
QUESTION [79 upvotes]: I am interested in how to select interesting yet reasonable problems for students to work on, either at Honours (that is, a research-based single year immediately after a degree) or PhD. 
By this I mean a problem that is unsolved but for which there is a good chance that a student can solve it either completely or partially and come out with a thesis either way.
There are a number of possible strategies that I see some of my colleagues use, but which are sadly not available to me:
(1) Be sufficiently brilliant that you already know roughly how something unsolved should be solved and guide the student accordingly, modifying the strategy on the fly.
(2) Have a major project, say classifying a big class of structures, that is amenable to attack with a big general theorem with many well-defined sub-cases that can be assigned individually to students. 
Personally I often work on problems that lead nowhere - I don't solve it, it's too hard, I only rediscover known examples etc - but provided at least some of the problems work out - it doesn't matter.
But for students, its more of a "one-shot" affair - they can't afford to work for a year or, worse, three years, and get nothing.
Are there any general principles that will help in the selection of problems? Or is just a situation where you've either "got the knack" or you haven't?

REPLY [2 votes]: I have advised just one honors student, and in the process of advising another one. However, in my field, (combinatorics) there are some problems that are great for students:

Anything with computers and computation. Students are great at finding patterns. This approach paired with OEIS is a nice combination.
My current student is looking at $q$-analogues of certain counts. Pick your favorite family of combinatorial objects, and see if it has a natural $q$-analogue. This is perfect for a small thesis.
If you feel like the student can handle it and you are lucky with your $q$-analogue, it might have a cyclic sieving phenomena, which one can explore.

Even computer experiments can be reported on, and enumerative problems can usually be "tuned" where special cases can be attacked first.
In my concrete example, we started looking at border-strip decompositions, and counting these. My student quickly came up with a nice family which admits a different interpretation, and fulfills some $q$-analogue properties.<|endoftext|>
TITLE: Why do Delta-sets not allow quotients?
QUESTION [7 upvotes]: A $\Delta$-set is a contravariant functor from the category $\Delta'$ of order-preserving injections to the category of sets (this is essentially what Allen Hatcher calls a $\Delta$-complex).
A main reason for working with simplicial sets instead of $\Delta$-sets should be that they allow quotients (see e.g. Allen Hatcher's nice appendix "CW complexes with simplicial structure" to his Algebraic Topology book: "A major disadvantage of $\Delta$-complexes is that they do not allow quotient constructions"), How does this go well with the fact that the category of functors $\Delta'op\to Sets$ has colimits?
(This question was already asked in a comment on Allen Hatcher's answer to this question on the definition of simplicial complexes. I apologize for asking it twice but there has been no answer given and I am afraid that the reason is - if it's not the silliness of my question - that the comment appears only after pressing the "more comments" button. However, I apologize.)

REPLY [6 votes]: The basic issue is that not every function that we would like to describe between $\Delta$-complexes can be realized by a natural transformation between functors.  The lack of degeneracy maps means that no map $X \to Y$ of $\Delta$-complexes that sends any simplex down to a degenerate simplex can be realized by a natural transformation of functors.  For example, if $X$ is a $\Delta$-complex interval realizing $[0,1]$ and $Y$ is a $\Delta$-complex realizing $[0,1]^2$, then there is no natural transformation of functors realizing the projection maps $p_i:[0,1]^2 \to [0,1]$.
As a consequence, the category of $\Delta$-complexes does not have enough immediately-available maps between objects to construct the kinds of colimit diagrams one would like to realize.<|endoftext|>
TITLE: Do symmetric spaces admit isometric embeddings as intersections of quadrics?
QUESTION [5 upvotes]: While preparing a seminar I gave today, the following question arose.  I asked the seminar participants, but nobody knew the answer.  Hence I'm asking it here in MO.
Background
Recall that a complete, connected and simply connected pseudoriemannian manifold $(M,g)$ is a symmetric space if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection: $\nabla R = 0$.
Typical examples are the (simply-connected) spaces of constant curvature: sphere, hyperbolic space, (anti) de Sitter spacetimes,... all of which admit local isometric embeddings as quadrics in some  flat pseudoriemannian manifold $\mathbb{R}^{p,q}$.  Recall that the flat metric on $\mathbb{R}^{p,q}$ is given in flat coordinates by
$$\sum_{i=1}^p (dx_i)^2 - \sum_{i=1}^q (dx_{p+i})^2.$$
For example, the sphere with unit radius of curvature embeds in $\mathbb{R}^{n+1}$ as the quadric
$$x_0^2 + x_1^2 + \cdots + x_n^2 = 1,$$
whereas the hyperbolic space embeds in $\mathbb{R}^{1,n}$ as one sheet of the quadric
$$-x_0^2 + x_1^2 + \cdots + x_n^2 = -1,$$
again for unit radius of curvature.
Similarly, and again for unit radii of curvature, $n$-dimensional de Sitter spacetime is the universal covering space of the quadric
$$-x_0^2 + x_1^2 + \cdots + x_n^2 = 1$$
in $\mathbb{R}^{1,n}$, whereas $n$-dimensional anti de Sitter spacetime is the universal covering space of the quadric
$$-x_0^2 + x_1^2 + \cdots + x_{n-1}^2 - x_n^2 = -1.$$
This continues to be the case for other spaces of constant curvature in other signatures.
Other riemannian symmetric spaces, such as the grassmannians, can also admit isometric embeddings, this time in projective spaces, whose image is the intersection of a number of quadrics.  This is the celebrated Plücker embedding.  Notice that grassmannians do not (generally) have constant sectional curvature.
The remaining nontrivial lorentzian symmetric spaces -- the $n$-dimensional Cahen-Wallach spaces -- can also be locally embedded isometrically in $\mathbb{R}^{2,n}$ as the intersection of two quadrics.
In particular this shows that all the indecomposable lorentzian symmetric spaces (in dimension $>1$, at least), which are the (anti) de Sitter and Cahen--Wallach spacetimes, can be locally embedded isometrically as the intersection of quadrics in some pseudoeuclidean space.
Question

Is this also the case for the other simply-connected (pseudo)riemannian symmetric spaces?

Perhaps asking about quadrics is too strong, so perhaps a weaker question is

Are simply-connected symmetric spaces always (locally) algebraic?

Here by locally algebraic I mean that they are the universal covering space of an algebraic space.

REPLY [4 votes]: Not so much an answer to the original question as a very late (5 months—sorry: only just noticed!) response to José's request for more on symmetric $R$-spaces as intersections of quadrics.
(1) Flag manifolds (thus $G/P$ for $G$ semisimple and $P$ parabolic) arise in lots of ways as projective highest weight orbits: this is a generalisation of the Plücker embedding of Grassmannians, for example.  If $V$ is an irrep of $G$ with highest weight $\lambda$ and ($1$-dimensional) highest weight space $L\in\mathbb{P}V$, the orbit $G\cdot L$ is a copy of $G/P$ for some parabolic $P$, and all $G/P$ arise this way for many $\lambda$.
(2) These orbits are intersections of quadrics: first note that $L^2$ is a highest weight space for the weight $2\lambda$ in $S^2 V$ so that the image under the Veronese map $\mathbb{P}V\to \mathbb{P}(S^2V)$ of $G\cdot L$ lies in $\mathbb{P}W$ where $W$ is the irreducible submodule of $S^2V$ with that highest weight, $W$ is called the Cartan square of $V$.  Kostant's result (in J. Alg, I think) is that the converse is true: the image of the Veronese map intersects $\mathbb{P}W$ in exactly this orbit.  Otherwise said, the flag manifold is cut out by quadratic equations $p(v^2)=0$ for $v\in V$ and $p:S^2 V\to S^2 V/W$ the natural projection.
(3) Example: The Grassmannian of $2$-planes in $\mathbb{C}^4$ is the highest weight orbit in $\mathbb{P}(\bigwedge^2\mathbb{C^4})$.  Meanwhile $S^2\bigwedge^2\mathbb{C}^4=W\oplus\bigwedge^4\mathbb{C}^4=W\oplus\mathbb{C}$ so that the Grassmannian is cut out by one quadric equation and we see the celebrated identification with the Klein quadric.<|endoftext|>
TITLE: Do rational numbers admit a categorification which respects the following "duality"?
QUESTION [22 upvotes]: I need to give a lot of quite basic background to this question because I think (at least from conversing with fellow graduate students) that most mathematicians have not really thought about fractions  for a long time. I think that there is an interesting germ of an idea in here somewhere, but I cannot exactly pinpoint it. Essentially there seems to be two canonical ways to solve division problems and there does not seem to be a "natural isomorphism" relating the two ways. I am interested in framing this duality formally: is there a "categorification" of the rational numbers where this duality can be precisely framed?
I TA a class for future elementary school teachers. The idea is to go back and really understand elementary school mathematics at a deep level. Hopefully this understanding gets passed on to the next generation.
We were discussing division of fractions. Rather than say "Yours is not to wonder why, just invert and multiply", we try to make sense of this question physically, and then use reasoning to solve the problem. Take (3/4) / (2/3). When doing this there seems to be two reasonable interpretations:
1) 3/4 of a cup of milk fills 2/3 of a container. How much milk (in cups) does it take to fill the entire container?
This is a "How many in each group" division problem, analogous to converting 6/3 into the question "If I have six objects divided into three equal groups, how many objects will be in each group?"
The solution that stares you in the face if you draw a picture of this situation is the following: 3/4 of a cup fills 2 thirds of a container. That means there must be 3/8 cups of milk in each third of the container. One container must have 9/8 cups of milk then, because this is 3 of these thirds. Note that the solution involved first dividing and then multiplying.
2) I have 3/4 cups of milk, and I have bottles which each hold 2/3 of a cup. How many bottles can I fill?
This is a "How many groups" division problem, analogous to converting 6/3 into the question "If I have six objects divided into groups of two, how many groups do I have?"
The solution suggested by this situation is the following: 3/4 of a cup of milk is actually 9 twelfths of a cup . Each twelfth is an eighth of a bottle. So I have 9/8 of a bottle. This solution involved first multiplying and then dividing.
Now I come to my question. This pattern persists! Every real world example of a "how many in each group" division problem suggests a solution by first dividing and then multiplying, whereas each "How many groups" division problem involves first multiplying and then dividing. It seems that solving the problem in the other order does not admit a conceptual realization in terms of the original problem. This is interesting to me! It suggests that the two solution methods are fundamentally different somehow. The standard approach to rational numbers (natural numbers get grothendieck grouped into integers, which get ring of fractioned into rational numbers) ignores this kind of distinction. Is there a "categorification" of the rational numbers which preserves the duality between these two types of question?
UPDATE 1:  In the category of sets, if you wanted to express $(\frac{6}{2})(3) = \frac{(6)(3)}{2}$ you would have to do something like this:
Let $A$ be a set with 6 elements, $B$ a set with 3 elements, $\sim$ an equivalence relation on A where each equivalence class has 2 elements, and $\cong$ an equivalence relation on $A \times B$ where each equivalence class has 2 elements. Then there is no canonical morphism from $(A/ \sim) \times B \to (A \times B)/ \cong$.  This seems to explain things somewhat on the level of integers, but we are talking fractions here.
UPDATE 2:  Qiaochu points out in a comment to his answer that the order of operations is not the most essential thing here.  You can solve the first problem by observing that 9/4 cups of milk fill 2 containers, so 9/8 must fill one.  Torsors give a formal distinction between the two situations, but it still feels like UPDATE 1 should go through in a suitable category of "fractional sets".
UPDATE 3:  For a very nice discussion of the ideas related to Theo's answer see http://golem.ph.utexas.edu/category/2008/12/groupoidification_made_easy.html

REPLY [5 votes]: Another way of categorifying $\mathbb{Q}_+$ is as follows:
Let $\mathcal{F}$ be the category of finite non-empty sets and bijections. This is a symmetric monoidal category with respect to $\times$. Let $\mathbb{N}$ be viewed as a symmetric monoidal category, such that the objects are $\mathbb{N}$ and we only have identity morphism. The product is given by the usual product $\cdot$ in $\mathbb{N}$.
Then there is an obvious symmetric monoidal functor $G \colon \mathcal{F} \to \mathbb{N}$, which simply counts the elements in the set. Taking Grayson Quillen construction on both sides, we get a symmetric monoidal functor:
$G' \colon \mathcal{F}^{-1}\mathcal{F} \to \mathbb{Q}_+$,
whic may be viewed as a categorification (note that this is also the $\pi_0$ functor). The Grayson Quillen construction may be thought of as a categorified version of the Grothendeick construction. Objects in $\mathcal{F}^{-1}\mathcal{F}$ are pairs of objects in $\mathcal{F}$ which I suggestively write as $A/B$
If we only use the skeletal category of $\mathcal{F}$ given by objects $s_n=\{1,\dots,n\}$ for each $n\in \mathbb{N}$ we get that $G$ is an isomorphism on the set of objects. This is not true for $G'$, because we have implicitly divided out by isomorphisms in the image (i.e. $p/q=np/(nq)$ in $\mathbb{Q}_+$ not just isomorphic objects as they are in the Grayson Quillen construction on $\mathbb{N}$). We cannot do this on the domain of $G'$ because $s_p/s_q$ is NOT isomorphic to $s_{np}/s_{nq}$. However, there are many maps from $s_p/s_q$ to $s_{np}/s_{nq}$, which identifies them when taking $\pi_0$.
I should note that I believe (but am not sure) that this can be made bi-monoidal by taking the category of finite sets and bijections as a bimonoidal category and localizing with respect to $\times$ over the full subcategory of finite and non-empty sets.<|endoftext|>
TITLE: Wanted: A constructive version of a theorem of Furstenberg and Weiss
QUESTION [5 upvotes]: Let $f(x)$ be a polynomial with real coefficients, and let $||\cdot||$ be the distance-from-the-nearest-integer function. It is  known  that for any $ \epsilon > 0 $, the set $S$ of positive integer solutions of the inequality $||f(x)|| < \epsilon$ has bounded gaps. This  means that if $x_1 < x_2  < \ldots$ are the elements of $S$ in increasing order, then the set of differences $x_{n+1}-x_n$ is bounded.  This is proved in   "Simultaneous Diophantine Approximation and IP Sets" by Furstenberg and Weiss, Acta Arith. 1988.
I would like to know how to compute the maximum gap, i.e. the maximum of the values $  x_{n+1}- x_n $, given a specific polynomial and a specific value of $\epsilon$. I can do this for linear polynomials, but not for quadratics. The proof of Furstenberg and Weiss appears to be  non-constructive. 
To take a particular example, what is the maximum possible gap between successive solutions of the inequality $||2^{1/2}x^2|| < .01$?  Can anyone suggest a method, no matter how impractical, that would eventually lead to an answer? Could it be that, even in this particular case, no one knows how to find the maximum gap? 
Note: Experimentation suggests that the maximum gap is 627, which occurs for the first time following the solution $x=1115714$.

REPLY [5 votes]: Dear RJS,
I think Tim Gowers is right - the problem seems too hard. Reasonably good bounds are known on (for example) the least $n \geq 1$ for which $\Vert n^2 \sqrt{2} \Vert \leq \epsilon$; one can find such an $n$ with $n \leq \epsilon^{-7/4 + o(1)}$. This is a result of Zaharescu [Zaharescu, A; Small values of $n^2\alpha\pmod 1$. Invent. Math. 121 (1995), no. 2, 379--388.] Zaharescu in fact obtains this result for any $\theta$ in place of $\sqrt{2}$. From a cursory glance at the paper I see that he uses the continued fraction expansion for $\theta$ and so it may be that one can slightly improve his bound in the particular case of $\sqrt{2}$.
It is an old conjecture of Heilbronn that the right bound here should be $\epsilon^{-1 + o(1)}$. I don't know off the top of my head whether any more precise conjectures have been made based on sensible heuristics either for this or for your original problem.
To get an explicit upper bound for your problem one can proceed quite straightforwardly using arguments due to Weyl. I don't think this is the right place to describe an argument in detail: there are several variants, and I first learnt this from a Tim Gowers course at Cambridge. See Theorem 3.10 of these notes:
http://www.math.cmu.edu/~af1p/Teaching/AdditiveCombinatorics/notes-acnt.pdf
If you really had to show that 627 is the answer to your specific problem, probably the best bet would be to inspect all the quadratics $n^2\sqrt{2} + \theta n + \theta'$ for $\theta,\theta'$ in some rather dense finite subset of $[0,1]^2$ and show using a computer that each takes (mod 1) a value less than 0.009999 for some $n \leq 627$. Painful!
The argument of Furstenberg and Weiss uses ergodic theory and so will not directly lead to an effective bound.
There are quite detailed conjectures about the fractional parts of $n^2\sqrt{2}$ (and other similar sequences) due to Rudnick, Sarnak and Zaharescu, essentially encoding the fact that this sequence of fractional parts is expected to behave like a Poisson process. I don't think those conjectures are likely to be helpful in your context since, taken too literally, they would seem to suggest that there are arbitrarily long intervals without a number such that $\Vert n^2 \sqrt{2} \Vert < 0.01$ - contrary to Furstenberg-Weiss.
Nonetheless let me point out a recent paper of Heath-Brown which is very interesting in connection with these matters:
http://arxiv.org/pdf/0904.0714v1.
One more point perhaps worthy of mention: sequences with bounded gaps are usually known as syndetic.<|endoftext|>
TITLE: Reference for base change of cohomology pull-push for clean intersections.
QUESTION [6 upvotes]: Let $X$ be a compact oriented manifold, and $A$ and $B$ closed oriented submanifolds intersecting cleanly.  Then I've always been under the impression that pushing forward a cohomology class from $A$ to $X$ and then pulling back from $B$ should have a base change formula where instead one pulls back to $A\cap B$ and pushes forward to $B$.
Of course, this couldn't possibly be right if $A$ and $B$ aren't transverse.  I think in the non-transverse case, one should correct by the Euler class of the excess bundle $T_{A\cap B}X/(T_{A\cap B}A+T_{A\cap B}B)$.
All of my intuition for algebraic topology tells that this true and easy to prove, but of course, one can't write that in a paper.

Does anyone know a convenient reference for this fact?  I've tried to find it via Google, but apparently can't find the right keywords, and a quick scan of Hatcher came up negative.

REPLY [5 votes]: Apologies if this is too late, but the canonical reference for this is Quillen's seminal paper 
"Elementary proofs of some results of cobordism theory using Steenrod operations"  Advances in Math.  7  1971 29--56 (1971).
 The proof given there is for complex cobordism and is entirely geometric. Presumably Quillen learned this from Bott, who gives a clean intersection formula in his paper
 "On the iteration of closed geodesics and the Sturm intersection theory"
Comm. Pure Appl. Math. 9 (1956), 171--206. 
If you are interested in the generalisation to immersions, see 
F. Ronga, "On multiple points of smooth immersions"  Comment. Math. Helv.  55  (1980), no. 4, 521--527.<|endoftext|>
TITLE: scheme-theoretic description of abelian schemes
QUESTION [7 upvotes]: Let $S$ be a locally noetherian scheme, $C$ the category of proper smooth $S$-schemes with geometrical connected fibres and $C_*$ the category of pointed objects of $S$, i.e. objects of $C$ together with a morphism $S \to C$. Also denote $A$ the category of abelian schemes over $S$. There is a well-known rigidity result stating that a pointed morphism between $X,Y \in A$ is already a group morphism. In other words, the inclusion functor
$A \to C_*$
is fully faithful. Is there a nice description for the image? In other words, which purely scheme-theoretic properties do abelian schemes have and are there enough to characterize them? For example, $X \in A$ is "homogeneous".

REPLY [6 votes]: How about smooth proper morphisms $X \to S$ with connected fibers, a section $S \to X$, such that the sheaf of Kähler differentials $\Omega_{X/S}$ is a pullback from $S$, and such that the group scheme $\underline{\rm Aut}_S X$ acts transitively on the fibers? The essential point is that the hypothesis on the differentials insures that no geometric fiber can contain a rational curve, so no affine algebraic group can act non-trivially. The result should follow from Chevalley's structure theorem for algebraic groups, with some fairly standard arguments (I haven't checked the details, though).<|endoftext|>
TITLE: Etale coverings of certain open subschemes in Spec O_K
QUESTION [14 upvotes]: Let $U$ be an open subscheme of $\textrm{Spec} \ \mathbf{Z}$. The complement of $U$ is a divisor $D$ of $\textrm{Spec} \ \mathbf{Z}$.
Q. Can we classify the etale coverings of $U$ of a given degree?  
Given a finite etale morphism $\pi:V\longrightarrow U$, the normalization of $X$ in the function field $L$ of $V$ is $\textrm{Spec} \ O_L$. Can we also say what $V$ itself should be?
Of course, we can complicate things by replacing $\textrm{Spec} \ \mathbf{Z}$ by $\textrm{Spec} \ O_K$. 
Example. Take $U= \textrm{Spec} \ \mathbf{Z} -\{ (2)\}$ and let $V\rightarrow U$ be an finite etale morphism. Suppose that $V$ is connected and that let $K$ be its function field. The normalization of $\textrm{Spec} \ \mathbf{Z}$ in $K$ is of course $O_K$. The extension $\mathbf{Z}\subset O_K$ is unramified outside $(2)$ and (possibly) ramified at $(2)$. Can one give a description of $V$ here?
EDIT. I just realized one can also ask themselves a similar question for $\mathbf{P}^1_{\mathbf{C}}$. Or even better, for any Riemann surface $X$.

REPLY [2 votes]: @Ariyan and BCnrd: Cf. Exercise 1.9, Section 4.1 of Q. Liu's wonderfully written book, "Algebraic Geometry and Arithmetic Curves".
@Ariyan: Exercise: re-write the one-dimensional (i.e. classical) idele class group as a certain complement of $Spec\mathcal{O}_K$...cf. the intro to "Global class field theory" by Kato-Saito.<|endoftext|>
TITLE: Noncontractible connected topological rings ?
QUESTION [12 upvotes]: Are there any non-contractible connected topological rings?

Of course, such a thing cannot be a (topological) algebra over the reals.
(I have a vague memory of having a glance at an erticle by Lurie in which some (for me) rather esoteric theory of higher categorical structures gave rise to topological rings that would have some very nontrivial topology, but I know nothing about that field(s) and, well, I just don't remember... Maybe someone can provide less "esoteric" examples! :) )

REPLY [21 votes]: Here is a method for manufacturing such topological rings. 
The main technical ingredient is a product-preserving functor 
$$\Theta: \mathrm{Set}^{\Delta^{op}} \to \mathrm{CGHaus}$$ 
from the category of simplicial sets to the category of compactly generated Hausdorff spaces that is not, however, the usual geometric realization functor. This will almost undoubtedly be unfamiliar, and so will require some preface. The basic idea though is that while the usual geometric realization uses for its topological input the usual interval $I = [0, 1]$, the formal properties of the realization functor, particularly the fact it preserves finite products, still hold upon replacing $I$ by any compact topological interval $L$ and replacing ordinary affine simplices by $L$-valued simplices. This $L$-based realization $\Theta$, being product-preserving, takes simplicial rings to ring objects in $\mathrm{CGHaus}$. By choosing an appropriate $L$ that is connected but not path-connected, we can construct a topological ring that is connected but not path-connected, hence not contractible. 

We define an interval to be a linearly ordered set with distinct top and bottom elements, and an interval map as an order-preserving map that preserves the top and bottom. Observe that the usual affine simplex $\sigma_{n-1}$ of dimension $n-1$ can be described as the space of $(n-1)$-tuples $0 \leq x_1 \leq \ldots \leq x_{n-1} \leq 1$ (topologized as a subspace of $I^n$), or in other words as the space of interval maps $[n+1] \to I$ from the finite interval with $n+1$ points to $I$. Meanwhile, the category $\mathrm{FinInt}$ of finite intervals $[n+1]$ is equivalent to $\Delta^{op}$ (where $\Delta$ is the category of finite nonempty ordinals); indeed we have a functor $\hom(-, [2]): \Delta^{op} \to \mathrm{FinInt}$, where the set of order-preserving maps $\hom([n], [2])$ from the $n$-element ordinal $[n]$ to $[2]$ is given the pointwise order, thus inheriting an interval structure from the interval structure on $[2]$, where we have $[n+1] \cong \hom_{\Delta}([n], [2])$ as intervals). 
The usual geometric realization $R(X)$ of a simplicial set $X$, from a categorical point of view, is a tensor product $X \otimes_\Delta \sigma$ of a "right $\Delta$-module" $X: \Delta^{op} \to \mathrm{Set}$ with a left $\Delta$-module $\sigma: \Delta \to \mathrm{CGHaus}$ (the affine simplex functor): 
$$\sigma: \Delta \simeq \mathrm{FinInt}^{op} \stackrel{\hom(-, I)}{\to} \mathrm{CGHaus}$$ 
$$[n] \mapsto [n+1] \mapsto \hom_{\mathrm{Int}}([n+1], I).$$ 
This tensor product is often described by a coend formula 
$$R(X) = X \otimes_\Delta \sigma = \int^{[n] \in \Delta} X([n]) \cdot \hom_{\mathrm{Int}}([n+1], I).$$ 
As is well-known, $R$ is product-preserving. What is perhaps less well-known is that the only thing we need from $I$ to prove this fact is that it's compact Hausdorff and the interval order $\leq$ is a closed subset of $I \times I$. Complete details may be found in the nLab here. Therefore, if we replace $I$ with another compact Hausdorff topological interval $L$ (so that $\leq_L$ is a closed subset of $L \times L$), we get the same result, that the functor $\Theta = R_L$ defined by the formula 
$$R_L(X) = \int^{[n] \in \Delta} X([n]) \cdot \hom_{\mathrm{Int}}([n+1], L)$$ 
is also product-preserving. 

Let us take our compact topological interval $L$ to be the end-compactification of the long line (so, adjoin points $-\infty$ and $\infty$ to the ends of the long line). This is connected, but not path-connected because for example there is no path from $\infty$ to any other point. Now we just turn a crank: start with any denumerable non-trivial ring $R$ in $\mathrm{Set}$ -- I'll take $R = \mathbb{Z}/(2)$ -- and apply a sequence of product-preserving functors, 
$$\mathrm{Set} \stackrel{K}{\to} \mathrm{Cat} \stackrel{N}{\to} \mathrm{Set}^{\Delta^{op}} \stackrel{R_L}{\to} \mathrm{CGHaus}.$$ 
(Here $K$ is the functor that takes a set $S$ to the category such that $\hom(x, y)$ is a singleton for any $x, y \in S$; this is right adjoint to the forgetful functor $\mathrm{Cat} \to \mathrm{Set}$ that remembers only the set of objects, and being a right adjoint, $K$ preserves products. The nerve functor $N$ also preserves products.) Since ring objects can be defined in any category with finite products, we have that product-preserving functors transport ring objects to ring objects. One should draw a picture of the category $K(\mathbb{Z}/(2))$; it's pretty clearly connected, and its nerve will be a connected simplicial set, or indeed a connected simplicial ring. The $L$-based realization of that will thus be a connected colimit of (connected) $L$-based simplices $\sigma_L(n) = \hom([n+1], L)$ (see the nLab here for connected colimits of connected spaces), and so it too will be a connected ring object in $\mathrm{CGHaus}$. 
At this point, the overall idea should be pretty clear, and the rest is just some technical mopping-up. 

One technical point is that products in $\mathrm{CGHaus}$ need not be usual topological products (as shown by a famous example of Dowker), so one might object that we could end up not with a topological ring, but some kind of funny ring object in $\mathrm{CGHaus}$. However, in many cases of interest, topological products do coincide with $\mathrm{CGHaus}$ products. This is particularly the case for colimits of countable increasing sequences of compact Hausdorff spaces: their product in $\mathrm{CGHaus}$ is the usual topological product. (The same proof as given by Allen Hatcher for Theorem A.6 here will do.) Thus, what counts here is that $N(K(\mathbb{Z}/(2)))$ is a simplicial set with finitely many cells in each dimension, and $R_L$ applied to this involves taking a countable union of compact Hausdorff spaces, so we are okay here. 
A second technical point involves showing that $X = (R_L \circ N \circ K)(\mathbb{Z}/(2))$ is not path-connected, which is intuitively clear, but an idea of proof would be nice. $X$ can be described as a union of nondegenerate simplices, where there are two such simplices in each dimension $n$ (corresponding to paths of length $n$ of the form $0 \to 1 \to 0 \to \ldots$ and $1 \to 0 \to 1 \to \ldots$), and a point in the interior of each such simplex has coordinates given by an increasing chain of length $n$ in a dictionary order, say $(j_1, t_1) < (j_2, t_2) < \ldots < (j_n, t_n)$ where the $j_k$ belong to the order type $-\omega_1 \cup \omega_1$ ($\omega_1$ being the first uncountable ordinal, and $-\omega_1$ is of opposite order type, extending in the "negative" direction), and the $t_k$ belong to $[0, 1)$. Every point of $X$ is an interior point of some unique $n$-simplex. Now if $\alpha: I \to X$ is a path connecting a point in the interior of an $n$-simplex, $n > 0$, to a 0-simplex, then let $(a, b) \subset I$ be a connected component of the open set of $t \in I$ such that $\alpha(t)$ is interior to an $n$-simplex with $n > 0$. Since $(a, b)$ has countable cofinality, there is a countable ordinal $\kappa$ such that for every $t \in (a, b)$, the maximum ordinal $|j_k|$ occurring in the coordinate description of $\alpha(t)$ is bounded above by $\kappa$. But $\alpha(a)$, being a 0-cell, has a neighborhood $U$ where every point $p \in U$, $p \neq \alpha(a)$, has a maximum $|j_k|$ (in its coordinate description) greater than $\kappa$, and we have reached a contradiction.<|endoftext|>
TITLE: Serre type vanishing theorem of coherent sheaves on quasi-projective variety?
QUESTION [5 upvotes]: For a projective variety $X$, Serre's vanishing theorem says that $H^i(X, \mathcal{F}(n))=0$ for any coherent sheave, $i>1$ and sufficiently large $n$. I am wondering, is there a similar type of vanishing theorem on quasi-projective varieties, namely, let $Y$ be a quasi-projective variety, what can we say about the vanishing of $H^i(Y, \mathcal{F}(n))$ under the same setting as projective case.
Or is there a similar type of theorem for local cohomology, say, when is $H^i_{pt}(X, \mathcal{F}(n))$ vanishing?

REPLY [4 votes]: Another comment on the local cohomology part of the question: Besides Karl's perfect answer one might also point out that the twisting does nothing to the cohomology group. The twisting line bundle is trivial in a neighbourhood of the point, so locally near the point $\mathcal F(n)\simeq \mathcal F$. 
On the other hand, your question is exactly about local cohomology: If $Y$ is quasi-projective and its projective closure is $X$, then the groups $H^i(X,\mathcal F(n))$ and $H^i(Y,\mathcal F(n))$ are part of a long exact cohomology sequence with the third vertex being local cohomology supported on $X\setminus Y$.
Finally, ploughshare's example is actually just an explicit manifestation of Karl's comment via the (equivalent of the) above sequence: $\mathbb A^2$ is affine, so the cohomology of $\mathbb A^2\setminus {0}$ is the same as the local cohomology at the point $0$. There is a shift of indexes, so the non-vanishing of $H^1(\mathbb A^2\setminus 0, \mathcal O)$ is equivalent to the non-vanishing of $H^2_0(\mathbb A^2, \mathcal O_0)$ which follows from the theorem of Grothendieck Karl mentioned.<|endoftext|>
TITLE: Fields with trivial automorphism group
QUESTION [42 upvotes]: Is there a nice characterization of fields whose automorphism group is trivial? Here are the facts I know.

Every prime field has trivial automorphism group.
Suppose L is a separable finite extension of a field K such that K has trivial automorphism group. Then, if E is a finite Galois extension of K containing L, the subgroup $Gal(E/L)$ in $Gal(E/K)$ is self-normalizing if and only if L has trivial automorphism group. (As pointed out in the comments, a field extension obtained by adjoining one root of a generic polynomial whose Galois group is the full symmetric group satisfies this property).
The field of real numbers has trivial automorphism group, because squares go to squares and hence positivity is preserved, and we can then use the fact that rationals are fixed. Similarly, the field of algebraic real numbers has trivial automorphism group, and any subfield of the reals that is closed under taking squareroots of positive numbers has trivial automorphism group.

My questions:

Are there other families of examples of fields that have trivial automorphism group? For instance, are there families involving the p-adics? [EDIT: One of the answers below indicates that the p-adics also have trivial automorphism group.]
For what fields is it true that the field cannot be embedded inside any field with trivial automorphism group? I think that any automorphism of an algebraically closed field can be extended to any field containing it, though I don't have a proof) [EDIT: One of the answers below disproves the parenthetical claim, though it doesn't construct a field containing an algebraically closed field with trivial automorphism group]. I suspect that $\mathbb{Q}(i)$ cannot be embedded inside any field with trivial automorphism group, but I am not able to come up with a proof for this either. [EDIT: Again, I'm disproved in one of the answers below]. I'm not even able to come up with a conceptual reason why $\mathbb{Q}(i)$ differs from $\mathbb{Q}(\sqrt{2})$, which can be embedded in the real numbers.

ADDED SEP 26: All the questions above have been answered, but the one question that remains is: can every field be embedded in a field with trivial automorphism group? Answering the question in general is equivalent to answering it for algebraically closed fields.

REPLY [8 votes]: In the paper below Shelah, among many other things, gives constructions of real closed fields with no nontrivial automorphisms that are not subfields of the reals.  
S. Shelah, Models with second order properties. IV. A general method and eliminating diamonds -- Annals Pure and Applied Logic 25 (1983) 183-212<|endoftext|>
TITLE: Non-isomorphic groups with isomorphic nth powers (and similarly in other categories)
QUESTION [7 upvotes]: What's the simplest example (if any) of two non-isomorphic groups G and H such that $G \times G \cong H \times H$? A similar question can be asked for $n^{th}$ powers for fixed $n > 1$.
The Krull-Remak-Schmidt theorem guarantees a "unique factorization" into direct products of directly indecomposable groups. It applies to finite groups, and more generally for groups that satisfy both the ascending and descending chain conditions on normal subgroups. 
Wherever such a unique factorization holds, it would follow that isomorphic $n^{th}$ powers implies isomorphic groups (by the usual expedient on counting the multiplicities of each directly indecomposable factor on both sides). So, counterexamples must be infinite at the very least.
However, I'm not able to come up with any counterexample nor can I see an easy proof. [EDIT: Resolved now in one of the answers. The question below still seems open.]
Also related: can we have groups G,H such that $G^2 \cong H^3$ but G is not the cube of any group and H is not the square of any group? Again, this fails in the cases where we have a unique factorization, because we can count multiplicities of indecomposables. $2$ and $3$ can be replaced by any two relatively prime numbers above.
Incidentally, this is somewhat related to an earlier question on Math Overflow: When is A isomorphic to A^3?
A similar question can probably be asked in many other categories.

REPLY [7 votes]: Concerning a "similar question ... in ... other categories": this is not an uncommon phenomenon for modules over many familiar rings. I note that the example A above is an abelian group, i.e. a Z-module, albeit a fairly involved one; replacing Z with a larger ring leads to more mundane (and, in particular, finitely generated) examples. 
The first Weyl algebra $A_1=k[x,y]/(xy-yx-1)$ (k is a field of characteristic zero) is a Noetherian simple noncommutative domain which is not a principal ideal ring: e.g. the right ideal $I=(x^2, 1+xy)$ is not principal, hence not isomorphic to $A_1$ as a module. Weyl algebra $A_1$ has two more interesting properties: (1) it is hererditary, i.e. every right ideal is a projective module; (2) every projective module over $A_1$ of uniform dimension $r\geq 2$ is free. It follows that $I\oplus I$ is isomorphic to $A_1\oplus A_1,$ although $I$ and $A_1$ aren't isomorphic. To get a commutative example with this flavor, take $R$ to be your favorite Dedekind domain whose class group has an even order (e.g. $Z[\sqrt{10}]$) and $I$ an ideal of order 2 in $Cl(R)$. Then $I\oplus I\simeq R\oplus I^2\simeq R\oplus R$, yet $I$ isn't principal. For details of these examples (and much more), see McConnell and Robson, Noncommutative Noetherian Rings, especially Chapter 11 dealing with stable freeness.<|endoftext|>
TITLE: Is there a version of König's theorem for tripartite 3-graphs?
QUESTION [13 upvotes]: I would like to know if there exists a version of König's theorem for tripartite $3$-graphs.
In other words, let $G = (V,T)$ be a tripartite $3$-graph.  That is, $V$ is a set of vertices (with $V$ equal to the union of three disjoint subsets $A,B,C$) and $T$ is a set of hyperedges $\{v_1,v_2,v_3\}$, with $v_1 \in A, v_2 \in B, v_3 \in C$. A matching is a collection of hyperedges such that no two of them share a vertex.  A cover is a set of vertices that meets every hyperedge. 

Is there a relationship between the size of a maximum matching and the size of a minimum cover in $G$.

Any suggestions or references?
Thanks in advance!

REPLY [18 votes]: This is a special case of Ryser's Conjecture, which states that in 
an $r$-partite, $r$-uniform hypergraph (with $r>1$) 

$\tau \leq (r-1) \nu$,

where $\tau$ is the size of a minimum cover and $\nu$ is the size of a maximum matching. Note that the case $r=2$ is simply König's theorem.
You are interested in the case $r=3$, which was settled by Aharoni.  The reference is 
Ryser's conjecture for tripartite 3-graphs.  Combinatorica 21 (2001), no. 1, 1--4, 
and you can find a copy of the paper here.<|endoftext|>
TITLE: Confusion about how the first cohomology classifies torsors
QUESTION [9 upvotes]: This question is inspired by, but is independent of: Sheaf Description of G-Bundles
Line bundles are classified by $H^1(X,\mathcal{O}^\times_X)$. We also know that in general that $H^1(X,G)$, where $G$ is a sheaf from open sets in $X$ to $Grps$, classifies $G$-torsors over X.
With this insight in mind: $\mathcal{O}^\times_X$-torsors should correspond to line bundles. Indeed, if $P$ is one, then $\mathcal{O} _X \times _{\mathcal{O} _X^\times}P$ gives the desired line bundle, and all line bundles are achieved this way (see the question I linked to).
My question is about the more mundane $H^1(X,\mathbb{C})$, which can be thought of as $H^1(X,\underline{\mathbb{C}})$ where $\underline{\mathbb{C}}$ is the constant sheaf $\mathbb{C}$ (which I think of as a sheaf going to $Grps$). These should supposedly correspond to $\mathbb{C}$-bundles over $X$. Which appears to be line bundles. But of course there's no reason for $H^1(X,\mathbb{C})$ to equal $H^1(X,\mathcal{O}^\times_X)$... My intuition is that this should correspond to the more naive version of fiber bundles that doesn't involve a structure group. Do you have any thoughts?

REPLY [4 votes]: I think the right thing to look at is $H^1(X, \mathbb C^\*)$. This classifies line bundles with  a flat connection, or equivalently, line bundles with locally constant transition functions.
Now the natural embedding $\mathbb C^\* \to \mathbb O_X^\*$ induces on a map on cohomology $H^1(X, \mathbb C^\*) \to H^1(X, \mathbb O_X^\*)$ which is forgetting the flat connection.<|endoftext|>
TITLE: Does anyone want a pretty Maass form?
QUESTION [34 upvotes]: A few months ago, I was curious about some properties of Maass cusp forms, of nonabelian arithmetic origin.  As a result, I went through a somewhat predictable process of finding a totally real $A_4$ extension of $Q$, lifting the resulting projective Galois representation to an honest Galois representation, and writing a short program to compute as many coefficients of the Artin L-function (thus coefficients of the Maass form) as needed.
Well, as often happens, I didn't find anything particularly surprising in the end.  
But now I "have a Maass form".  Its a pretty Maass form -- the simplest one of eigenvalue 1/4, of "nonabelian" origin (not arising from a dihedral Galois representation).  Its conductor is 163 -- a very attractive prime number (though its appearance here seems coincidental).  Some class number 1 coincidences make the computation of its coefficients extremely quick and simple.  
So, does anyone want the Maass form (i.e. code to output coefficients quickly)?  It's fun to play with, and doesn't take up too much space.  I guarantee its modularity.  If not, any suggestions where to put it (a little journal that publishes such cute examples)?

REPLY [3 votes]: Here it is from Dokchitser in Magma:
> L:=LSeries(HilbertClassField(QuadraticField(145)) : Method:="Artin");
> L`prod;
[
  <L-series of Riemann zeta function, 1>,
  <L-series of Artin representation of Number Field with defining polynomial x^8 - 
    636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with
    character ( 1, 1, 1, -1, -1 ) and conductor 5, 1>,
  <L-series of Artin representation of Number Field with defining polynomial x^8 - 
    636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with
    character ( 1, 1, -1, -1, 1 ) and conductor 145, 1>,
  <L-series of Artin representation of Number Field with defining polynomial x^8 - 
    636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with
    character ( 1, 1, -1, 1, -1 ) and conductor 29, 1>,
  <L-series of Artin representation of Number Field with defining polynomial x^8 - 
    636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with
    character ( 2, -2, 0, 0, 0 ) and conductor 145, 2>
]
> L5:=L`prod[5][1];
> CheckFunctionalEquation(L5); // LCfRequired(L5) demands 161 terms
1.57772181044202361082345713057E-30
> [<p,-Integers()!Coefficient(EulerFactor(L5,p),1)> : p in PrimesUpTo(100)];
[ <2, 0>, <3, 0>, <5, -1>, <7, 0>, <11, 0>, <13, 0>, <17, 0>, <19, 0>, <23, 0>, 
  <29, -1>, <31, 0>, <37, 0>, <41, 0>, <43, 0>, <47, 0>, <53, 0>, <59, -2>, <61, 0>,
  <67, 0>, <71, -2>, <73, 0>, <79, 0>, <83, 0>, <89, 0>, <97, 0> ]

These are the same as your ap(p), essentially.<|endoftext|>
TITLE: Exactness of 2nd-Order Differential Equations via Differential Forms
QUESTION [14 upvotes]: This (probably very elementary) question came up the last time I taught differential equations, and I've been toying with it for a while with no success:
A 1st-order differential equation $M(x,y)dx+N(x,y)dy=0$ is exact if $$M(x,y)dx+N(x,y)dy=f_x(x,y)dx+f_y(x,y)dy$$ for some differentiable function $f(x,y)$ defined on the domain of $\omega$.  In this case, we easily arrive at an implicitly-defined solution to the differential equation.  Importantly, there is a nice test for exactness stemming from Clairaut's theorem -- for everywhere smooth $M$ and $N$ (for simplicity/laziness...obvious generalizations abound), the differential equation is exact iff $N_y=M_x$.  Of course, this procedure is easily re-interpreted as saying that by the triviality of $H^1(\mathbb{R}^2)$, a one-form is closed if and only if it is exact.
Now let's move one degree higher.  Boyce and Di Prima define a 2nd-order differential equation $P(x)y''+Q(x)y'+R(x)y=0$ to be exact if there exists a differentiable function $f(x,y)$ such that the differential equation can be written
$$P(x)y''+Q(x)y'+R(x)y=[P(x)y']'+[f(x)y]'=0.$$
The analogous expression to Clairaut's theorem seems to be that (again, for sufficiently smooth inputs) an equation of that form is exact iff $P''(x)-Q'(x)+R(x)=0.$  Of importance is that such forms can be integrated once to leave us with a 1st-order differential equation.  So we've successfully lowered the degree of our problem.
This feels to me very much like an analogous $H^2$ calculation.  We have a condition on some coefficients that very much looks like an alternating sum coming from a $d$ map on forms, and lets us conclude that the equation "comes from" a one-degree-smaller differential equation.  
But! (and here's the question) I can't seem to fit any 2-forms into this picture that would explain this analogy.  Presumably there's some big story here linking the two notions of exactness about which I'd love to be enlightened.
Side remark:  I once received a partial response that there might be a link with Cartan tableau, which I've been unsuccessful in pursuing, if that helps spark an idea.

REPLY [15 votes]: What you are looking for nowadays goes by the name of the Rumin complex and is defined on any contact manifold.  Moreover, there is a vast generalization of this that sometimes goes by the name of 'the variational bicomplex' and sometimes by the name 'characteristic cohomology'.  Here is a brief description that is suited for the question you asked:
On $M = \mathbb{R}^3$ with coordinates $x,y_0,y_1$, consider the differential ideal $\mathcal{I}\subset\Omega^\ast(M)$ generated by the $1$-form $\omega = dy_0 - y_1\ dx$, i.e., $\mathcal{I}$ is the set of linear combinations of all multiples of $\omega$ and $d\omega$.  Note that $\mathcal{I}$ is a homogeneous ideal and equals $\Omega^\ast(M)$ in degrees 2 and 3.  Because $\mathcal{I}$ is closed under exterior derivative, it is a sub-complex of $\bigl(\Omega^\ast(M),d\bigr)$. Thus, there is a graded quotient complex, call it $\bigl(\mathcal{Q},\bar d\bigr)$, that vanishes in degrees above $1$.  Note that $\mathcal{Q}^0 = \Omega^0(M)= C^\infty(M)$, since $\mathcal{I}$ vanishes in degree $0$.
Now, say that an element $\phi \in \mathcal{Q}^1$ is exact if $\phi = \bar d f$ for some $f\in \mathcal{Q}^0 = \Omega^0(M)= C^\infty(M)$.  Unfortunately, unlike $\bigl(\Omega^\ast(M),d\bigr)$, the complex $\bigl(\mathcal{Q},\bar d\bigr)$ is not locally exact in positive degree.  In fact, $\bar d \phi =0$ for all $\phi\in\mathcal{Q}^1$, even though $\bar d: \mathcal{Q}^0\to \mathcal{Q}^1$ is not onto.
Let me pause just a moment to explain how this fits into your question.  Your equation $P(x)y'' + Q(x)y' +R(x)y = 0$ is encoded as the $1$-form $\phi = P(x) dy_1 + (Q(x)y_1 + R(x) y_0) dx$ (which represents the same class as the $1$-form $P(x) dy_1 + Q(x)dy_0 + R(x) y_0 dx$ in $\mathcal{Q}^1$), and you are asking when there is a function $f(x,y_0,y_1)$ such that $\phi = \bar d f$.  (You should verify that $f = P(x) y_1 + (Q(x)-P'(x))y_0$ works when your equation is satisfied and that, otherwise, nothing does.)
Now, how can we test for exactness in this sense?  This is where the Rumin complex (aka the variational bicomplex, etc.) comes in.  It turns out that there is a way to embed the operator $\bar d:\mathcal{Q}^0\to\mathcal{Q}^1$ into a complex that provides a fine resolution of the constant sheaf, the same way that the exterior derivative does for the full complex of exterior differential forms.  
What you do is this:  Let $\mathcal{E}^2\subset\Omega^2(M)$ be the set of multiples of $\omega$ and let $\mathcal{E}^3=\Omega^3(M)$.  
We now want to define a complex
$$
0\longrightarrow \mathcal{Q}^0
\buildrel{\bar d}\over\longrightarrow \mathcal{Q}^1
\buildrel{D}\over\longrightarrow \mathcal{E}^2
\buildrel{d}\over\longrightarrow \mathcal{E}^3
\longrightarrow 0.
$$
The map from $\mathcal{E}^2$ to $\mathcal{E}^3$ is the usual exterior derivative, so the only thing left to define is the map $D:\mathcal{Q}^1\to \mathcal{E}^2$.  To do this, we first define a (first-order) operator $\delta: \mathcal{Q}^1\to\Omega^1(M)$, by requiring that $\delta(\phi)$ be a 1-form representing $\phi$ in the quotient complex and that $d\bigl(\delta\phi\bigr)$ lie in $\mathcal{E}^2$, i.e., that it be a multiple of $\omega$.  (I'll let you write down the formula for $\delta$ in local coordinates.)  Now, define $D\phi$ to be $d\bigl(\delta\phi\bigr)$.  (Sounds almost trivial doesn't it?)  The operator $D$ is easily verified to be second order and linear.
Now, it is not hard to verity that this complex is locally exact in positive degrees.  (It also gives a fine resolution of the constant sheaf, so its cohomology on $M$ is canonically isomorphic to the deRham cohomology of $M$.)  In particular, the local condition that $\phi\in\mathcal{Q}^1$ be exact is that $D\phi=0$.  
You should verify (after you have defined $D$) that this reproduces your condition precisely in the linear case you asked about.<|endoftext|>
TITLE: Is the chain homotopy category, K(Ab), an Abelian category?  By Ab, I mean the category of Abelian groups.
QUESTION [13 upvotes]: Let A be an Abelian category.
From this category, we can form the chain complex category Ch(A).  The objects of Ch(A) are chain complexes of objects of A.  The morphisms of Ch(A) are chain maps.  Ch(A) is an Abelian category for every Abelian category A.
Now from Ch(A), we can form the chain homotopy category K(A).  The objects of K(A) are just objects of Ch(A), but the morphisms of K(A) are chain homotopy classes of chain maps. Thus, K(A) is a quotient of Ch(A).
It turns out that K(A) is an additive category for any Abelian category A.  From asking people, I seem to get the impression that K(A) is not always abelian, but I've had a hard time showing this.  If all objects of A are projective (e.g. if A is the category of vector spaces over some field k), then K(A) will be Abelian.
I've been trying to show that K(Ab) is not Abelian (where Ab is the category of Abelian groups).  More specifically, I've been trying to show this by showing the following (which may or may not be true):
Let X be a chain complex with the group of integers in dimension 0, and zero in every other dimension. Let f be the chain map from X to X, that sends each integer x to 2x.  I've been trying to show that in K(Ab), the homotopy class of this chain map does not have a cokernel.
Any answers, suggestions, hints, or comments would be greatly appreciated!

REPLY [11 votes]: You might want to have a look at my answer to this question.<|endoftext|>
TITLE: Computing the Galois group of a polynomial
QUESTION [38 upvotes]: Does there exist an algorithm which computes the Galois group of a polynomial 
$p(x) \in \mathbb{Z}[x]$? Feel free to interpret this question in any reasonable manner. For example, if the degree of $p(x)$ is $n$, then the algorithm could give a set of permutations $\pi \in Sym(n)$ which generate the Galois group.

REPLY [2 votes]: There is another way that doesn't seem to be mentioned here. This is just something that occurred to me a few days ago; if anyone knows whether this has been done before I would greatly appreciate your comments.
It is well known that, given a field $K_0$ and a polynomial $p \in K_0[x]$, the following process will eventually give us a field $K_n$ which is a splitting field for $p$:

choose a monic irreducible factor of $p$ of degree $> 1$ in $K_0[x]$. Call this factor $q_1$.
Let $K_1 = K[r_1] / (q_1(r_1))$.
choose a monic irreducible factor of $p$ of degree $> 1$ in $K_1[x]$. Call this factor $q_2$.
Let $K_2 = K_1[r_2] / (q_2(r_2))$.
choose a monic irreducible factor of $p$ of degree $> 1$ in $K_2[x]$, etc.

So we have a splitting field $K_n$, explicitly constructed as a quotient of $K_0[r_1, \ldots r_n]$. Let $I$ be the kernel of the obvious map from $K_0[r_1, ... r_n]$ to $K_n$. The algorithm above also gives us a Gröbner basis for $I$: for each of the polynomials $q_i$, with $2 \leq i \leq n$ let $q'_i$ be a lift of $q_i$ to a monic polynomial with coefficients in the polynomial ring $K_0[r_1, ... r_{i - 1}]$. Then it is easy to see that $B:=\{q_1(r_1), q'_2(r_2), ... q'_n(r_n)\}$ is a Gröbner basis for $I$ with the lexicographic monomial ordering with $r_n > r_{n-1} > ... > r_1$.
In general, if we have a ring $R$ with an ideal $J$, an automorphism $f: R \rightarrow R$ will induce an automorphism of $R/J$ iff $J$ is $f$-invariant, i.e. $f(x) \in J$ whenever $x \in J$. In particular, if $\sigma$ is a permutation of $\{ r_1 \ldots r_n \}$, and $f_\sigma$ is the corresponding automorphism of $K_0[r_1, ... r_n]$, we have that $f_\sigma$ induces an automorphism of $K_n$ iff $I$ is $f_\sigma$-invariant, or equivalently, $f_\sigma(b) \in I$ for each $b \in B$. Furthermore, we can test if $f_\sigma(b) \in I$ with multivariate division, which is convenient as $B$ is already a Gröbner basis for $I$.
In summary, we can check if a permutation $\sigma$ of the roots of $p$ is in the Galois group by checking if $f_\sigma(b) \in I$ for each $b \in B$, and this can be done with multivariate division.<|endoftext|>
TITLE: Why worry about the axiom of choice?
QUESTION [213 upvotes]: As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory.  Yet I still see people fuss about whether or not theorem X depends on it, and I don't see the point.  Yes, one can prove some pretty disturbing things, but I just don't feel like losing any sleep over it if none of these disturbing things are in conflict with the rest of mathematics.  The discussion seems even more moot in light of the fact that virtually none of the weird phenomena can occur in the presence of even mild regularity assumptions, such as "measurable" or "finitely generated".
So let me turn to two specific questions:

If I am working on a problem which is not directly related to logic or set theory, can important mathematical insight be gained by understanding its dependence on the axiom of choice?
If I am working on a problem and I find a two page proof which uses the fact that every commutative ring has a maximal ideal but I can envision a ten page proof which circumvents the axiom of choice, is there any sense in which my two page proof is "worse" or less useful?

The only answer to these questions that I can think of is that an object whose existence genuinely depends on the axiom of choice do not admit an explicit construction, and this might be worth knowing.  But even this is largely unsatisfying, because often these results take the form "for every topological space there exists X..." and an X associated to a specific topological space is generally no more pathological than the topological space you started with.
Thanks in advance!

REPLY [3 votes]: Another useful example is the statement that  $M$ and $M^2$ have the same power for  any infinite set $M$. This is proved with the use of the axiom of choice.
This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators  of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. (See paper of mine in: 
Siegel W., Duplij S. and  Bagger J., eds. (2004), Concise Encyclopedia of Supersymmetry And Noncommutative Structures in Mathematics and Physics, Berlin, New York: Springer-Verlag; math-ph/0009006.) To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.<|endoftext|>
TITLE: Extra automorphisms of curves and definability over \bar Q
QUESTION [12 upvotes]: A generic elliptic curves over C has automorphism group of order 2. The elliptic curves with extra automorphisms are C/Z[i] (automorphism group of order 4) and C/Z[w] where w is a primitive 3rd root of unity (automorphism group of order 6). One can use their extra automorphisms to prove that they can be defined over Q
Similarly, the Klein quartic (a genus 3 curve with 168 automorphisms, the maximum possible number for genus 3 curves) can be defined over Q.

Suppose X is a genus g curve over C.
  Is there a condition on the group
  Aut(X) that guarantees that X is
  defined over a number field?

I feel like maybe one could try to approach this question by passing to the Jacobian of X and arguing that extra automorphisms of X give extra automorphisms of J(X) which means that J(X) has "large" endomorphism ring and hence must be defined over a number field (just as CM elliptic curves are defined over number fields), but I don't know whether "large" is large enough.

REPLY [11 votes]: One has the notion of a compact Riemann surface with "many automorphisms", a concept for which there exist many equivalent definitions.  You have correctly identified the "genus one Riemann sufaces with many automorphisms" -- a situation which is slightly complicated by the fact that, strictly speaking, the automorphism group of any complex elliptic curve as a Riemann surface is infinite -- so let's concentrate on the case of Riemann surfaces $X$ of genus $g \geq 2$, in which case the automorphism group is always finite.
The following three conditions on $X$ are equivalent:
(1) On the moduli space $\mathcal{M}_g$ of genus $g$ complex algebraic algebraic curves, the function $C \mapsto \# \operatorname{Aut}(C)$ has a strict local maximum at $X$ -- explicitly, there exists a neighborhood of $U$ of $X$ such that every curve in $U \setminus \{X\}$ has smaller automorphism group than $X$.  
(2) The natural map $X \rightarrow X/\operatorname{Aut}(X)$ is a Belyi map -- i.e., the quotient has genus zero and the map is ramified only over three points.  By the "easy" direction of Belyi's theorem, it follows that $X$ can be defined over some number field, and it is interesting to study the field of moduli of such curves $X$ (which in this case is known to be the minimal field of definition).
(3) $X$ is uniformized by a finite index torsion free subgroup $\Gamma$ of a hyperbolic triangle group $\Delta(a,b,c)$.  This makes clear that these curves are a significant generalization of the Hurwitz curves, which correspond to the "first" hyperbolic triangle group $\Delta(2,3,7)$.  At times I have expressed the opinion that it is strange that there has been so much research done on the $\Delta(2,3,7)$ case -- i.e., when the automorphism group is numerically as large as it could possibly be -- rather than on the more general case of $\Delta(a,b,c)$, i.e., the case where the automorphism group is sufficiently large to have the above very interesting analytic / topological / arithmetic implications.  
For more information on these Riemann surfaces, see the bibliography of 
http://math.uga.edu/~pete/triangle-091309.pdf
Especially highly recommended is [50], a paper by J. Wolfart.  In general Wolfart is one of the pioneers of this subject and has written many interesting papers, mostly from the perspective of complex function theory.<|endoftext|>
TITLE: A question of Erdős on equidistribution
QUESTION [36 upvotes]: In his book Metric Number Theory, Glyn Harman mentions the following problem he attributes to Erdős:
Let $f(\alpha)$ be a bounded measurable function with period 1. Is it true that
$$\lim_{N\rightarrow\infty} \frac{1}{\log N} \sum_{n=1}^N \frac{1}{n}f(n\alpha) = \int_0^1 f(x) dx$$
for almost all $\alpha$,
writing "so far as the author is aware, this question remains open."
Harman's book is from 1997. Does anyone know the current status of the problem?
Motivation, for the curious
We lose no generality in assuming $f$ has mean $0$. The rough idea is that for almost all $\alpha$, $n\alpha$ will be equidistributed $(\mod 1)$ in a strong enough way to cause a great deal of cancellation in the sum, so in particular we might guess the sum is $o(\log N)$. It is a weaker version of a more classical conjecture of Khintchine that 
$$\lim_{N\rightarrow\infty} \frac{1}{N} \sum_{n=1}^N f(n\alpha) = \int_0^1 f(x) dx$$
for almost all $\alpha$, where $f$ is as above. This is known to be false. (Of course, if $f$ is continuous it is true, for all irrational $\alpha$ even.)

REPLY [23 votes]: The statement was shown to be false by J. Bourgain in a paper published in 1988 (Almost Sure Convergence and Bounded Entropy, doi:10.1007/BF02765022). Well before either Harman's 1997 book and the 2003 paper Gjergji mentioned in the comment, which both say that it is an open problem!
From page 2 of Bourgain's paper

As further application of our method, a problem due to A. Bellow and a question raised by P. Erdős are settled.

And, further down (using ${\bf T}={\bf R}/{\bf Z}$),

The problem of Erdős mentioned above deals with weaker versions of the Khintchine problem. In particular he raised the question whether given a measurable subset $A$ of $\bf T$, then for almost all $x$ the set $\lbrace j\in{\bf Z }_+\mid jx\in A\rbrace$ has a logarithmic density, i.e.
  $$
\frac{1}{\log n}\sum_{\substack{j\le n\\\\ jx\in A}}\frac1j\to\vert A\vert.
$$
  We will disprove this fact.

I can't vouch that Bourgain's paper is free of errors, as I have only just found it now and haven't read through it all in detail. However, Bourgain's result is also (very briefly) referred to in this 2004 paper, http://arxiv.org/abs/math/0409001v1, so I assume it is considered to be valid.<|endoftext|>
TITLE: why isn't the mobius band an algebraic line bundle?
QUESTION [16 upvotes]: When I hear the phrase "line bundle" the first thing that pops into my head is a mobius band. But this is a bad picture from an algebraic point of view since any line bundle on an affine variety is trivial. Anyway, my question is: is there a way of seeing more concretely what "goes wrong" when you try to construct the mobius band as an algebraic line bundle over ℝ, and what changes when you move to analytic line bundles?

REPLY [26 votes]: Consider the real algebraic line bundle $\mathcal{O}(-1)$ over the real algebraic variety $\mathbb{R}\mathbb{P}^1$. It is nontrivial hence continuously isomorphic to the "Moebius" line bundle (there are only 2 line bundles on the circle, up to continuous isomorphism), so its total space is homeomorphic o the "Moebius strip".
By the way, it is false that line bundles on affine algebraic varieties are trivial. One example is the above universal line bundle over $\mathbb{R}\mathbb{P}^1$. If you want an example over $\mathbb{C}$, consider the complement of a point in an elliptic curve: it's an affine variety but its Picard group is far from being trivial.<|endoftext|>
TITLE: distribution of coprime integers
QUESTION [12 upvotes]: Let $0 < a < 1$ be fixed, and integer $n$ tends to infinity. It is not hard to show that the number of integers $k$ coprime to $n$ such that $1\leq k\leq an$ asymtotically equals $(a+o(1))\varphi(n)$.
The question is: what are the best known estimates for the remainder and where are they written?
Many thanks!

REPLY [3 votes]: In a similar question Bound the error in estimating a relative totient function , Alan Haynes notes in an answer that Vijayraghavan in 1951 had published a result showing many $n$ for which (for certain a) the error approaches $2^{\omega(n) - 1}$.  Further, Lehmer in 1955 showed that for certain values of a (namely q/k where k divides p-1 and p divides n) that the error was 0.<|endoftext|>
TITLE: What are the motivations for studying Cherednik (symplectic reflection, graded Hecke) algebras?
QUESTION [14 upvotes]: Several times I have come across these algebras and I wonder why any of these are interesting; I'm very sure they are, but I could not find an answer in the literature. 
For example (the very general version of) a graded Hecke algebra for a finite group $G$ acting on a finite dimensional $\mathbb{C}$-vector space $V$ is defined as an algebra $A_\kappa := (T(V) \sharp \mathbb{C}G)/\langle vw - wv - \kappa(v,w) \mid v,w \in V \rangle$ satisfying the PBW-property $S(V) \sharp \mathbb{C}G \cong \mathrm{gr}(A_\kappa)$, where $\kappa:V \times V \rightarrow \mathbb{C}G$ is an alternating bilinear form. For these algebras I've seen an explanation like "we want to study the geometry of the action of $G$ on $V$, but the commutative algebraic geometry, i.e. $S(V)^G$, is bad, so we better study $S(V) \sharp \mathbb{C}G$"; but then I don't know why I'm interested in those $A_\kappa$. Can anybody explain this and make this precise? 
I was also told that graded Hecke algebras (for Weyl groups) were introduced by Lusztig in order to study affine Hecke algebras which in turn are important in the representation theory of semisimple split $p$-adic groups. Do the general graded Hecke algebras above have a similar use?
The Cherednik and symplectic reflection algebras are special cases of graded Hecke algebras. If I would have a nice motivation for graded Hecke algebras, I would also have one for them; but I am pretty sure that there are different motivations for them!?
I am thankful for any explanations and hints to literature.

REPLY [11 votes]: In addition to many excellent answers posted so far, I would like to explain another way in which the relations of the "graded" or degenerate affine Hecke algebras arise in representation theory, which I find most helpful in understanding them (and is how I would recover them for myself if stranded on a desert island).  This introduction has the advantage that it could be explained to an undergraduate who knew what a surface and a fundamental group was.  All of this is in Cherednik's book in one form or another.
In this point of view, the "full" DAHA and AHA are the primary objects, and one recovers the degenerate versions by a process similar to the degeneration from $U_q(\mathfrak{g})$ to $U(\mathfrak{g})$.  Let me just discuss $A_n$, although an analogous construction works for any Weyl group (and parts of the discussion for any complex reflection or symplectic reflection group).
The Artin braid group $B_n$ is $\pi_1(C_n(\mathbb{C}))$, the fundamental group of configurations of $n$ points in $\mathbb{C}$.  It's easy to identify this with the braid group associated to the root system of type $A_{n-1}$, since the reflection hyperplanes are precisely what impose the distinctness of the points in the configuration.  Denote the usual generators of $B_n$ by $T_i$.
One may instead consider the "double affine braid group" $DB_n$, $\pi_1(C_n(E))$, which is the configuration space of $n$ points on an elliptic curve (or what matters topologically is that it's a $S^1\times S^1$).  One finds generators $B_n$ corresponding to loops which are contained in a contractible ball, and new generators $X_1,\ldots, X_n, Y_1,\ldots, Y_n$ corresponding to taking the $i$th point of the configuration and running it around the the inside our outside ring of the torus.  One computes relations that the $X_i$ commute, the $Y_i$ commute, and $T_iX_iT_i=X_{i+1}$, $T_iY_{i+1}T_i=Y_i$, and (Y_2,X_1):=Y_2X_1Y_2^{-1}X_1^{-1}=T_1^{-2}$.
Now, how is this related to the algebras in your question?  Let $\tilde{A}$ denote the group algebra of $DB_n$ with coefficients in $\mathbb{C}[[\hbar]]$, completed in the $\hbar$-adic topology.  Let $A$ denote the quotient by the additional relations
$(T_i-q)(T_i+q^{-1})$, where $q:=e^{\hbar/2}$.
Now suppose that $V$ is some representation of $A$ such that $Y_i$ acts as $1$ modulo $\hbar$.  In this case, it makes sense to define $s_i$ and $y_i$ in $A$ by the relations:
$Y_i=e^{\hbar y_i}$, $T_j:=s_je^{k\hbar s_j}$.
Where I'm evaluating in the representation $V$ so that the first equation makes sense, but I'm not writing that in explicitly.
One now would like to check what relations are imposed on the generators $X_i$, $y_i$ and $s_j$ so defined.  Let us just check what relations we get by considering the relations of $A$ up to first non-trivial order in $\hbar$. We find:
$(T_i-q)(T_i+q^{-1})=0 \Rightarrow s_i^2=1$.
Braid relations for $T_i \Rightarrow$ braid relations for $s_i$. 
$X_i$'s commute (as before).  $\tilde y_i:=y_i + \sum_{i `< j} s_{ij}$.
$T_iX_iT_i=X_{i+1}\Rightarrow s_iX_is_i=X_{i+1}$
$T_iY_{i+1}T_i=Y_i\Rightarrow s_iy_is_i=y_{i+1}+s_i$
$(Y_2,X_1):=Y_2X_1Y_2^{-1}X_1^{-1}=T_1^{-2}\Rightarrow [y_2,X_1]=ks_1X_1$
which are (one form of) the relations of the trigonometric Cherednik algebra.
Further writing $X_i=e^{\hbar x_i}$, one recovers the so called rational Cherednik algebra.
So there is a hierarchy of degenerations.  The top and bottom of the hierarchy are essentially symmetric in the variables $X$ and $Y$, in the precise sense that there is a "Fourier transform" automorphism swapping the variables, in both cases.  Note that at the top of the hierarchy, the Fourier transform is just the order four automorphism of the elliptic curve which is the matrix $((0,1),(-1,0))$ in $PSL_2(Z)$, the mapping class group of the torus, and which can be seen in various elementary ways.  At the bottom of the hierarchy, the Fourier transform is just swapping $x_i$ and $y_i$ and is related to Fourier transform of differential operators on an abelian group.  In the middle, there isn't really  a Fourier transform, because the symmetry was broken by degenerating the $Y_i$, but leaving the $X_i$ unscathed.   

I'm not an expert on any of these things, and I can't say I've worked through the topological construction for all the cases, but I think, to answer Stephen's question, we can simply choose any lattice in $\mathbb{C}$, of rank $2$ (say $\Lambda=\mathbb{Z}1 \oplus \mathbb{Z}\mathbf{i})$, and consider $\pi_1(\mathfrak{h}_{reg}/W\ltimes\Lambda^r)$, where $r$ is the rank of $\mathfrak{h}$.  Then one quotients by the relations that loops around singular points of this quotient have order 2.  For instance, let me explain a nice thing that happens for BC_n type.
BC_n non-affine braid group means we configurations of 2n distinct points in $\mathbb{C}\backslash\{0}$, such that $x$ is in the configuration if and only if $-x$ is in the configuration.  We don't allow zero because that we want pairs of matched points.  It's not hard to see that $\pi_1$ of that configuration space is the braid group of type $BC_n$, because you can choose the repesentative of each pair lying in the upper half plane, and you get the upper half plane, except that the $r$ and $-r$ are identified for all real points, and zero is excluded.  This is a punctured plane.
Now the prescription above should lead you to consider $2n$ distinct points on the elliptic curve (or rather $S^1\times S^1$; I only say elliptic curve because people sometimes mean $\mathbb{C}^\times$ by torus...).  However, now you have more points to remove.   Not only zero, but all half integer points would correspond to a place where $x$ and $-x$ collide.  So you get the usual picture of the torus from a first course in topology, except with half-integer points removed.  A very fun exercise is to work out that you can again choose the representative of each pair which lies in, say, the lower left corner of the torus, where we cut the torus in half along the diagonal from upper left to lower right.  But now again you have to identify some edges, and you get ....  $\mathbb{CP}^1$ with four punctures (corresponding to the four half integer points of your real torus.)  Now you get to choose five parameters:  one parameter for non-affine hyper planes, meaning the $T_i$, and then one for each of the new poles you've introduced.  Imposing the Hecke relations on all those you get the double affine Hecke algebra of type $BC_n$ (sometimes called $C^\vee C_n$ for historical reasons) with those five parameters.  Notice that this process leads you to a very different presentation of the DAHA, where you de-emphasize the lattices, and emphasize the loops around the singularities.  I understand that S. Sahi gave a similar presentation for all DAHA's coming from root systems, but with probably deeper motivations than these drawings on surfaces.  I'm not sure of the reference where Sahi did this.
At least in this example, one again sees the Fourier transform as just the obvious move on the real torus.<|endoftext|>
TITLE: Bounded and weakly bounded sets in top. vector spaces
QUESTION [6 upvotes]: Consider a locally convex topological vector space V over the complex numbers. Is it true that every weakly bounded subset of V is indeed bounded? If not, what additional requirements are needed for this to hold? Perhaps someone has a reference, I was not able to find something in the literature.
Thanks for your help.
Cheers,
Ralf

REPLY [3 votes]: This is direct consequence of the Mackey Theorem: Having a dual pair (V,V') with V' as the dual of the locally convex space V, the bounded sets on V under any dual topology are identical. A dual topology on V is a locally convex topology $\tau$ such that (V,$\tau$)' = V'.
As the original and the weak topology give the same dual, the bounded sets are identical.<|endoftext|>
TITLE: Is every (finite) group action on R^n by diffeomorphisms conjugate to a linear action?
QUESTION [10 upvotes]: I want to know if every smooth (finite)group action on $\mathbb{R}^n$ is conjugate to some linear action.Thank you!

REPLY [10 votes]: Take a contractible manifold $C$, multiply it by $\mathbb R^n$, and let the finite group act trivially on $C$, and linearly  on $\mathbb R^n$ such that $0$ is the unique fixed point.  Then $C\times 0$ is the fixed point set. If $C$ is not diffeomorphic to $\mathbb R^n$, the action is not linear, but for sufficiently large $n$ the product $C\times\mathbb R^n$ will be diffeomorphic to a Euclidean space, by Stallings's characterization of Euclidean space as the contactible space that is simply-connected at infinity. In fact, Craig Guilbault proved that $n=1$ suffices (except possibly when $\dim(C)=3$, which I do not quite understand at the moment). See here  for Craig's paper.<|endoftext|>
TITLE: How to demonstrate $SO(3)$ is not simply connected?
QUESTION [12 upvotes]: A quote from Wikipedia's article on the Rotation group: 

Consider the solid ball in $\mathbb{R}^3$ of
  radius $\pi$ [...].
  Given the above, for every point in
  this ball there is a rotation, with
  axis through the point and the origin,
  and rotation angle equal to the
  distance of the point from the origin.
  The identity rotation corresponds to
  the point at the center of the ball.
  Rotation through angles between $0$ and
  $-\pi$ correspond to the point on the same axis and distance from the origin but
  on the opposite side of the origin.
  The one remaining issue is that the
  two rotations through $\pi$ and through $-\pi$
  are the same. So we identify [...] antipodal points on the
  surface of the ball. After this
  identification, we arrive at a
  topological space homeomorphic to the
  rotation group.

So far, so good. This illustrates $SO(3)\cong \mathbb{RP}^3$.

These identifications illustrate that
  $SO(3)$ is connected but not simply
  connected. As to the latter, in the
  ball with antipodal surface points
  identified, consider the path running
  from the "north pole" straight through
  the center down to the south pole.
  This is a closed loop, since the north
  pole and the south pole are
  identified. This loop cannot be shrunk
  to a point, since no matter how you
  deform the loop, the start and end
  point have to remain antipodal, or
  else the loop will "break open".

I believe that $SO(3)$ is connected but the "intuitive argument" for $\pi_1(SO(3))\neq 0$ is not clear to me: The starting point at the "north pole" is a rotation of $\pi$ counterclockwise around the $z$ axis. This agrees with the "south pole", a rotation of $\pi$ clockwise around the $z$ axis. So the described path is a full $2\pi$ rotation counterclockwise around the $z$ axis, stating not in the identity position. Why isn't this homotopic to the trivial path? Antipodal points are identified, so what does "start and end point have to remain antipodal, or else the loop will "break open"" mean?

REPLY [2 votes]: You have two different proofs (a.k.a. the cup proof and the belt proof) right here 
http://www.youtube.com/watch?v=CYBqIRM8GiY
There are several different demonstrations of the second one (Harrison Brown linked to this video in a comment to an answer I submitted - the cup proof - in the question asking for "proofs w/o words").<|endoftext|>
TITLE: Parallel translation in Lie groups
QUESTION [7 upvotes]: Let G be a Lie group with a left invariant metric. If X and Y are left invariant vector fields and [X,Y]=0, then it is easy to show that Y is parallel to exp(tX). 
But if [X,Y] is not zero, what is the parallel translate of Y along exp(tX)?

REPLY [4 votes]: For a general $Y$, it will be matrix-exponential in $t$ with initial conditions determined by $Y$.  Here's an explicit computation.  Pick a left-invariant global frame $(E_1, \ldots E_n)$ for the group, and define structure constants
$[E_i, E_j]=\sum c_{ij}{}^kE_k$.
The covariant derivative of $E_i$ along the geodesic $\exp(tX)$ from 0 is the constant
$\frac{1}{2}[X, E_i]=\frac{1}{2}\sum X^jc_{ji}{}^kE_k$
(see eg Lee "Riemannian Manifolds" problem 5-11).  Therefore a vector field
$t\mapsto \sum f^i(t)E_i$
along this geodesic is parallel if it is a solution to
$0=D_t\left(\sum f^i(t)E_i\right)=\sum_k\left[(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k\right]E_k$,
i.e., to
$\forall k \ 0=(f^k)'(t)+\frac{1}{2}\sum_{i,j} f^i(t)X^jc_{ji}{}^k$.
The parallel transport of $Y$ will be the solution to this linear system with initial value $Y$.  Perhaps there's a nice basis-invariant way of expressing this? I can't think offhand.<|endoftext|>
TITLE: The kernel of the map from the handlebody group to Outer automorphisms of a free group
QUESTION [18 upvotes]: Let $K$ be a compact oriented 3-dimensional handlebody of genus $g$.  The group $H_g$ of isotopy classes of diffeomorphisms of $K$ is called the handlebody group. (It embeds as a subgroup of the mapping class group of the genus $g$ surface $\partial K$.)  The fundamental group of $K$ is a free group of rank $g$, so there is a homomorphism
$H_g \to Out(F_g).$
I've been thinking about this homomorphism and its kernel, and I've come to suspect that the kernel is generated by Dehn twists around curves in $\partial K$ that bound discs in $K$.  These elements are all clearly contained in the kernel, but do they generate the entire kernel?
Does anyone know of a reference, proof, or counter example?

REPLY [12 votes]: This result is due to Luft; see  "Actions of the homeotopy group of an orientable 3-dimensional handlebody".
McCullough, in "Twist groups of compact 3-manifolds", proves that the twist group is not finitely generated and gives further references. 
EDIT. Ninja'ed (well, at least the first sentence).<|endoftext|>
TITLE: Choice vs. countable choice
QUESTION [21 upvotes]: This question arose after reading the answers (and the comments to the answers) to Why worry about the axiom of choice?.
First things first.  In my intuitive conception of the hierarchy of sets, the axiom of choice is obviously true. I mean, how can the product of a family of non-empty sets fail to be non-empty? I simply cannot fathom it. Now, I understand that there are people who disagree with me; a mathematician of a (more) constructive persuasion would reply that mathematical existence is constructive existence. Well, we can agree to disagree. And besides, the distinction between constructive and non-constructive proofs is very much worth having in mind. First, because constructive proofs usually give more information and second, there are many contexts where AC is not available (e.g. topoi).
A second (personal) reason for championing AC is a pragmatic one: it allows us to prove many things. And "many things" include things that physicists use without a blink. Analysis can hardly get off the ground without some form of choice. Countable choice (ACC) or dependent countable choice (ACDC) is enough for most elementary analysis and many constructivists have no problem with ACC or ACDC. For example, ACC and the stronger ACDC are enough to prove that the countable union of countable sets is countable or Baire's theorem but it is not enough to prove Hahn-Banach, Tychonoff or Krein-Milman (please, correct me if I am wrong). And this is where my question comes in. In one of the comments to the post cited above someone wrote (quoting from memory) that the majority of practicing mathematicians views countable choice as "true". I have seen this repeated many times, and the way I read this is that while the majority of practicing mathematicians views ACC as "obviously true", a part of this population harbours, in various degrees, some doubts about full AC. Assuming that I have not misread these statements, why in the minds of some people ACC is "unproblematic" but AC's validity is not? What is the intuitive explanation (or philosophical reason, if you will) why making countably infinite choices is "unproblematic" but making arbitrarily infinite choices is somehow "more suspicious" and "fraught with dangers"? I for one, cannot see any difference, but then again I freely confess my ignorance about these matters. Let me stress once again that I do not think for a moment that denying AC is "wrong" in some absolute sense of the word; I just would like to understand better what is the obstruction (to use a geometric metaphor) from passing from countably infinite choices to arbitrarily infinite ones.
Note: some rewriting and expansion of the original post to address some of the comments.

REPLY [5 votes]: I apologize for writing the following from memory, but I don't have (and won't for a while have) access to the book I want to refer to; I trust that others on MO will correct whatever I mess up here.  The book is Gregory Moore's "Zermelo's Axiom of Choice: Its origins, development and influence," and the reason I want to refer to it is not Moore's excellent history of the subject but some letters reprinted at the end of the book.  The letters contain a discussion of the axiom of choice, not very long after Zermelo introduced it.  They were written by Baire, Borel, Hadamard, Lebesgue, and (I think) one other prominent analyst of that period.  If I remember correctly, only Hadamard favored the full axiom of choice; the others wanted only countable choice.  Why did they consider countable choice more plausible than full choice?  A cynical answer would be that countable choice is needed for lots of results in analysis, including the Baire category theorem (in general complete metric spaces) and the countable additivity of Lebesgue measure. But the reasons actually given in the letters were, to the best of my recollection, rather close to what David Speyer wrote in his answer here.<|endoftext|>
TITLE: Is there a good account of D-affinity and localization theorem for partial flag varieties?
QUESTION [8 upvotes]: Recall that a topological space is called $A$-affine for a sheaf of algebras $A$ if taking global sections of coherent sheaves of $A$-modules is an equivalence of categories to finitely generated $\Gamma(A)$-modules.  (for example, an affine scheme is one which is affine for the structure sheaf).
It seems to be a "well-known fact" that the variety $G/P$ for any simple complex algebraic group $G$ and parabolic $P$ is $D$-affine where $D$ is the sheaf of differential operators (and that more generally, one can quite explicitly describe the set of TDO's which are affine).  I've found this stated in several books and papers (Beilinson and Bernstein's original paper, "Algebra V: homological algebra", this paper of Alexander Samokhin) but have yet to find an actual proof.  One place one might guess it would be that it seems to not be is the book of Hotta, Takeuchi and Tanisaki.

Does anyone know a published source where this is proved?

I'll emphasize, what I want is not a proof of the theorem in the answers here; that's easy once you understand Beilinson and Bernstein's original argument.  What I'm looking for in a place in the literature where this result is clearly and precisely stated, with a proof or clear reference to a proof.

REPLY [2 votes]: The answer is now yes, I think 
http://arxiv.org/PS_cache/arxiv/pdf/1011/1011.0896v2.pdf
Edit: as requested: http://arxiv.org/abs/1011.0896<|endoftext|>
TITLE: Analogue of Shimura curves in the symplectic case?
QUESTION [12 upvotes]: My understanding is this: one can attach 2-d Galois representations to classical modular eigenforms because one can look in the etale cohomology of modular curves. For Hilbert modular forms the naive analogy breaks down, because the middle cohomology of the Hilbert modular varieties will (conjecturally at least, and possibly this is known in this case) be built up from tensor products of the 2-dimensional Galois representations attached to the automorphic representations contributing to the cohomology, and one can't unravel the prodands from the product.
One way of resolving this problem is to instead use Shimura curves. By Jacquet-Langlands, cuspidal automorphic forms on GL_2 over a totally real field (of odd degree over Q say) biject
with cuspidal automorphic forms on a quaternion algebra ramified at all but one infinite place. And we have the happy coincidence that the associated algebraic group satisfies Deligne's axioms for a Shimura variety, and we can again look in the cohomology of a curve to construct the Galois representation.
This trick relies on two things, one local and one global: the local thing is that GL_2(R) has an inner form which is compact mod centre, and the global thing is that the quaternion algebra satisfies Deligne's axioms for a Shimura variety.
Now let's try and generalise all of this to the symplectic case, so G=GSp_{2g} with g>1. If the base field is Q then Weissauer and others constructed the Galois representations attached to a Siegel modular form in the case g=2 by looking in the etale cohomology of a Siegel modular 3-fold. Now what about if the base is bigger? Can one pull off the same trick?
Local question: does GSp_4(R) have an inner form which is compact mod centre?
Global question: if so, does GSp_4(F) (F totally real) sometimes have an inner form which is compact mod centre at all but one infinite place, and for which Deligne's axioms hold? If so, might one hope to see the Galois representations attached to Hilbert-Siegel modular forms over F here?
[Edit: the local question is solved below by Hansen. I thought that the papers he linked to would deal with the global question too, but now I suspect they don't. I've put a 150-point bounty on for the global question.]
[Edit: because of bounty daftness I can now no longer accept any answer for this question.]

REPLY [4 votes]: Local question: Yes, $GU2(\mathbb{H})$, where $\mathbb{H}$ is the Hamilton quaternions.
Global question: Not an answer, but perhaps useful to you - There are two quite relevant papers of  Claus Sorenson which can be found here and here. The first paper constructs the Galois representations attached to Hilbert-Siegel modular forms over totally real fields. The second paper concerns level-lowering for GSp4 -  along the way, he proves a Jacquet-Langlands transfer to an inner form of GSp4 compact at all infinite places and split at all finite places (at least when F has even degree and pi satisfies the usual conditions).<|endoftext|>
TITLE: Understanding different Q-models of a curve over C
QUESTION [5 upvotes]: These are parametrized by $H^1(Gal(\mathbb{Q}), Aut X)$, where X is some $\mathbb{Q}$-model of the curve.
It was established in Confusion about how the first cohomology classifies torsors
that fiber bundles over $B$ with fiber $F$, structure group $G$ and transition maps with property $P$ are classified by $T$-torsors, where $T$ is the sheaf on $B$ of functions to $G$ with property $P$. $T$-torsors, in turn, are classified by $H^1(B, T)$.
Is there a way to interpret the aforementioned classification of $\mathbb{Q}$-models of a curve in these terms?

REPLY [4 votes]: Yes.
Roughly, the idea is that once your base object $X/B$ is fixed then for any object $Y$ you can consider the sheaf $Iso(X,Y)$ of isomorphisms from $X$ to $Y$.  This has an action of the sheaf $Aut(X)$ and on any open cover $U$ where $Y|_U \cong X|_U$, making a choice of such an isomorphism gives an isomorphism of $Aut(X)(U)$-sets $Iso(X,Y)(U) \cong Aut(X)(U)$.
The difference from the standard argument in this case is that you need to use the etale topology instead of the Zariski topology.  The etale covers of $\mathbb{Q}$ are finite field extensions $F$ and so the statement that $Y$ is "locally" equivalent to $X$ means that $Y_F \cong X_F$ for a large enough field extension $F$.<|endoftext|>
TITLE: Failure of Fin. Presented and Fin. Generated Modules to be Abelian Categories? 
QUESTION [12 upvotes]: Let R be a ring. I'm trying to understand when the categories of finitely presented R-modules and finitely generated R-modules can fail to be abelian categories. 
Poking around on the internet has lead me to believe that these categories will agree and be abelian if R is a (left?) Noetherian ring, but I'm interested in the more general case. I've found warnings that these categories can fail to be abelian but haven't found much discussion of what goes wrong. I feel like there should be some really good illuminating examples.
I think it's not too hard to show that the cokernel of a map of finitely presented modules is again finitely presented. So one of the things I'm really looking for is an example of a map of between finitely presented modules in which the kernel is not finitely presented. It would be even better if the kernel was not finitely generated. Is this possible?

REPLY [5 votes]: [Background to the following can be found in Lam's Lectures on Modules and Rings, section 4.G.]
Definition: A finitely generated (f.g.) right module $M_R$ is coherent if every f.g. submodule of $M$ is finitely presented (f.p.).  
Now let $M_R$ be a finitely presented module that is not coherent.  There exists some submodule $N_R\subseteq M$ that is f.g. but not f.p.  Pick a surjection $R^n\twoheadrightarrow N$, and consider the composition $f\colon R^n \twoheadrightarrow N\hookrightarrow M$. Because $N\cong R^n/\ker(f)$ is not f.p., it must be the case that $\ker(f)$ is not f.g.  
So all we need is an explicit example of a f.p. module that is not coherent.  For this, see Jack Schmidt's answer to this question.  Using my approach in his case, instead of looking at a surjection $R\twoheadrightarrow R/I$, we would look at the composition $R^n\twoheadrightarrow I\hookrightarrow R$.<|endoftext|>
TITLE: If the tensor product of two representations are crystalline, are the original representations crystalline?
QUESTION [15 upvotes]: Let $K$ be a finite extension of the $p$-adic numbers. Suppose that $V$ and $W$ are two (finite dimensional, $p$-adic) continuous representations of $G_K$. Suppose that $V \otimes W$ is crystalline. Is $V$ crystalline up to twist by a character of $G_K$?

REPLY [17 votes]: I'm indeed pretty sure that the answer is "yes". I'd prefer not to post the idea of the proof here because I asked one of my PhD students to write it down with all the details.<|endoftext|>
TITLE: Graphs of maps between manifolds as cycles and intersection theory
QUESTION [6 upvotes]: I'm guessing that the answer to this question is well-known, but I'm struggling to find anything to help me.
Let $X,Y$ be compact manifolds of dimension $n,m$ respectively. Let $f:X \to Y$ be a smooth map. Then one can consider the graph $\Delta_f$ of $f$ as a cycle in $X \times Y$. 
Firstly what is "known" about $\Delta_f$ considered as a homology class? (I appreciate that this is a little vague). There might need to be some extra conditions placed of $f$, as clearly if for example $f$ maps everything to a point then there is nothing to be said.
Secondly (and related to the first question), suppose that $n=m$, and $f$ is an immersion. Then the self intersection $\Delta_f^2$of $\Delta_f$ with itself is well-defined. Is there a simple expression for this in terms of the basic properties of $f$?

REPLY [2 votes]: We have a short exact sequence:
$$
0\to T_{\Delta_f} \to T_{X\times Y}|_{\Delta_f}
\to N_{\Delta_f/Y} \to 0
$$
We also have that $T_{X\times Y}\simeq p_X^*T_X\oplus p_Y^*T_Y$ (where $p_X$ and $p_Y$ are the projections) and if $\Gamma_f: X\to X\times Y$ is the induced embedding, then $T_{\Delta_f}\simeq \Gamma_f^*T_X$. 
This implies that then
$$
{\rm deg}\ N_{\Delta_f/Y}={\rm deg}\ (p_Y^*T_Y)|_{\Delta_f}
$$
and hence (in the case $m=n$)
$$
\Delta_f^2={\rm deg} \left((p_Y^*T_Y)|_{\Delta_f}\right) = {\rm deg} \left(f^*T_Y\right) = {\rm deg} (f^*c_1(Y))= {\rm deg}\ f\cdot {\rm deg}\ c_1(Y).
$$
By the way, I would not say that "if for example f maps everything to a point then there is nothing to be said". Granted, it is simple, but one can still say that, and also that $\Delta_f^2=0$.<|endoftext|>
TITLE: Why does the arXiv produce a messed-up DVI when the PDF is fine?
QUESTION [8 upvotes]: Historically, I've checked the quality of arXiv output by reading the PDF it creates; I learned today that this actually a serious mistake.  If you look at the PDF and DVI for my most recent article, you'll see that the PDF looks fine, and the DVI is essentially unusable from a mathematical perspective.  The tikz pictures don't come out at all, and there seem to be serious font issues.
I'll note that this file compiles fine as a DVI on my home computer.

What's causing this?  Is it a problem a user can fix by include font/style files with his/her upload?  is there someone at the arXiv I should be telling to install something?  Why is the arXiv less competent than a home installation of MikTeX?


EDIT: So, the take-away seems to be that your DVIs from the arXiv will only look right if you use Computer Modern.  Hooray for all math papers looking the same!

REPLY [2 votes]: Please review section
9.2.1 Selecting the Backend Driver
of the pgf/tikz manual. arXiv dvi is generated with the dvips backend driver, and it must be post-processed with dvips to render in a useful way. In addition all the fonts referenced in the dvi need to be available to dvips when post-processing the dvi.
There is also section 
9.2.5 Producing Perfectly Portable DVI Output
which seems like a choice of last resort.
Cheers
T.<|endoftext|>
TITLE: How can I understand the "groupoid" quotient of a group action as some sort of "product"?
QUESTION [6 upvotes]: Recall the notion of groupoid (Wikipedia, nLab).  An important construction of groupoids is as "action groupoids" for group actions.  Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms.  Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x \overset g \to gx$ for each $g\in G$ and $x\in X$.  The composition of morphisms is well-defined if the action is by groupoid automorphisms.  (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${\rm Aut}(\bar x) = {\rm Stab}_G(x)$.)
(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)
Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms.  Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$.  In general, although $X\times 1 \cong X$, we do not have $X \times (1//G) \cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial.  (Here $\times$ denotes the groupoid product, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X \times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X \times (1//G)$.

Question: Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G \cong X \rtimes (1//G)$?  By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X \rtimes Y$?

I see that the page semidirect product in nLab defines $X \rtimes G$ as something closely related to $X//G$.  But clearly this ought to be called $X\rtimes (1//G)$, but then I do not know what the right definition for $X\rtimes Y$ is, hence the question.  And really I'd like to know about a "double crossed product" $X\bowtie Y$.
My motivation for this question is from my answer to Do rational numbers admit a categorification which respects the following “duality”?.

REPLY [7 votes]: The notion of semidirect product $\Gamma \rtimes G$ where $G$ is a group acting on a groupoid $\Gamma$ is set up in Chapter 11, Section 11.4, of my book "Topology and Groupoids".
It is used there in connection with studying orbit groupoids, and their relevance to the fundamental groupoid of an orbit space by a group action. 
One nice point is that this semidirect product includes the case $\Gamma$ is a discrete groupoid, i.e. essentially a set, when you get what is commonly called the action groupoid. In this case the morphism $p: \Gamma \rtimes G \to G$ is known as a covering morphism of groupoids, and  all covering morphisms of $G$ arise in this way. 
I feel the use of covering morphisms of groupoids makes for a nice exposition, base point free, of the theory of covering spaces. Such an idea was pointed out for the simplicial case in the 1967 book on simplicial theory by Gabriel and Zisman, was used in the first 1968 edition of my book, and is partially used in Peter May's 1999 book "A concise  course in algebraic topology". 
Update: I should also add that the notion of action of a groupoid on a groupoid is given, following C. Ehresmann, in my paper 
[11]  ``Groupoids as coefficients'', Proc. London Math Soc.
(3) 25 (1972)  413-426.
available here. The aim was cohomology with coefficients in a groupoid. One of the methods exploited is fibrations of groupoids. 
A  notion of "double product" for groups which act on other "compatibly" is discussed in paper 22 of this list. Another relevant paper on groupoids and actions  is this paper.<|endoftext|>
TITLE: The egg and the chicken
QUESTION [9 upvotes]: After posting this question (in particular after Carl's and Peter's answers) I have realized that the answer seems to depend on a basic problem in foundations.
Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets. These are treated at an intuitive level. Using set theory they define languages, axioms, theories, models and all the logic toolbox. Then they define (formalized) set theory again, using this language.
The second point of view is typical of logicians. They realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead. Then set theory is formalized in this framework.
I always thought that the points of view were interchangeable, as far as one was interested in the mathematical consequences. But comparing Carl's and Peter's answers it seems that actual (but still foundational) mathematic may depend on the point of view accepted. I'd like to understand this better.

Are there any mathematical consequences of choosing one of the two points of view?

REPLY [9 votes]: I would like to question two statements you make because they paint an oversimplified picture, which unfortunately is alluring to mathematicians who do not want to think about foundations (and they should not be blamed for it anymore than I should be blamed for not wanting to think about PDEs).

"Most mathematicians accept as given the ZFC (or at least ZF) axioms for sets." This is what mathematicians say, but most cannot even tell you what ZFC is. Mathematicians work at a more intuitive and informal manner. High party officials once declared that ZFC was being used by everyone, so it has become the party line. But if you read a random text of mathematics, it will be equally easy to interpret it in other kinds of foundations, such as type theory, bounded Zermelo set theory, etc. They do not use the language of ZFC. The language of ZFC is completely unusable for the working mathematician, as it only has a single relation symbol $\in$. As soon as you allow in abbreviations, your exposition becomes expressible more naturally in other formal systems that actually handle abbreviations formally. Informal mathematics is informal, and thankfully, it does not require any foundation to function, just like people do not need an ideology to think. If you doubt that, you have to doubt all mathematics that happened before late 19th century.
"They [logicians] realize that in order to talk of logic they don't need the full power of set theory, so they take logic as God-given instead." I do not know of any logicians, and I know many, who would say that logic is "God-given", or anything like that. I do not think logicians are born into a life rich with the "full power of set theory" which they throw away  in order to become ascetic first-order logicians. That is a nice philosophical story detached from reality. The logicians I know are usually quite careful, skeptical, and inquisitive about foundational issues, reflect carefully on their own experiences, and almost never give you a straight answer when you ask "where does logic come from?" Your view is naive and inaccurate, if not slightly demeaning.

If I understand your question correctly, you are asking whether there is a difference between the following two views:

We start with naive set theory and on top of it we formalize set theory.
We start with first-order logic and immediately formalize set theory.

Well, we are proceeding from two different meta-theories. The first one allows us a wide spectrum of semantic methods. We can refer to "the standard model of Peano arithmetic" because we "believe in natural numbers", and we can invent Tarskian model theory without worrying where it came from.
The second method is more restricted. It will lead to syntactic and proof-theoretic methods, since the only thing we have given ourselves initially are syntactic in nature, namely first-order theories. There will be careful analysis of syntax. For advanced methods, however, we will typically resort to at least some amount of "naive mathematics". Ordinals will come into play, it will be hard to live without completeness theorems (which involve semantics), etc.
However, this is not how real life works. The dilemma you present is not really there. A working mathematician does not concern himself with these issues, anyhow, while a logician will likely refuse to be categorized as one or the other breed.
That is my guess, based on the experience that my fellow logicians are complicated animals and it is hard to get to the bottom of their foundational guts.<|endoftext|>
TITLE: Set theory and Model Theory
QUESTION [43 upvotes]: This question probably doesn't make any sense, but I don't see why, so I ask it here hoping someone will illuminate the matter:
There is this whole area of study in Set Theory about the consistency, independence of axioms, etc. In some of these you use model theory (e.g. forcing) to prove results about set theory. 
My question is: What is the foundation of this model theory we are using? We are certainly using sets to talk about the models, what some may call sets in the "meta"-mathematics, that is to say, the "real" mathematics.
But then, all these arguments in the end in are all about the theory of sets as a theory, and not the theory of sets as a foundation of math, since we are using these sets in the meantime. So our set theory is not about the foundation of math. 
Am I right?

REPLY [8 votes]: First of all, I would like to say that your question is not only justified, but actually most welcome: it shows the King naked. Who is the King? You can call him Semantics,or the mathematical theory of Truth (notice the capital letter)  if you wish.
There is a widespread usage among logicians of sentences such as " the true universe of sets", the "set of natural numbers", and so forth, as if these entities were crystal clear and thus indisputable. They aren't, at least to me. But even if they are, even if there is such a thing as mathematical intuition which informs us on the world of platonic math (what my teacher aptly named PLATO's ATTIC), the fact is this: as soon as we SPEAK about them, they just become syntax, no more no less (ie part of a frame syntactic theory which grounds them). 
Model theory is a great branch of mathematics, no doubts about it. But it is formalized mathematics, and grounded in ZFC. Does that mean that all its results are empty, especially in the model theory of ZFC? I say absolutely no. Instead, it bespeaks of the uncanny capability of a powerful theory such as ZFC to "reflect" upon itself and the rest of mathematics. The king is naked, but a king nevertheless. Long live the king!
PS I just discovered the recent work of Joel D. Hamkins and a few others on the Multiverse: finally some people from the set theory ranks are moving away from a dogmatic  and monolitic notion of truth  (the ghost-like "true universe of sets" and its equally ghostly eternal properties) towards a new dynamic and contextual one. I think this is a beginning of a new era, if I am not mistaken....<|endoftext|>
TITLE: Real representation of finite groups 
QUESTION [6 upvotes]: I want to know if there is complete theory on real representation of finite
groups.
Say, given a finite group G, can we know all the injections from G to 
GL(n,R) for a particular n?

REPLY [3 votes]: Some of the other answers omit several details and I had already written this on another forum.  At any rate, perhaps the example would be useful:
More or less yes, though for larger groups it is not very practical.
Calculate the character table of G. Use the Frobenius-Schur indicator to
fix the complex and quaternionic irreducible characters to be characters of
irreducible real representations. Take all non-negative integer combinations
of these characters of real reps whose total degree is n and whose kernel is
trivial (the degree and the kernel of a character are very easy to calculate).
There is a 1-1 correspondence between injections of G into GL(n,R) up to GL(n,R)
conjugacy and characters of real reps whose total degree is n and whose kernel
is trivial.
Except for cyclic groups G and n=1, there will be uncountable many injections
into GL(n,R), so you'll want to only consider them up to GL(n,R) conjugacy in
order to get a finite number.
Also notice that if f1:G→GL(m1,R) is an injection and f2:G→GL(m2,R) is any group
homomorphism, then another injection is "f1⊕f2":G→GL(m1+m2,R):g→[f1(g),0;0,f2(g)],
the function that takes the two matrices from f1 and f2 and makes them the blocks
of a block diagonal matrix. For this reason the number of injections, even up to
conjugacy, can get unwieldy.

Example: The non-cyclic group of order four:
For G the Klein four-group the irreducible characters are all real, and are
[1,1,1,1], [1,-1,-1,1], [1,-1,1,-1,], [1,1,-1,-1]. The degree of each of these
is 1, the first number. The kernel of each of these are the positions where
the value and the degree are equal, so all of them have a kernel. If n was 10,
then I'd look at all the ways of choosing non-negative integers a,b,c,d such
that a+b+c+d=10 (so that the degree was 10) and such that at least 2 of b,c,d
were non-zero (so that the kernel was trivial). There are 282 such combinations,
one of which is a=5, b=0, c=2, d=3. This corresponds to the character
5*[1,1,1,1]+0*[1,-1,-1,1]+2*[1,-1,1,-1]+3*[1,1,-1,-1] = [ 10, 6, 4, 0 ]. It has
degree 10 (the first and largest positive number) and trivial kernel (all the
other positive numbers are strictly smaller). It corresponds to the representation
g1 → diag( 1,1,1,1,1, -1,-1, 1,1,1 ) and g2 → diag( 1,1,1,1,1, 1,1, -1,-1,-1 )
and of course g1*g2 → diag( 1,1,1,1,1, -1,-1, -1,-1,-1 ). This is 5 copies of
the representation g→1, g2→1, followed by two copies of g1→-1, g2→1, and
finally three copies of g1→1, g2→-1. Since b=0 we did not include any copies
of g1→-1, g2→-1.
By some silly counting, the number of injections of the Klein four-group into
GL(n,R) up to GL(n,R) conjugacy is (n^3+6*n^2+11*n-18)/6, so it grows somewhat
quickly but not too crazily.

For non-abelian groups G, the conversion of the characters into the matrices
is a little harder (for abelian groups, the matrices are all 1x1 and really
are just the character values), but at some point the matrices are too big
anyways, and the characters are a more efficient data type.<|endoftext|>
TITLE: Is there an obvious way for showing singularities are quotient?
QUESTION [6 upvotes]: I'm stuck on a technicality concerning singularities. 
Basically, I have to show that the singularities of a $\mathbf{certain}$ normal projective variety over $\mathbf{C}$ are rational. (I won't bother you with the exact set-up. For, it's possible that I'm even wrong.) 
My idea was to use a theorem of Viehweg and show that all singularities are quotient. But I don't have a clue of how to do the latter. What are the standard techniques used in such a proof? That is, say you want to show that all singularities are quotient. What would be your first idea to apply? Do problems like this become easier in low dimensions or does it really not matter?
Note. Forgive me if the question is ill-posed/vague.
Added later.
In view of Karl's remark, I decided to give the set-up.
Let $X$ be a smooth projective variety over $\mathbf{C}$ (of any dimension). Let $\pi:Y\longrightarrow X$ be a finite morphism, where $Y$ is a normal projective variety over $\mathbf{C}$. We also are given a flat morphism $h:X\longrightarrow C$, where $C$ is a smooth projective curve. Of course, it can happen that $Y$ has singularities that are not rational. (Example?) But I would like to show that the singularities of $Y$ are rational in the situation I will describe now.
Let $V\longrightarrow U$ be a connected finite etale covering of $U=X-D$, where $D$ is a simple normal crossings divisor on $X$. Define $Y$ to be the normalization of $X$ in the function field of $V$.
So the set-up is quite general, i.e., I don't have any equations. Maybe one could try to apply arguments based on fundamental groups to show that the singularities of $Y$ are quotient? I think I can show that the singularities of $Y$ occur in the inverse image under $\pi$ of $D^{sing}$, where $D^{sing}$ is the singular locus of $D$.
And Karl, what are the techniques in characteristic $p>0$ that you mention below?

REPLY [6 votes]: This is essentially Abhyankar's lemma.
What VA says is correct. However, one can simply remark that the subgroups $(d\mathbb Z)^n$ for  $d>0$ form a cofinal system of subgroups of finite index, and the normalization $Y_d$ of $X$ in the corresponding covering of $X \smallsetminus D$ is smooth. Since the fundamental group is abelian, the covering $Y \to X$ is Galois. If $G$ is the Galois group, the kernel $K$ of $\pi_1(X\smallsetminus D) \to G$ contains some $(d\mathbb Z)^n$; and then $Y$ is a quotient of $Y_d$ by the action of $K/(d\mathbb Z)^n$.<|endoftext|>
TITLE: Class Field Theory for Imaginary Quadratic Fields
QUESTION [8 upvotes]: Let $K$ be a quadratic imaginary field, and $E$ an elliptic curve whose endomorphism ring is isomorphic to the full ring of integers of $K$. Let $j$ be its $j$-invariant, and $c$ an integral ideal of $K$. Consider the following tower:
$$K(j,E[c])\ \ /\ \ K(j,h(E[c]))\ \ /\ \ K(j)\ \ /\ \ K,$$
where $h$ here is any Weber function on $E$. (Note that $K(j)$ is the Hilbert class field of $K$). 
We know that all these extensions are Galois, and any field has ABELIAN galois group over any smaller field, EXCEPT POSSIBLY THE BIGGEST ONE (namely, $K(j,E[c]) / K$). 
Questions:

Does the biggest one have to be abelian? Give a proof or counterexample.

My suspicion: No, it doesn't. I've been trying an example with $K = \mathbf Q(\sqrt{-15})$, $E = C/O_K$, and $c = 3$; it just requires me to factorise a quartic polynomial over $\bar{ \mathbf Q}$, which SAGE apparently can't do.

What about if I replace $E[c]$ in the above by $E_{tors}$, the full torsion group?

REPLY [8 votes]: Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.
> K<s>:=QuadraticField(-23);                                                    
> jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2);                             
> jrel:=PowerRelation(jinv,3 : Al:="LLL");                                      
> Kj<j>:=ext<K|jrel>;                                                           
> E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]);
> HasComplexMultiplication(E);
true -23                 
> c4, c6 := Explode(cInvariants(E)); // random twist with this j                
> f:=Polynomial([-c6/864,-c4/48,0,1]);                                          
> poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic                 
> R:=Roots(poly);                                                               
> Kj2:=ext<Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])>;                          
> KK:=ext<Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])>;                          
> assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here         
> f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1];  
> GaloisGroup(f); /* not immediate to compute */                                
Permutation group acting on a set of cardinality 12
Order = 48 = 2^4 * 3
> IsAbelian($1);                                                                
false

This group has $A_4$ and $Z_2^4$ as normal subgroups, but I don't know it's name if any.
PS. 5-torsion is too long to compute most often.<|endoftext|>
TITLE: method of finding roots of polynominal equations with arithmetic operations and roots and other functions
QUESTION [8 upvotes]: Lets recall Platonic construction in plane geometry. It is impossible to square a circle using only ruler and callipers. But is also known that it is possible to do it with ruler which has a mark on it.
So we know by Gaois theory that we can't write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots. 
Question: is it possible to write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots and something else in finite numbers of factors? For example adding logarithm to the set of operations? ( or other function, or class of functions...)? 
If Yes: Do they form any kind of algebraic structure?

REPLY [4 votes]: Eisenstein in 1844 found a very simple hypergeometric series solution of the
quintic $x^5+x=a$, namely
$x=a-a^5+10\frac{a^9}{2!}-15\cdot 14 \frac{a^{13}}{3!}+20\cdot 19\cdot 18\frac{a^{17}}{4!}-\cdots$
Since it was shown by Bring in 1786 that the general quintic is reducible by
radicals to this form, Eisenstein's solution is probably the simplest solution
possible for the quintic.
For more on this story, see S.J. Patterson, Eisenstein and the quintic equation,
in Historia Mathematica 17 (1990), pp.132--140.<|endoftext|>
TITLE: What is known about the MMP over non-algebraically closed fields
QUESTION [14 upvotes]: I would like to know how much of the recent results on the MMP (due to Hacon, McKernan, Birkar, Cascini, Siu,...) which are usually only stated for varieties over the complex numbers, extend to varieties over arbitrary fields of characteristic zero or to the equivariant case. 
I assume that the basic finite generation results hold for any such field, by base extending to an algebraic closure, so I would guess that most results should extend without too much difficulty. The particular questions that I am really interested in are: 
1) Given a smooth projective rationally connected variety X over a field k of charaterisitic zero, can we perform a finite sequence of divisorial contractions and flips to obtain a Mori fibre space?
and 
2) The equivariant version of 1) for the action of a finite group on X.

REPLY [8 votes]: Below you will find some references. I would also add a warning that these questions can become very difficult very soon.
Of course, one can base extend to the algebraic closure, and run the mmp there, but what guarantees that even the first step can be performed such that the resulting objects over the algebraic closure is the base extension of something reasonably related to the original object over the original field?
In particular, take a surface defined over $\mathbb R$ and let $X$ denote the corresponding surface over $\mathbb C$. Now if $X$ contains a $(-1)$-curve that has no real points, then when this curve is contracted, the real points, that is the original surface $X_{\mathbb R}$ does not change at all. That itself is of course not a problem and also one may argue that that $(-1)$-curve has to have a conjugate pair and contracting that results in a surface that should be again defined over $\mathbb R$ (In a different space but with the same set of real points as before). You can imagine that in higher dimensions even more complicated problems arise. 
Another issue to keep in mind is that in real algebraic geometry people care about keeping the topology the same. Doing a flip screws that up, so the preferred way to do it is without flips. This seems like an impossible proposition, but Kollár's ingenious series of papers does exactly that. Well, actually what he does is that he performs flips on the complexified space but proved that it happens on the imaginary part and hence it does not mess with the topology of the real points. However, you still need to do it in order to keep the mmp going.
Anyway, to learn more about this, look at these papers:
MR1677128 (2000c:14078) Kollár, János Real algebraic threefolds. I. Terminal singularities. Dedicated to the memory of Fernando Serrano. Collect. Math. 49 (1998), no. 2-3, 335--360. (Reviewer: Mark Gross) 14P25 (14E30)
MR1639616 (2000c:14079) Kollár, János Real algebraic threefolds. II. Minimal model program. J. Amer. Math. Soc. 12 (1999), no. 1, 33--83. (Reviewer: Mark Gross) 14P25 (14E30) 
MR1703903 (2000h:14049) Kollár, János Real algebraic threefolds. III. Conic bundles. Algebraic geometry, 9. J. Math. Sci. (New York) 94 (1999), no. 1, 996--1020. (Reviewer: Mark Gross) 14P25 (14E30) 
MR1760882 (2001c:14087) Kollár, János Real algebraic threefolds. IV. Del Pezzo fibrations. Complex analysis and algebraic geometry, 317--346, de Gruyter, Berlin, 2000. (Reviewer: Mark Gross) 14P25 (14J30) 

Another issue is the rational connectivity and its relation to Mori fiber spaces. 
For this, again, one has to be very careful and state exactly what one wants. What notion of rational connectivity are you using? Are the curves defined over $\mathbb R$ or $\mathbb C$? Do you require $X_{\mathbb R}$ or $X$ (or both) be rationally connected?
To illustrate the difficulties, here is a conjecture of Nash (yes, that Nash): Let $Z$ be a smooth real algebraic variety. Then $Z$ can be realized as the real points of a rational complex algebraic variety. 
This, actually, turns out to be false. Kollár calls it the shortest lived conjecture as it was stated in 1954 and disproved by Comessatti around 1914 (I don't remember the exact year). However, even if the statement is false, just the fact that it was made and no one realized for 50 years that it was false should show that these questions are by no means easy. (Comessatti's paper was in Italian and I have no idea how Kollár found it.)
Kollár showed more systematically the possible topology types of manifolds that can satisfy this statement. In particular, Kollár shows that any closed connected 3-manifold occurs as a possibly non-projective real variety birationally equivalent to ${\mathbb P}^3$. In other words the way Nash's conjecture fails is on the verge of the difference between projective and proper again showing that these questions are not easy.
Here are some references for Kollár's work on the Nash conjecture.
MR1641168 (99k:57038) Kollár, János The Nash conjecture for threefolds. Electron. Res. Announc. Amer. Math. Soc. 4 (1998), 63--73 (electronic). (Reviewer: A. Tognoli)
MR1865090 (2003a:14033) Kollár, János The topology of real and complex algebraic varieties. Taniguchi Conference on Mathematics Nara '98, 127--145, Adv. Stud. Pure Math., 31, Math. Soc. Japan, Tokyo, 2001. (Reviewer: Sándor J. Kovács) 
MR1882536 (2002m:14046) Kollár, János The topology of real algebraic varieties. Current developments in mathematics, 2000, 197--231, Int. Press, Somerville, MA, 2001. (Reviewer: Grigory B. Mikhalkin) 14P25 
MR1941579 (2004c:14117) Kollár, János The Nash conjecture for nonprojective threefolds. Symposium in Honor of C. H. Clemens (Salt Lake City, UT, 2000), 137--152, Contemp. Math., 312, Amer. Math. Soc., Providence, RI, 2002. (Reviewer: Sándor J. Kovács)<|endoftext|>
TITLE: Applications of pattern-free continued fractions
QUESTION [12 upvotes]: Questions about continued fractions reminded me about a related diophantine problem. I am not quite sure that diophantine equations are still in fashion but
$$
1^k+2^k+\dots+(m-1)^k=m^k,
$$
the Erdős--Moser equation, is quite special. This seems to be the only known equation in two unknowns ($k$ and $m$ are assumed to be positive integers) for which it is not proven whether there are finitely many solutions. Presumably there is only one solution, $1^1+2^1=3^1$. In a joint work with Yves Gallos and Pieter Moree (the preprint is available from arxiv.org) we show that if a solution exists than $m>10^{10^9}$ "by showing that $2k/(2m-3)$ is a convergent of $\log 2$ and making an extensive continued fraction digits calculation of $(\log 2)/N$, with $N$ an appropriate integer." Trying to 
build up some history on the use of continued fractions in diophantine equations we could not find any other example, besides the continued fractions of real quadratic irrationalities with their ultimate relation to Pell's equation (which in turn appears in studying some other equations). I am definitely interested if there are some, but also in other "serious" applications of "generic" (presumably pattern-free) continued fractions of mathematical constants (or functions) outside number theory. Please suggest "generic" cases where no rule for partial quotients is known.

REPLY [13 votes]: Continued fractions are used in the effective solution of Thue equations. Given such an equation, say $x^4-3x^3y+7x^2y^2-y^4 = 1$ to be specific, we can brute force find all small $x,y$ solutions, and linear forms in logarithms results can be used to eliminate solutions with $x,y$ astronomically large. In the large intermediate range (which typically can mean $x,y$ with between 50 and $10^{50}$ digits, depending on the specifics of the equation) LLL is often helpful. The small intermediate range (say $x,y$ having between 3 and 50 digits) is where continued fractions make their appearance.
An example of this is the following. The equation given above is the same as
 $$r^4 - 3r^3 + 7 r^2 -1 = 1/y^4$$
where $r$ is the rational $x/y$. This factors (for some complex numbers $\alpha_i$) as
 $$(r-\alpha_1)(r-\alpha_2)(r-\alpha_3)(r-\alpha_4) = \frac{1}{y^4}.$$
Since the right hand side is small, one of the factors on the left must be small, say it's the first that is smallest. Letting $C=(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_1-\alpha_4)/8$, our equation gives the inequality:
 $$\left| \alpha_1 - \frac xy \right| < \frac{2/(C y^2)}{2y^2}.$$
If $y$ is large enough (typically, say, 3 digits), then $2/(C y^2)<1$, and a theorem from a couple centuries ago states that if
 $$\left| \alpha - \frac xy \right| < \frac{1}{2y^2},$$
then $x/y$ is a convergent of the continued fraction of $\alpha$. This means that every solution to the original equation (except for the small ones) is a convergent to one of the roots of the equation, and numerators and denominators of convergents grow exponentially. So, as a practical matter, we can find all solutions of a Thue equation with between 3 and 50 digits essentially instantaneously.
Disclaimer: all statements here about how many digits constitutes large or small depend on the specific equation (among other things, on the embeddings into $\mathbb{C}$ of the group of fundamental units of the relevant extension of $\mathbb{Q}$), and the numbers I've given here derive from my personal experience. Some years ago, I worked at Wolfram Research to help bring diophantine equation solving to Mathematica, it will solve the above equation if you input:
Reduce[x^4 - 3 x^3 y + 7 x^2 y^2 - y^4 == 1, {x, y}, Integers]<|endoftext|>
TITLE: Is there any geometry where the triangle inequality fails?
QUESTION [27 upvotes]: We know from elementary school that the triangle inequality holds in Euclidean geometry. Some where in High School or in Univ., we come across non-Euclidean geometries (hyperbolic and Riemannian) and Absolute geometry where in both the inequality holds.
I am curious whether the triangle inequality is made to hold in any geometry( from the beginning) or is  a consequence of some axioms. Presumably, the denial of the inequality would create havoc in that conceivable geometry.
Thanks.

REPLY [8 votes]: The reverse triangle inequality|x+y|>|x|+|y| holds in the Minkowski Space of Special Relativity for two timelike vectors in the same direction. So geometries that deny the triangle inequality in some signficant way can be very important I guess.<|endoftext|>
TITLE: sums involving the nearest-integer function
QUESTION [5 upvotes]: Let $r$ be a positive real number.  For each positive integer $n$, let $y = m_n x + b_n$ be the line that best approximates (in the least-squares sense) the set of $n$ points $(k,{\rm nint}(kr))$ with $1 \leq k \leq n$, where ${\rm nint}$ is the nearest integer function.  Does the difference between $m_n$ and $r$ go to zero like $O(1/n^2)$?
Failing an answer to that question, I'd welcome pointers to any known bounds on the difference between $\sum_{k=1}^n {\rm nint}(kr)$ and $\sum_{k=1}^n kr$ as well as bounds on the difference between $\sum_{k=1}^n k \: {\rm nint}(kr)$ and $\sum_{k=1}^n k^2r$, since these could be used to answer the above question.

REPLY [3 votes]: The error term depends on continued fraction expansion of $r$. It is Khinchine's theorem. In the simple case (when $r=[q_0;q_1,q_2,...]$ is a Liouville number with convergents $P_k/Q_k$) for $n=Q_k$ the error term will has the following form
$$\frac{q_1+\cdots+q_k}{Q_k}\asymp \frac{q_k}{Q_k}.$$ It can be $\gg Q_k^{-\varepsilon}=n^{-\varepsilon}$ for some fast growing $q_k$.<|endoftext|>
TITLE: Local form of a real-analytic function taking values in a Banach space
QUESTION [5 upvotes]: Let $B$ be an infinite-dimensional Banach space, and let $M\subset\mathbb{R}^n$ be a neighborhood of the origin in $\mathbb{R}^n$.
Suppose that $I:M\to B$ is a real-analytic function with $I(0)=0$ and such that the derivative of $I$ at $0$ has maximal rank.
Is it true that there exist neighborhoods $U,V\subset B$ of $0$ and a real-analytic diffeomorphism $\phi:U\to V$ such that $\phi\circ I$ is the restriction of a linear map $\mathbb{R}^n\to B$? If so, what is a good reference?
EDIT: I asked this question in the real-analytic setting, but might as well have done so in the complex-analytic case.

REPLY [2 votes]: I cannot give you a reference, but the answer ought to be yes.
To simplify notation, identify $\mathbb{R}^n$ with its image in $B$ under the derivative $I'(0)$. That image, being finite-dimensional, is the range $pB$ of a finite rank, bounded projection $p$. Replacing $M$ by a smaller neighbourhood if necessary, we can assume that $p\circ I$ is a diffeomorphism of $M$ onto a neighbourhood $N$ of $0$ in $pB$. Let $h\colon N\to M$ be its inverse, and define $\phi\colon p^{-1}(N)\to p^{-1}(M)$ by $$\phi(w)=x+w-I(x),\qquad x=h(pw).$$
If $w\in pB$ then $h(p(I(w)))=w$, so $\phi(I(w))=w+I(w)-I(w)=w$ – i.e., $\phi\circ I$ is the identity on $pB$.
To find the inverse of $\phi$, note that if $y=\phi(w)$ and $x=h(pw)\in pB$ then $py=x+pw-p(I(h(pw))=x$, so $w=y-x+I(x)=y-p(y)+I(py)$, i.e., $$\phi^{-1}(y)=y-py+I(py).$$
Edit: On second thought, it would have been more natural to define $\phi^{-1}$ first, with the requirement that its restriction to $pB$ be $I$. Letting its restriction to $(1-p)B$ be the identity is the simplest way to make it a diffeomorphism.<|endoftext|>
TITLE: What theorem constructs an initial object for this category? (Formerly "Integrability by abstract nonsense")
QUESTION [30 upvotes]: Tom Leinster has a note here about how you can realize L^1[0,1] as the initial object of a certain category. You should really read his note because it is only 2.5 pages and is much more charming than what I am going to write below as background, but if you don't want to click on the link here is the idea:
We work in the category of Banach spaces with contractive maps, where we are defining $X \oplus Y$ to have the norm $|| (x,y) || = \frac{1}{2}(||x|| + ||y||)$. Consider triples $(X, \xi, u)$ where $X$ is a Banach space, $u \in X$ has norm at most 1, $\xi:X \oplus X \to X$ is a map of Banach spaces with $\xi(u,u) = u$.  A morphism of such triples is a map of Banach spaces commuting with all structure in sight.  The it turns out that the initial object in this category is $(L^1[0,1], \gamma, 1)$ where $\gamma(f,g)$ smushes $f$ and $g$ by a factor of two horizontally and then puts them side by side.  Essentially this is because once you know where the constant function 1 goes, you can determine where any piecewise constant function whose discontinuities are at dyadic rationals goes, and then by density you get a unique map out of $L^1$.
Leinster mentions that there are some abstract results which actually construct an initial object for a category like this. I have looked through Barr and Wells, but I do not see what exactly I should be using here. The only general initial object construction I know (The one at the beginning of the adjoint functor theorem chapter of Categories for the Working Mathematician), doesn't seem to apply (maybe it does but I am not seeing it).  Does anyone know such a general construction which applies here?  How does that construction look when you apply it to this situation?  Does it look anything like the usual construction of $L^1[0,1]$?

REPLY [21 votes]: I hope this answers the question a bit more explicitely as the other one. The general theorem which applies here is the following:

Theorem: Let $C$ be a category which as an initial object and colimits of $\omega$-chains. Then for every functor $F : C \to C$ which preserves these colimits, there exists an initial $F$-algebra. Namely, you can take the colimit of $0 \stackrel{i}{\rightarrow} F(0) \stackrel{F(i)}{\rightarrow} F^2(0) \stackrel{F^2(i)}{\rightarrow} \cdots$.

Remark the similarity to other fixed point theorems, such as by Banach (for Banach spaces) or Tarski (for CPOs). The proof is easy and straight forward. In our case, we have the category of Banach spaces $B$ and consider the comma category $C=\mathbb{K} / B$. It has products and also colimits of $\omega$-chains (take the completion of the colimit of the underlying normed spaces) and the functor $F : C \to C,~ X \mapsto X \times X$ preserves them. Thus we can apply the Theorem.
The sequence $\mathbb{K} \to \mathbb{K}^{\{0,1\}^1} \to \mathbb{K}^{\{0,1\}^2} \to \dotsc$ identifies with the sequence of inclusions $X_0 \subseteq X_1 \subseteq X_2 \subseteq \dotsc$, where $X_n$ is the space of step functions $[0,1[ \to \mathbb{K}$ which are constant on every intervall of the $n$-th dyadic subdivision of the interval. The norm becomes the usual $\mathrm{L}^1$-norm. The union $\cup_n X_n$ consists of those step functions which are locally constant with respect to some dyadic subdivision and is endowed with the $\mathrm{L}^1$-norm. All step functions can be approximated by these dyadic step functions. The completion is thus, indeed, $\mathrm{L}^1[0,1]$. Also some inspection gives us that the algebra structure consists of the constant function $1$ and the map $(f,g) \mapsto f \star g$, which squeezes and juxtaposes.
So basically this construction is very similar to the usual one of $\mathrm{L}^1[0,1]$, but we don't have to care about any ambiguity of choices since all the involved universal properties ensure this automatically. Besides, it avoids the general measure theory.<|endoftext|>
TITLE: When does a "representable functor" into a category other than Set preserve limits?
QUESTION [6 upvotes]: This might be a dumb question.  If $C$ is an ordinary category, then for any $c \in C$ the covariant representable functor $\text{Hom}(c, -) : C \to \text{Set}$ preserves limits.  However, it can happen that $c$ can be equipped with extra structure which in turn gives the morphisms out of $c$ extra structure, so that there is a "representable functor" $\text{Hom}(c, -) : C \to D$ where $D$ is a category equipped with a forgetful functor $F : D \to \text{Set}$ such that composing with the above gives the original representable functor.  
In this situation, when does the functor into $D$ still preserve limits?  How is this situation formalized?  (Assume that $C$ is not enriched over $D$ in any obvious way.)  
There are several examples of this coming from algebra, but the one that got me curious is the following.  Let $C$ denote the homotopy category of pointed (path-connected?) topological spaces and let $S^1$ denote the circle with a distinguished point.  I believe I am correct in saying that if the fundamental group functor $\pi_1 : C \to \text{Grp}$ is composed with the forgetful functor $U : \text{Grp} \to \text{Set}$, then $S^1$ represents the resulting functor $U(\pi_1(-))$.  (The extra structure on $S^1$ that makes this possible is, if I'm not mistaken, a cogroup structure internal to $C$.)  Can I conclude that $\pi_1$ preserves limits?  

Edit:  I've been told that the above example is problematic, so here's a simpler one.  Let $C = \text{Set}$ and suppose that $c \in C$ is equipped with a morphism $f : c \to c$.  Then by precomposition $\text{Hom}(c, d)$ is also equipped with such a morphism, so $\text{Hom}(c, -)$ has values in the category of dynamical systems.  Does it preserve limits?  Another example is my attempted answer to question #23188.

REPLY [6 votes]: [Collecting my sporadic comments into one (hopefully) coherent answer.]
A more general question is as follows: For functors
$C\stackrel{F}{\to}D\stackrel{U}{\to}E$ and for an index category
$J$ such that $UF$ preserves $J$-limits, when does $F$ preserve $J$
limits?
A useful sufficient condition is that if $U$ creates $J$-limits, then
in the above situation $F$ preserves $J$-limits.  Proof:  Let $T\colon
J\to C$ be a functor, and suppose that $\tau\colon
\ell\stackrel{\cdot}{\to} T$ is a limiting cone in $C$.  Since $UF$
preserves $J$-limits, $UF\tau\colon UF\ell\stackrel{\cdot}{\to} UFT$ is a limiting cone in
$E$. As $U$ creates $J$-limits, there is a unique lifting of $UF\tau$ to a cone in
$D$, and this cone is a limiting cone.  But $F\tau\colon
F\ell\stackrel{\cdot}{\to} FT$ is such a lift, and hence we're done.
This condition is quite useful, because many forgetful functors are
monadic, and monadic functors create all limits (by combining their definition
on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by
Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a
comment of Tom Leinster from which I learned this :)).
For example, consider the category of all small algebraic systems of
some type.  From the AFT, we know that the forgetful functor to
$\mathbf{Set}$ has a left adjoint, and it is the content of Theorem
6.8.1, p. 156 of Mac Lane that this forgetful functor is monadic.
Returning to the original question, this means that whenever the
category $D$ is one of $\mathbf{Grp}$, $\mathbf{Rng}$,
$\mathbf{Ab}$,... and $U\colon D\to \mathbf{Set}$ is the forgetful
functor, then for any $J$, $UF$ preserves $J$-limits implies $F$ preserves $J$
limits.  In particular, if $UF$ is a representable functor
(and hence preserves all limits), then $F$ preserves all limits.
Next, let me try to comment on your motivating examples (the one from
Q. 23188 and the one from the 'Edit' part of the current question.)
Regarding your example in Q. 23188:  Unfortunately I know nothing of
Hopf algebras, so I can't understand all the details of your
construction. If I understand correctly, you construct a functor
$F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the
forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is
representable. If this is indeed the case, then by the above $F$
itself preserves all limits.
[EDIT: corrected the part concerning the last example.] 
Finally, regarding your example in the edited question: 
While I
know nothing of dynamical systems, from a quick glance at Terence
Tao's blog
it seems that the category of dynamical systems is the
category whose objects are pairs $\langle X,f\colon X\to X\rangle$ with
$X$ a (small) set and whose arrows
$\phi\colon\langle X, f\rangle\to\langle Y, g\rangle$ are those
functions $\phi\colon X\to Y$ with $g\circ\phi =\phi\circ f$. 
To show that the above sufficient condition works in this case, we
would like to show that the forgetful functor to $\mathbf{Set}$ crates
limits.  More generally, we will show that if $C$ is a
category and $D$ is the category whose objects are pairs $\langle
x,f\colon x\to x\rangle$ (where $x\in\operatorname{obj}(C)$,
$f\in\operatorname{arr}(C)$), and whose arrows $\phi\colon \langle
x,f\rangle\to \langle y,g\rangle$ are those arrows $\phi\colon x\to y$
with $g\circ\phi =\phi\circ f$, then the forgetful functor $U\colon
D\to C$ creates limits.
[I'm sure that this follows from some well-known result, but since I
don't see it, I'll just continue with a direct proof.]
So, let $J$ be an index category, let
$F\colon J\to D$ be a functor, and suppose that $\tau\colon
x\stackrel{.}{\to} UF$ is a limiting cone in $C$.  We would like to
show that there exists a
unique cone $\sigma\colon L\stackrel{.}{\to} F$ in $D$ such that
$U\sigma=\tau$, and that this unique cone is a limiting cone.
For uniqueness, suppose that $\sigma\colon L\stackrel{.}{\to} F$
satisfies $U\sigma = \tau$. Write $F_j:=\langle y_j,f_j\rangle$.  Then
we must have for all $j$
$$
\sigma_j=(x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j)
$$
for some $f\colon x\to x$ (hence we immediately see that $\sigma$ is
determined up to $f$). Now, since by the above we see that $\tau_j$
must be an arrow
$$
(x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j)
$$
of $D$, the following diagram must be commutative for all $j$:
$$
\begin{matrix}
x & \stackrel{\tau_j}{\longrightarrow} & y_j =UF_j\\
f\downarrow & & f_j\downarrow\\
x&\stackrel{\tau_j}{\longrightarrow} & y_j = UF_j.
\end{matrix} \quad \text{(Diagram 1)}
$$
Now we claim that the $\to\downarrow$ part of the above diagram
forms a cone to $UF$, that is, we claim that the family
$\{f_j\tau_j\}$ forms a cone $x\stackrel{.}{\to} UF$.  Indeed, for an
arrow $g:j\to j'$ of $J$, consider the following diagram:
$$
\begin{matrix}
&&&&x\\
&&&\stackrel{\tau_j}{\swarrow}&&\stackrel{\tau_{j'}}{\searrow}\\
&&y_j && \stackrel{UFg}{\longrightarrow} && y_{j'}\\
&\stackrel{f_j}{\swarrow} &&&&&&\stackrel{f_{j'}}{\searrow}\\
y_j&&&&\stackrel{UFg}{\longrightarrow}&&&&y_{j'}
\end{matrix}
$$
The upper triangle is commutative because $\tau$ is a cone to the base
$UF$, and the lower trapezoid is commutative because $F$ is a
functor, and hence $Fg$ is an arrow $F_j\to F_{j'}$ in $D$. Hence the
outer triangle commutes, as required. From the universality of $\tau$,
it follows that there is a unique $f$ for which Diagram 1 is
commutative, and we have uniqueness.
For existence, we can take $f$ to be the unique arrow $x\to x$ for
which Diagram 1 is commutative, and we get a cone 
$$
\sigma=\{\sigma_j=\tau_j\colon (x\stackrel{f}{\to}x)\to F_j=(y_j\stackrel{f_j}{\to}y_j)\} 
$$
with $U\sigma=\tau$.  We claim that this is a limiting cone.
To see this, let $\alpha\colon(z\stackrel{g}{\to}z)\stackrel{.}{\to}F$
be a cone, so that 
for all $j$ the following diagram is commutative:
$$
\begin{matrix}
z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\
g\downarrow & & f_j\downarrow\\
z &\stackrel{\alpha_j}{\longrightarrow} & y_j.
\end{matrix} \quad\text{(Diagram 2)}
$$
Then $U\alpha$ is a cone $z\stackrel{.}{\to} UF$ in $C$, and by the
universality of $\tau$ there exists a unique arrow $h\colon z\to x$
for which the following diagram is commutative for all $j$:
$$
\begin{matrix}
z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\
h\downarrow& \stackrel{\tau_j}{\nearrow}\\
x& 
\end{matrix}\quad\text{(Diagram 3)}
$$
If this $h$ is an arrow $(z\stackrel{g}{\to}z)\to
(x\stackrel{f}{\to}x)$ in $D$, then we're done.  In other words, all
that remains to do is to show that the outer rectangle of the
following diagram is commutative:
$$
\begin{matrix}
z && \stackrel{h}{\longrightarrow} && x\\
& \stackrel{\alpha_j}{\searrow} && \stackrel{\tau_j}{\swarrow}\\
&& y_j\\ 
g\downarrow&& \downarrow f_j && \downarrow f\\
&& y_j\\
& \stackrel{\alpha_j}{\nearrow} && \stackrel{\tau_j}{\nwarrow}\\
z && \stackrel{h}{\longrightarrow} && x\\
\end{matrix}
$$
Now, the left trapezoid is just Diagram 2, the upper
and lower triangles are just Diagram 3, and the
right trapezoid is commutative for all $j$ by the definition of $f$.
It follows that both paths of the outer rectangle have the same
composition with the limiting cone $\tau$, and hence the outer rectangle
is commutative, as required.<|endoftext|>
TITLE: A non-commutative Radon-Nikodym derivative.
QUESTION [6 upvotes]: In this classic paper, Sakai proves the following Radon-Nikodym theorem:

Let $M$ be a von Neumann algebra, and let $\phi$ and $\psi$ be two normal positive linear functionals on $M$. If $\psi \leq \phi$, then there is a positive operator $t_0\in M$ such that $0 \leq t_0 \leq 1$, and $\psi(x) = \phi(t_0 x t_0)$ for all $x \in M$.

The paper provides no uniqueness result. One would naively expect that any two such operators $t_0$ and $t_1$ would satisfy $\phi((t_1-t_0)^2)=0$. I can find no such statement in the literature. Is this true?
Please note that $\phi$ is not assumed to be faithful.

REPLY [6 votes]: Such t_0 is unique if its support is at most p, where p is the support of ϕ.
Note that we can replace t_0 by pt_0p and the support of pt_0p is at most p.
Without this additional condition t_0 is highly non-unique, because
we can replace t_0 by t_0 + q,
where q is an arbitrary self-adjoint element with support at most 1-p such that t_0 + q ≥ 0.
Using simple algebraic manipulations one can show that all solutions can be obtained in this way.
See Lemma 15.4 (page 104) in
Takesaki's book “Tomita's theory of modular Hilbert algebras and its applications”. [Note that Takesaki implicitly assumes that φ_0 is faithful, hence you need to introduce
an additional condition on the support of h.]

One would naively expect that any two such operators t_0 and t_1 would satisfy ϕ((t_1−t_0)^2)=0. 

This is a trivial corollary of the above statement characterizing all possible solutions.<|endoftext|>
TITLE: Mathematically mature way to think about Mayer–Vietoris
QUESTION [36 upvotes]: This question is short but to the point: what is the "right" abstract framework where Mayer-Vietoris is just a trivial consequence?

REPLY [11 votes]: Here's an answer somewhat different from those already given. 
Associated to any homotopy pullback square, there's a long exact sequence of homotopy groups often called the Mayer-Vietoris sequence.  It comes from weaving together the long exact sequences for, say, the two vertical maps in the square, which have homotopy equivalent homotopy fibers.  (This weaving is a standard homological algebra exercise, and appears somewhere in Hatcher's book...)
Now, to build the Mayer-Vietoris sequence in cohomology for a CW complex X written as a union of subcomplexes $X = A\cup B$, just note that the homotopy pushout square formed by $A\cap B$, A, B, and X becomes a homotopy pullback square after applying Map(-, K(G, n)), where G is the coefficient group you're using. The Mayer-Vietoris homotopy sequence is now precisely the M-V sequence in cohomology.
(Annoyingly, for a fixed value of n this only gives you some of the sequence.)
It would be interesting to see a variant of this for homology, maybe using the infinite symmetric product?  I suppose the place to look would be the book by Aguilar-Gitler-Prieto, where homology is introduced entirely in terms of symmetric products.  The relevant bit seems to be missing from the Google preview.<|endoftext|>
TITLE: explicit big linearly independent sets
QUESTION [65 upvotes]: In the following, I use the word "explicit" in the following sense: No choices of bases (of vector spaces or field extensions), non-principal ultrafilters or alike which exist only by Zorn's Lemma (or AC) are needed. Feel free to use similar (perhaps more precise) notions of "explicit", but reasonable ones! To be honest, I'm not so interested in a discussion about mathematical logic. If no example is there, well, then there is no example. ;-)
Can you give explicit large linearly independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? For example, $\{\ln(p) : p \text{ prime}\}$ is such a set, but it's only countable and surely is no basis. You can find more numbers which are linearly independent, but I cannot find uncountably many. AC implies $\dim_\mathbb{Q} \mathbb{R} = |\mathbb{R}|$. Perhaps $ZF$ has a model in which every linearly independant subset of $ \mathbb{R}$ is countable?
The same question for algebraically independent subsets of $ \mathbb{R}$ over $\mathbb{Q}$? Perhaps the set above is such a subset? But anyway, it is too small.
Closely related problems: Can you give an explicit proper subspace of $ \mathbb{R}$ over $\mathbb{Q}$, which is isomorphic to $ \mathbb{R}$? If so, is the isomorphism explicit? Same question for subfields.
That would be great if there were explicit examples. :-)

REPLY [24 votes]: Here's an answer that's similar in spirit to Pietro Majer's, but not quite the same.
As a first step, choose an uncountable family of infinite subsets of positive integers such that any two distinct sets in the family have finite intersection. This can be done explicitly in many ways. One I like is as follows. Since one can take an explicit bijection between $\mathbb{N}$ and $\mathbb{Z}^2$, it's good enough to create a family of subsets of $\mathbb{Z}^2$ instead. And to do that, for each real number $\alpha\in[0,\pi)$ take the set of all points in $\mathbb{Z}^2$ that are within a distance 2 (say) of the line that makes an angle $\alpha$ with the x-axis.
Once we have such a family F, we define a real number $r_X$ for each X in F as follows. It is a number between 0 and 1 that has only 0s and 1s in its decimal expansion. And it has a 1 at the nth place if and only if $n=m^2$ for some $m\in X$. (The reason for restricting to the squares is simply that we want the gaps between successive places where there might be a 1 to get larger and larger, so that we can ignore carrying problems.)
The numbers $r_X$ are linearly dependent over $\mathbb{Q}$ only if we can find a non-zero integer combination of finitely many of them that gives zero. But we can't: if we've got a non-zero coefficient t, then after a while the gaps will be longer than the number of digits of t (or even of the sum of the absolute values of the coefficients, say), and we'll be able to find an element of the corresponding set $X\in F$ that belongs to none of the other sets, thereby proving that that integer combination is not zero.<|endoftext|>
TITLE: Nonexistence of determinantal functional equation for $\arccos$
QUESTION [7 upvotes]: Suppose I have distinct real numbers $a_i \in [-1,1]$,  $i \in [k]$. I want to choose real numbers  $b_j, j\in [k]$ such that the matrix $(\arccos(a_i b_j))_{i,j \in [k]}$ is nonsingular. 

Is this always possible?  Equivalently, does $\arccos$ not satisfy a functional equation of the form  $\det(\arccos(a_i x_j)) = 0$ for indeterminates  $x_j, j \in [k]$?

It seems like this should hold for any sufficiently ``non-algebraic'' function in place of  $\arccos$. Is there a theory that handles such questions?  Does functional analysis deal with these questions? I know very little functional analysis, so some general references may be helpful.

REPLY [12 votes]: A quick counter-example to the question as stated is $a_0=0$, $a_1=1$ $a_2=-1$. Since $2arccos(0)-arccos(b)-arccos(-b)=0$ for all $b$, we have $2M_1-M_2-M_3=0$ where $M_1, M_2, M_3$ are the rows of the matrix.
So it is better to assume that the $a_i$ are nonnegative. In this case, the answer is yes. More generally, consider this problem for a function $f$. You want to choose $n$ linearly independent vectors from the set
$$
X := \{ (f(a_1x), f(a_2x),\dots, f(a_nx)) \mid x\in\mathbb R \} \subset \mathbb R^n
$$
This is not possible if and only if $X$ lies in an $(n-1)$-dimensional subspace. This means that there are constants $c_1,\dots,c_n$ (not all zero) such that $c_1v_1+\dots+c_nv_n=0$ for all $v\in X$. Or, equivalently,
$$
 c_1 f(a_1x) + \dots c_n f(a_nx) = 0
$$
for all $x\in\mathbb R$ (such that all $a_ix$ belong to the domain of $f$). In other words, the functions $x\mapsto f(a_ix)$ are linearly dependent over $\mathbb R$.
This cannot happen if $f=arccos$. Indeed, assuming that $a_n$ is the maximum of $a_i$ such that $c_i\ne 0$, the above sum is well-defined and analytic on $[0,1/a_n)$ but its derivative goes to infinity as $x$ approaches $1/a_n$. Hence it is not constant on $[0,1/a_n)$, and hence non-constant in any neighborhood of 0.
UPDATE. A similar argument shows that the answer is the same for any non-polynomial function analytic near the origin (assuming $a_i>0$). Indeed, if $f(x)=\sum_{j=1}^\infty q_j x^{k_j}$ where $q_j\ne 0$, then Taylor expansion of the identity
$$
 c_1 f(a_1x) + \dots c_n f(a_nx) = 0
$$
implies that $\sum_i c_i a_i^{k_j}=0$ for all $j$. This cannot happen because, if $a_n$ is the maximum of $a_i$, the term $a_i^{k_j}$ grows faster (or decays slower) than all other terms as $k_j\to\infty$.<|endoftext|>
TITLE: Does category theory help understanding abstract algebra?
QUESTION [15 upvotes]: I'm studying category theory now as a "scientific initiation" (a program in Brazil where you study some subjects not commonly seen by a undergrad), but as I've never studied abstract algebra before, so it's hard to understand most examples and to actually do most of the exercises. (I'm using Mac Lane's Categories for the Working Mathematician and Pareigis Categories and functors.)
To solve this, my advisor recommended me to get S. Lang's Algebra as a reference, but I don't know if that's the most appropriate book and if it's better to get Lang and study algebra through category theory or to study (with a different book and approach, maybe Fraleigh) algebra and then category theory.
PS: I'll have to study by myself (with my advisor's help), as I can't enroll in the abstract algebra course without arithmetic number theory.

REPLY [9 votes]: There is the book Algebra: Chapter 0 by Paolo Aluffi that might fit your needs.
It is a textbook on algebra (as the title says), but it uses the language of category theory from the beginning. Category theory is mostly used to motivate definitions using universality properties.
It is only in the last two chapters that the author introduces more advanced concepts from category theory (functors, abelian categories, etc.).<|endoftext|>
TITLE: Nonseparable Hilbert spaces as quotients of spaces of bounded functions
QUESTION [10 upvotes]: Is the following result true: the Hilbert space $\ell^{2}\left(2^{\Gamma}\right)$ is a quotient of $\ell^{\infty}\left(\Gamma\right)$ for any
uncountable $\Gamma$ ? [I think it is, but cannot remember where I saw it, long time ago.] I would be very grateful for any (freely available, if possible) reference (Pelczynski ? Rosenthal ?).

REPLY [7 votes]: I don't know who first observed this (maybe Archimedes?) but it is true because $C(\{0,1 \}^\Gamma)$ is a quotient of $\ell_1^\Gamma$ and hence $\ell_1(2^\Gamma)$ embeds into $\ell_\infty(\Gamma)$.  
@Ady  
Here is a more serious answer to your question.  Take a quotient map $Q$ from $\ell_1(2^\Gamma)$ onto $C([0,1]^{2^\Gamma})$ and extend to a norm one mapping $T$ from $\ell_\infty(\Gamma)$ into some injective space $Z$ that contains $C([0,1]^{2^\Gamma})$ (you cannot extend $Q$ to an operator from  $\ell_\infty(\Gamma)$ into  $C([0,1]^{2^\Gamma})$ because, e.g., $C([0,1]$ is not a quotient of $\ell_\infty$).  Use partitions of unity to get a net $(P_a)$ of norm one finite rank projections on $Z$ taking values in $C([0,1]^{2^\Gamma})$ and whose restrictions to $C([0,1]^{2^\Gamma})$ converge strongly to the identity.  A weak$^*$ cluster point of $(P_a^* T^*)$ gives an isometric embedding of the dual of $C([0,1]^{2^\Gamma})$ (which contains $L_1([0,1]^{2^\Gamma})$) into the dual of $\ell_\infty(\Gamma)$.  Thus if $Y^*$ is any reflexive subspace of  $L_1([0,1]^{2^\Gamma})$, such as $\ell_2(2^\Gamma)$, then $Y$ is isometric to a quotient of $\ell_\infty(\Gamma)$.<|endoftext|>
TITLE: half-plane percolation clusters
QUESTION [9 upvotes]: Consider critical edge-percolation in the induced subgraph of the square grid with vertex set {$(i,j) \in Z \times Z:\ i+j \geq 0$}, and let $p_n$ be the probability that the cluster containing $(0,0)$ has size $> n$.  How quickly does $p_n$ fall as $n \rightarrow \infty$?

REPLY [2 votes]: In the case of oriented percolation the following regarding your
question are rigorously known in any dimension. Presumably these
results are also known for non-oriented percolation in half-spaces because in
this case it is also known that there is no infinite cluster at
criticality (Barsky, Grimmett and Newman 1991) and furthermore similar arguments apply  for a continuous time percolation model
where there is no orientation (i.e. time axis, see, Bezuidenhout and Grimmett 1991).
Let $\theta(p) = P(|C| = \infty)$,  where $C$ is the
cluster of the origin in oriented percolation with supercritical retention
parameter $p > p_{c}$. Further, let $\theta(p; \gamma) = 1 -
E_{p}(e^{\gamma |C|})$, $\gamma>0$. It is known that
(Aizenman and Barsky 1987, Bezuidenhout and Grimmett 1991 or Aizenman and Jung 1991) there exist
$a,b>0$ such that
\begin{equation}
\theta(p; \gamma) \geq a \gamma^{1 / 2}, \mbox{ for } 0<\gamma<b
\end{equation}
By means of emulating Tauberian theory arguments (due to Aizenmann and
Barsky (1989), see Grimmett's v. 1991 book, Proposition 10.29) and
using that $\theta(p_{c}) = 0$ (Bezuidenhout and Grimmett (1990)),
the last display above is tantamount to 
\begin{equation}
P( |C| \geq m) \approx m^{-\frac{1}{\delta}},\mbox{ } \delta \geq 2,
\end{equation}
where $\approx$ is meant in the logarithmic sence here ($a_{m} \approx b_{m}$ denotes $\frac{\log a_{m}}{\log b_{m}} \rightarrow 1$, as $m \rightarrow \infty$). Note that this implies that $P( |C| \geq m) \geq m^{-1 / 2}$, for all $m$ sufficiently large. 
In dimension 2 in particular the critical exponent for the probability of  $\{0 \leftrightarrow \partial B(n)\}$, that is the origin being connected to the boundary of a box of size $n$, is furthermore known. Durrett (1988) proves that $\lim_{n \rightarrow \infty} \sqrt{n}P(0 \leftrightarrow \partial B(n)) = +\infty$. 
Hope this helps.<|endoftext|>
TITLE: Importance of Log Convexity of the Gamma Function
QUESTION [23 upvotes]: The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies:
1) f(x+1) = x*f(x)
2) f(1) = 1
3) ln(f(x)) is convex
The Gamma function is meant to interpolate the factorial function, so I can see the importance of the first two properties. But why is log convexity important? How does it affect the Gamma function's applicability in other areas of mathematics?

REPLY [15 votes]: First, let me mention that log convexity of a function is implied by an analytic property, which appears to be more natural than log convexity itself. Namely, if $\mu$ is a  Borel measure on $[0,\infty)$ such that the $r$th moment 
$$f(r)=\int_{0}^{\infty}z^r d\mu(z)$$
is finite for all $r$ in the interval $I\subset \mathbb R$, then $\log f$ is convex on $I$.
Log convexity can be effectively used in derivation of various inequalities involving the gamma function (particularly, two-sided estimates of products of gamma functions).  It is linked with the notion of Schur convexity which is itself used in many applications.
An appetizer. Let $m=\max x_i$, $s=\sum x_i$, $x_i > 0$, $i = 1,\dots,n$, then
$$[\Gamma(s/n)]^n\leq\prod\limits_{1}^{n}\Gamma (x_i)\leq \left[\Gamma\left(\frac{s-m}{n-1}\right)\right]^{n-1}\Gamma(m).\qquad\qquad\qquad  (1)$$
(1) is trivial, of course, when all $x_i$ and $s/n$ are integers, but in general the bounds do not hold without assuming log convexity.
Edit added: a sketch of the proof. Let $f$ be a continuous positive function defined on an interval $I\subset \mathbb R$. One may show that the function $\phi(x)=\prod\limits_{i=1}^{n}f(x_i)$, $x\in I^n$ is Schur-convex on $I^n$ if and only if $\log f$ is convex on $I$. Thus the function 
$$\phi(x)=\prod\limits_{i=1}^n \Gamma(x_i),\quad x_i>0,\qquad \quad\qquad\qquad\qquad\qquad\qquad\quad   (2)$$ 
is Schur-convex on $I^n=(0,\infty)^n$. Since $x_i\le m$, $i=1,\dots,n$, and $\sum x_i=s$, it is easy to check that 
$$x \prec \left(\frac{s-m}{n-1},\dots,\frac{s-m}{n-1},m\right).$$
The latter majorization and the fact that $\phi(x)$ defined by (2) is Schur-convex imply the upper bound (1). The lower bound follows from the standard majorization $x\succ (s/n,\dots,s/n)$.

Have a look at the recent short article by  Marshall and Olkin concerning this and related inequalities for the gamma function.<|endoftext|>
TITLE: Is there a non-Gorenstein ring but locally Gorenstein?
QUESTION [7 upvotes]: A commutative noetherian ring $R$ is Gorenstein if $R$ has finite injective dimension.
Obviously, if $R$ is Gorenstein, then $R$ localized at any prime ideal $P$ is also Gorenstein. But I don't know whether the converse holds?

REPLY [15 votes]: In all of the standard references I know, Gorenstein is either only defined for local rings or a not necessarily local ring is defined to be Gorenstein if all of its localizations at maximal ideals are Gorenstein (which implies that its localizations at all prime ideals are Gorenstein).  So I am guessing you really want to ask: is there a commutative Noetherian ring $R$ all of whose localizations have finite injective dimension but such that $R$ itself does not?
I believe that the answer is "yes" and that a counterexample is given by Nagata's (in)famous example of a Noetherian ring of infinite Krull dimension.  See (5.96) in Lam's Lectures on modules and rings for an explanation why Nagata's example is regular, hence locally Gorenstein, i.e., locally of finite injective dimension.  
Furthermore, the proof of the Theorem at the bottom of p. 7 of 
http://www.math.hawaii.edu/~lee/homolog/Goren.pdf
states that the injective dimension of a ring is the supremum of the injective dimensions of its local rings $R_{\mathfrak{m}}$.   In this case each $R_{\mathfrak{m}}$ is regular, hence Gorenstein, hence its injective dimension is simply equal to its Krull dimension, i.e., to the height of $\mathfrak{m}$.  It follows that the injective dimension of $R$ itself is infinite.  
Conversely, Lee's handout contains a proof that the answer is "no" if $R$ is Noetherian of finite Krull dimension.<|endoftext|>
TITLE: Peculiar examples with Axiom of Countable Choice ?
QUESTION [12 upvotes]: I've been going over the extremely interesting discussions about Axiom of Choice.
It looks to me like all the "weird" consequences of AC (Banach-Tarski etc) come from using it on uncountable collections of sets.
If, instead, we only believe the Axiom of Countable Choice, do we still get unintuitive consequences in the same sense ?
Apologies in advance if the question is vague.

REPLY [13 votes]: It seems that many people who express a willingness to use countable AC are often actually thinking of arguments that use a somewhat stronger principle, the principle of Dependent Choices (DC). Both involve making only countably many choices, but the difference is that the DC principle allows one to make countably many choices in succession, so that the later choices can depend on the earlier choices. So the two principles are similar, but it turns out that DC is provably stronger than countable choice. It is DC and not just countable choice that one uses when choosing, for example, a nested sequences of closed balls in a metric space, since the choice of the later balls depends on the earlier-selected larger balls. It is the DC principle rather than mere countable AC that makes many arguments in analysis and measure theory work out. For example, I believe that many of the remarks made in favor of countable choice in the previous questions you mention can be construed as equally supportive of DC.
For this reason, it may be sensible to replace countable choice in your question with DC.
But it turns out that for statements about countable objects, and more generally, for statements expressible in L(R), the consequences of DC and AC are exactly the same. The reason is that if DC holds, then there is a forcing extension of L(R) adding no new reals, but adding a well-ordering of the reals. Thus, the forcing extension L(R)[G] satisfies ZFC with full AC, and has the same L(R) as the original universe. Another way to say it is that the theory ZF + AC is conservative over ZF + DC for statements about L(R), which includes all projective statements (e.g. those involving only quantification over the reals) and more. I give some fuller details in this related MO answer, to a question about DC and countable objects.
In particular, it follows from this that for countable objects, and more generally for statements about L(R), the principles DC and AC have exactly the same paradoxical statements. 
There is no need for large cardinals in this argument. Simon's point was that when there are large cardinals, then L(R) has highly regular features indeed, and so the classically paradoxical consequences of AC will not be found among statements about L(R). So this answer is a lower bound in the sense that Simon's is an upper bound.<|endoftext|>
TITLE: What is the earliest definition given by a universal mapping property?
QUESTION [7 upvotes]: As I study category theory, I'm finding the use of universal mapping properties in defining basic concepts to be both simple and clever.  Yet, the idea seems non-obvious enough that I expect quite a bit of mathematics had been done before the discovery of the technique.
What is the chronologically earliest abstract definition given by a universal mapping property?
Note that this question is not intended to be restricted to category theory.
Thank you!

REPLY [9 votes]: I very much agree with Pete's comment that we will find incipient instances of the universal mapping property among many classical constructions in mathematics, even if the original users of those concepts would not describe the idea in those terms. Indeed, I believe that these instances will stretch back through the whole of mathematics, and for this reason, there may be no definitive answer to the question. 
But let me anyway introduce a very early classical construction that we might agree has the hallmarks of a universal mapping property. To my mind, one of the fundamental essences of the UMP definitions is that they specify an object $U$ that relates to given objects $A,B,\ldots$ in a certain way, such that any other object $V$ relating to $A,B,\ldots$ in that same way then stands in a certain relation to $U$. This is the sense in which $U$ is universal with that property, and the particular details of the relations determine the nature of the universal property. 
My proposal is to consider the ancient idea of commensurability of line segments, appearing in Euclid's Elements and used earlier. Commensurability is of course intimately connected with the concept of greatest common divisor arising in the Euclidean algorithm, appearing in Books VII and X of Euclid's Elements. 
Specifically, two line segments $K$ and $L$ are comensurable when there is another line segment $R$ such that $K$ and $L$ are common multiples of $R$. The ancients knew not only that there was a largest such common measure $R$, but also that this largest common measure has a universal property: if $S$ is any other common measure of $K$ and $L$, then $R$ is a multiple of $S$. Thus, the largest common measure of commensurable line segments  is characterized by a universal property, known in antiquity. 
This fact is of course related to the fact, also known to the ancients, that the greatest common divisor $d$ of two natural numbers $a$ and $b$ is not only the greatest common divisor of $a$ and $b$, but has the universal property that any other common divisor of $a$ and $b$ is a divisor of $d$. Thus, the gcd of two integers is characterized by a universal property, also known in antiquity.<|endoftext|>
TITLE: Are $C$ and $\bar{Q_p}$ isomorphic?
QUESTION [16 upvotes]: There is a famous passage on the third page of Deligne's second paper on the Weil conjectures where he expresses his dislike of the axiom of choice, as manifested in the isomorphism between $C$ and $\bar{Q_p}$. The proof of said isomorphism runs as follows. Both $C$ and $\overline{Q_p}$ have transcendence bases, $S$ and $T$. Then $C\simeq \overline{Q(S)}$ and $\overline{Q_p}\simeq \overline{Q(T)}$. But $C$ and $\overline{Q_p}$ have the same cardinality, and hence, so do $S$ and $T$. Therefore, $Q(S)\simeq Q(T)$ and, from there, $C\simeq \overline{Q_p}$. 
For myself, this proof is quite convincing. Recently, Torsten Ekedahl expressed his opinion to the contrary, and this led to
the following exchange:
Why worry about the axiom of choice?
So I wondered about other expert opinions on this matter. Do you find the isomorphism unbelievable and, if so, why?

REPLY [11 votes]: I don't think one should try to determine whether to accept or reject the axiom of choice or any other independent axiom by appealing to "believability" of some consequences of it. With infinite sets, our intuition is just too often misleading. We get used to certain "paradoxa" like Hilbert's hotel because we see them very early in our mathematical life, but nobody should ever claim that he has a complete intuition for set theory.
As for the example, $\bar{\mathbf{Q}}_p$ and $\mathbf{C}$ are isomorphic if the axiom of choice is true, and that's that. Both are constructed using a completion, which makes them topological fields, and they are not homeomorphic or normed isomorphic, that's probably why it feels a bit wrong to some of us.<|endoftext|>
TITLE: Non-examples of model structures, that fail for subtle/surprising reasons?
QUESTION [44 upvotes]: An often-cited principle of good mathematical exposition is that a definition should always come with a few examples and a few non-examples to help the learner get an intuition for where the concept's limits lie, especially in cases where that's not immediately obvious.
Quillen model categories are a classic such case.  There are some easy rough intuitions—“something like topological spaces”, “somewhere one can talk about homotopy”, and so on—but various surprising examples show quite how crude those intuitions are, and persuade one that model categories cover a much wider range of situations than one might think at first.
However, I haven't seen any non-examples of model structures written up, or even discussed—that is, categories and classes of maps which one might think would be model structures, but which fail for subtle/surprising reasons.  Presumably this is because, given the amount of work it typically takes to construct an interesting model structure, no-one wants to write (or read) three-quarters of that work without the payoff of an actual example at the end. Has anyone encountered any interesting non-examples of this sort?

Background on my motivations: I'm currently working with Batanin/Leinster style weak higher categories, and have a problem which seems amenable to model-theoretic techniques, so I'm trying to see if I can transfer/adapt/generalise the model structures defined by Cisinski et al, Lafont/Métayer/Worytkiewicz, etc. in this area.  So I have some candidate (cofibrantly generated) classes of maps, and am trying to prove that they work; and there are lots of good examples around of how to prove that something is a model structure, but it would also be helpful to know what kinds of subtleties I should be looking out for that might make it fail to be.

REPLY [6 votes]: I've recently stumbled across another example, from 2017. It's in this paper by Wojciech Chacholski, Amnon Neeman, Wolfgang Pitsch, and Jerome Scherer. https://arxiv.org/abs/1702.05357v1
They are studying unbounded chain complexes over an abelian category $A$. For every injective class $I$ you can define a class of weak equivalences $f$ such that $A(f,W)$ is a quasi-isomorphism for all $W \in I$. You can then try to build resolutions, following Spaltenstein's work, via infinite towers where you truncate further and further to the left (i.e. more and more negative). If you have a model structure, you'll get these resolutions, but the authors get at them via "model approximations" instead, and give an example where even that fails.
In this paper, the authors formalize Spaltenstein's idea using a model category Tow($T$) of towers of successive truncations $\tau_{\leq -n}$. In Theorem 8.4 they give an example, using Nagata's "bad Noetherian ring," where the towers do not form a model approximation of $Ch(R)$. So, for this ring $R$, and for the choice of injective class in Section 8, $Ch(R)$ doesn't have a fibrant replacement functor built in the usual way. (and very likely, not one at all with respect to that injective class). The failure has to do with infinite Krull dimension, which I hope qualifies as "subtle and surprising."<|endoftext|>
TITLE: Tensor product of abelian categories
QUESTION [10 upvotes]: It seems the notion of tensor product of abelian categories exists naturally. 
Does someone know the reference of the construction?

REPLY [3 votes]: Yes Julian you are exactly right. In the paper you cite the Deligne product is extended to general (bi-)module categories in a way that reduces to the product of abelian categories in the case that the monoidal category involved is $Vec$. There it is also shown that certain classical formulas (adjunction of hom and tensor, for example) hold on the level of module categories. 
In an upcoming article the tensor product of module categories is studied from the perspective that it is a special case of a generalization of the center of a module category as introduced by Drinfeld and studied to great effect by Müger et. al., the so called relative center.<|endoftext|>
TITLE: Historical transition from classical homotopy to modern homotopy theory....... 
QUESTION [8 upvotes]: I plan to begin seriously studying model categories and thier applications to homotopy theory this summer. But I was hoping the topologists and historians in here could help me with something related:I was hoping to begin a push to get George Whitehead's ELEMENTS OF HOMOTOPY THEORY republished in a nice expensive edition for students. But looking at Whitehead's opus, it's amazing how alien most of it looks compared to the model categoric approach used today-long calculations with complexes and spectral sequences-and many problems were simply too difficult to attack directly.
My problem is if this work was republished, there really should be some historical context attached to it so students could transition from it to the more abstract methods today. (Sadly, Whitehead himself was planning a second volume detailing the model categorical approach-which was just beginning to become widely used in research at that point-and apparently he gave up attempting to compose it before be passed away.) 
  Does anyone know a good historical account of the transitional works between classical homotopy theory and the modern approach? I was hoping Whitehead's own "50 Years Of Homotopy Theory" would do the job and it would be perfect to bookend with the treatise,but it's not really about that. None of the review articles on model categories-like Dwyer,et.al.-really do this either. 
Can anyone outline historically this development for me?

REPLY [10 votes]: I am surely not a historian of topology, but I might try a few words.
That the usual literature concerning model categories is quite far away from traditional homotopy as presented in Whitehead's classic, is no wonder. Indeed, model categories are abstracted from homotopy theory, but not really that of the classical flavour. Quillen's lecture notes are not without reason entitled 'Homotopical Algebra'. As discussed in its introduction, its main object is to present an abstract framework where one can consider simplicial objects in categories of relevance for algebra. This leads to a theory of "non-additive derived functors", e.g. Andre-Quillen homology. 
In particular, the example of the model structure on topological spaces inducing the classical homotopy category is not presented in Quillen's book - only the one using Serre fibrations, generalized CW-complexes and weak homotopy equivalences. The model structure with Hurewicz fibrations/cofibrations and homotopy equivalences had to wait until Strom's The homotopy category is a homotopy category . As a consequence, the first absorbers of the theory of model categories were more simplicial minded guys. See for example Bousfield and Kan's Homotopy limits, completions and localizations. 
One reason, why the notion of a model category is today so omnipresent in algebraic topology is that they provided a very good framework to discuss the homotopy theory of spectra and it was important to work both simplicially and in topological spaces. But the model structure on topological spaces used here was again the Quillen model structure.
I think, it is only in the last years that topologists are caring more again to reunion classical homotopy theory and model categories. One important work for this is Cole's Mixing model structures . Here, a model structure on topological spaces is constructed, where the weak equivalences are again the weak homotopy equivalences, but fibrations are now the Hurewicz fibrations. This leads to a theory, where the cofibrant objects are all spaces homotopy equivalent to a CW-complex. This model structure interacts rather well with more classical homotopy theory (using Hurewicz cofibrations and so on) as is seen e.g. here or in (section 8 of) this, which is also used in the five-author paper Units of ring spectra and Thom spectra. The reason, why the latter needs the connection to more classical homotopy theory is that the theory of $E_\infty$-spaces stems from classical homotopy theory and is simultaneously deeply linked to modern stable homotopy theory.<|endoftext|>
TITLE: Isometric embedding of a positively curved polyhedral surface
QUESTION [5 upvotes]: Suppose you have a 2-dimensional polyhedral surface with specified lengths for the edges so that the vertices all have positive curvature. I believe this has a unique isometric embedding into 3-dimensional Euclidean space as the boundary of a convex polyhedron. Could someone confirm this? If so, is there a reasonable algorithm for finding the isometric embedding computationally?

REPLY [11 votes]: Yes, this is the historically first version of the Alexandrov embedding theorem which was earlier discussed here.  See the original Alexandrov's monograph (reviewed here), or see my book if you like downloadable stuff (at least for now).  
The algorithmic issue is difficult.  If you want the usual complexity, in my paper with Fedorchuk we showed that this requires finding roots of polynomials with exponential degrees.  The practical algorithm (based on a new completely different proof of the Alexandrov theorem) was found by Bobenko and Izmestiev here and further analyzed here.<|endoftext|>
TITLE: Textbooks on SINGULAR and Macaulay 2
QUESTION [7 upvotes]: Recently, I'm tired of those theoretical parts on commuative algebra. So I hope that someone could recommend me some good textbooks on SINGULAR and Macaulay 2. And I'm wondering whether SINGULAR is better that  Macaulay 2?
Thank you in advance!

REPLY [10 votes]: It is not true that Singular and Macaulay 2 use the same "computational engine (singular)"
As far as books are concerned - 
"Computational Algebraic Geometry'', by Hal Schenck is fantastic. Cambridge University Press, (2003).
http://www.amazon.com/Computational-Algebraic-Geometry-Mathematical-Society/dp/0521536502<|endoftext|>
TITLE: monodromy and global cohomology
QUESTION [6 upvotes]: Let $C$ be a compact Riemann surface, and let $U$ be a Zariski open subset in $C.$ Let $L$ be a local system (with coefficients $\mathbb C$ or $\mathbb Q_{\ell}$) on $U.$ For each point $z_i\in C-U,$ let $M_i$ be the monodromy matrix of $L$ around $z_i.$ If we identify $L$ as a representation of $\pi_1(U),$ and let $\gamma_i$ be a small loop in $U$ around $z_i,$ then $M_i$ is the image of $\gamma_i$ under the representation $\rho_L:\pi_1(U,a)\to GL(L_a)$ (where $a$ is a base point and $L_a$ is the fiber of $L$ at $a$). 
Here's my question. 
It seems to me that there should be some relation between (the traces of) these monodromies (as well as the monodromies around loops in $\pi_1(C)$) and (Betti numbers of) the cohomology groups $H^i_c(U,L)$ of $L$ with compact support, but I don't know the precise formulation or reference. 
As an example, if $C$ has genus $g$ and $L=\mathbb C$ is constant of rank 1, then $H^1_c(U,\mathbb C)$ has dimension $2g+n-1,$ where $n$ is the number of points at infinity. This number is the rank of $\pi_1(U),$ and each ``canonical generator" of $\pi_1(U)$ has trace 1.

REPLY [9 votes]: Suppose that $C = \mathbb P^1$ and $U = \mathbb P^1\setminus \{0,\infty\}.$  Then
$\pi_1(U)$ is cyclic, freely generated by a loop around $0$.
The local system $L$ is thus given by the vector space $L_a$, equipped with an invertible operator,
call it $T$, corresponding to the action of the generator of $\pi_1(U)$.  This operator $T$ is the monodromy matrix.
Now $H^1_c(U,L) = $ the space of $T$-invariants of $L_a$, while $H^2_c(U,L) = $ the space
of $T$-coinvariants of $L_a$.  If you think of the possible Jordan decompositions of 
$T$, and the fact that the trace is insensitive to the unipotent part, but just depends on
the semi-simple part, you'll see that the it's going to be hard to find any interesting relation of the type that you want.  (E.g. if $L_a$ is $n$-dimensional, and $T$ acts by the
identity, or by a maximally non-trivial unipotent element, in both cases the trace of $T$ is equal to $n$, but in the first case the Betti numbers are also both equal to $n$,
while in the second, they are both equal to $1$.)
You might also wonder about the Euler characteristic $H^2_c(U,L) - H^1_c(U,L)$, but this is always
equal to (rank $L$) $\cdot \chi(U)$, and so is insensitive to the monodromy matrices.<|endoftext|>
TITLE: How to compute irreducible representation of Lie algebra in the framework of BBD
QUESTION [8 upvotes]: We know Beilinson-Bernstein established the following famous equivalence:
$D-mod_{G/B}\rightarrow U(g)-mod_{\lambda}$,where $G$ is algebraic group and $B$ is Borel subgroup, $G/B$ is flag variety of finite dimensional Lie algebra $g$, $\lambda$ is central character.
This equivalence means that one can study representations of Lie algebra $g$ via D-modules. But How? 
My question:
Is there machinery in the framework of BBD to construct irreducible representations of $g$  explicitly? 
I am aware that there is Riemann-Hilbert correspondence to describe the correspondence between Perverse sheaves and holonomic D-modules. It seems that it is possible to know the irreducible objects in category of Perverse sheaves.(I guess in this case,we will know the irreducible representations corresponding to holonomic modules) But even in this case, I did not find appropriate reference. I wonder whether somebody compute some concrete examples such as flag variety of $sl_2$($P^1$). 
Further question: I also want to know the answers for affine Lie algebra case(Frenkel-Gaitsgory established analogue of BB-equivalence for critical level affine Lie algebra). Does this work give new class of irreducible representations of affine Lie algebra?
REMARK: What I want to know is the advantage to use D-module theory to construct representations.(if we can)For example, for some general Lie algebra $g$, we consider flag variety $X$. Then we consider $D-mod_{X}$, how to use algebraic geometric machinery on this category to construct irreducible representations of $g$ explicitly?(I would like to know is there any construction(in BBD)which can describe representations)

REPLY [6 votes]: Hello, I want to try to answer the first question you pose (at least I can answer the question I believe you are asking) -- sorry it's so far past when you asked it, but perhaps (if you know the answer yourself now) it will be useful to someone else to just have this information up anyway?  It sounds like you are asking "what is the D-module version of IC-sheaves?" In the BGG setting, Beilinson-Bernstein localization gives an equivalence between $B$-equivariant $D_{G/B}$-modules and highest weight $\mathcal{U}(\mathfrak{g})$-modules with trivial infinitesimal character. We can twist the D-module picture I am about to describe to obtain modules with other infinitesimal characters.  Irreducible $B$-equivariant $D_{G/B}$-modules are constructed as follows:  Let $Q$ be a $B$-orbit on $G/B$ and $i: Q\to G/B$ its inclusion.  Let $\tau$ be the structure sheaf of $Q$ (we can replace $B$ by more general groups in which case $\tau$ must range over all irreducible connections on $Q$, by which I mean specifically "line bundle with connection" so that we are working with $D_Q$-modules).  The D-module direct image $i_+ \tau$ has a unique irreducible submodule $L(\tau)$, and this construction in fact establishes a bijection between orbits $Q$ (more generally, pairs $(Q,\tau)$) and irreducible objects $L(\tau)$. Under Riemann-Hilbert, the $L(\tau)$ are precisely the IC sheaves, and upon taking global sections we recover irreducible representations.  The sheaves $i_+\tau$ have the co-Verma modules as their global sections and you can even recover the (co-)BGG resolution using D-module constructions alone (it's the Cousin resolution -- see Hartshorne's Residues and Duality). 
A good reference which has this all written down in the context of Harish-Chandra modules (and for arbitrary twists) is the paper "Localization and standard modules for real semi-simple Lie groups I: The duality theorem" -- the BGG version of the entire paper should work exactly the same but with $B$ instead of $K$ and "co-Verma" instead of "co-standard."<|endoftext|>
TITLE: Conservation of Hyperarithmetic Sentences over AC and CH.
QUESTION [5 upvotes]: I know that arithmetic sentences are conserved under the addition of the axiom of choice and the continuum hypothesis to ZF (i.e. ($ZF+AC \vdash \phi$ iff $ZF \vdash \phi$) and ($ZF+CH \vdash \phi$ iff $ZF \vdash \phi$) for arithmetic sentences $\phi$) Does this result extend to analytic sentences or other hyperarithmetic sentences? If so, how far does it extend up the hyperarithmetic hierarchy?

REPLY [7 votes]: First, let me remark that in your question, you can combine AC and CH together, rather than having two separate conservation results as you did. In fact, you can ramp CH up to GCH and more, including such principles as $\Diamond$ or others, without any problem. That is, the conservation result is that ZFC + GCH  proves $\varphi$ if and only if ZF proves $\varphi$, for a large class of statements $\varphi$, including the arithmetic statements, as you mentioned, but much more.
The phenomenon extends completely up the hyperarithmetic hierarchy and beyond, beyond even the analytic sentences up into the projective hiearchary at the level of $\Sigma^1_2$. (In this hiearchy, the hyperarithmetic statements are $\Delta^1_1$ and the analytic statements $\Sigma^1_1$.) 
This absoluteness result is the content of the Shoenfield Absoluteness Theorem, which asserts that any $\Sigma^1_2$ statement is absolute between between any two transitive models of set theory $V\subset W$ having the same ordinals. In particular, a $\Sigma^1_2$ statement holds in the universe if and only if it holds in the constructible universe $L$, where both AC and GCH hold. 
Thus, the $\Sigma^1_2$ statements provable in ZFC+GCH are the same as those provable in ZF.<|endoftext|>
TITLE: What are the interesting cases of the generalized Korteweg-de Vries equation?
QUESTION [6 upvotes]: The generalized Korteweg-de Vries equation is
$u_t + u_{xxx} + (u^p)_x=0$
for integer $p$.  (The original Korteweg-de Vries equation is the case $p=2$.)  I need to understand solutions for $p=1$, but I haven't been able to find this case addressed in the literature despite extensive searching.  Is there a reason why $p=1$ isn't interesting?  Does it reduce to a simpler form?  Is it intractable?
If not, what the best way to learn about this case?
Thanks!

REPLY [8 votes]: For $p=1$ your equation indeed reduces to a simpler form. Put $X=x-t, T=t$. Then $u_t=u_T-u_X$, $u_x=u_X$, and hence in the new independent variables $X,T$ your equation (with $p=1$) becomes a well-studied linear third-order equation
$$
u_{T}+u_{XXX}=0,
$$
which is, inter alia, a linearized version of the "standard" KdV equation with $p=2$,
see e.g. the discussion in this book by Ablowitz and Segur (and cf. also Willie's answer).
Note that the case of $p=3$ (the modified KdV equation) is also quite interesting. It is integrable by the inverse scattering transform just as the "usual" KdV and, in fact, is related to it through the Miura transformation (see e.g. the above book for details).<|endoftext|>
TITLE: A decision problem concerning polynomial rings
QUESTION [13 upvotes]: Let $f_1,f_2, \ldots ,f_n$ be polynomials in any number of variables with algebraic coefficients. Is there algorithm to determine whether the ring $\mathbb{Z}[f_1,f_2,\ldots ,f_n]$ contains a non-constant  polynomial with integer coefficients? 
To illustrate what the game is here, I'll give some examples:

$\mathbb{Z}[\sqrt{2}x-y]\cap \mathbb{Z}[x,y]=\mathbb{Z}$. Proof-sketch: Suppose that $H\in\mathbb{Z}[u]$ and $H(\sqrt{2}x-y)\in\mathbb{Z}[x,y]$. Choose a sequence of distinct points $(x_n,y_n)\in\mathbb{Z}^2$ such that $\sqrt{2}x_n-y_n$ converges. Then $H(\sqrt{2}x_n-y_n)$ is eventually an integer constant $c$. Thus the equation $H(u)=c$ has infinitely many solutions. It follows that $H$ is identically $c$
$\mathbb{Z}[\sqrt{2}x-y, 2\sqrt{2}xy-z]$ contains the polynomial $2x^2+y^2-z$. (Take the square of the first polynomial plus the second.) Note that the generators are algebraically independent.
$\mathbb{Z}[\sqrt{2}x-y, \sqrt{3}xy-z]\cap \mathbb{Z}[x,y,z]=\mathbb{Z}$. (I know only a rather lengthy and ad hoc proof)

REPLY [3 votes]: Let $\alpha \in \overline{\mathbb{Q}}$ be such that all $f_i$ have coefficients in $\mathbb{Q}(\alpha)$ and $k \in \mathbb{N}$ such that the $f_i$ are in 
$\mathbb{Q}(\alpha)[y_1, \ldots, y_k]$. Then the ideal the generated by the $f_i$ in this ring corresponds to an ideal $J$ in $\mathbb{Q}[x,y_1, \ldots, y_k]$ which is generated by lifts of the $f_i$ together with the minimal polynomial $f$ of $\alpha.$
Now clearly a neccessary condition for your problem is that $J \cap \mathbb{Q}[y_1 \ldots, y_k]\neq \{0\}.$ This can be checked algorithmically by computing a Groebner basis with respect to an elimination term ordering, c.f. e.g. Kreuzer/Robbiano, Computational Commutative Algebra. 
For a complete solution to the original question, you are not interested in the ideal generated by the polynomials but in the subalgebra they generate. Again, sometimes it"s possible to compute the gadgets corresponding to Groebner bases for ideals; these then are called SAGBI bases, which you would need to compute with respect to an elimination term ordering. In contrast to Groebner bases, these do not neccessarily exist, though.
To summarise: SAGBI basis is the notion you're looking for, if I'm not mistaken.<|endoftext|>
TITLE: Translation of Goldbach's 1742 letter to Euler
QUESTION [17 upvotes]: This ought to be a simple one to answer. Does anyone know of, or can anyone provide, an accurate English translation of the marginal remarks in Goldbach's letter to Euler
http://upload.wikimedia.org/wikipedia/commons/1/18/Letter_Goldbaxh-Euler.jpg
in which a statement equivalent to the Goldbach conjecture is first stated?

REPLY [14 votes]: I have an English translation of the letter; you can find my email address on my homepage. 
Here is the relevant part:
"By the way, I take it to be not useless to note down also such propositions
that are very probable, even if a real proof is lacking; for even if
afterwards they were found to be erroneous, they could all the same give
occasion for the discovery of a new truth. Thus Fermat's idea that all the
numbers $2^{2^{n-1}}+1$ yield a series of prime numbers cannot hold up, as
you already demonstrated, Sir; but it should still be remarkable if this
series were composed only of numbers that could be split into two squares in
a unique way.  I should like to risk another conjecture of that kind: any number 
composed from two primes is the sum of as many prime numbers (including $1$) as one wishes, right down to the sum that consists just of ones.
After reading this through again, I see that the conjecture can be proved 
quite rigorously for the case $n+1$ if it holds for the case $n$ and 
$n+1$ can be split into two prime numbers. The proof is very easy;
and at least it appears to be true that every number greater than 
$2$ is the sum of three prime numbers."
The translation was done by Martin Mattmüller.

REPLY [2 votes]: I do not have the reputation to comment on the answer, so I have to start a new one.
Freely translating page 127 of http://www.math.dartmouth.edu/~euler/correspondence/letters/OO0765.pdf :

I deem it to be advantageous to note
  the following conjecture, even though
  it lacks a proof, as a counterexample
  could provide further insights.
  Fermat's idea that every number of the
  form $2^{2^{n-1}}+1$ is prime can, as
  you have shown, not be true; but it
  would be strange if this series
  yielded a lot of "numeros unico modo
  in in duo quadratis divisibiles"
  (numbers that can be divided into two
  squares???). I, too, would like to
  hazard a conjecture: that every number
  that is the sum of two primes is the 
  sum of arbitrary numbers of primes (or 1),
  except the "congierem omnium unitatum" 
  (the collections of all in one???; the footnote
  was already translated by Mark), for example...<|endoftext|>
TITLE: An explicit example of a finitely presented group containing a subgroup isomorphic to $(\mathbb Q,+)$.
QUESTION [37 upvotes]: A theorem (I do unfortunately not remember to whom it is due) states that there exists a finitely presented group containing a subgroup isomorphic to the additive group of rational numbers. Can somebody give an explicit construction?

REPLY [6 votes]: Here's an answer after reading Jim's answer. Indeed, given the previous knowledge on finite order elements in Thompson's group $T$ it seems we were very close to the answer.
First, a general fact:
 Proposition. Let $G$ be a torsion-free group with central cyclic subgroup $Z$. Suppose that $G/Z$ is isomorphic to $\mathbf{Q}/\mathbf{Z}$. Then $G$ is isomorphic to $\mathbf{Q}$.
Proof, since $G$ is central-by-(locally cyclic), it is abelian. It necessarily has $\mathbf{Q}$-rank 1, so has an injective homomorphism $i$ into $\mathbf{Q}$. This induces an injection $\bar{i}$ from $G/Z\simeq\mathbf{Q}/\mathbf{Z}$ into $\mathbf{Q}/i(Z)$, which is itself isomorphic to $\mathbf{Q}/\mathbf{Z}$. Since every injective endomorphism of $\mathbf{Q}/\mathbf{Z}$ is an automorphism, we deduce that $\bar{i}$ is an isomorphism, and hence so is $i$.
Now let us consider finite order elements. First recall that in $G=\mathrm{Homeo}^+(\mathbf{R}/\mathbf{Z})$, every element of finite order is conjugate to a (finite order) rotation (=translation $r\mapsto r+\theta$); the rotation number $\theta$ is a conjugacy invariant. Thus there are $\varphi(n)$ conjugacy classes of elements of order $n$, and single conjugacy class of cyclic subgroups of order $n$.
I find in this 2012 unpublished paper of A. Fossas "On the number of conjugacy classes of torsion elements on Thompson's group T" that the same counting holds in $T$ (she asserted it for elements, but the simpler assertion for cyclic subgroups immediately follows). Let me emphasize this:
 Theorem. In Thompson's group $T$, there is exactly one conjugacy class of cyclic subgroup of order $n$, for every integer $n\ge 1$.
An immediate corollary is that if $m|n$, every cyclic subgroup of order $m$ is contained in a cyclic subgroup of order $n$. In this way, we can produce an ascending sequence $(H_n)$ of cyclic subgroups with $H_n$ of order $n!$. Hence its union is isomorphic to $\mathbf{Q}/\mathbf{Z}$. Now the inverse image $\tilde{T}$ of $T$ in the group of $\mathbf{Z}$-equivariant self-homeomorphisms of $\mathbf{R}$ is a torsion-free central extension of $T$ and the proposition applies:
Corollary. The finitely presented group $\tilde{T}$ contains an isomorphic copy of $\mathbf{Q}$.
Moreover when we produce the embedding $H_{n}\subset H_{n+1}$ with $n\ge 1$, we readily see that there exists an element in $T$ centralizing $H_n$ but not normalizing $H_{n+1}$. Conjugating by this element, we see that we have at least two choices for $H_{n+1}$. Hence we obtain $2^{\aleph_0}$ choices of such sequences, and hence $2^{\aleph_0}$ subgroups of $\tilde{T}$ isomorphic to $\mathbf{Q}/\mathbf{Z}$, and, pulling back, $2^{\aleph_0}$ subgroups of $T$ isomorphic to $\mathbf{Q}$ (as obtained by Belk, Hyde and Matucci). In particular, these copies are not all conjugate.
In addition let me mention that it is easy to find one copy of $\mathbf{Q}$ with solvable membership problem, just being careful when choosing roots (of course at most countably many of those copies have solvable membership problem).

Note: Fossas' short paper on uniqueness up to conjugation is unpublished. I have double checked (in my own way) and am convinced it's true.
I'm far from an overview of the past knowledge on classifying finite order elements in Thompson's group $T$. Let me mention this 2008 paper by Liousse "Rotation numbers in Thompson-Stein groups and applications", which already includes (see Cor. 1) the fact that $T$ admits elements of all possible orders.
All this retrospectively looks very easy. But I have to confess that I hadn't even realized so far (before Jim's answer) that $T$ has a element of order $3$...!<|endoftext|>
TITLE: Does the Grothendieck group depend on the embedding?
QUESTION [6 upvotes]: This might turn out to be a silly question, but here goes.
Let $\mathcal{C}$ be a full additive subcategory of an abelian category $\mathcal{A}$. I'm wondering if the Grothendieck group $K(\mathcal{C})$ (defined below) depends on $\mathcal{A}$. I can't think of any examples, and I believe that it does not. Something with Yoneda embeddings (see Weibel) comes to mind, but I can't really get it precise. 
Idea. Given two embeddings $\mathcal{C}\subset\mathcal{A}_1$ and $\mathcal{C}\subset \mathcal{A}_2$, I'm guessing it suffices to find a third one containing these.
Def. 
Let $\textrm{Ob}(\mathcal{C})$ denote the class of objects in $\mathcal{C}$ and let $\textrm{Ob}(\mathcal{C})/\cong$ be the set of isomorphism classes. Let $F(\mathcal{C})$ be the free abelian group on $\textrm{Ob}(\mathcal{C})/\cong$. To any sequence $$(E) \ \  0 \longrightarrow  M^\prime \longrightarrow M \longrightarrow M^{\prime\prime} \longrightarrow 0 .$$ in $\mathcal{C}$, which is exact in $\mathcal{A}$, we associate the element $Q(E) = [M] - [M^\prime] - [M^{\prime\prime}]$ in $F(\mathcal{C})$. Let $H(\mathcal{C})$ be the subgroup generated by the elements $Q(E)$, where $E$ is a short exact sequence. We define the Grothendieck group, denoted by $K(\mathcal{C})$, as the quotient group $$K(\mathcal{C}) = F(\mathcal{C})/H(\mathcal{C}).$$
(All categories are assumed to be at least skeletally small in this definition, but ok.)

REPLY [9 votes]: The abelian category $\mathcal A$ is inducing extra structure on $\mathcal C$, namely, 
it singles out a certain class of short exact sequences in $\mathcal C$ (namely the sequences that are short exact in $\mathcal A$), which (under additional mild assumptions) give
$\mathcal C$ the structure of an exact category.  The $K_0$ is then an invariant of the exact category (i.e. $\mathcal C$ together with this chosen class of exact sequences).<|endoftext|>
TITLE: Uniformization in Descriptive Set Theory
QUESTION [6 upvotes]: I'm a beginner in descriptive set theory.
There is a series of connected exercises (1C.6, 1C.7, 1C.8) in Moschovakis' classic text (new edition) on uniformization. They are simple case uniformization, reduction (by uniformization), and separation of sets (via reduction).
It seems like with such layout, the point of uniformization is to give us a way to separate sets.
My question is, why are we interested in uniformization and reduction? What I can see is that uniformization can always be carried out with AC, pretty obvious. So, has it anything to do with Choice?
What about reduction?

REPLY [4 votes]: One reason to be interested in uniformization is a very nice application of the Lusin–Novikov Uniformization Theorem to the theory of countable Borel equivalence relations. Here an equivalence relation $E$ on a Polish space is said to be Borel if $E$ is a Borel subset of $X \times X$ and $E$ is said to be countable if every $E$-class is countable. For example, if a countable discrete group $G$ has a Borel action $(g,x) \mapsto g \cdot x$ on a Polish space $X$, then the corresponding orbit equivalence relation $E_{G}^{X}$
is countable Borel. Using the Lusin–Novikov Uniformization Theorem, Feldman-Moore proved that every countable Borel equivalence relation arises in this manner; namely, if $E$ is a countable Borel equivalence relation on the Polish space $X$, then there exists a countable group $G$ with a Borel action on $X$ such that $E = E^{X}_{G}$.
Of course, if $E$ is an arbitrary equivalence relation on $\mathbb{R}$ with countable classes, then the Axiom of Choice implies that there is an action of $\mathbb{Z}$ on $\mathbb{R}$ which realizes $E$ as its orbit equivalence relation. The point of the Feldman-Moore Theorem is that if $E$ is Borel, then $E$ can be realized by a Borel action. It is interesting to note that there are countable Borel equivalence relations which cannot be realized by Borel actions of $\mathbb{Z}$.<|endoftext|>
TITLE: A split short exact sequence of algebraic fundamental groups
QUESTION [7 upvotes]: If we have a variety, $X$, over a field, $k$, and $x$ is a geometric point of $X$, and let $\bar x$ be a geometric point of $X_{k^s} := X \times_k k^s$ above $x$ then we have the following short exact sequence:
$1 \rightarrow \pi_1(X_{k^s}, \bar x) \rightarrow \pi_1(X,x) \rightarrow Gal(k) \rightarrow 1$
Implicit in this is a choice of $k^s$ (if you want, this is a choice of geometric point, $z$, on $Spec(k)$; $\pi_1(Spec(k), z)=Gal(k)$).
I'm wondering how to interpret the splitting of this short exact sequence, and more specifically: what is the significance of choosing different splittings? I'm having a hard time picturing intuitively how to think of this splitting.

REPLY [2 votes]: For me, philosophically, the splitting of the short exact sequence means that etale coverings of $X$ basically come in two flavors: Geometric coverings (classified by $\pi_1(X_{\bar{k}})$ which is sometimes also called "geometric fundamental group of $X$") and arithmetic coverings (classified by $Gal(k)$). All coverings can be obtained by "combining" geometric and arithmetic coverings.
Another similar interpretation is the following: By passing to the limit over all galois coverings of $X$ (more precisely, over the system of pointed galois coverings of $(X,x)$) one obtains a universal covering scheme $\hat{X}$. As a set, the fiber over the base point is the profinite set $\pi_1(X,x)$! Similarly one can construct the universal covering of $X_{\bar{k}}$ and of $Spec(k)$ (which is just $Spec(k^{sep})$. The fibers over the fixed base point of these covering schemes are $\pi_1(X_{\bar{k}})$ and $Gal(k)$ respectively (as sets). The splitting of the short exact sequence now gives information about the fiber of the universal covering of $X$ in terms of points coming from the fibers of $X_{\bar{k}}$ and $Spec(k)$.<|endoftext|>
TITLE: What is known about this plethysm?
QUESTION [31 upvotes]: Let $S^{\lambda}$ be a Schur functor. Is there a known positive rule to compute the decomposition of $S^{\lambda}(\bigwedge^2 \mathbb{C}^n)$ into $GL_n(\mathbb{C})$ irreps?

In response to Vladimir's request for clarification, the ideal answer would be a finite set whose cardinality is the multiplicity of $S^{\mu}(\mathbb{C}^n)$ in $S^{\lambda}(\bigwedge^2 \mathbb{C}^2)$. As an example, the paper Splitting the square of a Schur function into its symmetric and anti-symmetric parts gives such a rule for $\bigwedge^2(S^{\lambda}(\mathbb{C}^n))$.
Formulas involving evaluations of symmetric group characters, or involving alternating sums over stable rim hooks, are not good because they are not positive.
And, yes, it is easy to relate the answers for $\bigwedge^2 \mathbb{C}^n$ and $\mathrm{Sym}^2(\mathbb{C}^n)$, so feel free to answer with whichever is more convenient.

REPLY [7 votes]: Let $V = \mathbb{C}^n$ where $n$ is sufficiently large. This paper by Melanie de Boeck and Rowena Paget determines the constituents of $S^\lambda (\mathrm{Sym}^2 V)$ when $\lambda$ has either two rows, or two columns or is a hook partition of the form $(k-r,1^r)$. Since $S^\mu (V)$ appears in $S^\lambda (\mathrm{Sym}^2 V)$ if and only if $S^{\mu'}$ appears in $S^\lambda (\bigwedge^2V)$, these results apply to the question.
Explicit positive formulae are given for the multiplicities of irreducible consituents of $S^{(k-1,1)}(\mathrm{Sym}^2 V)$, $S^{(2,1^{k-2})}(\mathrm{Sym}^2V)$, $S^{(k-2,2)}(\mathrm{Sym}^2 V)$ and $S^{(k-2,1^2)}(\mathrm{Sym}^2 V)$. These results give a complete answer to the question in three new cases.
For example, Corollary 3.2 states that if $\mu$ is a partition of $2k$ then $S^\mu V$ appears in $S^{(k-1,1)}(\mathrm{Sym}^2 V)$ if and only if either $\mu$ has  only even parts, or $\mu$ has exactly two odd parts of distinct sizes. In the latter case the multiplicity is $1$, in the former case the multiplicity is one less than the number of distinct part sizes of $\mu$.
Edit. Say that $S^\lambda(V)$ is a minimal constituent of a polynomial $\mathrm{GL}(V)$-module $W$ if $S^\lambda(V)$ appears in $W$ and $\lambda$ is minimal with this property. Define maximal constituent analogously. Let $m \in \mathbb{N}$. This paper by Rowena Paget and me characterizes, in terms of certain tuples of families of $m$-subsets of $\mathbb{N}$, all partitions $\mu$ such that $S^\mu$ is a minimal  constituent of $S^\lambda(\mathrm{Sym}^m(V))$. 
There is an analogous characterization of the maximal constituents of $S^\lambda(\mathrm{Sym}^m(V))$ by replacing sets with multisets.
To give a very small example, the minimal constituent $S^{(4,3,1)}(V)$ of $S^{(1^4)}({\mathrm{Sym}^2(V)})$ corresponds to the family of $2$-sets $\bigl\{ \{1,2\}, \{1,3\}, \{2,3\}, \{1,4\} \bigr\}$ of multidegree $(4,3,1)' = (3,2,2,1)$.
These results give a practical sufficient condition on a partition $\nu$ for $S^\nu(V)$ to have multiplicity zero in $S^\lambda(\mathrm{Sym}^2(V))$, so are also relevant to the question.<|endoftext|>
TITLE: Set Theory and V=L
QUESTION [10 upvotes]: From http://en.wikipedia.org/wiki/Analytical_hierarchy
"If the axiom of constructibility holds then there is a subset of the product of the Baire space with itself which is $\Delta^1_2$ and is the graph of a well ordering of the Baire space. If the axiom holds then there is also a $\Delta^1_2$ well ordering of Cantor space."
Can someone (give here or point me to) a (sketch or thorough description) of a $\Delta^1_2$ set that does this for (Baire space or Cantor space)?
I can see how V=L implies there is a definable well order, but I can't see how it would be in the analytical hierarchy.

REPLY [16 votes]: In the constructible universe $L$, there is a definable well-ordering of the entire universe. This universe is built up in transfinite stages $L_\alpha$, and the ordering has $x\lt_L y$ when $x$ is constructed at an earlier stage, or else they are constructed at the same stage, but $x$ is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. I also explain this in this MO answer.
One may extract from this definition a rather low-complexity definable well-ordering of the reals by capturing the countable pieces of the $L$ hieararchy by reals. That is, if $x$ is a real number of $L$, then it appears at some countable stage $L_\alpha$ for a countable ordinal $\alpha$, and the entire structure $L_\alpha$ is countable, and hence itself coded by a real. Here, we code a set by a real in any of the standard ways, for example, by coding a well-founded extensional relation on $\omega$ whose Mostowski collapse is the given set. Furthermore, the $L$-order is absolute to any $L_\alpha$, since $L_\alpha$ knows about the $L_\beta$-heirarchy for $\beta<\alpha$. Also, if a countable structure is well-founded and thinks $V=L$, then it is $L_\alpha$ for some $\alpha$. Note that if a real $z$ codes a first order structure $M$, then the question of whether $M$ satisfies a first order assertion is an arithmetic statement in $z$, since we need only quantify over the coded elements, which are coded by natural numbers. 
Putting all this together, we get that the following are equivalent for any two reals $x$ and $y$:

$x\lt_L y$ in the $L$ order. 
There is some countable ordinal $\alpha$ such that $L_\alpha$ satisfies $x\lt_L y$.
For every countable ordinal $\alpha$, if $x$ and $y$ are reals in $L_\alpha$, then $L_\alpha$ satisfies $x\lt_L y$. 
There is a real $z$ coding a well-founded structure that thinks $V=L$ (and so this structure must be some $L_\alpha$) in which $x$ and $y$ are reals and the structure satisfies $x\lt_L y$. 
All reals $z$ coding well-founded structures $L_\alpha$ in which $x$ and $y$ are reals satisfy $x\lt_L y$. 

The fourth statement has complexity $\Sigma^1_2$, since being-well-founded is $\Pi^1_1$. Similarly the fifth statement has complexity $\Pi^1_2$, so overall the ordering is $\Delta^1_2$. 
The end result is that in the universe $L$, there is a low-complexity definable well-ordering of the reals. In this universe, therefore, all of the supposedly non-constructive applications of AC turn out to be completely definable.<|endoftext|>
TITLE: Applications of algebraic geometry over a field with one element
QUESTION [23 upvotes]: I would like to understand at least one of the several existing approaches to algebraic geometry over $\mathbb{F}_1$ (the field with one element). Is there an example of an "interesting" theorem that can be formulated purely in the language of ordinary schemes, but which can be proved using algebraic geometry over $\mathbb{F}_1$?
Of course, the interpretation of the word "interesting" is entirely up to your own taste. An example in which the theorem cannot be proved using "classical" methods would be most desirable, but examples where (one of) the theories of schemes over $\mathbb{F}_1$ gives an alternative proof of an already known result would also be very much appreciated.
[On a related note, perhaps there should be a tag "naive-question" for situations like this one.]

REPLY [11 votes]: To elaborate on the possible connection with the Riemann hypothesis:
One of the parts of the Weil conjectures (which were proved by Deligne, and which also follow from Grothendieck's "standard conjectures") states:

If $X$ is a smooth projective variety of dimension $n$ over $\mathbb{F}_q$, then its zeta function $Z_X(t)$ has the form $$\frac{P_1(t) P_3(t) \cdots P_{2n-1}(t)}{P_0(t)P_2(t) \cdots P_{2n}(t)},$$ where each $P_i(t)$ has the form $\prod_j (1-\alpha_{ij}t)$, and such that each $\alpha_{ij}$ is an algebraic integer such that $|\alpha_{ij}| = q^{i/2}$.

The zeta function $Z_X(t)$ of a variety $X$ is defined to be $\exp(\sum N_r t^r/r)$, where $N_r$ is the number of $\mathbb{F}_{q^r}$-points of $X$.
Write $\zeta_X(s) = Z_X(q^{-s})$. Then the above statement implies that the zeros and poles of $\zeta_X(s)$ lie on the lines given by $\mathfrak{Re}(s) = i/2$ with $i=1, \dots, 2n$. The function $\zeta_X(s)$ is also equal to $$\prod_{x \in X} \frac{1}{1-|k(x)|^{-s}}$$ where $|k(x)|$ is the order of the residue field of $x$ (which is always finite).
Now if $X = \operatorname{Spec} \mathbb{Z}$ were somehow a smooth projective variety of dimension 1 over $\operatorname{Spec} \mathbb{F}_1$, then $\zeta_X(s)$ would be the Riemann zeta function, and...<|endoftext|>
TITLE: A problem on finiteness of Ext
QUESTION [5 upvotes]: If $R$ is a commutative noetherian ring and $I$ is an ideal of $R$, $M$ is an $R$-module. Does $Tor_i^R(R/I, M)$ is finitely generated for $i\ge 0$ imply $Ext^i_R(R/I, M)$ is finitely generated for $i\ge 0$?

REPLY [9 votes]: In fact, the two are equivalent.  My apologies for the length of this argument - if someone else has a shorter one, I'd be happy to hear it.
Let $(a_1,\ldots,a_k) = I$, and let $K_\bullet$ be the Koszul complex  associated to this set of generators.  Note that its zero'th homology group is $R/I$, and all the homology groups are finitely generated $R/I$-modules because $R$ is Noetherian.
Let C be a Serre class of R-modules (i.e. one such that for any $0 \to A' \to A \to A'' \to 0$ exact, $A$ is in $C$ if and only if $A'$ and $A''$ are both in C).  The result you ask is obtained by letting C be the class of finitely generated $R$-modules (which is only a Serre class because $R$ is Noetherian).  We have that the following are equivalent for an R-module M:

$Tor_i(R/I,M)$ is in C for all values of i.
$Tor_i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.
$H_i(P \otimes_R M)$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.
$H_i(K \otimes_R M)$ is in C for all i.

The implication 1 => 2 follows inductively by writing $N$ in a short exact sequence $0 \to J \to \oplus R/I \to N \to 0$ and applying the long exact sequence of Tor.
The implication 2 => 3 follows from a hyperhomology spectral sequence.
The implication 3 => 4 is immediate from the definition of the Koszul complex.
The implication 4 => 1 is proved inductively.  The hyperhomology spectral sequence
$$
E^2_{p,q} = Tor_p(H_q(K), M) \Rightarrow H_{p+q}(K \otimes_R M)
$$
first shows $E^2_{0,0} = Tor_0(R/I,M)$ is in C.  If $Tor_i(R/I,M)$ is in C for $0 \leq i \leq m$, the above argument implies that $Tor_i(N,M)$ is in C for all finitely generated $N$, which forces $E^2_{p,q}$ to be in C for all $p \leq m$.  As the abutment is in C, this forces $E^2_{m+1,0} = Tor_{m+1}(R/I,M)$ to be in C.
Now, there is an exactly analogous string of implications in Ext.  The following are equivalent:

$Ext^i(R/I,M)$ is in C for all values of i.
$Ext^i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.
$H^i(\underline{Hom}_R(P,M))$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.
$H^i(\underline{Hom}_R(K,M))$ is in C for all i.

However, the Koszul complex has self-duality; the tensor product complex $K \otimes_R M$ visibly has the same homology groups as a shift of the Hom-complex $\underline{Hom}_R(K,M)$.  Therefore, the two versions of statement (4) are equivalent.<|endoftext|>
TITLE: Reference for the Gelfand duality theorem for commutative von Neumann algebras
QUESTION [36 upvotes]: The Gelfand duality theorem for commutative von Neumann algebras states that the following three categories are equivalent:
(1) The opposite category of the category of commutative von Neumann algebras;
(2) The category of hyperstonean spaces and hyperstonean maps;
(3) The category of localizable measurable spaces and measurable maps.
[Apparently one more equivalent category can be defined using the language of locales.
Unfortunately, I am not familiar enough with this language to state this variant here.
Any help on this matter will be appreciated.]
While its more famous version for commutative unital C*-algebras is extensively covered in the literature,
I was unable to find any complete references for this particular variant.
The equivalence between (1) and (2) follows from the Gelfand duality theorem for commutative C*-algebras
via restriction to the subcategory of von Neumann algebras and their morphisms (σ-weakly continuous morphisms of unital C*-algebras).
Takesaki in his Theory of Operator Algebras I, Theorem III.1.18,
proves a theorem by Dixmier that compact Hausdorff spaces
corresponding to von Neumann algebras are precisely hyperstonean spaces
(extremally disconnected compact Hausdorff spaces that admit sufficiently many
positive normal measures).
Is there a purely topological characterization of the last condition (existence
of sufficiently many positive normal measures)?  Of course we can require that
every meager set is nowhere dense, but this is not enough.
I was unable to find anything about morphisms of hyperstonean spaces in Takesaki's book
or anywhere else.
The only definition of hyperstonean morphism that I know is a continuous map between hyperstonean spaces such that
the map between corresponding von Neumann algebras is σ-weakly continuous.
Is there a purely topological characterization of hyperstonean morphisms?
I suspect that it is enough to require that the preimage of every nowhere dense set is nowhere dense.  Is this true?
To pass from (2) to (3) we take symmetric differences of open-closed sets and nowhere dense sets as measurable subsets and nowhere dense sets as null subsets.
Is there any explicit way to pass from (3) to (2) avoiding any kind of spectrum construction (Gelfand, Stone, etc.)?
Any references that cover the above theorem partially or fully and/or answer any of the three questions above will be highly appreciated.

REPLY [3 votes]: As shown in the paper Gelfand-type duality for commutative von Neumann algebras, the following categories are equivalent.

The category CSLEMS of compact strictly localizable enhanced measurable spaces,
whose objects are triples $(X,M,N)$, where $X$ is a set, $M$ is a σ-algebra of measurable subsets of $X$,
$N⊂M$ is a σ-ideal of negligible subsets of $X$
such that the additional conditions of compactness (in the sense of Marczewski) and strict localizability are satisfied.
Morphisms $(X,M,N)→(X',M',N')$ are equivalence classes of maps of sets $f:X→X'$ such that $f^*M'⊂M$ and $f^*N'⊂N$
(superscript $*$ denotes preimages)
modulo the equivalence relation of weak equality almost everywhere: $f≈g$ if for all $m∈M'$ the symmetric difference $f^*m⊕g^*m$
belongs to $N$.
The category HStonean of hyperstonean spaces and open maps.
The category HStoneanLoc of hyperstonean locales and open maps.
The category MLoc of measurable locales, defined as the full subcategory of the category of locales
consisting of complete Boolean algebras that admit sufficiently many continuous valuations.
The opposite category CVNA^op of commutative von Neumann algebras,
whose morphisms are normal *-homomorphisms of algebras in the opposite direction.

The paper contains an extensive discussion with counterexamples why this particular definition of CSLEMS is necessary.
In particular, the choices of strictly localizable vs localizable, weak equality almost everywhere vs equality almost everywhere,
and the property of compactness are all crucial.
Measurable spaces commonly encountered in analysis are typically compact, strictly localizable, and countably separated.
The latter property guarantees that weak equality almost everywhere implies equality almost everywhere.
Notice a curious property of the category MLoc of measurable locales:
it is a full subcategory of the category of locales.
Thus, measure theory quite literally is part of (pointfree) general topology.<|endoftext|>
TITLE: Intuitive explanation for the Atiyah-Singer index theorem
QUESTION [77 upvotes]: My question is related to the question Explanation for the Chern Character to this question about Todd classes, and to this question about the Atiyah-Singer index theorem.
I'm trying to learn the Atiyah-Singer index theorem from standard and less-standard sources, and what I really want now is some soft, heuristic, not-necessarily-rigourous intuitive explanation of why it should be true. I am really just looking for a mental picture, analogous somehow to the mental picture I have of Gauss-Bonnet: "increasing Gaussian curvature tears holes in a surface".
The Atiyah-Singer theorem reads $$\mathrm{Ind}(D)=\int_{T^\ast M}\mathrm{ch}([\sigma_m(D)])\smile \mathrm{Td}(T^\ast M \otimes \mathbb{C})$$
What I want to understand is what the Chern character cup Todd class is actually measuring (heuristically- it doesn't have to be precisely true), and why, integrated over the cotangent bundle, this should give rise to the index of a Fredholm operator. I'm not so much interested in exact formulae at this point as in gleaning some sort of intuition for what is going on "under the hood". The Chern character is beautifully interpreted in this answer by Tyler Lawson, which, however, doesn't tell me what it means to cup it with the Todd class (I can guess that it's some sort of exponent of the logarithm of a formal group law, but this might be rubbish, and it's still not clear what that should be supposed to be measuring). Peter Teichner gives another, to my mind perhaps even more compelling answer, relating the Chern character with looping-delooping (going up and down the n-category ladder? ), but again, I'm missing a picture of what role the Todd class plays in this picture, and why it should have anything to do with the genus of an elliptic operator. I'm also missing a "big picture" explanation of Fei Han's work, even after having read his thesis (can someone familiar with this paper summarize the conceptual idea without the technical details?). Similarly, Jose Figueroa-O'Farrill's answer looks intriguing, but what I'm missing in that picture is intuitive understanding of why at zero temperature, the Witten index should have anything at all to do with Chern characters and Todd classes.
I know (at least in principle) that on both sides of the equation the manifold can be replaced with a point, where the index theorem holds true trivially; but that looks to me like an argument to convince somebody of the fact that it is true, and not an argument which gives any insight as to why it's true.
Let me add background about the Todd class, explained to me by Nigel Higson: "The Todd class is the correction factor that you need to make the Thom homomorphism commute with the Chern character." (I wish I could draw commutative diagrams on MathOverflow!) So for a vector bundle $V\longrightarrow E\longrightarrow X$, you have a Thom homomorphism in the top row $K(X)\rightarrow K_c(E)$, one in the bottom row $H^\ast_c(X;\mathbb{Q})\rightarrow H_c^\ast(E;\mathbb{Q})$, and Chern characters going from the top row to the bottom row. This diagram doesn't commute in general, but it commutes modulo the action of $\mathrm{TD}(E)$. I don't think I understand why any of this is relevant.
In summary, my question is

Do you have a soft not-necessarily-rigourous intuitive explanation of what each term in the Atiyah-Singer index theorem is trying to measure, and of why, in these terms, the Atiyah-Singer index theorem might be expected to hold true.

REPLY [22 votes]: I am going to give a topologically biased answer, which will proceed by restating what the Index Theorem says so that its plausibility (though not its truth) is more immediate.  [Oops - I see that Paul Siegel said much of this in the comments following his answer - oh well.]

Philosophy: It is a good thing to take a homology and cohomology theory and find geometric/ analytic models for them as well as the linear and Poincare-Lefshetz duality between them, when the spaces in questions are manifolds.

The most basic example is of course ordinary homology and cohomology, which are quite familiar, but worth revisiting since there are actually a few different models of this philosophy. The standard geometric model for homology is chains, and then cohomology can be obtained as a "formal linear dual."  But in an oriented manifold, one can show that differential forms provide functionals on chains and go on to establish the de Rham theorem.  Or one can take proper submanifolds as partially-defined cochains and get an intersection-theoretic interpretation of cohomology.  Or one can decide that de Rham theory is so wonderful that it should be the basic theory, and then passing to linear duals from there leads to the theory of currents to represent homology.
Another example which works very easily but is not as well known is bordism and cobordism.  By definition bordism groups are defined by maps from manifolds.  Cobordism is typically defined by maps to Thom spectra, but standard transversality shows that such maps are represented by proper submanifolds.  The linear/Poincare/Lefshetz duality between bordism and cobordism is given by intersection theory.
The Index Theorem manifests this philosophy for K-theory.  K-theory cohomology classes are of course represented by formal sums (or complexes) of vector bundles.  K-homology, as developed in Higson and Roe's book, is represented by Fredholm operators.  The pairing between them is roughly "counting the dimensions of spaces of solutions of an operator on a vector bundle."  (On some planet with very advanced topologists but relatively weak other flavors of math, one could envision the topologists "inventing" this kind of analysis just so they could have more fun with K-theory.)  While the analysis (pseudodifferential operators and the like) which shows that Fredholm operators "pair" with vector bundles is standard enough, showing that there is such a homology theory is of course involved, and takes up much of the Higson-Roe book.
Thus, the Index Theorem follows immediately from the stronger statement that one can define K-homology theory using differential operators and that the pairing between that and K-theory is given by the Index. In order to get the cohomology statement, as Paul Siegel mentions above, one uses the Chern character to translate from K-theory to cohomology but then must multiply by the Todd genus because the Chern character does not preserve Thom classes.
I realize that this stronger statement while perhaps feeling plausible is more of a formal answer than the geometric answer (like for Gauss-Bonnet) that you sought.  But when I studied the Index Theorem, mostly on the topological side, years ago I found that this formalism did help me develop some "pictures" for some of the arguments (for example the Fredholm theoretic proof Bott periodicity etc).
By the way, while we're mentioning this philosophy, there is one case which is very much a hot/interesting topic, namely for elliptic cohomology where objects like conformal nets, higher categories and "stringy" topology of manifolds all need to be further developed to tell the story.<|endoftext|>
TITLE: How to Compute the coordinate ring of flag variety?
QUESTION [12 upvotes]: Let $G$ be algebraic group over $\mathbb(C)$(semisimple), $B$ be Borel subgroup. consider flag variety $G/B$. It is known that $G/B$ is projective variety. So one consider the homogeneous coordinate ring $ \mathbb{C}[G/B]$, it is known that it is a graded polynomial ring quotient by some homogeneous ideal. For example, consider flag variety of $sl_2$, it is $\mathbb{P}^{1}$, so the coordinate ring is $\mathbb{C}[x_1,x_2]$ as graded ring.
I want to know for general case, i.e. flag variety of finite dimensional Lie algebra $g$($G/B$), how to compute $\mathbb{C}[G/B]$? But we know that from Borel-Weil, we can write it as
$\bigoplus_{\lambda\in P_+}$ $R_\lambda$, where $R_\lambda$ is irreducible highest dominant weight representations of $g$
My question
How to compute $\mathbb{C}[G/B]$? explicitly?
How to build explicit ring isomorphism from $\bigoplus_{\lambda\in P_+}$ $R_\lambda$ to some concrete ring?
This question might be elementary. Thanks anonymous and Ben pointing out the stupid mistake I made

REPLY [7 votes]: The ring in question may be identified with the ring of $N$-invariant functions on $G$. (Where $N$ is the unipotent radical of $G$.) 
In type $A$, the group $G$ is $n \times n$ matrices with determinant $1$. A function is $N$ invariant if it is preserved by rightward column operations. (That is, adding $a \times (\mbox{column} \ i)$ to $\mbox{column} \ j$, for $i<j$.) It is obvious that those minors which use the left $k$ columns for some $k$ have this invariance. 
In fact the ring is generated by these minors, which are called Plucker coordinates. The relations between these minors, called Plucker relations, are also extremely classical and well known. A good low level reference to this topic is Chapter 14 of Miller and Sturmfels' book Combinatorial Commutative Algebra.<|endoftext|>
TITLE: Fundamental groups of topoi
QUESTION [32 upvotes]: Just yesterday I heard of the notion of a fundamental group of a topos, so I looked it up on the nLab, where the following nice definition is given:
If $T$ is a Grothendieck topos arising as category of sheaves on a site $X$, then there is the notion of locally constant, locally finite objects in $T$ (which I presume just means that there is a cover $(U_i)$ in $X$ such that each restriction to $U_i$ is constant and finite). If $C$ is the subcategory of $T$ consisting of all the locally constant, locally finite objects of $T$, and if $F:C\rightarrow FinSets$ is a functor ("fiber functor"), satisfying certain unnamed properties which should imply prorepresentability, then one defines $\pi_1(T,F)=Aut(F)$.
Now, if $X_{et}$ is the small étale site of a connected scheme $X$, then it is well known the category of locally constant, locally finite sheaves on $X$ is equivalent to the category of finite étale coverings of $X$, and with the appropriate notion of fiber functor it surely follows that the étale fundamental group and the fundamental group of the topos on $X_{et}$ coincide. 
Similarly, as the nlab entry mentions, if $X$ is a nice topological space, locally finite, locally constant sheaves correspond to finite covering spaces (via the "éspace étalé"), and we should recover the profinite completion of the usual topological fundamental group.
Before I come to my main question: Did I manage to summarize this correctly, or is there something wrong with the above?

My question:
Has the fundamental group of other topoi been studied, and in what context or disguise might we already know them? For example, what is known about the fundamental group of the category of fppf sheaves over a scheme $X$?

REPLY [33 votes]: The profinite fundamental group of $X_{fppf}$ as you define it is again the etale fundamental group of X.  More precisely, the functor (of points)
$f : X_{et} \to \mathrm{Sh}_{fppf}(X)$
is fully faithful and has essential image the locally finite constant sheaves (image clearly contained there, as finite etale maps are even etale locally finite constant, let alone fppf locally so).  Proof in 3 steps:

It is fully faithful by Yoneda (note also well-defined by fppf descent for morphsisms).
Both sides are fppf sheaves (stacks) in $X$, by classical fppf descent.
Combining 1 and 2, it suffices to show that a sheaf we want to hit is just fppf locally hit, which is obvious since locally it's finite constant.

Note that the same proof also works for $X_{et}$ or anything in between -- once your topology splits finite etale maps it doesn't really matter what it is.  So we usually just work with the minimal one, the small etale topology.  As Mike Artin said to me apropos of something like this, "Why pack a suitcase when you're just going around the corner?"<|endoftext|>
TITLE: What is an example of a non-regular, totally path-disconnected Hausdorff space?
QUESTION [5 upvotes]: I need this for a counterexample: the multiplication in the fundamental group $\pi_1(\Sigma X_+)$, when it is equipped with the topology inherited from $\Omega \Sigma X_+$, fails to be continuous for the sort of space in the question, by a result from

J. Brazas, The topological fundamental group and hoop earring spaces, 2009, arXiv:0910.3685

but the author doesn't supply an explicit example of such a space.

REPLY [6 votes]: One of the easiest examples is the rational numbers with the subspace topology of the real line with the K-topology. Total path disconnectedness is not entirely necessary for multiplication of $\pi_{1}(\Sigma X_{+})$ to fail to be continuous. It just makes the path component space of $X$ equal to $X$, greatly simplifying complications.<|endoftext|>
TITLE: Are the q-Catalan numbers q-holonomic?
QUESTION [20 upvotes]: The generating function $f(z)$ of the Catalan numbers which is characterized by $f(z)=1+zf(z)^2$  is D-finite, or holonomic, i.e. it satisfies a linear differential equation with polynomial coefficients. 
The generating function $F(z)$ of the $q$-Catalan numbers is analogously characterized by the functional equation $F(z)=1+z F(z) F(qz)$.  I suspect that  $F(z)$ is not $q$-holonomic, i.e. does not satisfy a linear $q$-differential equation with polynomial coefficients. But I have no proof.  Is there a proof in the literature or references which may lead to a proof?
Since there were some misunderstandings I want to clarify the situation.
A power series  $F(z)$ is called $q - $holonomic if there exist polynomials $p_i (z)$ such that $\sum\limits_{i = 0}^r {p_i (z)D_q^i } F(z) = 0$ where $D_q $ denotes the $q - $differentiation operator  defined by $D_q F(z) = \frac{{F(z) - F(qz)}}{{z - qz}}.$  Equivalently if there exist (other) polynomials such that $\sum\limits_{i = 0}^r {p_i (z)F(q^i } z) = 0.$
Let $f(z)$ be the generating function of the Catalan numbers  $\frac{1}{{n + 1}}{2n\choose n}$ . Then $f(z) = 1 + zf(z)^2 $ or equivalently  $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$
There are 3 simple $q - $analogues of the Catalan numbers: 
a) The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$.  My question is about these polynomials. Their generating function satisfies a simple equation, but there is no known formula for the polynomials themselves.
b) The polynomials  $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function.
c) The $q - $Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$  is a  $q - $analogue of  $ \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$  Let $h(z)$ be the $q - $analogue of $sqrt {(1 + z)}$ defined by $h(z)h(qz)=1+z$.  Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function. But they are  not polynomials in $q.$ 
Both b) and c) are $q$-holonomic.  My question is a proof that a) is not $q$-holonomic.

REPLY [14 votes]: The problem is to show that the function $F(z)=1+z+(1+q)z^2+O(z^3)$
satisfying the functional equation $F(z)=1+zF(z)F(qz)$ does not satisfy
$\sum_{j=0}^{n-1}P_j(z)F(q^jz)+Q(z)=0$ identically in $z$ for some $n$;
here $P_j$ and $Q$ are polynomials in both $z$ and $q$.
(Although the original question assumes the homogeneous equation,
$Q(z)=0$, the limiting case $q\to1$ suggests to consider $Q(z)$ more
generally.) In what follows we show that such a functional equation
implies the algebraicity of $F(z)$; this is known to be false.
First of all, switch to the function $G(z)=zF(z)$ which satisfies
$$
G(qz)G(z)=q(G(z)-z).
$$
The problem is then to show that the newer function does not satisfy
$\sum_{j=0}^{n-1}\tilde P_j(z)G(q^jz)+\tilde Q(z)=0$ for some $n$.
By applying $z\mapsto q^{-k}z$ we can assume that $\tilde P_0(z)\ne0$
in this relation. The substitution $z\mapsto qz$ results in the relation
$\sum_{j=1}^n\hat P_j(z)G(q^jz)+\hat Q(z)=0$ where $\hat P_1(z)\ne0$.
The next step is to show, by iterating the functional equation for $G(z)$, that
$$
G(q^nz)\dots G(qz)G(z)=X_n(z)G(z)-Y_n(z).
$$
Indeed, we have $X_0=1$, $Y_0=0$, and
$$
X_n(z)G(z)-Y_n(z)=\bigl(X_{n-1}(qz)G(qz)-Y_{n-1}(qz)\bigr)G(z)
$$
implying
$$
X_n(z)=qX_{n-1}(qz)-Y_{n-1}(qz), \quad Y_n(z)=qzX_{n-1}(qz)
\qquad\text{for}\quad n\ge1.
$$
By means of the formula we see that $\deg Y_n$ does not decrease with $n$,
so that $Y_n\ne0$ for $n\ge1$. Note that our computation implies
$$
G(q^nz)=\frac{X_n(z)G(z)-Y_n(z)}{X_{n-1}(z)G(z)-Y_{n-1}(z)}
\qquad\text{for}\quad n\ge1.
$$
Substitute the above finding into the equation
$\sum_{j=1}^n\hat P_j(z)G(q^jz)+\hat Q(z)=0$ with $\hat P_1(z)\ne0$.
We obtain
$$
\sum_{j=1}^n\hat P_j(z)\frac{X_j(z)G(z)-Y_j(z)}{X_{j-1}(z)G(z)-Y_{j-1}(z)}
+\hat Q(z)=0.
$$
The term corresponding to $j=1$ is equal to
$$
\hat P_1(z)\frac{X_1(z)G(z)-Y_1(z)}{X_0(z)G(z)-Y_0(z)}
=\hat P_1(z)\frac{qG(z)-qz}{G(z)}.
$$
Note that the denominator $G(z)$ in this expression is not
canceled by the other denominators in the former relation
because $X_{j-1}(z)G(z)-Y_{j-1}(z)$ is never divisible by $G(z)$
as $Y_{j-1}(z)\ne0$ for $j\ge2$. In other words, after multiplying
$$
\sum_{j=1}^n\hat P_j(z)\frac{X_j(z)G(z)-Y_j(z)}{X_{j-1}(z)G(z)-Y_{j-1}(z)}
+\hat Q(z)=0
$$
by $G(z)$ we get an algebraic relation
$$
-qz\hat P_1(z)+G(z)\cdot\text{rational function in }z,q,G(z)
=0
$$
where the rational function does not involve $G(z)$ as a multiple in the denominator.
Since $\hat P_1(z)\ne0$, this gives us a nontrivial algebraic relation for $G(z)$.
Thus the function $G(z)$, hence $F(z)$ as well, are algebraic.
Addition.
I have never seen Carlitz's function $F(z)$ before, so I was pretty sure that
things like its transcendence had been already established. Especially, since
it is not hard (although required some time from me). Any way, this explains
my reasons for not having a reference to this fact.
If we write
$$
F(z)=F_q(z)=\sum_{n=0}^\infty a_n(q)z^n=\sum_{n=0}^\infty a_nz^n=1+z+(1+q)z^2+\dots,
$$
the functional equation $F(z)=1+zF(z)F(qz)$ implies
$$
a_{n+1}=\sum_{k=0}^nq^ka_ka_{n-k} \quad\text{for}\; n\ge0,
\qquad a_0=1.
$$
The clear induction on $n$ shows that $a_{n+1}(q)$ is a polynomial from $\mathbb Z[q]$
with leading term $q^{n(n-1)/2}$. In particular, the denominator of $a_{n+1}(1/2)$ is
exactly $2^{n(n-1)/2}$.
If the function $F_q(z)$ were algebraic then its specialization $F_{1/2}(z)$ should be
algebraic. This would imply that for a certain integer $A$ the $z$-expansion of
$F_{1/2}(Az)$ has integral coefficients. But no $A\in\mathbb Z$ with the property
$A^{n+1}/2^{n(n-1)/2}\in\mathbb Z$ for all $n$ could be given. Therefore, $F_{1/2}(z)$ and
$F_q(z)$ in general are transcendental.<|endoftext|>
TITLE: Reference request for relative bordism coinciding with homology in low dimensions
QUESTION [8 upvotes]: It's a standard fact that, for finite CW complexes, the relative (edit: oriented) bordism group $\Omega_n(X,A)$ coincides with the homology $[H_\ast(X,A;\Omega_\ast(pt))]_n\simeq H_n(X,A)$ for $n<5$ (edit: this should be $n<4$). One can prove this using the Atiyah-Hirzebruch spectral sequence, and all papers I've seen seem to just state it as a fact without citation. I really want to find the original reference for the above isomorphism, but have wasted much time and found nothing. 
What is the original reference for the above proof (and the fact itself) that relative bordism and relative homology coincide in low dimensions?

REPLY [4 votes]: Yuli Rudyak gave a perfect answer to this question in May, but it was by e-mail. Because his might be of interest to others, I reproduce it in full.

Look Theorem IV.7.37 of my book "On Thom spectra, orientability, and cobordism", Corrected reprint, Springer, 2008.
There is proved explicitly that the map $E_i(X,A) \to H_i(X,A)$ is an isomorphism for $i<4$ and epimorphism for $i<7$, where $E$ denotes the ORIENTED bordism  group.
This is important to cite 2008 Corrected reprint: In previous 1998 edition I did not make an explicit claim (although is follows easily from what has been done), and many people asked about explicit citation. I included the reference in corrected reprint.
By the way, you mention that $\Omega_n^O(X,A)$ is isomorphic to $[H_\ast(X,A; \Omega^O_*(pt.))]_n$. This is correct for NON-oriented bordism, but it is wrong for oriented one.
For non-oriented bordism, $\Omega_n\ne H_n$ even  in dimension 2.

I subsequently bought Rudyak's book. The Amazon page had a mix-up and they kept sending me the first edition, but it was resolved eventually by getting Springer to intervene. I hope it's sorted out- otherwise, I recommend ordering the book directly from Springer.<|endoftext|>
TITLE: On The Convergence of Ergodic Measures
QUESTION [9 upvotes]: I would like to know an example (not using the Gibbs measure Theory) of a sequence of measures $\mu_n:\mathcal B\to[0,1]$ , where $\mathcal B$ is the $\sigma$-algebra of the borelians of a compact space $X$ such that :
1) $\mu_n$ is ergodic, with respect to a fixed continuous function $T:X\to X$, for all $n\in\mathbb N$;
2) $\mu_n\to \mu$ in the weak-$*$ topology and $\mu$ is not ergodic.

REPLY [4 votes]: consider the torus map $(x,y) \mapsto (x+y, y)$. For every $y$, this gives a rotation on the circle $S^1 \times y$. That is, Lebesgue measure on any "horizontal" circle is preserved. For $y$ rational it is not ergodic, but for $y$ irrational it is ergodic. Well, that's it. Lebesgue on a circle with rational rotation is weak* approximated by Lebesgue on circles with irrational rotation.<|endoftext|>
TITLE: Existence of Limiting Distribution for Moving Regions in Stat. Phys. Models
QUESTION [7 upvotes]: As the title (hopefully) suggests, I've been trying to prove (or disprove) the existence of a limiting distribution for a certain projection in a statistical physics model. I'll give the details of the model below, but if you are reading this, I'm particularly interested in proofs that seem 'quantifiable' (e.g. maybe implicitly has some estimate of distance to the limiting distribution). I think it is unlikely that anyone reading this has familiarity with the below model, so reasonable-sounding references would also be great.
The model is the infinite 'East Model', discussed in e.g. 'The Asymmetric One-Dimensional Constrained Ising Model: Rigorous Results'. For a short description, the process is on (0,1) labellings of the positive integers, and at time 0, all labels are 0 except for 1, which is labeled 1 (and for convenience we write that 0 is also labeled 0). Each positive integer k has its own Poisson-1 clock, and when the clock goes off, the label does nothing (if the label of k-1 is 0) or flips with probability 0.5 (if the label of k-1 is 1). 
At each time, there is a highest non-zero entry. I am interested in the vector consisting of the labellings of the n entries below the highest non-zero entry. This vector isn't by itself a Markov chain, and if it has a limiting distribution, it isn't quite a Bernoulli process (which is the limit for the East process itself). Any insight into existence of a limit, convergence, and how it might be affected by starting in some different initial state with finitely many 1's would be welcome.

REPLY [4 votes]: you might get more insight with pictures:
the whole process (at time when something happened at the east-end)
whole process http://www.freeimagehosting.net/uploads/aefaf14d4f.png
and the last $50$ non-zero sites<|endoftext|>
TITLE: Connected Graph
QUESTION [6 upvotes]: Suppose $\Delta(G) + \delta(G) \geq n-1$. Can we conclude that $G$ is connected?

REPLY [11 votes]: Yes. Suppose that $V(G)$ can be split into two sets $A$ and $B$ of respective sizes $a \geq b \geq 1$, and no edges between the two sets.  Since vertices in $A$ (resp. $B$) have degree at most $a-1$ (resp. $b-1$), we see that $\Delta(G) \leq a-1$ and $\delta(G) \leq b-1$. So $\Delta(G) + \delta(G) \leq a+b-2 = n-2$.

REPLY [7 votes]: We can do this without inducting on the size of the graph: choose a vertex $v$ of degree $\Delta$ and any other vertex $w$ of degree $d \geq n - 1 - \Delta$.  Either $w$ is adjacent to $v$ or, by the pigeonhole principle, they have a common neighbor among the other $n - 2$ vertices.  So not only is the graph connected but it's actually of diameter at most 4.

REPLY [5 votes]: Take a vertex $v$ of maximum degree, then delete a vertex $w$ not adjacent to $v$.<|endoftext|>
TITLE: What is exactly the (singularity) confinement property ?
QUESTION [8 upvotes]: This property seems to be used both in the context of differential equations and several kinds of discrete equation systems or automata.
It seems to be related in certain case to the Painlevé Property first discovered for Painlevé equations and their solutions.
I have seen several definitions, notations, criteria, which do not always seem to match each other or to be easily applicable to all context.  I would appreciate some clarification about it.
I am mostly interested in the discrete applications but all explanations about the continuous cases are welcome especially if there are links to the discrete setting.

REPLY [9 votes]: The singularity confinement property refers to a property of discrete integrable systems. I am unaware of this property in the context for continuous systems. I can understand why you might have difficulty in getting a definition, since it is rather oddly defined all too often. Since the paper of Goriely and La Fortune have been referenced I will assume that you are interested in this property for maps rather than partial difference equations. 
Let me start with an example that hopefully illustrates what is going on, before generalizing: Let us take a simple second order difference equation, which we right as a sequence, satisfying the recurrence relation:
$x_{n+1}x_{n-1} = 1+ x_n$
If we let $x_1 = -1$, then $x_2 = 0$, regardless of what $x_0$ is. Then $x_3 = -1$ and $x_4 = \infty$. Worst of all, you have also lost your initial conditions, namely $x_0$. Suppose, instead of $x_1 = -1$, we take $x_1 = -1+e$, where e is assumed very small. Then
$x(2) = \frac{e}{x_0}$
$x(3) = \frac{e+x_0}{(e-1)x_0}$
$x(4) = \frac{1+x_0}{e-1}$
Notice that in the limit as $e \to 0$, $x_2 \to 0$, $x_3 \to -1$, $x_4 \to -1 -x_0$. Somehow, in the limit around a "bad" point, or "singularity", we recover initial conditions, namely $x_0$. We say that the singularity is confined because, despite a loss of information at $x_2$ and $x_3$ in the limit, we can regain the information again at $x_4$ in the same limit. We say the equation has the singularity confinement property if we can always regain the initial conditions in the limit somewhere along the line. This is the usual, however you may want something more general. 
The generalization of the above is a map 
$f: (x_n, x_{n-1}) \to (x_{n+1},x_n) = \left(\frac{x_n+1}{x_{n-1}},x_n\right)$, 
however, we generally take a map of $R^n$ and then take the corresponding projective space, $P^n$, then we consider a map
$f: P^n \to P^n$.
This function now describes a discrete dynamical system, via $y_{n+1} = f(y_n)$. A singularity in this context is where the function, $f$, fails to be invertible. For example,  in the above, we had all initial conditions of the for $(x_0,-1)$ being sent to $(-1,0)$. However, we say the singularity is confined if some power of $f$ is invertible. Note that in the above example, the third power of the map can be continuously extended to the map $(x_0,-1) \to (-1,-1-x_0)$, hence the singularity, $(-1,0)$ is confined. A system has the singularity confinement property if all singularities are confined.<|endoftext|>
TITLE: Examples of common false beliefs in mathematics
QUESTION [979 upvotes]: The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.
Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are
(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.
Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.
A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.
Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

REPLY [5 votes]: "A closed unit-ball (in a Banach space) is compact!"

It only true in finite-dimensional spaces.<|endoftext|>
TITLE: Finite groups with a character having very few nonzero values?
QUESTION [7 upvotes]: A number theorist I know (who studies Galois representations) raised a question recently:

Which finite groups can have an irreducible character of degree at least 2 having only $n=2, 3$, or 4 classes where the character takes nonzero values? 

He has learned about a few special examples involving nonabelian groups which are very close to being abelian.    My limited intuition about the question, based on finite groups of Lie type, suggests that the sort of group he is looking for will be far from simple.   But maybe there is no reasonable characterization  of these groups for a given $n \leq 4$?   In any case, there should be some relevant literature out there.   The question itself belongs to a familiar genre in finite group theory:  What does the character table tell me about the structure of a group? 
Appropriate references: S. , Gagola, Pacific J. Math 108 (1983), 363-385,
Berkovich-Zhmud', Characters of Finite Groups, Part 2, Chapter 21.

REPLY [8 votes]: For n=2, your question is addressed in:
Gagola, Stephen M., Jr.  "Characters vanishing on all but two conjugacy classes."
Pacific J. Math. 109 (1983), no. 2, 363–385.
MR721927
euclid.pjm/1102720107
In it, he shows that if such a nearly-zero character exists, it is unique, and the unique faithful irreducible character of G (similar to the extra-special 2 and 3 groups mentioned by Kevin Buzzard).  A doubly transitive Frobenius group has such a character, but the non-solvable doubly transitive Frobenius groups are in short supply (five or so).  The possible forms of more general non-solvable examples are restricted in Theorem 5.6 (basically SL2).

Berkovich and Zhmud have results for more general cases.  See chapter 27 of their book on character theory (volume 2) which answers a broader question for n=2 and n=3:

Which groups have all irreducible non-linear characters taking on only n distinct values?

The list of groups is fairly short, but I haven't had time to verify it; they are all far from simple, usually with a normal Sylow and nilpotent quotient.  They have a result for n=4 too, on p. 239, but it is much less complete.
Since your character is nonzero except on n classes, it can take on at most n nonzero values.  It is fairly likely to be faithful, since every class contained in the kernel is a wasted class that cannot be zero.  This allows some of the lemmas to be used.  If Gagola's case is indicative, then the "all" versus "one" will not actually be a huge difference, at least mod the kernel of your single character.
Here are the paper references:
Berkovich, Yakov; Chillag, David; Zhmud, Emmanuel. "Finite groups in which all nonlinear irreducible characters have three distinct values."
Houston J. Math. 21 (1995), no. 1, 17–28.
MR1331241
Zhmudʹ, È. M. "On finite groups, all of whose irreducible characters take at most two nonzero values."
Ukraïn. Mat. Zh. 47 (1995), no. 8, 1144–1148;translation in 
Ukrainian Math. J. 47 (1995), no. 8, 1308–1313 (1996)
MR1367729
DOI:10.1007/BF01057720

Berkovich and Zhmud also pose an exercise that if all the irreducible characters of degree coprime to p take on only 3 values, then the group is p-solvable.  The same is true if "coprime to" is replaced by "divisible by".  This is page 238.

REPLY [8 votes]: For the case of n=2 see Gagola's paper "Characters vanishing on all but two conjugacy classes" MR0721927.  Some improvements on Gagola's results can be found in Kuisch and van der Waall's papers "Homogeneous character induction [I and II]" MR1172440/MR1302857 and in my paper "Groups with a Character of Large Degree".  
See Theorem 4.1 in my paper for what I think is the strongest result in this direction (as far as I know):

Gagola showed that the character of
  $V$ vanishes on all but one nontrivial
  conjugacy class if and only if there
  exists a subgroup $N$ of $G$ such that
  $N$ acts trivially on each simple
  $\mathbb{C}[G]$-module except $V$.
Let
  $G$ be a group of order $n$ and $V$ a
  simple $\mathbb{C}[G]$-module of
  dimension $d$.  Define $e$ such that
  $n = d(d+e)$.   Assume that there
  exists a normal subgroup $N \neq
> \{1\}$ such that $N$ acts trivially on
  each simple $\mathbb{C}[G]$-module
  other than $V$.  Let $x$ be a
  nontrivial element of $N$ and $C$ be
  the centralizer of $x$ in $G$.  Then
  there exist a prime number $p$, a
  positive integer $k$ and a
  non-negative integer $m$ such that:

$N$ is elementary abelian of order $p^k$,
$C$ has order $p^k e^2$, and $d = e(p^k-1)$, and $n = e^2 p^k (p^k-1)$,
$C$ is a Sylow $p$-subgroup of $G$ and $e = p^m$,
if $H$ is any group such that $N \subseteq H \subseteq C$ and $|H| >
> p^{k+m}$, then $N \subseteq [H,H]$.

The above conditions are also sufficient, see Thm 4.3

For n bigger than 2 there may be some results in MR1487363 Chillag "On zeroes of characters of finite groups."<|endoftext|>
TITLE: (When) does Schur's lemma break?
QUESTION [8 upvotes]: Edit: I wrote the following question and then immediately realized an answer to it, and moonface gave the same answer in the comments.  Namely, $\mathbb C(t)$, the field of rational functions of $\mathbb C$, gives a nice counterexample.  Note that it is of dimension $2^{\mathbb N}$.
The following is one statement of Schur's lemma:

Let $R$ be an associative unital algebra over $\mathbb C$, and let $M$ be a simple $R$-module.  Then ${\rm End}_RM = \mathbb C$.

My question is: are there extra conditions required on $R$?  In particular, how large can $R$ be?
In particular, the statement is true when $\dim_{\mathbb C}R <\infty$ and also when $R$ is countable-dimensional.  But I have been told that the statement fails when $\dim_{\mathbb C}R$ is sufficiently large.
How large must $\dim_{\mathbb C}R$ be to break Schur's lemma?  I am also looking for an explicit example of Schur's lemma breaking for $\dim_{\mathbb C}R$ sufficiently large?

REPLY [8 votes]: It all depends on what you call "Schur's Lemma". If M is a simple module over a ring R then D=EndRM is always a division ring (think of it as a weak Schur's lemma). The question is, can the endomorphism ring be pinned down more concretely, for example, if R is an algebra over a field k? In the usual Schur's lemma for finite groups, k=C and D=C. More general versions of Schur's lemma assert that D is algebraic over k (so D=k if k is algebraically closed).
If R is an affine (i.e. finitely-generated) commutative algebra over a field k and I is a maximal ideal of R then Hilbert's Nullstellensatz is asserting that R/I is algebraic over k. Since M=R/I is a simple module and R/I=EndRM, you may interpret the statement as a version of Schur's lemma: R/I=EndRM is algebraic over k. It is also known that every prime ideal of R is an intersection of maximal ideals.
Now if R is a noncommutative algebra over k one can ask whether the analogous properties hold: the endomorphism ring of a simple module is algebraic over k ("endomorphism property", implying the usual Schur's lemma when k is algebraically closed) and every prime ideal of R is an intersection of primitive ideals ("R has radical property"). This is true in a range of situations and constitutes noncommutative Nullstellensatz. Duflo proved that the endomorphism property for R[x] implies Nullstellensatz (endomorphism + radical) for R. 

As Kevin pointed out, if the dimension of M over k is smaller than the cardinality of k then the endomorphism property holds for R, and by Duflo, full Nullstellensatz holds for R.
In general, some extra assumptions are necessary, but Nullstellensatz holds for the Weyl algebra An(k), the universal enveloping algebra U(g) of a finite-dimensional Lie algebra g, and the group algebra k[G] of a polycyclic by finite group. (The first two cases use Quillen's lemma also mentioned by Kevin). 

This is described in chapter 9 of McConnell and Robson, Noncommutative Noetherian rings.<|endoftext|>
TITLE: Minimal Backtracking Proof Tree
QUESTION [5 upvotes]: When trying to prove that a particular instance of a problem like graph coloring or SAT is unsatisfiable, generally one explores the search tree using an algorithm like DPLL and the proof of unsatisfiability consists of the exhaustive exploration of the search tree - i.e. try an partial assignment of values to variables, do propagation and if the domain of a variable becomes empty, you know the assignment does not satisfy the instance so you backtrack. By propagation I mean, one looks at every constraint and then for every variable in the constraint, one prunes the values in the domain of the variable which are not part of any model satisfying that particular constraint.
My question is that how hard in terms of computational complexity is to find a search tree proving infeasibility of minimal size, i.e. if the size of the search tree is $S$, then how hard is it to find the minimal such search tree given that the size of the input is taken as $S$ (Note - that can be much larger than the size of the original problem, exponential in the worst case).

REPLY [2 votes]: I found a reference which proves that minimum propositional proof length is NP-hard to linearly approximate for a variety of systems from resolution proofs to Frege proofs.
Minimal Propositional Proof Length is NP-hard to Linearly Approximate<|endoftext|>
TITLE: What are trig classes like within a universe that's "noticeably" hyperbolic?
QUESTION [9 upvotes]: [I want to think that this question has an answer, but it may be more a "community wiki" discussion. Feel free to re-tag.]

What are trig classes like within a universe that's "noticeably"[*] hyperbolic?

Using an appropriate model to peek into such a universe from our Euclidean one shows us that all the fundamental trig relations (Law of Sines, Law of Cosines, etc) involve transcendental functions in both angle measures and lengths. But, then, in our earliest mathematical days, similar triangles allowed us to build a theory of (circular) trig functions based on ratios of lengths of sides, and eventually our mathematics became sophisticated enough to include hyperbolic functions and to be comfortable enough with them and how they might apply to our non-Euclidean models. (We also had the benefit of being able to interpret products as areas of rectangles whose side-lengths correspond to factors.) What if you don't --can't-- draw from our experience base?
How do you even get started developing trigonometry (or explaining it to your hyperbolic trig students[**]) without a concept of similar triangles?
Obviously(?), the Unit Circle is out.
It makes some sense that the Angle of Parallelism would be the fundamental bridge between angle-information and distance-information; one can imagine that hyperbolic people would be "aware" of the phenomenon on some level, and we know that it's a "universal" property. Even so, the most-concise representations of the AoP relationship are transcendental in both angle measure and length. How insightful would a hyperbolic mathematician have to be to discern the equations from tables of observed measurements? And is there a clear path from those equations to, say, (what we know as) the Law of Sines and the Laws of Cosines?
Or perhaps the fundamental figure in an "intrinsic" hyperbolic trig class is the Equilateral Triangle. This idea actually exploits non-similarity to set up a bridge between angle-information and distance-information (with area-information thrown in as a bonus); and it seems that it might be more likely to provide a path to the Laws of Sines and Cosines, since it already relates angles and sides of triangles. But, is it really a particularly helpful starting point? Can you get from there to the Angle of Parallelism relation?
Something else?
[*] For instance, anyone can pull out a protractor and easily see that triangles have an angle-sum less than 180 degrees.
[**] "hyperbolic" modifying both "trig" and "students". :)

REPLY [3 votes]: Chapter V of Harold E. Wolfe's book:  Introduction to Non-Euclidean Geometry (Holt, Rinehart, and Winston), 1945 (and reprinted, 1966) is entitled: Hyperbolic Plane Trigonometry, and has a systematic treatment of this topic.<|endoftext|>
TITLE: Is a subspace with a certain property dense in the dual of a vector space?
QUESTION [7 upvotes]: Suppose we have a normed vector space $V$ and its dual $V^*$, and suppose that $X \subseteq V^*$ has the property that for every $v \in V$, there is some $\phi \in X$ with $\Vert \phi \Vert = 1$ such that $\phi(v) = \Vert v \Vert$.  Is $X$ dense in $V^*$ (in the operator norm)?  Note that this is a stronger property than $\Vert v \Vert = \sup_{\phi\in X} \frac{\phi(v)}{\Vert \phi \Vert}$, since we are assuming that the supremum is realized.
I think the answer is probably "no."  A nice example (passed to me originally made up by Terry Tao) showing that the second condition (the supremum over $X$ gives the norm) does not imply dense is the following: consider $l^1$ and $(l^1)^* = l^\infty$.  Then the space of eventually zero sequences in $l^\infty$ is sufficient for the norm: given $f\in l^1$, let $\phi_n$ be a truncation of the sign function of $f$ to the first $n$ indices.  Then $\lim_{n\to \infty} \phi_n(f) = \Vert f \Vert$.  However (for $f$ with infinite support), there is no finite sequence $\phi$ with $\phi(f) = \Vert f \Vert$.

REPLY [11 votes]: The answer is negative. Since the linear span of the Dirac masses is not a dense subspace of the dual of $C[0,1]$.<|endoftext|>
TITLE: How canonical are localization maps in Galois cohomology?
QUESTION [8 upvotes]: The setup for my question is as follows: $k$ is a field, $K$ a Galois extension of $k$ with group $G$, $k^\prime$ an arbitrary extension of $k$, and $K^\prime/k^\prime$ another Galois extension of fields with group $G^\prime$ (the notation, as well as the setup, is essentially verbatim from Chapter X, Section 4 of Serre's Local Fields). Given a $k$-monomorphism $f:K\rightarrow K^\prime$, we get a homomorphism $\overline{f}:G^\prime\rightarrow G$ (sending $s^\prime\in G^\prime$ to the unique $s\in G$ with $f\circ s=s^\prime\circ f$), and if we have another $k$-monomorphism $g:K\rightarrow K^\prime$, then there is some unique $s\in G$ with $f\circ s=f^\prime$ (both these facts are consequences of the normality of $K$ over $k$, which implies that every $k$-monomorphism of $K$ into $K^\prime$ has the same image), so the homomorphism $\overline{g}:G^\prime\rightarrow G$ induced by $g$ is just that induced by $f$ followed by a conjugation of $G$.
Now, if $A$ is a commutative group scheme over $k$, then such an $f$ (as above) gives a map $A(K)\rightarrow A(K^\prime)$ that is compatible (in the sense of group cohomology) with $\overline{f}:G\rightarrow G^\prime$, so we get maps $H^n(G,A(K))\rightarrow H^n(G^\prime,A(K^\prime))$. The key fact which comes from the considerations in the previous paragraph is that if we use a different $k$-monomorphism $g:K\rightarrow K^\prime$, then the induced maps on cohomology differ by the map $H^n(G,A(K))\rightarrow H^n(G,A(K))$ induced by the pair $x\mapsto sxs^{-1}$ and $p\mapsto s^{-1}p$, which is the identity.
My question regards a particular case of this situation but with (what appears to me) to be a twist. $k$ is a global field, $k^\prime=k_v$ is the completion of $k$ at some place $v$, $K=k^s$ is a separable closure of $k$, and $K^\prime=k_v^s$ is a separable closure of $k$. Any two embeddings of $k^s$ into $k_v^s$ over $k\rightarrow k_v$ differ by an element of the absolute Galois group $G_k=Gal(k^s/k)$ of $k$ (these are just two choices of places of $k^s$ extending $v$). As before, with such an embedding, I get a map $G_{k_v}=Gal(k_v^s/k_v)\rightarrow G_k$ which is (by Krasner's lemma) an injection with image the decomposition group in $G_k$ of the corresponding place. Any two such injections differ by conjugation in $G_k$. So, if I have, for instance, an abelian variety $A$ over $k$, I get canonical "localization" maps $H^n(k,A(k^s))\rightarrow H^n(k_v,A(k_v^s))$, i.e., they are independent of the embedding of $k^s$ into $k_v^s$. But now if I start with an arbitrary $G_k$-module $M$, the homomorphism $G_{k_v}\rightarrow G_{k}$ makes $M$ into a $G_{k_v}$-module and is compatible with the identity $M\rightarrow M$ (where the target is regarded as a module over the local Galois group), so I get a map on cohomology $H^n(k,M)\rightarrow H^n(k_v,M)$. My question is how can this map be independent of the embedding $k^s\rightarrow k_v^s$? It's true that any two embeddings give rise to conjugate injections of the local Galois group into the global one, but the action of the local group on $M$ is defined in terms of the map $G_{k_v}\rightarrow G_k$. There is no canonical choice of $G_{k_v}$-action on $M$ as there is in the case of a commutative $k$-group scheme. As far as I can tell, if I use a different embedding, I get a conjugate homomorphism $G_{k_v}\rightarrow G_k$, and this gives a DIFFERENT action of $G_{k_v}$ on $M$. So, strictly speaking, the map on cohomology induced by a second embedding shouldn't even have the same target as the one induced by the first. Even in dimension zero, shouldn't the Galois invariants of $G_{k_v}$ with respect to the different actions differ by  an element of $G_k$ (i.e. if $a$ is fixed by $G_{k_v}$ with respect to the first action then $sa$ is fixed by $G_{k_v}$ with respect to the second action, where $s\in G_k$). I would greatly appreciate if somebody could clarify what I'm not getting right here. I know these maps have to be canonical, and everywhere this is stated it's attributed to the fact that conjugation induces the identity on cohomology, but when I've tried to apply this, it doesn't seem to work out if my Galois module isn't already endowed with an action of $G_{k_v}$. 
I apologize for this post being so long. I suspect there's something simple I've overlooked that will warrant maybe a sentence or two, but I felt like I needed to try to motivate this question somewhat.

REPLY [5 votes]: As you note, if we choose two different embeddings $k^s \to k_v^s$, say $\imath_1$
and $\imath_2$, then we get two different $G_{k_v}$-module structures on $M$, call
them $M_1$ and $M_2$, and two different restriction maps $r_1:H^n(k,M) \to H^n(k_v,M_1)$
and $r_2:H^n(k,M) \to H^n(k_v,M_2)$.
The point is that there will also be a canonical isomorphism
$i:H^n(k_v,M_1) \cong H^n(k_v,M_2)$ such that $i\circ r_1 = r_2,$
given by a formula analogous to the one you gave in the abelian variety context.
Namely, if $\imath_2 = \imath_1\circ g,$ then the isomorphism $i$ will be induced by
$m \mapsto g\cdot m$ (if I have things straight; you can easily check if this is correct,
of if I should have $g^{-1}$ instead).  The fact that $i\circ r_1 = r_2$ will then depend
on the fact that the automorphism of $H^n(k,M)$ induced by conjugation by $g^{-1}$ and the
map $m\mapsto g\cdot m$ is the identity.  So one does not have a literal independence of
the embedding, but rather, the restriction is defined up to a canonical isomorphism independent of the embedding, and this latter fact does depend upon conjugation inducing the
identity on cohomology (which is why people often summarize it in that way).
Note also that your abelian variety example is actually a special case of this, because
the $M$ is $A(k^s)$, and the $G_{k_v}$-action on $M$ does depend on the embedding
of $k^s$ in $k_v^s$.  But the natural map $A(k^s) \to A(k_v^s)$ also depends on this embedding,
in such a way that, when you compose the restriction from $G_k$ to $G_{k_v}$ (with coefficients in
$A(k^s)$) with the map on $G_{k_v}$-cohomology induced by the embedding $A(k^s) \hookrightarrow
A(k_v^s)$, you do obtain a map on cohomology that is independent of the embedding.
But it is not that in this case $M$ has a well-defined action of $G_{k_v}$ independent
of the choice of embedding $k^s \hookrightarrow k_v^s$.  It is rather that $M$ embeds into
a bigger module $M_v$ (in a way that also depends on the embedding) so that the
composite $H^n(k,M)\to H^n(k_v,M) \to H^n(k_v,M_v)$ is independent of the embedding.
It is the embeddings $M \hookrightarrow M_v$ that are missing in the more general context
(i.e. when $M$ is not of the form $A(k^s)$).<|endoftext|>
TITLE: Does pi contain 1000 consecutive zeroes (in base 10)?
QUESTION [41 upvotes]: The motivation for this question comes from the novel Contact by Carl Sagan.  Actually, I haven't read the book myself.  However, I heard that one of the characters (possibly one of those aliens at the end) says that if humans compute enough digits of $\pi$, they will discover that after some point there is nothing but zeroes for a really long time.  After this long string of zeroes, the digits are no longer random, and there is some secret message embedded in them.  This was supposed to be a justification of why humans have 10 fingers and increasing computing power.  
Anyway, apologies for the sidebar, but this all seemed rather dubious to me.  So, I was wondering if it is known that $\pi$ does not contain 1000 consecutive zeroes in its base 10 expansion?  Or perhaps it does?  Of course, this question makes sense for any base and digit.  Let's restrict ourselves to base 10.  If $\pi$ does contain 1000 consecutive $k$'s, then we can instead ask if the number of consecutive $k$'s is bounded by a constant $b_k$.
According to the wikipedia page, it is not even known which digits occur infinitely often in $\pi$, although it is conjectured that $\pi$ is a normal number.  So, it is theoretically possible that only two digits occur infinitely often, in which case $b_k$ certainly exist for at least 8 values of $k$.    
Update. As Wadim Zudilin points out, the answer is conjectured to be yes.  It in fact follows from the definition of a normal number (it helps to know the correct definition of things).  I am guessing that a string of 1000 zeroes has not yet been observed in the over 1 trillion digits of $\pi$ thus computed, so I am adding the open problem tag to the question.  Also, Douglas Zare has pointed out that in the novel, the actual culprit in question is a string of 0s and 1s arranged in a circle in the base 11 expansion of $\pi$.  See here for more details.

REPLY [6 votes]: I do not know what kind of effect this has on your question, but it might be a good thing to look at in this context. In 1995, a formula for the $n$-th digit of $\pi$ written in hexadecimal was found. See the wikipedia article, http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula.<|endoftext|>
TITLE: "Nice" definition of discriminant as alluded to in an answer of Qing Liu
QUESTION [14 upvotes]: In his answer 
    here
 Qing Liu mentioned "the 'discriminant' of X which measures the defect of a functorial isomorphism which involves powers of the relative dualizing sheaf of X/R." 
Could somebody give a reference for or explain this?
I ask because there is a natural definition of conductor of a proper morphism to a normal scheme (for example as in Serre's "Algebraic groups and class fields"), and some relationship between the conductor and discriminant, but the definitions of discriminants I have seen always seem situation-specific and ad-hoc. 
Thanks.

REPLY [7 votes]: I want to mention that there is a published version of Deligne's letter online, namely the famous paper "Le déterminant de la cohomologie". See here:
https://publications.ias.edu/node/404
I am trying to type and translate the paper on my own, but so far I have only translated bits of it here and there as it is lengthy. I heard from Soule that the exponent $12$, is ultimately coming from the second term contribution of the expansion of the Todd class. 
In light of Brian Conrad's comment ("the oracle status of the Deligne reference is an obstruction to saying more"), I want to point out a possible mistake in Deligne's paper. Here I quote from Gerard Freixas i Montplet and Anna von Pippich's paper (see footnote of page 2):

" To be rigorous, Deligne mistakenly applies Bismut–Freed’s curvature
  theorem for Quillen connections. The subtle point is the compatibility
  of the Quillen connection with the holomorphic structure of the
  Knudsen–Mumford determinant. Nevertheless, the Riemann–Roch isometry
  he claims can be established by appealing instead to the results of
  Bismut–Gillet–Soule, where this compatibility is addressed. Notice
  however these results came later in time."

I am not sure if the issue was noticed by other people, but this may be a subtle point worth mentioning as the paper is a canonical reference in Arakelov intersection theory.<|endoftext|>
TITLE: Is "all categorical reasoning formally contradictory"?
QUESTION [30 upvotes]: In the December 2009 issue of the newsletter of the European Mathematical Society there is a very interesting interview with Pierre Cartier.  In page 33, to the question

What was the ontological status of categories for Grothendieck?

he responds

Nowadays, one of the most interesting points in mathematics is that, although all categorical reasonings are formally contradictory, we use them and we never make a mistake.

Could someone explain what this actually means?  (Please feel free to retag.)

REPLY [17 votes]: With regard to what Darij said, if Haskell is viewed as a formal system, it's quite easy to derive contradictions; they correspond to infinite loops:
loop :: forall a. a
loop = loop

However, when viewed in this light, the quote becomes something like:

Nowadays, one of the most interesting things in computer programming is that, although most languages allow one to write infinite loops, most programs that people write aren't infinite loops.

But this isn't surprising at all, because most people construct programs with some productive activity in mind, rather than just spinning in a vicious loop. And similarly, most mathematicians presumably choose theorems and proofs that they find to be similarly productive, rather than a masquerading paradox, even if the formal system would allow the latter. Just because the formal system allows paradox doesn't mean that there aren't non-paradoxical ways to prove things in the system, or that most proofs people write wouldn't be the latter.
Now, one might say, "people do write infinite loops." And this is true. But, the analogy with Haskell breaks down a bit at that point, because Haskell makes it easy to write infinite loops. It's just a given in the language, like mine above. With inconsistent formal systems, it's typically not as easy to get a proof of false. People did a lot of naive set theory without noticing that you could do it for instance. Russel's paradox is probably the easiest of the bunch, but it's quite easy to break it. For instance, you could have a naive set theory with more typing restrictions, such that propositions like x ∉ x aren't well-formed. It will still be inconsistent, but you'll have to do more work to construct a paradox. As another example, a type theory with Type : Type (that is, there's a type of all types) is inconsistent, but proving false is a lot of work.
So, to posit a reason for why people never make mistakes with their categorical proofs, despite it being a possibility given that people typically work in a naive system: it may well be much harder to construct circular proofs (in category theory) for most theorems that people are interested in than it is to write good proofs. For one, it might be hard to prove false. For two, proving false and then inferring whatever you want is obviously bad, so the paradoxical logic would have to be disguised as a legitimate argument. And that's unlikely to be the sort of reasoning a mathematician has in mind to prove things that aren't fairly related to paradoxes.<|endoftext|>
TITLE: Möbius and projective 3-manifolds
QUESTION [9 upvotes]: A projective 3-manifold is a smooth manifold that admits an atlas with values in the real projective 3-space such that all transition maps are restrictions of projective transformations. A Möbius 3-manifold is defined similarly, with the projective space replaced with the standard 3-sphere and projective transformations replaced with Möbius transformations. (Recall that a Möbius transformation of the sphere $S^n$ is a self-diffeomorphism of the sphere that preserves the angles of the standard metric; such transformations form a Lie group isomorphic to $SO_{n+1,1}(\mathbf{R}))/\pm I$).
Both projective and Möbius manifolds are particular cases of manifolds admitting an $(M,G)$-structure in the sense of W. Thurston.
Every closed (=compact, orientable and without boundary) 2-surface admits both a Möbius structure and a projective one. 
I vaguely remember having been to a talk some time ago where the speaker said that (conjecturally?) the situation in dimension 3 is similar. But I don't remember the details at all. So I would like to ask if anyone knows whether either of the statements (each closed 3-manifold admits a Möbius, resp. projective structure) is a theorem, a conjecture or becomes one or the other after eliminating some counter-examples.
A related question: if memory serves, in the same talk it was mentioned that the $PGL_{n+1}(\mathbf{R})$ and the Möbius group are (conjecturally?) the maximal groups that can act faithfully on an $n$-manifold. I was wondering if anyone knows a reference for this.

REPLY [7 votes]: The questions about manifolds admitting flat conformal or projective structures have been answered, so I'll just address your final question, which is whether $\mathrm{PGL}(n{+}1,\mathbb{R})$ and $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$ are 'maximal' finite dimensional groups acting on $n$-manifolds.
The answer depends on what you mean by 'maximal'.  They are maximal in the sense that any finite dimensional Lie subgroup of $\mathrm{Diff}(\mathbb{RP}^n)$ that contains $\mathrm{PGL}(n{+}1,\mathbb{R})$ must be equal to $\mathrm{PGL}(n{+}1,\mathbb{R})$ and any finite dimensional Lie subgroup of $\mathrm{Diff}(S^n)$ that contains $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$ must be equal to $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$.  This is essentially due to Élie Cartan, who proved the corresponding statement that the corresponding Lie algebras of vector fields are maximal finite dimensional Lie algebras in $\mathrm{Vect}(M)$ in his series of papers on so-called 'infinite groups' of diffeomorphisms.  For a more modern treatment, see Singer and Sternberg's "The infinite groups of Lie and Cartan, Part I (the transitive case)".
However, if by maximal, you mean 'highest (finite) dimensional Lie subgroup of $\mathrm{Diff}(M^n)$', then this is false, as there is no upper bound for trivial reasons:  Take $k$ disjoint open sets $U_j\subset M$ each with compact closure, and let $X_j$ be a nonzero vector field on $M$ that is supported in $U_j$.  Then the flows of the $X_j$ commute and generate an effective action of $\mathbb{R}^k$ on $M^n$.  Since $k$ can be arbitrary, it follows that there is no upper bound on the dimension of Lie subgroups of $\mathrm{Diff}(M)$.
Of course, this example is not satisfying, and you well might ask whether there is an upper bound for transitive actions.   Even then, the answer is 'no'.  For example, for a fixed integer $m$, consider the transformations of $\mathbb{R}^2$ that are of the form
$$
\Phi(x,y) = (\ ax{+}b,\ cy + d_m x^m + d_{m-1} x^{m-1} + \cdots + d_1 x + d_0\ ) 
$$
where $a,b,c, d_0,\ldots, d_m$ are constants with $ac\not=0$.  These constitute a Lie group of dimension $m{+}4$ acting effectively and transitively on $\mathbb{R}^2$.  A similar example can obviously be constructed on $\mathbb{R}^n$ for any $n>1$.  One can construct transitive examples for any $n>1$ on compact manifolds as well.
The key missing condition necessary for classification is to ask that the group act both transitively and primitively, i.e., that the group action not preserve any nontrivial foliation of $M$.  Then there is a classification, due in large part to Élie Cartan, but finally completed to modern standards of rigor by Ochiai in the 70s.  The conformal and projective transformation groups are indeed maximal among primitive, transitive actions, but there are many others.  For example, the action of $\mathrm{PGL}(n,\mathbb{R})$ on the Grassmannian of $k$-planes in $\mathrm{R}^n$ is maximal, primitive, transitive for every pair $(k,n)$.  For a given  dimension $n$ of the underlying manifold, the projective group, at dimension $n^2{+}2n$, is the largest primitive transitive Lie subgroup in overall dimension.<|endoftext|>
TITLE: Number fields with same discriminant and regulator?
QUESTION [17 upvotes]: Are there non-isomorphic number fields (say of the same degree and signature) that have the same discriminant and regulator? I'm guessing the answer is no - why?
And focusing on fields of small degree (n=3 and n=4), what about a less restrictive question: can we find two such fields that have the same regulator (no discriminant restrictions)?

REPLY [21 votes]: Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator.  [Edited to add examples where $K$ and $K'$ do not have the same roots of unity.] 
Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.((-\epsilon)^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$.  Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator.
For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let
$K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$.  Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.
The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator.  More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$.
[The requirement that $K$ have no roots of unity other than $\pm 1$ is used for this conclusion $O_K^* = O_F^*\phantom.$, but is not needed for equality of regulators.
EDIT Indeed it may happen that in such a pair of quartic fields $K$ had more roots of unity than $K'$: if $\epsilon \equiv 1 \bmod 4$ then ($\epsilon$ is totally positive and) $K=F\phantom.(\sqrt{-3})$ has sixth roots of unity while $K'=F\phantom.(\sqrt{-3\epsilon})$ does not.  The regulators are still the same unless $K = F\phantom.((-\epsilon)^{1/2})$, that is, unless $3\epsilon$ is a square in $F$, in which case the regulator of $K'$ is twice that of $K$.  For example we can take $F = {\bf Q}(\sqrt{203})$, which has $\epsilon = 57 + 4 \sqrt{203}$,  but not $F = {\bf Q}(\sqrt{39})$ because then $\epsilon = 25 + 4 \sqrt{39} = (6+\sqrt{39})^2/3$ so $K$ contains the square roots of $-\epsilon$.  TIDE]
Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and as for degree $3$, while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.<|endoftext|>
TITLE: Do morphisms locally decompose into finite surjective followed by smooth? (update: Is every projective variety over a finite field a finite cover of $\mathbb{P}^d$ for some $d$?)
QUESTION [5 upvotes]: I have looked through all my standard algebraic geometry texts and tried many tricks using Zariski's main theorem and Noether normalization, but remain stuck by the following:
Let $\pi:X\to S$ be a morphism of finite type between integral, Noetherian schemes and let $x$ be a point of $X$. Does there exist an open neighbourhood of $X$ which admits a finite, surjective morphism onto a smooth $S$-scheme?
In this generality I think that the answer is 'no', though I do not have a counterexample. What if we impose additional assumptions such as $\pi$ being flat or proper (or even projective)?
A related question, an affirmative answer to which would imply the same for the previous question in the projective case, is the following: if $X$ is a projective scheme over a local ring $A$, then does $X$ admit a finite surjection to $\mathbb{P}_A^d$ for some $d\ge 1$?
(I am imagining that $A=\mathbb{Z}_p$, so please do not assume that the residue field of $A$ is infinite!)
Thank you!

Update
With Brian's help (thank you), the interesting remaining problem is the following: does every projective variety $V$ over a finite field $k$ admit a finite surjective morphism to $\mathbb{P}_k^d$ for some $d$? I have a gnawing suspicion that the answer is 'no'.
It is useful to remember Noether normalisation in this case: if $I$ is a non-zero ideal of $k[X_1,\dots,X_n]$, then one can find a finite morphism $k[Y_1,\dots,Y_{n-1}]\to k[X_1,\dots,X_n]/I$ by sending $Y_i$ to $X_i-X_n^e$ for some big enough $e\ge 1$. Unfortunately, projectivising this construction produces a morphism $\mathbb{P}_k^n\setminus C\to\mathbb{P}_k^{n-1}$ where $C$ is quite a large closed subscheme of $\mathbb{P}_k^n$ (unless I have made a mistake); so if $V$ is our variety inside $\mathbb{P}_k^n$, then it is difficult to ensure that $V$ doesn't meet $C$. Therefore we can't successively project down to smaller dimensional spaces. (In contrast with the case when $k$ is infinite, for then we use changes of variables looking like $Y_i=X_i-\alpha_iX_n$ and the resulting morphism between projective spaces is defined everywhere except for one point, which we can assume doesn't lie on $V$).

REPLY [6 votes]: I claim that every projective scheme $V$, over a finite field $k$, all of whose components have dimension $\leq d$, admits a finite morphism to $\mathbb{P}^d_k$. Let $q=|k|$.
Key Lemma: Let $V_1$, $V_2$, ... $V_s$ be a finite collection of subvarieties of $\mathbb{P}_k^N$. Then there is a homogenous polynomial $f$ in $k[x_0, x_1, \ldots, x_N]$ such that $f|_{V_i}$ is nonzero for every $i$.
Proof: (improved thanks to comments below)
Choose a closed point $v_i$ on each $V_i$; we will force $f$ not to vanish at any of the $v_i$. Choose $M$ large enough that every $V_i$ is in $\mathbb{P}^N(\mathbb{F}_{q^M})$. Let $a_0$, $a_1$, ... $a_N$ be a basis for $\mathbb{F}_{q^{(N+1)M}}$ over $\mathbb{F}_{q^M}$. Then the linear form 
$$a_0 x_0 + a_1 x_1 + \cdots a_N x_N$$
does not vanish on any $Vvi$. This, of course, does not have coefficients in $k$. But the product
$$\prod_{i=0}^{(N+1)M-1} \left( a_0^{q^i} x_0 + a_1^{q^i} x_1 + \cdots a_N^{q^i} x_N \right)$$
is similarly nonzero, and does have coefficients in $k$. This concludes the proof of the key lemma.

Now, let $V$ be the projective variety, for which we want to prove the result. By the key lemma, there is some polynomial $f_0$, of degree $r_0$, which does not vanish on $V$. Every irreducible component of $V \cap \{ f_0 =0 \}$ is of dimension $\leq d-1$. Apply the lemma again to these irreducible components to find a polynomial $f_1$, of degree $r_1$, so that every component of $V \cap \{ f_0 = f_1 = 0 \}$ has dimension $\leq d-2$. Continuing in this manner, we construct polynomials $f_0$, $f_1$, ..., $f_d$ such that $V \cap \{ f_0 = f_1 = \cdots = f_d = 0 \}$ is empty.
Set $R=\prod r_i$. Our map $V \to \mathbb{P}^d_k$ will be given by 
$$(f_0^{R/r_0} : f_1^{R/r_1} : \cdots : f_d^{R/r_d})$$.
By our construction, this map has no base points, and is thus well defined. 

By the "standard argument", this map is finite. I now recall the standard argument.
Since $V$ is proper, we just need to check that the map has finite fibers. 
Suppose, for the sake of contradiction, that the the fiber above
$$(a_0: a_1 : \cdots a_{i-1} : 1 : a_{i+1} :  \cdots : a_d)$$
has positive dimension. 
Let $C$ be a curve in this fiber, so $f_j^{R/r_j} - a_j f_i^{R/r_i}$ vanishes on $C$ for all $j \neq i$. 
The curve $C$ must meet the hypersurface $f_i=0$, say at $z$. Then all the $f_i$ vanish at $z$, contradicting our construction.<|endoftext|>
TITLE: What are the oldest illustrations of "Venn" diagrams?
QUESTION [6 upvotes]: Graphical representations of intersection of sets as logical combinations are much older than Venn.
Euler and Leibniz are often quoted and the current Wikipedia article also quotes Ramon Llull but I do not really find the illustrations provided in the Wiki Commons for Llull very compelling.
I expect that these kind of ideas can be found in many other places and even older times, perhaps in disguise.
In this context I find the heraldic uses of theological diagrams such as shown here quite fascinating as a kind of medieval fashion statement.
Do you know of older examples of graphical representation of logical and/or set relations, for instance of Chinese, Arabic and Greek origin ?
(ps: at least one of the tags is a joke)

REPLY [3 votes]: You may already be familiar with Ruskey and Weston's "A Survey of Venn Diagrams," which includes a discussion of Borromean rings. Such rings are similar to the valknut and the triskelion, of which the gankyil is a type. All of these figures are quite old.
Of course these observations don't answer your question about the use of graphical representations of logical and/or set relations in antiquity. 
Plato refers to diagrams, for example, in his discussion of the double-divided line in Book 6 of Republic and in Meno when Socrates questions Meno's slave about a problem in geometry -- how to find a square double in area to any given square. I imagine more examples can be found.
An interesting project would be to find examples of "visual" language in the works of Plato, Aristotle, and Euclid.
While writing this post I came across the following two references:
Edwards, Anthony W. F. Cogwheels of the Mind: The Story of Venn Diagrams. Baltimore, Maryland: The John Hopkins University Press, (2004). 
Kuehni, Rolf. "On the Source of d’Aguilon’s Arc Color Mixture Diagram." Unpublished manuscript, 2003.<|endoftext|>
TITLE: self-avoidance time of random walk
QUESTION [15 upvotes]: How many steps on average does a simple random walk in the plane take before it visits a vertex it's visited before?
If an exact formula does not exist (as seems likely), then I'm interested in good approximations.
I'd also like to know the standard nomenclature associated with the question, if any exists.

REPLY [12 votes]: [I've corrected a stupid mistake below and added an upper bound... Please check the numerical values!]
Well, I doubt that an explicit expression exists. However, it should be possible to get good bounds. The lower bound is easy: observe that
$$
E(T) = \sum_{k\geq 1} P(T> k-1) = 1+\sum_{k\geq 1} c_k 4^{-k},
$$
where $c_k$ is the number of self-avoiding paths of length $k$ (and, of course, $4^k$ is the number of all paths of length $k$), so that we get a lower bound by truncating this series. Using the (known) values for $c_k$, $k=1,\ldots,71$, we get
$$
E(T) > 4.58607909
$$
Now, you can get an upper bound by bounding the neglected part of the series using $c_k\leq 4 \cdot 3^{k-1}$. This gives you a very narrow interval containing the right value: if I have made no mistake ;) , we get
$$
4.58607909 < E(T) < 4.58607911
$$<|endoftext|>
TITLE: Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form.
QUESTION [66 upvotes]: OK so let's see if I can use MO to explicitly compute an example of something, by getting other people to join in. Sort of "one level up"---often people answer questions here but I'm going to see if I can make people do a more substantial project. Before I start, note
(1) the computation might have been done already [I'd love to hear of a reference, if it has]
(2) the computation may or may not be worth publishing
(3) If it is worth publishing, there may or may not be a debate as to who the authors are.
I personally don't give a hang about (2) or (3) at this point, but others might. Let me get on to the mathematics. Oh---just a couple more things before I start---this project is related to the mathematics at this question, but perhaps pushes it a bit further (if we can get it to work). I had initially thought about these issues because I was going to give them to an undergraduate, but the undergraduate tells me today that he's decided to do his project on the holomorphic case, and it seemed a bit daft to let my initial investment in the problem go to waste, so I thought I'd tell anyone who was interested. If no-one takes the bait here, I'll probably just make this another UG project.
OK so here's the deal. Say $K/\mathbf{Q}$ is a finite Galois extension, and $\rho:Gal(K/\mathbf{Q})\to GL(2,\mathbf{C})$ is an irreducible 2-dimensional representation. General conjectures in the Langlands philosophy predict that $\rho$ comes from an automorphic form on $GL(2)$ over $\mathbf{Q}$. The idea is that we are going to "see" this form in an explicit example where general theory does not yet prove that it exists.
Now the determinant of $\rho$ is a 1-dimensional Galois representation, and it makes sense to ask whether $det(\rho(c))$ is $+1$ or $-1$, where $c$ is complex conjugation. It has to be one of these, because $c^2=1$. The nature of the automorphic form predicted to exist depends on the sign. 
If the determinant is $-1$ then the form should be holomorphic, and a classical weight 1 cusp form. In this case the existence of the form is known, because it is implied by Serre's conjecture, which is now a theorem of Khare and Wintenberger.
If the determinant is $+1$ then the conjectural form should be a real-analytic function on the upper half plane, invariant under a congruence subgroup, and satisfying a certain differential equation (which is not the Cauchy-Riemann equations in this case). If the image of $\rho$ is a solvable group then the existence of this form is known by old work of Langlands and Tunnell. 
So in summary then, the one case where the form is not known to exist is when the determinant of complex conjugation is $+1$ and the image of Galois is not solvable.
Here is an explicit example. The polynomial
g5=344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5

has splitting field $L$, an $A_5$-extension of $\mathbf{Q}$ ramified only at the prime 1951. Now $A_5$ is isomorphic to the quotient of $SL(2,\mathbf{F}_5)$ by its centre $\pm 1$, and $L$ has a degree two extension $K_0$, also unramified outside 1951, with $Gal(K_0/\mathbf{Q})$ being $SL(2,\mathbf{F}_5)$. Turns out that $K_0$ can be taken to be the splitting field of the rather messier polynomial
g24 = 14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 -   1381768039105642956*x^6 + 4291028045077743465*x^8 -   2050038614413776542*x^10 + 287094814384960835*x^12 - 9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 +   152929135*x^20 - 50726*x^22 + x^24

I should perhaps say that David Roberts told me these polynomials in Jan 2008; they're in a paper by him and John Jones---but they learnt about them from a paper of Doud and Moore. 
Now $SL(2,\mathbf{F}_5)$ has two faithful 2-dimensional complex representations; the traces of each representation take values in $\mathbf{Q}(\sqrt{5})$ and one is of course the conjugate of the other. The determinant of both representations is trivial so in fact they are $SL(2,\mathbf{C})$-valued. Oh---also, all the roots of g24 are real---and hence $K_0$ is totally real. So what we have here is a representation
$$\rho_0:Gal(K_0/\mathbf{Q})\to GL(2,\mathbf{C})$$
which is conjectured to come from automorphic forms, but, as far as I know, the conjecture is not known in this case.
Unfortunately the conductor of $\rho_0$ is $1951^2$, which is a bit big. In fact let me say something more about what is going on at 1951. In the $A_5$ extension the decomposition and inertia groups at 1951 are both cyclic of order 5. If my understanding of what David Roberts told me is correct, in the $SL(2,\mathbf{F}_5)$ extension the decomposition and inertia groups are both cyclic of order 10 (in fact I just got magma to check this). But the upshot is that $\rho_0$ restricted to a decomposition group at 1951 is of the form $\psi+\psi^{-1}$ with $\psi$ of order 10. Which character of order 10? Well there are four characters $(\mathbf{Z}/1951\mathbf{Z})^\times\to\mathbf{C}^\times$ of order 10, and two of them will do, and two won't, and which ones will do depends on which 2-dimensional representation of $SL(2,\mathbf{F}_5)$ you chose.
The key point though is that if you get $\psi$ right, then the twist $\rho:=\rho_0\otimes\psi$ will have conductor 1951, which is tiny for these purposes.
Now, as Marty did in an $A_4$ example and as I did and Junkie did in a dihedral example
in the Maass form question cited above, it is possible to figure out explicitly numbers $b_1$, $b_2$, $b_3$,..., with the property that 
$$L(\rho,s)=\sum_{n\geq1}b_n/n^s.$$
If one had a computer program that could calculate $b_n$ for $n\geq1$, then there are not one but two ways that one could attempt to give computational evidence for the predictions given by the Langlands philosophy:
(A) one could use techniques that Fernando Rodriguez-Villegas explained to me a few months ago to try and get computational evidence that $L(\rho,s)$ had analytic continuation to the complex plane and satisfied the correct functional equation, and
(B) one could compute the corresponding real analytic function on the upper half plane, evaluate it at various places to 30 decimal places, and see if the function was invariant under the group $\Gamma_1(1951)$.
I don't know much about (A) but I once tried, and failed, to do (B), and my gut feeling is that my mistake is in the computer program I wrote to compute the $b_n$. But as junkie's response in the previous question indicates, there now seem to be several ways to compute the $b_n$ and one thing I am wondering is whether we can use the methods he/she indicated in this question.
Let me speak more about how I tried to compute the $b_n$. The character of the Maass form is $\psi^2$, the determinant of $\rho$. General theory tells us that $b_n$ is a multiplicative function of $n$ so we only need compute $b_n$ for $n$ a prime power. Again general theory (consider the local $L$-functions) says that for $p\not=1951$ one can compute $b_{p^n}$ from $b_p$. If $p=1951$ then $b_{p^n}=1$ for all $n$, because the decomposition and inertia groups coincide for 1951 in $K_0$ [EDIT: This part of the argument is wrong, and it explains why my programs didn't work. I finally discovered my mistake after comparing the output of mine and Junkie's programs and seeing where they differed. It's true that decomposition and inertia coincide in $K_0$ but when one twists by the order 10 character this stops being true. In fact $b_{1951}$ is a primitive 5th root of unity that I don't know how to work out using my method other than by trial and error.]. Finally, if the $L$-function of $\rho_0$ is $\sum_n a_n/n^s$ then $b_n=\psi(n)a_n$ for all $n$ prime to 1951, so it suffices to compute $a_p$ for $p\not=1951$.
To compute $a_p$ I am going to compute the trace of $\rho_0(Frob_p)$. I first compute the GCD of the degrees of the irreducible factors of g24 mod $p$. If $p$ doesn't divide the discriminant of g24 then this GCD is the order of $Frob_p$ in $SL(2,\mathbf{F}_5)$. If $p$ does divide the discriminant of g24 then rotten luck, I need to factorize $p$ in the ring of integers of the number field generated by a root of g24. I did this using magma. Here are the results:
                                                   prime   order
                                                     2       6
                                                     3      10
                                                     5       4
                                                   163       5
                                                 16061       1
                                                889289      10
                                          451400586583       2
                             1188493301983785760551727       2
120450513180827412314298160097013390669723824832697847       1

Now unfortunately just computing the order of the conj class of Frobenius is not enough to determine the trace of the Galois representation, because $SL(2,\mathbf{F}_5)$ contains two conj classes of elements of order 5, and two of order 10. However, in both cases, the conj classes remain distinct in $A_5$ and so it suffices to have an algorithm which can distinguish between the two conjugacy classes in $A_5$. More precisely, we have to solve the following problem: we label the conj classes of elements of order 5 in $A_5$ as C1 and C2, and we want an algorithm which, given a prime $p$ for which g5 is irreducible mod $p$, we want the algorithm to return "C1" or "C2" depending on which class $Frob_p$ is in. Here is a beautiful way of doing it, explained to me by Bjorn Poonen: if g5 is irreducible mod $p$ then its roots in an alg closure of $\mathbf{F}_p$ are $x,x^p,x^{p^2},...$. Set $x_i=x^{p^i}$ and compute $\prod_{i&lt;j}(x_i-x_j)$. This product is a square root of the discriminant of g5. Choose once and for all a square root of the discriminant of g5 in the integers; if the product is congruent to this mod $p$ return "C1", else return "C2".
That's it! I implemented this. I had a program which returned a bunch of $a_n$'s, and hence a bunch of $b_n$'s. I built the function on the upper half plane as the usual sum involving Bessel functions and so on, but it computationally did not come out to be invariant under $\Gamma_1(1951)$. 
If anyone wants to take up the challenge of computing the $b_n$ that would be great. I have explained one way to do it above, but I am well aware that there might be other ways to compute the $b_n$ analogous to junkie's approach from the previous question.

REPLY [8 votes]: I have accepted Junkie's answer to this question. But on analysing his working program and comparing it with my program-that-didn't-work I could easily find the bug in my program: I erroneously thought that the coefficient of $q^{1951}$ was 1 and it's not, it's a primitive 5th root of unity. My code works now, and is written in PARI-GP, so I thought I would post it so that people who don't have access to magma can play with the miracle that is a cuspidal Maass eigenform that (conjecturally) corresponds to a non-solvable even Galois representation. I make my post community wiki because my post says nothing that Junkie's doesn't so I don't want any credit for it.
\\ nonsolvmaass.g: works out a maass form associated to an explicit
\\ nonsolvable finite image even irred 2-d Galois rep.
\\ Note: rho(c)=-1 so we use sines.

\\ N is number of coefficients of the L-function we will compute.

N=2000000;

\\ g5 has splitting field an A_5 field.

g5=344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5

\\ g24 has splitting field a field with Galois group SL(2,5).

g24 = 14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 -   1381768039105642956*x^6 + 4291028045077743465*x^8 -   2050038614413776542*x^10 + 287094814384960835*x^12 -   9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 +   152929135*x^20 - 50726*x^22 + x^24

\\ The character table of SL(2,5) starts like this:
\\
\\
\\ -------------------------------------------
\\ Class |   1  2  3  4    5    6  7    8    9
\\ Size  |   1  1 20 30   12   12 20   12   12
\\ Order |   1  2  3  4    5    5  6   10   10
\\ -------------------------------------------
\\ X.1   +   1  1  1  1    1    1  1    1    1
\\ X.2   -   2 -2 -1  0   Z1 Z1#2  1  -Z1-Z1#2
\\ X.3   -   2 -2 -1  0 Z1#2   Z1  1-Z1#2  -Z1
\\
\\...
\\
\\ # denotes algebraic conjugation, that is,
\\ #k indicates replacing the root of unity w by w^k
\\
\\ Z1     = (CyclotomicField(5)) ! [ RationalField() | 0, 0, 1, 1 ]
\\
\\ If we let K be the splitting field of g24 then K contains a splitting
\\ field for g5. If we fix an isomorphism of Gal(K/Q) with SL(2,5) and
\\ use X.2 to define a 2-dimensional representation of Gal(K/Q), then
\\ the resulting representation has conductor 1951^2. 
\\
\\ Here is a pari script to compute the order of Frob_p for the resulting
\\ representation (for p not 1951). First a function that only works for the primes
\\ not dividing disc(g24).

ordfrob0(p)=local(m);m=factormod(g24,p);lcm(vector(#m~,i,poldegree(m[i,1])))

\\ For the primes dividing disc(g24) other than 1951 I used magma to work
\\ out the answer by hand [computing a maximal order etc].

ordfrob(p)=if(p==1951,0,if(p==2,6,if(p==3,10,if(p==5,4,if(p==163,5,if(p==16061,1,if(p==889289,10,if(p==451400586583,2,if(p==1188493301983785760551727,2,if(p==120450513180827412314298160097013390669723824832697847,1,ordfrob0(p)))))))))))

\\ If the order of Frob_p in SL(2,5) is 5 or 10 then we have to decide which
\\ of the two conjugacy classes p is in! Classes 5 and 8 reduce to the same
\\ class in A_5, and classes 6 and 9 reduce to the other. So we have to
\\ be able to distinguish between the classes in A_5. Bjorn Poonen told
\\ me how to do this. The trick is to compute the product whose square
\\ is the discriminant mod p and to see which of the square roots of the
\\ discriminant that the product itself is isomorphic to.
\\ This will not work for any p dividing the discriminant of g5.
\\ Fortunately I checked explicitly that no prime p dividing disc(g5)
\\ has Frob_p of order 5 or 10. Hooray for good fortune!
\\ The function returns 0 or 1.

whichclass(p)=local(t,conjs,pro);t=Mod(x,Pol(Mod([1, -1, -780, -1795, 3106, 344],p)));conjs=[t];for(i=1,4,conjs=concat(conjs,conjs[i]^p));pro=1;for(i=1,5,for(j=i+1,5,pro=pro*(conjs[i]-conjs[j])));pro==518348075378;

\\ Now finally an algorithm to compute a_p.

zet10=exp(2*Pi*I/10);Z1=zet10^4+zet10^6;Z1h2=zet10^2+zet10^8;

ap(p)=if(p==1951,0,orp=ordfrob(p);if(orp==1,2,if(orp==2,-2,if(orp==3,-1,if(orp==4,0,if(orp==6,1,whp=whichclass(p);if(orp==5,if(whp==0,Z1,Z1h2),if(whp==0,-Z1,-Z1h2))))))))

\\ The problem with the Maass form associated to these eigenvalues is that, although it
\\ has trivial character, it has conductor 1951^2. At a prime above 1951, the
\\ inertia subgroup has order 10, as does the decomposition group. The representation
\\ sends a generator of this inertia subgroup to something with eigenvalues zet10
\\ and zet10^9. So we need to twist by a Dirichlet character of conductor 1951
\\ and order 10 and we will get a representation of conductor 1951.
\\ But which one?! Well, there are only four, and two will work, so we have
\\ a sporting chance! If we get it wrong we can just try another one.
\\ Note that 3 is a primitive root mod 1951.

twist=vector(1950,i,0);z0=zet10^3;z=z0;n=Mod(3,1951);for(i=1,1950,twist[lift(n)]=z;n=n*3;z=z*z0);

\\ The character of the form we are after is the square of "twist".

\\ Here's the trace of Frobenius on the conductor 1951 twist.
\\ Note that the value at p=1951 was found by trial and error.
\\ The error I made, that Junkie fixed, was that I thought that "pure thought" gave
\\ that bp(1951)=1.

bp(p)=if(p==1951,zet10^6,ap(p)*twist[p%1951])

\\ Here's the character of the form.

chi(p)=if(p==1951,0,twist[p%1951]^2);

\\ Now v will be the vector of Hecke eigenvalues.

v=vector(N,i,0);
v[1]=1;
for(i=2,N,if(i%1000==0,print(N-i));fac=factor(i);k=matsize(fac)[1];\
   if(k>1,v[i]=prod(j=1,k,v[fac[j,1]^fac[j,2]]),\
   if(fac[1,2]==1,v[i]=bp(i),\
   p=fac[1,1];e=fac[1,2];v[i]=v[p]*v[p^(e-1)]-chi(p)*v[p^(e-2)]))\
);

\\ Now glue them together with the Maass form analogue of a q-expansion.
\\ The function q^n is replaced by sqrt(y)*K(2.Pi.n.y)*sin(2.Pi.n.x)
\\ where K=K_0 is a certain Bessel function.

F(z)=local(x,y,M);x=real(z);y=imag(z);M=ceil(11/y);if(M>N,error("y too small."));sqrt(y)*sum(n=1,M,if(v[n]==0,0,v[n]*besselk(1e-30*I,2*Pi*n*y)*sin(2*Pi*n*x)))

If you cut and paste that into a pari session then it will either run out of memory, or spend a long time counting down from 2,000,000 to 0 and then stop. If it runs out of memory then type something like "allocatemem(2000000000)" or some such thing. If you can't be bothered to wait for the countdown to finish then change the "$N=$" line to something smaller at the beginning. It's computing the first $N$ coefficients of the Maass form. 
The form itself is the function $F$ defined on the last line. The larger you let $N$ be, the better precision $F$ will be defined to, and hence the nearer to the real line you'll be able to evaluate $F$. The point is that if $z$ is in the upper half plane but very close to the real line then you need a lot of terms of $F$ to evaluate $F(z)$ accurately.
As in Junkie's post, we may try a random verification, to see whether $F$ looks like it's invariant under $\Gamma_1(1951)$:
(22:47) gp > zz4=0.0001+0.0001*I
%7 = 0.0001000000000000000000000000000 + 0.0001000000000000000000000000000*I
(22:47) gp > F(zz4)
%8 = 0.000000008720947535106531893030584189 - 0.00000001730002410267070596763764527*I
(22:48) gp > F(zz4/(1951*zz4+1))
%9 = 0.000000008720947535106531893415409754 - 0.00000001730002410267070596828184932*I
(22:51) gp > %8-%9
%10 = -3.848255644 E-28 + 6.442040498 E-28*I

The point is that the difference between the two values of $F$ is a tiny number, so $F$ is, to within computational error, seemingly invariant under $\Gamma_1(1951)$. People can play with other examples, but be warned: the smaller $N$ is (i.e. the less time you could be bothered to wait at the beginning) the more careful one has to be (i.e. the more likely the program is to refuse to compute $F$ where you want it to compute it).<|endoftext|>
TITLE: Monoidal structures on von Neumann algebras
QUESTION [18 upvotes]: My question is based on the following vague belief, shared by many people: It should be possible to use von Neumann algebras in order to define the cohomology theory TMF (topological modular forms) in the same way one uses Hilbert spaces in order to define topological K-theory. More precisely, one expects hyperfinite type $\mathit{III}$ factors to be the analogs of (separable) infinite dimensional Hilbert spaces.
Now, here is a fundamental difference between Hilbert spaces and type $\mathit{III}$ factors: The category of Hilbert spaces has two monoidal structures: direct sum $\oplus$, and tensor product $\otimes$, and both of them preserve the property of being an infinite dimensional Hilbert space.
In von Neumann algebras, the tensor product of two hyperfinite type $\mathit{III}$ factors is again a hyperfinite type $\mathit{III}$ factor, but their direct sum isn't (it's not a factor).
Hence my question: are there other monoidal structures on the category of von Neumann algebras that I might not be aware of? More broadly phrased, how many ways are there of building a new von Neumann algebra from two given ones, other than tensoring them together?
Ideally, I would need something that distributes over tensor product, and that preserves the property of being a hyperfinite type $\mathit{III}$ factor... but that might be too much to ask for.

REPLY [2 votes]: Suppose that there was such a category. Then, all objects would isomorphic to $R$ anyway and the question is how sum and tensor product act on the morphisms. The natural choice of morphisms is the set of endomorphisms of the hyperfinite factor. Now, pick a unital inclusion $\iota :R \oplus R \subset R$ and and isomorphism $\mu \colon R \cong R \otimes R$ and define
$$(\phi \oplus \psi)(x) := \iota(\phi(x) \oplus \psi(x)) \quad \mbox{and} \quad
(\phi \otimes \psi)(x) := \mu( (\phi \otimes \psi)(\mu^{-1}(x))).$$
The question is now how sum and tensor product behave. This of course requires a bit of work (and may depend on the choice of $\mu$ and $\iota$). 
However, let us look more concretely. Assume for the moment that $R = \otimes_{n \in {\mathbb N}} M_2 {\mathbb C}$ (with some fixed state on $M_2 {\mathbb C}$). Then, it seems that the inclusion ${\mathbb C}^{\oplus 2} \subset M_2 {\mathbb C}$ in the first factor gives some $\iota$ and a bijection between ${\mathbb N}$ and ${\mathbb N} \times {\mathbb N}$ yields some map $\mu$. It seems straightforward that this should give a bimonoidal category with the required distributivity isomorphisms.
Added August 15 to incorporate Andre's remarks:
The choice of $\iota$ is basically a choice of an infinite projection $p \in R$ and isomorphisms $\phi : R \to pRp$ and $\psi : R \to (1-p)R(1-p)$. Since all projections in $R$ are equivalent (we are in type $III$) i.e. $p \sim 1-p$, $\phi(p) \sim p$ and $\psi(p) \sim p$ the sum should be coherently associative. I think that it is a first step to provide an (inner) isomorphism of $R$ which maps the triple $p, \psi(p),\psi(1-p)$ to $\phi(p),\phi(1-p),1-p$. I have to really think about it, but each triple is a triple of pairwise orthogonal projections in $R$ which sums up to one. I think, that up to unitary conjugacy, there is only one such triple (in the type $III$-setting).<|endoftext|>
TITLE: Is -1 a sum of 2 squares in a certain field K?
QUESTION [13 upvotes]: Consider the field of fractions $K$
of the quotient algebra $\mathbb{R}[x,y,z,t]/(x^2+y^2+z^2+t^2+1)$,
where $\mathbb{R}$ is the field of real numbers and $x,y,z,t$ are variables.
Clearly $-1$ is a sum of 4 squares in $K$.
How can one prove that $-1$ is not a sum of 2 squares in $K$?
Serre mentions without proof this (probably known or easy) fact
in a letter to Eva Bayer of May 1, 2010,
and I am stuck: I cannot prove it.

REPLY [22 votes]: This is a special case of a theorem of A. Pfister.  It is well known to quadratic forms specialists.  See e.g. Theorem XI.2.6 in T.Y. Lam's Introduction to Quadratic Forms over Fields.  
I believe the original paper is 

Pfister, Albrecht,
  Zur Darstellung von $-1$ als Summe von Quadraten in einem Körper. (German)
  J. London Math. Soc. 40 1965 159--165. 

In this same paper Pfister defines the "stufe" (which Lam has successfully campaigned to be called the "level") of a non-formally real field, namely the least positive integer $n$ such that $-1$ is a sum of $n$ squares.  Among his other achievements, he proves that the level is always a power of $2$ (so that Kevin Buzzard's recollection is correct).  It is also worth remarking that his work is an insightful and rapid response to previous work of J.W.S. Cassels.<|endoftext|>
TITLE: Math History books
QUESTION [12 upvotes]: I'm teaching a course over the summer (it's a sort of make-your-own course for non-majors) and I'm planning on organizing it as a math history course, hitting on major threads through about 1900, and focusing on the evolution of ideas and on people, rather than on the details of proofs.  I've also been having a lot of trouble finding a good book covering this material (none finding books on ancient mathematics, but I want to focus on Renaissance to 19th Century, if possible), and so, here's my question:

What would be a good textbook for a course of this nature? Specifically, for a math history course targeted at non-science majors.

REPLY [3 votes]: The question might use a bifurcation: do you want a systematic treatise or a collection of episodes?
In the latter case, there is Gindikin's Tales of Physicists and Mathematicians. It is a collection of (mostly independent, if my memory doesn't cheat me - I have read parts of the Russian original) "tales" on some particular people and/or developments, mostly pre-1900. History is interlaced with mathematics - but almost all of the latter should be on high school level. Probably this is better used in a student seminar rather than in a lecture course, though.

REPLY [3 votes]: I think Development of mathematics in the 19th century by Felix Klein would be useful.<|endoftext|>
TITLE: Non-diagonalizable complex symmetric matrix
QUESTION [19 upvotes]: This is a question in elementary linear algebra, though I hope it's not so trivial to be closed.
Real symmetric matrices, complex hermitian matrices, unitary matrices, and complex matrices with distinct eigenvalues are diagonalizable, i.e. conjugate to a diagonal matrix.

I'd just like to see an example of a complex symmetric $n\times n$ matrix that is not diagonalizable.

REPLY [38 votes]: $$\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}.$$
How did I find this? Non-diagonalizable means that there is some Jordan block of size greater than $1$. I decided to hunt for something with Jordan form $\left( \begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix} \right)$. So I want trace and determinant to be zero, but the matrix not to be zero. The diagonal entries made sure the trace vanished, and then the off diagonal entries were forced.<|endoftext|>
TITLE: Books about history of recent mathematics
QUESTION [79 upvotes]: I draw on this question to ask something that has always been a pet peeve of mine. It is very easy to find books about the history of mathematics, much less so if one wants books about the recent (say > 1850) one.
Of course I know that this is difficult because not so many people would understand what's going on; to learn about the history of a subject, one should better know the subject beforehand. On the other hand, my feeling is that more or less all mathematics I know has been developed after 1850, and the growth, like in many other sciences, has been exponential. So the amount of mathematics which appears in history book seems negligible to me.
Can you point me to any good resources about the recent history of maths?

REPLY [7 votes]: I didn't see this mentioned elsewhere: the AMS has a whole series devoted to the history of mathematics, much of it fairly recent.<|endoftext|>
TITLE: Do Berkovich homogenous spaces exist?
QUESTION [10 upvotes]: Let G be a k-analytic group, and let H be a closed subgroup of G. Then does there exist a k-analytic space, which can be reasonably called the quotient G/H?
Commentary: I realise that I am not being overly precise here. This is partly because I am not sure in exactly what form I want to ask this question. In lieu of this, I will mention an example, that I suspect may produce an interesting quotient space.
Let Spec(A) be an affine group scheme of finite type over ko. Let G be the analytification of the generic fibre of Spec(A). On $A\otimes_{k^\circ} k$, one can define the seminorm where ||a|| is defined to be the infimum of all |t| where $t\in k^\times$ is such that $a\in tA=t(A\otimes_{k^\circ} 1)\subset A\otimes_{k^\circ} k$. Take the completion of this tensor product with respect to this seminorm, and let H be its spectrum (which is then naturally an affinoid subgroup of G).

REPLY [6 votes]: There is an old and forgotten paper of Z. Bosch on homogeneous spaces for affinoid groups (in German), where he construct rigid analytic quotients for affinoid groups and also proves some properties. The same arguments should work in the case of Berkovich spaces.<|endoftext|>
TITLE: Why is the intersection of complex submanifolds always positive.?
QUESTION [5 upvotes]: Hi, everyone:
    I was finally able to show that all complex manifolds are orientable, by
    generalizing to many variables the fact that , for a single complex variable,
    the Jacobian matrix is of the form (after using Cauchy-Riemann to substitute).
     ( This is my first post here. I read the FAQ's, but I apologize if I am not following protocol correctly.Please let me know if so.)
(a b)
(-b a) 

which has non-zero determinant a2+b2 . We can induct, to
    show something similar holds for higher dimensions, i.e., the Jacobian ( of overlapping charts will necessarily be positive.
    Now, a couple of questions, please:
1)Is there a more topological proof of the orientability.?.I thought of using Lie theory, that Gl(n;C) is connected, may work, but I don't see how to rigorize this argument; in Milnor and Stasheff's book , it is stated that(paraphrase) this path connectedness allows one basis to be deformed into another homotopically, so that orientation is preserved. But, AFAIK, the bases are just elements in Cn. Any ideas on this direction.?. I know that in Cn, connectedness implies path-connectedness. I also know that Gl(n;C) can be embedded in Gl+(2n;IR),
one of the connected components of Gl(2n;IR). I think this helps, but I don't know
how to "rigorize" this idea. Any suggestions, please.?   
I also wonder if one can generalize the CW-decomposition of CPn, where we can see that there is only one cell in the top dimension, to other complex manifolds.
2)Any suggestions, please, for showing that complex submanifolds S1,S2( of the "right dimensions, to make sure they can intersect) of a manifold M ,have positive intersection (self- or otherwise).? . I understand that, at every point p of intersection, we append the
tangent spaces, defining:
Tp(S1/\S2)=Tp(S1)(+)Tp(S2)
and then the intersection is positive if there is an even permutation from the basis
of TpM to the basis of Tp(S1/\S2). But I have no idea of how to show that, for complex
submanifolds, the intersection is always positive.
Thanks For Any Help/Suggestions.

REPLY [6 votes]: You can answer both your questions with the following remark: if $x_1,\ldots,x_n$ is a complex basis of $\mathbb{C}^n$, then $x_1,ix_1,x_2,ix_2,\ldots,x_n,ix_n$ is a real basis of $\mathbb{C}^n$ whose orientation does not depend upon $x_1,\ldots,x_n$ (in particular it does not depend upon their order).
Now you can give an orientation to a complex manifold $M$ by locally choosing a complex basis and use the subsequent local orientation (which will be globally defined thanks to the remark), and if you endow two complex submanifolds $N_1,N_2$ of complementary dimensions with the orientations defined in the same way, then automatically at each (transversal) intersection point $x$ the orientation of $T_x N_1\oplus T_xN_2$ obtained by concatenation is the same than that of $T_x M$. This exactly means that the intersection is positive.<|endoftext|>
TITLE: Fourier transform of Analytic Functions
QUESTION [12 upvotes]: Forgive me if this question does not meet the bar for this forum. But i would really appreciated some help.
I'm trying to construct a function according to some conditions in the frequency domain of the Fourier transformation. I want the function to be analytic and real when I transform it back to the time domain. The Fourier transformation of $f$ has of course some symmetry criteria to make $f$ real. But what about the Analytic property. As an analytic function imply some convergent power series expansion, and the Fourier transform of a polynomial is a sum of derivatives of Delta functions, I assume that there is a corresponding criteria of the Fourier transformation.
So the question is:
If a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is assumed to be analytic, what is the corresponding criteria for the Fourier transform of the function $\mathcal{F}[f] (k)$?
Edit: what I am trying to construct is probability distribution with the following condition
$f(x/\mu)/\mu=\frac{2}{3} f(x) + \frac{1}{3} (f\ast f)(x)\quad$
where $\ast$ mark the convolution, and $\mu=\frac{4}{3}$. $f$ is positive and real for $x\in [0,\infty)$
Taking the fourier transformation make the condition simpler:
$\tilde f(\mu k) = \frac{2}{3}\tilde f(k) + \frac{1}{3}\tilde f^2(k)$
So my problem is to construct $f$ (I am in particular interested in the tail behavior) and I try to use the properties of $\tilde f$. I posted a similar problem a while ago (see here). Julián Aguirre answered how to construct $\tilde f$ if it is analytic. But the inverse transformation of the power expansion is an infinite sum of derivatives of Delta functions, and is of little help.

REPLY [4 votes]: The following basic result needs to be quoted on these matters of analyticity: the Paley-Wiener-Schwartz theorem gives a characterization of distributions with compact support. Let $u$ be a tempered distribution ; $u$ is compactly supported in a ball of center $0$ and radius $R$ if an only if $\hat u$ is an entire function such that
$$\exists C_0, \exists N_0,\forall \zeta\in \mathbb C^d,\quad
\vert\hat u(\zeta)\vert\le C_0(1+\vert \zeta\vert)^{N_0}e^{R\vert\Im\zeta\vert}.
$$
Something analogous allows a characterization of $C^\infty$ functions with compact support.
This leads to the following characterization of the analytic wave front set, due to Bros and Iagolnitzer.
Let $v\in \mathcal E'(\mathbb R^d)$.
We define the Fourier-Bros-Iagolnitzer transform 
$Tv$
of $v$ by the following  formula, where the integral is in fact a bracket of duality,
$$
(Tv)(z,\lambda)=\int_{\mathbb R^d} e^{-\pi\lambda(z-x)^2} v(x) dx,\qquad z\in \mathbb C^d, \lambda >0.
$$
Let $\Omega$ be an open subset of $\mathbb R^d$;
let us note $\Omega\times(\mathbb R^d\backslash
\{0\})$ by $\dot T^*(\Omega)$ and by $dL(z)$ the Lebesgue measure on 
$\mathbb C^d$.
Let $u\in \mathcal D'(\Omega)$.
The analytic wave-front-set of $u$, denoted by $WF_{A}(u)$,
is the complement in $\dot T^*(\Omega)$
of the set of points $(x_{0},\xi_{0})$
such that
$$
\exists W_{0}\in \mathscr V_{x_{0}-i\xi_{0}},
\exists \chi_{0}\in C^\infty_c(\Omega), \chi_{0}(x)=1\  \text{near $x_{0}$},
\exists \epsilon_{0}>0\quad \text{with}
$$
$$
\sup_{\lambda\ge 1, z\in W_{0}}e^{\epsilon_{0}\lambda}
\vert{(T\chi_{0} u)(z,\lambda)}\vert e^{-\pi\lambda(\Im z)^2}
<+\infty.
$$
The first projection of $WF_A(u)$ is the analytic singular support.<|endoftext|>
TITLE: representability of consecutive integers by a binary quadratic form
QUESTION [14 upvotes]: I have two related questions on the representability of integers 
by quadratic forms in two variables :
(1) Let $f: {\mathbb Z} \times {\mathbb Z} \to {\mathbb Z}  $ be such a quadratic
form, i.e. we have $f(x,y)=ax^2+bxy+cy^2+dx+ey+g$ for some integer constants
$a,b,c,d,e,g$. Suppose that $f$ is not surjective, i.e. some integer is not represented
by $f$. Is it true that there is an integer constant $C$ such that in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
(2) If the answer to (1) is yes, is there a uniform bound ? In other words,
is there a uniform constant $C$ such that for any non-surjective $f$, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
Update : Good answers to my original questions appeared quickly. It seems the only
interesting subquestion left is the one asked by fedja : 
(2') is there a universal $C$ such that for any $f$ with positive definite  quadratic part, in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?
One may also ask, 
(3)  is there a universal $C$ such that for any non-surjective and irreducible $f$,  in any block of
$C$ consecutive integers, at least one of them is not represented by $f$ ?

REPLY [4 votes]: Seems worth recording my thoughts so far. I am sticking with positive forms, which means homogeneous. My guess of $C=31$ was probably larger than necessary, but we will see. Certainly I think there is a finite uniform $C$ and it can be taken quite small. 
It turns out to be quite difficult to have a positive binary represent any string of 7 consecutive positive integers. I finally managed it, $x^2 + x y + 18 y^2$ represents the numbers from 1,953,139 to 1,953,145. Of course, even for this form, most intervals of length 7 contain a nonrepresented number. 
If the discriminant is even there is a congruence obstruction on represented numbers mod 4 or mod 8, of a nature that prevents strings longer than 5. The form $x^2 + 2 y^2$ does well, many strings of length 5 as high as I could check.
If the discriminant is 5 mod 8 no numbers 2 mod 4 are represented. 
So we have explicit bounds of length no more than 5 unless $ \Delta \equiv 1 \pmod 8.$ Even here, for composite discriminant, say $ 3 | \Delta $ any form of that discriminant is restricted to all quadratic residues or all quadratic nonresidues $ \pmod 3 ,$ maximum length 3. Similar for  $ 5 | \Delta $ or  $ 7 | \Delta .$
The way to get big lengths is prime discriminants with either a long set of consecutive residues or a long set of consecutive nonresidues. Residues seem a better bet, as if 2,3,5 are residues then so are 1,2,3,4,5,6. And, indeed we have $ 1953139 \equiv 0 \pmod {71}, \ldots, \;  1953145 \equiv 6 \pmod {71} .$ Note that 7 is a nonresidue mod 71, the first nonresidue is always prime (for a prime modulus). 
Note that Fedja's argument also concerns the first nonresidue $q.$ Whatever else may or may not matter, the maximum length is $2 q - 1,$ for a possible interval from $ k q^2 - q + 1$ through to $ k q^2 + q - 1.$<|endoftext|>
TITLE: What are the normal subgroups of a direct product?
QUESTION [11 upvotes]: Let $N$ be a normal subgroup of $G \times H$, and let $\pi_1: G \times H \to G$ and $\pi_2: G \times H \to H$ be the canonical projections. Then $\pi_1(N)$ is normal in $G$ and $\pi_2(N)$ is normal in $H$. What else can we say? I know that it is not true, in general, that $N \simeq \pi_1(N) \times \pi_2(N)$.
I'm particularly interested in the case where $G$ and $H$ are simple. In that case, $N \simeq \pi_1(N) \times \pi_2(N)$ except possibly in the case where $\pi_1(N) = G$ and $\pi_2(N) = H$. In that case, what do we know?

REPLY [5 votes]: The following corollary is from Normal Subgroup Growth of Linear Groups: the $(G_2, F_4, E_8)$-Theorem, by Michael Larsen and Alex Lubotzky.

Corollary 1.4:
  Let $G = A \times B$ be the product of two groups. If for every (finite index) normal subgroup $M \lhd A$, $Z(A/M) = \{ 1 \}$, then every (finite index) normal subgroup $N \lhd G$ is of the form $N = (N \cap A) \times (N \cap B)$.<|endoftext|>
TITLE: Does this approach for the Poincaré conjecture work?
QUESTION [21 upvotes]: Several months ago a paper was posted at
http://arxiv.org/abs/1001.4164
called "Another way of answering Henri Poincaré's fundamental question." The author gave a talk on it today at my institution. If it's correct, it is a major breakthrough in terms of proof length (~10 pages). However, it is very outside my specialty. There's apparently been very little feedback, but the author is ok with public discussion. Therefore, can anyone say whether they have read the paper? Whether it is correct, missing details, clearly flawed, or what have you?

REPLY [15 votes]: I also Had a quick look (maybe a little less quick), and although I very much like the other answer, which illustrates that it may be difficult to fix, I may have found a more specific error, which may be more helpfull as an answer to the question.
Firstly, I am a little confused as to what constitutes a stratification. I see two possibilities:
1) The one which is actually defined which allows the following stratification: $S_1=S_2=D^2\times [0,1]$ and they are glued along a closed disc in the interior of $D^2\times \{1\}$ of $S_1$ and the same disc in the interior of $D^2 \times \{0\}$ in $S_2$.
2) The one which I think is implied at some points:
$S_i$ and $S_{i+1}$ may only be identified such that $U(S_i) \cup L(S_{i+1})$ is in fact a sub-surface in the 3-manifold.
I will describe my problems related to both definitions:
In the proof of prop 5.8 parts (2-3-4) he attaches "3-cell"s (I would write 3-disc as to avoid confusion with CW complex attachments of cells, or attach both a 2-cell and a 3-cell) $W$, and extends the stratification.
If we work under definition 2) above then this seems generally impossible because you would often also have to attach it at the top of $S_{i+1}$ to get the extra surface assumption in 2).
If we work under definition 1) above then this doesn't even make the new $F_{i+1}$ a surface in the simple example described above.<|endoftext|>
TITLE: What can we say about center of rational absolute Galois group?
QUESTION [14 upvotes]: Well the question is in the title.
I asked myself this question while thinking about something in Grothendieck-Teichmüller theory. I guess class field theory gives some insight into this, or I am missing something absolutely obvious..

REPLY [36 votes]: The proof of triviality is a step in the famous Neukirch-Uchida theorem of anabelian geometry, which says a number field is characterized by its absolute Galois group, even functorially, in an appropriate sense. The key elementary fact is the following:
Let $k$ be a number field, $K$ an algebraic closure, and $G=Gal(K/k)$. Let $P_1$ and $P_2$ be two distinct primes of $K$ with corresponding decomposition subgroups $G(P_i)\subset G$. Then
$G(P_1)\cap G(P_2)=1.$
Once this is stated for you, it's essentially an exercise to prove. 
Determining the center of $G$ becomes then completely straightforward: Suppose $g$ commutes with everything. Then for any prime $P$, $G(gP)=gG(P)g^{-1}=G(P)$. So $g$ must fix every prime, implying that it's trivial.
I think this is spelled out in the book Cohomology of Number Fields, by Neukirch, Schmidt, and Wingberg. Unfortunately, I left my copy on the plane last year.<|endoftext|>
TITLE: Simplicial complexes vs. geometric realization of abstract simplicial complexes
QUESTION [7 upvotes]: A finite abstract simplicial complex is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.
For $n\geq 0$ the topological space $\Delta^n=\{(x_0,...,x_n)\in\mathbb{R}^{n+1}\mid x_i\geq 0, \sum x_i =1\}$ is called the standard $n$-simplex. A topological space homeomorphic to the standard $n$-simplex is called an $n$-simplex. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.
A finite topological simplicial complex is a pair $(X,F)$ where $X$ is a topological space and $F=(F_1,...,F_m)$ is a finite sequence of embeddings $F_k:\Delta^{i_k}\to X$ such that

$X=\cup_k F_k(\Delta^{i_k})$
$F_k\neq F_l$ if $k\neq l$ 
for every $1\leq k\leq m$ with $i_k\geq 1$ and for every face $A$ of the $i_k$-simplex $F_k(\Delta^{i_k})$ there is a $1\leq l\leq m$ with $F_l(\Delta^{i_l})=A$
for every $1\leq k\neq k'\leq m$ the simplex $F_k(\Delta^{i_k})\cap F_{k'}(\Delta^{i_{k'}})$ is a face of each of them.

I hope, the definitions are correct.
There is the notation of a geometric realization of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.
If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.
My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?
I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:
A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the same directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in opposite directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?

REPLY [3 votes]: To answer just a small part of your question, three small picture-examples that can help see why "edges match up in opposite directions" is the right condition for "orientability":

Analogy with how vertices match up when you glue paths together along vertices.  Each path has one vertex oriented as "source", and another oriented "target".  To glue paths together in a consistently oriented way, we do it like this: $ \quad \cdot \longrightarrow \cdot \longrightarrow \cdot \quad $ where a "source" vertex is matched to a "target" vertex; in other words, a vertex should have opposite orientations in the edges that it joins.  (Matching them in the same orientation would give $ \quad \cdot \longrightarrow \cdot \longleftarrow \cdot \qquad $ ...ouch!)
(Working from the picture in David Sayer's example.) Each triangle in his picture has the edges in its boundary oriented clockwise, which makes them match up consistently at their vertices, in the sense of the previous point.  When we paste the triangles together along an edge, that edge is no longer part of the boundary of the resulting quadrilateral.  So looking at just the resulting boundary, we want that still to be consistently oriented, with each vertex occurring as a source of one edge and a target of another.  Having the glued edge matched in opposite directions is what will make that work...  contemplate how the vertices of the "gluing edge" occur in the remaining boundary edges, as sources/targets, and what will make sure that they still cancel out.
Now, imagine you're standing somewhere on the surface of either triangle; the notion of orientation you can see, locally, is the idea of "clockwise around you".  What should it mean to say the edges are consistent with this?  It seems reasonable to say that as you walk up to any edge of the triangle, its direction should agree with your idea of "clockwise around you".  So let's you're on the right-hand triangle and you sidle up to the "gluing edge"; it's on your left, and it points forward — that's good, that's clockwise!  Now you step over it; but now it's on your right, so to still agree with your orientation, it should point backwards, seen from this side... and it does!  Awesome!

As I see it, these (and a few other such examples) are the intuitively natural seeds from which the various abstractions and generalisations of orientation grow, and whenever I have trouble seeing why some choice is made the way it is, chasing it back to something like this usually resolves the question.
Looking outward: the ideas in 1 and 2 are the first steps down the path of combinatorial approaches to orientation, where the orientaion is on entire simplices, as seen in simplicial complexes like you sketch.   No. 3 points down a different path, to more geometric notions of local orientation at a point, which can be expressed in terms of  local homology groups, among other ways; Hatcher's excellent free book has a nice treatment of this, with really good examples/exercises, in Section 3.3 under "Orientations and Homology".<|endoftext|>
TITLE: How hard is it to get tenure in mathematics?
QUESTION [42 upvotes]: My curiousity provoked by this question and Igor's answer, I'd like to know: how many mathematics PhD's who want to get a tenured job as a research mathematician actually get that job?  
As a first approximation, I'd like to know how what percentage of math PhD's who get a research-oriented postdoc wind up as tenured professors in research-oriented jobs (pick a number of years $n \geq 6$, and determine the percentage of math PhD's who went on to research postdocs that had tenured jobs within $n$ years; the more data, the better).
(Of course I'd be happy with data for any particular region---e.g. US or Europe or Japan or Australia, and it'd be great to have it all; but maybe the US is the project to start with)
Added in response to Joel's answer: I haven't been able to extract what I want from the AMS data.  If someone else can do that, great!  But even though I haven't search it exhaustively, I don't think the information I want can be deduced from it.

REPLY [4 votes]: In the UK, impossible: tenure was abolished by Margaret Thatcher's government in 1988.<|endoftext|>
TITLE: Covers of the projective line over Z and arithmetic Grauert-Remmert
QUESTION [8 upvotes]: This question is the two-dimensional analogue of Etale coverings of certain open subschemes in Spec O_K
There I asked if one could characterize in a way certain covers of $\textrm{Spec} \ O_K$. As Cam Mcleman answered, this is basically done by the Galois group of the maximal extension unramified outside $D$. A covering of $U$ is of the form $O_L[\frac{1}{D}]$, where $L$ is any extension of $K$. 
Here I would like to ask the same question, only now for $X=\mathbf{P}^1_{\mathbf{Z}}$. 
Let $D$ be a normal crossings divisor on $\mathbf{P}^1_{\mathbf{Z}}$ and let $U$ be the complement of its support.
Q1. Is there an "equivalence of categories" as Georges Elencwajg mentions in his answer for the analytic case. (See above link.) Basically, is there an arithmetic Grauert-Remmert theorem? 
Q2. What is known about the etale fundamental group in this case? Is it "finitely generated"? Has anybody studied the maximal pro-p-quotients of these groups? 
Q3. The analytic analogue would be to consider the same question for $\mathbf{P}^1_{\mathbf{C}} \times  \mathbf{P}^1_{\mathbf{C}}$.  
Q4 Lars (see above link) mentions a result for $\mathbf{P}_{\mathbf{Q}}^1$. Is there something similar for $\mathbf{P}^2_{\mathbf{Q}}$?

REPLY [4 votes]: This is a question which is too long to put in a comment box: What exactly do you mean by a simple normal crossings divisor in $\mathbb P^1_{\mathbb Z}$? 
Let me recall that an irreducible divisor in $\mathbb P^1_{\mathbb Z}$ consists either of a Galois conjugacy class of $\overline{\mathbb Q}$ points of $\mathbb P^1$ (closed up to make a divisor in
$\mathbb P^1_{\mathbb Z}$ (these are the horizontal divisors), or else one of the closed
fibres $\mathbb P^1_{\mathbb F_p}$ of the map $\mathbb P^1_{\mathbb Z} \to \mathbb Z$ (these are the vertical divosors).  
The vertical divisors are mutually disjoint, while the horizontal divisors meet each vertical divisor in a point (if the horizontal divisor is the point
$\alpha \in \overline{\mathbb Q} \cup \{\infty\}$, and the vertical divisor is $\mathbb P^1_{\mathbb F_p}$, then they intersect in the point $\overline{\alpha}$ obtained by specializing $\alpha$ into char. $p$).  Two horizontal divisors, say corresponding to the points $\alpha, \beta \in \overline{\mathbb Q}\cup \infty$, may or may not intersect; they meet in a point lying over the point $p \in $ Spec $\mathbb Z$ if and only if $\alpha$ and $\beta$ have the same specialization to char. $p$, and one can define a multiplicity of intersection, which reflects the precise power of $p$ modulo which they are congruent.
I'm not exactly sure in this arithmetic context what the s.n.c. condition is.<|endoftext|>
TITLE: What are some good group theory references?
QUESTION [36 upvotes]: I'm curious about what books people use for a group theory reference. I don't currently own a dedicated group theory book, and I think I'd find such a book very helpful in my research. What is your favorite book on group theory? Please tell us why you like it — and what sort of groups it focuses on (finite? discrete? finitely generated? etc.)
(For my part, I'm interested mainly in discrete, finitely generated groups, but I enjoy the "flavor" of general group theory books more than combinatorial group theory books.)

REPLY [3 votes]: The book "Linear Algebraic Groups" by Armand Borel and "Linear Algebraic Groups" by James Humphreys are great (and standard) references for the theory of linear algebraic groups. In both of these books, the structure theory of linear algebraic groups uses some algebraic geometry and representation theory.<|endoftext|>
TITLE: Finding zeroes of classical modular forms
QUESTION [14 upvotes]: There are several papers which compute zeroes of modular forms for genus 0 congruence subgroups, such as "Zeros of some level 2 Eisenstein series" by Garthwaite et al published in Proc AMS and work of Shigezumi and others in levels 3,5 and 7. However, there don't seem to be generalizations of this to higher genus subgroups.
I know a few examples of modular forms for higher genus subgroups where one can compute all the zeroes; for instance, the unique normalized cusp form of weight 1 and level $\Gamma_0(31)$ with character the Legendre character modulo 31 has simple zeroes at the two cusps and the two elliptic points because the valence formula forces them to be there. Similar ideas work for levels 17, 19, 21 and 39.
My question is this: is there a more general way to find the zeroes of modular forms in an explicit way for congruence subgroups?

REPLY [7 votes]: If you are looking for examples of modular forms whose zeros can be described explicitly, then you probably want the zeros to be cusps or imaginary quadratic irrationals. In this case the Gross-Kohnen-Zagier theorem implicitly gives lots of examples, by describing the relations between Heegner points on modular elliptic curves. (Heegner points are closely related to imaginary quadratic numbers in the upper half plane.) Many examples of modular forms with zeros at imaginary quadratic irrationals can also be constructed explicitly as automorphic products.<|endoftext|>
TITLE: Published results: when to take them for granted?
QUESTION [49 upvotes]: Two kinds of papers.  There are two kinds of papers: self-contained ones, and those relying on published results (which I believe are the vast majority).
Checking the result. Of course, one should check carefully other's results before using them. There are several incentives to do that: become a real specialist; expand one's knowledge of concepts and techniques; find and mention a gap in the proof should that happen; get the ability to interact with more people ("I read your paper..."). So ideally, in a sense, checking a result before using it should always be the case.
Trusting peer-review. Yet, the very idea of academic peer-reviewed publications is to allow readers to locate results deemed trustable. The implied degree of trustability varies among scientific disciplines, but one would expect mathematics to have to most stringent one: a proof is either correct or it is not.
Given this, it is sometimes very tempting to use a result as a kind of "useful axiom", especially if that result has been proven with concepts very far from one's own area(s) of expertise, or if it is the culmination of several long papers: in those cases it would require a substantial amount of time, maybe even years, to personally check the results in their own right. Someone wanting to move forward quickly (or with a short-term position) may not want to go into this.
How to decide? Some cases are clear-cut (e.g. most people would accept the classification of finite simple groups), while others are borderline. 
My questions on that matter are:


Are there rules of thumb that you have come up with when deciding between checking a result, and taking it as an axiom ? 
When accepting without checking, how do you phrase it?
Has it ever occured to you that taking a result for granted actually backfired: what happened, and what would you do differently (job interview, retraction of publication)?



EDIT (friday 7 may):  many thanks to those who have replied, very interesting comments! (Also, please note that since there is no "best answer" to that kind of question I will not single out one over the others.)

REPLY [4 votes]: It is the life's work of many people to make this question obsolete.  More precisely, we aim to make mathematical assistants which are both pervasive and easy-to-use so that, in say 50 year's time, all high-quality journals would immediately reject a paper which has not been checked by one of these systems.  [There are sub-fields of computer science where the time horizon for this seems closer to 5 years, with the 'best' papers already being machine-checked today].
Note that some people mis-interpret such statements.  In the past, it is true that formal verification was extremely difficult and it intruded too much into the actual results and their write-up.  But this has changed tremendously of late, where 'modern' verified papers (through the liberal use of literate programming tools and ideas) look the same as non-verified papers, they just come with attachments which contain the fully formal parts.  Mathematical papers can then retain their human-oriented aspects of communicating the crucial ideas and insights, while allowing a certain lightening of the formalism in the text.
This is coming.  Current mid-career mathematicians don't have to worry about this too much, but I would certainly advise the younger generation to keep an eye on these developments.<|endoftext|>
TITLE: A Reference for Schubert's Theorem
QUESTION [9 upvotes]: Schubert's Theorem in Knot Theory says that any knot can be uniquely decomposed as the connected sum of prime knots. 
Unfortunately the original paper is in German. 
Does anyone know a good english reference for this. Or just the special case of the unknot. (i.e. that the unknot can't be written as the connected sum of two knots which aren't the unknot.)

REPLY [2 votes]: This is more general than what you ask for, but the following paper by Ryan Budney is deeply relevant:
 JSJ-decompositions of knot and link complements in the 3-sphere. L'enseignement Mathe'matique (2) 52 (2006), 319--359 math/0506523.
By looking at the JSJ decomposition of knot complements, Ryan shows, among other things, that any knot can be constructed via "satellite operations" from hyperbolic knots and torus knots (in particular these are prime knots) in an essentially unique way. Connect-sums are an example of a satellite operation. As he mentions, this theorem (in some form) is also in an unpublished manuscript of Bonahon and Seibenmann; and Schubert proved some of it (is this right?). The paper is quite readable, and in English, and the result is much stronger than Schubert's Theorem.<|endoftext|>
TITLE: Why are normal crossing divisors nice?
QUESTION [19 upvotes]: This question is going to be extremely vague.
It seems that wherever I go (especially about Grothendieck's circle of ideas) the higher-dimensional analogue of a curve minus a finite number of points is a scheme minus a normal crossing divisor.
Why is that? What's so special about a normal crossing divisor that it simulates a curve minus a finite number of points better?

REPLY [15 votes]: A related point is that if say $X\subset Y$ and you want an embedded resolution of $X$, then you can of course ask that you want a birational morphism $\pi:Z\to Y$ such that the strict transform of $X$ is smooth, but a better thing to ask is the the entire pre-image of $X$ is as nice as possible. Unfortunately (in general) you cannot make the preimage of $X$ smooth as the exceptional set will add additional components and where they meet is going to be a singular point. So, you can ask for the next best thing: normal crossings. You could even say that normal crossings is the reducible analogue of smooth. 
Anyway, this is the result of Hironaka, JS Milne referred to above: for any $X\subset Y$ (plus some reasonable assumptions) there exists a projective birational $\pi$ such that $Z$ is smooth and $\pi^{-1}X$ is a normal crossing divisor. If $Y\setminus X$ is smooth, then you may even require that $\pi$ is an isomorphism outside $X$.
The compactification result is a simple consequence of this: if $U$ is open (say quasi-projective), pick a projective compactification $Y$ and let $X=Y\setminus U$. Perform Hironaka's embedded resolution of singularities and you get $U\subset Z$ with the complement being a normal crossing divisor.

One, slightly independent word on normal crossings. There seems to be some confusion in the literature about what normal crossings mean. Or rather what the difference is between normal crossings and simple normal crossings. Well, the point is that (nowadays) the latter is understood in the Zariski topology while the former in the analytic or formal topology. In other words, simple normal crossings mean that each irreducible component is smooth and they meet transversally, while normal crossings allows for a component to meet itself transversally. In particular, a nodal curve has normal crossings but not simple normal crossings. 
In the above discussion and in the other answers before this one, you can always put simple normal crossings in place of normal crossings and the statements remain true. It is possible that back when Hironaka proved his famous theorem, this distinction had not been made so in older texts the meaning might be different. At the same time, according to Miles Reid, it was the Japanese who invented the term simple normal crossings.<|endoftext|>
TITLE: Nonuniqueness of maps of representing spaces
QUESTION [5 upvotes]: In Rudyak's On Thom Spectra, Orientability, and Cobordism, two variants of Brown's representability theorem are presented: given a natural transformation $f^*: E^* \to F^*$ of cohomology theories, Brown's representability theorem asserts that we can lift $f^*$ uniquely to a map $f: E \to F$ of spectra, and we can also lift $f^*$ to map $f^i: \Omega^\infty \Sigma^i E \to \Omega^\infty \Sigma^i F$ of representing spaces -- but he does not assert that this second kind of lift is unique.
My feeling is that this second map becomes unique after stabilization.  Is there an accessible example of such a natural transformation with two nonhomotopic representing maps on classifying spaces, illustrating nonuniqueness?

REPLY [6 votes]: The actual work being done here is by the Yoneda lemma.  Brown's representability theorem tells you that these spaces represent the cohomology theory, turning natural transformations into morphisms is done by Yoneda.
That said, uniqueness holds.
Of course, one has to be a little bit more careful about what one means by "natural transformation" in this case.  Natural transformation of what?  If your natural transformation is of the whole cohomology theory in one go (aka a stable operation) then, indeed, you get a morphism of spectra.  If your natural transformation acts on one particular level (aka an unstable operation) then you get a morphism of spaces.
Where you do not get uniqueness is if you have a family of unstable operations (aka a family of morphisms of spaces) which look as if they fit together to give a stable operation.  You can get "phantom" morphisms, and there's a "$\lim^1$" term that controls this.  A nice place to read about all of this is the papers by Boardman and Boardman, Johnson, and Wilson on stable and unstable cohomology operations (Handbook of algebraic topology, also available via Steve Wilson's homepage).

Update: To expand on that last point (I was deliberately vague because I didn't have Boardman's paper in front of me and couldn't remember which way the maps went): a stable operation (morphism of spectra) defines a compatible (under suspension) family of unstable operations (morphisms of spaces, indeed, morphisms of infinite loop spaces) but this assignment may not be injective (it is always surjective).  There is a short exact sequence (attributed to Milnor, and (9.7) in Boardman's paper):
$$
0 \to \lim_n{}^1 E^{k-1}(\underline{E}_n,o) \to E^k(E,o) \to \lim_n E^k(\underline{E}_n,o) \to 0
$$
The "o" means "pointed" and $\underline{E}_n$ is the $n$th component in the spectrum $E$.  So if the $\lim^1$ term vanishes, you get an isomorphism (whence injectivity) but if not, then there may be "phantom" stable operations that are non-trivial but can't be detected in the unstable realm.

Update: And now I've been shown a copy of the book references in the question and have read what may well be the statements themselves.  The second one is subtly different to the summary above.  In the book, the statement is that a natural transformation of cohomology theories of spaces can be lifted to a morphism of the representing spectra.  This is not necessarily unique.  The basic reason being that the Yoneda lemma does not apply to this case because the representing object is in a different category.  So cohomology of spaces represented by a spectrum is the composition of $\Sigma^\infty$ and the cohomology theory and thus not an honest representable functor.  In general, there's no way (that I can think of) to pull-back or push-forward natural transformations along a functor.
Nonetheless, because spectra and spaces are closely related, it is possible in this case to construct a natural-transformation-of-cohomology-of-spaces from one of spectra.  This works because the pieces of the cohomology theory are individually representable as spaces, and so again the Yoneda lemma comes into play.  Then one can ask about this map, and the $\lim{}^1$ sequence says that it is always surjective but not necessary injective.
So, in summary:

$E^\ast(R)$: both spectram: Yoneda => $Nat(E^\ast,F^\ast) \cong \{E,F\}$
$E^k(X)$: both spaces: Yoneda => $Nat(E^k,F^k) \cong [\underline{E}_k, \underline{F}_k]$
$E^\ast(X)$: spectrum and space: Milnor => $\{E,F\} \to Nat(E^\ast,F^\ast)$ surjective<|endoftext|>
TITLE: Reducing ACA₀ proof to First Order PA
QUESTION [13 upvotes]: According to the Wikipedia ACA0 is a conservative extension of First Order logic + PA.
http://en.wikipedia.org/wiki/Reverse_Mathematics
First of all I have a few questions about the proof:
a - What is the general sketch of this proof, is it based on models?
b - Consider the theorem that ACA0 is a conservative extension of First Order + PA, and the proof of that theorem is proven in a formal system, what kind of logic is needed? If the proof is based on models, then it requires second order logic. However, the theorem itself is a ∏02 question as far as I understand, and can be expressed in First Order logic + PA. Is there also a proof in First Order logic + PA?
Then I am interested in the following:
c - Given an ACA0 formal proof that ends in a theorem that is part of First Order logic + PA, is there an algorithm that reduces the ACA0 proof to First Order + PA proof?
One could just do a breath first search on First Order logic + PA and given the fact that ACA0 is a conservative extension, it is guaranteed to end. So, the answer to question c is definitely "yes", but I am looking for something more clever.
I am struggling with this algorithm for months. In general an ACA0 proof, with a First Order + PA end theorem reduces rather easier. However, there are some non-trivial cases. If the answer to question b is "yes", then that proof might give hints for constructing the algorithm.
I want to use this algorithm to reduce proofs of full second order, such that the reduced proof is First Order logic + PA, or contains the use of the induction scheme with a second order induction hypothesis.
In many cases the use of second order induction hypothesis, can be reduced by using the "Constructive Omega Rule". I want to understand the limitations of this (if any).
Thanks in advance,
Lucas

REPLY [6 votes]: To elaborate on robin-adams' answer, the proof of the conservation using cut elimination produces an algorithm running in superexponential time (i.e., $t(n)$ is $n$-times iterated exponentiation; that's the complexity of cut elimination), and it can be formalized in $I\Delta_0+SUPEXP$. This is essentially optimal, as a result of Solovay states that $\mathrm{ACA}_0$ has non-elementary speed-up over PA (i.e., a finite tower of exponentials won't do).<|endoftext|>
TITLE: Kan extensions and the yoneda embedding.
QUESTION [6 upvotes]: [Edit: For a category $C$ let $C^\wedge$ denote the category of presheaves on $C$.]
Let $f:C\to D$ be a functor. Precomposition with $f^{op}$ induces a functor
$f^\wedge:D^\wedge \to C^\wedge$. This functor has both a left- and a right adjoint, called  left- and right kan extension:
$f_\wedge \dashv f^\wedge \dashv f_+$.
Now for $c\in C$ we get $D^\wedge(D(-,fc),Y)=Y(fc)=f^\wedge Y(c)=C^\wedge(C(-,c),f^\wedge Y)$. This gives us the restriction of $f_\wedge$ to $C$ along the yoneda embedding: It is $f$ (composed with the yoneda embedding).
Now here's my question:

What is the restriction of $f_+$ to $C$ along the yoneda embedding?

It seems not to agree with $f$ but:

Is there a nice connection between $f_+C(-,c)$ and $D(-,fc)$?

REPLY [4 votes]: I know its good "explanation".
Let $X, Y$ be sets, and $f:X\to Y$ be a map. Then, it induces a inverse image map $f^{-1}:P(Y)\to P(X)$. Regarding $P(X),P(Y)$ as categories by its inclusion order, $f^{-1}$ can be regard as a functor. Then, this functor has a left and right adjoint. The left adjoint functor is just an image $f_*:P(X)\to P(Y)$, and right adjoint $f_!:P(X)\to P(Y)$ is called "dual image, small image" defined by $f_!(U)=Y-f(X-U)$. It is easy to check they are actually adjoint.
This is a special case of Kan extensions. Because by taking $2=\{0\to 1\}$ and regarding $X$ as discrete category, we can describe $P(X)$ as just a functor category $2^X$. And the "composite with $f$ functor" is nothing but the inverse image map. 
Furthermore, these categories are $2$-enriched, so the contravariant Hom-functor $Hom(-,x):X^{op}=X\to 2$ is seen as an object of $2^X$. But except for $y=x$, $Hom(y,x)$ is empty. This means it's a "characteristic functor" of $\{x\}\in P(X)$. So, $\{-\}:X\to P(X), x\mapsto \{x\}$ is seen as a "$2$-enriched yoneda embedding".
In this case, the restriction of $f_*$ to X along the yoneda embedding is $f$, composed with the yoneda embedding. However, it is clear that there is no nice connection between $f_!$ and the yoneda embedding. I think this is a good analogy to explain the advantage of left Kan extension in combination with yoneda embedding.<|endoftext|>
TITLE: A question about open induction
QUESTION [10 upvotes]: An old theorem of A. J. Wilkie (Some results and problems on weak systems of arithmetic, Logic Colloquium '77) asserts that  a discretely ordered ring $R$ can be extended to a model of open induction if and only if  for all $n>1$, there is a homomorphism from $R$ onto $\mathbb{Z}/n\mathbb{Z}$. 
Wilkie's proof proceeds by adjoining transcendental elements to $R$, but it is not clear that this is ever necessary:
Does every  ring that extends to a model of open induction have an algebraic extension to a model of open induction?
Does anyone know anything about this? I know of no place in the literature where the question is even mentioned, although it has come up more than once in conversation.
Here is simple test-case: Let $R$ be the ring $\mathbb{Z}[t,\sqrt{2}t-r]$, where $r$ is a real transcendental and $t$ is an indeterminate. Order $R$ by declaring $t$ positive infinite. It is not hard  to show that $R$ extends to a  model of true arithmetic. I don't know if $R$ has an algebraic extension to a model of open induction.

REPLY [6 votes]: I think that the answer in general is “no”, and that this follows from results by D’Aquino, Knight, and Lange [1].
Namely, by Theorem 3.5 and Remark 3, there exists a real-closed field $R$ with an element $r\in R$ and a $\mathbb Z$-ring $I\subseteq R$ such that no discretely ordered subring $I\subseteq J\subseteq R$ intersects $[r,r+1)$, but $b/i\in[r,r+1)$ for some $b,i\in I$. Then $I$ (being a $\mathbb Z$-ring) can be extended to a model of $\mathit{IOpen}$, but no such extension can be found in its real closure $\operatorname{rcl}(I)\subseteq R$ (since no discretely ordered $I\subseteq J\subseteq \operatorname{rcl}(I)$ can contain an integer part of $b/i$).
Reference:
[1] Paola D’Aquino, Julia Knight, Karen Lange: Limit computable integer parts, Archive for Mathematical Logic 50 (2011), no. 7, pp. 681–695, doi:10.1007/s00153-011-0241-z. Erratum in: Archive for Mathematical Logic 54 (2015), no. 3, pp. 487–489, doi:10.1007/s00153-015-0418-y<|endoftext|>
TITLE: Fundamental domain for symmetric group $S_n$ acting on $\mathop{Gr}(k,n)$?
QUESTION [5 upvotes]: Is there a nice fundamental domain for the symmetric group $S_n$ acting on the Grassmannian of $k$-planes in $\mathbb{R}^n$?  (The action of $S_n$ is by permuting the coordinates, of course.)
I'm looking for a way to efficiently test whether two subspaces of $\mathbb{R}^n$ are related by permuting and/or negating the coordinates.  (So I really care not about $S_n$, but about $(\mathbb{Z}/2)^n \rtimes S_n$, the Weyl group of the B/C series.)  I'm somewhat skeptical this is easy, but has anyone thought about it?

REPLY [2 votes]: An observation: If you ask for an exact decision procedure for whether to points of $G(k,n)$ are in the same $S_n$ orbit, this problem includes the problem of bipartite graph isomorphism. Namely, if $A$ and $B$ are the adjacency matrices of bipartite graphs on $k$ white and $n-k$ black vertices, then the row spans of
$$\begin{pmatrix} 2 \cdot \mathrm{Id} & A \end{pmatrix} \ \mbox{and} \ \begin{pmatrix} 2 \cdot \mathrm{Id} & B \end{pmatrix}$$
are in the same orbit if and only if the graphs are isomorphic.
Since you really care about generic points of $G(k,n)$, and since graph isomorphism is quite fast in practice, this shouldn't be considered to rule out the possibility of a good practical algorithm, but it does rule out certain strategies.<|endoftext|>
TITLE: Images of action of Galois on the Tate module of Elliptic Curve,
QUESTION [9 upvotes]: Let E be an elliptic curve over the rationals, and let $TE = \lim_\leftarrow E[n]$ be the Tate module of the elliptic curve. The action of the Galois group of $\bf Q$ gives rise to a representation $\rho_E : G_{\bf Q} \rightarrow GL_2(\widehat{\bf Z})$. 
My first question is why is the image of this representation not surjective? I seem to recall there is a very easy argument for it, but I can't remember what that is.
My second question is can one use this to give a level structure to all rational elliptic curves? Specifically, can I find a collection of modular curves with some level structure $\{X_i/{\bf Q}\}$ such that $\cup j(X_i({\bf Q})) = X(1)({\bf Q})$, where $j:X_i \rightarrow X(1)$ is the natural forgetful map? (I'm sure the answer to this is no, but it never hurts to be overly optimistic.)

REPLY [22 votes]: Let $\Delta$ be the discriminant of $E$.  Then the action of $G_{\mathbf{Q}}$ on $E[2]$ determines the action on $\sqrt{\Delta}$.  On the other hand, the action of $G_{\mathbf{Q}}$ on $E[n]$ determines the action on a primitive $n$-th root of unity $\zeta_n$, via the Weil pairing.  The Kronecker-Weber theorem implies that $\sqrt{\Delta} \in \mathbf{Q}(\zeta_n)$ for some $n$, and this forces a compatibility between the actions on $E[2]$ and $E[n]$, which forces the image of $G_{\mathbf{Q}}$ into an index-2 subgroup of $\operatorname{GL}_2(\hat{\mathbf{Z}})$.  But this index-2 subgroup varies with $E$, so you do not get a rational point on a nontrivial modular curve corresponding to any one kind of level structure.
Remarks: 
1) Nathan Jones in his Ph.D. thesis at UCLA, building on earlier work of William Duke, proved that in a precise sense, asymptotically 100% of elliptic curves are such that the image of Galois is of index 2.  
2) Over higher number fields $K$, quadratic extensions are not necessarily contained in cyclotomic ones, and in fact there exist elliptic curves over certain number fields $K$ other than $\mathbf{Q}$ for which $G_K \to \operatorname{GL}_2(\hat{\mathbf{Z}})$ is surjective.  The first example was given by Aaron Greicius in his Ph.D. thesis.  
3) David Zywina then proved that under mild necessary conditions on a number field $K$, asymptotically 100% of elliptic curves are such that $G_K \to \operatorname{GL}_2(\hat{\mathbf{Z}})$ is surjective.<|endoftext|>
TITLE: Reporting all faces in a planar graph
QUESTION [8 upvotes]: Hi, I was looking to traverse a planar graph and report all the faces in the graph (vertices in either clockwise or counterclockwise order). I have build a random planar graph generator that creates a connected graph with iterative edge addition and needed a solution to report all the faces that were created in the final graph. I was contemplating several strategies such as doing a sweep line algorithm and tracking the areas between lines or tracking the faces as I generate the graph however I have not been able to find much material regarding this matter and was wondering where I could find some assistance/ideas how to do this.
So far I found a Boost algorithm here http://www.boost.org/doc/libs/1_36_0/boost/graph/planar_face_traversal.hpp however I am having a lot of trouble decrypting the boost library to determine how it is done. 
Any thoughts, ideas or existing algorithms would be welcome.

REPLY [5 votes]: Hi, there is one such routine in SAGE (http://www.sagemath.org/)
see here:
http://www.sagemath.org/doc/reference/sage/graphs/generic_graph.html#sage.graphs.generic_graph.GenericGraph.trace_faces
def trace_faces(self, comb_emb):
    """
    A helper function for finding the genus of a graph. Given a graph
    and a combinatorial embedding (rot_sys), this function will
    compute the faces (returned as a list of lists of edges (tuples) of
    the particular embedding.

    Note - rot_sys is an ordered list based on the hash order of the
    vertices of graph. To avoid confusion, it might be best to set the
    rot_sys based on a 'nice_copy' of the graph.

    INPUT:


    -  ``comb_emb`` - a combinatorial embedding
       dictionary. Format: v1:[v2,v3], v2:[v1], v3:[v1] (clockwise
       ordering of neighbors at each vertex.)


    EXAMPLES::

        sage: T = graphs.TetrahedralGraph()
        sage: T.trace_faces({0: [1, 3, 2], 1: [0, 2, 3], 2: [0, 3, 1], 3: [0, 1, 2]})
        [[(0, 1), (1, 2), (2, 0)],
         [(3, 2), (2, 1), (1, 3)],
         [(2, 3), (3, 0), (0, 2)],
         [(0, 3), (3, 1), (1, 0)]]
    """
    from sage.sets.set import Set

    # Establish set of possible edges
    edgeset = Set([])
    for edge in self.to_undirected().edges():
        edgeset = edgeset.union( Set([(edge[0],edge[1]),(edge[1],edge[0])]))

    # Storage for face paths
    faces = []
    path = []
    for edge in edgeset:
        path.append(edge)
        edgeset -= Set([edge])
        break  # (Only one iteration)

    # Trace faces
    while (len(edgeset) > 0):
        neighbors = comb_emb[path[-1][-1]]
        next_node = neighbors[(neighbors.index(path[-1][-2])+1)%(len(neighbors))]
        tup = (path[-1][-1],next_node)
        if tup == path[0]:
            faces.append(path)
            path = []
            for edge in edgeset:
                path.append(edge)
                edgeset -= Set([edge])
                break  # (Only one iteration)
        else:
            path.append(tup)
            edgeset -= Set([tup])
    if (len(path) != 0): faces.append(path)
    return faces

I also have my own implementation which is an adaptation from the SAGE lib:
  def Faces(edges,embedding)
   """
   edges: is an undirected graph as a set of undirected edges
   embedding: is a combinatorial embedding dictionary. Format: v1:[v2,v3], v2:[v1], v3:[v1] clockwise ordering of neighbors at each vertex.)

    """

    # Establish set of possible edges
    edgeset = set()
    for edge in edges: # edges is an undirected graph as a set of undirected edges
        edge = list(edge)
        edgeset |= set([(edge[0],edge[1]),(edge[1],edge[0])])

    # Storage for face paths
    faces = []
    path  = []
    for edge in edgeset:
        path.append(edge)
        edgeset -= set([edge])
        break  # (Only one iteration)

    # Trace faces
    while (len(edgeset) > 0):
        neighbors = self.embedding[path[-1][-1]]
        next_node = neighbors[(neighbors.index(path[-1][-2])+1)%(len(neighbors))]
        tup = (path[-1][-1],next_node)
        if tup == path[0]:
            faces.append(path)
            path = []
            for edge in edgeset:
                path.append(edge)
                edgeset -= set([edge])
                break  # (Only one iteration)
        else:
            path.append(tup)
            edgeset -= set([tup])
    if (len(path) != 0): faces.append(path)
    return iter(faces)<|endoftext|>
TITLE: Combinatorial proof of (a special case of) the spectral theorem?
QUESTION [13 upvotes]: The spectral theorem for a real $n \times n$ symmetric matrix $A$ says that $A$ is diagonalizable with all eigenvalues real.  If $A$ happens to have non-negative integer entries, it can be interpreted as the adjacency matrix of an undirected graph $G$, and the spectral theorem gives us information about how the sequences $A_{ij}^n$ behave, which count the number of walks of length $n$ from vertex $i$ to vertex $j$.  In particular, it says that $A_{ij}^n$ has the form $\sum_{k=1}^{n} a_{ijk} \lambda_k^n$ for some real $\lambda_k$ and some $a_{ijk}$.  
If $A$ is not symmetric, on the other hand, two things can happen that don't happen in the above case:

The $\lambda_k$ may fail to be real.  In other words, there can be "periodicity" of period greater than $2$ in the sequences $A_{ij}^n$.  This happens, for example, if $A$ is a directed cycle graph.
The coefficients $a_{ijk}$ may be polynomials in $n$.  This happens, for example, if $A$ is a directed path graph with loops based at each vertex.

Is it possible to prove "combinatorially" that neither of these things can happen when $A$ is symmetric?  (What I mean is that, given only that you know what $A_{ij}^n$ looks like in terms of eigenvalues, what can you prove just by looking at $G$?)  For example, it is straightforward to prove that the coefficient of the largest positive eigenvalue is constant by a path-counting argument which I described here.  I think a path-counting argument can also in principle prove that the $\lambda_k$ are real via some kind of mixing argument which could show that the only constraint on the length of a long walk from a vertex to itself is its length mod 2 (due to the presence of even cycles).  I think such an argument can at least show that the eigenvalues of maximum absolute value must be real, but I don't know how easy it is to deduce information about the other eigenvalues.
If you have a solution with "adjacency matrix" replaced by "Laplacian," I'd also be interested in that.

REPLY [3 votes]: Not an answer, but perhaps the basis for a plan of attack ...
A consequence of the spectral theorem as it applies to the adjacency matrix of a graph $G$ is that, for every (presumably real) eigenvalue --say, with multiplicity $m$-- there exists a geometric realization, $R$, of $G$ in $\mathbb{R}^m$ such that each automorphism of $G$ induces a rigid isometry of $R$. I call these "spectral realizations" (which I discuss at length in my note "Spectral Realizations of Graphs" (PDF)). A spectral realization has this eigenic property: moving each vertex to the vector sum of its neighbors has the same effect as scaling the realization by the associated eigenvalue.
For example, these are the spectral realizations of the 15-cycle and the (skeleton of the) Truncated Octahedron (uniform polyhedron $U_8$).
Spectral Realizations of the 15-cycle http://daylateanddollarshort.com/math/pdfs/mathoverflow-spectralpolygons.png
(Whoops! Typo: Caption should read "Note that (b), (c), (e), and (h) are faithful".)
Spectral Realizations of the Truncated Octahedron http://daylateanddollarshort.com/math/pdfs/mathoverflow-spectralpolyhedra.png
(The caption seems to imply that there are other, higher-dimensional, realizations of the Truncated Octahedron; in this case, there are none. The caption is a template used for over 70 figures in my note, and I haven't gone back to edit the individual cases.)
One way to attack your problem, then, would be to attempt to reverse the implication: Use the combinatorics of the graph to prove independently that sufficiently many spectral realizations exist, "using up" all the eigen-dimensions; the associated eigenvalues must be real as they serve as scale factors for the eigenic property of these realizations.
It's not clear to me how a combinatorial approach to finding these realizations would proceed; they fall so neatly out of the spectral theorem, after all. However, I can demonstrate that sometimes there's a nice interplay between the combinatorics and the geometry; here's an example I left as a comment to an answer of another MO question:

Proposition: If $\lambda$ is a (real) eigenvalue of a bipartite graph, then so is $-\lambda$.

Proof: If $\lambda$ is a (real) eigenvalue, then there exists a corresponding spectral realization of the graph, such that the realization has the eigenic property with scale factor $\lambda$. If the graph is bipartite, take a two-coloring of its vertices; for each vertex of a chosen color, move the corresponding realized vertex to its reflection in the origin. The result is a realization with eigenic scale factor $-\lambda$, which must therefore be an eigenvalue of the graph. QED. (Note: The Truncated Octahedron, pictured above, is bipartite.)
I'd love to see the geometry of these realizations explained completely from the combinatorics of the graphs. (I'm personally trying to investigate the combinatorial conditions under which two spectral realizations may share the same vertex coordinates, as with pairs of the Truncated Octahedra realization shown, or the Dodecahedron and Great Stellated Dodecahedron, etc.) I hope you keep us posted on your progress in this area.<|endoftext|>
TITLE: Solutions to the Continuum Hypothesis
QUESTION [134 upvotes]: Related MO questions: What is the general opinion on the Generalized Continuum Hypothesis? ; Completion of ZFC ; Complete resolutions of GCH How far wrong could the Continuum Hypothesis be? When was the continuum hypothesis born?
Background
The Continuum Hypothesis (CH) posed by Cantor in 1890 asserts that $ \aleph_1=2^{\aleph_0}$. In other words, it asserts that every subset of the set of real numbers that contains the natural numbers has either the cardinality of the natural numbers or the cardinality of the real numbers.  It was the first problem on the 1900 Hilbert's list of problems. The generalized continuum hypothesis asserts that there are no intermediate cardinals between every infinite set X and its power set.
Cohen proved that the CH is independent from the axioms of set theory. (Earlier Goedel showed that a positive answer is consistent with the axioms).
Several mathematicians proposed definite answers or approaches towards such answers regarding what the answer for the CH (and GCH) should be.
The question

My question asks for a description and explanation of the various approaches to the continuum hypothesis in a language which could be understood by non-professionals.

More background
I am aware of the existence of 2-3 approaches.
One is by Woodin described in two 2001 Notices of the AMS papers (part 1, part 2).
Another by Shelah (perhaps in this paper entitled "The Generalized Continuum Hypothesis revisited "). See also the paper entitled "You can enter Cantor paradise" (Offered in Haim's answer.);
There is a very nice presentation by Matt Foreman discussing Woodin's approach and some other avenues. Another description of Woodin's answer is by Lucca Belloti (also suggested by Haim).
The proposed answer $ 2^{\aleph_0}=\aleph_2$ goes back according to François to Goedel. It is (perhaps) mentioned in Foreman's presentation. (I heard also from Menachem Magidor that this answer might have some advantages.)
François G. Dorais mentioned an important paper by Todorcevic's entitled "Comparing the Continuum with the First Two Uncountable Cardinals".
There is also a very rich theory (PCF theory) of cardinal arithmetic which deals with what can be proved in ZFC.
Remark:
I included some information and links from the comments and answer in the body of question. What I would hope most from an answer is some friendly elementary descriptions of the proposed solutions.

There are by now a couple of long detailed excellent answers (that I still have to digest) by Joel David Hamkins and by Andres Caicedo and several other useful answers. (Unfortunately, I can accept only one answer.)
Update (February 2011): A new detailed answer was contributed by Justin Moore.
Update (Oct 2013) A user 'none' gave a link to an article by Peter Koellner about the current status of CH:
Update (Jan 2014) A related popular article in "Quanta:"  To settle infinity dispute a new law of logic
(belated) update(Jan 2014) Joel David Hamkins links in a comment from 2012 a very interesting paper Is the dream solution to the continuum hypothesis attainable written by him about the possibility of a "dream solution to CH." A link to the paper and a short post can be found here.
(belated) update (Sept 2015) Here is a link to an interesting article: Can the Continuum Hypothesis be Solved? By Juliette Kennedy
Update A videotaped lecture The Continuum Hypothesis and the search for Mathematical Infinity by Woodin from January 2015, with reference also to his changed opinion. (added May 2017)
Update (Dec '15): A very nice answer was added (but unfortunately deleted by owner, (2019) now replaced by a new answer) by Grigor. Let me quote its beginning (hopefully it will come back to life):
"One probably should add that the continuum hypothesis depends a lot on how you ask it.

$2^{\omega}=\omega_1$
Every set of reals is either countable or has the same size as the continuum.

To me, 1 is a completely meaningless question, how do you even experiment it?
If I am not mistaken, Cantor actually asked 2..."
Update A 2011 videotaped lecture by Menachem Magidor: Can the Continuum Problem be Solved? (I will try to add slides for more recent versions.)
Update (July 2019) Here are slides of 2019 Woodin's lecture explaining his current view on the problem. (See the answer of Mohammad Golshani.)
Update (Sept 19, 2019) Here are videos of the three 2016 Bernay's lectures by Hugh Woodin on the continuum hypothesis and also the videos of the three 2012 Bernay's lectures on the continuum hypothesis and related topics by Solomon Feferman.
Update (Sept '20) Here are videos of the three 2020 Bernays' lectures by Saharon Shelah on the continuum hypothesis.
Update (May '21) In a new answer, Ralf Schindler gave a link to his 2021 videotaped lecture in Wuhan, describing a result with David Asperó that  shows a relation between two well-known axioms. It turns out that Martin's Maximum$^{++}$ implies Woodin's ℙ$_{max}$ axiom. Both these axioms were known to imply the $\aleph_2$ answer to CH. A link to the paper: https://doi.org/10.4007/annals.2021.193.3.3

REPLY [8 votes]: In Aug. 2020, I gave a talk at Wuhan with the title "How many real numbers are there?", taking into account my result with D. Asperó on MM++ => (*). There is a recording: https://m.bilibili.com/video/BV1TV411h714 , and there is also a set of notes: https://ivv5hpp.uni-muenster.de/u/rds/Notes_Wuhan.pdf . Comments are very welcome.<|endoftext|>
TITLE: Drawing conclusions by NOT using AC.
QUESTION [12 upvotes]: The existence of non-measurable subsets and functions on $\mathbb{R}$ require the use of the axiom of choice. That is, there exist models of ZF in which all subsets of (and hence all functions defined on) $\mathbb{R}$ are measurable. Does this mean that if I can define a subset or function on $\mathbb{R}$ without invoking the axiom of choice, it must be measurable?
Let's say I fully accept AC and am just looking for a quick and dirty way of proving a function is measurable.

REPLY [3 votes]: Solovay showed in the 1960's the consistency of  ZF + axiom of Dependent Choice (for countable sets) + "every set of reals is Lebesgue measurable".  .
This is a formal justification of the idea that any set of reals that can be proved to exist in ZFC without making uncountably many choices ---- i.e., a set that can be constructed in ZF + DC ---- cannot be proved to be non-measurable.   
There are also theorems of Shelah, coming out of the analysis of determinacy (AD, Woodin cardinals, etc) to the effect that "any reasonably defined set is Lebesgue measurable".
So it is impossible to produce a Lebesgue non-measurable set without specifically looking for it in the wilds of uncountable AC.
It would be interesting to know what the state of the art is for metatheorems on how broad a range of constructions can be used in ZF+DC, that are guaranteed to stay within the world of provably Lebesgue measurable sets.<|endoftext|>
TITLE: When does the Torelli Theorem hold?
QUESTION [18 upvotes]: The Torelli theorem states that the map $\mathcal{M}_g(\mathbb{C})\to \mathcal{A}_g(\mathbb{C})$ taking a curve to its Jacobian is injective.  I've seen a couple of proofs, but all seem to rely on the ground field being $\mathbb{C}$ in some way.  So:

Under what conditions does the Torelli Theorem hold?

Is algebraically closed necessary? Characteristic zero? It is known if there are any other base rings/schemes over which it is true?

REPLY [23 votes]: The Torelli theorem holds for curves over an arbitrary ground field $k$ (in particular, $k$ need not be perfect).  A very nice treatment of the "strong" Torelli theorem may be found in the appendix by J.-P. Serre to Kristin Lauter's 2001 Journal of Algebraic Geometry paper Geometric methods for improving the upper bounds on the number of rational points on algebraic curves over finite fields.  It is available on the arxiv:
http://arxiv.org/abs/math/0104247
Here are the statements (translated into English):
Let $k$ be a field, and let $X_{/k}$ be a nice (= smooth, projective and geometrically integral) curve over $k$ of genus $g > 1$.  Let $(\operatorname{Jac}(X),\theta_X)$ denote the Jacobian of $X$ together with its canonical principal polarization.  Let $X'_{/k}$ be another nice curve.
Theorem 1: Suppose $X$ is hyperelliptic.  Then for every isomorphism of polarized abelian varieties $(\operatorname{Jax}(X),\theta_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta_{X'})$, there exists a unique isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ such that $F = \operatorname{Jac} f$.  
Theorem 2: Suppose $X$ is not hyperelliptic.  Then, for every isomorphism $F: (\operatorname{Jax}(X),\theta_X) \stackrel{\sim}{\rightarrow} (\operatorname{Jac}(X'),\theta_{X'})$ there exists an isomorphism $f: X \stackrel{\sim}{\rightarrow} X'$ and $e \in \{ \pm 1\}$ such that $F = e \cdot  \operatorname{Jac} f$.  Moreover, the pair $(f,e)$ is uniquely determined by $F$.<|endoftext|>
TITLE: What does the classifying space of a category classify?
QUESTION [31 upvotes]: A finite group $G$ can be considered as a category with one object. Taking its nerve $NG$, and then geometrically realizing we get $BG$ the classifying space of $G$, which classifies principal $G$ bundles.
Instead starting with any category $C$, what does $NC$ classify? (Either before or after taking realization.) Does it classify something reasonable?

REPLY [24 votes]: Ieke Moerdijk has written a small Springer Lecture Notes tome addressing this question:"Classifying Spaces and Classifying Topoi" SLNM 1616.
Roughly the answer is: A $G$-bundle is a map whose fibers have a $G$-action, i.e. are $G$-sets (if they are discrete), i.e. they are functors from $G$ seen as a category to $\mathsf{Sets}$. Likewise a $\mathcal C$-bundle for a category $\mathcal C$ is a map whose fibers are functors from $\mathcal C$ to $\mathsf{Sets}$, or, if you want, a disjoint union of sets (one for each object of $\mathcal C$) and an action by the morphisms of $\mathcal C$ — a morphism $A \to B$ in $\mathcal C$ takes elements of the set corresponding to $A$ to elements of the set corresponding to $B$.
There is a completely analogous version for topological categories also.

REPLY [11 votes]: It's one level up the categorical ladder, but you may find this paper interesting:
http://arxiv.org/abs/math/0612549
Two-Categorical Bundles and Their Classifying Spaces
Authors: Nils. A. Baas, Marcel Bokstedt, Tore August Kro<|endoftext|>
TITLE: Covers of Riemann surfaces which become arbitrary close in Teichmuller space
QUESTION [7 upvotes]: Suppose $S$ and $S'$ are two compact Riemann surfaces of genus $g$.  Does there exist a sequence of genera $g_i \to \infty$ and covers $S_i, S_{i}'$ of $S,S'$, both of genus $g_i$, such that $d(S_i,S_{i}')\to 0$?  Here $d$ a "natural" distance function on Teichmuller space, of which I suppose there are many, but for definiteness let's take it to be induced by the Teichmuller metric.
This question was asked to me by Rick Kenyon last year, and some brief thought on it got me nowhere.

REPLY [10 votes]: This is the Ehrenpreis Conjecture, and is still open.
Jeremy Kahn and Vlad Markovic have made some progress recently.
UPDATE: Kahn and Markovic have now announced a proof of the entire conjecture.  See http://arxiv.org/abs/1101.1330

REPLY [10 votes]: At least for the Weil-Petersson metric on Teichmuller space, this is a well-known open problem known as the Ehrenpreis conjecture.  It has a rather fearsome reputation.<|endoftext|>
TITLE: Tannaka formalism and the étale fundamental group
QUESTION [33 upvotes]: For quite a while, I have been wondering if there is a general principle/theory that has
both Tannaka fundamental groups and étale fundamental groups as a special case.
To elaborate: The theory of the étale fundamental group (more generally of
Grothendieck's Galois categories from SGA1, or similarly of the fundamental group of a
topos) works like this: Take a set valued functor from the category of finite étale
coverings of a scheme satisfying certain axioms, let $\pi_1$ be its automorphism group and
you will get an equivalence of categories ( (pro-)finite étale coverings) <-> ( (pro-)finite cont.
$\pi_1$-sets).
The Tannaka formalism goes like this: Take a $k$-linear abelian tensor category $\mathbb{T}$ satisfying certain axioms (e.g. the category of finite dim. $k$-representations of an abstract group), and a $k$-Vector space
valued tensor functor $F$ (the category with this functor is then called neutral Tannakian), and let $Aut$ be its tensor-automorphism functor, that is the functor that assigns to a $k$-algebra $R$ the set of $R$-linear tensor-automorphisms $F(-)\otimes R$. This will of
course be a group-valued functor, and the theory says it's representable by a group scheme $\Pi_1$,
such that there is a tensor equivalence of categories $Rep_k(\Pi_1)\cong \mathbb{T}$.
Both theories "describe" under which conditions a given category is the (tensor) category of
representations of a group scheme (considering $\pi_1$-sets as "representations on sets" and $\pi_1$ as constant group scheme). Hence
the question: 

Are both theories special cases of some general concept? (Maybe, inspired by
  recent questions, the first theory can be thought of as "Tannaka formalism for
  $k=\mathbb{F}_1$"? :-))

REPLY [5 votes]: I stumbled upon this very old question and I had some observations to make, I hope they are not considered entirely unuseful after all this time. I want to highlight two points.

Even if we restrict to profinite groups, Tannakian categories contain strictly more information than Galois categories.
There is (almost) no reason to search for a concept more general than Tannakian categories.

If $k$ is not separably closed, the classical étale fundamental group should not be thought as a group scheme: the reason is that if we want to work with group schemes over $k$, everything should be relative to the base $\text{Spec}~k$, while the classical étale fundamental group is absolute. For example, the only reasonable fundamental group scheme of $\text{Spec}~k$ is is the trivial one, while the étale fundamental group is $\text{Gal}(k_s/k)$.
If one has a group scheme $G$, the right way of constructing a group out of it is to take $G(k_s)\rtimes\text{Gal}(k_s/k)$: the group scheme contains more information, since it remembers the projection to $\text{Gal}(k_s/k)$. By the way, an usual profinite group should not be thought as a constant group scheme: there is a natural way of giving it an algebraic structure such that the associated topology is the profinite one, not the discrete one. 
Let us now restrict ourselves to Tannakian categories whose associated group (or gerbe) is profinite, since Galois categories can only "see" profinite groups. By what we have said above, even in this case a Tannakian category contains more information than a Galois category: the Tannakian category determines a Galois category plus a Galois subcategory isomorphic to $\text{Ét}~k$. The Galois category associated to a Tannakian category $T$ can be constructed by taking the étale covers of the gerbe associated to $T$.
As already said by others, one can compare the two concepts if the base field is algebraically closed. In this case, Galois categories and "profinite" Tannakian categories are essentially the same thing, i.e. Tannakian categories do generalize Galois categories. 
One could try to take categories with fibre functors in other categories different from $\text{Set}$ or $\text{Vect}_k$. But the point now is that there is (almost) no reason to do this: if, in the end, one wants to obtain an affine group scheme, Tannakian categories already detect them all. Hence, one may try to do this only to get groups which are not affine.<|endoftext|>
TITLE: Lower bounds on zeta(s+it) for fixed s
QUESTION [11 upvotes]: This is most probably widely known and discussed here many times, so I am preliminay sorry.
Does Riemann conjecture imply some lower estimates on values, say $|\zeta(3/4+it)|$ for real $t$, when $|t|$ tends to infinity?
Are any such results known without assuming Riemann conjecture (many doubts here)?
Thanks!

REPLY [12 votes]: Yes, such conditional results are covered in Chapter 14 of the standard reference - the second edition of The Theory of the Riemann Zeta-Function by E. C. Titchmarsh. This edition has end-of-chapter notes by D. R. Heath-Brown bringing it up to date as of 1986.
In particular a lower bound
$$
|\zeta(3/4 + it)| \gg e^{-c\sqrt{\log(t)}/\log\log(t)}
$$
holds with some $c > 0$, conditionally on RH. See page 384 of the cited reference.
Such results are only known unconditionally for a region to the left of the line $\sigma = 1$ that 
narrows to zero width as $t \rightarrow {\pm}\infty$. Not coincidentally, the best zero-free region known is also of this form. See page 135 of the cited reference for the best result of this kind.<|endoftext|>
TITLE: Slices of presheaf categories
QUESTION [10 upvotes]: Apparently it's 'well known' that if $P$ is a presheaf on $C$ then there is an equivalence $\widehat{C}/P \simeq \widehat{\int P}$, where $\int P$ is the usual category of elements and $\widehat{C} = [C^{\rm op},{\rm Set}]$.  (I've seen a reference to Johnstone's Topos Theory for this, but I don't have easy access to the book.  It's also Exercise III.8(a) in Mac Lane--Moerdijk)
Now, I've come across comma categories $\widehat{C}/H$ for functors $H \colon D \to \widehat{C}$ (mainly when $H = D(F-,-)$ for $F \colon C \to D$), and I'd like to have a similarly useful/interesting result for them.  The equivalence doesn't generalize to $\widehat{C}/H \simeq \widehat{\int H}$, or at least I can't see any way to make that work.  So I want to understand the first equivalence from a more abstract-nonsensical point of view, to figure out what's really going on.
Is there a way to see the above equivalence as living in a fibrational cosmos or something similar?  If not, is there any other kind of machinery that might help me understand categories of the form $\widehat{C}/H$?

REPLY [14 votes]: Yes, you can see this as happening in a "fibrational cosmos."  I'll describe how it goes, but then we'll see that the description of $\widehat{C}/H$ that comes out could also be deduced pretty naively.
The universal property of $\widehat{C}$ is that the category of functors $A\to \widehat{C}$ is equivalent to the category of discrete fibrations from $A$ to $C$, via pullback of the universal such.  (A discrete fibration from $A$ to $C$ is a span $A\leftarrow E \to C$ such that $E\to C$ is a fibration, $E\to A$ is an opfibration, the two structures are compatible, and $E$ is discrete in the slice 2-category over $A\times C$.)
Now the slice category $\widehat{C}/P$, for $P\in\widehat{C}$, also has a universal property: it is the comma object of $\mathrm{Id}_{\widehat{C}}$ over $P:1\to \widehat{C}$.  Thus a functor $A\to \widehat{C}/P$ is equivalent to a functor $A\to \widehat{C}$ and a natural transformation from it to the composite $A \to 1 \overset{P}{\to} \widehat{C}$.  By the universal property of $\widehat{C}$, this is the same as giving a discrete fibration $A\leftarrow E \rightarrow C$ together with a map to the discrete fibration classified by $P:1\to \widehat{C}$, which is just $1\leftarrow \int P \rightarrow C$.
Next, (discrete) fibrations have the special property that not only is the composite of two (discrete) fibrations again a (discrete) fibration, but if $g$ is a discrete fibration and $g\circ f$ is a fibration, then $f$ is a fibration.  Therefore, given a discrete fibration from $A$ to $C$ together with a map $E\to \int P$ over $C$, the map $E\to \int P$ is itself a fibration, and we can actually show that $A\leftarrow E \to \int P$ is itself a discrete fibration from $A$ to $\int P$, and that this is an equivalence.  Hence, the slice category $\widehat{C}/P$ has the same universal property as $\widehat{\int P}$, so they are equivalent.
Now we can ask about replacing $P:1\to \widehat{C}$ with a more general functor $H:D\to \widehat{C}$.  Here the fibrational-cosmos argument fails, because in the discrete fibration $D \leftarrow \int H \to C$ it is no longer true that the solitary map $\int H \to C$ is itself a discrete fibration, so cancellability no longer holds and $E\to \int H$ is no longer necessarily a fibration.  However, at least in the 2-category $Cat$, we can still use it to figure out what $\widehat{C}/H$ should look like, by looking at the case $A=1$.  In this case, to give a map $1 \to \widehat{C}/H$ means to give a map $d:1\to D$ (i.e. an object of $D$) along with a discrete fibration $E\to C$ and a map from $E$ to $D \leftarrow \int H \to C$ over $d$ and $\mathrm{Id}_C$.  This is the same as a map from $E$ to $d^*(\int H) = \int H(d)$ over $C$, and we can then apply the previous argument here since both are discrete fibrations over $C$.
Thus, an object of $\widehat{C}/H$ consists of an object $d\in D$ together with a presheaf on $\int H(d)$.  This is not a surprise, though, since we're just applying the original fact $\widehat{C}/P \simeq \widehat{\int P}$ objectwise.  Since $d$ can vary between objects, what we're saying is really that $\widehat{C}/H$ is the category of elements of the functor $D\to Cat$ defined by $d\mapsto \widehat{\int H(d)}$, or more evocatively
$$ \widehat{C}/H = \int_d \widehat{\int H(d)} $$
as a "double integral".<|endoftext|>
TITLE: When is the push-forward of the structure sheaf locally free
QUESTION [11 upvotes]: Let $f:X\longrightarrow Y$ be a morphism of noetherian schemes. Under what conditions is $f_\ast \mathcal{O}_X$ a locally free $\mathcal{O}_Y$-module?
Example 1. Suppose that $f$ is affine. Then $f_\ast\mathcal{O}_X$ is a quasi-coherent $\mathcal{O}_Y$-module.
Example 2. Suppose that $f$ is finite. Then $f_\ast \mathcal{O}_X$ is even coherent.
Example 3. Suppose that $f:X\longrightarrow Y$ is a finite morphism of regular integral 1-dimensional schemes. Then $f_\ast \mathcal{O}_X$ is coherent and locally free. (The local rings $\mathcal{O}_{Y,y}$ are discrete valuation rings.)
In view of the above examples, I'm basically looking for a higher-dimensional analogue of Example 3. But, I don't require quasi-coherence in my question. (Although this is quite unnatural.)
Idea. For any finite morphism $f:X\longrightarrow Y$, we have that $f_\ast \mathcal{O}_X$ is locally free. Only what are the precise conditions on $X$ and $Y$?

REPLY [9 votes]: One should probably also mention the "miracle flatness" theorem: If $f: X \to Y$ is finite, $X$ and $Y$ have the same dimension, $X$ is Cohen-Macaulay and $Y$ is regular, then $f$ is flat. As everyone has mentioned above, finite and flat implies locally free, so this theorem can be one useful way to get the flatness hypothesis.<|endoftext|>
TITLE: Is a non-abelian free group fully residually a finite non-abelian simple group?
QUESTION [5 upvotes]: It is well known that a non-abelian free group is residually a finite simple group.  Katz and Magnus proved, in fact, that non-abelian free groups are residually alternating and residually $PSL_{2}$.  S. J. Pride has some nice results along these lines as well.  The best result that I know of is the theorem of Weigel that can be formulated as follows.  If $\mathfrak{X}$ is a group-theoretic class containing an infinite set of pairwise non-isomorphic finite non-abelian simple groups, then every non-abelian free group is residually an $\mathfrak{X}$-group.

My question is this:

Is a non-abelian free group fully residually a finite non-abelian simple group?

It seems likely that the answer to such an obvious question is known, but I have not been able to find it in the literature.
I should probably add that I suspect we can probably replace "finite non-abelian simple" with "alternating", but I haven't yet given any thought to the other infinite series.  I'd like to learn whether anything is known before spending more time on this.

REPLY [4 votes]: Okay I have found an online version of the paper by Dixon, Pyber, Seress, and Shalev at
http://mathstat.carleton.ca/~jdixon/Residual.pdf so I was able to check things. We need the following:
Theorem 3: Let $S$ be a finite simple group and let $w$ be a non-trivial element of the free group $F_2$ on $X,Y$. Then the probability that two randomly chosen elements $x$ and $y$ of $S$ satisfy both that $x$ and $y$ generate $S$ and $w(x,y) \neq 1$ tends to $1$ as $|S|$ tends $\infty$.
Now, clearly replacing $w$ by any finite set of words, the theorem is still true. Therefore, the fully residually case is true.
Now, the pro-$p$ case is more difficult because while the random generation has probability $1$ the not satisfying an identity has only positive probability. But it may be that the proof itself still works for finite number of words.<|endoftext|>
TITLE: What's the difference between a real manifold and a smooth variety?
QUESTION [42 upvotes]: I am teaching a course on Riemann Surfaces next term, and would like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties (over, say, $\mathbb{C}$).  I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.
Here are some examples that I thought of:
1.  Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$.  By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$.  Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.
2.  Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$.  Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that
agrees with $f$ on $V$ and is identically zero outside of $U$.
By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function.  It may be impossible find a global regular function $g$ that agrees with $f$ on $V$.  When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.
3.  If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian.  The local ring of a variety at a point is always noetherian.  
What are some more examples?
Answers illustrating the difference between real manifolds and complex manifolds are also welcome.

REPLY [6 votes]: Some embedding statements.
A compact complex subvariety of ${\mathbb{C}}^n$ is a point. However, every compact real manifold of dimension $n$ can be realized as a submanifold of some ${\mathbb{R}}^{2n}$.
There are compact complex manifolds that cannot be embedded into complex projective space. An example most often quoted in textbooks is the Hopf manifold, which is not even Kahler. On the other hand, I heard that embedding into real projective space is not often considered in differential geometry.<|endoftext|>
TITLE: Shoenfield's Absoluteness Theorem
QUESTION [7 upvotes]: From wikipedia: http://en.wikipedia.org/wiki/Absoluteness#Shoenfield.27s_absoluteness_theorem
"Shoenfield's theorem shows that if there is a model ZF in which a given $\Pi^1_3$  statement $\phi$ is false, then $\phi$ is also false in the constructible universe of that model."
The problem is that the only proofs I can find seem to use "a model of ZF+DC" or "a model of ZFC".
Is wikipedia right?
If so, is there a proof from just "a model of ZF" online or short enough to sketch here?

REPLY [13 votes]: Wikipedia is correct in that the Shoenfield Absoluteness Theorem holds for plain ZF.
Since the proof of the theorem relies heavily on the absoluteness of well-foundedness, it is tempting to assume DC. However, since the trees that occur in the usual proof of the theorem are canonically well-ordered, DC is not necessary to prove that the well-foundedness of these trees is absolute. For a different approach, see the proof given by Barwise and Fisher in The Shoenfield Absoluteness Lemma. [Israel J. Math. 8 1970, 329-339, MR278934]<|endoftext|>
TITLE: Geneology of survivors in a critical discrete Galton-Watson process
QUESTION [6 upvotes]: Hello. After flipping through a few textbooks on birth-death processes, I can't seem to find anything about genealogical distribution of survivors (conditioned on non-extinction). What I am looking for is a statement roughly of the type, "in generation n, there is probability at least P(j,k,n) that descendants have survived from at least k distinct members of generation j".
I'm interested in the critical case of the Galton-Watson process, where the number of descendents is i.i.d. with mean 1. If necessary, assume the distribution is Bin(r,1/r).
Also, my sample question about P(j,k,n) is just the easiest thing I could think to write down. But I would like to know how likely it is for there to exist a very uniformly distributed subpopulation of size roughly n in generation n.

REPLY [4 votes]: Take a look at
Elementary New Proofs of Classical Limit Theorems for Galton-Watson Processes by Jochen Geiger, Journal of Applied Probability, Vol. 36, No. 2 (1999), pp. 301-309.  It looks to me like the construction there could be helpful to you.<|endoftext|>
TITLE: Why free topological groups on Tychonoff spaces?
QUESTION [9 upvotes]: This is a question of the motivation for a common assumption found in the literature.
The free topological group $F(X)$ on a space $X$ exists for all spaces $X$ (It seems this was first shown by Katutani and Samuel). I mean "free topological group" in the sense that $F:Top\rightarrow TG$ is left adjoint to the forgetful functor $U:TG\rightarrow Top$ from the category of topological groups to the category of topological spaces.
$F(X)$ is well studied when $X$ is a Tychonoff space. This permits the application of pseudometrics which seems to be a powerful tool for describing the complicated topological structure of $F(X)$. Also, it seems to be a useful fact that the canonical map $\sigma:X\rightarrow F(X)$ is an embedding when $X$ is Tychonoff. 
These two conveniences do seem to make it convenient to study $F(X)$ when $X$ is Tychonoff but it seems almost no one is interested in $F(X)$ when $X$ is not Tychonoff. Why is this? Are these uninteresting for some reason?

REPLY [9 votes]: Let $X$ be a topological space. If $F(X)$ is $T_0$ then I think $F(X)$ is isomorphic (as topological groups) to $F(Y)$, where $Y$ is the Tychonofficiation (see below) of $X$. So it is enough to study topological free groups on a Tychonoff space.
Explanation:
First let me remind myself about some notation. Completely regular means that any point can be separated from a closed set not containing it by a continuous real-valued function. Tychonoff then means completely regular and $T_2$(=Hausdorff). Any topological group is completely regular; and for topological groups $T_0$ is equivalent to $T_2$.
Suppose $X$ is an arbitrary topological space, and let $Y$ be its "Tychonoffication" (!). That is, set theoretically $Y$ is the quotient of $X$ by the equivalence relation $x\sim x'$ if and only $f(x)=f(x')$ for all continuous $f:X\to\mathbb{R}$; each such $f$ descends to $Y$ and we give $Y$ the weak topology induced by all these real-valued maps. This makes $Y$ into a Tychonoff space which I think satisfies the following universal property: any continuous map from $X$ to a Tychonoff space factors uniquely through $Y$. Also, the natural map $X\to Y$ induces an isomorphism $C(Y)\stackrel{\cong}{\to} C(X)$, where $C(-)$ denotes the ring of real-valued continuous functions.
Assuming $F(X)$ is $T_0$, then the natural map $F(X)\to F(Y)$ seems to be an isomorphism of topological groups. It is enough to construct an inverse. Since $F(X)$ is $T_0$, it is even $T_2$, and therefore it is Tychonoff. So the natural map $X\to F(X)$ factors through $Y$ and induces $F(Y)\to F(X)$, which surely does the trick?
What about that $T_0$ assumption?
Lots of people are only interested in Hausdorff topological groups, so it seems reasonable to only study spaces $X$ for which $F(X)$ is $T_0$ (hence $T_2$). Otherwise you could replace $Y$ by the "complete-regularization" of $X$ (i.e. $X$ equipped with the weak topology induced by $C(X)$) and repeat the argument, but it doesn't work so nicely.
Edit:
While I was typing my answer, you asked about this Tychonoffication business!<|endoftext|>
TITLE: Can a positive binary quadratic form represent 14 consecutive numbers?
QUESTION [59 upvotes]: NEW CONJECTURE: There is no general upper bound.
Wadim Zudilin suggested that I make this a separate question. This follows
representability of consecutive integers by a binary quadratic form
where most of the people who gave answers are worn out after arguing over indefinite forms and inhomogeneous polynomials. Some real effort went into this, perhaps it will not be seen as a duplicate question.
So the question is, can a positive definite integral binary quadratic form
$$ f(x,y) = a x^2 + b x y + c y^2 $$
represent 13 consecutive numbers?
My record so far is 8: the form $$6x^2+5xy+14y^2 $$
 represents the 8 consecutive numbers from 716,234 to 716,241. Here we have discriminant $ \Delta = -311,$ and 2,3,5,7 are all residues $\pmod {311}.$ I do not think it remotely coincidental that $$6x^2+xy+13 y^2 $$ represents the 7 consecutive numbers from 716,235 to 716,241. 
I have a number of observations. There is a congruence obstacle $\pmod 8$ unless, with 
$ f(x,y) = a x^2 + b x y + c y^2 $  and $\Delta = b^2 - 4 a c,$ we have $\Delta \equiv 1 \pmod 8,$ or 
$ | \Delta | \equiv 7 \pmod 8.$ If a prime $p | \Delta,$ then the form is restricted to either all quadratic residues or all nonresidues $ \pmod p$ among numbers not divisible by $p.$ 
In what could be a red herring, I have been emphasizing $\Delta = -p$ where $p \equiv 7 \pmod 8$ is prime, and where there is a very long string of consecutive quadratic residues $\pmod p.$ Note that this means only a single genus with the same $\Delta = -p,$ and any form is restricted to residues. I did not anticipate that long strings of represented numbers would not start at 1 or any predictable place and would be fairly large. As target numbers grow, the probability of not being represented by any form of the discriminant grows ( if prime $q \parallel n$ with $(-p| q) = -1$), but as the number of prime factors $r$ with $(-p| r) = 1$ grows so does the probability that many forms represent the number if any do.  Finally, on the influence of taking another $\Delta$ with even more consecutive residues, the trouble seems to be that the class number grows as well. So everywhere there are trade-offs.
EDIT, Monday 10 May. I had an idea that the large values represented by any individual form ought to be isolated. That was naive. Legendre showed that for a prime $q \equiv 7 \pmod 8$ there exists a solution to $u^2 - q v^2 = 2,$ and therefore infinitely many solutions. This means that the form $x^2 + q y^2$ represents the  triple of consecutive
numbers $q v^2, 1 + q v^2, u^2$ and then represents $4 + q v^2$ after perhaps skipping $3 + q v^2$.
Taking $q = 8 k - 1,$ the form $ x^2 + x y + 2 k y^2$ has no restrictions $\pmod 8,$ while an explicit formula shows that it represents every number represented by   $x^2 + q y^2.$ Put together, if $8k-1 = q$ is prime, then $ x^2 + x y + 2 k y^2$ represents infinitely many  triples. If, in addition, $ ( 3 | q) = 1,$ it seems plausible to expect infinitely many  quintuples. It should be admitted that the recipe given seems not to be a particularly good way to jump from length 3 to length 5, although strings of length 5 beginning with some $q t^2$ appear plentiful.
EDIT, Tuesday 11 May. I have found a string of 9, the form is $6 x^2 + x y + 13 y^2$ and the numbers start at $1786879113 = 3 \cdot 173 \cdot 193 \cdot 17839$ and end with 
$1786879121$ which is prime. As to checking, I have a separate program that shows me the particular $x,y$ for representing a target number by a positive binary form. Then I checked those pairs using my programmable calculator, which has exact arithmetic up to $10^{10}.$ 
EDIT, Saturday 15 May. I have found a string of 10, the form is $9 x^2 + 5 x y + 14 y^2$ and the numbers start at $866988565 = 5 \cdot 23 \cdot 7539031$ and end with 
$866988574 = 2   \cdot  433494287.$
EDIT, Thursday 17 June. Wadim Zudilin has been running one of my programs on a fast computer. We finally have a string of 11, the form being $ 3 x^2 + x y + 26 y^2$ of discriminant $-311.$ The integrally represented numbers start at  897105813710 and end at   897105813720. Note that the maximum possible for this discriminant is 11. So we now have this conjecture: For discriminants $\Delta$ with absolute values in this sequence
http://www.oeis.org/A000229
some form represents a set of $N$ consecutive integers, where $N$ is the first quadratic nonresidue. As a result, we conjecture that there is no upper bound on the number of consecutive integers that can be represented by a positive quadratic form.

REPLY [4 votes]: Here's a heuristic that suggests why arbitrarily large strings of consecutive 
numbers should be representable by some binary quadratic form.  For simplicity 
consider a prime $p$ that is $3\pmod 4$ so that $-p$ is a fundamental discriminant.  Suppose that $p$ has been chosen in such a way that all the primes $\le k$ are 
quadratic residues $\pmod p$.  Suppose there are $h$ quadratic forms of discriminant $-p$ (and note that there is only one genus).  
Recall that a number $n$ is represented by some form of discriminant $-p$ if 
every prime factor $\ell$ of $n$ satisfies $\chi(\ell)=1$.  As discussed in my 
answer to Achieving consecutive integers as norms from a quadratic field we should expect to find many strings of $k$ consecutive numbers, each of which is representable by some form of discriminant $-p$.  
In my answer to that question, I focused on such strings of (almost) prime numbers, 
but the same Hardy-Littlewood heuristics would predict lots of strings $n+1$, $\ldots$, 
$n+k$ where each $n+j$ is divisible only by primes that are quadratic residues 
$\pmod p$, and each $n+j$ has a typical number of prime factors.  Under the 
restriction that all prime factors of $n$ are quadratic residues $\pmod p$, if $n$ is 
large then typically it will have about $\frac 12 \log \log n$ such prime factors.  Moreover we may expect these prime factors to be roughly equally distributed 
in the class group (which is of fixed size $h$).  Thus since there are $2^{\omega(n)}$ 
factorizations of $n$ as a product of two ideals, we would expect that typically there are many such factorizations with one of the factors lying in a prescribed 
ideal class.  
Summarizing one would expect that there are many strings $n+j$ 
($1\le j\le k$) with each $n+j$ composed of about $\frac 12\log \log n$ primes 
that are all quadratic residues $\pmod p$, and (typically) each such $n+j$ would 
be represented by every form of discriminant $-p$.  
It should be possible with a little effort to turn this into a precise 
Hardy-Littlewood type conjecture, but I don't see any hope of a proof. 
The heuristic described above is probably classical.  One place where this 
heuristic is described is a paper of Blomer and Granville: see pages 9 and 10 of 
http://www.dms.umontreal.ca/~andrew/PDF/quadraticforms.pdf<|endoftext|>
TITLE: How does $f_* O_X$ measure ramification and Grothendieck-Riemann-Roch
QUESTION [10 upvotes]: Let $f:X\longrightarrow Y$ be a finite morphism of smooth projective varieties over a field $k$ of characteristic zero, where $\dim X=\dim Y$. Then $f$ is flat. Hence $f_\ast \mathcal{O}_X$ is a coherent locally free sheaf on $Y$.
Now, my question is based on the following example. (For simplicity, take $k=\overline{k}$.)
Example. Suppose that $\dim X =\dim Y = 1$ (i.e., curves). Apply Grothendieck-Riemann-Roch to $f$ and $\mathcal{O}_X$. In degree 0, we get the fact that $\textrm{deg} \ f = \textrm{rk} \ f$. In degree 1 we get a "Hurwitz theorem". In fact, with little effort the formula reads $$2c_1(f_\ast \mathcal{O}_X) =\deg f \cdot K_Y -  f_\ast(K_X) = f_\ast(-R), $$ where $R$ is the ramification divisor on $X$.
Now for my two questions that are based on this formula.
Q1. The divisor $R$ is not called the ramification divisor for nothing. Its support is the set of ramification points and the multiplicity of $R$ at a point $P$ is precisely $e_P-1$. So in my opinion, it "measures" the ramification. What about $c_1(f_\ast \mathcal{O}_X) = c_1(\det f_\ast \mathcal{O}_X)$? How does he "measure" the ramification? (I'm probably missing something really elementary here.)
Q2. In higher-dimensions, if I understand correctly, one should get a "higher-dimensional" Hurwitz formula:  $$2c_1(f_\ast \mathcal{O}_X) =f_\ast(\textrm{td}(X/Y)).$$ I doubt that this "measures" all the ramification. And, to be frank, I don't really know what it "measures". Can anyone provide some insight?

REPLY [9 votes]: The following does not exactly answer your question, but you may find it interesting. It is the Riemann-Hurwitz formula for surfaces.
Let $\phi:S_1\to S_2$ be a finite morphism between smooth, projective surfaces (over an algebraically closed field of characteristic zero) of degree $n$, and let $B\subseteq S_2$ be the set of $y\in S_2$ such that $\phi^{-1}(y)$ does not contain $n$ points (i.e. $B$ is the ramification locus). Zariski's purity theorem states that $B$ is pure of dimension one; let $B_1,\dots,B_r$ be its irreducible components, and let $n_i$ be the degree of the morphism $\phi|_{\phi^{-1}(B_i)}:\phi^{-1}(B_i)\to B_i$. Then
$$\chi(S_1)=\chi(S_2)\deg \phi-\sum_{i=1}^r(n-n_i)\chi(B_i)+\sum_{y\in B}\left(|\phi^{-1}(y)|-n+\sum_{i=1}^r(n-n_i)m_i(y)\right)$$
where $m_i(y)$ denotes the number of local branches of $B_i$ at $y$. Here $\chi$ is the $\ell$-adic Euler characteristic of the surface ( topological Euler characteristic if $k=\mathbb{C}$), which can be translated into a Chern class if you prefer.
The proof is B. Iversen, 'Numerical invariants and multiple planes', Amer. J. Math. 92 (1970), 968-996. When $k=\mathbb{C}$, you can prove it by thinking of the topological Euler characteristic as a measure on constructible sets (e.g. O. Ya. Viro, Some integral calculus based on Euler characteristic); then the formula is equivalent to Fubini's theorem ($\int\int dxdy=\int\int dydx$) for the graph of $\phi$.<|endoftext|>
TITLE: Automorphisms of $\pi_1$ induced by pseudo-Anosov maps
QUESTION [5 upvotes]: Suppose $X$ is an orientable surface with non-empty boundary and $f:X\to X$ is a pseudo-Anosov automorphism that acts identically on $H_1(X,\mathbf{Z})$. Let $x$ be a fixed point of $f$.
For any $\gamma\in\pi_1(X,x)$ we have $\gamma^{-1}f(\gamma)\in [\pi_1(X,x),\pi_1(X,x)]$, the commutant of $\pi_1(X,x)$. More generally, we have $\gamma\cdot g^{-1}f g(\gamma)\in [\pi_1(X,x),\pi_1(X,x)]$ where $g$ is an automorphism of $X$ that fixes $x$.
I would like to ask what one can say about the normal closure in $\pi_1(X,x)$ of the set of all elements $\gamma\cdot g^{-1}f g(\gamma)$ where $\gamma$ runs through $\pi_1(X,x)$ and $g$ runs through the set of all diffeomorphisms $X\to X$ that fix $x$. In particular, does this closure  coincide with the commutant of $\pi_1(X,x)$?

REPLY [8 votes]: No. Let $\Gamma_i$ be the lower central series defined by $\Gamma_1=\pi_1(X,x)$, $\Gamma_{i+1}=[\Gamma_1,\Gamma_i]$. The Johnson filtration $\text{Mod}_g(k)$ is the descending filtration of the mapping class group relative to $x$ defined by:
$f\in \text{Mod}_g(k)\iff f$ acts trivially on $\Gamma_1/\Gamma_k$
The first term $\text{Mod}_g(2)$ is the Torelli group, consisting of diffeomorphisms acting trivially on homology. The next term $\text{Mod}_g(3)$ is the Johnson kernel. By a beautiful theorem of Johnson, this is the subgroup generated by Dehn twists around separating curves.
By residual nilpotence of surface groups, we have $\bigcap \text{Mod}_g(k)=\{1\}$, but every individual term in the filtration is nontrivial. It is not hard to see that every term of the Johnson filtration contains pseudo-Anosovs. Indeed every normal subgroup of the mapping class group contains pseudo-Anosovs (see Lemma 2.5 of Long, "A note on the normal subgroups of mapping class groups") from which Long concluded that any two normal subgroups intersect nontrivially!
Thus since $\text{Mod}_g(k)$ is normal, if we take $f\in\text{Mod}_g(k)$ we have $\gamma^{-1}\cdot g^{-1}fg(\gamma)\in \Gamma_k$ for all $g$ and all $\gamma$.<|endoftext|>
TITLE: Earliest diagonal proof of the uncountability of the reals.
QUESTION [12 upvotes]: I cited the diagonal proof of the uncountability of the reals as an example of a `common false belief' in mathematics, not because there is anything wrong with the proof but because it is commonly believed to be Cantor's second proof. The stated purpose of the paper where Cantor published the diagonal argument  is to prove the existence of uncountable infinities, avoiding the theory of irrational numbers. I have no problem believing that Cantor himself realized that a diagonal proof of the uncountability of R was possible but I have not even found an allusion to this in his collected works. The earliest appearance in print that I know is on page 43 of The theory of sets of points by W. H. Young and Grace Chisholm Young (1906). I would be very grateful for any reference to some scrap of paper where Cantor himself mentions the possibility of using the diagonal method to prove the set of reals uncountable.

REPLY [2 votes]: The evidence that you look for can already be found in the second paragraph of Cantor's original paper. There he states "Es läßt sich aber von jenem Satze ein viel einfacherer Beweis liefern, der unabhängig von der Betrachtung der Irrationalzahlen ist.", meaning that his diagonal argument supplies a much simpler proof of the theorem proved in his first paper on the uncountability of the real numbers. This does not only mention but declare that his method can be applied to real numbers.
Probably in order to emphasize its independence of numbers, Cantor did not use numerals 0 and 1, but m and w which (and this answers your last remark) in German are abbreviations of male and female. So he had a good substitution for 0 and 1 or up and down or yes and no - and he deliberately or unconsciously circumvented the problem that this proof without reservations, as stated in his original text, would fail on binary sequences.<|endoftext|>
TITLE: Thom's seminal cobordism paper in English?
QUESTION [9 upvotes]: In Quelques proprietes globales des varietes differentiables, Thom classifies unoriented manifolds up to cobordism.  I've been struggling a bit to understand this paper, and while Stong's cobordism notes have helped a bit, I was wondering if an English translation (of the entire paper or just parts) exists.  Thank you in advance.

REPLY [4 votes]: An English translation is also available online here:
http://math.mit.edu/~hrm/kansem/thom-some-global-properties.pdf<|endoftext|>
TITLE: How big can the Hausdorff dimension of a function graph get?
QUESTION [7 upvotes]: This question is inspired by How kinky can a Jordan curve get?
What is the least upper bound for the Hausdorff dimension of the graph of a real-valued, continuous function on an interval? Is the least upper bound attained by some function?
It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. However, this does not rule out the possible existence of a function graph of dimension two.

REPLY [3 votes]: It seems that most of the continuous functions over the unit interval have a graph of Hausdorff dimension 2. To be more precise, it is proved in https://projecteuclid.org/euclid.rae/1366030629 (The Hausdorff Dimension of Graphs of Prevalent Continuous Functions by Jonathan M. Fraser and James T. Hyde) that the set of continuous function whose graph has Hausdorff dimension 2 is prevalent.<|endoftext|>
TITLE: What's known about the stalks of Lusztig's perverse sheaves on quiver varieties?
QUESTION [11 upvotes]: Lusztig has defined a category of perverse sheaves on the moduli space of representations of a Dynkin quiver (see his paper) corresponding to canonical basis vectors.  
I'm interested in the stalks of these perverse sheaves, in particular, the stalk at 0.  I believe I've reduced a (seemingly unrelated) conjecture to the question

Is the stalk of any one of Lusztig's perverse sheaves at 0 1-dimensional? (A weaker claim, which I think would make me roughly as happy is that the Euler characteristic of a stalk is 1.)

(The conjecture is that it is 1-dimensional, but I'm fairly agnostic on this point.  I wouldn't be surprised either way, though if I had to choose I'd say this conjecture sounds a little unlikely).

Is there a proof or counterexample of this claim in the literature? 

After a quick read-through of some the literature (Lusztig's paper, Reineke's paper on resolutions, Kashiwara and Saito's paper on crystal structures on components, etc.)  I'm not feeling any closer to understanding these stalks, and am having trouble finding anywhere else to look.  

EDIT: This seems like an even longer shot, but what would be even better would be to understand the cohomology of one of these sheaves on a "quiver orbital variety,'' the space of quiver representations (not modulo isomorphism!) which preserve a particular flag.  Is there anything about this in the literature?

REPLY [7 votes]: The dimensions of the stalks of Lusztig's sheaves give the coefficients when a canonical basis element is expanded in a PBW basis. These stalks satisfy a parity vanishing condition.
For these positive results, see Corollary 10.7 of Lusztig's "Canonical Bases Arising from Quantized Enveloping Algebras".
The first example of a 2-dimensional stalk at zero occurs in type A2, in dimension vector (2,2). If X is the variety of 2x2 matrices of rank at most 1, then it is not hard to see (e.g. using the Deligne construction) that IC(X) is one of Lusztig's sheaves and has a 2-dimensional stalk at zero. Alternatively the canonical basis computation can be carried out, which is implemented in GAP via the package QuaGroup.<|endoftext|>
TITLE: How to find more (finite almost simple) groups with a given Sylow subgroup
QUESTION [16 upvotes]: I'm looking for some examples of actions on Sylow p-groups, and often those actions appear in the case of finite almost simple groups.  Given a finite almost simple group, I understand in principle how to calculate its Sylow p-subgroups (here p is usually 2 or 3), but perhaps I am just too slow at doing it.  In particular, I am familiar with the papers of Weir and Carter-Fong.
I am not sure how to do the reverse calculation: given a p-group P (possibly described as "the Sylow p-subgroup of the almost simple group X", and X is something explicit like "PGL(3,19)" or "M11"), find all of the almost simple groups that have a Sylow p-subgroup isomorphic to P.
I am pretty sure some people know how to do this, but it's not really clear to me how to go about it.  For instance, it would have never occurred to me that PSU(3,8), PSL(3,19), and 3D4(2) have isomorphic Sylow 3-subgroups.
Is there a description of how this is done?
I think there is likely to be a finite answer to the question: Obviously only finitely many (and probably O(p)) alternating groups could work for a given P.  We take for granted that only finitely many sporadic groups could work.  It seems that, similarly to the alternating case, there are only finitely many ranks (again probably O(p)) of Lie groups that could work, and hopefully for each Lie type (and rank), there are just some congruences on "q" that indicate which ones work and which don't.
However, I've not had much luck doing this calculation in examples, and so I am looking for papers or textbooks where this has been done.  I have found some that state the result of doing something like this (post CFSG), and I have found several that do this in quite some detail, but pre-CFSG so they spend hundreds of pages eliminating impossible groups obscuring what should now be an easy calculation.  I'm looking for something with the pedagogical style of the pre-CFSG papers, but that doesn't mind using the standard 21st century tools.
Alternatively: does anyone know of a vaguely feasible approach to construct all groups with given Sylow subgroups?  Blackburn et al.'s Enumerating book has some upper bounds, but they are pretty outrageous and don't seem adaptable to a feasible algorithm for my problem.

REPLY [4 votes]: There is a sense in which, for primes p >3, "most" finite p-groups can NOT occur as Sylow
p-subgroups of finite simple groups (I know you asked about almost simple groups).
For example, for such primes p, George Glauberman proved that a finite p-group P
whose outer automorphism group is a p-group can not be the Sylow p-subgroup of 
a finite simple group. There are various theorems which show that "most"
finite p-groups have no outer automorphism of order prime to p.
 But I agree with previous comments. Your last question sounds very very general.<|endoftext|>
TITLE: Are the norms of graphs dense in any interval?
QUESTION [32 upvotes]: It is known that there is a gap between 2 and the next largest norm of a graph.
Is there an interval of the real line in which norms of graphs are dense?

REPLY [38 votes]: I found a reference that seems to answer your question:
Shearer, James B.
On the distribution of the maximum eigenvalue of graphs, 1989.
The theorem in this paper is that the set of largest eigenvalues of adjacency matrices of graphs is dense in the interval $\left[\sqrt{2+\sqrt{5}},\infty\right)$.   Here's an online version.

Here's a related paper:
Hoffman, Alan J.
On limit points of spectral radii of non-negative symmetric integral matrices, 1972.
In this paper limit points less than $\sqrt{2+\sqrt{5}}$ are described. In particular, they form an increasing sequence starting at 2 and converging to $\sqrt{2+\sqrt{5}}$.  Here's an online version.    The author also posed the problem that led to Shearer's paper.<|endoftext|>
TITLE: Question about computing group cohomology using cochains
QUESTION [6 upvotes]: In Milne's notes on Class Field Theory (http://www.jmilne.org/math/CourseNotes/CFT.pdf), he initially defines group cohomology in terms of injective resolutions, then he talks about computing cohomology using cochains. I don't see him mention anywhere that the group has to be finite in order for cochains to work, but this seems to be the case?
Later, he discusses profinite groups, in which he says that cohomology of profinite groups can be computed using continuous cochains. What isn't clear is the following: is the cohomology using continuous cochains a modified cohomology theory, different from the one using injective resolutions? In this case, then, do we get the cohomology theory using injective resolutions if we use all cochains, not just continuous ones? Or do the continuous cochains give the same cohomology as injective resolutions, and cochains which are not necessarily continuous only give cohomology in the case of finite groups? I.e., is there only one such cohomology theory? At the very least, using general cochains versus continuous cochains in the case of infinite profinite groups is different, for in one case $H^1$ is $\mathrm{Hom}(G,M)$ when M is trivial, and in the other case $H^1$ is $\mathrm{Hom}_{\mathrm{cts}}(G,M)$.
Assuming that set-theoretic cochains only work for finite groups, why is it the case? It seems that the proof that cochains compute cohomology (i.e. by looking at a projective resolution of $\mathbb{Z}$) fails because the modules used in the case of finite groups, i.e. tensor powers of $\mathbb{Z}[G]$, aren't necessarily projective when $G$ is infinite (in the case when $G$ is finite, they are free). Is this correct?

REPLY [9 votes]: You should take a look at the beginning of chapter 2 of Serre's Galois Cohomology.
He explains there that if G is a profinite group, then the category of discrete abelian groups with a continuous action of G has enough injectives (but not enough projectives in general), and that cohomology can be "computed" as a direct limit of cohomology of a finite group.
To sum up:
-if G is discrete (i.e. no topology), then the category has enough injectives and projectives (it is the category of left $\mathbb{Z}[G]$-modules), and using a projective resolution for $\mathbb{Z}$ gives you the equivalence between the derived functor definition and the cochains definition (using the fact that Ext can be computed two ways). You can find this in Serre's Local Fields.
-if G is profinite, and we consider the category of discrete modules with a continuous action of G, then there are enough injectives (this can be seen quite easily from the discrete case), but not enough projectives. Luckily though, the two definitions agree (thanks to the "direct limit computability"). You can find this in Serre's Galois Cohomology.
-if G is an arbitrary topological group, there is not much left. There aren't enough injectives nor projectives in general, and if you define cohomology with cochains, you don't get an homological functor (only the beginning of the long exact sequence exists). However, see the end of J.-M. Fontaine and Yi Ouyang's book (it's a pdf, I found it on Fontaine's web page) about p-adic representations, they mention that if you have a continuous set-theoretic section in your short exact sequence, you get a long exact sequence. I haven't read the reference they provide, though.<|endoftext|>
TITLE: What are the merits of the different finiteness conditions on quasi-coherent sheaves?
QUESTION [16 upvotes]: It's my understanding that there's no disagreement about the right way to define a quasi-coherence for a sheaf $F$ of $O_X$-algebras (over a scheme, locally ringed space, or even locally ringed topos). It means that, after passing to some cover, it's isomorphic to a cokernel of a map of free $O_X$-modules.
But now there are several different finiteness conditions you can put on these.

$F$ is of finite type if, after passing to some cover, it is isomorphic to a quotient of a finite free module.
$F$ is of finite presentation if, after passing to some cover, it is isomorphic to a cokernel of a map of finite free modules.
$F$ is coherent if it is of finite type and for every open $U$, every integer $n>0$, and every map $O_X^n\to F$, the kernel is of finite type.

Then 3 ==> 2 ==> 1. They are all equivalent if $X$ is Spec of a noetherian ring.
The first two conditions seem very natural and are of the standard kind in sheaf theory: there exists a cover over which some property exists. But the definition of coherence is very different, and purely on formal grounds, we might expect that the class of coherent sheaves would be not so well behaved. (For instance, probably 1 and 2 can be expressed in terms of some allowable syntax in topos theory, but 3 can't.) Sure enough, the early sections of EGA are a mess when they talk about coherent sheaves, with noetherian hypotheses all over. For instance, if $X=\mathrm{Spec}(R)$, then $O_X$ is proved to be coherent only when $R$ is noetherian, as far as I can tell, whereas it's obviously finitely presented. Also, I think a quasi-coherent sheaf on an affine scheme is finitely presented if and only if the corresponding module is. And finite presentation is stable under pull back, but coherence isn't (e.g. $X\to\mathrm{Spec}(\mathbf{Z})$, where $O_X$ isn't coherent).
So coherence seems like a bad condition in the absence of some other hypotheses which make it collapse into one of the good ones. My feeling is that such foundational things should be very formal and tight, and if they're not, it's probably because we're using approximations of the right concepts.
Question #1: What is coherence actually good for? Suppose we tried to replace it with finite presentation everywhere. Would anything go wrong? (Is there difference between algebraic and analytic geometry here?)
Question #2: If finite presentation has its problems (which it does, I think, but I can't remember them now), are there any known variants that are better behaved? For instance, what about this condition: For an integer $n\geq 0$, let's say that $F$ is of $n$-finite type if, after restricting to some cover, there exists an exact sequence of $O_X$-modules $$M_n\to M_{n-1} \to \cdots \to M_0 \to F \to 0,$$ where each $M_i$ is $O_X^{r_i}$ for some integer $r_i\geq 0$. So finite type = 0-finite type, and finite presentation = 1-finite type. Then say $F$ is of $\infty$-finite type if there exists an $n$ such that $F$ is of $n$-finite type. Is there any chance that being of $\infty$-finite type is well behaved?

REPLY [8 votes]: Of course, the correct definition of coherence is that in your Question 2. It just so happens that for a sheaf of modules on a scheme it is equivalent to the easier one.
As far as a I know, the notion of coherence is mostly used when one has a sheaf of ring (often non-commutative), different from the structure sheaf. So, for example, the sheaf of D-modules on a smooth variety is not noetherian, but it is a coherent sheaf of rings; this is a very important fact.
There are schemes whose structure sheaf is not coherent, and those are a bit of a mess; for example, the locally finitely presented quasi-coherent sheaves do not form an abelian category. However, in most cases one is not usually bothered by them. 
For example, in setting up a moduli problem, it is quite useful to consider non-noetherian base schemes, because the category of locally noetherian schemes has problems (for example, is not closed under fibered products). For example, if $X$ is a projective scheme over a field $k$, and $F$ is a coherent sheaf on $X$, one defined the Quot functor from schemes over $k$ to sets by sending each (possibly non-noetherian scheme) $k$-scheme $T$ into the set of finitely presented quotients of the pullback $F_T$ of $F$ to $T$ that are flat over $T$. Of course, when you actually prove something, one uses that fact that locally on $T$ any finitely presented sheaf comes from one defined a finitely generated $k$-algebra, and works with that, free to use all the results that hold in the noetherian context. Thus, in practice most of the time you don't need to do anything with non-noetherian schemes.<|endoftext|>
TITLE: Is there a knotted torus in 4-sphere whose complement's fundamental group is infinite cyclic ?  
QUESTION [5 upvotes]: I am reading the book 'surface in 4-space' about the unknotting conjecture (Page 97): a 2-knot (2-sphere in 4-sphere) is trival if and only if the fundamental group of the exterior is infinite cyclic. 
It said that in TOP category, Freedman proved the statement is true. I don't know why it is also true for general surface. in top category?

REPLY [3 votes]: Anthony Conway and Mark Powell give a proof that a locally flat embedded closed orientable surface in the 4-sphere whose complement has infinite cyclic fundamental group is topologically unknotted, provided that the genus is greater than or equal to 3 (the genus 0 case is a result of Freedman and Quinn as mentioned in the question).
Apart from Kawauchi's controversial work, the topological unknotting conjecture for genus 1 and 2 appears to be an open question.<|endoftext|>
TITLE: Multiplicative structure on spectral sequence
QUESTION [20 upvotes]: Let $E$ be a spectral sequence and assume that there is a product
$E^{r}_{p_1,q_1} \times E^r_{p_2,q_2} \to E^r_{p_1+p_2,q_1+q_2}$
which satisfies the Leibniz rule (for all $p_i,q_i$, but $r$ fixed). Then it extends to a product
$E^{r+1}_{p_1,q_1} \times E^{r+1}_{p_2,q_2} \to E^{r+1}_{p_1+p_2,q_1+q_2}$.
In the errata for Weibel 5.2.13 it is suggested that the Leibniz rule does not hold automatically in $E^{r+1}$. I've convinced myself of this through lengthy calculations with the conclusion: Ok, nothing is compatible, and there is no reason why the product has something to do with the isomorphisms $H(E^r) \cong E^{r+1}$.

Is there an easy insightful example where the Leibniz rule fails?
Is there a simple-to-check criterion which makes the Leibniz rule valid?

Just to clarify: Of course there are lots of special cases where you can write down a product on $E^r$ for each $r$. For example if $E$ is the spectral sequence coming from a filtered differential graded algebra. But I'm rather interested in the general case. I think often you only know $E^2$ well and it would be interesting if you can lift the product to $E^{\infty}$ without actually make your hands dirty.

REPLY [11 votes]: This is an expansion of John Rognes' answer. I have filled in a few details in Douady's seminare notes and noticed that one gets away with slightly weaker axioms. If there is already a reliable reference for all this, please let me know.
Recall that a Cartan-Eilenberg system $(H,\eta,\partial)$ (see here, chapter XV.7) consists of modules $H(p,q)$ for each $p\le q$, morphisms $\eta\colon H(p',q')\to H(p,q)$ for all $p\le p'$, $q\le q'$, and boundary morphisms $\partial\colon H(p,q) \to H(q,r)$ for all $p\le q\le r$, such that

$\eta=\mathrm{id}\colon H(p,q)\to H(p,q)$,
$\eta=\eta\circ\eta\colon H(p'',q'')\to H(p',q')\to H(p,q)$,
$\eta$ and $\partial$ commute,
there are long exact sequences $\cdots\to H(q,r)\stackrel\eta\to H(p,r)\stackrel\eta\to H(p,q)\stackrel\partial\to H(q,r)\to\cdots$.

The conditions needed for convergence have been omitted. A typical example is $H(p,q)=\tilde h_\bullet(X_p/X_q)$, where $\cdots\supset X_{-1}\supset X_0\supset X_1\supset\cdots$ is a decreasing sequence of cofibrations and $\tilde h_\bullet$ is some generalised homology theory. The grading is suppressed in the following, but you can easily fill it in.
To set up a spectral sequence from $(H,\eta,\partial)$, one defines
$$Z^r_p=\mathrm{im}\bigl(H(p,p+r)\stackrel\eta\to H(p,p+1)\bigr)\;,$$
$$B^r_p=\mathrm{im}\bigl(H(p-r+1,p)\stackrel\partial\to H(p,p+1)\bigr)\;,$$
$$E^r_p=Z^r_p/B^r_p\;,$$
$$d^r_p\colon Z^r_p/B^r_p\twoheadrightarrow Z^r_p/Z^{r+1}_p\cong
B^{r+1}_{p+r}/B^r_{p+r}\hookrightarrow Z^r_{p+r}/B^r_{p+r}\;.$$
Details are in Switzer's book, chapter 15. In particular
$$\ker(d^r_p)=Z^{r+1}_p/B^r_p\qquad\text{and}\qquad
\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p\;.$$
For $a=\eta(a_0)\in H(p,p+1)$, $a_0\in H(p,p+r)$, one has
$$d^r_p([a])=[\partial a_0]\in E^r_p\qquad\text{with}\qquad\partial a_0\in H(p+r,p+r+1)\;.$$
Definition (Douady, II.A)
Let $(H,\eta,\partial)$, $(H',\eta',\partial')$ und $(H'',\eta'',\partial'')$ be Cartan-Eilenberg systems. A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ is a sequence of maps
$$\mu_r\colon H'(m,m+r)\otimes H''(n,n+r)\to H(m+n,m+n+r)$$
such that for all $m$, $n$, $r\ge 1$, the following two diagrams commute.
$\require{AMScd}$
\begin{CD}
  H'(m,m+r)\otimes H''(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\
  @V\eta'\oplus V\eta''V@VV\eta V\\
  H'(m,m+1)\otimes H''(n,n+1)@>\mu_1>>H(m+n,m+n+1)\rlap{\;,}
\end{CD}
\begin{CD}
  H'(m,m+r)\otimes H''(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\
  @V\partial'\otimes\eta''\oplus V\eta'\otimes\partial''V@VV\partial V\\
  {\begin{matrix}H'(m+r,m+r+1)\otimes H''(n,n+1)\\\oplus\\H'(m,m+1)\otimes H''(n+r,n+r+1)\end{matrix}}@>\mu_1+\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{\;.}
\end{CD}
The first diagram is weaker than in Douady's notes.
The second can be read as a Leibniz rule.
Theorem (Douady, Thm II)
A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ induces products
$$\mu^r\colon E^{\prime r}_m\otimes E^{\prime\prime r}_n\to E^r_{m+n}\;,$$
such that

$\mu^1=\mu_1$
$d^r_{m+n}\circ\mu^r=\mu^r\circ(d^{\prime r}_m\otimes\mathrm{id})\pm\mu^r\circ(\mathrm{id}\circ d^{\prime\prime r}_n)$,
$\mu^{r+1}$ is induced by $\mu^r$.


Proof.
Assume by induction that $\mu^r$ is induced by $\mu_1$. In particular,
$$Z^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to Z^r_{m+n}\;,$$
$$B^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;,$$
$$Z^{\prime r}_m\otimes B^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;.$$
This is clear for $r=1$ if we put $\mu^1=\mu_1$ because $E^1_p=Z^1_p=H(p,p+1)$
and $B^1_p=0$.
Let $[a]\in Z^{\prime r}_m$, $[b]\in Z^{\prime\prime r}_n$ be represented by $a=\eta'(a_0)\in H'(m,m+1)$, $b=\eta''(b_0)\in H''(n,n+1)$ with $a_0\in H'(m,m+r)$, $b_0\in H''(n,n+r)$.
Using the first diagram and the construction of $d^r_{m+n}$, we conclude that
$$(d^r_{m+n}\circ\mu^r)([a]\otimes[b])=d^r_{m+n}[\mu_1(a\otimes b)]=d^r_{m+n}[\eta(\mu_r(a_0\otimes b_0))]=(\partial\circ\mu_r)(a_0\otimes b_0)\;.$$
From the second diagram, we get
$$(\partial\circ\mu_r)(a_0\otimes b_0)=\mu_1(\partial'a_0\otimes\eta''b_0)\pm\mu_1(\eta'a_0\otimes\partial''b_0)=\mu^r(d^{\prime r}_m[a]\otimes[b])\pm\mu^r([a]\otimes d^{\prime\prime r}_n[b])\;.$$
This proves the Leibniz rule (2).
From the Leipniz rule and the facts that $\ker(d^r_p)=Z^{r+1}_p/B^r_p$ and $\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p$, we conclude that $\mu^r$ induces a product on $E^{r+1}_p\cong\ker(d^r_p)/\mathrm{im}(d^r_p)$, which proves (3). Because $\mu^r$ is induced by $\mu_1$, so is $\mu^{r+1}$, and we can continue the induction.<|endoftext|>
TITLE: Dimension of module
QUESTION [5 upvotes]: Does dimension of a module (say, dimension of its support) have anything to do with the supremum length of chains of prime submodules like rings?
Let's restrict to finitely generated modules over Noetherian ring.
Prime submodules are defined analogously to primary submodules: a submodule P in M is prime if P$\neq$M and $M/P$ has no zero divisors, i.e. $am\in P$ implies $m\in P$ or $a \in \mbox{Ann}(M/P)$.

REPLY [2 votes]: Let $R$ be an integral domain, then for the module $R^n$ its maximal length of chains of prime submodules is much larger than its dimension (for $n>>0$).<|endoftext|>
TITLE: Can Cantor set be the zero set of a continuous function?
QUESTION [39 upvotes]: More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?
Some days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.

REPLY [4 votes]: André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where
$\alpha$ is the supremum of all elements $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).<|endoftext|>
TITLE: Ultrafilters and automorphisms of the complex field
QUESTION [168 upvotes]: It is well-known that it is consistent with $ZF$ that the only automorphisms of the complex field $\mathbb{C}$ are the identity map and complex conjugation. For example, we have that $\vert\operatorname{Aut}(\mathbb{C})| = 2$ in $L(\mathbb{R})$. But suppose that we are given a nonprincipal ultrafilter $\mathcal{U}$ over the natural numbers $\mathbb{N}$. Is there any way to use $\mathcal{U}$ to define a third automorphism of $\mathbb{C}$?
Some background ... the "obvious" approach would be to note that the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ has lots of automorphisms arising as ultraproducts of Frobenius automorphisms. Of course, working in $ZFC$, this ultraproduct is isomorphic to $\mathbb{C}$ and hence we obtain many "strange " automorphisms of $\mathbb{C}$. However, the isomorphism makes heavy use of the Axiom of Choice and these fields are not isomorphic in $L(\mathbb{R})[\mathcal{U}]$. So a different approach is necessary if we are to find a third automorphism of $\mathbb{C}$ just in terms of $\mathcal{U}$ ...
Edit: Joel Hamkins has reminded me that I should mention that I always assume the existence of suitable large cardinals when I discuss properties of  $L(\mathbb{R})$ and $L(\mathbb{R})[\mathcal{U}]$. For  example, if $V = L$, then $L(\mathbb{R}) = L= V$ and so $L(\mathbb{R})$ is a model of $ZFC$. Of course, nobody would dream of studying $L(\mathbb{R})$ under the assumption that $V = L$ ...

REPLY [32 votes]: It seems not. 
It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), then after forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ (the infinite subsets of $\omega$, ordered by mod-finite containment) to produce a selective ultrafilter $U$, there is no selector (i.e., set meeting each equivalence class in exactly one point) for the equivalence relation $E_{0}$ (mod-finite equivalence on $\mathcal{P}(\omega)$) in the inner model $L(\mathbb{R})[U]$.
It also seems to follow from ZF + DC$_{\mathbb{R}}$ (which holds in $L(\mathbb{R})[U]$) that the existence of a discontinuous homomorphism from either of $(\mathbb{R}, +)$ or $(\mathbb{C}, +)$ to itself implies the existence of an $E_{0}$ selector, as we will show below. Since a discontinuous automorphism of $(\mathbb{C}, +, \times)$ restricts to one for $(\mathbb{C}, +)$, this answers the question. 
The proof is the same for each of $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$; moreover, the existence of each type of homomorphism implies the existence of the other. I haven't tried writing it up this way, but it seems that the argument can be carried out over an arbitrary complete additive metric group satisfying the triangle inequality. The existence of a discontinuous homomorphism of $(\mathbb{R}, +)$ easily gives one for
$(\mathbb{C}, +)$; we give a proof of the reverse direction at the end of this answer.  
So, let $h$ be a discontinuous homomorphism from $(\mathbb{R}, +)$ (or $(\mathbb{C}, +)$) to itself. 
As shown in the proof of Theorem 1 of a 1947 paper by Kestelman, for each positive real number $\delta$, $h$ is unbounded on $\{ x : |x| < \delta \}$. The same proof shows that the same fact holds for $(\mathbb{C}, +)$  (moreover, the fact follows easily from the definition of "discontinuous homomorphism").  Applying DC$_{\mathbb{R}}$, we may find $\{ x_{i} : i < \omega \}$ such that (1) each $|x_{i}|$ is more than $\sum \{ |x_{j}| : j > i\}$
and such that (2) for each $i$, $|h(x_{i})| - \sum \{ |h(x_{j})| : j < i \} > i.$
Let $X = \{ x_{i}: i < \omega \}$ and let $Y$ be the set of reals (or complex numbers) which are sums of (finite or infinite) subsets of $X$ (note that all the infinite sums converge).
By condition (1) on $X$, each $y \in Y$ is equal to $\sum \{ x_{i} : i \in S_{y}\}$ for a unique subset $S_{y}$ of $\omega$. 
Let $F$ be the equivalence relation on $Y$ where $y_{0} F y_{1}$  if and only if $S_{y_{0}}$ and $S_{y_{1}}$ have finite symmetric difference. 
By condition (2) on $X$, the $h$-preimage of each bounded subset of $\mathbb{R}$ ($\mathbb{C}$) intersects each $F$-equivalence class in only finitely many points (since if the bounded set is contained in an interval of length $i$, then for every $y$ in the intersection $S_{y} \setminus i$ is the same, which can be seen be consideration of the maximum point of disagreement between the sets $S_{y}$). It follows then that there is an $F$-selector : for each equivalence class, let $n \in \mathbb{Z}^{+}$ be minimal such that the $h$-preimage of $[-n, n]$ intersects the class, and then pick the least element of this intersection. Since $Y/F$ is isomorphic to $\mathcal{P}(\omega)/E_{0}$ via the map $y \mapsto S_{y}$, there is then an $E_{0}$-selector.
As for getting a discontinuous homomorphism of $(\mathbb{R}, +)$ from one on $(\mathbb{C}, +)$ : Suppose that $h$ is a homomorphism of $(\mathbb{C}, +)$. Define $f_{0},\ldots,f_{3}$ on $\mathbb{R}$ as follows:
(1) If $h(x) = a + bi$, then $f_{0}(x) = a$. 
(2) If $h(x) = a + bi$, then $f_{1}(x) = b$. 
(3) If $h(iy) = a + bi$, then $f_{2}(y) = a$.
(4) If $h(iy) = a + bi$, then $f_{3}(y) = b$.
Then each of $f_{0},\ldots,f_{3}$ is a homomorphism of $(\mathbb{R}, +)$. 
Since $h(x + iy) = h(x) + h(iy) = f_{0}(x) + if_{1}(x) + f_{2}(y) + if_{3}(y),$ if all of $f_{0},\ldots,f_{3}$ are continuous 
then $h$ is.<|endoftext|>
TITLE: Microwaving Cubes
QUESTION [44 upvotes]: First a little background. Microwaves do not heat uniformly. To help overcome this, your food is rotated, however this is not usually sufficient to produce totally uniform heating. Informally, this is the question: Is there a way of moving our food in order to heat it uniformly throughout?
Let $f : \mathbb{R}^n \to R$ be our heat function. Let $I^n = [-0.5,0.5] \times \cdots \times [-0.5,0.5]$ be the unit n-dimensional cube centered at the origin; it will be our food. Let $\gamma : [0,1] \to \mathbb{R}^n \times SO(n)$ be a map specifying a path along which to translate and rotate $I^n$. If $x \in I^n$ then let $h(x)$ denote the total heat absorbed by $x$ as it travels along $\gamma$. 
Note that if $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ then $h(x) = \int_0^1 f(\gamma_2(t)(x) + \gamma_1(t)) dt$.
We will call a curve $\gamma$ 'uniformly heated' iff $\forall x,y \in I^n$, $h(x) = h(y)$.
How sufficiently nice must our heat function $f$ be in order to guarantee that there exists a uniformly heated curve? Do these requirements change if we consider a different food to heat, for example, if we heat $I^m \times 0^{n-m}$ in $\mathbb{R}^n$?
Note that in $\mathbb{R}^1$, as $SO(1) = 1$, if $f$ is a strictly monotonic function then there cannot exist any uniformly heated curves as (assuming wlog $f$ is increasing) $h(-0.5) < h(0.5)$.

REPLY [23 votes]: You can uniformly cook the cube if $f$ is harmonic, i.e. $\Delta f=0$. Note that, if e.g. $\Delta f>0$ everywhere, then the center of the cube will always receive less heat than  the average over a sphere with the same center. Thus if $\Delta f$ happens to have constant sign, it must be zero.
To achieve uniform cooking in the harmonic case, let the center of the cube stay at the origin and rotate the cube using a Peano-like curve $\gamma:[0,1]\to SO(n)$ such that the push-forward measure $\gamma_*m$ (where $m$ is the Lebesgue measure on $[0,1]$) is the normalized Haar measure on $SO(n)$. Then the heat received by a point $x\in I^n$ is the average of $f$ over the sphere of radius $r=|x|$ centered at the origin. Since $f$ is harmonic, this average value equals $f(0)$.
To construct such a curve $\gamma$, follow the standard procedure for the Peano curve:  partition $SO(n)$ into reasonable sets (connected and with piecewise smooth boundaries) and visit all of them by a continuous path; this path is the first approximation. Then subdivide the partition and change the path so that it visits all sub-parts but do not make new intersections with boundaries of old parts. And so on. At each step, choose the parametrization so that the time spent in each piece equals its Haar volume. Let the diameters go to zero, then the paths will converge to a Peano-like curve $\gamma$. The push-forward measure $\gamma_*m$ coincides with the Haar measure on all elements of the partitions and hence on all Borel sets.
Remark.
A similar trick works if $f$ has compact support (more precisely, its support should be separated away from the microwave walls by distance at least 1). Just move around so as to realize a Haar measure on the relevant subset of the translation group rather than $SO(n)$.<|endoftext|>
TITLE: Is the absolutely continuous image of a nowhere dense set is also nowhere dense?
QUESTION [5 upvotes]: Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an  absolutely continuous  map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?
Remarks:

The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the  Cantor function . 
If $f$ is assumed to be $C^1$, then the answer is yes - a nice proof can be found at this page. Essentially the same proof works if it is assumed that $f$ is not differentiable at at most countably many points. 
 Edit:  I would like to retract the previous sentence. Now I don't see why it should be true.

REPLY [5 votes]: I think this is a  counter-example.
Let $C$ be a cantor set of positive measure, so $C$ is nowhere dense, perfect and is the countable decreasing intersection of sets $C_{n}$ each of which are a finite union of closed  disjoint intervals in $[0,1]$. 
Let $f(x)=\int_{0}^{x}\chi_{C}(t)\mbox{ }dt,$ so $f$ is certainly absolutely continuous. 
Assuming I did this right $f(C)$ contains $[0,m(c)).$ Consider each $C_{n},$ since $f$ is increasing and continuous $f([0,1])$ contains $[0,m(C)].$ Also $[0,1]\setminus C_{n}$ is a union of open intervals each of which are in the compliment of $C$ so it follows that $f$ is constant on each such interval. Since $f([0,1])=[0,m(C)]$ it follows that for any $x\in [0,m(C))$ we can find $t\in C_{n}$ so that $f(t)=x.$ Indeed we already know we can do this with $t\in [0,1]$ but since $f$ is constant on the intervals in the complement of $C_{n}$ we can force $t\in C_{n}$ (for instance if $x$ is in some interval $I$ in $[0,1]\setminus C_{n}$ then its left endpoint is in $C_{n}$ and since $f$ is constant on $I$ we have that $f$ has the same value at the left-endpoint of $I$ as on $I$.)
Now fix $y\in [0,m(C)).$ Since $\lbrace x\in C_{n}:f(x)=y\rbrace$ is non-empty and these sets are decreasing, (since the $C_{n}$ are decreasing) by compactness we can find $x\in C$ so that $f(x)=y.$ Thus $f(C)=[0,m(C))$ and we have found a nowhere dense set which is mapped to a set which is not nowhere dense.<|endoftext|>
TITLE: When do the Galois reps of modular forms have open image?
QUESTION [8 upvotes]: Suppose f is a newform (with coefficients generating some number field E), and $\rho_{f,\lambda}: {\rm Gal}(\overline{\mathbb{Q}} / \mathbb{Q}) \to {\rm GL}_2(E_\lambda)$ the associated Galois rep (for some prime $\lambda$ of E). When does $\rho$ have open image in ${\rm GL}_2(E_\lambda)$?
This clearly isn't the case if f has weight 1, or if f is of CM type; and I gather that it's a theorem of Serre that if f is attached to an elliptic curve, then not having CM is sufficient. What's known about this question in general?

REPLY [4 votes]: The answer of TSG is not correct in all cases. For a complete and detailed answer see this later question and the answer of David Loeffler (the very OP of the current question).<|endoftext|>
TITLE: Are there sigma-algebras of cardinality $\kappa>2^{\aleph_0}$ with countable cofinality?
QUESTION [8 upvotes]: A standard homework in measure theory textbooks asks the student to prove that there are not countably infinite $\sigma$-algebras. The only proof that I know is via a contradiction argument which yields no estimate on the minimum cardinality of an infinite $\sigma$-algebra.
Given an a set $X$ of infinite cardinality $\kappa$, the $\sigma$-algebra of all co-countable subsets of $X$ is of cardinality $2^\kappa$ $\kappa^{\aleph_0}$. This example doesn't tell me whether there are $\sigma$-algebras of cardinality below $2^{\aleph_0}$, if I don't assume the Continuum Hypothesis.
My question is as the title says: Are there $\sigma$-algebras of every uncountable cardinality? 
Edit: The combined answer with Stephen, Matthew proves that the cardinality of a $\sigma$-algebra is necessarily at least $2^{\aleph_0}$. Further, for each cardinality $\kappa\ge 2^{\aleph_0}$ with uncountable cofinality, the $\sigma$-algebra of countable (or cocountable) subsets of a set $X$ with cardinality $\kappa$, is of cardinality $\kappa$.
What is left is whether for $\kappa\ge 2^{\aleph_0}$ with $cf(\kappa)=\aleph_0$ are there $\sigma$-algebras of cardinality $\kappa$. (I changed the title to reflect this.)
Thanks Stephen, Matthew, Apollo, for the combined work!

REPLY [9 votes]: A Boolean algebra is $\sigma$-complete if every countable subset has a least upper bound
and a greatest lower bound.  Every $\sigma$-algebra is a $\sigma$-complete Boolean algebra.
Every (infinite) $\sigma$-complete Boolean algebra $B$ satisfies $|B|^{\aleph_0}=|B|$.
(I am almost certain that a proof of this fact is in the Handbook of Boolean Algebras, Volume 1.)
Since for every infinite cardinal $\kappa$, $\kappa<\kappa^{\text{cf}(\kappa)}$,
there is no infinite $\sigma$-algebra of some size $\kappa$ of cofinality $\aleph_0$.  
Something stronger is actually true:
By a result of Koppelberg [Boolean algebras as unions of chains of subalgebras, Algebra Universalis, Vol. 7 (1977), 195-203], no $\sigma$-complete Boolean algebra is the union of
a countable increasing chain of proper subalgebras.  This also implies that the
size of a $\sigma$-algebra cannot be of countable cofinality.<|endoftext|>
TITLE: Solvable transitive groups of prime degree
QUESTION [7 upvotes]: Is the following true ?

Every solvable transitive subgroup
  $G\subset\mathfrak{S}_p$ (the symmetric group on
  $p$ letters, where $p$ is a prime)
  contains a unique subgroup $C$ of
  order $p$ and is contained in the
  normaliser $N$ of $C$ in $\mathfrak{S}_p$.  The
  quotient $G/C$ is cyclic of order
  dividing $p-1$.  If $G$ is not cyclic,
  then it has exactly $p$ subgroups of
  index $p$.

I need such a result for an arithmetic application.  A reference or a short argument will be appreciated.
Addendum. For those interested in the arithmetic application, see http://arxiv.org/abs/1005.2016

REPLY [4 votes]: Happened to come across the following Satz in Huppert, Endliche Gruppen I, [S. 163].<|endoftext|>
TITLE: What are the prime ideals of k[[x,y]]?
QUESTION [16 upvotes]: Let $k$ be a field. Then $k[[x,y]]$ is a complete local noetherian regular domain of dimension $2$. What are the prime ideals?
I've browsed through the paper "Prime ideals in power series rings" (Jimmy T. Arnold), but it does not give a satisfactory answer. Perhaps there is none. Of course you might think it is more natural to consider only certain prime ideals (for example open/closed ones w.r.t. the adic topology), but I'm interested in the whole spectrum.
A first approximation is the subring $k[[x]] \otimes_k k[[y]]$. If we know its spectrum, perhaps we can compute the fibers of $\text{Spec } k[[x,y]] \to \text{Spec } k[[x]] \otimes_k k[[y]]$. Now the spectrum of the tensor product consists of $(x),(y),(x,y)$ and $\text{Spec } k((x))  \otimes_k k((y))$. The latter one is still very complicated, I think. For example we have the kernel of $k((x)) \otimes_k k((y)) \to k((x))$. Also, for every $p \in k[[x]]$, we have the prime ideal $(y - p)$.

REPLY [13 votes]: I really come long after the battle on this one, but I have something to add to Angelo's answer. To rephrase what Angelo said, there is only one prime ideal of height 2, namely the maximal ideal $(x,y)$, and one of height $0$. The problem is to classify the prime ideals of height one which (as in any UFD) are exactly the principal ideals generated by an irreducible element $f(x,y)$.
Now there are several possible meaning for "classify", but one reasonable sense is
to describe all such prime ideal up to an automorphism of $k[[x,y]]$ (automorphism of $k$-algebra, I mean, of course.) 

Theorem: if $k=\mathbb C$, then up to a an automorphism of $k[[x,y]]$ any prime ideal of height one is of the form $(f(x,y))$ where $f(x,y)$ is an irreducible polynomial in two variables.

A proof for this theorem (that is, a combination of two references that do the job) can be found in my answer in the following question. I think that the theorem is true for any field $k$, but one of the references used there use $k=\mathbb C$.
EDIT: I give a little bit more explanation as requested. Let start with $f(x,y)$ an irreducible element. One has $f(0,0)=0$ otherwise $f$ would be a unit. Then a theorem
of Artin (references in the question linked above) tells that up to a change of variable
$x \rightarrow x'$, $y \rightarrow y'$, one can assume that $f(x,y)=g(x',y')$ will be analytical, that is the power series $g$ converges on some neighborhood of $(0,0)$. 
In algebraic terms, the change of variables defines an automorphism $\psi$ if $k[[x,y]]$ sending $x$ to $x'$ and $y$ to $y'$, and $g = f \circ \psi^{-1}$. This theorem is actually much more general: it works in any dimension and over any field, with analytic replaced then by "in the Henselization of the subring of polynomials". In the case of dimension $2$ over $\mathbb C$, i would not be surprised if it was much older (Puiseux? Newton?). 
The second step is proving that analytic $g(x,y)$ can be changed, up to a second change of variable to an $h(x,y)$ which is simply a polynomial. The reference given in the linked question is to a book of complex geometry, hence is over $\mathbb C$, but the result 
may be true over any field. However it is not true in three and more variable, as explained in Moret-Bailly's answer to this question. 
So we have an automorphism $\psi$ such that $h = f \circ \psi^{-1}$. Since $(f)$ is prime, so is $h$, hence $h$ is irreducible as a power series, hence as a polynomial.
To make this more concrete, let me gives two examples. if $f(x,y) =a x + b y$ + higher terms, with $a$, or $b$, or both, non zeros (This condition is equivalent to $A/(f)$ being regular, or a discrete valuation ring.) Say $a \neq 0$, then one can take for new variables $x'=f(x,y)$, $y'=y$, and $h(x,y)$ is just $x$.
If on the contrary $a=b=0$, looking at the quadratic terms of $f$ gives a quadratic forms over $k$. If this quadratic form $h$ is non-degenerate, then by the theory of Morse function $f$ can be transformed to $h$ (I guess $\operatorname{char} k \neq 2$ may be necessary).<|endoftext|>
TITLE: reversible Turing machines
QUESTION [12 upvotes]: Hello,
Let T be a Turing machine such that
1) it operates on the alphabet {0,1}, 
2) its set of states is A
3) the language it accepts is $L$ . 
Does there exists a Turing machine S which also operates on the alphabet {0,1} and such that the language it accepts is L (the set of states might be different though) and such that, crucially, S is reversible? 
By reversible I mean "the computational paths of S are disjoint". More precisely, the transition table of S gives rise to a map $K_S: \text{Tapes}\times B \to \text{Tapes} \times B$, where Tapes is the subset of the infinite product $\{0,1\}^Z$ consisting of those sequences which have a finite number of 1's, and B is the set of states of S. S is reversible iff, by definition, $K_S$ is injective on the set
$$
 \bigcup_{i=0}^\infty K_S^{i}(\text{Tapes}\times \{Initial \} ),
$$
where $Initial\in B$ is the initial state of S.
If the answer to the above question is "no" then what if we allow S to operate on an alphabet which is larger then {0,1}?

REPLY [5 votes]: Well there are universal reversible Turing machines (RTMs), for example [1].
So you can simulate T with a universal RTM.
The resulting machine would be reversible and accept L.

[1] Kenichi Morita and Yoshikazu Yamaguchi, "A Universal Reversible Turing Machine" MCU2007<|endoftext|>
TITLE: What is so "spectral" about spectra?
QUESTION [25 upvotes]: What is the background of the terminology of spectra in homotopy theory? In what extend does the name "spectrum" fit to the definition and the properties? Also, are there relations to other spectra in mathematics (algebraic geometry, operator  theory)?
PS: The title is an allusion to this question ;-)

REPLY [6 votes]: It seems reasonable to me that in operator theory the term "spectrum" comes from the Latin verb spectare (paradigm: specto, -as, -avi, -atum, -are), which means "to observe". After all in quantum mechanics the spectrum of an observable, i.e. the eigenvalues of a self adjoint operator, is what you can actually see (measure) experimentally.
Edit: after having a look to an online etymological dictionary, it seems the relevant Latin verb is another: spècere (or interchangeably spicere)= "to see", from which comes the root spec- of the latin word spectrum= "something that appears, that manifests itself, vision". Furthermore, spec- = "to see", -trum = "instrument" (like in spec-trum). Also the term "spectrum" in astronomy and optics has the same origin.
In algebraic geometry, I believe the term "spectrum", and the corresponding concept, has been introduced after the development of quantum mechanics became well known. In this context, the concept of spectrum as a space made of ideals is perfectly analogous of that in operator theory (think of Gelfand-Naimark theory, and that the Gelfand spectrum of the abelian C-star algebra generated by one operator is nothing but the spectrum of that operator).
I wouldn't be surprised if the term "spectral sequence" had something to do with "inspecting" [b.t.w. also "to inspect" comes from in + spècere...] step by step the deep properties of some cohomological constructions.
Maybe the term "spectrum" in homotopy theory and generalized (co)homology -but I don't know almost anything about these- has to do with "probing", "testing", a space via maps from (or to?) certain standard spaces such as the Eilenberg-MacLane spaces or the spheres. Does it sound reasonable?
Edit: The following paragraph from the wikipedia article on "primon gas" seems to support my guess:

"The connections between number theory and quantum field theory can be somewhat further extended into connections between topological field theory and K-theory, where, corresponding to the example above, the spectrum of a ring takes the role of the spectrum of energy eigenvalues, the prime ideals take the role of the prime numbers, the group representations take the role of integers, group characters taking the place the Dirichlet characters, and so on"<|endoftext|>
TITLE: What does "kernel" mean in integral kernel?
QUESTION [11 upvotes]: In functional analysis, there is the term "integral kernel". Examples are Possion kernel, Dirichlet kernel etc. 
In algebra, the term kernel of a homomorphism refers to the inverse image of the zero element.
Are these two terms related? If not, where did the word "kernel" in the term "integral kernel" come from?

REPLY [3 votes]: An integral kernel is, of course, an integrable generalization $K(x,y)$ of a matrix $M_{j,k}$.  You could very loosely call this a "kernel" in the sense of the "core" of the formula for a integral linear operator.  For comparison, Wiktionary tells me that in German, a Kerngehäuse is an apple core, while Kernphysik is nuclear physics.
But mainly, I think that the two uses of kernel, one for the null space and one for an integration matrix, is just a terrible collision of terminology that became standard by accident.  It's nice when a mathematical term is an inspired metaphor or neologism.  For instance the word "spectrum" for the set of eigenvalues of an operator was not just inspired, but also prescient and profound.  (As I understand it, the term was chosen by mathematicians, by analogy with spectral lines in chemistry, shortly before the development of quantum mechanics.)  But sometimes we're just unlucky, or maybe collectively stupid.
As Jan Kolar points out, the kernel of an operating system is a third metaphorical use of the word that makes vastly more sense.<|endoftext|>
TITLE: "Natural" generating sets for symmetric groups
QUESTION [15 upvotes]: The symmetric group on $n$ letters has 
many sets of generators. Some of them are more natural than others, eg the
set $(i,i+1)$ of adjacent transpositions (natural with respect to the type A Weyl group), the set of all shuffles (permutations corresponding to "card-shuffles", ie $\sigma(1),\sigma(2),\dots,$ contains at most two increasing subsequences) perhaps also sets consisting of conjugacy
classes (preferably of signature $-1$ in order to avoid a stupid mistake).
Which other sets of generators of symmetric groups occur in a natural way?

REPLY [2 votes]: Another answer from a combinatorial point of view: the set of all transpositions is a generating set for $S_n$. Obviously this is a huge overkill. But a nice thing about including all transpositions is now your generating set is closed under conjugation. In fact, this generating set is used to define the so-called absolute order or reflection order (which applies more generally to any Coxeter group); the absolute order has as its Hasse diagram the Cayley graph with respect to the transpositions as generating set. Absolute order comes up for instance in Coxeter-Catalan combinatorics, a currently popular topic.<|endoftext|>
TITLE: Historically first uses of mathematical induction
QUESTION [10 upvotes]: I'm interested in find out what were some of the first uses of mathematical induction in the literature. 
I am aware that in order to define addition and multiplication axiomatically, mathematical induction in required. However, I am certain that the ancients did their arithmetic happily without a tad of concern about induction.
When did induction get mentioned explicitly in the mathematical literature? Definitely this places before about 1800 when the early logicians started formulating axioms for arithmetic.

REPLY [19 votes]: There are several questions here, so my answer overlaps with some of the
others.

First use of induction in some form. I would nominate the "infinite
descent" proof that $\sqrt{2}$ is irrational -- suppose that $\sqrt{2}=m/n$, 
then show that $\sqrt{2}=m'/n'$ for smaller numbers $m',n'$
-- which probably goes back to around 500 BC.
First published use of induction in some form. Euclid's infinite descent
proof that every natural number has a prime divisor, in the Elements.
First use of induction in the "base step, induction step" form. I
suggest Levi ben Gershon and (more definitely) Pascal, as mentioned in
danseetea's answer.
First mention of "induction". The one suggested by Gerald Edgar is
the earliest I know of.
First realization that induction is fundamental to arithmetic:
Grassmann's Lehrbuch der Arithmetik of 1861, where he defines addition
and multiplication by induction, and proves their ring properties by 
induction. This idea was rediscovered, and built into an axiom system by 
Dedekind, in his Was sind und was sollen die Zahlen? of 1888. It
became better known as the Peano axiom system, when Peano redeveloped
it a couple of years later.<|endoftext|>
TITLE: Potential modularity and the Ramanujan conjecture
QUESTION [26 upvotes]: A little background: Let $f(z)=\sum_{n=1}^{\infty} a(n) e^{2\pi i nz}$ be a classical holomorphic cuspidal eigenform on $\Gamma_1(N)$, of weight $k \geq 2$ normalized with $a(1)=1$.  The Ramanujan conjecture is the assertion that $|a(n)|  \leq n^{\frac{k-1}{2}}d(n)$.  This is a theorem, due to several people, but the main steps in its proof are the following two:

Show that $f$ can be "realized" in the etale cohomology of a suitable variety.  In fact $f$ can be found in $H^{k-1}$ of a $k-2$-fold self-product of the universal elliptic curve over $X_1(N)$.  This step is due to Eichler, Igusa, Kuga/Sato, and Deligne (who showed how to desingularize the variety).  This reduces the Ramanujan conjecture to the Weil conjectures.
Prove the Weil conjectures.  This is of course due originally to Deligne, although if I understand correctly, there are now several other proofs (a p-adic proof by Kedlaya, etc.)

Now, there is another approach to the Ramanujan conjecture, essentially through Langlands functoriality.  In particular, if we knew for all $n$ that the L-functions $L(s,\mathrm{sym}^n f)$ were holomorphic and nonvanishing in the halfplane $Re(s)\geq 1$, the Ramanujan conjecture would follow.  This observation is, I believe, due to Serre Langlands.
Nowadays these analytic properties are known, by the potential modularity and modularity lifting results of Barnet-Lamb/Clozel/Gee/Geraghty/Harris/Shephard-Barron/Taylor.  However, the proofs of potential modularity and modularity lifting seem to utilize the Ramanujan conjecture.  Hence my question:

Can the recent proofs of potential modularity for symmetric powers of $GL2$ modular forms be modified so they do not assume the Ramanujan conjecture, hence giving a new proof of the Ramanujan conjecture?

(If I got any of the history or attributions wrong, please correct me!)

REPLY [4 votes]: Leaving aside the original question, it seems that this new paper https://arxiv.org/abs/1812.09999 might be relevant. They claim to prove the Ramanujan conjecture for a class of automorphic forms as a consequence of automorphy results for symmetric powers and not (directly) by using Deligne's theorem, since the corresponding Galois representations in this case apparently don't come in any obvious way from geometry.<|endoftext|>
TITLE: Subgroups of direct product of groups 
QUESTION [7 upvotes]: I am interested in the following question on products of finite groups. Let $\Gamma$ be a subgroup of $U_1\times U_2$ such that the compositions with the canonical projections $\Gamma \subset U_1\times U_2 \rightarrow U_1$ and $\Gamma \subset U_1\times U_2 \rightarrow U_2$ are both surjective.
Does it follow that there is a group $G$ such that $\Gamma$ is isomorphic to the fiber product $U_1 \times_G U_2$? This means that there are surjections $\pi_1:U_1\rightarrow G$ and $\pi_2:U_2\rightarrow G$ such that $\Gamma$ is the set of pairs $(u_1,u_2)$ with $\pi_1(u_1)=\pi_2(u_2)$.
Goursat's Lemma mentioned in this question proves the statement in the case $\Gamma$ is a normal subgroup of $U_1\times U_2$.
If  the statement is not true without the normality assumption, then what would be a general characterization of these subgroups $\Gamma$?

REPLY [11 votes]: Goursat's Lemma provides a complete characterization of subgroups of a direct product of two groups as fiber products.  In the language I am used to: subgroups correspond to the graphs of isomorphisms between isomorphic sections of the two factors.  Some subgroup embedding properties can be read from the embedding of the sections in the factors (for instance being normal in the factors, or being central in the factors), but there are no embedding properties required to use the lemma.
Goursat's lemma appears in Roland Schmidt's Lattice of Subgroups book in chapter 1.6 (google books), and as an exercise in several textbooks.<|endoftext|>
TITLE: Is the semigroup of nonnegative integer matrices with determinant 1 finitely generated?
QUESTION [19 upvotes]: The group of $n\times n$ matrices with integer entries and determinant equal to 1, $SL(n,Z)$, is a finitely generated group (in fact, it is generated by 2 matrices). I am interested to know if the semigroup of the matrices in $SL(n,Z)$ where all the entries are nonnegative is also finitely generated. This is true at least in dimension $n=2$.

REPLY [16 votes]: The question has been amply answered, but perhaps it's worth explaining the geometry of the situation. The positive monoid acts  by projective transformations on the $n-1$-simplex (the view of the positve orthant as seen from the origin, and the partially order on the monoid is the order by inclusion of the image under this action.  In $R^n$, the image of the simplex spanned by the origin together with the basis vectors cannot contain any interior lattice points.  This implies that in the projective action, no proper image can contain its barycenter. This, with its ramifications, gives a strong limitation on images of the entire simplex under the positive monoid.
On the other hand, it's not hard to prove that for $n \ge 3$, the set of images of any face are dense.  In fact, using elementary and well-known theory, there is a positive density of $(n-1)$-tuples of lattice points that extend to a basis for $Z^n$, and any such $n$-tuple of positive elements can be extended to a positive basis.
This implies that there is no finite set of generators, because a single image of the $n-1$-simplex can contain at most simplices near the barycenter over a small range of angles. Here's a picture for the 2-dimensional case:
alt text http://dl.dropbox.com/u/5390048/MatrixSemigroup.jpg
The line segment  is the projection of the pair of lattice points 
$L_1=(285, 684, 112)$ and $L_2 = (764, 318, 949)$ found, by pseudorandom selection from lattice cubes that projected near the targeted segment, subject to the condition that they extend to a free basis.
A third element of a basis can be transformed to be project arbitrarily close to any desired point on the line through the image of $L_1$ and $L_2$ by adding $N*L1 + M*L2$, for suitable $N$ and $M$.
Moreover, this characterize the closure (with respect to the Hausdorff topology) of the set of images of the triangle under the positive monoid:  it consists of all triangles spanned by positive free bases, together with set of all line segments contained in the triangle.<|endoftext|>
TITLE: What are examples of mathematical concepts named after the wrong people? (Stigler's law)
QUESTION [34 upvotes]: It's a common observation in Lie theory that Cartan matrices and the Killing form are named after the wrong people; they were discovered by Killing and Cartan, respectively.  I remember learning about many other examples of this phenomenon, but can't think of too many at the moment.  Wikipedia has some examples here and here, but I'm curious about more obscure examples.  
Bonus points for an interesting story behind why the concept was incorrectly named.  Concepts that were deliberately named in honor of another mathematician don't count.

REPLY [3 votes]: A favorite of mine is l'Hôpital's rule. l'Hôpital paid Johann Bernoulli a retainer to keep him updated on developments in calculus and to solve problems he had. Correspondence shows that Bernoulli stated and proved the rule, which l'Hôpital then published.
Heine-Borel was first published by Borel, not Heine. In fact, Heine's name was attached because he was using similar methods to solve related problems. Too bad for both of them that it was actually Dirichlet who was the first recorded to have proved it, but his notes were published posthumously and after Borel's proof. 
Cramer's Rule was published first by MacLaurin, and some believe MacLaurin knew the proof some 20 years before Cramer's publication.<|endoftext|>
TITLE: Generalizing the notion of Farey neighbors to the algebraic numbers
QUESTION [36 upvotes]: "The Beauty of Roots" is about plots of roots of polynomials—specifically, those with degree less than a given number and height less than another given number. As you can see, these plots are really pretty:
(source)
Looking on the inside and the outside of that glowing ring, you can see some neat fractal patterns. But today, I'm not interested in those; I'm interested in those holes.
If you look at the roots of unity (and some other places, presumably the roots of other simple polynomials), you'll see that near each one, the algebraic numbers are especially sparse. As the page describes, for any algebraic number that's particularly "simple", its surroundings are relatively vacant of other algebraic numbers, and the "simpler" the algebraic number is, the more its fellow algebraic numbers tend to keep away. (I imagine that every algebraic number has such a "circle of emptiness" surrounding it, but for all but the simplest ones, this circle is tiny.)
I know of one other set that has this property, and that's the set of rational numbers. It's a theorem that given two fully reduced rational numbers $a/$b and $c/d$, the closest they can possibly be to each other is $1/bd$; thus, if a rational number is "simple" in the sense of having a small denominator, other rational numbers will tend to be far away from it.
Rational numbers $a/b$ and $c/d$ that differ by exactly $1/bd$ are called Farey neighbors; if two rational numbers are Farey neighbors, then they have exactly one Farey neighbor in common that lies between them, $(a+c)/(b+d)$. For more information, see Farey sequences on Wikipedia.
So, algebraic numbers that are "simple" are never close to each other. The rational numbers exhibit the same phenomenon; here, "simple" refers to the denominator, and you can determine exactly what the minimum distance is (1 over the product of the denominators). Is it possible to extend the notion of denominators and Farey neighbors to the algebraic numbers in general, thereby explaining the "holes" in the picture?

REPLY [3 votes]: Inversion with respect to a circle of radius $\sqrt 2$ centered at $i$ exchanges the unit circle and the real numbers (union infinity). One can apply this inversion to the set of roots of
a palindromic polynomial (palindromic: $\xi$ and $1/\xi$ are simultaneous roots) getting the
roots of another palindromic polynomial. This transformation yields an involution on the set of palindromic polynomials with rational coefficients (and roots in $\mathbb C^*\setminus\lbrace i,-i\rbrace$) which exchanges the role of the unit circle and the real numbers. This is thus a sort of geometrical explanation of the phenomenon.<|endoftext|>
TITLE: Composing left and right derived functors
QUESTION [17 upvotes]: I would appreciate either an explanation or a reference for what is going on here.
Motivation:
Let $f : X \rightarrow Y$ be a morphism of algebraic varieties. The derived projection formula implies that for a sheaf $\mathcal{F}$ on $Y$, we have
$$Rf_\ast Lf^\ast \mathcal{F} \cong Rf_\ast \mathcal{O}_X \otimes^L \mathcal{F}.$$
Suppose for example that $R^i f_\ast \mathcal{O}_X = 0$ when $i>0$, and is $\mathcal{O}_Y$ when $i=0$. It's natural to guess that the cohomology of the right hand side is just $\mathcal{F}$ in degree 0 and 0 otherwise.
Question:
Is this true? More generally, is there a spectral sequence calculating the cohomology of the composite of a right derived- with a left derived functor?
Is there an exact sequence of terms of low degree, as there is for the composite of two right derived functors?

REPLY [28 votes]: The answer is yes. Now, concerning the general question - it is not important here that one functor is right derived and the other is left derived. You just have triangulated functors between triangulated categories. Also you have t-structures on all categories which give you filtrations on all objects. And the spectral sequence controls their interaction.
To be more precise, let $F:T_1 \to T_2$ and $G:T_2 \to T_3$ be two triangulated functors. You want to apply $G\circ F$ to a pure (with only one nontrivial cohomology) object $X \in T_1$ and compute the cohomology of $G(F(X))$. First, let us denote by $Y_i$ the $i$-th cohomology of $F(X)$. In other words, there is a filtration on $F(X)$ with factors $Y_i$. Now we apply the functor $G$. We obtain an object $G(F(X))$ with a filtration with factors $G(Y_i)$. Now we want to compute the cohomology of this object. These are computed by the spectral sequence. So, you see, there is nothing special about the functors, the question is about filtered objects in triangulated categories.<|endoftext|>
TITLE: Is there an uncountable, non-discrete, Hausdorff Toronto space?
QUESTION [11 upvotes]: We call a topological space $X$ a Toronto space if for any subspace $Y \subseteq X$ such that $Y$ and $X$ have the same cardinality it follows that $Y$ is homeomorphic to $X$. 
Does anybody know what is known about the following question?:
Is there an uncountable, non-discrete, Hausdorff Toronto space?
It is not hard to show that if $X$ is countable, Hausdorff and Toronto then $X$ has the discrete topology. I have been thinking about the uncountable case for a while and it turns out it is a much harder question.

REPLY [8 votes]: You can read "The Toronto problem" by William Rea Brian (February 2014) to learn pretty much everything that is known about this problem. The article includes proofs of the folklore facts mentioned by Apollo and some other very interesting facts (e.g. Kunen´s result: An uncountable Hausdorff Toronto space contains no non-trivial convergent sequences).
But to sum up, the problem is now as open as it was 24 years ago: Under $GCH$ the only Hausdorff Toronto spaces (of any cardinality) are the discrete spaces. Also under $PFA$ there are no $T_3$ non-discrete Toronto spaces of size $\aleph_1$. We don´t know if the existence of a Hausdorff non-discrete Toronto space is consistent.<|endoftext|>
TITLE: Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope?
QUESTION [5 upvotes]: In this link, Corollary 3.2.2, page 59 the author claims that: The Euler characteristic of the toric variety $X_K$ associated to a convex polytope $K$ is the number of vertices of $K$.
I want to see how it works.  Could someone please illustrate this for me by using this method to compute the Euler characteristic of $\mathbb{P}^{2}$ and $\mathbb{P}^{1}\times \mathbb{P}^{1}$?  Thanks so much.

REPLY [16 votes]: Merely observe that a toric variety is the union of torus orbits $(\mathbb C^\*)^r$ for various dimensions $r$, and that the Euler characteristic of $(\mathbb C^\*)^r$ is zero if $r>0$ and $1$ if $r=0$. 
Vertices of a polytope correspond to 0-dimensional orbits, $r$-dimensional faces -- to $r$-dimensional orbits.
$\mathbb P^2$ corresponds to a triangle, $\mathbb P^1\times\mathbb P^1$ to a square. It is not very hard to count their vertices.<|endoftext|>
TITLE: Representations of Pin vs. Representations of Clifford
QUESTION [11 upvotes]: This may be total nonsense. But I need to know the answer quickly and I am too tired to think about it thoroughly. Let $k$ be a positive integer. Roe's "Elliptic Operators" claims that there is a 1-to-1 correspondence between:

representations of the Clifford algebra $\mathrm{Cl}\mathbb R^k$ of the vector space $\mathbb R^k$ with the standard inner product;
representations of the Pin group of this vector space (i. e., of the subgroup of the multiplicative group of $\mathrm{Cl}\mathbb R^k$ generated by vectors from $\mathbb R^k$) on which the element $-1$ of the Pin group acts as $-\mathrm{id}$;
representations of the subgroup $\left\lbrace \pm e_1^{i_1}e_2^{i_2}...e_k^{i_k} \mid 0\leq i_1,i_2,...,i_k\leq 1 \right\rbrace$ of the Pin group (where $\left(e_1,e_2,...,e_n\right)$ is the standard orthogonal basis of $\mathbb R^k$) on which the group element $-1$ acts as $-\mathrm{id}$.

I do see how representations restrict from the above to the below, and also how there is a 1-to-1 correspondence between the first and the third kind of representations. But is it really that obvious that there are no "strange" representations of the second kind? I mean, why is a representation of the Pin group uniquely given by how it behaves on $-1$, $e_1$, $e_2$, ..., $e_k$ ?
Any help welcome, I'd already be glad to know whether it's really that obvious or not.

EDIT: This seems to have caused some confusion. Here is the core of the question:
Assume that we have a representation $\rho$ of the Pin group $\mathrm{Pin}\mathbb R^k$ such that $\rho\left(-1\right)=-\mathrm{id}$. This, in particular, means an action of each unit vector. By linearity, we can extend this to an action of every vector. Is this always a representation (i. e., does the sum of two vectors always act as the sum of their respective actions)?

REPLY [12 votes]: Thanks to everyone who posted here.  It is not "obvious" to me what I was thinking of here, and I'm embarrassed that this argument has stood unchanged in the book since the first edition in 1988 or so.  I appreciate your pointing the issue out.  There aren't any plans currently for a new edition of the book - I just dragged the old TeX files out and found that I can no longer compile them - but if one does come to be I will ensure that appropriate corrections are incorporated.<|endoftext|>
TITLE: Can an arbitrary collection of circles of total area 1/2 fit into a circle of area 1?
QUESTION [53 upvotes]: Assume the circles are actually open disks, otherwise two circles each of area $\frac{1}{4}$ wouldn't fit into the circle of area 1. 
This seems like it should be true, thinking about packing density, but I've not been able to find an algorithm that works in all cases.

REPLY [4 votes]: After weeks of sweating through computations without success, I think I have an auxillary result which I can use to tackle the problem posted above.  However, the path to the conclusion is so surprising that I invite verification, in particular to point out any show-stopper mistake that may or may not be lurking.
The auxillary result involves a region that is the "suitcase" circle with a segment removed.  Let C be a disk of radius 1, and draw a chord of C which has distance 1/2 from the center of C.  Let D be that part of C (including the chord itself) that is on the same side of the chord as the center of C, so D is a disk with a segment chopped off so that the area of D is less than 5/6 the area of C.  Let F and G be two closed disks with unequal radii whose areas sum to half the area of C.
Result:  F and G pack inside D.
Note that if F and G have the same radius, they will not pack inside D for the same reason they won't pack inside C: they have to share a point which is the circle center.
The proof of the result involves placing F and G so that they touch the chord and otherwise are as far apart inside D as possible.  The analogous problem for packing inside C involves placing F and G inside C so that the centers of F, G, and C are on the same line and F and G are as far apart as possible, then use the area sum condition to show the radii of F and of G sum to less than 1, and so they can be pushed together just enough to remain disjoint and keep away from the region boundary.
For the result about D, I orient the coordinates to have the chord parallel to the x axis, and then I show that the x coordinates of the centers of F and G are greater than distance 1 apart.  It took me weeks to realize that was what was needed to be shown.  Even so, the algebra involved led to a surprise, which I want someone else to confirm or deny.
Now for the algebraic verification.  Let $0 \leq s \lt 1/2 \lt r \leq \sqrt{1/2}$ where $s$ and $r$ are the lengths of the radii of F and G.  The area sum condition yields $r^2 + s^2 = 1/2$, and the inequality to be shown is inequality (A): $\sqrt{3/4 - r} + \sqrt{3/4 - s} \gt 1$.  Since this inequality on the x coordinates of the centers of F and G imply that F and G are disjoint, inequality (A) will yield the result.
There may be a slick analytic way to show (A), but I use repeated squaring and rearrangments to yield a quartic polynomial in $r$.  I start by writing $s$ as $\sqrt{1/2 - r^2}$ and isolate the radical with $s$ on one side, and square both sides. I end up with an inequality to prove that has $\sqrt{1/2 - r^2}$ as a subexpression, and so I isolate the summand containing that subexpression on one side and square again.
I collect and rearrange and square again to get the following inequality to be shown:  $49/4 - 42r + 50r^2 - 24r^3 + 4r^4 \gt (1 - 2r + r^2)(12 - 16r)$.  I then collect terms and decide to factor out a term of $(r - 1/2)$ in hopes of getting an easier polynomial to handle to show that it is positive for $r$ in the range given above.
I know the relations above yield 0 when $r$ is $1/2$, so it seems like a nice simplification.
The surprise comes when I find out the above inequality is equivalent to (B): $4(r - 1/2)^4 \gt 0$.  I was not expecting that at all!
If I have it right, then the result is proved, and I can move on after weeks of fruitless labor involving delta and epsilons and sketches that had no $r$'s or $s$'s.  Can someone rederive or otherwise prove inequality A above?  Even better, can someone tell me how I could know inequality B was coming?
Gerhard "Surprised After All These Weeks" Paseman, 2012.05.07<|endoftext|>
TITLE: Sum of subset of geometric series: a^2^n
QUESTION [8 upvotes]: The formula for 1 + a + a^2 + .... where 0 < a < 1 is $\frac{1}{1-a}$, but what if you wanted to sum only those where the exponent is a power of 2? That is,
$S = a + a^2 + a^4 + a^8 + \cdots$ 
I feel like this is an easy one but I just can't seem to find a closed expression for it, nor search for it on Google.

REPLY [16 votes]: Mahler proved in the 1930s that the values of $f(z)=\sum_{n=0}^\infty z^{d^n}$, $d>1$ is an integer, are transcendental for any algebraic $z$ satisfying $0<|z|<1$. A related problem of transcendence of the function $f(z)$ was discussed in this question. This motivates nonexistence of simple formula like $1/(1-z)$ for $f(z)$.<|endoftext|>
TITLE: The Poincare Bundle(s) on C \times J
QUESTION [5 upvotes]: The set up is $C$ is a curve and $J$ is its Jacobian.  On the $C \times J$ there is the Poincare bundle $P$ which is the universal family of degree zero line bundles on $C$.  For every integer $d$ there is also a line bundle $P(d)$ on $C \times J$ which is a family of line bundles of degree $d$ on $C$.
I've seen a construction of $P$ and given $P$ an example of a $P(d)$ would be $P^L:= q_1^*L\otimes P$ where $q_1 \colon C \times J \to C$ is the projection and $L$ is a line bundle of degree $d$ on $C$.  If $L,L'$ both have degree $d$ you can form either $P^L$ or $P^{L'}$.  I've seen $P(d)$ defined as the inverse limit of $P^L$ as $L$ ranges over all degree $d$ line bundles.  I don't really know how to think of such an inverse limit.  Is there a more concrete way to describe $P(d)$?  It seems up to isomorphism, $P(d) = P^L$, but this is "very" non canonical which seems bad.  For example does $P^L$ have a universal property like $P$ does?
I have a more specific question related to this.  This question is coming from Prop. 21.6 of Polishchuk's book on Abelian varieties and the Fourier Mukai tranform, if anyone is curious.  If $q_2 \colon C \times J \to J$ is the projection, then apparently $q_{2*}P(g-1) = 0$.  Why is this so?
One argument for this (that I don't understand) is the following.
1) Embed $P(g-1) \to F$ with $F$ flat over $J$ and $R^1q_{2*}F = 0$.
2) From $0 \to P(g-1) \to F \to F/P=: E \to 0$ and cohomology we get
$0 \to q_{2*}P(g-1) \to q_{2*}F \to q_{2*}F \to R^1q_{2*}P(g-1) \to 0$
3)The restriction of $R^1q_{2*}P(g-1)$ to $a \in J$ is $H^1(C \times a, P(g-1)|_{C \times a})$ which is $H^1$ of a line bundle of degree $g-1$.  This is zero outside the theta divisor and Riemann-Roch says $h^1(L) = h^0(L)$ for degree $g-1$ line bundles, so  
$q_{2*}F \to q_{2*}F$
is an isomorphism outside a divisor.  And hence
4) $q_{2*}F \to q_{2*}F$ is injective.
Part 4) is the part I don't see.
I don't necessarily want to understand this line of reasoning.  But I don't see why $q_{2*}P(g-1) = 0$ since it seems to be supported on the theta divisor.

REPLY [4 votes]: For your first question, if you want a proper universal property it is defined
by a variety $J^{(d)}$ and a line bundle $L^{(d)}$ on $C\times J^{(d)}$ which is
of degree in the $C$-direction, i.e., of degree $d$ on each fibre $C\times
x$. The universality then says that for every $X$ and every line bundle $M$ on
$C\times X$ of degree $d$ in the $C$-direction there is a unique morphism $f\colon
X\rightarrow J^{(d)}$ such that $(\mathrm{id}\times f)^(L^{(d)})$ and $M$ differ
by a line bundle from $X$. Picking any line bundle of degree $d$ on $C$ (which
may not exist if the base field is not algebraically closed) allows you to
construct such a line bundle on $J$ but different choices will give different
line bundles on $J$ and hence an automorphism of $J$. This automorphism is then
explicitly given by a translation.
As for your second question $q_{2\ast}P(g-1)$ is a torsion free sheaf on $J$ whose
restriction to the complement of the theta divisor is $0$ and must therefore be
$0$. What happens is that the base change formula makes $H^0(C\times
a,P(g-1)_{|C\times a})$ come from $R^1q_{2\ast}P(g-1)$. An algebraic model is the
exact sequence $0\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$
where $k[t]\rightarrow k[t]$ is multiplication by $t$ (you should think of the
first $k[t]$ as $q_{2\ast}F$, the second as $q_{2\ast}E$ and $k$ as
$R^1q_{2\ast}P(g-1)$). When you tensor this with $k=k[t]/(t)$ the map
$k[t]\rightarrow k[t]$ grows a kernel, i.e., we have an exact sequence
$0\rightarrow k\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$. This
new kernel is $\mathrm{Tor}^1(k,k)$ which models $H^0(C\times
a,P(g-1)_{|C\times a})=\mathrm{Tor}^1(k(a),R^1q_{2\ast}P(g-1))$ (which comes from
the base change formula).<|endoftext|>
TITLE: Injective dimension of cyclic modules
QUESTION [6 upvotes]: Let $R$ be a non-Noetherian ring. Is its left global dimension ${\rm{lD}}(R)$ equal to $\sup \{ {\rm{id}}(M) \mid M \text{ is a cyclic $R$-module} \}$? Here $\rm{{id}}(M)$ denotes the injective dimension of $M$.

REPLY [3 votes]: The answer is no (in general) according to B. L. Osofsky, Global dimension of valuation rings, Corollary 3. For any $1 \leq n \leq \infty$, there are examples where $$\sup \{ {\rm{id}}(M) \mid M \text{ is a cyclic $R$-module} \}=1$$ and ${\rm{lD}}(R)=n$.<|endoftext|>
TITLE: Why is it so difficult to write complete (computer verifiable) proofs?
QUESTION [29 upvotes]: For example I have read that is agony to give a complete proof of the Jordan curve theorem. Since all statements are meant to be justified by the postulates, where does the difficulty lie?

REPLY [17 votes]: There are probably many reasons for the difficulty, but there might be one particularly difficult problem: Mathematicians may be using meta-induction instead of induction leading to erroneous proofs.
Bundy et al. in "What is a proof?" claims that S. Baker in "Aspects of the constructive omega rule within automated deduction" claims that often when mathematicians claim to do induction they actually do meta-induction.  This means that instead of proving $\forall n. P(n)$, they instead end up proving $\forall n:{\mathbb N}. Provable(P(term(n))$ (i.e. $P(0), P(1), \ldots$).  This result is, in general, weaker that the result $\forall n. P(n)$ that they claim to be proving.  The paper proofs are so informal that it is hard to tell that this is what they are doing, and this coupled with the fact that the theorems they are claiming to prove are actually true (and provable), means that this subtle error goes unnoticed.  This leave formalizers with the difficult task of providing an actual inductive proof.
Thanks to schropp for pointing this out.  Sorry for stealing your Mathoverflow karma.<|endoftext|>
TITLE: Reference request for conceptual numerical analysis
QUESTION [10 upvotes]: I am interested in clean algorithms for approximating solutions and so I am interested in numerical analysis, but most of the books I have seen get bogged down in error analysis or they spend a lot of time and effort in squeezing an additional 2% efficiency from a classical algorithm. What are some works which just have the fundamental ideas and methods?

REPLY [4 votes]: To augment Timur's answer:

Claes Johnson's introductory book on FEM 
Braess's book on FEM
Iserles's book on numerical analysis of DE
Gottlieb and Orzsag's book on spectral methods 
Nick Trefethen's book on spectral methods for spectral collocation ideas.
Quarteroni, Sacco, Saleri  on numerical methods. 
From 'the horse's mouth', the Cleve Moler book on numerical computing using Matlab.

I've picked these books for their balance of important algorithms and key insights, delivered with clear prose. 
I also like Strikwerda's book on finite difference methods, and the Hairer-Wanner books on numerical methods for ODE. But these focus a lot on error analysis, which may not be what you wish.<|endoftext|>
TITLE: Are these notions of strongly equivariant D-modules equivalent?
QUESTION [7 upvotes]: It seems that there are two notions of strongly equivaraint $D_X$- Modules and I would like to know if they are equivalent, or at least how they are related.
Let $\rho: G\times X \rightarrow X$ be an action of an algebraic group on a smooth variety over the complex numbers.
The first definition goes like this:
An equivariant $D_X$ Module is just a $D_X$ module $M$ together with an isomorphism 
$$\rho^* M\rightarrow \pi^* M$$ of $D_{G\times X}$ -modules. That isomorphism has to satisfy some cocycle condition.
The other definition is a bit more cumbersome to write down. First it requires just an  isomorphism of $O_{G\times X}$ modules, not necessarily of $D_{G\times X}$-modules $$\rho^* M\rightarrow \pi^* M$$  modules, which again satisfies the cocycle conditon. 
In addition it requires the action map $$D_X\otimes M \rightarrow M$$ to be equivariant.
Finally there is another condition to be satisfied:
Observe that we get two operations of the liealgera on $M$:
One operation, by directly differentiating the action of $G$ on $M$.
Another operation in the following way: First we differentiate the action of $G$ on $X$, and get a map
$$Lie(G)\rightarrow Der_X$$
from the liealgebra into vectorfields on $X$. Because $M$ is a $D_X$ module we can compose this map with the action of vectorfields on $M$ and get our second operation.
We require these operations to coincide.
A more precise definition of the second kind is given here on pages 48-49:
http://www.math.harvard.edu/~gaitsgde/267y/catO.pdf
So the question is, are these two notions equivalent?
Edit: If anybody else needs these facts, I found a reference which gives a proof:
http://alpha.uhasselt.be/Research/Algebra/Publications/Geq.ps

REPLY [12 votes]: Yes. They are both discussed in Chapter 7 (Hecke Patterns) of Beilinson-Drinfeld's Quantization of Hitchin Hamiltonians, accessible (like most things in the area) off Dennis Gaitsgory's page you quote above. 
More precisely, there are two ways to think of D-modules on a stack $X/G$ (aka $G$-equivariant D-modules on $X$). First, you can just describe them by descent from $X$ -- the descent data is exactly your first description - the two pullbacks have to agree, in a way that's associative. The same description works for the (dg) derived category of D-modules on the stack $X/G$ -- you make the simplicial scheme with simplices $G\times G\times\cdots\times G\times X$, and take the (homotopy) limit of the dg categories of $D$-modules on the simplices. Again this is explicitly in BD chapter 7, as well as in the appendix to the long paper by Frenkel-Gaitsgory and in my Character Theory paper with Nadler.
The second description is quantum hamiltonian reduction (aka BRST). You can write the cotangent of $X/G$ as the quotient by $G$ of the zero fiber of the moment map on $T^* X$. Now quantize, you can write the quantum version of $T^*(X/G)$ --- ie D-modules on $X/G$ - by first taking G-equivariants as O-modules (this gives the quantization of $(T^*X)/G$) and then imposing the zero value of the moment map, which means the compatibility of Lie algebra actions you wrote. This picture is also in BD..and again if you take invariants and coinvariants (ie impose 0 moment value) in a derived way the statement holds on the derived level.
One way Nadler and I like to say this is by thinking about D-modules as O-modules on the de Rham space $X_{dR}$ (quotient of X by formal neighborhood of the diagonal). Then D-modules on $X/G$ means O-modules on the de Rham space of $X/G$, which is $X_{dR}/G_{dR}$ -- spelling that out gives the first description you wrote. To get the second picture, you first impose $G$-equivariance as an $O$-module, which means looking at $X_{dR}/G$. But $G_{dR}=G/\widehat{G}$ (formal group of $G$) is a very useful realization -- ie Lie groups DO have canonical normal "subgroups"!).. the second step, fixing the action of the Lie algebra (or equivalently the formal group) tells you which sheaves on $X_{dR}/G$ actually come from $X_{dR}/G_{dR}$, ie are strongly equivariant..<|endoftext|>
TITLE: Functoriality of fundamental group via deck transformations
QUESTION [8 upvotes]: Problem
I'm trying to understand this with a view towards the etale fundamental group where we can't talk about loops. What I'm missing is how the fundamental group functor should work on morphisms, without mentioning loops.
Naive attempt
Let's say we have a map $X \rightarrow Y$ (of topological spaces, schemes, what have you). Let's say $\tilde Y$ is $Y$'s universal cover (in the case of schemes, this only exists as a pro-object, and only in some cases, but for simplicity assume it exists) and $\tilde X$ is $X$'s fundamental group.
My first, naive, approach was the following: take $X \times_Y \tilde Y$. This is a cover of $X$ (etale is invariant to base change. Again $\tilde Y$ isn't really etale over $Y$ because it's not finite, but once we have the topological case down, ironing out the arithmetic details should be easy). So we have a map $\tilde X$ to $X \times_Y \tilde Y$.
Now, since $\tilde Y$ to $Y$ was Galois (- normal for the topologists; with group of deck transformations $\pi_1(Y, y)$) then so is $X \times_Y \tilde Y$ over $X$. With what group? It seems (and correct me if I'm wrong) that this will always be some quotient of $\pi_1(Y,y)$ (meaning that the group action of $\pi_1(Y,y)$ on $X \times_Y \tilde Y$ as a map $\pi_1(Y,y) \times X \times_Y \tilde Y \rightarrow (X \times_Y \tilde Y) \times_X (X \times_Y \tilde Y)$ is surjective but not nec. an immersion).
Since $\tilde X$ maps to $X \times_Y \tilde Y$, we get a natural map $\pi_1(X,x) \twoheadrightarrow Aut_X(X \times_Y \tilde Y)$, where, as we said, $Aut_X(X \times_Y \tilde Y)$ is a quotient of $\pi_1(Y,y)$.
This is not going to work. What is the right definition of how the fundamental group functor acts on morphisms, via a deck-transformations approach?

REPLY [5 votes]: Your spaces are not pointed, so the fundamental group is not a functor with values in groups.
It's a functor with values in groupoids: The objects of $\pi_1(X)$ are the universal covers of $X$, and the morphisms of $\pi_1(X)$ are the isomorphisms between different universal covers.
For completeness, I recall that a universal cover of $X$ is a cover $\tilde X\to X$, such that $\tilde X$ is connected and simply connected (i.e. admits no non-trivial covering maps from other spaces).
Given a map $f:X \to Y$, and a universal cover $\tilde X\to X$, you get a universal cover $\tilde Y$ of $Y$ uniquely defined up to unique isomorphism by the property that it admits a map $\tilde X\to \tilde Y$ making the following square diagram commute:
$\tilde X\to \tilde Y$
$\downarrow\qquad\downarrow$
$X\to Y$. $\qquad\qquad$ We then let $\pi_1(f)$ be the functor sending $\tilde X\in\pi_1(X)$ to $\tilde Y\in\pi_1(Y)$.
In topology, a concrete construction of $\tilde Y$ involves paths in $Y$ starting from the image of some point $p\in \tilde X$. I don't know how to do this construction in your arithmetic context.

The tricky thing in the above argument is the difference between objects that are uniquely defined up to isomorphism, and objects that are uniquely defined up to unique isomorphism.
• A universal cover is uniquely defined up to isomorphism: "it's not really well defined".
• The space $\tilde Y$ in the above diagram is uniquely defined up to unique isomorphism: "it's truly well defined".

REPLY [2 votes]: Alright, this answer is a rephrasing of all the comments:
Personally I think the algebro-geometric language is completely superfluous here. This question is topological should have a completely topological answer. You start with a $\tilde Y$ to $Y$ (pt'd, whatever); take fiber product with $X$ and get a cover of $X$. Then $\pi_1(X,x)$ acts on the preimages of $x$ in each connected component of this fiber product, and therefore on all the preimages of $x$. Notice that over $x$ the fiber of this cover is naturally iso. to the fiber over $y$ in $\tilde Y$. Okay, good. So $\pi_1(X,x)$ acts on the fiber of $y$ in $\tilde Y$, so that gives a map to $\pi_1(Y,y)$. As Scott mentioned, the reason it's not onto is that $\tilde Y \times_Y X$ is not nec. connected.<|endoftext|>
TITLE: Procreation with several genders
QUESTION [15 upvotes]: Most mathematicians are aware that our species consists of two genders, 
denoted for simplicity by the multisets $\lbrace X,X\rbrace$ and $\lbrace X,Y\rbrace$, with 
offspring given by $\lbrace A,B\rbrace$ for $A\in\lbrace X,X\rbrace$ 
and $B\in \lbrace X,Y\rbrace$.
I am asking for the existence of combinatorial structures generalizing this construction. 
More precisely, given a finite set $\mathcal C$ of $n$ distinct elements
($n=2$ and $\mathcal C=\lbrace X,Y\rbrace$ in the above example) and an integer $k$, we denote by $\mathcal M_k$ the set of all multisets containing exactly 
$k$ not necessarily distinct elements of $\mathcal C$. 
A gender partition of a subset $\mathcal S\subset \mathcal M_k$ is a partition of $\mathcal S$ into $k$ non-empty parts $\mathcal G_1,\dots,\mathcal G_k$ called genders such that the following two conditions hold: 
(i) Given $(g_1,\dots,g_k)\in\mathcal G_1\times \dots
\times \mathcal G_k$, every element $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ gives rise to a multiset $\lbrace x_1,\dots,x_k\rbrace$ which is in $\mathcal S$.
(ii) Every multiset $g\in\mathcal S$ is of the form $\lbrace x_1,\dots,x_k\rbrace$
for $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ where $g_i\in
\mathcal G_i$ are suitable elements.
Examples with $\mathcal C=\lbrace X,Y\rbrace$ are:
(a) $\mathcal S= \lbrace X,X\rbrace\cup \lbrace X,Y\rbrace$ with $\mathcal G_i$ 
given by singletons.
(b) $\mathcal S= \lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace
\cup \lbrace X,Y\rbrace$ with $\mathcal G_1=\lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace$ and $\mathcal G_2$ consisting of $\lbrace X,Y\rbrace$.
(c) An example with $k=3$ (easily generalizable to arbitrary values of $k$)
is given by $\mathcal S=\lbrace \lbrace X,X,X\rbrace,\lbrace X,X,Y\rbrace,
\lbrace X,Y,Y\rbrace\rbrace$ with $\mathcal G_i$ given by singletons.
More examples of gender partitions $\mathcal S=\mathcal G_1\cup
\dots\cup \mathcal G_k$ are fairly easy to construct. (And there are fairly easy
notions for "products", "quotients", one can split an element of $\mathcal C$ 
into several new elements, etc.)
The following additional condition is more difficult to satisfy:
Call a gender partition $\mathcal S=\mathcal G_1\cup \dots\cup\mathcal G_k$
balanced if $\mathcal S$ admits a stationary probability measure 
$\mu$ giving equal weight $\frac{1}{k}$ to all genders $\mathcal G_i$.
A probability measure $\mu$ on $\mathcal S$ is stationary if 
the probabiliy $\mu(\lbrace x_i,\dots,x_k\rbrace)$ of every offspring 
of $(g_1,\dots,g_k)$ (with respect to uniform choices for $x_i\in g_i$)
is proportional to $\prod_{i=1}^k\mu(g_i)$. (Stationary probability 
measures exist always and are unique if $\mathcal S$ is
minimal in some sense.)
Example: The examples (a) and (b) above are balanced, (c) is not balanced.
Question: Produce other examples of balanced gender partitions.
Is there for example a balanced gender partition for $k=3$?
Remark: One can also consider probabilities on offsprings 
which depend on the choice of $x_i\in g_i$. Example (c) is not balanced
even in this more general framework.
Variation: Instead of working with multisets, one can also work with 
sequences of length $k$. An offspring of $k$ sequences
$g_1=(g_1(1),\dots,g_1(k)),\dots,g_k=(g_k(1),\dots,g_k(k))$ with 
$g_i\in\mathcal G_i$ is then given by 
$g_1(\sigma(1)),\dots,g_k(\sigma(k))$ where $\sigma$ is a (not necessarily arbitrary) permutation of 
$\lbrace 1,\dots,k\rbrace$.

REPLY [6 votes]: So, I hope I understand the definitions correctly.  Here's a way to construct an example with $k = 3$ genders (say A, B and C) using $n = 9$ sex chromosomes, which I will take to be the elements of $\mathbb{Z}/9\mathbb{Z}$: start with all 165 multisets of chromosomes unused.  Choose any unused multiset $\{x, y, z\}$, and add it, together with the multisets $\{x + 3, y + 3, z + 3\}$ and $\{x + 6, y + 6, z + 6\}$, to gender A, add the multisets $\{x + 1, y + 1, z + 1\}, \{x + 4, y + 4, z + 4\}, \{x + 7, y + 7, z + 7\}$ to gender B and add the multisets $\{x + 2, y + 2, z + 2\}, \{x + 5, y + 5, z + 5\}, \{x + 8, y + 8, z + 8\}$ to gender C.  Continue until every multiset of chromosomes is used up.  The resulting genders (each consisting of 55 multisets of chromosomes) are symmetric and so (if I'm not mistaken) are balanced.
It's easy to see how to construct a wide variety of similar gender partitions for given $k$ if we may choose $n$ appropriately.  These partitions have nice symmetry and use all possible multisets of chromosomes.  This says nothing at all about, say, constructing gender partitions for $n = 2$ (though I believe that I've confirmed by case analysis that there are no balanced gender partitions for $k = 3$ and $n = 2$).<|endoftext|>
TITLE: A question about the Osgood curve
QUESTION [9 upvotes]: Does every sub-arc of the Osgood curve (with distinct end-points) have positive two-dimensional 
Lebesgue measure? If not, do there exist Jordan curves whch have this property?

REPLY [2 votes]: Another example: in a 1961 paper, Besicovitch and Schoenberg constructed Jordan arcs with no negligible subarcs. More precisely, according to (1), for every $\varepsilon > 0$, there exists a Jordan arc $J$ such that for $v \neq v'$ in $J$, denoting $J(v,v')$ the subarc having $v$ and $v'$ as endpoints, $$0 < m_2\big(J(v,v')\big) < C_{\varepsilon} |v'-v|^{2-\varepsilon}.$$
$ $
References 

$\ \ $ (1)$\ \ $ On Jordan Arcs and Lipschitz Classes of Functions Defined on Them<|endoftext|>
TITLE: What is the limit of gcd(1! + 2! + ... + (n-1)! ,  n!) ?
QUESTION [61 upvotes]: Let $s_n = \sum_{i=1}^{n-1} i!$ and let $g_n = \gcd (s_n, n!)$.  Then it is easy to see that $g_n$ divides $g_{n+1}$. The first few values of $g_n$, starting at $n=2$ are $1, 3, 3, 3, 9, 9, 9, 9, 9, 99$, where $g_{11}=99$.  Then $g_n=99$ for $11\leq n\leq 100,000$.
Note that if $n$ divides $s_n$, then $n$ divides $g_m$ for all $m\geq n$.  If $n$ does not divide $s_n$, then $n$ does not divide $s_m$ for any $m\geq n$.
If $p$ is a prime dividing $g_n$ but not dividing $g_{n-1}$ then $p=n$, for if $p<n$ then $p$ divides $(n-1)!$ and therefore $p$ divides $s_n-(n-1)!=s_{n-1}$, whence $p$ divides $g_{n-1}$.
So to show that $g_n\rightarrow \infty$ it suffices to show that there are infinitely many primes $p$ such that $1!+2!+\cdots +(p-1)! \equiv 0$ (mod $p$).

REPLY [11 votes]: The current state of search is that $g_n=99$ for $11\leq n\leq 10,000,000,000$. I'll mention that Kurepa's conjecture, which means that $s_p\neq -1\pmod p$ holds for odd primes $p$, is true for $p<10^{10}$. It is interesting that $s_p=1\pmod p$ holds for $p=6,855,730,873$, and this is the first such prime after $p=31$ and $p=373$.
Searching for a counterexample of Kurepa's conjecture, arXiv:1409.0800 [math.NT]<|endoftext|>
TITLE: A number encoding all primes
QUESTION [10 upvotes]: This may be a soft question, but it's just something I thought of one night before sleeping. It's not my field at all, so I am just asking out of curiosity. Has anyone studied the number which is the sum over primes $\sum{ 2^{-p}}$? Its binary expansion (clearly) has a 1 in each prime^th "decimal place", and a zero everywhere else, so, it should be important in number theory I would guess.

REPLY [6 votes]: See http://oeis.org/A051006 and http://mathworld.wolfram.com/PrimeConstant.html which cover this particular sequence.<|endoftext|>
TITLE: Current status of Bloch Constant and Landau Constant bounds
QUESTION [7 upvotes]: The Bloch constant B (based on a theorem introduced by André Bloch in 1925 on the maximum radius of a one-to-one disk in the image of a normalized analytic function of the unit disk, see for instance Remmert Funktionentheorie II or Steven Finch marvelous "Mathematical Constants") was conjectured by Ahlfors to be
$$ \frac{1}{\sqrt{1+\sqrt{3}}}\frac{\Gamma(\frac{1}{3})\Gamma(\frac{11}{12})}{\Gamma(\frac{1}{4})}$$
(This value, if I remember well Ahlfors' article corresponds to a particular function that he constructed for this purpose).
The Bloch Constant $B$ is currently known to be at least slightly greater than  $\frac{\sqrt{3}}{4}$ (several articles improving upon each other by Mario Bonk, Chen and Gauthier, Xiong).
Has there been some progress since 1998 on the lower bound ?
Same question for the closely related (univalent) Landau constant (quite often called Bloch-Landau constant, sometimes seen as $B_\infty$) ?
The conjectured upper bound is 
$$\frac{\Gamma(\frac{1}{3})\Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})}$$

What can be said of the various adaptations or specializations of this constant to various class of functions, and extensions of these constants to several complex variables or other functional spaces ?

I give as background the original article from Bloch, Ahlfors and Grunsky.
(1) A. Bloch, Les théorèmes de M. Valiron sur les fonctions entières et la théorie de l'uniformisation, Ann. Fac. Sci. Univ. Toulouse, vol. 17, (1925), pp1-22.
(2) L. V. Ahlfors and H. Grunsky, Über die Blochsche Konstante, Math. Zeitschrift 42 (1937), pp671–673.
(3) L. V. Ahlfors, An extension of Schwarz's lemma, Trans. Amer. Math. Soc. 43 (1938), pp359–364.
(these two are reprinted in Ahlfors Works vol 1)

Ahlfors life and works are evocated in an AMS Notices of 1998.

REPLY [6 votes]: The world record on the Bloch's constant seems to be MR1690898 by C. Xiong, who proved
$B\geq \sqrt{3}/4+3.10^{-4}$. I recall that $\sqrt{3}/4$ is the Ahlfors estimate, then Heins
proved that $B$ is strictly greater than that, and Bonk was the first to prove this with
a specific constant. Then this constant was slightly improved, first by Gauthier and Chen
MR1428103 and then by Xiong.
This is far from the conjectured value. There are also results which show that
the Ahlfors - Grunsky conjectured extremal function gives a local extremum for certain
variations. But the classes of variations considered are narrow.
For example, Baernstein II and Vinson proved that the Ahlfors Grunsky function gives
a local extremum for the class of ramified coverings which are ramified 
only over some lattice
points. (So the hexagonal lattice is locally extremal for such restricted problem).<|endoftext|>
TITLE: Geometry for Anderson's motives?
QUESTION [14 upvotes]: Anderson's $t$-motives satisfy most of what is expected of a reasonable category of mixed motives, except of course that everything is in positive characteristic. For instance, it is a linear category with a tensor product, we have a well behaved weight filtration, and there are realisation functors with good properties. (After all, the name "motive" was not chosen at random.)
However, a classical motive is something that one wants to associate with a scheme or variety, thinking of it as its universal cohomology, and now it bugs me that I have no geometric objects at hand to which I could associate $t$-motives. I don't even know whether these, if any, should be varieties or something else.
To make this a halfway real question: Let $C$ be a smooth proper curve over $\mathbb{F}_p(t)$ of genus $g$. Is there a natural, functorial way of associating with $C$ a pure $t$-motive $M(C)$ of rank $g$, in such a way of course that the cohomology of $C$ is related, via a functor morphism, to the realisations of $M(C)$? So I'm asking here for some kind of Jacobian construction.
But as I said, I don't even know whether varieties over $\mathbb{F}_p(t)$ are the right geometric objects to look at.

REPLY [7 votes]: One object with a very geometric flavor associated with Anderson's $t$-motives are $\tau$-sheaves, which are sheaves on the curve with a certain Frobenius structure.  For Drinfeld modules (so Anderson $t$-motives of dimension 1), these are Drinfeld shtukas, which were then generalized to higher dimensions by several authors.  There is a nice treatment of them in the recent book by Böckle and Pink, Cohomological Theory of Crystals over Function Fields.  These are probably what you are looking for.
Pink has also defined Hodge structures for $t$-motives in the paper "Hodge structures over function fields," which is available on his web page: http://www.math.ethz.ch/~pink/.

To make this a halfway real question: Let C be a smooth proper curve over Fp(t) of genus g. Is there a natural, functorial way of associating with C a pure t-motive M(C) of rank g, in such a way of course that the cohomology of C is related, via a functor morphism, to the realisations of M(C)? So I'm asking here for some kind of Jacobian construction.

What you are proposing will have trouble working in some generality.  For example, the characteristic polynomials of Frobenius acting on the Tate module of the Jacobian of the curve will have coefficients in ℤ, while those acting on the Tate module of the $t$-motive itself will have coefficients in the function field.<|endoftext|>
TITLE: Raising and lowering operators for SL(n,K) on homogenous polynomials
QUESTION [8 upvotes]: The short version:

Can the theory of weights for SL(n,C) be explained concretely in terms of raising and lowering operators on spaces of polynomials?

A deleted question asked how to prove SL(3,C) acts irreducibly on the space of homogenous polynomials of fixed degree.  I think it was deleted as a homework question, but it has never been a homework question at my university (at least not in the past decade), and so I thought it might be reasonable to try and work out these ideas in a setting that makes sense to me:
Let K be a field of characteristic 0 and let SL(n,K) be the group of n×n matrices with entries from K and determinant 1.  Let V be the K-vector space with basis the monomials xi of total degree w, where i = (i1,i2,...,in) is a multi-index with n entries, all non-negative integers whose sum is w.  V has dimension w+1.  SL(n,K) operates on V in the one true manner, by substitution of variables.  In particular, for j = 2,3,...,n, the group SL(n,K) contains the elementary substitutions Ej which takes xj to x1+xj and takes xk to xk when j≠k.
Define the lowering operator Lj as Ej−1, so it takes a polynomial f to the difference between Ej(f) and f itself.  The L are called lowering operators because they lower the degree of f when considered as a polynomial in xj, though they hold the total degree constant.  This can be checked by applying it to monomials.  For any individual monomial, factor it as xja⋅xi where ij=0, then Lj(xja⋅xi) = (x1+xj)a.xi − xja⋅xi = ( (x1+xj)a−xja )⋅xi.  The leading term, when considered as a polynomial in xj is then ax1xja−1xi, so it has degree a−1 in xj.
If W is a submodule of V containing some nonzero polynomial f, then it must have nonzero degree in one of the variables xj.  If any of these j are greater than 1, then apply Lj to transfer the degree from j to 1.  Repeat until one is left with a⋅x1w.  Since each Lj takes W to W and W is closed under division by nonzero scalars, we must have that any non-empty SL(n,K) invariant subspace of V contains x1w.  In fact, we only use that it is closed under the maximal unipotent subgroup generated by the Ej.
For instance, if n=3, then we can more simply write x1=x, x2=y, and x3=z.  If w=4, and W contains xyyz+2xyzz, then apply L3 = Lz to get xyy(x+z)+2xy(x+z)(x+z) − xyyz+2xyzz = xyyx + 2xyxx + 4xyxz = 2xxxy + xxyy + 4xxyz.  The total degree remained the same, but shifted from z to x, at the cost of increasing the leading coefficient.  Applying Lz again, one gets the first two terms vanish, and one is left with 4xxy(x+z) − 4xxyz = 4xxxy.  Applying Ly one gets the pure 4xxx(x+y)−4xxxy = 4xxxx.
Notice that there were multiple paths we could have taken to move from f to xw.  I think these paths are basically what are called "weights", and the lowering operators are called (simple, positive) "roots".
Now we must recover all other monomials from this monomial, and so we define the raising operators.  We begin with the elementary Fj in SL(n,K) defined by taking x1 to x1+xj and xk to xk for k≠j.  The raising operator Rj is then Fj−1 which takes a polynomial f to the difference between Fj(f) and f itself.  Rj sacrifices a degree in x1 to provide a degree in xj.   Applying each Rj operator ij times to x1w we obtain a polynomial with nonzero xi term, but with a great many other terms that seem difficult to control.

How does one use the raising operator to recover the other monomials?

For instance, to show xxyz is in W, I imagine that we should apply Rz to xxxx to get (x+z)(x+z)(x+z)(x+z) − xxxx = 4xxxz + 6xxzz + 4xzzz + zzzz.  But then, how does one isolate 4xxxz?  Here is my current method, which seems overly complex compared to lowering:  Applying Lz 3 times and subtract 60xxxx (already known to be in W) takes zzzz to a multiple of xxxz, but it does so by zero-ing out the other terms.  Applying Ry to xxxz gives (x+y)(x+y)(x+y)z−xxxz = 3xxyz + 3xyyz + yyyz.  Now apply Ly twice to zero out the first two terms and take yyyz to 12xxxz + 6xxyz.  Since xxxz is known to be in W, we can remove it and rescale to get xxyz in W.
This random sort of zig-zag makes it hard to keep track of which "roots" we've been applying (that is, which raising and lowering operators).  In the Lie algebra case, I thought things were a bit cleaner.  At any rate, successfully applying these operators should have the side-effect of detecting in a very concrete way the "highest weight", if I understand correctly.  However, I cannot yet check, since I cannot yet successfully keep track of how I am applying these operators.

Is there a clear relationship between the paths one takes using these operators and the theory of weights?  For instance, is x1w a highest weight vector?

It seems to me that SL(n,K) should have other irreducible representations other than just these polynomial representations, just because the paths defined by the operators are so symmetric.  I think for SL(2,K) this is basically all there is, perhaps allowing for field automorphisms to be applied first.

Can all of the finite dimensional irreducible representations of SL(n,K) be explained in terms of actions on polynomials by substitutions and field automorphisms?

Assuming that this stuff makes sense, it would be handy to try it out on a different weight lattice.

Is there a version of this sort of description for any of the other classical groups or for G2?

It would be nice if say the symplectic group operated by substitutions on some subspace of homogenous polynomials, possibly restricted in some way, and that one could then find the raising and lowering operators.  Any group where the action is natural and is likely to work well in the setup of Dickson's Linear Groups is fine, but I suspect the symplectic group might be the one with the least complications.

REPLY [6 votes]: I am a bit reluctant to post a reply, since this is not even a homework question: you really ought to learn theory first from a textbook on representations of Lie algebras, and several of them have been published recently, instead of "discovering" it on your own and asking other people to check your understanding.
Answers (K has characteristic zero):

The Lie algebra sl(n,K) and the algebraic group SL(n,K) act on the polynomial algebra in $n$ variables so that the generator $E_{ij}$ of the Lie algebra sl(n) acts by the differential operator $x_i\partial_j$. This representation is irreducible of highest weight (N,0,...,0)=$N\omega_1$ in standard enumeration, with the highest weight vector given by the monomial $x_1^N$.
All simple finite-dimensional modules over a semisimple Lie algebra (equivalently, simple algebraic representations of a semisimple algebraic group) are classified in terms of their highest weights. For sl(n), a weight is an (n-1)-tuple of nonnegative integers and only very special n-tuples arise from the polynomials in n variables.
For the classical Lie algebras sl(N), so(N), sp(N), the theory of standard monomials (due to Hodge, Pedoe, Laskshmibai, Musili, Sheshadri, Sundaram) gives an explicit realization of these modules on certain spaces of polynomials. For G2 Greg Kuperberg found a combinatorial analogue. The Littelmann path model provides a uniform description for all types that encompasses a lot of previous work including also Kashiwara's crystal bases.<|endoftext|>
TITLE: Locus of equal area hyperbolic triangles
QUESTION [17 upvotes]: Henry Segerman and I recently considered the following question:
Given a fixed area $A < \pi$ and two fixed points in the upper half-plane model for hyperbolic $2$-space, what is the locus of points which give rise to a hyperbolic triangle of the given area?
We found it a fun exercise in hyperbolic geometry to show that the answer is a Euclidean straight line, or an arc of a Euclidean circle. As this requires only elementary properties of hyperbolic geometry, we strongly suspect it should be known, but have thus far been unable to find a reference for it. Does anyone know whether it's known, and if so, where one can find it?

REPLY [8 votes]: Nice problem!  After googling "hyperbolic triangles of equal area on a fixed base" I found the paper "Extremal properties of the principal Dirichlet eigenvalue for regular polygons in the hyperbolic plane" by Karp and Peyerimhoff.  Their Theorem 7 looks like the result you ask for.  In a footnote KP refer to a 1965 book of Fejes Toth, "Regulare Figuren", as containing the same result.  There appears to be an English version "Regular Figures" published in 1964. 
This question seems very interesting in spherical geometry, as well.<|endoftext|>
TITLE: Dual Curves in Fancy Language
QUESTION [8 upvotes]: I decided to spend this summer working through exercises in Hartshorne, and I found myself frustrated by the way I was solving one of them, specifically IV.2.3 on pp. 304-305. No, I'm not asking for a solution to the problem --- I almost have the whole thing solved, as far as I can tell --- what worries me is that I think I'm thinking about this stuff the wrong way.
The problem is about the map from a projective plane curve in characteristic 0 to its dual curve, and it has 8 parts, which ask you to prove things like the fact that bitangents on the original curve correspond to nodes on the dual. I managed to solve all of them except the one about ordinary inflection points on X giving ordinary cusps on X*, but all my solutions involved picking affine charts, finding an ugly formula for the map in question, and computing first and second partial derivatives, and it's all very long and messy and doesn't give any clue as to what's going on until you get to the end and see that the thing you have is 0 or whatever if and only if the condition in the problem is met for some magical reason.
I feel like there must be a more "high-brow" way to approach this object, and the fact that I haven't been able to come up with one seems to speak to the sort of backward way I've been learning about the subject (I just took a class that was very good and covered a lot but was almost completely devoid of examples). There must be some approach to this that actually uses all the machinery that's developed in the rest of the book. Is there a satisfying, pretty way to deal with this thing that I'm missing, something that would tell me why these relationships hold and not just that they hold?

REPLY [2 votes]: You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve.  For example, try to identify the line bundle on your curve which embeds it into the dual plane.  E.g.:  let $X \subset \mathbb{P}^2$ be your curve.  Then define $X' \subset X \times \mathbb{P}^{2*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$.   When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N_{X/\mathbb{P}^2}$.  
Then think about the map $X' \rightarrow X \times \mathbb{P}^{2*} \rightarrow \mathbb{P}^{2*}$.  This is the dual curve embedding.  The morphism comes from the inclusion of $N^* \subset \Omega_{\mathbb{P}^2}|_X \subset \mathcal{O}(-1)^3$.  
At some point, you have to do computations in local coordinates - but setting it up like this may help a little.
Example of local computation:
Locally, the curve looks like $(x(t), y(t), z(t))$.  The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$).  If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.<|endoftext|>
TITLE: Extending properties of commutative rings to schemes
QUESTION [12 upvotes]: I'm trying to pin down the various ways we can extend a property of commutative rings to a corresponding property for schemes. Let $P$ be a property of commutative rings. We could define a scheme $(X,\mathcal{O}_X)$ to have property $P$ in one of the following ways:

$\mathcal{O}_X(U)$ has $P$ for every open subset $U\subset X$.
$\mathcal{O}_X(U)$ has $P$ for every affine open subset $U\subset X$.
there exists an affine open cover ${U_i}$ of $X$ such that $\mathcal{O}_X(U_i)$ has $P$ for all $i$.
for each $x\in U\subset X$ with $U$ an open subset, there exists an affine open $V\subset X$ with $x\in V\subset U$ such that $\mathcal{O}_X(V)$ has $P$.
$\mathcal{O}_{X,x}$ has $P$ for all $x\in X$.

Evidently, $(1)\Rightarrow (2)\Rightarrow (3)$. If the property $P$ is stable under inversion of single elements (that is, $A$ has $P$ $\Rightarrow$ $A[1/s]$ has $P$ for any element $s\in A$) then $(3)\Rightarrow (4)$. Furthermore, if the property $P$ is stable under arbitrary localizations (that is, $A$ has $P$ $\Rightarrow$ $S^{-1} A$ has $P$ for any multiplicative subset $S$ of $A$) then $(4)\Rightarrow (5)$.
Thus, for many properties of commutative rings $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)\Rightarrow (5)$.
Now we need to consider going in the other direction. I believe it can be shown that if $P$ is a local property in the sense that $A$ has $P$ iff $A_{\mathfrak{p}}$ has $P$ for each $\mathfrak{p}\in$ Spec$(A)$ then $(5)\Rightarrow (2)$.
Now, it seems to me that if $P$ is a local property in this sense then the property is stable under localization by arbitrary multiplicative subsets. Thus if $P$ is a local property in this sense then $(2) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5)$.
Finally, here is another notion of a property being local. Suppose that $P$ is such that $A$ has $P$ implies that $A[1/s]$ has $P$ for each $s\in A$ and that on the other hand, if $s_1,\ldots,s_n \in A$ are such that Spec$(A)=D(s_1)\cup D(s_2) \cup \cdots \cup D(s_n)$ then $A[1/s_i]$ has $P$ for all $i=1,..,n$ implies that $A$ has $P$. Then I believe it can be shown that $(4)\Rightarrow (2)$ and thus for such a property we have that $(2) \Leftrightarrow (3) \Leftrightarrow (4)$.
Does this seem right to you? I haven't seen any books on algebraic geometry discuss this question to my satisfaction and I am nervous that there may be some holes in my proofs, so if anyone knows off the top of their head that what I have described seems right then I would be happy to hear from you. Do you have any further comments to make about this process of extending a property of commutative rings to schemes?

REPLY [5 votes]: The $U$-sections funtctors $\Gamma_U: Sheaves(X, Ring)\to Ring: \mathscr{O} \mapsto  \mathscr{O}(U)  $ preserve (and reflect)  only limits ($Ring$ mean commutative rings by unit).
The Stalk functors $\Gamma_x: Sheaves(X, Ring)\to Rings: \mathscr{O} \mapsto  \mathscr{O}_x $ $x\in X$  preserve (and reflect)  also colimits.
Then the question is what kind of propriety is preserved (and reflexed) by the funtors  $\Gamma_U$ or $\Gamma_x$?
Formally:
1) Is: $\mathscr{O}$ is $P$ $\Leftrightarrow$  $\Gamma_U(\mathscr{O} )$ is $P$ $\forall U\in \tau_X$ ?
2) Is: $\mathscr{O}$ is $P$ $\Leftrightarrow$  $\Gamma_x(\mathscr{O} )$ is $P$ $\forall x\in \tau_X$ ?
These questions  involving the “logic  inside a category” concept (for formalizing what you mean about property  $P$). 
The (1) has the answere: for all $P$ definible in terms of  “Cartesians formulas”
The (2) has the answere: for all $P$ definible in terms of  “geometrical  formulas” 
The fist step in this direction was the cap.III of  Monique Hakim's  book "Schemas relatifs et Topos anelles”, for a more easy lectures can read 
Saunders MacLane, Ieke Moerdijk
Sheaves in Geometry and Logic: A First Introduction to Topos Theory 
But for example if $P$ is the $local-ring$ proprierty the this means that $\mathscr{O}_{X,x} $ a unique maximal ideal but dont means that $\mathscr{O}_X(U) $ has a unique maximal ideal, but a more general condiction  on the sheaf $\mathscr{O}$ : for any  $ U\tau_X$  and $s\in \mathscr{O}(U) $ we have $U=U_s \cup U_{1-s}$ where $U_t \subset  U$
($t\in \mathscr{O}(U) $) is the maximal open  of $U$ where $t$ is invertible (this definitions coincides by the usal  for sheaves on thee trivial one-point space, i.e. and the funtor $\Gamma_x$ is just the restriction the the space {x}.<|endoftext|>
TITLE: Emptiness and determinization of NFAs
QUESTION [7 upvotes]: Consider an NFA on n states. Is it possible to determine whether it accepts all strings in poly(n) time?
Suppose the NFA above has an equivalent DFA on d states. Is it possible to construct this DFA in poly(n,d) time?

REPLY [2 votes]: Your second question is a little ambiguous, because it admits the following cheating affirmative answer, which is probably not what you intend. 
Namely, every NFA with n states is equivalent to a DFA with d states for d sufficiently large. And for large $d$, allowing poly(n,d) steps is plenty of time. The well-known equivalent DFA, such as the one provided in Sipser's book, has $2^n$ states (plus a constant), and this example can be constructed in poly(n,2^n) steps, simply because $d=2^n$ is already so large here. More generally, for even larger $d$, we can simply pad this one with extra irrelevant states, and build it also in poly(n,d) steps. 
Perhaps you mean to ask about the optimal $d$? Or do you want to ask for all $d$ for which there is an equivalent DFA with $d$ states? Or do you also want to ask whether the optimal $d$ itself is poly(n)?<|endoftext|>
TITLE: What does it mean for a mathematical statement to be true?
QUESTION [12 upvotes]: As I understand it, mathematics is concerned with correct deductions using postulates and rules of inference. From what I have seen, statements are called true if they are correct deductions and false if they are incorrect deductions. If this is the case, then there is no need for the words true and false. I have read something along the lines that Godel's incompleteness theorems prove that there are true statements which are unprovable, but if you cannot prove a statement, how can you be certain that it is true? And if a statement is unprovable, what does it mean to say that it is true?

REPLY [9 votes]: Part of the reason for the confusion here is that the word "true" is sometimes used informally, and at other times it is used as a technical mathematical term.
Informally, asserting that "X is true" is usually just another way to assert X itself.  When I say, "I believe that the Riemann hypothesis is true," I just mean that I believe that all the non-trivial zeros of the Riemann zeta-function lie on the critical line.  (Note in particular that I'm not claiming to have a proof of the Riemann hypothesis!)  This insight is due to Tarski.  If you know what a mathematical statement X asserts, then "X is true" states no more and no less than what X itself asserts.  Now, there is a slight caveat here: Mathematicians being cautious folk, some of them will refrain from asserting that X is true unless they know how to prove X or at least believe that X has been proved.  So in some informal contexts, "X is true" actually means "X is proved."  As we would expect of informal discourse, the usage of the word is not always consistent.
The word "true" can, however, be defined mathematically.  Truth is a property of sentences.  If you have defined a formal language $L$, such as the first-order language of arithmetic, then you can define a sentence $S$ in $L$ to be true if and only if $S$ holds of the natural numbers.  So for example the sentence $\exists x: x > 0$ is true because there does indeed exist a natural number greater than 0.  Here it is important to note that true is not the same as provable.  The formal sentence corresponding to the twin prime conjecture (which I won't bother writing out here) is true if and only if there are infinitely many twin primes, and it doesn't matter that we have no idea how to prove or disprove the conjecture.
Now, perhaps this bothers you.  Is it legitimate to define truth in this manner?  Some people don't think so.  However, note that there is really nothing different going on here from what we normally do in mathematics.  When we were sitting in our number theory class, we all knew what it meant for there to be infinitely many twin primes.  Why should we suddenly stop understanding what this means when we move to the mathematical logic classroom?  If we understand what it means, then there should be no problem with defining some particular formal sentence to be true if and only if there are infinitely many twin primes.  It is as legitimate a mathematical definition as any other mathematical definition.
Now, how can we have true but unprovable statements?  And if we had one how would we know?  Joel David Hamkins explained this well, but in brief, "unprovable" is always with respect to some set of axioms.  Therefore it is possible for some statement to be true but unprovable from some particular set of axioms $A$.  In order to know that it's true, of course, we still have to prove it, but that will be a proof from some other set of axioms besides $A$.<|endoftext|>
TITLE: Cutting convex sets
QUESTION [19 upvotes]: Any bounded convex set of the Euclidean plane can be cut into two convex pieces of equal area and circumference.
Can one cut every bounded convex set of the Euclidean plane into an arbitrary number $n$ 
of convex pieces having equal area and circumference?
The solution of this problem for $n=2$ is generically unique. Are there other values of $n$
(assuming that the problem is possible) where this happens?
More generally, given a $d-$dimensional bounded convex set $C$ in the Euclidean space $\mathbf E^d$ of dimension $d$. for which values of $n$ can one cut $C$ into $n$ convex pieces of equal area with boundaries of equal $(d-1)-$dimensional area? ($n=2$ is again easy, but the solution is no longer generically unique if $d>2$). Are there values for $(n,d)$ for which the solution always exists in a generically unique way (or for which the number of solutions is generically finite)?

REPLY [2 votes]: There is a generalization of this question, where "area" and "circumference" are replaced by arbitrary "nice" measures (for the purpose of this answer, say absolutely continuous measures) $\mu$ and $\nu$ on $\mathbb{R}^2$. Bárány and Matoušek have a nice paper on the subject. 
Even more generally, fix nice probability measures $\mu_1,\ldots,\mu_i$ on $\mathbb{R}^2$. 
A $k$-fan in $\mathbb{R}^2$ consists of $k$ rays (semi-infinite lines) $r_1,\ldots,r_k$ emanating from a point, listed in some clockwise order. (In fact $k$-fans are also allowed to emanate from the point at infinity, i.e., a set of $k$ parallel lines is considered to be a $k$-fan.) Write $C_k$ for the region proceeding $r_k$ in the clockwise order. 
Given a vector $\alpha=(\alpha_1,\ldots,\alpha_k)$ with non-negative entries summing to one, 
say that $\mu_1,\ldots,\mu_r$ can be simultaneously $\alpha$-partitioned if 
there exists a $k$-fan such that $\mu_i(C_j)=\alpha_j$ for each $i=1,\ldots,r$ and $j=1,\ldots,k$. (If $\alpha_1=\ldots=\alpha_k=1/k$ say that the measures can be simultaneously equipartitioned. This case, with $k=2$, is closest to the original )
Bárány and Matoušek have a whole host of results about when such partitions exist and do not exist. Here are just a couple: 

For any $k \geq 5$ and any $\alpha$, there are two measures that can not be simultaneously $\alpha$-partitioned. 
For any $\alpha=(\alpha_1,\alpha_2)$, any two measures can be simultaneously $\alpha$-partitioned, even if the center of the fan is specified in advance. 

No one knows, for example, if any two measures can be simultaneously equipartitioned into four parts. Karasev seems to have a paper where he proves that any two measures can be simultaneously equipartitioned into $q$ convex parts, whenever the number of parts is a prime power. (This was first  achieved for three parts -- this is the result by Bárány et al that Joseph O'Rourke mentioned.) I am unclear on the relation between this and the result of Hubard and Aronov, mentioned by Joseph O'Rourke in his answer.
Higher-dimensional versions have also been considered but much is open. For example, for any three measures in $\mathbb{R}^3$ can one always find a convex $3$-partition of space so that each measure has measure $1/3$ on each part? (I heard Bárány say in a seminar that the version with $3$ replaced by a power of $2$, is known to be true; but I didn't note down a reference.)<|endoftext|>
TITLE: Are two sheaves that are locally isomorphic globally isomorphic ?
QUESTION [6 upvotes]: Let $X$ be a topological space and let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over $X$.
Of course, if one has a morphism $f : \mathcal{F} \to \mathcal{G}$ such that for all $x\in X$, $f_x : \mathcal{F}_x \to \mathcal{G}_x$ is an isomorphism, then it is known that $f$ itself is an isomorphism.
My question is the following: if we don't have such a morphism $f$, but if we know that for all $x\in X$, $\mathcal{F}_x$ and $\mathcal{G}_x$ are isomorphic, is it true that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic ?

REPLY [2 votes]: No: think of non-isomorphic vector bundles over $X$. They are stalkwise isomorphic even as modules (over $\mathcal{O}_{X,x}$ at various $x \in X$, where $\mathcal{O}_X$ is the sheaf of continuous functions on $X$), hence as abelian groups.<|endoftext|>
TITLE: If G is monadic and the comparison functor is an equivalence that is not an isomorphism, does G create limits?
QUESTION [6 upvotes]: Background
Recall that a functor $G\colon A\to X$ is called
monadic if it has a
left adjoint $F$ for which the Eilenberg--Moore comparison functor $K\colon A\to
X^{\mathbb{T}}$ is an equivalence of categories, where $\mathbb{T}$ is
the monad in $X$ defined by the adjunction $\langle
F,G,\ldots\rangle\colon X\rightharpoonup A$, and $X^{\mathbb{T}}$ is
the category of $\mathbb{T}$-algebras in $X$.
This means that a monadic functor is the forgetful functor
$G^{\mathbb{T}}\colon X^{\mathbb{T}}\to X$ up to composition with an
equivalence of categories (the comparison functor $K$). Now, it can be
verified that $G^{\mathbb{T}}$ creates
limits (Ex. 6.2.2 of Mac Lane).  If the comparison functor is an
isomorphism, then it is straightforward to verify that $G$ creates
limits.  In fact, I think that even if it is only assumed that $K$ is an
equivalence for which the object function is surjective, then $G$
creates limits. 
However, in Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, it is
stated that any monadic functor creates limits.  The proof starts
with the following words (with minor omissions):

Let $G$  be monadic. Then by definition, $G$ is the 
  forgetful functor $G^\mathbb{T}$ up to an equivalence of
  categories. It thus suffices to show that such a forgetful functor
  $G^\mathbb{T}$ creates limits.

I simply do not understand this statement:  In general, the composition
of an equivalence and a functor that creates limits need not create
limits. For example, the identity $\mathbf{Set}\to\mathbf{Set}$
creates limits, and for any skeleton $X$ of $\mathbf{Set}$ the inclusion
$X\subseteq \mathbf{Set}$ is an equivalence. Let $X$ be some skeleton
of $\mathbf{Set}$ (for which I am happy to assume any necessary axiom
of choice), and take a one-element set $1$  that is not in $X$. Then
$1$ is a limit of the functor obtained by composing the unique functor
from the empty category to $X$ with
$X\subset\mathbf{Set}\stackrel{\operatorname{Id}}{\to}\mathbf{Set}$,
but $1$ has no lifting in $X$. 
So it seems that there are 4 possibilities:

The above Proposition 4.4.1, as stated, is wrong.  There is a
counter example where a monadic functor (for which the comparison
functor is not an isomorphism) does not create limits.  
The proof in ML-M covers just some of the cases, and for the other
cases it is not known if the assertion is true (namely, for a monadic
functor $G$ for which the comparison functor is not an isomorphism, it
is not known whether in general $G$ creates limits).
The proposition is correct because the comparison functor has some
additional special property (e.g., its object function must be
surjective whenever it is an equivalence).
(Most likely) I am wrong, and the quoted argument from Mac Lane--Moerdijk is
correct.

I would like to note that in Theorem 3.4.2 on p. 105 of Barr-Wells, it
is only claimed that monadic functors reflect limits.
Question
Which one of the above 4 possibilities is true?  In essence, my
question is:  If $G$ is monadic and the comparison functor is an equivalence
that is not an isomorphism, does $G$ create limits?

REPLY [2 votes]: Mac Lane-Moerdijk slipped up; they really should have said "reflects limits". Now it's true that the forgetful functor from the literal category of algebras to the underlying category creates limits (according to the definition of "creates" in CWM), but this notion doesn't transfer across equivalences, as you have observed. 
An example where the distinction is important is the statement that for a topos E, the power object functor P: E^{op} --> E is monadic. It wouldn't make much sense to say that this creates limits in Mac Lane's sense of the term.<|endoftext|>
TITLE: Characterizing the Local Langlands Correspondence
QUESTION [17 upvotes]: In the p-adic case, is there any hope for a set of conditions on the local Langlands correspondence which would make it unique? In the case of GL(n) this is provided by L and epsilon factors. For classical groups, can one make a precise statement (even conjecturally), using lifting to GL(n)?

REPLY [6 votes]: This answer will necessarily be full of conjecture, but I'll try to make things more concrete for some classical groups.
For a reductive group $G$, and smooth irreducible representation $(\pi, V)$ of $G$, and L-group homomorphism $\sigma: {}^L G \rightarrow {}^L GL_n$, one may conjecturally associate a local L-function and $\epsilon$-factor $L(\pi, \sigma, s)$ and $\epsilon(\pi, \sigma, s)$.
As $\sigma$ varies over all L-group representations of ${}^L G$, this family of L-functions and $\epsilon$-factors will uniquely determine the (conjectural) Langlands parameter 
$$\phi_\pi: W' \rightarrow {}^L G,$$
where $W'$ is the Weil-Deligne group of the $p$-adic field.  Namely, there will be a unique $\phi_\pi$ such that
$$L(\sigma \circ \phi_\pi, s) = L(\pi, \sigma, s), \epsilon(\sigma \circ \phi_\pi, s) = \epsilon(\pi,\sigma,s)$$
where the L and $\epsilon$ on the left sides are Artin L-functions (adapted to the Weil-Deligne group instead of Galois group in a simple way).
In this way, the L- and $\epsilon$-factors determine the Langlands parameter of a smooth irrep of $G$.  This partitions the smooth irreps of $G$ into the "L-packets".  
Now, for classical groups, one can get away with less information (or at least more computable information) in many cases.  I refer to the recent paper of Wee Teck Gan and Takeda, "The Local Langlands Correspondence for Sp(4)" (avaialble on Wee Teck's home page, for example), where they prove that the Langlands parameterization is uniquely determined by L- and $\epsilon$-factors (and $\gamma$-factors) of pairs (i.e. twists by $GL(n)$ with $n \leq 3$) for generic representations and a fact about Plancherel measure for certain nongeneric representations (this fact is analogous to a fact about Shahidi-type L-functions, which are only understood in the generic case).
For any classical group, I imagine one (probably not me) could come up with an analogous set of twists, etc.. to find a reasonable set of L-functions and $\epsilon$-factors, and Plancherel measures to determine the Langlands parameterization.  Perhaps this is done by others who prove functoriality for classical groups (P.S., Cogdell, Rallis, Jiang, Kim, Shahidi, and many others). 
Refining the local Langlands conjectures (I appreciate Vogan's survey most here), one should be able to parameterize the contents of an L-packet (with parameter $\phi: W' \rightarrow {}^L G$) by the representations of a finite group $S_\phi$.  This finite group has a general description, which in some nice cases (split, adjoint perhaps, I don't recall), is that $S_\phi$ is the component group of the centralizer of the image of $\phi$ in the dual group $\hat G$.)
So to pin down the Langlands correspondence, one must characterize this parameterization of the elements of an L-packet as well.  For this, it seems that characters are key -- there is certainly no telling the difference with L-functions or $\epsilon$-factors.  I believe this parameterization of the elements of an L-packet should be uniquely determined by the stability of an associated sum of character distributions of elements of the L-packet.  I tend to avoid characters, but maybe the work of DeBacker and Reeder -- who prove such a stability result for depth zero representations -- is a good place to start.<|endoftext|>
TITLE: Equality vs. isomorphism vs. specific isomorphism
QUESTION [49 upvotes]: This question prompted a reformulation: 
What is a really good example of a situation where keeping track of  isomorphisms leads to tangible benefit?
I believe this to be a serious question because it actually is oftentimes a good idea casually to identify isomorphism classes. To bring up an intermediate-level example I've alluded to often, consider  the classification of topological  surfaces. When I explain it to students, I do somewhat consciously write equalities as I manipulate one shape into another homeomorphic one. I even do it rather quickly to encourage intuitive associations that are likely to be useful. In any case, for arguments of that sort, it would be really tedious, and probably pointless, to write down isomorphisms with any precision.
Meanwhile, at other times, I've also  joined in the chorus of criticism that greets the conflation of equality and isomorphism.
The problem is it's quite challenging to come up with really striking examples where this care is rewarded. Let me start off with a somewhat specialized class of examples. These come from descent theory. The setting is a map $$X\rightarrow Y,$$ which is usually submersive, in some sense suitable to the situation. You would like criteria for an object $V$ lying over $X$, say a fiber bundle, to arise as a pull-back of an object on $Y$. There is a range of formalism to deal with this problem, but I'll just mention two cases. One is when $Y=X/G$, the orbit space of a group action on $X$. For $V$ to be pulled-back from $Y$, we should have $g^*(V)\simeq V$ for each $g\in G$. But that's not enough. What is actually required is that there be a collection of isomorphisms $$f_g: g^*(V)\simeq V$$ that are compatible with the group structure. This means something like $$f_{gh}=f_g\circ f_h,$$ except you have to twist in an obvious way to take into account the correct domain. So you see, I have at least to introduce notation for the isomorphisms involved  to formulate the right condition. In practice, when you want to construct something on $Y$ starting from something on $X$, you have to specify the $f_g$  rather precisely. 
Another elementary case is when $X$ is an open covering $(U_i)$ of $Y$. Then an object on $Y$ is typically  equivalent to a collection $V_i$ of objects, one  on each $U_i$, but with additional data. Here as well, $V_i$ and $V_j$ obviously have to agree on the intersections. But that's again not enough. Rather there should be a collection of isomorphisms  $$\phi_{ji}: V_i|U_i\cap U_j\simeq V_j|U_i\cap U_j$$
that are compatible on the triple overlaps:
$$\phi_{kj}\circ \phi_{ji}=\phi_{ki}.$$ Incidentally, for something like a vector bundle, since any two of the same rank are locally 'the same,' it's clear that keeping track of isomorphisms will be the key to the transition from collections of local objects to a global object. The formalism is concretely applied in situations where you can define some objects only locally, but would like to glue them together to get a global object. For a really definite example that comes immediately to mind, there is the determinant of cohomology for vector bundles on a family of varieties over a parameter space $Y$. Because a choice of resolution is involved in defining this determinant, which might exist only locally on $Y$, Knudsen and Mumford struggled quite a bit to show that the local constructions glue together. Then Grothendieck suggested the remedy of defining the determinant provisionally as a signed line bundle, which then allowed them to nail down the correct $\phi_{ji}$. These days, this determinant is a very widely useful tool, for example, in generating line bundles on moduli spaces.
I apologize if this last paragraph is a bit too convoluted for non-specialists. Part of my reason for writing it down is to illustrate that my main examples for bolstering the 'keep track of isomorphisms' paradigm are a bit too advanced for most undergraduates.
So, to conclude, I'd be quite happy to hear of better examples. As already suggested above, it would be nice to have them be accessible but substantively illuminating. If you would like to discuss, say, different bases for vector spaces, it would be good if the language of isomorphism etc. clarifies matters in a really obvious way, as opposed to a  sets-and-elements exposition.
Added: Oh, if you have advanced examples, I would certainly like to hear about them as well.
Added: I see now there are three levels at least to distinguish:
Regarding objects as equal vs. regarding them as isomorphic vs. paying attention to  specific isomorphisms.
I somehow conflated the two transitions in the course of asking the question. Of course I'm happy to see good examples illustrating the nature of either, but I'm especially interested in the second refinement.
Added yet again:: I'm grateful to everyone for contributing nice examples, and to Urs Schreiber who put in some effort to instruct  me over at the n-category cafe. As I mentioned to Urs there, it would be especially nice to see examples of the following sort.

One usually thinks $X=Y$;
A careful analysis encourages the view $X\simeq Y$;
This perspective leads to genuinely new insight and benefit. 

Even better would be if some specific knowledge of the isomorphism in 2. is important. Of course, more than two objects might be involved. I was initially hoping for some input from combinatorics, with the emphasis on 'bijective proofs' and all that. Anything?
Added, 14 May:
OK, I hope this will be the last addition. Because this question flowed over to the n-category cafe, I ended up having a small discussion there as well. I thought I'd copy here my last response, in case anyone else is interested.
n-cafe post:
I suppose it's obvious by now that I'm using a specific request to drive home the need for 'small but striking examples' in favor of category theory.
Last fall, Eugenia Cheng told me of a visit to some university to give a colloquium talk. The host greeted her with the observation that he doesn't regard category theory as a  field of research. OK, he was probably a bit extreme, but milder versions of that view are quite common. Now, one possible response is to regard all such people as unreasonable and talk just to friends (who of course are the reasonable people!). This is not entirely bad, because that might be a way to buy time and gain enough stability to eventually prove the earth-shattering result that will show everyone! Another way is to take up the skepticism as a constructive everyday challenge. This I suppose is what everyone here is doing at some level, anyways.
Other than the derived loop space, which is not exactly small, Urs' examples are all of the simple subtle sort that can, over time, contribute to a really important change in scientific outlook and maybe even the infrastructure of a truly glorious theory. For example, I agree wholeheartedly about the horrors of the old tensor formalism. But it's not unreasonable to ask for more striking accessible evidence of utility when it comes to the current state of category theory.
The importance of small insights and language that gradually accumulate into  the edifice of a coherent and powerful theory is the usual interpretation of Grothendieck's 'rising sea' philosophy. However, the process is hardly ever smooth along the way, especially the question of acceptance by the community. I'm not a historian, but I've studied arithmetic geometry long enough to have some sense of the changing climate surrounding etale cohomology theory, for example, over the last several decades. The full proof of the Weil conjectures took a while to come about, as you know. Acceptance came slowly with many bits and pieces sporadically giving people the sense that all those subtleties and abstractions are really worthwhile. Fortunately, the rationality of the zeta function was proved early on. However, there was a pretty concrete  earlier proof of that as well using $p$-adic analysis, so I doubt it would have been the big theorem that convinced everyone. One real breakthrough came in the late sixties when Deligne used etale cohomology to show that Ramanujan's conjecture on his tau function could be reduced to the Weil conjectures. There was no way to do this without etale cohomology and the conjecture in question concerned something very precise, the growth rate of natural arithmetic functions. This could even be checked numerically, so impressed people in the same way that experimental verification of a theoretical prediction does in physics. Clearly something deep was going on. Of course there were many other indications. The construction of entirely new representations of the Galois group of $\mathbb{Q}$ with very rich properties, the unification of Galois cohomology and topological cohomology, a clean interpretation of arithmetic duality theorems that gave a re-interpretation of class field theory, and so on.
For myself, being a fan of you folks here, I believe this kind of process is going on in category theory. But I don't think you have to be too unreasonable to doubt it. In a similar vein, I don't agree with Andrew Wiles' view that physics will  be irrelevant for number theory, but also think his pessimism is perfectly sensible.
I think I'm trying to make the obvious point that the presence of pessimists can be very helpful to the development of a theory, in so far as the optimists interact with them in constructive ways. I haven't been coming to this site much lately, because the bit of internet time I have tends to be absorbed by Math Overflow. But I did catch David's recent post on Frank Quinn's article, which ended up as a catalyst for my MO question.
At the Boston conference following the proof of Fermat's last theorem, I've been told Hendrik Lenstra said something like this: 'When I was young, I knew I wanted to solve Diophantine equations. I also knew I didn't want to represent functors. Now I have to represent functors to solve Diophantine equations!' So should we conclude that he was foolish to avoid  representable functors for so long? I wouldn't.
This response to the MO question brings up the importance of knowing the specific isomorphism between some Hilbert spaces given by the Fourier transform. This is an excellent example, especially when we consider how it relates to the different realizations of the representations of the Heisenberg group and the attendant global issues, say as you vary over a family of polarizations. But I couldn't resist recalling Irving Segal's insistence that 'There's only one Hilbert space!' Obviously, he knew, among many other things, the different realizations of the Stone-Von-Neumann representation as well as anyone, so you can take your own guess as to the reasoning behind that proclamation. He certainly may have lost something through that kind of philosophical intransigence. But I suspect that he, and many around him, gained something as well.

REPLY [5 votes]: I'm surprised no one gave this answer. Every Banach space isometrically embeds in its second dual by a natural map. If this map is surjective, the Banach space is said to be reflexive.
James gave an example of a Banach space which is isometrically isomorphic to its second dual but is not reflexive.<|endoftext|>
TITLE: Simple inequality in C*-algebras
QUESTION [7 upvotes]: Sorry the title is a bit vague.  Let A be a C*-algebra, and let x and y be positive elements in A.  Is it true that $$ \|x-y\|^2 \leq \|x^2-y^2\|? $$
Well, yes.  But the proof I have is a bit of a hack, so I wonder if anyone has a "nice" proof, or a reference?
Aside: if $A=C_0(X)$ then this reduces to the inequality $(a-b)^2 \leq |a^2-b^2|$ for non-negative real numbers a and b.
Update: Jonas points me to http://www.springerlink.com/content/j4756m418220644r/ where Kittaneh has a proof pretty similar to what I had in mind (unpack the proof of Theorem 1).  I guess I was interested in whether this sort of thing was standard (if I looked in the right textbook) or if it was a bit of a curiosity.  I think the latter seems more likely...

REPLY [5 votes]: Okay, so actually, decoding Phillip's work gives what I think is a very nice proof.
Let x and y be positive.  Let $\epsilon=\|x-y\|$ so as $x-y$ is self-adjoint, it follows that
$x \leq y+\epsilon 1$.  What seems to be a very standard inequality is that then $x^{1/2} \leq (y+\epsilon 1)^{1/2}$.  We then claim that $(y+\epsilon 1)^{1/2} \leq y^{1/2} + \epsilon^{1/2} 1$.  This follows by working in the commutative C*-algebra generated by y and 1, and using that $(s+t)^{1/2} \leq s^{1/2} + t^{1/2}$ for positive real numbers s and t.  So $$x^{1/2} - y^{1/2} \leq \epsilon^{1/2} 1$$and by symmetry, also $y^{1/2}-x^{1/2}\leq\epsilon^{1/2}1$.  Thus $\|x^{1/2}-y^{1/2}\|\leq\epsilon^{1/2}=\|x-y\|^{1/2}$ as required.<|endoftext|>
TITLE: Splitting of a division algebra with an involution of second kind
QUESTION [5 upvotes]: Let $k$ be a field, $K/k$ a separable quadratic extension,
and $D/K$ a central division algebra of dimension $r^2$ over $K$
with an involution $\sigma$ of second kind
(i.e. $\sigma$ acts non-trivially on $K$ and trivially on $k$).
Does there exist a field extension $F/k$ such that $L:=K\otimes_k F$
is a field, and $D\otimes_K L$ splits (i.e. is isomorphic to the matrix algebra $M_r(L)$ over $L$)?
Motivation: Let $h\in D$ be a Hermitian element ($h^\sigma =h$), and let
 $G$ be the $k$-group with
$G(k)=${$g\in D^\times\ | \ ghg^\sigma=h$}.
I want to find a field extension $F/k$ such that $G\times_k F$
is a unitary group over a field $L$ (and not over a division algebra over $L$).

REPLY [3 votes]: You find the answer to the question in "The Book of Involutions". More precisely, in the first  5 lines of the proof of Lemma 10.27.<|endoftext|>
TITLE: Quantitative questions about the size of a finite epsilon net
QUESTION [5 upvotes]: Let $X$ be a metric space, and let $U \subset X$ be any set.  A finite set $N = N(\epsilon) \subset U$ is called a finite $\epsilon$-net of $U$ if every point of $U$ is at most a distance of $\epsilon$ from some point of $N$.
It is easy to show that if $U$ is compact, then for any $\epsilon>0$, a finite $\epsilon$-net exists.  I am interested in the behavior of the function $|N(\epsilon)|$ as $\epsilon$ goes to zero.  If there are answers in this generality, great.  I am mostly interested in the particular case where $U$ is also convex, and where $X$ is also an infinite dimensional topological vector space, but any answers are of course welcome.
Edit: I'm also interested in weakening the notion of $\epsilon$-net, so instead of requiring every point of $U$ to be close to a point in $N$, we could require every point of $U$ to be close to a point in the convex hull of $N$.  The motivation for this comes from looking at objects which are "convexly compact"; this means (in a metric space) that given any sequence $(f_n)$ in $U$, there exist $g_j \in \text{conv}(f_j,f_{j+1},\ldots)$ such that $g_j \rightarrow g$ in $U$.

REPLY [5 votes]: This is a huge subject. The minimum sizes of $\epsilon$-nets of compacts in linear spaces 
were studied by Kolmogorov and his school. They showed that in general there are no good bounds for this quantity for convex compacts in infinite-dimensional Banach spaces. However,
reasonable estimates can be obtained in some interesting specific cases.
Let $|N_\epsilon(U)|$ be the minimum size of an $\epsilon$-net for the set $U$. The value 
$\mathcal N_\epsilon(U)=\log_2 |N_\epsilon(U)|$ is called the $\epsilon$-entropy of the set $U$.
The following results are due to Kolmogorov and Tikhomirov.
Theorem 1. Let $\phi(\epsilon)$ goes to $+\infty$ (and monotonically increases) as
$\epsilon\to 0$. Assume that $X$ is an infinite-dimensional Banach space. Then there is a compact $K\subset X$ such that
$$ \mathcal N_\epsilon(K)\succeq\phi(\epsilon).$$ 
Theorem 2. Assume that $X$ is an infinite-dimensional Banach space. Let $K\subset X$ be a convex set which is not contained in any finite-dimensional subspace of $X$. Then
$$ \mathcal N_\epsilon(K)\succeq\left(\frac{1}{\epsilon}\right)^n$$
for any $n\in \mathbb N$.<|endoftext|>
TITLE: A variation on "Hearing the shape of a drum" for polytopes.
QUESTION [8 upvotes]: Let $\varphi:\mathcal S^{d-1}\longrightarrow \mathbb R_{>0}$ be a strictly positive function describing the boundary $\varphi(\mathbf x)\mathbf x,\mathbf x\in\mathbb S^{d-1}$ of a $d-$dimensional closed convex set $C$ with barycenter at the origin.
We consider the decomposition $\varphi=\oplus_{\lambda\in \mathbf{Spec}(\Delta)}\varphi_\lambda$ into eigenvectors associated 
to distinct eigenvalues of the Laplacian.
To what extend does the sequence $\parallel \varphi_\lambda\parallel_{\lambda\in\mathbf{Spec}(\Delta)}$ determine $C$?
(This is certainly the case for a sphere of radius $\rho$ since then $\varphi=\varphi_0=\rho$.) 
Are there examples of two non-isometric convex sets (or even two non-isometric polytopes) for which the two corresponding sequences of norms coincide?
Moreover, the convexity of $C$ gives probably constraints for the norms $\parallel \varphi_\lambda\parallel$ since spherical harmonics associated to high eigenvalues wiggle a lot
and have thus to be involved with coefficients that are small with respect to
$\parallel \varphi_0\parallel$. What are these constraints?

REPLY [4 votes]: The short answer is that there are no particular constraints on the spectral decomposition of the function $\varphi$, as long as a basic convexity condition is satisfied. 

Lemma..Assume that $\varphi\in C^2(\mathbb S^{d-1},\mathbb R)$ satisfies for all $x\in \mathbb S^{d-1}$, $i$, $j=1,\dots,d$, and some $r>0$ the inequalities 
  $$|\varphi(x)|+\sqrt{2d^3}|D^i\varphi(x)|+2d^3|D^{ij}\varphi(x)|< r, \qquad\qquad (1)$$
  where
  $$D^i\varphi(x)=\left(\frac{\partial \varphi(u/|u|)}{\partial u_i}\right)_{u=x},\quad
D^{ij}\varphi(x)=\left(\frac{\partial^2 \varphi(u/|u|)}{\partial u_i\partial u_j}\right)_{u=x},\quad u\in \mathbb R^d,\ x\in \mathbb S^{d-1}.$$
  Then $\varphi_r(x)= r+\varphi(x)$ is the support function of a compact convex set in $\mathbb R^d$.

Condition (1) can be translated into constraints on the harmonic expansion of the support function.

Theorem. Let $P_i=P_i(x)$, $x\in \mathbb S^{d-1}$  denote a spherical harmonic of order i.

There exists a constant $c_0$ such that for all $c\geq c_0$
  $$c+P_{n_1}+\dots+P_{n_m} $$
  is the support function of a compact convex set in $\mathbb R^d$.
The subset of those compact convex sets whose support functions are finite sums of spherical harmonics is dense (with respect to the Hausdorff metric) in the set of all convex bodies in $\mathbb R^d$.
If $C\subset \mathbb R^d$ is a compact convex set whose principal radii of curvature exist and srtictly positive and whose support function
  $\varphi=\sum\limits_{n=1}^{\infty} P_n $
  belongs to the class $C^k(\mathbb S^{d-1},\mathbb R)$ with $k>\frac{d+4}{2}$, then there is 
  an $n_0$ such that the partial sum
  $$P_{0}+P_1\dots+P_{n}  $$
  is the support function of a convex body in $\mathbb R^d$ for any $n>n_0$.


Reference. Geometric Applications of Fourier Series and Spherical Harmonics by  H. Groemer.<|endoftext|>
TITLE: sub-tori of a torus, generated by 1-dimensional subgroup
QUESTION [10 upvotes]: Ok the question is pretty dumb: suppose you have a torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ and a vector $\bar{v}=(v_1,\ldots,v_n)\in\mathbb{R}^n$.
Consider the torus $T_{\bar{v}}$ given by the closure of the one parameter group in $T^n$ generated by $\bar{v}$:
$T_{\bar{v}}=\overline{ \{t\cdot\bar{v}\mod\mathbb{Z}^n|\phantom{a}t\in\mathbb{R}\}}$
My questions are:

what is the dimension of $T_{\bar{v}}$?
How can i find a basis of vectors spanning the tangent space of $T_{\bar{v}}$ at the origin?

My guess for question 1. is $\dim T_{\bar{v}}=\dim_{\mathbb{Q}}\langle v_1,\ldots, v_n\rangle$, but i don't know what the answer to question 2. can be.
Thanks!

REPLY [5 votes]: The key to solving both problems is the use of the following two facts: 1) Any
closed subgroup of $T^n$ is the intersection of the kernels of characters of
$T^n$, i.e., continuous group homomorphisms $T^n \rightarrow S^1$. 2) Any
continuous homomorphism $T^n \rightarrow S^1$ is of the form $(\overline
x)\mapsto e^{2\pi i x\cdot m}$ for a unique $m\in\mathbb Z^n$. Hence, the
closure of $T_{\overline \nu}$ is the intersection of the kernels of the
characters corresponding to $m$ for which $\nu\cdot m\in\mathbb Z$. Picking a
basis $m_1,\dots,m_k$ of the group of such $m$ gives a surjection $T^n
\rightarrow T^k$ for which $T_{\overline\nu}$ is the kernel ($T_\nu$ is not
necessarily a torus as it might not be connected but it doesn't change
anything). By the above this map is just given by an $n\times k$ integer matrix
specified by $m_1,\dots,m_k$. The tangent map at the origin is then obtained by
regarding this matrix as a real matrix and thus the tangent space of
$T_{\overline \nu}$ is the null space of this matrix.
In particular this gives that the dimension of $T_{\overline \nu}$ is equal to
$\dim_{\mathbb Q}\langle1,\nu_1,\nu_2,\ldots,\nu_n\rangle -1$. This is off by one
from your guess if $1$ is in the span of of the $\nu_i$ but equal to it if it isn't.
[[Added]]
I misread the question and the above is for the closed subgroup generated by $\overline\nu$ while the question was about the closure of the $1$-parameter subgroup generated by it. To answer the question everything works the same only the condition is that $r\nu\cdot m\in\mathbb Z$ for all real $r$ which gives $\nu\cdot m=0$ and indeed the dimension is $\dim_{\mathbb Q}\langle\nu_1,\nu_2,\ldots,\nu_n\rangle $.<|endoftext|>
TITLE: Ordinary cohomology of stacks
QUESTION [14 upvotes]: Let $\mathbf{X}$ be a stack over $Top$ (a lax sheaf of groupoids, or some such thing). If it admits a surjective representable map $F \to \mathbf{X}$ then one can form the iterated fibre product to get a simplicial space $F_\bullet$, and its realisation $X$ is the homotopy type of $\mathbf{X}$, which comes with a (homotopy class of?) map $X \to \mathbf{X}$.
It seems that the reasonable thing to call the singular cohomology of $\mathbf{X}$ is the singular cohomology of its homotopy type, because the homotopy type is sufficiently unique for this to be well-defined.
On the other hand, consider the composition of functors
$$\mathbf{h}^i : Top \overset{H^i(-;\mathbb{Z})}\to AbGp \to Set \overset{inc}\to Gpd$$
where the middle functor is forgetful, and the last one is the inclusion of sets as groupoids with no non-identity morphisms. 
The stack $\mathbf{X}$ gives another functor $Top \to Gpd$, and one may consider the set $H^i$ of natural transformations $\eta : \mathbf{X} \to \mathbf{h}^i$. This set has the structure of an abelian group as one may add pointwise (as $\mathbf{h}^i$ factors through abelian groups).
Given any such $\eta$, we may apply it to the unique homotopy class of maps $X \to \mathbf{X}$ to obtain $\eta(X \to \mathbf{X}) \in H^i(X;\mathbf{Z})$ a cohomology class on the homotopy type. Any $Y \to \mathbf{X}$ factors up to homotopy through $X$, and so $\eta(Y \to \mathbf{X})$ is obtained by pullback from $\eta(X \to \mathbf{X})$. Thus it seems to me that the group $H^i$ is naturally isomorphic to $H^i(X;\mathbf{Z})$.
My first question, at last, is: is the group $H^i$ the correct notion of the cohomology of the stack $\mathbf{X}$, which happens to coincide with the cohomology of its homotopy type? Would it still be a reasonable definition on some terrible stack that does not admit an atlas?
Secondly: how does one do homology like this?
Thirdly: one can do the above over $Diff$ instead of $Top$, and use de Rham cohomology in the definition of $\mathbf{h}^i$. Then it seems the group one produces, call it $H^i_{dR}$ now, still makes sense, but the homotopy type of a stack over $Diff$ is not necessarily itself a manifold and does not necessarily have a de Rham theory. Is this a reasonable thing to do? Why is it not done this way (for example by Behrend)?
Fourthly: if, as I suspect, all this is known, where can I find it?

REPLY [5 votes]: Does this work for de Rham cohomology? 
Sure.  While the homotopy type of a stack in $Diff$ doesn't exactly have a de Rham complex, it does have a couple of things that work perfectly as substitutes.  Behrend uses the Cech-de Rham double complex from a particular choice of atlas.  The homotopy type of the stack is the geometric realisation of a simplicial manifold, and the Cech-de Rham double complex is made from the simplicial object in cochain complexes by taking the de Rham complex levelwise, so it really is the de Rham model for the simplicial manifold.
Alternatively, the homotopy type of the stack has a real homotopy type (which can be defined, for example, by taking Sullivan's PL forms with real coefficients).  On manifolds the  cohomology of the real homotopy type is of course naturally isomorphic to the de Rham cohomology.  Thus on stacks over $Diff$ the Cech-de Rham complex is a model for the real homotopy type.
What about K-theory?
This wasn't one of your questions, but as long as we are varying the cohomology theory, it's worth mentioning this.  The K-theory of a stack really is different from the K-theory of its homotopy type.  The simplest example is the stack $*/G$.  The K-theory of the stack is the G-equivariant K-theory of a point, but K-theory of the homotopy type is K(BG), which is of course the completion at the augmentation ideal by the Atiyah-Segal theorem.
I think that defining K-theory in terms of natural transformations $hom(-,\textbf{X}) \to K(-)$ just gives you the K-theory of the homotopy type.<|endoftext|>
TITLE: Correspondence between functions on a set and "states" on its power set
QUESTION [6 upvotes]: Let $L$ be the poset (ordered by set inclusion) that is the power set of some set $X$.
A state is a function $s:L \rightarrow [0,1]$ satisfying
i) for {$p_1,p_2,...$}, $p_i \in L$ a
    pairwise orthogonal (i.e. $p_i \leq p_j'$
    where $a'$ is the complement of $a$)
    countable sequence, $\bigvee_i p_i$
    exists, and $s(\bigvee_i p_i) =
    \sum_i s(p_i)$.
ii) s(X) = 1.
also consider functions $f:X \rightarrow [0,1]$.
Then for $X$ countably infinite, every state $s$ is in one to one correspondance to a function $f$ by the following argument (skip the next three paragraphs if you are OK with this):

As $L$ is atomistic, we have for arbitrary $p \in L$, $p = \bigvee_i a_i$ ($= \bigcup_i a_i$ as the join of the poset corresponds to the union of subsets) where $a_i \in L$ are atoms (which are pairwise orthogonal). 
The atoms $a_i$ are in one to one correspondence to the elements of $X$ such that, given $f:X \rightarrow [0,1]$, we can define $f$ on the set of atoms via $f(a) = f(x_a)$ where $x_a$ is the element of $X$ associated to the atom $a$.
Then for a given state $s$ and arbitrary $p \in L$, $s(p) = s(\bigvee_i a_i) = \sum_i s(a_i)$. So $s$ is determined by its values on the atoms and we can associate a state to a function $f$ by setting for all atoms $a$, $s_f(a) = f(a)$. This is bijective. (End of argument.)

Questions:
1) Now for $X = R^n$ (or uncountable) is there a similar correspondence? 
2) Or do I need to adapt condition i) ? 
What I fail to show in the uncountable case, is whether condition i) is strong enough to ensure that any state on the power set of such an $X$ is uniquely determined by its values on the atoms. 
I hope this question is worthy of a response, it is my first one and I hesitated for the last 4 days.

REPLY [3 votes]: Your correspondence is equivalent to the existence of a real-valued measurable cardinal, a large cardinal concept equiconsistent with the existence of a measurable cardinal. 
First, note that if $\kappa$ is a measurable cardinal, then there is a 2-valued measure $\mu$ on $P(\kappa)$ which is not only countably-additive but $\kappa$-additive, in the sense the measure of the union of fewer than $\kappa$ many disjoint sets is the sum of the measures. For this measure, every set gets measure either 0 or 1, and so there are no disjoint sets of positive measure, and also every singleton gets measure 0. So it is an instance of a violation of your correspondence. 
More generally, if $\kappa$ is a real-valued measurable cardinal, then there is a real-valued $\kappa$-additive measure $\mu$ on $P(\kappa)$ giving measure 0 to singletons. In particular, such a measure would be countably additive, and it would not correspond to function in your sense. 
Conversely, suppose that there were a countably additive real-valued measure $\mu$ on $P(X)$ for some set $X$. If this measure does not correspond to a function, let's subtract from it the sum measure on singletons, to arrive without loss of generality at a measure that gives measure zero to singletons, but positive measure to the whole space. In this case, let $\kappa$ be the additivity of $\mu$, the largest cardinal such that the $\mu$ measure of any less-than-$\kappa$ sized disjoint union is equal to the sum of the measures individually. In this case, there is a set $Y\subset X$ of positive measure and a $\kappa$ partition of $Y=\cup_{\alpha\lt\kappa} Y_\alpha$ such that each $Y_\alpha$ has measure $0$. We may now define a $\kappa$-additive measure on $P(\kappa)$ by $\mu_0(I)=\Sigma_{\alpha\in I}\mu(Y_\alpha)$. Thus, $\kappa$ is a real-valued measurable cardinal. 
So your question is equivalent to the existence of a real-valued measurable cardinal. Such a hypothesis is equiconsistent with the existence of a measurable cardinal. 
The particular case when the set $X$ has size continuum $c$ corresponds to the situation where $c$ is a real-valued measurable cardinal. This implies a strong failure of the Continuum Hypothesis, since in this case $c$ would be weakly inaccessible. It is equivalent to the existence of a countably-additive extension of Lebesgue measure measuring all sets. (Such an extension cannot be translation invariant by Vitali.)<|endoftext|>
TITLE: Products of Baire spaces
QUESTION [13 upvotes]: I could not find any references about this fact. I apologize if this is completely trivial, but is the product of two Baire spaces, or for that matter of finitely many of them a Baire space? Now is a countable product of Baire spaces a Baire space? 
What about an uncountable product of Baire space? This fact seems to be treated in an article I can't access.
It seems to work for the Sorgenfrey line: $S$ is a Baire space and $SxS$ is a Baire space since if you consider the diagonal $A$={($-x$,$x$): $x\in S$ and $x\in$ℚ} then this is a closed discrete subspaces which is the union of countably many closed nowhere dense sets but its interior is empty.
Is that true?
Thx

REPLY [2 votes]: There are two metric Baire spaces, whose product is not Baire; a counterexample is given in the slides An absolute barely Baire space, L.F. Aurichi and G.A.A. Medina, 2020. We present Fleissner and Kunen's proof with applications of Clubs and stationary sets in $\omega_{1}$.
Fleissner, W. G.; Kunen, K., Barely Baire spaces, Fundam. Math. 101, 229-240 (1978). ZBL0413.54036.<|endoftext|>
TITLE: What do people mean by "subcategory"?
QUESTION [13 upvotes]: Mac Lane defines a subcategory as a subset of objects and a subset of morphisms that form a category. But the first rule of category theory is that you do not talk about equality of objects. Up to equivalence, the definition becomes a faithful functor. This is a useful concept, but I don't think it fits the name. I don't want groups to be a subcategory of sets!
This is not a question about aesthetics, but about usage. I don't think people tend to use Mac Lane's definition. Maybe they're just wrong, but I'd like to know if there is another definition which fits the usage better. All the time I see people say things like "we may assume that our subcategory contains every object isomorphic to an object of the subcategory." I guess we can expand to an equivalent subcategory to achieve this, but we probably have to choose how the objects are isomorphic (though it may be easier if to change the ambient category). This is a much more natural thing to do (and safer) if the subcategory is full, or at least contains all the automorphisms of its objects. This leads me to suspect that people are assuming or thinking of some stronger definition than faithful.
Do people tend to mean the official definition? or do they also require full? containing all the automorphisms? Are there other useful intermediate notions?

REPLY [3 votes]: Upon request, I will clarify -- and partially take back! -- my earlier comment.
What I find completely unobjectionable is Mac Lane's definition of a subcategory $\mathcal{D}$  of a category $\mathcal{C}$ as being given by subclasses of objects and morphisms which forms a category under the induced composition.  I don't see what else you would want a subcategory to be.  I do agree that the notion of "subcategory" is not one of the more useful categorical concepts I know, and it even has some potential to be evil in the sense that modern categorists use the word.  (Surely it would be more in the spirit of things to talk about a functor from $\mathcal{C}$ to $\mathcal{D}$ which satisfies certain "injectivity" properties.)  
Now let's return to the statement "I don't want Groups to be a subcategory of Sets".  In my comment I said that I did want this, but I don't now know why I said that: I think I must simply have been confused.  Indeed, it is not obvious to me that this definition makes Groups a subcategory of Sets, at least not in any unique or benign way.  
If you asked me to spell out the most evident categorical relationship between sets and groups, I would first of all point to the category NonEmptySets -- now that's a subcategory of Sets! -- and then the "forgetful" functor from Groups to NonEmptySets.  This functor is (at least assuming the Axiom of Choice) surjective: every nonempty set is the underlying set of some group.  But most sets can be endowed with a group law in multiple (usually nonisomorphic) ways, so this is not an "inclusion functor".  
(Even the other way around, namely the free group functor from Sets to Groups seems not to quite make Sets into a subcategory of groups, because the class of sets is not a subclass of the class of groups.)
Maybe you are thinking of doing something tricky: defining a group to be an ordered pair [identifying ordered pairs with sets in one of the usual -- silly! -- ways] $(S,\circ)$ where $S$ is a set and $\circ$ is a subset of $S \times S \times S$ satisfying certain axioms.  (Note that this is definitely incompatible with the above way of thinking about groups as having -- but not being -- an "underlying set".)  But isn't this especially evil?
Comments more than welcome.<|endoftext|>
TITLE: Why are rational Cherednik algebras so... rational?
QUESTION [7 upvotes]: The real question is both more serious and somewhat longer than the title.
For the definition of the rational Cherednik algebra attached to a complex reflection group $W$, see for instance 5.1.1 of Rouquier's paper here.  It is a flat family of algebras depending on some parameters $h_{H,j}$ indexed by pairs consisting of a $W$-orbit of reflecting hyperplanes $H$ and an integer $0 \leq j \leq e_H-1$, where $e_H$ is the order of the pointwise stabilizer of $H$.  
Various features of the structure of the Cherednik algebra turn out to be governed by systems of linear equations with rational coefficients in the parameters.  For instance, in this paper Dunkl and Opdam show that the polynomial representation is irreducible exactly if the parameters avoid a certain locally finite system of rational hyperplanes, and in this paper a couple of guys show that it is Morita equivalent to its spherical subalgebra off a certain set of rational hyperplanes.  In this paper, Etingof shows (for real reflection groups) that the set of parameters for which the irreducible head of the polynomial representation is finite dimensional is a set described by linear conditions with rational coefficients on the parameters.
Every time someone discovers the set of parameters for which the rational Cherednik algebra satisfies some reasonable properties, it winds up being described linearly in the parameters with rational coefficients.  Why?
(I'm asking for a conceptual reason---in each case I mentioned I know the proof.  For instance, I'd love to know an a priori proof that the set of parameters where the RCA is not Morita equiv. to its spherical subalgebra is a finite union of rational hyperplanes, without necessarily giving the union explicitly)

REPLY [6 votes]: The phenomenon seems the same as for affine Kac-Moody algebras, where rational level is where everything special happens, or quantum groups, where roots of unity are the exceptional locus (and of course these examples are related). The parameter for Cherednik algebras is an additive/Lie algebra type parameter, like the KM level, as opposed to the exponentiated version as in the quantum groups case. If you mod out by the action of translation functors, you're asking why finite order points of the parameter are special. I don't know that I can give a completely uniform answer, but it certainly seems reasonable.
For example in type A, modules for Cherednik algebras are realized using twisted D-modules on some stack. The parameter for twisting is an additive one, but the abstract category of twisted D-modules depends only on this parameter mod integral translations (hence translation functors). At integral points (or rather "the" integral points) many special things happen -- there are geometric obstructions to the existence of objects with various supports, and these obstructions vanish integrally -- which is why say category O for Lie algebras becomes much bigger integrally. More generally twisted D-modules can be described as sheaves on a gerbe, which depends only on the twist mod integers. If (and certainly only if) the parameter is rational - ie the twist is a torsion element - then you might expect to represent your gerbe by an Azumaya algebra, or equivalently to have finite rank twisted sheaves. I would imagine this general type of phenomenon is behind the results you mention, though I haven't thought about the specifics. But in any case this is a geometric phenomenon about categories of twisted D-modules in general, and as we know "basically all interesting categories of representations are some categories of twisted D-modules" so this is quite general.<|endoftext|>
TITLE: Generalized quadratic Gauss sums
QUESTION [6 upvotes]: I was wondering whether anyone knows how to approach the following two
generalizations of the quadratic Gauss sum:
Given integers r,s with gcd(r,s)=1 and integers a,b,N
$F(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $
$G(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{r})(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $
Note that removing the sine terms and the sign, setting a = 2, N = 4, r = 1 and s = prime gives the classical quadratic Gauss sum.
Some experimentation suggests that
$F(r,s,N,a,b) = 0$ for all integers b,N, r,s if a is even and (r,s) =1 and
$G(r,s,N,a,b) = 0$ for all a,b,N and r,s with (r,s) =1
Is there a good reason for these sums to vanish? Or a clean proof/reference?
Is it possible to evaluate F in the case a = 1? It seems to be non-zero then.
I tried reducing to the original Gauss sum by completing the square but 
this seems to get quite ugly.
More generally, do such Gauss-like sums have a more natural generalization
that turns up somewhere?
Thanks

REPLY [2 votes]: I think this should just be computations. Define as usual $e(x) = \exp(2 \pi i x)$. Define
$$
 f(r,s,N,a,b) = \sum_{w=1}^{rsa} e\left(\frac{N}{4 rs} w^2 + \left(\frac{1}{2s}+\frac{b}{2} \right) w\right)
$$
Then $F(a,r,s,N,a,b) = \frac{1}{2i } (f(r,s,N,a,b) - f(r,s,N,a, -b))$, at least if I didn't make any computational mistakes. 
Now, start with $a = 1$. Then you can use, the method described in http://en.wikipedia.org/wiki/Quadratic_Gauss_sum#Generalized_quadratic_Gauss_sums . 
Next, one has to understand what happens if one passes from $a$ to $a + 1$. For this compute
$$
 f(r,s,N,a+1,b) - f(r,s,N,a,b)
$$
I guess, the result should be of the form $z f(\hat r,\hat s, \hat  N, 1, \hat b)$ with $|z| = 1$.
Note: I began summing at $1$ on purpose, so one sums over the group $\mathbb{Z}_{rsa}$, which seems like the correct choice ...<|endoftext|>
TITLE: Borel(X) = \sigma(X') for X non-separable 
QUESTION [7 upvotes]: Let $X$ be a Banach space, $X' = \mathcal{L}(X, \mathbb{K})$ its dual space. Denote by $\mathcal{B}(X)$ the $\sigma$-algebra of Borel sets and denote by $\sigma(X')$ the $\sigma$-algebra which is generated by all sets of the form $u^{-1}(C)$ for $u \in X'$ and $C \in \mathcal{B}(\mathbb{K})$.
For $X$ separable we have that
$\mathcal{B}(X) = \sigma(X')$  (*)
see e.g. "Gaussian measures in Banach spaces" by Hui-Hsiung Kuo, p. 74 - 75.
Now the author of this book does not bother to discuss the case of $X$ non-separable.
In [1] is a halfway believable counterexample for $X = \ell^2(\mathbb{R})$.
I'm specifically interested in the case $X = \ell^{\infty}$. 
Does (*) hold in this case and why or why not?
Thanks.
[1] http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst;task=show_msg;msg=1533.0001.0001.0001

REPLY [8 votes]: If $I$ is uncountable, then in space $l^2(I)$ no countable set of functionals separates points.  Consequently, for any set $A$ in the sigma-algebra generated by these functionals [the Baire sets for the weak topology, see reference below], if $0 \in A$, then an entire subspace is contained in $A$.  So all elements of this sigma-algebra are unbounded.  Thus this sigma-algebra is not all of the norm-Borel sets.
My papers on measurability in Banach space:
Indiana Univ. Math. J. 26 (1977) 663--677
Indiana Univ. Math. J. 28 (1979) 559--579
edit
For gaussian measures in Banach space, you really want the example
of Fremlin and Talagrand, "A Gaussian measure on $l^{\infty}$". Ann. Probab. 8 (1980), no. 6, 1192--1193.  This gaussian measure on $l^\infty$ with the cylindrical sigma-algebra has total mass 1, yet every ball of radius 1 has measure 0.<|endoftext|>
TITLE: Dissecting a tetrahedron into orthoschemes
QUESTION [6 upvotes]: Is there a way to dissect any tetrahedron into a finite number of orthoschemes?
I know that for a tetrahedron which only has acute angles, one can take the center of the inscribed circle and project the center on all the faces and edges and connect it with the vertices to get the orthoschemes. This however does not work when the tetrahedron is allowed to have obtuse angles since the projection of the center of the inscribed circle on the plane containing a face for instance may fall outside of the tetrahedron.

REPLY [8 votes]: Yes, this is known (12 is always enough).  Interestingly, in higher dimension it is open whether every simplex in $\Bbb R^d$ can be dissected into finitely many orthoschemes (also called path-simplices). This is called Hadwiger's conjecture.  See this survey for results and refs to proofs of the conjecture for $d \le 5$.  
P.S.  In 1993, Tschirpke showed that 12,598,800 orthoschemes suffices in $\Bbb R^5$.<|endoftext|>
TITLE: How does singular homology H_n capture the number of n-dimensional "holes" in a space?
QUESTION [8 upvotes]: This is a foundational doubt I have. How does singular homology H_n capture the number of n-dimensional holes in a space?
We disregard the case of $H_0$ as it has the very satisfactory explanation that it is the direction sum of $\mathbb Z$ over the path-connected components of the space. 
Now, handwaving aside, we consider the most important example of this "detecting hole" phenomenon, viz,, the fact that for $i \geq 1$ $H_i(S^n) = \mathbb(Z)$ if and only if $i = n$. For this we use Mayer-Vietoris and a decomposition of $S_n$ into a union of two open sets which are the complements of the north pole and south pole. And the intersection deformation retracts to $S^{n -1}$ and from the long exact sequence we get the isomorphisms $H_i \cong H_{i -1}$. 
Now, by the above computation, it seems that the "hole detection" is achieved via Mayer-Vietoris and going up from the dimension below, using the long exact sequence. Mayer-Vietoris on the other hand depends on the snake lemma, which is very un-geometric and difficult to visualize.
So I would be most grateful for a more intuitive explanation of this hole capturing phenomenon. I can see that it is very natural that boundaries should be cancelled out as the solid simplices can be contracted to the central point. I can also "feel" that a hollow $n$-simplex, there should be a nontrivial $n$-chain which is not a boundary of an $n+1$-chain. But I am still left with a feeling of partial understanding. I hope this fundamental vagueness of understanding of mine can be cleared here.

REPLY [10 votes]: Rather than thinking directly about "holes", I suggest you think about how a circle (in the guise of the boundary of a triangle) is obtained by gluing three intervals at their endpoints, or how a 2-sphere (in the guise of the surface of a tetrahedron) is obtained by gluing together four triangles along their edges. In general, the boundary of an $n+1$ simplex, which is topologically a sphere, is the sum of $n+1$ different $n$-simplices, and this sum is the non-trivial $n$-cycle giving the top homology of the sphere.
If you have trouble connecting this picture with the definitions of singular homology, then try
learning simplicial homology first. The computations are then much more explicit,
and you can compute the homology of a sphere directly from the preceding triangulations,
rather than from a diagram-chasing interpretation of Mayer--Vietoris.
Once you are comfortable with computations in that context, return to the singular theory.<|endoftext|>
TITLE: Computing Bruhat Order Covering Relations
QUESTION [11 upvotes]: To put this in context:  I am in the process of developing a package for Macaulay 2 (a commutative algebra software,) called "Permutations", which will add permutations as a type of combinatorial object that M2 will handle, hopefully integrating nicely with the (still in development) Posets package, the (still in development) Graphs package, and various current M2 functions.
One of the first things that I'd wanted to put in was functions which compute the poset of the Bruhat order on $S_n$ and compute when one permutation covers another in the Bruhat order.  This was all coded and "working" - but has been producing a poset which is decidedly NOT the desired poset (among other things, the graph of the Hasse diagram isn't regular.)  I'd like to ask whether the (probably somewhat naive) algorithm I was using to check covering relations seems reasonable (so the problem is just in the coding of it, not in the theory behind it) or not.
To see if $P\leq R$ in the Bruhat order:
Given a pair of permutations P and R, compute their lengths (the number of simple transpositions in their decomposition, or [as implemented right now] the sum of entries in their inversion vectors.)  If length(P)=length(R)+1, then we compute
$(P^{-1})*R$.
If R covers P in the Bruhat order, then length$((P^{-1})*R)=1.$
Am I missing some subtlety of the Bruhat order?  I thought one permutation covered enough exactly when they differed by a single, simple transposition.  This seemed to capture that - but is giving me an incorrect poset.
Coding error or theory error?  I'd love to hear it.

REPLY [3 votes]: For  Stephen Griffeth and others:
download: http://math.univ-lyon1.fr/~ducloux/coxeter/coxeter3/english/coxeter3_e.html
To tell the software what group you want to work with, get yourself to the "type" prompt, either by pressing Enter at the main prompt, or by entering "type" as a command. The names of the groups are:

A-I, corresponding to the finite Coxeter groups
a-g, corresponding to the affine Coxeter groups
X or x, for input from a file
Y or y, for interactive input

For (1) (2) and (4) you will next be presented with the "rank" prompt, which refers to the size of the generating set of the Coxeter group, so for example the affine Weyl group of SL(3) should be "a" followed by "3". For (4) you will also be asked for the entries of the Coxeter matrix. For (3) you will be asked for a file.
When you need to enter a group element, you merely type a sequence of integers 1 through rank separated by periods, e.g. 1.2.1. If the rank is at most 9 then you can (must?) omit the periods, e.g. 121. If you need to input the identity element, you simply type nothing and press Enter. 
I don't remember how the simple reflections are numbered, but I seem to remember that the affine generator is always the last one. There is a prefix and postfix command to change these naming conventions (so you could call the 1st reflection "s(1)" if you wanted), which is useful when you are doing input from the output of some other program or script.
You can see a list of commands by typing "help", and they should be mostly self-explanatory knowing the above.<|endoftext|>
TITLE: Measurable subgroups.
QUESTION [6 upvotes]: Let $G$ be a compact connected topological group and let $H$ be a subgroup of $G$. Suppose that $H$ is measurable with respect to the normalised Haar measure $\mu$ on $G$. Do we necessarily have $\mu(H)=0$ or $\mu(H)=1$?
Maybe this is well--known, I ask it just out of curiosity. The question is related to this one: If you provide a measurable subgroup $H$ of $\mathbb R/\mathbb Z$ of measure not 0 or 1, then the characteristic function of $H$ violates the conjecture stated there.

REPLY [10 votes]: Don't we still have this:  if $A$ is measurable of positive measure, then $A A^{-1}$ contains a neighborhood of the identity...?  So: a measurable subgroup of positive measure itself contains a neighborhood of the identity, and thus by connectedness is all of $G$.<|endoftext|>
TITLE: Non-absolute convergence of series with asymtotically  equal coefficients
QUESTION [5 upvotes]: The following seems to be a question related to standard calculus, but I am not quite sure
where to look for an answer.
Suppose $f,g:\mathbb{N} \to \mathbb{C}$ are such that the have the same asymptotical behaviour, i.e. $f(n)/g(n) \to 1$ as
$n \to \infty$. Of course, suppose that one of the sums $\sum_{n=0}^\infty f(n)$ and $\sum_{n=0}^\infty g(n)$ converges absolutely,
then so does the other. This can be proven by a standard estimate. However this standard estimate fails
if we do not have absolute convergence. I do not see how to prove convergence of one of sums implies the convergence of the other.
I feel that it may be actually false.
So the first question is:

$1$. Is it true that one series converges iff the other does?

If this is not the case, however, in the problem I am studying, I want to prove convergence for both series. For my application in mind, you may assume that $f(n)/g(n)$ is always in $\mathbb{R}$. So the second question is  

$2$. Under which additional conditions (which do not! imply absolute convergence) can
     we deduce both series have the same behaviour. Are there books treating such topics?

REPLY [2 votes]: Following Theo Johnson--Freyd's suggestion, I am making my above comment an answer:
Just subtracting one series from the other, it seems that you need $\sum_{n=0}^{\infty} (f(n)−g(n))$  to converge.  Writing $$f(n)/g(n)=1+\delta_n,$$  $$\text{so that } \qquad \qquad\sum_{n=0}^{\infty} (f(n)−g(n))=\sum_{n=0}^{\infty} \delta_n g(n),$$  you see need control over the signs of the $\delta_n$, or (as Theo notes in his comment, on their
rate of growth).
E.g. if they are all of the same sign, you are okay, while if the sign of $\delta_n$ is always the same as, or always opposite to, that of $\delta_n$, then you could be in bad shape. (This is what goes wrong in Xandi Tuni's example.) 
As Theo notes in his comment, you are also okay if $\sum_{n = 0}^{\infty} \delta_n$
converges absolutely.  Whether this
applies in your case will depend on how closely $f$ and $g$ approximate one another.
(This is illustrated by the example in Julian Aguirre's answer.)<|endoftext|>
TITLE: Contracting divisors to a point
QUESTION [17 upvotes]: This is quite possibly a stupid question, but it is pretty far from what I normally do, so I wouldn't even know where to look it up.
If $X$ is a projective variety over an algebraically closed field of arbitrary characteristic and $Y\subset X$ a smooth divisor. Under which conditions can I contract $Y$ to a point, i.e. under which conditions is there a projective (smooth!?) variety $V$, and a morphism $f:X\rightarrow V$, such that $f$ is an isomorphism away from $Y$, and $Y$ is mapped to a point.
What can one say if $Y$ is a strict normal crossings divisor? 
Hints and references are very appreciated!

REPLY [29 votes]: For a smooth $Y$, a necessary condition for contractibility is that the conormal line bundle $N_{Y,X}^\*$ is ample. It is also sufficient for contracting to an algebraic space. The reference is Algebraization of formal moduli. II. Existence of modifications. by M. Artin. 
$Y$ can be contracted to a point on an algebraic (projective) variety if in addition $Y=\mathbb P^{n-1}$, $n=\dim X$. You can prove this easily by hands. Start with an ample divisor $H$ and then prove that an appropriate linear combination $|aH+bY|$ is base point free and is zero exactly on $Y$. You will find the argument in Matsuki's book on Mori's program for example.
So if $X$ is a surface and $Y=\mathbb P^1$ with $Y^2<0$ then it is contractible to a projective surface. For a reducible divisor $Y=\sum Y_i$ a necessary condition (which is also sufficient in the category of algebraic spaces) is that the matrix $(Y_i.Y_j)$ is negative definite. The strongest elementary sufficient condition for contractibility to a variety is that $\sum Y_i$ is a rational configuration of curves. This is contained in On isolated rational singularities of surfaces by M. Artin. 
This paper also contains an example of an elliptic curve $Y$ with $Y^2=-1$ which is not contractible to an algebraic surface. The surface $X$ is the blowup of $\mathbb P^2$ at 10 sufficiently general points lying on a smooth cubic, $Y$ is the strict preimage of that cubic.
Finally, for an irreducible divisor $Y$ the resulting space $V$ is smooth iff $Y=\mathbb P^{n-1}$ and $N_{Y,X}=\mathcal O(-1)$. Indeed, $X\to V$ has to factor through the blowup of $V$ at a point by the universal property of the blowup. But then $X$ has to coincide with this blowup by Zariski main theorem. And on the blowup at a point the exceptional divisor is $\mathbb P^{n-1}$ with the normal bundle $\mathcal O(-1)$.

REPLY [3 votes]: If you ask V to be smooth and if the divisor Y is irredicible then Y should be isomorphic to the projective space and an intersection on X of the divisor Y with a line L on Y have to be equal to $-1.$<|endoftext|>
TITLE: Ultrafilters arising from Keisler-Shelah ultrapower characterisation of elementary equivalence
QUESTION [14 upvotes]: In model theory, two structures $\mathfrak{A}, \mathfrak{B}$ of identical signature $\Sigma$ are said to be elementarily equivalent ($\mathfrak{A} \equiv \mathfrak{B}$) if they satisfy exactly the same first-order sentences w.r.t. $\Sigma$.  An astounding theorem giving an algebraic characterisation of this notion is the so-called Keisler-Shelah isomorphism theorem, proved originally by Keisler (assuming GCH) and then by Shelah (avoiding GCH), which we state in its modern strengthening (saying that only a single ultrafilter is needed):
$\mathfrak{A} \equiv \mathfrak{B} \ \iff \ \exists \mathcal{U} \text{ s.t. } (\Pi_{i\in\mathcal{I}} \ \mathfrak{A})/\mathcal{U} \cong (\Pi_{i\in\mathcal{I}} \ \mathfrak{B})/\mathcal{U},$
where $\mathcal{U}$ is a non-principal ultrafilter on, say, $\mathcal{I} = \mathbb{N}$.  That is, two structures are elementarily equivalent iff they have isomorphic ultrapowers.
My question is the following (admittedly rather vague): Does anyone know of constructions in which an ultrafilter is chosen by an appeal to this characterisation and then used for other means?  An example of what I have in mind would be something like this (using the fact that any two real closed fields are elementarily equivalent w.r.t. the language of ordered rings): In order to perform some construction $C$ I ``choose'' a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$ by specifying it as a witness to the following isomorphism induced by Keisler-Shelah:
$\mathbb{R}^\mathbb{N}/\mathcal{U} \cong \mathbb{R}_{alg}^\mathbb{N}/\mathcal{U},$ 
where $\mathbb{R}_{alg}$ is the field of real algebraic numbers.  So the construction $C$ should be dependent upon the fact that $\mathcal{U}$ is a non-principal ultrafilter bearing witness to the Keisler-Shelah isomorphism between some ultrapower of the reals and the algebraic reals, resp.
Also, a follow-up question: Let's say I'd like to ``solve'' the above isomorphism for $\mathcal{U}$.  Are there interesting things in general known about the solution space, e.g., the set of all non-principal ultrafilters bearing witness to the Keisler-Shelah isomorphism for two fixed elementarily equivalent structures such as $\mathbb{R}$ and $\mathbb{R}_{alg}$?  What machinery is useful in investigating this?

REPLY [10 votes]: As you might expect, things are consistently much more interesting if $CH$ fails. This has been explored by Shelah in a fascinating series of papers "Vive la difference I - III". For example, it is consistent that there is a nonprincipal ultrafilter $\mathcal{U}$ on $\omega$ such that if $(R_{n})$ and $(S_{n})$ are sequences of discrete rank 1 valuation rings having countable residue fields, then any isomorphism $\varphi: \prod_{\mathcal{U}}R_{n} \to \prod_{\mathcal{U}}S_{n}$ is an ultraproduct of isomorphisms $f_{n}: R_{n} \to S_{n}$. In particular, $\mathcal{U}$-almost all $R_{n}$ are isomorphic to the corresponding $S_{n}$ and so the Ax-Kochen isomorphism theorem doesn't hold with respect to $\mathcal{U}$.
If you are only interested in ultraproducts of fixed structures $A$, $B$, then I should mention that it is also consistent that there exists an ultrafilter $\mathcal{A}$ on $\omega$ such that if $A$ and $B$ are countable structures which satisfy the strong independence property, then the corresponding $\mathcal{A}$-ultraproducts are isomorphic iff $A \cong B$.<|endoftext|>
TITLE: Sequence that converge if they have an accumulation point
QUESTION [9 upvotes]: I am looking for classes of sequence, that converge iff they contain a converging sub-sequence.

The basic example of such sequences are monotone sequences of real numbers.
A more interesting examples comes from metric fixed point theory:
Let $B$ be a Banach space and $f\colon B \to B$ be a continuous mapping that is non-expansive (i.e. $\lVert f(x) - f(y)\rVert \le \lVert x -y\rVert$).
Define $x_{n+1} := \frac{1}{2} x_n + \frac{1}{2} f(x_n)$ for any startingpoint $x_0\in B$.
This is the so called Krasnoselski iteration.
One can show that any accumulation point $\tilde{x}$ of $(x_n)$ is a fixed point of $f$.
Since $f$ is non-expansive, it follows that
$\lVert x_{n+1}-\tilde{x}\rVert = \frac{1}{2}\lVert (x_{n}-\tilde{x}) + (f(x_{n}) - f(\tilde{x}))\rVert\le \lVert x_n -\tilde{x}\rVert$.
Hence $(x_n)$ converges iff it contains a converging sub-sequence.  
This is a special case of Ishikawa's fixed point theorem. (The Krasnoselski-Mann iteration - a generalization of the Krasnoselski iteration - also has this property.)

I am interested in this sequence because they provide very nice applications of the Bolzano-Weierstrass principle.
Do you know of any other examples of sequences with this property?
Do you know other proofs that uses this property together with the Bolzano-Weierstrass principle to prove the convergence of a sequence?

REPLY [2 votes]: The following version of the mean ergodic theorem is taken from the book of Krengel, "ergodic theorems".
Let T be a bounded linear operator in a Banach space X. The Birkhoff averages are denoted by $A_n = {1\over n} \ \Sigma_{k=0}^{n-1} \ T^k$.
Assume that the sequence  of operator norms $||A_n||$ is bounded independently of $n$. Then for any x and y in B, the following is equivalent :
-- y is a weak cluster point of the sequence $(A_nx)$,
-- y is the weak limit of the sequence $(A_nx)$,
-- y is the strong limit of the sequence $(A_nx)$.
(note that we talk about cluster points instead of converging subsequences because we didn't assume B separable. Hence the weak topology is not necessarily metrizable.)
This theorem implies e.g. the ergodic theorem for Markov operators on $C(K)$ (sequential compactness follows from Azrela-Ascoli), or the ergodic theorem for power bounded operators defined on reflexive Banach spaces (sequential compactness follows from Eberlein-Smulian). 
There is a whole set of theorems in ergodic theory along these lines.  Let me mention the convergence of the one sided ergodic Hilbert transform, discussed in Cohen and Cuny (see Th 3.2) as another example.<|endoftext|>
TITLE: Maximal extension almost everywhere unramified and totally split at one place
QUESTION [11 upvotes]: Fix a finite set of primes $S$ and an additional prime $p$. Let $K$ be the maximal extension of $\mathbb{Q}$ that is unramified outside $S$ and $\infty$ and totally split at $p$. Is the extension $K$ finite?
My intuitive guess would be no, but the simple constructions (based on class field theory) I tried so far do not prove this, at least for the ground field $\mathbb{Q}$. In contrast, for imaginary quadratic fields, there are extensions with Galois group $\mathbb{Z}_{\ell}^2$ ramified only above $\ell$, and any place not above $\ell$ splits completely in an infinite subextension, as $\mathbb{Z}_{\ell}^2$ has no procyclic subgroups of finite index.

REPLY [11 votes]: Nope.  
I'm lacking a reference in front of me at the moment (see NSW's Cohomology of Number Fields, or Gras's Class Field Theory -- I'll update with a precise reference later), but there are remarkably clean formulas for the generator and relation ranks for the Galois group of the maximal $\ell$-extension of $\mathbb{Q}$ unramified outside $S$ and completely split at $T$, for finite sets of primes $S$ and $T$.  Throwing out some silly cases, these depend only on $|S|$ and $|T|$ (and, in your problem, maybe even just $|S|-|T|$).  In your case, where $|T|=1$, it's just a matter of making $S$ big enough (again, a reference will say how big, but right now, I think $|S|=4$ does the trick.)
Edit to add in in a precise reference (though the above book references certainly contain the results as well):  Christian Maire's "Finitude de tours et p-tours T-ramifiees moderees, S-decomposees".<|endoftext|>
TITLE: Can one extend a morphism of commutative triangles to a morphism of  octahedral diagrams?
QUESTION [6 upvotes]: Consider two (distinct) octahedral diagrams i.e. diagrams mentioned in the octahedron axioms of triangulated categories (with four 'commutative triangular faces' and four 'distinguished triangular faces'). Is it true than one can extend to a morphism of such diagrams:
1. any morphism of one of the 'commutative faces' of the octahedron
2 any morphism of the pair of morphisms whose target is the upper vertex of the octahedron (i.e. a morphism of commutative triangles not lying on the faces of the octahedrons)?
Is there any text where I could look for various facts of this sort?
P.S. It seems that the answer is 'no' in general. Having a morphism of 'commutative faces', one can extend it to a morphism of three neighbouring 'triangulated faces'. Thus one obtains morphisms of each of six vertices. Yet (all possible) compositions of edges of the 'first' commutative triangles and the neigbouring distinguished faces do not yield all edges of the octahedron; two of the edges (in the 'lower hat') are missing.
Yet it would be very interesting to know which additional conditions are needed in order for the morphism of the octahedrons desired to exist. I would be deeply grateful for any comments!!

REPLY [4 votes]: The identity of commutative triangles does not lift to a morphism of Verdier octahedra in general, for it may happen that there exist two mutually nonisomorphic Verdier octahedra on the same commutative triangle.
Cf. http://www.math.rwth-aachen.de/~kuenzer/counterexample.pdf . This has been observed already by Amnon Neeman in the 90s (unpublished). 
(The situation is better if one uses 3-triangles, i.e. "distinguished octahedra", in the sense of Heller/oo-triangulated categories. Cf. http://www.math.rwth-aachen.de/~kuenzer/heller.pdf , Lem. 3.2. So there exists a choice of octahedra such that morphisms of commutative triangles lift to morphisms of these particular octahedra.)<|endoftext|>
TITLE: Are centrally extended p-adic groups defined over F_1?
QUESTION [12 upvotes]: Let G be a semisimple algebraic group.
Following work of Matsumoto [1], Brylinski and Deligne [2] constructed a central extension of the functor G : Rings → Groups by the second algebraic K-theory functor.
Plugging in ℂ((t)) into those functors, we get the well known central extension $\widetilde{G\big(\mathbb C((t))}\big)$ of
the loop group G(ℂ((t))) by the multiplicative group ℂ*=K2(ℂ((t))).
It is interesting to note that the above group comes from an algebraic 
group defined over the subfield ℂ of ℂ((t)). Namely, $\widetilde{G\big(\mathbb C((t))}\big)$ = $\widetilde{LG}(\mathbb C)$. 
Doing all this with ℚp instead of ℂ((t)),
we get a central extension $\widetilde{G(\mathbb Q_p)}$ of G(ℚp) by the group K2(ℚp) = Fp*.
Now, here's an idea: maybe that central extension is defined over... the subfield 
F1 of ℚp?...
My questions:
• Has this been considered before?
• If yes, among all the exitsing notion of "defined over F1", 
which one(s) make this possible?
• If no: is my heuristic argument is convincing?

References:
[1] Matsumoto,
"Sur les sous-groupes arithmétiques des groupes semi-simples déployés".
[2] Brylinski, Deligne, "Central extensions of reductive groups by $K_2$".

REPLY [10 votes]: First a small thing. I am pretty sure we don't have $K_2(\mathbb C((t)))=\mathbb
C^*$, we have a surjective residue homomorphism $K_2(\mathbb C((t)))\rightarrow
\mathbb C^*$ but, I believe, with a non-trivial kernel. In any case, we can look
at the induced central extension and then the rest of what you say is
OK. Similarly, we have a surjective map $K_2(\mathbb Q_p)$.
Disrergarding this, there is a much simpler analogy between the two cases which on the
one hand, I think, makes the analogy that you want less likely and on the other
hand can be proven... To begin with it is not quite true that even $G(\mathbb
C((t)))$ is defined over $\mathbb C$ at least not as a group scheme. What
happens is that $G(\mathbb C[[t]])$ is a group scheme, it is the inverse limit
of the $G(\mathbb C[t]/(t^n))$ and these have a natural structure of algebraic
group over $\mathbb C$ (through the Greenberg functor). then $G(\mathbb C[[t]])$
as the inverse limit of algebraic groups is a group scheme (it is not of finite
type hence convention forces us to call it a group scheme rather than algebraic
group). Now, if we try to pass to $G(\mathbb
C((t)))$ we get into trouble. It is an infinite union of schemes (bound the
valuations of the entries of the elements of $G(\mathbb
C((t)))$ in some faithful linear representation of $G$) but an infinite union of
schemes does in general not have a scheme structure. There are ways of extending
the scheme notion to cover this case and what we get is what is called an
ind-group scheme over $\mathbb C$. Also the loop group type extension of $G(\mathbb
C((t)))$ by $\mathbb C^*$ has such an extension (as does every Kac-Moody type
group).
The situation for $G(\mathbb Q_p)$ is almost identical; $G(\mathbb Z/p^n)$ are
the $\mathbb Z/p$-points of a $\mathbb Z/p$-algebraic group, $G(\mathbb Z_p)$
are the $\mathbb Z/p$-points of a group scheme over $\mathbb Z/p$ and $G(\mathbb
Q_p)$ are the $\mathbb Z/p$-points of an ind-group scheme over $\mathbb Z/p$. I
think that the same thing is true for the central extension. The upshot is that
there is a close analogy to the $\mathbb C$ case but in that analogy $\mathbb C$
is replaced by $\mathbb F_p$ not by $\mathbb F_1$.
Note that in the Connes-Consani version of $\mathbb F_1$ $G$ is defined over $\mathbb F_{1^2}$ so perhaps that is the place to look for a version of the Brylinski-Deligne result.
Addendum: Just to add even more concreteness to George's answer about the explicit form of Greenberg's functor for $\mathbb G_m$. We have that $W_n(B)$ is just $B^n$ with a funny multiplication and addition. They are however given by polynomials (which are independent of $B$). The units in this ring are the tuples of the form $B^\ast\times B^{n-1}$ and multiplication is given by polynomials. This means that the algebraic group associated to this is just $\mathbb G_m\times\mathbb A^{n-1}$ as scheme but with a funny product structure. In particular its $\mathbb F_p$-points are just $(\mathbb Z/p^n)^\ast$.<|endoftext|>
TITLE: How to prove that a set of facets are all the facets of a convex polytope.
QUESTION [6 upvotes]: Say that you know all the vertices of a polytope P, and a set of facet defining hyperplanes that you guess give all the facets of P. What are some good ways to try to prove that the guess is right?
A common idea seems to be to find a way to linearly optimize over the polytope defined by the facets and then show that you always end up in one of the vertices of P. Another type of argument involves the volume of the polytopes. There is also algorithms that involve constructing a simplicial complex of the given data and then compute homology. Are there other common teqniques?
Are there any good ways to prove that the guess is wrong?

REPLY [4 votes]: Hi,
The question itself and all answers seem to be "practice-oriented" but as far as the complexity status of the problem is concerned then you are speaking about famous L.Lovasz's "polytope- polyhedron question". Namely, to check whether a polytope given by vertices coincides with a polyhedron given by facets. Note, that the lengths of the vertex and, respectively, the facet descriptions may differ exponentially and thus we should speak about incremental polynomial algorithms. Say, if a polyhedron is given by facets you should find a new vertex in polynomial time with respect to the input  and the length of the list of already calculated vertices. In this format the question is still open for (bounded) polytopes but is proved to be NP-hard for (unbounded) polyhedrons (see, http://portal.acm.org/citation.cfm?id=1109640 and references therein)<|endoftext|>
TITLE: What is the geometry behind psi classes in Gromov-Witten theory?
QUESTION [10 upvotes]: Intuitively, Gromov-Witten theory makes perfect sense. Via Poincare duality, we look at the cohomology classes $\gamma_1, \ldots, \gamma_n$ corresponding to geometric cycles $Z_i$ on a target space $X$, pull them back and then the integral
$$\langle \gamma_1 \cdots \gamma_n\rangle=\int_{\overline{\mathcal{M}}_{g,n}(X,\beta)^{vir}}ev_1^*\gamma_1 \smile \cdots \smile ev_n^*\gamma_n$$
should count the number of curves whose intersection with the given cycles is non-empty.
However, we also have the ψ-classes (or "gravitational descendants") arising from the moduli space $\overline{\mathcal{M}}_{g,n}$ which are the chern classes of the $i$-th cotangent line bundle to a given $(C, x_1, \ldots, x_n) \in \overline{\mathcal{M}}_{g,n}$.
So what, geometrically, do these represent? The fact that they arise from $\overline{\mathcal{M}}_{g,n}$ means that the inclusion of a ψ-class places restriction on the geometry of the curves which we count; that much is clear. What is this restriction?
The reason that I am curious is that I am trying to evaluate the GW-invariants corresponding to maps which have components collapsing to an A1 singularity (i.e. a $B\mathbb{Z}/2$), but such that not all of the curve collapses. It has been mentioned in passing that including a ψ-class could help with this, and while the little I understand makes this sound plausible, I don't exactly see why.
So what are ψ-classes? Can I use them to split my curve up into parts so that a fixed component lands on my stacky point, while the rest of it does whatever else curves do?

REPLY [12 votes]: The following answer is unfortunately not quite correct, but it may be useful anyway. I will of course be ignoring any virtual fundamental class issues.
Imagine that you are computing a Gromov-Witten invariant where you require the i-th marked point to land at a specific point (i.e. your i-th insertion γi is the class of a point), and now lets add aditionally the i-th psi-class as an insertion. You can restrict to the subspace of maps with $f(x_i) = x$ for some generic choice of $x \in X$. Fixing an arbitrary non-trivial map $\Phi \colon T_x \to k$ gives you by composition a map from the relative tangent bundle of the universal curve over $M_{g, n}(X)$ at the section xi to the trivial line bundle, in other words a section $\phi$ of the relative cotangent bundle of the universal curve. It will vanish on curves which are tangent to a hypersurface through x with tangent direction matching the zero-locus of the map $\Phi$.
So you can think of Gromov-Witten invariants with psi-classes as counting maps which additionally satisfy tangency conditions at the marked points.
Why is this not correct? The zero locus of $\phi$ computes the Chern class of the relative cotangent bundle at $x_i$ over Mg, n(X), which is not the same as the pull-back of the $\psi$-class from Mg, n. Insertions of the former are sometimes called "gravitational ancestors", and the difference to gravitational descendants is described explicitly in alg-geom/9708024.<|endoftext|>
TITLE: What interesting/nontrivial results in Algebraic geometry require the existence of universes?
QUESTION [46 upvotes]: Brian Conrad indicated a while ago that many of the results proven in AG using universes can be proven without them by being very careful (link).  I'm wondering if there are any results in AG that actually depend on the existence of universes (and what some of the more interesting ones are).
I'm of course aware of the result that as long as we require that the classes of objects and arrows are sets (this is the only valid approach from Bourbaki's perspective),  for every category C, there exists a universe U such that the U-Yoneda lemma holds for U-Psh(C) (this relative approach makes proper classes pointless because every universe allows us to model a higher level of "largeness"), but this is really the only striking application of universes that I know of (and the only result I'm aware of where it's clear that they are necessary for the result).

REPLY [22 votes]: There is a paper of Solomon Feferman from 1969 in which he proposes a conservative extension of ZFC adequate for formulating much of category theory.  Since the need for universes in algebraic geometry seems to arise mainly via the use of category theory, I conjecture that Feferman's approach may be useful in algebraic geometry.  The paper in question is "Set-theoretical foundations of category theory" (in "Reports of the Midwest Category Seminar III", Springer Lecture Notes in Math 106, edited by Saunders Mac Lane, pages 201-247).  Feferman proposes to add to the language of ZFC a new constant symbol $\kappa$ and to add to the axioms the schema saying that $V_\kappa$ is an elementary submodel of the universe $V$ of all sets, with respect to the original language of ZFC.  That is, for each formula $\phi(x_1,\dots,x_n)$ in the original language (i.e., not involving $\kappa$), there is an axiom saying that, for each $x_1,\dots,x_n\in V_\kappa$, the statement $\phi(x_1,\dots,x_n)$ is equivalent to the same statement with all quantified variables restricted to range over $V_\kappa$.  The idea is that this $V_\kappa$ can play the role of a Grothendieck universe, even though it isn't really one.  Furthermore, in situations where one would expect to use two or more universes, one can often get by with one, by a judicious use of the "elementary submodel" axioms.  
Note that one cannot express the notion of "elementary submodel of the universe" as a single formula, since there is no uniform way to express "truth in the universe" (Tarski's theorem).  But one can express it one formula at a time, as an infinite axiom scheme, since there's no problem expressing truth of a single formula $\phi$ (just say $\phi$).  And this is what Feferman does.
Feferman's axiom system is conservative over ZFC.  That is, if you can prove, in Feferman's system, a sentence that doesn't involve $\kappa$, then you can prove the same sentence in ZFC.  (This follows from Levy's reflection principle plus the compactness theorem of first-order logic.)
Many years ago, I taught a category theory course, using Feferman's axioms as the foundation.  For the most part, this foundation worked well, but, if I remember correctly, I ran into a problem proving the existence of Kan extensions.  I don't remember what I did to overcome the problem (nor do I even remember exactly what the problem was).  
I believe Feferman has done additional work improving this foundational system, but I have not yet absorbed that work.<|endoftext|>
TITLE: Factorial Rings and The Axiom of Choice
QUESTION [9 upvotes]: It is shown in Lang's Algebra (and many other books I assume) that:
if A if a principal entire ring, then A is a factorial ring.
The proof uses Zorn's Lemma. Is this theorem equivalent to the axiom of choice?

REPLY [6 votes]: This is an extended comment on Keith's answer:
Another way to think about Keith's argument (or the more general argument for the existence of minimal primes in Noetherian rings) is that one constructs a tree as follows: beginning 
with an ideal $I$, if it not prime then we may find $J_1, J_2$ properly containing $I$
such that $J_1 J_2 \subset I$, and we make a "factorization tree" joining $I$ to $J_1$
and $J_2$, and then iterate this procedure.
Noetherianness says that each branch of the tree has finite length, and since each
edge has finite valence, the whole tree is finite, hence we eventually must reach prime
ideals, and we find a finite collection of prime ideals $P_i$ whose product is contained
in $I$.
The fact about trees being used can be phrased (in the contrapositive) as:
an infinite tree with every vertex of finite valence has an infinite branch.  (This has a name, which I forget; it is also what underlies the proof of Bolzano--Weierstrass.)
Now the proof of this fact uses DC (as far as I can tell), and it's probably well-known
to logicians exactly whether or not it is equivalent to DC (or, at any rate,
exactly how strong it is).
The question then becomes whether we can model any tree (whose internal vertices have
valence 3) as a factorization tree (in a PID, say, although the possibility of allowing
more general Noetherian rings seems reasonable, at least if it helps to give a positive
answer).
Added: The result on trees is called Koenig's lemma. (Thanks to Artie Prendergrast-Smith for this.)  
I should also note that my original movitation for formulating the above argument was similar to that of Keith: the usual proof of the existence of minimal primes in Noetherian rings uses Zorn's lemma and has a non-constructive feel, whereas the above reformulation feels more concrete.<|endoftext|>
TITLE: Seeking reference for the enumerative "mass formula" concept
QUESTION [8 upvotes]: I am teaching a combinatorics class in which I introduced the notion of a "mass formula".  My terminology is inspired by the Smith–Minkowski–Siegel mass formula for the total mass of positive-definite quadratic forms of a given size and genus.  That famous mass formula is much too fancy of an example for my class.  All that I really do is define the concept of the "mass" of a combinatorial object to be $1/|G|$ if $G$ is its automorphism group, and then argue that it can be easier to find the total mass of a collection of objects than to count them straight (using Polya counting theory).  For example, the total mass of unlabeled trees of order $n$ is $n^{n-2}/n!$, because there are $n^{n-2}$ labeled trees.
So I have two questions for which a quick answer (i.e. sooner than two weeks) would be most convenient:

Is "mass formula" a standard name for this concept?  Is there a standard name?
Can someone suggest a free on-line reference, comparable to a Wikipedia page or a little longer?  The class textbook doesn't have a discussion.

REPLY [3 votes]: This doesn't qualify as a free reference, but "Graphs on surfaces and their applications" by Lando and Zvonkin has some nice examples. On p.46, after stating a theorem enumerating trees with a given "passport", the authors remark:
 We will often encounter enumerative formulas where the objects are not counted one by one but a weight us assigned ti each object, and this weight is equal to 1/|Aut|, where the denominator means the order of the automorphism group of the object. Formulas of this kind are often called mass-formulas. (Footnote: The first mass-formula was proposed by H.J.S. Smith in 1867. Mass-formulas are also called Siegel–Minkowski formulas.)<|endoftext|>
TITLE: What's wrong with compact-open topology on the space of maps?
QUESTION [10 upvotes]: Given a smooth vector bundle $E$ with non-compact base, let
$\Gamma(E)$ be the space of $C^\infty$ sections equipped with compact-open $C^\infty$-topology. 

I have heard that $\Gamma(E)$ is not locally-contractible. Why not?
Is $\Gamma(E)$ contractible? Visibly any section can be joined to the zero section by "straight line", doesn't this prove that $\Gamma(E)$ is contractible?
Is it true that every convex subset of $\Gamma(E)$ is contractible? The argument of 2 seems to apply, but then it seems plausible that each section has an arbitrary small convex neighborhood, contradicting 1.

CLARIFICATION: One source of "rumor 1" is the book "The Convenient Setting of Global Analysis" freely available 
 here. On page
429 one reads: "Unfortunately, for non-compact $M$, the space $C^\infty(M, N)$ is not locally contractible in the compact-open $C^\infty$-topology". Another source is the discussion in Hirsch's book in the beginning of Chapter 2, which says "It can be shown that
$C^\infty(M, N)$ has very nice features, e.g. it has a complete metric, and a countable base; if $M$ is compact, it is locally contractible and $C^r(M, \mathbb R^n)$ is a Banach space for $2\le r<\infty$".
Thus I assumeed that in general, if $M$ is non-compact, then the space $\Gamma(E)$ is not (or maybe just need not be?) locally contractible. Also I am uncertain whether $C^\infty(M, \mathbb R^n)$ or $\Gamma(E)$ is a topological vector space, is it really? There seems to be a sequence of semi-norms giving these spaces a structure of Frechet spaces, but then they must be locally convex, hence locally contractible. Obviously, I am missing something. 
In response to comments I ask a more specific question. 
Question. Let $T_{r,s}(M)$ denote the space of $C^\infty$-smooth $(r,s)$-tensors on a connected non-compact $C^\infty$ manifold $M$. For $k$ with $2\le k\le \infty$, give $T_{r,s}(M)$ the weak $C^k$-topology as in Hirsch's book (roughly for $k$ finite we require that given $\epsilon>0$ and compact subset $K$ all derivatives up to $k$ are $\epsilon$-close over $K$, and for $k=\infty$ we take the union of all $C^k$-topologies for all finite $k$ under the inclusions $C^\infty\to C^k$). Now I ask
Is  $T_{r,s}(M)$ a Fréchet space with respect to the weak $C^k$-topology?
In particular, I want to conclude that $T_{r,s}(M)$ is locally contractible, and any convex subset of $T_{r,s}(M)$ is contractible; I think Fréchet spaces must have this property.

REPLY [5 votes]: (For the more specific question)
Yes for $k = \infty$ if $M$ can be exhausted by a countable number of compact sets, no otherwise.  However, the failure is due to a lack of completeness (for $k \ne \infty$) or size issues (if $M$ can't be exhausted) rather than anything else and thus the local contractibility still holds.  Indeed, contractibility holds simply by contracting the vector bundle itself down to the image of the zero section.
The reason is due to the fact that the topology can be described by a family of semi-norms, as Sergei indicates in his comment to his earlier answer, so you get a locally convex topological vector space.  I recommend that you read about these spaces; Schaefer's book is a good place to start (as in Jarchow's but that doesn't seem to be available any more).
(In particular, be wary of saying "union of all $C^k$-topologies"; actually you are taking a projective limit here which means that the space has a $0$-neighbourhood base which is a union of the $0$-neigbourhood bases from each of the $C^k$ topologies, but that doesn't mean that the final topology is the union of all of the $C^k$ topologies.  Simply take a set of point $x_n$ that are "far apart" and put a set $U_n$ about each one so that $U_n$ is open in the $C^n$-topology but not in $C^{n-1}$.  Then $\bigcup U_n$ is open in the $C^\infty$ topology but not in any of the $C^n$-topolgies.)<|endoftext|>
TITLE: Drawing of the eight Thurston geometries?
QUESTION [60 upvotes]: Do you know of a picture, drawing, or other concise visual representation of the eight three-dimensional Thurston geometries?

I am imagining something akin to the standard picture (of a sphere, plane, and saddle) used to illustrate the three constant curvature geometries in dimension two.  Of course, it takes more doing to illustrate representative three-manifolds, and there are more choices for natural examples, but I was surprised when I couldn't find such a picture.  Another option would be to  depict or indicate some of the geometries in less direct ways, for instance via the structure of stabilizers.

REPLY [18 votes]: We have recently started working on visualizing Sol.
Sol is defined by the following metric in $\mathbb{R}^3$: $ds^2 = (e^zdx)^2 + (e^{-z}dy)^2 + dz^2$
I think it is quite easy to see what is going on there: there is a Z coordinate and moving along this Z coordinate makes your Y steps larger (double each $\log(2)$ moved on the Z axis) while making your X steps smaller (double each $-\log(2)$ moved). In a hyperbolic space both would expand together (that's how the half-space model works if you replace the Z coordinate by its logarithm).

SolvView by MagmaMcFry gives a native perspective visualization of both Sol and Nil. By native perspective I mean that the view you see here is the view you would get if you were inside the space, assuming that the light travels on geodesics.
We have added Sol as a playable geometry in the current beta of HyperRogue (viewable both in the native perspective projection and in projection of the simple model above). Here is a video of a camera rotating in Solv, looking at some surfaces of constant Z.
UPDATE: we have implemented all Thurston geometries. See the release post and the geometry page.<|endoftext|>
TITLE: Is there a nice proof of the fact that there are (p-1)/24 supersingular elliptic curves in characteristic p?
QUESTION [36 upvotes]: If $k$ is a characteristic $p$ field containing a subfield with $p^2$ elements (e.g., an algebraic closure of $\mathbb{F}_p$), then the number of isomorphism classes of supersingular elliptic curves over $k$ has a formula involving $\lfloor p/12 \rfloor$ and the residue class of $p$ mod 12, described in Chapter V of Silverman's The Arithmetic of Elliptic Curves.  If we weight these curves by the reciprocals of the orders of their automorphism groups, we obtain the substantially simpler Eichler-Deuring mass formula: $\frac{p-1}{24}$.  For example, when $p=2$, the unique supersingular curve $y^2+y=x^3$ has endomorphisms given by the Hurwitz integers (a maximal order in the quaternions), and its automorphism group is therefore isomorphic to the binary tetrahedral group, which has order 24.
Silverman gives the mass formula as an exercise, and it's pretty easy to derive from the formula in the text.  The proof of the complicated formula uses the Legendre form (hence only works away from 2), and the appearance of the $p/12$ boils down to the following two facts:

Supersingular values of $\lambda$ are precisely the roots of the Hasse polynomial, which is separable of degree $\frac{p-1}2$.
The $\lambda$-line is a 6-fold cover of the $j$-line away from $j=0$ and $j=1728$ (so the roots away from these values give an overcount by a factor of 6).

Question:  Is there a proof of the Eichler-Deuring formula in the literature that avoids most of the case analysis, e.g., by using a normal form of representable level?
I suppose any nontrivial level structure will probably require some special treatment for the prime(s) dividing that level.  Even so, it would be neat to see any suitably holistic enumeration, in particular, one that doesn't need to single out special $j$-invariants.
(This question has been troubling me for a while, but Greg's question inspired me to actually write it down.)

REPLY [22 votes]: You can also do this "topologically".  The idea is to separately calculate the (l-adic etale) Euler characteristics of the stack $Y_0(p)$ in $\mathrm{char}$ $0$ and in  $\mathrm{char}$ $p$, then see how they must relate.  Here we go:
Over $\mathbb{C}$, and hence over $\bar{\mathbb{Q}}_p$ as well, the Euler characteristic of $Y_0(p)$ is $(p+1)\times(-1/12)$, since $Y_0(p)$ is a $(p+1)$-fold cover of $Y=M_{ell}$.
On the other hand, over $\bar{\mathbb{F}}_p$, up to "homeomorphism" $Y_0(p)$ is two copies of Y glued at the supersingular points, so the Euler characteristic is $2\times(-1/12) - S$ (where $S$ is the "number" of supersingular elliptic curves).
However, since $Y_0(p)$ has semi-stable reduction (and is constant at infinity) over $\mathbb{Z}_p$, the special fiber is gotten from the generic fiber by contracting a bunch of "circles" to points, one for each nodal point of the special fiber; thus our second ($\mathrm{char}$ $p$) Euler characteristic is equal to our first ($\mathrm{char}$ $0$) Euler characteristic plus $S$.  Comparing and solving for $S$ gives the formula pretty quickly.<|endoftext|>
TITLE: Convergence of $\sum(n^3\sin^2n)^{-1}$
QUESTION [159 upvotes]: I saw a while ago in a book by Clifford Pickover, that whether $\displaystyle \sum_{n=1}^\infty\frac1{n^3\sin^2 n}$ converges is open. 
I would think that the question of its convergence is really about the density in $\mathbb N$ of the sequence of numerators of the standard convergent approximations to $\pi$ (which, in itself, seems like an interesting question). Naively, the point is that if $n$ is "close" to a whole multiple of $\pi$, then $1/(n^3\sin^2n)$ is "close" to $\frac1{\pi^2 n}$.
[Numerically there is some evidence that only some of these values of $n$ affect the overall behavior of the series. For example, letting $S(k)=\sum_{n=1}^{k}\frac1{n^3\sin^2n}$, one sees that $S(k)$ does not change much in the interval, say, $[50,354]$, with $S(354)<5$. However, $S(355)$ is close to $30$, and note that $355$ is very close to $113\pi$. On the other hand, $S(k)$ does not change much from that point until $k=100000$, where I stopped looking.]
I imagine there is a large body of work within which the question of the convergence of this series would fall naturally, and I would be interested in knowing something about it. Sadly, I'm terribly ignorant in these matters. Even knowing where to look for some information on approximations of $\pi$ by rationals, or an ad hoc approach just tailored to this specific series would be interesting as well.

REPLY [30 votes]: There is an even bigger reduction that can be done:
Theorem: The Flint Hills series converges if and only if the series
$$
\sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n\pi - p_n)^2}
\qquad{(1)}
$$
converges, where $(p_n/q_n)_1^\infty$ is the sequence of convergents of $\pi$.
Proof: Let
$$
S = \sum_{q = 1}^\infty \frac{1}{q^3 (q\pi - p)^2},
\qquad{(2)}
$$
where $p\in\mathbb N$ is chosen to minimize $|q\pi - p|$. As Wadim Zudilin argued, the Flint Hills series converges if and only if $S$ converges. Now consider the unimodular lattice $\Lambda = \{(q,q\pi - p) : p,q\in\mathbb Z\}$. We can rewrite $S$ as
$$
S = \sum_{\substack{(q,r)\in\Lambda^* \\ q > 0 \\ -1/2 < r < 1/2}} \frac{1}{q^3 |r|^2}\cdot
$$
Here $\Lambda^* = \Lambda\setminus\{\mathbf 0\}$. Next, using the identity
$$
\frac{1}{q^3 r^2} = \int_{s > q} \int_{t > r} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \;dt\;ds
= \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} [s > q][t > r] \;dt\;ds
$$
we get
$$
S = \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \sum_{\substack{(q,r)\in\Lambda^* \\ q > 0 \\ -1/2 < r < 1/2}}  [s > q][t > |r|] \;dt\;ds\\
= \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \#\big\{(q,r)\in\Lambda^* : 0 < q < s,\; |r| < \min(t,1/2)\big\}  \;dt\;ds.
\qquad{(3)}
$$
We can bound the integrand in two different ways, depending on whether or not
$$
N_{s,t} := \#\big\{(q,r)\in\Lambda^* : 0 < q < s,\; |r| < \min(t,1/2)\big\} \leq \max(0,3st - 1/2).
\qquad{(4)}
$$
If (4) holds, then it can be used to bound the entire integral; I leave it to the reader to verify that the resulting integral converges. So let us consider the cases where (4) fails.
Fix $s > 1$ and $t > 0$, and let $D_{s,t} = (-s,s)\times(-t,t)$. If $D_{s,t}$ contains two linearly independent elements of $\Lambda$, then $D$ contains a fundamental domain for $\Lambda$, say $F$; we have
$$
1 + 2N_{s,t} = \#(\Lambda\cap D_{s,t}) = \sum_{\mathbf x\in\Lambda\cap D_{s,t}} m(\mathbf x + F)
= m\left(\bigcup_{\mathbf x\in\Lambda\cap D_{s,t}}(\mathbf x + F)\right)
\leq m(2D_{s,t}) = 4st,
$$
which implies that (4) holds. Similarly, if $D_{s,t}\cap\Lambda = \{\mathbf 0\}$, then (4) holds.
So if we assume that (4) fails for some pair $(s,t)$, then we have $D_{s,t}\cap\Lambda = D_{s,t}\cap \mathbb Z\mathbf x$ for some $\mathbf x = (q,r)\in D_{s,t}\cap\Lambda^*$. It follows that
$$
\max(1,3st - 1/2) \leq N_{s,t} = \left\lfloor \min\left(\frac sq,\frac t{|r|}\right)\right\rfloor \leq \min\left(\frac sq,\frac t{|r|}\right)
$$
and thus
$$
\frac{st}{q|r|} \geq \min\left(\frac sq,\frac t{|r|}\right)^2 \geq \max(1,3st - 1/2)^2 \geq \max(1,3st - 1/2) \geq 2st,
$$
so $q|r| = q|q\pi - p| \leq 1/2$. A well-known theorem now implies that $p/q$ is a convergent of $\pi$, i.e. $(q,p) = (q_n,p_n)$ for some $n\in\mathbb N$. So if we let
$$
\Lambda_c = \{(k q_n, k(q_n\pi - p_n)) : n,k\in\mathbb N\}
$$
then
$$
N_{s,t} = \#(\Lambda_c\cap D_{s,t}).
$$
In other words, the only points which are contributing to the integrand of (3) are points which come from $\Lambda_c$. Reversing the argument of (3) now gives
$$
S \leq C + \sum_{\substack{(q,r)\in\Lambda_c \\ q > 0 \\ -1/2 < r < 1/2}} \frac{1}{q^3 |r|^2},
$$
where $C < \infty$ is a constant describing an upper bound on the contribution to the integral (3) of pairs $(s,t)$ satisfying (4). Thus,
$$
S \leq C + \sum_{n = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(k q_n)^3 (k(q_n \pi - p_n))^2}
= C+\zeta(5)\sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n \pi - p_n)^2}\cdot
$$
It follows that (1) converges if and only if (2) converges.

If this proof was too technical to follow, I'll try to summarize the main ideas: First of all, any rational number $p/q$ which is not a convergent of $\pi$ must satisfy $q|q\pi - p| > 1/2$ (this is a well-known fact). By itself this fact isn't enough to guarantee that the terms coming from non-convergents won't make the series (2) diverge, since you end up comparing it with the harmonic series, which (just barely) diverges. But that's just the crudest possible bound: most rationals $p/q$ will satisfy $q|q\pi - p| \gg 1$. Since (2) involves a summation over all $q$, there will be a lot of "averaging", and so the "spikes" which occur when $q|q\pi - p|$ is small will be washed out in the long run. In order to formalize this you need to talk about lattices and fundamental domains - basically, the idea is that the number of intersection points of a lattice with a convex centrally symmetric region is about the same as the area of the region except for certain exceptional cases; these exceptional cases turn out to correspond to the convergents of $\pi$.
Corollary:
If the exponent of irrationality of $\pi$ is strictly less than $5/2$, then the Flint Hills series converges.
Proof: If $\mu(\pi) < 5/2$, then there exists $\varepsilon > 0$ such that for all but finitely many $n$, we have
$$
|q_n \pi - p_n| \geq q_n^{-3/2 + \varepsilon}.
$$
This gives the following upper bound for (1):
$$
\sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n^{-3/2 + \varepsilon})^2} = \sum_{n = 1}^\infty \frac{1}{q_n^{2\varepsilon}}\cdot
$$
But since the sequence $(q_n)_1^\infty$ must grow at least exponentially fast, this series converges.<|endoftext|>
TITLE: If Erdős is published as Erdös in a paper, which do I cite?
QUESTION [15 upvotes]: There seems to be a few papers around with Erdős written as Erdös.  For example:
MR0987571 (90h:11090) Alladi, K.; Erdös, P.; Vaaler, J. D. Multiplicative functions and small divisors. II. J. Number Theory 31 (1989), no. 2, 183--190. (Reviewer: Friedrich Roesler) 11N37

Would it be incorrect to cite such papers using Erdős instead?

REPLY [5 votes]: This is a matter of convention.  One guideline is in paragraph 17.20 of the Chicago Manual of Style (15th ed):
Authors' names are normally given as they appear in the title pages of their books. Certain adjustments, however, may be made to assist correct identification (unless they conflict with the style of a particular journal or series.  First names may be given in full place of initials.  If an author uses his or her given name in one cited book in and initials in another (e.g., "Mary L. Jones" versus "M. L. Jones"), the same form, preferably the fuller one, should be used in all references to that author.
I would err on the side of consistency.  Some bibliography styles in LaTeX/bibtex replace subsequent references to the same author with an em dash.  If you use many different spellings of the same author's name, this behavior will break.<|endoftext|>
TITLE: Rallis inner product formula for U(2,2) and U(3)
QUESTION [5 upvotes]: Victor Tan has a couple of papers on a regularized Siegel-Weil formula for U(2,2) and U(3).  The papers I'm talking about are:

"A Regularized Siegel-Weil Formula on U(2,2) and U(3)", Duke, 1998.
"An Application of the Regularized Siegel-Weil Formula on Unitary Groups to a Theta Lifting Problem", Proceedings of the AMS, 1999.

One natural thing to look for after obtaining such a result is a Rallis inner product formula.  Tan doesn't prove such a result in either of the two references mentioned above, though he seems to come close at the end of the second one.
Does anyone know if such a formula is written down anywhere?

REPLY [2 votes]: See recent preprint of Harris-Li, which is base on Ichino`s S-W formula for unitary groups<|endoftext|>
TITLE: Survey articles on homotopy groups of spheres
QUESTION [15 upvotes]: Are there general surveys or introductions to the homotopy groups of spheres? I'm interested especially in connections to low-dimensional geometry and topology.

REPLY [32 votes]: While my Algebraic Topology book and my unfinished book on spectral sequences (referred to in other answers to this question) contain some information about homotopy groups of spheres, they don't really qualify as a general survey or introduction.  One source that fits this bill more closely is Chapter 1 of Doug Ravenel's "green book" Complex Cobordism and Stable Homotopy Groups of Spheres, from 1986.  This introductory chapter starts at a reasonably accessible level, with increasing prerequisites in the later sections of the chapter. More recent surveys ought to exist, although at the moment I can't recall any.  With the recent solution of the Kervaire invariant problem by Hill-Hopkins-Ravenel, this would be a good time for an updated survey.
Connections between homotopy groups of spheres and low-dimensional geometry and topology have traditionally been somewhat limited, with the Hopf bundle being the thing that comes most immediately to mind.  A fairly recent connection is Soren Galatius' theorem that the homology groups of $Aut(F_n)$, the automorphism group of a free group, are isomorphic in a stable range of dimensions to the homology groups of "loop-infinity S-infinity", the space whose homotopy groups are the stable homotopy groups of spheres.<|endoftext|>
TITLE: Why are modular forms interesting?
QUESTION [126 upvotes]: Well, I'm aware that this question may seem very naive to the several experts on this topic that populate this site: feel free to add the "soft question" tag if you want... So, knowing nothing about modular forms (except they're intrinsically sections of powers of the canonical bundle over some moduli space of elliptic curves, and transcendentally differentials on the upper half plane invariant w.r.t. some specific subgroup of $SL(2,\mathbb{Z})$), I have the curiosity -that many other non experts might have- to understand a bit why that is considered a so vast and important topic in mathematics. The wikipedia page doesn't help: on the contrary, it makes this topic appear as quite narrow and merely technical.
I would roughly divide the question into three (though maybe not neatly distinct) parts:

1) Why are modular forms per se interesting?

That is, do they "generate" some piece of rich self-contained mathematics? To make an analogy: cohomology functors were born as applied tools for studying spaces, but have then evolved to a very rich theory in itself; can the same be said about m.f.'s?

2) How are modular forms deeply related to other, possibly quite distant, mathematical areas?

For example: I've heard about deep relations to some generalized cohomology theories (elliptic cohomology) via formal group laws coming from elliptic curves; and I've heard about the so called moonshine conjecture; there should also be some more classical relations to the theory of integral quadratic forms and diophantine equations, and of course to elliptic curves; and people here always mention Galois representations...

3) Why are modular forms useful as "applied" technical tools?

In this last question I'm ideally expecting indications of cases (or actual theorems) in which some questions that do not involve modular forms are asked about some mathematical objects, and an answer that does not involve m.f.'s is given, but the method used to obtain that answer/proof makes consistent use of m.f.'s.

REPLY [4 votes]: Elsewhere commented. However because of the importance of the only known proofs of optimality of sphere packing in higher dimensions posting here. https://www.quantamagazine.org/sphere-packing-solved-in-higher-dimensions-20160330/ covers the solution to optimality of known sphere packings in dimensions $8$ and $24$ using modular forms and this is one reason why modular forms are interesting as they show up naturally to solve a central problem in an area of mathematics.<|endoftext|>
TITLE: Cyclic extensions
QUESTION [8 upvotes]: Hi. Are there nice/simple examples of cyclic extensions $L/K$ (that is, Galois extensions with cyclic Galois group) for which $L$ cannot be written as $K(a)$ with $a^n\in K$?
Thanks.

REPLY [5 votes]: Take $\zeta = e^{2\pi i/p}$ for a prime number $p\equiv1$ (mod 3), e.g. $p=7$.
Then  $Q(\zeta+\bar\zeta)$ is a totally real cyclic Galois extension of $\mathbf{Q}$ of degree a multiple of 3, hence contains a cubic extension $L$ that is Galois with cyclic Galois group.
Being totally real it cannot be the splitting field of a polynomial of the
form $X^3-a$. (or use David Loeffler's argument above).
Dirichlet's theorem on primes in arithmetic progressions assures us that  we have infinitely many such examples over the base $\mathbf{Q}$.<|endoftext|>
TITLE: Are there primes p, q  such that p^4+1 = 2q^2 ?
QUESTION [5 upvotes]: $\exists p, q \in \mathbb{P}: p^4+1 = 2q^2$? I suspect there is some simple proof that no such p, q can exist, but I haven't been able to find one.
Solving the Pell equation gives candidates for p^2=x and q=y, with x=y=1 as the first candidate solution and subsequent ones given by x'=3x+4y, y'=2x+3y; chances of a prime square seem vanishingly unlikely as x increases, but I don't have a proof.
Meta: how do you search for a question like this? I looked for a searching HOWTO here and on meta, and couldn't find one. That the search appears to strip '^' and '=' makes it all the harder.

REPLY [20 votes]: This is not my solution, but I don't remember where I learned it.
Square both sides, subtract $4p^4$, and divide by 4 to obtain
$({p^4-1\over 2})^2=q^4-p^4$.
However, $z^2=x^4-y^4$ has no solutions in non-zero integers.
This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem.
The proof uses the representation of Pythagorean triples and infinite
descent. 
So you must have $p=\pm 1$.<|endoftext|>
TITLE: Characterizing a tumbling convex polytope from the surface areas of its two-dimensional projections
QUESTION [9 upvotes]: My general question concerns what we can learn about an arbitrary, three-dimensional convex polytope (or convex hull of an arbitrary polytope) strictly from the surface areas of its two-dimensional projections on a plane as it 'tumbles' in 3-space (i.e. as it rotates along an arbitrary, shifting axis).  
If it's helpful, please imagine the following physical set-up:
We take an arbitrary three-dimensional convex polytope, and fix the center of mass to a coordinate in 3-space, $(x_0, y_0, z_0)$, located at some distance, $D$, above a flat surface.  While this prohibits translation of the center of mass, the polytope is still allowed to tumble freely (i.e. it is allowed rotation around an arbitrary axis centered at the fixed coordinate).
There is no 'gravity' or other force to stabilize the tumbling polytope in a particular orientation.  Over time it will continue to tumble randomly.  (The 'physical' set-up is only meant for descriptive reasons.)
We shine a beam of coherent light on the tumbling polytope, larger than the polytope's dimensions, and continually record the area of the resulting shadow, or two-dimensional projection on the surface.  To be clear, the area of the two-dimensional projection is the only information we are allowed to observe or record, and we are allowed to do so over an arbitrary length of time.

My question is - From observing the area of the tumbling polytope's shadow, or two-dimensional projection over time, what can we learn about it's geometry?  To what extent can we characterize and/or reconstruct the polytope from its changing shadow (extracting the surface area for example - hat tip to Nurdin Takenov)? 
Do we gain anything by watching the evolution of the convex polytopes shadow as it tumbles (part of the point for the physical example), as opposed to an unordered collection of two-dimensional projections?
Update - Nurdin Takenov (and Sergei Ivanov in later comments) nicely points out that we can use the average surface area of the two-dimensional projection to find the surface area of the tumbling convex polygon.  Might we be able to find it's volume?  
(Addendum - I would be really neat if somebody could point me to any algorithms in the literature... or available software.... that let's me calculate and characterize two-dimensional surface projections of convex polytopes!)

REPLY [6 votes]: Hej, 
there seem to be a number of variations of the original question out there. 
Let me tell you what is known (to me): 
Let us assume that an (unknown) convex polytope $P$ in $R^3$ is given. 
I assume that we can observe orthogonal projections onto planes (think of shadows 
that we see on a screen that is orthogonal to the direction of the incoming parallel light). 
If the direction of the incoming light is a unit vector u, then the information we have 
is the data 
$
F(P,u)=area(proj_u(P))
$
where $proj_u(P)$ is the orthogonal projection of $P$ along the direction $u$. 
This function is called projection function or brightness function in convex geometry. 
What do we know about $P$, when we know the projection function for all directions u?
1) The surface area of $P$ is proportional to the averaged function $F(P,u)$, averaged over all directions (was remarked before, and is a consequence of Cauchy's projection formula, 
even true in all dimensions and when the polytope is some convex set). 
2) If $P$ is translated then $F(u)$ does not change, so the best we can hope for is determination up to translations. In general $P$ is not uniquely determined up to translations by $F(P,u)$. Also this was remarked earlier. It is also clear. 
Take some polytope $P$ which does not have a centre of symmetry. Then its 
reflection at the origin $\hat P$ is not a translation of the original $P$, but 
$F(P,u)=F(\hat P,u)$ for all $u$. 
3) Amazingly, the central symmetry is the crucial property: If $P$ is central symmetric
the $F(P,u)$ (known for all u) determines $P$ uniqely up to translations. 
This was already shown by Alexandrov and is often called Alexandrov's projection theorem. 
It holds in all dimensions and for arbitrary central symmetric convex bodies. See Richard Gardner's book on geometric tomography. 
4) Under the central symmetry assumption, there is an algorithm to determine $P$ from 
$F(P,u)$, see the article RECONSTRUCTION OF CONVEX BODIES FROM BRIGHTNESS FUNCTIONS
by R. Gardner and Peyman Milanfar. 
Best wishes, Markus Kiderlen<|endoftext|>
TITLE: What is the coordinate ring of symmetric product of affine plane?
QUESTION [5 upvotes]: The symmetric product of a variety $M$ is the quotient of $M^n/S_n$ where $S_n$ is the symmetric group permuting components of n-fold product $M^n$. IF $M$ is an affine plane $C^k$ over complex numbers, the coordinate ring of the symmetric product is the invariant polynomials in $R:=C[x^1_1,...,x^1_k, x^2_1,...,x^2_k,... ,x^n_1,...,x^n_k]$ under the action of $S_n$ where $S_n$ permutes the variables $x_i^1,...,x_i^n$ simultaneously for $i=1,...,k$. I want to know the invariant subring $R^{S_n}$ in terms of generators and relations. Could anybody help me?

REPLY [2 votes]: The relations might be complicated.  The multisymmetric functions of degree up to n generate the ring, but very redundantly.  In Lemma 2.2 of 
http://annals.princeton.edu/annals/2006/163-2/p11.xhtml
Venkatesh and I show that you can get by with using many fewer of these multisymmetric functions, if you are content to generate a subring of R^{S_n} whose fraction field is finite-index in the fraction field of R^{S_n}.<|endoftext|>
TITLE: Banach spaces with a certain separability property
QUESTION [7 upvotes]: In Ledoux and Talagrand's "Probability in Banach Spaces", for technical reasons they frequently assume that a Banach space $B$ has the property that the unit ball of $B^*$ contains a countable subset $D$ such that
$$ \Vert x \Vert = \sup_{f\in D} \vert f(x) \vert$$
for every $x\in B$.  Examples of such spaces $B$ include both all separable Banach spaces, and all duals of separable Banach spaces (e.g. $\ell_\infty$).
My question is, is there a standard name or alternative characterization of spaces with this property?  The authors decline to discuss this at all, except to point out that separable spaces and $\ell_\infty$ have this property; they don't even point out that it extends to duals of separable spaces.

REPLY [4 votes]: A subset D of $B^{*}$ with the property you describe is called "norming" (wether or not it is countable).
Recall that a subset D is called "total" if if $d(x) = 0$ for all $d \in D$ implies $x = 0$.
Every norming set is total.
Maybe a partial answer to your question is given by proposition 5.17 in the "Handbook of the Geometry of Banach Spaces" volume 1, chapter 15 "Infinite dimensional convexity", it says when the converse is true:
Every total subset of $B^{*}$ 
is norming iff B is quasireflexive, that is $\frac{dim(B^{**})}{dim B} \lt \infty$.<|endoftext|>
TITLE: Quotients of Abelian varieties by finite groups
QUESTION [9 upvotes]: Let $A$ be an Abelian variety (over an algebraically closed field).  The group $\mathbb{Z}/2\mathbb{Z}$ acts on $A$ and the quotient is called the Kummer variety of $A$.  These are well studied and, as I understand it, their geometry is understood and a classical subject (especially for surfaces).
What happens when you take the quotient of an Abelian variety by other finite groups?  Is the geometry of the quotient understood in any sense (e.g., Kodaira dimension)?  Are the quotients somehow classified?  Is anything else known about them?

REPLY [2 votes]: If you are still interested at the moment there is a complete classification in 3 and 4 dimension when the group acts freely on the abelian varieties (they are called hyperelliptic varieties). Some authors are Catanese and Demleitner: if you search their papers you can find others (and better) references.
Maybe you are also interested in the so called generalized kummer.<|endoftext|>
TITLE: construct scheme from quivers?
QUESTION [12 upvotes]: I heard from some guys working in noncommutative geometry talking about the idea that one can construct the noncommutative space from quivers. I feel it is rather interesting. However, I can not image how one can construct a non-affine scheme or space. For affine case, one might consider the path algebra of a quiver $Q$, say, $KQ$. It is a hereditary algebra(or quasi-free algebra, formally smooth algebra). How can one construct a non-affine space from quiver?
Is it possible to construct a noncommutative space using different interesting quivers as "local data"?
If it is possible, the algebraic geometry machine working on this space will be helpful to study the representations of quiver?
Thanks

REPLY [15 votes]: It is possible to construct more general schemes (even 'projective' schemes) starting from quiver-representations.
The idea is to consider a stability structure and consider the corresponding moduli spaces of semi-stable quiver representations as introduced by Alastair King (Moduli of representations of finite dimensional algebras, Quat. J. Math. Oxford 45 (1994) 515-530.)
If one starts with a quiver without oriented cycles, these moduli spaces are projective and so it makes sense to take the collection of all these moduli spaces (or multiples of a fixed dimension vector compatible with the stability structure) as a noncommutative projective space. This point of view was advocated in 'Noncommutative compact manifolds constructed from quivers'.
More generally, allowing cycles, the collection of moduli spaces (or multiples of fixed dim vector) for all dim vectors compatible with the stabilitu structure can be viewed as a noncommutative scheme.
Here the main idea is that polynomial semi-invariants of quivers are known to be generated by so called determinental semi-invarinats (a result by Aidan Schofield and Michel Van den Bergh 'Semi-invariants of quivers for arbitrary dimension vectors'.
What this says is that these collections of moduli spaces have an affine cover determined by universal localizations of the path algebra. This point of view is expressed here.
EDIT : Here's an example to illustrate all of this : noncommutative projective n-space. Consider the quiver with 2 vertices and n+1 arrows, all from the first to the second vertex, Take stability structure (-1,1), then we consider only representations of dimension vector (m,m). A semi-stable representation is one given by the n+1 mxm matrices X0,...,Xn such that there are numbers ai with det(a0X0+a1X1+...+anXn) is non-zero. For example, consider the open piece where Xi is invertible, then one can use Xi to identify the two vertex-spaces and the other arrows now become loops in that common vertex. That is, this open piece gives all representations of the free algebra in n variables. Hence, we can cover the moduli space by affine pieces, all iso to the reps of a one-vertex quiver with n loops.
Clearly one can generalize this procedure to glue certain representations of two very different quivers. Formally invert one arrow in your quiver, this comes down to taking a stability structure with 0's at all other vertices and (-1,1) at start and end-point of your arrow. Then the 'Zariski open piece' of repQ consists the representations of a new quiver Q' in which one identifies the two vertices, and turns all other arrows between these two vertices into loops at the new vertex. In this way to quivers can be 'glued' to form more general schemes provided they are isomorphic after doing this 'invert one arrow' trick on each of them a finite number of times.
Universal localizations corresponding to 'determinental semi-invariants' are a generalization of this idea, allowing paths (or even collections of linear combinations of paths) to be 'inverted'.  (End EDIT)
A recent incarnation of this strain of ideas (restricting to stable rather than semi-stable representations) is taken by Markus Reineke in his study of 'noncommutative Hilbert schemes' which have been used by Markus recently in 'Cohomology of quiver moduli, functional equations, and integrality of Donaldson-Thomas type invariants'.
As to the local structure of these noncommutative schemes. Their 'local rings' are indeed again path algebras of 'local quivers' which can be determined from the stable components of the semi-stable representation. The details are here.
If you want to extend all of this to general formally smooth algebras (rather than just path algebras) a place to start is with 'One quiver to rule them all'.
Finally, Ive tried to collect all of this in a book as Daniel mentioned in his answer (one might replace 'blablabla' by 'noncommutative geometry'...). The latest version of it can be downloaded here.<|endoftext|>
TITLE: Why do the homology groups capture holes in a space better than the homotopy groups?
QUESTION [38 upvotes]: This is a follow-up to another question. 
A good interpretation of having an $n$-dimensional hole in a space $X$ is that some image of the sphere $\mathbb{S}^n$ in this space  given by a mapping $f: \mathbb{S}^n \rightarrow X$ cannot shrink down to a point. The matter of "shrinking to a point" is best expressed by being $f$ homotopic to some constant map. Next, the homotopy groups $\pi_n$ can be defined as the homotopy classes of base-point preserving maps from $\mathbb{S}^n$ to $X$. In this way it might be argued that the homotopy groups $\pi_n$ should best capture the holes in $X$. 
But this is not so. One has the most satisfying result that for $i \geq 1$ the homology $H_i (\mathbb{S}^n) $ is nontrivial iff $n = i$. But the higher homotopy groups of spheres are very complicated.
Why does this complication occur? Why are homology groups far better for capturing the holes than the homotopy groups, which are intuitively better suited, but are not actually so? In the case of $1$-dimensional holes, the homology $H_1$ and $\pi_1$ captures the holes equally well; but of course in this case the former is the abelianization of the latter.

REPLY [12 votes]: For another example that homology and homotopy capture holes differently: note that the curve $C$ in the bitorus below is homologous to zero but not homotopic to zero.
 
Also, a remarkable difference between them is that while an homology chain can be subdivided into smaller chains, a map from a sphere cannot be subdivided into smaller maps from a sphere. Historically, higher homotopy groups were only pursued after Hopf's fundamental example of a map $S^3 \to S^2$ not homotopic to zero, which was already commented above (I recommend the reading of H. Samelson "$\pi_3(S^2)$, H. Hopf, W. K. Clifford, F. Klein", In: History of Topology, p.575-578, Elsevier, Amsterdam, 1999.)<|endoftext|>
TITLE: Perturbations of an operator that disconnect the spectrum 
QUESTION [9 upvotes]: The following question came to me while working on a technical matter about transversality in infinite dimension, and I'm really curious to know whether it has an affirmative answer at least under extra hypotheses.

Let A be a bounded linear operator on
  a Banach space E. Does it exist a
  bounded linear operator  S such that 0
  and 1 do not belong to the same
  connected component of the  spectrum
  of the operator AS:= A + A(I-A)S?

That is, S is OK if either 0 or 1 is not in the spectrum of AS, or if they both are in the spectrum, they should belong to different connected component of it. Thus it may be assumed that 0 and 1 belong to the same component of spec(A), otherwise S=0 trivially solves the problem.
The first idea is to look for S of the form f(A), but this can't work if f is continuous,
since then spec(AS) is the continuous image of spec(A) with a map that fixes 0 and 1. However, if A admits a discontinuous functional calculus (e.g. a normal operator on Hilbert space), the trick does work. 
I do not know the answer to the question even on Hilbert spaces. In a general Banach space the problem seems even harder, due the difficulty of building operators.
I'd very grateful of any suggestion! (Pietro Majer).
edit (17/11/2011). 
Here are a few more or less trivial facts that I know.

For a Banach space $X$, the set $\mathcal{A}$ of all $A\in L(X)$ such that there exists  $S\in L(X)$ such that no connected component of $\operatorname{spec}(A_S)$ contains both $0$ and $1$, is an open set; 
If $A\in \mathcal{A}$, then there is $S$ such that $A_S$ is even a linear projector (thus satifying the condition on the spectrum ad abundantiam); 
If $A\in L(X)$ and   $A_S\in \mathcal{A}$ for some $S\in  L(X)$, then $A$ itself is in $\mathcal{A}$;
$A\in \mathcal{A}$ if and only if there are closed subspaces $V$ and $W$ of $X$ such that $V\times W\ni (v,w) \mapsto Av + (I-A)w \in X$ is bijective; 
if $AX$ is a closed subspace and $(I-A)^{-1}(AX)$ is a complemented subspace of $X$, then $A\in \mathcal{A}$.

REPLY [6 votes]: For Hilbert spaces, the conjecture follows from fact 4 and the answer to question Complement of a subspace which is a cartesian product applied to the kernel of the map $H\times H\ni (v,w) \mapsto Av + (I-A)w \in X$ .<|endoftext|>
TITLE: "negative" vs "minus"
QUESTION [8 upvotes]: Not too long ago, whenever I was confronted with the expression, -x, and I was in a position where I needed to communicate it to someone verbally, I would say, "negative x", as opposed to "minus x", probably because "negative x" sounded more professional to me.  This went on until a professor pointed out to me some of the problems with this usage of the word negative.
After all, whatever the object "negative x" is, it should at the very least be negative, right?  However, objects written down as $-x$ (or $-g$ or whatever) are usually elements of a vector space or an abelian group, where there is usually no concept of negative element.  Even when $x$ is a real number where the concept of negative element exists, the terminology is still bad (in fact it is probably worse, since in the case where $x$ is a negative number, "negative x" is not negative, but positive!).
I suspect that people regard the expression "negative x" as somehow emphasizing the fact that $-x$ is the additive inverse of $x$.  This has other harmful effects.  Just last year, acting under these pernicious influences, I caught myself telling linear algebra students about the "existence and uniqueness of negatives in a vector space"!
On the other hand the alternative, "minus x", is a straightfoward and unambiguous verbal description of the written form of the expression.
One might argue that this is a trivial issue which is hardly worth the effort of discussion, but mathematicians prize clarity, and if one alternative is clearly better than the other, why not stick to it?  Another reason for using better terminology is that even though the underlying issues are trivial to us, they may not be clear to others.  I distinctly recall being frustrated to tears at one point as a child, trying to understand how to add integers, and the phrase "negative one", etc. very well may have been a serious source of confusion.  At the very least this discussion may be helpful for teaching mathematics to five year olds.
The expression "negative x" is not some fictitious straw man of my own construction.  To the contrary, in my experience, it is the dominant verbal alternative.  This turn of phrase even turns up in mathematical writing when authors decide not to assign a symbol to the mathematical object they are discussing.  For example: "The 1-form is the negative of the differential of the function". It seems to me that the case for "minus x" is very strong, but I have nevertheless had a lot of difficulty winning people over to the "minus" side of the debate. Some argue that saying "negative x" is logical because it describes the process of obtaining -x from x by mutliplying x by -1, which is a negative number.  However this strikes me as a convoluted way of constructing terminology, and the argument does nothing to address the potential for confusion.  
I'll admit, there may be some arguments for "negative x" or against "minus x" which I haven't considered.
So what do you think? 

"negative x" or "minus x"?

REPLY [2 votes]: As far as I can check, in french this issue does not arise. The only available description is "moins x" or "moins 3", corresponding to "minus x" or "minus 3". The closest equivalent of "negative x" would be "x négatif" and it would only be used in phrases such as "pour x négatif", meaning "for x < 0" and not as a description of part of such a formula. "Négatif" seems to me mostly used in french in a categorical way "températures négatives", "nombres négatifs".
It would be informative if native german, russian, chinese, etc. speakers could comment on this.<|endoftext|>
TITLE: Why “syntomic” if “flat, locally of finite presentation, and local complete intersection” is already available?
QUESTION [15 upvotes]: Dear everyone,
(i) Who is the father of the adjective “syntomic” in algebraic geometry?
(ii) And why did he choose to introduce a new term for what we already know from EGA IV.19.3.6 and SGA 6.VIII.1.1 as “flat, locally of finite presentation, and local complete intersection”? 
Thanks!

REPLY [25 votes]: Mazur gives the following beautiful justification, which explains the “syn-” in “syntomic” as well.

Dear Thanos,
Thanks for your question.  I'm thinking of “ local complete
  intersection” as being a way of cutting out a (sub-) space from an
  ambient surrounding space; the fact that it is flat over the parameter
  space means that each such "cutting" as you move along the  parameter
  space, is---more or less---cut out similarly. I'm also thinking of the
  word "syntomic" as built from the verb temnein (i.e., to cut) and the
  prefix "syn" which I take in the sense of "same" or "together".  So I
  think it  fits.
Best wishes,
Barry<|endoftext|>
TITLE: If two fields are elementarily equivalent, what can we say about their Witt rings?
QUESTION [6 upvotes]: The question is in the title exactly as I want to ask it, but let me provide some background and motivation.
Many of the properties of fields studied in the algebraic theory of quadratic forms are manifestly elementary properties in the sense of model theory: that is, if one field has this property, then any other field which has the same first-order theory in the language of fields has that same property.  Examples:
being quadratically closed, being formally real, being real-closed, being Pythagorean (sum of two squares is always a square), for any fixed positive integer n, having I^n = 0 (follows from the Milnor conjectures!), the u-invariant, the level, the Pythagoras number...
These properties imply that at least for some fields $K$, if $L$ is any field elementarily equivalent ot $K$, then $W(L) \cong W(K)$: e.g. $K$ is quadratically closed, $K$ is real-closed, $K = \mathbb{C}((t))$.  Is it always the case that $K \equiv L$ implies $W(K) \cong W(L)$?  I am pretty sure the answer is no because for instance if $\operatorname{dim}_{\mathbb{F}_2} K^{\times}/K^{\times 2}$ is infinite, I think it is not an elementary invariant.  And if you take a field with vanishing Brauer group, then $W(K)$ is, additively, an elementary $2$-group of dimension $\operatorname{dim}_{\mathbb{F}_2} K^{\times}/K^{\times 2} + 1$.
But are there known positive results in this direction?

REPLY [8 votes]: As you point out, one cannot hope that the Witt ring, up to isomorphism, be an elementary invariant of a field. The strongest statement which I might conjecture would be that if $K \preceq L$ is an elementary extension of fields, then $W(K) \to W(L)$ is an elementary extension of rings.  If this statement were true, then the theory of the Witt ring would be an elementary invariant as any two elementarily equivalent fields have a common elementary extension.
It is true that if $K \preceq L$ is an elementary extension of fields, then map $W(K) \hookrightarrow W(L)$ is an inclusion [Why?  Being zero in the Witt ring is defined by an existential condition.]   One might try to prove that $W(K) \hookrightarrow W(L)$ is elementary by induction where the key step would be to show that if $W(L) \models (\exists x) \phi(x;a)$ where $a$ is a tuple from $W(K)$, $x$ is a single variable, and $\phi$ is a formula in the language of rings, then $W(K) \models (\exists x) \phi(x;a)$.  The witness in $W(L)$ would be represented by a quadratic space of some finite dimension $n$.  One would like to argue that the set defined by $\phi(x;a)$ in the space of $n$-dimensional quadratic forms is definable in the field language in $K$ in which case a witness could be found in $K$ via elementarity.  This last part of the argument is delicate as it would require knowing bounds for checking equalities in the Witt ring.
The Witt ring construction is an example of an ind-definable set modulo an ind-definable equivalence relation.  These are discussed in some detail in Hrushovski's paper on approximate groups ( arXiv:0909.2190 ).   With Krajiceck, I considered similar issues (how does the Grothendieck ring of a first-order structure depend on its theory) in  Combinatorics with definable sets: Euler characteristics and Grothendieck rings. Bull. Symbolic Logic 6 (2000), no. 3, 311--330.<|endoftext|>
TITLE: Relative K-theory and split exact sequences of C* algebras
QUESTION [10 upvotes]: Let $A$ be a C* algebra, $J$ an ideal, $\pi: A \to A/J$ the quotient map.  Recall that the relative K theory group $K_0(A, A/J)$ consists of equivalence classes of triples $(p,q,x)$ where $p$ and $q$ are projections over $A$ and $\pi(x)$ implements a Murray-Von-Neumann equivalence between $\pi(p)$ and $\pi(q)$.  One has the standard equivalence relations involving homotopies and direct sums, together with the relation $[p,q,x] = 0$ if $x$ implements a Murray-Von-Neumann equivalence between $p$ and $q$.
One can check that the sequence $K(A, A/J) \to K(A) \to K(A/J)$ is exact in the middle, where the first map is given by $[p,q,x] \mapsto [p] - [q]$.  I am trying to understand how the excision theorem for K-theory of C* algebras works, and it would help if I had a direct proof of the following lemma: if the exact sequence $0 \to J \to A \to A/J \to 0$ splits then the map $K(A, A/J) \to K(A)$ is injective.  It would even be helpful to understand what is going on in the simple case where $A$ is the unitalization of a nonunital C* algebra $J$ (where the relevant short exact sequence necessarily splits). 
Can anyone help?

REPLY [2 votes]: Although this question was posted and answered a very long time ago, I just stumbled upon it and I thought it might be worthwhile if an answer is provided near the post (it actually seems like a full answer is not yet available). 
First I will assume $A$ is untital. Second I will use the fact that given $[(p',q',x')]\in K_0(A,A/J)$ I can write it as $[(\ell_k,q,\ell_k)]$ where 
$\pi(q)=\pi(\ell_k)=\ell_k$ where $\ell_k$ is the standard (bigger than $k\times k$) matrix over the complex numbers with ones on the first $k$ entries of the diagonal and zeros elsewhere. A sketch of the proof is: 
-note that 
$$[(p',q',x')]=[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]$$
-note that 
$$[(p'\oplus 1-p',q'\oplus 1-p', x'\oplus 1-p')]=[(\ell_k,q'',x'')]$$ 
- Use the fact that given a partial isometry $y$ we can find a path of unitaries in matrices of quadruple the size from the identity to $V$ such that 
$V^*yy^*V=y^*y$ and $yy^*Vy^*y=y$ (where we include partial isometries and projections in the upper left hand corner in bigger matrices).
-So let $u_t$ this path of unitaries for the partial isometry $\pi(x)$ and let $v_t$ a lift starting at the identity. 
-Note that 
$$[(\ell_k,q'',x'')]=[(\ell_k, q, x)],$$
where $\ell_k=\pi(\ell_k)=\pi(q)=\pi(x)$ (use $(\ell_k,v_t^*q''v_t,v_t^*x'')$). 
-Note that 
$$[(\ell_k, q, x)]=[(\ell_k,q,\ell_k)]$$ by a linear path $t\ell_k+(1-t)x$ since 
$\pi(x)=\pi(\ell_k)$.  
Denote by $\sigma:A/J\rightarrow A$ the section which exists by assumption. Now suppose $r=[(\ell_k, q, \ell_k)]$ is a general element in $K_0(A,A/J)$ such that $[\ell_k]-[q]=0$ in $K_0(A)$. Then there must exist a path of unitaries $u_t$ starting at the identity such that $u_1^*qu_1=\ell_k$. So we have 
$r=[(\ell_k,\ell_k, u_1^*\ell_k)]$. Now denote $U:=\sigma(\pi(u_1))$ then 
$\pi(U)=\pi(u_1)$ and thus by a linear path $r=[(\ell_k,\ell_k, U^*\ell_k)]$. Now note that $U^*\ell_k U=\sigma\pi(u_1^*\ell_ku_1)=\ell_k$ which shows that $(\ell_k,\ell_k,U^*\ell_k)$ is degenerate thus $r=0$. So we find that the kernel of $K_0(A,A/J)$ is trivial. Which was to be shown.<|endoftext|>
TITLE: Efficiently sampling points uniformly from the surface of an n-sphere
QUESTION [15 upvotes]: Is there an efficient way to sample uniformly points from the unit n-sphere? Informally, by "uniformly" I mean the probability of picking a point from a region is proportional to the area of that region on the surface of the sphere. Formally, I guess I'm referring to the Haar measure.
I guess "efficient" means the algorithm should take poly(n) time. Of course, it's not clear what I mean by an algorithm since real numbers cannot be represented on a computer to arbitrary precision, so instead we can imagine a model where real numbers can be stored, and arithmetic can be performed on them in constant time. Also, we're given access to a random number generator which outputs a real in [0,1]. In such a model, it's easy to sample from the surface of the n-hypercube in O(n) time, for example. 
If you prefer to stick with the standard model of computation, you can consider the approximate version of the problem where you have to sample from a discrete set of vectors that $\epsilon$-approximate the surface of the n-sphere.

REPLY [6 votes]: There is a completely different way of doing this based on Hurwitz's 1897 "factorization of measures".  Essentially, you make sequential rotations in $(12)$-plane, then in $(13)$-plane,..., and finally $(n-1,n)$-plane.  You can calculate explicitly the $(ij)$-distribution in terms of beta functions.  The result is a random rotation in $SO(n)$ which in particular gives you a uniform point in the sphere.  See section 2 in this paper by Diaconis and Saloff-Coste for explicit formulas and references.

REPLY [3 votes]: In case you don't want to worry about dividing by small norms (admittedly, less and less of an issue for higher $n$...)
Generate a point $\mathbf{x}$ on the $(n-1)$-sphere; and generate a number $y\in [-1,1]$ with density $k (1-y^2)^{(n-2)/2}$ for the appropriate constant $k$.  Letting $\mathbf{x}'=(y,(1-y^2)^{1/2} \mathbf{x})$ is a uniform point on the $n$-sphere.

Alternatively, in the unlikely event that you don't mind funny correlations between consecutively generated points, you might prefer randomly generating elements of the orthogonal group, for instance letting $A_n$ be sufficiently random${}^1$ antisymmetric matrices, $O_n=(I-A_n)^{-1}(I+A_n)$ is orthonormal, and then the sequence of matrices $U_0=I$, $U_{n+1} = O_n U_n$ is uniformly distributed w.r.t. Haar measure.  If later you want to kill consecutive correlations, generate a bunch of them and then sample the sequence at random.

${}^1$: "Sufficiently random": Leveraging the mean-value theorem for all it's worth, we might even take the $A_n$s concentrated on a "generic curve in general position".  For sure this isn't efficient, but it gives an idea of how little is needed to get quantitative convergence to uniformity.<|endoftext|>
TITLE: Neutral tic tac toe
QUESTION [24 upvotes]: I heard this puzzle from Bob Koca.  Suppose we play misere tic-tac-toe (a.k.a. noughts and crosses) where both players are X.  Who wins?
That particular puzzle is easy to solve, but more generally, has $n \times n$ impartial tic tac toe, in both normal and misere forms, been studied before?

EDIT: Thane Plambeck's paper, mentioned at the end of his answer below, coined the term Notakto for this game. That name seems to have caught on; for example, there is now a Wikipedia article on Notakto.

REPLY [10 votes]: It's possible to give a complete theory of 3x3 misere "X-only" tic-tac-toe disjunctive sums by introducing the 18-element commutative monoid $Q$ given by the presentation 
$Q = \langle\ a,b,c,d\ | \ a^2=1,\ b^3=b,\ b^2c=c,\ c^3=ac^2,\ b^2d=d,\ cd=ad,\ d^2=c^2 \rangle\ $.
Such a "disjunctive" game of 3x3 neutral tic-tac-toe is played not just with one tic-tac-toe board (as has been previously discussed in this thread), but more generally with an arbitrary (finite) number of such boards forming the start position.  On a player's move, he or she selects a single one of the boards, and makes an X on it (a board that already has a three-in-a-row configuration of X's is considered out-of-play).  Play ends when every board has a three-in-a-row configuration, and the player who completes the last three-in-a-row on the last available board is the loser.  
The game analyzed already in this thread corresponds to play on a 3x3 single board.
The monoid Q arises as the misere quotient of the impartial game 
G = 4 + {2+,0} 
{2+,0} is the canonical form of the 3x3 single board start position, and "4" is the nim-heap of size 4, which also happens to occur as a position in this game.  I'm using the notation of John Conway's On Numbers and Games, on page 141, Figure 32.  
One way to think of Q is that it captures the misere analogue of the "nimbers" and "nim addition" that are used in normal play disjunctive impartial game analyses, localized to the play of this particular impartial game, neutral 3x3 tic-tac-toe.
I performed these calculations partly using Mathematica, and partly using Aaron N. Siegel's "MisereSolver" program.
See also 
http://arxiv.org/abs/math/0501315
http://arxiv.org/abs/math/0609825
http://arxiv.org/abs/0705.2404 
http://www.miseregames.org
It's possible to build a dictionary that assigns an element of Q to each of the conceivable 102 non-isomorphic positions in 3x3 single-board neutral tic-tac-toe. (I mean "non-isomorphic" under a reflection or rotation of the board.  In making this count, I'm including positions that couldn't be reached in actuality because they have too many completed rows of X's, but that doesn't matter since all those elements are assigned the identity element of Q). To determine the outcome of a multi-board position (ie, whether the position is an N-position -- a Next player to move wins in best play, or alternatively, a P-position-- second player to move wins), what a person does is multiply the corresponding elements of Q from the dictionary together, and reduce them via the relations in the presentation Q that I started with above, arriving at a word in the alphabet a,b,c,d.
If that word ends up being one of the four words {a, b^2, bc, c^2 }, the position is P-position; otherwise, it's an N-position.
I'm guessing the the 4x4 game does not have a finite misere quotient, but I don't know for sure.
If people want more details, I'm happy to send them.  Google my name for my email address.
Best wishes
Thane Plambeck
Postscript (added 8 Jan 2013)  Here's a paper http://arxiv.org/abs/1301.1672 I just put up in the arXiv that has more details.<|endoftext|>
TITLE: Ways of formulating homological algebra without diagram chasing
QUESTION [12 upvotes]: The beginning of homological algebra involves lots of diagram chasing, for proving most of the theorems. This gets repetitive after a while. To make things more interesting and satisfactory, one would like to remove the mechanical dredge work as far as possible. 
One possible approach, as advocated in Lang's Algebra book, is to first prove the snake lemma by diagram chasing, and then reduce everything else as much possible to the snake lemma. This is mostly satisfactory. But some proofs, such as that of five lemma(as far as I am able to reconstruct now), require a small amount of diagram chasing in addition to using snake lemma.
There seems to be an alternate approach using the salamander lemma, as given in the references in this question. Salamander lemma can be used to prove the other important lemmas. But according to the description in Anton Geraschenko's post, it is like chasing salamanders around, rather than chasing elements.
Are there other possible approaches to founding homological algebra avoiding extensive diagram chasing? That is, is it possible to base homological algebra on some other lemma/proposition, than the two possibilities mentioned above?

REPLY [10 votes]: One other way you can prove all diagram chasing results (that I know of) is to first prove that there is a spectral sequence associated with a double complex (actually, there are two!). This involves some diagram chasing, but you only do it once. I guess you could also argue that you still have to chase the spectral sequences.
As an example, here's how you'd prove that a short exact sequence of chain complexes induces a long exact sequence in homology. We start with the double complex with exact rows

    ↑      ↑      ↑
0 → Ai+1 → Bi+1 → Ci+1 → 0
    ↑      ↑      ↑
0 → Ai  →  Bi  →  Ci  → 0
    ↑      ↑      ↑

(I'm assuming the columns are bounded below, but it doesn't matter since this result is "local": you can truncate above and below the point you're interested in and then prove the result)
We can regard this as the $E_0$ page of two different spectral sequences, depending on whether we decide to start with the vertical or horizontal arrows as our differential. Both spectral sequences have to abut to the homology of the total complex associated to this double complex.
First, use the horizontal arrows. Since the rows are exact, when we take homology, we get zero everywhere, so the $E_1$ page is identically zero. Nothing interesting is going to happen now: we've gotten to $E_\infty$. So the homology of the total complex is zero.
Now what if we use the vertical arrows? We get the $E_1$ page

0 → Hi+1(A) → Hi+1(B) → Hi+1(C) → 0

0 → Hi(A)  →  Hi(B)  →  Hi(C)  → 0

We'd like to prove that the rows are exact in the middle and that kernel of $H^{i+1}(A)\to H^{i+1}(B)$ is isomorphic to the cokernel of $H^i(B)\to H^i(C)$. Let's flip to the $E_2$ page and see what happens. Let $K^i=\ker(H^{i+1}(A)\to H^{i+1}(B))$, $M^i=\mathrm{cok}(H^i(B)\to H^i(C))$, and let $L^i$ be the homology of the above row at $H^i(B)$. We get the $E_2$ page

0   Ki+1   Li+1   Mi+1   0
   ↘     ↘     ↘
       ↘      ↘      ↘
0   Ki     Li     Mi   0

(pardon my ascii art: those are meant to be arrows going two spots to the right and one spot down)
Note that on the $E_{\ge 3}$ pages, the differentials will be too long to connect anything in these three columns to anything other than zero. Since we must abut to zero, this $E_2$ page is the "last chance" for the homology to vanish. It follows that the sequences $0\to L^i\to 0$ and $0\to K^{i+1}\to M^i\to 0$ must be exact, so $L^i=0$ and $K^{i+1}\cong M^i$, which is exactly what we wanted to show.

Note that I was able to completely ignore the question of what the differentials on higher pages were; I just had to know that they exist.
Diagram-chasing results almost always assume that the rows are exact and then make assertions about the homology groups you get from the columns, so you can always run one spectral sequence and immediately get that the homology of the total complex is zero, then run the other spectral sequence and note when the spectral sequence has it's "last chance" to cancel all homology groups.

REPLY [9 votes]: But isn't doing homological algebra without diagram chasing akin to doing geometry without circles, or doing algebra without multiplication? It's just the nature of the subject, a big part of it (and a fairly pleasant, I might add).
A standard solution is to use Freyd's Embedding theorem for abelian categories, which turns on diagram chase for all abelian categories. Taking it on faith and proving separately later is not ideal, but to me it seems better than the alternatives.

REPLY [2 votes]: Checkout chapter "Abstract Homological Algebra" on the book "Basic Homological Algebra", by M. Scott Osborne (http://books.google.com/books?id=0_3swQG0hpsC). Everything is done very close to the axioms of Abelian Category, whithout chasing. Morphisms are constructed only through factorization properties.<|endoftext|>
TITLE: Are operators with trivial spectrum nilpotent in a sense?
QUESTION [10 upvotes]: Being far from analysis, I recently learned about the Invariant subspace problem and came up with the following (perhaps simple or well-known) question.
Let $H$ be a separable complex Hilbert space and $T:H\to H$ a bounded operator. Assume that the spectrum of $H$ is $\{0\}$, i.e. $T-\lambda I$ has a bounded inverse for every $\lambda\in\mathbb C\setminus\{0\}$. In finite dimensions, this would imply that $T$ is nilpotent ($T^n=0$ for some $n$). I wonder if there is something similar in the infinite dimensional case. The precise formulation I have in mind is the following.
It is easy to see that there is a maximal subspace $X\subset H$ such that $T(X)=X$. This is a purely set-theoretic fact, a sort of explicit construction is the intersection of the images of iterations of $T$ up to and beyond infinity (via transfinite induction).
Question: can it happen that $X\ne\{0\}$?
Here are some observations that I made:

$T$ cannot be onto. (I derived this from some random Wikipedia quotes so there are high chances of error; please correct me if I am wrong.) It follows that a nontrivial $X$ cannot be closed.
If there is an example, then there is one where $X$ is dense. Just take the closure of $X$ for $H$.
It is possible that $T(H)$ is dense. My example is the shift in $\ell^2(\mathbb Z)$ composed with a mutiplication by a positive function (sequence) that goes to zero at both ends. Again, please correct me if I am wrong.

REPLY [2 votes]: Just some remarks. A natural example of a quasinilpotent operator, which may be of interest to you, is 
Tu(x):=∫0xu(s)ds 
on C[0,1], or on Lp[0,1], for 1≤p≤∞. In all these cases T is compact. However note that a noncompact example can be easily defined e.g. on a Hilbert sum of infinitely many Hilbert copies of a Hilbert space H, as the direct sum of infinitely many copies of a non-zero quasinilpotent T:H→H (that is, just taking {hj} to {Thj}).<|endoftext|>
TITLE: Why are order-k differential forms sections of the kth exterior power of the cotangent bundle?
QUESTION [8 upvotes]: The question I ask is in the title. This should be quite well-known, and in fact probably I am going to get the response that it is the definition. To convey my confusion, I have to convey my understanding of what is a differential form and what is the contangent bundle. To simplify things, we assume that our whole setup is immersed in the Euclidean space. 
$1$. Differential forms.
We take the definitions from Rudin, Principles of Mathematical Analysis. This is for an open set in $\mathbb R^n$.
Suppose $E$ is an an open set in $\mathbb R^n$. A $k$-surface in $E$ is a differentiable mapping $\Phi$ from a compact subset $D \subset \mathbb R^k$ into $E$. $D$ is called the parameter domain of $\Phi$ consisting of points $\mathbf u = ( u_{i_1}, \cdots , u_{i_k} )$. 
A differential form of order $k \geq 1$ in $E$ is a function $\omega$, symbolically represented by the sum
$$\omega = \sum a_{i_1, \cdots , i_k}(\mathbf x) dx_{i_1}\wedge \cdots \wedge dx_{i_k}$$
where the indices $i_1, \cdots , i_k$ range independently from $1$ to $n$, and so that $\omega$ assigns to each $k$-surface $\Phi$ in $E$ a number$\omega(\Phi) = \int_\Phi \omega$ 
, according to the rule
$$\int_\Phi \omega = \int_D \sum a_{i_1, \cdots , i_k}(\Phi((\mathbf{u})) \frac{\partial ( x_{i_1}, \cdots , x_{i_k})}{\partial ( u_{i_1}, \cdots , u_{i_k})}d\mathbf u $$
where $D$ is the parameter domain of $\Phi$, and the functions $a_{i_1}, \cdots, a_{i_k}$ are assumed to be real and continuous in $D$. 
So in the above definition the differential $k$-form is a certain integral for functions on compact $k$-surfaces. Thus a differential form can be treated as a measure for the $k$-surfaces, which can be integrated.
$2$. Cotangent bundle
We take this from wikipedia.
Let $M \times M$ be the Cartesian product of $M$ with itself.  The diagonal mapping $\Delta$  sends a point $p$ in $M$ to the point $(p,p)$ of $M \times M$.  The image of $\Delta$ is called the diagonal.  Let $\mathcal{I}$ be the sheaf of germs of smooth functions on $M \times M$ which vanish on the diagonal.  Then the quotient sheaf $\mathcal{I}/\mathcal{I}^2$  consists of equivalence classes of functions which vanish on the diagonal modulo higher order terms.  The cotangent sheaf $\Omega$ is the pullback of this sheaf to $M$.
Now, Def 2: A differential form $k$-form $\omega$ is a section of $\wedge^k\ \Omega$.
Question.
We consider an open set in the Euclidean space and look at the two definitions. A priori, to my eyes, both appear to be different things. It needs to be proved that they are the same. Please help me out with a reference with the required proofs.

REPLY [8 votes]: You have to work a bit to get those two definitions to agree, but it is all standard lore in differential geometry.  Both of the references in the comments - Spivak and Madsen & Tornehave - are good and should have what you need, the latter a bit more useful in my opinion.  But I am writing this answer because in neither text (or virtually any other introductory differential geometry text) will you encounter explicitly your definition of the cotangent bundle, or for that matter words like "sheaf" and "germ".  Such notions are useful in differential geometry, but it is not so crucial to incorporate them into the foundations the way it is in algebraic geometry.
A definition that you will see in books (in some form) proceeds as follows.  Given a point $p$ in $M$, define the tangent space $T_p M$ to be the vector space of point derivations of $C^\infty(M)$ at $p$.  If $(x_1, \ldots, x_n): U \subseteq M \to \mathbb{R}^n$ is a local coordinate system near $p$ then the directional derivative derivations $\frac{\partial}{\partial x_i}|_{p}$ form a basis of $T_p M$.  Construct a vector bundle $TM$ whose fiber over a point $x$ in $M$ is $T_x M$, and form its dual $T^*M$.  This is the cotangent bundle, and it takes only some basic techniques with sheaves to prove that it is the same as what you defined.
The equivalence of this definition with your first definition is just a bunch of coordinate calculations, complicated by the fact that Rudin defines surfaces extrinsically (in contrast to the intrinsic definitions that you and I produced).  From the intrinsic point of view, a differential k-form is a smooth section of the bundle $\wedge^k T^*M$.  As above, a local coordinate system $(x_1 \ldots x_n)$ on an open neighborhood $U$ of $M$ yields a trivializing frame $\frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n}$ for $TM$ over $U$, and the corresponding dual frame for $T^*M$ over $U$ is denoted by $dx_1 \ldots dx_n$.  Thus a trivializing frame for $\wedge^k T^*M$ is given by $\{ dx_{i_{1}} \wedge \ldots \wedge dx_{i_{k}}: 1 \leq i_{1} \leq \ldots \leq i_{k} \leq n\}$, which more or less explains the local formula in your question.  The details of your integral formula are explained by noting that Rudin is defining a $k$ form on $\mathbb{R}^n$ with its standard coordinate system and pulling it back to a $k$ form on the surface along the embedding of the surface in $\mathbb{R}^n$.  Thus it is necessary to sort out how an intrinsic differential form behaves under coordinate change (from $\mathbb{R}^n$ coordinates to coordinates on the surface), and the whole point of the theory is that they change in a way which makes integration coordinate invariant.
One last comment.  Rudin defines differential k-forms for k-dimensional surfaces in $\mathbb{R}^n$ which comes naturally equipped with a notion of integration.  Of course most interesting manifolds (e.g. the sphere) are not the images of compact domains in $\mathbb{R}^k$ embedded in $\mathbb{R}^n$; this is the right LOCAL picture, but not the right global picture.  So to properly define the integral of a k-form over a k-manifold, it is necessary to define the global integral in terms of patched-together local integrals via a partition of unity.  Most books set up the theory this way and prove that the integral doesn't depend on the choice of partition of unity, but no books explain why, for example, the intrinsic integral of an n-form $f dx_1 \wedge \ldots \wedge dx_n$ over an open set in $\mathbb{R}^n$ (involving partitions of unity) agrees with standard Lebesgue integral of $f$ over that open set.  I suspect that one would have to sort out this kind of issue in order to REALLY prove that your two notions of differential form agree.  It's possible to work this out on your own, but Brian Conrad sorted it out in his "How to compute integrals" handout here:
http://math.stanford.edu/~conrad/diffgeomPage/handouts.html
You might find some of the other handouts helpful too.<|endoftext|>
TITLE: What do you use categorical glueing/sconing/Freyd covers for?
QUESTION [15 upvotes]: In the theory of programming languages and structural proof theory, one of the handiest techniques we have available is a method called "logical relations", in which you can prove properties of languages by taking the type structure of the language, and defining families of predicates or relations by induction on the type structure. 
For example, if you want to prove that all definable functions in a language halt, you start by giving a predicate $P_X$ picking out the terms of base type $X$ which halt. Then, at higher type (say $A \to B$) you define $P_{A \to B}$ to pick out those terms which halt, and which additionally take elements of $P_A$ to elements of $P_B$. (You need this extra strength because otherwise a term $t$ of type $A \to B$ may evaluate to a function value, which diverges when given an input.) Then you prove a theorem showing that every definable term is in the predicate, and there you go.
This is a very natural technique, and I've often wondered where it shows up in the rest of mathematics. Recently, I learned that category theorists know it too, and call it by various names -- I listed the ones I was able to find in the question. Since (a) categories seem to serve as an interlingua for mathematics, and (b) lots of mathematicians hang out here, I figured I could ask: what other branches of mathematics has this technique been used in, and for what?

REPLY [8 votes]: I think the terminology "gluing" comes from the following example.  Let X be a topological space, let U be an open subset of X, and let $K=X\setminus U$ be its complementary closed subset.  Then the inclusions of U and K into X induce geometric morphisms $p:Sh(U)\to Sh(X)$ and $q:Sh(K)\to Sh(X)$.  The composite $q^* p_* :Sh(U) \to Sh(K)$ is then left exact, and the "gluing construction" applied to it (i.e. the comma category $(Sh(K) / q^* p_* )$) recovers $Sh(X)$.  In other words, the functor $q^* p_*$ tells you how to glue together $U$ and $K$ to get back $X$, and the comma category is what does the gluing.  This generalizes to complementary open and closed subtoposes of any topos, not just of sheaves on a topological space.<|endoftext|>
TITLE: Singular homology of a graph.
QUESTION [7 upvotes]: By a graph I will understand an undirected graph without multiple edges or loops. By a morphism of graphs I will understand a map $f$ between the underlying sets of vertices, such that if $x$ and $y$ are adjacent, then $f(x)$ and $f(y)$ are either adjacent or equal.
Let $G$ be a finite graph. One can realise $G$ as a CW-complex $|G|$ and look at topological invariants, such as singular homology. But this captures only very little information about $G$, because except from $H_0(|G|)$ and $H_1(|G|)$ all homology groups are zero.
Consider the following alternative construction: Let us write $\Delta_n$ for the complete graph on $n$ vertices, and let us re-baptise this graph by the name "standard $n$-simplex". There are obvious codegeneracy and coface maps between standard simplices, so that we obtain a cosimplicial object $\Delta_\bullet$ in the category of graphs. Now, proceed as usual: Morphisms $\Delta_\bullet \to G$ form a simplicial set, applying the free group construction yields then simplicial group, and the associated chain complex is the one whose homology $H_i^{\mathrm{sing}}(G)$ I shall call "singular homology of $G$".
Obvious properties of $H_i^{\mathrm{sing}}(G)$ are: It is a finitely generated commutative group ($G$ is finite), covariantly functorial in $G$. In particular, if we work with coefficients in a field, we obtain representations of the automorphism group of $G$. The homology of the point is $\mathbb Z$ in degree $0$ and trivial in higher degrees. We can define singular cohomology accordingly, and get then a natural pairing between homology and cohomology.
The list of all natural questions one must ask after making such a definition is long, so I will not ask everything. 
(a) Is there a comparison map $H_i(|G|) \to H_i^{\mathrm{sing}}(G)$, maybe even on the level of chain complexes? Is there some more elaborate CW-complex $||G||$ one can naturally associate with $G$ such that $H_i(||G||)$ gives back singular homology of $G$? In that case, one would ask for a natural map $|G| \to ||G||$.
(b) Given a graph $G$, is there a largest integer $i$ such that $H_i^{\mathrm{sing}}(G)$ is nonzero? Assuming yes, is this integer less or equal the size of the largest complete subgraph of $G$.
(c) Is there a Künneth morphism in singular cohomology? --is there a natural ring strucure on cohomology?
(d) What is a homotopy between morphisms of graphs? Given an answer to that, do homotopically equivalent morphisms induce the same maps in homology?
(z) Can you give an example of a graph with nontrivial $H_2^{\mathrm{sing}}(G)$?

REPLY [3 votes]: In response to (d), you may find some of the papers by Anton Dochtermann interesting. In particular, the paper Hom complexes and homotopy theory in the category of graphs might contain something you are looking for. I've seen two different notions of graph morphism: one where you allow $f(x) = f(y)$ if $x$ and $y$ are adjacent as you have, and one where you don't. Not allowing equality is what Dochtermann does, but your definition can be recovered if you put a loop at each vertex, for example.
If you use this definition (so the complete graphs form a simplicial object given by the different ways of embedding), then homology is not a homotopy invariant if my old notes are correct: the line graph on 3 vertices and the line graph on 2 vertices are homotopic but $H_1$ for the first is rank 2 while for the second it is rank 1. This should probably be double-checked (sorry that I am not doing this).
Maybe I should also mention that homology defined this way has been studied in the literature. It usually goes by the name of the "clique complex".
Hope that's of some help. I wrote some notes exploring some of these ideas which you can have if you like (but looking back, it looks like I was confused and I don't remember what I was trying to do), just send me an email.<|endoftext|>
TITLE: Suggestions for good books on class field theory
QUESTION [26 upvotes]: Recently I tried to learn class field theory, but I find it is difficult. I have read the book "Algebraic Number Theory" by J. W. S. Cassels and A. Frohlich. In the book, the approach to class field theory is cohomology of groups. Although I have learned cohomology of groups, I find that those theorems in the book are complicated and can not form a system.
I'm wondering what are people's opinions of the book above, can you give me some suggestions on learning class field theory, and could you recommend some good books on class field theory?

REPLY [4 votes]: I'm a beginner with basically no background in algebraic number theory, and am close to finishing  Number Theory 2: Introduction to Class Field Theory by Kato, Kurokawa, Saito. 
I love this book - it does a good job explaining the big pictures of number theory (e.g. the local-global philosophy) and providing motivation. It begins with concrete examples of what class field theory says - this is one thing I wanted. Throughout the book, they state theorems and their consequences and postpone the proofs, but still give proofs eventually. I really like this style, as it prevents getting bogged down in proofs. The last section claims to outline the proof of the main theorems of class field theory, but I haven't read it yet. I have not read Volume 1, and was able to get through Volume 2 without any difficulty.<|endoftext|>
TITLE: Motives versus Motifs
QUESTION [17 upvotes]: I was in Paris recently for a meeting about motives or motifs, and since I'm too jet lagged
for real work let me ask the following somewhat frivolous question. The word "motif" is
usually translated as "motive" in English.  However, I wonder if this is really the best choice. "Motive" has, for me, a primarily psychological meaning, whereas "motif" -- which is
a perfectly good English word -- means pattern or theme. I guess my question is which word better captures the intended meaning?
Incidentally, it appears that this usage of "motive" goes back to Grothendieck himself, cf."Standard conjectures on algebraic cycles". So perhaps,  one should allow him to have the last word and not question his motives, which have wonderful if unintended consequences.

REPLY [16 votes]: Dear Donu, here are Grothendieck's own words:
"Contrary to what occurs in ordinary
topology, one finds oneself confronting
a disconcerting abundance of different
cohomological theories. One has the
distinct impression (but in a sense that
remains vague) that each of these theories
“amount to the same thing”, that
they “give the same results”. In order to
express this intuition, of the kinship of
these different cohomological theories,
I formulated the notion of “motive” associated
to an algebraic variety. By this
term, I want to suggest that it is the
“common motive” (or “common reason”)
behind this multitude of cohomological
invariants attached to an
algebraic variety, or indeed, behind all
cohomological invariants that are a
priori possible"
They can be found in his autobiographical "Récoltes et Semailles", where there is also an allusion to a musical meaning of "motif". 
The translation is Barry Mazur's in his article "What is... a Motive?" which is, needless to say, a fascinating short survey (plus bibliography) .Here is the reference:
http://www.ams.org/notices/200410/what-is.pdf<|endoftext|>
TITLE: Non-real constants
QUESTION [6 upvotes]: Constants are usually real numbers e.g. e, pi, gamma etc. Can you give examples of special constants that are not real? e.g. complex or p-adic constants. 
A real number in base10 can be viewed as the coefficients of a power series evaluated at x=1/10, so I suppose a constant in another context such as a complete ring could just be some value of a function evaluated at some point. Can you give examples of such a value that could be considered as a special mathematical constant.
More generally a whole object such as a set or group could be thought of as a constant if it appeared in many formulae relating such objects.

REPLY [3 votes]: Since all the known non-trivial zeroes of Riemann Zeta function are on the Re(z)=1/2 line we only give their imaginary parts, but in fact their are complex,
$$1/2+ i*14.1347251417346937904572519835624702707842571156992431756855674601499...$$
being the first above the real line.  If Riemann Hypothesis is true we will never have to mention a different real part.

REPLY [3 votes]: Since we do not want to restrict ourselves to values in traditional number systems...

Each remarquable/exceptional finite algebraic structure, graph, can be described/encoded as a specific value in a numbering system.  For instance as a series of generating matrices, multiplication table, representation tables, etc.
We can consider each sporadic group as a remarquable constant. To me the Monster Group is a mathematical attraction point that can be compared to $\pi$ or $e$.  And it is well hidden in the armies of soluble groups around it.<|endoftext|>
TITLE: Hopf fibration inside the retraction of R^4 minus line -> S^2?
QUESTION [7 upvotes]: This was inspired by this question.
Let $Y = {\mathbb R}^4 \setminus$a coordinate line, which retracts to ${\mathbb R}^3 \setminus$a point, which retracts to $S^2$.
What is an explicit immersion $S^3 \to Y$, whose composition with the above retraction gives the Hopf fibration? 
My idea being, perhaps this would make clearer in what sense the $S^3$ is surrounding "a hole" in $S^2$.

REPLY [2 votes]: First, note that $\mathbb R^4 \setminus \mathbb R \simeq S^2\times \mathbb R^2$.
So you are asking for an immersion $S^3\to S^2\times \mathbb R^2$
representing the generator $\eta$ of $\pi_3(S^2\times \mathbb R^2)=\pi_3(S^2)=\mathbb Z$.
I'm guessing that your immersion doens't exist, and that you need to consider maps $S^3\to S^2\times \mathbb R^n$ with larger $n$, in order to represent $\eta$ by an immersion.
But if you are willing to go a little bit up in dimension, and consider maps $S^3\to S^2\times \mathbb R^4$, then you can even find an embedding representing $\eta$.
It is given by $(H,I):S^3\to S^2\times \mathbb R^4$, where $H:S^3\to S^2$ is the Hopf map, and $I: S^3 \rightarrow {\mathbb R}^4$ is the standard inclusion.<|endoftext|>
TITLE: (Co)homology of the Eilenberg-MacLane spaces K(G,n)
QUESTION [27 upvotes]: Let $(G, n)$ be a pair such that $n$ is a natural number, $G$ is a finite group which is abelian if $n \geq 1$. It is well-known that $\pi_n(K(G,n)) = G$ and $\pi_i (K(G,n)) = 0$ if $i \neq n$.
Also it is known that these spaces $K(G,n)$ play a very important role for cohomology. For any abelian group $G$, and any CW-complex $X$, the set $[X, K(G,n)]$ of homotopy classes of maps from $X$ to $K(G,n)$ is in natural bijection with the $n^{\mathrm{th}}$ singular cohomology group $H^n(X; G)$ with coefficients in $G$.
But what is known about the cohomology of the $K(G,n)$ themselves? It is interesting in the light of the above. Here I mean the singular cohomology with integral coefficients.

REPLY [8 votes]: In Q-subalgebras, Milnor basis, and cohomology
of Eilenberg – Mac Lane spaces Tamanoi gives explicit polynomial generators of $H^*(K(\mathbb Z/p^k,n);\mathbb Z/p)$ and $H^*(K(\mathbb Z,n);\mathbb Z/p)$ for all primes $p$, using Milnor basis of the dual Steenrod algebra.<|endoftext|>
TITLE: Advice on changing topic for thesis problem
QUESTION [19 upvotes]: I am starting to find my thesis problem not meaningful nor interesting and too technical.  As I learn more mathematics I am finding myself attracted to other topics and only started on my intital problem with my advisor because I could understand it at first.
My problem is a very "good" one, fulfuling many criteria mentioned by mathematicians:

It is pitched at exactly the right difficult.
It involves learning a big machine
It solution leads to further work and questions. 

Should I:

Find a new problem? (I'm sure I can't think of such a "good" problem as the one given to me by my advisor.)
Stick with the same problem and only do new problems after graduation?
Anything else?

REPLY [10 votes]: My advice: find someone who knows you in person, and knows more about your situation to talk to about this.  The best would be if you felt comfortable discussing it with your advisor, but another professor you know, or some sort of graduate subchair, or even an older graduate student would also be good.  You just aren't going to get anything but the most vague advice that's not very well-suited to your situation from strangers on the internet.<|endoftext|>
TITLE: Why do categorical foundationalists want to escape set theory?
QUESTION [38 upvotes]: This is a question that I have seen asked passively in comments relating to the separation of category theory from set theory, but I haven't seen it addressed in full.
I know that it's possible to formulate category theory within set theory while still being albe to construct the useful things one would want from category theory. So as far as I understand, all normal mathematics that involves category theory can be done as long as a little caution is taken. 
I also know that some people (categorical foundationalsists) would still like to formulate category theory without use of or reference to set theory. While I admit that I am curious about this for curiosity's sake, I'm not sure if there are any practical motivations for doing this. The only reason for wanting to separate category theory from set theory that I have read about is for the sake of `autonomy of category theory'.
So my question is twofold: What other reasons might categorical foundationalists have for separating category theory from set theory, and what practical purposes might it serve to do this?

REPLY [5 votes]: Todd Trimble is right: not all categoriests who are interested in foundations reject set theory. It seems to me that category and set theory are deeply interrelated. Categorical problems, as the existence of some adjoint functors depends on some completions of ZFC. For example the statement every subcategory closed under limits of a locally presentable category is reflective (that is, the inclusion functor has a left adjoint) is equivalent to a set theoretical statement called Vopenka principle (see J. Adamek, J. Rosicky, Locally presentable and accessible categories, London Math. Soc. Lect. Notes, Cambridge University Press, 1994). There are lots of such problems, which depend on extensions of ZFC, both in category theory but also in other mathematical areas which are formulated categorical e.g. the Whitehead problem in abelian group theory: Is it true that every abelian group which has no nontrivial extensions through $\mathbb Z$ is projective (free)?<|endoftext|>
TITLE: no lines/conics on a degree 4/5 surface?
QUESTION [8 upvotes]: Are there explicit examples of smooth surfaces of degree 4 and 5 in P^3, which 
contain no lines or conics (i.e. smooth rational curves of degrees 1 and 2)? An example over a finite field would be better, but if not - I will take one over complex numbers :)

REPLY [3 votes]: A quartic surface containing a line or a conic has Picard number at least 2. There are explicit examples of Ronald van Luijk of quartic surfaces defined over $\mathbb{Q}$ such that the (geometric) Picard number is 1. (See http://www.math.leidenuniv.nl/~rvl/ps/picone.pdf)
The reductions of these surfaces mod 2 and 3 contain a line resp, a conic, so they do not work for you. 
However, this paper hints at algorithm how to find examples in positive characteristic:
Take a smooth quartic surface $S$ over a finite field with $q$ elements. By point counting (or more efficient methods) you can determine the characteristic polynomial $P_2(T)$ of Frobenius on the second (etale) cohomology of $S$. 
Let $r$ be the number reciprocal zeros of $P_2$ the form $q\zeta$ with $\zeta$ a root of unity. This number $r$ has the same parity as the degree of $P_2$, which is $22$ for quartic surfaces. It is well-known that $r$ is at most the geometric Picard number of $S$.
Suppose now that you find a surface such that $r=2$. Then either the geometric Picard number $\rho(S)$ equals 1, and you are done, or the geometric Picard number equals 2. In the latter case the Tate-conjecture holds. Milne proved that the Tate-conjecture implies the Artin-Tate conjecture. Now the Artin-Tate conjecture enables you to determine the discriminant $D$ of the intersection pairing on the Neron-Severi group up to squares from $P_2(T)$. (See, e.g. my paper on elliptic K3 surfaces with geometric Mordell-Weil rank 15.) Now if $\rho(S)=2$ and $S$ contains a line then $D=-k^2$ for some integer $k$, and if $S$ contains a conic then $D=-3k^2$ for some integer $k$ (this is an easy exercise). Hence you have a criterion to check whether your quartic surface contains a line or a conic. 
For quintics something similar works, only the degree of $P_2$ differs, and the values of $D$ in the case that $S$ contains a line or a quintic differ.<|endoftext|>
TITLE: Beginning a sentence with a mathematical symbol
QUESTION [19 upvotes]: This is more of a mathematician question than a mathematics question, but I hope it is still appropriate.
Several times now, when friends have been editing my mathematical writing, they have pointed out instances where I began a sentence with a mathematical symbol, such as $X$, or $a\in A$, etc.  This is the sort of thing that never struck me as inherently bad writing, since it is often exactly how I would speak the sentence aloud.  However, I can also see their point, in that a sentence beginning with a mathematical symbol can look a little weird (especially when it starts with a lowercase letter), and doesn't quite read the same as a sentence beginning with an English word.
However, now that I have been trying to avoid doing this, more and more I find myself writing awkward sentences to avoid the most natural way of speaking, and I can't quite decide if I am increasing or decreasing the quality of my writing.
Do people have strong opinions on this?  Do people have tricks for dealing with this, besides a succession of empty and synonymous clauses ("Thus", "Then", "Therefore", "We find", "Looking at"...)?

REPLY [14 votes]: If one stoops to starting sentences with a symbol, then one
soon descends to finishing a sentence with a symbol and starting
the next with a symbol. Then one is liable to finish a sentence
with a symbol and start a sentence with the same symbol.
To the reader (if you still have one) madness beckons.<|endoftext|>
TITLE: The mathematical theory of Feynman integrals
QUESTION [91 upvotes]: It is well known that Feynman integrals are one of the tools that physicists have and mathematicians haven't, sadly.
 Arguably, they are the most important such tool. Briefly, the question I'd like to ask is this: what are the existing proposals for defining Feynman integrals rigorously and why don't they work?
I should admit that my interest in all of this comes from attempts to understand Witten's definition of invariants of links in 3-manifolds (Comm Math Phys 121, 1989).
Here are some comments on some of the existing approaches and a few specific questions. (And please let me know if I've missed any important points!)
Here is a general framework. Suppose $E$ is a topological vector space of some kind and $L,f:E\to\mathbf{R}$ are functions. These are usually assumed continuous and moreover $L$ is (in all examples I know of) a polynomial of degree $\leq d$ ($d$ is fixed) when restricted on each finite-dimensional subspace. The Feynman integral is $$Z=\int_E e^{i L(x)}f(x) dx$$ where $dx$ is the (non-existent) translation-invariant measure on $E$ (the Feynman measure).
The way I understand it, this supposed to make sense since $L$ is assumed to increase sufficiently fast (at least, on the "most" of $E$) so that the integral becomes wildly oscillating and the contributions of most points cancel out.
The above is not the most general setup. In some of the most interesting applications there is a group $G$ (the gauge group) acting on $E$. The action preserves $L$ and $f$ and then what one really integrates along is the orbit space $E/G$, but still the "integration measure" on $E/G$ is supposed to come from the Feynman measure on $E$. Moreover, there is a way (called Faddeev-Popov gauge fixing) to write the resulting integral as an integral over $E'$ where $E'$ is a $\mathbf{Z}/2$-graded vector space (a super vector space), which is the direct sum $E\oplus E_{odd}$ with $E$ sitting in degree $0$ and $E_{odd}$ in degree 1.
There are two naive approaches to Feynman integrals -- finite-dimensional approximations and analytic continuation. Both are discussed in S. Albeverio, R. Hoegh-Krohn, S. Mazzucchi, Mathematical theory of Feynman path integrals, Springer LNM 523, 2 ed. As far as I understand, both work fine as long as $L$ is a non-degenerate quadratic function plus a bounded one. Moreover, in the 1st edition of the above-mentioned book Albeverio and Hoegh-Krohn give yet another definition, which works under the same hypotheses. 
However, one can try to do e.g. finite dimensional approximations for any $L$ (and probably renormalize as divergencies occur). So here is the first naive question: has anyone tried to find out what are the functions $L$ for which this procedure can be carried out and leads to sensible results?
Since it is so difficult to define Feynman integrals directly, people have resorted to various tricks. For example, suppose $0$ is a non-denenerate critical point of $L$, so $L$ can be written $L(x)=Q(x)+U(x)$ where $Q$ is a non-degenerate quadratic function and $U$ is formed by the higher order terms. Introducing a parameter $h$ one can write (in finite dimensions and for a sufficiently nice $U$) the Taylor series for $\int_E e^{i (Q(x)+hU(x))}f(x) dx$ at $h=0$ in terms of Feynman diagrams. This is explained in e.g. in section 2 of the paper "Feynman integrals for pedestrians" by M. Polyak, arXiv:math/0406251. When one tries to mimic this in infinite dimensions, one runs into difficulties: the coefficients of Feynman diagrams are given by finite-dimensional but divergent integrals (e.g. if $E$ is the space of functions of some kind on some manifold, then the integrals are taken along the configuration spaces of that manifold).
A. Connes and D. Kreimer have proposed a systematic way to get rid of the divergencies (Renormalization in quantum field theory and the Riemann-Hilbert problem I: the Hopf algebra structure of graphs and the main theorem, arXiv:hep-th/9912092). I think I understand how it works in the particular case of the Chern-Simons theory, but I was never able to understand the details of the above paper. In particular, I'd like to ask: what are precisely the conditions on $Q$ and $U$ in order for this procedure to be applicable? What happens when we take the resulting series in $h$? If it diverges, is it still possible to deduce the value of the Feynman integral from it when the latter can be computed by other means (as it is the case e.g. with the Chern-Simons theory on a 3-manifold)?
Finally, let me mention yet another approach to Feynman integrals: the "white noise analysis" (see e.g. Lectures on White noise functionals by T. Hida and S. Si). It was used by S. Albeverio and A. Sengupta (Comm Math Phys, 186, 1997) to define rigorously the Chern-Simons integrals when the ambient manifold is $\mathbf{R}^3$ (they have to use gauge-fixing though). Since I don't know much about this I'd like to ask a couple of naive questions: Is there a way this could help one handle the cases other approaches can't? Can this help eliminate the necessity to start with a non-degenerate quadratic function?

REPLY [53 votes]: Most of this is standard theory of path integrals known to mathematical physicists so I will try to address all of your questions.  
First let me say that the hypothesis you list for the action $S$ to make the path integral well defined, ie that $S=Q+V$ where $Q$ is quadratic and non-degenerate and $V$ is bounded are extremely restrictive.  One should think of $V$ as defining the potential energy for interactions of the physical system and while it certainly true that one expects this to be bounded below, there are very few physical systems where this is also bounded above (this is also true for interesting mathematical applications...).  Essentially requiring that the potential be bounded implies that the asymptotic behavior of $S$ in the configuration space is totally controlled by the quadratic piece.  Since path integrals with quadratic actions are trivial to define and evaluate, it is not really that surprising or interesting that by bounding the potential one can make the integral well behaved.    
Next you ask if anyone has studied the question of when an action $S$ gives rise to a well defined path integral: $ \int \mathcal{D}f \ e^{-S[f(x)]}$
The answer of course is yes. The people who come to mind first are Glimm and Jaffe who have made whole careers studying this issue.  In all cases of interest $S$ is an integral $S=\int L$ where the integral is over your spacetime manifold $M$ (in the simplest case $\mathbb{R}^{n}$) and the problem is to constrain $L$.  The problem remains unsolved but nevertheless there are some existence proofs.  The basic example is a scalar field theory, ie we are trying to integrate over a space of maps $\phi: M \rightarrow \mathbb{R}$.  We take an $L$ of the form:
$ L = -\phi\Delta \phi +P(\phi)$
Where in the above $\Delta$ is the Laplacian, and $P$ a polynomial.  The main nontrivial result is then that if $M$ is three dimensional, and $P$ is bounded below with degree less than seven then the functional integral exists rigorously.  Extending this analysis to the case where $M$ has dimension four is a major unsolved problem.
Moving on to your next point, you ask about another approach to path integrals called perturbation theory.  The typical example here is when the action is of the form $S= Q+\lambda V$ where $Q$ is quadratic, $V$ is not, and $\lambda$ is a parameter.  We attempt a series expansion in $\lambda$.  The first thing to say here, and this is very important, is that in doing this expansion I am not attempting to define the functional integral by its series expansion, rather I am attempting to approximate it by a series.  Let me give an example of the difference.  Consider the following function $f(\lambda)$:
$f(\lambda)=\int_{-\infty}^{\infty}dx  \ e^{-x^{2}-\lambda x^{4}}$ 
The function $f$ is manifestly non-analytic in $\lambda$ at $\lambda=0$.  Indeed if $\lambda<0$ the integral diverges, while if $\lambda \geq 0$ the integral converges.  Nevertheless we can still be rash and attempt to define a series expansion of $f$ in powers of $\lambda$ by expanding the exponential and then interchanging the order of summation and integration (illegal to be sure!).  We arrive at a formal series:
$s(\lambda)=\sum_{n=0}^{\infty}\frac{\lambda^{n}}{n!}\int_{-\infty}^{\infty}dx  \ e^{-x^{2}}(-x^{4})^{n}$
Of course this series diverges.  However this expansion was not in vain.  $s(\lambda)$ is a basic example of an asymptotic series.  For small $\lambda$ truncating the series at finite order less than $\frac{1}{\lambda^{2}}$ gives an excellent approximation to the function $f(\lambda)$
Returning to the example of Feynman integrals, the first point is that the perturbation expansion in $\lambda$ is an asymptotic series not a Taylor series.  Thus just as for $s(\lambda)$ it is misguided to ask if the series converges...we already know that it does not!  A better question is to ask for which actions $S$ this approximation scheme of perturbation theory itself exists.  On this issue there is a complete and rigorous answer worked out by mathematical physicists in the late 70s and 80s called renormalization theory.  A good reference is the book by Collins "Renormalization."  Connes and Kreimer have not added new results here; rather they have given modern proofs of these results using Hopf algebras etc.
Finally I will hopefully answer some of your questions about Chern-Simons theory. The basic point is that Chern-Simons theory is a topological field theory.  This means that it suffers from none of the difficulties of usual path integrals.  In particular all quantities we want to compute can be reduced to finite dimensional integrals which are of course well defined.  Of course since we lack an independent definition of the Feynman integral over the space of connections, the argument demonstrating that it reduces to a finite dimensional integral is purely formal.  However we can simply take the finite dimensional integrals as the definition of the theory.  A good expository account of this work can be found in the recent paper of Beasley "Localization for Wilson Loops in Chern-Simons Theory."
Overall I would say that by far the currently most developed approach to studying path integrals rigorously is that of discretization.  One approximates spacetime by a lattice of points and the path integral by a regular integral at each lattice site.  The hard step is to prove that the limit as the lattice spacing $ a  $ goes to zero, the so-called continuum limit, exists.  This is a very hard analysis problem.  Glimm and Jaffe succeeded in using this method to construct the examples I mentioned above, but their arguments appear limited.  Schematically when we take the limit of zero lattice size we also need to take a limit of our action, in other words the action should be a function of $ a$.  We now write $S(a)=Q+\lambda V+H(a,\lambda)$  Where as usual $Q$ is quadratic $V$ is not an $\lambda$ is a parameter. Our original action is $S=Q +\lambda V$ 
The question is then can we find an $H(a,\lambda)$ such that a suitable $a\rightarrow 0$ limit exists?  A priori one could try any $H$ however the arguments of Glimm and Jaffe are limited to the case where $H$ is polynomial in $\lambda$.  Physically this means that the theory is very insensitive to short distance effects, in other words one could modify the interactions slightly at short distances and one would find essentially the same long distance physics.  It seems that new methods are needed to generalize to a larger class of continuum limits.<|endoftext|>
TITLE: Integer matrices with no integer eigenvalues
QUESTION [21 upvotes]: Let $$A = \begin{pmatrix} 3&1 \\  0&1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 1&0\\ 1&2 \end{pmatrix}$$ I want to show that the only elements of the semigroup generated by $A$ and $B$ that have integer eigenvalues are elements of the form $A^n$ and $B^n$, where $n \in \mathbb{N}$. I have tried every way that I can think of. Is it possible that a problem like this is undecidable?

REPLY [6 votes]: I just found this problem. If you try the matrix $A^nB^m$, then your question for such matrices is equivalent to this number theory question: Can 
$9^n+2\cdot 9^n\cdot 2^m-12\cdot 3^n\cdot 2^m+2\cdot 3^n+4^{m}\cdot 9^n+4^{m}+2\cdot 2^m+9$ 
be a square provided $m,n\ne 0$. Note that if we denote $3^n$ by $x$, $2^m$ by $y$, we get a quartic polynomial in $x,y$. I hope number theorists here can say something about this exponential Diophantine equation. 
The answer to problem with question mark is "obviously NO". To be undecidable, you should have a mass problem. For given $A,B$, you have the following problem: 
given a product $W(A,B)$ is it true that the matrix has an integer eigenvalue. That problem is obviously decidable. The question of whether this is true for every word $W$ requires answer "yes" or "no" and is not a mass problem. You can still ask whether it is independent from ZF or even ZFC (or unprovable in the Peano arithmetic). What Bjorn had in mind is a completely different and much harder problem when you include $A, B$ in the input and ask if for this $A$, $B$ some product $W(A,B)$ not of the form $A^n, B^m$ has an integer eigenvalue. This is a mass problem which could be undecidable (although he, of course, did not prove it). But this has nothing to do with the original question.<|endoftext|>
TITLE: Any paradoxical theorems arising from large cardinal axioms?
QUESTION [15 upvotes]: If we accept the axiom of choice, we take the responsibility of living in a world in which, e.g., a ball in Euclidean 3-space is equiscindable to two isometric copies of itself (Banach-Tarski). So we have a very natural and seemingly harmless axiom (that most matehmaticians are probably willing to accept and use) on one hand, and a quite paradoxical and counterintuitive phenomenon arising from accepting that axiom on the other.
I would like to know if the assumpion of some large cardinal axiom is known to produce some sort of paradoxical phenomena in the "everyday world", as it happens in the case of the axiom of choice.

REPLY [11 votes]: (This was a reply to John that grew beyond the alloted limit.)

large cardinals have consequences that were initially surprising to set theorists.

But, John, this is simply because our intuitions about large cardinals took a bit to form, just as one would expect with any relatively recent theory. Now we understand significantly better how far from $L$ the universe ought to be in the presence of large cardinals, and find the puzzlement at Scott's result curious. We know, for example, that much much much less than a measurable already contradicts $V=L$. 
Similarly, when our understanding of large cardinals was relatively poor (early 80s), it was the common belief that for any large cardinal there was a nice inner model with a $\Delta^1_3$ well-ordering of ${\mathbb R}$, and it was by refuting this that we eventually arrived at Woodin cardinals and our current view of the set theoretic universe, where (as Simon mentioned) large cardinals make such well-orderings impossible, since all projective sets are measurable. 
(This significant shift in understanding started with the Martin's Maximum paper by Foreman-Magidor-Shelah, Annals of Mathematics, 127 (1988), 1-47, and is very nicely explained in 'Iteration trees' by Martin-Steel, Journal of the American Mathematical Society, 7(1):1–73, 1994.) 
This also led, by the way, to the proof that large cardinals imply determinacy in nice inner models, again shifting the prior poorly developed intuitions that had us expect much larger consistency strength for determinacy than it turned out to be the case.<|endoftext|>
TITLE: Explicit cocycle for the central extension of the algebraic loop group G(C((t)))
QUESTION [14 upvotes]: Let $G$ be a simple Lie group and let $G(\mathbb{C}((t)))$ be its loop group.
The Lie algebra $\mathfrak{g}[[t]][t^{-1}]$ has a well known central extension
(see e.g. 
Wikipedia) given by the cocycle
$c(f,g) = Res_0\langle f,dg\rangle$. Here, $\langle\ ,\ \rangle \colon \mathfrak{g}\otimes \mathfrak{g}\to\mathbb{C}$ denotes some invariant bilinear form on $\mathfrak{g}$, and $f dg$ is the $\mathfrak{g}\otimes \mathfrak{g}$-valued differential given by multiplying $f$ and dg.

Question: It there a similarly concrete cocycle for the central extension of $G(\mathbb{C}((t)))$ by $\mathbb{C}^\ast$?

To give you an idea of what I'm looking for, let me show
you a cocycle for central extension by $S^1$ of the smooth loop group $LG = \mathop{Map} _ {C^\infty} (S^1,G)$ of a compact Lie group $G$.
Pick a bounding disc $D_\gamma$ : $D^2 \to G$ for each element $\gamma\in LG$. The cocycle is then given by
$$
c(\gamma,\delta) = \exp\left(i\int \langle D_\gamma^*\theta_L,D_\delta^*\theta_R\rangle
+i\int H^*\eta\right)
$$
where $\theta_L,\theta_R\in\Omega(G,\mathfrak{g})$ are the Maurer-Cartan 1-forms, $\eta\in\Omega^3(G)$ is the Cartan 3-form, and $H:D^3\to G$ in a homotopy between $D_\gamma D_\delta$ and $D _ {\gamma\delta}$.

References: The cocycle for the smooth loop group can be found on page 19 of the paper From Loop groups to 2-groups, by Baez, Crans, Schreiber, and Stevenson, and also on page 8 of Mickelsson's paper From Gauge anomalies to Gerbes and Gerbal actions.

REPLY [12 votes]: The Kac-Moody central extension can be described in terms of algebraic $K_2$. This was first discovered I think by Spencer Bloch in the early '80s. There is a scattered literature that spells this out in different contexts - the main published references I can think of are by Deligne-Brylinski (Central extensions of groups by $K_2$) and the papers it cites by Deligne (in particular Le Symbole Modéré), the papers by Brylinski-Mclaughlin on the Segal-Witten reciprocity law and symbols etc. (I learned of this from the famous unpublished manuscript of Beilinson-Kazhdan, I think it appears also in later published works of these two individually). Anyway this gives a formula for the Kac-Moody central extension in terms of the tame symbol. Actually one place where the whole story is spelled out beautifully is Kapranov's paper on Eisenstein series and S-duality.
To summarize briefly: $H^4(BG,Z)$ actually consists of algebraic cycle classes, i.e. it's equal to $Chow^2 (BG)$. Bloch showed (in the 70s) that this is the same as $H^2(BG,K_2)$ and used this to give a beautiful picture for second Chern classes. Anyway this can be interpreted as central extensions of $G$ by $K_2$. Now if you're over a local field (Laurent series say) the tame symbol is a kind of residue map, taking $K_2$ of the local field to $K_1$ (i.e. units) in the residue field. So you can push out the $K_2$ extension to get a $C^*$ extension of the loop group, as desired. While $K_2$ is an intimidating beast, this gives an explicit formula I think since the tame symbol is explicit... but I'm the wrong person to give you that formula. 
BTW for the multiplicative group this ends up giving a POV on Weil reciprocity, that
was spelled out by Witten in his gorgeous paper on Grassmannians, QFT and Algebraic Curves, and is explicated in a paper by Brylinski with a related title (Central Extensions and Reciprocity Laws) and most recently in a very pretty paper of Takhtajan.<|endoftext|>
TITLE: (Path) connected set of matrices?
QUESTION [12 upvotes]: Let $N \in \mathfrak{M}_n(\mathbb{C})$ nilpotent, such that there exists $X \in \mathfrak M_n(\mathbb{C})$ with $X^2=N$ (take for instance $n>2$ and $N(1,n)=1$; $N(i,j)=0$ otherwise).
Denote by $\mathcal{S}_N$ the set of $X \in \mathfrak M_n(\mathbb{C})$ such that $X^2=N$.
Is $\mathcal{S}_N$ connected or path-connected ? What happens when we change $2$ by $3,4,\ldots $?

REPLY [2 votes]: This is a memorial to an incorrect solution that used to be here. Unfortunately, I can't delete it since it was accepted.<|endoftext|>
TITLE: Almost orthogonal vectors
QUESTION [23 upvotes]: This is to do with high dimensional geometry, which I'm always useless with.  Suppose we have some large integer $n$ and some small $\epsilon>0$.  Working in the unit sphere of $\mathbb R^n$ or $\mathbb C^n$, I want to pick a large family of vectors $(u_i)_{i=1}^k$ which is almost orthogonal in the sense that $|(u_i|u_j)| < \epsilon$ when $i\not=j$.  I guess I'm interested in how the biggest choice of $k$ grows with $n$ and $\epsilon$.
For example, we can let $\{u_1,\cdots,u_n\}$ be the usual basis, and then choose $u_{n+1} = (1,1,\cdots,1)/n^{1/2}$, which works if $n^{-1/2} < \epsilon$.  Then you can let $u_{n+2} = (1,\cdots,1,-1,\cdots,-1)/n^{1/2}$ and so forth, but it's not clear to me how far you can go.

REPLY [2 votes]: This was too long to be a comment, my apologies:
Further to Tim Gowers' comment and the following discussion, Chapter 6 of Bollobas' lovely white book "Combinatorics - set systems hypergraphs, families of vectors, and combinatorial probability" studies exactly this kind of question,. 
In fact Tim's comment corresponds to Theorem 6 in that chapter of the book, I believe. The language of that theorem uses symmetric difference of subsets of $\{1,2,..,n\}$, which can be converted to inner products of normalized $n$ dimensional vectors with entries $\pm 1/\sqrt{n}$.
Let $\epsilon>0$ be small. The second case of Theorem 6 essentially states that if the normalized inner product is allowed to be in $[-1,\epsilon]$, equivalently, if the pairwise symmetric differences of the sets in the family are all slightly less than $n/2$ in relative terms, then the number of sets in the family, and hence the number of vectors with entries $\pm 1/\sqrt{n}$ can be as large as $2^{\epsilon n}.$
The proof of this part of the Theorem is given as an exercise with a nice hint in the book: 
Focus on subsets of size $k=\lceil n/2 \rceil$ and do sphere packing in the Hamming space.
The variation with the OPs question is that the allowed range for the normalized inner product is $[-\epsilon, +\epsilon]$ in the OPs question. From Jelani Nelson's answer, it seems that the penalty for restricting the inner product to a band is perhaps not as severe as one might expect. We go from $\epsilon n$ down to $\epsilon^2 \log(1/\epsilon)n$ in the exponent.<|endoftext|>
TITLE: What is the reverse mathematics of first-order logic and propositional logic?
QUESTION [8 upvotes]: Suppose one tries to formalize first-order logic. How much "strength" is required to do this?
Strength can mean in various senses:

The fragment of ZFC needed to codify first-order logic.
Which system of 2nd-order arithmetic is needed to codify first-order logic. (reverse mathematics)
The fragment of PA needed to codify first-order logic.
Categorial logic?

Codify means be able to (i) represent formulae and sentences, (ii) recognize proofs, (ii) represent structures and models.
(I'm also interested to the answer for this question in the case of propositional logic.)

REPLY [11 votes]: In general, "reverse mathematics" refers to work with subsystems of second-order arithmetic only; it does not include ZFC.  Assuming this is actually what you meant, everything is in Stephen Simpson's book Subsystems of second-order arithmetic.
All the basic syntactics of formulas can be done in a primitive recursive way, and can be directly formalized into RCA0. It is well known that Goedel's incompleteness theorems can be formalized in PRA, and so they can also be formalized in RCA0.
The restriction of Goedel's completeness theorem to countable theories is equivalent to WKL0 over RCA0, as is the similar restriction of the compactness theorem.  The corresponding theorems for propositional logic are also equivalent to WKL0 over RCA0. 
The underlying reason that WKL0 is needed is that RCA0 is not strong enough to prove that the deductive closure of an arbitrary set of sentences exists. If one adds a hypothesis that the input theory is already deductively closed, RCA0 will do. This is all made precise in Simpson's book. 
There has been some interesting work [1] on the reverse mathematics strength of the atomic model theorem of elementary model theory, which turns out to be very weak in the sense of reverse mathematics while still being stronger than RCA0.
[1] Denis R. Hirschfeldt; Richard A. Shore; Theodore A. Slaman. "The atomic model theorem and type omitting". Trans. Amer. Math. Soc.  361  (2009), 5805-5837.

REPLY [3 votes]: Just as in any area of mathematics, different parts of the subject have different strengths. The theory of Reverse Mathematics investigates these strengths by determining for a large number of mathematical theorems, including theorems arising in the development of first order logic, the weakest axioms that are able to prove them. In order to do this, one reverses the usual direction of mathematical proof, by proving the axioms from the theorems, over an extremely weak base theory. 
The remarkable fact of Reverse Mathematics is that many theorems of classical mathematics fall into five large equivalence classes, consisting of provably equivalent theorems. Many instances of this are listed on the Wikipedia page I link to above. 
It appears that all the most trivial parts of logical syntax can be developed in the weak base system $RCA_0$, and this may be the answer to your question. For more powerful parts of the theory, one needs stronger axioms. 

A weak version of the Goedel Incompleteness Theorem (for countable theory, already closed under conseqeunce) is also provable already in the base theory $RCA_0$. (As is the Baire Category theorem and the existence, but not uniqueness, of an algebraic closure for any countable field.)
The Completeness Theorem for countable languages is equivalent to $WKL_0$. (As is the Heine-Borel theorem, the Jordan curve theorem, and many others.)
the sequential completeness of the reals and various forms of Ramsey's theorem are equivalent to $ACA_0$.
Comparability of well-orders and Open Determinacy are equivalent to $ATR_0$.
the Cantor-Bendixon theorem is equivalent to $\Pi^1_1-CA_0$.<|endoftext|>
TITLE: A Sketch of "Esquisse d'un Programme"
QUESTION [17 upvotes]: I'm refering, of course, to Grothendieck's ambitious program available fully here: http://www.math.jussieu.fr/~leila/grothendieckcircle/EsquisseFr.pdf
The text is, as described in its title, a sketch, and is supposed to be written in terms one can easily follow. However, for a non-expert like me it is still, to say the least, difficult. What I've done so far is read some things that were written about it, and only glanced at "esquisse", which still strikes me as an intimidating text. My understanding is still less than preliminary.
This is going to be a series of related questions, whose answers will hopefully sketch to me the program described in "esquisse", or at least put me in better position to sketch it to myself:

As I understand it, $Gal( \mathbb{Q})$ acts on $\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty$ in the obvious way. Now $\mathcal{M}_{0,n}$ (the moduli space of genus $0$ $\bar {\mathbb{Q}}$-curves with $n$ marked points) can be identified with $(\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty)^{n-3}-\Delta$, generalizing just $\mathbb{P}^1 _ {\bar {\mathbb{Q}}}- 0,1,\infty = \mathcal{M}_{0,4}$. So $Gal( \mathbb{Q})$ acts on $\mathcal{M} _ {0,n}$. Here I start getting confused. On the one hand it seems like the generalization that is described is an action of $Gal( \mathbb{Q})$ on all $\mathcal{M}_{g,n}$, and on the other it seems that the action described is on its universal cover, the Teichmüller space, $\mathcal{T} _{g,n}$.
Okay. Now the assertion seems to be, we have a map from $Gal( \mathbb{Q})$ to each fundamental group (fundamental groupoid? I imagine that we may view $Gal( \mathbb{Q})$ as a baseless fundamental groupoid if we choose) of each $\mathcal{M}_{g,n}$. We can in fact view this as mapping simultaneously to all of them, respecting certain operations between the $\mathcal{T} _{g,n}$. What are these actions? Can you restate this last paragraph rigorously?
This map (sketchily sketched in question 2) is then conjectured to be an imbedding (I think this has been proven, am I right?). According to a paper by Pierre Lochak and Leila Schneps ($\hat {GT}$ and braid groups): "A very important suggestion of Grothendieck is that the tower of Teichmüller groupoids ... should be entirely constructable from its first two levels". What does "first two levels" mean? (first two $n$'s for every $g$?) And why should that be true? (what would behoove anyone to think it's true?)
How do braid groups have to do with any of this? Is the assertion that $\pi_1 ( \mathcal{M} _{0,n},basept)$ is isomorphic to $B_n$ (the braid group on $n$ generators). Is that even true? I'm confused about this point.
This question is regarding notations for profinite groups, used generously anywhere related to this topic. $\hat {F_2}$, the profinite completion of the free group on two generators, is ubiquitous in such discussions because it is the fundamental group of $\mathbb{P}^1 _{\bar {\mathbb{Q}}}-0,1,\infty$. It is "generated" by two elements, meaning that it is the closure of a group generated by two elements. While defining the Grothendieck-Teichmüller group, Lochak and Schneps write "if $x, y$ are the generators of $\hat {F_2}$, we write an element of $\hat {F_2}$ as a 'profinite word' $f(x,y)$, although it is not generally a word in $x$ and $y$. This notation allows us to give a meaning to the element $f(\bar x, \bar y)$ where $\bar x$ and $\bar y$ are arbitrary elements of a profinite group". Well... then, if $f(x,y)$ is not an actual word in $x$ and $y$, then what does $f(\bar x,\bar y)$ mean?
At the same place, $\hat {GT}_0$ is defined as the subgroup of $\hat{ \mathbb{Z}}^{\times} \times \hat {F_2} '$ of elements $(\lambda,f)$ such that, among other things, $f(x,y)f(y,x)=1$. What does that mean? What is $f(x,y)$, and how is it different from $f(y,x)$ in its definition? Is it that somehow I can "express" any element of $\hat {F_2}$ as a "word" in $x$ and $y$, and then $f(y,x)$ is just flipping them? How do you make that rigorous?
Do I understand correctly that $\hat {GT}$ is a subgroup of the automorphism group of this tower of Teichmüller spaces, and that it is conjectured that $Gal( \mathbb{Q})$ surjects onto it (as well as imbeds into it)?
Do any of these things have to do with the anabelian geometry that is sketched later in "esquisse"? If so, how? And in general, how do these sets of ideas fit into esquisse? Are they the essence, one idea of many, or a piece in a bigger picture I can't see (combining dessin d'enfants, the Teichmüller tower, and anabelian geometry)? If so, very roughly, what is that sketch?

I wasn't sure whether to make this community wiki.

REPLY [4 votes]: For question 5, I think the motivation for thinking of elements of $\hat{F_2}$ as "profinite words" is that $\hat{F_2}$ enjoys the same universal mapping properties as $F_2$, but in the profinite setting.
Indeed, any map from $F_2$ to a profinite group $G$ extends uniquely (at least if we require continuity) to a map from $\hat{F_2}$ to $G$. So let $f=f(x,y)\in \hat{F_2}$. Then if $\overline{x},\overline{y}$ are elements of some profinite group $G$, define $f(\overline{x},\overline{y})$ to be the image of $f$ under the unique map extending $x\mapsto \overline{x}, y\mapsto\overline{y}$.
This should also clear up number 6... the "flip" map is the unique continuous map from $\hat{F_2}$ to $\hat{F_2}$ sending $x$ to $y$, $y$ to $x$.<|endoftext|>
TITLE: Algorithms for the Lakes of Wada
QUESTION [5 upvotes]: The Lakes of Wada partitions the unit square in to three regions, all of whom share a common boundary. The Wikipedia entry (http://en.wikipedia.org/wiki/Lakes_of_Wada) gives a construction approach, and a picture of a first few steps of the construction.
Is there a good algorithm available somewhere to explicitly list the partition up to some specific iteration? Or a closed form expression for the boundary in the limit of the process? 
I'm hoping for something I can implement myself, or a piece of software that already does it, and in the end get a picture to arbitrary high levels of detail and arbitrary high iterations of the construction.

REPLY [8 votes]: Wada lakes can be obtained following a recipe given by Plykin.
An algorithm is explained in an online article in the Notices of the AMS. You only need to iterate a single explicit function to get the picture. So, this can be done using any fractal generator (chaospro,fractint,...). Note that zooming on
the boundary of the lake is not so interesting. You only get straight interlaced color bands after a few zooms.
The result can be seen online (scroll to the section about Wada Lakes). There is also a movie.
Actually there are four regions on the pictures/movie. You can merge two of them if three is enough for your need.
Cheers<|endoftext|>
TITLE: Quick proofs of hard theorems
QUESTION [112 upvotes]: Mathematics is rife with the fruit of abstraction.  Many problems which first are solved via "direct" methods (long and difficult calculations, tricky estimates, and gritty technical theorems) later turn out to follow beautifully from basic properties of simple devices, though it often takes some work to set up the new machinery.  I would like to hear about some examples of problems which were originally solved using arduous direct techniques, but were later found to be corollaries of more sophisticated results.
I am not as interested in problems which motivated the development of complex machinery that eventually solved them, such as the Poincare conjecture in dimension five or higher (which motivated the development of surgery theory) or the Weil conjectures (which motivated the development of l-adic and other cohomology theories).  I would also prefer results which really did have difficult solutions before the quick proofs were found.  Finally, I insist that the proofs really be quick (it should be possible to explain it in a few sentences granting the machinery on which it depends) but certainly not necessarily easy (i.e. it is fine if the machinery is extremely difficult to construct).
In summary, I'm looking for results that everyone thought was really hard but which turned out to be almost trivial (or at least natural) when looked at in the right way.  I'll post an answer which gives what I would consider to be an example.
I decided to make this a community wiki, and I think the usual "one example per answer" guideline makes sense here.

REPLY [5 votes]: The alternating sign matrix conjecture was first proved by Doron Zeilberger; the proof was enormously computational.  Later, Greg Kuperberg gave a much shorter proof using results from statistical mechanics.  Kuperberg's proof is still not trivial, but it is more conceptual.
(It's worth mentioning that very recently, Fischer and Konvalinka have announced a bijective proof of the same theorem, which is pleasing to a combinatorialist, but far from trivial, and hence probably not an answer to the question.)<|endoftext|>
TITLE: Galois orbits of newforms with prime power level
QUESTION [9 upvotes]: Let me start with a simple observation.  Suppose $f$ is a weight two newform of level $p^3$.  Write $d$ for the size of the Galois orbit $f^\sigma, \sigma \in \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$.  Then $d \geq (p-1)/2$.   The proof is quite simple: associated to $f$ is an abelian variety $A_f$ of dimension $d$ and conductor $p^{3d}$.  If a prime $p$ divides the conductor of an abelian variety of dimension $d$, and $p>2d+1$, then the maximal power of $p$ dividing the conductor is $p^{2d}$ (I believe this is a theorem of Serre-Tate).  Hence if we had $d < (p-1)/2$, this would immediately imply the absurd inequality $3d<2d$, so contradiction.  Of course this also works for newforms of weight two and higher prime power level.
My questions:

Is this written down in the literature somewhere?  It seems like such a simple observation that I want to guess that it is, but I cannot find a reference.
Is the same result true for higher weight newforms?

REPLY [4 votes]: See Section 4 of my paper The Hecke algebra T_k has large index, joint with Frank Calegari, where this sort of argument is applied.  Lemma 4.1 is a variant of the statement you prove; see also the last sentence.  
In any case, the argument does generalize to higher weight newforms; it is just a statement
about $p$-torsion elements in $GL_n(\mathbb Z_p)$ (there are none when $p > n + 1$).  This
in turn implies that a Galois rep. into $GL_n(\mathbb Z_p)$ must be tamely ramified at $p$
if $p > n + 1$, and hence has conductor at $p$ bounded by $p^n$ (because the exponent of the conductor is then just the codimension of the fixed subspace under tame inertia).  Note also
(see our paper) that you can replace $\mathbb Z_p$ by any unramified extension, and thus
that the argument controls not only the degree of the coefficient field of a newform, but also the ramification of $p$ in the coefficient field (the larger the degree of the coeff.
field, the more ramified $p$ has to be).<|endoftext|>
TITLE: Differentiable structures on R^3
QUESTION [9 upvotes]: This is for the sake of completeness(for references, understanding). 
I ask for references for proofs that:
-There is exactly one differentiable(ie $C^\infty$) structure on $\mathbb R$, upto diffeomorphism.
-Ditto for $\mathbb R^2$.
-Ditto for $\mathbb R^3$.

REPLY [16 votes]: UPDATE : I found some precise references that answer the OP's question and fill in some details in my original answer.  See the end for them.
I don't have a precise reference for the cases of $n=2$ or $n=3$, though I suspect that they could be found in Moise's "Geometric Topology in Dimensions 2 and 3".
However, I do have a nice reference for the amazing theorem (due to Stallings) that $\mathbb{R}^n$ has a unique $C^{\infty}$ structure for $n \geq 5$.  It is written up very nicely in Steve Ferry's Geometric Topology Notes.  See Chapter 10 starting on page 56.  What he proves is that $\mathbb{R}^n$ has a unique PL structure, but that is the key to the result.  The rest of the notes are also wonderful.
This theorem is surprising for a number of reasons.  For instance, it says that if $\Sigma$ is an exotic $n$-sphere for $n \geq 5$, then $\Sigma \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^n$ for any point $p \in \Sigma$.  In other words, the "exoticness" is "concentrated at a point".  Of course, this also follows from the usual proof of the high dimensional Poincare conjecture using the $h$-cobordism theorem, which constructs a homeomorphism between a homotopy sphere and the usual sphere which is differentiable except at one point (that one point giving trouble due to the "Alexander trick").
Another remark that should be made is that Freedman and Donaldson proved that $\mathbb{R}^4$ has uncountably many $C^{\infty}$ structures.
UPDATE : OK, here are some precise references.  In
MR0121804 (22 #12534)
Munkres, James
Obstructions to the smoothing of piecewise-differentiable homeomorphisms.
Ann. of Math. (2) 72 1960 521--554. 
it is proven in Corollary 6.6 that two PL-isomorphic differentiable manifolds that are homeomorphic to $\mathbb{R}^n$ are actually diffeomorphic.  To make sense of this, recall that Cairns proved that differentiable manifolds have canonical PL structures; the original reference for this is his paper
MR0017531 (8,166d)
Cairns, Stewart S.
The triangulation problem and its role in analysis.
Bull. Amer. Math. Soc. 52, (1946). 545--571. 
I think JHC Whitehead might have also proven this, but I don't have a reference for that.
This reduces us to showing that $\mathbb{R}^n$ has a unique PL structure. The cases $n=2$ and $n=3$ can be found in Moise's book "Geometric Topology in Dimensions 2 and 3".  As far as the original results go, the 2-dimensional case is classical, while the 3-dimensional case was originally proven by Moise in his paper
MR0048805 (14,72d)
Moise, Edwin E.
Affine structures in $3$-manifolds. V. The triangulation theorem and Hauptvermutung.
Ann. of Math. (2) 56, (1952). 96--114. 
For dimensions at least $5$, the original reference is the following
MR0149457 (26 #6945)
Stallings, John
The piecewise-linear structure of Euclidean space.
Proc. Cambridge Philos. Soc. 58 1962 481--488.
I hope this is helpful!<|endoftext|>
TITLE: Why are parabolic subgroups called "parabolic subgroups"?
QUESTION [51 upvotes]: Over the years, I have heard two different proposed answers to this question.

It has something to do with parabolic elements of $SL(2,\mathbb{R})$. This sounds plausible, but I haven't heard a really convincing explanation along these lines.
"Parabolic" is short for "para-Borelic," meaning "containing a Borel subgroup."

Which answer, if either, is correct?
A related question is who first introduced the term and when.  Chevalley perhaps?

REPLY [15 votes]: The naming (attributed by Borel as quoted by @GjergjiZaimi) happened quite publicly, in Roger Godement : Groupes linéaires algébriques sur un corps parfait, Sémin. Bourbaki 13 (1960/61), Exp. No. 206, 22 p. (1961). ZBL0119.27206, first page (my bold & link):

When $G_A/G_{\mathbf Q}$ is not compact, it is equally easy to conjecture that one must be able to define something like Poincaré’s classical “parabolic cusps”, which must correspond to nontrivial unipotent subgroups of $G_{\mathbf Q}$ (...) We shall, in this talk, define and study “parabolic subgroups” by methods of algebraic group theory.

Godement describes this at length in his Analyse mathématique IV, e.g. p. 441: in the Poincaré upper half-plane,

a parabolic cusp with vertex $\xi$ is the part of a horocycle with center $\xi$ comprised between two arcs of circle orthogonal to the real axis at $\xi$; see the figure (...)
$\hspace{9em}$<|endoftext|>
TITLE: What is the physical meaning of a Lie algebra symmetry?
QUESTION [16 upvotes]: The physical meaning of a Lie group symmetry is clear: for example, if you have a quantum system whose states have values in some Hilbert space $H$, then a Lie group symmetry of the system means that $H$ is a representation of some Lie group $G$.  So you want to understand this Lie group $G$, and generally you do it by looking at its Lie algebra.  At least initially, this is understood as a way to make the problem of classifying Lie groups and their representations easier.
But there are Lie algebras which are not the Lie algebra of a Lie group, and people are still interested in them.  One possible way to justify this perspective from a physical point of view is that Lie algebras might still be viewable as ("infinitesimal"?) symmetries of physical systems in some sense.  However, I have never seen a precise statement of how this works (and maybe I just haven't read carefully enough).  Can anyone enlighten me?

REPLY [6 votes]: I would like to provide you with a non-trivial example (and a reference) of a Lie algebra of symmetries
which is not a Lie algebra of a Lie group within the framework of conventional quantum mechanics.
This example relies on the notion of "dynamical groups" (which you can find a lot of literature about in the net).
I think that the most precise definition of a dynamical group of a quantum system would be a Lie group
which the system's phase space is a coadjoint orbit of. In the cases known to me
the system's Hamiltonian belongs to the Lie algebra of the dynamical group, but I don't think that
this is an essential requirement (The Hamiltonian can be a member of the universal enveloping algebra).
The main application of dynamical groups is to provide algebraic solution to the quantum mechanical problem. 
The spectrum of the systems can be obtained from the representation of the dynamical group associated
with the coadjoint orbit instead of solving the Shroedinger equation.
The Lie algebra of the dynamical group generally consists of the usual space time symmetries 
such as the angular momentum su(2) in addition to more generators which originally required
a lot of ingenuity to come up with, such as the Runge-lenz generators of the Hydrogen atom problem, sometimes
referred to as the Kepler problem,which close together with the angular momentum generators to o(4) for elliptical motion and o(3,1) for hyperbolic motion.
Another known example in which a dynamical group formulation is used is the harmonic oscillator 
with the dynamical group SU(1,1).
Returning to the Kepler problem. The treatment described above considers only a fixed energy subspaces. In the following article , by C. Dabul, J. Dabul, P. Slodowy,
a twisted Kac-Moody dynamical algebra is constructed which is a simultaneous dynamical Lie algebra
of the full problem (corresponding to elliptic, parabolic and hyperbolic trajectories).
This is an example of a dynamical Lie algebra of symmetries which is not a Lie algebra of 
a Lie group. 
Regrettably, I didn't see a followup of this work in terms of algebraic solution of the 
full Kepler problem in terms of representations of this algebra, nor a treatment of this problem
in terms of coadjoint orbits of (the non-Lie) Kac-Moody groups which correspond to this algebra and are 
subjects of active research. I think that these would be interesting research problems.<|endoftext|>
TITLE: Exotic differentiable structures on R^4?
QUESTION [37 upvotes]: This was going to be a comment to Differentiable structures on R^3, but I thought it would be better asked as a separate question.
So, it's mentioned in the previous question that $\mathbb{R}^4$ has uncountably many (smooth) differentiable structures.  This is a claim I've certainly heard before, and I have looked a little bit at the construction of exotic $\mathbb{R}^4$s, but it's something that I really can't say I have an intuitive understanding of.  
It seems reasonable enough to me that a generic manifold can have more than one differentiable structure, just from the definition; and is in fact, a little surprising to me that manifolds have only one differentiable structure for dimension $d \le 3$.
But it's very odd to me that $\mathbb{R}^d$ has exactly one differentiable structure, unless $d=4$, when it has way too many!  
Naively, I would have thought that, since $\mathbb{R}^4 = \mathbb{R}^2 \times \mathbb{R}^2$, and $\mathbb{R}^2$ has only one differentiable structure, not much can happen.  Although, we know $\text{Diff}(M\times N)$ cannot generically be reasonably decomposed in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$ in general, I would not have expected there to be obstructions for this to happen in this case.
I would have also thought, that since $\mathbb{R}^5$ has only one differentiable structure, and $\mathbb{R}^4$ is a submanifold of $\mathbb{R}^5$, and $\mathbb{R}^3$ is a submanifold of $\mathbb{R}^4$ with only one differentiable structure, this would be fairly restrictive on the differentiable structures $\mathbb{R}^4$ can have.  
Although it seems that this only restricts the "inherited" differentiable structure to be a unique one, it still seems odd to me that the there are "non-inherited" structures in $d=4$ from $d=5$, and somehow all of these non-inhereted structures are identical on the submanifold $\mathbb{R}^3$!
Anyway, can anyone provide a intuitively sensible explanation of why $\mathbb{R}^4$ is so screwed up compared to every other dimension?  Usually I would associate multiple differentiable structures with something topologically "wrong" with the manifold.  Is something topologically "wrong" with $\mathbb{R}^4$ compared to every other dimension?  Or is this a geometric problem somehow?

REPLY [20 votes]: This started as a comment on Greg's post, but my comments are getting too long...
The additional topological condition you need for an open contractible 4-manifold to be homeomorphic to $\mathbb{R}^4$ is that it be "simply connected at infinity" (this is definitely necessary, and I would bet that it is sufficient, though I am definitely not an expert at $4$-manifold topology). This excludes interiors of contractible manifolds w/ boundary a homology 3-sphere. By the way, the result Igor quoted is a deep theorem of Friedman.
Here's something else that will blow your mind.  There actually exist exotic $\mathbb{R}^4$'s $U$ that embed as open submanifolds of $\mathbb{R}^4$ (the so-called "small" exotic $\mathbb{R}^4$'s).  We thus get that $U \times \mathbb{R}$ embeds as an open submanifold of $\mathbb{R}^5$.  By Stallings's theorem, $U \times \mathbb{R}$ is thus diffeomorphic to $\mathbb{R}^5$ even though $U$ is only homeomorphic to $\mathbb{R}^4$.
This is actually a very common type of thing -- often bizarre examples become nice when you "stabilize them" by, say, crossing with $\mathbb{R}$ or some other related operation.<|endoftext|>
TITLE: Practical applications of algebraic number theory?
QUESTION [35 upvotes]: I'm interested in learning about any applications, the more worldly the better*.
Pointing to a nice reference on the number field sieve, for example, would be fine.
However, let me mention one direction I would be especially grateful to learn about.
In my introductory course, I like to spend some time on the perspective that algebraic number theory
is the study of sophisticated multiplications on $\mathbb{Q}^n$ (an algebraic number field $F$ of
degree $n$) and on $\mathbb{Z}^n$ (the ring of algebraic integers in $F$).
This is in part because I still find it amazing that a little bit of abstract algebra (of irreducible polynomials)
enables us to construct such things systematically**. 
But I also  believe at least half-seriously that this is the view through which the general public will
gradually learn about algebraic numbers, until the time they're taught in primary schools several thousand years hence.
After all, we have ourselves witnessed the remarkable ascent of multiplication on $\mathbb{F}_2^n$, a set whose initial practical use was entirely devoid
 of algebraic content, as a powerful tool
for information processing.
After such grandiose reflection, I can't help but wonder: are multiplicative structures on $n$-tuples of integers provided
by algebraic number theory already of some practical use? A superficial google search uncovered nothing.
But surely, there must be something? I would love to be able to mention some examples to my students.
As I write, one class of examples occurs to me. Algebraic integers can be used to construct arithmetic groups, which I understand can be applied in a number of ways. Perhaps someone can comment knowledgeably on that. But something direct that could
 at least vaguely be explained in an undergraduate course would be even better.
Added: 
Such was the depth of my ignorance that I didn't even know about number field codes until Victor Protsak pointed to them in his answer. Thanks to him, I stumbled upon a short survey by Lenstra. To get the gist of it, one need only read this quote: 
'The new codes are the analogues, for number fields, of the codes constructed
by Goppa and Tsfasman  from curves over finite fields.'
The time-worn analogy continues to prosper.
Added again: 
In order to avoid misleading people with the word 'prosper,' I should say that Lenstra has many negative things to say about these codes. For example,
'If the generalized
Riemann hypothesis is true our codes are, asymptotically speaking, not as
good as those of Goppa and Tsfasman Also, the latter codes are linear and
non-mixed.'
My original question still stands.

*I do not wish, however, to give the impression of a firm belief in the division
between pure and applied mathematics.
** To appreciate this, one need only spend a little time
on a direct approach using the multi-linear algebra of  structure constants.

REPLY [4 votes]: The Bellows Conjecture was not named after someone named Bellows, but after the physical device for pumping air. There are polyhedra that can flex if their faces are rigid but you let them bend along their edges, so the dihedral angles are not fixed. The Bellows Conjecture was that a polyhedron that can flex does not change its volume.  
This was proved by Sabitov for spherical polyhedra, and by Connelly, Sabitov, and Walz in general in three dimensions, by showing that the volume is integral over an extension of the rationals by the edge lengths by considering the places of that field.<|endoftext|>
TITLE: Spectral graph theory: Interpretability of eigenvalues and -vectors
QUESTION [38 upvotes]: I thought "Wow!" when I learned that the eigenvector of the adjacency matrix of a cycle graph $C_n$ corresponding to the second largest eigenvalue gives the coordinates of the vertices when equally distributed on the unit cycle: the $n$-th roots of unity (from complex analysis) come up in a completely discrete context! What's more: The coordinates of the eigenvector can be "interpreted" straight forwardly when assigned properly to the vertices.
There's another straight-forward interpretation of an adjacency eigenvector: the eigenvector corresponding to the largest eigenvalue gives the relative importances of the vertices, being proportional to the sum of the relative importances of its neighbors.

Question: Can more - or eventually all - adjacency eigenvectors be sensibly
  "interpreted"?

Or does it in general depend on the type of graph, whether and how the eigenvectors can be interpreted, and the first example above is just a curio? 
What about the interpretation of the eigen-values? Do at least some of them "mean" something conceivable?

REPLY [3 votes]: A definite reference is Lovasz' Eigenvalues of graphs, a paper I wasn't aware of when asking this question.<|endoftext|>
TITLE: A place to find original papers
QUESTION [11 upvotes]: I currently use scholar.google.com to find papers in cases like Sophus Lie's original papers on "Transformation Groups". Does anyone know of other places that collect original works like this, i.e. works by those such as Weierstrass, Riemann or Lie describing the ideas that they originated. scholar.google.com isn't "bad" per se, I'd just like to look up just those papers and not every paper on the same topic.
I have access to many journals so that isn't so much the problem as knowing where to look.

REPLY [4 votes]: I have mentioned this link before. You can find links to almost everything that is freely and legally available on the web as far as mathematical publications are concerned.<|endoftext|>
TITLE: Is there an interesting definition of a category of test categories?
QUESTION [8 upvotes]: Given a pair of test categories $C_1$ and $C_2$ (in the sense of Grothendieck - weak or strict or otherwise), has anyone defined an interesting notion of morphism between them? Or are ordinary functors enough? Given that, has anyone studied the resulting category? In particular, what effect do properties of the functors have on the resulting presheaf categories, $Set^{C_1^{op}}$ and $Set^{C_2^{op}}$? Is there a characterisation of functor between test categories which induces a Quillen equivalence of presheaf categories, using the model structure as in Cisinski's Asterisque volume?

REPLY [7 votes]: I guess that it might be slightly more interesting to look for a notion of morphism of local test categories. A first natural candidate is given by the notion of locally constant functor: a functor $u:A\to B$ is locally constant is it satisfies the assumpions of Quillen's theorem B, namely that, for any map $b\to b'$ in $B$, the induced functor on the comma categories $A/b\to A/b'$ is a weak equivalence. If $A$ and $B$ are local test categories, then a functor $u:A\to B$ is locally constant if and only if the inverse image functor $u^\star : Set^{B^{op}}\to Set^{A^{op}}$ is a left Quillen functor (and such a $u^\star$ is moreover a Quillen equivalence if and only if $u$ is a weak equivalence); see prop 6.4.29 in Astérisque 308.
If we want to look at something which looks like some theory of bimodules, in order to produce a bicategory of local test categories, we might start with spans of shape 
$$A\overset{w}{\leftarrow} C \overset{u}{\to} B$$
where $w$ is aspherical (i.e. satisfies the assumptions of Quillen's theorem A), while $u$ is locally constant (with $C$ a local test category as well). From such data, we obtain a left Quillen functor 
$$Set^{B^{op}}\overset{u^\star}{\longrightarrow} Set^{C^{op}}$$
as well as a left Quillen equivalence
$$Set^{C^{op}}\overset{w^\star}{\longleftarrow} Set^{A^{op}}.$$
The trouble is that such spans may not compose (because neither aspherical functors nor locally constant functors are stable under pullbacks in general). We may correct this defect by asking furthermore that $w$ is smooth (e.g. that $C$ is fibred over $A$): a reformulation of Quillen's theorem B is that the pullback of a locally constant functor by a smooth map is always homotopy cartesian (see prop 6.3.39 and thm 6.4.15 in Astérisque 308); furthermore, for any smooth functor $p:X\to Y$, if $Y$ is a local test category, so is $X$ (in fact, for such a property, we need only $p$ to be locally aspherical); see thm 7.2.1 in loc. cit. In other words, we obtain a bicategory of local test categories, for which the $1$-cells are the couples $(w,u)$, with $w$ an aspherical smooth functor, while $u$ is locally constant (e.g. any smooth and proper functor is locally constant). This bicategory is in fact another model for homotopy types: if we denote by $S(A,B)$ the category of spans $(w,u)$ from $A$ to $B$ as above, then the classifying space of $S(A,B)$ has the homotopy type of the mapping space from the classifying space of $A$ to the classifying space of $B$ (and any homotopy type is the classifying space of some local test category).
Here is some explanation of the meaning of this bicategory: this has to do with the description of homotopy types as $\infty$-groupoids. Let us start with the toposic description of $1$-groupoids. There is a $2$-functor from the $2$-category of $1$-groupoids to the $2$-category of (Grothendieck) topoi which associates to a $1$-groupoid $G$ the category of presheaves of sets over $G$. This functor is fully faithful (i.e. induces equivalences of categories at the level of Hom's). Its essential image may be characterized as the $2$-category of topoi which are generated (as cocomplete categories) by their locally constant objects. This embedding of $1$-groupoids into topoi is the starting point of the Grothendieck version of Galois theory (unifying classical Galois theory of fields and topological Galois theory). If we consider homotopy types as $\infty$-groupoids (in the spirit of Toën, Lurie, and others), then one can consider the classifying space of a small category $A$ as the $\infty$-groupoid obtained from $A$ by inverting all the arrows in $A$. The $\infty$-topos associated to this $\infty$-groupoid may be described as the one associated to the model category obtained from the (injective) model category of simplicial presheaves over $A$, by considering the left Bousfield localization by the arrows between representable presheaves. The latter model category is precisely the model structure corresponding to the local test category $A\times \Delta$ (recall that the product of any small category with a (local) test category is a local test category). There is a higher version of higher Galois theory whose first fundamental statement is that the $(\infty,2)$-category of $\infty$-groupoids is embedded fully faithfully in the $(\infty,2)$-category of $\infty$-topoi; see Toën's Vers une interprétation Galoisienne de la théorie de l'homotopie Cahiers de Top. et de Geom. Diff. Cat. 43, No. 4 (2002), 257-312". The bicategory of local test categories as described above is a convenient presentation to see this canonical embedding of homotopy types into $\infty$-topoi explicitely: one associates to a span $(w,u)$ as above the map induced by the left Quillen functors ${w^{\star}}$ and ${u^{\star}}$  (inverting ${w^{\star}}$); see the last paragraph of my paper Locally constant functors, Math. Proc. Camb. Phil. Soc. 147 (2009), 593-614, for a similar description of this embedding. In other words, given a local test category $A$, the corresponding model category on $Set^{A^{op}}$ defines an $\infty$-topos, which is Galois (in the sense that it is generated, as a cocomplete $(\infty,1)$-category, by its locally constant objects), and the corresponding $\infty$-groupoid is simply the one associated to $A$ by inverting all the arrows in $A$. This is how I understand the meaning of these model category structures associated to local test categories.<|endoftext|>
TITLE: Gaining intuition for how submodules behave
QUESTION [5 upvotes]: I'm studying elementary commutative algebra this semester, largely following Atiyah-MacDonald. I often find myself in a situation where I'm interested in whether some property of an R-module M is inherited by its submodules (e.g. the property of being finitely generated over R) and I feel like I am lacking the necessary intuition to decide whether something is the case or not. In the case of finite-generation, for example, I think that my mental picture of a finitely-generated R-module is still too close to that of a finite-dimensional vector space to be able to intuit the fact that this simply shouldn't hold true in general. 
So here's my question - when you run across some property for a module and you want to know whether this is inherited by its submodules, how do you begin to think about the problem? Do you have a standard stock of counterexamples (or a procedure of sorts to concoct a counterexample)? Or do you have a more nuanced way of informally thinking about modules that captures more of their behavior? As I only have a semester of commutative algebra under my belt (think Atiyah-MacDonald), I'd appreciate answers that tend towards the more elementary end of the subject, although if you think that it's not possible to gain a good feel for the way that modules behave without diving deeper, I'd like to hear that too.

REPLY [14 votes]: Doing all exercises in Atiyah-MacDonald, like BCnrd suggested, is surely the ideal way to learn about this and much more. Let me offer a couple of practical tips to get you started:
A surprisingly effective example to keep in mind when you deal with any question about submodules of a module $M$ is to take $M=R$. Then the submodules of $R$ are just the ideals of $R$, which are concrete enough to check your intuition, but still possess a very rich structure so that not much is lost. 
Also, since many properties of modules fail to pass to submodules in higher dimension, it usually suffices to consider some small example, say $R= k[x,y]$.
As an example, let says you are trying to understand the following question: Over what Noetherian ring $R$ is a submodule of any  free module free? (this is of course true for vector spaces). 
If you  take $M=R$, it follows that  all ideals $I$ have to be free. If $R=k[x]$, this is true, and already an interesting exercise, but if $R=k[x,y]$, just take $I=(x,y)$. $I$  is not free because the generators have a non-zero relation: $xy-yx=0$. This example also suggests that all ideals in $R$ have to be principal, otherwise similar counter-examples can be found. So you naturally gets to principal ideal rings. 
If you want to play with it a bit more, since $R/I$ fits into an exact sequence:
$$0 \to I \to R \to R/I \to 0 $$
This says that $R/I$ has projective dimension at most $1$ for any ideal $I$. This leads you to  some serious restriction on $R$, which will point you to the right condition, from a different perspective.
You can replace "free" by "locally free" and play the same game, it will naturally leads you to all sort of interesting things worth learning about commutative rings, for examples projective modules or Quillen-Suslin theorem, etc.  
(There are, of course, other ways to approach this particular question, my point is by considering $M=R$ you can already get quite far). I hope you will have some fun!<|endoftext|>
TITLE: Degrees of a polynomial $p$ such that $x\mapsto p(x)$ is one-to-one
QUESTION [10 upvotes]: Given a finite field $K$, what are the possible degrees of a polynomial $p\in K[x]$ such that $x\longmapsto p(x)$ is one-to-one?
Such a polynomial has clearly not degree $0$ and it cannot be of degree two except for
$x\longmapsto (\alpha(x))^2$ for $\alpha$ an affine bijection of a field of characteristic $2$.
Are there many examples of degree $3$ (except for the stupid $x\longmapsto (\alpha(x))^3$
with $\alpha$ an affine bijection of a field of characteristic $3$)?
I guess that the degrees of such polynomials (except for affine bijections and their
composition with the Frobenius map)
are generically fairly high (the interpolation polynomial for a "random" permutation
of a finite field with $q$ elements should typically be of degree $q-1$).
What can for instance be said on the smallest degree $>1$ of a non-affine polynomial
inducing a bijection of $\mathbb Z/p\mathbb Z$?

REPLY [23 votes]: I only saw this a decade late, but just to record answers to these questions in case someone else comes across this post in the future:

The only degree-3 examples are as follows, up to composing on both sides with degree-$1$ polynomials: $x^3$ with $q\not\equiv 1\pmod{3}$, and $x^3-cx$ with $3\mid q$ and $c$ a nonsquare in $\mathbb F_q$.  This was proved by Dickson in his 1896 Ph.D. thesis The analytic representations of substitutions on a power of a prime number of letters with a discussion of the linear group.

Yes, the degrees are generically quite high.  As in the statement of the question, we could start with any of the $q!$ permutations $\pi$ of $\mathbb F_q$, and use Lagrange interpolation to produce a unique polynomial $f_\pi(x)$ of degree at most $q-1$ which acts on $\mathbb F_q$ as $\pi$.  If $q>2$ then $f_\pi(x)$ cannot actually have degree $q-1$, for instance because $f_\pi(x)=\sum_{\alpha\in\mathbb F_q} \pi(\alpha)\cdot\frac{(x-\alpha)-(x-\alpha)^q}{x-\alpha}$ and visibly the coefficient of $x^{q-1}$ is $-\sum_{\alpha\in\mathbb F_q}\pi(\alpha)$, which equals $-\sum_{\alpha\in\mathbb F_q} \alpha$ since $\pi$ is a permutation, and finally this last summation equals $0$ when $q>2$.  However, most choices of $\pi$ do yield polynomials $f_\pi(x)$ of degree very close to $q$.  Specifically, Konyagin and Pappalardi proved in 2002 (Enumerating permutation polynomials over finite fields by degree) that the number $N$ of permutation polynomials of degree less than $q-2$ satisfies
$$\lvert N-(q-1)!\rvert\le\sqrt{2e/\pi}q^{q/2}.$$
So roughly $(q-1)/q$ of the $q!$ permutations $\pi$ yield permutation polynomials $f_\pi(x)$ having degree $q-2$, and roughly $1/q$ of the $\pi$'s yield lower-degree $f_\pi$'s.  Note that this is what one would naïvely expect if the coefficient of $x^{q-2}$ in $f_\pi(x)$ were randomly distributed among elements of $\mathbb F_q$.  In a 2006 followup paper (Enumerating permutation polynomials over finite fields by degree II), Konyagin and Pappalardi extended their result to degrees $q-3$, $q-4$, ..., $0.96017q$, yielding the answers one would have naïvely expected.

It is known that if a permutation polynomial $f(x)\in\mathbb F_q[x]$ has degree $n$, where $n\le q^{1/4}$, then $f(x)$ has the very unusual property that there are infinitely many $k$ for which $f(x)$ permutes $\mathbb F_{q^k}$ (see Rem. 8.4.20 of this paper).  Polynomials with the latter property are called "exceptional".  I interpret this $n\le q^{1/4}$ result as suggesting that, since exceptionality seems very unlikely to ever happen "at random", therefore any permutation polynomial of degree at most $q^{1/4}$ should have a good causal reason for existing.  There are partial classification results for exceptional polynomials: for instance, it is known that the exceptional polynomials with degree coprime to $q$ are precisely the compositions of degree-one polynomials, $x^r$ with $\gcd(r,q^2-q)=1$, and degree-$r$ Dickson polynomials with $\gcd(r,q^3-q)=1$ (Dickson polynomials are defined by the functional equation $f(x+a/x)=x^r+(a/x)^r$ for some $a\in\mathbb F_q^*$; in characteristic not $2$ they are quadratic twists of Chebyshev polynomials) -- this was shown in a hard-to-find 1975 paper by Klyachko (Monodromy groups of polynomial mappings, in Russian, pp. 82-91 of vol. 6 of "Studies in Number Theory", Izdat. Saratov. Univ., Saratov, 1975), and also in the appendix of a 1997 paper by Müller
(A Weil-bound free proof of Schur's conjecture).
More generally, any exceptional polynomial of degree at least $2$ can be written as the composition of indecomposable exceptional polynomials, meaning polynomials which cannot be written as compositions of lower-degree polynomials in $\mathbb F_q[x]$.  There are only a few known families of indecomposable exceptional polynomials, and it is known that any as-yet-unknown example would have degree $p^k$ where $k\ge 4$ and $p:=\operatorname{char}(\mathbb F_q)$; but it is not expected that there actually exist any as-yet-unknown examples.
The known examples are listed in my brief survey paper
Exceptional polynomials.

It is unknown whether the bound $n\le q^{1/4}$ in item 3 can be improved substantially.  This is especially interesting when $q$ is a prime $p$, where it is unknown whether every permutation polynomial over $\mathbb F_p$ of degree less than $p/(2\log p)$ must be exceptional, and hence (by Klyachko's result mentioned in item 3) must be a composition of degree-one polynomials, polynomials of the form $x^r$, and Dickson polynomials.  (The bound $p/(2\log p)$ is natural to consider because that is roughly the value at which naïve heuristics would predict that at random there would not exist any permutation polynomials with degree below that bound.)  When $q$ is a square, there are various constructions of non-exceptional permutation polynomials over $\mathbb F_q$ of degree roughly $\sqrt{q}$.  For some infinite classes of $q$'s, there are constructions of non-exceptional permutation polynomials of degree roughly $\sqrt[3]{q}$.  But to date there is no known example of a non-exceptional permutation polynomial over $\mathbb F_q$ of degree less than $\sqrt[3]{q}$, for any $q$.

Roland Bacher mentioned in a comment that the affine bijections and the power permutations $x^n$ probably generate all permutations of $\mathbb F_q$.  That is true — in fact, Carlitz showed in 1953 that if $q>2$ then all permutations are generated by the permutations of the form $ax+b$ and $x^{q-2}$ (Permutations in a finite field).  But I don't think this says much about degrees of permutation polynomials, since compositions of several copies of $ax+b$ and $x^{q-2}$ typically have huge degree with loads of terms, and it's hard to control the degrees of the polynomials you'll get by reducing these gigantic polynomials mod $x^q-x$.<|endoftext|>
TITLE: Counting submanifolds of the plane
QUESTION [7 upvotes]: After thinking about this question and reading this one I am led to ask for an uncountable collection of homeomorphism types of boundaryless connected path-connected submanifolds of the plane.  
My guess is that is suffices to consider complements of Cantor sets.  However, I do not know how to distinguish ends (up to homeomorphism) sufficiently well to ensure that this works.   Are there other, easier, invariants?

REPLY [9 votes]: See a theorem of Richards, which implies that homeomorphism types of planar surfaces are in 1-1 correspondence with homeo. types of compact subsets of the Cantor set. I think there should be uncountably many homeo. types of totally disconnected compactums, but I don't know a reference or an argument off the top of my head. I think this should be related to the ordinalities of the accumulation points, but I'm not sure which ordinals can occur. 
Addendum: Googling, I found references to a result of Markiewicz-Sierpinski classifying
countable compact metric spaces up to homeomorphism by their Cantor-Bendixson rank (see section 3 of this paper for a statement). 
The CB-rank must be a countable ordinal $\zeta$, and the space is homeomorphic to the
ordinal $\omega^\zeta\cdot n+1$ with the order topology for some $n\in \mathbb{N}$. These may all be realized as
compact subsets of the line. This gives uncountably many non-homeomorphic compacta, 
which by Richards' theorem implies that there are uncountably many planar surfaces.<|endoftext|>
TITLE: Choosing postdocs
QUESTION [25 upvotes]: In this period I have been applying for some postdoc positions. One of the main difficulties is that there is no central organization, so the times and practices for applying vary considerably even in the same country. Even being able to start a postdoc exactly when the previous one ends is not always simple.
Here I have a question about etiquette. What is people expected to do when moving from one postdoc to another one? A typical example would be a position which starts a few months before the previous one is finished. Or it is conceivable that one may have to accept a position in a short time, only to find later that he has won another position (for which he had already applied) which is "better" for various reasons, for instance it lasts longer or is in a place which is more favourable for personal reasons. An extreme case would be that on getting a permanent position while still doing a postdoc.
What is the expected etiquette in such cases? I'm interested mainly in what happens in Europe, if the opinion changes worldwide, but it is probably useful to hear from anyone about this matter.

REPLY [5 votes]: Just so we have a complete picture here: all of the other answers I've seen describe a type of postdoc which is common in Europe (I'm not really sure about the rest of the world), but unheard of in the US.  In the US, basically any postdoc will fit into the standard academic calendar (with a few exceptions that are 1 semester); sometimes you can have flexibility about starting a year later (this is actually quite common with tenure-track positions).
I think the generally accepted standard in the US is that you should only accept one job in a given year (i.e. if you accept one job, you shouldn't just drop it and take another one a month later), though I think people violate this rarely enough that I'm not sure we have reliable data on how it really affects one's career.  I think it's generally considered fine to look for jobs after that (say after 1 or 2 years of a 3 year position).<|endoftext|>
TITLE: What is the universal enveloping algebra?
QUESTION [13 upvotes]: Let ${\mathfrak g}$ be a Lie algebra in a symmetric monoidal category enriched over $K$-vector spaces, i.e., in particular, hom-s are $K$-vector spaces (where $K$ is a field of characteristic zero). What is its universal enveloping algebra?
As one can talk about associative and Lie algebras there, I can imagine the definition in terms of the universal property but I am interested in its existence, a construction, if you may. Completing the category appropriately (direct sums and direct summands) could give familiar tensor and symmetric algebras $T({\mathfrak g})$ and $S({\mathfrak g})$ (i.e. they are objects in a certain completion of the original category). Is there a way to quotient $T({\mathfrak g})$ or to deform $S({\mathfrak g})$ at this point?

REPLY [5 votes]: I missed this question but I still want to have my say as I think this deserves to be better known. Perhaps this would make a good topic for a blog post? This is, I think, Poincare's proof of a strong form of the PBW theorem.
Birkoff and Witt proved a weaker result a couple of decades later. This story is told in this reference:
MR1793103 (2001f:01039)  Ton-That, Tuong ;  Tran, Thai-Duong . Poincaré's proof of the so-called Birkhoff-Witt theorem.
 Rev. Histoire Math.  5  (1999),  no. 2, 249--284 (2000).
The set-up is given in Torsten's answer. We have a symmetric monoidal category enriched in the category of vector spaces over a field of characteristic zero. We also assume we can form countable direct sums and that idempotents have images (this is no loss of generality as the original category can be formaly enlarged if necessary). This gives the structure and there is a long list of compatibility conditions most of which should be obvious. I am not sure if we require the tensor product to be distributive over countable direct sums.
Just to be clear I do not assume we have cokernels and I have in mind examples where cokernels do not exist.
However let's start in the category of vector spaces. I was given a copy of notes taken at a talk by Kostant in France in the 1975. The only references I know of are the following (both of which make the construction seem obscure). 
MR2301242 (2008d:17015)  Durov, Nikolai ;  Meljanac, Stjepan ;  Samsarov, Andjelo ;  Škoda, Zoran . A universal formula for representing Lie algebra generators as formal
 power series with coefficients in the Weyl algebra.
 J. Algebra  309  (2007),  no. 1, 318--359.
http://arxiv.org/abs/math/0604096
MR1991464 (2004f:17026)  Petracci, Emanuela . Universal representations of Lie algebras by coderivations.
 Bull. Sci. Math.  127  (2003),  no. 5, 439--465.
Anyway the basic idea is that we take the symmetric algebra $S(g)$ and define an action of $g$. This then generates the action of $U(g)$ and so constructs $U(g)$. In order to define the action of $g$ it is sufficient to define $x*y^n$ since we obtain the action of $x$ by polarisation. The key is that this is given by:
$$x*y^n = \sum_{j=0}^n \binom{n}{j}B_j ad^j(y)(x)y^{n-j}$$
where the $B_j$ are the Bernoulli numbers with generating function $x/(e^x-1)$.
Once you unwind this you find that you have constructed maps $S^r(g)\otimes S^s(g)\rightarrow S^{r+s-j}(g)$ for $r,s,j\ge 0$ by universal formulae. These
formulae then make sense in the abstract setting.
I would be delighted to see a good exposition of this.
Edit: The following reference looks as though it should be relevant but I didn't get much from it.
MR1894038 (2003b:17014)  Cortiñas, Guillermo . An explicit formula for PBW quantization.
 Comm. Algebra  30  (2002),  no. 4, 1705--1713.<|endoftext|>
TITLE: tetrahedron edges probability 
QUESTION [8 upvotes]: If 6 numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the edges of a tetrahedron? 
I wrote some code and simulate this probability and I suppose the answer is 1/3.
As for 2-dimensional question it is trivial to find that:
If 3 numbers are chosen at random, uniformly and independently, from the interval [0,1], the probability that they are the lengths of the sides of a triangle is 1/2.
3-dimensional case seems to be very difficult to prove.
Moreover is this true (n-dimensional case):
If n(n+1)/2 numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the edges (1-faces) of a n-dimensional simplex?
Inspired by the 2 and 3-dimensional cases I suspect the answer is 1/n but this one seems to be a real chestnut.

REPLY [8 votes]: There are four ways to interpret your question:
1) compute the probability exactly when each length $\ell_{ij}$ is chosen independently
2) compute the probability exactly when 6 numbers are chosen independently and you have a freedom to assign them to edges in any way
3) you want some (say, 0.001) lower bound in either case
4) you want a 1/3 lower bound in either case
In case 1), consider the set of possible tetrahedron 6-tuples of edge lengths as a points in $\Bbb R^6$.  You basically want to compute the volume of this set.  Unfortunately, this set is non-convex and is defined by rather nasty inequalities (see this Rivin's paper which I already mentioned on this MO answer).  To get convexity, Rivin shows you need to consider squares of edge lengths.  In other words, the desired volume is a volume of an algebraic body and is likely be non-algebraic itself.  It being 1/3 is doubtful.  
For 2), the problem is much harder as you have various permutations to consider.  For 3), this is easy - a small enough perturbation of lengths of a regular simplex will work.  For 4), this may or may not be true.  I sort of doubt it if you don't allow permutations, but with permutations I have no intuition.  In principle, you can simply approximate the volume of a body in $\Bbb R^6$, there are better ways for doing that than sampling random points and checking if it's in there.  Either way, it is hard to imagine how you would later extend this to simplices in higher dimension.<|endoftext|>
TITLE: Can we select a rainbow matching if each degree is 6 and each colorclass is a C_6?
QUESTION [10 upvotes]: Suppose that we have a 2d-regular graph whose edges are colored such that the edges of each color form a cycle of length 2d. (So if the graph has 2n vertices, then there are n colors.) Is it true that there always is a perfect matching containing one edge of each color?
Remarks. For d=2 there is a simple proof by Zoltan Kiraly who also invented the above formulation of the problem. I even do not know the answer for d=3.

REPLY [2 votes]: This is a little embarrassing, but it turned out that not even a (non-rainbow) matching is guaranteed to exist. The problem was solved on this workshop by a number of people, presented by Tamas Terpai. They raised the same question for bipartite graphs, for which a matching must always exist.<|endoftext|>
TITLE: Packing twelve spherical caps to maximize tangencies
QUESTION [10 upvotes]: Suppose that $v_i$, for $i \in \{1, 2, \ldots 11, 12\}$, are twelve unit length vectors
  based at the origin in $R^3$.  Suppose that $|v_i - v_j| \geq 1$ for all $i
\neq j$.  What arrangement of the $v_i$ maximizes the number of pairs $\{i,j\}$
  so that $|v_i - v_j| = 1$?

If C is a cube of sidelength $\sqrt{2}$ centered at the origin then we can
place the $v_i$ at the midpoints of the twelve edges.  Taking the convex
hull of the $v_i$ gives a cube-octahedron of edge-length one. See here
for a picture.  If you cut the cubeoctahedron along a hexagonal equator and 
rotate the top half by sixty degrees you get another polyhedron.  Both of 
these have 24 edges.  Are these the unique maximal solutions to the above 
problem?
Notice that if you place the $v_i$ at the arguably nicer vertices of a 
icosahedron then the $v_i$ become too widely separated.  It is easy to 
check this by making a physical model!
I spent some time thinking about areas of spherical polygons and restrictions on the graph of edges (and its dual graph) coming from the Euler characteristic.  However, I don't think I got very far - in particular ruling out pentagons seems to be a crucial point that I couldn't deal with.  Finally, to explain the problem title: instead of thinking of unit vectors with spacing restrictions, consider the (equivalent) problem of placing twelve identical spherical caps, of radius $\pi/12$, on the unit sphere with disjoint interiors in such a way as to maximize the number of points of tangency. 
This question was asked of me by an applied mathematician.  It comes from a problem involving packing balls in three-space, minimizing some quantity that is computed by knowing pairwise distances.  The solution to the kissing problem thus justifies the "twelve" appearing in the problem statement.  The projection of surrounding balls to a central one gives the spherical caps.

REPLY [9 votes]: Interesting question. I can find answer using my program, which was made for solving Tammes problem for 13 points. But I need some time for answer.
UPD: I wrote program. Result: 24 is a maximal number of edges.
I did in three steps.
First, I enumerated planar graphs with 12 vertices with at least 25 edges, at most 5 edges in a vertex and at most hexagonal faces.
Total number of suc graphs is 67497.
Second, I eliminated by linear programming by considering values of face angles as variables. 
My constrains was:
1. angle in triangle is ~1.2310
2. each angle no less than 1.2310
3. sum of angles around vertex is 2*pi
4. opposite angles of rectangle are equal
5. sum of non-opposite angles in rectangle between 3.607 and 3.8213
I solve feasibility of this LP problem (with some tolerance)
After this step all graph were eliminated.<|endoftext|>
TITLE: Examples of undergraduate mathematics separation from what mathematicians should know
QUESTION [43 upvotes]: I'm looking for examples of four kinds of things:

Material that is usually covered in standard undergraduate mathematics courses and/or in first-year graduate work (or tested in qualifying examinations) but that most mathematicians aren't really expected to know/remember: Some things that come to my mind are Sylow's theorems and their applications (for mathematicians outside of group theory and geometric group theory) and point set topology (except perhaps logicians and some algebraic geometers). If there are other examples of this kind of stuff, why is it taught in undergraduate courses? I can think of three explanations: (a) it is useful to learn (either the content or the techniques) at least once, even if people forget; (b) it is so important for people who go into that area of mathematics that it's worth subjecting everyone else to it; (c) inertia.
Material that is not taught or covered in undergraduate courses and/or in most first-year graduate work, but that professional mathematicians across multiple specialties are supposed to be comfortable with. Things that might fit the bill (but I'm not sure) are various techniques in combinatorics and elementary number theory, and ideas from category theory. But I'm not really sure. 
On a related note to (1), mathematical skills that undergraduates get good at while studying the courses but that most of them forget even if they become mathematicians. Examples include all the tricks and techniques for integration, Sylow's theorem tricks.
In contrast to (3), skills that people get better at in general as they do more and more mathematics. This probably includes things like a better understanding of quotients, asymptotic behavior, universal properties, product spaces, multiple layers of abstraction (like a norm on a space of operators on a space of linear functionals on a space of functions on a topological space, or one of those typical things in category theory).

All the things above are guesses and I'm curious to hear what items others have in mind and whether people think there exists any notable divide or difference of the kind I've suggested above between what undergraduates learn/get good at and what mathematicians are expected to be good at.

REPLY [2 votes]: Lagrange multipliers. Not adequately covered in my Princeton math-major curriculum. (I did take some physics and economics courses, where they were used as a magic formula without any explanation beyond the heuristic "shadow price", etc.) By grad school though (CS at CMU), we were expected to feel fully comfortable with them.<|endoftext|>
TITLE: Examples of computing Ext and Tor functors?
QUESTION [10 upvotes]: So I understand in theory the definition of Ext and Tor, but when it comes to actually computing them, I'm stuck.  For example, could someone show me how to compute $\text{Ext}(\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$?  I tried this by taking an injective resolution ($0 \to \mathbb{Z}/m\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \to 0$?) and I got an exact sequence with $\text{Ext}^1(\mathbb{Z}/m\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) = 0$, but I don't see what to do next?
P.S. No, this is not a homework question.

REPLY [10 votes]: $\mathbb{Q}/\mathbb{Z}$ is a pretty terrible abelian group, or a rather hard one, there may be better injective resolutions to work with. It would certainly be easier to do the projective resolution, use $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/n \to 0$. this will surely be easier to work through than the one involving $\mathbb{Q}/\mathbb{Z}$. Then compute the appropriate tensor product or hom group.
I started learning this stuff on more interesting modules as Schmidt suggests. For example, modules over the group ring of some cyclic group, or maybe an exterior algebra on two generators (if you make the generators of different gradings, in particular 1 and 3). This happens to be the category of modules you need to understand in order to compute complex connective k-theory! 
this should help get you going. These computations are very fun!<|endoftext|>
TITLE: Basic question about differential forms and physics
QUESTION [9 upvotes]: When, if ever, can we view a differential form, e.g. like $dx \wedge dy$, as the similar looking expression used in physics to represent the product of "infinitesimals" e.g. $dx$ $dy$? In particular, I'm wondering why differential forms are anti-symmetric, e.g. $dx \wedge dy=-dy \wedge dx$, whereas in physics we often are happy to write $dx$ $dy=dy$ $dx$. Am I misunderstanding something basic?

REPLY [2 votes]: This question is long dead, but I think there's one aspect of it that wasn't really addressed, which is how there can be such a direct relationship between the antisymmetric object $\mathrm{d}x \wedge \mathrm{d}y$ and the symmetric object $dxdy$.
The reason is that ever-important proviso when integrating a differential form: the result depends on an arbitrary choice of orientation. So for a manifold $M$ admitting an orientation $o$, the relationship is actually $\int_{(M,o)} dx \wedge dy = \begin{cases} \int_M dxdy = \int_M dydx & \text{if } [\mathrm{d}x \wedge \mathrm{d}y] = o\\
-\int_M dxdy = -\int_M dydx & \text{if } [\mathrm{d}x \wedge \mathrm{d}y] \neq o\end{cases}$
Here the integral on the left is integration of forms and the integral on the right is a measure-theoretic integral. Also, the brackets denote taking the orientation-equivalence classes, where $\omega \sim \omega'$ iff $\omega = f\omega'$ for some everywhere-positive function $f$. Actually, this only makes sense if $\omega$ is nonvanishing; for a general 2-form $\omega = f dx\wedge dy$ we extend by locality of the integral / linearity of the measure and have $\int_{(M,o)} f dx\wedge dy = \int_{M_+} f dxdy - \int_{M_-}fdxdy$ where $M_{\pm} = \{p \in M \mid [f_p \mathrm{d}x \wedge \mathrm{d}y] = \pm o_p\}$ (and $M_0$ gets thrown out).
Now, what is going on when we swap the order of symbols under an integral sign? Well, from the equation above, we see that when we swap $\mathrm{d}x \wedge \mathrm{d}y$ with $\mathrm{d}y \wedge \mathrm{d}x$, we're multiplying our integrals by $-1$ (assuming we intend on keeping the same orientation). On the other hand, when we swap $dxdy$ for $dydx$, we're just using a notational variant as far as the product measure is concerned. But there's an ambiguity in standard integral notation between integration over a product measure and "Fubini'ed" integration, and usually if we swap $dxdy$ for $dydx$, what we mean is that we're changing the order in which we want to Fubini! What is the counterpart in terms of integrals of differential forms? Well, the equations, with $\omega = f \mathrm{d}x\wedge \mathrm{d}y$, will be
$\int(\int fdx)dy = \int fdxdy = \int (\int fdy)dx$
$\int(\int i_{\partial_{y|x}} \omega) \mathrm{d}y = \int \omega = \int(\int i_{\partial_{x|_y}} \omega) \mathrm{d}x$
Here $i_X$ is the insertion operator for the vector field $X$, which feeds $X$ into one of $\omega$'s input slots (first or last, by convention, I'm not sure which), lowering the degree by 1. Also, $\partial_{x|y}$ is short for $\frac{\partial}{\partial x}|_y$. Orientations: suppose that $\int \omega$ is oriented like $\mathrm{d}x \wedge \mathrm{d}y$. Then we can take the other integrals to be oriented like $\mathrm{d}x$ or $\mathrm{d}y$ as appropriate, as long as we take the insertion operator to be inserting in the last slot on the LHS, and the first slot on the RHS... 
Independently of what happens when you swap the order of symbols, we can talk about the weird role of orientation in integration of forms. A form like $\mathrm{d}x \wedge \mathrm{d}y$ doesn't integrate to volume exactly, because its integral changes sign with orientation whereas volume doesn't. A quantity like this is usually called (somewhat pejoritively) a pseudoform.  Whereas a "true form" requires a choice of orientation on its domain of integration, a pseudoform requires a choice of orientation on the normal bundle of its domain of integration (despite what it sounds like, this does NOT require a metric). One reference for this material is in Theodore Frankel's Geometry of Physics, which Steve Huntsman (cryptically) linked to in his comment above.<|endoftext|>
TITLE: Are schemes that "have enough locally frees" necessarily separated
QUESTION [20 upvotes]: Let me motivate my question a bit. 
Thm. Let $X$ be a locally noetherian finite-dimensional regular scheme. If $X$ has enough locally frees, then the natural homomorphism $K^0(X)\longrightarrow K_0(X)$ is an isomorphism.
A locally noetherian scheme has enough locally frees if every coherent sheaf is the quotient of a locally free coherent sheaf, $K^0(X)$ denotes the Grothendieck group of vector bundles on $X$ and $K_0(X)$ denotes the Grothendieck group of coherent sheaves on $X$.
The above theorem is shown as follows.
By the regularity (and finite-dimensionality!) of $X$, we can construct a finite resolution by a standard procedure. (Surject onto the kernel at each stage with a vector bundle.) Then the "Euler characteristic" associated to this resolution is inverse to the natural morphism.
Now, I was looking through the literature (Weibel's book basically) and I saw that this theorem appears with the additional condition of separatedness. (Edit: This is not necessary. The point is that noetherian schemes that have enough locally frees are semi-separated.)
Example. Take the projective plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}^3$ whereas $K_0(X) \cong \mathbf{Z}^4$. 
Example. Take the affine plane $X$ with  a double origin. Then $K^0(X) \cong \mathbf{Z}$, whereas $K_0(X) \cong \mathbf{Z}\oplus \mathbf{Z}$.
So I figured I must be missing something...Thus, I ask:
Q. Are locally noetherian schemes that have enough locally frees separated? 
EDIT.
The answer to the above question is "No" as the example by Antoine Chambert-Loir shows. 
From Philipp Gross's answer, we conclude that a noetherian scheme which has enough locally frees is semi-separated. This means that, for every pair of affine open subsets $U,V\subset X$, it holds that $U\cap V$ is affine. Note that separated schemes are semi-separated and that Antoine's example is also semi-separated.
Taking a look at Totaro's article cited by Philipp Gross, we see that a regular noetherian scheme which is semi-separated has enough locally frees. (Do  regular semi-separated and locally noetherian schemes have enough locally frees?)
This was (in a way) also remarked by Hailong Dao. He mentions the result of  Kleiman and independently Illusie. Recently, Brenner and Schroer observed that their proof works also with $X$ noetherian semi-separated locally $\mathbf{Q}$-factorial. See page 4 of Totaro's paper. In short, separated is not really needed but semi-separated is.
Thus, we can conclude the following.
Suppose that $X$ is a regular and finite-dimensional scheme.
If $X$ has enough locally frees, then $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. For example,  $X$ is noetherian and semi-separated.
Anyway, thanks to everybody for their answers. They helped me alot!

REPLY [15 votes]: The property that every coherent sheaf admits a surjection from a coherent locally free sheaf is also known as the resolution property.
The theorem can be refined as follows:
Every noetherian, locally $\mathbb Q$-factorial scheme with affine diagonal (equiv. semi-semiseparated) has the resolution property (where the resolving vector bundles are made up from line bundles).
This is Proposition 1.3 of the following paper:
Brenner, Holger; Schröer, Stefan
Ample families, multihomogeneous spectra, and algebraization of formal schemes. 
Pacific J. Math. 208 (2003), no. 2, 209--230. 
You will find a detailed discussion of the resolution property in
Totaro, Burt. 
The resolution property for schemes and stacks. 
J. Reine Angew. Math.  577  (2004), 1--22
Totaro proves in Proposition 3.1. that every scheme (or algebraic stack with affine stabilizers) has affine diagonal if it satisfies the resolution property.
The converse is also true for smooth schemes:
Proposition 8.1 : Let $X$ be a smooth scheme of finite type over a field. Then the following are equivalent:

$X$ has affine diagonal.
X has the resolution property.
The natural map $K_0^{naive} \to K_0$ is surjective.<|endoftext|>
TITLE: Examples of non-abelian groups arising in nature without any natural action 
QUESTION [25 upvotes]: It's said that most groups arise through their actions. For instance, Galois groups arise in Galois theory as automorphisms of field extensions. Linear groups arise as automorphisms of vector spaces, permutation groups arise as automorphisms of sets, and so on.
On the other hand, abelian groups often arise without any natural (or at least obvious) action -- the "class groups" such as the ideal class group and Picard group, as well as the various homology groups and higher homotopy groups in topology are examples. [ADDED: One (sloppy?) way of putting it is that abelian groups arise quite often for "bookkeeping" purposes, where we think of them simply as more efficient ways to store invariants, and their actions are not obvious and not necessary for most of their basic applications.]
What are some good examples of non-abelian groups that arise without any natural action? Or, where the way the group is defined doesn't seem to indicate any natural action at all, even though there may be an action lurking somewhere? The only prima facie example I could think of was the fundamental group of a topological space, but as we know from covering space theory, for nice enough spaces (locally path-connected and semilocally simply connected), the fundamental group is the group of deck transformations on the universal covering space.
This might be somewhat related to the question raised here: Why do groups and abelian groups feel so different?.
To clarify: There are surely a lot of ways of constructing groups within group theory (or using the tools of group theory, which includes various kinds of semidirect and free products, presentations, etc.) where there is no natural action. These examples are of interest, but what I'm most interested in is cases where such groups seem to arise fully formed from something that's not group theory, and there is at least no immediate way of seeing an action of the group that illuminates what's happening.

REPLY [8 votes]: The Weil group is an extension of the absolute Galois group of a number field by the connected component of the identity of its idele class group. Of course, the quotient given by the Galois group acts on stuff. Whether the bigger group naturally acts on some space is a big open problem in number theory that some people think holds the key to the Riemann hypothesis.
Tate, J. Number Theoretic Background, Proc. Symp. Pure Math. 33 (1979) 3-26.<|endoftext|>
TITLE: Cardinality: Why is there no "ℵ½"?
QUESTION [7 upvotes]: A wikipedia page/paragraph on ℵ₁ states: 

"The definition of ℵ₁ implies (in
ZF, Zermelo-Fraenkel set theory
without the axiom of choice) that no
cardinal number is between ℵ₀ and
ℵ₁."
"If the axiom of choice (AC) is
used, it can be further proved that
the class of cardinal numbers is
totally ordered, and thus ℵ₁ is the
second-smallest infinite cardinal
number."

Can someone point me at a lay explanation of this please?
Is it simply saying that ℵ½'s existence is up to definition, or choice?  And this has been shown by the axiom of choice?

REPLY [39 votes]: The point is that without the Axiom of Choice, cardinalities are not linearly ordered, and it is possible under $\neg AC$ that there are additional cardinalities to the side of the $\aleph$'s. Thus, the issues is not additional cardinalities between $\aleph_0$ and $\aleph_1$, but rather additional cardinalities to the side, incomparable with these cardinalities. 
Let me explain. We say that two sets $A$ and $B$ are equinumerous or have the same cardinality if there is a bijection $f:A\to B$. We say that $A$ has smaller-or-equal cardinality than $B$ if there is an injection $f:A\to B$. It is provable (without AC) that $A$ and $B$ have the same cardinality if and only if each is smaller-or-equal to the other (this is the Cantor-Shroeder-Bernstein theorem). 
Under AC, every set is bijective with an ordinal, and so we may use these ordinals to select canonical representatives from the equinumerosity classes. Thus, under AC, the $\aleph_\alpha$'s form all of the possible infinite cardinalities. 
But when AC fails, the cardinalities are not linearly ordered (the linearity of cardinalities is equivalent to AC). Let me mention a few examples:

It is a consequence of the Axiom of Determinacy that there is no $\omega_1$ sequence of distinct reals. Thus, in any model of AD, the cardinality of the reals is uncountable, but incomparable to $\aleph_1$. Thus, in such a model, it is no longer correct to say that $\aleph_1$ is the smallest uncountable cardinal. One should say instead that $\aleph_1$ is the smallest uncountable well-orderable cardinal.
A more extreme example is provided by the Dedekind finite infinite sets. These sets are not finite, but also not bijective with any proper subset. It follows that they can have no countably infinite subsets. In particular, they are uncountable sets, but their cardinality is incomparable with $\omega$. Thus, in a model of $\neg AC$ having a Dedekind finite infinite set, it is no longer correct to say that $\aleph_0$ is the smallest infinite cardinal. 

Thus, the issue isn't whether there is something between $\aleph_0$ and $\aleph_1$, but rather, whether there are additional cardinalities to the side of these cardinalities.<|endoftext|>
TITLE: The characteristic (indicator) function of a set is not in the Sobolev space H¹
QUESTION [7 upvotes]: Is it true that the characteristic
(indicator) function of a subset of
Euclidean space with finite positive
measure is never in the Sobolev space
$H^1 = W^{1,2}$? And if so, what is the best/easiest/most elementary way to see this?

Context:
I have this on good authority (it is stated in a decent textbook). However, I have had no joy in showing this to be the case myself. Due to its placement in the aforementioned textbook as the first exercise at the end of a chapter about $H^1$, it feels like it oughtn't be difficult to show, but a group of my friends and I had no luck. It is bugging me now.
[Though I am a student taking a course based on the textbook, this is not ‘homework’; I will not be graded on it in any way and I have attempted the problem myself.]

REPLY [4 votes]: A not-so-elementary way to see it is to use the theorem that if $f$ is an $H^1$ function, then for almost every line segment, the restriction of $f$ to that line segment gives an absolutely continuous function.  (In fact, this is essentially sufficient as well as necessary.) Wikipedia cites Maz'ya's book Sobolev spaces, and I also found a proof in Ziemer.

REPLY [2 votes]: The answer posted by Tom, as written is actually not true. A function in $H^1$ will not in general be differential almost everywhere; it depends on the dimension. In one dimension however it is indeed true that $H^1$ functions are differentiable almost everywhere (they are in fact absolutely continuous). There are two ways of seeing it is not in $H^1$. The simple answer is that if you differentiate the characteristic function of say $[0,\infty)$ then you will get the Dirac measure. However let me just answer your question first:
Answer 1:
Take any smooth compactly supported $\phi:\mathbb{R} \to \mathbb{R}$. By definition of  weak derivative we have
$\int \phi\, g^{\prime} \,\mathrm dx = - \int \phi^{\prime} g \,\mathrm dx$ where I've set $g=1_{[0,\infty)}$. This would have to be true for all such $\phi$ if the weak derivative existed. Now take $\phi^{\epsilon}$ to be supported in a neighborhood $(-\epsilon,\epsilon)$ of $0$. We are making the crucial assumption that $g^{\prime}$ is an integrable and hence it follows that $\int \phi^{\epsilon} g^{\prime} \to 0$ as $\epsilon \to 0$. However, $\phi^{\epsilon}$ is smooth and so
$\int \partial_x\phi^{\epsilon}(x)g(x)\,\mathrm dx = \phi^{\epsilon}(0)$ since $\phi$ was assumed to have compact support in $(-\epsilon,\epsilon)$. Now just fix $\phi^{\epsilon}(0)=1$ and we have that $\phi^{\epsilon}(0) \to 0$ by the first integral equality. This is a clear contradiction.
Notice that in fact that this really shows that $g' \,\mathrm dx = \delta(x)$.
Answer 2:
Take $1_{[0,1]}$ instead so that it is an $L^2([0,1])$ function. This is in fact the Fourier transform of a "sinc" function, $\sin(k)/k$ up to some normalization constants. If we consider the $H^1$ norm in frequency space we would need $\int_0^{\infty} |k|^2\frac{\sin(k)^2}{|k|^2} \,\mathrm d k < \infty$ which is clearly false. This requires being at ease with the Fourier transform so if you're not, answer 1 is probably best.
It is true in $\mathbb{R}^n$ that if $u \in W^{1,p}$ for $p > n$ then $u$ is a.e. differentiable and equals a.e. its weak gradient (see Evans chapter 5). This is to correct what Tom had said although perhaps we was thinking about the $n=1$ case in which case $2 > 1$.
Hope this helps!<|endoftext|>
TITLE: Why is Lebesgue integration taught using positive and negative parts of functions?
QUESTION [80 upvotes]: Background:  When I first took measure theory/integration, I was bothered by the idea that  the integral of a real-valued function w.r.t. a measure was defined first for nonnegative functions and only then for real-valued functions using the crutch of positive and negative parts (and only then for complex-valued functions using their real and imaginary parts).  It seemed like a strange starting point to make the theory dependent on knowledge of the nonnegative function case when this certainly isn't necessary for Riemann integrals or infinite series: in those cases you just take the functions or sequences as they come to you and put no bias on positive or negative parts in making the definitions of integrating or summing.  
Later on I learned about integration w.r.t a measure of Banach-space valued functions in Lang's Real and Functional Analysis.  You can't break up a Banach-space valued function into positive and negative parts, so the whole positive/negative part business has to be tossed aside as a foundational concept.  At the end of this development in the book Lang isolates the special aspects of integration for nonnegative real-valued functions (which potentially could take the value $\infty$).  Overall it seemed like a more natural method.
Now I don't think a first course in integration theory has to start off with Banach-space valued functions, but there's no reason you couldn't take a cue from that future generalization by developing the real-valued case in the same way Banach-space valued functions are handled, thereby avoiding the positive/negative part business as part of the initial steps.  
Finally my question:  Why do analysts prefer the positive/negative part foundations for integration when there is a viable alternative that doesn't put any bias on which function values are above 0 or below 0 (which seems to me like an artificial distinction to make)?  
Note:  I know that the Lebesgue integral is an "absolute" integral, but I don't see that as a justification for making the very definition of the integral require treatment of nonnegative functions first.  (Lang's book shows it is not necessary. I know analysts are not fond of his books, but I don't see a reason that the method he uses, which is just copying Bochner's development of the integral, should be so wildly unpopular.)

REPLY [5 votes]: The question has been exhaustively answered, I'd like to add a remark. The way I view the Lebesgue integral is: to every positive measurable function you can associate a meaningful integral (i.e. stable by all natural operations and limit procedures), which might be infinite. Now if you have a sign-changing measurable function, you can assign an integral to its positive and its negative part. If one of the two is finite, then you can associate a meaningful integral to that function too. If both are infinite, there are ways to define an integral in some cases, but much less meaningful and stable; some natural operations become impossible to define or unstable in general. I find this a quite natural approach: in most theorems you can replace the assumption 'integrable' with 'whose integral is defined'. Also, it is pretty intuitive that if both areas above and below the $x$-axis are infinite, there is little point in defining the signed area of the whole region.
Consider for instance the Fubini theorem (for $L^1$ functions) and its counterpart for positive functions, which we call the Tonelli theorem here in Italy. You can actually merge the two results and just say: if the integral of a measurable $f(x,y)$ is defined, then both iterated integrals are defined and give the same result as the double integral.<|endoftext|>
TITLE: Global stability for dynamical systems in $R^n$
QUESTION [6 upvotes]: Suppose we have a smooth dynamical system on $R^n$ (defined by a system of ODEs).
Assume that:
(1) The system has an absorbing ball, that is every trajectory eventually enters this ball
and stays in it. 
(2) The system has a unique stationary point, and this stationary point is locally
asymptotically stable.
(2) The system has no period orbits.
Can we conclude that the stationary point is in fact globally stable?

REPLY [4 votes]: As the questioner notes in a comment, the answer is Yes for n<3. 
One way to create counterexamples for larger n is to use the work on the Seifert Conjecture. Start with a vector field pointing inward to the origin, and replace a little piece of it with an "aperiodic plug." This "plug" looks from the outside like a constant flow, has no periodic orbits in the interior, but there is at least one orbit that goes in and never comes out.
For details on various plug constructions, this note from the Geometry Center is very readable and also has references to the original papers of Wilson and Kuperberg.<|endoftext|>
TITLE: Texts on the General History of Contemporary Combinatorics
QUESTION [11 upvotes]: I am looking for some core texts (books, book chapters, papers) about the general history of contemporary combinatorics, starting, say, from the interwar period up to today. 
Texts about the history of computer science & the theory of computation might be ok, but I guess I want to focus more on the general history of the 'pure' mathematics side.
Most of the main works I have found so far seem to talk about the earlier periods. I am familiar with some of Gian-Carlo Rota, Timothy Gowers and Terence Tao's expository pieces, and also the many biographies of Paul Erdos. I am also familiar with Norman Biggs and Edward Brian Davies's work.

REPLY [5 votes]: At George Andrews' web site there is a link to an article "Partitions".  He refers to this as chapter eight in History of Combinatorics, edited by Robin Wilson.  Robin Wilson has a list of Book Projects in which he describes this book as follows:
"History of Combinatorics, Mathematical Association of America; an edited collection of articles on topics in the history of combinatorics: target handover date, end-2008."
But I can't find anybody who's saying that they've written some other chapter of this book. Nor can I find any reference to it at the MAA web site. Perhaps someone else reading this knows more about the status of this project.
Edited 31 May 2011: I came across this old answer of mine and decided to dig around a little more. John Watkins, recently retired from Colorado College, is apparently working on this project with Wilson. This June 2010 Mathematics Magazine article about Euler squares includes the following citation for another chapter: 
L. D. Andersen, History of latin squares, Department of Mathematical Sciences, Aalborg University, Research 
Report Series R-2007-32, 2007. To appear in The History of Combinatorics, R. Wilson and J. Watkins, eds. 
Edited 5 May 2016: This book appears to have been published as Combinatorics Ancient and Modern.<|endoftext|>
TITLE: How thinly connected can a closed subset of Hilbert space be?
QUESTION [19 upvotes]: Let H be a separable (and infinite-dimensional) Hilbert space. Is it known whether there exists an infinite
subset C of H with the following properties.? (1) C is connected and closed in H. (2) No infinite proper 
subset of C is both connected and closed in H.
Perhaps the answer depends upon whether or not the Axiom of Choice is assumed to hold. In any case, no 
finite-dimensional Euclidean space-because it is locally compact-can contain a subset with these properties.

REPLY [6 votes]: Problem has now been solved.
There is a Polish space that is widely-connected. 
A space is widely-connected if it is connected and  each connected subset with more than one point is dense (so every connected subset is “wide”).  In the posts and comments above there are several equivalent definitions. 

Surely continuum theorists have thought about this question, haven't they?

Yes, they have. It appeared in various problem books which is how I became aware of it. The earliest mention I could find was actually by Paul Erdős, according to Mary Ellen Rudin.  It probably goes back to the 1940's, when Erdős was publishing on connected sets and dimension theory.<|endoftext|>
TITLE: A Non-Commutative Nullstellensatz
QUESTION [15 upvotes]: In studying presentations of pro-$p$-groups via generators and relations, one is led (via the so-called Magnus embedding) to questions involving power series in non-commuting variables.  Results from local algebraic geometry occasionally shed some insight on how to make progress, but more often that not, I find myself lacking appropriate analogs of major theorems from the commutative case.  I haven't had much luck in books on non-commutative ring theory or non-commutative algebraic geometry -- the focus seems to be on completely different ideas (though I'll happily stand corrected).  In any case, here's an important and seemingly basic question that I don't know how to answer.

Let $\mathbb{F}_p\langle\langle x,y\rangle\rangle$ be the ring of formal power series over $\mathbb{F}_p$ in two non-commuting variables $x$ and $y$.  This ring has a unique two-sided maximal ideal $I=(x,y)$.  Suppose $f,g\in I$.  Can anything be said about the smallest $n$, if one exists, such that $I^n\subset (f,g)$?  Namely, when does this quantity exist?  Is this quantity computable?  Boundable?

It's trivial to come up with examples for which there is no $n$, e.g., $(xy,yx)$, since no $x^n$ is contained in this ideal.  I'm not sure how exactly to quantify this observation.  Is there some kind of non-commutative resultant at play here?
Edit:  I think it might be helpful for me to update with some examples as we go along.  Here's one that I thing captures at least some of the interesting parts of this question.
Take $p=3$, $f=x+y$, and $g=x^3$.  Then the inclusion $I^3\subset (f,g)$ can be seen by taking each of the 8 monomials in $I^3$ verifying that they are in $(f,g)$, e.g., $yxy=yfy-f^3+g\in (f,g)$.  The same argument applies with the same $f$ and taking $g=x+y+x^3$.  This seems to me evidence that this question can't be answered only by looking at the leading monomials (though admittedly it might be easy enough to exclude these trivial counter-examples).

REPLY [4 votes]: Let $F$ be a field, and let $f_1,f_2,\ldots, f_k\in R:=F\langle\langle x,y\rangle\rangle$ with $k\in \mathbb{N}$.  Order monomials in $R$ by degree, and then lexicographically.  Since the question concerns computability, assume that there is an algorithm which spits out the coefficients of the monomials in $f_i$ (in order).  I claim that:
Proposition: Given any fixed $n\in \mathbb{N}$, there is an algorithm to decide whether $I^n\subseteq (f_1,f_2,\ldots, f_k)$.
To prove this, first we need a lemma and some notation.  Given $r\in R$ we write $r[n]$ for the homogeneous component of $r$ in degree $n$.
Lemma: $I^n\subseteq (f_1,f_2,\ldots, f_k)$ iff for each of the $2^n$ monomials $m$ of total degree $n$ there is a power series $g_m\in (f_1,f_2,\ldots, f_k)$ such that $g_m[n]=m$.
Proof of the lemma: The forward direction is obvious.  For the reverse, let $m_1,m_2,\ldots, m_{2^n}$ be the list of all monomials of degree $n$, and let $g_i:=g_{m_i}$.  Fix $h\in I^n$.  We can write $h[n]=\sum m_i a_{i,n}$ for some $a_{i,n}\in R$ (all of degree $0$).  Setting $h':=h-\sum g_i a_{i,n}$, we see that $h'$ has zero homogeneous components in degree $\leq n$.  We can write $h'[n+1]=\sum m_i a_{i,n+1}$ for some $a_{i,n+1}\in R$ (all of degree $1$).  Set $h''=h-\sum g_i(a_{i,n}+a_{i,n+1})$.  Repeating this process, we obtain power series $a_i=\sum_{m\geq n}a_{i,m}$ such that $h=\sum_i g_i a_i\in (f_1,f_2,\ldots, f_k)$.$\qquad \blacksquare$
Proof of the proposition:  By the lemma, it suffices to decide for each monomial $m\in I^n$, whether there is a power series $g_m\in (f_1,f_2,\ldots, f_k)$ with $g_m[n]=m$.  We may as well work in the quotient ring $R/I^{n+1}$.  The image of the ideal $(f_1,f_2,\ldots, f_k)$ modulo $I^{n+1}$ is a finite-dimensional $F$-vector space, and so the question is easily decided by a row-reduction style argument.  [More concretely, one can modify the set $\{f_1,f_2,\ldots, f_k\}$ modulo $I^{n+1}$ so that no leading terms are linear combinations of others, and the set is closed under left and right multiplication by $x$ and $y$.  Then, the leading terms of degree $n$ either do or do not generate all of the appropriate monomials.]$\qquad \blacksquare$

To give an interesting example, consider $f_1=yx-y^2+x^3$, $f_2=xy-x^2$.  One can show directly (using the methodology above) that $I^{5}\subseteq (f_1,f_2)$.  If we modify $f_1=yx-y^2+x^k$ (with $k\geq 3$), then we get $I^{k+2}\subseteq (f_1,f_2)$.

On the question of whether it is decidable, given as input only the algorithms describing the power series $f_1,\ldots, f_k$, whether an $n$ exists for which $I^n\subseteq (f_1,f_2,\ldots, f_k)$, I don't know the answer.<|endoftext|>
TITLE: Resources for graphical languages / Penrose notation / Feynman diagrams / birdtracks?
QUESTION [31 upvotes]: There is an idea I've recently gotten interested in that doesn't seem to have a good agreed-upon name ("diagrammatic algebra?").  It centers around the use of two-dimensional diagrams of dots, possibly boxes, and arrows, and is related to (in no particular order) knot theory, braided monoidal categories, quantum groups and Hopf algebras, subfactors, planar algebras, and (topological) quantum field theory.  However, it also has a more accessible aspect: it can be used as an elegant notation for working with $\text{Vect}$ (a particularly ubiquitous braided monoidal category; see question #6139), and at least one textbook has used a variant of it to develop the basics of Lie theory.  There is also John Baez's Physics, Topology, Logic, and Computation: a Rosetta Stone, and another accessible introduction to some of these ideas is Kock's Frobenius Algebras and 2D Topological Quantum Field Theories.  These ideas have also been used to understand quantum mechanics.
This is all pretty fascinating to me.  These are elegant and beautiful ideas, and it seems to me that they are badly in need of unification and accessible exposition (something like Selinger's A survey of graphical languages for monoidal categories, but maybe with a more historical and/or expository bent).  Beyond Baez's paper, does anyone know of any resources like that?  Where can I learn more about what you can do with these diagrams that doesn't necessarily require a lot of background?  
Related: how should I TeX these diagrams?

REPLY [3 votes]: Here is a recent talk by Micah McCurdy on how to extend string diagrams to express (monoidal) functors, and monads.
Since monads and functors are ubiquitous in computer science (e.g. denotational semantics, functional programming etc.) McCurdy's work has immediate practical applications.<|endoftext|>
TITLE: Does every finitely presentable group have a presentation that simultaneously minimizes the number of generators and number of relators?
QUESTION [27 upvotes]: This should probably be an easy question, but I don't know how to answer it: Suppose G is a finitely generated presentable group. Suppose a is the absolute minimum of the sizes of all generating sets for G and b is the absolute minimum of the number of relations over all presentations of G. Question: Is it necessary that G has a presentation that simultaneously has a generators and b relations?
The case b = 0 is just the fact that a free group cannot be generated by fewer elements than its free rank.
The problem could probably be interpreted in terms of CW-complexes (where the generators give rise to 1-cells and the relators give rise to 2-cells) but, because of my lack of familiarity with CW-complexes, I don't immediately see how to use these to solve the problem.
It also seems to be related to the notion of "deficiency" of a group, which is the (maximum possible over all presentations) difference #generators - #relations (under the opposite sign convention, the minimum possible difference #relations - #generators).

REPLY [17 votes]: A stronger question, is the deficiency of $G$ realized for a presentation with
the minimal number of generators (rank($G$))? This question is asked in a paper of Rapaport, and proved to be true for nilpotent and 1-relator groups. 
Addendum:
The question appears as Question 2, p. 2, of a book by Gruenberg. Lubotzky
has answered the analogous question affirmatively in the category of profinite
groups (Corollary 2.5). (Note though that this is in the category of profinite presentations, so it does not imply an affirmative answer even for finite groups).<|endoftext|>
TITLE: A subset of Baire space Wadge incomparable to a Borel set?
QUESTION [6 upvotes]: Let $\omega^\omega$ be Baire space. If $A,B\subseteq\omega^\omega$ we say that $A$ is Wadge reducible to $B$ (written $A\leq_w B$) if there is a continuous function $f:\omega^\omega\rightarrow\omega^\omega$ with $x\in A$ if and only if $f(x)\in B$. (In other words $A$ is a continuous preimage of $B$). By identifying sets with $A\leq_w B$ and $B\leq_w A$ we induce a partial ordering on the set of corresponding equivalence classes, or Wadge degrees. Under the axiom of determinacy AD, it turns out by the so-called Wadge lemma that this hierarchy is almost linearly ordered, meaning we get a linear order if we identify a degree $a$ and the degree consisting of complements of members of $a$. Clearly the Borel sets will form an initial segment of this hierarchy. What I am curious about is if this is still the case if we drop the AD assumption.
Even without AD, determinacy holds for the Borel sets and so the Borel Wadge degrees will still be almost linear ordered - indeed almost well-ordered. For non-Borel degrees using AC it is possible to get a lot of bad behavior; for example many incomparable Wadge degrees. But all the ways I can figure to do such things is to enumerate all continuous functions and build simultaneously the incomparable sets by diagonalizing against the continuous functions. An argument like this is (I think) rather unlikely to produce a Borel set.
So specifically my question is: is it possible for there to be a non-Borel set which is Wadge-incomparable to some Borel set? This is the same as asking if it is possible for there to be a Borel set $B$ and a non-Borel set $A$ with $B\not\leq_w A$.

REPLY [6 votes]: The same type of diagonalization should allow you to do this.
Suppose $B$ is a Borel set that is not $F_\sigma$.
Let $(f_\alpha:\alpha<2^{\aleph_0})$ list all continuous functions.
At any stage $\alpha<\aleph_0$ we will have $A_\alpha$ and $C_\alpha$ disjoint
subsets of $\omega^\omega$ of size less than $2^{\aleph_0}$.  Elements in $A_\alpha$ will
end up in our set $A$ while elements of $C_\alpha$ will not.  That is $A=\bigcup A_\alpha$
and $A\cap C_\alpha=\emptyset$ for all $\alpha$.
At stage $\alpha$ we make sure $f_\alpha$ will not be a reduction of $B$ to $A$.
case 1: there is $y$ in the image of $f_\alpha$ but not in $A_\alpha\cup C_\alpha$.
Say $y=f(x)$.  If $x\in B$, put $y\in C_{\alpha+1}$, otherwise put $y\in A_{\alpha+1}$.
In either case, we have insured $f_\alpha$ is not a reduction.
case 2: there is $x$ in B with $f_\alpha(x)\in C_\alpha$ or $x\not\in B$ with
$f_\alpha(x)\in A_\alpha$.
In this case we already see $f_\alpha$ is not a reduction and do nothing.
case 3: otherwise  $f_\alpha$ maps $B$ into $A_\alpha$ and the complement into $C_\alpha$.
But the continuous image of a Borel is either countable or has size continuum.  Thus the images of $B$ and its complement both must be countable.  But then $B$ is $F_\sigma$
and its complement is $F_\sigma$, a contradiction.<|endoftext|>
TITLE: Example sought of an atomic domain R such that R[t] is not atomic
QUESTION [6 upvotes]: Recall that an integral domain $R$ is atomic if every nonzero nonunit admits at least one factorization into irreducible elements.  (Indeed, hard-core factorization theorists have replaced the word "irreducible" by "atom".)
From prior reading, I happen to know that there exist atomic integral domains $R$ such that the univariate polynomial ring $R[t]$ is not atomic.  This is a somewhat surprising pathology, because the implication is true if both instances of "atomic" are replaced by "UFD", "Noetherian" or "Ascending Chain Condition on Principal Ideals".
But I don't know a precise example or a reference, and I would like one for an expository article I'm writing.  Of course, the chronologically earlier and logically simpler the example, the better.

REPLY [5 votes]: According to the book "Non-Noetherian commutative ring theory" by S.T. Chapman and S. Glaz the question was first asked in "Factorization of integral domains" by D.D. Anderson, D.F. Anderson, M. Zafrullah, Journal of Pure and Applied Algebra 69 (1990) 1-19 (question 1). An answer was given here by M. Roitman.
There it was conjectured that $R[X]$ atomic $\implies$ $R[X,Y]$ is also atomic.<|endoftext|>
TITLE: Reconstruction conjecture: Can other decks do the job?
QUESTION [10 upvotes]: The standard reconstruction conjecture states that a graph is determined by its deck of vertex-deleted subgraphs.

Question: Have other decks been investigated, finding out
  that only vertex-deleted subgraphs can do the job? If so: Which property of vertex-deleted subgraphs makes them exceptional?

I have three candidates in mind, others are conceivable. (For the sake of simplicity I consider only simple connected graphs $G$.)

the deck of sub-maximal neighbourhoods: Let the sub-maximal neighbourhood of $v$ be the $v$-rooted graph constructed from $G$ by deleting all vertices with maximal distance from $v$.
the deck of distinguishing neighbourhoods: Let the distinguishing neighbourhood of $v$ be the smallest $n$-neighbourhood of $v$ which distinguishes it from all vertices not conjugate to it ($n$-neighbourhood = the $v$-rooted induced subgraph containing  all vertices $w$ with distance $d(v,w) \leq n$).
the deck of crossref-deleted subgraphs: Let the crossref-deleted subgraph with respect to $v$ be the $v$-rooted graph constructed from $G$ by deleting all edges between vertices that have the same distance from $v$.

Note that the vertex-deleted subgraph with respect to $v$ is nothing but the $v$-rooted graph constructed from $G$ by deleting all edges between $v$ and its neighbours.
I am not good in systematically constructing counterexamples, and I do not have very much intuition about general graphs. So, any counterexample to one of the candidates above would be very welcome.
What I do know is that a) trees are trivially reconstructible from their deck of crossref-deleted subgraphs, that b) graphs with one node of which the distinguishing neighbourhood is the whole graph are trivially reconstructible from their deck of distinguishing neighbourhoods, and that c) reconstructing (very) small graphs from one of the decks above is fun.

REPLY [8 votes]: I suspect that there are counterexamples to all three of your proposals.  For example, you clarified that the distinguishing neighborhood of a vertex-transitive graph is just its 1-neighborhood.  Then the dodecahedral graph and the Desargues graph will have the same decks; in each case you'll just have twenty copies of $K_{1,3}$.  Similarly, although I don't have an explicit counterexample offhand, I suspect that if you examine strongly regular graphs with the same degree and number of vertices, you will find plenty of pairs of graphs with the same decks of sub-maximal neighborhoods.  (Strongly regular graphs have diameter 2 so again you're just looking at 1-neighborhoods.)
As for your question of what other reconstruction conjectures there are, the most famous is the edge reconstruction conjecture.  There is also the vertex-switching reconstruction conjecture and the $k$-reconstruction conjecture (see Bondy's Graph Reconstructor's Manual for definitions).  For more examples, you could try contacting Mark Ellingham at Vanderbilt, who keeps a list of papers on the reconstruction conjecture.<|endoftext|>
TITLE: Why is Riemann-Roch for stacks so hard?
QUESTION [29 upvotes]: First some indication that it really is a difficult problem: Both Vistoli and Gillet in their classics on intersection theory on stacks remark that their should be a Riemann-Roch theorem for proper representable morphisms, but that they are not able to prove it. I think thats more than enough evidence. 
The two existing proofs by Toen and Joshua both involve using not the naive Chow ring, but a modified version. This makes both proofs quite heavy on K-Theory, and I don't really get them. 
So what makes the proof using the naive Chow-Ring so difficult? If I remember correctly from reading Fulton-Langs "Riemann-Roch Algebra", the basic technique is to factor a morphism as a regular imedding followed by a projection. The cases of a regular imbeddings and projections are treated by a hands-on methods. Here are some reasons I can think of why this might not work for stacks:

Its hard to find such a factorization.

    There's a problem with identifying the Chow-Ring of a stack with the K-Group equipped with the gamma-filtration.

    The factorization exists, but the hands-on part is too difficult.
    Maybe the K-Group doesn't have a lamda-ring structure?

REPLY [13 votes]: Ben gave much of the answer, but I'll try to make it precise. Toen says there is no Riemann-Roch for the naive (rational) chow ring (Remark 4.3 in Theoremes de R-R) (EDIT: unless you also take the naive K-theory).
He says the problem is the chow rings missing the stack structure (p.1), like Ben says. Like you point out, the morphism to a point won't be representable. If a stack has a representable morphism to a scheme, it must be an algebraic space. (Just pull-back by the identity. Representable means this will be an algebraic space.) There's also a problem with the G-theory. The problems are:

The morphism from a DM-stack to its coarse moduli space $p : F \to M$ induces an isomorphism of rational Chow rings,
$p_{*} : CH(F) \otimes \mathbf{Q} \to CH(M) \otimes \mathbf{Q}$
That's Thm 0.5 from Gillet's intersection theory paper.
The same morphism induces a weak equivalence (now we're working with simplicial objects)
$p_{*} : G(M) \otimes \mathbf{Q} \to H(F_{et}, G \otimes \mathbf{Q})$
of the cohomology of the G-theory sheaf with the G-theory of the coarse moduli space. (Corollary 3.8 of Toen's R-R paper). 
Note that G-theory is formed from the K-groups of the category of coherent sheaves, K-theory from the category of vector bundles. The natural morphism $K \to G$ gives a "Poincare" duality. It is an isomorphism in the case of algebraic spaces, but not in general for stacks (Prop 2.2 of Toen's thesis).

The workaround considers the ramification stack (aka. classifying stack of cyclic subgroups) of F, denoted $I_F$, and has been known in the case of complex orbifolds (V-varieties) since Kawasaki wrote on it in 1979. Toen's proof seems to center on proving
$G_{*}(F) \otimes \mathbf{Q}(\mu_\infty) \cong H^{-*}(I_F, G \otimes \mathbf{Q}(\mu_\infty))$.
I read the left side as the K-theory and the right side as the Chow ring.
He eventually reduces to the known case $F = [X / H]$ of equivariant K-theory with a smooth projective variety $X$ quotiented by a finite group $H$:
$\mathbf{K}_{*}(X, H) \otimes \mathbf{C} \cong \bigoplus_{h \in c(H)} \mathbf{K}_{*}(X^h)^{Z(h)}$,
where the sum is over the set $c(H)$ of conjugacy classes of $H$, $X^h$ is the fixed point subscheme, and $Z(h)$ is the centralizer of $h$ in $H$. (That's Vistoli 1991, maybe also Angeniol, Lejeune-Jalabert 1985)<|endoftext|>
TITLE: when are algebras quiver algebras ?
QUESTION [14 upvotes]: Good Morning from Belgium,
I'm no stranger to the mantra that quiver-algebras are an extremely powerful tool (see for example the representation theory of finite dimensional algebras). But what is a bit unclear to me is what is known about what kind of algebras are quiver algebras. The first case I can prove is for graded rings with finite dimensional semisimple $A_0$. But what about the ungraded case (I'm pretty sure that it is also true for finite-dimensional algebras over an algebraically closed field for example).
Is there a larger generality possible ?
Answers (and references) are as always much appreciated !

REPLY [13 votes]: Louis, do you mean by a 'quiver-algebra' the path algebra of a quiver, or do you mean a quotient of a path algebra?
If the first, then I do not understand your comments. k[x,y] is graded with semi-simple part of degree zero but not a path algebra. 
If you mean by quiver-algebra a quotient of a path algebra then the answer is simple : any finitely generated C-algebra will do as they are quotients of free algebras (path algebra of one vertex multiple loop quiver).
If you mean by quiver-algebra really the path algebra of a quiver, the answer is trickier. 
If your algebra is finite dimensional (I'm always working over C) then the classification is : hereditary and basic (that is, all simples are one-dimensional). In that case, any hereditary is Morita equivalent to a path algebra. All this goes back to Gabriel.
If your algebra is infinite dimensional one has to be careful. Surely it must be formally smooth (that is, it has the lifting property for algebra maps through nilpotent ideals) and have a finite number of isoclasses of one-dimensional representations.
EDIT : Oops, if one has loops then there are of course infinitely many 1-dmls. I should have said that there are only finitely many components of 1-dml representations, all parametrized by affine spaces and such that one can pick one rep in each component to perform the trick with the structural morphism described below.
But that is not enough, take e.g. the groupalgebra of the modular group PSL(2,Z). It has 6 one-dimensionals but is not isomorphic to the 'obvious' quiver one would construct out of these 6 (arrows corresponding to extensions between the simples).
What one needs is that the structural morphism A --> End(SS) (where SS is the semi-simple on the finite number of one-dimls) splits so that A becomes a C^k-algebra. Then one can use formal smoothness of A and the path algebra with semi-simple part End(SS) and arrow part determined by M/M^2 (where M is the kernel of the structural morphism above) to prove that they are isomorphic. An argument like this appears in the paper by Cuntz and Quillen on noncommutative smoothness (they call a formally smooth algebra 'quasi-free').<|endoftext|>
TITLE: Using the multiverse approach to decide the law of the exluded middle?
QUESTION [9 upvotes]: Recently, in response to deciding the Continuum Hypothesis $CH$, Hamkins and Gitman have proposed consider a multiverse of set-theoretic universes, some in which $CH$ is true, some in which $\neg CH$ is true (and some in which $CH$ is not a relevant hypothesis?).
In formulating logical languages, there has been an outstanding problem: that of deciding the Law of the Excluded Middle $p\vee\neg p$. In classical logic $p\vee\neg p$ is true, but in intuitionistic logic this is not the case. With $CH$, the pragmatic mathematician tries to avoid invoking $CH$; if he assumes $CH$ or $\neg CH$, he will state it clearly. In everyday mathematics practice, the mathematician does use $p\vee\neg p$, but we do see an effort to give  effective constructions, and hard estimates in analysis.
It seems that people are using classical logic basically because it's a core logic, in analogy to the constructible universe $L$ in which $CH$ is true. It is the logic first discovered (and often wrongly attributed as the Platonic choice) and the consistency and strength of other languages proven in terms of this core logic. Afterwards people come up with other logics, like Brouwer coming up with intuitionistic logic, in which a fundamental principle, the Law of the Excluded Middle, does not hold. It seems to me that this debate regarding the Law of Excluded Middle can be formalized using the multiverse approach.
So has anyone tried to use the multiverse approach towards considering these plurality of languages? Perhaps by considering a multiverse of topoi?
PS. The difficulty for arithemetizable syntax to describe continuous properties of geometric/measure-theoretic space is central (may I say?) to the difficulty of deciding $CH$. In a like fashion, the difference between classical and intuitionistic logic plays up in comparing the Dedekind-reals and the Cauchy-reals. In other words, continuous properties have different descriptions in these two logics. I'm hoping that work on the Law of the Excluded Middle will shed some light on trying to use discrete languages to model continuous properties. There's more where this comes from, but it is enough for now...

REPLY [2 votes]: I offer a different perspective on the question, which I hope will be found
useful.
One view of logic is strongly syntactic:  develop a language, and develop
rules for manipulating this language and classifying parts of this language
(terms, well-formed formulae, sentences, derivation, proofs, theories, and so on).
One of course would like to arrange things so that from a syntactic perspective
one can easily define such subsets so as to say "for each x in this collection
of objects, I can interpret x in this environment so that the interpretation is
at the same time true, meaningful, and tells me something about the environment." .
Indeed, such a viewpoint helped in various areas of scientific and technological
development.  Compiler theory, software development, fuzzy logic are just three
areas of "real-world" applications.  In this "real-world", the focus is often on 
solving a relatively narrow class of problems with a small set of tools, 
as opposed to considering many alternative problems with a wide variety of tools.
I call this an "engineering" perspective.
In my view, what you propose is a comparison of several logical systems in the
context of some unifiable framework, loose enough to encompass all of the
possibilities that you want to consider, but tight enough that you can resolve
some of the potential questions you see.  I call this a "scientific" perspective.
Speaking from ignorance, the closest I see in attempting such things are in fields
like modal logic, algebraic logic, and certain forms of general proof theory. I don't know enough of modal logic to comment intelligently, but I see various 
logics being compared and contrasted in some papers on modal logic.  I haven't 
worked in algebraic logic, but relational algebras, cylindric algebras, 
Heyting algebras, and more occurred in my studies in universal algebra: these 
structures were meant to put a strong algebraic component in studying and modeling 
logical systems.  From my naive perspective, one of the intentions was to provide 
an algebraic way to "define, true, meaningful, and useful subsets of a language". 
I know (almost) less than nothing of proof theory, but I recall Kleene's 
"Introduction to Metamathematics" where he develops formalisms for both classical 
and intuitionistic logic in parallel, using notation similar to those found
in proof theory, and I dimly recall Lindstroem's result on first-order logic being the 
strongest logic system in a sense (something about having both compactness and 
downward L-S theorems). So attempts have been made which are (as Gerry Myerson 
might say) "not a million miles away" from your proposal.
Of course, people who have spent time studying such things should weigh in and
clarify the situation.  Andrej Bauer has given a nice philosophical perspective,
but I think he missed the aspects above.  In particular, I think the "engineering"
perspective can be brought more in line with the "scientific" perspective, in that
while a "full" multiverse of logics could be studied, emphasis should be placed on
the "favored" multiverse of logics that are currently or soon will be applied to give
us some of the technological benefits mathematical logic can give society.
Gerhard "Ask Me About System Design" Paseman, 2015.10.08<|endoftext|>
TITLE: Properties of the n-dimensional Stereographic Projection
QUESTION [5 upvotes]: Hello,
I'm looking for an argument that the n-dimensional stereographic projection maps circles (intersections of affine two-dimensional subspaces with S^n) to circles in R^n. I've looked around and the only argument I saw for the n-dimensional case is a generalization of the geometric proof for n = 2 (with the tangent cone) which I don't really feel comfortable with, even when n = 2. Is it possible to reduce it to the n = 2 case somehow or give a "direct", algebraic, proof?

REPLY [6 votes]: I don't know what is the "tangent cone" argument you mention, but anyway here is my favorite  proof of the fact.
The stereographic projection is the restriction of an inversion $I$ from $\mathbb R^{n+1}\setminus \{p\}$ to itself, where $p$ is the pole from which you project (or maybe $I$ is a map from $\mathbb R^{n+1}\cup\{\infty\}$ to itself if you prefer it this way). So it suffices to prove that $I$ sends circles to circles and lines.
First observe that $I$ sends $n$-spheres and hyperplanes to $n$-spheres and hyperplanes. (This follows from the planar case with circles and lines and rotation symmetry. The planar case is elementary). In $\mathbb R^3$, every circle is an intersection of two spheres. Hence its image is an intersection of two sets each of which is a sphere or a plane. Such set is a circle again.
In higher dimensions, either restrict to a suitable 3-dimensional subspace, as Agol said in his answer, or represent a circle as an intersection of several spheres.<|endoftext|>
TITLE: Database of Steiner triple systems
QUESTION [11 upvotes]: Can anyone point me to an online database of Steiner triple systems?
My Google-fu is only getting me to descriptions of the few smallest ones, mostly Google book scans (which are rather useless to process using a computer...)

REPLY [18 votes]: The first answer is not a database of the Steiner triple systems, but rather how many there are up to isomorphism.  It is known that they there are 2 of order 13, 80 of order 15, and 11084874829 of order 19.  The last number was computed by Kaski and Ostergard, and I suppose that the best approach is to ask them for their data.  It's more or less the end of the story, because it is easy to make larger Steiner triple systems, but impossible to compile  them into a complete database.
Actually this paper describes a compressed 39-gigabyte file with the Steiner triple systems of order 19, and says that it is available by e-mail request from three of the authors (including Kaski and Ostergard).
It looks like a number of people have the Steiner triple systems of order 15, but I didn't find a paper that simply lists them.  
Update 1: It seems that everyone works from the paper "Small Steiner triple systems and their properties", by Mathon, Phelps, and Rosa.  This paper is basically an encyclopedia of the 80 Steiner triple systems of order 15 and many of their properties.  It also introduces a somewhat standard numbering.  The thing to do at this point would be to transcribe the data in this widely cited paper into a file.  Google seems to indicate that no such file has been posted to the web.
Update 2: It was done!  See file data/steiner.tbl in this GAP package by Nagy and Vojtechovsky.  For some reason, a Steiner triple system of order $n$ is also called a Steiner loop of order $n+1$, and that is the terminology that they use.  They copied the data from Colbourn and Rosa, Triple Systems, which presumably is the same as in Mathon, Phelps, and Rosa.
How I found it:  I Googled one of the hexadecimal strings used to describe one of the STS(15)s.  None of the Google's heuristics worked for me, so instead I used the old-fashioned trick of searching for a very specific keyword.  It also shows up in a LaTeX PHD thesis, but the GAP file, which has almost the same syntax as Python, is better.<|endoftext|>
TITLE: Most squares in the first half-interval
QUESTION [20 upvotes]: It is well known that if $p$ is an odd prime, exactly one half of the numbers $1, \dots, p-1$ are squares in $\mathbb{F}_p$. What is less obvious is that among these $(p-1)/2$ squares, at least one half lie in the interval $[1, (p-1)/2]$.
I remember reading this fact many years ago on a very popular book in number theory, where it was claimed that this is an easy consequence of a more sophisticated formula of analytic number theory.
Sadly I forgot both the formula and the book. So the purpose of the question is double:

1) Has any simple way been found to derive the fact mentioned above?
2) Does anybody know a reference for the analytic number theory route to the proof?

REPLY [28 votes]: There is a method of explaining this without using analytic methods.  I will get to that at the end of this answer.  
First, if $p \equiv 1 \bmod 4$ then this result is clear since -1 is a square mod $p$.  So here exactly half the squares mod $p$ lie in the first half of $[1,p-1]$.  The real problem is for $p \equiv 3 \bmod 4$, where analytic methods show there are more squares mod $p$ lying in the first half of that interval than in the second half because there is a formula for the class number of ${\mathbf Q}(\sqrt{-p})$  that is 1 or 1/3 times $S - N$, where $S$ is the number of squares mod $p$ in $[1,(p-1)/2]$ and $N$ is the number of nonsquares mod $p$ in $[1,(p-1)/2]$. Class numbers are positive integers, so $S > N$, which means in $[1,(p-1)/2]$ the squares mod $p$ outnumber nonsquares mod $p$. Since there are as many squares as nonsquares mod $p$ on $[1,p-1]$, the square vs. nonsquare bias on the first half of this interval forces there to be more squares mod $p$ on the first half than squares mod $p$ on the second half.
For a proof by analytic methods, see Borevich-Shafarevich's "Number Theory", Theorem 4 on p. 346.  It is not true that no non-analytic derivations of this bias are known.  For instance, Borevich and Shafarevich say on p. 347 that Venkov gave a non-analytic proof for some cases in 1928 (which came out in German in 1931: see Math Z. Vol. 33, 350--374). I should clarify this point since there is a bad typo in Borevich and Shafarevich here. What Venkov did was give a non-analytic proof of Dirichlet's class number formula for imaginary quadratic fields having discriminant $D \not\equiv 1 \bmod 8$.  Here the book unfortunately has $D \equiv 1 \bmod 8$. (It's clear from the book that something is wrong because shortly after saying Venkov treated $D \equiv 1 \bmod 8$ by non-analytic methods they say the case $D \equiv 1 \bmod 8$ still awaits a non-analytic proof.) The class number formula only gets an interpretation about squares or nonsquares for the fields ${\mathbf Q}(\sqrt{-p})$, but Venkov was working on non-analytic proofs of the class number formula for imaginary quadratic fields without having this restrictive case as the only one in mind. R. W. Davis (Crelle 286/287 (1976), 369--379) made simplifications to Venkov's argument.
What cases for ${\mathbf Q}(\sqrt{-p})$ are covered by Venkov? When $p \equiv 3 \bmod 4$, the discriminant of ${\mathbf Q}(\sqrt{-p})$ is $-p$. If $p \equiv 3 \bmod 8$ then $-p \equiv 5 \bmod 8$, while if $p \equiv 7 \bmod 8$ then $-p \equiv 1 \bmod 8$, so Venkov had non-analytically proved the formula when $p \equiv 3 \bmod 8$.  The case $p \equiv 7 \bmod 8$ remained open. 
In 1978 the whole problem was solved.  Davis, in a second paper (Crelle 299/300 (1978), 247--255), handled some but not all cases of imaginary quadratic fields with discriminant $1 \bmod 8$ (corresponding to $p \equiv 7 \bmod 8$ for the fields ${\mathbf Q}(\sqrt{-p})$) by non-analytic methods and in the same year H. L. S. Orde settled everything by non-analytic methods.  See his paper "On Dirichlet's class number formula", J. London Math. Soc. 18 (1978), 409--420.<|endoftext|>
TITLE: Random Walks and Lyapunov exponents
QUESTION [10 upvotes]: Given a sequence $Y_1, Y_2, \dots$ of i.i.d. matrices in $GL_n(\mathbb R)$, there is a theorem of Furstenberg and Kesten which says that if $\mathbb E(\log||Y_1||)$ is finite, there exists a constant $\gamma$ (the Lyapunov exponent) such that
$$\lim_{n\rightarrow\infty}\frac{1}{n}\log||Y_n\dots Y_1|| = \gamma$$
There are also versions of central limit theorems for this scenario.  I'm pretty sure this is also known in a more general case (e.g. suppose we have a sequence of matrices $Y_i$ of order 2, and I don't want to consider sequences of length $n$ in which $\dots Y_i Y_i\dots$ appears).    I am wondering if anyone knows a good reference for theorems regarding Lyapunov exponents and central theorems in this case.

REPLY [2 votes]: I believe the paper
  Almost all cocycles over any hyperbolic system have non-vanishing Lyapunov exponents
  Annals of Math. 167 (2008), 643-680.

available for free at http://www.preprint.impa.br/Shadows/SERIE_A/2005/384.html contains positivity results for the Lyapunov exponent over hyperbolic dynamical systems.
Also the paper by C. Bonatti, X. Gomez-Mont, and M. Viana cited as [7] in there should be of interest.
I am not sure if they treat central limit theorems in these works.<|endoftext|>
TITLE: Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components?
QUESTION [6 upvotes]: Do you know how to construct a compact hyperbolic 3-manifold with three or four totally geodesic boundary components?  The only constructions I could find have one boundary component.  A reference would be appreciated.

REPLY [9 votes]: A beautiful construction (relying on 2-dimensional hyperbolic circle packings) of hyperbolic 3-manifolds with many totally geodesic boundary components can be found in the paper
Totally geodesic boundaries are dense in the moduli space.
J. Math. Soc. Japan 49 (1997), no. 3, 589--601.
by Fujii and Soma.<|endoftext|>
TITLE: How to compute the (co)homology of a compact Riemann surface?
QUESTION [5 upvotes]: The situation is the following. 
A finite-index subgroup $\Gamma$ of $SL_2(\mathbb Z)$ acts on the upper-half plane $\mathcal H$. It has a fundamental domain, obtained by a union of translates of the fundamental domain for $SL_2(\mathbb Z)$. The quotient $\mathcal H/\Gamma$ can be compactified by adding a finite number of cusps(I have checked that this can indeed be done, in some book). We call it $X$. 
Now I want to compute the de Rham cohomology or singular (co)homology of $X$. I am unable to do it in the general case. Any hints on how to proceed would be appreciated.
The difficulty I am facing is that I am given a group to work with, and the standard examples of computations are with simple spaces through Mayer-Vietories. I do not a priori have a nice Mayer-Vietories decomposition of the space. Or perhaps the best method is not through Mayer-Vietories? 
More generally, if $\Gamma$ is a discrete subgroup of $SL_2(\mathbb R)$, firstly, 1) How would one construct a fundamental domain? and, 2) How would one compute the homology?
Re to Sam Nead: I had only the computation of $H_1$ or $H^1$ in mind.

REPLY [2 votes]: It is hard to answer the question without actually knowing what the group is. Here are some remarks:

For classical congruence subgroups of $PSL_2(\mathbf{Z})$ there is a formula for the number of cusps and the genus. It can be found in many books on modular forms and related things. If memory serves, it is given in Diamond-Shurman (and probably in Shimura's book too).
For a general $\Gamma$ of finite index or not the quotient $H/\Gamma$ is homotopy equivalent to the graph $X_{comb}$ constructed as follows. Let $\Gamma\setminus PSL_2(\mathbf{Z})$ be the set of the right cosets; there is a natural right action of $PSL_2(\mathbf{Z})$ on it. The set of the vertices of $X_{comb}$ is the disjoint union of the sets of orbits of the standard elements of order 2 and 3 (tried to put the matrices here, but these won't show properly). Two vertices of $X_{comb}$ are joined with an edge iff  the corresponding orbits intersect (and there are as many edges joining the vertices as there are elements in the intersection). This gives the cohomology and more.<|endoftext|>
TITLE: Are there applications of algebraic geometry into algebraic topology?
QUESTION [13 upvotes]: It is described in many sources that algebraic topology had been a major source of innovation for algebraic geometry. It is said that the uses of cohomology, sheaves, spectral sequences etc. in algebraic geometry were motivated by algebraic topology. Moreover it is said that Weil conjectures arose out of inspiration from algebraic topology.
So it seems a very clear thing that algebraic topology tremendously influenced algebraic geometry, at least historically.
But are there influences in the other way? Did it ever happen that the modern developments in algebraic geometry were ever taken back to algebraic topology and led to developments over there?
Edit: Topological K-theory is one application. Are there more?

REPLY [6 votes]: A nice conceptual link between derived algebraic geometry and algebraic topology is established in
Bertrand Toen, Champs affines ,
http://arxiv.org/abs/math/0012219
An impression of the central conceptual picture developed there is given at
nLab: rational homotopy theory in an (oo,1)-topos.
The main point is, in slight paraphrase of what Toen writes, that he shows that the relation between dg-algebras and rational homotopy theory factors through derived algebraic stacks. This allows to closely connect algebraic topology with derived algebraic geometry and improve on ordinary rational homotopy theory by admitting algebraic models for spaces with arbitrary fundamental groups.
This is supposed to go back to and provide a solution to Grothendiecks "schematization problem" in terms of "schematic homotopy types".<|endoftext|>
TITLE: Introduction to the Podles Sphere
QUESTION [6 upvotes]: I am just looking for a basic introduction to the Podles sphere and its topology.  All I know is that it's a q-deformation of $S^2$.

REPLY [4 votes]: First of all some terminology. One usually talks about Podles spheres, since they are a one parameter family. If you say the Podles sphere you probably mean the one that is often referred to as the standard one. My indications will refer to the whole family.
You do not clarify your background and your directions. The family of Podles spheres has been used as an example in all possible meaning of the word quantization, I guess you could easily find a list of hundred of references about it. I'll stick to some paper that can be considered as foundational in various aspects.
First: as NC C*-algebra the first reading should be Podles original paper, which is simple and well written P.Podles Quantum spheres, Lett.Math.Phys. 14, 193-202 (1987).
From the point of view of quantum homogeneous spaces, i.e. -coideal subalgebras in Hopf--algebras I strongly suggest M.Dijkhuizen and T. Koornwinder Quantum homogeneous spaces, duality and quantum 2-spheres, Geom.Dedicata 52, 291-315 (1994).
In both these approaches, more or less evidently, q-special functions pop up at some point.
M. Noumi and Mimachi K., Quantum 2-spheres and big q-Jacobi polynomials, Comm. Math. Phys. 128, 521-531 (1990) is the one not to avoid.
Last but not least you may be interested in understanding Podles spheres as deformations of Poisson structures (say à la Rieffel), which is beautifully explained in A.Sheu, Quantization of the Poisson SU(2) and its Poisson homogeneous space - the 2 sphere, Comm. Math. Phys. 135, 217-232 (1991). I suggest that here you first read the appendix by Lu and Weinstein where the Poisson structures are explained neatly and simply, and then go the quantization part (some of which may result rather technical at first).
Then, of course, as mathphysicist mentioned, the whole issue of putting spectral triples opens up; the literature there is much more scattered (several attempts and several choices as well) and I guess one should just simply dive into open sea and see what happens...
ADDED:
Personally I would start with the paper by Dijkhuizen and Koornwinder that settles the algebraic part (generators and relations) and has a down-to-earth approach, without technicalities (if you know a little about Hopf algebras). I would not dismiss the paper by Noumi and Mimachi if you're interested in spectral triples. q-special functions means harmonic analysis on the sphere: it shouldn't surprise it is important if you look for Dirac operators satisfying some invariance condition.<|endoftext|>
TITLE: References regarding a connection between recursion theory and sheaves
QUESTION [14 upvotes]: In Manin's A Course in Mathematical Logic for Mathematicians, he defines (p.201) a structure $(\mathcal{E},R)$ given an enumerable set $E \subset (\mathbb{Z}^+)^n$ by:

$\mathcal{E}$ is the set of all enumerable subsets of $E$.
For each $E' \in \mathcal{E}$, $R(E')=\{f|\text {domain}(f)=E', f:E'\rightarrow (\mathbb{Z}^+) \text{ is recursive}\}.$

He then demonstrates (cumulating on p. 205-6) that there is an analogy between $(\mathcal{E},R)$ and (his quotes) "a topological space together with a sheaf", and a way to define "recursive Cech cohomology of groups" of some complexes that arise from $(\mathcal{E},R)$. He then claims that "it would be interesting to study such cohomology groups".
My question: is this a well known construction/analogy? Has it been studied further? Are there any relevant references?

REPLY [2 votes]: Here is a somewhat positive answer in terms of
germs and equivalence modulo finite differences.
There has been a good deal of impressive lattice-theoretical work on the lattice of recursively enumerable sets. (This topic is a subtopic of recursively enumerable sets and degrees (AMS classification 03D25); these are also known as r.e. sets, computably enumerable sets, and c.e. sets.)
This work in particular has dealt with the lattice of c.e. subsets of a given c.e. set $E$, $\mathcal L(E)$, which has different properties depending on properties of $E$ such as hypersimplicity and creativity.
About half of the work concerns the quotient structures modulo finite differences. So in a sense they are studying germs and there is some analogy with what you are asking for. That is, one studies $\mathcal E$, the lattice of r.e. sets, and $\mathcal E^*$, the lattice of equivalence classes of elements of $\mathcal E$, where $A$ and $B$ are equivalent if
$$
\{n : A(n)\ne B(n)\}
$$
is a finite subset of $\omega=\mathbb N$.
The focus is on the automorphism group, its orbits, and first order definability.
See for instance

Cholak, Peter A.; Downey, Rodney; Harrington, Leo A. On the orbits of computably enumerable sets. J. Amer. Math. Soc. 21 (2008), no. 4, 1105–1135.<|endoftext|>
TITLE: Dual Spaces of Sobolev Spaces
QUESTION [5 upvotes]: I will consider Sobolev spaces with $p=2$, only, so that they are Hilbert spaces. Hence the Sobolev inner product identifies each Sobolev space with its dual.  In other words, I have an isomorphism $H^m\to (H^m)^\ast$ given by $x\mapsto \langle x,\cdot\rangle_m$
Now, if $\sigma>0$, then I have an embedding $H^{m+\sigma}\hookrightarrow H^m$.  Under the above isomorphism, how can I describe the image of $H^{m+\sigma}$ inside $(H^m)^\ast$?  In particular, is there a $\tau>0$ such that $H^{m+\sigma}$ it identified with $(H^{m-\tau})^\ast$?

REPLY [12 votes]: I am going to stick with the standard terminology $H^m$ here. Taking Fourier transforms one finds that $$\langle u,v\rangle_m=\int\hat u(\xi)\bar{\hat v}(\xi)(1+|\xi|^2)^m\,d\xi$$ (give or take the odd multiplicative constant), where $H^m$ consists precisely of those $u\in L^2$ for which $\langle u,u\rangle<\infty$. This works even for $m<0$, if you allow distributions whose Fourier transforms are functions. Everything follows from this, including the fact that $H^{-m}$ acts as the dual of $H^m$ simply by the distribution $u$ acting on the function $v$, which corresponds to the integral $$\langle u,v\rangle=\int \hat u(\xi)\bar{\hat v}(\xi)\,d\xi=\int \hat u(\xi)(1+|\xi|^2)^{-m/2}\cdot\hat{\bar v}(\xi)(1+|\xi|^2)^{m/2}\,d\xi$$ where I have split up the integrand into a product of two $L^2$ functions.
For this reason, it seems more natural to identify $H^{-m}$ with the dual of $H^m$ than to identify $H^m$ with its own dual. However, you can go ahead and identify any Sobolev space with the dual of any other just by inserting a suitable power of $1+|\xi|^2$ in the integral defining the pairing between the two.
Rather than coming straight out and answering your question, I'll leave it to you to ponder the consequences of the above. In particular, note that you we embed and identify you have to keep careful track of what space you have identified with whose dual, or you will be endlessly befuddled.
Addendum: To spell out a more direct answer to your question, $\langle\cdot,\cdot\rangle_m$ can identify $H^{m+\sigma}$ with the dual of $H^{m-\sigma}$, since we can write $$\langle u,v\rangle_m=\int \hat u(\xi)(1+|\xi|^2)^{(m-\sigma)/2}\cdot\bar{\hat v}(\xi)(1+|\xi|^2)^{(m+\sigma)/2}\,d\xi$$ where I have split the integrand into a product of two $L^2$ functions.
Edit: Changed a couple $\hat{\bar v}$ into $\bar{\hat v}$.<|endoftext|>
TITLE: Groups with all subgroups normal
QUESTION [28 upvotes]: Is there any sort of classification of (say finite) groups with the property that every subgroup is normal?
Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary.
If there isn't a classification, can we at least say the group must be of prime power order, or even a power of two?

REPLY [4 votes]: If we replace “(say finite)” by “(say connected Lie)”, then the fact that such groups are abelian may be first observed in:
Ahrens, W., Über Transformationsgruppen, deren sämtliche Untergruppen invariant sind, Hamb. Mitt. 4, 72-78 (1902). ZBL33.0162.02.<|endoftext|>
TITLE: Number of spanning trees of a quotient graph
QUESTION [11 upvotes]: Let $G$ be a finite connected graph on a $2m$-element vertex set $V$. For any graph with vertices $u,v$, let $\mu(u,v)$ denote the number of edges between $u$ and $v$. Suppose that $G$ has an automorphism $f$ that is a fixed-point free involution on the vertices. We can define a quotient graph $G/f$ by letting the vertices of $G/f$ be the orbits $[u]=\left\lbrace u,f(u)\right\rbrace$ of $f$, and setting
 $$ \mu([u],[v]) = \mu(u,v) + \mu(f(u),v). $$
Let $\kappa(H)$ denote the number of spanning trees of a graph $H$. By the Matrix-Tree Theorem and some simple linear algebra, one can show that $2\kappa(G)$ is divisible by $\kappa(G/f)$. My question is whether the factor of 2 is necessary, i.e., is it always true that $\kappa(G)$ is divisible by $\kappa(G/f)$? Similar questions can be asked for more complicated automorphism groups of $G$.

REPLY [8 votes]: Let us group the vertices as $U=\{u_1,u_2,\dots,u_n\}$ and $V=\{v_1,v_2,\dots, v_n\}$ where $f(u_i)=v_i$. Let $L_0$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, let $L _1 = \operatorname{diag}\left( \sum _{j=1}^n \mu(u _i, v _j)\right)$ and $L=L _0+L _1$,
also let $A$ be the symmetric matrix whose $a _{ij}$ entry is $-\mu(u _i,v _j)$. Clearly the Laplacian of $G$ is
$M=\left( \begin {array} {cc} L & A \\\ 
A & L \end {array} \right)$. Let $M^*$ stand for the matrix $M$ with deleted first row and column. We have
$$\kappa(G)=\det \left( \begin {array} {cc} L & A \\\ 
A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\\ 
D & (L+A)^ * \end {array} \right)$$
for some block matrices $B,C,D$ of size $n\times n,n\times (n-1),$ and $(n-1)\times n$ where this second matrix was obtained by adding the $i$th row of $M^{*}$ to it's $n+i$th row for $1\le i\le n-1$ and then adding the first (or last) $n-1$ columns to the $n$th column. So $D$ is the matrix $(L+A)^*$ together with a last column of zeros, making $\left((L+A)^*\right)^{-1}D$ with integer entries.
Next we factor it using one of these identities
$$\kappa(G)=\det(L+A)^ * \det(B-C\left((L+A)^*\right)^{-1}D)$$
and observe that $L+A$ is the Laplacian of $G/f$ so $\det(L+A)^ *=\kappa(G/f)$, and since the second factor is an integer we get the desired divisibility.<|endoftext|>
TITLE: Finitely many arithmetic progressions
QUESTION [25 upvotes]: A few years ago, somebody told me a lovely problem. I suspect there may be more to it (which I would be interested in learning), and would very much like to find a reference, it makes me uncomfortable to use it in class without being able to point to its source. 
The problem is as follows. I'll post the solution I know, which is the reason I like it, as an answer, to give a bit of a chance to people who read it and want to think about it without being spoiled.
Assume the natural numbers are partitioned into finitely many arithmetic progressions. Then two of these progressions must have the same common difference.

REPLY [10 votes]: Chapter one of the Mathematical Coloring Book  discuss  this problem. There you will
 found some of its history. Apparently it was conjecture by Erdös in 1950 and  proved (but not published) a few months later by Donald Newman and Leon Misrky.<|endoftext|>
TITLE: Picard Groups of Moduli Problems
QUESTION [14 upvotes]: First, yes, I've seen Mumford's paper of this title.  I'm actually interested in specific ones, and looking for really the most elementary/elegant proof possible.
I'm told that for $g\geq 2$ it is known that the Picard groups of $\mathcal{M}_g$ and $\mathcal{A}_g$ (the moduli spaces of curves of genus $g$ and abelian varieties of dimension $g$) are both isomorphic to $\mathbb{Z}$ (at least, over $\mathbb{C}$).  What's the most efficient way to compute this? In fact, for $\mathcal{M}_g$, it's even generated by the Hodge bundle, I'm told.  Ideally I want to avoid using stacks (though if stacks give an elegant proof, I'm open to them) and also would like to be able to calculate the degrees of some natural bundles, though I get that that's going to be a bit harder, so I want to focus this question on the computation of the Picard group.

REPLY [8 votes]: The fact that the Picard group of the moduli variety (not stack) A(g) is of rank 1, is sketched in a footnote of Mumford's paper on the Kodaira dimension of A(g), in LNM 997.  This footnote is elaborated (over Z) in a paper of Smith-Varley: in LNM 1124.  Another reference is Freitag's paper in Arch. Math. 40 (1983), pp.255-259.  The whole point of Mumford's argument was that it follows from Borel's computation of the rank of the second cohomology group of the symplectic group (i.e. rank one).<|endoftext|>
TITLE: Lifting varieties to characteristic zero.
QUESTION [16 upvotes]: If you want to compute crystalline cohomology of a smooth proper variety $X$ over a perfect field $k$ of characteristic $p$, the first thing you might want to try is to lift $X$ to the Witt ring $W_k$ of $k$. If that succeeds, compute de Rham cohomology of the lift over $W_k$ instead, which in general will be much easier to do. Neglecting torsion, this de Rham cohomology is the same as the crystalline cohomology of $X$.
I would like to have an example at hand where this approach fails: Can you give an example for

A smooth proper variety $X$ over the finite field with $p$ elements, such that there is no smooth proper scheme of finite type over $\mathbb Z_p$ whose special fibre is $X$.

The reason why such examples have to exist is metamathematical: If there werent any, the pain one undergoes constructing crystalline cohomology would be unnecessary.

REPLY [3 votes]: In general the obstruction to lift a scheme $X$ in characteristic zero is in $H^2(X,T_X)$. For examples of $3$-folds in positive characteristic that connot be lifted in characteristic zero you may look at Theorem 22.4 in Hartshorne's "Deformation Theory".<|endoftext|>
TITLE: uncountable algebraically closed field other than C ?
QUESTION [9 upvotes]: Is there any "well-known" algebraically closed field that is uncountable other than $\mathbb{C}$ ?
The algebraic closure of $\mathbb{C}(X)$ would work, but is it meaningful, is this field used in some topics ?
Have you other examples ?
Thank you.

REPLY [23 votes]: The algebraic closure of $\mathbb{F}_p((t))$ is uncountable of characteristic $p$.  It comes up naturally in number theory and algebraic geometry.
For every characteristic $p \geq 0$ and uncountable cardinal $\kappa$, there is up to isomorphism exactly one algebraically closed field of characteristic $p$ and cardinality $\kappa$.  The examples of $\mathbb{C}$ and closures of Laurent series fields as above give you the ones of continuum cardinality and all characteristics.  Indeed I do not know any specific reason to consider algebraically closed fields of larger than continuum cardinality.

REPLY [7 votes]: The algebraic closure of the p-adic field $Q_p$ is also of interest. One may even want to consider the completion (with respect to the p-adic absolute value) of this algebraic closure.
The resulting field is both complete and algebraically closed. It is denoted by $C_p$, and is considered as an p-adic analog of $C$.<|endoftext|>
TITLE: For which rings does a projectivization of modules exist?
QUESTION [6 upvotes]: Let $R$ be a ring. Consider the inclusion functor from the category of finitely generated, projective $R$-modules to the category of all finitely generated $R$-modules. For which rings does it have a left adjoint.
For example for any principal ideal domain $R$, we have the structure theorem of f.g. $R$-modules. The upper left adjoint is given by $M\mapsto M/torsion$.
In general there is no reason, why such a construction should exist. I tried to check the assumptions  for Freyds adjoint functor theorem. But it seems quite hard to check whether, they are satisfied for a given ring $R$.
So does anyone know a counterexample, where such a adjoint doesn't exist or a weaker property than PID, that implies the existence ?

REPLY [6 votes]: A necessary condition is that, for any finitely generated R-module M, the Hom-object $Hom_R(M,R)$ is a finitely generated projective right R-module.  This is because if a left adjoint existed, this would be isomorphic to the set $Hom_R("Proj(M)",R)$, and this is a summand of a free right module.
Consider the case $R = \mathbb{Z}/p^2$, $M = \mathbb{Z}/p$, $P = R$.  Then $$Hom_{\mathbb{Z}/p^2}(\mathbb{Z}/p, \mathbb{Z}/p^2) \cong \mathbb{Z}/p$$
and so this ring admits no projectivization.
Conversely, if this condition is satisfied then we can define $D_R M = Hom_R(M,R)$, and $DM$ is always finitely generated projective with a natural double-duality map $M \to D_{R^{op}} D_R M$ which is an isomorphism for projective modules.  This provides the desired adjoint, because any map $M \to P$ produces a natural factorization
$$M \to DDM \to DDP \leftarrow^{\sim} P$$
which is the desired adjoint.<|endoftext|>
TITLE: Constructing coherent sheaves on Shimura varieties.
QUESTION [15 upvotes]: Let me first run through the setting of my question in an example I understand well; that of modular curves. If $Y_1(N)$ denotes the usual modular curve over the complexes, the quotient of the upper half plane by the congruence subgroup $\Gamma=\Gamma_1(N)$, then there are two kinds of sheaves that one often sees showing up in the theory of automorphic forms in this setting:
1) Locally constant sheaves. The ones showing up typically come from representations of $\Gamma$ of the form $Symm^{k-2}(\mathbf{C}^2)$, with $\Gamma$ acting in the obvious way on $\mathbf{C}^2$. These sheaves---call them $V_k$---are related to classical modular forms of weight $k$ via the Eichler-Shimura correspondence. They only exist for $k\geq2$ (weight 1 forms are not cohomological) and representation-theoreticically the sheaves are associated to representations of the algebraic group $SL(2)$ (the reason one starts in weight 2 rather than weight 0 is that there is a correction factor of "half the sum of the positive roots").
2) Coherent sheaves. The ones showing up here are powers $\omega^k$ of a canonical line bundle $\omega$ coming from the universal elliptic curve. The global sections of $\omega^k$ (which are bounded at the cusps) are classical modular forms of weight $k$. Although there are no classical modular forms of negative weight, the sheaf $\omega^k$ still makes sense for $k<0$ (in contrast to case 1 above). I am much vaguer about what is conceptually going on here. I have it in my mind that here $k$ is somehow a representation of the group $SO(2,\mathbf{R})$.
Now my question: what is the generalisation of this to arbitrary, say, PEL Shimura varieties? Part (1) I understand: I can consider algebraic representations of the reductive group I'm working with and for each such gadget I can make a locally constant sheaf. But Part (2) I understand less. I am guessing I can construct a big vector bundle on my moduli space coming from the abelian variety. Now, given some representation of some group or other, I can build coherent sheaves somehow, possibly by "changing the structure group" somehow. For which representations of which group does this give me a coherent sheaf on the moduli space?
Basically---what is the general yoga for supplying natural coherent sheaves on Shimura varieties, which specialises to the construction of $\omega^k$ in the modular curve case, and which explains why $\omega^k$ exists even for $k<0$?

REPLY [11 votes]: As others have pointed out, the word you are looking for is automorphic vector bundle. The holomorphic automorphic forms are exactly the sections of these bundles. To define them, you start with a homogeneous bundle on the (compact) dual hermitian symmetric space. Initially, the construction of the automorphic vector bundle attached to the homogeneous bundle is analytic, but it can be made algebraic by the introduction of the standard principal bundle. It is known that the standard principle bundle has a canonical model over the reflex field, so this all works arithmetically. [The definition and proof of the existence of a canonical model of the standard principal bundle is in my 1988 Inventiones paper, after earlier work of Shimura, Harris, and others; there is a discussion of such things in Chapter III of my article in the Proceedings of the 1988 Ann Arbor conference, which are available on my website.]<|endoftext|>
TITLE: Convergence of L-series
QUESTION [13 upvotes]: I remember to have read that the L-function of an elliptic curve, which a priori only converges for $\Re s > \frac{3}{2}$ also converges at $s=1$ provided that the $L$-function
satisfies the functional equation.
I always thought that this is due to the fact that in this case the L-function is also the L-function of a modular form and in this case we have better convergence. However, the modular forms which correspond to these curves are cusp forms of weight 2 and so have a priori even worse convergence properties, namely convergence for $\Re s > 2$.
So I now wonder, whether I remember correctly and the the claim above is indeed correct?
If so, I would like to see a reference for the proof.
What is the reason for this fact. Does some non-trivial fact about elliptic curves play a role. It will (almost surely) not hold for arbitrary modular forms or cusp forms, since they always satisfy the functional equation. What do we need? An Euler product?
EDIT: The $L$-function extends to an entire function. But what I am interested in is the original series representation. Is it true that the series representation for the $L$-function of an elliptic curve which converges a priori for $\Re s > \frac{3}{2}$, is valid also for a bigger strip  $\Re s > c$ with $c<\frac{3}{2}$.

REPLY [23 votes]: Wood, see the article by Kumar Murty in Seminar on Fermat's Last Theorem.  He shows how the L-series converges (conditionally!) on Re(s) > 5/6, thus in particular at s = 1.  You can find the book on Google books and do a search on "5/6" to find the page.  OK, I just did that and will tell you: it's on page 15.  He proves the theorem for any Dirichlet series converging abs. for Re(s) > 3/2 and having an analytic continuation and suitable functional equation relating values at s and 2-s.  (Thus in practice it is a theorem about L-functions of suitable weight 2 modular forms, certainly nothing directly about elliptic curves!) He also says that what he describes is a special case of a more general result, with citation.
Where the Dirichlet series converges, it still represents the L-function that may have been analytically continued to the wider region by some other method, since Dirichlet series are analytic on the half-plane to the right of any point where they converge and moreover at any point where they converge the value is the limit of the function taken along the line to the right of the point (Abel's theorem for power series on discs works for Dirichlet series on right half-planes).  So we are assured that if you happen to know the series itself converges somewhere new it still equals the orthodox analytic continuation (whatever that means). 
On the other hand, the Euler product has surprises.  Goldfeld discovered that if the Euler product for an ell. curve over Q converges at s = 1, in the natural sense of partial Euler products over primes up to x as x goes to $\infty$, and the value of the Euler product is nonzero, then this value is not L(1) but rather is off from this by a factor of $\sqrt{2}$.  Of course there was no real input about elliptic curves directly: Goldfeld was assuming the ell. curve was modular (he was writing in the 1980s) and used that right away. 
It turns out exactly the same thing happens for quadratic Dirichlet L-functions at s = 1/2: if the partial Euler product at s = 1/2 converges to a nonzero value then again you're off by a factor of $\sqrt{2}$ from L(1/2), but in one setting the factor is $\sqrt{2}$ and in the other it's $1/\sqrt{2}$.  For non-quadratic Dirichlet L-functions there's no funny business: if the Euler product converges at s = 1/2 to a nonzero value (which, by the way, it always should since nobody expects Dirichlet L-functions vanish at 1/2) then the value will be L(1/2).  I first heard about Goldfeld's result at a talk by Karl Rubin (he gave the usual heuristic for BSD conjecture by looking at the Euler product at s = 1 and some wiseguy in the audience asked if it really did converge at s = 1 and Rubin mentioned there was a paper of Goldfeld on that), but when I read Goldfeld's paper I was confused by part of it, so in trying to work it out in the simpler example of Dirichlet L-functions I wound up seeing I could prove the same kind of theorem for any Euler product over a global field having the properties everyone expects it should have. This turns out to be related to properties of symmetric and exterior square Euler products and is morally the same quadratic bias (called Chebyshev's bias) that Sarnak and Rubinstein found when they worked out comparative statistics on the number of primes up to x in different congruence class mod $m$: classes of squares or non-squares exhibit different fine growth rates compared to one another.  For more details on the partial Euler products, including some numerical examples, see my paper http://www.math.uconn.edu/~kconrad/articles/eulerprod.pdf.  
By the way, one can definitely observe this $\sqrt{2}$ business happening numerically but we'll never expect to prove it happens since it actually implies the Riemann hypothesis for the relevant L-function.  If the product converges to a nonzero value at a point on the critical line it is no real surprise that you could prove the Dirichlet series for the log of the Euler product converges everywhere to the right of the critical line, which implies the Euler product itself converges to a nonzero value everywhere to the right of the critical line, hence Riemann hypothesis. In fact, as I show in that paper, this (suitable) Euler product convergence at a point on the critical line is actually equivalent to something which at present appears to lies deeper than the Riemann hypothesis but is still plausible. There's no lack of results which imply RH but are themselves false, you see.  This probably isn't one of them.
In summary, the Dirichlet series for ell. curve L-functions (over $\mathbf Q$) can provably be shown to converge on a wider region than they are usually said to converge and still equal the L-function there, including s = 1, while the Euler product probably does converge at s = 1 too but if the value is not 0 then it's not going to converge to what you expect... unless you think the Riemann hypothesis is false.<|endoftext|>
TITLE: Au revoir, law of excluded middle?
QUESTION [13 upvotes]: In what way and with what utility is the law of excluded middle usually disposed of in intuitionistic type theory and its descendants? I am thinking here of topos theory and its ilk, namely synthetic differential geometry and the use of topoi in algebraic geometry (this is a more palatable restructuring, perhaps), where free use of these "¬⊨P∨¬P" theories is necessarily everywhere--freely utilized at every turn, one might say. But why and how are such theories first formulated, and what do they look like in the purely logical sense?
You will have to forgive me; I began as a student in philosophy (not even that of mathematics), and the law of excluded middle is something that was imbibed with my mother's milk, as it were. This is more of a philosophical issue than a mathematical one, but being the renaissance guys/gals that you all are, I thought that perhaps this could generate some fruitful discussion.

REPLY [2 votes]: A version of Russell's paradox provides a failure of "law of excluded middle". I don't know if this directly answers your question, but it illustrates the constraints on getting this "law" to work.
Define a "logical function" to be one that returns 'yes' or 'no' when given an input. "law of excluded middle" asserts that if the 'yes' inputs are specified then assigning 'no' to everything else defines a logical function. Consider  

$R[f]$:= If( $f$ is a logical function) then not($f[f]$), else 'no')

But $R$ cannot be a logical function, for if it were we would have $R[R]= not(R[R]$. The error must be that "($f$ is a logical function)" is not a logical function. If it is not 'true' then it is 'wrong' in the sense that the reasoning is invalid. Usually this is seen as invalidating "unlimited comprehension" (naive set theory). 
Most mathematics requires excluded-middle arguments, so we interpret this problem as showing inputs have to be restricted for these arguments to work. In a sense the job of set theory is to provide safe contexts for excluded-middle arguments. This also suggests why set theory is complicated: we cannot describe "safe contexts" using logical functions.<|endoftext|>
TITLE: Why are integrals over cycles called periods?
QUESTION [13 upvotes]: In the statement of de Rham theorem, a pairing is defined as follows:
$H_i(X, \mathbb R) \times H^i_{\mathrm{de Rham}}(X) \rightarrow \mathbb R$
It is given by
$\left( \left( \sum a_i \gamma_i \right) , \omega \right) \mapsto \sum a_i \int_{\gamma_i} \omega $.
Here $a_i$ takes real values, and $\gamma_i$ are integral homology cycles, and $\omega$ is an $i$-form.
The integral given above is called the period of the above integral, and the isomorphism given by this pairing is sometimes called the period isomorphism.
Question:

Why is the above integral of a closed form over a cycle called a period?

My peeve with this terminology of "period" is that it does not agree at all with anything else I know about this word. They are the following: 1) The period of a periodic function. 2) Periods as generalizations of algebraic numbers, as integrals of algebraic(rational) expressions over domains in Euclidean spaces defined by algebraic inequalities.
Indeed, this strange use of the word "period" is used even by Ahlfors in his book on complex analysis, for integrals of holomorphic(rather, meromorphic? depending on the domain ...) functions over loops.
I do not understand why on earth the word period appears in this setting of integrating on abstract manifolds. True, the cycles capture some underlying geometry of the space. But why is something a "period" when you integrate a form over a cycle? Why is the integral of a $1$-form over a line segment not a period?(Or is it also period, in some definition I am not aware of?)

REPLY [12 votes]: I must admit that this is the first time I heard the suggestion that
the word "period" came into complex analysis/algebraic geometry from
the period of the pendulum. The word was certainly used in astronomy
before the theory of the pendulum was known -- e.g. Kepler stated his
third law in 1618 as "The squares of the periodic times are to each
other as the cubes of the mean distances" -- and uniform circular motion
gives a "period" that was known before the period of the pendulum.
In any case, the concept of "period" is not a big deal until one
discovers functions with two (or more) periods, which first happened
when Gauss discovered this property of the lemniscate sine function in
1797. The lemniscatic sine, $sl(u)$,is the inverse function of the 
elliptic integral
$
u=\int^{x}_{0} \frac{dt}{\sqrt{1-t^4}},
$
and Gauss called its real period $2\varpi$, using the variant form of
the Greek letter pi. Presumably this was to stress its analogy with the
period $2\pi$ of the sine function.<|endoftext|>
TITLE: Chopping up Dynkin diagrams
QUESTION [6 upvotes]: Suppose I have a simple, simply connected (linear) algebraic group $\mathcal{G}$ over an algebraically-closed field $k$, which could have any characteristic. In fact, to keep things simple, let's imagine $\mathcal{G}$ is simply-laced. Let $\mathcal{B}$ be a choice of Borel, and $\mathcal{T}$ be a maximal torus for $\mathcal{B}$ (and hence for $\mathcal{G}$). Our $\mathcal{B}$ gives a choice of simple positive roots. I'll write $\mathbb{G}_M$ for the multiplicative group of $k$.
Now, suppose we look at the Dynkin diagram for $\mathcal{G}$, and imagine removing a single node from the diagram; for simplicity let's imagine this doesn't disconnect the diagram. Then I believe we should be able to find a subgroup $\mathbb{G}_M\times\mathcal{G}'$ in $\mathcal{G}$ and a Borel subgroup $\mathcal{B}'$ in $\mathcal{G}'$, such that the inclusion $\mathbb{G}_M\times\mathcal{B}'\hookrightarrow \mathcal{G}$ factors through $\mathcal{B}$; such that the Dynkin diagram for $\mathcal{G}'$ is just the Dynkin diagram for $\mathcal{G}$ with our chosen node removed; and indeed such that when we use the inclusion $\mathcal{G'}\hookrightarrow\mathcal{G}$ to map the simple roots of $\mathcal{G}'$ determined by $\mathcal{B'}$ into roots of $\mathcal{G}$, and hence map the Dynkin diagram of $\mathcal{G}'$ into the Dynkin diagram of $\mathcal{G}$, we get the obvious inclusion of 'the diagram with the node removed' into 'the original diagram'.
Moreover, if instead I did disconnect the diagram, I should get a similar picture, with a subgroup $\mathbb{G}_M\times\mathcal{G}_1\times\dots\times\mathcal{G}_k$ where the groups $\mathcal{G}_k$ have Dynkin diagrams corresponding to the components of the picture I get by removing the node from the original Dynkin diagram.
My question is: a) am I right that this kind of operation is all legal and above board, b) are there any caveats one should be aware of, and c) what are good references to appeal to to make this kind of thing rigorous.

REPLY [8 votes]: Brian's comment does what you want, and describes the almost direct product caveat.
A standard and excellent reference for all things of this nature is Demazure's "Sous-groupes Paraboliques des groupes réductifs," Exposè XXVI of SGA3.  I believe that Borel and Tits earlier work "Groupes Réductifs" has similar results covered, in the case of a base field instead of base scheme.  Maybe this would be better for your question.
There's no need for the assumptions simple, simply-connected, simply-laced, nor the assumption that you're working over an algebraically closed field.  The general setting is a reductive group $G$ over a base scheme $S$.
For a (connected) reductive group $G$ over an algebraically closed field, once you fix $T \subset B \subset G$ as you wish, there's a bijection between subsets of the vertex-set of the Dynkin diagram and parabolic subgroups $P$ of $G$ containing $B$ (so-called standard parabolics).  Each parabolic subgroup has a standard Levi decomposition $P = LU$, where $U$ is the unipotent radical of $P$, and $L$ is a reductive group containing $T$.
When $G$ is a simply-connected group, the group $L$ has simply-connected semisimple derived subgroup $L'$.  The Dynkin diagram of $L'$ is the subdiagram induced by subset of the vertex-set of the Dynkin diagram of $G$.  
Getting the isogeny class right -- e.g. identifying a precise surjective map from $G_m^r \times L' \rightarrow L$ with finite kernel -- is kind of a pain, in my experience.  Typically, I just "know the answer" for classical groups (e.g., Jim's example, in $SL_3$, one finds a Levi $L \cong GL_2$) and for certain exceptional groups.  Occasionally, I have to do the tedious work of determining the root datum (in the sense of Borel) of the Levi to figure it out (or more commonly looking through the literature to find someone else who's done it).  Oh how I wish that this were implemented in SAGE.<|endoftext|>
TITLE: duplicate detection problem
QUESTION [17 upvotes]: Restated from stackoverflow:
Given:

array a[1:N] consisting of integer elements in the range 1:N

Is there a way to detect whether the array is a permutation (no duplicates) or whether it has any duplicate elements, in O(N) steps and O(1) space without modifying the original array?
clarification: the array takes space of size N but is a given input, you are allowed a fixed amount of additional space to use.

The stackoverflow crowd has dithered around enough to make me think this is nontrivial. I did find a few papers on it citing a problem originally stated by Berlekamp and Buhler (see "The Duplicate Detection Problem", S. Kamal Abdali, 2003)

REPLY [5 votes]: This is still an open and interesting problem. The best deterministic algorithm that I know of takes $O(n \log n)$ time and $O(\log n)$ words of space by Munro, Fich and Poblete in Permuting in place. This paper doesn't explicitly mention the problem of detecting if there is a duplicate but the method they develop for permuting in place is directly applicable. It is still possible that there is a true linear time and $O(1)$ words of space solution (either randomised or deterministic).
If you simply increase the alphabet size from $n$ the situation changes drastically. Even if you change it to $2n$ the complexity of finding if there is a duplicate is unknown and in particular no near linear time solution is known for small space. The most obvious randomised approach is to hash the elements down to the range $[1,\dots,n]$.  You are then left with the problem of trying to distinguish real duplicates from ones created by hash collisions.  With full independence it seems you can most likely do this in something like $O(n^{3/2})$ time  but I am not sure if this has ever been formally analyzed in published work. However, we can't actually use a hash function with full independence without also using linear space so the problem as before is to show that a hash function family whose members can be represented in small space and which has the desired properties actually exists.
For even larger alphabets of size $n^2$ there is an existing lower bound for small space algorithms given in Time-space trade-off lower bounds for randomized computation of decision problems.  With space $O(\log n)$ bits (or $O(1)$ words) it simplifies to approximately $\Omega(n \sqrt{\log n/\log{\log{n}}})$. This means that no linear time solution is possible in this case.
COMMENT: This should be a comment to David Eppstein's answer but I don't have the points for that.  The function $h(x) = 2^x \bmod p$ with $p$ a prime with $O(\log n)$ bits is very interesting. Although it is clear that it takes $\Theta(\log n)$ time to evaluate the hash function once (by repeated squaring, assuming constant time operations on words),  is it obvious that it can't be done faster on average when evaluating at $n$ points by some clever method? Consider, for example, an array with the elements in increasing order. In this case it takes only $O(n)$ time to compute all the hash values.<|endoftext|>
TITLE: AC in group isomorphism between R and R^2
QUESTION [12 upvotes]: Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups.  In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of these two vector spaces of cardinalities $c$ and $c\times c = c$, so they are isomorphic as vector spaces over $\mathbb{Q}$.
Is there a way to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups which does not use AC?  Are there models of set theory in which these groups are not isomorphic?  I'm also curious whether there is a proof (perhaps using AC) which does not make use of this vector space machinery, though I'm fairly sure the above proof is the simplest.

REPLY [12 votes]: You cannot prove that $\mathbb{R}$ and $\mathbb{R}^{2}$ are isomorphic in $ZF$. To see this, note that the map $(x,y) \mapsto (x,0)$ is a nontrivial non-surjective additive endomorphism of $\mathbb{R}^{2}$. Assuming the existence of suitable large cardinals, every map $f: \mathbb{R} \to \mathbb{R}$ is measurable in $L(\mathbb{R})$ and it follows that the only additive endomorphisms of $\mathbb{R}$ in $L(\mathbb{R})$ are the maps $x \mapsto rx$ for some $r \in \mathbb{R}$. Thus $\mathbb{R} \not \cong \mathbb{R}^{2}$ in $L(\mathbb{R})$.
There is an interesting open problem related to your question of whether an isomorphism can be found which does not make use of the vector space machinery.
If we add a Ramsey ultrafilter $\mathcal{U}$ to $L(\mathbb{R})$, then we obtain an interesting model $L(\mathbb{R})[\mathcal{U}]$ which contains the ultrafilter $\mathcal{U}$ while still retaining some of the "nice" properties of $L(\mathbb{R})$. In particular, neither $\mathbb{R}$ nor $\mathbb{R}^{2}$ has a basis as a vector space over $\mathbb{Q}$ in $L(\mathbb{R})[\mathcal{U}]$ . However, it is unknown whether or not $\mathbb{R} \cong \mathbb{R}^{2}$ in $L(\mathbb{R})[\mathcal{U}]$. In other words, does an ultrafilter help in trying to construct such an isomorphism?

REPLY [6 votes]: Here's the way I like to do it:  
A group homomorphism with the property of Baire from one Polish group to another is automatically continuous.  Of course no bijection between your two groups is a homeomorphism.  Solovay provided a model (AC fails) in which every subset of a Polish space has the property of Baire, so in Solovay's model these two groups are not isomorphic.  
A more down-to-earth consequence of the result says: no map you can actually write down is a group isomorphism between these two groups.<|endoftext|>
TITLE: Cauchy-Crofton formula for curvature
QUESTION [7 upvotes]: The Cauhy-Crofton formula relates the length of a curve to an integral over lines:
$$
\text{length} (\gamma) = \frac14\iint n_\gamma(\varphi, p)\; d\varphi\; dp,
$$
where $\gamma$ is a curve and $n_\gamma(\varphi, p)$ is the number of times the line defined by the angle $\varphi$ and the distance to the origin $p$ intersects the curve.
This formula has had a major impact on image analysis in the last decade. It gives a method to discretize functionals used in e.g. image segmentation and solve them with graph methods.
My question is: is there a similar formula, but for curvature instead? I.e.
$$
\int_\gamma |\kappa(x)|^p dx = ?
$$
Now using lines would not be enough, bur perhaps circles?

REPLY [5 votes]: I doubt it. Let's first dissect your Cauchy-Crofton formula. Up to minor technical assumptions, what we have is that 
$$\mbox{length}(\gamma) = \int_{\mathbb{R}^2} \delta_\gamma(x) d^2x$$
where $\delta_\gamma(x)d^2x$ represents the measure concentrated on $\gamma$. Locally you can think of its as the pull back of the Dirac delta via the charaterization function of $\gamma$ as a submanifold. 
Next we do a change of coordinates. Observe that (not being too precise here) $\mathbb{R}^2 = TS^1 \times \mathbb{R} / S^1$ (basically you can rotate a standard coordinate system; in other words, the set of all lines on the plane can be identified with $TS^1$, and we take its Cartesian product with $\mathbb{R}$ to measure along each of the lines). So let $\pi$ be the canonical projection map from $TS^1\times \mathbb{R}$ to $\mathbb{R}^2$. Then your integral can be re-written as
$$ \mbox{length}(\gamma) = \frac{1}{|S^1|}\int_{TS^1\times \mathbb{R}} \pi^*\delta_\gamma(y) dy$$
where $\pi^*\delta_\gamma(y) dy$ is the pull back measure. Now you integrate out the fibers $\mathbb{R}$ first and by simple geometry you see that 
$$ \int_{TS^1\times\mathbb{R}}\pi^*\delta_\gamma(y) dy = \int_{TS^1} n(s) ds $$
where $n(s)$ is the number of times the line indexed by the coordinate $s$ intersects $\gamma$. So up to some normalization constants we recover the Cauchy-Crofton formula. 
Now, if you want to get an integral of the curvature $k$ along the curve, you can write it as
$$ \mbox{curvature integral} = \int_{\mathbb{R}^2} |k(x)|^p \delta_\gamma(x) d^2x$$
the procedure of lifting to the space of lines is no problem, so you can again get an integral over $TS^1\times \mathbb{R}$. The problem is that I can't see any obvious way of integrating out the additional factor of $\mathbb{R}$. In the Cauchy-Crofton case, each time the line hits your curve it picks up a unit bundle of mass. In the curvature case, you pick up some number which depends on the curvature at the point of intersection. 
Using circles won't help either. You can consider a foliation (or almost a foliation) of $\mathbb{R}^2$ by some family of curves, each of which can be identified with some curve $m$. Then you can do the same thing as above: let $M$ be the parameter space of this foliation (the foliation can be moved under rigid motion transformations to get a new foliation) then $\mathbb{R}^2$ can be identified as $M\times m / G$ by $G$ being some subgroup of the Euclidean symmetries. So formally your integral can be re-written as
$$ \frac{1}{|G|}\int_{M\times m} \mbox{something} dy $$
but to simplify down to just an integral over the parameter space $M$, you need to integrate out along the fibre $m$, and to get a simple expression at the end you almost certainly need that the family of curves $m$ in your foliation must be very special. Unless $m$ is adapted to $\gamma$ such that the integral along $m$ of $|k|^p\delta_\gamma$ can be easily evaluated, you have no hope of arriving at a simple integral expression.<|endoftext|>
TITLE: Maximal ideal in polynomial ring
QUESTION [9 upvotes]: Is it true that the intersection of a maximal ideal in $A[x]$ with $A$ is a maximal ideal in $A$?
Let's say A is Noetherian. I would be surprised if it isn't true but somehow I can't seem to show it. Any help or tip will be appreciated. Thanks!

REPLY [2 votes]: See also my sci.math post [1] for some further discussion and references
[1] sci.math, 22 Apr 2009
http://groups.google.com/group/sci.math/msg/b00f5e1f7397842f
http://google.com/groups?selm=y8zfxg12ckl.fsf%40nestle.csail.mit.edu<|endoftext|>
TITLE: Is a conceptual explanation possible for why the space of 1-forms on a manifold captures all its geometry?
QUESTION [16 upvotes]: Let $M$-be a differentiable manifold. Then, suppose to capture the underlying geometry we apply the singular homology theory. In the singular co-chain, there is geometry in every dimension. We look at the maps from simplexes, look at the cycles and go modulo the boundaries. This has a satisfying geometric feel, though I need to internalize it a bit more(which matter I tried to address in other questions).
Now on the other hand, let $\Omega^1$ be the space of $1$-forms on the space. The rest of the de Rham complex comes out of this object, wholly through algebraic processes, ie by taking the exterior powers and also the exterior derivative. After getting this object in hand, the journey upto getting the de Rham cohomology ring is entirely algebraic.
And by the de Rham theorem, this second method is equally as good as the more geometric first method. In the second method no geometry is explicitly involved anywhere in any terms after the first term. So the module of $1$-forms somehow magically capture all the geometry of the space $M$ without need of any explicit geometry. This is amazing from an algebraic point of view since we have less geometric stuff to understand.
This makes me wonder for the conceptual reason why this is true. I know that one should not look a gift horse in the mouth. But there is the need to understand why there is such a marked difference in the two approaches to capturing the geometry in a manifold, viz, through de Rham cohomology in differential topology, and through singular homology in algebraic topology. I would be grateful for any explanations why merely looking at all the $1$-forms is so informative.
Edited in response to comments: I meant, the de Rham cohomology is as good as singular homology for differentiable manifolds. What is "geometry of rational homotopy type"? And what is "geometry of real homotopy type"?

REPLY [17 votes]: We should distinguish between thinking locally and thinking globally. Locally, any differential $k$-form is built out of wedging together $1$ forms and taking linear cmobinations. Globally, a closed $k$-form need not be a wedge of closed $1$-forms. So the statement that everything is built out of $1$-forms is only a local statement, whereas cohomology is precisely about the difference between the local and the global. (Indeed, cohomology measures the extent to which a locally exact $k$-form need not be exact.) For this reason, I don't like your statement that the de-Rham complex is all built from $1$-forms.
I want to say that there should be a similar story in singular cohomology: that any closed $k$-co-chain should be locally a wedge of closed-$1$-co-chains. The picture I want to give is that any $k$-simplex "is" the wedge of the $k$ edges coming out of a single vertex, and so singular cohomology is locally built from $1$ dimensional objects to the same extent that deRham cohomology is. But I am hazy as to how to make this precise.<|endoftext|>
TITLE: Is the Green-Tao theorem true for primes within a given arithmetic progression?
QUESTION [57 upvotes]: Ben Green and Terrence Tao proved that there are arbitrary length arithmetic progressions among the primes.
Now, consider an arithmetic progression with starting term $a$ and common difference $d$. According to Dirichlet's theorem(suitably strengthened), the primes are "equally distributed" in each residue class modulo $d$. Therefore we imagine that the Green-Tao theorem should still be true if instead of primes we consider only those positive primes that are congruent to $a$ modulo $d$. That is, Green-Tao theorem is true for primes within a given arithmetic progression.
Question: Is something known about this stronger statement?

REPLY [133 votes]: The Green-Tao is true for any subset of the primes of positive relative density; the primes in a fixed arithmetic progression to modulus $d$ have relative density $1/\phi(d)$.<|endoftext|>
TITLE: Matsumura: "Commutative Algebra" versus "Commutative Ring Theory"
QUESTION [32 upvotes]: There are two books by Matsumura on commutative algebra.  The earlier one is called Commutative Algebra and is frequently cited in Hartshorne.  The more recent version is called Commutative Ring Theory and is still in print.  In the preface to the latter, Matsumura comments that he has replaced a section from a previous (Japanese?) edition because it "did not substantially differ from a section in the second edition of my previous book Commutative Algebra." This suggests that Matsumura considered the two books complementary, and certainly did not intend Commutative Ring Theory to replace Commutative Algebra.
Basically, my question is

Why are there two books, by the same author, apparently on the same subject?

A more practical question is whether  both books are equally appropriate as references when reading a book like Hartshorne.  However, I am curious to know the answer to the broader question as well.

REPLY [38 votes]: By comparing the tables of contents, the two books seem to contain almost the same material, with similar organization, with perhaps the omission of the chapter on excellent rings from the first, but the second book is considerably more user friendly for learners.  There are about the same number of pages but almost twice as many words per page.  The first book was almost like a set of class lecture notes from Professor Matsumura's 1967 course at Brandeis.  Compared to the second book, the first had few exercises, relatively few references, and a short index.  Chapters often began with definitions instead of a summary of results. Numerous definitions and basic ring theoretic concepts were taken for granted that are defined and discussed in the second.  E.g. the fact that a power series ring over a noetherian ring is also noetherian is stated in the first book and proved in the second.  The freeness of any projective modules over a local ring is stated in book one, proved in the finite case, and proved in general in book two.  Derived functors such as Ext and Tor are assumed in the first book, while there is an appendix reviewing them in the second.  Possibly the second book benefited from the input of the translator Miles Reid, at least Matsumura says so, and the difference in ease of reading between the two books is noticeable.  Some arguments in the second are changed and adapted from the well written book by Atiyah and Macdonald.   More than one of Matsumura's former students from his course at Brandeis which gave rise to the first book, including me, themselves prefer the second one. Thus, while experts may prefer book one, for many people who are reading Hartshorne, and are also learning commutative algebra, I would suggest the second book may be preferable.
Edit:  Note there are also two editions of the earlier book Commutative algebra, and apparently only the second edition (according to its preface) includes the appendix with Matsumura's theory of excellent rings.<|endoftext|>
TITLE: How do undergrad students write papers by themselves?
QUESTION [5 upvotes]: Can a student write a paper and send it to a professor review?

REPLY [3 votes]: Permit me to draw attention to a relatively new journal,
Involve—A Journal of Mathematics. Snippets from About the Journal:

"Involve is dedicated to showcasing and encouraging high quality mathematical research involving students (at all levels). ... Submissions in all mathematical areas are encouraged. However, each manuscript should include a minimum of 1/3 student authorship. ...
  Involve is a publication in between the extremes of purely undergraduate research journals, which in general are aimed at undergraduate audiences, and mainstream research journals."

I would guess that nearly all of the papers that have appeared in its first nine issues reflect work supervised by a faculty mentor, and a good percentage are coauthored with that faculty member (and not infrequently with more than one student).<|endoftext|>
TITLE: The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
QUESTION [13 upvotes]: I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with 
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$ A natural question to ask is whether such a formula exists for $(R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle$. An obvious guess would be to take the inverse of the matrix $[R_{js}^{ir}]_{i,r,j,s}$. 
That is, to guess that $$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$ 
is equal to $$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that
$$
\delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}.
$$
The matrix inverse is easy to calculate and gives us the formula
$$\langle R, u^i_j \otimes u^r_s \rangle  = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.
$$
Now let's try and test this result: As is very well known $(S \otimes $id)$R = R^{-1}$. Thus, $$\langle R^{-1}, u^i_j \otimes u^r_s \rangle = \langle R, S(u^i_j) \otimes u^r_s \rangle.$$ In the case of $N=2$, $i=j=r=s=1$, we have $S(u^1_1) = u^2_2$, and so,
$$
(R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,S(u^1_1) \otimes u^1_1 \rangle = \langle R,u^2_2 \otimes u^1_1 \rangle  = R^{21}_{21} = 1.
$$ 
But the formula above gives us that $(R^{-1})^{11}_{11}$ is equal to
$$ q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1},$$ 
Moreover, performing the analogous calculations for the other possible values of $i,j,r,s$ we get two different matrices: Using the general formula we get
$$
\left(
\begin{array} {cccc}
                q^{-1} & 0              & 0 & 0     \\\
                0      & 1              & 0 & 0     \\\
  0      & -\lambda             & 1 & 0  \\\
  0      & 0              & 0 & q^{-1} \\\
\end{array}
\right);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
                1  &  0  & 0 & -\lambda     \\\
                0  &  q  & 0 & 0           \\\
  0  &  0  & q & 0           \\\
  0  &  0  & 0 & 1           \\\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.
I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text

REPLY [9 votes]: I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U_q(sl_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).
First, I get $(R^{-1})^{21}_{12} = -q^{-1} R^{21}_{12}$,
because $\langle(S\otimes id)(R),a^2_1\otimes a^1_2\rangle = \langle R,S(a^2_1)\otimes a^1_2 \rangle = -q^{-1} \langle R,a^2_1\otimes a^1_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen.  So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.
Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.
I hope this helps!<|endoftext|>
TITLE: What is special about polylogarithms that leads to so many interesting identities and applications?
QUESTION [30 upvotes]: I have heard that Polylogarithms are very interesting things. The wikipedia page shows a lot of interesting identities. These functions are indeed supposed to have caught the attention of Ramanujan. Moreover, they seem to be important in physics for various purposes like Bose-Einstein integrals, which I am not really knowledgeable enough to understand. These are all things I have heard from people after I queried "why polylogarithms are interesting".
So this function
$$Li_s (z) = \sum_{k=1}^\infty \frac{z^k}{k^s}$$
is very interesting and has a lot of useful properties as I can see. Especially for integral values of $s$.
What is bothering me is the following.
Let $f(z) = \sum a_n z^n$ be analytic inside the unit disc. That is, the radius of convergence at least $1$. For simplicity, we assume it is $1$, i.e.,
$$r = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}} = 1$$
implying that inside the unit disc, term-by-term differentiation and integration are possible. Since $\left( a_n \right)^{1/n}$ goes to $1$ as $n$ goes to $\infty$, it is also true that $\left(na_n \right)^{1/n}$ goes to $1$. Similarly, $\left(\frac{a_n}{n} \right)^{1/n}$ also goes to $1$. Now, defining 
$$f(z,s) = \sum_{n=1}^\infty \frac{a_n z^n}{n^s}$$
it is the case that $f_{2}(z) = f(z,2)$ and in general iterating, $f_n(z) = f(z, n)$ are analytic inside the unit disc. In fact, uniform bounds hold true for compact sets contained in the domain  $|z| \leq 1$ and $|s| \leq K$ for arbitrary $K$, and so $f(z,s)$ is analytic in the domain $D^\circ \times \mathbb C$, where $z \in D^\circ$, the open unit disc, and $s \in \mathbb C$.
So any analytic function in the unit disc can be extended just like the logarithm is extended to polylogarithms. But there is a rich theory about polylogarithms, and I haven't heard of any such theories about other functions analytic inside the unit disc.
So, what makes polylogarithm so amenable to an extended theory of polylogarithms yielding so many results? Why is this not giving such an interesting theory for just any other complex functions?

REPLY [4 votes]: Polylogarithms have interesting connections in the Theory of Partitions.  This is mainly because a class of generating functions for bivariate partition statistics can be approximated by polylogarithms.  
Consider for example the class of generating functions with $s>1,$
$\displaystyle \sum_{n=0}^\infty \sum_{m=0}^\infty p(n,m)x^n u^m =\prod_{k=1}^\infty \frac{1}{(1-xu^k)^{k^{s-2}}}. $
When $s=2,$ $p(n,m)$ is the number of partitions of $m$ with $n$ parts.  When $s=3,$ $p(n,m)$ represents the number of plane partitions of $m$ with trace $n.$
For simplicity assume $1>x>0,$ as $u\to ^-1,$
$\displaystyle \prod_{k=1}^\infty \frac{1}{(1-xu^k)^{k^{s-2}}} \thicksim\sqrt[ - \frac{1}{\zeta(2-s)}]{1-x}\exp(\Gamma(s-1)Li_{s}(x)/(\ln u^{-1})^{s-1}). $
When used in conjunction with other techniques like the Circle Method, this estimate serves as a foundation for many facts about $p(n,m).$
Off the top of my head here are three papers that use this kind of estimate:
http://www.math.drexel.edu/~rboyer/papers/partitions_experimental.pdf
http://plms.oxfordjournals.org/content/s2-36/1/117.full.pdf
http://www.springerlink.com/content/d65348128235x7m1/<|endoftext|>
TITLE: Is every G-invariant function on a Lie algebra a trace?
QUESTION [17 upvotes]: I am in the (slow) process of editing my notes on Lie Groups and Quantum Groups (V Serganova, Math 261B, UC Berkeley, Spring 2010.  Mostly I can fill in gaps to arguments, but I have found myself completely stuck in one step of one proof.  One possibility that would get me unstuck is a positive answer to the following (which may be obviously false or trivial, but I'm not thinking well):

Question: Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb K$, and if necessary you may assume that $\mathbb K = \mathbb C$ and that $\mathfrak g$ is semisimple.  Then $\mathfrak g$ acts on itself by the adjoint action, and on polynomial functions $f : \mathfrak g \to \mathbb K$ via derivations.  A polynomial $f: \mathfrak g \to \mathbb K\,$ is $\mathfrak g$-invariant if $\mathfrak g \cdot f = 0$.  For example, let $\pi: \mathfrak g \to \mathfrak{gl}(V)$ be any finite-dimensional reprensentation.  Then $x \mapsto \operatorname{tr}_V \bigl(\pi(x)^n\bigr)$ is $\mathfrak g$-invariant for any $n\in \mathbb N$.  Is every $\mathfrak g$-invariant function of this form?  Or at least a sum of products of functions on this form?

When $\mathfrak g$ is one of the classical groups $\mathfrak{sl},\mathfrak{so},\mathfrak{sp}$, or the exceptional group $G_2$ the answer is yes, because we did those examples in the aforementioned class notes.  But I have no good grasp for the $E$ series, and I don't know if the statement holds for non-semisimples.
What I'm actually trying to prove is a weaker statement, but I figured I'd ask the stronger question, because to me the answer is not obviously "no".  The weaker statement:

Claim: Let $\mathfrak g$ be a finite-dimensional semisimple Lie algebra over $\mathbb C$.  Then every $\mathfrak g$-invariant function is constant on nilpotent elements of $\mathfrak g$.  (Recall that $x\in \mathfrak g$ is nilpotent if $\operatorname{ad}(x) = [x,] \in \mathfrak{gl}(\mathfrak g)$ is a nilpotent matrix — some power of it vanishes.)

It's clear that the spectrum of any nilpotent matrix is $\{0\}$, and for a semisimple Lie algebra, any nilpotent element acts nilpotently in all representations.  For the classical groups, in the notes we exhibited generators for the rings of $\mathfrak g$-invariant functions as traces of representations, and so we can just check the above claim.  But we did not do the $E$ series or $F_4$.

REPLY [5 votes]: If the representation is fixed as the fundamental representation, then in the case of $\mathfrak{so}(2n)$, you need Pfaffians as well as traces.<|endoftext|>
TITLE: Hamiltonian displaceability of tori in symplectic balls
QUESTION [12 upvotes]: Here is my first try at a question, which is a really easy to state question
about displaceability:
Let $D$ be the unit disk in the complex plane $D = \{ |z| \leq 1 \}$ equipped
with its standard symplectic form and for $r \in (0,1)$ let $S(r) \subset D$ be the Lagrangian circle of radius $r$ centered at $0$, enclosing area $\pi r^2$. 
Question:  for which $r_1,\ldots, r_n$ is the Lagrangian torus
$S(r_1) \times \ldots \times S(r_n)$ Hamiltonian displaceable in $D^n$?
Conjecture:  for a given $r_1,\ldots, r_n$, the Lagrangian is non-displaceable iff
$r_j^2 \ge 1/2$ for all $j$. 
Known: Using McDuff's probes, one can see that if some $r_j^2 < 1/2$, then the Lagrangian is displaceable.   I think I can show that if each $r_j^2$ lies in the set $\{ 1/2, 2/3,3/4, \ldots \}$ then the Lagrangian is non-displaceable, by embedding D^n in a product of weighted projective lines and showing the Floer homology of the Lagrangian is non-vanishing. But there must be a way of doing better. 
Sub-question: (which came out of a discussion with Abouzaid): is there a Floer-theoretic
way of seeing the non-displaceability of $S(r)$ for $r^2 \ge 1/2$?   This is easy
to prove by elementary means.

REPLY [5 votes]: I think Thomas is right:  taking a different area form on $S^2$, but still taking the Lagrangian to be the equator is equivalent to changing r, since the unit disk still embeds symplectically as long as the two-sphere has area at least \pi.   Then non-vanishing Floer homology of the product of equators in the product of $S^2$'s implies non-displaceability in the product of unit disks.  Thanks Thomas, I missed that!
Here is a similar problem that this technique solves:  take the ball $B_r$ of radius r in $C^2$ centered at zero, and a product of circles $S(r_1) \times S(r_2)$ contained in $B_r$.  When is $S(r_1) \times S(r_2)$ displaceable?  Thomas' suggestion, applied to the embedding in $CP^2$, gives non-displaceability for $r_1^2 = r_2^2 \ge 1/3$ for $r = 1$. Using probes this is the complete answer.

Chris<|endoftext|>
TITLE: Fundamental group of the line with the double origin.
QUESTION [14 upvotes]: In the simplest cases, the fundamental group serves as a measure of the number of 2-dimensional "holes" in a space. It is interesting to know whether they capture the following type of "hole". 
This example may look pathological, but one must understand where one gets stuck, when one tries to study pathological spaces. It helps one in understanding where exactly all the extra nice conditions are used, and hopefully this type of approach will help in minimizing the number of false beliefs we unconsciously have.
The line with the double origin is the following space. In the union $\{0\} \times \mathbb R \cup \{ 1 \} \times \mathbb R$, impose the equivalence relation $(0, x) \sim (1, x) $ iff $x \neq 0$. 
This space is locally like the real line, ie a $1$-manifold in everything except the Hausdorff condition. It is connected, path-connected and semilocally simply connected. Just the sort of nice space you study in the theory of fundamental group and covering spaces, except for the (significant) pathology that it has one inconvenient extra point violating the Hausdorff-ness.
It seems that the usual methods of computing the fundamental group are not working for this space. Van Kampen's theorem in particular does not apply. Also the covering spaces are weird, just like this space. In fact this space would have been a covering of $\mathbb R$, were it not for the condition that the preimage of every point is a disjoint union of open sets.
So, what if we try to compute the fundamental group of this space? I would be satisfied to know whether it trivial or not. Say, is the collection of homotopy classes of loops based at $1$ nontrivial? It is possible to speculate that a certain loop based at $1$ which passes through both the origins on this special line, in such a way that it passes through the "upper" origin, ie $(1,0)$ on the way left, and it passes through the lower origin, ie $(0,0)$, ought not to be homotopic to the constant loop based at $1$. But how to go about proving/disproving this statement?

REPLY [30 votes]: The earlier answers showing that the fundamental group of this space is infinite cyclic by determining its universal cover or by constructing a fiber bundle over it with contractible fibers are very nice, but it's also possible to compute $\pi_1(X)$ by applying the classical van Kampen theorem not to $X$ itself but to the mapping cylinder of a map from the circle to $X$ representing the supposed generator of $\pi_1(X)$, namely the map that sends the upper and lower halves of $S^1$ to arcs in $X$ from $+1$ to $-1$ in the two copies of $\mathbb R$ in $X$. Decompose the mapping cylinder into the two open sets $A$ and $B$ which are the complements of the two "bad" points in $X$ (regarding $X$ as a subspace of the mapping cylinder). Taking a little care with the point-set topology, one can check that $A$, $B$ and $A\cap B$ each deformation retract onto the circle end of the mapping cylinder. Then van Kampen's theorem says that $\pi_1$ of the mapping cylinder, which is isomorphic to $\pi_1(X)$, is isomorphic to the free product of two copies of $\mathbb Z$ amalgamated into a single $\mathbb Z$.
An interesting fact about $X$ is that it is not homotopy equivalent to a CW complex, or in fact to any Hausdorff space. For if one had a homotopy equivalence $f:X \to Y$ with $Y$ Hausdorff then $f$ would send the two bad points of $X$ to the same point of $Y$ so $f$ would factor through the quotient space of $X$ obtained by identifying these two bad points. This quotient is just $\mathbb R$ and the quotient map $X \to \mathbb R$ is not injective on $\pi_1$, so the same is true for $f$ and $f$ can't be a homotopy equivalence.<|endoftext|>
TITLE: Isolated quadratic residues in integers mod p
QUESTION [8 upvotes]: For prime p sufficiently large, there is always an integer q such that q is a residue mod p, but neither q−1 nor q+1 are; the number of such residues scales like p/8 (and similarly for any sequence of residues/non-residues in three consecutive integers). 
What are the best lower bounds on primes p, for which such "isolated" residues are guaranteed to exist? Do they exist, for instance, for all p ≡ 3 (mod 4) aside from p = 3?
(If this is a typical homework problem, please point me to a textbook for which it is an exercise.)

REPLY [3 votes]: In addition to the character sum approach explained by Charles and Robin, which gives you a precise formula for the number of NRN patterns $\mod p$ for any $p$, in the case $p\equiv 3(\mod 4)$ there is an elementary proof that avoids elliptic curves or Gauss sums. (Notation: N is quadratic non-residue, R is quadratic residue.)
I claim that it is sufficient to find an arithmetic progression $\mod p$ of the form $-a^2,1,-b^2$ with non-zero $a,b.$ Indeed, $-a^2$ and $-b^2$ are N and 1 is R. Choose the sign so that $n=\pm(1+a^2)$ is R: this is possible since $-1$ is a quadratic non-residue $(\mod p),$ by the assumption on $p$. Dividing by $n$, we'll get three consecutive non-zero numbers $\mod p$ in pattern NRN. Thus it remains to show that $a^2+b^2=-2 (\mod p)$ has a solution with $a,b\ne 0,$ which is a special case of the well known fact that for $p\geq 7,$ any non-zero element of $\mathbb{Z}/p\mathbb{Z}$ is the sum of two non-zero squares. Conversely, all NRN patterns are obtained in this way and with a tiny bit of extra work, one can count that there are $(p+1)/8$ or $(p-3)/8$ of them, depending on whether $p\equiv 7(\mod 8)$ or $p\equiv 3(\mod 8).$<|endoftext|>
TITLE: Primes in a (commutative) Jacobson ring
QUESTION [8 upvotes]: Recall that a commutative ring is Jacobson if every prime ideal is the intersection of the maximal ideals that contain it. 
In the exercises of a commutative algebra course I gave I asked the students to show that a commutative ring is Jacobson if and only if every non-maximal prime ideal is the intersection of the prime ideals that strictly contain it. I now suspect that somewhere in the back of my mind I had imposed the condition that the ring should be Noetherian without actually saying this. Of course, Jacobson rings will always have this other property, and the converse is straightforward to prove if there is no strictly ascending chains of prime ideals. But is the result true in general?

REPLY [11 votes]: The result is true in general. 
We may assume a counterexample is given in the form of a domain $R$ satisfying the second property but with nontrivial Jacobson radical, i.e. the closed points of Spec $R$ are not dense. Let D be an affine open neighborhood of $(0)$ in Spec $R$ which contains no closed points. Since D is affine, there exists some $x\in$ D which is closed in D. That is,
$\overline{\lbrace x\rbrace}\setminus x\subset\text{Spec }R\setminus D$.
Since $D$ is open, this implies
$x\not\in \overline{\overline{\lbrace x\rbrace}\setminus x}$,
but this contradicts the requirements of the second property.<|endoftext|>
TITLE: Is there a "Grothendieckification" functor from elementary toposes to Grothendieck toposes?
QUESTION [15 upvotes]: One of my friends asked me whether or not the inclusion of the category of Grothendieck toposes into elementary toposes has a left adjoint.  We are looking at the categories of geometric morphisms.  I am not really sure how to start but nothing seems to rule it out immediately.

REPLY [23 votes]: No, it doesn't.  If it did, then it would preserve limits.  But the category of Grothendieck toposes and geometric morphisms has a terminal object, namely the category of sets, while there are elementary toposes not admitting any geometric morphism to Set (for instance, any small elementary topos).<|endoftext|>
TITLE: Moduli of Extensions
QUESTION [7 upvotes]: Given two modules $M$ and $N$ there is a nice scheme parametrizing extensions
$0 \rightarrow M \rightarrow E \rightarrow N \rightarrow 0$
namely $Ext^1(N,M)$ or, leaving out the trivial extension, the projective space $P(Ext^1(N,M))$.
There are (at least) two natrual generalizations

n-step extensions
$M \rightarrow E_1 \rightarrow E_2 \rightarrow \dots \rightarrow E_n \rightarrow N$
between $N$ and $M$.
Filtered modules: Parametrize modules $E$ which admit a filtration 
$0 \subset F_1 \subset F_2 \dots F_n=E$
with fixed graded objects $E_i=F_i/F_{i-1}$.

I suppose in the first case one can use the group Ext^n(M,N), although I never
saw a construction of a universal family. Is there a good reference?
In the second case, I do not have a clue. So my main question is:
Is there a nice moduli space of filtered objects?

REPLY [3 votes]: Hi Heinrich,
in the situation you have in mind (sheaves on an algebraic variety),
such spaces are not too difficult to construct as Artin stacks. If you omit the condition that the i-th filtration quotient is isomorphic to a given one, then such a universal Artin stack is e.g. constructed in Bridgeland's introduction to Hall-algebras (arXiv:1002.4372, he calls them $\mathcal M^{(n)}$), but of course also in earlier articles by Joyce. Basically it follows from the existence of relative quot schemes.
These universal extension stacks have evaluation morphisms to $\mathcal M$, the stack of all sheaves, sending the filtration to its i-th quotients, so you can take a base change via the map from $\Spec k \to \mathcal M \times \dots \times \mathcal M$ given by your set of objects $E_i$, and the fiber product will be the Artin stack you are looking for.
If you want a scheme instead of an Artin stack - then I would ask back "why?" :) Nevertheless, it would be useful to understand this fiber product better when $n > 2$.<|endoftext|>
TITLE: Genealogy of the Lagrange inversion theorem
QUESTION [27 upvotes]: A wonderful piece of classic mathematics, well-known especially to combinatorialists and to complex analysis people, and that, in my opinion, deserves more popularity even in elementary mathematics, is the Lagrange inversion theorem for power series (even in the formal context). 
Starting from the exact sequence
$0 \rightarrow \mathbb{C} \rightarrow  \mathbb{C}((z)) \xrightarrow{D} \mathbb{C}((z)) \;\xrightarrow{ \mathrm{Res}    }\; \mathbb{C} \rightarrow 0,$
and using the simple rules of the formal derivative D, and of the formal residue Res (that by definition is the linear form taking the formal Laurent series $f(z)=\sum_{k=m}^{\infty}f_k z^k \in \mathbb{C}((z))$ to the coefficient $f_{-1}$ of $z^{-1}$) one easily proves:

(Lagrange inversion formula): If $f(z):=\sum_{k=1}^{\infty}f_k z^k\in \mathbb{C}[[z]]$ and $g(z):=\sum_{k=1}^{\infty}g_k z^k\in\mathbb{C}[[z]]$ are composition inverse of each other, the
  coefficients of the (multiplicative) powers of $f$ and $g$ are linked
  by the formula $$n[z^n]g^k=k[z^{-k}]f^{-n},$$ and in particular (for
  $k=1$), $$[z^n]g=\frac{1}{n} \mathrm{Res}( f^{-n} ).$$

(to whom didn't know it: enjoy computing the power series expansion of the composition inverse of $f(z):=z+z^m$, 
or of $f(z):=z\exp(z),$ and of their powers).
My question: what are the generalization of this theorem in wider context. I mean, in the same way that, just to make one example, the archetypal ODE $u'=\lambda u$ procreates the theory of semigroups of evolution in Banach spaces. 
Also, I'd be grateful to learn some new nice application of this classic theorem.
(notation: for $f=\sum_{k=m}^{\infty}f_k z^k \in \mathbb{C}((z))$ the symbol $[z^k]f$ stands, of course, for the coefficient $f_{k}$)

REPLY [11 votes]: This theorem has been proven by many people and in the context of many diverse fields of mathematics since Lagrange presented "Nouvelle methode pour resoudre les equations litterales par le moyen des series."
It is so ubiiquitous that I've often thought that getting all the proofs together in one place would capture in a unique and compelling way the ebb and flow of the styles, habits, aesthetics and interests of the mathematics community over time. Indeed, I started on this project and quickly discovered that it was more than I could possibly do on my own. Just starting the project however gave me an even deeper appreciation of works such as Dickson's History of the Theory of Numbers and Fletcher's Index of Mathematical Tables.
I've got a list of 50+ proofs, a large collection of books, papers and articles and a bibtex bibliography if anybody is intrigued.  
IMHO, one of the most general statements is Krattenhaler's "Operator Methods and Lagrange Inversion: A Unified Approach to Lagrange Formulas."  One of the most unique is Abdesselam's "A Physicist's Proof of the Lagrange-Good Multivariable Inversion Formula."
Cheers, Scott (sbg at acw dot com)<|endoftext|>
TITLE: Primes in quasi-arithmetic progressions?
QUESTION [8 upvotes]: Suppose $\alpha > 1$ is irrational.  Are there infinitely many primes of the form $\left\lfloor \alpha n \right\rfloor$?  Is the number of $p \leq X$ of this form $\sim \alpha^{-1} X (\log{X})^{-1}$?  I know this is the kind of thing the circle method was born to do, but I cannot for the life of me find a reference for this!

REPLY [8 votes]: I think the uniform distribution mod1 of $\{p/\alpha\}$ is due to Vinogradov, and the asmptotic for primes in a Beatty sequence $\sim \frac{\pi(x)}{\alpha}$ is an immediate consequence. Indeed for $p$ to be equal to some $\lfloor k\alpha\rfloor$ it is equivalent to $1-\frac{1}{\alpha}<\frac{p}{\alpha}-\lfloor \frac{p}{\alpha}\rfloor<1$. So you just need the fractional part of $p/\alpha$ to be on a fixed interval of length $\alpha$ mod1.
On a related note this paper discusses the general sequence $q\lfloor \alpha n+\beta\rfloor +a$.<|endoftext|>
TITLE: how to construct a spherical dodecahedron?
QUESTION [7 upvotes]: using only a spherical ruler (to construct great lines) and a pair of compasses, how can you construct a regular dodecahedron on the surface of the sphere? thank you very much.

REPLY [2 votes]: Not an answer to the question as stated but relevant:
I don't seem to have a reference, but the quickest description of possible constructions on, say, the unit sphere, is by the angles made by intersecting curves. The constructible angles are the same as the constructible angles in the Euclidean plane. Thus the constructible lengths are those arclengths $\alpha$ for which $\cos \alpha$ or $ \sin \alpha$ or $\tan \alpha$ (the conditions are equivalent) are in the "constructible field," the smallest extension of the rationals in which the square root of any positive element is still in the field. One might also wish to require  $\alpha \leq \pi.$ 
Actually, let me make that a request. If anybody knows of a reference on the constructible lengths and angles on the surface of the sphere, please let me know.
This is strictly analogous to (and presumably far, far older than)  the situation in the hyperbolic plane,  I will try to make a working link:
http://zakuski.math.utsa.edu/~jagy/papers/Intelligencer_1995.pdf 
See also Marvin Jay Greenberg, "Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries,"  The American Mathematical Monthly (an M.A.A. journal), Volume 117, number 3, March 2010, pages 198-219. I have a pdf of that as well if anyone cannot find it.
There is a bit of a story. The results on constructibility in $H^2$ were in a string of papers in Russian and Ukrainian in the 1930's and 1940's. I found, and used, the simple conclusions. I later sent my paper to Greenberg, so that material is in the M.A.A. paper mentioned and in the fourth edition of his book. Meanwhile, Robin Hartshorne (yes, that Hartshorne) heard of this result from Marvin and came up with his own proof using the Hilbert Field of Ends formalism, expressing regret that such a pretty result did not make it into his own book on the subject, "Geometry: Euclid and Beyond."<|endoftext|>
TITLE: Zariski closed sets in C^n are of measure 0
QUESTION [12 upvotes]: This is related to another question in which it is proved that Zariski open sets are dense in analytic topology.
But it is intuitive that something more is true. Namely, that they are the sets where some polynomials vanish, and consideration of a few examples in $\mathbb R^n$ where they are of Lebesgue measure $0$, suggest strongly that the Zariski-closed sets(except the whole affine space) are of measure $0$ in $\mathbb C^n$ as well. This should be quite simple; but I am unable to prove it due to inexperience in measure theory.
The nice thing about proving this is that once this is done, then we are able to claim safely that so-and-so statement is true almost everywhere, if it is true on a Zariski-open set.
So, in a more measure theoretic formulation:

Let $X$ be a set in $\mathbb C^n$ contained in the zero locus of some collection of polynomials. How to show that $X$ is of measure $0$?

In fact my feeling is that more should be true, ie, we can replace polynomials by analytic functions at least, and get the same result.

REPLY [4 votes]: And if you want to take this to an extreme... for a function on a domain in ${\bf R}^n$, it's enough to assume that at every $x$ there's a ball $B_x$ centered at $x$ and a multiindex $\alpha$ for which $\partial^{\alpha} f$ is nonzero and continuous on $B_x$. 
To see this, first note that it suffices to show that the zeroes of $f$ in a given $B_x$ have measure zero. This is proven by induction on $|\alpha|$. If $\alpha = 0$ it's trivial, and if $\partial^{\alpha'}f(x) \neq 0$ for any $\alpha '$ with $|\alpha '| < |\alpha|$ it follows by the induction hypothesis, shrinking $B_x$ if necessary. Otherwise we can write $\partial^{\alpha} f = \partial_{x_i}\partial^{\beta} f$ for some $\beta$, where we can assume $\partial^{\beta} f(x) = 0$. By the implicit function theorem, if $B_x$ is small enough the zeroes of $\partial^{\beta} f$ in $B_x$ form a $C^1$ hypersurface with measure zero. For each $y$ off this surface, $\partial^{\beta} f$ is nonzero and then you can apply the inductive hypothesis to find an appropriate $B_y$. A simple compactness argument shows you need only countably many $B_y$.. so you're done.<|endoftext|>
TITLE: Funky congruences
QUESTION [6 upvotes]: Suppose we have the remainders: {$(a^0_1, a^1_1), \ldots, (a^0_n, a^1_n)$} and the moduli {$c_1, \ldots, c_n$}. We want to know if there exists $b_1, \ldots, b_n \in$ {0,1} and $m \in \mathbb{N}$ such that:
$$
\begin{array}{lcl}
m & = & a^{b_1}_1 \quad (\mod c_1) \newline \\
  & = & a^{b_2}_2 \quad (\mod c_2) \newline \\
  & = & \ldots \newline \\
  & = & a^{b_n}_n \quad (\mod c_n) \newline \\
\end{array}
$$
That is basically a generalization of a system of linear congruences with an OR operator. Is there a nice necessary and sufficient condition for such an $m$ to exists?

REPLY [5 votes]: Here is an alternative approach, by reducing the problem to 2-SAT. It makes me ask: what if there were three values $(a_i^0, a_i^1, a_i^2)$ for each $i$ (and $b_i$'s in $\{0, 1, 2\}$)? It analogously can be reduced to some type of 2-SAT with ternary-valued variables. I don't know immediately whether either of these more general problems is NP-complete.
We need the following
Theorem: Suppose we have a collection of modular constraints on $m$, of the form $\{m \equiv v_i \; (\mod{c_i})\}_{i=1\ldots n}$. Then there is no $m$ meeting all constraints iff there are two particular constraints $1 \le j < k \le n$ so that no $m$ simultaneously satisfies those two constraints.
The proof is given later below. But, assuming the theorem, we just need to determine whether $b$ exists such that all pairs of two constraints $m \equiv a_j^{b_j} \; (\mod{c_j}), m \equiv a_k^{b_k} \; (\mod{c_k})$ are pairwise satisfiable. This can be done with 2-SAT: it is easy to check which pairs of $b^j, b^k$ allow a mutual solution (see Wadim's note in his first 2 lines), then for each pair $(j, k)$ we get up to 4 constraints depending on which pairs of values for $b_j, b_k$ can hold at the same time. For example if $a_j^0 = 5, c_j^0 = 20, a_k^0 = 6, c_k^0 = 42$, we see $b_j=0, b_k=0$ cannot both hold, since $m$ would have to be both even and odd. Thus we add the clause $(b^j = 1$ OR $b^k = 1)$ to our 2-SAT instance. Repeating this gives $O(n^2)$ clauses, then we run a linear-time 2-SAT algorithm to see if $b$ exists meeting all clauses, and are done.
Proof: $\Leftarrow$ is clear, so it suffices to prove the contrapositive of $\Rightarrow$. Thus, assume we can satisfy any given two constraints; we'll constrct an $m$ that satisfies all constraints. 
We focus on one prime $p$ at a time. Let $p^t$ be the maximum power of $p$ dividing any $b_i$, and let $\ell$ (which depends on $p$) be the maximizer, so $p^t | b_\ell$. The $m$ we hope to construct must satsify $m \equiv b_\ell \; (\mod{p^t})$, call this a crazy constraint. We get one crazy constraint for each prime dividing $c_1c_2\cdots c_n$. Moreover, for any $m$ that satisfies this crazy constraint, using Wadim's note twice, we deduce that $m \equiv v_i \; (\mod{\gcd(p^t, c_i)})$ for each $i$. But $\gcd(p^t, c_i)$ is just the highest power of $p$ dividing $c_i$. Thus by CRT, an $m$ which satisfies all crazy constraints (i.e. for all $p$), satisfies $m \equiv v_i \; (\mod{c_i})$, and hence all $n$ of the "input" constraints. But the crazy constraints are for relatively prime moduli, hence they  all simultanouesly hold for some $m$, and we are done.<|endoftext|>
TITLE: An elementary proof that the degree of a map of spheres determines its homotopy type
QUESTION [13 upvotes]: I'm helping to teach an undergraduate algebraic geometry course (out of Hatcher's textbook).  We have recently defined the degree of a map of spheres using homology, and the professor and I thought it would be nice if we could give some kind of argument that such a map is determined up to homotopy by its degree.  I know two proofs of this: one using the Freudenthal suspension theorem, and the other using the Pontryagin correspondence between homotopy classes of [smooth] maps and framed cobordism classes of framed submanifolds (see Milnor, Topology from a Differential Viewpoint).  Unfortunately, neither of these arguments would be accessible to our students, who have only seen the fundamental group and homology (no higher homotopy theory) and who are not necessarily expected to know any differential topology.
Thus, I ask the following question:

Is there an elementary argument (i.e., that can be understood by someone who only knows about homology and the fundamental group) that the degree of a map of spheres determines its homotopy type?

More precisely, what we have (or will have) available is most of the material in the first two chapters of Hatcher, not including the "additional topics."
If necessary, I'm willing to make plausible assumptions that the students may not know how to prove, such as
-Replacing $f \colon S^n \to S^n$ by a homotopic map if necessary, we may assume that there exists points with only finitely many preimages, such that $f$ is a homeomorphism locally about each preimage. (i.e., regular values)
-Every map of CW complexes is homotopic to a cellular map.
-The degree map $\pi_n(S^n) \to \mathbb{Z}$ is a group homomorphism. [This reduces us to showing that a degree-0 map is nullhomotopic.]

REPLY [19 votes]: Take a look at Exercise 15 in Section 4.1, page 359 of the book you're referring to.  This outlines an argument that should be the sort of thing you're looking for.  The main step is to deform a given map to be linear in a neighborhood of the preimage of a point, using either simplicial approximation or the argument that proves the cellular approximation theorem.  Once this is done, the rest is essentially the Pontryagin-Thom argument (in a very simple setting), plus the fact that $GL(n,\mathbb R)$ has just two path-components.<|endoftext|>
TITLE: What does the Fourier transform of an L-infinity function look like locally?
QUESTION [7 upvotes]: Question:
What does an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ look like locally?
As formulated, the question might be a bit difficult to answer since the Fourier transform of a function f ∈ L∞(ℝ) is a distribution, and it is not easy to "write down" a distribution.
So let me first illustrate the situation at hand with an easy example:

Example:
  The Fourier transform of the Heaviside function H(x) (i.e. the characteristic function of the positive reals) is given by
  a linear combination of the function 1/x and of the Dirac delta function
  (see this Wikipaedia entry for the exact formula, as well as for the meaning of the distribution "1/x").

The formalism of distributions is bit overkill for talking about measures, and things that look like 1/x.
For example, the primitive of an element of $\mathcal F \big( L^\infty(\mathbb R)\big)$ is
always a function (well defined outside of a set of measure zero).
Using the above observation, we get the following
Reformulation of the question:
Let f(x) ∈ L∞(ℝ) be a function, and let g(x) be a primitive of its Fourier transform.
• What can g(x) look like locally?
• What local conditions must g satisfy in order to have a chance of coming from some
f ∈ L∞(ℝ)?
• On what kind of sets can g fail to be continuous?

REPLY [8 votes]: It is pretty much the same as to describe the class $G$ of functions $g$ on the circle whose Fourier coefficients decay as $O(|k|^{-1})$. There is no nice "space side" property $P$ that would characterize them but for every nice "space side" property $P$ one can figure out in finite time if it holds for all such functions or not.
As to your particular questions, the answers are
1) On the circle being in this class it is a local property (this needs compactness of the circle) because if $g\in G$, then the product of $g$ and any sufficiently smooth function is in $G$ and you can do partitions of unity.
2) There are obvious inclusions $BV\subset G\subset BMO$ ("bounded variation" and "bounded mean oscillation"). If you need something tighter than that, tell the family of comparison spaces you want to use.
3) Since $\sum_{k\ge 1} \frac 1kz^k$ is unbounded at $1$ and continuous everywhere else, we can move such spikes around to create a function that is locally unbounded on any closed set we want and discontinuous on any $F_\sigma$ set we want including the entire circle.<|endoftext|>
TITLE: Is there an intuitive explanation for an extremal property of Chebyshev polynomials?
QUESTION [18 upvotes]: Consider the following optimization problem:
Problem: find a monic polynomial $p(x)$ of degree $n$ which minimizes $\max_{x \in [-1,1]} |p(x)|$. 
The solution is given by Chebyshev polynomials:
Theorem: Let $T_n(x) = cos (n \cdot cos^{-1} x)$. Then $(1/2^{n-1}) T_n$ is a monic polynomial of degree $n$ which achieves the above minimum.
The proof of this fact is short and surprisingly free of messy calculations. From the definition of $T_n$ you derive a recurrence relation expressing $T_n$ in terms of $T_{n-1}, T_{n-2}$, which shows that $T_n$ are indeed polynomials. Then
you argue that $(1/2^{n-1}) T_n$ is monic and achieves its extrema $\pm 1/2^{n-1}$ at least $n+1$ times in $[-1,1]$, from which the above theorem easily follows. If you'd like a nice exposition of this argument which does not skip any steps,  this is short and clear.
However, I don't get much enlightenment from this proof: it feels pulled out of a 
hat. For example, it gives me no clue about which other polynomial optimization problems
have similar solutions. 
My question: Is there a natural and motivated sequence of steps which, starting from the above optimization problem, leads to Chebyshev polynomials?
Update: I changed the title to better reflect the question I am asking.

REPLY [8 votes]: Well, let's try to avoid the hat.
Consider the dual (and obviously equivalent) problem: find the polynomial $p(x):[-1,1]\rightarrow [-1,1]$ of degree $n$ with the greatest possible leading coefficient. We have some information on values of $p$, and need something about its coefficient. Let's try Lagrange's interpolation. Take some $n+1$ values $t_1 < t_2 < \dots < t_{n+1}$ from $[-1,1]$ and write down (for $u(x)=(x-t_1)\dots(x-t_{n+1})$) the formula
$$
p(x)=\sum p(t_i) \frac{u(x)/(x-t_i)}{u'(t_i)}.
$$
Then take a look on coefficient of $x^n$. It equals
$$
\sum \frac{p(t_i)}{u'(t_i)}.
$$
We know that $|p(t_i)|\leq 1$, so the leading coefficient does not exceed
$
\sum 1/|u'(t_i)|.
$ 
Ok, when does equality occur? The answer is: $p$ should take values $(-1)^{n-i+1}$ in $t_i$. That is, we have to find a polynomial of degree $n$ with $n+1$ extremal values $\pm 1$ on $[-1,1]$. This may hold only if $t_1=-1$, $t_{n+1}=1$, and $t_2$, $\dots$, $t_n$ are roots of $p'$. So, $1-p^2(x)$ should be divisible by $(1-x^2)p'(x)$. Hereon the trigonometic substitution $x=\cos t$, $p=\cos f$ is very natural, as we know that $1-f^2$ is divisible by $f'$ for $f=\cos$. So we invent Chebyshev's polynomials.
Also, it is seen from Lagrange formula that they are extremal in many other problems with restrictions $|p(x)|\leq 1$ on $[-1,1]$. For example, the value in each specific point $x_0>1$ is maximized also for Chebyshev polynomial, it is proved by exactly the same way.<|endoftext|>
TITLE: Is geometric realization of the total singular complex of a space homotopy equivalent to the space?
QUESTION [6 upvotes]: Let $X$ be a topological space and let $|Sing(X)|$ be the geometric realization of the total singular complex of $X$. 
Then $|Sing(X)|$ is a CW complex with one cell for each non-degenerate singular simplex. There's a natural map $f:|Sing(X)|\to X$ and there's a theorem that says that $f$ is a weak homotopy equivalence. That is, $f$ induces isomorphisms of homotopy groups.
Then it seems that Whitehead theorem applies and gives that $f$ is homotopy equivalence as long as $X$ is homotopy equivalent to a CW complex (i.e. $X$ is m-cofibrant). Is that correct?
Is there an example when $f$ is not homotopy equivalence? Any examples that come up in "real life"?

REPLY [14 votes]: The map from the (realization of the) singular complex of a space $X$ to $X$ is a homotopy equivalence if and only if $X$ is homotopy equivalent to a CW complex, so to get examples where the map is not a homotopy equivalence you just need spaces that are not homotopy equivalent to CW complexes.  There are plenty of these that come up all the time in various places outside algebraic topology. For example, a compact space with infinitely many path-components isn't homotopy equivalent to a CW complex. For an explicit example, take the Cantor set, or more simply just a convergent sequence with its limit point. Suspending these examples gives path-connected examples, and iterated suspensions give examples with higher connectivity. More generally, any compact space having a nonfinitely generated homology group is an example. There are lots of other ways that spaces not homotopy equivalent to CW complexes arise too.<|endoftext|>
TITLE: Real interpretations of Discontinuities in Floer homology
QUESTION [10 upvotes]: This question is motivated by the answer in this question (you don't have to read it to understand the following).
I am not that proficient in calculating Floer homology, and I held back on answering and just commenting on the question because of the following.
Let $L$ be the equator on $S^2$ then it is obvious that no "Hamiltonian" symplectomorphism can take $L$ to another Lagrangian not intersecting $L$. Indeed, $L$ cuts $S^2$ into two equally sized pieces and it is not difficult from there.
It is quite a different thing to assert that the intersection Floer homology $FH(L,L)$ is non-trivial, which is used in the solution to the problem (together with a Kunneth formula).
The thing that initially bugged me about this was that arbitrarily close to $L$ there are Lagrangians whose Floer homology intersection is $0$ both with $L$ and themselves. I have later come to realize that I myself when considering exact Lagrangians in cotangent bundles have encountered this discontinuity frequently, and should really not be surprised. Here the exactness is very important for non-triviality. However, for me this comes up in a different way because I am considering finite reductions of the loop spaces using Chaperons broken geodesics (as Viterbo does).
I, however, realized that the discontinuity does not seem that bad and this leads me to my question:
Question: Is there a natural Floer homology with real coefficients, which resolves these discontinuities in the sense that the differentials are continuous.
A little more motivation: In the above case it seems that with coefficients in the Novikov ring $\mathbb{Z}[t]$ we can realize the Floer complex with two generators, and then for all Lagrangians not dividing $S^2$ into two equally volumed parts the differential is an isomorphism. But if the Lagrangians are both of that particular type the differential is $0$. This could be described continuously with real coefficients as:
$ d \colon \mathbb{R}[t] \to \mathbb{R}[t]$
where $d$ is the differential from the $\mathbb{Z}[t]$ coefficient complex multiplied by the scalar $(A_1-A_2)^2+(A_1'-A_2')^2$ where $A_i$ are the two volumes for the first Lagrangian and $A_i'$ are the volumes for the second Lagrangian.

REPLY [11 votes]: The brief answer is yes, using ideas from Novikov homology. 
Here's an example of the discontinuity and how it can be fixed.  Take $L=S^1\times y$ as a Lagrangian in standard symplectic $T^2=S^1\times S^1$. Then $HF_\ast(L,L)$ (which means $HF_\ast(L,\phi L)$ for $\phi$ Hamiltonian) is isomorphic to $H_\ast(L)$, while $HF(L,S^1\times y')=0$ for $y' \neq y$. But if we couple the Floer complex to a local system $l$ on the space of paths from one Lagrangian to another which restricts to a non-trivial local system $\alpha$ on $L$, then we have $HF_\ast(L,S^1\times y';l)=0$ for $y'=y$ (because there are no intersections) and $HF_\ast(L,L;l)\cong H_\ast(L;\alpha) = 0$ as well.
Now here's some detail.
There's a loose principle in Floer theory which says that if you have Floer modules $HF(a_t)$ for data $a_t$ depending smoothly on a parameter $t$, all defined over the same coefficient ring, then you can expect them to vary continuously provided a certain vector space $K_t$ does not vary with $t$. Let $C$ be the ambient configuration space of the theory. Then we have an index (or spectral flow) class $I\in H^1(C;\mathbb{Z})$, and for each $t$ an action class $[A_t]\in H^1(C;\mathbb{R})$. Then $K_t = \ker I \cap \ker [A_t] \subset H_1(C;\mathbb{R})$. The point is that $\ker I$, the group of periods of the action functional $A_t$, shows up in the Floer-theoeretic differential, and if the behaviour of the action functional on $\ker I$ changes then unexpected cancellations could occur. Note that this does indeed jump in the circle example above (I get $K_t=\mathbb{R}$ for $t\neq 0$, $K_t=0$ for $t=0$).
The principle can be proved in particular instances either using the homotopy method (bifurcation analysis), which is very general but analytically hard, or Floer's continuation method (in which case you'll need good energy estimates).
We get  bit more flexibility using a local system $l$ on $C$. If our coefficient ring is $\Lambda$, this assigns an automorphism of $\Lambda$ to each homotopy class of paths from $c_1$ to $c_2$, where $c_1,c_2\in crit(A_t)$, compatibly with concatenation of paths. We can build this system into the Floer differential. The local system defines a class $[l]\in H^1(C;\Lambda)$. If $R$ comes with a homomorphism $f\colon \mathbb{R}\to \Lambda$ (think Novikov ring!)  then a local system $l_\mathbb{R}$ with coefficients in $\mathbb{R}$ defines one with coefficients in $\Lambda$. If contributions to the differential are weighted by $f\circ A_t$, then for such local systems we should modify the definition of $K_t$, since the contribution of periods to the differential has changed: now the definition should be 
$$K_t= \ker I \cap \ker ([A_t]+l_{\mathbb{R}}).$$ 
The continuity principle is as before.
As an example, one can see this way that if $L_t$ is a Lagrangian isotopy with flux $F\in H^1(L_0;\mathbb{R})$ then the local Floer homology $HF(L_0,L_1)_{loc}$ (taken in a suitable small neighbourhood of $L_0$ so as to avoid "instanton corrections") should be isomorphic to $H_\ast(L_0; \Lambda_F)$, where $\Lambda_F$ is the local system of Novikov rings determined by $F$.
Sample references: 
Le Hong Van, Kaoru Ono, "Symplectic fixed points, the Calabi invariant and Novikov homology", MR1308493. 
Kaoru Ono, "Floer-Novikov cohomology and the flux conjecture", MR2276532.
Yi-Jen Lee, "Reidemeister torsion in Floer-Novikov theory and counting pseudo-holomorphic tori. I", MR2199540
Mihai Damian, "Constraints on exact Lagrangians in contangent bundles of manifolds fibered over the circle". MR2534477<|endoftext|>
TITLE: limiting behaviour of converging loops on a torus
QUESTION [8 upvotes]: Suppose (Ln) is a sequence of loops in a torus  S1 × S1 converging in the Hausdorff metric to some set L in the torus. Suppose also that for each loop Ln the projection map p:S1 × S1 -> S1 defined by p(x,y) =x when restricted to Ln is not null-homotopic, then can we conclude that the restriction of the projection map p to L is also not null homotopic? Or are there counter examples?

REPLY [4 votes]: Your intuition is correct: the map p : L → S1 is not nullhomotopic.
If p : L → S1 was nullhomotopic, then it would factor through the universal cover ℝ of S1.
The inclusion ι : L → S1 × S1 would therefore factor through ℝ × S1.
Let ι' : L → ℝ × S1 denote a lift of ι,
and let L(0) := ι'(L).
The subset L(0) ⊂ ℝ × S1 is then one of the connected components of the preimage of L in ℝ × S1.
Let us call the other components L(n) for n ∈ ℤ.
Let C0 ⊂ ℝ × S1 be a loop in that
separates L(0) from L(1).
We may also assume that C0 is not nullhomotopic in ℝ × S1 [this still needs a small argument...].
The projection C ⊂ S1 × S1 of C0 is then a loop in the complement of L, and
represents the element (0,1) of ℤ×ℤ = π1(S1 × S1).
Since L lies in the complement of C and Ln → L in the Hausdorff metric, there exists an n such that Ln lies in the complement of C.
This contradicts your assumtion that p : Ln → S1 is nullhomotopic.

REPLY [3 votes]: EDITED. The following example should not be conisdered as a contre-example, 
it is just a well known example of a pathology that can  happen.
EXAMPLE. $L$ is composed of a union of the vertical circle $(0, S^1)$ and a disjoint $R^1$ that is emdedded in $T^2$ in such a way, that it accumulates to $(0, S^1)$ from both sides. This $R^1$ projects one-to one to the horisontal circle without a point and can be represented a as graph (function of $x$), than if is given by the following formula:
$y= (sin(1/x))$
The point is that such $L$ can be Hausodrff approximated by a sequence of circles $s_n$ that are not null-homotopic ($s_n$ wiggles more an more near the vertical circle $(0, S^1)$ as $n\to \infty$).  
The problem with this example, is that the topology on the union of $S^1$ and $R^1$ that we should take (I guess) is the toplogy induced from $T^2$. And for this induced topology, I guess the projection to the horisontal cicle is non null-homotopic...<|endoftext|>
TITLE: What role does the "dual Coxeter number" play in Lie theory (and should it be called the "Kac number")?
QUESTION [47 upvotes]: While trying to get some perspective on the extensive literature about highest weight modules for affine Lie algebras relative to "level" (work by Feigin, E. Frenkel, Gaitsgory, Kac, ....), I run into the notion of dual Coxeter number but am uncertain about the extent of its influence in Lie theory.   The term was probably introduced by Victor Kac and is often denoted by $h^\vee$ (sometimes by $g$ or another symbol).   It occurs for example in the 1990 third edition of his book Infinite Dimensional Lie Algebras in Section 6.1.   (The first edition goes back to 1983.)   It also occurs a lot in the mathematical physics literature related to representations of affine Lie algebras.   And it occurs in a 2009 paper by D. Panyushev in Advances which  studies the structure of complex simple Lie algebras.   

Where in Lie theory does the dual Coxeter number play a natural role (and why)?

A further question is whether it would be more accurate historically to refer instead to the Kac number of a root system, since the definition of $h^\vee$ is not directly related to the work of Coxeter in group theory.
BACKGROUND: To recall briefly where the Coxeter number $h$ comes from, it was introduced by Coxeter and later given its current name (by Bourbaki?).   Coxeter was studying a finite reflection group $W$ acting irreducibly on a real Euclidean space of dimension $n$: Weyl groups of root systems belonging to simple complex Lie algebras (types $A--G$), these being crystallographic, together with the remaining dihedral groups and two others.   The product of the $n$ canonical generators of $W$ has order $h$, well-defined because the Coxeter graph is a tree.    Its eigenvalues are powers of a primitive $h$th root of 1 (the "exponents"): $1=m_1 \leq \dots \leq m_n = h-1$. Moreover, the $d_i = m_i+1$ are the degrees of fundamental polynomial invariants of $W$ and have product $|W|$.  
In the Weyl group case, where there is an irreducible root system (but types $B_n, C_n$ yield the same $W$), work of several people including Kostant led to the fact that $h$ is 1 plus the sum of coefficients of the highest root relative to a basis of simple roots.  On the other hand, the dual Coxeter number is 1 plus the sum of coefficients of the highest short root of the dual root system.   For respective types $B_n, C_n, F_4, G_2$, the resulting values of $h, h^\vee$ are then $2n, 2n, 12, 6$ and $2n-1, n+1, 9,4$.  This gets pretty far from Coxeter's framework.   
One place where $h^\vee$ clearly plays an essential role is in the study of a highest weight module for an affine Lie algebra, where the canonical central element $c$ acts by a scalar (the level or central charge).   The "critical" level $-h^\vee$ has been especially challenging, since here the theory seems to resemble the characteristic $p$ situation rather than the classical one.

REPLY [7 votes]: Let $\mathfrak{g}$ and $\mathfrak{h}$ be semisimple Lie algebras corresponding to connected simply connected compact groups $G,H$. Any map $\mathfrak{g} \to \mathfrak{h}$ has a Dynkin index, which is the induced map $\mathrm{H}^4(BH) \to \mathrm{H}^4(BG)$. When $\mathfrak{g}$ and $\mathfrak{h}$ are simple, $\mathrm{H}^4(BH)$ and $\mathrm{H}^4(BG)$ are both isomorphic to $\mathbb Z$ (and this isomorphism can be chosen canonically by using the generator which maps to a positive-definite element under $\mathrm{H}^4(BG) \to \mathrm{H}^4(BG) \otimes \mathbb{R} \cong \mathrm{Sym}^2(\mathfrak{g})^W$), and the Dynkin index is then just a number.
The dual Coxeter number of $\mathfrak{g}$ arises as the Dynkin index of the map $\mathrm{adj} : \mathfrak{g} \to \mathfrak{so}(\dim\mathfrak{g})$. (Almost. When $\dim \mathfrak g \leq 4$, this fails. What is correct is to look at the stablized adjoint map $\mathfrak{g} \to \mathfrak{so}(\infty) = \varinjlim \mathfrak{so}(n)$.)
What's called "the Dynkin index of a representation" $V$ is the Dynkin index of the map $\mathfrak{g} \to \mathrm{sl}(\dim V)$. Note that $\mathfrak{so}(n) \to \mathfrak{sl}(n)$ has Dynkin index $2$ when $n \geq 5$, explaining the factor of two in the usual formulas about $2h^\vee$ and Killing forms.<|endoftext|>
TITLE: Functoriality of Poincaré duality and long exact sequences
QUESTION [7 upvotes]: Hi all,
Today I was playing with the cohomology of manifolds and I noticed something intriguing. Although I might just have been caught out by a couple of enticing coincidences, it feels enough like there might be something going on that I thought I'd put it out here.  
Let $M$ be an $n$-manifold with boundary $\partial M$. We write out the long exact homology sequence for the pair $(M, \partial M)$:
$$\cdots \to H_k(M) \to H_k(M, \partial M) \to H_{k - 1}(\partial M) \to \cdots$$
Let's apply Poincaré duality termwise, and keep the arrows where they were out of sheer faith. What we get is
$$\cdots \to H^{n - k}(M, \partial M) \to H^{n - k}(M) \to H^{n - k}(\partial M) \to \cdots$$
Surprisingly, this is the long exact cohomology sequence for the pair $(M, \partial M)$! To my mind, two things here are weird. The first is that intuitively, any functoriality properties Poincaré duality possesses should be arrow-reversing. The second is that we have a shift - but not a shift by a multiple of 3. So the boundary map in the homology sequence 'maps' to something that doesn't change degree in the cohomology sequence.  
Let's play the same game with the Mayer-Vietoris sequence. For simplicity, suppose now $M$ is without boundary. Write $M = A \cup B$ where $A$ and $B$ are $n$-submanifolds-with-boundary of $M$ and $A \cap B$ is a submanifold of $M$ with boundary $\partial A \cup \partial B$. Then we have
$$\cdots \to H_k(A \cap B) \to H_k(A) \oplus H_k(B) \to H_k(M) \cdots$$
Hitting it termwise with Poincaré duality, and cruelly and unnaturally keeping the arrows where they are once again, we get
$$\cdots \to H^{n - k}(A \cap B, \partial A \cup \partial B) \to H^{n - k}(A, \partial A) \oplus H^{n - k}(B, \partial B) \to H^{n - k}(M) \to \cdots$$
This looks unfamiliar, but by looking at cochains it's not hard to see that there actually is a long exact sequence with these terms. However, this time we don't have the weird shift.
Now is there anything going on here, or just happenstance? Is there really a sense in which Poincaré duality is functorial with respect to long exact sequences? If so, what's the 'Poincaré dual' of the long exact sequence of the pair $(M, N)$ where $N$ is a tamely embedded submanifold of $M$? 
Edit: Realised that in the final l.e.s. the arrows should actually go the other way, which is slightly less impressive. Even so...

REPLY [5 votes]: One of the standard proofs of Poincaré duality, at least for those manifolds that have handle decompositions, provides a reason for some of these naturality properties.  Every piecewise linear manifold, or every smooth manifold, has a handle decomposition, and many but not all topological manifolds also do.  (Amazingly enough, the only exceptions are in 4 dimensions.)  A handle decomposition gives rise to two different CW cellulations on the manifold, one using cores and the other using co-cores.  Then this proof of Poincaré duality posits that the CW chain complex of one cellulation is identical to the CW cochain complex of the other cellulation.
You can extend this coincidence of chain complexes to both of your examples, the Mayer-Vietoris sequence and the exact sequence of a pair.  Obtaining identical chain complexes also gives you other information, for instance that the Bockstein maps are the same.<|endoftext|>
TITLE: Do continuous maps give continuity in the 'topology' of Hausdorff distance?
QUESTION [5 upvotes]: I was reading this question:
limiting behaviour of converging loops on a torus
And I wanted to be able to give an argument along the lines of: "If your loops are converging in your torus, their projections must converge in your $S^1$", but a quick google search gives me no results along these lines- do they exist? If not why not?
I am aware that if either of your spaces are unbounded then a sensible topology isn't particularly forthcoming, but is there a situation in which a result of this form can make sense? As a starting point let's set the bar at:

Do compact fibrations induce maps on their subsets that are continuous wrt Hausdorff distance?

Can we do better? Can we do a little worse? Or does none of this make sense?

REPLY [5 votes]: Any uniformly continuous map $f$:X→Y between metric spaces induces a uniformly continuous map $C\mapsto \overline{f(C)}\ $  between the spaces of closed subsets wrto the Hausdorff distances; in fact with the same modulus of continuity. (Just recall that the Hausdorff distance between A and B is less than δ if and only if for any a∈A there is some b∈B with d(a,b)<δ and for any b∈B, there is some a∈A with d(a,b)<δ).
PS: As to the topologic side of the question (stability of topological or homotopical properties of subsets of a space under perturbations in Hausdorff distance). As far as I know these things usually work well in an ANR metric space X, because of the homotopy extension property. For instance, any closed subset of X that is contractible in X (meaning that the inclusion map $C\to X$ is null homotopic) has a contractible nbd in X. If X is also compact, the nbd is a uniform nbd, so if a sequence of closed sets $C_n$ converges in the Hausdorff sense to C, the $C_n$ are eventually contractible in X, too. Generalizing a bit, we may also say that the Lusternik-Schnirelman category of C (minimum cardinality of a contractible closed covering of C) is lower semicontinuous wrt the Hausdorff distance (the category of the limit is larger or equal to the limit of the category along the sequence). Reference (for ANR's and homotopy): e.g. the first 2 chapters of Spanier book.<|endoftext|>
TITLE: Ample line bundles, sections, morphisms to projective space
QUESTION [19 upvotes]: This concerns a number of basic questions about ample line bundles on a variety $X$ 
and maps to projective space. I have searched related questions and not found answers, but I apologize if I missed something. I'll work with schemes of finite type over a field $k$ for simplicity.
Background
A quasi-coherent sheaf $F$ on a $k$-scheme $X$ is globally generated if the natural map $H^{0}(X,F)\otimes \mathcal{O}_{X} \rightarrow F$ is a surjection of sheaves. Basically, this says that for any point on $X$, there is at least one section of $F$ that doesn't vanish at that point, so there are enough sections of $F$ to see all the points of $X$. (EDIT: As pointed out in the comments below, this last sentence does not describe a situation equivalent to being globally generated. Perhaps it is better to say that
globally generated means that for each point $x \in X$, $F$ has some rank $r$ at $x$ and globally generated means that there are at least $r$ sections of $F$ that are linearly independent over $x$.)
The notion of globally generated is especially useful when $F=L$ is a line bundle on $X$.
If $V$ is a finite dimensional subspace of $H^{0}(X,L)$ such that $V \otimes \mathcal{O} \rightarrow L$ is surjective, then we get a morphism $\varphi_{V}:X \rightarrow \mathbb{P}(V)$ 
by the universal property of the projective space $\mathbb{P}(V)$ of hyperplanes in $V$. 
Essentially, given a point $x \in X$, we look at the fibre over $x$ of the surjection 
$V \otimes \mathcal{O} \rightarrow L$ to get a quotient $V \rightarrow L_{x}$. The kernel
is a hyperplane in $V$, and the morphism $\varphi_{V}$ sends $x$ to that hyperplane as a point in $\mathbb{P}(V)$.
So how to build globally generated sheaves? A line bundle $L$ is called ample 
if for every coherent sheaf $F$, $F \otimes L^{\otimes n}$ is globally generated for all large $n$. The smallest $n$ after which this becomes true can depend on $F$.
Finally, a line bundle is called  very ample if $L$ is globally generated and $\varphi_{V}$ is an embedding
for some subspace of sections $V$.
There are various properties of and criteria for ample line bundles, which can be found in Hartshorne, for example. What we need for the below questions are the following: $L$ is ample if and only if $L^{m}$ is ample for some $m$ if and only if $L^{n}$
is very ample for some $n$; if $L$ is ample, eventually $L^{k}$ will have sections, be globally generated, be very ample, and have no higher cohomology.
Questions

Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?
Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^{k}$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?
If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X \subset \mathbb{P}^{N}$, I can eventually get a finite morphism $X \rightarrow \mathbb{P}^{d}$, where $d$ is the dimension of $X$. But what if
I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb{P}^{d}$?

REPLY [11 votes]: The answers already given seem to settle your questions, but I'd like to emphasise one point that hasn't been mentioned.
Suppose X is a projective variety, and $\pi: X' \rightarrow X$ any morphism which contracts at least one curve in X'. (For example, it could be the blowup of any subvariety of X of codimension at least 2.) Then for any very ample line bundle L on X, the pullback $\pi^*L$ is globally generated but not ample, because it has degree zero on any curve contracted by $\pi$. (So to get a simple example for the first part of your first question, let $\pi$ be the blowup of P^2 in a point, and L the class of a line.)
The point I want to make is that this is not just a class of examples, but in fact the general case. To see this, suppose X is a projective variety and L a line bundle which is globally generated but not ample. Then L defines a morphism $f_L: X \rightarrow Y \subseteq \mathbb P^n$ to some subvariety of projective space with $L = f_L^* H$, the pullback of the hyperplane class (which is very ample on Y). Now since L is not ample, the morphism $f_L$ must contract some curve on X: in other words, the Stein factorisation of $f_L$ consists of a nontrivial contraction followed by a finite morphism. (Here a contraction means a projective morphism with connected fibres.) 
The moral is that for X any projective variety, globally generated but not ample line bundles on X are the "same thing" as nontrivial contraction morphisms $\pi: X \rightarrow Y$. (The correspondence can be made more precise by considering the face structure of the nef cone of X, but that's not really relevant to your question.)
Finally, a word on terminology: a line bundle L on X of which some power $L^k$ is globally generated is called (by some people at least) semi-ample. For some purposes this turns out to be a better notion than globally generated, just as ample is a better notion than very ample.<|endoftext|>
TITLE: Making N (set of all positive integers) a group
QUESTION [5 upvotes]: Can anybody please give me an example of a binary operation under which N forms a group? More generally, how to find some operations to make possibly any set a group?

REPLY [4 votes]: So far, all the answers have taken the following form: you just biject N with some countably infinite group. But what if one regarded that as "cheating" and asked for a truly "natural" group operation on N? So far, I think Pietro Majer's answer comes closest, since his bijection with an infinite product of cyclic groups of order 10 is fairly natural (at least in the sense of "familiar"). But can one do any better than this?
Here is an attempt to argue that there is no 100% natural group structure on N. (I don't claim that this is surprising.)
First, we need to choose an identity element. The identity element is a privileged element of the group, so for the choice to be natural it should be a privileged element of N. The only privileged element of N is 1 (for me the natural numbers start at 1) so we are forced to choose that. Now let's consider the following question: what is the inverse of 2? I think there are two possible natural answers, namely 2 itself and 3. (One is the smallest element you can choose, and the other is the smallest element that is not equal to 2.) If we choose 2 and continue, then perhaps we'll end up with the binary version of PM's answer (except that everything would be shifted by 1 because of my insistence on making 1 the smallest natural number -- perhaps that was silly). So perhaps, contrary to my original intention, that is an argument that PM's answer is (apart from taking base 10) the most natural thing one can do, and not too unnatural.
If we go for 3, then there doesn't seem to be any natural choice for the order of 2 other than infinity. So then we need to keep powers of 2 away from powers of 3. The natural way of doing that looks like actually using the multiplicative structure of N. But then we get into trouble later, since 2*3 will be 1, so our treatment of 6 becomes a bit arbitrary. Or at least, I think it does. An implicit question here: is there some optimally natural way of putting a group structure on N in such a way that every element has infinite order? Perhaps the bijection-with-Z answer is the best one can do.<|endoftext|>
TITLE: Applications of Group Theory Which Motivate Theoretical Questions?
QUESTION [8 upvotes]: I'm going to be a teaching assistant for an undergraduate class in abstract algebra next semester, for students who have not taken abstract algebra before. It will deal with group theory and linear algebra, but the students have already had a semester of linear algebra so I'm thinking about how to deal with group theory.
I'd like to present some examples of applications of group theory that motivate the theoretical questions the course will deal with. For instance, "symmetries of three-dimensional objects form groups. Crystals have symmetrical structure. If we could get some bounds on possible forms of three-dimensional symmetry, this would give limits on the sorts of crystals that could form, which would be interesting to a chemist." Another example could be, "The roots of a polynomial can sometimes be permuted in ways that do not change the value of polynomial functions of the roots, and these permutations form a group. [insert explanation of Galois theory here] If we could determine whether this group is trivial, we could see whether solving the polynomial is possible."
My goal is to persuade the students that group theory is useful and therefore interesting, but since they will most likely be interested in different things, I'd like to have a big list in the hope of being able to find something for each of them. Can people suggest either applications of group theory or places to find such applications?

REPLY [3 votes]: An obvious one is proof of Fermat's little theorem from Lagrange's theorem.  This has immediate application in public-key cryptography, which computer people will care about.<|endoftext|>
TITLE: Geodesics on a hyperbolic paraboloid
QUESTION [8 upvotes]: Given any two points on a hyperbolic paraboloid ($xy = z$ or $z = (x^2 - y^2)/2$) how does one find the geodesic between them?
I know that since the hyperbolic paraboloid is doubly ruled, some of the geodesics are lines. However, I have very little idea of how to find the geodesics between arbitrary points. 
If an exact answer cannot be given, then I would be interested in a technique of approximating the geodesics. 
Finally, I'd like to add that this is outside my area of expertise (I'm an algebraist!) so you'll have to explain it to me.
Thanks!

REPLY [7 votes]: I was working on answering another question involving integrating the geodesic equations on a surface, and the links there lead me back to this question, which I hadn't noticed before.  In case anyone is interested, there is a more direct way to get to the geodesics of the induced metric on the hyperbolic paraboloid, so I thought that would go ahead and input this approach.
As Greg pointed out, the geodesic equations in the graphing coordinates are
$$
x'' = -\frac{2yx'y'}{(1+x^2+y^2)}\qquad\text{and}\qquad y'' = -\frac{2xx'y'}{(1+x^2+y^2)}.
$$
These equations have two first integrals that are quadratic in velocity:
$$
Q_1 = (1+y^2)\,x'^2 + 2xy\,x'\,y' + (1+x^2)\,y'^2,
$$
i.e., the induced metric itself, which is, of course, the statement that geodesics have constant speed, and
$$
Q_2 = 2(1+x^2+y^2)\,x'\,y',
$$
which is easily verified.  (It's a classical fact that the geodesic equations on any (non-flat) surface of degree $2$ in Euclidean $3$-space have a second first integral that is quadratic in velocities, namely $Q_2 = |K|^{-3/4}I\!I$ in addition to the obvious first integral $Q_1 = I$.)
Since there are two independent first integrals quadratic in velocity, this is a Liouville metric and hence can be put in Liouville form.  This is an algorithmic procedure; following it leads to the result that, if one establishes new coordinates $z$ and $w$ on the surface
by 
$$
 x = \sinh \left(\frac{z+w}2\right)\quad\text{and}\quad 
 y = \sinh\left(\frac{z-w}2\right),
$$ 
then we have the first integrals
$$
\begin{aligned}
Q_1 &= \tfrac14\bigl(\cosh(z)+\cosh(w)\bigr)\bigl(\,\cosh(z)\,z'^2 +\cosh(w)\,w'^2\,\bigr)\\
Q_2 &= \tfrac14\bigl(\cosh(z)+\cosh(w)\bigr)\bigl(\,\cosh(w)\cosh(z)\,z'^2 -\cosh(z)\cosh(w)\,w'^2\,\bigr)
\end{aligned} 
$$
(The actual Liouville coordinates would be $(u,v)$ where $\mathrm{d}u = \sqrt{\cosh z}\,\mathrm{d}z$ and $\mathrm{d}v = \sqrt{\cosh w}\,\mathrm{d}w$, but these are elliptic integrals, and it seems pointless to change to these coordinates.) 
In particular, for a unit speed geodesic, i.e., one for which $Q_1 = 1$ and $Q_2 = c$ for some constant $c$, we have $Q_2-c\,Q_1 =0$, so
$$
(\cosh(w)-c)\cosh(z)\,z'^2 -(\cosh(z)+c)\cosh(w)\,w'^2,
$$ 
and one can thus separate variables, yielding
$$
\frac{\cosh(z)\,\mathrm{d}z^2}{(\cosh(z)+c)} = \frac{\cosh(w)\,\mathrm{d}w^2}{(\cosh(w)-c)}.
$$
Now, $c=0$ corresponds to $z\pm w$ being constant, i.e., $x$ or $y$ is constant, which are straight lines in the surface.  When $c\not=0$, the geodesic has to stay in the region where $\cosh z + c$ and $\cosh w - c$ are non-negative, and we have an equation that can be integrated 'by quadratures' to yield two foliations by geodesics of the region where $\cosh z + c$ and $\cosh w - c$ are positive.
$$
\sqrt{\frac{\cosh z}{\cosh z + c}}\,\mathrm{d}z = \pm\, \sqrt{\frac{\cosh w}{\cosh w - c}}\,\mathrm{d}w
$$
This will include a family that envelopes either $\cosh z + c  = 0$ (if $c<-1$) or $\cosh w - c = 0$ (if $c>1$).  (The curves $z = 0$ and $w=0$ are geodesics.)
In order explicitly to compute the distance between two points using these formulae, one would need to compute the corredponding $z$ and $w$ coordinates of the two points (easy), find the $c$ belonging to the (unique) geodesic joining those two points (nontrivial), and then compute the elliptic integrals as above.  
I expect that the likelihood that one could actually carry this out and find a 'closed form' expression for the distance function on the surface is rather low.<|endoftext|>
TITLE: Reference request: representation theory of the hyperoctahedral group
QUESTION [17 upvotes]: I was wondering if someone knows a good reference for the representation theory of the hyper-octahedral group $G$. The hyper-octahedral group $G$ is defined as the wreath product of $C_2$ (cyclic group order $2$) with $S_n$ (symmetric group on $n$ letters). 
I understand that the representations of $G$ are in bijection with bi-partitions of $n$. I am looking for a reference which explains the details of why the representations of $G$ are in bijection with bi-partitions of $n$, and constructs the irreducible representations of $G$ (I imagine this is vaguely similar to the construction of Specht modules for $S_n$). 
So far, the only reference I have is an Appendix of MacDonald's "Symmetric functions and Hall polynomials" (2nd version), which deals with the representation theory of the wreath product of $H$ with $S_n$ (for $H$ being an arbitrary group, not $C_2$).

REPLY [2 votes]: A paper dealing with the hyperoctahedral group :
"Branching rules for the hyperoctahedral group",J.Math.Phys.30,11,1989,pp2469-2475.<|endoftext|>
TITLE: Major mathematical advances past age fifty
QUESTION [62 upvotes]: From A Mathematician’s Apology, G. H. Hardy, 1940:
"I had better say something here about this question of age, since it is particularly important for mathematicians. No mathematician should ever allow himself to forget that mathematics, more than any other art or science, is a young man's game. ... I do not know an instance of a major mathematical advance initiated by a man past fifty. If a man of mature age loses interest in and abandons mathematics, the loss is not likely to be very serious either for mathematics or for himself."
Have matters improved for the elderly mathematician? Please answer with major discoveries made by mathematicians past 50.

REPLY [3 votes]: George Pólya (1887-1985) wrote the wonderful paper
Pólya, George, On the eigenvalues of vibrating membranes, Proc. Lond. Math. Soc., III. Ser. 11, 419-433 (1961). ZBL0107.41805.
at the age of 73. This paper motivated large chunk of research known as the Pólya conjecture of the eigenvalues of the Laplacian. See for example this MO-Question.<|endoftext|>
TITLE: Telling group algebras apart
QUESTION [7 upvotes]: It's a big, famous, hard problem in operator algebras to determine if the von Neumann algebras $L(F_2)$ and $L(F_3)$ are isomorphic, or not.  Here $F_n$ is the free group on n generators and $L(F_n)$ is the weak-operator-topology closure of the group algebra $\mathbb C[F_n]$ acting naturally on the Hilbert space $\ell^2(F_n)$.
I presume it must be known if the algebras $\mathbb C[F_2]$ and $\mathbb C[F_3]$ are isomorphic or not.  But from casually asking a few algebraists, I've never had any luck in finding this out (I admit to not working very hard on this!)  I'm guessing some (co)homology theories must help...?  What about for replacing $\mathbb C$ by a more general ring?

REPLY [5 votes]: One has $Hom({\mathbb C}[F_n],{\mathbb C}) = ({\mathbb C}^{\times})^n$ with the obvious topology. (Here, $Hom$ denotes the space of $\mathbb C$-linear homomorphisms.) This of course uses a little bit more than only the algebra structure, but every ${\mathbb C}$-linear isomorphism would preserve the topology on the space of $\mathbb C$-linear representations. Since the spaces $({\mathbb C}^{\times})^n$ are not homeomorphic for different $n$, the claim follows. The same applies to the maximal group $C^{\star}$-algebra of $F_n$. One has $Hom(C^{\star}(F_n),{\mathbb C}) = (S^1)^n$, where one considers only $\star$-homomorphisms.<|endoftext|>
TITLE: Why do modules with small support have high Exts?
QUESTION [15 upvotes]: Let $M$ be a module over a ring $R$. In nice situations (though I don't know what exactly nice means...) the following two numbers are equal:
1.) The codimension of the support of $M$
2.) The biggest $k$ such that $\text{Ext}^k(M,.)$ doesn't vanish
Why do we expect this intuitively? Why should lengths of injective/projective/flat resolutions have anything to do with the support?

REPLY [14 votes]: To understand what "nice" is in your sense has been a very interesting question in commutative algebra.
In the following discussion I will assume, unless otherwise notice, that $(R,m,k)$ is Noetherian local, and $M$ is finitely generated. Let $(1)$ be the codimension of support of 
$M$ and $(2)$ be the biggest non-vanishing index of $\text{Ext}(M,-)$.
First, the number (2) is finite forces $M$ to have finite projective dimension by taking $N=k$ the residue field. So we will assume $\text{pd}\ M <\infty$. Then, as BCrd pointed out:
$$ (2)  = \text{pd} \ M = \text{depth} \ R -  \text{depth} \ M$$
The first inequality is easy by computing Ext via a projective res. of M + Nakayama lemma. The second is the Auslander-Buchsbaum theorem.
On the other hand:
$$(1) = \text{dim} \ R - \text{dim} \ M $$ 
So 
$$(1) - (2) = (\text{dim} \ R - \text{depth} \ R) -  (\text{dim} \ M-\text{depth} \ M) $$
Thus, if both $R,M$ are Cohen-Macaulay (which by def. means dim=depth) and $\text{pd}\ M <\infty$ then $(1) = (2)$. If $R$ is "more Cohen-Macaulay" then $M$, we will have $(1)<(2)$. 
The situation described in Emerton's answer is also very interesting. In general, the smallest index for which $\text{Ext}^i(M,N) \neq 0$ is the biggest length of an $N$-regular sequence inside the annihilator of $M$. When $N=R$ this number is called the grade of $M$, which I will call (3). 
It is easy to see that $(3) \leq (1)$ in general. One can prove that $(1) = (3)$ if $R$ is Gorenstein as follows: By Local Duality, $\text{Ext}^i(M,R)$ is Matlis dual to the local cohomology module $\text{H}_{m}^{d-i}(M)$, here $d= \text{dim}\  R$. It is not hard to show that local cohomology vanish beyond $\text{dim}\ M$, QED. 
Amazingly, it has been an open conjecture for 50 years that $(1)=(3)$ whenever $M$ has finite projective dimension!   
For "intuitive" understanding, I would offer the following: often when study modules of finite projective dimension one draw inspirations from those of the form $R$ modulo a regular sequence (so the resolution is a Koszul complex). In such case one can easily see that $(1) = (2) =(3)$.
EDIT: I got too caught up in the results and forgot your main question: why bigger codimension implies bigger projective resolution? A very low-tech way to see it is: bigger codimension means bigger annihilator of $M$. Now each element of the annihilator of $M$ gives a non-trivial relation on elements of $M$, namely $ax=0$, so it is not surprising that the modules with bigger annihilators  have more complicated resolutions.

REPLY [13 votes]: One intuitive explanation (in the case $R$ regular that Emerton discusses) is that on a derived level a module knows all about the formal neighborhood of its support, and the higher ext's (up to the codimension) account for the missing directions. 
Let me assume for simplicity that $M$ is the structure sheaf of a subscheme. Then the functor $Ext(M,-)$ is derived restriction (more precisely restriction-with-supports) to the support of $M$. This functor sets up a derived equivalence (by the derived form of the Barr-Beck theorem, or a fancy form of Koszul duality) between the completion of the regular ring $R$ along the support and a derived refinement of $M$. The idea is that transversally to the support we are using Koszul duality to identify modules over a symmetric algebra and an exterior algebra (ie if the support of $M$ is a smooth variety of codim k, we enhance the structure sheaf of the support by an exterior algebra on a k-dim vector space). 
(edit:) More generally, for any $M$ coherent or a perfect complex, ($R$ still regular) $Ext(M,-)$ preserves all limits and colimits on the derived level (in the dg derived catetgory of $R$), and sets up an equivalence between $Ext(M,M)$-modules and whichever $R$-modules are not annihilated by the functor $Ext(M,-)$. For $M$ a structure sheaf this means the formal completion along the support, shouldn't be hard to describe in general. I think it should be easy to deduce from this that we have to have Exts up to the codimension, since we have to account for all the homological complexity of an "open nbhd" of the support using a sheaf of dg algebras supported on the subscheme.<|endoftext|>
TITLE: tangent sphere bundle over sphere
QUESTION [7 upvotes]: are there some general description about tangent sphere bundle over sphere?
(it is a special $S^{n-1}$bundle over $S^n$)
say for n=1,it is trivial,$S^0\times S^1$,for n=2,it is $SO(3)\cong \mathbb{R}P^3$,
for n=3,it is trivial again,so it is for n=7.
how about other cases.

REPLY [8 votes]: If you like clutching maps descriptions of bundles the sphere has a nice one.   Think of $S^n$ as the union of two discs corresponding to an upper and lower hemi-sphere.  Then the tangent bundle trivializes over both hemispheres.  You can write down the trivializations explicitly with some linear algebra constructions.  Think of the intersection of the two hemi-spheres as an $S^{n-1}$, this allows you to think of the tangent bundle as a union $D^n \times \mathbb R^n \cup D^n \times \mathbb R^n$ along the common boundary $S^{n-1} \times \mathbb R^n$. The clutching (gluing) map is then a map of the form:
$$ c: S^{n-1} \to SO_n $$
and it is explicitly the map $c(v) = M(v)M(x_{n+1})$
where $v \in S^{n-1} \subset \mathbb R^n \subset \mathbb R^{n+1}$ where we think of $\mathbb R^n$ as the orthogonal complement of the $(n+1)$-st coordinate vector $x_{n+1}$ in $\mathbb R^{n+1}$.  $M(v)$ denotes mirror reflection fixing the orthogonal complement to $v$. 
The basic idea in this construction is that if one takes a geodesic between two points on a sphere, parallel transport from one point to the other can be written as a composite of two reflections, the latter reflection corresponding to the mid-point of the geodesic, the initial reflection corresponding to the initial point of the geodesic.<|endoftext|>
TITLE: Elementary proof wanted: every local principal ideal ring is a quotient of a PID
QUESTION [17 upvotes]: I am looking for a more elementary proof of the following result:
Theorem (Hungerford, 1968): Let $R$ be a principal ideal ring.  Then $R \cong \prod_{i=1}^n R_i$, where each $R_i$ is a homomorphic image of a principal ideal domain (PID).
Hungerford's article is available free online at:
http://projecteuclid.org/euclid.pjm/1102986148
What do I mean by "more elementary"?  Hungerford uses the Cohen structure theory of complete local rings, which I would like to avoid (because I have notes on commutative algebra which do not discuss such things).  
Note that Hungerford's theorem is a refinement of a previous result of Zariski and Samuel, 
which asserts that a principal ideal ring is isomorphic to a finite direct product of rings, each of which is either a PID or a "special principal ideal ring", i.e., a local Artinian principal ideal ring.  The proof of this result uses primary decomposition, which is acceptable to me (in fact I put a section on primary decomposition into my notes for exactly this application).  
Given the theorem of Zariski-Samuel, Hungerford's result is plainly equivalent to the fact that every Artinian local principal ideal ring is the quotient of a PID.  Now doesn't that sound like you should be able to prove it without invoking the structure theory of complete local rings?

REPLY [5 votes]: Theorem 5.2 in http://www.emis.de/journals/BAG/vol.46/no.1/b46h1her.pdf gives an answer
(take the projective limit. The paper has a related one with corrections,
but not for the part that is related to your question). This is for a non-commutative
case, and the theorem has a non-commutative extension: a PIR is a finite direct product 
of prime and artinian indecomposable cases, which are matrix rings over CPU rings
(Faith, Algebra II should contain all the needed references)<|endoftext|>
TITLE: Axiom of Computable Choice versus Axiom of Choice
QUESTION [6 upvotes]: What would be the consequence of requiring that any choice function be computable; i.e. using as the foundational basis ZF + ACC? Does it make a difference if we admit definable functions?
I guess I am sometimes bothered by the thought that any random choice over an uncountable set by definition would seem to almost certainly return a non-computable member. This seems impractical and perhaps even problematic, considering that major branches of mathematics such as for example analysis, with only few notable exceptions, mainly operate within the computable or definable realm.
Presumably an immediate consequence would be that the Banach–Tarski paradox and similar theorems related to unmeasurable sets would fail. But would there be more fundamental consequences?

REPLY [21 votes]: As others have observed before, you cannot simply say "computable" and "ZF" in the same sentence without explaining what you mean. But I can tell you what your options are.
In order to speak about computability you have to provide some sort of an axiomatization or a model of computability. ZF is not such a model, but there are many others. Let us look at some of them and what happens to the axiom of choice. In what follows I mean by "computable model" a model of set theory or type theory in which all (global) maps are computable in some sense. In particular, choice functions happen to be computable in such models, so far as they exist.
Intuitionistic set theory IZF is an intuitionistic variant of set theory. It has many different models, some of which are computable. In IZF we can prove that the axiom of choice implies the law of excluded middle, so this kind of destroys the I in IZF. But restricted forms of choice are ok, notably countable and dependent choice are fine (in the sense that they are consistent with IZF and are validated by various computable models of IZF).
Higher-order intuitionistic logic (internal language of a topos) is essentially the same as IZF with regards to choice.
Martin-Löf type theory is a formulation of constructive mathematics in which choice is valid, in fact it is easy to prove it. The caveat here is that the interpretation of logic is a bit unusual because a proposition is equated with the collection of its proofs (as opposed to with its extension).
Brouwerian intuitionism accepts some choice but not all. More precisely, it accepts countable choice and $AC_{1,0}$, which is choice for families indexed by the set $\mathbb{N}^\mathbb{N}$. There are computable models of Brouwerian intuitionism (certain kinds of realizability models).
Bishop-style constructive mathematics accepts countable and dependent choice but not more. It has many computable and classical models because it is agnostic with respect to the law of excluded middle.
Russian constructivsm is another form of constructivism which accepts countable and dependent choice, but not more. The effective topos is a model.
Realizability toposes provide a rich class of models of computability. In fact, they are so general that the topos of (classical) sets is a special case. I should also point out that realizability toposes are larger than classical sets because they contain the category of sets as a subtopos of sheaves (for the double negation topology). Therefore, they provide the sort of setup that is needed to make sense of your question. In those realizbility toposes that are built from reasonable computational models, i.e., those that are based on the standard notion of Turing computability, choice is never generally valid. This is so because general choice implies the law of excluded middle (as mentioned above), and the law of excluded middle allows us to define the Halting oracle. Nevertheless, countable choice is always valid, which is one reason why various "schools of computability" accept it. In some realizability toposes you get more choice, but never a lot.
Let me make one last remark. There is a very general principle that "computable maps are continuous", where of course we have to look at "correct" topologies for this to make sense. (Ask a MO question if you want to know why.) Applied to choice this says that computable choice functions are continuous. But it is quite easy to come up with examples where the choice function cannot be continuous, for example we cannot choose continuously for each $x \in \mathbb{R}$ an integer $k \in \mathbb{Z}$ such that $x < k$. So you need not get into computability to see why choice has to fail in certain contexts. This might be helpful if you are familiar with topological sheaves.<|endoftext|>
TITLE: Canonical fundamental domain for a discrete subgroup Γ of SL₂(R) acting on hyperbolic plane
QUESTION [8 upvotes]: Let a discrete subgroup $\Gamma$ of $SL_2(\mathbb R)$ act on the hyperbolic plane by Möbius transformations. Is there a "best" or "most canonical" fundamental domain for this action? 
Some (mostly unhelpful) observations:

For the action of $SL_2(\mathbb Z)$ the usually taken fundamental domain is this one. Any particular reason?
Unfortunately the hyperbolic triangle with vertices $0, 1, \infty$ also won't do for $SL_2(\mathbb Z)$, as it is not (quite) a fundamental domain.
If we are considering a finite-index sugroup of $SL_2(\mathbb Z)$, then we can form the fundamental domain as a finite union of the fundamental domain of  $SL_2(\mathbb Z)$. This is not very nice-looking, but it can be determined effectively given a description of $\Gamma$.
Let the action be specified for $\Gamma$. Then, for any point $z_0 \in \mathbb H$, there is the standard polygon around it. This was pointed out to me by sigfpe. Thanks a lot, sigfpe. This is a very nice and very canonical construction, once you have a starting point $z_0$. But this is very theoretical. We do not know what are the vertices, in a computable way. We just know that they exist. And then again, what would be a canonical choice for the point $z_0$? If any point inside the hyperbolic plane is canonical, it is $i$. But the two examples considered above for  $SL_2(\mathbb Z)$ does not arise as a fundamental polygon or standard polygon for $i$. In fact $i$ is on the edge of both of these fundamental domains.

REPLY [3 votes]: I found interesting the question of why we normally use "that" fundamental domain for the action of $PSL_2(\mathbb{Z})$. I thought it worthy of its own answer, though I make no claims that the reasons I will list are exhaustive (and I'd be interested to see others!).

Once we have a fundamental domain, and as long as we keep track of the ways the sides get identified by the group, the Poincaré Polyhedron Theorem allows us to recover a presentation for the group, where side-pairings correspond to generators, and vertex cycles to relations. If, say, we were interested in finding minimal generating sets, that would correspond to finding fundamental domains with minimal numbers of sides. Since $PSL_2(\mathbb{Z})$ has rank 2, the "cleanest" fundamental domains we can find (in this sense) are those with 4 sides. (Here we regard the two halves of the bottom circle arc as distinct sides, since they are identified by the linear fractional transformation $z \rightarrow \frac{-1}{z}$, and we regard $i$ as a vertex.) Further, the two matrices which correspond to the side-pairings of the usual domain are $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right)$, which are frequently taken as generators for $PSL_2(\mathbb{Z})$. So, in this regard, the domain "fits in" with other existing conventions.
As Greg Kuperberg mentions, $PSL_2(\mathbb{Z})$ is a subgroup of index 2 in the $(2,3,\infty)$ triangle reflection group, and it's easy to see how our domain is just one copy of this triangle unioned with its reflection along a vertical side.
As coudy mentions, our domain is a Dirichlet domain, and the proof of this fact is fairly straightforward (and elegant, in my opinion). Certainly, it's easier to see with this particular domain than with other domains one might try to construct.
I suspect that one reason this domain is used so frequently is that short, elementary proofs exist that it is indeed a fundamental domain, which avoid any of the discussion listed above. In what little I have glanced at on automorphic forms, this type of proof is given at an introductory stage, so that one can move on to more content. I like to see the simplest arguments as being tied up with the previously mentioned Ford domains, though I'm not sure how useful others might find it. Essentially, the unit circle is the isometric circle of the map $z \rightarrow \frac{-1}{z}$, which (because this particular element has order 2, or is its own inverse) means the circle is sent to itself, and its exterior and interior are interchanged. This observation is part of showing that no two points of our domain are equivalent under the action of the group. Using other domains involves using more than one (isometric) circle, so our domain is simplest in this sense.<|endoftext|>
TITLE: Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective?
QUESTION [9 upvotes]: Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0$  forall $I$ then $P$ is projective? 
Similar statements are true for flat and injective modules, but I'm beginning to suspect that projective modules cannot be characterized solely by ideals.

REPLY [2 votes]: I think the answer is no. I found the following counterexample.
Let $K$ be the field of complex Hahn series with real exponents, i.e.
$$K = \left\{ f = \sum_{r \in \mathbb{R}} a_r X^r ; \, \operatorname{supp}(f) \textrm{ is a well-ordered subset of } (\mathbb{R}, \ge) \right\},$$
where $\operatorname{supp}( \sum_{r \in \mathbb{R}} a_r X^r ) := \{r \in \mathbb{R} ; \, a_r \neq 0\}$.
Let $R$ be the subring of $K$, defined as $R = \{f \in K ; \, \operatorname{supp}(f) \subseteq \mathbb{R}_{\ge 0}\}$. This is a local ring, with maximal ideal $\mathfrak{m} = \{f \in K ; \operatorname{supp}{f} \subseteq \mathbb{R}_{>0}\}$. The residue field $k= R/m \cong \mathbb{C}$ is an $R$-module.
I claim that $M = R/\mathfrak{m}$ is a counterexample, i.e. $\operatorname{Ext}_R^1(M,R/I) = 0$ for all ideals $I$ of $R$, yet $M$ is not a projective $R$-module.
Proof:
$M$ is not projective: If $M$ were projective, then $0 \to \mathfrak{m} \to R \to M \to 0$ would split, hence $\mathfrak{m}$ would be a quotient of $R$, so in particular it could be generated by one element. However $\mathfrak{m}$ is not a finitely generated ideal.
$\operatorname{Ext}_R^1(M,R/I) = 0$:
The short exact sequence $0 \to \mathfrak{m} \to R \to M \to 0$ gives the long exact sequence
$$
\begin{split}
0 &\to \operatorname{Hom}(M, R/I) \to \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I) \to \\
 &\to \operatorname{Ext}^1(M, R/I) \to \operatorname{Ext}^1(R,R/I) \to \dotsm
\end{split}
$$
Here $\operatorname{Ext}^1(R,R/I) = 0$ since $R$ is projective. So it is enough to prove that $R/I = \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I)$ is surjective.
The ideals of $R$ are easy to describe: If $c \in \mathbb{R}_{\ge 0}$, then let $I_{\ge c} = (X^c)$ and $I_{>c} = (X^r ; \, r > c)$. Then every nonzero ideal is of the form $I_{\ge c}$ or $I_{>c}$. In particular $\mathfrak{m} = I_{>0}$.
If $I=0$:
we need that $R = \operatorname{Hom}(R,R) \to \operatorname{Hom}(\mathfrak{m}, R)$ is surjective (in fact it is bijective).
This is not hard to check (for $\varphi \in \operatorname{Hom}(\mathfrak{m}, R)$,
show that $\varphi(X^r) \in I_{\ge r}$, and $X^{-r} \varphi(X^r) \in R$ is independent of $r>0$).
If $I=I_{\ge c}$:
Let $\varphi \colon \mathfrak{m} \to R/I_{\ge c}$ be a homomorphism.
We need to show that there is an $h \in R$ such that $\varphi(f) = f h + I_{\ge c}$.
To prove this, look at $\varphi(X^r)$ and take $r \to 0$.
If $r<c$, then $\varphi(X^r)$ will determine $h$ modulo $I_{\ge c-r}$.
To finish the proof, we need that if $E \subseteq [0, c)$ such that $E \cap [0,d)$ is well-ordered for all $d<c$, then $E$ is also well-ordered.
For $I=I_{> c}$, the proof is almost the same.<|endoftext|>
TITLE: Some models for random graphs that I am curious about
QUESTION [25 upvotes]: G(n,p)
We are familiar with the standard notion of random graphs where you fixed the number n of vertices and choose every edge to belong to the graph with probability 1/2 (or p) independently. This model is referred to as G(n,1/2) or more generally G(n,p).
Random graphs with prescribed marginal behavior
Now, suppose that rather than prescribe the probability for every edge, you presecribe the marginal probability for every induced subgraph H on r vertices. (So r should be a small integer, 3,4,5, etc.) You make the additional assumptions that 
a) the probability $p_H$ does not depend on the identity of the r vertices, 
and 
b) It depends only on the isomorphism type of $H$. 
So, for example: for r=3 you can think about the case that $p_H = 1/9$ if H is a triangle or an empty graph and $p_H=7/54$ otherwise. 
Once these $p_H$ are assigned you consider among all the probability distributions with these marginal behavior (if there are any) the one with maximal entropy. (But this choice is negotiable; if there is something different worth doing this is fine too.) 
My questions:
1) Are these models studied in the literature? 
2) When are such $p_H$'s feasible? 
3) Given such feasible $p_H$'s say on graphs with 4 vertices, is there a quick algorithm to sample from the maximal-Entropy distribution which will allow to experiment with this model?
Background
This question is motivated by a recent talk by Nati Linial in our "basic notion" seminar on extremal graph theory. (Maybe these are well studied models that I simply forgot, but I don't recall it now.)

REPLY [16 votes]: The Lovasz-Szegedy theory of graphons is likely to be relevant.  Every measurable symmetric function $p: [0,1] \times [0,1] \to [0,1]$ (otherwise known as a graphon) determines a random graph model, in which every vertex v is assigned a colour c(v) uniformly at random from the unit interval [0,1], and then any two vertices v, w are connected by an edge with an independent probability of p(c(v),c(w)).  These are in some sense the only models of large graphs in the sense that any sequence of increasingly large graphs has a subsequence that converges to a graphon (in the sense that the $p_H$ statistics converge).
Each $p_H$ (in the asymptotic limit $n \to \infty$) can be read off from the graphon as an integral.  For instance, the density of triangles is $\int\int\int_{[0,1]^3} p(x,y) p(y,z) p(z,x)\ dx dy dz$.
If the $p_H$ are specified for all finite graphs H, then this determines p up to change of variables (measure-preserving bijections on [0,1]) outside of a set of measure zero.  But if one only specifies the $p_H$ for a finite number of H then there are multiple choices for p (and in some cases, no choices at all) and it is not obvious to me how to find a solution or even to determine whether when a solution exists.  (Note that even for just two choices of H, one being an edge and the other being a bipartite graph, the question of determining the possible values of $p_H$ is essentially Sidorenko's conjecture, which is still not fully resolved.)  But perhaps numerical methods (annealing, gradient descent, etc.) may be able to find solutions some of the time (though they will hardly be "canonical").<|endoftext|>
TITLE: Intuition for a formula that expresses the class number of an imaginary quadratic field by counting quadratic residues
QUESTION [8 upvotes]: If $p$ is a prime of the form $4n+3$, the class number $h$ of $Q[\sqrt{-p}]$ can be expressed using the number $V$ of quadratic residues and $N$ nonresidues in the interval $[1,\frac{p-1}{2}]$: 

If $p=8n+7$ then $h=V-N$
If $p=8n+3$ then $h=\frac{1}{3}(V-N)$

This result seems so simple and elegant, but its proof (which I saw in Number Theory by Borevich & Shafarevich - p. 346), while definitely beautiful, is not very short and is based on the analytic class number formula. And so I didn't feel that the proof really helped "demystify" the result.
This leads me to ask: Can someone see intuition for this result? Any short heuristic argument that would lead to it? Any obvious meaning of $h=V-N$? In short, "why" is it true?

REPLY [5 votes]: Update: more thoughts, including a shorter and more nonsensical "proof", on my blog. 

Orde's paper, "On Dirichlet's Class number formula", gives a beautiful nonsense proof, before giving a rigorous one. Thanks to KConrad for pointing out Orde's paper to me. Orde is a little terse, so let me expand:
Let $R$ be the ring of integers in $\mathbb{Q}[\sqrt{-p}]$. For simplicity, take $p>3$. For any positive integer $N$, let $S(N)$ be the number of ideals in $R$ with norm $N$. By unique factorization into prime ideals and a little thought, we have
$$S(N) = \sum_{d|N} \left( \frac{-p}{d} \right). \quad (*)$$
This formula is correct for $N>0$. Orde explains how to extend this formula to be correct for $N \neq 0$. The formula at $N=0$ will then be the class number formula!
Let $C$ be the class group of $R$. Let $Q$ be the set of integral quadratic forms with discriminant $-p$, modulo equivalence. Note that $Q$ is the disjoint union of $Q^{+}$ and $Q^{-}$; the positive definite forms and the negative definite ones. For most purposes, we discard $Q^{-}$, but today we want it around. There is a standard bijection between $C$ and $Q^{+}$. 
For $c \in C$ and $N>0$, let $S_c(N)$ be the number of ideals of $R$ of class $c$ and norm $N$. So $S(N)=\sum_{c \in C} S_c(N)$. Let $q$ be the corresponding positive definite form and let $T_q(N)$ be the number of representations of $N$ be the form $q$. By the standard relationship between quadratic forms and ideals, $T_q(N)=2 S_c(N)$. (That $2$ is because $R$ has $2$ units.) Also, since $N>0$, we have $T_{-q}(N)=0$. So
$$\frac{1}{2} \sum_{q \in Q} T_q(N) = S(N) = \sum_{d|N} \left( \frac{-p}{d} \right) \quad (**).$$
The left and right hand sides of $(**)$ are symmetric in exchanging $N$ and $-N$, so $(**)$ is also valid for $N<0$.
Now, consider $(**)$ for $N=0$. For any $q \in Q$, we have $T_q(0)=1$, since $q$ is either positive or negative definite. So the left hand side is $(1/2) |Q|=|C|$. 
Everything divides $0$, so the right hand side is $\sum_{d>0} \left( \frac{-p}{d} \right)$. That doesn't converge, but its Cesaro sum is $(1/p) \sum_{d=1}^{p} (p-d) \left( \frac{-p}{d} \right).$ (If we were doing the case that $p \equiv 1 \mod 4$, that average would be over $4p$ terms, instead of just $p$ of them.) So we "derive" that
$$|C| = (1/p) \sum_{d=1}^{p} (p-d) \left( \frac{-p}{d} \right).$$
This is easily shown to be equivalent to the class number formula.<|endoftext|>
TITLE: The set of non-smooth points of a convex function is (m - 1)-rectifiable
QUESTION [6 upvotes]: I am looking for a reference to the following result. 

Let $f:\mathbb R^m\to\mathbb R$ be a convex function.
  Then $f$ is differentiable at all points of outside of a countable union of $(m-1)$-rectifiable sets.

Comments:

$n$-rectifiable set is an image of Lipschitz map from bounded domain in $\mathbb R^n$
I checked Federer's "Geometric Measure Theory", but I might miss the right place.
Extract from the Greg's answer (for those who are lazy to read the paper): In this paper, it is given a complete characterization of subsets of nondifferatiable points of a convex function. Namely, it is proved that $A$ is a such a set if and only if it can be covered by countably many graphs of DC-functions. ("DC-function" = "Difference of Convex functions".)

REPLY [5 votes]: The paper On the differentiation of convex functions in finite and infinite dimensional spaces by Zajíček primarily deals with the general question in Banach space, but it looks like it has a summary of the situation (as of 1979) that tells you everything that you might want  to know.  It has references to closer papers by Anderson-Klee and Besicovitch.
I just Googled around and eventually got to this paper.<|endoftext|>
TITLE: Dolbeault cohomology of Hopf manifolds
QUESTION [16 upvotes]: This should be straightforward; I'm sorry if it's too much so.  Can someone point me to a reference which computes the Dolbeault cohomology of the Hopf manifolds?
Motivation:  I'd like to work through a concrete example of the Hodge decomposition theorem failing for non-Kähler manifolds.  The textbook I have handy (Griffiths & Harris) doesn't treat this, and the obvious Google search was unhelpful.

REPLY [3 votes]: There are some references where it is computed
D. Mall, {\em The cohomology of line bundles on Hopf manifolds},
Osaka J. Math. {\bf 28} (1991), 999--1015.
D. Mall, {\em Contractions, Fredholm operators and the cohomology
of vector bundles on Hopf manifolds}, Arch. Math., {\bf 66} (1996), 71--76.
M. Ise, {\em On the geometry of Hopf manifolds}, Osaka J. Math.
{\bf 12} (1960), 387--402.
A. Libgober, {\em Cohomology of bundles on homological Hopf mani
folds},  Sci. China Ser. A {\bf 52} (2009), no. 12, 2688--2698.
A. Haefliger: Deformations of transversely holomorphic flows on spheres and deformations
of Hopf manifolds. Compositio Mathematica,55 (1985), 241-251.<|endoftext|>
TITLE: Is there a purely algebraic criterion which characterizes the real algebraic numbers?
QUESTION [7 upvotes]: By the fundamental theorem of algebra, the algebraic closure $\mathbb{K}$ of $\mathbb{Q}$ decomposes as $\mathbb{K} = F \oplus i F$ where $F = \mathbb{R} \cap \mathbb{K}$ (the intersection is in $\mathbb{C}$).  I want to know if there is a purely algebraic way to characterize $F$, i.e. without invoking any analysis, topology, or transcendental number theory.  I am asking this because I noticed that it is often convenient when working with examples in characteristic 0 algebraic number theory to give preference to the real roots of a polynomial, and I am wondering if there is a canonical algebraic way to formulate this preference.  It doesn't seem like an object built out of lots and lots of transcendental extensions should be so fundamental to purely algebraic examples.
Here are some specific questions that I have been playing with.
Is there a purely algebraic way to distinguish between the splitting fields of $x^2 + 2$ and $x^2 - 2$?
Is there a purely algebraic way to distinguish the real root among the three roots of $x^3 - 2$ in a splitting field?
Of course, the relevant algebraic structures can't be invariant under $\mathbb{Q}$ automorphisms.  But I don't see why one can't just be a little bit imaginative.
(Examples edited, changing 1 to 2)

REPLY [2 votes]: As other correspondents have pointed out there is no algebraic way
to distinguish the elements of $\mathbb{Q}^{alg}\cap\mathbb{R}$ but there
is an algebraic way of distinguishing totally real algebraic numbers
(those whose conjugates are all real). An algebraic number $\alpha$
is totally real iff $n-r=1$ where $n$ is the degree of $K=\mathbb{Q}(\alpha)$
and $r$ is the rank of the unit group of the integral closure of
$\mathbb{Z}$ in $K$. This follows from Dirichlet's units theorem.<|endoftext|>
TITLE: Is there a non-trivial knot with trivial Homfly polynomial?
QUESTION [13 upvotes]: We have no topologists on our faculty, and from time to time I get to teach our topology course. I know that there are examples of inequivalent knots with the same Homfly polynomial, and I know that there are non-trivial knots with trivial Alexander polynomial, but I don't know whether the question has been settled as to whether there is a non-trivial knot (or link?) with a trivial Homfly polynomial. I'd like to give my students up-to-date information on this, and being outside the area I don't know quite where to look.

REPLY [6 votes]: All I know is that in their 2003 paper Eliahou, Kauffman and Thistlethwaite
write that they did not find any links with trivial HOMFLY-PT. Although they do find links
with both trivial Jones and Alexander.
http://www.math.uic.edu/~kauffman/ekt.pdf
My guess would be there exist links with trivial HOMFLY-PT but no such knots.
Although as Qiaochu mentions it is still open whether there are knots with trivial Jones, Joergen Andersen claims on his website that there are no knots with trivial
Colored Jones. (a.k.a Jones of all the cables of the knot)
http://home.imf.au.dk/andersen/<|endoftext|>
TITLE: Residues of $1/\zeta$
QUESTION [6 upvotes]: Are there any bounds on residues of $1/\zeta$ in roots of $\zeta$ in critical strip, which may use RH, but do not use the conjecture on simplicity of roots or something similar? I did not find such resuts in Titchmarsh, but I could miss something. Thanks!

REPLY [7 votes]: This is actually a very difficult problem, and currently most results are highly conjectural. It essentially comes down to finding useful bounds on discrete moments of the Riemann zeta function of the form
$$J_k(T) = \sum_{0 < \Im(\rho) < T}{|\zeta'(\rho)|^{2k}},$$
as one can then choose the correct value of $k$ and apply partial summation.
For positive $k$, recent results of Milinovich and Milinovich and Ng show under the Riemann Hypothesis that
$$T(\log T)^{(k+1)^2} \ll J_k(T) \ll T(\log T)^{(k+1)^2 + O(1/\log \log \log T)};$$
and slightly stronger results are known for $k = 0,1,2$ (with the latter two being under the Riemann Hypothesis).
For negative $k$, the situation is much more difficult. A conjecture of Gonek-Hejhal suggests that for all $k > -3/2$,
$$J_k(T) \asymp T(\log T)^{(k+1)^2}$$
and this has been refined significantly by Hughes, Keating, and O'Connell. But it seems out of reach to prove anything significantly useful in this area; the best result so far has been by Gonek, who proved that $J_{-1}(T) \gg T$ assuming the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$. But this isn't useful for most applications, where an upper bound is needed. I believe it is possible to show $J_{-1} \ll T^{2+\varepsilon}$ under the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$, though I don't have a reference for this. Also it is quite possible that this result also holds if we replace $|\zeta'(\rho)|^{-2}$ by $\left|\mathrm{Res}_{s = \rho} \zeta(\rho)^{-1}\right|^2$, though again I don't know of a reference for this.<|endoftext|>
TITLE: Seeking Noetherian normal domain with vanishing Picard group but not a UFD
QUESTION [15 upvotes]: Once again, the question says it all.
My motivation is the article on factorization I am writing.  I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater than one, the obstruction to factoriality is the nonvanishing of the (Weil) divisor class group $\operatorname{Cl}(R)$, not the Picard group $\operatorname{Pic}(R)$ (equivalently, the Cartier divisor class group).
My understanding is that it is equivalent to find a normal Noetherian domain with vanishing Picard group which is not locally factorial.  I would be especially happy to see an example among affine domains, i.e., in which the domain is finitely generated as an algebra over some field.

REPLY [2 votes]: These results hold for any field and for many semigroup rings. The first reference is a paper of mine from 1978 or so in the Canada. Math. J  (I think).
      David F. Anderson<|endoftext|>
TITLE: Consolidation: Aftermathematics of fads
QUESTION [22 upvotes]: From Frank Quinn's THE NATURE OF CONTEMPORARY CORE MATHEMATICS: "Mathematics has occasional fads, but for the most part it is a long-term solitary
activity. 
In consequence the community lacks the customs evolved in physics to deal with
the aftermathematics of fads. If mathematicians desert an area no one comes in
afterwards to clean up. 
Lack of large-scale cleanup mechanisms makes mathematical areas vulnerable
to quality control problems. There are a number of once-hot areas
that did not get cleaned up and will be hard to unravel when the developers
are not available. Funding agencies might watch for this and sponsor
physics-style review and consolidation activity when it happens."
Can you give examples of such once-hot areas in need of consolidation ?

REPLY [2 votes]: It may be unpopular to say this, but the theory of subfactors needs a consolidating account. (I don't think of this theory as a fad, but I do think it may be in danger of being difficult to unravel without some consolidating effort.)
This is a major reason why I asked the question here.
If I'm wrong about this, someone please point me to a reference!!!<|endoftext|>
TITLE: Where can we find Deligne's paper " Theorie de Hodge I"?
QUESTION [6 upvotes]: Where can we find Deligne's paper " Theorie de Hodge I"?

REPLY [25 votes]: Amazingly, ALL the ICM talks since the beginning of time can now be found online at http://mathunion.org/ICM/<|endoftext|>
TITLE: Precise relation between prime number theorem and zero-free region
QUESTION [7 upvotes]: I was wondering about the following, and I was hoping that some expert here could answer, rather than me indulging in a search for a needle in the haystack of formulas in books like Titchmarsch.
Notation:

$\zeta(s)$ is the Riemann zeta function.
$f : \mathbb R^+ \rightarrow (0,1/2)$ is such that $\zeta(s)$ does not vanish between $s = 1+it$ and $s=1 - f(t) + it$.
$\pi(x)$, $Li(x)$ as in wikipedia.

Assuming the above data, suppose the version of the prime number theorem that can be proven is:
$$ \pi(x) = Li(x) + O\left(G(x)\right) $$
Question:

Can G(x) be given a closed form expression showing its precise(if and only if) dependence on $f(t)$?

Heuristics: When $f = 0$, $G(x) = x \mathrm{e}^{-a\sqrt{\ln x}}$ and when $f = 1/2$, $G(x) = \sqrt x \ln x$. So possibly there would be a term like $x^{1-f(x)}$ in a putative expression for $G(x)$.

REPLY [7 votes]: Your heuristic is wrong: $G(x)=x\exp{(-a\sqrt{\log{x}}})$ follows from $f=\frac{c}{\log{(|t|+3)}}$ for some fixed real $c>0$.
I really don't want to tell you the answer, because this is a great exercise!  A big hint: use the "approximate explicit formula"
$\psi(x)=x-\sum_{|\rho|\leq T} \frac{x^{\rho}}{\rho}+O(T^{-1} x \log^2{x}),$ 
bound the sum over zeros trivially given what you know about $f$, and then choose $T$ so that the two error terms balance.<|endoftext|>
TITLE: the hopf invariant of the hopf construction
QUESTION [7 upvotes]: I'm having some trouble with a problem about the Hopf construction, in the exercises for Ch. 4 of Mosher & Tangora.  Given a map $g : S^{n-1} \times S^{n-1}  \rightarrow S^{n-1}$, we get a map $h(g) : S^{2n-1} \rightarrow S^n$ by considering
$S^{2n-1} = S^{n-1} * S^{n-1} = S^{n-1} \times I \times S^{n-1} / \sim$
$S^n = S(S^{n-1}) = S^{n-1} \times I / \sim$
and by putting $h(g)(a,t,b) = (g(a,b),t)$.  There's an easy homotopy invariant $(a,b)$ of the map $g$, the degree of the map when restricted to either factor (times any point of the other $S^{n-1}$).  The problem asks me to show that the Hopf invariant of the map $h(g)$ is $H(h(g))=ab$.  This is defined by $s^2=H(h(g))t$ for generators $s\in H^n(X)$ and $t\in H^{2n}(X)$, where $X=S^n \cup_{h(g)} S^{2n}$ (here $h(g)$ is the attaching map).
I'm trying to mimic a proof in the chapter, which constructs a map with Hopf invariant 2 for even $n$ which is really just the Whitehead square of the identity map.  They use the following diagram ($f$ is the folding map, $F$ is induced from $f$, $g$ is the attaching map to get from $S^n \vee S^n$ to $S^n \times S^n$, the vertical maps are inclusions, and $K=S^n \cup_{fg} e^{2n}$ is the complex in which we need to calculate cup products):
          g                     f
S^{2n-1}  ---->   S^n V S^n ------------> S^n
                      |                    |
                      |                    |
                      |                    | i
                      |                    |
                      V              F     V
S^n x S^n = ( S^n V S^n ) U_g e^{2n} ----->  K

[Edit: Uh oh, how do I get it to put stuff in a uniformly spaced font?  I'm too illiterate to understand the "formatting reference" -- any help??]
Note that $F$ exists because the composition $ifg:S^{2n-1} \rightarrow K$ is nullhomotopic.  We know that $f^*s=s_1 + s_2$, where $s$ and $s_i$ represent the obvious cohomology generators.  Denoting 2n-dim. cohomology generators of $K$ and $S^n \times S^n$ by $t$ and $r$ respectively, since $F$ induces an isomorphism in $H^{2n}$, (we can choose orientations so that) $F^* t=r$.  And by the Kunneth formula, in $H^*(S^n \times S^n)$ we have $s_1 s_2=r$ and $s_1^2=s_2^2=0$.  So now we calculate:
$F^*(s^2) = (F^*s)^2 = (f^*s)^2 = (s_1 + s_2)^2 = s_1^2 + s_1 s_2 + s_2 s_1 + s_2^2 = 2r$
(assuming $n$ is even; otherwise $F^*(s^2)=0$).
So somehow they're using things they know about the cohomology of $S^n \times S^n$ to compute cup products in $H^*(K)$.  This seems like it's possible because the attaching map $fg$ factors through $S^n \vee S^n$.  In the problem I'm stuck on, I can't figure out what should be the analogous factorization.  Alternatively, I was thinking that I could possibly use Poincare duality and just try to find the self-intersection of the copy of $S^n$ inside of $K$, but that seem too silly/gimmicky.  Another fact I know is that $A * B \simeq \Sigma(A \wedge B)$.  But suspension doesn't preserve cup products, so I don't think this could help either...

REPLY [6 votes]: This can be seen as follows in terms of the older geometric definitions of degree and Hopf invariant.
The degree of a smooth map $S^{n-1}\to S^{n-1}$ is the number of points mapping to a given point. More precisely, for a generic point in the codomain the map is transverse to it so it has finitely many points in its preimage and these can be assigned signs according to orientation. The degree is the sum of the signs.
The Hopf invariant of a smooth map $S^{2n-1}\to S^n$ is the linking number of the preimages of two points. More precisely, for a generic point in the codomain the map is transverse to it, making the preimage a closed $(n-1)$-manifold with a trivialized normal bundle and therefore an orientation. The Hopf invariant is the linking number of any two of these.
In the case at hand, you can think of $S^{2n-1}$ as the union of two pieces, $S^{n-1}\times D^n$ and $D^n\times S^{n-1}$, glued along $S^{n-1}\times S^{n-1}$. These two doughnuts are linked once, and I believe you can arrange for the point preimages to be a manifold in $S^{n-1}\times D^n$ that is homologically $a$ times $S^{n-1}\times point$ and a manifold in $D^n \times S^{n-1}$ that is homologically $b$ times $point \times S^{n-1}$.
Added: To connect this with the definition of Hopf invariant of $f:S^{2n-1}\to S^n$ in terms of cup product in $S^n\cup_fD^{2n}$: As intermediary use the following cochain-level description: Let $\alpha$ be a cocycle of $S^n$ whose value on the fundamental class is $1$. Then $f*\alpha$ is $\delta \beta$ for some $(n-1)$-cochain of $S^{2n-1}$. Evaluate the cocycle $\beta\cup f*\alpha$ on the fundamental class of $S^{2n-1}$. The result is independent of choices and equal to the Hopf invariant of $f$. It's also OK to choose $\beta$ so that $\delta\beta$ is a different cocycle in the class of $\alpha$. Now to get from here to the linking number let the $n$-cocycle $\alpha$ be given by a $0$-cycle, a point in $S^n$, so that $f*\alpha$ is given by an $(n-1)$-cycle, the preimage of the point, and the cochain $\beta$ corresponds to an $n$-chain with boundary is that preimage. Cup product corresponds to intersecting one point's preimage with something whose boundary is the other point's preimage, and that's the linking number.<|endoftext|>
TITLE: Shortest/Most elegant proof for $L(1,\chi)\neq 0$
QUESTION [64 upvotes]: Let $\chi$ be a Dirichlet character and $L(1,\chi)$ the associated L-function evaluated at $s=1$. What would be the 'shortest' proof of the non-vanishing of $L(1,\chi)$?
Background: The non-vanishing of $L(1,\chi)$ plays an essential role in the proof of Dirichlet´s theorem on primes in arithmetic progressions. In his "Introduction to analytic number theory", T. M. Apostol gives an elementary proof of the above fact estimating various sums in a few lemmas in the context of a proof of the aforementioned Dirichlet theorem. While his approach has the advantage of being self-contained and not requiring much of a background, it is quite lenghty. In their "Analytic number theory", H. Iwaniec and E. Kowalski remark that in Dirichlet´s original proof the non-vanishing of $L(1,\chi)$ for real Dirichlet characters is a simple consequence of Dirichlet´s class number formula. However, in both approaches it is necessary to distinguish between real and complex Dirichlet characters. Hence my two "sub"-questions:
1) Is there a proof that avoids the distinction between the complex and real case?
2) Are there in general other proof strategies for $L(1,\chi)\neq 0$ that can be considered shorter and/or more elegant than the two mentioned above?

REPLY [3 votes]: An algebraic number theory argument? At least when $\chi(n)$ is the quadratic character $\bmod p$ an odd prime.
Up to a constant it is its own discrete Fourier transform:
$$\chi(n) = \frac{G(\chi)}{p} \sum_{k=1}^p \chi(k) e^{2i\pi kn/p}$$ so that
$$L(1,\chi) =\sum_{n\ge 1}\frac{\chi(n)}{n}= -\frac{G(\chi)}{p}\sum_{k=1}^p \chi(k) \log(1-e^{2i\pi k/p}) $$
$L(1,\chi)=0$ is the same as $$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})= \prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi a k/p})\tag{1}$$
for a quadratic non-residue $a$.
$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi a k/p})$ is the only Galois conjugate of $\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})$.
Whence $(1)$ is the same as $$\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})\in \Bbb{Q}$$
This is impossible because the $p$-adic valuation of $1-e^{2i\pi k/p}$ is $\frac1{p-1}$ so $$v_p\left(\prod_{k=1,\chi(k)=1}^p (1-e^{2i\pi k/p})\right)=\frac12$$<|endoftext|>
TITLE: Finding Two Rainbow Spanning Trees
QUESTION [10 upvotes]: Suppose we have a graph whose edges are coloured. It's not necessarily a proper colouring: a given node may have 0, 1, or several incident edges of a given colour.
Is the following problem NP-complete? Determine whether there are two edge-disjoint spanning trees, such that in each individual tree, no colour appears twice.
I am curious because the variant "determine whether there are two edge-disjoint spanning trees, such that in the union of the trees, no colour appears twice" is solvable in polynomial time, for example using matroid theory.

REPLY [2 votes]: Here are some nice lemmas that you can use:
http://www.cs.elte.hu/egres/qp/egresqp-10-04.pdf
Dave pointed out my mistake, the subgraphs of spanning trees do not have to be trees. So I have no clue about the answer.
Espacially Problem 4 (or 10) seems promising. Take the graph from their construction such that all of its edges have a different color, suppose it has e edges. Then if we allow multigraphs, adding every edge with multiplicity 2n-2-e, all of a different color but same for each edge (thus in total we have 2n-2 colors) shows that your question solvable in P is NP-hard for multigraphs. Am I right? I would guess that with some further tricks you can make a simple graph from this for the variant you asked.<|endoftext|>
TITLE: Origami Constructions: Intersecting two Circles
QUESTION [16 upvotes]: It is well known that every construction that can be performed with compass and straightedge alone can also be performed using origami, see:
R. Geretschlager. Euclidean Constructions and the Geometry of Origami.
Mathematics Magazine 68 (1995), no. 5, 357–371.
If one checks the paper by Geretschlager above, one sees that the construction for intersecting two circles is quite complex. Is there an easier construction?
The axioms for origami can be seen at:
http://origami.ousaan.com/library/conste.html
Background: I teach a course in geometry for future teachers. With this construction (and other easy constructions), it would be fairly clear that every construction that can be performed with compass and straightedge alone can also be performed using origami.
I only have found two origami constructions for intersecting two circles - one in the above reference and another in a forgotten (!!!) reference. It was a strange book written in the 1950s about various geometric constructions.

REPLY [8 votes]: Have you found the papers by Roger Alperin?  He has some very nice articles, especially in regards to relating various construction systems -- in addition to origami and compass-straightedge constructions, there are a variety of others, like the Vieten constructions, Pythagorean constructions,  and even some variants on the origami constructions.  Basically, each additional axiom potentially provides an entirely new class of constructions (which may or may not actually enlarge the set of valid constructions).  Some of these questions end up being very geometric in nature, and others have some fantastic ties to the structure of algebraic numbers.
In any case, you can find several papers of his via Google.  I believe your specific question about intersecting two circles is answered in Section 6 of Alperin's "Mathematical Origami:  Another View of Alhazen's Optical Problem."
Hope that helps.<|endoftext|>
TITLE: Why is this algebra called the q-Weyl algebra?
QUESTION [7 upvotes]: Recall that the (first) Weyl algebra over $\mathbb{C}$ is the algebra generated by $x,y$ with the relation $yx-xy=1$. It can be realized as the algebra of polynomial differential operators in 1 variable, i.e. $\mathbb{C}[x]$ is a faithful representation, where $x$ acts by multiplication by $x$ and $y$ acts by $\frac{\partial}{\partial x}$.
In many places I've seen the q-Weyl algebra defined as the algebra generated by $x,y$ with relations yx-qxy=1, where q is some fixed nonzero scalar. This seems like a natural way to do it, and seems to be quite common.
Now in the notes Introduction to representation theory Etingof et al. define an algebra which they call the q-Weyl algebra. This is the $\mathbb{C}$-algebra generated by $x,x^{-1},y,y^{-1}$ with the relations $xx^{-1} = x^{-1}x = 1$, $yy^{-1}= y^{-1}y = 1$ and $xy=qyx$ where q is some fixed nonzero scalar.
My question then is: What is the reason for the name 'q-Weyl algebra' for the algebra defined by Etingof et al.?

REPLY [18 votes]: Suppose that $x$ and $y$ obey $xy-yx=h$, where $h$ is a central element. Set $X$ and $Y$ to be $e^x$ and $e^y$. For now, don't worry too much about what this exponentiation means. Then $XY=q YX$, where $q=e^h$. 
If we interpret $X$ and $Y$ as operators on functions then $(Xf)(x)=e^x f(x)$ and $(Yf)(x) = f(x+h)$. You can check that $XY$ does indeed equal $q YX$.  So the $q$-commutation relation can be seen as an exponentiation of the standard Weyl relation.<|endoftext|>
TITLE: Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance
QUESTION [9 upvotes]: Problem: Consider a random walk on the lattice $\mathbb{Z}^2$ where on each iteration a particle either stays at its current location or moves to a neighboring vertex with probability 1/5. We start the random walk with one particle at the origin. For each $n \geq 1$ and $x \in \mathbb{Z}^2$ let $p_n(x)$ be the probability of finding the particle at $x$ after $n$ iterations. For two points $x,y \in \mathbb{Z}^2$ let $|\cdot|$ denote the Euclidean distance of $x$ and $y$ via the standard embedding $\mathbb{Z}^2 \subset \mathbb{R}^2$. 
For what $n$ is it true that $|x| \leq |y| \Rightarrow p_n(x) \geq p_n(y)$? What kind of techniques are available to prove statements like this? Barring arithmetic mistakes I have verified this up to n=6 via explicit computation.
Please forgive me if this is actually a trivial question (I know very little about random walks). I would also be very happy with suggested approaches or references.
A Little Motivation/Another Problem: Suppose we list the elements of $\mathbb{Z}^2$ is ascending order by Euclidean distance from the origin, $z_1 \leq z_2 \leq \cdots$, and then set $D_n = \cup_{i=1}^n z_i$. For various reasons I have been dealing with these $D_n$ and would like to consider analogues in other groups. Hence I would very much like to have a "$\mathbb{Z}^2$-intrinsic" characterization of these $\{D_n\}$, i.e. it would be nice to have a characterization of $D_n$ that only used group or graph theoretic statements about $\mathbb{Z}^2$. Most importantly I do not want to mention the specific embedding of $\mathbb{Z}^2$ into $\mathbb{R}^2$.
Note: The $D_n$ are not exactly well defined since there are choices involved in the list $z_1 \leq z_2 \leq \cdots $. So I am actually interested in characterizing them up to the forced ambiguity.

REPLY [3 votes]: As written the statement is false for $n=3$: note that $p_3(2,2) = 0$ but $p_3(3,0) > 0$, while $|(2,2)| < |(3,0)|$.  Similar counterexamples exist for all $n\geq 5$.  So for larger $n$ you would at least need some extra condition about $L^1$ norms to guarantee that you can't have $|x|<|y|$ with $p_n(x)=0$ and $p_n(y)>0$.  I would guess that this would still be too weak, however.<|endoftext|>
TITLE: Shortest Paths on fractals
QUESTION [8 upvotes]: How can one find shortest paths between 2 specified points on fractals, or (since I'm pretty sure this is quite complicated) make useful generalizations about them?
Since the above question is broad, how about this one: What is the general formulation (in a direct equation, recursive formulation, or other form) for distance between 2 points on the sierpinski carpet?
Obviously for some fractals all points are infinite. Identifying these is often easy, but are there any edge cases where it's hard to decide whether all paths are infinite length? And if so, how does one decide?
Edit: This question was inspired, by the way, by this thread on a different website (where it became clear that it was beyond the average math knowledge there).
http://echochamber.me/viewtopic.php?f=3&t=40348#p1618494
That particular post shows paths(whose presence is recursive) in the carpet.

REPLY [2 votes]: For the Sierpinski carpet, using the metric induced by its embedding in $\mathbb{R}^2$ is quite difficult computationally. There are alternative metrics: for fractals arising from hyperbolic iterated function systems (of which the Sierpinski carpet is one), one can identify points via its tops code space, which involves determining which subcopies of the Sierpinski carpet a point lies in. It's not difficult to place a metric on the tops code space of such a fractal, and this in some sense tells us how "close" two points are, in terms of how many iterations of the function system have to occur before the two points are sent to different subcopies of the fractal. It's a bit difficult to make this more precise without going into a fair bit of detail; I'm pretty sure this is all covered in Michael Barnsley's textbook Fractals Everywhere though. Note also of course that this only works for fractals arising from hyperbolic iterated function systems, which includes other fractals like the Koch curve and the Menger sponge.<|endoftext|>
TITLE: Can we make rigorous the 'obvious' characterisation of singular homology?
QUESTION [8 upvotes]: It is a well known and often touted fact that the singular homology groups 'count the k- dimensional holes' in a space (see: How does singular homology H_n capture the number of n-dimensional "holes" in a space? for an explanation), but has anyone succeeded in making this heuristic rigorous?
I am aware of the work of Rene Thom on a related subject (see: Cohomology and fundamental classes), but am dubious that this comes close to what I am asking.
Specifically what I am asking is whether there is or indeed ever could be a theorem of the form:

Suppose $M^n \subset \mathbb{R}^m$ is a hausdorff [insert category here] manifold of dimension n: let $Y_k$ be the set of generators of $H_k(M^n)\otimes \mathbb{Q}$ with $k\neq 0,n$ then for each $[x]\in Y_k$ $\exists x \in [x]$ and a point of $x$, $\hat{x}$, s.t. $\exists$ a convex neighbourhood $U$ of $\hat{x} $ with $U \cap M $ homotopic to $\mathbb{R}^{k+1} \setminus 0$.

(The torsion has been killed in an effort to make this a reasonable request)
If not, why not? (I am aware that there could be an arbitrary number of twists in the codimension of the fundamental class for example, but surely this could be sorted out!) Are there counter examples? Is there a situation where this is true?

Edit: After a counter example to the original conjecture, rather than retreating all the way back to manifolds Let's attempt to defend the intermediate bridge of $U \cap M $ homotopic to $S^{i_1} \times... \times S^{i_p}$ with $\Sigma_j i_j=k$
Edit2: $H_2(\mathbb{R}^3 \setminus T \wedge T)$ scuppers that. Compact orientable smooth manifolds it is then...

REPLY [3 votes]: I think that your theorem is "almost" true if we restrict ourselves to a neighborhood of x in (some quotient of) the k-skeleton. And indeed, this is a way to give a precise meaning to "holes" in singular homology. Let me be more specific.
Let's consider a finite CW-complex. The space is built starting with a finite number of points, and attaching a finite number of cells of various dimensions. The k-skeleton $X_k$ is the union of all cells of dimension less than or equal to k. 
The smash $X_k|X_{k-1}$ is obtained from $X_k$ by identifying all points in $X_{k-1}$. If X is a CW-complex, that "smash" is homeomorphic to a bouquet of k-spheres. These spheres are the "holes" we are looking after. From the standard identification
$H_k(X_k|X_{k-1})\simeq H_k(X_k,X_{k-1})$ , we see that each sphere gives rise to an element in the relative homology group 
$H_k(X_k,X_{k-1})$. There are two ways these elements may fail to give an element in $H_k(X)$. 
-->  Instead of capturing a k-dimensional "hole", the cell may in fact "fill" a (k-1)-dimensional "hole". That's what happens when, for example, we cap a cylinder with a disk. So, we are only interested in elements in the kernel of the boundary operator $\delta_k : H_{k}(X_k,X_{k-1})\rightarrow H_{k-1}(X_{k-1})$.
-->  The k-dimensional "hole" may be filled by some $k+1$-dimensional cell. So we should quotient $H(X_k,X_{k-1})$ by the image of the operator $H_{k+1}(X_{k+1},X_k)\rightarrow H_{k}(X_k)\rightarrow H_{k}(X_k,X_{k-1})$. Let us denote that image by $E_k$.
And it works. The group $H_k(X)$ is actually isomorphic to the quotient $ker\ \delta_k / E_k$. So there is a subset of the "holes" in $X_k|X_{k-1}$ that provide a generating family for $H_k(X_k)$. Arguably, the spherical neighborhood you are looking for exists in the smash, not in X, but still, I think it succeeds in making our intuition rigorous. As a reference, I may point to Greenberg "Algebraic topology, an introductory course" (21.8 ff).<|endoftext|>
TITLE: Naive questions about "matrices" representing endomorphisms of Hilbert spaces.
QUESTION [15 upvotes]: This is a very basic question and might be way too easy for MO. I am learning analysis in a very backwards way. This is a question about complex Hilbert spaces but here's how I came to it: I have in the past written a paper about (amongst other things) compact endomorphisms of $p$-adic Banach spaces (and indeed of Banach modules over a $p$-adic Banach algebra), and in this paper I continually used the notion of a "matrix" of an endomorphism as an essential crutch when doing calculations and proofs. I wondered at the time where more "conceptual" proofs existed, and probably they do, but I was too lazy to find them.
Now I find myself learning the basic theory of certain endomorphisms of complex separable Hilbert spaces (continuous, compact, Hilbert-Schmidt and trace class operators) and my instinct, probably wrong, is to learn the theory in precisely the same way. So this is the sort of question I find myself asking.
Say $H$ is a separable Hilbert space with orthonomal basis $(e_i)_{i\in\mathbf{Z}_{\geq1}}$. Say $T$ is a continuous linear map $H\to H$. Then $T$ is completely determined by its "matrix" $(a_{ij})$ with $Te_i=\sum_ja_{ji}e_j$. But are there completely "elementary" conditions  which completely classify which collections of complex numbers $(a_{ij})$ arise as "matrices" of continuous operators?
I will ask a more precise question at the end, but let me, for the sake of exposition, tell you what the the answer is in the $p$-adic world.
In the $p$-adic world, $\sum_na_n$ converges iff $a_n\to 0$, and life is easy: the answer to the question in the $p$-adic world is that $(a_{ij})$ represents a continuous operator iff 
(1) For all $i$, $\sum_j|a_{ji}|^2<\infty$ (equivalently, $a_{ji}\to 0$ as $j\to\infty$), and 
(2) there's a universal bound $B$ such that $|a_{ij}|\leq B$ for all $i,j$.
[there is no inner product in the $p$-adic case, so no adjoint, and the conditions come out being asymmetric in $i$ and $j$].
See for example pages 8--9 of this paper of mine, although of course this isn't due to me---it's in Serre's paper on compact operators on $p$-adic Banach spaces from the 60s---see Proposition 3 of Serre's paper. In particular, in the $p$-adic world, one can identify the continuous maps $H\to H$ (here $H$ is a $p$-adic Banach space with countable ON basis $(e_i)$) with the collection of bounded sequences in $H$, the identification sending $T$ to $(Te_i)$.
In the real/complex world though, the analogue of this result fails: the sequence $(e_1,e_1,e_1,\ldots)$ is a perfectly good bounded sequence, but there is no continuous linear map $H\to H$ sending $e_i$ to $e_1$ for all $i$ (where would $\sum_n(1/n)e_n$ go?). 
Let's consider the finite rank case, so $T$ is a continuous linear map $H\to H$ with image landing in $\mathbf{C}e_1$. Then by Riesz's theorem, $T$ is just "inner product with an element of $H$ and then multiply by $e_1$". Hence we have an additional condition on the $a_{ij}$, namely that $\sum_j|a_{ij}|^2<\infty$. Furthermore a continuous linear map is bounded, as is its adjoint.
This makes me wonder whether the following is true, or whether this is still too naive:
Q) Say $(a_{ij})$ $(i,j\in\mathbf{Z}_{\geq1})$ is a collection of complex numbers satisfying the following:
There is a real number $B$ such that
1) For all $i$, $\sum_j|a_{ij}|^2\leq B$
2) For all $j$, $\sum_j|a_{ij}|^2\leq B$
Then is there a unique continuous linear map $T:H\to H$ with $Te_i=\sum_ja_{ji}e_i$?
My guess is that this is still too naive. Can someone give me an explicit counterexample? Or, even better, a correct "elementary" list of conditions characterising the continuous endomorphisms of a Hilbert space?
On the other hand, it clearly isn't a complete waste of time to think about matrix coefficients. For example there's a bijection between Hilbert-Schmidt operators $T:H\to H$ and collections $(a_{ij})$ of complexes with $\sum_{i,j}|a_{ij}|^2<\infty$, something which perhaps the experts don't use but which I find incredibly psychologically useful.

REPLY [5 votes]: So maybe the following is a counterexample to Kevin's original post. (It was created by computer scientist and friend Erik Vee as a "counterexample" to that exercise 3.14 in Zimmer (which says that a bounded operator is compact iff $a_{ij}$ goes to 0 as i and j go to $\infty$); and it was Robert Pollack who figured out that the reason it's not a counterexample is that it doesn't represent a continuous operator. So my role here is transcriber only.)
Define the matrix as follows. The first column is (1, 0, 0, ...), the second is (1/2, 1/2, 1/2, 1/2, 0, 0, ...), the nth has $\frac 1 n$ appearing $n^2$ times, followed by all zeros. Then $\ell_2$-norm of each column is exactly 1, and the $\ell_2$-norm of each row is bounded by $\sqrt{\frac{\pi^2}{6}}$. So this matrix has bounded $\ell_2$ rows and columns as necessary. 
But this matrix cannot represent a continuous operator. If it did, then, since it satisfies Kevin's/Zimmer's criterion, this operator -- call it $A$ -- would be compact, and hence a uniform limit of the operators $A_n$ given by the first n rows of $A$. But the operator $A - A_n$ has, in its nth column, $\frac 1 n$ appearing $n^2 - n$ times, which means that that column's $\ell_2$ norm is $1 - \frac 1 n$, which is bounded away from zero.
It's still unclear to me if this matrix represents an unbounded linear map, or if it doesn't represent a well-defined map at all.<|endoftext|>
TITLE: Torsors for monoids
QUESTION [6 upvotes]: Torsors are defined as a special kind of group action. I am wondering whether the analogous notion exists for monoid actions. Some references would be helpful.
In general I'm interesting in the notion of 'subtraction/division' induced by having a torsor. My application is in computer science. A monoid is used to capture modifications to a computer program and the monoid action corresponds to performing the modification on the program. If I have a torsor-like entity, I can take two software entities and produce the modification required to convert one into the other.
The answer is that any such monoid will automatically be a group. In my application, it seems that I will only get close to the notion of torsor if my modifications have inverses, which they do not.

REPLY [9 votes]: This will be an expansion on Ben Steinberg's answer (which unfortunately I saw only after I was in the middle of compiling this answer). 
For a group $G$, another point of view on $G$-torsors is that they are classified by $BG$, which here denotes the topos of $G$-sets $Set^G$. (This should resonate with the meaning of the classifying space $BG$ in algebraic topology, which classifies $G$-principal bundles.) The meaning of this in topos theory is that geometric morphisms $Set \to Set^G$ ("points" of $BG$) are equivalent to torsors over $G$, and more generally that a $G$-torsor as interpreted in any topos $E$ (not just $Set$) is equivalent to a geometric morphism $E \to Set^G$. 
Equivalently, left exact left adjoints $Set^G \to Set$ are equivalent to torsors; by a result known as Diaconescu's theorem (see Mac Lane and Moerdijk's book on topos theory), these in turn are equivalent to flat functors $G \to Set$, i.e., a torsor as $G$-set is essentially the same as a filtered colimit of copies of the representable $G$-set. 
We could if we like expand the meaning of "torsor" by adopting this as a definition: if $C$ is a category, then a $C$-torsor is a functor $C \to Set$ obtained as a filtered colimit of representables $\hom_C(c, -): C \to Set$. 
I realize this may seem highly abstract, so it's worth seeing what this means for a monoid and bringing this back down to earth. As an example, consider the monoid $\mathbb{N}$. We are trying to understand filtered colimits of copies of $\mathbb{N}$ as $\mathbb{N}$-sets. An example is the colimit of the filtered diagram 
$$\mathbb{N} \stackrel{s}{\to} \mathbb{N} \stackrel{s}{\to} \ldots$$ 
where $s$ is the successor function. Here the colimit is $\mathbb{Z}$ with the "standard" action $\mathbb{N} \times \mathbb{Z} \to \mathbb{Z}$ given by addition. This is isomorphic to the "non-standard" action $\mathbb{N} \times \mathbb{Z} \to \mathbb{Z}$ given by subtraction, $(n, m) \mapsto -n+m$. 
For commutative monoids at least, I think this is a somewhat typical example: filtered colimits of copies of the representable are isomorphic to those obtained by formally inverting some set of elements. (Cf. the fact that localizations of commutative rings are flat.) I am less sure about the general case of non-commutative monoids.<|endoftext|>
TITLE: The word "torsion" and its connection to geometry and homology
QUESTION [8 upvotes]: In an $R$-module $M$, an element $m \in M$ is said to be torsion if $am = 0$ for some $a \in R$ with $a \neq 0$.
Also, for a non-orientable (closed) surface such as the projective plane or the Klein bottle, there is a $\mathbb Z/2$ term in the first homology. This part is said to detect the "twisting" in the surface. 
So this leads to the plausible notion that the origin of the word "torsion" in algebra is related to this "torsion" or "twisting" from topology.
Unfortunately the difficulty is that this is not exactly true. For example, the Möbius band is surely a twisted object. But it deformation retracts to the circle, and therefore its homology is very normal.
I hope somebody can shed more light on this terminology.

REPLY [6 votes]: See Stillwell's Classical Topology and Combinatorial Group Theory, pp. 170--171 for a discussion of the historical origins of torsion, which of course is in algebraic topology, not abstract algebra. Find the title on Google books and search for the word torsion.<|endoftext|>
TITLE: Geometric interpretation of filtered rings and modules
QUESTION [14 upvotes]: Let $A$ be a commutative algebra, say over $\mathbb{C}$. 
Giving a grading on $A$ corresponds at least morally to giving a $\mathbb{C}^*$ action on spec(A): $A_i$ can be thought of as  those functions on which $t$ acts by multiplication with $t^i$. Similary a graded $A$ module is just a $\mathbb{C}^*$ equivariant sheaf.
Now I want to know, if there is also a geometric interpretation of filtered rings/modules.

REPLY [23 votes]: To a filtered algebra $(A,F)$ one can assign its Rees algebra $R=\bigoplus_i F_iA$.  It is a graded algebra containing the algebra of polynomials in one variable $\mathbb{C}[t]$ naturally embedded as the subalgebra generated by the element $t\in R_1$ corresponding to the element $1\in F_1A$.  So the algebra $R$ defines a $\mathbb{C}^*$-equivariant quasi-coherent sheaf of algebras $\mathcal{R}$ over the affine line $\operatorname{Spec}\mathbb{C}[t]$.  The algebra $A$ can be recovered as the fiber of $\mathcal{R}$ at the point $t=1$, and the associated graded algebra $\operatorname{gr}_FA$ is the fiber of $\mathcal{R}$ at $t=0$.  Filtered $A$-modules correspond to $\mathbb{C}^*$-equivariant quasi-coherent sheaves of modules over $\mathcal{R}$.
The algebra $R$ is a torsion-free $\mathbb{C}[t]$-module, so the quasi-coherent sheaf $\mathcal{R}$ over $\operatorname{Spec}\mathbb{C}[t]$ has to be torsion-free.  This description does not take into accout the issue of completeness of the filtration $F$ (in case it extends also in the decreasing direction), which requires a separate consideration.<|endoftext|>
TITLE: What is the minimal size of a partial order that is universal for all partial orders of size n?
QUESTION [40 upvotes]: A partial order $\mathbb{B}$ is universal for a class $\cal{P}$ of partial orders if every order in $\cal{P}$ embeds
order-preservingly into $\mathbb{B}$.
For example, every partial order
$\langle\mathbb{P},\lt\rangle$ maps order-preservingly into
its power set by the map 
$$p\mapsto\{q\in\mathbb{P}\mid q\leq p\}$$ that sends each element $p$ to its
lower cone.
Thus, the power set order $\langle
P(\{1,2,\ldots,n\}),{\subseteq}\rangle$ is universal for
the class of partial orders of size $n$. This provides an
order of size $2^n$ that is universal for orders of size
$n$.
Question. What is the minimal size of a partial
order that is universal for orders of size $n$?
In particular, is there a polynomial upper bound? 
One can make at least slight improvements to the $2^n$
upper bound, by observing that the emptyset was not needed,
as it never arises as a lower cone, and we don't need all
the atoms, since if they are needed, then one can use the
co-atoms instead. I suspect that there is a lot of waste in
the power set order, but the best upper bound I know is
still exponential.
For a lower bound, my current knowledge is weak and far
from exponential. Any order that is universal for orders of
size $n$ will contain a chain and an antichain, making it
have size at least $2n-1$. (That bound is exact for $n\leq
3$.) A student in my intro logic course extended this to
$n\log(n)$ by considering $k$ chains (and antichains) of size
$n/k$.
Can one find better lower bounds?
Interestingly, the same student observed that we cannot in
general expect to find unique smallest universal orders,
since he found several orders of size 5 that are
universal for orders of size 3 and which are minimal with
that property. So in general, we cannot expect a unique
optimal universal order. Does this phenomenon occur for
every $n$? (He also found minimal universal orders of size larger than the minimal size universal order.)

REPLY [6 votes]: We proved in this paper https://arxiv.org/abs/2012.01764 that the
answer is $2^{n/4+o(n)}$. As observed in the other answers, a counting
argument shows that this is optimal (up to the lower order term).

Update: May 7, 2021. Unfortunately there was a flaw in our proof and we have to take back this answer. The main result of the paper is still valid but we cannot obtain the universal poset as a direct consequence of our comparability labelling scheme for posets.
Very sorry, feel free to down-vote this answer !<|endoftext|>
TITLE: Relevance of the complex structure of a function algebra for capturing the topology on a space.
QUESTION [11 upvotes]: This question is the outcome of a few naive thoughts, without reading the proof of Gelfand-Neumark theorem.
Given a compact Hausdorff space $X$, the algebra of complex continuous functions on it is enough to capture everything on its space. In fact, by the Gelfand-Neumark theorem, it is enough to consider the commutative C*-algebras instead of considering compact Hausdorff spaces.
The important thing here is that $C*$-algebras have a complex structure. The real structure is not enough. Given the algebra $C(X, \mathbb R) \oplus C(X, \mathbb R)$ of real continuous functions on $X$, the algebra $C(X, \mathbb{C})$ is simply the direct sum $C(X, \mathbb R) \oplus C(X, \mathbb R)$, as a Banach algebra(and this can be given a complex structure, (seeing it as the complexification...)). But to obtain a C*-algebra, we need an additional C*-algebra, and the obvious way, ie, defining $(f + ig)$* $= (f - ig)$ does not work out. More precisely, the C* identity does not hold.
So one cannot weaken (as it stands) the condition in the Gelfand-Neumark theorem that we need the algebra of complex continuous functions on the space $X$, since we do need the C* structure. Of course, this is without an explicit counterexample. Which brings us to:

Qn 1. Please given an example of two non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that the function algebras $C(X, \mathbb R)$ and $C(Y, \mathbb R)$ are isomorphic(as real Banach algebras)?

(Here I am hoping that such an example exists).
Then again,

Qn 2. From the above it appears that the structure of complex numbers is involved when the algebra of complex functions captures the topology on the space. So how exactly is this happening?

(The vague notions concerning this are something like: the complex plane minus a point contains nontrivial $1$-cycles, so perhaps the continuous maps to the complex plane might perhaps capture all the information in the first homology, etc..)..
Note : Edited in response to the answers. Fixed the concerns of Andrew Stacey, and changed Gelfand-Naimark to Gelfand-Neumark, as suggested by Dmitri Pavlov.

REPLY [6 votes]: Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff.  For each closed subset $K$ of $X$, define $\mathcal{I}_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$.  The map $K\mapsto\mathcal{I}_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$.  Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required.  I remember doing this as an exercise in Douglas's Banach algebra techniques in operator theory in the complex setting, but the same proof works in the real setting.

Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof.  Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)
I will take it for granted that each $\mathcal{I}_K$ is a closed ideal.  This doesn't require that the space is Hausdorff (nor that $K$ is closed).  Suppose that $K_1$ and $K_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K_2\setminus K_1$.  Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K_1$ but $f(x)=1.$  Thus, $f$ is in $\mathcal{I}_{K_1}\setminus\mathcal{I}_{K_2}$, and this shows that $K\mapsto \mathcal{I}_K$ is injective.  The work is in showing that it is surjective.
Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K_\mathcal{I}=\cap_{f\in\mathcal{I}}f^{-1}(0)$, so that $K_\mathcal{I}$ is a closed subset of $X$.  Claim: $\mathcal{I}=\mathcal{I}_{K_\mathcal{I}}$.  
It is immediate from the definition of $K_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K_\mathcal{I}$, so that $\mathcal{I}\subseteq\mathcal{I}_{K_\mathcal{I}}.$  Let $f$ be an element of $\mathcal{I}_{K_\mathcal{I}}$. Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|_\infty<3\epsilon$.  Define $U_0=f^{-1}(-\epsilon,\epsilon)$, so $U_0$ is an open set containing $K_\mathcal{I}$.  For each $y\in X\setminus U_0$, because $y\notin K_\mathcal{I}$ there is an $f_y\in \mathcal{I}$ such that $f_y(y)\neq0$.  Define $$g_y=\frac{f(y)}{f_y(y)}f_y$$ and $U_y=\{x\in X:|g_y(x)-f(x)|<\epsilon\}$.  Then $U_y$ is an open set containing $y$.  The closed set $X\setminus U_0$ is compact, so there are finitely many points $y_1,\dots,y_n\in X\setminus U_0$ such that $U_{y_1},\ldots,U_{y_n}$ cover $X\setminus U_0$.  Relabel: $U_k = U_{y_k}$ and $g_k=g_{y_k}$.  Let $\varphi_0,\varphi_1,\ldots,\varphi_n$ be a partition of unity subordinate to the open cover $U_0,U_1,\ldots,U_n$.  Finally, define $g=\varphi_1 g_1+\cdots+\varphi_n g_n$.  That should do it.
In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points.  (Maximal ideals are actually always closed in a Banach algebra, real or complex.)<|endoftext|>
TITLE: Representing repeated structure in graphs
QUESTION [9 upvotes]: Apologies for not knowing exactly what I'm looking for, but I'd appreciate any general pointers to get me started.
I'm interested in efficient representations (graphical and otherwise) of finite graphs with repeated structure --- something like coding theory for graph structures.  For example, an $N$ by $N$ grid graph where each non-boundary vertex has four neighbors (up, right, left, down) could be represented with far fewer than O($N^2$) parameters.  A $N$ by $N$ grid where each non-boundary vertex has 8 neighbors (up left, up, up right, ...) should take more parameters to represent, but not many more.  I'm imagining a notation or diagram that uses a base template, then defines a local connectivity pattern and a repetition structure.
I'd like to be able to ask (and start to answer) questions like:

What is the minimum number of parameters needed to describe a particular graph structure?
Are there some problems that become fixed-parameter tractable for graphs that can be described with $k$ parameters?

Is there a branch of graph theory that studies these types of questions?  I'd be grateful for pointers to starting points or related work.

REPLY [5 votes]: Once you have found the smallest description of a graph (in your favored scheme), then a philosophically minded person will say, Yes, but how about the smallest description-of-a-description of a graph? And how about the smallest description-of-a-description-of-a-description? Why stop?
These ideas will lead you to the concept of Kolmogorov complexity, where one measures the complexity or information content of a mathematical object by the smallest computational description of it, essentially the size of the smallest program that generates the object. The subject is a part of the emerging theory of computational concepts of randomness, an extremely active current area of research in computability theory. The theory is usually thought of as applied to strings: a string is incompressible if the shortest computational description of it has the same size as itself (thus, the most efficient way to describe it is by explicitly listing it out). Compressible strings, in constrast, have comparatively low information density, since they are describable by a much smaller object. Thus, the graphs you seek to describe are exactly the graphs that are compressible with respect to Kolmogorov complexity. 
Thus, I propose that we measure the complexity of a graph on vertices $\{1,2,\ldots,n\}$ by the size of the smallest program able to compute the edge relation. (Let us fix for this purpose a notion of computability, such as Turing machines.) This would correspond to the Kolmogorov complexity, and it will be an extremely robust notion of the measure of the complexity of description of your graph. The class of graphs you have in mind are those that are compressible with respect to this measure, computed by a program that is strictly smaller than the program that simply stores the edge relation in state memory.
There is a small paradox in the subject of computational randomness, since ordinarily one might think of a random string as containing very little information, but in this subject, such strings are incompressible, as it is difficult to describe them exactly except by listing them out explicitly. In this sense, therefore, random strings contain a huge amount of information. Similarly, random graphs are hard to describe in any other way other than by listing the edge relation explicitly.
Because of these ideas, your question (perhaps in extreme form) may ultimately have more to do with logic than with graph theory. Almost any mathematical object can be coded into a graph (in a precise sense, every mathematical structure is interpretable inside a graph), and all such mathematical objects can be ultimately described by strings, to which the Kolmogorv ideas apply. You are asking about are the graphs that are compressible in the sense of Kolmogorov complexity. On cardinality grounds, of course, most graphs are not like this. In general, one should expect Turing noncomputability issues to arise, since the question of whether a given string is compressible is undecidable. Similarly, the question of whether a given finite graph is incompressible is undecidable.<|endoftext|>
TITLE: Classification of adjoint orbits for orthogonal and symplectic Lie algebras?
QUESTION [8 upvotes]: This might be standard, but I have not seen it before:
Let $K$ be an algebraically closed field (of characteristic 0 if necessary). Let $G$ be the orthogonal group ${\bf O}(m)$ or the symplectic group ${\bf Sp}(2n)$, and let $\mathfrak{g}$ be its Lie algebra. Is there a classification of the adjoint orbits of $G$ acting on $\mathfrak{g}$? An answer for the special orthogonal group ${\bf SO}(m)$ is also welcome. I am aware of the classification of nilpotent and semisimple orbits.
I am looking for an answer kind of like Jordan normal form in the case that $G$ is the general linear group. My guess is that the classification is Jordan normal form plus some other invariant.

REPLY [4 votes]: There's a paper by Burgoyne and Cushman in the Journal of Algebra which answers this for the real field, the complex field and finite fields. My recollection is that there is a paper by Milnor for perfect fields.
Here are the references:
MR0432778 (55 #5761)  Burgoyne, N. ;  Cushman, R.  Conjugacy classes in linear groups.
 J. Algebra  44  (1977),  no. 2, 339--362.
MR0249519 (40 #2764)  Milnor, John . On isometries of inner product spaces.
 Invent. Math.  8  1969 83--97.<|endoftext|>
TITLE: Creating high quality figures of surfaces
QUESTION [25 upvotes]: I am not sure if this question is suitable for mo, it is more about visualization than math. Anyway, here it is:
What is the best way to visualize a 2-surface in Euclidean space with high quality?
Of course Maple or Matlab produce some grapical output but if one is interested in high quality figures, these methods are insufficient.
I am currently using the following procedure (POV-Ray is a free rendering software based on C): 

produce the surface with Matlab (or C) and store the surface as a triangle mesh.
write out the triangle mesh to a Povray file.
produce parameter curves with Matlab (or C)
write out the parameter curves (as a union of cylinders) to a Povray file.

This produces very nice figures but suffers from a lack of interactivity. For instance the camera position has to be specified a-priori in Povray.
My question: what do you use? Is there a better method?

REPLY [4 votes]: Please excuse me in offering as a possible answer a software that I developed/am developing myself, namely 
asxp 
which stands for "algebraic surface explorer". 
It can 

create living views of an algebraic surface under parameter change, 
produce Floyd-Steinberg dithered grayscale images of surfaces
create images which are cross hatched along the principal directions of curvature
contour images
triangularize surfaces, smoothen and reduce the triangularization
produce STL models ready for 3D printing (with the help of an auxiliary program, "renderstl" that I wrote too)

The program uses QT, CGAL, GNU GTS and CUDA. At the moment it is in late prototypical state and will probably soon be published, maybe as open-source-software.
More information and a demo video is on
http://www.aviduratas.de/asxp.html
Below are two cross-hatched images:
A certain quartic surface:

A torsal algebraic surface (note the straight lines in the cross hatch):

The curves on the cross-hatched surfaces are available in vector form and would therefore be usable for engraving or pen-plotting.<|endoftext|>
TITLE: Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance Part II
QUESTION [7 upvotes]: For some context see Random Walks in $Z^2$/$Z^2$-intrinsic characterization of Euclidean distance
As per Noah's answer and JBL's comment this was false as stated. However, I think the following reformulation is interesting. 
As before we consider a random walk on $\mathbb{Z}^2$ where a particle either stays at its vertex or moves to a neighbor with probability 1/5. We start the process with a particle at the origin.  For $x \in \mathbb{Z}^2$ we let $p_n(x)$ denote the probability that we find the particle at $x$ after $n$ iterations. Let $\left|\cdot\right|$ denote the Euclidean distance of two points in $\mathbb{Z}^2$ via the standard embedding of $\mathbb{Z}^2 \subset \mathbb{R}^2$. 
Now for the reformulated question: For each $n$, let $C_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| \leq \left|y\right| \Rightarrow p_n(x) \geq p_n(y)$. }
$$
Does $\lim_{n\to\infty} C_n = \infty$? If so, how fast does this diverge? 
EDIT: As per George Lowther's comment, I now find it quite probable that $\lim\inf_{n\to\infty} C_n \leq 5$ if not $C_n = 5$ for all large $n$.
A natural attempt to salvage the question is the following: For each $n$, let $\tilde{C}_n$ be the supremum over all $C > 0$ so that for all $x,y \in \mathbb{Z}^2$ we have 
$$
\text{$\left|x\right|, \left|y\right| \leq C$ and $\left|x\right| < \left|y\right| \Rightarrow p_n(x) > p_n(y)$. }
$$
Again we ask if $\lim_{n\to\infty} \tilde{C}_n = \infty$ and if so, how fast this diverges.

REPLY [8 votes]: As I mentioned in my comment - what you are suggesting implies that the probability of being at (3,4) is the same as being at (5,0) for all large $n$. That seems unlikely, and would guess that $C_n=5$ for $n$ large.

The answer to your modified question is yes! $\tilde C_n$ tends to infinity as $n$ goes to infinity. (Phew! It took me a couple of revisions to prove this, but hopefully the calculations below are now correct). 
In fact, $\tilde C_n\ge c\sqrt{n}$ for some positive constants c. I think that you can also show that $\tilde C_n\le C\sqrt{n}$ for some other constant $C$ but I'm not completely sure yet, although it should follow from a closer examination of my expression below for $p_n(x)$.
You can derive an asymptotic expansion for $p_n(x)$ in 1/n. Evaluating this to second order is enough to answer your question. After $n$ steps the distribution of the particle will be approximately normal with variance 2n/5 in both dimensions, so we expect to get $p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}$ to leading order.
The idea is to note that you are repeatedly applying a linear operator,
$$
p_{n+1}=Lp_n,\ Lp(x) \equiv (p(x)+p(x-e_1)+p(x+e_1)+p(x-e_2)+p(x+e_2))/5
$$
where $e_1=(1,0)$, $e_2=(0,1)$. In finite dimensional spaces, you would solve this by decomposing $p_0$ into a sum of eigenvectors and for large n, the dominant term of $L^np_0$ will be that corresponding to the largest eigenvalue. In this case, the infinite dimensional operator $L$ has a continuous spectrum, and is diagonalized by a Fourier transform.
$$
p_0(x)=1_{\lbrace x=0\rbrace}=\int_{-[\frac12,\frac12]^2}e^{2\pi ix\cdot u}\,du.
$$
Noting that $e^{2\pi ix\cdot u}$ (as a function of $x$) is an eigenvector of $L$,
$$
Le^{2\pi ix\cdot u}=\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)e^{2\pi ix\cdot u}
$$
gives the following for $p_n$,
$$
p_n(x)=L^np_0(x)=\int_{[-\frac12,\frac12]^2}\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)^ne^{2\pi ix\cdot u}\,du.
$$
The term inside the parentheses is less than 1 in absolute value everywhere away from the origin, so looks like a Dirac delta when  raised to a high power n. Using a Taylor expansion to second order,
$$
\left(\frac15+\frac25\cos(2\pi u_1)+\frac25\cos(2\pi u_2)\right)^n
=e^{-\frac45\pi^2n\vert u\vert^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u_1^2u_2^2)+O(n\vert u\vert^6)\right).
$$
This expansion is valid over any domain on which $n\vert u\vert^6$ is bounded. Say, $\vert u\vert\le n^{-1/6}$. Outside of this domain, the integrand above is bounded by $e^{-cn(n^{-1/6})^2}=e^{-cn^{2/3}}$ for a constant c, which is much smaller than O(1/n^3) and can be neglected. Then,
$$
p_n(x)=\int_{\mathbb{R}^2}\left(1+\frac{8\pi^4n}{75}(7\vert u\vert^4-20u_1^2u_2^2)+O(n\vert u\vert^6)\right)e^{-\frac45\pi^2n\vert u\vert^2+2\pi ix\cdot u}\,du.
$$
Here I not only substituted in the second order approximation to the integrand, but also extended the range of integration out to infinity. This is fine, because it can be shown that the value of this integral over $\vert u\vert\ge n^{-1/6}$ has size of the order of no more than $e^{-cn^{2/3}}$, so vanishes much faster than $O(1/n^3)$. Substituting in $v=\sqrt{\frac{8n}{5}}\pi u$ also shows that the $O(nu^6)$ term in the integrand vanishes at rate $1/n^3$, giving the following.
$$
p_n(x)=\frac{5}{8\pi^2n}\int_{\mathbb{R}^2}\left(1+\frac{1}{24n}(7\vert v\vert^4-20v_1^2v_2^2)\right)e^{-\frac12\vert v\vert^2+i\sqrt{\frac{5}{2n}}x\cdot v}\,dv+O(n^{-3}).
$$
This integral can be computed,
$$
p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{1}{24n}\left(36-\frac{90}{n}\vert x\vert^2+\frac{175}{4n^2}\vert x\vert^4-\frac{125}{n^2}x_1^2x_2^2\right)\right)+O(n^{-3}).
$$
This is a bit messy, but the exact coefficients are not too important. What matters is the general form of the expression.
The leading order term also agrees with the guess above based on it being approximately normal.
Also, for any fixed $\vert x\vert \lt\vert y\vert$, the leading order term in $p_n(x)-p_n(y)$ will dominate for large n, giving $p_n(x)\gt p_n(y)$. So, $\tilde C_n\to\infty$.
Consider $\vert x\vert\le c\sqrt{n}$ for some $c\le1$. Then,
$$
p_n(x)=\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+\frac{3}{2n}\right)+O(c^2n^{-2}).
$$
If $\vert x\vert\lt\vert y\vert\le c\sqrt{n}$ then $\vert y\vert^2-\vert x\vert^2\ge 1$ (as it is a nonzero integer)
$$
\begin{align}
p_n(x)-p_n(y)&=\frac{5}{4\pi n}\left(1+\frac{3}{2n}\right)e^{-\frac{5}{4n}\vert x\vert^2}\left(1-e^{-\frac{5}{4n}(\vert y\vert^2-\vert x\vert^2)}\right)+O(c^2n^{-2})\\
&\ge\frac{5}{4\pi n}e^{-\frac{5}{4n}\vert x\vert^2}(1-e^{-\frac{5}{4n}})+O(c^2n^{-2})\\
&=\frac{25}{16\pi n^2}e^{-\frac{5}{4n}\vert x\vert^2}\left(1+O(c^2)\right).
\end{align}
$$
As long as $c$ is chosen small enough that the $O(c^2)$ term is always greater than -1, this expression will be positive. So $p_n(x)\gt p_n(y)$ for all $\vert x\vert\lt\vert y\vert\le c\sqrt{n}$, giving $\tilde C_n\ge c\sqrt{n}$.<|endoftext|>
TITLE: Is the Euler characteristic a birational invariant
QUESTION [13 upvotes]: Suppose that $X$ and $Y$ are smooth projective varieties which are birationally equivalent. I would like to have  that $$\textrm{deg} \ \textrm{td}(X) = \textrm{deg} \ \textrm{td}(Y).$$ Invoking the Hirzebruch-Riemann-Roch theorem, this boils down to showing that $$ \chi(X,\mathcal{O}_X) = \chi(Y,\mathcal{O}_Y).$$ 
This is probably a basic fact. A stronger statement is apparently shown in Birationale Transformation von linearen Scharen auf algebraischen Mannigfaltigkeiten by van der Waerden. The only problem is that I can't seem to find it in that article (probably because I don't read that well German).
For $\dim X =2$ one can prove this as Hartshorne does as follows.
Any birational transformation of nonsingular projective curves can be factored into a sequence of monoidal transformations and their inverses. For such a monoidal transformation, the result follows from Proposition 3.4 in Chapter V of Hartshorne. 
Does this work in the general case?

REPLY [17 votes]: The dimensions $h^i(\mathcal O_X)$ of the cohomology groups of $\mathcal O_X$, and thus the Euler characteristic, are birational invariants of smooth proper varieties in positive characteristic as well, by a recent work of Andre Chatzistamatiou and Kay Rülling. It is not published yet but a preprint is available.<|endoftext|>
TITLE: Introducing Cryptology to Undergraduates
QUESTION [11 upvotes]: This summer I am going to give some lectures to some REU students. I am still tossing around ideas for what I am going to talk about, but one thing I would at least like to give one or two lectures on, is Cryptology.
I had a fairly standard undergraduate course on Number Theory where we learned basic cyphers and some things about encryption. However, I am hoping to talk about the relationship of elliptic curves to encryption. Is there an appropriate level book that covers this relationship? Many of the students are strong, but lack some background. Many will have some experience with number theory, but may lack Abstract Algebra and Advanced Calculus.
In the absence of a nice book talking about elliptic curves relation to cryptology, I will probably talk about the excellent book by Ash and Gross.
I was just hoping to add this topic into my mix of lectures so I thought the MO community could offer some suggestions.
Thanks in advance! 
EDIT: I wanted to add that Diffe-Hulman will definitely be covered as one of the main research projects will focus on it. The elliptic curves comes by request of the students, who have heard cool things about them :D

REPLY [4 votes]: If it's not too late, I'll mention A Course in Number Theory and Cryptography, by Neil Koblitz.  It has a good, undergraduate-level treatment of elliptic curves.<|endoftext|>
TITLE: U(3) Sato-Tate measure.
QUESTION [10 upvotes]: An undergraduate is performing some computations, related to a Sato-Tate conjecture of $U(3)$ type (a curve over $Q$, for which the roots of local L-functions look like eigenvalues of a random matrix in $U(3)$ and their complex conjugates, scaled appropriately).
There are well-known results in the literature, on random matrices from $U(3)$, with respect to Haar measure.  For example, the expected value of the trace is zero (obviously), and the expected value of the square of the trace is one (not-so-obviously, but this is what I recall).  The Weyl integration formula allows one to express other expectations as definite integrals.  Results of Diaconis and others give a few moments, but most results I find look at $U(N)$ as $N$ grows large.
Does anyone know a precise elementary formula for the distribution of traces of a random matrix in $U(3)$?  Or a good reference?
For example, the trace of a random matrix $g$ in $SU(2)$ is between $-2$ and $2$.  The probability distribution of $Tr(g) = t$ is given, up to some normalization, by:
$$P(t) = \sqrt{1 - \left( \frac{t}{2} \right)^2 },$$
The real part of the trace of a random matrix $g$ in $U(3)$ is between $-3$ and $3$.  Can anyone write the probability distribution of $Tr(g) = z$, in as simple a fashion?  Of course, here the traces can be complex -- I'd be happy also to find the probability that $Re(Tr(g)) = t$, as an elementary function of $t$.

REPLY [3 votes]: If the literature is such a poor state, I'm a little puzzled as to why. 
Maybe for SU(3) it should be clear in principle what is happening? The trace is an invariant of conjugacy classes, and the Weyl integration formula should reduce any issue to an integral over a two-dimensional region. Namely the maximal torus T is acted on by the Weyl group W, and apart from stuff of Haar measure zero the quotient can be represented by an explicit fundamental domain. The push forward of Haar measure on SU(3) has a density with respect to Lebesgue measure on the fundamental domain that was written down by Weyl (if not before). The trace is also a function on this fundamental domain. Points in T can be represented by three angles summing to zero mod 2pi. 
Am I being slow? The generic conjugacy class is not a difficult thing to understand, once having thrown out something of measure zero. This looks like an exercise in multiple integration over sets x + y + z = constant.<|endoftext|>
TITLE: Fastest Rolling Shape?
QUESTION [16 upvotes]: The following questions occurred to me.
This is not research mathematics, just idle curiosity.
Apologies if it is inappropriate.

Suppose you have a fixed volume V of maleable material, 
perhaps clay.
The goal is to form it into a shape S (convex or nonconvex) that would roll down an inclined
plane as fast as possible.
The plane is tilted at θ with respect to the horizontal.
The race track that quantifies "as fast as possible"
is of length L.  The shape S must be
entirely behind a starting plane orthogonal to the inclined plane,
and its race is finished when it is entirely ahead of a finishing plane,
L distant, again orthogonal to the inclined plane.
So if S is a disk of radius r, its center of
gravity will have to travel a total distance of L + 2r
to complete the race.  S is released at rest, and rolls
under gravity.  Assume that the materials of S and of the
inclined plane have a sufficient coefficient of friction μ 
such that there is no slippage, just pure rolling.

   The stipulation that S start behind and end
ahead of the orthogonal planes suggests that the optimal shape
is an arbitrarily thin (and therefore arbitrarily long) cylinder, to minimize r.
But this does not accord with (my) physical intuition.
What am I missing here?  
A variation on the same problem replaces 
the inclined plane with an inclined half-cylinder of radius R, like a rain gutter.
It may be necessary to assume a relation between R and the volume V of material
to make this problem reasonable (so that S fits in the gutter). But certainly this version's
answer could not be an arbitrarily long cylinder!

REPLY [2 votes]: I am going to post a bunch of assumptions which will impose further constraints on the 2nd problem. I think these assumptions maybe natural in view of the thought experiment. 

Without loss of generality, we can assume that the gutter has width 1, so the width of the gutter is parametrized by $[0,1]$. 
The cross-section of the gutter itself is parametrized by some convex function $g: [0,1] \to \mathbb{R}$ with $g(0) = g(1)$. 
The object is not allowed to stick out of the gutter side-ways (else we can just take a heavy rod that rests on the top of the gutter...).
The object is assumed to have uniform density and total mass 1.
The object is obtained by rotation (since it is to rotate) around an axis. 

First, by energy conservation. similar to described in Victor Protsak's answer, to minimize the total time of travel it suffices to minimize
$$ \mathcal{E} = (L + 2 R_0)( 1 + I / R_1^2 ) $$
where $R_0$ is the maximum radius of the rotating body and $R_1$ is radius at which it is rolling, in other words the maximum distance from the axis to a point on the surface of the object that is in contact with the gutter. (Note that it is impossible for a ball of radius 1 to roll inside a gutter of radius 1 without slipping!!! We'll just conveniently forget frictional dissipation for our computations.)
Now, by the definition of the angular momentum and simple re-arrangement argument, it is clear that the cross-section of the object can be obtained as a function over its axis. More precisely, in a cylindrical coordinate system where $z$ is along the axis of rotation, the object 
$$ S = \{ r \leq f(z) \} $$
(If there are any "holes" near the axis, by moving the mass inward closer to the axis it reduces both $I$ and $R$.)
It remains to write $I$ and $R$ as a functional of $f$. Working in the cylindrical coordinate still, the volume of $f$ is
$$ V(f) = \pi \int_0^1 f(z)^2 dz $$
By assumption this is a constant number.
The moment of inertia is 
$$ I(f) = \int_0^1 \int_0^{f(z)} 2\pi r^3 \mu dr dz = \frac{\mu\pi}{2}\int_0^1 f(z)^4 dz $$
where $\mu = 1 / V$ the uniform mass density. 
Lastly, $R_0 = \sup f$ by definition. But $R_1$ is a bit tricky to compute. A bit of thought tells you that $R_1 = \sup_{P} f(z)$, where $P$ is the set of points satisfying $f(z) + g(z) = \sup (f+g)$. 
Observe that $R_1 \leq R_0$. 
To summarize, the task then becomes: given a function $g(z)$ and constants $\alpha,\beta,V$, minimize the functional 
$$\mathcal{E} = (\alpha + \|f\|_\infty)(\beta + \frac{\|f\|^4_4}{R_1(f;g)^2} )$$
under the constraint that $\|f\|_2^2 = V$, where $\|\cdot\|_p$ denotes the Lebesgue $p$-norm over $[0,1]$. 
If we impose the constraint that $R_1 = R_0$, then the problem reduces to the first one which Victor examined, where the solution is a Rolling cylinder. I rather suspect that the cylinder is also the solution in the general case. (I am pretty sure that to minimize $L^4$ while fixing $L^2$ on a compact interval one needs to make the function as flat as possible.)<|endoftext|>
TITLE: Does every retraction of free groups arise from projection to a subset of a freely generating set?
QUESTION [8 upvotes]: Suppose $F_1$ and $F_2$ are free groups, and suppose $\alpha:F_1 \to F_2$ is a surjective homomorphism. Then, because $F_2$ is free, the homomorphism splits, and we get a subgroup $H$ of $F_1$ isomorphic to $F_2$ and a retraction of $F_1$ onto $H$, i.e., a surjective map to $H$ that restricts to the identity on $H$ (with kernel a normal complement to $H$).
Question: Can we find a freely generating set $A$ for $F_1$ and a freely generating set $B$ for $H$ such that $B$ is a subset of $A$ and the retraction sends all elements of $B$ to themselves and sends all elements of $A \setminus B$ to the identity element?
The corresponding statement for free abelian groups is true: simply pick a (free abelian) generating set for the retraction image and the kernel and take their union to get a freely generating set for the whole group. [Note: Any subgroup of a free abelian group is free abelian.] But this technique of taking a freely generating set for the kernel fails in the non-abelian case, because the kernel is too big.

REPLY [12 votes]: No. This is explicitly stated in the paragraph above Theorem 1 of:
Turner, Edward C, Test words for automorphisms of free groups.
Bull. London Math. Soc. 28 (1996), no. 3, 255--263.
The author refers to Proposition 1.
EDIT:
Let's give an explicit example.  Let $F=\langle a,b\rangle$ and let $g=a[b,a]=ab^{-1}a^{-1}ba$.  Clearly $\langle g\rangle$ is a retract of $F$.  But the Whitehead graph of $g$ is two triangles glued along an edge.  This is Whitehead-reduced, so $g$ is not part of a free basis for $F$.<|endoftext|>
TITLE: Positivity of $L(1,\chi)$ for real Dirichlet's character
QUESTION [9 upvotes]: Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.
In my
answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples
of the real characters modulo 3 and 4. The proof goes as follows: one takes
$$
f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j
$$
and uses Abel's theorem to write
$$
L(1,\chi)=\int_0^1f(x)dx;
$$
since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.
Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of
the polynomial
$$
g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n
$$
on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works
perfectly further; for example,
$$
g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5
$$
or
$$
g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11.
$$
Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is
always holds for small values $m\le30$.
Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$,
is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for
which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed
in the literature?

REPLY [4 votes]: As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).<|endoftext|>
TITLE: What is the most simple non-planar Gorenstein curve singularity?
QUESTION [11 upvotes]: Let $R$ be a reduced curve singularity over an algebraically closed field $k$ and $\tilde{R}$ its integral closure in its total ring of fractions.  
The $k$-dimension of $\tilde{R}/R$ is finite.  If we assume $R$ is non-planar and Gorenstein, then how small can this number be?
The ring $R = k[[x,y,z]]/(xy = z^2, z x = y^2)$ is a complete intersection, hence Gorenstein, and the dimension of $\tilde{R}/R$ is $4$.  The question is thus "is $2$ or $3$ possible?"
For the sake of concreteness, let's say that a curve singularity is a $1$-dimensional quotient of $k[[x_1, \dots, x_n]]$ for some $n$.
Edit: I had thought that the $k$-dimension of $\tilde{R}/R$ was widely known as the $\delta$-invariant; I think this the notation Serre uses in Algebraic Groups and Class Fields.  From the comments, it seems this is non-standard, and I have edited accordingly.
As Graham points out, the number $\operatorname{dim}(\tilde{R}/R)$ is also the colength of the conductor ideal.  The number also comes up in computing the (arithmetic) genus of a singular curve.

REPLY [3 votes]: Here is a short geometric proof that if $R$ is Gorenstein and $\tilde{R}/R$ has dimension $\delta \le 3$, then $R$ is planar. 
We can realize $R$ as the local ring of a rational curve $X$ of genus $\delta$.  If $X$ is hyperelliptic (i.e. admit a degree $2$ morphism $f$ to $\mathbb{P}^{1}$), then $X$ embeds into a smoth surface: the ruled surface $\mathbb{P}(\mathcal{E})$ for $\mathcal{E}=f_{*}\mathcal{O}_{X}$.  In particular, the singularities of $X$ are planar.
Otherwise, $X$ is non-hyperelliptic of genus $3$.  But then the canonical map embeds $X$ as a plane quartic curve. In particular, $X$ again embeds in a smooth surface and hence has planar singularities.<|endoftext|>
TITLE: Calculus on rationals
QUESTION [11 upvotes]: Choosing for every prime number $p$ an arbitrary rational value
$\alpha_p$ for its formal derivation $p'$, one gets
a derivation on the rationals by extending the definition
with the Leibniz rule given by
$\left(\pm\prod_{p}p^{\nu_p}\right)'=\pm\sum_{p}\alpha_p\frac{\nu_p}{p}
\prod_{p}p^{\nu_p}$ where sums and products are over all primes 
(and one can even consider rational exponents $\nu_p$ but let us 
stick to integers for simplicity). We have then $(ab)'=a'b+ab'$ 
and $(a^n)'=na^{n-1}a'$ but $(a+b)'\not=a'+b'$ in general.
Are there choices of the values $\alpha_p$ giving rise to interesting
dynamical systems $x\longrightarrow x'$ on the rationals?
(The word "interesting" is of course not well-defined but examples are: 
only finitely many orbits (in the sense that $a,b$ are considered 
in the same orbit if $a^{(n)}=b^{(m)}$ for natural integers $n,m$), 
almost all orbits $a^{(\mathbb N)}$ have infinitely many different elements,
almost all orbits $a^{(\mathbb N)}$ are finite, ...)
Natural candidates are perhaps $\alpha_p=1$ for all $p$, $\alpha_p$
a (linear combination of) Dirichlet character(s) with rational values 
or for example $\alpha_p=p$ (this choice fixes all primes). I guess that 
in general one cannot say much and the study is messy, except 
if all $\alpha_p$ except one are zero. A first perhaps non-trivial case is
given by $\alpha_p=0$ for all primes $p\geq 5$ but $\alpha_2$ and
$\alpha_3$ are both non-zero rationals.
Which choices of values for $\alpha_2,\dots$ give rise to an
interesting differential calculus?
More precisely, we want a canonical primitive in $\mathbb Q[l_2,l_3,l_5,l_7,l_{11},\dots]$
for every rational number where
$l_p$ is a symbol for the logarithm $\int p^{-1}$ of each prime $p$. 
Are there such choices and if yes has the
map $x\longrightarrow \int x$ an interesting dynamical behaviour?
I guess that this can be extended to general number fields
(where one has probably to work with ideals instead of numbers).
Motivation: none except fun and perhaps the existence of
a good exercise on the Leibniz rule and the chain rule for undergraduates.

REPLY [4 votes]: Yes, you get some interesting dynamics out.  The derivative with all $\alpha_p=1$ goes by the name "the arithmetic derivative," and there are a few references around (Ufnarovski's being the most complete).  The short version of the dynamics story is the following:  There are many numbers $n$ (e.g., primes, or twice a twin prime) whose higher order derivatives $n^{(k)}$ are eventually zero.  There are an easily-described set of numbers (those of the form $n=p^p$ with $p$ prime) which satisfy $n'=n$, and so $n^{(k)}=n$ for all $n$.  Finally, there are many numbers (e.g., non-trivial multiples of $p^p$) with $n^{(k)}\rightarrow\infty$.  A fairly major open problem is whether or not there are any other possible orbits (i.e., non-trivial cycles).
A comment on importance:  Though it's not clear to me that there's any way these links are genuinely helpful, there are some amusingly sneaky ties relating statements about arithmetic derivatives to statements about other classical number theory problems.  Ufnarovski gives links to Goldbach's conjecture and the twin primes conjecture, and some undergraduate research I (and colleague Ben Levitt) supervised extends this to Sophie Germain primes and Cunningham chains.
As to the number fields case, one can certainly still play some analogous games, though it's impossible, even in nice cases, to "extend" the arithmetic derivative.  Even if the class group is trivial (e.g., $K=\mathbb{Q}(\sqrt{2})$), one has the obvious problem that one would like to have the prime element $\sqrt{2}$ have derivative one, but this is inconsistent with the product rule (if one wants to maintain that $2'=1$.)
I haven't thought much about other values of $\alpha_p$ (though in many cases, I'd imagine you'd get exactly the same dynamics), but as an aside, let me also reference you to Buium's notion of a derivative (also going by the name arithmetic derivative) which is a little fancier, but currently seems to be of more theoretical significance.<|endoftext|>
TITLE: How to understand the concept of compact space
QUESTION [16 upvotes]: the definition of compact space is: A subset K of a metric space X is said to be compact if every open cover of K contains finite subcovers. What is the meaning of defining a space is "compact". I found the explanation on wikipedia : "In mathematics, more specifically general topology and metric topology, a compact space  is an abstract mathematical space in which, intuitively, whenever one takes an infinite number of "steps" in the space, eventually one must get arbitrarily close to some other point of the space. " I can not understand it, or at least I can not get this by its definition. Can anybody help me?

REPLY [15 votes]: Terry Tao has a nice explanation in the Princeton companion to maths. The article's also on his website: 
http://www.math.ucla.edu/~tao/preprints/compactness.pdf<|endoftext|>
TITLE: Intuitive crutches for higher dimensional thinking
QUESTION [235 upvotes]: I once heard a joke (not a great one I'll admit...) about higher dimensional thinking that went as follows-

An engineer, a physicist, and a mathematician are discussing how to visualise four dimensions:
Engineer: I never really get it
Physicist: Oh it's really easy, just imagine three dimensional space over a time- that adds your fourth dimension.
Mathematician: No, it's way easier than that; just imagine $\mathbb{R}^n$ then set n equal to 4.

Now, if you've ever come across anything manifestly four dimensional (as opposed to 3+1 dimensional) like the linking of 2 spheres, it becomes fairly clear that what the physicist is saying doesn't cut the mustard- or, at least, needs some more elaboration as it stands.
The mathematician's answer is abstruse by the design of the joke but, modulo a few charts and bounding 3-folds, it certainly seems to be the dominant perspective- at least in published papers. The situation brings to mind the old Von Neumann quote about "...you never understand things. You just get used to them", and perhaps that really is the best you can do in this situation.
But one of the principal reasons for my interest in geometry is the additional intuition one gets from being in a space a little like one's own and it would be a shame to lose that so sharply, in the way that the engineer does, in going beyond 3 dimensions.
What I am looking for, from this uncountably wise and better experienced than I community of mathematicians, is a crutch- anything that makes it easier to see, for example, the linking of spheres- be that simple tricks, useful articles or esoteric (but, hopefully, ultimately useful) motivational diagrams: anything to help me be better than the engineer.
Community wiki rules apply- one idea per post etc.

REPLY [2 votes]: This is more of a joke. We see the world in 3D by moving our eyeballs. All movements of an eye are locally described by the Lie group SO(3). Similarly, to "see" things in 4D we need to get info from SO(4)-parametrized set of directions. But SO(4) is locally isomorphic to SO(3) x SO(3), so it is enough to teach our eyes to move independently! (Chameleons do that!)<|endoftext|>
TITLE: Sets with positive Lebesgue measure boundary
QUESTION [23 upvotes]: Consider a compact subset  $K$ of $R^n$ which is the closure of its interior. Does its boundary $\partial K$ have zero Lebesgue measure ?
I guess it's wrong, because the topological assumption is invariant w.r.t homeomorphism, in contrast to being of zero Lebesgue measure. But I don't see any simple counterexample.

REPLY [15 votes]: Let $D_0,D_1,\ldots$ enumerate a sequence of disjoint intervals in the unit interval with $\bigcup_n D_n$ open dense and having measure less than $1$. For example, place a very tiny interval around each rational number, so that the sum of the intervals is less than $1$. Now, let $E=\bigcup_n D_{2n}$ be the union of the even intervals and $O=\bigcup_n D_{2n+1}$, the union of the odd intervals. The entire interval is the union of $E$, $O$ and their boundaries, so one of these boundaries must have positive measure. So we may take $K$ to be the closure of $E$ or $O$.<|endoftext|>
TITLE: Are the rationals homeomorphic to any power of the rationals?
QUESTION [52 upvotes]: I asked myself, which spaces have the property that $X^2$ is homeomorphic to $X$. I started to look at some examples like $\mathbb{N}^2 \cong \mathbb{N}$, $\mathbb{R}^2\ncong \mathbb{R}, C^2\cong C$ (for the cantor set $C$). And then I got stuck, when I considered the rationals. So the question is:
Is $\mathbb{Q}^2$ homeomorphic to $\mathbb{Q}$ ?

REPLY [19 votes]: As Gerald Edgar points out in his comment on Xandi Tuni's answer, the continued fraction trick works for the irrationals, not for the rationals.  But this turns out to be enough:
1) An irrational number has a unique continued fraction expansion.  Therefore the irrationals are isomorphic to an infinite direct product of ${\mathbb Z}$ with itself.  Therefore (writing ${\mathbb I}$ for the irrationals) we have ${\mathbb I}={\mathbb I}^2$.  
2)  Therefore ${\mathbb I}^2$ imbeds as a dense subset of the reals.  Now map ${\mathbb Q}\times {\mathbb Q}$ to ${\mathbb I}\times {\mathbb I}$ by (say) adding $\sqrt{2}$ to each component.  This imbeds ${\mathbb Q}\times {\mathbb Q}$ as a countable dense subset of the reals.  
3)  Now use the fact that every countable dense subset of the reals is homeomorphic to ${\mathbb Q}$.<|endoftext|>
TITLE: What are "variational crimes" and who coined the term?
QUESTION [33 upvotes]: I just caught sight on arXiv a paper by Holst and Stern titled Geometric Variational Crimes. Apparently a Variational Crime is an approach to solve problems using a finite element method (e.g. Galerkin) where certain assumptions about approximate solutions are violated. As you can see, even after doing a brief search in the literature my understanding of the "problem" is still rather superficial and limited (and very possibly wrong). 
So, my questions are

Can someone give me a quick and dirty explanation of what Variational Crimes is about (and how it is useful)?
Does anyone know where the name comes from?

This is just out of curiosity into what other people are doing; I am by no means a numerical analyst.

REPLY [43 votes]: Thanks for your interest in the paper!  (It's also nice to see something on Math Overflow that I know something about.)  Your summary of variational crimes is actually pretty close to the mark: it refers to certain "abuses" of the Galerkin method, where some of the assumptions are violated, and thus the standard error estimates (e.g., Céa's lemma) are no longer valid.
Since you have the basic idea right, let me try to provide some context and motivation.  As a simple example, consider Poisson's equation on some domain $ U \subset \mathbb{R}^n $ with Dirichlet boundary conditions,
$$ - \Delta u = f \text{ on } U, \quad u \rvert _{\partial U} = 0 .$$  
This can be written as a variational problem on $ V = \{ v \in H^1(U) : v \rvert_{\partial U} = 0 \} $: Find $ u \in V $ such that
$$ \int _U \nabla u \cdot \nabla v \; dx = \int _U f \thinspace v \; dx , \quad \forall v \in V .$$
(You can see, using integration by parts, that any classical solution solves this variational problem.)  If we define the bilinear form $ B(u,v) = \int _U \nabla u \cdot \nabla v \; dx $ and functional $ F(v) = \int _U \thinspace f \thinspace v \; dx $, then this problem can be written in the usual abstract form: Find $ u \in V $ such that
$$ B(u,v) = F(v) ,\quad \forall v \in V .$$
To apply the Galerkin method, we need to take a subspace $ V _h \subset V $ (e.g., the span of some finite element basis) and solve the Galerkin variational problem: Find $ u _h \in V _h $ such that 
$$ B(u_h, v ) = F(v), \quad \forall v \in V_h .$$
The problem is that, for many practical purposes, this is impossible to compute.  First, the bilinear form $ B(\cdot, \cdot) $ requires us to calculate an integral exactly.  In practice, this is usually not possible, so people instead approximate the integral using numerical quadrature.  However, this is a variational crime, since using numerical quadrature replaces $ B (\cdot, \cdot) $ by some $ B_h (\cdot, \cdot) \approx B(\cdot, \cdot) $ in the Galerkin variational principle; likewise, numerical quadrature also replaces $ F(\cdot) $ by some $ F_h(\cdot) \approx F(\cdot) $.  (This is aside from the fact that computers only use finite-precision arithmetic, so even if we had an closed formula for these integrals, there would always be some floating-point error involved.)
Moreover, if $U \subset \mathbb{R}^n $ is polyhedral, then it can be triangulated exactly, so we can get $ V_h \subset V $ to be some finite element space supported on this piecewise-linear mesh.  However, if $U$ has a curved boundary, then a piecewise-linear (or piecewise-polynomial, in the case of isoparametric elements) mesh only approximates the actual domain.  Since the functions in $ V_h$ are defined on a slightly different domain than those in $V$, in this case $ V_h \not\subset V $.
Now, in the real world of numerical computation (engineering, etc.), people didn't worry too much about using these approximations instead of the exact Galerkin variational problem; it was a practical necessity, and the approximations seemed to converge just fine.  However, these "variational crimes" meant that the abstract Galerkin error analysis was no longer valid for the modified methods.  Strang pointed this out, and his lemmas quantify the additional errors introduced by these "crimes."
As far as the history/terminology: to the best of my knowledge, Strang himself coined the term "variational crime."  The earliest reference I know is 
[Strang, G. (1972), Variational crimes in the finite element method. In The mathematical foundations of the finite element method with applications to partial differential equations (Proc. Sympos., Univ. Maryland, Baltimore, Md., 1972), pages 689–710. Academic Press, New York.], although I haven't been able to find an electronic copy.  He followed this up with a more easily-located article for a wider audience: Strang, G. (1973), Piecewise polynomials and the finite element method. Bull. Amer. Math. Soc., 79, 1128–1137.  Finally, an excellent account is given in the book Brenner, S. C., and L. R. Scott (2008), The mathematical theory of finite element methods, volume 15 of Texts in Applied Mathematics. Springer, New York, third edition.; Chapter 10 is entirely about variational crimes.<|endoftext|>
TITLE: Periodic orbits and polynomials
QUESTION [16 upvotes]: There are two simple and classic enumerations that still I'm puzzled about. Let's start with a simple counting problem from a well-known dynamical system.
fact 1 Consider the "tent map" f:[0,1]→[0,1] with parameter 2, that is 

f(x):=2min(x,1-x).

Clearly, it has 2 fixed points, and more generally, for any positive integer n, there are 2n periodic points of period n (it's easy to count them as they are fixed points of the n-fold iteration of f, which is a piecewise linear function oscillating up and down between 0 and 1 the proper number of times). To count the number of periodic orbits of minimal period n, a plain and standard application of the Moebius inversion formula gives

Number of n-orbits of (I,f) =
  $\frac{1}{n}\sum_{d|n} \mu(d)2^{n/d}.$

(rmk: any function with similar behaviour would give the same result, e.g. f(x)=4x(1-x),...&c.)
Now let's leave for a moment dynamical systems and consider the following enumeration in the theory of finite fields.
fact 2 Clearly, there are 2n polynomials of degree n in $\mathbb{F}_2[x]$. With a bit of field algebra it is not hard to compute the number I(n) of the irreducible ones. One can even make a completely combinatorial computation, just exploiting the unique factorization, expressed in the form:
$\frac{1}{1-2x}=\prod_{n=1}^\infty (1-x^n)^{-I(n)}.$ 
One finds:

Number of irreducible polynomials of
  degree n in $\mathbb{F} _ 2[x]$ =
  $\frac{1}{n}\sum_{d|n} \mu(d)2^{n/d}.$

Question: it's obvious by now: is there a natural and structured bijection between periodic orbits of f and irreducible polinomials in $\mathbb{F}_2[x]$? How is interpreted the structure of one context when transported ni the other?
(rmk: of course, analogous identities hold for any p > 2)

REPLY [11 votes]: How I see it is every fixed point comes from an equation $T^nx=x$. Denoting $f(x)=2x$ and $g(x)=2(1-x)$ we see that this corresponds to solving equations 
$$h_1\circ h_2\circ\cdots \circ h_n (x)=x$$ where $h_i\in \{f,g\}$. This is geometrically the intersection of two lines and thus gives a unique solution, and therefore a correspondence between periodic points of period $n$ and binary strings of length $n$. Now since concatenating a binary string $L$ with itself clearly gives the same $x$ with $x=L(x)=L\circ L(x)$ and you have a shift operator $L\circ h(x)=x \implies h\circ L(y)=y$ where $y=h(x)$ you get a bijection between periodic orbits and aperiodic cyclic sequences of zeros and ones, which are well known to be in bijection with the irreducible polynomials of $\mathbb F_2 [x]$.<|endoftext|>
TITLE: Topological homotopy category as derived category 
QUESTION [7 upvotes]: In the Introduction of his
Derived Categories for the working mathematician 
Richard Thomas mentions the following theorem of Whitehead.
Suppose that $X,Y$ are simplicial complexes, then
the underlying topological spaces |X|, |Y| (both simply connected) are homotopy equivalent 
if and only if there are maps of simplicial complexes
$X \leftarrow Z \rightarrow X$ inducing quasi-isomorphims 
$C_X \leftarrow C_Z \rightarrow C_Y$ of simlicial-chain complexes.
This suggest that there is a relation between the homotopy category of
simplicial complexes and the derived cagtegory of abelian groups.
E.g. the functor
$ Ho( SimplicialSpaces ) \longrightarrow \mathcal{D}(Mod-\mathbb{Z}), X \mapsto C_X$.
induces an injection on the level of isomorphy classs when restricted to simply-connected
spaces.
Does this functor have any good properties? 
What about homomorphisms (fullness/faithfulness)? 
How does it help to study topological spaces?
More generally one can ask: 
What kind of topological homotopy categories are related to "algebraic" derived-categories?
The above example can be interpreted as $C_x = R\pi_* (\mathbb{Z}_X)$
where $\pi: X \rightarrow \{pt\}$ is the projetion to a point, and $\mathbb{Z}-Mod$ occures as abelian sheaves on $pt$.
One could try to generalize using sheaved spaces $X \rightarrow B$ over a scheme $B$.
This seem to be quiet obvious questions. However I have seen nobody elaborating
on this so far. Or maybe I am missing a link to something well known?

REPLY [7 votes]: More generally one can ask: What kind of topological homotopy categories are related to "algebraic" derived-categories?

First, to have a triangulated category analogous to $\mathcal{D}(Mod-\mathbb{Z})$, you should replace $Ho( SimplicialSpaces )$ by $Ho( Spectra )$, i.e. the stable version of the former. This said, there are intrinsic obstructions to express $Ho( Spectra )$ as the derived category of an algebraic one (even a "stable" category). Fo full details, see
Muro, F.; Schwede, S.; Strickland, N.: 
Triangulated categories without models. Invent. Math. 170 (2007), no. 2, 231–241.
and
Schwede, S.:
Algebraic versus topological triangulated categories. Triangulated categories, 389–407, 
London Math. Soc. Lecture Note Ser., 375, Cambridge Univ. Press, Cambridge, 2010.<|endoftext|>
TITLE: Evaluation of the following series... $S = 1/(2\times3) + 1/(5\times6) + 1/(7\times8) + 1/(10\times11) + ...  $
QUESTION [12 upvotes]: EDIT, Will Jagy, December 8, 2010: to anyone considering working on this, please first see  http://mathoverflow.tqft.net/discussion/817/could-a-few-moderators-please-remove-one-of-my-questions/#Item_9 
which gives the story behind this peculiar sum. Note that the OP is no longer interested in the results, as they arose from one kind of error and cannot be applied because of a different sort of misunderstanding. The double sum version below was provided recently by Harald Hanche-Olsen.
ORIGINAL. I'm curious one of you is able to find the exact evaluation of 
the following series: 
$$\begin{aligned} S &= 1/(2\times3) +1/(5\times6) + 1/(7\times8) + 1/(10\times11) + \cdots 
  \\\\&= \sum_{n=1}^\infty\sum_{k=1}^{n}\frac1{(n^2+2k-1)(n^2+2k)} \end{aligned}$$ 
I'm not exactly sure on how to state the 'general term' of the series. Perhaps I can illustrate it with an example:
$ 1/(1\times2) + 1/(3\times4) + 1/(5\times6) + 1/(7\times8) + \ldots + 1/((2n - 1) \times 2n) + \ldots = \log(2)$. 
Now, to answer Nate Eldredge: let $a_0=2$ and $a_{k+1}=a_{k} + 1 $ unless $ a_{k} + 1$ is a square, in which case let $a_{k + 1} = a_{k} + 2$. Now, multiply $a_{k}$ with $a_{k+1}$. That's a term. Let me show the first few terms:
$ S = 1/(2\times3)$ [now skip 4] $ +  1/(5\times6) + 1/(7\times8)$ [now skip 9] $ + 1/(10\times11) + 1/(12\times13) + 1/(14\times15)$ [now skip 16] $ + 1/(17\times18) + \ldots $ 
So all the squares (1,4,9,16,25, etc) are 'skipped' in the terms.
I hope this clarifies it a bit...
Thanks a lot in advance,
Max Muller
PS: If someone has any ideas as to how the general term of this series can be written in a more concise manner, please let me know! For the Meta-users, see also the relevant discussion on this question.

REPLY [6 votes]: As already mentioned by Mariano, the sum in question is a simple logarithmic sum minus
$$
\sum_{n>1}\biggl(\frac1{n^2(n^2-1)}+\frac1{n^2(n^2+1)}\biggr)
=2\sum_{n>1}\frac1{n^4-1}
=\frac74-\frac\pi2\coth\pi.
$$
The closed form evaluation is mentioned in Michel Waldschmidt's
"Open Diophantine Problems" (Moscow Math. Journal 4:1 (2004), 245-305, 312) together with "which is a transcendental number since $\pi$ and $e^\pi$ are algebraically independent over $\mathbb Q$ (Yu.V. Nesterenko)."
EDIT. I was confused by too many answers and edits. On using
$$
\frac1{n(n+1)}=\frac1n-\frac1{n+1},
$$
one can write the wanted sum as
$$
\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\sum_{k=1}^\infty\frac1{k^2}
=\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\frac{\pi^2}6.
$$
I can hardly imagine that the remaining single sum has a closed form... But this leaves a nice problem in 1st year analysis: Show that the series
$$
\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]}}n
$$
converges.<|endoftext|>
TITLE: Ramsey multiplicity
QUESTION [22 upvotes]: Given a positive integer $a$, the Ramsey number $R(a)$ is the least $n$ such that whenever the edges of the complete graph $K_n$ are colored using only two colors, we necessarily have a copy of $K_a$ with all its edges of the same color.
For example, $R(3)= 6$, which is usually stated by saying that in a party of 6 people, necessarily there are 3 that know each other, or 3 that don't know one another; but there is a party of 5 people without this property. This is probably known to everybody.
Slightly less known is the fact that any such coloring of $K_6$ in fact contains 2 monochromatic triangles.
The Ramsey multiplicity $m(a)$ (there does not seem to be a standard notation) is the largest number $m$ of monochromatic copies of $K_a$ that we can guarantee in any 2-coloring of $K_{R(a)}$. For example, $m(3)=2$, and Piwakowski and Radziszowski showed around 1999 that $m(4)=9$.
I have a couple of questions (please forgive me if they are trivial, I'm just beginning to form my intuitions in this field) :

Is it known that $m(n)$ is monotonically increasing?

Do we know anything about the rate of growth of the function $m(n)$?


I suspect that the answer to both questions is yes and that reasonable bounds for $m(n)$ are known, but haven't been able to locate any references. The best I know is that $$ m(n)\le \frac{\binom{r(n)}n}{2^{\binom n2-1}}, $$
(proved by Burr and Rosta in 1980), which is probably too high, and a recent result of Conlon suggests that
$$  m(n)\ge C\frac{\binom{r(n)}n}{2^{n(3n-1)/2}} $$
for some appropriate $C$. I say "suggests" because Conlon's results carry some additional implicit constants that I haven't checked can be absorbed this way. (Please let me know if I am completely off the mark here.) [Edit: Unfortunately, Conlon's bounds (in his paper "On the Ramsey Multiplicity of Complete Graphs") do not apply here. No lower bound beyond $m(n)\ge1$ seems known.]

"Update": William Gasarch's Open Problems column in the June 2020 issue of the ACM SIGACT News is devoted to Ramsey multiplicity.

REPLY [12 votes]: I emailed David Conlon about this question. He agreed to let me share his answer. In short, the problem very much seems to be open (I've added the relevant tag). As Thomas mentions, the upper bound I cite is straightforward. And nothing better is known!
If one looks for papers on Ramsey multiplicity, a few come up, but they deal with a different concept, that I explain below. The quotes are from Conlon's emails.

Unfortunately, the concept you're talking about is also known as the Ramsey
  multiplicity! There are very few references as far as I know. The only one I
  can think of offhand is the Piwakowski and Radziszowski paper which you
  quoted. Perhaps there's something in the references to that paper, but I
  doubt it somehow.

Indeed, in the papers I have seen (included P-R, where $m(4)=9$ is proved), there are no arguments about $m(n)$ for general $n$ (or even $n=5$).

The function you're interested in is rather amorphous, I'm afraid. My result
  will imply that if $n \ge 4^t$ you must have at least $n^t/2^{3t^2/2}$ copies of
  $K_t$ or thereabouts. But when your number is below $4^t$ it implies nothing. 
In general, because we don't understand the Ramsey function, I find it hard
  to imagine how we might be able to say anything at all about $m(n)$. Unless
  there's an elementary argument which gives something. It reminds me of
  estimating the difference between successive Ramsey numbers like $r(n,n)$ and
  $r(n,n+1)$, where, though the difference is almost certainly exponential, the
  largest difference that can be guaranteed is tiny (I think linear or
  quadratic even, though I can't remember exactly).

Here, $r(m,n)$ are the usual Ramsey numbers (what I called $R(n)$ in the question, is $r(n,n)$ in this notation). In general, $r(m,n)$ is the smallest $k$ such that any coloring of the edges of $K_k$ with blue and red either contains a blue copy of $K_m$ or a red copy of $K_n$.

The only thing that appears clear to me is an upper bound following from the
  probabilistic method, namely $\displaystyle \frac{\binom{r(n)}{n}}{2^{\binom n2}}$. It's not even
  obvious how one would approach showing that the multiplicity is at least 2!

Finally, as to the question of how to call this concept:

I'd suggest that this be called the critical multiplicity or something like
  that, just to distinguish it from the usual multiplicity function.

The usual Ramsey multiplicity is defined as follows. It is significantly better understood than $m(n)$.
Let $k_t(n)$ be the minimum number of monochromatic copies of $K_t$ within a two-coloring of the 
edges of $K_n$, and let $$ c_t(n)= \frac{k_t(n)}{\binom nt} $$ be the minimum proportion of monochromatic copies of $K_t$ in such a two-coloring.
It is known that the numbers $c_t(n)$ increase with $n$. The Ramsey multiplicity of $t$ (or of $K_t$) is $\displaystyle c_t\lim_{n\to\infty} c_t(n)$.
(Relevant references can be found in Conlon's paper mentioned in the question.)<|endoftext|>
TITLE: Topological conjugacy between homeomorphisms and diffeomorphisms
QUESTION [10 upvotes]: Consider a compact differentiable manifold $M$. We say that $f:M\to M$ and $g: M \to M$ are topologically conjugated if there exists $h:M\to M$ a homeomorphism such that $f\circ h= h \circ g$. The conjugacy class of a homeomorphism $f$ is the set of all $g$ such that $g$ is topologically conjugated to $f$.
If a homeomorphism $f: M\to M$ has infinite topological entropy (which is an invariant under topological conjugacy), then the conjugacy class of $f$ has no diffeomorphisms.
Is there any other known obstruction for a homeomorphism not to have diffeomorphisms in its conjugacy class? I would guess that yes, but I could not find one.  
Is there a restriction on the dimension of the manifold? 
(Remark: In dimension one, that is, in the circle, every homeomorphism has a diffeomorphism-of class $C^1$ in its conjugacy class. However, Denjoy counterexamples have no $C^2$ diffeomorphisms in theirs).

REPLY [13 votes]: Here is an example. Consider a map $f:\mathbb R^2\to\mathbb R^2$ given by $(r,\varphi)\mapsto (r,\varphi+\sin(1/r))$ in polar coordinates $(r,\varphi)$, $0<r\le 1/\pi$. For $r>1/\pi$, let $f$ be identity, then close up the plane to make a compact manifold.
This map has zero topological entropy but has no conjugate $C^1$ diffeomorphism. Indeed, we may assume that 0 is mapped to itself by the conjugation. Then the circles centered at the origin are mapped to Jordan curves winding around the origin. There are arbitrarily small circles made of fixed points, hence the derivative of the diffeomorphism at the origin is the identity. On the other hand, there are arbitrary small circles on which our map is periodic with rotation number, say, 1/10. It is easy to see that this cannot happen on a Jordan curve winding around the origin, unless some point changes is angular coordinate by $2\pi/10$. This contradicts the fact that the derivative at 0 is the identity.<|endoftext|>
TITLE: Example of Noetherian group (every subgroup is finitely generated) that is not finitely presented
QUESTION [14 upvotes]: A Noetherian group (also sometimes called slender groups) is a group for which every subgroup is finitely generated. (Equivalently, it satisfies the ascending chain condition on subgroups).
A finitely presented group is a group with a presentation that has finitely many generators and finitely many relations.
Flipping through some search results and references, I get the impression that there should be examples of Noetherian groups that are not finitely presented (because I can locate references to "finitely presented Noetherian group", a name that shouldn't exist if being Noetherian implies being finitely presented). However, I'm not able to get an explicit reference or example. I would be grateful if somebody could point out a reference or example.
For a solvable group, being Noetherian is equivalent to being polycyclic (i.e., having a subnormal series where all the successive quotients are cyclic groups), and polycyclic groups are finitely presented. Hence, any counterexample must be a non-solvable group.
[Note: My standard example of a finitely generated group that is not finitely presented is a wreath product of the group of integers with itself. But this is far from Noetherian.]

REPLY [11 votes]: Tarski monsters provide examples of 2-generator noetherian groups that is not finitely presented.

Edit (YCor): Tarski monsters, as defined in the link (infinite groups of prime exponent $p$ in which every nontrivial proper is cyclic) exist for large $p$ and all currently known constructions of Tarski monsters are known to yield groups that are not finitely presented. However, it is unknown whether there exists a finitely presented Tarski monster.<|endoftext|>
TITLE: Derivative of a random variable
QUESTION [14 upvotes]: Hi,
If I have two i.i.d random variables $X,Y$ and a parameter $a$. If I define a new random variable $Z(a)=aX+(1-a)Y$. 
Does it makes sense to talk about first, second derivative of the random variable $Z(a)$ respect to $a$. Then $Z^{\prime}(a)=(X-Y)$ and so on.
Thanks

REPLY [10 votes]: Yes, it makes sense if for example your random variables are in $L^1$. 
Your map $Z(a)=aX+(1-a)Y$ is a well-defined map from an open set in a Banach space to a Banach space, $Z: R \mapsto L^1$.
In such situation, you can talk about the Frechet derivative of Z, and it satisfies the usual properties you can expect from a derivative. 
If you are dealing with random variables living in non-locally convex topological vector spaces (e.g. in $L^p$, $0\leq p < 1$), then I think you run quickly into several problems. 
The standard procedure to prove results from calculus for a vector-valued function Z is to go back to a real-valued function just by replacing Z with $\lambda(Z)$, where $\lambda$ is a continuous linear functional on the vector space. That is, we are just looking at the "coordinates" of Z. But if there are no non-zero linear functionals on the vector space (e.g. $L^p$, $0\leq p<1$), then there is not much that can be done.<|endoftext|>
TITLE: Comparison of Local and Global Duality
QUESTION [6 upvotes]: Hello! 
I'd like to understand the relation between the following two theorems:

The "global" duality for projective schemes, as explained in [Hartshorne]: If $X$ is an equidimensional projective Cohen-Macaulay scheme of dimension $n$ over an algebraically closed field with dualizing sheaf $\omega_X$, then for all $i$ there is a natural isomorphism $Ext^i({\mathcal F},\omega_X)\cong H^{n-i}(X,{\mathcal F})^{\ast}$.
The "local" duality theorem Cohen-Macaulay rings, as explained in [Bruns, Herzog]: If $(R,{\mathfrak m},k)$ is a complete local ring of dimension $d$, then for all finite $R$-modules $M$ and all $i$ there is a natural isomorphism $\text{Ext}_R^i(M,\omega_R)\cong\text{Hom}_R(H_{\mathfrak m}^{d-i}(M),E(k))$, where $\omega_R$ is the canonical module of $R$ and $E(k)$ the injective hull of the residue field $k$.

The isomorpisms are strikingly similar, but I don't know if there is a rigorous way to deduce, say, the global duality from the local one. Can somebody explain this to me or give references?
Thank you!

REPLY [5 votes]: In the case of varieties over a perfect field this question is explained with great detail in the book by J. Lipman:
Dualizing sheaves, differentials and residues on algebraic varieties. (French summary) 
Astérisque No. 117 (1984)
known by some people as "Lipman's blue book". If you want to get rid of the base field then you should look at Greenlees May duality for schemes and Grothedieck duality for formal schemes, see joint work by Alonso, Jeremías & Lipman.<|endoftext|>
TITLE: Characterization of Kleisli adjunctions
QUESTION [12 upvotes]: There's a well known theorem due to Beck that characterizes when an adjunction is monadic, that is, if $F$ is left adjoint to $G$, $G:D \to C$, $GF:=T$ is always a monad on $C$, and the adjunction is called monadic, essentially, when $D$ is the Eilenberg–Moore category $C^T$ of $T$-algebras and $G$ is the forgetful functor. (For the precise definition see http://ncatlab.org/nlab/show/monadic+adjunction). I was wondering if there was a similar characterization to determine when $D$ is the Kleisli category of FREE $T$-algebras?

REPLY [18 votes]: There is a unique functor $\mathbf{Kl}(GF) \rightarrow \mathbf{D}$ commuting with the adjunctions from $\mathbf{C}$, since the Kleisli category is initial among adjunctions inducing the given monad; and this functor is always full and faithful, since $\mathbf{Kl}(GF)(A,B) \cong \mathbf{C}(A,GFB) \cong \mathbf{D}(FA,FB)$.
So this functor will be an equivalence iff it is essentially surjective, and an isomorphism iff it is bijective on objects.  But its object map is just the object map of $F$.
So $\mathbf{Kl}(FG)$ is equivalent to $\mathbf{D}$ compatibly with the adjunctions from $\mathbf{C}$ precisely when $F$ is essentially surjective, and isomorphic just when $F$ is bijective on objects.<|endoftext|>
TITLE: Shortest-path Distances Determining the Metric?
QUESTION [14 upvotes]: The metric of a Riemannian manifold determines the shortest
distance between any two points.

I assume the reverse holds?  That is, if you are given the
shortest distance d(x,y) between every pair of points
of a manifold M, the metric for M is determined?
I am mainly interested in compact, connected, closed 2-manifolds,
but the most general answer would be appreciated.
(Apologies in advance for my naivete.)
Assuming I am correct above, is there some natural subset
S of M such that knowing d(x,y) between every
pair of points of S uniquely determines the metric (up to isometry)?
For example, suppose S is a simple closed geodesic on
M?  Perhaps some assumptions on M are necessary:
genus zero, convex, ... ?

REPLY [4 votes]: Nice paper, I was not aware of it. There is this paper: 
I.G. Nikolaev, A metric characterization of riemannian spaces, Siberian Advances in Mathematics, 1999, v. 9, N4, 1-58 
see the mathscinet page: 
http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=Nikolaev&s5=metric%20characterization&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq
Citing from the Abstract: "We present a metric characterization of Riemannian spaces: a locally compact metric space with intrinsic metric in which geodesics are locally extendable, 
and which has Holder-continuous curvature as a metric space, is isometric with a C^2-Riemannian manifold."
Edit [by J.O'Rourke]. Following Willie's advice (in the comments), here is the MathSciNet link.<|endoftext|>
TITLE: Möbius-invariant triangle center?
QUESTION [7 upvotes]: Given any two points x and y on a circle O, one can form four different lenses (regions between two circles, one of which is O) that have corners at x and y and make angles of 2π/3 at their corners. For three points x, y, and z, with O the circumcircle of triangle xyz, two of these lenses are outside O and two of them have z on the boundary, so there is a unique lens L(x,y) that is inside O and disjoint from z, and lenses L(y,z) and L(x,z) defined symmetrically.
These three lenses L(x,y), L(y,z) and L(x,z) intersect in a unique point c contained in O, defining c as a triangle center for the triangle xyz. There's also a closely related center c' that you can get by inversion of c through O (the intersection point of three lenses outside O). For an equilateral triangle xyz, c is (as for any triangle center) at the centroid of xyz, while c' is at the point at infinity in the one-point completion of the plane. {c,c'} is equivariant under Möbius transformations, an unusual property for a triangle center.
Another equivalent way of defining c, I think, is that it's the point for which angles xcy = π/3 + xzy, xcz = π/3 + xyz, and ycz = π/3 + yxz. This is similar to the fact that the circumcenter C is the point for which xCy = 2xzy etc.
Does anyone recognize these centers c and c'? Do they have entries in the Encyclopedia of Triangle Centers? The definitive way to answer this would be to compute trilinear coordinates and look them up using ETC's search feature, but I'm hoping that this looks familiar enough to someone here that I can skip that step. I tried searching for Möbius and inversion in the text of ETC but found nothing that way.

REPLY [6 votes]: Unless I'm mistaken you seem to describe the isogonal conjugate of the Fermat point X(13). This is the first isodynamic point, or X(15) in ETC.<|endoftext|>
TITLE: Analytic continuation of holomorphic functions
QUESTION [11 upvotes]: Analytic/meromorphic continuation is a difficult problem in general. For "motivic L-functions", the idea of proving their analytic continuations by first proving their modularity goes back, I guess, to Riemann. 
Here I just want to ask a purely complex-analytic question. Let's restrict ourselves to the case of one variable functions. Let $U$ be a region in the complex plane, and let $f$ be a holomorphic function on $U.$ Is there any criterion for $f$ to have analytic continuation to a larger region? And what is this "maximal domain of regularity"? 
Feel free to assume $U$ and $f$ to have the shape you like, e.g. a power series on an open disk or a Dirichlet series on some half plane. I guess even if $U$ is an annulus or a punctured disk, where one can compute (theoretically or numerically) the value of the extended function (if exists) at the points inside the inner loop by Cauchy's formula, it is still difficult to decide if this extension is continuous or analytic.

REPLY [3 votes]: To answer your specific questions, a) a maximal (by inclusion) domain does not have to exist. Consider $\sqrt{1-z}$ in the unit disk. That is actually the reason why Riemann introduced Riemann surfaces.
But there are maximal domains in various other senses. Suppose you have an analytic germ at infinity. Then there is a unique (!) set $K$ in the plane of
of minimal logarithmic capacity, such that our germ has a (single-valued) analytic continuation to $C\backslash K$. This is due to H. Stahl. 
b) The answer depends on what you exactly mean by a criterion. There are necessary and sufficient conditions for existence of an analytic continuation
(multiple valued) to a given region. Not surprisingly they are difficult to check. See, for example
 MR1711032 
Atzmon, A., Eremenko, A., Sodin, M.
Spectral inclusion and analytic continuation. 
Bull. London Math. Soc. 31 (1999), no. 6, 722–728.
There is a necessary and sufficient condition for a function defined in the unit disk to have an analytic continuation to a given point on the circumference (Euler).
And there are many separate necessary (but not sufficient) and sufficient (but not necessary) conditions in terms of coefficients in various special cases. This is a very vast subject. A good survey of these conditions can be found in Bieberbach, Analytische Fortsetzung, Springer, 1955, and
V. Bernstein, Lecons sur les progres recents de la théorie des séries de Dirichlet, Paris: Gauthier- Villars. XIV, 320 S. (1933).<|endoftext|>
TITLE: A topologically mixing subshift with multiple measures of maximal entropy
QUESTION [5 upvotes]: Let Σp={1,...,p}ℤ be the full shift on p symbols, and let X ⊂ Σp be a subshift -- that is, a closed σ-invariant subset, where $\sigma\colon \Sigma_p\to \Sigma_p$ is the left shift.  Then σ is expansive, and hence there exists a measure of maximal entropy (an mme) for (X,σ).
It is well known that if X is a subshift of finite type on which σ is topologically mixing, then there is a unique mme.  (See, for example, Rufus Bowen, Equilibrium states and the ergodic theory of Anosov diffeomorphisms, 1975.  In fact, Bowen proves uniqueness of equilibrium states for any Hölder continuous potential φ, but let's stick with the case φ≡0 for now.)
If X is not a subshift of finite type, then less is known.  For example, if X is a β-shift, then it has a unique mme, but it is not known if this holds for subshifts that are factors of a β-shift.
Question:  Does anybody know of a subshift that is topologically mixing but does not have a unique mme?  (That is, a subshift that has multiple measures of maximal entropy despite being topologically mixing.)

REPLY [3 votes]: There is a book by Denker, Grillenberger and Sigmund which deals extensively with this topic: they prove a whole range of theorems which construct subshifts whose invariant measures have specified properties. In particular they construct a mixing subshift with multiple invariant measures of maximal entropy, but this is just one of many interesting constructions.
Subshifts which are not of finite type are, as a general class, quite well understood. In terms of their measurable ergodic theory, they can do pretty much anything which is not prevented by entropy. An example is the following theorem of Jewett. Suppose that $T$ is an ergodic measure-preserving transformation of a probability space $(Z,\mathcal{F},\mu)$, and the entropy of $T$ with respect to $\mu$ is finite. Let $p$ be an integer such that $\log p$ is strictly greater than the entropy of $T$ with respect to $\mu$. Then there is a subshift $X$ of $\Sigma_p$, having a unique invariant measure $\nu$, such that the measure-preserving transformation $(X,\mathcal{B},\nu,\sigma)$ is measurably isomorphic to the transformation $(Z,\mathcal{F},\mu,T)$. (Here $\mathcal{B}$ of course denotes the Borel sigma-algebra of $X$.)<|endoftext|>
TITLE: How many conjugacy classes of subgroups does GL(2,p) have?
QUESTION [14 upvotes]: How many conjugacy classes of subgroups does $\mathrm{GL}(2,p)$ have?

For instance the dihedral group of order $2n$ has $\tau(n)$ cyclic normal subgroups and $\sigma(n)$ "dihedral" subgroups (as in, containing a reflection), but they fall into $\gcd(2,n) \tau(n) + \tau(n/\gcd(2,n))$ conjugacy classes. Here $\tau(n)$ is the number of divisors of $n$, and $\sigma(n)$ is their sum.
The formula is relatively compact and can be explicitly evaluated for $n$ in the millions without much work.  The description is nice because it even indicates the structure of the subgroups.
The subgroups of $\mathrm{GL}(2,p)$ whose order is divisible by $p$ either have a normal Sylow $p$-subgroup or contain $\mathrm{SL}(2,p)$.  The former types have conjugacy classes indexed by the subgroups of $(p-1) × (p-1)$, and the latter by subgroups of $(p-1)$.  The number of the first type has some reasonable formulas at OEIS: A060724 and the latter is just $\tau(p-1)$ again.
Again the description is compact and can be explicitly evaluated for numbers into the millions without any real effort: $\mathrm{GL}(2,1000003)$ has $1000008$ conjugacy classes of subgroups of order divisible by $1000003$ and $\mathrm{GL}(2,10000019)$ has $10000024$ conjugacy classes of subgroups of order divisible by $10000019$, each number computed in under 1ms.  Again the description is especially nice because it even indicates the structure of the subgroups.

What about the conjugacy classes of subgroups of $\mathrm{GL}(2,p)$ whose order is coprime to $p$?

Is there a similarly compact and easily evaluated description of their number, and even more nicely, does it also indicate the structure of the subgroups?

REPLY [4 votes]: Another way to look at these is that there are three kinds of subgroup: those which contain 
a trivial Sylow $p$-subgroup, those which contain exactly one Sylow subgroup of order $p,$ ad those which contain more than one Sylow $p$-subgroup. The last type is easy to deal with: any such subgroup contains all $p+1$ Sylow $p$-subgroups of ${\rm GL}(2,p,$ so contains ${\rm SL}(2,p)$ (and is, in particular, normal). Every subgroup between ${\rm SL}(2,p)$ and ${\rm GL}(2,p)$ occurs, and the number  (of conjugacy classes of) such subgroups is the number of divisors of $p-1.$ The second type is also reasonably straightforward: any such subgroup is conjugate to one and only one subgroup of the group of invertible upper triangular matrices and that group of upper triangular matrices contains all upper unitriangular matrices. The first type consists of subgroups of order prime to $p.$ Since matrix representations over finite fields which are conjugate over an extension field are already conjugate over the field of realizability, the conjugacy class of such a group is uniquely determined by its Brauer character. The isomorphism types which can occur are described already in earlier answers.<|endoftext|>
TITLE: Examples of the moduli space of X giving facts about a certain X
QUESTION [14 upvotes]: What is a good example of a fact about the moduli space of some object telling us something useful about a specific one of the objects?
I am currently learning about moduli spaces (in the context of the moduli space of elliptic curves). While moduli spaces do seem to be fascinating objects in themselves, I am after examples in which facts about a moduli space tell us something interesting about the specific objects that they parametrise. For example, does the study of $\mathbb RP^n$ tell us anything we don't already know about some given line through the origin (say, the $x_1$-axis) in $\mathbb R^{n+1}$?

REPLY [2 votes]: Earlier today I thought of something that may be an example of this. If not, at least it's very elementary and has something of the same spirit. I don't really know what any kind of moduli space is in any technical sense.
Suppose you know the formula for volume of a sphere, but you do not know the formula for surface area. Parameterize the set of all spheres in $ \mathbf{R}^3 $ at the origin by their radius $ r $, forming the space $ X = ( 0, \infty ) $. Assignment of volume $ V(r) $ to the sphere with radius $ r $ defines a differentiable function $ V $ on $ X $. It is clear from geometry that if $ r $ is changed by $ \Delta r $ then its volume should change by approximately $ A ( r ) \cdot \Delta r $, for $ A ( r ) $ the surface area of the sphere with radius $ r $. In other words, $ A ( r ) = V^{\prime} ( r ) = 4 \pi r^2 $.
I guess the point is that even if you are interested in only one specific sphere S, you can compute the number that you want by forming a continuum of spheres and using analysis on that parameter space. Until you have the space, there is no derivative. Does that make any sense?<|endoftext|>
TITLE: Connections between ultrafilters in topology and logic
QUESTION [22 upvotes]: I have a some-what vague question. It seems to me that there are two main ways in which ultrafilters (on a set) can be used. One is in topology. The notion of an ultrafilter converging to a point is very useful since, in particular, knowing the limit points of every ultrafilter on a space is equivalent to knowing its topology. The other use is in logic (a subject about which I admittedly know very little). For instance, ultraproducts (and more generally ultralimits) can be used to construct non-standard models etc. I'm just curious about any connections that exist between these two uses of ultrafilters. For example, is there any logical interpretation of ultrafilter convergence on a topological space, etc? Is there a connection to the internal logic of its topos of sheaves for example? I'm really a beginner with this stuff, so any interesting connections, even trivial ones, would be most interesting to me. If this question is too open-ended, feel free to change this to community wiki.

REPLY [2 votes]: Ultrafilters appear naturally as ultraproducts in the space of marked groups as introduced by V. Guirardel and C. Champetier in Limit groups as limits of free groups. Indeed, if $(G_k,\phi_k)$ is a sequence of marked groups converging to some marked group $(G,\phi)$, then, for every nonprincipal ultrafilter $\omega$ on $\mathbb{N}$, $G$ embeds into the ultraproduct $\prod\limits_{n \in \mathbb{N}} G_k / \omega$. This key observation links convergence in the space of marked groups and universal theory of groups. For example, it can be proved that limit groups are exactly the finitely generated groups with the same universal theory as a free groups of finite rank, thanks to a embedding into $^{\omega}\mathbb{F}_2$. (See also Non-standard free groups, written by I. Chiswell, for an approach based on real trees.)
In Structure and rigidity in hyperbolic groups, E. Rips and Z. Sela combined ultralimits of metric spaces and Rips theory to show their shortening argument. For instance, they applied it to show that the automorphism group of a torsion-free hyperbolic group is finitely generated. The shortening argument is also a key result for Makanin-Razborov diagrams, and more generally for Sela's solution of Tarski problem.<|endoftext|>
TITLE: Cohomology of orthogonal and symplectic groups 
QUESTION [9 upvotes]: Hello,
in their book  Cohomology of Finite Groups Adem and Milgram investigate the cohomology of the finite orthogonal and symplectic groups only in case $\mathbb{F}_2$. 
Let $p$ be a prime dividing the order of $\text{O}_n(q)$, $\text{Sp}_n(q)$ and $q$ a prime power. 
I am wondering if anything is known about $\text{H}^\ast(\text{O}_n(q),\mathbb{F}_p)$ or $\text{H}^\ast(\text{Sp}_n(q),\mathbb{F}_p)$.
I am also interested in the maximal elementary abelian $p$-subgroups of these groups.
In light of Quillen's stratification theorem these two questions are, of course, related to each other.
I would be grateful for any kind of information.

REPLY [8 votes]: Direct finite group computations of cohomology of the finite groups of Lie type tend to be very sparse.   The case $p=2$ has special interest for topologists and does provide some explicit results.    More generally, some indirect results of interest have been found in recent decades by systematically comparing cohomology of the finite groups (in the defining characteristic $p$) with rational cohomology of the ambient algebraic groups.    The main work in this direction is found in papers by some combination of Cline, Parshall, Scott, van der Kallen (see their joint 1977 paper in Invent. Math.), Friedlander.   But a full description of the cohomology rings is elusive.   Benson and Carlson have done a lot of work in the traditional setting as well.   There are lots of papers, but not many definitive results.   
For maximal elementary abelian $p$-subgroups, some of the work by Avrunin, Parshall, Scott would be relevant, as well as the older LMS Lecture Notes by Kleidman and Liebeck.   Perhaps the most helpful source is the third volume of an ongoing series of books on the classification of finite simple groups.   See especially section 3.3 and the summary table there for groups of Lie type:MR1490581 (98j:20011) 20D05 (20-02)
Gorenstein, Daniel; Lyons, Richard (1-RTG); Solomon, Ronald (1-OHS)
The classification of the finite simple groups. Number 3. Part I. Chapter A. 
Almost simple K-groups.
Mathematical Surveys and Monographs, 40.3.
American Mathematical Society, Providence, RI, 1998. xvi+419 pp.
ISBN 0-8218-0391-3<|endoftext|>
TITLE: Binomial supercongruences: is there any reason for them?
QUESTION [17 upvotes]: One of the recent questions, in fact
the answer
to it, reminded me about the binomial sequence
$$
a_n=\sum_{k=0}^n{\binom{n}{k}}^2{\binom{n+k}{k}}^2,
\qquad n=0,1,2,\dots,
$$
of the Apéry numbers. The numbers $a_n$ come as
denominators of rational approximations to $\zeta(3)$ in Apéry's
famous proof of the irrationality of the number.
There are many nice properties of the sequence, one of these
is the observation that for primes $p\ge5$,
$$
a_{pn}\equiv a_n\pmod{p^3},
\qquad n=0,1,2,\dots.
$$
The congruence was conjectured for $n=1$ by S. Chowla, J. Cowles and M. Cowles
and proved in the full generality by I. Gessel (1982).
There is probably nothing strange in the congruence
(which belongs, by the way, to the class of supercongruences as it happens
to hold for a power of prime higher than one). But already the classical
binomials behave very similarly: for primes $p\ge5$,
$$
\binom{pm}{pn}\equiv\binom mn\pmod{p^3},
$$
the result due to G.S. Kazandzidis (1968). There are many other examples
of modulo $p^3$ congruences, most of them explicitly or implicitly related
to some modular objects, but that is a different story. My question is:
what are the grounds for the above (very simple) congruence for binomial
coefficients to hold modulo $p^3$? Not modulo $p$ or $p^2$ but $p^3$.
I do not ask you to prove the supercongruence but to indicate a general
mechanism which provides some kind of evidence for it and can be used
in other similar problems.
My motivation rests upon my own research on supercongruences; most of them
are just miracles coming from nowhere...

REPLY [5 votes]: Andrew Granville, in his beautiful article Binomial coefficients modulo prime powers (1997), attributes to Jacobsthal the congruence
$$
\binom{np}{mp} \Big/ \binom{n}{m} \equiv 1 \mod{p^{3+\mathrm{ord}_p\{nm(n-m)\}}} \quad \quad \quad (p \geq 5).
$$
In Chapter 7 of Alain's Course in $p$-Adic Analysis (GTM 198), the same congruence is attributed to Kazandzidis. 
This is a (mild?) generalization of Wolstenholme's theorem, $\binom{np}{mp} \equiv \binom{n}{m} \mod{p^3}$.
As both references explain, the general congruence is best understood via the $p$-adic analog of Stirling's asymptotic formula for the $\Gamma$-function. The $p$-adic logarithm of the left-hand side admits a convergent power series expansion involving the $p$-adic Bernoulli numbers - just as the higher order terms in the usual Stirling formula involve the usual Bernoulli numbers. This is covered by formulas (35) and (38) in Granville's article. When $p \geq 5$, all terms in the expansion are seen to have at least the stated valuation $\delta := 3+\mathrm{ord}_p\{nm(n-m)\}$, and the only term with valuation possibly $\delta$ has the form $B_{p-3} \times p^3 \times nm(n-m) \times u_p$, where $u_p \in \mathbb{Z}_p^{\times}$ is a $p$-adic unit. 
Therefore the congruence holds modulo $p^{4 + \mathrm{ord}_p(nm(n-m))}$ if and only if $p \mid B_{p-3}$. Those are the Wolstenholme primes referred to by Greg Kuperberg in his answer.
[I came to this old question via the "Related" column for the new post A congruence involving binomial coefficients by Richard Stanley, which proposes a different generalization of Wolstenholme's $\binom{2p}{p} \equiv 2 \mod{p^3}$. ]<|endoftext|>
TITLE: Finite number of minimal ideals
QUESTION [10 upvotes]: What is the necessary condition on a ring that guarantees the number of minimal non-zero ideals to be finite? Neither Noetherian or Artinian condition seems sufficient, and the ring being semisimple seems too strong.

REPLY [3 votes]: I'm not sure how to answer your exact question ("the necessary condition that guarantees..."?), but here are a few minor observations.   A direct product of commutative rings has only finitely many minimal (by definition nonzero) ideals if and only if each component ring has only finitely many and all but finitely many of the component rings have none.   So it suffices to consider indecomposable commutative rings with only finitely many ideals.   There are all sorts of indecomposable commutative rings with no minimal ideals.   Now suppose the indecomposable commutative ring has a positive finite number of minimal ideals.   The socle of such a ring has square zero; thus, the socle is a nonunital subring with the structure of an additive abelian group with zero multiplication on it (of course, this additive abelian group need not have only finitely many minimal subgroups).   One can construct all sorts of examples of this sort.   For instance, let $A$ be an indecomposable commutative ring with no minimal ideals, and let $M$ be an $A$-module with only finitely many simple submodules.   (For example, one might take $M$ to be a uniserial $A$-module.)   Let $R = A \oplus M$ as an additive group, with multiplication given by $(a_1, m_1) (a_2, m_2) = (a_1 a_2, a_1 m_2 + a_2 m_1)$.   Then $R$ is an indecomposable commutative ring with only finitely many minimal ideals.<|endoftext|>
TITLE: What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ?
QUESTION [8 upvotes]: Let $P,Q$ be $n$ by $n$ invertible matrices. Suppose further that $P$ and $Q$ satisfies the following equation :
$$(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$$
where $\circ$ denotes the Hadamard matrix product, which is simply the entrywise product. 
Then what can be said about $P$ and $Q$? More precisely, I want to know if there are additional relations between $P$ and $Q$. For example, one can show that the condition $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ implies
$$tr(P^{-1}DPE) = tr(Q^{-1}DQE)$$ for all diagonal matrices $D$ and $E$.
References in the litterature about matrices of the form $(P^{-1})^T \circ P$ would help too.
Thank you,
Malik

REPLY [3 votes]: Actually, if $D, E$ are diagonal and $Q=EPD$, then $(P,Q)$ is such a pair. A natural question is whether every pair such that $P^{-T}\circ P=Q^{-T}\circ Q$ is of the form above. But even this is false, because if $P$ is triangular, then $P^{-T}\circ P=I_n$.
I take the occasion to mention an open question: define $\Phi(P):=P^{-T}\circ P$ (the gain array). What are the matrices $P$ such that $\Phi^{(k)}(P)\rightarrow I_n$ as $k\rightarrow+\infty$ ? According to Johnson & Shapiro, this is true at least for

Strictly diagonally dominant matrices
Symmetric positive definite matrices

On the contrary, $\Phi$ has fixed points, for instance the mean $\frac12(P+Q)$ of two permutation matrices. See Exercises 335, 336, 342, 343 of my blog about Exercises on matrix analysis.<|endoftext|>
TITLE: Unipotency in realisations of the motivic fundamental group
QUESTION [19 upvotes]: Deligne, in his 1987 paper on the fundamental group of $\mathbb{P}^1 \setminus \{0,1,\infty\}$ (in "Galois Groups over $\mathbb{Q}$"), defines a system of realisations for a motivic fundamental group. Namely, Tannakian categories are defined of unipotent lisse $\ell$-adic sheaves, unipotent vector bundles with a flat connection and the crystalline equivalent given by unipotent overconvergent isocrystals, whereas the automorphism groups of their fibre functors define the $\ell$-adic, de Rham and crystalline fundamental groups respectively. Now, I do not claim to have a deep understanding of these constructions, nor of the paper as a whole (or to even have read it fully in detail, given its size!), but I was left with a question: why does the word unipotent appear in all the constructions?
Having read parts of Minhyong Kim's work it is understandable that in this context, these subcategories generated by the unipotent objects are not only enough, but seemingly exactly what is needed. In other words, one still acquires the Tannakian structure (which if I'm not mistaken exists on the full categories as well) and thus has fundamental groups which are affine group schemes and can start playing the game of using unipotency and quotients in the lower central series to construct moduli spaces of path torsors, moving gradually away from abelianness and deeper into Diophantine information. This however doesn't answer the original motivation.
I apologise if the answer is hidden in Deligne's original paper or if the context makes it obvious to everyone but me.

REPLY [16 votes]: Essentially because the Tannakian theory gives in the unipotent case (and only in
that case) a reasonably sized answer with an easy motivic interpretation.
For the size you should be aware that already in the topological situation the
group scheme associated by Tannaka theory to the fundamental group of $\mathbb
P^1$ minus three points (i.e., the free group on two generators) is
huge. For one thing each irreducible representation gives rise to a
reductive quotient and there are continuous families (i.e., positive dimensional
varieties) of such representations. From this one can see that the group scheme
maps onto a product of reductive groups where the index set are the points of
some algebraic variety (see
What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?
for the case of $1$-dimensional representations).
Added to this is the fact that most of the topological representations do not
have geometric origin and hence have no motivic interpretation. If one looks at
$\ell$-adic representations of the fundamental group over $\mathbb Q$ of $\mathbb
P^1$ minus three points (or of suitable germs if one wants a theory over
$\mathbb C$) and adds mixedness assumptions, then the Tannakian category should
have a motivic interpretation which also should be independent (in some suitable
sense) of $\ell$ and should be comparable to its cristalline equivalent. This
however all depends on the Langlands program and hence is currently beyond our
reach.
If one sticks to unipotent representations then essentially all these problems
disappear. A unipotent representation (over some field $k$ of characteristic
$0$) of the free group $F$ on two elements factors through a nilpotent quotient $\Gamma$ of
$F$ and such a nilpotent quotient has a Malcev completion, a unipotent algebraic
group $G$ over $\mathbb Q$ of dimension the rank of $\Gamma$, such that the
Tannakian category of unipotent representations of $\Gamma$ over $k$ is
equivalent to the category of $k$-representations of $G$. Passing to the limit
gives us a pro-unipotent algebraic group $G_\infty$ over $\mathbb Q$ whose category of
$k$-representations is equivalent to the category of unipotent
$k$-representations of $F$. Furthermore, the Lie algebra of $G_\infty$ has a
nice cohomological description; it is the free Lie algebra generated by
$H_1(X,\mathbb Q)$, where $X$ is $\mathbb P^1$ minus three points.
The motivic side of things now comes along very gracefully: For one of several
natural categories that has an appropriate $H_1(X)$ in it there is a
corresponding relative Tannakian description of unipotent families of objects
over $X$. As examples we have unipotent variations of rational Hodge structures,
geometrically unipotent $\mathbb Q_{\ell}$-adic sheaves over $\mathbb Q$ and
successive extensions of constant $F$-iso-crystals. In all these cases these
categories are described by representations of a pro-unipotent algebraic group
object in the appropriate base category (rational Hodge structure, $\mathbb
Q_{\ell}$-adic sheaves over $\mathbb Q$ and $F$-isocrystals) and in all the
cases its Lie algebra is the free Lie algebra on $H_1(X)$.

REPLY [14 votes]: I'm sure other people can, and will, give more detailed and insightful answers, so I'll be
brief. From my point of view, the unipotent part of the motivic fundamental group is
an important approximation to the true object. Nevertheless it is an approximation. The
full motivic fundamental group should be the group associated the Tannakian category of
all motivic local systems (once it's constructed!), and this would include motives associated
to smooth projective families which are not unipotent. 
One had a similar situation in Hodge theory, where most of the initial effort was on the unipotent side, but in the last few years people have started looking at the deeper structure. More information about this can be found in
http://arxiv.org/abs/0902.4252<|endoftext|>
TITLE: Uses of bisimulation outside of computer science.
QUESTION [11 upvotes]: Bisimulation is one of the most important ideas of theoretical computer science. I was wondering whether bisimilarity is used/known outside of computer science/modal logic? I am aware that it corresponds more or less to back and forth techniques from model theory, but are there any other areas where it finds application?
For those not in the know, here is a definition:

Given a labelled transition system $(S,\Lambda,\to)$,
  a bisimulation relation is a binary relation $R$ over $S$ (that is, $R\subseteq S\times S$) such that for all pairs of elements $p,q\in S$ with $(p,q)\in R$, and for all $\alpha\in\Lambda$, we have 

$p\to^\alpha p'$ implies that there is a $q'$ such that $q\to^\alpha q'$ and $(p',q')\in R$, and
symmetrically for $q$, namely, $q\\to^\alpha q'$ implies that there is a $p'$ such that $p\to^\alpha p'$ and $(p',q')\in R$.


Applications collected thus far in answers include:

process equivalence in concurrency theory
model logic: expressiveness characterisations, modal correspondence theory
coinduction, for example in Game Theory
non-well founded set theory
algebraic set theory
geometric topology

REPLY [2 votes]: Bisimularity of graphs arises in geometric topology.  See "Quasi-isometric classification of graph manifold groups" by Behrstock and Neumann.<|endoftext|>
TITLE: Decide how many non-negative solutions a set of multivariate quadratic equations have
QUESTION [5 upvotes]: Given a set of multivariate, quadratic, non-homogeneous equations, is there a way to decide how many non-negative roots it have?
Some explanations:

All the coefficients are real numbers.
The number of variables is the same as the number of equations, and all the equations are independent.
Some of the equations might actually be linear equations.
By "non-negative" solution, I mean a solution in which all the variables take non-negative values.
By some "physical" reasoning, we know it must have at least one non-negative solution.

Further explanations:

I know generally a set of n quadratic equations with n variables has at most $2^n$ distinct roots.
The background of this problem is: the set of quadratic equations is the right-hand side of the chemical rate equation. By equating is to zero, the steady-state case is being considered. As we only take one-body or two-body reactions into account, the degree of the equations are at most two. As the abundance of the molecules cannot be negative, we only care about the non-negative solutions.
The number of variables can be up to 1000, so simple numerical test is not practical.

I am not a math student, and I am not sure whether this kind of question is allowed here.

REPLY [4 votes]: Not efficiently, at least not unless the problem has some additional structure which can be exploited.  The set of mixed Nash equilibria of a two-player game can be written as the nonnegative solutions of such a polynomial system.  In general it is #P-hard to count the equilibria of such a game (Conitzer and Sandholm).  It is even PPAD-hard (still thought to be polynomial time unsolvable) to compute a single equilibrium, although one is guaranteed to exist by Nash's Theorem.
You may want to investigate connections with Nash equilibria further -- if your problem is somehow equivalent to this one then you will have a lot of established results and techniques to build on.  Also people would probably find a connection between game theory and chemistry surprising.  Or if your problem is easier, looking at Nash equilibria might help you figure out what extra features your problem has that makes it solvable efficiently.  If it is harder or incomparable it might still be interesting to know that.<|endoftext|>
TITLE: Floer homology and status of the Arnold conjecture
QUESTION [13 upvotes]: The Arnold conjecture on a closed symplectic manifold $(M,\omega)$ says in the weakest version that for a non-degenerate Hamiltonian there are at least $k$ 1-periodic orbits where $k$ is the sum of the Betti numbers of $M$. It is easy to show that one can assume that $\omega$ is integral, so I do so in the following.
On wikipedia it says that the Arnold conjecture is solved in many cases using Floer homology. 
However, I was given the impression that this version of the Arnold conjecture has been solved in all cases, but are scattered around in several papers - due to several different complications.
Question: What is the current status of this weak Arnold conjecture precisely, and what are the refferences for these results?
Added: I know most of the details of the monotone case, so I am mostly interested in the more exotic cases.

REPLY [21 votes]: V. I. Arnol'd, June 12, 1937 - June 3, 2010.
The very sad news of his death is reported today here.

After Floer, the main difficulty in solving the weak Arnol'd conjecture on a compact symplectic manifold $M$ lies in defining a Floer chain complex generated by 1-periodic orbits of an arbitrary non-degenerate Hamiltonian $H: S^1\times M \to \mathbb{R}$, in such a way that the homology is independent of $H$. Once one has that, the remaining step (proving an isomorphism with Morse homology) can be done either by a computation with small autonomous Hamiltonians, or by a "PSS" isomorphism.
When $M$ is monotone, the crucial compactness theorems for solutions to Floer's equation (used to define the candidate-differential on the Floer complex, to prove that it squares to zero, and, in a variant, to prove the invariance of the theory) can be proved using index considerations. When $M$ is Calabi-Yau, compactness needs an additional idea, that holomorphic spheres generically don't hit cylinders solving Floer's equation. This is beautifully worked out in

Hofer, H.; Salamon, D. A. "Floer homology and Novikov rings." The Floer memorial volume, 
483--524, Progr. Math., 133, Birkhäuser, Basel, 1995; MR1362838.

In general, where there may be holomorphic spheres with small negative Chern number, one has little choice but to allow "stable trajectories" consisting of broken Floer trajectories with holomorphic bubble-trees attached. Transversality is proved by introducing multi-valued perturbations to the equations, and this forces one to use rational coefficients. References:

Fukaya, Kenji; Ono, Kaoru. "Arnold conjecture and Gromov-Witten invariant". Topology 38 (1999), no. 5, 933-1048. MR1688434


Liu, Gang; Tian, Gang, "Floer homology and Arnold conjecture", J. Differential Geom. 49 (1998), no. 1, 1-74. MR1642105

[Edit: both these references offer proofs of the weak Arnol'd conjecture with rational coefficients.] For a detailed introduction to these "virtual transversality" methods, see

Salamon, Dietmar, "Lectures on Floer homology". MR1702944

The technical complications of virtual transversality theory are notorious, and one could wish for a fully detailed textbook account.
What's left?
So far as I know, there is no proof for general manifolds that the number $h$ of 1-periodic orbits of a non-degenerate Hamiltonian is at least the sum of the mod $p$ Betti numbers. The strong Arnol'd conjecture for non-degenerate Hamiltonians, that $h$ is at least the minimum number of critical points of a Morse function, is wide open.<|endoftext|>
TITLE: What is the best way to peel fruit?
QUESTION [34 upvotes]: A mango made me wonder about this. (See also this question, which is in a similar spirit.)
Fix $L >0$ and a smooth body (possibly nonconvex—pears or bananas are fair game!) $B \subset \mathbb{R}^3$ (and assume w/l/o/g below that $L$ is sufficiently large since we can dilate $B$). For $\gamma:[0,L] \rightarrow \mathbb{R}^3$ smooth and parametrized by arclength and $\theta:[0,L] \rightarrow S^1 $ smooth, let $k(\gamma, \theta,s)$ denote a copy of the unit interval centered at $\gamma(s)$ and in the plane orthogonal to $\dot \gamma(s)$, and at the angle $\theta(s)$ in that plane (we require $k(\gamma,\theta,0)$ to be tangent to $B$, say, and w/l/o/g that this sets $\theta(0) = 0$; angles in planes away from $s=0$ can be sensibly defined via parallel translation). Let $K(\gamma,\theta):= \{ k(\gamma, \theta,s) \cap B : s \in [0,L] \} $. If $K$ contains the boundary of a body $C_K \subset B$ then say that $(\gamma, \theta)$ is a peeling of $B$.
For $L$ fixed, is there an effective way to determine a peeling that minimizes $\mbox{vol}(B \backslash C_K)$? 
Followup: can the best peeling of the unit ball for a given value of $L$ be explicitly constructed?

REPLY [11 votes]: Sorry, I'm still not allowed to comment. So I use the "Answer" window...
I'm not completely sure to understand your formulation, but for the case of a 2-dimensional sphere and some fixed width of the pealing, you may find your answer in the sphere-filling ropes of Gelrach and von der Mosel. These are ropes with a certain fixed width that are going on a sphere and trying to cover the greatest area. For some width, it is possible to cover everything. 

Heiko von der Mosel et Henryk Gerlach On sphere-filling ropes. Amer. Math. Monthly 118 (2011), no. 10, 
Heiko von der Mosel et Henryk Gerlach : What are the longest ropes on the unit sphere ? Arch. Ration. Mech. Anal. 201 (2011), no. 1, 303–342.<|endoftext|>
TITLE: How do researchers carry out computational experiments in Graph Theory?
QUESTION [5 upvotes]: There are packages like, Combinatorica in Mathematica, GA, SA, PSO can be comparatively easily done in MATLAB, C++ has boost, Java has JGraphT and so on and on. My question is,

How do Graph Theorists carry out large computational experiments?
What languages or packages or libraries do they usually use? Is their any general preferences?
Do they use a combination of many sporadic resources? Like use something that is good at plotting, then use some other thing that is good for numerical solutions etc?
Where from you get Hard-Instances ?

Background:
I'm assigned to work on a GT problem named 'Degree Constrained Minimum Spanning Tree'. I want to study, implement and compare current algorithms. As I've been studying, I come to observe most of the algorithms are : Heuristics, Distributed Algorithms, Genetic Algorithms, Linear Programming (Narula-Ho) etc. Some uses methods called 'Particle Swarm Optimization', 'Simulated Annealing' and 'Ant Colony Optimization'. As far as I know MATLAB has built in routines for GA, SA, PSO, ACO etc but don't have any graph theory package. Combinatorica seems very good package but I don't have any access to it's accompanying book 'Computational Discrete Mathematics'. I don't know MATLAB or Mathematica.
Update:
A comprehensive list on GT packages/systems can be found here: http://wiki.sagemath.org/graph_survey
Update
It seems Mathematica 7 has all the above, mostly as built in functions/Commands.

REPLY [4 votes]: As TheMadman mentionned in a comment, Sagemath has an impressive Graph library. I have been working on it for quite a while, and I expect you wouldn't find anywhere else the list of methods it has.
I don't think I already wrote a solver for a degree-constrained spanning tree, but I definitely implemented a degree-constrained subgraph there. As Mixed Integer Linear Programming is easily available, something like 20 lines of code are enough to get an exact solver for the min degree spanning tree problem anyway.
You will find help of its methods (and so the list of them) at this address :
http://www.sagemath.org/doc/reference/graphs.html
be sure to check both "generic graphs" and "Graph" :-)
Nathann<|endoftext|>
TITLE: The harmonic (series) beetle: live illustrations of mathematical theorems
QUESTION [8 upvotes]: In my analysis class I use the following problem to illustrate the divergence
of the harmonic series (consider this as a hint for solving it).
Exercise.
A beetle creeps along a 1-meter infinitely elastic tape with constant velocity.
Every hour the tape is lengthened out by 1 meter, and the beetle remains at the
same rate of the tape it has already reached to the moment. Will the beetle ever reach
the end of the tape?
This is not a paradox but a calculation of a mathematically idealised model
"from life". Do you have some other, probably nicer examples which illustrate
some standard but deep results in analysis, algebra, probability, geometry,
and so on, and so on, and so on. Please keep in mind an average undergraduate
student as the audience for your example and allow others to use it in his/her
teaching.
Thank you!

REPLY [2 votes]: How do you hold a pizza (with your hands) so that the toppings don't slide off? That is, how do you keep the pizza "straight"? You bend it through the middle and use one of your fingers to support the bend. Why does it work? Because a pizza is inherently "flat." You would expect any reasonable definition of curvature to give a (flat) pizza slice a 0. It turns out this expression (Gaussian curvature) is the product of two numbers, $\kappa_1$ and $\kappa_2$, corresponding to the two orthogonal curvatures. Thus, one of the two curvatures must be 0 -- so if you're bending one direction, the other must necessarily be "straight" (no curvature).<|endoftext|>
TITLE: Asymptotic approximation of $x^\alpha$ by entire functions
QUESTION [15 upvotes]: Given a non-integral real $\alpha$, is there an entire (see http://en.wikipedia.org/wiki/Entire_function) function $h(x)$ such that $x^{-\alpha}h(x)\longrightarrow 1$
for $x\rightarrow+\infty$ (with $x$ real non-negative)?
Clearly, such a function if it exists is not unique since $h(x)+e^{-x}$ and similar functions
work also.

REPLY [24 votes]: As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in  0, with real coefficients and radius of convergence infinite) are dense in $C^0({\mathbb R}, \mathbb{R})$ in the sense of the order, that is:

Theorem (T.Carleman, 1927). For any two continuous real valued
  functions f < g there exists a real
  entire function $\phi$ in between:
$f(x)<\phi(x) < g(x)$ for all $x\in\mathbb{R}$.

So in particular, an entire function may be asymptotic to any continuous real function, and also, it may grow as fast as any continuous function.<|endoftext|>
TITLE: Gauss-Bonnet Theorem for Graphs?
QUESTION [23 upvotes]: One can define the Euler characteristic χ for a graph as the number of vertices minus the number of edges.  Thus an $n$-cycle has $\chi = 0$ and    $K_4$ has $\chi=-2$.
Is there an analog for the Gauss-Bonnet theorem for graphs, something akin to:
 
     [total turn angle] $+$ [enclosed curvature] 
$= \tau + \omega = 2 \pi \chi$ ?

Certainly if one embeds the graph on a manifold, then an interpretation is possible via Gauss-Bonnet on the manifold.  But is there a more purely combinatorial interpretation?
Addendum. (27Nov11).
A new paper on this topic just appeared on the arXiv:
Oliver Knill (who answered below back in March),
"A graph theoretical Gauss-Bonnet-Chern Theorem."
arXiv:1111.5395v1.  Here is Knill's first figure:

REPLY [6 votes]: There are different type of curvatures for graphs. In two
dimensions its not the degree of the point which matters but
the length of the circles at the point like in differential geometry.
This is different from the degree if graphs with boundary are considered.
The simplest curvature for two dimensional graphs 
(I defined the dimension for graphs in
http://arxiv.org/abs/1009.2292). For the curvature
K(x) = 6-|S(x)|, any two dimensional graph G has total curvature
6 chi(G). More interesting and subtle is a second order curvature 2|S1(x)|-|S2(x)| 
which is differential geometrically motivated because curvature is a 
second order difference notion. Gauss-Bonnet is now not true in general
and it seems that some smoothness condition needs to be satisfied. It
is true for all two dimensional graphs non-negative curvature and for graphs
which are "smooth" enough in some sense which I still struggle to define.
I myself got more interested in Gauss-Bonnet-Chern for n-dimensional graphs
and have a result there. There is a natural Chern-Euler form on n-dimensional
graphs those total sum is the Euler characteristic for an n-dimensional graph. 
This form is defined also in odd dimensions but seems to be zero (it is in
3 dimensions but I still struggle to verify this in higher dimensions).
Note that all this is purely graph theoretical. No additional structure on
the graph is necessary. No embedding in an ambient Euclidean space
necessary for example.
The inductive dimension as defined in the above mentioned paper (revised here:
http://www.math.harvard.edu/~knill/graphgeometry/papers/1.pdf)
is a natural way to define polyhedra and polytopes as graphs which become
n-dimensional graphs after some truncation or stellation procedure.
The inductive dimension for graphs looks similar to inductive dimension
usually considered in topology but is utterly different. 
With the usual inductive dimension, any graph would be one dimensional.<|endoftext|>
TITLE: Non-symmetric Braiding on finite group Representation Categories
QUESTION [6 upvotes]: Do the fusion categories $Rep(S_4)$ and $Rep(A_5)$ admit non-symmetric braidings?  All the other rep. cats. of finite subgroups of $SU(2)$ do (in the McKay correspondence).  My guess is no.

REPLY [6 votes]: You can also answer the question using the classification of R-matrices over group algebras. Indeed, the R-matrices in the Hopf algebra $ kG $, are classified by pairs $ (A, \rho) $ where $ A \subset G $ is an abelian normal subgroup and $ \rho: A \times A \to k^* $ is a bilinear form ad $ G $-invariant. It is easy to see that the associated R-matrix is symmetric if and only if the bilinear form is skew-symmetric.
It is now clear that $ kA_5 $ does not have R-matrices in addition to the trivial. On the other hand, the only normal abelian subgroup of $ S_4 $ is formed by the union of the conjugacy classes of $(12)(34)$ and the identity of $S_4$. This normal subgroup is isomorphic to $C_2 \times C_2 $. As was mentioned by Ostrik, one example could be the next: the quadratic form that takes values -1 on nontrivial elements. That bilinear form satisfies the conditions and defines a non-symmetric R-matrix.<|endoftext|>
TITLE: When is it true that the ring of global regular functions on a projective variety is just the base ring?
QUESTION [9 upvotes]: Apologies in advance if this is too elementary.
The following is well known when $A$ is an algebraically closed field:

Let $X$ be an integral closed subscheme of $P^n_A$. Then $\Gamma(X, \mathcal{O}_X) = A$. 

My question is: For what other rings does the above statement hold?
There are two  proofs of this (for $A$ algebraically closed) in Hartshorne. Both seem to use the fact that the integral closure of $A$ in its quotient field is just $A$ itself in a key way.
So I suspect that having $A$ integrally closed  will be crucial, but I do not know.  In particular, does the proof in Hartshorne still work, and if so, does it apply to nonnormal domains?

REPLY [5 votes]: The statement is true if and only if $A$ is an algebraically closed field.
Assume the statement is true.
Let $\mathfrak{m}$ be a maximal ideal of $A$.  Then $A \to A / \mathfrak{m}$ is certainly finite.  Thus, as I explain in my comment above, $\mathbb{P}^n_{A / \mathfrak{m}}$ is an integral subscheme of some $\mathbb{P}^N_A$.  Thus, $A = A / \mathfrak{m}$; since $\mathfrak{m}$ was maximal, $A$ is a field.
Suppose $A$ is not algebraically closed.  Then $A$ has a finite extension $B$; and again, as remarked in the comment, this implies that $B$ is the ring of global regular functions of some integral closed subscheme of $\mathbb{P}^N_A$, a contradiction.<|endoftext|>
TITLE: Who first proved that there are at least n^(1-ε) primes up to n?
QUESTION [15 upvotes]: It's well-known that Hadamard and de la Vallée-Poussin independently proved the Prime Number Theorem in 1896: that $\pi(n)=n/\log n+o(n/\log n)$.  I'm curious as to a weaker result: that for any $\varepsilon>0$, $\pi(n)\gg n^{1-\varepsilon}$.
Chebyshev famously proved that if $\lim \pi(n)\log n$ exists it must be equal to 1, but I seem to remember that he also proved bounds on that value, pushing the date back to 1850 or so in that case.  But were there earlier results in this direction?

REPLY [17 votes]: The following is an MO-adopted extract from my lectures.
Euclid's theorem about the infiniteness of primes
can be written as
$$
\pi(x)\to+\infty \qquad\text{as}\quad x\to+\infty.
$$
However, this theorem tells us nothing about how fast the function
$\pi(x)$ increases with $x$.
In 1808 Legendre published a discovered empirical formula, namely,
$$
\pi(x)\approx\frac x{\ln x-1.08366};
$$
some years later Gauss noted that $x/(\ln x-1)$ and even
$$
\int_2^x\frac{dt}{\ln t}
$$
is a much better approximation to $\pi(x)$ for larger $x$.
In 1850 Chebyshev published his work
containing the following result.
Theorem (Chebyshev).
There exist two constants $A$ and $B$,
$0< A< 1< B$, such that for all $x\ge2$ the following bounds are valid:
$$
\frac{Ax}{\ln x}< \pi(x)< \frac{Bx}{\ln x}.
\qquad(1)
$$
Chebyshev actually proved the theorem by showing that one can take
$A=0.921$ and $B=1.106$ for $x\ge x_0$, so that
$A$ and $B$ are very close to $1$. In our simplified proof
we will get more modest estimates for $A$ and $B$.
Note however that for the proof of the prime number theorem
one needs the existence result for
these constants. Chebyshev's theorem states the correct
order of growth of the function $\pi(x)$
as $x\to+\infty$.
We need some properties of the function
$$
T(x)=\sum_p\ln p\biggl(\biggl[\frac xp\biggr]
+\biggl[\frac x{p^2}\biggr]
+\biggl[\frac x{p^3}\biggr]+\dots\biggr),
$$
where $[\cdot]$ denotes the integer part of a real number.
Lemma 1.
For positive integers $n$ we have the equality
$T(n)=\ln n!$.
Proof.
Factorise $n!$ into product of primes,
$n!=\prod_pp^{\nu_p}$, and use
the formula
$$
\operatorname{ord}_p n!=\biggl[\frac np\biggr]
+\biggl[\frac n{p^2}\biggr]
+\biggl[\frac n{p^3}\biggr]+\dots
$$
(the sum is in fact finite!) to see that
$$
\nu_p=\biggl[\frac np\biggr]
+\biggl[\frac n{p^2}\biggr]
+\biggl[\frac n{p^3}\biggr]+\dots
\qquad\text{for each prime $p$}.
$$
Taking the logarithm of the factorisation of $n!$
we deduce the required equality.
Lemma 2.
The function $T(n)$ (of natural argument $n$) is nondecreasing.
Proof.
It follows from Lemma 1 that
$$
T(n+1)-T(n)=\ln(n+1)!-\ln n!=\ln(n+1)> 0.
$$
Lemma 3.
If $[x]=n$, then $T(x)=T(n)$.
Proof.
Let $n=[x]$, $\alpha=\{x\}$ (the fractional part of $x$, hence $0< \alpha< 1$),
and let $m$ be a power of prime $p$.
If $[n/m]=q$, then $n=mq+r$ where $0\le r\le m-1$.
Therefore, $x=n+\alpha=mq+(r+\alpha)$ where
$0\le r+\alpha< m$; consequently,
$$
\biggl[\frac xm\biggr]
=\biggl[q+\frac{r+\alpha}m\biggr]
=q+\biggl[\frac{r+\alpha}m\biggr]
=q=\biggl[\frac nm\biggr].
$$
From this inequality and from the definition of $T(x)$
the required property follows immediately.
We will also need the following elementary inequality:
for $n\ge3$,
$$
(n+1)^3\le2^{2n}.
$$
Lemma 4.
For $x\ge6$ the following estimates take place:
$$
\frac{\ln2}2\cdot x< T(x)-2T\biggl(\frac x2\biggr)
< \frac{4\ln2}3\cdot x.
$$
Proof.
Denote $n=[x/2]\ge3$, so that $2n\le x< 2n+2$. Then by Lemma 3,
$$
T(x)-2T\biggl(\frac x2\biggr)
=T(2n)-2T(n) \quad \text{if $2n\le x< 2n+1$}, \qquad\text{and}\qquad
=T(2n+1)-2T(n) \quad \text{if $2n+1\le x< 2n+2$},
$$
hence with the help of Lemma 2 we obtain
$$
T(2n)-2T(n)
\le T(x)-2T\biggl(\frac x2\biggr)
\le T(2n+1)-2T(n).
\qquad(2)
$$
By Lemma 1 for $n\ge3$ we find that
$$
T(2n)-2T(n)
=\sum_{k=1}^{2n}\ln k-2\sum_{k=1}^n\ln k
=\sum_{k=n+1}^{2n}\ln k-\sum_{k=1}^n\ln k
$$
$$
=\ln\frac{n+1}{1}+\ln\frac{n+2}{2}+\dots+\ln\frac{n+n}{n}
\ge(n+1)\ln2
\gt \frac{x}{2}\ln 2,
\qquad (3)
$$
so that
$$
\ln\frac{n+1}1\ge\ln4=2\ln2,
\qquad
\ln\frac{n+k}k=\ln\biggl(1+\frac nk\biggr)\ge\ln2
\quad\text{for } k=2,3,\dots,n,
$$
and $x/2< n+1$.
On the other hand, for $n\ge3$,
$$
\frac{(2n+1)!}{n!^2}
=(n+1)\cdot\frac{(2n+1)!}{n!(n+1)!}
=\frac{n+1}2\cdot\biggl(\binom{2n+1}n+\binom{2n+1}{n+1}\biggr)
$$
$$
< \frac{n+1}2\sum_{k=0}^{2n+1}\binom{2n}k
=\frac{n+1}2\cdot(1+1)^{2n+1}=2^{2n}(n+1)
$$
$$
\le2^{2n+2n/3}=2^{8n/3}
$$
(on the last step we use the above elementary inequality).
Thus, by Lemma 1
$$
T(2n+1)-2T(n)=\ln\frac{(2n+1)!}{n!^2}
< \frac{8n}3\ln2\le\frac{4x}3\ln2.
\qquad(4)
$$
Substituting the inequalities (3) and (4)
into (2), we arrive at the desired claim.
Lemma 5.
For a real number $\alpha$, the difference
$[2\alpha]-2[\alpha]$ is either $0$ or $1$.
The proof of the lemma is straightforward.
Proof of Chebyshev's theorem.
Lower bound.
Using the definition of $T(x)$ we obtain
$$
T(x)-2T\biggl(\frac x2\biggr)
=\sum_{p\le x}\ln p\biggl(
\biggl(\biggl[\frac xp\biggr]-2\biggl[\frac x{2p}\biggr]\biggr)
+\biggl(\biggl[\frac x{p^2}\biggr]-2\biggl[\frac x{2p^2}\biggr]\biggr)
+\dots\biggr),
\qquad(5)
$$
since all the terms on the right-hand side vanish for $p> x$.
The groups
$$
\biggl(\biggl[\frac x{p^k}\biggr]-2\biggl[\frac x{2p^k}\biggr]\biggr)
$$
also vanish for $p^k> x$, whence the coefficient of $\ln p$
on the right-hand side of (5) can be written as
$$
\sum_{k=1}^s
\biggl(\biggl[\frac x{p^k}\biggr]-2\biggl[\frac x{2p^k}\biggr]\biggr),
$$
for $p^s\le x$, that is, for $s\le\dfrac{\ln x}{\ln p}$.
By Lemma 5 each term in parenthesis on the right-hand side of (5)
is at most $1$, and the number of such terms is at most $s$,
hence
$$
T(x)-2T\biggl(\frac x2\biggr)
\le\sum_{p\le x}\ln p\cdot\frac{\ln x}{\ln p}
=\ln x\cdot\sum_{p\le x}1
=\pi(x)\ln x.
$$
Consequently, by Lemma 4
$$
\pi(x)\ln x> \frac{\ln2}2x,
$$
and the desired lower estimate is deduced with $A=\frac12\ln2$.
Upper bound.
Since all the terms on the right-hand side of (5)
are non-negative, leaving those for which $p> x/2$ does not
increase the whole sum. For $x/2< p\le x$ we have
$$
\biggl(\biggl[\frac xp\biggr]-2\biggl[\frac x{2p}\biggr]\biggr)
+\biggl(\biggl[\frac x{p^2}\biggr]-2\biggl[\frac x{2p^2}\biggr]\biggr)
+\dots
=1,
\qquad(6)
$$
because for such $p$ and $x\ge6$ there holds $2p> x$ and $p^2> x^2/4> x$,
that is, the left-hand side of (6) involves a unique nonzero term,
$[x/p]$. Thus, from (5) we obtain
$$
T(x)-2T\biggl(\frac x2\biggr)
\ge\sum_{x/2< p\le x}\ln p
\gt \ln\frac x2\cdot\sum_{x/2< p\le x}1
=\biggl(\pi(x)-\pi\biggl(\frac x2\biggr)\biggr)\ln\frac x2
$$
$$
=\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2-\pi(x)\ln2
$$
$$
\ge\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2-x\ln2
$$
(where we applied the trivial bound $\pi(x)\le x$),
whence
$$
\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2< T(x)
-2T\biggl(\frac x2\biggr)+x\ln2< \biggl(\frac{4\ln2}3+\ln2\biggr)x< 2x
\qquad(7)
$$
for $x\ge6$. For any $x> 0$ there exists a positive integer $s$
such that $x/2^s< 2$, so that $\pi(x/2^s)=0$.
Successively applying the inequalities (7) we find that
$$
\pi(x)\ln x
< 2x+\pi\biggl(\frac x2\biggr)\ln\frac x2
< 2x+2\cdot\frac x2+\pi\biggl(\frac x4\biggr)\ln\frac x4
< \dots
$$
$$
< 2x+2\cdot\frac x2+\dots+2\cdot\frac x{2^{s-1}}
+\pi\biggl(\frac x{2^s}\biggr)\ln\frac x{2^s}
$$
$$
< 2x\biggl(1+\frac12+\frac14+\dots\biggr)
=4x,
$$
which gives us the required estimate with $B=4$.
Remarks.
In order to get $A$ and $B$ closer to $1$,
Chebyshev considered the expression
$$
T(x)-T\biggl(\frac x2\biggr)
-T\biggl(\frac x3\biggr)
-T\biggl(\frac x5\biggr)
+T\biggl(\frac x{30}\biggr)
$$
instead of $T(x)-2T(x/2)$; in addition, he first
deduced estimates for the functions
$$
\vartheta(x)=\sum_{p\le x}\ln p
\qquad\text{and}\qquad
\psi(x)=\sum_{p\le x}\biggl[\frac{\ln x}{\ln p}\biggr]\ln p,
$$
and then translated them for $\pi(x)$.
The functions $\vartheta(x)$ and $\psi(x)$ are now known as
Chebyshev's functions.<|endoftext|>
TITLE: Which finite groups are generated by n involutions?
QUESTION [8 upvotes]: One of the interesting problems in abstract polytope theory is to determine, for a given finite group, when that group is the automorphism group of a regular abstract polytope. This is equivalent to the following question: Given a finite group G, when is G generated by involutions $\rho_0, \ldots, \rho_n$ such that $(\rho_i \rho_j)^2 = 1$ if $|i - j| \geq 2$ and such that for all $I, J \subset \{0, \ldots, n\}$ we have $\langle \rho_i \mid i \in I \rangle \cap \langle \rho_i \mid i \in J \rangle = \langle \rho_i \mid i \in I \cap J \rangle$?
The last property can be difficult to check, so let's relax that requirement for now. If a finite group G is generated by n involutions such that non-adjacent generators commute, what can we say about the structure or size of G? Of particular interest: what if G is simple?
Here are a few simple observations:

For each n, the smallest (abstract) n-polytope has an automorphism group that is isomorphic to the direct product of n copies of $C_2$, corresponding to the trivial Coxeter diagram on n nodes. So a finite group G cannot be the automorphism group of an abstract regular n-polytope for $n > \log_2(|G|)$. 
A (nontrivial) group generated by involutions has even order.
The abelianization of a group generated by involutions such that nonadjacent generators commute is a quotient of the group in (1), the direct product of n copies of $C_2$.

REPLY [14 votes]: Really quite a few finite simple groups are generated by three involutions, two of which commute (n=3).
For instance, this papers provides a (revised) proof that almost all sporadic groups have such a generating set:
Mazurov, V. D. "On the generation of sporadic simple groups by three involutions, two of which commute."
Sibirsk. Mat. Zh. 44 (2003), no. 1, 193–198; translation in Siberian Math. J. 44 (2003), no. 1, 160–164
MR1967616
DOI:10.1023/A:1022028807652
Its references provide a large list of other simple groups with this property:

almost all groups of lie type in char 2: MR1131150
almost all alternating groups: MR1172472
almost all groups of lie type in odd char: MR1454692 (low rank exceptions) and MR1601503 (all large rank)

Since one of the reviews was inaccurate, I quickly checked through the sources I had access to, and the following is probably quite close to accurate: Every finite simple group other than:

A6, A7, A8, S4(3)=U4(2), M11, M22, M23, McL
L3(q), U3(q) for all prime powers q
L4(q) for even prime powers q

has a generating set consisting of three involutions, two of which commute.  In particular the Monster group does have such a generating set (a short proof in the first paper, an earlier proof due to Simon Norton in a letter).
I (quickly) verified that A6, A7, A8, S4(3), M11, M22, M23 have no generating set of involutions where non-adjacent involutions commute (even for more than three generators).  You can bound n by the 2-rank of the group: having lots of commuting involutions means you have a large elementary abelian subgroup, and so n ≤ 5 for these groups.<|endoftext|>
TITLE: Is there a generalization of Burnside's theorem for compact Lie groups?
QUESTION [7 upvotes]: Recall the following theorem due to Burnside:Let $G$ be a finite group and let $V$ be its  irreducible complex representation of dimension greater than 1, then the character
of this representation is $0$ on some element of $G$. Is this statement still correct
if $G$ is any compact Lie group? Thanks.

REPLY [7 votes]: The answer is yes--use the Weyl character formula, for example.
See: Patrick X. Gallagher, Zeros of group characters. Math. Z. Volume 87 (1965), Number 3.<|endoftext|>
TITLE: Two fancy ways of defining differential forms: How does one show that they are equivalent?
QUESTION [13 upvotes]: Given a smooth manifold M, the following procedures yield the differential graded algebra (Ω*(M),ddR) of differential forms:

Procedure 1 (synthetic geometry).
For each n, consider the object of infinitesimal n-simplices in M. This is a ringed space whose underlying space is the diagonal M ⊂ Mn+1, and whose structure sheaf is $C^\infty (M^{n+1}) / I^2$, where $I$ is the ideal defining the diagonal.
The objects of infinitesimal n-simplices assemble into a simplicial ringed space, and by taking global function one gets a cosimplicial algebra. The normalized Moore complex of that cosimplicial algebra is then canonically isomorphic to the deRham complex of M.
Procedure 2 (supergeometry).
Letting ℝ0|1 denote the odd line, we may consider the internal hom
$Map(\mathbb R^{0|1} , M)$ in the category of supermanifolds. The global functions $C^\infty( Map(\mathbb R^{0|1} , M))$ then form a ℤ/2-graded algebra. Noting that the supergroup of automorphisms of ℝ0|1 acts on $Map(\mathbb R^{0|1} , M)$, one can then upgrade that ℤ/2-graded algebra to a ℤ-graded algebra with differential d. Once again, that procedure recovers the deRham complex of M.


Why do these two recipes produce the same outcome?

More generally, one can imagine applying Procedures 1 and 2 to objects M that are more general than manifolds (supermanifolds, singular spaces, schemes, derived manifolds, differential stacks, infinity stacks, ...). Are those two procedures always going to agree?
If not, when do they agree, when do they disagree, and why?

REPLY [2 votes]: The first construction is an explicit realization of the oo-categorical derived de Rham space functor, which identifies all infinitesimal neighbour points, up to higher equivalence. So it's an explicit realization of the derived dR functor in Simpson-Teleman, I think. http://ncatlab.org/schreiber/show/path+%E2%88%9E-groupoid#Infinitesimal<|endoftext|>
TITLE: Is there a definition of analogue Weyl group for Lie super algebra?
QUESTION [12 upvotes]: I heard from some people working in Lie super algebra that there was no proper definition for Weyl group of Lie super algebra. I do not know Lie super algebra at all. But When I searched on Google, I found that it seems there still exists some definitions of Weyl group. 
I wonder whether there is a well-accepted definition for it. 
The reason I want to ask this question is that I need Weyl group for Lie super algebra to play with the geometry related to super Lie algebra. 
Another question is that I heard from some experts in Lie super algebra that there was no well-accepted super geometry related to Lie super algebra. 
However, It seems that one of the students of Manin, who is Dimitry Leites ever developed supergeometry.

REPLY [2 votes]: As it is clear from the other answers there are several viewpoints on Weyl group in super case.
Let me mention papers by Sergeev and Veselov, who also stands on the point of the Weyl groupoid in the super case. 
As far as I understand such viewpoint agrees with their works on 
Calogero-Moser integrable system.
https://arxiv.org/abs/1504.08310
"Orbits and invariants of super Weyl groupoid"
https://arxiv.org/abs/0704.2250
"Grothendieck rings of basic classical Lie superalgebras"
Abstract:
"The Grothendieck rings of finite dimensional representations of the basic classical Lie superalgebras are explicitly described in terms of the corresponding generalised root systems. We show that they can be interpreted as the subrings in the weight group rings invariant under the action of certain groupoids called Weyl groupoids."

PS
By the way, GL(n,F_q) for q=1 "is" Weyl group (see e.g. field with one element). I wonder is there some kind of super analog of that heuristics ?
Is there some kind of "bijection" between "irreps" and "conjugacy classes"
for Weyl grouppoid (see  MO270916) ?<|endoftext|>
TITLE: Deformations of sheaves via automorphisms. How to express $Ext^1$?
QUESTION [5 upvotes]: Let $X$ be a complex manifold (for example $\mathbb CP^n$), let $v$ be a holomorphic vector field on $X$, and let $F$ be a coherent sheaf (for example a vector bundle or a structure sheaf of a point). Then $v$ defines an element in $Ext^1(F,F)$. Indeed $v$ generates an action of $\mathbb C$ on $X$ and taking pull-backs of $F$ under this action we get a deformation of $F$, hence an element of $Ext^1(F,F)$.
Question. Is there any fancy (or not fancy) way to express the corresponding element of $Ext^1(F,F)$ in terms of $v$ and $F$? Maybe there is some construction with jets? (I understand, this is a bit vague)
Added. Two equally nice and far leading answers were given to this question. I would like summarise here what I understood from David's answer in down to earth terms. So, we want to associate an extension  $F\to E\to F$ to a vector field $v$. Suppose (just for the sake of been very much down to earth), that $F$ is a vector bundle, and $v$ has no zeros. Then we can consider $1$-jets of sections of $F$ in the direction of $v$ (or if you want, along trajectories of $v$). It is not hard to see that this is a bundle of rank $2rank(F)$, and this is exactly $E$, that we are looking for.

REPLY [7 votes]: Here is a not so fancy description.
There is a general principle (in algebraic geometry but applicable to some
neighbouring disciplines) that says that anything that is functorial and
commutes with base change admits an action by vector fields. The reason is that
a vector field on $X$ gives an automorphism on $X[\delta]$, the base extension
of $X$ to the ring of dual numbers, that is the identity modulo $\delta$. 
As an example that is unrelated to the question but illustrative take the
example of a vector bundle $E$ and its symmetric algebra $S^*E$. The fact that
the symmetric algebra is defined by a universal property and commutes with base
extension immediately gives an action of derivations of $\mathcal O_X$ on $S^*E$
(the usual proof of this uses these properties implicitly but is less
transparent as it doesn't use them explicitly).
In any case turning to the problem at hand, given a vector field $v$ we get an
automorphism $\varphi$ of $X[\delta]$ and hence an $\mathcal
O_{X[\delta]}$-module which is the pullback by $\varphi$ of the constant
extension $F[\delta]$ of $F$ to $X[\delta]$. Concretely it means that we
consider $F[\delta]=F\bigoplus F\delta$ with action of $f\in\mathcal O_X$ given
by $f\cdot e=fe+v(f)e\delta$ (and $\delta$ acting in the obvious way). The
identification of first order deformations of $F$ with $\mathrm{Ext}^1(F,F)$
is the done by considering this as an extension of $F=F\delta$ by
$F[\delta]/F\delta=F$.
This can be elaborated as follows (which should give a closer relation with the
answer provided by David): Instead of taking the pullback of $F[\delta]$ by
$\varphi$ we can instead consider the vector field as a morphism $\varphi\colon
X[\delta]\rightarrow X$ and pullback $F$ along it. We of course also have the
constant morphism $X[\delta]\rightarrow X$ and the coincide on the closed
immersion $X\hookrightarrow X[\delta]$. Hence the induced map
$X[\delta]\rightarrow X\times X$ factors through the first order neighbourhood
of the diagonal $X^{(2)}:=\mathrm{Spec}\mathcal O_{X\times X}/\mathcal I^2_\Delta
\hookrightarrow X\times X$, i.e., a map $X[\delta]X^{(2)}$ such that the
composite of the first projection $X^{(2)}\rightarrow X$ is $\varphi$ and with
the second is the constant map. (Concretely, we have $\mathcal I_\Delta/\mathcal
I^2_\Delta=\Omega^1_X$ and $v$ is a map $\Omega^1_X\rightarrow\mathcal O_X$ and
the map is just the pushout along this map.) Hence the pullback of $F$ along $\varphi\colon
X[\delta]\rightarrow X$ is obtained by first taking the pullback $F'$ of $F$ along
the first projection $X^{(2)}\rightarrow X$ and the pullback further on to
$X[\delta]$. However, $F'$ as an extension $F\bigotimes\Omega^1_X$ by $F$ is the
Atiyah extension and we the extension of $F$ by $F$ associated to $v$ is just
the pushout of it along the map $F\bigotimes\Omega^1_X \rightarrow F$ induced by
$v$. Hence, the extension is the pushout of the universal Atiyah extension.<|endoftext|>
TITLE: A Counterexample to the HIrsch Conjecture
QUESTION [5 upvotes]: Recently Francisco Santos has announced that he has a counterexample to the Hirsch conjecture. The last I heard it was circulating among several people and there would be a public version of it available soon. I am curious how close it is to release. Also has there been any progress in the attempt to find the vertices of the counterexample. The last I heard to find the vertices a series of steps had to be done and each step increased the complexity of the problem by a geometric factor making it difficult to complete the computation.

REPLY [3 votes]: The public version is now out. It is available here<|endoftext|>
TITLE: Is a weak monoidal category a monoid object in some category?
QUESTION [7 upvotes]: A monoid in the Category Cat is a strict monoidal category according to Wikipedia. Is it possible to weaken the monoid so that its realisation in Cat is a weak monoidal category? Do we shift up a dimension, and throw in a 2-morphism associator for the monoid satisfying the analogue of Mac Lane's Pentagon, and similarly for the identities? 
Or is it still possible to get a weak monoidal category using a monoid by changing the category we realise it in from Cat to some modification of it?

REPLY [7 votes]: Yes, you can define a pseudomonoid in any monoidal 2-category, such that a pseudomonoid in the 2-category Cat is precisely a non-strict monoidal category.  The definition of monoidal category, interpreted in terms of functors and natural isomorphisms, i.e. in terms of the 2-category Cat, tells you exactly how to define a pseudomonoid.<|endoftext|>
TITLE: Points on a sphere
QUESTION [18 upvotes]: Wonder whether any of you know where it was that the following pearl of topology first appeared:

Prove that at any instant of time you can find three isothermal points on the surface of the Earth that correspond to the vertices of an equilateral triangle.

According to Léo Sauvé, all problems appeared once in the Monthly. Does that dictum apply to the above teaser, too?

REPLY [23 votes]: Let me just elaborate a little on the references that Charlie Frohman listed (so this isn't really a separate answer, but it's too long for a comment).
The theorem for equilateral triangles is due to S. Kakutani (1942 Annals). He stated it just for triangles formed by orthonormal bases for $\mathbb R^3$, but the argument applies for all sizes of equilateral triangles. He deduced the interesting corollary that any compact convex set in $\mathbb R^3$ has a circumscribing cube, answering a question posed by Rademacher. Charlie Frohman's proof is similar in spirit to Kakutani's.
The n-dimensional generalization of Kakutani's theorem (and corollary) is due to H. Yamabe and Z. Yujobo (1950 Osaka Math J).
Returning to 3 dimensions, the corresponding result for 4 points at the vertices of a square inscribed in a great circle was shown by F. Dyson (1951 Annals), and G.R. Livesay generalized this to rectangles inscribed in a great circle (1954 Annals). The case of arbitrary triangles was done by E.E. Floyd (1955 PAMS).
It's interesting that this little topic produced three Annals papers, though each one was short -- just 3 pages.  (When was the last 3-page Annals paper?)
The techniques used to prove the later theorems varied considerably. It might be interesting to see which ones can be proved by basic algebraic topology arguments as in Kakutani's theorem (which certainly deserves to be included in algebraic topology textbooks!).<|endoftext|>
TITLE: To what extent is it true that "number theory = mathematics"?
QUESTION [19 upvotes]: In a thought-provoking answer to this MO question, Kevin Buzzard
and several commentators have described a multitude of ways in which
number theory is related to other parts of mathematics. It seems that,
in practice, to know number theory you have to know all mathematics.
But what is "all mathematics"? The usual description is top-down -- that is, 
give a high-level theory, such as category theory, 
that includes nearly everything we currently consider to be important. 
Alas, there is no telling whether such a theory will continue to be a 
good description; category theory has only been around for a few decades.
Another way to describe "all mathematics" is from the bottom up -- give a 
basic form of mathematics that has always existed and which keeps growing 
and ramifying in all mathematical directions. Elementary number theory is 
very tempting bottom-up answer, because of the connections with other parts 
of mathematics  already noted, and because it will satisfy our non-mathematical
friends who think that mathematicians are people who are good with numbers.
So my basic questions are:

Is number theory a good bottom-up description of all mathematics? And
  if so, why?

Answers can be anything from general theories about the universality of
number theory to examples of unexpected appearance of number theory in other
branches of mathematics. And if you are not convinced that number theory rules:

Is there any good bottom-up description of all mathematics (one you
  can explain to a non-mathematical friend), and if so what?

REPLY [3 votes]: You really don't need PDE's and propagation of singularities (just to name an example) in order to do number theory. It simply never comes up. However PDE's are a part of mathematics, a big one at that. Therefore this disproves your equation (I'm a number theorist and in honesty I do find the equation mathematics = number theory borderline offensive).<|endoftext|>
TITLE: Integer-valued factorial ratios
QUESTION [45 upvotes]: This historical question recalls
Pafnuty Chebyshev's estimates for the prime distribution function. In his derivation
Chebyshev used the factorial ratio sequence
$$
u_n=\frac{(30n)!n!}{(15n)!(10n)!(6n)!}, \qquad n=0,1,2,\dots,
$$
which assumes integer values only. The latter fact can be established with the help of
$$
\operatorname{ord}_p n!
=\biggl\lfloor\frac{n}{p}\biggr\rfloor+\biggl\lfloor\frac{n}{p^2}\biggr\rfloor
+\biggl\lfloor\frac{n}{p^3}\biggr\rfloor+\dots
$$
and routine verification of
$$
\lfloor 30x\rfloor+\lfloor x\rfloor-\lfloor 15x\rfloor-\lfloor 10x\rfloor-\lfloor 6x\rfloor\ge0.
$$
Other Chebyshev-like examples of integer-valued factorial sequences are known;
the complete list of such
$$
u_n=\frac{(a_1n)!\dots(a_rn)!}{(b_1n)!\dots(b_sn)!}
$$
in the case $s=r+1$ was recently tabulated in
[J.W. Bober, J. London Math. Soc. (2) 79 (2009) 422--444].
A motivation for this classification problem is in relation with
a certain approach to Riemann's hypothesis, but I would prefer to
refer everybody interested in to Bober's paper (which could be
found in the arXiv as well). The proofs of $u_n\in\mathbb Z$
make use of the above formula for $\operatorname{ord}_p n!$
There are three 2-parameter families in Bober's list, namely,
$$
\frac{(n+m)!}{n!m!}, \qquad
\frac{(2n)!(2m)!}{n!(n+m)!m!}, \qquad\text{and}\qquad
\frac{(2n)!m!}{n!(2m)!(n-m)!} \quad (n>m);
$$
the first one includes the binomial coefficients, while some
properties of the second family are mentioned in
this question.
For the binomial family, a standard way to establish integrality
purely combinatorially amounts to interpreting the factorial
ratio as coefficients in the expansion
$$
(1+t)^{n+m}=\sum_{k=0}^{n+m}\binom{n+m}{k} t^k,
$$
that is, as the number of $m$-element subsets of an $(n+m)$-set.
There is lack of similar interpretation for the other two 2-parametric
families, although Ira Gessel indicates in
[J. Symbolic Computation
14 (1992) 179--194] that the inductive argument together with identity
$$
\frac{(2n)!(2(n+p))!}{n!(n+(n+p))!(n+p)!}
=\sum_{k=0}^{\lfloor p/2\rfloor}2^{p-2k} \binom{p}{2k}
\frac{(2n)!(2k)!}{n!(n+k)!k!}
\qquad (p\geq 0)
$$
allows one to show that the numbers in question are indeed integers.
A slight modification of the formula can be used for showing that
the third 2-parametric family is integer valued. In these cases
one uses a reduction to binomial sums for which the integrality is
already known. But what about the 1-parametric families, like Chebyshev's
or, say,
$$
\frac{(12n)!n!}{(6n)!(4n)!(3n)!}?
$$
Is there any way to establish the integrality without referring to
the $p$-order formula?
My own motivation is explained in the joint recent
preprint
with Ole Warnaar, where we observe a $q$-version of the integrality
in a "stronger form".

REPLY [5 votes]: Although this isn't an answer to the question, it's worth pointing out that the second and third families are essentially binomial coefficients.
We have
$$U_2(m,n):=\frac{(2m)!\,(2n)!}{m!\, n!\, (m+n)!} = (-1)^m 2^{2m+2n}\binom{n-\frac12}{m+n}$$
and
$$U_3(m,n):=\frac{(2n)!\,m!}{n!\,(2m)!\,(n-m)!}=(-1)^{n-m}2^{2n-2m}\binom{-m-\frac12}{n-m}.$$
It follows that
$U_2(m,n)$ is $(-1)^n$ times the coefficient of $x^{m+n}$ in $(1-4x)^{n-1/2}$ and
$U_3(m,n)$ is the coefficient of $x^{n-m}$ in $(1-4x)^{-m-1/2}$. Thus these numbers are integers since they are coefficients of (odd) integer powers of $(1-4x)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n} x^n$. (And in a sense they are really just one family.) It also follows that there is a simple combinatorial interpretation for $U_3(m,n)$ since it is a coefficient of a positive integer power of $(1-4x)^{-1/2}$, but we don't get an interpretation for $U_2(m,n)$ since there is cancellation in expanding positive powers of $(1-4x)^{1/2}$<|endoftext|>
TITLE: Triangulations of exotic 4-spheres
QUESTION [8 upvotes]: Are there explicit examples of triangulations of exotic 4-spheres?

REPLY [2 votes]: The current status of the smooth Poincare conjecture in dimension 4 is presented in the paper:
Michael Freedman, Robert Gompf, Scott Morrison and Kevin Walker "Man and machine thinking about the smooth 4-dimensional Poincare conjecture" in Quantum Topology, Volume 1, Issue 2 (2010), pp. 171–208 (arXiv)
Thus the Cappel-Shaneson approach seem to fail by Akbuluts work. Now there is only possible construction via the Gluck twist with a real 2-knot (i.e. knotted 2-sphere), i.e. a knot not coming from a 3-dimensional (classical) knot.<|endoftext|>
TITLE: Chebyshev's approach to the distribution of primes
QUESTION [10 upvotes]: This is motivated by a recent  question by Wadim.
The negative answer should be known, since t is very natural, in this case I would be happy to see any reference.
May Pafnuty Lvovich Chebyshev's approach to distribution of primes lead to PNT itself, if we replace $\frac{(30 n)! n!}{(15 n)! (10 n)! (6 n)!}$ to other integer ratios of factorials? If not, what are the best constants in asymptotic relation
$$
c_1 \frac{n}{\log n}< \pi(n)< c_2 \frac{n}{\log n}
$$
which may be obtained on this way?

REPLY [8 votes]: Erdős and Diamond proved in [1] that Chebyshev could have achieved sharper bounds for the asymptotic behavior of the prime counting function. Nevertheless, their proof does not shed any light on the first question that you posed because they took the PNT for granted throughout their note.
References:
[1] H. G. Diamond; P. Erdős. On sharp elementary prime number estimates,  Enseign. Math. (2)  26 (1980) 313-321.

REPLY [5 votes]: According to the notes in fifth edition of Niven, Zuckerman and
Montgomery's An Introduction to the Theory of Numbers for each
$\epsilon\in(0,1)$ there is
a series of parameters in Chebyshev's method that proves
$$(1-\epsilon)\frac{\log x}{x} < \pi(x) < (1+\epsilon)\frac{\log x}{x}$$
for all large enough $x$ but that the proof of this uses PNT
so that it doesn't provide an alternative proof of PNT.
They cite a paper of H. G. Diamond and P. Erdos:
"On sharp elementary prime estimates", L'Enseignment Math.
26 (1980), 313-321.<|endoftext|>
TITLE: Can we color Z^+ with n colors such that a, 2a, ..., na all have different colors for all a?
QUESTION [45 upvotes]: For example for n=2 coloring odd numbers red, numbers of the form 4k+2 blue and so on works.
This problem was posed in the KoMaL for n+1 prime, by Peter Pach Pal. I verified it for all n<30, I think with a computerprogram one can easily verify it for much bigger numbers by trying certain periodic colorings.
Ideas. As there is a lot of discussion going on, I thought I share here my attempts as Ewan and Gowers took similar paths. First of all, if we denote the primes by $p_i$ and the largest prime not bigger than $n$ by $p_d$, then it is sufficient to color the numbers of the form $\Pi p_i^{\alpha_i}$. This is equivalent to coloring $\mathbb Z^d$ with n colors such that the translates of a special poliomino are all rainbow colored, meaning they contain all $n$ colors. This is also equivalent to tiling the space with translates of this poliomino. The easiest way to give a coloring is if we have some nice periodicity, eg. if n+1 is prime, then $\Pi p_i^{\alpha_i} \mod (n+1)$ is such, whenever we go in a direction, it corresponds to an authomorphism of $Z_{n+1}^*$. Another possibility is to give a "linear" coloring using the addition $\mod n$, for example for $n=5$, one can take $x+3y+4z \mod 5$. So far I could always find such a linear coloring but I cannot prove that it always exist, eg. we have too many constraints to use the combinatorial nullstellensatz.

REPLY [19 votes]: This is  not an answer, but an update (with side remarks) indicating limitations of approaches suggested in other answers. 
I suggested to my Master student Thomas Chartier to work on this problem. You can find a copy of his thesis here, together with a short description of its contents. 

Given $n$, let $K_n$, the $n$-core,
  denote the set of positive integers
  whose prime factorization only
  involves primes less than or equal to
  $n$.

It should be clear that the question has a positive answer for $n$ iff it has a positive answer when ${\mathbb Z}^+$ is replaced with $K_n$, so I will focus on $K_n$ in what follows. In fact, if there is a coloring of $K_n$ as in the question (a satisfactory $n$-coloring), then there are as many such colorings of ${\mathbb Z}^+$ using $n$ colors as there are real numbers.

Given a group $G$ of order $n$, say
  that a group structure
  $(\{1,\dots,n\},\cdot)$ is
  $G$-satisfactory iff it is isomorphic to $G$ and extends the
  partial graph of multiplication, i.e.,
  whenever $i,j$ and $ij$ are all of
  size at most $n$, then the group
  operation satisfies $ij=i\cdot j$. (In
  particular, $1$ is the identity of the
  group.)

Suppose that, for some abelian $G$ of order $n$, there is a $G$-satisfactory group. Then there is a satisfactory $n$-coloring of $K_n$: In effect, we can assign to $p_1^{m_1}\dots p_k^{m_k}\in K_n$ the 'color' $p_1^{m_1}\dots p_k^{m_k}$ where now exponents and products are computed in the sense of the $G$-satisfactory group. 
This generalizes Ewan Delanoy's suggestion (where $G={\mathbb Z}/n{\mathbb Z}$). (It is a true generalization, in that we have examples where $G\not\cong{\mathbb Z}/n{\mathbb Z}$.) 
In turn, Ewan's approach generalizes Victor Protsak's. This is because if $p=nk+1$ is prime, then the nonzero $k$-th powers modulo $p$ form a group isomorphic to ${\mathbb Z}/n{\mathbb Z}$, and if $1^k,\dots,n^k$ are all different modulo $p$, then the identification $i\mapsto(i^k\mod p)$ induces a ${\mathbb Z}/n{\mathbb Z}$-satisfactory group structure on $\{1,\dots,n\}$. (Again, this is a true generalization, in that we have examples of ${\mathbb Z}/n{\mathbb Z}$-satisfactory groups that are not induced by any $p$.)
Some comments:

Of course, in order to define the coloring as above, we need $G$ to be abelian. Independently of the problem at hand, the question of whether the partial graph of multiplication on $\{1,\dots,n\}$ can be extended to a group structure (regardless of whether it is abelian or not) is interesting in its own right. It turns out that if $n$ is odd, any such group structure must be abelian. This is proved in "Groups formed by redefining multiplication", by K. Chandler. Canadian Mathematics Bulletin, 31 (4) (1988), 419-423. As far as I know, it is open whether this must always be the case. (I would appreciate any updates in this regard.)
Not every ${\mathbb Z}/n{\mathbb Z}$-satisfactory group is induced by a prime $p=nk+1$ as in Protsak's condition. (In Chartier's thesis, we call such a prime $p$ a strong representative.) The question of when there are such primes is also interesting in its own right. This is essentially addressed by a theorem of Mills that in turn extends a result of Kummer, see "Characters with preassigned values", by W.H. Mills. Canadian Journal of Mathematics, 15 (1963), 169-171. Mills characterization shows that, in particular, not every $n$ admits a strong representative $p$. The proof of Mills's theorem makes strong use of Chebatorev's theorem. As a consequence of this result, we know (for example) that for $n=34$ there must be such a strong representative $p$, but extensive computer search has not found it. As an indication of the difficulties of the search, the smallest $p$ for $n=32$ is $p=5209690063553$, and we expect any $p$ for $34$ to be orders of magnitude larger.
If $n+1$ or $2n+1$ is prime, then Protsak's condition is automatically satisfied. For other values of $k$, this is no longer the case. In fact, $k$ can never be 3, for example, and for any $k$ there are only finitely many possible $n$, see this question.

As mentioned by Fedor Petrov, domotorp's question immediately implies the Balasubramanian-Soundararajan theorem (formerly, Graham's conjecture). Sure enough, Delanoy's and Protsak's suggestions had been previously considered in connection with Graham's conjecture. In 

"What is special about 195? Groups, $n$th power maps and a problem of Graham", by R. Forcade and A. Pollington. Proceddings of the First Conference of the Canadian Number Theory Association, Ban, 1988, R.A. Mollin, ed., Walter de Gruyter, Berlin, 1990, 147-155,

it is shown that $n=195$ is the least integer for which there is no $G$-satisfactory structure for any $G$. Forcade provided us with the list of all $n\le 500$ for which this is the case. In Chartier's thesis, these $n$ are called groupless. The existence of groupless numbers unfortunately shows that Greg Kuperberg's probabilistic suggestion cannot be formalized.
The first few are listed below:

195, 205, 208, 211, 212, 214, 217, 218, 220, 227, 229, 235, 242, 244, 246, 247, 248, 252, 253, 255, 257, 258, 259, 263, 264, 265, 266, 267, 269, 271, 274, 275, 279, 283, 286, 287, 289, 290, 291, 
  294, 295, 297, 298, ...

(The sequence is not currently listed in OEIS.)
We do not know if there is a satisfactory $n$-coloring for any/some groupless $n$. In particular, the first open instance of domotorp's question is when $n=195$.
In Chartier's thesis a coloring induced by a $G$-satisfactory group is called multiplicative (so, of course, any coloring for $n=195$ would be non-multiplicative). We know that non-multiplicative colorings exist (at least, for $n=6$), see this question. However, though my examples are non-multiplicative, one can define multiplicative colorings from them. The situation for $n=195$ is very much open.
To close, let me remark that it has been suggested, for example, in

"Constructing $k$-radius sequences", by S. Blackburn and J. McKee. Mathematics of computation, to appear,

that (at least for $n$ large enough) it is the case that $n$ is groupless iff neither $n+1$ nor $2n+1$ is a prime number. This also seems interesting to investigate on its own.<|endoftext|>
TITLE: Semisimple Hopf algebras with commutative character ring
QUESTION [5 upvotes]: Suppose that $A$ is a semisimple Hopf algebra with a commutative character ring. Does it follow that $A$ is quasitriangular, i.e $\mathrm{Rep}(A)$ is a braided tensor category?
I think I 've seen this statement in a paper without a proof long time ago. It might be obvious although I don't see how to construct a braiding just knowing non-functorial commutativity of the tensor products.

REPLY [7 votes]: Sebastian,
No, it does not follow.
In this paper (Example 6.14) we proved that if a Tambara-Yamagami fusion category 
admits a braiding then its dimension is a power of 2. Note that a Tambara-Yamagami
category
has a commutative Grothendieck ring. Hopf algebras whose representation
category is of Tambara-Yamagami type are classified by Tambara (Representations of tensor categories with fusion rules of self-duality for abelian groups, Isr. J. Math. 118 (2000), 29-60). For example, there is a Hopf algebra $A = k^9 \oplus M_3(k)$ (so-called Kac-Paljutkin algebra) with commutative chracater ring and $Rep(A)$ admitting no braiding.<|endoftext|>
TITLE: For which classes of topological spaces Euler characteristics is defined?
QUESTION [10 upvotes]: I would like to know something more than what is written on wikipedia http://en.wikipedia.org/wiki/Euler_characteristic
What would be some large (largest?) class of topological spaces for which $\chi$ is defined, so that all standard properties hold, for example that $\chi(X)=\chi(Y)+\chi(Z)$ if $X=Y \cup Z$, ($Y\cap Z=0$).
ADDED.  The answer of Algori indicates that a reasonably large class of spaces for which Euler characteristics can be defined are locally compact spaces $X$, whose one point compactification $\bar X$ is a  CW complex. Then we can define $\chi(X)=\chi(\bar X)-1$. For example, the Euler characteristics of an open interval according to this definition is $-1$.
This definition rases a second (maybe obvious) question. 
Question 2. Suppose $X$ is a locally compact space whose 1 point compactification is a $CW$ complex, and $Y$ is a subspace of $X$ such that both $Y$ and $X\setminus Y$ have this property. Is it ture that $\chi(X)=\chi(Y)+\chi(X\setminus Y)$?
Also, I was thinking, that Euler characteristics is more fundamental then homology.But can it be defined for spaces, where homology is not defined? 
Finally, Quiaochu pointed out below that a very similar question was already discussed previously on mathoverflow.

REPLY [2 votes]: The Euler Characteristic can be defined for a larger class of spaces. For example, Euler Characteristic is defined for definable sets in an o-minimal system such as: semi-linear sets, semi-algebraic sets, algebraic varieties, compact smooth manifolds... and not all of them are locally compact. 
However, when they are not locally compact, then it is not necessarily true that the different definitions of the characteristic give the same result. If you are interested in this point of view you can read:

L. Van den Dries. Tame topology and o-minimal structures. Vol. 248. London Mathematical Society Lecture Notes Series. Cambridge University Press, 1998.
J. Curry, R. Ghrist y M. Robinson. “Euler calculus with applications to signals and sensing”. Proceedings of Symposia in Applied Mathematics. Vol. 70. 2012, pages: 75-146. (they have a pdf freely and legally available online here)<|endoftext|>
TITLE: Showing an Ext^2 element is zero
QUESTION [5 upvotes]: If we have an extension of bundles $0 \to E \to F \to G \to 0$ on $X$,  then to show that this is the zero element in $Ext^1_X(G,E)$, we need to show that this sequence splits. To produce a splitting is a concrete question in homological algebra. I understand that in general if we have an extension $ 0 \to E \to E_1 \to \cdots \to E_k \to G \to 0$, short of showing $Ext^k_X(G,E)=0$, there is no  simple recipe for showing that this particular extension is zero. But are there any known examples where such things have been shown? I am basically interested only in the case $k=2$. Is there a simple recipe in this case?

REPLY [10 votes]: There is a simple recipe to show that a product of two $Ext^1$ is zero (and it is clear that any $Ext^2$ can be represented as such a product). Namely, let $0 \to E_0 \to E_{01} \to E_1 \to 0$ and $0 \to E_1 \to E_{12} \to E_2 \to 0$ are two exact triples.
The product of the corresponding elements $e_{01} \in Ext^1(E_1,E_0)$ and $e_{12} \in Ext^1(E_2,E_1)$, that is $e_{01}\circ e_{12} \in Ext^2(E_2,E_0)$, is zero, if and only if there is an object $E$ with a filtration of length 3 on it $F_0E \subset F_1E \subset F_2E = E$ such that the first triple is isomorphic to $0 \to F_0E \to F_1E \to F_1E/F_0E \to 0$ and the second triple is isomorphic to $0 \to F_1E/F_0E \to F_2E/F_0E \to F_2E/F_1E \to 0$.
Here is a sketch of a proof. Consider the long exact sequence obtained by applying $Hom(E_2,-)$ to $0 \to E_0 \to E_{01} \to E_1 \to 0$:
$$
\dots \to Ext^1(E_2,E_{01}) \to Ext^1(E_2,E_1) \to Ext^2(E_2,E_0) \to \dots
$$
It is clear that the class $e_{12}$ is mapped by the second arrow to the class $e_{01}\circ e_{12} = 0$, hence there exists a class $e \in Ext^1(E_2,E_{01})$ which is mapped to $e_{12}$ by the first arrow. Let $E$ be the corresponding extension. Then we have an exact sequence $0 \to E_{01} \to E \to E_2 \to 0$ and a morphism from this
to the sequence $0 \to E_1 \to E_{12} \to E_2 \to 0$ such that the map $E_{01} \to E_1$ is the given one and the map $E_2 \to E_2$ is the identity (you can write down a commutative diagram here). It follows that the induced map $E \to E_{12}$ is a surjection and its kernel is $E_0$. This is precisely what was claimed ($F_1E$ is the image of $E_{01}$ in $E$ and $F_0E$ is the image of $E_0 \subset E_{01}$).<|endoftext|>
TITLE: When factors may be cancelled in homeomorphic products?
QUESTION [37 upvotes]: It is easy to see that if $A\times B$ is homeomorphic to $A\times C$ for topological spaces $A$, $B$, $C$, then one may not conclude that $B$ and $C$ are homeomorphic (for example, take $C=B^2$, $A=B^{\infty}$). The question is: for which $A$ such conclusion is true? I saw long ago a problem that for $A=[0,1]$ it is not true, but could not solve it, and do not know, where to ask. Hence am asking here. The same question in other categories (say, metric spaces instead topological) also seems to have some sense.

REPLY [10 votes]: It is easy to see that if $A\times B$ is homeomorphic to $A\times C$ for topological spaces $A$, $B$, $C$, then one may not conclude that $B$ and $C$ are homeomorphic (for example, take $C=B^2$, $A=B^∞$). The question is: for which $A$ such conclusion is true? 

Witold Rosicki has a lot of results of this sort (usually under some conditions on $B$ and $C$). For instance,
On decomposition of polyhedra into a Cartesian product of 1-dimensional and 2-dimensional factors
On uniqueness of decomposition of 4-polyhedron into Cartesian product of the 2-dimensional factors
On uniqueness of Cartesian products of surfaces with boundary
(with J. Malešič, D. Repovš, A. Zastrow)
There also exist papers of a different flavor on this subject
All lens spaces have diffeomorphic squares (S. Kwasik, R. Schultz)
Non-cancellation and a related phenomenon for the lens spaces (A. J. Sieradski)
As for nice examples, there exist manifolds $M$ such that $M\times I$ is homeomorphic a ball. For instance, Mazur's 4-manifold, as described by Zeeman:

Start with $S^1\times I^3$. In the boundary $S^1\times S^2$, choose a knotted $S^1$ homologous to the first factor. Knotted means that $S^1$ is not isotopic to a 1-sphere $S^1\times y$, $y\in S^2$. Form $M^4$ from $S^1\times I^3$ by attaching a handle to
  $S^1$ (i.e., attach a disk to $S^1$ and then fatten the disk so that its fattened
  boundary is identified with some chosen tubular neighbourhood of $S^1$ in
  $S^1\times S^2$). Form the cube $I^4$ by the same process, only omitting the knotting. The knotting ensures that $M^4\not\cong I^4$. But one extra dimension permits unknotting 
  $M^4\times I\cong I^4\times I$ (by just untwisting the handle).

Zeeman also notes a parallel construction of Whitehead's example with surfaces $\times I$ (mentioned above by Sergei Ivanov): `Start with $S^0\times I^2$. In the boundary $S^0 \times S^1$, choose three linked $S^0$'s, each homologous to the first factor', etc.
A really cool cancellation theorem is about joins of polyhedra, rather than products (H. Morton):

If $A*B\cong A*C$, then  either  $B\cong C$ or else $A\cong pt*A'$, $B\cong pt*X$ and
  $C\cong S^0*X$ for some polyhedra $A'$ and $X$.<|endoftext|>
TITLE: Eigenvectors of a certain big upper triangular matrix
QUESTION [9 upvotes]: I'm looking at this matrix:
$$
\begin{pmatrix}
1 & 1/2 & 1/8 & 1/48 & 1/384 & \dots \\
0 & 1/2 & 1/4 & 1/16 & 1/96 & \dots \\
0 & 0 & 1/8 & 1/16 & 1/64 & \dots \\
0 & 0 & 0 & 1/48 & 1/96 & \dots \\
0 & 0 & 0 & 0 & 1/384 & \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
$$
The first row contains the reciprocals of the double factorials
$$
2, \qquad 2 \cdot 4, \qquad 2 \cdot 4 \cdot 6, \qquad 2 \cdot 4 \cdot 6 \cdot 8, \qquad \dots
$$
Each row is a shift of a scalar multiple of the first row, and the scalar multiple is in each case itself a reciprocal of a double factorial, so that the main diagonal is the same as the first row.  A consequence is that each column is proportional to the corresponding row of Pascal's triangle.  E.g. the last column shown is proportional to
$$
1, 4, 6, 4, 1.
$$
This matrix is the matrix of coefficients in the "inversion formulas" section of this rant that I wrote.
I found the first three eigenvectors:
$$
\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix},
\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix},
\begin{pmatrix} 5 \\ -14 \\ 21 \\ 0 \\ 0 \\ \vdots \end{pmatrix}
$$
Meni Rosenfeld pushed this through some software and found that up to the 40th eigenvalue, the signs of the components of the eigenvectors alternate.
Can anything of interest be said about the eigenvectors?
Can anything of interest be said about this matrix?

REPLY [3 votes]: Might be a wild intuition, I'd say the eigenvalues are the entries of the first row, and that the eigenvector coresponding to the $n$th eigenvalue, $k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n\times n$ minor of the matrix.
Example:
for eigenvaue $1$ we take the matrix 
$ \begin{pmatrix} 1 \end{pmatrix}$ ,the eigenvetor corresponding to $1 $ is $\begin{pmatrix} 1  \end{pmatrix}$ , so we obtain   $\begin{pmatrix} 1 \cr 0 \cr 0 \cr  0 \cr 0 \cr \vdots \end{pmatrix} $ as the first eigevector.
for eigenvaue $1/2$ we take the matrix 
$ \begin{pmatrix} 1& 1/2 \cr  0 & 1/2\end{pmatrix}$ ,the eigenvetor corresponding to $1/2$ is $\begin{pmatrix} 1 \cr -1\end{pmatrix}$ , so we obtain  $\begin{pmatrix} 1 \cr -1 \cr 0 \cr  0 \cr 0 \cr \vdots \end{pmatrix} $ as the second eigevector . 
The eigenvector coresponding to $1/8$ for $ \begin{pmatrix} 1& 1/2 & 1/8 \cr  0 & 1/2 & 1/4 \cr 0 & 0  & 1/8 \end{pmatrix}$ is $\begin{pmatrix} 5 \cr -14 \cr 21 \cr \end{pmatrix}$, you get the ideea .Also , the eigenvectors span the entire space , ie if a possibly infinite (but convergent)  sum of eigenvectors is $\vec 0$ then the coefficients of those vectors are $0$.
Here is an explicit formula for the eigenvectors: first select $M_n$, the $n\times n$ truncation of the matrix and calculate $M_n - I v_n $, the $n$th eigenvalue. Example:
for $n=3$, we obtain 
\begin{pmatrix} 7/8 & 1/2 & 1/8 \cr  0 & 3/8 & 1/4 \cr 0 & 0  & 0 \end{pmatrix}. Now let $S$ be the $(n-1)\times(n-1)$ truncation of that , ie \begin{pmatrix} 7/8 & 1/2 \cr  0 & 3/8 & \end{pmatrix}
Calculate $S^{-1} = \begin{pmatrix} 8/7 & -32/21 \cr  0 & 8/3 & \end{pmatrix}$, now multiply  $S^{-1}$ with the truncation of the last column of $M_n$, \begin{pmatrix} 1/8 \cr 1/4 \end{pmatrix}
You obtain $$\begin{pmatrix} -5/21 \\ 2/3 \end{pmatrix}.$$ Concatenating $-1$ to that, you obtain $$\begin{pmatrix} -5/21 \\ 2/3 \\ -1  \end{pmatrix},$$ the third eigenvector, or the $n$th eigenvector in the general case.<|endoftext|>
TITLE: Is the desingularization of a normal variety with only quotient singularities projective
QUESTION [6 upvotes]: The base field will be the field of complex numbers. I have a slightly technical problem concerning the resolution of singularities of a certain variety. Basically, I want to to know if it is projective.
Let $Y$ be a normal projective variety with only quotient singularities. Let $C$ be a smooth projective curve.
Let $p:Y\longrightarrow C$ be a flat projective morphism.
Let $f:X\longrightarrow Y$ be a resolution of singularities. Is $X$ necessarily projective over the base field? I know this is true in dimension 2. Are there any counterexamples in dimension 3?

REPLY [2 votes]: Here is an explicit example that this does not work for arbitrary resolutions. The assumptions below are only needed to get the flat morphism over the curve and they are not really necessary but make life a little easier. I think the example Karl is referring to is Hartshorne, GTM 52, Appendix B, Example 3.4.1. Anyway, here we go.
Let $Y=C\times S$ where $C$ is a smooth projective curve and $S$ is a smooth projective surface. The projection $\pi:Y\to C$ is not only flat, but smooth and projective. Assume further that there exists a morphism $\alpha: C\to S$ such that it is an embedding everywhere except at two points, whose images are the same point $P\in S$. Consider the following curves: $C_1=C\times \{P\}$,$C_2=\{(x,\alpha(x)\mid x\in C \}$. By the previous assumptions, $C_1$ and $C_2$ meet in exactly $2$ points. Now perform Hironaka's trick to produce a non-projective smooth $3$-fold: blow-up $C_1$ and $C_2$ in the two possible orders, but in different ways near the two intersection points (need to do it locally and then glue). See Hartshorne, GTM 52, Appendix B, Example 3.4.1 for details on this construction. Let $\sigma: X\to Y$ be the $3$-fold obtained this way. It is easy to see that this is not projective using intersection numbers. See Hartshorne for the computation. 
Perhaps the more interesting point about this example is that it shows that a morphism being projective is not local on the target. (Like being proper is, [Hartshorne, II.4.8(f)]). 
Obviously $\pi\circ\sigma$ is projective locally on $C$, since it is a combination of the original projective morphism $\pi$ and two (projective) blow-ups, but $\pi\circ\sigma$ cannot be projective, since then so would be $X$ (over the base field, since $C$ is). 
Finally, to answer your question in the comments, whether there is "a" resolution that
's projective, the answer is certainly "yes". If you pick one that's a sequence of blow-ups, then the morphism is projective and hence so is $X$. In fact, using Chow's Lemma, you can "correct" any resolution that may accidentally be non-projective: by Chow's Lemma, you can always dominate birationally your non-projective (but proper) variety by a projective one and then applying a projective resolution of that will save the day.<|endoftext|>
TITLE: Are there arbitrarily sparse "lattices" in negatively curved symmetric spaces?
QUESTION [8 upvotes]: Let $X$ be a negatively curved symmetric space. In other words, $X$ is one of the four examples: a hyperbolic space, a complex hyperbolic space, a quaternionic hyperbolic space or the hyperbolic Cayley plane.
Let $B\subset X$ be a compact set (wlog, a large ball). Does there always exist a subgroup $\Gamma$ of the isometry group of $X$ such that $X/\Gamma$ is a compact manifold and the images of $B$ under $\Gamma$ are disjoint (i.e. $B\cap\gamma B=\emptyset$ for every nontrivial $\gamma\in\Gamma$)?
Remark: It would suffice to find just one (for each $X$) residually finite discrete co-compact group of isometries. I am sure there is one but I don't know where to look. (Or maybe they all are residually finite?)
Meta-proof: If the answer is negative, then, for some of the model geometries (in some dimension), all compact manifolds with this local geometry are "uniformly thin" - all injectivity radii are bounded above by some universal constant. Such a wonderful fact would be widely known.

REPLY [11 votes]: The examples are arithmetic groups, constructed in general by Borel and Harish-Chandra. 
See also Dave Witte Morris' preliminary book. 
However, examples in hyperbolic and complex hyperbolic spaces probably go back further to the study of quadratic forms. 
For hyperbolic lattices, one can take a quadratic form over a quadratic number field (such as $\mathbb{Q}(\sqrt{2})$), which
is Lorentzian at one place, and definite at the other places (such as $x_1^2+\cdots +x_n^2-\sqrt{2} x_{n+1}^2$), and take the group of matrices $\Gamma$ in $GL(n+1,\mathbb{Z}[\sqrt{2}])$ which preserve this quadratic form. Then Mahler's compactness theorem (cf. Witte Morris) implies 
that the quotient $\mathbb{H}^n/\Gamma$ is compact. Then by Selberg's Lemma and residual finiteness (as Greg points out, Malcev's Theorem), you may find a torsion-free subgroup 
of finite-index with as large injectivity radius as you like. 
For hyperbolic and complex-hyperbolic spaces, there are other examples which don't come from the arithmetic construction (in fact, most hyperbolic surfaces and 3-manifolds are not arithmetic). These are attributable to Gromov and Piatetski-Shapiro in the hyperbolic case in all dimensions, and there are finitely many examples in the complex hyperbolic case (at least for $\mathbb{C}$-dim >1) going back to Deligne and Mostow (see also Thurston's paper). However, Gromov and Schoen have shown that quaternionic and Cayley-hyperbolic lattices are all arithmetic, so the Borel Harish-Chandra construction is complete in these cases.  
Addendum: Related to Protsak's comment, there is a simple example of a negatively curved homogeneous space which has no lattice action. This is Thurston's "9th geometry", which is excluded as a geometry precisely for this reason. One can take the double warped product metric $$ dr^2 + e^{2a r} dx^2 + e^{2b r} dy^2, $$ for $a,b >0$. When $a=b$, this gives hyperbolic space. But when $a\neq b$, the sectional curvatures are $-a^2, -b^2, -ab$. This has a solvable transitive group of isometries, so is homogeneous. But using the solvability, one may see that there is no cocompact action. (Remark:  when $a=-b$, this gives sol geometry).<|endoftext|>
TITLE: Triangle centers as power sum minimizing points: Which is the locus of all these points? 
QUESTION [9 upvotes]: Some of the classical triangle centers can be expressed as solutions to minimization problems: Given a triangle $A_1, A_2, A_3$ define $d_i, i=1,2,3$ to be the distance of a given point $P$ to $A_i$, and $f_q$ as the sum of the $q$-th power of these distances:$f_q = \sum_{i=1}^3 d_i^q$. I'm looking for the point $P$ which minimizes $f_q$. For $q=1$ this is the Steiner point, for $q=2$ the centroid, for $q \to \infty$ the circumcenter, for $q \to 0$ the point where the product of distances is minimized. An obvious question is to find the curve of all these points for reasonably general $q$ (e.g. $q \in \mathbb{R}_{>0}$). However, in the ressources for triangle centers, as e.g. http://faculty.evansville.edu/ck6/encyclopedia/ETC.html this problem seems not to be considered. 
EDIT: 
1) I would like to restrict the problem to $q \ge 1$ since then $f_q$ is convex and a unique minimum is guaranteed. 
2) I would like to add a generalization of the question: Consider all continuous functions $f(d_1,d_2,d_3)$ that a) are invariant under permutations of $d_1, d_2, d_3$ and b) have a unique minimum. What can be said about the locus of all these minima?

REPLY [3 votes]: I posted an animation showing approximations of this curve for a sequence of triangles, and a range of q values, here.<|endoftext|>
TITLE: {transcendental numbers} \ {computable transcendental numbers}
QUESTION [12 upvotes]: I know Chaitin's constant Ω is not computable (and therefore transcendental).  Are there other specific, known noncomputable numbers?  I am trying to understand what distinguishes a computable transcendental number, such as π, from a noncomputable transcendental number, such as Ω.  Is there anything revealing that can be said about the set difference {transcendental numbers} \ {computable transcendental numbers}?
     I ask this as a novice.  I am re-visiting a wonderful book that sadly can no longer be updated by Victor Klee, in which he and Wagon pose this as an open problem: If an irrational number is real-time computable, is it then necessarily transcendent? [Problem 23] 
Update (19Jun12). There is an illuminating discussion under the title
"Why The Hartmanis-Stearns Conjecture Is Still Open" at the Lipton-Regan blog.
The Hartmanis-Stearns Conjecture is the open problem mentioned above:
If a number is real-time computable, it is either rational or transcendental.
If true, this has what strikes me as a counterintuitive consequence: that algebraic irrationals like $\sqrt{3}$
are in some sense "more complicated" than transcendentals.

REPLY [5 votes]: Let me consider your question: Are there other specific, known
non-computable numbers?
There are indeed numerous specific, known definable real numbers
that we know are not computable. Many of these real numbers arise
when considering various notions of definability in various
stronger-than-computable systems, such as the arithmetic
hierarchy, the projective hierarchy and other hierarchies of
definability and complexity, some of which I describe in my answer
to I. J. Kennedy's question Are some numbers more
irrational than others? In my
answer to Paul Budnik's question Is there a well-defined subset
of the integers that cannot be defined as a property of a
recursive process or Turing machine?, I
mention several specific numbers that are definable, but not
computable. In this information-theoretic context, the difference
between a real number and a set of natural numbers is often not so
great, and every set of natural numbers (or sentences in a given
language) naturally corresponds to a real number, whose digits
indicate membership or non-membership in that set. Let me list
several definable non-computable reals by their names:

The real $0'$, pronounced "$0$-jump". This is just the halting
problem, which you can think of as the real number whose $n^{th}$
digit is $1$ if the $n^{th}$ Turing machine program halts on
trivial input, and was mentioned already in some of the other
answers.
Kleene's $O$.
$0''$, $0'''$, the double jump, triple jump and so on, which
relativizes the halting problem to an oracle.
$x'$ for any definable $x$, we have the halting problem relative
to Turing machines with oracle $x$, which is strictly harder than
$x$ to compute.
Tot, the set of programs computing total functions. This has
complexity $\Pi^0_2$.
Fin, the set of programs computing a finite function. This has
complexity $\Sigma^0_2$.
TA, or "true arithmetic", is the set of arithmetic sentences
true in $\langle\mathbb{N},+,\cdot,0,1,\lt\rangle$. It is Turing
equivalent to $0^{(\omega)}$, the $\omega^{th}$ jump.
WO, the set of programs computing a well-ordered relation on
$\mathbb{N}$. This has complexity complete $\Pi^1_1$, just beyond
the hyperarithmetic hierarchy.
Th(HC), the set of statements true for hereditarily countable
sets. This is an analogue of TA for the projective hiearchy.
One can define other specific real numbers, which are not
computable, such as the real number whose $n^{th}$ binary digit
records whether or not $2^{\aleph_n}=\aleph_{n+1}$. This real
number is definable, but not necessarily computable. In fact, for
any real number, there is a forcing extension of the universe in
which this definition defines that number. So in this sense, any
number you like can be made to be a specific definable number in some set-theoretic
universe.
$0^\sharp$, pronounced "$0$-sharp", is a real number whose existence is a kind of large
cardinal assertion, equivalent to the existence of a proper class
of order-indiscernible ordinals for the constructible hierarchy.
The real $0^\sharp$ is the theory of these indiscernibles.
One can iterate this with $0^{\sharp\sharp}$ and $x^\sharp$ for
any $x$.
$0^\dagger$, pronounced "$0$-dagger", is a similar theory for indiscernibles over a richer
inner model $L[\mu]$, with a measurable cardinal. This is a
specific real number whose existence has consistency strength
greater than the existence of a measurable cardinal.
There are other similar reals, such as $0$-hand-grenade and
others, which carry a stronger large cardinal strength.<|endoftext|>
TITLE: When is a conjugacy class of matrices an embedded submanifold?
QUESTION [7 upvotes]: Let $M_{n\times n}$ denote the set of $n\times n$ real matrices and let $GL_n$ be the subgroup of invertible matrices.  $GL_n$ acts on $M_{n\times n}$ smoothly by conjugation, which means that each conjugacy class (which is an orbit of this action) is an immersed submanifold of $M_{n\times n}$.  However, the action is not proper (e.g. the isotropy groups are not compact) so the orbits may not be embedded submanifolds.
My question is if there are nice conditions on a matrix that guarantee that its conjugacy class is or is not an embedded submanifold.  My interest in this question actually comes from trying to understand the space of all complex structures on a real vector space:  it can be shown that the set of all complex structures is the conjugacy class of the block matrix $\begin{pmatrix} 0 & -I \\\ I & 0\end{pmatrix}$ and I was wondering if this is an embedded submanifold.  So an answer to this question (if the above doesn't have a nice answer) would also be appreciated.

REPLY [5 votes]: If $G$ is an algebraic group (say over $\mathbb C$), acting on an algebraic variety $X$, the orbits are always locally closed smooth algebraic subvarieties of $X$. This is standard, and easy to prove. If $x$ is a point of $X$, and $G \to X$ is the morphism sending $g$ to $gx$, consider the orbit $Gx$ as a subset of its Zariski closure $\overline{Gx}$. By Chevalley's theorem, $Gx$ contains a dense open subset of $\overline{Gx}$. Hence, by homogeneity it is open in $\overline{Gx}$, and smooth.
From this it is easy to deduce that all the orbits of $\mathrm{GL}_n(\mathbb R)$ in $\mathrm{M}_n(\mathbb R)$ are embedded submanifolds. The point is that if two real matrices are conjugate as complex matrices, then they are conjugate as real matrices. These means that the orbits of the action of $\mathrm{GL}_n(\mathbb R)$ in $\mathrm{M}_n(\mathbb R)$ are of the form $\Omega(\mathbb R)$, where $\Omega$ is an orbit of  $\mathrm{GL}_n(\mathbb C)$ in $\mathrm{M}_n(\mathbb C)$; since $\Omega$ is a smooth algebraic variety, it follows from the implicit function theorem that $\Omega(\mathbb R)$ is an embedded submanifold of $\mathrm{M}_n(\mathbb R)$.<|endoftext|>
TITLE: models of PA which are isomorphic but not elementarily equivalent?
QUESTION [7 upvotes]: On page 164 of his book Models of Peano Arithmetic, Kaye states Friedman's Theorem:
Let $M{\vDash}PA$ be nonstandard and countable, let $a\in M$ and let $n\in {\mathbb N}$.  Then there is a proper initial segment $I\subseteq_c M$ ($\subseteq_c$ means "cofinal in") containing $a$ such that $I\cong M$ and $I<_{\Sigma_n} M$.
However, on the next page he writes "Nor can we expect in general to get initial segments $I$ with $M\cong I < M$ and $M\neq I$, i.e., elementary for all formulas.  For example if $M=K_T$ (where $T\neq Th({\mathbb N})$ is a complete extension of PA) then $M$ has no proper elementary substructures, and so certainly has no proper elementary initial segments!"
I am confused.  If two models in the same language are isomorphic, are they not elementarily equivalent?  Of course the converse need not be true.
An isomorphism of models is a bijective homomorphism of the language's algebraic portion (constants and functions) which preserves and reflects all relations of the language (Hodges, Model Theory, p5).  Therefore by induction on the structure of any ${\mathcal L}_{\omega,\omega}$ formula, the isomorphism will both preserve and reflect it.  So an isomorphism preserves all formulas (Hodges, Theorem 2.4.3(c)).
If the proper initial segment is isomorphic as a model to the entire model, how could any first-order sentence possibly be true in one and not in the other?
Thanks,

a

REPLY [10 votes]: There is a major difference between elementary equivalence and elementary embedding. Moreover, in this case, the actual embedding is somewhat ambiguous. First, let me recap some often confused terminology.
Two models are elementary equivalent if they satisfy the same first-order sentences. Any two isomorphic models are always elementary equivalent. An elementary embedding is a map j:A→B such that, for all first-order formulas φ(v1,...,vk) and all a1,...,ak ∈ A, A ⊧ φ(a1,...,ak) iff B ⊧ φ(j(a1),...,j(ak)). An isomorphism is always an elementary embedding.
The notation A ≺ B means that A is an elementary submodel of B, i.e. the inclusion map A  ⊆ B is an elementary embedding from A into B. In your context, the isomorphism (or its inverse) is not the proposed elementary embedding, it is the inclusion map which is in question: it is elementary for Σn formulas, but not elementary for all first-order formulas.<|endoftext|>
TITLE: Uniform solutions to Post's problem for axiomatizable theories
QUESTION [11 upvotes]: The Second Incompleteness Theorem says that if $T$ is a consistent (computably) axiomatizable theory which extends IΣ1, then $\mathrm{Con}(T)$ is not provable from $T$. By analogy with computability theory, the stronger theory $T + \mathrm{Con}(T)$ can be thought of as the "jump" of $T$. To abuse this analogy, I will use $T'$ to denote the theory $T + \mathrm{Con}(T)$. I will write $T \leq S$ when $S$ proves every axiom of $T$; I will also write $S \equiv T$ (resp. $T < S$) when $T \leq S$ and $S \leq T$ (resp. $S \nleq T$). 
It is well-known that if $T$ is consistent there are plenty of axiomatizable theories $S$ such that $T < S < T'$. In the following questions $H$ will denote an operator (like $\mathrm{Con}$) that uses the computable axiomatization of $T$ to produce a sentence $H(T)$. I will write $T^H$ for the theory $T + H(T)$.

Is there a computable operator $H(T)$ such that $T < T^H < T'$ for every consistent  axiomatizable theory $T$ extending IΣ1? Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?
Is there a computable operator $H(T)$ such that $(T^H)^H \equiv T'$ for every consistent  axiomatizable $T$ extending IΣ1?  Is there such an operator which moreover satisfies that $T \equiv S$ implies $T^H \equiv S^H$?

Question 1 asks for a uniform solution to the analogue of Post's Problem for axiomatizable theories. Question 2 asks for a uniform "half-jump" operator.

REPLY [5 votes]: The premise in your question that the Con operator itself
has the desired property and serves as a jump operator is
not universally true among the theories you consider.
Specifically, you seem to assume that because
$\text{Con}(T)$ is not provable in $T$, that
$T+\text{Con(T)}$ is consistent. But this is not correct,
because perhaps $T$ actually proves $\neg\text{Con}(T)$.
One easy instance of this is the theory
$T=PA+\neg\text{Con}(PA)$, which is consistent by the 2nd
Incompleteness Theorem, but clearly proves
$\neg\text{Con}(PA)$ and hence also $\neg\text{Con}(T)$.
Thus, as weird as it sounds, $T$ is a consistent theory
that proves its own inconsistency. In this case your theory
$T'$ is inconsistent and the jump failed. Carl's theory
$T^H$ in this case is consistent, but upon inspection you
will find that it is equivalent to $T$. So for this theory
$T$, your theory $T'$ jumped into inconsistency, and his
theory didn't jump at all.
One can similarly replace $PA$ here with any representable
theory $T_0$ and arrive at similar counterexamples, densely above any theory.
You can fix the question by
considering only the case where $T'$ is consistent, which
is surely what you had in mind. In this event, you would
only apply the jump when it happens to arrive at a
consistent theory. Since this question is not decidable
from a presentation of the theory, however, even from a
finite axiomatization, it may affect your motivation for
considering computable versions of the half-jump, since even the
full jump is not computable. 
For this reason, and also because there is something a
little arbitrary about having the jump only partially
defined, it may be that a more robust jump arises from the
Rosser sentence---there is no proof of me without a
shorter proof of my negation---instead of $\text{Con}(T)$?
This would put you back into the universal domain of all
representable consistent theories.<|endoftext|>
TITLE: Are there any sofware packages for computing Picard numbers?
QUESTION [7 upvotes]: Are there any computer algebra systems (e.g. Macaulay2 og singular) that allows one to compute the Picard number (i.e. the rank of the Neron-Severi group) of a given variety?

REPLY [11 votes]: A more basic question is whether there even exists an algorithm to compute this number.
I've wondered this for a long time, and I honestly don't know what to expect. 
Any algorithm would have to be quite subtle. In the early 1980's Shioda had to work quite hard to construct explicit examples of surfaces in $\mathbb{P}^3$, defined over $\mathbb{Q}$, with Picard number 1. 
Added: Of course, I should have said Shioda's example have degree >1. As further
evidence of subtlety of the problem: one can decide whether an elliptic curve $E$
has CM by computing the Picard number of $E\times E$.<|endoftext|>
TITLE: Minimal basis of set of positive integers
QUESTION [12 upvotes]: Let $A$  be a set of positive integers and $A+A = \{a_1 + a_2 | a_1,a_2 \in A \}$. If $A+A$ contains all positive integers, $A$ is called a basis (of order 2) of the set of positive integers. A basis $A$ is called a minimal basis, if no proper subset of $A$ is a basis. 
E.g. the set of all numbers with only "0"s and "1"s in ternary representation is a (even minimal) basis. This is somehow the discrete analogon to the well known fact that the sumset of two Cantor ternary sets is the closed interval [0,2].
I'm looking for a 'smallest' minimal basis of the set of positive integers ('smallest' in the sense of the lexikographic order - also other orders would be interesting, e.g. the order induced by sum of inverse squares of the elements of the basis). 
There are a lot of articles about minimal asymptotic basis of the set of positive integers ("asymptotic" meaning that only every \textit{sufficiently large} natural number is the sum of two elements of the basis). But I'm struggling to get or find my question above answered.  
ADDENDUM: Sorry for not having been clear, and thank you for your comments. I hope the following remarks will clarify things a bit: a) Yes, I consider only order 2. b) To make my intention clearer, lets consider the finite set $M_n = \{0, 1, 2, 3, \ldots, n-1\}$ and ask for a basis $A$ such that $M_n \subset A + A$. E.g. $A = \{0,1,3,4,9,10,12,13\}$ is a basis of $M_{27}$ (this is just the ternary construction mentionned above). In the finite case I'm looking for the basis with the smallest number of elements and in case of a tie (same number of elements), I prefer the basis which comes later in lexicographic order (i.e. I would prefer $\{0,1,3,4,10\}$ to $\{0,1,3,4,9\}$. My intention with the question was to ask this problem not for finite $M_n$, but for $M_{\infty} = \mathbb{N}$ (sorry also for the confusion whether to include "0" or not). 
Since the cardinality of the "ternary basis" scales with $n^{2/3}$ and the cardinality ofthe proposed basis of odd numbers is $n/2$, I regard the "ternary basis" as a better one. 
Perhaps the problem is still not well defined and/or there is no such "best" basis. This would also be helpful for me to know.

REPLY [8 votes]: The exponent can be arbitrarily close to $\frac{1}{2}$: If $n=r^2$ then $A+B=\lbrace 0,1,2,\cdots n-1 \rbrace$ where $A=\lbrace 0,1,2,\cdots r-1\rbrace$ and $B=\lbrace 0,r,2r,\cdots,(r-1)r\rbrace$. This means that $S=A \cup B$ is a set of size $2r=2\sqrt{n} $  with $S+S \supset \lbrace 0,1,2,\cdots n-1 \rbrace$. So by iteration (as described in the old answer) we can have exponent $\frac{1}{2}+\log_n(2)$.
Modifying this a bit gives a construction I probably have seen before: Let $A=\lbrace 0,1,4,5,16,17,20,21\cdots\rbrace$ be the (infinite) set of non-negative integers whose binary expansion is $0$ in all the odd positions and $B=2A$ the set of those with $0$ in all the even positions. Here both have exponent $\frac{1}{2}$ and $A+B=\mathbb{N}_0$. Then $S=A \cup B$ has about $2\sqrt{N}$ members less than $N$ and $S+S=\mathbb{N}_0.$

Old Answer Here is a nice ad hoc start: The set $S=\lbrace 0, 1, 3, 5, 6, 13, 15, 16, 18, 25, 26, 28, 30, 31 \rbrace$ has $S+S=\lbrace 0,1,2,\cdots,62 \rbrace$ (It is the start of a sequence in the OEIS with next term 63,  but I'm not sure if that helps). Then $T=S+63S$ has 196 members the largest being $1984$ and $T+T=\lbrace 0,1,2,\cdots,3968 \rbrace$. Now $14=31^{0.63944}$ and $196=3968^{0.63699}$ and further iterating will continue to give examples sparser than $k^{2/3}$. The same trick can be done with any finite example. The densities decrease but only very very slightly.
LATER the set $S'=\lbrace 0, 1, 3, 5, 6, 13, 14,17, 18, 25, 26, 28, 30, 31 \rbrace$ also has $S'+S'=\lbrace 0,1,2,\cdots,62 \rbrace.$
Observe that the places where $S$ has a jump from $j$ to $2j+1$ are at $0,1,6,31$ (numbers of the form $\frac{5^j-1}{4}$) and that there is a central symmetry for $0,1$ and for  $0,1,3,5,6$ and for $S$. This suggests a treasure hunt for an symmetric extension with largest member $1+5+25+125=156$.
UPDATE For any one of the four choices $(a,b)=(68,69),(68,72),(69,70),(70,71)$ the set  $V=S \cup \lbrace 63, 64, a,b,156-b,156-a, 92, 93  \rbrace \cup (S+125)$ has $36$ members, the largest being $156$. and $V+V=\lbrace 0,1,2,\cdots,312\rbrace$ There is only one way to similarly extend $S'$ to a set of size $36$ with the same properties, it has lower half $S' \cup \lbrace 63, 65, 67, 69 \rbrace$. These are better than the examples above since $36=312^{0.62398}$. Also, these can be iterated as above.
Maybe an even lower density is possible for $W=V \cup \lbrace ?? \rbrace \cup(V+625)$ where the set in the middle is made of several pairs $q,781-q$. I'd guess 8 pairs leading to $88=1562^{0.60885}$ but that is pure speculation.  Of course there might be better examples which are not symmetric (or even ones which are). There turns out to be no advantage at this stage to putting the exact middle of $78$ in $V$ (for any of the 5 choices). I don't think it would help get a better $W$ but I am not sure.<|endoftext|>
TITLE: References for complex analytic geometry?
QUESTION [20 upvotes]: I'm looking for references on the "algebraic geometry" side of complex analytis, i.e. on complex spaces, morphisms of those spaces, coherent sheaves, flat morphisms, direct image sheaves etc. A textbook would be nice, but every little helps.
Grauert and Remmert's "Coherent analytic sheaves" seems to contain what I want, but it is very dense reading. You could say I'm looking for sources to read on the side as I work through G&R, to get different points of views and examples. For example, B. and L. Kaup's "Holomorphic functions of several variables" talks about the basics of complex analytic geometry, but doesn't go into much detail.
My motivation is twofold. First, I'm studying deformation theory, which necessarily makes use of complex spaces, both as moduli spaces and objects of deformations, so while I can avoid using complex spaces at the moment they're certain to come in handy later. Second, I want to be able to talk to the algebraic geometers in my lab, so I should know what their schemes and morphisms translate to in the analytic case. I like reading as much as I can about what I'm trying to learn, so:
Do you know of other sources (anything: textbooks, lecture notes, survey articles, historical overviews, comparisons with algebraic geometry ...) that talk about complex spaces and their geometry?

REPLY [8 votes]: For complex geometry,which really is fundamental in analytic deformation theory,I strongly suggest 2 sources besides the classical source by Griffiths and Harris: Complex Geometry:An Introduction by Daniel Huybrechts,which has rapidly become the standard text on the subject,and the online text draft of a comprehensive work by Demially. The Demailly text is much more comprehensive and more advanced,with an emphasis on algebraic and differential geometry.But you may find it more helpful as it contains a great deal more near the research level. It can be found here: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf<|endoftext|>
TITLE: How should one present calculations?
QUESTION [18 upvotes]: It is often necessary to present calculations, or at least their outlines, in a proof.  However, when I need to do it, if the calculation takes more than one line (or perhaps two at the most), I feel a bit uneasy, and have wrestled with trying to present it in a better way.  On friday, I was browsing a journal, and came across a paper, in a subject that I was interested in, which contained a number of calculations which were at least a half a page of nothing but formulas.  This made me shudder.  Ultimately the reason why we should write papers is to foster insight, and I think that such presentations fail that test.
So, how do people deal with this problem?

REPLY [9 votes]: I think that in all writing, and especially in mathematical writing, signposting is extremely important.  So a calculation-heavy proof should include a short summary, and then the calculations should be very clearly marked so that the reader knows that "here are the nitty gritties of the calculations, and here's where they stop, so you can skip them and pick up at the end".
For short papers, I like the following format.  Begin with a very short introduction, setting the paper in context and giving a couple-sentence outline of the paper.  In the second section, state all definitions and theorems precisely, but do not provide any proofs that are more than one or two sentences.  Then the rest of the document, in as many sections as necessary, provides the detailed proofs, with all non-immediate calculations.  The point is that most readers can read sections 1 and 2 and get everything out of the paper, and the only people who will read sections 3+ are: the reviewers (one hopes), and anyone trying to generalize the actual proof to a different setting.<|endoftext|>
TITLE: Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology?
QUESTION [12 upvotes]: Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth closed $n$-dimensional manifolds such that the induced map
$$f^* \colon H^*(N,G) \to H^*(L,G)$$
is an isomorphism.
Question: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.
This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?
Motivation: This came up because (relative) spin is important in defining Floer homology with $\mathbb{Z}$ coefficients. So I am in fact mostly interested in the following sub-question.
Question: In particular what about the second Stiefel-Whitney class in the case where both $N$ and $L$ are also assumed to be oriented? and if the answer is negative: what extra conditions do I need to make it positive?
The idea is that I apriori have to use $G$ coefficients, but can prove that it is a $G$-cohomology equivalence, and want to use that to start the argument over again with other coefficients, but for that I need this property of the second Stiefel-Whitney class. 
This sub-question and the relation to Floer homology is related to orientations in real $K$-theory and delooping in the following sense: take a map $h\colon X \to U/O$ by delooping we get a map $\Omega h \colon \Omega X \to \Omega U/O \simeq \mathbb{Z}\times BO$ which classifies a virtual bundle over the loop space of $X$. This bundle is oriented iff the original map composed with the canonical map $U/O \to BO$ classified a virtual $0$-dimensional bundle with vanishing second Stiefel-Whitney class. This is the main point of why orientations in Floer homology is initimitely linked with spin! In the case of a Lagrangian sub-manifold $L\subset T^*N$ the difference of the tangent bundles precisely defines such a map $L \to U/O$ ($U(n)/O(n)$ classifies Lagrangians in $\mathbb{R}^{2n}$) such that the composition to $BO$ classifies the virtual bundle $TN-TL$. So in fact you may add this lifting property as an extra condition to the sub-question if you like, and then I would lose no generality. I believe that this condition implies that all the relative Prontryagin classes vanishes, which may be helpfull.
ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.

REPLY [5 votes]: The most conceptual way of understanding the relation between the mod 2 Wu and Stiefel-Whitney classes of a manifold and the action of the Steenrod algebra on the mod 2 cohomology is to use the homotopy theory of Poincare duality spaces and the Spivak normal fibration, and also the chain homotopy theory of chain complexes with symmetric Poincare complexes and the normal chain bundle expounded in my 1980 paper The algebraic theory of surgery (Part I, Part II). A map $f:L\to N$ of $n$-dimensional manifolds which induces isomorphisms in $Z_2$-coefficient cohomology also induces a chain equivalence of $n$-dimensional symmetric Poincare complexes over $Z_2$. Such a chain equivalence automatically preserves the Spivak normal chain bundles. The mod 2 Wu and Stiefel-Whitney classes of the manifolds are preserved by $f$ because they only depend only on the underlying chain homotopy structure. It is also worth reminding ourselves that Atiyah's 1960 paper Thom complexes established the $S$-duality between the Thom space of the normal bundle of a manifold $X$ and $X_+=X \cup \{*\}$, and so proved a conjecture of Milnor and Spanier: the stable fibre homotopy type of the tangent sphere bundle of a differentiable manifold $X$ depends only on the homotopy type of $X$.<|endoftext|>
TITLE: Cohomological dimension
QUESTION [8 upvotes]: Cohomological dimension arises in many things I read, but my familiarity with it is superficial. What's a good source for understanding cohomological dimension on its various examples, fascets, theorems, and applications, basically providing the maximal intuition-building narative?

REPLY [3 votes]: Cohomological dimension arises in different contexts, algebraic topology, algebraic geometry, group theory... So I think that it would be difficult to give a single reference for this broad topic. 
It was already mentioned that Brown "cohomology of groups" has a chapter on it, and I think this is a good reference for cohomological dimension of groups. However cohomological dimension  also arises in the context of topology, 
and when the space is well-behaved, say a manifold, it just coincides with the usual topological dimension. So it is perhaps better to start building intuition in that context.
Topological dimension is defined with covers, so Cech cohomology, which is also defined using covers, is perhaps the best cohomology to start with in order to understand the relation between the two notions. Cech cohomology is used for practical computations in algebraic geometry, so it will be useful if you are interested in that subject.
There is a chapter on Cech cohomology in Hartshorne, but there may be better references more focused on dimension in the algebraic setting. If you are interested in the differential viewpoint, then Bott-Tu "Differential forms in algebraic topology" is a good reference.
Bredon, "sheaf theory", has  a chapter on dimension from the topological viewpoint. Here is what I learnt from it. You may expect that if X is a well-behaved subset of Y, the cohomological dimension of X (that is, the greatest integer, or $\infty$) is less than or equal to the dimension of Y. This is true for some cohomologies, like the sheaf cohomology with constant coefficients (X,Y compact), but it is false for others. In particular, singular cohomology behaves a bit erratically with respect to cohomological dimension. Barratt and Milnor built in 62 an example of a compact manifold in R^3 (infinitely spheres of radius 1/n tangent to xy plane at the origin) with infinite singular cohomological dimension. This is definitely surprising !<|endoftext|>
TITLE: Is there a ring of integers except for Z, such that every extension of it is ramified?
QUESTION [34 upvotes]: This is probably common knowledge, alas I have to confess my ignorance.
In simpler more abstract language, does $\mathcal{O}_K$ being simply connected (having trivial etale $\pi_1$) imply $\mathcal{O}_K=\mathbb{Z}$?

REPLY [21 votes]: Yes, there are many such fields.
(Edit:  Let me put up here that "many" is still finite, and only in the hundreds if my list below represents most of the known examples.  Note that "state-of-the-art" is still pretty primitive here -- infinitely many examples would give infinitely many number fields with class number one, which we don't know to exist (but strongly strongly suspect to!).)
As Torsten brings up (and what is behind the exercise in Washington that dke mentions), the Odlyzko bounds are really the key here.  Briefly, the remarkable result that comes from these bounds is:

The ring of integers in any number field with sufficiently small root discriminant has trivial fundamental group, i.e., admits no non-trivial unramified extensions.

The "sufficiently small" is large enough, with the Odlyzko bounds, to generate fairly large examples of what you're looking for:

  All quadratic imaginary number fields with class number 1 (of which there are 9)
  All cyclotomic fields with class number 1 (of which there are 30)
  All real quadratic fields with prime-power conductor $\leq 67$ (I count 9).
  The top of the class field tower over a quadratic imaginary number field $K$ for many many $K$, including at least all such number fields with conductor $\leq 1000$ (a couple hundred).

The last two of these are work of Yamamura (results spread out, but see "Maximal Unramified Extensions of Imaginary Quadratic Fields of Small Conductors", parts I and II).
While I'm here, this last one answers affirmatively the question brought up by Minhyong in the comments.  For example, $\mathbb{Q}(\sqrt{-771})$ has a ring of integers with fundamental group $S_4\times \mathbb{Z}/3\mathbb{Z}$, for this is precisely the Galois group of the Hilbert class field tower over this base.  Finite but non-trivial.  Many other examples of fundamental groups that can show up in this way are found in Yamamura.
I think the natural follow-up (and incredibly interesting) question is to classify exactly which groups can show up in this fashion.  This is completely intractable at the moment, modulo some easy partial results in both directions.  Edit:  For example, by class group considerations, it is impossible to have a quadratic imaginary number field with etale fundamental group isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.<|endoftext|>
TITLE: software for computations on flag varieties in arbitrary characteristic
QUESTION [5 upvotes]: Is there any software that will compute cohomology of vector bundles (or just line bundles) on flag manifolds?  
The only one I know of is Macaulay2, via the Schubert2 package, but it works with what it calls "abstract varieties", which are really just the intersection rings over $\mathbb{Q}$, so it's explicitly limited to characteristic zero.  I'm interested in (among other things) bad behavior at small prime characteristics.

REPLY [5 votes]: To answer the original question explicitly, there seems to be no relevant software in prime characteristic.   Nor is there any on the horizon, unless the theory developed so far becomes much more definitive.   In the setting of flag varieties, general principles show that Euler characters in characteristic $p$ are the same as in characteristic 0 for line bundles (etc.) because the objects involved have compatible $\mathbb{Z}$-forms.  Kempf even showed that for a dominant line bundle, sheaf cohomology vanishes except for degree 0; so the formal character and dimension of the global section module are given by Weyl's formula.   Similarly for the Serre dual, but there are some systematic patterns (based on alcoves for an affine Weyl group relative to $p$) in which some other line bundles have nonvanishing cohomology in multiple degrees.   This appears only for weights "close to" Weyl chamber walls and is conjecturally due to failure of "cancellation"  in Jantzen-Andersen filtrations near walls.
A moral of the existing work on line bundles and some other vector bundles is that module structure seems needed to understand the vanishing behavior of cohomology.   For small $p$ one lacks analogues of Lusztig's conjectures on characters of the simple modules, which may be an added obstacle.   Even small calculations are very difficult, for example for small primes in rank 2.  Some have occurred in the literature, but there is no mechanical method to generate them.  
Most work was done in the 1980s, following a thesis by Mumford's student W.L. Griffith Jr. showing a few counterexamples to the Borel-Weil-Bott picture for some flag varieties and small $p$, as well as Henning Andersen's MIT thesis around the same time.   Full references before 1990 appear in my short conference survey: 
MR1131312 (92k:20084) 20G10 (14M17 20G05)
Humphreys, J. E. (1-MA),
Cohomology of line bundles on flag varieties in prime characteristic.
Proceedings of the Hyderabad Conference on Algebraic Groups (Hyderabad, 1989), 193–204,
Manoj Prakashan, Madras, 1991.
A couple of recent papers by Steve Donkin focus mostly on SL(3):
MR1958906 (2004f:20083) 20G05 (14M15 20G10)
Donkin, Stephen (4-LNDQM)
Anote on the characters of the cohomology of induced vector bundles on G/B in
characteristic p.
J. Algebra 258 (2002), no. 1, 255–274.
MR2275364 (2008a:20077) 20G10 (14L30 14M15)
Donkin, Stephen (4-YORK),
The cohomology of line bundles on the three-dimensional flag variety.
J. Algebra 307 (2007), no. 2, 570–613.<|endoftext|>
TITLE: Maximal Ideals in the ring k[x1,...,xn ]
QUESTION [20 upvotes]: Hi. From one of the forms of Hilbert's Nullstellensatz we know that all the maximal ideals in a polynomial ring $k[x_1, \dots, x_n]$ where $k$ is an algebraically closed field, are of the form $(x_1 - a_1, \dots , x_n - a_n)$. So that any maximal ideal in this case is generated by polynomials $g_j \in k[x_1, \dots, x_n]$ for $j = 1, \dots , n$, where $g_j$ only depends on the variable $x_j$ (Obviously by taking $g_j = x_j - a_j$). Now, I'm interested in the case of a polynomial ring $k[x_1, \dots, x_n]$ where $k$ is an arbitrary field (i.e., I can't make use of the Nullstellensatz). I suppose this may no longer be the case, i.e., I don't expect any maximal ideal to be generated by n polynomials, each of them only dependent on one of the variables $x_j$, but my question is if maybe the maximal ideals can be generated by polynomials $g_j$ that only depend on the first $j$ variables $x_1, \dots , x_j$? If so, does anybody know how to prove this or can anyone suggest me some references that may help me?. Thanks.

REPLY [6 votes]: In response to the OP's request for an explicit reference:
see Theorems 128 and 130 on pages 82-83 of
http://math.uga.edu/~pete/integral.pdf<|endoftext|>
TITLE: A condition on finite groups
QUESTION [5 upvotes]: While reading a paper, I came across the following peculiar condition:
Let $1 \rightarrow H \rightarrow G \rightarrow G/H \rightarrow 1$ be a short exact sequence, and let $H$ be abelian. We require that any automorphism, $\sigma$, of $G$ that preserves $H$ pointwise and such that $\sigma(g)H=gH$ (preserves cosets pointwise) is trivial.
This doesn't give me a good clue as to what we're talking about. For example: is it true for any semi-direct product of two groups of coprime order (the first one being abelian, as required)? How about for a semi-direct product of cyclic groups of coprime order? Prime coprime orders? What is this condition, and what are some canonical examples of it?
I apologize if this is simple, but my finite group theory is pretty shoddy.

REPLY [10 votes]: Short answer: a typical example is G=SL(2,5), H = Z(G) = Z/2Z.  If G/H and H are coprime and satisfy the condition, the G = G/H × H is quite dull.

I'll assume you find this interesting, and want to read about it:
Let G be a group (finite is good), H be an abelian normal subgroup of G, and Q be the quotient group.
If σ is an automorphism of a group G such that σ(H) = H, then σ induces automorphisms on H and G/H=Q.
If σ(h)=h for all h in H, then σ(H) = H, so we are interested in those σ that are "invisible" as automorphisms both of H and of Q.
The paper's condition is that no automorphism is invisible, which should seem a reasonable crutch if you want to talk about automorphisms of G in terms of those of H and Q (here it helps if H is also characteristic).
What do invisible automorphisms look like?
Well every element of G can be written as a product q*h for some q in Q and h in H.  (q1*h1)*(q2*h2) = (q1*q2)*(h1^q2 * h2 * ζ(q1,q2)) where ζ:Q×Q→H is a (set-theoretic) function called a 2-cocycle.  If you are only interested in semidirect products, then ζ(q1,q2) = 1H can be ignored.
What does σ do to q*h? Well it takes products to products, and h to h, but it only takes q to another element of the same coset, q*δ(q) where δ:Q→H is another set-theoretic function.  Hence σ(q*h) = q*h*δ(q).
A good example to keep in mind here are the extra-special groups, like the dihedral group of order 8.  They have lots of invisible automorphisms.  For instance (x,y)→(x,yz) where x,y are the main generators and z=[x,y] generates the center.
Now clearly not all δ can work, since surely δ(q1*q2) is related somehow to δ(q1) and δ(q2).  Indeed σ(q1*q2) = (q1*q2)*δ(q1*q2), but it is also equal to σ(q1)*σ(q2) = (q1*δ(q1))*(q2*δ(q2)) = (q1*q2)*( δ(q1)^q2 * δ(q1) * ζ(q1,q2) ).
Ignoring ζ for a moment, one gets the equation δ(q1*q2) = δ(q1)^q2 * δ(q2), which expresses the fact that δ:Q→H is a derivation of the Q-module H.  It is not too hard to see that the implications are reversible, and derivations help to define automorphisms fixing H and Q.
Not ignoring ζ does not change things very much, as instead of the subgroup of derivations inside the abelian group of all functions from Q to H, you just take a coset of this subgroup determined by ζ.
At any rate, so in your semi-direct products the condition is that there are no non-identity derivations from Q to H; the only derivation should have δ(q)=1H for all q in Q.
Now of course derivations can exist even when Q and H are coprime:  Take G to be the non-abelian group of order 6, H to be its subgroup of order 3.  Then δ:Q→H takes the non-identity element of Q to any one of the three elements of H, giving three derivations δ.  Checking the automorphism group of G, it is easy to see that every automorphism must take H to H, and must act as the identity on Q, since Aut(Q) = 1.  Of the six automorphisms, three act as inversion on H, so are not inivisible, but three must be invisible, one for each δ.
In particular, being cyclic of prime and coprime order is not sufficient.  Your coprime condition does, however, severely limit the variety of invisible homomorphisms that are available: they must all be conjugations by elements of H.
An automorphism of G induced by conjugation by an element of H must act trivially on H, since H is abelian.  It must act trivially on Q, since hH = 1Q in Q.  Hence every automorphism induced by an element of H is invisible.
If G is to have no invisible automorphisms, then H must be central, since q^h = h^-1 * q * h = q* (h^q)^-1 * h is only the identity on G if h^q = h for all q.
In other words, the paper's condition implies H is central, so you are looking for a group Q with a trivial module H whose first cohomology vanishes, but whose second does not.  I think this is reasonably rare.  Probably a very standard example is a perfect group Q that is not superperfect, and H to be its Schur multiplier.
For instance, take G=SL(2,5), Q = Alt(5), H = Z/2Z.<|endoftext|>
TITLE: What are the limits of non-halting?
QUESTION [9 upvotes]: It's easy enough to build Turing Machines that don't halt. But how complex can we make these?  For example, suppose a machine has access to its state transition table and can write to it like a C program could point to its own code page in RAM and poke around. The motivation for the question should clear up the particulars:
Imagine that we've build an intelligent (but deterministic) autonomous robot that can completely self-repair from the environment. Imagine that it's a space probe.  We don't want it to shut itself off.  Because it can change itself physically, it can also change its own programming.  We have no control over that once we launch the thing. It's within the realm of possibility it will go through a series of changes that result in it halting and becoming space junk.  
Is there any way to understand the topology of a self-modification "trajectory" so that we could minimize the risk of halting?  For example, maybe there's some kind of "attractor" where halting is rare.  
Or do we just have to assume that Chaitin's Omega constant applies, and there's an unknown constant probability that the thing will halt?

Update: Thanks for the comments--they sent me in new directions.  Here is some additional background.

Microsoft has an active research project along these lines.  


Turing proved that, in general, proving program termination is undecidable.
  However, this result does not preclude the existence of
  future program-termination proof tools that work 99.9 percent of
  the time on programs written by humans. This is the sort of tool
  that were aiming to make. --Byron Cook, the project leader


Usually we want programs to halt and give us some output.  But for the example I gave, we want it to run forever.  Can we build an AI that won't spontaneously turn itself off with high probability, like Shannon's "Ultimate Machine"?  Supposing that a civilization is effectively computable (a big if, but somewhere to start), is there any way to guard against self-halting? Peter Suber studied this idea, limited to legislative systems, and created the game Nomic.  Paul Krugman gives an example of a government that actually did self-halt.  My own thoughts about this are in this paper, where I assumed Chaitin's Omega would "tax" survival probability of any computable system.  This is not very satisfying, however.  It implies that we can't do any better than randomly selecting an algorithm.

REPLY [6 votes]: Total functional programming allows considerable freedom to program with a guarantee of termination. You don't get unbounded loops but you can still use structural recursion.
Such a computer would be connected to the outside world via sensors. If we allow guarded recursion then we get a nice framework for writing algorithms to process the data that can guarantee that from time to time, the robot will produce an output, rather than disappear into its own thoughts.
I don't know if this answers your question, but it seems to fit. Such a machine wouldn't poke its own state transition table, but it would be able to "program itself" by making use of higher order functions to build new functions.<|endoftext|>
TITLE: Categorical construction of the category of schemes?
QUESTION [30 upvotes]: The answer to the following question is probably well known or the question itself is well known not to have a reasonable answer. In the latter case could you please let me know what the "right" question may be (rather then stating that the answer is 42;))
Is there a purely categorical procedure that takes the category of commutative rings as input and produces the category of schemes (over $\mathbf{Z}$) as output?
A possible place to start would be to consider a scheme $X$ as a functor from the category $CommRing$ of commutative rings to the category of sets: $A\mapsto Hom_{Sch}(Spec(A),X)$ where $A$ a commutative ring. If we instead of $Spec(A)$'s we consider all schemes, then we simply get the Yoneda embedding. But some questions arise.

Does this give a fully faithful functor from schemes to functors from commutative rings to sets? Or loosely speaking, do $Spec(A)$-valued points ($A$ a commutative ring) suffice to determine a scheme? (My guess is that the answer is yes and this is classical.)
Is there a way to characterize those functors that actually come from schemes? For example one can introduce a Grothendieck topology on $CommRing$ (or its opposite) and require that the functor should be a sheaf in that topology. But in that case, can one describe the topology without referring to the fact that the objects of $CommRing$ are commutative rings? (Here my guess is the first question is probably too complicated but there are some necessary conditions.)
Regardless whether the answer to 2. is positive or negative, is there a way to describe algebraic spaces or stacks as presheaves on $CommRing^{op}$ that satisfy some conditions?

REPLY [3 votes]: It's worth pointing out Zhen Lin Low's thesis

Low, Z. L. (2016). Categories of spaces built from local models (doctoral thesis). doi:10.17863/CAM.384

with abstract

Many of the classes of objects studied in geometry are defined by first choosing a class of nice spaces and then allowing oneself to glue these local models together to construct more general spaces. The most well-known examples are manifolds and schemes. The main purpose of this thesis is to give a unified account of this procedure of constructing a category of spaces built from local models and to study the general properties of such categories of spaces. The theory developed here will be illustrated with reference to examples, including the aforementioned manifolds and schemes. For concreteness, consider the passage from commutative rings to schemes. There are three main steps: first, one identifies a distinguished class of ring homomorphisms corresponding to open immersions of schemes; second, one defines the notion of an open covering in terms of these distinguished homomorphisms; and finally, one embeds the opposite of the category of commutative rings in an ambient category in which one can glue (the formal duals of) commutative rings along (the formal duals of) distinguished homomorphisms. Traditionally, the ambient category is taken to be the category of locally ringed spaces, but following Grothendieck, one could equally well work in the category of sheaves for the large Zariski site—this is the so-called ‘functor of points approach’. A third option, related to the exact completion of a category, is described in this thesis. The main result can be summarised thus: categories of spaces built from local models are extensive categories with a class of distinguished morphisms, subject to various stability axioms, such that certain equivalence relations (defined relative to the class of distinguished morphisms) have pullback-stable quotients; moreover, this construction is functorial and has a universal property.

That last sentence is perhaps stronger than other answers here, which in all likelihood give the construction that Zhen Lin does, albeit in the special case of only schemes.<|endoftext|>
TITLE: Do Lie algebroids pull back (along submersions)?
QUESTION [5 upvotes]: There are more general definitions, but for my purposes a Lie algebroid on a smooth manifold $X$ is a vector bundle $A \to X$, a map $\rho: A \to {\rm T}X$ of vector bundles over $X$, and a bracket $[,]$ making $\Gamma(A)$ into an $\mathbb R$-Lie algebra, such that $\rho$ induces is a map $\Gamma(A) \to \Gamma({\rm T}X)$ of Lie algebras and such that the Leibniz rule $[a,fb] = f[a,b] + (\rho(a)f)b$ is satisfied for $f\in \mathcal C^\infty$ and $a,b\in \Gamma(A)$.
So, suppose I happen to have a Lie algebroid $A\to X$ lying around, and also a smooth map $\phi: Y\to X$.  (In my case, $Y \to X$ happens to be a vector bundle, so you can assume some fairly strong properties of the map.)  Then I can certainly pull back the vector bundle $A\to X$ to $\phi^*A \to Y$.  Is there a natural Lie algebroid structure I can put on this pullback?  The answer is probably "how natural do you want it?": if $Y = \{{\rm pt}\}$ and $\phi({\rm pt}) = y\in X$, then the anchor map for $\phi^*A \to \{{\rm pt}\}$ must be trivial, but the only part of the fiber $A_y$ with a Lie algebra structure is $\ker \{A_y \overset\rho\to {\rm T}_y X\}$.
So the real question is:

Along what type of maps do Lie algebroids pull back?  In particular, can I always pull back a Lie algebroid along a submersion?

REPLY [6 votes]: As your example suggests, the vector bundle pullback is perhaps not the right thing to consider when wanting defining the notion of a pullback Lie algebroid. You have to go one step further and take the  pullback of $\phi^*\rho: \phi^* A \to \phi^*TX$ along $d\phi: TY \to \phi^*TX$. The resulting bundle $\phi^{**}A$ (which exists when $d\phi - \phi^*\rho: TY \oplus \phi^*A \to \phi^*TX$ has constant rank, so in particular when $\phi$ is a submersion) then has a Lie algebroid structure with gives the right universal property you would want, if that's what you want. In the case where $Y$ is a point, this gives exactly your example.
This is all detailed in the paper "Algebraic Constructions in the Category of Lie Algebroids" by Higgins and Mackenzie. This may even have something to say about just trying to do something with $\phi^*A$ itself, but I haven't checked.<|endoftext|>
TITLE: Rational congruence of binomial coefficient matrices
QUESTION [13 upvotes]: Skip Garibaldi asks if there is an elementary proof of the following fact that "accidentally" fell out of some high-powered machinery he was working on.
Say that two matrices $A$ and $B$ over the rationals are rationally congruent if
there exists a nonsingular matrix $S$ over the rationals such that $S^t A S = B$.
Theorem (Garibaldi).  Suppose $n \equiv 0 \pmod 4$. Then the diagonal matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 2}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 1}\right]$$
are rationally congruent. Similarly, suppose $ n \equiv 2 \pmod 4$. Then the matrices
$$A = diag\left[\binom{n}{0}, \binom{n}{2}, \binom{n}{4}, \ldots, \binom{n}{n/2 - 1}\right]$$
and
$$B = diag\left[\binom{n}{1}, \binom{n}{3}, \binom{n}{5}, \ldots, \binom{n}{n/2 - 2}, \frac{1}{2}\binom{n}{n/2}\right]$$
are rationally congruent.

REPLY [3 votes]: There is a problem with this approach (I refer to Wadim Zudilin's answer). At least I don't see how to get from the first displayed equation (involving the $P_i$) to the second (involving the $L_i$) in his post.
Here is an example:
$(t^2+t-2)^2 + (3t+2)^2 + 1^2 = (t^2+t+2)^2 + (t+2)^2 + 1^2$
but 
$(t_2+t_1-2t_0)^2 + (3t_1+2t_0)^2 + t_0^2 \ne (t_2+t_1+2t_0)^2 + (t_1+2t_0)^2 + t_0^2$,
by considering the coefficient of $t_0t_2$ or of $t_1^2$.
Clearly, if the second equation were true it would imply the first. But the problem with the converse is that the first involves 5 degrees of freedom (the coefficients of $t^i$ with $0 \le i \le 4$), whereas the second has 6 (the coefficients of the $t_it_j$ with $i \le j$).
Edit: Here is a better example:
$(t^2+1)^2 -2 (t)^2 + 1^2 = (t^2-1)^2 + 2 (t)^2 + 1^2$,
but the two forms $diag(1,-2,1)$ and $diag(1,2,1)$ don't even have the same signature.<|endoftext|>
TITLE: Coiling Rope in a Box
QUESTION [21 upvotes]: What is the longest rope length L of radius r that can fit into a box?
The rope is a smooth curve with a tubular
neighborhood of radius r, such that the rope does not
self-penetrate.  For an open curve, each endpoint is surrounded
by a ball of radius r. For a box of dimensions
$1{\times}1{\times}\frac{1}{2}$ and rope with $r=\frac{1}{4}$,
perhaps $L=\frac{1}{2}+\frac{\pi}{4} \approx 1.3$,
achieved by a 'U':
           

I know packing circles in a square is a notoriously difficult problem,
but perhaps it is easier to pack a rope in a cube, because the
continuity of the curve constrains the options?
(I struggle with this every fall, packing a gardening hose in
a rectangular tub.) I am more interested in general strategies for how to best coil the rope, rather than specific values of L. It seems that if r is large w.r.t. the box dimensions
(as in the above example), no "penny-packing" cross-sectional structure is possible, where one layer nestles in the crevices of the preceding layer.

   This is a natural question and surely has been explored, but I didn't find much.

Edit 1. It seems a curvature constraint is needed to retain naturalness:
The curve should not turn so sharply that the disks of radius r orthogonal to the curve
that determine the tubular neighborhood interpenetrate.
Edit 2 (26Jun10).  See also the MO question concerning decidability.
Edit 3 (12Aug10).
Here is an observation on the 2D version, where a $1 {\times} 1 {\times} 2r$ box may only
accommodate one layer of rope. If $k=\frac{1}{2r}$ is an even integer, then I can see two natural strategies
for coiling the rope within the box:


$\qquad \qquad \qquad \qquad \qquad$ Red is rope core curve, blue marks the rope boundary.

Interestingly, if I have calculated correctly, the length of the red rope curve
is identical for the two strategies:
$$L = 2 (k-1)[r \pi/2] + 2(k-1)^2 r \;. $$
For $r=\frac{1}{16}$, $k=8$ as illustrated, $L=\frac{7\pi}{16} + \frac{49}{8} \approx 7.5$.
(As a check, for $r=\frac{1}{4}$, $k=2$, and $L$ evaluates to $\frac{\pi}{4}+\frac{1}{2}$ as in the
first example above.)

REPLY [11 votes]: Part of this question has an ad hoc nature that, in my opinion, weakeans it as a math question.  How much spaghetti can you fit into a Volkswagen Beetle?  Remark:  Slightly more than otherwise if you remember to open the glove compartment.  Okay, the question is not quite that bad, but it is clearly sensitive to boundary behavior.  If you allow the box to be an arbitrary convex shape (say), then who knows if it can ever be completely solved.
So let's look at the boundary-independent part of the question.  In 2 dimensions, if you have a convex box of any shape with a very large inradius, then it is not hard to show that the asymptotic density of the rope is 1.  You can go back and forth across the box in a boustrophedonic pattern (like the diagram on the left).
In 3 dimensions, one can speculate that the asymptotic density of the rope is $\sqrt{\pi}/12$, the same as the circle packing density in the plane.  There is a theorem of Andras Bezdek and my dad, Włodzimierz Kuperberg, that the maximum density space packing with congruent circular cylinders of infinite length is attained when the cylinders are parallel.  Their theorem includes the special case of the rope question in which most of the rope is parallel to itself.  (On the other hand, it is not quite obvious that the entire theorem is a special case.  Given a pile of stiff, straight spaghetti, can you always efficiently connect the ends to make one long noodle?)  In any case, I asked my dad this asymptotic rope question, and in his opinion, it is an open problem.  A rope seems much harder to control than straight cylinders.
Also, as far as I know, the same questions concerning either straight, round cylinders or one long rope in higher dimensions are also open.
It's my philosophy that an argument that a problem is open is a valid MO answer to a MO question.<|endoftext|>
TITLE: Connectedness and the real line
QUESTION [7 upvotes]: It is fundamental to topology that $\mathbb{R}$ is a connected topological space. However, all the topology books that I have ever looked in give the same proof. (the proof I am thinking of can be seen in Munkres's topology or Lee's Introduction to topological manifolds)
This seems strange to me, because for other fundamental results such as the Compactness of $[0,1]$, I can think of several proofs. 
Does anyone know any different proofs of the connectedness of $\mathbb{R}$?

REPLY [2 votes]: If $\mathbb{R} = A \uplus B$, for $A,B$ nonempty open subsets, then each of them is a countable union of disjoint open intervals (to prove this, one has to use the completeness of $\mathbb{R}$). Let $(a_1,a_2)$ be an interval which appears in this decomposition of $A$. Then $a_2$ can't be in $A$, but also can't be in $B$ since $B$ is open. Contradiction.<|endoftext|>
TITLE: 4900, a particularly square number
QUESTION [7 upvotes]: I read in "Letters to a young mathematician" that 4900 is the only square integer that is the sum of consecutive squares (I believe he meant by that "starting from 1", but maybe that's not even necessary). I did a quick run through with a python script and of course this seems totally devoid of a computational pattern. Why is 4900 (and 1 of course) the only number such that this works?
I did find out that the sum of squares is the following...
$\sum^{n}_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}$

REPLY [12 votes]: I remember vaguely that the equation $1^2+\dots+24^2=70^2$ gives a construction of the Leech lattice (one has to consider one of the two even neighbours  of the lattice $\mathbb Z \frac{w}{70}+\Lambda'$ where $w=(1,2,\dots,24)$ and $\Lambda'=\lbrace z\in\mathbb Z^{24}|\langle z,w\rangle\in70\mathbb Z\rbrace$) which is a rather unique and exceptional structure.<|endoftext|>
TITLE: Is there much difference between Kronecker's and Dedekind's methods in algebraic number theory and commutative algebra?
QUESTION [13 upvotes]: Edwards, in his book "Divisor theory" says that Kronecker's methods are quite different to Dedekind's and those of today. Is there really much of a difference apart from Kronecker's methods being more constructive?

REPLY [7 votes]: For a good introduction to the different approaches of Kronecker and Dedekind you should read the relevant sections in Weyl's "Algebraic theory of numbers", Princeton UP 1940. He strongly
favors Kronecker's approach, at least if the goal is saving unique factorization. In fact, Kronecker's methods do what they're supposed to do in the polynomial ring $R = {\mathbb Z}[X]$ (in particular, they show that $R$ is a UFD), whereas Dedekind's ideal theory fails miserably in this respect since $R$ is a UFD yet its ideal theory is complicated as $R$ is not even a Dedekind domain.
Taking up a comment made by Victor Protsak, the divisor theory developed by Borevich and Shafarevich (which I also neglected until very recently) isn't that foreign at all: their "rings with a divisor theory" coincide, unless I am mistaken, with the modern notion of Krull domains. I do not (yet) know any serious and readable account in English (B & S have superfluous axioms; see Gundlach's book below). For those comfortable in German I suggest 

Olaf Neumann, Was sollen und was sind Divisoren? (What are divisors and what are they good for?),  Math. Semesterber. 48, No. 2, 139-192 (2001)
F. Lucius, Ringe mit einer Theorie des grössten gemeinsamen Teilers
(Rings having a theory of greatest common divisor), Diss. Goettingen 1996
K.-B. Gundlach, Einführung in die Zahlentheorie, BI 1972<|endoftext|>
TITLE: Why were Abelian functions so important in the 19th century?
QUESTION [17 upvotes]: Felix Klein, when discussing how the popularity of areas in mathematics rises and falls, mentions that in his youth Abelian functions were at the summit of mathematics, and that later on their popularity plummeted. I could hardly find anything on the net on Abelian functions and Wikipedia thinks that they are barely worth mentioning. What did the 19th century theory of these functions consist of, and why was it so important?

REPLY [13 votes]: For a really detailed answer to your question, see The Legacy of Niels Henrik
Abel, edited by O.A. Laudal and R. Piene (Springer 2004). In particular, there is a long introductory article by Christian Houzel, most of which can be viewed here.
Addendum. The complete article "The Work of Niels Henrik Abel" by
Christian Houzel may be found here.<|endoftext|>
TITLE: Are spaces of holomorphic maps manifolds?
QUESTION [10 upvotes]: Hello,
Let $X$ and $Y$ be two smooth (probably projective) algebraic varieties defined over $\mathbf{C}$.
What is known in general about the (topological) space of holomorphic maps $\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ from $X(\mathbf{C})$ to $Y(\mathbf{C})$?
In particular, I would be interested to know under what hypothesis on $X$ and $Y$ the space 
$\mathrm{Hol}(X(\mathbf{C}),Y(\mathbf{C}))$ is a smooth manifold (and to have a formula to compute the dimension of its connected components).
One example I have in mind is : if $X=Y=\mathbf{P}^1$ (the projective line) then $\mathrm{Hol}(\mathbf{P}^1(\mathbf{C}),\mathbf{P}^1(\mathbf{C}))$, in which case one gets the space of complex rational functions (which has connected components indexed by the positive integers (the degree), but each is a smooth algebraic variety).
I think this might be an (easy?) application of the theory of Grothendieck $\mathrm{Hom}$-schemes, but I don't feel very at ease with this. 
I would also be interested to know when $\mathrm{Hom}(X,Y)$ is a smooth algebraic variety.
Many thanks,
K.

REPLY [8 votes]: Let $X,Y$ be analytic spaces with $X$ compact reduced and $Y$ arbitrary (i.e. maybe with nilpotents in its structure sheaf). Then Douady showed in his thesis
that the set of holomorphic maps $Hol(X,Y)$ can be endowed with the structure of an analytic space whose underlying topology is the compact-open topology. If $X,Y$ are compact manifolds, the Zariski tangent space at $f:X\to Y$ is a subspace of the finite-dimensional vector space of sections of the tangent bundle to $Y$ pulled-back to $X$ viz. $T_f(Hol(X,Y))\subset \Gamma(X,f^\star TY)$.
I don't know any good general criterion  for $Hol(X,Y)$ to be smooth at $f$.
Edit Here is a class of examples which might interest you, where  smoothness occurs. Let $X$ be a Riemann surface of genus $g$. The space of ramified covers  $f:X\to \mathbb P^1$  of degree $d$ is non-empty and smooth of dimension $2d+1-g$ as soon as $d\geq g+1$. But there are explicit cases for smaller $d$ where the corresponding space is singular. You can read about these results in this article by Akaohori and Namba.<|endoftext|>
TITLE: Why is the Hahn-Banach theorem so important?
QUESTION [56 upvotes]: Every time I hear it mentioned it is praised in the highest possible terms, and I remember one of my old lecturers saying that it is one of the 3 most important theorems in analysis. Yet the only consequences of it that I have read is that it proves that there are lots of functionals and that separating hyperplanes exist. Are those 2 consequences really that spectacular, or are there other ones that I don't know of?

REPLY [46 votes]: I have used it (sometimes with coauthors) several times in the following general context. I have wanted to prove that a function f can be decomposed as a sum g+h, where g has certain properties and h has certain properties. It has been possible to show that the set of acceptable g is convex, as is the set of acceptable h. So I'm trying to show that f belongs to a sum of two convex sets K+L. But a sum of convex sets is convex, so if f cannot be written in such a way, then it can be separated from K+L by a functional. It is often possible to derive a contradiction from this.
The result is that one can prove the existence of the desired decomposition under circumstances where explicitly defining a decomposition would be difficult.
Incidentally, the applications I am alluding to are of the finite-dimensional Hahn-Banach theorem. Some people call it the minimax theorem, and still others would call it duality for linear programming or something like that.<|endoftext|>
TITLE: Is there a category of topological spaces such that open surjections admit local sections?
QUESTION [21 upvotes]: The class of open surjections $Q \to X$ is a Grothendieck pretopology on the category $Top$ of spaces, and includes the class of maps $\amalg U_\alpha \to X$ where $\{U_\alpha\}$ is an open cover of $X$. There are large classes of spaces for which these two pretopologies aren't equivalent (any non-locally contractible space, for example). What I'd like to know is if there are any spaces for which they are. My question can be split into two parts:

Is there a full subcategory of $Top$ such that every open surjection admits local sections?
Is there a non-full subcategory, like that of finite CW complexes and cellular maps (not that I'm claiming this is), in which every open surjection - in this category - admits local sections?

Clearly the non-full subcategory part needs to include enough maps to be sensible, e.g. every continuous map is homotopic to one in the subcategory, and enough objects to also be considered nontrivial. 

Edit: I'm bumping this question because it received little interest, and I thought I'd explain the example which brought me to this idea.
Consider the path fibration $P_xX \to X$, the total space of which is the space of based paths with the compact-open topology. If $X$ is locally contractible then this is fibre homotopy trivial, and in particular admits local sections. As $P_xX$ is contractible, it can be seen as a sort of 'free resolution' of the space $X$ - some sort of 'cover'. (If we work in the smooth setting, and let $X=G$ a Lie group, then $P_eG \to G$ is even a locally trivial $\Omega G$-bundle.) However, going to the other extreme and only asking for $X$ to be path connected and locally path connected then $P_xX \to X$ is an open surjection. There are lots of other maps which are open surjections which admit local sections (such as principal bundles)
I'm happy for open surjections to be covers (i.e. form a Grothendieck pretopology) but then I want to be able to specify when open surjections and open covers generate the same Grothendieck topology, and this involves finding a category of spaces (which I'd like to be enough to model all homotopy types and mapping spaces correctly) where open surjections admit local sections.

REPLY [5 votes]: Maybe this is not quite satisfactory answer, but the category of zero-dimensional Polish spaces and their continuous maps has the required property: each open continuous map between Polish zero-dimensional spaces has a section (this follows from a suitable Michael's Selection Theorem, see e.g. http://www.renyi.hu/~descript/papers/Michael_monthly.pdf).<|endoftext|>
TITLE: Applications of connectedness
QUESTION [23 upvotes]: In an «advanced calculus» course, I am talking tomorrow about connectedness (in the context of metric spaces, including notably the real line).

What are nice examples of applications of the idea of connectedness?

High wow-ratio examples are specially welcomed... :)

REPLY [4 votes]: This is probably not as impressive as other answers, but since I am quite fond of real induction, I could not resist.
Real induction can be stated as follows:

Let $a < b$ be real numbers. Let a subset $S \subseteq [a,b]$ be inductive, i.e.:
  (RI1) $a\in S$.
  (RI2) If $a\le x<b$, then $x\in S$ $\implies$ $[x,y]\subseteq S$ for some $y > x$.
  (RI3) If $a < x \le b$ and $[a,x)\subset S$, then $x \in S$.
  Then $S=[a,b]$.

This formulation is taken from this Pete L. Clark's answer where also his text on this topic is linked.1
If we look at the set $S'=\{x\in[a,b]; [a,x]\subseteq S\}$, then the above conditions simply say that $S'$ is non-empty clopen subset of $[a,b]$. So in this way, we can view this as a consequence of connectedness.
The linked paper contains several theorems shown using real induction. Usually the proof goes by choosing a suitable set $S$ and showing that this set fulfills the conditions (RI1), (RI2), (RI3). At least for some of them it seems that there is not much difference between difficulty of the proof that $S$ is a non-empty clopen subset and the proof that $S$ fulfills these conditions. (Of course, it is more elegant to use real induction if we want to illustrate this method as unifying principle for proofs of several basic theorems. And also in this way similarity to the usual mathematical induction is highlighted. But for the purposes of this question, this approach might be a way how to view proofs by real induction as a source of applications of connectedness in analysis.)

1 Pete L. Clark: The Instructor's Guide to Real Induction, https://arxiv.org/abs/1208.0973, http://math.uga.edu/~pete/realinduction.pdf<|endoftext|>
TITLE: Can one bound the Todd class of a 3-dimensional variety polynomially in c_3
QUESTION [10 upvotes]: This question is on bounding the degree of the Todd class on a complex threefold.  
Let $X$ be a smooth compact connected complex surface. Let $c_i=c_i(TX)$ be its $i$-th Chern class. Recall the following two facts. These allow one to bound the degree of the Todd class on a surface in terms of $c_2$.
1. If $X$ is not of general type, we have that $c_1^2$ is bounded absolutely from above by 9. See Table 10 of Chapter VI of Compact complex surfaces by Barth, Hulek, Peters and van de Ven.
2. If $X$ is of general type, then the Bogomolov-Miyaoka-Yau inequality states that $$c_1^2 \leq 3c_2.$$ 
Now, I am   interested in similar results for 3-dimensional smooth projective connected   varieties over $\mathbb{C}$.
In this case,  the degree of the Todd class of $X$ is the degree of $$\frac{c_1 c_2}{24}.$$  

Question. For 3-dimensional smooth projective connected varieties over $\mathbb{C}$, do there exist any absolute upper bounds on $c_1c_2$ (or any bounds for that matter) which are polynomial in $c_3$?

REPLY [17 votes]: The case of complex manifolds of higher dimension is very different from the case of complex surfaces. So the answer to the question about complex $3$ folds is no, there exists a real 6-dimensional simply connected manfiold with integrable complex structures $J_m$ for all $m\in \mathbb Z^+$ such that $c_1c_2=48m$. This is a theorem A from the acticle of LeBrun. Though, the manifolds that he constructs are not algebraic
Topology versus Chern Numbers for Complex 3-Folds
http://arxiv.org/PS_cache/math/pdf/9801/9801133v1.pdf
The question for algebraic manifolds was studied by Kotschick, you may be interested this the following two articles:
TOPOLOGICALLY INVARIANT CHERN NUMBERS OF PROJECTIVE VARIETIES
http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.1587v1.pdf
CHERN NUMBERS AND DIFFEOMORPHISM TYPES OF PROJECTIVE VARIETIES
http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.2857v2.pdf
Finally, it should be added that a systematic attempt to construct various complex 3-fold is given in a very nice article of Okonek and Van de Ven "CUBIC FORMS AND COMPLEX 3-FOLDS" of Okonek, Ch. / Van de Ven, A, L'Enseignement Mathématique Volume / Année: 41 / 1995. The link is given in the comments<|endoftext|>
TITLE: complexity of counting homomorphisms
QUESTION [11 upvotes]: This is a question I have thought about and asked a number of people, but have never got an answer beyond "It should be true that..."
Given a finitely generated group $G$ (eg. a link group $G_L:=\pi_1(S^3-L)$ for a link $L$) and a finite group $H$ we want to count homomorphisms from $G$ to $H$.  For link groups as above, this is an invariant of $L$.  
My question: (for which $H$) is there a polynomial-time algorithm (in the number of generators and relations for $G$) for computing $N(G,H):=|Hom(G,H)|$ (particularly for $G_L$)?
Some things I know: 
1) If $L$ is a knot and $H$ is nilpotent then $N(G_L,H)$ is constant (M. Eisermann)
2) D. Matei; A. I. Suciu, have an algorithm for solvable $H$, but the complexity is not clear.
3) The abelianization of $G_L$ is just $Z^c$, $c$ the number of components, so for $H$ abelian it is easy.
A wild conjecture is that it should always be "FPRASable" i.e. there exists a fully polynomial randomized approximation scheme for the computation.

REPLY [3 votes]: For $G$ a knot group, and for $H$ a dihedral group, there should be a simple algorithm for counting the number of homomorphisms. The meridians of $G$ normally generate, and are all conjugate, so they must be sent to conjugate elements in $H$. If they are sent to the cyclic subgroup of index 2, then the image is cyclic, and this is easy to count. 
If a meridian is sent to an involution, then an index 2 subgroup of $G$ is sent to a cyclic group. This amounts to computing the homology of the 2-fold branched cover of the knot, together with the action of the involution on this homology. This is certainly polynomial-time computable, and I'm pretty sure one can determine its dihedral quotients easily. In any case, at least this reduces it to the problem of finding dihedral quotients of abelian-by-$\mathbb{Z}/2$ groups.<|endoftext|>
TITLE: The fibers of M_{g,n} \to M_g and the Fulton-MacPherson compactification
QUESTION [12 upvotes]: Let $g \geq 2$, and consider the moduli space $\bar M_{g,n}$ of stable n-pointed curves of genus g. There is a natural forgetful map to $\bar M_g$, which forgets the markings and contracts any resulting unstable component. I am thinking about what the fibers of this map look like. Consider first the n-fold fibered product of the universal curve over $\bar M_g$ -- this would be a moduli space of curves of genus g with n not necessarily distinct markings. We know that this product is singular at all points where more than one of the markings coincide, or when a marking is placed on a node of a singular base curve. Moreover, we know that $\bar M_{g,n}$ is a desingularization. 
Of course, the difference between the two spaces is that on $\bar M_{g,n}$, whenever two markings or a marking and a node come together, one ''bubbles off'' a rational curve, and this can be seen as a blow-up of an ambient space in which the curve is embedded. My question is: can one in a similar way explicitly realize the total space $\bar M_{g,n}$ as a sequence of blow-ups of the n-fold fibered product of the universal curve, or a similar construction?
And a related naive question, which may or may not be equivalent to the one above: given a point [C] in the interior of $\bar M_g$, the fiber over it in $\bar M_{g,n} \to \bar M_g$ is a compactification of the configuration space of n distinct ordered points on C. Is this fiber isomorphic to the Fulton-MacPherson compactification of this configuration space?

REPLY [2 votes]: It may be worth mentioning briefly that the special case of $\overline{M_{0,n}}$ can be constructed fairly explicitly by blow-ups. The Fulton-MacPherson construction works. There are at least two additional constructions: an inductive one by Sean Keel in "Intersection theory on moduli space of stable N-pointed curves of genus zero" and the construction of Kapranov using what he calls Veronese curves towards the end of this paper and in an earlier paper not on the arXiv that he sites. Kapranov's construction expresses $\overline{M_{0,n}}$ as a composition of blow-ups of $\mathbb{P}^{n-3}$ at an explicit sequence of (strict transforms of) linear subspaces.<|endoftext|>
TITLE: A specific Dedekind-esque sum
QUESTION [13 upvotes]: Let $\tau$ be $(-1+\sqrt{5})/2$, let $f(x)$ be $\lfloor (x+1)\tau \rfloor$, let
$s_n$ be $\tau n (n+1) (n+2) / 6$, and
let $S_n$ be $$\sum_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1)
+ (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0).$$
Is $(S_n-s_n)/(n \log n)$ bounded for $n > 1$?
(It stays between -0.35 and +0.30 for all $n$ between 2
and $10^6$.)
This is a specific instance of the question
Dedekind-esque sums
that I posted a few weeks ago.  It may be an atypical instance in some
ways (since $\tau$ is a pretty atypical real number for Diophantine
approximation problems) but it's the one that interests me most
right now.  An affirmative answer to my question would have
implications concerning the "Goldbug machine" described in

Michael Kleber, Goldbug Variations, Mathematical Intelligencer 27 #1 (Winter 2005), pp. 55–63, https://arxiv.org/abs/math/0501497.

The plot at the bottom of http://www.cs.uml.edu/~jpropp/Phi-short.pdf
is a histogram of $(S_n - s_n)/n$ for $n$ going from 1 to a million.
As you can see, it doesn't stray very far away from 0.
So perhaps that $\log n$ in the denominator could be replaced by
something smaller, like $\sqrt{\log n}$ or even $\log \log n$ (or
maybe even 1, though I doubt it).

REPLY [3 votes]: I can't help thinking that the answer ought to come out of a combination of standard techniques. First, indeed, reduce to fractional parts: but with a caveat, that it would be sensible to do some manipulation first. I'm thinking about "summation by parts". 
Edit: There is a classic reference: G. H. Hardy & J. E. Littlewood, "Some problems of Diophantine approximation: The lattice points of a right-angled triangle," (1st memoir), Proc. London Math. Soc. (2), v. 20, 1922, pp. 15-36. They consider the modified fractional part sum, with {x} set as x - [x] - 1/2, of the ${k\theta}$ up to n, where to be compatible with their notation $\theta$ would be the reciprocal of $\tau$, not that this matters at all. The bound they get is O(log n) (Hardy's Works vol. I p. 145), which depends only on the continued fraction having bounded partial quotients. The particular case relevant to $\tau$ is worked out in detail over the next few pages. The result is sharp. Off the top of my head this looks enough to get the error term O(nlog n) for the sum as posted, by breaking into at most n sums of this type.<|endoftext|>
TITLE: Papers that debunk common myths in the history of mathematics
QUESTION [111 upvotes]: What are some good papers that debunk common myths in the history of mathematics?
To give you an idea of what I'm looking for, here are some examples.
Tony Rothman, "Genius and biographers: The fictionalization of Evariste Galois," Amer. Math. Monthly 89 (1982), 84-106.  Debunks various myths about Galois, in particular the idea that he furiously wrote down all the details of Galois theory for the first time the night before he died.
Jeremy Gray, "Did Poincare say 'set theory is a disease'?", Math. Intelligencer 13 (1991), 19-22.  Debunks the myth that Poincare said, "Later generations will regard Mengenlehre as a disease from which one has recovered."
Colin McLarty, "Theology and its discontents: The origin myth of modern mathematics," http://people.math.jussieu.fr/~harris/theology.pdf .  Debunks the myth that Gordan denounced Hilbert's proof of the basis theorem with the dismissive sentence, "This is not mathematics; this is theology!"
Some might say that my question belongs on the Historia Matematica mailing list; however, besides the fact that I don't subscribe to Historia Matematica, I think that the superior infrastructure of MathOverflow actually makes it a better home for the list I hope to create.  Still, maybe someone should let the Historia Matematica mailing list know that I'm asking the question here.

REPLY [4 votes]: There are stories about work of Banach being written up by people other than Banach. Details and debunking links available on another MO thread, Who wrote up Banach's Thesis?<|endoftext|>
TITLE: Free splittings of one-relator groups
QUESTION [22 upvotes]: Roughly speaking, I want to know whether one-relator groups only have 'obvious' free splittings.

Consider a one-relator group $G=F/\langle\langle r\rangle\rangle$, where $F$ is a free group.  Is it true that $F$ splits non-trivially as a free product $A * B$ if and only if $r$ is contained in a proper free factor of $F$?

Remarks

One direction is obvious.  It is clear that if $r$ is contained in a proper free factor then $G$ splits freely.  (We think of $\mathbb{Z}\cong\langle a,b\rangle/\langle\langle b\rangle\rangle$ as an HNN extension of the trivial group, so it's not really a counterexample, even though it might look like one.)
A quick search of the literature suggests that the isomorphism problem for one-relator groups is wide open.  (I'd be interested in any details that anyone may have.)
There is no decision-theoretic obstruction.  Magnus famously solved the word problem for one-relator groups. Much more recently, Nicholas Touikan has shown that, for any finitely generated group, if you can solve the word problem then you can compute the Grushko decomposition.  So one can algorithmically determine whether a given one-relator group splits.  If the answer to my question is 'yes' then one can use Whitehead's Algorithm to find this out comparatively quickly.
When I first considered this question, it seemed to me that the answer was obviously 'yes' - I don't see how there could possibly be room in a presentation 2-complex for a 'non-obvious' free splitting.  But a proof has eluded me, and of course many seemingly obvious facts about one-relator groups are extremely hard to prove.

REPLY [12 votes]: I just came across the following, which is Prop. II.5.13 of Lyndon-Schupp.
Proposition.  Let $G = \langle x_1, \ldots , x_n: r\rangle$ where $r$ is of minimal length under $Aut(\langle x_1, \ldots , x_n\rangle)$ and contains exactly the generators $x_1,\ldots , x_k$ for some $0 \leq k \leq n$.  Then $G \cong G_1*G_2$ where $G_1 = \langle x_1, \ldots , x_k:r\rangle$ is freely indecomposable and $G_2$ is free with basis $x_{k+1}, \ldots ,x_n$.
Unless I'm missing something, up to isomorphism you can assume that your relator has minimal length.  If it is not contained in a free factor of $G$, then $k=n$ in the Proposition, hence $G = G_1$ is freely indecomposable.<|endoftext|>
TITLE: Is there a simple method to test a local ring to be Cohen Macaulay?
QUESTION [16 upvotes]: Hi, everybody. I'm recently reading W.Bruns and J.Herzog's famous book-Cohen-Macaulay Rings. I personally believe that it would be perfect if the authors provide for readers more concrete examples. After reading the first two sections of this book, I have two questions.

Given a non-negative integer n, how can we construct a non-Gorenstein Cohen-Macaulay ring with dimension n?
Except by the definitions of Cohen-Macaulay rings, is there a more efficient way to check the Cohen-Macaulayness of local rings?

REPLY [6 votes]: For me personally, the whole theory started to take shape (and make sense) once I learned about the graded case and understood connections with combinatorics.
For a graded (sometimes called $*$-local) ring, a basic technique for establishing the Cohen-Macaulay property is "Gröbner degeneration": using a Gröbner basis, deform the ring to a quotient of a polynomial ring by a monomial ideal. Another approach is to deform a ring to a multigraded ring (=an affine semigroup ring) by exhibiting a SAGBI basis. This is known as "toric degeneration". The question then may be decided by combinatorial techniques. The commutative algebra bit is that if $R_t$ is a flat deformation with a CM special fiber $R_0$ and general fiber $R$ then $R$ is also CM. 
A quotient $k[x_1,\ldots,x_n]/I$ of a polynomial ring by a square-free monomial ideal is a Stanley-Reisner ring of a simplicial complex $\Delta$ and CM property of the ring can be decided at the level of homology of $\Delta$ by the Reisner criterion. The corresponding simplicial complexes $\Delta$ are also called Cohen-Macaulay and have been much studied by people in algebraic combinatorics. 
The Cohen-Macaulayness of determinantal rings mentioned in Hailong's answer can be established using the strategy I outlined (I think that Bruns and Herzog actually do it in a later chapter; I can't verify it since I don't have the book). "Combinatorial commutative algebra" by Miller and Sturmfels is well worth looking at for a more encompassing view. Stanley's "Combinatorics and commutative algebra" is older, but retains much of its appeal: it is very explicit and to the point. You can find many examples there.<|endoftext|>
TITLE: Incompleteness and nonstandard models of arithmetic
QUESTION [9 upvotes]: The following are a collection of doubts, some of which shall have concrete answers while others may have not. Any kind of help will be welcome.
Reading Peter Smith's "Gödel Without (Too Many) Tears", particularly where he gives a nonstandard model of Q, I began wondering if the reason for the existence of nonstandard models of arithmetic has anything to do with incompleteness theorems.
I do not know if categoricity implies completeness (in the sense of every sentence being decidable by proof), but anyway, it seems reasonable, when one is formalizing a given (informal) theory, to try to "force" somehow the formal theory to talk "almost exclusively" about the intended interpretation. So I started thinking if some axiom (or axiom schema) could be added to PA in order to forbid its most obvious nonstandard models.
The first idea in this line was: ok, we have our class of terms 0, S0, SS0, etc. So, if we found a way to tell that for every x there is some term to which it is equal, we would be done.
But then I realized that our terms are defined inductively and that we are making implicitly the assumption: “and nothing else is a term”, very similar to the desired “and nothing else is a number” we would like to add to PA. This thought sort of worried me: every metatheoretic concept (terms, formulas, and even proofs!) is based on assumptions like these! (I have not still found a way out of these worries).
Leaving that apart. What if we move on to a stronger theory (with different axioms, but with an extension by definitions that proves every axiom of PA), for example ZFC? Natural numbers become then 0 (the empty set) plus the von Neumann ordinals (obtained by Pair and Union) that contain no limit ordinal. The set of natural numbers is obtained from Infinity, just selecting them by Comprehension. Kunen says in page 23 of his “The Foundations of Mathematics” that the circularity in the informal definition of natural number is broken “by formalizing the properties of the order relation on omega”. Could nonstandard models survive this formalization?
Well, I think I've read somewhere that being omega is absolute, so forcing would not be a way to obtain such nonstandard models. Also, I am not sure if (the extension by definitions from) ZFC set theory is a conservative extension of PA, but then it would not be able to prove anything about natural numbers (expressible in the language of arithmetic) that PA alone cannot prove. So somehow it looks like nonstandard models must manage to survive! Maybe due to the notion of being a subset of a given set not being particularly clear (although it looks like it should not be problematic with hereditarily finite sets).
Thank you in advance.

REPLY [3 votes]: That's the blue pill. Forgive me if I can't resist mentioning the red pill. 
One way out is (constructive) Church's Thesis, from which it follows there are no nonstandard models of first-order arithmetic. 
http://www.jstor.org/pss/2274603
http://en.wikipedia.org/wiki/Extended_Church%27s_thesis<|endoftext|>
TITLE: Fundamental group as topological group
QUESTION [48 upvotes]: Background
Let $(X,x)$ be a pointed topological space. Then the fundamental group $\pi_1(X,x)$ becomes a topological space: Endow the set of maps $S^1 \to X$ with the compact-open topology, endow the subset of maps mapping $1 \to x$ with the subspace topology, and finally use the quotient topology on $\pi_1(X,x)$. This topology is relevant in some situations. A very interesting paper dealing with this topology is:

[1] Daniel K. Biss, A Generalized Approach to the Fundamental Group, The American Mathematical Monthly, Vol. 107

You can find this online. This is somehow an introduction to

[2] Daniel K. Biss, The topological fundamental group and generalized covering spaces , Topology and its Applications, Vol. 124

Question
How can we prove that $\pi_1(X,x)$ is a topological group? Clearly the inversion map $\pi_1(X,x) \to \pi_1(X,x)$ is continuous, since $S^1 \to S^1, z \mapsto \overline{z}$ is continuous and induces this map. But I don't know how to attack the continuity of the multiplication. It's not hard to see that the multiplication on $map((S^1,1),(X,x))$ is continuous, since it is induced by a fold map $S^1 \to S^1 + S^1$. In order to carry this over to $\pi_1(X,x)$, there are at least two problems which I encounter:

The quotient map $map((S^1,1),(X,x)) \to \pi_1(X,x)$ may be not open.
The product of the quotient maps $map((S^1,1),(X,x))^2 \to \pi_1(X,x)^2$ may be not a quotient map.

In [1] it is claimed that $\pi_1(X,x)$ is always a topological group, and this should be proven in [2], but I have no acecss to [2].
An example that products of quotient maps don't have to be quotient maps can be found here. Remark however that this is true in the category of compactly generated spaces.

REPLY [2 votes]: What has not been said is that if $X$ is path connected and has a universal cover, then the fundamental groupoid $\pi_1(X)$ can be given a topology so that the source map $s:\pi_1 (X) \to X$  is a bundle over $X$ with fibre over $x \in X$ the universal cover of $X$ based at $x$: see Chapter 10 of Topology and Groupoids. This has been known for a long time.<|endoftext|>
TITLE: Can an integral equation always be rewritten as a differential equation?
QUESTION [19 upvotes]: Given an integral equation is there always a differential equation which has the same (say smooth) solutions?
It seems like not but can one prove this in some example?
Edit: Naively I'm hoping for some algorithm which takes an integral equation and applies some operations like taking derivatives, substituting variables for some new ones, adding additional differential equations etc... such that after this procedure you have made all integral signs vanish and obtained a differential equation which has the same solutions as the integral eqution. (maybe similarly to how one can transform any system of PDEs into a system of first order equations)

REPLY [3 votes]: Well, the answer, if there is any, must be subtle. Take the following example. Let $H$ be the Hilbert transform. Then
$$\frac{d}{dx}Hu=f$$
is an integral equation, where the kernel is the Fourier transform of $\xi\mapsto|\xi|$. It cannot be rewritten as a set of differential equations (ODEs) within $\mathbb R$. Nevertheless, it can be recast in terms of PDEs, using an harmonic extension in the upper half-plane:
$$-\Delta_{x,y}\phi=0\hbox{ in }\{y>0\},\qquad \frac{\partial\phi}{\partial y}(x,0)=f(x),\quad u(x)=\phi(x,0).$$<|endoftext|>
TITLE: Double Category of Topological Stacks
QUESTION [9 upvotes]: There are two equivalent ways of describing topological stacks. 

One is the "stacky" definition, that is, a topological stack is a stack $\mathbb{X}$ on $Top$ (a Grothendieck universe thereof, if you'd like) equipped with the topology generated by open covers, such that $\mathbb{X}$ admits an atlas (representable epimorphism) $X \to \mathbb{X}$ from a topological space. This is equivalent to saying that $\mathbb{X}$ is 2-iso to the stackification of a pseudofunctor $Hom(blank,G)$ for some topological groupoid $G$. Topological stacks are then the full sub-2-category all stacks on $Top$ consisting of those stacks with an atlas.
One is a "groupoidy" definition. The bicategory $BunGpd$ has topological groupoids as objects, and a morphism $H \to G$ is a principal $G$-bundle over $H$ (and biequivariant maps of as 2-cells). This bicategory is equivalent to that of topological stacks.

However, 2.) can be naturally strengthened to a weak double category by declaring vertical morphisms to be continuous functors and horizontal arrows to be principal bundles.
Now, I have a preference to working in 1.) as the stacky-language is quite useful. However, 2.) manifestly has "more structure". My question is, is there a way to "beef up" topological stacks into a weak double category in a natural way? I'm not really satisfied with taking the objects to be topological stacks with a preferred atlas and declaring the vertical arrows to be maps which factor through these atlases.

REPLY [3 votes]: A quick answer regarding 2. This could be approached by using a proarrow equipment, of which I know nothing bar their existence and the easy example of Prof - a weak double category with vertical arrows honest functors, and horizontal arrows profunctors. Given the connection (tenuous, perhaps) between the bicategory of (saturated) anafunctors/generalised morphisms between ordinary small groupoids and profunctors, I would think that a lot of this would go across without too much difficulty.<|endoftext|>
TITLE: Examples of locally ringed spaces
QUESTION [10 upvotes]: I want to know more classes of examples of locally ringed spaces. The reason is that when I want to prove/disprove something about locally ringed spaces, my examples are often not eclectic enough. Besides, it is interesting to what extent special theories can be generalized to locally ringed spaces.
Here are the classes of examples I know:

Local rings (underlying topological space is a point)
Schemes
Algebraic Varieties in the classical sense (just closed points)
Manifolds with smooth/analytic/holomorphic functions
Topological spaces with continuous functions
Various subcategories of the above examples
The category of locally ringed spaces is complete and cocomplete. Thus you can build new locally ringed spaces out of the above ones.

Which substantial different examples do you know? Please no fancy Grothendieck topologies ;).

REPLY [4 votes]: The following is BCnrd's answer, posted as requested by Martin:  "Complex/real-analytic spaces, formal schemes, topological space equipped with locally constant functions valued in a local ring. Strictly speaking, examples with exotic Grothendieck topologies yield a locally ringed topos but are not locally ringed spaces (e.g., etale or crystalline of fppf topos of a scheme equipped with "structure sheaf", rigid-analytic spaces, Berkovich spaces, adic spaces, simplicial objects of various sorts, etc.). A nifty related fact is that there are rigid-analytic spaces admitting abelian sheaves having a nonzero global section whose stalks at all points vanish."<|endoftext|>
TITLE: Is the tensorproduct of a triangulated category with a ring again triangulated?
QUESTION [7 upvotes]: $\underline{Background}$ : 
Suppose $\tau$ is a preadditive category and $R$ a ring. Then one may form a new preadditive category $\tau \otimes R$ in the following way: 
$\tau \otimes R$ has the same objects as $\tau$ and for objects $A$ and $B$, set $\tau \otimes R(A,B) := \tau(A,B)\otimes_{\mathbb{Z}} R$. Composition is given by $\tau(A,B)\otimes_{\mathbb{Z}} R\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}} R \cong \tau(A,B)\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}}R\otimes_{\mathbb{Z}}R \rightarrow \tau(A,C) \otimes_{\mathbb{Z}}R$, where the last arrow is given by the composition in $\tau$ and multiplication in $R$.
$\tau \otimes R$ is additve if $\tau$ is additive.
$\underline{Question}$ : Suppose $\tau$ is triangulated, is there a (canonical) structure of a triangulated category for $\tau \otimes R$?
Actually, I only need the case $R := \mathbb{Z}[\frac{1}{n}, \theta] \subset \mathbb{C}$, where $\theta$ is a primitive root of unity of order $n$ for some $n \in \mathbb{N}$, but I would also be interested in a general statement/counterexample to the general case.
Thank you for reading.

REPLY [8 votes]: It seems unreasonable to expect tensoring the morphisms in a triangulated category by a ring to be a meaningful operation - after all these morphisms are typically cohomology groups of natural complexes, and we know there are universal-coefficient-theorem type corrections to tensoring cohomology with a ring as opposed to the more natural operation of tensoring a complex.
On the other hand, if we enrich our triangulated categories everything works fine.
For example on the level of (pretriangulated) differential graded categories or $A_\infty$ categories, or if you prefer, stable $\infty$-categories, there's a perfectly nice
operation of tensor product by a ring - in fact one can tensor by another such category (generalizing the case you're asking about, which is tensoring with R-mod). Even more, there's a structure of symmetric monoidal $\infty$-category on such categories (developed in Lurie's Derived Algebraic Geometry series (an exposition and references can be found eg here).<|endoftext|>
TITLE: Does smooth target space and smooth fibers imply smooth total space?  
QUESTION [6 upvotes]: Suppose that $f: X \rightarrow Y$ is a morphism between algebraic varieties.  If $Y$ is smooth, and the fibers of $f$ over closed points of $Y$ are proper and nonsingular, does it follow that $X$ is smooth?
Update:
The answer to the question as posed, is NO.  See a comment by Karl Schwede below for a counterexample.  
Modified question:
Let $f$ be a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field).  Let $x \in X$ be a closed point and let $y = f(x)$.  Just because the fiber $f^{-1}(y)$ is smooth does not mean $X$ is smooth at $x$.  What if $X \times_Y Spec \mathcal{O}_y/m^n$ is smooth over $Spec \mathcal{O}_y/m^n$ for every positive integer n - is $X$ smooth at $x$?  Here $m$ is the maximal ideal of the local ring at $y$.  
Is there any condition on $f$ or the fibers which will guarantee smoothness of the total space?  Flatness plus smooth fibers is one, is there anything weaker?

REPLY [2 votes]: I apologize for answering such an old question, but it seems fundamental.  A classical counterexample occurs for the abel map of a Prym variety with exceptional singularities on the theta divisor.  The point is that the fibers of the abel prym map  X-->Y of the double cover C'-->C are included among those for the abel map of C', hence are all smooth.  (A map obtained by restricting another map over a subvariety of the target has the same fibers.)
Nonetheless X is singular at any exceptional divisor. (see lemma 2.13 of A Riemann singularities theorem for Prym theta divisors, with applications).
The point of the previous paper was that generalizing the Riemann - Kempf singularity theorem to prym varieties is easy when X is smooth.  But when X is singular it is considerably harder:
A necessary and sufficient condition for Riemann's singularity theorem to hold on a Prym theta divisor
Singularities of the Prym theta divisor
For a detailed discussion of the case of the abel prym map for a prym variety isomorphic to the intermediate jacobian of the cubic threefold, see:
On parametrizing exceptional tangent cones to Prym theta divisors
The answer is yes however if the target Y is a smooth curve, since X is smooth at any point lying on a smooth cartier divisor,  (compare Mumford, chap.7, Prop. 2, redbook.)<|endoftext|>
TITLE: Polarizations on intermediate Jacobians
QUESTION [6 upvotes]: Let $X$ be a Kahler variety of dimension $n$. For each odd number $2k-1 \leq n$ one can consider the $k$-th intermediate Jacobian, that is, the complex torus $$J^{k}X := \frac{H^{2k+1}(X, \mathbb{C})}{F^k H^{2k+1}(X, \mathbb{C}) + H^{2k+1}(X, \mathbb{Z})},$$ where $F^{\cdot}$ is the Hodge filtration. In general $J^k X$ is not an abelian variety, except for the extreme cases of the Picard and Albanese varieties (when $X$ itself is algebraic).

Are there any criteria to determine whether $J^K X$ is polarized? Or have some nontrivial cases where $J^k X$ is polarized been worked out?

REPLY [4 votes]: If enough Hodge numbers vanish so that the Hodge structure $H^{2k+1}(X)$ has level one,
then $J^kX$ should be an abelian variety. This applies to Fano (e.g. cubic) 3-folds for example.
Later that day: Partly in response to Charles Siegel's comment/question, let me
sketch a proof of a slightly more general statement. Suppose X is a projective rather
than just Kaehler (which I forgot to say before), so $H$ has a polarization $Q$. Assume
further that 
$$H^{2k+1}(X) = H= H^{pq}\oplus H^{qp}$$
has only two terms.
Let $G$ be the same thing as $H$ viewed as a weight one structure. More precisely,
the lattices $G_Z=H_Z$ are the same, and $G^{10}=H^{pq}$.
Then one sees 
that $J^kH= G^{01}/G_Z$, and that $\pm Q$ gives a polarization on $G$. So this is abelian variety.

REPLY [2 votes]: The best known examples are the cubic threefolds and the quartic double-solids (that is, double covers of $\mathbb{P}^3$ branched along a quartic).  In general, this works for quadric bundles, see Beauville's paper.<|endoftext|>
TITLE: Navigating Z/pZ
QUESTION [30 upvotes]: Let's consider a silly-looking question first. Consider Z/pZ. Say I am allowed the two operations x->x+1 and x->2*x. Then, starting from 0, I can express every given element y of Z/pZ in O(log p) steps; moreover, I can figure out how to do it in time O(log p).
The answer is trivial: just lift y to an integer, and express it in base 2.
Now, what happens if we consider x->r*x instead of x->2*x?
(Assume that r has order at least >> log p, as otherwise things don't work.)
That is: can one express every given element y of Z/pZ in O(log p) steps by starting from 0 and using the operations x->x+1 and x->r*x? If so, can one figure out how to express such an element in that way in O(log p) (or O((log p)^c)) steps?

REPLY [4 votes]: Let $r$ be the "base" and $x$ the number to represent.
Let $m = \log_{2} (p) + \epsilon$. Construct the matrix $L$:
$$\begin{pmatrix}
x & \lambda & 0 & & ... & & 0 \\\\
1 & 0 & \lambda & &  & & 0 \\\\
\bar{r} & & & \lambda & & & \\\\
... & & & & & & \\\\
\bar{r^m} & & & & & & \lambda \\\\
p & & & & & & 0 \end{pmatrix}$$
where $\bar{r^i}$ is $(r^i \mod p)$.
Given a representation:
$$x = \sum_{i = 0}^{m} a_i r^i (\mod p)$$
or more precisely:
$$x = \sum_{i = 0}^{m} a_i \bar{r^i} - tp$$
where $a_i \in \{0,1\}$, we multiply the matrix from the left by the row:
$$\left( \begin{array}{ccc} 1 & -a_0 & -a_1 & ... & -a_m & t \end{array} \right)$$
and get the short row:
$$\left( \begin{array}{ccc} 0 & \lambda & -a_0 \lambda & -a_1 \lambda & ... & -a_m \lambda \end{array} \right)$$
since we expect the number of non zero $a_i$'s to be around $\log_2(p)/2$, we expect the norm of this row to be:
$$\sqrt{\lambda^2 \sum_{i = 0}^{m} a_i^2} \sim \lambda \sqrt{\log_2(p)/2}$$
Note that:
$$|\det(L)|^{1/(m+3)} = (p \lambda^{m+2})^{1/(m+3)}$$
While we want to keep the row short, we want to make other rows long. So we choose:
$$\lambda \sim p \sqrt{\log_2(p)/2}$$
Going the other way around, applying the LLL algorithm to $L$, for a few small $\epsilon$, should produce a representation.
This algorithm is heuristic and I am not sure how to prove anything stronger. Try looking for papers on the knapsack problem, since it is quite similar.<|endoftext|>
TITLE: What, if anything, makes homogeneous polynomials so great?
QUESTION [21 upvotes]: It should be obvious from the question that I am not any kind of algebraic geometer, so if there are definitions of hom-polys as comonoidal dyadic functors or whatnot, let's leave that to one side for the purposes of this question. I really mean hom-polys in the most pedestrian sense possible.
From the outside, it seems that homogeneous polynomials get a lot of attention in alg-geom. (Perhaps in other areas as well?) I know that they are, well, homogeneous with respect to dilations, and that this allows one to look at their zeros in projective space in a natural way.
I usually like to keep a safe distance between myself and projective space, and have always looked at hom-polys as "merely" a technical tool. But, just the other day, I was able to quickly solve a small, elementary number-theoretic problem by converting it into "homogeneous" form. (Some vague memories of homogeneous problem-solving heuristics prompted this.) To be precise, the problem was that of computing how many solutions there are to m^2 + n^2 = 1 in Z_p, where p is a prime congruent to 3 mod 4. (Of course this is trivial by a change of variables when p = 1 mod 4.) The problem seemed unfriendly at first, but passing to the related question of the solutions to m^2 + n^2 - t^2 = 0 revealed a lot of symmetry and made it quite trivial. (Of course, solving small cases of the first problem showed a lot of symmetry, but it wasn't obvious how to get a handle on it.)
My question is: do hom-polys help to solve a lot of seemingly unrelated problems via some process of homogeneization of the problem? Do you have examples? Is this one reason for their popularity? I'm looking for heuristics mostly, but if you think an actual theorem makes precise or helps formulate a heuristic, go nuts.

REPLY [3 votes]: As you already noticed,  thinking in terms of homogeneous coordinates in a complex projective space $\mathbb{P}^n$, in order to define a function (any old complex-valued function, not necessarily holomorphic or even continuous) on $\mathbb{P}^n$ we must require that $f$ is homogeneous of order zero, i.e., $f(\lambda x)=f(x)$ for any $\lambda \in \mathbb{C}\setminus 0$ and any $x \in \mathbb{P}^n$. (I am cutting some corners with this notation.) But in order to define a zero set of a function on   $\mathbb{P}^n$ , we have a little more leeway: requiring that the the ``function" be homogeneous of order $k \geq 0$, i.e.,  $f(\lambda x)=\lambda^kf(x)$ for any $\lambda \in \mathbb{C}\setminus  0$ and any $x \in \mathbb{P}^n$, will do. A more profound statement (=Chow's theorem) is true (sort of converse to the above): every analytic set in $\mathbb{P}^n$ is the (common) zero set of finitely many homogeneous polynomials. 
Aside 1: This point of view generalizes to toric varieties, since they admit an analog of homogeneous coordinates (first introduced in the smooth case in 
MR1106194 Audin, Michèle: The topology of torus actions on symplectic manifolds. Translated from the French by the author. Progress in Mathematics, 93. Birkhäuser Verlag, Basel, 1991. 181 pp. ISBN: 3-7643-2602-6, then developed for the general case in 
MR1299003 Cox, David A.: The homogeneous coordinate ring of a toric variety. J. Algebraic Geom. 4 (1995), no. 1, 17–50)
Aside 2:  the interpretation of sections of a line bundle over a complex manifold as homogeneous functions has nothing to do with projectivity and/or Serre's theorem (as an analyst, I'll stick to the language of complex manifolds and line bundles instead of schemes and sheaves). It is more general and follows  from the relation between the $k$-th tensor power of the line bundle $\pi: L \to X$ over an $n$-dimensional complex manifold $X$ and the dual bundle $\pi': L' \to X$. Namely (as can be checked using respective definitions, taking into account relations between trivializations), every holomorphic function $f$ on $L'$ which is homogeneous of degree $k$ on the fibers is a section in $L^k$ (not just a ''stand-in").  In the specific case of $X=\mathbb{P}^n$ and $L=\mathcal{O}(1)$, the space of sections $\Gamma(\mathbb{P}^n, \mathcal{O}(k))$ is isomorphic to the space of homogeneous polynomials of degree $k$ in the variables $z_0,...,z_n$.<|endoftext|>
TITLE: Best exposition of the Proof of the Hilbert Syzygy Theorem by Eilenberg-Cartan
QUESTION [6 upvotes]: Where can I find a comprehensive treatment of this important result at the level of a very advanced undergraduate/beginning graduate student? What works develop the relevant material in a cohesive and readable way? Assume only a familiarity with basic homological algebra and ring theory on the part of the reader in assessing this question, please. Thank you!

REPLY [7 votes]: The place I learned it from is Chapter 19 of Eisenbud's Commutative Algebra. Most of the proofs in that section do not use material from previous chapters if I recall correctly. The route (which I think is what you are looking for) is to construct the Koszul complex of the residue field of a regular (graded) local ring and also prove the symmetry of the Tor functor, and then use these two facts to get finite global dimension which implies Hilbert's syzygy theorem.<|endoftext|>
TITLE: Shuffling decks of cards where not all cards are distinguishable
QUESTION [6 upvotes]: Suppose a deck of cards consists of $a_1+a_2+\cdots+a_k$ cards of $k$ types, where there are $a_i$ indistinguishable cards of each type. How many shuffles does it take, on average, to randomize the deck?  For example, $a_i=4$ and $k=13$ gives a standard deck of playing cards; $a_i=4$ for $1\le i\le9$, $i_{10}=16$, $k=10$ gives the cards for blackjack.
In general I would expect that this would be an easier task than shuffling a deck where all cards are indistinguishable.  A standard deck has about 226 bits of entropy, while the same deck without suits has only 166 bits of entropy.

Consider a standard deck of 52 playing cards.  Bayer & Diaconis (famously) showed that, under a certain model, it takes 7 riffle shuffles to sufficiently randomize the deck. Salem applies a different methodology and gets 6 idealized riffle shuffles. Mann uses a much stricter measurement and determines 11.7 as the expected stopping time for the riffle unshuffle (and thus the riffle shuffle) to randomize the deck.
In particular, Mann gives a formula:
$$E(\mathbf{T})=\sum_{k=0}^\infty\left[1-{2^k\choose n}\frac{n!}{2^{nk}}\right]$$
which lends itself to generalization nicely.
I'm partial to Mann's method, but my question applies broadly.

[1] David Aldous and Persi Diaconis, "Shuffling cards and stopping times", The American Mathematical Monthly 93:5 (1986), pp. 333-348.
[2] Dave Bayer and Persi Diaconis, "Trailing the dovetail shuffle to its lair", The Annals of Applied Probability 2:2 (1992), pp. 294-313. JSTOR: 2959752
[3] Brad Mann, How many times should you shuffle a deck of cards?
[4] Michael P. Salem, How many shuffles are necessary to randomize a standard deck of cards?

REPLY [7 votes]: A place to start is the recent preprint:
Riffle shuffles of a deck with repeated cards
Sami Assaf, Persi Diaconis, K. Soundararajan.
They also have another preprint about reducing this to a "rule of thumb", and cite some earlier work of Conger and Viswanath, available here.
Pulling carelessly from tables in the different papers without checking exactly what I'm saying, it seems that in the total variation distance, (the measure which yields seven shuffles for a standard deck), four shuffles are already sufficient for blackjack, while for the separation distance (the measure yielding eleven shuffles for a standard deck), it still takes nine shuffles.<|endoftext|>
TITLE: Does the p-adic Tate module of an elliptic curve with ordinary reduction decompose?
QUESTION [13 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}_p$ and $E$ an elliptic curve over $K$ with good ordinary reduction.
The p-adic Tate module $T_p(E)$ is (after tensoring with $\mathbb{Q}_p$) a 2-dimensional $\mathbb{Q}_p$-representation of $\mathop{\mathrm{Gal}}(\bar{K}/K)$.
It is reducible: the kernel of reduction to the residue field is an invariant line.
Does $T_p(E) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ contain another invariant line?

REPLY [12 votes]: If you go to the maximal unramified extension of $K$ (so the residue field is algebraically closed) then you can write $T_p(E)$ as an extension of $\mathbb{Z}_p$ by $T_p(\mu)$. The class of this extension is the Serre-Tate parameter and can be viewed as a one-unit in the ground field. The Serre-Tate parameter parametrizes the set of elliptic curves with a fixed ordinary reduction.
To answer your question, you get another invariant line if and only if the Serre-Tate parameter is a root of unity (since the extension of groups splits up to isogeny). As in David's answer, this only happens if the elliptic curve has CM. The curve with Serre-Tate parameter equal to one is the canonical lift of the reduction ($T_p(E)$ is a direct sum) and the other CM curves are called quasi-canonical lifts.<|endoftext|>
TITLE: Galois theory timeline
QUESTION [27 upvotes]: A recent question on the history of Galois theory wasn't the most satisfactory. But the historical issues do seem quite attractive. They relate to innovation, and to exposition. There is a perspective (which based on past teaching I don't entirely share) that the periods worth considering are pre-Artin, classic Artin treatment, and post-Artin. To make the point explicitly, that is to do with the influence of Artin's Galois Theory Notre Dame notes, copyright dates 1940 and 1942. 
My issues with this periodisation are primarily to do with a wish to have a proper view of innovation, starting with Galois (admitting pre-history evident in Gauss and Abel, solution of the quartic, group theory and other contributions in Lagrange). There is something like this:
*Galois
*Liouville writes up the theory
*French school of group theory and treatment by Camille Jordan
*Riemann surface theory in general, and isogenies of elliptic curves in particular, develop in parallel
*Presumably Hurwitz knew how to connect the dots
*Algebraic number theory uses abelian extensions and Kummer theory extensively
*Hilbert lays conjectural foundations for class field theory, post-Kronecker Jugendtraum and complex multiplication theory, using a version of Galois theory that seems to be much influenced by Hurwitz/Riemann surfaces
*Steinitz, abstract theory of fields, idea of separable extensions clarified
*New expositions from Emmy Noether and Artin in the 1920s (are these documented, though?), against the background of completing proofs of class field theory, and Artin L-functions
*The Inverse Problem for Galois groups is stated and leads to work on invariant theory
*1930s: Galois theory for infinite extensions is enunciated
*C.1940: Tensor products of fields.
This takes us just about to 1940. I think it is a trap to assume Artin in 1940 was lecturing on Galois theory in the precise terms he would have used in the 1920s. 
I'd be grateful for help making this tentative timeline more solid. Further interesting things did happen after 1942, but that seems enough for one question.
[Edit:The older question was What was Galois theory like before Emil Artin? - treat my remarks there as tentative.]
Edit: Dedekind's contribution should have been on the list. See hss.cmu.edu/philosophy/techreports/184_Dean.pdf about what Dedekind did in his Vorlesungen. That article credits Artin with the formulation of the Fundamental Theorem in abstract terms, while crediting Dedekind with the theory for subfields of the complex numbers. In that context it becomes clearer, I think, why "innovation" goes to Dedekind on the lattice-theoretic way of thinking about Galois theory, while Artin might reasonably have thought he was doing exposition (bringing those ideas explicitly into the post-Steinitz era).

REPLY [2 votes]: There is a very interesting paper 
van der Waerden, B.L.
Die Galois-Theorie von Heinrich Weber bis Emil Artin. 
Arch. Hist. Exact Sci. 9, 240-248 (1972); DOI: 10.1007/BF00327305, jstor.
In this paper van der Waerden criticizes a paper the below paper
Kiernan, Melvin
The development of Galois theory from Lagrange to Artin. (English) Zbl 0231.01003
Arch. Hist. Exact Sci. 8, 40-154 (1971); doi: 10.1007/BF00327219, jstor.
He says the Kiernan's treatment from Lagrange to Webber is quite good but from Weber to Artin is not adequate (to him).
There is an entire section on Hilbert's contribution (and his school) on Galois theory. In particular he writes in Zahlberichtes Hilbert proves the normal basis theorem for abelian extensions and Noether (1931) and Deuring (1932) treated the general case of  the normal bassis theorem.<|endoftext|>
TITLE: deformation to the normal cone
QUESTION [12 upvotes]: What is the idea behind deformation to the normal cone and what are examples of its applications?

REPLY [20 votes]: Here's a place to see the normal cone side-by-side with other familiar constructions, that I learned from Fulton's "Intersection Theory". Here $X \subset Y$.
Start with the space $Y \times {\mathbb P}^1$, thought of as a trivial family over ${\mathbb P}^1$. Blow up the subscheme $X \times \infty$. Now we still have a flat family over ${\mathbb P}^1$, in which all the fibers except the one over $\infty$ are still copies of $Y$. The fiber over $\infty$ is reducible: one piece $Z_1$ is the blowup of $Y$ along $X$, and the other is the projective completion $Z_2$ of the normal cone to $X$ inside $Y$. These intersect along the projectivization of the normal cone, which appears in $Z_1$ as the exceptional divisor, and in $Z_2$ as the stuff added in projective completion.
(An example, for people who like polytopal pictures of toric varieties: let $Y$ be ${\mathbb P}^2$, pictured as a triangle, and $X$ be a point, pictured as a vertex. Then $Y \times {\mathbb P}^1$ is pictured as a triangular prism, and its blowup by cutting a corner off that prism. The fiber over $\infty$ is then pictured as cutting that triangle into a trapezoid $Z_1$ union a small triangle $Z_2$, glued along an interval, matching the decomposition above.)
So, if perhaps you don't like degenerating $Y$ (which may be complete) to something noncomplete, you can complete it by including $Z_1$. The normal cone is $Z_2 \setminus Z_1$.
Two other comments. On $Y \times {\mathbb P}^1$ there is a circle action, dilating the ${\mathbb P}^1$. It acts trivially on the $0$ and $\infty$ fibers, moving the rest around. When we blow up $X \times \infty$, the circle action on the new $\infty$ fiber $Z_1 \cup Z_2$ is nontrivial on $Z_2$; it is the dilation of the cone fibers. This is one place to lay blame for the existence of this "cone" structure.
Finally, there's a conformally equivalent way to think about the $Y$ to $Z_1 \cup Z_2$ degeneration, at least when $X$ and $Y$ are smooth. Pick a small tubular neighborhood (nonalgebraic!) around $X$, with boundary $S$, a sphere bundle over $X$. Let the metric on $Y$ get very looong nearby $S$, in directions passing through $S$. You might say that $X$ is falling into a black hole, with $S$ the Schwarzschild boundary. In the limit, it gets infinitely long, and $X$ has bubbled off into its own universe.<|endoftext|>
TITLE: Degree of sum of algebraic numbers
QUESTION [58 upvotes]: This is an elementary question (coming from an undergraduate student) about algebraic numbers, to which I don't have a complete answer.
Let $a$ and $b$ be algebraic numbers, with respective degrees $m$ and $n$. Suppose $m$ and $n$ are coprime. Does the degree of $a+b$ always equal $mn$?
I know that the answer is "yes" in the following particular cases (I can provide details if needed) :
1) The maximum of $m$ and $n$ is a prime number.
2) $(m,n)=(3,4)$.
3) At least one of the fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ is a Galois extension of $\mathbf{Q}$.
4) There exists a prime $p$ which is inert in both fields $\mathbf{Q}(a)$ and $\mathbf{Q}(b)$ (if $a$ and $b$ are algebraic integers, this amounts to say that the minimal polynomials of $a$ and $b$ are still irreducible when reduced modulo $p$).
I can also give the following reformulation of the problem : let $P$ and $Q$ be the respective minimal polynomials of $a$ and $b$, and consider the resultant polynomial $R(X) = \operatorname{Res}_Y (P(Y),Q(X-Y))$, which has degree $mn$. Is it true that $R$ has distinct roots? If so, it should be possible to prove this by reducing modulo some prime, but which one?
Despite the partial results, I am at a loss about the general case and would greatly appreciate any help!
[EDIT : The question is now completely answered (see below, thanks to Keith Conrad for providing the reference). Note that in Isaacs' article there are in fact two proofs of the result, one of which is only sketched but uses group representation theory.]

REPLY [8 votes]: For those who want to read Isaac's proof --- mentioned in Pete's answer --- in the language of Molière and Bourbaki, there is François Brunault's exposé.
Addendum (30/01/2011).  Today I came across the following related article by Weintraub in the Monatshefte.<|endoftext|>
TITLE: Enumerating representations of an integer as a sum of squares
QUESTION [6 upvotes]: Let $p$ be an odd prime.  An old theorem of Jacobi asserts that $p$ has exactly $8(p+1)$ representations as a sum of four squares of integers (solutions counted with order and sign).  What is the most effective way to enumerate these solutions computationally?  Can it be done in time $p^{1+\varepsilon}$, or even in time $p (\log{p})^A$?

REPLY [8 votes]: Form the set $S$ of all squares less than $p$. This has $O(\sqrt{p})$ elements, and writing them down takes $O(\sqrt{p} \log p)$ time. (You don't have to implement fast multiplication to do this; just compute the list of squares by successively adding odd numbers.)
Let $T$ be the set of all integers expressible as the sum of two elements of $S$. This has $O(p)$ elements, and takes $O(p \log p)$ steps to write down.
Sort $T$ and sort $p-T$. This is $O(p \log p)$ steps each. Find all duplicates between the lists $T$ and $p-T$; this takes $O(p)$ steps because they are already sorted. 
All in all, $O(p \log p)$ steps, the same size as the output.<|endoftext|>
TITLE: How thick is the reciprocal of the squares
QUESTION [26 upvotes]: There is a unique nonempty set $B$ of nonnegative integers such that
every positive integer can be written in the form
$$b + s^2, b\in B, s\ge0$$
in an even number of ways.
$B = \{0, 1, 2, 3, 5, 7, 8, 9, 13, 17, 18, 23, 27, 29, 31, 32, 35, 37,$
$ 39, 41, 45, 47, 49, 50, 53, 55, 59, 61, 63, 71, 72,$
$ 73, 79, 81, 83, 87, 89, 91, 97, 98, 101, 103, 107,$
$ 109, 113, 115, 117, 121, 127, 128, 137, 139, 149,$
$ 151, 153, 157, 159, 162, 167, 171, 173, 181, 183,$
$ 191, 193, 197,\dots\}$
Does the set $B$ have positive density?
Now for some context. Every set $A$ of nonnegative integers that contains 0 has a unique set $B$ of nonnegative integers so that
$$\left( \sum_{a\in A} q^a \right) \, \left( \sum_{b\in B} q^b \right) = 1$$
in the ring ${\mathbb F}_2[[q]]$ of binary power series. We call $B$ the reciprocal of $A$.
As a consequence of a Euler's pentagonal number theorem, the reciprocal of the set $\{n(3n+1)/2 \colon n \in \mathbb{Z}\}$ is the set $\{ n \colon p(n)\equiv 1 \bmod 2\}$, where $p(n)$ is ordinary partition function. Almost nothing interesting is known about the parity of the partition function, but computationally it seems to be even and odd with equal frequency. This question arises out of an effort to put the parity of the partition function into some context.
In this article (arxiv, Int. J. Number Theory 2 (2006), no. 4, 499--522), Josh Cooper, Dennis Eichhorn and I investigated the properties of $A$ that lead to $B$ having positive density, and all of our data and partial results can be summed up in the following conjecture:

Conjecture: If $A$ contains 0, 
  is not periodic, and is uniformly 
  distributed in every congruence class
  modulo every power of 2, then $B$ has
  positive density.

Letting $A$ be the set of squares, we were able to prove that the even numbers in $B$ are exactly $\{2k^2 \colon k\ge 0\}$, and we were able to classify the $1\mod 4$ elements of $B$. 
Update Greg Kuperberg's answer concerning the conjecture displayed above is, while not quite a disproof, utterly convincing. So convincing, I can no longer understand how I thought the conjecture could plausibly be true. In our paper, we described it as "the strongest conjecture that is consistent with our theorems, our experiments, and Conjecture 1.1", so I see we weren't too enthusiastic about its truth. We should have been even less so!
The question directly asked, the density of the reciprocal of the squares, remains unanswered. Paul Monsky has introduced a new (to me, at least) approach, and has made striking progress both in the answer below and in his answer to this question. 
I love Greg's answer to the question I didn't dare ask, and want to accept it, but Paul's is more directly relevant to the question I did ask.
Here are some computational counts of the number of elements of $B\cap[0,2^{23}]$ in particular congruence classes.
(1 mod 4, 371867), (3 mod 4, 760697)
(1 mod 8, 185336), (5 mod 8, 186531), (3 mod 8, 294045), (7 mod 8, 466652)
(1 mod 16, 92703), (5 mod 16, 93236), (9 mod 16, 92633), (13 mod 16, 93295),
(3 mod 16, 147232), (11 mod 16, 146813),
(7 mod 16, 204808), (15 mod 16, 261844)
(7 mod 32, 102487), (23 mod 32, 102321), 
(15 mod 32, 130895), (31 mod 32, 130949)

Since there was a specific request for 15 mod 32 data, here are the first 10 such numbers in $B$: (47,79,271,559,623,687,719,815,879,911). Here are the last 10 that I've computed: (8388539, 8388551, 8388559, 8388563, 8388567, 8388571, 8388581,
8388591, 8388593, 8388603, 8388607)

REPLY [3 votes]: O'Bryant et. al. showed that the elements of B in 12 of the congruence classes mod 16
form a set of density 0. In my answer to this question and 2 other questions I used
a result of Gauss to show that the k in B that are 3, 7 or 11 mod 16 also form sets of
density 0. So only the case of k congruent to 15 mod 16 remains.
NEVERTHELESS, I believe that the above results are somewhat accidental, and that a k
congruent to 15 (16) is "just as likely to be in B as out of it" so that the density of
B is 1/32. Here are the back-up data, provided by Kevin, for k<2^23.
k=15  (64):       2*(65536) elements,     65446 in B
k=31  (64):       2*(65536) elements,      65247 in B
k=47  (64):       2*(65536) elements,      65449 in B
k=63  (64):       2*(65536) elements,      65702 in B
Readers are invited to make further calculations, but the above are highly suggestive.<|endoftext|>
TITLE: Looking for an interesting problem/riddle involving triple integrals.
QUESTION [17 upvotes]: Does anyone know some good problem in real analysis, the solution of which involves triple integrals, and which is suitable for second semester Analysis students?
Thanks!

REPLY [2 votes]: This riddle is better phrased in term of double integral but can work for cubes instead of rectangles.
Here is the riddle: Call a rectangle integral if the length of (at least) one of its edges is an integer.
Prove that a rectangle that can be tessellate by integral  rectangles if also integral rectangle.
It is a huge hint to say that it has something to do with double integral so you may want to avoid it in the beginning.<|endoftext|>
TITLE: Gluing perverse sheaves?
QUESTION [8 upvotes]: It might be a stupid question. 
How to glue perverse sheaves? I am considering the following example. Flag variety of $sl_2$, which is $P^1$. Consider category of perverse sheaves on $P^1$. Denoted by $Perv(P^1)$. The big cells of flag variety of $sl_2$ give the affine cover for it. In this case, they should be two $A^1$. We can also consider category of perverse sheaves on $A^1$. My question is:

How to glue two pieces of perverse sheaves on 1-dimensional affine spaces to that of projective spaces? More general, is there any gluing machinery which can globalize the perverse sheaves? It seems that Beilinson had a paper talking about this, but what I preferred is some expository notes explaining with some examples.
Maybe I need to ask this in another question. How can one define perverse sheaves on noncommutative space. I am aware that there is a paper by Amnon Yekutieli, James J. Zhang talking about perverse sheaves on noncommutative space. However, what they considered was not really a noncommutative space from my understanding.(If I made mistake or bullshit, point out please). They consider quasi coherent sheaves of (not necessarily)commutative algebra on commutative scheme. Which does not fit my need. I am considering the following example: Quantized flag variety of $sl_2$ i.e. $Proj(O_q(G/N))$ in the sense of Lunts-Rosenberg(see also Erik Backelin and Kobi Kremnitzer and Tanisaki). It is a noncommutative scheme. I wonder whether James Zhang has also defined dualizing complexes for this case. How to define category of perverse sheaves on quantized flag variety ? 
The motivation for this question is I think there should be quantum version of Riemann-Hilbert correspondence. Which should describe the categorical equivalence:

$Perv(Proj(O_q(G/N))$ and category of quantum holonomic D-modules on quantized flag variety.
At present, I have more interest to know the answer of question 1. Thank you!

REPLY [6 votes]: As others have said, to glue perverse sheaves given on two open copies of $\mathbb{A}^1$ into a sheaf on $\mathbb{P}^1$, it is necessary (and sufficient) simply to give an isomorphism of these sheaves on the intersection.  This is the same procedure as for gluing anything usually called a "sheaf".  Beilinson's paper concerns a different kind of "gluing" where the covering is not open, but in the form of a two-piece stratification.<|endoftext|>
TITLE: Explicit ordering on set with larger cardinality than R
QUESTION [17 upvotes]: Is it possible to construct (without using Axoim of Choice) a totally ordered set S with cardinality larger than $\mathbb{R}$?
Motivation: A total ordering is often called a “linear ordering”. I have heard the following explanation: “If you have a total ordering on a set S, you can plot the set on the real line such that elements to the right are greater than elements to the left”. Formally this means that there exist a function $\phi:S\rightarrow \mathbb{R}$ such that for all $ a$ ,$b\in S$, x < y $ \Leftrightarrow \phi(x)$ < $\phi(y)$. This is of course correct if the set is finite or countable (and it gives a good intuition on what a total ordering is), but obviously not if $|S|>|\mathbb{R}|$, and using the axiom of choice it is easy to “construct” a total ordering on, say, the power set of $\mathbb{R}$. But I would prefer to have a more concrete counterexample, and this is why I asked myself this question.
Later I realized that is was possible to construct a total ordering on a set $|S|=|\mathbb{R}|$, such that no such function $\phi$ exist, but I still think that the above question is interesting.

REPLY [16 votes]: This is a response to Joel's comment about whether $2^{\mathbb R}$ can be linearly ordered without choice. In general, no. There is a concrete obstacle, actually: Vitali's equivalence relation. Recall that this relation is defined by $x\sim y$ iff $x-y\in{\mathbb Q}$. Now consider ${\mathbb R}/\sim$, the collection of equivalence classes. This is a concrete subset of $2^{\mathbb R}$ that in general cannot be linearly ordered without some appeal to choice. 
For example, under determinacy, this set is not linearly orderable, so in $L({\mathbb R})$ there is no linear ordering of it in the presence of large cardinals. In short, under reasonable assumptions, there is no way of linearly order this set without appealing to choice.
Things get interesting. For example, in $L({\mathbb  R})$ (the smallest model of ZF that contains all the reals), in the presence of large cardinals, a set is linearly orderable iff ${\mathbb R}/\sim$ does not inject into it, and a set is well-orderable iff ${\mathbb R}$ does not inject into it.

Here are some details, it is not a complete argument, it requires knowing some descriptive set theory (and there may be some typos), but the sketch should give a decent idea. 
I'll actually work with $2^\omega/E_0$ (which is another manifestation of Vitali's relation). I learned this from Benjamin Miller, by the way, and it immediately became key for some results Richard Ketchersid and I have been working on. The result itself, that under AD this quotient does not admit a linear ordering, has been known to descriptive set theorists for ages, I am not sure who first noticed it.
Recall that $x\mathrel{E_0}y$, for $x,y\in2^\omega$, iff there is some $n$ such that for all $m\ge n$ we have $x(m)=y(m)$. It suffices to assume that all sets of reals have the Baire property.
Suppose $R$ is a linear ordering of $2^\omega/E_0$. Then the pullback $\hat R$ of $R$ is a quasi-ordering of $2^\omega$. Begin by noticing that $\hat R$ is not meager. Otherwise, $2^\omega$ itself would be meager, being the union of $\hat R$ and $\hat R^{-1}$ (its ``flip'').
Note that the set 
{$x \mid${ $y \mid x \mathrel{\hat R} y$} is non-meager } 
is itself non-meager, by the Kuratowski-Ulam theorem, so we can fix some $s \in 2^{<\omega}$ such that 
{$ x \mid${ $y \mid x \mathrel{\hat R} y $} is co-meager in $N_s$}
is non-meager, where $N_s$ is the basic neighborhood consisting of sequences in $2^\omega$ that begin with $s$.
The key point is that if a set has the Baire property and $E_0$ restricted to that set is smooth (i.e.,there is a Borel reduction to the identity on that set), then the set is actually meager (this follows from the Glimm-Effros dichotomy of Harrington-Kechris-Louveau).
Note that $E_0$ is smooth on the set 
{$x\mid$ there are $y$, $z$ such that $x \mathrel{E_0} y \mathrel{E_0} z$, and exactly one of {$y'\mid y' \mathrel{\hat R} y$}, {$z'\mid z \mathrel{\hat R} z'$ } is co-meager in $N_s$}.
This is not hard, but needs a tiny bit of thought. The point is that any $E_0$-class admits a natural ${\mathbb Z}$-ordering, and on the set above we can pick representatives from each class, since we actually have a way of ``assigning an origin'' to this ordering.
It follows that that the set
{ $x \mid$ for all $x' E_0 x$ the set { $y \mid x' R y$ } is co-meager in $N_s$ }
is non-meager. 
Now: This set is $E_0$-invariant, and therefore it must actually be co-meager.
But then $\hat R$ itself is co-meager in $N_s \times N_s$. Now let $E$ be the equivalence relation $\hat R\cap \hat R^{-1}$. Then $E$ is also co-meager in $N_s \times N_s$. But then it admits an equivalence class which is co-meager in $N_s$. Since $E$ actually contains $E_0$, we then have that it is co-meager in all of $2^\omega$.
But then $R$ cannot be a linear order, as it cannot distinguish between co-meager many $E_0$-classes.
[As a final remark: One can of course organize the whole thing using Lebesgue measurability rather than the property of Baire, and Fubini's theorem rather than Kuratowski-Ulam. But the argument using the Baire property shows that this is equiconsistent with ZF (by Shelah), while using measurability would in consistency require an inaccessible.]<|endoftext|>
TITLE: Hecke operators acting as correspondences?
QUESTION [16 upvotes]: This question is inspired by Relation between Hecke Operator and Hecke Algebra
I remember having heard of yet another way of looking at Hecke operators acting on the spaces of modular forms for classical congruence subgroups of $SL_2(\mathbf{Z})$, and I would like to ask if anyone knows a reference for that. Here are some details and a related question.
There is a universal elliptic curve $U$ over $\mathbf{H}$, the upper half-plane. It is an analytic manifold obtained by taking the quotient of $\mathbf{C}\times\mathbf{H}$ by the action of $\mathbf{Z}^2$ given by $(n,m)\cdot (z,\tau)=(z+n+m\tau,\tau)$ where $n,m\in\mathbf{Z},z\in \mathbf{C}$ and $\tau\in\mathbf{H}$.
A torsion free finite index subgroup $\Gamma$ of $SL_2(\mathbf{Z})$ (congruence or not) acts on $U$. Here is the first question: for which $\Gamma$ is the quotient $U/\Gamma$ algebraic?
In any case $U/\Gamma$ is algebraic if $\Gamma=\Gamma(N)=\ker (SL_2(\mathbf{Z})\to SL_2(\mathbf{Z}/N))$ and $N\geq 3$. Denote the corresponding quotient $U/\Gamma(N)$ by $U(N)$. This is the universal elliptic curve with a level $N$ structure. (The notation $U(N)$ may be non-standard, in which case please let me know.)
There is a natural map $U(N)\to Y(N)=\mathbf{H}/\Gamma(N)$. Now take the $n$-th fibered cartesian power $U^n(N)$ of $U(N)$ over $Y(N)$ and let $p:U^n(N)\to Y(N)$ be the projection.
All derived direct images of the constant sheaf under $p$ decompose as direct sums of the Hodge local systems, which correspond to the symmetric powers of the standard representation of $\Gamma(N)$. Using the Eichler-Shimura isomorphism one can construct classes in $H^1(Y(N),V_k)$ from modular forms of weight $k+2$ for $\Gamma(N)$ where $V_k$ is the local system on $Y(N)$ that comes the $k$-th symmetric power of the standard representation. (In fact these classes can be interpreted in terms of the Hodge theory, see Eichler-Shimura isomorphism and mixed Hodge theory) So modular forms give elements in the $E_2$ sheet of the Leray spectral sequence for $p$.
I would like to ask: can one interpret the Hecke operators acting on modular forms as correspondences acting on $U^n(N)$ over $Y(N)$ (and hence, on the Leray spectral sequence for $p$)?

REPLY [11 votes]: The answer to your last question is yes, modulo my own misunderstandings. In Scholl's "Motives for modular forms", §4 (DigiZeitschreiften), the Hecke operators are defined in this way, and I think the equivalence of these definitions is implied by the diagram (3.16) and proposition 3.18 of Deligne's "Formes modulaires et representations l-adiques" (NUMDAM).
Let p be a prime not dividing N, and let Y(N,p) be the moduli scheme parametrizing elliptic curves with full level N structure and a choice of a cyclic subgroup C of order p. There are two natural "forgetful" maps q1 and q2 from Y(N,p) to Y(N) -- the former forgets the cyclic subgroup, the latter takes the induced level N structure on the quotient by C. 
From these two maps, one gets two different families of elliptic curves over Y(N,p). Explicitly, the pullback of Un(N) along q1 is isomorphic to the n:th fibered power of the universal elliptic curve over Y(N,p); let us denote it Un(N,p). Let Q(N,p) be the quotient of U(N,p) by the cyclic subgroup C, and take its n:th fibered power as well. Then similarly Qn(N,p) is the pullback of Un(N) along q2. Finally, the quotient map gives us $\phi : U^n(N,p) \to Q^n(N,p)$. 
But this data gives us the right correspondence on Un(N) over Y(N), namely, one takes the composite $q_{1\ast}  \phi^\ast q_2^\ast$ (where I use q1 and q2 also for the induced maps on fibered powers of universal curves).<|endoftext|>
TITLE: How do you find out the latest news in fields other than your own?
QUESTION [37 upvotes]: Most of us are at least somewhat curious about what's going on in areas of mathematics outside our own area of research.  If a significant breakthrough is made, and can be stated in language that we understand, then we would enjoy hearing about it.  Now, for truly fantastic breakthroughs like Fermat's Last Theorem or the Poincaré conjecture, the mathematical grapevine functions quite well and we all hear about it pretty quickly.  For anything below that level, however, I for one feel that there is no good way to find out what is going on in fields other than my own.  Every four years I seem to have the following conversation: "Did you hear?  So-and-so won the Fields Medal!"  "No.  And, uh, who in the world is so-and-so?"  Shameful.
What I'm wondering is, am I doing something wrong?  Does anyone else have a better way of keeping abreast, even at a superficial level, of what major advances are happening in other areas of mathematics?  At one time, Mathematical Reviews would select some articles or books for "Featured Reviews."  I really enjoyed Featured Reviews and learned about many interesting results this way.  However, MR stopped doing Featured Reviews after a while.  No official reason was given, but I have heard that one reason was that some people were treating Featured Reviews as judgments as to which papers/books were "the best," and that MR did not want to accept the responsibility for making such judgments, especially if they were going to be used for tenure and promotion decisions.  This is understandable, but unfortunate (from my point of view).
The series of books What's Happening in the Mathematical Sciences is also excellent.  However, producing an expository article of that quality takes a lot of time, and so What's Happening is necessarily limited in scope.  I would like to know what's happening in between issues of What's Happening.  If there are resources out there that others have found useful for keeping abreast of mathematical research news, I would like to hear about them.

REPLY [3 votes]: Here https://theorems.home.blog/theorems-list/ is the website you are asking for.
It covers all recently solved mathematical problems, which are important (for example, published in a top journal) but at the same time can be understood with not too much background.<|endoftext|>
TITLE: How to count symmetry factors of Feynman diagrams? 
QUESTION [21 upvotes]: I have enough fears that this question might get struck down. Still let me try. 
I shall restrict myself to $\frac{\lambda \phi^4}{4!}$ perturbed real scalar quantum field theory and call as "symmetry factor" of a Feynman diagram to be the eventual number by which the power of $\lambda$ is divided in the final integral representation of the diagram. 
In that way the symmetry factor of the figure of eight vacuum bubble is 8 and of the "tadpole diagram" it is 2. 
One way to get this factor right is to count the number of ways the free arms in the "pre-diagram" can be contracted. But this is more like a cook-book rule than an understanding of how the factor comes. 
I believe the most conceptually correct way is to count for every diagram the number of  terms in the representation as a functional derivative of the path-integral which give that diagram. 
In that picture one has to argue that there were precisely $4!$ terms produced by the functional derivative which produced that figure of eight vacuum bubble. Which I can argue.
But for the tadpole diagram and the product of the vacuum bubble with the free-propagator, I can't find an argument. Like one has to be able to show that in the functional derivative picture there are $4!\times 3! \times (2!)^2$ terms corresponding to the tadpole diagram. 
Any help regarding how this counting is done or any general framework which helps compute these symmetry factors correctly?

REPLY [6 votes]: I think this question is mostly about mentally internalizing the interpretation of generating functions and the orbit-stabilizer theorem.  That is, if you can understand how the rules arise from a method that seems more natural to you, then the enumeration of automorphisms should appear less like arbitrary cookbook rules.
There is a good exposition of the symmetry factors in section 5 (page 25) of Alex Barnard's notes from Richard Borcherds's 2001 QFT class.  In order to distill the combinatorics of the diagrams, it gives a treatment of $\phi^4$ theory in zero spacetime dimensions, where the path integral is just an ordinary one dimensional integral: $$Z = \int_\mathbb{R} e^{-\phi^2/2 - \lambda \phi^4/4!}d\phi.$$
You can expand this integral as a series in polynomials times Gaussians, and observe that the $\frac{(2n)!}{n!2^n}$ in the identity $\int_\mathbb{R} \phi^{2n} e^{-\phi^2/2} d\phi = \frac{(2n)!}{n!2^n}\int_\mathbb{R} e^{-\phi^2/2}d\phi$ enumerates matchings of $2n$ labeled vertices.  If you ask for $k$ ($=\frac{n}{2}$) vertices to have valence 4, then there is a group of order $k!(4!)^k$ that permutes the vertices and their inputs, and this order divided by the order of the stabilizer of a particular isomorphism type is the order of the automorphism group.  By comparing coefficients, you find that a sum over graph isomorphism types, weighted by their automorphism groups, yields exactly the terms that come up in the path integral.
I imagine there may be a high-level explanation using species theory, but I'm not the right person to give it.<|endoftext|>
TITLE: Normal Ordering with Vertex Operators in Conformal Field Theory 
QUESTION [12 upvotes]: The "definition" of the normal ordering in CFT looks a bit vague to me. 
I found the definition in terms of exponentiated functional derivative pretty opaque.
Also in this context it might help if someone can give a reference or if there is a short explanation to understand how the Operator Product Expansion is derived using products of normal ordered operators.
I don't see the conceptual framework in which these ideas fit together. 
Some of the books I looked at gave a very disparate view as a collection of some complicated formulas. 
Let me give a precise example of the kind of calculation that I am stuck with,
Refer to these lecture notes
I can understand equation 4.26 of this but not the next 4 equations that seem to follow from it leading to 4.28.
It would be helpful if someone can decrypt the calculation. 

In light of the kinds of references that came in as responses, I think it would help if I make the problematic calculation a little more explicit.
This has to do with what are called "Vertex Operators" in CFT given as $:e^{ikX(z)}:$ where $::$ is the notation for normal ordering and $k$ is some scalar and $X$ is a conformally invariant free Bosonic field. Then I would like to understand the derivation of this equality,
(all expressions are understood to be valid under the Feynman Path Integral)
$:\partial X(z)\partial X(z)::e^{ikX(w)}:  = -\frac{k^2\alpha ^2}{4}\frac{:e^{ikX(w)}:}{(z-w)^2}-ik\alpha\frac{:\partial X(z) e^{ikX(w)}:}{(z-w)}$
where we have $X(z)X(w) = -\frac{\alpha}{2}ln \vert z - w \vert$
and what would be the similar simplification of 
$:e^{ikX(z)}::e^{ikX(w)}: = ?$

Some more elaboration on what about normal ordering I am concerned about. 
The problem is that I can't these books give an honest definition of what it means to "normal order" operators in CFT. Like there is a very clean definition in rest of QFT whose relation to time-ordering is given by the Wick's Theorem. Here in CFT one is supposed to understand that while normal ordering a string of operators inserted at different points on the space-time one is subtracting away from the product every possible way in which one or more pairs of insertion points can coincide and produce a singularity
Like if A,B,C,D are 4 different Bosonic operators say inserted at 4 different space-time points. Then one would define normal ordering as,
$:ABCD: = ABCD - (AB):CD: - (AC):BD: - (AD):BC:-(BC):AD:-(BD):AC:$
$$-(CD):AB:-(AB)(CD)-(AC)(BD)-(AD)(BC)$$
where () denotes the correlation function of the operators.
Now the point is whether one is supposed to take the above kind of equations as being just well-motivated definition or is there is anything more fundamental from which it is derivable? 
There is definitely an issue about defining difference of two divergent expressions here.

REPLY [3 votes]: The field $L=:\partial X\partial X:$ is Virasoro and your first formula says that a vertex operator is a primary field (highest weight vector) with respect to the Virasoro action, with appropriate weight. This is a basic computation with vertex operators done in nearly any textbook on conformal field theory (Kaku, Di Francesco, etc). You may also want to look at one of the first mathematical treatments of vertex operators, very clearly written:
Frenkel, I. B.; Kac, V. G.
Basic representations of affine Lie algebras and dual resonance models.
Invent. Math. 62 (1980/81), no. 1, 23--66
By the way, the proper way to think about OPE for bosonic fields is 
$$\partial X(z) \partial X(w) = \frac{1}{(z-w)^2}.$$
The "raw" $X(z)$ occurs only as a part of the exponential expression defining the vertex operator, never by itself.<|endoftext|>
TITLE: When are entire functions surjective?
QUESTION [10 upvotes]: Is there some useful criterion to determine whether or not an entire function is surjective?

REPLY [7 votes]: Maybe Picard's theorem is of help
http://en.wikipedia.org/wiki/Picard_theorem<|endoftext|>
TITLE: How do you know when a reflective subcategory of Top is quotient-reflective?
QUESTION [5 upvotes]: A subcategory $\mathcal{C}$ of the category $Top$ of topological spaces is a reflective subcategory if the inclusion functor $i:\mathcal{C}\hookrightarrow Top$ has a left adjoint $R:Top\rightarrow \mathcal{C}$. In other words, for every space $X$, there is a space $RX\in \mathcal{C}$ and a map $r:X\rightarrow RX$ such that for each map $f:X \rightarrow Y$ with $Y\in \mathcal{C}$, there is a unique map $\tilde{f}:RX\rightarrow Y$ such that $\tilde{f}\circ r=f$.
If the reflection map $r$ is always a quotient map, $\mathcal{C}$ is said to be a quotient-reflective subcategory. For example, it seems that the subcategories of $T_0$ spaces and $T_2$ spaces are quotient-reflective but the subcategories of completely regular spaces and compact Hausdorff spaces are reflective but not quotient-reflective. I'd like to hear of other examples of quotient-reflective subcategories as well.
There seem to be conditions fully characterizing when a subcategory is reflective. For example, some necessary conditions are mentioned in the answer to this MO question. Are there also conditions characterizing subcategories which are quotient-reflective?

REPLY [4 votes]: The following theorem (16.8 in Abstract and Concrete Categories, Adamek-Herrlich-Strecker) could be of use.
Let E and M be subclasses of epis and monoes respectively, closed under composition with isomorphisms.
If A is a full subcategory of an (E,M)-factorisable category B, then the following conditions are
equivalent:
(1) A is E-reflective in B.
(2) A is closed under the formation of M-sources in B.
In the case that B has products and is E-co-wellpowered, the above conditions are equivalent to:
(3) A is closed under the formation of products and M-subobjects in B.
In the case of TOP, E would be extremal epis, and M monoes. In topological constructs extremal epis are exactly quotients (final and surjective). TOP is extremal epi-mono factorisable, has products and is extremally co-wellpowered. So a full subcategory is quotient reflective in TOP if and only if it is closed under products and (not necessarily initial) subobjects.
The subcategories of completely regular spaces is for example not closed for "adding more opens", so it is, as you said, not quotient reflective.<|endoftext|>
TITLE: Real analysis has no applications?
QUESTION [13 upvotes]: I'm teaching an undergrad course in real analysis this Fall and we are using the text "Real Mathematical Analysis" by Charles Pugh. On the back it states that real analysis involves no "applications to other fields of science. None. It is pure mathematics." This seems like a false statement. My first thought was of probability theory. And isn't PDE's sometimes considered applied math? I was wondering what others thought about this statement.

REPLY [4 votes]: I have a good undergraduate analysis book, "Real Analysis with Real Applications," by Kenneth R. Davidson and Allan P. Donsig.  The book is divided into two parts.  Part A deals with "Abstract Analysis" which includes theory, proofs, examples, and problems found in most undergraduate analysis books.  Part B deals with "Applications" which include polynomial approximations, discrete dynamical systems, diff. eqns, fourier series and physics, wavelets, and optimization.<|endoftext|>
TITLE: How to prove that a projective variety is a finite CW complex?
QUESTION [13 upvotes]: Let $X$ be a (singular) projective variety, in other words something given by a collection of polynomial equations in $\mathbb CP^n$ or $\mathbb RP^n$. How can one prove it is a finite $CW$ complex? 
Similar question: Suppose that $X$ affine (i.e. given by polynomial equations in $\mathbb C^n$, or $\mathbb R^n$). How can one prove its one point compactification is a finite $CW$ complex? 
These questions are sequel to the discussions here:
For which classes of topological spaces Euler characteristics is defined?

REPLY [5 votes]: The following thesis seems to be more general and worth referencing at this post:
Triangulation of Locally Semi-Algebraic Spaces. by K.R. Hofmann.
I quote from the abstract:

"We give necessary and sufficient conditions for a locally
  semi-algebraic space to be homeomorphic to a simplicial complex. Our
  proof does not require the space to be embedded anywhere, and it
  requires neither compactness nor projectivity of the space. A
  corollary is that every real or complex algebraic variety is
  triangulable, a result which does not seem to be available in the
  literature when the variety is neither projective nor real and
  compact."<|endoftext|>
TITLE: Making sure that you have comprehended a concept
QUESTION [29 upvotes]: I have a question that I've been thinking about for a long time.

How can you assure yourself that you've fully comprehended a concept or the true meaning of a theorem in mathematics?

I mean how can you realize that you totally get the concept and it's time to move on to the next pages of the book you're reading?

Thanks in advance for your responses.

REPLY [3 votes]: I usually consider myself understanding something, if I can implement it on a computer. I do algebraic combinatorics, so it is fairly easy to verify identities and bijections.
Basically, if I can teach my computer a concept - I must be understanding it myself.<|endoftext|>
TITLE: Can the continuum be a singular cardinal?
QUESTION [8 upvotes]: Martin's Axiom implies that $2^{\aleph_0}$ is a regular cardinal. But can $2^{\aleph_0}$ be a singular cardinal? 
By Konig's Lemma, it can never be $\aleph_{\omega}$ since cf($2^{\aleph_0}$)>$\aleph_0$ but under what conditions can it be  $\aleph_{\omega_1}$? It is even possible it can be $\aleph_{\omega_1}$?

REPLY [13 votes]: Yes, but it must have uncountable cofinality. So if it is to be singular, the smallest possibility is $\aleph_{\omega_1}$. 
The basic fact is that if $\kappa$ is any cardinal such that $\kappa^\omega=\kappa$, then there is a forcing extension $V[G]$ in which $2^\omega=\kappa$. The forcing to achieve this is $\text{Add}(\omega,\kappa)$, which consists of finite partial functions from $\kappa\times\omega$ to $2$.
In particular, if you start with GCH in the ground model, and add $\aleph_{\omega_1}$ many Cohen reals, then you will have $2^\omega=\aleph_{\omega_1}$ in the forcing extension. The proof uses the following facts: (1) adding any number of Cohen reals is c.c.c. and therefore preserves all cardinals. (2) The forcing clearly adds at least that many reals. (3) A nice name argument shows that every real in the extension has a nice name in the ground model, and there are only $\aleph_{\omega_1}$ many such names. So the continuum of the extension is exactly $\aleph_{\omega_1}$. The same ideas work for any $\kappa$ for which $\kappa^\omega=\kappa$.<|endoftext|>
TITLE: How to become a good mathematician?
QUESTION [5 upvotes]: HOW to BECOME a GOOD  THEORETICAL PHYSICIST by Gerard 't Hooft (Nobel Prize Winner)
Is there similar "How to become a good mathematician by __"?
Humble Suggestion : Why not build it here?

REPLY [2 votes]: There is: How to Solve it, by George Polya (reprinted by Princeton University Press, 2004).<|endoftext|>
TITLE: Zariski tangent spaces to representation varieties
QUESTION [11 upvotes]: In Bill Goldman's paper "The Symplectic Nature of the Fundamental Groups of Surfaces" (Advances, 54, 200-225, '84) it is stated that the "Zariski tangent space" to a representation space Hom$(\pi, G)/G$ at a representation $\rho$ is the cohomology group $H^1 (\pi; Ad\ \rho)$.  Here G is a Lie group, and Ad $\rho$ is the representation of $\pi$ on the Lie algebra of G induced by the adjoint action of G.  In the context of Goldman's paper, maybe this is just meant to refer to the case when $\pi = \pi_1 S$ with $S$ a Riemann surface.
My question is: is there some sense in which this is true for other discrete groups $\Gamma$?  In general, $Hom(\pi, G)/G$ is only a semi-algebraic set, and not a variety, so maybe the question is not really meaningful.  But I'd like to know whether these first cohomology groups behave like tangent spaces in some useful way.  
Here are two specific questions.  I'm most interested in the case when G = U(n).  
Edit:
Let me emphasize that I'm really talking about the topological quotient Hom(G, U(n))/U(n), which is a reasonably nice space.  Since U(n) is compact, general nonsense implies that this space is Hausdorff, and even better, it's a semi-algebraic set.  So in particular, it's homeomorphic to a simplicial complex.

Say $[\rho]\in$ Hom$\left(\Gamma, U(n)\right)/U(n)$ has an open neighborhood homeomorphic to $\mathbb{R}^m$ for some $m$.  Is it then true that dim $H^1 (\Gamma; Ad\ \rho) = m$?  In other words, does the cohomology group give the "topological" dimension at "smooth" points in Hom$\left(\Gamma, U(n)\right)/U(n)$?  When $H^1 (\Gamma; Ad\ \rho) = 0$, a theorem of Weil (Ann. of Math. (2)  80  1964 149--157) says that $[\rho]$ is an isolated point in Hom$\left(\Gamma, U(n)\right)/U(n)$.  This is the converse statement for m=0. 
If $[\rho]$ is not a smooth point in the above sense, then in any triangulation of Hom$\left(\Gamma, U(n)\right)/U(n)$, we see that $[\rho]$ must not lie in the interior of a maximal simplex.  If $\sigma$ is a maximal simplex of dimension m containing $[\rho]$ (in its boundary), is it true that dim $H^1 (\Gamma; Ad\ \rho) > m$?  In other words, does the dimension of the "tangent space" jump up at non-smooth points?

Any ideas, references, examples, or counterexamples would be welcomed!

REPLY [5 votes]: It is false in general. A counter-example was constructed by J.Huebschmann in section 6 of 
his paper "Singularities and Poisson geometry of certain representation spaces". However, if, say, $\rho(\Gamma)$ has trivial centralizer, then the Zariski tangent space to $Hom(\Gamma, G)//G$ (you need to use Mumford quotient for the question to make sense) at $[\rho]$ is indeed isomorphic to $H^1(\Gamma, Ad \rho)$. The reason is that the $G$-action at such $\rho$'s admits a local cross-section and then everything works, as it was explained by A.Weil in his 1964 paper "Remarks on cohomology of groups".<|endoftext|>
TITLE: Ergodic splitting in L_p
QUESTION [7 upvotes]: I have a curiosity on the Ergodic decomposition given by the von Neumann's theorem:
$$L^2(X,\Sigma,\mu)=L^2(X,\Sigma_T,\mu)\oplus\overline{\{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu)\}},$$
that occurs for a measure-preserving map $T$ of a probability space $(X,\Sigma,\mu)$, $\Sigma_T$ being the sub-σ-algebra of all $T$-invariant measurable sets, and the (orthogonal) projector being the conditional expectation $E(\cdot|\Sigma_T)$. For all $1\leq p \leq\infty$,  the conditional expectation is well-defined as a linear projector of norm 1 on $\textstyle L^p(X,\Sigma,\mu)$, with range the closed subspace $L^p(X,\Sigma_T,\mu) \subset L^p(X,\Sigma,\mu)$. Therefore it's quite natural to consider the analogue decomposition of the $L^p$ spaces given by the $L^p$ projector , that ''should'' be:  
$$L^p(X,\Sigma,\mu)=L^p(X,\Sigma_T,\mu)\oplus\ {\overline{\{ f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)}\} }^{L^p}.$$
Now if $1\leq p\leq 2$, this splitting actually holds true, and it is easily obtained with a Lp-closure starting from the $L^2$ splitting. For $2\leq p \leq\infty$, a splitting is obtained by restriction to $L^p(X,\Sigma,\mu)$. And here it comes the problem: this way I get the right first factor (the range of the projector) $L^p(X,\Sigma_T,\mu)=L^2(X,\Sigma_T,\mu)\cap L^p(X,\Sigma,\mu)$, but I can't see why the kernel of the projector,
$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)$$
should be equal to (and not larger than)
$$\overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}.$$
Maybe it's not a fundamental point (it does not enter in the proof of the main ergodic theorems) but I think that if a complete analogy holds true, it would be nice to state it, and if it doesn't, one would like to know what goes wrong. I checked the main texts of ergodic theory on this point, and found nothing. 
Summarizing: 
is there a (hopefully quick) way to see whether for $2\leq p \leq\infty$ there is an inclusion (hence equality)
$$\overline{ \{f-f\circ T\ :\ f\in L^2(X,\Sigma,\mu) \} }^{L^2} \cap\ L^p(X,\Sigma,\mu)\ \subset  \ \overline{\{f-f\circ T\ :\ f\in L^p(X,\Sigma,\mu)\}}^{L^p}$$
$$\mathbf{?}$$

REPLY [2 votes]: The mean ergodic theorem on L^p spaces is due to F. Riesz (1938) and S. Kakutani (1938). For p in $[1,\infty[$, this is theorem 1.2 ff in the book of Krengel, "Ergodic theorems".
If $p=\infty$, you get the splitting only on the set of functions for which the Birkhoff sums are converging, which, I think is not everything in general. In general Banach spaces, you always have the splitting in restriction to the space of vectors with converging averages (see Krengel, theorem 1.3).
The decomposition is more or less explicit. For all measurable function g, we can write 
$g - {1\over n} S_n(g) = g_n - g_n\circ T$
with $S_n(g)=\Sigma_0^{n-1}g\circ T^k$ and $g_n={1\over n}\ \Sigma_0^{n-1} S_k(g)$.
If g is in the $L^2$ closure of coboundaries, we know that its conditional expectation w.r.t the invariant $\sigma$-algebra is zero. If moreover g is in $L^p$, $1\leq p < \infty$, we know that ${1\over n} S_n(g)$ goes to zero in $L^p$ norm, using the $L^p$ ergodic theorem. Also, the functions $g_n$ are in $L^p$, from their very definition. So g is in the closure of $L^p$ coboundaries.
I am not sure that the result holds for $p=\infty$.<|endoftext|>
TITLE: Research statement in PhD applications--how much is too much?
QUESTION [11 upvotes]: I intend, in the somewhat near future, to engage preparing my graduate school applications for next year. I have worked hard to secure a solid application as far as coursework, grades, recommendations, etc., etc., though the statement of my research plan is very important to me (and, as a friend of mine who is on an admissions committee at a top school has informed me, it is far more important than most students believe it to be). 
However, I find myself at an impasse; I have research interests which lie at the intersection of a broad array of wider mathematical disciplines (algebraic geometry, algebraic topology, a bit of number theory, representation theory, categorical algebra, and even model theory--I always try utilizing my mathematical toolkit in assessing problems in mathematical physics, as well). 
If I just go on in my application listing these disciplines, I won't be taken seriously. Though if I am too particular, I risk appearing too specialized for the research being conducted at school X (and I am broadly interested, though this can be a boon if not taken too far). 
So rather than expressing my interests (and potential interests) in the following way:
--algebraic geometry
--algebraic topology
--arithmetic geometry/algebraic number theory
--n-categories/topoi
--representation theory
--etc., etc.
I would like say something like 
--motivic cohomology, etale homotopy, Hodge theory, stacks, D-modules (algebraic geometry/topology)
--braided monoids and algebras (representation theory, category theory)
--model-theoretic proofs of Mordell-Lang and geometric stability theory (model theory/arithmetic geometry)
--n-categories, higher constructions with topoi (this ties in with my interest in etale homotopy).
What would be a good strategy here? I don't want to seem unfocused or naive, but I don't want to leave out any of the many things in which I have some degree of interest? (Many of the subjects listed here are things which I have actively pursued outside of the classroom to some degree, some of them at an advanced level--e.g., motivic cohomology and etale homotopy).

REPLY [10 votes]: Some random thoughts:
1) I recommend that you discuss this with a professor who knows you well and show him drafts of your statement.
2) There is no reason why you need to submit the same research statement to every school. You can focus on different research topics, depending on the strengths of each department.
3) Your list of interests above is way too long and broad. I doubt it will be taken seriously. Focus on only one or two and discuss them in enough depth to show that you really know more than just the terminology.
4) I agree with everybody else that the statement should be no longer than two pages, no matter what.
5) Nobody expects an undergraduate to have much breadth or depth in their knowledge of mathematics. What you want to demonstrate is your desire and commitment to building greater depth in your knowledge of mathematics. Although you don't want to appear too narrow (and this does not seem to be a problem for you anyway), demonstrating breadth or an interest in breadth is far less important than showing the desire for depth.<|endoftext|>
TITLE: Lowest Unique Bid
QUESTION [20 upvotes]: Each of n players simultaneously choose a positive integer, and one of the players who chose [the least number of [the numbers chosen the fewest times of [the numbers chosen at least once]]] is selected at random and that player wins.
For n=3, the symmetric Nash equilibrium is the player chooses m with probability 1/(2^m).
What is the symmetric Nash equilibrium for n=4?  Is it known for general n?

REPLY [3 votes]: There are non symmetric Nash equilibria. For example with 3 players: (1,1,2) is a silly solution.<|endoftext|>
TITLE: Is it possible to classify all Weil cohomologies?
QUESTION [6 upvotes]: Weil cohomologies seem to be "natural" and useful cohomology theories. Wikipedia lists Betti, De Rham, l-adic etale, and crystalline cohomologies as examples of Weil cohomology. Do we have more of them? and is it plausible to classify all or some Weil cohomologies? Or more generally, classify these really good Grothendieck sites?

REPLY [4 votes]: If we believe in the standard conjectures (or something similar) so that the category of motives is Tannakian, then a Weil cohomology theory is just a fibre functor and as such twists of each other (that does not quite give the multiplicative structure however) which gives, more or less, a classification. Note that your question on Grothendieck sites is only vaguely related, the whole point of the notion of Weil cohomology theory is that there is no site in sight (pun intended).
Addendum: This is a little bit simple minded as the requirements for a Weil cohomology theory to be a fibre functor go beyond the standard conjectures I think. Also one would need to define motives using cohomological equivalence. However, I think the statements I made are philosophically OK and one cannot hope to get anything in the way of a more precise classification.<|endoftext|>
TITLE: Non-degeneracy of ground state in quantum mechanics
QUESTION [12 upvotes]: In non-relativistic quantum mechanics, what are the necessary conditions on the potential (or on the hamiltonian in general) for the ground state to be non-degenrate?

REPLY [2 votes]: A stronger result is true: the ground state of $H=-\Delta+V(x)$ (if one exists, that is, if the spectrum is bounded below and its minimum is an eigenvalue) is positive pointwise.
For a sketch of a proof, recall that an eigenfunction $y(x)$ with eigenvalue $\min\sigma$ minimizes the quadratic form $Q(y)=\int (|\nabla y|^2+V|y|^2)$. We can in fact assume that $y$ is real valued (take the real or imaginary part otherwise).
Notice that if $y\in H^1$ minimizes $Q$, then so do $|y|$ and $y\pm |y|$. If
we now repeat Riemann's error from his proof of the Riemann mapping theorem and just assume that a minimizer has enough smoothness so that it will be a classical solution of the Euler-Lagrange equation, which here is just the original eigenvalue equation $-\Delta y+Vy=E y$, then we're already done (or close at least), because $y\pm |y|$ will not have second derivatives if the original function took values of both signs. One can make a rigorous proof out of this, based on the maximum principle for second order elliptic equations; see Theorem 6.5.2 of Evans's PDE book.
Finally, to make the connection to the original question more explicit, let me state more clearly what a rigorous version of this argument proves: any minimizer $y$ of $Q$ satisfies $y(x)>0$ after multiplication by a suitable constant. Since minimum energy eigenfuctions are minimizers, we can't have more than one, or we could take a linear combination that violates this condition.<|endoftext|>
TITLE: Pushing convex bodies together
QUESTION [13 upvotes]: Given two convex bodies $A$ and $B$, in $\mathbb R^3$ let's say. We define $A(t)$ and $B(t)$ as $A+xt$ and $B+yt$ where $x,y$ are two arbitrary points. (That is the Minkowski sum, so the two bodies are moving at constant velocity in the $x$ and $y$ directions, and $t$ is the time variable.) Can one show that the function
$$f(t)=\operatorname{Vol}(A(t)\cap B(t))$$ is unimodal? That is nondecreasing up to some point, and then nonincreasing.

REPLY [15 votes]: The sets $\{ (A(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ and $\{ (B(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ are convex, their intersection $K$ is a bounded convex set, and $f(t)$ is the volume of the slice of $K$ at height $t$. By Brunn-Minkowski inequality, this is log-concave, so definitely unimodal.<|endoftext|>
TITLE: Green's formula for nonorientable manifolds
QUESTION [8 upvotes]: Usually in differential geometry one proves the Stokes theorem and then obtains divergence theorem and Green's formulas as corollaries. However, divergence theorem is also valid for nonorientable riemannian manifolds when one replaces forms with densities. But then the Green's formulas should also be valid. Am I missing something? I haven't found a discussion about where precisely the orientability is needed. 
Further supposing Green formulas in nonorientable manifolds one could define variational formulation for example for elliptic PDEs there. I wonder if the (non)orientability has some effect on the solutions of such PDEs?

REPLY [4 votes]: So it seems that Bott & Tu are saying that orientability doesn't really matter: Stokes theorem remains valid if one replaces forms by densities. But then it's curious why this fact is not clearly stated in differential geometry books. For example some projective spaces are nonorientable and obviously projective spaces are fundamental in all mathematics. So it's funny why there is no explicit discussion of these matters in projective spaces.<|endoftext|>
TITLE: A variation of Minkowski sum
QUESTION [7 upvotes]: I have to work with the following variation of Minkowski sum:

Let $\mathbb E$ be a Euclidean space and $K$ be a convex set in $\mathbb E\times \mathbb E$.
  Set 
  $$K^+=\{\\,x+y\in\mathbb E\mid(x,y)\in K\\,\}.$$

Note that if $K=K_x\times K_y$ for some convex sets  $K_x$ and $K_y$ in $\mathbb E$ then $K^+$ is the usual Minkowski sum of $K_x$ and $K_y$.
Questions: 

Did anyone consider this construction? 
Does it have a name?

REPLY [5 votes]: In additive combinatorics, we call the Minkowski sum the sumset, and write it as ${\mathbb E}+{\mathbb E}$. We call what you're talking about the "sumset along a graph", and write it as ${\mathbb E}+_K{\mathbb E}$, where $K$ is any graph (you call it a subset of ${\mathbb E}\times {\mathbb E}$ and I call it a graph, but it's the same thing!).
For an example of this terminology in use, check out this paper of Alon, Angel, Benjamini, and Lubetzky. Also, a google scholar search shows the terminology in action.<|endoftext|>
TITLE: Are the underlying undirected graphs of two mutation-equivalent acylic quivers isomorphic?
QUESTION [10 upvotes]: Quiver mutation, defined by Fomin and Zelevinsky, is a combinatorial process. It is important in the representation theory of quivers, in the theory of cluster algebras, and in physics. 
We consider a finite directed graph Q (aka a quiver) without loops and without 2-cycles which may contain parallel edges. Then Fomin-Zelevinsky associated with every vertex k of Q a new quiver μk(Q) which has again neither loops nor 2-cycles. Every μk is involutive in the sense that μk(μk(Q)=Q for every  quiver Q and every vertex k. The precise definition may be found in Fomin-Zelevinsky's works on cluster algebras. In the theory of cluster algebras the quivers encode exchange relations.
You might check Keller's java program to see quiver mutation in action.
Although quiver mutation has an easy combinatorial definition, many natural questions are either open or rely on sophisticated techniques. For example, there is no recipe (simple algorithm) to decide whether two given quivers can be obtained from each other by a sequence of mutations. 
Question: Suppose I start with an acyclic quiver Q. (That is, Q contains no oriented cycles.) After a sequence of mutations I get another acyclic quiver Q'. Does it follow that the underlying undirected graphs of Q and Q' obtained by ignoring the orientation of the edges are isomorphic as (undirected) graphs?

REPLY [9 votes]: Two acyclic quivers which are mutation equivalent are always related by a sequence of mutations at sinks and sources. This is proved in Caldero-Keller's 2006 paper.<|endoftext|>
TITLE: Is perfect play possible in continuous rock-paper-scissors? game "step size" vs. "acceleration"
QUESTION [6 upvotes]: The first part of my question is simple: Is every game continuous in time and strategy-space also a game of perfect information with a good equilibrium? For example, consider rock-paper-scissors. The discrete version has no nash equilibrium - a perfectly uniform random mixed strategy is the best option.
Continuous rock-paper-scissors, by contrast, allows players to move at some limited velocity (consider 2 cases, acceleration is limited and acceleration is infinite) through a "strategy space" s.t. R+P+S = 1, and (0.5, 0.5, 0) vs. (0,1,0) returns 0.5 to player one and -0.5 to player 2, while (1,0,0) returns 1 to player one, -1 to player 2. To avoid the "go directly to the middle" strategy, it's fine to remove (1/3,1/3,1/3) or some disk around it from the strategy plane.
So, is continuous RPS effectively a game of perfect information?
For a more dramatic example, consider the stock market as a game. If it were continuous, would randomness essentially be removed? Would a player also need to explicitly know the strategies of all other players as individuals, or only the end result of those strategies (i.e. value of stocks at a given point in time) in order to play perfectly?
For a more realistic example, consider a hunt between a dog and hare. Strategies for them are the direction they choose to run in the pursuit. The dog has reflexes r, the time it takes him to notice the hare's change in direction. The rabbit has acceleration a. Ignore the dog's acceleration for now. As r*a becomes extremely small (i.e. the dog's reflexes are swift relative to the hare's acceleration), does this effectively converge to a continuous, perfect-information game (specifically the game of the homicidal chaffeur), or is the difference still important? Specifically, suppose that the dog can only make decisions on pursuit directions at increments equal to r - I don't want it to be a continuous game with a lagging signal.

REPLY [4 votes]: If I interpret your question correctly, continuous RPS isn't a game of perfect information.  I'm assuming what I know at any instant is A) my strategy at any past instant, and B) the outcome of our two strategies matched up against each other at any past instant.  
But I can't deduce your strategy from knowing mine and the outcome; consider my strategy of (1,0,0) and an outcome of 0 for both players.  I can't tell if you played (1,0,0) or (0,.5,.5) at that moment.  In the continuous game, you could be moving anywhere you please on the line connecting those two strategies while I stayed at (1,0,0), and I would be none-the-wiser.
It is, however, true that in any continuous game where your opponent's strategy can be completely deduced from the outcome, you have perfect information.  This follows by reverse-engineering his strategy at all previous moments, and taking the limit.  Unfortunately, if the outcome is represented by a single real number, then the strategy space for such a game has to be essentially 1-dimensional, which means they are rather boring.<|endoftext|>
TITLE: Why does undergraduate discrete math require calculus?
QUESTION [14 upvotes]: Often undergraduate discrete math classes in the US have a calculus prerequisite.
Here is the description of the discrete math course from my undergrad:

A general introduction to basic
  mathematical terminology and the
  techniques of abstract mathematics in
  the context of discrete mathematics.
  Topics introduced are mathematical
  reasoning, Boolean connectives,
  deduction, mathematical induction,
  sets, functions and relations,
  algorithms, graphs, combinatorial
  reasoning.

What about this course suggests calculus skills would be helpful?
Is passing calculus merely a signal that a student is ready for discrete math?
Why isn't discrete math offered to freshmen — or high school students — who often lack a calculus background?

REPLY [3 votes]: This has been dormant for a while, but it's worth pointing out the ACM recommendations, which essentially say what J W says - but I don't have enough rep to vote up that answer or comment on it, so I provide the link here for those searching for info.  The ACM also recommended calculus in this set of recs, whereas the update is more about the core CS curriculum. It's also worth mentioning that the ACM is focused more on "sound reasoning", not "formal symbolic proof", in its guidelines.  That doesn't necessarily mean less mathematical, from what I can tell.<|endoftext|>
TITLE: Criteria for boundedness of power series
QUESTION [26 upvotes]: Consider a power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real 
x, thus defining a function $f: \mathbb{R} \to \mathbb{R}$. 
Can one give necessary and sufficient criteria the sequence of the coefficients $(a_n)$ has to meet in order for $f$ to be bounded on $\mathbb{R}$? (Let's disregard the trivial case that $a_0$ is the only non-zero coefficient and let's call a sequence "function-bounded" if the power series is bounded.) Criteria for boundedness seem to be far more difficult to obtain than the usual criteria for convergence of a power series, here some remarks: 
a)  A necessary condition for $\sum_n a_n x^n$ to be bounded is that there are infinitely many non-zero coefficients which change sign infinitely many times. 
b) The boundedness of $f$ is an "unstable" property of the sequence of coefficients: any non-zero change in any finite subset (except $a_0$) will destroy boundedness. Thus the linear subspace of all function-bounded sequences is rather "sparse" in the vector space of all sequences representing convergent power series.  
c) On the other hand, the linear subspace of all function-bounded sequences contains at least all power series of functions that can be written as $\cos \circ h$ with $h$ an entire, real-analytic function, and the algebraic span of these functions. One could conjecture that this space is already the space of all bounded functions that can be represented as power series[EDIT: seems to be refuted, cf. comment below]. And perhaps this could be a starting point for deducing the criteria. 
EDIT (conjecture added): 
Is is true, that every power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad  \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?
Example: One can modifify the signs of the power series of the exponential function $\sum_{n=0}^{\infty} x^n/n!$ pretty easily to a bounded power series by $\epsilon_n = +1$ for $n = 0 or 1 \mod 4$ and $\epsilon_n = -1$ for $n = 2 or 3 \mod 4$, yielding the function $\sin(x) + \cos(x)$. (One can modify the signs pretty easily a bit further such that the power series is not only bounded on the real axis, but also on the imaginary axis - but this is not the question here). 
I have neither succeeded in finding a counterexample nor in prooving this conjecture. 
EDIT2: 
Thanks for the nice counterexample. I would like to improve the conjecture as follows: Define a power series $\sum_{n=0}^{\infty} a_n x^n$ as nondominant, if for all $x \in \mathbb{R}$ the absolute value of every term $a_kx^k$ is smaller or equal than the sum of the absolute values of all the other terms: $|a_kx^k| \le \sum_{n \neq k} |a_n x^n|$. The improved conjecture is: Is is true, that every nondominant power series $\sum_{n=0}^{\infty} a_n x^n$ that is convergent for all real $x$ can be modified only by changing the signs of the terms to a convergent power series $\sum_{n=0}^{\infty} \epsilon_n a_n x^n, \quad  \epsilon_n \in \{\pm1\}$ that is bounded for all real $x$?

REPLY [7 votes]: I think the nice added conjecture goes to the core of the problem, nevertheless it has to be modified because it doesn't hold as it is.
Consider for instance an entire function
$f(x):=\sum_{n=0}^\infty a_n x^n$ with 
$$|a_n|=3^{-n^2}.$$
then $f$ is unbounded on $\mathbb{R}$ since we have, for all $a\in\mathbb{N}:$
$$\left|\  f(3^{2a})\right| \geq 3^{a^2} -\sum_{n\neq a} 3^{n(2a-n)}\geq 3^{a^2}\left( 1-2\sum_{k>0 }3^{-k^2}\right)\ge \frac{3^{a^2}}{4}.$$<|endoftext|>
TITLE: Interpreting the Famous Five equation
QUESTION [32 upvotes]: $$e^{\pi i} + 1 = 0$$
I have been searching for a convincing interpretation of this. I understand how it comes about but what is it that it is telling us? 
Best that I can figure out is that it just emphasizes that the various definitions mathematicians have provided for non-intuitive operations (complex exponentiation, concept of radians etc.) have been particularly inspired. Is that all that is behind the slickness of the Famous Five equation?
Any pointers?

REPLY [3 votes]: A way to introduce the exponential is by the ODE:
$$f' = \alpha f, \ \ f(0) = 1,$$
whose unique solution is $t \mapsto exp(\alpha t)$. This is common when $\alpha$ is natural, but what if $\alpha$ is complex? And, in particular, what if $\alpha = i$? Of course, $f$ must be complex-valued instead of real-valued.
Then $f' = if$, that is, $f'$ is the vector $f$ rotated by $\pi/2$ in the trigonometric direction. If we forget a moment about the complex structure, we get:
$$y' = x,  \ x'=-y, \ \ f(0) = (x,y)(0) = (1,0).$$
A solution of this system is a trajectory whose speed is always tangential to its position vector. But then, the function $(x,y) \mapsto x^2+y^2$ is a first integral, so it is constant along trajectories. In particular, the trajectory starting from $(1,0)$ stays on the circle of radius $1$ centered on $0$. But then, since $|f|$ is constant, so is $|f'|$: the trajectory moves at unit speed along this circle of radius $1$.
Since the trajectory starts from $1$, after a time $\pi$, it will have traveled along a half-perimeter, so it will be in $(-1,0)$. Whence $exp(i \pi) = -1$.
@Andrey Rekalo's interpretation of this equation is Euler's method applied to this ODE.<|endoftext|>
TITLE: Involutions in GL_n(Z)
QUESTION [12 upvotes]: Is there a classification of involutions in $\text{GL}_n(\mathbb{Z})$?
Here's some more details about what I mean.  Consider $f \in \text{GL}_n(\mathbb{Z})$ such that $f^2=1$.  Regard $f$ as an automorphism of $\mathbb{Z}^n$.  Extend $f$ to an automorphism $g$ of $\mathbb{Q}^n$.  Then we can write $\mathbb{Q}^n = E_1 \oplus E_{-1}$, where $g$ acts as the identity on $E_1$ and as $-1$ on $E_{-1}$.  Restricting this decomposition to $\mathbb{Z}^n$, we obtain a finite-index subgroup $A$ of $\mathbb{Z}^n$ and a decomposition $A = F_1 \oplus F_{-1}$ such that $f$ acts as the identity on $F_1$ and as $-1$ on $F_{-1}$.
However, we definitely cannot assume that $A = \mathbb{Z}^n$.  For instance, the matrix whose first row is $(0 1)$ and whose second row is $(1 0)$ (by the way, I can't figure out how to get my matrices to display correctly) is an involution in $\text{GL}_n(\mathbb{Z})$ that can be diagonalized over $\mathbb{Q}$ but not over $\mathbb{Z}$.
What else can be said here?

REPLY [15 votes]: The problem is equivalent to classifying isomorphism classes of $n$-dimensional integral representations of the cyclic group $C_2$ of order 2, or $\mathbb{Z}[C_2]$-modules on $\mathbb{Z}^n$. This group has exactly 3 isomorphism classes of indecomposable free $\mathbb{Z}$-modules: 
(1) trivial 
(2) sign representation 
(3) 2-dimensional with matrix $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}.$
Every $n$-dimensional $\mathbb{Z}[C_2]$-module is a direct sum of (1), (2), (3) with uniquely determined multiplicities. Thus any involution is conjugate over $\mathbb{Z}$ to a block diagonal matrix with blocks [1], [-1], $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ whose sizes are uniquely determined.<|endoftext|>
TITLE: Name of amateur who gave a new proof of the Ramanujan-Nagell theorem?
QUESTION [33 upvotes]: In an article by George Johnson in the New York Times back in 1999, it says that an amateur mathematician from India once sent Ian Stewart a proof of the Ramanujan-Nagell theorem that the Diophantine equation $x^2 + 7 = 2^n$ is solvable if and only if $n = 3, 4, 5, 7, 15$.  The proof "was badly typed on strange paper and cast in an idiosyncratic style that would have given any journal editor the impression that the writer was a crank."  However, it was correct, and after getting some help cleaning it up, the man published the proof.
To me, this is an inspiring story, and I would like to know the name of this man and to see the paper.  I asked Ian Stewart but he said that he remembers the incident but not the identity of the man in question.  I would try asking George Johnson but I am not sure how to contact him.  I searched MathSciNet but was not able to guess which paper it was.
Does anyone know more details?

REPLY [3 votes]: I'm going to argue that the story is apocryphal; there was no such proof by an amateur.  Here's why
The linked story is not about the amateur and the proof; it's about mathematical cranks.  The amateur is not named; all the article says is

Several years ago Dr. Stewart heard from a man -- in India again --
  who had found a new, simpler proof for an obscure, pointless theorem
  in number theory written by Ramanujan and a collaborator. According to
  the Ramanujan-Nagell theorem, the only numbers one can square and add
  7 to, yielding an answer that is a power of 2, are 1, 3, 5, 11 and
  181. For example, squaring 3 and adding 7 gives 16, which is the fourth power (the square of the square) of 2.
Dr. Stewart was surprised to realize that the proof was correct, but
  it was badly typed on strange paper and cast in an idiosyncratic style
  that would have given any journal editor the impression that the
  writer was a crank. Dr. Stewart advised the writer to find an Indian
  number theorist who could teach him how to present a proper paper.
  Several years later the result was published, and soon after came
  another publication from the same man. ''It is worth reading these
  things occasionally,'' Dr. Stewart said.

A search on MathSciNet for any article between 1998 and 2003 containing both the words Ramanujan and Nagell turns up only 19 hits.  None are the article in question.
A Google Scholar search with the same parameters turns up only 4 pages of hits.  None are plausibly an article of a proof by one or two authors, (at least one from India)
The 2002 survey article Relevance of Srinivasa Ramanujan at the dawn of the new millennium by KS Rao talks about the Ramanujan-Nagell equation does not mention a new proof - one might expect that Rao would have known about this.

So what did happen?  Where did the story come from?  One could imagine that Stewart had some conversation with Johnson about mathematical cranks, and misremembered  a story, which Johnson did not do enough checking to correct.  (side note: Johnson writes regularly about science but has only written a couple of math articles)<|endoftext|>
TITLE: Why didn't Vladimir Arnold get the Fields Medal in 1974?
QUESTION [82 upvotes]: As you all probably know, Vladimir I. Arnold passed away yesterday. In the obituaries, I found the following statement (AFP)

In 1974 the Soviet Union opposed Arnold's award of the Fields Medal, the most prestigious recognition in work in mathematics that is often compared to the Nobel Prize, making him one of the most preeminent mathematicians to never receive the prize.

Since he made some key results before 1974, it seems that the award would have been deserved. Knowing that the Soviets sometimes forced Nobel laureates not to accept their prizes, I thought at first that the same happened here - but noticing that Kantorovich received his Nobel prize the next year, and that Fields laureates both in 1970 and 1978 were Russians (Novikov and Margulis, respectively), I cannot understand why did the Soviets oppose it in case of Arnold. Can someone shed some light?
EDIT: I googled the resources online in English before asking this question here and found no answers. But after posting it here, I googled it in Russian, and found the following:

Владимир Игоревич Арнольд был номинирован на медаль Филдса в 1974 году. Далее — изложение рассказа самого Арнольда; надеюсь, что помню его правильно. Всё было на мази, Филдсовский комитет рекомендовал присудить Арнольду медаль. Окончательное решение должен был принять высший орган Международного математического союза — его исполнительный комитет. В 1971 — 1974 годах вице-президентом Исполнительного комитета был один из крупнейших советских (да и мировых) математиков академик Лев Семёнович Понтрягин. Накануне своей поездки на заседание исполкома Понтрягин пригласил Арнольда к себе домой на обед и на беседу о его, Арнольда, работах. Как Понтрягин сообщил Арнольду, он получил задание не допустить присуждение тому филдсовской медали. В случае, если исполком с этим не согласится и всё же присудит Арнольду медаль, Понтрягин был уполномочен пригрозить неприездом советской делегации в Ванкувер на очередной Международный конгресс математиков, а то и выходом СССР из Международного математического союза. Но чтобы суждения Понтрягина о работах Арнольда звучали убедительно, он, Понтрягин, по его словам, должен очень хорошо их знать. Поэтому он и пригласил Арнольда, чтобы тот подробно рассказал ему о своих работах. Что Арнольд и сделал. По словам Арнольда, задаваемые ему Понтрягиным вопросы были весьма содержательны, беседа с ним — интересна, а обед — хорош. Не знаю, пришлось ли Понтрягину оглашать свою угрозу, но только филдсовскую медаль Арнольд тогда не получил — и было выдано две медали вместо намечавшихся трёх. К следующему присуждению медалей родившийся в 1937 году Арнольд исчерпал возрастной лимит. В 1995 году Арнольд уже сам стал вице-президентом, и тогда он узнал, что в 1974 году на членов исполкома большое впечатление произвела глубина знакомства Понтрягина с работами Арнольда.

Translation of the text:

Vladimir Arnold was nominated for the Fields Medal in 1974. The following is Arnold’s own version of the story; I hope I’m remembering it correctly. The matter seemed settled: the Fields Medal Committee recommended to award a medal to Arnold. The final decision was to be taken by the top administrative body of the International Mathematical Union, the Executive Committee. In 1971–1974 the vice-president of the Executive Committee was one of Soviet Union’s (even world’s) greatest mathematicians, Lev Semenovich Pontryagin (also a member of the Soviet Academy of Sciences). On the eve of his visit to the meeting of the EC, Pontryagin invited Arnold to his home for lunch and to talk about Arnold’s work. Pontryagin told Arnold he was instructed not to allow the Fields Medal to be awarded to him. In case the executive committee wouldn’t agree and would still try to award the medal to Arnold, Pontryagin was authorized to threaten that the Soviet delegation would boycott the next International Congress of Mathematicians in Vancouver, or even that the USSR would leave the IMU. However, in order for Pontryagin’s assertions about Arnold’s work to be convincing, Pontryagin said, he would need to know that work very well. That’s why he invited Arnold in order that he describe his work in detail. Arnold did. According to Arnold, the questions Pontryagin asked him were profound, the talk with him interesting, and the meal good. I do not know if Pontryagin had to make his threat known, but Arnold did not receive the medal—and only two medals were awarded instead of the intended three. By the next time the medals were being awarded, Arnold, born in 1937, was over the age limit. In 1995, Arnold himself became vice-president and learned that in 1974 the depth of Pontryagin’s familiarity with Arnold’s work made a great impression on the members of the EC.

REPLY [3 votes]: Arnold himself confirmed what you wrote in your question:
V.I.Arnold "Yesterday and long ago", Springer, 2007 (translated from Russian), p. 94:
"...while having dinner he (L.S.Pontryagin) explained to me that he had to go to an international meeting and talk against me..."
So basically what you cited seems to be true, though Arnold did not mention that "...Pontryagin was ordered not to allow the award of Fields Medal to Arnold" (in Russian "had to" not necessary means "was ordered to").<|endoftext|>
TITLE: Replacing logician-constructive with combinatorist-constructive?
QUESTION [13 upvotes]: Logicians interpret the word "constructive" in a very well-defined way: they take it to mean, more or less, "computability". Taking constructivity seriously and working in a world where everything must be constructive, leads to intuitionistic logic, which has been a very productive and fascinating subfield of logic.
On the other hand, combinatorists use "constructive" in a different sense. They use it to mean "better than brute force". For example, Ramsey's theorem is non-constructive from the POV of a combinatorist, since its proof offers no method better than just enumerating the subgraphs until you find a complete monochromatic one. On the other hand, from a logician's POV, it is constructive -- just enumerate the subgraphs until you find a complete monochromatic one! (Or even more simply, the pigeonhole principle has the same flavor.) 
So: 

Has anyone looked at logics in which only combinatorist-constructive methods are ok? 
If not, has anyone done a formal analysis of what "better than brute force" means? (This seems different than the questions typically asked in algorithmics, but I would not be shocked if they've thought about it too.)

REPLY [5 votes]: I think the closest thing to what you are looking for are the logical systems studied in Cook and Nguyen's recent book Logical Foundations of Proof Complexity.  These are systems in which the provably total functions lie in certain well-defined computational complexity classes.  In particular, existence proofs in these systems imply that the object whose existence is asserted can be computed "easily."
This line of research goes back at least to Buss (as mentioned by Andrej Bauer), who defined systems of bounded arithmetic that are closely related to the levels of the polynomial hierarchy (a hierarchy of complexity classes whose lowest levels are $P$ and $NP$).  More generally, the field known as "proof complexity" is devoted to studying the relationship between computational complexity classes (particularly circuit complexity classes) and formal systems for arithmetic with suitably weakened induction axioms.
This all assumes that you are satisfied with the idea that "better than brute force" means something like "polytime solvable."  There are limitations with the concept of polynomial time solvability, notably its emphasis on asymptotic behavior and its focus on worst-case complexity.  (Although average-case complexity has been studied, the natural questions there are very difficult to answer and the theory is much less developed.)  Still, a lot of interesting insights have emerged from studying proof complexity and I think it is a very promising avenue for further research.<|endoftext|>
TITLE: Nilradicals without Zorn's lemma
QUESTION [26 upvotes]: It's well known that the nilradical of a commutative ring with identity $A$ is the intersection of all the prime ideals of $A$.
Every proof I found (e.g. in the classical "Commutative Algebra" by Atiyah and Macdonald) uses Zorn's lemma to prove that $x \notin Nil(A) \Rightarrow x \notin \cap_{\mathfrak{p}\in Spec(A)} \mathfrak{p}$ (the other way is immediate).
Does anybody know a proof that doesn't involve it?

REPLY [16 votes]: Since you asked for a proof, let me complement Chris Phan's answer by outlining a proof that relies only on the Compactness Theorem for propositional logic, which is yet another equivalent to the Ultrafilter Theorem over ZF. 
Let A be a commutative ring and let x ∉ Nil(A). To each element a ∈ A associate a propositional variable pa and let T be the theory whose axioms are

p0, ¬p1, ¬px, ¬px2, ¬px3,...
pa ∧ pb → pa+b for all a, b ∈ A.
pa → pab for all a, b ∈ A.
pab → pa ∨ pb for all a, b ∈ A.

Models of T correspond precisely to prime ideals that do not contain x. Indeed, if P is such an ideal, then setting pa to be true iff a ∈ P satisfies all of the above axioms, and conversely. So it suffices to show that T has a model.
Since xn ≠ 0 for all n, one can verify using ideals over finitely generated subrings of A that the theory T is finitely consistent, i.e. any finite subset of T has a model. (What I just swept under the rug here is a constructive proof of the theorem for quotients of Z[v1,...,vn].) The Compactness Theorem for propositional logic then ensures that T has a model.<|endoftext|>
TITLE: Why is Fourier analysis so handy for proving the isoperimetric inequality?
QUESTION [33 upvotes]: I have just completed an introductory course on analysis, and have been looking over my notes for the year.  For me, although it was certainly not the most powerful or important theorem which we covered, the most striking application was the Fourier analytic proof of the isoperimetric inequality.  I understand the proof, but I still have no feeling for why anyone would think to use Fourier analysis to approach this problem. I am looking for a clear reason why someone would look at this and think "a Fourier transform would simplify things here".  Even better would be a physical interpretation.  Could this somehow be related to "hearing the shape of a drum"? Is there any larger story here?

REPLY [10 votes]: I'd like to add a few words on what happens in higher dimensions. First, a convexity assumption becomes essential (as in the second proof of Hurwitz which works only for convex domains). The isoperimetric inequalities in $\mathbb R^n$, $n>2$, are much easier to deal with in case of convex bodies, and the whole problem in some sense looks most natural under the convexity assumption. Second, there are many different isoperimetric inequalities in higher dimensions. And Fourier analysis (or rather harmonic analysis on a sphere) can be successfully applied to prove at least some of them.  
There is a classical approach to isoperimetric problems based on Steiner's theorem. Let $K$ be a convex body in $\mathbb R^n$ and let $K_r$ denote the "parallel" body
$$K_r=\{x\in\mathbb R^n|\ dist(x, K)\leq r \},\quad r>0.$$ 
Then, by Steiner's theorem, there exist $n+1$ numbers $W_0^n(K),W_1^n(K),\dots,W_n^n(K)$, such that 
$$V(K_r)=Vol(K_r)=\sum\limits_{i=0}^{n}{n \choose i}W^n_i(K)r^{i}.$$
It can be shown that 
$$W^n_0(K)=V(K),\quad W^n_1(K)=\frac{S(K)}{n},\qquad(1)$$
where $S(K)$ is the surface area of $\partial K$. Moreover, $W^n_n(K)$ is equal to the volume $\pi_n$ of the unit ball in $\mathbb R^n$ and
$$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),\qquad\qquad\qquad\quad (2)$$
where $w(K)$ is the mean width of $K$. Note that for $n=2$ the perimeter $P(K)$ equals
$\pi w(K).$ The numbers $W_i^n(K)$ give some information on how the convex body $K$ is different from a ball (for the unit ball in $\mathbb R^n$, obviously $W^n_i(K)=\pi_n$ for all $i$.)
A convex body is completely determined by its support function
$$h(x)=\sup\{x\cdot y|\ y\in K\}$$
which measures the directed distance of the origin to the tangent plane of $K$ at direction $x\in S^{n-1}.$  Now, the second proof of Hurwitz deals with the Fourier decomposition of the support function of a convex 2D domain. The problem is that in dimension $n>2$ the formulas for volume and surface area in terms of the support function
cannot be expressed nicely by means of spherical harmonics. However, it is still possible
to derive an isoperimetric inequality for the numbers $W^n_{n-2}$ and $W^n_{n-1}$ via harmonic analysis, namely
$$W^n_{n-1}\geq\sqrt{\pi_n W^n_{n-2}}. \qquad\qquad\qquad(3)$$
When $n=2$ this is the standard isoperimetric inequality $P^2\geq 4\pi A$. 
If $n=3$ $(3)$ gives the isoperimetric inequality between the mean width and surface area of a convex body $$\pi [w(K)]^2\geq S(K).\qquad\qquad\qquad (4)$$ 
The proof is a straightforward extension of the second Hurwitz proof (using a decomposition of the support function into a series of spherical harmonics) and can be found here. 

Update (concerning the question in Victor's comment below). If we assume as known the inequality 
$$W_{1}^3\geq \sqrt{W_{0}^3W_{2}^3},$$
then  together with (1),(2) and (4) it implies that $S^3\geq 36\pi V^2$. ("Known" means that I don't know how to obtain the inequality using only harmonic analysis. It follows from the Alexandrov-Fenchel inequality for mixed volumes.)<|endoftext|>
TITLE: Representation theory over Z
QUESTION [9 upvotes]: In his answer to my question here, Victor Protsak quoted the following result:
Let $C_2$ be a finite cyclic group of order $2$.  Then every $\mathbb{Z}[C_2]$ structure on $\mathbb{Z}^n$ is isomorphic to a direct sum of representations of the following types:

The trivial representation
The sign representation
The representation on $\mathbb{Z}^2$ that permutes the two factors.

This strikes me as a very beautiful result!  I know a lot of nice sources for representation theory over various kinds of fields, but a bit of searching does not turn up any books or surveys on representation theory over the integers.  Does anyone have anything they recommend?

REPLY [17 votes]: See Curtis–Reiner's textbook on the Representation Theory of Finite Groups and Associative Algebras (MR 144979), Theorem 74.3, page 507, and especially the introduction starting on page 493.
The result for cyclic groups of prime order, and for order 4 was originally done in:

Diederichsen, Fritz-Erdmann.
"Über die Ausreduktion ganzzahliger Gruppendarstellungen bei arithmetischer Äquivalenz"
Abh. Math. Sem. Hansischen Univ. 13, (1940). 357–412. MR2133.

However, Reiner has written quite a few nice papers on similar subjects.  One of his earlier ones is on the same topic:

Reiner, Irving.  "Integral representations of cyclic groups of prime order."
Proc. Amer. Math. Soc. 8 (1957), 142–146.
MR83493
doi:10.2307/2032829

One can also consult texts on things called "crystallographic groups", "space groups", and "p-adic space groups".  Plesken has written several nice books using this sort of thing.  These give infinite families of nicely related finite groups and of course help crystallographers.
Be careful to distinguish these sorts of representations from ZG-modules.  ZG-modules are basically incomprehensible, so instead lots of people focus on ZG-lattices, where the underlying module is projective.  This means the idea of using matrices still makes some sense.  There is a lot of literature on modules over group rings over nice rings (like Z or Dedekind domains), but a fair amount of it is not applicable to questions about GL(n,Z).
Roughly speaking, even for G=1, ZG modules are too difficult to understand, and adding a G just makes it worse.  Another common tack is to look at $\hat {\mathbb{Z}}_p$ modules, p-adic modules.  Again the results are nicest for lattices, but things do not get so bad near so fast there.  Reiner's Maximal Orders textbook describes some of the beautiful and well-behaved things you can see there.<|endoftext|>
TITLE: Can the Jacobi-Trudi identity be understood as a BGG resolution?
QUESTION [12 upvotes]: The thought process that led me to this question is that the identity 
$$ \left(\prod_i \frac1{1-x_i}\right)\left(\prod_i {1-x_i}\right)=1$$
can be understood as expressing exactness of the Koszul complex.
This identity is rewritten by taking $\left(\prod_i \frac1{1-x_i}\right)$
as the generating function for the complete symmetric functions $h_n$
and $\left(\prod_i {1+x_i}\right)$ as the generating function for the elementary symmetric functions $e_n$.
Next we have the Jacobi-Trudi identity which expresses a Schur function as the determinant of a matrix whose entries are complete (or elementary) symmetric functions. Also the Specht module is sometimes constructed as a quotient (or submodule) of the trivial representation of the Young subgroup induced to a representation. This suggests that this is the start of a BGG resolution.
I imagine that if this works then it is well-known. Could I have some references? and where does line of thought lead?

REPLY [12 votes]: Look at the short paper MR902299 (89a:17012) 17B10 (20C30)
Zelevinski˘ı, A.V. [Zelevinsky, Andrei] (2-AOS-CY),
Resolutions, dual pairs and character formulas. (Russian)
Funktsional. Anal. i Prilozhen. 21 (1987), no. 2, 74–75, as well as the independent work by Kaan Akin (a former student of David Buchsbaum) including MR1194310 (94e:20059) 20G05
Akin, Kaan (1-OK),
On complexes relating the Jacobi-Trudi identity with the Bernstein-Gel0fand-Gel0fand
resolution. II.
J. Algebra 152 (1992), no. 2, 417–426.   A further refinement is given in MR1379204 (97b:20066) 20G05
Maliakas, Mihalis (1-AR),
Resolutions and parabolic Schur algebras.
J. Algebra 180 (1996), no. 3, 679–690.<|endoftext|>
TITLE: Stone Spaces, Locales, and Topoi for the (relative) beginner
QUESTION [7 upvotes]: I am currently reading Vickers' text "topology via logic" and Peter Johnstone's "stone spaces", and I understand the material in both of these texts to pertain directly to constructions in elementary topos theory (by which I do not mean 'the theory of elementary topoi). However, these things do not seem to be mentioned explicitly in these texts, at least not to great extent. Where might I avail myself of material which really 'brings home' the notion of topoi as 'generalized spaces' in the context of stone spaces and locales as alluded to in Vickers and Johnstone? I understand that Borceaux's third volume in the 'handbook of categorical algebra' is probably a good place to start...

REPLY [5 votes]: My article "Locales and toposes as spaces", in the Handbook of Spatial Logics, serves as a readers' guide to Mac Lane and Moerdijk to show how the results there can be used to create Grothendieck's concept of toposes as generalized spaces.<|endoftext|>
TITLE: Local complete intersections which are not complete intersections
QUESTION [29 upvotes]: The following definitions are standard:
An affine variety $V$ in $A^n$ is a complete intersection (c.i.) if its vanishing ideal can be generated by ($n - \dim V$) polynomials in $k[X_1,\ldots, X_n]$. The definition can also be made for projective varieties.
$V$ is locally a complete intersection (l.c.i.) if the local ring of each point on $V$ is a c.i. (that is, quotient of a regular local ring by an ideal generated by a regular sequence).
What are examples (preferably affine) of l.c.i.  which are not c.i. ? I have never seen such one.

REPLY [2 votes]: Varieties in an affine space.

in characteristic zero any lci $X\subseteq\mathbb{A}^n$ is a set-theoretic complete intersection.
In characteristic $p$ any lci curve $C\subseteq\mathbb{A}^n$ is a set-theoretic complete intersection.

Projective varieties 
Let $X\subseteq\mathbb{P}^n$ be a smooth non-degenerate degree $p$ (a prime number) variety of codimension $c$. Then $X$ is not a scheme-theoretic complete intersection. Indeed, if $X = H_1\cap H_2\cap...\cap H_c$, then $deg(H_2)=...=deg(H_c) = 1$ and $deg(H_1) = p$ by Bezout's theorem because $p$ is prime. Therefore $X$ would be degenerate. An example is again the twisted cubic $C\subset\mathbb{P}^3$. However $C$ is a set-theoretic complete intersection. There exist a quadric surface $Q$ and a cubic surface $S$ such that $Q\cap S = 2C$ (i.e. $Q$ and $S$ are tangent along $C$). 
Hartshorne Conjecture: If $X\subseteq\mathbb{P}^N$ is a smooth variety of dimension $n$, codimesnion $c$ and $c\geq 2n+1$ then $X$ is a scheme-theoretic complete intersection. 
Hartshorne Conjecture has been proven for Fano varieties of codimension two and quadratic varieties (i.e. varieties that can be defined just by quadratic polynomials).
Thanks to Barth’s result: Barth, W.: ”Transplanting cohomology classes in complex-projective
space”, Amer. J. Math., 92, 951-967 (1970),  and since no indecomposable rank
two vector bundle on $\mathbb{P}^N$, $N\geq 5$, is known, it is generally believed that any
smooth, codimension two subvariety of $\mathbb{P}^N$, $N \geq 6$, is a complete intersection. The main results for codimension two subvarieties can be summarized as follows: let $\omega_X\cong \mathcal{O}_X(e)$, $d$ the degree of $X$ and $s$ the minimal
degree of an hypersurface containing $X$. if $e \leq N + 1$ or if $d < (N − 1)(N + 5)$ or if $s \leq N − 2$, then $X$ is a complet intersection. For $N = 5,6$ we can something more: let $X \subset \mathbb{P}^6$ be a smooth, codimension two subvariety, if $s\leq 5$ or if $d \leq 73$, then $X$ is a complete intersection. Let $X \subset \mathbb{P}^5$ be a smooth, subcanonical threefold. If $s \leq 4$, then $X$ is a complete intersection. This is Theorem 1.1 of http://arxiv.org/abs/math/9909137.<|endoftext|>
TITLE: Can equivalences be strictified to isomorphisms?
QUESTION [24 upvotes]: In category theory there are lots of examples of isomorphisms that cannot be strictified to become identities.  For instance, every monoidal category is equivalent to a strict monoidal category, where the associativity and unit isomorphisms are identities, but not every braided monoidal category is equivalent to a strictly braided one where the braiding is an identity.
In higher category theory, constraints are in general (adjoint) equivalences rather than isomorphisms.  For instance, the associativity and unit 2-cell constraints for a tricategory (weak 3-category) are equivalences, but the 3-cell constraints (such as the interchanger) are still isomorphisms, since there is "no room" for anything weaker (there being no 4-cells in a tricategory).  Every tricategory is equivalent, not to a strict 3-category where all constraints are identities, but to a Gray-category where the associators and unitors are identities but the interchanger is not.
I am looking for an example of a higher-categorical structure containing constraint equivalences which cannot be strictified even to become isomorphisms (not necessarily identities).  Since the only nontrivial constraints in a Gray-category are top-dimensional and hence isomorphisms, the first place to look for this would be in some sort of 4-category.  But we can also make it more manageable by being somewhat degenerate.  A triply degenerate 4-category (exactly one 0-, 1-, and 2-cell) would be (by the delooping hypothesis) a symmetric monoidal category, with no room for any equivalences that aren't isomorphisms, so the next level of complexity seems the first place to look.
A doubly degenerate 4-category should be the same as a braided monoidal bicategory, and by the coherence theorem for tricategories, everyone of those is equivalent to a braided Gray-monoid (a Gray-monoid being a Gray-category with one object).  Howver, the braiding in a braided Gray-monoid is, a priori, still only an equivalence, so one way to make this question precise would be:

Is every braided Gray-monoid equivalent to one whose braiding is an isomorphism, rather than merely an equivalence?

REPLY [3 votes]: It appears that the answer to the corresponding question for symmetric Gray-monoids is yes; this was shown by Schommer-Pries and cleanly reformulated by Gurski-Johnson-Osorno.  Nick says that he thinks the braided case should work similarly.<|endoftext|>
TITLE: Pointed vs. unpointed homotopy colimits
QUESTION [6 upvotes]: Let $C$ be a category with a zero object, i.e. an object 0 which is both initial and terminal.  Then $C$ is automatically (and uniquely) enriched over the category $Set_\star$ of pointed sets with smash product, where the basepoint $0\in C(x,y)$ is the unique map which factors through the object 0.  Also, a functor between such categories is $Set_\star$-enriched (i.e. preserves the basepoints of homsets) if and only if it preserves the zero object.  Call these pointed categories and pointed functors.  (This is the "zero-ary" version of additive categories and biproducts.)
Now let $G:C^{op} \to Set_\star$ and $F:C\to Set_\star$ be pointed functors; we can then consider the tensor product $G \otimes_C F$ (otherwise known as the $G$-weighted colimit of $F$) either (1) as ordinary unenriched functors, or (2) as $Set_\star$-enriched functors.  The first is a quotient of $\bigvee_c G(c)\times F(c)$, while the second is a quotient instead of $\bigvee_c G(c) \wedge F(c)$.  (I'm writing $\vee$ for the coproduct in $Set_\star$, since it is a "wedge" which identifies basepoints in the disjoint union.)  I believe, however, that these two tensor products turn out to be the same, since any pair $(x,0)$ in $G(c)\times F(c)$ is equal to $(x,F(t)(0))$ where $t:0\to c$ is the unique map from 0 to $c$ in $C$, and hence gets identified in the tensor product with $(G(t)(x),0) = (0,0)$, which is the basepoint of $\bigvee_c G(c)\times F(c)$.  Thus when $C$ has a zero object, the ordinary tensor product over $C$ has the effect of automatically performing the smash product as well.
My question is: does this remain true for homotopy tensor products?  Suppose we consider $Set_\star$ as sitting inside the category $sSet_\star$ of pointed simplicial sets, and instead of the tensor product we take its homotopical replacement, which can be described as a two-sided bar construction.  We thus get two pointed simplicial sets $B^u(G,C,F)$ and $B^p(G,C,F)$ (for unpointed and pointed), defined by
$$B^u_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \times C(c_{n-1},c_n)\times\dots\times C(c_0,c_1)\times F(c_0) $$
and
$$B^p_n(G,C,F) = \bigvee_{c_0,\dots,c_n} G(c_n) \wedge C(c_{n-1},c_n)\wedge\dots\wedge C(c_0,c_1)\wedge F(c_0) $$
There is a canonical quotient map $B^u(G,C,F) \to B^p(G,C,F)$, and the above observation (assuming it is correct) means that this map induces an isomorphism on $\pi_0$.  Is it a weak homotopy equivalence?
Edit: Reading over the answers, I realized that there are actually two different questions here.  Greg answered a question which isn't quite what I asked, but fortunately the question he answered is the one that I meant to ask, which is comparing $B^p(G,C,F)$ with the homotopy tensor product when considering $G$ and $F$ as functors landing in unpointed simplicial sets, so that the $\bigvee$s would actually become disjoint unions $\coprod$.

REPLY [6 votes]: I'll sketch an alternative approach. I've been on the lookout lately for ways to argue about homotopy (co)limits using universal properties instead of explicit constructions, so I like this question.
Let $Fun(C,Set_* )$ be the category of functors and let $Fun_* (C,Set_* )$ be the category of pointed functors. For a given choice of $G$, the weighted colimit functor $G\otimes -$ in the enriched sense is a composition $Fun_* (C,Set_* ) \rightarrow Fun(C,Set_* )\rightarrow Set_* $ , forgetful functor followed by the unenriched $G\otimes -$. This follows from the fact that its right adjoint is the composition of right adjoints $ Set_* \rightarrow  Fun(C, Set_* ) \rightarrow Fun_* (C,Set_* )$, $X\mapsto (c\mapsto Set_* (G(c),X))$ followed by $F\mapsto (c\mapsto (fiber(F(c)\rightarrow F(0))))$. 
Now make a derived-functor version of the same argument.<|endoftext|>
TITLE: Polynomials with rational coefficients
QUESTION [26 upvotes]: Long time ago there was a question
on whether a polynomial bijection $\mathbb Q^2\to\mathbb Q$ exists. Only one attempt
of answering it has been given, highly downvoted by the way. But this answer isn't obviously
unsuccessful, because the following problem (for case $n=2$) remains open.
Problem. Let $f$ be a polynomial with rational (or even integer!) coefficients
in $n$ variables $x_1,\dots,x_n$. Suppose there exist two distinct points
$\boldsymbol a=(a_1,\dots,a_n)$ and $\boldsymbol b=(b_1,\dots,b_n)$ from $\mathbb R^n$
such that $f(\boldsymbol a)=f(\boldsymbol b)$. Does this imply the existence
of two points $\boldsymbol a'$ and $\boldsymbol b'$ from $\mathbb Q^n$ satisfying
$f(\boldsymbol a')=f(\boldsymbol b')$?
Even case $n=1$ seems to be non-obvious.
EDIT. Just because we have a very nice counter example (immediately
highly rated by the MO community) by Hailong Dao in case $n=1$ and because
for $n>1$ there are always points $\boldsymbol a,\boldsymbol b\in\mathbb R^n$
with the above property, the problem can be "simplified" as follows.
Is it true for a polynomial $f\in\mathbb Q[\boldsymbol x]$ in $n>1$ variables
that there exist two points $\boldsymbol a,\boldsymbol b\in\mathbb Q^n$ such
that $f(\boldsymbol a)=f(\boldsymbol b)$?
The existence of injective polynomials $\mathbb Q^2\to\mathbb Q$ is discussed
in B. Poonen's preprint
(and in comments to this question).
What can be said for $n>2$?
FURTHER EDIT.
The expected answer to the problem is in negative. In other words, there exist injective polynomials
$\mathbb Q^n\to\mathbb Q$ for any $n$.
Thanks to the comments of Harry Altman and
Will Jagy, case $n>1$ is now fully
reduced to $n=2$. Namely, any injective polynomial $F(x_1,x_2)$ gives rise to the injective
polynomial $F(F(x_1,x_2),x_3)$, and so on; in the other direction, any $F(x_1,\dots,x_n)$ in more
than 2 variables can be specialized to $F(x_1,x_2,0,\dots,0)$.
In spite of Bjorn Poonen's verdict
that case $n=2$ can be resolved by an appeal to
the Bombieri--Lang conjecture for $k$-rational points on surfaces of general type
(or even to the 4-variable version of the $abc$ conjecture), I remain with a hope
that this can be done by simpler means. My vague attempt (for which I search
in the literature) is to start with a homogeneous form $F(x,y)=ax^n+by^n$,
or any other homogeneous form of odd degree $n$, which has
the property that only finitely many integers are represented by $F(x,y)$ with
$x,y\in\mathbb Z$ relatively prime. In order to avoid this finite set of "unpleasant"
pairs $x,y$, one can replace them by other homogeneous forms $x=AX^m+BY^m$ and
$y=CX^m+DY^m$ (again, for $m$ odd and sufficiently large, say),
so that $x$ and $y$ escape the unpleasant values.
Then the newer homogeneous form $G(X,Y)=F(AX^m+BY^m,CX^m+DY^m)$ will give
the desired polynomial injection. So, can one suggest a homogeneous form $F(x,y)$
with the above property?

REPLY [32 votes]: Let $f(x)=x^3-5x/4$. Then for $x\neq y$, $f(x)=f(y)$ iff $x^2+xy+y^2=5/4$ or $(2x+y)^2+3y^2=5$. The last equation clealy have real solutions. But if there are rational solutions, then there are integers $X,Y,N$ such that $(2X+Y)^2+3Y^2=5N^2$. This shows $X,Y,N$ all divisible by $5$, ...<|endoftext|>
TITLE: Is there an i.c.c. nonamenable simple group that is inner amenable?
QUESTION [7 upvotes]: A finitely presented, countable discrete group $G$ is amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gX)=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$ 
A countable discrete group $G$ is inner amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gXg^{-1})=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$
The growth $b:\mathbb{N} \rightarrow \mathbb{N}$ of $G$ (with respect to a given word length metric on $G$) is defined as the number of elements $b(n)$ in $G$ lying inside the ball of radius $n$ around $e$. 
It is possible to detect the amenability of $G$ in terms of the growth of G (c.f. R. I. Grigorchuk, “Symmetric random walks on discrete groups”, UMN, 32:6(198) (1977), 217–218).
Can the growth of G detect inner amenability?
I'd like to know if there is an i.c.c. discrete nonamenable simple group that is inner amenable? 
On a related note, what about an answer to Owen's question below?

REPLY [6 votes]: The group $G:=SL_{\infty}(\mathbb Q) = \cup_n SL_n(\mathbb Q)$ is a concrete example. It is obviously simple and non-amenable. Let $g_n \in SL_{\infty}(\mathbb Q)$ be the matrix which is 
$$g_n:= 1_n \oplus \left(\begin{matrix} 0 & 1 \newline -1 & 0 \end{matrix}\right) \oplus 1_{\infty}.$$
and let $$m_{n}(A) := \begin{cases} 1 & g_n \in A \newline 0 & g_n \not \in A \end{cases}.$$
be the finitely additive probability measure associated with $g_n$. Now, for any non-principal ultrafilter $\omega \in \beta \mathbb N \setminus \mathbb N$,
$$m(A) := \lim_{n \to \omega} m_n(A) \in [0,1]$$
is a conjugation invariant finitely additive probability measure on $G \setminus \{e\}$. Conjugation invariance follows since the each element in $G$ commutes with $g_n$ for $n$ large enough.<|endoftext|>
TITLE: Reasons for success in automated theorem proving
QUESTION [14 upvotes]: It seems to me (at least according to books and papers on the subject I read) that the field of automated theorem proving is some sort of art or experimental empirical engineering of combining various approaches, but not a science which tries to explain WHY its methods work in various situations and to find classes of situations for which they work. Or may be I am not aware of some results or the problem is too hard.
Has it been clarified in a rigorous way why, for example, Robinson's resolution works better than Gilmore's saturation? 
(intuitive reason of course is that most general unifier already contains all information about infinitely many terms from Herbrand's universe, but it is only intuition, not a concrete reason, as well as the fact that you can not know when to stop searching for a proof of a formula A in the case A and not(A) are both unprovable is not a rigorous explanation of undecidability of first order logic). 
I even saw a paper shows that there are cases in which London museum algorithm (brute force search, essentially) performs better than resolution! 
Maybe somebody have found some large classes of formulas on which resolution works well?
Maybe some probabilistic analysis of proving methods?
Thanks in advance.

REPLY [12 votes]: You may be interested in the wonderful little book ``The Efficiency of Theorem Proving Strategies: A Comparative and Asymptotic Analysis'' by David A. Plaisted and Yunshan Zhu.  I have the 2nd edition which is paperback and was quite cheap.  I'll paste the (accurate) blurb:
``This book is unique in that it gives asymptotic bounds on the sizes of the search spaces generated by many common theorem-proving strategies. Thus it permits one to gain a theoretical understanding of the efficiencies of many different theorem-proving methods. This is a fundamental new tool in the comparative study of theorem proving strategies.''
Now, from a critical perspective: There is no doubt that sophisticated asymptotic analyses such as these are very important (and to me, the ideas underlying them are beautiful and profound).  But, from the perspective of the practitioner actually using automated theorem provers, these analyses are often too coarse to be of practical use.  A related phenomenon occurs with decision procedures for real closed fields.  Since Davenport-Heinz, it's been known that general quantifier elimination over real closed fields is inherently doubly-exponential w.r.t. the number of variables in an input Tarski formula.  One full RCF quantifier-elimination method having this doubly-exponential complexity is CAD of Collins.  But, many (Renegar, Grigor'ev/Vorobjov, Canny, ...) have given singly exponential procedures for the purely existential fragment.  Hoon Hong has performed an interesting analysis of this situation.  The asymptotic complexities of three decision procedures considered by Hong in ``Comparison of Several Decision Algorithms for the Existential Theory of the Reals'' are as follows:
(Let $n$ be the number of variables, $m$ the number of polynomials, $d$ their total degree, and $L$ the bit-width of the coefficients)
CAD: $L^3(md)^{2^{O(n)}}$
Grigor'ev/Vorobjov: $L(md)^{n^2}$
Renegar: $L(log L)(log log L)(md)^{O(n)}$
Thus, for purely existential formulae, one would expect the G/V and R algorithms to vastly out-perform CAD. But, in practice, this is not so.  In the paper cited, Hong presents reasons why, with the main point being that the asymptotic analyses ignore huge lurking constant factors which make the singly-exponential algorithms non-applicable in practice.  In the examples he gives ($n=m=d=L=2$), CAD would decide an input sentence in a fraction of a second, whereas the singly-exponential procedures would take more than a million years.  The moral seems to be a reminder of the fact that a complexity-theoretic speed-up w.r.t. sufficiently large input problems should not be confused with a speed-up w.r.t. practical input problems.
In any case, I think the situation with asymptotic analyses in automated theorem proving is similar.  Such analyses are important theoretical advances, but often are too coarse to influence the day-to-day practitioner who is using automated theorem proving tools in practice.
(* One should mention Galen Huntington's beautiful 2008 PhD thesis at Berkeley under Branden Fitelson in which he shows that Canny's singly-exponential procedure can be made to work on the small examples considered by Hong in the above paper.  This is significant progress.  It still does not compare in practice to the doubly-exponential CAD, though.)<|endoftext|>
TITLE: How many tacks fit in the plane?
QUESTION [17 upvotes]: Call a tack the one point union of three open intervals.  Can you fit an uncountable number of them on the plane?  Or is only a countable number?

REPLY [4 votes]: While the result of Young mentioned in the answer of Greg Kuperberg seems to be of purely topological interest, quite surprisingly, it has applications to study of the structure of the support of Sobolev functions: 
T. Bagby, P. M. Gauthier,
Note on the support of Sobolev functions.
Canad. Math. Bull. 41 (1998), no. 3, 257–260. 
(MathSciNet  review). 
The paper of Bagby and Gauthier contains a proof of Young's result (Lemma 2 in the paper). The authors knew the result of Moore, but they were not aware of the generalization obtained by Young.<|endoftext|>
TITLE: What does the ample cone look like?
QUESTION [6 upvotes]: For a variety $X/k$, consider the monoid $A$ of classes of ample line bundles in $NS(X)$. What does $A \otimes_\mathbf{Z} \mathbf{R} \subset NS(X)_\mathbf{R}$ look like?

REPLY [9 votes]: I guess I should take this opportunity to mention a personal hobby-horse: the Cone Conjecture of Morrison--Kawamata. (See e.g. http://arxiv.org/abs/0901.3361 for precise statements, history, and known cases.)
To give a little background, note that (as mentioned in Michael Thaddeus' answer) for Fano varieties the nef cone (i.e. closure of the ample cone) is polyhedral, and spanned by rational rays. That's the nicest possible answer one can hope for. 
Once we leave the Fano realm, however, things get worse. Already for Calabi--Yau varieties, the cone of curves can fail to be polyhedral, and even be a completely "round" cone. (This happens e.g. for Abelian surfaces of Picard number at least 3.)
That seems pretty bad, but the Morrison--Kawamata conjecture predicts that there is still something we can say about the nef cone, in good cases such as Calabi--Yau varieties. The key observation is that the automorphism group Aut(X) of any variety acts on the nef or ample cone, and one can hope that this group action simplifies the picture somehow.
Morrison's form of the conjecture is roughly the following.
Cone conjecture: Let X be a Calabi--Yau variety. Then there is a rational polyhedral cone Π inside the nef cone Nef(X) which is a fundamental domain for the action of Aut(X) on Nef(X): in particular, Aut(X).Π = Nef(X).
(To give a more precise statement, one has to consider what happens with non-rational rays on the boundary on the nef cone. Those can never be the image of a rational ray, hence we have to throw out all points of the boundary which are not in the convex hull of the rational rays. There is also an issue of what "fundamental domain" means. But the abbreviated statement above gives the essential flavour, I think.)
So the conjecture predicts that although the nef cone may be very far from being a polyhedral cone, it is "generated" by a polyhedral cone when we take into account the action of automorphims (and, as mentioned, throwing away some boundary points.) If true, that would be a very satisfying description. 
The conjecture has been generalised to Calabi--Yau fibre spaces and then klt Calabi--Yau pairs. In its most general form, it includes all Fano varieties, all Calabi--Yau varieties, and many in between.<|endoftext|>
TITLE: Skellam distribution: Deep connection between Poisson distributions and Bessel function?
QUESTION [8 upvotes]: The probability mass function for the Skellam distribution for a count difference $k=n_1-n_2$ from two Poisson-distributed variables with means $\mu_1$ and $\mu_2$ is given by:
$$
  f(k;\mu_1,\mu_2)= e^{-(\mu_1+\mu_2)}
  \left({\mu_1\over\mu_2}\right)^{k/2}I_{|k|}(2\sqrt{\mu_1\mu_2})
$$
where $I_k(z)$ is the modified Bessel function of the first kind.
My question: Is it just for convenience to get to grips with the resulting infinite summation terms that the Bessel function appears in this formula or is there a deeper mathematical reason connecting Poisson distributions and Bessel functions or even the Poisson distribution with Bessel differential equations? Is there perhaps even some physical interpretation or intution?

REPLY [12 votes]: A mathematical reason is as follows.
On the one hand, the Laurent series for the modified Bessel functions of the first kind $I_k$ can be deduced from the Laurent series for the Bessel functions of the first kind $J_k$ given here. It reads 
$$
\sum_{k\in\mathbb{Z}}I_k(x)t^k=\mathrm{e}^{(x/2)(t+1/t)}.
$$
On the other hand, the characteristic function of a Poisson random variable $Y$ with mean $\mu$ is $E(z^{Y})=\mathrm{e}^{-\mu(1-z)}$, at least for every complex number $z$ of modulus $1$. Hence the characterization, which you recalled in your post, of Skellam distribution as the distribution of $X=Y_1-Y_2$ for independent Poisson random variables $Y_1$ and $Y_2$ with means $\mu_1$ and $\mu_2$ shows that 
$$
E(z^X)=\mathrm{e}^{-(\mu_1+\mu_2)}\mathrm{e}^{\mu_1 z+\mu_2/z}.
$$
Solving $(x/2)(t+1/t)=\mu_1 z+\mu_2/z$ for $x$ fixed and $t$ depending on $z$ yields $$
x=2\sqrt{\mu_1\mu_2},\qquad
t=z\sqrt{\mu_1/\mu_2}.
$$
The value of $\mathbb{P}(X=k)$ for every integer $k$ follows, which involves $I_k(2\sqrt{\mu_1\mu_2})$. 
Finally, one should not worry too much about the appearance of $I_k$ in this answer versus $I_{|k|}$ in the OP's post because $J_{-k}(-x)=(-1)^kJ_k(x)$ for every integer $k$, hence $I_{-k}(x)=I_k(x)$ (a relation which is also a consequence of the invariance by $t\to1/t$ of the Laurent series for the functions $I_k$).<|endoftext|>
TITLE: Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)
QUESTION [15 upvotes]: I recently read the statement "up to conjugacy there are 4 nontrivial finite subgroups of ${\rm SL}_2(\mathbb{Z})$." They are generated by 
$$\left(\begin{array}{cc} -1&0 \\\ 0&-1\end{array}\right),
 \left(\begin{array}{cc} -1&-1 \\\ 1&0\end{array}\right),
 \left(\begin{array}{cc} 0&-1 \\\ 1&0\end{array}\right),
 \left(\begin{array}{cc} 0&-1 \\\ 1&1\end{array}\right) $$
and are isomorphic to $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_4$, and $\mathbb{Z}_6$, respectively. Does someone know a reference for this statement? (Or, is it easy to see?) My attempt at a Google search turned up this statement, but I wasn't able to find a reference.

REPLY [3 votes]: Perhaps the following explicit calculations clarify Ekedahl's argument: consider, for example, the case where the group is cyclic of order 3. Let $w$ be a generator of the group. The characteristic polynomial of w is then necessarily $X^2 + X + 1$. Thus the action of $w$ on $M = \mathbf{Z}\oplus\mathbf{Z}$ makes $M$ into a $\mathbf{Z}[X]/(X^2 + X + 1)$-module. This module is torsion free (since it is torsion-free as a $\mathbf{Z}$-module). Consequently, it is locally free (necessarily of rank $1$) and hence free since $\mathbf{Z}[X]/(X^2+X+1)$ is a PID.
The freeness means that one can find a module generator $m\in M$. Since $(-X,1)$ is a $\mathbf{Z}$-module basis for $\mathbf{Z}[X]/(X^2+X+1)$, one finds that that $(-wm,m)$ is a $\mathbf{Z}$-module basis for $M$.
If the basis $(-wm,m)$ has the same orientation as the standard basis for $M$, then the change-of-basis matrix to $(-wm,m)$ is in ${\rm SL}_2(\mathbf{Z})$. Since the matrix of $w$ with respect to $(-wm,m)$ is the standard matrix of the original problem statement, this construction finishes the argument in this case.
If, on the other hand, the basis $(-wm,m)$ has the opposite orientation from the standard basis for $M$, then $(-w^2 m, m)$ has the same orientation as the standard basis of $M$ (algebra omitted). The change-of-basis matrix to $(-w^2 m, m)$ is thus in ${\rm SL}_2(\mathbf{Z})$. The matrix of $w^2$ with respect to $(-w^2 m, m)$ is the standard matrix of the original problem statement, which finishes the argument in this case.
Let me state this last part of the calculation abstractly: since the question involves ${\rm SL}_2(\mathbf{Z})$ conjugacy, one should consider locally free $\mathbf{Z}[X]/(X^2+X+1)$-modules of rank $1$ equipped with an orientation. We use the orientation $(-X,1)$ on $\mathbf{Z}[X]/(X^2+X+1)$. Given such a module $N$, we can obtain a new conjugate module $N'$ by changing the $\mathbf{Z}[X]/(X^2+X+1)$ action by the (orientation-reversing) automorphism $X\mapsto X^2$. For any oriented $N$, either $N$ or $N'$ is module-isomorphic to $\mathbf{Z}[X]/(X^2+X+1)$ (with orientation). Corresponding to these two cases, one can conjugate (in ${\rm SL}_2(\mathbf{Z})$) either $w$ or $w^2$ to the desired standard form. 
Note that it is straightforward to find a module generator $m$ (and hence to find a standardizing basis for $M$): the function $$ q(m):m\mapsto \det(-wm \; m) $$ is a quadratic form on $\mathbf{Z}\oplus\mathbf{Z}$. It is either positive definite or negative definite. (One can see that easily in the abstract using a module generator for $M$.) An $m\in M$ is a module generator precisely when $q(m) = \pm 1$. We know that such an $m$ exists, and since $q$ is definite, it is easy to search for one.<|endoftext|>
TITLE: Is there an index for solutions to American Mathematical Monthly problems?
QUESTION [31 upvotes]: There is a lot of good stuff contained in the Problems section of the American Mathematical Monthly.  One difficulty with extracting that information, however, is that if I see an old Monthly problem, it is not always so easy to locate the solution.  Sometimes the solution does not appear until many years later, and in some cases the solution has never appeared at all.  For those at an academic institution with access to JSTOR, sometimes an electronic search will turn up the solution, but if the search comes up empty, I am left wondering whether the solution was really never published or whether I just did something wrong with my search.  Furthermore, even with JSTOR access, there is no easy way to list (for example) all unsolved Monthly problems.
Does anybody have, or know of, an index of all the Monthly problems and solutions that would address this issue?  Failing that, is there any interest in creating such a publicly available resource?  If the work is divvied up among a few dozen volunteers it shouldn't take that much work.  As I envisage it, the index would (at least initially) not contain the mathematical content of the problems but would just give the bibliographic information for the problem and the solution (if any), and indicate which parts (if any) remain unsolved.

REPLY [9 votes]: In May 2017 I proposed an MAA pdf ebook series organized by subject.  My proposal included a sample covering the years 2000-2016, a subsample of which can be found here.
The proposal was received well, but it ultimately got rejected because nobody at the MAA knows whether it has the authority to reprint such a collection without obtaining new permissions from all the thousands of published proposers and solvers.  It probably does not help that Taylor & Francis now owns the journals too.
After receiving the rejection I sent each lead editor of the three problem columns a full version covering 2000-2016 (including the Putnam), organized by subject.  One of them replied: "Thank you very much for this incredible work!  This is very useful. 
Is this open to everybody out there?"
Sadly, no.
Earlier attempts include an ambitious database project spearheaded by Ivars Peterson during his tenure as Director of Publications (2009-2014). It eventually met with powerful resistance and died.
The MathPro Press collection (see Jonas's answer) was once available in a free searchable database.  Snapshots of the 1999-2005 and 2005-2010 versions are here and here, respectively.
I am currently putting together a free pdf of Index to Mathematical Problems 1975-1979 for my ResearchGate page.  Will put a link here when it's finished.  Stanley Rabinowitz is interested in doing the same for the 1980-1984 index.
============================ added 2018 Aug 16 ============================
Update:
The 1975-1979 index is now freely available as a pdf on my ResearchGate page and on my website, here:
http://www.mathematrucker.com/problems_1975-1979.pdf<|endoftext|>
TITLE: The definition of "proof" throughout the history of mathematics
QUESTION [12 upvotes]: It is widely believed that mathematicians have a uniform standard of what constitutes a correct proof. However, this standard has, at minimum, changed over time. What are some striking examples where controversies have arisen over what constitutes a correct proof?
Examples of this include: 

The acceptability of the use of the axiom of choice
The acceptability of proofs that rely on assuming that a computer has performed a certain computation correctly
The debate over intuitionistic logic versus classical logic
Hilbert's re-examination of Euclid's axioms and his discovery of unstated assumptions therein
Debates over the use of infinitesimals in calculus, culminating in Weierstrass's epsilons and deltas. There are of course many others.

Edit:
The above re-formulation of the question was provided by Timothy Chow and copied directly from the meta.mathoverflow thread about this question.

REPLY [4 votes]: Appel and Haken's proof of the Four Color theorem.<|endoftext|>
TITLE: Is this problem solvable in polynomial time?
QUESTION [8 upvotes]: Let's start with a picture: http://i.imgur.com/P1k8T.png
What you see here are boxes and circles inside the boxes. Each circle is connected to zero or more boxes. One box is the primary box, it's the grey one. Here is another example (the top box is the primary box): http://i.imgur.com/ibOIO.png (I apologize for the crappy drawings).
The goal is to select a subset of all circles such that:

If a circle is selected, then all the boxes it is connected to must be selected.
If a box is selected then one of the circles in it must be selected.
The primary box is selected.
The sum of the numbers in the selected circles is minimal.

In the top picture the optimal selection of circles has been colored blue. In the bottom picture there are two optimal solutions: select the 3 in the top box and the 1 in the right box. Other solution: select the 2 in the top box, the 1 in the left box and the 1 in the right box.
My question is: is this solvable in polynomial time, or if not is it possible to approximate it? If the graph is tree structured then dynamic programming can solve the problem, but diamond structures seem to complicate the problem.
I have asked this problem on stackoverflow.com, but that doesn't seem to get very far and I thought maybe this is a more appropriate place?

REPLY [12 votes]: There is a straightforward encoding of 3-SAT into this problem, which means that unless P=NP, no, there cannot be a polynomial-time algorithm that solves it. The encoding can be constructed as follows. Given a formula $\phi=\psi_1\wedge\dots\wedge\psi_k$ in conjunctive normal form over propositional variables $v_1,\dots,v_n$, you have the following boxes:

For each $v_i$, two boxes $A_i, B_i$ representing the literals $v_i$ and $\neg v_i$. Both contain a single unit-weight circle. 
Also for each $v_i$, a box $C_i$ representing the law of the excluded middle: This box contains two circles with weight 0, one connected to $A_i$, the other to $B_i$.
For each clause $\psi_j=l_1\vee l_2 \vee l_3$, a box $D_j$ containing three weight-0 circles connected to the corresponding $A_i$ or $B_i$.
The root box, containing a single weight-0 circle connected to all the $C_i$ and $D_j$.

This can be done in time linear in the size of $\phi$.
Then a set of circles satisfying your constraints must contain at least one of $A_i,B_i$ for all $i$, and has weight $n$ if and only if it corresponds to a valid valuation of the $v_i$ which satisfies $\phi$. In particular, satisfiability of $\phi$ can be decided by checking if the minimum weight solution has weight $n$.<|endoftext|>
TITLE: A binomial generalization of the FLT: Bombieri's Napkin Problem
QUESTION [35 upvotes]: This is an extract from Apéry's biography
(which some of the people have already enjoyed in
this answer).

During a mathematician's dinner in
  Kingston, Canada, in 1979, the
  conversation turned to Fermat's last
  theorem, and Enrico Bombieri proposed
  a problem: to show that the equation
  $$ \binom xn+\binom yn=\binom zn \qquad\text{where}\quad n\ge 3 $$ has
  no nontrivial solution. Apéry left the
  table and came back at breakfast with
  the solution $n = 3$, $x = 10$, $y =
 16$, $z = 17$. Bombieri replied
  stiffly, "I said nontrivial."

What is the state of art for the equation above? Was it seriously studied?
Edit. I owe the following official name of the problem to Gerry,
as well as Alf van der Poorten's (different!) point of view on this story and
some useful links on the problem (see Gerry's comments and response).
The name is Bombieri's Napkin Problem. As the OEIS link suggests,
Bombieri said that

"the equation $\binom xn+\binom yn=\binom zn$
  has no trivial solutions for $n\ge 3$"

(the joke being that he said "trivial" rather than "nontrivial"!).
As Gerry indicates in his comments, the special case $n=3$ has a long history
started from the 1915 paper [Bökle, Z. Math. Naturwiss. Unterricht 46 (1915), 160];
this is reflected in
[A. Bremner, Duke Math. J. 44 (1977) 757--765].
A related link is [F. Beukers, Fifth Conference of the Canadian Number Theory Association, 25--33]
for which I could not find an MR link.
Leech's paper indicates
the particular solution
$$
\binom{132}{4}+\binom{190}{4}=\binom{200}{4}
$$
and the trivial infinite family
$$
\binom{2n-1}n+\binom{2n-1}n=\binom{2n}n.
$$

REPLY [7 votes]: My first instinct is to say it seems unlikely there's been serious progress on this problem for general n.  Unlike the Fermat equation, this one is not homogeneous of degree n, which means that it's really a question about points on a surface rather than points on a curve.  We don't have a giant toolbox for controlling rational or integral points on surfaces as we do for curves.
In fact, I can't think of any example of a family of surfaces of growing degree where we can prove a theorem like "there are no nontrivial solutions for n > N."  OK, I guess one knows this about the symmetric squares of X_1(n) by Merel...<|endoftext|>
TITLE: entropy and flatness of densities
QUESTION [15 upvotes]: I was reading C.R Rao's Linear Statistical inference. Rao presents the entropy of a continuous distribution (expectation of -log density) as a measure of closeness to the uniform distribution, and shows that of all distributions whose densities which have a bounded interval [a,b] as its support the uniform distribution on [a,b] has the highest entropy. Can this connection be formalized? For example if I restrict myself to probability distributions which put all mass on a bounded interval and have sufficiently smooth densities, is there a measure of the flatness of the density based on geometry which is connected to the entropy of the distribution.

REPLY [4 votes]: Two great answers above - I just wanted to comment on the "measure of closeness to the uniform distribution"-part. Now I do not have Rao's book in my office so I can't check if this is already in it, my apologies if that's the case.
There is a concept in information theory called the relative entropy, or Kullback-Leibler divergence, which measures the distance between two probability distributions Wiki: Kullback-Leibler.  
Since we are now only interested in continuous distributions, let $P$ and $Q$ be two such distributions with densities $p$ and $q$ respectively. The Kullback-Leibler divergence from $P$ to $Q$ is
$\mathcal{H}(P|Q) = \int \log \frac{dP}{dQ}dP = \int \log \frac{p(x)}{q(x)}p(x)dx $ 
Now take p to have support $[a,b]$ and let $Q$ be the uniform distribution on this interval (i.e., $q = \frac{1}{b-a}$ on $[a,b]$). Inserting this into the expression for $\mathcal{H}(P|Q)$ yields
$\mathcal{H}(P|Q) = \int \log p(x) dP(x) + \log(b-a), $
where the first term is the entropy of $P$ sans the minus sign. Thus, since the Kullback-Leibler divergence in this case is precisely a measure of the closeness of $P$ to the uniform distribution $Q$, the entropy of $P$ will have this interpretation - the term $\log (b-a)$ is only a constant and will be present for every distribution $P$. 
It holds that $\mathcal{H}(P|Q) = 0$ iff $P=Q$ and this explains why the uniform distribution will also be the one with the highest entropy, namely $\log (b-a)$<|endoftext|>
TITLE: Feynman Kac Formula as appears in Krzysztof Gawedzki's Lectures on conformal field theory
QUESTION [5 upvotes]: The lecture notes appeared in the second volume of "Quantum Fields and Strings, a course for mathematicians". I would like to understand the derivation of (1.3), the 2-point correlation function:
$$ \int_{C_{\rm{per}}([0,L])} \phi(x_1) \phi(x_2) d\mu_G(\phi) = \text{tr } e^{-x_1 H} \phi e^{(x_2 - x_1)H} / \text{tr }  e^{-LH} $$
which is listed as a problem, under the heading of Feynman kac's formula. Specifically I would like to know how exactly $\mu_G$ is defined and how it relates to Wiener measure. 
On a more practical note, I would like to know what's a good source (if not here) for obtaining solutions to these sporadic exercises in high level lecture notes. Since time is limited, those of us who do not want to specialize in an area, but only want to get a taste of a subject, but do not want to sacrifice rigor, might find this kind of information very useful.

REPLY [5 votes]: As explained in Gawedzki's lectures, in the line preceding equation (3), the measure $d\mu_G$ differs from the Wiener measure by the multiplicative density $e{-\frac{\beta m^2}{2\pi}\int_0^L\phi(x)^2 dx}$. The exponent is proportional to the Potential energy of the Harmonic oscillator (This example treats the case of the harmonic oscillator).
The equation of the correlation functions is sometimes called the Feynman-Kac-Nelson formula.
It describes a probabilistic evaluation of the trace formula of the correlators in Hilbert space .This is similar to the usual Feynman-Kac formula which is a probabilistic description of the solution of the diffusion equation.
A full proof of the Feynman-Kac-Nelson formula can be found in appendix D of Arai's paper<|endoftext|>
TITLE: Separable and algebraic closures?
QUESTION [21 upvotes]: I have no intuition for field theory, so here goes. I know what the algebraic and separable closures of a field are, but I have no feeling of how different (or same!) they could be.
So, what are the differences between them (if any) for a perfect field? A finite field? A number field?
Are there geometric parallels? (say be passing to schemes, or any other analogy)

REPLY [48 votes]: Geometrically there is a very big difference between separable and algebraic
closures (in the only case where there is a difference at all, i.e., in positive
characteristic $p$). Technically, this comes from the fact that an algebraically
closed field $k$ has no non-trivial derivations $D$; for every $f\in k$ there is
a $g\in k$ such that $g^p=f$ and then $D(f)=D(g^p)=pg^{p-1}D(g)=0$. This means
that an algebraically closed field contains no differential-geometric
information. On the other hand, if $K\subseteq L$ is a separable extension, then
every derivation of $K$ extends uniquely to a derivation of $L$ so when taking a
separable closure of a field a lot of differential-geometric information remains.
Hence I tend to think of a point of a variety for a separably closed field as a
very thick point (particularly if it is a separable closure of a generic point)
while a point over an algebraically closed field is just an ordinary (very thin)
point.
Of course you lose infinitesimal information by just passing to the
perfection of a field (which is most conveniently defined as the direct
limit over the system of $p$'th power maps). Sometimes that is however exactly
what you want. That idea first appeared (I think) in Serre's theory of
pro-algebraic groups where he went one step further and took the perfection of
group schemes (for any scheme in positive characteristic the perfection is the limit,
inverse this time, of the system of Frobenius morphisms) or equivalently
restricted their representable functors to perfect schemes. This essentially
killed off all infinitesimal group schemes and made the theory much closer to
the characteristic zero theory (though interesting differences remained mainly
in the fact that there are more smooth unipotent group schemes such as the Witt
vector schemes). Another, interesting example is for  Milne's flat cohomology
duality theory which needs to invert Frobenius by passing to perfect schemes in
order to have higher $\mathrm{Ext}$-groups vanish (see SLN 868).<|endoftext|>
TITLE: Hyperbolic structure on surfaces with boundary
QUESTION [5 upvotes]: I have following two questions
1) Let $S$ be a compact oriented surface with (non-empty) boundary. Also assume that the Euler characteristic of $S$ is negative (Thus, $S$ is not disk or annulus). Then is it possible to put a hyperbolic metric on $S$  such that every boundary component becomes a geodesic? 
2) In above case, what is the universal cover? Is it a subset of the unit disk with hyperbolic metric?

REPLY [9 votes]: The answers to both questions are positive. 
1) If the Euler characteristic of $S$ is negative, then you can find a pants decomposition (where some of the pants boundary components are glued together, while the others are boundary components of $S$). Then you can put hyperbolic metrics on each pair of pants, while prescribing the lengths of the boundary component, so that you can glue them into a hyperbolic metric of $S$ (and you have parameters left for the gluing, just like for a closed surface).
2) You can always, by gluing additional pants, consider $S$ endowed with a hyperbolic metric constructed as above, as a subsurface with geodesic boundary of a closed hyperbolic surface $\bar S$. Then the universal cover of $\bar S$ is the Poincaré disc, and the boundary components of $S$ lift to complete geodesics. The universal cover of $S$ is then one of the connected components of the complement of these complete geodesics.<|endoftext|>
TITLE: Norms of Commutators
QUESTION [77 upvotes]: If an $n$ by $n$ complex matrix $A$ has trace zero, then it is a commutator, which means that there are $n$ by $n$ matrices $B$ and $C$ so that $A= BC-CB$. What is the order of the best constant $\lambda=\lambda(n)$ so that you can always choose $B$ and $C$ to satisfy the inequality $\|B\|\cdot \|C\| \le \lambda \|A\|$?
Added June 10:  Gideon Schechtman showed me that for normal $A$ you can take $B$ a permutation matrix and $\|C\|\le \|A\|$ s.t. $A=BC-CB$.

REPLY [7 votes]: In a recent paper ([1]), Ravichandran and Srivastava (RS) study pavings for collections of matrices. Their main theorem claims to yield an improvement to the bound obtained by Johnson, Ozawa, and Schechtman (JOS). However, as noted by YCor in a comment, RS [1] cite the JOS work as satisfying a bound on $\max(\|B\|,\|C\|)$, instead of a bound on the product $\|B\| \|C\|$ as in Bill Johnson's answer above. 
But as YCor notes, we can scale $B$ by $\|A\|$ (or both $B$ and $C$ by suitably, e.g., $\sqrt{\|A\|}$), to recover the inequality for the case noted in the OP and in the JOS paper.
In particular, Ravichandran and Srivastava's results imply the following:

Corollary (Corollary 3 in [1]). Every zero trace matrix $A \in M_n(\mathbb{C})$ may be written as $A=[B,C]$ such that $\|B\|$, $\|C\| \le K\log^2(n)\|A\|$ for some universal constant $K$.

(By suitable scaling, this translates into $\|B'\|\|C'\| \le K^2\log^4(n)\|A\|$, for $[B',C']=A$). 
[1]. M. Ravichandran and N. Srivastava. Asymptotically Optimal Multi-Paving. arXiv. Jun 2017.<|endoftext|>
TITLE: Has anyone implemented a recognition algorithm for totally unimodular matrices?
QUESTION [16 upvotes]: One of the consequences of Seymour's characterization of regular matroids is the existence of a polynomial time recognition algorithm for totally unimodular matrices (i.e. matrices for which every square sub-determinant is in {0, 1, -1}).
But has anyone actually implemented it?

REPLY [10 votes]: EDIT. Walter and Trümper have announced on arXiv their implementation, with source code available, of two methods for testing total unimodularity. Their paper describes the technical details of the implementation / algorithm, and also provides several experimental results.

I found the following link for an implementation in R, where they claim to have a function for testing whether a matrix is totally unimodular. I have not checked which particular algorithm they use.
Link: R package<|endoftext|>
TITLE: A question about fields of real numbers
QUESTION [6 upvotes]: Assume that the continuum hypothesis holds. If $F$ is an uncountable field of real numbers, does $F$ always contain a proper uncountable subfield? Are there many specific uncountable fields of real
numbers whose existence can be proved without assuming the axiom of choice?

REPLY [16 votes]: Take a compact Cantor set $K \subseteq \mathbb{R}$ of Hausdorff dimension zero.  Actually we need all cartesian powers $K^n$ of dimension zero as well.  The field $\mathbb{Q}(K)$ generated by it is uncountable, but still of Hausdorff dimension zero, so it is a proper subfield.
edit
That field consists of the values of rational functions $w(x_1,\dots,x_n)$ of many variables with rational coefficients, where the variables range over $K$.  There are countably many such things, so you just have to show any one of them has dimension zero.  The domain of any such $w$ (that is, the set where the denominator does not vanish) consists of an increasing countable union $\bigcup_k A_k$ of sets where the gradient is bounded, so that $w$ is Lipschitz continuous on each $A_k$.  So the image of $w$ on $K^n$ is again a countable union of sets of dimension zero.
plug
G. A. Edgar & Chris Miller, Borel subrings of the reals.  Proc. Amer. Math. Soc. 131 (2003) 1121-1129
LINK
Borel sets that are subrings of $\mathbb R$ either have Hausdorff dimension zero as described, or else are all of $\mathbb R$.<|endoftext|>
TITLE: What are your favorite finite non-commutative rings?
QUESTION [19 upvotes]: When you are checking a conjecture or working through a proof, it is nice to have a collection of examples on hand. 
There are many convenient examples of commutative rings, both finite and infinite, and there are many convenient examples of infinite non-commutative rings. But I don't have a good collection of finite non-commutative rings to think about. I usually just think of a matrix ring over a finite field.
Do YOU have other examples that you particularly like/find easy to use/find to be a good source of counterexamples?

REPLY [5 votes]: The path algebra of an acyclic quiver (no loops or closed circuits) over a finite field. These rings are hereditary, i.e. they have projective dimension $\leq 1$ and occupy an intermediate position between Greg's semisimple and wild examples. If a quiver is "arboreal" (underlying undirected multigraph is in fact a tree) then its path algebra may be identified with the incidence algebra of the poset of vertices in which $i\leq j$ $\iff$ there is a directed path from $i$ to $j$ (by the assumption, there is at most one).
For any ring $R$ you can form matrix algebra $M_n(R)$ and they will be Morita equivalent. In particular, matrix rings of path algebras are also hereditary, but their simple modules are not all one-dimensional.
By the way, I think that in the noncommutative setting, when Frobenius morphism is not available, this is a wrong question: all finite-dimensional algebras over a field should be considered on a similar footing, regardless of whether the field is finite or infinite.<|endoftext|>
TITLE: Do actual Sudoku puzzles have a unique rational solution?
QUESTION [36 upvotes]: Here is a question in the intersection of mathematics and sociology.  There is a standard way to encode a Sudoku puzzle as an integer programming problem.  The problem has a 0-1-valued variable $a_{i,j,k}$ for each triple $1 \le i,j,k \le 9$, expressing that the entry in position $(i,j)$ has value $k$.  The Sudoku rules say that four types of 9-sets of the variables sum to 1, to express that each cell is filled with exactly one number, and that each number appears exactly once in each row, column, and $3 \times 3$ box.  And in a Sudoku puzzle, some of these variables (traditionally 27 of them) are preset to 1.
It is known that generalized Sudoku, like general integer programming, is NP-hard.  However, is that the right model for Sudoku in practice?  I noticed that many human Sudokus can all be solved by certain standard tricks, many of which imply a unique rational solution to the integer programming problem.  You can find rational solutions with linear programming, and if the rational solution is unique, that type of integer programming problem is not NP-hard, it's in P.  Traditionally Sudoku puzzles have a unique solution.  All that is meant is a unique integer solution, but maybe the Sudoku community has not explored reasons for uniqueness that would not also imply a unique rational solution.
Are there published human Sudoku puzzles with a unique solution, but more than one rational solution?  Is there a practical way to find out?  I guess one experiment would be to make such a Sudoku (although I don't know how difficult that is), and then see what happens when you give it to people.

REPLY [3 votes]: Towards a weak proof of existence, I would go through Gordon Royle's database of Sudoku with a minimum number of givens, and (program a computer to) check for multiple rational solutions on any of the puzzles.  If all of them had unique rational solutions, I would
take that as strong evidence that all proper Sudoku have unique rational solutions.  If
one had multiple rational solutions, add some givens until you have a maximal desired
puzzle, and publish it (or send it to Nick Baxter et. al. for the sociological answer).
Gerhard "Ask Me About System Design" Paseman, 2010.06.08<|endoftext|>
TITLE: Good reference for homology of $K(\mathbb{Z}, 2n)$?
QUESTION [14 upvotes]: The homology algebra $H_*( K(\mathbb{Z},2n); \mathbb{Z})$ contains a 
divided polynomial algebra on a generator $x$ of dimension $2n$.
I suppose I could read through the Cartan seminar for a proof, but I'm hoping
someone knows of a nice simple argument for this fact.

REPLY [5 votes]: For a reference, you might see this paper of Birgit Richter.  A rough outline follows:
Since $X = K(Z,n)$ can be made a commutative topological monoid per Ben Wieland's answer, its singular chain complex $C_*(X)$ is made into a commutative and associative differential graded algebra via the commutative associative Eilenberg-Zilber shuffle product $C_*(X) \otimes C_*(X) \to C_*(X \times X)$ composed with the multiplication on $X$.
The formula for the shuffle product is slightly involved when you iterate it, but essentially: to multiply $\alpha_1 \cdots \alpha_n$ with $\alpha_i$ of degree $k_i$, you sum over all ways to divide a set of size $\sum k_i$ into subsets of size $k_1, \cdots, k_n$ of a product of certain degeneracy operators, depending on the subdivision, applied to the chains $\alpha_i$.  (With signs.)
If all the $\alpha_i$ are equal and in positive even degree $k$, then the signs don't interfere and we are summing over all ways to divide $nk$ into $n$ equal pieces of size $k$.  However, because the chains are now equal we get the same term if we simply permute the pieces of size $k$, and so each term appears in the sum at least $n!$ times.  (Note that $k > 0$ is necessary here in order for these permutations to actually give different subivisions.)  As a result, the n-fold product chain is divisible by $n!$.<|endoftext|>
TITLE: Geometric interpretation of group rings?
QUESTION [20 upvotes]: For a group $G$, is there an interpretation of $\mathbb C[G]$ as functions over some noncommutative space? 
If so, what does this space "look like"? What are its properties? How are they related to properties of $G$?

REPLY [26 votes]: The noncommutative space defined by $C[G]$ is (by definition) the dual $\widehat{G}$ of G. 
There are as many ways to make sense of this space as there are theories of noncommutative geometry. 
(Edit: In particular if G is not finite you have lots of possible meanings for the group ring, depending on what kind of regularity and support conditions you put on G, or equivalently, what class of representations of G you wish to consider, and I will ignore all such issues - which are the main technical part of the subject - below.)
One basic principle is that noncommutative geometry is not about algebras up to isomorphism, it's about algebras up to Morita equivalence -- in other words, it's in fact about categories of modules over algebras (the basic invariant of Morita equivalence). You can think of these (depending on context) as vector bundles or sheaves of some kind on the dual. In this case we're looking at the category of complex representations of G, which are sheaves on the dual $\widehat{G}$. You can think of this as a form of the Fourier transform (modules for functions on G with convolution = modules for functions on the dual with multiplication), though obviously at this level of detail it's a complete tautology. Coarser invariants such as K-theory of group algebras, Hochschild homology etc give invariants, eg K-theory and cohomology, of the dual, noncommutative as it may be. There are many conjectures about this noncommutative topology, most famously the Baum-Connes conjecture relating the K-theory of this "space" to that of classifying spaces associated to G.
As to what the dual looks like, this is of course highly dependent on the group. 
For completely arbitrary groups I don't know of anything meaningful to say beyond structural things of the Baum-Connes flavor, so you have to pick a class of groups to study.
If G is abelian, the dual is itself a group (the dual group). The formal thing you can say in general is that the dual of G fibers over the dual of the center of G -- this is a form of Schur's lemma, saying irreducible reps live over a particular point of the dual of the center (ie the center acts by evaluation by a character). You might get some more traction by looking at the "Bernstein center" or Hochschild cohomology --- endomorphisms of the identity functor of G-reps. This is a commutative algebra and the dual fibers over its spectrum. In many cases this is a very good approximation to the dual -- ie the "fibers are finite" (this is what happens for say real and p-adic groups).
The orbit method of Kirillov says that for a nilpotent or solvable group, the dual looks like the dual space of the Lie algebra, modulo the coadjoint action. So again that's quite nice.
Very very roughly the Langlands philosophy says that for reductive groups G (in particular over local or finite fields) the dual of G is related to conjugacy classes in a dual group $G^\vee$. This is if you'd like a way to make meaningful the observation that conjugacy classes and irreps are in bijection for a finite group -- you roughly want to say they're in CANONICAL bijection if the two groups are "dual".
Rather than say it this coarsely, it's better to think in terms of the Harish-Chandra / Gelfand philosophy, which (again whittled down to one coarse snippet) says that the dual of a reductive group (over any field) is a union of "series", ie a union of subspaces each of which looks like the dual of a torus modulo a Weyl group. In other words, you look at all conjugacy classes of tori in G, for each torus you construct its dual (which is a group now!), and mod out by the symmetries inherited by the torus from its embedding in G, and this is the dual of G roughly. (This is also very close to saying semisimple conjugacy classes in the dual group of G, which is where the Langlands interpretation comes from). 
Anyway this is saying that the dual is a very nice and manageable, even algebraic, object.
Kazhdan formulated this philosophy as saying that the dual of a reductive group is an algebraic object --- the reps of the group over a field F are something like the F-points of one fixed variety (or stack) over the algebraic closure.
Anyway one can go much further, and that's what the Langlands program does.

REPLY [5 votes]: I hope that people with more expertise also answer, and give more information than I can.  My memory is that $\mathbb C[G]$ is something like the functions on $\{{\rm pt}\}/G$.  More generally, Connes says that when $G$ acts on a (nice, say locally compact Hausdorff) space $X$, then the functions on the noncommutative space $X/G$ are given by the semidirect product $\mathcal C_0(X) \rtimes G$, where by $\mathcal C_0(X)$ I mean the functions that are smaller than $\epsilon$ outside of a compact set, and the semidirect product is as a vector space (a C-star completion of) the tensor product $\mathcal C_0(X) \otimes \mathbb C[G]$, and the algebra structure is such that $\mathcal C_0(X)$ and $\mathbb C[G]$ are subalgebras.  Connes encourages this way of thinking about bad quotients; the typical example being the $\mathbb R$ action on a torus given by an irrational line.
I'm not entirely sure that I like this answer, however.  First of all, when $G$ happens to be commutative, then $\operatorname{Spec}(\mathbb C[G])$ is the dual group, which seems to me very different from how I think about $\{{\rm pt}\}/G$.  And there are, to my mind, better ways to think about bad quotients, namely through the language of groupoids and stacks.  But I am not an expert on NCG.<|endoftext|>
TITLE: number fields generated by units of number fields
QUESTION [8 upvotes]: Which number fields are generated by the units of some number field? That is, if $K$ is a number field and $U(K)$ its group of units, the field $k = \mathbb{Q}(U(K))$ is a subfield of $K$. But which number fields $k$ occur in this way as $K$ varies over all number fields?
This is an unmotivated question that struck me while making up exercises. It does not seem usable for an exercise, but who knows?

REPLY [9 votes]: The rank of the unit group of $K$ is $r+s-1$ where $r$ and $2s$
are the numbers of real and complex embeddings of $K$. As $r+2s=n$
this rank is less than $n=r+2s$, the degree of $K$. It follows that the
degree of $k$ is at least $r+s\ge n/2$. Thus if $K\ne k$ then the degree
of $k$ is $n/2$, $r=0$ and $s=n/2$. Hence $K$ is a totally complex
quadratic extension of the totally real field $k$, and so is a so-called
CM field.
When we have a CM field, the unit groups of $K$ and $k$ have the same
rank (where $k$ is the real subfield of $K$) but the unit group of $K$
may still be strictly larger (as an example the $p$-th cyclotomic field
where $p$ is prime). I suspect that there must be examples where the
unit groups are the same, but I don't have any to hand.
Added (8/6/2010) I claim that each totally real number field $k$
has infinitely many quadratic extensions $K$ with the same unit
group. Such an extension must be a CM field: $K=k(\sqrt a)$ where $a\in k$
is totally negative.
First I claim that $k$ has infinitely many CM quadratic extensions.
Given a prime ideal $P$ in $k$, by weak approximation there is
a totally negative $a\in K$ whose valuation at $P$ is $1$. Then
$k(\sqrt{a})$ is a CM field ramified over $P$. So there must be infinitely
many such fields.
I now claim that only finitely many CM quadratic extensions of $k$ have
larger unit group. Such an extension $K$ may be a cyclotomic extension,
got by adjoining a root of unity. But the number of roots of unity in $K$
is bounded by a number depending on the absolute degree of $K$, so there
are only finitely many cyclotomic quadratic extensions of $k$.
Assume that $K/k$ is a CM quadratic extension, not cyclotomic and
with larger unit group. Then $K=k(\epsilon)$ where $\epsilon$ is a unit.
As the ranks of the unit groups coincide, then some power of $\epsilon$
lies in $k$. We may assume that $\epsilon^p=\eta\in k$ where $p$
is prime. If $\epsilon^\ast$ is the Galois conjugate of $\epsilon$ over
$k$ then $\epsilon^\ast{}^p=\eta$ and so $(\epsilon^\ast/\epsilon)^p=1$.
Therefore $\epsilon^\ast/\epsilon$ is a nontrivial $p$-th root of unity.
If $\epsilon^\ast/\epsilon\notin k$ then $K/k$ is cyclotomic. So
$\epsilon^\ast/\epsilon\in k$, and $k$ contains all $p$-th roots of unity.
This is only possible if $p=2$ since $k$ is totally real.
Hence $K=k(\sqrt{\eta})$ where $\eta$ is a unit. As the unit group is finitely
generated, there are only finitely many such extensions $K$.<|endoftext|>
TITLE: Category Theory / Topology Question
QUESTION [10 upvotes]: Let me begin by noting that I know quite little about category theory.  So forgive me if the title is too vague, if the question is trivial, and if the question is written poorly.
Let $\mathcal{C}$ and $\mathcal{D}$ be categories.  Say that a functor $T: \mathcal{C} \to \mathcal{D}$ has property X (maybe there is a real name for this property?) if a morphism $f: A \to B$ between objects of $\mathcal{C}$ is an isomorphism whenever $T(f): T(A) \to T(B)$ is an isomorphism.  For example, the obvious forgetful functor $CH \to Set$ where $CH$ is the category of compact Hausdorff spaces has property X because a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.
Here is my question.  Is there a (nontrivial) functor $T: LCH \to \mathcal{D}$ from the category of locally compact Hausdorff spaces to some category $\mathcal{D}$ with property X?  Even better, can we assume that $LCH$ is a subcategory of $\mathcal{D}$ and that $T$ is a forgetful functor?
I don't care to specify what I mean by "nontrivial", except that the "identity" functor from $LCH$ to itself doesn't count.  I want it to be genuinely easier to decide whether or not a morphism is an isomorphism in $\mathcal{D}$.  If there happen to be lots of ways to do this, perhaps it will help to know that my interest comes from some problems in analysis.
Thanks in advance!

REPLY [6 votes]: The property $X$, as you call it, is well-known. A functor with this propery is said to "reflect isomorphisms". Another example of such a functor is the geometric realization functor from simplicial sets to compactly generated Hausdorff spaces. There are all sorts of ways of building a category $D$ and a functor $T$ with the properties you want, however, depending on what you want to do, different answers can be more suiting. For example, you can let $D$ be the category $Sh(CH)$ of sheaves on the site of compact Hausdorff spaces and $T$, be "Yoneda": if $Y$ is a $LCH$ space, then $T(Y)$ is the sheaf that assigns each compact Hausdorff space $X$ the set $Hom(X,Y)$. Then, it is a simple exercise to verify that $T$ is fully-faithful and that $T(f)$ is an isomorphism implies that $f$ is (SINCE every locally compact Hausdorff space is compactly generated).<|endoftext|>
TITLE: When is a stack (NOT) geometric?
QUESTION [8 upvotes]: Following the terminology of $n$-Lab, a geometric stack $\mathcal{X}$ on a site $\mathcal{(C,J)}$ is a stack for which there exists a representable epimorphism $X \to \mathcal{X}$ from an object $X$ of $\mathcal{C}$ (viewed as a representable presheaf). Equivalently, $\mathcal{X}$ is geometric if and only if there exists a (nice enough) groupoid object $\mathcal{G}$ in $\mathcal{C}$ such that $\mathcal{X}$ is (2-iso to) the stackification of the strict presheaf of groupoids $Hom(blank,\mathcal{G})$ (where nice enough essentially means that you can take enough iterated pullbacks in $\mathcal{C}$ to form a $\mathcal{C}$-enriched nerve).
My question is, is there a more intrinsic definition of geometric stack? By "more-intrinsic" I mean a definition that does not use the existential quantifier. For example, if our site is topological spaces, we know a presheaf is representable if and only if it sends colimits in $Top$ to limits in $Set$. Since geometric stacks are in some sense a natural generalization of representable presheaves, it would seem natural to expect a similar characterization of geometric stacks (at least in the case when our site is nice enough, like $Top$).
I ask this mostly because, although in some circumstances there is a natural atlas or a natural choice of representing groupoid object around to try to prove that something is a geometric stack, proving that a stack is NOT geometric becomes very difficult when the definition involves the EXISTENCE of a nice atlas.
If someone only knows the answer for certain sites, this is still interesting to me.

REPLY [4 votes]: One problem -- or at least one characteristic aspect -- of the notion of geometric stack is of course that it makes explcit reference to a fixed chosen site. Different sites may give rise to equivalent toposes and still to different notions of geometric stacks.
One approach is to make that extra information an explicit extra piece of data in a controlled way. This is effectively what is achieved by the notion of geometry for a structured topos. In terms of this one can then characterize geometric stacks fairly intrinsically. For instance in Structured Spaces it is shown that with a standard choice for "geometry" a Deligne-Mumford stack is precisely a "2-scheme" in a suitable sense.
Going beyond that, one could ask which "geometries" in this sense are naturally associated to a given topos, without choosing them by hand, such that the corresponding 2-schemes are the natural notion of geometric stack.
I think a big step in that direction is achieved in Bertrand Toen's work Champs affine. As reviewed at rational homotopy theory in an (oo,1)-topos, Toen there shows that for stacks or higher stacks on the algebraic site, one can characterize "affine stacks" intrinsically, as the objects of the reflective sub-(oo,1)-category on objects that are local with respect to morphisms that induce isomorphisms in "rational cohomology", where "rational" is as seen by the ground field. 
Using that intrinsic notion of "affine stack", Toen then gives in section 4 a definition of geometric oo-stacks.
This may or may not be exactly what you are asking for, but I think it does provide some noteworthy indications of the kind of approach that  one should think about.<|endoftext|>
TITLE: Defining Quotient Bundles
QUESTION [15 upvotes]: This is an extremely elementary question but I just can't seem to get things to work out.  What I am looking for is a natural definition of the quotient bundle of a subbundle    $E'\subset E$ of     $\mathbb R$ (say) vector bundles over a fixed base space    $B$.
Every source I find on this essentially leaves the construction to the reader.  I would like to glue together sets of the form    $U\times E_x/E'_x$ where    $x\in U$ is a locally trivial neighborhood by some sort of transition function derived from those corresponding to    $E$ and    $E'$, but this doesn't actually make sense in any meaningful way.
While I am tagging this as differential geometry, I would like a construction that works in the topological category (i.e., does not invoke Riemannian metrics) and avoids passing to the category of locally free sheafs.  
Sorry if this is a repost (I'm sure it is, but I can't seem to find anything) and thanks in advance.

REPLY [10 votes]: (I was going to leave this as a comment but decided that it's a bit long for that)
A couple of remarks:

You express an aversion to Riemannian metrics because you want to be able to apply this in the topological category.  That's fine, except for two things: firstly, Riemannian metrics would not be explicitly involved in this construction as it is a general construction that applies to all vector bundles, not just tangent bundles.  Secondly, having inner products on the fibres of a vector bundle is not something that is special to the smooth category.  Using a partition of unity argument (assuming you're working over a sufficiently nice space, or your vector bundle is a pull-back of a universal one - look up "numerable cover" for more on this - but note that all the answers to this question tacitly assume this), any finite dimensional vector bundle admits a continuous choice of inner product on its fibres.  So the standard argument: "choose an orthogonal structure and take the orthogonal complement" works equally well in the continuous category as the smooth one.
What is really going on here is a reduction of structure group.  The structure group of the big bundle is $Gl(n)$.  The inclusion of the subbundle implies that it reduces to the subgroup that preserves $\mathbb{R}^k \subseteq \mathbb{R}^n$.  At this point, you should work out what this subgroup consists of - think in terms of matrices if you don't see it immediately.  General Nonsense (although for this case, the more junior Lieutenant Nonsense will do) implies that this subgroup is homotopy equivalent to $Gl(k) \times Gl(n-k)$.  A reduction to this defines an isomorphism $E \cong E' \oplus E''$, where $E''$ is the required quotient bundle.  The two previous answers can be viewed as constructing this reduction.  The "standard orthogonality" argument rests on the observation that $Gl(m) \cong O(m)$ for any $m$ so we can reduce everything to the corresponding orthogonal group.  Thus we start with $O(n)$, reduce to the subgroup that preserves $\mathbb{R}^k$ and then ... but there is no "and then" because this subgroup is already $O(k) \times O(n-k)$.  So either way, you are doing one "reduction of structure group", the difference between the two methods is simply a choice of doing $Gl(n) \to O(n)$ at the start, or doing $Gl(n;k) \to Gl(k) \times Gl(n-k)$ in the middle.<|endoftext|>
TITLE: Free resolution dimension?
QUESTION [9 upvotes]: Is there a notion of a dimension associated to free resolutions like projective and injective dimensions associated to projective and injective resolutions? My guess is that it coincides with projective dimension but couldn't find any reference talking about this. Any help or tips will be appreciated!

REPLY [8 votes]: When requiring finitely-generatedness of the resolution, then the free dimension of a projective module can be infinite.
As a simple example, take the ring $R=k\oplus k$, and let $e_1=(1,0)$ and $e_2=(0,1)$ be the obvious idempotents.  Then the module $ke_1$ is projective.  However, any surjection
$$ R^{\oplus n}\rightarrow ke_1 $$
will have a kernel $S_1$ which is isomorphic to $R^{n-1}\oplus ke_2$.  Therefore, any surjection
$$ R^{\oplus m}\rightarrow S_1 $$
will have a kernel $S_2$ which is isomorphic to $R^{m-n}\oplus ke_1$.  Thus, any f.g. free resolution of $ke_1$ must infinite.
Of course, as noted above, $ke_1$ has an infinitely-generated free resolution of length 1, as it is projective.<|endoftext|>
TITLE: When are Ehrhart functions of compact convex sets polynomials?
QUESTION [13 upvotes]: Given a lattice $L$ and a subset $P\subset \mathbb R^d$, we define for each positive integer $t$ $$f_P(L,t)=|(tP\cap L)|$$ the number of lattice points in $tP$. Let's say $P$ is nice if $f_P(L,t)$ is a polynomial. We know that if $P$ is a convex polytope with vertices in $L$ then $P$ is nice and $f_P(L,t)$ is its Ehrhart polynomial. My question is about some converse of this statement.

Are there some mild assumptions (for example convexity etc.) on $P$, under which if $f_P(L,t)$ is a polynomial with respect to at least some lattice $L$ then $P$ must be a convex polytope? Or a weaker question: Is any polynomial arising this way also the Ehrhart polynomial of some polytope?

P.S. I haven't thought much about this question so I apologize if it is well-known or it has an obvious negative answer. Also feel free to retag.

Richard Stanley suggested the following in the comments (edited to take into account a trivial family of counter-examples):

Could the following be true? It seems more in line with the question. Let $P$ be a compact convex $n$-dimensional set in $\mathbb R^n$. Suppose that the Ehrhart function $f_P(t)$ is a polynomial for positive integers $t$. Then $P$ is a translation of a rational polytope.


Edit: I would also be interested in a slightly weaker statement: Suppose a convex set has positive curvature almost everywhere, must the Ehrhart function necessarily be non-polynomial?
For example given an arbitrary lattice, what would be the easiest way to see that a circle doesnt have a polynomial Ehrhart function?

REPLY [3 votes]: I believe that the strong form of the conjecture is false. In lieu of a simple counterexample, let me point you towards a centrally symmetric 10-gon $\hat P$ in arXiv:0801.2812, Figure 6. It is a bit of a mess to explain exactly what it is, but it has something to do with Picard lattice of a toric DM stack. It need not be rational or a translate of rational.
The key properties of $\hat P$ are 
(1) It is centrally symmetric.
(2) The midpoints of all the sides are lattice points. As a result, opposite sides are 
lattice translates of each other.
As a result, generic translates of $\hat P$ have the same number of lattice points. Indeed,
as you move the polytope in a plane along a general curve as soon as a point appears on one side of it, another point exits from the opposite side. This implies that the opposite sides of $\hat P$ glue together to give a "no-gaps" cover of the torus $\mathbb R^2/L$ (preimage of a generic point has the same cardinality $k$). Then if one takes a $t$-multiple of it, one gets a "no-gaps" cover of $\mathbb R^2/tL$, and will thus have $kt^2$ points in $t(\hat P+ c)$ for a generic shift $c$.
I assume that this construction can be simplified to give something more explicit and palatable, so long as the property that the opposite sides are lattice translates of each other is satisfied.
It clearly requires flat sides to be able to glue them together on the torus, so this idea is not going to work for the positive curvature problem.<|endoftext|>
TITLE: A ring such that all projectives are stably free but not all projectives are free? 
QUESTION [19 upvotes]: This question is motivated by this recent question. Suppose $R$ is commutative, Noetherian ring and $M$ a finitely generated $R$-module. Let $FD(M)$ and $PD(M)$ be the shortest length of free and projective resolutions (with all modules f.g.)  of $M$ respectively. 
We will be interested in two conditions on $R$:
(1) For all f.g. $M$, $PD(M)<\infty$ if and only if $FD(M)<\infty$.
(2) For all f.g. $M$, $PD(M)=FD(M)$. 
It is not hard to see that one is equivalent to (3): all f.g. projective modules are stably free. Also, (2) is equivalent to (4): all f.g projectives are free. It is natural to ask whether (1) and (2) are equivalent. I don't think they are, but can't find a counter-example. Thus:

Can we find a commutative Noetherian ring which satisfies (3) but not (4)?  

Some thoughts: (3) is equivalent to the Grothendick group of projective $K_0(R)=\mathbb Z$. Well-known class of rings satisfying (3): local rings or polynomial rings over fields. 
Of course, there are well known rings which fails (4): coordinate rings of $n$-spheres $R_n=\mathbb R[x_0,\cdots,x_n]/(x_0^2+\cdots+x_n^2-1)$ for $n$ even. Unfortunately, for those rings we have $K_0(R_n)=\mathbb Z^2$. (except for $n=8r+6$ which do provide examples, see below)
UPDATE: it seems to me that Tyler's answer suggests $R_5$ may work, but one needs to check some details. 
UPDATE 2: I think one can put together a more or less complete solution, based on Tyler and Torsten's answers, the comments and some digging online. 
Claim: $R_n$ satisfies (3) but not (4) if $n=5,6$ or $n=8r+k$ with $r>0$, $k\in \{3,5,6,7\}$ 
Proof: Let $S^n$ be the $n$-sphere and $C(S^n)$ be the ring of continuous functions.
To show that $R_n$ satisfies (3), it suffices to show $K_0(R_n)=\mathbb Z$ (Lemma 2.1, Chapter 2 of Weibel's K-book) or the reduced group $\tilde K_0(R_n)=0$. But one has isomorphism (known to Swan, see for example 5.7 of this paper):
$$\tilde K_0(R_n) \cong \tilde KO_0(S^n) $$
the reduced $K$-groups of real vector bundles, and it is well-known that $\tilde KO_0(S^n) =0$ if $n\equiv 3,5,6,7 (\text{mod} 8)$.  
It remains to show our choices of $R_n$ fail (4). Let $T_n$ be the kernel of the surjection $R_n^{\oplus n+1} \to R$ defined by the column of all the $x_i$s. $R$ is projective, so the sequence splits, thus $T_n$ is stably free. 
On the other hand, $R_n$ embeds in $C(S^n)$ and tensoring with $C(S^n)$ turns $T_n$ into the tangent bundle of $S^n$. If $T_n$ is free, said bundle has to be trivial. But $S_n$ has trivial tangent bundle if and only if $n=1,3,7$.
While it is not surprising that topological machinery is helpful, I would still be interested in seeing a pure algebraic example, so if you have one, please post.

REPLY [19 votes]: The more canonical example probably is the standard universal example for such a question. So, let $R_n=k[x_i,y_i]/\sum x_iy_i=1$  where $k$ is any field and there are $2n$ variables. By localization one easily checks that $K_0(R_n)=\mathbb{Z}$ for any $n$. But the projective module given by the presentation,
$$0\to R_n\stackrel{(x_i,\ldots,x_n)}{\to} R_n^n\to P\to 0$$
is clearly stably free but not free if $n\geq 3$. An algebraic proof (given by myself and Madhav Nori) can be found in the article of Swan (Annals of Math studues, vol 113, pp 432-522).

REPLY [3 votes]: Here is an example of a non commutative ring which satisfies (3) but not (4): the integral group ring  $\mathbb{Z}[C_\infty \times Q_8]$ where $Q_8$ is the quaternion group of order 8 and $C_\infty$ is the infinite cyclic group. Using the methods developed by Swan in (Projective resolutions over binary polyhedral groups, Journal fur die Reine und Ang. Math. 342 66---172 (1983)), Frank Johnson at UCL has shown:
(i) $K_0(\mathbb{Z}[C_\infty \times Q_8])\cong K_0(\mathbb{Z}[Q_8])\cong \mathbb{Z}$ (which is (3));
(ii) $\mathbb{Z}[C_\infty \times Q_8]$ has infinitely many isomorphically distinct stably free modules of rank 1. So (4) fails quite badly. 
The proof involves decomposing the ring as a fibre product and then analyzing the module structure a la chapter 2 of Milnor's Algebraic K-Theory.
Swan's paper gives examples of integral group rings of finite groups with stably free, non-free modules. However there are no finite commutative groups whose integral group rings have non-trivial stably frees. T.Y. Lam's excellent book on Serre's problem has lots of information on this kind of thing.<|endoftext|>
TITLE: How does the Lefschetz-Poincare dual torsion linking pairing on manifolds with boundary interact with the maps of the long exact sequence of the manifold-boundary pair?
QUESTION [9 upvotes]: I'm wondering if anyone can point me to a reference on how the various 
Lefschetz-Poincare dual torsion pairings of a manifold with boundary fit 
together.
To explain in more detail, consider a connected compact oriented 
n-manifold $M$ with boundary. Then we have the various dualities with 
rational coefficients $H_i(M;Q)=H_{n-i}(M, \partial M; Q)$ and
$H_i(\partial M;Q)=H_{n-i-1}(\partial M; Q)$, but we also obtain, from some easy 
tinkering with the long exact sequence of the pair and other basic 
observations that, e.g. $im(H_i(M)\to H_i(M, \partial M))$ is dual to 
$im(H_{n-i}(M)\to H_{n-i}(M, \partial M))$ (in fact this is how we define 
signatures on manifolds with boundary) and similar results hold for the 
other "shared" terms in the long exact sequence.
I'd like to know more about the generalization of this over Z and, in 
particular, about what torsion pairings to Q/Z exist on the various 
images, cokernels, etc. of the long exact sequence and which are 
nonsingular. Does anyone know of any place in the literature where this 
is worked out?
Thanks!
Updated:
Thanks, Tom, for your response. I've been thinking about it, and while I agree that the situation is much murkier over Z, I think there are still some things that can be said.
For example, it's true that the map $H_n(M)\to H_n(M, bd M)$ (let's assume $n$ is the middle dimension to simplify the discussion), only yields a nondegenerate pairing on the image (which you call $A$) mod torsion  and not a perfect (nonsingular) pairing. But now since $H_n(M)\to A/\text{torsion}$ has a free group as its image, we have a (non-unique) splitting that lets us consider $A/\text{torsion}$ as a direct summand of $H_n(M)$. Let's fix this summand for now (the choice turns out not to matter). Since $H_n(M)/\text{torsion}$ and $H_n(M, bd M)/\text{torsion}$ are (perfectly) Z-dual, $A/\text{torsion}$ must be dual to something in $H_n(M, bd M)/\text{torsion}$, and I claim that the thing it's dual to is isomorphic to the kernel of $H_n(M,bd M)\to H_{n-1}(bd M)/\text{torsion}$ (mod torsion). Notice that any non-torsion element of this kernel has a multiple represented by a cycle in $M$, and so it must have non-zero intersection number with any cycle in the boundary, and this shows that as far as intersection pairings are concerned, the choice of $A/\text{torsion}$ as a summand of $H_n(M)$ doesn't matter. I don't really want to get into the details of the rest of my claim here, but the basic idea should be that these are the things that should be dual once we tensor with Q but over Z we need to make sure we have all the appropriate Z-primitives, which I'm pretty sure this does (since if a multiple of $x\in H_n(M,bd M)$ is in $A$, then $x$ becomes torsion in $H_{n-1}(M)$).
So what's the point of all this? Well, even over Z we can say that there is a nonsingular pairing between a certain cokernel and a certain kernel, as long as we mod out the torsion in the right places. I suspect that there's a more subtle version of this for the torsion linking pairings where instead of quotienting out all torsion we just kill certain torsion subgroups. I think I've seen things somewhat of this nature in papers relating Witt groups of Z pairings to Witt groups of Q/Z pairings.

REPLY [9 votes]: I don't know a reference, but I have some thoughts. I'm going to tackle this using just homology. I'll ignore gradings in the name of legibility.
There are three kinds of homology in this story: $H(M)$, $H(M,b M)$, and $H(b M)$. The last of these is self-dual; the other two are dual to each other. What I just said is literally true over $Q$, with 'dual to' meaning 'canonically isomorphic to the vector-space dual of'. Over $Z$ of course it's only true in a derived sense. One consequence of the derived statement is that, of the free abelian groups $H(M)/{torsion}$, $H(M,b M)/{torsion}$, and $H(b M)/{torsion}$, the last one is self-dual in the $Hom(-,Z)$ sense while the other two are dual to each other. Another is that the torsion part of $H(b M)$ is self-dual in the $Hom(-,Q/Z)$ sense and the torsion parts of the others are dual to each other in the same sense.
Then there are the other three players: the images of the three maps $H(M)\rightarrow H(M,b M)\rightarrow H(b M)\rightarrow H(M)$. Call them $A$, $B$, and $C$. (Each of the three can also be described as a kernel, or as a cokernel.) Here the rational story is very nice: $A$ is self-dual (more precisely the perfect pairing between $H(M)$ and $H(M, b M)$ when restricted to be a pairing between $H(M)$ and $A$ yields a pairing between $H(M)/ker=A$ and $A$ which is perfect). And $B$ and $C$ are dual to each other (more precisely, in the perfect self-pairing of $H(b M)$ the subspace $B$ is its own orthogonal complement, so that $B$ and $C=H(b M)/B$ are dual).
Over $Z$ things become murkier for $A$, $B$, and $C$. It's true that we get a pairing of $A$ with itself, and that this yields a nondegenerate pairing between the free abelian group $A/{torsion}$ and itself. But it's not perfect; it just injects $A$ into $Hom(A,Z)$. The same goes for pairing $B/{tors}$ with $C/{tors}$. And as for pairings of the torsion parts into $Q/Z$, the images just don't seem to behave well in general.
Update in response to Greg's update to the question: I think the following is basically what you're saying, but it can be said just algebraically and has nothing to do with choosing a splitting (i. e. identifying a quotient of $H(M)/tors$ with a subgroup). The sequence of free abelian groups $H(M)/tors\to H(M,bM)/tors\to H(bM)/tors$ is not exact. The image of $H(M)/tors\to H(M,bM)/tors$, which is the same as my $A/tors$, is not a summand in general; the kernel of $H(M,bM)/tors\to H(bM)/tors$ is a summand and consists of all elements such that some multiple is in $A/tors$; call this $A'$. So $A'$ contains $A/tors$ with finite index. And we get a perfect Z-pairing between $A/tors$ and $A'$. Likewise we get $B'$ and $C'$ containing $B/tors$ and $C/tors$ with finite index, and Z-dual to $C/tors$ and $B/tors$. I don't know what to say about all of this in relation to pairings into $Q/Z$.<|endoftext|>
TITLE: Simple explicit example of local Jacquet-Langlands theorem for inner forms of GL(n), and consequences
QUESTION [11 upvotes]: This one will be very easy for the experts.
Let $F$ be a nonarch local field, let $n\geq1$ be an integer, choose $0\leq d<n$ and let $D$ be the central simple algebra over $F$ with invariant $d/n$ in the Brauer group [EDIT: and with $F$-dimension $n^2$, so e.g. if $gcd(d,n)>1$ then $D$ will not be a division algebra but rather a matrix algebra over the division algebra with invariant $d/n$]. Let $G$ be the algebraic group $D^\times$ and let $\pi_0$ denote the trivial 1-dimensional representation of $G(F)$.
The local Jacquet-Langlands theorem guarantees us the existence of a smooth irreducible 
representation $\pi=JL(\pi_0)$ of the group $GL(n,F)$ canonically associated to $\pi_0$. This construction gives a completely canonical map from the set $\{0,1/n,2/n,\ldots,(n-1)/n\}$ to the set of smooth irreducible representations of $GL(n,F)$, sending $d/n$ to $\pi$.
If you put a gun to my head and asked me to guess what $\pi=\pi(d/n)$ was in this situation, I would probably go for the following construction: set $h=gcd(d,n)$, Let $P$ be the standard parabolic in $GL(n)$ whose Levi is $GL(h)^{n/h}$, and (non-normally) induce the trivial 1-dimensional representation of $P$ up to $G$; such a representation will, I suspect, have a canonical "biggest" irreducible subquotient, corresponding on the Galois side to an $n$-dimensional representation of the Weil-Deligne group of $F$ which is a direct sum of $h$ representations of degree $n/h$ each of which is Steinberg (in the sense that $N$ is maximally unipotent). I only envisage this because I can't imagine any other such map which agrees with what I know in the $GL(2)$ case!
Here is a consequence of my guess: if $d$ is coprime to $n$ then the trivial 1-dimensional representation of $D^\times$ corresponds to the Steinberg representation of $GL(n)$, whatever $d$ is. This makes me wonder whether the following is true: say $d_1$ and $d_2$ are both coprime to $n$ and let $G_i$ be the group of units of the central simple algebra over $F$ with invariant $d_i/n$. Are the smooth irreducible representations of $G_i$ canonically in bijection with one another? Does this remain true if I relax the condition that the $d_i$ are coprime to $n$ but instead only demand that $gcd(d_1,n)=gcd(d_2,n)$? 
More generally is it true that if $gcd(d_1,n)$ divides $gcd(d_2,n)$ then there's a canonical injection from the irreps of $G_1$ to the irreps of $G_2$, which is a bijection iff the gcd's coincide?
I don't know where in the literature to look for such statements :-/ so I ask here.

REPLY [2 votes]: Sorry Kevin, I read your question too quickly. For non essentially square integrable representations you have to give a meaning to the Jacquet-Langlands correspondence. This is done in Badulescu's article:
"Jacquet-Langlands et unitarisabilité", Journal de l'Institut de Mathématiques de Jussieu, 2007, 
where the correspondence is generalized to the Grothendieck group of finite length representations. Moreover Badulescu proves that the extended correspondence is compatible with the Zelevinsky-Aubert duality and one knows that the dual of the trivial representation is a (generalized ???) Steinberg representation.  So you're right : one should obtain a (generalized) Steinberg representation.<|endoftext|>
TITLE: profinite spaces are the pro-completion of finite sets
QUESTION [5 upvotes]: The title sounds tautological, right? Perhaps I'm missing something completely trivial here ...
Assume $X$ is a compact totally disconnected hausdorff space. It is known that $X$ can be written as directed inverse limit of finite discrete spaces $X_i$ with surjective transition maps (i.e. $X$ is profinite). How do you prove that every map from $X$ to a finite discrete space factors through some projection $X \to X_i$?
I know that the fibers of the projections are a basis of the topology of $X$ (not only a subbasis). The corresponding result for profinite groups is true, but I cannot adopt the proof.
Of course you could use Stone duality to reduce the assertions to a completely trivial one (a finite boolean ring in a directed limit of boolean subrings lies in some of these boolean subrings), but I want a direct topological proof.

REPLY [7 votes]: Let $f:X \to Z$ be a map to a finite discrete space. Note that each fiber, $f^{-1}(z)$, is both open and closed in $X$. Let $p_i: X \to X_i$ be the projection maps. 
Fix some $z \in Z$. Since $f^{-1}(z)$ is open, and the fibers of the maps are a basis, there is an open cover of $f^{-1}(z)$ by sets of the form $p_i^{-1}(x)$, for $x$ in various $X_i$. 
Since $f^{-1}(z)$ is closed in a compact space, it is compact. So we can take a finite subcover of this cover. Thus, there is some single index $i$ for which $f^{-1}(z)$ is covered by sets of the form $p_i^{-1}(x)$, $x \in X_i$. 
Since $Z$ is finite, there is a single $i$ such that, for every $z \in Z$, the fiber $f^{-1}(z)$ is covered by sets of the form $p_i^{-1}(x)$, $x \in X_i$. The map $f$ factors through $X_i$.

REPLY [3 votes]: The key to generalizing the proof for groups is to argue that the space is a uniform space. In the case of a group, everything can be translated into neighborhoods of the identity. To do the same in the general case, we work with neighborhoods of the diagonal Δ = {(x,x) : x ∈ X}.
When X is an inverse limit of finite discrete spaces pi:X→Xi, then the preimages of the finite diagonals Ei = pi-1(Δi) form a fundamental system of entourages for X; this follows directly from the universal property of inverse limits. Now consider a partition U1,...,Uk of X into pairwise disjoint clopen sets, then U1×U1 ∪ ... ∪ Uk×Uk is a clopen neighborhood of the diagonal Δ. This neighborhood must contain one of the entourages Ei, which gives the required factorization.
See also my answer to question 15440 where I similarly characterize spaces that are inverse limits of discrete spaces.<|endoftext|>
TITLE: Existence of a pair of matrices in SL(2,Z) satisfying certain constraints on the spectral radius
QUESTION [10 upvotes]: Some background: my coauthors and I are working on a problem which deals with the exponential growth rates of certain infinite products of matrices. One of the sub-problems which arises in this project is to prove that a particular function from the unit interval to the reals, which describes the growth rates of a family of related infinite matrix products, does not have any line segments in its graph. Now, the existence of a line segment would imply the existence of a pair of matrices some of whose characteristics coincide in a precise way. 
We have found an analytic proof that the graph cannot contain line segments, but it is long and displeasingly ad-hoc, and we would rather like to replace it with a shorter argument! Hence, we have found ourselves wondering whether there is an algebraic obstacle which would prevent these coincidences from occurring. 
The question which arises from this research, then, is:

Do there exist matrices $A,B \in SL(2,\mathbb{Z})$ which satisfy the following constraints: the matrices $A$ and $B$ have only positive entries, do not commute, and satisfy
  $$\rho(AB)=\rho(A)\rho(B),$$
  $$\mathbb{Q}\left(\rho(A)\right)=\mathbb{Q}\left(\rho(B)\right) \neq \mathbb{Q},$$
  where $\rho$ denotes spectral radius? 

The non-existence of such a pair of matrices would imply the non-existence of the line segments mentioned above. 
One of my coauthors suspects that the spectral radius constraints on $A$ and $B$ cannot be met unless $A=C^k$, $B=C^\ell$ for some $C \in SL(2,\mathbb{Z})$ and $k,\ell \in \mathbb{N}$, which would contradict the hypothesis that $A$ and $B$ do not commute. Any solutions which end up being used in our paper will of course be gratefully acknowledged!

REPLY [10 votes]: This is true: if one has two matrices $A, B \in SL_2 C$, and one knows the three traces $tr(A),tr(B),tr(AB)$,
then this uniquely determines the matrices $A,B$ up to conjugacy if $A$ and $B$ generate a non-elementary discrete group (except in 
a few degenerate cases which won't occur for positive matrices). See for example Maclachlan-Reid. Since $tr(X) = \rho(X)+\rho(X)^{-1}$ for a positive matrix $X$, this implies that the three matrices have the same traces as two commuting matrices with the same traces,
and therefore $A,B$ must be conjugate to a pair of commuting matrices. 
In fact, as your coauthor suspects, then they must be powers of the same matrix. 
Addendum: I'll give some details of the argument, and I forgot a condition ($A$ and $B$ might generate a solvable group otherwise, although this is not possible for discrete groups). 
Assume that $A, B \in SL_2 Z$. We want to show that $tr(A),tr(B),tr(AB)$
uniquely determine $A$ and $B$ up to conjugacy in $SL_2 C$. 
Since $A$ is a positive matrix in $SL_2 Z$, we may conjugate it in $SL_2 R$ to be of the form
$$
A=\begin{pmatrix} \lambda & 0 \\\ 0 & \lambda^{-1} \end{pmatrix}
$$
where $\lambda >1$ (because $tr(A)>2$). We conjugate $B$ by the same matrix, and relabel so that 
$$
B=\begin{pmatrix} a & b \\\ c & d \end{pmatrix}.
$$
If $A$ and $B$ commute, then we check that $b=c=0$, so in particular $ad=1$. Conversely, if $ad=1$, then we conclude that $bc=0$, and therefore $b=c=0$ since $A,B$ cannot generate a noncommutative solvable group since $SL_2 Z$ is discrete (see below).
From now on, let's assume that $A$ and $B$ don't commute. 
The centralizer of $A$ in $SL_2 C$ consist of matrices of the
form 
$$
\begin{pmatrix} z & 0 \\\ 0 & z^{-1} \end{pmatrix}.
$$
We may conjugate $B$ by the centralizer of $A$ to be of the form 
$$
B=\begin{pmatrix} a & 1 \\\ ad-1 & d \end{pmatrix}.
$$
Here I'm making use of the fact that $A,B$ are conjugate into  $SL_2 Z$ to conclude that they 
don't generate a solvable group. This follows from the Margulis lemma (or an easy exercise). 
If $B$ had off-diagonal entries which were
zero, then $\langle A, B\rangle$ would be solvable. The point is that then we may
conjugate by elements of the centralizer of $A$ in $SL_2 C$ to arrange the upper
right entry to be 1. 
Thus, we have 
$$tr(A)=\lambda+\lambda^{-1}, tr(B)=a+d, tr(AB)=\lambda a +\lambda^{-1} d.$$
Now we compute 
$$a= \frac{\lambda tr(AB)-tr(B)}{\lambda^2-1} , d= \frac{\lambda^{-1} tr(AB)-tr(B)}{\lambda^{-2}-1}.$$
Thus, $tr(A), tr(B), tr(AB)$ canonically determine $A,B$ up to conjugacy (of course $\lambda$ is uniquely determined by $tr(A)$ given $\lambda >1$). 
Now, we need to check that $\rho(AB) \neq \rho(A)\rho(B)$, where $\rho(A)=\lambda, \rho(B)=\mu$. If not, $tr(AB)=\lambda\mu +(\lambda\mu)^{-1}$ (again since $AB$ is positive). 
Plugging into the above equations, we conclude that $a=\mu, d=\mu^{-1}$, a contradiction. 
I'm sure this argument can be simplified, but I wanted to carry through the sketch of the argument I gave the first go around. 
Also, one may see that any subgroup of commuting matrices in $SL_2 Z$ is either powers of a parabolic element, a hyperbolic element, or a finite-order element.<|endoftext|>
TITLE: Geometric interpretation of the fundamental groupoid
QUESTION [6 upvotes]: Motivation
The common functors from topological spaces to other categories have geometric interpretations. For example, the fundamental group is how loops behave in the space, and higher homotopy groups are how higher dimensional spheres behave (up to homotopy in both cases, of course). Even better, for nice enough spaces the (integral) homology groups count $n$-dimensional holes.

A groupoid is a category where all morphisms are invertible. Given a space $X$, the fundamental groupoid of $X$, $\Pi_1(X)$, is the category whose objects are the points of $X$ and the morphisms are homotopy classes of maps rel end points. It's clear that $\Pi_1(X)$ is a groupoid and the group object at $x \in X$ is simply the fundamental group $\pi_1(X,x)$. My question is:

Is there a geometric interpretation $\Pi_1(X)$ analogous to the geometric interpretation of homotopy groups and homology groups explained above?

REPLY [2 votes]: Thanks Andrew for the nice comments! 
In relation to the comment on G-spaces by Donu, I should point out that Chapter 11 of "Topology and groupoids" is on  "Orbit spaces, orbit groupoids". But I doublt many topologists are aware of the latter concept! 
My new jointly authored book `Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids' is published in August 2011 by the European Math. Soc and distributed in the Amercas by the AMS from October 2011. It is a kind of sequel to the above book, exploring some major uses of groupoids in higher homotopy theory.  A kind of new foundation for algebraic topology at the border between homotopy and homology, and applications of some higher order Seifert-van Kampen Theorems. In particular, the 2-dimensional version allows some quite explicit calculations of homotopy 2-types.  
See my web site for more infomration.<|endoftext|>
TITLE: Separable sigma-algebra: equivalence of two definitions
QUESTION [13 upvotes]: The two definitions alluded to in the title can be found here: http://en.wikipedia.org/wiki/Separable_sigma_algebra (one is that the $\sigma$-algebra is countably generated, the other is pretty much the standard usage of the word separable wrt the semi-metric given by the measure). Why are they equivalent?

REPLY [8 votes]: [Edited 10/22/15: there were embarrassing errors in my original answer, which were pointed out in another question; I've corrected these, and I've written up further details in a blog post.]

I'm late to the party, but here's my two cents.  References in what follows are to

[Ha] P.R. Halmos, Measure Theory, Springer, 1950.
[Ro] V.A. Rokhlin, On the fundamental ideas of measure theory, Transl. AMS, Series 1, No. 10 (1952), 1-54.  (The original Russian article is from 1949.  A pdf of the English translation is presently available here.)
[JDH] will denote the answer already given to this question by Joel David Hamkins.

Consider a measured $\sigma$-algebra $(S,\mu)$.  Assume that $\mu$ is normalised to have total weight 1, and that $S$ is complete (contains all subsets of null sets).
In [Ha], $(S,\mu)$ is said to be separable if it has a countable subset that is dense w.r.t. the metric $\rho(A,B) = \mu(A\bigtriangleup B)$.  We denote this property by (S).
In [Ro], $(S,\mu)$ is said to be separable if it has a countable subset $\Gamma$ such that for every $A\in S$, there exists $B\in \sigma(\Gamma)$ such that $A\subset B$ and $\mu(B\setminus A) = 0$.  Here $\sigma(\Gamma)$ is the $\sigma$-algebra generated by $\Gamma$.  (In fact, Rokhlin's definition is given for a measure space, not just a measured $\sigma$-algebra, and requires that $\Gamma$ separate points of the space.)  Since we're already using the word "separable" for (S), let's say that in this case $(S,\mu)$ is one-sided countably generated mod zero, and denote this property by (CG0+).  
So that's two conditions.  Another natural condition would be that $S$ itself is countably generated, that is, that $S = \sigma(\Gamma)$ for some countable $\Gamma$; call this (CG), and not that it applies to the Borel $\sigma$-algebra on $[0,1]$, but not the Lebesgue $\sigma$-algebra.  The latter satisfies the weaker condition (CG0+), and hence is also countably generated mod zero, meaning that there is a countable $\Gamma \subset S$ such that for every $A\in S$, there exists $B\in \sigma(\Gamma)$ such that $\mu(A \bigtriangleup B) = 0$.
(In the previous version of this answer I got carried away and also defined (S1), one-sided separability, to be the property that every $A\in S$ and $\epsilon>0$ there exists $B\in \Gamma$ such that $A\subset B$ and $\mu(B\setminus A) < \epsilon$.  As was correctly pointed out by Rina Shora and Nik Weaver on another question, this fails to hold for even the most standard examples.)
It is immediate that (CG) $\Rightarrow$ (CG0+) $\Rightarrow$ (CG0), and [JDH] shows that (CG0) $\Rightarrow$ (S) (it is written with (CG) in mind, but works just as well for (CG0)).  In fact one can also show that (S) $\Rightarrow$ (CG0) (details are in this blog post) -- that is, separability is equivalent to being countably generated mod zero -- and that the first two implications above are strict (the Lebesgue $\sigma$-algebra satisfies (CG0+) but not (CG), and the $\sigma$-algebra of Lebesgue subsets of $[0,1]$ with measure 0 or 1 satisfies (CG0) but not (CG0+)).
Finally, another related condition would be that $S$ is contained in the completion of a countably generated $\sigma$-algebra; this property lies somewhere between (CG) and (CG0), but its relation to (CG0+) is not clear to me.
The example in [JDH] involving the club filter shows that (S) (and hence (CG0)) is not enough to imply this condition.<|endoftext|>
TITLE: When does a unitary Hilbert space rep of a reductive Lie group decompose into a direct sum of irreps with finite multiplicities?
QUESTION [8 upvotes]: I'm giving some lectures on the trace formula. Here's something I proved in the last lecture. Let $G$ be a locally compact Hausdorff unimodular topological group (e.g. a reductive Lie group), let $\Gamma$ be a discrete subgroup with $\Gamma\backslash G$ compact, and let $H$ be $L^2(\Gamma\backslash G)$, considered as a representation of $G$ via the right regular action.
Then $H$ is a Hilbert space direct sum $H=\oplus m_\pi\pi$ with $\pi$ running through irreducible unitary Hilbert space reps of $G$ and each $m_\pi<\infty$.
Although I don't think I need it for my course, it seems to me that all I really used about $H$ was that it was a unitary Hilbert space rep of $G$ and that if $f\in C_c(G)$ (continuous functions on $G$ with compact support) then (fixing a Haar measure on $G$) the induced action of $f$ on $H$ is a Hilbert-Schmidt operator. I hope that's right because I'm no expert! The point is that you can use a kernel function argument to prove that if $H=L^2(\Gamma\backslash G)$ as above then each $f$ acts via a Hilbert-Schmidt operator, but that's all you seem to need in the proof, which now goes through for any $H$ with this property.
I am minded to actually revisit the proof and remark that the arguments all (seem to) go through in this generality---not least because I'm not sure in what generality I'll need this decomposition later.
I am also minded to make the following definition:
Definition: a unitary Hilbert space rep of $G$ is Hilbert-Schmidt if each $f\in C_c(G)$ acts via a Hilbert-Schmidt operator.
Then the theorem is that Hilbert-Schmidt reps decompose into direct sums of irreps each occurring with finite multiplicity.
I am not at all sure that this is standard terminology though! Is there standard terminology to describe such reps? That's the question! Even less mathematical: is $L^2(\Gamma\backslash G)$ a representation which has some standard property $P$ (I want $P$ to be "Hilbert-Schmidt!") and the real theorem is that every rep with property $P$ decomposes into irreps each occurring with finite multiplicity?
Finally a technical point: if $H$ is a unitary rep of $G$ and $f\in C_c(G)$ then I want to define an action of $f$ on $H$ by $fv=\int_Gf(g)gv dv$. The question is how to give this integral a formal definition. Initially I had got the following definition in my mind: $fv$ is the unique element of $H$ with the property that $(fv,w)=\int_Gf(g)(gv,w) dg$ for all $w$. But now I realise there might be a "genuine" definition of $fv$ involving approximating $f$ by step functions and so on and so on. Am I right?

REPLY [4 votes]: A bit belatedly, but perhaps marginally usefully: first, yes, indeed, all you needed was that $f\in C^o_c(G)$ gave a compact operator, not even necessarily Hilbert-Schmidt or trace-class, tho' the latter does lead to further interesting things.
Second, about integrals, Gelfand and Pettis (c. 1928) effectively created a very nice notion of "integral" which works wonderfully for continuous, compactly-supported functions with values in any locally-convex, quasi-complete tvs. (These integrals are sometimes called "weak", but this is misleading in several ways.) The characterization is "weak" in the sense that $\int_X f\in V$ is uniquely determined by the fact that, for every continuous linear functional $\lambda$ on $V$, $\lambda(\int_X f)=\int_X \lambda\circ f$, where the scalar-valued integral of continuous, compactly-supported is unambiguous. One also proves that the integral is in the closure of the convex hull of the image $f(X)$, which gives a grip on "estimates". One immediate corollary is that for any continuous linear $T:V\rightarrow W$, the integral of $Tf$ is $T$ of the integral of $f$, which leads to justification of differentiation under integrals, and such. Further, giving operators on a Hilbert space the "strong operator topology" (not norm...) by $p(T)=\sup_{|x|\le 1} |Tx|$, which is what makes $G\times V\rightarrow V$ continuous, actually the operator-valued integral $\int_G f(g)\,T(g)\,dg$ makes sense, for $f\in C^o_c(G)$. E.g., see  here . Various natural continuations of this, such as vector-valued holomorphic functions, were treated by Schwartz and Grothendieck. E.g., see  here . Very handy on occasion.
Third, about repns "decomposing discretely, with finite multiplicities". I think for practical purposes it's not so clear what a "decomposition" would mean outside the Hilbert space context, although notions of compact or trace-class or nuclear operators have senses in larger contexts. I've heard gossip about concerted efforts at Yale in the 1950s to be able to talk about "spectral theory" in more general contexts, but it seems that it just doesn't work very well beyond Hilbert spaces, and the Fredholm alternative for special operators on Banach spaces. That such a decomposition succeeds for $L^2(compact-quotient)$ was arguably known to several people in the 1950s already: probably Gelfand et alia, but also Selberg and others, and certainly Langlands by the early 1960s. Probably those people would say that the compact quotient case was "obvious", and that the issue of serious interest was the non-compact quotient case, where one has to do some serious work to prove that the operators are still trace-class on cuspforms. Probably Gelfand-PS gave the first more-or-less proof of that, although the reader has pretty substantial responsibilities there.
The "continuous" decompositions we know for general unitary repns of type I groups are not very useful, unfortunately, in that they give no particular information. Indeed, one can execute the proof that Eisenstein series span various bits of continuous spectra... literally decomposing $L^2$ functions... without even formulating the general notion, somewhat like we can prove Fourier inversion without explaining how to view $L^2(\mathbb R)$ as the Hilbert direct integral of one-dimensional repns...
Edit: after @pm's answer, I realized I was not clear in what I wrote, at best:
First, yes, I was thinking only of $G$ unimodular.
Second, for many applications one wants to take trace, and trace class is a proper subset of Hilbert-Schmidt is a proper subset of "compact", altho' the composition of two Hilbert-Schmidt is trace class (perhaps by definition). 
Again, the most difficult issue is proving trace class _on_cuspforms_ for integral operators attached to not-necessarily very smooth functions on a Lie group (e.g.), which Langlands did in the 1960s, I think roughly the same time as Gelfand-PS, tho' perhaps much worse documented. Adele-group-or-not is not the key point, I think. I am not aware of any systematic approach to not-co-compact $\Gamma$ in general topological groups $G$, only for reductive Lie or adele or similar.
Re: chronology, one should note that Harish-Chandra proved sharp versions of admissibility of unitaries of reductive Lie groups in the 1950s, tho' I do not know whether his results explicitly mentioned "trace class" issues.
And, yes, since the composition of two Hilbert-Schmidt ops is trace-class, the Cartier/Dixmier-Malliavin result that all smooth functions are finite sums of convolutions of smooth functions certainly crushes certain issues. 
Many interesting things here! :)<|endoftext|>
TITLE: What kind of line bundles have Chern class of Hodge type (2,0) or (0,2)?
QUESTION [5 upvotes]: If $L$ is a complex line bundle on a topological space $X$, let $c_1(L)$ denote the image of its Chern class in $H^2(X;C)$.  A complex manifold structure on $X$ [ok which is also compact and say algebraic] splits $H^2(X,C)$ into $H^{2,0}$, $H^{1,1}$, and $H^{0,2}$.  If $L$ can be given a compatible complex structure, then $c_1(L)$ belongs to $H^{1,1}$.  It's rewarding to study the moduli space of isomorphism classes of compatible complex structures on a given $L$--for instance, this is another finite-dimensional complex manifold in a natural way.
One not-very-specific way to explain what a "compatible complex structure" is, is to say that it's a tensor on the total space of L, satisfying a differential equation.  Is there a natural kind of differential-geometric structure like this that one can put on a line bundle that guarantees that c_1(L) belongs to $H^{2,0}$ or $H^{0,2}$?  Is there an interesting moduli space of such structures?
Tim points out that the question is silly, because $H^{2,0}$ and $H^{0,2}$ do not intersect H^2(X;Z) (which classifies line bundles per Donu), but that there might be some integral classes in $H^{2,0} + H^{0,2}$, or the part of it that's stable by complex conjugation.  Is there a structure on a line bundle that lands its Chern class there?

REPLY [3 votes]: When $(X,g,\omega)$ is a compact Kaehler surface, the integral classes representable as the real part of a (2,0)-form can be represented as the curvatures of anti-instantons.
One has two splittings of the $g$-harmonic 2-forms $\mathcal{H}^2_g(X)$: the Hodge-theoretic one, and the one determined by the Hodge star,
$$  \mathcal{H}^2_g(X) = \mathcal{H}^+_g(X) + \mathcal{H}^-_g(X), $$
into self-dual and anti-self-dual harmonic forms. 
The relation with the Hodge decomposition is that
$$ \mathcal{H}^+ = \mathbb{R}\omega \oplus (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}} $$
and
$$ \mathcal{H}^- = (\mathcal{H}^{1,1}_0)_{\mathbb{R}}$$
(trace-free part of the real (1,1)-forms).  We can check this pointwise, where it's linear algebra. The wonderful thing about Hodge theory is that it then implies the corresponding cohomology-level statement. (Note that $\mathcal{H}^\pm$ are maximal positive-definite (+) and negative-definite (-) subspaces of the wedge-product form on $H^2(X;\mathbb{R})$. So as a by-product we  get the Hodge index theorem.) 
Suppose that there's an integral class $c$ in $ (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}}$. It's represented by a hermitian line bundle $L_c$, and (by Hodge theory again) we can choose a unitary connection in $L_c$ whose curvature is harmonic and hence self-dual. 
Corrected: This connection, which is called an abelian anti-instanton, is not quite unique, because we could add to it $i$ times any closed 1-form; but if this closed form is exact, or represents $2\pi$ times an integral cohomology class, then the new connection is gauge-equivalent to the original one. Hence, the space of anti-instantons mod gauge is an torsor for the torus $H^1(X;S^1)$. The catch, I think, is that for generic Kaehler metrics in a fixed Kaehler class, there will be no such integral classes unless $\mathcal{H}^{1,1}_0=0$.<|endoftext|>
TITLE: Factors of p-1 when p is prime.
QUESTION [17 upvotes]: Hi,
I have no idea where to look for, so I'm hoping you can give me some pointers.
I'm interested by numbers of form $p-1$ when $p$ is a prime number. Do they have a name, so that I can google them?
More precisely, I'm interested in their factors. Ok, obviously 2 is a factor, but what about the others? Are there a lot of small factors? Do we know the rate of the growth of its larger factor, is it linear, logarithmic?
Thanks.

REPLY [6 votes]: You should look at the paper by Erdos and Odlyzko "On the density of odd integers of the form $(p − 1)/2^n$ and related questions" in Journal of number theory, vol. 11 (1979) pp 257-263.  Among other things they prove the the set of odd divisors of $p-1$ (where $p$ runs over primes) has a positive density.
As far as the largest prime factor of $p-1$ there is a result of Fouvry "Theoreme de Brun-Titchmarsh: application au theoreme de Fermat", Invent. Math. v. 79 (1985) pp 383-407, in which he proves that the set of primes $p$ for which the largest prime factor of $p-1$ is $\ge p^{.6687}$ has positive relative density in the set of all primes.<|endoftext|>
TITLE: Is a simple loop in a spine of a strongly irreducible Heegaard splitting primitive in the fundamental group?
QUESTION [5 upvotes]: Let $\gamma$ be a simple loop in a spine of a strongly irreducible Heegaard splitting of a closed 3-manifold $M$ with torsion-free fundamental group.  Does $\gamma$ necessarily correspond to a primitive element of the fundamental group of $M$, or is it possible for $\gamma$ to be a power of some other element?
I suspect that $\gamma$ is not necessarily primitive in the fundamental group of $M$, but I do not know of any examples.
Edit: I added torsion-free fundamental group after Charlie's comment.  In particular, I am most interested in this question for hyperbolic 3-manifolds.

REPLY [4 votes]: New Answer: Take a 2-bridge knot, and perform hyperbolic
Dehn filling (so that the core of the Dehn filling is geodesic),
and so that the filling slope has intersection number $>1$ with
the meridian. Then the meridian will not be primitive, since it
will be a multiple of the core of the Dehn filling. 2-bridge
knots have a genus two Heegaard splitting, which has a spine
for the handlebody which is a wedge of two meridians
at the bottom. This remains a spine in the
Dehn filling, so the loops represented by the meridians are not
primitive. This also works for the (2,n) torus knots (which are
2-bridge), so I think Charlie's answer is right (at least for
many small Seifert fibered-spaces). 
Old (non)Answer:
Here's almost an example. All punctured torus bundles have
Heegaard genus $\leq 3$, and many have Heegaard genus 3 (I discussed
this once in my defunct blog). One may find a genus 3 Heegaard splitting
of any once punctured torus bundle by taking two copies of a
fiber, tubing them together along the boundary on one side,
and adding a handle to the other side (drill out discs from
both fibers, and glue an annulus in). By a theorem of 
Moriah-Rubinstein, most Dehn fillings will also have
Heegaard genus 3 if the punctured torus bundle does. 
The Heegaard splitting of the punctured torus bundle 
has one side which is a handlebody,
and the other side a compression body. We may think of the
handlebody as a product neighborhood of the fiber (which is
a punctured torus) with a 1-handle attached. We may find a
spine for the handlebody which consists of a wedge of two
loops which is a spine for the punctured torus, together with
another loop going through the 1-handle. 
Now, the peripheral curve of the punctured torus is not primitive
in Dehn fillings along curves which intersect the longitude
multiple times. This curve is represented in the spine not
as an embedded curve, but has multiplicity two (since it is
a commutator of the generators). If we choose a small punctured torus bundle
of genus three, most Dehn fillings will be small (non-Haken) of
Heegaard genus 3, and so this Heegaard surface will be strongly
irreducible. But the peripheral curve will not be primitive,
since it will be a multiple of the core of the Dehn filling. However,
it is not embedded in the spine. 
Even though this doesn't answer the question, it gives a strategy
for trying to find an example. Namely, if one can find a 1-cusped
hyperbolic 3-manifold which is small, and contains an incompressible
surface with boundary, such that the boundary slope is a generator in
the surface,
and such that a tubular neighborhood of the surface (or a slight modification by
drilling a hole in a tubular neighborhood)
is a minimal genus Heegaard splitting, then many Dehn fillings will
have the desired property.<|endoftext|>
TITLE: minimize the sum of absolute eigenvalues
QUESTION [6 upvotes]: Hi,
We have a real symmetric matrix M with diagonal elements 0's, the eigenvalues and eigenvectors of M are computed.
Now we wish to change its diagonal elements arbitrarily to minimize the sum of absolute eigenvalues. Does there exist a way to find such modifications?
If we add a constraint : keep Tr(M)=0, would that become easier?
Is there some topics about these?
Thank you for your help
Zhi Ming

REPLY [3 votes]: Interesting question. Nuclear norm minimization is getting much attention right now as it relates directly to compressed sensing.
Some software for minimization with this constraint that I've used:
http://perception.csl.illinois.edu/matrix-rank/sample_code.html
A fun related problem:
http://www-stat.stanford.edu/~candes/papers/MatrixCompletion.pdf<|endoftext|>
TITLE: How do I describe the GL_n torsor attached to a smooth morphism of relative dimension n?
QUESTION [10 upvotes]: Edit: It seems I had two different constructions mixed up in my head, namely the frame torsor and the automorphism bundle of a vector bundle.  This made the main question a bit confusing.  The first two auxiliary questions were about the frame torsor, and the last one was about the automorphism bundle.  If anyone knows a published reference for either construction, I would still be most appreciative.
The original question is below the line:

I feel like I should have learned this in grad school, but I never encountered a construction.
Let $n$ be a positive integer, and let $f: Y \to X$ be a smooth morphism of schemes of relative dimension $n$.  The sheaf $\Omega_{Y/X}$ is then a rank $n$ locally free $\mathcal{O}_Y$-module.  Using the symmetric algebra functor, we can form the associated rank $n$ vector bundle $\mathbf{V}(\Omega_{Y/X}) = \operatorname{Spec}_Y \operatorname{Sym}_{\mathcal{O}_Y} \Omega_{Y/X}$ (cf. EGA2 1.7.8).  I've heard it called the bundle of 1-jets, which ought to mean tangent bundle, but I'm always confused by this, so maybe it's the cotangent bundle.
Main question: Is there a reference for the construction of the commutative $\mathcal{O}_Y$-algebra $A$ for which $\operatorname{Spec}_Y(A)$ is the $GL_{n,Y}$-torsor $P$ of automorphisms of $\mathbf{V}(\Omega_{Y/X})$?  Specifically, I'd like the torsor to satisfy the property that I can retrieve 1-jets by the associated bundle construction: $\mathbf{V}(\Omega_{Y/X}) \cong P \overset{GL_{n,Y}}{\times} \mathcal{O}_Y^{\oplus n}$
This can be viewed as a question about constructing the automorphism torsor of any bundle, but 1-jets seem to have specific structural features that may make a more specialized construction possible.  For example, it should be a quotient of some canonical infinite-dimensional torsor of coordinates coming from the Gelfand-Kazhdan formal geometry theory.
Auxiliary questions (not as important):

Is there a concise description of the functor the torsor represents, e.g., are $S$-points on the torsor equal to $S$-points $g:S \to Y$ equipped with isomorphisms $\mathcal{O}_S^{\oplus n} \to g^*\Omega_{Y/X}$?
Is there a nice way to describe the $GL_{n,Y}$-action (since writing an explicit comodule structure sounds like it could be a mess)?
I would be interested in seeing how the torsor can be cut out of the rank $n^2$ bundle of endomorphisms by inverting determinants.

REPLY [5 votes]: If $V$ is a vector bundle of rank $n$, the corresponding universal algebra $A$ which makes $V$ trivial (i.e. $V \otimes A \cong A^n$), or equivalently the algebra of the corresponding $\mathrm{GL}_n$-torsor, is given by
$$A = \mathrm{Sym}(V^n) \otimes_{\mathrm{Sym}(\Lambda^n V)} \mathrm{Sym}^{\mathbb{Z}}(\Lambda^n V).$$ 
Here, we define $\mathrm{Sym}^{\mathbb{Z}}(\mathcal{L})=\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z}$ for the line bundle $\mathcal{L}=\Lambda^n V$, and the tensor product is taken with respect to the morphism $\delta : \mathcal{L} \to \mathrm{Sym}^n(V^n)$ which maps $v_1 \wedge \dotsc \wedge v_n$ to $\sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} \iota_i(v_{\sigma(i)})$. This description is global in nature and actually generalizes to arbitrary cocomplete linear tensor categories. Some details can be found in my thesis, Section 4.9.
The idea of the construction of $A$ is the following: $\mathrm{Sym}(V^n)$ is the universal algebra $B$ with a morphism of $B$-modules $V \otimes B \to B^n$. Then we construct $B \to A$ so that the determinant of this morphism becomes invertible over $A$, so that $V \otimes A \cong A^n$. 
We could also construct $A$ as a quotient of $\mathrm{Sym}(V^n) \otimes \mathrm{Sym}((V^*)^n)$, which introduces morphisms $V \otimes A \to V^n$ and $A^n \to V \otimes A$, and the quotient should be made in such a way that these morphisms become inverse to each other.<|endoftext|>
TITLE: A recent talk by Kollar on cohomology of structure sheaves
QUESTION [10 upvotes]: Recently I noticed an intriguing talk by  Kollár at the MAGIC conference. The abstract says:

Title: Cohomology groups of structure sheaves
Abstract: I will discuss the behavior of cohomology groups of the structure sheaf and of the dualizing sheaf under deformations and birational maps.

This sounds very interesting to me and can be potentially helpful for my research. However, I could not find any further information about the talk, except for some pictures, from which I shamefully failed to reconstruct the materials!
Would any expert who happened to be there please point me to some references for which the talk is based on, or better yet, explain briefly the motivations and key results? Many thanks.

REPLY [7 votes]: I would guess, he was lecturing about this article: http://arxiv.org/abs/0902.0648<|endoftext|>
TITLE: For which values of $N$ is known the Lieb-Simon Inequality for $Z_N$ Models ?
QUESTION [5 upvotes]: Background:
Let $\mathbb Z^d$ denote the $d$-dimensional integer lattice with norm $|x|=\sum_i|x_i|$. 
For each $x\in\mathbb Z^d$ we associate a spin variable, $\sigma_x$ taking values on the set 
$\{\exp(2\pi i j/N);1\leq j\leq N\}$ with uniform a priori distribution. 
Consider the formal Hamiltonian given on the lattice $\mathbb Z^d$ by
$$
H_{ \Lambda }(\{\sigma\}) = -\sum_{\langle x,y\rangle} J_{xy} \vec\sigma_{x} \vec\sigma_{y}    
$$
where $J_{xy}$ are nonnegative constants and $\vec\sigma_x \vec\sigma_y$ is inner product in $\mathbb R^2$. 
The sum is taken over all pair of first neighbors $\langle x,y\rangle$ means that $|x-y|=1$. The Partition function on a finite
$\Lambda\subset \mathbb Z^d$ is given by
$$ 
Z_\Lambda = \int\exp\Big( \beta \sum_{\langle x,y\rangle \in \Lambda }J_{xy} \vec\sigma_x \vec\sigma_y\Big)\;d\sigma 
$$
where the integral is taken over all sites of $\Lambda$. The two point correlations are given by
$$
\langle \vec\sigma_x \vec\sigma_y \rangle_{\Lambda}=
Z_{\Lambda}^{-1}\int\vec\sigma_x \vec\sigma_y\exp\Big( \beta \sum_{\langle x,y\rangle \in \Lambda }J_{xy} \vec\sigma_x \vec\sigma_y\Big) d\sigma 
$$
 Question:
For which values of $N$ is known that the Lieb-Simon Inequality is true or false ?
 Lieb-Simon Inequality
$$
\langle \vec\sigma_x \vec\sigma_y \rangle_{\Lambda} \leq  \sum_{b \in \partial B} \bigl<\vec\sigma_x \vec\sigma_b \bigr>_B  \bigl<\vec\sigma_b \vec\sigma_y  \bigr>_\Lambda,
$$
where $B\subset\Lambda\subset\mathbb Z^d$ are finite, $x,y\in\Lambda$, $\partial B=\{z\in B; d(z,B^c)=1\}$ and $\partial B$ separates $x$ and $y$ ($i.e.$ any path from $x$ to $y$ must intercept $\partial B$).

REPLY [2 votes]: As far as I know, it has only been proved for $N=1$ (Lieb) and $N=2$ (Lieb+Rivasseau) in the form you want. With an additional prefactor $\beta/N$ and with the infinite-volume measure in the RHS, it has been extended to $N=3$ and $N=4$ by Aizenman and Simon (see also Spohn and Zwerger). All these proofs rely on various correlation inequalities that, to my knowledge, have not been extended to general $O(N)$ models.<|endoftext|>
TITLE: Spanning polytopes
QUESTION [10 upvotes]: Hamiltonian cycles (seen as spanning polygons) are interesting for several reasons (only a few of which I am aware of), but especially because not every connected graph has a Hamiltonian cycle (is Hamiltonian), so the characterization of Hamiltonian graphs becomes interesting (see wikipedia article on Hamiltonian paths). 
Side note: Each platonic and archimedian solid is Hamiltonian. 
What about spanning polytopes, as
one possible generalization of
Hamiltonian cycles = spanning polygons?
(By "spanning polytope" I mean a spanning subgraph that is the 1-skeleton of a polytope of arbitrary dimension.)
There are connected graphs without spanning polytopes (trees obviously), but there are non-Hamiltonian graphs that have a spanning polytope of dimension d>2, e.g. the Herschel graph.
A google search for "spanning polytope" yields only very few and unrelated results, so my question is:

Is there research on this or a related
  topic, only under another name?

If not so, does this have an obvious - or not so obvious - reason?

REPLY [4 votes]: Unfortunately, there is no known good characterization of which graphs are the 1-skeleton of some polytope in some dimension d. There are several interesting properties of polytope graphs that may be of interest to you:
$\bullet$ In $\mathbb{R}^3$, Steinitz's theorem characterizes all polytopal graphs as the planar 3-connected graphs.
$\bullet$ In d > 3, the cyclic polytopes have $K_n$ as a 1-skeleton. A result of Perles implies that every graph $G$, $G$+$K_n$ is a d-polytopal graph for some n and some d, although it is unknown what the minimum n is for a given $G$.
$\bullet$ It is a theorem of Balinski that the graph of a polytope in $\mathbb{R}^d$ is d-connected, so graphs with polytopal spanning subgraphs must have a highly connected subgraph.
There are also results about how to determine whether a specific graph $G$ is a polytopal graph, but they do not extend into a classification of all polytopal graphs.
There have been studies into classes of abstract polytope graphs, classes which contain all the 1-skeletons of polytopes but likely contain graphs with no polytopal realization, for example in this paper of Eisenbrand, Haehnle, and Rothboss Diameter of Polyhedra: Limits of Abstraction 
For more information, check out Gil Kalai's chapter 19 in the Handbook of Discrete and Computational Geometry by Jacob E. Goodman and Joseph O'Rourke (or Gil Kalai's excellent blog gilkalai.wordpress.com/ ) or Grunbaum's Convex Polyopes.<|endoftext|>
TITLE: Twistings for other cohomology theories
QUESTION [22 upvotes]: Twistings in cohomology theories have a long history and have been used to great effect. The classical example is cohomology with local coefficients. Using this one can formulate Poincaré duality and the Thom isomorphism for unorientable manifolds. More recently, twisted K-theory has been linked to string theory (as B-fields) and the representation theory of loop groups (the Freed-Hopkins-Teleman theorem).
However, cohomology and K-theory are the only examples I know of where twistings are used regularly. What is known for twistings of other cohomology theories, e.g. the one associated to cobordism? In this case, one would expect a similarity to K-theory, from the Conner-Floyd theorem. Do these have geometric interpretations? What are the applications of such twistings?

REPLY [18 votes]: Twisted forms exist for all multiplicative generalized cohomology theories. A nice paper which discusses a modern point of view for twists of homology, K-theory, and TMF is the following paper Twists of K-theory and TMF by Ando-Blumberg-Gepner.
If E is a generalized cohomology theory, represented by a spectrum also denoted E, then the E-cohomology of X coincides with the homotopy classes of maps 
$$[ \Sigma^{-i} X, E] $$
i.e. the "E-valued functions on X". Morally, if E is a ring spectrum then we can talk about E-module spectra, and about E-lines: those E-module spectra which are equivalent to E, but not necessarily canonically so. With a sufficiently robust theory, we should be able to talk about bundles of spectra over X, and in particular E-line bundles over X. Then the usual E-cohomology of X can be thought of as the sections of the trivial E-line bundle over X. An E-twist is a possibly non-trivial E-line bundle over X, and twisted E-cohomology consists of the sections of this line bundle. 
The tricky part is making this philosophical picture into something mathematically precise. The above paper is one way to do this. 
In general the E-lines are classified by the space $BGL_1(E)$, which is the classifying space of the $A_\infty$-space $GL_1(E)$. This space is defined by the pull-back diagram

GL_1(E) --> Ω∞E
   |        |
   v        v
 π0(E)x --> π0(E)

From this you can read off the homotopy groups of $BGL_1(E)$ and you see that for $n \geq 2$ they agree with those of E, but are shifted in degree. 
More generally, when E is a commutative ring spectrum, one can study the larger class of "E-lines" which are invertible E-modules. This requires a robust theory of spectra where you have a good notion of smash product over E. This leads to a larger classifying space of E-lines whose zeroth homotopy group is the Picard group Pic(E). Even more generally, you could consider bundles of general E-modules (not necessarily invertible) to be twists. There are probably applications of this, but I don't recall any off-hand. 
As far as geometric descriptions go, you might be asking for too much. Even for K-theory it is only the simplest kinds of twists corresponding to the bottom few homotopy groups of $BGL_1(K)$ which appear to have a clear geometric description (e.g. in terms of super gerbes and clifford algebras). The higher twists of K-theory are more subtle and it is not a priori clear that they have a purely geometric description.<|endoftext|>
TITLE: Are there complexity classes with provably no complete problems?
QUESTION [24 upvotes]: A problem is said to be complete for a complexity class $\mathcal{C}$ if a) it is in $\mathcal{C}$ and b) every problem in $\mathcal{C}$ is log-space reducible to it.  There are natural examples of NP-complete problems (SAT), P-complete problems (circuit-value), NL-complete problems (reachability), and so on.
Papadimitrou states that "semantic" complexity classes like (I think) BPP are less likely to admit complete problems. Then again, we do not actually know whether P = BPP or not, and if so, there would be BPP-complete problems. (There exist IP-complete problems, since IP=PSPACE, and determining whether a quantified boolean formula is satisfiable is PSPACE-complete.)

Question: Are there any natural complexity classes that can be shown not to have complete problems?

I think this question has to be modified slightly, because I would imagine $Time(n)$ has no complete problems because log-space reductions can introduce a polynomial factor. 
So, in the word "natural," I include the assumption that the complexity class should be invariant under polynomial transformations (I don't know how to make this precise, but hopefully it is clear).
(Also, time or space bounds should be at least $\log n$, of course.)
Edit: A commenter has pointed me to an interesting result of Sipser that $\mathrm{BPP}^M$ for $M$ a suitable oracle does not have complete problems.  Is the same true for a (less fancy) class of the form $\bigcup_f TIME(f)$, where the union is over a class of recursive functions $f$ that are all polynomially related to each other? (Same for $\bigcup_f NTIME(f)$, and ditto for space.

REPLY [4 votes]: This is more of a comment than an answer, but the comment go too long. From this thread, there seem to be two different themes to coming up with classes without complete problems. 
Completeness is defined using two properties. L is X-complete if
(1) L is in X
(2) L is X-hard (under some suitable notion of efficient reductions)
The first theme involves classes which have hard problems, but if the hard problem were a member of the class itself, it would cause problems. The examples of POLYLOGSPACE and ELEMENTARY fall in this category. Both have hard problem, of course, but if the hard problem were a member of the class, some hierarchy theorem would be violated. (Space hierarchy and time hierarchy theorems, respectively.) Similarly one could come up with more examples of this kind.
The second theme involves classes which have no hard problems, such as ALL or P/poly. These classes don't have a complete problem for a fundamentally different reason than the previous case.
It would be interesting to see if there are other classes which fail to have complete problems for completely different reasons.<|endoftext|>
TITLE: Rational numbers with dense  orbits in [0,1] under iteration by f(x)=4x(1-x)
QUESTION [11 upvotes]: Let $f(x)=4x(1-x)$.
For which rational numbers $r\in [0,1]$ is the sequence $f^n(r)$, $n\in \mathbb N$, dense in $[0,1]$ ?
$(f^n(r)=f\circ f\circ ...\circ f(r)$ n times)
I would be happy to find a single rational number with dense orbit in $[0,1]$, but my guess is that all rational numbers different from k/2^n should work. The numbers 1/3, 1/5, 1/10 are candidates (at least numerically).
It is known that a.e. point x in [0,1] has a dense orbit (w.r.t the Lebesgue measure). This is shown by conjugating f to x->2x mod 1 with the map $x\rightarrow sin^2(\pi x/2)$. So the question can be rephrased as: for which rational r does ${1\over \pi} Arcsin(r)$ have a dense orbit in [0,1] under the action of x->2x mod 1 ? (this does not seem simpler though)
EDIT: from W. Zudilin answer, it seems that the question is open in full generality. But maybe there is a chance of finding just one rational $r$ such that $f^n(r)$, or $Frac({2^n\over \pi} Arcsin\ r ) $,  is dense in [0,1] ?

REPLY [5 votes]: Ian Morris has it essentially correct in his comment.  If you solve the equation $x = (\sin \pi y/2)^2$ for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under $y \mapsto 2y$ is dense if and only if every finite binary string appears in the binary expansion of $y$.  Now, this is weaker than being normal in base 2, because that requires that every binary string appears equally often, not just that it appears.  Let's call such a number "topologically 2-normal" (or 2-dense could be another name), because the 2-normality condition is equivalent to saying that the orbit of $y$ is not just dense, but ergodic.  My impression is that not much more is known about topologically normal numbers than about normal numbers.  For instance, you can conjecture that any irrational algebraic number is topologically normal in base 2 (or in any other base), but it doesn't look like it is known.  In any case, topological normality is the heart of the question.<|endoftext|>
TITLE: When is a TQFT the dimensional reduction of a higher dimensional TQFT?
QUESTION [23 upvotes]: In Lurie's framework for TQFT's, a TQFT is a symmetric mondoial functor from $Cob_n(n)$ to some symmetric monoidal $n$-category $\mathcal{C}$. One can construct an $(n-1)$-dimensional TQFT from an $n$-dimensional TQFT by taking the product of a cobordism with a circle first: $Red(F)(M) = F(M \times S^1)$. 
My first question is: what are necessary or sufficient conditions for a TQFT to be the dimensional reduction of a higher-dimensional TQFT?
As an illustration, consider the following. For example, the 2-dimensional TQFT $K^\tau_G(G)$ for a compact simple Lie group $G$ arises as the dimensional reduction of Chern-Simons theory. In Consistent Orientation of Moduli Spaces, Freed-Hopkins-Teleman remark that the fact that $K^\tau_G(G)$ can be refined to be defined over $\mathbb{Z}$ instead of $\mathbb{C}$ "reflects that the theory is a dimensional reduction". Can this statement be made precise?

REPLY [14 votes]: Asking for sufficient conditions that an $n$-dimensional TQFT lifts to $n+1$ dimensions is like asking for sufficient conditions that an egg will hatch into a bird.  You might think of necessary conditions.  The only reasonable sufficient condition is to see the egg hatch, i.e., to see the lift or categorification of the TQFT into the next dimension.  Lest that sound like a fatuous pronouncement, the reason is that an $(n+1)$-dimensional TQFT is often vastly more complicated than its $n$-dimensional predecessor.  Certainly a Chern-Simons TQFT in 3 dimensions is vastly more complicated than its dimensional reduction.
As for necessary conditions, that depends on what kind of bird you expect.  Suppose that you are set on a specific target category $\mathcal{C}$.  Then you know that $F(M \times S^1)$ is the trace of the identity functor on $M$, and that gives you a lot of information.  For example, in the dimensional reduction of a Chern-Simons TQFT, you always assign a non-negative integer to a closed surface.  You have to, because its lift is vector space.
The problem is that $\mathcal{C}$ might be negotiable.  Suppose that you have a 2D TQFT in which you assign a negative integer to a surface.  If know that that should lift to a vector space, that's impossible.  But it could instead lift to a supervector space (a graded vector space with its signed dimension), or a chain complex, or something else.  If the value is not even an integer, that might not be a show-stopper either:  Its lift could be a graded space with a Poincaré-Hilbert series, and there could be a formal sum of that series with a non-integer answer.  One important moral example is Khovanov homology, even though it is not a complete TQFT.  Khovanov homology categorifies the Jones polynomial.  A Laurent polynomial with both signs is an inauspicious candidate for categorification, but the categorification exists.<|endoftext|>
TITLE: Why do non-equioriented A<sub>n</sub> quivers have singularities identical to the singularities of Schubert varieties?
QUESTION [7 upvotes]: In the general case, quiver cycles are of the form of orbit closures of $GL\cdot V_{\vec{r}}$, where $GL= \prod_{i=0}^n GL_{r_i}$ is the possible changes of basis on all of the vector spaces on each of the vertices and $V_{\vec{r}}$ is any representation of the quiver with fixed dimension vector $\vec{r}$.  In the equioriented An case, these are well understood by Zelevinsky and Lakshmibai-Magyar by showing them isomorphic to open sets in Schubert varieties.  Bobinski and Zwara claim to reduce the non-equioriented case to the equioriented case, but I don't see how they are doing that.  
In the introduction to ``Normality of Orbit Closures for Dynkin Quivers" (manuscripta math. 2001), Bobinski and Zwara say that they will generalize the result that equioriented An quivers have the same singularities as Schubert varieties to non-equioriented An quivers.  They claim that they will do this by reducing the non-equioriented case to the equioriented case.  So far, so good.  But then, they say that this result follows from the proposition that they will prove, which I don't see has to do with the theorem at all.  
The proposition is about a Dynkin quiver, Q, of type Ap+q+1 with p arrows in one direction and q arrows in the other and Q' an equioriented Dynkin quiver of type Ap+2q+1, their respective path algebras B=kQ and A=kQ', and respective Auslander-Reiten quivers &GammaB and &GammaA over the category of finite dimensional left modules over A and B.  The proposition says ``Let A=kQ' and B=kQ be the path algebras of quivers Q' and Q, respectively, where Q and Q' are Dynkin quivers of type A.  Assume there exists a full embedding of translation quivers $F: \Gamma_B \to \Gamma_A$.  Then there exists a hom-controlled exact functor $\mathcal{F}: \text{mod }B \to \text{mod }A$."  
Can anyone tell me how (or if) their results translate into a result that tells me a recipe for constructing a Kazhdan-Lustzig variety from my non-equioriented quiver?  (By K-L variety, I mean a Schubert variety intersect an opposite Bruhat cell.)  Alternately, is there a way to see which particular sub-variety of the representation variety of equioriented Ap+2q+1 I get out of this theorem and how that is (maybe a GIT quotient away from) a Kazhdan-Lustzig variety?
Thanks,
Anna

REPLY [2 votes]: The relevance of hom-controlled functors comes from Zwara's paper "Smooth morphisms of module schemes" (Theorem 1.2). The definition there is that two schemes with basepoints $(X,x)$ and $(Y,y)$ have identical singularities if there is a smooth morphism $f \colon X \to Y$ such that $f(x) = f(y)$. 
Let $F$ be a hom-controlled functor. He shows that when we're dealing with module varieties, and $X$ is an orbit closure $\overline{O}_M$ with basepoint $x$ some closed point of $O_N$ (so $x$ represents the isomorphism class of a module $N$), then $(\overline{O}\_M, x)$ has identical singularities as $(\overline{O}\_{FM}, y)$ where $y$ is a closed point of $O_{FN}$.
So if one starts with non-equioriented ${\rm A}_n$ and picks an orbit closure $\overline{O}_M$ together with a closed point in it, then one knows that there is a smooth morphism to some orbit closure in a bigger ${\rm A}_m$. The orbit closure is just the image of $M$ under the hom-controlled functor constructed in Bobinski and Zwara's paper (though this construction is long and I don't remember the details). Then one can use Lakshmibai–Magyar to get a smooth morphism from this orbit closure to some Schubert variety.
So it's enough to understand how to construct $F$, which I remember being explicit but requiring quite a few steps, if we just want the varieties together with the singularities, but constructing the smooth morphism itself would take a lot more digging to construct explicitly.<|endoftext|>
TITLE: Is the smallest primitive root modulo p a primitive root modulo p^2?
QUESTION [15 upvotes]: Let $p \ne 2$ be a prime and $a$ the smallest positive integer that is a primitive root modulo $p$. Is $a$ necessarily a primitive root modulo $p^2$ (and hence modulo all powers of $p$)? I checked this for all $p < 3 \times 10^5$ and it seems to work, but I can't see any sound theoretical reason why it should be the case. What is there to stop the Teichmuller lifts of the elements of $\mathbb{F}_p^\times$ being really small?

REPLY [23 votes]: The key term here is: Wieferich prime base $a$.
What you observed can be presented to children in the following form:  if $p$ is a prime greater than 5 and the fraction $1/p$ has decimal period $d$, numerical tables show $1/p^2$ has decimal period $dp$, $1/p^3$ has decimal period $dp^2$, and generally the decimal period  of $1/p^k$ is $dp^{k-1}$. For example, 1/13 has decimal period 6, 1/169 has decimal period $78 = 6 \cdot 13$, and 1/2197 has decimal period $1014 = 6 \cdot 13^2$.
This works for primes below 100, but if you search far enough you will find a counterexample.  The first one is $p = 487$: 1/487 and $1/487^2$ both have decimal period 486. The second counterexample is $p = 56,598,313$. (!!) This list has been Sloaned: http://oeis.org/A045616.
For a general article about this business, see http://www.jstor.org/stable/3219294.
Within algebraic number theory, this phenomenon appears when you compute the ring of integers of ${\mathbf Q}(\sqrt[n]{2})$, which turns out to be ${\mathbf Z}[\sqrt[n]{2}]$ for all $n \leq 1000$.  With that evidence you might guess the ring of integers is always ${\mathbf Z}[\sqrt[n]{2}]$, just like the ring of integers of ${\mathbf Q}(\zeta_n)$ is always ${\mathbf Z}[\zeta_n]$.  But in fact it's not always true.  There are $n > 1000$ such that ${\mathbf Z}[\sqrt[n]{2}]$ is not the full ring of integers of ${\mathbf Q}(\sqrt[n]{2})$. If you search for Wieferich primes to base 2 you will find them.<|endoftext|>
TITLE: Process equivalent to conditional probability
QUESTION [5 upvotes]: Hi,
Having a random variable $X$ I am trying to find a stochastic process $Z_t$ such that:
$$P[Z_t>T] = P[X > T | X > t]$$
for all $T>t$, or a proof that such a process does not exist.
Please note that this question is not related to any homework and that I actually need this result for my research in financial maths.
Edit
I haven't really mentioned it, but what I am really after is some sort of closed form formula for $Z_t$, ideally as a function of $X$ and $t$.

REPLY [2 votes]: Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct. 
Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is
$$
Z_t=t+(1-t)X. 
$$
In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, 
hence a (pathwise nondecreasing) solution in the general case is
$$
Z_t=G^{-1}(G(t)G(X)). 
$$
Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula
$$
G^{-1}(u)=\inf\{x \vert G(x)\le u\},
$$ 
for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.
A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. A conjugate example is when $X$ follows a power law in the sense that $G(x)=(x_0/x)^a$ for every $x\ge x_0$, for given positive $x_0$ and $a$, then $Z_t=tX/x_0$ for every $t\ge x_0$ and $Z_t=X$ for every $t\le x_0$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.<|endoftext|>
TITLE: Why is $ \frac{\pi^2}{12}=\ln(2)$ not true?
QUESTION [25 upvotes]: This question may sound ridiculous at first sight, but let me please show you all how I arrived at the aforementioned 'identity'.
Let us begin with (one of the many) equalities established by Euler:
$$ f(x) = \frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) $$
as $(a^2-b^2)=(a+b)(a-b)$, we can also write: (EDIT: We can not write this...)
$$ f(x) = \prod_{n=1}^{\infty} \Big(1+\frac{x}{n\pi}\Big) \cdot \prod_{n=1}^{\infty} \Big(1-\frac{x}{n\pi}\Big) $$ 
We now we arrange the terms with $ (n = 1 \land n=-2)$, $ (n = -1 \land n=2$), $( n=3 \land -4)$ , $ (n=-3 \land n=4)$  , ..., $ (n = 2n \land n=-2n-1) $ and $(n=-2n \land n=2n+1)$ together.
After doing so, we multiply the terms accordingly to the arrangement. If we write out the products, we get:
$$ f(x)=\big((1-x/2\pi + x/\pi -x^2/2\pi^2)(1+x/2\pi-x/\pi - x^2/2\pi^2)\big) \cdots $$
$$ 
\cdots \big((1-\frac{x}{(2n)\pi} + \frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(n-1))^2\pi^2})(1+\frac{x}{2n\pi} -\frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(2n-1))^2\pi^2)})\big) $$
Now we equate the $x^2$-term of this infinite product, using 
 Newton's identities (notice that the '$x$'-terms are eliminated) to the $x^2$-term of the Taylor-expansion series of $\frac{\sin(x)}{x}$. So,
$$ -\frac{2}{\pi^2}\Big(\frac{1}{1\cdot2} + \frac{1}{3\cdot4} + \frac{1}{5\cdot6} + \cdots + \frac{1}{2n(2n-1)}\Big) = -\frac{1}{6} $$
Multiplying both sides by $-\pi^2$ and dividing by 2 yields
$$\sum_{n=1}^{\infty} \frac{1}{2n(2n-1)} = \pi^2/12 $$
That (infinite) sum 'also' equates $\ln(2)$, however (According to the last section of  this  paper).
So we find $$ \frac{\pi^2}{12} = \ln(2) . $$
Of course we all know that this is not true (you can verify it by checking the first couple of digits). I'd like to know how much of this method, which I used to arrive at this absurd conclusion, is true, where it goes wrong and how it can be improved to make it work in this and perhaps other cases (series). 
Thanks in advance, 
Max Muller
(note I: 'ln' means 'natural logarithm)
(note II: with 'to make it work' means: 'to find the exact value of)

REPLY [28 votes]: Maybe I'm too late to be of much use to the original question-asker,
but I was surprised to see that all of the previous answers seem to 
not quite address  the real point in this question.


While it is important to be aware of the dangers of rearranging conditionally convergent series, it not true that any rearrangement is invalid, in terms of  changing the value of the sum.

Namely, any finite rearrangement of terms will obviously leave the sum unchanged.
So will any collection of disjoint finite rearrangements.
For example, by a standard Taylor series argument the following sum converges conditionally to $\ln (2)$:
$$ H_{\pm} = \sum_{m\geq 1}(-1)^{m+1}\frac1{m} = 1 - \frac12  + \frac13 - \frac14 + \cdots = \ln(2).$$
(Note that by grouping terms, 
$ H_{\pm} = \sum_{n\geq 1} \left( \frac{1}{2n-1} - \frac{1}{2n}\right)= \sum_{n\geq1} \frac1{2n(2n-1)}$.) 
The following ''rearranged'' sum also converges to the same value:
$$ H_{\pm}^* = \sum_{n\geq 1} \left(  - \frac{1}{2n} + \frac{1}{2n-1}\right)
 = -\frac12 + 1 - \frac14 + \frac13 - \cdots .$$
The partial sums of $H_{\pm}$ and $H_{\pm}^*$ share a subsequence in common, 
the partial sums indexed by even numbers,
so the two series must  both converge to the same value.
This is essentially the same type of rearrangement that Max is considering in the question.
The product of 
$$ \textstyle\left(1 + \frac1{2n-1}\frac{x}{\pi}\right)
      \left(1 -\frac1{2n-1} \frac{x}{\pi} \right) 
\qquad\text{and}\qquad
\left(1 + \frac1{2n}\frac{x}{\pi}\right)
      \left(1 - \frac1{2n} \frac{x}{\pi} \right) $$
can be rearranged as the product
$$ \textstyle\left(1 + \frac1{2n-1}\frac{x}{\pi}\right)
      \left(1 -\frac1{2n} \frac{x}{\pi} \right) 
\qquad\text{and}\qquad
\left(1 - \frac1{2n-1}\frac{x}{\pi}\right)
      \left(1 +\frac1{2n} \frac{x}{\pi} \right) .$$
This does not change the value of the (conditionally convergent) infinite product
for $\frac{\sin x}{x}$.
So if the error in this ''proof'' of $\pi^2/6 = \ln(2)$ is not in rearranging terms,
 where is the actual mistake?


The  mistake is in  leaving out a term when "foiling" the product of two polynomials.
(Or, in a misapplication of Newton's identities.)

The valid infinite product expression
$$ \frac{\sin x}{x} = \prod_{n\geq 1}
\textstyle\left(1 + \frac1{2n-1} \frac x{\pi}\right)\left(1 -\frac1{2n} \frac{x}{\pi}\right) \left(1 - \frac1{2n-1}\frac{x}{\pi}\right)\left(1 +\frac1{2n}\frac{x}{\pi} \right) $$
$$\qquad\qquad \qquad\quad= \prod_{n\geq 1}
\textstyle\left(1 + \frac1{2n(2n-1)} \frac{x}{\pi} - \frac1{2n(2n-1)}\frac{x^2}{\pi^2} \right) \left(1 - \frac1{2n(2n-1)} \frac{x}{\pi} - \frac1{2n(2n-1)}\frac{x^2}{\pi^2} \right) $$
simplifies to
$$\frac{\sin x}{x} = \prod_{n\geq 1}
\textstyle\left(1 - \frac{2}{2n(2n-1)}\frac{x^2}{\pi^2} - \frac{1}{(2n)^2(2n-1)^2} \frac{x^2}{\pi^2}  + O(x^4)\right)  .$$
The coefficient of $x^2$ in this product is the series
$$ -\frac1{\pi^2}  \sum_{n\geq 1}\left( \frac{2}{2n(2n-1)} + \frac{1}{(2n)^2(2n-1)^2} \right)$$
which does, in fact, converge to $ -\frac1{\pi^2}\zeta(2) = -\frac1 6$.
This can be checked through some algebra, or by asking WolframAlpha.
(This explains why
$ \sum_{n\geq 1} \frac{1}{(2n)^2(2n-1)^2} = \frac{\pi^2}{6}-2\ln(2),$
which I would not have known how to evaluate otherwise.)<|endoftext|>
TITLE: An example of a non-amenable exact group without  free subgroups.
QUESTION [20 upvotes]: A countable discrete group $\Gamma$ is said to be exact if it admits an amenable action on some compact space. 
So clearly amenable groups are exact, but large familes of non-amenable groups are as well.
For many of the families that I know of (ex. linear groups, hyperbolic groups) that are exact, they also satisfy the von Neumann conjecture (i.e. that if they are non-amenable then they have subgroup isomorphic to a free group.)
So my questions is:
Are there examples of exact groups that are non-amenable and do not contain free subgroups?

REPLY [3 votes]: It has been some years since this question was posted, but maybe (if you haven't seen it yet) you'll enjoy reading the new geometric solution for the von Neumann problem.
http://www.math.cornell.edu/~justin/Ftp/vN_fp.pdf<|endoftext|>
TITLE: Character sums over prime arguments
QUESTION [14 upvotes]: Let $f$ be a monotone decreasing, continuously differentiable function with $\lim_{x\rightarrow \infty}f(x)=0$.  Let $\chi$ be a non-principal Dirichlet character.  It is standard to show that $\sum_{n\geq x}\chi(n)f(n)=O(f(x))$, where the big-O constant is easily computable and depends only on $\chi$.  In particular, we have $\sum_{n\leq x}\chi(n)f(n)=A+O(f(x))$ where $A=\sum_{n\in \mathbb{N}}\chi(n)f(n)$ is a constant.
When $f(x)=\log(x)/x$, Mertens used the fact that $\log(ab)=\log(a)+\log(b)$ to re-express the sum in terms of a sum over primes.  He showed that $\sum_{p\leq x}\chi(p)\log(p)/p$ is bounded, in absolute value, by a computable constant.  Then, by partial summation techniques, he removed the $\log(p)$ from the numerator and obtained bounds of the form $$\left|\sum_{p\leq x}\chi(p)/p- C \right| < D/\log(x)$$ where $C$ and $D$ are easily computable constants (possibly depending on $\chi$).
My question is two-fold.  First, what conditions on a function $f$ (satisfying any of the nice properties above, or more) guarantees that $\sum_{p\in \mathbb{N}}\chi(p)f(p)$ exists?
Second, since $-L'(s;\chi)/L(s;\chi)$ is analytic in a neighborhood of $s=1$, we know that $\sum_{p\in\mathbb{N}}\chi(p)\log(p)/p$ converges, say to a constant E.  Is there an easy way to obtain explicit bounds of the form  $$\left|\sum_{p\leq x}\chi(p)\log(p)/p - E \right| < o(1)$$ where the $o$-function is fairly simple, etc...?
The reason I ask is that I want to find an effective formula for $\sum_{p\equiv a\pmod{k},\, p\leq x}\log(p)/p$, where the error term is small.  If anyone has such a reference that would also be appreciated.

REPLY [3 votes]: An update on this problem:
I found out how to compute effective (and asymptotically accurate) bounds for $\sum_{p\leq x,\, p\equiv a\pmod{k}}\log(p)/p$.  Basically it boils down to the usual analytic techniques [see Rosser and Schoenfeld's paper "Formulas for some functions of prime numbers"], and effective bounds on $\theta(x;k,a)$ [e.g. see a 2002 paper of Pierre Dusart], which can be found by transferring information about zero-free regions of Dirichlet L-functions to this context.  I'm still interested in the original question though.<|endoftext|>
TITLE: Transcription of an interview of Kazuya Kato
QUESTION [10 upvotes]: On this page, the interview is here. Can someone provide an English translation?

REPLY [14 votes]: I can write a short summary of the nice interview, sacrifising lots of details, if that is really what you want. But, you can probably ask the video archive stuff video@ms.u-tokyo.ac.jp? They should be happy to receive your responce.<|endoftext|>
TITLE: Given $v,w$ primes of $k$, is there $K/k$ so $K_v\cap\Bbb Q^{cycl}=K_w\cap\Bbb Q^{cycl}=K\cap\Bbb Q^{cycl}$?
QUESTION [7 upvotes]: For any field $k$, let $\mu(k)$ denote the roots of unity in $k$.  Now let $k$ be a number field and let $v, w$ be non-archimedean primes of $k$ with distinct residual characteristics.  Does there exist a finite Galois extension 
$K/k$, with $v',w'$ primes of $K$ lying over $v,w$, such that $\mu(K)=\mu(K_{v'})=\mu(K_{w'})$?
For example, if $k=\mathbb{Q}$ and you're looking at the primes $3$ and $5$, then you can take $K=\mathbb{Q}(\zeta_{24})$ as your galois extension.

REPLY [2 votes]: Okay, I'm giving up.  I hadn't realized local fields had large subfields of cyclotomic extensions.  Thanks to BCnrd - I learned something interesting today.
Second try (also wrong): Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension.  By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$.  Let ${k_v}'$ be the subfield of $k_v$ consisting of elements that are algebraic over $k$ and lie in $\mathbb{Q}^{ab}$ (under any choice of embedding), and let $m_v$ be a positive integer such that ${k_v}' \subset \mathbb{Q}[\zeta_{m_v}]$.  Let $m_w$ be defined similarly.  Let $n$ be the least common multiple of $m$, $m_v$, and $m_w$, and let $K = k[\zeta_n]$.
$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$.  The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over the subfield of $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$.  Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$.  Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively.  $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$.  Is there a subtlety I'm missing?<|endoftext|>
TITLE: Indices for math texbooks/monographs
QUESTION [5 upvotes]: There are a number of mathematical books/monographs that do not have indices.  In some cases, this is no huge deal; for instance, it is often easy to find something in Bourbaki using the table of contents.  However, sometimes it can be incredibly frustrating.  For instance, in Mumford's Red Book, if you want to know what it means for a prescheme to be a scheme (in more recent terminology, what it means for a scheme to be separated), you have to look in the section titled "The functor of points of a prescheme."  Pedagogically, it works, but who would ever think to look there?
Thus, my "question" has two parts:  
1) Are there any good resources (online or otherwise) that provide indices for such works as the Red Book?
2) Assuming the answer to 1) is no, could we, as an online community, produce such a resource in the answers to this question (which I am making a community wiki for this purpose)?
To try to jumpstart 2), in case the answer to 1) is no, I am including as an answer a partial index I have produced for the Red Book.  This is also to show I am not asking others to contribute to something that I myself have not put time into.

REPLY [2 votes]: Mumford's "Red Book" is not really a book or monograph in the usual sense but started life as mimeographed lecture notes (bound with red paper covers).  The later photographic reproduction as Springer Lecture Notes 1358 doesn't change the format or add indexing.   This is still a useful resource, but isn't as reader-friendly as some of the more recent textbooks on algebraic geometry.    It's always a useful educational exercise for an individual to add index entries or notes to such a volume, though even a careful group indexing effort would still be invisible to future readers if they didn't suspect its existence.    
Other widely used lecture notes pose similar problems.  For example, Steinberg's often-cited 1967-68 lectures on Chevalley groups at Yale have rich content in spite of some rough spots in the write-up but exist only in cumbersome versions with no index and no page headings for easy location of material.    (Publishers like AMS have tried unsuccessfully to extract a more durable version.)    
As noted, a few actual books lack indexes of any kind.   An example is the otherwise valuable Birkhauser monograph Complex Geometry and Representation Theory by Chriss and Ginzburg, which I believe still lacks an index.   But the individual chapters or multi-chapters of Bourbaki I own (in several of their "books") do have an "index des notations" (in order of appearance), an "index terminologique" (alphabetical), and finally a table of contents.     
Much more commonly, advanced books often have incomplete or erratic indexes, partly due to older technology and reliance on authors to make up their own lists after the fact.  More elementary textbooks or books written for a general audience tend to be indexed by professionals, one of whom is a sibling of mine and has lots to say about the fine points.    Current software makes it much easier for authors to add index entries as they go along, but even so there is wide scope for individual judgment about inclusion/exclusion or how elaborate an index to construct.   Aside from that, I prefer books to have a full table of contents, descriptive section headings, logical internal numbering, and page headers that make it easy to find your way around.  But as I pointed out, the "Red Book" is not really a "book".<|endoftext|>
TITLE: Piecewise-smooth manifolds?
QUESTION [11 upvotes]: We know about the existence of topological (Top), differentiable (Diff) and piecewise-linear (PL) manifolds, and such things that, say, in four dimensions PL=Diff, but $\ne$Top.
The question is: do there exist piecewise-smooth manifolds? Are they equivalent to something in some dimensions?

REPLY [20 votes]: A homeomorphism $h:U\rightarrow V$ between open subsets of $\mathbb R^n$ is called piecewise differentiable (PD) -- you could also say piecewise smooth -- if there is a triangulation of $U$ by linear simplices such that the restriction of $h$ to each simplex is a smooth embedding. This class of maps is invariant under composition with diffeomorphisms on one side, and under composition with piecewise linear homeomorphisms on the other. They play an essential role in defining the good notion of compatiblility of a PL structure and a smooth structure. It is a nontrivial result of J H C Whitehead that for every smooth structure there is a compatible PL structure, unique up to a certain equivalence relation. The composition of PD maps is not in general PD, though, so this does not lead (by using "PD atlases") to a notion of PD manifold.<|endoftext|>
TITLE: Is this a definition of equivariant derived category?
QUESTION [7 upvotes]: Let $X$ be a topological space and $G$ be a topological group acting on $X$, both locally compact Hausdorff. Denote by $D^b(X)$ the derived category of sheaves (say of abelian groups) on $X$. We define a new category in the following way:
An object is an object $M$ of $D^b(X)$, together with an isomorphism 
$$\pi^*M\leftrightarrow \rho^*M$$
that satisfies the cocycle condition, where $\pi,\rho\colon G\times X\to X$ are the projection and the action maps. A morphism is a morphism of objects in $D^b(X)$ which is compatible with the action isomorphisms.
Note that this category is a special case of the category of equivariant objects:
http://ncatlab.org/nlab/show/equivariant+object

Is this category equivalent to the equivariant derived category? If not, why is the equivariant derived category a better construction in this case?

PS: Bernstein and Lunts give among others the following definition of the equivariant derived category in their book "Equivariant sheaves and functors" (p.32):
Denote by $[X/G]$ the "action simplex" $\Delta(n)=G^{n-1}\times X$.
By a simplicial sheaf we mean a collection of sheaves $F^n$ on $\Delta(n)$ together with maps
$\alpha_h: h^*F^m \rightarrow F^n$ for each map $h$ in $[X/G]$, which satisfy the cocycle condition:
$$\alpha_{h' h}=\alpha_h \circ h^* \alpha_{h'}$$ 
Denote by $Sh([X/G])$ the category of simplicial sheaves and by $Sh_{eq}([X/G])$ the full subcategory where all $\alpha_h$ are isomorphisms.
The derived equivariant category $D^b_G(X)$ is then the full subcategory of $D^b(Sh([X/G]))$ consisting of objects, which have cohomology in $Sh_{eq}([X/G])$.

REPLY [7 votes]: There's a related discussion here. The idea that the equivariant version of the derived category consists simply of complexes with equivariance structure is correct, once you work in a refined enough setting such as differential graded, $A_\infty$ or stable $\infty$-categories. As was explained it's certainly not true just at the homotopy level. But at this enhanced level, the equivariant derived category is indeed just the derived invariants of the group acting on the derived category of X.  The Bernstein-Lunts definition is the same, just writing down invariants using a standard resolution (ie the simplicial model for BG). A good reference for such things is chapter 7 of the manuscript of Beilinson-Drinfeld on Quantization of Hitchin's Hamiltonians.<|endoftext|>
TITLE: Splitting the determinant polynomial into linear factors - a Dedekind problem
QUESTION [23 upvotes]: Here's the question in a nutshell. For some $n\in\mathbb N$, we consider the polynomial
$\det\left(\left(X_{i,j}\right) _ {1\leq i\leq n,\ 1\leq j\leq n}\right)\in\mathbb Z\left[X_{i,j}\mid 1\leq i\leq n,\ 1\leq j\leq n\right]$
in $n^2$ indeterminates $X_{i,j}$. This is known to be irreducible over $\mathbb Z$, but is there a "nice" ring in which $\mathbb Z$ embeds and where this polynomial splits into linear factors? The ring needs not be commutative, but the variables $X_{i,j}$ are still supposed to commute with everything from this ring. For instance, if $n=1$, then the ring can be taken to be $\mathbb Z$, and if $n=2$, then it can be taken to be $M_2\left(\mathbb Z\right)$, since
$\det\left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right) = \left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right)\mathrm{adj}\left(\begin{array}{cc}X&Y \newline Z&W\end{array}\right)$,
and both factors on the right hand side are linear in $X$, $Y$, $Z$, $W$. For $n>2$, however, the adjoint is not linear anymore. Can we extend $M_n\left(\mathbb Z\right)$ further to make it split? (This is what I mean by "nice" - it should be a kind of natural generalization of $M_n\left(\mathbb Z\right)$. Although I have troubles constructing even a non-nice splitting ring...)

Here is the actual source of the question:
Keith Conrad, in his expository paper The Origin of Representation Theory, discusses an apparently forgotten problem that goes back to Dedekind: Given a finite group $G$, the polynomial $\det\left(\left(X_{gh^{-1}}\right)_{g\in G,\ h\in G}\right)\in\mathbb Z\left[X_g\mid g\in G\right]$ (this is a generalization of the circulant, which is obtained if $G$ is a cyclic group) is known to split into a product of irreducible factors as follows:
$\det\left(\left(X_{gh^{-1}}\right) _ {g\in G,\ h\in G}\right) = \prod\limits_{\rho\text{ is an irrep of }G\text{ over }\mathbb C} \det\left(\sum\limits_{g\in G}X_g\rho\left(g\right)\right)^{\dim\rho}$.
(Okay, apparently Keith writes $\deg \rho$ instead of $\dim \rho$, but otherwise this is in his Section 5.)
Now, some of the factors on the right hand side - those corresponding to representations of dimension $> 1$ - are nonlinear, and Dedekind tried to split them into linear factors by extending the base field. Two examples are given, and both times the extension of the field is more or less the endomorphism ring of the representation $\rho$ - but this is not surprising, because both times $\dim \rho=2$, and we have the adjoint decomposition I gave above for the case $n=2$. The actually interesting problem seems to be the $n > 2$ case. Since any irrep $\rho$ over an algebraically closed field like $\mathbb C$ has the property that the $\rho\left(g\right)$ for all $g\in G$ span the whole endomorphism ring of the irrep (this is called the density theorem, I believe), we can actually forget about the irrep and try to split the determinant of the general matrix. That's the problem above.

REPLY [2 votes]: Your question extends to other homogeneous polynomials, and is interesting in Partial Differential equations. For instance, 
$$X^2-Y^2-Z^2-W^2=(X+Ya+Zb+Wc)(X-Ya-Zb-Wc)$$
where $a,b,c$ are the Pauli matrices
$$\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix},\qquad
\begin{pmatrix} 0 & -i \\\\ i & 0 \end{pmatrix},\qquad
\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \end{pmatrix}.$$
This gives a factorisation of the wave operator $\partial_t^2-\partial_x^2-\partial_y^2-\partial_z^2$ into first-order operators. Im am not sure whether there is a characterization of those hyperbolic operators that can be split into first-order factors.<|endoftext|>
TITLE: Non-canonicity of skeleta
QUESTION [5 upvotes]: I was mulling on this previous question of mine, and I think I'd like to play the devil's advocate a bit more. I am now convinced that skeleta do not make category theory any simpler, and this is mostly due to the fact that there is no canonical way to construct a skeleton for a given category.
I am now wondering whether this can be made into a precise theorem. I was thinking of something about this lines (this may not be the right formulation):

Let $Cat$ be the $2$-category of small categories. There is no pseudofunctor $S \colon Cat \to Cat$ such that for every $C \in Cat$, $S(C)$ is a skeletal category equivalent to $C$.

Is this true? I have no clue how to prove this.
And if not, is there some variation which puts the intuition  that skeleta involve arbitrary choices on a sound basis?

REPLY [4 votes]: As David says, such pseudofunctors do exist.  Moreover, any such pseudofunctor is pseudonaturally equivalent to the identity.  However, I think that's the wrong question to be asking, because working with pseudofunctors and pseudonatural transformations means that equivalent categories will be essentially indistinguishable.  Just as any functor can be modified by replacing each of its values by an isomorphic object along a specified isomorphism, any pseudofunctor can be so modified along specified equivalences.  If you want to be able to distinguish between a category and its skeleton, then you need to talk about something stricter.  For instance, here's one statement which may be more along the lines of what you're looking for.
Claim: There is no pseudofunctor $skel:Cat\to Cat$ such that each category $skel(C)$ is skeletal and there is a strict 2-natural transformation $\alpha:skel\to Id$ whose components are equivalences.
Proof  Let $C$ be the terminal category and let $D$ be the walking isomorphism with two objects $x,y$.  Then $skel(C)\cong C$ and $skel(D)\cong C$, and so the inclusion $\alpha_D : C \to D$ must either pick out $x$ or $y$.  Let $f:C \to D$ pick out the other one; then $skel(f)=id$ and so $\alpha_D \circ skel(f) \neq f \circ \alpha_C$, i.e. $\alpha$ is not strictly natural.
One could also ask about strictifying things in other ways, such as looking for a 2-natural transformation $Id\to skel$, or asking whether $skel$ could be made a strict 2-functor.<|endoftext|>
TITLE: Hanner's inequalities: the intuition behind them
QUESTION [6 upvotes]: Hanner's inequalities in the theory of $L^p$ spaces (see http://en.wikipedia.org/wiki/Hanner's_inequalities) look hard to come-up with at the first glance. Their proof (say, the one in Lieb & Loss "Analysis", Theorem 2.5.) gives no intuition (at least for me) how they come about. How does one see that these inequalities turn up naturally? Do you know a proof which at the same time hints to how one starts considering Hanner inequalities. I hope this is not too vague of a question. Both Wikipedia and Lieb & Loss mention that Hanner's inequalities have to do with uniform convexity of $L^p$ spaces, but from that alone I cannot see how they arise "naturally".

REPLY [7 votes]: First, a probably unsatisfactory answer to your first question.  I believe that Hanner proved his inequalities because he was investigating the modulus of convexity of $L^p$.  So possibly he formed a guess on whatever basis, saw that guess would be correct if what have come to be known as Hanner's inequalities were true, then found that he could prove the inequalities.  (I haven't actually looked at Hanner's paper so I might be wrong about the history, and in any case this is the kind of thought process that is seldom recorded in papers.)
A better answer is that Hanner's inequalities generalize the parallelogram identity in a very natural way.
As for your second question, I don't know that this really fits what you're asking for either, but this paper contains the most natural-looking proof of Hanner's inequalities that I've seen.
Added: On further reflection, the best answer to the first question combines my first two paragraphs above.  The parallelogram identity very precisely expresses the uniform convexity of $L^2$.  So in investigating the modulus of convexity of $L^p$, it is natural to look for some inequality that generalizes the parallelogram identity to $L^p$.<|endoftext|>
TITLE: Capitalization of theorem names
QUESTION [53 upvotes]: I hope this question is suitable; this problem always bugs me. It is an issue of mathematical orthography.
It is good praxis, recommended in various essays on mathematical writing, to capitalize theorem names when recalling them: for instance one may write "thanks to Theorem 2.4" or "using ii) from Lemma 1.2.1" and so on.
Should these names be capitalized when they appear unnumbered? For instance which of the following is correct?
"Using the previous Lemma we deduce..." versus "Using the previous lemma we deduce..."
"The proof of Lemma 1.3 is postponed to next Section." versus "The proof of Lemma 1.3 is postponed to next section."

REPLY [2 votes]: The beauty of (La)TeX is that you can, for example,
\usepackage{xspace}
\providecommand{\Thmref}[#1]{Theorem~\ref{#1}\xspace}
\providecommand{\thmref}[#1]{theorem~\ref{#1}\xspace}
%%\providecommand\thmref\Thmref

and choose between the last two according to your editor's whims (remember to use the capitalized macro at the beginning of sentences, or wherever capitalization is required by other reasons)
Even easier, consider
\usepackage{cleveref}<|endoftext|>
TITLE: What is the Gromov-Witten potential associated to String Topology?
QUESTION [5 upvotes]: Kevin Costello's article on the Gromov-Witten potential associated to a TCFT constructs for each TCFT, i.e. a functor from chains on Riemann surfaces with boundary to chain complexes satisfying certain conditions, a canonical formal power series $D$ with coefficients in a certain Fock space $\mathcal{F}$, which is constructed from the chain complex $V$ that a TCFT associates to the circle. 
When one can construct the Gromov-Witten invariants for a manifold, we get a TCFT from Gromov-Witten theory. In that case a certain choice of polarization allows us to identify this potential $D$ with the Gromov-Witten potential. This potential encodes the intersection numbers of $\Psi$-classes and the fundamental class.
According to Costello's earlier article on TCFT's and $A_\infty$ Calabi-Yau categories the constructions of string topology allow us to define a TCFT for each oriented closed manifold $M$ (basically the chain level operations of Godin's operations in homology). One also gets a Gromov-Witten potential in this case. Is there an easier expression known for this potential? What geometric information does it encode?

REPLY [4 votes]: Here's how I understand the situation(I'm not very deep and might well be wrong) --- Costello's paper explains how to construct the Gromov Witten potential from a compact A(infinity) Calabi Yau category given two little additional conditions 1) the Hodge to de-Rham spectral sequence degenerates and 2) the induced pairing as defined in his first paper on $HH_*$   is non degenerate. This is explained for example on the bottom of page 9... 
In the case of an affine (Z- graded) dg-category--- e.g. C= the category of perfect modules over a dg-algebra A, these conditions are guaranteed by A being homologically smooth. The reference for the first being implied is a famous paper of D. Kaledin and the second is a paper of D. Shklyarov http://arxiv.org/abs/math/0702590 (I think anyways). 
None of these conditions are satisfied for string topology of a manifold(probably never somehow). The smoothness for example breaks pretty clearly --- think about $S^n$, C*(X) is $Q[x]/x^2$ with no higher operations.) The calculation of the HH to cyclic spectral sequence is a tad easier for odd spheres--- it is the case of a free commutative algebra on an odd variable which you can find in say Loday's book. HH is differential forms on the superspace $R(0,1)$ and normally the Connes operator acts by de Rham d extended to superdifferential forms(I haven't checked this little part but it's the only thing that makes any sense...)
What you do have from degenerate theories like this is an ideal in a certain Weyl algebra (see page 10 of the paper) I am pretty confident you could compute it at least for the case of spheres without too many problems, but I tend to be optimistic about these sort of things so... Hopefully some of this is right.<|endoftext|>
TITLE: Proof of PCT theorem for Haag-Kastler nets in QFT
QUESTION [7 upvotes]: This question is about a theorem in the Haag-Kastler axiomatic approach to quantum field theory (QFT), also known as axiomatic or algebraic or local QFT.
PCT stands for parity, charge and time, a "PCT theorem" says roughly that if a quantum field theory describes a universe, then after reversing the parity, charges and the arrow of time in the theory, the resulting theory still describes the same universe. 
It is one of the fundamental symmetries of today's theoretical physics.
Various versions of PCT theorems in different QFT frameworks have been around at least since the 1950ties.
My question is: What versions of the PCT theorem exist that are stated and proven using some version of the Haag-Kastler axioms?
There are three papers that I am aware of:
H.J. Borchers, J. Yngvason: On the PCT--Theorem in the Theory of Local Observables
H.J. Borchers: “On Revolutionizing of Quantum Field Theory with Tomita’s Modular Theory”
available free for download here.
Longo, Guido: An Algebraic Spin Statistics Theorem
My motivation is twofold: 

The "conceptional proofs" of physicists tend to be rather simple and general, but are not rigorous. The proofs in the papers above are rigorous, but rather involved and not as general as I expected (which could simply be a misunderstanding or too big expectations on my part). I would like to know if there is a proof that is either simpler (does use less mathematical machinery, e.g. does not use modular theory) or more general (does not need one of the assumptions or weakens one of the assumptions).
I would like to know who constructed the first proof of a PCT theorem in the Haag-Kastler approach and when.

BTW: Any information about the twin of the PCT symmetry, the spin-statistics theorem, in the Haag-Kastler approach would be welcome, too.

REPLY [2 votes]: Jens Mund has some papers on spin-statistics and PCT in the case of massive particles in d=2+1 (e.g. here and here), but as far as I recall he also uses modular theory, so this might not provide a full answer to your question.<|endoftext|>
TITLE: Homologically trivial submanifolds
QUESTION [15 upvotes]: Unuseful prequel
Let $M$ be a (compact, oriented, differentiable) manifold. Before knowing anything about homology theory a naif but clever mathematician may want to measure the holes in $M$ by the following procedure. One lets $B_k(M)$ be the abelian group generated by embedded submanifolds of $M$ of dimension $k$ with border. There is an obvious border operator $\delta \colon B_{k}(M) \to B_{k-1}(M)$ and trivially $\delta^2 = 0$ since the border of a manifold with border is a closed manifold. So one may define $H^K_{naif}(M)$ by taking the quotient of cycles modulo boundaries.
Ok, this does not really work, for at least two reasons. The first there is no way to make this functorial without considering more degenerate objects (singular simplices or currents being two possible choices). The second is that even if one adjusts the definition to get, for instance, the bordism groups of $M$, these will be highly nontrivial for differential geometric reasons, rather than for the topology of $M$ (for instance they will be nontrivial for a point).
Anyway, one finally chooses a working (co)homology theory, for instance singular, and then can use the fundamental class of submanifolds to link this theory to the naif one. Two questions naturally arise:

Are fundamental classes of manifolds enough to generate the cohomology? This is nicely answered here.
Are submanifolds with border enough to generate homology relations? This is the purpose of the present question.

The actual question
Let $M$ be a (compact, oriented, differentiable) manifold and let $N \subset M$ be a (closed) submanifold. The problem is to test whether $N$ is the boundary of submanifold with boundary of $M$. There are two obvious obstructions:

$N$ should be bordant as an abstract variety
The class $[N] \in H_{*}(M, \mathbb{Z})$ should be $0$.


Assume 1 and 2 hold. Are there any conditions on $M$, $N$ or the embedding $N \to M$ which guarantee that $N$ is the boundary of submanifold with boundary of $M$?

There are a few classical cases which come to mind:

If $N$ and $M$ are spheres, 1 and 2 are vacuous and the thesis is true by the (generalized) Jordan curve theorem.
If $N = S^1$ and $M = \mathbb{R}^3$, the thesis is still true by the existence of Seifert surfaces.
If $\dim N = 1$ and $\dim M = 2$ the result seems true to me by the classification of surfaces as quotients of polygons and the usual Jordan curve theorem (but I did not check the details).

On the other hand I'm sure there are plenty of negative example even in $3$-dimensional topology. Is there anything nontrivial which can be said about this question, whether on the positive or negative side?

REPLY [4 votes]: (A similar question has since been asked here.)
This is a supplement to the answers of Jeff and Oscar, who note that first you must ask if the embedding $f\colon M \to N$ is the boundary of a smooth map $F\colon W\to N$, then ask if $F$ can be made an embedding. The bordism problem can be attacked using characteristic numbers of the map $f$. This is described in section 17 of the book "Differentiable Periodic Maps" by Conner and Floyd. In particular, their Theorem 17.5 states that

If the torsion of $H_*(N;\mathbb{Z})$ consists only of elements of order two, then maps $f_i\colon M_i\to N$, $i=1,2$ of oriented closed manifolds represent the same oriented bordism class if and only if all their Stiefel-Whitney and Pontrjagin numbers agree.

So, under these homological hypotheses, a necessary and sufficient condition for $f\colon M\to N$ to be null-bordant is that all the numbers $\langle w_I\cup f^*(x),[M]_2\rangle$ and $\langle p_J\cup f^*(x),[M]\rangle$ are zero, where $x$ ranges over the cohomology of $N$ with appropriate coefficients, and the $w_I$ and $p_J$ are monomials in the Stiefel-Whitney and Pontrjagin classes of $M$.<|endoftext|>
TITLE: formal completion
QUESTION [20 upvotes]: When I study formal completion and formal schemes, on p.194 of Hartshorne's "Algebraic Geometry", he said "One sees easily that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings." 
Notice that here $\mathcal{O}_{\hat{X}}$ is not the structure sheaf of X, there is a "hat" on the symbol $X$.
But I can't see the reason for that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings.
Could someone explains this for me, thanks.

REPLY [29 votes]: Here is a self-contained explanation (hopefully without any blunder): 
Locally, $\hat{X}$ is an affine formal scheme, so each point has a neighbourhood basis admitting of open sets $U$ admitting the following description:
there is a ring
$A$, with ideal $I$, such that the underlying topological space is $U_0 :=$ Spec $A/I$, and the structure sheaf is the projective limit of the sheaves $\mathcal O_{U_n},$
where $U_n :=$ Spec $A/I^{n+1}$.  
(Note that the underlying topological space of all the $U_n$ coincide, so as topological spaces $$U = U_0 = \cdots = U_n = \cdots .$$
The sheaves $\mathcal O_{U_n}$ are all sheaves on this same underlying topological space,
which form a projective system with obvious transition maps, corresponding to the surjections of rings $A/I^{n+1} \to A/I^n.$
Now if $x$ is a point of $\hat{X}$, and if we choose a neighbourhood $U$ of $x$ as above,
then the natural map on sheaves $\mathcal O_{\hat{X}} \to \mathcal O_{U_0}$ induces
a natural map on stalks $\mathcal O_{\hat{X},x} \to \mathcal O_{U_0,x}$.
The target is a local ring.  Let $\mathfrak m_x$ denote the preimage in $\mathcal O_{\hat{X},x}$ of the maximal ideal
in $\mathcal O_{U_0,x}$;  I claim that it is the unique maximal ideal of
$\mathcal O_{\hat{X},x}$.
To see this, suppose that $f$ is an element of $\mathcal O_{\hat{X},x}$ which does not
lie in $\mathfrak m_x$.  Then by definition of the stalk, $f$ extends to a section
of $\mathcal O_{\hat{X}}$ over some neighbourhood of $x$, which (by shrinking $U$ as necessary) we may as well
assume is our affine neighbourhood $U$.
Thus we may think of $f$ as a section of the projective limit of $\mathcal O_{U_n}(U)$,
i.e. the projective limit of the rings $A/I^{n+1}$.
The assumption that $f$ is not in $\mathfrak m_x$ says that its image in $\mathcal O_{U_0}(U)$ is not in the maximal ideal at $x$, and so shrinking $U$ further,
if necessary, we may assume that $f$ is a unit in $\mathcal O_{U_0}(U) = A/I$.
Thus $f$ is an element of the projective limit of $A/I^{n+1}$ which is a unit
in $A/I$.  One easily verifies that $f$ is then a unit in every $A/I^{n+1}$,
and hence in the projective limit.  Thus $f$ is a unit in the ring 
$\mathcal O_{\hat{X}}(U)$, and so in particular in the stalk $\mathcal O_{\hat{X},x}$.
I've shown that every element of the stalk $\mathcal O_{\hat{X},x}$ not lying
in $\mathfrak m_x$ is a unit, which implies that $\mathcal O_{\hat{X},x}$ is local
with maximal ideal $\mathfrak m_x$.<|endoftext|>
TITLE: Does there exist a comprehensive compilation of Erdos's open problems?
QUESTION [37 upvotes]: Fan Chung and Ron Graham's book Erdos on Graphs: His Legacy of Unsolved Problems (A. K. Peters, 1998) collects together all of Erdos's open problems in graph theory that they could find into a single volume, complete with bounties where applicable.  Of course Erdos posed many other open problems in combinatorics and number theory that do not appear in this book.  I once heard a rumor that some people were working on a project to publish a similar but more comprehensive book or series of books, covering all of Erdos's open problems, but I don't know if the rumor is true.  Does such a compilation exist?  If not, is there anything else like this besides Chung and Graham's book?

REPLY [7 votes]: For graph theory problems, I find this the most comprehensive resource: http://www.math.ucsd.edu/~erdosproblems/<|endoftext|>
TITLE: Do these conditions on a semigroup define a group?
QUESTION [17 upvotes]: As is well known, if $S$ is a semigroup in which the equations $a=bx$ and $a=yb$ have solutions for all $a$ and $b$, then $S$ is a group. This question arose when someone misunderstood the conditions as requiring that the solution to both equations be the same element of $S$. He suggested instead replacing one of the equations with a cancellation condition (he was thinking along the lines of trying to specify that the Cayley table would be a latin square). It is easy to see that there are semigroups that are not groups in which every equation of the form $a=xb$ has a solution and you can cancel on the right (the standard example that sets $ab=a$ for all $a,b$ works). What is not clear to me is what happens if the equations and cancellations are on the same side. That is:

Suppose $S$ is a semigroup in which the following two conditions hold:

 For all $a,b\in S$ there exists $x$ such that $a=xb$.
 For all $a,b,c\in S$, if $ab=ac$ then $b=c$.
 
Is $S$ a group?

It is easy to see that if $S$ contains an idempotent, then $S$ will be a group: if $e^2=e$, then for all $a\in S$ we have $e^2a=ea$, so $ea=a$ for all $a$; then solving $e=xa$ shows $S$ has a left identity and left inverses, hence is a group. In particular, $S$ will be a group if at least one cyclic subsemigroup of $S$ is finite, and also in particular if $S$ is finite.
I suspect that the answer will be "no" in full generality (that is, there are examples of semigroups $S$ that satisfy 1 and 2 above but are not groups), but I have not been able to construct one. Does any one have an example, a proof that $S$ will always be a group under these conditions, or a reference?

REPLY [16 votes]: I am a victim of timing... I had asked this of a colleague a few days ago and had received no answers, but today at lunch he gave me a counterexample and reference (Clifford and Preston's The Algebraic Theory of Semigroups, volume II, pp. 82-86). The example is the Baer-Levi semigroup: the semigroup of all one-to-one mappings $\alpha$ of a countable set $X$ into itself such that $X\setminus \alpha(X)$ is not finite. 
Left cancellation follows trivially; if $a$ and $b$ are such mappings, then to construct $c\in S$ such that $a=cb$ proceed as follows: if $y=b(x)\in b(X)$, set $c(y)=c(b(x))=a(x)$. Now let $\delta$ be a one-to-one map from $X\setminus b(X)$ to $X\setminus a(X)$ with $(X\setminus a(X))\setminus\delta(X\setminus b(X))$ infinite. Then define $c(y)$ for $y\notin b(X)$ by $c(y)=\delta(y)$.<|endoftext|>
TITLE: What are normalized singular chains good for?
QUESTION [14 upvotes]: One of the common definitions of homology using the singular chains, i.e. maps from the simplex into your space. The free abelian group on these can be made into a chain complex and one can take the homology of this. The result is usually called singular homology.
However, one can use a smaller chain complex instead, by taking the quotient with the degenerate chains (those that are the image of a degeneracy map $\sigma$). This will give the same result in homology.
In some articles, I've seen authors replace the singular chains with the normalized singular chains, often claiming that this is "for technical reasons" (a example is Costello's article on the Gromov-Witten potential associated to a TCFT). What are the important technical differences between these two functors? In which situations is there a preferred one?

REPLY [7 votes]: The question was about chains on a space, presumably singular chains. If what you're up to is, for example, learning or teaching the basic facts about singular homology, like homotopy-invariance and excision, then the switch to normalized chains seems like an unnecessary complication. In that context, you could say that non-normalized chains have the advantage.
To expand on Ben's comments: 
For A and B simplicial abelian groups and NA and NB the associated normalized chain complexes, there is a quasi-isomorphism $NA\otimes NB\rightarrow N(A\otimes B)$, where the tensor product of simplicial groups is defined levelwise $(A\otimes B)_n=A_n\otimes B_n$ and the tensor product of chain complexes is the usual $(C\otimes D)_n=\oplus_p (C_p\otimes D_{n-p})$. An associative simplicial ring leads in this way to an associative differential graded ring. 
Another technical advantage of the normalized complex is its role in proving that homotopy groups of a simplicial abelian group are homology groups (Dold-Thom Theorem).<|endoftext|>
TITLE: Does this sequence span $L^2$?
QUESTION [13 upvotes]: Consider the following sequence of functions in $L^2[0,\infty)$:
$$f_n(x)=e^{-x/n}x^n,\;\;n\geq 1$$
Does this sequence span $L^2[0,\infty)$ (that is, is the set of finite linear combinations
of these functions dense)?
(My guess is that it doesn't).

REPLY [2 votes]: This is a very interesting problem.  An intermediate generalization between that stated problem and that proposed by Zen Harper is to look at the sequence
$$f_n(x) = x^n e^{-a_n x}, \quad n=1,\ldots$$
for some given sequence $a_n$ of positive numbers.  As in Zen Harper's post, the Laplace transform of $f_n$ is the function $F_n(z)= (-1)^n \frac{1}{(z+a_n)^n}$, defined for $z$ in the right half plan $\{ \mathrm{Re}z >0\}$, and a function $f$ is orthogonal to the span of $\{f_n\}$ if and only if its Laplace transform $F$ satisfies
$$ F^{(n)}(a_n)=0.$$
Thus, if $a_n$ is a constant sequence then the span of this sequence is dense, since anything orthogonal to the span would vanish to infinite order at a point.
On the other hand if $\sum_{n=1}^\infty n \frac{a_n}{(1+a_n)^2} <\infty$ then using a bit of complex analysis one can find a non-zero function $g$ orthogonal to the span of $\{f_n\}$.  Specifically, we form the Blaschke product $B(z)$ with $n$ zeros at $a_n$ -- this is an analytic function in the right half plane, everywhere bounded by one and with zero of order $n$ at $a_n$ for each $n$.  So $B(z)$ satisfies $B^{(n)}(a_n)=0$ and in fact
$$B^{(k)}(a_n)=0,\quad k=1,\ldots,n.$$
Multiplying $B$ by an non-zero $H^2$ function $F$ produces a function $BF \in H^2$ and perpendicular to the given sequence.  Taking the inverse Laplace transform gives the desired function $g$.
But the original problem has $n a_n/(1+a_n)^2\rightarrow 1$ so this does not produce a counterexample.<|endoftext|>
TITLE: Seeking examples or proof: injectivity of Cartan homomorphism for commutative rings? 
QUESTION [5 upvotes]: This question is motivated by some issue raised by David Speyer in this question. 
Let $R$ be a ring. Let $K_0(R)$ and $G_0(R)$ be the Grothendieck groups of f.g. projective modules and f.g. modules over $R$, respectively (you just kill all relations generated by short exact sequences). There is a natural map, called the Cartan homomorphism (see Serre's "Linear reps of finite groups", Chapter 15) $$c: K_0(R) \to G_0(R)$$ given by forgetting a module is projective. 
In general, $c$ needs not be injective nor surjective. For non-surjectivity, take $R$ to be a local ring, then $K_0(R)=\mathbb Z$ but $G_0(R)$ can be huge (in particular, if $R$ is normal, $\mathbb Z\oplus \text{Cl}(R)$ is a quotient of $G_0(R)$). Examples of non-injectivity can be found by taking $R$ to be some group rings, as the Cartan matrix is not always invertible, see for example Section 4 of this paper by Martin Lorenz . But I don't know any commutative example of non-injectivity.
Is $c$ always injective if $R$ is commutative? How about if $R$ is commutative and Noetherian?
(If this is true, one can prove the original question quoted above with the assumption $G_0(R)=\mathbb Z$)

REPLY [9 votes]: No, it is not injective in general, unless $R$ is regular notherian. There are many counterexamples; for a simple one you can take the ring $R := \mathbb C[t^2, t^3] \subseteq \mathbb C[t]$, compute that $G_0(R) = \mathbb Z$, while $K_0(R)$ maps onto the Picard group of $R$, which is the additive group $\mathbb C$.<|endoftext|>
TITLE: What are some correct results discovered with incorrect (or no) proofs?
QUESTION [76 upvotes]: Many famous results were discovered through non-rigorous proofs, with
correct proofs being found only later and with greater difficulty.  One that is well
known is Euler's 1737 proof that
$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$
in which he pretends that the power series for $\frac{\sin\sqrt{x}}{\sqrt{x}}$
is an infinite polynomial and factorizes it from knowledge of its roots.
Another example, of a different type, is the Jordan curve theorem. In this case,
the theorem seems obvious, and Jordan gets credit for realizing that it requires
proof. However, the proof was harder than he thought, and the first rigorous proof
was found some decades later than Jordan's attempt. Many of the basic theorems of topology are like this.
Then of course there is Ramanujan, who is in a class of his own when it
comes to discovering theorems without proving them.
I'd be interested to see other examples, and in your thoughts on what the examples reveal about the connection between discovery and proof. 
Clarification.  When I posed the question I was hoping for some explanations
for the gap between discovery and proof to emerge, without any hinting from me. Since this hasn't happened much yet, let me suggest some possible
explanations that I had in mind:
Physical intuition. This lies behind results such as the Jordan curve theorem,
Riemann mapping theorem, Fourier analysis.
Lack of foundations. This accounts for the late arrival of rigor in calculus, topology,
and (?) algebraic geometry.
Complexity.  Hard results cannot proved correctly the first time, only via a
series of partially correct, or incomplete, proofs. Example: Fermat's last theorem.
I hope this gives a better idea of what I was looking for. Feel free to edit your
answers if you have anything to add.

REPLY [2 votes]: The (sharp) bound on the number of non-repelling cycles of a rational map of a Riemann sphere, sometimes called Fatou-Shishikura inequality, is such an example. It says that a rational map $f: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ of degree $d \geq 2$ has at most $2d-2$ non-repelling (i.e., attracting or neutral) cycles. 
This bound was first stated (without proof or even any particular motivation) by Lucjan Emil Boettcher, in his paper ''Zasady rachunku iteracyjnego (czesc pierwsza i czesc druga) [Principles of iterational
calculus (part one and two)]", Prace Matematyczno Fizyczne, vol. X (1899-1900),
pp. 65-86, 86-101. In 1920 it was formulated independently by Pierre Fatou. He managed to prove a weaker estimate, by $4d-4$. Later Adrien Douady and John Hamal Hubbard proved the conjectured estimate in the case when $f$ is polynomial, and finally Mitsuhiro Shishikura proved it  in the general case, using the theory of quasiconformal surgery.
(Shishikura, Mitsuhiro: Surgery of complex analytic dynamical systems. In: Dynamical systems and nonlinear oscillations (Kyoto, 1985), 93-105, World Sci. Adv. Ser. Dynam. Systems, 1, World Sci. Publishing, Singapore, 1986). Subsequently, another proof was given by Adam L. Epstein: http://arxiv.org/pdf/math/9902158.pdf<|endoftext|>
TITLE: Cardinality of the permutations of an infinite set
QUESTION [21 upvotes]: If you have an infinite set X of cardinality k, then what is the cardinality of Sym(X) - the group of permutations of X ?

REPLY [15 votes]: Since the comments on Steve0078's answer raise issues concerning the axiom of choice, let me point out that John Dawson and Paul Howard have shown that, in choiceless set theory, the number of permutations of an infinite set $X$ can consistently be related to the number of subsets of $X$ by a strict inequality in either direction; the two numbers can also be incomparable; and of course they can be equal as in the presence of choice.  (Slogan: Without choice, nothing can be proved about those two cardinals.)  The reference for this is "Factorials of Infinite Cardinals" in Fundamenta Mathematicae 93 (1976) pp. 186-195 (Math Reviews volume 55 #7779).<|endoftext|>
TITLE: association schemes, infinite schemes, semi-schemes, quasi-schemes
QUESTION [5 upvotes]: Some questions about possibly nonsensical ideas:
1) Can you come up with a definition of an infinite association scheme ?
2) Would infinite association schemes relate to infinite groups the way association schemes relate to finite groups ?
(see http://www.ams.org/bull/2006-43-02/S0273-0979-05-01077-3/S0273-0979-05-01077-3.pdf)
3) Can you define objects that relate to semigroups and quasigroups as schemes relate to groups ? 
4) What combinatorial, statistical or other properties would these infinite schemes, semi-schemes and quasi-schemes have ?

REPLY [3 votes]: A (probably not very brilliant) partial answer to the first question.
The following definition of infinite association scheme semi-rings makes sense: A set $A_i,i\in\mathcal I$ (with $\mathcal I$ not necessarily finite) of infinite matrices (indexed by $\mathbb N$ or $\mathbb Z$) with coefficients in $\{0,1\}$ containing the infinite identity matrix such that $\sum_{i\in \mathcal I}A_i=J$ where $J$ denotes the infinite all $1$ matrix and
$A_iA_j=\sum_{k\in\mathcal I}\gamma_{i,j}^kA_k$ with $\gamma_{i,j}^k$ in
$\mathbb N\cup\lbrace\infty\rbrace$ and where the last sum is finite. 
We require moreover the equalities $\gamma_{i,j}^k=\gamma_{j,i}^k$.
All operations are then well-defined on
the semiring $\sum_{i\in\mathcal I}\lambda_i A_i$ of finite sums with coefficients in
$\mathbb R_{\geq 0}\cup\{\infty\}$ and can even be extended to infinite sums (this is
useful since the Hadamard product identity, $J$ is an infinite sum). 
Negative coefficients should be avoided. 
I am not convinced of the interest of such a structure.<|endoftext|>
TITLE: Making 'circles' on a lattice/ Making distinct fractions from partitions of a number
QUESTION [5 upvotes]: First formulation: discrete geometry
Pick your favourite 2D square lattice (I'm sure we all have one...) and try to place n points 'in a circle' (that is: in general position [no 3 should be colinear] and forming the vertices of a convex polygon). Easy-peasy (assuming your favourite lattice is infinite- I know mine is). But now make your lattice smaller- finite even- say, $K \times K$- can you still do it? What is K is smaller? Bigger? Clearly the answer depends on n, so we ask:

What is the smallest $K=K(n)$ such that this can be done?

Motivation?
This little problem- way out of my comfort zone in terms of field- was sort of inspired by this question. It initially looked like an interesting problem but, sadly, turned out to be rather trivial- putting points in a circle spelled trouble, it seemed, and I idly suggested a finite lattice to stop it happening- idle suggestion turned to idle speculation and I began to wonder how small the lattice had to be to stop the circle forming- beyond that, intrigue is the only reason why I am pursuing it.
So far my progress has been scant, but I have found a rather interesting reformulation: by focusing on the most densely populated 'quadrant' of the lattice and observing that the gradients of the edges within it must all differ. Modulo some suitable considerations of n mod 4 we can get-
Reformulation: number theory

Given $\hat{n}$ what is the smallest $\hat{K}$ such that we may find $\hat{n}$ distinct fractions $\frac{p_1}{q_1},....\frac{p_n}{q_n}$ with $\Sigma_i q_i \leq \Sigma_i p_i = \hat{K}$?

Here $\hat{n}$ would be the number of points in the quadrant $\hat{K} \times \hat{K}$ the smallest size of that quadrant.
If we set $\hat{n}=\lceil \frac{n}{4} \rceil$, assume $\hat{K}=\frac{K}{2}$ and let the $p_i$, $q_i$ be the first $\hat{n}$ natural numbers we get a [really crappy] upper bound of $K \leq \lceil \frac{n}{4} \rceil (\lceil \frac{n}{4} \rceil +1)$- this fails to be sharp by the time n=5 for obvious reasons- 1) There has been no real consideration of the mod 4 behaviour and 2) By no means is this the most efficient way to make differing fractions.
Fixing 1) seems pretty easy (we can look at 2 different quadrants), but I didn't want to clog up the question with numerics when 2) is the real toughie. As mentioned above, I am neither a discrete geometer nor a number theorist, so this may unwind to be as trivial as the question that inspired it- still, to me it seems intriguing- and I am sufficiently invested to be hankering for an answer.
Edit: Gjergji's answer seems to be pretty great as a problem posed for polytopes, but I haven't accepted it yet because I am curious- can the number theory formulation do better? Is there a sharper result in that context?

REPLY [5 votes]: I believe your question (and many other variations) are answered here. Also see this survey by I. Barany about extremal problems for convex lattice polytopes. 
Actually the problem for the convex polygon inside the square was considered earlier than that. See here for an article that seems to be written more in the lines of your question. The result is that
$$K(n)\sim 2\pi \left(\frac{n}{12}\right)^{2/3}.$$<|endoftext|>
TITLE: Is there a mathematically precise definition of turbulence for solutions of Navier-Stokes?
QUESTION [28 upvotes]: Given a solution $S$ of the Navier-Stokes equations, is there a way to make mathematically precise a statement like: "$S$ is turbulent in the spacetime region $U$"? 
And if such a definition exists, are there any known exact solutions of Navier-Stokes exhibiting turbulence?
Reading Wikipedia makes me also want to ask the following related questions: 

Does a single vortex count as turbulence?
Is the appearance of vortices a necessary feature of turbulence?
What's the difference between a vortex and an eddy?

Pointers to (mathematically rigorous) literature are much appreciated.

REPLY [4 votes]: I understand turbulence to be defined as a chaotic patch of vorticity. This definition comes from http://books.google.co.uk/books?id=rkOmKzujZB4C&lpg=PP1&dq=davidson%20turbulence&pg=PP1#v=onepage&q=davidson%20turbulence&f=false 
This definition implies that vorticity is a necessary feature of turbulence. I suppose the vortex patch would tend towards a stable structure such as a vortex.
A vortex often refers to a mathematically definable point vortex. An eddy is a less rigorous term for an overturning motion that often implies a vortex.<|endoftext|>
TITLE: Morse-Bott homology for infinite-dimensional manifolds
QUESTION [5 upvotes]: Is there any work on Morse-Bott homology for infinite-dimensional manifolds (e.g. Hilbert manifolds). I am particularly interested in the case where we have a locally trivial fiber bundle and the Morse-Bott function is the function on the total space which is a pulled back Morse function on the base.

REPLY [3 votes]: See Atiyah and Bott's papers on Yang-Mills for Riemann surfaces.
The essence of it is that the Yang-Mills functional acts like a
perfect Morse functional after quotienting by gauge transformations.
So, look at it before the quotient to get your Morse-Bott function.<|endoftext|>
TITLE: Derived Physics
QUESTION [12 upvotes]: Hello to all,
This question will probably be closed down as being off-topic faster than one can say "string theory", but here it goes: I've noticed that the problems I'm working on -the structure of derived categories of some interesting surfaces- have a very important physical meaning. Unfortunately, I have in no way any idea as to why. Is there anyone out there who could give a mathematical explanation -or a link to a paper- as to why physicists would be interested in such highly abstract gizmos like derived categories, mutation, orbifolding, tilting...(the list goes on and on)

REPLY [13 votes]: The short but ahistorical answer is that topological string theories turn out to be examples of $(\infty,1)$-categories. The mathematical formulation of this statement is in Lurie's classification of topological field theories http://www.math.harvard.edu/~lurie/papers/cobordism.pdf (building on work of Atiyah, Segal, Getzler, Costello, Baez-Dolan, Kontsevich and probably a bunch more I'm forgetting.)
The content of this statement is that when you write down the axioms for a topological string theory, the collection of "boundary conditions" or "D-branes" look like the collection of objects in an $(\infty,1)$ category.
Of course, you can ask why the derived category of coherent sheaves. Historically, the answer to that is that it is very easy to write down a boundary condition for a holomorphic vector bundle in the topological B-model. It's not a huge leap from there to coherent sheaves, and if you start mumbling words like tachyon condensation, you can get to the derived category with a fair bit of hand waving.
That's from the physics side of things. On the math side, Kontsevich got there first, possibly by noting that the space of closed string states in the B-model ($H^\bullet(\wedge^\bullet TX)$) is exactly the Hochschild coohomology of the derived category of coherent sheaves. He then followed up by associating the (still not yet defined?) Fukaya category with the A-model and conjecturing that mirror symmetry is an equivalence of the two (with some Hodge structure goodies thrown in). Subsequently, it looks like you have to add in some things called coisotropic branes to cover all your bases, but the basic idea is right.
Kontsevich formulated all this in terms of $A_\infty$ categories which in the Lurie language turn into $(\infty,1)$ categories which are just TQFTs in disguise. So, Kontsevich's homological mirror symmetry is then the statement that two TQFTs are the same, just like mirror symmetry in string theory.
From the physics side of things, this was all a bit of a mess, but we now understand that the derived category really arises via Block's construction of the derived category (I'm being intentionally vague as to which version of the derived category) as arising from integrable super-connections of graded smooth vector bundles http://www.math.upenn.edu/~blockj/papers/BottVolume.pdf. You can see this explicitly in the physics from a few sources, particularly Kapustin, Rozansky and Saulina, and Herbst, Hori and Page, but I'm rather fond of my own contribution http://arxiv.org/abs/0808.0168.<|endoftext|>
TITLE: Algebraicity of holomorphic representations of a semisimple complex linear algebraic group
QUESTION [9 upvotes]: Let $G$ be a complex linear algebraic group, given to us as a closed subgroup of some $\mathrm{GL}(n,\mathbb{C})$.  Suppose moreover that $G$ is semisimple.  Then it's a fact that every finite-dimensional holomorphic representation of the complex Lie group $G(\mathbb{C})$ is actually an algebraic representation (i.e., given by polynomials in the matrix entries, together with $\mathrm{det}^{-1}$).
One can certainly deduce this from the highest-weight theory (that is, by provably constructing all the representations, and noting that everything you've constructed is in fact algebraic).  But this isn't remotely satisfying.
In another direction, I was led by notes of Milne to a book and some papers by Dong Hoon Lee, and from those to a series of papers by Hochschild and Mostow.  But those authors want to do something harder: classify the Lie groups (not necessarily semisimple!) that can be given an algebraic group structure such that all the holomorphic representations of the Lie group are algebraic representations of the algebraic group.  It seems to me that if you start out with an algebraic group, and then assume that the group is semisimple, then most of the complications should go away.
So my question is, is there a satisfying and reasonably elementary proof of the fact in the first paragraph?  I should say something about my motivation.  I'll be teaching a Lie groups/algebras course next year, and when we talk about representation theory we'll observe this phenomenon; so it would be nice to explain it if there's a reasonable way to do so. Given that I can't expect my students to have had an algebraic geometry course, I'd want to minimize the algebraic geometry in the argument, possibly at the cost of making more serious use of the structure theory of Lie groups/algebras.  
Here's an approach that I would find especially clarifying if it can be made to work. We're handed a faithful representation of $G(\mathbb{C})$ (the inclusion into $\mathrm{GL}(n,\mathbb{C})$) which is certainly algebraic.  Its tensor powers are algebraic.  Then the claim would be immediate by semisimplicity if one can show that every irreducible representation of $G(\mathbb{C})$ (or perhaps of Lie groups in some more general class than these) occurs as a subquotient of a tensor power of a faithful one.  How might one prove the latter?  (Can one prove the latter for compact real groups in a manner similar to the proof for finite groups, and then pass to semisimple complex groups by the unitary trick?)

REPLY [4 votes]: I also want to add another, much more elementary, answer in the case $G=\mathbb{C}^*$. Let $\rho(t)$ be a holomorphic map $\mathbb{C}^* \to \mathrm{GL}_n(\mathbb{C})$. Then we can write $\rho$ as a convergent sum $\sum_{i=-\infty}^{\infty} P_i t^i$. Write out the equation $\rho(tu) = \rho(t) \rho(u)$ and look at the $t^i u^i$ term to deduce that $P_i^2=P_i$. So $\mathrm{Tr} \ P_i$ is a nonnegative integer, equal to its rank. But $\sum P_i = \rho(1) = \mathrm{Id}_n$, so $\sum \mathrm{Tr} P_i =n$. We deduce that all but at most $n$ of the $P_i$ must be $0$, so $\rho(t)$ is a polynomial.
I've been trying to think of a clever way to extend this to the general case, using that a generic matrix is diagonalizable, but I haven't found one yet.

REPLY [2 votes]: (Edit: removed application of Peter-Weyl's --- thanks to Victor Protsak for pointing out it is unnecessary.)
As long as you don't mind passing to compact subgroups, it can be done using Weyl's unitary trick. Here's a sketch: let $K\subset G({\mathbb C})$ be the maximal compact. Consider in $L^2(K)$ the matrix elements of irreducible representations. They form an orthogonal set.
On the other hand, the subset consisting of matrix elements of irreducible algebraic representations spans $L^2(K)$ (by the Stone-Weierstrass Theorem --- any polynomial in matrix elements and $det^{-1}$ is a linear combination of those). Hence the two sets are the same. 
P.S. If you prefer, you can work with characters instead of matrix elements and with class functions instead of all functions.<|endoftext|>
TITLE: Polynomials having a common root with their derivatives
QUESTION [54 upvotes]: Here is a question someone asked me a couple of years ago. I remember having spent a day or two thinking about it but did not manage to solve it. This may be an open problem, in which case I'd be interested to know the status of it.
Let $f$ be a one variable complex polynomial. Supposing $f$ has a common root with every $f^{(i)},i=1,\ldots,\deg f-1$, does it follow that $f$ is a power of a degree 1 polynomial?
upd: as pointed out by Pedro, this is indeed a conjecture (which makes me feel less badly about not being able to do it). But still the question about its status remains.

REPLY [25 votes]: The strongest result in this direction that I've heard of is Sudbery's theorem (which was 
originally conjectured by Popoviciu and Erdös).

Theorem. Let $P(z)$ be a polynomial of degree $n\geq 2$ and let 
  $\Pi(z)=\prod\limits_{k=0}^{n-1}P^{(k)}(z)$ where $P^{(k)}$ is the $k$th derivative of $P$.
  Then either $\Pi(z)$ has exactly one distinct root or $\Pi(z)$  has at least $n+1$ distinct roots.

See the original paper by Sudbery.<|endoftext|>
TITLE: Infinite dimensional unitary representations of SU(2) for non-half-integer j?
QUESTION [5 upvotes]: The finite dimensional irreducible unitary representations of $SU(2)$ are labelled by $j$ which needs to be half-integer, the dimension of the representation is $2j+1$. This is well-known, all is good.
If we do not require finite dimension for the representation, is it possible to make sense of representations with an arbitrary real number $j$? They will, presumably, be infinite dimensional but hopefully still unitary.
In the half-integer case when represented on at most $2j$ degree holomorphic polynomials, the 3 basis elements of the Lie-algebra in representation $j$ act as
$e_1 = \frac{1-z^2}{2}\frac{d}{dz} + jz$
$e_2 = \frac{1+z^2}{2i}\frac{d}{dz} + ijz$
$e_3 = -z\frac{d}{dz} + j$
Clearly, if $j$ is not half-integer and we start from $f(z) = z$ and start acting on it with $e_i$, it will generate an infinite dimensional space.
This kinda gives me the feeling that perhaps non-half-integer $j$ representations are still meaningful and are infinite dimensional. But I'm not sure, is this really the case or something will go wrong?
Basically what I'm asking is whether analytic continuation in $j$ makes any sense.

REPLY [4 votes]: You have to distinguish between representations of a compact group $SU(2)$ and of its Lie algebra $su(2)$, which can be complexified to $sl(2,\mathbb{C}).$ By Peter-Weyl theorem, every irreducible representation of the group $SU(2)$ is finite-dimensional. However, you can let Lie algebra $sl(2)$ act on various infinite-dimensional spaces by differential operators, e.g.
$$e\to z^2 d/dz, h\to zd/dz, f\to d/dz,$$ 
$$\text{ where as usual}\quad 
e=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix},
h=\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix},
f=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix},
$$
$$\text{acts on }\quad V=z^\mu\mathbb{C}[z,z^{-1}]$$ (this is probably equivalent to your formulas after some change of coordinates).  They will not exponentiate to an action of $SU(2)$.
Nonetheless, as Scott indicated in the comments, some of these representations do arise from bona fide representations of the group $SL(2,\mathbb{R})$, which has the same complexified Lie algebra. This group is noncompact and some of them are unitary and some are not. You can use algebraic methods to describe a large class of representations of $SL(2,\mathbb{R})$ in this way. I've omitted some technical details, but all this is beautifully explained in Howe and Tan's book.<|endoftext|>
TITLE: Does every automorphism of G come from an inner automorphism of S_G?
QUESTION [12 upvotes]: I feel sort of silly asking this question. Unless I'm very much mistaken the paper I'm reading assumes the following statement:
Let $G$ be a finite group. We may embed it via the Cayley embedding into an ambient permutation group $G \leq S_{|G|}$. Then any automorphism of $G$ comes from conjugation by an element in $N_{S_{|G|}}(G)$.
Is this statement true?

REPLY [16 votes]: The statement is true. Let $g \in G$ and $\pi \in Aut(G)$. Let $\lambda_{g}$ be the corresponding left translation by $g$. Regard $\pi$ and $\lambda_{g}$ as elements of $Sym(G)$. Then for all $x \in G$,
$(\pi \lambda_{g} \pi^{-1})(x) = (\pi \lambda_{g}) ( \pi^{-1}(x)) = 
\pi( g \pi^{-1}(x)) = \pi(g) x = \lambda_{\pi(g)}(x)$.
Thus $\pi \lambda_{g} \pi^{-1} = \lambda_{\pi(g)}$.<|endoftext|>
TITLE: Hodge standard conjecture in positive characteristic
QUESTION [8 upvotes]: In the Wikipedia article on the Hodge Standard Conjecture it is written (note [Oct. 2015]: it has since been fixed):

In characteristic zero the Hodge standard conjecture holds, being a consequence of Hodge theory. In positive characteristic the Hodge standard conjecture is known only for surfaces and abelian varieties.

I have three questions: 
(1) Is the characteristic zero version the Hodge Index Theorem? 
(2) If so, what is a good reference for an algebraic geometry proof? I know it can be proved for a surface using the Riemann-Roch Theorem. Does this continue to be true for higher dimensions?
(3) Is the conjecture really only known in positive characteristic for surfaces and abelian varieties? Surely it should be possible to at least compute a result for projective $n$-space.

REPLY [4 votes]: (1) No, the Hodge index theorem, valid in arbitrary characteristic, is the Hodge standard conjecture for divisors on surfaces. Taking 2-dimensional linear  sections, one can deduce the Hodge standard conjecture for divisors on arbitrary smooth projective varieties. 
The Hodge standard conjecture over $\mathbf{C}$ is not a consequence of the Hodge index theorem but rather of the fact that the space of algebraic cycles is contained in the space of Hodge classes $H^{p,p}(X)\cap H^{2p}(X, \mathbf{Q})$, and by Hodge theory (Riemann bilinear relations), the Lefschetz pairing is definite on the primitive parts of those a priori bigger spaces. 
(2) I do not think there is an algebraic proof, since the above arguments requires some "positivity" (a definite pairing on a $\mathbf{Q}$-vector space remains definite after restricting to any subspace). 
Speculation: Maybe the Weil cohomology theory for varieties over $\overline{\mathbf{F}}_p$ with values in the Kottwitz category ${\rm Kt}_{\mathbf{R}}$ whose existence was recently conjectured by Scholze could make such an argument possible in positive characteristic.
(3) Tetsushi Ito in "Weight-monodromy conjecture for $p$-adically uniformized varieties" (Invent. Math. 2005) crucially proved the Hodge standard conjecture for the varieties obtained from $\mathbf{P}^n$ over $\mathbf{F}_q$ by successively blowing up all $\mathbf{F}_q$-points, the strict transforms of the lines between those points etc. The cohomology of these varieties is generated by algebraic cycles, but they do not lift to characteristic zero. 
(Obviously, if a variety in characteristic $p$ whose cohomology is generated by algebraic cycles lifts to characteristic zero "together with all cycles", then the Hodge standard conjecture holds.)<|endoftext|>
TITLE: Periodic Automorphism Towers
QUESTION [23 upvotes]: In Scott's classic textbook on Group Theory, he asks:
Suppose that $G$ is a finite group. Is the sequence of isomorphism types
of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?
Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence
$|Aut^{(n)}(G)|$ always bounded above?
It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:
Does there exists a finite group such that $Aut(G) \not \cong G$ but
$Aut^{(n)}(G) \cong G$ for some $n \geq 2$?
Edit: Joel has pointed out that my question is perhaps even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.

REPLY [8 votes]: As a follow up to my previous answer, I have computed all the automorphism series for groups of order up to 24 within the limits of what I could do with GAP. I will share some of what I found, throughout I will use the notation $(n,d)$ for the group of order $n$ with GAP library id $d$. For large groups not contained in the library, I will use $<r,n>$ to denote a group with minimum generating set of size $r$ and of order $n$, if I know only a bound for $r$, I will use $<\leq r,n>$ to denote that fact. We will say that a group $G$ stabilizes (at $k$) if $Aut^{(k-1)}(G) \not\simeq Aut^{(k)}(G) \simeq Aut^{(k+1)}(G)$. We will call a group $G$ stable if $G \simeq Aut(G)$.
I have verified that all groups of order up to 24 stabilize except for the following groups:
$(16,5)$, $(16,6)$, $(16,7)$, $(16,8)$, $(16,9)$, $(16,10)$, $(16,11)$, $(24,4)$, $(24,6)$, $(24,7)$, $(24,9)$, and $(24,10)$.
Of note is that many of these groups have the same automorphism group, hence their series is identical from $k=1$ onwards. Specifically, $(16,5)$, $(16,6)$, $(16,8)$, $(24,9)$, and $(24,10)$ all have $(16,11)$ as their automorphism group. The groups $(16,7)$, $(16,9)$ have the same automorphism group and so do the groups $(24,4)$, $(24,6)$, and $(24,7)$.
The following is a list of the groups I know to be stable, it is complete only up to order 24, and I give a Structure Description of the ones of order up to 24:
$(1,1) \simeq \mathbb{Z}_1$, $(6,1) \simeq S_3$, $(8,3) \simeq D_8$, $(12,4) \simeq D_{12}$, $(20,3) \simeq \mathbb{Z}_5 \rtimes \mathbb{Z}_4$, $(24,12) \simeq S_4$.
These are still in the library: $(40,12)$, $(42,1)$, $(48,48)$, $(54,6)$, $(110,1)$, $(144,183)$, $(336,208)$, $(384,5678)$, $(432,734)$, $(1152,157849)$.
These are too big to be in the library: $<2,40320> \simeq S_8$, $<4,442368> \simeq Aut^{(6)}((16,3))$.
These results give me two (general) ideas on how to attack this problem: One, analyze those groups for which I haven't been able to determine stabilization to see if I can find anything they have in common and use it show the conjecture is false. Two, Analyze the stable groups to see what causes them to be stable and use that knowledge to (somehow) show that every group must eventually stabilize. Both will likely require a detailed analysis of how $Aut(G)$ arises from $G$, to see how $G$ controls the properties of $Aut(G)$.
Edit:
I now have a complete list of stable groups of order up to 511, their GAP structure descriptions already reveal some very interesting patterns:
$(1,1)\simeq\mathbb{Z}_{1}$
$(6,1)\simeq S_{3}$
$(8,3)\simeq D_{8}$
$(12,4)\simeq D_{12}$
$(20,3)\simeq\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$
$(24,12)\simeq S_{4}$
$(40,12)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(42,1)\simeq(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(48,48)\simeq\mathbb{Z}_{2}\times S_{4}$
$(54,6)\simeq(\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(84,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(108,26)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(110,1)\simeq(\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2}$
$(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
$(252,26)\simeq S_{3}\times(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(272,50)\simeq\mathbb{Z}_{17}\rtimes\mathbb{Z}_{16}$
$(312,45)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(320,1635)\simeq((\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2)\rtimes\mathbb{Z}_5)\rtimes\mathbb{Z}_4$
$(324,118)\simeq S_{3}\times(\mathbb{Z}_9\rtimes\mathbb{Z}_3)\rtimes\mathbb{Z}_2)$
$(336,208)\simeq PSL(3,2)\rtimes\mathbb{Z}_2$
$(342,7)\simeq (\mathbb{Z}_{19}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_2$
$(384,5677)\simeq((((\mathbb{Z}_{4}\times\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(384,5678)\simeq((((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(432,520)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{2}$
$(432,523)\simeq(((\mathbb{Z}_{6}\times\mathbb{Z}_{6})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(432,533)\simeq\mathbb{Z}_{2}\times((((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(432,734)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(480,1189)\simeq(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})\times S_{4}$
$(486,31)\simeq(\mathbb{Z}_{27}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{2}$
$(500,18)\simeq(\mathbb{Z}_{25}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{4}$
$(506,1)\simeq(\mathbb{Z}_{23}\rtimes\mathbb{Z}_{11})\rtimes\mathbb{Z}_{2}$<|endoftext|>
TITLE: How would one even begin to try to prove that a simple number-theoretic statement is undecidable?
QUESTION [42 upvotes]: This question is closely related to this one: Knuth's intuition that Goldbach might be unprovable. It stems from my ignorance about non-standard models of arithmetic. In a comment on the other question, Chandan Singh Dalawat reproduced the following interesting quotation:

There are very many old problems in
  arithmetic whose interest is
  practically nil, i.e. the existence of
  odd perfect numbers, the iteration of
  numerical functions, the existence of
  infinitely many Fermat primes $2^{2^n}+1$,
  etc. Some of these questions may well
  be undecidable in arithmetic; the
  construction of arithmetical models in
  which questions of this type have
  different answers would be of great
  importance." (Bombieri, 1976)

What I'd like to know is, what might such a model look like, even roughly? For example, suppose one wanted to construct a model in which there were only finitely many Fermat primes. Would one do something like adjoin a nonstandard integer N, add the statement that all Fermat primes were less than N, and somehow demonstrate that that did not lead to a contradiction? (For all I know, that is an obviously flawed or even ridiculous suggestion.) A slightly more general question is this: is it conceivable that a number-theoretic independence proof might be achieved without one going too deeply into the number theory? (I ask that in the expectation, but not strong expectation, that the answer is no: that you can't get something for nothing.)

REPLY [2 votes]: I would like to extend the above question: It seems that most previous comments are dealing with independence of number theoretic statements from PA. I would like to know your input regarding the possible independence of number theoretic statements from ZFC. Unlike the Paris-Harrington theorem, in which we got the unprovability of a true statement, I would like to know whether some work has been done on proving the (symmetric/CH-like) independence of number theoretic statements from ZFC.
It seems that forcing is not very useful in this situation. Is it likely that such independence results may be somehow connected to possible (non-trivial) independence results from ZFC+V=L?<|endoftext|>
TITLE: Non-existence of algorithm converting NP algorithm to P algorithm?
QUESTION [9 upvotes]: [Edit: in the light of Nate Eldredge's answer below I rephrase the question]
P=NP is equivalent to the existence of a map of the following form:

Input: a polynomial-time non-deterministic Turing machine which accepts some language (call the language L) [Edit: we are not to assume these NDTMs come with any certificate proving they run in polynomial time -- Ryan requested this clarification, below]
Output: a polynomial-time deterministic Turing machine which accepts the language L

Is it known that if such a map exists then it cannot be computable?

REPLY [3 votes]: The question has basically been answered via the comments but it may help to summarize the conclusion.  If you insist that the input be unclocked NP machines then nothing useful can possibly be computed from the input, as explained in the answer to this related MO question by Joel David Hamkins.  But this kind of uncomputability result is, I would argue, completely uninteresting and irrelevant to your intended question, because it has absolutely nothing at all to do with P or NP.  It just amounts to the fact that arbitrary Turing machines are intractable objects.  On the other hand, if the input is a clocked NP machine, then Cook's reduction shows how to construct a P machine that solves your problem (assuming P = NP).  This is really what we care about in practice.  If I have a problem that I know is in NP, then I want a mechanical way of producing a polytime algorithm for it (assuming P = NP).  It's really irrelevant that there are all kinds of other, bizarre NP machines that accept the same language, and that it's an uncomputable task to sift through them.<|endoftext|>
TITLE: Who is the last mathematician that understood all of mathematics.
QUESTION [7 upvotes]: We were discussing this question at dinner this evening: Who is the last mathematician who had an understanding of a large proportion of mathematics (at the time they were alive)? 
I think it is safe to say that the mathematician lived and worked before the secong world war. After that period it has become impossible to aquire knowledge of such a large and rapidly growing subject. 
Picard (1856-1941) was the best suggestion we have come up with because he published research papers and wrote text books in a vast range of different mathematical subjects.

REPLY [6 votes]: Many people have mentioned Jean Dieudonné (1906-92) in this regard.

REPLY [6 votes]: By the way, facetiously one could add "Bourbaki" to that list.<|endoftext|>
TITLE: Non-negative integer solutions of a single Linear Diophantine Equation
QUESTION [6 upvotes]: Consider the following linear Diophantine Equation::
  ax + by + cz = d  ------------ (1)

for all, a,b,c and d positive integers, and relatively prime, and assume a>b>c without loss of generality.
Can we find a lower bound on d which ensures at least one non-negative solution to this equation?
I know we can solve this problem easily for 
     ax+by = c. -------- (2)

The answer is   
c>=ab, ------------(3)
which derives from the fact that the distance  between two consecutive solutions  of this equation is
     $D = (\sqrt(a^2 + b^2))$  ----------(4)

and c>=ab ensures that the length of line in x-y plane is large enough to include at least one solution).
Since equation (1) is a plane in the xyz coordinate system, and the distance between consecutive solution can be shown to be DD = sqrt(b^2+c^2) (though this may not be smallest distance between solutions). I was thinking that if we can show that an inscribed circle with diameter DD can be enclosed within the triangle formed by x,y and z intercepts of Eq.(1), i.e. (c/a,0,0), (0,c/b,0) and (0,0,c/a), then we have at least one non-negative solution. But the in-circle radius has an inconvenient relationship with the original variables (a,b,c,d), and may not be a monotonic function of d. 
Is there a smarter way to do this? and if such a bound exists, can it be extended to higher dimensions?
Thanks.

REPLY [4 votes]: There's a whole book on this problem; J L Ramirez Alfonsin, The Diophantine Frobenius Problem, Oxford University Press, 2005.<|endoftext|>
TITLE: Generalization of Hamiltonian cycles to "Hamiltonian spheres"
QUESTION [8 upvotes]: One possible generalization of a Hamiltonian cycle in a triangulated plane graph is what could be
called a Hamiltonian sphere: a collection of triangles within a simplicial complex in $\mathbb{R}^3$
that forms a surface homeomorphic to a sphere and which includes every vertex.  For example, the triangulated surface of a cube is a Hamiltonian sphere for any one of the tetrahedralizations of the cube interior. And the notion could be generalized to arbitrary dimensions.
Has this concept been studied?  I can imagine there are results specifying properties of the simplicial complex that guarantee it is Hamiltonian in the sense above.  A trivial example is that the convex hull of points in convex position constitute a Hamiltonian sphere for a triangulation of the hull interior into simplices. (Here convex position means that all points are on the hull.)

REPLY [3 votes]: Amos Altshuler studied a related notion in his Ph. D. thesis and the paper "Altshuler, Amos Manifolds in stacked $4$-polytopes. J. Combinatorial Theory Ser. A 10 1971 198--239." In his version he allows manifolds and not only spheres and demand that the manifold contains every edge. (He consider simplicial 4-polytopes but thos can be realized as simplicial complexes in 3-space.)
I think I saw some subsequent papers on similar notions of "hamiltonian manifolds and spheres" but I could not track them. (I think this is Schulz ' paper)
Other related papers (still not those I thought about)  Effenberger, Felix; Kühnel, Wolfgang, Hamiltonian submanifolds of regular polytopes. Discrete Comput. Geom. 43 (2010), 242--262. 
Schulz, C.: Polyhedral manifolds on polytopes, Proc. Conf. Palermo 1993. Rend. Circ. Mat. Palermo (2) Suppl. 35, 291--298 (1993) 
The Mathscinet review by David Walkup reads:
M is said to be a “good” manifold in L if L is a simplicial complex, M is a subcomplex of L
containing all the edges of L, and |M| is a closed 2-manifold. The case when |L| is a 3-sphere,
especially if L is the boundary complex of a simplicial 4-polytope, is of interest. A “stack” is
defined inductively as follows: A 4-simplex with its faces is a stack, and the union of two stacks
is a stack if their intersection is the closure of a common 3-simplex. If K is a stack, then |K| is a
4-ball, the boundary complex Bd(K) is well-defined, and Bd(K) can be realized as the boundary
complex of a simplicial 4-polytope. Any good manifold in a stack K must lie entirely in Bd(K).
A “star” is a stack obtained by stacking a 4-simplex on each of the five 3-faces of a central 4-
simplex. Theorem 7: If K is a star, then there are exactly 6 good manifolds in Bd(K), each is
a torus, and all 6 imbeddings are isomorphic. Theorem 14: For any n >= 1 there is a stack K
of 6n 4-simplices which can be obtained by stacking together n stars so that Bd(K) contains a
good manifold of genus n. Conversely, Theorems 16 and 4: If M is a good manifold of genus
n in a stack K, then K must be the result of stacking together exactly n stars. Other theorems
characterize those ways of stacking stars so that the result admits a good manifold. Theorem 23:
The maximum number of different good manifolds in K as K ranges over all stacks with 6n 4-
simplices is 6, 8, 12, 24, 40, 80, or 2n according as n = 1, 2, 3, 4, 5, 6, or n >= 7. Theorem 26: For
every g >= 1 there exists a 3-sphere which cannot be realized as the boundary of a stack but does
contain a good manifold of genus g. Many other nice results and fruitful ideas are developed.
The MR for Schulz paper: The author gives a survey on polyhedral manifolds which are contained in the boundary complex of convex polytopes. Manifolds with certain extremal properties (e.g. minimal number of vertices) are of particular interest. Another fruitful field is Hamiltonian $2$-manifolds (i.e. $2$-manifolds containing the $1$-skeleton of the polytope). The author gives a wealth of material by several authors from Möbius, Coxeter, Ringel, Jungerman to Altshuler, Brehm, Bokowski, Kühnel, McMullen, Schulz and others.<|endoftext|>
TITLE: Diophantine equation: Egyptian fraction representations of 1
QUESTION [14 upvotes]: According to the OEIS (A002966) there are 294314 solutions in positive integers to the equation
$$\sum_{i=1}^7\frac{1}{x_i}=1$$ assuming $x_1\leq x_2\leq\cdots\leq x_7$.
Similarly for 8 summands there are 159330691 solutions.
My question: What are they? Is there a way of counting them without knowing them?
The bound for $x_n$ for $n$ summands is double exponential and I could only compute the solutions up to $n=6$ with Maple.

REPLY [2 votes]: A related interesting problem is  the number of partitions of 1 into n distinct positive Egyptian fractions, the first few terms of which are given in A006585.
These values were counted laboriously by brute force in the early 1990's, and although they have been reconfirmed (by Jud McCranie) the sequence has only been extended one further term (by John Dethridge) in 2004.
Additional terms, cleverer algorithms and/or a generating function, would be most welcome!<|endoftext|>
TITLE: Examples of statements that provably can't be proved using a promising looking method 
QUESTION [43 upvotes]: Motivation: In Razborov and Rudichs article "Natural proofs" they define a class of proofs they call "natural proofs" and show that under certain assumptions you can't prove that $P\neq NP$ using a "natural proof". I know that this kind of results is common in complexity theory, but I don't know any good examples from other fields. This is why I ask:
Question: Can you give an example of a statement S that isn't known to be unprovable (it could be an unsolved problem or but it could also be a theorem), a promising-looking class of proofs and a proof that a proof from this class can't prove S.
I'm interested in both famous unsolved problems and in elementary examples, that can be used to explain this kind of thinking to, say, freshmen.

REPLY [3 votes]: Terry Tao has a great post on his blog explaining the limitations of the circle method with respect to the binary Goldbach conjecture.<|endoftext|>
TITLE: Why Is $\frac{163}{\operatorname{ln}(163)}$ a Near-Integer?
QUESTION [36 upvotes]: The $j$-function and the fact that 163 and 67 have class number 1 explain why:
$\operatorname{exp}(\pi\cdot \sqrt{163}) = 262537412640768743.99999999999925$,
$\operatorname{exp}(\pi\cdot \sqrt{67}) = 147197952743.9999987$.
But is there any explanation for these?:
$\frac{163}{\operatorname{ln}(163)} = 31.9999987 \approx 2^5$,
$\frac{67}{\operatorname{ln}(67)} = 15.93 \approx 2^4$,
$\frac{17}{\operatorname{ln}(17)} = 6.00025$.
These numbers seem too close to integers to occur by chance.

REPLY [9 votes]: What make us confident about some mystery in these observations?
1st note:
"An example discovered by Srinivasa Ramanujan around 1913 is $\exp(\pi\sqrt{163})$,
which is an integer to one part in $10^{30}$, and has second continued fraction term
$1,333,462,407,511$. (This particular example can be understood from the fact
that as $d$ increases $\exp(\pi\sqrt{d})$ becomes extremely close to
$j((1 + \sqrt{-d})/2)$, which turns out to be an integer whenever there
is unique factorization of numbers of the form $a + b \sqrt{-d}$ --- and $d=163$
is the largest of the 9 cases for which this is so.) Other less spectacular examples
include $e^{\pi}-\pi$ and $163/\log(163)$."
2nd note:
"Any computation involving 163 gives an answer that is close to an integer:
$$
163\pi = 512.07960\dots, \quad
163e = 443.07993\dots, \quad
163\gamma = 94.08615\dots\text{"}
$$
and
$$
\text{"}67/\log(67)=15.9345774031\dots, \quad
43/\log(43)=11.432521184\dots
$$
...nah, with class number 1 it's not connected.
It's just the same 163. $\ddot\smile$"
A synthetic example of my own:
$$
\root3\of{163}-\frac{49,163}{9,000}
=0.0000000157258\dots
$$
(note the double appearance of 163).
So, let's feel that the prime 163 is a supernatural number. $\ddot\smile$
EDIT. Another interpretation the original question is related to
the observation of Kevin O'Bryant who computed the first successive maxima
of the sequence $\|n/\log(n)\|$ where $\|\ \cdot\ \|$ denotes the distance
to the nearest integer. The existence of infinitely many terms
is guaranteed by the following
Problem.
For any $\epsilon>0$, there exists an $n$ such that $\|n/\log(n)\|<\epsilon$.
See solution by Kevin Ventullo to this question. I hope that this fact demystifies the original problem in full.<|endoftext|>
TITLE: In what ways is physical intuition about mathematical objects non-rigorous?
QUESTION [29 upvotes]: I'm asking this question as a mathematician who is very far removed from the Physics world, and has little to no knowledge of what math goes into it, and what math comes out of it. What I do hear is that people have "physical intuition" about mathematical objects (especially in algebraic geometry), and that they then try to prove it mathematically.
So out of curiousity, my question, then, is this. Which combination of the following is true for why "physical intuition" isn't already rigorous:

They assume the existence of objects that they don't construct.
Their logic is flawed.
They experiment (with particles and such) and assume that if it works enough times then it is true.
They assume that "reasonable" mathematical conjectures are true without bothering to be sure.
They don't have axiomatized definitions, and rely on vague notions.

REPLY [7 votes]: I think some of what's been said is a little misleading.  A lot of people have implied that because physicists have the ability to rely on experiments, they don't have to worry as much about formal proofs.  But this isn't really right, and although ultimately if a particular theory is "correct" or not depends on this, physical intuition can still lead to good ideas and good math, even if it leads to empirically incorrect ideas (in fact, many good theoretical physicists care as little for experiments as they do for proofs!).  The history of physics is actually full of "recycling" good ideas from places where they didn't work, to new places where they do!  These ideas are so re-usable not because they were based in experiment, but because the kind of reasoning behind them was qualitatively good, even if it did not end up being quantitatively good (compare to similar instances in math!).
I'm a theoretical physicist, but have an undergrad degree in math, and that's caused me to put a lot of thought into exactly how my thinking in my physics training differs from my thinking in my math training.  And I really think that the kind of qualitative thinking that I see a lot of mathematicians use to reason with before building a formal argument is almost exactly the kind of thinking physicists use.
I know one of the things made me first realize this, was a few years ago, reading a blog post by Terry Tao. The post was about some analysis topic that I wasn't familiar with (which I no longer recall specifically) and I had stopped to think carefully about what he was saying for a few minutes after seeing a confusing statement, and tried to reason through it using my physics intuition.  After getting some idea of what was going on, I finished reading the article, and, after finishing it, I realized that the logic behind the article as a whole (as opposed to behind each individual statement) was basically identical to what my physics explanation was.
In terms of the points you mention above, "physics intuition" would correspond roughly to 1, 5, and a lesser extent 4.  But this (from my point of view) seems to correspond pretty well to exactly how mathematicians think both before they formalize an argument carefully, and in the "big picture" point of view (which is really partly inherited from the former).  
So in a sense, physical intuition is everything you do in math, up to, but not including the final step when you make your arguments careful.  Although we usually go "most" of the way to making an argument careful, ultimately we do have to bring things to the level of being able to make a calculation which one could compare with experiment, and this requires being fairly careful about the reasoning we use being mathematically sensible (although, from the point of view of most theorists, this is not the interesting part).
We also like to break our arguments up into "fundamental" pieces, but not in the same way as mathematicians do, in terms of axioms/definitions/theorems/lemmas, but in terms of "physically reasonable" pieces, since they are easier to get a handle on in terms of theory-building.  But the problem is that, while these physically reasonable pieces usually correspond to simple physical statements, they usually correspond to very complicated statements when spelled out axiomatically, which makes that form of them too cumbersome to work with.  
It's difficult to explain specifically what the similarities I'm thinking about are, so if you want to see some specific examples, that would be more amicable from a mathematician's point of view, it could be valuable to grab a text on the calculus of variations and go through some of the proofs of things that you already know through other means (e.g., geodesics) since this specifically is one kind of reasoning that's used all over physics.  There're also a number of such books written from a solving physics problems point of view.  
There is also "quantum fields and strings: a course for mathematicians" which is a bit tougher, but written by actual physicists, and I think could give a good deal of insight into how we actually think.  I would avoid anything called "quantum mechanics for mathematicians" for this because they tend to not be written by people who are primarily physicists.
You could also go back to Euler or Gauss or Riemann, since a lot of their arguments are very "physical" and are highly recognizable for physicists.  I believe Spivak's volumes on differential geometry contain some of Riemann's papers, along with discussion translating them into modern language which could be useful to see.  The MAA also has a "how Euler did it" column that could be interesting in this regard, too.<|endoftext|>
TITLE: Do there exist chess positions that require exponentially many moves to reach?
QUESTION [50 upvotes]: By "chess" here I mean chess played on an $n\times n$ board with an unbounded number of (non-king) pieces.  Some care is needed if you want to generalize some of the subtler rules of chess to an $n\times n$ board, but I will not dwell on this point because the answer to the question I'm interested in should be the same under any reasonable generalization.  Namely, does there exist an infinite sequence $(A_n, B_n)$ of pairs of chess positions on an $n\times n$ board such that the minimum number of legal moves required to get from $A_n$ to $B_n$ is exponential in $n$?  Here I allow any legal moves and not only strategically intelligent moves.
Technically this question might be classified as an "open problem" (which is illegal on MO) because it was implicitly asked by A. Fraenkel and D. Lichtenstein in "Computing a perfect strategy for $n\times n$ chess requires time exponential in $n$," J. Comb. Th. A 31 (1981), 199–214.  However, I think it is fair game for MO because I'm pretty confident that this has not been looked at much.  Fraenkel and Liechtenstein showed that determining whether a given chess position is won for White (with best play) is EXPTIME-complete and asked for the computational complexity of the chess reachability problem ("is $B_n$ reachable from $A_n$?").  Clearly chess reachability is in NPSPACE = PSPACE, and Hans Bodlaender has shown that it is NP-hard.  If the answer to the question I've posed above is "No, it can always be done with polynomially many moves" then it would solve this problem by showing that chess reachability is NP-complete, because exhibiting the sequence of moves yields a short certificate.
If you have some experience with retrograde chess problems and if you've read Hearn and Demaine's lovely book Games, Puzzles, and Computation then you may get the intuition that shuffling chess pieces around is reminiscent of other rearrangement puzzles that have been shown to be PSPACE-complete.  However, I've asked both Demaine and Hearn and neither of them saw immediately how to show that chess reachability is PSPACE-complete.
[Edit: Searching more carefully through Hearn and Demaine's book, I see that they list this problem in their list of open problems at the end of the book under the name "Retrograde Chess."  I didn't notice it before because for some reason that page is not listed in the index under "chess."  I can perhaps be blamed for using the name "Retrograde Chess" for this problem because that's what I called it when I first posted this question to USENET way back when.  I think that "reachability" is a better name for it.]

REPLY [16 votes]: Here is a summary of solution proposed in this arXiv paper.
A pair of positions on the $n\times n$ chessboard is constructed
for which (1) there is a sequence $\sigma$ of legal chess moves leading
from $P$ to $P'$; (2) the length of $\sigma$ cannot be less than $\exp\Theta(n)$.
The idea is to construct a position which consists essentially of $m$ tracks each of which is in some state in every moment. The set of possible states of $i$th track is a cycle group of order equal to $(i+3)$rd prime, and the position is defined uniquely by states of all tracks. A "move" increases every state by $1$ or decreases every state by $1$. Transforming $(0,0,\ldots,0)$ to $(1,0,\ldots,0)$ is possible by Chinese remainders but requires at least $p_5\ldots p_{m}=\exp(p_m+o(p_m))$ "moves".

  (source)

The chess positions representing tracks use the above pattern, which is a specific example corresponding to $m=3$. The dots denote pawns, and we assume that the kings are located somewhere else on the board. In the paper it is explained (in different terms) why we can think of dark cycles as tracks, the positions of white bishops on them as states, and "moves" as minimal sequences of legal chess moves whose initial and resulting positions coincide up to a color of pieces.<|endoftext|>
TITLE: When does replacement (accidentally) hold in amenable sets?
QUESTION [9 upvotes]: A set $M$ is called amenable if it is transitive and satisfies the following conditions:

For all $x,y\in M$,  $\{x,y\}\in M$
For all $x\in M$, $\bigcup x \in M$
$\omega \in M$
For all $x,y \in M$, $x\times y \in M$
($\Sigma_0$ comprehension) Whenever $\Phi$ is a $\Sigma_0$ formula of one free variable with parameters from $M$, then for all $x\in M$, $\{z\in x | \Phi(z)\}\in M$

Although the definition of an amenable set does not include replacement, some very limited amount of replacement follows from the axioms given.  For example, for all $x,y\in M$, it must be that $\{\{z,w\}|z\in x, w\in y\}\in M$ and $\{\{z\}|z\in x\}\in M$.  So just how limited is the replacement in amenable sets?  In particular,

If $M$ is an amenable set and $x\in M$, does it follow that $\{\bigcup z | z \in x\} \in M$?

REPLY [5 votes]: I agree with Andreas Blass's solution. The problem, or difficulty, with the definition
of amenable is highlighted with this example: $\Sigma_0$-comprehension in this questioner's
scheme is not really adequate.
For this reason it is sometimes replaced with $\Sigma_0$- (or `rudimentary') -closure for
$\Delta_0$ formulae $\varphi$: 
$$\forall x \exists w \forall \vec v \in x \exists t\in w \forall u
   (u \in t \leftrightarrow u \in x \wedge \varphi[u, \vec v])  $$
This is more useful, implies $\Sigma_0$-Comprehension, and rules out the undesirable effect of the example.<|endoftext|>
TITLE: Decidability of chess on an infinite board
QUESTION [59 upvotes]: The recent question Do there exist chess positions that require exponentially many moves to reach? of Tim Chow reminds me of a problem I have been interested in. Is chess with finitely many men on an infinite board decidable? In other words, given a position on an infinite board (say $\mathbb{Z}\times \mathbb{Z}$, though now pawn promotion is not possible) with finitely many men, say with White to move, is there an algorithm to determine whether White can checkmate Black (or prevent Black from checkmating White) against any Black defense?

REPLY [36 votes]: There is a positive solution for the decidability of the mate-in-$n$ version of the problem. 
Many of us are familiar with the White to mate in 3
variety of chess problems, and we may consider the natural analogue
in infinite chess. Thus, we refine the winning-position problem,
which asks whether a designated player has a winning strategy from
a given position, to the mate-in-$n$ problem, which asks whether
a designated player can force a win in at most $n$ moves from a
given finite position. (And note that as discussed in Johan
Wästlunds's question checkmate in $\omega$
moves?, there are finite winning positions in infinite
chess which are not mate-in-$n$ for any finite $n$.) Even so, the
mate-in-$n$ problem appears still to be very complicated, naturally
formulated by assertions with $2n$ many alternating quantifiers:
there is a move for white, such that for every black reply, there
is a countermove by white, and so on. Assertions with such quantifier complexity are 
not generally decidable, and one cannot expect to search an
infinitely branching game tree, even to finite depth. So one might
naturally expect the mate-in-$n$ problem to be undecidable.
Despite this, the mate-in-n problem of infinite chess
is computably decidable, and uniformly so. Dan Brumleve, myself and Philipp Schlicht have just submitted
an article establishing this to the CiE 2012, and I hope to speak on it there
in June.

D. Brumleve, J. D. Hamkins and P. Schlicht, "The mate-in-n problem of infinite
  chess is decidable," 10 pages, arxiv pre-print, submitted to CiE 2012.
Abstract. Infinite chess is chess played on an infinite
  edgeless chessboard. The familiar chess pieces move about according
  to their usual chess rules, and each player strives to place the
  opposing king into checkmate. The mate-in-$n$ problem of infinite
  chess is the problem of determining whether a designated player can
  force a win from a given finite position in at most n moves. A
  naive formulation of this problem leads to assertions of high
  arithmetic complexity with $2n$ alternating quantifiers---there is
  a move for white, such that for every black reply, there is a counter-move
  for white, and so on. In such a formulation, the problem does not
  appear to be decidable; and one cannot expect to search an
  infinitely branching game tree even to finite depth. Nevertheless,
  the main theorem of this article, confirming a conjecture of the
  first author and C. D. A. Evans, establishes that the mate-in-$n$
  problem of infinite chess is computably decidable, uniformly in the
  position and in $n$. Furthermore, there is a computable strategy
  for optimal play from such mate-in-$n$ positions. The proof
  proceeds by showing that the mate-in-$n$ problem is expressible in
  what we call the first-order structure of chess $\frak{Ch}$, which
  we prove (in the relevant fragment) is an automatic structure,
  whose theory is therefore decidable. Unfortunately, this resolution
  of the mate-in-$n$ problem does not appear to settle the
  decidability of the more general winning-position problem, the
  problem of determining whether a designated player has a winning
  strategy from a given position, since a position may admit a
  winning strategy without any bound on the number of moves required.
  This issue is connected with transfinite game values in infinite
  chess, and the exact value of the omega one of chess $\omega_1^{\rm
chess}$ is not known.

The solution can also be cast in terms of Presburger arithmetic, in a manner close to Dan Brumleve's answer to this question. Namely, once we restrict to a given collecton of pieces $A$, then we may represent all positions using only pieces in $A$ as a fixed-length tuple of natural numbers, and the elementary movement, attack and in-check relations are expressible for this representation in the language of Presburger arithmetic, essentially because the distance pieces---rooks, bishops and queens---all move on straight lines whose equations are expressible in Presburger arithmetic. (There is no need to handle sequence coding in general, since the number of pieces does not increase during play.) Since the mate-in-$n$ problem is therefore expressible in Presburger arithmetic, it follows that it is decidable.<|endoftext|>
TITLE: Why is there no Cayley's Theorem for rings?
QUESTION [58 upvotes]: Cayley's theorem makes groups nice: a closed set of bijections is a group and a group is a closed set of bijections- beautiful, natural and understandable canonically as symmetry. It is not so much a technical theorem as a glorious wellspring of intuition- something, at least from my perspective, that rings are missing; and I want to know why.
Certainly the axiom system is more complicated- so there is no way you're going to get as simple a characterisation as you do with groups- but surely there must be some sort of universal object for rings of a given cardinality, analogous to the symmetric group in group theory. I would be surprised if it was a ring- the multiplicative and additive properties of a ring could be changed (somewhat) independently of one another- but perhaps a fibration of automorphisms over a group? If so is there a natural(ish) way of interpreting it?
Perhaps it's possible for a certain subclass of rings, perhaps it's possible but useless, perhaps it's impossible for specific reasons, in which case: the more specific the better. 
Edit: So Jack's answer seems to have covered it (and quickly!): endomorphisms of abelian groups is nice! But can we do better? Is there a chance that 'abelian' can be unwound to the extent we can make this about sets again- or is that too much to hope for?

REPLY [45 votes]: You can think of Cayley's theorem as the special case of Yoneda's lemma, where the category has only one object. If you take the additive version of Yoneda's lemma, and you plug in an additive category with one object then you get the desired statement for rings that's in Jack's answer. Viewing Cayley's theorem like this let's you generalize it to many other structures then groups.<|endoftext|>
TITLE: Relatively ample line bundles
QUESTION [18 upvotes]: Is it true that a line bundle is relatively ample iff its restsriction to fibers is? If so, what would be the reference?

REPLY [2 votes]: EDIT : my previous answer was wrong. Thanks to BConrad for pointing it out.
Here is a counterexample if the map is not proper. Let $X$ be the plane, let $Y$ be the blow-up of the plane in one point, and $U$ be $Y$ with one point of the exceptionnal divisor removed. Let $f|_U:U\to X$ be the projection and consider the line bundle $\mathcal{O}_U$. 
Since the fibers of $f|_U$ are affine (either points or the affine line), $\mathcal{O}_U$ becomes ample when restricted to fibers of $f|_U$. However, $\mathcal{O}_U$ is not $f|_U$-ample.
Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$, but we can compute :
$$H^0(U,\mathcal{O}_U^N)=H^0(U,\mathcal{O}_U)=H^0(Y,\mathcal{O}_Y)=H^0(X,\mathcal{O}_X),$$
where the second equality comes from property $S2$ and the third holds because $f_* \mathcal{O}_Y=\mathcal{O}_X$. Hence, $H^0(U,\mathcal{O}_U^N)$ cannot distinguish between two points of the exceptionnal divisor, and $\mathcal{O}_U$ cannot be ample on $U$.

Warning : what follows is false. I kept it here so that the comment below remains understandable.
Here is a counterexample if the map is not proper. Consider the inclusion $f$ of the plane minus a point $U$ in the plane $X$. The line bundle $\mathcal{O}_U$ is ample restricted to the fibers of $f$ (they're points...). However, it is not $f$-ample. Indeed, if it were, $\mathcal{O}_U$ would be ample on $U$. Choosing $N>>0$, we would get $$H^1(U,\mathcal{O}_U)=H^1(U,\mathcal{O}_U^N)=0.$$
But a simple computation via Cech cohomology shows that this cohomology group is not trivial (in fact, infinite).<|endoftext|>
TITLE: Does the free resolution of the cokernel of a generic matrix remain exact on a Zariski open set?
QUESTION [5 upvotes]: "Random" modules of the same size over a polynomial ring seem to always have the same Betti table. By a "random" module I mean the cokernel of a matrix whose entries are random forms of a fixed degree. 
For example if we take the quotient of the polynomial ring in three variables by five random cubics:  
$S = \mathbb{Q}[x1,x2,x3]$ 
$M$ = coker random( S^1, S^{5:-3} ) 
then Macaulay2 "always" (e.g. 1000 out of 1000 times) gives the following Betti table
total: 1 5 9 5 
0: 1 . . . 
1: . . . . 
2: . 5 . .
3: . . 9 5
It seems that the behavior can be explained by the fact that we can resolve the cokernel of a generic matrix of the given form and this resolution remains exact when specializing to any point in a Zariski open subset of some affine space. My question is whether anyone knows a slick proof of this fact.
To elaborate: we can adjoin a new variable to our original ring for each coefficient appearing in each entry of the matrix. So in the above example we would adjoin 10*5 = 50 new variables to $S$, say $y1..y50$. Call the new ring $T$. Consider the $1x3$ matrix $N$ over $T$ whose entries are cubic in the $x_i$ and linear in the $y_i$. Resolve the cokernel of $N$ over $T$ to get a complex $F$. We can then substitute any point in $\mathbb{Q}^{50}$ into the maps of $F$ to get a complex over the original ring $S$. The claim is that this complex is exact on a Zariski open set of the affine space $\mathbb{Q}^{50}$.
It seems like this must be well known but I'm having trouble finding references.

REPLY [6 votes]: It sounds like the question you mean to ask is the following: if $X$ is an integral noetherian scheme with generic point $\eta$ and $C^{\bullet}$ is a finite complex of coherent sheaves on $X$ such that $C^{\bullet}_ {\eta}$ is exact, then does there exist a dense open $U$ in $X$ such that $C^{\bullet}|_U$ is exact and has exact fibers?  If that is your question, then the answer is yes.  It is instructive to make your own proof by using generic flatness considerations for finite modules over noetherian domains (applied to various kernels, cokernels, images, etc.).  Alternatively, look at EGA IV$_3$, 9.4.2 and 9.4.3 for vast generalizations without noetherian hypotheses (e.g., 9.4.3 concerns compatibility of formation of homology sheaves with respect to passage to fibers over some dense open, not assuming exactness at the generic fiber). Even in this hyper-generality, the principle of using "generic flatness" remains the same.<|endoftext|>
TITLE: Splitting of the Universal Coefficients sequence
QUESTION [8 upvotes]: The really beautiful way to prove the Universal Coefficients theorem, to my taste, 
is to use the fibration sequence $K(\mathbb{Z}, n) \to K(\mathbb{Z}, n) \to 
K(\mathbb{Z}/k, n)$ (I'm using $\mathbb{Z}/k$ coefficients for simplicity)
to give a long exact sequence of cohomology, and split it into short exact sequences.
(I'm pretty sure I learned this from Adams' Generalized Homology...).
Anyway, this gives the desired exact sequence, but I don't know how to get
the (nonnatural) splitting that the homological algebra derivation provides.

REPLY [4 votes]: (This is a corrected version of my original, off-the--cuff answer).
If R has proj dim 1 and C is a flat chain complex, then you get the
UCT sequence.  To get the splitting, you need to also assume that
C is projective, so that the map C -->> B from the complex to the
subcomplex of boundaries has a splitting.  
A spectrum level construction of the splitting seems unlikely since you
don't have kernels and cokernels, but only fibers and cofibers there.<|endoftext|>
TITLE: Computing only the order of Galois group (not the group itself).
QUESTION [8 upvotes]: My question is related to  this one: Computing the Galois group of a polynomial.
I was wondering if there is a faster algorithm just to compute the order of the group rather than the group itself.
Also, has anybody compared the performance of GAP and Magma in computing Galois groups? I just heard Magma is very good at it.
I asked this question because I encounter every so often new bug with Magma's implementation and I wanted to see if I can implement something similar. But at this time I'm just interested in the exponent at the first place. This is the last annoying error that I get for basically any deg 5 poly that has Gal group $S_5$.
k := FiniteField(2);
kx<x> := RationalFunctionField(k);
kxbyb<y> := PolynomialRing(kx);
MinP :=  y^5 + y + x^2 + x;
print GaloisGroup(MinP);

The result is:
Runtime error: too much looping

Which I don't understand what it means (Magma Ver 2.16-8).
To be more clear, my ultimate goal is to check a lot of polynomials and throw out those with $S_n$ Gal group and focus on those which are not such. As you see even an upper bound over the exponent is enough for me.

REPLY [9 votes]: I am actually one of the authors of the Galois package in Magma. Firstly, the "too much looping" error does not happen anymore (for this example at least) in the current Magma version (2.16-13). Secondly, the way Sn/An is recognized in general is through the use of
factorisation as suggested. More precisely, the polynomial is factored modulo several primes and the resulting factors (well their degrees) are noted. Those give possible cycle
types of the Galois group. If cycle types of certain patterns happen, we know the group
is An/Sn. Those types are very frequent, hence this is trivial. However, then we hit a problem. In order to distinguish An and Sn usually one looks at the discriminant of the polynomial with the idea that the group is An (or in general contained in An) iff the discriminant is a square. This unfortunately breaks down in characteristic 2 and the currently employed test is slow. (And caused the "too much looping" message).
Unfortunately, we don't have an interface like IsAnOrSn(f) which would be sufficient here.
In general, looking at cycle types or even at types and their frequency, will not determine the group nor the group size. All one gets from here are a lower bounds. However
for small degrees (and 5 is small) this would work.
The connection between Kash and Magma here is difficult: Magma used to rely on Kash for the Galois groups, but the algorithm was limited to degree <= 23. This is the PhD of
Katharina Geissler, her thesis can be found on the Kash page in Berln. The current Magma implementation of Galois groups is independent and does not share any code with Kash.<|endoftext|>
TITLE: At what point in history did it become impossible for a person to understand most of mathematics?
QUESTION [12 upvotes]: Disclaimer:
I am asking this question as an improvement to this question, which should be community wiki.  This is in line with the actions taken by Andy Putman in a similar case (cf. meta).
See the relevant meta thread about the previous question.  
Edit: If it wasn't already obvious, I only asked this question to prevent the other one (which was not made community wiki) from being reopened.  
Question:
The scope of mathematics has grown immensely since ancient times.  At what point in time did it become impossible for a single person to understand the majority of mathematics enough to keep current with contemporary research?  
Edit: Clarified the wording.

REPLY [11 votes]: The world's output of scientific papers increased exponentially from 1700 to 1950. 
One online source is this article (which is concerned with what has happened since then). The author displays a graph (whose source is a 1961 book entitled "Science since Babylon" by Derek da Solla Price) showing exponential increase in the cumulative number of scientific journals founded; an increase by a factor of 10 every 50 years or so, with around 10 journals recorded in 1750.
Perhaps someone can locate similar statistics specific to mathematics, but it's reasonable to expect the same pattern. If so, it is a long time since any individual could follow the primary mathematical literature in anything close to its entirety. 
But then, gobbling papers is not how leading mathematicians (or scientists) actually operate. 
By making judicious choices of what to pursue when, and with sufficient brilliance and vision, it is possible even today to make decisive contributions to many fields. Serre has done so in, and between, algebraic topology, complex analytic geometry, algebraic geometry, commutative algebra and group theory, and continues to do so in algebraic number theory/representation theory/modular forms.<|endoftext|>
TITLE: Differential of the Torelli morphism at the boundary
QUESTION [9 upvotes]: Let consider the Torelli morphism $T:\mathcal{M}_g \rightarrow \mathcal{A}_g$, from the moduli space of curves of genus $g$ to the moduli space of principal polarized abelian varieties of dimension $g$, that maps a curve to its Jacobian. The differential of $T$ at a point $[C]$ is the natural map
$$H^1(C, T_C) \rightarrow Sym^2H^1(C, \mathcal{O}_C).$$
I know that $T$ can be extended to a map 
$$T:\bar{\mathcal{M}_g} \rightarrow \bar{\mathcal{A}_g}$$
from the Deligne-Mumford compactification of $\mathcal{M}_g$ to some compactification of $\mathcal{A}_g$.
I would like to know if there is a way to describe the differential of $T$ at a point representing a nodal curve. More specifically, how can we describe the deformations space of a semi-abelian variety and in particular of a generalized Jacobian variety?
Over $\mathbb{C}$, by computing the period matrix, one can show that the differential of $T$ has maximal rank at each point representing a nodal curve with non-hyperelliptic  normalization. I'm wondering if, perhaps, there is a more algebraic way to see it.

REPLY [2 votes]: As mathwonk said, Friedman and Smith has what you want although is still over C. For a more general result, stated algebraically and a connection with Serre-Tate deformation theory, see my paper with Coleman, also in Inv. Math. (1992).<|endoftext|>
TITLE: Clarification of Gödel's second incompleteness theorem
QUESTION [16 upvotes]: I am sorry for the following question, because the actual answer to this question is in the beautiful works of Feferman and Jeroslow, but, unfortunately, I havn't any time to go into that specific field right now and maybe some of you is aware of the answer.
The situation with Gödel's second incompleteness theorem is quite delicate. Let Pf(a, b) means that a is the Gödel number of proof of statement with Gödel number b and Neg(a, b) means that b is the Gödel number of negation of statement with Gödel number a. The main result of Gödel's work is representability of predicates Pf and Neg in formal arithmetic. Knowing that, we can define the Consistency of formal arithmetic in formal arithmetic as follows:
" it is not the case that there exist a statement A such that A and (not) A are provable in formal arithmetic " or in the language of formal arithmetic:
$$\forall x_1, x_2, x_3, x_4 [\neg (\operatorname{Pf}(x_1, x_3) \wedge \operatorname{Pf}(x_2, x_4) \wedge \operatorname{Neg}(x_3, x_4))].$$
Let's denote last proposition by W. Gödel's second incompleteness theorem says that W is not provable in formal arithmetic, i. e. the consistency of formal arithmetic is not provable in formal arithmetic.
But Solomon Feferman in his remarkable paper of 1960 "Arithmetization of metamathematics in general setting" have found other formalization of consistency of formal arithmetic W' that is provable in formal arithmetic (see notes by E. Mendelson to the second theorem in formal arithmetic chapter of his "Intoduction to Mathematical logic"). Jeroslow ("Consistency statements in formal theories"), thereafter, studied consistency statements in a broader way.
Can someone explain the nature of W'?
(obviously, $W\Leftrightarrow W'$ can not be proven in PA).
Can we interpret the provability of W' as the proof of consistency of formal arithmetic in formal arithmetic?
How well are various consistency statements grasped today?  
Thanks in advance.

REPLY [4 votes]: Reading between the lines, I think your question may be answered in David Auerbach's article "How to Say Things With Formalisms."  Regarding whether these exotic "consistency" statements may be interpreted as "expressing" consistency, I would say that the answer is no.  The crux of the matter may be put more simply.  Let's suppose we want to express "$x$ is even" in the first-order language of arithmetic.  A natural attempt is $\exists y: y+y=x$.  However, you could be sneaky and try something like $\exists y: [y+y=x \wedge \neg G(y,"1=0")]$ where $G(a,b)$ says "$a$ is not the Goedel number of a proof of $b$."  If the theory in question is consistent then your exotic evenness predicate will be true of precisely the even integers.  However, it's pretty clear that there's something fishy about this predicate.  It doesn't really say only that $x$ is even; it says that $x$ is even and $x/2$ doesn't witness a proof of a contradiction.  Similarly, the exotic consistency predicates don't just say that the theory is consistent.
The issue is necessarily somewhat philosophical and not strictly mathematical, because when deciding whether $\exists y: y+y=x$ correctly "expresses" the statement that $x$ is even, we have to resort to our informal understanding of what it really means for $x$ to be even.  But any reasonable person would agree that $\exists y: y+y=x$ is the right way to express "$x$ is even" and not the exotic alternative above.  Similarly, when deciding whether a consistency predicate correctly expresses what we really mean by consistency, we have to look at each step of the construction and check that all the formalizations are "reasonable."  If someone digs in their heels and insists that the exotic predicate is the reasonable one then it's impossible to prove them wrong; however, any reasonable person can see that the standard consistency predicate is the right one.<|endoftext|>
TITLE: How to define a differential form on a fractal?
QUESTION [30 upvotes]: It is well known how to construct a Laplacian on a fractal using the Dirichlet forms (see e.g.
the survey article by Strichartz). This implies, in particular, that a fractal can be "heated", i.e. one can write (and solve) the heat equation on the fractal.
The question is, can one run a fluid flow through a fractal set? In other words, is there
a proper way to write the Navier-Stokes equations on a fractal? In order to do this, it seems that  we need a "correct" notion of divergence at least.
More generally, is there a "correct" way to define a differential form on a fractal?

REPLY [6 votes]: There is an example of Hassler Whitney that is a lot of fun.

Whitney, Hassler, "A function not constant on a connected set of critical points." 
  Duke Math. J. 1 (1935), no. 4, 514–517. 

He constructs a curve $C$ (this is your fractal) in the plane...



and a $C^1$ function $f : \mathbb R^2 \to \mathbb R$ such that $f$ is not constant
on the curve, yet the gradient of $f$ vanishes at every point of the curve.
(Time-travelers may attend my lecture at the undergraduate math club...)



The relevance here: if you want
$$
\int_C df = f(b) - f(a) ,
$$
then that "differential" $df$ must involve more than just the partial derivatives of $f$ at the points of $C$...<|endoftext|>
TITLE: De Rham decomposition theorem, generalisations and good references 
QUESTION [23 upvotes]: De Rham decomposition theorem states that every simply-connected Riemannian manifold $M$ 
that admits complementary sub-bundles $T'(M)$ and $T''(M)$ of its tangent bundle parallel with respect to the levi-Chevita connection is isometric to the direct product of two Riemanninan manifolds $M'\times M''$.  
Question 1. In the first place I would like to have a good reference for a clear "modern" and complete proof of this theorem, if it exists (more recent than Kobayshi-Nomizu pp. 187-193) (Note, that Besse 10.44 claimed that no simple proof exists yet).
Eddited.
Question 2.  Secondly it seems to me that there should be some statement much more general than de Rham theorem. Namely, suppose we have a metric space $X$ that is locally decomposable as an isometric product of two in such a way that this decomposition is "coherent" in a appropriate sense, i.e. forms something like a presheaf. When will we be able to say that $X=Y\times Z$? (I am interested only in the cases when this will work, not when this will fail). As corollary of such a general statement one should be able to deduce de Rham theorem for example, for Finsler of polyhedral manifold, ect.

REPLY [4 votes]: The following theorem is cloasely related to your second question.
It is proved in The de Rham decomposition theorem for metric spaces by T. Foertsch and A. Lytchak.<|endoftext|>
TITLE: Some constructive versions of the Continuum Hypothesis are false. Are any true, or open?
QUESTION [13 upvotes]: Background
In constructive set theory (say based on CZF) there are inequivalent ways of stating the continuum hypothesis. Some of them are easily if not trivially refutable with common anti-classical assumptions. For instance:
Theorem: The class of all subsets of $\mathbb{N}$ is not in bijection with the class of all hereditarily transitive countable sets. (Clearly classically equivalent to ~CH)
Proof (CZF+REA+SC): The class of all hereditary transitive countable sets is a set. But the class of all subsets of N is a proper class (because SC: all sets are subcountable -- the anti-classical assumption). 
QED
But the same subcountability argument shows the class of all subsets of $\mathbb{N}$ is not in bijection with the set $2^\mathbb{N}$, nor with the set of real numbers. And anything that implies UZ (continuum is undecomposable) will imply that the set of real numbers is not in bijection with $2^\mathbb{N}$ either. So there are some classically equivalent but constructively inequivalent natural variants on CH there.
And of course CH can be rephrased without $\aleph_1$. We can ask whether there is a set which embeds into the continuum and into which the integers can be embeded, but neither converse holds. Constructively this is a weaker hypothesis, giving another "dimension" of natural variants.
It seems plausible to me that all of these variants are false under SC or UZ or intuitionistic continuity principle, or common assumption like that. 
Question
Does anyone know of any simple statements that are classically equivalent but constructively inequivalent to CH, and which are true, or open problems, under common anti-classical assumptions like subcountability or unzerlegbarkeit?

REPLY [6 votes]: It is unclear to me what formulation of CH you have in mind that might make sense intuitionistically. Could you clarify that? I think I can still answer usefully the question about embeddings $\mathbb{Z} \to {?} \to \mathbb{R}$ which cannot be inverted intuitionistically.
By an embedding I mean an injective map $e$, i.e., $e(x) = e(y)$ implies $x = y$ for all $x$ and $y$ in the domain of $e$. (Incidentally, please please don't teach people that a map is injective or 1–1 when $x \neq y \implies e(x) \neq e(y)$, you're just making them addicted to negation, and it takes years to get rid of the habit.)
There is obviously a sequence of embeddings

$\mathbb{Z} \to \mathbb{Z}^\mathbb{N} \to \mathbb{R}$.

(I will leave it to you to find them.) There is no embedding from $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$ by the usual diagonalization argument. Classically, there is an embedding $\mathbb{R} \to \mathbb{Z}^\mathbb{N}$, but we cannot construct such an embedding intuitionistically. More precisely, Markov principle implies that such an embedding gives a decomposition of the reals. Since in the effective topos Markov principle is valid and the reals are indecomposable, intuitionistic logic alone cannot prove the existence of such an embedding.
Suppose $e : \mathbb{R} \to \mathbb{Z}^\mathbb{N}$ is an embedding. Because $e(0) \neq e(1)$, by Markov principle there exists $k$ such that $e(0)(k) < e(1)(k)$ or $e(0)(k) > e(1)(k)$. Let $t = \min(e(0)(k), e(1)(k))$ and consider the sets

$A = \lbrace x \in \mathbb{R} \mid e(x)(k) \leq t \rbrace$ and
  $B = \lbrace x \in \mathbb{R} \mid e(x)(k) > t \rbrace$.

They are inhabited, disjoint and their union is $\mathbb{R}$. Therefore $\mathbb{R}$ is decomposable. (It seems to me that we should be able to get rid of Markov principle in this argument.)
We cannot expect to show intuitionistically the stronger result, namely the existence of embeddings $\mathbb{Z} \to {?} \to \mathbb{R}$ such that intuitionistic logic proves that neither embedding can be reversed. That would consitute a proof of $\lnot CH$ which is classically valid and it would be quite surprising indeed.<|endoftext|>
TITLE: Osculating conics and cubics and beyond
QUESTION [7 upvotes]: The osculating circle at a point of a smooth plane curve can be obtained by considering three points on the curve and the circle defined by them. When the three points approach $P$, the circle becomes the osculating circle at $P$. 
Generalizing this method to a conic defined by five points on a curve, one obtains an 'osculating conic' at each point of a smooth plane curve. It is pretty straightforward to show that this osculating conic is an ellipse (resp. parabola, hyperbola) if $y'' y^{(4)} > \frac{5}{3}y'''^2$ (resp. =, <). 
Thus one can classify points on a plane curve, e.g. all points on $y=e^x$ are hyperbolic (which is not obvious to the 'naked eye'). The aforementioned criterion is an affine differential invariant. 
I'm wondering why these 'osculating conics' seem to be relatively unknown (they would give a nice textbook example connecting elementary calculus and linear algebra - you can find the criterion above in one handwritten page, starting with the Taylor expansion) and whether there are any applications (generalization of evolute; can you get the curve back from the focal curve of the osculating conics)?
Furthermore it would be interesting to know whether there are any results for 'osculating cubics' (nine points define a cubic plane curve) or 'osculating quartics' (fourteen points define a quartic plane curve) or for 'osculating quadrics' (would in general yield another classification of points on a surface beyond the usual elliptic/parabolic/hyperbolic one).

REPLY [3 votes]: My REU project was about this topic.
We found some interesting things about when osculating cubics are unique (plot twist: not always), as well as a formula for the osculating conic of a smooth plane curve (at a non-flex point). Here is the formula (copied from a software package I wrote, sorry for the mess),

Given F(x,y):=(higher order terms) +a*x^4+b*x^3*y+c*x^2*y^2+d*x*y^3+e*y^4+f*x^3+g*x^2*y+h*x*y^2+i*y^3+j*x^2+k*x*y+l*y^2+m*x+n*y
The osculating conic of V(F) at (0,0) is given by
OscConic(a,b,c,d,e,f,g,h,i,j,k,l,m,n)=m*x+n*y+(j*x^2+k*x*y+l*y^2)+(-((-f)*n^3+g*m*n^2+(-h)*m^2*n+i*m^3)^2)/(j*n^2-k*m*n+l*m^2)^3*(m*x+n*y)^2+(a*n^4-b*m*n^3+c*m^2*n^2+(-d)*m^3*n+e*m^4)/(j*n^2-k*m*n+l*m^2)^2*(m*x+n*y)^2+(((x*(k*m-2*j*n)+y*(2*l*m-k*n))*((-f)*n^3+g*m*n^2+(-h)*m^2*n+i*m^3)-(x*(3*f*n^2-2*g*m*n+h*m^2)+y*(g*n^2-2*h*m*n+3*i*m^2))*(j*n^2-k*m*n+l*m^2))*(m*x+n*y))/(j*n^2-k*m*n+l*m^2)^2

The formula comes from the paper we wrote (look at Lemma 2.22 and 2.24) here
Points of Ninth Order on Cubic Curves<|endoftext|>
TITLE: The Green-Tao theorem and positive binary quadratic forms
QUESTION [14 upvotes]: Some time ago I asked a question on consecutive numbers represented integrally by an integral positive binary quadratic form. It has occurred to me that, instead, the Green-Tao theorem may  include a result on arithmetic progressions represented by a positive binary form. So my question is whether that is the case, do we already know that a positive binary form represents arbitrarily long arithmetic progressions? These would be primes in this general setting, thus quite different from consecutive integers of course.
My main reference is
David A. Cox,
Primes of the form $x^2 + n y^2.$ He defines the Dirichlet density on page 169. Then he states the 
Chebotarev Density Theorem (8.17) on page 170. Finally he gives the Dirichlet density of primes represented by a positive binary form on page 188, Theorem 9.12. EDIT::: Not difficult to state: with discriminant $ \Delta < 0$ and class number $ h(\Delta),$ if the form is ambiguous (such as the principal form) the Dirichlet density of the set of primes it represents is $$ \frac{1}{2 h(\Delta)},$$
while if the form is not ambiguous the Dirichlet density is
$$ \frac{1}{h(\Delta)}.$$ On page 190 he does the example $ \Delta = -56.$ Here $x^2 + 14 y^2$ represents a set of primes with Dirichlet density $1/8,$ while $2 x^2 + 7 y^2$ also gets density $1/8,$ but in the other genus $ 3 x^2 + 2 x y + 5 y^2 $ and  $ 3 x^2 - 2 x y + 5 y^2 $ each represent the same set of primes with density $1/4.$ Note on page 195 we have Exercise 9.17, that the sum of these densities for any discriminant must be $1/2.$ A little fiddling, not mentioned in the book, shows that each genus (of a fixed discriminant $\Delta$) represents the same total density, something we really want because of the relationship between genera and arithmetic progressions of primes.
From an earlier answer by David Hansen it would appear that the only missing ingredient is a comparison between Dirichlet density and "relative density." I have never been sure on this point, does a positive binary form represent the same relative density of primes as its Dirichlet density? Anyway, see:
Is the Green-Tao theorem true for primes within a given arithmetic progression? 
My earlier question, about which I should say that I have come to believe there is no upper bound on the length of intervals represented, despite the great difficulty finding examples: 
Can a positive binary quadratic form represent 14 consecutive numbers?
On the Green-Tao theorem itself:
http://arxiv.org/abs/math.NT/0404188 
http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem 
same-day EDIT: I looked up Dirichlet density on wikipedia. As relates to the earlier David Hansen answer I liked, wikipedia stops short of saying that the Dirichlet density of primes in an arithmetic progression is identical to the relative density. Personally, I cannot see how the relative density could be anything else, but that is just my opinion. So I think I am also asking for references that prove the relative density is equal to the Dirichlet density in some naturally-occurring situations. 
http://en.wikipedia.org/wiki/Dirichlet_density
Later on the same day EDIT: There is stronger language in this next wikipedia page, so I think we can conclude that for primes in an arithmetic progression the Dirichlet density and the relative density are equal, but I would still like a more substantial reference. I'm the nervous type. I worry.
http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem

REPLY [15 votes]: Edit: {The answer to your question, "...do we already know that a positive binary form represents arbitrarily long arithmetic progressions?" is yes. See the second paragraph below.}
If the relative density exists, so does the Dirichlet density and they are equal. The converse is not true in general. For primes in a given arithmetic progression, both densities exist. See Lang's Algebraic Number Theory, Ch. VIII.4 and XV. Given those facts, one approach to the problem would be trying to show that the relative density of the set of primes represented by a positive binary quadratic form actually exists (I have no idea how hard this might be). 
On the other hand, if you only want to know about a.p.'s of primes represented by a positive quadratic form, a better approach might be answering the question, "Does Green-Tao still hold for sets of primes with positive Dirichlet density?" The answer is yes since G-T only requires that the limsup of the relative density be positive, and positive Dirichlet density implies positive limsup (if the limsup were 0, the lim would be 0).<|endoftext|>
TITLE: Group structures on the cartesian product of two groups
QUESTION [7 upvotes]: This question probably has a simple and immediate answer which escapes me now. (And, I should admit, it's more my curiosity than anything else.) The only natural way to construct a group structure on the cartesian product $G\times H$ of two groups $G$ and $H$ (in particular, ``natural'' to me means that on each factor the group product should be the original one) is the semi-direct product in the case when one group acts on another one by automorphisms. Are there any natural constructions of a group structure on $G\times H$  where neither factor is a normal subgroup?
Update: I was pointed out that the notion of Zappa-Szep product that appears in the answer given by Steven Gubkin  is also mentioned in an earlier MO discussion; I thought I'd link it here for some sort of connectivity.

REPLY [10 votes]: Wikipedia to the rescue!
http://en.wikipedia.org/wiki/Zappa-Szep_product<|endoftext|>
TITLE: "Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?
QUESTION [21 upvotes]: The group $Diff(S^n)$ ($C^\infty$-smooth diffeomorphisms of the $n$-sphere) has many interesting subgroups.  But one question I've never seen explored is what are its "big" finite-dimensional subgroups? 
For example, $Diff(S^n)$ contains a finite-dimensional Lie subgroup of dimension $n+2 \choose 2$, the subgroup of conformal automorphisms of $S^n$.  Similarly it contains a compact Lie subgroup of dimension $n+1 \choose 2$, the isometry group of $S^n$. 
Is it known that:
1) A finite-dimensional Lie subgroup of $Diff(S^n)$ having dimension at least $n+2 \choose 2$ is conjugate to a subgroup of the conformal automorphism group of $S^n$ ? (Answer, no, see Algori's answer below). Modified question: As Algori notes, $GL_{n+1}(\mathbb R) / \mathbb R_{>0}$ acts on $S^n$ and has dimension $n^2+2n$. So is a finite-dimensional Lie subgroup of $Diff(S^n)$ of dimension $n^2+2n$ (or larger) conjugate to a subgroup of this group? 
2) A compact Lie subgroup of $Diff(S^n)$ having dimension at least $n+1 \choose 2$ is conjugate to a subgroup of the isometry group of $S^n$ ? (Answer: Yes, see Torsten Ekedahl's post below)
For example, arbitrary compact subgroups of $Diff(S^n)$ do not have to be conjugate to subgroups of the above two groups -- perhaps the earliest examples of these came from exotic projective and lens spaces.  But I have little sense for how high-dimensional these "exotic" subgroups of $Diff(S^n)$ can be.

REPLY [32 votes]: You can make big Lie groups act effectively on small manifolds by cheating: make the group a product of groups, with each factor acting by compactly supported diffeomorphisms on a different disjoint open subset. So the additive group $\mathbb R^N$ becomes a subgroup of $Diff(S^1)$ by flowing along $N$ commuting vector fields supported in $N$ disjoint arcs.
(added later) A similar cheat: let $a_1,\dots ,a_N$ be linearly independent functions of one variable. Then $a_1(x)\frac{\partial}{\partial y},\dots ,a_N(x)\frac{\partial}{\partial y}$ are independent commuting vector fields in the $x,y$ plane. Modify this example to make it compactly supported if you like.
(added still later) My proposed extension of the first cheat to a semisimple group (comment thread of Torsten's answer) is doomed: Choose a point on the circle and choose an element of SL_2(R) that fixes this point and acts on the tangent space there with eigenvalue c>1. Lifting the group element to the universal covering group in the right way, you get an element g of the latter group that fixes all the points above the given point in the universal covering space of the circle, in each case with eigenvalue c. But now if this line with this action could be embedded in a longer line with trivial action outside then there would be a sequence of fixed points of g with eigenvalue c converging to a fixed point of g with eigenvalue 1, contradiction.<|endoftext|>
TITLE: When is an Anosov diffeomorphism mixing?
QUESTION [6 upvotes]: Let $M$ be a compact Riemannian manifold and let $T : M \rightarrow M$ be Anosov. I have read here that it is an open problem to prove that $T$ is topologically mixing if $M$ is connected. Katok and Hasselblatt point out that if $(T, \mu)$ is mixing, then the restriction of $T$ to the support of $\mu$ is topologically mixing. Yet I have also read here (among other places) that Axiom A diffeomorphisms (and hence Anosov diffeomorphisms) satisfy exponential decay of correlations, which seems like it implies mixing (take characteristic functions). Still, I see references in the physics literature (which is where I am coming from) to "mixing Anosov maps". 
While the Anosov alternative for flows is clear enough, I haven't found a characterization of mixing properties for Anosov diffeomorphisms that sorts out these apparently contradictory statements.
With that in mind, my question is (sort of): under what circumstances is an Anosov diffeomorphism guaranteed to be mixing?

REPLY [2 votes]: I believe it's true that an Anosov diffeomorphism on a nilmanifold or infranilmanifold has a full non-wandering set. In combination with Vaughn and coudy's answers, the question of whether every Anosov diffeomorphism of a connected compact Riemannian manifold is mixing is therefore linked to the conjecture that every Anosov diffeomorphism takes place on an (infra-)nilmanifold. If you have a specific example in mind, then either it's defined on an infranilmanifold and is mixing, or it's not defined on an infranilmanifold, and hence its very existence is a shoo-in for Annals of Mathematics... 
I don't know a good general survey on Anosov diffeomorphisms, but there are quite a few references to papers on this topic in Gorodnik's survey paper "Open problems in dynamics and related fields" (the conjecture mentioned above is Conjecture 3 in Gorodnik's paper).<|endoftext|>
TITLE: Highly connected, compact complex manifolds
QUESTION [14 upvotes]: Here are four remarks about the homology and homotopy type of a compact, complex manifold $M$:

If $M$ is Kähler, then it is symplectic and thus $H^2(M,\mathbb{R}) \ne 0$.  (Also, as explained in a blog posting by David Speyer, you still have $H^2(M,\mathbb{R}) \ne 0$ even if $M$ is non-projective but algebraic.)
An interesting first example of a non-Kähler manifold is a Hopf manifold, by definition $(\mathbb{C}^n\setminus 0)/\Gamma_r$, where $\Gamma_r$ is a rescaling by $r$ with $|r| \ne 0,1$.  This example has $H^1(M,\mathbb{R}) \ne 0$.
On the other hand, even-dimensional, compact Lie groups have left-invariant complex structures.  If $M$ is such a manifold and is simply connected, then it is also 2-connected. $H^1(M,\mathbb{Z}) = H^2(M,\mathbb{Z}) = 0$ and $M$ is manifestly not Kähler.  On the other hand, no such example is 3-connected and you always have $H^3(M,\mathbb{R}) \ne 0$.
There is (or was) a long-standing conjecture that no even-dimensional sphere other than $S^2$ has a complex structure.

So, question:  Is there for each $n$, a compact, complex manifold $M$ which is $n$-connected?

REPLY [20 votes]: E. Calabi, B. Eckmann, A class of compact, complex manifolds which are not algebraic.
Ann. of Math. (2) 58, (1953). 494–500. 
From Chern's MR review (MR0057539): 

This paper defines on the topological product $S^{2p+1} \times S^{2q+1}$ of two spheres of dimensions $2p+1$ and $2q+1$ respectively, $p$ > 0, a complex analytic structure. The complex manifold so obtained ... admits a complex analytic fibering, with two-dimensional tori as fibers and having as base space the product $\mathbb{P}^p \times \mathbb{P}^q$ of complex projective spaces of (complex) dimensions $p$ and $q$ respectively.

REPLY [8 votes]: As shown by Calabi and Eckmann, products of odd-dimensional spheres admit complex structures. See Anns of Maths 58, 1953, 494-500.<|endoftext|>
TITLE: Are periods of rigid Calabi-Yau threefolds over $Q$ algebraic?
QUESTION [11 upvotes]: Let $X$ be a (smooth) compact complex manifold, and suppose that $H^1(X, \Theta_X) = 0$, where $\Theta_X$ is the tangent sheaf.  In other words, suppose that $X$ is rigid.
Suppose moreover that $X$ arises as the complex points of a smooth projective variety over $Q$.
Is it known or expected that the periods of $X$ are algebraic numbers?  If $X$ were not rigid, then the periods would be values (at zero) of functions satisfying Picard-Fuchs equations.  But the rigidity suggests (to my intuition) that the periods should not be transcendental.
Is anything known?  Expected?  Written?  How about the specific case when $X$ is a rigid Calabi-Yau 3-fold?  Has anyone computed such periods?  Could one compute them easily?

REPLY [4 votes]: For rigid, at least in the modular case (known in many events), you can compute the periods of the form, though this supposes you can explicitly write down the weight 4 newform. For instance, Schutt ( http://arxiv.org/pdf/math/0311106 ) gives examples of level 73, and using Magma you can compute the periods as
> M:=NewformDecomposition(NewSubspace(CuspidalSubspace(ModularSymbols(73,4))))[1];
> Periods(M,100);
[ (0.902834199842382836695960181248 + 0.0526923557275574794028757363126*i),
  (0.285105536792331422114513708795 + 0.0175641185758524931342918404798*i) ]

Here $L$-functions are not applicable, as the $L$-functions vanishes at the central point. I extended the above to a few hundred digits and found nothing with PowerRelation.<|endoftext|>
TITLE: Iterated Circumcircle
QUESTION [20 upvotes]: Take three noncollinear points (a,b,c), compute the center of their circumcircle x, and replace a random one of a,b,c with x.  Repeat. It seems this process may converge to a point, assuming no collinearities ever develop along the way.  But the dynamics seem complicated, with intermediate growth before ultimate convergence.  And I am by no means certain of ultimate convergence, even generically.
Here is an example, iterated 1000 times:

Has this or similar processes been studied?  Might one establish conditions under which initial triangles definitely lead to convergence?
Reformulation. In light of the many cogent comments, let me attempt a reformulation and sharpening of the question.  Restrict attention to isosceles triangles, represented by their apex angle θ and incident side length L. Let T represent the space of all (θ,L) pairs. Color a point of T red if almost surely the process converges to a point, and blue otherwise.  What does the coloring of T look like?  Are the red points
dense in T?
Edit(10Sep10). 
A knowledgeable colleague believes that the scaling limit of the evolution
of $a_n$ in Victor's answer (the isosceles side length)
should be Brownian motion on the line. In light of this opinion,
I (belatedly) accept Victor's answer.  Thanks again for the interest shown in my question!

REPLY [3 votes]: This is also similar to a different technique for generating the Sierpinski Triangle through an iterated method which neither converges nor diverges but chaotically stays in a particular set.

Given any triangle $ABC$ with non-collinear endpoints $A,B,C$,
select a random point $P_0$ within the center of $ABC$
then iterate the following
pick any one of the vertices, $A,B,$ or $C$ at random
generate $P_{i+1}$ as the midpoint of the line-segment $P_i$ and the randomly selected vertex

Iterated multiple times, this generates a Sierpinski triangle with a few extra points thrown in at the beginning.  If your initial point is definitely on the Sierpinski triangle (say you start with one of the vertices as your initial point $P_0$), then all of the subsequent points are definitely in the Sierpinski triangle.  The Sierpinski triangle has Hausdorff dimension $log(3)/log(2)$ ≈ $1.585$ (copied from wikipedia)
I remember writing this as a program in BASIC on the Apple ][, but I cannot recall the source of the question that led me to the program.  Most likely it was an article in Byte or Creative Computing.
This is similar to your selecting a new point defined as the circumcenter of your three points, and then using that new point along with two of the vertices of your current triangle to generate then next triangle to find the circumcenter of. Your iterated method leads to points outside of the triangle, leading to a wandering triangle in most cases, leading to your second question, in which cases of initial triangles do you end up with convergence or divergence, which seems to have been addressed with some of the earlier answers.  I'll think about that a bit more before commenting on that.
This "iterated line-segment midpoint" technique definitely does not converge.  It leads to selecting points within the set of points contained within the Sierpinski Triangle.  The new points also do not diverge away; the points always stay within the confines of the triangle $ABC$, and if the initial point is in the Sierpinski triangle, then the set of points generated are all also within the set of points in Sierpinski triangle.<|endoftext|>
TITLE: Points on binary hemispheres of the n-sphere
QUESTION [9 upvotes]: Let $\mathbb{S}^{n-1}=${$ x\in \mathbb{R}^n| \sum_{k=1}^n x_k^2 =1 $} be the $n-1$ sphere and $n_i\in\mathbb{R}^n$ with components $n_{ij}\in${$-1,1$}$\ \forall\ j=1,2,\dots,n$. There are obviously $2^n$ such distinct "binary" vectors $n_i$. Now define for every binary vector $n_i$ a Hemisphere $H_i=${$x \in \mathbb{S}^{n-1}\ |\ (x,n_i)\geq0$} where $(.,.)$ means the usual inner product $(x,y)=\sum_{k=1}^n x_k y_k$. 
What is the Probability that $p$ uniformly distributed vectors on $\mathbb{S}^{n-1}$ are all on the same of those $H_i$?
EDIT, Will Jagy: evidently these really are hemispheres, Tom Goodwillie pointed out that for $n=3$ they are 
$$ x + y + z \geq 0, \; \; \; x + y -z \geq 0, \; \; \; x - y + z \geq 0, \; \; \; x - y - z \geq 0 , $$ and their complements.
$$ -x - y - z \geq 0, \; \; \; -x - y +z \geq 0, \; \; \; -x + y - z \geq 0, \; \; \; -x + y + z \geq 0 . $$
EDIT,problem poser: 
Here is a solution for $n=3$: 
The problem is much easier, if you see the similarity of the sphere with the great arcs that define the $8$ hemispheres and a cuboctahedron (www.wikipedia.org/wiki/Cuboctahedron). 
As you can see the cuboctahedron actually consists only of $2$ distinct kinds of faces, these are $8$ triangles and $6$ squares. It is not difficult to see, that the intersections of the hemispheres consists of only these triangles and squares projected on the sphere and that these spherical triangles and “spherical” squares are the smallest parts you can get by the intersection of some hemispheres (except the empty set of course). Unfortunately the area is not preserved under this projection. But what is preserved is of course the fact that one hemisphere consists of $3$ spherical triangles and $4$ “spherical” squares (Lets call the triangles $T$ and the squares $S$ from now on).
It is also pretty obvious that the intersection of two hemispheres with hamming distance $2$ (that means two hemispheres with binary vectors with only two different components) consists of two $T$ and one $S$. That gives us the two equations $2T+S=I_2$  and $4T+3S=2\pi$  where $I_2$  is the intersection of two hemispheres with hamming-distance $2$.
To calculate $I_2$ we use the fact that the angle between two binary vectors of hamming distance $2$  is given by $\varphi_2=arccos−\frac{1}{3}$ and therefor $I_2=\frac{\pi-\varphi_2}{2\pi}4\pi=2(\pi-\varphi_2)$. 
Now we have from the equations above $$T=\frac{3I_2-2\pi}{2}=2\pi-3\varphi_2$$ and $$S=2(\pi-I_2))=4\varphi_2-2\pi.$$Now, that we know the smallest parts $T$ and $S$ we have to understand what kinds of intersections we have and how they consist of $T$ and $S$. To do that we only need to care wlog about the hemisphere $H_1$ with binary vector $n_1=(1,1,1)^T$. Lets call $I_{ijk}$ the intersection of $H_1$ and hemispheres with hamming distance $i,j,k$(relatively to $n_1$). It is obvious, that there cant be more than intersections of $4$ hemispheres. (If there would be more, there would be a pair of disjoint hemispheres with vanishing intersection.) So there are the non vanishing intersections $I_1,I_2,I_{11},I_{12},I_{22},I_{111},I_{211},I_{221}$ (note that i mean here that you can find hemispheres with hamming distances $i,j,k$ so that $I_{ijk}$ doesnt vanish. For $n=3$ I think it is obvious, that in this case the value of the intersection is either $0$ or $I_{ijk}$ depending on which hemispheres of the given hamming distances you chose.). As you can easily check we have $I_{221}=I_{111}=I_{22}=T$, $I_{211}=S$, $I_{12}=I_{11}=S+T$, $I_2=2T+S$ and $I_1=2T+2S$. I think everyone sees now, that we just have to use the inclusion exclusion principle to get the desired probability $P(3,p)$. I won't do this now because i think that this doesnt give more insights.
The big question that arises now is, whether there are similarities in higher dimensions. If we could find the smallest parts of intersections on $\mathbb{S}^3$ without too much effort (like having to integrate) we could use a similar strategy I used for $n=3$. Then, if we could find some further symmetries, we may even be able to get some kind of recursion and solve it for arbitrary dimensions.
EDIT, problem poser: 
Here is a little summary of the progress I made for n>3:
My aim is to find the smallest pieces of intersections of the hemispheres. If you look at the case n=3, you recognize that one of the smallest pieces is a spherical triangle with equal angles (the equal angles come from the symmetry of the problem). The good thing is, as Will Jagy pointed out, that you can get the area  by the method Marco Radeschi describes in his comment of Proofs without words . Even more great is the fact, that we can always apply this method if we have a set of hemispheres, that all have the same distance to each other (that means that all have binary vectors with same hamming distance to each other) and for that the intersections of all these hemispheres do not vanish. For example for n=3 the intersection of the hemispheres with binary vectors (1,1,1), (1,-1,-1) and (-1,1,-1) is such a spherical triangle mentioned above. If you add the hemisphere (-1,-1,1) now to that set, all binary vectors still have the same distance, but the intersection of all four vanishes. That shows that the vanishing property is important to apply the method.
Now it seems logical to search for such a set for n=4 and to hope that the intersection (that you could easily calculate by the method mentioned above) is a smallest piece. 
What really concerns me is the fact that there is no such set for n=4. For example a set with the maximum number of hemispheres that all have hamming distance 2 to each other and for that the intersection of all hemispheres does not vanish is {(1,1,1,1), (1,-1,-1,1), (1,1,-1,-1), (1,-1,1,-1)}. (It is easy to prove this, because there are only 3 possible hemispheres left and each of them is the negative one of a hemisphere in the set. ) Now you can add for example the hemisphere (1,-1,-1,-1) and you will still have a non vanishing intersection of all hemispheres. This proves that for n=4 the intersection of hemispheres that all have hamming distance 2 to each other and that have a non vanishing intersection is not a smallest piece. The same argumentation proves that there also can't be such sets with hamming distance 1 and 3. 
But I also found a positive result: For all dimensions the intersection of hemispheres with the same given component, is a smallest piece.(To avoid confusion I mean hemispheres that for example all have the same first component) For example for n=3 these intersections are the spherical squares. The problem for n>3 is to find the volume of such intersection. 
Any ideas?

REPLY [2 votes]: A good way to think about a uniformely random point on the sphere is to think of it as the normalization of a vector of normally distributed i.i.d random variables. In fact, you don't even need to normalize. Consider hemisphere $v_i$ and $v_j$, form the $2 \times N$ matrix $V$, $v_i$ and $v_j$ being the rows of $V$. Let $X$ be a random Gaussian vector with covariance matrix $I$ (identity). $V.X$ is a vector with two components, whose sign represent membership in the hemisphere $i$ and $j$... $V.X$ follows a multivariate normal distribution with mean $\left(\begin{array}{cc}0\\\0\end{array}\right)$ and covariance matrix $n \left[\begin{array}{cc}1 & 1-2h/n\\\1-2h/n & 1\end{array}\right]$ where $h$ is the hamming distance. Notice that any covariance matrix can be approximated, up to a multiplicative factor, for $n$ large enough. The measure of the intersection of the two hemispheres is the integral of the the probability density over an orthant.
For the intersection of two hemispheres, the orthant is a quartant and the intersection is given as
$$\frac{1}{4} + \frac{\sin^{-1}(1-2h/n)}{2\pi}$$
(note that this is not the area but the fraction of the $(n-1)$ sphere area that is in the intersection)
There is also a formula for the trivariate case, i.e. if you're considering the intersection of three hemispheres
$$\frac{1}{8} + \frac{1}{4\pi}( \sin^{-1}(1-2h_{12}/n) + \sin^{-1}(1-2h_{23}/n) + \sin^{-1}(1-2h_{13}/n) )$$
Unfortunately, it is known that there is no closed-form expression of the probability of the orthants of a general multivariate normal distribution above the trivariate case. In your case, the covariance matrix have a certain structure, so it is not definite evidence that there is no formula for the area of the intersection of more than three hemispheres, but it is pretty good evidence.<|endoftext|>
TITLE: Number-theoretic congruences with geometry and topology?
QUESTION [5 upvotes]: There are many examples of $q$-series identities being proven by interpreting them as generating series of geometric invariants like the Donaldson invariants.  I would like to know if there are ways of establishing congruences among the coefficients of theta functions or other $q$-series using geometric invariants.  As an example, here's one relating $\mathrm{SO}(3)$ Donaldson invariants and mock theta functions.  Here's another one involving McMahon's enumeration of plane partitions.  These deal with infinite products or series.  I'd like to know if physical or geometric methods have been used to prove congruences
My naive opinion is that since physics deals mostly with "real" spaces, there's no way to get the "torsion" involved in number theory.
EDIT: Perhaps what I ask asking for is much simpler.  Are there number-theoretic congruences among invariants in topology or geometry?  Examples might be Bott periodicity or Adams' theorem on the number of independent vector fields on spheres.

REPLY [13 votes]: I came onto mathoverflow for a reference on an unrelated topic, but since I noticed this question I thought I would chime in:  the image of the J homomorphism in the stable homotopy groups of spheres, by the work of Adams, is dictated by the denominators of Bernoulli numbers $-B_n/n$ (i.e. $\zeta(1-2n)$) and it is these orders that are what are fundamentally related to the vector fields on spheres problem.
The image of $J$ is the $n = 1$ instance of some general $v_n$-periodic families in the stable stems called Greek letter elements (See Miller-Ravenel-Wilson's Annals paper "Periodic phemonena in the Adams-Novikov spectral sequence").  The n=2 analog is called the divided beta family.
Adams's identification of the Image of J with denomenators of Bernoulli numbers was through his "e"-invariant, which took values in Q/Z equal to the images of the Bernoulli numbers.
Gerd Laures, in his paper "The topological q-expansion paper" introduced a higher form of the "e" invariant called the "f"-invariant, which takes values in the quotient of the Katz's ring of divided congruences of modular forms by the sum of the integral q-expansions, and the classical modular forms over Q.  (A kind of higher analog of Q/Z)  He showed that his f invariant gave an injection from the divided beta family to this quotient, but did not characterize the image.
In my paper "congruences of modular forms and the divided beta family in homotopy theory" I showed that the divided beta family was actually in bijective correspondence with a set of congurence conditions between modular forms.  Gerd and I later showed in another paper that these congruences I wrote down were precisely the ones coming from his f invariant.
But what I do not know, and would be curious if any user had an idea concerning this, is if there is a natural family of modular forms, kind of like a higher form of Eisenstein series, which would realize these beta elements in its image in the quotient of the ring of divided congruences.
The phenomenon I describe is not unique to n = 2, abstractly there is some kind of "congruence condition" amongst holomophic automorphic forms on U(1,n-1) that describes the nth Greek letter family in the stable homotopy groups of spheres - I just don't know how to express this congruence condition in clasical terms.<|endoftext|>
TITLE: When is $n/\ln(n)$ close to an integer?
QUESTION [19 upvotes]: As usual I expect to be critisised for "duplicating"
this question. But I do not! As Gjergji immediately
notified, that question was from numerology. The one I ask you here
(after putting it in my response) is a mathematics
question motivated by Kevin's (O'Bryant) comment to the earlier post.
Problem.
For any $\epsilon>0$, there exists an $n$ such that 
$\|n/\log(n)\|<\epsilon$ where $\|\ \cdot\ \|$ denotes the
distance to the nearest integer.
In spite of the simple formulation, it is likely that
the diophantine problem is open. I wonder whether it follows
from some known conjectures (for example, Schanuel's conjecture).

REPLY [13 votes]: [Edited mostly to extend the computation from $1.5 \cdot 10^{13}$
to a bit over $2^{50} > 10^{15}$ and give the heuristics for expected number of
records for $\| r(n) \|$ vs. $\log n \cdot \| r(n) \|$]
Just ran across this.  I see that Kevin's answer completely settles the
original question, but meanwhile Will Jagy raised the question of
finding new record lows for
$$
\log n \cdot \left\| \frac{n}{\log n} \right\|
$$
and proving their infinitude.  I next outline a proof that there are
infinitely many such record lows, and then report on a computation of
all such $n$ up to $1.5 \cdot 10^{13}$.
For the infinitude: Since $r(n) := n / \log n$ can never be an exact integer,
it is enough to prove that for each $\epsilon > 0$
there exist infinitely many solutions of $\| r(n) \| < \epsilon/\log n$.
In fact it's not hard to show that $\| r(n) \|$ can get as small as some
negative power of $n$, because $r(n)$ is almost linear
(its second derivative is $o(n^{-1})$ as $n \rightarrow \infty$)
and we can choose $n_0$ to make $r'(n_0)$ as far as possible from
any rational number.  If I did this right, we can find intervals
$|n - n_0| \leq h$ in which $\min_n \| r(n) \| \ll h^{-1}$
where $h^{-1} = |r''(n_0)|^{1/3} \sim (n_0 \log^2 n_0)^{-1/3}$.
For instance, we may choose $n_0$ so that
$r'(n_0) = 1 / (k + \sqrt 2)$ for $k = 1, 2, 3, \ldots$
[that is, so that $\log n_0$ solves the quadratic equation
$\lambda^2 = (k+\sqrt2) (\lambda-1)$].
On such an interval, $r(n)$ is approximated by $r(n_0) + r'(n_0)(n-n_0)$
to within $O(r''(n_0) (n-n_0)^2) = O(h^2/h^3) = O(h^{-1})$, and
(since $h$ grows much faster than $k$)
the arithmetic sequence with common difference $r'(n_0)$
is close enough to being equidistributed that it comes within
$O(1/h)$ of an integer.  [We probably expect that $\| r(n) \|$
is random enough that it gets as small as $c/n$ or even $o(1/n)$,
but proving such a result must be well out of reach.]
For the numerical search, the problem is quite similar to
MO.19170
on nearly-integral values of $\log_{10} n!$ (since $n/ \log n$,
like $\log_{10} n!$, is nearly linear in $n$).  Again it takes time
only $\tilde O(N^{2/3})$ to find all examples with $n < N$ using
a linear-approximation technique such as described at the bottom of
page 15 of
Lefèvre's slides.
This is actually the same idea as in the previous paragraph:
partition $[1,N]$ into intervals $|n-n_0| \leq h \sim (n_0 \log^2 n_0)^{1/3}$;
in general $r'(n_0)$ might be so close to a rational number that
equidistribution fails, but we can still use continued fractions
to find all $n$ in that interval for which $\|r(n)\| \ll h^{-1}$.
I ran this with $N = 2^{50} > 10^{15}$ on ten alhambra heads.
Most finished in under two days; two took an extra day or two,
probably spending most of them on $n_0$ for which
$r'(n_0)$ was nearly rational (in this case one can do much better than
trying every $n \in [n_0-h,n_0+h]$ for which
$\| r(n_0) + r'(n_0)(n-n_0) \|$ is small,
but I didn't take the extra time to implement that refinement).
The computation found fourteen new records beyond the
12 initial terms 2, 17, 163, 715533, 1432276, 6517719, 11523158,
11985596, 24102781, 254977309, 451207448, 1219588338
of OEIS sequence A178806, namely

2048539023, 10066616717, 42116139191, 47657002570,
  73831354169, 122478947521, 143949453227, 3152420311977,
  5624690531099, 14964977749017, 25999244327633, 92799025313425,
  164330745650026, and 604329910739082.

There is also a new example, namely $n = 3040705645816$, of a number
that is not in this sequence but does belong in the closely
related OEIS sequence A178805,
which consists of $n$ that achieve record low values of $\| r(n) \|$
instead of $\log n \cdot \| r(n) \|$.  In general a
$\log n \cdot \| r(n) \|$ record is automatically also an
$\| r(n) \|$ record, but the converse can fail on occasion.
If we imagine that the $\| r(n) \|$ are independent random numbers
uniformly distributed on $(0,1/2)$ then the probability that
$\| r(n) \|$ is a new record is $1/n$, so we expect $\log N + O(1)$
record values with $n \leq N$.  The same question for
$\log n \cdot \| r(n) \|$ is trickier, but if I did this right
the probability that $\| r(n) \|$ is a new record but
$\log n \cdot \| r(n) \|$ is not one is approximately $1 / n \log n$,
so we expect only $\log\log N + O(1)$ examples such as
$n = 3040705645816$ up to $N$, and might never see another one
even though there should be infinitely many more.
Here is a table of the values of $n < 2^{50}$
for which $\| r(n) \|$ attains a new record low, together with
the signed fractional part of $r(n)$, and $\log n$ times that fractional part:
$$
\begin{array}{rrrc}
2 & -0.1146099 & -0.0794415 & \\
5 & 0.1066747 & 0.1716863 & ! \\
9 & 0.0960765 & 0.2111017 & ! \\
13 & 0.0683262 & 0.1752532 & ! \\
17 & 0.0002541 & 0.0007199 & \\
163 & -1.26 \cdot 10^{-6} & -6.43 \cdot 10^{-6} & \\
53453 & 1.22 \cdot 10^{-6} & 1.33 \cdot 10^{-5} & ! \\
110673 & 6.68 \cdot 10^{-7} & 7.76 \cdot 10^{-6} & ! \\
715533 & 3.84 \cdot 10^{-7} & 5.17 \cdot 10^{-6} & \\
1432276 & 2.33 \cdot 10^{-7} & 3.30 \cdot 10^{-6} & \\
6517719 & -2.00 \cdot 10^{-7} & -3.14 \cdot 10^{-6} & \\
11523158 & -9.95 \cdot 10^{-8} & -1.62 \cdot 10^{-6} & \\
11985596 & -7.26 \cdot 10^{-8} & -1.18 \cdot 10^{-6} & \\
24102781 & 4.43 \cdot 10^{-9} & 7.53 \cdot 10^{-8} & \\
254977309 & 9.12 \cdot 10^{-10} & 1.76 \cdot 10^{-8} & \\
451207448 & 3.68 \cdot 10^{-10} & 7.33 \cdot 10^{-9} & \\
1219588338 & -2.57 \cdot 10^{-10} & -5.38 \cdot 10^{-9} & \\
2048539023 & -5.89 \cdot 10^{-11} & -1.26 \cdot 10^{-9} & \\
10066616717 & 4.85 \cdot 10^{-11} & 1.12 \cdot 10^{-9} & \\
42116139191 & -4.47 \cdot 10^{-11} & -1.09 \cdot 10^{-9} & \\
47657002570 & -2.43 \cdot 10^{-11} & -5.97 \cdot 10^{-10} & \\
73831354169 & 1.35 \cdot 10^{-11} & 3.38 \cdot 10^{-10} & \\
122478947521 & 7.53 \cdot 10^{-13} & 1.92 \cdot 10^{-11} & \\
143949453227 & -5.50 \cdot 10^{-13} & -1.41 \cdot 10^{-11} & \\
3040705645816 & 5.18 \cdot 10^{-13} & 1.49 \cdot 10^{-11} & ! \\
3152420311977 & -3.36 \cdot 10^{-13} & -9.67 \cdot 10^{-12} & \\
5624690531099 & 1.28 \cdot 10^{-13} & 3.76 \cdot 10^{-12} & \\
14964977749017 & -7.15 \cdot 10^{-14} & -2.17 \cdot 10^{-12} & \\
25999244327633 & -2.02 \cdot 10^{-14} & -6.25 \cdot 10^{-13} & \\
92799025313425 & 6.01 \cdot 10^{-15} & 1.93 \cdot 10^{-13} & \\
164330745650026 & -1.00 \cdot 10^{-15} & -3.28 \cdot 10^{-14} & \\
604329910739082 & -4.59 \cdot 10^{-16} & -2.27 \cdot 10^{-14} &
\end{array}
$$
the "!"'s mark the $\| r(n) \|$ records that aren't $\log n \cdot \| r(n) \|$ records.<|endoftext|>
TITLE: What part do arguments from authority play in mathematical reasoning?
QUESTION [8 upvotes]: In forming your answer you may choose to address any or all of the following aspects of the question:

Descriptive.  What part do arguments from authority actually play in mathematical reasoning?
Normative.  What part do arguments from authority ideally play in mathematical reasoning?
Regulative.  What if any discrepancies exist between the actual and the ideal and what if anything should be done about them if there are?

REPLY [10 votes]: You could interpret your question outside of the context of mathematical proof. It seems like your question was phrased broadly-enough to allow this interpretation. 
There are what are perhaps the most subjective aspects of mathematical reasoning:
1) What questions do we consider important? 
2) What results do we consider worthy of publication? 
3) What do we consider worthy of teaching our students? 
I suppose this is a type of metamathematical reasoning.  But arguments from authority frequently get used in these situations.  
Especially in grad school I seemed to encounter far too many people with fantastical images of certain mathematicians, frequently fields-medalists, sometimes with little  understanding of what these people have accomplished.  The further I get from grad school the less of this I see, thankfully.<|endoftext|>
TITLE: Feasibility of linear programs
QUESTION [5 upvotes]: It's known that finding the intersection of n halfplanes in 2-d takes $\Omega(n\log n)$ time. Does the lower bound apply if we change the question to deciding whether the intersection is non-empty?

REPLY [2 votes]: It seems this can be done in linear time. Algorithms that solve linear programs are also capable of deciding whether the LP is feasible or not and 2-d linear programs can be solved in linear time (linear in terms of the number of constraints). So to decide whether a set of n halfplanes is non-empty or not, just solve the LP that has those halfplanes as its constraints with any objective function.<|endoftext|>
TITLE: Does the image of a differential operator always contain an ideal?
QUESTION [11 upvotes]: Let $\delta$ denote a non-zero complex algebraic differential operator in a single variable x.  That is, it can be written as a sum
$$ \delta = \sum_i f_i\partial_x^i$$
where there $f_i$ are complex polynomials in x.
Let $R=\mathbb{C}[x]$, and consider the image of $\delta$ as a map on R.  As a subspace of $R$, does $Im(\delta)$ always contain an non-trivial ideal?
It does in every case I can think of where there is some trick I can use to understand the image better:

When $\delta$ is a function.
When $\delta$ is a constant coefficient differential operator.
When $\delta$ has order 1.
When $\delta$ is homogeneous for the Euler grading; that is, it takes monomials to monomials.

It seems like it should be related to the simpler fact that $\delta$ is zero if $\delta$ kills functions of unboundedly high degree, which can be shown from the Formal Continuity of differential operators.
Remark. For more than one variable, the above question is false.  If $\delta=x\partial_x-y\partial_y$, then $\delta$ is homogeneous for the Euler bigrading (it takes monomials to monomials), but it kills all monomials of the form $x^iy^i$.  Since any monomial ideal in $\mathbb{C}[x,y]$ must contain some monomial of this form, the image of this $\delta$ contains no ideal.

REPLY [14 votes]: No. Let $\delta=x-\partial$ and $L=Im(\delta).$ I claim that $L$ does not contain any non-zero ideal of $\mathbb{C}[x].$ Indeed, $x^k\equiv (k-1)x^{k-2}\ (\mod L)$ and, by induction, 
$$x^{2n+1}\equiv (2n)!!x\equiv 0(\mod L),\ x^{2n}\equiv (2n-1)!!\ (\mod L).$$
Thus $L$ contains all odd powers of $x$ and has codimension 1 in $R.$ 
Suppose that $L$ contains a principal ideal $(f)$. Let $f=f_0+f_1$ be the decomposition of $f$ into the even and odd parts ($f_0$ is the span of the even degree monomials of $f$). Then $x^{2N}f_0\in L$ and $x^{2N+1}f_1\in L$ for any $N\geq 0.$ 
At least one of $f_0$ and $xf_1$ is non-zero and has the form $g=\sum_n a_{n}x^{2n}.$ 
Then for any $N\geq 0,$
$$x^{2N}g=\sum_n a_nx^{2(n+N)}\equiv \sum_n (2n+2N)!!a_n\ (\mod L)\quad 
\text{ and }\quad \sum_n (2n+2N)!!a_n=0.$$
However, this is impossible: for sufficiently large $N,$ the term involving $a_n\ne 0$ with the largest $n$ clearly dominates the rest of the sum.<|endoftext|>
TITLE: Diameter of reduction graph of a curve over a complete discrete valuation ring
QUESTION [6 upvotes]: Let $R$ be a complete discrete valuation ring with field of fractions $K$ and algebraically closed residue field $k$, and let $X$ be a proper, smooth, geometrically connected curve over $K$. Take a finite extension $L/K$ and a regular proper model $\widetilde{X}$ of $X$ over the ring of integers of $L$ whose special fibre $\widetilde X_k$ is semi-stable.  Let $e$ be the ramification index of $L/K$.  The reduction graph of $X$ is a metrised graph obtained as follows. Take a set of intervals of lengh $1/e$ indexed by the singular points of $\widetilde X_k$.  For every singular point $x$ of $\widetilde X_k$, label the endpoints of the corresponding edge by the two irreducible components (possibly the same) on which $x$ lies. For every irreducible component $C$ of $\widetilde X_k$, identify all the endpoints labelled $C$. The result (as a metric space) is independent of the choice of $L$.
Question: Suppose we have a regular proper model of $X$ over $R$ whose special fibre $X_k$ is reduced, but not necessarily semi-stable.  Suppose furthermore that we know the irreducible components and singular points of $X_k$ and their intersection multiplicities in this model.  What can be said about the reduction graph of $X$?
It seems reasonable to ask this question in this generality, but I am actually interested in the modular curves X1(n) over Wp[ζp2], where p is a prime number dividing n exactly twice and Wp is the ring of Witt vectors of an algebraic closure of Fp.  In this situation non-semi-stable models as above were found by Katz and Mazur.  What I would like specifically is an upper bound on the diameter of the reduction graph for these curves.

REPLY [6 votes]: This is a sequel to the above comments. 
Consider an elliptic curve $E$ over $K$ with additive reduction over $K$ and multiplicative reduction over some extension $L/K$. Then we can find a quadratic extension $L/K$, and the Kodaira symbole of $E$ over $K$ is $I^*_m$ for some $m$ (see below), and the group of components of $E_L$ is $Z/nZ$, where $n=[L:K]\nu_K(\Delta)$ and $\Delta$ is the minimal discriminant of $E$ over $K$. 
Now how to compute $n$ from data over $K$ ? I mean from data that you can read from the special fiber of the minimal regular model over $K$ ? The link between $n$ and $m$ is $n/2=(2+\delta) + (m+4)-1$ (Ogg's formula), and $\delta$ is the Swan conductor of $E$. If the residue characteristic is different from $2$, then $\delta=0$ and $n=2(m+5)$. Perfect. 
But if the residue characteristic is 2 (so wild ramification happens), then $\delta$ can be arbitrarily big (if the absolute ramification index of $K$ is big). So $n$ is not a function of data from the special fiber over $K$.<|endoftext|>
TITLE: Does the norm of a normed linear space determine the form of its dual spaces elements?
QUESTION [5 upvotes]: Hello everybody,
As an introductory example, suppose $U \subset R^n$ is open and bounded, let $p = 2$. Then there is a constant $c>0$ s.t. $\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}_0} \le c \Vert \vert \nabla u \vert\Vert_{L^p(U)}$. This implies that the latter expression defines an equivalent norm on ${W^{1,p}_0}$.
Let $f \in L^2(U), g \in C^1(\bar{U})$. Then there exists an unique solution $u \in W^{1,p}$ to the system $ \triangle u = f $ over $U$, $u = g$ over $\partial U$ - or equivalently, there exists an unique solution $u \in W^{1,p}_0$ to the system $ \triangle u = f - \triangle g$ over $U$, (in the distributional sense).
Proof: $W^{1,2}_0$ is a Hilbert space, hence self-dual. The rhs $f - \triangle g$, defines an element of $D'$, which by density can extended to $W^{1,2}_0$. On the other side, the equivalent norm as introduced above is defined the inner product $(u,v) = \int \nabla u \cdot \nabla v$, by riesz' representation theorem, there is an $u \in W^{1,2}_0$ s.t. the induced form $(u, \cdot)$ coincides with the form defined by the rhs. But then this $u$ is a weak solution to $ \triangle u = f - \triangle g$.
So far, so good. I would like to ask some questions on this.
i) Can this be extended to other dual exponents $p$, $q$ ?
ii) The equivalent norm that regards first derivatives only is not only an equivalent norm for $p=2$, but also for $1 \leq p < \infty$. In the above case, it seems the norm imposes a form the dual vectors are subject to. I wonder whether in general - not only in the case of $L^p$ and its friends - there is some way how the form of linear functionals on some normed space $X$ are determined by the norm attached to the vector space $X$.
I hope this questions ain't too vacuous and there are interesting answers. In either case, thanks.

REPLY [5 votes]: The argument you have presented is an adaptation of the Lax-Milgram theorem which is essentially equivalent to the Riesz representation theorem (and generally speaking both of these results hold only in the Hilbert space framework). The Lax-Milgram theorem fails for the Laplace equation in $L^p$-spaces with $p\neq2$. Instead, some analogous results based on the ideas of coercivity, duality and monotonicity can be obtained in any reflexive Banach space.
The Dirichlet problem for the $p$-Laplace operator
$$-\nabla(|\nabla u|^{p-2}\nabla u)=f,\quad x\in\Omega,\qquad (*)$$
$$u=0,\qquad x\in\partial\Omega,$$
might be a "correct" $L^p$-analogue of the problem described in the question.
The right hand side of $(*)$ gives rise to the mapping $A: W_{0}^{1,p}\to(W_{0}^{1,p})^{*}$
defined by the identity
$$\langle Au,v\rangle=\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla v\ dx\quad \mbox{for all }  v\in W_{0}^{1,p}.$$
A straightforward check shows that $A$ satisfies the conditions of the following theorem
(which might be viewed as an $L^p$-analogue of the Lax-Milgram theorem).

Theorem. Let $A$ be a strictly monotone, coercive operator from a reflexive Banach space $E$ to its dual $E^{* }$.   If $A$ is continuous on finite-dimensional subspaces of $E$ then for every $f\in E^{*}$ there exists a unique solution to the problem
  $$Au=f.$$  

Have a look at the textbook by Chipot or  the free monograph by Showalter where the approach is explained in detail.<|endoftext|>
TITLE: Why the roots of unity are the analogs of constants ?
QUESTION [10 upvotes]: Hello,
Joel Dogde, in a comment on his question "Roots of unity in different completions of a number field", says the following, about the analogy between number fields and function fields :  
Number of roots of unity in number fields is something like the size of the constant field for function fields.
Could anyone explain that ?
Thanks.

REPLY [3 votes]: The number of roots of unity in a local field (of order prime to the characteristic) plus 1 is the cardinality of the residue class field (i.e. constant field in the function field case).  This follows because the $q$-th roots for $q$ prime to the characteristic are distinct in the residue class field, because $\prod_i (1 - \zeta_q^i) = q$.  In addition, each element $\bar{x}$ of the residue class field must satisfy $\bar{x}^r= 1$ for some $r$ prime to the characteristic, and $\bar{x}$ can be lifted to the local field by Hensel's lemma to an $r$-th root of unity.  The plus one was added to account for zero (thanks to KConrad for pointing this out).
An unramified extension of local fields is obtained by adjoining a root of unity (prime to the characteristic of the residue class field).  In the case when the local fields are completions of function fields, the additional roots of unity correspond precisely to increasing the field of constants (because constants are roots of unity, as Pete Clark as mentioned).<|endoftext|>
TITLE: Trigonometric identities
QUESTION [7 upvotes]: In the rant I wrote at
http://ncatlab.org/nlab/show/trigonometric+identities+and+the+irrationality+of+pi
I asked: Are these four identities the first four terms in a sequence that continues?
This referred to the identities in the last bullet point above that question.
While we're at it, is there any intuitive geometric interpretation of the identity involving $f_2$?
OK, here are the functions involved:
$$
\begin{align}
f_0(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 0} (-1)^{n/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_1(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 1} (-1)^{(n-1)/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_2(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 2} (-1)^{(n-2)/2} n \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_3(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 3} (-1)^{(n-3)/2} (n-1) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_4(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 4} (-1)^{(n-4)/2} n(n-2) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_5(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 5} (-1)^{(n-5)/2} (n-1)(n-3) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_6(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 6} (-1)^{(n-6)/2} n(n-2)(n-4) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
f_7(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 7} (-1)^{(n-7)/2} (n-1)(n-3)(n-5) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\
& \,\,\,\vdots
\end{align}
$$
In each function the coefficient kills off the terms involving values of $n$ smaller than the index, so that for example we could have said "$\text{odd }n \ge 1$" instead of $\text{odd }n \ge 7$ and it would be the same thing.
Now some facts:

Each $f_k$ is a symmetric function of $\theta_1,\theta_2,\theta_3,\dots$.

$0$ is an identity element for each of these functions, in the sense that $f_k(0,\theta_2,\theta_3,\dots) = f_k(\theta_2,\theta_3,\dots).$

Let $\lfloor k\rfloor_\text{even}= 2\lfloor k/2\rfloor$ be the "even floor" of $k,$ i.e. the largest even integer not exceeding $k.$ Then \begin{align}
& f_k(\theta_1,\theta_2,\theta_3,\dots) - f_k(\theta_1+\theta_2,\theta_3,\dots) \\[8pt]
= {} & \begin{cases}
\lfloor k \rfloor_\text{even} \cdot \sin\theta_1\sin\theta_2 f_{k-2}(\theta_3,\theta_4,\dots) & \text{if } k\ge2, \\[6pt]
\quad 0 & \text{if } k = 0\text{ or } 1.
\end{cases}
\end{align}


Now the sequence of identities:
$$
\begin{align}
f_0 & = \cos(\theta_1 + \theta_2 + \theta_3 + \cdots) \\
f_1 & = \sin(\theta_1 + \theta_2 + \theta_3 + \cdots) \\
\text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then }
f_2 & = \sum_{i=1}^\infty \sin^2\theta_i=\frac{1}{2} \sum_{i=1}^\infty (1-\cos(2\theta_i))\\
\text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then }
f_3 & = \frac{1}{2} \sum_{i=1}^\infty \sin(2\theta_i)
\end{align}
$$
The QUESTION is whether these are the first four identities in a sequence that continues beyond this point.

REPLY [2 votes]: With binomial theorem, the products on the right have closed form $$ \sum_{|A|=n} \prod_{i \in A} \sin \theta_i \prod_{i \notin A} \cos \theta_i = \prod_{i=1}^n ( \sin \theta_i + \cos \theta_i )^n = \prod_{i=1}^n \sqrt{2} \sin (\theta_i + \pi/4)^n$$
So we'll let $x = \sin \theta + \cos \theta$ so the first sum looks like
$$ \sum_{n \geq 0,\text{ even}} (-1)^{n/2} \prod_{i=1}^n x_i = 1 - x_1x_2 + x_1x_2x_3x_4 - \dots$$
This is not symmetric in the x's.<|endoftext|>
TITLE: What is a flop (and when are they conjectured to give derived equivalences)?
QUESTION [13 upvotes]: (1)  Is the definition of flop given by Wikipedia the industry standard?
(2)  Regardless of the answer to (1), when is it expected that a birational transformation gives rise to a derived equivalence?
References to places where precise conjectures are recorded will be very much appreciated!
The reason I'm asking: apparently it is conjectured that different crepant resolutions are derived equivalent.  On page 40 of this paper of Bondal-Orlov, they conjecture that flops induce derived equivalences.  Apparently "flop" is sometimes used to mean birational transformation preserving canonical classes (without specifying the type of surgery actually being performed).  So I'm interested to know whether such transformations are expected to be factorizable into (Wikipedia) flops, or produce derived equivalences for other reasons.

REPLY [5 votes]: I'm not any kind of expert on this stuff and I'm not sure what the current state of this conjecture is, but Kawamata has conjectures in this paper and this paper regarding when two birational varieties have equivalent derived categories. He also discusses flops in the first paper.
He has partial results, including: if $X$ is general type and $\mathcal{D}^b(X) \cong \mathcal{D}^b(Y)$ as triangulated categories then $X$ and $Y$ are K-equivalent. This generalizes the famous theorem of Bondal-Orlov that the bounded derived category of a Fano variety determines the variety. IIRC, in the proof of his theorem he takes the kernel of the Fourier-Mukai transform that gives the equivalence, shows that the support of the kernel (meaning the union of the supports of the cohomology sheaves of the kernel) has a component $Z$ dominating both varieties and uses $Z$ for the "roof" of the K-equivalence. The assumption that $X$ is general type is used to show that the projections from $Z$ are birational.<|endoftext|>
TITLE: how can I minimise (n * y) (mod x) for known x and y, and for a given range of n?
QUESTION [10 upvotes]: How can I minimise (n.y) (mod x), for known x and y, and for a given range of n?
($x$ and $y$ are actually the components of a 2D vector for a line for which I'm trying to generate a set of bounding integer points)
So, for example, if x = 61, y = 17, and n must be in the range 0 < n < 12, then minimum value of the modulo operation is at n = 11, i.e. (11 * 17) (mod 61) = 4.
If we changed the range to 0 < n < 9, the minimum value is then at n = 4,  i.e. (4 * 17) (mod 61) = 7.
I need to be able solve this for arbitrary values, but within a known range (around +/- 3000000).
This is a practical question so if there is no direct solution (or if a direct solution is very complicated) then a numerical method may be preferrable.

REPLY [4 votes]: What I'm going to say is somewhat similar to Roland's answer but more precise in the case when the range for $n$ is given in the form of upper bound, i.e., $0 < n < N$. 
Notice that $ny\bmod x = ny - kx$ for some integer $k$. We want to minimize $ny-kx$ that, if we disregard for a moment the sign, can be formulated as minimizing 
$$\left|n\frac{y}{x} - k\right|$$
over integer $n$ in the given range and arbitrary integer $k$.
It is known that if some $n,k$ give better approximation (in the sense of the above absolute value) than any other $n',k'$ with $n' < n$, then $\frac{k}{n}$ with necessity represents a convergent to $\frac{y}{x}$.
Therefore, a first good candidate for the anticipated $n$ is the largest denominator of a convergent $\frac{k}{n}$ for $\frac{y}{x}$ that fits the given range (i.e., $n < N$).
For such $n$, if we have $n\frac{y}{x} - k > 0$ (equivalently, $\frac{k}{n}<\frac{y}{x}$), then it is indeed a solution. 
However, if $n\frac{y}{x} - k < 0$ (equivalently, $\frac{k}{n}>\frac{y}{x}$), then the solution is given by largest allowed denominator of a semi-convergent located between the preceding and subsequent convergents of $\frac{k}{n}$. That is, if $\frac{k'}{n'}, \frac{k}{n}, \frac{k''}{n''}$ are consecutive convergents, then $\frac{k'}{n'} < \frac{k''}{n''} < \frac{y}{x}$ and $n' < N \leq n''$. Then one needs to find a semi-convergent between $\frac{k'}{n'}$ and $\frac{k''}{n''}$ with the largest denominator smaller than $N$.<|endoftext|>
TITLE: Why do we care about $L^p$ spaces besides $p = 1$, $p = 2$, and $p = \infty$?
QUESTION [139 upvotes]: I was helping a student study for a functional analysis exam and the question came up as to when, in practice, one needs to consider the Banach space $L^p$ for some value of $p$ other than the obvious ones of $p=1$, $p=2$, and $p=\infty$.  I don't know much analysis and the best thing I could think of was Littlewood's 4/3 inequality.  In its most elementary form, this inequality states that if $A = (a_{ij})$ is an $m\times n$ matrix with real entries, and we define the norm
$$||A|| = \sup\biggl(\left|\sum_{i=1}^m \sum_{j=1}^n a_{ij}s_it_j\right| : |s_i| \le 1, |t_j| \le 1\biggr)$$
then
$$\biggl(\sum_{i,j} |a_{ij}|^{4/3}\biggr)^{3/4} \le \sqrt{2} ||A||.$$
Are there more convincing examples of the importance of "exotic" values of $p$?  I remember wondering about this as an undergraduate but never pursued it.  As I think about it now, it does seem a bit odd from a pedagogical point of view that none of the textbooks I've seen give any applications involving specific values of $p$.  I didn't run into Littlewood's 4/3 inequality until later in life.
[Edit: Thanks for the many responses, which exceeded my expectations!  Perhaps I should have anticipated that this question would generate a big list; at any rate, I have added the big-list tag.  My choice of which answer to accept was necessarily somewhat arbitrary; all the top responses are excellent.]

REPLY [2 votes]: One answer that I don't see explicitly mentioned here is that $L^\infty = \lim_{p\to\infty} L^p$, so to speak.  So if you in fact care about $L^\infty$ but there are easier theorems regarding $p<\infty$, then do what we do in analysis, and get a sequential approximation.<|endoftext|>
TITLE: Picard-Fuchs equations for modular functions
QUESTION [14 upvotes]: Hello, MathOverflow community!
Suppose we have a modular curve of genus $0$, whose rational function field is generated by the modular function $f$. We can view $f$ as the parameter for some pencil of elliptic curves over $\mathbb{C}$. Under certain conditions, it is possible to express $f$ as the inverse function of the ratio of two linearly independent solutions of a second-order linear differential equation. The prototypical example is the case of the period integrals of the Legendre elliptic curve $y^2=x(x-1)(x-\lambda)$, which satisfy the Fuchsian equation $$\lambda(1-\lambda)D^2y + (1-2\lambda)Dy - y/4=0,$$ where $D=d/d\lambda$. We can interpret this differential equation as measuring the variation of the periods of an elliptic curve, as the parameter $\lambda$ changes. 
Now my question is : have other Picard-Fuchs equations been calculated for modular functions? In principle, there should be many such equations; the Picard-Fuchs equation for Klein's $j$ function, without the calculation, is given in (Harnad, McKay). I have seen the calculation for the $\lambda$ case carried out in a few books. But I have not seen such equations for the Hauptmodul associated to the other genus $0$ modular curves.
Any thoughts, comments, questions or references are much appreciated.
Please be kind, as I am only an undergraduate. (There seems to be much "tough love" here!)

REPLY [2 votes]: See the following papers as well...

Linear Ordinary Differential
Equations Satisfied by Modular
Forms by Xiaolong Ji and Yujie
Ma
Modular Functions and Their
Differential Equations by Joshua
L. Wiczer See also
http://arxiv.org/PS_cache/math/pdf/0611/0611291v1.pdf
MR2031441 (2005b:11049)  Yang, Yifan
. On differential equations
satisfied by modular forms. Math. Z.
246  (2004),  no. 1-2, 1--19.<|endoftext|>
TITLE: Intersection homology for toric varieties
QUESTION [5 upvotes]: is there any algorithm known for computing (middle perversity)intersection homology of complex toric varieties based on their combinatorial data? I'm not looking for a computer program.
Regards,
Peter

REPLY [4 votes]: There is a simple and beautiful description in terms of commutative algebra (repeatedly calculating global sections and taking a projective cover). The work of Braden-MacPherson cited by Alexander is relevant, but only for certain toric varieties (those admitting affine pavings). Also, the Braden-MacPherson paper is really aimed at handling the case of flag varieties etc., which is more complicated than toric varieties.
I think the first combinatorial description was given by Bernstein and Lunts at the end of their book on equivariant sheaves:
Bernstein, Joseph; Lunts, Valery
Equivariant sheaves and functors. LNM 1578. Berlin: Springer-Verlag.
This was then abstracted to arbitrary (perhaps non-rational) polytopes here:
Bressler, Paul and Lunts, Valery, Intersection Cohomology on Nonrational Polytopes,
Compositio Mathematica, Volume 135, Issue 3, pp 245-278.
http://arxiv.org/abs/math/0002006
There is parallel work by BBFK:
Gottfried Barthel, Jean-Paul Brasselet, Karl-Heinz Fieseler, and Ludger Kaup
Combinatorial intersection cohomology for fans, Tohoku Math. J. (2) Volume 54, Number 1 (2002), 1-41.
All of this is summarized quite nicely in Kirwan and Wolf, An introduction to Intersection Cohomology Theory, Second Edition, Chapman and Hall, 2006.<|endoftext|>
TITLE: Dimension of central simple algebra over a global field "built using class field theory".
QUESTION [17 upvotes]: If $F$ is a global field then a standard exact sequence relating the Brauer groups of $F$ and its completions is the following:
$$0\to Br(F)\to\oplus_v Br(F_v)\to\mathbf{Q}/\mathbf{Z}\to 0.$$
The last non-trivial map here is "sum", with each local $Br(F_v)$ canonically injecting into $\mathbf{Q}/\mathbf{Z}$ by local class field theory.
In particular I can build a class of $Br(F)$ by writing down a finite number of elements $c_v\in Br(F_v)\subseteq \mathbf{Q}/\mathbf{Z}$, one for each element of a finite set $S$ of places of $v$, rigging it so that the sum $\sum_vc_v$ is zero in $\mathbf{Q}/\mathbf{Z}$.
This element of the global Brauer group gives rise to an equivalence class of central simple algebras over $F$, and if my understanding is correct this equivalence class will contain precisely one division algebra $D$ (and all the other elements of the equiv class will be $M_n(D)$ for $n=1,2,3,\ldots$).
My naive question: is the dimension of $D$ equal to $m^2$, with $m$ the lcm of the denominators of the $c_v$? I just realised that I've always assumed that this was the case, and I'd also always assumed in the local case that the dimension of the division algebra $D_v$ associated to $c_v$ was the square of the denominator of $c_v$. But it's only now, in writing notes on this stuff, that I realise I have no reference for it. Is it true??

REPLY [2 votes]: The fact that the Schur index of an element of the Brauer group of
a number field equals its order in the Brauer group (and also
is a cyclic algebra) is Theorem 18.6 in Richard Pierce's
Associative Algebras (GTM88 Springer-Verlag 1982). The proof
uses the Grunwald-Wang theorem.<|endoftext|>
TITLE: Universal property of X//G?
QUESTION [8 upvotes]: Given an operation of say a topological group on a topological space, one can form the quotient stack X//G: 
the stack associated to the action groupoid.
Does this stack satisfy some kind of universal property?

REPLY [9 votes]: Probably the simplest example is when the space $X$ is a single point. Then $pt//G$ classifies principal $G$-bundles. Here, the action groupoid is just $G$ considered as a one-object groupoid. The one object, $pt$ becomes an atlas for the stack, so we have a representable epimorphism $pt \to pt//G$. This is in fact the universal principal $G$-bundle! Given any map $X \to pt//G$, the 2-fibered product $X \times_{pt//G} pt\to X$ is a principal $G$-bundle. What we have in general is, given any topological groupoid $H$, its associated stack (stackification of $Hom(blank,H)$ ) is equivalent to $Bun_H:T \mapsto Bun_H(T)$, where $Bun_H(T)$ is the groupoid of principal $H$-bundles over $T$. So, by Yoneda, we have for all spaces $T$, an equivalence of groupoids $Hom(T,Bun_H) \cong Bun_H(T)$. Moreover, you can show that for any map $T \to Bun_H$, the principal bundle $P$ to which the map corresponds is the 2-fibered product $T \times_{Bun_H} H_0 \to T$. Applying this to the case when $H$ is the group $G$, we get what I claimed. Now, if instead $H$ is the action groupoid of $G$ on a space $X$, we get that $X//G$ classifies principal $G \ltimes X$-bundles. This answers the question as far as what maps INTO the stack yield. 
As Chris suggested, for any topological groupoid, you can take its enriched-nerve to get a simplicial space. Applying Yoneda on each component, you get a simplicial stack. The weak 2-colimit of this, is equivalent to the stack associated to the groupoid. (In fact, you only need to consider the 2-truncation of this diagram). Spelling this out, you get Proposition 3.19 out of Noohi's Foundations of Topological Stacks I. This roughly says that maps from the stack associated to a groupoid $H$ into another stack $Y$ are the same as maps $f:H_0 \to Y$ (or by Yoneda elements $f \in Y(H_0)_0$) together with an isomorphism $f \circ s \to f \circ t$ satisfying some obvious coherencies, where $s$ and $t$ are the source and target map of the groupoid $H_1 \rightrightarrows H_0$. In the case where $H$ is the action groupoid $G \ltimes X$, you get maps $f:X \to Y$, together with isomorphisms $f \circ proj \to f \circ \rho$, where $\rho:G \times X \to X$ is the action. This becomes more concrete, if for example, $Y$ is not some arbitrary stack, but instead the stack of principal $K$-bundles for some group $K$. Then $Hom(X//G,Bun_K)$ is the groupoid with objects principal $G$-bundles $P$ over $X$ together with an isomorphism $proj^*P \cong \rho^*P$ (satisfying a cocycle condition), with the obvious arrows.
If you were wondering why $X//G$ was the "like the ordinary quotient but better", then the following heuristics should help. The 2-truncated guy whose colimit equals $X//G$ is "like the colimit expressing $X/G$ but with the action built in more". In fact, we we took the UNtruncated guy, so, the simplicial space associated to the action groupoid, then it "IS" the Borel construction (or rather its geometric realization). In particular, this implies the homotopy type of $X//G$ is the same as the homotopy quotient $EG \times_G X$. Now consider one more thing- if $G$ is acting on $X$ we have the commutative square with $G \times X$ on the upper left, $X$ on bottom left and on the upper right, and $X/G$ (the coarse quotient) on the bottom right, with map up top being the action, and the map down below being the projection, then this is a pullback diagram if and only if the action is faithful. If you replace $X/G$ with $X//G$, then this is ALWAYS a (2-)pullback diagram. Hence, going to stacks "makes all actions faithful up to homotopy" (more accurately, stacks allow us to encode the isotropy data that would otherwise be lost in such a way that any action becomes as good as a faithful one).

REPLY [6 votes]: Yes. It is the (homotopy) colimit in stacks of the simplicial diagram:
$$\cdots X \times G^2 \Rightarrow X \times G \rightrightarrows X $$
(here $\Rightarrow$ is a substitute for three arrows). This simplicial object is built from the action of G on X.  Since stacks form a 2-category, the natural thing is a "weak" or "homotopy" colimit. See the n-lab pages. The inclusion from spaces into stacks doesn't preserve colimits, so even though this is a diagram of ordinary spaces, the colimit is an interesting stack. 
Since it is a colimit, there is a clear associated universal property for maps out of this stack. 
It is also the stackification of the presheaf of groupoids associated to the action groupoid, hence from this description it satisfies another additional universal property for maps out of it, and there is a concrete description for maps into it. It is a fun (but tedious) exercise to see that the two universal properties are in fact the same.<|endoftext|>
TITLE: Checking whether the image of a smooth map is a manifold
QUESTION [9 upvotes]: I have a specific problem, but would also like to know how to tackle the general case. I will first state the genral question. Let $M$ be an embedded submanifold of $\mathbb{R}^n$ and let $F: \mathbb{R}^n \to \mathbb{R}^n $ be a smooth map. How do I go about checking whether $F(M)$ is a smooth embedded submanifold of $\mathbb{R}^n$ or not? The specific problem I have is the following :- Let $F:\mathbb{C}^2\to \mathbb{C}^2$ the map $(z_1,z_2) \mapsto (z_1+z_2,z_1z_2)$ and let $M$ be the unit sphere in $\mathbb{C}^2$, i.e., $\lbrace (z_1,z_2) : |z_1|^2 + |z|^2 = 1 \rbrace $. Is $F(M)$ an embedded submanifold of $\mathbb{C}^2$ (considered as $\mathbb{R}^4$)?

REPLY [9 votes]: The specific $F(M)$ is not a smooth submanifold. Here is an argument.
To simplify formulas, I renormalize the sphere: let it be the set of $(z_1,z_2)\in\mathbb C^2$ such that $|z_1|^2+|z_2^2|=2$ rather than 1. Then, as Gregory Arone pointed out, $F(M)$ is the set of $(b,c)\in\mathbb C^2$ such that the roots $z_1,z_2$ of the equation $z^2-bz+c$ satisfy $|z_1|^2+|z_2^2|=2$. I claim that it is not a smooth manifold near the point $p:=(2,1)\in F(M)$.
Let us intersect $F(M)$ with two planes: $\{b=2\}$ and $\{c=1\}$. If it is is a smooth submanifold, at least one of the intersections should be locally (near $p$) a 1-dimensional smooth submanifold of the respective plane (otherwise both planes are tangent to $F(M)$ at $p$, but this is impossible since they span the whole space). But this is not the case:
If $b=2$, the equation is $z^2-2z+c=0$, hence $z_1+z_2=2$, then $|z_1|+|z_2|\ge 2$ and therefore $|z_1|^2+|z_2|^2\ge 2$. Equality is attained only for $z_1=z_2=1$, thus the intersection is a single point $c=1$, not a 1-dimensional submanifold.
If $c=1$, the equation is $z^2-bz+1$, hence $z_1z_2=1$, then $|z_1|\cdot|z_2|=1$ and therefore $|z_1|^2+|z_2|^2\ge 2$. The equality is attained if and only if $|z_1|=|z_2|=1$, so $z_1$ and $z_2$ are conjugate to each other. The set of $b$'s for which this happens is the real line segment $[-2,2]$ which is not a submanifold near 2.<|endoftext|>
TITLE: Proving Hodge decomposition without using the theory of elliptic operators?
QUESTION [35 upvotes]: In the common Hodge theory books, the authors usually cite other sources for the theory of elliptic operators, like in the Book about Hodge Theory of Claire Voisin, where you find on page 128 the Theorem 5.22:

Let P : E → F be an elliptic differential operator on a compact manifold. Assume that E and F are of the same rank, and are equipped with metrics. Then Ker P is a finite-dimensional subspace of the smooth sections of E and Im P is a closed subspace of finite codimension of the smooth sections of F and we have an L²-orthogonal direct sum decomposition:
smooth sections of E = Ker P ⊕ Im P*
(where P* is the formal L²-adjoint of P)

In the case of Hodge Theory, we consider the elliptic self-adjoint operator Δ, the Laplacian (or: Hodge-Laplacian, or Laplace-Beltrami operator).
A proof for this theorem is in Wells' "Differential Analysis on Complex Manifolds", Chapter IV - but it takes about 40 pages, which is quite some effort!

Now that I'm learning the theory of elliptic operators (in part, because I want to patch this gap in my understanding of Hodge Theory), I wonder if this "functional analysis" is really always necessary.
Do you know of any class of complex manifolds (most likely some restricted class of complex projective varieties) where you can get the theorem above without using the theory of elliptic operators (or at least, where you can simplify the proofs that much that you don't notice you're working with elliptic operators)? Maybe the general theorem really requires functional analysis (I think so), but the Hodge decomposition might follow from other arguments.
I would be very happy to see some arguments proving special cases of Hodge decomposition on, say, Riemann surfaces. I would be even happier to hear why this is implausible (this would motivate me to learn more about these fascinating elliptic differential operators).
If this ends up being argumentative and subjective, feel free to use the community wiki hammer.

REPLY [12 votes]: The Hodge decomposition can be proved, in a very nice, abstract-functional-analysis setting, on so-called Hilbert complexes.  Brüning and Lesch wrote an excellent paper on the topic in J. Funct. Anal., first developing the theory on arbitrary Hilbert complexes, and then discussing the application to elliptic complexes.<|endoftext|>
TITLE: Differential operators preserving the space of harmonic functions (aka higher symmetries of the Laplacian)
QUESTION [6 upvotes]: The article http://arxiv.org/abs/hep-th/0206233 (published in Ann. of Math. (2)  161  (2005),  no. 3) deals with linear differential operators $D$ for which there exists another linear differential operator $\delta$ such that $\Delta D = \delta \Delta$. Obviously these operators preserve the kernel of $\Delta$, i.e. the space of harmonic functions. The mentioned article finds essentially all such operators $D$. The result is that up to trivial operators $D = P\Delta$ all the operators $D$ have polynomial coefficients and are generated by sums of compositions of first order operators of this kind.
First question: Let $D$ be any differential operator preserving the space of harmonic functions. It is easy to see that the operator $\delta = \Delta D (\Delta)^{-1}$ is well defined and satisfies $\Delta D = \delta \Delta$. Is $\delta$ also a differential operator?
Second question: Is it true that all differential operators, which preserve the space of harmonic functions, are generated by first order ones with this property?
One can also ask these questions only for linear differential operators or for operators from the Weyl algebra (i.e. linear differential operators with polynomial coefficients). For example, by a theorem of Peetre, the answer to the first question is affirmative if the operator $\delta = \Delta D (\Delta)^{-1}$ is local (i.e. the support of $\delta u$ is contained in the support of $u$).
Third question: What makes the linked article so interesting that it was published in Annals?

REPLY [2 votes]: The answer to your second question (unless I somehow misread it) is yes precisely because of the result of the paper you refer to (you may also wish to look at this paper and the preprint math-ph/0506002 which address the same subject). This is the case because if $D$ is a differential operator that preserves the space of harmonic functions then there indeed exists a differential operator $\delta$ such that $\Delta D = \delta \Delta$. The latter holds (see e.g. the discussion at p.290 near Eq.(5.5) of the book Applications of Lie groups to Differential Equations by P.J. Olver) because the equation $\Delta f=0$ is totally nondegenerate in the sense of Definition 2.83 of the same book. In spite of the rather technical language the idea behind all this is very simple: if you have a submanifold $N$ of an manifold $M$ defined by the equations $F_1=0, \dots, F_k=0$ with smooth $F$'s and $k<\mathrm{dim}\ M$, then a smooth function $h$ vanishes on $N$ iff there exist smooth functions hj on $M$ such that 
$$h=h_{1} F_1+\cdots+h_k F_k$$ provided $dF_1\wedge \dots \wedge dF_k\neq 0$ on $N$ (see Proposition 2.10 of the same book). In a sense, this is a smooth counterpart of the famous Hilbert's Nullstellensatz in the form stated e.g. here. This result is then applied to the case when $M$ is a jet bundle and $N$ is a submanifold thereof defined by a system of differential equations and all its differential consequences (more precisely, one should rather consider the consequences only up to a certain order, to avoid dealing with infinitely many equations), et voila.<|endoftext|>
TITLE: Does a positive binary quadratic form represent a set of primes possessing a natural density
QUESTION [9 upvotes]: In his answer to my question 
The Green-Tao theorem and positive binary quadratic forms
                Kevin Ventullo answers my initial question in the affirmative. What remains is the title question here, of separate interest to me. 
Any integral positive binary quadratic form integrally represents a set of primes with known Dirichlet density. This is an application of the Cebotarev density theorem (or Chebotarev or Tchebotarev), in particular it is Theorem 9.12 in David A. Cox, Primes of the form $x^2 + n y^2,$ with the example $\Delta = -56$ on page 190. I typed this out in the previous question.
Now, Jurgen Neukirch "Class Field Theory" points to Serre "A Course in Arithmetic," and on page 76 Serre says that the set of primes p such that a fixed polynomial has a root $\pmod p$ has a natural density, and refers to K. Prachar "Primzahlverteilung" chapter 5 section 7. By results (theorem 9.2, page 180) in the Cox book, this means that the principal form 
$x^2+ny^2$ or $x^2+xy+ky^2$ does represent a set of primes with a natural density, therefore equalling the Dirichlet density. And by the result on arithmetic progressions, a full genus of forms has a natural density.
Combining observations, the principal form always has a natural density of primes, any full genus does, therefore we are done for one class per genus, and in the case of two classes per genus we are done with the principal genus and any genus with two distinct opposite forms. So we have natural densities for Cox's $\Delta = -56$ example. We are also done with the principal genus when it has three classes.
So, (and I would love a reference), does every positive binary quadratic form represent a set of primes for which the natural density exists?

REPLY [6 votes]: If $Q$ is a positive binary quadratic form whose discriminant $D$ is fundamental, then the number of primes $\leq X$ represented by $Q$ is given asymptotically as
$\pi_{Q}(X)=\frac{1}{2h(D)} \mathrm{Li}(X)+O(X \exp{(-c_{Q}\sqrt{\log{X}})})$. 
Here $\mathrm{Li}(X)$ is the usual logarithmic integral. My reason for giving this as an answer is to point out that this was proven by de la Vallee Poussin himself, in the same paper where he proved the usual prime number theorem!  Dating from the 1890s, this definitely predates Chebotarev, and de la V.P.'s contribution deserves (IMHO) to be better known than it is.  Hadamard's work on the PNT is a little easier on the reader than de la V.P.'s, but de la V.P.'s results were both stronger and more general, which sometimes is forgotten.<|endoftext|>
TITLE: Tensor products of Weyl modules in positive characteristic
QUESTION [5 upvotes]: Let $G$ be a simple algebraic group over a field $k$, and let $U$ be the unipotent radical of a Borel subgroup $B$. Because $B$ normalises $U$, the group $H = B/U$ acts on the coordinate ring $\mathcal{O} = k[X]$ of the basic affine space $X = G/U$ via $(h.f)(x) = f(xh)$. We get a decomposition of $\mathcal{O}$ into a direct sum 
$\mathcal{O} = \oplus_{\lambda \in \Lambda^+} \mathcal{O}^\lambda$
where the Weyl module $\mathcal{O}^\lambda$ is the set of all $f \in \mathcal{O}$ such that $h.f = \lambda(h)f$ for all $h \in H$. Because the action of $H$ commutes with the action of $G$ on $k[X]$ given by $g.f(x) = f(g^{-1}x)$, each $\mathcal{O}^\lambda$ is a $G$-submodule of $\mathcal{O}$. We can also identify $\mathcal{O}^\lambda$ with the space of global sections $H^0(G/B, \lambda)$.
Next, multiplication in $\mathcal{O}$ induces a $G$-module map $\mathcal{O}^\lambda \otimes \mathcal{O}^\mu \to \mathcal{O}^{\lambda + \mu}$ for any $\lambda, \mu \in \Lambda^+$. Since $\mathcal{O}$ has no zero-divisors, this map is non-zero. 
Now if the base field $k$ has characteristic zero, it is well-known that the $G$-modules $\mathcal{O}^\lambda$ for $\lambda \in \Lambda^+$ are irreducible, so the multiplication map above must be surjective. Does this remain true when the characteristic of $k$ is positive, when the Weyl modules $\mathcal{O}^\lambda$ are no longer irreducible in general?

REPLY [7 votes]: The question has an affirmative answer and a fairly long history as well, but the
proof uses some nontrivial ideas.   The notation used here is nonstandard relative to that found in Jantzen's book Representations of Algebraic Groups (second edition, AMS, 2003).   Also, a "Weyl module" (in the usual sense)  of a given highest weight is the dual of the module of global sections for a related line bundle on the flag variety, using Kempf's vanishing theorem (1976).   The term "Weyl module" was coined by Carter and Lusztig in their paper on special linear groups, partly because the formal character is given by Weyl's formula.
A Weyl module has a unique simple quotient, while the corresponding module has this module as its unique simple submodule.   
There was a series of papers by Lakshmibai-Musili-Seshadri on the geometry of flag varieties in prime characteristic, in which they stated along the way that the tensor product of
these dual Weyl modules maps onto the one specified by the sum of highest weights.  (Their proof may not be rigorous.  In any case, Kempf's theorem comes into play here.)  A focused reference is the paper in J. Algebra 27 (1982) by Jian-pan Wang, "Sheaf cohomology on $G/B$ and tensor product of Weyl modules".   That paper followed up a suggestion of mine that such a tensor product should have a filtration with appropriate Weyl modules as subquotients.   The paper by Olivier  Matthieu in Duke Math. J. 59 (1989) used Frobenius splitting techniques to prove this in full generality after the partial results by Wang and then by Steve Donkin in Springer Lecture Notes 1140 (1985).  Eventually all of this gets folded into the general theory of "tilting modules" for reductive algebraic groups (Chapter G in Jantzen).
[ADDED] As Ekedahl just pointed out, a treatment is given in the more recent and more extensive book by Brion and Kumar along with history.

REPLY [5 votes]: Yes, it is true in general.
I found it as Thm 3.1.2 of Brion, Michel(F-GREN-IF); Kumar, Shrawan(1-NC)
Frobenius splitting methods in geometry and representation theory.
Progress in Mathematics, 231. Birkhäuser Boston, Inc., Boston, MA, 2005. x+250 pp. ISBN: 0-8176-4191-2. The result itself is earlier (see historical remarks at the end of Chapter 3).<|endoftext|>
TITLE: Salem Inequality
QUESTION [8 upvotes]: I have come across this inequality in the paper "Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type" http://www.math.msu.edu/~fedja/Published/paper.ps by Nazarov and he calls it by the name of Salem Inequality (which according to him is well known, but I can't find a reference).
If I have understood it correctly, the Inequality says that if $p$ is an exponential polynomial whose exponents are well separated, then the average value of square of the modulus of $p$ over a sufficiently large interval dominates the sum of the square of the modulus of its coefficients.
Let $p(t) = \Sigma_{k=1}^n c_k e^{ i \lambda_k t}$, where $ \lambda_1<\lambda_2\dots<\lambda_n \in \mathbb R$ and $\lambda_k$'s satisfies a separation condition i.e., $\lambda_{k+1}-\lambda_k \geq \Delta >0$. Let $I$ be an interval of length bigger than $4\pi / \Delta$, then  $$\sum_{k=1}^{n} |c_k|^2 \leq \frac{4}{|I|} \int_I |p(t)|^2 dt. $$
How can one prove this Inequality? This surely would have a lot of applications (and as he says must be well known – maybe under a different name?). I would appreciate some references to such inequalities in general.  Also I find it curious that the length of the interval does not seem to depend on $n$ and depends only on $\Delta$.

REPLY [8 votes]: Following a cue from Wadim, this inequality is Theorem 9.1 in Chapter 5 of Zygmund's Trigonometric series, vol 1. Note that although the book is mostly dealing with trigonometric series, the proof is given for general lacunary $\lambda_k.$ (Salem was a good friend of Zygmund's; see the preface to the book.)<|endoftext|>
TITLE: Analog of Chebyshev's inequality for higher moments
QUESTION [12 upvotes]: I have a positive random variable $X$ with $E[X] = 1$ and a small number $k$ more moments bounded by constants:
$$E[(X-1)^i] = O(1) \forall i \in \{2, ..., k\}.$$
I'd like to bound the average of $n$ independent samples of $X$.  Markov's inequality only uses the first moment to get:
$$\Pr[\frac{1}{n}\sum_{i=1}^n x_i \geq c] \leq 1/c$$
Chebyshev's inequality also uses the second moment to get:
$$\Pr[\frac{1}{n}\sum x_i \geq 1 + c] \leq \frac{O(1)}{nc^2}$$
which is better asymptotic behavior in $n$.  If infinite moments converged, I could use Azuma/Chernoff/Hoeffding/McDiarmid to get
$$\Pr[\frac{1}{n}\sum x_i \geq 1 + c] \leq e^{-f(c)n}$$
for some function $f(c)$.  But what if I have somewhere between 2 and infinite moments, say 10 moments.  Is there a theorem with better asymptotic behavior for intermediate moments?

REPLY [12 votes]: Tom is right: the proof of Chebyshev's inequality can be easily adapted to every nondecreasing nonnegative function. The proof of this generalization that I prefer has a principle worth remembering:


First find an inequality between random variables, then integrate it.


To apply the principle, let $g$ denote a nondecreasing nonnegative function defined on $[0,+\infty)$ and $Z$ a nonnegative random variable. Let $z\ge0$ such that $g(z)>0$. Then,
$$
g(z)\mathbf{1}_A\le g(Z)\ \mbox{with}\ A=[Z\ge z].
$$
Proof: if $\omega\notin A$, the assertion reduces to $0\le g(Z(\omega))$, which holds because $g$ is nonnegative everywhere; if $\omega\in A$, the assertion reduces to $g(z)\le g(Z(\omega))$, which holds because $Z(\omega)\ge z$ and $g$ is nondecreasing. 
Integrating the inequality yields
$$
g(z)P(A)\le E(g(Z)),
$$
and, dividing both sides by $g(z)$, we are done. 
The usual case is when $Z=|X-E(X)|$ and $g(z)=z$. The case mentioned by Thomas is when $Z=|X-E(X)|$ and $g(z)=z^p$, for every positive $p$ (and not only for $p\ge2$). Another case, mentioned by Tom and at the basis of the whole field of large deviations principles, is when $Z=\mathrm{e}^{rX}$ for a nonnegative $r$ and $g(z)=z$. But to deal with the $u\log(u)$ case mentioned by Tom requires to be more careful because the function $u\mapsto u\log(u)$ is non monotonous on $[0,1]$ and not of a constant sign on $[0,+\infty)$ (but everything works fine for $g(z)=z\log(z)$ on $[1,+\infty)$, that is, if $Z\ge1$ almost surely).<|endoftext|>
TITLE: Vertex connectivity of random graphs?
QUESTION [8 upvotes]: Consider simple, undirected Erdős–Rényi graphs $G(n,p)$, where $n$ is the number of vertices and $p$ is the probability for each pair of vertices to form an edge. Many properties of these graphs are known - in particular, $G(n,p)$ is almost surely connected when $p \gt (1 + \epsilon)\frac{\log(n)}{n}$, and the largest clique in $G(n, \frac{1}{2})$ is almost surely about $2\log_2(n)$.
What is known about the vertex connectivity number $\kappa(G)$, $G\in G(n,p)$, the minimum number of vertices that one must remove in order to disconnect the graph?
It is known that for fixed $k$ and fixed $p\in (0,1)$, almost every graph in $G(n,p)$ is $k$-connected, but what is the expected connectivity as a function of $p$ and $n$?

REPLY [5 votes]: The expected connectivity cannot be higher than the expected minimal degree, which jumps to roughly $pn$ after getting into the range $p\gg\frac{\log n}{n}$. On the other hand, sloppily counting potential clusters of size $m < n/2$ that have boundaries of less than $k$ vertices gives a probability of $\binom{n}{m}\binom{n-m}{k}(1-p)^{m(n-m-k)}$, which is for $k \ll n$ decreasing in $m$ up to $m\approx \frac{n-k}{2}$ and increasing after that value, so we can get an estimate by considering only $m=1$ (checking for vertices with at most $k$ neighbours) and $m=\frac{n}{2}$:
$$
\binom{n}{n/2}\binom{n/2}{k}(1-p)^{n(n-2k)/4} < \exp(n \log 2+k \log n - pn(n-2k)/4) <
$$
$$
< \exp(n \log 2 - pn(\frac{n}{4}-\frac{k}{2}-\log n)) < \exp(-\frac{n \log n}{4} + n \log 2 +2(\log n)^2),
$$
this latter number tending to $0$ fast enough to ignore it. So, the expected connectivity is the expected minimal degree and is roughly $pn$ once $p$ exceeds $\log n/n$. Do you need the behaviour of expected connectivity specifically in this region?<|endoftext|>
TITLE: Solenoid of a continuous map of a ball, is it contractible?
QUESTION [5 upvotes]: Let $B$ be the closed unit ball in $\mathbb R^n$ and $f\colon B\to B$ a continuous map.
Consider the infinite product $B^{\mathbb Z}$ equipped with the product topology. Consider the solenoid
$$
S_f=\{\{x_n; n\in\mathbb Z\}: x_{n+1}=f(x_n)\}
$$
equipped with the induced topology.
Question: Is $S_f$ contractible? If yes, is there a deformation retraction?
Motivation is here: Two commuting mappings in the disk

REPLY [9 votes]: No, it is not even path-connected in general, already for $n=1$.
Consider the folding map $f:[0,1]\to[0,1]$, namely $f(t)=2t$ for $t\le 1/2$ and $f(t)=2(1-t)$ for $t\ge 1/2$. There is no path connecting the orbits of the two fixed points: 0 and 2/3.
Indeed, suppose there is a continuous path $t\mapsto \{x_n(t):n\in\mathbb Z\}$ in $S_f$ such that $x_n(0)=0$ and $x_n(1)=2/3$ for all $n\in\mathbb Z$. Consider $x_0(t)$ and $x_{-n}(t)$ where $n>0$. While $x_{-n}(t)$ travels from 0 to 1/2, $x_0(t)$ will take the value 0 at $t_0=0$, then the value 1 at some $t_1>t_0$, then 0 at some $t_2>t_1$, then 1 again, and so on, $2^n$ times. Since $t\mapsto x_0(t)$ is a continuous function, it can have only finite number of alternating 0 and 1 values, but $n$ can be arbitrarily large, a contradiction.<|endoftext|>
TITLE: A nontrivial surface on which any two curves intersect
QUESTION [13 upvotes]: One interesting property of the projective plane is that any two plane curves intersect.  (More generally, if $V$ and $W$ are subvarieties of any projective space, and codim $V$ + codim $W \geq 0$, then $V$ and $W$ intersect.) However, the same does not seem to hold for most other easy examples of surfaces.  For instance, any ruled surface $S \to C$ has non-intersecting curves (take the fibers over any two distinct points of $C$).  Furthermore, any surface $S$ obtained from $\mathbb{P}^2_k$ by blowing up points $p_1, \ldots, p_n$ has two non-intersecting curves: take two lines that intersect transversely at $p_1$ and avoid $p_2, \ldots, p_n$, and lift them to curves on $S$.
Thus, my question:

Is there any nonsingular algebraic surface other than the projective plane such that any two curves on the surface have nontrivial intersection?

(Note: assume the base field is algebraically closed.)

REPLY [11 votes]: Here is another way to construct such examples. This way one can get examples with Picard number at least up to $4$.
Notation: Let $S$ be a smooth projective surface, $\overline{NE}(S)$ the closed cone of effective curves, $Q_{\text{tot}}(S)$ be the set of numerical classes with positive self-intersection, and $Q(S)$ the closure of the connected component of $Q_{\text{tot}}(S)$ containing an ample class.
Fact (a simple consequence of Riemann-Roch): $Q(S)\subseteq \overline{NE}(S)$.

Claim Let $S$ be a smooth projective surface such that every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$. Then for any two proper effective curves $C_1,C_2\subset S$ we have $C_1\cdot C_2>0$ and in particular $C_1\cap C_2\neq\emptyset$.

Remark: If the Picard number is not $1$, then the condition that "every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$" is equivalent to assuming that
$Q(S)=\overline{NE}(S)$ and that the boundary of $Q(S)$ does not contain any effective classes.

Proof
By the nature of the statement we may assume that the Picard number of $S$ is at least $2$ and that $C_1$ and $C_2$ are irreducible.
By the assumption $C_1^2>0$ and if it is irreducible, then this implies that it is nef. Hence the linear functional induced by $C_1\cdot(\quad)$ is non-negative on $Q(S)$, but since the boundary of $Q(S)$  does not contain any effective classes, it follows that $C_1\cdot(\quad)$ is actually positive on every effective class and hence the statement follows.

To see that there exists surfaces satisfying the condition in the Claim simply consider K3 surfaces that do not contain smooth rational or elliptic curves. These exist with Picard number 1-4. I suspect that one can find examples satisfying the condition in the Claim with higher Picard numbers, too.<|endoftext|>
TITLE: diameter of a graph with random edge weights
QUESTION [6 upvotes]: Given a weighted directed graph $G=(V,E, w)$, suppose we generate a new graph $G'=(V,E,w')$ with the same vertices and edges, but now letting the weight of edge $(i,j)$ be
an exponential random variable with mean $w_{ij}$. My question is: what is the
expected diameter of $G'$?
Why I'm interested in this: I was intrigued by the observation that the expected diameter of $G'$ can be
quite different from the diameter of $G$. Indeed, consider the following example: define $G$ by taking the complete graph $K_{n+1}$, picking an arbitrary vertex $a$, and assigning weight $n$  to any edge incident on $a$, and weight $1$ to 
every other edge. Then, the diameter of $G$ is $n$. On the other hand, the expected diameter
of $G'$ is O(1) since we can expect one of the edges incident on $a$ to have small weight.

REPLY [4 votes]: For the special case of the complete graph $K_n$ which you mention in your post, Svante Janson answered your question in this paper; the answer is that the weighted diameter grows like $3 \log n$ in probability. 
There is also some very nice work by Bhamidi et. al on this question when the underlying graph is the giant component of Erdos--Renyi random graph $G_{n,c/n}$ with $c>1$ fixed, although they only prove lower bounds. Amini et. al. (link is to a PDF) have found the asymptotics of the weighted diameter for random graphs with a given degree sequence, under some conditions, for degree sequences which in particular result in graphs which are with high probability connected. Ding et.al. (Theorems 3.7 and 3.8 of the linked paper) prove quite refined estimates, and tail bounds, for the weighted diameter of random $d$-regular graphs, for $d \geq 3$. (Since random regular graphs are a special case of random graphs with a given degree sequence, the results of Amini et. al. and Ding et.al. have some overlap). 
There is also related work, on the hopcount of randomly weighted graphs. The hopcount is what you get if you count the number of edges on the smallest-weight path. The primary interest of Bhamidi et. al. in fact seems to be hopcounts rather than weighted path lengths.<|endoftext|>
TITLE: How should the Math Subject Classification (MSC) be revised or improved?
QUESTION [18 upvotes]: Most of us are familiar with the Math Subject
Classification
(MSC),
a coded index attempting to classify all mathematical
research areas by topic. The MSC, devloped jointly by the Math Reviews and Zentralblatt, is used by most journals and many grant institutions,
such as the US National Science Foundation, as a way of
grouping mathematical work into topic categories. The MSC
codes were recently updated from the year 2000 codes to the
current 2010 Mathematics Subject
Classification. These codes are organized hierarchically, first
dividing into broad research areas, then into sections and
finally into more specific research categories.
Question. How well do these codes describe the natural divisions of
research in mathematics? Could they be improved in some
way? How should they be revised?
Most of us, when submitting a research article for
publication, have to decide on the most appropriate codes
for that particular work. My own experience is that usually
there there is a natural code or two codes that fit very
well, which aptly describe the research topic of the
article. Sometimes I use two or more codes in a situation
where the work doesn't really fit well into either of them
alone, so that it isn't really a primary/secondary
classification for me, but rather a classification into the
union of two categories. Increasingly, however, I find
myself stymied by the classification scheme, frustrated in
my newest projects that perhaps four or five subcategories
are involved, with none of them truly apt, except for the
unhelpful "None of the above, but in this section"
category. In such cases, I feel that the MSC has failed
me.
I recognize that this may simply mean that I sometimes
favor offbeat topics, and so perhaps this is my problem
rather than the MSC's problem. Or perhaps my problem is
that I would like my research to be categorized by the
bottom level of the hierarchy, but I should be content just
with using the middle level of the hierarchy.
At the same time, I recognize that the mathematical
community has a specific interest in encouraging research
that crosses the boundaries between established areas,
perhaps cross-pollinating or unifying them or at least
transferring methods and techniques from one area to
another. In time, therefore, we expect subject
classification boundaries to migrate or split in various
ways. Indeed, perhaps some of the most valuable
mathematical work tends to destroy the old classification
scheme for precisely this kind of reason. Presumably, this
is part of the reason why the MSC is somewhat regularly
updated (every ten years I think).  So I suspect that there may be many people who share my frustration. 
How would you revise the MSC?
Let's have standard community-wiki rules; please provide
just one group of changes per post.

REPLY [2 votes]: First, we should ask what purpose MSC type codes serve.  After all modern papers can be indexed in full text and, even if paywalls make full text search impractical one can do a full text search on abstracts at mathscinet.  It seems to me they offer two important features.

MSC codes let one track all papers in a given are even when no particular search terms would be sufficiently specific without leaving anything out.   This allows paper archives to present their contents in a hierarchical manner or searches to be restricted to a particular subject.
MSC codes (at least in theory, I've never used them for this purpose) distinguish papers working with particular mathematical objects/approaches that can't easily be identified using keyword searches.   For instance even if I restrict my attention to papers in computability a search for "admissible set" is likely to turn up to much (admissible, like good, is overused) and miss some instances that use other terminology.

The first usage calls for a basic hierarchical description of major areas of mathematical research.  Each area should be well populated and each paper should fall into only one or at most a handful of areas.  These areas should be developed enough that they won't disappear with changes in research focus.  
The second usage calls for a plentiful list of canonical tags for particular objects, approaches or questions.  Each paper might fall under arbitrarily many such tags, the more the better.  As far as this use is concerned there is no real harm if research directions change and papers falling under some tag stop being published.  Codes should be generously added for every conceivable object/approach/question constrained only by the requirement that every such concept have a canonical code (no codes that are synonyms).  Search engines could then maintain a list of terms associated with each such code so one could do a search in computability theory for papers that mention both "model" (in the model theory sense) and admissible ordinal.
I submit that this naturally suggests two systems of codes.  One hierarchical that roughly corresponds to the first two digits + letter, e.g., 03A, of MSC2010 and the other a much more numerous database of tags and their synonyms.<|endoftext|>
TITLE: Who is the Youngster in the Automorphic Room?
QUESTION [6 upvotes]: Iwaniec and Friedlander wrote a short survey article for the notices of the AMS, entitled "What is the Parity Phenomenon?"
http://www.ams.org/notices/200907/rtx090700817p.pdf
At the end of the article they refer to a young mathematician:
"Sometimes it almost seems as though there is a
ghost in the House of Prime Numbers. Perhaps that
will be ruled out some day. There are suggestions
of a youngster who might do this, one who will
come from the Automorphic Room of the house."
Does anyone know who this mathematician is?
Thanks!

REPLY [13 votes]: They wrote a mixed technical summary and allegory, something of a prose poem. The allegorical part is concentrated in three paragraphs. These are the second paragraph, the last paragraph, and one in the middle in which people in the "Analytic Room" regard their methods as recent in that Euler is only about three hundred years old. The youngster is evidently anyone from the "Automorphic Room." So the suggestion is that progress is likely to come from automorphic methods, and no specific mathematician is indicated.  My best guess on the "ghost" is the possibility of an "exceptional" character whose $L$-function could possess a bad zero. This is from the next to last paragraph, after mention of the "Algebraic Room." 
Overnight my impression of the piece clarified a bit, from a pretty vague sense that the inhabitants in the Rooms were not specific people, even from the distant past. I am close to Unreasonable Sin's comment of just an hour ago. I think the inhabitants of the Rooms are ideas more than specific theorems, techniques, particular papers. So the Youngster is an idea. Note Victor Protsak's  comment of ten hours ago, I think he got it exactly right. Victor is a smart man. Heed him.
For examples of this style, see the fantasy short story collections of Lord Dunsany,
http://en.wikipedia.org/wiki/Edward_Plunkett,_18th_Baron_of_Dunsany
especially, from 1905, The Gods of Pegana
http://en.wikipedia.org/wiki/The_Gods_of_Peg%C4%81na 
evidently available online in full, see for instance
http://en.wikisource.org/wiki/The_Gods_of_Peg%C4%81na#The_Chaunt_of_the_Priests<|endoftext|>
TITLE: Howe duality for exceptional algebras
QUESTION [5 upvotes]: There is a nice tool in representation theory, the Howe duality, which as I know works for certain pairs of classical Lie algebras (the reference to the complete list of Howe dual pairs is appreciated very much). Is there any extension of the Howe duality for exceptional algebras?

REPLY [4 votes]: Yes, and it was studied both from the point of view of the local theory (correspondence of the infinitesimal characters) and automorphic forms. The following review of Jian-Shu Li's paper contains a nice selection of references:
MR1682229 (2000b:22014) Li, Jian-Shu, The correspondences of infinitesimal characters for reductive dual pairs in simple Lie groups. Duke Math. J. 97 (1999), no. 2, 347--377.
More recent work was done by Gordan Savin and Wee Teck Gan.<|endoftext|>
TITLE: Factoring operators $L_\infty \longrightarrow L_2$ as the composition of $n$ strictly singular operators, $n\in \mathbb{N}$
QUESTION [7 upvotes]: Motivation and background This question is motivated by the problem of classifying the (two-sided) closed ideals of the Banach algebra $\mathcal{B}(L_\infty)$ of all (bounded, linear) operators on $L_\infty$ $(=L_\infty[0,1])$. As far as I am aware, the only known nontrivial ideals in $\mathcal{B}(L_\infty)$ are the ideals $\mathcal{K}(L_\infty)$ of compact operators and $\mathcal{W}(L_\infty)$ of weakly compact operators. Most of the other well-known closed operator ideals not containing the identity operator of $L_\infty$ seem to coincide with one of these two operator ideals on $L_\infty$. Let me mention explicitly the following further relevant pieces of background information:

Any nontrivial closed ideal $\mathcal{J}$ of $\mathcal{B}(L_\infty)$ must satisfy $\mathcal{K}(L_\infty)\subseteq \mathcal{J} \subseteq \mathcal{W}(L_\infty)$.
We have $\mathcal{S}(L_\infty) = \mathcal{W}(L_\infty)$, where $\mathcal{S}$ denotes the (closed) operator ideal of strictly singular operators. Moreover, the ideal of operators $L_\infty \longrightarrow L_\infty$ that factor through $L_2$ $(=L_2[0,1])$ is a norm dense subset of $\mathcal{W}(L_\infty)$. With regards to this latter fact, we have that if $\mathcal{J}$ is any closed ideal of $\mathcal{B}(L_\infty)$ satisfying $\mathcal{K}(L_\infty)\subsetneq \mathcal{J} \subsetneq \mathcal{W}(L_\infty)$, then there exists an element of $\mathcal{W}(L_\infty)\setminus \mathcal{J}$ that factors through $L_2$.

One possible avenue towards discovering more closed ideals in $\mathcal{B}(L_\infty)$ is to consider products of closed operator ideals, and my question below concerns but one approach along these lines. We recall now that for operator ideals $\mathcal{A}$ and $\mathcal{B}$, their product $\mathcal{B}\circ\mathcal{A}$ is the class of all operators of the form $BA$, where $A\in\mathcal{A}$, $B\in\mathcal{B}$ and the codomain of $A$ coincides with the domain of $B$ (so that the composition is defined). It is well-known that $\mathcal{B}\circ\mathcal{A}$ is an operator ideal and that $\mathcal{B}\circ\mathcal{A}$ is a closed operator ideal whenever $\mathcal{A}$ and $\mathcal{B}$ are, the latter fact being due to Stefan Heinrich. For $n$ a natural number, one may define the power of an operator ideal $\mathcal{A}^n$ as the product $\mathcal{A}\circ \ldots \circ \mathcal{A}$ with $n$ factors, and $\mathcal{A}^n$ is closed for all $n$ whenever $\mathcal{A}$ is closed. Moreover, we may define $\mathcal{A}^\infty:= \bigcap_{n\in\mathbb{N}}\mathcal{A}^n$, with $\mathcal{A}^\infty$ being a closed operator ideal whenever $\mathcal{A}$ is.
It is well-known that every operator from $L_\infty$ to $L_2$ $(=L_2[0,1])$ is strictly singular. However, the answer to the following question is not so clear to me:
Question: Is $\mathcal{B} (L_\infty, L_2) = \mathcal{S}^\infty (L_\infty, L_2)$? If no, what is the least $n$ for which $\mathcal{B} (L_\infty, L_2) \neq \mathcal{S}^n (L_\infty, L_2)$?
Further comments: A negative answer to my question above will yield through easy arguments that $\mathcal{K}(L_\infty) \subsetneq \mathcal{S}^\infty (L_\infty) \subsetneq \mathcal{W}(L_\infty)$ (note that $\mathcal{K}(L_\infty) \subsetneq \mathcal{S}^\infty (L_\infty)$ in any case since there are noncompact operators $L_\infty \longrightarrow L_\infty$ that have the formal inclusion operator $\ell_2 \hookrightarrow \ell_\infty$ as a factor, and this inclusion operator belongs to $\mathcal{S}^\infty$).
It seems to me that the best chance of obtaining an element of $\mathcal{B} (L_\infty, L_2) \setminus \mathcal{S}^\infty (L_\infty, L_2)$ would be to consider a surjective element of $\mathcal{B} (L_\infty, L_2)$, for instance the adjoint of an isomorphic embedding of $L_2$ into $L_1$. A further possibility along these lines would be to factor such an embedding as the product of $n$ operators whose adjoints are strictly singular, for each $n\in \mathbb{N}$. However, it seems to me that such an approach will require a far deeper knowledge of the subspace structure of $L_1$ than I have at the present time. Interpolation methods may work too, but I am also too ignorant of that theory to know whether it's a genuinely feasible approach.

REPLY [5 votes]: I think this is an outline of a proof, Phil. It is enough to factor $I_{\infty,2}$  as the product of strictly singular operators, where $I_{\infty,2}$ is the identity from $L_\infty(\mu)$  to $L_2(\mu)$  with $\mu$ a probability (since every operator from $L_\infty$ to $L_2$ is $2$-summing). I guess it can be assumed that $\mu$ is a separable measure (if not, I think using Maharam's theorem works) and certainly it is enough to look at $\mu$ purely non atomic. But then $I_{\infty,2}$ can be regarded as the identity from $L_\infty([0,1])$ to $L_2([0,1])$ since every separable non atomic probability space is measure isometric to $[0,1]$ with Lebesgue measure. Now $I_{\infty,2}$ factors through $L_\Phi([0,1])$ for every fast growing  Orlicz function $\Phi$. Consider $\Phi_p(t)=e^{t^p} -1$ for $p>$. Rodin and Semenov showed that the identity from $L_{\Phi_p}$ to $L_2$ 
is strictly singular and the injection from $L_\infty$ to $L_\Phi$ is strictly singular
always.  This takes care of the case $n=2$.  
Schechtman suggests that to do general $n$ to show that the injection from $L_{\Phi_p}$ to $L_{\Phi_r}$ is strictly singular for 
$2<r<p$.  At least this map is not an isomorphism on the span of the Rademachers.<|endoftext|>
TITLE: Compact simple simply connected algebraic groups over $Q_p$ or other local non-archimedean fields
QUESTION [9 upvotes]: My motivation is to understand the following situation: Given absolutely and almost simple algebraic group $G$ defined over a number field $k$ and a finite valuation $v$ on $k$, when $G(k_v)$ can be compact (with respect to the $p$-adic topology)?
I more or less understand that if $G=SL_1(D)$ where $D$ is a division ring of dimension $n^2$ and of order $n$ in the Brauer group over $Q_p$ then $G(Q_p)$ is compact. I also understand that $Spin(q)$ when $q$ has more than $5$ variable cannot be compact over local non-archimedean fields.
Are there more examples? I think that one can classify all the examples but I don't manage to do it or find a reference for it.
Can someone outline a route to take in order to understand it thoroughly (for someone with basic understanding of algebraic groups and Galois cohomology)? 
Thanks a lot!

REPLY [13 votes]: Here are some remarks on the answers of Charles Matthews, Kevin Buzzard, and Victor Protsak. For justification of Kevin Buzzard's claim, see G. Prasad, "An elementary proof of a theorem of Bruhat-Tits-Rousseau and of a theorem of Tits" from Bull. Soc. Math. France 110 (1982), pp. 197--202, for an incredibly elegant and short proof that over any henselian valued field $F$, a connected reductive $F$-group $G$ is $F$-anisotropic if and only if $G(F)$ is "bounded" (a property defined in terms of a choice of closed immersion of $G$ into an affine space over $F$, the choice of which doesn't matter; this is meaningful for any affine $F$-scheme of finite type and equivalent to compactness when $F$ is locally compact). Platanov-Rapinchuk has a universal assumption that all fields of characteristic 0 (except when they're finite), so unfortunately that reference is insufficient for uniform arguments over all non-archimedean local fields. I suppose (near?-)circularity (suggested by Victor Protsak) is a more serious issue. :)
There remains the matter of determining, for locally compact non-archimedean $F$ and connected reductive $F$-groups, precisely when the $F$-anisotropic case can actually occur. As Victor Protsak mentioned, via Bruhat-Tits theory one sees that for connected semisimple $F$-groups which are absolutely simple and simply connected over a non-archimedean local field $F$ (i.e., $G$ a simply connected $F$-form of a Chevalley group), such forms never exist away from type A, and in type A the $F$-anisotropic examples are precisely the $F$-groups of norm-1 units of central simple algebras over $F$.  (Note the contrast with the case $F = \mathbb{R}$, for which there's always a "compact form" of any Chevalley type.) 
Let me now briefly explain why this handles the general connected reductive case, by a standard kind of argument with central isogenies and separable Weil restriction. (This is explained also in the article [2] of Tits referenced in Charles Matthews' answer.) If $f:G' \rightarrow G$ is a (possibly inseparable) central $F$-isogeny between connected reductive $F$-groups then the preimage of an $F$-torus of $G$ is an $F$-torus of $G'$ (since maximal tori in $G'$ are their own functorial centralizers, so $\ker f$ is of multiplicative type, nothing funny happens when $f$ is not separable).  Since $F$-anisotropcity of an $F$-torus is invariant under $F$-isogenies (as we see using the $F$-rational character group, or more direct arguments), it follows that $G$ is $F$-anisotropic if and only if $G'$ is. (This argument has the advantage of working over any field $F$, in contrast with a direct attack on the topology of rational points by using finiteness theorems for Galois cohomology of connected reductive groups.)  Thus, by considering an arbitrary connected reductive $F$-group $G$ and letting $G'$ denote the product of its maximal central $F$-torus and the simply connected central cover of the derived group $D(G)$, we see that the problem comes down to the simply connected case.  
But in the simply connected semisimple case, the general structure of connected semisimple groups over fields (in terms of central isogenous quotient of direct product of commuting simple "factors") implies that $G$ is uniquely a direct product of commuting $F$-simple connected semisimple $F$-groups, each of which is simply connected, so we may assume $G$ is $F$-simple.  Then by an elementary result of Borel and Tits (6.21 in "Groupes reductif", IHES), $G = {\rm{Res}}_ {F'/F}(G')$ for a finite separable extension $F'/F$ and a connected semisimple $F'$-group $G'$ that is absolutely simple and simply connected.  By the good behavior of Weil restriction with respect to the formation of the topological group of rational points, it follows that the equality ${\rm{Res}}_ {F'/F}(G')(F) = G'(F')$ of abstract groups is a homeomorphism, so we can replace $(G,F)$ with $(G',F')$ to reduce to the case when $G$ is also absolutely simple, the case addressed by Victor Protsak above.
(A more algebraic argument with Galois descent relating maximal $F'$-tori in $G'$ and maximal $F$-tori in its Weil restriction to $F$ shows the equivalence of anisotropicity for $G'$ and its Weil restriction through the finite separable $F'/F$, where $F$ can be taken to be any field at all.) 
Conclusion: for non-archimedean local $F$, the $F$-anisotropic connected reductive $F$-groups are precisely the central quotients of products of an $F$-anisotropic torus and groups of norm-1 units of central division algebras over finite separable extensions of $F$.<|endoftext|>
TITLE: units in distinct division algebras over number fields---are they definitely not isomorphic as abstract groups?
QUESTION [18 upvotes]: This is really an irrelevant question in the sense that the answer isn't remotely "logically crucial for the Langlands programme" or whatever---it's just something that occurred to me when writing some lectures on the subject.
Let $F$ be a number field and let $D$ and $D'$ be division algebras over $F$ with centre $F$. Let me put myself in the following simplified situation: assume $D$ and $D'$ both have dimension $p^2$ over $F$ (so are forms of $M_p(F)$) with $p$ an odd prime, and let me also assume that $D$ and $D'$ are both ramified at precisely the same finite places of $F$ (and at no infinite places of $F$) but that $D$ and $D'$ are not isomorphic.
The question: Are $D^\times$ and $(D')^\times$ sometimes non-isomorphic as groups?
Let me put this slightly strange question into context. Experts will already know what is coming: $D^\times$ and $(D')^\times$ are group-theoretically similar in certain ways; this is what I'm about to explain. This group-theoretic similarity would be trivially explained if the groups were isomorphic! But I think the whole point should be that the groups aren't isomorphic, although I don't know an instance where I can prove this.
OK so here are the details. Given $D$ and $D'$ as above, there is a link between automorphic forms on $D^\times$ and $(D')^\times$. When proving this link using the trace formula, one has to check that two complicated formulae for traces, one associated to $D$ and one to $D'$, coincide. The formulae are of the form "sum over conjugacy classes of certain orbital integrals". The strategy I understand to check these sums coincide is first to use general theory of central simple algebras to explicitly list the conjugacy classes in the groups $D^\times$ and $(D')^\times$, and then to write down an explicit natural bijection between them, and then to check that things add up in the sense that each term on the $D$ side is equal to the corresponding term on the $D'$ side, and then to deduce that certain traces are equal, and then you follow your nose to the answer (modulo a technicality that has to be dealt with using a Plancherel measure argument but which isn't relevant here).
Crucial in this strategy is the identification of the conjugacy classes in $D^\times$ with the conjugacy classes in $(D')^\times$. The way this works in this situation is that a conjugacy class in $D^\times$ is either an element of $F^\times$ (the central classes) or contains some $e\not\in F$; in this case $E:=F(e)$ is a field extension of $F$ of degree $p$ which splits $D$, and the fields that split $D$ of this form are precisely the degree $p$ extensions of $F$ which have one prime above $v$ for all $v\in S$ our bad set. The miracle that has occurred here is that this argument (when fleshed out) shows that the conjugacy classes in $D^\times$ are parametrised by a set that depends only on $F$, $p$ and $S$, and in particular this set is the same for $D$ and $D'$, which have the same degree and which ramify at the same primes.
I will present this argument in detail in class today, and my instinct is to stress that a miracle has occurred, because $D^\times$ and $(D')^\times$ are non-isomorphic groups whose conjugacy classes are completely naturally in bijection with one another. Of course if $D$ and $D'$ are isomorphic then then $D^\times$ and $(D')^\times$ will be isomorphic. Moreover, if $D'=D^{opp}$ then again $D^\times$ and $(D')^\times$ will be isomorphic as groups. In both cases it is hardly surprising that I have managed to canonically biject the conjugacy classes of $D^\times$ and $(D')^\times$! But in general I am still convinced that a miracle has occurred. However to really convince the audience of this I want to assert confidently that the groups $D^\times$ and $(D')^\times$ really are not always isomorphic. Is this definitely true??
{\bf EDIT}: Here's an even stronger question, which is even less likely to be true, and given that I'm hoping/guessing that such things aren't true, it seems worth formulating (the stronger it is, the easier it should be to falsify).
OK so same set-up: $D$ and $D'$ division algebras of dimension $p^2$, $p$ prime, and ramified at the same set of finite primes $S$, but $D$ and $D'$ not isomorphic. Let $G$ be the form of $PGL_p$ associated to $D^\times/Z(D^\times)$ and let $G'$ be the form corresponding to $D'$. If $\mathbf{A}^S$ denotes the adeles of $F$ away from $S$ then choosing isomorphisms $D_v=D'_v$ for all $v\not\in S$, sending $O_v$ to $O'_v$ for two chosen maximal orders $O$ and $O'$ in $D$ and $D'$ when $v$ is finite, gives us an induced isomorphism $G(\mathbf{A}^S)=G'(\mathbf{A}^S)$. Call this group $X$. Now $\Gamma:=D^\times/F^\times$ and $\Gamma':=(D')^\times/F^\times$ are two discrete subgroups of $X$ and using the trace formula one can prove that not only do the conjugacy classes of $D^\times$ and $(D')^\times$ match up, but (if my understanding is correct) the covolumes of $\Gamma$ and $\Gamma'$ are the same. This would be explained by the highly unlikely statement that $\Gamma$ and $\Gamma'$ were actually conjugate within $X$! This is perhaps even stronger than just being abstractly isomorphic, in this setting? One might also ask to give an explicit example where this cannot possibly be the case?
I mention this stronger statement because people might find it easier to use high-powered methods to refute it.

REPLY [6 votes]: I know this is a very old question but I saw it just a few days ago. Super-rigidity can indeed be applied not for "higher rank lattices"  but for the global points $D^*/F^*$ (which form a lattice in the adelic group $ PGL_1(D\otimes_{\mathbb{Q}} {\mathbb A})$ (the latter is certainly a "higher rank group". 
First of all, as was observed in the comments (by Emerton?), the groups $D^*$ and $(D^{opp})^*$ are isomorphic under the map $x\mapsto x^{-1}$. The result is then as follows.
Proposition:  Suppose $\theta: D^* \rightarrow D'^*$ is an isomorphism of groups. Then one of the following two alternatives hold.
[1] There is an isomorphism $\sigma: D \rightarrow  D'$ of central division algebras over $F$, and a homomorphism $\phi: D^*\rightarrow F^* \subset D'^*$ such that the isomorphism $\theta $ is given by 
$$\theta (x)=\sigma (x)\phi (x) \quad \forall  \quad x\in D^*.$$  
[2] There is an isomorphism $\tau : D^{opp} \rightarrow D'$ of division algebras and a homomorphism $\psi :(D^{opp})^*\rightarrow F^*$ such that 
$$\theta (x)= \tau (x^{-1})\psi (x) \quad \forall x\in D^*.$$ Since $x\mapsto x^{-1}$ maps $D^* $ into $(D^{opp})^*$, the rhs makes sense. 
So $D^*$ and $D'^*$ are isomorphic if and only if $D'$ is isomorphic as a division algebra either to $D$ or to $D^{opp}$. 
The proposition follows from superrigidity (see (C) on page 259 of Chapter VIII (title of the chapter: Normal Subgroups and "Abstract" homomorphisms of semi-simple algebraic groups over global fields) of Margulis 's book "discrete subgroups of semi-simple Lie groups", where he states the superrigidity for an arbitrary set $S$ (not necessarily finite) of finite places of $F$ (If you take $S$ to be all finite places, then you get global points).
Let me state the (consequence of) superrigidity that I am using. If $\theta: G(F)\rightarrow G'(F)$ is an isomorphism of groups over $F$, with $G,G'$ absolutely simple and adjoint, then there exists an automorphism $a$ of $F$ and a morphism of algebraic groups $\alpha :^a (G)\rightarrow G'$ such that $\theta =\alpha \circ a$. This is deduced in Margulis' book from the "usual" superrigidity.  
The isomorphism from $D^*$ to $(D')^*$ gives an isomorphism from $PGL_1(D)$ (a lattice in a semi-simple adelic group) to $PGL_1(D')$. Then superrigidity applies for this case. To go from the theorem stated in Margulis' book to the proposition  needs some work; the essential point is that the only algebraic automorphisms of $PGL_p({\mathbb C})=GL_1(D\otimes {\mathbb C})$ are inner conjugations (then we get into the case of [1] of the proposition), or else are inner conjugate to the map $x\mapsto (^tx)^{-1}$, and then we land in case [2].<|endoftext|>
TITLE: Positivity of "harmonic" summation
QUESTION [6 upvotes]: The settings for the problem are as follows. Given
a real number $\alpha\in[0,1]$, consider
a sequence of real (positive, negative and zero) numbers
$a_1,a_2,\dots,a_n,\dots$ satisfying
(1) $a_1=1$,
(2) $|a_n|\le n^\alpha$ for all $n=1,2,\dots$, and
(3) $\displaystyle\max_{1\le k\le n}\lbrace a_1+a_2+\dots+a_k\rbrace
+\min_{1\le k\le n}\lbrace a_1+a_2+\dots+a_k\rbrace\ge0$ again for all $n=1,2,\dots$.
Is it true that
$$
s_n=\sum_{k=1}^n\frac{a_k}k>0
$$
for any $n$?
The answer is no for $\alpha=1$, as the choice $a_k=(-1)^{k-1}k$ shows
that we can only achieve a nonstrict inequality in this case. So, what
are the conditions on $\alpha$ to ensure $s_n>0$ for any $n$?
I spent some time trying to construct a counterexample (for $\alpha=0$
and $\alpha=1/2$) but with no result. Let me note that one can
consider a finite sequence $a_1,a_2,\dots,a_n$ (but of arbitrary
length $n$, of course) which corresponds to the choice
$a_{n+1}=a_{n+2}=\dots=0$. A tedious analysis shows that $\alpha<1$
implies $s_n>0$ for $n=1,2,3,4$ but sheds no light on how to proceed
further.
Any ideas?!
EDIT. The problem has finally got a
solution in negative in the most interesting case $\alpha=0$. (This is
automatically a solution for any $\alpha\ge 0$.)

REPLY [3 votes]: You can get counter examples for various $\alpha$ of the following form.  Let
$$a_1=1, a_2=-2 + 2 \epsilon, a_3=0, a_4=a, a_5 = -b.$$
Here $0<\epsilon <1$ and we take $a$ is big enough so that $a + 2\epsilon -1 \ge 1$. Here is one such counterexample
$$a_1 = 1, a_2 = -\frac{7}{4}, a_3=0, a_4=3, a_5= -\frac{71}{16}.$$
The sequence of partial sums is $ 1, -\frac{3}{4},-\frac{3}{4}, \frac{9}{4}, -\frac{35}{16}$ which satisfies your min-max requirement.  The harmonic sum is
$$1 -\frac{7}{8} + \frac{3}{4} - \frac{71}{80} = \frac{-1}{80}.$$
More generally, the constraints on $a,b$, $\epsilon$ are
$$ 2(a + 2\epsilon -1) - b >0$$
(from the max-min restriction), and 
$$ \epsilon + \frac{a}{4}- \frac{b}{5} <0 $$
(so we get a counter example). In other words,
$$ 5 \epsilon + \frac{5a}{4} < b <2a + 4 \epsilon -1.$$
For $a> \frac{8}{3} +  \frac{4}{3} \epsilon$ these are consistent and leave us room to choose $b$. (And $a$ automatically satisfies our previous requirement that $a>2+2\epsilon$.)
We also need $a <4$ and $b<5$ to fit your requirement that $|a_j| \le j^\alpha$.  The requirement $a<4$ only asks that $\epsilon <1$, which we already assumed, however $b<5$ forces $\epsilon <1/4$.  Thus only for $\epsilon <1/4$ can you find $a$ and $b$ which make the harmonic sum at order $5$ negative and satisfy all your constraints.<|endoftext|>
TITLE: A canonical and categorical construction for geometric realization 
QUESTION [20 upvotes]: There is a very intimate connection between categories, simplicial sets, and topological spaces. On one hand, simplicial sets are the presheaf category on the category $\Delta$ and $\Delta$ is a canonically defined "invariant" of the theory of categories. (e.g. the machinery of Mark Weber "spits out" $\Delta$ when you "plug in" the free category monad: http://golem.ph.utexas.edu/category/2008/01/mark_weber_on_nerves_of_catego.html)
However, $\Delta$ is also linked with topological spaces. The key to this link is the functor $\Delta \to Top$ which assigns the category $[n]$ the standard n-simplex $\Delta^n$. It is this functor which produces the adjunction between the geometric realization functor and the singular nerve functor which allow you to transfer the model structure on $Top$ to $Set^{\Delta^{op}}$ so that this adjunction becomes a Quillen equivalence.
My question is the following:
Is there a deep categorical justification for the functor $\Delta \to Top$ being defined exactly how it is? If we didn't know about the standard n-simplices, how could we "cook up" such a functor? I would like a construction of this functor which is truly canonical.
The closest to an answer I've found is Drinfeld's paper http://arxiv.org/abs/math/0304064. However, this doesn't quite "nail it home" to me. First of all, the definition is just made, but not motivated. The definition shouldn't be a "guess that works", but something canonical. Moreover, if you unwind it enough, it is secretly using the fact that finite subsets of the interval with cardinality $n$ correspond to points in (the interior of) the $(n+1)$-simplex. Plus, there's some funny business going on for geometric realization of non-finite simplicial sets. (Don't get me wrong- I think it's a great paper. It just doesn't totally answer my question).
EDIT: A possible lead:
$Set^{\Delta^{op}}$ is the classifying topos for interval objects and the standard geometric realization functor $Set^{\Delta^{op}} \to Top$ is uniquely determined by its sending the generic interval to $[0,1]$. This reduces the question to "why is [0.1] the canonical interval?". Is it perhaps the unique interval object whose induced functor $Set^{\Delta^{op}} \to Top$ is both left-exact and conservative?
EDIT: I've proposed a partial answer to this below, along the lines of the above lead. I would love any feedback that anyone has on this.

REPLY [10 votes]: As to "why is the unit interval the canonical interval?", there is an interesting universal property of the unit interval given in some observations of Freyd posted at the categories list, characterizing $[0, 1]$ as a terminal coalgebra of a suitable endofunctor on the category of posets with distinct top and bottom elements. 
There are various ways of putting it, but for the purposes of this thread, I'll put it this way. Recall that the category of simplicial sets is the classifying topos for the (geometric) theory of intervals, where an interval is a totally ordered set (toset) with distinct top and bottom. (This really comes down to the observation that any interval in this sense is a filtered colimit of finite intervals -- the finitely presentable intervals -- which make up the category $\Delta^{op}$.) Now there is a join $X \vee Y$ on intervals $X$, $Y$ which identifies the top of $X$ with the bottom of $Y$, where the bottom of $X \vee Y$ is identified with the bottom of $X$ and the top of $X \vee Y$ with the top of $Y$. This gives a monoidal product $\vee$ on the category of intervals, hence we have an endofunctor $F(X) = X \vee X$. A coalgebra for the endofunctor $F$ is, by definition, an interval $X$ equipped with an interval map $X \to F(X)$. There is an evident category of coalgebras. 
In particular, the unit interval $[0, 1]$ becomes a coalgebra if we identify $[0, 1] \vee [0, 1]$ with $[0, 2]$ and consider the multiplication-by-2 map $[0, 1] \to [0, 2]$ as giving the coalgebra structure. 
Theorem: The interval $[0, 1]$ is terminal in the category of coalgebras. 
Let's think about this. Given any coalgebra structure $f: X \to X \vee X$, any value $f(x)$ lands either in the "lower" half (the first $X$ in $X \vee X$), the "upper" half (the second $X$ in $X \vee X$), or at the precise spot between them. Thus, you could think of a coalgebra as an automaton where on input $x_0$ there is output of the form $(x_1, h_1)$, where $h_1$ is either upper or lower or between. By iteration, this generates a behavior stream $(x_n, h_n)$. Interpreting upper as 1 and lower as 0, the $h_n$ form a binary expansion to give a number between 0 and 1, and therefore we have an interval map $X \to [0, 1]$ which sends $x_0$ to that number. Of course, should we ever hit $(x_n, between)$, we have a choice to resolve it as either $(bottom_X, upper)$ or $(top_X, lower)$ and continue the stream, but these streams are identified, and this corresponds to the identification of binary expansions 
$$.h_1... h_{n-1} 100000... = .h_1... h_{n-1}011111...$$ 
as real numbers. In this way, we get a unique well-defined interval map $X \to [0, 1]$, so that $[0, 1]$ is the terminal coalgebra. 
(Side remark that the coalgebra structure is an isomorphism, as always with terminal coalgebras, and the isomorphism $[0, 1] \vee [0, 1] \to [0, 1]$ is connected with the interpretation of the Thompson group as a group of PL automorphisms $\phi$ of $[0, 1]$ that are monotonic increasing and with discontinuities at dyadic rationals.)<|endoftext|>
TITLE: The cohomology of a product of sheaves and a plea.
QUESTION [14 upvotes]: The question Consider a topological space $X$ and a family of sheaves (of abelian groups, say)  $\; \mathcal F_i \;(i\in I)$ on $X$. Is it true that 
$$H^*(X,\prod \limits_{i \in I} \mathcal F_i)=\prod \limits_{i \in I} H^*(X,\mathcal F_i)  \;?$$
According to Godement's and to Bredon's monographs this is correct if the family of sheaves is locally finite (In particular if $I$ is  finite). [Bredon also mentions in an exercise that equality holds for spaces in which every point has a smallest open neighbourhood.]
What about the general case?
A variant Same question for $\check{C}$ech cohomology: is it true that
$$\check{H}^*(X,\prod \limits_{i \in I} \mathcal F_i)=\prod \limits_{i \in I} \check{H}^*(X,\mathcal F_i)  \;?$$
(Of course, $\check{C}$ech cohomology often coincides with derived functor cohomology but still the question should be considered independently)
A prayer Godement's book Topologie algébrique et théorie des faisceaux was published in 1960 and is still, with Bredon's, the most complete book on the subject.  I certainly appreciate the privilege of working in a field where a book released half a century ago is still relevant:  programmers and molecular biologists are not so lucky. Still I feel that a new treatise is due, in which naïve/foundational questions like the above would be addressed, and which would take the research and shifts in emphasis of half a century into account: one book on sheaf theory every 50 years does not seem an unreasonable frequency. So might I humbly suggest to one or several of the awesome specialists on MathOverflow  to write one? I am sure I'm not the only participant here whose eternal gratitude they would earn.

REPLY [8 votes]: The answer to the first question is almost always no, see Roos, Jan-Erik(S-STOC)
Derived functors of inverse limits revisited. (English summary)
J. London Math. Soc. (2) 73 (2006), no. 1, 65--83. .
Addendum: The crucial point is that infinite products are not exact. The most precise counterexample statement is Cor 1.11 combined with Prop 1.6 which identifies the stalks of the higher derived functors of the product with what you are interested in. Formally, it doesn't give a counter example for a single $X$ but Cor 1.11 shows that for any paracompact space with positive cohomological dimension there is some open subset for which your question has a negative answer. It seems clear that one could examples for specific $X$.<|endoftext|>
TITLE: Connection between the axiom of universes and Tarski's axiom
QUESTION [5 upvotes]: Grothendieck's axiom states that any set is a member of a Grothendieck universe (i.e. of a set that is closed under the subset, powerset, pairing and family-union relations), or equivalently, that there is a proper class of inaccessible cardinals. Tarski's lemma states that any set is a member of a set that is closed under the subset and the powerset relations and that contains each of its subsets of smaller cardinality than itself.
So the axioms seem to state similar things, but I can't really work out the precise connection between them. Does one imply the other? If not, is it known which one has stronger consistency strength?
Is Tarski's axiom as useful as the axiom of universes for category theory? Does the axiom of universe, like Tarski's axiom, imply AC (over ZF)?

REPLY [12 votes]: Over ZFC, the theories are equivalent. Both Grothendieck's Universe Axiom and Tarski's Axiom are equivalent to the assertion that there are unboundedly many inaccessible cardinals.
Grothendieck universes are exactly the sets $H_\kappa$ for an inaccessible cardinal $\kappa$, consisting of all sets hereditarily of size less than $\kappa$. They can also be described as $V_\kappa$ for inaccessible $\kappa$, using the cumulative von Neumann hierarchy. The Tarski universes, in contrast, needn't be transitive, and so the notions are not equivalent.
The issue is that over ZF, they lose their equivalence. Solovay explained on the FOM list that this is due to the way that cardinalities are treated in TG, making ZF+TG imply AC, whereas ZF+GU does not imply AC. You can read Solovay's interesting post here.

REPLY [5 votes]: Both axioms are equivalent with ZFC.
They are not equivalent with only ZF, because Ac is a cosequence of ZF+Grothendieck, but not of ZF+ tarski A.
See the message "AC and strongly inaccessible cardinals" (29/02/2008) of Robert SOLOVAY on the list FOM (Foundation of Mathematics).<|endoftext|>
TITLE: The Jacobi Identity for the Poisson Bracket
QUESTION [22 upvotes]: It is well known that if $M, \Omega$ is a symplectic manifold then the Poisson bracket gives $C^\infty(M)$ the structure of a Lie algebra.  The only way I have seen this proven is via a calculation in canonical coordinates, which I found rather unsatisfying.  So I decided to try to prove it just by playing around with differential forms.  I got quite far, but something isn't working out and I am hoping someone can help.  Forgive me in advance for all the symbols.
Here is the setup.  Given $f \in C^\infty(M)$, let $X_f$ denote the unique vector field which satisfies $\Omega(X_f, Y) = df(Y) =  Y(f)$ for every vector field $Y$.  We define the Poisson bracket of two functions $f$ and $g$ to be the smooth function $\{f, g \} = \Omega(X_f, X_g)$.  I can show that the Poisson bracket is alternating and bilinear, but the Jacobi identity is giving me trouble.  Here is what I have.
To start, let's try to get a handle on $\{ \{f, g \}, h\}$.  Applying the definition, this is given by $d(\Omega(X_f, X_g))X_h$.  So let's try to find an expression for $d(\Omega(X,Y))Z$ for arbitrary vector fields $X, Y, Z$.
Write $\Omega(X,Y) = i(Y)i(X)\Omega$ where $i(V)$ is the interior product by the vector field $V$.  Applying Cartan's formula twice and using the fact that $\Omega$ is closed, we obtain the formula
$$d(\Omega(X,Y)) = (L_Y i(X) - i(Y) L_X) \Omega$$
where $L_V$ is the Lie derivative with respect to the vector field $V$.  Using the identity $L_V i(W) - i(W) L_V = i([V,W])$, we get:
$$(L_Y i(X) - i(Y) L_X) = L_Y i(X) - L_X i(Y) + i([X,Y])$$
Now we plug in the vector field $Z$.  We get $(L_Y i(X) \Omega)(Z) = Y(\Omega(X,Z)) - \Omega(X,[Y,Z])$ by the definition of the Lie derivative, and clearly $(i([X,Y])\Omega)(Z) = \Omega([X,Y],Z)$.  Putting it all together:
$$d(\Omega(X,Y))Z = Y(\Omega(X,Z)) - X(\Omega(Y,Z)) + \Omega(Y, [X,Z]) - \Omega(X, [Y,Z]) + \Omega([X,Y], Z)$$
This simplifies dramatically in the case $X = X_f, Y = X_g, Z = X_h$. The difference of the first two terms simplifies to $[X_f, X_g](h)$, and we get:
$$
\begin{align}
\{\{f, g\}, h\} &= [ X_f, X_g ](h) + [ X_f, X_h ](g) - [ X_g, X_h ](f) - [ X_f, X_g ](h)\\
&= [ X_f, X_h ](g) - [ X_g, X_h ](f)
\end{align}$$
However, this final expression does not satisfy the Jacobi identity.  It looks at first glance as though I just made a sign error somewhere; if the minus sign in the last expression were a plus sign, then the Jacobi identity would follow immediately.  I have checked all of my signs as thoroughly as I can, and additionally I included all of my steps to demonstrate that if a different sign is inserted at any point in the argument then one obtains an equation in which the left hand side is alternating in two of its variables but the right hand side is not.  Can anybody help?

REPLY [7 votes]: Here's the way it's done in John Lee's Smooth Manifolds book:
$${\small
\iota_{X_{\lbrace f,g\rbrace}}\omega = d\lbrace f,g \rbrace=d(X_gf) = d(\mathcal{L}_{X_g}f) =\mathcal{L}_{X_g}df=\mathcal{L}_{X_g}(\iota_{X_f}\omega)=\iota_{[X_g,X_f]}\omega + \iota_{X_f}\mathcal{L}_{X_g}\omega=\iota_{[X_g,X_f]}\omega}
$$
which by nondegeneracy of $\omega$ implies the desired result.<|endoftext|>
TITLE: Does the Poincare series of a Coxeter group always describe a "flag variety"?
QUESTION [13 upvotes]: Let $W$ be a Coxeter group and let $P_W(q) = \sum_{w \in W} q^{\ell(w)}$ be its Poincare series.  When $W$ is the Weyl group of a simple algebraic group $G$ (hence $W$ is finite), $P_W(q)$ is the generating function describing the cells in the Bruhat decomposition of the flag variety $G/B$, $B$ a Borel.  
What happens when $W$ is infinite, e.g. when $W$ is an affine Weyl group or a hyperbolic Coxeter group?  Can $P_W$ still be associated to an infinite-dimensional "flag variety"?

REPLY [6 votes]: I think that Shrawan Kumar's book "Kac-Moody groups, their flag varieties, and representation theory" will contain the flag varieties (which are really ind-varieties in the non-finite case) that you are looking for.
A crystallographic Coxeter group is one of the form $\langle s_1,\ldots,s_n| s_i^2=(s_i s_j)^{m_{ij}}=1\rangle$ where each mij is equal to 2,3,4,6 or infinity. Such Coxeter groups are precisely the Weyl groups associated to arbitrary Kac-Moody algebras. In this case, there is a corresponding Kac-Moody group, together with an associated flag variety and Schubert cells, which seems to me to be what you want.
It is this geometry that is the starting point of the geometric interpretation of Kazhdan-Lusztig polynomials in the crystallographic case. As is to be expected, the finite case is easier than the affine case, which again is easier than the arbitrary KM case.<|endoftext|>
TITLE: Convergence of Fourier Series of $L^1$ Functions
QUESTION [25 upvotes]: I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e.  Also, Wikipedia informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$.  Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.
My first question is about the $p = 1$ case.  That is:

If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?


I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere.  Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.
My second question is similar:

Do there exist functions $f \in L^1$ whose Fourier series converge pointwise a.e., yet diverge in the $L^1$ metric?


(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system.  I am also referring only to the Lebesgue measure on [0,1].  Of course, if anyone knows any more general results, that would be great, too.)

REPLY [16 votes]: The answer to your first question is no. There is an $L^1$ function with Fourier series not converging in measure.
In the Kolmogorov example of an $L^1$ function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure $E$ and a subsequence $n_k$ such that for all $x$ in $E$, the absolute values of the partial sums $S_{n_k}$ of the Fourier series goes to infinity with $k$.
$$\forall x\in E,\ \ |S_{n_k}f(x)|\rightarrow \infty$$
This can be checked from the construction of $f$ in the original article of Kolmogorov, in its selected works.
If $S_nf$ converges in measure, then $S_{n_k}f$ must also converges in measure. This implies that there is a subsequence $n_{k_l}$ such that $S_{n_{k_l}}f(x)$ converges a.e. $x$, a contradiction.<|endoftext|>
TITLE: Mathematics and autodidactism
QUESTION [20 upvotes]: Mathematics is not typically considered (by mathematicians) to be a solo sport; on the contrary, some amount of mathematical interaction with others is often deemed crucial. Courses are the student's main source of mathematical interaction. Even a slow course, or a course which covers material which one already knows to some level, can be highly stimulating. However, there are usually a few months in the year when mathematics slows down socially; in the summer, one might not be taking any courses, for example. In this case, one might find themselves reduced to learning alone, with books.
It is generally acknowledged that learning from people is much easier than learning from books. It has been said that Grothendieck never really read a math book, and that instead he just soaked it up from others (though this is certainly an exaggeration). But when the opportunity does not arise to do/learn math with/from others, what can be done to maximize one's efficiency? Which process of learning does social interaction facilitate?
Please share your personal self-teaching techniques!

REPLY [3 votes]: I don't know if this counts as a correction, but in Pursuing Stacks Grothendieck mentions reading Quillen's book on model categories and was able to correct a mistake he made by reading a paper about model structures on the category of categories.<|endoftext|>
TITLE: Random geometric graphs and spanners
QUESTION [10 upvotes]: I would grateful to learn of work mixing
random geometric graphs with random graphs under
the
Erdős-Renyi model, and in particular concerning spanners.
Select $n$ points uniformly at random from the unit square,
and then form a graph $G=G(n,p)$ by connecting points
by adding edges with probablity $p$.
If $p_1$ is the threshhold
for the formation of the giant component $C$, is $C$
almost surely a geometric spanner for the points it connects
for $p=p_1 + \epsilon$?
(My guess is: No.)
One could ask a similar question about the threshold
$p_2$ for complete connection of the point set.
(Here perhaps the answer is: Yes?)
A geometric spanner has the property that, between any pair of points, 
there is path whose length
is not much longer than (no more than some constant times)
the Euclidean
distance between those points.

REPLY [7 votes]: I assume you meant $p=(1+\varepsilon)p_1$ in the first question, in which case the answer seems to be "no".
To see this, we note that, with high probability:
(i) The size of $C$ is $\Theta(n)$ (this is the classical result);
(ii) The average graph-theoretic distance $d_G(x,y)$ between two randomly chosen vertices $x,y\in C$ is concentrated around $c\ln n$ for some $c>0$ (this is proven in eg. Durrett's Random Graph Dynamics);
(iii) All points $p$ in the square are such that the ball of radius $r$ around $p$ contains $\Theta(r^2 n)$ points of $C$, simultaneously for all $r\gg \ln^2n/n$. (To see this, notice that the positions of points in the square are independent from their being or not in the giant component, then apply a VC dimension argument + part (i)).
Now let $\varepsilon>0$ be small (and fixed) and let $n$ grow. By item (ii), there is a high probability that one can find a point $p\in C$ such that the set
$$P(p)\equiv \{\mbox{all points $q$ in $C$ with }\frac{d_G(p,q)}{c\ln n}\in [1/2,1]\}$$
has size $\geq (1-\varepsilon^2)|C|$. By part (iii), there is a high probability that at least one point $q_0\in P(p)$ with $|p-q_0|\leq \varepsilon$ and at least one point $q_1\in P(p)$ with $q_1\geq 1/3$. Since $d_G(p,q_0),d_G(p,q_1)=O(\ln n)$, this shows that:
$$\frac{d_G(p,q_0)}{|p-q_0|}\geq \Omega\left(\frac{1}{\varepsilon}\right)\frac{d_G(p,q_1)}{|p-q_1|}.$$<|endoftext|>
TITLE: Why do generic polynomials work in reality?
QUESTION [8 upvotes]: I understand that a generic $G$-polynomial $f(t_1,...,t_n)[X]$ over field $k$ has Galois group $G$ over $k(t_1,...,t_n)$. And basically any $G$ extension of $k$ should be generated by a realization of $f$.(even a bit stronger but that is not the point here).
Now as much as I understand, our motivation for hunting these polynomials is that in real  (constructive) life, we would like to plug random elements of $k$ into $t_1,...,t_n$ and get a $G$-extension. However, it's obvious that the definition doesn't guarantee it. For example as a trivial failure, we know that $X^n + t_1X^{n-1} + \cdots + t_n$ is generic for $S_n$, but not all values for $t_1, ..., t_n$ (basically all polynomials) lead to an $S_n$-extension. 
So, basically, my question is this: what is the constructive value of the definition of generic polynomial. Is there any (although I know I'm saying nonsense) high probabilistic/statistic success rate in getting a $G$-extension when a random realization is chosen. Is there some kind of definition of "odd" that says those times that we don't get a $G$-extension are somehow odd and not normal?

REPLY [6 votes]: Adding unto Boyarsky's answer: Stephen Cohen has given quantative bounds for how often generic polynomials work. If I've skimmed his paper correctly, when the coefficients are integers chosen from the interval $[-N, N]$, the probability that the Galois group comes out wrong is $O(N^{-1/2} \log N)$, with an explicitly computable constant which depends on the group and the precise parameterization being used.<|endoftext|>
TITLE: How to implement Horner’s scheme for multivariate polynomials?
QUESTION [9 upvotes]: Background
I need to solve polynomials in multiple variables using Horner's scheme in Fortran90/95. The main reason for doing this is the increased efficiency and accuracy that occurs when using Horner's scheme to evaluate polynomials.
I currently have an implementation of Horner's scheme for univariate/single variable polynomials. However, developing a function to evaluate multivariate polynomials using Horner's scheme is proving to be beyond me.

An example bivariate polynomial would be: $12x^2y^2+8x^2y+6xy^2+4xy+2x+2y$ which would factorised to $x(x(y(12y+8))+y(6y+4)+2)+2y$ and then evaluated for particular values of x & y.

Research
I've done my research and found a number of papers such as:
staff.ustc.edu.cn/~xinmao/ISSAC05/pages/bulletins/articles/147/hornercorrected.pdf
citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.40.8637&rep=rep1&type=pdf
www.is.titech.ac.jp/~kojima/articles/B-433.pdf  
Problem
However, I'm not a mathematician or computer scientist, so I'm having trouble with the mathematics used to convey the algorithms and ideas.
As far as I can tell the basic strategy is to turn a multivariate polynomial into separate univariate polynomials and compute it that way.
Can anyone help me? If anyone could help me turn the algorithms into pseudo-code that I can implement into Fortran myself, I would be very grateful.

REPLY [6 votes]: I implemented this in Python: multivar_horner
You can look at the approach used there and port it to Fortran.
NOTE: In contrast to the one dimensional case there are multiple possible Horner factorisations of multivariate polynomials. One can allow a search over the possible factorisations to find a minimal representation as described HERE. In most cases however it suffices to use a heuristic to find a single "good" factorisation. multivar_horner implements the greedy heuristic described in "Greedy Algorithms for Optimizing Multivariate Horner Schemes".
The authors of some related publications (including the above mentioned) claim to have an implementation of their proposed algorithms, but I was not able to find any publicly available ones.<|endoftext|>
TITLE: Why are there usually an even number of representations as a sum of 11 squares
QUESTION [20 upvotes]: Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?
Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.
The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.
If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.
The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps:
  $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$
If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.
This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form:
 $$n = x_0^2 + 2x_1^2 + 8 x_3^2 \; , \; \; \mbox{with} \; \; x_0, x_1, x_3 \geq 0 \;?$$
Some play with the parities of binomial coefficients reduces this to the question I led with.

REPLY [4 votes]: This is not an answer to your stated question, but a relevant remark.  Suppose that $a(x)$ is a unipotent formal power series in the formal power series ring $(\mathbb{Z}/p)[[x]]$.  (Here unipotent just means constant term 1, so you could say topologically or adically unipotient.)  Then obviously $a(x)^n$ is well-defined for any integer $n$.  What is somewhat less obvious, but not hard and an interesting general principle, is that $a(x)^d$ is well-defined for any $p$-adic integer $d$.  Namely, if $d$ has digits $\ldots d_2d_1d_0$, then 
$$a(x)^d := a(x)a(x^{d_1p})a(x^{d_2p^2})\cdots$$
It's not hard to check that this formula is (a) correct when $d$ is an integer using the Frobenius map, (b) convergent when $a(x)$ unipotent, and (c) continuous in $d$ and $a$.  Therefore (d) it satisfies $a^{d+e} = a^d a^e$ and $(ab)^d = a^db^d$.
You use this formula twice in your question.  You use it with $d=-1$ when you call your power series "nontrivially the same" as your other question.  You use it again with $d=11$ with the phrase "some play with parities of binomial coefficients".  Of course in both cases you're using the power series from your paper,
$$a(x) = 1+x+x^4+x^9 + x^{16} + \cdots \in (\mathbb{Z}/2)[[x]].$$
I hadn't thought of this style of $p$-adic exponentiation and I think that it's cute, and it could be a unifying principle for some of what you are doing.<|endoftext|>
TITLE: Hilbert 90 for algebras
QUESTION [14 upvotes]: Let $L\diagup K$ be a Galois extension of fields satisfying $\left[L:K\right] < \infty$. Let $B$ be a finite-dimensional (as a $K$-vector space) $K$-algebra. Then, the Galois group $G$ of $L\diagup K$ acts on the $L$-algebra $L\otimes_K B$ (although not by $L$-linear homomorphisms), thus also on its unit group $\left(L\otimes_K B\right)^{\times}$. Is it true that $H^1\left(G,\left(L\otimes_K B\right)^{\times}\right)$ is the one-element set?
For $B=K$, this would be Hilbert 90. More generally, for $B$ being a matrix algebra over $K$, this would be an extension of Hilbert 90 Milne claims to hold in his CFT. My main reason for generalizing to arbitrary $B$ is to prove the following fact, known as Noether-Deuring theorem:
Let $A$ be a $K$-algebra, and let $U$ and $V$ be two finite-dimensional representations of $A$ over $K$. Then, $U$ and $V$ are isomorphic representations if and only if the representations $L\otimes_K U$ and $L\otimes_K V$ are isomorphic representations of the algebra $L\otimes_K A$. Note that this holds not only for Galois extensions $L\diagup K$ but for arbitrary field extensions $L\diagup K$, but the (Galois) case of finite fields is the hardest. This generalizes the fact that two matrices over some field are similar if and only if they are similar over a field extension, which, in turn, is a particular case of "Conjugacy rank" of two matrices over field extension .

REPLY [17 votes]: It's actually easier to go the other way around. Finite dimensional modules over
an algebra $A$ fulfils the Krull-Remak-Schmidt theorem of being isomorphic to a
direct sum of indecomposable modules with the indecomposable factors unique up
to isomorphism. If now $L\bigotimes_KU$ and $L\bigotimes_KV$ are isomorphic as
$L\bigotimes_KA$-modules they are also isomorphic as $A$-modules but as
$A$-modules they are isomorpic to $U^n$ resp. $V^n$ where $n=[L:K]$ and by the
KRS-theorem this implies that $U$ and $V$ are isomorphic. The generalised Thm 90
now follows as the cohomology set classifies $B$-modules whose scalar extension
to $L$ are free of rank $1$.

REPLY [6 votes]: Dear darij: yes, your $H^1$ is the one element set. A reference is Exercise 2 in Chapter X, §1 of Serre's Local Fields (page 152 in my edition). [The exercise is provided with hints!]
And, yes, the case of finite fields is indeed the hardest.<|endoftext|>
TITLE: Book on Hochschild (co)homology
QUESTION [37 upvotes]: There is currently no treatise treating Hochschild (co)homology systematically. There is a chapter in Weibel's book, there's parts of Loday's and a few others...

What should be covered by such a mythical treatise?

REPLY [2 votes]: There is so much on this subject coming from different perspectives. 
 I would add to all of the above  a complete, up to date, exposition regarding the relationship of Hochschild homology with loop spaces (for example including the non-simply connected case) and string topology (the structure of the hochschild chain complex of an algebra with a homotopy version of Poincaré duality, etc...)<|endoftext|>
TITLE: Integral points on varieties
QUESTION [19 upvotes]: I recently came across an interesting phenomenon which confused me slightly, concerning integral points on varieties.
For example, consider $X = \mathbb{A}_{\mathbb{Z}}^{n+1} \setminus \{0\}$, affine $n$-space over $\mathbb{Z}$ with the origin removed. Naively, one would guess that $X(\mathbb{Z})$ is the set of integers $\{ (x_0,x_1,\ldots,x_n) \in \mathbb{Z}^{n+1} \setminus \{0\}\}$.
However, some work that I have been doing on recently with universal torsors has in fact led me to believe that $X(\mathbb{Z})$ should equal $\mathbb{P}^n(\mathbb{Z})$, at least modulo the action of $\mathbb{G}_m$. That is $X(\mathbb{Z})$ is actually the set of integers $\{ (x_0,x_1,\ldots,x_n) \in \mathbb{Z}^{n+1}\setminus \{0\}\}$ such that $\gcd(x_0,x_1,\ldots,x_n)=1$.
Is there a simple explanation for why this is the case?
Thanks!
Dan

REPLY [19 votes]: Dear Daniel, here is a detailed explanation, respectfully following the sacred texts (EGA or Hartshorne).
a) First of all, $\mathbb A^{n+1}_{\mathbb Z}$ has no origin, despite our classical intuition! As a substitute, it has the prime (but not maximal) ideal $\mathcal P=(X_1,X_2,...,X_{n+1})$ and corresponding to it the integral subscheme $V=V(\mathcal P)$. And what you want to calculate is the set of $\mathbb Z$-points of
 $U=\mathbb A^{n+1}\setminus V(\mathcal P)$, i.e. the set of morphisms $Spec(\mathbb Z)\to U$. Let's do that.
b) A morphisms $f: Spec \mathbb Z \to \mathbb A^{n+1}$ corresponds to a morphism of rings 
$ev_a: \mathbb Z[X_1,X_2,...,X_{n+1}] \to \mathbb Z$ , evaluation of integral polynomials at a tuple $a=(a_1,a_2,...,a_{n+1}) \in {\mathbb Z}^{n+1} $. Call $f=f_a : Spec \mathbb Z \to \mathbb A^{n+1}$ the corresponding morphism.
c) We must ensure that the image of $f$ lies in $U$ i.e. that it is disjoint from 
$V=V(\mathcal P)$. But the points of $V$ are its generic point $(X_1,X_2,...,X_{n+1})$ and its closed points $\mathcal M_p=(X_1,X_2,...,X_{n+1}, p)$, $p$ a prime. It is enough to show that these closed points are not in the image of $f$. Equivalently, we must show that the fibre of $f$ at $\mathcal M_p$ is empty. Since this fibre is the spectrum of
$\mathbb Z / (a_1,a_2,...,a_{n+1},p)$ our condition is that this ring be zero or equivalently that the ideal $(a_1,a_2,...,a_{n+1},p)$ be zero: this will happen exactly if $p$ does not divide all ot the $a_i$'s. Since this must hold for all primes, we get:
Final result The $\mathbb Z$-points of $U=\mathbb A^{n+1}\setminus V(\mathcal P)$ are given by $(n+1)$-tuples of integers whose g.c.d. is 1.
Reminder I  have used that the fibre of a morphism of affine schemes $f:SpecB\to Spec A$ at $\mathcal M \in Specmax A$ is $Spec ( B/\mathcal M B) $.<|endoftext|>
TITLE: What should be learned in a first serious schemes course?
QUESTION [178 upvotes]: I've just finished teaching a year-long "foundations of algebraic
geometry" class.  It
was my third time teaching it, and my notes are gradually converging.
I've enjoyed it for a number of reasons (most of all the students, who
were smart, hard-working, and from a variety of fields).  I've
particularly enjoyed talking with experts (some in nearby fields, many
active on mathoverflow) about what one should (or must!) do in a first
schemes course. I've been pleasantly surprised to find that those who
have actually thought about teaching such a course (and hence who know
how little can be covered) tend to agree on what is important, even if
they are in very different parts of the subject.  I want to raise this
question here as well:

What topics/examples/ideas etc. really really should be learned in a 
  year-long first serious course in schemes?

Here are some constraints.  Certainly most excellent first courses
ignore some or all of these constraints, but I include them to focus
the answers.  The first course in question should be
purely algebraic.  (The reason for this constraint: to avoid a 
debate on which is the royal road to algebraic geometry --- this is
intended to be just one way in.  But if the community thinks that a
first course should be broader, this will be reflected in the voting.)
The course should be intended for people in all parts of algebraic
geometry.  It should attract smart people in nearby areas.  It should
not get people as quickly as possible into your particular area of
research.  Preferences: It can (and, I believe, must) be hard.  As
much as possible, essential things must be proved, with no handwaving
(e.g. "with a little more work, one can show that...", or using
exercises which are unreasonably hard).  Intuition should be given
when possible.
Why I'm asking:   I will likely edit the notes further, and hope to post 
them in chunks over the 2010-11 academic year to provoke further debate.  Some hastily-written thoughts are
here, 
if you are curious.
As usual for big-list questions: one topic per answer please.  There
is little point giving obvious answers (e.g. "definition of a
scheme"), so I'm particularly interested in things you think others
might forget or disagree with, or things often omitted, or things you
wish someone had told you when you were younger.  Or propose dropping
traditional topics, or a nontraditional ordering of traditional topics.  Responses
addressing prerequisites such as "it shouldn't cover any commutative
algebra, as participants should take a serious course in that subject
as a prerequisite" are welcome too.  As the most interesting
responses might challenge (or defend) conventional wisdom, please give
some argument or evidence in favor of your opinion.
Update later in 2010:  I am posting the notes, after suitable editing, and trying to take into account the advice below, here.  I hope to reach (near) the end some time in summer 2011.  Update July 2011: I have indeed reached near the end some time in summer 2011.

REPLY [3 votes]: Being a differential geometer, it might be nice if you can point out analogies (perhaps even make them rigorous ?) to differential geometry. Like a scheme being flat over another is perhaps akin to a fibre bundle. A scheme itself is like a manifold, etc. This might make the subject slightly less scary for geometric analysts.<|endoftext|>
TITLE: The functoriality of group C* algebra structure
QUESTION [6 upvotes]: Let $G$ and $H$ be discrete groups and $f:G \rightarrow H$ be any homomorphism of these groups. I have three questions about it:
1) How to prove the functoriality of the construction of universal $C^*$-algebra of discrete
group (the existence of induced homomorphism $C^*(G) \to C^*(H)$)?
2) How to prove that the construction of reduced $C^*$-algebra of discrete group is not functorial (I am especially interested in counterexamples) and in which case (I mean conditions for group homomorphism) it'll be functorial? 
3) Let us consider the case when $G = \mathbb{Z}$ (integers) and $H = \mathbb{Z}/ n\mathbb{Z}$. How to describe the kernel and the image of the induced homomorphism of group $C^*$-algebras?

REPLY [7 votes]: Regarding Q2: the reduced $C^\star$-algebra is functorial wrt homomorphisms with amenable kernels. Indeed, let $N$ be a normal, amenable subgroup of $G$; since the trivial representation of $N$ is weakly contained in the regular representation of $N$ (by amenability), by continuity of induction the regular representation of $G/N$ is weakly contained in the regular representation of $G$, which means that the reduced $C^\star$-algebra of $G$ maps onto the one of $G/N$.<|endoftext|>
TITLE: References for "modern" proof of Newlander-Nirenberg Theorem
QUESTION [24 upvotes]: Hi,
I'm starting to prepare a graduate topics course on Complex and Kahler manifolds for January 2011. I want to use this course as an excuse to teach the students some geometric analysis. In particular, I want to concentrate on the Hodge theorem, the Newlander-Nirenberg theorem, and the Calabi-Yau theorem.
I have many excellent references (and have lectured before) on the Hodge and CY theorems. However, for the Newlander-Nirenberg theorem, I am finding it hard to find a "modern" treatment. I recall going through the original paper in my graduate student days, but I hope that there is a more streamlined version floating around somewhere. (I want to consider the general smooth case, not the easy real-analytic version). Besides the original paper, so far I can only find these references:
J. J. Kohn, "Harmonic Integrals on Strongly Pseudo-Convex Manifolds, I and II" (Annals of Math, 1963)
and
L. Hormander, "An introduction to complex analysis in several variables" (Third Edition, 1990)
Both are easier to follow than the original paper. But my question is: are there any other proofs in the literature, preferably from books rather than papers? The standard texts on complex and Kahler geometry that I have looked at don't have this.

REPLY [21 votes]: Hi Spiro:  I have had much the same difficulties as you, but I now know a modern proof.
At heart, the original proof is an application of the implicit function theorem. More specifically, let $U$ be a polydisk in $C^n$ consider the sequence of Banach manifolds
$Diff^{k,\alpha}(U,C^n) \to AC^{k-1,\alpha}(U) \to (A^{0,2})^{k-2,\alpha}(U,TU).$
These are respectively the diffeomorphisms $U\to C^n$ of class $(k,\alpha)$, the almost complex structures on $U$ of class $(k-1,\alpha)$ and the $(0,2)$ forms on $U$ with values in the holomorphic tangent bundle, of class $(k-2,\alpha)$.  The first map is the pullback of the standard complex structure, and the second is the Frobenius integrability form 
$\phi \mapsto \overline \partial \phi - \frac 12 [\phi\wedge \phi].$ 
The object is to show that the first map is locally surjective onto the inverse image of $0$ by the second. These spaces are Banach manifolds, and if you can show that the sequence  of derivatives (respectively at the identity, at the standard complex structure and at 0) is split exact, the result follows from the implicit function theorem.
This sequence of derivatives is the Dolbeault sequence on $U$ (in the appropriate class), and it is split exact, though this is NOT obvious.  There is an error in the original paper, or rather in the paper of Chern's that it depends on, but the result is true.  The remainder of the mess in the original proof is due to the authors writing out the Picard iteration in the specific case, rather than isolating the needed result.
I am working on getting this written up with Milena Pabiniak, a graduate student here at Cornell.  Write me at jhh8@cornell.edu  if you are interested in seeing details.
John Hubbard<|endoftext|>
TITLE: Hodge numbers of compactifications
QUESTION [9 upvotes]: Let $X$ be a smooth complex quasi-projective variety. We can find good compactification: a smooth proper variety $\bar{X}$ such that ${\bar X} \setminus X$ is a divisor with normal crossing. The variety $\bar{X}$ is then stratified by the singulartities of the divisor. And one can compute the mixed Hodge structure on $H^{\bullet}(X)$ in terms of the pure Hodge structures $H^{\bullet}(S_\alpha)$ of the smooth closed strata using a spectral sequence.   
Let's say a variety $Y$ is Hodge-Tate if $h^{p,q}(Y) = 0$ for $p\neq q$. 
If all the closed strata of $\bar{X}$ are Hodge-Tate then $X$ is Hodge-Tate. 
Question: Let $X$ be a smooth complex quasi-projective variety. Assume $X$ is Hodge-Tate. 

Can one find a good compactification $\bar{X}$ with Hodge-Tate strata? 
Are all good compactifications of $X$ of this type? (Edit: Answer is no, see Torsten's elementary example).

REPLY [7 votes]: I asked the question to Claire Voisin and she immediatly gave me the following counter example:
Consider Fermat's cubic of dimension 3 $X$. There are 5 cones on elliptic curves $E_i$ inside $X$ and one can verify that $Y=X\setminus U_iE_i$ is Hodge-Tate. But $Y$ doesn't admit a Hodge-Tate compactification as that would imply that $X$ is birational to a Tate variety and this would contradict Clemens-Griffiths' theorem saying that the intermediate jacobian of $X$ is not a direct sum of jacobians of curves as a polarized variety.<|endoftext|>
TITLE: "Industry"/Government jobs for mathematicians
QUESTION [42 upvotes]: Suppose that you graduate with a good PhD in mathematics, but don't necessarily want to go into academia, with the post-doc years that this entails. Are there any other options for continuing to do "real math" professionally?
For example, how about working at the NSA? I don't know much of what is done there -- is it research mathematics? Are there other similar organizations? Perhaps corporations that contract with the federal government? Companies like RSA?
Other areas of industry? Is there research mathematics done in any sort of financial or tech company?
I've made this a community wiki, since there aren't any right answers...

REPLY [37 votes]: I have worked in academia, at the research center of a telecommunications company (Tellabs), and at two different FFRDCs (MIT Lincoln Laboratory and IDA).  At all of the non-academic jobs, I have done "real math," published papers, attended conferences, given talks, etc.  So it is certainly possible to continue doing "real math" outside of academia.
You should be aware, however, that in almost any non-academic job, there is pressure on you to produce results that are "useful" for the company or the government.  The amount of such pressure varies, but it always exists, because ultimately that is the main justification for your paycheck.  In academia, the corresponding fact is that in almost any academic job, there is pressure on you to teach, since that is usually the justification for a significant portion of your salary.  Finding a non-academic job where there is no pressure on you to do anything "useful" is akin to finding an academic job where you have no teaching responsibilities.
Certain high-tech companies and certain FFRDC's recognize that a good way to attract top talent is to give their employees the freedom to pursue their own research interests, whatever that may be.  All the non-academic jobs I had were like this.  They actively encouraged me to spend some amount of my time doing "real math" regardless of whether the results were of any "use."  How much time?  Well, if the company was doing well, and if I was doing a good job of producing "useful" results that they liked, then they would give me more freedom.  But if the company was doing poorly then they would start to squeeze.  During the telecom industry meltdown in the late 1990s, Tellabs eventually eliminated its research center entirely, along with my job; Bell Labs (more famously) suffered a similar fate.
So far I have been drawing a dichotomy between "what the company finds useful" and "real math," and maybe you don't find that satisfactory.  After all, if you're sufficiently motivated, you can do "real math" on your own time regardless of what your "day job" is.  Maybe what you want is a job where providing what is useful to the company involves doing real math.  This is a taller order; for example, at Lincoln Labs I found that there was almost no real math involved in the work they wanted me to do, and I eventually left that job for that reason even though it was a great job in almost every other respect.  However, it is still possible to find such jobs, depending on what area of math you are interested in.  If you are interested in large cardinals and are hoping for a job where your theorems about large cardinals will be "useful" then you are probably out of luck.  However, if your interests lean towards areas with known relevance to computer science or various branches of engineering then your chances are much better.  The NSA scores pretty well in this regard since it is no secret that number theory and various other branches of so-called "pure" mathematics are relevant to cryptology.
In summary, jobs where you do "real math" do exist.  When considering such a job, though, you should first ask yourself, will I enjoy producing what this company considers to be "useful" results?  If the answer is no, then you will probably not be happy at the job even if they give you some freedom to do "real math."  However, if the answer is yes, and the company gives you some amount of freedom to do "real math," then it will probably be an excellent fit for you.<|endoftext|>
TITLE: Classical mechanics motivation for poisson manifolds?
QUESTION [24 upvotes]: Suppose I want to understand classical mechanics. 
Why should I be interested in arbitrary poisson manifolds and not just in symplectic ones?
What are examples of systems best described by non symplectic poisson manifolds?

REPLY [7 votes]: The following  article by J. Butterfield suggests 3 reasons motivating Poisson manifolds (over symplectic manifolds)(page 81)

Parameters and stability. Sometimes it is easier to analyze stability of the dynamics on a Poisson manifold than on its symplectic leaves.
Naturality: The rigid rotator problem is more natural to analyze on SO(3) than on its coadjoint orbit.

3.Reduction: When the configuration space is a Lie group, it is natural to use T*G/G as the reduced phase space. This is a Poisson manifold isomorphic to the dual of the Lie algebra and having a Lie Poisson structure.<|endoftext|>
TITLE: Extra principal Cartier divisors on non-Noetherian rings? (answered: no!)
QUESTION [17 upvotes]: On the way to defining Cartier divisors on a scheme $X$, one sheafifies a presheaf base-presheaf of rings $\mathcal{K}'(U)=Frac(\mathcal{O}(U))$ on open affines $U$ to get a sheaf $\mathcal{K}$ of "meromorphic functions".1
(ETA: See Georges Elencwajg's answer for Kleiman's article on why $Frac(\mathcal{O}(U))$ doesn't define an actual presheaf.  The correct base-free way to make a presheaf $\mathcal{K}'$ is to let $S(U)$ be the elements of $\mathcal{O}(U)$ which are "stalk-wise regular", i.e. non-zerodivisors in $\mathcal{O}_p$ for every $p\in U$, and define $\mathcal{K}'(U)=\mathcal{O}(U)[S(U)^{-1}]$.  This agrees with the base-presheaf above on affines.)
Have you ever wondered what this sheaf does on affine opens?  That's how I usually grasp what a sheaf "really is", but Hartshorne's Algebraic Geometry (Definition 6.11-, p. 141) doesn't tell us.  The answer is non-trivial, but turns out to be nice for lots of nice rings.  Q. Liu's Algebraic Geometry and Arithmetic Curves shows that:

If $A$ is Noetherian, or reduced with finitely many mimimal primes (e.g. a domain), then 
$\mathcal{K}(Spec(A))=Frac(A)$.  (Follows from Ch.7 Remark 1.14.)

If $A$ is any ring, then $Frac(A)$ is a subring of $\mathcal{K}(Spec(A))$.  (Follows from Ch.7 Lemma 1.12b.)

So for $A$ non-Noetherian, we could be getting some extra elements, and presumably, they could be units.  In other words, we could have principal Cartier divisors that don't come from $Frac(A)$.  
Is there an example where this happens? 

Follow-up: Thanks to BCnrd's proof below, the answer is "no": even though $\mathcal{K}(Spec(A))$ can be strictly larger than $Frac(A)$, it can't contain additional units, so there are no such extra principal divisors!

Footnotes:
1 Here "Frac" means inverting the non-zero divisors of the ring; I'm not assuming anything is a domain.

REPLY [21 votes]: In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra principal Cartier divisors". This is what I will prove cannot happen. The argument is a correction on an earlier attempt which had a bone-headed error. [Kleiman's construction from Georges' answer is not invertible, so no inconsistency. Kleiman makes some unfortunate typos -- his $\oplus k(Q)$ should be $\prod k(Q)$, and more seriously the $t$ at the end of his construction should be $\tau$, for example -- but not a big nuisance.]
For anyone curious about general background on meromorphic functions on arbitrary schemes, see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is a little subtle error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of locally regular sections of $O_X$; this is the issue in the Kleiman reference mentioned by Georges. The content of EGA works just fine upon making that little correction. There are more hilarious errors elsewhere in IV$_4$, all correctable, such as fractions with infinite numerator and denominator, but that's a story for another day.) Also, 20.2.12 there is the result cited from Qing Liu's book in the setup for the question. 
The first step in the proof is the observation that for any scheme $X$, the ring $M(X)$ of meromorphic functions is naturally identified with the direct limit of the modules Hom($J, O_X)$ as $J$ varies through quasi-coherent ideals which contain a regular section of $O_X$ Zariski-locally on $X$.  Basically, such $J$ are precisely the quasi-coherent "ideals of denominators" of global meromorphic functions. This description of $M(X)$ is left to the reader as an exercise, or see section 2 of the paper "Moishezon spaces in rigid-analytic geometry" on my webpage for the solution, given there in the rigid-analytic case but by methods which are perfectly general.   
Now working on Spec($A$), a global meromorphic function "is" an $A$-linear map $f:J \rightarrow A$ for an ideal $J$ that contains a non-zero-divisor Zariski-locally on $A$.
Assume $f$ is an invertible meromorphic function: there are finitely many $s_i \in J$ and a finite open cover {$U_i$} of Spec($A$) (yes, same index set) so that $s_i$ and $f(s_i)$ are non-zero-divisors on $U_i$; we may and do assume each $U_i$ is quasi-compact. Let $S$ be the non-zero-divisors in $A$. Hypotheses are preserved by $S$-localizing, and it suffices to solve after such localization (exercise). So without loss of generality each element of $A$ is either a zero-divisor or a unit. If $J=A$ then $f(x)=ax$ for some $a \in A$, so $a s_i=f(s_i)$ on each $U_i$, so all $a|_{U_i}$ are regular, so $a$ is not a zero-divisor in $A$, so $a$ is a unit in $A$ (due to the special properties we have arranged for $A$). Hence, it suffices to show $J=A$. 
Since the zero scheme $V({\rm{Ann}}(s_i))$ is disjoint from $U_i$ (as $s_i|_ {U_i}$ is a regular section), the closed sets $V({\rm{Ann}}(s_i))$ and $V({\rm{Ann}}(s_2))$ have
intersection disjoint from $U_1 \cup U_2$.  In other words, the quasi-coherent ideals ${\rm{Ann}}(s_1)$ and ${\rm{Ann}}(s_2)$ generate the unit ideal over $U_1 \cup U_2$.
A quasi-coherent sheaf is generated by global sections over any quasi-affine scheme, such as $U_1 \cup U_2$ (a quasi-compact open in an affine scheme), so we get $a_1 \in {\rm{Ann}}(s_1)$ and $a_2 \in {\rm{Ann}}(s_2)$ such that $a_1 + a_2 = 1$ on $U_1 \cup U_2$.  Multiplying both sides by $s_1 s_2$, we get that $s_1 s_2 = 0$ on $U_1 \cup U_2$.  But
$s_1$ is a regular section over $U_1$, so $s_2|_ {U_1} = 0$.  But $s_2|_ {U_2}$ is a regular section, so we conclude that $U_1$ and $U_2$ are disjoint.  This argument shows that the $U_i$ are pairwise disjoint.   
Thus, {$U_i$} is a finite disjoint open cover of Spec($A$), so in fact each $U_i = {\rm{Spec}}(A_i)$ with $A = \prod A_i$.  But recall that in $A$ every non-unit is a zero-divisor.  It follows that the same holds for each $A_i$ (by inserting 1's in the other factor rings), so each regular section $s_i|_ {U_i} \in A_i$ is a unit. But the preceding argument likewise shows that $s_i|_ {U_j} = 0$ in $A_j$ for $j \ne i$, so each $s_i \in A$ has a unit component along the $i$th factor and vanishing component along the other factors.  Hence, the $s_i$ generate 1, so $J = A$.  QED<|endoftext|>
TITLE: A question about tilings of the plane
QUESTION [8 upvotes]: Let C be a compact subset of the Euclidean plane E whose boundary is a Jordan curve J. If C tiles the 
plane, can J be such that it has a unique tangent line at each point and none of its sub-arcs is a 
straight line segment with distinct end-points? If so, can you give an example? J does not need to be
convex and the tiling need not be regular. The only requirement is that the plane E be a countable union
of congruent copies of C, no two of which have a common interior point.

REPLY [6 votes]: I have a way to do it, but read my note at the bottom about why you may not find it to be valid.
Use an eye-shaped figure, but where the ends of the eyes meet at angle 0. This will allow unique tangents at the "corners".
alt text http://www.freeimagehosting.net/uploads/1e7dbd258e.jpg
The eyes will fit together in a standard brick pattern. 
In the image, I used a sin function, but you can do it also using circle fragments. In this case, the tiling corresponds to a standard penny tiling of the plane, but apportioning the empty space to adjacent pennies in order to make the eye shapes. 
Since my Jordan curves turn directly around in the opposite direction at those cusps, however, you may not consider this to be a valid example, since perhaps you regard this as two tangent lines at those points, pointing in opposite directions.
I think someone will show up and prove that you cannot do it without any cusps. I would like to know whether you can do it with only one cusp.<|endoftext|>
TITLE: How do I convert a uniform value in [0,1) to a standard normal (Gaussian) distribution value?
QUESTION [13 upvotes]: I have uniform value in [0,1). I'd like to transform it into a standard normal distribution value, in a deterministic fashion.
What I'm confused about with the Box-Muller transform is that it takes two uniform values in [0, 1), and transform them into two normal random values.
However, I only have one uniform value. How do I apply Box-Muller over a single value?

REPLY [2 votes]: The best way to obtain the inversion from U[0, 1] to Normal distribution is by using an algorithm presented in a famous short paper of Moro (1995). Moro presented a hybrid algorithm: he uses the Beasley & Springer algorithm for the central part of the Normal distribution and another algorithm for the tails of the distribution.
He modeled the distribution tails using truncated Chebyschev series.....
..................
http://marcoagd.usuarios.rdc.puc-rio.br/quasi_mc.html<|endoftext|>
TITLE: Can't find Segal's papers in CFT
QUESTION [5 upvotes]: At various places I have seen people referring to Segal's papers in CFT as the "standard definition" of the subject. These seem to have become classics in this field. But I can't locate them on the net. I searched on arxiv,spires,Google/Scholar,AMS etc. 
The papers I am referring to are (especially the fist one),

G. Segal, The definition of conformal field theory, in: Differential geometrical methods in theoretical physics (Como, 1987), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 250, Kluwer Acad. Publ., Dordrecht, 1988, 165-171
G. Segal, Two-dimensional conformal field theories and modular functors, in: Proceedings of the IXth International Congress on Mathematical Physics, Swansea, 1988, Hilger, Bristol, 1989, 22-37.
G. Segal, The definition of conformal field theory, preprint, 1988; also in: Topology, geometry and quantum field theory, ed. U. Tillmann, London Math. Soc. Lect. Note Ser., Vol. 308. Cambridge University Press, Cambridge, 2004, 421-577.

and on similar lines this paper by Moore and Seiberg 

G. Moore and N. Seiberg, Lectures on RCFT, in: Physics, geometry, and topology (Banff, AB, 1989), ed. H.C. Lee, NATO Adv. Sci. Inst. Ser. B Phys., 238, Plenum, New York, 1990, 263-361.

One possibility is that these are only available in the proceedings mentioned and hence only as books in some library. Are there no online copies of these?

REPLY [3 votes]: http://www.math.upenn.edu/~blockj/scfts/segal.pdf<|endoftext|>
TITLE: Jordan Form Over a Polynomial Ring
QUESTION [9 upvotes]: Let $X$ be the set of $k\times k$ matrix with entries in $\mathbb{C}$, and let $M\in X$.  The group $GL(k,\mathbb{C})$ acts on $X$ by conjugation, and according to the Jordan decomposition theorem (see e.g wikipedia) somewhere in the orbit containing $M$ is a block diagonal matrix with non-zero entries only on the diagonal and superdiagonal.
Suppose now we consider $k\times k$ matrices whose entries lie in the polynomial ring $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ and we study the action by conjugation of $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$.  Then the Jordan decomposition theorem, as formulated above, clearly no longer holds.  For example consider the matrix:
$ M=\left(
{\begin{array}{cc}
0 & 1 \\\
z^{p}_{1} & 0 
\end{array}}
\right)$, 
where in the above $p$ denotes a positive integer.  If $p $ is odd, then $M$ cannot be diagonalized since the ring $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ does not contain the eigenvalues of $M$.  On the other hand, if $p$ is even we still cannot diagonalize $M$ since when $z_{1}=0 $,  $M$ is not diagonalizable.
My question is then what, if anything, remains of the Jordan decomposition in this case?  Or equivalently given a $k\times k$ matrix $M$ with entries in $\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}]$ are there any particularly simple matrices related to $M$ via conjugation by an element of $GL(k,\mathbb{C}[z_{1},z_{2}, \ldots ,z_{n}])$?

REPLY [12 votes]: The short answer is "no". It is not too difficult to construct *invariants, but the canonical forms are hard.a What follows is not a full answer, but a useful way to think about the question. 
The problem of classifying $k\times k$ matrices over a commutative ring $R$ up to conjugacy is equivalent to the problem of classifying all $R[\lambda]$-module structures on $R^k$, the free $R$-module of rank $k,$ up to isomorphism. Given $A\in M_k(R)$, let the variable $\lambda$ act on $R^k$ via $A$ and conversely, given an $R[\lambda]$-module structure on $R^k,$ the action of $\lambda$ is $R$-linear, hence yields a matrix. It's a good exercise to see that conjugacy of matrices $\leftrightarrow$ isomorphism of modules.
Now, when $R$ is a field, $R[\lambda]$ is a principal ideal domain, and all finitely-generated modules can be completely classified using the theory of elementary divisors.b However, if $R$ is even a bit more complicated, such as $\mathbb{Z}$ or $K[X]$, the question involves modules over a ring of Krull dimension $2$ or larger, and one cannot hope for an explicit easy solution (except $k=1$). Already for $n=1, R=K[x]$ we are looking at the classification of $K[x,y]$-modules.  See van der Waerden for classical treatment.

Footnotes 
a For example, the characteristic polynomial and the Fitting invariants of the matrix in b.
b In the context of the conjugacy problem, you can replace a $k\times k$ matrix $A$ over $R$ modulo conjugation by $GL_k(R)$ with a $k\times k$ matrix $A-\lambda I_k$ over $R[\lambda]$ modulo left and right multiplication by $GL_k(R).$ The "elementary" part refers to the fact that when $R=K$ is a field, the general linear group is generated by elementary transformations, and  "divisors" refers to the form of the answer, where the canonical form is diagonal and $d_i$ divides the entries in rows $1$ through $i$. Neither fact is true for a more general $R.$<|endoftext|>
TITLE: Is there a Morse theory proof of the Bruhat decomposition?
QUESTION [27 upvotes]: Let $G$ be a complex connected Lie group, $B$ a Borel subgroup and $W$ the Weyl group. The Bruhat decomposition allows us to write $G$ as a union $\bigcup_{w \in W} BwB$ of cells given by double cosets.
One way I have seen to obtain cell decompositions of manifolds is using Morse theory. Is there a way to prove the Bruhat decomposition using Morse theory for a certain well-chosen function $f$? More generally, are there heuristics when obtaining a certain cell decomposition through Morse theory is likely to work?

REPLY [8 votes]: Let me give  a very elementary description of the Morse-theoretical interpretation of the Bruhat decomposition in a simple case, of thecomplex Grasmannians. $\DeclareMathOperator{\Gr}{Gr}$ Denote by $\Gr_{k,n}$ the Grassmannian of complex $k$-dimensional  subspaces of $\mathbb{C}^n$.
We can identify $\Gr_{k,n}$  with a submanifold of the vector space $\mathscr{S}_n$ of hermitian $n\times n$ matrices by associating to a subspace $L$  the orthogonal  projection  $P_L$. Let $A\in\mathscr{S}_n$ a hermitian matrix  with distinct eigenvalues. Define $\DeclareMathOperator{\tr}{tr}$
$$f_A:{\Gr}_{k,n}\to\mathbb{R},\;\;L\mapsto  \tr(AP_L).$$
Denote by $X_A$ the (negative) gradient of $f_A$ with respect to the metric on $\Gr_{k,n}$ induced by the Euclidean metric on $\mathscr{S}_n$. Then the unstable  manifolds  of  the flow  generated by $X_A$ are the Schubert cells    on $\Gr_{k,n}$ with respect to a flag determined by the eigenvectors of $A$.

REPLY [8 votes]: My answer has some overlap with Lucas', but a different point of view.
In the appendix to our paper 


Cycles of Bott-Samelson type for taut representations 


G. Thorbergsson and I considered generalized real flag manifolds $G/P$ where $G$ is a noncompact real semisimple Lie group and $P$ is a parabolic subgroup. It is well known that those arise as orbits of representations of isotropy representations of symmetric spaces. In a celebrated paper in the 1950's, Bott and Samelson had constructed concrete cycles for those spaces which represent a basis for the $\mathbb Z_2$-homology. We wrote 
in the appendix an explicit (Lie-theoretic) proof that 


the images of the cycles of Bott and Samelson equal the closures of the Bruhat cells


(such a proof had been previously given by Hansen in the case $G$ is a complex Lie group
in "On cycles in flag manifolds", Math. Scand. 33 (1973), 269-274).
As an application of the cycles of Bott and Samelson, (squared) distance functions to orbits of isotropy representations of symetric spaces are perfect in the sense that the Morse inequalities are in fact equalities; submanifolds with such property are usually called taut. Because of their tautness, the Bruhat decomposition of he spaces $G/P$ is minimal in the sense that the number of cells in dimension $k$ is equal to the 
$k$th $\mathbb Z_2$-Betti number of $G/P$. Notice that in the case $G$ is a complex Lie group, the cells are all even-dimensional which makes the minimality of the Bruhat decompostion trivial, but such easy arguments do not apply otherwise. Another 
Morse-theoretic interpretation of the Bruhat cells is their appearence and the unstable manifolds of suitable height functions on $G/P$ seen as orbits of isotropy representations of symmetric spaces, see M. Atiyah, "Convexity and commuting Hamiltonians", Bull. London. Math. Soc. 14 (1982), 1-15 and  R. R. Kocherlakota, "Integral homology of real flag manifolds and loop spaces of symmetric spaces", Adv. Math. 110 (1995), 1-46.<|endoftext|>
TITLE: I am searching for the name of a partition (if it already exists)
QUESTION [11 upvotes]: I derived this definition by searching for a representation of a family of sets. I am quite sure that someone should have thought to this before, because it seems to be quite straightforward given a family of sets. However, I did not find anything suitable on google or on wikipedia.
Let a family of sets, say $A_1, \ldots, A_n$, be given. To avoid misunderstanding I will call them modules. This family induces a unique partition on the union set $A = \bigcup_{i=1}^{n} A_i$ in the following way: I call building block a maximal subset $B$ of $A$ such that do not exist 2 different modules $A_i$ and $A_j$ with:
$B \cap A_i \not= \emptyset$,
$B \cap A_j \not= \emptyset$ and
$B\not\subseteq A_i \cap A_j$.
For example, if my family consists of 2 different overlapping modules $A_1$ and $A_2$, I can partition the set $A = A_1 \cup A_2$ as:
(elements in $A_1 \cap A_2$);
(elements in $A_1$ but not in $A_2$);
(elements in $A_2$ but not in $A_1$).
I know that in logic there is something similar, but I am searching for something in set theory. Moreover, I want to underline the dependency of this uniquely derived partition from the family of sets I am given.
See also http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Given_a_family_of_sets.2C_I_can_generate_a_partition_of_their_union_set.2C_by_looking_at_their_overlapping...
Thanks to all!
A newcomer

REPLY [19 votes]: Your building blocks are known as the atoms in the Boolean algebra or field of sets generated by the $A_i$. Each building block will consist of points that have the same pattern of answers for membership in the various $A_i$. To see this, observe first that by maximality any building block will respect this equivalence and therefore be a union of such atoms. Conversely, if a set has points with two different patterns of answers, then it will contain points from two $A_i$ without being contained in their intersection. (Specifically, if $x,y\in B$ and $x\in A_i$ but $y\notin A_i$, then pick $j$ such that $y\in A_j$, and observe that $B$ meets both $A_i$ and $A_j$, buit is not contained in $A_i\cap A_j$.) So the building blocks are the atoms.
If your family is finite as you indicated, then the Boolean algebra generated by the $A_i$ consists precisely of the unions of blocks. This is the representation theorem showing that every finite Boolean algebra is isomorphic to a finite power set---the power set of the atoms. 
In the infinite case, however, the Boolean algebra generated by the $A_i$ may be atomless---it may have no atoms at all, and this is a fascinating case. Nevertheless, your blocks still form a partition, and are precisely the atoms in the infinitary-generated field of sets, still determined by my argument above by the patttern-of-answers to membership in the $A_i$.<|endoftext|>
TITLE: Probability of a Point on a Unit Sphere lying within a Cube
QUESTION [14 upvotes]: Suppose we have a ($n-1$ dimensional) unit sphere centered at the origin: $$ \sum_{i=1}^{n}{x_i}^2 = 1$$
Given some some $d \in [0,1]$, what is the probability that a randomly selected point on the sphere, $ (x_1,x_2,x_3,...,x_n)$, has coordinates such that $$|x_i| \leq d$$  for all $i$?
This is equivalent to finding the intersection of the $(n-1)$-hypersphere with the $n$-hypercube of side $2d$ centered at origin, and then taking the ratio of that $(n-1)$-volume over the $(n-1)$-volume of the $(n-1)$-hypersphere.
As there are closed-form formulas for the volume of a hypersphere, the problem reduces to finding the $(n-1)$-volume of the aforementioned intersection.
All attempts I've made to solve this intersection problem have led me to a series of nested integrals, where one or both limits of each integral depend on the coordinate outside that integral, and I know of no way to evaluate it. For example, using hyperspherical coordinates, I have obtained the following integral: $$ 2^n n! \int_{\phi_{n-1}=\tan^{-1}\frac{\sqrt{1-(n-1)d^2}}{d}}^{\tan^{-1}1} \int_{\phi_{n-2}=\tan^{-1}\frac{\sqrt{1-(n-2)d^2}}{d}}^{\tan^{-1}\frac{1}{\cos\phi_{n-1}}}\ldots\int_{\phi_1=\tan^{-1}\frac{\sqrt{1-d^2}}{d}}^{\tan^{-1}\frac{1}{\cos\phi_2}} d_{S^{n-1}}V$$
where
$$d_{S^{n-1}}V = \sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin(\phi_{n-2})\ d\phi_1 \ d\phi_2\ldots d\phi_{n-1}$$  is the volume element of the $(n-1)$–sphere. But this is pretty useless as I can see no way of integrating this accurately for high dimensions (in the thousands, say).
Using cartesian coordinates, the problem can be restated as evaluating: $$ \int_{\sum_{i=1}^{n-1}{x_i}^2\leq1, |x_i| \leq d} \frac{1}{\sqrt{1-\sum_{i=1}^{n-1}{x_i}^2}}dx_1 dx_2 \ldots dx_{n-1}$$ which, as far as I know, is un-integrable.
I would greatly appreciate any attempt at estimating this probability (giving an upper bound, say) and how it depends on $n$ and $d$. Or, given a particular probability and fixed $d$, to find $n$ which satisfies that probability.
Edit: This question leads to two questions that are slightly more general:

I think part of the difficulty is that neither spherical nor Cartesian coordinates work very well for this problem, because we're trying to find the intersection between a region that is best expressed in spherical coordinates (the sphere) and another that is best expressed in Cartesian coordinates (the cube). Are there other problems that are similar to this? And how are their solutions usually formulated?

Also, the problem with the integral is that the limits of each of the nested integrals is a function of the "outer" variable. Is there any general method of solving these kinds of integrals?

REPLY [15 votes]: Denote the median of $\max_{i=1,\dots,n}|x_i|$ on the sphere by $M_n$. It is known that the ratio between $M_n$ and $\sqrt{\log n/n}$ is universally bounded and bounded away from zero. If you take $d=M_n$ then the quantity you are looking for is exactly $1/2$. It is also known that the $\infty$-norm ($\max_{i=1,\dots,n}|x_i|$) is ``well concentrated" on the sphere meaning in particular that for any $\epsilon>0$, if you take $d<(1-\epsilon)M_n$ the probability you're interested in tends to zero and if you take $d>(1+\epsilon)M_n$, it tends to 1. Quite precise estimates are known.
The way these things are estimated is by relating the uniform distribution on the sphere to the distribution of a standard Gaussian vector:
 If $g_1,\dots,g_n$ are independent standard Gaussian variables then the distribution of 
 $$
 \frac{(g_1,\dots,g_n)}{(\sum g_i^2)^{1/2}}
 $$ is equal to the uniform distribution on the sphere. So the quantity you're looking for is the probability that $\max_{i=1,\dots,n}|g_i|\le d (\sum g_i^2)^{1/2}$.
 Since $(\sum g_i^2)^{1/2}$ is very well concentrated near the constant $\sqrt n$, this is asymptotically the same as the probability that $\max_{i=1,\dots,n}|g_i|\le d \sqrt n$. 
For general reference in a much more general setting you can look here:  Milman, Schechtman, Asymptotic theory of finite-dimensional normed spaces. For a finer treatment of the special case of the $\infty$-norm, look here: http://www.wisdom.weizmann.ac.il/mathusers/gideon/papers/ranDv.pdf.<|endoftext|>
TITLE: Surfaces all of whose geodesics are both closed and simple
QUESTION [8 upvotes]: The Zoll surfaces have the property that all of their geodesics are closed.
If one futher stipulates that all geodesics are also simple, i.e., non-self-intersecting,
does this leave only the sphere?
Apologies for the simplicity of this question, but I am not finding an answer in the literature,
and I suspect many just know this off the top of their head.  Thanks!

REPLY [12 votes]: From Guillemin's "The Radon transform on Zoll surfaces", it follows that there are deformations of $S^2$ which keep all geodesics closed AND simple.<|endoftext|>
TITLE: Reference for Grothendieck-Riemann-Roch for $Ext$.
QUESTION [9 upvotes]: A reference to the Riemann-Roch theorem can be found here: http://en.wikipedia.org/wiki/Riemann-Roch
and here: http://en.wikipedia.org/wiki/Grothendieck%E2%80%93Riemann%E2%80%93Roch_theorem 
Would you know any reference that introduces Grothendieck-Riemann-Roch for $Ext$?  (If it includes a reference for $Ext$ of elements in a derived category, all the better.  But, if not, then that's also pretty good.)  
I am looking for a reference of this formula because in my current work I would like to apply the following argument.  I have a bunch of $Ext$ groups:  $Ext^0(I^\bullet,F), Ext^1(I^\bullet,F), Ext^2(I^\bullet,F)$, where $F$ is a sheaf over a surface $X$ and $I^\bullet$ is a chain complex of sheaves over $X$ (if that is confusing, just think $I^\bullet$ is a sheaf.)  In my case, one can show that $Ext^2(I^\bullet, F)$ vanishes and that the dimension of $Ext^0(I^\bullet,F)$ is constant.  I would like to conclude that $Ext^1(I^\bullet,F)$ also has constant dimension.

REPLY [7 votes]: One has $Ext^\bullet(I,F) = H^\bullet(I^*\otimes F)$, where the dualization and the tensor product are derived and $H^\bullet$ stands for the hypercohomology. The Euler characteristic can be computed via Riemann--Roch, so if you know that all $Ext$'s with exception of two vanish, and one of this two has constant dimension, then you can conclude that the other one has constant dimension as well.<|endoftext|>
TITLE: Is it possible to partition $\mathbb R^3$ into unit circles?
QUESTION [51 upvotes]: Is it possible to partition $\mathbb R^3$ into unit circles?

REPLY [13 votes]: A very nice, explicit, elementary partition (without the Axiom of Choice) of $\mathbb{R}^3$ into geometric circles of variable radii is given in: [MR0719756] Szulkin, Andrzej. $\mathbb{R}^3$ is the union of disjoint circles. Amer. Math. Monthly 90 (1983), no. 9, 640–641.<|endoftext|>
TITLE: How many Hamiltonians Paths there are in almost regular graph ?
QUESTION [5 upvotes]: Let be $G=(V,E)$, where $V=\{1,\ldots,n\}$ and $E=\{\{i,j\}\subset V;|i-j|\leq k\}$ and $k<n$.
For which values of $k\geq 2$, can we count explicitly the number of Hamiltonian paths in $G$ ?

REPLY [5 votes]: Explicit values for $k\leq 9$ and small $n$ are given in the OEIS:
k=2: http://oeis.org/A003274 (contains some references and a generating function)
k=3: http://oeis.org/A174700
k=4: http://oeis.org/A174701
k=5: http://oeis.org/A174702
k=6: http://oeis.org/A177278
k=7: http://oeis.org/A177279
k=8: http://oeis.org/A177280
k=9: http://oeis.org/A177281<|endoftext|>
TITLE: When Have Numerology and Computational Experimentation Been Successful?
QUESTION [10 upvotes]: When has numerology been successfully used in math and science?  The Monstrous Moonshine conjecture led to a Fields medal for Borcherds.  Balmer's formula for hydrogen spectra led to the Bohr model of the atom.
We could extend this to general computational experimentation.  For example, the Birch-Swinnerton-Dyer conjecture was originally formulated based on sketchy computational results.  Gauss guessed the law of quadratic reciprocity and the prime number theorem from his calculations too.  Are there other interesting or instructive examples?

REPLY [5 votes]: There is this MO thread Experimental Mathematics about experimental mathematics, with more than 40 answers.

REPLY [4 votes]: Perhaps it would be useful to record some of the failures of numerology as well, e.g., Kepler's attempt to model the solar system by inserting the five Platonic solids among the six planets that were known in his day. Kepler's eventual success with ellipses could be viewed as numerology as well, as the theoretical basis for it had to wait for Newton.<|endoftext|>
TITLE: Existence of a vector field with a finite number of limit cycles.
QUESTION [8 upvotes]: The following question is related to the Seifert conjecture.
Let $M$ be a closed manifold with zero Euler characteristic. Is it true that each homotopy class of nowhere-zero vector fields on $M$ contains a vector field with a finite number of (stable) limit cycles (closed trajectories)? Is it easy to construct?

REPLY [7 votes]: In any dimension bigger or equal to $4$, the answer is yes. See here.
In dimension 3, the question is adressed here. In fact, in this paper neccessary and sufficient conditions to be homotopic to a non singular Morse-Smale flow are given. Morse-Smale means to have finitely many non degenerate closed orbits and that those are al the non-wandering set. In dimension 3 there are restrictions to satisfy that property, however, this paper proves that one has in the homotopy class vector fields whose minimal sets consist of finitely many periodic orbits (maybe degenerate). 
The reference is K. Yano, The homotopy class of non singular Morse Smale vector fields on 3 manifolds, Inventiones Math. (1985).<|endoftext|>
TITLE: Explicit invariants (under change of basis) of maps $V \to V \otimes V$.
QUESTION [7 upvotes]: It is standard to construct numbers associated to a linear transformation $f: V \to V$  of a finite-dimensional vector space which are invariant under change of basis.  The coefficients of the characteristic polynomial are such, and it is quite simple to see for example that the trace is invariant by equating it with the map which takes $\sum v_i \otimes w_i \in V \otimes V^* \cong Hom(V, V)$ and then evaluates $\sum w_i(v_i)$.
My question is: can one explicitly associate quantities to maps $V \to V \otimes V$ which are invariant under change of basis?  A simple dimension count indicates there should be plenty of invariants, since the orbits will have dimension roughly $d^3 - d^2$, where $d$ is the dimension of $V$, but I have yet to construct a single one.

REPLY [5 votes]: This is essentially a rephrazing of Victor Protsak's reply (which I read too late).
In characteristic zero the answer is in principle given by classical invariant theory. Let $(e^i)$, $i=1,...,n$ be a basis for V. Then
coordinates on the space $Hom(V,V \otimes V)$ are given by tensors $c^i_{jk}$. Such a
tensor sends the vector $\lambda_i e^i$ to $c^i_{jk}\lambda_i e^j \otimes e^k$ (using the summation condition for repeated indices).
In addition let $\epsilon^{i_1,...,i_n}$ be the tensor which is $1$ if the $i_1,...,i_n$
form an even permutation of $1,...,n$, which is $-1$ for an odd permutation and which is zero otherwise. Similarly for $\epsilon_{i_1,...,i_n}$. 
SL(V) invariants are now given by expressions in $c,\epsilon$ which have no free indices
(it is possible to do this via graphs). One has to be careful since many expressions
are zero for symmetry reasons.
There are clearly no linear invariants. If $dim(V)=2$ then I believe $\epsilon^{kl}c^i_{kj}c^j_{il}$ is an honest quadratic invariant. Another one
is $\epsilon^{kl}c^i_{ik}c^j_{lj}$. I.e. the determinant of the left partial
trace with the right partial trace.
EDIT:
The two dimensional case can also be viewed in a less abstract way. If $dim V=2$ then
$V\cong V^\ast$ as $SL(V)$ representation. Furthermore by Clebsch Gordan we have $V^* \otimes V\otimes V\cong
S^3V\oplus V\oplus V$. This is the problem of finding the generating concomitants for binary cubic forms. The answer can be found in Grace and Young. In this setting I
see only one quadratic invariant, namely the determinant between the two copies of V.
This means that the two quadratic invariants identified above are linearly
dependent which does not appear obvious. But they are indeed. If you write them
out explicitly you get (up to sign)
$c^1_{11}c^1_{12}-c^1_{21}c^1_{11}+c^1_{12}c^2_{12}-c^2_{21}c^1_{21}+c^2_{12}c^2_{22}
-c^2_{22}c^2_{21}$

REPLY [2 votes]: I've been thinking about the case when $V$ is $2$-dimensional. Here, I claim, is a parametrization of the $GL(V)$-orbits of 'generic' linear maps $F:V\to V\otimes V$:
To numbers $p, a, b, c$ we associate a map which acts on a basis $v, w$ by:
$F(v)= -p(v\otimes v)+ w\otimes w$
$F(w)= (v\otimes w-w\otimes v)+ a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$.
To put a generic $F$ in this form, write it as $F_++F_-$, symmetric plus antisymmetric. $F_-$ is given by $x\mapsto v\otimes x-x\otimes v$ for a unique $v\in V$. Assume $v\ne 0$. 
Choose $w$ such that $v,w$ is a basis, but remember we are free to replace $w$ by $sv+tw$ for any scalars $s$ and $t\ne 0$. Think of the symmetric tensor $F_+(v)$ as a homogeneous quadratic polynomial in indeterminates $v,w$. Assume it has a $w^2$ term. Replacing $w$ by suitable $tw$ we can make that term $w^2$. Now replacing $w$ by suitable $sv+w$ ("completing the square") we can eliminate the $vw$ term. So $F_+(v)= -pv^2+w^2=-p(v\otimes v)+ w\otimes w$ for some $p$. And $F_+(w)= av^2+2bvw+cw^2=a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$ for some $a,b,c$. This gives the formulas above for $F$.
I used up all the choices I had except that $w$ can still be changed to $-w$, which would change $p,a,b,c$ to $p,-a,b,-c$. So the "parametrization" is actually two to one.<|endoftext|>
TITLE: Computational complexity of the word problem for semi-Thue systems with certain restrictions
QUESTION [6 upvotes]: The word problem (from wikipedia).

Given a semi-Thue system T: = (Σ,R)
  and two words , can u be transformed
  into v by applying rules from R?

This problem is undecidable, but with a certain restriction, it is decidable.
The Restriction:
All the rules in R are of the form A->B where A and B are string of the same length.
What is the computational complexity of this problem?

REPLY [6 votes]: Joel David Hamkins has done most of the work in his answer, but just to wrap it up, the problem is PSPACE complete. As Joel has explained, the given
semi-Thue problem can be simulated by a space-bounded, nondeterministic
Turing machine, and conversely. So the problem is (nondeterministic)
PSPACE complete. But Savitch's theorem says that this is the same as
PSPACE complete.<|endoftext|>
TITLE: Learning Algebra & Group Theory on my own
QUESTION [12 upvotes]: I'm learning Algebra & Group Theory, casually, on my own.  Professionally, I'm a computer consultant, with a growing interest in the mathematical and theoretical aspects.  I've been amazed with the applications of Algebra to CS things like cryptography, coding theory, and combinatorial search.
I'm using Artin's book, and, as high quality as it is, I find it leaves me dry.  A lot of names for new concepts, and proving that one thing follows from others.  But I don't see the powerful tool I envisioned that I'd gain taking shape in my hand.
Perhaps because I'm learning independently and casually, I need a book which will give me tools, showing me how they can be used to solve concrete problems outside of algebra.
What book(s) do you recommend?

REPLY [2 votes]: First, my background - a math major and a retired software engineer.
It takes time and efforts to go from theory to application. You need to convert yourself from a computer scientist to a mathematician. Then convert back to computer scientist. A hard process.  Please be patient and give yourself more time.
All the books the answerers recommend seem to be good although I did not read all of them. I am reading Dummit and Foote’s Abstract Algebra because I am studying for math, not for computers. Please note the word I use – ‘for’.
To answer your particular question, I would recommend P.M. Cohn’s two volumes – Basic Algebra and Further Algebra and Applications.  These two books contain enough theory and also pay attention to applications. Cohn tried to tell the reader how to apply the theory. You may not find too much number theory in those two books. For that, you need a good number theory book.
EDIT:
If I were you, I would go to a university library and take a glance at all those books people here recommend before I buy any book. Everybody has his own taste.<|endoftext|>
TITLE: Singmaster's conjecture
QUESTION [19 upvotes]: Has any work been done on Singmaster's conjecture since Singmaster's work?
The conjecture says there is a finite upper bound on how many times a number other than 1 can occur as a binomial coefficient.
Wikipedia's article on it, written mostly by me, says that

It is known that infinitely many numbers appear exactly 3 times.
It is unknown whether any number appears an odd number of times where the odd number is bigger than 3.
It is known that infinitely many numbers appear 2 times, 4 times, and 6 times.
One number is known to appear 8 times. No one knows whether there are any others nor whether any number appears more than 8 times.
Singmaster reported that Paul Erdős told him the conjecture is probably true but would probably be very hard to prove.

REPLY [22 votes]: There is an upper bound of $O\left(\frac{(\log n)(\log \log \log n)}{(\log \log n)^3}\right)$  due to Daniel Kane: see "Improved bounds on the number of ways of expressing t as a binomial coefficient," Integers 7 (2007), #A53 for details.<|endoftext|>
TITLE: Good differential equations text for undergraduates who want to become pure mathematicians
QUESTION [49 upvotes]: Alright, so I have been taking a while to soak in as much advanced mathematics as an undergraduate as possible, taking courses in algebra, topology, complex analysis (a less rigorous undergraduate version of the usual graduate course at my university), analysis, model theory, and number theory. That is, I have taken enough 'abstract' (proof-based) mathematics courses to fall in love with the subject and decide to pursue it as a career. 
However, I have been putting off taking a required ordinary differential equations course (colloquially referred to as 'calc 4', though this seems inappropriate) which will likely be very computational and designed to cater to the overpopulation of engineering students at my university. 
So my question is, for someone who might have to actually concern themselves with the theory behind the 'rules' and theorems which will likely go unproven in this low-level course (likely of questionable mathematical content), what might be a decent supplementary text in ODE? That is, something substantive to counter-balance the 'ODE for students of science and engineering'-type text I will have to wade through. I want to study algebraic geometry further (I have gone through Karen Smith's text and the first part of Hartshorne), so something which goes from basic material through differential forms and related material would be nice. 
Thanks! (and yes, it's embarrassing that I still haven't taken the 200-level ODE course, but I have been putting it off in favor of more interesting/rigorous courses... but now there's that whole graduation requirements issue). 
--Lambdafunctor

REPLY [4 votes]: A classic is Coddington and Levinson. There is also a much simpler and smaller book by Coddington alone.<|endoftext|>
TITLE: Group homomorphisms and maps between function spaces
QUESTION [8 upvotes]: Let G and H be locally compact groups, and let $\theta:G\rightarrow H$ be a continuous group homomorphism.  This induces a *-homomorphism $\pi:C^b(H) \rightarrow C^b(G)$ between the spaces of bounded continuous functions on H and G.
If $\theta$ is an injection with closed range, then as locally compact groups are normal, you can use the Tietze extension theorem to show that $\pi$ is a surjection.
Conversely, if $\pi$ surjects, then $\theta$ must be an injection.  Need $\theta(G)$ be closed in H??
(If G and H are just locally compact spaces, and $\theta$ just a continuous map, then no: you could let G be non-compact and $H=\beta G$ the Stone-Cech compactification, with $\theta$ being the canonical inclusion.  The resulting map $\pi$ is just $C^b(H) = C(\beta G) \rightarrow C^b(G) = C(\beta G)$, which is the identity, once suitably interpreted.  Of course, here $\theta$ has open range, and in a topological group, an open subgroup is closed, so maybe there's hope... hence my question).
More thoughts: As in my comment, we can extend $\theta$ to a map $\tilde\theta:\beta G\rightarrow\beta H$ between the Stone-Cech compactifications: this induces the map $\pi:C(\beta H)\rightarrow C(\beta G)$.  As these are compact, it follows that $\pi$ is surjective if and only if $\tilde\theta$ is injective.  By replacing $H$ with the closure of $\theta(G)$, we may suppose that $\theta$ has dense range: this forces $\tilde\theta$ to be a bijection, and hence a homeomorphism.  So is it possible for $\theta$ to be an injection with dense range, and $\tilde\theta$ a homeomorphism, but without $\theta$ being onto?  For example, certainly H cannot be compact, as then $\beta G$ would be a topological group, which is possible only if $G$ is compact (I think).

REPLY [2 votes]: Inspired by Victor's idea:
Setup: Let $\theta:G\rightarrow H$ be a continuous dense range map between locally compact (Hausdorff) spaces.  Let $G'=\theta(G)$ with the subspace topology from H.  Let $\pi:C^b(H) \rightarrow C^b(G)$ be the pull-back of $\theta$.
Lemma: The collection of sets of the form $f^{-1}((1/2,\infty))$, where $f$ is a continuous map $G\rightarrow [0,1]$ vanishing at infinity, is a base for the topology on $G$.
Proof: Let $G_\infty$ be the one-point compactification, let $s\in G$, and let $U\subseteq G$ be open with compact closure (which is okay, as $G$ is locally compact) with $s\in U$.  By Urysohn, there exists a continuous $f:G_\infty\rightarrow [0,1]$ with $f(s)=1$ and $f|_{G_\infty\setminus U} \equiv 0$.  Then $f|_G$ vanishes at infinity, and $s\in f^{-1}((1/2,\infty)) \subseteq U$.  Clearly every open set can now be written as a union of these special open sets.  QED.
Claim: Suppose that $\pi$ is surjective (so $\pi$ is infact a bijection).  Let $\phi:G\rightarrow G'$ be $\theta$, considered as having codomain $G'$.  Then $\phi$ is a homeomorphism, and so $\theta(G)$ is open in $H$.
Proof: We show that $\phi$ is open.  Let $f\in C^b_{\mathbb R}(G)$, and let $g=\pi^{-1}(f)\in C^b(H)$, so that $g(\theta(s)) = f(s)$ for $s\in G$.  Let $U=f^{-1}((1/2,\infty))$ so $\theta(U) = \{ t\in H : \exists s, \phi(s)=t, f(s)>1/2 \}$ $= \{ t\in G' : f(\phi^{-1}(s))>1/2\}$ $= \{ t\in G' : g(t)>1/2 \} $ $= G' \cap g^{-1}((1/2,\infty))$.  Thus $\phi(U)$ is open, as $G'$ has the subspace topology.  By the lemma, this does show that $\phi$ is open.  Then $G'$ is itself locally compact, and so as $G'$ is dense, it must be open in $H$.  QED.
Claim: If additionally $G$ and $H$ are groups and $\theta$ a homomorphism, then $\theta$ is a surjection.
Proof: An open subgroup is closed.  QED.
If this is all correct, then I'd be a little surprised if this wasn't known (say, the stuff not about groups).  Any ideas???<|endoftext|>
TITLE: A plausible positivity
QUESTION [5 upvotes]: After getting stuck with the
previous positivity
(it probably sounds too complex),
I would like to give a version of the problem which is of most interest to me.
Consider a sequence of real numbers
$a_1,a_2,\dots,a_n,\dots$ with absolute values bounded above by the first term $a_1=a>0$,
which satisfies, for all $n=1,2,\dots$,
$$
|A_n|\le A \qquad\text{where}\quad A_n=a_1+a_2+\dots+a_n.
$$
In addition, assume that infinitely many terms of the sequence are nonzero.
These settings and Dirichlet's convergence test
guarantee that the series
$$
\sum_{n=1}^\infty\frac{a_n}n
$$
converges.
Assume, in addition, that
$$
\max_{1\le k\le n}A_k+\min_{1\le k\le n}A_k\ge0
\qquad\text{for all}\quad n=1,2,\dots.
$$
The problem is to show that
$$
\sum_{n=1}^\infty\frac{a_n}n>0
$$
and to provide, in terms of $a$ and $A$, a lower (strictly positive) bound for the series.
(The latter is optional, as I am not sure that such a bound exists.)

REPLY [8 votes]: The sum $\sum a_n/n$ can be negative. Below I construct a finite sequence; one can always add a negligibly small tail to get infinitely many non-zeroes.
Begin with $a_1=1$ and $a_2=-1$.
This gives $A_2=0$ and the partial sum of the main series is $1-1/2=1/2$.
Then, repeat 100 times the following procedure:
Pick an integer $k$ larger than the length of the sequence so far.
Extend $(a_n)$ by zeroes up to $n=10k-1$.
Then set $a_n=1$ for all $n$ from $10k$ to $11k-1$ and $a_n=-1$ for all $n$ from $11k$ to $13k-1$. The $k$ ones contribute less than $1/10k$ each to the main series $\sum a_n/n$, and this is less than $1/10$ in total. The $2k$ negative ones contribute absolute value at least $1/13k$ each, this sums up to at least $2/13$ of negative amount. So the partial sum of the main series went down by at least $2/13-1/10>1/20$.
But we have $A_n=-k$ now (for $n=13k-1$). To fix this, extend $(a_n)$ by a huge amount of zeroes, followed by $k$ ones, so that the contribution of these ones to the main sum is less than $1/100$. 
Now we extended the sequence so that the last $A_n$ is zero again but the partial sum of the main series went down by at least $1/30$. Choose the next $k$ and repeat (finitely many times!).
[Edit] Since the sequence is finite, one can define $A=\max|A_n|$ to satisfy the condition $|A_n|\le A$. The $\max+\min$ condition is immediate from the construction.<|endoftext|>
TITLE: Bourbaki theory of isomorphism, examples of untransportable formulas
QUESTION [5 upvotes]: In their book "Theory of sets" Bourbaki suggested a general theory of isomorphism.
(See also http://www.tau.ac.il/~corry/publications/articles/pdf/bourbaki-structures.pdf )
The example of an untransportable relation (i.e. formula) in the book involves 2 principal base sets.
Are there examples of untrasportable formulas when we allow only one principal base set?

REPLY [3 votes]: An example of untrasportable sentence, when there is only one principal base set X, may be the following one: 
All elements of the set X are finite sets,
Because, by definition, the truth value of a transportable sentence must be preserved under all bijections from the set X. Obviously, there  exists a bijection from X to a set Y, where not all elements of Y are finite sets.
A simpler example is "the set X contains the empty set". 
There is a paper "Sentences of type theory: the only sentences preserved under isomorphisms" by Victoria Marshall and Rolando Chuaqui - see The Journal of symbolic Logic, vol 56, #3, Sep 1991.<|endoftext|>
TITLE: Are all Galois cohomology groups also étale cohomology groups?
QUESTION [21 upvotes]: Let $K$ be a field and $K^s$ a separable closure of $K$, and let $\mathcal{F}$ be a sheaf on $\mathrm{Spec}(K)$ (in the étale topology).
By Grothendieck's Galois Theory, we have the isomorphism
$$H_{\text{et}}^k(\mathrm{Spec}(K),\mathcal{F}) \cong H^k(\mathrm{Gal}(K^s/K),\mathcal{F}_{\mathrm{Spec}(K^s)})$$
i.e. the étale cohomology groups of $\mathrm{Spec}(K)$ with coefficients in $\mathcal{F}$ correspond to the Galois cohomology groups of the absolute Galois group of $K$ acting on the stalk of $\mathcal{F}$ at the (separably closed) geometric point $\mathrm{Spec}(K^s) \to \mathrm{Spec}(K)$.
Question: Is there any similar interpretation of more general Galois cohomology groups in terms of étale cohomology groups? For instance, can we write the cohomology groups $H^k(\mathrm{Gal}(L/K),L^\times)$ directly in terms of étale cohomology groups?
I do not see how this would be possible, given that topos-theoretically we can just restrict our attention to points with separably closed residue field for the geometric points of the relevant étale topos.
Is there any chance we can describe more Galois cohomology groups if we consider other topologies on a category of schemes (Edit: preferably to obtain a uniform description using a single site)?
Finally, if this is not possible, are there any "higher dimensional" appearances of this discrepancy? This is deliberately vague, but I am thinking about some sort of cohomology groups that restrict to étale cohomology groups when considering just separably closed geometric points (via some sort of higher dimensional version of Grothendieck's Galois Theory).

REPLY [24 votes]: If $L/K$ is a Galois extension, then one can define a site in which the objects are intermediate fields $K \subseteq E \subseteq L$ which are finite over $K$. Or, you can take finite étale algebras $A$ over $K$ with a homorphism of $K$-algebras $A \to L$. The topology is the étale topology. These two sites define the same topos. There is an equivalence between sheaves on either of these sites and discrete $\mathrm{Gal}(L/K)$-modules, hence cohomology coincides with Galois cohomology.
Does this answer your question? I am not sure I understand what you have in mind.<|endoftext|>
TITLE: Uppercase Point Labels in High-School Diagrams: from Euclid?
QUESTION [17 upvotes]: I wonder if the convention of labeling points in geometric
diagrams with uppercase symbols ultimately derives from 
Greek mathematics, which was originally written in
"majuscule" (uppercase) Greek script (in contrast to the "minuscule"
script that was introduced much later (9th century?)).
Certainly Euclid and Archimedes used only uppercase,
and all of Descartes diagrams in La Geometrie (1637)
follow the same convention.
It seems that middle- and high-school textbooks continue
to use uppercase labels (is this only in the U.S.?), 
but college texts do not follow
this as rigidly.  This was brought home to me when I wrote
a chapter for high-school teachers and the editors changed
all my lowercase vertex labels to uppercase.
I much prefer lowercase for point labels, although
I do not quite know why I have this preference.
(Maybe because uppercase seems like SHOUTING?)
But when writing for an audience accustomed to
a particular convention, it seems prudent to follow that convention.
My questions are: 
(1) Is the Greek majuscule script the origin of the uppercase
diagram-labeling convention?
(2) In so far as I am correct that the uppercase convention
is followed up to high school but dissolves at more advanced
levels, why does it persist to one level but dissolve beyond?

REPLY [4 votes]: As mentioned by the OP in the comments, Reviel Netz is an academic who thinks about these kinds of questions.  For example, he has a great chapter in "The Archimedes Codex" (by Netz and his coauthor William Noel) on the topic of how the Greeks labelled their figures.  He explains how the philological method - comparing and contrasting different manuscripts to reconstruct a common origin manuscript - can be applied to the various figures in the extant codices of the Archimedean tradition.  He applies the philological method to construct what the figures drawn on the beaches of Syracuse in the third century BCE might have looked like.
However, Netz also notes the innovations of the medieval scribes.  He states (p. 111):

Medieval scribes, at this point, took some crucial steps in the invention of more effective typesetting interfaces.  The use of several cases - an uppercase alongside a lower case - is of great value.  It allows us to separate the letters referring to the diagram (which remains in upper case) from the rest (which is now in lower case).

Thus, Netz appears to argue that the real innovation is not so much writing in majuscule for the figures, but in writing a mix of majuscule and miniscule at the same time.  As pointed out by others, this is a medieval construct.  
(Netz states in passing that the ancient Chinese used different colors to identify the components of their figures.)<|endoftext|>
TITLE: Status of Beal, Granville, Tijdeman-Zagier Conjecture
QUESTION [10 upvotes]: The Beal, Granville, Tijdeman-Zagier Conjecture, i.e.
If $A^x+B^y=C^z$ , where $A, B, C, x, y,z$ are positive integers and $x,
y$ and $z$ are all greater than $2$, then $A, B$ and $C$ must have a common prime
factor.
... and its associated $1,000,000 prize for proof or disproof seems to have gone largely unnoticed in the mathematics community. Please answer with (A) references to past or ongoing research or (B) references to equivalent forms of this conjecture known prior to Andrew Beal posing it in 1993.

REPLY [11 votes]: At present there is no real strategy for the general problem.  But progress on individual cases, or families of cases, keeps moving along.  For instance, Poonen, Schaeffer, and Stoll handled the case x^2 + y^3 + z^7 in 2005; last year, Mike Bennett, Nathan Ng and I finished off the case x^2 + y^4 = x^p and David Brown did x^2 + y^3 + z^10.<|endoftext|>
TITLE: Galois groups via cohomology
QUESTION [6 upvotes]: I would like to know about references for the following result (point 3): 
Let $K/k$ be a normal extension (I am interested in number fields, but everything should work in fields of characteristic $\ne 2$) with Galois group $G$, and let 
$L = K(\sqrt{\mu}\,)$ be a quadratic extension.

$L/k$ is normal if and only if for every $\sigma \in G$ there is an
$\alpha_\sigma \in K$ such that $\mu^{\sigma-1} = \alpha_\sigma^2$.
Let $L/k$ be normal. Then we can define an element $[\beta]$ in the second cohomology
  group $H^2(G,\mu_2)$ with values in $\mu_2 = \{-1,+1\}$ by setting
  $$ \beta(\sigma,\tau) = 
           \alpha_\sigma^\tau \alpha_\tau \alpha_{\sigma\tau}^{-1}. $$
If $L/k$ is normal, then $[\beta]$ is the element of the second cohomology group 
attached to the group extension
$$  1 \rightarrow \mu_2 \rightarrow  Gal(L/k) \rightarrow  Gal(K/k) \rightarrow   1.  $$

REPLY [4 votes]: I don't have a reference, but it does not seem too hard.

Assume $L/k$ normal, and take $\sigma \in G$, which can be extended to an element of $Gal(L/k)$. Then $\sigma(\sqrt{\mu})/\sqrt{\mu} \in L$, and if $\gamma$ is the nontrivial element of $Gal(L/K)$, $\sigma^{-1} \gamma \sigma$ is trivial on $K$, and being nontrivial on $L$ it has to be equal to $\gamma$. So $\gamma\left( \sigma(\sqrt{\mu})/\mu \right) = \sigma(\gamma(\sqrt{\mu}))/\gamma(\sqrt{\mu}) = \sigma(\sqrt{\mu})/\sqrt{\mu}$, so we can take $\alpha_{\sigma} = \sigma(\sqrt{\mu})/\sqrt{\mu}$, it is an element of $K$.
Now for the other way, take $\tilde{\sigma} \in Gal(\bar{L}/k)$. Denote the restriction of $\tilde{\sigma}$ to $K$ by $\sigma$. Then $\sigma(\sqrt{\mu})/\sqrt{\mu}= \pm \alpha_{\sigma}$ (this equality is in $\bar{L}$), so $\sigma(\sqrt{\mu}) = \pm \alpha_{\sigma} \sqrt{\mu} \in L$, so $L$ is normal.
Consider a set-theoretic section $\sigma \mapsto \tilde{\sigma}$ for the surjective morphism $Gal(L/k) \rightarrow G$. Then the (up to a coboundary) 2-cocycle $\beta_0$ associated to the group extension is given by the formula $\tilde{\sigma} \tilde{\tau} = \beta_0(\sigma,\tau) \widetilde{\sigma \tau}$. Evaluating at $\sqrt{\mu}$ gives the equality between $\beta$ and $\beta_0$, if for every $\sigma \in G$, $\alpha_{\sigma} = \tilde{\sigma}(\sqrt{\mu})/\sqrt{\mu}$. You can always choose your section so that it is the case (change of section = associating a sign to each $\sigma \in G$).

EDIT: There's a left/right action problem, because $\beta_0(\sigma,\tau) = \sigma(\alpha_{\tau}) \alpha_{\sigma} \alpha_{\sigma \tau}^{-1}$ with what I wrote. I think it has to do with the fact that you use exponential notation, so somehow your action is on the right? Maybe you define $x^{\sigma} = \sigma^{-1}(x)$? Otherwise the definition of $\beta$ doesn't make it a 2-cocycle, with the definitions I know. Could you clarify?<|endoftext|>
TITLE: When can boundedness be characterized topologically in Metric spaces?
QUESTION [5 upvotes]: Let H be a separable and infinite-dimensional Hilbert space. Is every closed subset of H homeomorphic 
to some closed and bounded subset of H?

REPLY [11 votes]: It is an old result of Klee saying that the infinite-dimensional Hilbert space is homeomorphic with both its unit sphere and its closed unit ball. See e.g. http://www.ams.org/journals/bull/1961-67-03/S0002-9904-1961-10589-2/S0002-9904-1961-10589-2.pdf , and the references therein.<|endoftext|>
TITLE: Does (the ideal class of) the different of a number field have a canonical square root?
QUESTION [37 upvotes]: A theorem of Hecke (discussed in this question)
shows that if $L$ is a number field, then the image of the
different $\mathcal D_L$ in the ideal class group of $L$ is a square.
Hecke's proof, and all other proofs that I know, establish this essentially by
evaluating all quadratic ideal class characters on $\mathcal D_L$ and showing
that the result is trivial; thus they show that the image of $\mathcal D_L$
is trivial in the ideal class group mod squares, but don't actually exhibit a square root
of $\mathcal D_L$ in the ideal class group.
Is there any known construction (in general, or in some interesting cases) of an ideal
whose square can be shown to be equivalent (in the ideal class group) to $\mathcal D_L$.
Note: One can ask an analogous question when one replaces the rings of integers by
Hecke algebras acting on spaces of modular forms, and then in some situations I know
that the answer is yes.  (See this paper.)  This gives me some hope that there might be a construction in
this arithmetic context too.  (The parallel between Hecke's context (i.e. the number
field setting) and the Hecke algebra setting is something I learnt from Dick Gross.)
Added: Unknown's very interesting comment below seems to show that the answer is "no",
if one interprets "canonical" in a reasonable way.   In light of this,  I am going to ask another question which is a tightening of this one. 
On second thought: Perhaps I will ask a follow-up question at some point, but I think I need more time to reflect on it.  In the meantime, I wonder if there is more that one can say about this question, if not in general, then in some interesting cases.

REPLY [9 votes]: Recently the following was explained to me by Melanie Wood.  (What follows is a summary of some results from her preprint here; see in particular the discussion at the top of page 3.)
If $f$ is a primitive irreducible binary $n$-ic form with coefficients in $\mathbb Z$, then we can consider the locus $S_f$ in $\mathbb P^1_{\mathbb Z}$ that $f$ cuts out.  This will be finite over $\mathbb Z$, and hence is equal to Spec $R$, where $R = H^0(S_f, \mathcal O)$.  The finite $\mathbb Z$-algebra $R$ will be an order in a number field of degree $n$. Not all orders in all number fields arise this way (if $n >3$), but certainly some do!
Now it turns out that restriction of $\mathcal O(n-2)$ to $S_f$ (which is an invertible sheaf on $S_f$, and hence gives an ideal class of $R$) is canonically isomorphic to the inverse different of $R$, and hence if $n$ is even then we can consider the restriction to $S_f$ of $\mathcal O((n-2)/2)$, and so obtain a square root of the inverse different of $R$.
As is explained in the above linked preprint, $R$ alone does not suffice to determine $f$; rather $f$ is determined by $R$ together with some extra ideal-theoretic strucure. So (as far as I know) this construction of the square root does not depend on $R$ alone, but on extra data.
Nevertheless, it gives a very interesting answer to my question in those cases where it applies.<|endoftext|>
TITLE: Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules
QUESTION [8 upvotes]: Let $A$ be an algebra over an algebraically closed field $k.$ Recall that if $A$ is
a finitely generated module over its center, and if its center is a finitely generated
algebra over $k,$ then by the Schur's lemma all simple $A$-modules are finite dimensional
over $k.$ 
Motivated by the above, I would like an example of a $k$-algebra $A,$ such that:
1) $A$
has a simple module of infinitie dimension over $k,$ 
2) $A$ contains
a commutative finitely generated subalgebra over which $A$ is a finitely generated 
left and right module. 
Thanks in advance.

REPLY [4 votes]: Doc, this is a stinker. Your condition (2) forces your algebra to be finitely generated PI, and every little hare knows that simple modules over such algebras are finite-dimensional. See 13.4.9 and 13.10.3 of McConnell-Robson...<|endoftext|>
TITLE: What is the proper initiation to the theory of motives for a new student of algebraic geometry?
QUESTION [35 upvotes]: A preliminary apology is in order: I realize that most of my contributions to this site are in the form of reference requests. I understand that this makes it seem as though I do nothing more than sit around most of the time, soaking in as much advanced mathematics as possible, despite my position as a lowly undergraduate. In all actuality, this couldn't be more accurate; I really do just sit around reading maths most of the time. 
Alright, now that I put that out there, I am curious as to where I might find a coherent treatment of the theory of motives; one which is below the level of a professional mathematician and roughly suited for readers of Hartshorne or Eisenbud/Harris's wonderful scheme theory text. That is, I want to understand the discipline which I hear extolled as beautiful and complex by researchers in the field, but which is notoriously abstruse and difficult to learn/understand. I wonder if expositions of the theory of motives are necessarily highly technical, or if it is approachable to the ambitious advanced undergraduate. 
Thank you again, MO community, for imparting your wisdom regarding good references. It is very much appreciated =)

REPLY [3 votes]: Here is a very understandable introductory article by R. Sujatha. For a beginning student this is good.
In my case, after that article, my next encounter with motives was with the more precise definition of a motive from the initial parts of Deligne's monograph “Le Groupe Fondamental de la Droite Projective Moins Trois Points”. It even sort of defines a mixed motive; in fact it is the only definition of mixed motive that I know.
Read the Mathscinet review and also, Jordan Ellenberg's opinion on this remarkable paper of Deligne. I myself was astonished when I first looked into it and saw how much stuff was contained in it.
Deligne's paper "Formes modulaires et représentations $l$-adiques" proving that the Weil conjectures imply the Ramanujan conjecture, is almost close to the theory of motives even though it does not explicitly mention motives. Here the representations of the absolute Galois group on the étale, or rather on the $\ell$-adic, cohomology is considered. This might give some starting insight into the Galois representations approach to motives.<|endoftext|>
TITLE: Why is Cohen's result insufficient to settle CH?
QUESTION [6 upvotes]: OK, Cohen has constructed a model in which both ZFC and ~CH are true. Isn't this model an answer to the continuum problem? Hasn't he showed that it is indeed possible to construct a set with cardinality between that of the integers and that of the reals? Why is it still not considered sufficient to settle CH? Why is one model not enough? Why for all models? In other words, why do we have to answer whether "ZFC |- CH" instead of just "CH" itself?

REPLY [3 votes]: CH has not been "settled" (and there are obstacles to settling it) in any of the following senses:
$\quad 1.$ Finding a compellingly natural extension of standard set theory (more natural than ZF+CH) that decides CH, i.e., proves CH or proves its negation.
Here the main approach is blocked, because large cardinal axioms don't directly decide CH.
$\quad 2.$ Finding compelling arguments for replacing set theory, wherever it is used (e.g., as a foundation or formalization scheme), with set-theory-plus-CH.
This approach is blocked by the lack of "material consequences" of CH.  For example, the set of true first-order sentences of arithmetic is not affected by assuming CH, so there would be no concrete statement such as the Twin Prime Conjecture that could be proved only with the use of CH.  For similar reasons, it is unlikely that there exists a proof of any concrete statement that is much shorter or easier with CH than without it.
$\quad 3.$ Finding a compellingly natural alternative to standard axiomatic set theory (one whose theorems are not a subset or superset of the theorems in ZFC, and which comes to be preferred over ZFC) that can formulate and decide CH.  
This development would be a lot more significant than deciding CH, and would presumably affect a large number of other questions.  So to the extent that this possibility is relevant it should be discussed directly, and CH itself is immaterial.  More on this argument in the earlier thread: Knuth's intuition that Goldbach might be unprovable
(The same comments also apply to the negation of the Continuum Hypothesis; everything above is phrased in terms of CH only to avoid clunky qualifiers in the sentences.) 
EDIT:  I am not counting another possibility, where partial answers to CH are accepted as the best that can be done, or the original problem comes to be seen as the wrong formulation (but better formulations are decidable in ZFC).   For example, there are theorems to the effect that "any reasonably defined set of real numbers satisfies CH", and PCF theory that tries to capture the ZFC content of set theoretic cardinality questions while avoiding independence phenomena.   For purposes of this answer I refer only to approaches that would "settle" CH by exhibiting a formal system that is strong enough to derive CH or its negation, and is also adequate in other respects, such as being extremely psychologically or pragmatically compelling compared to systems (such as ZFC) that don't decide CH.<|endoftext|>
TITLE: Structure Theorem for finitely generated commutative cancellative monoids?
QUESTION [8 upvotes]: Is there a Structure Theorem for finitely generated commutative cancellative monoids?
Of course they can be densely embedded into a finitely generated abelian group, whose structure is known. Also, in the book of J. C. Rosales and P. A. García-Sánchez there are some special embedding theorems: If the monoid is torsionfree, it even embeds to some free abelian group, and if the monoid is also reduced, it embeds in some free commutative monoid.
But I want to know if it is possible to give a complete classification (for example, in terms of generators and relations, as in the case of groups).

REPLY [4 votes]: Finitely generated commutative semigroups have decidable elementary theories and the isomorphism problem for them is decidable. This means that some sort of description does exist. For example, with every pair of finitely generated commutative monoids, one can associate two tuples of integer matrices such that the monoids are isomorphic iff the tuples are conjugate (Taiclin), and then one can use a result of Grunewald and Segal or Sarkisyan. For references look at the survey http://www.math.vanderbilt.edu/~msapir/ftp/pub/survey/survey.pdf (the part about semigroups there).<|endoftext|>
TITLE: How does one use the Poisson summation formula?
QUESTION [55 upvotes]: While reading the answer to another Mathoverflow question, which mentioned the Poisson summation formula, I felt a question of my own coming on. This is something I've wanted to know for a long time. In fact, I've even asked people, who have probably given me perfectly good answers, but somehow their answers have never stuck in my brain. The question is simple: the Poisson summation formula is incredibly useful to many people, but why is that? When you first see it, it looks like a piece of magic, but then suddenly you start spotting that people keep saying "By Poisson summation" and expecting you to fill in the details. In that respect, it's a bit like the phrase "By compactness," but the important difference for me is that I can fill in the details of compactness arguments.
What I would like to know is this. What is the "trigger" that makes people think, "Ah, Poisson summation should be useful here"? And is there some very simple example of how it is applied, with the property that once you understand that example, you basically understand how to apply it in general? (Perhaps two or three examples are needed -- that would obviously be OK too.) And can one give a general description of the circumstances where it is useful? (Anyone familiar with the Tricki will see that I am basically asking for a Tricki article on the formula. But I don't mind something incomplete or less polished.)
For reference, here is a related (but different) question about the Poisson summation formula: Truth of the Poisson summation formula

REPLY [5 votes]: This is really a comment that could be added on to existing answers, but I can't add comments. As a particularly simple example that follows Ben's answer, as well as an example of SandeepJ's observation about changing the 'size' of the sum, consider counting solutions to some linear equation L = 0 in a subset A of a finite abelian group G. Let's say $L = x + y + z$. To count solutions to $L(x,y,z) = 0$ with $x,y,z \in A$ one could look at the sum $\sum_{(x, y, z) \in H} F(x, y, z)$, where $F(x, y, z) = 1_A(x)1_A(y)1_A(z)$ and $H \leq G^3$ is the subgroup of solutions $(x,y,z) \in G^3$ such that $L(x,y,z) = 0$. Up to some normalization factors, Poisson summation then says something like 
$$\sum_{\mathbf{x} \in H} F(\mathbf{x}) = \sum_{\chi \in H^\perp} \widehat{F}(\chi) = \sum_{\gamma \in \widehat{G}} \widehat{1_A}(\gamma)^3,$$
since $H^\perp \cong \widehat{G}$. So the 'two-dimensional' sum on the left has become a 'one-dimensional' sum on the right.
This is of course the basis of a lot of arguments in additive combinatorics, and as Ben said one might fruitfully consider the term $\widehat{F}(0)$ in the sum on the right-hand side above.<|endoftext|>
TITLE: Natural morphism appearing in Grothendieck spectral sequence
QUESTION [5 upvotes]: Assume we are in the setting of the Grothendieck spectral sequence (Weibel, 5.8): $G : A \to B, F : B \to C$ are left exact functors such that $G$ sends injective objects to $F$-acyclic objects. Now the edge maps should be some natural maps $(R^p F)(GA)  \to R^p(FG)(A)$ and $R^q(FG)(A)  \to F((R^q G)(A))$. But how are they defined? 
I've managed to write down a definition in the special case of the Leray spectral sequence, there $G : Sh(X)  \to Sh(Y)$ is a direct image functor of a map $f : X \to Y$ and $F : Sh(Y)  \to Ab$ is a global section functor: The first map is $H^q(Y,f_* A) \to H^q(X,A)$ and the second map is $H^q(X,A)  \to H^0(Y,(R^q f_*) A)$. For the first one, you may use the inverse image functor and define $H^q(Y,-) \to H^q(X,f^{-1} -)$ using universal $\delta$-functors, and then compose this with the adjunction morphism $f^{-1} f_* A \to A$. For the second one, you may use an injective resolution $I^*$ of $A$ and use the canonical maps $H^0(Y,Z)/H^0(Y,B) \to H^0(Y,Z/B)$ for a subsheaf $B \subseteq Z$.
But this method does not generalize.
Also, I want to know why these maps are natural with respect to $F$ and $G$. For example in the special case above, if $f' : X' \to Y$ is another map and $g : X \to X'$ is a map over $Y$ and $A'$ is a sheaf on $X'$, why is the diagram [Feel free to edit this!]
$H^q(X',A') \to H^q(X,g^{-1} A')$
$\downarrow ~~~~~~~~~~~~~~~~~~~~~~~~~~ \downarrow$
$H^0(Y,(R^q f'_*) A') \to H^0(Y,(R^q f_*) g^{-1} A')$
commutative? Motivation: This is needed to show that the morphism $Pic(X)  \to Pic(X/Y)$ is natural, where $f$ is a morphism of locally ringed spaces.

REPLY [2 votes]: The second edge morphism $\kappa_2: R^p(FG)(X) \to FR^pG(X)$ is induced by universality of $R^*(FG)$ from the identity on $FG(X)$.
The first edge morphism $\kappa_1: R^pF(GX) \to R^p(FG)(X)$ can be described as follows:  Assume $G$ has an exact adjoint functor $T$.  Choose an injective resolution $X[0] \to J^\bullet$ in $\mathcal{A}$. Apply $G$ and choose an injective resolution $GX[0] \to I^\bullet$ in $\mathcal{B}$.  Apply $T$ and use the composition of the natural map $TGX[0] \to X[0] \to J^\bullet$ and the theorem on the natural extending of morphisms between an exact complex and an injective complex to obtain a morphism $TI^\bullet \to J^\bullet$.  Adjoint to this is $\phi: I^\bullet \to GJ^\bullet$, which gives us $\kappa_1$ after applying $F$ and taking cohomology. (see also commutative diagram with Yoneda pairing, Weil pairing and edge morphism)<|endoftext|>
TITLE: Approximation by analytic functions
QUESTION [6 upvotes]: Dear all.
Let 
$$
 f(x) = \sum_{k \in \mathbb{Z}} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
be a function given by usual fourier series. 
Since my original question hasn't got any answer yet, and I came across another related question, I am just adding it. Denote by $T_n$ the set of all trigonometric polynomials of degree $n$, that is $g\in T_n$  if 
$$
 g(x) = \sum_{k=-n}^{n} \hat{g}(k) \exp(2\pi \mathrm{i} kx).
$$
So now what is $\min_{g \in T_n} \|f - g\|_{\infty}$ and what is the optimal $g$?
Since the Fourier series of a continuous function must not converge, I expect that the answer isn't $g(x) = \sum_{k=-n}^{n} \hat{f}(k) \exp(2\pi \mathrm{i} kx)$ but something else. However, the other choice the Fejer kernel
$$
 g(x) = \sum_{k=-n}^{n} \frac{n - |k|}{n} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
seems to give worse estimates on $\min_{g \in T_n} \|f - g\|_{\infty}$ once $\hat{f} \in \ell^2$.
Thanks,
Helge
Original question:
I am interested in the question of how well one can approximate $f$ by functions that are analytic in some strip. The naive approach yields for example that if one sets
$$
 f_R(x) = \sum_{|k|\leq R} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
and assumes $f \in C^{n+1}$ then $f_R(x)$ has an extension to a strip of width $\frac{n \log(k)}{2\pi k}$ on which $f_R$ is bounded by $\|\hat{f}\|_{\ell^1}$.
This seems like a pretty natural question so I expect it to be well studied, but I don't know where... Does anybody has references? I am also interested in stronger regularity assumptions than $C^n$...

REPLY [2 votes]: The answer to the modified question is given by Jackson-type theorems.
The classic book by N.I. Akhiezer which is quoted in the Wikipedia article contains a number of specialised results on optimal approximation by trigonometric polynomials. 
A typical optimal result improves the approximation by finite Fourier sums by a logarithmic factor.

Theorem. Let $f$ be a periodic function on $\mathbb R$ of class $C^{m}$. Then for any $n\in\mathbb N$
   $$\inf\limits_{g \in T_n} \|f - g\|_{L^\infty}\leq \frac{K_m}{n^m}\|f^{(m)}\|_{L^\infty},$$
  where the constant $K_m$ is sharp (and can be written in a closed form).

A result of S. Bernstein says roughly that the order of approximation $n^{-m}$ cannot be improved.
To find the trigonometric polynomial $g$ which minimizes $\|f - g\|_{L^\infty}$ for a given $f$ is a difficult problem. I am not sure if it has been solved.<|endoftext|>
TITLE: Eilenberg–Moore algebras in terms of Kleisli ones
QUESTION [6 upvotes]: Suppose I know what the category of free algebras for a particular monad look like. Can I then describe what the category of Eilenberg–Moore algebras look like? E.g. Suppose that I have a good handle on a category $D$ and that I have two functors $F$ and $G$, with $G:D \to C$ and $F$ left-adjoint to $G$ and $F$ essentially surjective- this guarantees that this adjunction exhibits $D$ as the Kleisli-category for the monad $T:=GF$ on $C$ (see Characterization of Kleisli adjunctions). Is there a way to exhibit the Eilenberg–Moore category $C^T$  as "generalized objects of $D$"? This is somewhat vague of a question I realize, but I'm not sure how to make it more precise. (Feel free to help me do so).
EDIT: To be slightly more specific, in "Toposes, Triples, and Theories" it is said that the Eilenberg-Moore category "is in efect all the possible quotients of objects in Kleisli's category". Can anyone make this precise in a way that answers my question??

REPLY [11 votes]: One nice result is Street's theorem 14 in The formal theory of monads, generalized in Elementary cosmoi, which says that $C^T$ is isomorphic to the full subcategory of $[(C_T)^{\mathrm{op}}, \mathrm{Set}]$ containing those presheaves that become representable when precomposed with the inclusion $C \to C_T$.  That is, $C^T$ is the pullback of $[F^{\mathrm{op}},\mathrm{Set}]$ along the Yoneda embedding.  So at least informally you can think of T-algebras as being represented by certain formal colimits of free algebras.
I think it's standard, too, that every algebra has a canonical presentation -- its (2-truncated) bar resolution -- as the coequalizer of maps between free objects.  Can't think of a good reference at the moment, but Mac Lane has a section on the bar construction, and there's some good stuff on nlab.  Maybe try Kelly & Power, Adjunctions whose counits are coequalizers, and presentations of finitary enriched monads, JPAA 89 (1993).
I don't know if there's a relationship between these two ideas, but I'd be very interested to find out!  (Can't do it myself, I'm supposed to be on holiday.)<|endoftext|>
TITLE: Definitions of real reductive groups
QUESTION [25 upvotes]: There are several definitions of real reductive groups, sometimes subtly inequivalent. The following come to my mind:

A closed subgroup of $GL(n,\mathbb C)$ closed under conjugate transpose.

The set of real points $G(\mathbb R)$ of a real algebraic group such that $G(\mathbb C)$ is reductive.

Knapp's  definition i.e. a Lie group $G$ with a reductive Lie algebra $\mathfrak g$, a Cartan decomposition $\mathfrak g =\mathfrak k +\mathfrak p$, a maximal compact subgroup $K$ such that $G = K\cdot\mathrm{exp}(\mathfrak p)$ and such that every automorphism of $\mathfrak g$ of the form $\operatorname{Ad}(g)$, $g \in G$, is in $\operatorname{Int}(\mathfrak g^{\mathbb C})$.

Harish-Chandra definition i.e. 3. plus the condition that the connected component of the identity of a semisimple factor of $G$ has finite center.


Maybe there are others too. What are the relations between these different definitions?

REPLY [14 votes]: This is an old question, probably abandoned because its answer would require a short article (complete with references).   While I can't supply such an article, I can point to some of Borel's writings which involve a definition of "real reductive group".   In an old comment, Brian Conrad already referred to the added survey in Section 24C of Borel's second edition of Linear Algebraic Groups (GTM 126, Springer, 1991).  There is also a set of his lecture notes intended for a 2003 Chinese summer school, published (along with other lectures) as: Lie groups and linear algebraic groups. I. Complex and real groups.  Lie groups and automorphic forms, 1–49, AMS/IP Stud. Adv. Math., 37, Amer. Math. Soc.,
Providence, RI, 2006.  (See Sections 5-6.)
The definition he uses is in some ways the simplest and most natural, I think.  In the general framework of finite dimensional Lie algebras over an arbitrary field $K$ of characteristic 0, there are elementary definitions of semisimple and reductive Lie algebras.   When you take $K = \mathbb{R}$ and work in the classical framework of real Lie groups, it's then natural to define a connected Lie group $G$ to be reductive if its Lie algebra is.   Of course, the Lie algebra only sees local behavior, so one could leave it at that.  However, disconnected Lie groups come up immediately when Lie group theory is combined with linear algebraic groups in the study of representations, automorphic forms, etc.   So Borel adds in this case the extra requirement that $G$ have only finitely many connected components in the euclidean topology.
Where does this condition come from?  Starting with a connected linear algebraic group (scheme) $H$ over $\mathbb{R}$, the resulting group $G:= H(\mathbb{R})$ is a real Lie group but need not be connected.   An obvious example is the multiplicative group.   But a basic theorem states that this Lie group has only finitely many components in the euclidean topology.  The theorem comes, for instance, from Whitney's older work on real affine varieties but is also a consequence of a finiteness theorem in Galois cohomology proved in the work of Borel-Serre.   (I'm not sure what the best modern proof of the theorem is, but that's another question.)
Borel's notes were intended partly to prepare for Wallach's related lectures.  In any case, his approach is closely related to some of the other proposed definitions of real reductive group in the question.   But it strikes me as a more straightforward starting point.<|endoftext|>
TITLE: How to associate a Dirichlet character to a Tate character?
QUESTION [8 upvotes]: A Dirichlet character is a multiplicative map from (Z/N)* to $S^1$.
A Tate character is a continuous map from I/Q to $S^1$, where I is the Idele group of Q.
It is always claimed that they are equivalent but I never see a convincing source.
There is something involved called $F^N/P^N$ where $F^N$ is the group of fraction ideals not involving primes dividing $N$ and $P^N$ is the group of fraction ideals generated by $a$ such that a=1 mod N.  
I cannot see why a map on $F^N/P^N$ can be associated to a Dirichlet character, either.

REPLY [3 votes]: I think it easier to see this from an adelic point of view. I prefer to use $I =\mathbb{A}^\times$ for the ideles.
Using strong approximation: 
$$ 
\mathbb{A}^\times  = \mathbb{Q}^\times \times \prod\limits_p \mathbb{Z}_p^\times \times \mathbb{R}^\times
$$
Accordingly, we can factor a Hecke character $\chi$ on $\mathbb{Q}^\times \backslash \mathbb{A}^\times $ to character $\chi_p$ for all $p$ (all but finitely many trivial by Tychonoff's theorem) and $\chi_\infty$.
Now a continuous character $\chi_p$ on $\mathbb{Z}_p^\times$ must factor through $\mathbb{Z}_p^\times /(1 + p^k\mathbb{Z}_p) \cong ( \mathbb{Z} / p^k)^\times$.
In the end, we use the Chinese remainder theorem (this is essentially the approximation property above)
$$ (\mathbb{Z}/N)^\times = \prod\limits_{p^k || N }(\mathbb{Z}/p^k)^\times.$$<|endoftext|>
TITLE: Haar Measure Existence/A problem with Borel sets and regularity.
QUESTION [6 upvotes]: In Paul Halmos' Measure Theory book, section 53, he defines a content on a locally compact Hausdorff space to be a set function, $\lambda$ that is additive, subadditive, monotone, and $0\le\lambda(C)<\infty$ for all $C$ compact.  The "Borel sets" he considers(section 52) in this book are the smallest $\sigma$-ring generated by the class of compact sets. The difference between $\sigma$-ring and $\sigma$-field is that the superset need not be contained. It can be shown, and is given as an exercise by Halmos, that the smallest $\sigma$-ring generated by compact sets is the same as the smallest $\sigma$-ring generated by open bounded sets.  A bounded set is one that is contained in a compact set.  Halmos then shows that a content induces a Borel Regular measure. He introduces the inner content and an outer measure in order to do so. Regular is the important word for this question.  In fact, later in this chapter he proves the Riesz Representation Theorem, proving the existence of a Regular measure induced from positive linear functional of the continuous functions of compact support.  In the next chapter, Halmos provides an existence proof of the Haar Measure by creating a left-invariant content, which then induces a Regular measure. 
The issue I am having is that in almost every other source(books, wikipedia), the Haar Measure --and the Reisz-Rep measure which is related because one can prove the existence via a left or right invariant functional--is a Radon measure.  A Radon measure is outer regular on Borel sets, finite on compact sets, and inner regular on open sets and Borel sets with finite measure. I do realize that in the other books, the Borel sets are the smallest $\sigma$-algebra generated by open sets.
The questions I have are:
1) Is the Borel $\sigma$-ring generated by compact sets in a locally compact Hausdorff space the same as the Borel $\sigma$-field generated by the open sets.  I think the answer is no. One reason is a $\sigma$-ring and and the other is a $\sigma$-field.  I don't even think that the $\sigma$-ring generated by compact sets is the same as the $\sigma$-ring generated by open sets.  I think the best you can say is bounded open sets(mentioned above). Perhaps they are the same when the space is separable. Is this true? Is the $\sigma$-algebra generated by compact sets the same as the $\sigma$-algebra generated by the open sets the same. 
2)Although this questions is more about basic measure theory than the Haar measure, in the end I want to show that the measure Halmos created is the same as the one constructed in most other books. In order to do this, I need to go from a measure defined on the $\sigma$-ring generated on compact sets to the $\sigma$-field generated by open sets.  How do I do this?
Thank you in advance for all the help.

REPLY [4 votes]: The situation is indeed a delicate one and one needs to carefully check the conventions before transferring a result from one context to another. The situation is summarized in Royden's Real Analysis, though with just a few examples.
For a given space $X$, the main players are:

$\mathcal{B}a$ — the σ-algebra generated by the compact Gδ sets. (The Baire algebra.)
$\mathcal{B}c$ — the smallest σ-algebra with respect to which all continuous real-valued functions are measurable.
$\mathcal{B}k$ — the σ-algebra generated by the compact sets.
$\mathcal{B}o$ — the σ-algebra generated by the closed sets. (The Borel algebra.)
$\mathcal{S}$ — the smallest σ-ring containing the compact sets.
$\mathcal{R}$ — the smallest σ-ring containing the compact Gδ sets.

In general, we have the inclusions
$$\mathcal{B}a \subseteq \mathcal{B}k \subseteq \mathcal{B}o\quad\mbox{and}\quad\mathcal{B}a \subseteq \mathcal{B}c \subseteq \mathcal{B}o,$$
but $\mathcal{B}c$ and $\mathcal{B}k$ are not necessarily related. Moreover, $\mathcal{B}a = \mathcal{B}c \cap \mathcal{B}k$ and $\mathcal{B}o$ is generated by $\mathcal{B}c \cup \mathcal{B}k$. The σ-rings $\mathcal{S}$ and $\mathcal{R}$ consist of the σ-bounded elements of $\mathcal{B}a$ and $\mathcal{B}o$, respectively.

When $X$ is σ-compact and locally compact, then 
$$\mathcal{R} = \mathcal{B}a = \mathcal{B}c
\quad\mbox{and}\quad\mathcal{S} = \mathcal{B}k = \mathcal{B}o.$$
A compact example showing the strict inequality is $\beta\mathbb{N}$.
When $X$ is metrizable, more generally when closed sets are Gδ, we have 
$$\mathcal{R} = \mathcal{S} \subseteq \mathcal{B}a = \mathcal{B}k \subseteq \mathcal{B}c = \mathcal{B}o.$$
An example showing strict inequality is the space of irrational numbers.
When $X$ is locally compact and separable, then we have
$$\mathcal{R} = \mathcal{S} = \mathcal{B}a = \mathcal{B}k = \mathcal{B}c = \mathcal{B}o.$$

That said, the uniqueness (up to normalization) of Haar measure and it's (inner) regularity guarantee that all constructions will agree on their common domain of definition. Proving this from scratch does require some work depending on where you start and end.

REPLY [3 votes]: If your space X is sigma-compact (can be written as the union of a countable family of compact sets) then everything is okay, I think.  This follows as X is in the Borel sigma-ring, and hence the Borel sigma-ring is a sigma-field (or sigma-algebra).  If U is open, then $X\setminus U$ is closed, and so can be written as the union of a countable family of compacts (intersect each member of your initial countable family with $X\setminus U$).  Thus $X\setminus U$, and so U, is in the Borel sigma-ring/field.  So the two possible definitions of the Borel sigma-algebra agree.
Now, groups are rather nice in this regard.  If G is a locally compact group, then pick some compact neighbourhood of the identity, say A.  We may suppose that $A=A^{-1}$.  Then let H be the union $\bigcup_{n\geq 1} A^n$ where $A^2 = \{ab:a,b\in A\}$ and so forth.  Then H is an open subgroup of G, so H is also closed.  Each $A^n$ is compact, so H is also sigma-compact.  Thus the above applies to H.  What's really nice is that the coset space $G/H$ carries the discrete topology, and so the Haar measure on G can be reconstructed from that of H and the counting measure on $G/H$.  So, for example, $L^1(G) \cong \ell^1(L^1(H),G/H)$.  In essential, for measure theory issues, you need only worry about H.
Note: Francois posted an answer as I was typing: he makes some similar points to my first paragraph.<|endoftext|>
TITLE: How to think like a set (or a model) theorist.
QUESTION [6 upvotes]: Kenneth Kunen in his “The Foundations of Mathematics” writes:

‘Set theory is the study of models of ZFC’ (p. 7) 
‘Set theory is the theory of everything’ (p. 14)

With (1) Kunen is pointing to a change in the intended use of the axioms of ZFC: ‘there are two different uses of the word “axioms”: as “statements of faith” and as “definitional axioms”.’ (p. 6).
With (2) he means ‘set theory is all-important. That is

All abstract mathematical concepts are set-theoretic.
All concrete mathematical objects are specific sets.’ (p. 14)

According to (1), to be a set is to be any of the individuals of the universe of a particular model of ZFC, just like being a numeral (standard or not) is being any of the individuals of the universe of a particular model of PA (here I am using Shoenfield’s terminology in “Mathematical Logic”, p. 18).
But, according to (2), models are sets too, as any other objects dealt with in the metatheory.
What's more, models of set theory are defined in terms of relative interpretations of set theory into itself, a syntactical concept. (See Kunen’s “Set Theory. An Introduction to Independence Proofs”, p. 141), which makes the whole thing a bit more confusing.
The view of the axioms of set theory as "definitional axioms" is appealing. And more in regard of (2) since then they pretend to define all that there is. The study of models of set theory has an intrinsic interest, but why reduce the study of set theory to it? Or stated another way, why abandoning the old view?
I would like to know if set theorists do stick to one view or another or shift comfortably between both at need, and the reasons they have to do so.

REPLY [2 votes]: I think the more fundamental question to ask is why set theorists insist that the axioms of set theory be strictly first-order in nature (*).  I claim that you can't really explain the motivations of set theorists until you address this phenomenon.
In light of this near-universal assumption, I think it is fair to say that:

First-order logic is the foundation of set theory (which is in turn the foundation of other things)
The axioms selected for set theory are those which enable as much model theory as possible, without risking inconsistency.

The fact that one can then turn around and use model theory to study set theory itself is a major bonus, but not really a part of the foundations argument.  I've heard some quasi-mystical tales about sets existing in some alternate universe out there, but I feel far more comfortable using "it lets me do model theory and hasn't led to contradictions" as a justification for the axioms.
(*) Excluding, for example, second-order quantification, the logic $L_{\omega_1,\omega}$ of countable conjunctions, or the "exists uncountably many" quantifier.<|endoftext|>
TITLE: What is Borel-de Siebenthal theory?
QUESTION [14 upvotes]: What is Borel-de Siebenthal theory?

REPLY [11 votes]: In my head at least, part of it is this...
Let G be reductive. Consider the following algorithm:

Extend one component of the Dynkin diagram to its affine diagram, by attaching the lowest root.
From that component, remove one or more vertices. (If you want to stay semisimple, remove only one.)
Repeat to taste. (If you want to stay semisimple, and you're at a disjoint union of $A_m$ diagrams, then you're definitely done.)

The result is the Dynkin diagram of a subgroup H of G of the same rank. Every such subgroup (up to finite factors) arises this way.<|endoftext|>
TITLE: von dyck groups and solvability
QUESTION [7 upvotes]: A von Dyck group is a group with presentation $< a,b | a^m=b^n=(ab)^p=1 >$ with m,n,p natural numbers.  Is it known which of these groups are solvable and which are not?  Is there a reference for this?  Thanks.

REPLY [5 votes]: You might try Generators and Relations for Discrete Groups by Coxeter and Moser. 
Specifically for 1/m + 1/n + 1/p = 1 there are only 3 cases up to permutation, (2,3,6), (2,4,4) and (3,3,3). Map a and b to an appropriate root of unity to get a homomorphism onto C_6, C_4, or C_3, respectively. The kernel of the map is in all three cases isomorphic to Z^2.<|endoftext|>
TITLE: What can be tiled by T-tetrominoes?
QUESTION [24 upvotes]: The T-tetromino is a T-shaped figure made of four unit squares.
An $m\times n$ rectangle can be tiled by T-tetrominoes if and only if both $m$ and $n$ are multiples of 4. This was proved in a 1965 paper by D.W.Walkup, and the proof was "hands on".
Some "algebraic" tricks like colouring or tiling groups can prove that $mn$ must be a multiple of 8, but they do not seem to rule out the cases like $99\times 200$ and $100\times 102$.
I wonder whether a better proof of D.W.Walkup's theorem is known today. By "better" I mean applicable to non-rectangular regions as well. For example, is there a way to determine what 6-gons (8-gons, ...) admit tiling by T-tetrominoes?

REPLY [24 votes]: There are only partial answers to this question.  First, one can prove that Walkup's result cannot be proved using coloring arguments (I think I did this in New horizons paper, but the setting is formalized in the Ribbon tile invariants paper). Second, Walkup's proof uses an easy induction argument, and it extends to regions with sides multiples of 4.  Third, I am pretty sure you can classify all 6- and 8-gons tileable by T-tetrominoes.  This won't be conceptual.  Why do it then?  
Now, motivated by the quest to find a better proof, I made a "local move connectivity" conjecture saying that every two T-tetromino tilings of a simply connected region are connected by a series of moves involving either two T-tetrominoes or four T-tetrominoes (forming a $4\times 4$ square).  Usually, the "conceptual proof" comes from some kind of height function argument which also proves the local move connectivity.  Now, Mike Korn in his thesis disproved this by a simple construction.  One can ask if the Conway group approach in full generality can prove something like what you are asking.  You need to compute $F_2/\langle tile~words\rangle$ (see Conway-Lagarias paper, "New horizons" or Korn's thesis).  We did not do that, but I won't be very optimistic - it is a bit of a miracle when this approach works out.  
Mike and I were still able to prove the conjecture (by a height function argument) for rectangles and the above mentioned 4-multiple regions, but that proof assumes Walkup's theorem.  Independently this was established by Makarychev brothers, using a related but somewhat different argument (in Russian, based on connection to the six-vertex model).  In fact, in a followup paper we use Walkup's theorem as a definition of the graphs in which the number of claw partitions is "nice".  Anyway, hope this helps.  
UPDATE:  I just remembered that Michael Reid also did the T-tetromino computation (as well as many other computations) here.<|endoftext|>
TITLE: Which pair of mathematicians has the most joint papers?
QUESTION [16 upvotes]: I was searching on MathSciNet recently for a certain paper by two mathematicians.  As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list.  My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!
So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers?  
A good meta-question would be: can MathSciNet search for this sort of thing automatically?  The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers.  
Trying this, to my surpise I couldn't beat the 80 joint papers I've already found.  Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos).  Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).
I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.

Addendum: All right, multiple people seem to want to know.  The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams.

REPLY [15 votes]: How about Ravi Agarwal and Donal O'Regan with 445 joint journal publications on MathSciNet?<|endoftext|>
TITLE: Concentration of Measure for Power Law
QUESTION [5 upvotes]: I have a power law distribution $X$ with exponent $c$: $$p(X=t) = \left\\{\begin{array}{cl}(c-1)/t^{c} & t \geq 1 \\\\ 0 & t < 1\end{array}\right.$$
From $X$ I take $n$ independent samples $X_1, \ldots, X_n$ and would like to give a tail bound for the mean; i.e. a good bound on $\Pr[\frac{1}{n}\sum X_i > t]$.
Unfortunately, my $X$ is unbounded and much of the work on concentration of measure assumes bounded variables.  However, I do know that $X$ has bounded moments, so I can bound the variance and get a weak concentration.  But if $c$ is larger, say $10$, I ought to be able to use the higher moments to get a stronger result.  In particular it seems like the moment method outlined in http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/ should be applicable, but the treatment there assumes bounded variables.
I can work through the modifications to that treatment myself and get a reasonable result, but it's kind of messy.  So my question is: is there any existing published theorem that does what I want, so I don't have to add an ugly proof to my paper's appendix?

REPLY [3 votes]: This topic is examined by Olivier Catoni in his recent paper: High confidence estimates of the mean of heavy-tailed real random variables. Denote the mean, variance and kurtosis of your distribution by $m$, $v$ and $\kappa$, respectively. Then, by Catoni's proposition 7.1, the following holds with probability at least $1-2\epsilon$:
$$\frac{|(\frac{1}{n}\sum_{i=1}^n X_i) - m|}{\sqrt{v}} \leq \frac{2\log (\frac{3}{2}\epsilon^-1)\sqrt{\kappa}}{5n} + \sqrt{\frac{2\log (\frac{3}{2}\epsilon^-1)}{n}} + \left(\frac{3\kappa}{2\epsilon n^3}\right)^{1/4}\left(1 + \frac{3^5(n-1)\log (\frac{3}{2}\epsilon^-1)^2\kappa}{2500n^2} + \frac{12\sqrt{2}\log(\frac{3}{2}\epsilon^-1)^{3/2}\sqrt{\kappa}}{25n^{3/2}}\right)^{1/4}$$
There is no mention of boundedness but of course you will want to check his proof.
Are you using the empirical average as an estimate of the mean? If so, you may prefer Catoni's estimator, which is conceptually similar to the truncation trick described in Tao's notes. The raw empirical average performs poorly because a single $X_i$ can throw it wildly off course. Instead, blunt the incoming $X_i$'s and use them to update a guess of $m$. Of course, an initial guess is needed, so the scheme is naturally iterative.<|endoftext|>
TITLE: MathSciNet vs Google Scholar
QUESTION [25 upvotes]: What are the pros and cons of the MathSciNet database vs Google Scholar?
I don't have access to Mathscinet so this question is out of curiosity, and also this question where MathSciNet is used to find paper counts. I reckon Google Scholar will almost always be the more comprehensive of the two with higher paper counts for any particular author and includes papers that don't use the MSC codes as well papers from other subjects that may be of mathematical interest but aren't included in mathscinet.
One definitely annoying thing about Google is that in the advanced search it doesn't have a mathematics only category, but lumps it in with computer science and engineering so you sometimes need to add something like -"computer science" -"engineering" mathematics to your search term to filter out unwanted results, which isn't ideal.

REPLY [5 votes]: One disadvantage of Google Scholar is that it is not quite reliable in recognizing the same paper coming from different sources (arXiv, multiple preprint servers, journal website, the author's homepage, etc). This leads to a citation overcounting and unreliable metrics.
On the other hand, the author can update his own profile and merge all these duplicates. In this case, Google Scholar is quite accurate and powerful. Since not a lot of authors do this, unless Google implements some automatic merging algorithm, the metrics it provides are not very reliable (for what they matter, anyway).<|endoftext|>
TITLE: Covering a circle with red and blue arcs
QUESTION [22 upvotes]: We have a circle and two families of $n$ red arcs and $n$ blue arcs, positioned on the circle so that every two arcs of different colors intersect. Can one show that there is a point in the perimeter which is part of at least $n$ arcs?
(The statement sounds very simple. It makes me think the answer should be very simple too, but I've been struggling with this one for a bit and got nowhere.)
After reading Suresh's answer below, I can't help but think that there must be some colorful Helly theorem on manifolds, of which the question above is a special case. At the moment I don't even have a meaningful formulation of what this theorem could be, has it been treated before?

REPLY [12 votes]: This is the second half of a proof started by Peter Shor.
I assume that the set of arcs is already in a position as in Peter's answer: the red arcs $(L_1,R_i)$ are cyclically ordered and all blue arcs are of the form $(R_i,L_{i-1})$. For convenience, I also assume that no two of red arcs coincide (and hence their $2n$ endpoints are distinct). This is easy to achieve by perturbation. Let $S$ denote the set of the red endpoints and $r:S\to\mathbb N$ denote the covering multiplicity by red arcs: $r(p)$ is the number of red arcs containing $p$.
Fix the red arcs. Then a configuration is determined by a vector of $n$ multiplicities $T=(t_1,\dots,t_n)$ where $t_i$ is how many copies of a blue arc $(R_i,L_{i-1})$ we have. For such $T$ and each $p\in S$, let $b_T(p)$ be the number of blue arcs (in the configuration defined by $T$) containing $p$. Note that $b_T$ is linear in $T$. By Peter's observation, we have $b_{T_0}(p)=n-r(p)$ for all $p\in S$ where $T_0$ is the standard vector: $T_0=(1,\dots,1)$.
We have $\sum t_i=n$ for every admissible vector $T$. I claim that none of the resulting  functions $b_T:S\to\mathbb R_+$ strictly majorizes any other. The result follows from this claim applied to $T=T_0$. To prove the claim, it suffices to find a collection  $w:S\to\mathbb R_+$ of nonnegative weights such that, for every (potentially) blue arc, the sum of weights of points covered by this arc equals 1. Indeed, if such weights are found, we have $\sum_{p\in S} w(p) b_T(p)=n$ for any $T$, and the claim follows.
To construct $w$, consider the standard covering map $f:\mathbb R\to S^1$. On the line, we have a discrete set $\tilde S=f^{-1}(S)$ and a collection of segments corresponding to pre-images of the potentially blue arcs (both structures are $2\pi$-periodic). The endpoints of the segments are in $\tilde S$, and the segments are ordered from left to right (none of them is contained in another). Let us first solve the weight problem on the line. Begin with one of the segments and mark is right endpoint. Then take the first segment contained in the open half-line after the marked point, and mark its right end. Repeat for the new marked point, and so on. Then every segment to the right of the first one contains exactly one marked point. Let the marked points have weight 1 and all other points have weight 0. We have solved the weigh problem on a half-line.
Observe that the set of marked points is eventually $2\pi k$-periodic for some integer $k$. This follows from the deterministic nature of the construction: once some $x$ and some $x+2\pi k$ are both marked, the pattern will repeat itself. Hence there is a $2\pi k$-periodic weight function on the whole line. We can make it $2\pi$-periodic by averaging over $2\pi m$-translations, $0\le m<k$. Once it is $2\pi$-periodic, it projects back to the circle.<|endoftext|>
TITLE: An optimization problem for points on the sphere (master's dissertation)
QUESTION [9 upvotes]: First, by means of a disclaimer, some background. I am entering the fourth and final year of an undergraduate master's degree in maths, and a quarter of the maximum credit for this year will be for a project on the subject of my choice, which I will describe below. In this project, I am allowed to cite and quote the work of others to any extent so long as it is correctly referenced. My current understanding is that final-year projects in pure maths tend to include little or none of the author's original research, but rather take the form of a summary with perhaps some relatively minor original investigation. I have checked that asking about my chosen subject on this website is within regulations, so long as I cite the thread, although I'm less clear on Math Overflow's attitude to this sort of thing. Assuming it's alright, please bear in mind that I might quote you unless you ask not to be quoted.
Moving on to the subject of the project, I am interested in "sphere-like" polyhedra. More specifically: given a polyhedron with a fixed number n of sides, I would like to minimize its surface area relative to its volume. Some preliminary research turned up Michael Goldberg's paper "The Isoperimetric Problem For Polyhedra", from Tohoku Mathematical Journal vol. 40 (1934), which can be found at http://staff.aist.go.jp/d.g.fedorov/Tohoku_Math_J_1934_40_226-236.pdf
This is the only paper in English that I was able to find directly addressing the problem. It cites a number of previous non-English papers, which it summarizes as not having established very much, the main result being that a solution polyhedron can be considered as a set of n points on the unit sphere, representing the centres of mass of the polyhedron's faces which are tangent to the sphere at this point. Then, if we define sphericity as inverse to the surface area of such a polyhedron circumscribed about a unit sphere, we are looking to maximize the sphericity given n.
Goldberg also makes some further progress on the problem, and conjectures that the solutions fall into a particular class of polyhedra, which he terms "medial". He published a later paper on the subject of medial polyhedra, which can be found at http://staff.aist.go.jp/d.g.fedorov/Tohoku_Math_J_1936_43_104-108.pdf
As an undergraduate making his first foray into mathematical research, I have little intuition for the directions I should be going in (and hence, apologies if any of the tags I've added are inappropriate); but I have considered looking at what happens if I add the additional constraint that the polyhedron should have a nontrivial symmetry group, as the few known results do (tetrahedron, triangular prism, cube, pentagonal prism, dodecahedron). To this end I thought using the stereographic projection, fixing one of the n points to be at infinity, might make it easier to study the symmetry and perhaps even view the points as roots of a polynomial. Then I may be able to provide new proofs for the known solutions, which might generalize so that I can investigate unknown solutions for small values of n, at least those which have nontrivial symmetry.
In the other direction, I am interested in the asymptotic behaviour of the solution set. Goldberg places some bounds on its behaviour in his first paper; I've been wondering about possible bounds on the minimum or maximum area of a single face in terms of n, or on the maximum number of edges it can have (I've got a gut feeling that most of the faces are going to be hexagons for large enough n). An important question is of course how sphericity behaves asymptotically with respect to n. While I expect the solution set will eventually become quite irregular, is it possible to find a well-behaved family of polyhedra for which sphericity behaves similarly, if not quite as well?
Finally, how if at all does this problem relate to other optimization problems for points on the sphere? The known solutions at least coincide, for example, with those for minimizing the energy of n point charges on a sphere (the Thomson problem), though I suspect the coincidence only occurs because n is small.
In summary, my question is this: are there any English-language publications, or translations, on this subject of which I am currently unaware? Has any further progress been made? If not, what avenues sound like they may be worth exploring? If anyone is interested in pursuing the problem themselves and would be willing to give me permission to cite their work I would of course be delighted.
Many thanks,
Robin

REPLY [2 votes]: You might be  interested in the recent article by Böröczky and Csikós concerning polytopes $P_n$ of $n$ facets with minimal surface area. They studied  best approximations of a convex body $K\subset \mathbb E^d$  with $C^2$ boundary by circumscribed polytopes with respect to the surface area. Their main results can be roughly summarized as follows.

Theorem. Given a convex body $K$ in $\mathbb E^d$, $d \geq 2$, with $C^2$ boundary, let $P_n$ be a circumscribed polytope with $n$ facets that has minimal surface area. Denote 
  by $\Xi_n$ a family of points where the $n$ facets of $P_n$ touch $\partial K$. Then

$\{\Xi_n\}$ is uniformly distributed with respect to a density defined in terms of the 
  the second fundamental form of $\partial K$;
$$S(P_n) − S(K)\sim \frac{C_K}{n^{2/(d-1)}}\quad \mbox{as }\ n\to\infty,$$
  where $S(K)$ ($S(P_n)$) is the surface area of $K$ ($S(P_n)$).


This work extends (and builds on) some of the earlier results by Fejes Tóth, Gruber, Ludwig and many others who studied optimal polytopal approximations of convex bodies in terms of volume and mean width. For a modern survey of these results and related references check out "Convex and Discrete Geometry" by Peter Gruber.<|endoftext|>
TITLE: Division algebras in which every proper subfield is maximal
QUESTION [10 upvotes]: I have a (noncommutative) division algebra D which is finite dimensional over its center F. I know that every subfield of D which contains F properly is a maximal subfield of D. What can we say about D? 
Is there any characterization of such division algebras? 
Does anybody know any book or paper that discusses this? 
By the way, the set of such division algebras is obviously not empty because, for example, the (real) quaternion algebra (or any division algebra of prime degree) is in that set.
(The degree of D is the square root of the dimension of D over F.)
I hope my question is not too trivial! Thanks.

REPLY [11 votes]: The short answer is that not too much is known about this situation, beyond the easy observations that I will now list.  I will call $D$ irreducible if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.

If $D$ has prime degree, then $D$ is irreducible.  This is obvious because every subfield is contained in a maximal subfield, and the maximal subfields all have the same dimension over $F$.
If the degree of $D$ has at least two prime factors, then $D$ is reducible.  In this case you can factor $D$ as a tensor product of two division algebras of relatively prime degrees.  Then you just take a maximal subfield in one of the two factors.  This reduces us to considering algebras $D$ whose degree is a prime power.
If $D$ has composite degree and $D$ is a crossed product, then $D$ is reducible.  Recall that $D$ is a crossed product if it has a maximal subfield $L$ that is Galois over $F$.  So suppose that the Galois group is $G$, necessarily of composite order.  Then there is a nonzero proper subgroup of $G$, hence $D$ is reducible.  (This deduction sounds foolish, because the theorem that something is a crossed product is much stronger than what you are asking about.  But asking if something is a crossed product is a standard question, so in this way you can connect your question to standard results.)
As a consequence of #3, every $D$ of degree 4 is reducible, and every $D$ of degree 8 and exponent 2 is reducible.  That is because such algebras are cross products under $Z/2 \times Z/2$ (Albert) and $Z/2 \times Z/2 \times Z/2$ (Rowen) respectively.
If $D$ has degree $p^r > p$ for some $p$ prime and it happens that every finite extension of $F$ has dimension a power of $p$, then $D$ is reducible.  Indeed, by Galois theory every maximal subfield contains proper subfields.

So the first open case is where $D$ has degree 8 and exponent at least 4 in the Brauer group, and the base field has extensions of degree not a power of 2.
Translation in terms of algebraic groups
Your question is closely related to the question of whether the group $SL_1(D)$ has nonzero, proper connected subgroups.  Well, $SL_1(D)$ always contains maximal tori.  So the question is: Are there others?  (If your field has nonzero characteristic, probably one should only consider reductive subgroups.)  Subfields of $D$ correspond to tori in $SL_1(D)$, so your question is the same as asking: For what $D$ are maximal tori the only nonzero, proper reductive subgroups of $SL_1(D)$?
These sorts of questions are addressed in my joint paper with Philippe Gille Algebraic groups with few subgroups, J. London Math. Soc., vol 80 (2009), 405-430. http://dx.doi.org/10.1112/jlms/jdp030 See especially section 4.
Also, the paper Irreducible tori in semisimple groups by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587 discusses a similar question for tori.  Your maximal subfield has no proper intermediate fields if and only if the corresponding torus is irreducible in their sense.  This is why I called your $D$ irreducible above.<|endoftext|>
TITLE: Infinite groups which contain all finite groups as subgroups
QUESTION [16 upvotes]: There are infinite graphs which contain all finite graphs as induced subgraphs, e.g. the Rado graph or the coprimeness graph on the naturals.

Are there infinite groups which
  contain all finite groups as
  subgroups?

REPLY [10 votes]: My favorite is Thompson's group $V$.  
My favorite picture of $V$ is to take a set $X$ which is a disjoint union of subsets $L$ and $R$ having fixed bijections $l:X\rightarrow L$ and $r:X\rightarrow R$.  Finite words in $l$ and $r$ map $X$ to "fragments" of $X$.  Two "fragments" $W$ and $U$ that are the images, respectively, of words $w$ and $u$ in $l$ and $r$ are connected by the bijection $uw^{-1}:W\rightarrow U$.  The two "fragments" will be disjoint iff neither of $w$ nor $u$ is a prefix of the other.  If this condition holds, let $(w,u)$ represent the permutation on $X$ that is $uw^{-1}$ on $W$, is $wu^{-1}$ on $U$, and is the identity elsewhere.  The group $V$ is generated by all such $(w,u)$.  It is finitely presented and contains all finite groups.
Other f.p. groups containing all finite groups are known as Houghton groups.  See the section on Houghton groups K. S. Brown "Finiteness properties of groups" in Journal of Pure and Applied Algebra, 44 (1987), 45-75.  I forget how they are indexed, but from about n=3 on up, they are all f.p.

REPLY [2 votes]: I can't resist mentioning that the group of unitary elements in the hyperfinite $II_{1}$ factor (The group von Neumann algebra of Bruce Westbury's "locally finite" permutation group above) in fact contains every countable discrete amenable group as a subgroup.<|endoftext|>
TITLE: Gromov's list of 7 constructions in differential topology
QUESTION [77 upvotes]: At the 2010 Clay Research Conference, Gromov explained that we know of only 7 different methods for constructing smooth manifolds.  Working from memory, and hence not necessarily respecting the order he used:

Algebraic geometry (affine and projective varieties, ...)
Lie groups (homogeneous spaces, ...)
General position arguments (Morse theory, Pontryagin-Thom construction, ...)
Solutions to PDE (Moduli spaces in gauge theory, Floer theory, ...)
Surgery (Cut and paste techniques, ...)
Markov processes

I realise that I only gave 6 constructions; this was the number of separate items listed on his slides, and since he failed to discuss this part, I am left to guess that he either listed two different constructions on one line, which I interpreted to be variants of the same construction, or that failed to include one altogether.
Question How does one construct a smooth manifold from Markov processes?
I asked Gromov after the talk for explanation, but due to the rudimentary nature of my Gromovian, I was unable to understand the answer.  The only word I managed to parse is "hyperbolic," though I wouldn't put too much stock in that.

REPLY [4 votes]: There is an interview with Gromov: http://www.ihes.fr/~gromov/PDF/rtx100300391p.pdf
Q: (...) an you describe your involvement and how your mathematical and geometric insights can be useful for problems in biology?
Gromov: I can explain how I got involved in that. Back in Russia, everybody was excited by ideas of René Thom on applying mathematics to biology. My later motivation started from a mathematical angle, from hyperbolic groups. I realized that hyperbolic Markov partitions were vaguely similar to what happens in the process of cell division. So I looked in the literature and spoke to people, and I learned that there were so-called Lindenmayer systems. (...)
And his paper on the subject: "Cell Division and Hyperbolic Geometry" http://www.ihes.fr/~gromov/PDF/16%5B71%5D.pdf
I was just reading: Visions in Mathematics: GAFA 2000 Special Volume, Part I.
Gromov's article in the collection, titled: "Spaces and Questions"
has a subsection: "Symbolization and Randomization" which you might find interesting, he discusses "random manifolds" at length and even touches on one of the questions in his talk: assembling combinatorial manifolds out of simplices (i.e. how many triangles).<|endoftext|>
TITLE: Are there natural choices of $\sqrt{-1}$ in $\mathbb Z/p\mathbb Z$ for a prime $p\equiv 1\pmod 4$
QUESTION [14 upvotes]: The definition of "natural" is of course somewhat personal and involves perhaps aesthetic aspects and/or interesting properties. I illustrate it with two examples:
The integer $((p-1)/2)!$ is such a square-root of $-1$ (modulo $p$ for $p$ a prime congruent to $1$ modulo $4$). 
Another choice which is perhaps natural is given by $2a/b\pmod p$ where $p=4a^2+b^2$ with $a,b\in\mathbb N$.
Are there other nice formulae?
An example of a not very natural choice (in my eyes) is $a^{(p-1)/4}\pmod p$ where $a$ is the smallest natural integer generating the multiplicative 
group of invertible elements modulo $p$. However, I accept this answer gladly, if there is some generator $a$ of $(\mathbb Z/p\mathbb Z)^*$ given by a "natural" formula. 
(The motivation for this question was Does (the ideal class of) the different of a number field have a canonical square root? )

REPLY [3 votes]: Even $((p-1)/2)!\pmod p$ is a "computationally difficult" square root of $-1\pmod p$, this seems to be the simplest uniform formula. Less universal is $a^{(p-1)/4}\pmod p$ for a quadratic nonresidue $a$ modulo $p$, which is easy $2^{(p-1)/4}\pmod p$ for $p\equiv 5\pmod 8$ as Victor mentions. For $p\equiv1\pmod8$, you simply search for the least quadratic nonresidue $a$ modulo $p$; as far as I remember, there is a bound of the form $a<\operatorname{const}\cdot \log^2p$ assuming GRH. Computing the Legendre--Jacobi symbol $(a/p)$ has polynomial complexity.
I cannot be original in what I indicated above, so let me add some "complicated" (but curious) stuff on a discrete analogue of the formula
$$
(1-x)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}\biggl(\frac x4\biggr)^n.
$$
This is the subject of Lemma 3 in arXiv:1004.4337.
Denote $q(x)=q_p(x)=(x^{p-1}-1)/p$ the Fermat quotient of $x\in\mathbb Z_p^*$.
For a prime $p>3$, let $x$ be a rational number such that both
$x$ and $1-x$ are not divisible by $p$ and $1-x$ is a quadratic residue modulo $p$. Take $y$ such
that $y^2\equiv 1-x\pmod p$. Then
$$
\sum_{n=0}^{p-1}\binom{2n}n\biggl(\frac x4\biggr)^n
\equiv 1+\frac{px}{2(1-x)}\bigl(-q(x)+q(y+1)(y+1)-q(y-1)(y-1)\bigr)\pmod{p^2}.
$$
Taking $p\equiv1\pmod 4$ and $x=2$ one obtains a closed form for the left-hand side by means of $y$ which is your square root of $-1$ modulo $p$. Isn't it a curious result?<|endoftext|>
TITLE: Examples of compact hyperbolic surfaces/manifolds with very small or very large eigenvalues
QUESTION [5 upvotes]: Hello,
Is there any general ways to construct compact hyperbolic 2-manifolds with very small or very large eigenvalues ?  Also, as a special case, can we construct a sequence of compact hyperbolic 2-manifolds with sequence of genus $g_n$  and varying hyperbolic metrics $d_n$ such that its ($2g_n - 3$)rd eigenvalue $\lambda_n$ is of the order $o(g_n ^3)?$
Any answer or any reference would be appreciated.
P.S. : in a paper by Schoen-Wolpert-Yau, they stated a relation between $\lambda_n$ and $l_n,$ of the form $c(g) l_n \leq \lambda_n \leq d(g) l_n,$ which I do not want to use, since I do not understand fully the proof of $\lambda_n \geq c(g)l_n\ldots$.I might ask a question about it later. The quantity l_n is defined below : consider all the family F of sets of disjoint closed geodesics on the compact hyperbolic surface M such that they divide M into ( n+1 ) components. Then l_n = infimum of sum of all the closed such geodesics in all such sets in the family F.

REPLY [3 votes]: Dear SPal,
while I didn't try to answer your main question, let me make a few comments around the first eigenvalue. In general, there is a nice trick (of taking cyclic coverings) in order to get $\lambda_1$ arbitrarily small. More precisely, given $\varepsilon>0$ and any compact hyperbolic surface $S$, there is a finite (cyclic) covering $S'$ of $S$ such that $0<\lambda_1(S')\leq\varepsilon$. 
A (quick) proof can be found in the excellent book (in French) of Nicolas Bergeron (see this link here, Corollaire 3.39) 
Just as a side remark, this construction goes back to Selberg (which obtained this result in connection with his lower bounds on the first eigenvalue of arithmetic hyperbolic surfaces). Another observation is that one can use the Cheeger-Buser inequalities to give lower and upper bounds on the first eigenvalues in terms of appropriate isoperimetric inequalities (see Bergeron's book). 
Furthermore, the problem of getting large eigenvalues was discussed here  .
In any case, I believe that there are similar results for higher eigenvalues, but I don't remember any references right now.
Best,
Matheus<|endoftext|>
TITLE: A characterization of convexity
QUESTION [15 upvotes]: While doing some research on polytopes I came to the following question. Maybe it's already somewhere but anyway I'll post it here.
Let $X\subset \mathbb{R}^3$ be such that, for every plane $P$, $P\cap X$ is simply connected. Is $X$ convex?
I'm not sure, but I think maybe it's necesary to assume some well behaving like local simple connectedness. Anyway I think this is true with the apropiate asumptions. I would not be surprised if it was true just as stated.
Probably this is true even in greater dimensions.

REPLY [17 votes]: A version of this is well known:

Let $X\subset \Bbb R^3$ be a compact
  set and suppose every intersection of
  $X$ by a plane is contractible. Then
  $X$ is convex.

This is due to Schreier (1933) in $\Bbb R^3$, and Aumann (1936) generalized this to higher dimensions.  See this and other related results in Ju.D. Burago and V.A. Zalgaller, Sufficient criteria of convexity, J. Math. Sci. 10 (1978). 395–435.  Incidentally, this a (hard) exercise in my book (Exc. 1.25).<|endoftext|>
TITLE: colimits of spectral sequences
QUESTION [5 upvotes]: I'm looking for some references about colimits of spectral sequences.
More precisely: let $X : I \longrightarrow \cal{C}$ be a functor from a filtered category $I$ to the category of double cochain complexes of an abelian category $\cal{C}$, in which  filtered colimits exist and commute with cohomology.
Let $E_2(X_i)$ be the second page of the first filtration ss associated to $X_i$. Assuming that the $X_i$ are right-half plane double complexes, it weakly converges to $H^*(\mbox{Tot}^\prod X_i)$ for all $i$ (Weibel, "An introduction to homological algebra", page 142):
$$
E_2(X_i) \Longrightarrow H^*(\mbox{Tot}^\prod X_i)\ ,
$$
where $\mbox{Tot}^\prod$ is the total product complex,
$$
(\mbox{Tot}^\prod X)^n = \prod_{p+q=n} X^{pq} \ .
$$
For the same reason:
$$
E_2(\underset{i}{\lim_\longrightarrow} X_i) \Longrightarrow H^*(\mbox{Tot}^\prod \underset{i}{\lim_\longrightarrow} X_i )\ .
$$
Then, because of the exactness of $\displaystyle \lim_\longrightarrow$, we have
$$
\underset{i}{\lim_\longrightarrow} E_2 (X_i) = E_2(\underset{i}{\lim_\longrightarrow} X_i) \ .
$$
Then my question is: under which conditions can I assure that I have a comparison theorem like
$$
\underset{i}{\lim_\longrightarrow} H^* (\mbox{Tot}^\prod X_i) = H^*(\mbox{Tot}^\prod \underset{i}{\lim_\longrightarrow} X_i) \quad \mbox{?}
$$
Any hints or references will be appreciated.

REPLY [5 votes]: This is a nice question, so I'm not sure why it was never answered- maybe my answer is wrong and this question is harder than I thought? In any event here's my attempt:
Under fairly mild hypotheses on your spectral sequences (i.e. that it converges in the sense of Weibel 5.2.11), we have a comparison theorem which says that if a map of convergent spectral sequences is an isomorphism for any $r$, then it induces an isomorphism on the abuttments. In particular, in this case I think that you definitely have a map of spectral sequences $\text{colim } E(X_i) \rightarrow E(\text{colim } X_i)$, and since it's an isomorphism at the $E_2$ page then it's an isomorphism from then on, so the theorem applies (at least in the case when the double complexes in question are, say, right half-plane or something). The comparison theorem is in Weibel, 5.2.12.<|endoftext|>
TITLE: $p$-torsion in the Mordell-Weil group of Abelian varieties injecting in reduction
QUESTION [7 upvotes]: Let $K$ be a number field and $\mathfrak{p}$ be a place of good reduction.  It is easy to see that the reduction map on prime-to-$p$ torsion $A(K)[p'] \hookrightarrow A_{\mathfrak{p}}(\kappa(\mathfrak{p}))$ is injective.
But if $p > e(\mathfrak{p}/p) + 1$, the reduction map is even injective on $p$-torsion. This can be seen by showing that the formal group has no $p$-torsion.
Now I ask if one can show this without using the formal group. By a theorem of Raynaud in "Schémas en groupes de type $(p, \ldots, p)$", the inequality for $p$ implies that for a $p^n$-torsion commutative finite flat group scheme the generic fibre can be spread out uniquely over the special fibre and $\mathrm{Hom}(G,H) = \mathrm{Hom}(G(\bar{K}),H(\bar{K}))$. Can this be used somehow?

REPLY [6 votes]: One has the finite flat group scheme $\mathbb Z/p$ over $\mathcal O_{K_{\mathfrak p}}$
(I write $K_{\mathfrak p}$ for the $\mathfrak p$-adic completion of $K$, and 
$\mathcal O_{K_{\mathfrak p}}$ for its integer ring), 
as well as
the finite flat group scheme $A[p]$.  Giving a $p$-torsion point over $K_{\mathfrak p}$ (and hence in particular over $K$) is the same as giving a closed embedding
on generic fibres:
$(\mathbb Z/p)\_{/ K_{\mathfrak p}} \hookrightarrow A[p]\_{/K_{\mathfrak p}}.$
Raynaud's results imply that this extends to a closed embedding over $\mathcal O_K$:
$\mathbb Z/p \hookrightarrow A[p],$
which is another way of saying the that the non-zero $p$-torsion point has non-zero
reduction.
Just to see concretely what can happen in the situation when $e \geq p-1$, suppose
that $K = \mathbb Q$ and $p = 2$.  Then we could have a map
$(\mathbb Z/2)\_{/\mathbb Q_2} \hookrightarrow A[2]_{/\mathbb Q_2}$
which extends to a closed immersion
$\mu_2 \hookrightarrow A[2].$
This would correspond to having a 2-torsion point in the kernel of the reduction map.
(Note that $\mu_2$ has a non-trivial point in char. zero, which collapses down to
the identity in char. two.)
[Added in response to unkwown's comment:]
The point is that one can form the scheme-theoretic closure in $A[p]$ of the image
of $(\mathbb Z/p)\_K,$ which is some finite flat subgroup scheme over $\mathcal O_K$
which is embedding as a closed subgroup scheme of $A[p]$ (by construction: we formed it as a scheme-theoretic closure).  And it
has $(\mathbb Z/p)\_K$ as its generic fibre (again by construction).
Now when $e < p-1$, Raynaud's results show that this finite flat group scheme has no
choice but to be $\mathbb Z/p$, and so we get a copy of $\mathbb Z/p$ embedding into $A[p]$,
extending the original embedding of generic fibres.  Thus the order $p$ point in $A[p](K)$
reduces mod $p$ to an order $p$ point.
But if e.g. $p = 2$, then this scheme-theoretic closure could be $\mu_2$.  Now the non-trivial point ($-1$) of $\mu_2(K)$ specializes to the trivial point in char. 2,
and so when we have a copy of $\mu_2$ inside $A[2]$, the non-trivial point of $\mu_2(K)$
lies in the kernel of the reduction mod 2 map.<|endoftext|>
TITLE: Characterizing convex polynomials
QUESTION [5 upvotes]: Let $p=\sum_{i=0}^{n}a_ix^i$. Under what conditions on the coefficients $a_i$  is $p$ convex? Strictly convex?

REPLY [8 votes]: To add some more stuff to @coudy's answer. We need to check whether $p''$ is nonnegative.
To that end let us recall some basic results (copied from another MO answer of mine), discussed in these lecture notes; these results are helpful for checking if a univariate polynomial is nonnegative.
Theorem A univariate polynomial is nonnegative if and only if it is a sum of squares.
Theorem Let $\mathbf{x}=[1,x,x^2,\ldots,x^m]^T$, and let $P(x)$ have degree $2m$. Then, $P(x)$ is nonnegative if and only if there exists a $m+1 \times m+1$ positive semidefinite matrix $Q$ such that $P(x) = \mathbf{x}^TQ\mathbf{x}$.
Let $P(x)=\sum_{i=0}^{2m} p_ix^i$. It can be further shown that the matrix $Q$ must satisfy
\begin{equation*}
 p_i = \sum_{j+k=i} Q_{jk},\qquad i=0,1,\ldots,2m.
\end{equation*}
Theorem If we can find a positive semidefinite matrix $Q$ that satisfies the above linear constraints (this can be done using regular semidefinite programming software), then the polynomial $P(x)=\sum_{i=0}^{2m} p_ix^i$ is nonnegative or SOS, otherwise not.<|endoftext|>
TITLE: A question about Iwasawa Theory
QUESTION [13 upvotes]: I am just reading about Iwasawa theory about Coates and Sujatha's book on Iwasawa Theory. I was wondering that since Iwasawa thought about the whole theory from the analogy of curves over finite fields, so what should be the analog of the module $U_\infty$/$C_\infty$ in the curve case (if there is any) where $U_\infty$ is the inverse system of local units and $C_\infty$ is the cyclotomic units.

REPLY [9 votes]: There is a very close analogy but to unravel it requires some work.
So take $X$ a smooth curve over $\mathbb F_{\ell}$ (more generally you could take $X$ a scheme over $\mathbb F_{\ell}$) and let $\mathscr F$ be a smooth sheaf of $\mathbb Q_{p}$-vector spaces on $X$ (you could be much more general in your choice of coefficient ring, and indeed, I think you might need to consider more general coefficient rings in order to really grasp the analogy, but that will do for the moment). Moreover, we will assume for simplicity that $\mathscr F$ comes from a motive over $X$, a sentence which will remain vague but aims to convey the idea that $\mathscr F$ has geometric origin.
Then the cohomology complex $R\Gamma(X,\mathscr F)$ is a perfect complex so it has a determinant $D$. This complex fits in a exact triangle
$$ R\Gamma(X,\mathscr F)\rightarrow R\Gamma(X\otimes\bar{\mathbb F}_{\ell},\mathscr F)\rightarrow R\Gamma(X\otimes\bar{\mathbb F}_{\ell},\mathscr F)$$
Here the very important fact to understand in order to grasp the analogy is that the second arrow is given by $Fr(\ell)-1$. This exact triangle induces an isomorphism $f$ of $D$ with $\mathbb Q_{p}$.
There is conjecturally another such isomorphism. Assume that the action of the Frobenius $Fr(\ell)$ acts semi-simply on $H^{i}(\bar{X},\mathscr F)$ for all $i$ (this is widely believed under our hypothesis on $\mathscr F$). Then degeneracy of a the spectral sequence $H^{i}(\mathbb F_{\ell},H^{j}(X\otimes\bar{\mathbb F}_{\ell},\mathscr F))$ gives an isomorphism $g$ between $D$ and $\mathbb Q_{p}$. 
Now, consider $gf^{-1}(1)$. This happens to be the residue at 1 of the zeta function of $X$. 
What has all this to do with units in number fields? Change setting a bit and take $X_{n}=\operatorname{Spec}\mathbb Z[1/p,\zeta_{p^{n}}]$ and $\mathscr F=\mathbb Z(1)$. We would like to carry the same procedure as above but we can't, because we are lacking crucially the exact triangle involved in the definition of the isomorphism $f$. Nonetheless, there is a significantly more sophisticate way to construct a suitable $f_{n}$ for all $n$ and it turns out that this construction will crucially involve $U_{\infty}/C_{\infty}$.
So to sum up, the analog of $U_{\infty}/C_{\infty}$ for curves over finite fields is none other than the Frobenius morphism $Fr(\ell)-1$. You may know that the cyclotomic units form an Euler system, that is to say that they satisfy relations involving corestriction and the characteristic polynomial of the Frobenius morphisms. This fact is I believe what led K.Kato to describe the analogy above.
You can read about all this in much much greater details in the contribution of Kato in the volume Arithmetic Algebraic Geometry (Springer Lecture Notes 1553).<|endoftext|>
TITLE: Does anyone know an intuitive proof of the Birkhoff ergodic theorem?
QUESTION [47 upvotes]: For many standard, well-understood theorems the proofs have been streamlined to the point where you just need to understand the proof once and you remember the general idea forever.  At this point I have learned three different proofs of the Birkhoff ergodic theorem on three separate occasions and yet I still could probably not explain any of them to a friend or even sit down and recover all of the details.  The problem seems to be that they all depend crucially on some frustrating little combinatorial trick, each of which was apparently invented just to service this one result.  Has anyone seen a more natural approach that I might actually be able to remember?  Note that I'm not necessarily looking for a short proof (those are often the worst offenders) - I'm looking for an argument that will make me feel like I could have invented it if I were given enough time.

REPLY [6 votes]: I like an answer that was shown to me by Mate Wierdl.
You restate the maximal inequality this way:
Let
$$
Mf(x) = \sup_N (1/N)\sum_{n < N}f(T^nx).
$$
and first prove that $\|Mf\|_{1,\infty}\le\|f\|_1$.
Here by $\|\cdot\|_{1,\infty}$
I mean the "weak $L^1$ norm" (which is not actually a norm at all) defined by
$\|g\|_{1,\infty}=\sup_a a \cdot m( \{ x\colon |g|(x)\le a \}) $
(the area of the biggest rectangle that fits under $|g|$).
This has a beautiful (and simple and intuitive) proof if you are working on the integers and $T$ is the map $T(n)=n+1$. There is then a technique called transference that allows you to take the proof in your favourite system and transfer it to any other system.
You then show that this implies that when you have a maximal inequality, the set of functions for which you get almost everywhere convergence is closed in $L^1$. (Suppose the ergodic theorem works for $f_1,f_2,\ldots$ and $\|f_n-f\|_1\to 0$. Use the maximal inequality to deduce that the ergodic theorem holds for $f$).
At this stage you know that the set of $f$'s for which the ergodic theorem holds is a closed set, so you're done if you can find a dense set of $f$'s for which this works. Since you're looking for a dense set you're OK to work with $L^2$. The result is obvious for coboundaries and the orthogonal complement of the space of coboundaries is the set of invariant functions (for which the result is even more obvious). QED.<|endoftext|>
TITLE: Is there a universal countable group? (a countable group containing every countable group as a subgroup)
QUESTION [41 upvotes]: This recent MO
question,
answered now several times over, inquired whether an
infinite group can contain every finite group as a
subgroup. The answer is yes by a variety of means.
So let us raise the stakes: Is there a countable group
containing (a copy of) every countable group as a subgroup?
The countable random graph, after all, which inspired the
original question, contains copies of all countable graphs,
not merely all finite graphs. Is this possible with groups?
What seems to be needed is a highly saturated countable group.

An easier requirement would insist that
the group contains merely all finitely generated groups as
subgroups, or merely all countable abelian groups.
(Reducing to a countable family, however, trivializes the question via the direct sum.)
A harder requirement would find the subgroups in particularly nice ways: as direct summands or as normal subgroups. 
Another strengthened requirement would insist on an
amalgamation property: whenever
$H_0\lt H_1$ are finitely generated, then every copy of $H_0$ in the universal group $G$
extends to a copy of $H_1$ in $G$. This property implies
that $G$ is universal for all countable groups, by adding
one generator at a time. This would generalize the
saturation property of the random graph.
If there is a universal countable group, can one find a
finitely generated such group, or a finitely presented
such group? (This would lose amalgamation, of course.)
Moving higher, for which cardinals $\kappa$
is there a universal group of size $\kappa$? That is, when is there
a group of size
$\kappa$ containing as a subgroup a copy of every group of size
$\kappa$?
Moving lower, what is the minimum size of a finite group
containing all groups of finite size at most $n$ as subgroups?
Clearly, $n!$ suffices. Can one do better?

REPLY [9 votes]: Hall's universal group is a countable locally finite group that contains every countable locally finite group (see these lecture notes).<|endoftext|>
TITLE: Volumes of sets of constant width in high dimensions
QUESTION [92 upvotes]: Background
The $n$-dimensional Euclidean ball of radius $1/2$ has width $1$ in every direction. Namely, when you consider a pair of parallel tangent hyperplanes in any direction the distance between them is $1$.
There are other sets of constant width $1$. A famous one is the Reuleaux triangle in the plane. The isoperimetric inequality implies that among all sets of constant width $1$ the ball has largest volume. Let's denote the volume of the $n$-ball of radius $1/2$ by $V_n$.
The question
Is there some $\varepsilon >0$ so that for every $n>1$ there exist a set $K_n$ of constant width 1 in dimension n whose volume satisfies $\mathrm{vol}(K_n) \le (1-\varepsilon)^n V_n$.
This question was asked by Oded Schramm who also asked it for spherical sets of constant width r.
A proposed construction
Here is a proposed construction (also by Schramm). It will be interesting to examine what is the asymptotic behavior of the volume. (And also what is the volume in small dimensions 3,4,...)
Start with $K_1=[-1/2,1/2]$. Given $K_n$ consider it embedded in the hyperplane of all points in $R^{n+1}$ whose $(n+1)$-th coordinate is zero.
Let $K^+_{n+1}$ be the set of all points $x$, with nonnegative $(n+1)$-th coordinate, such that the ball of radius $1$ with center at $x$ contains $K_n$.
Let $K^-_{n+1}$ be the set of all points $x$, with nonpositive $(n+1)$-th coordinate, such that $x$ belongs to the intersection of all balls of radius $1$ containing $K_n$.
Let $K_{n+1}$ be the union of these two sets $K^-_{n+1}$ and $K^+_{n+1}$.
Motivation
Sets of constant width (other than the ball) are not lucky enough to serve as norms of Banach spaces and to attract the powerful Banach space theorist to study their asymptotic properties for large dimensions. But they are very exciting and this looks like a very basic question.
References and additional motivation
In the paper: "On the volume of sets having constant width" Isr. J Math 63(1988) 178-182, Oded Schramm gives a lower bound on volumes of sets of constant width. Schramm wrote that a good way to present the volume of a set $K \subset R^n$  is to specify the radius of the ball having the same volume as $K$, called it the effective radius of the set $K$ and denote it by $\operatorname{er} K$. Next he defined $r_n$ as the minimal effective radius of all sets having constant width two in $R^n$. Schramm proved that $r_n \ge \sqrt {3+2/(n+1)}-1$. He asked if the limit of $r_n$ exists and if $r_n$ is a monotone decreasing sequence.
Our question is essentially whether $r_n$ tends to 1 as $n$ tends to infinity.
In the paper: O. Schramm, Illuminating sets of constant width. Mathematika 35 (1988), 180--189, Schramm proved a similar lower bound for the spherical case and deduced the best known upper bound for Borsuk's problem on covering sets with sets of smaller diameter.

REPLY [4 votes]: Warning: This is not an answer to the question as posed, just an explanation of my comment at Gil's request. The question itself remains as open as it was!
Let $L$ be the intersection of the unit (radius $1$) ball with the positive orthant $\{x\in\mathbb R^n: x_i\ge 0\  \forall i\}$. Note that the support function of $L$ with respect to the origin is $h_L(\theta)=|\theta_+|$ where $\theta$ is a unit vector and $\theta_+$ is its "positive part" (i.e., all negative coordinates get replaced by $0$). Take small $r>0$ and let $K$ be the convex hull of $L\cup (-rL)$.
Then $h_K(\theta)=\max(|\theta_+|,r|\theta_-|)$, so the width of $K$ in the direction $\theta$ is $h_K(\theta)+h_K(-\theta)|=\max(|\theta_-|,r|\theta_+|)+\max(|\theta_+|,r|\theta_-|)$. Since $|\theta_+|^2+|\theta_-|^2=1$, this is at least $\frac{1+r}{\sqrt{1+r^2}}\ge 1+\frac r2$ for small $r$.
Now let us estimate the volume of $K$ up to a polynomial in $n$ factor. $K$ is the union of the sets $K_{\rho,m}$ consisting of points $x$ with $m$ negative and $n-m$ positive coordinates such that $|x_-|\le\rho$, $|x_+|\le 1-\frac \rho r$. After some usual mumbo-jumbo about $n$ choices of $m$ and a polynomial net in $\rho$, we see that we just need to bound $\max_{m,\rho}|K_{\rho,m}|$.
Now, $|K_{\rho,m}|={n\choose m}\omega_m\omega_{n-m}2^{-n}\rho^m(1-\frac\rho r)^{n-m}$. We want to compare it to the volume of the ball of diameter $1+\frac r2$, which is $2^{-n}\omega_n(1+\frac r2)^n$ ($\omega_k$ is the volume of the $k$-dimensional unit ball). Note that $\frac{\omega_m\omega_{n-m}}{\omega_n}\approx {n\choose m}^{1/2}$ (up to a polynomial factor), so we want to bound
$$
{n\choose m}^{3/2}\rho^m(1-\tfrac\rho r)^{n-m}\le\left[ 
{n\choose m}(\rho^{2/3})^m(1-\tfrac 23\tfrac\rho r)^{n-m}
\right]^{3/2}
\\
\le (1+\rho^{2/3}-\tfrac 23\tfrac{\rho}{r})^{\frac 32n}\le (1+r^2)^{\frac 32 n}\,
$$
so the ratio is essentially $\left[\frac{(1+r^2)^{3/2}}{1+\frac r2}\right]^n$, which is exponentially small in $n$ for fixed small $r>0$.
I re-iterate that it, probably, doesn't say absolutely anything about the original question, but Gil was interested in details, so here they are. BTW, I'll not get surprised if an even simpler construction is possible. I just needed some bounds for self-dual cones and this particular example is merely a byproduct of one failed attempt to prove something decent in another problem...<|endoftext|>
TITLE: Characterisation of parabolic subalgebras: reference sought
QUESTION [8 upvotes]: Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{p}$ a subalgebra.  As we all know, $\mathfrak{p}$ is parabolic if it contains a Borel (thus maximal solvable) subalgebra.  In this case, with $\mathfrak{p}^\perp$ the orthocomplement of $\mathfrak{p}$ with respect to the Killing form of $\mathfrak{g}$, $\mathfrak{p}^\perp$ is the nilradical of $\mathfrak{p}$.
There is a handy converse to this statement which goes as follows: a subalgebra $\mathfrak{p}$ of $\mathfrak{g}$ is parabolic if $\mathfrak{p}^\perp$ is a nilpotent (thus central descending series terminates) subalgebra of $\mathfrak{g}$.  Note that there is no a priori demand that $\mathfrak{p}^\perp$ is even contained in $\mathfrak{p}$ (though that is, of course. part of the conclusion).
My question: does anyone know a reference for this (not difficult to prove) fact?  (I have, in the past, incorrectly attributed it to Grothendieck.)

REPLY [2 votes]: If $\mathfrak{p}^\perp$ is a nipotent subalgebra then it must consist of nilpotent elements. Indeed, let $h\in \mathfrak{p}^\perp$ and suppose ${\rm ad}\,h$ is not nilpotent.
Let $\mathfrak{g}^0(h)$ be the Fitting null-component of  ${\rm ad}\,h$ (it consists of all
$x\in\mathfrak{g}$ such that $({\rm ad}\,h)^N(x)=0$ for $N\gg 0$). Since $\mathfrak{p}^\perp$ is nipotent and $[h,\mathfrak{p}]\subseteq \mathfrak{p}^\perp$ we have that $\mathfrak {p}+\mathfrak{p}^\perp\subseteq \mathfrak{g}^0(h)$. If $\mathfrak{g}^0(h)$ is a proper Lie subalgebra of $\mathfrak {g}$, there is a nonzero $a\in \mathfrak{g}^0(h)^\perp$. Then $a\in \mathfrak{p}^\perp\subset \mathfrak{g}^0(h)$ forcing $a\in \mathfrak{g}^0(h)\cap \mathfrak{g}^0(h)^\perp$. However, decomposing $\mathfrak{g}$ into the direct sum of generalised eigenspaces for ${\rm ad}\,h$ it is easy to see that the restriction of the Killing form of $\mathfrak g$ to $\mathfrak{g}^0(h)$ is non-degenerate. So ${\rm ad}\,h$ has to be nilpotent for every $h\in\mathfrak{p}^\perp$.<|endoftext|>
TITLE: Birkhoff ergodic theorem for dynamical systems driven by a Wiener process
QUESTION [6 upvotes]: At the risk of asking a stupid question I have the following problem.
Suppose I have a measure preserving dynamical system $(X, \mathcal{F}, \mu, T_s)$, where 

$X$ is a set 
$\mathcal{F}$ is a sigma-algebra on $X$,
$\mu$ is a probability measure on $X$,
$T_s:X \rightarrow X$, is a group of measure preserving transformations parametrized by $s \in \mathbb{R}$.

Suppose that this dynamical system is ergodic, so that for any $f \in L^1(\mu)$,
$\lim_{t\rightarrow \infty}\frac{1}{2t}\int_{-t}^t f(T_s x) ds = \int f(x)d\mu(x)$.
Now let $B_s$ be a real valued Wiener process such that $B_0 = 0$, then I can define the following process:
$\frac{1}{t}\int_{0}^t f(T_{B_s} x) ds$
Does anybody know how this process would behave as $t\rightarrow \infty$?  Intuitively I would expect it to converge to a similar constant for a.e realisation of the brownian motion, but I can't find a convincing argument.
Thanks for your help.

REPLY [5 votes]: Not a stupid question, but I think the answer is no. 
The paper Random Ergodic Theorems with Universally Representative Sequences
by Lacey, Petersen, Wierdl and Rudolph gives a counterexample in the case where the system is being driven by a simple symmetric random walk (based on an application of Strassen's functional law of the iterated logarithm). I'm pretty sure the same technique would give a counterexample here.
The paper can be found online at: http://www.numdam.org/item?id=AIHPB_1994__30_3_353_0<|endoftext|>
TITLE: Ulam spiral coordinate system
QUESTION [6 upvotes]: Inspired by a Project Euler problem, I recently started playing around with Ulam spirals. My first thought was that an Ulam spiral could be a (rather useless) coordinate system, and how I might be able to convert from "Ulam coordinates" (i.e. just the number on the spiral at a given point) to rectangular coordinates and vice versa.
So the question is: Is there a single function that would return the number on the Ulam spiral given (x,y)? And also (and perhaps more of a challenge), one to convert back? 
I'm just a sophomore computer science student, and not terribly competent in math compared to the people on these site, sadly (I'm in Calc III right now). I was able to work out four separate equations to find the number on the spiral given an ordered pair, but that's one for each of four sections in between the 'diagonal axes' (I'm not sure what to call them). Unfortunately, I have yet to make this more simple. Any ideas?

REPLY [3 votes]: A while ago, Dan Pearcy approached me with a similar problem, asking for the inverse formula to convert $n$ into coordinates $(x,y)$. Using the floor function, it was not too difficult to provide an explicit closed form. He wrote a blog post about this, including my extremely messy closed form:
https://web.archive.org/web/20141202041502/https://danpearcymaths.wordpress.com/2012/09/30/infinity-programming-in-geogebra-and-failing-miserably/<|endoftext|>
TITLE: Books you would like to see retranslated.
QUESTION [10 upvotes]: As a follow on to this question, what books would you like to see retranslated or rewritten as the original translation wasn't very good, or can you give examples of books that have been translated more than once into the same language.

REPLY [8 votes]: I nominate Felix Klein's Lectures on the Icosahedron and the Solution
of Equations of the Fifth Degree as a book that deserves retranslation. 
The present English translation was made in 1888, and it contains a lot of 
archaic terminology, such as "permutable" for "commuting," "transformation" 
for "conjugation," and "associates" for "conjugates."  Also confusing, though 
in principle a good idea, a normal subgroup is called "self-conjugate."
Best of all, a new edition would give an opportunity to 
introduce some pictures, which are incredibly absent from Klein's original text.<|endoftext|>
TITLE: Classification of finite complex reflection groups
QUESTION [6 upvotes]: Background:
Let $K$ be a field and let $V$ be a finite-dimensional $K$-vector space. A pseudoreflection (or usually imprecisely just reflection) in $V$ is an element $1 \neq s \in \mathrm{GL}(V)$ fixing a hyperplane. A reflection representation of a group $W$ over $K$ is a $K$-linear representation $\rho:W \rightarrow \mathrm{GL}(V)$, such that $\rho(W)$ is generated by reflections. A group $W$ is called a reflection group over $K$ if it admits a reflection representation over $K$.
Shephard-Todd classified (see below) the finite irreducible reflection groups over $\mathbb{C}$ (i.e. those finite groups admitting an irreducible reflection representation over $\mathbb{C}$). 
Question:
Is there also a classification of the finite irreducible reflection representations over $\mathbb{C}$? 
Edit: This question is very imprecise as indicated in the comments below. I should say what "classification of representations" means, and I have to admit: I don't know. A few ideas in this direction are:

determine the isomorphism classes of finite irreducible reflection representations over $\mathbb{C}$, where an isomorphism between two reflection representations $\rho:W \rightarrow \mathrm{GL}(V)$, $\rho':W' \rightarrow \mathrm{GL}(V')$ is a vector space isomorphism $f:V \rightarrow V'$ such that $f \rho(G) f^{-1} = \rho'(G)$. (I think the Shephard-Todd classification is a classification relative to this notion!?)
the same as above but an isomorphism is a vector space isomorphism $f:V \rightarrow V'$ and a group isomorphism $\varphi:W \rightarrow W'$ such that $f \rho(g) f^{-1} = \rho'( \varphi(g) )$ for all $g \in W$.
consider pairs $(W,T)$ consisting of a finite irreducible reflection group over $\mathbb{C}$ and a subset $T$ which are generating reflections of some irreducible reflection representation of $W$ and then determine isomorphism classes of such pairs.
[Insert your idea here].

My motivation for this question is something like this: A Cherednik-Algebra is defined for any finite irreducible reflection representation over $\mathbb{C}$. In what sense does the algebra depend on the group alone and not on the choice of a particular reflection representation?

REPLY [3 votes]: Answering the first question, if the field has characteristic zero then the classification will be reduced to Shephard-Todd. By this I mean that every finite reflection group will be on Shephard-Todd list. In the opposite direction, finite Weyl groups will appear over every field and the rest of them will need some algebraic integers to be present in your field... 
In positive characteristic, the life gets tough as pseudoreflections can be unipotent. I do not know whether classification is known but I can say that the list will get much longer...<|endoftext|>
TITLE: Minimal elements of minimal R^k actions
QUESTION [6 upvotes]: C. Pugh and M. Shub showed in 1971 that, given an ergodic action of $G=\mathbb{R}^k$ on some separable finite measure space $(X,\mu)$, then all elements of $G$ , off a countable family of hyperplanes, are ergodic.
Is there an analogous statement in the topological setting, with ergodicity replaced by minimality (i.e. all orbits are dense) and $X$ assumed to be compact ?

REPLY [4 votes]: A colleague pointed out the following counterexample.
Let $h_t$ be the horocyclic flow on a negatively curved compact surface S. This R action is known to be minimal. Now Consider the $R^2$ action on $S\times S$ given by $(s,t)\rightarrow (h_s,h_t)$. This action is again minimal. 
The action of the diagonal $\{(s,s), s\in R\}$ is not minimal since the orbit of any point (x,x) stays in the diagonal. 
Let $\theta\in R$. The action of the line $\{(s,\theta s), s\in R\}$ is not minimal because it is conjugated to the diagonal action. This comes from the fact that the two actions $h_{\theta s}$ and $h_s$ are conjugated by the geodesic flow. 
As a result, there are no elements in $R^2$ acting minimally, although $R^2$ itself acts minimally.<|endoftext|>
TITLE: Conjugate cocharacters in a maximal torus
QUESTION [5 upvotes]: Let $G$ be a linear algebraic group over an algebraically closed field $k$, and $T$ a maximal torus of $G$.
Suppose we have two cocharacter $\mu, \mu' : \mathbb{G}_m \to T$, which are conjugate under $G$ i.e. there exists $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$.
Question. Can we always choose $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$ and such that $g$ normalizes $T$?
If $G = \mathrm{GL}_n$ then this is true: suppose that $T$ is the diagonal torus.  Then a cocharacter of $T$ is just determined by its weights on the basis vectors of the natural representation of $G$ (i.e. integers $k_i$ such that $\mu(z)e_i = z^{k_i} e_i$).
Since $\mu'$, $\mu$ are $G$-conjugate, the weights of $\mu$ are a permutation of the weights of $\mu'$.  We can choose an element of $G$ which induces this permutation on the basis vectors.
For general $G$, we can embed $G$ in $\mathrm{GL}_n$ for some $G$, such that $T$ is the intersection of $G$ with the diagonal torus.  The weights of $\mu$ on the basis vectors of the corresponding representation are a permutation of the weights of $\mu'$.  The question is whether we can choose this permutation (since the weights need not be distinct, there may be several choices for the permutation) so that it is induced by $N_G(T)$.
For my application, to show that the number of $G$-conjugates of $\mu$ in $T$ is finite, it is enough to know this for $\mathrm{GL}_n$, but it would be nice to know if it is true in general.

REPLY [8 votes]: In fact something better is true (properly formulated!) over any field $k$, using $k$-rational conjugacy and maximal $k$-split $k$-tori.  I will give a precise statement and proof below, with $G$ any affine $k$-group of finite type (no reductivity or smoothness hypotheses, though lack of smoothness is a red herring, as will be explained below; lack of reductivity is more serious with general $k$). In the end it will come down to the same idea as in the  argument which is given by an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta, but we need preparations to make it work over any field and for any $G$ (even with smoothness restrictions, there is work to do).  To stick with a single reference, below I refer to things in "Pseudo-reductive groups". 
Let $G$ be an affine group scheme of finite over a field $k$, and $S$ a maximal $k$-split $k$-torus in $G$ (these exist, for dimension reasons). I first claim that any two such $S$ are $G(k)$-conjugate, which is really the whole point.  We can replace $G$ with the identity component of the smooth Galois descent of the Zariski-closure of $G(k_s)$ in $G_{k_s}$ (see Lemma C.4.1 in "Pseudo-reductive groups" for why this is a smooth $k$-subgroup of the original $G$) to reduce to the case when $G$ is smooth and connected, and then it is very well-known when $k$ is algebraically closed (the case of interest in the question), less widely-known but due to Grothendieck (in SGA3) by a nice short inference from the algebraically  closed case when $k = k_s$, and not at all widely known but announced (without published proof) by Borel & Tits; a proof of the Borel-Tits result is given in Theorem C.2.3 of the book "Pseudo-reductive groups".  The proof in the general case involves a lot of work (but not the main results of the book). 
With the $G(k)$-conjugacy settled, consider the group $W := N_ {G(k)}(S)/Z_ {G(k)}(S)$.  I claim that this is always a finite group, even with $S$ replaced by any $k$-torus in $G$ and without smoothness hypotheses on $G$ (a red herring, by the trick in the preceding paragraph). 
This can be proved in several ways. Here is one.  The functorial normalizer and centralizer of $S$ are defined in Definition A.1.9 of "Pseudo-reductive groups", immediately after which it is proved that they are represented by closed $k$-subgroup schemes $N_G(S)$ and $Z_G(S)$, with $Z_G(S)$ normal in $N_G(S)$ from the definitions. The quotient $W(G,S) := N_G(S)/Z_G(S)$ then makes sense as a finite type $k$-group, and (as is also shown in SGA3 by the same method) the proof of Lemma A.2.9 shows that $W(G,S)$ is \'etale and hence $k$-finite (the smoothness of $G$ which is assumed there isn't relevant to that part of the proof).  Thus, $W(G,S)(k)$ is finite, and $W$ is a subgroup of this (since $N_ {G(k)}(S) = N_G(S)(k)$ and $Z_ {G(k)}(S) = Z_G(S)(k)$). 
Remark: By Proposition C.2.10, if $G$ is smooth and connected then $W(G,S)$ is a constant $k$-group and $W \hookrightarrow W(G,S)(k)$ is an isomorphism. That's nice to know, but we make no use of it. 
OK, so finally we can state the general result: 
Theorem: Let $G$ be an affine group of finite type over a field $k$, and $S$ a maximal $k$-split $k$-torus.  Define $W$ as above, let ${\rm{X}}_ {\ast}(G) = {\rm{Hom}}_ k(\mathbf{G}_ m, G)$, and likewise
for ${\rm{X}}_ {\ast}(S)$.  
Under the natural $G(k)$-action on ${\rm{X}}_ {\ast}(G)$ and the natural $W$-action on ${\rm{X}}_ {\ast}(S)$, the natural map
$$W\backslash {\rm{X}}_ {\ast}(S) \rightarrow G(k) \backslash {\rm{X}}_ {\ast}(G)$$
is bijective.  In particular, $G(k)$-orbits on ${\rm{X}}_ {\ast}(G)$ have finite non-empty intersection with ${\rm{X}}_ {\ast}(S)$. 
Proof: Letting $G' \subseteq G$ denote the Galois descent of the Zariski closure of $G(k_s)$ in $G_ {k_s}$, so $G'$ is a smooth closed $k$-subgroup of $G$ (see Lemma C.4.1 once again), clearly ${\rm{X}}_ {\ast}(G') = {\rm{X}}_ {\ast}(G)$, $G'(k) = G(k)$, and $S$ is contained in $G'$. Thus, we can replace $G$ with $G'$ without affecting what is to be proved, and so we can assume $G$ is smooth (but maybe not connected).
We now adapt the argument from the proof of Lemma C.3.5 (which treats a variant for pseudo-reductive $k$-groups and maximal $k$-tori).  Any $k$-homomorphism $\mathbf{G}_ m \rightarrow G$ has image which is a $k$-split $k$-torus, so it lands in a maximal $k$-split $k$-torus.  But all such $k$-tori are $G(k)$-conjugate to our friend $S$, as explained above (even without smoothness), so we get the surjectivity.
Now consider injectivity (which was the original MO question!). This will go exactly as in  an$\overline{\rm{a}}$dimadhy$\overline{\rm{a}}$nta's answer, using what was done above, but for convenience of a reader who may be nervous about general $k$ we now give the argument. For cocharacters $\mu$ and $\mu'$ of $S$ over $k$, suppose $g \in G(k)$ satisfies $g.\mu' = \mu$ in ${\rm{X}}_ {\ast}(G)$ (where $g.\lambda: t \mapsto g \lambda(t) g^{-1}$ denotes the $G(k)$-action on $k$-rational cocharacters $\lambda$ of $G$).  Thus, $\mu'$ is valued in the commutative $g^{-1} S g$, so $g^{-1} S g$ and $S$ are both contained in the centralizer $Z_G(\mu')$ of $\mu'$ (which is the same as the centralizer of the image of $\mu'$, so it is represented by a closed $k$-subgroup of $G$).  Hence, $g^{-1} S g$ and $S$ are contained in $H := Z_G(\mu')$. Clearly $g^{-1} S g$ and $S$ are visibly maximal $k$-split $k$-tori in $H$, 
so there exists $h \in H(k)$ such that $h^{-1} (g^{-1} S g) h = S$.  It is harmless to replace $g$ with $gh$, yet this brings us to the case $g \in N_ {G(k)}(S)$. QED<|endoftext|>
TITLE: Is there a notion of a "perfectly regular" topological space?
QUESTION [9 upvotes]: The separation axioms have exploded a little since the original list of four!  Amongst them can be found "completely regular" spaces and "perfectly normal" spaces.  The former is well-known: a point can be separated from a disjoint closed subset by a continuous real-valued function.  The latter may be less well-known, but is fairly simple: given two disjoint closed sets then there is a continuous real-valued function such that the first closed set is the preimage of {0} and the second closed set is the preimage of {1}.  This is equivalent to the condition that every closed set be the set of zeros of some continuous real-valued function.
Urysohn's lemma says that there's little point in looking for a completely normal space: in a normal space, any disjoint closed sets can be separated by a continuous real-valued function.  Similarly, if we imposed the obvious exactness condition from perfectly normal onto completely regular then we'd actually end up with T6 (perfectly normal and T0).
However, there's room in the middle for something else and that's what I'm interested in: spaces which are completely regular and are such that the function separating a point from a closed set can be chosen so that the point is the preimage of its value.  No assumption is made on the interaction of the closed set and the function (beyond that already given by the "completely regular" condition).
So my question is simple:

Has anyone encountered this notion before, and if so, where?

If it helps, I think that this condition is equivalent to the space being completely regular and singleton sets being Gδ sets (intersection of a countable number of open sets).  Note that this isn't the same as first countable.

Motivation: As one might expect, my motivation comes from Froelicher spaces.  Any Froelicher space comes equipped with two topologies: generated by either the curves or the functions.  I'm trying to find conditions under which they are equal.  Since the identity is continuous from the curvaceous topology to the functional topology, theorems like "a continuous bijection from a compact space to a Hausdorff space is a homeomorphism" spring to mind.  Hausdorff is no problem - I can easily assume that both topologies are Hausdorff.  In another context, I decided that I quite liked it when the curvaceous topology was sequentially compact.  I think that I can prove that a continuous bijection from a sequentially compact space to a "perfectly regular" space is a homeomorphism, so it's a bit like a weakening of one condition at the expense of strengthening the other.
For a purely topological motivation, consider the following.  In a completely regular space, one can test whether or not a net converges to a particular point by looking at its image under all continuous functions.  In a "perfectly regular" space, one can test whether a convergent net converges to a particular point by using just one function and that function depends only on the proposed limit and not on the net.
So for both of these reasons, it's a nice condition to have and I wondered if it was known about so that, hopefully, I could build on others' work rather than having to invent it myself (hey, I've invented this really cool round thing!).

REPLY [8 votes]: An answer is: completely regular plus countable pseudocharacter, the latter means that points are $G_\delta$-sets. In completely regular spaces a point is a $G_\delta$-set iff it is the zero-set of a continuous function, the proof is just like that for arbitrary closed $G_\delta$-sets in normal spaces: if $\lbrace x\rbrace=\bigcap_u U_n$ then pick continuous functions $f_n:X\to[0,1]$ that are zero at $x$ and $1$ outside $U_n$; then $\sum_n2^{-n}f_n$  has $\lbrace x\rbrace$ as its zero-set. 
See Engeling's General Topology, proof of Corollary 1.5.12.<|endoftext|>
TITLE: What is happening to Martin Gardner's files?
QUESTION [11 upvotes]: Martin Gardner kept voluminous correspondence with amateur and professional mathematicians worldwide throughout his career.  His files are a treasure trove of information about all areas of recreational mathematics.  Does anybody here know whether any portion of those files will be made available to the public for research purposes?

REPLY [18 votes]: According to a simple Google search, the papers were donated to Stanford.
http://www.oac.cdlib.org/data/13030/6s/kt6s20356s/files/kt6s20356s.pdf<|endoftext|>
TITLE: A question on curved algebras, papers by Positselski and E. Segal
QUESTION [6 upvotes]: I am trying to understand something about curved dg algebras as studied by Positselski, E. Segal. These come up in mirror symmetry and when one wants to study Kozsul duality for algebras that are more general then those that fit into the classical framework. 
Let C denote the complex numbers. Suppose we take a cdg-algebra B that is C[x] with a curving x^2, where x is now a variable in odd degree and so does not fit into the matrix factorization framework.  As I understand the situation, then the Koszul dual of this should be an algebra A which is C[y]/(y^2-1), where y is in even degree.In the framework of Positselski, as I understand it one can compute A as Ext(C,C) and pass back to B by taking the Cobar construction. The curving arises from the fact that the coalgebra dual to A is no longer co-augmented. I think that in this simple example one equivalently can also take a more down to earth approach, as was taken by Dyckerhoff in his paper on matrix factorizations, using explicit Koszul resolutions. 
My question is about a confusion I have about the Hochschild cohomology of this dg-category. As I understand it, the Hochschild cohomology of D(B) should be isomorphic to the center of A which is A. Yet if one uses the complex defined in Segal's paper, I believe one gets C[y]/y^2, with y an even variable. The complex he claims should compute HH*(D(B)) is given by the usual Hochschild complex with differential on the algebra B + Gerstenhaber bracket with x^2. To compute, I first computed ordinary HH*(C[x]) and the Gerstenhaber bracket on HH* and then concluded the answer using a spectral sequence. 
Segal's justification seems to be that this is what you get when you regard B as a curved A-infinity algebra, but as I understand Positselski, there seems to be some subtelties with this and that one can often end up with no objects in the category. So I was made a little bit nervous by the justification. Most likely everything is ok and I am missing something stupid, but it would be great if someone would be so kind as to point out where my mistake is.

REPLY [4 votes]: From what you describe my guess is that you may have made a mistake in the argument with the spectral sequence.  When you have a multiplicative spectral sequence, assuming that it converges, the term E_\infty is still not the limit, but rather, the associated graded algebra to a filtration on the limit.  The algebra C[y]/(y^2) is the associated graded algebra to an increasing filtration on the algebra C[y]/(y^2-1).
I would not venture to discuss here the much more delicate issues raised in Ed Segal's answer, which require a careful consideration.  I do have a point of view on these issues; those desiring to get a taste of this point of view can just read the beginning of the introduction to my paper, where the story of two kinds of derived functors and derived categories is told.<|endoftext|>
TITLE: If every point is contained in at most 3 sets and all sets are big, then is the discrepancy zero?
QUESTION [7 upvotes]: Suppose we have a finite, 100-uniform system of sets such that any point is contained in at most 3 sets. Is it true that we can color the points such that every set contains 50 red and 50 blue points?
The question is by Thomas Rothvoss. A positive answer would solve the Three permutations problem of Beck, so a simple answer would be a counterexample...

REPLY [6 votes]: No. 
Given sets 
$$
a_1,a_2,\dots,a_{99},b_1{\rm\ and\ }a_1,a_2,\dots,a_{99},b_2
$$
we see that $b_1$ and $b_2$ must be the same color, say, red. Then from 
$$
b_1,b_2,c_1,c_2,\dots,c_{98}{\rm\ and\ }d_1,d_2,c_1,c_2,\dots,c_{98}
$$
we see $d_1$ and $d_2$ must both be red. Then from 
$$
d_1,d_2,e_1,e_2,\dots,e_{98}{\rm\ and\ }f_1,f_2,e_1,e_2,\dots,e_{98}
$$
we see that $f_1$ and $f_2$ must both be red. Dot, dot, dot. You wind up with as many elements as you like, all of which must be red, and none of them are in more than two of the sets. Once you have more than 50 of them, you can put them in another set which will then have more than 50 red points. 
Obviously, we can take 100 to be a variable in this problem and solution, provided we restrict its range to the positive even integers and understand 50 to be 100/2.<|endoftext|>
TITLE: How long are the certificates produced by the Zeilberger and WZ methods for solving combinatorial sums (A=B)?
QUESTION [7 upvotes]: In the book "A = B" by Petkovesk, Wilf, and Zeilberger, (downloadable here), the authors provide several algorithmic methods for finding closed forms or recurrences for sums involving e.g. binomial coefficients.  Even more exciting, their methods provide seemingly short certificates for the truth of these computer-verified claims.  In particular, the WZ method prints a single rational function as such a certificate.
In more detail, here is the broad outline of the WZ method, where I directly quote from page 25 of "A = B":


Suppose that you wish to prove an identity of the form $\sum_k t(n, k) = rhs(n)$,
  and let’s assume, for now, that for each n it is true that the summand $t(n, k)$
  vanishes for all $k$ outside of some finite interval.
Divide through by the right hand side, so the identity that you wish to prove
  now reads as $\sum_k F (n, k) = 1$, where $F (n, k) = t(n, k)/rhs(n)$.
Let $R(n, k)$ be the rational function that the WZ method provides as the proof
  of your identity (this is described in Chapter 7 of "A=B"). Define a
  new function $G(n, k) = R(n, k)F (n, k).$
You will now observe that the equation
                   $$F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k)$$
  is true. Sum that equation over all integers $k$, and note that the right side
  telescopes to 0. The result is that
                             $$\sum_k F (n + 1, k) =    \sum_k F (n, k),$$
  hence we have shown that         $\sum_k F (n, k)$ is independent of $n$, i.e., is constant.         
Verify that the constant is $1$ by checking that        $F (0, k) = 1$.


What I want to know is:

Are there known bounds on the length of these certificates $R(n, k)$, in terms of the length of the description of the combinatorial sum in question?  If so, what are they?

REPLY [6 votes]: Already for the case of Gosper summation (single-variable), it is known that things can get exponentially larger than the input, because the 'answer' fundamentally depends on the dispersion of the input term.  You will find much more comprehensive answers in the papers of Sergei Abramov as well as those of Marko Petkovsek (especially so in their joint papers!)
The disperson of a polynomial $p(x)$ is the largest integer $n$ such that $p(x)$ and $p(x+n)$ have a non-trivial gcd.  A good understanding of how dispersion enters the picture can be gotten from the paper Shiftless decomposition and polynomial-time rational summation.<|endoftext|>
TITLE: A toy example of a tensor triangulated category?
QUESTION [13 upvotes]: I've been reading Paul Balmer's paper about constructing a "spectrum of prime ideals" on an (essentially small) tensor triangulated category in order to then classify thick subcategories. This is all done to generalize work done in various fields throughout mathematics (e.g. Devinatz, Hopkins, and Smith's work in stable homotopy theory, and Pevtsova and Friedlander's work in finite group schemes). The classic examples of tensor triangulated categories that Balmer talks about are the category of spectra, the category of $G$-modules for some finite group scheme $G$, or the perfect derived category associated to a (topologically Noetherian) scheme (this is related to Thomason's work reconstructing a scheme from the aforementioned category).
But I can't, for the life of me, think of more examples of tensor triangulated categories! (I'm new at all of this...) Can anyone give me a "toy" example of a tensor-triangulated category that is not an example of any of the ones I just listed? By "toy" example I mean that it should be relatively simple with an easy to understand structure. The purpose will be so that I can do Balmer's construction on the toy category to get a better understanding of what's going on.

REPLY [4 votes]: Take a finite dimensional Hopf algebra $H$, the category of $H$-modules is Frobenius (projectives=injectives and there is enough of both); e.g. take $H$ to be the group algebra of a finite group.   So the stable category is triangulated (see Happel's book). In these examples, the usual tensor product of $H$-modules is well defined in the stable category.
This is explained in Khovanov's paper here.    and developed further by You Qi. I recently upload a preprint considering co-Frobenius Hopf algebras (i.e. not necesarily finite dimensional Hopf algebras admitting a (left or right) integral $\Lambda\in H^*$, https://arxiv.org/abs/1904.10430)
The category of comodules over a co-Frobenius Hopf algebra is Frobenius, the stable category is triangulated and the standard tensor product of comodules is well defined in the stable category.<|endoftext|>
TITLE: Relation between Tate's thesis and Class Field Theory
QUESTION [8 upvotes]: Class Field Theory states the correspondence between abelian extensions of k and congruence divisor class.  In idelic language, there is a surjective map from $J_k/k^*$ to $Gal(k^{ab}/k)$ with its kernel unkonwn.
Tate's Thesis proved some functional equations and analytic continuity(with a finite character of $J_k/k^*$). 
Question: Why Tate's thesis contributed to class field theory?

REPLY [3 votes]: Boyarsky has already answered the question in the comments section.  These are a couple of expositions which put John Tate's contribution in perspective.

Stephen Kudla's chapter on Tate's
thesis
Stephen Gelbart's article on Elementary
introduction to Langlands'
program.  See the section beginning page 194<|endoftext|>
TITLE: What is the relationship between modular forms and the Rogers-Ramanujan identities?
QUESTION [6 upvotes]: Let G(q) be the generating function for partitions such that if k is a part, then it occurs once and k+1 is not a part.
Let H(q) be the generating function for partitions with the same condition plus that 1 is not a part.
These are the left-hand sides of the Rogers-Ramanujan Identities.

$G(q)=\displaystyle\sum_{n=0}^\infty\frac{q^{n^2}}{(q;q)_n}=\frac{1}{(q;q^5)_\infty(q^4;q^5)_\infty}$
$H(q)=\displaystyle\sum_{n=0}^\infty\frac{q^{n^2+n}}{(q;q)_n}=\frac{1}{(q^2;q^5)_\infty(q^3;q^5)_\infty}$

I am intrigued by the following unreferenced statement in the wikipedia page:

If q = e2πiτ, then q−1/60G(q) and q11/60H(q) are modular functions of τ.


Do modular forms shed any light on the Rogers-Ramanujan Identities, or is the connection (as far as we know) a curious coincidence?

Is there some class of modular forms whose Fourier series count natural collections of partitions such as those counted by the left-hand sides of the Rogers-Ramanujan Identities? In particular I have in mind the seemingly "non-local" condition that if k is part, then it is distinct and also k+1 is not parts.

In general, how does one tell if a certain generating function (that counts partitions of a certain type, say) is related (by a multiplicative factor like above) to a modular form of some weight for some group (maybe even with some character)?

REPLY [3 votes]: It's hard to compete with Berndt's former student and Berkovich's active collaborator in providing an exhaustive link of references. I can only indicate my own modest contribution, joint with Ole Warnaar (who is an expert in the business), in which you can find links to further literature as well as discussion of other (not originally expected!) aspects of Rogers-Ramanujan identities.
As for the original question,

What is the relationship between Modular Forms and the Rogers-Ramanujan Identities?

the answer is straightforward: whenever you see Rogers-Ramanujan-type identities, both sides are modular forms. It doesn't however work in the opposite direction: there are plenty of modular forms for which an RR-style interpretation isn't known.<|endoftext|>
TITLE: Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$
QUESTION [26 upvotes]: I am trying to prove $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}\binom{2k+1}{j}E^j$;
These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2P_{k-1}(E) =$ some binomial in $E$;
The original sum can be expressed as $P_k(E)x^{(2k-2)} |_{x=k} $;
Try to derive a recurrence for $P_k(E)x^{(2k-2)}$ before plugging in
$x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested by clue 4.

REPLY [20 votes]: This is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.
The Eulerian numbers are expressible as 
$$\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+1\choose i}(m+1-i)^n.$$
View them as functions of $m$ and let $\Delta$ be the backward difference operator, 
$$\Delta f(m)=f(m)-f(m-1).$$
Claim  The $r$th iterated backward difference of the Eulerian number is given by the formula
$$\Delta^r\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+r+1\choose i}(m+1-i)^n.$$
Proof This is proved by induction in $r$ using the binomial identity $${n+r\choose i}+{n+r\choose i-1}={n+r+1\choose i}. \quad\square$$
Setting $m=k-1$ and comparing with the definition of the sequence, we see that 
$$A(k)=\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \choose j}(k-j)^{2k-2}=-\Delta^2\left\langle {n\atop k-1}\right\rangle\ \text{evaluated at }\ n=2k-2.$$
Thus 
$$A(k) = -\Delta\left\langle {n\atop k-1}\right\rangle + \Delta\left\langle {n\atop k-2}\right\rangle = \Delta\left\langle {n\atop k-2}\right\rangle\ \text{evaluated at }\ n=2k-2$$
and the first summand vanishes due to the symmetry of the Eulerian numbers, $\left\langle {n\atop m}\right\rangle=\left\langle {n\atop n-1-m}\right\rangle$, which implies that $\left\langle {2k-2\atop k-1}\right\rangle=\left\langle {2k-2\atop k-2}\right\rangle.$ 
Now the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\Delta\left\langle {n\atop m}\right\rangle\geq 0$ for $m\leq n/2.$ Explicitly, 
$$A(k)=\left\langle {2k-2\atop k-2}\right\rangle-\left\langle {2k-2\atop k-3}\right\rangle > 0\ \text{for}\ k\geq 2.$$<|endoftext|>
TITLE: Is there any literature on multivariable theta functions?
QUESTION [5 upvotes]: The theta function of a lattice is defined to be
$$ \vartheta_\Lambda = \sum_{v\in\Lambda} q^{{\Vert v\Vert}^2}$$
which yields as a coefficient of qk the number of vectors of norm-squared k.
On the other hand, the Jacobi theta function is given by
$$ \vartheta(u,q) = \sum_{n=-\infty}^\infty u^{2n}q^{n^2}$$
and we have the obvious fact that if $\Lambda = \mathbb{Z}$ with its usual intersection form, then $\vartheta(1,q)$ is the theta function for that lattice.
We also have the fact that $\vartheta_{\Lambda_1 \oplus \Lambda_2} = \vartheta_{\Lambda_1}\vartheta_{\Lambda_2}$, and so we can decompose our theta functions into products of theta functions of primitive lattices.
Combining these facts, it is not entirely ridiculous to hope that there is some way to write, for a lattice of rank k, a ``theta function'' of the form
$$\vartheta(u_1, \ldots, u_k, q)$$
such that $\vartheta(1, \ldots, 1, q)$ is the ordinary theta function of the lattice. In some sense, the u-variables keep track of the basis elements of the lattice which immediately raises the question as to well-definedness of such an idea; it is worth noting that for the lattice $\Lambda = \bigoplus_i\mathbb{Z}$ that this definition does make sense.
So is there any literature on such objects? Do they make sense for lattices which are not just sums of copies of ℤ? Do they have nice relations akin to those of normal theta functions?

REPLY [6 votes]: There are three ways to view theta functions

as classical homomorphic functions in
vector z and/or period matrix T
as matrix coefficients of a representation of the
Heisenberg and/or Metaplectic grp
as sections of Line bundles on the Abelian variety
and/or moduli space of the abelian variety

Ram Murty's Theta functions - from the classical to the modern discusses Weil's representation-theoretic interpretation of theta functions.  See chapter 3 by Hoffstein on Eisenstein series and theta functions on the metaplectic group.  It is the connection to the metaplectic group which gives rise to the functional equation of the multivariable theta function, which you will also find in the chapter on the Metaplectic group in vol 3 of Mumford's Tata Lectures
Bellman's Brief introduction to Theta functions Section 61 alludes to theta functions in several complex variables.
You may also want to search for material in books on Abelian varieties.  For example, Baker's Abelian functions Chapter X develops the theory based on the period matrix.  Also Murty's book on Abelian varieties and Polishchuk's Abelian Varieties, Theta Functions and the Fourier Transform 
Tyurin's Quantization, Classical and Quantum Field Theory, and Theta Functions might also be a useful reference, which I haven't browsed.
See also: Springer Encyclopedia of Math entry on theta functions<|endoftext|>
TITLE: Symmetric colorings of regular tessellations
QUESTION [5 upvotes]: Given a regular tessellation, i.e. either a platonic solid (a tessellation of the sphere), the tessellation of the euclidean plane by squares or by regular hexagons, or a regular tessellation of the hyperbolic plane.
One can consider its isometry group $G$. It acts on the set of all faces $F$. I want to define a symmetric coloring of the tessellation as a surjective map from $c:F\rightarrow C$ to a finite set of colors $C$, such that for each group element $G$ there is a permutation $p_g$ of the colors, such that $c(gx)=p_g\circ c(x)$. ($p:G\rightarrow $Sym$(C)$ is a group homomorphism).
Examples for such colorings are the trivial coloring $c:F\rightarrow \{1\}$ or the coloring of the plane as an infinite chessboard.
The only nontrivial symmetric colorings of the tetrahedron, is the one, that assigns a different color to each face. For the other platonic solids there are also those colorings that assign the same colors only to opposite faces. 
So my question is: Does every regular tessellation of the hyperbolic plane admit a nontrivial symmetric coloring?
I wanted to write a computer program that visualizes those tessellations, but I didn't find a good strategy which colors should be used. So I came up with this question.

REPLY [5 votes]: The answer is yes. Moreover, for every two different faces $A$ and $B$ there is a symmetric coloring assigning different colors to $A$ and $B$.
The isometry group $G$ is residually finite, hence here is a normal finite index subgroup $H$ of $G$ that contains no elements (except the identity) sending $A$ to itself or to $B$. Assign a unique color to each orbit of $H$.
The coloring symmetry condition is essentially he following: if $f\in G$ and faces $X$ and $Y$ are of the same color, then so are $f(X)$ and $f(Y)$. Since $X$ and $Y$ are of the same color, there exists $h\in H$ such that $h(X)=Y$. Since $H$ is normal, $h_1:=f^{-1}hf\in H$. But $h_1(F(X))=F(Y)$, hence $F(X)$ and $F(Y)$ are of the same color.<|endoftext|>
TITLE: Local root number
QUESTION [6 upvotes]: I am reading about the L-functions of elliptic curves and I was thinking about the root number as the product of local root numbers. So my question is how to think about the local root numbers geometrically or arithmetically. I have also read that even though the functional equation is conjectural (in different cases) but root number is a well-defined concept. Is it somehow related to the l-adic representation attached to my objects?

REPLY [12 votes]: Yes, if $X$ is a variety over an extension $K$ of $\mathbb Q_p$, then the $\ell$-adic cohomology spaces
$H^i(X,\mathbb Q_{\ell})$ are $\ell$-adic representations of $G_{K}$,
which give rise to Weil--Deligne representations.  (See Tate's Corvallis article,
for example.)  The resulting Weil--Deligne representation is conjectured to be
independent of the choice of $\ell$ (as long as $\ell \neq p$).  This is known
when $X$ has a smooth proper model over the ring of integers of $K$ (although
from the point of view of root numbers this case is not so interesting; in
this good reduction situation the associated Weil--Deligne representation
is unramified, so the local root number is 1, for the right choice of additive 
character).   It is also known when $X$ is an elliptic curve (in which case
$i = 1$ is the interesting choice; this gives the contragredient of the
Tate module).  
Actually computing root numbers is a non-trivial business, especially for
instances of very bad reduction at small primes.  For examples, see recent
work by Mazur and Rubin, and by the Dokchitsers.
[Added in response to Arijit's comment below:]
One of the most interesting applications of root numbers is when you can prove that the product of the local root numbers (which is the global root number) of an elliptic curve is -1.  Then BSD predicts that there will be a rational point of infinite order.
In the Heegner point situation studied by Gross--Zagier, one is in this context (the global root number of the elliptic curve over a quad. imag. field is -1), and if the order of vanishing is precisely one, they produce a point of infinite order.  In general we don't have the technology yet to control Mordell--Weil groups, but can control Selmer groups (which morally should be the same thing, since, following Shafarevic and Tate, one conjectures that Sha has finite order, and hence that the Selmer rank and Mordell--Weil rank coincide).
In the work of Mazur and Rubin, and also of Nekovar, and of the Dokchitsers, the goal is,
under assumptions which makes the global root number -1, to produce the predicted rank in
the Selmer group.  (They succeed in doing in this in many cases!)  Perhaps developing some understanding of this question, and some sense of what people are doing about it, will help give you motivation.   After all, it's pretty surprising that just by computing a sign,
you can predict whether a Diophantine equation will have an interesting solution, and even
more surprising that you can prove non-trivial results in this direction!<|endoftext|>
TITLE: Subposets of small Dushnik-Miller dimension
QUESTION [18 upvotes]: The Dushnik–Miller dimension of a partial order $(P,{\leq})$ is the smallest possible size $d$ for a family ${\leq_1},\ldots,{\leq_d}$ of total orderings of $P$ whose intersection is ${\leq}$, i.e. $x \leq y$ iff $x \leq_i y$ holds simultaneously for all $i = 1,\ldots,d$. Equivalently, the dimension is the smallest $d$ such that $P$ embeds in $L^d$ of some total ordering $L$, where $L^d$ is endowed with the coordinatewise partial ordering.
Since chains have dimension 1 and antichains have dimension 2, Dilworth's Theorem guarantees that every poset of size $n$ contains a $2$-dimensional subposet of size at least $\sqrt{n}$. Is this optimal? In general, what can we say about subposets of dimension $d$?

Tom Goodwillie's argument below shows that for sufficiently large $n$, every poset of size $n$ either has an antichain of size $\sqrt{dn}$ or a $d$-dimensional subposet of size $\sqrt{dn}$. This result is optimal for $d = 1$; stated this way, this could also be optimal for $d > 1$ too. For $d = 2$, this improves my lower bound $\sqrt{n}$ above by a factor of $\sqrt{2}$.
In view of this, let me reformulate the question as follows. Let $F_d(n)$ be the largest integer such that every poset of size $n$ has a $d$-dimensional subposet of size $F_d(n)$. Note that $F_1(n) = 1$ for all $n$ and, when $d > 1$, $F_d(n) \geq \sqrt{dn}$ for large enough $n$.

Is $F_2(n) \leq C\sqrt{n}$ for some constant $C$? In general, what is the asymptotic behavior of $F_d(n)$?

REPLY [3 votes]: Together with Grzegorz Guśpiel and Piotr Micek, we have come up with another upper bound construction. Our example is simply an $n^{d+1}$ grid with natural product order. Obviously it has a $d$-dimensional subposet of size $n^d$, and we show that this is asymptotically the largest such subposet. This gives $F_2(n) = O(n^{2/3})$, and in general $F_d(n) = O(n^\frac{d}{d+1})$, which improves the upper bound by Ben Reiniger and Elyse Yeager for $d \leq 7$.
Our proof uses a multidimensional version of Marcus-Tardos theorem, and for $d=2$ we also provide a slightly different example, which can be analysed with bare hands.
You can find the details on arXiv: https://arxiv.org/abs/1705.00176<|endoftext|>
TITLE: Club sets and substructures
QUESTION [11 upvotes]: This was of course motivated by this question. 
Suppose $\kappa<\theta$ are uncountable regular cardinals. Given a structure ${\mathcal A}=(H_\theta,\in,<,\dots)$ where < is a well-ordering, let $C_{\mathcal A}=${$\sup(M\cap\kappa)\mid M\prec{\mathcal A}\mbox{ and }|M|<\kappa$}. Then $C_{\mathcal A}$ contains a club. Under what (reasonably general) circumstances can we guarantee that $C_{\mathcal A}$ actually is (respectively, is not) a club?

REPLY [12 votes]: The idea in my previous answer can, I think, be upgraded to solve the whole problem, as follows.  Again, fix Skolem functions for $\mathcal A$ as given by the well-ordering $<$, and again let $D$ be a set of fewer than $\kappa$ ordinals $\delta$, each of which is $\sup(\kappa\cap M_\delta)$ for some $M_\delta\prec\mathcal A$ with $|M_\delta|<\kappa$.  I need to show that $\sup(D)$ is also in $C_{\mathcal A}$.  For each $\delta\in D$, define $N_\delta$ to be the Skolem hull of $\kappa\cap\bigcup_{\xi\in D, \xi\leq\delta}M_\xi$.  With appropriate gratitude for the hypothesis that $\kappa$ is regular, note that $N_\delta$ is an elementary submodel of $\mathcal A$ of size $<\kappa$ and that the sequence $\langle N_\delta\rangle$ is an elementary chain.  Let $N$ be its union, and note that it, too, is an elementary submodel of $\mathcal A$ of size $<\kappa$.  So it suffices to show that $\sup(D)=\sup(\kappa\cap N)$.  The $\leq$ direction here is obvious, as $N$ includes $\kappa\cap M_\delta$ for each $\delta\in D$.  To complete the proof, suppose the $\geq$ direction failed.  Then we would have $\sup(D)<\sup(\kappa\cap N)$, so there would be an ordinal $\alpha\in\kappa\cap N$ with $\sup(D)\leq\alpha$.  By construction, we would have some $\delta\in D$ with $\alpha\in N_\delta$, and so $\alpha$ would be of the form $f(\vec\beta)$ for some Skolem function $f$ and some ordinals $\beta_i$ in $\kappa\cap M_{\xi_i}$ for certain $\xi_i\leq\delta$.  For each $i$, we have $\beta_i<\xi_i\leq\delta$, and, since there are only finitely many $i$ (as Skolem functions are finitary), we can find $\gamma<\delta$ with all $\beta_i<\gamma$.  Increasing $\gamma$ if necessary, we can arrange that $\gamma\in M_\delta$.  In $\mathcal A$, we can define the function $g$ sending each ordinal $\nu<\kappa$ to the supremum of all $f(\vec\eta)<\kappa$ for $\eta$ bounded by $\nu$; the values of this function are $<\kappa$ by regularity.  As an elementary submodel of $\mathcal A$, $M_\delta$ is closed under $g$ and, in particular, contains $g(\gamma)$.  But (again by elementarity) $g(\gamma)$ majorizes $f(\vec\beta)=\alpha>\sup(D)\geq\delta$.  That contradicts the fact that $\delta$ is the supremum of $\kappa\cap M_\delta$, and this contradiction completes the proof.<|endoftext|>
TITLE: Leray-Hirsch principle for étale cohomology
QUESTION [10 upvotes]: Let $p:E\to B$ be a continuous map of topological spaces and set $F_x=p^{-1}(x)$ for an $x\in B$. Take a commutative ring  $A$ and assume for simplicity that each $H^\*(F_x,A)$ is a free $A$-module. Let $a_1,a_2,\ldots \in H^\*(E,A)$ be classes that give a basis of $H^\*(F_x,A)$ when restricted to any $F_x$. Assume that the direct image $R^0p_\ast \underline{A}_E$ of the constant sheaf on $E$ is constant. The Leray-Hirsch principle says that $H^\*(E,A)$ is a free $H^\*(B,A)$-module generated by the $a_i$'s.
I would like to ask if anyone knows a reference for a similar result for étale cohomology. Ideally I would like to have a statement for $E,B$ varieties over an algebraically closed field $k$ and finite coefficients of order prime to $char (k)$.

REPLY [11 votes]: [[ I have added a discussion of when $p$ is smooth or has quotient singularities. ]]
[[ I added a discussion on the cohomology of $[X/G]$. ]]
The étale case follows in a way that is altogether analogous to the topological
case. Let me give a proof that gives a teeny bit of extra information. I assume
that $\alpha_i$ is homogenous (with respect to cohomological degree) of degree
$d_i$. Then $\alpha_i$ gives a map in the derived category $A[-d_i]\to
Rp_\ast A$ and combining them a map $\bigoplus_iA[-d_i]\to Rp_\ast A$. If we can
show that this map is an isomorphism then we get an isomorphism
$\bigoplus_iR\Gamma(B,A)[-d_i]\to R\Gamma(E,A)$ which on taking
cohomology gives the L-H theorem. That this is an isomorphism can be checked
fibrewise and if the natural map $(R^ip_\ast A)_x\to H^i(F_x,A)$ is an isomorphism for
all geometric points $x$ we are through.
This condition is true under one of the following conditions:

 $p$ is proper (by the proper base change theorem).

 $p$ is locally trivial by the Künneth formula.

 The second case covers the case of a $G$-torsor. If $G$ acts on $X$ with
finite stabiliser scheme (a condition slightly stronger than having finite
stabilisers but which guarantees that $X/G$ exists) and the orders of the
stabilisers are invertible in the ring of coefficients it is still true. This
can be seen by looking at the stack quotient $X\to[X/G]$ which is a
$G$-torsor (though with base a stack) and at $[X/G]\to X/G$ which induces an
isomorphism in cohomology fulfilling the condition. It should also be possible
to do directly imitating Deligne's proof (in SGA 4 1/2 I think) that the
cohomology of $G$ is indendent of the characteristic (it use the sequence of
fibrations $G\to G/U\to G/B$ where $B$ is a Borel subgroup and $U$ its unipotent radical).

Addendum: Here are, as requested below by algori, some details on the fact that $\pi\colon[X/G]\to X/G$ induces an isomorphism for coefficients $A$ for which the order of the (group of connected components of the) stabilisers are invertible. (This of course is well-known, so well-known in fact that I don't know if there is a proper reference for it.) I will not use that the stack is a global quotient so we may as well consider $\pi\colon\mathcal X\to X$ where $\mathcal X$ is a stack with finite stabiliser scheme and $X$ is its spatial quotient. For simplicity I will assume that the automorphism groups are reduced (i.e., that $\mathcal X$ is a Deligne-Mumford stack). The general case can be proved along the same lines but would be longer and more technical. What we are going to show is that $R\pi_*A=A$. As the construction of the spatial quotient commutes with étale localisation on $X$ we may assume that $X$ is local strictly Henselian and then by the local structure theory of DM-stacks (to be found for instance in Laumon-Moret-Bailly) $\mathcal X$ has the form $[Y/G]$, where $G$ is a finite group which can be assumed to be the stabiliser of a point of $Y$ and hence has order invertible in $A$ and $Y$ is also local strictly Henselian. Now using the usual simplicial resolution $T_n=G^n\times Y$ of $[Y/G]$ we get that $H^*([Y/G],A)=H^*(G,A)=A$ as $H^*(T_n,A)=A^G$.

 $p$ is smooth. This is proved as follows: For any $A$-complex $K$ on $B$ we
get a map $\bigoplus_iK[-d_i]\to Rp_\ast p^\ast K$ which we want to show is an
isomorphism. This map is functorial so in particular if is an isomorphism for
two complexes in a distinguished triangle it is so for the third. By Noetherian
induction (assuming for simplicity $B$ is Noetherian) we may assume that $B$ is
local Henselian and that the statement is true for $B$ replaced by the
complement $U$ of the closed point. We start by showing that that implies that
if $K$ is an $A$-complex on $U$ and if $j\colon U\to B$ is the inclusion, then the
result is true for $Rj_\ast K$. Indeed, this follows directly from smooth base
change which implies that $p^\ast Rj_\ast K=Rj'_\ast p^\ast K$, where $j'\colon p^{-1}U\to E$ is the
inclusion, and then the result follows from the induction hypothesis. On the
other hand, if $K$ is supported on the closed point $x$ of $B$, then the map is
$\bigoplus_iK_x[-d_i]\to R\Gamma(E_x,K_x)$ which is an isomorphism as it is for
$K=A$. Now, the mapping cone of $K\to Rj_\ast j^\ast K$ has support at $x$ so the statement
follows.

 The only thing that is used about a smooth map is that $p$ is universally
locally acyclic for the torsion primes of $A$ (SGA IV: Exp. XVI,
Thm. 1.1). Unless I am mistaken this is true for $p$ that locally are of the
form $E\times_GU\to B\times_GU$, where $E\to B$ is a smooth $G$-map, $U$ a $G$-scheme and $|G|$
is invertible in $A$.

Another way of dealing with the $(G,X)$ case which I think should be more
efficient and general is, following Deligne, to split $X \to X/G$ up into $X\to
X\times_GG/U\to X\times_GG/B\to X/G$ (where $U$ is the unipotent radical of a Borel subgroup
$B$). Then $X\times_GG/B\to X/G$ is proper and in fact the Leray-Hirsch argument
applies provided a large enough integer is invertible in the coefficients (it is
in general not enough to invert the $|H|$ but one also needs to invert some
primes intrinsincally defined by $G$) and $X\to X\times_GG/U$ has more or less affine
spaces as fibres and induces an isomorphism if the $|H|$ are invertible. Finally
$X\times_GG/U\to X\times_GG/B$ is essentially a torus bundle and the cohomology of $X\times_GG/U$
can be analysed in terms of the cohomology of $X\times_GG/B$ and the characteristic
classes of the bundle.<|endoftext|>
TITLE: Dirichlet's theorem for number fields
QUESTION [8 upvotes]: I'd like to see a formulation of Dirichlet's theorem for number fields, i.e. some analogue of the assertion: 
The number of primes less than $N$ congruent to $a \pmod{m}$ where $(a,m)=1$ is 
$\frac{1}{\phi(m)}\pi(N) + o(\pi(N))$
I am really only specifically concerned with such a theorem for the Eisenstein integers  $\mathbb{Z}[\omega]$, but I am curious about the general case as well. Is this true, and if so can somebody provide me with a reference? Thanks!

REPLY [11 votes]: To expand on the excellent comments a bit, one needs both a bit
more and a bit less than Chebotarev's density theorem. :-)
Let's take a number field $K$, and a nonzero ideal $\mathfrak{N}$
of the ring $\mathfrak{O}$ of $K$. We divide the nonzero ideals coprime to $\mathfrak{N}$
into a finite number of classes depending on $\mathfrak{N}$.
There is an equivalence relation on ideals
defined by $\mathfrak{a}\sim\mathfrak{b}$
if $b\mathfrak{a}=a\mathfrak{b}$ where $a$, $b\in\mathfrak{O}$,
$a\equiv b\equiv1$ (mod $\mathfrak{N}$) and $a$ and $b$ are totally positive
(positive in each embedding of $K$ in $\mathbb{R}$). The equivalence classes
are called ray classes modulo $\mathfrak{N}$.
The analogue of Dirichlet's theorem for $K$ is that the prime ideals
of $\mathfrak{O}$ are equidistibuted amongst the ray classes. The strong form
states that if $\pi_{\mathfrak{a}}(N)$ is the number of prime ideals of norm
$\le N$ in the ray class of $\mathfrak{a}$ and 
$\pi_{\mathfrak{O}}(N)$ is the number of prime ideals of norm
$\le N$ in $\mathfrak{O}$ then
$$\lim_{N\to\infty}\frac{\pi_{\mathfrak{a}}(N)}{\pi_{\mathfrak{O}}(N)}
=\frac1m$$
where $m$ is the number of ray classes modulo $\mathfrak{N}$.
To prove this we need less than Chebotarev's theorem, as we need
to apply that only to an abelian extension of $K$, but we need more,
namely a suitable extension $L/K$ to apply it to. This extension
$L/K$ has abelian Galois group $G$ which is in natural correspondence
with the set of ray classes. In detail the Frobenius element attached
to a prime ideal $\mathfrak{p}$ is the element of $G$ corresponding to
the ray class of $\mathfrak{p}$. This extension exists by the
existence theorem of class field theory, quite a deep result.
Let's consider particular examples. For $K=\mathbb{Q}$ all
ideals are principal so take $\mathfrak{N}=(N)$ for a positive integer $N$.
Then for positive integers $r$ and $s$, the ideals $(r)\sim(s)$
iff there are positive integers $a$ and $b$ congruent to $1$ modulo $N$
with $br=as$. This condition is equivalent to $r\equiv s$ (mod $N$).
So ray classes correspond to congruence classes and so we recover
Dirichlet's theorem.
Now let $K=\mathbb{Q}(\omega)$. In this case all ideals are principal.
Take an ideal $\mathfrak{N}=(\nu)$ of $\mathfrak{O}$.
and we find this time that $(\alpha)\sim(\beta)$ if $\alpha\equiv\eta\beta$
(mod $\nu$) where $\eta=\pm \omega^j$ is a unit in $\mathfrak{O}$. (As
$K$ has no real embeddings, the condition of total positivity is redundant.)
So the analogue of Dirichlet here is that for $\alpha$ coprime to
$\nu$ the density of prime ideals $\pi$ with $\pi\equiv\pm\omega^j\alpha$
(mod $\nu$) is indepdendent of the choice of $\alpha$.<|endoftext|>
TITLE: Non-computable but easily described arithmetical functions
QUESTION [15 upvotes]: I have read about the existence of functions of the kind described in the title in several places, but never seen an instance of them. Sorry if this is too much an elementary question to be posted here.

REPLY [11 votes]: Joel and John have given lots of important examples; here is a curiosity:
John Conway invented the Turing complete language FRACTRAN. In this language, a program is simply a a finite sequence of fractions $p_1/q_1$, $p_2/q_2$, ..., $p_r/q_r$. Given an integer $n$, FRACTRAN multiplies $n$ by the first $p_i/q_i$ for which $q_i$ divides $n$. It does this repeatedly until no such $i$ exists, then it halts.
For any FRACTRAN program $(p_1/q_1, \ldots, p_r/q_r)$, consider the function $f$ which returns $1$ if the program halts and $0$ if it does not. This function can be rewritten as a simple piecewise-linear recurrence:

$f(n)=1$ if none of the $q_i$ divides $n$. 
  If $q_i$ is the first $q_i$ which divides $n$, then $f(n)=f(n p_i/q_i)$.
  If $f(n)$ is not forced by the above conditions, then $f(n)=0$.

Now take a program for which it is undecidable, given an input $n$, whether or not it will halt. Would you agree that this is an easily described arithmetic function, if it were written out as a recurrence like this?<|endoftext|>
TITLE: The invariant 3-form on a compact Lie group
QUESTION [19 upvotes]: Let $G$ be a compact Lie group. We have the well-known Maurer-Cartan left-invariant and right-invariant $1$-forms $\theta$ and $\bar\theta$ in $\Omega^1(G, \mathfrak{g})$, probably discussed in every Lie theory lectures.
However the canonical bi-invariant closed $3$-form $\chi = \frac{1}{12} (\theta, [\theta, \theta]) = \frac{1}{12} (\bar\theta, [\bar\theta, \bar\theta])$ in $\Omega^3(G)$ may be a little bit less-known. And when I heard of it I had some questions in mind...
1) Are there canonical invariant $5$-forms and higher invariant forms on a $G$?
2) How are they related to Lie algebra cohomology and equivariant cohomology?
3) The construction looks a little bit like $tr(A \wedge dA)$, if we regard $\theta$ as a connection $A$ on the frame bundle of $G$, and use the Maurer-Cartan equation $d\theta = -\frac{1}{2}[\theta, \theta]$. 
Now there is another famous $3$-form: the Chern-Simons form $tr(A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$, and I wonder if these two are somehow related. Does the $tr(A \wedge A \wedge A)$ part vanish here, due to Jacobi identity? 
(and, note there are Chern-Simons 5-forms etc.)
Thank you very much.

REPLY [8 votes]: The 3-form in question is the extension to de Rham cohomology of the group of the canonical semisimple Lie algebra 3-cocycle $\langle - [-,-] \rangle$. 
A Lie algebra cocycle is the same as a closed element in the Chevalley-Eilenberg algebra. If any Lie algebra cocycle is in transgression with an invariant polynomial on the Lie algebra, then this is witnessed by the existence of a corresponding Chern-Simons element in the Weil algebra. The detailed construction is recalled here.
From this simple cohomological construction result most of the phenomena in higher Chern-Simons theory relating these algebraic entities. I have once written up some notes on this with Jim Stasheff and Hisham Sati (from section 6.3 on it discusses what I just mentioned).<|endoftext|>
TITLE: Does a "Chern character" exist for any generalized cohomology theory?
QUESTION [23 upvotes]: The Chern character is a ring homomorphism from complex K-theory to the usual cohomology.
1) I wonder if there are "Chern character"-like ring homomorphisms from other generalized cohomology theories to the usual cohomology. Are they related with Atiyah-Hirzebruch?
2) And if there are such nice homomorphisms, what is the "Todd genus" in these cases, making the generalization of that famous diagram in Grothendieck–Hirzebruch–Riemann–Roch commute?
When I think about it, I cannot even recall seeing anything like this in real K-theory, but that is probably because I don't really know real K-theory at all.

REPLY [30 votes]: In Oscar Randall-Williams' answer "connective" is unnecessary. Also, there is no need to choose that isomorphism (from a rational theory to the ordinary theory with the same coefficient groups); it is canonical.
The generalization of the Todd genus or Todd class arises when the multiplicative theory $E$ has a "complex orientation": a multiplicatively well-behaved way of producing Thom isomorphisms in $E^*$-theory for all complex vector bundles.<|endoftext|>
TITLE: The unification of Mathematics via Topos Theory
QUESTION [50 upvotes]: In her paper The unification of Mathematics via Topos Theory, Olivia Caramello says "one can generate a huge number of new results in any mathematical field without any creative effort". Is this an exaggeration, and if not is this a new idea or has it always been thought that topos theory could enable automatic generation of theorems ?

REPLY [21 votes]: It would be presumptuous on my part to attempt to answer this question, but I want to share with other MOers this recent paper 
http://www.ihes.fr/~lafforgue/math/TheorieCaramello.pdf
of Laurent Lafforgue and this video 
https://sites.google.com/site/logiquecategorique/Contenus/20130227_Lafforgue
of one of his recent lectures, [dont] le but [] est de poser cette question (inspirée par la théorie de Caramello) : l'indépendance de $l$ de la cohomologie $l$-adique et la correspondance de Langlands sont-elles des équivalences de Morita entre topos classifiants ?
Here is a quote from the paper : La théorie de Caramello... offre déjà un très grand nombre d'exemples d'équivalences
de Morita et de leurs applications. Ces exemples sont étonnament divers et
ils apparaissent presque toujours comme surprenants. Beaucoup d'énoncés
auraient été très difficiles à démontrer, et plus encore à imaginer, sans les
topos et sans les méthodes de calcul que la théorie des topos classifiants et
des équivalences de Morita rend possibles et naturelles. Quand on songe
que la correspondance de Langlands ressemble beaucoup à une equivalence
de Morita et qu'elle en est peut-être une, on se dit que le champ ouvert à
cette théorie est immense.<|endoftext|>
TITLE: How large a subset do you need to uniquely determine a 2-cocycle?
QUESTION [5 upvotes]: Suppose A and B are abelian groups. I want to find subsets D of $A \times A$ such that any 2-cocycle $c:A \times A \to B$ for the trivial action is uniquely determined by what it does on D. (Equivalently, if a 2-cocycle is trivial on all of D it is the zero cocycle).
I'm not looking for a characterization of all such subsets, just some convenient way of describing some such subsets that makes it easier to list all possible 2-cocycles.
By the way, the 2-cocycle condition states that, for all $g,h,k \in A$:
$$c(g,h + k) + c(h,k) = c(g,h) + c(g + h,k)$$
My first thought was to take a generating set S for A and set $D = S \times S$, but this doesn't seem to work. The problem appears to be that, say, something like $c(g,h + k)$ cannot be described purely in terms of $c(g,h)$, $c(g,k)$, and $c(h,k)$.
I'm also interested in the corresponding question when we are restricted to the subgroup of the group of 2-cocycles by one or both of the following conditions: (i) it is skew symmetric, i.e., $c(x,y) = -c(y,x)$, and (ii) $c(a,b) = 0$ whenever $a,b$ generate a cyclic subgroup.
For simplicity, please feel free to assume, for instance, that both A and B are p-groups for the same prime p, or even that they are both vector spaces over the field of p elements. My actual reason for asking this question involves trying to compute 2-cocycles in the case $p = 2$, but I thought this question might be of general interest.

REPLY [5 votes]: See section 8.7.2 p307ff of Holt, Eick, O'Brien's Handbook of Computational Group Theory.  Also see TwoCohomology in the GAP manual.
Roughly speaking normalized cocycles are determined by their values on rule overlaps, and those values are independent of how you overlap.  So for instance if A was cyclic of order 8 with generator a, then ζ( a^i, a^(8-i) ) has a constant value for each 0 < i < 8, namely the element b of B such that in your extension G, the image α of a satisfies α^8 = b.  It is important to normalize the cocycles, otherwise the formulas get messy.  In other words, in the extension G you take a sane transversal of B consisting of reduced words, so there is no loss of generality, just encouragement to keep good hygiene.
If A is a general finite group, then you can replace the confluent polycyclic rewriting system by any confluent rewriting system and still have a very effective algorithm.  You can also replace it by a general finite presentation, but then determining values of the cocycle become "word problems" in a finitely presented group, and are not very suited to algorithmic determination (though as long as A and B are finite, the problems are only practical).
Roughly 1-cohomology corresponds to the bridge between generators and relations, 2-cohomology between relations and associativity (known as "overlaps" when the relations are a rewrite system).<|endoftext|>
TITLE: Catenary curve under non-uniform gravitational field
QUESTION [15 upvotes]: The catenary curve is the shape of a chain hanging between two equal-height poles under the influence of gravity.  But the derivation of the (hyperbolic cosine) curve equation from the physics traditionally assumes a uniform gravitational field.  Suppose instead one uses the non-uniform gravitational field that
diminishes with distance from the center of the Earth.  (Perhaps this would be relevant for
a very long chain that sags significantly.)  Does this lead to an interesting curve, known in some
closed form?  Or just to a differential equation that can only be solved numerically?
I ask this primarily out of curiosity, so please interpret in that spirit!

REPLY [3 votes]: Have a look at 

J.Denzler, A.M.Hinz, Catenaria Vera---the True Catenary, Exposition. Math.
17 (1999), 117--142. (MR 2000e:49006, ZB 943.49001).

—A.M.Hinz (Maribor, Slovenia)

(Image from the cited paper added by J.O'Rourke:)<|endoftext|>
TITLE: Combining DAGs into an acyclic tournament
QUESTION [6 upvotes]: I have a vertex set $V$ and a collection of disjoint arc sets $E_1, \ldots, E_t$ such that $$G_i = (V, E_i),\quad\forall i = 1, \ldots t,$$ are directed acyclic graphs (DAGs) and $$G = (V, E_1 \cup \ldots \cup E_t)$$ is a tournament. We note that the individual DAGs may be disconnected and that $G$ may not be acyclic. However, suppose there exists a bipartition of the arc set indices $\alpha \cup \beta$ such that $$G' = (V, E_\alpha\cup E_\beta^T)$$ is an acyclic tournament where $$E_\alpha = E_{\alpha_1} \cup \ldots \cup E_{\alpha_p}$$ and $$E_\beta = E_{\beta_1} \cup \ldots \cup E_{\beta_q}$$ and $E^T$ is the transpose of $E$ (all the arcs are reversed).
Does anybody know of any results relating to the above? In particular, does anybody know of a method of determining a bipartition $\alpha \cup \beta$, given that at least one exists, other that enumerating all possible bipartitions and checking if the resulting $G'$ is acyclic?

REPLY [2 votes]: The problem you pose, of finding a bipartition if one exists, is of polynomially equivalent difficulty to the decision problem of determining whether a bipartition exists.  The decision problem in turn is NP-complete, by reduction from 3-SAT (and the fact that a solution is easily checked.)
Given an instance of 3-SAT with $n$ clauses, we construct a family of DAGs on $4n$ vertices.  All edges in the complement of $n$ disjoint $4$-cycles will be singleton DAGs.  One "universal" DAG consists of a single edge in each $4$-cycle, and establishes a potential (forbidden) orientation on each $4$-cycle.  Then for every variable in the 3-SAT instance we define a DAG consisting of an edge in each of the $4$-cycles corresponding to the clauses in which that variable appears, with the direction depending on whether the variable appears negated in the clause, in such a way that the forbidden orientation imposed by the universal DAG is achieved in a given $4$-cycle if and only if no literal in the corresponding clause is true, where a variable is considered true when its DAG lies on the same side of the bipartition as the universal DAG and is considered false otherwise.  Then an acyclic bipartition of the DAGs exists if and only if the instance of 3-SAT has a satisfying assignment.<|endoftext|>
TITLE: Algebraic geometry used "externally" (in problems without obvious algebraic structure).
QUESTION [21 upvotes]: This is a request for a list of examples of problems (or other mathematical situations) that are not initially of algebro-geometric nature, but can be solved or understood by using algebraic geometry.  
Here are some applications that are not of the kind sought:

Diophantine equations or other problems whose basic data are specified in algebraic terms, or have an immediate translation into such terms.
GAGA or reduction to finite characteristic arguments, but applied to problems that are clearly already within (or very near) the algebro-geometric sphere, involving varieties or moduli spaces, or cohomology of such spaces.

REPLY [2 votes]: There are several papers by Michael Atiyah where he studies partial differential equations and distributions by methods of algebraic geometry - especially the Hironaka resolution of singularities. See eg Resolution of Singularities and Division of Distributions or the two articles with Bott and Garding about Lacunas for Hyperbolic Differential Operators with Constant Coefficients.<|endoftext|>
TITLE: Achieving consecutive integers as norms from a quadratic field
QUESTION [14 upvotes]: This question is inspired by my inability to make any progress on Will Jagy's question. 
Giving a positive answer to this question should be strictly easier than proving Jagy's conjectures.
Suppose that $K/\mathbb{Q}$ is an imaginary quadratic extension. Let $\chi$ be the corresponding quadratic character. Suppose that there exist $k$ consecutive integers such that $\chi(a)=\chi(a+1)=\ldots=\chi(a+k-1)=1$. Do there necessarily exist infinitely many integers $b$ such that $b$, $b+1$, ... and $b+k-1$ are all norms of ideals in $\mathcal{O}_K$?
For example, the first interesting case is to determine whether there are infinitely many $b$ such that, in the prime factorizations of both $b$ and $b+1$, those primes which are $3$, $5$ or $6$ modulo $7$ all occur an even number of times. 
The motivation here is that Jagy's questions seem to mix a "sieve" question and a "class group" question. My question aims to isolate the sieve problem as its own challenge.

REPLY [7 votes]: The answer to Speyer's question as stated is no, this need not be 
the case.  To see this let $p\equiv 3\pmod 4$ be a prime and consider 
the associated imaginary quadratic field ${\Bbb Q}(\sqrt{-p})$.  Note that 
the associated quadratic character is simply the Legendre symbol $(\frac{n}{p})$.  
Then as observed by David Hansen, if $p$ is sufficiently large, there will 
certainly exist $k$ consecutive quadratic residues $\pmod p$ so that the 
hypothesis in Speyer's question is satisfied.  But on the other hand if 
say $3$ is a non-residue $\pmod p$ then among any six integers there would 
be an integer divisible by $3$ and not $9$ and this integer cannot be 
the norm of an ideal.   
The right conjecture is that if the primes below $k$ are split in 
the number field $K$ then there must be infinitely many strings of $k$ 
consecutive numbers all of which are norms of ideals in $K$.   This does not 
seem easy to prove, and I think (for $k\ge 3$) is comparable in strength 
to the Hardy-Littlewood prime $k$-tuples conjecture.  One can also deduce this 
result from the (generalized) Hardy-Littlewood conjectures:  If $D$ is 
such a discriminant, consider the $k$-tuple $|D|k! n +1, \ldots, |D|k! n+k$. 
These will all satisfy $\chi(|D|k! n+\ell) =1$ (since $\chi(\ell)=1$ for all $\ell \le k$ by assumption), and it should be possible (by Hardy-Littlewood) to arrange all 
the $|D|k!n/\ell +1$ to be primes.  That does the job.  
Alternatively, one can argue as in Hardy-Littlewood and write down conjectures 
for the number of consecutive $k$-tuples that are norms of ideals. One would also 
guess that these could be made to lie in arbitrary classes of the class group, 
and that would answer the motivating question of Jagy on quadratic forms.<|endoftext|>
TITLE: Which Steiner systems come from algebraic geometry?
QUESTION [11 upvotes]: This question is motivated by the ongoing discussion under my answer to this question.  I wrote the following there:

A $(p, q, r)$ Steiner system is a collection of $q$-element subsets $A$ (called blocks) of an $r$-element set $S$ such that every $p$-element subset of $S$ is contained in a unique element of $A$.  Good examples come from considering as blocks the set of hyperplanes in $\mathbb{A}^n$ or $\mathbb{P}^n$ over a finite field.  For example, $\mathbb{A}^2$ over $\mathbb{F}_3$ gives a $(2, 3, 9)$ Steiner system:  it contains $9$ ($\mathbb{F}_3$-rational) points, and let the blocks be the lines, each of which consists of $3$ points.  Then any $2$ points are contained in a unique line.  This is the unique $(2, 3, 9)$ Steiner system.

In general, considering lines in $\mathbb{A}^n$ or $\mathbb{P}^n$ gives an analogous Steiner system, and papers such as this one contain similar constructions.
Loosely, my question is:  which Steiner systems come from similar constructions?  I'll make this more precise in a bit.
An artificial construction allows us to realize any Steiner system as the points of a variety in $\mathbb{A}^n$ over $\mathbb{F}_2$, the blocks of which are given by the intersection of the variety with some specified hyperplanes, as follows.  (This construction is due to Jeremy Booher.)  Say we have a $(p,q, r)$ Steiner system with $k$ blocks; consider the subvariety of $\mathbb{A}^k$ containing the point $y_j=(a_i)_{1\leq i\leq k}$ with $a_i=0$ if the $j$-th element in our Steiner system is in block $i$ and $1$ otherwise.  Then the intersections with the hyperplanes $x_i=0$ give our blocks.  And this subset is a variety as any subset of $\mathbb{A}^k$ is a variety, as it is finite.

So let us try for something harder:  Which Steiner systems $(p, q, r)$ come from a subvariety $X$ of $\mathbb{A}^n$ or $\mathbb{P}^n$ containing $r$ ($\mathbb{F}_s$-rational) points, with the blocks given as the intersections with all $p+1$-dimensional hyperplanes?  A slightly weaker version: when is there a subvariety $X$ of $\mathbb{A}^n$ or $\mathbb{P}^n$ with $r$ $\mathbb{F}_s$-rational points such that every $p+1$-plane intersects $X$ at $q$ points?  In particular, this requires that any $p$ points in $X$ be in general position, so things like rational normal curves are natural candidates.

This is probably too hard, so perhaps the simpler question is tractable:

Can you prove that some Steiner system does not come from this construction?

EDIT:  One way of doing this might be to find a Steiner system with $b$ blocks, where $b$ is not a $q$-binomial coefficient with $q$ a prime power; such systems exist for $p=1$ but I am looking for a non-trivial example.

REPLY [5 votes]: It seems that no Steiner System of the form $(2, 3, 25)$ can be represented in this fashion---many such systems do exist; see here.  In particular, such a system would contain $100$ blocks; but no Grassmannian of lines in $\mathbb{A}^n$ or $\mathbb{P}^n$ over a finite field contains $100$ points.
However, this leaves the possibility of the following (less appealing) construction.  Say a scheme $S$ is $k$-rigid for a scheme $Y$ if for any $k$ points of $Y$, $S$ embeds uniquely in $Y$ (up to automorphisms of $S$) so that it passes through all $k$ points.  (This seems like a natural definition; does it already have a name?)  Then we may ask for varieties $S; T\subset U$ with $S$ $p$-rigid for $U$ such that any embedding of $S$ in $U$ intersects $T$ at $q$ points, where $T$ has $r$ points.<|endoftext|>
TITLE: What's wrong with the surreals?
QUESTION [80 upvotes]: Of all the constructions of the reals, the construction via the surreals seems the most elegant to me.
It seems to immediately capture the total ordering and precision of Dedekind cuts at a fundamental level since the definition of a number is based entirely on how things are ordered.  It avoids, or at least simplifies, the convergence question of Cauchy sequences.  And it naturally transcends finiteness without sacrificing awareness of it.
The one "rumor" I've consistently heard is that it is hard to naturally define integrals and derivatives in the surreals, although I have yet to see a solid technical justification of that.
Are there known results that suggest we should avoid further study of this construction, or that show limitations of it?

REPLY [19 votes]: This is more of an extended footnote to Nombre’s answer than an answer itself. As Nombre’s observations would suggest, I heartily agree that the algebraico-tree-theoretic simplicity hierarchy is critical to the surreals. $\mathbf{No}$ is not just a monster ordered field containing the reals and the ordinals.
The following is a list of some recent papers on the surreals that make critical use of the simplicity hierarchy, and thereby lend credence to Nombre's observations. It is only the beginning of a new wave of work presently being done by model theorist, order algebrists and analysts that take advantage of $\mathbf{No}$’s simplicity-hierarchical structure.
Berarducci, A. and Mantova, V. (2018): Surreal numbers, derivations and transseries, Journal of the European Mathematical Society 20, pp. 339-390. arixv:1503.00315.
Berarducci, A. and Mantova, V. (forthcoming): Transseries as germs of surreal functions, Transactions of the American Mathematical Society, arXiv:1703.01995.
Aschenbrenner, M., van den Dries, L. and van der Hoeven, J. (2018): Numbers, germs and transseries, Proceedings of the International Congress of Mathematicians, Rio De Janeiro, 2018, arXiv:1711.06936.
Aschenbrenner, M., van den Dries, L. and van der Hoeven, J. (forthcoming): Surreal numbers as a universal $H$-field, Journal of the European Mathematical Society arXiv:1512.02267.
Ehrlich, P. and Kaplan, E.: Number systems with simplicity hierarchies: a generalization of Conway's theory of surreal numbers II, The Journal of Symbolic Logic 83 (2018), No. 2, pp. 617-633, arXiv:1512.04001.
Kuhlmann, S. and Matusinski, M.
The exponential-logarithmic equivalence classes of surreal numbers, Order 32 (2015), no. 1, 53–68. arXiv:1203.4538.
Costin, O., Ehrlich, P. and Friedman, H. (24 Aug 2015): Integration on the surreals: a conjecture of Conway, Kruskal and Norton, preprint, arXiv:1505.02478.
The last paper is a rather old version of a paper now in the process of being revised and will eventually be two separate papers.
Edit. May 17, 2020.
The following recent paper by Elliot Kaplan and myself adds further credence to the idea that the algebraico-tree-theoretic simplicity hierarchy is of critical importance to the surreals.
Surreal ordered exponential fields: (https://arxiv.org/abs/2002.07739)
Abstract: In (Ehrlich, J Symb Log, 66, 2001: pp. 1231-1266), the algebraico-tree-theoretic simplicity hierarchical structure of J. H. Conway's ordered field $\mathbf{No}$ of surreal numbers was brought to the fore and employed to provide necessary and sufficient conditions for an ordered field (ordered $K$-vector space) to be isomorphic to an initial subfield ($K$-subspace) of $\mathbf{No}$, i.e. a subfield ($K$-subspace) of $\mathbf{No}$ that is an initial subtree of $\mathbf{No}$. In this sequel to (Ehrlich, J Symb Log, 66, 2001: pp. 1231-1266), piggybacking on the just-said results, analogous results are established for ordered exponential fields. It is further shown that a wide range of ordered exponential fields are isomorphic to initial exponential subfields of $(\mathbf{No}, \exp)$. These include all models of $T(\mathbb{R}_W, e^x)$, where $\mathbb{R}_W$ is the reals expanded by a convergent Weierstrass system $W$. Of these, those we call trigonometric-exponential fields are given particular attention. It is shown that the exponential functions on the initial trigonometric-exponential subfields of $\mathbf{No}$, which includes $\mathbf{No}$ itself, extend to canonical exponential functions on their surcomplex counterparts. This uses the precursory result that trigonometric-exponential initial subfields of $\mathbf{No}$ and trigonometric ordered initial subfields of $\mathbf{No}$, more generally, admit canonical sine and cosine functions. This is shown to apply to the members of a distinguished family of initial exponential subfields of $\mathbf{No}$, to the image of the canonical map of the ordered exponential field $\mathbb{T}$ of transseries into $\mathbf{No}$, which is shown to be initial, and to the ordered exponential fields $\mathbb{R}((\omega))^{EL}$  and $\mathbb{R}\langle\langle\omega\rangle \rangle$, which are likewise shown to be initial.<|endoftext|>
TITLE: When does a submersion have connected fibers?
QUESTION [7 upvotes]: Can we characterize when a submersion $F:M \to N$ between two smooth manifolds has connected fibers? If this is too hard, what are some sufficient conditions?

REPLY [8 votes]: If $M$ and $N$ are both compact, then the submersion $F$ can be thought of as a fiber bundle map with fiber $F^{-1}(p)$ for any $p\in N$.  Then one can apply the long exact sequence of homotopy groups of a fiber bundle to learn that if, for example, $M$ is connected and $N$ is 1-connected, that the fibers must be connected.
These sufficient conditions may be too specific, though.<|endoftext|>
TITLE: Reasons to believe Vopenka's principle/huge cardinals are consistent
QUESTION [28 upvotes]: There are a number of informal heuristic arguments for the consistency of ZFC, enough that I am happy enough to believe that ZFC is consistent.  This is true for even some of the more tame large cardinal axioms, like the existence of an infinite number of Grothendieck universes.  
Are there any such heuristic arguments for the existence of Vopenka cardinals or huge cardinals?  I'd very much like to believe them, mainly because they simplify a great deal of trouble one has to go through when working with accessible categories and localization (every localizer is accessible on a presheaf category, for instance).  
For Vopenka's principle, the category-theoretic definition is that every full complete (cocomplete) subcategory of a locally presentable category is reflective (coreflective).  This seems rather unintuitive to me (and I don't even understand the model-theoretic definition of Vopenka's principle).  
What reason is there to believe that ZFC+VP (or ZFC+HC, which implies the consistency of VP) is consistent?  Obviously, I am willing to accept heuristic or informal arguments (since a formal proof is impossible).

REPLY [33 votes]: Most of the arguments previously presented take a set-theoretic/logical point of view and apply to large cardinal axioms in general.  There's a lot of good stuff there, but I think there are additional things to be said about Vopěnka's principle specifically from a category-theoretic point of view.
One formulation of Vopěnka's principle (which is the one that I'm used to calling "the" category-theoretic definition, and the one used as the definition in Adamek&Rosicky's book, although there are many category-theoretic statements equivalent to VP) is that there does not exist a large (= proper-class-sized) full discrete (= having no nonidentity morphims between its objects) subcategory of any locally presentable category.  I think there is a good argument to be made for the naturalness of this from a category-theoretic perspective.
To explain why, let me back up a bit.  To a category theorist of a certain philosophical bent, one thing that category theory teaches us is to avoid talking about equalities between objects of a category, rather than isomorphism.  For instance, in doing group theory, we never talk about when two groups are equal, only when they are isomorphic.   Likewise in doing topology, we never talk about when two spaces are equal, only when they are homeomorphic.  Once you get used to this, it starts to feel like an accident that it even makes sense to ask whether two groups are equal, rather than merely isomorphic.  And in fact, it is an accident, or at least dependent on the particular choice of axioms for a set-theoretic foundation; one can give other axiomatizations of set theory, provably equivalent to ZFC, in  which it doesn't make sense to ask whether two sets are equal, only whether two elements of a given ambient set are equal.  These are sometimes called "categorial" set theories, since the first example was Lawvere's ETCS which axiomatizes the category of sets, but I prefer to call them structural set theories, since there are other versions, like SEAR, which don't require any category theory.
Now there do exist categories in which it does make sense to talk about "equality" of objects.  For instance, any set X can be regarded as a discrete category $X_d$, whose objects are the elements of X and in which the only morphisms are identities.  Moreover, a category is equivalent to one of the form $X_d$, for some set X, iff it is both a groupoid and a preorder, i.e. every morphism is invertible and any parallel pair of morphisms are equal.  I call such a category a "discrete category," although some people use that only for the stricter notion of a category isomorphic to some $X_d$.  So it becomes tempting to think that one might instead consider "category" to be a fundamental notion, and define "set" to mean a discrete category.
Unfortunately, however, what I wrote in the previous paragraph is false: a category is equivalent to one of the form $X_d$, for some set X, iff it is a groupoid and a preorder and small.  We can just as well construct a category $X_d$ when X is a proper class, and it will of course still be discrete.  In fact, just as a set is the same thing as a small discrete category, a proper class is the same thing as a large discrete category.  However, this feels kind of bizarre, because the large categories that arise in practice are almost never of the sort that admit a meaningful notion of "equality" between their objects, and in particular they are almost never discrete.  Consider the categories of groups, or rings, or topological spaces, or sets for that matter.  Outside of set theory, proper classes usually only arise as the class of objects of some large category, which is almost never discrete.  The world would make much more sense, from a category-theoretic point of view, if there were no such things as proper classes, a.k.a. discrete large categories --- then we could define "set" to mean "discrete category" and life would be beautiful.
Unfortunately, we can't have large categories without having large discrete categories, at least not without restricting the rest of mathematics fairly severly.  This is obviously true if we found mathematics on ZFC or NBG or some other traditional "membership-based" or "material" set theory, since there we need a proper class of objects before we can even define a large category.  But it's also true if we use a structural set theory, since there are a few naturally and structurally defined large categories that are discrete, such as the category of well-orderings and all isomorphisms between them (the core of the full subcategory of Poset on the well-orderings).
Thus Vopěnka's principle, as I stated it above, is a weakened version of the thesis that large discrete categories don't exist: it says that at least they can't exist as full subcategories of locally presentable categories.  Since locally presentable categories are otherwise very well-behaved, this is at least reasonable to hope for.  In fact, from this perspective, if Vopěnka's principle turns out to be inconsistent with ZFC, then maybe it is ZFC that is at fault!  (-:<|endoftext|>
TITLE: Why is the fibered coproduct of affine schemes not affine?
QUESTION [5 upvotes]: I am confused about the following issue:
Let $X=SpecS$, $U_1=SpecR_1$, $U_2=SpecR_2$. and suppose we have maps $S \rightarrow R_1$, $S \rightarrow R_2$. Let $U_3=Spec (R_1 \otimes_S R_2)$. We have scheme maps $U_1 \rightarrow X$, $U_2 \rightarrow X$, $U_3 \rightarrow U_1$, $U_3 \rightarrow U_2$. The particular situation I have in mind is when $U_1$ and $U_2$ are distinguished (corresponding to localization of S at some element) open subschemes of $X$. Intersection of $U_1$ and $U_2$ is $U_3$, and the inclusion of $U_3$ in $X$ corresponds to $S$-algebra structure on $(R_1 \otimes_S R_2)$.
The category of affine schemes (ASch) is the opposite category of commutative rings (CRing).
In CRing kernels (equalisers) of pairs of maps and products exist, so by a lemma from category theory limits should exist, in particular fibered coproducts should exist, so union of two affine schemes $U_1$ and $U_2$ over $U_3$ should be affine scheme $U_4$! But we know that in general it is not so! Maybe the problem is that abstractly it is an affine scheme but what is it's inclusion map into X? Actually there exists an obvious map on the ring side from $S$ to kernel of a pair of maps $R_1 \rightarrow (R_1 \otimes_S R_2)$, $R_2 \rightarrow (R_1 \otimes_S R_2)$.
Thank you!

REPLY [5 votes]: From the categorical point of view the situation is the following. We have the category of affine schemes $AffSch$, the category of all schemes $Sch$, and the inclusion functor $i:AffSch \to Sch$. The functor $i$ has a left adjoint functor $i^*:Sch \to AffSch$, $S \mapsto Spec \Gamma(S,{\mathcal{O}}_S)$, which is sometimes called the affine envelope. 
Now, the coequalizer (the pushout) by definition is the object which corepresents a certain contravariant functor to $Sets$. Note that whenever we have a functor $i:C \to D$ having a left adjoint and a contravariant functor $F:D \to Sets$, if $F$ is corepresentable by an object $X$, then $F\circ i$ is corepresentable as well and the corepresenting object is $i^{\ast}(X)$. Indeed, if $F(Y) = Hom(X,Y)$ then $$F(i(Z)) = Hom(X,i(Z)) = Hom(i^{\ast}(X),Z).$$ Applying this to the situation of the first paragraph we see that the restriction of a corepresentable (by an scheme $S$) functor from $Sch$ to $AffSch$ is correspresentable by $i^{\ast}(S)$. In particular, the coequalizer in the category of affine schemes is the affine envelope of a coequalizer in the category of all schemes.<|endoftext|>
TITLE: Smoothness of frame bundle of (global) orbifolds [reference request]
QUESTION [5 upvotes]: Background
Let $(M,g)$ be a riemannian manifold and let $G$ be a finite group acting effectively and isometrically on $M$.  Recall that this means that for all $x \in G$, the diffeomorphism $\gamma_x$ is such that $\gamma_x^* g = g$ and that if $\gamma_x(p) = p$ for all $p \in M$, then $x$ is the identity element.
If $G$ acts freely, then $M/G$ is a smooth manifold, otherwise let's call it a (global) orbifold.
Years ago I was told that even when the action of $G$ on $M$ is not free, its lift to an action on the orthonormal frame bundle $F(M)$ is free. This means that the quotient $F(M)/G$ is smooth and is the orthonormal frame bundle of $M/G$.
I can see this when $M = \mathbb{R}^n$ with the standard euclidean inner product, so that the action of $G$ is via  orthogonal (hence in particular linear) transformations.  Indeed, suppose that $x\in G$ fixes a point in the frame bundle.  Such a point consists of a  pair $(p,f)$ where $p$ is a point in $M$ and $f$ is a frame for the tangent space $T_pM$ to $M$ at $p$.  The action of $x$ on the pair $(p,f)$ is given by
$$(p,f)\mapsto (\gamma_x(p),(D\gamma_x)_p f),$$
where $(D\gamma_x)_p$ is the derivative (i.e., the push-forward) of $\gamma_x$ at $p$.
Now if $x$ fixes $(p,f)$, then $\gamma_x(p)=p$ and $(D\gamma_x)_p$ is the identity endomorphism of $T_pM$.  But since $\gamma_x$ is linear, it agrees with its derivative, which means that $\gamma_x$ itself is the identity.  Finally, since $G$ acts effectively, we conclude that $x$ is the identity.
Question

Is this still true for $M/G$, where $M$ is a riemannian manifold?  And if so, can someone point me in the direction of a reference where this is proved?

Many thanks in advance.

REPLY [3 votes]: First, one can clearly assume $M$ is connected by simply applying the argument to each componenet of $M$.
The key fact is a generalization of your argument for $M=\mathbb{R}^n$: that if $f:M\rightarrow M$ is an isometry with $M$ connected and if there is a point $p\in M$ with $f(p) = p$ and $d_pf = Id$, then $f$ itself is the identity map.
Assuming this fact for the moment, then if $\gamma_x(p,f) = (p,f)$ one concludes $\gamma_x = Id$, and then since the action is effective, x itself must have been the identity element.  Thus, the only element of $G$ which fixes any element of the frame bundle is the identity, so the action on the frame bundle is free.
Now, why is the fact true?
Set $X = \{p\in M| f(p) = p$ and $D_p f = Id\}$.  $X$ is nonempty by assumption and clearly closed (by, say, a continuity argument).  If we can show it's open, then we'll have $X=M$ by connectedness of $M$.  Thus, in particular, $f(p) = p$ for all $p\in M$.
To see $X$ is open, let $q\in X$.  Let $U_q$ be a normal neighborhood around $q$.  Normal means that every pair of points in $U_q$ has a unique minimal geodesic between them.  Normal neighborhoods always exist (any sufficiently small neighborhood is totally normal), though I don't immediately remember how to prove it, only that it uses the Gauss lemma.  Do Carmo's Riemannian Geometry book proves it, if you need a reference for this.
I claim that $U_q\subseteq X$.  For, if $r\in U_q$, there is a unique minimal geodesic $c$ with $c(0) = q$ and $c(1) = r$.  Since $d_q f = Id$, we must have $f(c(t)) = c(t)$, so in particular, $f(r) = f(c(1)) = c(1) = r$.  In short, $f$ fixes all points in $U_q$
So, we must simply establish that $d_r f = Id$.  But, if there is a $v\in T_r M$ with $d_r f v\neq v$, then the we have $f(exp(tv))\neq f(exp(t d_r f v))$ for all sufficiently small $t$.  But for all sufficiently small $t$, both curves lie in $U_q$ and we know $f$ fixes all points in $U_q$.  Thus, $d_f v = v$ and the result follows.

REPLY [3 votes]: I recommend looking at Moerdijk and MrCun's "Introduction to Foliations and Lie Groupoids". On page 42, they define the frame bundle and orthonormal frame-bundle of an orbifold and show it's a smooth manifold. To do this, they first establish this on orbifold chart $(U,G,\varphi)$ by essentially your proof. Then you need to take the filtered colimit of the smooth manifolds resulting from each chart. This becomes the frame bundle of the orbifold, and is of course, a manifold.
(So, applying this to canonical orbifold charts on $M/G$, this gives you the proof).

REPLY [2 votes]: Jason has already explained the proof. The point is that, given a connected Riemannian manifold, any isometry is uniquely determined by how it acts on a single point and the tangent space at that point. The core reason, as explained by Jason, for this is the existence and uniqueness theorem for second order ODE's applied to geodesics radiating from the point.
As for references, I would be surprised if it is not in Kobayashi-Nomizu or Kobayashi's book, Transformation Groups in Differential Geometry. I'm sure it's in other places, as well. Look in any proof that the isometry group is finite-dimensional.<|endoftext|>
TITLE: Geometric (or intuitive) interpretation of additional derivatives in characteristic p > 0
QUESTION [9 upvotes]: In characteristic $p > 0$ there are "extra" differential operators, i.e., ones that are outside the algebra generated by first-order derivations.
Is there any interpretation of these operators in terms of Grothendieck (or other) style geometric ideas such as thickenings, formal schemes, lambda rings, or crystalline cohomology?
A related and possibly equivalent question is whether there is any precise sense in which the "extra" material in divided power algebras (compared to polynomial or power-series rings) is dual to the "extra" differential operators in positive characteristic(s).

REPLY [5 votes]: In a nutshell, observe that $\partial_x^p(x^p)=0$ in characteristic $p$. This means that you can divide either $\partial_x^p$ or $x^p$ by $p!$ and get a sensible object, i.e., the second object will interact with the divided object and you get $=1$ in the formula. Dividing $\partial_x^p$ you get your missing differential operators. Dividing $x^p$ you lose it but discover "crystalline world", "sickening", etc.
To put it bluntly, No, in crystalline world, differential operators on a smooth scheme are generated in degree 1, but yes, your missing operators will live on a formal scheme...
What was your question exactly, Doc?<|endoftext|>
TITLE: K_2 of rings of algebraic integers
QUESTION [14 upvotes]: Let $R$ be the ring of integers in an algebraic number field.  There are beautiful descriptions of $K_0(R)$ and $K_1(R)$.  Namely, $\tilde{K}_0(R)$ is the class group of $R$ and $K_1(R)$ is the group of units of $R$.  Question : Is there a nice description of $K_2(R)$ (or at least some reasonable conjectures)?  I couldn't find much about this in Milnor's or Rosenberg's books on algebraic K-theory, so I expect that the answer is pretty complicated.  Is it maybe at least known in some special cases (say, for $R$ the integers in a quadratic extension of $\mathbb{Q}$)?

REPLY [4 votes]: I think Kolster's survey is a good introduction to questions related to 
arithmetic interpretations of higher K-groups.<|endoftext|>
TITLE: Can we hope to solve all Diophantine equations?
QUESTION [6 upvotes]: According to Godel result, neither ZFC nor other particular theory is strong enough to resolve all questions about, say, Diophantine equations. But maybe we can hope that a sequence of theories will help? It is known that ZFC-1 theory (ZFC + Cons(ZFC) ) is much stronger than ZFC, in sense that now there are theorems with extremely shorter proofs, and many new theorem are now decidable. If we continue this to ZFC-2, .., ZFC-n, ... then ZFC-w which is union of all, ZFC-(w+1) and so on, we can continue to extremely large sets of theories, about all of them we have no doubts, and maybe now for every natural Diophantine equation we can choose a theory witch resolve it? Moreover, if we would be able to imagine non-enumerable set of such theories, may be we could hope that for EVERY Diophantine equation has a corresponding theory from this set in which it can be resolved? Or this is trivially incorrect “conjecture”? It seems that this does not contradict to Godel Theorem, which consider one theory, not a sequence of theories.
Another way of thinking about the same idea is to take only axioms from ZFC but add a new derivation rule, which would say that "from any set of axioms A it follows that A is consistent". With this derivation rule we would derive Cons(ZFC) in one step! So, for some reasons (by the way, I do not understand why) Godel theorem is not applicable here. May we hope that with ZFC extended with such a new derivation rule we can, say, solve all Diophantine equations?

REPLY [23 votes]: On the one hand, no consistent theory $T$ that we can
describe by giving a computable enumeration of its axioms
can settle all Diophantine equations. The reason is that
for any such theory $T$, we can construct an integer
polynomial $p_T(\vec x)$ that has a solution in the
integers if and only if $T$ is inconsistent. Thus, since
$T$ is consistent, $p_T$ will have no solutions, but $T$
will not prove this.
One may construct $p_T$ by understanding the MRDP solution
to Hilbert's 10th problem. In that argument, for any Turing
machine program, one may construct a polynomial whose
solutions correspond to halting instances of the program.
But consider the program that searches for a proof of a
contradiction in $T$, halting only when one is found. The
corresponding polynomial $p_T$ for this program will have a
solution if and only if the program halts, which is if and
only if $T$ is inconsistent, as desired.
So we cannot hope to settle all Diophantine equations with
respect to one computably enumerable theory.
Nevertheless, we can describe theories that solve all
Diophantine equations in other ways. For example, the
theory TA known as True Arithmetic, extends PA and consists precisely of
the first order assertions that are true in the standard
model $\langle \mathbb{N},+,\cdot,0,1,\lt\rangle$. (One could use $\mathbb{Z}$ in place of $\mathbb{N}$ here, or just realize that $\mathbb{Z}$ is interpretable in $\mathbb{N}$.) Our
background theory ZFC proves that TA is consistent and
complete, and it certainly correctly settles all the
Diophantine equations, proving of exactly those polynomials
that have solutions that they do have solutions and of the others that they do not. So this
theory is the kind of limit theory you requested. But the
difficulty with TA and with any of the non-enumerable limit
theories that you seek, is that it is too difficult to
recognize the axioms. We cannot tell if a proof from TA
is legitimate, because we cannot even recognize the axioms.
For the purpose of settling Diophantine equations, it would
suffice to use $TA_1$, consisting just of the true
$\Pi_1$ assertions. But recognizing whether a given
$\Pi_1$ assertion is true or not (that is, recognizing it as an axiom of $TA_1$) is exactly as difficult
as recognizing whether a given Diophantine equation has solutions in the integers or not. 
This last difficulty is inherent in the problem, since any theory correctly proving whether the polynomials have solutions or not will prove all the instances of $TA_1$. So $TA_1$ a minimal instance of the theory you seek, but it is not clear how useful it is for your purpose, since it is as difficult to recognize the axioms of this theory as it is to solve the problems you intend to solve with it.<|endoftext|>
TITLE: Minimal-length embeddings of braids into R^3 with fixed endpoints
QUESTION [13 upvotes]: (Apologies in advance for any imprecision in the following; I am a computer scientist and regret never having taken an actual course on topology.)
One way to define the pure braid group $P_n$ is as follows: consider a pure braid to be a set of $n$ non-intersecting arcs in $x,y,t$-space which are monotone in the $t$ direction, such that the $i$th arc connects $(i,0,0)$ to $(i,0,1)$. Sets of arcs which can be deformed continuously into each other without any arcs intersecting are considered equivalent pure braids.
Consider a generalization where the endpoints are not integer positions on two lines, but arbitrary fixed positions in the $t = 0$ and $t = 1$ planes. It seems clear to me that this does not change the topology of the space [1], so I will call such a collection of arcs an embedded braid. I'm looking at characterizing "optimal" embedded braids that minimize a certain functional $F$. For concreteness, let $F$ be the total length of the arcs (the actual functional I need to use is slightly different, but this should be close enough to carry over the results).
If the requirement that arcs do not intersect is removed, the space of embedded braids can be given an affine structure over which $F$ is convex, and it is immediately apparent that there is a single local minimum which is the global minimum: connect each pair of endpoints with a straight line. What can we say about the local minima of $F$ if we retain the non-intersection property? My intuition says that each topologically distinct braiding (corresponding to a particular element of $P_n$) forms a connected component with a unique local minimum, but I cannot tell how to begin proving this. It would be true if one could show that the connected components are convex subsets of the space; this does not hold under the "obvious" affine structure where we treat each arc and each $t$ independently, but that leaves the possibility of some other choice of affine structure which works. Are there some nice proof techniques that would help proving uniqueness of local minima in this context? Would any ideas or analogies from the theory of minimal surfaces help?
As Andrew Stacey mentions, the connected components are open subspaces; I believe existence of local minima can be guaranteed by considering the closure of the component seen as a subset of the space of embeddings that allow intersection. (Following Kevin Walker's comment, I realize this is called compactification.) This would include embeddings "on the boundary", whose arcs can intersect but which are only an infinitesimal displacement away from non-intersecting embeddings with the right topology. As a concrete example, the following braid,

\   /
 \ /
  \
 / \
 \ /
  \
 / \
/   \

on being pulled tight, approaches a configuration where the arcs are infinitesimally close together in the middle; the minimal length is attained by the intersecting embedding where the two arcs share a common point.
[1] Since the endpoints are fixed, we can associate each such generalized braid uniquely with a pure braid by shrinking the $t$ dimension slightly and composing each arc $i$ with a fixed arc from its endpoints to $(i,0,0)$ and $(i,0,1)$ respectively.

REPLY [7 votes]: UPDATE.
I revisited the question and realized that verifying the local CAT(0) property is not that easy. When I wrote the original answer, I was under impression that removing any collection of codimension 2 subspaces (more precisely, their tubular neighborhoods) from $\mathbb R^n$ leaves one with a locally CAT(0) space. This is true in $\mathbb R^3$ but there are counterexamples in $\mathbb R^4$. This particular subset might satisfy the condition but this does not seem to follow from any "generic" argument.
Further, there are some discouraging examples. First, approximating by arcs by "ropes" with fixed-size square sections does not actually work: in this case there are non-unique minimal configurations. So one really needs to deal with zero-width ones. Second, if you consider 3 arcs and allow two of them intersect while deforming the braid (but still disallow intersections with the 3rd one), then again, you can have two distinct minimal configurations in the same equivalence class.
Below is the original answer (and I suggest that it is unaccepted).

I can prove uniqueness of a local minimum for another length-like functional
(similar but not equal to the sum of lengths). I believe that it should work the same way
for the sum of lengths, but unfortunately the underlying geometric theory does not
seem to exist (yet?).
First let me reformulate the problem. Let $X$ denote the set of possible horisontal cross-sections of braids.
This is the set of $n$-tuples of distinct points in $\mathbb R^2$. Geometrically this is $(\mathbb R^2)^n$
minus a collection of codimension 2 subspaces corresponding to positions where some two points coincide.
Actually I prefer another formalization: the arcs forming the braid are ropes of nonzero width
and with square cross-sections.
More precisely, the horizontal section of every rope is an $\varepsilon\times\varepsilon$ square (with sides parallel
to the coordinate axes), and these sections should not overlap. So $X$ is $\mathbb R^{2n}$ minus a union of polyhedral
heighborhoods of codimension 2 subspaces. This formalization makes the local structure simpler,
and the original one is the limit as $\varepsilon\to 0$.
An embedded braid is a path $f:[0,1]\to X$. We want to minimize the functional
$$
 L = \int_0^1 \left(\sqrt{(df_1/dt)^2+1}+\dots+\sqrt{(df_n/dt)^2+1}\right) dt,
$$
over all paths between two given positions $a,b\in X$, in a given connected component
of the space of such paths. Here $f_i=f_i(t)$ are 2-dimensional coordinates.
Another functional,
$$
 L' = \int_0^1 \sqrt{(df_1/dt)^2+\dots+(df_n/dt)^2+1}\ \  dt,
$$
is easier to deal with, because this is the Euclidean length of the corresponding
path $(f(t),t)$ in $X\times\mathbb R\subset \mathbb R^{2n+1}$. Let us work with $L'$.
The connected component of the set of paths is the homotopy class. Fixing the homotopy class of a path is the same as
fixing endpoints of its lift to the universal cover of the space.
(In the case of zero-width ropes, you first take the universal cover and then the completion in order to add "boundary cases".)
And local length minimizers are called geodesics. So we want to show that, in the universal cover of $X\times\mathbb R$,
every pair of points is connected by a unique geodesic.
Our space $X$ (and hence $X\times\mathbb R$) is a locally CAT(0) space.
This can be shown using standard curvature tests for polyhedral spaces.
A globalization theorem says that a complete, simply connected, locally CAT(0) space is globally CAT(0),
in other words, it is a Hadamard space. So the universal cover is a Hadamard space.
Hadamard spaces feature uniqueness of geodesics, hence the result.
Now what about the original functional $L$? It is also a length functional, but for a non-Euclidean norm on $\mathbb R^{2n+1}$.
So, to carry over the proof, one needs to develop an analog of CAT(0) spaces modelled after Finsler spaces rather than Riemannian. I think it should be possible - after all, all this CAT(0) business is about convexity of distance, and this convexity is there in all normed vector spaces.
But I have not heard of such generalizations (maybe this is just my ignorance).<|endoftext|>
TITLE: Sheaf cohomology question
QUESTION [8 upvotes]: For a topological space $X$ and a sheaf of abelian groups $F$ on it, sheaf cohomology $H^n(X,F)$ is
defined.
Singular cohomology of $X$ can be expressed as sheaf cohomology if $X$ is locally contractible and $F$ is the sheaf of locally constant functions.
I have two related questions.
For an algebraic scheme $X$, one uses the sheaf cohomology with the structure sheaf $F=\mathcal{O}_X$. What happens if $X$ is a topological manifold and $F$ is the sheaf of continuous functions to the real numbers? Or differentiable manifolds? Does this cohomology have a special name under which I can search for literature?
Doesn't one get many interesting cohomology theories besides singular cohomology for a topological space $X$ from sheaf cohomology? I mean for a classical topological space, not a scheme, which is Hausdorff and so forth. Is there a list or an overview in the literature? Thank you.
I like to restate the second question. Does any reasonable cohomology theory of topological spaces come from sheaf cohomology?

REPLY [8 votes]: Sheaves of continuous or smooth functions with values in a vector bundle have zero cohomology groups because of the existence of partitions of unity. 
Still they provide interesting resolutions for the locally constant sheaf, e.g. by considering the bundle built from exterior algebras on the tangent space, and that's a way to prove that De Rham cohomology coincides with singular or Cech cohomology for paracompact manifolds.
Of course, we can look at the De Rham complex obtained by tensoring the exterior forms by an arbitrary bundle on the manifold. In which case we obtain an interesting cohomology with "values" in a vector bundle instead of a coefficient ring.
This is similar to singular cohomology with twisted coefficient.
And I think that's the kind of examples that gave rise to the concept of sheaf.<|endoftext|>
TITLE: Convergence of orthogonal polynomial expansions
QUESTION [6 upvotes]: "Everyone" knows that for a general $f\in L^2[0,1]$, the Fourier series of $f$ converges to $f$ in the $L^2$ norm but not necessarily in most other senses one might be interested in; but if $f$ is reasonably nice, then its Fourier series converges to $f$, say, uniformly.
I'm looking for similar results about orthogonal polynomial expansions for functions on the whole real line.  What I specifically want at the moment is sufficient conditions on a bounded function $f:\mathbb{R} \to \mathbb{R}$ so that the partial sums of its Hermite polynomial expansion are uniformly bounded on compact sets, but I'm also interested in learning what's known about pointwise/uniform/etc. convergence results for Hermite and other classical orthogonal polynomials.
Possibly such results follow trivially from well-known basic facts about Hermite polynomials, but I'm not familiar with that literature and I'm having trouble navigating it. So in addition to precise answers, I'd appreciate literature tips (but please don't just tell me to look at Szegő's book unless you have a specific section to recommend).

REPLY [5 votes]: I too was looking into this recently and found good citations difficult to find.  Here are two I found which I can recommend:
.1. In the book "Special Functions and their Applications" by Lebedev, he gives the following clear theorem statement (with a moderately technical but not too bad-looking proof): (Theorem 2) "Assume $f \in L^2(\gamma)$ (i.e., is square-integrable w.r.t. the standard Gaussian measure) and is piecewise-$\mathcal{C}^1$ on every finite interval $[-a,a]$.  Then the Hermite expansion of $f$ converges pointwise at every point of continuity of $f$", and further, (Remark 1) "converges to $(f(x^+)+f(x^-))/2$ at any jump discontinuity $x$."
It would be nice to also know that one has uniform convergence on any interval $[-a,a]$, but I didn't immediately see how to read that out.
.2. The book "Gaussian measures" by Bogachev gives pretty careful statements about the domains under which formal operations (e.g., differentiation of Hermite expansions) hold.<|endoftext|>
TITLE: Is Soergel's proof of Kazhdan-Lusztig positivity for Weyl groups independent of other proofs?
QUESTION [16 upvotes]: Let $(W, S)$ be a Coxeter system.  Soergel defined a category of bimodules $B$ over a polynomial ring whose split Grothendieck group is isomorphic to the Hecke algebra $H$ of $W$.  Conjecturally, the image of certain indecomposable (projective?) bimodules in $B$ is the well-known Kazhdan-Lusztig basis of $H$.  Assuming the conjecture, Soergel showed that the coefficients of the Kazhdan-Lusztig polynomials of $W$ are given by the dimensions of certain Hom-spaces in $B$.  It follows that these coefficients are non-negative, which was already known by work of Kazhdan-Lusztig in the Weyl group case by linking these coefficients to intersection cohomology of the corresponding Schubert varieties.
Soergel proved this conjecture in 1992 for $W$ a Weyl group, and Härterich proved it in 1999 for $W$ an affine Weyl group.  Unfortunately, I can't access the first paper, and the second paper is in German, so I don't know anything about either of these proofs.  
Question:  Do these proofs depend on the relationship of the coefficients of the K-L polynomials to intersection cohomology, or are they independent of the corresponding machinery?
(The reason I ask is that I am potentially interested in relating known combinatorial proofs of positivity to Soergel's work, and I want to get an idea of how much machinery I would need to learn to do this.)
Edit:  Soergel's 1992 paper is here, if only I had the appropriate journal access.  If anybody does and would like to send me this paper, that would be excellent - my contact information is at a link on my profile.

REPLY [23 votes]: Perhaps I can supplement Jim's answer a little.
In the paper "Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln uber Polynomringen" Soergel shows that there are certain graded indecomposable bimodules over a polynomial ring (now known as Soergel bimodules) which categorify the Hecke algebra. (Note that these are not projective!!)
That is, the indecomposable objects are classified (up to shifts and isomorphism) by the Weyl group, and one has an isomorphism between the Hecke algebra and the split Grothendieck group of the category of Soergel bimodules.
As a consequence, one obtains a basis for the Hecke algebra which is positive in the standard basis (as follows from the construction of the isomorphism of the Grothendieck group with the Hecke algebra) and has positive structure constants (because it is a categorification).
Soergel conjectures that this basis is in fact the Kazhdan-Lusztig basis, which would imply positivity in general.
Up until now there are only two cases when one can verify Soergel's conjecture:

when one has some sort of geometry (in which case one can show that the indecomposable Soergel bimodules are the equivariant intersection cohomology of Schubert varieties, and then use old results of Kazhdan and Lusztig). This shows Soergel's conjecture for Coxeter groups associated to Kac-Moody groups (in particular finite and affine Weyl groups).
when the combinatorics is very simple (i.e. for dihedral groups (Soergel) or universal Coxeter groups (Fiebig, Libedinsky)).

Hence, up until now there are no examples where Soergel's conjecture has yielded positivity when it was not known by other means. Also note that in the vast majority of cases, the proof using Soergel bimodules is strictly more complicated than the geometry proof, as one needs an extra step to get from geometry to Soergel bimodules.
Soergel's conjecture would however have more far reaching consequences than a proof that Kazhdan-Lusztig polynomials have positive coefficients. For example it provides a natural "geometry" for arbitrary Coxeter groups. For example, generalising some sort of Soergel bimodules to complex reflection groups would yield a natural setting for the study of "spetses" (unipotent characters associated to complex reflection groups).
One should also note that Dyer has developed a very similar conjectural world associating commutative algebra categories to Coxeter groups. He instead considers modules over the dual nil Hecke ring (which is the analogue of the cohomology of a flag variety), and has many nice results and conjectures. (Much of his work considers more general orders than the Bruhat order, and so will probably come in handy soon ...!)
While I am at it I should mention Peter Fiebig's theory of Braden-MacPherson sheaves on moment graphs. This is (in a sense made precise in one of Peter's papers) a local version of Soergel bimodules, and hence many questions become more natural on the moment graph.
Finally, one should mention the recent work of Elias-Khovanov and Libedinsky, which give generators and relations for the monoidal category of Soergel bimodules for certain Coxeter groups. (Elias-Khovanov in type A and Libedinsky in right-angled type.) These are very interesting results, but it is unclear to what extent they can be used to attack Soergel's conjecture.<|endoftext|>
TITLE: Non-split extensions of $GL_n(F_q)$ by $F_q^n$ ?
QUESTION [19 upvotes]: A very naive question :
I just learned that there is a non-split extension of $GL_3(F_2)$ by $F_2^3$ (with standard action). It can be realized as the subgroup of the automorphism group $G_2$ of Cayley-Graves octaves (edit: octonions) that preserve up to sign the basis $e_i$, $i=1..7$of imaginary octaves. Does this happen for other values of $(n,q)$ (as in the title) ?

REPLY [10 votes]: There's an elementary (non-cohomological) group-theoretic explanation for why $GL(n,q)$ must split over its natural module in odd characteristic. Suppose $N \triangleleft G$, where
$G/N \cong GL(n,q)$ with $q$ odd and $N$ is isomorphic to the natural module for $G/N$. The negative of the identity matrix in $GL(n,q)$ is a central element of order $2$ that acts on $N$ with no nonidentity fixed points. Thus there exists $T \triangleleft G$ with
$N \subseteq T$ such that $T/N$ has order $2$, and if $S$ is a Sylow $2$-subgroup of $T$ then ${\bf C}_N(S) = 1$. Now let $K = {\bf N}_G(S)$. By the Frattini argument,
$G = TK = NSK = NK$. Also, $N \cap K = 1$ since $T$ must centralize $N \cap K$. Thus $K$ is the desired complement for $N$ in $G$.<|endoftext|>
TITLE: Difference between represented and singular cohomology?
QUESTION [19 upvotes]: Ordinary cohomology on CW complexes is determined by the coefficients.  There are (more than) two nice ways to define cohomology for non-CW-complexes:  either by singular cohomology or 
by defining $\widetilde H^n(X;G) = [X, K(G,n)]$.  Are there standard/easy examples where these
two theories differ?
One idea that comes to mind is the paper by Milnor and Barratt (about Anomolous Singular
Homology) which says that the  $n$-dimensional Hawaiian earring $H^n$ has nontrivial singular
homology in arbitrarily high dimensions.  But I don't see an easy way to compute
$[H^n, K(G, m)]$.

REPLY [7 votes]: Here are two examples similar to Ben's answer, followed by an example with a path-connected space.
Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.
Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve
$$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$
Its zeroth singular cohomology is the finite product
$$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$
Its zeroth naive cohomology is
$$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$
A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.
Example 2. Let $X \subset \mathbb{R}$ be
$$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$
Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product
$$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$
The zeroth naive cohomology is the subgroup
$$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$
which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.
Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map
$$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$
which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space
$$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does.
Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map
$$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$
is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $E' \stackrel{\sim}{\to} E$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.
Edit. For the record, I previously thought that a similar argument would work with integral cohomology $H^1(E;\mathbb{Z})$, but I had forgotten Specker's theorem $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$.
In fact, it turns out that said cohomology group is $H^1(E;\mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$, as explained here.
[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.<|endoftext|>
TITLE: Noncombinatorial proofs of Ramsey's Theorem?
QUESTION [13 upvotes]: I know of 2(.5) proofs of Ramsey's theorem, which states (in its simplest form) that for all $k, l\in \mathbb{N}$ there exists an integer $R(k, l)$ with the following property: for any $n>R(k, l)$, any $2$-coloring of the edges of $K_n$ contains either a red $K_k$ or a blue $K_l$.
Both the finite and the infinite versions (the latter being--a 2-coloring of the edges of $K_\mathbb{N}$ contains an infinite monochrome $K_\mathbb{N}$) are proven on Wikipedia, and one may deduce the finite version from the infinite one by compactness, or equivalently using Konig's lemma.  The infinitary proof does not give effective bounds on $R(k, l)$, but can be converted to one that does as follows (this is the .5 proof): 
Consider a $2$-coloring of the edges of a complete graph on $N=2^{k+l}$ vertices, $v_1, ..., v_N$.  Let $V_0$ be the set of all vertices, and let $V_i$ be the largest subset of $V_{i-1}$ connected to $v_i$ by edges of a single color, $c_i$.  After $k+l$ steps, at least $k$ of the $c_i$ are read or $l$ of the $c_i$ are blue by pigeonhole; let the set of indices for which this happens be denoted $S$.  Then $(v_i)_{i\in S}$ is the desired subgraph.
My question is:

Does anyone have a fundamentally different proof of this theorem?  In particular, I am curious to know if there are any of a less combinatorial flavor.

REPLY [14 votes]: One can prove Ramsey's theorem by using a minimal ultrafilter on the infinite permutation group to make a graph monochromatic, see
http://terrytao.wordpress.com/2008/01/21/254a-lecture-5-other-topological-recurrence-results/
(This proof was first discovered by Neil Hindman.)  But this proof is somewhat idiosyncratic and might not be to most people's taste.  (Also, on some level it is equivalent to the usual iterated pigeonhole proof, though heavily disguised through several applications of the axiom of infinity and axiom of choice.)  It is also closely related to the proof that Pietro mentioned.<|endoftext|>
TITLE: Proofs of Lower Bounds for Ramsey Numbers?
QUESTION [5 upvotes]: As a sort of dual question to this question, I am wondering what proofs people know of lower bounds on Ramsey numbers $R(k, k)$.  I know of two proofs:  there is Erdos's beautiful probabilistic argument, given here for example, as well as the following:
(This is a sketch; it's worth working out the details.)  Represent a two-coloring of the edges of a complete graph on $n$ vertices as the upper triangle (strictly above the diagonal) of an $n\times n$ matrix of zeroes and ones (that is, ${n\choose 2}$ bits).  We may rewrite this representation by noting which vertices are contained in our monochrome subgraph and what color it is, as well as including all the remaining edge data, using some special characters to block off this data.  If Ramsey numbers are small, this sends each string of bits under an appropriate encoding to a smaller string of bits, which is impossible by pigeonhole.  (I am being purposely vague about the encoding--pick your own, anything goes essentially--because it's a bit boring).  The bound this argument gives is essentially the same as the probabilistic one, and indeed it seems to me to be essentially a "derandomization" of that argument.
My question is:

Does anyone know a proof of a similarly good lower bound using a fundamentally different method?

REPLY [5 votes]: There have been a two recent improvements on the Barak-Rao-Shaltiel-Wigderson result result mentioned by Ryan Williams. There are now explicit construction of a $2^{(\log\log N)^C}$-Ramsey graph over $N$ vertices for some constant $C$. Both results are to be presented at STOC 2016 in June.
The first is due to Gil Cohen: Two-Source Dispersers for Polylogarithmic Entropy and Improved Ramsey Graphs. 
The second is a corollary to a widely publicized breakthrough in random number generation by Eshan Chattopadhyay and David Zuckerman: Explicit Two-Source Extractors and Resilient Functions.<|endoftext|>
TITLE: L-S category versus Betti numbers
QUESTION [7 upvotes]: Is there a sequence of topological spaces $X_n$ (manifolds ideally), where the sum of the Betti numbers of $X_n$ remains bounded but the Lusternik–Schnirelmann category is unbounded, as $n \to \infty$?  What about vice versa?
One might think of both of these numbers as very rough measures of the "complexity" of a space, and it is well known in particular that both quantities are lower bounds on the number of critical points of a Morse function.  But it would be nice to hear about any facts governing the relations between them.

REPLY [4 votes]: The dumb answer is infinite projective space, which has zero Betti numbers as classically computed (over $\mathbb{Q}$).  But what you really want is very small 
cohomology and very large L-S category.
Let $X$ be a noncontractible 
acyclic space.  Then all Betti numbers are zero (no matter what coefficients you use), 
and the same is true for the 
$n$-fold product $X^n$.  But a theorem of Hilton says (as I recall) that an 
$n$-fold product of noncontractible spaces must have category at least $n$.
On the other hand, if $X$ is simply-connected and $X$ has category $n$, then $X$
must have cohomology in at least $n$ different dimensions.<|endoftext|>
TITLE: PDEs, boundary conditions, and unique solvability
QUESTION [14 upvotes]: I'm interested in a criterion that determines whether a linear scalar PDE (arbitrary order) has a unique solution given vanishing boundary conditions at spatial infinity. I'll try to formulate the question more precisely below.
Consider a PDE of the form $L[u]=0$ where $u(t,x,y,z)$ is a scalar function of one time $(t)$ and three spatial variables $(x,y,z)$, though this choice of dimensionality is not central to the question. The function $u$ is required to vanish "sufficiently" fast if the $(x,y,z)$ variables are taken to infinity, keeping $t$ fixed. [If that's not enough, it can also be required to vanish at infinity along any hyperplane that is space-like with respect to the Lorentzian metric $\mathrm{diag}(-1,1,1,1)$.] However, no requirements are put on the behavior of $u$ as $t\to\pm\infty$ for fixed $(x,y,z)$. The linear differential operator $L$ can be assumed to have constant coefficients, but could by of any order. Though, I'd also like to know how the answer generalizes to the case when the coefficients and the background Lorentzian metric are no longer constant.

So, my question is this: for which operators $L$ does the equation $L[u]=0$ have a unique solution?

Let me give some examples.

Equation $\partial_z u=0$ has a unique solution. An arbitrary solution comes from integrating the rhs wrt to $z$ and adding any function that's constant wrt to $z$. From the boundary conditions, it is easy to see that both pieces must be zero. Hence, $u=0$ is the unique solution.
The same argument does not work for $\partial_t u=0$. For any given solution, I can get another solution by adding a function of $(x,y,z)$ only that vanishes at infinity, and there are plenty of those.
The equation $(\partial_x^2+\partial_y^2+\partial_z^2)u=0$ is uniquely solvable: ignore $t$ dependence and invert the Laplacian, with uniqueness given by the same argument as in the first example.
The equation $(-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2)u=0$ is not uniqely solvable: solutions are parametrized by Cauchy data on, say, the $t=0$ hyperplane, and so are definitely not unique.

These examples make me think that the answer is some version of an ellipticity condition. Unfortunately, I'm only aware of how to formulate this condition for second order systems. Any help appreciated!
Status Update: Willie Wong provided some good, relevant information below. Let me summarize my understanding of it here and then sharpen my question in light of it.
If the partial differential operator (PDO) $L$ contains no time derivatives, then the Fourier transform $\hat{u}$ of a nontrivial solution $u$ must be supported on the zero set of $P(\xi)$, the symbol of $L$. If this set is of measure zero, then $\hat{u}$ can only be a distribution. On the other hand, local regularity of $\hat{u}$ is controlled by the decay of $u$ at infinity. In particular, if $u$ is in the Schwarz space, then $\hat{u}$ cannot be sufficiently singular to be a distribution. Hence, $L[u]=0$ would admit only the trivial solution, and hence be uniquely solvable.
In principle, I can use the same argument for any $L$ that is expressible only in terms of derivatives parallel to a given space-like hyperplane, by appealing to the fact that I've imposed the same boundary conditions at infinity (say Schwarz) for each such hyperplane.
In principle, the above reasoning gives a nice large space of PDOs that satisfy my criteria. But are there any more? I think there are (see below).
Now, suppose that I cannot ignore time derivatives. Willie's suggestion is to write the equation $L[u]=0$ in evolution form $\partial_t v + Av=0$ and exclude $L$'s for which the evolution equation is well posed as an initial value problem. But not all such $L$ can be excluded, since not all initial data generates solutions that satisfy the boundary conditions (decay at infinity along any space-like direction). I'm thinking that there should be a geometric condition involving the background null cone and the characteristics of $L$ or the zero set of $P(\xi)$. For instance, if $L=-c^{-2}\partial_t^2+\partial_x^2$ with $c>1$, then the corresponding equation has infinitely many solutions parametrized by Cauchy data on $t=0$. However, these solutions would correspond to waves pulses propagating along the $x$-axis at a speed faster than the background speed of light. On the other hand, the solution $u(t,ct)$ must vanish for large $t$ as $(1,c)$ is a space-like vector. I believe this is enough to show that this $L$ also admits only trivial solutions. My reasoning here is based on the exact representation of solutions via the D'Alambert formula, but that doesn't generalize easily. Any idea what kind of geometric condition could be used for more general operators?
Ultimately, I'd like to know something about the geometry of the space of operators that satisfy my criteria. Say I fix the maximal order to make things easier. The space is clearly not linear, but could it be convex or the complement of a convex set? (These last guesses are probably not right. I'm just throwing out ideas.) I'd be happy if I could understand this space for just first and second order operators, preferably with hints of how this understanding could generalize to higher orders.

REPLY [3 votes]: Hi, I am adding another answer because this suggests a rather different approach then what I have outlined before, and this is targeted at the fact you are willing to grant smooth with compact support on any space-like hyperplane. 
If you are willing to let your solutions vanish in such a large set, then the proper tool for the analysis is the theory of unique continuation for solutions to linear differential equations. For example, solutions to elliptic equations tend to have the property of strong unique continuation: if the solution vanish on any open set it must vanish everywhere. Using the geometry you can weaken ellipticity to have cases where no derivatives are taken in certain directions ($\partial_x^2 + \partial_y^2$, for example). 
Similarly you can address parabolic type operators. (They have infinite speeds of propagation as well.)
For hyperbolic type, if the coefficients of the operator are real analytic, then you can use Holmgren's Uniqueness theorem if the inner-most characteristic cone of the operator remains outside of the Minkowskian null cone: the compact-support criterion on space-like (to back ground metric) slices will guarantee that there will be a time-like surface (relative to the differential operator, not to the background metric) to one side of which the solution vanishes, then Holmgren kicks in and tells you that the solution must vanish everywhere. 
(A version of Holmgren's also applies to ultrahyperbolic operators, with suitable changes in the geometry.)
Outside of these trivial cases, you need to actually consider the geometry of the characteristic cones and compare them against the geometry of the support set of your function. The keywords you will be looking for are Carleman estimates, unique continuation, and pseudo-convexity. There's a nice, comprehensive treatment in Lars Hormander's Analysis of Linear Partial Differential Operators (I think volume III, may be IV?); there are also some nice notes by Daniel Tataru on his website.<|endoftext|>
TITLE: What is the opposite category of the category of modules (or Hopf algebra representations)?
QUESTION [30 upvotes]: This question is, in some sense, a variant of this, but for certain cases.
The opposite category of an abelian category is abelian. In particular, if $R-mod$ is the category of $R$-modules over a ring $R$ (say left modules), its opposite category is abelian.  The Freyd-Mitchell embedding theorem states that this opposite category can be embedded in a category of modules over a ring $S$. This embedding is usually very noncanonical though.

Question: Is there any way to choose $S$ based on $R$?

My guess is probably not, since these notes cite the opposite category of $R-mod$
as an example of an abelian category which is not  a category of modules. I can't exactly tell if they mean "it   is (for most $R$) (provably) not equivalent to a category of modules over any ring" or "there is no immediate structure as a module category." If it is the former, how would one prove it?
I also have a variant of this question when there is additional structure on the category.
The module category $H-mod$ of a Hopf algebra $H$ is a tensor category.  A finite-dimensional Hopf algebra can be reconstructed   from the tensor category of finite-dimensional modules with a fiber functor  via Tannakian reconstruction.   Now $(H-mod)^{opp}$ is a tensor category as well satisfying these conditions (namely, the hom-spaces in this category are finite-dimensional).  The dual of the initial fiber functor makes sense and becomes a fiber functor from $(H-mod)^{opp} \to \mathrm{Vect}$ (since duality is a contravariant tensor functor on the category of vector spaces).  In this case, $(H-mod)^{opp}$ is the representation category of a canonical Hopf algebra $H'$.

Question$\prime$ What is $H'$ in terms of $H$?

REPLY [8 votes]: I am three months late for the party, but I'll still add an answer (to the first question):
There is the notion of "locally finitely presentable category" which can be described equivalently as
(i) a category such that there exists a set of finitely presentable objects such that every object is a directed colimit of these.
(Explanation: $X$ being finitely presentable means that $Hom(X,-)$ commutes with directed colimits, for modules it is equivalent to the usual notion of finitely presented, intuitively because for a given morphism into the colimit each of the finitely many generators has to go into some finite stage of the colimit diagram, then one can form a cone over these finitely many objects, as the diagram is directed, and the morphism factors through there)
(ii) a category of all models in SET of a finite limit sketch
(Explanation: A finite limit sketch is a small diagram, or small category, with distinguished cones, a model in SET is a functor from this small category to sets which maps the distinguished cones to limit cones - example: the limit sketch of groups is a category $C$ with an object $G$, a map $G \times G \rightarrow G$ a commutative diagram stating associativity etc and the cones which e.g. say that the object $G \times G$ is the product of $G$ with itself. A group is a functor from this to SET which takes $G \times G$ to the product, i.e. maps the distinguished cones to limit cones. Likewise for a ring $R$ the limit sketch for $R$-modules is given by an abelian group object $G$ as above plus one group endomorphism of $G$ for each element of $R$, behaving as dictated by the multiplication in $R$)
By either description you see that the category of $R$-modules is a locally finitely presentable category and now there is a theorem (e.g. Adamek/Rosicky Thm 1.64) saying that the opposite of a locally finitely presentable category is never itself locally presentable unless we are in a trivial case where our category is a poset.
The cool thing about this result is that it not only answers your question about modules, but also applies to all other algebraic structures (the theory of locally presentable cats gives a characterization of categories of algebraic structures) and I like to read it as is a provable mathematical statement which reflects the duality between algebra and geometry: The dual of an algebraic category is not itself algebraic, but geometric...<|endoftext|>
TITLE: Work of ICM 2010 plenary speakers (and other humans)
QUESTION [38 upvotes]: The ICM is approaching. It would be nice for everybody who feels qualified to give a brief overview of the work of one of the plenary speakers. If anything, this would serve to make all of us a little more cultured. On that same note - I would be especially interested in getting to know more about the work of anybody who has been invited to speak in more than one section. (For one thing, the probability of understanding some of the work of any person satisfying that property might be higher than the mean.)
(There could be some real-life follow-up on this on the part of those of us who are going to be in India in August. Perhaps we could organise some very informal seminars?)

REPLY [5 votes]: R. Balasubramanian is an expert in the field of analytic number theory, especially the theory of the Riemann Zeta function. He, along with his colleagues, is well known for solving the Waring's problem for the 4th power i.e. $g(4)=19$. More about his works and publications can be found here:
http://www.imsc.res.in/~balu/balulist.htm<|endoftext|>
TITLE: How many surjections are there from a set of size n?
QUESTION [39 upvotes]: It is well-known that the number of surjections from a set of size n to a set of size m is quite a bit harder to calculate than the number of functions or the number of injections. (Of course, for surjections I assume that n is at least m and for injections that it is at most m.) It is also well-known that one can get a formula for the number of surjections using inclusion-exclusion, applied to the sets $X_1,...,X_m$, where for each $i$ the set $X_i$ is defined to be the set of functions that never take the value $i$. This gives rise to the following expression: $m^n-\binom m1(m-1)^n+\binom m2(m-2)^n-\binom m3(m-3)^n+\dots$.
Let us call this number $S(n,m)$. I'm wondering if anyone can tell me about the asymptotics of $S(n,m)$. A particular question I have is this: for (approximately) what value of $m$ is $S(n,m)$ maximized? It is a little exercise to check that there are more surjections to a set of size $n-1$ than there are to a set of size $n$. (To do it, one calculates $S(n,n-1)$ by exploiting the fact that every surjection must hit exactly one number twice and all the others once.) So the maximum is not attained at $m=1$ or $m=n$. 
I'm assuming this is known, but a search on the web just seems to lead me to the exact formula. A reference would be great. A proof, or proof sketch, would be even better.
Update. I should have said that my real reason for being interested in the value of m for which S(n,m) is maximized (to use the notation of this post) or m!S(n,m) is maximized (to use the more conventional notation where S(n,m) stands for a Stirling number of the second kind) is that what I care about is the rough size of the sum. The sum is big enough that I think I'm probably not too concerned about a factor of n, so I was prepared to estimate the sum as lying between the maximum and n times the maximum.

REPLY [2 votes]: Many people may be interested in the asymptotics for $n=cm$ where $c$ is constant (say $c=2$).
That is, how likely is a function from $2m$ to $m$ to be onto?
I couldn't dig the answer out from some of the sources and answers here, but here is a way that seems okay.
Bender (Central and local limit theorems applied to asymptotics enumeration) shows
$$\begin{eqnarray*}{n\brace m}&\sim&\frac{n!e^{-\alpha m}}{m!\rho^{n-1}(1+e^\alpha)\sigma\sqrt{2\pi n}},\\
\frac{n}m &=& (1+e^\alpha)\ln(1+e^{-\alpha}),\\
\rho&=&\ln(1+e^{-\alpha}),\\
\sigma^2&=&\left(\frac{m}n\right)^2[1-e^\alpha\ln(1+e^{-\alpha})].\end{eqnarray*}$$
Notice that for constant $n/m$, all of $\alpha$, $\rho$, $\sigma$ are constants.
For $c=2$, we find $\alpha=-1.366$
and then $\rho=1.59$
and $$\sigma^2=\frac14 [1-e^\alpha\ln(1+e^{-\alpha})]=\frac14 [1-e^{-1.366}(1.59)]=.149$$
Thus the probability that our function from $cm$ to $m$ is onto is
$$\Pr(\text{onto})=\frac1{m^n}m!{n\brace m}=\frac1{m^n}m!\frac{n!e^{-\alpha m}}{m!\rho^{n-1}(1+e^\alpha)\sigma\sqrt{2\pi n}}$$
$$=\frac1{m^n}\frac{n!e^{-\alpha m}}{\rho^{n-1}(1+e^\alpha)\sigma\sqrt{2\pi n}}$$
Does it go to 0? Well, $\rho=1.59$ and $e^{-\alpha}=3.92$, so up to polynomial factors we have
$$\frac1{m^n}\frac{n!e^{-\alpha m}}{\rho^{n}}\approx\frac{n!(3.92^m)}{(1.59)^n(n/2)^n}$$
$$\approx \frac{(3.92^m)}{e^{n}(1.59)^n(1/2)^n}=\left(\frac{2^2\times 3.92}{1.59^2\times e^2}\right)^m=0.839^m$$
So phew... it goes to 0, but not as fast as for the case $n=m$ which gives $(1/e)^m$.<|endoftext|>
TITLE: Coordinates on Flag Manifolds
QUESTION [6 upvotes]: Suppose you want to work with complete flags $\mathbb{F}_3$ on $\mathbb{C}^3$.  Given a flag
$$ \{0\}\leq V_1\leq V_2 \leq \mathbb{C}^3$$
you can think of $V_1$ as the span of a vector $\vec{u}$, and then you can choose a vector
$\vec{v}$ that is Hermitian orthogonal to $\vec{u}$ so that $V_2=<\vec{u},\vec{v}>$.  Finally you
can choose $\vec{w}$ so that it is Hermitian orthogonal to $<\vec{u},\vec{v}>$.  This gives an embedding
$$\mathbb{F}_3\rightarrow \mathbb{C}P(2)^3 .$$
Since the Hermitian inner product involves complex conjugates, this embedding cannot possibly be
holomorphic.  For instance if the first line in homogeneous coordinates is $[a,b,c]$ and the second
is $[d,e,f]$ then they satisfy $a\overline{d}=b\overline{e}+c\overline{f}=0$ where the overline
indicates complex conjugate.  Is there some way of playing around with complex structures to fix
this? Is there a similar map, that is better behaved?

REPLY [7 votes]: I think you can use wedge products. Choose $v \in V_1$, then $u \in V_2$, which is linearly independent. Map the flag to $([v], [v \wedge u]) \in (CP^{2})^2$. This should be well defined and holomorphic.<|endoftext|>
TITLE: Is there a central limit theorem for bounded non identically distributed random variables?
QUESTION [15 upvotes]: I have a sequence of centered independent random variables $X_i$ that are all
bounded by one in absolute value. They are not identically distributed, though.
I would like to know if the central limit theorem is still true 
for such a sequence. Putting $S_n= X_1+...+X_n$, do we have 
$$
c_n = P(\ {S_n\over\sigma(S_n)} \in [a,b] ) - 
{1\over \sqrt{2\pi}}\int_a^b exp(-t^2/2) dt \ \rightarrow \ 0\ ?
$$
(let's assume $\sigma(S_n)$ goes to infinity with n).
I guess it is true but I can't find a reference.
Also, what can be said from the rate of convergence of $c_n$ ?
Since the $X_i$ are uniformly bounded, does $c_n$ goes to zero
exponentially fast ?

REPLY [10 votes]: Theorem (Billingsley, "probability and measure", example 27.4)
Let X_i a sequence of independent, uniformly bounded random variables with zero mean, such that $\sigma(S_n)$ goes to infinity with n. Then $S_n/\sigma(S_n)$ converges in law to the normalized Laplace-Gauss distribution.
This follows from the Lindeberg triangular array theorem. As pointed out in the other answers, the convergence can be slow. The Bernstein inequality may be used to bound the tail. Under the assumption of the previous theorem, for all n, we have
$$P(S_n>t) \leq exp(-t^2/(\sigma^2(S_n)+Ct/3))$$
where C is a bound for the |X_i|'s.<|endoftext|>
TITLE: When is a sheaf of smooth functions acylic?
QUESTION [6 upvotes]: Let $G$ be a Lie group, and let $\underline{G}$ denote the sheaf of smooth $G$-valued maps, i.e. for a smooth manifold $M$ we have $G(M) = C^\infty(M,G)$. 
What are conditions on $G$ that imply that $\underline{G}$ is acyclic, i.e. the sheaf cohomology $H^n(M,\underline{G})=0$ for all smooth manifolds $M$ and all $n>0$? 
It is clear that soft, flabby or fine sheaves are acyclic. I am interested in concrete conditions on the group $G$, e.g. like smooth contractibility. 
EDIT: Daniel's answer below answers my question in the case that $G$ is abelian, using the classification of abelian Lie groups. So let us concentrate on the case that $G$ is non-abelian. The condition I am looking for is supposed to imply the vanishing of the set $H^1(M,\underline{G})$. This set can be defined for example via Cech cohomology. Its geometrical meaning is that it classifies principal $G$-bundles over $M$ up to isomorphism.

REPLY [6 votes]: For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).
Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.

Added (7/7/2010):  Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem. 
Theorem.  Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi_1(G)\neq 0, \mathbb{Z}$.  Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.
Proof.  Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$.  Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi_1(U(n))=\mathbb{Z}$ wheareas $\pi_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption.  That is, $U(n)\not\simeq G\times M$ as $\pi_1(U(n))\not\simeq \pi_1(G)\times \pi_1(M)$.           $\square$
This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.

Added (7/9/2010):  I don't know why I didn't mention it before, but replacing "unitary" with "complex" in the theorem above gives the same result for e.g. complex connected semisimple Lie groups, by an identical proof.  In this case the manifold $M$ constructed in the proof cannot be guaranteed to be compact however.<|endoftext|>
TITLE: Why does the triangle groups not include a tiling by 30-30-120 triangles?
QUESTION [7 upvotes]: Looking at http://en.wikipedia.org/wiki/Triangle_group I begin to wonder why the definition explicitly excludes the tessellation of the Euclidean plane by 30-30-120 triangles? In terms of the Wallpaper groups, I am thinking of the group p6 ( http://en.wikipedia.org/wiki/Wallpaper_group#Group_p6 ). 
Is it just an exceptional case? The definition of the triangle group asks that the "order" at each vertex to be even, which is natural, as for odd orders only isosceles triangles can "close" under reflections. So if one of the angles of a tessellating triangle is $2\pi / k$ for $k$ odd, it is necessary that there exists some integer $l$ such that $$\frac{1}{k} + \frac{2}{l} = \frac12$$ 
and the only solution is $k = 3$ and $l = 12$.
But for less rigid geometries (say hyperbolic), this seems to introduce a large number of additional tessellations. (Though not really more groups, I think, since geometrically replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined cases.) 
Motivation: I am trying to describe an arts/craft project for demonstrating hyperbolic geometry. As such it is more natural to start from the tessellation picture, rather than the reflection group picture. Therefore it was a bit surprising to me that despite the introductory paragraph on Wikipedia, the two points of view are not exactly the same.
My apologies for the somewhat muddy wording of the question.

REPLY [6 votes]: The answer is already contained in your question.
You do not describe more reflection groups, since replacing the obtuse triangle by two congruent acute triangles leads to one of the already defined symmetry groups.
Not directly related to your question: here are two cool java applet having to do with hyperbolic tesselations.
Applet 1. 

Applet 2.<|endoftext|>
TITLE: How to calculate partition function of a QFT by summing over irreducible representations of the symmetry group? 
QUESTION [6 upvotes]: By definition computing the partition function of a QFT amounts to doing a Feynman Path Integral exactly. At a schematic level I can see why this can become a  question of summing/integrating over characters of the irreducible representations of the symmetry group. 
But I don't understand this precisely enough to do a calculation of this kind. 
In QFT books I have never come across such a calculation. There one is always bothered about doing a perturbative evaluation of that using Feynman diagrams. One runs into this kind of a calculation in only papers like this. 
I would like to know if there is an expository/introductory reference about this translation and computational technology. Something which say explains this representation theory approach of exact computation of partition function starting from simple field theories which will eventually help me understand the exotic ones seen in the papers like the above.

REPLY [2 votes]: The following article by A.B. Balantekin describes in section III the structure of thermal grand canonical partition functions (i.e., with chemical potentials) for systems with Lie group symmetries (which is the partition function of the type described in the reference given in the question).
The sum over the group charcters originates from the presence of the chemical potentials, 
which are the (analytical continuations of the) coordinates of the maximal torus of the group. 
The Balantekin article describes the "sum over states" formulas of the partition functions in which the summation over the group representations is explicit, in contrast to the Feynman's "sum over paths" (mentioned in the question) in which the summation over the representations is implicit.
I'll describe here a simple example of an explicit calculation of the grand canonical partition function over the two-sphere.
Let $-H$ be the scalar Laplacian on the two sphere. The state Hilbert space is spanned the spherical representations (with multiplicity 1), 
corresponding to integer spins only. The symmetry group $SU(2)$ is generated by the usual set of generators $[J_i, J_j] = \epsilon_{ijk} J_k$ 
The grand canonical partition function is given by:
$Z = \textrm{Tr}(exp(-\beta H + \mu J_3)) = \sum_{j=0}^{\infty}\chi_j(i \mu)exp(-\beta j(j+1))$
$= \sum_{j=0}^{\infty}\frac{sinh((2j+1) \mu)}{sinh(\mu)}exp(-\beta j(j+1))$
$= \frac{e^{\frac{\beta}{4}}}{2sinh(\mu) }\theta_1(\mu, \beta)$
The actual computation (of the N=4 SYM) referred to in the question needs much more work and it relies on several assumptions and approximations.
The introduction section of the following article by Yamada and Yaffe and references therein may be helpful.<|endoftext|>
TITLE: covering a square with unit squares
QUESTION [15 upvotes]: Can some square of side length greater than $n$ be covered by $n^2+1$ unit squares?  (The unit squares may be rotated. The large square and its interior must be covered.)

REPLY [7 votes]: $\Pi(3)>10$ according to "A note on covering a square with equal squares", Januszweski, The American Mathematical Monthly Vol. 116, No. 2 (Feb., 2009), pp. 174-178.<|endoftext|>
TITLE: The product of n radii in an ellipse
QUESTION [13 upvotes]: I sent the following question to another forum more than a week ago but haven't got any responses. The moderator of that forum suggested that I pose the following question here:
Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the 
origin). Let $n>4$. There are $n$ rays going out of the origin, at angles 
$0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x_1,y_1),...,(x_n,y_n)$ be 
the points of intersection of the rays and the ellipse. The product 
from $k=0$ to $n-1$ of $(x_k)^2 + (y_k)^2$ is equal to one. Can $a$ and $b$ be 
rational? Note that this is obviously possible if $a=b=1$, since then the 
ellipse becomes a circle of radius 1. But what about if $a$ is not equal 
to $b$? Can $a$ and $b$ still be rational? 
Craig

REPLY [13 votes]: So, let's finish this. Starting from Qiaochu's formula,
$$\prod_{k=0}^{n-1} (r^2 \cos^2 (2 \pi k/n) + s^2 \sin^2 (2 \pi k)/n))=1$$.
Each factor is
$$\left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) + i s \sin(2 \pi k/n) \right) \left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) - i s \sin(2 \pi k/n) \right) =$$ 
$$\left( \frac{r+s}{2} e^{2 \pi i k/n} + \frac{r-s}{2} e^{-2 \pi i k/n} \right) \left( \frac{r+s}{2} e^{2 \pi i k/n} - \frac{r-s}{2} e^{-2 \pi i k/n} \right).$$ 
Putting $u=(r+s)/2$, $v=(r-s)/2$ and $\zeta=e^{2 \pi k/n}$, we have 
$$\prod_{k=0}^{n=1} (u + v \zeta^{2k}) (u - v \zeta^{2k})$$
If $n$ is odd, this is $(u^n+v^n)(u^n - v^n)= (u^{2n} - v^{2n})$. If $n$ is even, I get $(u^{n/2} - v^{n/2})^2 (u^{n/2} + v^{n/2})^2 = (u^n-v^n)^2$. So either $u^{2n} - v^{2n}=1$ or $u^n-v^n=1$ or $u^n-v^n=-1$. These are all forms of FLT.<|endoftext|>
TITLE: Whitening a random bit sequence
QUESTION [5 upvotes]: Given an (infinite) stream of uncorrelated random bit with a known "reasonable" bias (say 15-85% 1's) I want to whiten it, e.i. produce a shorter stream of bits that has no bias. The restriction is that the output must be usable as a cryptographically secure ransom bit stream.
The proposal is to compress the stream with a Huffman code constructed from a table of theoretic frequencies of bit sequences (say 10 bits at a time). As the number of bits used increases, will this approach ideal performance?
Clearly, the ratio of bits consumed to bits produced will be nearly ideal, but what about the other interesting properties?

Edit1: A while back I looked around a bit on this topic and found some methods but didn't see this approach used and I'm wondering if it has some sort of hidden flaw.
Edit 2: I'm only interested in the performance of this device for uncorrelated input, that is (to make sure I'm using the term correctly) where the bias of any given element is independent of any and all other values. This happens to make the frequency of any given sequence a function only of it's length and sum.
Edit 3: Assume the input is not a bottle neck, that it can generate bits as fast as I need them.

REPLY [3 votes]: If your goal is to create an unbiased truly random stream, this page has an algorithm which produces results much faster than the usual technique (the one mentioned by qwerty1793 in the comments above).
However, if your concern is simply real-world cryptographic security, since random generators like the one you describe usually produce unbiased streams very slowly, you are better off whitening your stream once using the usual method and using that to seed a provably-secure cryptographic PRNG.
In fact, if you are able to produce your truly-random biased bits as quickly as your PRNG, you could even introduce true randomness to the result by XOR'ing the biased stream with the PRNG output.  This works regardless of the stream's bias (due to the fact that random data XOR'ed with non-random data produces random data).<|endoftext|>
TITLE: Can string topology be a open-closed TCFT with the full set of branes?
QUESTION [7 upvotes]: String topology studies the algebraic structure of the homology of the free loop space $LM = Map(S^1,M)$ of a oriented closed manifold. One aspect of this structure is that the pair $(H_\ast(LM;\mathbb{Q}),H_\ast(M;\mathbb{Q}))$ forms a open-closed HCFT with positive boundary (work of Godin). This means that there are operations coming from the homology of moduli space of Riemann surfaces with each connected component having a non-empty outgoing or free boundary. The conjecture (although Blumberg-Cohen-Teleman claim it is a theorem) is that in fact $H_\ast(M;\mathbb{Q})$ should be seen has $H_\ast(P(M,M),\mathbb{Q})$, where $P(M,M)$ is the space of paths starting and ending in $M$, and the HCFT-structure can be extended to include open-closed cobordisms which have open boundaries labelled with a closed oriented submanifold $N$ of $M$. This then gives operations not only for $H_\ast(LM;\mathbb{Q})$ and $H_\ast(M;\mathbb{Q})$, but also on $H_\ast(P(N_1,N_2);\mathbb{Q})$ for any two closed compact oriented $N_1$, $N_2$ in $M$. This is called the full set of branes. For $N_1 = N_2$ a single point and $M$ connected, $P(N_1,N_2) = \Omega M$, the based loop space. In general $H_\ast(P(N_1,N_2);\mathbb{Q})$ will therefore not be finite-dimensional.
On the other hand, Costello has proven a classification theorem of open-closed TCFT. In this we don't work with homology, but chains in the moduli space of Riemann surfaces, and chain complexes instead of (graded) vector spaces. Costello has proven that a open-closed TCFT can be constructed from an open TCFT (cobordism without incoming or outgoing boundary components equal to the circle) and that an open TCFT is equivalent to a Calabi-Yau $A_\infty$ category. One of the properties of a Calabi-Yau $A_\infty$ category is that all hom-spaces are finite-dimensional, forced by a certain non-degenerate pairing.
One can construct a HCFT from a TCFT by applying homology everywhere. I think this HCFT will in fact be positive (or negative?) boundary, because the TCFT is defined from open-closed cobordisms where each connected component has at least one incoming boundary component. Is this correct? 
Furthermore, Costello conjectures in his paper that string topology (with the full set of branes) can be constructed as a TCFT, and applying homology then reduces to the HCFT given by Godin. But I can think of two reasons which make this conjecture seems false:
1) The naive choice of $C_ast(P(N_1,N_2);\mathbb{Q})$ as Calabi-Yau $A_\infty$ category is impossible, because these spaces will certainly be infinite-dimensional.
2) But no choice will work, because the homology of a finite-dimensional cell complex is finite-dimensional and we know some branes must be assigned infinite-dimensional spaces in the HCFT structure.
So my question is: Is this reasoning enough to make my naive interpretation of Costello's conjecture false? If not, what is the mistake?

REPLY [4 votes]: No, the conjecture is not false. The mistake is the following: I said "On the other hand, Costello has proven a classification theorem of open-closed TCFT." He proves a classification of open TCFT's and gives a construction of an open-closed TCFT from an open TCFT. String topology can't be obtained by this specific construction.
However, Lurie(-Hopkins) gives a more general construction in section 4.2 of the article on the classification of TQFT's. This can deal with string topology. However, the Calabi-Yau objects in this case only require a non-degenerate cotrace, and string topology fits in now.<|endoftext|>
TITLE: recommendation letter for teaching
QUESTION [7 upvotes]: I need a recommendation letter on my teaching. I want to ask the instructor in last semester for which I was a TA, but I don't know how his impression for my teaching. So do I need to ask him for his opinion about my teaching before letting him write the recommendation letter?

REPLY [9 votes]: People usually do not feel comfortable to say no when they are been asked to write a recommendation letter as they do not like to imply that they have negative views about your work. However, most decent people would try to avoid writing you a letter if they feel they cannot write a good letter. So when you ask for a letter always try to leave as much room for a polite refusal as possible. Therefore, phrase your request as a question, something like: if you feel you know my work well enough, would you be able to write me a letter of recommendation? If the answer is not yes or if they express too much hesitation, you probably better off asking someone else.<|endoftext|>
TITLE: A condition that implies commutativity
QUESTION [53 upvotes]: Let $R$ be a ring. A notable theorem of N. Jacobson states that if the identity $x^{n}=x$ holds for every $x \in R$ and a fixed $n \geq 2$ then $R$ is a commutative ring.
The proof of the result for the cases $n=2, 3,4$ is the subject matter of several  well-known exercises in Herstein's Topics in Algebra. The corresponding proofs rely heavily on "elementary" manipulations. For instance, the proof of the case $n=3$ can be done as follows:
1) If $a, b \in R$ are such that $ab= 0$ then $ba=0$.
2) $a^{2}$ and $-a^{2}$ belong to $\mathbf{Z}(R)$ for every $a \in R$.
3) Since $(a^{2}+a)^{3} = (a^{2}+a)^{2}+(a^{2}+a)^{2}$ it follows that
$a=a+a^{2}-a^{2} = (a+a^{2})^{3}-a^{2} = (a^{2}+a)^{2}+(a^{2}+a)^{2}-a^{2}$
and whence the result. ▮
Certainly, the mind can't but boggle at the succinctness of the above solution. Actually, it is the conciseness of this argument that has prompted me to pose the present question: is an analogous demonstration of the general theorem possible? The one that appears in [1] depends on some non-trivial structure theorems for division rings.
As usual, I thank you in advance for your insightful replies, reading suggestions, web links, etc...
References
[1] I. N. Herstein, Noncommutative rings, The Carus Mathematical Monographs, no. 15, Mathematical Association of America, 1968.
[2] I. N. Herstein, Álgebra Moderna, Ed. Trillas, págs. 112, 119, and 153.

REPLY [31 votes]: For fixed $n \in \mathbb{N}$, Birkhoff's completeness theorem implies that such a proof must exist in the first-order equational theory of rings - as I mentioned here in a recent post. Many years ago Stan Burris told me that John Lawrence discovered such an equational proof that works uniformly for all $n$ (possibly also for Jacobson's form $x^{n(x)} = x$). I don't know if the proof is published yet, but some clues as to how it may proceed might be gleaned from their earlier joint work [1]
1 S. Burris and J. Lawrence, Term rewrite rules for finite fields.
International J. Algebra and Computation 1 (1991), 353-369. 
http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf<|endoftext|>
TITLE: Sticks and thread
QUESTION [8 upvotes]: In this recent question Math puzzles for dinner we had a nice time as we were asked to provide new maths puzzles for dinners. I suggested the following:

Given three equal sticks, and some
  thread, is it possible to make a rigid
  object in such a way that the three
  sticks do not touch each other? (all
  objects are 1 dimensional; sticks are
  straight and rigid, and the thread is
  inestensible).

I'm not particularly fond of maths puzzles, and I found this one in order to satisfy the party in a dinner of  non-mathematic people, when I'm asked for a puzzle. Indeed, it's suitable for a dinner, and people get to the right conclusion in reasonable time, amusing themselves and disputing a bit. 
Possibly as a consequence of the fact that I always proposed it to non-mathematicians, who are not interested in the proof, now I realized that I do not have, or I forget, the proof for the answer (a bit embarassing indeed). Can somebody find a quick proof for the answer, without too many technical computations?

REPLY [9 votes]: Instead of a proof, I will provide references. It's called a "tensegrity prism". See especially sections 1.4, 3.5 and 3.6 of Dynamics and Control of Tensegrity Systems. Also see "Review of Form-Finding Methods for Tensegrity Structures" and the MS thesis Kinematic Analysis of Tensegrity Structures.<|endoftext|>
TITLE: Non-isomorphic graphs of given order.
QUESTION [6 upvotes]: It is well discussed in many graph theory texts that it is somewhat hard to distinguish non-isomorphic graphs with large order. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. So, it follows logically to look for an algorithm or method that finds all these graphs. 
A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately $\sqrt{T_n}$ non-isomorphic graphs of order n.
( $\{T_n}$ being the number of labeled graphs of order n.)
So, I have the followings to ponder over: 
(1) Are there such algorithms or has there been an improvement on the aforementioned algorithm?
(2) Where can I find a collection of non-isomorphic graphs of a given order?  
If you allow me, I would also like to extend my question to connected graphs. 
Many thanks. 
(I am a beginner in Graph theory, so please give answers in not-very-specialized terms.)

REPLY [3 votes]: See:
Combinatorial algorithms: an update, 
Herbert S. Wilf, Albert Nijenhuis
SIAM, 1989. Chapter 8: Generating Random Graphs.
The chapter gives an algorithm for producing an undirected graph uniformly over all graphs of size $n$. It is based on Polya counting. Computing the enumerating polynomial depends on some group theory that is time consuming (I don't know the complexity class, but I'll just conjecture it is most likely exponential space on $n$). But it is a guarantee of uniform distribution. Unfortunately I don't know of a way (I haven't heard of a way) to derandomize this to create an unranking algorithm (to give a mapping from the naturals to the set of unlabeled graphs). 
The algorithm presented in your link (by de Wet) is cute (I mean that in the sense that it is cleverly simple, does not lie, but doesn't really give the meat of it, what it means to have a list of non-isomorphic graphs). The graphs created there have a very particular structure (two paths with an arbitrary subset of edges between the paths, plus some small widgets on one end of each path to break symmetry. All subsets is a good trick but having two paths is pretty uncommon and $\sqrt{T_n}\over T_n$ goes to 0 as $n$ grows.
As to practicality, in addition to the suggestions of nauty and Sage, there's also Mathematica (commercial) which has a list (that you can manipulate) of graphs up to size 11.<|endoftext|>
TITLE: Treating Differential Operators as Numbers
QUESTION [7 upvotes]: In Penrose's book (The Road to Reality, chapter 21) he gives an example of Oliver Heaviside's observation that you can treat differential operators like numbers:
The differential equation $(1+D^2)y = x^5$ can be solved by dividing by $(1+D^2)$ then taking the power series expansion: $$y = (1-D^2+D^4-D^6+\cdots)x^5$$ which evaluates to $y = x^5 - 20x^3 + 120x$.
Apparently this can be made perfectly rigorous!
How is this done? and do you know where I can read more details about this idea?

REPLY [12 votes]: This is just a fact from linear algebra: if $T$ is a nilpotent transformation of a vector space $V$, then $(1-T)^{-1} = 1  + T + T^2 + \dots$.  More generally, the same is true in any commutative Banach algebra (such as the endomorphism ring of a normed complex vector space) if $T$ is of norm less than 1.
In your case, the differential operator $D$ is a nilpotent operator on the vector space of polynomials, and consequently this formula applies.
As people have pointed out in the comments, you may want to look at the Fourier transform, which realizes differentiation as multiplication, which allows you to treat differentiation kind of like a number. (More precisely, there is an isomorphism $F:L^2 \to L^2$ such that $F D F^{-1}$ is equal to multiplication by $x$ on sufficiently nice (e.g. Schwarz) functions.) 
In higher dimensions, for instance, this means that if $\Delta$ is the Laplacian, then multiplication by $(1 + |x|^2)^k$ corresponds to applying $(I-\Delta)^k$, and it is thus possible to define an operator $(I - \Delta)^r$ for any real $r$.<|endoftext|>
TITLE: An Operation on Multisets
QUESTION [7 upvotes]: Define a 2 x n array of positive integers where the first row consists 
of some distinct positive integers arranged in increasing order, and the second row consists of any positive 
integers in any order.  Create a new array where the first row consists of all the integers that occur in the first array,
arranged in increasing order, and the second row consists of their multiplicities.
Repeat the process.   For example, starting with the 
2 x 1 array [1; 1], the sequence is:
[1; 1] -> [1; 2] -> [1, 2; 1, 1] -> [1, 2; 3, 1] -> [1, 2, 3; 2, 1, 1]
-> [1, 2, 3; 3, 2, 1] -> [1, 2, 3; 2, 2, 2] -> [1, 2, 3; 1, 4, 1] -> 
[1, 2, 3, 4; 3, 1, 1, 1] -> [1, 2, 3, 4; 4, 1, 2, 1] -> [1, 2, 3, 4; 3, 2, 1, 2]
-> [1, 2, 3, 4; 2, 3, 2, 1], and we now have a fixed point (loop of one array).
Does the process always result in a loop of 1, 2, or 3 arrays?

REPLY [6 votes]: But I don't have a clear proof that the sequence always terminates in a loop. – Martin Erickson 

Here is a proof that the sequence always terminates in a loop. 
Let $A, B$ be consecutive arrays in the sequence, and c(X) denote the number of columns in array X.
Claim 1. $c(A) \leq c(B)$.
Proof. Trivial.
Claim 2. $$\sum_{i=1}^{c(B)} B[2,i] = 2 c(A).$$
Proof. By the sequence definition, the second row of each subsequent array contains the multiplicities of elements of the preceding array, and thus the sum of multiplicities equals the total number of elements in the preceding array. QED
Now, let $A', A, B$ be three consecutive arrays in the sequence.
Claim 3. Let $m\geq 5$ be an odd integer such that no element $B$ exceeds $m$. Then no element larger than $m$ can appear in subsequent arrays (while their size may eventually grow up to $2\times m$).
Proof. Assume that a larger element $m'\geq m+1$ appears. Without loss of generality suppose that this happens in (the bottom row of) array $C$ that immediately follows $B$. By Claim 1, we have $c(A)\leq c(B)\leq c(C)\leq m$. 
By Claim 2 and since $C$ contains $m'\geq m+1$ (while the other elements are at least 1), 
$$m + c(C) \leq m' + (c(C)-1) \leq \sum_{i=1}^{c(C)} C[2,i] = 2c(B) \leq 2c(C)\leq 2m$$
implying that $c(B)=c(C)=m$ and $m'=m+1$. 
Therefore, the top row of both $B$ and $C$ contains all integers from 1 to $m$. 
The bottom row of $C$ consists of one number $m'=m+1$ and $m-1$ ones. If $m'$ appear under the number $k$, then bottom row of $B$ contains at least $m$ numbers $k$, whose sum must not exceed $2c(A)\leq 2m$, implying that $k\leq 2$. Consider two cases:
If $k=1$ then the bottom row of $B$ consists of all ones, implying that the elements of $A$ are the integers from $1$ to $m$ without repetitions, and hence $m$ is even, a contradiction proving that no element larger than $m$ may appear. 
If $k=2$ then the bottom row of $B$ consists of all twos, implying that $c(A)=m$ and $A$ contains in each row all integers from 1 to $m$ so that the sum of its bottom row equal $1+2+\dots+m = m(m+1)/2.$
The inequality $m(m+1)/2 \leq 2c(A') \leq 2c(A) = 2m$ then implies that $m\leq 3$, that is not the case.
QED
Claim 4. The sequence always terminates in a loop. 
Proof. By Claim 3, there exists an integer $m$ such that elements of the arrays in the sequence do not exceed $m$. By Claim 1 and since $c(X)\leq m$ for all $X$ in the sequence, the size (and hence the top row) of arrays stabilizes to a certain $c(X)=n$. Then by Claim 2, the sum of the bottom row stabilizes to $2n$. Since, there are only a finite number of compositions of $2n$ into the sum of $n$ positive integers (namely, $\binom{2n-1}{n}$), there exists only a finite number of distinct arrays that may appear after the size stabilization, implying that the sequence loops. QED
Claim 5. The length of the terminal loop is bounded by $\binom{2m-1}{m}$, where $m$ is defined as in Claim 3.
Proof. See proof of Claim 4.<|endoftext|>
TITLE: A Game on a Finite Projective Plane
QUESTION [7 upvotes]: Two players Oh and Ex alternately choose points of a finite projective plane.
The first player (if any) to make a line in his/her chosen points is the winner.
Using the Erdos-Selfridge theorem, we can see that the game is a draw if the order of the projective plane is 5 or greater.
  The game is a trivial Oh win if the order is 2.  Does Oh win if the order is 3 or 4?

REPLY [8 votes]: See "Tic-Tac-Toe on a Finite Plane", Maureen T. Carroll and Steven T. Dougherty,
Mathematics Magazine, Vol. 77, No. 4 (Oct., 2004), pp. 260-274. (Preprint here: http://academic.scranton.edu/faculty/carrollm1/tictac.pdf) The second player can force a draw on the 3-by-3 and 4-by-4 projective planes.<|endoftext|>
TITLE: What does the expression count?
QUESTION [8 upvotes]: Let $q \geq 2$.  What does the expression $(q^n-1)(q^n-q)(q^n-q^2)(q^n-q^3)\ldots(q^n-q^{n-1})/n!$ count?  If $q$ is a prime power, then this is
the number of bases of an $n$-dimensional vector space over a field with $q$ elements.

REPLY [4 votes]: This is a partial answer...perhaps someone will improve on it!  
Historically, most of the $q$-analog formulae (beginning from Euler) were derived based on the assumption that  $|q|< 1$ (to ensure series convergence) or $q=p^k$ for a prime $p$.   John Baez in one of his weekly finds (week184) discusses the geometric interpretation of $q=1$ (counting over $\mathbb CP^n$), $q=-1$ (counting over $\mathbb RP^n$) and $q=$a prime power (counting over PG($\mathbb F_q$)).  There is no discussion for other values of $q$.
However, in Gasper and Rahman's Basic Hypergeometric Series, there is an inversion identity on page 4 which can be used when $|q| > 1$:  

$(a; q)_n = (a^{-1}; p)_n (-a)^n p^{-n(n-1)/2} $ where $p=1/q$.

This returns a new expression in base $|1/q| < 1$.   You can see some examples of the identity being applied in Gasper's Lecture Notes on q-series (Exercise 1.1, Exercise 2.3, pg 14)
I have no idea how to interpret the result geometrically.  Could it be relevant to Buildings, Buekenhout geometry or $p$-adic geometry?<|endoftext|>
TITLE: A Transversal Achievement Game on a Grid
QUESTION [7 upvotes]: Two players alternately write O's (first player) and X's (second player) in the unoccupied cells of an n x n grid.
The first player (if any) to occupy some transversal (i.e., a set of n cells having no two cells in the same row or column) is the winner.  What is 
the outcome of the game given best possible play by both players?

REPLY [3 votes]: As Michael Albert mentioned, an interesting variant is when X wins by building a permutation, while O wins only by preventing X from building a permutation.  (The standard term for such a game is maker-breaker.)
According to my coding (which should not be regarded as infallible), X wins this game on a $4\times 4$ grid both when X goes first and when O goes first.  By a straight-forward inductive argument, this implies that X wins the maker-breaker game on an $n\times n$ grid for all $n\ge 4$.
The question remains of whether this has anything to do with the version of the game posed by Martin, where both X and O are trying to build permutations.<|endoftext|>
TITLE: Square Achievement Game on a Grid
QUESTION [8 upvotes]: Two players alternately write O's (first player) and X's (second player) in the unoccupied cells of an n x n grid.
The first player to occupy the vertices of a square with horizontal and vertical sides
is the winner.  What is the smallest n such that the first player has a winning strategy?
Note: Roland Bacher and Shalom Eliahou proved that every 15 x 15 binary matrix contains four equal entries (all 0's or all 1's) at the vertices of a square with horizontal and vertical sides.  So the game must result in a winner (the first player) when n=15.

REPLY [3 votes]: I believe the first player has a winning strategy for N=7.
In the configuration below (with "o" having the next move), "o" can force a win by making a "cross".  Note that the "cross" can also be applied to two non-adjacent "o"s in the same row or column.

The first player plays in the center.  The second player has 9 unique responses.  I'll consider the 3 most interesting responses here.
Below, the first player places the second "o" next to the first, threatening to make a cross.  So, the second player has 2 unique moves available.  However, both moves allow the first player to force a few moves and achieve the "cross", as illustrated.  (The two starting pieces for the "cross" are marked with a square around them.)

Another possible position for the initial "x".  This is also a win for "o".

Another possible position for the initial "x".  This is also a win for "o".  (I omitted 2 branches to save space, but they follow the same pattern).

I didn't consider the other 6 initial moves for "x", but I think the same techniques can be used to solve them.
FWIW, I'm pretty sure N=5 is a draw.  I haven't considered N=6, yet.<|endoftext|>
TITLE: How many orders of infinity are there?
QUESTION [84 upvotes]: Define a growth function to be a monotone increasing function $F: {\bf N} \to {\bf N}$, thus for instance $n \mapsto n^2$, $n \mapsto 2^n$, $n \mapsto 2^{2^n}$ are examples of growth functions.  Let's say that one growth function $F$ dominates another $G$ if one has $F(n) \geq G(n)$ for all $n$.  (One could instead ask for eventual domination, in which one works with sufficiently large $n$ only, or asymptotic domination, in which one allows a multiplicative constant $C$, but it seems the answers to the questions below are basically the same in both cases, so I'll stick with the simpler formulation.)  
Let's call a collection ${\mathcal F}$ of growth functions complete cofinal if every growth function is dominated by at least one growth function in ${\mathcal F}$.  
Cantor's diagonalisation argument tells us that a cofinal set of growth functions cannot be countable.  On the other hand, the set of all growth functions has the cardinality of the continuum.  So, on the continuum hypothesis, a cofinal set of growth functions must necessarily have the cardinality of the continuum.
My first question is: what happens without the continuum hypothesis?  Is it possible to have a cofinal set of growth functions of intermediate cardinality?
My second question is more vague: is there some simpler way to view the poset of growth functions under domination (or asymptotic domination) that makes it easier to answer questions like this?  Ideally I would like to "control" this poset in some sense by some other, better understood object (e.g. the first uncountable ordinal, the nonstandard natural numbers, or the Stone-Cech compactification of the natural numbers).
EDIT: notation updated in view of responses.

REPLY [4 votes]: I would like to make two points addressing the topological aspect of the OP by Terence Tao. The cardinals $\frak b$ and $\frak d$, although defined "arithmetically," are in fact equivalent to some incompleteness parameters of the Boolean Algebra $P (\omega) / fin$, equivalently of the space $\omega^* = \beta\omega \setminus \omega$.
1A) Indeed: $\frak b$ is the smallest cardinal of a family $\mathcal B$ of clopen subsets of $\omega^*$ such that, for some countable family $\mathcal A$ "disjointed" from $\mathcal B$ (and this means that for all $A$ in $\mathcal A$ and all $B$ in $\mathcal B$ the meet $A \wedge B = 0$), $\mathcal A$ and $\mathcal B$ cannot be separated by a clopen set.
This is well-defined because $P (\omega) / fin$ is not countably complete, i.e. countable subsets need not have the least upper bound.
On the other hand, it is well-known that  $P (\omega) / fin$ possesses the so called the du Bois-Reymond separation property (this means that disjointed  $\mathcal B$, $\mathcal A$ can be separated when both families are countable). A compact space with this property is called an $F$-space (even if it is  not Boolean, i.e. zero-dimensional), and the Boolean Algebra is called (by Eric van Douwen) weakly countably complete. 
1B) Now, regarding the cardinal $\frak d$, which is our main concern: $\frak d$ is the smallest cardinal for a $\mathcal B$ so that, for some countable $\mathcal A$ disjointed from $\mathcal B$, $\mathcal A$ and $\mathcal B$  cannot be "weakly left-separated." This means that there is no clopen set $C$ such that $\forall \  A \in \mathcal A$, $A \leq C$ and $\forall \  B \in \mathcal B$, $B - C \neq 0$.
These facts are, practically, obvious (so that I could not find a formal reference even in the excellent monograph of Andreas Blass he mentions in his answer).
2) My second remark is that the $\frak b$ and $\frak d$ so re-defined for  $P (\omega) / fin$  are (despite their "arithmetical" origin) perfectly applicable to any Boolean Algebra/Boolean space which is not countably complete, and that they measure the rough cardinal degree of such an incompleteness. 
(3) I also have some mild puzzlement about the a-symmetricity in the (revised) definition of $\frak d$.<|endoftext|>
TITLE: What determines a model structure?
QUESTION [18 upvotes]: It is easy to prove that a model structure is determined by the following classes of maps (determined = two model structures with the mentioned classes in common are equal).

cofibrations and weak equivalences
fibrations and weak equivalences

The second statement follows immediately from the first by duality.
What about the following classes of maps/objects (A short argument would be very helpful)?

cofibrations and fibrations
cofibrant objects and weak equivalences
cofibrant objects and fibrations
cofibrant objects and fibrant objects

I think each of these classes determine the structure respectively. For the last one I suppose that one has to use framings but I cannot see how to do it.
Edit: Thank you all for the illuminative answers.

true
?
true
false

REPLY [15 votes]: Mark's answer explains why (1) cofibrations and fibrations do determine the model structure, and Charles' and Tom's examples show that (4) cofibrant objects and fibrant objects do not.
For (2), any weak factorization system (L,R) on a category determines a model structure in which L is the cofibrations, R is the fibrations, and all maps are weak equivalences.  Thus, it suffices to find a category C admitting two wfs (L,R) and (L',R') with the same cofibrant objects, which is pretty easy; take for instance C=Set with (all functions, isos) and (monos, epis).
Finally, for (3) it is true, surprisingly, that the cofibrant objects and the fibrations determine a model structure.  In fact, merely the cofibrant+fibrant objects together with the fibrations also determine the model structure.  This observation is due to Joyal and can be found on his Catlab (currently Proposition 4).<|endoftext|>
TITLE: Operad terminology - Operads with and without O(0). 
QUESTION [8 upvotes]: In the Markl, Schneider and Stasheff text, topological operads are an indexed collection of spaces $O(n)$ for $n \in \{1,2,3,\cdots\}$ satisfying some axioms.  In May's text, the index set is allowed to include zero.  
1) Is there a standard terminology for operads with and without $O(0)$?  
2) Is there standard terminology for topological operads where $O(0)$ is a point, vs. $O(0)$ not being a point? 
Although it's less important I'd be curious if people have examples where these distinctions are interesting. 
Since any operad acts on its $O(0)$ part perhaps the $O(0)$ part should be called something like its "base"?  But then "baseless operad" would sound kind of pejorative.

REPLY [3 votes]: It is interesting to note that the general theory of operads with constants and that of operads without constants (here I refer to $O(0)$ as $\it{constants}$ showing my personal preference for terminology) admit the following distinct difference. Just for simplicity, let's consider operads enriched in sets and let's allow all (coloured) operads instead of just the monochromatic ones. Thus, let $\mathbf{Ope}$ be the category of all small coloured operads (symmetric or not, does not matter for this example) in $\mathbf{Set}$. Let $\mathbf{cfOpe}$ be the full subcategory of $\mathbf{Ope}$ consisting of the constant-free operads (that is those operads in which no $0$-ary arrows exist). 
Now consider the obvious functors $j:\mathbf{Cat}\to\mathbf{Ope}$ and $l:\mathbf{Cat}\to\mathbf{cfOpe}$. It is rather simple to show that each of these functors has a right adjoint so we get $j':\mathbf{Ope}\to\mathbf{Cat}$ and $l': \mathbf{cfOpe}\to\mathbf{Cat}$. However, $l'$ has again a right adjoint while $j'$ does not. 
Not much changes if one considers operads enriched in some symmetric monoidal category $V$.<|endoftext|>
TITLE: Enumerating ways to decompose an integer into the sum of two squares
QUESTION [29 upvotes]: The well known "Sum of Squares Function" tells you the number of ways you can represent an integer as the sum of two squares.  See the link for details, but it is based on counting the factors of the number N into powers of 2, powers of primes = 1 mod 4 and powers of primes = 3 mod 4.
Given such a factorization, it's easy to find the number of ways to decompose N into two squares. But how do you efficiently enumerate the decompositions?
So for example, given N=2*5*5*13*13=8450 , I'd like to generate the four pairs:
13*13+91*91=8450
23*23+89*89=8450
35*35+85*85=8450
47*47+79*79=8450
The obvious algorithm (I used for the above example) is to simply take i=1,2,3,...,$\sqrt{N/2}$ and test if (N-i*i) is a square. But that can be expensive for large N. Is there a way to generate the pairs more efficiently?   I already have the factorization of N, which may be useful.
(You can instead iterate between $i=\sqrt{N/2}$ and $\sqrt{N}$ but that's just a constant savings, it's still $O(\sqrt N)$.

REPLY [2 votes]: Another point of view, which might be easier from an algorithmic point of view (in my opinion), is to look at the decomposition $n = a^2+b^2$ using complex numbers $n = (a+bi)(a-bi)$. Suppose you know how to find the solutions $a,b$ when $n$ is prime. Then note that the identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ can be seen as the equality of the product of modules of two complex numbers. 
In order to find all possible decompositions of a number $n$ into a sum of squares just write the factorization of $n$ in $\Bbb{Z}[i]$: $n = \prod (a_j+ib_j)$ and note that in this product every factor comes with its conjugate. First, ignore the powers of $2$ and the primes of the form $4k+3$. Now, in order to find all factorizations, just split all factors into two columns with conjugate pairs being on different columns. Doing this, when taking the product on each column we'll get a pair of conjugate numbers whose product equals to $n$, so we find a solution of the representation $n=a^2+b^2$. The number of ways to split the factors in the two columns with conjugate pairs on different sides will be equal the number of divisors of $n$ in $\Bbb{Z}[i]$ (divided by 4, since you can multiply factors by $1,-1,i,-i$) so all factors will be generated. 
If $n$ contains primes of the forms $4k+3$ or powers of $2$ look at Will Jagy's answer. Note that if you reduce all powers of $4$ and you still have a $2$ left in $n$ you can write $2 = (1+i)(1-i)$ and split this on different sides of the two columns.<|endoftext|>
TITLE: What are the most elegant proofs that you have learned from MO?
QUESTION [84 upvotes]: One of the things  that MO does best is provide clear, concise
answers to specific mathematical questions. I have picked up ideas
from areas of mathematics I normally wouldn't touch, simply because
someone posted an eye-catching answer on MO.
In particular, there have been some really elegant and surprising proofs. 
For example, this one  by villemoes, when the questioner asked for a simple proof that there are uncountably many permutations of $\mathbb{N}$.

The fact that any conditionally convergent series [and that such exists] can be rearranged to converge to any given real number x proves that there is an injection P from the reals to the permutations of $\mathbb{N}$.

Or this one  by André Henriques, when the questioner asked whether the
Cantor set is the zero set of a continuous function:

The continuous function is very easy to construct: it's the distance to the closed set.

There must many such proofs that most of us have missed, so I'd like to see a list,
an MO Greatest Hits if you will. Please include a link to the answer, so that the author
gets credit (and maybe a few more rep points), but also copy the proof, as
it would nice to see the proofs without having to move away from the page.
(If anyone knows the best way to copy text with preservation of LaTeX, please advise.)
I realize that one person's surprise may be another person's old hat, so that's why I'm
asking for proofs that you learned from MO. You don't have to guarantee that the
proof is original.

REPLY [5 votes]: My candidate is Jim Belk's one-line answer to the question about the existence of functions from $\Bbb{R}$ to $\Bbb{R}$ whose range is $\Bbb{R}$ on every open interval. 
I do wonder, however, if Jim Balk's solution was known to founders of classical set theory (Cantor, Bernstein, Hausdorff, ...).<|endoftext|>
TITLE: Generalizations of the nonabelian group of order $pq$
QUESTION [8 upvotes]: If $p < q$ are primes then there is a nonabelian group of order $pq$ iff $q = 1 \pmod p$, in which case the group is unique. If $p = 2$ we obtain the dihedral group of order $2q$, which generalizes first to the dihedral group of order $2n$ and then even further to the "generalized dihedral group" where the cyclic group of order $n$ is replaced with any abelian group.
What if $p > 2$? Is there a natural generalization of the groups of order $pq$ to a family of groups of order $pn$? Maybe more than one possible generalization? Is it maybe even meaningful to talk about the "generalized $p$-hedral group"?

REPLY [7 votes]: The nonabelian group of order $pq$ is given by generators $a$, $b$, with relations $a^p=1$, $b^q=1$, $a^{-1}ba=b^r$, where $r$ is chosen so $r^p$ is 1 mod $q$. If there is an element $r$ of order $p$ mod $n$, then there is a nonabelian group of order $pn$ with generators $a$, $b$, and relations $a^p=1$, $b^n=1$, $a^{-1}ba=b^r$.<|endoftext|>
TITLE: When sticks fall, will they weave?
QUESTION [12 upvotes]: Imagine $n$ $z$-vertical sticks uniformly spaced around a unit-radius circle in the $xy$-plane.
At $t{=}0$, each is randomly $\epsilon$-perturbed from the vertical, and they fall under
the influence of gravity. Will some sticks form a "teepee" suspended above the $xy$-plane?
Let us assume the sticks are one-dimensional segments of height $h$, perhaps $h{=}2$ so they
span the diameter,
and that their base points are pinned to the plane via universal joints.
It seems possible that a subset of sticks could fall to form a weaving with a cyclic
on-top-of graph, as illustrated below.



Assuming a sufficient coefficient of friction $\mu$ between pairs of sticks, it seems conceivable that such a
structure would not collapse to the plane.
What is your intuition here?  For sufficiently large $n$, and adequate $h$ and $\mu$,
would some sticks form a woven structure above the plane?
Or would all sticks ultimately flatten to the plane?
If the latter, are there natural conditions that would lead to formation of a "teepee"?
I am less interested in probability calculations than in a qualitative assessment.
I've tagged this 'recreational' because it is only a peripheral spinoff of my research.
Addendum.  It is now clear, from the comments of Scott Morrison and Rahul Narain, that
the random perturbations should be random both in direction and magnitude.
Otherwise, as Scott incisively observed, the sticks, with high probability, all fall without touching
one another until they reach the plane (which to me is already rather remarkable!).

REPLY [5 votes]: Couldn't accomplish this in a comment to Rahul's #1, but here is a physical realization of my drawing with colored pencils. Compare:


The pencils are in contact, but the lead inside could serve as lines in space without contacts.
I don't know how to reconcile this with your quote from Whiteley...<|endoftext|>
TITLE: Is there a name for a family of finite sequences that block all infinite sequences?
QUESTION [26 upvotes]: Let ${\bf N}^\omega = \bigcup_{m=1}^\infty {\bf N}^m$ denote the space of all finite sequences $(N_1,\ldots,N_m)$ of natural numbers.  For want of a better name, let me call a family ${\mathcal T} \subset {\bf N}^\omega$ a blocking set if every infinite sequence $N_1,N_2,N_3,N_4,\ldots$ of natural numbers must necessarily contain a blocking set $(N_1,\ldots,N_m)$ as an initial segment.  (For the application I have in mind, one might also require that no element of a blocking set is an initial segment of any other element, but this is not the most essential property of these sets.)
One can think of a blocking set as describing a machine that takes a sequence of natural number inputs, but always halts in finite time; one can also think of a blocking set as defining a subtree of the rooted tree ${\bf N}^\omega$ in which there are no infinite paths.  Examples of blocking sets include

All sequences $N_1,\ldots,N_m$ of length $m=10$.
All sequences $N_1,\ldots,N_m$ in which $m = N_1 + 1$.
All sequences $N_1,\ldots,N_m$ in which $m = N_{N_1+1}+1$.

The reason I happened across this concept is that such sets can be used to pseudo-finitise a certain class of infinitary statements.  Indeed, given any sequence $P_m(N_1,\ldots,N_m)$ of $m$-ary properties, it is easy to see that the assertion

There exists an infinite sequence $N_1, N_2, \ldots$ of natural numbers such that $P_m(N_1,\ldots,N_m)$ is true for all $m$.

is equivalent to

For every blocking set ${\mathcal T}$, there exists a finite sequence $(N_1,\ldots,N_m)$ in ${\mathcal T}$ such that $P_m(N_1,\ldots,N_m)$ holds.

(Indeed, the former statement trivially implies the latter, and if the former statement fails, then a counterexample to the latter can be constructed by setting the blocking set ${\mathcal T}$ to be those finite sequences $(N_1,\ldots,N_m)$ for which $P_m(N_1,\ldots,N_m)$ fails.)
Anyway, this concept seems like one that must have been studied before, and with a standard name.  (I only used "blocking set" because I didn't know the existing name in the literature.)  So my question is: what is the correct name for this concept, and are there some references regarding the structure of such families of finite sequences?  (For instance, if we replace the natural numbers ${\bf N}$ here by a finite set, then by Konig's lemma, a family is blocking if and only if there are only finitely many finite sequences that don't contain a blocking initial segment; but I was unable to find a similar characterisation in the countable case.)

REPLY [31 votes]: Intuitionists use the name "bar" for what you called a blocking set.  The relevant context is "bar induction," the principle saying that, if (1) a property has been proved for all elements of a bar and (2) it propagates in the sense that, whenever it holds for all the 
one-term extensions of a finite sequence s then it holds for s itself, 
then this property holds of the empty sequence.  (I'm omitting some technicalities here that distinguish different versions of bar induction.)  
There's also a closely related notion in infinite combinatorics, called a "barrier"; this is a collection $B$ of finite subsets of $\mathbf N$ such that no member of $B$ is included in another and every infinite subset of $\mathbf N$ has an initial segment in $B$.  This is the subject of a partition theorem due to Nash-Williams: If a barrier is partitioned into two pieces, then there is an infinite $H\subseteq\mathbf N$ such that one of the pieces includes a barrier for $H$ (meaning that every infinite subset of $H$ has an initial segment in that piece).

REPLY [14 votes]: If you assume that no element of the family is an initial segment of the other, it is called a barrier in wqo (well quasi order) theory. See mostly Nash-Williams.<|endoftext|>
TITLE: Is there a solution for the equation $x^m-y^n=k$ in which $k > 1$?
QUESTION [6 upvotes]: The Catalan conjecture state that $x^m-y^n=1$ has only the solution $x=3, m=2, y=2, n=3$. This conjecture was proved by Preda Mihailescu in 2004, but I want to know about the equation mentioned above. Is there a asolution of this?

REPLY [14 votes]: Pillai's conjecture is that for each $k$, there are only finitely many solutions.
The ABC conjecture implies Pillai's conjecture as follows. First, I state a form of the ABC conjecture. Given three relatively prime positive integers $A+B=C$, the quality of the triple $(A,B,C)$ is $\log(C)/\log(R)$, where $R$ is the product of the primes that divide $ABC$. For example, the quality of $(5,27,32)$ is $\log(32)/\log(30)$. One strong form of the ABC conjecture is that there are only finitely many triples (of relatively prime positive integers) with quality greater than $1.001$.
Now a solution to $x^m-y^n=k$ with $\gcd(x,y)=1$ has $(A,B,C)=(k,y^n,x^m)$. Observe that $R\leq k y x$. Thus, the quality of the triple is at least 
  $$\frac{m\log(x)}{\log(k)+\log(x)+\log(y)}\approx \frac{m\log(x)}{\log(k)+(\frac mn +1)\log(x)}.$$ 
As $x\to\infty$, this gives a quality approaching $\frac{mn}{m+n}$, and for $m n>4$ this is $\geq 1.2$, which can only happen finitely many times by ABC. Also, $\frac{mn}{m+n}=1$ if $m=n=2$, so that case has to be handled separately. If $x$ and $y$ are not relatively prime, as KConrad commented below, set $d=\gcd(x^m,y^n)$ and apply ABC to the triple $(k/d, y^n/d,x^m/d)$.

REPLY [4 votes]: There is a conjecture that for every positive natural number k there are just finitely many solutions of the above equation. As far as I know, this is open for k>1. In fact, Erdős conjectured that the difference between a full power x and the next full power is at least xc for some positive constant c.<|endoftext|>
TITLE: Primitive element theorem without building field extensions
QUESTION [12 upvotes]: Is there are nice way to prove the primitive element theorem without using field extensions?
The primitive element theorem says that if $x$ and $y$ are algebraic over $F$ and $y$ is separable over $F$, then there exists a $z \in F(x,y)$ such that $F(x,y) = F(z)$.  In the case where $F$ is infinite, $z$ can be expressed in the form $x + {\lambda}y$ with $\lambda \in F$.  In fact, almost any $\lambda$ will do.  There are only a finite number of exceptions.  These exceptions are $\frac{\alpha_i - x}{y - \beta_j}$ where $\alpha_i$ and $\beta_j$ range over the (other) roots of the minimal polynomials of $x$ and $y$ respectively.
But in order to talk about $\alpha_i$ and $\beta_j$ we need to build a field extension where the minimal polynomials of $x$ and $y$ split.  This is the step I'm hoping to avoid.
Perhaps we can build a polynomial in $F[x]$ whose roots are $\frac{\alpha_{i_1} - \alpha_{i_2}}{\beta_{j_1} - \beta_{j_2}}$ and simply avoid picking $\lambda$s which are roots of this polynomial, and then use whatever properties this polynomials has to prove that this works.
Or maybe there is another completely different way of proving the primitive element theorem while avoiding building field extensions.
I found a nice proof that $x$ is separable over $F$ if and only if every $y \in F(x)$ is separable over $F$ that avoids building field extensions.  It uses derivations instead.  Now I'm hoping to do the same with the primitive element theorem.
Edit: I'll try to give some motivation. Adding roots of polynomials to fields in constructive mathematics is more difficult than in classical mathematics (because irreducibility is undecidable). It only works for countable fields, and then you have no guarantee that the original field will be a decidable subset of the new field. Yes, it can be done, but it seems like a pain. There is another way too using double negation translations, but it also seems like a pain. Instead I'd rather avoid the whole issue of building splitting fields, if I can, and it seems tantalizingly close to possible.  After all, the only reason splitting fields are used here is to build a finite set of elements in the base field to avoid.

REPLY [7 votes]: OK, I have a proof which meets your conditions. I relied on this write up of the standard proof as a reference.
Lemma 1: Let $K/F$ be an extension of fields, and let $f(x)$ and $g(x)$ be polynomials in $F[x]$. Let $d_F(x)$ be the GCD of $f$ and $g$ in $F[x]$ and let $d_K(x)$ be the GCD of $f$ and $g$ in $K[x]$. Then $d_F(x)$ and $d_K(x)$ coincide up to a scalar factor.
Proof: Since $d_F(x)|f(x)$ and $d_F(x)|g(x)$ in $K[x]$, we have $d_F(x)|d_K(x)$. Now, there are polynomials $p(x)$ and $q(x)$ in $F[x]$ such that $f(x) p(x) + g(x) q(x) = d_F(x)$. So $d_K(x) | d_F(X)$. Since $d_F(x)$ and $d_K(x)$ divide each other, they only differ by a scalar factor.
Lemma 2: Let $f(x)$ and $g(x)$ be polynomials with $g(0) \neq 0$. Then, for all but finitely many $t$, the polynomials $f(tx)$ and $g(x)$ have no common factor.
Proof: Let $f(x) = f_m x^m + \cdots + f_1 x + f_0$ and $g(x) = g_n x^n + \cdots + g_1 x + g_0$. If $f(tx)$ and $g(x)$ HAVE a nontrivial common factor, then there are polynomials $p(x)$ and $q(x)$, of degrees $\leq n-1$ and $\leq m-1$, such that 
$$f(tx) p(x)=g(tx) q(x).$$ 
This is $m+n$ linear equations on the $m+n$ coefficients of $p$ and $q$. 
Writing this out in coefficients, the matrix
$$\begin{pmatrix} 
f_m t^m & \cdots & f_1 t & f_0 & 0 & 0 & \cdots & 0 \\
0 & f_m t^m & \cdots & f_1 t & f_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & f_m t^m & \cdots & f_1 t & f_0 \\
g_n & \cdots & g_1 & g_0 & 0 & 0 & \cdots & 0 \\
0 & g_n & \cdots & g_1 & g_0 & 0 & \cdots & 0 \\
\ddots \\
0 & 0 & \cdots & 0 & g_n & \cdots & g_1 & g_0
\end{pmatrix}$$
has nontrivial kernel. The determinant of this matrix is a polynomial in $t$, with leading term $(f_m)^n (g_0)^n t^{mn} + \cdots$. (Recall that $g_0 \neq 0$.) So, for all but finitely many $t$, this matrix has nonzero determinant and $f(tx)$ and $g(x)$ are relatively prime. QED.
Now, we prove the primitive element theorem (for infinite fields). Let $\alpha$ and $\beta$, in $K$, be algebraic and separable over $F$, with minimal polynomials $f$ and $g$. We will show that, for all but finitely many $t$ in $F$, we have $F(\alpha - t \beta) = F(\alpha, \beta)$.
Let $f(x) = (x -\alpha) f'(x -\alpha)$ and $g(x) = (x - \beta) g'(x - \beta)$. Since $\beta$ is separable, we know that $g'(0) \neq 0$. Note that $f'$ and $g'$ have coefficients in $K$. By Lemma 2, for all but infinitely many $t$ in $F$, the polynomials $f'(ty)$ and $g'(y)$ have no common factor. Choose a $t$ for which no common factor exists. Set $F' = F(\alpha - t \beta)$; our goal is to show that $F'=F(\alpha, \beta)$.
Set $h_t(x) = f(tx + \alpha-t \beta)$. Note that $h_t(x)$ has coefficients in $F'$. We consider the GCD of $h_t(x)$ and $g(x)$. 
Working in $K$, we can write $h_t(x) =  t (x - \beta) f'(t (x- \beta))$ and $g(x) = (x-\beta) g'(x-\beta)$. By the choice of $t$, the polynomials $f'(t (x- \beta))$ and $g'(x-\beta)$ have no common factor, so the GCD of $h_t(x)$ and $g(x)$, in the ring $K[x]$, is $x-\beta$.
By Lemma 1, this shows that $x - \beta$ is in the ring $F'[x]$. In particular, $\beta$ is in $F'$. Clearly, $\alpha$ is then also in $F'$, as $\alpha= (\alpha - t \beta) + t \beta$. We have never written down an element of any field larger than $K$. QED.<|endoftext|>
TITLE: Path continuity for (closed) martingales?
QUESTION [6 upvotes]: Take a time interval $[0,T]$, and a filtered probability space $(\Omega,P,\mathcal{F},\mathcal{F}_t)$.  If $X \in L^1(\mathcal{F}_T)$, then $M_t = E [X \ | \ \mathcal{F}_t]$ is a martingale.  If I want the martingale $M$ to have continuous or right continuous paths, is there a condition I can impose on the filtration to ensure this?
A standard result says that if the filtration is right-continuous, meaning that $\cap_{s>t} \mathcal{F}_s = \mathcal{F}_t$, then there exists a modification of $M$ with right continuous paths (in fact right continuous with left limits).  However, I want to say something about the original process, and not a modification.

REPLY [2 votes]: Hi,
The thing is that both your original process and the càldlàg modification of it, are equivalent with respect to your probability measure (if the usual hypothsesis hold). 
Probabilisticly speaking, there is nothing that can be said about your original process that cannot be said from your modification (if the usual hypothsesis hold) and this is why we work with the nice one.
Why don't you ask the real question you have in mind about your original process ?
(which is not continuous even for a Brownian Motion look for example at the construction by Karatzas and Shreve !!! )
Best Regards<|endoftext|>
TITLE: Older editions of which books were better than the new ones?
QUESTION [5 upvotes]: When choosing some mathematics book to study, is it always the case that one should look for the current edition of the book. Are there any examples when the older edition of some book is clearly better than the latest version?

REPLY [5 votes]: Hausdorff's book Mengenlehre in the first edition had an appendix, omitted in subsequent editions,  on the Banach paradox.  (Later made into the Banach-Tarski paradox by Tarski...)  Someone once told me this was the best, most elementary, presentation of it -- I haven't compared different versions of the proof myself.<|endoftext|>
TITLE: A Model Category of Segal Spaces?
QUESTION [15 upvotes]: So in Julie Bergner's work on $(\infty, 1)$-categories arXiv:0610239, she considers several model categories which model $(\infty, 1)$-categories, which are known to be equivalent. I'm guessing that there is another model which is equivalent to these. It is probably known to experts, and probably exists for easy reasons, but I'm not seeing it so I'm asking here. 
Background
In Julie's survey, which reviews some of her own work in arXiv:0504334, the work of Joyal and Tierney arXiv:0607820, and the work of others, she compares four models of $(\infty, 1)$-categories. They are model categories of Segal Categories, Complete Segal spaces, Quasi-categories, and Simplicial Categories. 
All of these models are related in multiple ways, but some are more closely aligned than others. In particular the Complete Segal spaces and the Segal categories are very similar objects. The underlying categories for these two model categories are almost the same. They are the category of simplicial spaces and the category of simplicial spaces whose zeroeth space $X_0$ is discrete, respectively. 
Among the simplicial spaces we have those which satisfy the Segal condition. These are the (Reedy fibrant) simplicial spaces such that the Segal map
$$ X_n \to X_1 \times_{X_0} \cdots \times_{X_0} X_1 $$
is a weak equivalence. These are called Segal spaces. There is a model structure on the category of simplicial spaces such that the Segal spaces are the fibrant objects. It is a localization of the Reedy model structure. But in this model structure the weak equivalences between Segal spaces are just the level-wise weak equivalences.
The model category of complete Segal spaces is a further localization of this. The weak equivalences between fibrant objects (complete Segal spaces) in this model category are pretty easy. They are the level-wise weak equivalences. More generally the weak equivalences between Segal spaces in this new model structure can also be identified, without too much trouble. They are called Dwyer-Kan equivalences or DK-equivalences for short. 
A Segal category is a Segal space where the space of objects $X_0$ is discrete. There is a Quillen equivalent model structure on the category of those simplicial spaces whose zeroeth space is discrete. 
In this model category, a map between two Segal categories is a weak equivalence if and only if it is a DK-equivalence. 
Question
I'm wondering if there is an intermediary model category which is equivalent to both the above model categories (hopefuly in an obvious way) which has the following properties: 

It should be a model category on the category of simplicial spaces.
The fibrant objects should be the Segal spaces. (Not necessarily complete).
The weak equivalences should be the DK-equivalences. 

Does such a model category exist? is it well known?

REPLY [9 votes]: I'm not aware that the model category you want has been constructed.  But it seems like an interesting question.  You should ask Julie Bergner if she has thought of anything along these lines.
I don't have an answer, but I will think out loud for a bit.
(I'll be a little vague by what I mean by "space", but I probably mean "simplicial sets" here.)
I will assume that your property 3 should say: "The weak equivalences between fibrant objects (i.e., between Segal spaces) are the DK-equivalences."   I would also like to throw in an additional property:
4.  The trivial fibrations between Segal spaces are maps $f:X\to Y$ which are DK-equivalences, Reedy fibrations, and such that the induced map $f_0:X_0\to Y_0$ on $0$-spaces is surjective.  (Note: since $f$ is a Reedy fibration, $f_0$ is a fibration of spaces.)
Yes, this comes out of thin air ... but it's modelled on the trivial fibrations in the "folk model structure" on Cat.  You could go further, and posit that fibrations between Segal spaces are Reedy fibrations such that $f_0$ is surjective.
Given a space $U$, let $cU$ denote the "$0$-coskeleton" simplical space, with $(cU)_n=U^{\times (n+1)}$.  If $U$ is a fibrant space, then $cU$ is Reedy fibrant; if $U\to V$ is a fibration, $cU\to cV$ is a Reedy fibration.  Furthermore, $cU$ clearly satisfies the Segal condition.
Thus, if $g:U\to V$ is a surjective fibration of spaces, $cg: cU\to cV$ should be a trivial fibration in our model category, according to 4.
The functor $c$ is right adjoint to $X\mapsto X_0$: that is, maps of simplicial spaces $X\to cU$ are naturally the same as maps $X_0\to U$ of spaces.
Putting all this together, we discover that, if such a model category exists, a cofibration $f: A\to B$ should have the following properties: the map $f_0 : A_0\to B_0$ is a cofibration of spaces, and $B_0=B_0'\amalg B_0''$ so that $f_0$ restricts to a weak equivalence $A_0\to B_0'$, and such that $B_0''$ is homotopy discrete (i.e., has the weak homotopy type of a discrete space). 
In particular, a necessary condition for $B$ to be cofibrant is that $B_0$ is homotopy discrete.
This is a pretty restrictive condition on cofibrations, but it does not seem impossible.  If there actually was a model category with all these properties, it appears that the class of fibrant-and-cofibrant objects would be what you might call the quasi Segal categories.  These are the Segal spaces $X$ such that $X_0$ is homotopy discrete.  Cofibrant replacement of a Segal category would give a DK-equivalent quasi-Segal category.
That would be a pleasing outcome, and probably along the lines of what you're looking for.<|endoftext|>
TITLE: Must the set of lines through the origin on which a nonconstant entire function is bounded be finite?
QUESTION [40 upvotes]: If an entire function is bounded for all $z \in \mathbb{C}$, than it's a constant by Liouville's theorem. Of course an entire  function can be bounded on lines through the origin $z=r \exp(i \phi), \phi= \text{const.}, r \in \mathbb{R}$ without being constant (e.g. $\cos(z^n)$ is bounded on $n$ lines).
What is the maximum cardinality of the set of "directions" $\phi$ for which an entire function can be bounded without being constant?
From intuition I would expect only finitely many directions. Is this correct?
(Picard's second theorem says that in any open set containing $\infty$ every value with possibly a single exception is taken infinitely often by an entire non-constant function. Here I'm asking a somehow "orthogonal" question, looking for lines through $\infty$ where an entire non-constant function is bounded.)

REPLY [59 votes]: Newman gave an example in 1976 of a non-constant entire function bounded on each line through the origin in "An entire function bounded in every direction".  
I like the second sentence of the article:

This is exactly what is needed to confuse students who have just struggled to comprehend the meaning of Liouville's theorem.

Armitage gave examples in 2007 of non-constant entire functions that go to zero in every direction in "Entire functions that tend to zero on every line".  For this I have only seen the MR review.  (If you don't have MathSciNet access, the link should still give you the publication information to find the article.)

Update: I just decided to take a look at the Armitage paper, and the introduction was enlightening:

Although every bounded entire (holomorphic) function on $\mathbb{C}$
  is constant (Liouville’s theorem), it has been known for more than a hundred years
  that there exist nonconstant entire functions $f$ such that $f(z) → 0$ as $z →∞$ along
  every line through 0 (see, for example, Lindelöf’s book [10, pp. 119–122] of 1905). And it has been known for more than eighty years that such functions can tend to 0
  along any line whatsoever (see Mittag-Leffler [11], Grandjot [8], and Bohr [4]). Further
  references to related work are given in Burckel’s review [5] of Newman’s note [12].
  Entire functions with radial decay are used by Beardon and Minda [3] and Ullrich [14]
  in studies of pointwise convergent sequences of entire functions.

Armitage goes on to mention that Mittag-Leffler and Grandjot also gave explicit constructions, but states, "The examples given in what follows may nevertheless
be of some interest because of their comparative simplicity."  The examples are 
$$F(z)=\exp\left(-\int_0^\infty t^{-t}\cosh(tz^2)dt\right) - \exp\left(-\int_0^\infty t^{-t}\cosh(2tz^2)dt\right)$$ and
$$G(z)=\int_0^\infty e^{i\pi t}t^{-t}\cosh(t\sqrt{z})dt\int_0^\infty e^{i\pi t}t^{-t}\cos(t\sqrt{z})dt .$$

REPLY [14 votes]: The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector 
$$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$
In particular, $e^z=E_{1,1}(z)$ is bounded in
$$\frac{\pi}{2}< \arg z<\frac{3\pi}{2}.$$<|endoftext|>
TITLE: Understand Cech Cohomology
QUESTION [13 upvotes]: I am currently trying to understand Cech cohomology. Five questions arised and I would be glad for help. In what follows $X$ is a topological space.

I really like Dugger's and Isaksen's paper "Topological Hypercovers and ...". They prove for an arbitrary open cover $U=(U_a)_{a\in A}$ of $X$ the weak equivalence
$$
\operatorname{hocolim} ~C(U)\to X
$$
with $C(U)$ is the usual simplicial space of Cech namely
$$
\dots\to\coprod_{(b,c)\in A\times A}U_b\cap U_c\to\coprod_{a\in A}U_a
$$
Related statements are due to Segal. If $U$ is now a cover with every iterated intersection contractible we have a weak equivalence
$$
\operatorname{hocolim} ~C(U)\to \operatorname{hocolim} ~\pi_0C(U)
$$
The space $\pi_0C(U)$ is a simplicial set. You can find such a cover for every locally contractible space for instance. Dugger and Isaksen build the category $OpCov(X)$. This has open covers of $X$ as objects and morphisms are mappings between the index sets $f:A\to B$ and for every $a$ in $A$ a mapping $U_a\to V_{f(a)}$. One can also build the category $Cov(X)$ with same objects but morphisms only strict containments of covers. Is then
$$
\operatorname{hocolim} ~C(U)\sim \lim_{U\in Cov(X)}~\operatorname{hocolim} ~\pi_0C(U)\sim \lim_{U\in OpCov(X)}~\operatorname{hocolim} ~\pi_0C(U)
$$
all weakly equivalent for locally contractible $X$?
Why does one define Cech cohomology as $H^n(X)=\operatorname{colim} H^n(\operatorname{hocolim} C(U))$ with colimit over $Cov(X)^{op}$? Is this the right definition? Why not the limit over $Cov(X)$ instead of colimit over the opposite category?
What is the problem with Cech Homology? One defines $\hat H^n(X)=\operatorname{colim} ~H^n(C(U))$ as Cech cohomology. Then for a locally contractible $X$ it coincides with singular cohomology. Why not define $\hat H_n(X)=\lim ~H_n(C(U))$. I have read that it is because the limit functor does not respect exact sequences. Does it mean $\hat H_n(X)$ coincides not with singular homology for locally contractible $X$?
Can one see directly without going through sheaf cohomology that singular cohomology and Cech cohomology are the same for locally contractible $X$? I have written down complexes for the boundary of the two simplex with a nice cover and yes the cohomology is the same. Is there a kind of MayerVietoris argument?
Here is my last question. $X$ is a good space now. Where is the mistake in the following equality. I consider Top as a topologgical model category and the crucial step is perhaps the one which looks like a smallness argument on the homotopy category.
$$
\begin{array}{rcl}
\pi_0(X)&=&[S^0, \operatorname{hocolim} ~C(U)]\\
&=& \operatorname{hocolim}[S^0, C(U)]\\
&=& \operatorname{hocolim} \pi_0(C(U))\\&=&X
\end{array}
$$
Here $U$ is a cover with everything contractible as above.

Thank endurance reading and for help.

REPLY [4 votes]: On 3rd question: Čech homology is not a homology theory in the sense of Eilenberg-Steenrod: the exactness axiom (long exact sequence in homology) does not hold, exactly because of the problem with limits. There is however a sophisticated method, discovered by Sibe Mardešić, to correct this, by modifying slighly the Čech definition. The resulting "strong homology theory" agrees with singular homology on the spaces having homotopy type of CW complexes, and does give long exact sequence of pairs $(X,A)$ where $X$ is paracompact and $A$ closed; moreover for metric compacta it satisfies not only all the axioms of Eilenberg-Steenrod, but also the relative homeomorphism axiom and the wedge axiom. The only homology theory on the metric compacta satisfying not only the Eilenberg-Steenrod but also the wedge axiom is the Steenrod-Sitnikov homology theory, hence the strong homology agrees with it.<|endoftext|>
TITLE: Drinfeld's 1988 letter to Schechtman: translation request
QUESTION [9 upvotes]: Inspired by an old question by Kevin Lin and the communal translation of an answer by Laurent Fargues, I am proposing a communal effort to translate from Russian to English Drinfeld's famous 1988 letter to Schechtman: typeset and in the handwritten original.
I am starting a blank, communal answer below in which to whip up the translation. Thanks in advance for any contributions!

REPLY [11 votes]: Keith Conrad has kindly produced a translation available here. It can be cited as

Vladimir Drinfeld, A letter from Kharkov to Moscow. EMS Surv. Math. Sci. 1 (2014), 241-248. doi:10.4171/EMSS/5<|endoftext|>
TITLE: Semisimple-ish rings!
QUESTION [8 upvotes]: Let S be the class of all rings R which have 1 and satisfy this condition:
for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.)
All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings.
My questions:
Will this condition by itself force an element of S to have any (known, interesting) structure?
A more important question:  
What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!)

REPLY [16 votes]: By Zorn's lemma, each right ideal is contained in a maximal right ideal,
therefore if $I+J = R$ then $I+M = R$ where $M$ is a maximal right ideal.
If $I+M\ne R$ for all maximal right ideals $M$ then $I\subseteq M$ for
all maximal ideals $M$. Thus $I\subseteq J(R)$, the Jacobson radical of $R$
which is the intersection of all maximal right ideals of $R$. Hence
condition $S$ is equivalent to $J(R)=0$.
A ring with vanishing Jacobson ideal is called semiprimitive.
As $J(R)$ is also the intersection of the maximal left ideals of $R$
then the property of semiprimitivity is left-right symmetric.
There are plenty of examples of semiprimitive rings which are not
semisimple. For instance every simple ring is semiprimitive
and every subdirect product of semiprimitive rings is semiprimitive
($\mathbb{Z}$ is a subdirect product of finite fields).
As a reference see Section 10.4 of P. M. Cohn
Algebra (2nd ed. vol 3) Wiley 1991.<|endoftext|>
TITLE: Is it true that if the pushforward of a coherent sheaf is locally free, then the original sheaf is locally free?
QUESTION [11 upvotes]: I think the title says it all.  If I have a finite map $p:X\to Y$ between schemes, and $F$ is a coherent sheaf on $X$ such that $p_*F$ is locally free, can I conclude that $F$ is locally free?
Assumptions I would be happy to make:

The map $p$ is flat.
$X$ and $Y$ are both $\mathbb{A}^n$.


I would be much more pleased with a reference than a proof, since I would like to use this result in a paper.

REPLY [8 votes]: Since you asked for reference, here are some references that may be helpful  (the question is local):
1) (EDITED: Thanks to BCnrd for keeping me honest here!) If $(R,m)\to (S,n)$ is a map of  Cohen-Macaulay local rings of same dimensions, $N$ a finite $S$-module which is $R$-free flat then $${\rm{depth}}_ S N ={\rm{depth}} R = {\rm{depth}} S.$$
This follows from Prop 1.2.16 and Theorem A.11 of Bruns-Herzog "Cohen-Macaulay rings". Namely, take $M=R$ in both results, one get from A.11 that $\dim_SN = \dim_RN + \dim_SN/mN$, so $\dim_SN/mN=0$, then 1.2.16 gives  ${\rm{depth}}_SN = {\rm{depth}} R$.
2) If $S$ is regular, then f.g modules with maximal depth are free.
This is well-known. It follows from Auslander-Buchsbaum formula, as Boyarsky pointed out.
EDIT: Just to be clear, it is enought to assume: $f$ is finite, $X$ regular, and $Y$ Cohen-Macaulay scheme (certainly true if $Y$ also regular or smooth, but is a much weaker condition). For example the map induced by $k[x,y]/(xy) \to k[x]$ by killing $y$ works.<|endoftext|>
TITLE: Pushouts of noetherian rings
QUESTION [7 upvotes]: Does the category of noetherian commutative rings have pushouts?
Background: If $X/S$ is an abelian scheme, then the relative Picard functor $\mathrm{Pic}_{X/S}$ is only defined on the category of locally noetherian $S$-schemes (as far as I know). It is a group functor and in some situations it is representable. We then get a group object in the category of locally noetherian $S$-schemes, and I ask myself if it has a multiplication morphism. [Edit: Boyarsky has mentioned in the comments how to deal with this.]
Observe that the tensor product of noetherian commutative rings does not have to be noetherian (isn't this ugly?). Even for fields there is a counterexample: Let $L/K$ be a purely transcendental field extension of infinite transcendence degree. Then $\Omega^1_{L/K}$ is infinite-dimensional, from which you can concluce that the kernel of $L \otimes_K L \to L, a \otimes b \mapsto ab$  is not finitely generated. Thus $L \otimes_K L$ is not noetherian.
Of course, this does not disprove that $L \leftarrow K \rightarrow L$ has a pushout in the category of noetherian commutative rings. How can this be done? The question has a similar spirit as this one.

REPLY [7 votes]: Let $k$ be a field, and let $\ell = k(x_1,x_2,\ldots)$ be the fraction field of $k[x_1,x_2,\ldots]$. Then $\ell \otimes_k \ell$ is the localisation of $k[x_1,x_2,\ldots][y_1,y_2,\ldots]$ at the multiplicative set
$$S = \left\{fg\ \bigg|\ \begin{array}{ll}f \in k[x_1,x_2,\ldots]\setminus\{0\},\\g \in k[y_1,y_2,\ldots] \setminus\{0\}\end{array}\right\}.$$

Claim. The pushout $\ell \coprod_k \ell$ in the category of Noetherian rings does not exist.

Proof. If it does, it has natural maps $\iota_i \colon \ell \to \ell \coprod_k \ell$, hence a natural map $\phi \colon \ell \otimes_k \ell \to \ell \coprod_k \ell$. Consider the ideal $I = (x_1-y_1,x_2-y_2,\ldots) \subseteq \ell \otimes_k \ell$. Since $\ell \coprod_k \ell$ is Noetherian, the ideal in $\ell \coprod_k \ell$ generated by $\phi(I)$ is finitely generated. But if $\phi(I) (\ell \otimes_k \ell)$ is generated by $n$ elements, then the same is true for $\psi(I)A$ for any morphism $\psi \colon \ell \otimes_k \ell \to A$ for a Noetherian ring $A$. On the other hand, consider the ring 
\begin{align*}
A &= (\ell \otimes_k \ell)/(x_{n+2}-y_{n+2},x_{n+3}-y_{n+3},\ldots)\\
&\cong k(z_{n+2},\ldots)[x_1,\ldots,x_{n+1},y_1,\ldots,y_{n+1}][T^{-1}]\\
&\cong k(z_{n+2},\ldots)(x_1,\ldots,x_{n+1}) \underset{{k(z_{n+2},\ldots)}}{\otimes} k(z_{n+2},\ldots)(y_1,\ldots,y_{n+1}),
\end{align*}
where
$$T = \left\{fg\ \bigg|\ \begin{array}{ll}f \in k(z_{n+2},\ldots)[x_1,\ldots,x_{n+1}]\setminus\{0\},\\g \in k(z_{n+2},\ldots)[y_1,\ldots,y_{n+1}] \setminus\{0\}\end{array}\right\}.$$
Then $\mathfrak p = \psi(I)A = (x_1-y_1,\ldots,x_{n+1}-y_{n+1})$ is a prime ideal with $\operatorname{ht}(\mathfrak p) \geq n+1$, since we have a chain of prime ideals
$$0 \subsetneq (x_1-y_1) \subsetneq \ldots \subsetneq (x_1-y_1,\ldots,x_{n+1}-y_{n+1}).$$
But now Tag 0BBZ (1) (a form of Krull's Hauptidealsatz) shows that $\mathfrak p$ cannot be generated by $n$ elements. $\square$<|endoftext|>
TITLE: List of recently solved mathematical problems
QUESTION [8 upvotes]: I'm looking for a news site for Mathematics which particularly covers recently solved mathematical problems together with the unsolved ones. Is there a good site MO users can suggest me or is my only bet just to google for them?

REPLY [2 votes]: Here https://theorems.home.blog/theorems-list/ is the website you are asking for. 
It covers all recently solved mathematical problems, which are important (for example, published in a top journal) but at the same time can be understood with not too much background.<|endoftext|>
TITLE: Differential between tangent sheaves
QUESTION [5 upvotes]: Let $M$ be a smooth manifold. Its structure sheaf $\mathcal{O}_M$ is the sheaf of smooth real-valued functions. Together they form a ringed space $(M,\mathcal{O}_M)$. The tangent sheaf $\mathcal{T}_M$ is a sheaf of modules over the structure sheaf. It can be defined as the sheaf of derivations of the structure sheaf.
A smooth map of manifolds $f: M \rightarrow N$ induces a morphism $df: \mathcal{T}_M \rightarrow f^*(\mathcal{T}_N)$ of $\mathcal{O}_M$-modules, where $f^*(\mathcal{T}_N)$ is the inverse image of $\mathcal{T}_N$. It is called the differential or pushforward of the map $f$.
Does anyone have a reference for the definition of the differential? In which kind of textbook would this be explained? It seems to be somewhere between differential geometry and algebraic geometry but I could not find it in any textbook in neither of these areas.
(I am not looking for the differential between tangent bundles which is explained in detail in every basic book on differential geometry.)
Thanks in advance!

REPLY [3 votes]: If you are interested in the real $C^{\infty}$ case there are also: Jet Nestruev, Smooth Manifolds and observables, Springer Graduate Texts as well as Gonzales, Salas, C-differentiable spaces, Lecture notes in Mathematics, Springer.
The algebraic definition of the differential consist in composing a derivation $X:\mathcal{O}_M \to \mathcal{O}_M$ with  $f^*:\mathcal{O}_N \to \mathcal{O}_M$ which gives you a derivation from $\mathcal{O}_N$ to $\mathcal{O}_M$. It remains to show that this sheaf of $\mathcal{O}_M$ modules is isomorphic to $f^*(\mathcal{T}_N)$. But maybe you did not ask for that.<|endoftext|>
TITLE: Undiscovered for a long time before it is realised it is the same concept developed under different names. 
QUESTION [10 upvotes]: Mathematics has been described as the giving of the same name to different things,
but sometimes different names are given to the same thing.
Can you give examples of concepts where researchers in different areas have used the same concept under different names for a long time before it is discovered they are talking about the same thing ?
Wavelets might be an example.
EDIT: In response to Willie Wong's comment. I was thinking of the book "The World According to Wavelets. The Story of a Mathematical Technique in the Making" by Barbara Burke Hubbard. 
Here's some quotes from page 26:
"I have found at least 15 distinct roots of the theory, some going back to the 1930s", Meyer said. "David Marr, who worked on artificial vision and robotics at MIT, had similar ideas. The physics community was intuitively aware of wavelets dating back to a paper on renormalization by Kenneth Wilson, in 1971".
Littlewood and Paley developed wavelet-like techniques.
Alberto Calderon developed a continuous version of wavelets.
Yet other researchers developed wavelets-which they called "self-similar Gabor functions"-to model the visual system.
Jean Morlet developed wavelets as a tool for oil processing.
From page 40:
"Multiresolution approximation and wavelets", Mallat. The paper made it clear that work that existed in many different guises and under many different names -- were at heart all the same.

REPLY [3 votes]: Denis-Charles Cisinski mentioned in an answer that Heller's homotopy theory of homotopy theories is the same theory as Grothendieck's theory of derivators.

REPLY [2 votes]: The concept of an $r$-cover-free family of sets was studied independently in at least three different communities: combinatorics, group testing, and information theory, and of course was called by different names (superimposed codes, $ZFD_r$ codes, etc.).  I myself independently rediscovered the concept and was inclined to call it a "$k$-Sperner family" before I discovered Ruszinkó's paper (J. Combin. Theory Ser. A 66 (1994), 302–310).  Ruszinkó did a fine job of tracking down the literature in all the different fields, and proved one of the basic theorems in the subject.<|endoftext|>
TITLE: Binary Sequences of Length 2n
QUESTION [19 upvotes]: I had posted an urn probability problem that didn't have good motivation.  I'd like to try to explain the motivation here, and reintroduce the problem.
Consider binary sequences of length $2n$.  Let's say we put a marker in such a sequence as soon as we see a total of $n$ 0's or $n$ 1's, reading left to right.  For example, if $n=4$, then the sequence 00101011 would receive a marker thus:  001010|11.  Now write down the bits to the right of the marker.  In the case of our example, this would be 11.  Do this for every binary sequence of length $2n$.  We observe that we have written down 
$2n\binom{2n}{n}$ bits, half 0's and half 1's.  It is possible to prove this observation using binomial coefficient identities, but I wonder whether there is a simple bijective proof.  
The previous urn problem was an equivalent probabilistic formulation of the case $n=5$.

REPLY [4 votes]: Given the symmetry between 0 and 1 past the end of the first "run" of n equal digits, it does not matter whether we count zeros or ones after a "run" of n zeros. That is, if we only count zeros after n zeros and ones after n ones, we should get $n\binom{2n}{n}$. This count is the total number of excess (over n) digits in all the bit patterns of length 2n, and these are split evenly between zeros and ones, denote by $S=\sum_{k=0}^{n}k\binom{2n}{n+k}$ this half-count.
We have $\binom{2n}{n+k}(n+k) = \binom{2n}{n+k-1}(n-k+1)$ -- both are the number of partitions of 2n elements into sets of size n+k-1, n-k and 1. Hence
$$S+\sum_{k=0}^{n} n\binom{2n}{n+k} = \sum_{k=0}^{n} \binom{2n}{n+k}(n+k) = \sum_{k-1=-1}^{n-1} \binom{2n}{n+(k-1)}(n-(k-1)) =$$
$$= \sum_{k-1=-1}^{n} \binom{2n}{n+(k-1)}(n-(k-1)) = \binom{2n}{n-1}(n+1) + \sum_{k=0}^{n} n\binom{2n}{n+k} - S$$
and we have $2S=(n+1)\binom{2n}{n-1} = n\binom{2n}{n}$ as required.
This is essentially counting all zeros in the patterns which have an excess of zeros with weight -1 and all ones in the patterns which have at least as many ones as zeros with weight +1, and noticing that we get $n\binom{2n}{n}$ by straight counting (only the n+n patterns are not pairwise annihilated with their complementary sequence) on one hand and $2S$ on the other hand (we can pair a bit pattern with a selected majority digit with a pattern with a selected minority digit and count each excess majority digit twice in doing so).<|endoftext|>
TITLE:  Weyl Character Formula for Quantum Groups
QUESTION [5 upvotes]: How much is known about the Weyl character formula for quantum groups? More specifically, has the formula been generalized to the general setting of deformed coordinate algebras $\mathbb{C}[G_q]$ of semi-simple Lie groups and their associated flag varieties? I am most interested in the non-root of unity case.

REPLY [5 votes]: The question still feels a bit vague to me, but this started to get too long to be a comment. There are a number of issues: 
There's the question of the definition of $\mathbb C[G_q]$, and say an analogue of the Peter-Weyl theorem, and there is also the issue of doing this at a root of unity case, or better studying things integrally. For this say, a recent paper of Lusztig gives a definition of a quantum coordinate ring for any (finite type) root datum, which specializes to the Kostant-Chevalley form. 
Andersen-Polo-Wen and others have studied studied quantum induction functors which correspond to taking global sections on the classical flag variety, and these might be what you want (Ryom-Hansen also proved a version of Kempf vanishing in this context for example). This was also more recently taken up by Kumar and Littelmann in the context of studying Frobenius splitting. Finally there's the issue of understand quantum flag varieties as noncommutative spaces as in the previous comment, for which along with the Lunts-Rosenberg paper, there is also more recent work of Backelin and Kremnizer.<|endoftext|>
TITLE: Grothendieck's mathematical diagram
QUESTION [19 upvotes]: I was going through this article (Who Is Alexander
Grothendieck?)(Wayback Machine) which appeared in the Notices of the AMS, and in it, there's a picture(page 936) which shows a mathematical diagram drawn by Grothendieck. I would be delighted if anyone could explain what that is.
Thanks in advance.
EDIT: The comments below suggest that the diagram is a dessin d'enfant. And even though I am by no means an expert, I somehow feel that it may not be one. I'd love some clarification on how one may call it so. Questions I want to ask now, are

Is the diagram really a dessin? Or is it something else?

Do the shaded parts of the diagram signify something? Are there any special types of dessins which have a similar a structure?

REPLY [3 votes]: Actually, this diagram appears in Malgoire's presentation Alexander Grothendieck à Montpellier (1973–1991) at 38:54.
It appears to be a diagram illustrating Grothendieck's ideas on "pseudo-droites", or pseudo-lines.  This idea is mentioned in the Esquisse, but I do not understand the mathematics behind it at all and I don't know if there has been anything published on this topic.<|endoftext|>
TITLE: Do semisimple algebraic groups always have faithful irreducible representations?
QUESTION [21 upvotes]: For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.
The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations.  The fundamental theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.

My question is whether such a representation does, in fact, exist.
(Answered: iff the center is cyclic.)

This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}_m^r \to \mathbb{G}_m$, which therefore have nontrivial kernels.  More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.
The exceptional case $r = 1$ does have an example, namely $\operatorname{GL}_n$, whose standard representation is faithful and irreducible and whose center has rank 1.  A more general version of this question might be, then:

Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?
(Answered: when not semisimple, iff the center is connected.)

Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial).  A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:

Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?
(Not yet answered!)

This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."
Finally, suppose the answer is "no".

What is the obstruction to such a representation existing?
(Answered: for $Z$ the center, it is the existence of a generator for $X^*(Z)$.)

REPLY [16 votes]: This is an elaboration of George's answer (and my own deleted semi-answer), along with a comment on the further questions raised.  George has already addressed the basic question about existence of faithful irreducible representations for semisimple groups.   So this is just a supplement, not a separate answer.
For a connected semisimple algebraic group $G$ (over any algebraically closed field), the Chevalley classification determines $G$ up to isomorphism in terms of the way the weight lattice of a maximal torus of $G$ lies between the full weight lattice (of a simply connected group) and the root lattice.   The related classification of irreducible representations of $G$ by highest weights is also given in terms of dominant integral weights lying in this weight lattice.  In particular, if such a weight can be chosen to lie in no smaller intermediate lattice, it will produce a faithful irreducible representation of $G$: otherwise it gives a faithful irreducible representation of a proper quotient group and thus has highest weight in a proper sublattice.   
Such a weight exists unless the "co-fundamental group" (quotient of weight lattice by root lattice) is noncyclic.   For an irreducible root system this happens just for even type D, where you get a Klein 4-group (for Spin groups); in other cases the co-fundamental group (= center of $G$ in characteristic 0) is cyclic.   In general $G$ is an almost-direct product of groups with irreducible root systems; so one has to be careful about
factors of even type D.
ADDED: For any connected semisimple $G$ (and any characteristic), the argument just outlined leads routinely to the conclusion that a faithful irreducible representation exists precisely when no simple factor of $G$ is simply connected of even type D.
That exception was pointed out in earlier comments/answer and uses only Schur's Lemma.   To treat all simple types, the (Killing-Cartan/Chevalley) classification is used to avoid cases when the co-fundamental group is noncyclic.   Then the proof for a general $G$ reduces to this one: $G$ is an almost-direct product of simple groups, so you just have to find suitable highest weights for the individual
factors to get a suitable highest weight for $G$.  (In effect, you find an irreducible representation of the simply connected covering group, a direct product of simple groups, which induces a faithful irreducible representation of $G$ just in case $G$ has no Spin factors.) 
As Ryan points out there is a problem for reductive groups if the center is a torus of dimension $>1$.   Otherwise the situation for general linear groups can probably be imitated: start with a faithful irreducible representation of the derived group (if available), then throw in the scalars (which may of course already overlap the image of the derived group in a finite subgroup).
The further question about "construction" of $G$ (in a suitable representation) from knowledge of its adjoint representation depends on what
"construction" means in this context.   For semisimple groups, there are indirect ways to construct each irreducible representation of a given highest weight (in characteristic 0) by working with tensor products of fundamental representations.   This does not give direct information of the sort contained in Weyl's character and dimension formulas, however.   Anyway, if there exists a suitable highest weight (as discussed above) giving a faithful representation of $G$, this rather abstract procedure does "construct" the corresponding faithful irreducible representation.   But I think you are asking for a more "natural" construction based on the adjoint representation, which I can't visualize.

REPLY [12 votes]: Edit: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^*(T)$ for its group of characters.  Write $R$ for the
subgroup of $X$ generated by the roots of $G$.  Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$. 

Claim: $G$ has a faithful irreducible representation if and only if the character 
  group $X/R$ of $Z$ is cyclic.

Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is  (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$
of its irreducible components. There is an isogeny
$$\pi:\prod_i G_{i,sc} \times T \to G$$
where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group
with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.
The key fact is this: 
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in  X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant.  After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(*) \quad \text{for each $i$, there is $\beta_i \in \Phi_i$ with $\langle \lambda,\beta_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.<|endoftext|>
TITLE: Cohomology Theories on The Stone Space of Complete n-types
QUESTION [10 upvotes]: Just a random thought here: Can cohomology theories (e.g. sheaf cohomology) on the Stone space $S_n(T)$ (the space of complete n-types) of a first-order theory $T$ tell us anything interesting (e.g. the classification of theories)? Is there any result in model theory that is obtained (probably most easily) by this kind of application of cohomology theories? Thanks!

REPLY [4 votes]: I can not really inform you about this since I don't know, but I can point you to some notes of Angus Macintyre, 
http://modular.math.washington.edu/swc/notes/files/03MacintyreNotes.pdf
Here are some excerpts: 
"For me personally, the main surprise arising from the discovery of ACFA
was how much there was to be done in terms of a model-theoretic reaction
to the development of etale cohomology and its relatives."
"Again, in a different direction, one begins to see cohomological ideas coming
up all over applied model theory, for example in o-minimality."
I hope that you find this useful.<|endoftext|>
TITLE: A (known?) hypergeometric identity
QUESTION [5 upvotes]: Incidentally I've obtained a hypergeometric identity that I've not seen before:
$${}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2$$
So, I wonder if it is well-known and possibly represents a particular case of something more general?
P.S. I've tried to simplify() the l.h.s. in Maple but it did not succeed, giving a hope that the identity is not completely trivial. ;)
EDIT: There seems to be a bug in formula rendering, so I'm repeating it below in plain LaTeX:
{}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2

REPLY [14 votes]: Your relation is a particular case of the Karlsson--Minton relations (see Section 1.9 in the $q$-Bible by Gasper and Rahman). It's also a contiguous identity to Pfaff--Saalschütz.
EDIT.
First of all I apologise for giving insufficient comments on the problem.
I learned from Max a very nice graph-theoretical interpretation of the identity
which makes good reasons for not burring it in the list of "ordinary" problems.
The hypergeometric series (function)
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} a_ 0,\ a_ 1,\ \dots,\ a_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};x\biggr)
= \sum_ {k=0}^\infty \frac{(a_ 0)_ k(a_ 1)_ k\dots (a_ p)_ k}{(b_ 1)_ k\dots (b_ p)_ k}\frac{x^k}{k!},
$$
where
$$
(a)_ 0=1 \quad\text{and}\quad
(a)_ k =\frac{\Gamma(a+k)}{\Gamma(a)}= a(a+1)\dots (a+k-1) \quad\text{for } k\in \mathbb Z_ {>0}
$$
(I consider the ones with finite domain of convergence $|z|<1$), have very nice
history and links to practically everything in mathematics. There are many
transformation and summation theorems for them, both classical and contemporary.
There are very efficient algorithms and packages for proving them, like
the algorithm of creative telescoping (due to W. Gosper and D. Zeilberger)
and the package HYP
which allows one to manipulate and identify binomial and hypergeometric series
(due to C. Krattenthaler). An example of classical summation theorem is
the Pfaff--Saalschütz sum
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ 1+a+b-c-m\end{matrix};1\biggr)
=\frac{(c-a)_ m(c-b)_ m}{(c)_ m(c-a-b)_ m}
$$
where $m$ is a negative integer, with a generalisation
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} a,\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=0
\quad\text{if } \operatorname{Re}(-a)>m_ 1+\dots+m_ p
$$
and
$$
{}_ {p+1}F_ p\biggl(\begin{matrix} -(m_ 1+\dots+m_ p),\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr
b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=(-1)^{m_ 1+\dots+m_ p}
\frac{(m_ 1+\dots+m_ p)!}{(b_ 1)_ {m_ 1}\dots (b_ p)_ {m_ p}}
$$
due to B. Minton and Per W. Karlsson (here $m_ 1,\dots,m_ p$ are nonnegative integers).
Max's original identity is not a straightforward particular case but a linear combination
of three contiguous Pfaff--Saalschütz-summable hypergeometric series.
(Two hypergeometric functions are said to be contiguous if they are alike except
for one pair of parameters, and these differ by unity.) Because of having three
hypergeometric functions, I do not see any fun in writing the corresponding details
but indicate a simpler hypergeometric derivation.
Applying Thomae's transformation
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr
c,\ d\end{matrix};1\biggr)
=\frac{(d-b)_ m}{(d)_ m}\cdot{}_ 3F_ 2\biggl(\begin{matrix} -m,\ c-a,\ b \cr
c,\ 1+b-d-m\end{matrix};1\biggr)
$$
the problem reduces to evaluation of the series
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr).
$$
Writing
$$
\frac{(n+1)_ k}{(n)_ k}=\frac{n+k}{n}=1+\frac kn
$$
the latter series becomes
$$
{}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr
1,\ n\end{matrix};1\biggr)
={}_ 2F_ 1\biggl(\begin{matrix} -m,\ m+n \cr 1 \end{matrix};1\biggr)
+\frac{(-m)(m+n)}{n} {}_ 2F_ 1\biggl(\begin{matrix} -m+1,\ m+n+1 \cr 2 \end{matrix};1\biggr)
$$
and the latter two series are summed with the help of the Chu--Vandermonde summation
(a particular case of the Gauss summation theorem).
As for general forms of Max's identity, I can mention that there is no use of the integrality of $n$
in the last paragraph, and I could even expect something a la Minton--Karlsson in general.<|endoftext|>
TITLE: When is the $(F_!,F^*)$ counit a natural isomorphism?
QUESTION [5 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and suppose $F\colon\mathcal{C}\to\mathcal{D}$ is a functor.  It induces two adjoint pairs between $Set~^{\mathcal{C}}$ and $Set~^{\mathcal{D}}$; one is denoted $(F^\star,F_\star)$ and one is denoted $(F_!,F^\star)$.  One proves easily that the counit to $(F^\star,F_\star)$ is a natural isomorphism of functors $\mathcal{C}\to Set$ if and only if $F$ is fully faithful.
I am interested in the counit of the other adjunction $F_!:Set~^{\mathcal{C}}\Longleftrightarrow Set^{\mathcal{D}}:F^*$.  Lets denote it by $$\epsilon_F\colon F_!F^*\to \operatorname {id}_{Set^{\mathcal{D}}}.$$
Question: Under what conditions on $F$ is $\epsilon_F^~$ a natural isomorphism?

REPLY [5 votes]: In general, the counit of an adjunction is an isomorphism if and only if the right-adjoint is fully faithful (dually the unit is an iso iff the left-adjoint is fully-faithful). So, your question is easily seen to be equivalent to asking "When is $F^{*}$ fully-faithful? In topos-theory lingo, when is the induced geometric morphism $\mathbf{F}:Set^{C^{op}} \to Set^{D^{op}}$ satisfies $F^*$ is faithful, then $\mathbf{F}$ is said to be a SURJECTION of topoi. In this setting, this is equivalent to every object in $D$ being a retract of an object of the form $F(C)$.
Ok, so how about asking for $F^*$ to also be full? $F^*$ being faithful AND full means you are looking at what is called a CONNECTED geometric morphism of topoi. What properties $F$ do we need to ensure this? This is in general a hard problem. However, there are at least sufficient conditions. Given $F$, you first construct the category $Ext_{F}$ of "F-extracts"- these are quadruples $(U,V,r,i)$ with $U \in C$, $V \in D$, $r:FU \to B$, and $i:V \to FU$ such that $ri=1$, with the evident morphisms. There is a canonical functor $\tilde F:Ext_{F} \to D$ which sends $(U,V,r,i) \mapsto V$. Denote by $Ext_F(V)$ the fiber over $V$ of this functor. Then if $\tilde F$ is full and each $Ext_F(V)$ is a connected category, then $\mathbf{F}$ is a connected morphism.
This is in "Sketches of an Elephant" C.3.3.<|endoftext|>
TITLE: Homology of lens spaces using Morse theory?
QUESTION [5 upvotes]: To test Morse theory, one day I tried to compute the homology of Lens spaces.  You can build a lens space out of $S^3 = \{ |z|^2 + |w|^2 = 1: z, w \in \mathbb{C} \}$ using the quotient by $z \mapsto e^{2\pi i /p}z, w \mapsto e^{2\pi i q/p}w$.  The quotient by this action L(p,q) is a lens space.
With Morse theory estimates the Betti numbers of a manifold by counting the critical points of a specified index.  For any real-valued function $f: L(p,q) \to \mathbb{R}$, we look for critical points where $\nabla f(p_0) = 0$.  Expanding around p0 $f(p) = f(p_0) + (p - p_0)^T A (p - p_0) + O(|p - p_0|^2)$ for some matrix $A$, the Hessian.  If the eigenvalues of A are all real and nonzero and the critical values of f are all different then $f$ is called "Morse".  The Morse theory says L(p,q) is homotopic to a CW-complex with a cell complex of dimension k for each critical point of index k.
The Morse function I chose is $h = r \cos p\theta$ where $r,\theta$ come from $z = r e^{i \theta}$.  Also let $w = \rho e^{i \phi}$.  This function is well-defined on the lens space L(p,q) and its critical points are (+/-1,0,0,0) and its images under the deck transformation.  To see this, notice the gradient in this coordinate system is $$\nabla =  \left( \frac{\partial }{\partial r}, \frac{1}{r}\frac{\partial }{\partial \theta},, \frac{\partial }{\partial \rho}, \frac{1}{\rho} \frac{\partial }{\partial \phi}\right) $$ and you look for points where the gradient is normal to S3.  However, $H_1[L(p,q)]= \mathbb{Z}/p$ suggesting we should have a saddle point of index 1.
I am told that Morse theory only gives you the singular $\mathbb{R}$ homology.  Did I find all the critical points correctly? Is there a way to get the $\mathbb{Z}$ homology using Morse homology or other piece of differential topology?

REPLY [2 votes]: I tried a computation, I hope it's correct. I considered the map h = Re $z^p$ + Re $w^p$ on the sphere, and I found four sets of critical points: (1) ($\mu_{1}^{+} / \sqrt{2}$, $\mu_{2}^{+} / \sqrt{2}$), (2) ($\mu_{1}^{-}/ \sqrt{2}$, $\mu_{2}^{-} / \sqrt{2}$), (3) ($\mu_{1}^{+}$, 0) and (0, $\mu_{2}^{+}$), (4) ($\mu_{1}^{-}$, 0) and (0, $\mu_{2}^{-}$), where $\mu_{j}^+$ denotes a p-root of 1 and $\mu_{j}^-$ denotes a p-root of -1. 
The first set gives maxima (where h = 2), the second set gives minima (h=-2) and the other two give index 1 and 2 points (I think). The first and second set contribute to p points in the lens space, the other two contribute 2 points each. Perhaps doing Morse homology as Pietro suggested counting connecting flow lines gives you the correct result.... but I have not tried.<|endoftext|>
TITLE: An obstruction theory for promoting homotopy equivalences that are equivariant maps to equivariant homotopy equivalences?
QUESTION [7 upvotes]: Say I have a map of $G$-spaces $f : X \to Y$ and I know it is a homotopy-equivalence in the plain sense that there exists a map (maybe not equivariant) $g : Y \to X$ such that the two composites are homotopic to the identity $f\circ g \simeq Id_Y$, $g\circ f \simeq Id_X$. 
Is there something of an obstruction theory that tells you when you can promote $f$ to a homotopy-equivalence in the category of $G$-spaces? 
Ideally the level of generality I care for is $X$ and $Y$ manifolds and $G$ a compact Lie group.  But anything in that ballpark interests me.

REPLY [3 votes]: A facetious answer would be: sure, but one of the ingredients is going to be the group of homotopy equivalences between $X^G$ and $Y^G$. Since $G$-spaces are diagrams of fixed spaces, the Dwyer-Kan obstruction theory of diagrams (c 1984) probably describes the space of maps, in principle allowing you to answer this question and decide which maps of the given fixed sets are realized. This doesn't sound remotely practical to me.
If you don't understand the fixed points to begin with, Smith theory or bust. The map on fixed sets is going to be an isomorphism on homology with coefficients (possibley $0$) depending on the group and exotic fixed sets satisfying that condition are possible. If you're willing to assume everything is simply connected and the group is the circle, it's probably automatically an equivalence. On the opposite extreme is (binary?) $A_5$ acting on high dimensional manifolds. It has PL actions on the disk with the fixed set arbitrary simplicial complexes. I forget what happens in the smooth category; I think empty fixed set is possible, but I forget how. In between are $p$-groups, where you'll keep control of the mod $p$ homology but not the general homology.

REPLY [2 votes]: The simplest interesting case would be when G is Z/p, p prime. In this case the main issue is that by Smith Theory (applied to the mapping cylinder rel domain, say) you will only know that the induced map of fixed point sets is a Z/pZ equivalence, and definitely not necessarily a homotopy equivalence. If the G-map that is a homotopy equivalence induces a homotopy equivalence of fixed point sets, then I believe one can prove the desired result by bare hands, assuming, say the spaces are G-CW complexes, or smooth G-manifolds.  For more general groups G one would need to assume or arrange that the induced maps on fixed point sets for all subgroups are themselves homotopy equivalences. The idea is the same, but the details seem more daunting.<|endoftext|>
TITLE: Presenting a paper: Do's and Don'ts?
QUESTION [38 upvotes]: I was wondering what would be the best way to present your paper at a conference, if your paper is selected for "short communication", lasting for about 15 minutes?
Should you concentrate on the main results or the proofs?
And what should a first-time presenter be wary of?
Thanks in advance.

REPLY [2 votes]: Some general points I find important:

focus on the main results
don't spend time with proofs (people can read those them self)
don't give any hand-outs (this just distracts people; wastes paper)
put the important points on the slides, talk about what is not on the slides
don't put too much into the slides, make the fonts big and readable
humor is useful to keep people paying attention. this might be more difficult with a math paper than for example with a paper from computer science, but there are still posibilities to use humor.
talk loud and clearly, don't turn around to look at the slides or text.
use the blackboard to answer questions and to give examples (don't put the examples in the presentation)<|endoftext|>
TITLE: Density of smooth functions under "Hölder metric"
QUESTION [19 upvotes]: This question came up when I was doing some reading into convolution squares of singular measures. Recall a function $f$ on the torus $T = [-1/2,1/2]$ is said to be $\alpha$-Hölder (for $0 < \alpha < 1$) if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\alpha(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)|$. This behaves much like a metric, except functions differing by a constant will not differ in $\omega_\alpha$. My primary question is this: 
1) Is it true that the smooth functions are "dense" in the space of continuous $\alpha$-Hölder functions, i.e., for a given continuous $\alpha$-Hölder $f$ and $\varepsilon > 0$, does there exists a smooth function $g$ with $\omega_\alpha(f-g) < \varepsilon$?
To be precise, where this came up was worded somewhat differently. Suppose $K_n$ are positive, smooth functions supported on $[-1/n,1/n]$ with $\int K_n = 1$. 
2) Given a fixed continuous function $f$ which is $\alpha$-Hölder and $\varepsilon > 0$, does there exist $N$ such that $n \geq N$ ensures $\omega_\alpha(f-f*K_n) < \varepsilon$? 
This second formulation is stronger than the first, but is not needed for the final result, I believe.
To generalize, fix $0 < \alpha < 1$ and suppose $\psi$ is a function defined on $[0,1/2]$ that is strictly increasing, $\psi(0) = 0$, and $\psi(t) \geq t^{\alpha}$. Say that a function $f$ is $\psi$-Hölder if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\psi(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)|$. Then we can ask 1) and 2) again with $\alpha$ replaced by $\psi$.
I suppose the motivation would be that the smooth functions are dense in the space of continuous functions under the usual metrics on function spaces, and this "Hölder metric" seems to be a natural way of defining a metric of the equivalence classes of functions (where $f$ and $g$ are equivalent if $f = g+c$ for a constant $c$). Any insight would be appreciated.

REPLY [11 votes]: Smooth functions are not dense in the space of Hölder continuous functions, but it is possible to characterize those functions that can be approximated. This is done below.

Definition. Let $(X,d)$ be a metric space and $0<\alpha\leq 1$. The space $C^{0,\alpha}(X)$ of  Hölder continuous functions is a space
  of bounded real valued functions such that $$
 [f]_{C^{0,\alpha}}:=\sup\left\{\frac{|f(x)-f(y)|}{d(x,y)^\alpha}:\,
 x\neq y\right\}<\infty. $$

$C^{0,\alpha}(X)$ is a Banach space with respect to the norm
$$
\Vert f\Vert_{C^{0,\alpha}}=\Vert f\Vert_\infty+[f]_{C^{0,\alpha}}.
$$

Definition. Let $(X,d)$ be a metric space. For $0<\alpha\leq 1$ we define the space $C^{0,\alpha+}(X)$ to be a subspace of
  $C^{0,\alpha}(X)$ (equipped with the  same norm) that consists of
  functions $f\in C^{0,\alpha}(X)$ such that for every compact set $K$, $$
 \lim_{t\to 0+}\ \sup\left\{\frac{|f(y)-f(x)|}{d(x,y)^\alpha}:\ x,y\in
 K,\ d(x,y)\leq t,\ x\neq y\right\}= 0. $$

In other words $C^{0,\alpha+}(X)$ is a subspace of $C^{0,\alpha}(X)$ consisting of functions that, on every compact set, 
have a slightly better modulus of continuity than 
that one in the definition of the $C^{0,\alpha}$ norm.

Theorem. Let $0<\alpha<1$. Then a function  $f\in C^{0,\alpha}(\mathbb{R}^n)$  can be approximated by a sequence smooth
  functions $f_k\in C^\infty$  so that for every compact set
  $K\subset\mathbb{R}^n$, $\Vert f_k-f\Vert_{C^{0,\alpha}(K)}\to 0$ as
  $k\to\infty$ if and only if $f\in C^{0,\alpha+}(\mathbb{R}^n)$.

Remark. The result is not true for $\alpha=1$. Lipschitz functions that are not $C^1$ cannot be approximated by smooth functions in the Lipschitz norm, because the Lipschitz norm is the same as $C^1$ norm.

Example. $|x|^{1/2}\in C^{0,1/2}(-1,1)\setminus C^{0,1/2+}(-1,1)$ and hence it cannot be approximated by smooth functions in the
  $C^{0,1/2}$ norm (see also the accepted answer).

Proof of the theorem.
Suppose that $f\in C^{0,\alpha}(\mathbb{R}^n)$ can be approximates by smooth functions. We need to show that $f\in C^{0,\alpha+}(\mathbb{R}^n)$.
Let $B_R$ be a ball of any radius. Let $\varepsilon>0$ be given. Then for a sufficiently large $k$,
$$
|(f_k-f)(y)-(f_k-f)(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha
\quad
\text{for all $x,y\in B_R$.}
$$
Let $M=\sup_{B_R}|\nabla f_k|$. Hence the mean value theorem yields
$$
|f(y)-f(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha +|f_k(y)-f_k(x)|\leq
\left(\frac{\varepsilon}{2}+M|x-y|^{1-\alpha}\right)|x-y|^\alpha
$$
so
$$
|f(y)-f(x)|\leq\varepsilon |x-y|^\alpha
\quad
\text{for all $x,y\in B_r$ satisfying $|x-y|<(\varepsilon/2M)^{1/(1-\alpha)}$.}
$$
This proves that $f\in C^{0,\alpha+}$.
Suppose now that
$f\in C^{0,\alpha+}(\mathbb{R}^n)$. We will show that the approximation by mollification $f_t$ has the desired property i.e., 
for every ball $B_R$, $\Vert f_t-f\Vert_{C^{0,\alpha}(B_R)}\to 0$ as $t\to 0$. 
Since $f_t\to f$ uniformly, it remains to estimate the constant in the H\"older estimate of the difference $f_t-f$.
Let $\varepsilon>0$ be given. It follows from the definition of $C^{0,\alpha+}$
that there is $R>\tau>0$ such that if $x,y\in B_{2R}$, $|x-y|<\tau$, then $|f(x)-f(y)|\leq \frac{1}{2}\varepsilon|x-y|^\alpha$. Hence for $0<t<R$, 
$|f_t(x)-f_t(y)|\leq\frac{1}{2}\varepsilon|x-y|^\alpha$ for $x,y\in B_R$ satisfying $|x-y|<\tau$. This easily follows from the definition of $f_t$, because $f_t(x)$ 
is a weighted average of $f$ on the ball $B(x,t)\subset B_{2R}$.
Therefore
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha 
\quad
\text{for all $x,y\in B_R$ satisfying $|x-y|<\tau$.}
$$
Let $0<\delta<R$ be such that 
$\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha/2$ for $0<t<\delta$. 
If $x,y\in B_R$, $|x-y|\geq\tau$, then 
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq 2\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha\leq\varepsilon |x-y|^\alpha.
$$
We proved that if $0<t<\delta$, then
$$
|(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha 
\quad
\text{for all $x,y\in B_R$}
$$
as desired. This completes the proof. 
$\Box$<|endoftext|>
TITLE: Hochschild cohomology and A-infinity deformations
QUESTION [7 upvotes]: When we are dealing with ordinary things or dg things (where thing = algebra or category), I think I understand how HH^2 corresponds to 1st order deformations and HH^3 corresponds to obstructions.
One often hears (or at least I often hear) that HH^* corresponds to A-infinity deformations. I am wondering whether there is any reference which works this out precisely. EDIT: This seems to be incorrect (depending on what we mean by "deformation"). See Damien's answer. And see David Ben-Zvi's comment.

REPLY [11 votes]: Well. Even in the case of a DG (or $A_\infty$) algebra $A$, infinitesimal (i.e. 1st order) deformations are classified by $HH^2(A,A)$. Namely, the structure maps (a-k-a Taylor components) of an $A_\infty$-algebra, viewed as elements of the Hochschild cochain complex, do have total degree $2$. 
I think that one recovers the full Hochschild cohomolgy $HH^*(A,A)$ by considering "derived" infinitesimal deformations (namely, deformations for which the deformation parameter is allowed to have non zero degree). 
In other words, and making use of funny words, $HH^*(A,A)$ is the tangent to the derived stack of associative (better, $A_\infty$) algebras at the point $A$. While $HH^2(A,A)$ can be viewed as the tangent to the coarse moduli space. As an indermediate statement between those two, in his PhD thesis Mathieu Anel computed the tangent complex to the 2-stack of associative algebras (not in the derived context): he found that it is precisely a 2 step complex, obtain as a truncation of the Hochschild complex. See http://arxiv.org/abs/math/0607385 (in french, sorry).<|endoftext|>
TITLE: Spaces with both "simple homology" and "simple homotopy" at the same time
QUESTION [14 upvotes]: Maybe every algebraic topology student, at some moment, will ask himself/herself the question: why are $\pi_*$ so difficult and mysterious, especially when compared with (co)homology? Think about the weird connections between $\pi_n(S^k)$ and number theory... it is insane!
But one day I realize this idea may be wrong and biased.
Homology, is probably not really easier than homotopy.
One tends to think $H_*$ is easy, only because most of us only care about finite-dimensional manifolds in our daily life. And then one have that nice vanishing result for higher-dimensional $H_*$.
Consider the infinite-dimensional Eilenberg−MacLane space $K(\mathbb{Z}, n)$ when $n>2$. Its $\pi_*$ is surely as easy as one could hope, but how about its $H_*$, especially the torsions?
It appears to me, the better (?) statement might be
"Spaces with simple $\pi_*$ tends to have complicated $H_*$.
Spaces with simple $H_*$ tends to have complicated $\pi_*$."
This gives one some strange feeling. It is almost like some kind of "Fourier transformation", some kind of duality.
And this makes some spaces particularly interesting: spaces with both simple $H_*$ and simple $\pi_*$ at the same time.
Let me start with some simple examples:
1) $S^1 \simeq K(\mathbb{Z}, 1)$.
2) $\Sigma^g$, that is, Riemann surfaces.
3) $K(\mathbb{Z}, 2) \simeq CP^\infty$.
4) $K(G, 1)$ when $G$ is a finite group.
And I am looking forward to your examples of such spaces, especially comments from AT experts. 
Thank you very much.

REPLY [5 votes]: For the nicest answers to this kind of question, you should fix a prime $p$ and look for $p$-local spaces $X$ such that 

$X$ is simply connected and admits an $H$-space structure
Each homotopy group $\pi_k(X)$ is free and finitely generated over $\mathbb{Z}_{(p)}$.
Each homology group $H_k(X)$ is also free and finitely generated over $\mathbb{Z}_{(p)}$.

These are called Wilson spaces, because they were first studied in Steve Wilson's paper "The $\Omega$-spectrum for Brown-Peterson homology II".  He showed that

For each $k\geq 2$ there is an indecomposable Wilson space $Y_k$ whose first homotopy group is in dimension $k$.
This $Y_k$ is unique up to (non-canonical) weak equivalence.  It can be constructed as a space in the $\Omega$ spectrum for $BP\langle n\rangle$ for suitably chosen $n$.
Any Wilson space can be written as a product $\prod_iY_{k_i}$, where the indices $k_i$ form a nondecreasing sequence that tends to infinity.<|endoftext|>
TITLE: Which tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields?
QUESTION [6 upvotes]: Consider a connected symplectic manifold $(M, \omega)$ of dimension $m=2n$. A few preliminary reminders (mostly to fix the notation): A vector field $X$ is symplectic if its flow preserves the symplectic form, ie. $L_X \omega = 0$, where $L_X$ denotes Lie derivative with respect to $X$. The Cartan formula shows that this is equivalent to the 1-form $i_X\omega = \omega(X, -)$ being closed. The Hamiltonian vector field associated to a smooth function $f$ is the vector field determined by $\omega(X_f, -) = df$; any symplectic vector field is locally Hamiltonian. The questions I'm interested in are of local nature, so we don't have to worry about the distinction.
Question 1: Which differential forms are invariant under all Hamiltonian flows (meaning $L_{X_f}\alpha = 0$ for all smooth functions $f$)?
Clearly, the symplectic form itself generates a truncated polynomial algebra (isomorphic to $\mathbb{R}[x]/(x^{n+1})$) inside $\Omega^*(M)$ which is invariant under all Hamiltonian flows. But is it possible that there are other than those? I believe I have shown that there are no invariant 1-forms using a horrible calculation in local (Darboux) coordinates, but I'm not sure if this method is suitable for higher degrees. In the even degrees, we know that the answer is not 0, and I can't see how to prove that an invariant $2d$-form is necessarily a constant multiple of $\omega^d$.
Question 2: What can one say about more general tensor fields on $M$? I am especially interested in the sections of the symmetric powers of $TM$ (ie. symmetric multi vector fields).
The proof that no $1$-forms are invariant is easily adapted to proving that no vector fields are invariant, but again, I'm not sure if this generalizes.
Question 3: Suppose we have a subalgebra $A\subset C^\infty(M)$ with the property that for each $p\in M$, $\{df_p \mid f\in A\} = T^*_pM$ (in other words, the Hamiltonian vector fields associated to the functions in $A$ realize every tangent vector on $M$). Do the answers to Questions 1 and 2 change if we only insist that the forms/tensor fields should be invariant under the Hamiltionian vector fields associated to the elements of $A$?
It is certainly important that we still have a whole algebra of functions available; on $\mathbb{R}^{2n}$, any constant coefficient tensor field is invariant under the Hamiltonian vector fields associated to the coordinate functions $x_j, y_j$ (which, up to a sign, are just the corresponding coordinate vector fields $\partial/\partial y_j, \partial/\partial x_j$). The proof that no invariant vector fields exists also requires one to consider the Hamiltonians associated to $x_j^2$ and $y_j^2$.

REPLY [14 votes]: Any symplectic linear transformations in $T_xM$ is locally realizable as a Hamiltonian vector field, thus for questions 1 and 2, one can profitably use representation theory of the symplectic group.
FACT (Lefschetz decomposition) Let $W$ be a $2n$-dimensional symplectic vector space, $\bigwedge^\ast W$ its exterior algebra, and $\omega\in\bigwedge^2 W$ the invariant two-form. Exterior multiplication by $\omega$ and the contraction with $\omega$  define a pair of $Sp(W)$-equivariant graded linear transformations $L, \Lambda$ of $\bigwedge^\ast W$ into itself of degrees $2$ and $-2,$ and let $H=\deg-n$ be the graded degree $0$ map acting on $\bigwedge^k$ as multiplication by $k-n.$ Then $L,H,\Lambda$ form the standard basis of the Lie algebra $\mathfrak{sl_2}$ acting on $\bigwedge^\ast W$ and the actions of $Sp(W)$ and $\mathfrak{sl_2}$ are the commutants of each other.
See, for example, Roger Howe, Remarks on classical invariant theory.
Corollary Every homogeneous $Sp(W)$-invariant element of $\bigwedge^\ast W$ is a multiple of $\omega^k$ for some $0\leq k\leq n.$
Since, conversely, every polynomial in $\omega$ is invariant under the Hamiltonian vector fields, this gives a full description of the invariant differential forms.
For question 2, locally every invariant tensor must reduce to an $Sp(W)$-invariant element of the tensor algebra. For the special case of symmetric tensors, the answer is trivial.
FACT Under the same assumptions, the $k$th symmetric power $S^k W$ is a simple $Sp(W)$-module (non-trivial for $k>0$).
General case can be handled using similar considerations from classical invariant theory. A more involved question of describing the invariant local tensor operations on symplectic manifolds (an analogue of the well-known problem of invariant local operations on smooth manifolds, such as the exterior differential or Schoutens bracket) was considered in an old article by A.A.Kirillov.<|endoftext|>
TITLE: What's the notation for a function restricted to a subset of the codomain?
QUESTION [27 upvotes]: Suppose I have a function f : A → B between two sets A and B. (The same question applies to group homomorphisms, continuous maps between topological spaces, etc. But for simpicity let's restrict ourselves to the case of arbitrary maps between sets.)
If I have subset A' ⊆ A of the domain A, then I can restrict f to A'. There is the function f|A' : A' → B which is given by f|A'(a) = f(a) for all a ∈ A'. 
Suppose now that I have subset B' ⊆ B of the codomain B that contains the image of the map f. Similarly I can restrict f to B', meaning that there is a function g : A → B' which is given by g(a) = f(a) for all a ∈ A. 
In general, it might be useful to consider such functions g and have a name for them, for example if I have a function B' → C that I want to apply afterwards. Unfortunately, I haven't seen a name or symbol for this function g in literature. 

Is there notation for the
  restriction g of f to a subset of the
  codomain similar to the notation
  f|A' for a restriction to a
  subset of the domain?

REPLY [3 votes]: Here's some notation.
Given a morphism $f : A \rightarrow B$ and a subobject $b': B' \rightarrow B$, the astriction of $f$ to $B' \subseteq B$ can be denoted $b'\backslash\, f : A \rightarrow B'$, pronounced "$b'$ under $f$." It satisfies $b'\circ (b'\backslash\, f).$ For example, if we have a function $f : A \rightarrow B$ and subsets $X$ of $A$ and $Y$ of $B$ such that $f(X)\subseteq Y$, then the relevant map $X \rightarrow Y$ can be denoted $Y\,\backslash\, (f \circ X).$
Similar notation can be used in linear algebra to denote solution sets. For example; to denote the set of all vectors $x$ satisfying a linear equation $Ax=b,$ I tend to write $A \backslash b.$ This has an intuitive look about it: $$Ax = b \iff x \in A \backslash b.$$
The connection basically is that in the category of matrices, $Ax=b$ can be seen as a factorization of $b$ through the tail of $A$ (because $A$ is on the left). So, the general principle is that:

In any category, $g \backslash \,f$ is good notation for the set of all factorizations of $f$ through the tail of $g$, or for the unique such factorization just in case a unique factorization exists.

Dually, if we have a function $f : A \rightarrow B$ and a quotient object $a' : A \rightarrow A'$, we can write $f/a'$ for the relevant morphism $A' \rightarrow B$ when it exists, or else use this notation to denote the set of all such morphisms.<|endoftext|>
TITLE: How ugly is the isomorphism R[GxH] = R[G] (X) R[H] for groups G, H?
QUESTION [13 upvotes]: Clearly, when $G$ and $H$ are two finite groups, and $V$ and $W$ are two representations of $G$ and $H$, respectively, then $V\otimes W$ is a representation of the group $G\times H$. It is a well-known fact that over an algebraically closed field of characteristic zero, every representation of $G\times H$ is a direct sum of such $V\otimes W$'s if the groups $G$ and $H$ are finite. Here are some things I am wondering about:
1) How canonical can these $V$'s and $W$'s be chosen?
EDIT: This was discussed at decomposition of representations of a product group as I see. Yet the other questions are new.
2) Do we actually need all the conditions? What if our groups are not finite, or the characteristic of the field is nonzero? The latter may mean different things - we can work in the representation groups, we can work in the Grothendieck groups and we can work in the Grothendieck groups of the projective $k\left[G\right]$-modules (i. e. in the K-theory).
Note that we cannot lift the condition that $k$ be algebraically closed.
3) Does anything improve if $G=S_a$ and $H=S_b$ for integers $a$ and $b$ ? After all, symmetric groups have the nice property that all representations over $\mathbb C$ are defined over $\mathbb Q$, and this gives us hope that the tensorands $V$ and $W$ have some meaning.
Here is why I care:
The famous Hopf algebra $R\left(S\right)=\bigoplus\limits_{n\geq 0}R\left(S_n\right)$ (where $R\left(G\right)$ denotes the (Grothendieck) group of representations of a group $G$) has its product defined by
$U\cdot V = \mathrm{Ind}_{S_a\times S_b}^{S_{a+b}} U\otimes V$
(for $U\in R\left(S_a\right)$ and $V\in R\left(S_b\right)$) and its comultiplication (I hesitate to say coproduct) defined by
$\Delta\left(U\right) = \sum\limits_{k=0}^n \mathrm{Res}_{S_k\times S_{n-k}}^{S_n} U$
(for $U\in R\left(S_n\right)$). Now, $\mathrm{Res}_{S_k\times S_{n-k}}^{S_n} U$ is not an element of $R\left[S_k\right]\otimes R\left[S_{n-k}\right]$ per se, but an element of $R\left[S_k\times S_{n-k}\right]$, and one wishes to have a canonical isomorphism $R\left[S_k\times S_{n-k}\right]\to R\left[S_k\right]\otimes R\left[S_{n-k}\right]$ here. (Of course, it is canonical on the $R\left(S\right)$ level, but it would be great if it would also work out that nicely on the level of representations - after all we're being constructive. Note that multiplication in $R\left(S\right)$ is canonical on the level of representations.)
Summary of the question: Given a Young diagram $\lambda$ with $n$ boxes, is there a "canonical" (as in, explicit formulae or nice deterministic algorithm) way to decompose the restriction of the corresponding Specht module $R_{\lambda}$ to $S_a\times S_b$ (where $a+b=n$) into tensor products of the form $\left(\text{Specht module for }S_a\right)\otimes\left(\text{Specht module for }S_b\right)$ ?
Actually this is part of a bigger question, in case anyone is willing to answer that:
4) Is there a systematic text-book like account of representation theory of $S_n$ which actually uses the modern approaches (Hopf algebras, the Okounkov-Vershik constructive theory avoiding characters, algebraic combinatorics of Young tableaux, Liulevicius' K-theoretical interpretation), doesn't shy away from difficult parts (such as plethysms) and gives modern proofs of classical results rather than just refer to them as well-known (I am not really satisfied with the Schur-Froebnius era proofs, they are rather clumsy and intricate, some of them require working over $\mathbb C$ and character theory and they are just long).
I am aware of Goldschmidt (very nice but too basic), Zelevinsky (seems to become tough reading) and Liulevicius (alas, only journal articles). Is there more?

REPLY [7 votes]: You can somewhat lift the algebraic closedness assumption: You have to allow an auxillary ring (actually, division algebra) to act equivariantly on both representation and tensor over it.
Such a decomposition should hold whenever one of the groups has semi-simple representation category (the division rings in 1 are endomorphisms of simples).  Then, the decomposition can be made canonical precisely up to choosing representative simple objects.  If $V$ is a $G \times H$-rep, and $\rho$ are representative simples for $G$, then the natural map
$$ \bigoplus_{\rho} \rho \otimes_{D_\rho} Hom_G(\rho, V) \to V $$ with $D_\rho = End_G(\rho)$ will be an isomorphism of $G \times H$-modules.
(Conversely, given applying such a decomposition to $k[G]$ viewed as $G \times G$-module one would have to recover a representative set of simples.)
For symmetric groups (in char. $0$), the endomorphism rings of simples are just the field (i.e., the simples remain irreducible over the alg. closure), so in particular you get such decompositions.  Moreover, there are explicit representative simples that one can write down (the Specht modules).  I don't know of the combinatorial theory to say if this gives any sort of satisfactory answer to your question 3.<|endoftext|>
TITLE: Are the banded versions of a positive definite matrix positive definite?
QUESTION [5 upvotes]: Consider $M$,  a positive definite matrix. Let $M^{(1)}$ be the diagonal matrix which agrees with $M$ on the diagonal ($M_{ii}=M^{(1)}_{ii}$). We have that $M^{(1)}$ is positive definite because it is diagonalizable and it has non-negative eigenvalues.
What about general bands? Let  $M^{(b)}$ be the restriction of $M$ on a band: $M_{ij}^{(b)}=M_{ij}$ when  $i$  and $j$ differ by less than $b$ in absolute value and  $M^{(b)}_{ij}=0$ for otherwise. Is $M^{(b)}$ positive definite for all $b$?

REPLY [10 votes]: No.  The matrix
$M = \begin{bmatrix}5 & 4 & 4 \\\\ 4 & 5 & 4 \\\\ 4 & 4 & 5\end{bmatrix} = \begin{bmatrix}2 & 2 & 2\end{bmatrix}\begin{bmatrix}2 \\\\ 2 \\\\ 2\end{bmatrix} + I$
is positive definite, but 
$\begin{bmatrix}1 & -\sqrt{2} & 1\end{bmatrix}M^{(2)}\begin{bmatrix}1 \\\\ -\sqrt{2} \\\\ 1\end{bmatrix} = 20 - 16\sqrt{2}<20 - 22.4 < 0$,
so $M^{(2)}$ is not.

REPLY [6 votes]: Wouldn't that mean that the quadratic form $x^2+y^2+z^2+2xy+2yz$ must be nonnegative definite (as it is a band restriction of the quadratic form $x^2+y^2+z^2+2xy+2yz+2zx$, which is clearly nonnegative definite), which contradicts its value at $x=1$, $y=-1$, $z=1$ ?
(Note that I replaced your "positive definite" by "nonnegative definite" - feel free to add $\epsilon\left(x^2+y^2+z^2\right)$ to the form for some $\epsilon>0$ to keep everything positive.)
EDIT: There's a bit more to this:
Let us denote by $A\ast B$ the Hadamard product of two $n\times n$ matrices $A$ and $B$ (defined by
$A\ast B=\left(a_{i,j}b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$,
where
$A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$
and $B=\left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$).
Let $A$ be a symmetric matrix. Then, (the matrix $A\ast B$ is nonnegative definite for every nonnegative definite matrix $B$) if and only if the matrix $A$ is nonnegative definite. The $\Longrightarrow$ direction is more or less trivial (just take $B$ to be the matrix $\left(1\right)_{1\leq i\leq n,\ 1\leq j\leq n}$) and disproves your conjecture (by taking $A$ to be the matrix whose $\left(i,j\right)$-th entry is $1$ if $\left|i-j\right|\leq d$ and $0$ otherwise). The $\Longleftarrow$ direction is interesting and most easily proven by decomposing the matrix $A$ in the form $u_1u_1^T+u_2u_2^T+...+u_nu_n^T$, where $u_1$, $u_2$, ..., $u_n$ are appropriate vectors. Another proof reduces it to Corollary 2 in my answer to MathOverflow #19100 - do you see how?<|endoftext|>
TITLE: Strange boundary-like map on tensor algebra: what is its kernel?
QUESTION [9 upvotes]: Let $k$ be a commutative ring and $L$ a $k$-module. The tensor algebra $\otimes L$ is $\mathbb{Z}$-graded and $\mathbb{Z}_2$-graded (an element of $L^{\otimes n}$ has degree $n$ and $\mathbb{Z}_2$-degree $n\mod 2$); hence it is a superalgebra over $k$. This canonically induces a supercommutator $\left[\cdot,\cdot \right]_{\mathrm{s}}$ on $\otimes L$, which is simply given by
$$\left[U,V\right]_{\mathrm{s}}=UV-\left(-1\right)^{nm}VU$$
for any $U\in L^{\otimes n}$ and any $V\in L^{\otimes m}$.
Let $T:\otimes L\to \otimes L$ be the $k$-linear map which acts on pure tensors according to the formula
$$T\left(u_1\otimes u_2\otimes \ldots\otimes u_k\right) = \sum\limits_{i=1}^{k} \left(-1\right)^i u_i \otimes u_1 \otimes u_2 \otimes \ldots \otimes u_{i-1} \otimes u_{i+1} \otimes \ldots \otimes u_k$$
(this is clearly well-defined). It is easy to see that $L^{\otimes 0}\subseteq \operatorname{Ker} T$, that $\left[L, L\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$, and that
$$\operatorname{Ker} T\cdot \operatorname{Ker} T\subseteq \operatorname{Ker} T$$
(where multiplication is the multiplication in the tensor algebra $\otimes L$), so that $\operatorname{Ker} T$ is a subalgebra of $\otimes L$. (Thus, in particular, $\left[\operatorname{Ker} T,\operatorname{Ker} T\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$.) Also, $\left[L, \operatorname{Ker} T\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$.
Consequently, by induction, any nontrivial tree of supercommutator brackets decorated by elements of $L$ must evaluate to an element of $\operatorname{Ker} T$, and so must any tensor product of such trees (including empty products). I am wondering: do these generate (over $k$) all of $\operatorname{Ker} T$ or is there more? I am mostly interested in the case when the characteristic of $K$ is zero, but experimentation with Sage (see comments) suggests that the answer is characteristic-independent as long as $K$ does not have characteristic $2$. In characteristic $2$, one has to additionally consider the elements $x \otimes x \in \operatorname{Ker} T$ for all $x \in L$ (in all other characteristics, this follows from $\left[L, L\right]_{\mathrm{s}}\subseteq \operatorname{Ker} T$).

Here is a bit of motivation (I said this is a curiousity question, but in fact the original curiousity question was about bilinear forms):
Let $f:L\times L\to k$ be a bilinear form. We define a bilinear map $U:L\times\left( \otimes L\right)\to \otimes L$ by $U\left(u,v_1\otimes v_2\otimes ...\otimes v_k\right)=\sum\limits_{i=1}^k\left(-1\right)^{i-1}f\left(u,v_i\right)\cdot v_1\otimes v_2\otimes ...\otimes \hat{v_i}\otimes \ldots\otimes v_k$,
where $\hat{v_i}$ means that the factor $v_i$ is omitted from the product. (Of course, we have just defined $U\left(u,V\right)$ for pure tensors $V$ only, but the rest is clear by linearity.)
This bilinear map $U$ is rather natural; people use to denote the similarly defined map $L\times\left( \wedge L\right)\to \wedge L$ (where all $\otimes$ signs have been replaced by $\wedge$ signs) as the "interior product" (the "exterior product" is just the wedge product) - best known in the context of differential forms.
Now I'm wondering what tensors $V\in \otimes L$ satisfy ($U\left(u,V\right)=0$ for every $u\in L$ and every bilinear form $f$). This is equivalent to $V\in\operatorname{Ker} T$.

EDIT: This seems closely related to Sections V and VI in Wilhelm Specht, Gesetze in Ringen I, Mathematische Zeitschrift (1950), Volume: 52, page 557-589. Essentially, Specht proves my conjecture for multilinear tensors (= tensors spanned by pure tensors whose tensorands are a permutation of the basis vectors). Apparently, Specht's motivation was understanding PI-algebras.

REPLY [3 votes]: My conjecture was correct. This, and more, is now proven in The signed random-to-top operator on tensor space (draft) (aka arXiv preprint arXiv:1505.01201).
(Some questions do remain, such as those in §9.)<|endoftext|>
TITLE: $3^n - 2^m = \pm 41$ is not possible. How to prove it?
QUESTION [10 upvotes]: $3^n - 2^m = \pm 41$ is not possible for integers $n$ and $m$. How to prove it?

REPLY [23 votes]: The congruence $3^n - 2^m \equiv 41\pmod{60}$ has no solutions.
The congruence $3^n - 2^m \equiv -41\pmod{72}$ has no solutions.<|endoftext|>
TITLE: Schemes as a model category
QUESTION [9 upvotes]: I'm just learning some basics of model categories, so please forgive me if my question turns out to be trivial. I hope it does at least make sense.
A natural temptation is to relate this machinery to birational geometry; in particular one would like to find a model category structure having the birational morphisms as weak equivalences. More precisely it would be nice to have such a model structure on the category $Sch_k$ of schemes of finite type over a field $k$.
A natural problem arises: a model category is required by definition to have all small limits and colimits, and $Sch_k$ does not satisfy this. For limits the situation is not that bad. I believe the original work of Quillen required only the existence of finite limits and colimits. Since $Sch_k$ has finite products and fiber products, it has all finite limits.
On the other hand finite colimits need not exist. A simple way to see this is to realize that categorical quotients by equivalence relations do not always exist in $Sch_k$, and these are just some coequalizers. So my questions are:

Is there a canonical way to enlarge a category to add finite limits?
If this is the case, what do we obtain when applying this to $Sch_k$? The resulting category would have to contain algebraic spaces, as these arise as quotients of schemes by étale equivalence relations. How much bigger would it be?
Assuming one has a decent notion of birational morphism for these objects: is there a model structure on the enlarged category such that birational morphisms are the weak equivalences?

REPLY [5 votes]: A canonical way would be to embed $Sch_k$ into the category of (pre)sheaves.<|endoftext|>
TITLE: Forcing as a tool to prove theorems
QUESTION [20 upvotes]: It is often mentioned the main use of forcing is to prove independence facts, but it also seems a way to prove theorems. For instance how would one try to prove Erdös-Rado, $\beth_n^{+} \to (\aleph_1)_{\aleph_0}^{n+1}$ (or in particular that $(2^{\aleph_0})^+ \to (\aleph_1)_{\aleph_0}^2$) by using forcing? Is it simpler than the combinatorial proof? Which partial order would one use? 
Or can one try to show that forcing the negation of the Erdös-Rado is impossible or inconsistent?

REPLY [5 votes]: One use of forcing as a tool to prove theorems that has not been mentioned in the answers is the method of generic ultrapowers, where we take an ideal $I$ on an uncountable regular cardinal $\kappa$ (in the sense of $M$), and consider the poset $P,\leq$ of those subsets of $\kappa$ that has positive measure (the ordering is by subset). A generic filter $F$ of that poset will end up being an $M$-ultrafilter in $M[G]$, which extends the filter dual to $I$. And if $I$ is {$\kappa$-complete, normal} in $M$, $G$ will be {$\kappa$-complete, normal}.
If $I$ indeed has the requisite properties, then in $M[G]$ we can form the ultrapower of $M$ by $G$, and an elementary embedding $j:V\to Ult(M,G)$. This way, we allow ourselves the liberty of "pretending" to have an elementary embedding $j:V\to M$ without being necessarily committed to the existence of large cardinals.
The generic ultrapower is not necessarily well-founded (an ideal is precipitous (which, incidentally, means příkrý in Czech) iff the generic ultrapower obtained is well-founded). But having such a construction at hand can be useful. Example 22.15 in Jech illustrates one such use.
Theorem. Let $\kappa$ be a singular cardinal of uncountable cofinality, and assume GCH holds below $\kappa$, then $2^\kappa=\kappa^+$.
The proof is to use the nonstationary ideal on $(cf(\kappa))^M$, then in the extension we obtain a normal $M$-measure on $(cf(\kappa))^M$, which we use to construct the ultrapower. One key observation is that the poset of positive measure subsets will have size $2^{cf(\kappa)}<\kappa$, so its cc property implies that cardinals above $\kappa$ are preserved. By studying the combinatorial properties of objects in $Ult$, we may infer that $(|\mathcal{P}^M(\kappa)|)^{M[G]}\leq(\kappa^+)^{M[G]}$, which can be transferred back to $M$ because of the observation.<|endoftext|>
TITLE: Applications of topological and diferentiable stacks
QUESTION [22 upvotes]: What are some examples of theorems about topology or differential geometry that have been proven using topological/differentiable stacks, or, some examples of proofs made easier by them? I'm well aware of several statements made more beautiful in the language of stacks, but, I'm looking for a concrete application.

REPLY [4 votes]: I should update with a mention of some of my own results in http://arxiv.org/abs/1504.02394: 

There is a proof of Segal's theorem that the classifying space $B\Gamma^q$ of Haefliger's foliation groupoid is homotopy equivalent to classifying space of the discrete monoid of embeddings of $\mathbb{R}^n$ into itself $B\mathbf{Emb}\left(\mathbb{R}^n\right)$ using differentiable stacks (and higher topos theory). (This is theorem 3.7)
You can also use the same machinery to prove the following theorem (Theorem 4.1):

Let $G$ be a Lie group acting almost freely on a manifold $M$. Then the homotopy type of the Borel construction $M\times_G EG$ is the same as the the classifying space of a certain discrete category, whose objects are smooth tranversals, i.e. maps $f:\mathbb{R}^n \to M,$ with $n=\dim M - \dim G$ which are transverse to the $G$-orbits.<|endoftext|>
TITLE: What is the shortest program for which halting is unknown?
QUESTION [38 upvotes]: In short, my question is:

What is the shortest computer program for which it is not known whether or not the program halts?

Of course, this depends on the description language; I also have the following vague question:

To what extent does this depend on the description language?

Here's my motivation, which I am sure is known but I think is a particularly striking possibility for an application to mathematics:
Let $P(n)$ be a statement about the natural numbers such that there exists a Turing machine $T$ which can decide whether $P(n)$ is true or false.  (That is, this Turing machine halts on every natural number $n$, printing "True" if $P(n)$ is true and "False" otherwise.)  Then the smallest $n$ such that $P(n)$ is false has low Kolmogorov complexity, as it will be printed by a program that tests $P(1)$, then $P(2)$, and so on until it reaches $n$ with $P(n)$ false, and prints this $n$.  Thus the Kolmogorov complexity of the smallest counterexample to $P$ is bounded above by $|T|+c$ for some (effective) constant $c$.  
Let $L$ be the length of the shortest computer program for which the halting problem is not known.  Then if $|T|+c < L$, we may prove the statement $\forall n, P(n)$ simply by executing all halting programs of length less than or equal to $|T|+c$, and running $T$ on their output.  If $T$ outputs "True" for these finitely many numbers, then $P$ is true.
Of course, the Halting problem places limits on the power of this method.
Essentially, this question boils down to:  What is the most succinctly stateable open conjecture?
EDIT:  By the way, an amazing implication of the argument I give is that to prove any theorem about the natural numbers, it suffices to prove it for finitely many values (those with low Kolmogorov complexity).  However, because of the Halting problem it is impossible to know which values!  If anyone knows a reference for this sort of thing I would also appreciate that.

REPLY [5 votes]: May I suggest the use of Binary Lambda Calculus [1] for writing programs and measuring their size in bits?
There are many BLC programs of only a few dozen bits, comparable in complexity to 5 state TMs that need nearly 50 bits to describe, whose halting behavior is unknown.
Beyond that, there are programs like this 215 bit one for computing Laver tables [2], whose halting behavior is related to existence of large cardinals. A counterexample to Goldbach's conjecture can be found with a 267 bit program.
I decided to pose a question on mathoverflow [3] addressing the specific form of the question.
[1] https://tromp.github.io/cl/Binary_lambda_calculus.html
[2] https://codegolf.stackexchange.com/questions/79620/laver-table-computations-and-an-algorithm-that-is-not-known-to-terminate-in-zfc
[3] What's the smallest lambda calculus term not known to have a normal form?<|endoftext|>
TITLE: What is the etymology of the term "perverse sheaf"?
QUESTION [46 upvotes]: Grothendieck famously objected to the term "perverse sheaf" in Récoltes et Semailles, writing "What an idea to give such a name to a mathematical thing!  Or to any other thing or living being, except in sternness towards a person—for it is evident that of all the ‘things’ in the universe, we humans are the only ones to whom this term could ever apply.”  (Link here, in an excellent article by Allyn Jackon.)  But a google search for '"perverse sheaf" etymology' gives only nine hits, none of which seem informative.  

What is the etymology of the term "perverse sheaf"?

REPLY [10 votes]: When I was a grad student, a classmate commented about the odd term "perverse" in IH but I defended it. I thought I had figured out the rationale, which I denied had anything to do with perversity in the ordinary sense of that word. My theory was that the "verse" was from "transverse," which is a generic kind of intersection of objects of complementary dimension. By replacing "trans" with "per" we get a more general notion of intersection. The "per" here meant excess or abundance, following the way chemists use that prefix in words like peroxide or permanganate. A perversity in IH in fact specifies how far in excess of the generic expectation the dimension of an intersection with a stratum is allowed to be. My classmate was skeptical, but this theory seemed so plausible to me that I never really doubted it. I am genuinely surprised to be finally refuted.<|endoftext|>
TITLE: Generalized smooth spaces and infinite dimensional manifolds
QUESTION [11 upvotes]: There is a theorem due to Losik which shows that the category of Frechet manifolds embeds fully-faithfully into diffeological spaces. (Diffeological spaces are concrete sheaves on the site of (Euclidean) manifolds http://ncatlab.org/nlab/show/diffeological+space). Diffeological spaces are a complete and cocomplete quasitopos, so, in particular are Cartesian-closed. Froelicher spaces are also complete, cocomplete and Cartesian-closed: http://ncatlab.org/nlab/show/Froelicher+space#hausdorff. Do Frechet manifolds also embedd fully-faithfully into Froelicher spaces? If so, if we "cut out a submanifold" of a Frechet space, does it correspond to the sub-Froelicher space when embedded? How about for diffeological spaces?

REPLY [6 votes]: In Jordan Watt's thesis it is shown that Frölicher spaces form a full subcategory of diffeological spaces, so if the functor from Fréchet manifolds to diffeological spaces is fully faithful (as seems to be the case by the result of Losik), and if this functor factors through Frölicher spaces (as shown in Kriegl-Michor), then Fréchet manifolds embed fully faithfully in Frölicher spaces.<|endoftext|>
TITLE: Generalizing square wheels rolling on inverted catenaries
QUESTION [12 upvotes]: It is not uncommon to see in a science museum a bicycle with
square wheels that rides smoothly over a washboard-like
surface made from inverted catenary curves (e.g., at the Münich museum).
The square wheel may be generalized to any regular polygon (except the triangle),
which rolls on a similar curve without slippage.
Here, for example, is a nice Mathematica demo.
My question is: For which wheel shapes does there exist
a matching road shape that permits the wheel to
roll over it without slippage so that:
(a) the wheel center remains
horizontal throughout its motion,
(b) the wheel can turn at constant angular velocity,
and (c) if possible, the wheel center also moves at
constant horizontal velocity?
The square satisfies (a) and (b), but only regular hexagons
and beyond satisfy (c).  If you've experienced a
square-wheel bicycle ride, you can feel it jerk because (c)
fails to hold.
It would be interesting to know the class of closed wheel curves
that satisfy (a) and (b), and also those that in addition
satisfy (c).  For example, must all (a,b) curves be star-shaped from the wheel center $x$?
(star-shaped: every point of the curve is visible from $x$).
This is probably all known, so an appropriate reference
may suffice.
Addendum1 (1July10).
The delightful Hall-Wagon paper that user abel found (below) answers many of my questions, and may be
the last word (or the most recent work) on the topic.  However, it does not seem to address the broader question I posed: For which class of wheel shape curves is a such a wheel-road construction
possible?  I'll update further if anything comes to light.
Addendum2 (8June11).
A paper just appeared in the Amer. Math. Monthly (Vol.118, No.6, 2011),
"Roads and Wheels, Roulettes and Pedals," by Fred Kuczmarski, which seems to
establish that a wheel-road construction is possible for every

continuously differentiable plane curve
such that the angle of rotation of its tangent lines, as measured relative to some initial
position, is a strictly monotonic function of arc length.
We call such curves rollable.
The monotonic condition implies that rollable curves have no inflection points, while
the strictness of the monotonicity precludes rollable curves from containing line segments.

Certainly this is not the full class (as he mentions), but he has a nice theorem that
constructs a road for any rollable-curve wheel.
Added from mathcurve.com, as
cited by @J.M.isntamathematician: Ellipses on sinusoid.
Animation by Alain Esculier.

REPLY [4 votes]: The link between Wheel and Ground is general : to any curve (W) in polar (rho, theta) is associated a ground (G) in cartesian orthonormal frame (x, y) and conversely. Gregory's transformation direct and inverse give parametric equations with one integration. 
James Gregory in "Geometriae pars universalis 1668" invented a direct transformation  GT equivalent :

for a given wheel in polar coordinates (rho, theta)  it gives the ground (x,y) in orthonormal coordinates y=rho  and dx=Integral rho.d theta  Inverse transformation GT-1 defines, 
for a given ground (x,y), the associated wheel : rho=y and theta=Integral dx/y if y<>0. 
GT gives the ground if we know the wheel (rho, theta) and 
GT-1 gives the wheel if we know the ground (y, x). 

In each case there is only one integration. Cesaro in NAM 1886 has given many examples and properties of these associated curves which have same arc length. The theory is linked with integration and area. The area of the wheel is half of the one under the ground. When the polar curve rolls on the ground (with initial conditions) the pole O runs along the x-axis (called base-line). When you fix the Wheel then the base line pass through the fixed pole if the ground rolls on the Wheel. The problem was much studied about 1845-1920 In NAM, Mathesis,JMPA, etc.
There is identity of arc length between the polar curve (rho, theta)and (x,y). 
A theorem of Steiner-Habich is important in the theory (pp 3-4 of the paper I Gregory's transformation). Apply the theory to special family of curves as wheels "sinusoidal spirals" for which pedals are in the same family gives examples: line-Catenary , Circle-double circle, parabola-parabola, Cardioid-Cycloid, Tractrix spiral-Tractrix, etc. 
You can view examples here http://christophe.masurel.free.fr/#s9 All papers are open-access. 
There are also many informations in "Nouvelles annales de mathematiques" (1842-1927) 
-but in french language- http://www.numdam.org/numdam-bin/feuilleter?j=nam 
or on Gallica.fr and also in Mathesis (Google books on line).
C. Masurel<|endoftext|>
TITLE: Fourier transforms of compactly supported functions
QUESTION [12 upvotes]: One manifestation of the uncertainty principle is the fact that a compactly supported function $f$ cannot have a Fourier transform which vanishes on an open set.  As stated, this phenomenon applies when $f$ lives on the integers, or on Euclidean space, but it is false for, say, the p-adic rationals, on which the characteristic function of the p-adic integers provides a counterexample.
One normally proves this phenomenon on Euclidean space through complex analysis, although it does follow from the real-variable argument on this post of Tao on Hardy's Uncertainty Principle.
For $f$ on the integers, it's even easier because only the zero polynomial has infinitely many roots (and this appears to me to be more or less the kind of input which goes into the other proofs as well).
What I want to know is whether there is a proof of this phenomenon -- namely, the refusal of a compactly supported functions F.T. to vanish on an open set --  which directly relates to the connectedness of the frequency space, if that is what is behind it.
Here is a failed effort in this direction...
You know that $f = f \chi$ for any function $\chi$ which is $1$ on the support of $f$.  Then $\hat{f}$ should remain unchanged when convolved with the Fourier transform of $\chi$, but since $\hat{f}$ lives on the reals you would like to think that this convolution would partially fill up some open set connected to the boundary of the original support of $\hat{f}$ and thus enlarge the support of $\hat{f}$.  This argument goes through as long as you don't get an absurd cancellation; but while $\hat{f}$ may be assumed positive, one has less freedom to renormalize $\chi$.
Does anyone know if there is a version of this argument which actually goes through?  As stated it's no different than saying "$\hat{f} \ast \hat{\chi} = 0$ whenever $\chi$ lives away from the support of $f$; how weird..."  It's possible my intuition here is all wrong and it's really the regularity of $\hat{f}$ which is completely behind the phenomenon, but I am at a loss for other examples of connected, locally compact abelian groups (are there any?) and I don't know how connectedness behaves with respect to Pontrjagin duality.

REPLY [7 votes]: For $\mathbb{R}$. Suppose f is our compactly supported function and g(x) is its Fourier transform. Since f is compactly supported, $\hat{f} = g$ is the restriction to $\mathbb{R}$ of an entire function g(z) by the Paley-Wiener theorems. Since g is entire and vanishes on an open set, $g \equiv 0$. The proof of this last fact (weakening the assumption to vanishing on a set with an accumulation point) uses that $\mathbb{C}$ is connected which is of course directly related to $\mathbb{R}$ being connected.
I expect that you knew this proof, but maybe you accidentally overlooked where connectedness was used. Or more likely, this proof didn't explain what you had in mind and you want a more general proof for $\mathbb{R}^n$. I can't currently do that. Instead, I have another idea which focuses on a different aspect than connectedness, but seems to be related.
In connection with the analogous statement for polynomials. A polynomial can only have finitely many zeroes over a field is proved via a complexity argument using that infinity > finite. Analytic functions, i.e. the completion of polynomials over $\mathbb{C}$ can have infinitely many zeroes, but uncountably many zeroes implies the analytic function is identically 0. So it seems that a set that has a limit point is more complex (in terms of complexity) than a countable set. I'm thinking the complexity argument should be interpreted in terms of density in topology - no finite subset of a $\mathbb{N}$ is dense in the discrete topology or any open subset of the co-finite topology on $\mathbb{N}$. Similarly for $\mathbb{R}$ and $\mathbb{C}$.
I hope this is helpful. This is an interesting question and I'll think more about it.<|endoftext|>
TITLE: Rank of a module
QUESTION [28 upvotes]: What's wrong with defining the rank of a finitely generated module over any (commutative) ring to be just the smallest number of generators? All books I know define rank only locally this way. But why not define it globally?

REPLY [42 votes]: Since your profile says you are interested in Algebraic Geometry, here are geometric considerations that might appeal to you.
Consider a projective module $P$ of finite type over a commutative ring $A$. It corresponds to a locally free sheaf $\mathcal F $ over $X=Spec(A)$. The rank of $\mathcal F $ at the prime ideal $\mathfrak p$ is that of the free $A_{\mathfrak p}$-module $\mathcal F_{\mathfrak p}$.
The rank is then a locally constant function on $X$ and if $X$ is connected (this means that the only idempotents in $A$ are $0$ and $1$) it may be seen as an integer. 
If  $A$ is a domain, then $X$ is certainly connected and has a generic point $\eta$ whose local ring is the field of fractions $\mathcal O_\eta=K=Frac(A)$. The rank of $\mathcal F $  or of $P$ is then simply the dimension of the $K$ vector space $P\otimes_A K$. 
Actually, if $A$ is a domain, this formula can be used to define the rank of any $A$-module $M$ (projective or not, finitely generated or not) : $rank(M)=dim_K ( M\otimes_A K) $ .
This is the definition given by Matsumura in his book Commutative Rings, page 84.
 It corresponds to the maximum number of elements of $M$ which are linearly independent over $A$. 
The minimum number of generators of $M$  (which started this discussion) is quite a different, but interesting invariant, which has been studied by Forster, Swan, Eisenbud, Evans,...
Geometrically it corresponds to the minimum numbers of global sections of $\tilde{M}$ which generate this sheaf at each point of $Spec(A)$.
Elementary example:   Every non-zero ideal of a Dedekind domain is of rank one, can be generated by at most two elements and can be generated by one element iff it is principal.
If the Dedekind domain is not a PID there always exist  non free ideals which thus cannot be generated by less than two elements.
Bibliography
 Ischebeck and Rao have published a monograph Ideals and reality: projective modules and number of generators of ideals on exactly this theme<|endoftext|>
TITLE: Do you understand SYZ conjecture
QUESTION [17 upvotes]: The aim of this question is to understand SYZ conjecture ("Mirror symmetry is T-Duality").
I don't expect a full and quick answer but to find a better picture from answers and comments.
The whole idea is to construct the Mirror C.Y $Y$ from $X$ intrinsically as follows.
One considers the moduli of special Lagrangian tori with a flat $U(1)$ bundle on it in $X$.
Then we put a metric on this moduli (plus corrections coming from J-holomorphic disks) and expect that this moduli and the metric given on it is the mirror C.Y we were looking for.
Here are the things I can not understand:
---What is the metric given in the paper "Mirror symmetry is T-Duality"?
Where does it come from? (I can not understand the formulation of metric there).
and more importantly
--How do we deform the metric using J-holomorphic disks(instantons)?

REPLY [18 votes]: Hi-
Just saw this thread.  Maybe I should comment.  The conjecture
can be viewed from the perspective of various categories:
geometric, symplectic, topological.  Since the argument is
physical, it was written in the most structured (geometric)
context -- but it has realizations in the other categories
too.
Geometric:  this is the most difficult and vague, mathematically,
since the geometric counterpart of even a conformal field theory
is approximate in nature.  For example, a SUSY sigma model with
target a compact complex manifold X is believed to lie in the
universality class of a conformal field theory when X is CY,
but the CY metric does not give a conformal field theory on
the nose -- only to one loop.  Likewise, the arguments about
creating a boundary conformal field theory using minimal (CFT) +
Lagrangian (SUSY) are only valid to one loop, as well.
To understand how the corrections are organized, we should
compare to (closed) GW theory, where "corrections" to the classical
cohomology ring come from worldsheet instantons -- holomorphic
maps contributing to the computation by a weighting equal
to the exponentiated action (symplectic area).  The "count"
of such maps is equivalent by supersymmetry to an algebraic
problem.  No known quantity (either spacetime metric or
Kahler potential or aspect of the complex structure) is
so protected in the open case, with boundary.  That's why
the precise form of the instanton corrections is unknown,
and why traction in the geometric lines has been made
in cases "without corrections" (see the work of Leung, e.g.).
Nevertheless, the corrections should take the form of 
some instanton sum, with known weights.  The sums seem
to correspond to flow trees of Kontsevich-Soibelman/
Moore-Nietzke-Gaiotto/Gross-Siebert, but I'm already running
out of time.
Topological:  Mark Gross has proven that the dual torus
fibration compactifies to produce the mirror manifold.
Symplectic:  Wei Dong Ruan has several preprints which
address dual Lagrangian torus fibrations, which come
to the same conclusion as Gross (above).  I don't know
much more than that.
Also-
Auroux's treatment discusses the dual Lagrangian
torus fibration (even dual slag, properly understood)
for toric Fano manifolds, and produces the mirror
Landau-Ginzburg theory (with superpotential) from this.
With Fang-Liu-Treumann, we have used T-dual fibrations
for the same fibration to map holomorphic sheaves
to Lagrangian submanifolds, proving an equivariant version of
homological mirror symmetry for toric varieties.
(There are many other papers with similar results
by Seidel, Abouzaid, Ueda, Yamazaki, Bondal, Auroux,
Katzarkov, Orlov -- sorry for the biased view!)
Reversing the roles of A- and B-models, Chan-Leung
relate quantum cohomology of a toric Fano to the
Jacobian ring of the mirror superpotential via T-duality.
Help or hindrance?<|endoftext|>
TITLE: Galois Group as a Sheaf
QUESTION [25 upvotes]: I've noticed that the Galois groups associated to Galois field extensions $L$ of a given field $K$ seem remarkably like a sheaf, with the field extensions taking the place of open set, and the Galois group $\mathrm{Gal}(L/K)$ of a field extension $L/K$ is the sections over $L/K$. This makes sense because we have a natural restriction map $\mathrm{Gal}(M/K) \to \mathrm{Gal}(L/K)$ when $M/L/K$. Furthermore, we can define an open covering of an extension $L/K$ as a collection of subextensions whose compositum is $L/K$, and then a collection of elements of the Galois group in each subextension which match up under restriction gives an element of $\mathrm{Gal}(L/K)$. This is a key ingredient in proving that the Galois group of an infinite extension is an inverse limit of finite Galois groups. Is there a way to formalize this?
It also has interesting ramifications for understanding the topology of the Galois groups. Two elements of an infinite Galois group of $L/K$ are close if they agree on larger and larger subextensions of $L/K$, i.e. if they agree on larger "open sets" in this topology. So there's some kind of interesting inverse relation between the topology on the infinite Galois group and the "topology" of the subextensions.

REPLY [14 votes]: You might introduce a Grothendieck topology on the Galois subextensions of $L/K$; then the Galois group is indeed a sheaf.
I just want to remark the following: There is a natural homeomorphism between $Gal(L/K)$ and $Spec(L \otimes_K \overline{K})$. Namely, if $\sigma \in Gal(L/K)$, then the kernel of $L \otimes_K \overline{K} \to L \otimes_K \overline{K} \subseteq \overline{K} \otimes_K \overline{K} \to \overline{K}$ is a prime ideal of $L \otimes_K \overline{K}$. This is easily checked to be a homeomorphism if $L/K$ is finite, and then also in general. In particular, you can endow the profinite space $Gal(L/K)$ with a sheaf so that we get an affine scheme. If $L'/K$ is a Galois subextension of $L/K$, then the restriction $Gal(L/K) \to Gal(L'/K)$ comes from the inclusion $L \otimes_K \overline{K} \subseteq L' \otimes_K \overline{K}$. This provides another reason why $Gal(-/K)$ is a "sheaf", namely because $Spec$ commutes with filtered direct limits.<|endoftext|>
TITLE: Proving non-existence of solutions to $3^n-2^m=t$ without using congruences
QUESTION [23 upvotes]: I made a passing comment under Max Alekseyev's cute answer to this question and Pete Clark suggested I raise it explicitly as a different question. I cannot give any motivation for it however---it was just a passing thought. My only motivation is that it looks like fairly elementary number theory but I don't know the answer.
OK so one problem raised in the question linked to above was "prove there are no solutions to $3^n-2^m=41$ in non-negative integers" and Aleksevev's answer was "go mod 60". It was remarked afterwards that going mod 601 or 6553 would also nail it. For example, modulo 6553 (which is prime), 3 has order 39, 2 has order 117, but none of the 39 values of $3^n-41$ modulo 6553 are powers of 2 modulo 6553.
My question (really just a passing remark) is:
Is there an integer $t$ such that the equation $3^n-2^m=t$ has no solutions in non-negative integers $m$, $n$, but for which there are solutions modulo $N$ for all $N\geq1$? (By which of course I mean that for each $N\geq1$ the equation is satisfied mod $N$ for some integers $m,n\geq0$ depending on $N$; I am not suggesting that $m$ and $n$ be taken modulo $N$ or are independent of $N$).
This for me looks like a "Hasse principle" sort of thing---in general checking congruences doesn't give enough information about solvability of the polynomial in integers and there are many examples of such phenomena in mathematics. As exponential Diophantine equations are harder than normal ones I would similarly expect the Hasse Principle to fail here, but others seemed to be more optimistic.

REPLY [6 votes]: Just in case people were not aware, in the case of the single exponential problem $a^{n}=t$, if there is a solution modulo all prime powers, then there is an integer solution.  [A nice proof is given in Cojocaru and Murty's book "An introduciton to sieve methods and their applications".<|endoftext|>
TITLE: Some questions about scalar curvature
QUESTION [33 upvotes]: Recall that the scalar curvature of a Riemannian manifold is given by the trace of the Ricci curvature tensor.  I will now summarize everything that I know about scalar curvature in three sentences:

The scalar curvature at a point relates the volume of an infinitesimal ball centered at that point to the volume of the ball with the same radius in Euclidean space.
There are no topological obstructions to negative scalar curvature.
On a compact spin manifold of positive scalar curvature, the index of the Dirac operator vanishes (equivalently, the $\hat{A}$ genus vanishes).

The third item is of course part of a larger story - one can use higher index theory to produce more subtle positive scalar curvature obstructions (e.g. on non-compact manifolds) - but all of these variations on the compact case.
I am also aware that the scalar curvature is an important invariant in general relativity, but that is not what I want to ask about here.  This is what I would like to know:

Are there any interesting theorems about metrics with constant scalar curvature?  For example, are there topological obstructions to the existence of constant scalar curvature metrics, or are there interesting geometric consequences of constant scalar curvature?
Can anything be said about manifolds with scalar curvature bounds (other than the result I quoted above about spin manifolds with positive scalar curvature), analogous to the plentiful theorems about manifolds with sectional curvature bounds?  (Thus additional hypotheses like simple connectedness are allowed)
Is anything particular known about positive scalar curvature for non-spin manifolds?

REPLY [6 votes]: There is an important relation between variation of scalar curvature Kahler metric and Lichnerowicz operator. Lets explain it here
Let $(X,J,\omega)$ be Fano Kahler manifold with $\omega_0\in [\omega_0]=\kappa$, then by $\partial\bar\partial$-lemma any other Kahler 2-form can be written as $\omega_0+\sqrt{-1}\partial\bar\partial \varphi$. Now the set of Kahler metrics in $[\omega_0]$ can be identified with $$\mathcal H=\{\varphi \in C^\infty(X,\mathbb R)|\omega_0+\sqrt{-1}\partial\bar\partial \varphi>0\}/\mathbb R$$. Note that $T_\varphi\mathcal H=C^\infty(X,\mathbb R)/\mathbb R$
The Scalar curvature can de define as 
$$Sc:\mathcal H\to C^{\infty}(X,\mathbb R)$$
$$Sc(\varphi)=\Lambda_\varphi\sqrt{-1}\partial\bar\partial\log \omega^n$$
and for $\psi\in T_{\varphi}\mathcal H$ we have 
$$d \text{Sc}(\varphi).\psi=\Delta_{\omega_\varphi}^2\psi-Sc(\varphi)\Delta_{\omega_{\varphi}}\psi$$ and moreover by using following known relation of Lichnerowicz  
$$\mathcal D_\varphi^*\mathcal D_\varphi \psi=\Delta_{\omega_\varphi}^2\psi-Sc(\varphi)\Delta_{\omega_\varphi}\psi+\frac{1}{2}g(dSc(\varphi),d\psi)$$ so we can rewrite the derivative of scalar curvature metric as follows
$$d \text{Sc}(\varphi).\psi=\mathcal D_\varphi^*\mathcal D_\varphi \psi-\frac{1}{2}g(dSc(\varphi),d\psi)$$
See also Oliver Biquard expository paper<|endoftext|>
TITLE: Jordan decomposition in a classical group
QUESTION [6 upvotes]: Let $\mathfrak{g} \subset \mathfrak{gl}_n$ be one of the classical real or complex semisimple Lie algebras.  If $g \in \mathfrak{g}$, then $g$ has a Jordan decomposition $g = g_s + g_n$ with $g_s$ semisimple and $g_n$ nilpotent, and $[g_s,g_n]=0$.  
The elements $g_s,g_n$, which a priori are just in  $\mathfrak{gl}_n$, are both in $\mathfrak{g}$ again.  There are various middle-brow general ways to see this (for one, use that $\mathfrak{g}$ is algebraic), but for concrete choices of $\mathfrak{g}$ it's basically elementary, as follows.  One knows from the construction of the Jordan decomposition that $g_s,g_n$ are both polynomials in $g$ (different polynomials for different $g$, of course), and (EDIT) you can rig the construction so that these polynomials are odd.  The Lie algebra $\mathfrak{g}$ is the subspace of $\mathfrak{gl}_n$ cut out by conditions like $\mathrm{trace}(g)=0$,  or $Jg = -g^{t} J$ for some matrix $J$, and so forth.  The condition $\mathrm{trace}(g)=0$ is always true for $g_n$, so it's true for $g_s$ if true for $g$.  The condition $Jg=-g^t J$ is visibly true for odd $p(g)$ if true for $g$, so if true for $g$ then it's true for both $g_s$ and $g_n$.  Thus $g_s$ and $g_n$ visibly satisfy whatever conditions $g$ is required to satisfy, and so are contained in $\mathfrak{g}$.
(This might seem lowbrow but in fact I think this is basically the idea of the proof that Fulton-Harris give for general semisimple Lie algebras.)
Now suppose instead that $G$ is a real or complex linear Lie group with Lie algebra $\mathfrak{g}$.  This time the Jordan decomposition is $g = g_s g_u$ with $g_u$ unipotent, and indeed $g_s$ and $g_u$ are still in $G$.  But if you try to make the same lowbrow argument as in the Lie algebra case, it appears to die horribly (a condition like $g^t = g^{-1}$ certainly need not be preserved by taking a polynomial in $g$).  My question is, is there an elementary way to rescue it?  (In particular, something other than just the general argument for algebraic groups.) Obviously you're fine for elements $g$ in the image of the exponential map, so the issue is passing to the whole group.  A caveat is that I do $\textit{not}$ want to assume that $G$ is connected.

REPLY [2 votes]: The proof of the Jordan decomposition for algebraic groups over perfect fields has two parts:
(a) Linear algebra: An automorphism of a vector space has a unique multiplicative Jordan decomposition, which is compatible with maps and tensor products...
(b) Some baby Tannakian stuff.
Most proofs in the literature mix the two parts, making the proof seem more difficult than it is. If you accept the baby Tannakian stuff, which everyone should know anyway, one is left with some easy linear algebra (see, for example, I Section 9 of my online notes).<|endoftext|>
TITLE: Existence/Uniqueness of Nonnegative Solutions of Linear Systems of Equations
QUESTION [5 upvotes]: Suppose we have an $m$x$n$ matrix $A$, with $m\lt n$, and an $m$x$1$ vector $b$. Are there existence and uniqueness conditions characterizing nonnegative solutions of the system of linear equations $Ax=b$? i.e. When is there an $x\geq 0$ such that $Ax$=$b$?
I'm sure it is a very well-known result I'm after here but I can't seem to find the answer easily. Any references would be helpful. This (http://www.jstor.org/pss/1968384) looks related but it is not free.

REPLY [5 votes]: The following work seems to be exactly what you want.
M. Wang, and A. Tang. 
Conditions for a Unique Non-negative Solution to an Underdetermined System.
in Proc. of Allerton Conference , Monticello, Illinois, Sep. 2009.  
Here is the link:
http://networks.ece.cornell.edu/meng/pub/Allerton09.pdf<|endoftext|>
TITLE: profinite spaces coming from profinite groups
QUESTION [6 upvotes]: This is probably well-known:
Does every nonempty profinite space occur as the underlying space of a profinite group? If not, which conditions have to be imposed?
- Is every profinite group isomorphic to the Galois group of some Galois extension? (yes, see the comments)

REPLY [4 votes]: (Note: This was intended to be a comment to unknown (google)'s answer - but as I'm new here I can't post comments.)
As Pete L. Clark points out, unknown (google)'s answer is false as stated.
However, this is only because of the omission of the word "infinite".
A correct statement is:
An infinite profinite group $G$ is homeomorphic to $\{0,1\}^{w(G)}$,
where $\{0,1\}$ is the $2$-point discrete space,
and $w(G)$ is the weight of $G$.
This is Theorem 9.15 (pages 95-98) of "Abstract Harmonic Analysis I" by Edwin Hewitt and Kenneth A. Ross. (Hewitt and Ross actually state the result using the minimum cardinality of a local base at $1_G$, rather than the weight of $G$, but these are equal for infinite profinite groups.)
Notice that the case of countable weight is an immediate consequence of the usual characterisation of the Cantor set.<|endoftext|>
TITLE: Cardinality of maximal linearly independent subset
QUESTION [6 upvotes]: M a finitely generated module over a commutative ring A. I can't think of an example of two maximal linearly independent subsets of M having different cardinality. I know that they all have the same cardinality if A is integral domain. Any suggestions are welcome!

REPLY [9 votes]: I found an old paper by Lazarus (Les familles libres maximales d'un module ont-elles le meme cardinal?, Pub. Sem. Math. Rennes 4 (1973), 1-12) which contains the the following result: Let A be a commutative ring with unit and M an A-module. In the following situations, maximal linearly independent subsets of M have the same cardinality:

If M is a free A-module of infinite rank.
If A is reduced and has only finitely many minimal primes (e.g. integral domain, reduced Noetherian ring)
If A is Noetherian and M is a free A-module.
If A is Noetherian and M is a submodule of a free A-module of finite rank.
If A is Noetherian and M has an infinite linearly independent subset.
If A is Noetherian and M is a submodule of a flat A-module.
If A is Artin local and the zero ideal $(0)\subset A$ is irreducible.

And the examples given in the paper of modules not satisfying this same cardinality property are highly nontrivial.<|endoftext|>
TITLE: Efficient approximation of a matrix and its inverse
QUESTION [6 upvotes]: Assume that $ A $ is a real $ n\times n $ matrix whose rows constitute an orthonormal basis of $ \mathbb R^n $. 
Informal statement of question: Assume we want to approximate $ A $ by a rational matrix, such that each entry can be written efficiently (that is, has a small binary encoding), but we require also the inverse of the approximate matrix to have small representation. Is this possible? 
Formal statement of question: Let $ p(n) $ be some polynomial in $ n $. For a real number $ r $, we say that $ a/b $ is a polynomial approximation of $ r $, if $ a/b$ is a rational number (that is, $ a,b $ are integers) and both $ a $ and $ b $ are of size at most  $p(n) $ (e.g., their binary representation is of logarithmic size in $ n $), such that $ |r-a/b|\le 1/p(n) $. 
Question: Does there exist a rational matrix $ B$, such that $ B $ polynomially approximates $ A $ (that is, the entry $ B_{ij} $ in $ B $, is a polynomial approximation of the entry $ A_{ij} $ in $ A $, for all $ 1\le i,j\le n $), and such that $ B^{-1} $ is a rational matrix whose entries are all polynomially-bounded (that is, for any $ 1\le i,j\le n $, $ B^{-1}_{ij}=a/b$, where $ a,b $ are integers of size at most $ p(n) $) ?

REPLY [4 votes]: In $\mathbb{R}^3$, Milenkovic and Milenkovic give an alogrithm for efficiently approximating an orthogonal matrix by a rational orthogonal matrix. As lhf points out, the inverse of an orthogonal matrix is its transpose, so the inverse will also have short entries in this setting. 
Regarding $n>3$, here is a tentative thought, and a reference. I haven't put much effort into either :).
Let $v=(v_1, v_2, \ldots, v_n)$ be a nonzero vector. Define a linear operator 
$$s_v(u) := u - 2 \frac{\langle v,u \rangle}{\langle v,v \rangle} v.$$
This is the orthogonal reflection that negates $v$. Note that, if $v \in \mathbb{Q}^n$, then the entries of the matrix $s_v$ are rational. This is true even if $v$ does not have norm $1$.
Now, any rotation matrix can be written as a product of $\leq n$ reflections: $R=\prod_{i=1}^h s_{v_i}$ for some sequence of vectors $v_i$ in $\mathbb{R}^n$. A potential algorithm, then, is to find such a factorization and then approximate each $v_i$ by a rational vector $w_i$ which is roughly parallel to it. (There are plenty of standard algorithms for rational approximation of a vector.) Then take $\prod s_{w_i}$ as the approximation to $R$.
I got this strategy from a paper of Eric Schmutz. Schmutz follows this strategy, but he forces his approximating vectors $w_i$ to lie on the unit sphere. As far as I can see, this is a waste of effort, since $s_v$ is orthogonal with rational entries even if $v$ is not on the unit sphere. However, Schmutz has exact bounds, which you may find useful.<|endoftext|>
TITLE: roots of sum of two polynomials
QUESTION [10 upvotes]: I believe that there is no common theory for finding roots of polynomial sum. In my case I have 
$$P_{n}(x)+AQ_{n}(x)$$.
I am wondering how roots of this sum depend on $A$?

REPLY [9 votes]: If they are complex polynomials or can be treated as such, then you could apply Rouche's theorem, where the location of the zeros is determined by the dominant polynomial within the sum. ("Walk the dog on the leash")
Possibly related: you could use the Wronskian to determine the values of A that make $P_n(x)$ and $Q_n(x)$ linearly independent.
Your question is related to Mason's theorem.  There are a few papers which explore this specifically

MR1923392 (2003j:30012)  Kim, Seon-Hong . Factorization of sums of
polynomials.  Acta Appl. Math.  73 
(2002),  no. 3, 275--284.
MR2103113 (2005h:30011)  Kim,
Seon-Hong . On zeros of certain sums
of polynomials.  Bull. Korean Math.
Soc.  41  (2004),  no. 4, 641--646
MR1911767 (2003d:11036)  Pintér, Á. 
Zeros of the sum of polynomials.  J.
Math. Anal. Appl.  270  (2002),  no.
1, 303--305.<|endoftext|>
TITLE: Set theoretical realizations of the hidden variables program in quantum mechanics
QUESTION [12 upvotes]: The hidden variables program in quantum mechanics has been largely discredited by two powerful theorems, namely those of Bell and Kochen/Specker. Nonetheless, this program retains a certain philosophical appeal ("God does not play dice" and all that jazz) so I won't bother to motivate my interest in the topic.
More specifically, I am investigating recent efforts to construct alternative frameworks for QM in which hidden variables are possible, thereby making the theory deterministic/realistic, etc. In particular, there are two very intriguing papers, one by William Boos and the other by Robert Van Wesep, which make use of set theoretical tools to create (plausible?) hidden variable theories:

William Boos (1996) claims that random ultrafilters can provide a realization of the hidden variable program:
http://www.springerlink.com/content/n3gr194551472536/
Robert Van Wesep (2006) argues that the hidden variable program is entirely characterized by generic filters and uses forcing techniques on the algebra of quantum propositions:
http://arxiv.org/abs/quant-ph/0506040

Interestingly, both authors use related techniques (ultrafilters & forcing) which perhaps indicates that they are on to something... However, the papers are very technical and I do not fully understand their results; sadly, I could not locate any reviews of either paper online (which is surprising to me, considering how intriguing these papers are).
So my question is the following: has anyone read these papers, and if so, could you please comment on them?
(Although the question ultimately relates to physics, I feel that the highly mathematical nature of the methods used in these papers (and their beauty!) should appeal to the audience of Math Overflow, and indeed, I hope that someone here has already perused them...)
Thank you!
PS: Boos' paper is not freely available from the publisher, but a copy exists online:
http://uploading.com/files/37m88a11/boos%2Bultrafilters.pdf/

REPLY [9 votes]: There are also papers from the 80's by Itamar Pitowsky on hidden variables models using nonmeasurable sets.
http://edelstein.huji.ac.il/staff/pitowsky/             (papers #1 and #4 for example)
For the Kochen-Specker theorems there is a long line of later results but see especially the Conway-Kochen "Free Will Theorem", available on arxiv (quant-ph).
I don't think physicists take seriously the idea that set theory is (or even might be) relevant to quantum mechanics.   On the mathematical side the set-theoretic models can be seen as demonstrations that certain types of hidden variable models are logically consistent, and so cannot be ruled out without additional assumptions, such as Lebesgue measurability, that are implicit in ordinary physics reasoning.<|endoftext|>
TITLE: Origin of symbol  *l* for a prime different from a fixed prime?
QUESTION [18 upvotes]: I've never seen an authoritative explanation for the choice of the lower case letter $\ell$ or $l$ to denote an arbitrary prime different from a given prime $p$.  This now has its own LaTeX command \ell, but has been in use at least since the old work of Taniyama and Weil involving L functions.  That use of the upper case letter might have suggested the lower case here, I guess(?)  The letter q would seem more natural in elementary number theory.   The write-up of Serre's 1967 McGill lectures was published in 1968 by W.A. Benjamin under the title Abelian l-adic representations and elliptic curves.  There his convention is to denote prime numbers by $\ell, \ell', p, \dots$, stating: "we mostly use the letter $\ell$ for $\ell$-adic representations and the letter $p$ for the residue characteristic of some valuation".  
I've heard this question raised but not answered quite a few times.  For instance, after a colloquium talk in Hamburg given by Bhama Srinivasan on Deligne-Lusztig characters, the elderly Ernst Witt asked the non-technical question I've just raised.   (He had done impressive work in his youth but became a convert to the Nazi cause without apparently committing any war crimes.   Possibly he was the young man reported to have shown up once at Emmy Noether's seminar wearing a pro-Nazi uniform.  In old age he had retained some mental acuity but developed phobias about for example the flooring material in the math tower, which required talks like the ones Bhama and I gave to move to a remote building.)
[ADDED] Both Franz and quim point in the direction of how the symbol $l$ became common for prime numbers in Hilbert's development of Kummer's work.   There he considers an $l$th root of unity ($l$ an odd prime) instead of $\lambda$ used earlier by Kummer.  Later on I guess it became a default option for many people to use $l$ for a prime different from a given prime $p$, especially when $q$ became used commonly for a power of $p$.

REPLY [10 votes]: This elaborates quim's answer. Kummer did indeed use $\lambda$ for denoting primes (in connection with cyclotomic fields); he borrowed the notation from Jacobi's articles on cyclotomy as well as from his notes of the number theory lectures in 1836/37. When Hilbert rewrote Kummer's contributions in his Zahlbericht, he started the chapter on cyclotomic fields with "Let $l$ denote an odd prime number". The reason for switching from the Greek to the Latin alphabet was Hilbert's custom to use Latin letters for rational numbers. Hilbert also used ${\mathfrak l}$ for prime ideals above $2$.
 Edit.  For what it's worth: Euler used primes $\lambda n + 1$ in art. 92 of his article E449.<|endoftext|>
TITLE: Composite pairs of the form n!-1 and n!+1
QUESTION [25 upvotes]: It's well known that the numbers of the form $n!\pm1$ are not always prime. Indeed, Wilson's Theorem guarantees that $(p-2)!-1$ and $(p-1)!+1$ are composite for every prime number $p > 5$. 

Is there a proof, preferably an elementary proof, that there are infinitely many composite pairs of the form $n!\pm1$?

The motivation for this question comes from my answer to this recent question. There, I show that every nonstandard model of Peano Arithmetic has a $\mathbb{Z}$-chain consisting entirely of composite numbers. The example I gave is that of a $\mathbb{Z}$-chain contained in the infinite interval $[N!+2,N!+N]$, where $N$ is any nonstandard natural number. I wonder if I could have picked some $\mathbb{Z}$-chain centered at $N!$ instead. A positive answer to the above question would mean that this is indeed possible. Note that it is important in this context that the proof is elementary, but I will also accept beautiful analytic arguments.
Andrey Rekalo pointed out that $(N!)^3 \pm 1$ are both composite. This means that, if $N$ is a nonstandard integer, then the $\mathbb{Z}$-chain centered at $(N!)^3$ has only composite numbers all but two have standard factors. I don't know if it's possible to find a $\mathbb{Z}$-chain all of whose elements have a standard factor.

REPLY [3 votes]: Explicit constructions of infinitely many examples seem to be difficult. Looking at a table of factorizations of $N! \pm 1$ I noticed the following pattern (and now I see that this is essentially what Dror suggested in his comment):
Assume that $q \equiv 3 \bmod 4$ and $p = \frac{q+3}2$ are prime numbers. Then for $n = p-2$, we have $p \mid n!-1$ and $q \mid n!+1$ if $h(-q) \equiv 1 \bmod 4$, where $h(m)$ denotes the class number of ${\mathbb Q}(\sqrt{m})$. Probabilistically, the class number of $h(-p)$ should
be $\equiv 1 \bmod 4$ in half the cases.<|endoftext|>
TITLE: (Elementary?) combinatorial identity expressing binomial coefficients as an alternating sum over permutations.
QUESTION [9 upvotes]: Background
I came up with this while trying to find a sort of high-level exposition of the exterior algebra of a vector space.  Let $V$ be a vector space of dimension $n$ over $\mathbb{C}$, and let $k \in \mathbb{N}$.  One picture of $\Lambda^k(V)$, the $k^{th}$ exterior power of $V$, is as the space of totally antisymmetric tensors in $V^{\otimes k}$.
This can be constructed as follows.  Let
$$ \rho : S_k \to \mathrm{End}(V^{\otimes k})$$
be the representation given by
$$ \rho_\pi (v_1 \otimes \dots \otimes v_k) = v_{\pi(1)} \otimes \dots \otimes v_{\pi(k)},$$
and then let $\sigma$ be the alternating form of this representation, i.e. $\sigma_\pi = sgn(\pi) \rho_\pi$.  The total antisymmetrizer is the map
$$A_k = \frac{1}{k!} \sum_{\pi \in S_k} \sigma_\pi.$$
This is the projection onto the space of totally antisymmetric tensors, and so we can calculate the dimension of $\Lambda^k(V)$ simply by taking the trace of the map $A_k$.  It turns out that 
$$\mathrm{tr}(\rho_\pi) = n^{cyc(\pi)},$$
where by $cyc(\pi)$ I mean the number of cycles in the factorization of $\pi$ into disjoint cycles (including cycles of length 1).  This can be shown as follows.  Take a basis $\{ e_1, \dots, e_n \}$ of $V$ and then form the basis for $V^{\otimes k}$ consisting of all vectors $ e_{i_1} \otimes \dots \otimes e_{i_k}$ such that $1 \le i_1, \dots, i_k \le n $.  Then 
$$ \rho_\pi(e_{i_1} \otimes \dots \otimes e_{i_k}) = e_{i_{\pi(1)}} \otimes \dots \otimes e_{i_{\pi(k)}},$$
so this basis vector contributes 1 to the trace of $\rho_\pi$ if and only if $i_j = i_{\pi(j)}$ for all $1 \le j \le k$, i.e. if and only if all labels are constant over cycles of $\pi$.  Since there are $n$ choices for each label, this gives
$$\mathrm{tr}(\rho_\pi) = n^{cyc (\pi)},$$
and thus
$$ \mathrm{tr}(A_k) = \frac{1}{k!} \sum_{\pi \in S_k} sgn(\pi) n^{cyc (\pi)}.$$


Question: does anybody know a simple combinatorial proof that 
  $$ \frac{1}{k!} \sum_{\pi \in S_k} sgn(\pi) n^{cyc (\pi)} = \binom{n}{k},$$
  where (in case you didn't read the long-winded background that I wrote), $cyc(\pi)$ is the number of cycles in the disjoint cycle factorization of $\pi$.

REPLY [16 votes]: $n(n-1)...(n-(k-1))$ is the number of injective functions from a set of size $k$ to a set of size $n$.  We can count these using inclusion-exclusion: first include all such functions, of which there are $n^k$.  Then, for each transposition $(ij)$ in $S_k$, exclude all the functions such that $f(i) = f(j)$, of which there are $n^{k-1}$.  And so forth.  This alternating sum cancels out all functions which are invariant under a permutation of the domain, so the only ones left are the injective ones.
There's a much easier way to prove an equivalent identity, which is
$$\frac{1}{k!} \sum_{\pi \in S_n} n^{\text{cyc}(\pi)} = {n+k-1 \choose k}.$$
This identity is equivalent because the sign of a permutation is determined by the parity of its number of cycles, and it corresponds to replacing "antisymmetric" by "symmetric" everywhere in your question.  But this identity has an obvious proof by Burnside's lemma: the LHS and RHS both count the number of orbits of functions $[k] \to [n]$ under permutation of the domain.  (This is a special case of a result I call the baby Polya theorem.)
Both identities are in turn a special case of the exponential formula, which one can state as a generating function identity for the cycle index polynomials of the symmetric groups.  I explain some of this here.  The relevant specializations are
$$\frac{1}{(1 - t)^n} = \exp \left( nt + \frac{nt^2}{2} + ... \right)$$
and
$$(1 + t)^n = \exp \left( nt - \frac{nt^2}{2} \pm ... \right).$$<|endoftext|>
TITLE: Geometric model for classifying spaces of alternating groups
QUESTION [19 upvotes]: The classifying space of the nth symmetric group $S_n$ is well-known to be modeled by the space of subsets of $R^\infty$ of cardinality $n$.  Various subgroups of $S_n$ have related models.  For example, $B(S_i \times S_j)$ is modeled by subsets of $R^\infty$ of cardinality $i + j$ with $i$ points colored red and $j$ points colored blue.  More fun: the wreath product $S_i \int S_j \subset S_{ij}$ has classifying space modeled by $ij$ points partitioned into $i$ sets of cardinality $j$ (but these sets are not "colored").
My question: is there a geometric model, preferably related to these, for classifying spaces of alternating groups?  [Note: since any finite group is a subgroup of a symmetric group one wouldn't expect to find geometric models of arbitrary subgroups, but alternating groups seem special enough...]

REPLY [22 votes]: $n$ linearly independent points in $R^\infty$ together with an orientation of the $n$-plane which they span.

REPLY [12 votes]: Probably the right thing to do is to express the classifying space of $A_n$ as the non-trivial double cover of the classifying space of $S_n$.  A point in the classifying space is then a set of $n$ points in $\mathbb{R}^\infty$ with a "sign ordering".  A sign ordering is an equivalence class of orderings of the points, i.e., ways to number them from 1 to $n$, up to even permutations.  I coined the term "sign ordering" by analogy with a cyclic ordering. But that name aside, the idea comes up all the time in various guises.  For instance an orientation of a simplex is by definition a sign ordering of its vertices.
This is in the same vein as your other examples and you can of course do something similar with any subgroup $G \subseteq S_n$.  You can always choose an ordering of the points up to relabeling by an element of $G$.

A bit more whimsically, you could call the configuration space of $n$ sign-ordered points in a manifold "the configuration space of $n$ fermions".  Although a stricter model of the $n$ fermions is the local system or flat line bundle on $n$ unordered points, in which the holonomy negates the fiber when it induces an odd permutation of the points.  This local system is similar to the sign-ordered space in the sense that the sign-ordered space is the associated principal bundle with structure group $C_2$.<|endoftext|>
TITLE: Methods of showing an element is / is not in a field
QUESTION [5 upvotes]: Let $K$ be a field, $\alpha\in\bar{K}$, and $L/K$ a finite extension. How can we determine whether $\alpha\in L$, preferably in as much generality as possible?
Of course, there may be special cases where this is easy, e.g. $K\subset\mathbb{R}$ and $\alpha\in\mathbb{C}\setminus\mathbb{R}$. Another trick, using the field trace, is the described in exercise 16 of Chapter 2 of Marcus's Number Fields, though as far as I can tell this only works for the special case of radicals (since their traces are always 0).
There's only two general approaches I can think of at the moment: checking whether $\deg(\alpha)\mid [L:K]$, though this may not suffice; or somehow finding the minimal polynomial of $\alpha$ over $L$ (including proving that it is irreducible), which will be of degree 1 iff $\alpha\in L$ (not entirely sure how one would do this).
I'm wondering because I recently had the following messy situation: $K$ is the splitting field of a cubic $g\in\mathbb{Q}(t)[x]$, having $[K:\mathbb{Q}(t)]=3$, and $f_1$, $f_2\in K[x]$ are two cubics with splitting fields, $L_1$ and $L_2$ respectively, having $[L_1:K]=[L_2:K]=3$, and I wanted to know whether $L_1=L_2$. It would suffice to show $f_1$ has a root in $L_2$ or vice versa (since $L_1=K($any root of $f_1)$ and $L_2=K$(any root of $f_2)$, which led me to my question. Ultimately (and I still want to double-check my answer), I found that $L_1=L_2$, but I depended heavily on the specific properties of my $f_1$, $f_2$, and $g$.

REPLY [5 votes]: See section 5.4 "The Subfield Problem and Applications" in Henri Cohen's book "A Course in Computational Algebraic Number Theory". This is an excellent reference for computational algebraic number theory and the first place one should look when investigating such topics.<|endoftext|>
TITLE: How many model categories have the same weak equivalences?
QUESTION [19 upvotes]: There are many situations which arise where one might consider different Model categories with the same underlying category. For example in (left) Bousfield localization you start with a model category M and construct a new model category structure on M with the same cofibrations, but with more weak equivalences and fewer fibrations. In these settings the identity adjoint equivalence often serves as a Quillen pair relating the Model categories. 
One classic example is when the underlying category is a the category of functors from a (small) category C into a (nice) model category M. In that case, we have two naturally occuring model structures: the projective and the injective model structures. In some cases, when the source category C is a Reedy category, there is a third model structure on D=Fun(C,M) known as the Reedy model structure. All three of these model category structures are Quillen equivalent. In fact the weak equivalences are exactly the same. They are the level-wise weak equivalences. 
In some sense which I'm trying to make precise, the injective and projective model structures on on opposite sides of the spectrum. The Reedy model structure is something like a mix of these two. In fact the model structures with a fixed set of weak equivalences should form a poset. For example we can just look at the set of cofibrations. Afterall the cofibrations and the weak equivalences determine the model structure if it exists. Then the model category structures with a fixed set of weak equivalences are ordered by inclusion (of sets of cofibrations). In the above example we have the ordering: $$Proj \subseteq Reedy \subseteq Inj$$ 
So this raises the question, what is known about different model structures on a category with a fixed set of weak equivalences? Is there always a maximal/minimal model structure? If not, are there some conditions which ensure its existence? Are there properties which characterize these model structures? Since the weak equivalences are always the same, then the identity functor should induce an isomorphism on homotopy categories. Thus if two such model structure are comparable (so that the identity adjoint equivalence is a Quillen pair) then they are Quillen equivalent. So if there is a minimal or maximal model structure, they all these model structures are Quillen equivalent via a zig-zag of Quillen equivalences. Is this the case?

REPLY [6 votes]: Tibor Beke has some comments on this too.
http://faculty.uml.edu/tbeke/cofib.pdf<|endoftext|>
TITLE: Examples where physical heuristics led to incorrect answers?
QUESTION [86 upvotes]: I have always been impressed by the number of results conjectured by physicist, based on mathematically non-rigorous reasoning, then (much) later proved correct by mathematicians. A recent example is the $\sqrt{2+\sqrt{2}}$ connective constant of the honeycomb lattice, derived non rigorously by the physicist B. Nienhuis in 1982 and rigorously proved this year by S. Smirnov and H. Duminil-Copin.
I would be interested in knowing examples of results conjectured by physicists and later proved wrong by mathematicians.  Furthermore it would be interesting to understand why physical heuristics can go wrong, and how wrong they can go (for example, were the physicists simply missing an important technical assumption or was the conjecture unsalvagable).

REPLY [5 votes]: In the context of the Abelian sandpile model, in the paper ``Absorbing-state phase transitions in fixed-energy sandpiles'' (https://arxiv.org/abs/cond-mat/0003285) the physicists Vespignani, Dickman, Muñoz, and Zapperi predicted that the threshold density of the fixed energy sandpile should be the same as the stationary density of this model. But in fact this heuristic was refuted in some graphs by Fey, Levine and Wilson (https://arxiv.org/abs/1001.3401v2 and https://arxiv.org/abs/0912.3206). In particular while the two quantities in question appear to be close to one another in the cases of interest to physicists (torus/grid graphs), they aren't the same. The issue is that the threshold density depends on the initial configuration one starts from: the system remembers some of its pre-critical past at criticality. One then suspects that if the system is allowed enough time to mix, the threshold density will be the same as the stationary density. Hence, some more physicists (Poghosyan et al. https://arxiv.org/abs/1104.3548) conjectured that if the total number of grains of sand of the initial configuration goes to negative infinity, in the limit the threshold density and stationary density agree. Levine proved this is the case ("Threshold state and a conjecture of Poghosyan, Poghosyan, Priezzhev and Ruelle" https://arxiv.org/abs/1402.3283). Levine's paper is very well written and he goes over all this history in the introduction.<|endoftext|>
TITLE: What is the set of possible values of the degree of the sum of two algebraic numbers with fixed degrees?
QUESTION [6 upvotes]: This question is related to Degree of sum of algebraic numbers and algebraic numbers of degree 3 and 6, whose sum has degree 12. 
  In this last question I asked a very special case of the following problem : given two
algebraic numbers $\alpha$ and $\beta$ with degrees $a$ and $b$ respectively,
what can the degree of $\alpha+\beta$ be ?
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(1) $d \leq ab$ and $a \leq db$ and $b \leq da$. (this condition is obviously
necessary)
(2) $d$ divides $ab$, or $a$ divides $db$, or $b$ divides $da$.
The "if" part probably involves Galois theory as in Gerry's answer to the special case.  
EDIT 07/01/2010 : As Gerry noted, the conjecture above is grossly false. Below is a "corrected version"
of my conjecture.
I believe the answer is as follows : the degree of $\alpha+\beta$ can equal some
value $d$ iff
(*)  There is some integer $e$ divisible by all of $a,b,d$, and lower than or equal to
all of $ab,ad,bd$ (this is a necessary condition, as is seen by taking $e$ to be the degree
of the extension $k(\alpha,\beta)/k$, where $k$ is the base field).

REPLY [3 votes]: This isn't a solution, just a comment that got too long for the comment box: Assuming that all finite groups occur as Galois groups over $k$, the answer to this question should only depend on the characteristic of $k$.
Consider the more detailed question: 

For a finite group $G$, and subgroups
  $H_1$, $H_2$ and $H_3$, is there a
  Galois extension $L$ of $k$ with
  Galois group $G$, and elements $v_i
> \in L$ such that the stabilizer of
  $v_i$ is $H_i$ and $v_1+v_2+v_3=0$.

I claim that, for given $(G, H_1, H_2, H_3)$, assuming that there is some $G$-extension of $k$, the answer to this question only depends on the characteristic of $k$. Proof: As a $G$-representation, $L$ is the permutation representation on $X:=G/(H_1 \cap H_2 \cap H_3)$. The question, then, is whether we can find $L$-valued functions, $f_i$ on $X$, such that $f_i$ is constant on $H_i$ orbits (but not for any larger subgroup) and $f_1+f_2+f_3=0$. This is a collection of linear equalities and inequalities in $3 |X|$ variables, with integer coefficients. So whether or not they have a solution depends only on the characteristic of $k$ (we are using that $k$ is infinite). To see that the characteristic can matter, take $G=S_3$, $H_1$ and $H_2$ two different subgroups of order $2$ and $H_3$ the subgroup of order $3$. You should get a solution in characteristic $3$, and not otherwise.
Of course, answering the original question just means answering this question for all $(G, H_1, H_2, H_3)$ with $|G/H_i| = d_i$. (One can immediately make two reductions. First, a necessary condition is that $H_1 \cap H_2 = H_1 \cap H_3= H_2 \cap H_3$. Second, one can immediately reduce to the case that $H_1 \cap H_2$ contains no nontrivial normal subgroup. The latter means that $|G| \leq (d_1 d_2)!$, so the problem is finite.) 
I see no reason to believe that you will get a nicer answer by forgetting the groups and only remembering the degrees, but of course I haven't thought very hard about the problem.<|endoftext|>
TITLE: The ring of integers  looks  like the 3-dimensional sphere viewed as the Hopf fibration
QUESTION [6 upvotes]: This question is based on the following phrase:
"In a sense, $\textrm{Spec} \ \mathbf{Z}$ looks topologically like a 3-dimensional sphere viewed as the Hopf fibration over $\mathbf{S}^2$."
See page 88 of Algebraic Geometry II by Shafarevich.
I find this remark very interesting but I can't seem to parse it. 
I always just viewed $\textrm{Spec} \ \mathbf{Z}$ as an arithmetic analogue of $\mathbf{P}^1(\mathbf{C}) = \mathbf{S}^2$. This remark would add "something" to that in a sense.

REPLY [5 votes]: Various pieces of exposition and references are to be found - here, here, here, and here.<|endoftext|>
TITLE: Software for Tree-Decompositions
QUESTION [6 upvotes]: Does anybody know about software that exactly calculates the tree-width of a given graph and outputs a tree-decomposition? I am only interested in tree-decompositions of reasonbly small graphs, but need the exact solution and a tree-decomposition. Any comments would be great. Thanks!

REPLY [3 votes]: You could try the LibTW software, which is freely available from http://www.treewidth.com/ (also read their FAQ linked at the bottom of the page). It's written in Java so you can easily extend it with your own functionality.<|endoftext|>
TITLE: To differently gluing of two Riemann surfaces with boundary we get different surfaces
QUESTION [7 upvotes]: If $M,N$ are two Riemann surfaces with boundary, then we can glue them along one of each of their boundary component, which is $S$, to form a new Riemann surface with boundary, but for different gluing we may form different Riemann surfaces with boundary, for example, there may be a $S$ twist, intuitively, it is just we rotate one $S$ an angle then glue it with another surfaces, but my question is how can we show the resulting two surfaces (twisted gluing and untwisted gluing) are different Riemann surface with boundary (their differential structures are the same because we can regard it as a kind of connected sum)? Thanks!

REPLY [2 votes]: When you glue $M$ and $N$, for example, along some boundary component $C$. Choose a closed geodesic $\gamma$ on $S$ intersect with $C$. It can be shown that the hyperbolic length of $\gamma$ is changed when you twist along $C$. This means that different twist makes different hyperbolic surface.<|endoftext|>
TITLE: Existence of supercuspidal representations
QUESTION [6 upvotes]: Let $G$ be an unramified reductive group over $Q_p$. I want to prove that the group $G(Q_p)$ has a supercuspidal representation (complex coefficients).
I have been looking in many parts of the literature, and it seems that many people are convinced that it is true; however up to now I never saw it stated explicitly.
By the works of L. Morris I reduce to showing that a reductive group $M$ over $F_p$ has a cuspidal representation  (L. Morris, level zero G-types, p 140).
So then, I should prove that $M(F_p)$ has a cuspidal representation. The article of Deligne–Lusztig provides such a representation when given a minisotropic torus $T$ in $M$ and a character $\chi$ of $T(F_p)$ which is in general position.
Let me recall that "character in general position" means that the rational Weil group acts freely on the character.
So now comes my doubt and question. Is it true that such a pair $(T, \chi)$ can always be found for all reductive groups $M$ over $F_p$ ? 
I am "afraid" of "small" groups that have tori with "large" Weyl groups.
The supercuspidal representations that come from the above construction are of "level 0". 
In the book of Carter (Finite Groups of Lie Type) I found a result pointing in this direction. Lemma 8.4.2 p. 281 (with an easy proof) shows that for $T$ given and $q$ sufficiently large, the torus $T(F_q)$ has a character in general position.

REPLY [2 votes]: Various special constructions make the point for particular groups, which tends to convince one about the general case. E.g., for $G=SL(2,\mathbb Q)$, a well-known (Shalika, Jacquet, c. 1970) idea is that inducing "supercuspidals" from $K=SL(2,\mathbb Z_p)$ to $G$ is a finite direct sum of supercuspidals. The "level zero" cuspidals on $SL(2,\mathbb Z)$ factor through $SL(2,\mathbb Z/p)$. The irreducibles of the latter finite group are ascertained in various ways, and one finds that a significant fraction are "cuspidal", so compose to "supercuspidal" on $K$, and (with proof) induce to finite direct sums of supercuspidals on $G$.
The critical point is that induced repns (from Iwahori subgroups) do not exhaust the irreds of $K$. For $SL(2)$ this is easily proven by a counting argument, and/or by a theta-correspondence argument. For general groups, some of these choices persist.<|endoftext|>
TITLE: Integer values of a rational function
QUESTION [11 upvotes]: Suppose we are given a rational function with numerator and denominator being polynomials with integer coefficients. Is there an efficient algorithm for finding all integers arguments at which the function takes integer values?
In other words, for given polynomials $F(x)$ and $G(x)$ with integer coefficients, how to compute efficiently all such integers $m$ that $G(m)$ divides $F(m)$ ?
I've developed a rather straight forward approach to this problem at
http://list.seqfan.eu/pipermail/seqfan/2010-April/004339.html
but I suspect it is far from optimal.

REPLY [3 votes]: Some closely related (but not algorithmic) results are discussed in this paper of Corvaja and Zannier -- they have a number of other papers over the last few years on the same subject.<|endoftext|>
TITLE: The smallest Laplace-Beltrami eigenvalue on hyperbolic surfaces
QUESTION [9 upvotes]: For $g\geq 2$, let $M_g$ be the moduli space of genus $g$ hyperbolic surfaces, and let $\lambda_1(S_x): M_g \to \mathbb{R}$ be the smallest eigenvalue of the Laplace-Beltrami operator on the surface $S_x$ parametrized by a point $x\in M_g$.  Is there anything known about how the values of $\lambda_1$ are distributed when viewed as a function on moduli space?  For example, does the volume of the set of surfaces with  $\lambda_1(S_x)<\varepsilon$ go to zero rapidly as $\varepsilon \to 0$?

REPLY [5 votes]: Not anywhere near as much as known as one might like, but for enlightenment on your specific question see M. Mirzakhani's recent preprint on arXiv.org:
http://arxiv.org/abs/1012.2167v1<|endoftext|>
TITLE: Abstract thought vs calculation
QUESTION [59 upvotes]: Jeremy Avigad and Erich Reck claim that one factor leading to abstract mathematics in the late 19th century (as opposed to concrete mathematics or hard analysis) was the use of more abstract notions to obtain the same results with fewer calculations.
Let me quote them from their remarkable historical paper "Clarifying the nature of the infinite: the development of metamathematics and proof theory".

The gradual rise of the opposing viewpoint, with its emphasis on conceptual
reasoning and abstract characterization, is elegantly chronicled by Stein (Wayback Machine),
as part and parcel of what he refers to as the “second birth” of mathematics.
The following quote, from Dedekind, makes the difference of opinion very clear:



A theory based upon calculation would, as it seems to me, not offer
the highest degree of perfection; it is preferable, as in the modern
theory of functions, to seek to draw the demonstrations no longer
from calculations, but directly from the characteristic fundamental
concepts, and to construct the theory in such a way that it will, on
the contrary, be in a position to predict the results of the calculation
(for example, the decomposable forms of a degree).



In other words, from the Cantor-Dedekind point of view, abstract conceptual
investigation is to be preferred over calculation.

What are concrete examples from concrete fields avoiding calculations by the use of abstract notions? (Here "calculation" means any type of routine technicality.) Category theory and topoi may provide some examples.
Thanks in advance.

REPLY [10 votes]: I think that Gauss Theorem on constructible polygons fit this category.
For more than 2000 years the actual construction only lead to 4 classes: $2^n; 2^n\cdot 3; 2^n \cdot 5; 2^n \cdot 15$.
Gauss' abstract approach solved the problem. The interesting case $n=17$ becomes easy to understand, and easy to construct once one understands the abstract approach, but hard to attack otherwise. $n=257$ and especially $n=65537$ and the ones derived from these are the perfect examples of easy abstract proof versus extremely complicated calculations.<|endoftext|>
TITLE: Looking for figure of part of an A2 affine building
QUESTION [10 upvotes]: Somewhere, I  don't remember where, I saw a beautiful 3D figure of part a CAT(0) simplicial complex.  I am thinking and hoping that this was some finite piece of an affine building of type A2, presumably in characteristic 2.  But I'm very frustrated now that I just can't remember exactly what I saw or where I saw it.  It was something like part of the neighborhood of radius 1 or radius 2 of a vertex, with enough simplices removed so that the rest fits in $\mathbb{R}^3$.  It looked like Mathematica graphics, since (in my recollection) it had good colors to help show the 3-dimensional structure.  I'm thinking that I saw it in the AMS Notices, but it could also have been in an AMS calendar or elsewhere.  Does anyone remember seeing an image like this, and if so, where?
I'm asking because I'd like to have such an image in a paper that I'm working on, not necessarily the one that I saw but something similar.

I got some good answers to my question, both here and in private e-mail to Bill Casselman.  But in the end I decided to make my own diagram (with the aid of TikZ and Python SAGE).  Here it is.
alt text http://www.freeimagehosting.net/uploads/e38638d43e.png

For those who are interested in the TikZ and SAGE code, I combined them into one TeX document.  I posted both the TeX source and its PDF output (from pdflatex) on my web page.

REPLY [3 votes]: A picture similar to the one you've made can be found in Garrett's book on Buildings and classical groups.
http://www.amazon.com/Buildings-classical-groups-Paul-Garrett/dp/041206331X<|endoftext|>
TITLE: Constructing 4-manifolds with fundamental group with a given presentation.
QUESTION [13 upvotes]: Reading Princeton Companion I found out that every finitely presented group can be realized as the fundamental group of a 4-manifold. 
When starting to write this answer I found this related MO question. However, my question has to do with one of its answers (which is similar to the hint given in the Princeton Companion). 
The proceedure for constructing this manifold from a given presentation is first to construct a CW-complex with that fundamental group (by wedge sum of circles puting 2-cells to cover the relations) embedding it in $R^5$ and considering the frontier of a tubular neighborhood. 
Two questions come up to me (which are maybe trivial):
1- Why this cannot be done in $R^4$? Or can it be but the result is not the same?
2- Why the resulting manifold has the desired fundamental group?

REPLY [14 votes]: Taking the boundary of tubular neighborhood in codimension two doesn't preserve fundamental group -- the codimension must be at least three.
For a simple example, if you take the boundary of a tubular neighborhood of a single point in $R^2$, you get a circle, which has nontrivial fundamental group.  Similarly, the boundary of a tubular neighborhood of a 2-plane in $R^4$ is the product of a 2-plane with a circle, which again has nontrivial fundamental group.  In general, if you start with a 2-manifold in $R^4$, the boundary of a tubular neighborhood will be a circle bundle over the manifold, and will therefore not have the right fundamental group.
This problem is fixed in $R^5$, because you get a sphere bundle instead of a circle bundle, and the 2-sphere has trivial fundamental group.<|endoftext|>
TITLE: Is there a non-trivial group G isomorphic to Aut(G)?
QUESTION [11 upvotes]: The title basically says it all.
Is there a group with more than one element that is isomorphic to the group of automorphisms of itself?
I'm mainly interested in the case for finite groups,
although the answer for infinite groups would still be somewhat interesting.

REPLY [3 votes]: See also Problem 2015-4C in the Nieuw Archief voor Wiskunde.
A solution (of mine) can be found here:
http://www.nieuwarchief.nl/home/problems/pdf/uitwerking-2015-4.pdf<|endoftext|>
TITLE: Who wins this two-player game based on the sandpile model?
QUESTION [25 upvotes]: Given a connected graph $G$, two players, Blue and Green, play the following game: initially, all vertices are unclaimed.  Players alternate turns.  On her turn, Blue adds a token to either an unclaimed vertex (turning it blue) or a blue vertex, and similarly on his turn, Green adds a token to either an unclaimed vertex (turning it green) or a green vertex.
If a vertex of degree $d$ ever receives $d$ tokens, it topples, donating one token to each of its neighbors.  This may in turn cause some of its neighbors to topple, and so on, as in the sandpile model.  If Blue sets off a sequence of one or more topplings, she colors blue every vertex on which a token landed as a consequence of her move, and likewise for Green.  (Note that because the sandpile model is abelian, we don't need to specify an order of topplings.)  A player wins when she has managed to color the entire graph in her color.
Question: For which graphs $G$ is this game a first-player win?
I would be interested in any nontrivial statements for interesting classes of graphs, e.g., for paths, trees, cycles, complete bipartite graphs, grid graphs, whatever.  The game is trivially a first-player win on a complete graph $K_n$: the first player simply plays $n - 1$ times on the same vertex, and there is nothing the second player can do.  It's also not hard to see that it's a second-player win on the graph with two edges and three vertices: the first player cannot win on her move, and following her move there is always one token on the degree-2 vertex.  The second player places a token on the unclaimed leaf vertex, which topples; there are now two tokens on the degree-2 vertex, which topples, making player 2 the winner.
Motivation: This came up not in my research but rather in my relaxation -- there's a game installed on my math department Linux distribution called KJumpingCube, which is precisely this game on square grid graphs.
One could also ask the same question for directed graphs, of course.

REPLY [2 votes]: For an $n$-cycle, player 1 wins if $n$ is odd, else player 2 wins.
Since all vertices have degree 2, any vertex can have at most 1 token on it. Token-carrying vertices (of different colors) can be grouped into connected components. Call a component occupied by a player if it contains some vertex colored by the player, and that two components separated by a single vertex are adjacent. For example, the 6-cycle $(-*-1-2-*-1-*-)$ has 2 components, the first occupied by both players, the second occupied by player 1, and they are adjacent.
Lemma: If less than $n-2$ tokens have been played, it is player $i$'s turn to play, and player $i$ occupies a component, then she can ensure that she continues to occupy a component until her next turn.
To do so, she simply topples a component occupied by her. Toppling an occupied component results in two adjacent components occupied by her (you can see this by working out the result of toppling an isolated component). These may merge with adjacent neighbors of the original component, but this does not affect her occupancy. The opponent can in her turn topple at most one of these two components, leaving the other still occupied.
Neither player can lose before $n-2$ tokens have been played, if both follow the above strategy. On the $(n-2)$th play, any move results in a single $(n-1)$-vertex component. Suppose $n$ is even, so that it is player 1's turn, and player 2 on the previous turn created two distinct components occupied by her as above. Whether player 1 topples one of the components if she also occupies it, or joins the two by playing on an empty vertex, it leads to a single connected component occupied by player 2. Player 2 then wins by toppling the final component on the $n$th move. The same argument holds for Player 1 if $n$ is odd.
The strategy as such does not necessarily work for paths, because toppling a component adjacent to an end of the path does not create two separate components. My hunch is that for even $n$, starting from the middle then playing as above may still be a winning strategy for player 1, as it avoids the ends till the last turn. I don't know what the right strategy is for odd $n$, though experimentation leads me to believe player 2 should win.
I also can't see if there's any way to generalize this for general graphs, because it depends crucially on there being at most one token per vertex.<|endoftext|>
TITLE: Reference for Pic(G) and central extensions.
QUESTION [10 upvotes]: Let $G$ be a connected reductive group over a (perfect, why not) field $F$.  Let $m$, $pr_1$, $pr_2$ denote the multiplication, first, and second projection maps from $G \times G$ to $G$.
Then I'm pretty sure that I can prove the following fact:  if $L$ is a line bundle on $G$, then $m^\ast(L)$ is isomorphic to $pr_1^\ast(L) \cdot pr_2^\ast(L)$.  This (plus Hilbert's 90) implies that $Pic(G)$ classifies the central extensions of $G$ by the multiplicative group $G_m$, by some stuff in SGA 7, I believe.
The way that I can prove the above fact is by using Kottwitz's isomorphism, which describes $Pic(G)$ in terms of the dual group.  I'll probably include this Kottwitzish proof in something I'm writing, but I'm left with the following question:
Is there a proof in the literature that $m^\ast(L)$ is isomorphic to $pr_1^\ast(L) \cdot pr_2^\ast(L)$ for line bundles over reductive groups?  Someone must have written this up 30 years ago, right?  And the implication that $Pic(G)$ classifies central extensions by $G_m$?  Is this published somewhere?  It certainly shouldn't require passage to the dual group!
Of course, if I've messed something up, and the above fact is false, I'd appreciate such information too!

REPLY [3 votes]: I can give a reference only for the second part of the question, 
namely, about central extensions.
It was answered by Colliot-Thélène in 2008, not 30 years ago!
Colliot-Thélène's paper Résolutions flasques des groupes linéaires connexes, J. für die reine und angewandte Mathematik (Crelle) 618 (2008), 77--133,
contains the following corollary (I type it in English):
Corollary 5.7. Let $G$ be a connected linear algebraic group, assumed reductive if char $k > 0$. For any smooth $k$-group of multiplicative type $S$, the natural arrow
Ext$(G,S)\to$ ker$[H^1(G,S)\to H^1(k,S)]$ is an isomorphism.
Taking $S=\mathbf{G}_m$, we obtain $H^1(k,\mathbf{G}_m)=1$ (Hilbert 90) and $H^1(G,\mathbf{G}_m)=\mathrm{Pic}(G)$
(here $H^1$ means étale cohomology). We obtain an isomorphism Ext$(G,\mathbf{G}_m)\cong \mathrm{Pic}(G)$.<|endoftext|>
TITLE: Maximum number of mutually equidistant points in an n-dimensional Euclidean space is (n+1). Proof?
QUESTION [13 upvotes]: How to prove that the maximum number of mutually equidistant points in an n-dimensional Euclidean space is (n+1)?

REPLY [8 votes]: Assume one of the points is the origin, the others are the vectors a1,...,an+1. If the pairwise distance is 1, then $a^2_i=1$ and $(a_i-a_j)^2=1$ (i < j)(scalar multiplication). This gives $(a_i,a_j)=\frac{1}{2}$. Now show that $a_1,\dots,a_{n+1}$ are linearly independent: if $\lambda_1,\dots,\lambda_{n+1}$ are scalars and $\sum \lambda_ia_i=0$, then scalar multiplication by $a_i$ gives $\lambda_i+\frac{1}{2}\sum_{j\neq i}\lambda_j=0$, then we have $\lambda_i=-\Lambda$ for each $i$ where $\Lambda=\lambda_1+\cdots+\lambda_{n+1}$, summing gives $\Lambda=0$ so each $\lambda_i=0$.<|endoftext|>
TITLE: Random walk is to diffusion as self-avoiding random walk is to ...?
QUESTION [15 upvotes]: One can view a random walk as a discrete process whose continuous
analog is diffusion.
For example, discretizing the heat diffusion equation
(in both time and space) leads to random walks.
Is there a natural continuous analog of discrete self-avoiding walks?
I am particularly interested in self-avoiding polygons,
i.e., closed self-avoiding walks.
I've only found reference 
(in Madras and Slade,
The Self-Avoiding Walk, p.365ff)
to continuous analogs of "weakly self-avoiding walks"
(or "self-repellent walks") which discourage but
do not forbid self-intersection.
I realize this is a vague question, 
reflecting my ignorance of the topic.
But perhaps those knowledgeable could point me to the
the right concepts.  Thanks!
Addendum.
Schramm–Loewner evolution is the answer.  It is the conjectured scaling limit of the self-avoiding walk and several other related stochastic processes.  Conjectured in low dimensions,
proved in high dimensions, as pointed out by Yuri and Yvan.  Many thanks for your help!

REPLY [14 votes]: In 2D the scaling limit is believed to be SLE with parameter 8/3. This was conjectured by Lawler, Schramm and Werner and, to the best of my knowledge, still remains open.

REPLY [3 votes]: For SLE see http://en.wikipedia.org/wiki/Schramm%E2%80%93Loewner_evolution<|endoftext|>
TITLE: volume of big line bundles under finite morphisms
QUESTION [6 upvotes]: Let $X$, $Y$ be complex projective varieties of dimension $n$, let $f:X \rightarrow Y$ be a surjective finite morphism of degree $d$ and let $B$ be a big line bundle on $Y$.
Is that true that vol($f^*B$)=d $\cdot$ vol($B$)?
(I know that if $B$ is not only big but also nef then the formula is true by Lazarsfeld's Positivity in Algebraic geometry I, remark 1.1.13, using the well-known fact that if $B$ is nef then vol($B$)=$B^n$).

REPLY [5 votes]: Yes, this is true, even in a slightly more general context (the varieties can be defined over any field $k$, and the morphism only needs to be generically finite and surjective).
The main parts of the argument are given in the books of Lazardfeld (Positivity in Algebraic geometry) and Debarre (Higher-dimensional Algebraic Geometry), although this specific property is never stated there (as far as I know).
By the projection formula, we have
$H^0(X,f^* B^{\otimes m}) \cong  H^0(Y,f_* \mathcal O_X \otimes B^{\otimes m}),$
so we can restrict our attention to $f_* \mathcal O_X$ and its twists by $B$. There is an open dense subset $U$ of $Y$ such that $f_* \mathcal O_X$ is free of rank $d = \deg(f)$, so  $(f_* \mathcal O_X)|_U \simeq \mathcal O_U^d$. This isomorphism gives an injection $f_* \mathcal O_X \hookrightarrow \mathcal K_Y^d$, where $\mathcal K_Y$ is the sheaf of total quotient rings of $\mathcal O_Y$. Set $\mathcal G = f_* \mathcal O_X \cap \mathcal O_Y^d$ and define $\mathcal G_1$ and $\mathcal G_2$ by the exact sequences of sheaves
$0 \to \mathcal G \to f_* \mathcal O_X  \to \mathcal G_1 \to 0$ , 
$0 \to \mathcal G \to \mathcal O_Y^d\;  \to \mathcal G_2 \to 0.$
The supports of $\mathcal G_1$ and $\mathcal G_2$ do not meet $U$, hence have dimension less than $n$. Therefore,
$h^0(Y, \mathcal G_i\otimes B^{\otimes m}) := \dim_k H^0(Y, \mathcal G_i\otimes B^{\otimes m}) = O(m^{n-1})$ 
for $i=1,2$ (see e.g. Proposition 1.31 in Debarre's book). Using the long exact cohomology sequence for the two short exact sequences above twisted by $B^{\otimes m}$, this implies
$h^0(X,f^* B^{\otimes m}) = h^0(Y,(f_*\mathcal O_X)\otimes B^{\otimes m})= d \cdot h^0(Y, B^{\otimes m}) + O(m^{n-1}),$
from which the assertion follows.
See also Proposition 4.1 in this arXiv paper.<|endoftext|>
TITLE: Geometric vs Arithmetic Frobenius
QUESTION [20 upvotes]: If an algebraic variety $X$ over a field characteristic p is given by equations $f_i(x_1,...,x_k) = 0$, we can consider the variety $X^{(p)}$ obtained by applying p-th powers to all the coefficients of all $f_i$'s.
Frobenius morphism, as I understand it, is a morphism $X \to X^{(p)}$,
given on points as raising all coordinates to p-th power. 
Can anyone please explain me, what is the geometric Frobenius, as opposed to the arithmetic one?
EDIT: Thanks to Florian and George for the answers. I understand the difference now.
I accepted Florian's answer because he was first and also because I found the last link 
http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf he provided especially helpful.

REPLY [17 votes]: Brion & Kumar ["Frobenius splitting methods in geom. and rep. thy" Birkhauser 2005] call the absolute
Frobenius endomorphism the mapping $F_{abs}:X \to X$ which is the identity on $X$ and with
comorphism given when 
$X = \operatorname{Spec}(A)$ is affine by the rule $(f \mapsto F_{abs}^*(f) = f^p):A \to A$.
This is not a morphism "over $k$" since $F^*:A \to A$ is "semilinear" for the Frobenius
endomorphism of $k$ (= Frobenius automorphism in Galois group of $k$ if $k$ is perfect).
In Jantzen ["Representations of algebraic groups", 2nd edition] 9.1 and 9.2, he describes the absolute Frobenius map - it is "the same" as the one describe by B&K, except that
the codomain is "twisted" to make $F$ a morphism over $k$. For $X = \operatorname{Spec}(A)$
this twisting amounts to: $X^{(p)} = \operatorname{Spec}(A^{(p)})$ where
the $k$-algebra $A^{(p)}$ is $A$ as a ring but an element $a \in k$ acts on $A^{(p)}$ as
$a^{p^{-1}}= a^{1/p}$ does on $A$.
Geometric and arithmetic Frobenii have meaning only (I believe) when $X$ is "defined
over" a finite field; here I'll assume $X$ is defined over $\mathbf{F}_p$. And I'll even suppose $X$ arises by base change to $k$ from the affine $k_0 = \mathbf{F}_p$-scheme $X_0 = \operatorname{Spec}(A_0)$ (otherwise, patch!). 
Then $X = \operatorname{Spec}(A)$ where $A = A_0 \otimes_{k_0} k$.
The arithmetic Frobenius map on $X$ is the $k_0$-morphism $F_{arith}:X \to X$ whose
comorphism is given by $$(f \otimes a \mapsto f \otimes a^p):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k$$ for $f \in A_0$ and $a \in k$.
Thus the set of $k_0$-points $X_0(k_0)$ is the set of points in $X(k)$ fixed by the
arithemtic Frobenius $F_{arith}$; i.e. the action of $F_{arith}$ on points just gives the
"usual" action of the Frobenius element of the Galois group on rational points (here I must be supposing $k$ to be perfect...)
The geometric Frobenius of $X$ is the $k$-morphism $F_{geom}:X \to X$ whose comorphism
is given by $$(f \otimes a \mapsto f^p \otimes a):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k.$$ If you pick an embedding $X \subset \mathbf{A}^N$ defined
over $k_0$, then $F_{geom}$ is given on $k$-points in these coordinates by
$$F_{geom}(x_1,\dots,x_N) = (x_1^p,\dots,x_N^p)$$.
The arithmetic and geometric Frobenius are defined (briefly) in Jantzen (loc. cit.). 
Note that $F_{arith} \circ F_{geom} = F_{geom} \circ F_{arith}$ is the "absolute Frobenius" of B&K mentioned above.
Also see Milne's "Lectures on Etale Cohomology" 29.11 for some discussion reconciling
the number theorists with their action of the Frobenius automorphism $\phi=(x \mapsto x^p)$ on the Tate group $T_\ell E$ of an elliptic curve defined over $k_0$ and the algebraic geometers with their action of $F_{geom}$ on $H^1(E,\mathbf{Z}_\ell)$.<|endoftext|>
TITLE: Explanation why $x,y,z$ are always variables
QUESTION [23 upvotes]: I heard or have read the following nice explanation for the origin of the convention that one uses (almost) always $x,y,z$ for variables. (This question was motivated by question
Origin of symbol  *l* for a prime different from a fixed prime?)
It seems this custom is due to the typesetter of Descartes. Descartes used initially other letters 
(mainly $a,b,c$) but the typesetter had the same limited number of lead symbols
for each of the 26 letters of the Roman alphabet. The frequent use of variables exhausted his stock and he asked thus Descartes if he could use the last three letters $x,y,z$ of the alphabet (which occur very rarely in French texts). 
Does anyone know if this is only a (beautiful) legend or if it contains some truth? (I checked that Descartes uses indeed already $x,y,z$ generically for variables in his printed works.)

REPLY [13 votes]: I'm often amazed how legends propose completely preposterous "explanations". Why on earth would a typesetter have equal numbers of lead symbols for every letter, given that they have a very non-uniform distribution in ordinary text? And why would it take Descartes' math formulas to discover that the distribution is not balanced?
If on the other hand the typesetter had a distribution of lead characters that corresponds roughly to their use in non-math texts (which would seem to be a reasonable assumption), then surely it would disturb this balance much more profoundly to start systematically using the least-occurring letters in mathematical formulae than to stick to using those letters which are already in common use. I don't think such the theory put forward is even worth seriously investigating.<|endoftext|>
TITLE: What's an example of an intersection cohomology sheaf whose stalks are pure but not pointwise pure?
QUESTION [10 upvotes]: I'll freely admit that I have a rather hard time keeping straight different notions of purity of etale sheaves, and I think part of the problem is the lack of counterexamples.
For example, it's a theorem that the stalks of intersection cohomology sheaves (with coefficients in $\overline{\mathbb{Q}}_\ell$, say) are pure.  But here, pure means "looks like the cohomology of a possibly singular projective variety" i.e., the weights are bounded above, but not below.  
This is perhaps not so surprising: if, say, one takes constant coefficients, then the IC is a summand of the pushforward of a resolution of singularities, and so the stalks are summands of the cohomology of projective varieties.  
But, if you do geometric representation theory, a funny thing happens.  In practice, one seems to always get "pointwise purity" i.e. the stalk has the weights that one expects from the cohomology of a smooth projective variety.  There are various geometric reasons for this (for example, if your resolution of singularity is symplectic, or if the fibers of your map have a paving/$\alpha$-partition by affines or other pure varieties), but it makes it hard for me to imagine anything else.  I think my understanding of algebraic geometry would be improved by knowing an example of an intersection cohomology sheaf on (say) an affine variety which isn't pointwise pure.  

What is a good example of such an intersection cohomology sheaf?

REPLY [5 votes]: Here is an example along the lines of Donu's suggestion:
We work over a (large) finite field $k$. Let $C \subset \mathbb{P}^2$ be an irreducible  plane curve of degree $d$ and geometric genus $g > 0$ having at least one node. For example, the curve given by the equation
$$y^2 = (x^2-1)(x-a_1)\cdots(x-a_{d-2})$$
 with $d>4$, all $a_i$ distinct and nonzero and $char(k) \neq 2$.
Blow up $d^2 + 1$ distinct smooth points on $C$ to get a smooth surface $X$ and let $D$ be the strict transform of $C$. We have $D\cdot D = -1$ since $C\cdot C = d^2$. By Artin's contractability criterion, we can contract $D$ to obtain a normal surface $Y$ with a singular point $p$. Now $H^1(D, \mathbb{Q}_{\ell})$ has weights $0$ and $1$, the $0$ coming from the node and the $1$ since the geometric genus is positive. By proper base change and the decomposition theorem, it follows that $IC_Y$ is not pointwise pure (at $p$).<|endoftext|>
TITLE: Predicting if something is a code
QUESTION [8 upvotes]: I'm trying to help a non-mathematical friend by posting a question of his here.  He studies literature and has come across a book which is written in a made-up language.  The book is hundreds of pages, but apparently it is unknown if there is meaning behind what's written.  Is there a standard way of predicting whether something like this is written in a code?  Obviously we can't know for sure without actually decoding it, but are there some traits to look for?  I have no idea, so I thought I would ask here.  Thanks!
Here's the book in question:
http://en.wikipedia.org/wiki/Codex_Seraphinianus

REPLY [4 votes]: The technique to investigate whether documents contain coded information is called Steganalysis. It is the counterpart of Steganography, namely hiding messages such that no one except the recipient suspects a message. The problem is generally handled with statistical analysis includig Zipf's law mentioned in the answer of Dan Piponi. For an overview see http://en.wikipedia.org/wiki/Steganalysis. And here are some valuable tools in case your problem has not yet been solved http://stegsecret.sourceforge.net/.<|endoftext|>
TITLE: English translation of Riemann's Habilitation Thesis
QUESTION [10 upvotes]: Does anyone know where to find an English translation of Riemann's Habilitation Thesis concerning trigonometric series?  The German title of the work is "Über die Darstellbarkeit einer Function durch eine trigonometrische Reihe" and the English title is "On the representation of a function as a Trigonometric Series".  So far I unfortunately haven't been able to find the English translation, so any suggestions as to where I might find it (if it has been translated) would be greatly appreciated.

REPLY [3 votes]: Riemann's complete papers, including are available used on amazon, currently for $80. Search for ISBN-10: 0974042730.<|endoftext|>
TITLE: Is there a good argument for why you can't place 4 queens which cover a chessboard?
QUESTION [31 upvotes]: It has been known since the 1850's (or even much earlier) that 5 queens could be placed on an 8*8 
chessboard so that every square on the board lies in the same row, column, or diagonal as at least one
of the queens. It was also "known" that this could not be done with 4 queens. But I have not been able to
obtain or track down any rigorous mathematical proof of this that could be (or could have been) carried out
in a reasonable time by a human being with pencil and paper. There are altogether 635376 ways of placing
4 queens on an 8*8 chessboard. Does anyone know of a combinatorial algorithm, exploiting the symmetries of
the chessboard, which would reduce the number of cases to be considered to the hundreds (or, at most, the
low thousands.)?
                  This is, of course, a trivial problem for modern computers, which have many times since
verified that 5 queens are needed.

REPLY [5 votes]: I also wonder whether the formation came before or after 1990, because that is when Spencer came up with the lower bound for n=11.  I know first-hand that formations can be searched for by hand for say, up to n=20 with reasonable results (note that is much different than actual proof that the formation is minimal, for if we find the formation we've only found an upper bound).  So it's quite possible that someone found the formation first, and then Spencer might have been the first to observe it.  Of course it's also possible that Spencer's bound came first.
I found Dr Chatham's comment interesting enough to vote it up, although seeing as how I'm stuck at one reputation, I can't.<|endoftext|>
TITLE: What is an example of a compact smooth manifold whose K-theory and Cech cohomology are not isomorphic?
QUESTION [15 upvotes]: On page 283 of Max Karoubi’s book, “K-theory,” he states that for any compact Hausdorff space $X$, the Chern character determines an isomorphism from the group $K^0(X) \otimes Q$ to $H^{even}(X; Q)$, the direct sum of the even-dimensional Cech cohomology groups of X with rational coefficients.  In particular, this theorem implies that $K^0(X)$ and $H^{even}(X; Z)$ are isomorphic up to torsion. 

Does anyone know of an example (and also a reference, preferably) of a smooth compact manifold $X$ with the property that $K^0(X)$ and $H^{even}(X; Z)$ are not isomorphic?

REPLY [21 votes]: For $X=\mathbb RP^4$ the groups $K^0(X)$ and $H^{even}(X)$ respectively are $\mathbb Z\oplus \mathbb Z/4$ and $\mathbb Z\oplus \mathbb Z/2\oplus \mathbb Z/2$. More generally, $K^0(\mathbb RP^{2k})\cong \mathbb Z\oplus \mathbb Z/2^{k}$. These computations of real and complex $K$-groups of real and complex projective spaces can be found in an early paper of J F Adams, I believe the one about vector fields on spheres. Let $k\to\infty$ so that $\mathbb RP^k$ becomes $BG$ for $G$ of order $2$. The Atiyah-Segal Completion Theorem says that for a finite or more generally compact Lie group $G$ the ring $K^0(BG)$ is the completion of the complex representation ring $RG$ with respect to the augmentation ideal (kernel of rank homomorphism $RG\to\mathbb Z$).
Even when the homology groups are torsion-free, so that $K^0$ is abstractly isomorphic to $H^{even}$, there is in some sense a difference between $K^0$ and $H^{even}$; they are then two slightly different lattices in the same rational vector space. For example, if $x$ generates $H^2(\mathbb CP^k,\mathbb Z)$ then a $\mathbb Z$-basis for $H^{even}(\mathbb CP^k)$ is $1,x,\dots,x^k$ while a $\mathbb Z$-basis for (the image under the Chern character of) $K^0(\mathbb CP^k)$ is $1,e^x,\dots,e^{kx}$ (or $1, e^x-1,\dots, (e^x-1)^k$). When $X$ is a sphere $S^{2k}$, they are the same lattice.<|endoftext|>
TITLE: What is the Batalin-Vilkovisky formalism, and what are its uses in mathematics?
QUESTION [23 upvotes]: I checked Wikipedia, I know it is a powerful quantization in physics, but I am wondering what is its relation in mathematics (like mirror symmetry as in wikipedia). A related thing is quantum master equation, what's its use in mathematics? Any reference or background? Thanks!

REPLY [30 votes]: The BV formalism provides a (co)homological reformulation of several important questions of quantum field theory. The kind of problems that are usually addressed by the BV formalism are:

the determination of gauge invariant operators,
the determination of conserved currents, 
the problem of consistent deformation of a theory,
the determination of possible quantum anomalies (the  violation of the gauge invariance due to quantum effects). 

The BV formalism is specially attractive because it  does not require one to make a choice of a gauge fixing and it maintains a manifest spacetime covariance. It can also deals with 
situation that the traditional BRST formalism can not handle. This is for example the case of gauge theories admitting an open gauge algebra (  a gauge algebra that is closed only modulo the equations of motion). The typical example are supergravity theories.
The BV formalism also allows an elegant and powerful mathematical reformulation of certain questions of quantum field theories in the language of homomological algebra. 
Mathematically, the BV formalism is simply a clever application of  homological perturbation theory. In order to understand the relation, I will first review the  geometry of a physical model described by a Lagrangian $\mathcal{L}$  depending of fields $\phi^I$ and a finite number of their derivatives and admitting a gauge symmetry $G$. 
The starting point is the space $\mathcal{M}$ of all possible  configurations of fields and their derivatives. This can be formalized using the language of jet-spaces. The Euler-Lagrange equations give the equations of motion of the theory and together with their derivatives, they define a sub-space $\Sigma$ of $\mathcal{M}$ called the  stationary space. The on-shell functions are the functions relevant for  the dynamic of the theory, they are defined on the stationary space $\Sigma$, they can be described alegebraically as   $\mathbb{C}^\infty(\Sigma)=\mathbb{C}^\infty(\mathcal{M})/ \mathcal{N}$ where $\mathcal{N}$ is the ideal of functions that vanish on $\Sigma$.
Because of the gauge invariance, the Euler-Lagrange equations are not independent but they satisfy some non-trivial relations called Noether identities. One has to identify different configurations related by a gauge transformation. Indeed, 
 a gauge symmetry is not a real symmetry of the theory but a redundancy of the description. 
The two steps that we have just described (restriction to the stationary surface and taking the quotient by the gauge transformations) are respectively realized in the BV formalism  by the homology of the Koszul-Tate differential $\delta$ and the cohomology of the  longitudinal operator $\gamma$. 
The Koszul-Tate operator defines a resolution of the equations of motion in homology. 
This is done by introducing one antifield $\phi^*_I$ for each field $\phi^I$ of the Lagrangian. The antifields are introduced to ensure that  the equations of motion are trivial in the homology of the Koszul-Tate operator. 
The gauge invariance of the theory  is taking care of by the cohomology of the longitudinal differential $\gamma$. In the case of Yang-Mills theories, the cohomology of $\gamma$ is equivalent to the Lie algebra cohomology. 
The full BV operator is then given by
$$s=\delta + \gamma+\cdots,$$
where the dots are for possible additional terms required to ensure that the BV operator $s$ is nilpotent ( $s^2=0$). The construction of $s$ from $\delta$ and $\gamma$ follows a recursive pattern borrowed from  homological perturbation theory. One can trace the need for the antifields and the Koszul-Tate differential to this recursive pattern. 
For simple theories like Yang-Mills, we just have $s=\delta+\gamma$ because the gauge algebra closes as a group without using the equations of motion. In more complicate situation when the algebra is open there are additional terms in the definition of $s$. 
One can generates $s$ using the BV bracket $(\cdot ,\cdot)$ (under which a field and its associated antifields are dual) and a source $S$ such that the BV operator can be expressed as 
$$
s F= (S,F).
$$
 The classifical master equation is
$$(S,S)=0,$$
  and it is just equivalent to $s^2=0$. 
At the quantum level, the action $S$ is replaced by a quantum action $W=S+\sum_ i \hbar^i M_i$ where the terms $M_i$ are contribution due to the path integral measure. The gauge invariant of quantum expectation values of operators is equivalent to the quantum master equation :
$$
\frac{1}{2}(W,W)=i\hbar \Delta W,
$$
where $\Delta$ is   an operator similar to the Laplacian but defined in the space of  fields and their antifields. This operator naturally appears when one considers the invariance of the measure of the path integral under an infinitesimal BRST transformation.
When $\Delta S=0$, we can take $W=S$.
We will now review the BV  (co)homological interpretation of some important questions in  quantum field theory:

The observables of the theory are gauge invariant operators, they  are described by the cohomology group    $H(s)$ in ghost number zero. 
Non-trivial conserved  currents of the theory are equivalent to the so-called characteristic  cohomology $H^{n-1}_0(\delta |d)$ which is the cohomology of the Koszul-Tate operator $\delta$ (in antifield number zero) modulo total derivatives for forms of degree $n-1$, where $n$ is the dimension of spacetime. 
The equivalent class of global symmetries is equivalent to $H^n_1(\delta| d)$. 
The gauge anomalies are controlled by the group $H^{1,n}(s|d)$ (that is $H(s)$ in antifield number 1  and in the space of $n$-form modulo total derivative). The  conditions that define the  cohomology $H^{1,n}(s|d)$ are generalization of the famous Wess-Zumino consistency condition.
The group $H^{0,n}(s|d)$ controls the renormalization of the theory and all the possible counter terms. 
The groups $H^{0,n}(\gamma,d)$ and $H^{1,n}(\gamma, d)$ control the consistent deformations of the theory. 

References:

For a short review, I recommend the
preprint  by Fuster,  Henneaux and Maas: hep-th/0506098.  
The classical reference is the book  of
Marc Henneaux and    Claudio   Teitelboim  (Quantization of    Gauge
Systems).
For   applications there is also a standard   review by  Barnich, Brandt
and Henneaux: ``Local BRST cohomology in gauge theories,''  Phys. Rept.338, 439 (2000) [arXiv:hep-th/0002245].<|endoftext|>
TITLE: Approximating operators on Banach spaces by bounded operators on a proper dense subspace
QUESTION [14 upvotes]: While digging through old piles of notes and jottings, I came across a question I'd looked at several years ago. While I was able to get partial answers, it seemed even then that the answer should be known and in the literature somewhere, but I never knew where to start looking. So I thought I'd ask here on MO if anyone knows of a reference for this observation.
Here's the notation and background for the question.
Let $(E,\Vert\cdot\Vert)$ be a real, normed vector space (I think the complex case works out to be almost identical). We will see shortly that my question is vacuous unless $E$ is incomplete. Let $F$ be the completion of $E$.
Denote by $B(E,\Vert\cdot\Vert)$ the space of all linear maps $T:E\to E$ which ae bounded with respect to $\Vert\cdot\Vert$, i.e. there exists $C$ depending on $T$ such that
$$ \Vert T(x)\Vert \leq C\Vert x\Vert \;\;\hbox{for all $x\in E$.} $$
Clearly each $T\in B(E,\vert\cdot\Vert)$ extends uniquely to a bounded linear operator $F\to F$, and we thus get an injective algebra homomorphism $\imath:B(E,\Vert\cdot\Vert)\to B(F)$.
The question arises: when does $\imath$ have dense range?

It is not hard to show that if $E=c_{00}$ and $\Vert\cdot\Vert$ is the $\ell_\infty$ norm then $\imath$ does indeed have dense range.
On the other hand, if $E=\ell_1$ and $\Vert\cdot\Vert$ is the $\ell_\infty$ norm, then by considering "blocks" which have $\ell_\infty$-norm 1 and large $\ell_1$-norm, we can construct an isometry on $c_0$ which is not approximable by operators of the form $\imath(T)$; in particular, $\imath$ does not have dense range in this case.
I can't find the piece of paper where I wrote down the details, but I seem to recall that one obtains the same answer if we replace $\ell_1$ by $\ell_p$ for $1\leq p < \infty$ and take $\Vert\cdot\Vert$ to be the $\ell_r$ norm for any $p<r\leq\infty$.

So here are two explicit questions. I haven't looked at them properly since about 2004/5, so they may well have straightforward solutions.
Q1. Let $\Vert\cdot\Vert$ be any norm on $c_{00}$, and let $F$ be the completion of $c_{00}$ in this norm. Does $\imath: B(c_{00},\Vert\cdot\Vert)\to B(F)$ have dense range?
Q2. Let $(F,\Vert\cdot\Vert)$ be a Banach space with an unconditional basis. Let $E$ be a proper dense subspace of $F$ which is a Banach space under some norm that dominates $\Vert\cdot\Vert$. 
Does $\imath: B(E,\Vert\cdot\Vert)\to B(F)$ always have non-dense range?
If the answers to these are known, does anyone know where I might find references to these in the literature?
Update 5th July 2010: Q1 has a positive answer, as given by Bill Johnson below (a simmilar approach was also elaborated by Pietro Majer). As pointed out (ibid.) the question can be rephrased/generalized to the following:

given a separable Banach space $F$ and
  a dense linear subspace $E$ of
  countable dimension, can every bounded
  operator on $F$ be approximated by
  operators which take $E$ to $E$?

I'd still be interested to know the answer to Q2, even in the special cases where $F=\ell_p$ for some $1\leq p < \infty$.

REPLY [4 votes]: For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations. 
Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)
I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that  $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$
The sum
$$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$
is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form 
$$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$
and one has, on the subspace $X$ 
$$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$
By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$ 
The claim should be proved as suggested below by  Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:  
Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such  that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$
Applying this to $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted. 
Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense linear subspace; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $D'$. This way one sees where the assumptions are needed: countability is only relevant for $D$, density is only relevant for $D'$.<|endoftext|>
TITLE: Lower bound on # of nonzero digits in ternary expansions of powers of 2?
QUESTION [9 upvotes]: Does anyone know of any lower bounds on the number of nonzero digits that appear in powers of 2 when written to base 3?  (Other than the easy "If it's more than 8 it has to have at least 3.")  I know there's been some stuff done with powers of 2 written to base 3, but I can't seem to find anything that quite answers this question.
(Are there more general such bounds? Powers of a written to base b? (To avoid triviality, suppose that no power of a is a power of b... or are stricter conditions needed?) From what I can tell, it looks like even specific instances of these are hard, so I suppose I should stick to the specific version.)

REPLY [13 votes]: A nontrivial lower bound can be found in a paper of Cam Stewart (see http://www.math.uwaterloo.ca/PM_Dept/Homepages/Stewart/Jour_Books/J-reine-ange-Math-1980.pdf). He proves, more generally, for fixed bases a and b for which $\log a/\log b$ is irrational, that the sum of the number of nonzero digits in the base a and base b digits of an integer n exceeds (essentially) $$\log \log n/ \log \log \log n.$$<|endoftext|>
TITLE: Definition of Function
QUESTION [13 upvotes]: What is authoritative canonical formal definition of function?
For example,
According to Wolfram MathWorld,
$$isafun_1(f)\;\leftrightarrow\;
\forall a\in f\;(\exists x\exists y \;\langle x,y\rangle = a)
\; \wedge \;
\forall x\forall y_1\forall y_2\;((\langle x,y_1\rangle\in f\wedge\langle x,y_2\rangle \in f)\rightarrow y_1=y_2))$$
According to Bourbaki "Elements de Mathematiques, Theorie des Ensembles",
$$
isafun_2(f)\;\leftrightarrow\;
\exists d\exists g\exists c\;(\langle d,g,c\rangle=f
\;\wedge\;isafun_1(g)\;\wedge$$
$$\;\wedge\;
\forall x(x \in d\rightarrow \exists y(\langle x,y\rangle \in g))
\;\wedge\;
\forall x\forall y(\langle x,y\rangle \in g\rightarrow (x \in d\wedge y\in c)))
$$
How to make agree definition of function as triple with extensional equality
$$
\forall f\forall g\;[\;(isafun(f)\wedge isafun(g))
\; \rightarrow \;
[\;(\forall x(\;f(x)=g(x)\;))\leftrightarrow f=g\;]\;]
$$
?
Why such divergences in definitions exist?
Upd: Two additional questions:

Why function is not a pair in $isafun_2$? First component of triple is perfectly derivable from the second.
What word function exactly means if no underlying theory is specified in context? If I build fully formal knowledge base about mathematics for automated reasoning and want to add notion of contextless function -- how I must describe it?

REPLY [3 votes]: When we use the Bourbaki definition of function as a triple (domain, codomain, graph), then two functions are usually defined to be equal iff their domains and graphs are equal. So  equal functions can have different codomains. The problem is that the same sign "=" is used both for the  the equality of functions and the "universal" equality (In ZFC, for example, the "universal" equality is defined for all sets). That is, the sign "=" is overloaded. Normally, from the context one can determine what is the intended meaning.
But there is a more serious trouble (as Vag pointed out early) with the Bourbaki definition, when a function is an element of a set. So it seems that the  definition of function as a set of ordered pairs having the functional property is more preferable.<|endoftext|>
TITLE: An experiment on random matrices
QUESTION [15 upvotes]: A bit unsure if my use/mention of proprietary software might be inappropriate or even frowned upon here. If this is the case, or if this kind of experimental question is not welcome, please let me know and I'll remove it quickly.
I was experimenting on the distribution of eigenvalues of random matrices via the following Mathematica code
dim = 301;
decomplex = {z_Complex -> {Re[z], Im[z]}, A_Real -> {A, 0}}; b = 
 Eigenvalues[
   SparseArray[{i_, i_} -> 1, {dim, dim}] + 
    Table[RandomReal[{-.9, 1.}], {i, 1, dim}, {j, 1, dim}]] /. decomplex;
ListPlot[b, PlotStyle -> PointSize[0.015], AspectRatio -> Automatic]
This simply draws (or should draw) a picture of the complex eigenvalues of a 301 x 301 matrix, with entries uniformly distributed in the real interval [-0.9, 1]. If you try the code you will notice a disk of random eigenvalues, which is expected, plus a single positive eigenvalue at some distance of the disk. Decreasing the lower bound on the entries to -1 makes this phenomenon disappear, while increasing the lower bound makes it more evident.
Question: is this an artifact, a real phenomenon, or maybe even a well known and simple to explain phenomenon?

REPLY [18 votes]: The distribution of the bulk of the spectrum is an example of the circular law.  For the model you selected (where each entry is uniformly chosen at random from an interval), the law was first proven by Bai (at least in the case where the entries are normalised to have mean zero), building upon previous work of Girko; the non-central case (non-zero mean) was recently established by Chafai.  
The non-central case is a rank one perturbation of the central case (by the matrix whose entries are all equal to the mean) and should therefore cause one exceptional eigenvalue.
(In the central case it can be shown that the spectral radius is close to the radius of the disk, so there are basically no exceptional eigenvalues.)  If the mean of each entry is $\mu$, then the rank one perturbation has an eigenvalue at $\mu n$, so one expects the exceptional eigenvalue to linger near this number.  In your case, $\mu = 0.05$ and $n = 301$, so the exceptional eigenvalue should linger near $15.05$.
There is a paper of Silverstein which makes the above heuristics precise; see also the earlier work of Andrew.
There is quite a bit of recent literature on the circular law, see for instance this survey by Van Vu and myself, or my lecture notes on this topic.<|endoftext|>
TITLE: Derived Algebraic Geometry and Chow Rings/Chow Motives
QUESTION [29 upvotes]: I recently heard a talk about Chow motives and also read Milne's exposition on motives. If I understand it correctly, the naive definition of the Chow ring would be that it simply consists of all algebraic cycles, but to define a multiplication one needs to impose a certain equivalence relation, either rational or numerical equivalence. 
I wondered myself if an alternative definition using derived algebraic geometry would be possible. Regardless which framework of derived algebraic geometry you use, a feature should be that you can get always the correct intersection/fiber product. Therefore, one might try to define a derived Chow ring by considering 'derived algebraic cycles' (without any equivalence relation). One would probably get a space out of this instead of a set, but this wouldn't necessarily be a bad thing. Also, the associated category of 'derived Chow motives' would then be a simplicial category (or $(\infty,1)$-category).
What I would like to know is the following: Has somebody tried to build such a theory and if not, what are the problems of it or why is it perhaps a bad idea right from the start?

REPLY [17 votes]: I can't give you a complete answer apart from saying that this is definitely a hard problem!
If you have a derived scheme $X$, you can always truncate to get an every-day scheme $t_0 (X)$. On the level of rings, this corresponds to truncating a simplicial ring $A$ to $\pi_0 (A)$. The canonical morphism $A \to \pi_0(A)$ is obviously 0-connected. A result of Waldhausen then tells you that $A$ and $\pi_0(A)$ have isomorphic $K_0$ and $K_1$. 
Now the Chow-Ring is more or less the same as $K_0$ (after tensoring with Q and up to rational equivalence). So I guess what this means is that on the level of cycles it is impossible to tell the difference between a derived scheme and an ordinary scheme, because $CH(X) = CH(t_0 (X)).$  
I think the fancy way of saying this is that the etale topoi of $X$ and $t_0(X)$ are the same, but that is slightly over my head. This result can be found both in Luries thesis and in HAG II by Toen-Vezzosi.
The situation immediately changes though if you work with Bloch´s higher Chow groups! They see the difference starting from $K_2$, since the higher Chow groups are tied to higher $K$-groups via a Riemann-Roch map. So on the level of these cycles you could tell a difference.<|endoftext|>
TITLE: Parabolic envelope of fireworks
QUESTION [38 upvotes]: The envelope of parabolic trajectories from a common launch point is itself a parabola.
In the U.S. soon many will have a chance to observe this fact directly, as the 4th of July is traditionally celebrated with fireworks.
If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows that the parabola
$$
y = x \tan \theta  - [x^2 /(2 v^2)] (1 + \tan^2 \theta)
$$
and the envelope of all such trajectories, is another parabola:
$$
y = v^2 /2 - x^2 / (2v^2)
$$

                


These equations are not difficult to derive.
I have two questions.
First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations?
Is there a purely geometric argument?
Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.
Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity.
Is the envelope of these elliptical trajectories also an ellipse?
(I didn't try to work out the equations.)
Perhaps the same geometric viewpoint (if it exists) could apply, e.g. by slightly tilting the sections.

REPLY [3 votes]: I found another article to supplement those to which Andrey linked:
Eugene I Butikov, "Comment on 'The envelope of projectile trajectories',"
Eur. J. Phys. 24 L5-L9, 2003.
He also explains the expanding-circles viewpoint that is Andrey's second bullet.
He imagines first that there is no gravity, in which case the particles are
on the surface of an expanding sphere whose radius $r$ equals $v t$.
"With gravity, this uniformly expanding sphere is falling freely as a whole with the
acceleration of free fall."  He then finds the envelope of these falling, expanding circles.
Later in the note he considers water drops spun off a spinning, wet bicycle wheel.
To continue the 4th-of-July theme, these could be sparks from a spinning sparkler wheel.
He proves that again, the envelope of the drops/sparks is a parabola.<|endoftext|>
TITLE: separated schemes
QUESTION [24 upvotes]: Here's a dangerous question:
We all know that a variety is an integral scheme, separated and of finite type over an algebraically closed field. Now, if I remove the separated hypothesis, I get the class of schemes which are made by gluing together (finitely many) classical affine varieties and applying the famous fully faithful functor $Var_k \longrightarrow Schemes_k$. Thus, separatedness must relate somehow to the "way of gluing" together these affine varieties, (which is kind of confirmed when you look at the uber-classical example of the double line). Is there someone here that can explain what way of gluing we ban when restricting to separated schemes ?

REPLY [16 votes]: The necessary and sufficient condition on a schemes $\{U_i\}$ and gluing isomorphisms $\varphi_{ij}:U_{ij} \simeq U_{ji}$ between opens $U_{ij} \subset U_i$ ($i, j \in I$) that the gluing $X$ be separated is that the graph map $U_{ij} \rightarrow U_i \times U_j$ defined by $u \mapsto (u, \varphi_{ij}(u))$ is a closed immersion (or equivalently, has closed image) for all $i, j$.  This is seen by intersecting $\Delta(X)$ with the opens $U_i \times U_j$ that cover $X \times X$. 
(Taking $i=j$, this says that all $U_i$ are separated, which is automatic when all $U_i$ are affine; in such cases the closed immersion condition forces all $U_{ij}$ to also be affine, so in the context of the question one loses nothing by requiring all $U_{ij}$ to be affine open prior to stating the closed immersion condition.)

REPLY [7 votes]: Basically it's the sort of gluing that would violate the (discrete) valuative criterion for separatedness.  See Chapter II, Section IV of Hartshorne for example.
For example, if you are gluing $U$ to $V$ (say separated noetherian schemes) along a common open subscheme $W$, if you want the result to be separated, you presumably want to require that:
For every spec of a DVR, $Z =$ { generic Pt, closed Pt } with a map $Z \to U$ such that,
the generic point of $Z$ is sent into $W$, then the closed point of $Z$ is also sent to $W$ as well whenever the map {generic Pt} $\to W$ extends to a map $Z \to V$.<|endoftext|>
TITLE: On the cohomology ring of the Grassmannian
QUESTION [6 upvotes]: The basis of Schubert classes for the cohomology ring $H^*(\text{Gr}(m,N))$ of the Grassmannian of $m$-dimensional subspaces of $\mathbb{C}^N$ is indexed by $L(m,N-m)$, the poset of all partitions fitting inside a $m \times (N-m)$ box.  This is the quotient of the powerset $2^{m(N-m)}$ by the action of the wreath product $S(m) \wr S(N-m)$.  How does this come from the fact that $\text{Gr}(m,N) \cong U(N) / \big (U(m) \times U(N-m) \big )$?  Can this be extended to other homogeneous spaces?

REPLY [4 votes]: The Schubert classes on $G/P$ are the classes of the Schubert varieties, which are the closures of the Schubert cells, each of which contains a unique $T$-fixed point. The $T$-fixed points on $G/P$ are the images of $T$-fixed points on $G/B$ (since $T$ acts on the fiber, which is a projective variety, hence itself has a $T$-fixed point by Borel's theorem).
Up on $G/B$, the $T$-fixed points are exactly of the form $N_G(T)B/B$, so indexed by the Weyl group $W_G = N_G(T)/T$. Down on $G/P$, they group together by the Weyl group $W_P = N_P(T)/T$, so they're indexed by $W_G/W_P$. Which is exactly what you observed in the $G/P =$ Grassmannian case.
(Actually you asked about compact groups, so $K/L$ where $K$ is compact and $L$ is compact of the same rank, which includes some cases like $S^4 = SO(5)/SO(4)$ that is not of the form $G/P$ for $G$ complex and $P$ parabolic. Then there's still a basis of "Schubert classes", indexed by $W_K/W_L$ similarly.)<|endoftext|>
TITLE: Does the Tannaka-Krein theorem come from an equivalence of 2-categories?
QUESTION [17 upvotes]: Possibly the correct answer to this question is simply a pointer towards some recent literature on Tannaka-Krein-type theorems.  The best article I know on the subject is the excellent

André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, Category Theory, Lecture Notes in Math, 1991 vol. 1488 pp. 412–492

from which I'm getting most of my facts.
Let me fix, once and for all, a field $\mathbb K$.  I will denote by $\text{Vect}$ the usual category of all $\mathbb K$-vector spaces, and by $\text{FinVect}$ its full subcategory of dualizable objects.  The notions of (coassociative, counital) coalgebra and (right, say, and coassociative and counital) comodule are standard; briefly, if $(\mathcal V,\otimes)$ is a monoidal category, a coalgebra in $\mathcal V$ is an object $A$ along with a map $A \to A \otimes A$ satisfying some axioms, and an $A$-comodule is an object $X$ and a map $X \to X\otimes A$ satisfying some axioms.  Here's the fast way to say the axioms.  Without telling you anything, it's clear what should be a homomorphism of comodules, namely a map $f: X\to Y$ so that the maps $X \overset f \to Y \to Y\otimes A$ and $X \to X\otimes A \overset{f\otimes {\rm id}}\to Y\otimes A$ agree.  Then a coalgebra $A$ is coassociative iff the comultiplication $A\to A\otimes A$ is a homomorphism of $A$-comodules, and a comodule $X$ is coassociative iff $X\to X\otimes A$ is a homomorphism of comodules.  (In both cases, the comodule structure on the right is just the comultiplication in the second factor.)  I will abuse language so that all coalgebras and comodules are coassociative.  Oh, and I haven't said anything about counits, but I want them as well.  Given a coalgebra $A$, I will denote its category of right comodules by $\text{Comod}_A$, and of finite-dimensional right comodules by $\text{FinComod}_A$.
Let's say that a Tannakian category is a $\mathbb K$-linear (i.e. $\text{Vect}$-enriched) abelian category $\mathcal C$ along with a faithful exact $\mathbb K$-linear functor $F: \mathcal C \to \text{FinVect}$.  (If this is not the most standard definition, let me know.)  The fundamental theorem is the following:

Theorem (c.f. Op. cit.): Let $F: \mathcal C \to \text{FinVect}$ be any functor from any category.  Then there is a coalgebra $\operatorname{End}^\vee(F) \in \text{Vect}$ given by a certain natural colimit (take the definition of the vector space of natural transformations and turn all arrows around; use the fact that the endomorphism of a finite-dimensional vector space is naturally a coalgebra).  The functor $F$ factors through $\text{FinComod}_{\operatorname{End}^\vee(F)}$ (and its forgetful functor to $\text{FinVect}$), and $\operatorname{End}^\vee(F)$ is universal with respect to this property (if $F$ factors through $\text{forget}: \text{FinComod}_A \to \text{FinVect}$, then there is a map $\operatorname{End}^\vee(F) \to A$ inducing this).  If $(\mathcal C,F)$ is Tannakian, then the map $\mathcal C \to \text{FinComod}_{\operatorname{End}^\vee(F)}$ is an equivalence of categories.

It follows from the above theorem that there is an equivalence of categories between:
$$ \{\text{Tannakian categories}, \text{strictly commuting triangles}\} \leftrightarrow \{\text{coalgebras}, \text{homomorphisms}\}$$
By a "strictly-commuting triangle" of Tannakian categories $(\mathcal C,F) \overset{f}\to (\mathcal D,G)$, I mean a functor $f: \mathcal C \to D$ so that we have strict equality $F = G\circ f$ as functors $\mathcal C \to \text{Vect}$.
But when talking about functors, etc., it's evil to think about strict equality.  Rather, there should be some two-categories floating around.  There is a natural two-category whose objects are coalgebras: the one-morphisms are bicomodules, the two-morphisms are homomorphisms of bicomodules, and the one-composition is cotensor product of bicomodules.  There are, to my mind, a few different two-categories whose objects are Tannakian categories: perhaps the one-morphisms should be triangles that commute up to specified natural isomorphism ("strong"?), or up to specified natural transformation in one way or in the other ("lax" and "oplax"?).
All together, my question is as follows:

Question: What is the correct, non-evil two-category of Tannakian categories so that the last two sentences of the "Tannaka theorem" above expresses an equivalence of two-categories
$\{\text{Tannakian categories}\} \leftrightarrow \{\text{coalgebras}, \text{bicomodules}, \text{homomorphisms}\}$?
Is this two-category a full sub-two-category of some larger two-category whose objects are pairs
$(\mathcal C, F:\mathcal C \to \text{Vect})$ with no further conditions?  If so, does the full theorem represent some sort of "Tannakaization" of such pairs?

If modifications need to be made (not quite the right two-category of coalgebras, not quite the right definition of Tannakian category for the question to have an answer, etc.) feel free to answer the corresponding question.

REPLY [5 votes]: Your question is dealt with (in a slightly more general setting) in section 11 of Daniel Schäppi's paper, Tannaka duality for comonoids in cosmoi. Specializing to your setting, he shows that there is a biadjunction (a weak 2-categorical form of adjunction) between the 2-category of $k$-linear categories equipped with a functor to $\operatorname{FinVect}$, where morphisms are triangles commuting up to specified natural isomorphism, and the usual category of coalgebras (thought of as a 2-category with only identity 2-morphisms). I believe this biadjunction should restrict to a biequivalence on the sub-2-category of Tannakian categories (as you have defined them).
This biadjunction is useful because there is a nice tensor product on the 2-category of $k$-linear categories that turns this biadjunction into a monoidal biadjunction, which gives us a way of relating things like bialgebras and tensor categories.<|endoftext|>
TITLE: The Wedge Sum of path connected topological spaces
QUESTION [9 upvotes]: A definition of wedge sum can be found here:
http://en.wikipedia.org/wiki/Wedge_sum
My professor has claimed that wedge sums of path connected spaces X and Y are well-defined up to homotopy equivalence, independently of choice of base points x0 and y0.  Base point here means the points that are identified under the equivalence relation forming the wedge product out of the disjoint union topology of X and Y.
Recall homotopy equivalence of X and Y means that there is f:X->Y and g:Y->X continuous with gf and fg homotopic to the identity.
With these definitions, please prove my professor's claim, which I have failed to do for a week. (It is left as an exercise in his lecture.)
Thanks.

REPLY [19 votes]: A counterexample is shown on the cover of the paperback edition of the classic textbook Homology Theory by Hilton and Wylie.  This can be viewed on the amazon webpage for the book.  The example consists of the wedge of two copies of a cone, the cone on the sequence 1/2, 1/3, 1/4, ... together with its limit point 0.  With one choice of basepoints the wedge is not contractible, but with other choices it is.<|endoftext|>
TITLE: Strictly convex equivalent norm
QUESTION [9 upvotes]: Does any Banach space admit an equivalent strictly convex norm (i.e. such a norm, that a unit sphere does not contain segments)?

REPLY [10 votes]: Every separable Banach space has an equivalent norm which is both strictly convex and smooth. For certain nonseparable spaces, in particular, $\ell_{\infty}(\Gamma)$ with $\Gamma$ uncountable, there may be no equivalent strictly convex or smooth norm. 

Link.<|endoftext|>
TITLE: Counting the Groups of Order n Weighted by 1/|Aut(G)| 
QUESTION [28 upvotes]: Idle question:  
Let $g(n)$ be the sum, over all isomorphism classes of groups of order $n$, of $\frac{1}{|Aut(G)|}$ where $G$ is a group in the class. Thus $g(n)n!$ is the number of group laws on a fixed set of size $n$. Is anything known about the asymptotic behavior of this quantity? I could easily believe that abelian groups account for most of it. If we only count abelian groups, calling the analogous number $a(n)$, then the function $a$ is clearly multiplicative in the sense that $a(mn)=a(m)a(n)$ when $m$ an $n$ are relatively prime, and I believe that the function $a(p^k)$ can be written as an explicit function of $k\ge 0$ and the prime $p$: $a(p)=\frac{1}{p-1}$, $a(p^2)=\frac{p}{(p-1)(p^2-1)}$, $a(p^3)=\frac{p^3}{(p-1)(p^2-1)(p^3-1)}$.
ADDED: So it looks like $a(p^k)=p^{\frac{(k(k-1)}{2}}\prod_{1\le j\le k}(p^j-1)^{-1}$. (I checked it up to $k=4$.) If you let $m\ge k$ and use the fact that every abelian group of order $p^k$ is isomorphic to a subgroup of $(\mathbb Z/p^m)^k$ and the fact that every isomorphism between two such subgroups is induced by an automorphism of $(\mathbb Z/p^m)^k$, you can interpret this as saying that the sum, over all automorphisms $g$ of $(\mathbb Z/p^m)^k$, of the number of subgroups of order $p^k$ in the fixed set of $g$, is a certain power of $p$. But I can't think of a reason why that should be true.

REPLY [15 votes]: Higman and Sims showed that the number of groups of order pm is somewhere around p2m3/27, up to some smallish correction that I cant remember offhand. This is vastly bigger than the size of the typical automorphism group of such groups, so dividing by the automorphism group does not make much significant difference to this number.
The number of groups of order at most N is dominated by those of order a power of 2. (The next most common are those of order 3 times a power of 2).  The number of groups of given order N has been worked out exactly for N up to about 2000.<|endoftext|>
TITLE: Evaluation of a combinatorial sum (that comes from random matrices)
QUESTION [16 upvotes]: I'm looking for an elementary combinatorial/generating function/etc proof of the following result:
For nonnegative integers $r$,
$$\frac{1}{r!} = \sum_{p_0+p_1+\cdots = r} \frac{1}{(p_0!)^2(p_1!)^2\cdots{p_0+p_1+1\choose 1}{p_1+p_2+2\choose 2}{p_2+p_3+3\choose 3}\cdots}.$$
Here the sum is over all sequences of nonnegative integers $(p_0,p_1,...)$ that sum to $r$. (Only finitely many terms in each such sequence will be nonzero.) 
It is related to a result of Diaconis and Shahshahani that the trace of a random unitary matrix (with probability measure being given by the Haar measure) is distributed like a Gaussian variable, and indeed can be proven using this result, but I had initially hoped to proceed in the other direction. The above sum, after all, can be evaluated for specific $r$ by inspection (although this rapdily becomes a bit tedious for $r > 2$), and it ought to be possible to somehow summarize this information in a general.
Edit:
Alternatively phrased, we want
$$e^x = \sum_{p_0,p_1,.. = 0}^\infty \frac{x^{p_0+p_1+\cdots}}{\left(\prod_{j \geq 0}(p_j!)^2\right)\cdot\left(\prod_{k\geq 1}{p_{k-1}+p_k+k\choose k}\right)} = \lim_{\lambda \rightarrow \infty} \sum_{p_0,p_1,.. p_\lambda = 0}^\infty \frac{x^{p_0+p_1+\cdots + p_\lambda}}{\left(\prod_{j=0}^\lambda(p_j!)^2\right)\cdot\left(\prod_{k=1}^\lambda{p_{k-1}+p_k+k\choose k}\right){p_\lambda+\lambda+1\choose \lambda+1}}$$

REPLY [5 votes]: I may as well write down how I've been attacking this, although I don't have a solution yet. Sequences $(p_i)$ of nonnegative integers with sum $r$ are in bijection with weakly increasing $r$-tuples $(n_1, n_2, \ldots, n_r)$ of positive integers. Specifically, given the sequence $(p_0, p_1, p_2, \ldots)$, form the sequence of partial sums $(q_1, q_2, \ldots)$ given by $q_i:=p_0+p_1+\ldots+p_{i-1}$. Let $n_j$ be the minimal $i$ for which $q_i \geq j$. For example, $(1,0,0,1,0,0,2,0,0,\ldots)$ corresponds to $(1, 4, 7,7)$. So, we want to compute the sum over all weakly increasing $r$-tuples and prove it is equal to $1/r!$. 
For each weakly increasing $r$-tuple, let us sum instead over all $r!$ permutations of the $r$-tuple. So we can view our sum as being over all $r$-tuples of nonnegative integers, and we want to prove now that the sum is $1$. One difficulty is that some $r$-tuples will appear more than once. For example, $(1,4,7,7)$ will appear twice, because the permutation which switches the $7$'s stabilizes this quadruple. It turns out that the multiplicity of an $r$-tuples is precisely $\prod (p_i)!$. So, what we want to show is that
$$\sum_{n_1, n_2, \ldots, n_r \geq 0} \frac{1}{\prod (p_i)! \prod \binom{p_{k-1}+p_k+k}{k}} =1$$
Now, when none of the $n_i$ are equal to each other, and when none of them differ by $1$, the summand is
$$\frac{1}{n_1(n_1+1)n_2(n_2+1) \cdots n_r(n_r+1)} = \left( \frac{1}{n_1} - \frac{1}{n_1+1} \right) \cdot \left( \frac{1}{n_2} - \frac{1}{n_2+1} \right) \cdots \left( \frac{1}{n_r} - \frac{1}{n_r+1} \right).$$
This is set up beautifully to telescope. If I could just find a similar nice description for when the $n_i$ collide or are adjacent ...<|endoftext|>
TITLE: Particle Physics and Representations of Groups
QUESTION [28 upvotes]: This question is asked from a point of complete ignorance of physics and the standard model.
Every so often I hear that particles correspond to representations of certain Lie groups. For a person completely ignorant of anything physics, this seems very odd! How did this come about? Is there a "reason" for thinking this would be the case? Or have observations in particle physics just miraculously corresponded to representation theory? Or has representation theory of Lie groups grown out of observations in particle physics?
In short: what is the chronology of the development of representation theory and particle physics (with relation to one another), and how can one make sense of this relation in any other way than a freakish coincidence?

REPLY [4 votes]: As a physicist, I may be able to give a different perspective on this question.  In particular, many of the responses so far have been about quantum mechanics and quantum field theory (which involve Lie groups), but if the question is, "Why is the particle content of physics theories derived from Lie groups?" then the answer is not specifically about the theories' quantumness.  It's about their geometry, which can be discussed separately from quantum effects.
In 1926, Kaluza and Klein attempted to unify electromagnetism with gravity by proposing a theory of General Relativity with 5 dimensions (4 spatial).  Since we don't macroscopically experience this extra degree of freedom, they proposed that it is topologically like a cylinder with a small radius, so small that the extra degree of freedom can't be probed as a direction.  This degree of freedom does, however, allow us to encode classical electromagnetism as part of the geometry of space-time.  We'll see in a moment that while this formulation isn't exactly right, it does show how the differential geometry concepts of General Relativity can be used in particle physics theories, leading to a unification of all four forces at a classical level.  (It's the quantization of gravity that's the hard part.)
The Lagrangian of quantum electrodynamics (late 1940's) is just the Lagrangian of the Dirac equation with an additional requirement: that the spinor field $\psi$ has a local $\mathcal{U}(1)$ symmetry.  I'll use the same notation as the Wikipedia article, except that I'll use $c = \hbar = 1$.  The Dirac Lagrangian
$\mathcal{L_D} = m\bar{\psi}\psi - \frac{i}{2}\left(\bar{\psi} \gamma^\mu (\partial_\mu\psi) - (\partial_\mu\bar{\psi}) \gamma^\mu \psi \right)$ (1)
has a global $\mathcal{U}(1)$ symmetry in that the complex phases of components of $\psi$ cancel in the $\bar{\psi}\psi$ terms: multiplying all instances of $\psi$ by $e^{i\alpha}$ for some constant $\alpha$ would not change the value of $\mathcal{L}$.  The Dirac equation does not have a local $\mathcal{U}(1)$ symmetry, that is, invariance under
$\psi(x,t) \to e^{i\theta(x,t)} \psi(x,t)$ where everything is a function of 4-D space-time points $(x,t)$ (2).
If we want to create a new Lagrangian which does have a local $\mathcal{U}(1)$ symmetry, we find that we would need to replace the derivative operators $\partial_\mu$ with
$D_\mu = \partial_\mu - iqA_\mu$ (3)
where $A$ is a new field with the transformation property
$A_\mu(x,t) \to A_\mu(x,t) + \frac{1}{q}\partial_\mu \theta(x,t)$ (4).
The new theory has a Lagrangian
$\mathcal{L_{QED}} = m\bar{\psi}\psi - \frac{i}{2}\left(\bar{\psi} \gamma^\mu (D_\mu\psi) - (D_\mu\bar{\psi}) \gamma^\mu \psi \right) + \frac{1}{4}\left((\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu)\right)$ (5)
where the last term is required to preserve symmetry under Lorentz boosts (conservation of energy in the new $A$ field).  Just following the consequences of a local $\mathcal{U}(1)$ symmetry, we have turned freely-streaming Dirac Lagrangian into the interacting electromagnetic Lagrangian, where we can interpret $\psi$ as charged particle (e.g. electron) waves and $A$ as the vector potential of electromagnetism, which is to say, the photon waves.  The transformation of Eqns (2) and (4) is the gauge transformation of electromagnetism: we've learned that the electromagnetic gauge symmetry is fundamentally a local $\mathcal{U}(1)$ symmetry.
Getting back to Kaluza and Klein's theory, a 5th compactified dimension is a little like having a $\mathcal{U}(1)$ invariance at every point in 4-D space, since it's hard to see where we are in the loop of the 5th dimension.  It's not exactly the same thing: with an extra dimension, we should in principle be able to perform rotations in which spatial dimensions and the extra dimension mix, while that would not be possible in a 4-D space plus $\mathcal{U}(1)$ fiber bundle.  (This difference is perhaps related to the reason Kaluza and Klein's original theory didn't work...?)  If we generalize our notion of space-time to include the $\mathcal{U}(1)$ fibers, we can think about electromagnetism and General Relativity in the same terms.  For instance, the photon field $A$ plays the same role in the $\mathcal{U}(1)$ symmetry as the connection/covariant derivative in the local Lorentz symmetry of the space-time metric.  That is, the classical photon field is the "curvature" of the fiber bundle in the same sense that gravitation is the curvature of space-time.
Moreover, this picture unifying the geometry of electromagnetism with the geometry of gravity also works for all the other known forces.  In 1954, Yang and Mills generalized the "local $\mathcal{U}(1)$-to-electromagnetism" idea to work for any Lie group, including non-Abelian ones.  The Yang-Mills idea wasn't popular at first because it didn't seem to describe the nuclear strong force (but that was based on a wrong assumption that the nuclear force is a Yukawa interaction).  By the late 1960's, Weinberg derived a unified electro-weak theory from local $\mathcal{SU}(2)\times\mathcal{U}(1)$, and Han and Nambu derived a theory of nuclear strong force from $\mathcal{SU}(3)$.  (I'm skipping over many important contributions for brevity.)  By the mid-1970's or early 1980's, depending on who I ask, this became known as the Standard Model of particle physics because of its experimental success.
We can think about the Standard Model geometrically as an $\mathcal{SU}(3)\times\mathcal{SU}(2)\times\mathcal{U}(1)$ at every point in 4-D space-time, with the gluon, W and Z bosons, and photon being connections through groups at neighboring points of space-time, constantly arranging themselves to hide information about the components of matter fields in all of these "internal" degrees of freedom.  The structures of the groups are directly responsible for the charges and interactions of the matter fields (quarks and leptons), but the matter fields themselves are not derived from the groups (supersymmetry might change that part of the picture).  There is a direct analogy between these group connections (the gluon, W, Z, and photon) and the space-time connection in General Relativity (which we could call a graviton field, if you wish).  I have said nothing at this point about the quantization of all of these fields, which further complicates the picture, especially in the case of gravity!
By the way, I would love to know more about the curvature of fiber bundles, in order to understand the above at a deeper mathematical level.  If you have any suggested reading, I'm interested.  Thanks!<|endoftext|>
TITLE: F(0) = 0? F: additive functor
QUESTION [7 upvotes]: If I define an additive functor to be a functor on abelian categories such that the action of $F$ on ${\rm Hom}(A,B)$ is a group homomorphism, do I necessarily have that $F(\text{zero object}) = \text{zero object}$?

REPLY [16 votes]: Since the OP asked for a detailed answer:
Let $A$ be an object of an abelian (or additive) category. Then $A$ is a zero object if and only if the zero endomorphism is the identity endomorphism (and then $Hom(A,A)$ is the zero ring). If $F$ is any functor, it sends the identity morphism of $A$ to the identity morphism of $F(A)$. If in addition $F$ is additive (no pun intended), it also sends the zero morphism to the zero morphism. Thus $F(0)$ is a zero object.
Another exercise in the similar spirit: show that additive functor is additive on objects: it sends finite direct sums to direct sums. (Your question is the particular case of empty direct sum.)<|endoftext|>
TITLE: Why do Physicists need unitary representation of Kac-Moody algebra?
QUESTION [11 upvotes]: My advisor mentioned to me that he talked to Witten last summer on representation theory, and Witten told him that unitary representations of Kac-Moody algebra are important to working physicists. But he did not explain in details to me. 
My question is Why? 
Second question: I know there are a lot of people devoting to studying unitary representation of Lie group. But are there papers investigating unitary representations of Kac-Moody algebra?
I am not an expert, so this question might be naive. Thanks

REPLY [9 votes]: As others have mentioned, the reasons lie indeed in two-dimensional conformal field theory and in string theory.
The propagation of string on a compact Lie group $G$ is described by the Wess-Zumino-Witten model, whose dynamical variables are maps $g:\Sigma \to G$ from a riemann surface $\Sigma$ to $G$.  The quantisation of that model is difficult in terms of $g$ (although see the 1988 papers of Gawedzki and Kupiainen, and also Felder, for a functional integral approach) and one instead chooses to quantise their currents, roughly the (anti)holomorphic components of the pullbacks $g^*\theta_L$ and $g^*\theta_R$ of the Maurer-Cartan forms on $G$.   There is a natural action of two copies of the affine Kac-Moody algebra associated to $G$ on the WZW model which preserves the Poisson structure of the WZW model and gives rise to moment mappings which are, essentially, the currents.  In other words, the Poisson bracket of the currents is that of two copies of the affine Kac-Moody algebra of $G$.  Hence the quantisation naturally leads one to consider unitary, integrable representations of the affine Kac-Moody algebra.  The first "modern" reference for this is a 1986 paper of Doron Gepner and Edward Witten String Theory on Group Manifolds; although there are pioneering papers of Halpern, Bardakci,... in from the late 1960s and early 1970s.
At a more abstract level, we can substitute the group $G$ by any (unitary) two-dimensional (super)conformal field theory with the right central charge.  This idea of replacing the geometry by a conformal field theory used to be known as "strings without strings", since one loses the description of strings propagating in some geometry.  In this context, it is important to have at one's disposal a number of unitary two-dimensional conformal field theories.  The natural ones are those coming from from unitary representations of infinite-dimensional Heisenberg and Clifford algebras (so-called free fields) and unitary integrable representations of affine Kac-Moody algebras, but one can also consider constructions (e.g., orbifolds, coset constructions,...) which generate new unitary CFTs from these ones.  The first "modern" reference for the coset construction is perhaps the 1986 paper of Peter Goddard, Adrian Kent and David Olive Unitary representations of the Virasoro and superVirasoro algebras.
Finally, I should say that although it's the affine Kac-Moody algebras which seem to have played the more important rôles thus far, there is also the emergence (in the context of M-theory) of more general Kac-Moody algebras.  There's work on this in King's College London (West et al.), Brussels (Henneaux et al.) and Potsdam (Nicolai et al.).  I'm not very familiar with this, though.<|endoftext|>
TITLE: A puzzling question on real interpolation
QUESTION [6 upvotes]: Suppose an operator $T$ is bounded on $L^2$ and also bounded from $L^{1}$ to $L^{1}$-weak. Then by Marcinkewicz interpolation one gets that $T$ is bounded on every $L^{p}$ for p between 1 and 2. Precise versions of the theorem (see the book of L.Grafakos) give an estimate of the norm of $T$ on $L^{p}$, and of course the norm diverges as $p\to 1$. If not, by a simple argument one could obtain that $T$ is bounded also on $L^{1}$, which is clearly false in general (e.g. singular integrals). By the way, even a weaker assumption on the growth of the norm should allow one to conclude that $T$ is bounded on $L^{1}$, as in extrapolation theorems.
Now suppose you (me) are hard-headed and want to use the general machinery of real interpolation, say the K-method as detailed in the book of Bergh and Löfström. Then $T$ is bounded from $L^{1}$ to $L^{1}$-weak which means the Lorentz space $L^{1,\infty}$, with norm $M_{0}$, and also from $L^{2}$ to $L^{2}$ with norm $M_{1}$. Then $T$ is bounded on the corresponding real interpolation spaces with norm $M_{0}^{1-\theta}M_{1}^{\theta}$. Real interpolates of Lorentz spaces are Lorentz spaces, see Theorem 5.3.1 in B-L. We conclude that $T$ is bounded on every Lorentz space $L^{p,q}$ with p between 1 and 2, and $1\le q\le \infty$, and in particular on $L^{p,p}=L^{p}$ as expected.
Unfortunately, now I have a uniform bound on the norm of $T$ as $p\to 1$, which would allow me to conclude that $T$ is also bounded on $L^{1}$. Where is the mistake? There must be some inaccuracy in one of the steps, but which one exactly?

REPLY [8 votes]: (Note: I don't have my copy of B&L handy, so I'm sort of doing this from memory.)
The problem is that $(L_{p_0.q_0},L_{p_1,q_1})_{\theta,q} = L_{p_\theta,q}$ under equivalent norm. There's no saying what the constant of equivalency are. Now, if you look at $T: \{X_1,X_2\} \to \{Y_1,Y_2\}$ with norm $\{M_1,M_2\}$, then you have 
$$ T: (X_1,X_2)_{\theta,q} \to (Y_1,Y_2)_{\theta,q} $$
with norm $M_1^{(1-\theta)}M_2^\theta$. Now put in $X_1 = L_{1,1}$, $X_2 = L_{2,2}$, $Y_1 = L_{1,\infty}$ and $Y_2 = L_{2,2}$, while the interpolants $(X_1,X_2)_{\theta,q}$ may be equivalent to $L_{p_\theta,q}$, and $(Y_1,Y_2)_{\theta,q}$ to $L_{p_\theta,q}$, the constants of equivalency may be different. That's where the degeneracy is hiding.
Observe that if $Y_1$ were $L_{1,1}$ (or if $X_1 = L_{1,\infty}$) instead, then the two interpolation spaces are actually equal, which is what you expect.<|endoftext|>
TITLE: étale fundamental group of projective space
QUESTION [7 upvotes]: What is the étale fundamental group of projective space over an algebraically closed field?
In char = 0 it is trivial (Lefschetz principle), as well as in dimension 1 (Riemann-Hurwitz).

REPLY [11 votes]: It is a birational invariant (for smooth proper connected schemes over a field, ultimately due to Zariski-Nagata purity of the branch locus), and its formation is compatible with products (for proper connected schemes over an algebraically closed field), so we can replace projective $n$-space with the $n$-fold product of copies of the projective line to conclude.  Likewise, due to limit arguments and invariance of the etale site with respect to finite radiciel surjections (such as a finite purely inseparable extension of a ground field), it suffices to take the ground field to be separably closed rather than algebraically closed. This is all in SGA1.<|endoftext|>
TITLE: Is Li(x) the best possible approximation to the prime-counting function?
QUESTION [42 upvotes]: The Prime Number Theorem says that $\lim_{n \to \infty} \frac{\pi(n)}{\mathrm{Li}(n)} = 1$, where $\mathrm{Li}(x)$ is the Logarithm integral function $\mathrm{Li}(x) = \int_2^x \frac{1}{\log(x)}dx$. It is also the case that $\lim_{n \to \infty} \frac{\pi(n)}{n/\log(n)} = 1$, as $\mathrm{Li}(x)$ and $\log(x)$ are asymptotically equivalent. However it seems that $\mathrm{Li}(n)$ is a better approximation to $\pi(n)$ (the Mathworld article states that this has been proven, I don't know in what precise sense).
There are also results for the absolute difference between $\pi(n)$ and $\mathrm{Li}(n)$; for instance $\pi(n) - \mathrm{Li}(n)$ is known to change sign infinitely often. We also know that the Riemann Hypothesis is equivalent to
$$ \pi(n) - \mathrm{Li}(n) \in O \left (\sqrt{n} \log(n) \right ).$$
In addition, Riemann showed that we have
$$\pi(n) = \mathrm{Li}(n) - \frac{1}{2} \mathrm{Li}\left ( \sqrt{n} \right ) - \sum_{\rho} \mathrm{Li}(x^\rho) + \text{lower order terms}$$ where $\rho$ runs over all the nontrivial zeroes of the Riemann zeta function.
Question: Is there a sense in which $\mathrm{Li}(n)$ is the best possible approximation to $\pi(n)$? Ideally, there would be some Bohr-Mollerup type theorem: $\mathrm{Li}(n)$ is uniquely characterized as being a good approximation to $\pi(n)$ which has some properties, such as analyticity and negative second derivative. Probably $\mathrm{Li}(n)$ isn't the best possible, for instance $\mathrm{Li}(n) - \frac{1}{2} \mathrm{Li}\left ( \sqrt{n} \right )$ might be better? Riemann also suggested
$$\sum_{n \geqslant 1} \frac{\mu(n)}{n} \mathrm{Li} \left (x^\frac{1}{n}\right ),$$ where $\mu(n)$ is the Möbius function.

REPLY [10 votes]: Whether for a finite set $\mathcal{R}$ of roots the approximation
$$
\pi(x)\approx\mathrm{Li}(x)-\frac{1}{2}\mathrm{Li}(x^{1/2})-\sum_{\rho\in\mathcal{R}}\mathrm{Li}(x^\rho)
$$
is "on average" better then $\pi(x)\approx\mathrm{Li}(x)-\frac{1}{2}\mathrm{Li}(x^{1/2})$ depends on the precise notion of average. Since $x^\rho=x^\sigma e^{i\gamma\log x}$, it might appear more natural to consider $\pi(e^t)$ instead of $\pi(x)$. Things are somewhat easier for $\Psi$, and there we get under RH
$$
\lim_{T\rightarrow\infty}\frac{1}{T}\int_2^T\left(\Psi(e^t)-\Big(e^t-\sum_{\rho\in\mathcal{R}}\frac{e^{t\rho}}{\rho}\Big)\right)^2\frac{dt}{e^{t/2}} = \sum_{\rho\not\in\mathcal{R}}\frac{1}{|\rho|^2}
$$
(insert explicit formula, throw away terms of small order, square out, interchange integral and summation). So after the proper rescaling including some roots actually does reduce the mean error.<|endoftext|>
TITLE: Deligne's proof of Ramanujan's conjecture
QUESTION [18 upvotes]: I am trying to understand Deligne's proof of the Ramanujan conjecture and more generally how one associates geometric objects (ultimately, motives) to modular forms.
As the first step, which I understand more or less, one identifies the space of cusp forms $S_k(\Gamma)$ with the first cohomology group $W$ of $X(\Gamma)$ with coefficients in some sheaf (Deligne calls W parabolic cohomology). If we assume the weight is equal to 2, then this parabolic cohomology would probably just become $H^1(X(\Gamma), \mathbb C)$.
After that point I unfortunately understand practically nothing in Deligne's paper (modular forms and l-adic cohomology), so maybe someone could give an informal explanation:

What are the next steps in order to identify the Dirichlet series corresponding to a cusp form $f$ with a Hasse-Weil series of a motive (is this what we are doing?)
Also, what role does the Hecke action play here (in particular, how is some adelic gadget that Deligne calls the Hecke action related to the usual one)? 
And what role does the field $\mathbb Q_f$ constructed by adjoining the coefficients of $f$ play?

Thanks a lot.

REPLY [11 votes]: Deligne's construction works as follows. He identifies the space $GL_2(\mathbb{A})/GL_2(\mathbb{Q})$ with the $\mathbb{C}$-points of the variety* $\mathcal{M}_{ell,level}$ parameterizing elliptic curves with complete level structure, or equivalently, the moduli space of elliptic curves up to isogeny with complete (in the appropriately modified sense) level structure. The latter perspective has the advantage that there's an obvious $GL_2(\mathbb{A}_f)$-action. Furthermore, this moduli perspective has the advantage that both the variety and this action are defined over $\mathbb{Q}$.
Since this action is defined over $\mathbb{Q}$, there are commuting (!) actions of $\operatorname{Gal}(\mathbb{Q})$ and $GL_2(\mathbb{A}_f)$ on the (e.g., parabolic, with coefficients in $Sym^k$ of the Tate module of the tautological elliptic curves) cohomology of $\mathcal{M}_{ell,level}$. Therefore, for a prime $p$, there are commuting actions of the Hecke algebra at $p$ on the $G(\mathbb{Z}_p)$-invariants and the Galois group. Running this over all primes over which your eigenform $f$ is unramified, you get a commuting action of a big Hecke algebra and the Galois group. Finally, $f$ determines a maximal ideal of this big Hecke algebra being an eigenform, so you can take the "$f$-isotypic" component of this cohomology.
The first theorem of Eichler and Shimura tells you that this isotypic component has dimension $2$. Then the Eichler-Shimura relation tells you that the Hecke-eigenvalues correspond to the traces of Frobenius as predicted by Langlands et al. This matching of eigenvalues/traces of Frobenius exactly means that the L-function of this modular form match the Artin L-function of the representation.
Deligne also gives a modular-type description of the action of these Hecke operators because knowing that the Hecke operators are given by correspondences (so within the world of motives) allows him to apply the Weil conjectures on how e.g. constant sheaves behave under motivic-type operations. From here he deduces the Ramanujan conjecture.
* Actually, it's somewhat better to think of it as a provariety. Then notions of compactly supported etale cohomology are actually defined, and from this perspective cohomology of $Sym^k$ of the tautological $\mathbb{Z}_{\ell}$-sheaf on this space given as the $\ell$-adic Tate module of the canonical elliptic curve (which is trivial on the variety) is not just the direct sum of $k+1$ copies of cohomology of $\mathbb{Z}_{\ell}$.<|endoftext|>
TITLE: quasi-isomorphism
QUESTION [5 upvotes]: Is the distinction between quasi-isomorphism and `weak homotopy equivalence'
ONLY that the first means inducing an isomorphism in homology
and the second to an isomorphism of homotopy groups?

REPLY [4 votes]: This is a terminology question.
I think that "weak homotopy equivalence" is mostly used for maps between topological spaces (as opposed to "weak equivalence", that is used in the much broader context of model categories).
The term "quasi-isomorphism", on the other hand, is typically used for (co)chain complexes,
or (co)chain complexes equipped with extra structure.
So I would say that the main difference between those two terms is that they are used in different contexts.<|endoftext|>
TITLE: t-structures and higher categories?
QUESTION [30 upvotes]: I'm curious to find out where the viewpoint of higher categories may be useful so here is a somewhat vague question (which may or may not have a reasonable answer).
Given a triangulated category, one can consider the set of all possible t-structures on it. Simple examples where one can compute things by hand indicate that this is something complicated but not hopelessly so. See for example the paper http://arxiv.org/abs/0909.0552 by Jon Woolf which describes a three parametric family of t-structures on the constructible bounded derived category of $\mathbf{P}^1(\mathbf{C})$ stratified by a point and its complement. Some of these t-structures are more interesting than others and there is one that is the most interesting of them all since by taking the bounded derived category of its heart one gets back the triangulated category one started with. (For that t-structure the heart is the category of perverse sheaves on $\mathbf{P}^1(\mathbf{C})$.)
On the other hand, the set of t-structures on a triangulated category is interesting since there lurks somewhere the conjectural motivic t-structure whose existence implies Grothendieck's standard conjectures. See the recent paper http://arxiv.org/abs/1006.1116 by Beilinson.
On the triangulated categories page of the n-category lab website it says "Therefore, all the structure and properties of a triangulated category is best understood as a 1-categorical shadow of the corresponding properties of stable (infinity,1)-categories". See http://ncatlab.org/nlab/show/triangulated+category. Note that this is quite a strong statement, since it is referring to all, and not just some, properties and structure of a triangulated category.
So I'd like to ask: is there a higher categorical analog of a t-structure? More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable $(\infty,1)$ category?
upd: as Mike points out in the comments, the answer to the first question is yes and it is given by proposition 6.15 of Lurie's Stable Infinity Categories. The second, more "philosophical" question remains.

REPLY [22 votes]: I wanted to contribute something because nobody's really explained why a t-structure on a stable $\infty$-category is the same thing as a t-structure on its homotopy category. It's not just because someone defined it as such in Definition 1.2.1.4 of Higher Algebra; it's because the most natural generalization is no generalization at all.
Roughly speaking, a t-structure on a triangulated category should have the following three properties:

h1. It determines a full subcategory $\mathcal D_{\geq 0}$ closed under the shift functor $\Sigma$.
h2. $hom(X,Y) = 0$ if $X \in \mathcal D_{\geq 0}$ and $Y \in \mathcal D_{\leq -1}$.
h3. Any object fits into an exact triangle, sandwiched between an object in $\mathcal D_{\geq 0}$ and $\mathcal D_{\leq -1}$. 

If you think about the algebraic manipulations for which one utilizes t-structures, the natural generalization of a t-structure to a stable $\infty$-category ought to be as follows:

1$.$ It determines a full subcategory $\mathcal C_{\geq 0}$ closed under the shift functor $\Sigma$.
2$.$ $hom(X,Y)$ is contractible if $X \in \mathcal C_{\geq 0}$ and $Y \in \mathcal C_{\leq -1}$.
3$.$ Any object fits into an fiber sequence, sandwiched between an object in $\mathcal C_{\geq 0}$ and $\mathcal C_{\leq -1}$. 

If $\mathcal C$ is stable, one defines a triangulated structure on its homotopy category $\mathcal D = ho \mathcal C$ by demanding that its exact triangles are precisely those arising from fiber sequences. Then we see immediately that (1)-(3) imply (h1)-(h3). That is, we have a map from "oo-categorical" t-structures on $\mathcal C$ to t-structures on $ho \mathcal C$.
What might be surprising is that the other direction holds, but it's actually an easy exercise to show that any collection of data satisfying (h1)-(h3) uniquely determines data satisfying the properties (1)-(3). In a nutshell: (h1) implies (1) obviously. (h1) and (h2) together imply (2) because you can write a fiber sequence for $X[i]$ and $X[i+1]$, then apply the hom functor to get a fiber sequence of mapping spaces. The long exact sequence of homotopy groups will prove that every homotopy group of hom(X,Y) vanishes. Finally, by definition of exact triangles for $ho\mathcal C$, (h3) also implies (3).
So we have an interesting phenomenon where the set of t-structures can't distinguish between two oo-categories with equivalent homotopy categories. This is one of the main reasons that Bridgeland stability conditions don't have an obvious generalization to see higher-homotopical structures beyond the homotopy category.<|endoftext|>
TITLE: Is it possible a variety be a manifold with boundary
QUESTION [5 upvotes]: As a complex affine variety or projective variety, is it possible it is a manifold with boundary?

REPLY [7 votes]: No, this is not possible (unless you allow the boundary to be empty). If $X$ is a complex algebraic variety (affine or projective, this doesn't matter), there are two possibilities. If $X$ is smooth (as an algebraic variety), then $X$ is a smooth manifold with empty boundary. Otherwise let $Y$ be the singular locus of $X$. Then $X\setminus Y$ is a smooth variety and hence a smooth manifold with empty boundary. On the other hand, the \emph{real} codimension of $Y$ in $X$ is at least two (because the complex codimension is at least one). However, if $M$ is a (real) manifold with nonempty boundary, it is not possible to find a subset $Y$ of $M$ that has real codimension two such that $M\setminus Y$ has empty boundary. (The boundary $\partial M$ of $M$ is either empty or has real codimension one in $M$.) This implies that $X$ is not a manifold with boundary.<|endoftext|>
TITLE: Symmetric polynomials theorem
QUESTION [7 upvotes]: Hello all, I would appreciate comments on the following question:
A main theorem of symmetric functions might be formulated: Let k be a field of char. 0. Then  $k[x_1,...,x_n]^{S_n} = k[s_1,...,s_n]$, i.e. symmetric polynomials can be written as polynomials in the elementary symmetric polynomials. Moreover, $s_1,...,s_n$ satisfy no polynomial relations.
I want to see how to prove it using Galois theory. I thus consider the field $M=k(x_1,...,x_n)$ and its subfields $K=k(s_1,...,s_n)$ and $L$, the subfield of symmetric functions. Thus $K \subset L \subset M$. I then consider the polynomial $G(t)=(t-x_1)...(t-x_n)$. It has coefficients in $K$. $M$ is the splitting field of $G$ over $K$. Hence $[M:K] \leq n!$. From this we already see that $s_1,...,s_n$ satisfy no polynomial relations. On the other hand, $M$ has $n!$ different automorphism over $L$, which are permuting the $x_i$. Hence from Galois theory we can conclude that $L=K$.
My question is: How can I deduce the claim $k[x_1,...,x_n]^{S_n} = k[s_1,...,s_n]$ from the corresponding one for rational functions: $L=K$.
Thanks.

REPLY [6 votes]: I hope the following works. Let $A=k[x_1,\ldots,x_n]^{S_n}$, and let $B=k[s_1,\ldots,s_n]$. The polynomial algebra $k[x_1,\ldots,x_n]$ is an integral extension of $B$, and hence, a fortiori, $A$ is integral over $B$. I think your argument proves that $A$ and $B$ have the same fraction field. However, since $B$ is (isomorphic to) a polynomial algebra, it must be integrally closed, whence $A=B$.<|endoftext|>
TITLE: Grothendieck spectral sequence and Mayer-Vietoris sequence
QUESTION [7 upvotes]: Suppose $U'\cup U''=X$ is an open cover $U$ of a topological space $X$ and $F$ is a sheaf on $X$ with values in abelian groups. There is a special instance of the Grothendieck spectral sequence relating Cech to sheaf cohomology:
$$E_2^{p,q}=\tilde{H}^p(U,H^q(-,F))\Rightarrow H^{p+q}(X,F)$$
I would like to see, how this implies the Mayer-Vietoris sequence for this easy cover $U$. Drawing the $E_2$-page, I get so far that only the first two columns $p=0,1$ are non-zero. Therefore this page equals the $E_\infty$-page.

REPLY [4 votes]: Here is a slightly different argument than algori's, not using the construction of the Čech-to-derived functor spectral sequence and only using $E_2$ terms, not $E_1$ terms.
As you say, the spectral sequence is given by
$$
E_2^{pq} = \check H^p(\mathcal{U},H^q(-,F)) \Longrightarrow H^{p+q}(X,F).
$$
Since the covering $\mathcal{U}$ only consists of two open sets – $A$ and $B$, say – the $E_2$ page looks like this:

$E_2^{pq}$ is zero for $p \geq 2$.
$E_2^{0q}$ equals the kernel of the map $H^q(A,F) \oplus H^q(B,F) \to H^q(A \cap B,F)$ which sends $(s,t)$ to $t|_{A \cap B} - s|_{A \cap B}$. (Use the alternating Čech complex.)
$E_2^{1q}$ is the cokernel of that map.

Therefore the spectral sequence degenerates on the $E_2$ page. Now consider, for any $n \geq 0$, the canonical short exact sequence
$$ 0 \longrightarrow F^1 E_\infty^n \longrightarrow E_\infty^n \longrightarrow E_\infty^n/F^1 E_\infty^n \longrightarrow 0. $$
Since $F^2 E_\infty^n = F^3 E_\infty^n = \cdots = 0$ and $F^0 E_\infty^n = F^{-1} E_\infty^n = \cdots = E_\infty^n$ (by the vanishing of most columns, see for instance these notes by Matthew Greenberg), we can express the outer terms of this sequence in $E_2$ terms:
$$ 0 \longrightarrow E_2^{1,n-1} \longrightarrow E_\infty^n \longrightarrow E_2^{0,n} \longrightarrow 0, $$
i.e.
$$ 0 \longrightarrow \mathrm{cok}(H^{n-1}(A) \oplus H^{n-1}(B) \to H^{n-1}(A \cap B)) \longrightarrow H^n(X) \longrightarrow \mathrm{ker}(H^n(A) \oplus H^n(B) \to H^n(A \cap B)) \longrightarrow 0.$$
We can splice these short exact sequences to obtain the long exact Mayer–Vietoris sequence.<|endoftext|>
TITLE: Level sets of Morse functions
QUESTION [26 upvotes]: Every compact two dimensional manifold admits a Morse function such that any its regular level set is at most two circles. I am interested in a generalization of that phenomenon.
Does there exist a finite collection of compact manifolds of dimension $n$ such that any compact manifold of dimension $n+1$ admits a Morse (variant - strong Morse, that is all its critical values are pairwise distinct) function $f$ such that for any regular value $c$ the level set $f^{-1}(c)$ is a disjoint union of manifolds from the collection. 
I am especially interested in the case $n=2$.

REPLY [14 votes]: I posted a short paper to the arXiv, Group width,
 which answers this question at least for nonsimply connected manifolds. I think the same question for simply connected manifolds still deserves an answer.<|endoftext|>
TITLE: What is the automorphism group of the additive group of the p-adic integers?
QUESTION [8 upvotes]: Sorry if this is an easy one, I'm a little rusty on my group theory. My first guess was that it's simply the inverse limit of the Aut($\mathbb{Z}/p^i\mathbb{Z})$, with the map when $i\leq j$ given by taking $\sigma\in$ Aut$(\mathbb{Z}/p^j\mathbb{Z})$ to the map $\tilde{\sigma}:\mathbb{Z}/p^i\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ defined by solving $\phi\circ\sigma=\tilde{\sigma}\circ\phi$ where $\phi:\mathbb{Z}/p^j\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ is the reduction map, but that seems too optimistic - I couldn't think of any reason $\tilde{\sigma}$ would be well-defined, much less be an automorphism of $\mathbb{Z}/p^i\mathbb{Z}$.
Also, barring a full answer to my question, I would be interested in whether Aut$(\mathbb{Z}_p)$ is a $p$-group. If not, what can we say about the elements $\sigma\in$ Aut$(\mathbb{Z}_p)$ with order a power of $p$?

REPLY [11 votes]: First the $p$-adic integers are finitely generated (actually cyclic) pro-$p$ group therefore from a result of Serre all automorphisms are continuous. Now as it cyclic it is enough to see what happens to $1$. It has to go to another generator, i.e. any element of the form $a_0+a_1p+a_2p^2+\cdots$, where $0 \leq a_i < p$ for all $i$ and $a_0 \ne 0$. Hence, every element in $\mathbb{Z}_p$ is just multiplied by $a_0+a_1p+a_2p^2+\cdots$. Thus, the automorphism group is the multiplicative group of $\mathbb{Z}_p$.
It is of course not a pro-$p$ group, but it contains a subgroup of index $p-1$ which is pro-$p$ and is actually again a cyclic pro-$p$ group, i.e. isomorphic to $\mathbb{Z}_p$ and thus have no elements of finite order.

REPLY [5 votes]: Any automorphism of $\mathbb{Z}_p$ preserves whether an element is divisible by $p^k$, so it is Lipschitz (in particular, continuous) with respect to the $p$-adic norm.  On the other hand, any automorphism must preserve $\mathbb{Z}$, which is dense in $\mathbb{Z}_p$.
What I should've said is that any automorphism is determined by its behavior on $\mathbb{Z}$, hence by its behavior on $1$.  To make up for that mistake, let me offer a sketch of the description of the structure of $\mathbb{Z}_p^{\ast}$.  This group clearly splits up as the direct product of $(\mathbb{Z}/p\mathbb{Z})^{*}$ and the multiplicative group $U = 1 + p \mathbb{Z}_p$.  It is now an interesting exercise to show that the exponential map $x \mapsto u^x, x \in \mathbb{Z}, u \in U$ extends to a map from $\mathbb{Z}_p$ to $U$ which, given the right choice of $u$, is an isomorphism for $p > 2$.  For $p = 2$, Yiftach Barnea's otherwise excellent answer is slightly wrong and $U$ is in fact isomorphic to $\{ \pm 1 \} \times \mathbb{Z}_2$.<|endoftext|>
TITLE: Extension of valuation
QUESTION [10 upvotes]: Fix a prime number $p$. Suppose that I have a valuation $v_p: \mathbb{Q} \to \mathbb{Q}$ on the rationals $\mathbb{Q}$. That is, $v_p( p^n(\frac{a}{b})) = p^{-n}$ where each of $a,b$ is not divisible by $p$. 
How can I extend $v_p$ to $v$ on the reals $\mathbb{R}$ such that $v|_\mathbb{Q} = v_p$? I am looking for an explicit description of $v$, if that is possible. I know for a fact that one can extend valuation on any field extension.
Thank you,

REPLY [7 votes]: Just in case you don't know this, the extension of $v$ to $\overline{\mathbb{Q}_p}$ is done as follows: If $x$ is algebraic over $\mathbb{Q}_p$, let $x^n + a_{n-1} x^{n-1} + \cdots a_0=0$ be the minimal polynomial of $x$. Then $v(x) = (1/n) v(a_0)$. Turning this into a valuation on $\mathbb{C}$ requires noncanonical choices of two kinds. 
For the first kind of choice, consider let $p=7$. The equation $x^2-6x+7=0$ has two roots in $\mathbb{R}$: one is roughly $4.414\ldots $ and the other is roughly $1.585\ldots $. It also has two roots in $\mathbb{Q}_7$: one is roughly $(\ldots 60)_7$ and the other is roughly $(\ldots16)_7$. You have to decide which one will be identified with which. For the second kind, there are transcendental elements of $\mathbb{C}$, such as $\pi$. You can choose their valuation freely. (More precisely, you can choose the valuations of a transcendence base freely.)<|endoftext|>
TITLE: Cohomological dimension-doubling
QUESTION [14 upvotes]: I'm sure this is a question which has been asked many times, if not necessarily on this site:

Why does a (smooth, projective) scheme over a field, with dimension d, behave as though it were a complex manifold of real dimension 2d?

I am pretty ignorant of most details, so this can be taken to represent the underlying question: what is the "reason" that etale cohomology has dimension 2d?  I am aware of the comparison theorem with singular cohomology but (being ignorant) I do not know if the proof contains something of an answer to my question or if it just "is".  One thing which seems unlikely is that the comparison theorem holds (or can even be stated) for schemes in arbitrary characteristic, where the notion of the associated complex variety is undefined.  So this is not really a reason.
I have also been told that the dimensional properties of etale cohomology are consequences of the same facts for motivic cohomology.  Of course, this is not an explanation either.  The real mystery seems to be: how does cohomology know that $\mathbb{C}/\mathbb{R}$ has degree 2, and how does it know that we had that in mind when we were working with schemes over finite fields?
Possibly, this is related to the fact (?) that any finite extension of fields of which the larger is algebraically closed must have degree 2.  Since I only mention this because it is the single natural occurrence of the number 2 that leaps to mind, it is in fact grasping at straws.
I would appreciate any suggestions towards improving this question.

REPLY [10 votes]: I'm not sure I can give you a morally satisfying answer. To my mind, this sort of theorem should be true because it works over $\mathbb{C}$, for topological reasons. Of course, that isn't a proof over other fields. But, to my limited understanding, the intuition comes from $\mathbb{C}$ and the proofs are motivated by taking proofs which work over $\mathbb{C}$ and seeing whether we can generalize them to an arbitrary field.
That said, I can sketch two proofs of this theorem. The first proof is by induction on dimension. Write $X$ as a family over $Y$, with fibers of dimension $\dim X - \dim Y$. A spectral sequence shows that, if the cohomology of $Y$ vanishes is in degree $>i$ and the cohomology of the fibers vanishes in dimension $>j$, then the cohomology of $X$ vanishes in degree $>i+j$. This reduces us to showing that curves have no cohomology above $H^2$, which can be done by hand; I think this is in Chapter 14 of Milne's lectures.
The other method is for deRham-like cohomology theories (rigid, crystalline, etc.). Roughly speaking, those methods compute cohomology as the hypercohomology of a complex $\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$. If your ground field is characteristic zero, and your variety is smooth, these are actually the familiar sheaves of differentials. If one of these two conditions fails, you have to adjust in some manner; the details of this adjustment describe which of the theories you are working in. In any case (very roughly speaking) the $2n$ appears here as $n+n$: the complex has length $n$ and each of the terms in the complex has no cohomology above degree $n$. Again, a spectral sequence finishes the proof from here.<|endoftext|>
TITLE: Is there a progress on a solution of the inequality $\pi (m+n) \leq \pi (m) + \pi (n)$
QUESTION [17 upvotes]: in 1923 Hardy and Littlewood proposed the conjecture $\pi (m+n) \leq \pi (m) + \pi (n)$. Is there any progress towards solving this conjecture?

REPLY [7 votes]: As has already been pointed out, Montgomery and Vaughn have proven that
$$\pi(x+y) \leq \pi(x)+ 2 \frac{y}{\log(y)} $$
from which, with some care, one can derive that: $\pi(x+y)  \leq \pi(x)+ 2\pi(y)$ (this is worked out in the original paper of Montgomery and Vaughn). 
The natural question is thus to further refine the constant $2$. There seems to be no unrestricted result on this problem, however Friedlander and Iwaniec (in their recent book on sieve theory) have shown that
$$\pi(x+y) \leq  \pi(x) + (2-\delta) \frac{y} {\log(y)} $$
holds for $x^{\theta} < y < x$ where $\delta:= \delta(\theta)$ is a function of $\theta$. In other words, one can improve the constant $2$ as long as $y$ isn't too small compared to $x$.  
Recently Bourgain and Garaev have refined this result to give the quantitative relationship of $\delta \sim \theta^2$.<|endoftext|>
TITLE: What matrix groups can be embedded in $Sp_4$?
QUESTION [8 upvotes]: In a joint paper with Yifan Yang we constructed an "exotic" embedding
of $SL_2(\mathbb R)$ in $Sp_4(\mathbb R)$ (in fact, of $PSL_2(\mathbb R)$ in $PSp_4(\mathbb R)$),
namely,
$$
\iota\colon\begin{pmatrix} a & b \cr c & d \end{pmatrix}
\mapsto\begin{pmatrix}
a^2d+2abc & -3a^2c & abd+\frac12b^2c & \frac12b^2d \cr
-a^2b & a^3 & -\frac12ab^2 & -\frac16b^3 \cr
4acd+2bc^2 & -6ac^2 & ad^2+2bcd & bd^2 \cr
6c^2d & -6c^3 & 3cd^2 & d^3 \end{pmatrix}.
$$
An equivalent form of the embedding was independently discovered by Don Zagier,
and we could not find it in the literature.
Although the properties of the embedding (discussed in the preprint above) are nice by themselves,
I am interested in an exhaustive list of possibilities to embed
other matrix groups and their direct products in $Sp_4(\mathbb R)$ (or $PSp_4(\mathbb R)$).
For example, can the direct product of two copies of $SL_2(\mathbb R)$ be embedded?
As I am not a specialist in Lie groups, I would appreciate plainer sources.
Thank you for any help in advance!

REPLY [2 votes]: You can embed the direct product of two copies of $SL_2(\mathbb{R})$. One embedding sends the entries to the center $2\times2$ block, the other sends the entries to the corners with the $2\times2$ identity matrix in the center block. Here, the $J$ matrix is the standard anti-diagonal matrix
For embeddings of other groups, you could look at the Bruhat decomposition of Sp(4) and write a decomposition of each cell. Some explicit information is given on the decomposition for GSp(4) in a book by Ralf Schmidt and Brooks Roberts, which is available on Ralf's website.<|endoftext|>
TITLE: Sum of Log Normal random variables
QUESTION [8 upvotes]: I would be grateful to anyone who could provide me with some reference concerning the behavior of the sum of Log Normal random variables (need not independent) with respect to a Log Normal random variable.
It is obvious that such a sum has no reason to be log normal but what is less obvious to me is "is it very far from a log normal distribution really ? " or "under what conditions those two objects could be considered as close or very close?".
Regards
Edit : 
A note to explain the "general" formulation of the question :  
There are many metrics to evaluate distance between 2 random variables (Kullback-Leibler and entropic-like metrics, total variations, Hellinger, and so on), I have asked the question in this unspecified fashion because I had no "a priori" on the metric or any other indicator that could cope with the subject, on the contrary the more approaches I could get, the happier. Second aspect of the question, which is implicit in its general formulation, is the way in which a sum of log-normal could be approximated by a log-normal variable. At the time, I had no insight on the methodology to do so. In respect to this second aspect I still think that a general formulation of the question is for the better. Nevertheless I have extended the tags with "reference request" so that it is more clear to the reader I'd rather be pointed to relevant literature on the subject than get a direct answer in a post.   ((spelling of "Kullback-Leibler" corrected, is seen wrong, too often))

REPLY [4 votes]: you can see links for independent and correlated case: 
http://airccj.org/CSCP/vol4/csit43104.pdf
http://airccj.org/CSCP/vol4/csit43105.pdf
Thanks<|endoftext|>
TITLE: Euclid with Birkhoff
QUESTION [28 upvotes]: I'm looking for a short and elementary book which does Euclidean geometry with Birkhoff's axioms.
It would be best if it would also include some topics in projective (and/or) hyperbolic geometry.
About the course. The students suppose know some basic calculus, but they did not see real proofs.
Most of the students in my course want to become math teachers.
The course description says: "Euclidean and Hyperbolic geometries and their development from postulate systems".
I choose Birkhoff's axioms since they use real numbers as a building block.
This makes it possible to do intro without cheating and without boring details.
I know some good books for school students, but I am looking for something a bit more advanced.
P.S. I want to thank everyone for their comments and answers.
As I stated in the comments, I did not find an appropriate book and wrote the lecture notes myself: Euclidean plane and its relatives; they are also available on arXiv.

REPLY [8 votes]: If you're willing to use an unpublished manuscript, from the little I've looked at it, this book by Matthew Harvey looks pretty good.  However, he uses Hilbert's axioms rather than Birkhoff's.  Jack Lee at the University of Washington is writing another book, using a variant on the SMSG postulates, designed for a geometry course for math majors who are considering teaching high school.  His book spends several chapters on hyperbolic geometry, but doesn't have any projective geometry.  The book is not publicly available, but you could email him and ask him about it.<|endoftext|>
TITLE: Are Calabi-Yau manifolds in dimension >= 3 algebraic?
QUESTION [13 upvotes]: I believe that I once saw a statement that every compact, smooth Calabi-Yau manifold in dimension at least 3 is algebraic, but I can remember neither the reference nor the proof (which would have been quite short) and I might just be confusing this with something else. Is it true?

REPLY [16 votes]: It depends a little bit on your definition of CY. If you're using a good one, it will imply that the Hodge numbers $h^{0,p} = 0$ for $p \neq 0,d$ (see, for example, Prop. 5.3 of Joyce's http://arxiv.org/abs/math/0108088). This implies that $H^2(X) \cong H^{1,1}(X)$. Since the Kaehler cone is an open set in $H^{1,1}(X)$, it contains an rational class, and we can scale that to be an integral class. So, by Kodaira and Chow, we're done.<|endoftext|>
TITLE: Sneaky Recursive Non-Well-Orders
QUESTION [14 upvotes]: Background
An ordinal $\alpha$ is called a recursive ordinal if there is a recursive well-order $R$ on $\mathbb{N}$ such that ordertype($\mathbb{N},R) = \alpha$.  For example, $\omega\cdot 2$ is a recursive ordinal because the ordering of $\mathbb{N}$ as 0, 2, 4, 6, 8,  ... 1, 3, 5, 7, ... is computable and has order type $\omega\cdot 2$.
Kleene encoded the recursive ordinals in the natural numbers in a nifty way which is described at the Wikipedia page on Kleene's O.  Now Kleene's $\mathcal{O}$ is a fairly powerful set -- given a Turing machine index for a linear order, $\mathcal{O}$ can decide whether that ordering is a well-ordering or not.
Using Kleene's $\mathcal{O}$, it is possible to describe how to iterate the Turing jump through the recursive ordinals.  For each natural number $a\in\mathcal{O}$, we can define a set $H_a$ recursively as follows:

$H_a = \emptyset$ if $a=0$
$H_a = {H_b}'$ if $a=2^b$
$H_a = \{\langle n, x \rangle | x \in H_{\phi_e(n)} \}$ if $a = {3\cdot 5^e}$

For each $a\in \mathcal{O}$, we have $H_a$ <$_T \ \mathcal{O}$ (strict inequality), and no $H_a$ is powerful enough to decide which recursive orders are well-orders.
Question
Among recursive non-well-orders, some hide their descending chains better than others do.
For example, if we only wanted to flag the non-well-orders sporting a recursive descending chain, the full power of $\mathcal{O}$ would not be necessary -- $\emptyset'''$ would do $(\exists e [ \phi_e$ is total and $\forall n [\ \phi_e(n+1)$ <$_R\ \phi_e(n)\ ]]$?).  Thus there is a recursive linear non-well-order with no recursive descending chain.
In fact (by similar reasoning), for each $a \in \mathcal{O}$ there must be a recursive linear non-well-order with no recursive-in-$H_a$ descending chain.
I wonder whether we could effectively construct these sneaky recursive non-well-orders.

Is there a recursive function $f$ such that whenever $a\in\mathcal{O}$, $f(a)$ is a Turing index for a linear non-well-order with no $H_a$ -computable descending chain?

REPLY [14 votes]: In the classic paper Recursive pseudo-well-orderings, TAMS 131 (1968), 526–543, Joe Harrison showed that one can in fact do much better: there are computable linear orderings which are not wellordered but have no hyperarithmetic descending sequences. An index for such a linear ordering satisfies your requirements simultaneously for all $a \in \mathcal{O}$.
Here is a sketch of Harrison's argument. First note that the ordering of $\mathcal{O}$ has a c.e. extension ${\prec}$ to all of $\mathbb{N}$, namely the smallest relation such that 

$x \neq 0 \to 0 \prec x$,
$x \prec 2^x$,
$\phi_e(n){\downarrow} \to \phi_e(n) \prec 3\cdot5^e$, and
$x \prec y \land y \prec z \to x \prec z$.

Harrison's $\mathcal{O}^*$ is the intersection of all hyperarithmetic sets $X$ with the following properties.

$0 \in X$.
If $a \in X$ then $2^a \in X$.
If $\phi_e$ is total ${\prec}$-increasing and the range of $\phi_e$ is contained in $X$, 
then $3\cdot5^e \in X$.

Since Kleene's $\mathcal{O}$ is the intersection of all sets $X$ with the above properties, we have $\mathcal{O} \subseteq \mathcal{O}^*$. Also, every property of $\mathcal{O}$ translates to a property of $\mathcal{O}^*$ by restricting all second-order quantifiers to range over hyperarithmetic reals, provided that all axioms used in the proof of the property remain valid when all quantifiers are replaced in the same way. Since $\Sigma^1_1$ dependent choice holds in the hyperarithmetic world, this includes a lot of properties of $\mathcal{O}$ including the fact that the initial intervals $I_n = \{x : x \prec n\}$ are computable linear orderings that have no hyperarithmetic descending sequences.
Harrison shows that $\mathcal{O}^*$ is (complete) $\Sigma^1_1$. Since $\mathcal{O} \subseteq \mathcal{O}^*$ is complete $\Pi^1_1$, we must have some $n \in \mathcal{O}^* \setminus \mathcal{O}$. Then the initial interval $I_n = \{x : x \prec n\}$ is a computable non-wellfounded linear ordering without hyperarithmetic descending sequences. In fact, one can show that $\mathcal{O} \cap I_n$ is a path through $\mathcal{O}$ and that $I_n$ has order-type $\omega_1^{CK}(1+\eta) + \alpha$ where $\eta$ is the order type of the rationals and $\alpha < \omega_1^{CK}$.<|endoftext|>
TITLE: Checking locally whether a homomorphism is a localization
QUESTION [6 upvotes]: All rings below are commutative with $1$.  
Suppose $A\subset B$ is a subring and that $A\rightarrow A'$ is a faithfully flat ring homomorphism.  [You may assume the rings are actually ${\mathbb C}$-algebras if it helps.]  Suppose that $B' = A'\otimes_A B$ is a localization of $A'$, i.e. there is a multiplicatively closed subset $S$ of $A'$ such that $B' = S^{-1}A'$.  Must $B$ be a localization of $A$?  
I find it hard to believe that the answer is "yes."  But I'm having a mental block coming up with an example to show that it's "no."

REPLY [5 votes]: Let $A$ be the coordinate ring of a smooth affine curve $X$ over $\mathbb C$, and let $p$ be a point of infinite order in the class group of $A$. Let $B$ be the coordinate ring of $X \smallsetminus \{p\}$, and let $C$ be the coordinate ring of an open subscheme $U$ of $X$ containing $p$ such that $p$ is principal in $U$. Set $A' = B \times C$. Then it it is easy to see that $A' \otimes_A B$ is a localization of $A'$, while $B$ is not a localization of $A$.<|endoftext|>
TITLE: Checking local solubility of varieties at "bad" primes
QUESTION [6 upvotes]: Let $X$ be a smooth variety defined over $\mathbb{Q}$. If we want to check that $X$ is locally soluble at a prime $p$, then it suffices to find a non-singular $\mathbb{F}_p$-point, which can be lifted to a $\mathbb{Q}_p$-point by Hensel's lemma.
However, it might happen that $X$ does not have any non-singular $\mathbb{F}_p$-points. For example $X$ could be given by a quadratic form and we are interested in the prime $p=2$, in which case the reduction mod $2$ of $X$ is a non-reduced scheme and hence every point is singular (at least if I understand the situation correctly).
What general methods are there, if any, to check local solubility in this kind of situation?
A nice toy example is the equation $x^2 + y^2 + z^2=0$. This is locally soluble for all primes $p \geq 3$ (by Chevalley–Warning) and clearly not soluble for $p=\infty$, hence it is not soluble at $p=2$ (by the Hilbert symbol formulation of quadratic reciprocity). Is there a simple way to see this using general methods?

REPLY [8 votes]: As mention in the comments by Pete Clark,  it is a theorem that when $n$ is sufficiently large, you will either be able to apply Hensel's Lemma to get a solution in the valuation ring, or you will find that there are no (primitive) solutions $\pmod{p^n}$.
In practice, I will assume your variety $X$ is affine and it is given as the zero set of some polynomials $f_1(x), \dots, f_m(x)$, where $x=(x_1,\dots,x_r)\in \mathbb{Z}_p[x_1,\dots,x_r]$, which you can assume is saturated, meaning that $$\left((f_1,\dots,f_m)\mathbb{Q}_p[x]\right)\cap \mathbb{Z}_p[x]=(f_1,\dots,f_m)\mathbb{Z}_p[x]$$
There is a procedure to saturate an ideal which essentially divides by $p$ a set of generators until is not possible to do it again, even making a linear combination of the generators.
If you have only one equation, which is the case one needs to have in mind, to be saturated means your coefficients are not all multiple of $p$.
If the reduction $\overline{X}$ modulo $p$ of $X$, i.e. given by the polynomials
$\overline{f_1}(x), \dots, \overline{f_m}(x)\in \mathbb{F}_p[x]$ contains a non singular point, then you are done by Hensel, so $X$ has a point over $\mathbb{Z}_p$.
If not, compute the $\mathbb{F}_p$-points, and for each $P=(a_1,\dots,a_r)\in \overline{X}(\mathbb{F}_p)$, consider the polynomials
$$f_{i,P}(x):=f_i(a_1+p\; x_1,\dots,a_r+p\; x_r)$$
(where I consider $a_i\in \mathbb{Z}$ given as $0\le a_i<p$). Saturate the ideal $I_P:=(f_{1,P}(x),\dots,f_{m,P}(x))$ and consider the scheme $X_P$ over $\mathbb{Z}_P$ given by that saturated ideal $\widetilde{I_P}$. Repeat the procedure. So, look at the $\mathbb{F}_p$-points of the reduction $\overline{X_P}$: if you find some non-singular point, this will lift to a point in $X_P$, so in $X$, whose reduction is $P$. If there are no points in $\overline{X_P}$, then $P$ does not lift to a point in $X$. If there are points, and all singular, we start again with $X_P$.
Let's do it in practice with your equation (which it can be proved also using Hilbert symbol). Here $f:=x_1^2+x_2^2+x_3^2$. Since we want to work in the affine setting (I guess you are considering the projective curve, since as affine variety we have the global point $(0,0,0)$), we take the affine patch "$x_3=1$", so we get $f:=x_1^2+x_2^2+1$. The reduction modulo $2$ has two points (in the affine space). I take $P=(1,0)$, but with the other will be analogous.
Then $$f_P:=(1+2x_1)^2+(2x_2)^2+1=2+4x_1+4x_1^2+4x_2^2$$
and we saturate (so we divide by 2) to get
$$\widetilde{f_P}:=1+2x_1+2x_1^2+2x_2^2$$
whose reduction modulo $2$ is $\widetilde{f_P}\equiv 1$, so it has no zeros. This is essentially showing that there is no solution modulo 4.
Consider another may be more interesting example, coming from famous Selmer example of a genus one curve having points locally but not globally. Take $f=3x^3+4y^3+5$ (this is an affine version of the curve), and consider $p=3$. It has three points in $\mathbb{F}_3$, namely $(0, 1), (1, 1), (2, 1)$, but all three are singular. Take $P=(0,1)$, and compute
$$f_P=3(3x)^3+4(1+3y)^3+5=9\left(9x^3 + 12y^3 + 12y^2 + 4y + 1\right)$$
whose saturation is $$\widetilde{f_P}=9x^3 + 12y^3 + 12y^2 + 4y + 1.$$ The reduction of the corresponding scheme $X_P$ is $\widetilde{f_P}\equiv y+1$, which has three points, all non-singular. Each one lifts to $\mathbb{Z}_3$ by Hensel. So the affine curve has points over $\mathbb{Q}_3$.
Magma software has this function implemented (see in here). I followed a bit the explanation given by Nils Bruin in "Some ternary Diophantine equations of signature (n, n, 2).", a chapter in W. Bosma and J. Cannon, editors, Discovering Mathematics with Magma. Springer-Verlag, Heidelberg, 2004, but the algorithm it is well known.<|endoftext|>
TITLE: Are real numbers countable in constructive mathematics?
QUESTION [11 upvotes]: We are talking about ordinary reals in constructive mathematics.

Let represent each real number by infinite converging series:
$$r = [\;(a_0,b_0),(a_1,b_1),...,(a_i,b_i),...\;]$$
$$where\quad a_i \leq b_i\quad and \quad a_i \leq a_{i+1} \; and \; b_{i+1} \leq b_i$$
And interval $(a_i,b_i)$ converges: for any given rational $e > 0$ there is index $j$ such that $b_k - a_k < e$ for all $k \geq j$.
There are only one way to construct such a number: to build an algorithm that produces $ (a_{i+1},b_{i+1}) $ from (a,b) (or some nearly equivalent).
Let model algorithms by lambda terms (we are able to do so because lambda calculus is Turing complete).
It is easy to show that each lambda term may be represented by unique natural number (this is simple serialization/deserialization process, well known for every programmer).
So there is a one-to-one correspondence between real numbers and subset of natural numbers.
This imply that constructive reals and naturals are equipotent sets.

What are not ok with this reasoning and why?

REPLY [6 votes]: I'd like to emphasise a few relevant distinctions. First, as mentioned, is the various senses of "constructive". The important two here are Markov's (russian) tradition, where it is 
assumed that all functions are computable. The other is Bishop's tradition, where all 
assumptions are valid both classically and in Markov's tradition (and in other senses). So, 
Bishop's school does not assume that there are non-computable functions, but neither does it 
assume that all functions are computable.
Andrej's earlier proof is in Bishop's tradition. There are of course proofs that the computable real numbers are not computably enumerable, and other proofs that the classical real numbers are not classically enumerable. But the proof he posted is a unifying proof that the real numbers are not enumerable.
A second distinction is between the set of real numbers and the set of infinite binary sequences. A bijection exists classically, but is non-constructive. A third distinction is between these two sets, and the class of all subsets of $\mathbb{N}$. It is useful to talk about sets of natural numbers, even if their characteristic function is undecidable. For instance, the set of (codes for) lambda terms of convergent series (Cauchy sequences). This is a subset of $\mathbb{N}$, but only classically does its characteristic function exist, so only classically will it be an element of the set $2^\mathbb{N}$. 
Finally, there is the distinction between countable and subcountable. A set is subcountable if 
it is the image of a subset of $\mathbb{N}$. Classically, every subcountable set is countable, but not so constructively. There are even simple models of CZF (a version of ZF with intuitionistic logic) where every set is subcountable (even within the model), whereas the set of real numbers, as said above, is not countable.
So, with all these in mind:
The real numbers are not countable. But there is proof, in Markov's tradition, that the real numbers are subcountable - which is basically yours, since the set of valid lambda terms is a subset of $\mathbb{N}$.
There is a diagonal argument, valid in Bishop's tradition, that $2^\mathbb{N}$ is not countable, but similarly there is a Markovian proof that $2^\mathbb{N}$ is subcountable. Finally, there is a diagonal argument, valid in Bishop's tradition, that the class of all subsets of $\mathbb{N}$ is not even subcountable.<|endoftext|>
TITLE: How to locate the paper that established Robinson Arithmetic?
QUESTION [10 upvotes]: If I'm not mistaken, it was in his seminal paper “An Essentially Undecidable Axiom System”, published in 
Proceedings of the International Congress of Mathematics (1950), 729–730,
where R.M. Robinson proved that Gödel Incompleteness Theorem still applies to Peano Axioms if we drop the induction schema (hence showing that infinite axiomatization is not necessary for essential undecidability), in what we now call Robinson Arithmetic.
I would like to know:

Is actually this paper what I should be looking for?
Can it be found anywhere on the net? (I already tried on MathSciNet, SpringerLink, JSTOR and Google Scholar, without success)
Can anyone pinpoint to closely related, or at least similar, accessible papers?

(Note: I already have the book "Undecidable theories", which he published in collaboration with Tarski, but I'd prefer to locate papers about 'Robinson theory', specifically).

REPLY [3 votes]: Hi Jose,
it's in the British library collection:
http://snurl.com/z16ud
Haven't checked what the fees are, but you could order it from there.
Alternatively, you could try the LMS:
http://www.lms.ac.uk
A good chance they will have the procs in their library, and you can get photocopies for a nominal fee.
Several other similar alternatives too.<|endoftext|>
TITLE: HNN Embedding Theorem for Amenable Groups?
QUESTION [24 upvotes]: Does there exist an analog of the HNN Embedding Theorem for the class of countable amenable groups? In other words, is it true that every countable amenable group embeds into a 2-generator amenable group? Perhaps easier, is it true that every countable amenable group embeds into a finitely generated amenable group?

REPLY [18 votes]: If I am not mistaken, the answer is "yes".
Theorem. Every countable amenable (respectively, elementary amenable) group embeds into a $2$-generated amenable (respectively, elementary amenable) group.
The proof is based on the following lemma, which admits a quite elementary proof using wreath products (see [P. Hall, The Frattini subgroups of finitely generated groups, Proc. London Math. Soc. 11 (1961), 327-352]). Given a group $X$, we denote by $X^\omega$ the restricted direct product of countably many copies of $X$. 
Lemma (P. Hall).  Let $H$ be a countable group. Then there exists a short exact sequence
$$
1\longrightarrow M \longrightarrow G \longrightarrow \mathbb Z \longrightarrow  1,
$$
where $G$ is $2$-generated and $[M,M]=[H,H]^\omega$. 
In particular, the lemma implies the theorem when the countable group is perfect. To prove the theorem in the general case we use the following trick which goes back, I believe, to the paper by B. H. Neumann and H. Neumann cited by Mark.
Starting with a countable group $K$, consider the subgroup $H$ of the Cartesian (unrestricted) wreath product $K \, {\rm Wr}\, \mathbb Z$ generated by $\mathbb Z$ and the set of all elements of the base group $ f_k\colon \mathbb Z\to K$, $k\in K$, such that $f_k(n)=1$ for $n\le 0$ and $f_k(n)=k$ for $n> 0$. Let $t$ be a generator of $\mathbb Z$. For definiteness let $t=1$. Then $t^{-1}f_ktf_k^{-1}$, considered as a function $\mathbb Z\to K$, takes only one nontrivial value $k$ (at $0$). This obviously gives an embedding $K\le [H,H].$
Moreover, it is easy to see that the intersection of $H$ with the base $B$ of the wreath product consists of functions $f\colon \mathbb Z\to K$ with the following property: There exists $N_f\in \mathbb N $ such that $f(n)=1$ whenever $n\le -N_f$ and $f(n)=f(N_f)$ whenever $n\ge N_f$. Obviously the map $\varepsilon\colon H\cap B\to K$, which maps every function $f\colon \mathbb Z\to K$ as above to $f(N_f)$, is a homomorphism and $Ker\, \varepsilon$ is isomorphic to a subgroup of $K^\omega $. In particular, if $K$ is amenable (or elementary amenable), then so is $H\cap B$ and, consequently, so is $H$. Now applying the lemma to $H$ yields the theorem as $K\le [H,H]$.<|endoftext|>
TITLE: Birkhoff conjecture about integrable billiards
QUESTION [8 upvotes]: There is a conjecture by Birkhoff which claims that for a simple closed $C^2$ plane curve $C$, if the billiard ball map is integrable then the curve is an ellipse. 
Integrability here might be formulated as follows: there exists a neighbourhood of $C$ in the interior $Int(C)$ that is foliated by caustics (caustics being curves that are everywhere tangent to a given trajectory of the billiard ball). 
I would be interested to know the current status (and progresses, if there are) of this conjecture.

REPLY [7 votes]: For a recent progress see http://arxiv.org/pdf/1412.2853.pdf
a local version of this conjecture is proven.<|endoftext|>
TITLE: Establishing zeta(3) as a definite integral and its computation.
QUESTION [17 upvotes]: I am a 19 yr old student new to all these ideas. I made the transformation $X(z)=\sum_{n=1}^\infty z^n/n^2$. Therefore $X(1)=\pi^2/6$ as we all know (it is $\zeta(2)$). To calculate $X(1)$, I integrated
$$\frac{Y(z)}{z}=\sum_{n=1}^\infty \frac{z^{n-1}}{n} $$
between 0 to 1;
We all know $Y(z)=-\log(1-z)$; so doing integral of $Y(z)/z$ from 0 to 1, I get
$$X(1)=\int_0^1\ln\frac{z}{z-1}dz$$
that is $\pi^2/6$; Now integrating $X(z)/z$ between 0 to 1, I get $\sum_{n=1}^\infty 1/n^3$;
therefore performing that integral I get $\sum1/n^3$ as 
$$\int_0^1 \frac{(\log x)^2 dx}{1-x};$$
which Iam unable to do.
So can anyone give an idea of how to find 
$$\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)=\int_0^1 \frac{(\log x)^2 dx}{1-x};$$
given
$$\zeta(2)=\int_0^1 \frac{\log x\ dx}{1-x}=\frac{\pi^2}{6}.$$
Please take time to read all these things and help me out by giving some suggestion.

REPLY [2 votes]: Following your way:
Let $F(x) = \displaystyle\int_0^x{-\frac{\ln(1-t)}{t}}dt$, where $x  \in [0,1]$. Easy to see that $F(1)$ yields $\zeta(2)$. 
$F(x)$ is continuous and derivable on $(0,1)$ [The integrand being continuous on $(0,1)$] and $F'(x)= -\displaystyle\frac{\ln(1-x)}{x}$. Again, $F(x)=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+... $.
It can be readily seen that: $\displaystyle\int_0^x{\frac{F(x)}{x}}dx=\mathrm{Li}_3(x).$
Integrating by parts:$\displaystyle\int_0^x{\frac{F(x)}{x}}dx=$ $\displaystyle{[{F(x){\ln(x)}-\int{F'(x)\ln(x)dx}}]_0^x=[F(x)\ln(x)+\frac{\ln(1-x)\ln(x)}{x}]_0^x}$
$F(0)=0$ [follows from the continuity of integral functions] and $\lim_{x \to 0}\ln(1-x)\ln(x)=0$. 
Hence, we finally arrive at:  
$\zeta(3)= \displaystyle\sum_{n=1}^\infty{\frac{1}{n^3}}=\displaystyle\int_0^1{\frac{\ln(1-x)\ln(x)}{x}dx}$<|endoftext|>
TITLE: Relatively countably compact subsets without countably compact closure.
QUESTION [9 upvotes]: I am looking for a (Hausdorff or better) space $X$ and a subset $A$ of $X$ that is relatively countably compact (every sequence from $A$ has an accumulation point in $X$) such that the closure of $A$ is not countably compact.
It is known that in many "nice" spaces such examples do not exist (a classical case being normed spaces in their weak topology).
Edit: for $T_4$ spaces this cannot happen, as the closure of a relatively countably compact subset is pseudocompact (Suppose A is relatively countably compact.
If f from cl(A) to R is unbounded then f|A is unbounded as well, as A is dense in cl(A).
So we can find {x_n: n in N} in A such that |f(x)| >= n.
Let p in cl(A) be an accumulation point of this set, by A being relatively countably compact.
Then continuity at f implies that |f(x_n)| <= |f(p)| + 1, for all but finitely many n.
This contradicts the choice of the x_n, contradiction.)
So cl(A) is pseudocompact, and hence in a normal space, countably compact.
This explains the properties of the example given below.

REPLY [4 votes]: To fix Pietro's idea: take a Maximal Almost Disjoint family on $\omega$, i.e., a family $\mathcal{A}$ of infinite subsets such that any two distinct elements have a finite intersection and such that it is maximal with respect to this property. Topologize $\omega\cup\mathcal{A}$ by making each $n\in\omega$ isolated and have $\bigl\lbrace\lbrace A\rbrace\cup(A\setminus F): F$ finite$\rbrace$ as a local base at $A\in\mathcal{A}$.
By maximality $\omega$ is relatively countably compact; its closure is the whole space, which is not countably compact. 
(This is generally know as the/a Moore Mrowka space, a $\psi$-space, or $\Psi$-space ...)<|endoftext|>
TITLE: Uniform convergence of difference quotient
QUESTION [5 upvotes]: Let $\phi\in C^\infty_c(\mathbb R)$ be a smooth function with compact support.
For $h>0$ define the difference quotient $\phi_h\in C^\infty_c(\mathbb R)$ by $\phi_h(t)=\dfrac{\phi(t+h)-\phi(t)}{h}$.
By definition, for fixed $t\in\mathbb R$, we have $\phi_h(t)\to\phi'(t)$ as $h\to 0$.

Question: Can we conclude, that $\phi_h\to\phi'$ uniformly on $\mathbb R$ (as $h\to 0)$?

Motivation: This is used in a proof of Stone's Theorem on the existence of generators of operator groups I'm trying to understand.

REPLY [3 votes]: More generally, any function with uniformly continuous first derivative satisfies 
$$\|\phi_h-\phi'\|_\infty\leq \omega(|h|),$$
where $\omega$ is any modulus of continuity of $\phi'$ (just recall that $\phi_h(x)$ equals a value of $\phi'$ in a point within a distance $|h|$ from $x$).<|endoftext|>
TITLE: Banach spaces with few linear operators ?
QUESTION [21 upvotes]: Sometimes, dealing with the concrete and familiar Banach spaces of everyday life in maths, I happen nevertheless to ask myself about the generality of certain constructions. But, as I try to abstract such constructions up to the level of general Banach spaces, I immediately feel the cold air of wilderness coming from the less known side of this huge category. One typical difficulty is, that it's not at all obvious how to find, or to prove the existence, of (bounded linear) operators $T:E\to E.$
Certainly, there are always a lot of finite rank operators, furnished by the Hahn-Banach theorem, and their norm-limits. However, in an anonymous infinite dimensional Banach space, I do not see how to guarantee the existence of bounded linear operators different from a compact perturbation of $\lambda I:$ hence my questions:
Are there always other operators than $\lambda I+K $ on an infinite dimensional Banach space? 
(in other words, can the Calkin algebra be reduced to just $\mathbb{C}\\ $)? More genarally, starting from a operator $T$, is there always a way to produce new operators different from a compact perturbations of the norm-closed algebra generated by $T$? Is there a class of Banach spaces that are rich of operators in some suitable sense (spaces with bases, for instance, but more in general?).
Thank you!

REPLY [17 votes]: Examples were constructed (about two years ago?) by Argyros and Haydon. See this blog post for some non-technical discussion. It seems worth noting, as one is almost obliged to, that the space originally constructed by A & H is an isomorphic predual of $\ell_1$.<|endoftext|>
TITLE: How can we show the spaces $M_{g}(n)$ and $M_{g, n}$ are homotopy equivalent?
QUESTION [5 upvotes]: How can we prove that the moduli space,$M_{g}(n)$, of genus $g$ Riemann surface with $n$ boundary components is homotopy equivalent to $M_{g,n}$, that is ,the moduli space of genus $g$ Riemann surface with $n$ punctures? Thanks! (It is very intuitive, but it seems that I can't make it)

REPLY [8 votes]: A compact Riemann surface of genus $g$ with $n$ boundary components has a unique realization as a hyperbolic surface with geodesic boundary. One may see this by reflecting through the boundary and uniformizing. The uniqueness of the uniformization implies it is invariant under reflection, and therefore the fixed point set is geodesic. 
Thus, the moduli space of genus $g$ Riemann surfaces with $n$ boundary components is equivalent to the space of hyperbolic surfaces with totally geodesic boundary. One may now insert a punctured disk into each boundary component, to obtain a Riemann surface with punctures. I don't know of a canonical way to do this, but for example for a boundary component of length $l$, one may attach isometrically the boundary of a punctured Euclidean disk of circumference $l$. The important thing is that this gluing only depends on $l$, and that it induces a conformal structure on the punctured surface. This gives a map between the spaces. Since the mapping class groups are the same, it induces a homotopy equivalence (in the category of orbifolds). Of course, there are some technical details one must carry out to make this argument rigorous. There are several other ways to fill in a punctured disk.   
Another possible approach is to use the Weil-Petersson metric on moduli space. One can take the WP nearest point in the Deligne-Mumford compactification of moduli space, which is finite distance away since the WP metric is incomplete. Because the metric is CAT(0), a unique nearest point exists.<|endoftext|>
TITLE: A remark on Cohen's theorem
QUESTION [10 upvotes]: It is well known as Cohen's theorem that a commutative ring is Noetherian if all its
prime ideals are finitely generated. Is this statement true or false when prime ideals are replaced by maximal ideals?

REPLY [2 votes]: This is an old question, but Julian Rosen and I observed an example that is a bit simpler than those written above. I am posting it in case it's helpful to others.
It is a bit closer to algebraic geometry, in particular a one-dimensional subquotient of $k[x,y]$ where $k$ is an algebraically closed field, basically a thickening of $k[x]$ at the origin.
Let $\displaystyle{R = \frac{k[x,y,\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots]}{(y)}.}$
Claim 1. The maximal ideals of $R$ are $(x-c)$ for $c \in k$.
Claim 2. The only other prime ideal is the nilradical of $R$, which is $\displaystyle{(\tfrac{y}{x}, \tfrac{y}{x^2}, \ldots)}$.
Proof: By the description of the nilradical, $R / \mathrm{rad}(R) \cong k[x]$, which makes it easy to list all the prime ideals. It suffices to check their generators in $R$, i.e. each $(x-c) \subset R$ actually is maximal and that the nilradical is as described.
For the nilradical claim, $(\tfrac{y}{x^n})^2 = y \cdot \tfrac{y}{x^{2n}} = 0$ for all $n$. Quotienting by them leaves $k[x]$ which is reduced.
For the maximal ideals claim, set $x=c$. If $c \ne 0$, then $\tfrac{y}{x^n} \cdot c^n = \tfrac{y}{x^n} \cdot x^n = y = 0$, so $\tfrac{y}{x^n} = 0$ for all $n$, which leaves a field. On the other hand if $c = 0$, then $\tfrac{y}{x^n} = \tfrac{y}{x^{n+1}} \cdot x = 0$ for all $n$. Either way, we get a field.<|endoftext|>
TITLE: When did the career of 1 as a prime number begin and when did it end?
QUESTION [19 upvotes]: The old Greek did not consider 1 a number, so it was not a prime. The theorem of unique prime factorization excludes 1 to be a prime number. But in between probably at Euler's and Goldbach's times? Who can determine most precisely (probably by original papers) when 1 first became a prime number and when 1 had been called a prime number for the last time?

REPLY [14 votes]: My naive understanding is like this:  most, but not all, of the ancient Greeks excluded one from the category of numbers ($\alpha\rho\iota\theta\mu o\varsigma$), hence from the primes (others excluded two as well).  Speussippus (c.350BC) is a rare exception:

Speussippus, then, is exceptional among pre-Hellenistic thinkers in that he considers
  one to be the first prime number.  [L. Taran, Speusippus of Athens: a critical study with a collection of the related texts and commentary, Philosophia antiqua, vol. 39-40, E.J. Brill, 1981] 

This view held (mostly) until Stevin's argument that one was a number and his development of the reals (late 16th century)

In general, mathematics before Stevin is of one character and, after him, it
  is of another re
  ecting his contributions. In this regard, he is like Euclid: he
  stood at a watershed in the history of mathematics. And as with Euclid, he
  was so successful that, from our present day vantage point, it is hard to see
  the other side of that watershed. Over there, one was not a number; here and
  now, it is; even  is a number, and i, and aleph nul. [C. J. Jones, The concept of one as a number, Ph.D. thesis, University of Toronto, 1978.]

Now we enter a period of confusion.  The view of one as a non-number slowly begins to die out, and some (e.g., Brancker + Pell's table 1688) begin to list one as a prime.
Not a prime for Schooten 1657, Clerke 1682, Chales 1690, Ozanam 1691, Brunot 1723, Cortes 1724, Reyneau 1739, Euler 1770, Horsley 1772, ...  A prime for Wallis 1685, Goldbach 1742, Kruger 1746, Willich 1759, Lambert 1770, Felkel 1776, Warring 1782, ... (Not these folks may have used both views at times, as many of us alternate between ln and log for the natural log, depending on our audience)     
The beginning of the end comes with Gauss' Disquisitiones Arithmeticae as the Fundamental theorem of arithmetic, and especially the uniqueness of factorization, becomes central.  At about the same time number fields are introduced and the role of units becomes understood. I could give a long list (like the above) of yeses and nos in this period as well.  But the choice of excluding one from the primes gains superiority and is now essentially universal among mathematicians.
Was there a time one was almost universally considered a prime? Absolutely not.  Was there a time it was almost universally considered a non-prime: yes, much of history.    
Edited to add "last time" from comments: 
It might be that the last major mathematician "wrote" that one was a prime in print was G. H. Hardy, who lists one as a prime in his "A course of pure mathematics, "3rd ed., Cambridge University Press, 1921:

If there are only a finite number of primes let them be 1, 2, 3, 5, 7, 11, ... $N$.  [section 61, page 143-144]

He does is indirectly later in this text:

the decimal $.111\ 010\ 100\ 010\ 10\ldots$, in which the $n$th figure is $1$ if $n$ is prime, and zero otherwise, represents an irrational number [section 78, page 174]

By the 9th edition, 1944, Gerry Myerson notes the first reference to 1 as prime is removed (I'd bet it was changed by the 7th edition, 1938, and will try to check). The decimal (surely accidentally) was still present in the 10th edition that I checked.  
However, I am not convince Hardy personally thought (defined) that 1 was prime in 1921, I suspect he thought it was still not important enough to bother changing what he had written in the first edition, 1908.  His comments in the article "The Theory of Numbers"  in Science (New Series, Vol. 56, No. 1450, Oct. 13, 1922, pp. 401-405), appears to imply 1 is not prime.  E.g., he repeats that Mersenne listed $2^n-1$ was prime for 2, 3, ... without commenting about $1=2^1-1$ or altering Mersenne's statement to start at 1---which was commonly done.  (I admit this is this is weak evidence!)<|endoftext|>
TITLE: Limit probability
QUESTION [5 upvotes]: You start with a bag of N recognizable balls. You pick them one by one and replace them until they have all been picked up at least once. So when you stop the ball you pick has not been picked before but all the others have been picked once or more.
Let $P_N$ be the probability that all the others were actually picked twice or more.
Questions: does $P_N$ have a limit as  $N\to \infty$, is this limit 0 or can you compute it?
Note: This is a question that was asked at the Yahoo answers forum, by gianlino. Since no one has found an answer in that forum, I forwarded it here. Here is the original link. http://answers.yahoo.com/question/index;_ylt=ApZ_Q6sS137DnEcNoUBqn.vty6IX;_ylv=3?qid=20100704050924AAfLYYJ

REPLY [2 votes]: Also, here's a heuristic argument to show that the probability behaves like $1/N$ as $N$ becomes large.
The last ball to be chosen is chosen for the first time after roughly $N\log N$ steps. (It's the maximum of $N$ random variables which are geometric with mean $N$). By this stage, the number of times that any particular other ball has been selected is Poisson($\log N$). So the probability a particular other ball has been selected only once is roughly $\log N e^{-\log N}$ which is $\log N/N$. Hence the probability that no other ball has been selected only once is roughly
$(1-\log N/N)^{N-1}$, which is $\exp[-\log N + O(\log N/N)] \sim 1/N$.
Of course the previous paragraph ignores various dependencies, but as $N$ becomes large these will be negligible and no doubt one could formalise the argument if desired.<|endoftext|>
TITLE: Nilpotent group with ascending and descending central series different?
QUESTION [6 upvotes]: This may turn out to be a bit embarassing, but here it is. It is probably not true that the ascending and descending central series (*) of a nilpotent group have the same terms. (Or at least one of MacLane-Birkhoff, Rotman and Jacobson would have mentioned it.) However, I have been unable to find an example where they are different. I thought I had a sketch of proof that they are always equal, but there is a gap, of the kind where you feel it is not patchable.
I've proved it for a few nilpotent groups (dihedral of the square, any group of order p^3, the Heisenberg groups of dimension 3 and 4 over any ring -- I think the argument extends to any dimension), and checked a few more exotic examples in the excellent Group Properties Wiki. So,

What is the simplest (preferably finite) nilpotent group such that its a.c.s. and d.c.s. are different?

and

Do the a.c.s. and d.c.s. coincide in some interesting general case?

(*) For completeness, the ascending central series of a group G is defined by Z_0 = 1, Z_{i+1} = the pullback of Z(G/Z_i(G)) along the projection, and the descending central series by G_0 = G, G_{i+1} = [G,G_i]. The group G is nilpotent iff ever Z_m = G or G_m = 1. It turns out that if such m exists it is the same for both.

REPLY [3 votes]: Not only can the two central series be different, there are many qualitative properties that the descending central series (also called the lower central series) satisfies that the ascending central series (also called the upper central series) usually does not. For instance, the lower central series is strongly central (as pointed out by Tom Church in his comment, though he didn't use that word) but the upper central series (turned around) need not be. At least half the members of the lower central series are abelian, but this need not be true of the upper central series. All the members of the lower central series are verbal subgroups, but the members of the upper central series need not even be fully invariant.
For more related information, see this page:
http://groupprops.subwiki.org/wiki/Nilpotent_not_implies_UL-equivalent
[DISCLOSURE: I wrote the material on that page and in most of the linked-to pages]<|endoftext|>
TITLE: Example of a quasitopological group with discontinuous power map
QUESTION [5 upvotes]: A quasitopological group is a group $G$ with topology such that multiplication $G\times G\rightarrow G$ is continuous in each variable (i.e. all translations are continuous) and inversion $G\rightarrow G$ is continuous. Sometimes these are called semitopological or semicontinuous groups. What (if it exists) is an example of a quasitopological group such that at least one of the $n$-th power maps $g\mapsto g^{n}$ (for $n\geq 2$) is discontinuous?
I am pretty sure such an example exists but I am having a hard time finding one in the literature.

REPLY [2 votes]: Maybe, the following topology on the plane works: a base at 0 is formed by the usual neighborhoods at 0 in the plane minus a convenient subset of the diagonal, e.g. the sequence 1/3^n (and -1/3^n).<|endoftext|>
TITLE: Torsion in K-theory versus torsion in cohomology
QUESTION [13 upvotes]: Inspired by  this  question, I wonder if anyone can provide an example of a finite CW complex X for which the order of the torsion subgroup of $H^{even} (X; \mathbb{Z}) = \bigoplus_{k=0}^\infty H^{2k} (X; \mathbb{Z})$ differs from the order of the torsion subgroup of $K^0 (X)$, where $K^0$ is complex topological K-theory.  This is the same as asking for a non-zero differential in the Atiyah-Hirzebruch spectral sequence for some finite CW complex X, since this spectral sequence always collapses rationally.
Even better, is there an example in which X is a manifold?  An orientable manifold?
Tom Goodwillie's answer to the question referenced above gave examples (real projective spaces) where the torsion subgroups are not isomorphic, but do have the same order.
It's interesting to note that the exponent of the images of these differentials is bounded by a universal constant, depending only on the starting page of the differential!  This is a theorem of D. Arrlettaz (K-theory, 6: 347-361, 1992).  You can even change the underlying spectrum (complex K-theory) without affecting the constant.

REPLY [12 votes]: This paper by Volker Braun shows that the orientable 8-manifold $X=\mathbb{RP}^3\times \mathbb{RP}^5$ gives an example. One has $$K^0(X) \cong \mathbb{Z}^2 \oplus \mathbb{Z}/4\oplus (\mathbb{Z}/2)^2$$ 
and 
$$H^{ev}(X) \cong \mathbb{Z}^2 \oplus (\mathbb{Z}/2)^5.$$
Braun does the calculations using the Künneth formulae for the two theories, with the discrepancy in size arising because the order of the tensor product of finite abelian groups  is sensitive to their structure, not just their order.
One another remark is that Atiyah and Hirzebruch told us the $d_3$-differential in their spectral sequence. It's the operation $Sq^3 \colon H^i(X;\mathbb{Z})\to H^{i+3}(X;\mathbb{Z})$ given by $Sq^3 := \beta\circ Sq^2 \circ r$, where $r$ is reduction mod 2 and $\beta$ the Bockstein. As you say, Dan, if this is non-vanishing, K-theory has smaller torsion. This happens iff there's a mod 2 cohomology class $u$ such that $u$ admits an integral lift but $Sq^2 (u) $ does not. Can someone think of a nice example where this occurs?<|endoftext|>
TITLE: Rotation part of short geodesics in hyperbolic mapping tori
QUESTION [5 upvotes]: Otal [Sur le nouage des géodésiques dans les variétés hyperboliques.  C. R. Acad. Sci. Paris Sér. I Math.  320  (1995),  no. 7, 847--852.] showed that "short" simple closed geodesics in 3-dimensional hyperbolic mapping tori are unknotted and unlinked with respect to the fiber, where "short" depends only on the genus of the fiber.
Another fact about short geodesics in hyperbolic mapping tori is the following: Given $\delta > 0$, the rotation part of a loxodromic isometry corresponding to a "short" simple closed geodesic is less than $\delta$, where "short" depends only on the genus of the fiber and $\delta$.
I haven't been able to find this written down.  Does anyone know a reference for this fact?

REPLY [8 votes]: This should follow from Minsky's paper The classification of Kleinian surface groups, I: Models and bounds on a priori bounds for surface groups, which is used in the proof of the ending lamination conjecture.
The punctured torus case (The classification of punctured-torus groups, Annals) is simpler and more explicit (see Theorem 4.1 and equations 4.4 and 4.5).
Addendum: Once I thought about it for a bit, I think it follows from much more elementary
considerations (in fact, I'm pretty sure someone explained this to me before, but I forgot the argument). Let $\Sigma$ be a surface. Suppose one has a very short geodesic $\gamma\subset M$, where $M\cong \Sigma\times \mathbb{R}$ is a hyperbolic manifold, then Otal's argument proves it is unknotted (this was actually known to Thurston, and generalized to multiple components by Otal). In fact, one may find a pleated surface $f:\Sigma \to M$ so that $\gamma$ is a closed geodesic on the image of this surface. Then the  Margulis tube $V$ of $\gamma$ is of very large radius, and therefore its boundary $\partial V$ is very close to being a horosphere (i.e., its principle curvatures are very nearly $=1$) and is isometric to a Euclidean torus. The boundary slope $\gamma'\subset \partial V$ of the surface $\Sigma$ is a Euclidean geodesic of bounded length - this follows from an area estimate of a pleated annulus $A \subset \Sigma$ such that $f(A)$ cobounds $\gamma$ and $\gamma''$, where $\gamma''\sim \gamma'\subset \partial V$, which has  $\mathrm{Area}(A) \approx \gamma''$ by a Gauss-Bonnet argument (if $V$ were a horocusp, then this would be an equality). But
$$
\mathrm{length}(\gamma')\leq \mathrm{length}(\gamma'')\approx \mathrm{Area}(A) \leq \mathrm{Area}(f^{-1}(\Sigma)) =  -2\pi \chi(\Sigma).
$$
The meridian $\mu\subset \partial V$ is a curve intersecting $\gamma'$ once. We may assume that $\gamma',\mu\subset \partial V$ are chosen to be Euclidean geodesics. Then $\partial V \backslash (\gamma'\cup \mu)$ is a Euclidean parallelogram, with one pair of sides of bounded length corresponding to $\gamma'$. Since $V$ has very large radius, $\mu$ must be extremely long. The rotational part corresponds to the fraction of the offset between the two sides of the parallelogram corresponding to $\mu$.
But this implies that the rotational part of $\gamma$ is less than $$
2\pi \mathrm{length}(\gamma')/\mathrm{length}(\mu),
$$
which is very small, and approaches zero as $\mathrm{length}(\gamma)\to 0$.<|endoftext|>
TITLE: approximately linear functions
QUESTION [7 upvotes]: i suppose it's fairly well known that if a (continuous, real-valued) function $f$ on the real line satisfies
$f(x-y)=f(x)-f(y)+const$
then it is necessarily linear.
are there any general characterizations of "approximate" linearity? for example, what can be said if $|f(x-y)-f(x)+f(y)-f(0)|$ is bounded by some small $\epsilon$? or, more relevant to what i need, if the $L^2$ norm of this difference is bounded by a small constant? in particular, suppose,
$$E[(f(X-Y)-f(Y)+f(Y)-f(0))^2]\leq\epsilon$$
for a pair of independent random variables $X,Y$. is there a sense in which $f$ is approximately linear?

REPLY [14 votes]: Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy  the inequality
$$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$
for all $x, y \in E$,  are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers
(link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space  $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$. 

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.
Edit. As Yemon Choi indicated, finite dimensional versions of the result had been discovered earlier and independently. Check out, for instance, the Pólya and Szegö problem book (Ch 3, Problem 99):

Assume that the terms of the sequence $a_1,a_2,a_3,\dots$ satisfy the condition 
  $$a_m+a_n-1 < a_{m+n} < a_m+a_n+1.$$
  Then
  $$\lim\limits_{n\to\infty}\frac{a_n}{n}=\omega$$
  exists; $\omega$ is finite and we have
  $$\omega n-1 < a_n < \omega n +1.$$<|endoftext|>
TITLE: What are you using for symbolic computation?
QUESTION [7 upvotes]: What are the pluses and minuses of different software packages? Anything new worth checking out?
I'm especially interested in open source packages.

REPLY [2 votes]: I suggest you to try Reduce, it is OpenSource and there are RPM packeges. It looks like this: 
alt text http://static.itmages.ru/i/10/1012/h_1286887665_4af06aa049.png<|endoftext|>
TITLE: Sum f(p) over all primes convergent with sum f(n) over all natural numbers divergent?
QUESTION [9 upvotes]: The sum $\sum_{n=1}^{\infty} 1/n^{s}$ is convergent for all real $s>1$ and diverges for all real $s \le 1$. The same holds for the sum $\sum_{p \ prime} 1/p^{s}$. Thus, for the functions $f(n)= 1/n^s, s \in \mathbb{R}$ the sum $\sum_{n=1}^{\infty}f(n)$ shows the same convergence behaviour as the sum $\sum_{p \ prime}f(p)$.
The same holds, if I'm not mistaken, for the functions $f(n)= 1/(n (\ln n)^s),  s \in \mathbb{R}$ (both for $n \in \mathbb{N}$ and for primes convergence iff $s>1$). 
Question: Is there a real monotonic function $f$ such that $\sum_{n=1}^{\infty}f(n)$ diverges whereas sum $\sum_{p \ prime}f(p)$ converges? (The monotony requirement is for preventing 'artificial' solutions that single out the primes (as e.g. $f(n) = 2^n$ if $n$ is prime; $f(n)=n$ if $n$ is not prime)).

REPLY [5 votes]: Note also that if $\{a_n\}$ is an increasing sequence of natural numbers that presents arbitrarily large gaps between consecutive terms (e.g. a sequence with with density $0$),  there is always a positive decreasing function $f$ such that the sum of $f(a_n)$ converges and the sum of $f(n)$ diverges to positive infinity. The reason is that $\sum_n f(n)\geq\sum_n f(a_n)(a_n-a_{n-1}),$ and as the $a_n-a_{n-1}$ are unbounded, the claim reduces to the easy: for any unbounded sequence $u_n>0$ there is a convergent series with coefficients $\epsilon_n>0$ such that the series of the $u_n\epsilon_n$ diverges. Use this taking $u_n:=a_n-a_{n-1}$ to define a decreasing $f$ such that $f(a_n)=\epsilon_n$. 
(Just to clarify that if your question was really "is there a function" and not "which function", then it has a basic answer, not requiring the prime number theorem).<|endoftext|>
TITLE: Where does the "Hardy-Littlewood" conjecture that pi(x+y) < pi(x) + pi(y) originate?
QUESTION [18 upvotes]: The conjecture that $\pi(x+y) \leq \pi(x) + \pi(y)$, with $\pi$ the counting function for prime numbers, is customarily attributed to Hardy and Littlewood in their 1923 paper, third in the Partitio Numerorum series on additive number theory and the circle method.  For example, Richard Guy's book Unsolved Problems in Number Theory cites the paper and calls the inequality a "well-known conjecture ... due to Hardy and Littlewood".  
Partitio Numerorum III is one of the most widely read papers in number theory, detailing a method for writing down conjectural (but well-defined) asymptotic formulas for the density of solutions in additive number theory problems such as Goldbach, twin primes, prime $k$-tuplets, primes of the form $x^2 + 1$, etc.  This formalism in the case of prime $k$-tuplets, with the notion of admissible prime constellations and an asymptotic formula for the number of tuplets, came to be known as the "Hardy-Littlewood [prime k-tuplets] conjecture" and indeed is one of several explicit conjectures in the paper.
However, the matter of whether $\pi(x+y) \leq \pi(x) + \pi(y)$, for all $x$ and $y$, does not actually appear in the Hardy-Littlewood paper. They discuss the inequality only for fixed finite $x$ and for $y \to \infty$, relating the packing density of primes in intervals of length $x$ to the $k$-tuplets conjecture.  (The lim-sup density statements that H & L consider were ultimately shown inconsistent with the k-tuplets conjecture by Hensley and Richards in 1973).
The questions:

Are there other works of Hardy or Littlewood where the inequality on $\pi(x+y)$ is discussed, or stated as a conjecture?
Where does the inequality first appear in the literature as a conjecture?
Is there any paper that suggests the inequality (for all finite $x$ and $y$, not the asymptotic statement considered by Hardy-Littlewood) is likely to be correct?

REPLY [8 votes]: I took a look at Schinzel and Sierpinski, Sur certaines hypotheses concernant les nombres premiers, Acta Arith IV (1958) 185-208, reprinted in Volume 2 of Schinzel's Selecta, pages 1113-1133. In the Selecta, there is a commentary by Jerzy Kaczorowski, who mentions "the G H Hardy and J E Littlewood conjecture implicitly formulated in [33] that $\pi(x+y)\le\pi(x)+\pi(y)$ for $x,y\ge2$." [33] is Partitio Numerorum III. Schinzel and Sierpinski (page 1127 of the Selecta) define $\rho(x)=\limsup_{y\to\infty}[\pi(y+x)-\pi(y)]$, and point to that H-L paper, pp 52-68. They then write (page 1131), "$\bf C_{12.2}.$ L'hypothese de Hardy et Littlewood suivant laquelle $\rho(x)\le\pi(x)$ pour $x$ naturels $\gt1$ equivaut a l'inegalite $\pi(x+y)\le\pi(x)+\pi(y)$ pour $x\gt1,y\gt1$." It should be said that the proof that the first inequality implies the second relies on Hypothesis H, which essentially says that if there is no simple reason why a bunch of polynomials can't all be prime, then they are, infinitely often. 
Schinzel and Sierpinski express no opinion as to any degree of belief in the conjecture under discussion. 
I don't suppose this actually answers any of the questions, although Kaczorowski's use of the word "implicitly" may be significant.<|endoftext|>
TITLE: Finite simple groups with upper bound on order of elements
QUESTION [8 upvotes]: Given a natural number k, are there only finitely many finite simple groups with the property that all elements have order at most k?
This holds if I only look at the finite simple groups I understand 
(e.g. alternating groups and SL(k,finite field)), but it's not clear to me whether this holds for all finite simple groups, even using  their classification.

REPLY [4 votes]: Indeed, a stronger claim is true: there are only finitely many finite simple groups $G$ such that the largest prime dividing $|G|$ (which is also the largest prime order of an element) is at most $k$.  For a proof, see Section 5 of:
L. Babai, A. J. Goodman and L. Pyber. Groups without faithful transitive permutation representations of small degree. J. Algebra 195 (1997), no. 1, 1–29.<|endoftext|>
TITLE: Ranks of free submodules of free modules
QUESTION [11 upvotes]: Possible Duplicate:
Atiyah-MacDonald, exercise 2.11 

The following question came up during tea today.
Let $R$ be a commutative ring with an identity and let $M \subset R^n$ be a submodule.  Assume that $M \cong R^k$ for some $k$.  Question : Must $k \leq n$?
If $R$ is a domain, then this is obvious.  The obvious approach to proving the general result then is to mod out by the radical of $R$.  If the resulting map $M / \text{rad}(R) M \rightarrow (R / \text{rad}(R))^n$ were injective, then we'd be done.  However, I can't seem to prove this injectivity (I'm not even totally convinced that it's true).
Thank you for any help!

REPLY [7 votes]: For a proof using multilinear algebra, see Corollary 5.11 at 
http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf<|endoftext|>
TITLE: Pi and the primes: a pattern related to the Ulam spiral
QUESTION [19 upvotes]: I have recently been considering the spiral packing pattern (illustrated here) which often appears in plants, and arises by placing the nth point at distance n from the origin and angle 2nπ/φ where φ is the golden ratio. I wondered: what happens if we rotate by 1 radian each time instead of the golden angle? Evidently a less dense packing is obtained; but also, by numbering the points, we have a situation analogous to the Ulam spiral - perhaps more accurately the Sacks spiral - in which the primes may be highlighted.
The Sacks spiral plots the natural number n at an angle proportional to √n and thus is roughly equivalent to considering the square root of primes modulo some value k: for the Sacks spiral itself k = 1, while for the Ulam spiral k = 2. By instead plotting n at an angle proportional to n, we are effectively considering the primes themselves modulo some k: in this case, k = 2π. 

The result is striking (all graphs are plotted for the first 10000 primes; apologies for the shoddy Excel graphs). My guess is that the 2*10 equidistant parallel lines, plus 2 gaps, correspond to 22 being the numerator in a best rational approximation to π, but honestly this is just a rather shaky intuition based on the appearance of the Fibonacci sequence in the aforementioned golden angle spiral.
In fact the two sets of lines coincide mod π; Shown here 

are the primes mod π. Just in case π turned out not to be so special after all (though I suspected it was), I returned to the original golden spiral packing by taking the primes 

and mod φ-1

I also tried taking them mod e

and mod √2

(obviously if I took them mod some rational k I'd just get a bunch of horizontal lines). In case anyone has difficulty accessing those images, here's a quick summary: φ and φ-1 give nice sets of lines on a large scale, though the pattern is less clear for smaller numbers of primes (in contrast with π) and I suspect spacings will be related to φ; for e the picture is less clear but there appear to be narrow, regularly-spaced "empty bands"; and for √2 there are again empty bands, but not evenly-spaced. In all cases, the bands form parallel lines. No other irrational k seems to give the nice, clear pattern that π does.
To summarize: there is the general pattern of bands for irrational k; the particularly nice behaviour in the case of π; and the question of when the bands are spaced periodically and, when they are, with what period. Is there some spectacularly simple explanation that I'm missing? If not, have any of these ever been noticed or studied before? What about the more general case of studying pα mod k for primes p and other powers α? Any relevant literature / ideas?
Edit: in fact, for k = π the lines coincide mod π/11 so a prime p is p(1-7π/22) + nπ/22 (mod π) where n is an odd number between -9 and 9.

REPLY [8 votes]: I've disappointingly come to the same conclusion as Gerry, that it's got more to do with continued fractions than a special relationship between π and the primes. (Edit: in fact, it doesn't have a great deal to do with the primes at all; rather, the "empty" bands appear due to almost all primes being coprime to any given numerator, as explained below. The only special thing about the primes is that they are coprime to all such numerators, so the pattern of empty bands will appear at all scales.) Returning to the case k = π:
For p ∉ {2,11} we have:
7p ≡ an odd number n ≠ 11 (mod 22)
so p(7π/22) ≡ nπ/22 (mod π),
so p ≡ p(1-7π/22) + nπ/22 (mod π)
The key property is that 22/7 ≅ π, so that the slope (1-7π/22) is small enough to be visible over a large range. Given γ irrational and a/b ≅ γ, only finitely many p divide a, so in general we have bp ≡ n coprime to a (mod a). Then p(bγ/a) ≡ nγ/a (mod γ), so p ≡ p(1-bγ/a) + nγ/a a(mod γ).
What made π seem special was that a was small, so the number of possible choices for n was also small. By contrast, φ is difficult to approximate so in order for (1-bγ/a) to be small a must be large.
Finally, my guess is that the less evenly-spaced bands when k = √2 are a result of the numerators of the convergents being particularly composite.<|endoftext|>
TITLE: Arithmetic fixed point theorem
QUESTION [34 upvotes]: I want to understand the idea of the proof of the artihmetic fixed point theorem. The theorem is crucial in the proof of Gödel's first Incompletness theorem.
First some notation: We work in $NT$, the usual number theory, it has implemented all primitve recursive functions. Every term or formula $F$ has a unique Gödel number $[F]$, which encodes $F$. If $n$ is a natural number, the corresponding term in $NT$ is denoted by $\underline{n}$. The function $num(n):=[\underline{n}]$ is primitive recursive. Also, there is a primitive recursive function $sub$ of two variables, such that $sub([F],[t])=[F_v(t)]$, where $v$ is a free variable of $F$ which is replaced by a term $t$.
Now the theorem assertions the following:

Let $F$ be a formula with only one free variable $v$. Then there is a sentence $A$ such that $NT$ proves $A \Leftrightarrow F_v(\underline{[A]})$.

This may be interpreted as a self-referential definition of $A$, which is, as I said, crucial in Gödel's work. I understand the proof, I just repeat it, but I don't get the idea behind it:
Let $H(v)=F_v(sub(v,num(v)))$ and $A = H_v(\underline{[H]})$. Then we have
$A \Leftrightarrow H_v(\underline{[H]})$
$\Leftrightarrow F_v(sub(v,num(v)))_v(\underline{[H]})$
$\Leftrightarrow F_v(sub_1(\underline{[H]},num(\underline{[H]})))$
$\Leftrightarrow F_v(sub_1(\underline{[H]},\underline{[\underline{[H]}]}))$
$\Leftrightarrow F_v(\underline{[H_v(\underline{[H]})]})$
$\Leftrightarrow F_v(\underline{[A]}), qed.$
But why did we choose $H$ and $A$ like above?

REPLY [4 votes]: I think that the best way to capture the idea beyond the proof of the
fixed point theorem is to mirror it in an ordinary language
formulation and then translate it back to the first order language of
arithmetic (cf. J.N. Findlay, Goedelian Sentences: A Non-numerical
Approach}, Mind, Vol. 51, 1942, pp. 259-65.). Clearly,
what we seek is a sentence asserting that it has a given property,
that is, a sentence that says "I have the property p". But,
in order for it to be formalizable, our sentence should consist of
components with easily identifiable formal first-order 
counterparts. Therefore we cannot use such indexicals as `I'.
In order to circumvent the need for indexicals, we reformulate
Grelling's paradox applying it to open sentences instead of
adjectives:
(1)      "x is heterological" is heterological,
where an open sentence is called autological if the property it
attributes to x possessed by the sentence itself, otherwise it is called
heterological. For example, "x consists of five words", "x is
English", are autological, while "x is long", "x is German", are
heterological. On the other hand, both in formal languages and in informal
ones, the fact that an object has a property is expressed by a
substitution of the name of the object into the open sentence 
expressing that property. Consequently,
(2)    x is heterological just in case the sentence obtained by
       substituting the name of x for the variable in it is false.
Now (using the convention that he name of linguistic objects are the
object itself between quotation marks), if we replace "being false" by
"having property p", (1) and (2) together yield:
(3)      the sentence obtained by substituting the name of 
         "the sentence obtained by substituting the name of 
         x for the variable in it has property p"
         for the variable in it has property p.
This is the sentence we need. On the one hand, it does not use indexicals, on the other, it indeed says of itself
that it has property p (and says nothing else), since it is built up in
such a way that if we perform the substitution described in it, then
we obtain the sentence itself, which is stated to have property p. 
Now, let s denote the open sentence between the quotation marks
in (3), that is, let s be:
(4)  the sentence obtained by substituting the name of 
     x for the variable in it has property p.
Then, clearly, the whole sentence (3) is s("s"). In order to obtain
the fix point lemma, we should translate it into the language of formal 
arithmetic. Clearly, the formalization process consists of two main steps.  In the first step, we have to find the formal version $\eta$ of s, and then the second step is obvious: the desired sentence $\lambda$ will simply be $\eta(g(\eta))$ (where $g(\varphi)$ is the G\"odel number of $\varphi$ and plays, of course, the formal counterpart of name of $\varphi$, and, for simplicity, I leave out of consideration the difference between numbers and their formal counterparts in the language).
Now, that is all. That is the essence of the proof. What remains to do
is simply translate the ordinary language argument into
the formal language of arithmetic. That is a completely mechanical task.
Let us recall that what we should show is that,
for any arithmetical formula $\varphi$ with at most one free
variable (this fact will be denoted by $\varphi=\varphi(x)$), there is a sentence $\lambda$ such that 
$Q\vdash \lambda \longleftrightarrow
\varphi(g(\lambda)),$
where $Q$ is Robinson arithmetic (essentially Peano arithmetic without induction). 
Now, let the formula corresponding to the property p be $\varphi=\varphi(x)$.  Then, obviously, the formal version of s is
$\varphi(g[x(g(x)])$.  In order to continue the formalization process,
we should find a formula that can play the role of
$\varphi(g[x(g(x))])$, that is, a formula $\eta=\eta(x)$ such that
$\eta(g(\psi))$ is provably equivalent to $\varphi(g[\psi(g(\psi))])$
for every $\psi=\psi(x)$, or equivalently (denoting the inverse of $g$
by $g^{-1}$), for any $n \in N$,
$Q\vdash  \eta(n)\longleftrightarrow\varphi(g[g^{-1}(n)(n)]). $
In order to find the appropriate formula $\eta$, 
let us consider the expression
substituted into the formula $\varphi$, 
and define the function $f:\omega\longrightarrow \omega$ accordingly:
$f(n)=g[g^{-1}(n)(n)]$  if  $n \in N$ and $f(n)=0$ otherwise. 
Since this function is obviously recursive and hence representable,
and, up to provable equivalence, the result of substituting a
representable function into a formula can also be expressed by a
formula, there is a formula $\eta$ such that, for any $n\in N$,
(5) 
$Q\vdash\eta(n)\longleftrightarrow\varphi(f(n))$
Thus we have obtained what we need, we have shown that there exists an 
$\eta$ that can be considered to be the formal version 
of s. Now, all that remains to do is straightforward: 
it follows from (5) that, for every $\psi$,
$Q\vdash \eta(g(\psi))\longleftrightarrow
 \varphi\big(g[\psi(g(\psi))])$,  
which, in turn, choosing $\psi$ to be $\eta$, yields
$Q\vdash \eta(g(\eta))\longleftrightarrow
\varphi(g[\eta(g(\eta))])$,  
showing that the sentence 
$\lambda =\eta(g(\eta))$ indeed  has the desired 
property.<|endoftext|>
TITLE: How is K-theory defined for coherent sheaves?
QUESTION [6 upvotes]: If I take $\mathcal{A}$ =  coherent sheaves on $X$* up to isomorphism, then there are two things I could do which come to mind.
The first is noticing that $(\mathcal{A},\oplus)$ is a monoid and subsequently applying Grothendieck's $K$-functor to it obtaining a group $K(\mathcal{A})$.
The second is to take the free abelian group generated by $\mathcal{A}$ quotiented out by relations
$B = A + C$ for any exact sequence
$$0 \to A \to B \to C \to 0$$
obtaining a second group $L(\mathcal{A})$.
Are these two groups $K(\mathcal{A})$ and $L(\mathcal{A})$ isomorphic?
If not, which one is the 'correct' one?
*I'm not sure what $X$ should stand for. I'd be happy to assume smooth and projective variety over $\mathbb{C}$, or maybe just locally noetherian scheme, or maybe just locally ringed space or etc.

REPLY [6 votes]: The Grothendieck group on $X$ is defined (in your notation) as $K(\mathcal{B})$, where $\mathcal{B}$ is the monoid of f.g. locally free sheaves on $X$, (i.e. vector bundles on $X$.) The group $L(\mathcal{A})$ is usually denoted as $G(X)$, or $K'(X)$. If $X$ is smooth, the embedding $\mathcal{B} \subset \mathcal{A}$ induces an isomorphism from $K(\mathcal{B})$ to $L(\mathcal{A})$ (i.e. from $K_0(X)$ to $G(X)$.)
Chuck Weibel's notes for his K-theory book (which can be found on his website) is a good place to find some of this information.<|endoftext|>
TITLE: Applications of classical field theory
QUESTION [7 upvotes]: What are the applications (physical and mathematical) of classical field theory beyond electrodynamics and gravity?
By such applications, I mean that either the field theory viewpoint adds some genuinely new insight into the underlying physics or that it gives rise to interesting mathematical problems. So I'm not thinking about:
-field-theoretical description of something that is very well understood with other tools (for example, describing classical electrodynamics in language of fibre bundles, differential forms etc. is very nice and elegant, but doesn't add much to physics)
-quantum field theory (in QFT you always write down the classical lagrangian and then turn the fields into operators, but there is not much actual classical field theory here)
Of course, you can always write some lagrangian like phi^34 + 14*phi^8 + ..., and study the resulting PDE (existence and uniqueness of solutions etc.), but I guess that lacks real motivation.

REPLY [8 votes]: How about the study of minimal surfaces (physical applications in soap films etc.)? In fact one might argue the Lagrangian formulation of minimal surfaces (the problem of Plateau) is one of the oldest "classical field theory" problems, and led to the revival of calculus of variations in the early twentieth century (see esp. the works of Morrey). 
Slightly related is the general study of continuum mechanics and (non-linear) elasticity. Which is kind of like fluid mechanics except for deformations of solids. 
Another well-known application of the general frame work is the study of harmonic maps and wave maps (also known as non-linear sigma model in physics). The study of such systems led to developments of the techniques of compensated compactness and multilinear product estimates in partial differential equations (see, e.g. works of Helein, Klainerman, Tao, Krieger, and many others). The regularity properties of the harmonic maps are still under active study (Li and Tian, Nguyen, Weinstein, and others). And in physics, the sigma models find application from particle physics (as a model for equivariant Yang-Mills equation) to general relativity (stationary solutions in Einstein-vacuum or Einstein-Maxwell theories). 
The sigma models are also generalized by Tony Skyrme in his namesake quasilinear model (both hyperbolic and elliptic), which is not yet well understood. This model has found applications from nucleon physics to condensed matter, and now to topological material science. The study of the stationary problem (and its generalization in the Fadeev-Skyrme model) led to interesting developments in topology and geometry (since the model admits topological solitons), see for example the work of Kapitansky.<|endoftext|>
TITLE: Fibrations of Simplicial sets
QUESTION [6 upvotes]: Hello,
Maybe it is too vague a question, but I would like to ask if anybody could say some explanatory words about the importance (for infinity category study) of studying all the kinds of fibrations there are for simplicial sets (there are more than eight: right, left, mid, Kan, trivial, and so on..). In general, it might be believed that there are few "important" notions, so why do we have so many fibrations, all of which are studied?
Thank you

REPLY [8 votes]: I'm not an expert, but, here is my understanding. Right-fibrations are important because they are the infinity-version of a category fibered in groupoids (that is an infinity-category fibered in infinity-groupoids). In particular, given an infinity-category $C$,there is a model structure on $sSet/C$, called the contravariant model structure, such that the fibrant and cofibrant objects are precisely the right-fibrations over $S$, and this model structure is Quillen-equivalent (through a generalization of the Grothendieck construction) to the projective model-structure of simplicial presheaves over $w(C)$, where $w(C)$ is a simplicial category and $w$ is the left-adjoint to the homotopy-coherent nerve. Both of these (simplicial) model categories model $Fun(C^{op},\infty-Gpd)$- the infinity-categeory of "weak presheaves in infinity groupoids". So, the upshot is, right fibrations are the infinity-analogue of Grothendieck fibrations in groupoids and provide a model for weak presheaves. This presheaf infinity-topos is the starting point for higher topos theory; infinity topoi are just left-exact (accessible) localizations of such presheaf-infinity categories.
Now, dually, left-fibrations are a model for "infinity-categories COfibered in infinity-groupoids".
The next step, is Cartesian-fibrations. Cartesian-fibrations are the infinity-version of categories fibered in categories (not necessarily groupoids), i.e. they are "infinity categories fibered in infinity-categories". Nearly everything above goes through again, except we need to work with marked-simplicial sets, where we "mark the cartesian-edges".
Again, dually, CoCartesian fibrations model "infinity categories cofibered in infinity-categories". You may wonder why we need both notions? In fact, we need both notions TOGETHER in order to define adjunctions. An adjunction between two infinity-categories $C$ and $D$ is a functor $K \to \Delta[1]$ which is simultaneously a Cartesian-fibration and a CoCartesian fibration, together with Joyal-equivalences $K_{0} \cong C$ and $K_{1} \cong D$. This definition is a generalization to the infinity-world of a characterization of adjunctions using cographs.
Now, Kan-fibrations are a relic of homotopy theory. They are the fibrations on the Quillen-model structure on simplicial sets. In a similar spirit, categorical fibrations are the fibrations in the Joyal-model structure on simplicial sets. Other than that, they are not that well behaved; they don't really play a role in infinity-category theory.
Finally, inner fibrations, as far as I know, are only used in defining Cartesian fibrations. That is, a Cartesian fibration is defined to be an inner fibration satisfying extra properties.
I hope this helps.<|endoftext|>
TITLE: explicit formula for the j-invariant of binary quartic form
QUESTION [6 upvotes]: A binary quartic form
$aX^4+bX^3Y+cX^2Y^2+dXY^3+Y^4$
decomposes as a product of linear factors $Y-t_jX$, $j=1,...,4$. 
I would like to have an explicit formula for symmetrization of the crossratio of $t_j$.

REPLY [11 votes]: The $j$ invariant is
$j=\frac{S^3}{S^3-27T^2}$
where
$S=a-\frac{bd}{4}+\frac{c^2}{12}$
and
$T=\frac{ac}{6}+\frac{bcd}{48}-\frac{c^3}{216}-\frac{ad^2}{16}-\frac{b^2}{16}$
for more details see my article "A computational solution to a question by Beauville on the invariants of the binary quintic", J. Algebra 303 (2006) 771-788. The preprint version is here.<|endoftext|>
TITLE: Fixed point theorems and equiangular lines
QUESTION [12 upvotes]: I've been thinking about the equiangular lines (or SIC-POVM) conjecture, and my conclusion is that the best means of attack would be through some kind of fixed point theorem -- I'm thinking specifically of geometric fixed point theorems, like Brouwer's. So my (rather vague) questions are:
1) is there some good survey article or classification for fixed point theorems?
2) are there fixed-point theorems which are related to actions of groups on geometric spaces?
3) has anybody tried this idea?
Added: In response to Joe's comment below, let me note that while the motivation is from quantum information theory, the equiangular lines conjecture is a purely classical geometry problem (see my comment below). The conjecture is really intriguing: numerical constructions of sets of equiangular lines have been found up to dimension 67, at which point the computer time required exceeded the patience of the investigators. However, only a handful of these numerical solutions have been shown to be rigorously correct by finding corresponding algebraic numbers. See this recent paper.

REPLY [7 votes]: The book "Fixed point theory" by Dugundji and Granas is a nice reference. The headers of the sections in the book give some kind of classification of fixed point theorems.

results based on compactness
order theoretic results
results based on convexity
Borsuk theorem and topological transitivity
homology and fixed points
Leray-Shauder degree and fixed point index

Part VI of the bibliography is really extensive and contains a finer classification of fixed point theorems.<|endoftext|>
TITLE: Ways to prove an inequality
QUESTION [8 upvotes]: It seems that there are three basic ways to prove an inequality eg $x>0$.

Show that x is a sum of squares.
Use an entropy argument. (Entropy always increases)
Convexity.

Are there other means?
Edit: I was looking for something fundamental. For instance Lagrange multipliers reduce to convexity. I have not read Steele's book, but is there a way to prove monotonicity that doesn't reduce to entropy? And what is the meaning of positivity?
Also, I would not consider the bootstraping method, normalization to change additive to multiplicative inequalities, and changing equalities to inequalities as methods to prove inequalities. These method only change the form of the inequality, replacing the original inequality by an (or a class of) equivalent ones. Further, the proof of the equivalence follows elementarily from the definition of real numbers. 
As for proofs of fundamental theorem of algebra, the question again is, what do they reduce too? These arguments are high level concepts mainly involving arithmetic, topology or geometry, but what do they reduce to at the level of the inequality?
Further edit: Perhaps I was looking too narrowly at first. Thank you to all contributions for opening to my eyes to the myriad possibilities of proving and interpreting inequalities in other contexts!!

REPLY [9 votes]: I don't think your question is a mathematical one, for the question about what do all inequalities eventually reduce to has a simple answer: axioms. I interpret it as a metamathematical question and still I believe the closest answer is the suggestion above about using everything you know.
An inequality is a fairly general mathematical term, which can be attributed to any comparison. One example is complexity hierarchies where you compare which of two problems has the highest complexity, can be solved faster etc. Another one is studying convergence of series, that is comparing a quantity and infinity, here you find Tauberian theory etc. Even though you did not specify in your question which kind of inequalities are you interested in primarily, I am assuming that you are talking about comparing two functions of several real/complex variables. I would be surprised if there is a list of exclusive methods that inequalities of this sort follow from. It is my impression that there is a plethora of theorems/principles/tricks available and the proof of an inequality is usually a combination of some of these. I will list a few things that come to my mind when I'm trying to prove an inequality, I hope it helps a bit.
First I try to see if the inequality will follow from an equality. That is to recognize the terms in your expression as part of some identity you are already familiar with. I disagree with you when you say this shouldn't be counted as a method to prove inequalities. Say you want to prove that $A\geq B$, and you can prove $A=B+C^2$, then, sure, the inequality follows from using "squares are nonnegative", but most of the time it is the identity that proves to be the hardest step. Here's an example, given reals $a_1,a_2,\dots, a_n$, you want to prove that $$\sum_{i,j=1}^n \frac{a_ia_j}{1+|i-j|} \geq 0.$$
After you realize that sum is just equal to $$\frac{1}{2\pi}\cdot\int_{0}^{2\pi}{\int_{0}^{1}{\frac{1-r^{2}}{1-2r\cos(x)+r^{2}}\cdot |\sum_{k=1}^{n}{a_{k}e^{-ikx}}|^{2}dx dr}}$$ then, yes, everything is obvious, but spotting the equality is clearly the nontrivial step in the proof.
In some instances it might be helpful to think about combinatorics, probability, algebra or geometry. Is the quantity $x$ enumerating objects you are familiar with, the probability of an event, the dimension of a vector space, or the area/volume of a region? There is plenty of inequalities that follow this way. Think of Littlewood-Richardson coeficients for example.
Another helpful factor is symmetry. Is your inequality invariant under permuting some of its variables? While I don't remember right now the paper, Polya has an article where he talks about the "principle of nonsufficient reason", which basically boils down to the strategy that if your function is symmetric enough, then so are it's extremal points (there is no sufficient reason to expect assymetry in the maximal/minimal points, is how he puts it). This is similar in vein to using Langrange multipliers. Note however that sometimes it is the oposite of this that comes in handy. Schur's inequality, for example is known to be impossible to prove using "symmetric methods", one must break the symmetry by assuming an arbitrary ordering on the variables. (I think it was sent by Schur to Hardy as an example of a symmetric polynomial inequality that doesn't follow from Muirhead's theorem, see below.)
Majorization theory is yet another powerful tool. The best reference that comes to mind is Marshall and Olkin's book "Inequalities: Theory of Majorization and Its Applications". This is related to what you call convexity and some other notions. Note that there is a lot of literature devoted to inequalities involving "almost convex" functions, where a weaker notion than convexity is usually used. Also note the concepts of Schur-convexity, quasiconvexity, pseudoconvexity etc. One of the simplest applications of majorization theory is Muirhead's inequality which generalizes already a lot of classical inequalities and inequalities such as the ones that appear in competitions.
Sometimes you might want to take advantage of the duality between discrete and continuous. So depending on which tools you have at your disposal you may choose to prove, say the inequality $$\sum_{n=1}^{\infty}\left(\frac{a_1+\cdots+a_n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p \sum_{n=1}^{\infty}a_n^p$$ or it's continuous/integral version
$$\int_{0}^{\infty}\left(\frac{1}{x}\int_0^x f(t)dt\right)^p dx \le \left(\frac{p}{p-1}\right)^p \int_{0}^{\infty} f(x)^p dx$$ I've found this useful in different occasions (in both directions). 
Other things that come to mind but that I'm too lazy to describe are "integration preserves positivity", uncertainity principle, using the mean value theorem to reduce the number of variables etc. What also comes in handy, sometimes, is searching if others have considered your inequality before. This might prevent you from spending too much time on an inequality like $$\sum_{d|n}d \le H_n+e^{H_n}\log H_n$$ where $H_n=\sum_{k=1}^n \frac{1}{k}$.<|endoftext|>
TITLE: Weakest subsystems of second order arithmetic for mathematical logic
QUESTION [7 upvotes]: It is known that to prove completeness of first-order logic for countable languages WKL0 is enough. But, is it the weakest subsystem where one can prove it?
What about the incompleteness theorems? Is it known which are the weakest subsystems of second order arithmetic where one would be able to prove each of them?

REPLY [7 votes]: In fact, the incompleteness and completeness theorems can be proven in subsystems of second-order arithmetic weaker than RCA-0: incompleteness can be proven in EFA (first-order elementary arithmetic), which proves exponentiation total, but cannot prove iterated exponentiation to be total.  In fact, systems much weaker than EFA can prove incompleteness: Solovay has shown that any sane system of arithmetic (more or less, any first-order equational logic where there are reasonable definitions of zero and successor) strong enough to prove that multiplication is total can prove incompleteness.  But EFA is interesting because "Exponential Function Arithmetic is the weakest system in use for which the coding of finite objects by nonnegative integers is worry free" (Friedman 2010): EFA is a reasonable first-order base upon which to build reverse mathematics.
EFA can be usefully extended to the language of second-order arithmetic using the comprehension scheme ∀x (φ(x) ↔ ψ(x)) → ∃Y ∀x (x ∈ Y ↔ φ(x)), where where φ and ψ are Σ-0-1 and Π-0-1 predicates which may have free second-order variables (this definition is from Avigad 2003).  This language, call it ERCA-0, is then an analog of RCA-0-like that is a conservative extension of EFA.  Avigad shows how this base system can be considered as a weaker base theory for reverse mathematics, with a series of weaker analogs to other fixtures of the reverse mathematics landscape: in particular, EWKL-0, that analog of WKL-0, can prove the completeness theorem.
To summarise: ERCA-0 is weaker than RCA-0 and can prove the incompleteness theorems; EWKL-0 is weaker than WKL-0 and can prove the completeness theorem.  We can hope for weaker systems still, but Friedman's remark suggests that such systems will be more complex, and less suitable for reverse mathematics: there's a sense in which we might expect this to be around the best "weak" base system.
References 

Avigad, 2003, Number theory and elementary arithmetic.  NB. Avigad calls elementary arithmetic, EA.
Friedman, 2010, Concrete Incompleteness from EFA through Large Cardinals.<|endoftext|>
TITLE: Toric varieties as quotients of affine space
QUESTION [11 upvotes]: One way to define toric varieties is as quotients of affine $n-$space by the action of some torus. However, this is not strictly true as we need to throw away "bad points" which ruin this construction.
For example consider the construction of projective space as a toric variety. Let $\mathbb{G}_m$ act on $\mathbb{A}^n$ in the obvious way. Then the quotient of $\mathbb{A}^n$ by this action is a single point, as "everything is rescaled to the origin". More rigously the only functions invariant under this action are the constants, thus the quotient is the spectrum of the ground field. To fix this we of course we remove the origin and then take the quotient and we get projective space as required.
So given an action of some torus on affine space, how do we know which points to remove before we take the quotient to make sure we get a toric variety? 
My first naive guess is to remove the points which are fixed under the action, but Im wary it may be more subtle than that, as I know GIT can get quite technical.

REPLY [3 votes]: Let me advertise the polytopal point of view.
Start with $T^k$ acting on ${\mathbb A}^n$ linearly, i.e. we have a map
$T^k \to T^n$. The $T^n$-moment polytope of ${\mathbb A}^n$ is just the
orthant ${\mathbb R}_+^n$. The $T^k$-moment polytope is the projection
of that orthant to ${\mathbb R}^k = Lie(T^k)$* under the transpose 
$Lie(T^n){}^* \to Lie(T^k){}^*$ of the Lie algebra map $Lie(T^k) \to Lie(T^n)$.
To do GIT, you need to pick a point $\mu$ (technically, an integral point, but
it doesn't much matter) in $Lie(T^k)^*$. The fiber over $\mu$ will
be some (not necessarily compact) polyhedron in the orthant. Not every
face of the original orthant will meet that fiber: the ones that don't
are the images of the unstable set for the GIT quotient. (Well, this is
only true if the point chosen was generic, i.e., if $\mu$ is in the
interior of each face whose image contains $\mu$.)
The polyhedron can be thought of as living in $Lie(T^n/T^k)^*$, and
is the moment polytope for the GIT quotient, itself a toric variety.
Here's an example. Let $T^1$ act on ${\mathbb A}^2$ by 
$t\cdot (x,y) = (tx, t^{-1} y)$. Then the corresponding projection can
be pictured by turning the first quadrant $45^\circ$ degrees counterclockwise,
and dropping it onto the $x$-axis. The genericity condition says that
we shouldn't take $\mu=0$. Then depending on the sign of $\mu$, GIT says
we must leave out either all $\{ (x,0) \}$ or all $\{ (0,y) \}$ in ${\mathbb A}^2$.
Quotienting either open set obviously gives ${\mathbb A}^1$, whose polytope
is the half-line fiber over $\mu$.
Note that your naive guess would have left out just the origin, 
and the quotient would have had two origins. Some entire axis must be left out,
or some further identification (not just quotienting by the group)
must be made.<|endoftext|>
TITLE: Is Hartshorne's definition of the category of varieties natural?
QUESTION [10 upvotes]: Hartshorne's "Algebraic geometry" begins with the definition of (quasi-)affine and (quasi-)projective varieties over some fixed algebraically closed field. At a first glance, these seem to be quite different, so that I would have expected that one would pose questions either on quasi-affine or on quasi-projective varieties.
However, Hartshorne then defines a variety to be either a quasi-affine or a quasi-projective variety. These varieties (together with certain continuous and in some sense regular maps) then form the category of varieties.

Here is my question: Is the above definition natural in the sense that we really want to compare quasi-affine and quasi-projective varieties or at least study them both at the same time?

For instance, is there a (non-trivial) example of a quasi-affine variety which is isomorphic in the above category to a quasi-projective variety? If not, isn't this "unifying" definition a bit artificial?

REPLY [11 votes]: I would just comment, but I'm a new user so I can't.
I believe you are referring to Hartshorne's definition on p. 15:
"A variety over $k$ is any affine, quasi-affine, projective, or quasi-projective variety as defined above."
The reason he defines things this way is that in section I.1 he defined affine and quasi-affine varieties; in section I.2 he defined projective and quasi-projective varieties.  You are right that he hasn't made very clear the relation between the two.  Exercise I.2.9 on projective closures hints at the relation, but it hasn't been made completely precise (he just uses the terminology "identify").  To make this precise, we need to say what the morphisms between any two different varieties are.
Suffice it to say, the identification of Exercise I.2.9 given by projective closure is in fact an isomorphism in the sense of section I.3, so all varieties are isomorphic to quasi-projective varieties.
If this is your first encounter with algebraic geometry, however, I'm not sure I would recommend chapter I of Hartshorne very highly.  I made the mistake of thoroughly studying chapter I myself, and I think there are far better ways to spend your time.<|endoftext|>
TITLE: Relation between motivic cohomology and Quillen K-theory
QUESTION [5 upvotes]: What's the relation between Voevodsky's motivic cohomology and Quillen K-theory of a scheme?

REPLY [7 votes]: You should look at Marc Levine's preprint "K-theory and motivic cohomology of schemes, I".  The version on the UIUC K-theory server seems to be older than the version on his  website .
Roughly speaking, the motivic spectral sequence starts from motivic cohomology and converges to algebraic K-theory.  This spectral sequence was conjectured by Beilinson and first written down for nice fields by Bloch and Lichtenbaum (pre-print on the UIUC server), and was extended to more general schemes by Friedlander and Suslin (Ann. Sci. École Norm. Sup. (4)  35  (2002),  no. 6, 773--875).  I think these papers were written before the bulk of Voevodsky's work, and the E_2 term was described in terms of Bloch's higher Chow groups.  The correspondence between motivic cohomology and higher Chow groups is Theorem 1.2 in Levine's paper, and is due to Voevodsky  ("Motivic cohomology groups are isomorphic to higher Chow groups in any characteristic," Int. Math. Res. Not.  2002,  no. 7, 351--355).<|endoftext|>
TITLE: Geometric proof of Robinson-Schensted-Knuth correspondence?
QUESTION [8 upvotes]: Famous Robinson Schensted Knuth correspondence gives a correspondence between the matrices with non-negative integer entries and pair of semi standard tableaux. The proof that I have seen is highly combinatorial e.g. in Knuth's paper [Permutations, matrices, and generalized young tableaux]. Does there exist a geometrical proof of this correspondence?

REPLY [2 votes]: Not involved approach to RSK is in 
Danilov, V. I.; Koshevoy, G. A. The octahedron recurrence and RSK-correspondence. Sém. Lothar. Combin. 54A (2005/07), Art. B54An, 16 pp. (electronic)<|endoftext|>
TITLE: Structure of Kähler cone
QUESTION [11 upvotes]: Are there examples of Kähler manifolds whose Kähler cone can be described explicitly, say spanned by certain cohomology classes?  As far as I know, Hirzebruch Surface has a complete description for its Kähler cone.

REPLY [3 votes]: Explicit description of a Kahler cone for all hyperkahler manifolds is here:
http://arxiv.org/abs/1401.0479 (Rational curves on hyperkahler manifolds,
Ekaterina Amerik, Misha Verbitsky)<|endoftext|>
TITLE: Diffeomorphism groups of orbifolds
QUESTION [6 upvotes]: A lot is known about geometric and topological properties of diffeomorphism groups of surfaces (here, I am mainly thinking about the work of Smale and Eells-Elworthy). Is there anything known for orbisurfaces ? My first guess would be that many of these groups must have contractible components since singular points impose extra conditions somewhat similar to fixed points. Is there a good reference on this topic ?

REPLY [7 votes]: The result you want can be found in the following paper:
MR0955816 (89h:30028)
Earle, Clifford J.(1-CRNL); McMullen, Curt(1-MSRI)
Quasiconformal isotopies. Holomorphic functions and moduli, Vol. I (Berkeley, CA, 1986), 143--154,
Math. Sci. Res. Inst. Publ., 10, Springer, New York, 1988.
What they prove is actually pretty remarkable.  Namely, let $S$ be a hyperbolic surface.  Then there is a family $\phi_t$ of self-maps of $\text{Diff}^{0}(S)$ such that $\phi_0$ is the identity, such that $\phi_1$ is the constant map taking each diffeomorphism to the identity diffeomorphism, and such if $f \in \text{Diff}^0(S)$ commutes with a finite order diffeomorphism $g$ of $S$, then $\phi_t(f)$ also commutes with $g$ for all $t$.  In other words, you can contract $\text{Diff}^0(S)$ in way that doesn't break any symmetries.
Now assume that $\Sigma = S / \Gamma$ is a good hyperbolic orbifold, where $\Gamma$ is a finite group of diffeomorphisms of $S$.  The identity component of the orbifold diffeomorphism group of $\Sigma$ is then homeomorphic to 
$$\text{Diff}^{0}(S,\Gamma) := \langle f \in \text{Diff}^{0}(S)\ |\ gfg^{-1}=f\ \text{for all}\ g \in \Gamma \rangle \subset \text{Diff}^{0}(S)$$
The null-homotopy $\phi_t$ preserves $\text{Diff}^{0}(S,\Gamma)$, so it is contractible.
(EDIT : I made a slight fix to the definition of the orbifold diffeomorphism above.  It doesn't change the argument.  Thanks to Tom Church for pointing it out to me!).<|endoftext|>
TITLE: Kalinin's formulation of the Anosov closing lemma
QUESTION [8 upvotes]: I'm trying to read a paper of Boris Kalinin on the cohomology of dynamical systems for a project.  The material is geared towards topologically transitive Anosov diffeomorphisms (which is how the initial (abelian) results were proved by Livsic). However, he axiomatizes things for homeomorphisms of metric spaces.  It is necessary in the proof to use a version of the Anosov closing lemma, but it looks stronger than the one I've seen.  I'd like to know whether elementary techniques will suffice to prove it.
Background: The statement of the closing lemma that I learned initially uis as follows. Let $M$ be a compact manifold, $f$ an Anosov diffeomorphism. Put a metric $d$ on $M$, and fix $\epsilon>0$.  Then there is $\delta$ such that if $n \in \mathbb{N}$ and $d(f^n(x), x)<\delta$, then there is  $p \in X$ with $f^n(p)=p$ and $d(p,x)<\epsilon$.  In other words, "approximately periodic points" can be approximately closely by actual ones.
The property Kalinin stipulates in section 1 of his paper is that there is a type of exponential closeness.  In other words, Kalinin wants that $c,\delta, \gamma>0$ exist such that any $x \in X$ with $d(f^n(x),x)<\delta$ can have the whole orbit be exponentially approximated by a periodic orbit (of $p$ with $f^n(p)=p$).  More precisely, one has
$d(f^i(p), f^i(x))< c d(f^i(x),x) e^{-\gamma \min(i,n-i)}$ for each $i=0, \dots, n-1$.  This means that the orbits of $p$ and $x$ get even closer in the middle, and this is a strenghtening of the usual closing condition.  
One can motivate this fact for Anosov diffeomorphisms geometrically by considering hyperbolic linear maps and drawing a picture of the stable and unstable subspaces, and I am told that it is a straightforward (and "effective") generalization of the usual statement of the closing lemma.  This seems more like an intuitive aid rather than a rigorous proof for general Anosov diffeomorphisms, though.
However, Kalinin goes on to say more.
He assumes that there exists $y \in X$ such that $d(f^i(x), f^i(y)) \leq \delta e^{-\gamma i}$, $d(f^i(y)), f^i(p)) \leq \delta e^{-\gamma(n-i)}$. Immediately thereafter, he states that this is true for Anosov diffeomorphisms in view of the closing lemma.  
Questions: 
1) Can one prove the (stronger) version of the closing lemma in Kalinin's paper using the usual statement itself standard techniques (i.e. successive approximation, basic linear algebra for hyperbolic maps, or lemmas like this one)?  The books I have seen do not mention it, and certainly say nothing about a point $y$ as in the statement.
2) Does anyone know a good reference for this material (or for the general theory of Anosov diffeomorphisms, for that matter)?

REPLY [3 votes]: Besides the Hasselblatt and Katok bible, these are the references on Anosov diffeomorphisms that I found worth buying in the past six months: HK's Handbook of Dynamical Systems vol 1A, Gallavotti's books (available online and giving some great physical background and intuition, but probably not of any interest for you otherwise) and last but certainly not least Bowen's Equilibrium states and the ergodic theory of Anosov diffeomorphisms, which covers the closing lemma (3.8). There is a PDF of this online that should be easy to find.

REPLY [3 votes]: The closing lemma as stated by Kalinin can be found in many textbooks e.g. Katok-Hasselblatt "Introduction to the modern theory of dynamical systems", corollary 6.4.17.
The closing lemma really gives a periodic point close to x,
with iterates also close to the iterates of x until the orbit of x returns. That's not just the fact that periodic points are close to non-wandering points.
The point y is obtained by taking the intersection of the local stable set of x with the nth pull-back of the local unstable set of $f^n(p)$. Draw a picture to understand what's going on.
There are "geometric" proofs of the closing lemma that build y before p. And of course the original article of Livsic contains such a proof.<|endoftext|>
TITLE: How are the Conway polynomial and the Alexander polynomial different?
QUESTION [21 upvotes]: Background story:
I have just come out from a talk by Misha Polyak on generalizations of an arrow-diagram formula for coefficients of the Conway polynomial by Chmutov, Khoury, and Rossi. In it, he described how to obtain formulae for a Conway polynomial of string links, string links in open-closed surfaces etc. These formulae do not give generalizations of the Alexander polynomial. This is funny, because I always thought that the Alexander polynomial and the Conway polynomial were basically the same, but that turns out not to be the case at all, philosophically.
Background:
The Alexander polynomial is a topological invariant for knots. It is a palindromic polynomial in t and t-1, which can be viewed as representing deck transformations of the infinite cyclic covering of the complement. Knot Floer Homology categorifies it. There is a multivariable version for links.
The Conway polynomial of a knot is obtained from the Alexander polynomial by a change of variables (see the wikipedia page for details). It's an honest polynomial, satisfying a particularly satisfying skein relation. There is no analogue known for links, although Misha mentioned some recent thesis which gives partial results in this direction. It's categorification seems to be unknown, and Misha suggested that a solution to this problem would lead to a categorification of the linking number.
My question:
 How are these two knot invariants different, beyond what I said above? Why is one "better" than the other? What is the Conway polynomial supposed to be measuring?

REPLY [4 votes]: a proper (i.e. non-stupid) generalization of the Conway polynomial to some multi-variable
polynomial for links seems to be something people are after

How about this:

Sergey A. Melikhov, Colored finite type invariants and a multi-variable analogue of the Conway polynomial, https://arxiv.org/abs/math/0312007


What is the Conway polynomial supposed to be measuring?

Cochran, Levine and others gave various good answers - see references in the linked paper.
Thinking of what the multi-variable Conway polynomial is measuring seems to give a much clearer picture, at least in the 2-variable case. What is less clear is why it does so -
I mean that there should be more conceptual proofs.<|endoftext|>
TITLE: Kirby calculus and local moves
QUESTION [54 upvotes]: Every orientable 3-manifold can be obtained from the 3-sphere by doing surgery along a framed link. Kirby's theorem says that the surgery along two framed links gives homeomorphic manifolds if and only if the links can be related by a sequence of Kirby moves and isotopies. This is pretty similar to Reidemeister's theorem, which says that two link diagrams correspond to isotopic links if and only if they can be related by a sequence of plane isotopies and Reidemeister moves.
Note however that Kirby moves, as opposed to the Reidemeister moves, are not local: the second Kirby move involves changing the diagram in the neighborhood of a whole component of the link. In "On Kirby's calculus", Topology 18, 1-15, 1979 Fenn and Rourke gave an alternative version of Kirby's calculus. In their approach there is a countable family of allowed transformations, each of which looks as follows: replace a $\pm 1$ framed circle around $n\geq 0$ parallel strands with the twisted strands (clockwise or counterclockwise, depending on the framing of the circle) and no circle. Note that this time the parts of the diagrams that one is allowed to change look very similar (it's only the number of strands that varies), but still there are countably many of them.
I would like to ask if this is the best one can do. In other words, can there be a finite set of local moves for the Kirby calculus? To be more precise, is there a finite collection $A_1,\ldots A_N,B_1,\ldots B_N$ of framed tangle diagrams in the 2-disk such that any two framed link diagrams that give homeomorphic manifolds are related by a sequence of isotopies and moves of the form "if the intersection of the diagram with a disk is isotopic to $A_i$, then replace it with $B_i$"?
I vaguely remember having heard that the answer to this question is no, but I do not remember the details.

REPLY [40 votes]: There is a finite set of local moves. For instance, these:

(source: unipi.it)
In the second row, the number of encircled vertical strands is $n\leqslant 3$ on the left and $n\leqslant 2$ on the right move. So they are indeed finite. The bottom-right move is just the Fenn-Rourke move (with $\leqslant 2$ strands). The box is a full counterclockwise twist. We also add all the corresponding moves with $-1$ instead of $+1$.
We can prove that these moves generate the Fenn-Rourke moves as follows. Consider as an example the Fenn-Rourke move with 3 strands:

(source: unipi.it)
To generate this move, we first construct a chain of 0-framed unknots as follows:

(source: unipi.it)
Then we slide the vertical strands along the chain:

(source: unipi.it)
Now it is sufficient to use the Fenn-Rourke move with 2 strands, slide, and use another Fenn-Rourke with one strand:

(source: unipi.it)
Finally, iterate this procedure 3 times:

(source: unipi.it)
and you are done. The same algorithm works for the general Fenn-Rourke move with $n$ strands.
 EDIT  I have slightly expanded the proof and posted it in the arXiv as https://arxiv.org/abs/1102.1288<|endoftext|>
TITLE: Dirichlet series expansion of an analytic function
QUESTION [14 upvotes]: Let $F(s)=\sum_{n\geq 1}\frac{a_n}{n^s}$ be a Dirichlet series with (finite) abscissa of absolute convergence $\sigma_a$. It can be shown that $\forall \sigma >\sigma_a:$
$$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}F(\sigma+ it)n^{it}\mathrm{d}t=\frac{a_n}{n^{\sigma}}.$$ 
The natural question arises, given some function $F$ holomorphic in some half-plane, under what conditions does it have a representation as a Dirichlet series. I believe this is a very broad question, so I would actually like to make things a bit more specific. 
For fixed $c\geq 0$ let $H:=H_c:=\{z\in\mathbb{C}:\Re(z)>c\}$ be some half-plane and let $f\in\mathcal{O}(H)$. For $\sigma> c$ define the linear functional
$$\Phi_{n,\sigma}: f\mapsto\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}f(\sigma + it)n^{it}\mathrm{d}t$$
I have intentionally left out the actual domain of $\Phi_{n,\sigma}$ in $\mathcal{O}(H)$ (since it is rather part of the general question than a known fact). It can be easily seen though that $\Phi_{n,\sigma}$ is not well-defined on the whole $\mathcal{O}(H)$. Let $\sigma_0>c$ be fixed real number. 
(Q1) Provided $\{\Phi_{n,\sigma_0}(f)\} _{n\in\mathbb{N}}$ exists, does it follow that $\{ \Phi_{n,\sigma}(f)\} _{n\in\mathbb{N}}$ exists for all $\sigma>\sigma_0$?
(Q2) Provided $\{\Phi_{n,\sigma_0}(f)\} _{n\in I}$ exists, where $I\subset\mathbb{N}$ is some infinite subset, does it follow that $\Phi_{n,\sigma_0}(f)$ exists for all $n\in\mathbb{N}$? How about sufficiently large finite subset $I\subset\mathbb{C}$?
(Q3) Provided that $n^{\sigma_0}\Phi_{n,\sigma_0}(f)=:a_n$ exists for all $n\in\mathbb{N}$. Does it follow that the Dirichlet series $\sum_{n\geq 1}\frac{a_n}{n^s}$ is absolute(?) convergent in some half-plane? If it is convergent, does it represent $f$ in that half-plane? 
(Q4) And the more general question: Are there any known conditions when an analytic function admits expansion as an ordinary Dirichlet series?
I am also aware of the existence of a series of papers of A.F. Leont'ev on the representations of analytic functions as Dirichlet series, e.g.  "On the representation of analytic functions by Dirichlet series", A. F. Leont'ev 1969 Math. USSR Sb. 9 111 and "On conditions of expandibility of analytic functions in Dirichlet series", A. F. Leont'ev 1972 Math. USSR Izv. 6 1265, etc. English translations as well as some of the original are available at iopsciences. Unfortunately for me, I don´t have institutional access to those :-(
However, while Leontev´s papers appear to be fundamental for the subject, they all date back to the period 1969-1975. So I was hoping that there might be some good serveys or other types of good references summarizing the recent developments, respectively the most general results in the subject so far and that would be also "easier to have" than the aforementioned papers. Also, per Andrey Rekalo´s comment it seems that Leont'ev´s work is not really applicable to the more specific case of representation by ordinary Dirichlet series.
Thank you in advance for any input!

REPLY [9 votes]: A.F. Leont'ev continued to work on general Dirichlet series well into 1980s (until his death in 1987). Actually, he published three monographs on the subject from 1976 to 1983! He made a short summary of his earlier results for the 1974 ICM  in Vancouver (a free preview of the lecture is available here).
A.F. Leont'ev obtained in some sense final results on the representation of analytic functions by general Dirichlet series of the form
$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{-\lambda_n s},\quad s\in D\subset \mathbb C.$$
He studied Dirichlet series in bounded and unbounded convex domains (including half-planes). The problem is that his results may not be directly applicable to the `ordinary' Dirichlet series with $\lambda_n=\ln n$.    A typical Leont'ev's theorem for half-planes is as follows (link to the original article in Russian).

Theorem. For every $\rho>1$, there is a sequence $\lambda_n>0$, $n\in\mathbb N$, satisfying the condition
  $$\lim\limits_{n\to\infty}\frac{n}{\lambda_n^\rho}=\tau,\quad 0<\tau<\infty,$$
  such that any function $f$, which is analytic in the right half-plane $\Re z > 0$ ,  can be represented in the form
  $$f(z)=\sum\limits_{n=1}^{\infty}a_n e^{-\lambda_n z}+\Phi(z),\qquad \Re z > 0,$$
  where $\Phi$ is entire.

This obviously doesn't cover the case $\lambda_n=\ln n$, $n\in\mathbb N$.
Anyway, if you're interested and if you have a colleague who speaks Russian I can send you a couple of original articles by Leont'ev (PDF files). (Edit: sent.) By the way, the papers you've mentioned both deal with the case of convergence in a bounded domain $D$.<|endoftext|>
TITLE: Sums of two squares in (certain) integral domains
QUESTION [9 upvotes]: While giving the first of eight lectures on introductory model theory and its applications yesterday, I stated Hilbert's 17th problem (or rather, Artin's Theorem): if $f \in \mathbb{R}[t_1,\ldots,t_n]$ is positive semidefinite -- i.e., non-negative when evaluated at every $x = (x_1,\ldots,x_n) \in \mathbb{R}^n$ -- then it is a sum of squares of rational functions.  One naturally asks (i) must $f$ be a sum of squares of polynomials, and (ii) do we know how many rational functions are necessary?  The general answers here are no (Motzkin) and no more than $2^n$ (Pfister).  Then I mentioned that the case of $n=1$ is a very nice exercise, because one can prove in this case that indeed $f(t)$ is positive semidefinite iff it is a sum of two (and not necessarily one, clearly) squares of polynomials.  Finally I muttered that this was a sort of function field analogue of Fermat's Two Squares Theorem (F2ST).
So I thought about how to prove this result, and I was able to come up with a proof that follows the same recipe as the Gaussian integers proof of F2ST.  Then I realized that the key step of the proof was that a monic irreducible quadratic polynomial over $\mathbb{R}$ is a sum of two squares, which can be shown by...completing the square.
But then today I went back to the general setup of a "Gaussian integers" proof, and I came up with the following definition and theorem.
Definition: An integral domain $R$ is imaginary if $-1$ is a square in its fraction field; otherwise it is nonimaginary.  (In fact I will mostly be considering Dedekind domains, hence integrally closed, and in this case if $-1$ is a square in the fraction field it's already a square in $R$, so no need to worry much about that distinction.)  Note that nonimaginary is a much weaker condition than the fraction field being formally real.
(Definition: An element $f$ in a domain $R$ is a sum of two squares up to a unit if there exist $a,b \in R$ and $u \in R^{\times}$ such that $f = u(a^2+b^2)$.)
Theorem: Let $R$ be a nonimaginary domain such that $R[i]$ ($= R[t]/(t^2+1)$) is a PID.
a) Let $p$ be a prime element of $R$ (i.e., $pR$ is a prime ideal).  Then $p$ is a sum of two squares up to a unit iff the residue field $R/(p)$ is imaginary.
b) Suppose moreover that $R$ is a PID.  Then a nonzero element $f$ of $R$ is a sum of two squares up to a unit iff $\operatorname{ord}_p(f)$ is even for each prime element $p$ of $R$ such that $R/(p)$ is nonimaginary.
[Proof: Introduce the "Gaussian" ring $R[i]$ and the norm map $N: R[i] \rightarrow R$.  Follow your nose, referring back to the proof of F2ST as needed.]
Corollaries: 1) F2ST.  2) Artin-Pfister for $n = 1$.  3) A characterization of sums of two squares in a polynomial ring over a nonimaginary finite field (a 1967 theorem of Leahey).
4) Let $p \equiv 3,7 \pmod{20}$ be a prime number.  Then $p$ is a sum of two squares up to a unit in $\mathbb{Z}[\sqrt{-5}]$ but is not (by F2ST) a sum of two squares in $\mathbb{Z}$.
Finally the questions:

Have you seen anything like this result before?

I haven't, explicitly, but somehow I feel subconsciously that I may have.  It's hard to believe that this is something new under the sun.

What do you make of the strange situation in which $R$ is not a PID but $R[i]$ is?

Note that one might think this impossible, but $R = \mathbb{R}[x,y]/(x^2+y^2-1)$ is an example.  [Reference: Theorem 12 of Elliptic Dedekind domains revisited.] Do you have any idea about how one might go about producing more such examples, e.g. with $R$ the ring of integers of a number field (or a localization thereof)?

Addendum: As I commented on below, a good answer to the first question seems to be the paper

MR0578805 (81h:10028)
Choi, M. D.; Lam, T. Y.; Reznick, B.; Rosenberg, A.
Sums of squares in some integral domains.
J. Algebra 65 (1980), no. 1, 234--256.

In this paper, they prove the theorem above with slightly different hypotheses: $R$ is a nonimaginary UFD such that $R[i]$ is also a UFD.  Looking back at my proof, the only reason I assumed PID was not to worry about the distinction between $R/pR$ and its fraction field.  Just now I went back to check that everything works okay with PID replaced by UFD.  So the second question becomes more important: what are some examples to exploit the fact that $R[i]$, but not $R$, needs to be a UFD?

REPLY [8 votes]: Let $K$ be a complex quadratic number field such that $K(i)$ has class number $1$. If $K$ has class number $\ne 1$, then $K(i)$ must be the Hilbert class field of $K$, which, in this case, coincides with the genus class field of $K$. By genus theory, the discriminant of $K$ must have the form $d = -4p$ for a prime number $p \equiv 1 \bmod 4$.
In these cases, the ring of integers in $K$ is $R = {\mathbb Z}[\sqrt{-p}]$, and the ring of integers in $L$ is ${\mathbb Z}[i, (1 + \sqrt{p})/2] \ne R[i]$. 
Finding number fields of higher degree with this property seems to be an interesting problem; 
I can't think of an obvious approach in general, but if I find anything, I'll let you know.
Edit 1. The argument works for all imaginary number fields: if $R$ is the ring of integers in a number field $K$, and if $S = R[i]$ is the ring of integers in the extension $L = K(i)$, then disc$(L) = \pm 4$ disc$(K)^2$ (this is a simple determinant calculation: take an integral basis
$\{\alpha_1, \ldots, \alpha_n\}$ for $K$; then $\{\alpha_1, \ldots, \alpha_n, i\alpha_1, \ldots, i\alpha_n\}$ is an integral basis for $L$). 
On the other hand, if $L$ has class number $1$ and $K$ is not a PID, then $K$ has class number $2$ and $L$ is the Hilbert class field of $K$. This implies disc$(L) = \pm$  disc$(K)^2$.
The problem in the non-imaginary case is that $K(i)$ might be unramified at all finite primes, but not at infinity; in this case, $K$ has class number $2$ in the strict sense, yet its ring of integers is a UFD. 
Remark 2. By looking at $p = \alpha^2 + \beta^2$ modulo $4$ it follows (unless I did something stupid) that primes $p \equiv 3, 7 \bmod 20$ are not sums of two squares in 
${\mathbb Z}[\sqrt{-5}]$. 
Edit 2. Your suggestion to look at rings $R[\frac12]$, where $R$ is the ring of integers of a quadratic number field, seems to work for $K = {\mathbb Q}(\sqrt{-17})$, which has a cyclic class group of order $4$. The ring $R[\frac12]$ has class number $2$ because $2$ is ramified and so generates a class of order $2$, and $S = R[\frac12,i]$ is the integral closure of $R[\frac12]$ in the extension $L = K(i)$. The ring $R[\frac12]$ has class number $2$, and $S$ is a UFD since $L$ has class number $2$, and its class group is generated by one of the prime ideals above $2$ (the ramified prime above $2$ in $K$ splits in $L$).
Edit 3. The corollary concerning sums of two squares in $R[\frac12]$ shows that primes $p \equiv 3 \bmod 4$ splitting in $K$ are sums of two squares up to units. In fact it follows from genus theory that the prime ideals above such $p$ are not in the principal genus, hence lie in the same class as the prime above $2$ or in its inverse. This implies that either $2p = x^2 + 17y^2$ or $8p = x^2 + 17y^2$, giving a representation of $p$ as a sum of two squares up to a unit (remember $2$ is invertible). Thus everything is working fine.<|endoftext|>
TITLE: Computational complexity of computing homotopy groups of spheres
QUESTION [74 upvotes]: At various times I've heard the statement that computing the group structure of $\pi_k S^n$ is algorithmic.  But I've never come across a reference claiming this. 
Is there a precise algorithm written down anywhere in the literature?  Is there one in folklore, and if so what are the run-time estimates?  Presumably they're pretty bad since nobody seems to ever mention them. 
Are there any families for which there are better algorithms, say for the stable homotopy groups of spheres?  or $\pi_k S^2$ ?  
edit: I asked Francis Sergeraert a few questions related to his project.  Apparently it's still an open question as to whether or not there is an exponential run-time algorithm to compute $\pi_k S^2$.

REPLY [2 votes]: This is a bit aside from the question at hand, but I think it's worth making the observation.  Consider the function of two variables:
$$ (X,n) \mapsto \pi_n X.$$
As a function of $n$, the computational complexity is believed (for general $X$) to grow  exponentially.  But for fixed $n$, as a function of simply-connected $X$ (measured in terms of, say, number of simplices), the growth of the computational complexity is polynomial.  
(In fact, I'd guess that you could even specify the degree of the polynomial to be something like $n/c$ where $c$ is the connectivity of the $X$s.  I don't know if anyone has made that precise.)
So if you want to get in the game of making algorithms to compute homotopy groups, don't bother with high-dimensional spheres: it's a waste of effort.  Instead, compute low-dimensional homotopy groups of large spaces.   (This is basically what the "effective homotopy program" of Sergeraert, Rubio, Romero, and others, does.)<|endoftext|>
TITLE: Action of the group of isometries on a manifold
QUESTION [6 upvotes]: Hi guys,
I am able to prove that any symmetric manifold is complete (Consider a local geodesic and use the symmetry to flip it, effectively doubling the length of the geodesic, ad infinitum). I want to use a similar procedure to prove that a manifold whose isometries act transitively is complete, i.e there is always an isometry which maps the start point of a local geodesic to its end point, preserving the geodesic. I am, however, unable to ensure that it is not `rotated' in the process, i.e I want the pushforward of the initial tangent, by the isometry, to be the final tangent, ensuring the resultant doubled geodesic is smooth.
My Lie group theory is a bit scratchy but I assume there is a method which allows me to construct the correct pushforward using only transitivity.
Any ideas would be great,
regards,
MK

REPLY [2 votes]: Your original approach for extending the geodesic is in fact possible, after a slight modification. (Credint: I heard this Idea from Eran Assaf). 
(You do not pushforward the original geodesic, but instead you pushforward another geodesic which starts at $p$).
Let $p \in M$. It's known there exists a neighbourhood (of $p$) $V$ in $M$, and $\epsilon > 0$ such that for every $q \in V \, , w \in T_qM , |w|<\epsilon$  the (unique) geodesic which starts at $q$ with velocity $w$ is defined on $(-2,2)$.
Let $c$ be geodesic:  $c(0)=p,|\dot c(0)|<\epsilon$. By the above statement, we can assume $c$ is defined on $(-2,2)$,So we can define $q = c(1)$. 
By assumption there is an isometry $f:M \to M$ such that $f(p)=q$.
Now let $\gamma$ be the geodesic satisfying: $\gamma(0)=p,\dot \gamma(0)=(df_p)^{-1}(\dot c(1))$. Since $f$ is an isometry: $|\dot \gamma(0)|=|\dot c(1)|\overset{ c \, \text{geodesic} }{=}|\dot c(0)|<\epsilon$. Hence $\gamma$ is also defined on $(-2,2)$.
Now consider the curve $\beta = f \circ \gamma$: 
$\beta(0) = f(\gamma(0))=f(p)=q=c(1)$
$\dot \beta(0) = df_p(\dot \gamma(0))=\dot c(1)$
Since $f$ is an isometry and $\gamma$ is a geodesic, $\beta$ is also a geodesic. Hence, $\beta$ is the unique geodesic starting at $q = c(1)$ with velocity $\dot c(1)$. But so is $\tilde c(t)=c(t+1)$, hence they are equal, whenever both are defined. 
So, we can now concatenate, to obtain a geodsic $\alpha:(-2,3) \to M$,
$\alpha(t) =
\begin{cases}
c(t),  & t \in (-2,2)  \\
\beta(t-1), & t \in (-1,3)
\end{cases}$
$\alpha$ is well-defined, and a geodesic. Now we may repeat this procedure ad infinitum.<|endoftext|>
TITLE: Intuitive explanation for the use of matrix factorizations in knot theory
QUESTION [26 upvotes]: Hello!
I read through parts of Khovanov/Rozansky's paper on the categorification of the HOMFLY polynomial using Matrix Factorizations. Technically, I can follow (though it seems to me that quite a lot of details are missing and tedious to fill in) - intuitively, however, I have no idea why one is lead to consider matrix factorizations when studying knot theory, in particular the RT invariants obtained from interpreting colored tangles as morphisms between modules over the quantum group.  Until now, it feels quite mysterious to me why Khovanov and Rozansky choose particular potentials like $x_1^n+x_2^n-x_3^n-x_4^n$ in their construction, and why one should expect that in the end we get something invariant under the Reidemeister moves.
Can somebody explain to me the motivation behind this construction? What is the relation between the morphism of modules over the quantum group a wide edge represents and the matrix factorization associated to it?
Thank you!

REPLY [11 votes]: Many knot homologies are expected to have Floer-theoretic interpretations. 
However, in Floer theory often the chain "complex" $CF(L_0,L_1)$ is not a complex but 
rather an a-infty bimodule over a pair of curved a-infty algebras; matrix factorizations are more or less a special case of these, where the curvature of the ainfty algebras
is a multiple of the identity.   Under some nice (monotone or exact) assumptions $CF(L,L)$ has differential that squares to zero, since the curvatures of the a-infty algebras 
on both sides occur with opposite sign and so cancel if the Lagrangians are the same
(or related by a symplectomorphism). But even in 
the monotone or exact situations $CF(L_0,L_1)$ can have a differential which gives
 a matrix factorization.   Now Manolescu has proposed a symplectic interpretation
of Khovanov-Rozansky for links, which looks like $CF(L,\beta(L))$ where \beta
is a braid presentation of the link, but so far there doesn't seem to be a proposal
for graphs.  Probably for graphs the invariant would be the homotopy type of $CF(L_0,L_1)$
(or some more general quilted Floer chain group)  which under suitable monotonicity assumptions will be a matrix factorization. Note that Kamnitzer has a proposal for a Floer-theoretic version for arbitrary G;  it would be interesting to see if one could extend this proposal to the graph case, which one would probably need for an exact triangle.<|endoftext|>
TITLE: Deeper meanings of barycentric subdivision
QUESTION [15 upvotes]: I just want to ask if there is any deeper motivation or clear geometric "sense" behind the barycentric subdivision. Some friend asked me about this a few months ago, looking back the section at Hatcher, I still feel quite confused. I remember one friend told me combinatorically one can do this from posets back to posets, but this does not give me any way to "understand" it properly. In some books (Bredon, for example), the author use excision property as one of the axioms, I'm wondering "where they came from, why they make any sense?".

REPLY [3 votes]: The following paper includes a very detailed and elegant description of how to construct edgewise subdivisions that subdivide a $d$-simplex into $k^d\cdot d$-simplices all of the same volume and shape characteristics, for every integer $k\geq 1$
Edgewise Subdivision of a Simplex
H. Edelsbrunner and D. R. Grayson
DISCRETE AND COMPUTATIONAL GEOMETRY
Volume 24, Number 4, 707-719<|endoftext|>
TITLE: Surjective *-homs between multiplier algebras
QUESTION [5 upvotes]: Let A and B be C*-algebras, and let $\phi:A\rightarrow B$ be a surjective *-homomorphism.  Then $\phi$ is non-degenerate, and so we can extend it to *-homomorphism between the multiplier algebras: $\tilde\phi: M(A)\rightarrow M(B)$.  It's rather tempting to believe that then, surely, $\tilde\phi$ is also surjective.  But I cannot for the life of me think of a proof.  Any ideas...?
Background: The multiplier algebra $M(A)$ is the largest C*-algebra containing A as an essential ideal.  Concretely, pick some "large enough" representation of A (either $A\rightarrow B(H)$ a non-degenerate *-representation, or $A\rightarrow A^{**}$ say) and then $M(A) = \{ x : xa,ax\in A \ (a\in A)\}$ the idealiser of $A$ in our large ambient algebra.  As $\phi$ surjects, it's very easy to define $\tilde\phi$: we simply have that $$\tilde\phi(x) \phi(a) = \phi(xa), \quad \phi(a) \tilde\phi(x) = \phi(ax).$$
This is well-defined, for if $\phi(a)=0$, then given an approximate identity $(e_i)$ for A, we have that $\phi(xa) = \lim_i \phi(xe_i a) = \lim_i \phi(xe_i) \phi(a) = 0$, and so forth.  Indeed, if $B\subseteq B(K)$ say, then $\tilde\phi(x)$ is the limit (strong operator topology say) of the net $\phi(xe_i)$ in $B(K)$; then clearly this is in the idealiser of $B$, and so does define a member of $M(B)$.

REPLY [6 votes]: This is true if $A$ is $\sigma$-unital, and is sometimes called the "noncommutative Tietze extension theorem".  A good reference is Proposition 6.8 in Lance's Hilbert $C^*$-modules.  Proposition 3.12.10 in Pedersen's $C^*$-algebras and their automorphism groups covers the separable case, which was first proved by Akemann, Pedersen, and Tomiyama in a 1973 paper called "Multipliers of C*-algebras".
Pedersen points out in Section 3.12.11 that you can get counterexamples in the commutative case by considering non-normal locally compact Hausdorff spaces, so that Tietze's extension theorem doesn't apply.

Akemann, Pedersen, and Tomiyama are more explicit:

In fact let $X$ be a locally compact Hausdorff
  space which is not normal, and consider two disjoint closed sets $Y_1$
  and $Y_2$ such that the function $b$ which is one on $Y_1$ and zero on $Y_2$ has
  no continuous extension to $X$. The restriction map of $C_0(X)$ to
  $C_0(Y_2\cup Y_2)$ is surjective, and $b\in M(C_0(Y_1\cup Y_2))$, but $b$ is not the
  image of a multiplier of $C_0(X)$.<|endoftext|>
TITLE: Homomorphisms from $\mathbb{R}$ to $\mathrm{Homeo}^+(\mathbb{R})$, or "fractional iterations"
QUESTION [8 upvotes]: Let $G$ be the group of orientation-preserving homeomorphisms (or, if you prefer, diffeomorphisms) of the real line. Does there exist a natural way to associate, to each function $f \in G$, a homomorphism $\phi_f \colon \mathbb{R} \to G$ such that $\phi_f(1) = f$?

Motivation: Many of us, in high school or before, have wondered whether for a given function $f$, it is possible to find a function $f^{1/2}$ such that $f^{1/2} \circ f^{1/2} = f$. I have never seen a satisfactory answer. This is intended as a more "grown-up" version: Is there some natural definition of $f^r$ ($= \phi_f(r)$), for $r \in \mathbb{R}$, such that $f^1 = f$ and $f^r \circ f^{-s} = f^{r-s}$?

REPLY [8 votes]: This has some relation with this question, but it is obviously different.
In general, a homeomorphism of $\mathbb{R}$ which preserves the orientation may have or not fixed points. 
If it has no fixed points, then it is conjugated to a translation and thus, one can easily construct such $\phi_f$. 
The other case is not much more difficult, since one can consider the set of fixed points of $f$ (that is, such that $f(x)=x$) which is closed and then do the trick in the complement of that set and leave fixed the set of fixed points for every $t$ (when defining $\phi_f(t)$). Notice that in either case, there is in general no unique way to do this. 
It is also interesting that the diffeomorphism case is quite different, in particular, one can easyly construct a diffeomorphism $f:\mathbb{R}\to \mathbb{R}$ such that there is no diffeomorphism $g$ such that $g\circ g =f$. This can be seen in the paper provided by Helge (in fact it has to do with distortion and the fact that if you take one contracting point, there are restrictions to construct, for example a square root, see Section 1 of this paper). 
ADDED RELATED REFERENCE: In this paper, Palis gives a not so difficult proof that $C^1$-generic diffeomorphisms (which belong to a $G_\delta$-dense subset of $Diff^1(M)$) of a compact manifold, the diffeomorphisms are not the time one map of a flow.

REPLY [2 votes]: I commented earlier, without finishing reading google's reply:
http://www.math.northwestern.edu/~wilkinso/papers/announce04152008.pdf
shows the answer is no if you replace $\mathbb{R}$ by a compact manifold. Key is googling for "Is every diffeomorphism a time one map?". I'm not sure what the answer for $\mathbb{R}$ is, and if it has been looked at.<|endoftext|>
TITLE: The vanishing of Ramanujan's Function tau(n)
QUESTION [15 upvotes]: This is a problem I had a look at some years ago but always had the feeling that I was missing something behind its motivation.
D.H. Lehmer says in his 1947 paper, “The Vanishing of Ramanujan's Function τ(n),” that it is natural to ask whether τ(n)=0 for any n>0.
My question is: Why is it natural to wonder whether τ(n)=0 any n>0?
Are there any particular arithmetic properties among the many satisfied  by τ(n) that would lead one to ponder its vanishing? The problem is mentioned here, where it's stated that it was a conjecture of Lehmer, although it's not actually presented as a conjecture in his paper, more a curiosity.
Maybe there is no deep reason to ponder the vanishing of τ(n), in which case that would be a satisfactory answer too.

REPLY [5 votes]: A simple reason: this is a function of $n$ satisfying significant congruences. If it vanishes, that is further congruence information.<|endoftext|>
TITLE: Applications of non-reductive GIT
QUESTION [7 upvotes]: Geometric invariant theory works well when the algebraic group $G$ acting on a variety is reductive.  There has been recent work by Doran and Kirwan here and here to find a canonical method of constructing GIT quotients for non-reductive groups.  My question is what are potential applications for their work?  One specific application they mention is constructing moduli of hypersurfaces in toric varieties.  I would be interested in knowing of other applications.

REPLY [6 votes]: On 80's Atiyah conference Kirwan spoke about one application. Namely she stated in her talk that there is an application to Green-Griffits conjecture.
You can download the talk here http://www.maths.ed.ac.uk/~aar/atiyah80.htm
and the slides are here http://www.icms.org.uk/downloads/GandP/Kirwan.pdf
I am not sure if this was written down somewhere.<|endoftext|>
TITLE: Why is the identity element of the sandpile group self-similar?
QUESTION [12 upvotes]: I've been reading about the Abelian Sandpile Model and noticed the identity element of the sandpile group on the square has self-similar components.  

The sandpile group of the 198x198 square of height 4 above is a finite abelian group. However, the sandpile corresponding to elements of this group can be fractal.  Where is the complexity coming from?
Also, how does can you find the identity element (say, with a computer)?

REPLY [3 votes]: Another way to compute the identity is to do it iteratively: start with an empty grid (all cells contain 0 chips), then iterate the following:

perform a reverse firing from the edges: add 2 chips to each corner cell and one chip to each edge cell
if the resulting configuration is not stable, i.e. some cells contain at least 4 chips, then fire these cells until you obtain a stable configuration (some chips will disappear from the grid at corner and edge cells).

Once you obtain twice in a row the same stable configuration, you have the identity.
During my PhD I have made a video of the stable configurations that you obtain at each step of this algorithm:
          
(many initial steps are missing to speed up the video).
You can see how the self-similar patterns evolve at each step and the formation of the center square.<|endoftext|>
TITLE: Characterizing the rationalization of spaces.
QUESTION [5 upvotes]: In the category of rational spaces, loop spaces split as products of Eilenberg-Mac Lane
spaces and SUSPENSIONS split as wedges of (rational) spheres.  I wonder if anything of the following form is true:
(*)  Any functor $F$ from spaces to spaces which splits suspensions and loop spaces as above must factor through the rationalization.
EDIT 1:  Greg raises some fine questions, but I stand by my wording.  This is a question that arises from curiosity, not because I need it for anything, so I'd be happy with "anything like" the given statement.  
EDIT 2:  At least for simply-connected spaces, rationalization commutes with loop and suspension.  But, it seems to me that the power of the property is that the suspension
of any F-space splits and the loops of any F-space splits.  So I would go with:
the suspension of any rational space splits as a wedge of rational spheres and 
the loops of any rational space splits as a product of rational Eilenberg-Mac Lanes spaces.
Thus, we'd be looking for functors to some model-esque category with some relatively manageable list of objects whose products exhaust the homotopy types of loop spaces and whose wedges exhaust the homotopy types of suspensions.

REPLY [3 votes]: I think the following is a trivial counterexample, which may lead you to reflect about your question:
\begin{align*}
F\colon  Spaces & \longrightarrow Spaces\\\\
X&\;\mapsto\;\bigvee_{H_1(X,\mathbb{F}_2)}S^1
\end{align*}
This functor takes any space to a wedge of several circles, one circle for each element in the homology group ${H_1(X,{\mathbb{F}}_{2})}$. Such wedges are both suspensions and Eilenberg-MacLane spaces. Obviously this functor does not factor through rationalization, since there are spaces $X$ and $Y$ with $X\simeq _{\mathbb{Q}} Y$ but $|H_1(X,\mathbb{F}_2)|\neq |H_1(Y,\mathbb{F}_2)|$. 
Of course, you can replace $H_1$ with $H_n$ for any $n$ if you wish to work with simply connected spaces.<|endoftext|>
TITLE: Is there an explicit construction of a free coalgebra?
QUESTION [36 upvotes]: I am interested in the differences between algebras and coalgebras. Naively, it does not seem as though there is much difference: after all, all you have done is to reverse the arrows in the definitions. There are some simple differences: 
The dual of a coalgebra is naturally an algebra but the dual of an algebra need not be naturally a coalgebra.  
There is the Artin-Wedderburn classification of semisimple algebras. I am not aware of a classification even of simple, semisimple coalgebras.  
More surprising is: a finitely generated comodule is finite dimensional.
This question is about a more striking difference. The free algebra on a vector space $V$ is $T(V)$, the tensor algebra on $V$. I have been told that there is no explicit construction of the free coalgebra on a vector space. However these discussions took place following the consumption of alcohol. What is known about free coalgebras?

REPLY [2 votes]: My two cents on the topic: dealing with co-structures, conilpotency (sometimes called connectedness or even cocompleteness) is a necessary condition to talk about cofreeness. A known counterexample is given by the coalgebra $(\mathbb k, \Delta),$ with $\Delta(1)=1\otimes 1$, denoting by $\mathbb k$ a field. The coalgebra is clearly not cocomplete and does not satisfy the lifting property of coalgebra morphisms (considering reduced tensor coalgebras, which are cocomplete).
I would like to share two references:

The book Algebraic Operads  by Loday and Vallette describes the topic in much detail (see, in particular, pag.8-9 in Chapter I).
The first chapter of Hasegawa's PhD thesis contains the definition of cocomplete coalgebras and discusses lifting properties.<|endoftext|>
TITLE: Zagier's one-sentence proof of a theorem of Fermat
QUESTION [109 upvotes]: Zagier has a very short proof (MR1041893, JSTOR) for the fact that every prime number $p$ of the form $4k+1$ is the sum of two squares. 
The proof defines an involution of the set $S= \lbrace (x,y,z) \in N^3: x^2+4yz=p \rbrace $ which is easily seen to have exactly one fixed point. This shows that the involution that swaps $y$ and $z$ has a fixed point too, implying the theorem.
This simple proof has always been quite mysterious to me. Looking at a precursor of this proof by Heath-Brown did not make it easier to see what, if anything, is going behind the scenes. 
There are similar proofs for the representation of primes using some other quadratic forms, with much more involved involutions.
Now, my question is: is there any way to see where these involutions come from and to what extent they can be used to prove similar statements?

REPLY [7 votes]: A recent article gives another very elegant proof of the two-square theorem in the Zagier style, but is easier.  See

Stan Dolan, A very simple proof of the two-squares theorem,
The Mathematical Gazette, Volume 105, Issue 564 (2021) p. 511. doi:10.1017/mag.2021.120 (the whole article is visible in the free preview)

The proof is not one sentence but it is still well worth reading.  One starts with the set $S = \{(x,y,z) \mid (x+y+z)^2 = p + 4xz, \text{ and } 
x,y,z \in \mathbb{Z}_{>0} \}$.  Short, clever arguments show that $S$ is finite,
$|S|$ is odd, and that there is a positive integer solution with $y=z$.<|endoftext|>
TITLE: A question about the construction of Francia flip
QUESTION [5 upvotes]: Here is the construction.
Start with $U$ a normal variety of dimension 3 with a unique singular point locally isomorphic to the quotient of $(x, y, z) \to (-x, -y, -z)$. Also assume we have a small contraction which contracts a smooth rational curve $C$ through the singular point. 
It is known that we can blow up the singular point in $U$ and get a smooth variety $V$ whose exceptional divisor is $P^2$ with normal bundle $\mathcal{O}(-2)$. Now my question is what is the normal bundle of the strict transform of $C$ in $V$?
It is easy to see the intersection number with $K_V$ should be $0$. In the book Geometry of Higer dimensional Algebraic Variety by Miyaoka and Peternell, the author claims it is easy to see the normal bundle is $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$ (p.184 Example 7.10). But I do not know why.

REPLY [4 votes]: Let the strict transform of $C$ on $V$ be denoted by $\widetilde C$. Obviously, $\widetilde C\simeq \mathbb P^1$. Let $\mathcal O_{\widetilde C}(1):=\mathcal O_{\mathbb P^1}(1)$ via this isomorphism.
Writing down the short exact sequence corresponding to the restriction of the cotangent bundle of $V$ to $\widetilde C$,
$$
0 \to \mathcal N_{\widetilde C|V} \to \Omega_V\otimes \mathcal O_{\widetilde C} \to \Omega_{\widetilde C} \to 0
$$
combined with (what you already computed) $K_V\cdot \widetilde C=0$
implies that ${\rm det}\ \mathcal N_{\widetilde C|V}\simeq \mathcal O_{\widetilde C}(-2)$ (I guess I should have just said that the "adjunction formula" does that). Since $\widetilde C\simeq \mathbb P^1$, it follows that $\mathcal N_{\widetilde C|V}\simeq \mathcal O_{\widetilde C}(a)\oplus \mathcal O_{\widetilde C}(b)$ for some $a,b \in\mathbb Z$ such that $a+b=-2$. Since $C$ is contractible, both $a$ and $b$ has to be negative, but the only way that can happen is if they are both equal to $-1$.<|endoftext|>
TITLE: CM field to Torus to Abelian Variety?
QUESTION [11 upvotes]: Given a CM field we can use its maximal order (and a choice of CM type) to construct an abelian variety $\mathbb{C}^g/\Lambda$ with complex multiplication by the maximal order.
How do I (or where can I find information on) explicitly write down equations for a projective embedding of this variety, and the action of the CM order on points? Is this implemented anywhere?
For genus one we can use the Eisenstein series to find the coefficients of a Weierstrass model for the elliptic curve. So I'm looking for a generalization of this.

REPLY [3 votes]: This is not the answer. I am adding some relevant papers (some possessing good examples) which won't fit in the comments section.  
I have been wanting to know the answer to this question as well.  It seems one has to find an ample line bundle, and then calculate the Riemann theta relations which define the projective embedding.  The equation(s) of the embedding is decided by relation between the theta functions.  The wikipedia page on equations for abelian varieties was written Charles Matthews.  Perhaps he might have more to say on this question...
MR0946234 (89i:14038)  Barth, Wolf . Abelian surfaces with $(1,2)$-polarization.
 Algebraic geometry, Sendai, 1985,  41--84, Adv. Stud. Pure Math., 10, North-Holland, Amsterdam,  1987. 
MR1257320 (95e:14033)  Barth, W. ;  Nieto, I.  Abelian surfaces of type $(1,3)$ and quartic surfaces with $16$ skew lines. J. Algebraic Geom.  3  (1994),  no. 2, 173--222.
MR1336597 (96h:14064)  Barth, W.  Quadratic equations for level-$3$ abelian surfaces.
 Abelian varieties (Egloffstein, 1993), 
 1--18, de Gruyter, Berlin,  1995. 
MR1602020 (99d:14046)  Gross, Mark ;  Popescu, Sorin . Equations of $(1,d)$-polarized abelian surfaces.
 Math. Ann.  310  (1998),  no. 2, 333--377.
MR2194379 (2006m:14054)  Gunji, Keiichi . Defining equations of the universal abelian surfaces with level three
 structure.
 Manuscripta Math.  119  (2006),  no. 1, 61--96.
MR0611469 (82h:14028)  Sasaki, Ryuji . Some remarks on the equations defining abelian varieties.
 Math. Z.  177  (1981),  no. 1, 49--60.
MR1048533 (91e:14043)  Birkenhake, Ch. ;  Lange, H.  Cubic theta relations. J. Reine Angew. Math.  407  (1990), 167--177.<|endoftext|>
TITLE: Obstruction theory for non-simple spaces
QUESTION [14 upvotes]: I'm looking for a good reference that has a detailed treatment of obstruction theory in the case where the target space is not simple. The specific situation I am interested in involves lifting a map of 3-skeletons from a $K(G, 1)$ to an arbitrary homotopy 3-type $X$ to the 4-skeletons (and hence to a true map between the spaces); the obstruction to this "should" live in $H^4(G, \pi_3(X))$ (with the appropriate action of $G$ on $\pi_3(X)$) via a local coefficient system, but I've been having trouble hashing out the details. Experts I've asked have given answers ranging from saying that it's impossible to saying that they're certain that it's possible but they don't know a reference that does it. Books I've looked at tend to gloss over the details, but they seem to indicate that this should work. Can anybody set me straight on this?

REPLY [10 votes]: @Evans Jenkins: A comparison of the work of Whitehead and Olum is given in
Ellis, G.J. "Homotopy classification the J.H.C. Whitehead way". Exposition. Math. 6 (1988) 97--110.
He writes (and I leave the reader to find the citations):
``In view of the ease with which
Whitehead's methods handle the classifications of Olum and Jajodia,
it is surprising that the papers \cite{Olum53} and \cite{Jaj80}
(both of which were written after the publication of
\cite{W49:CHII}) make respectively no use, and so little use, of
\cite{W49:CHII}.
``We note here that B. Schellenberg, who was a student of Olum, has
rediscovered in \cite{Sch73} the main classification theorems of
\cite{W49:CHII}. The paper \cite{Sch73} relies heavily on earlier
work of Olum.''
Whitehead used what he calls "homotopy systems", which we now call "free crossed complexes"; the notion of crossed complex goes back to Blakers in 1948 (Annals of Math), and a full account is in the 2011 EMS Tract Vol 15 Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy grouopids. The relation between crossed complexes and chain complexes with operators is quite subtle; it was first developed by Whitehead, and in CHII he explains, in our terms,  that  crossed complexes have better realisation properties that chain complexes with operators.  For example, the latter  do not  model homotopy 2-types.
Section 12.3 of the above Tract is on the homotopy classification of maps, including the non simply connected case, and Section 12.4 is on local systems, but it may be that your example is out of reach of the "linear" theory of crossed complexes. The homotopy classification of $3$-types requires quadratic information, see books by Baues and also
Ellis, G.J. "Crossed squares and combinatorial homotopy". Math. Z. (214} (1993) 93--110.
So there is sill a lot of work to be done!<|endoftext|>
TITLE: What's the current state of the classification of not-fully-extended TQFTs?
QUESTION [24 upvotes]: Recall that a $(k,k+1,\dots,k+n)$-TQFT is (supposed to be) a functor from the $n$-category whose $j$-morphisms are (isomorphism classes of) compact $(k+j)$-dimensional manifolds with boundary to some target category, usually your favorite version of $n$-Vect.  When $k=0$, a full "classification" of TQFTs with a given target category is given in:

Lurie, Jacob.  On the classification of topological field theories. Current developments in mathematics, 2008, 129--280, Int. Press, Somerville, MA, 2009.  58Jxx (57Rxx) MR2555928.  arXiv:0905.0465.

Or, rather, Lurie first provides reasonable definitions for a number of things, end then proves that there is an equivalence of $n$-categories between the $(0,\dots,n)$-TQFTs with target $\mathcal V$ and the $n$-groupoid of ("fully") dualizable objects in $\mathcal V$.  (The classification is not particularly effective in two ways: given a dualizable object, which is the value the TQFT assigns to a point, it can be still very hard to understand the functor on complicated manifolds; and given a category, it can be still very hard to classify its dualizable objects.)  For a review, see nLab: cobordism hypothesis.
But Lurie's result does not describe all gadgets that deserve to be called "TQFT"s.  For example, it is a classical folk theorem that $(1,2)$-TQFTs are the same as commutative cocommutative Frobenius algebras.  I think that there are other similar results of this nature, but I don't know of any theory that puts them all into a single framework.  Hence:

Question: Is there a classification, similar to Lurie's, for $(k,\dots,k+n)$-TQFTs with a give target $n$-category?

REPLY [40 votes]: When n > 1 the paper that you cite can give you a little bit of traction: the sketch proof of the main result gives a generators-and-relations presentation of (k,k+1,...,k+n)-Bord relative to (k,k+1)-Bord. There are two caveats:
1) (k,k+1)-Bord must be interpreted as an (infty,1)-category (or at least as an (n,1)-category), rather than as an ordinary category. Consequently, this is a very complicated
object even when k=1 (to my knowledge, there is no concrete description of its representations along the lines of "commutative Frobenius algebras"). Fortunately it
is quite easy to understand when k < 0, which is exploited in the treatment of the case of fully extended field theories.
2) The presentation is more complicated than in the fully extended case. When increasing the dimension, you need to add generators and relations corresponding to handles and handle cancellations for all indices (in the fully extended case, there is a cancellation phenomenon which ends up telling you that the only data you need to supply is for a handle of index 0).<|endoftext|>
TITLE: Unbounded linear operator defined on $l^2$
QUESTION [15 upvotes]: Let $l^2$ be a Hilbert space of infinite sequences $(z_0, z_1, \cdots)$ with finite $\sum_{i=0}^{\infty} |z_i|^2$.
Are there any simple example of unbounded linear opearator $T: l^2 \to l^2$ with $D(T)=l^2$?

REPLY [15 votes]: No there aren't any simple, or even any constructive, examples of everywhere defined unbounded operators.   The only way to obtain such a thing is to use Zorn's Lemma to extend a densely defined unbounded operator.  Densely defined unbounded operators are easy to find.
Zorn's lemma is applied as follows. Let $A$ be an operator on a domain $\mathcal D$. Consider the set $E$ of all extensions of $A$, that is the collection of operators $A'$ on domains $\mathcal D' \supset \mathcal D$ that agree with $A$ when restricted to $\mathcal D$. Then $E$ is partially ordered by inclusion on domains.  Furthermore, any linear chain has an upper bound, by taking unions of domains. So there is a maximal element by Zorn.  Finally, suppose the maximal element $A$ is defined on a domain $\mathcal D'$ that is not all of $\ell^2$.  Let $v$ be any vector in the complement of $\mathcal D'$. Define an extension of $A'$ on $\mathcal D'+\{a v\}$ by, say, mapping $v$ to zero. This contradicts maximality, so any maximal element is globally defined.<|endoftext|>
TITLE: Localizations as free, finite rank modules
QUESTION [6 upvotes]: Let $K$ be a algebraic number field of degree $n$ over $\mathbb{Q}$, and $O$ its ring of integers. Let $P$ be a prime ideal of $O$ and $(p)=P \cap \mathbb{Z}$.
Is it true that the localization $O_{P}$ is a rank $n$ free module over $\mathbb{Z}_{(p)}$ (the localization of $\mathbb{Z}$ at $(p)$) if and only if $P$ is the only prime above $(p)$?

REPLY [6 votes]: Well, if $P$ is not the only prime above $p$, then $O_P$ cannot be a finitely-generated $\mathbb{Z}_{(p)}$-module for the following reason. Suppose $Q$ is another prime ideal above $p$ and select $\beta\in Q\setminus P$. Then $\beta^{-1}\in O_P$. If $O_P$ were finitely-generated as a module over $\mathbb{Z}_{(p)}$, then it would be integral over $\mathbb{Z}_{(p)}$, and hence would be contained in the integral closure of $\mathbb{Z}_{(p)}$ in $K$, which is $O_p$. But then $\beta^{-1}\in O_p$, so $1/\beta=\alpha/m$ for some integer $m$ not divisible by $p$. This means that $m=\alpha\beta\in Q$, whence $m\in Q\cap\mathbb{Z}=(p)$, a contradiction.<|endoftext|>
TITLE: Uniqueness of local Langlands correspondence for connected reductive groups over real/complex field.
QUESTION [19 upvotes]: In Langlands' notes "On the classification of irreducible representations of real algebraic groups", available at the Langlands Digital Archive page here, Langlands gives a construction which is now referred to as "the local Langlands correspondence for real/complex groups".
What Langlands does in practice in this paper is the following. Let $K$ denote either the real numbers or a finite field extension of the real numbers (for example the complex numbers, or a field isomorphic to the complex numbers but with no preferred isomorphism). Let $G$ be a connected reductive group over $K$. Langlands defines two sets $\Pi(G)$ (infinitesimal equivalence classes of irreducible admissible representations of $G(K)$) and $\Phi(G)$ ("admissible" homomorphisms from the Weil group of $K$ into the $L$-group of $G$, modulo inner automorphisms). He then writes down a surjection from $\Pi(G)$ to $\Phi(G)$ with finite fibres, which he constructs in what is arguably a "completely canonical" way (I am not making a precise assertion here). Langlands proves that his surjection, or correspondence as it would now be called, satisfies a whole bunch of natural properties (see for example p44 of "Automorphic $L$-functions" by Borel, available here), although there are other properties that the correspondence has which are not listed there---for example I know a statement explaining the relationship between the Galois representation attached to a $\pi$ and the one attached to its contragredient, which Borel doesn't mention, but which Jeff Adams tells me is true, and there is another statement about how infinitesimal characters work which I've not seen in the literature either.
So this raises the following question: is it possible to write down a list of "natural properties" that one would expect the correspondence to have, and then, crucially, to check that Langlands' correspondence is the unique map with these properties? Uniqueness is the crucial thing---that's my question.
Note that the analogous question for $GL_n$ over a non-arch local field has been solved, the crucial buzz-word being "epsilon factors of pairs". It was hard work proving that at most one local Langlands correspondence had all the properties required of it---these properties are listed on page 2 of Harris-Taylor's book and it's a theorem of Henniart that they suffice. The Harris-Taylor theorem is that at least one map has the required properties, and the conclusion is that exactly one map does. My question is whether there is an analogue of Henniart's theorem for an arbitrary connected reductive group in the real/complex case.

REPLY [6 votes]: Here is a slight refinement of Emerton's answer. 
Any L-homomorphism $\phi:W_\mathbb R\rightarrow \phantom{}^LG$ defines an infinitesimal character $\lambda$ and a central character $\chi$. If $\phi(W_\mathbb R)$ is not contained in a proper Levi this defines an L-packet: the set of relative discrete series with this infinitesimal and central character (relative = discrete series modulo center). Every relative discrete series L-packet arises this way.
Now compatibility with parabolic induction determines all L-packets as follows. If $\phi(W_\mathbb R)$ is contained in a proper Levi subgroup $\phantom{}^LM$, the preceding construction defines a relative discrete series L-packet for $M$. The L-packet for $G$ is defined to be the irreducible summands of $Ind_{MN}^G(\pi_M)$ as $\pi_M$ runs over the L-packet of $M$ (for the correct choice of $N$). This induction can be done in stages: a tempered step, which is completely reducible, followed by an induction which gives unique irreducible summands. (This incorporates the "twisting" mentioned above).
As Emerton says this is canonical since we don't need L or epsilon factors to define relative discrete series L-packets. I'd like to know if anything like this, however speculative, might hold for general p-adic groups; in particular whether epsilon factors really do come to the rescue outside of GL(n) and perhaps classical groups. See Characterizing the Local Langlands Correspondence.<|endoftext|>
TITLE: Counting  branched covers of the projective line and Spec Z
QUESTION [12 upvotes]: I've asked a question like this before, but now I'm more interested in counting the number of covers.
We suppose given the following data.

A positive integer  $d$ 
A finite set of closed points $B= ( b_1,\ldots,b_n )$ in $\mathbf{P}^1_\mathbf{C}$ 
Branch types $T_1,\ldots, T_n$.

Question. How many branched covers of $\mathbf{P}^1_\mathbf{C}$ exist which are branched only over $b_i$ (with branch type $T_i$ over each $b_i$)?
The answer lies within the Hurwitz number for $(T_1,\ldots,T_n)$. This translates the problem to combinatorial group theory.
Now, for my main question:
Q1. Can one ``count'' covers of $\textrm{Spec} \mathbf{Z}$ as above? That is, can one count the number of finite field extensions $$\mathbf{Q}\subset K$$ of given degree $d=[K:\mathbf{Q}]$ which are unramified outside a given set of prime numbers $p_1,\ldots,p_n$ with ramification types $T_1,\ldots,T_n$?
I know that one can use Minkowski's Geometry of Numbers to give some nontrivial bounds on the discriminant. Is this the best we can do?

REPLY [9 votes]: One thing to keep in mind is that the analogue of Spec Z is really P^1 over a finite field k, not P^1/C.  And here already one does not have a simple "Hurwitz-type formula" for the number of G-covers with given branching which are defined over k.
Just to give an example which may be illustrative; suppose that G = S_3, and you require that the inertia at the primes p_1, ... p_n is tame and maps to a transposition in G.  The extensions of Q of this kind are more or less in bijection with the etale Z/3Z covers of the quadratic field K = Q(sqrt(N)) where N = p_1....p_n.  (I am being careless about the real place here.)  In any event, to "count" the number of covers is in effect to compute the size of the 3-part of the class group of K.  There is not going to be a nice formula for this, and in particular it will depend unpredictably on the primes in question.  On the other hand, you can compute the average of this quantity over squarefree integers N, by Davenport-Heilbronn.
So I would say:
"No" to your Q1 as stated.
"Yes, for some choices of G," to your Q1 on average -- e.g. for G = S_3 (by Davenport-Heilbronn), for G = D_4 (Cohen-Diaz y Diaz-Olivier), for G = S_4, S_5 (by Bhargava, though perhaps some slight and presumably true refinement of Bhargava to squarefree discriminants is needed), for G = D_p when K is F_ell(T) by work of myself, Venkatesh and Westerland.<|endoftext|>
TITLE: Relation between indefinite quadratic forms and continued fractions
QUESTION [11 upvotes]: Let $D$ be a positive square free integer; for simplicity let's take $D$ to be $2$ or $3$ modulo $4$. Then ideal classes in $\mathbb{Z}[\sqrt{D}]$ are in bijection with matrices $\left( \begin{smallmatrix} a & b \\ b & c \end{smallmatrix} \right)$ such that $ac-b^2=-D$, module the action of $SL_2(\mathbb{Z})$. Given such a matrix, consider the element $\theta := \frac{b+\sqrt{D}}{a}$ in $\mathbb{Z}[\sqrt{D}]$. (Since $D$ is not a square, $a$ is not zero.) So matrices of the sort we are interested in are in bijection with numbers $\theta$ of the form $\frac{b+\sqrt{D}}{a}$ such that $a | (b^2-D)$.
When people discuss the continued fraction of $\sqrt{D}$, the set of such real numbers comes up, as it has the property that it is stable under translation by integers, and under inversion. (Which, now that I think about it, is just the $SL_2(\mathbb{Z})$ action. In fact,  the relation between  $\left( \begin{smallmatrix} a & b \\ b & c \end{smallmatrix} \right)$ and $\theta$ is that $(\theta \ 1)$ is isotropic for  $\left( \begin{smallmatrix} a & b \\ b & c \end{smallmatrix} \right)$. So it makes sense that we have related $SL_2(\mathbb{Z})$ actions.)
This makes me think there should be some way to relate the continued fraction behavior of $\sqrt{D}$ to the class group of $\mathbb{Z}[\sqrt{D}]$. This is probably some standard thing that every number theorist knows. Where do I read about it?

REPLY [10 votes]: Recently, I got obsessed with working out this story, to the detriment of my mathematical work. Here is a quick crib sheet for relating properties of continued fractions to properties of real quadratic fields. Warning: I haven't checked this against a standard reference, so there may be errors.
Preliminary notation: Continued fractions 
It is convenient to convert a continued fraction to a sequence of binary symbols; I've been using red and blue dots, which I'll represent here as $r$ and $b$. For example, $\sqrt{13}=[3,1,1,1,1,6,1,1,1,1,6,\cdots]$; I'll write this as 
$$r\ r\ r\ b\ r\ b\ r\ b\ b\ b\ b\ b\ b\ r\ b\ r\ b\ r\ r\ r\ r\ r\ r\ \cdots$$
This makes the continued fraction purely periodic instead of having that peculiar $3$ at the beginning; the period starts again in the middle of that last block of six $r$'s. 
Note also that the period of the fraction now appears to be twice what it was; when we get to the old period, we have switched colors. Finally, a sequence which starts with $b$'s should be thought of as a continued fraction for a number in $(0,1)$; the sequence starts with zero copies of $r$.
Preliminary notation: Real quadratic fields
Let $K$ be a quadratic field and let $\Lambda$ be a rank $2$ sublattice of $K$, with $\mathbb{Q} \Lambda =K$. Let $\mathrm{End}(\Lambda)$ be the ring of $\theta \in K$ such that $\theta \Lambda \subseteq \Lambda$. This is an order in $\mathcal{O}_K$.
Let $K$ be a real quadratic field, with fixed embedding $K \to \mathbb{R}$, and write $z \mapsto \overline{z}$ for the Galois symetry of $K$. We'll say that two lattices $\Lambda_1$ and $\Lambda_2$ with CM are "strictly equivalent" if there is an element $\alpha \in K$ such that $\Lambda_1 = \alpha \Lambda_2$ with $\alpha$ and $\overline{\alpha}$ both positive.
Summary of results:
$\bullet$ Take any periodic sequence $a_i$ of $r$'s and $b$'s and turn it into a continued fraction. Let $z$ be the value of that continued fraction. Then $z$ is a quadratic irrational, with $z>0$ and $\overline{z}<0$. We have $a_0=r$ if $z>1$ and $a_0=b$ if $z<1$. Extending the periodicity to negative indices, $a_{-1}=r$ if $\overline{z} < -1$ and $a_{-1} =b$ if $\overline{z} > -1$. 
$\bullet$ Switching the colors changes $z$ to $1/z$. Reversing the sequence changes $z$ to $-\overline{z}$. 
Strict ideal classes
$\bullet$ Let $K = \mathbb{Q}(z)$ and let $\Lambda = \langle 1, z \rangle$. Shifting the periodic sequence does not change the strict equivalence class of $\Lambda$.
$\bullet$ The above is a bijection between periodic sequences of $r$'s and $b$'s, up to shift, and strict equivalence classes of lattices in real quadratic fields.
$\bullet$ Switching the colors corresponds to multiplying by our lattice by an element of 
negative norm. So producing a lattice which is equivalent, but not strictly equivalent.
$\bullet$ Reversing the sequence sends $\Lambda$ to $\overline{\Lambda}$. If $\Lambda$ is a lattice in $K$, and $R = \mathrm{End}(\Lambda)$, then $\Lambda \overline{\Lambda}$ is a strictly principal fractional ideal for $R$. (This is a special property of quadratic fields, which I know of no generalization of in higher degree number fields.) So, with the understanding that we treat a fractional ideal as a fractional ideal for its full endomorphism ring, reversing the sequence sends $\Lambda$ to $\Lambda^{-1}$.
Units
$\bullet$ Let $R$ be the endomorphism ring of $\Lambda$. Let $p/q$ be the convergent obtained by truncating the continued fraction just before the first repetition of the block which contains $a_0$. Let $u=p-qz$. Then $u$ is a unit of $R$ with norm $1$, and is the fundamental generator of the group of such units.
For example, $\langle 1, \sqrt{13} \rangle$ has endomorphism ring $\mathbb{Z}[\sqrt{13}]$. We truncate the above sequence to
$$r\ r\ r\ b\ r\ b\ r\ b\ b\ b\ b\ b\ b\ r\ b\ r\ b$$
or
$$[3,1,1,1,1,6,1,1,1,1] = \frac{649}{180}$$
and $649-180 \sqrt{13}$ is the fundamental positive unit of $\mathbb{Z}[\sqrt{13}]$.
$\bullet$ The color reversal of our sequence is a shift of itself if and only if $R$ has units of norm $-1$. We can recover them by the same recipe, truncating before the color reversed copy of $a_0$.
For example,
$$[3,1,1,1,1] = \frac{18}{5}$$
and $18-5 \sqrt{13}$ is the fundamental unit of norm $-1$ in $\mathbb{Z}[\sqrt{13}]$. 
Note, by the way, that we have not yet seen the fundamental unit of $\mathbb{Q}(\sqrt{13})$, which is $(3-\sqrt{13})/2$. That's because $\langle 1, \sqrt{13} \rangle$ doesn't have CM by this unit.
We can obtain the same unit from two quite different looking continued fractions. For example
$$\sqrt{10} = r\ r\ r\ b\ b\ b\ b\ b\ b\ r\ r\ r \cdots \quad \sqrt{10}/2 = r\ b\ r\ b\ b\ r\ b\ r \cdots$$
where I have given a full period for each fraction. Truncating to before the middle block gives
$$r\ r\ r = \frac{3}{1} \quad r\ b\ r\  = \frac{3}{2}.$$
Both of these give the unit $3-\sqrt{10} = 3-2 \frac{\sqrt{10}}{2}$.
Fixing the endomorphism ring; working with triples $(a,b,c)$
$\bullet$ Let $R = \mathbb{Z}[\sqrt{D}]$, for $D>0$ and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ of integers with $D=b^2+ac$ and $a$, $c>0$, by the recipe $(a,b,c) \mapsto (b+\sqrt{D})/a$.
The corresponding ring is exactly $\mathbb{Z}[\sqrt{D}]$ if and only if $(a,2b,c)$ have no common factor.
$\bullet$ Let $R = \mathbb{Z}[(1+\sqrt{D})/2]$ with $D \equiv 1 \mod 4$, positive and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ with $D=b^2+ac$, $a$ and $c>0$, and the additional condition that $b$ is odd and $a$ and $c$ are even.
Example: If we want to get the ring $\mathbb{Z}[(1+\sqrt{13})/2]$, we need to pick $z$ so that $\langle 1, z \rangle$ has CM by this ring. An obvious choice is $z=(1+\sqrt{13})/2$, with $(a,b,c) = (2,1,6)$. The continued fraction is 
$$r\ r\ b\ b\ b\ r\ r\ r\ b\ b\ b\ r\ r\ r\ b\ b\ b\ $$
or $[2, 3,3,3,3,\ldots]$ in conventional notation.
$\bullet$ The corresponding ring is exactly $\mathbb{Z}[(1+\sqrt{D})/2]$ if and only if $(a,b,c)$ have no common factor.
$\bullet$ There are only finitely many $(a,b,c)$ for any $R$.
$\bullet$ Adding an $r$ at the beginning of the sequence changes $(a,b,c)$ to $(a,a+b,c-a-2b)$. Adding a $b$ at the beginning changes $(a,b,c)$ to $(a-c-2b,b+c,c)$. Reversing the sequence changes $(a,b,c)$ to $(a,-b,c)$; color switching the sequence sends $(a,b,c)$ to $(c,-b,a)$.
Continued fractions with special symmetry
$\bullet$ A continued fraction is a shift of its color switch if and only if $R$ contains a unit with norm $-1$; we have already described how to find this unit.
$\bullet$ A continued fraction is a shift of its reversal if and only if it is a $2$-torsion class in the strict ideal class group.
Consider continued fractions which equal their reversal, so the periodic sequence starts at the middle of an even block of $r$'s or $b$'s, like the sequence for $\sqrt{13}$ above. These correspond to $z = \sqrt{D}/a$ for some divisor $a$ of $D$.
Consider continued fractions which are off from one by a shift of their reversal, so the periodic sequence starts in the middle of an odd block of $r$'s or $b$'s. These correspond to $z = (b+\sqrt{D})/(2b)$. If $D$ is odd, then we can take $b$ to be any divisor of $D$. If $D$ is $2 \mod 4$, then there are no solutions to $b^2+2bc=D$. If $D$ is $0 \mod 4$, then we can take $b$ of the form $2 b'$, where $b'$ is a divisor of $D/4$.
$\bullet$ Let $(-)$ denote the strict ideal class of principal $\Lambda$ ideals generated by elements of negative norm. A continued fraction is a shift of its color switched reversal if and only if $\Lambda^2 = (-)$ in the strict ideal class group.
A continued fraction actually equals its color switched reversal if and only if $a=c$. In other words, such continued fractions for $R = \mathbb{Z}[\sqrt{D}]$ are in bijection with solutions to $a^2+b^2=D$ with $a$ and $b>0$, and $GCD(a,2b)=1$. Such continued fractions for $R=\mathbb{Z}[(1+\sqrt{D})/2]$ are in bijection with solutions to $a^2+b^2=D$ with $a$, $b>0$, such that $a$ even and $b$ odd. 
Example: We have $34=3^2+5^2$. So take $z=(3+\sqrt{34})/5$. Take $\Lambda$ to be the lattice $\langle 1, (3+\sqrt{34})/5 \rangle$, which is strictly equivalent to the ideal $I=\langle 5, 3+\sqrt{34} \rangle$ in $\mathbb{Z}[\sqrt{34}]$. This is a non-principal prime ideal dividing $5$. We have $I^2 = \langle 3+\sqrt{34} \rangle$, which is principal, but not strictly principal. 
The corresponding continued fraction is 
$$r\ b\ r\ r\ r\ b\ b\ b\ r\ b\ r\ b\ r\ r\ r\ b\ b\ b\ r\ b\ r\ b\ \cdots$$
or $[1,1,3,3,1,1,1,1,3,3,1,1,1,1,\cdots]$ in conventional notation. This sequence is its own color switched reversal, reflecting that $I^2=(-)$. However, it is not a shift of its own reversal, reflecting that $I^2$ is not strictly principal, and it is not a shift of its color switch, reflecting that $\mathbb{Z}[\sqrt{34}]$ does not have a unit of norm $-1$.<|endoftext|>
TITLE: Curvature of a Lie group
QUESTION [20 upvotes]: Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold [Edit: provided we fix a metric, for example an invariant or bi-invariant one] has some scalar curvature R.  

Question [Edited] Is there a nice formula which expresses the scalar curvature at a point of the manifold in terms of the lie algebra of the group?

REPLY [11 votes]: For left-invariant (or right-invariant) metrics, this paper of Arnold gives a formula for  the sectional and Riemannian curvatures, in terms of the adjoint of the Lie bracket operation in the metric.<|endoftext|>
TITLE: Inner products and Norms
QUESTION [6 upvotes]: Let $f:[n]\times [n] \rightarrow [0,1]$ be a function from pair of integers to the real interval [0.1]. I would like to find sets of complex vectors 
$X= \{x_i\}$ and $Y=\{y_j\}$ satisfying $x_i\cdot y_j=f(i,j)$, in such a way that 
the vectors in $X$ and in $Y$ are as small as possible. More precisely, 
set $m= max_{i,j} f(i,j)$ and $N=max_{i,j}[\|x_i\|,\|y_j\|]$.
1) What is the minimal $N$ such that $x_i\cdot y_j = f(i,j)$ for all $i,j\in [n]$? 
2) Is there an upper bound on $N$ purely in function of $m$, i.e., with no depenence 
on $n$? 
3) If the answer to question two is no, what is the best upper bound that we can 
give for $N$ in function of $n$ and $m$? A trivial upper bound is 
$$N \leq \max_i{\sum_{j} f(i,j) } \leq mn.$$ 
but I believe that the dependence on $n$ might be lowered.
cordially,
mateus

REPLY [10 votes]: Such questions have been dealt with. Note first that your $m$ is just the norm of the matrix $F$ (see Robin's comment) as an operator from $\ell_1^n$ to $\ell_\infty^n$. $M$ also has a name, it is the $\gamma_2$ norm of this operator (This is the minimal product of 
$$\|Y\|_{1\to 2}\|X\|_{2\to\infty}$$ over all  factorizations $F=XY$ . $\|Z\|_{p\to q}$ denotes the norm of Z as an operator from $\ell_p$ to $\ell_q$.)
It is not hard to see that $M=\gamma_2(F)\le \sqrt n \|F\|_{1\to\infty}=\sqrt n m$.
For a random $0,1$ matrix $F$ one gets that this estimate is tight, up to a universal constant.
You can look here http://www.springerlink.com/content/px2324p527n19xj2/ for details.
In particular Cor 5.2 there (it deals with random $\pm 1$ matrices but it is easy to go between those and random $0,1$ matrices).<|endoftext|>
TITLE: Intuitive "proof" or explanation of a result in Friedman's urn
QUESTION [5 upvotes]: Let $g, r, a, b$ be positive integers. In Friedman's urn model we have an urn with $r$ red and $g$ green balls in it. In each step we take one ball out of urn, register its color and return it to the urn. Additionally, we put $a$ more balls of this color and $b$ more balls of the other color.
Let $X_n$ be the relative amount of green balls in the box after $n$-th step. It can be proven (e.g. Richard Durrett, Probability: Theory and Examples, pages 254-255) that
$$\lim_{n\to\infty}X_n = \frac12\quad\text{a.e.}$$
If one thinks about this statement for a second it would most probably strike him as extremely counter-intuitive. The proof cited above make use of square integrable martingales, and unfortunately doesn't seem to give intuitive explanation of this phenomena. 
I'm looking for an explanation which would explain on some heuristic level why this result is in some sense logical. An idea for more intuitive proof would also definitely be helpful.
I would also like to note that if we take $b=0$ the model becomes well-known Polya-Eggenberger urn model for which we have
$$\lim_{n\to\infty}X_n\sim B\left(\frac{g}{a}, \frac{r}{a}\right).$$

REPLY [4 votes]: You might be interested in the article by David A. Freedman on Friedman's urn. He reports a simple and intuitive proof due to Ornstein, which only uses the strong law 
of large numbers.
In his notation the urn contains $W_n$ white balls and $B_n$ black balls at time $n$, $a$ and $b$ have the same meaning as in the question.

D. Ornstein has obtained this very intuitive proof that $(W_n + B_n)^{-1}W_n$ converges
  to $1/2$ with probability 1 for $b > 0$. Suppose first $a > b$. If $0 \leq  x \leq 1$ and
  $$\mathbb P\left\{\limsup \frac{W_n}{W_n + B_n} \leq x\right\} = 1,$$ 
  by an easy variation of the Strong Law, with probability 1, in $N$ trials there will be at most $Nx + o(N)$ drawings of a white ball; so at least $N(1 - x) - o(N)$ drawings of black. Therefore, with probability 1, $\limsup (W_n+ B_n)^{-1}B_n$ is bounded above by
  $$\lim\limits_{N\to\infty}\frac{a[Nx + o(N)] + b[N(1 - x) - o(N)]}{N(a + b)}=\frac{b+(a-b)x}{a+b}.$$
  Starting with $x = 1$ and iterating,
  $$\mathbb P\left\{\limsup \frac{W_n}{W_n + B_n} \leq \frac{1}{2}\right\} = 1$$
  follows. Interchange white and black to complete the proof for $a > b$. If $a < b$, and 
  $$\mathbb P\{\limsup (W_n + B_n)^{-1}W_n \leq x\} = 1,$$
  then a similar argument shows 
  $$\mathbb P\left\{\limsup\frac{B_n}{W_n + B_n} < \frac{a+(b-a)x}{a + b}\right\}=1$$
  The argument proceeds as before, except both colors must be considered simultaneously.<|endoftext|>
TITLE: $C^1$ isometric embedding of flat torus into $\mathbb{R}^3$
QUESTION [66 upvotes]: I read (in a paper by Emil Saucan) that the flat torus may be isometrically embedded
in $\mathbb{R}^3$ with a $C^1$ map by the Kuiper extension of the Nash Embedding Theorem,
a claim repeated in this Wikipedia entry. I have been unsuccessful in finding a description
of such a mapping, or an image of what the embedding looks like.  I'd be grateful to any pointers on this topic.  Thanks!
Addendum.
It seems Benoît Kloeckner's answer below is definitive.  What I asked for 
apparently does not yet exist, but is 
"in process" and will soon be available through the work of the Hévéa project.
[23Apr2012] This is taken from the link in DamienC's comment and Benoît's update in the latter's answer below:

REPLY [7 votes]: I just wanted to add that there is also a very detailed paper on the isometric embedding of a flat torus by the four authors of the Hevea project (namely, Borrelli, Jabrane, Lazarus, and Thibert), which also includes photographs of a 3D-printed model on p.67.
3D-printed model of flat torus
The paper (considerably longer than the PNAS one) is available here.
There is also a fantastic video showcasing the 3D model, also made by the Hevea project.<|endoftext|>
TITLE: In what setting does one usually define mixed sheaves and weights for them?
QUESTION [6 upvotes]: In BBD mixed sheaves and weights for them were only defined for ($\overline{\mathbb{Q}_l}$-)sheaves over a variety $X_0$ defined over a finite field $F$. Weights start to behave better when one extends coefficients from $F$ to its algebraic closure i.e. passes from $X_0$  to $X$.
Now, BBD was published in 1982. Are any significant improvements and/or generalizations known in this field now? There is a paper by Huber and a book by Jannsen where mixed sheaves and weights for them are mentioned. Yet these authors were not able to generalize the results of BBD (in fact, an example of Jannsen shows that this is probably impossible). They also didn't extend scalars. So, are there any other papers on this subject?
Upd. There seems to be two basic ways to define weights (for sheaves) explicitly. The first method uses weights of Hodge structures. It seems that this method can work only for something like the category of mixed Hodge modules. Possibly, I will study these categories in the future. Yet at the moment I study motives, and it seems that 'motivic' people usually do not understand mixed Hodge modules (and so did not relate them with motives). 
So, I am currently interested in the second method. It uses the eigenvalues of the  Frobenius action. So, was anything interesting done using THIS approach after 1982?

REPLY [6 votes]: I'm not sure if this is the kind of answer that you're looking for. This is an extension of Tom Nevin's answer. Saito's theory of mixed Hodge modules is modeled, to some extent, on BBD. The results are quite similar, but the constructions and proofs are entirely different (and unfortunately rather opaque). In a sense, this follows a general pattern. In the $\ell$-adic world, the weight filtration is determined naturally from the Galois action. In Hodge theory, one usually has to guess what it is based on analogy with it.
Perhaps I can say a few more words to make the formal structure of Saito's theory  a little clearer. A mixed Hodge module consists a filtered perverse sheaf $(K,W)$ over $\mathbb{Q}$, and a bifiltered regular holonomic $D$-module $(M,W,F)$ such that $(M,W)$ corresponds to $(K,W)$ under Riemann-Hilbert. One then needs to impose some axioms in order to get good theory. The first is that over point, this datum defines a mixed Hodge structure.
Before getting to the second, recall that
one knows from BBD (and G I guess) that mixed perverse sheaves are stable under vanishing cycles. Saito takes this as an axiom. However, making that last statement precise takes
over 100 pages.<|endoftext|>
TITLE: Hopf algebra duality and algebraic groups
QUESTION [15 upvotes]: Background:
Let $G$ be a linear algebraic group over an algebraically closed field $k$ and let $I \subseteq k[G]$ be the ideal of the identity element. The hyperalgebra $U(G)$ of $G$ is defined to be the subspace of the linear dual of $k[G]$ consisting of all $f$ such that $f(I^n) = 0$ for some $n > 0$; then there is a natural Hopf algebra structure on $U(G)$.
Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^*$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^*$ has a natural Hopf algebra structure.
EDIT: As pointed out in comments, I am using what might be nonstandard terminology -- perhaps I should write $A^\circ$ instead of $A^*$.
Questions:
I now have two related questions.
1) I assume that in general the hyperalgebra of $G$ is not the same as the Hopf algebra dual of $k[G]$. What is the Hopf dual of $k[G]$?
2) Although one obtains $U(G)$ from $k[G]$ by a dual construction, you can't in general obtain $k[G]$ from $U(G)$ -- for example, if $G$ is reductive, you can't get $k[G]$ from $U(G)$ by some sort of duality since the hyperalgebra doesn't see isogeny. Is there something one can say in the reductive case about the structure of the Hopf dual of $U(G)$? More generally, for which algebraic groups $G$ can we obtain $k[G]$ from $U(G)$ by some sort of duality? And when we can do so, is it as simple as just taking the Hopf dual?

REPLY [8 votes]: In prime characteristic (or for algebraic groups rather than Lie algebras in general), the comments already posted indicate a need for caution.   Jantzen's
Part I covers a lot of the ground, but he refers back at a few delicate points to Demazure-Gabriel.   
Duality for general Hopf algebras is discussed in section 3.5 of Cartier's 2006 notes 
http://inc.web.ihes.fr/prepub/PREPRINTS/2006/M/M-06-40.pdf 
Starting with a Hopf algebra $A$, its reduced dual Hopf algebra (denoted by him $R(A)$) lives in the linear dual (a coalgebra in its own right) but is often smaller.   So it's complicated to go back and forth.   On the other hand, dealing with algebraic groups rather than just Lie algebras makes life a little more complicated, as suggested in his earlier discussion of the algebra of representative functions on a group; see also:
"Remark 3.7.3. Let $k$ be an algebraically closed field of arbitrary characteristic.
As in subsection 3.2, we can define an algebraic group over $k$ as a pair
$(G,O(G))$ where $O(G)$ is an algebra of representative functions on $G$ with
values in $k$ satisfying the conditions stated in Lemma 3.2.1. Let 
$H(G)$ be the
reduced dual Hopf algebra of $O(G)$. It can be shown that $H(G)$ is a twisted
tensor product $G \times U(G)$ where $U(G)$ consists of the linear forms on $O(G)$ vanishing on some power $\mathfrak{m}^N$ of the maximal ideal $\mathfrak{m}$ corresponding to the unit element of $G$; here $\mathfrak{m}$ is the kernel of the counit 
   $\epsilon: O(G) \rightarrow k$.  If $k$ is of characteristic 0, $U(G)$ is again the enveloping algebra of the Lie algebra $\mathfrak{g}$ of
$G$. For the case of characteristic $p >0$, we refer the reader to Cartier [18] or
Demazure-Gabriel [32]."  
[Note that the \times symbol should be the LaTeX symbol \ltimes, which apparently won't print here.]
ADDED.  Given $G$, the hyperalgebra of $G$ is what Cartier denotes by $U(G)$; so it is not quite the reduced dual Hopf algebra of the algebra of regular functions on $G$ in general.    In case $G$ is a connected reductive algebraic group over an algebraically closed field of characteristic $p>0$, the hyperalgebra has an explicit "divided power" construction starting with Kostant's integral form of the universal enveloping algebra of the complex analogue of the Lie algebra of $G$ (in the crucial case $G$ semisimple and simply connected), then reducing mod $p$.    This is used heavily by Jantzen to investigate the rational representations of $G$ and relevant closed subgroups such as Borel subgroups, etc.  (A similar construction is used by Lusztig for the quantum enveloping algebra at a root of unity.)<|endoftext|>
TITLE: entropy of normal distribution
QUESTION [5 upvotes]: What is the entropy of a normal distribution with mean 0 and variance \sigma?
Thanks!

REPLY [6 votes]: I found here that "the negative differential entropy of the normal distribution"
(which may not be what you are asking for?) is:
$$-\frac{1}{2} [ \log (2 \pi \sigma^2 ) + 1 ] ,$$
independent of $\mu$.<|endoftext|>
TITLE: What is the precise relationship between groupoid language and noncommutative algebra language?
QUESTION [20 upvotes]: I have sitting in front of me two 2-categories.  On the left, I have the 2-category GPOID, whose:

objects are groupoids;
1-morphisms are (left-principal?) bibundles;
2-morphisms are bibundle homomorphisms.

On the right, I have the 2-category ALG, whose:

objects are algebras (over $\mathbb C$, say);
1-morphisms are (adjectives?) bimodules;
2-morphisms are bimodule homomorphisms.

And probably I should go through and add "in TOP" to every word on the left and "C-star" to every word on the right.
(I have the impression that ALG is two-equivalent to another category, which I will put on the far right, whose:

objects are cocomplete VECT-enriched categories (with extra adjectives?);
1-morphisms are cocontinuous VECT-enriched functors;
2-morphisms are VECT-enriched natural transformations.

In one direction, the functor takes an algebra to its full category of modules.  In the other direction, there might be extra adjectives needed, and I mean to appeal to the Mitchell embedding theorem; but I'm pretty sure an equivalence exists if I insist that every object on the far right comes with a cocontinuous faithful VECT-enriched functor to VECT, and my idea is that Mitchell says that every category admits such a functor.
So anyway, the point is that either category on the right or far right is a sort of "algebraic" category, as opposed to the more "geometric" category on the left.)
Then I've been told on numerous occasions that there is a close relationship between GPOID and ALG.  See, for example, the discussion at Geometric interpretation of group rings? — in fact, it's reasonable to think of the present question as a follow-up on that one.
The relationship is something like the following.  To each (locally compact Hausdorff) topological, say, groupoid, we can associate a C-star, say, algebra — the construction restricts in various special cases to: a (locally compact Hausdorff) topological space $X$ going to its algebra of continuous vanishing-at-infinity functions $C_0(X)$; a finite group $G$ going to its group ring $\mathbb C G$; etc.  The construction extends to 1- and 2-morphisms to build a (contra)functor.  At least if I get all the adjectives right, the functor should be a two-equivalence.

Question: What's the precisification of what I have said above?  What exactly is the two-functor from groupoids to algebras, and which adjectives make it into an equivalence of two-categories?  Groupoids have natural "disjoint union" and "product" constructions; these presumably correspond to Cartesian product and tensor product (?!? that's not the coproduct in the category of algebras, but maybe in this two-category it is?) on the algebraic side?

Let me end with an example to illustrate my confusion, which I brought up in Op. cit..  Let $G$ be a finite abelian group; then it has a Pontryagin dual $\hat G$.  Now, there is a canonical way to think of $G$ as a groupoid: it is the groupoid $\{\text{pt}\}//G$ with only one object and with $G$ many morphisms.  If I'm not mistaken, the corresponding algebra should be the group algebra $\mathbb C G$.  But $\mathbb C G$ is also the algebra $C_0[\hat G]$ of functions on the space $\hat G$.  And if there is one thing I am certain of, it is that the underlying space of $\hat G$ (a groupoid with no non-identity morphisms) and the one-object groupoid $\{\text{pt}\}//G$ are not equivalent as groupoids.  And yet their "function algebras" are the same.  So clearly I am confused.

REPLY [4 votes]: Both David Ben-Zvi and Jeffrey Giansiracusa have given excellent answers, so I will only fill in some bits that with Douglas Rizzolo we worked through based on their suggestions.
Moreover, for the purposes of this answer, I will avoid almost all analysis questions.  So although I think I know how to say some of what I want to say more generally, I will take as my "geometric" category the category $\text{FinSet}$ of finite sets.
The final disclaimer is that I will not mention the "inverse" map: rather than groupoids, I will describe the construction for category objects; rather than Hopf algebras, I will have bialgebras; there will not be a C-star structure.  Nevertheless, I will use the shorthand of "groupoid"/"Hopf"/... in place of "category object"/"bialgebra"/...
Groupoid algebras
So, abusing language, a groupoid is a finite set $C_0$, a finite set $C_1$, maps $i: C_0 \to C_1$ and $s,t: C_1 \to C_0$ such that $s\circ i = t\circ i = \operatorname{id}$, and a composition map $m: C_1 \underset{C_0}\times C_1 \to C_1$ satisfying a number of equations.  The short-hand way to say all this is to say "algebra in the category of spans from $C_0$ to $C_0$".  I will denote a generic groupoid by $C = \{C_1 \rightrightarrows C_0\}$.  I have worked through precisely three examples:

Any set $M$ is a groupoid $M = \{M \rightrightarrows M\}$, where all maps are identities.
Any group $G$ is a groupoid $\text{pt}/G = \{G \rightrightarrows \text{pt}\}$, where the maps are the only thing they can be.
The "equivalence relation" groupoid $n/n = \{n^2 \rightrightarrows n\}$ has $n$ objects and for each pair of objects a unique morphism.  It is equivalent to $\text{pt}$.

Fix a field $\mathbb K$, and let $\text{FinVect}$ denote the category of finite-dimensional $\mathbb K$-vector spaces.  Recall the "linearization" functor $\mathbb K: \text{FinSet} \to \text{FinVect}$, which takes a set $S$ to the vector space $\mathbb KS$, which has $S$ as a (distinguished) basis.  This functor preserves colimits (being adjoint to $\text{forget}$) and takes products to tensor products.
For any groupoid $C = \{C_1 \rightrightarrows C_0\}$, we define the algebra of functions on $C$ to be the vector space $\mathbb K C_1$, with the multiplication determined by the multiplication in $C$.  More precisely, if $a,b \in C_1$ are basis elements, we set their multiplication in $\mathbb K C_1$ to be: $$ab = \begin{cases} ab \in C_1, & s(a) = t(b), \\ 0 ,& \text{otherwise.} \end{cases}$$
In the three examples above:

The "algebra of functions on $M$" is the algebra of functions $M \to \mathbb K$ with pointwise multiplication.
The "algebra of functions on $\text{pt}/G$" is the "group algebra" $\mathbb K G$.
The "algebra of functions on $n/n$" is the matrix algebra $\operatorname{Mat}(n,\mathbb K)$.  It is Morita equivalent to $\mathbb K$.

A "vector bundle" or "sheaf" on a groupoid $C$ is precisely a representation of $\mathbb KC_1$.  My earlier objection was that as algebras, if $G$ is abelian with Pontryagin dual $\hat G$, then as algebras the algebras of functions on $\text{pt}/G$ and on $\hat G$ are the same, when these are different as sheaves.
Hopfish structure
David Ben-Zvi hinted that what distinguishes the algebras of functions over $\text{pt}/G$ and $\hat G$ is the tensor structure on their categories of sheaves.  The following notion is due to Xiang Tang, Alan Weinstein, and Chenchang Zhu, Hopfish algebras, 2008, arXiv:math/0510421v2.  (Actually, they decategorify, considering only isomorphism classes of bimodules; I will describe the correct version, which is presumably the version they secretly wanted but were afraid to write about.  Also, I will continue to ignore antipodes.)  A hopfish algebra is:

A $\mathbb K$-algebra $A$,
a bimodule ${_A \Delta _{A\otimes A}}$,
and an "associativity" isomorphism ${_A \Delta_{A\otimes A}} \underset{A\otimes A}\otimes ({_A A _A} \otimes {_A \Delta_{A\otimes A}}) \overset{\phi}\to {_A \Delta_{A\otimes A}} \underset{A\otimes A}\otimes ( {_A \Delta_{A\otimes A}}\otimes{_A A _A} )$,
such that $\phi$ satisfies a pentagon.

Coining some words, a hopfish algebra is commutatish if there is an isomorphism ${_A \Delta _{A\otimes A}} \to {_A \Delta^{\rm flip} _{A\otimes A}}$, where ${_A \Delta^{\rm flip} _{A\otimes A}}$ is the module $\Delta$ with the action by the two $A$s on the right flipped, which is an involution and satisfies (with $\phi$) two hexagons.
The idea is the following.  If $A$ is an algebra so that the category $_A\operatorname{mod}$ of left $A$-modules has a $\mathbb K$-linear cocontinuous monoidal structure, then this structure makes $A$ hopfish.  Given a hopfish structure, the corresponding monoidal structure is: $$_A\bigl((_A X) \underset{\Delta}\otimes (_AY) \bigr) \overset{\rm def}= {_A \Delta _{A\otimes A}} \underset{A\otimes A}\otimes ({_AX}\underset{\mathbb K}\otimes {_AY})$$
The three examples above are naturally hopfish.

Any commutative ring $R$ is hopfish, with the bimodule given as the "comodulation" of the multiplication map $R \otimes R \to R$ (an algebra homomorphism iff $R$ is commutative), i.e. the bimodule is $_R R _{R\otimes R}$, where the right action is obvious and the left action is "multiply and act".  When $R$ is the algebra of functions on a set $M$, then the corresponding monoidal structure on sheaves is the "pointwise" or "fiber-by-fiber" tensor product.
A group algebra $\mathbb K G$ is hopfish with the bimodule given by the "modulation" of the "duplication" algebra homomorphism $\mathbb K G \to \mathbb K G \otimes \mathbb K G$ given on a basis by $g \mapsto g\otimes g$.  I.e. the bimodule is $_{\mathbb K G} {\mathbb K G \otimes \mathbb K G} _{\mathbb K G \otimes \mathbb K G}$, where the left action is obvious and the right action is given by the duplication homomorphism.  The corresponding monoidal structure on sheaves is the usual tensor structure on $G$-representations.
Hopfish structures are designed to be well-behaved under Morita equivalence.  The Morita equivalence between $\operatorname{Mat}(n,\mathbb K)$ and $\mathbb K$ is given by $\mathbb K^{\oplus n}$ with the obvious actions.  Pushing the hopfish structure on $\mathbb K$ through this Morita equivalence gives the bimodule: $$_{\operatorname{Mat}(n)} {\operatorname{Mat}(n\times n^2)} _{\operatorname{Mat}(n^2) = \operatorname{Mat}(n) \otimes \operatorname{Mat}(n)}$$

In fact, all three examples follow from a general construction that I will now describe.
Let $C = \{C_1 \rightrightarrows C_0\}$ be a groupoid and $A = \mathbb K C_1$ the corresponding algebra of functions.  As a vector space, $\Delta = \mathbb K ( C_1 { \underset{^t{C_0}^t}\times} C_1)$ is spanned by pairs $(g,h)$ of morphisms with the same target.  Identifying the basis of $A\otimes A$ as all pairs of morphisms $(x,y)$, the bimodule structure is:
$$ a \cdot (g,h) \cdot (x,y) = (agx,ahy) $$
with the convention that if any multiplication is non-composable, the whole pair is $0$.  This determines an action.  The associativity isomorphism comes from identifying both sides with the space spanned by triples of arrows with the same target.  This hopfish algebra is always commutatish, by switching $(g,h) \mapsto (h,g)$.
In cartoons, $\mathbb K C_0 = \{\bullet\}$, $\mathbb K C_1 = \{\leftarrow\}$, and $\Delta = \{ \rightarrow \bullet \leftarrow\}$.
Extending to a 2-functor
Let $C = \{ C_1 \rightrightarrows C_0\}$ be a groupoid.  A left $C$-set is a functor $C \to \text{FinSet}$, or more precisely an arrow $C_0 \leftarrow S$ and an action $C_1 \underset{C_0}\times S \to S$ satisfying an obvious square.  If $C,D$ are two groupoids, a bibundle is a diagram $C_0 \leftarrow S \rightarrow D_0$ with commuting actions $C_1 \underset{C_0}\times S \underset{D_0}\times D \to S$.
It should be pretty much clear that if $S$ is a left $C$-set, then $\mathbb K S$ is a left $\mathbb K C_1$-module, and similar on the right, so that linearization takes bibundles to bimodules.  Moreover, it should be clear that $\mathbb K : \underset{C_0}\times \to \underset{\mathbb K C_1}\otimes$, and that morphisms of bibundles go to morphisms of bimodules.  Thus, linearization is a 2-functor from $\text{Gpoid}$ to $\text{Alg}$.  In fact, it lands within the sub-2-category of commutatish hopfish algebras.
In particular, the hopfish algebra corresponding to a groupoid, considered up to Morita equivalence, gives an invariant of the stack represented by said groupoid.
Further questions
There are largely two issues that I would like to understand in the construction.

If our groupoids consist of topological spaces, then taking "linear combinations of points" may not be the right thing, as it ignores the topology.  But it is no longer correct to identify, as vector spaces, $\mathbb K S$ with the algebra of, say, continuous functions on $S$.  In particular, if $G$ is a compact Lie group, then the algebra of functions on $\text{pt}/G$ should be some "convolution algebra" for $G$, and in particular the elements of the convolution algebra are not functions, but rather distributions.  I don't know how to say such words when I move far away from manifolds, but presumably the C-star algebraists do.
Nowhere did I use the antipode.  In Op. Cit., defining what I have called a "Hopfish algebra" (what they call "sesquialgebra") takes a page, whereas the antipode takes most of the paper.  The antipode must also come into play to define the $*$-involution in the C-star approach.

Finally, the question arises:

Is the functor from groupoids to commutative Hopfish algebras the right 2-notion of "full and faithful"?  I.e. it is a complete invariant of stacks?  I.e. is a stack recoverable from its commutatish Hopfish algebra up to Morita?  Is a morphism of stacks recoverable from a morphism of algebras?  Etc.
What is the essential image of the functor described above?  Is every commutative Hopfish algebra Morita-equivalent to one coming from a groupoid?  Does every bimodule thereof come from a bibundle?  The second is doubtful; the first is more promising.<|endoftext|>
TITLE: Order information enough to guarantee 1-isomorphism?
QUESTION [15 upvotes]: I define a 1-isomorphism between two groups as a bijection that restricts to an isomorphism on every cyclic subgroup on either side. There are plenty of examples of 1-isomorphisms that are not isomorphisms. For instance, the exponential map from the additive group of strictly upper triangular matrices to the multiplicative group of unipotent upper triangular matrices is a 1-isomorphism. Many generalizations of this, such as the Baer and Lazard correspondences, also involve 1-isomorphisms between a group and the additive group of a Lie algebra/Lie ring.
Consider the following function F associated to a finite group G. For divisors $d_1$, $d_2$ of G, define $F_G(d_1,d_2)$ as the number of elements of G that have order equal to $d_1$ and that can be expressed in the form $x^{d_2}$ for some $x \in G$.
Question: Suppose G and H are finite groups of the same order such that $F_G = F_H$. Does there necessarily exist a 1-isomorphism between G and H?
Note that the converse is obviously true: if there exists a 1-isomorphism between G and H, then $F_G = F_H$.
Incidentally, just knowing the orders of elements does not determine the group up to 1-isomorphism. There are many counterexamples of order 16, with two non-abelian groups (one being the direct product of the quaternion group and the cyclic group of order two, and the other a semidirect product of cyclic groups of order four) having the same statistics on orders of elements as $\mathbb{Z}_4 \times \mathbb{Z}_4$, but neither being 1-isomorphic to it because they don't have the same number of squares.
Similarly, just knowing how many elements are there of the form $x^d$ for each divisor d of the order is not sufficient to determine the group up to 1-isomorphism. Again, there are counterexamples of order 16.

REPLY [16 votes]: Here is a counterexample of order $32$. 
$G$ and $H$ will each have $3$ elements of order $2$ and $28$ elements of order $4$. In both cases all three elements of order $2$ will have square roots. That insures that $F_G=F_H$. But in $G$ one of them will have $4$ square roots while the others each have $12$, and in $H$ one of them will have $20$ square roots while the others each have $4$. That rules out a $1$-isomorphism. 
Let $Q$ be a quaternion group of order $8$ and let $C\subset Q$ be a (cyclic) subgroup of order $4$. Inside $Q\times Q$ there are three subgroups of index $2$ that contain $C\times C$. Let $G$ be $Q\times C$ and let $H$ be the one that is neither $Q\times C$ nor $C\times Q$.<|endoftext|>
TITLE: Prime numbers as knots: Alexander polynomial
QUESTION [53 upvotes]: A naive and idle number theory question from a topologist (but not a knot theorist):
I have heard it said (and this came up recently at MO) that there is a fruitful analogy between Spec $\mathbb Z$ and the $3$-sphere. I gather that from an etale point of view the former is $3$-dimensional and simply connected; from the same point of view the subschemes Spec $\mathbb Z/p$ are $1$-dimensional and very much like circles; and the Legendre symbols for two odd primes that figure in quadratic reciprocity are said to be analogous to linking numbers of knots. So, prompted by a recent MO question, I started thinking:
The abelianized fundamental group of the complement of Spec $\mathbb Z/p$ (the group of $p$-adic units) is not terribly different from the abelianized fundamental group of a knot complement (an infinite cyclic group). For nontrivial knots, there is a lot more to the fundamental group of a knot complement than its abelianization. The next little bit, the abelianization of the commutator subgroup (or $H_1$ of the infinite cyclic cover) has an action of that infinite cyclic group, and I recall that the Alexander polynomial of the knot may be created out of this action. 
So there must be some analogue of that in number theory, right? Like, some construction involving ideal class groups or idele class groups of $p$-power cyclotomic fields can be interpreted as the Alexander polynomial of a prime number?

REPLY [10 votes]: A whole book has recently appeared (Primes and Knots,
Edited by Toshitake Kohno, University of Tokyo, Japan, and Masanori Morishita, Kyushu University, Fukuoka, Japan).  
This volume deals systematically with connections between algebraic number theory and low-dimensional topology. Of particular note are various inspiring interactions between number theory and low-dimensional topology discussed in most papers in this volume. For example, quite interesting are the use of arithmetic methods in knot theory and the use of topological methods in Galois theory. Also, expository papers in both number theory and topology included in the volume can help a wide group of readers to understand both fields as well as the interesting analogies and relations that bring them together.<|endoftext|>
TITLE: How to combine linear constraints on a matrix and its inverse?
QUESTION [11 upvotes]: Suppose there exists a $(n \times n)$ matrix $A$ that is real and invertible (nothing unusual or special about $A$). We do not know the entries of $A$. However, we do have linear constraints, some of which are on the entries of $A$ and some of which are on the entries of its inverse $A^{-1}$. All constraints are assumed to be consistent with the true invertible matrix $A$, but the system may be underdetermined.
The general question is whether there is an efficient way to solve the system and determine $A$ or to characterize the remaining underdetermination in $A$. How can the linear constraints on $A^{-1}$ be converted into constraints on $A$ such that one can still solve for $A$ (when the system is determined)?
Simplest case: I know some entries of $A$ and some entries of $A^{-1}$. How can these constraints be combined to solve for $A$, if possible? Obviously, $AA^{-1} = I$, but in general this is a quadratic system in many variables, for which I am unaware of any solution procedure.
Even general pointers would be most welcome.

REPLY [4 votes]: Since the question suggests that the questioner is looking for an efficient algorithm for this problem, here is my attempt to answer the question from the complexity-theoretic perspective.  Unfortunately, the answer is pretty negative.
The following problem, which is one of the possible formulations of the question, is NP-complete.
Given: N∈ℕ, finitely many linear constraints (equations or inequalities) over ℚ on variables aij and bij (1≤i,j≤N), and N×N rational matrices A and B satisfying AB=I and all the given linear constraints.
Question: Is there another pair (A, B) of N×N rational matrices that satisfy AB=I and all the given linear constraints?
A proof is by reduction from the following problem (called “Another Solution Problem (ASP) of SAT”):
Given: An instance φ of SAT and a satisfying assignment to φ.
Question: Is there another satisfying assignment to φ?
The ASP of SAT is known to be NP-complete [YS03].
Note: The following reduction is much simplified compared to the first version posted.  See below for the first version, which proves a slightly stronger result.
We can construct a reduction from the ASP of SAT to the problem in question as follows.  Given an instance of SAT with n variables x1,…,xn, let N=n and constrain A to be a diagonal matrix such that A=A−1; these are easily written as linear equality constraints on the elements of A and A−1.  These constraints are equivalent to the condition that A is a diagonal matrix whose diagonal elements are ±1.  Now encode a truth assignment to the n variables by such a matrix by letting aii=1 if xi is true and aii=−1 otherwise.  Now it is easy to write down the constraints in SAT as linear inequalities.
With this encoding, the solutions to the given instance of SAT correspond one-to-one to the pairs (A, A−1) satisfying all the linear constraints.  This establishes a reduction from the ASP of SAT to the problem in question, and therefore the problem in question is NP-complete.
Remark.  This reduction can be viewed as an ASP reduction from SAT to the problem of finding a pair (A, B) of matrices satisfying given linear constraints.  For more about ASP reductions, see [UN96] and/or [YS03].  (The notion of ASP reductions was used in [UN96], where the authors treated it as a parsimonious reduction with a certain additional property.  The term “ASP reduction” was introduced in [YS03].)

In fact, the problem remains NP-complete even if we allow only linear constraints on the variables aij and linear constraints on the variables bij (but not a linear constraint which uses both aij and bkl).  The NP-completeness of this restricted problem can also be shown by reduction from the ASP of SAT.
The following lemma is a key to construct this version of a reduction.
Lemma.  Let A be a real symmetric invertible matrix.  Both A and A−1 are stochastic if and only if A is the permutation matrix of a permutation whose order is at most 2.
I guess that this lemma can be proved more elegantly, but anyway the following proof should be at least correct.
Proof.  The “if” part is straightforward.  To prove the “only if” part, assume that both A and A−1 are stochastic.  Note the following properties of A:

Because A is symmetric, A can be diagonalizable and all eigenvalues are real.
Because A is stochastic, all eigenvalues have modulus at most 1.
Because A−1 is stochastic, all eigenvalues have modulus at least 1.

Therefore, A can be diagonalizable and all eigenvalues are ±1, and therefore A is an orthogonal matrix.  Since both the 1-norm and the 2-norm of each row are equal to 1, all but one entry in each row are 0.  Therefore, A is a permutation matrix, and the only symmetric permutation matrices are the permutation matrices of some permutations whose order is at most 2.  (end of proof of Lemma 1)
It is easy to write down linear constraints which enforce A to be symmetric and both A and A−1 to be stochastic.  In addition, write down linear constraints which enforce A to be block diagonal with 2×2 blocks.  Given an instance of SAT with n variables x1,…,xn, we encode a truth assignment by a 2n×2n matrix which is block diagonal with 2×2 blocks so that the first block is $\pmatrix{1 & 0 \\ 0 & 1}$ if x1 is true and the first block is $\pmatrix{0 & 1 \\ 1 & 0}$ if x1 is false and so on.
Now that a truth assignment can be encoded as a matrix, the rest is the same: just verify that it is easy to write down the constraints in SAT as linear inequalities and that there is one-to-one correspondence between the solutions to a SAT instance and the pairs (A, A−1) of matrices satisfying the linear constraints.

References
[UN96] Nobuhisa Ueda and Tadaaki Nagao.  NP-completeness results for NONOGRAM via parsimonious reductions.  Technical Report TR96-0008, Department of Computer Science, Tokyo Institute of Technology, May 1996.
[YS03] Takayuki Yato and Takahiro Seta.  Complexity and completeness of finding another solution and its application to puzzles.  IEICE Transactions on Fundamentals of Electronics, Communications and Computer Sciences, E86-A(5):1052–1060, May 2003.<|endoftext|>
TITLE: Smallest dilation of a quadrilateral?
QUESTION [9 upvotes]: What is the smallest dilation of a quadrilateral in $\mathbb{R}^d$?
This may be an open problem;
my question is: Is this indeed open?
It will take me some time to explain the terms.
The notion of dilation derives from Gromov, as far as I know
(He defines a version in
Metric Structures for Riemannian and Non-Riemannian Spaces,
p.11,
although he called it distortion).
I came upon it myself via $t$-spanners.
The version in which I am interested is this.
Let $P$ be a polygon (its boundary, not its interior), and $x,y$ two points on
$P$.  You can think of $P$ in $\mathbb{R}^2$, but also $\mathbb{R}^3$ and $\mathbb{R}^d$
for $d>3$ are interesting.
Define $\delta(x,y)$ as the maximum (supremum) of $d_P(x,y) / | x y |$,
where $d_P(x,y)$ is the distance between $x$ and $y$ following
$P$ (the shortest path staying on the closed path that consitutes $P$),
and $|xy|$ is the Euclidean distance in $\mathbb{R}^d$.
Thus $\delta(x,y)$ measures how much $P$ dilates w.r.t. Euclidean distance.
I am interested in the minimum value $\delta(P)$
of $\delta(x,y)$ over all
$x,y \in P$, for all $n$-gons $P$, for fixed $n$.

Example 1.
If $P$ is a unit square, then $\delta(x,y)$
for $x,y$ opposite corners is $\sqrt{2}$, but
$\delta(P)=2$ because with $x,y$ midpoints of
opposite sides, $\delta(x,y)= 2/1$.
Example 2.
If $P$ is an equilateral triangle, $\delta(P)=2$, as shown in the figure.
In fact, the dilation of any triangle is $\ge 2$ [Lemma 7 in the 2nd paper below].
            

Example 3.
It is known the the dilation of any closed curve $C$
satisfies $\delta(C) \ge \pi/2$, with equality achieved
only by the circle. [Corollary 23 in the first paper below.] This is (apparently) due to Gromov.

So I finally come to my question.  By reading these two
papers,
"Geometric Dilation of Closed Planar Curves: New Lower Bounds,"
and
"On Geometric Dilation and Halving Chords,"
it appears to me that the minimum dilation of a quadrilateral
in $\mathbb{R}^2$ (and $\mathbb{R}^d$) is not known.
I had heard this was the case three years ago in a seminar
in Brussels, but (a) I didn't quite believe it,
(b) it was hearsay, and (c) it is now out of date.
I am trying to clarify with the authors of these papers, but in parallel
I would appreciate any information on the status of
this question.
The latter paper cited above proves a lower bound of $4 \tan(\pi/8) \approx 1.66$
(if I have interpreted it correctly).
Addendum. I don't want to close-out this question, but I have heard from one of the authors of
the above cited papers, and indeed it appears that the dilation of a planar quadrilateral is unknown
[as of July 2010, the original posting date].
So I have tentatively tagged this as an open-problem, and I will update if new information surfaces.  Thanks for everyone's interest and input!

REPLY [6 votes]: Updated on 25th July -- see below
Make an isosceles trapezium by starting with a rectangle of height 12 and width 66/13, and attaching Pythagorean (5,12,13) triangles to each side. Then the perimeter P is 600/13, and the height h is 12; and the numbers have been selected so that the width w at the equator is 12 too (where the equator is the horizontal line that divides the perimeter into two halves of equal length). So the dilation is P/2h = 25/13, which is less than 2. 

          
[]
 
We can try this with a general right-angled triangle. It is convenient to let the angle at the base of the trapezium be 2θ, so in our example we have sin 2θ = 12/13. We find that the dilation is equal to 1 + sin 2θ. However, if θ is too small, then we can achieve a larger dilation simply by cutting across the corner; this dilation is 1/sin θ. So we get the smallest dilation for an isosceles trapezium when 1/sin θ = 1 + sin 2θ. I had to resort to numerical methods to solve this; I got θ = 0.5555166235227462... radians, for a dilation of 1.89615765267304...
We can't improve on this by using a non-isosceles trapezium, but a smaller dilation might be achieved by a general non-trapezoidal quadrilateral.
Sorry if this all looks a bit provisional. If I get a full answer, I will put more effort into drawing some nice pictures and stuff.
Update
I have carried out a computer search for the smallest dilation, as follows.
Without loss of generality, we can restrict the search to quadrilaterals ABCD such that:
- A=(0,0), D=(1,0);
- B and C lie above the x-axis;
- all side lengths are <= 1.  
Step 1: Evaluate the dilation of all such quadrilaterals with x- and y-coordinates a multiple of 0.01. Save the 10000 quadrilaterals with the smallest dilation to file.
Step 2: For each B,C in the file, evaluate the dilation for the 10000 quadrilaterals with coordinates differing from B,C by a multiple of 0.001 between -0.005 and +0.004. (In other words, decrease the grid size by a factor of 10.) Of the resulting 100000000 quadrilaterals, save the 10000 with the smallest dilation to file.
Step 3: Repeat Step 2 with the grid size decreased by a factor of 10. And so on.
This procedure converges on the isosceles trapezium described above. So while this is not a proof, it is likely that the smallest possible dilation of a quadrilaterlal is indeed  1.89615765267304... (which is the real root of the polynomial $x^5 - x^4 - 4x - 4$).
Edit by J.O'Rourke (16Aug10). If I've followed Tony's description in the comment below
correctly,
here is his quadrilateral with the (conjectured) smallest dilation:
$h=0.896158$, $w=0.25552$:<|endoftext|>
TITLE: Which statements in section 5 of BBD will fail if we consider $\mathbb{Q}_l$-adic sheaves there?
QUESTION [7 upvotes]: A stupid question: which statements in section 5 of BBD will fail if we replace $\overline{\mathbb{Q}_l}$-sheaves by just $\mathbb{Q}_l$-ones? I am especially interested in Proposition 5.1.15.
BBD = Beilinson A., Bernstein J., Deligne P., Faisceaux pervers, Asterisque 100, 1982, 5-171.

REPLY [5 votes]: I think that all the statements are true, except for 5.3.9 (ii). Remark 5.3.10 says that all the statements in 5 up to and including 5.3.8 are true for $\mathbb{Q}_\ell$-coefficients with the same proof, and that 5.3.9 (i) is still true but with a slightly more complicated proof. I am pretty sure that the proof of corollary 5.3.10 works too for $\mathbb{Q}_\ell$-coefficients. I cannot see any reason why the proofs in 5.4 would not also work, but I have read them quickly. Anyway, proposition 5.1.15 is certainly fine.<|endoftext|>
TITLE: Approximation with continuous functions
QUESTION [10 upvotes]: Is it true that for every function $\mathbb{R} \to \mathbb{R}$ there exists a sequence of continuous functions $f_n(x): \mathbb{R} \to \mathbb{R}$ such that for any $x \in \mathbb{R}$ $f_n(x)$ converges to $f(x)$?
I started with characteristic function of rationals and tried to find corresponding sequence and got stuck. So additional question if this statement is true for this function.

REPLY [3 votes]: All the answers here rely on measure-theoretic ideas to give a negative answer. I feel that this does not exactly meet the point, instead I would say: You did not quite ask the right question, but if you did, the answer would in fact be "yes".
The space of all functions $\mathbb{R} \longrightarrow \mathbb{R}$ with the topology of pointwise convergence is a locally convex vector space, as its topology is induced by the system of seminorms
$$ \|f\|_x = |f(x)|.$$
However, since this system of norms in uncountable, sequences are not the right tool to describe convergence in this space.
Instead you should use nets. Using nets, you indeed have the following result: Every (yes every!) function $\mathbb{R} \longrightarrow \mathbb{R}$ is even the pointwise limit of a net of piecewise linear functions with finitely many knots. In particular, it is the limit of a net of continuous functions.
The set of finite subsets $\tau$ of $\mathbb{R}$ is ordered by inclusion and it is a directed set as the union of two finite sets is finite.
Now take a function $f:\mathbb{R} \longrightarrow \mathbb{R}$. For each finite subset $\tau$ of $\mathbb{R}$, let $f_\tau$ be the piecewise linear function with
$$f_\tau(t) = f(t)~~~~~\forall t \in \tau$$
that is constant left of the smallest element of $\tau$ and right of the biggest element of $\tau$. Then $(f_\tau)$ is a net that converges to $f$.<|endoftext|>
TITLE: Does Smith normal form imply PID?
QUESTION [27 upvotes]: Let $R$ be a nonzero commutative ring with  $1$, such that all finite matrices over $R$ have a Smith normal form.  Does it follow that $R$ is a principal ideal domain?
If this fails, suppose we additionally suppose that  $R$  is an integral domain?
What can we say if we impose the  additional condition that the diagonal entries be unique up to associates?

REPLY [7 votes]: For the sake of completeness, I summarize several important definitions in this subject (all close to the original question) and their main relations, as can be found in the literature (see the references below).
We consider all rings $R$ commutative and unital. All the theory that follows was inspired by the structure theorem for finitely generated modules over PIDs.
Firstly, given any $R$-module $M$, we have a free presentation (exact sequence)
$$F_1\rightarrow^g F_0\rightarrow^f M\rightarrow 0$$
with $F_0,F_1$ free modules. If $M$ is finitely presented we can pick $F_0\cong R^n$, $F_1\cong R^m$, and then $g$ can be represented by a matrix $A\in$Mat$_{n,m}(R)$ (for fixed bases of $R^m,R^n$), so that
$$M\cong R^n/AR^m.$$
Thus the study of (finite) matrices over $R$ is the same as the study of finitely presented $R$-modules (more in general, the study of infinite matrices is the study of all modules).
When $A$ is equivalent to a (nonsquare) diagonal matrix, $PAQ=$diag$(d_1,\ldots,d_r)$ with $P,Q$ invertible over $R$, we have
$$M\cong R^s\oplus R/d_1R\oplus\cdots\oplus R/d_rR,$$
so $M$ is a direct sum of cyclic modules, which are moreover cyclically presented. If in addition $d_i|d_{i+1}$ for all $i$ (i.e., if $A$ has a Smith normal form) then the presentation ideals satisfy $R\supseteq d_1R\supseteq d_2R\supseteq\cdots\supseteq d_rR$.
Now let us abstract the previous ideas with the following definitions. We will call a ring:

FGC if all its finitely generated modules are direct sums of cyclic modules.
FPC if all its finitely presented modules are direct sums of cyclic modules (this nomenclature is not usual).
Elementary divisor (ED) if all its finite matrices have a Smith normal form.
Hermite if all its $2\times1$ and $1\times2$ matrices are equivalent to a diagonal matrix. This implies that all matrices have a Hermite normal form (Kaplansky 1949).
Bezout if all its finitely generated ideals are principal.
A PIR if all its ideals are principal.
CF if all finite direct sums of cyclic modules $M$ have a canonical form, i.e., $M\cong R/I_1\oplus\cdots\oplus R/I_n$ with $I_1\supseteq I_2\supseteq\dots\supseteq I_n$.
CP if all its finitely presented cyclic modules $C$ have a cyclic presentation, i.e., $C\cong R/cR$ with $c\in R$ (this nomenclature is not usual).

We have the following immediate relations between the definitions:

FGC implies FPC by definition.
ED implies FPC by the exposition above. Moreover it implies CP and CF-only-for-finitely-presented-modules.
ED implies Hermite: All matrices have Smith normal form, in particular the $2\times1$ and $1\times2$ ones.
Hermite implies Bezout: The ideal $I$ generated by the elements $a,b\in R$ can be seen as the sum of the components of the image of the matrix $\begin{pmatrix}a&0\\b&0\end{pmatrix}:R^2\rightarrow R^2$, which can be invertibly reduced on the left to the form $\begin{pmatrix}c&0\\0&0\end{pmatrix}$, giving $I=cR$.

We have also less obvious relations between them:

Bezout implies CP: Let $C$ be a finitely presented cyclic module, $C\cong mR$. Then
$$0\rightarrow\text{Ann}_R(m)\rightarrow R\rightarrow C\rightarrow 0$$
is a short exact sequence with $C$ finitely presented and $R$ finitely generated. By a lemma of Bourbaki, Ann$(m)$ is finitely generated, so a finitely generated ideal, hence Ann$(m)=cR$ since $R$ is Bezout. This comes from (Larsen, Lewis, Shores 1974) (they just refer to Bourbaki's result).
FPC implies elementary divisor: This is one of the main results from (Larsen, Lewis, Shores 1974). Note that, as elementary divisor equals FPC+CP+CF-fin.presented, what this is saying is that once all finitely presented modules are direct sums of cyclic modules, said modules can be taken cyclically presented and in canonical form, so that the Smith normal form of the corresponding matrix exists.

So now we have 
$$\text{FGC}\Rightarrow \text{FPC}=\text{ED}\Rightarrow \text{Hermite}\Rightarrow \text{Bezout}\Rightarrow \text{CP}.$$
In the literature there are also answers to interesting questions:

The Noetherian setting: As has also been proved here, an elementary divisor ring is Noetherian if and only if it is a PIR (Uzkov 1963). Since for Noetherian rings all submodules of finitely generated modules are finitely generated (so finitely generated modules are finitely presented), we also have that the Noetherian FGCs are exactly the PIRs. Observe that the existence of zero divisors does not prevent the matrices from having a Smith normal form.
Diagonal matrices having SNF: The rings such that every diagonal matrix has a Smith normal form are precisely the Bezout rings (Larsen, Lewis, Shores 1974). So, in Bezout rings, the obstruction for matrices having an SNF is not being equivalent to a diagonal matrix.
Sufficient condition for ED: Hermite rings ask for a diagonal form of small matrices, the $2\times1$ and $1\times2$ vectors. We may ask which other sizes are needed to guaraantee that the ring is not only Hermite, but elementary divisor. It turns out that order $2$ is enough: a ring is ED if and only if all its $2\times 2$ matrices are equivalent to a diagonal matrix (Larsen, Lewis, Shores 1974).
Conditions for Bezout being Hermite: Not all Bezout rings are Hermite. Bezout domains are Hermite, and it is an open problem to determine if all Bezout domains are ED domains, a question coming from (Helmer 1943). Bezout rings with a finite number of minimal prime ideals are Hermite (Larsen, Lewis, Shores 1974). 
The classification of FGCs: Due to the work of several people, the (commutative) FGCs have been classified. A ring is FGC if and only if:


It is Bezout and fractionally self-injective (Vámos 1979).
It is a finite direct sum of maximal valuation rings, almost maximal Bezout rings, and torch rings.
It is a finite direct sum of rings $S$ satisfying: $S$ has a unique minimal prime $P$, $S/P$ is an h-local Bezout domain, the ideals contained in $P$ form a chain, and for each maximal ideal $M$ of $S$, $S_M$ is an almost maximal valuation ring.

FGC implies CF: As a consequence of the classification theorems it can be shown that all finitely generated modules over an FGC can be presented in canonical form (Wiegand, Wiegand 1977).
Classification of CF rings: CF rings have also been classified (Shores, Wiegand 1974). They are the finite direct sums of local rings, h-local domains, and rings $S$ with a unique minimal prime $P$ such that $R/P$ is an h-local domain, $P^2=0$ and every ideal of $S$ contained in $P$ is comparable with every ideal of $S$.

A detailed account on FGC rings can be found in (Brandal 1979).
We can also think about diagonal forms of infinite matrices. In this context it is more usual to speak of stacked bases: given two free $R$-modules $F\leq G$, we say that they have stacked bases if there is a basis $\{x_i\}_{i\in I}$ of $G$ such that $\{r_jx_j\}_{j\in J}$ is a basis of $F$ with $I\subseteq J$ and $r_j\in R$ for all $j\in J$, that is, if a basis of $F$ can be formed by taking multiples of some of the vectors of the basis of $G$. Observe that if $F,G$ have stacked bases then $G/F$ is a direct sum of cyclic modules.
Over a PID, the stacked bases theorem generalizes the structure theorem for not necessarily finitely generated modules: If $R$ is a PID and $F\leq G$ with $G$ free are such that $G/F$ is a direct sum of cyclic modules then $F,G$ have stacked bases (Cohen, Gluck 1970).
To further generalize to integral domains, (Fuchs, Salce 2000) broadens the definition of stacked bases so that the submodule needs not be free, building on the structure theorem for finitely generated modules over Dedekind domains: Given modules $H\leq F$ with $F$ free, they have stacked bases if we can write $F=\bigoplus_{i\in\Lambda_F} J_ix_i$, $H=\bigoplus_{i\in\Lambda_H} I_iJ_ix_i$ with $\Lambda_H\subseteq\Lambda_F$, $J_i$ invertible ideals of $R$ (i.e., the $J_ix_i$ are rank one projective modules), and $I_i$ finitely generated ideals of $R$.
Then Dedekind domains satisfy the stacked bases property for finitely presented modules $F/H$. (Fuchs, Salce 2000) shows that if $R$ is an h-local Prüfer domain and $M$ is a direct sum of cyclic $R$-modules of projective dimension one, then every presentation $0\rightarrow H\rightarrow F\rightarrow M\rightarrow 0$ of $M$ has stacked bases.
References:

The elementary divisor theorem for certain rings without chain condition (1943). Helmer.
Elementary divisors and modules (1949). Kaplansky.
On the decomposition of modules over a commutative ring into a direct sum of cyclic submodules (1963). Uzkov.
Stacked bases for modules over principal ideal domains (1970). Cohen, Gluck.
Elementary divisor rings and finitely presented modules (1974). Larsen, Lewis, Shores.
Rings whose finitely generated modules are direct sum of cyclics (1974). Shores, Wiegand.
Commutative rings whose finitely generated modules are direct sums of cyclics. Wiegand, Wiegand.
Commutative rings whose finitely generated modules decompose (1979). Brandal.
Sheaf-theoretical methods in the solution of Kaplansky's problem (1979). Vámos.
Stacked bases over h-local Prüfer domains (2000). Fuchs, Lee
Rings and things and a fine array of Twentieth Century associative algebra (2004). Faith. Elementary division rings. FGC rings.<|endoftext|>
TITLE: Does a crossed  product R⋊_α F_n of the hyperfinite factor of type II_1 and a free group have the QWEP?
QUESTION [7 upvotes]: Let $\mathcal{R}$ be the hyperfinite factor of type $\rm{II}_1$ and let $\mathbb{F}_n$ be a free group with $n$ generators. Let $\alpha$ be an action of $\mathbb{F}_n$ on $\mathcal{R}$.
Does the von Neumann crossed product $\mathcal{R}\rtimes_{\alpha}\mathbb{F}_n$ have the QWEP?
Remarks: Since $\mathbb{F}_n$ is a residually finite group, the group von Neumann algebra $\rm{VN}(\mathbb{F}_n)$ has the QWEP.  Moreover $\mathcal{R}$ has the QWEP.

REPLY [9 votes]: Yes.  If $a$ and $b$ are generators of $\mathbb F_2$ then $\mathcal R \rtimes_\alpha \mathbb F_2$ decomposes as an amalgamated free product of $(\mathcal R \rtimes_\alpha \langle a \rangle)$ and  $(\mathcal R \rtimes_\alpha \langle b \rangle)$ over $\mathcal R$, where each of these are hyperfinite.  Brown, Dykema, and Jung showed in http://arxiv.org/abs/math/0609080 that for separable finite von Neumann algebras being embeddable into $\mathcal R^\omega$ is stable under amalgamated free products over a hyperfinite von Neumann algebra.  Thus $\mathcal R \rtimes_\alpha \mathbb F_2$ is embeddable into $\mathcal R^\omega$, which is equivalent to QWEP.  Induction then gives the case when $2 \leq n < \infty$, and the case $n = \infty$ then follows since QWEP is preserved under (the weak-closure of) increasing unions.
Related to this, Collins and Dykema in http://arxiv.org/abs/1003.1675 have recently shown that the class of Sophic groups is stable under taking amalgamated free products over amenable groups.
I believe this is an open problem however if we consider arbitrary residually finite groups instead of only $\mathbb F_n$.<|endoftext|>
TITLE: Suggestions for wiki farm with good latex support
QUESTION [8 upvotes]: I've decided to start a wiki to do collaborative mathematics. However I don't have access or control over a server. So I need a wiki farm. I've tried out pbworks and wikidot, but their latex support is not as straightfoward as say wordpress. 
Do you have a suggestion of which wiki farm to use?

REPLY [3 votes]: If you're looking for a wiki that can handle LaTeX-style equations, then you should take a look at instiki.  Not only does it display mathematics properly, it can also export pages to LaTeX.
http://www.instiki.org/show/HomePage<|endoftext|>
TITLE: Let a function f have all moments zero. What conditions force f to be identically zero?
QUESTION [16 upvotes]: Throughout, let $f$ be a Lebesgue measurable function (or continuous if you wish, but this is probably no easier). (Questions with distributions etc. are possible also but I want to keep things simple here).
FINAL CLARIFICATION/REWRITE!!
Thanks to all who have commented so far. I will need some more time to digest it properly. The original forms of the questions are at the end; here I have rewritten the questions, hopefully  more clearly; sorry for my poor explanation before!
Definition $\mathcal{M}_a$, the space of absolutely null moment functions, is the set of all measurable $f$ satisfying
$$
\int_0^\infty t^n |f(t)| dt < \infty,
\quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots.
$$
$\mathcal{M}$, the space of null moment functions, is the set of all measurable $f$ satisfying
$$
\int_0^T |f(t)| dt < \infty  \quad \forall \, T>0, 
\qquad
\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \quad \forall \, n=0,1,2,\ldots
$$
Thus, trivially $0 \in \mathcal{M}_a \subseteq \mathcal{M}$, but $\mathcal{M}_a$ contains many other non-trivial functions. It seems certain that $\mathcal{M}_a \ne \mathcal{M}$ (I would be amazed if the spaces were equal), although constructing an explicit example seems tricky.
Definition Given a function $\psi \geq 0$, let $G(\psi)$ be the set of all $f$ such that $|f| \leq \psi$.
(Of course we're identifying functions equal a.e., so really we should consider equivalence classes etc. just as for $L^p$ spaces).
Rephrased Question 1 Find general simple necessary and/or sufficient conditions on $\psi$ with the property
$$
 G(\psi) \cap \mathcal{M}_a = \{ 0 \}. 
$$
Rephrased Question 2 The same as Question 1, but with $G(\psi) \cap \mathcal{M}$ instead.
Or, if $G(\psi)$ is not the appropriate space for these problems, consider  $\int_T^{2T} |f| dt \leq g(T) $ or $\int_n^{n+1} |f| dt \leq A_n$ or something similar instead. Finding the correct kind of restrictions on $f$ is part of the problem.
Thus, $G(\psi) \cap \mathcal{M}_a = \{ 0 \}$ for $\psi(t) = \exp(-\delta t)$, by the discussion below; and also for any compactly supported $\psi \in L^1$. 
But $G(\psi) \cap \mathcal{M}_a \ne \{ 0 \}$ for $\psi(t) = \exp(-t^{1/4}) |\sin(t^{1/4})|$.

Original QUESTION 1
If 
$$
\int_0^\infty t^n |f(t)| dt < \infty,
\quad \int_0^\infty t^n f(t) dt = 0, \qquad \forall \, n=0,1,2,\ldots,
$$
when must $f \equiv 0$ almost everywhere? 

EDIT: CLARIFICATION: this is really about classes of functions, expressed in terms of a growth/decay rate function $\phi$, which give unique solutions to the moment problem. I am NOT asking how to solve the moment problem itself! 
If possible, find necessary and sufficient conditions on functions $\phi \searrow 0$ with the property that
$$
\int_R^\infty |f(t)| dt \leq \phi(R) \qquad \Rightarrow \qquad f \equiv 0.
$$
So, $\phi(R) = \int_R^\infty \exp(-t^{1/4}) |\sin(t^{1/4})| dt$ is not enough (by Example 2 below); nor is $\phi(R) = \int_R^\infty |f(t)| dt$ with f given by "coudy" in his/her answer below.
But $\phi = \chi_{[0,b]}$ is enough (by Example 1 below); moreover $\phi(R) = \exp(-\delta R)$ would be enough for any fixed $\delta > 0$, by my discussion of Example 1 below, because the relevant Laplace transform $F = \mathcal{L}f$ is analytic on the half-plane $\{ \mathrm{Re}(z) > -\delta \}$. So we want to know about the gap between $\exp(-\delta R)$ and functions like that given by "coudy" below.
FURTHER CLARIFICATION: by analogy, maybe an example from PDEs will explain better (I'm not saying this is related to my problem; I'm saying that this is the same kind of result as what I want):
Definition A null temperature function is a continuous function $u = u(x,t) : \mathbb{R} \times [0, \infty) \to \mathbb{R}$ such that the heat equation is satisfied, i.e. $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$ in $\{ t>0 \}$, and $u(x, 0) = 0$ for all $x$.
Theorem There exists a null temperature function satisfying $|u(x,t)| \leq \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for all $t>0$.
Theorem Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta<1$. Then $u \equiv 0$.
So here the "critical" growth rate for null temperature functions is roughly $\exp(A/t)$. I am looking for a similar thing with "null moment functions".
Note that this is a totally different problem to: given $v$, find some $u$ satisfying the heat equation such that $u(x,0)=v(x)$.

Original QUESTION 2
If instead we have only $\int_0^R |f(t)| dt < \infty$ for each $R>0$, and
$$
\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \qquad n=0,1,2,\ldots
$$
when must $f \equiv 0$ almost everywhere? (I have very little idea about this).

I think these questions are clearly very natural, interesting, and important, but Googling etc. didn't work well (I tried "vanishing moments" and other phrases, but there's just too much stuff out there). Standard known examples/methods follow.
Example 1: if $f$ is compactly supported on $[a,b]$, say, then $f \equiv 0$ a.e. because polynomials are dense in $C[a,b]$.
Example 2: by taking imaginary parts of $\int_0^\infty t^{4n+3} \exp(-(1+i)t) dt \in \mathbb{R}$, we get
$$
\int_0^\infty t^{4n+3} e^{-t} \sin t \, dt = 0,
$$
and so by substituting $t = x^{1/4}$,
$$
\int_0^\infty x^n \exp(-x^{1/4}) \sin(x^{1/4}) dx = 0, \qquad n=0,1,2,\ldots
$$
Alternative method for Example 1: consider the Laplace transform $F(z) = \int_0^\infty e^{-zt} f(t) dt$. In Example 1, $F$ is an entire function such that $F^{(n)}(0) = 0$ for all $n$, so $F \equiv 0$ and thus $f \equiv 0$ a.e. as required.
So, any condition on $f$ forcing $F$ to be analytic on some disc with centre $0$ is enough; but can we do better? 
In Example 2, $f \in L^1(0,\infty)$ and so $F$ is bounded and analytic on $\{ \mathrm{Re}(z) > 0 \}$, and continuous on the boundary, with $\lim_{z \to 0} F^{(n)}(z) = 0$ for all $n$. But this is still not enough to force $F \equiv 0$.

REPLY [17 votes]: The answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2.
$$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}_{[0,\infty[}(x)$$
You can check that all the moments are zero using the change of variable $\ln x = s+k$.
This example is used to show that a probability measure is not always determined by its moments.<|endoftext|>
TITLE: Is Lebesgue's "universal covering" problem still open?
QUESTION [46 upvotes]: The following problem has been attributed to Lebesgue. Let "set" denote any subset of the Euclidean plane. What is the greatest lower bound of the diameter of any set which contains a subset congruent to every set of diameter 1? There are a number of interesting geometric problems of this type. Is it possible that some of them may be difficult to solve because the solution is a real irrational number which (when expressed in decimal form) is not even recursive-and so cannot be approximated in the usual way?

REPLY [5 votes]: Philip Gibbs wrote:

the area of the universal cover for an angle $\sigma=0.4$ degrees can be computed to be $0.8441177$.

I believe Philip later noticed an error in this area computation.  He also later noticed that there's a constraint that $\sigma$ needs to obey, which is not obeyed by this choice: the best choice of $\sigma$ is about 
$$ \sigma = 1.294389444703601012^\circ $$
and the area of the universal covering this gives is about
$$ 0.844115297128419059\dots .$$
We wrote a paper on this together with Karine Bagdasaryan, and Greg Egan did some high-precision calculations that give the above numbers.   Our paper explains the details:

John C. Baez, Karine Bagdasaryan and Philip Gibbs, The Lebesgue universal covering problem.

Abstract: In 1914 Lebesgue defined a "universal covering" to be a convex subset of the plane that contains an isometric copy of any subset of diameter 1. His challenge of finding a universal covering with the least possible area has been addressed by various mathematicians: Pal, Sprague and Hansen have each created a smaller universal covering by removing regions from those known before. However, Hansen's last reduction was microsopic: he claimed to remove an area of $6 \cdot 10^{−18}$, but we show that he actually removed an area of just $8 \cdot 10^{-21}$. In the following, with the help of Greg Egan, we find a new, smaller universal covering with area less than $0.8441153$. This reduces the area of the previous best universal covering by a whopping $2.2 \cdot 10^{−5}$.<|endoftext|>
TITLE: Provably intractable problems
QUESTION [7 upvotes]: Let f(n) be a space-constructible superpolynomial function.  Then BQP $\subseteq$ PSPACE $\subset$ SPACE(f(n)), so in particular, SPACE(f(n)) $\not\subseteq$ BQP.  Let L be a problem such that every problem in SPACE(f(n)) is BQP-reducible to L.  Then L $\notin$ BQP.
Are there any problems that have been proven to not be in BQP for which that is not known to be provable by the above method?

REPLY [5 votes]: There are few complexity class separations known which do not follow from some type of diagonalization (a complexity hierarchy theorem of some kind). I know of none for $\mathbf{BQP}$. One canonical example of a separation that doesn't seem to follow from a diagonalization argument is $\mathbf{AC}^0 \subsetneq \mathbf{NC}^1$, which instead follows from the Ajtai-Furst-Saxe-Sipser theorem that the parity of $n$ bits does not have polynomial size circuits of unbounded fan-in and constant depth. 
Now, if by "the above method" you meant something more specific than just diagonalization, then there is just a little something else you can say about $\mathbf{BQP}$. Adleman, DeMarrais, and Huang proved that $\mathbf{BQP} \subseteq \mathbf{PP}$:

Leonard M. Adleman, Jonathan DeMarrais, Ming-Deh A. Huang: Quantum Computability. SIAM J. Comput. 26(5): 1524-1540 (1997)

(Recall that $\mathbf{PP}$ consists of languages recognized by randomized polynomial time algorithms with "exponential precision". Without loss of generality, we may say that an input is "accepted" by such an algorithm if and only if the probability of outputting $1$ is strictly greater than $1/2$. Note this probability could be $1/2+1/2^{n^{\Omega(1)}}$. It is known that $\mathbf{PP} \subseteq \mathbf{PSPACE}$, but the other direction is unknown.) 
Just like $\mathbf{PSPACE}$ has a superpolynomial analogue $\mathbf{SPACE}(f(n))$, $\mathbf{PP}$ has a superpolynomial analogue $\mathbf{PTIME}(f(n))$, so in your above argument you can replace $\mathbf{SPACE}(f(n))$ with $\mathbf{PTIME}(f(n))$. Note the latter is contained in the former.<|endoftext|>
TITLE: A question about regularity of foliations
QUESTION [5 upvotes]: Let $F$ be a smooth foliation of a torus. Assume that $F$ can be mapped by a homeomorphism to an irrational-straight-line foliation $L$. Does it follow that $F$ can be mapped to $L$ by a diffeomorphism?
I am interested in 2 dimensional case and higher dimensional case.

REPLY [6 votes]: As far as I understand, the answer to the question is no, you can check this in
Handbook of Dynamical Systems, Volume 1, Part 1 By Boris Hasselblatt, Anatole Katok, page 173. This question is identical to the following -- suppose we have a diffeo of S^1 toplogically conjugate to an irrational rotation, can we make this conjugation a diffeo? Here is the sitation from the book:
For smooth and analytic circle 
diffeomrophisms with extremely well approximable rotation
number, the conjugacy to a roation and hence the invariant 
measure tend to be singular. Arnold's theorem 
exposed sigularity of the conjugacies as a generic phenomenon
in typical one-parameters families of real-analityc maps<|endoftext|>
TITLE: How do I fix someone's published error?
QUESTION [117 upvotes]: Paper A is in the literature, and has been for more than a decade.
An error is discovered in paper A and is substantial in that many 
details are affected, although certain fundamental properties 
claimed by the theorems are not.  (As a poor analogue, it would be 
like showing that certain solutions to the Navier-Stokes equations 
had different local properties than what were claimed, but that the 
global properties were not affected.  The error is not of the same
caliber as Russell's correction of Frege's work in logic.)
The author is notified, who kindly acknowledges the error.
Now what?
Should the remaining action lie fully on the author, or should
the discoverer of the error do more, such as contact the journal,
or publish his own correction to paper?  How long should one
wait before suitable action is taken?  And what would be 
suitable action if not done by the author?
Based on remarks from those who previewed this question
on meta.mathoverflow, I propose the following
Taxonomy:  There are various kinds of error
that could be considered.

typographical - An error where a change of a character or a
word would render the portion of the paper correct.  In some
cases, the context will provide enough redundancy that the
error can be easily fixed by the reader.  Addressing these
errors by errata lists and other means have their importance,
but handling those properly is meant for another question.
slip - (This version is slightly different from the
source; cf the discussion on meta for the source
http://mathoverflow.tqft.net/discussion/493/how-do-i-fix-someones-published-error/ )
This is an error in a proof which may be corrected, although
not obviously so.  In a slip, the claimed main theorem is either
true or can be rescued with little cost.  In my opinion, the degree
of response is proportional to the amount of effort needed to fix it
(and is often minor),
but there may be slips major enough to warrant the questions above.
miscalculation -  Often a sign or quantity error.  In some cases
the results are minor, and lead to better or worse results
depending on the calculation.  I've included some miscalculations
in some of my work to see if anyone would catch them.  I've
also prepared a response which shows the right calculation and
still supports the main claims of the work.  (See below on
impact as a factor.)
oversight or omission - This is stating a fact as true without
sufficient folklore to back up that fact.  In some cases the
author doesn't include the backup to ease (the reading of) the
paper and because the author thinks the audience can provide 
such backup.  More seriously,
the omission occurs because the author thought the fact was
true and that there was an easy proof, when actually the fact
may or may not be a fact and the author actually had a faulty
argument leading him to think it true.
major blunder - This is claiming a result which is true,
and turns out not to be true in a socially accepted proof
system.  Proofs of Euclid's fifth postulate from the other
four fall into this type.

The above taxonomy is suggested to help determine the type
of response to be made by the discoverer.  Also, degree of
severity is probably not capable of objective measure, but 
that doesn't stop one from trying.  However, there are at
least two other considerations:

Degree to which other theorems (even from other papers)
depend on the error in the result.  I call this impact.
Degree to which the error is known in the community.

The case that inspired this question falls, in my mind,
into the category of a miscalculation that invalidates a 
proposition and several results in paper A following from
the proposition.  However, as I alluded to above in the 
Navier-Stokes analogy, the corrected results have the same 
character as the erroneous results.  I would walk on a 
bridge that was built using the general characteristics of 
the results, and not walk on a bridge that needed the 
specific results.  In this case, I do not know to what 
degree impact the miscalculation has on other papers, nor 
how well known this miscalculation is in the community.
If someone thinks they know what area of mathematics my case 
lies (and are sufficiently experienced in the area), and they 
are willing to keep
information confidential, I am willing to provide more
detail in private.  Otherwise, in your responses, I ask that no
confidentiality be broken, and that no names be used
unless to cite instances that are already well-enough
known that revealing the names here will do no harm.  Also,
please include some idea of the three factors listed
above (error type, impact on other results, community awareness),
as well as other contributing factors.  
This feels like a community-wiki question.  Please, one
response/case per answer.  And do no harm. 
Motivation:  Why do I care about fixing someone else's error? 
Partly, it adds to my sense of self-worth that I made a 
contribution, even if the contribution has no originality.
Partly, I want to make sure that no one suffers from the
mistake.  Partly, I want to bring attention to that area
of mathematics and encourage others to contribute.  Mostly
though, it just makes an empty feeling when one reaches
the "Now What?" stage mentioned above.  Feel free to
include emotional impact, muted sufficiently for civil
discourse.
Gerhard "Ask Me About System Design" Paseman, 2010.07.10

REPLY [21 votes]: I apologize for posting before reading the long discussion; I did a cursory check however and I noticed that Ted Hill has not yet been mentioned. His text How to Publish Counterexamples in 1 2 3 Easy Steps is a first-hand account of dealing with submitting corrections to authors and editors in a rather high-profile case. The facts of the case are intricate enough that one wants to remain circumspect in one's conclusions, but one thing is overwhelmingly clear: the process is not for the fainthearted.<|endoftext|>
TITLE: What is the most general "two in one row for A & in one column for B" theorem?
QUESTION [6 upvotes]: Let $A$ and $B$ be two Young tableaux, i. e. Young diagrams filled with the numbers $1$, $2$, ..., $n$ for some $n$ (not necessarily the same $n$). (They need not be semistandard.)
(a) (Etingof's Lectures on Representation Theory, proof of Lemma 4.40): If $A$ and $B$ share the same Young diagram, then there exist two entries which lie in the same row of $A$ and in the same column of $B$, unless we can make $A$ and $B$ equal by permuting some elements inside their rows in $A$ and permuting some elements inside their columns in $B$.
(b) (Etingof's Lectures on Representation Theory, proof of Lemma 4.41): If the partition corresponding to $A$ is lexicographically larger than that corresponding to $B$ - but both have the same sum -, then there exist two entries which lie in the same row of $A$ and in the same column of $B$.
(c) (Serganova's Representation Notes, lecture 6, Lemma 1.4): If the Young diagram of $B$ has only one less square than $A$ but does not result from $A$ by removing one square, then there exist two entries which lie in the same row of $A$ and in the same column of $B$, or two entries which lie in the same row of $B$ and in the same column of $A$.
I am rather new to Young tableaux, and I haven't looked into Fulton, Stanley or Knuth, but maybe someone can answer on the spot whether there is a more general statement behind these three results?
Oh, and since this fits so nicely: This paper gives a wonderful proof of the Littlewood-Richardson rule, even generalized to the product of a Schur function of a Young diagram with that of a skew Young diagram. Is there a reasonable generalization to the product of two skew Young diagrams?
UPDATE: Claim (c) is wrong, as easily checked for $\lambda = \begin{array}{ccc} 5&1&3\\\\ 2& & \\\\ 4 & & \\\\ \end{array}$ and $\mu = \begin{array}{cc} 1& 2 \\\\ 4 & 3 \end{array}$.

REPLY [8 votes]: I can't give you your desired "most general" theorem, but I can say a little about this. In (b), the condition "shape(A) is lexicographically larger than shape(B)" is much stronger than it needs to be: "shape(A) is not dominated by shape(B)" will yield the same conclusion (recall the dominance order on partitions: $\lambda$ dominates $\mu$ if $\lambda_1+\cdots+\lambda_i\geqslant\mu_1+\cdots+\mu_i$ for each $i$).
To prove this: suppose all the entries in each row of $A$ are in different columns of $B$.  Replace each entry of $B$ with the number of the row in which it appears in $A$; then by assumption the entries in each column of (the modified) $B$ are distinct.  So if we sort the entries in this tableau into increasing order, all the entries less than or equal to $i$ will appear in the top $i$ rows.  Hence the number of positions in the top $i$ rows of $B$ is at least the number of entries in the top $i$ rows of $A$, i.e. $\lambda_1+\cdots+\lambda_i\geqslant\mu_1+\cdots+\mu_i$ (where $\lambda=\operatorname{shape}(B)$ and $\mu=\operatorname{shape}(A)$).
This is all assuming that $A$ and $B$ have the same size.  If $A$ is bigger than $B$, then obviously it goes wrong (because then $\lambda$ can't possibly dominate $\mu$, but the conclusion could easily be false).  A more general statement (I think) is the following:
if either ($|\lambda|\geqslant|\mu|$ and $\lambda\ntrianglerighteq\mu$) or ($|\lambda|\leqslant|\mu|$ and $\mu'\ntrianglerighteq\lambda'$) then there are two entries in the same row of $A$ and the same column of $B$.
(Here I'm still writing $\lambda=\operatorname{shape}(B)$ and $\mu=\operatorname{shape}(A)$, $\lambda'$ denotes the conjugate (=transpose) partition to $\lambda$, and $\trianglerighteq$ is the dominance order.)<|endoftext|>
TITLE: Is there a way to see a topological group as the "Cayley graph" of its "infinitesimal generators"?
QUESTION [10 upvotes]: At the time of writing, the most recent blog post over at What's new by Terrence Tao is Cayley graphs and the geometry of groups, and that (excellent, as with most of Tao's writing) post most immediately inspired this question.  Also, a disclaimer: Cayley graphs, and, indeed, abstract group theory generally, are not my area of expertise, so feel free to explain how my question is overly naive and/or needs revision.
It is a classical result that a connected topological group is generated by any open neighborhood of its identity element.  (Proof: the subgroup generated by the open neighborhood is a union of opens, and hence open; but then so are all its cosets; so this subgroup is both open and closed; hence it is everything if the group is connected as a topological space.)  So I am tempted to take some topological group $G$, and some very small neighborhood $S$ of the identity $e$, and construct the corresponding Cayley graph.  My intuition says that as $S$ gets smaller and smaller, the Cayley graph better and better approximates the topological space $G$.  Notice also that arbitrary small open neighborhoods $S\ni e$ know everything about the topology of $G$, because $G$ is homogeneous.
I don't know how to define a "formal neighborhood" of a topological space, unless it is actually a manifold or algebraic space or ....  So maybe I should restrict my attention to Lie or Algebraic groups for this question.  But anyway:

Is there a way to define the "infinitesimal generators" or "formal neighborhood" for a topological group in such a way that the corresponding "Cayley graph" is the group as a topological space (Edit: which of course doesn't make any sense; see the comments.  I mean something like "so that the geometry/topology of the group comes immediately from the graph)?  If not in this generality, does it at least work when the group is Lie? Algebraic? Or other regularity conditions on the topology?

REPLY [10 votes]: Several people have been working on what might be described as geometric group theory for compactly generated, totally disconnected topological groups.  
The main definitions as well as several nice results are in this paper:
Krön, Bernhard; Möller, Rögnvaldur G.
Analogues of Cayley graphs for topological groups. 
Math. Z. 258 (2008), no. 3, 637--675. 
and if you search for papers which reference this one you'll find some more recent work.<|endoftext|>
TITLE: What can be said about a group from its presentation?
QUESTION [12 upvotes]: This maybe a very general question.
If we have a group given by its presentation only, what kind of properties could be proven about it?
I know examples about non-amenability of some Burnside groups.
What kind of examples are there in literature where one proves some property "just" from a presentation?

REPLY [5 votes]: Given a group $\Gamma$ which you are interested in, given by a presentation, and an easy group $G$ which you understand (a dihedral group or a symmetric group, maybe), you can often use the presentation to determine whether or not there exists a surjective homomorphism $\Gamma\to G$, and maybe you can even count them. This is a strong tool for showing that two given presentations give rise to different groups, and for showing that your group must be "at least as complicated" as $G$. For instance, Tietze originally proved in 1908 that the trefoil is knotted by exhibiting a surjection from a Wirtinger presentation of the fundamental group of its complement onto the symmetric group $S_3$, while the fundamental group of the unknot is abelian and so can admit no such homomorphism.<|endoftext|>
TITLE: Can a mathematical definition be wrong?
QUESTION [94 upvotes]: This question originates from a bit of history. In the first paper on quantum Turing machines, the authors left a key uniformity condition out of their definition. Three mathematicians subsequently published a paper proving that quantum Turing machines could compute uncomputable functions. In subsequent papers the definition of quantum Turing machine was revised to include the uniformity condition, correcting what was clearly a mathematical error the original authors made.
It seems to me that in the idealized prescription for doing mathematics, the original definition would have been set permanently, and subsequent papers would have needed to use a different term (say uniform quantum Turing machine) for the class of objects under study. I can think of a number of cases where this has happened; even in cases where, in retrospect, the original definition should have been different.
My question is: are there other cases where a definition has been revised after it was realized that the first formulation was "wrong"?

REPLY [2 votes]: For me the more interesting question is what makes definitions "right" — like how the definition of e.g. function, manifold, or Lie group are definitions that are widely believed to be very unlikely to undergo any future revisions.*
But maybe I'm being fussy. When a mathematician learns a mature definition for a subtle concept, there is a feeling as if something is clicking into place that can be very satisfying.

As for manifolds, there are actually a number of definitions. Most obviously, depending on the pseudogroup of transition functions allowed for its atlas, most commonly continuous, merely homeomorphic to a simplicial complex, piecewise linear, differentiable of class Ck with 1 <= k <= infinity, real analytic, or complex analytic. And there is the question of whether infinite dimensions should be included. But also, there are diverging opinions on whether a manifold must be second countable, and in some quarters even on whether a manifold must be Hausdorff. Personally I favor removing all such restrictions but the requirement that it be locally Euclidean of some nonnegative dimension. But this can be terminologically inconvenient.<|endoftext|>
TITLE: Proving that a binary matrix is totally unimodular
QUESTION [9 upvotes]: I'm working on a set of problems for which I can formulate binary integer programs. When I solve the linear relaxations of these problems, I always get integer solutions. I would like to prove that this is always the case. 
I believe that this involves proving that the constraint matrix is totally unimodular. Is there any sufficent conditions for binary matrices to be totally unimodular that might be of use for this?

REPLY [12 votes]: Here are some common ways of proving a matrix is TU.

The incidence matrix of a bipartite graph and network flow LPs are TU; these are standard examples usually taught in every book on TU.
The consecutive-ones property: if it is (or can be permuted into) a 0-1 matrix in which for every row, the 1s appear consecutively, then it is TU. (The same holds for columns since the transpose of a TU matrix is also TU.)
Every "network matrix," defined as follows, is TU (and they are a fundamental building block of the set of all TU matrices, according to Seymour's theorem). The rows correspond to a tree $T = (V, R)$ each of whose arcs have an orientation (i.e. it is not necessary that exist a root vertex $r$ such that the tree is "rooted into $r$" or "out of $r$").The columns correspond to another set $C$ of arcs on the same vertex set $V$. To compute the entry at row $R$ and column $C = st$, look at the $s$-to-$t$ path $P$ in $T$, then the entry is:

+1 if arc $R$ appears forward in $P$
-1 if arc $R$ appears backwards in $P$
0 if arc $R$ does not appear in $P$

[You can see more in Schrijver's 2003 book.]

Ghouila-Houri showed a matrix is TU iff for every subset $R$ of rows, there is an assignment $s : R \to \pm 1$ of signs to rows so that the signed sum $\sum_{r \in R} s(r)r$ (which is a row vector the same width as the matrix) has all its entries in $\{0, \pm1\}$. 

There are other if-and-only-if conditions like Ghouila-Houri too (see Schrijver 1998) but the 4 conditions I gave above have been the most practical for me.<|endoftext|>
TITLE: Describing global sections of sheafifications
QUESTION [7 upvotes]: Recently on glancing through Hartshorne's description
of Cartier divisors I started pondering the definition of
sheafification which led me to a question I can't answer. Neither
can I find a discussion in the standard texts.
First let's set up some notation.
Let $\mathcal{F}$ be a presheaf (let's say of sets)
on the topological space $X$.
It has a sheafification $\mathcal{F}^+$. We want to describe
the global sections of $\mathcal{F}^+$.
A global section of $\mathcal{F}^+$ consists
of a map $f:x\mapsto f_x$ where $f_x\in\mathcal{F}_x$,
the stalk of $\mathcal{F}$ at the point $x\in X$. This map
$f$ also obeys a local compatibility condition: there is an open
covering $(U_i)_{i\in I}$ of $X$, and sections $g_i\in\mathcal{F}(U_i)$
with the property that whenever $x\in U_i$ then $f_x$
equals the germ of $g_i$ at the point $x$. Let's call such a covering
$(U_i)$ and sections $(g_i)$ a representing system of sections for
the global section. (Is there a standard term for this concept?)
My question is this:

For each global section of $\mathcal{F}^+$
  is there always a representing system of sections for it having the
  stronger compatibility property that
  $$g_i|_{U_i\cap U_j}=g_j|_{U_i\cap U_j}\in\mathcal{F}(U_i\cap U_j)$$
  for all $i$, $j\in I$?
  If not, is there some reasonable condition on the presheaf $\mathcal{F}$
  that will guarantee this?

Another motivation is to find a good "pointless" description of
the sheafification functor (in the sense of "pointless topology" or locale
theory). The definitions of presheaf and sheaf only use the complete
lattice structure on the collection of open sets of $X$ and so are
thoroughly "pointless", but the usual description of the sheafification
functor uses the definitely "pointy" notion of stalk.

REPLY [6 votes]: As Kevin points out, one can't ask for a representing family of sections that are equal on every intersection.  However, the description can certainly be restated pointlessly.
For $f,g \in \mathcal{F}(U)$, define $f \approx g$ if there's a cover $U = \bigcup_i U_i$ such that for each $i$, $f|_{U_i} = g|_{U_i}$.  Read this as “$f$ and $g$ are equal on a cover”.
In pointy terms, $f \approx g$ iff all their germs are equal: for every $x \in U$, $f_x = g_x$.
Now, every section $f \in \mathcal{F}^+(U)$ of the sheafification can be represented by a weakly matching family in $\mathcal{F}$: that is, a cover $U = \bigcup_i U_i$, and sections $f_i \in \mathcal{F}(U_i)$, such that for each $i,j$, we have $f_i|_{U_i \cap U_j} \approx f_j|_{U_i \cap U_j}$.  (This is immediate from the representing families you exhibit in the question.)  Similarly, two weakly matching families are $(f_i), (g_j)$ are equal as sections of the sheafification iff for each $i, j$, we have $f_i|_{U_i \cap V_j} \approx g_j|_{U_i \cap V_j}$.
So the sections of $\mathcal{F}^+$ can be described exactly as equivalence classes of weak matching families from $\mathcal{F}$.  This is a standard way of constructing sheafification on general sites, known as the “double plus-” or “$(-)^{++}$-construction”; see e.g. the Mac Lane and Moerdijk book Sheaves in Geometry and Logic for more context.<|endoftext|>
TITLE: Vanishing of Euler class
QUESTION [13 upvotes]: Given a real oriented vector bundle E over the base space B of rank n, such that the Euler characteristic class in the n-th cohomology group of B vanishes, is it true that there exists a global nowhere-vanishing section of the bundle? 
Any idea where to find a proof or a counterexample?
Thanks!

REPLY [5 votes]: Let $n$ be odd. Recall that $S^n$ is parallelisable if and only if $n = 1, 3, 7$. For every other $n$, there exists $k < n$ such that $S^n$ admits $k$ linearly independent vector fields, but not $k + 1$. As $n$ is odd, $\chi(S^n) = 0$ so $S^n$ admits a nowhere-zero vector field by Poincaré-Hopf; that is, $k > 0$. Therefore $TS^n = E\oplus\varepsilon^k$ where $E$ has rank $n - k$ and does not admit a nowhere-zero section. Note however that $e(E) \in H^{n-k}(S^n; \mathbb{Z}) = \{0\}$ as $0 < n-k < n$.
It should be noted that the precise value of $k$ is known for every $n$ by work of Radon, Hurwitz, and Adams. Namely $k = \rho(n+1) - 1$ where $\rho(n+1)$ denotes the $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$.<|endoftext|>
TITLE: Functions of several complex variables: book recommendations?
QUESTION [18 upvotes]: Can anyone recommend a good comprehensive introduction to functions of several complex variables that a) is fairly up to date, b) isn't a geometry or an algebra book only, but takes multiple viewpoints? I don't mind texts that require a lot of background as long as they present it intelligibly.
Just as good would be links to lecture courses by experts on the subjects that are accessible and do the job. 
I've seen Griffith/Harris, Grauert and Gunning's 3 volume treatise. Gunning looks the best of the three, but I was hoping for something more up-to-date. This is still a very active field. I haven't seen Krantz yet.

REPLY [16 votes]: I know this is somewhat late answer for the original question, but here goes.  I've taught Several Complex Variables twice at Oklahoma State, and I wrote a hopefully easy to read textbook for the purpose.  It is freely available online under a CC license:
http://www.jirka.org/scv/
It is not a reference work, it is a hopefully gentle introduction, something that would be covered in one semester by a beginning graduate student after at least one semester of one complex variable.  It doesn't cover anything too deeply, but I did try to cover several different viewpoints rather than focus on one specific thing.<|endoftext|>
TITLE: Does listing the prime factors always stop?
QUESTION [13 upvotes]: Take a natural number's prime factors and list them increasingly and repeating them according to multiplicity. Concatenate their decimal (or in any base) representation to get a new number and repeat the process. Does this always end in a prime number for any input?

REPLY [17 votes]: It's open problem, sequence A037274 from OEIS, so-called "home primes". Hm, the value for n=77 is even unknown.
P.S. On-Line Encyclopedia of Integer Sequences definitely should be included in FAQ.<|endoftext|>
TITLE: When is  a surface in a threefold  contractible to a curve?
QUESTION [12 upvotes]: Given  a threefold $Y$ containing a surface $S$. Under which conditions can I contract $S$ so that I still end up with a smooth variety?
In other words

what are the conditions for the
  existence of a smooth variety $X$ and
  a  morphism $Y\rightarrow X$ such that
  the image of $S$ under the morphism is
  a curve and the morphism is an
  isomorphism away from $S$?

What are the conditions when $Y$ is a fourfold and $S$ is still a surface?

REPLY [11 votes]: You want a divisorial contraction $Y \to X$ on a smooth 3-fold $Y$. Extremal divisorial contractions (i.e., contractions associated to a $K_Y$-negative extremal ray in the Mori cone of $Y$) have been classified by S. Mori in his paper 
3-folds whose canonical bundles are not numerically effective, Annals of Mathematics 116, No. 1 (1982), pp. 133-176. 
Looking at Theorem 3.3, we see that here are exactly the following possibilities:

the smooth blow-up of a point; in this case $S$ is isomorphic to $\mathbb{P}^2$ with normal bundle $\mathcal{O}(-1)$;
the smooth blow-up of a curve, in this case $S$ is a ruled surface whose normal bundle restricted to the ruling has degree $(-1)$; this is the situation described by Donu Arapura in his answer;
the contraction of a plane $S$ with normal bundle $\mathcal{O}(-2)$; in this case the 
surface  $X$ has an isolated singularity isomorphic to the quotient of $\mathbb{A}^3$ by the involution $(x,y,z) \to (-x, -y, -z)$;
the contraction of a smooth quadric $S$ whose rulings are numerically equivalent; in this case the image of $S$ is a single point, which is a singular point for $X$;
the contraction of a singular quadric $S$; again, the image of $S$ is a point in $X$.

Summing up, if you want that $X$ is smooth and the image of $S$ is a curve, the only possibility is 2. 
In the case where $Y$ is a smooth 4-fold and $S$ is a smooth surface, the answer can be found in the paper of Kawamata "Small contractions of four-dimensional algebraic manifolds": in this case the only possibility is that $S$ is the disjoint union of copies of $\mathbb{P}^2$, with normal bundle $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$<|endoftext|>
TITLE: The role of the mean value theorem (MVT) in first-year calculus
QUESTION [25 upvotes]: Should the mean value theorem be taught in first-year calculus?
Most calculus textbooks present the MVT just before the section that says that if $f'>0$ on an interval then $f$ increases on that interval. Most of the related exercises that books list are either about examples related to the necessity of the hypothesis of the theorem (i.e. just checking it) or about proving theoretical facts (e.g. general properties of functions, or derivatives of functions, inequalities, Taylor, ...) I can imagine a text that doesn't require the student to understand logical rigor, but that instructs them on the role of the MVT in the theory.
It is a fact that many times we fail in making students able to use MVT in these kind problems. 
Assume that we are considering a first introduction to calculus for students that mostly will use it in application, students that will work in Biology, Engineering, Chemistry,... 
Is it possible to remove MVT from the program and get a consistent exposition of the rest of the results and techniques in Calculus?
How does eliminating MVT from the curriculum affect students from specialties exemplified above (i.e. In what ways is MVT used in further studies that they have to take or that they need)?
Can the role of MVT be replaced by a more easy to use/easy to grasp result?
Are there other uses (exercises) more suitable for the uses that MVT has for students of this specialties?  
Note added later: "Franklin" appears to have altered the meaning of the question with his later edits.  I'll say more on this soon.......

REPLY [4 votes]: First I'll apologize for necrobumping this thread after well over a year, but, it seemed to me that there is a simple answer to the OP's question which was never directly addressed.
Students are interested in existence and uniqueness of solutions to equations. They like the fact that $x^2 = -1$ has no solutions (in the real numbers) but $x^2=1$ has two solutions. They like the fact that $2x + 5y = 1$ has infinitely many solutions (in the $(x,y)$-coordinate plane).
I like to ask students: what are the solutions of the equation $y'=1$? The format of the answer must be a function $y=f(x)$. And of course they can all come up with infinitely many different solutions.
Then I ask them: are there any other solutions? If not, why not? They don't have to be "interested" in abstract mathematics and proofs and rigor in order to appreciate this question and to take interest when I show them why the functions $x + C$ are, in fact, all the solutions. And then one easily continues on to solutions of equations like $y' = $ any other of your favorite functions.
So yes, MVT can and should be taught in calculus.<|endoftext|>
TITLE: Largest number of vectors with pairwise negative dot product
QUESTION [22 upvotes]: What is the largest $m$ such that there exist  $v_1,\dots,v_m \in \mathbb{R}^n$  such that for all $i$ and $j$, $1\leq i< j\leq m$, we have  $v_i \cdot v_j < 0$.
Also, the preview screen is not displaying set braces for LaTeX.  Is that just the preview, or does it mean the site wouldn't display them after the question had been posted either?  (I formatted this question without the braces in case it's the latter.)

REPLY [4 votes]: A proof from matrix theory
Let $V=(v_1,...,v_m)\in\Bbb R^{n\times m}$ be the matrix with the $v_i$ as columns (and let us assume that $\|v_i\|=1$).
The matrix $X:=V^\top V\in\Bbb R^{m\times m}$ has the following properties:

$X$ is positive semi-definite, that is, all eigenvalues are real and $\ge 0$. Furthermore, the multiplicity of the eigenvalue zero is equal $m-d$, wehre $d$ is the rank of $X$ and equals $d:=\dim \mathrm{span}\{v_1,...,v_m\}\le n$.
$X_{ij}=v_i\cdot v_j$, in particular, the diagonal of $X$ consists of 1-s, and the off-diagonal entries of $X$ are negative.

The matrix $Y:=2I-X$ has only positive entries, and by Perron-Frobenius, its maximal eigenvalue has therefore multiplicity one.
Hence, the minimal eigenvalue of $X$ has multiplicity one.
Thus, the multiplicity of zero as eigenvalue of $X$ is at most one (it could be that zero is not an eigenvalue at all, but if it is one, it is the smallest).
Consequently, $m-d\le1\implies m\le d+1\le n+1$.<|endoftext|>
TITLE: On linear independence of exponentials
QUESTION [38 upvotes]: Problem.
Let $\{\lambda_n\}_{n\in\mathbb N}$  be a sequence of complex numbers .  Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients
$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$
converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that 
$a_n=0$ for all $n\in\mathbb N$. 

Question. Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent?


Background and motivation.
A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev  (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).
The question is related to the problem of uniqueness of solutions to the so called gravity equation
$$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$
where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).
Titchmarsh proved that an arbitrary solution to the gravity equation has the form 
$$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$
where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation
$\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial.

REPLY [14 votes]: I have some partial answers.
I. It is not hard to construct a Dirichlet series 
$$f(z)=\sum_{n=1}^\infty a_ne^{\lambda_n z}$$
which converges to $0$ absolutely and uniformly on the real line but does not converge at some points
of the complex plane.
It is constructed as a sum of 3 series $f=f_0+f_1+f_2.$ Let $f_1$ be a series with imaginary
exponents $\lambda_n$ which converges to an entire function in the closed lower half-plane,
but not in the whole plane.
Such series is not difficult to construct, see V. Bernstein, page 34, (see the full reference below) and there are simpler examples,
with ordinary Dirichlet series. Then put $f_2=\overline{f_1(\overline{z})}$,
and $f_0=-f_1-f_2$. So all three functions are entire. Now, according to Leontiev, EVERY entire function
can be represented by a Dirichlet series which converges in the whole plane.
Thus we have a Dirichlet series $f_0+f_1+f_2$ which converges on the real line to $0$ but does not
converge in the plane.
A counterexample to the original question also requires real coefficients, this I do not know
how to do (for $f_0$).
II. It is clear from the work of Leontiev, that to obtain a reasonable theory,
one has to restrict to exponents of finite
upper density, $n=O(|\lambda_n|)$, otherwise there is no uniqueness in $C$. In the result I cited
above the expansion of $f_0$ is highly non-unique.
Assuming finite upper density I proved that if a series is ABSOLUTELY and uniformly
convergent on the real line to zero, then all coefficients must be zero.
http://www.math.purdue.edu/~eremenko/dvi/exp2.pdf
I don't know how to get rid of the assumption of absolute convergence.
But there is a philosophical argument in favor of absolute convergence: the notion of "linear
dependence" should not depend on the ordering of vectors:-)
III. The most satisfactory result on my opinion, is that of Schwartz. Let us say that
the exponentials are S-linearly independent if none of them belongs to the closure of
linear span of the rest. Topology of uniform convergence on compact subsets of the real line. Schwartz gave a necessary and sufficient conditon of this:
the points $i\lambda_k$ must be contained in the zero-set of the Fourier transform of a measure
with a bounded support in R. 
(L. Schwartz, Theorie generale des fonctions moyenne-periodiques, Ann. Math. 48 (1947) 867-929.)
A complete explicit characterization of such sets is not known, but they have finite upper density,
and many of their properties are understood. These Fourier transforms are entire functions of
exponenitial type bounded on the real line. The link I gave above contains Schwartz's proof
in English. S-linear dependence is also non-sensitive to the ordering of functions, which is good.
IV. Vladimir Bernstein's book is "Lecons sur les progress recent de la theorie des series de Dirichlet", Paris 1933. This is the most comprehensive book on Dirichlet series, but unfortunately
only with real exponents.
V. The application to the functional equation mentioned by the author of the problem is not a good
justification for the study of the problem in such generality. The set of exponentials there is
very simple, and certainly we have $R$-linear independence for SUCH set of exponentials. Besides
the theorem stated as an application has been proved in an elementary way.  
VI. Finally, I recommend to change the definition of $R$-linear independence by allowing complex
coefficients (but equality to $0$ on the real line). Again in the application mentioned in the original problem, THIS notion of $R$-uniqueness is needed: the function is real, but the exponentials
are not real, thus coefficients should not be real.<|endoftext|>
TITLE: Can an infinite commutative ring have a finite (but nonzero) number of non-nilpotent zero-divisors?
QUESTION [13 upvotes]: By a theorem of Ganesan, if a commutative ring not a domain has only finitely many zero-divisors, then the ring must be finite. (There are analogous results for non-commutative rings.)
There are plenty of examples of infinite rings with a finite number of nonzero nilpotents. There are also plenty of examples of infinite rings with an infinite number of zero-divisors, all of which are nilpotent.
However, I am unaware of any ring with an infinite number of zero-divisors, of which $0 < n < \infty$ are non-nilpotent.
Can anyone give an example or explain why this can't happen. I am mostly interested in the commutative case, but non-commutative examples would be interesting too.

REPLY [12 votes]: No, there is no such example.
Recall that the nilradical $N$ of $R$ is the ideal of nilpotent elements.  It equals the intersection of all prime ideals of $R$.
On the other hand, the set $D$ of zero-divisors of $R$ can be expressed as the union of the radicals of the annihilators of individual nonzero elements of $R$ (Atiyah-MacDonald Prop. 1.15):
$$D = \bigcup_{x\neq 0} \sqrt{(0:x)}$$
Here $(0:x)$ is an ideal, and its radical is the intersection of all the primes containing it.  Thus $D$ is a union of ideals, each of which contains the nilradical $N$.  If any of these ideals $I$ properly contains $N$, then if $N$ is infinite we conclude $I\setminus N$ is also infinite (since it contains a whole coset of $N$), and hence $D\setminus N$ is infinite.
EDIT:  Here's an easier proof in a different spirit, motivated by the preceding argument.
Suppose $x,y\in R$, such that $x$ is nilpotent and $y$ is a zerodivisor.  I claim $x+y$ is a zerodivisor.  Let $z\neq 0$ be such that $yz=0$.  If $xz=0$, we are done.  Otherwise, let $n$ be the smallest number such that $x^nz=0$ (which happens for some $n$ since $x$ is nilpotent).  Then $x^{n-1}z\neq 0$ but $x(x^{n-1}z)=0$, so $(x+y)x^{n-1}z=0$.  Thus $x+y$ is a zerodivisor.
Now if $y$ is not nilpotent, $x+y$ is not nilpotent since the nilradical $N$ is an ideal.  It follows that the coset $N+y$ consists entirely of nonnilpotent zerodivisors, so if $N$ is infinite then there are infinitely many nonnilpotent zerodivisors.<|endoftext|>
TITLE: The Worst Possible Winner
QUESTION [12 upvotes]: First a little background. In racing it is possible for a player to win a tournament without winning a single race, however, how bad can a tournament winner actually be? Can a player win a tournament without even doing better than coming third? Or even fourth? Obviously this depends on the scoring method used for awarding points for each race.
More formally, suppose $p$ players, named $\alpha_1, \alpha_2, \ldots, \alpha_p$, play a game consisting of $n$ races (with no possibility of ties for a position). 
Suppse that player $\alpha_i$ finishes race $j$ in position $\beta_{i,j} \in \lbrace 1, 2, \ldots p \rbrace$ (with $\beta_{i,j} = 1$ being the best possible result for player $\alpha_i$). And that for each race the points scored by a player are given by a non-negative, strictly decreasing function called a scoring function $f : \lbrace 1,2, \ldots, p \rbrace \to \mathbb{N}$, i.e. the player coming first receives $f(1)$ points, the player coming second receives $f(2)$ points and the player coming last receives $f(p)$ points.
Let $\text{score}(\alpha_i) = \sum_{j = 1}^{n} f(\beta_{i,j})$ be the total score obtained by player $\alpha_i$.
Let $\text{best}(\alpha_i) = \min_{1 \leq j \leq n} \lbrace \beta_{i,j} \rbrace$, be the best position that player $\alpha_i$ came in.
We say that player $\alpha_i$ is a winner iff $\forall j \in \lbrace 1, 2, \ldots, p \rbrace$ $\text{score}(\alpha_i) \geq \text{score}(\alpha_j)$, note there may be more than one winner of a game. 

Given a particular choice of scoring function $f$, if $\alpha_i$ is a winner what is the maximum value $\text{best}(\alpha_i)$ can possibly be?

Or alternatively:

For what $k \in \lbrace 1, 2, \ldots, p \rbrace$, is there a choice of scoring function $f$ such that it is possible for $\alpha_i$ to be a winner and $\text{best}(\alpha_i) \geq k$?

If the general case is too hard, how about when $f(x) = p + 1 - x$?

REPLY [14 votes]: With the "right" scoring function, it is possible that $best(\alpha_{i}) = p-1$: Suppose our winner is next-to-last in every race, that each of the other racers is last in at least one race, and the scoring function awards $100^{p}-k+1$ (or some other large enough number) points for position $k=1,\ldots, p-1$ and $0$ points for position $p$.   (EDIT: Also suppose $n \geq p-1$.) The scoring function gives roughly equal awards to the racers who don't come in  last and severely penalizes a racer who comes in last.  Our consistently next-to-last racer is the overall winner since he or she is the only one who never comes in last.<|endoftext|>
TITLE: The "sensitivity" of 2-colorings of the d-dimensional integer lattice
QUESTION [53 upvotes]: Consider the $d$-dimensional integer lattice, $Z^d$.  Call two points in $Z^d$ "neighbors" if their Euclidean distance is 1 (i.e., if they differ by 1 on exactly one coordinate).
Let $C$ be a two-coloring of $Z^d$, which makes each point either red or blue.  We'll assume $C$ has the following "nontriviality" property: the origin is colored red, but on each of the $d$ axes through the origin, there's a point on that axis that's colored blue.
Let the "sensitivity" of a point $x$ with respect to $C$, or $s_x(C)$, be the number of $x$'s neighbors that are colored differently from $x$.  Then let $s(C) = \max_{x \in Z^d} s_x(C)$.
QUESTION: Can you give me any decent lower bound on $s(C)$ in terms of $d$?  For example, that $s(C) \ge k \sqrt{d}$ for some constant $k > 0$?  
REMARK 1: If you prove a lower bound of the form $k d^l$ (for constants $k,l > 0$), then you'll have solved an old open problem in the study of Boolean functions, namely the "sensitivity versus block sensitivity" problem (posed by Noam Nisan in 1991).  But please don't let that discourage you!  My variant feels more approachable, and maybe something is even already known about it.
(I'll be happy to supply full details of the reduction on request.  But here's the basic idea: let $f : \lbrace 0,1 \rbrace ^n \rightarrow \lbrace 0,1 \rbrace$ be a Boolean function such that the block sensitivity $bs(f)$ is much much larger than the sensitivity $s(f)$. Then there must be an input $x$ of $f$ that has $bs(f)$ disjoint sensitive blocks.  Let $d=bs(f)$.  Then we can construct a two-coloring of $Z^d$ with the properties listed above, and such that $s(C) \le 2 s(f)$ where $s(f)$ is the sensitivity of $f$.  The input $x$ gets mapped to the origin of $Z^d$, while each of the $d$ sensitive blocks of $x$ gets mapped to one of the axes of $Z^d$.  To map a Boolean assignment to an integer, in a way that preserves the sensitivity, we use the Gray Code.  Finally, we color each point $y \in Z^d$ either red or blue, depending on whether $f(x)$ is 0 or 1 for the corresponding Boolean point $x$.)
REMARK 2: I can give an example of a coloring with $s(C) = O(\sqrt{d})$, meaning that $s(C) \ge k \sqrt{d}$ really is the best lower bound one can hope for.  This coloring can be obtained by starting from "Rubinstein's function" -- a Boolean function $f : \lbrace 0,1 \rbrace ^n \rightarrow \lbrace 0,1 \rbrace$ with $bs(f) = n/2$ and $s(f) = 2 \sqrt{n}$ -- and then applying the reduction sketched in Remark 1.
(For those who are interested, let me now go ahead and describe a coloring with $s(C) = O( \sqrt{d} )$ explicitly.  Assume for simplicity that $d$ is a perfect square.  Partition the $d$ coordinates of $x$ into $\sqrt{d}$ "blocks" of $\sqrt{d}$ coordinates each.  Then we'll color $x$ blue, if and only if at least one of the blocks has a single coordinate equal to $2$ and all other coordinates equal to $0$.  I'll leave it as an exercise for you to verify that $s(C) = 2 \sqrt{d}$.)
Note: I edited the above paragraph a little, to simplify the construction and insert a missing factor of 2.
REMARK 3: At the moment, I don't even have a proof that $s(C)$ has to grow with $d$ (!!).  But I suspect at least $s(C) \ge k \log d$ ought to be doable.
EDIT: Sorry to switch notations in the middle of the game, but I have a better one if you want to talk about low dimensions (per domotorp's question below)!  Let's let $r_x(C)$ be the number of axes (up/down, left/right, etc.) along which $x$ has a neighbor that's colored differently than $x$ is.  Then let $r(C) = \max_x r_x(C)$.  Clearly $r(C) \le s(C) \le 2r(C)$ for all $C$.
In fact, something even stronger than that is true: given any coloring $C$, one can create a new coloring C' that satisfies $s(C')=r(C)$, by simply "blowing up" each point $x$ into a cube of $2^d$ points, which are all colored the same way $x$ was colored in $C$.  The nontriviality and sensitivity properties will clearly be preserved; all this transformation does is to eliminate the problem of a point having two differently-colored neighbors along the same axis.  So without loss of generality, we can shift attention to $r(C)$.
Now let $r_d = \min_C r(C)$ over all nontrivial colorings $C$ of $Z^d$.
Then here's what I know:
$r_1 = 1$
$r_2 = 2$
$r_3 = 2$
$r_4 = 2$
$r_5 \in \lbrace 2,3 \rbrace$
UPDATE: I created an image that shows an explicit coloring of $Z^3$ that satisfies both the nontriviality condition and $r(C)=2$.  (That is, from every point, you can change color by moving along at most 2 different axes.)  As explained above, this can easily be converted into a coloring with $s(C)=2$ as well.
        (source)
domotorp is right that proving $r_5=3$ could be a great start...

REPLY [8 votes]: Just seen on twitter: Hao Huang at Emory University has given a very short argument in this preprint proving the following

Theorem 1.1. For every integer $n\geq 1$, let $H$ be an arbitrary $(2^{n-1}+1)$-vertex induced subgraph of $Q^n$, then
$$\Delta(H)\geq\sqrt{n}$$
Moreover this inequality is tight when $n$ is a perfect square.

which implies the sensitivity conjecture! This goes via the approach of Gotsman and Linial mentioned in Noam's answer.
(Here, $Q^n$ is the $n$-dimensional hypercube graph and $\Delta(H)$ denotes the maximum degree of $H$.)
The paper is very readable, and the "meat" of the argument takes up less than 2 pages and uses not much more than basic interlacing properties of eigenvalues!<|endoftext|>
TITLE: When does a ring surjection imply a surjection of the group of units?
QUESTION [39 upvotes]: The following might be a very trivial question.  If so, I don't mind it being closed, but would appreciate a reference where I could read about it.
Let $R$ and $S$ be commutative rings and let $R^\times$ and $S^\times$ denote their respective multiplicative groups of units.  Let $f:R \to S$ be a ring homomorphism and let $f^\times : R^\times \to S^\times$ denote the induced group homomorphism.  Finally, suppose that $f$ is surjective.

Under what conditions (if any) will $f^\times$ be surjective?

Thanks in advance!

REPLY [7 votes]: Addendum I. I am raising this addendum at the top because it is more relevant to the question than what I have written before.
We denote by $R^{\times}$ the unit group of a unital ring $R$.

Claim 1. Let $R$ be a commutative and unital ring. Then the following are equivalent:

The natural map $R^{\times} \rightarrow (R/I)^{\times}$ is surjective for every ideal $I$ of $R$.

The Bass stable rank of $R$ is $1$.



This claim was inspired by Theorem 5.1 of [5], a preprint which cannot be more on topic!
Let us define the Bass stable rank of an associative and unital ring $R$.
We call $\mathbf{r} \in R^n$ a unimodular row of $R$ if the components of $\mathbf{r}$ generate $R$. We denote by $\operatorname{Um}_n(R)$ the set of unimodular rows of size $n$.
We say that $n > 0$ lies in the stable range of $R$ if for every $(r_1, \dots, r_{n + 1}) \in \operatorname{Um}_{n + 1}(R)$, there is $r \in R$ such that $(a_1 + r_1a_{n + 1}, \dots, a_n + r_n a_{n + 1}) \in \operatorname{Um}_n(R)$. It is well-known and easily checked that the stable range has no gap, i.e., if $n$ lies in the stable range of $R$, then so does $m$ for every $m > n$. It is also well-known that $\operatorname{sr}(R/\mathcal{J}(R)) = \operatorname{sr}(R)$ where $\mathcal{J}(R)$ denotes the Jacobson radical of $R$, that is, the intersection of the maximal ideals of $R$.
The Bass stable rank $\operatorname{sr}(R)$ of $R$, or simply stable rank of $R$, is the smallest element in the stable range of $R$.
The proof of Claim $1$ is actually straighforward.

Proof of Claim $1$. Assume that $R^{\times} \rightarrow (R/I)^{\times}$ is surjective for every ideal $I$ of $R$. Let $(a_1, a_2) \in \operatorname{Um}_2(R)$. Then $a_1 + I$ is unit of $R/I$ with $I = Ra_2$. Therefore we can find $u \in R^{\times}$ such that $a_1 + I = u + I$, i.e., there is $r \in R$ verifying $a_1 + r a_2 = u$. As a result $\operatorname{sr}(R) = 1$.
Let us assume now that $\operatorname{sr}(R) = 1$. Let $I$ be an ideal of $R$ and let $a + I \in (R/I)^{\times}$. Then we can find $b \in I$ such that $(a, b) \in \operatorname{Um}_2(R)$. Because $R$ is of stable rank $1$, there is $r \in R$ such that $u \Doteq a + rb$ is a unit of $R$.

Semilocal rings, and Artinian rings in particular, are known to have stable rank $1$. More generally, so are the rings $R$ such that $R/\mathcal{J}(R)$ has Krull dimension $0$ (Justin Chen's semifields), e.g., von Neumann regular rings.
The reader should be bear in mind that $$\operatorname{sr}(R) \le \dim_{Krull}(R) + 1$$ by the Bass Stable Range Theorem and its generalisation by R. Heitmann [2].
The class of local-global rings encompass all the previous classes.

Definition. A commutative and unital ring $R$ is local-global if every (possibly multi-variate) polynomial $f$ with coefficients in $R$ whose values generate $R$ represents a unit of $R$, i.e., some value  of $f$ is a unit.

The following is well-known and immediate.

Claim 2. If $R$ is local-global then $\operatorname{sr}(R) = 1$.
Proof. Given $(a, b) \in \operatorname{Um}_2(R)$, consider the polynomial $f(X) = a + bX$.

Addendum II: I just discovered that the connection between surjections on unit groups and the stable rank 1 condition has been known for a long time. This is Lemma 6.2 of "Stable range in commutative rings" (1967) by D. Estes and J. Ohm.
See also Corollaries 6.3 and 6.4 for results on the kernel of the induced group homomorphism.
(The paper is wonderfully written.)

Former answer.
I would like to contribute in outlining two natural generalisations of the original question for which the Jacobson radical also comes into play. Eventually, I would like to quote two research results for which OP’s question is essential.
Considering that $R^{\times} = GL_1(R)$, it is natural to ask

When does the surjection $R \twoheadrightarrow R/I$ induce a surjective map $GL_n(R) \rightarrow GL_n(R/I)$?.

Here $I$ is an ideal of $R$ and the second map is the reduction of matrix coefficients modulo $I$. If $I$ is an ideal contained in the Jacobson radical $\text{rad}(R)$ of $R$, then $GL_n(R) \rightarrow GL_n(R/I)$ is surjective [5, Exercise I.1.12.iv].
Considering that $R$ is an $R$-module and that every element in $R^{\times}$ generates $R$, it is natural to ask

For $M$ an $R$-module, when do $R$-generating sets of $M/IM$ lift to $R$-generating sets of $M$?.

This holds if $I \subseteq \text{rad}(R)$  and $M/IM$ is finitely generated over $R$ by Nakayama’s lemma, see also the related concept of superfluous submodule.
I introduce now a result published in 2003 which directly relates to OP's question. Let $R$ be a Dedekind domain which is universal for $GE_2$. The latter property means that in the subgroup $GE_2(R) \subseteq GL_2(R)$ generated by the transvections and the diagonal matrices, the latter generators are only subject to the « obvious » or universal relations (equivalently $K_2(2, R)$ is generated by symbols as a normal subgroup of $St_2(2, R)$ [1, 4]).
Then we have:

If $p$ is a prime element of $R$ such that the natural $R 
\twoheadrightarrow R/(p)$ induces a surjective homomorphism between unit groups, then the localization $RS^{-1}$ of $R$ with $S = \{1, p, p^2, \dots\}$ is also universal for $GE_2$.

This was proved in [3] and the result was shown to be sharp in [4].
I can’t resist mentioning a humble result of mine, which I would like to believe is entertaining.
A finitely generated group $G$ of rank $n$ is said to satisfy the generalised Andrews-Curtis conjecture if every $n$-tuple of elements which normally generate $G$ can be transitioned to an $n$-tuple of generators of $G$ by means of finitely many $AC$-moves, i.e., applying transvections or replacing one component by a conjugate a finite number of times. (If $G$ is the free group on $n$ generators, this is just the Andrews-Curtis conjecture, still unsettled.) Then the result reads as:

The solvable Baumslag-Solitar group $\langle a, b \,\vert\, aba^{-1} = b^n \rangle$ with $n \ge 1$ satisfies the generalised Andrews-Curtis conjecture if and only if the natural map $R^{\times} \rightarrow (R/(n - 1)R)^{\times}$ is surjective  where $R = \mathbf{Z}[1/n]$.



[1] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.
[2] R. Heitmann, "Generating non-Noetherian modules efficiently", 1984.
[3] E. Abe and J. Morita, "Tits systems with affine Weyl groups in Chevalley groups over Dedekind domains", 1988.
[4] H. Yu and S. Chen, "Subrings in quadratic fields which are not $GE_2$", 2003.
[5] C. Weibel, "The K-book", 2013.
[6] J. Chen, "Surjections of unit groups and semi-inverses", 2017.<|endoftext|>
TITLE: 100 Prisoners, 100 Boxes: Proof of Optimality
QUESTION [10 upvotes]: There's a chestnut about 100 prisoners, labeled 1 through 100, and 100 boxes, each with a number 1 through 100 in it. Each prisoner, completely independently of the others, tries to find the box which has their label in it. If they all find their label, they win.
They each get to sequentially look inside 50 boxes, meaning they look in the first box, and based on the response, decide which box to look into next, etc. They can't alter the boxes or communicate with other prisoners in any way. Of course, they agree on a strategy beforehand.
There is a really lovely strategy based on cycle representations of permutations (I have a soft spot for puzzles which hinge on cycle representations of permutations), but is there a lovely proof that its optimal? Or any proof?

REPLY [8 votes]: To prove that the strategy described above by Qiaochu Yuan gives the best chances of winning, the idea is as follows: consider a more fair, and more trivial variant of the game, where the rules are as before, but all open boxes remain open, so that each prisoner takes advantage of the information gathered; one may think that the boxes are simply being opened in some order by the same person. This sort of relaxation of the game is quite trivial to analyse: it consists essentially in choosing in which order to open the boxes, and of course whether or not the chosen permutation is winning, it's just random (and a priori no choice of a permutation is better than another). It turns out that the winning permutations for this game are as many as the permutations with no cycle longer than 50. Indeed, if $\lambda(1),\dots,\lambda(100)$ is the corresponding sequence of the labels found, let's divide it in consecutive substrings so that each string ends with the lower not yet found label. This subdivision may be seen as the cycle decomposition of a permutation $\sigma$, and of course it is a lucky choice if and only if $\sigma$ has no cycle of length greater that 50. So one can't do better than that even in the original game.<|endoftext|>
TITLE: Is the category of affine schemes (over a fixed field) Cartesian closed?
QUESTION [6 upvotes]: This is probably a trivial question, but I don't see the answer, and I haven't found it on Wikipedia, nLab, nor MathOverflow.
Let $\text{ComAlg}$ denote the category whose objects are commutative algebras over a fixed field $\mathbb K$ and whose morphisms are homomorphisms of algebras, and let $\text{ComAlg}^{\rm op}$ denote its opposite category.  Given commutative algebras $A,B$, let $\operatorname{hom}(A,B)$ denote the set of algebra homomorphisms $A\to B$, so that $\operatorname{hom}$ is the usual functor $\text{ComAlg}^{\rm op} \times \text{ComAlg} \to \text{Set}$.  The short version of my question:

Is $\text{ComAlg}^{\rm op}$ Cartesian closed?

The long version of my question (if I've gotten all the signs right):

Is there a functor $[,] : \text{ComAlg} \times \text{ComAlg}^{\rm op} \to \text{ComAlg}$ such that there is an adjunction (natural in $A,B,C$, i.e. an isomorphism of functors $\text{ComAlg}^{\rm op} \times \text{ComAlg} \times \text{ComAlg} \to \text{Set}$) of the form:
  $$ \operatorname{hom}([A,B],C) \cong \operatorname{hom}(A,B\otimes C) ?$$

Recall: $\otimes$ is the coproduct in $\text{ComAlg}$, hence the product in $\text{ComAlg}^{\rm op}$.
Motivation: $\text{ComAlg}^{\rm op}$ is complete and cocomplete, and so many constructions that make sense in $\text{Set}$ and $\text{Top}$ transfer verbatim to the algebraic setting.  I would like to know how many.

REPLY [9 votes]: The existence of such an adjunction implies that $B \otimes -$ preserves limits, which doesn't seem very likely.
Here is a counterexample, though probably not the simplest one. Set $B = k[y]$ and consider the inverse limit of $k[x]/(x^{n+1})$. If we take the tensor products first, then we get $k[y][[x]]$ while if we take the limit first we obtain $k[[x]][y]$. These are distinct, since the first contains for example $(1-yx)^{-1} = \sum_{k \geq 0} y^k x^k$ and the second does not.<|endoftext|>
TITLE: Skeleton of a braided monoidal category
QUESTION [9 upvotes]: Does every (lax) braided monoidal category  have a braided monoidal skeleton? That is, I want to define  a (lax) braided monoidal structure on a skeleton so that it is braided monoidal equivalent to the original category. If this is always possible, how does one accomplish it?
An extreme case of this question (which I most interested in) is this: suppose the original braided monoidal category was strict, in the sense that the associators were trivial. I would like to replace this category with its skeleton, perhaps at the expense of making it nonstrict (generating nontrivial associators).

REPLY [12 votes]: Yes. This follows from two simple facts:

If $F:C \simeq D$ is an equivalence of categories, and $D$ has a braided monoidal structure, then there exists a braided monoidal structure on $C$ and an enhancement of $F$ to a braided monoidal functor such that $F$ is an equivalence of braided monoidal categories. 
Every category is equivalent to a skeletal category. 

Number 2 is standard, so I won't dwell on it. If you don't believe number 1, try proving it. It is a great exercise. Here is a sketch of how to go about proving it. 

Replace $F$ by an adjoint equivalence, i.e. pick an adjoint inverse equivalence to $F$. You'll need this to transfer the structure from D to C. 
Now transfer the structure from D to C. There is not much to say here. Just follow your nose. For example the tensor product in C is defined by
$$ x \otimes_C y := G(F(x) \otimes_D F(y))$$
There is an associator because we have chosen F and G to be adjoint equivalences to each other. The braiding is similar. It is given by,
$$G(\beta_{F(x), F(Y)}): G(F(x) \otimes_DF(y)) \to G(F(y) \otimes_DF(x)). $$
It is tedious to verify, but this does actually satisfy the hexagon identities. 
Thus we've constructed a braided monoidal structure on C. To show that this new braided monoidal category C is equivalent as a braided monoidal category to D, we need to augment F and G to braided monoidal functors. To do this we just keep playing the same game. For example we need an equivalence $$F(x) \otimes_D F(y) \to F(x \otimes_C y) = FG(F(x) \otimes_D F(y)).$$
This is given by the inverse of the unit/counit of the adjoint equivalence between F and G. Constructing the other structure is no different. 

This is morally the strategy advocated by Benjamin Enriquez. He tried to do this with less then an equivalence and ran into trouble. For the question as stated, you really only need the case that C and D are equivalent. Notice also that this still works when the monoidal structures are just lax. It still produces a strong monoidal equivalence between C and D.<|endoftext|>
TITLE: Does ZF prove that all PIDs are UFDs?
QUESTION [16 upvotes]: Main Question:
Does ZF (no axiom of choice) prove that every Principal Ideal Domain is a Unique Factorization Domain?
The proofs I've seen all use dependent choice.
Minor Questions:
Does ZF + Countable Choice prove all PIDs are UFDs?
Does ZF prove "If all PIDs are UFDs, then [some choice principle]"?
(If anyone knows how I could force line breaks to put the questions on their own lines, please tell me.)

REPLY [23 votes]: ZF alone does not prove that every PID is a UFD, according to this paper: Hodges, Wilfrid. Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289--297. MR 422022.
One result in this paper is the following:COROLLARY 10. Neither (a) nor (b) is provable from ZF alone:
(a) Every principal ideal domain is a unique factorization domain.
(b) Every principal ideal domain has a maximal ideal. 
By the way, I didn't know the answer to this question until today. To find the answer, I consulted Howard and Rubin's book Consequences of the Axiom of Choice. (Actually, I did a search for "principal ideal domain" of their book using Google Books.)<|endoftext|>
TITLE: When does a matrix define a convolution operator on a hypergroup?
QUESTION [6 upvotes]: Let $H$ be a discrete hypergroup. Suppose I have a matrix $A=(A_{x,y})$ indexed over $H$ with nonnegative entries which defines a bounded operator on $\ell^2(H)$. When does there exist $f\in\ell^1(H)$ such that $A_{x,y}=\langle f*\delta_x,\delta_y\rangle$, i.e., $A$ is the matrix of transition probabilities for a random walk given by convolution with $f$?
A necessary condition is that $A$ commutes with $\ell^1(H)$ convolution on the right. Is this sufficient?

REPLY [10 votes]: In case of discrete groups, it requires amenability of $H$. Indeed, $H$ is amenable if and only if $f\in\ell^1(H)$ for all $f\geq0$ such that $[f(xy^{-1})]_{x,y} \in B(\ell^2H)$. I just don't know what are hypergroups.<|endoftext|>
TITLE: Real number happens to be an integer
QUESTION [10 upvotes]: Sometimes you have a real number (with a rather complicated definition), and with some effort you can show that
1) this real number is, actually, an integer;
2) the distance of this real number to an integer, say $0$, is less than $1/2$.
Thus you can conclude that this real number is $0$! I think this is a very nice trick. Especially when the argument for 1) is so involved that you don't really see this a priori. However, I don't remember in which context I have seen this. But I guess that this trick works in various situations.
So my question is: Can you give nice, explicit instances of this trick?

REPLY [3 votes]: See Making Transcendence Transparent: An Intuitive Approach to Classical Transcendental Number Theory by Burger and Tubbs.  It is only a slight exaggeration to say that the entire book is devoted to showing that the whole subject of diophantine approximation is based on the paradigm you describe.<|endoftext|>
TITLE: Non-abelian class field theory and  fundamental groups
QUESTION [62 upvotes]: Over the years, I've been somewhat in the habit of asking questions in this vein to experts in the Langlands programme. 
As is well known, given an algebraic number field $K$, they propose to replace the reciprocity map 
$$A_K^\*/K^*\rightarrow Gal(K^{ab}/K)$$ of abelian class field theory by a correspondence between the $n$-dimensional representations $\rho$ of $Gal(\bar{K}/K)$ and certain automorphic representations $\pi_{\rho}$ of $GL_n(A_K)$. (We'll skip the Weil group business for this discussion.) Substantial arithmetic information is carried on either side by the $L$-functions, which are supposed to be equal.
This involves  deep and beautiful mathematics whenever something can be proved, and there are many applications, such as the Sato-Tate conjecture or this recent paper of Chenevier and Clozel:
http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf
(I mention this one because it is in some ways very close to the point of this question.)
However, there are elementary consequences of abelian class field theory that seem not to have obvious non-abelian analogues. The one I wish to mention today has to do with the fundamental group. Given a number field $K$ (assume it's totally imaginary to avoid some silly issues), how can we tell if it has non-trivial abelian unramified extensions? Class field theory says we can look at the class group, which is quite computable in principle, and even in practice for small discriminants. But now, suppose we go on to ask the non-abelian question:  which number fields have $$\pi_1(Spec(O_K))=0?$$ That is to say, when does $K$ have no unramified extension at all, abelian or not? As far as I know, there is no easy answer to this question. Niranjan Ramachandran has pointed out that there are at least ten examples, $K=\mathbb{Q}$ (oops, that's real) and $K$ an imaginary quadratic field of class number one. I know of no others. Of course I would be happy to collect some more, if someone else has them lying around. 
But the question I really wanted to ask is: Suppose  we are in a Langlands paradise where everything reasonably conjectured by the programme is a theorem. Does this give a way to algorithmically (as we run over fields $K$) resolve this question  as in the abelian case? Otherwise, is there a sensible refinement of the usual formulation that would subsume such applications?
Added: 
I'm embarrassed to admit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I've changed the title following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme good for?'
Regarding jnewton's very natural thought: in addition to other difficulties, one would also need to bound $n$.

Added, 13 July:
Here is one more remark concerning jnewton's suggestion.  Of course in the realm of classical holomorphic
cusp forms, there are infinitely many of level one. More generally, it is shown in the paper 
http://www.math.uchicago.edu/~swshin/Plancherel.pdf
that whenever $G$ is a split reductive group over $\mathbb{Q}$, there are infinitely many cuspidal automorphic
representations  that are unramified everywhere and belong to the discrete series at $\infty$. (I presume there are other results of this sort. This one I just happen to know from a talk
last Fall.) According to Clozel's conjecture as you might find in
http://seven.ihes.fr/IHES/Scientifique/asie/textes/Clozel-juil06.pdf
(conjecture (2.1)), algebraic ones among them should correspond to motivic Galois representations (after we choose a representation
of the dual group)*. 
I don't have the expertise to recognize algebraicity in such constructions, in addition to the danger that I'm misunderstanding
 something  more elementary. But it seems to me quite a task to show directly that there are none corresponding to Artin representations. (The only case I could do myself is the classical one.)
Now, I would like very much to be corrected on all this.
But such families do seem to indicate that  a 'purely automorphic' approach to
the the $\pi_1$ question is somewhat unlikely, at least within the current framework of the Langlands correspondence. 
I suppose I'm sabotaging my own question.

*Note that in  these situations, the
 Galois representations don't have to be unramified, since there is the choice of a coefficient field $\mathbb{Q}_p$. In general, they should only be crystalline at $p$.

Added 14 July:
Matthew: Since I didn't really expect a complete answer to my question, if you could write your extremely informative series of comments as an answer,  I will accept it. (Barring the highly unlikely possibility that someone will write something better between the time you submit your answer and the time I look at it.)

REPLY [31 votes]: I'm going to take the step of disagreeing slightly 
with the perspective offered in Professor Emerton's answer,
in a bold attempt to prise away the green for myself.
The question is, what does the Langlands program have to say about
the finiteness (or otherwise) of $\pi_1(\mathcal{O}_K)$?
Let's first revisit what one knows by ``classical'' methods.
Minkowski proved that if $K$ is an extension of degree $n$, then
$$\sqrt{|\Delta_K|} \ge \left(\frac{\pi}{4}\right)^{r_2} \frac{n^n}{n!},$$
which, asymptotically, implies that the root discriminant
$\delta_K = |\Delta_K|^{1/n}$ is at least $e^2 \pi/4 - \epsilon$ 
for large $n$, and in particular, that $|\Delta_K| > 1$
if $n > 1$. This is purely a geometry of numbers argument.
What does the Langlands program have to say about this question?
First, let me ask a related question: what does the Langlands program say about
elliptic curves $E/\mathbb{Q}$ with good reduction everywhere? 
In this case, it implies (and is known, by Wiles!)
that there exists a classical cuspidal modular
form 
$$f \in S_2(\Gamma_0(1)).$$
 The latter group vanishes, and so $E$
does not exist. Yet this latter fact seemingly requires an actual computation
- namely, that $X_0(1)$ has genus zero. This could be tricky if one wants
to replace $E$ by an abelian variety $A$ of (varying)
dimension $g$, or replace $\mathbb{Q}$ by another
field $F$. It turns out, however, that one can show that $f$ does not exist
purely from the existence of the relevant functional equation - more
on this later.
Let us return to our original question. Suppose we have a field
$K$ unramified everywhere over $\mathbb{Q}$. It has been suggested
that one should ponder the existence of algebraic automorphic forms $\pi$
for $\mathrm{GL}_n/\mathbb{Q}$ associated to
irreducible Artin representations $\rho$ of 
$\mathrm{Gal}(K/\mathbb{Q})$. However, it is more natural
to consider the regular representation. 
In this case, it is (of course) known by Hecke that 
$\zeta_K(s)$ has a meromorphic continuation that is entire away from
$s = 1$, and, moreover, that $\zeta_K(s)$ satisfies a functional
equation of a precise type. 
Now, purely given the existence of the functional equation for
$\zeta_K(s)$, Odlyzko gives a formula(!) for $\log |\Delta_K|$
as some function of $s$ involving the roots of $\zeta_K$ 
(the function is (obviously) constant, but is not obviously constant).
One then deduces a lower bound for
$\delta_K$ of the form $2 \pi e^{\gamma} - \epsilon$, which is
better than Minkowski's estimate. (Odlyzko's method can be refined
to give better bounds.)
What is easy to miss in this argument is
that role that the Langlands conjecture plays in this argument -
 in this case the theorem of Hecke - is already
known! One might claim that this argument uses "more" than
Langlands and ask for an argument that is purely algebraic and
geometric (and here by "one", I mean Brian Conrad or
Chevalley), but I think this is a little misguided. After all, I
think it would be hard to prove that there does not exist any
Maass form for $\mathrm{SL}_2(\mathbb{Z})$ of eigenvalue
$1/4$ without using some analysis.
Can one argue similarly to see that there are no abelian varieties
$A$ with good reduction everywhere? Indeed, Mestre gave such
an argument (before Fontaine!). Namely, if $A$ is an abelian variety,
and $L(A,s)$ is automorphic in the expected sense, then $A$ has
conductor at least $10^{g}$. Moreover, this is close to optimal
($X_0(11)^g$ has conductor $11^g$).
There are other arguments along these lines.
Stephen Miller (and Fermigier independently)
 proves that there
are no cusp forms
for $\mathrm{SL}_n(\mathbb{Z})$
of "weight zero" (cohomological with the same infinitesimal character as the trivial representation) and "level one" for all $n$ in the range $2 \ldots 23$ - another generalization
of the fact that $X_0(1)$ has genus zero - using Rankin--Selberg
L-functions. 
Here is a final reason why one should expect some analysis.
All of these arguments fundamentally require the discriminant
of $K$ to be small. Any ``algebraic''  method should be
expected to work more generally. Yet consider the question of whether
 $\pi_1(\mathcal{O}_K)$ is finite when $\delta_K$
is large.  The only known method for answering this question
is producing an unramified extension $L/K$ such that the GS-inequality
applies to
$\pi_1(\mathcal{O}_L)$. The question on whether these groups
are (always?) infinite when $\delta_K$ is sufficiently large is
wildly open, and I know no good heuristics on this question.<|endoftext|>
TITLE: Can a scheme be defined by gluing open affines such that the intersections are affine?
QUESTION [10 upvotes]: One way to think of a manifold is as a family of of open subsets $U_i \subset \mathbb{R}^n$, together with distinguished subsets $V_{ij} \subset U_i$ and isomorphisms $\psi_{ij}: V_{ij} \to V_{ji}$ that satisfy the cocycle condition.  This may not be useful practically, but occasionally it might be an intuitive crutch. Now, just as manifolds are obtained by gluing together subsets of $\mathbb{R}^n$, a scheme is obtained by gluing together affines.  In other words, we have open affines $U_i = \mathrm{Spec} A_i$ for suitable rings $A_i$, open subsets $V_{ij} \subset U_i$, and isomorphisms $\psi_{ij}$ (of locally ringed spaces) as before.
However, the open subsets $V_{ij}$ need not be themselves affine.
Question: Is it possible to formulate this definition such that the sets $V_{ij}$ are affine? I know this can be done if the scheme is separated (because the intersection of open affines is affine).
One of the nice things about this is that wouldn't have to worry about the isomorphisms $\psi_{ij}$ being isomorphisms of locally ringed spaces, just isomorphisms of the corresponding rings.
This is probably a bad way of thinking of schemes in general; the only reason I was interested in it was because then the fibered product could perhaps be thought of more "explicitly."
The following (related) question also occurred to me when I was thinking about this.
Question': Is there an easy way to tell when the complement of $V(\mathfrak{a}) \subset \mathrm{Spec} A$ is affine? Of course, this is true when $\mathfrak{a}$  is principal.  (Answered: see the comments of Matthew Emerton and David Speyer.)

REPLY [6 votes]: You can, if you use a slightly more general notion of gluing.  (The notion of gluing you present is "wrong", or at least simplistic, in roughly the same way that it is "wrong" to require that a basis for a topology be closed under intersections.  E.g., if you do this, then the set of open balls in $\mathbb{R}^n$ for $n > 1$ does not form a "basis.")
Let $X$ be a scheme.  Consider the diagram whose objects are open affine subschemes of $X$, and whose morphisms are inclusions $U \hookrightarrow V$ such that $U$ is a distinguished open subset of $V$.  Whenever $U$ and $V$ are two objects and $x \in U \cap V$, there exists an object $W \subset U \cap V$ such that $x \in W$ and $W \hookrightarrow U$, $W \hookrightarrow V$ are both morphisms: Since the distinguished open subsets of $U$ form a basis for the topology, there is a distinguished open $W'$ in $U$ such that $x \in W' \subset U \cap V$. Similarly, there is a section $f$ over $V$ such that $x \in V_f \subset W'$.  But then $V_f = W'_f$ is a distinguished open subset of both $U$ and $V$, so we let $W = W'_f$.
It is also not too hard to show that whenever we have a category as above, we can glue things together to form a scheme (i.e., the diagram has a unique colimit in the category of schemes).  If someone asks for a precise statement of this, I'll try to cook one up, but it's not particularly nice.  (Not quite horrendous, but not very nice either.)
In particular, the fiber product is obtained by gluing together schemes of the form $\mathrm{Spec} A \otimes_C B$, where $\mathrm{Spec} C$ contains the images of both $\mathrm{Spec} A$ and $\mathrm{Spec} B$, with "overlap inclusions" specified by morphisms
$A \otimes_C B \to A_f \otimes_C B_g$.  An important note here: if $C \to D$ is a ring epimorphism (e.g., corresponds to an open immersion), and $A, B$ are $D$-algebras, then $A \otimes_C B$ is naturally isomorphic to $A \otimes_D B$.<|endoftext|>
TITLE: Finite nonabelian groups of odd order
QUESTION [14 upvotes]: For every even $n$ there exists nonabelian group. As example of such group we can take dihedral group.
The question is about odd $n$. For some of them there are no nonabelian groups of order $n$ (for example, if $n$ is prime then the group of order $n$ is cyclic and hence abelian).
For what odd $n$ are there known examples of nonabelian finite groups of order $n$?

REPLY [7 votes]: You might be interested in the result that if n is odd, |G| = n for a finite group G, and if every subgroup of G is normal, then G is abelian. (This does not hold if the hypothesis that n is odd is ommitted as the quaternion group of order 8 demonstrates.) 
A group whose every subgroup is normal is called a Dedekind group. A non-abelian Dedekind group is called a Hamiltonian group. With this terminology the result simply states that a Dedekind group of odd order is abelian. 
The proof is not immediately obvious. It relies on a classification result that states that every Hamiltonian group is a direct product of the quaternion group of order 8, an elemetary abelian 2-group, and a periodic abelian group of odd order. Once this classification result is established, however, the result can be seen easily.<|endoftext|>
TITLE: Is the feedback vertex number bounded by the maximum number of leaves in a spanning tree?
QUESTION [7 upvotes]: I have a graph-theoretical conjecture which I think would have been studied before, but for which I cannot find anything in the literature.
Let G be a finite, simple, connected graph. Let the feedback vertex number $FVS(G)$ be the minimum number of vertices that have to be deleted from $G$ to break all cycles, so the minimum number of deletions needed to turn $G$ into a forest. Let the max leaf number $MaxLeaf(G)$ be the maximum number of leaves in any spanning tree for $G$. 
My conjecture is that $FVS(G) \leq MaxLeaf(G)$. 
The two numbers come close for complete graphs: a $K_t$ has a spanning tree with $(t-1)$ leaves, and $(t-2)$ deletions are needed to turn $K_t$ into a forest. Since a forest can have an arbitrary number of leaves and has FVS number 0, the MaxLeaf number cannot be bounded by a function of the FVS number.
I can prove that $FVS(G) \leq 6 \cdot MaxLeaf(G)$ through a lemma on spanning trees which says that for every connected graph G containing m vertices of degree $\neq 2$, there is a spanning tree for G with at least $m/6$ leaves. Since the deletion of the set of vertices of degree $\neq 2$ turns a graph into a forest if the graph is not a simple cycle, this shows that $FVS(G) \leq 6 \cdot MaxLeaf(G)$ when $G$ is not a simple cycle; and it is easy to see that the claim also holds when $G$ is a simple cycle since $MaxLeaf(C_n) = 2$ and $FVS(C_n) = 1$ for $n \geq 3$.
Since the complement of the leaves in a spanning tree form a connected dominating set, and since the complement of a feedback vertex set is a maximum induced forest, an alternative way to state the conjecture is:
For any connected graph $G$ the number of vertices in the largest induced subforest of $G$ is at least as large as the minimum size of a connected dominating set in $G$.
So my question is: is this conjecture true, and does anyone know of any research related to it?

REPLY [6 votes]: Bill Waller and I proved the stronger statement that for G a graph on n(G) > 1 vertices, the order of a largest induced linear forest is at least one plus the connected domination number. See our preprint. A linear forest is a forest in which each connected component is a path, and clearly a lower bound for the order of a largest forest.<|endoftext|>
TITLE: "Affine communication" for topological manifolds
QUESTION [32 upvotes]: There is a situation that comes up regularly in algebraic topology when giving proofs of facts about manifolds, like Poincare duality and the like.  The typical sequence goes like this:

Prove something for $\mathbb{R}^n$.
Then it follows for open disks.
Use a Mayer-Vietoris argument to prove it for finite unions of convex sets in $\mathbb{R}^n$.
Use a colimit argument to prove it for arbitrary open subsets of $\mathbb{R}^n$.
It then follows for open subsets of a manifold admitting a homeomorphism to an open subset of $\mathbb{R}^n$.
Use a Mayer-Vietoris argument to prove it for finite unions of such subsets of a manifold.
Use a colimit argument to prove it for arbitrary open subsets of a manifold.

Obviously there is some redundancy here, and it makes the technical details in these proofs overwhelm the underlying ideas.  In the smooth category one can do better, but usually only by appealing to machinery which is useful but takes extra time to prove.
The answers/comments in the following question point towards a very useful way to compare coordinate charts in different affine covers of a given scheme:
What should be learned in a first serious schemes course?
Namely, the intersection of any two affine opens has a cover by open sets which are simultaneously distinguished in both.
I feel like I should know a reference to whether this is true in the topological category - and I suspect that it's not - but I shamefully don't know.  One can phrase this in terms of continuous local homeomorphisms from $\mathbb{R}^n$ to itself, but I'll instead just ask:
Given two coordinate charts on a topological manifold M and a point in their intersection, is there a neighborhood of this point which is simultaneously a convex open set in both charts?  Is there a simple counterexample?

REPLY [26 votes]: There are piecewise linear counterexamples in dimension $2$. 
Arrange $2m$ evenly spaced rays $R_i$ around the origin, $m\ge 3$. If $C$ is a convex neighborhood of the origin, let $r_i$ be the reciprocal of the length of the portion of $R_i$ in $C$. For some number $K>0$ (depending on $m$), convexity implies $r_{i+1}+r_{i-1}\ge 2Kr_i$. Thus if $e(C)$ and $o(C)$ are the sums of $r_i$ over even and odd $i$ we have $e(C)\ge Ko(C)$ and $o(C)\ge Ke(C)$. Now apply a homeomorphism $h$ that linearly stretches even-numbered rays by $A>0$ and odd-numbered rays by $B>0$. If $h(C)$ is a convex set then you would have numbers $e(h(C))=e(C)/A$ and $o(h(C))=o(C)/B$, whence $e(C)/A\ge Ko(C)/B$, contradicting $o(C)\ge Ke(C)$ if $A/B$ is chosen big enough.<|endoftext|>
TITLE: Non-existence of small resolutions for the singularity $y^2=u^2+v^2+w^3$
QUESTION [14 upvotes]: In his paper "Smooth models for elliptic threefolds" (In: The Birational Geometry of Degenerations, Progress in Mathematics, v. 29, Birkhauser, (1983), 85-133), Rick Miranda mentions in the example of section 8 (page 101-102) that it is an unfortunate fact of life that there are no small resolutions for the singularity of the threefold defined by the equation 
$$y^2=u^2+v^2+w^3\quad \text{in}\quad \mathbb{C}^4.$$
He said that it is a theorem of  Brieskorn but he does not give a reference. Anyone has a reference to this theorem and its proof?
Updated (following some of the answers, the question can be generalized right away):
More generally the same question can be asked for 
$$y^2=u^2+v^2+w^{2k+1}\quad \text{in}\quad \mathbb{C}^4, \quad\text{where} \quad  k \in \mathbb{N}_0.$$

REPLY [5 votes]: This answer to a more general question might also be relevant for similar questions.<|endoftext|>
TITLE: Heuristic justification for Goldbach's conjecture
QUESTION [9 upvotes]: On the Wikipedia page of Goldbach's conjecture, a heuristic justification is given, which did not completely satisfy me.  It roughly goes as follows:


randomly define a subset integers in accordance with the prime number
  theorem
Let $K_n$ be the random variable, counting the number of ways the
  natural number $2n$, can be written as
  a sum of two members of this set.

Then $E[K_n]\rightarrow \infty$ .

The problem is that, although the mean goes to infinity, it still might be true that the probability that $K_n>0$ for all $n$ is zero. 
So I thought of a different heuristic, and I am curious about whether anything is known about it:

Let $\mathcal P$ be the collection of
  all subsets of odd numbers whose
  density agrees with the prime number
  theorem, and let $\mathcal G$ be the
  collection of subsets for which
  Goldbach's property holds (i.e. every
  even number can be written in at least
  one way with two members of the set). 
  Let $\mu$ be the uniform product
  measure of the space $\{0,1\}^{\mathbb
> N}$.  Then the quantity  $$
> \frac{\mu(\mathcal P \cap \mathcal
> G)}{\mu(\mathcal P)} $$ 
   is (significantly) greater than zero. 
Edit: As pointed out in the comments,
  $\mu(\mathcal P) = 0$, so this
  quantity is  meaningless as it is, but
  I think it can be formalized in some
  way.

I do not know if this is easy or almost as difficult as the original problem.  But it would be a very convincing heuristic for me in that, it would tell me how much of Goldbach's conjecture is already explained by the prime number theorem.
I would appreciate answers, or references to any known results, or reasons if this kind of heuristic is not relevant, if that is the case.

REPLY [5 votes]: Let me add a historical remark. J.J. Sylvester was probably the first to come up with a heuristic argument.  See his (somewhat naive) approach in this short paper and this Nature note. 
P.S.  I found these two while working on this mildly related paper.<|endoftext|>
TITLE: How would you solve this tantalizing Halmos problem?
QUESTION [110 upvotes]: $1-ab$ invertible $\implies$ $1-ba$ invertible has a slick power series "proof" as below, where Halmos asks for an explanation of why this tantalizing derivation succeeds. Do you know one?

Geometric series. In a not necessarily commutative ring with
unit (e.g., in the set of all $3 \times 3$ square matrices with real
entries), if $1 - ab$ is invertible, then $1 - ba$ is invertible. However
plausible this may seem, few people can see their way
to a proof immediately; the most revealing approach belongs
to a different and distant subject.
Every student knows that
$1 - x^2 = (1 + x) (1 - x),$
and some even know that
$1 - x^3 =(1+x +x^2) (1 - x).$
The generalization
$1 - x^{n+1} = (1 + x + \cdots + x^n) (1 - x)$
is not far away. Divide by $1 - x$ and let $n$ tend to infinity;
if $|x| < 1$, then $x^{n+1}$ tends to $0$, and the conclusion is
that
$\frac{1}{1 - x} = 1 + x + x^2 + \cdots$.
This simple classical argument begins with easy algebra,
but the meat of the matter is analysis: numbers, absolute
values, inequalities, and convergence are needed not only
for the proof but even for the final equation to make
sense.
In the general ring theory question there are no numbers,
no absolute values, no inequalities, and no limits -
those concepts are totally inappropriate and cannot be
brought to bear. Nevertheless an impressive-sounding
classical phrase, "the principle of permanence of functional
form", comes to the rescue and yields an analytically
inspired proof in pure algebra. The idea is to pretend
that $\frac{1}{1 - ba}$ can be expanded in a geometric series (which
is utter nonsense), so that
$(1 - ba)^{-1} = 1 + ba + baba + bababa + \cdots$
It follows (it doesn't really, but it's fun to keep pretending) that
$(1 - ba)^{-1} = 1 + b (1 + ab + abab + ababab + \cdots) a.$
and, after one more application of the geometric series
pretense, this yields
$(1 -ba)^{-1} = 1 + b (1 - ab)^{-1} a.$
Now stop the pretense and verify that, despite its unlawful
derivation, the formula works. If, that is, $ c = (1 - ab)^{-1}$, 
so that $(1 - ab)c = c(1 - ab) = 1,$ then $1 + bca$ is the inverse
of $1 - ba.$ Once the statement is put this way, its
proof becomes a matter of (perfectly legal) mechanical
computation.
Why does it all this work? What goes on here? Why
does it seem that the formula for the sum of an infinite
geometric series is true even for an abstract ring in which
convergence is meaningless? What general truth does
the formula embody? I don't know the answer, but I
note that the formula is applicable in other situations
where it ought not to be, and I wonder whether it deserves
to be called one of the (computational) elements
of mathematics. -- P. R. Halmos [1]
[1] Halmos, P.R. Does mathematics have elements?
Math. Intelligencer 3 (1980/81), no. 4, 147-153
http://dx.doi.org/10.1007/BF03022973

REPLY [79 votes]: The best way that I know of interpreting this identity is by generalizing it:
$$(\lambda-ba)^{-1}=\lambda^{-1}+\lambda^{-1}b(\lambda-ab)^{-1}a.\qquad\qquad\qquad(*)$$
Note that this is both more general than the original formulation (set $\lambda=1$) and equivalent to it (rescale). Now the geometric series argument makes perfect sense in the ring $R((\lambda^{-1}))$ of formal Laurent power series, where $R$ is the original ring or even the "universal ring" $\mathbb{Z}\langle a,b\rangle:$
$$ (\lambda-ba)^{-1}=\lambda^{-1}+\sum_{n\geq 1}\lambda^{-n-1}(ba)^n=\lambda^{-1}(1+\sum_{n\geq 0}\lambda^{-n-1}b(ab)^n a)=\lambda^{-1}(1+b(\lambda-ab)^{-1}a).\ \square$$
A variant of $(*)$ holds for rectangular matrices of transpose sizes over any unital ring: if $A$ is a $k\times n$ matrix and $B$ is a $n\times k$ matrix then   
$$(\lambda I_n-BA)^{-1}=\lambda^{-1}(I_n+B(\lambda I_k-AB)^{-1}A).\qquad\qquad(**)$$
To see that, let $a = \begin{bmatrix}0 & 0 \\ A & 0\end{bmatrix}$ and $b= \begin{bmatrix}0 & B \\ 0 & 0\end{bmatrix}$ be $(n+k)\times (n+k)$ block matrices and apply $(*).\ \square$

Here are three remarkable corollaries of $(**)$ for matrices over a field:

$\det(\lambda I_n-BA) = \lambda^{n-k}\det(\lambda I_k-AB)\qquad\qquad\qquad$ (characteristic polynomials match)
$AB$ and $BA$ have the same spectrum away from $0$
$\lambda^k q_k(AB)\ |\ q_k(BA)\qquad\qquad\qquad\qquad\qquad\qquad\qquad $ (compatibility of the invariant factors) 

I used a noncommutative version of $(**)$ for matrices over universal enveloping algberas of Lie algebras $(\mathfrak{g},\mathfrak{g'})$ forming a reductive dual pair in order to investigate the behavior of primitve ideals under algebraic Howe duality and to compute the quantum elementary divisors of completely prime primitive ideals of $U(\mathfrak{gl}_n)$ (a.k.a. quantizations of the conjugacy classes of matrices). 

Addendum
The identity $(1+x)(1-yx)^{-1}(1+x)=(1+y)(1-xy)^{-1}(1+x)$ mentioned by Richard Stanley in the comments can be easily proven by the same method: after homogenization, it becomes
$$(\lambda+x)(\lambda^2-yx)^{-1}(\lambda+y)= (\lambda+y)(\lambda^2-xy)^{-1}(\lambda+x).$$
The left hand side expands in the ring $\mathbb{Z}\langle x,y\rangle((\lambda^{-1}))$ as
$$1+\sum_{n\geq 1}\lambda^{-2n}(yx)^n+ \sum_{n\geq 0}\lambda^{-2n}(x(yx)^n+y(xy)^n)+ \sum_{n\geq 1}\lambda^{-2n}(xy)^n,$$
which is manifestly symmetric with respect to $x$ and $y.\ \square$<|endoftext|>
TITLE: Why do probabilists take random variables to be Borel (and not Lebesgue) measurable?
QUESTION [156 upvotes]: I've been studying a bit of probability theory lately and noticed that there seems to be a universal agreement that random variables should be defined as Borel measurable functions on the probability space rather than Lebesgue measurable functions. This is so in every textbook on probability theory which I consulted. In general, it seems to me that probability theory favors the Borel algebra more than the algebra of Lebesgue measurable sets. My question is: why?
In every course in measure theory, one is taught of the notion of a complete measure and completion of measures and I got the impression that a complete measure space is somewhat superior to a non-complete one (or at least that completeness makes life a bit easier on the technical level), so this preference of Borel sets puzzles me.

REPLY [9 votes]: On pages 209-210 of his book "Real Analysis, A Comprehensive Course in Analysis, Part I" Barry Simon gives an argument for sticking with Borel measurable sets and functions.
Edit:
Barry Simon argues that Lebesgue measurable functions are not closed under composition, that it complicates arguments such as constructing product measures, requiring an extra completion set, and that nothing is gained since every Lebesgue measurable function is equal a.e. to a Borel function, and equivalence classes that matter.<|endoftext|>
TITLE: Does negative Kodaira dimension imply uniruled?
QUESTION [12 upvotes]: There is a conjecture (often attributed to Mumford) I believe which states that if, on a smooth proper variety $X$ (over an algebraically closed field of characteristic zero), there are no pluricanonical forms (that is $H^0(X, mK_X) = 0$ for all positive $m$) then $X$ is uniruled.  
Apparently, this conjecture follows from a "well known" conjecture arising from the minimal model program.  I believe, but am not entirely sure, that this is the Abundance Conjecture, which says (in one formulation) that the Kodaira dimension of $X$ agrees with the so called numerical Kodaira dimension of $X$.  There are by now many well written introductions to the MMP, here is one.  
At the same time, Professor Siu has recently posted a sketch of the proof the Abundance Conejcture.  Unfortunately, I am not sufficiently equipped to read the proof which uses L2 estimates of d-bar equations.  Here are my questions.

Is it true that over the complex number Siu's result implies Mumford's conjecture?  He doesn't mention this in the preprint, but is there a reference?
Is anyone well versed enough in both the analytic techniques and algebraic geometry to explain what Siu did to someone more algebraically minded?
Do people have an opinion (vague or otherwise) as to whether techniques coming from analysis are just stronger than techniques coming from algebra?  An if so, why is that?  (An obvious example, example: Hodge decomposition, but also Siu's proof of invariance of plurigenera, and now the abundance conjecture).

REPLY [5 votes]: The conjecture you attribute to Mumford is also sometimes called weak non-vanishing, or just non-vanishing. As already mentioned, by BDPP one is reduced to proving that if $h^0(mK_X) = 0$ for all $m \geq 0$ then $K_X$ is not even pseudo-effective, that is, not numerically a limit of effective divisors. There is an alternative formulation of the abundance conjecture which is arguably easier to understand: if $K_X + \Delta$ is nef and $(X, \Delta)$ is klt then $K_X + \Delta$ is semi-ample. The point is that this says that running the MMP turns $K_X$ into a semi-ample divisor.
My understanding is that the answer to question to 2 is something like the following. Siu explains and gives references in his introduction to Part II of his paper that there is something called the numerically trivial fibration that is similar to the Kodaira-Iitaka fibration but works for the numerical dimension. Its construction is analytic in nature. Siu claims that, if $\pi : X' \to Y$ is a realization of numerically trivial fibration for $K_X$ with $X'$ a birational model of $X$, then $\pi_* O_{X'}(mK_{X'/Y})$ carries some kind of metric with strictly positive curvature on a general fiber of $\pi$. This, in particular, shows that $\pi$ is not the identity map, which I think is now known to imply the abundance conjecture, though I am not sure if this is how Siu proceeds. I am not familiar with the details of the argument.
As for question 3, it is hard to say. The famous example is Siu's proof of the deformation invariance of plurigenera for all Kodaira dimensions, this is currently a purely analytic argument that involves taking limits of pluri-subharmonic functions to obtain a semipositive singular metric on $O_X(mK_X)$ that may not have analytic singularities but has the right multiplier ideal: see Paun's "Siu's invariance of plurigenera: a one-tower proof." This has resisted attempts to prove it algebraically so far. On the other hand, the algebraic approach allows reduction to positive characteristic. It may be that, ultimately, anything you can do with one you can do with the other. Of course, "algebraic" here allows the Kodaira vanishing theorem, which can be deduced from the homological statement that the map $H^i(X, \mathbb{C}) \to H^i(X, O_X)$ induced by the natural inclusion of sheaves is surjective.<|endoftext|>
TITLE: What can be the analogue of Frobenius in complex geometry?
QUESTION [6 upvotes]: The  Frobenius map is usually only defined in char p, or when we are working with $\mathbb{Q}_p$, etc.
What can be the analogue of Frobenius in complex geometry?
As a vague guess, about which I probably heard some years ago somewhere, is it somehow "related" to the Laplacian or Dirac operators? Do we need some very heavy machineries for a better analogue? Will $\mathbb{F}_1$ say something about it?
Thank you very much.

REPLY [9 votes]: I've been hesitating about whether or not to answer this question. Not because it's uninteresting, but because I think it's too interesting. I'm glad you didn't ask what the influence of Frobenius is on characteristic $0$ geometry, because it's enormous and would be too much to get into. Since some of these interactions have been discussed in other people's answers and elsewhere, I'll just make a list.
These include: Mori's "bend and break", the theory of weights, the decomposition theorem for perverse sheaves, Frobenius splitting, Deligne-Illusie splitting, tight closure... 
I'd rather get off the beaten track a bit, and discuss some answers which are  more obscure,
but hopefully interesting. Actually, these are closer to answers to the question you did ask.

Some special classes of complex varieties (e.g. toric varieties, modular curves, CM elliptic curves) carry endomorphisms or self correspondences which are in fact lifts of the Frobenius. For $\mathbb{P}^n$, one can take $[x_0,\ldots x_n]\mapsto [x_0^p,\ldots x_n^p]$.
Unfortunately, such lifts usually don't exist. (I'd be happy with more examples, if people
know of any.)
On K-theory the Adams operations $\psi^p:K^0(X)\to K^0(X)$ behave like the action of Frobenius. Defining these is a long story, suffice to say that for a line bundle $\psi^p(L) = L^p$, exactly like the Frobenius.
This is a bit of a crazy thing to do, but you can take the ultraproduct of the mod $p$ reductions of your complex variety and you will get a nonnoetherian (!) scheme dominating $X$ with a  Frobenius like endomorphism.

As I said in my comment above, there are many answers. The best one really depends on what you want to do.<|endoftext|>
TITLE: Advantages of working with CW complexes/spaces over Kan complexes/simplicial sets?
QUESTION [17 upvotes]: Many topologists express a clear preference for working with CW complexes instead of simplicial sets.  
One of the reasons is that the cellular chain complex of a CW complex is often easier to work with than a simplicial chain complex.  However, simplicial sets have many nice features that spaces do not.  The category of simplicial sets has a proper and combinatorial (in the sense of Jeff Smith) model structure and is a presheaf topos, which makes the objects behave very much like sets.  Surely these make up for the problems with specifying combinatorial data?
The question: Why do many topologists and homotopy theorists prefer to work with spaces and CW complexes over simplicial sets and Kan complexes?  What are some other advantages that CW complexes enjoy over Kan complexes?

REPLY [2 votes]: It is always good to know both CW complexes and simplicial sets. Let me give a few examples.

To define cup products in some multiplicative cohomology theory, one needs the diagonal map $X\to X\times X$. It does not respect a CW structure in general, so one has to approximate it. One knows that such an approximation always exists. If you need a concrete one, you have to work. For simplicial sets on the other hand, the diagonal map is simplicial. But then it is harder to relate $h^\bullet(X\times X)$ with $h^\bullet(X)\otimes h^\bullet(X)$.
One of the possible solutions for ordinary cohomology leads to the well-known cup-product formula in singular cohomology. It is interesting to notice that this cup-product formula looks as if it came from an approximation of the diagonal in the CW-product of the geometric realisations.
Every topological space has an approximation by a weakly homotopy equivalent CW complex. If you are lucky, you find one with very few cells, for example a single point suffices for the polish circle. But there always is a natural choice, the realisation of the singular complex $\left|S_\bullet(-)\right|$, which is awfully large in most cases. However, in the case of the polish circle, you may argue that a weakly homotopy equivalent approximation loses too much information, and prefer to use some entirely different theory.
Starting from a smooth manifold $M$, a Morse function together with a gradient-like vector field gives a CW complex, which is again far from natural. But you immediately recover the dimension of $M$ from the cell structure (which you can then use to prove a cup-length estimate, for example). The singular simplicial complex has nondegenerate simplices in all dimensions, so you really have to work until you recover $\dim M$.
Yet another point. For any topological group $G$, Milnor's join construction gives a model for the classifying space $BG$ that is a simplicial space. It is "made" to classify $G$-bundles gives by $G$-cocycles. On the other hand, if $G$ is a classical Lie group, you can approximate $BG$ through Grassmannians. These classify vector bundles that are given as subbundles of trivial bundles (which is how you usually view vector bundles in noncommutative geometry). The construction is again less universal, but it connects better with some analytic methods. And one has Schubert cells to work with.

There may even be situations where one wants to combine the strength of both approaches in some hybrid object.<|endoftext|>
TITLE: Lie group actions and f-relatedness
QUESTION [8 upvotes]: Background
Let $f: M \to N$ be a smooth map between smooth manifolds.
Two vector fields $X$ in $M$ and $Y$ in $N$ are said to be $f$-related if for all $p \in M$, $(f_*)_p(X_p) = Y_{f(p)}$; equivalently, if for every smooth function $g: N \to \mathbb{R}$, one has
$$(Yg) \circ f = X(g \circ f).$$
One immediate consequence of this definition is that if $X_i$ and $Y_i$ are $f$-related, for $i=1,2$, then so are their Lie brackets $[X_1,X_2]$ and $[Y_1,Y_2]$.
A number of basic results about Lie groups in a differential-geometric context follow from this observation; for example,

the left-invariant/right-invariant vector fields form a Lie subalgebra,
the differential of a smooth map between Lie groups is a Lie algebra homomorphism,
the Lie algebra of a Lie subgroup is a Lie subalgebra (really a special case of the above),
etc

Now, perhaps unwisely, while I was preparing a problem sheet for an undergraduate summer project, I added a problem on f-relatedness, one of whose parts was to show the following.

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, acting on the left right on a smooth manifold $M$.  If $X \in \mathfrak{g}$, let $X'$ denote the corresponding fundamental vector field on $M$.  Show that $[X',Y'] = [X,Y]'$, where the left-hand side is the Lie bracket of vector fields on $M$ and the right-hand side is the vector field on $M$ corresponding to the Lie algebra bracket of $X,Y \in \mathfrak{g}$.

One can prove this directly, of course, but I (mistakenly?) thought there ought to be a slick proof using $f$-relatedness.  After one of my students complained that he could not find a proof using $f$-relatedness, I realised (somewhat embarrassingly) that neither could I!
The naive thing does not work: one can exhibit $X'_m = (\alpha_m)_* X$, where $\alpha_m : G \to M$ is the map sending $g$ to $g\cdot m$.  This does not work because the map used to relate $X$ and $X'$ depends on $m$.
Hence the following
Question

Is the vector field $X'$ on $M$ $f$-related to the left-invariant vector field on $G$ corresponding to $X \in \mathfrak{g}$?

In other words, the question is to exhibit $f$.  It does not seem that it's as simple as a map $G\to M$, but perhaps there is some trick using the action $\alpha: G \times M \to M$ in some way.

REPLY [17 votes]: I hate to throw cold water on the party, but surprisingly, the formula that the OP was trying to prove ($[X',Y']=[X,Y]'$) is actually false when $G$ acts on the left.  This formula is correct if $G$ acts on the right on $M$; but if $G$ acts on the left, then the correct formula is $[X',Y']=-[X,Y]'$.  
Here's how to prove it. Suppose first that $G$ acts smoothly on $M$ on the right.  Fix $p\in M$, and consider the orbit map $\alpha^{(p)}: G\to M$ defined by $\alpha^{(p)}(g) = p\cdot g$.  Then I claim that for each $X\in \mathfrak g$, the fundamental vector field $X'$ is $\alpha^{(p)}$-related to $X$.  To see this, note that the group law $p\cdot gg' = (p\cdot g)\cdot g'$ translates to $$\alpha^{(p)}\circ L_g (g') = \alpha^{(p\cdot g)}(g')$$
(where $L_g$ is left multiplication by $g$).
Let $g\in G$ be arbitrary and set $q=p\cdot g=\alpha^{(p)}(g)$.  Because $X$ is left-invariant, 
$$ X'_q = d\bigl(\alpha^{(q)}\bigr)_1(X_1) = d\bigl(\alpha^{(p)}\bigr)_g\circ d(L_g)_1(X_1) = 
d\bigl(\alpha^{(p)}\bigr)_g(X_g)$$
(where the first equality is essentially the definition of $X'$, and the second follows from the previous equation by taking differentials at the identity). 
This proves the claim.  
Because brackets of $\alpha^{(p)}$-related vector fields are themselves $\alpha^{(p)}$-related, it follows that $[X',Y']_p = ([X,Y]')_p$, and since this is true for every $p$ it is true globally.
Now if $G$ acts on $M$ on the left, the above argument doesn't work; if $\tilde\alpha^{(p)}$ denotes the orbit map for the left action, then $X'$ is not $\tilde\alpha^{(p)}$-related to $X$.  Instead, we can create a right action by setting $p\cdot g = g^{-1}\cdot p$.  Letting $\alpha^{(p)}$ denote the orbit map for this new right action, we have $\tilde\alpha^{(p)}=\alpha^{(p)}\circ\iota$, where $\iota: G\to G$ is inversion; and the argument above shows that $X$ and $X'$ are $\alpha^{(p)}$-related.  Since $d\alpha^{(p)}$ preserves brackets and $d\iota$ reverses them, it follows that $d\tilde\alpha^{(p)}$ reverses brackets.  
The problem with Eric's argument has to do with the identification between $\operatorname{Lie}(\operatorname{Diff}(M))$ and $\mathfrak X(M)$.  Essentially the same argument as above shows that the natural map $\operatorname{Lie}(\operatorname{Diff}(M)) \to \mathfrak X(M)$ is actually a Lie algebra anti-isomorphism (if we consider $\operatorname{Diff}(M)$ as an infinite-dimensional Lie group acting on $M$ on the left).  
The problem with Victor Protsak's argument is that it doesn't actually show that $(X,0)$ (considered as a vector field on $G\times M$) is $\alpha$-related to $X'$ -- to show this, you'd have to prove that $(d\alpha)_{(g,m)}(X,0)=X'$ for every $g$ in the group, not just $g=1$. 
I had to sort all this out recently because I'm adding a section on infinitesimal group actions in the second edition of my book Introduction to Smooth Manifolds.  When I tried to prove the same formula the OP was trying to prove, I was surprised to find that it led to contradictions.  Eventually I came up with the argument I sketched here, and then found at least one other book that confirms the result.  I'm at home now and don't have access to my books, but if you'd like I'll find the reference and post it tomorrow.<|endoftext|>
TITLE: Theorems which say "such and such method cannot possibly prove FLT"
QUESTION [19 upvotes]: After a previous question that I asked https://mathoverflow.net/questions/31565/request-for-comments-about-a-claimed-simple-proof-of-flt-closed was closed, someone suggested in the comments that I ask another question that is more suited for MO. That question is as follows:
Are there any nontrivial theorems of the form "Method X cannot possibly prove FLT."? 
The reason I am asking is because I'd like to know (by deductive reasoning and not relying on 350 years experience) if it is impossible to prove FLT using elementary methods.

REPLY [8 votes]: There is also the relative approach, which is to show that making Method X work is at least as difficult as solving some set of known hard problems, due to relatively simple reductions between the problems.  This is what has been done for the P=NP problem --- thousands of NP problems are computationally equivalent to each other --- as well as the Riemann Hypothesis which has a large number of known equivalents.  If you had the bright idea of replacing the non-elementary Riemann zeta function and statements about its zeroes by statements about the distribution of primes, or sums of the Moebius function, then the translation between those contexts is known and much easier than the theory of the zeta function itself, so there would have to be some elementary idea that people had missed in each and every one of those equivalent environments. 
For example, if Method X = "prove an effective form of the ABC conjecture" (the non-effective form implies FLT for large exponents using a few lines of algebra), then solutions of dozens of hard unsolved problems, and easier proofs of very hard theorems, would be assured as soon as one could carry out Method X.  Because the reductions of these problems to the ABC conjecture are often quite simple, any elementary idea used to carry forward Method X would immediately translate into elementary methods for the dozens of other problems, and it is less likely that nobody would have ever noticed this in any of the other actively studied problems.<|endoftext|>
TITLE: Does "Algebraic numbers coloured by degree" form a fractal?
QUESTION [15 upvotes]: This picture from Wikipedia's article on Algebraic numbers shows a visualization of Algebraic numbers coloured by degree.
I'm wondering if this is a fractal?

REPLY [4 votes]: If you consider the set of roots of polynomials whose coefficients are entirely $1$ or $-1$, and take the topological closure of that set, you get a fractal pattern closely related to the Dragon curve.
see: http://math.ucr.edu/home/baez/roots/beauty.pdf<|endoftext|>
TITLE: Is an English translation of Grothendieck's EGA available?
QUESTION [24 upvotes]: I have always wondered whether there is an English translation of Grothendieck's EGA (Elements de Geometrie Algebrique) available. Does anyone know whether there is and if so where I can find it? If not, are there English texts that cover similar material to the EGA that you would recommend? (My knowledge of French is very rudimentary, and while I can roughly make meaning out of some (non-mathematical) passages, it seems (from what I have heard) that some mastery of French is necessary to leisurely read the EGA.)

REPLY [41 votes]: Yes, ish. There is a community effort (https://github.com/ryankeleti/ega) to translate the EGA into English. I’m posting this now because we’ve just finished EGA I, and around 30% of the  combined EGA 0.<|endoftext|>
TITLE: Modern algebraic geometry vs. classical algebraic geometry
QUESTION [54 upvotes]: Can anyone offer advice on roughly how much commutative algebra, homological algebra etc. one needs to know to do research in (or to learn) modern algebraic geometry. Would you need to be familiar with something like the contents of Eisenbud's Commutative Algebra: With a View Toward Algebraic Geometry, or is less needed in reality? (I am familiar with more commutative algebra than that which is covered in Atiyah and MacDonald's *Introduction to Commutative Algebra", but less than that which is covered in Eisenbud's textbook.)
Also, is modern algebraic geometry concerned with abstractions such as schemes, sheaves, topological spaces, commutative and noncommutative rings etc., or is it just classical algebraic geometry in an abstract form? Perhaps more specifically, to do research in modern algebraic geometry, do you need to be familiar with classical algebraic geometry, or is it possible to think of algebraic geometry as an "abstract language" and do research based just on this perception? 
While I suspect that, as with other branches of mathematics, "abstraction was invented to analyze the concrete", with all the emphasis currently given to the understanding of abstract tools, for someone who is not very familiar with the subject (such as myself), it seems that algebraic geometry is a "mixture" of general topology and abstract algebra. Is this right? If not, succinctly my question is: how great an influence does classical algebraic geometry have on modern algebraic geometry today?

REPLY [18 votes]: For several beautiful and expert discussions of the contrasts and relations between classical and the evolving subject of abstract or modern algebraic geometry, I recommend the following ICM lectures:
O.Zariski, 1950, vol.2, p.77ff; 
B.Segre, 1954, vol.3, p.497ff; 
J.P.Serre, 1954, vol.3, p.515ff; 
A.Weil, 1954, vol.3, p.550ff; 
A.Grothendieck, 1958, p.103.  
(This falls obviously under the heading "reading the masters".)
Indeed the whole algebraic geometry session, 1954, vol.3, pp.445-560, has an incredible list of short talks, (Groebner, Hirzebruch, Kodaira, Neron, Rosenlicht, Van der Waerden,...).
the link is:    http://www.mathunion.org/ICM/
My apologies for such a brief answer.
The article by Zariski, THE FUNDAMENTAL IDEAS OF ABSTRACT ALGEBRAIC GEOMETRY,  points out the advances in commutative algbra motivated by the need to substantiate results in geometry.  “The past 25 years have witnessed a remarkable change in the field of algebraic geometry, a change due to the impact of the ideas and methods of modern algebra. What has happened is that this old and venerable sector of pure geometry underwent (and is still undergoing) a process of arithmetization. This new trend has caused consternation in some quarters. It was criticized either as a desertion of geometry or as a subordination of discovery to rigor. I submit that this criticism is unjustified and arises from some 
misunderstanding of the object of modern algebraic geometry. This object is not 
to banish geometry or geometric intuition, but to equip the geometer with the 
sharpest possible tools and effective controls.”
That by Segre argues for the preservation of geometric intuition in algebraic geometry for just this reason, for motivating and suggesting new questions to investigate.  It seems particularly articulate and impassioned as he is arguing for a tradition that seems threatened to be lost.
GEOMETRY UPON AN ALGEBRAIC VARIETY 
BENIAMINO SEGRE 
I. Algebraic geometry — that is to say, the branch of geometry which 
deals with the properties of entities represented by algebraic equations — has in 
recent years developed in two distinct directions, which in a sense are opposed 
to one another. One of these directions is called abstract in as much as it is 
concerned with algebraic equations defined over commutative fields subject 
only to slight restrictions; here the means employed are purely algebraic, 
including in particular ideal theory and valuation theory. The other direction 
may properly be called geometrical) this usually deals with algebraic equations 
in the complex domain, and from time to time appeals to ideas and methods of 
analytic and projective geometry, topology, the theories of analytic functions 
and of differential forms. 
The dualism between these two disciplines has close relationship and affi- 
nity with that which, three centuries ago, arose between l'esprit géométrique of 
Descartes and l'esprit de finesse of Pascal, and which, in the past century, on 
the one hand divided the geometers into analysts of the school of Plücker and 
synthesists of the school of Steiner and, on the other, the algebraists into 
purists à la Dedekind and arithmetizers à la Kronecker. However, this dualism, 
instead of proving harmful to geometry, offers undoubted advantages when 
the two lines of development, with their respective merits and possibilities, are 
regarded not as contrasting but as complementary. 
We cannot fail to recognise in the abstract method and its technique a 
peculiar elegance, an impeccable logical coherence, and to appreciate the im- 
portance of the results so far obtained by it, particularly in the study of the 
foundations of geometry and the difficult questions concerning the singularities 
of algebraic varieties. But equally we cannot fail to recognise that the geometr- 
ical approach, with its greater concreteness, lends itself better to the formula- 
tion and initial study of new concepts and problems; and that it presents an 
incomparable wealth and colour of its own, due to the interweaving of many 
diverse strands, to the subtle and perspicuous play of geometrical intuition, 
and to the possibility of readily constructing examples and investigating special 
cases. We may also point out that, in the geometrical discipline, corresponding 
to a more definite notion of algebraic variety, there is a much wider range of 
subjects and a far greater number of orientations and contacts with other 
important branches of mathematics, which have found, and are finding, 
therein inspiration and extensions beyond the purely algebraic field.
Weil’s article describes how arithmetic benefits as well from the algebraization of geometry.
ABSTRACT VERSUS CLASSICAL ALGEBRAIC GEOMETRY 
ANDRé WEIL 
The word "classical", in mathematics as well as in music, literature or 
most other branches of human endeavor, may be taken in a chronological sense; 
it then means anything which antedates whatever one chooses to consider as 
"modern", and may be used to describe remote antiquity or the achievements 
of yesteryear, according to the mood and the age of the speaker. Sometimes, too, 
it is purely laudatory and is applied to any piece of work which is thought to be 
of permanent value. 
Here, however, while discussing algebraic geometry, I wish to use the words 
"classical" and "abstract" in a strictly technical sense which will be explained 
presently. Until not long ago algebraic geometers did their work exclusively 
with reference to the field of complex numbers; at the same time they worked 
on non-singular models, or at any rate their concern with multiple points was 
merely in order to try to push them out of the way by suitable birational trans- 
formations. Thus transcendental and topological tools of various kinds were 
available, and it was merely a matter of individual taste, personal inclination or 
expediency whether to use them or not on any given occasion. The most deci- 
sive progress ever made in the theory of algebraic curves was achieved by 
Riemann precisely by introducing such methods. Later authors took consider- 
able pains to obtain the same results by other means. In so doing, they were 
motivated, at least in part, by the fact that Riemann had given no justification 
for Dirichlet's principle and that it took many years to find one. Similarly, the 
use of topological methods by Poincaré and Picard, not to mention some more 
recent writers, has often been such as to justify doubts about the validity of their 
proofs, while conversely it has happened that theorems which had merely been 
made plausible by so-called geometrical reasoning were first put beyond doubt 
by the transcendental theory. 
Now we have progressed beyond that stage. Rigor has ceased to be thought 
of as a cumbersome style of formal dress that one has to wear on state occasions 
and discards with a sigh of relief as soon as one comes home. We do not ask any 
more whether a theorem has been rigorously proved but whether it has been 
proved. At the same time we have acquired the techniques whereby our prede- 
cessors' ideas and our own can be expanded into proofs as soon as they have 
reached the necessary degree of maturity; no matter whether such ideas are
based on topology or analysis, on algebra or geometry, there is little excuse left 
for presenting them in incomplete or unfinished form. 
What, then, is the true scope of the various methods which we have learnt 
to handle in algebraic geometry? The answer is obvious enough. Let us call 
"classical" those methods which, by their very nature, depend upon the pro- 
perties of the real and of the complex number-fields; such methods may be 
derived from topology, calculus, convergent series, partial differential equations 
or analytic function-theory. As examples, one may quote the use of the differ- 
ential calculus in the proof of the Kronecker-Castelnuovo theorem, of theta- 
functions in the theory of elliptic curves and abelian varieties, of topology in the 
proof of the "principle of degeneracy". Let us call "abstract" those methods 
which, being basically algebraic, are essentially applicable to arbitrary ground- 
fields; this includes for instance the theory of differentials of the first, second 
and third kinds (but of course not that of their integrals) and the greater part of 
the "geometric" proofs of the Italian school. Thus it is plain that, in all cases 
where an abstract proof is available, it may be expected to yield more than 
any classical proof for the same result. No one could deny this unless he had 
made up his mind to ignore fields of non-zero characteristic and was prepared 
to maintain that a theorem in algebraic geometry which has been proved for the 
field of complex numbers can always be extended to any field of characteristic 0. 
There are indeed many cases where this is so; quite often, however, the exten- 
sion can only be made to algebraically closed fields. As to denying any existence 
to algebraic geometry of non-zero characteristic, not merely would this, in 
view of recent developments, amount to denying motion; it would also deprive 
algebraic geometry of a rich and promising field of possible applications to 
number-theory, where one cannot do without reduction modulo p.
Serre and Grothendieck describe the contribution of cohomology.
I cannot give a good account of this material in a few words, but I strongly advocate reading these articles which marked the introduction of abstract methods in algebraic geometry in its most fruitful period.<|endoftext|>
TITLE: Statistics for mathematicians
QUESTION [122 upvotes]: I'm looking for an overview of statistics suitable for the mathematically mature reader: someone familiar with measure theoretic probability at say Billingsley level, but almost completely ignorant of statistics.
Most texts I've come across are either too basic, or are monographs focused on a specific area or technique.
Any suggestions?

REPLY [2 votes]: Mathematical Statistics, by Jun Shao, the most mathematical statistics book that I know.
This book is mainly focused on classical statistics. If you want to proceed further into classical statistics, E.L. Lehmann's two books Theory of Point Estimation and Testing Statistical Hypothesis are bibles of classical statistics, worth your reading.
High-Dimensional Statistics: A Non-Asymptotic Viewpoint, by A.M. Wainwright, another popular book, also quite mathematical, mainly focused on high dimensional statistics.
Enjoy!<|endoftext|>
TITLE: Distinct numbers in multiplication table
QUESTION [34 upvotes]: Consider the multiplication table for the numbers $1,2,\dots, n$. How many different numbers are there? That is, how many different numbers of the form $ij$ with $1 \le i, j \le n$ are there?
I'm interested in a formula or an algorithm to calculate this number in time less than $O(n^2)$.

REPLY [6 votes]: Regarding the algorithmic question, a recent paper of Brent, Pomerance, Purdum, and Webster presents a subquadratic algorithm to compute the number of distinct products $M(n)$ of the $n \times n$ multiplication table.  They have implemented their results to compute $M(n)$ exactly for all $n \leq 2^{30}$.  They note that for larger values of $n$, exact algorithms become impractical, and so the paper also presents two Monte Carlo algorithms to approximate $M(n)$. Monte Carlo computations are presented for $n$ up to $2^{100000000}$.<|endoftext|>
TITLE: Any further applications of Freudenthal's 1936 Spectral Theorem?
QUESTION [14 upvotes]: Seemingly completely forgotten, back in 1936, the Dutch mathematician Freudenthal, quite well known at the time, proved his so called Spectral Theorem, see chapter 6 in Luxemburg & Zaanen : Riesz Spaces I. North-Holland, 1971. The amusing thing is that the theorem is formulated eclusively in terms of partial orders, and on top of it, its proof is also in the very same terms. Yet one of the rather direct consequences of it is the spectral representation of normal operators in Hilbert spaces. Another one is the Radon-Nykodim theorem in measure theory. And to aggravate things, it can also solve some Poisson PDEs.
Does anybody know about more recent applications of that theorem ? And how about having more appreciation for the concept of partial order ?

REPLY [3 votes]: I think that some aspects of the Daniell integral and the Riesz-Markov representation theorem are other applications, however, I'm sorry that I neither can give a good reference nor can I really proof this here. I'll try to sketch the idea, maybe I'll come back later and improve it. And you might have a look at the book "Integration - a functional approach" by Klaus Bichteler (it never mentions Riesz spaces, but it gave me some of the ideas here):
For the Daniell integral, you essentially start with a Riesz subspace $L\subseteq \mathbb{R}^X$ of all  functions on a set (maybe just bounded ones?) and positive linear functionals $\psi$ on $L$ which have the property ($\star$) that $\lim_{n\to\infty} \psi(f_n) = 0$ for every monotonely decreasing sequence $(f_n)_{n\in\mathbb{N}}$ in $L$ with pointwise limit $0$. You can then complete $L$ to a Dedekind-$\sigma$-complete Riesz subspace $L^c$ of $\mathbb{R}^X$ (using the Daniell mean from Bichteler's book) and extend the linear functionals $\psi$ to the completion by using ($\star$). The Freudenthal spectral theorem now tells you that $L^c$ is essentially generated by indicator functions on a $\sigma$-algebra $\mathfrak{a}$ on $X$, which allows you to relate this approach to the Lebesgue-integral: The measure $\mu$ associated to $\psi$ is simply defined as $\mu(S) = \psi(\chi_S)$ for every $S\in\mathfrak{a}$, where $\chi_S\in L^c$ is the indicator function of $S$.
For the Riesz-Markov theorem, you start with $L=\mathcal{C}(X)$, the Riesz space of continuous real-valued functions on a compact Hausdorff space $X$, and an arbitrary $||\,\cdot\,||_\infty$-continuous linear functional $\psi\colon \mathcal{C}(X)\to\mathbb{R}$. Here $||\,\cdot\,||_\infty$ is the usual Banach-norm of $\mathcal{C}(X)$ and the $||\,\cdot\,||_\infty$-continuous linear functionals are just the order-bounded ones, which are a Riesz space themselves, so $|\psi|$ is defined and is a $||\,\cdot\,||_\infty$-continuous positive linear functional. Now one only has to show that $\lim_{n\to\infty}|\psi|(f_n)=0$ actually holds for every monotonely decreasing sequence $(f_n)_{n\in\mathbb{N}}$ in $\mathcal{C}(X)$ with pointwise limit $0$, because $X$ is compact. So we can construct the Daniell integral, which by the above gives rise to a Lebesgue integral.
PS: I'm all for more appreciation for the concept of partial order.<|endoftext|>
TITLE: Lie algebras and complements
QUESTION [6 upvotes]: I have some elementary questions about Lie algebras and vector space complements.
Let  $(\mathfrak{g},[.,.])$  be a finite-dimensional Lie algebra and $\mathfrak{g}_1$ a Lie ideal in $\mathfrak{g}$.
1) Is it possible to choose $\mathfrak{g}$ and $\mathfrak{g}_1$ in such a way that there is no vector space complement $\mathfrak{g}_2$ of $\mathfrak{g}_1$ in $\mathfrak{g}$ which is additionally a sub-Lie algebra of $\mathfrak{g}$?
2) Is there an example of $\mathfrak{g}$ and $\mathfrak{g}_1$ such that there is a vector space complement $\mathfrak{g}_2$ of $\mathfrak{g}_1$ in $\mathfrak{g}$ with $[\xi_1,\xi_2] = 0$ for all $\xi_1 \in \mathfrak{g}_1$ and $\xi_2 \in \mathfrak{g}_2$ but such that no such complement is a Lie ideal in $\mathfrak{g}$?
In the situation I need, all Lie algebras occur as Lie algebras of some Lie group. For every Lie algebra $\mathfrak{g}$ there exists a Lie group $G$ with $\mathrm{Lie}(G) = \mathfrak{g}$, but that's rather abstract, so it would be nice if $\mathfrak{g}$ and $\mathfrak{g}_1$ were the Lie algebras of some well-known Lie groups.
I'm not an expert in Lie algebra theory, so if some things described in 1) or 2) have special names in the literature it would be nice if you could mention that or give some references.

REPLY [12 votes]: Take $\mathfrak{g}$ to be the Lie algebra of the Heisenberg Lie group (i.e., the nilpotent Lie algebra of strictly upper triangular $3\times 3$ matrices) and let $\mathfrak{g}_1$ be the centre.  This gives you examples to (1) and (2).
Added
Now that I have more time, I can perhaps say something else about (1).  (I will work in zero characteristic, just in case.)
Every ideal $\mathfrak{g}_1$ of $\mathfrak{g}$ gives rise to an exact sequence of Lie algebras
$$\begin{matrix} 0 & \longrightarrow & \mathfrak{g}_1 & \longrightarrow & \mathfrak{g} & \longrightarrow & \mathfrak{h} & \longrightarrow & 0 \end{matrix}$$
where $\mathfrak{h} = \mathfrak{g}/\mathfrak{g}_1$.  This says that $\mathfrak{g}$ is an extension of $\mathfrak{h}$ by $\mathfrak{g}_1$.
Your condition (1), that there should not be a Lie subalgebra of $\mathfrak{g}$ complementary to $\mathfrak{g}_1$, is precisely the statement that the sequence does not split.  Hence, in particular, $\mathfrak{g}$ is not isomorphic to a (semi)direct product of $\mathfrak{g}_1$ and $\mathfrak{h}$.
If $\mathfrak{g}_1$ were abelian, it becomes in a natural way an $\mathfrak{h}$-module and you can classify such extensions by the Lie algebra cohomology group $H^2(\mathfrak{h},\mathfrak{g}_1)$.  Then your condition  (1) can be rephrased as saying that the corresponding extension class in $H^2(\mathfrak{h},\mathfrak{g}_1)$ is nonzero.  (I do not know what classifies extensions in the general case.)<|endoftext|>
TITLE: Euclidean function of Euclidean domain defined at 0
QUESTION [5 upvotes]: In a few places where I have looked the Euclidean Function of a Euclidean Domain is only being defined for non-zero elements. I am teaching an undergraduate course and I am trying to make things as simple as possible. Is there any good reason why not to define it as $0$ at $0$?

REPLY [3 votes]: Old thread, but nobody wrote the simple practical reason, so I'll do it now. In $K[T]$, we want to use $\deg$ as a Euclidean function, so $\deg(c) = 0$ for $c \in K - \{0\}$. But $\deg(0)$ is best thought of as $-\infty$, which breaks the usual definitions of "Euclidean function". It's common to sidestep this by simply not requiring the Euclidean function to be defined at 0 at all, which is fine since you never need it anyway.<|endoftext|>
TITLE: Examples of non-diffeomorphic smooth manifolds with diffeomorphic tangent bundle
QUESTION [27 upvotes]: Given a smooth manifold $M$, we define the differentiable structure on $TM$ in the usual way.
I would like to know examples of two smooth manifolds which are non-diffeomorphic, but their tangent bundles are.
Which is the smallest dimension in which one can find such examples?
What if I ask the same question for $k$ pairwise non-diffeomorphic manifolds?
Can we have $k=\infty$?

REPLY [43 votes]: Here are examples of non-diffeomorphic closed manifolds with diffeomorphic
tangent bundles:

3-dimensional lens spaces have trivial tangent bundles, which are diffeomorphic if and only if the lens spaces are homotopy equivalent, e.g. $L(7,1)$, $L(7,2)$ are not homeomorphic, but their tangent bundles are diffeomorphic. This follows from proofs in [Milnor, John, Two complexes which are homeomorphic but combinatorially distinct.
Ann. of Math. (2) 74 1961 575--590]. 
Tangent bundle to any exotic $n$-sphere is diffeomorphic to $TS^n$ as proved in [De Sapio, Rodolfo, Disc and sphere bundles over homotopy spheres, Math. Z. 107 1968 232--236].
In dimensions $n\ge 5$ one can attack this question via surgery theory. For example, let $f:N\to M$ be a homotopy equivalence of closed $n$-manifolds that has trivial normal invariant (which is a bit more than requiring that $f$ preserves stable tangent bundle). Multiply $f$ by the identity map of $(D^n, S^{n-1})$, where $D^n$ is the closed $n$-disk. Then Wall's $\pi-\pi$ theorem implies that $M\times D^n$ and $N\times D^n$ are diffeomorphic, so if tangent bundles of $M, N$ are trivial, this gives manifolds with diffeomorphic tangent bundles.  
To illustrate the method in 3, here is a particular example of infinitely many non-homeomorphic closed manifolds with diffeomorphic tangent bundles. Fix any closed $(4r-1)$-manifold $M$ where $r\ge 2$ such that $TM$ is stably trivial and $\pi_1(M)$ has (nontrivial) elements of finite order. Then results of Chang-Weinberger imply that there are infinitely many closed $n$-manifolds $M_i$ that are simply-homotopy equivalent to $M$ and such that  the tangent bundles $TM_i$ are all diffeomorphic (I am not quite sure how to get them be diffeomorphic to $M\times\mathbb R^n$ even though this should be possible). I know how to deduce this from [On invariants of Hirzebruch and Cheeger-Gromov, Geom. Topol. 7 (2003), 311--319].
I have been thinking extensively of related issues, so you might want to look at my papers at arxiv, e.g. this one.<|endoftext|>
TITLE: Inverting the totient function
QUESTION [20 upvotes]: For what values of $n$ does the equation $\phi(x) = n$ have at least one solution? Is there any efficient way to check it for a given $n$?
It obviously has no solutions for odd $n$. And the smallest even number for which it has no solutions is $14$.

REPLY [4 votes]: I recently answered this related question about the Carmichael function on math.SE.  The algorithm uses an unconditional lower bound so it should work just as well for the totient function because $\lambda(x) \le \phi(x)$.  My answer (the only one) has not been accepted and the question has a bounty which expires tomorrow.  I should not like to receive a bounty by default for an incorrect answer, so I am posting this here now as an invitation for you to correct me on math.SE.  It is not an efficient algorithm as this MO question demands, but I proffer it because no algorithm has yet been given to answer it.
Also related is Carmichael's totient function conjecture which is that there are no unique solutions to this equation.<|endoftext|>
TITLE: Best strategy for small resolutions
QUESTION [17 upvotes]: I would like to know if there is a standard technique to check if a singular variety admits a small resolution. What are the main references for these types of questions? 
I am mostly interested in threefolds and fourfolds with singularities in codimension 2 or higher.
(By a small resolution, I mean a proper birational transformation $Y\rightarrow X$ such  that Y is smooth and the exceptional locus does not contain any divisors.)

REPLY [21 votes]: I assume that by a small morphism you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a small resolution if $Y$ is smooth.
There is a nice criterion to check that a variety does not admit a small resolution:
Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f_*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^*(m(f_*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D. 
There is also something one can say for the reverse direction:
If $X$ has klt singularities, then it is possible to construct a morphism $f:Y\to X$ such that $f$ is small and $Y$ is $\mathbb Q$-factorial. I don't know a very easy proof of this. The essence is to take a resolution and then use a well-chosen directed mmp (in the sense of BCHM) to contract all the exceptional divisors.
This implies that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ has klt singularities and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution is then decided on whether or not there exists a directed mmp so the "$\mathbb Q$-factorial model" obtained by the above method is smooth. It is possible that all choices lead to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.
EDIT: Removed previous statement about the reverse direction as that was not true as stated. At this time I am not sure how to fix that statement. I replaced it with a different one that I know how to prove, but the margin is not wide enough to include a proof, so it is just stated without proof. Sorry. However, the current statement is probably pretty close to what one might hope for.<|endoftext|>
TITLE: Strengthening the induction hypothesis
QUESTION [49 upvotes]: Suppose you are trying to prove result $X$ by induction and are getting nowhere fast.  One nice trick is to try to prove a stronger result $X'$ (that you don't really care about) by induction.  This has a chance of succeeding because you have more to work with in the induction step.  My favourite example of this is Thomassen's beautiful proof that every planar graph is 5-choosable.  The proof is actually pretty straightforward once you know what you should be proving.  Here is the strengthened form (which is a nice exercise to prove by induction),
Theorem.
Let $G$ be a planar graph with at least 3 vertices such that every face of $G$ is bounded by a triangle (except possibly the outer face).  Let the outer face of $G$ be bounded by a cycle $C=v_1 \dots v_kv_1$.  Suppose that $v_1$ has been coloured 1 and that $v_2$ has been coloured 2.  Further suppose that for every other vertex of $C$ a list of at least 3 colours has been specified, and for every vertex of $G - C$, a list of at least 5 colours has been specified.  Then, the colouring of $v_1$ and $v_2$ can be extended to a colouring of $G$ with the specified lists.  
Question 1. What are some other nice examples of this phenomenon?
Question 2. Under what conditions is the strategy of strengthening the induction hypothesis likely to work?

REPLY [2 votes]: Euler's formula $V-E+F=2$ for convex polytopes has a trivial inductive proof after we replace polytopes to connected planar graphs (remove edges until you get a tree), and even more trivial if we replace 2 to $1+C$, where $C$ is a number of connected components of arbitrary planar graph (remove edges until you get a graph without edges). Note that such operations on a set of polytopes are not well defined. This is probably a reason why Euler did not prove this fact.<|endoftext|>
TITLE: Is Tor always torsion?
QUESTION [10 upvotes]: Question: Is the following statement true?

Let $R$ be an associative, commutative, unital ring. Let $M$ and $N$ be $R$-modules. Let $n\geq 1$. Then $Tor_n^R(M,N)$ is torsion.

By " $Tor_n^R(M,N)$ is torsion" I mean that every of its elements is a torsion element. Maybe I want to assume that $R$ is an integral domain.
Remark: The above statement is true if $R$ is a principal ideal domain (then $Tor_n^R$ vanishes for $n\geq 2$) and $M$ and $N$ are finitely generated (then we can apply the structure theorem).

REPLY [2 votes]: Yes. Set $K = \mathrm{Frac} \ R$.
Lemma: Let $\ldots \to C_2 \to C_1 \to C_0$ be a complex of $R$-modules. Suppose that $C^{\bullet} \otimes_R K$ is exact (but not necessarily surjective at $C_0$). Then $H_k(C_{\bullet})$ is $R$-torsion for $k>0$.
Proof: Let $v \in C_k$ with  $dv=0$. So $d(v \otimes 1)=0$. By the exactness of $C^{\bullet} \otimes_R K$, there is $u \in C_{k+1} \otimes_R K$ with $du=v$. Write $u=\sum w_i \otimes (f_i/g_i)$, with $f_i/g_i \in K$ and $w_i \in C_{k+1}$. Set $g=\prod g_i$ and $w=\sum \left( \prod_{j \neq i} g_j \right) f_i w_i$. Then $dw=gv$, so $[v]$ is $g$-torsion in $H_k(C_{\bullet})$. QED
Take resolutions $A_{\bullet} \to M$ and $B_{\bullet} \to N$ by free $R$-modules. Then $\mathrm{Tor}_{\bullet}(M,N)$ is the homology of the complex formed by collappsing the double complex $A_{\bullet} \otimes_R B_{\bullet}$. Note that $\left( A_{\bullet} \otimes_R B_{\bullet} \right) \otimes_R K \cong (A_{\bullet} \otimes_R K) \otimes_K (B_{\bullet} \otimes_R K)$.
Since $A^{\bullet}$ is exact, so is $A^{\bullet} \otimes_R K$. Thus $A_{\bullet} \otimes_R K$ breaks up as a direct sum of complexes of the form $\ldots \ldots 0 \to K \to K \to 0 \ldots$, and the complex $\ldots \to 0 \to K$, with the $K$ in the last position. (This uses the Axiom of Choice; I suspect you should be able to avoid it, but I don't see how right now.)  The complex $B \otimes_R K$ breaks up into pieces of the same kind. So the double complex breaks up into squares $\begin{smallmatrix} K & \to & K \\ \downarrow & & \downarrow \\ K & \to & K \end{smallmatrix}$, vertical strips $\begin{smallmatrix} K \\ \downarrow \\ K \end{smallmatrix}$, horizontal strips $\begin{smallmatrix} K & \to & K \end{smallmatrix}$ and, in position $(0,0)$, some isolated copies of $K$.
Only summands of the last type contribute to the cohomology of the double complex, so the double complex obeys the hypotheses of the lemma and we are done.<|endoftext|>
TITLE: Largest pair of homometric Golomb rulers?
QUESTION [11 upvotes]: A Golomb ruler is a set of $n$ integers that determines $\binom{n}{2}$ distinct differences.
Two sets are homometric if they determine the same (multiset) of differences.
For example,
$$\{0,1,4,10,12,17\} \;,\; \{0,1,8,11,13,17\}$$
are a homometric pair of Golomb rulers, determining 15 distinct differences (excluding only 
$\{14,15\}$).
Although there are arbitrarily large Golomb rulers, and arbitrarily large pairs of homometric
sets (allowing multipiclity), it is unclear (from my searching) if there are arbitrarily large pairs of Golomb rulers.
In the 1994 paper, "There Are No New Homometric Golomb Ruler Pairs with 12 Marks or Less,"
the authors say that they "are divided on whether any additional nontrivial homometric rulers are to be found." The nice paper "Reconstructing sets from interpoint distances" does not seem to
attend to the special case where all distances are distinct.  Nor does the Rosenblatt-Seymour paper "The structure of homometric sets" (inferring from secondary sources—I don't have that paper yet).
My question is: What is the largest pair of homometric Golomb rulers known? Is it still open whether or not there are arbitrarily large pairs?  Thanks for any pointers on this topic!
Addendum. Thanks to Yota Otachi for uncovering the 2007 paper by Bekir and Golomb he cites below.
As he says, it proves that there are no homometric Golomb rulers of more than six marks.
The proof uses Golomb's "polynomial method."

REPLY [7 votes]: I think "There are no further counterexamples to S. Piccard's theorem" by A. Bekir and S.W. Golomb is the answer. I skimmed the paper and if I understand correctly they proved that there are no homometric Golomb ruler pairs with 7 marks or more.<|endoftext|>
TITLE: The probabilistic method - reference to less challenging questions
QUESTION [13 upvotes]: I am teaching a course in combinatorics and large part of it is dedicated to the probabilistic method especially in the case of graphs. The course is an undergraduate level (almost none of the students will pursue a PhD in mathematics).
Last year when I though the course I had very hard time finding appropriate questions for the students. So for instance problems in Alon and Spencer’s book are way too hard.  I am looking for textbooks or online notes which contain less challenging problems. For example, problems which are direct implication of a method or problems which are broken down to small parts, etc. If the reference actually has solutions or hints, that will be a nice bonus.  
Clarification: After seeing Ravi’s answer I think I should be clearer. Olympiads problems are probably the complement of what I am looking for. They tend to be elementary but tricky. I, on the other hand, look for problems where the method might not be very elementary, but the answer is direct. So for instance, near the end of the course I teach them the Local Lemma. I would like to find 2-3 problems where I can more or less just ask the students to plug in the numbers and use the lemma to deduce the existence of some graph. I then will be happy to give them something more sophisticated but maybe with some hints or broken down to small parts.

REPLY [5 votes]: We dealt with the probabilistic method in our undergrad randomized algorithms class at Berkeley, and used Mitzenmacher's book.  Chapter 6 is about the probabilistic method and has a bunch of exercises that don't require as much computational yoga as Olympiad problems or Alon and Spencer.  
The chapter is mostly about combinatorics, and not about algorithms.  For instance, one of the questions in Chapter 6 is to show if $4 \binom{k}{2} \binom{n}{k-2} 2^{1-\binom{k}{2}} \leq 1$, then it's possible to two-color the edges of $K_n$ so there's no monochromatic $K_k$ subgraph, which is a direct application of Lovasz.<|endoftext|>
TITLE: Monotonic maximal chains in a Coxeter group
QUESTION [7 upvotes]: Let $(W, S)$ be a Coxeter system, and let $T = \bigcup_{w \in W, s \in S} wsw^{-1}$.  Associated to every element $t \in T$ is a unique positive root $\alpha_t \in \Phi^{+}$ considered as a vector in the standard geometric representation $V$ of $W$.  A total order on $T$ is a reflection order if, whenever $\alpha_{t_1} < \alpha_{t_2}$, it follows that $\alpha_{t_1} < x \alpha_{t_1} + y \alpha_{t_2} < \alpha_{t_2}$ whenever the middle term is a positive root with $x > 0, y > 0$.  (See, for example, Bjorner and Brenti's book.)
Fix a reflection order and let $[u, v]$ be a Bruhat interval.  A maximal chain $u = w_0 \to w_1 \to ... \to w_m = v$ in the Bruhat order is what I'll call monotonic if $w_i w_{i-1}^{-1} > w_{i+1} w_i^{-1}$ in the reflection order.  
There is a nonrecursive formula for the Kazhdan-Lusztig polynomials $P_{u,v}(q)$ which implies that $P_{u,v}(0)$ is equal to the number of monotonic maximal chains in $[u, v]$.  This number is known by other means to be equal to $1$, so I know that there is a unique monotonic maximal chain; however, I can't prove this directly.  So far all I've been able to do is use the greedy algorithm to prove that at least one monotonic maximal chain exists.
Does anyone have a direct proof of this fact?  
Edit:  No progress, but now I have a more general conjecture which I no longer know by other means is true.  Fix a sequence $a_1, ... a_m$ of odd positive integers such that $\sum_i a_i = \ell(v) - \ell(u)$.  Then there exists at most one monotonic chain (not necessarily maximal) such that $w_i w_{i-1}^{-1} \in T$ and such that $\ell(w_{i-1}) - \ell(w_i) = a_i$.

REPLY [6 votes]: Hi, you may want to try to peruse these two papers:
Dyer, M. J. Hecke algebras and shellings of Bruhat intervals. Compositio Math. 89 (1993), no. 1, 91--115. 
Dyer, M. J. Hecke algebras and shellings of Bruhat intervals. II. Twisted Bruhat orders. Kazhdan-Lusztig theory and related topics (Chicago, IL, 1989), 141--165, Contemp. Math., 139, Amer. Math. Soc., Providence, RI, 1992.
Dyer proves that Bruhat intervals are EL-shellable, and this gives the answer to your first question. As to your more general question, there may be answers to those as well (indirectly of course).<|endoftext|>
TITLE: Retraction of a Riemannian manifold with boundary to its cut locus
QUESTION [11 upvotes]: This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.
Recall that the cut locus of a set $S$ in a geodesic space $X$
is the closure of the set of all points $p \in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$.
http://en.wikipedia.org/wiki/Cut_locus
A simple lemma shows that, for a disk $D^2$ with a Riemannian metric and piecewise smooth generic boundary, the cut locus of $D^2$ with respect to its boundary is a tree. 
A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is white.
http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf
For an ellipse on the 2-plane, the tree is the segment that joins its focal points.
More generically for a Riemannian manifold $M^n$ with boundary, the cut locus of $\partial M$ should be a deformation retract of $M$.  (I guess it is a $CW$ complex of dimension less than $n$.) To prove this lemma, notice that $M^n\setminus\operatorname{cut-locus}(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.
I wonder if this lemma has a name or maybe is contained in some textbook on Riemannian geometry?

REPLY [6 votes]: The 'simple lemma' certainly isn't always true.  Let's consider just open disks in the Euclidean plane with a flat metric. Fremlin (Proc. LMS, 1997) calls the set of points without a unique nearest boundary point the "skeleton".  He gives an example where the skeleton is somewhere dense (that is, its closure, which you call the "cut locus", has nonempty interior).  The skeleton will always be an R-tree.<|endoftext|>
TITLE: What is the right notion of self-dual (two-dimensional) percolation in R^4?
QUESTION [27 upvotes]: For a lattice in $\mathbb{R}^2$, if we include each edge independently with probability $p$ (i.e. bond percolation), it is well known that there is a critical probability $0 < p_c < 1$ depending on the lattice, such that if $ p > p_c$ then there is almost surely an infinite component (i.e. with probability $1$), and if $p < p_c$ then almost surely all components are finite.
The square lattice is special because it is self-dual.  This allows one to guess that in this particular case $p_c = 1/2$, although the proof of this fact, originally due to Harry Kesten, is fairly subtle.
One wonders if there is an analogous fact for "two-dimensional" percolation in $\mathbb{R}^4$.  In particular, consider the hypercubic lattice as a $1$-dimensional cubical complex, and then attach two-dimensional square "plaquettes" independently with probability $p$.  Since the hypercubic lattice is self-dual, we might expect some kind of phase transition when $p=1/2$.  My hope is that topology would give us the right language to talk about this.
One possibility, and perhaps the nicest answer I can imagine, was suggested to me by Russ Lyons:  If $p > 1/2$ then there are almost surely embedded planes, and if $p< 1/2$ then there are almost surely not.  Here the embedded plane should be the union of closed $2$-cells.  It turns out that once $p > 1/2$ in $\mathbb{R}^2$, there are not only infinite components, which implies embedded half-lines, but embedded lines as well.
Another possibility is that once $p > 1/2$ there are almost surely bounded $1$-dimensional cycles in homology $H_1 ( - , \mathbb{Z} / 2\mathbb{Z})$, which are boundaries of unbounded two-dimensional complexes, and that when $p < 1/2$ there are almost surely not.  (I believe that this is similar to the "plaquette percolation" studied by Jennifer Chayes in her Ph.D. thesis, but the work I know of is in $\mathbb{R}^3$.)
I am not an expert in percolation theory, and would really like to know if anyone knows about any previous work in this area, or any standing conjectures.  (Or is there any obvious reason that either of the possibilities I suggested could be ruled out?)

REPLY [5 votes]: The paper "PLAQUETTES, SPHERES, AND ENTANGLEMENT" by GEOFFREY R. GRIMMETT AND ALEXANDER E. HOLROYD does not deal with the self-dual problem, but nevertheless could be of interest for you. If shows that "The high-density plaquette percolation model in d dimensions contains a surface that is homeomorphic to the (d − 1)sphere and encloses the origin." Here's the link: http://www.statslab.cam.ac.uk/~grg/papers/sphere8.pdf<|endoftext|>
TITLE: Finite groups with elements of order n
QUESTION [14 upvotes]: Consider a finite group where all elements have the same order $n$.
What could be said about such groups?
For $n=2$ it could be proved that such group is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$.
Could it be somehow generalized on case $n>2$?
EDIT: Surely the identity has order 1, so we have to exclude it.

REPLY [6 votes]: This is not really on point, but it may be of some interest. The finite groups of prime exponent are exactly those finite groups that are the set-theoretic union of a collection of pairwise trivially intersecting proper subgroups, all of which have the same order. (See my paper in the Pacific J, of Math. 49, 1973.)<|endoftext|>
TITLE: Nonconvex manhole covers
QUESTION [32 upvotes]: One common reason given for the circularity of manhole covers is that they can't fall through the manhole. For convex manhole covers, this property is equivalent to having constant width — if you have different widths, just orient the cover so that the shorter width slides through the larger one. Since convex polygons can't have constant width, this rules them out for manhole covers. However, for nonconvex shapes, a longer width does not necessarily give you a longer hole.
Is it possible to have a nonconvex polygon that cannot be moved through a hole of the same shape?
Some Clarifications:

The hole has zero thickness.
I was thinking of the polygon being simply connected. But if this matters I would like to know the answer both ways.

REPLY [5 votes]: What about a square with four thin slices taken out of it?
 (source)<|endoftext|>
TITLE: Feit-Thompson theorem: the Odd order paper
QUESTION [42 upvotes]: For reference, the Feit-Thompson Theorem states that every finite group of odd order is necessarily solvable. Equivalently, the theorem states that there exist no non-abelian finite simple groups of odd order. 
I am well aware of the complexity and length of the proof. However, would it be possible to provide a rough outline of the ideas and techniques in the proof? More specifically, the sub-questions of this question are: 

Are the techniques in this proof purely group-theoretic or are techniques from other areas of mathematics borrowed? (Such as, for example, other branches of algebra.) In the same vein, how great an influence do the techniques (if any) from number theory and combinatorics have on the proof? (Here "combinatorics" is of course not very specific. I should emphasize that I mean "tools from combinatorics that are pure and solely derived from techniques within the area of combinatorics and that do not require "deep" group theory to derive". Similarly for "number theory".)
What sorts of "character-free" techniques and ideas exist in the proof? Does a character-free proof of this result exist? (Since I suspect the answer to the latter is in the negative, I am primarily interested in an answer to the former.)
What are the underlying "intuitions" behind the proof? That is, how does one come up with such a proof, or at least, certain parts of it? This is a rough question of course; "coming up" with things in mathematics is very difficult to describe. However, since the argument is so long, I suspect some sort of inspiration must have driven the proof.
I have observed in group theory that many arguments naturally divide into "cases" and often the individual cases are easy to tackle and the arguments naturally "flow". Of course, here I speak of arguments whose lengths are no more than a few pages. Does the proof of the Feit-Thompson theorem share the same "structure" as smaller proofs, or is the proof structurally unique?
How often do explicit "elementwise computations" arise in the proof?
Is there any hope that one day someone might discover a considerably shorter proof of the Feit-Thompson Theorem? For example, would the existence of a proof of this theorem less than 50 or so pages be likely? (A proof making strong use of the classification of finite simple groups, or any other non-trivial consequence of the Feit-Thompson Theorem, does not count.) If not, why is it so difficult in group theory to provide more concise arguments?

While I have Gorenstein's excellent book entitled Finite Groups at hand, I did not go far enough (when I was reading it) to actually get into the "real meat" of the discussion of the Feit-Thompson theorem; that is, to actually get a sense of the mathematics used to prove the theorem. Nor do I intend to do so in the near future. (Don't get me wrong, I would be really interested to see this proof, but it seems too much unless you intend to research finite group theory or a related area.)
Thank you very much for any answers. I am aware that some aspects of this question are imprecise; I have tried my best to be as clear as possible in some cases, but there might still be possible sources of ambiguity and I apologize if they are. (If there are, I would appreciate it if you could try to look for the "obvious interpretation".) Also, I have a relatively strong background in finite group theory (but not a "research-level" background in the area) so feel free to use more complex group-theoretic terminology and ideas if necessary, but if possible, try to give an exposition of the proof that is as elementary as possible. Thanks again!

REPLY [10 votes]: I won't presume to attempt a precis of the Feit-Thompson proof. But I would suggest that your
question about the hope of finding a much shorter proof is impossible to answer meaningfully.
The current answer, backed up by almost 50 years of recent history,  is probably ``with currently available techniques, there appears to be little prospect of any dramatic shortening of the
length of the proof of the odd order theorem." It should also be remembered that many of the
currently available accepted techniques of finite group theory were developed to attack this
problem, and proved later to be very powerful in a wider context. Many of the techniques are
such an integral part of the weaponry of many modern group theorists that they implicitly impose
an inevitability and naturality to the structure of the proof of the odd order theorem, complex and forbidding though the details are. But had the question been asked, say in 1955, "Is there any prospect of proving  the solvability of finite groups of odd order in the near future?", the answer likely to be given at the time can only be a matter of speculation (for most of us at any rate), but with the benefit of hindsight we can see at present that to make the prospect of such a proof a reality, many new and innovative techniques had to be developed, and profound new insights brought to bear. 
However, it would be a rash mathematician (and one who took little account of the history of the 
subject) who would  pronounce it impossible to find a significantly shorter proof at some point in the future. It might be a safer bet to suggest that a significantly shorter proof would require
some genuinely new insights and ideas, but even a statement such as that might eventually be 
proved to be presumptuous.<|endoftext|>
TITLE: Checking whether given binary operation is a group operation
QUESTION [14 upvotes]: Given a binary function $f: [1..n] \times [1..n] \to [1..n]$ how to check that this operation is a group operation on $[1..n]$?
It's obvious that this can be done in $O(n^3)$ time just by checking all group properties. The most time-expensive property is associativity. Also it's clear that it could not be done faster than $O(n^2)$ time since you should at least examine all values $f(i,j)$.
The question is if there is any algorithm to solve this problem in time faster than $O(n^3)$?

REPLY [13 votes]: See this entry - http://rjlipton.wordpress.com/2010/06/03/an-amplification-trick-and-stoc-2010/ for a nice discussion on the question you have posed and other related ones. The article also has links to the original papers.

REPLY [10 votes]: Indeed, S. Rajagopalan and L. Schulman prove that this can be done in $O(n^2\log n)$ time in the paper "Verifying identities" from the proceedings of the 37th annual symposium on foundations of computer science. Their algorithm is applied to operations with cancellative properties, but checking if a binary operation has the cancellative property can be done in $O(n^2)$ time.<|endoftext|>
TITLE: Problems known to be in both NP and coNP, but not known to be in P
QUESTION [43 upvotes]: One such problem I know is integer factorization.
What are other interesting cases?

REPLY [4 votes]: Since you mentioned integer factoring, an analogous problem is the discrete log problem.
Given the cyclic group $G = \mathbb{Z}^*_p$ for a prime $p$ and any generator $g$ of $G$ along with another $h \in G$ (which will also be a generator), the discrete log asks to find $x \in \mathbb{Z}_{p-1}$ such that $g^x = h$. To convert to decision problem, let $k$ be any positive integer $< \log_2(p-1)$. The problem is then: 
Is the $k^{th}$ bit of $x$ true? 
This is also in NP $\cap$ coNP but not known to be in P.<|endoftext|>
TITLE: Is the Riemann Hypothesis equivalent to a $\Pi_1$ sentence?
QUESTION [56 upvotes]: 1) Can the Riemann Hypothesis (RH) be expressed as a $\Pi_1$ sentence?

More formally,

2) Is there a $\Pi_1$ sentence which is provably equivalent to RH in PA?


Update (July 2010):
So we have two proofs that the RH is equivalent to a $\Pi_1$ sentence. 

Martin Davis, Yuri Matijasevic, and Julia Robinson,
"Hilbert's Tenth Problem. Diophantine Equations: Positive Aspects of a Negative Solution", 1974.
Published in "Mathematical developments arising from Hilbert problems", Proceedings of Symposium of Pure Mathematics", XXVIII:323-378 AMS.
Page 335
$$\forall n >0 \ . \ \left(\sum_{k \leq \delta(n)}\frac{1}{k} - \frac{n^2}{2} \right)^2 < 36 n^3 $$

2.
Jeffrey C. Lagarias, "An Elementary Problem Equivalent to the Riemann Hypothesis", 2001
$$\forall n>60 \ .\  \sigma(n) < \exp(H_n)\log(H_n)$$
But both use theorems from literature that make it difficult to judge if they can be formalized in PA. The reason that I mentioned PA is that, for Kreisel's purpose, the proof should be formalized in a reasonably weak theory. So a new question would be:

3) Can these two proofs of "RH is equivalent to a $\Pi_1$ sentence" be formalized in PA?


Motivation:
This is mentioned in P. Odifreddi, "Kreiseliana: about and around George Kreisel", 1996, page 257. Feferman mentions that when Kreisel was trying to "unwind" the non-constructive proof of Littlewood's theorem, he needed to deal with RH. Littlewood's proof considers two cases: there is a proof if RH is true and there is another one if RH is false. But it seems that in the end, Kreisel used a $\Pi_1$ sentence weaker than RH which was sufficient for his purpose.
Why is this interesting?
Here I will try to explain why this question was interesting from Kreisel's viewpoint only.
Kreisel was trying to extract an upperbound out of the non-constructive proof of Littlewood. His "unwinding" method works for theorems like Littlewood's theorem if they are proven in a suitable theory. The problem with this proof was that it was actually two proofs: 

If the RH is false then the theorem holds.
If the RH is true then the theorem holds.

If I remember correctly, the first one already gives an upperbound. But the second one does not give an upperbound. Kreisel argues that the second part can be formalized in an arithmetic theory (similar to PA) and his method can extract a bound out of it assuming that the RH is provably equivalent to a $\Pi_1$ sentence. (Generally adding $\Pi_1$ sentences does not allow you to prove existence of more functions.) This is the part that he needs to replace the usual statement of the RH with a $\Pi_1$ statement. It seems that at the end, in place of proving that the RH is $\Pi_1$, he shows that a weaker $\Pi_1$ statement suffices to carry out the second part of the proof, i.e. he avoids the problem in this case.
A simple application of proving that the RH is equivalent to a $\Pi_1$ sentences in PA is the following: If we prove a theorem in PA+RH (even when the proof seems completely non-constructive), then we can extract an upperbound for the theorem out of the proof. Note that for this purpose, we don't need to know whether the RH is true or is false.
Note: 
Feferman's article mentioned above contains more details and reflections on "Kreisel's Program" of "unwinding" classical proofs to extract constructive bounds. My own interest was mainly out of curiosity. I read in Feferman's paper that Kreisel mentioned this problem and then avoided it, so I wanted to know if anyone has dealt with it.

REPLY [33 votes]: I realized that none of the answers present what I consider to be the most straightforward $\Pi^0_1$ expression for the Riemann hypothesis, namely bounds on the error term in the prime number theorem. I will write it in terms of Chebyshev’s $\psi$ function as I find it more natural, but it works for $\pi$ just the same. The following are equivalent:

The Riemann hypothesis.
$\psi(x)-x=O(x^{1/2+\epsilon})$ for all $\epsilon>0$.
$|\psi(x)-x|\le\frac1{8\pi}\sqrt x\log^2 x$ for all $x\ge74$.

The equivalence of 1 and 2 is classical, the explicit bound in 3 is due to Schoenfeld. Now, the large leeway between 2 and 3 allows one to write the bound as a $\Pi^0_1$ sentence, even though we cannot compute exactly all the logarithms involved: let $\mathrm{psi}(n)$, $\mathrm{sqrt}(n)$, and $\mathrm l(n)$ be computable functions that provide rational approximations within distance $1$ of $\psi(n)$, $\sqrt n$, and $\log n$, respectively. Then RH is equivalent to
$$\forall n\,|\mathrm{psi}(n)-n|\le42+\mathrm{sqrt}(n)\,\mathrm l(n)^2.$$
The beauty of this is not only that it is in line with the form of RH most likely to be useful in elementary number theoretic arguments, but perhaps more importantly, it easily generalizes to extensions of the RH to other $L$-functions.
For a specific formulation, Section 5.7 of Iwaniec and Kowalski’s Analytic number theory states for a large class of $L$-functions (basically, functions in the Selberg class with a polynomial Euler product; the assumptions are somewhat negotiable, in particular I’m confident one can eliminate the Ramanujan–Petersson hypothesis at the expense of somewhat worse bounds) the equivalence of

The Riemann hypothesis for $L(s)$.
$\psi_L(x)-n_Lx=O(x^{1/2+\epsilon})$ for all $\epsilon>0$.
$|\psi_L(x)-n_Lx|\le c\sqrt x\,(\log x)\log(x^dq_L)$.

Here $c$ is an absolute constant that can (in principle) be extracted from the proof, $d$ is the degree of the Euler product, $n_L$ is the order of the pole of $L(s)$ at $s=1$, $q_L$ is a conductor of sorts, and
$$\psi_L(x)=\sum_{n\le x}\Lambda_L(n),$$
where $\Lambda_L(n)$ is a “von Mangoldt” function of $L$ extracted from the expansion of the logarithmic derivative of $L$ as a Dirichlet series:
$$-\frac{L'(s)}{L(s)}=\sum_{n=1}^\infty\Lambda_L(n)\,n^{-s}.$$
The upshot is that the RH for a class of $L$-functions is $\Pi^0_1$, provided the class is “recursively enumerable”: we can parametrize the class as $L(s,a)$ where the $a$’s are finite objects (including basic data like $d,n_L,q_L$) in such a way that the set of valid $a$’s is r.e., and given $a$, $n$, and $\epsilon>0$, we can compute an approximation of $\Lambda_L(n)$ within distance $\epsilon$ (or equivalently, if we can approximately compute terms of the Euler product).
For example, each of the following can be expressed as a $\Pi^0_1$ sentence:

The RH for Dirichlet $L$-functions.
The RH for Dedekind zeta-functions.
The RH for Hecke $L$-functions.

(The first two classes can be enumerated in a straightforward way. Finite-order Hecke characters are also easily enumerable, as ray class groups are finite and computable. The case of general Hecke characters needs a bit more work, but basically, one can enumerate a basis of suitably normalized infinity types using an effective version of Dirichlet’s unit theorem.)
I can’t tell (but would be interested to hear from someone more knowledgeable) whether the RH for standard automorphic $L$-functions is also $\Pi^0_1$, that is, whether these functions are recursively enumerable. (There are certainly only countably many up to normalization, and polynomially many of bounded analytic conductor, so conceivably this may be true.)<|endoftext|>
TITLE: Characteristic surface for systems of PDE
QUESTION [5 upvotes]: Despite the title, this is probably actually a question in linear algebra or algebraic geometry. Let me write the question(s) first, before I explain the background. 
Problems
Let $h^{\mu\nu}_{ij}$ represent a map from $\mathbb{R}^4\otimes\mathbb{R}^4$ to $\mathbb{R}^2\otimes\mathbb{R}^2$ (here $\mu\nu$ are indices in the $\mathbb{R}^4$ directions, and $ij$ are in $\mathbb{R}^2$ directions). We shall assume that $h$ is symmetric swapping the $\mu,\nu$ indices and also symmetric swapping the $i,j$ indices. Then for any $\xi_\mu$ in $\mathbb{R}^4$, the object $H(\xi):= h^{\mu\nu}_{ij}\xi_\mu\xi_\nu$ is a symmetric bilinear form on $\mathbb{R}^2$. We say that $\xi$ is characteristic if $H(\xi)$ is degenerate. In other words, $\xi$ is characteristic if $\det(H(\xi)) = 0$. 
Since $H(\xi)$ is quadratic in $\xi$, the determinant is an 8th degree homogeneous polynomial in $\xi$. Furthermore, by definition if $\xi$ is characteristic, so is $-\xi$. Observe also that in general the characteristic set will have multiple sheets. 
Question 1, very specific
Does there exist an $h$ such that the characteristic surface is given by $\xi_1^4 + \xi_2^4 + \xi_3^4 - \xi_4^4 = 0$? 
Question 2, slightly more general
In general are there any obstructions to having a sheet of the characteristic surface described by the zero set of an irreducible (over the reals) polynomial of degree strictly higher than 2?
Question 3, even more general
What if we relax the condition on $h$ so that it is a map from $\mathbb{R}^m\otimes\mathbb{R}^m$ to $\mathbb{R}^d\otimes\mathbb{R}^d$ with the same symmetric properties. Define $H(\xi)$ analogously. Can a sheet of the characteristic surface have algebraic degree more than 2?
I'm particularly interested in concrete examples.  
Motivation
This comes from the study of hyperbolic systems partial differential equations. Recall that a second degree partial differential equation 
$$ h^{\mu\nu}_{ij} \partial_\mu\partial_\nu u^i = 0 $$
is said to be strictly hyperbolic in the direction of $e_\mu$ if the characteristic polynomial (a polynomial in $t$) $\det(H(x_\mu - te_\mu))$ is hyperbolic for any fixed $x_\mu$ linearly independent from $e_\mu$ and that the roots are distinct (it is enough that the second condition only holds for all by finitely many $x_\mu$ modulo $e_\mu$). 
The classical examples for strictly hyperbolic systems (wave equation, crystal optics, etc) all have the sheets of the characteristic surfaces being linearly transformed versions of the standard quadratic double cone: in other words there exists a basis of $\mathbb{R}^m$ such that a sheet is given by $\sum_{i = 1}^{m-1} e_i^2 - e_m^2 = 0$. 
I am guessing that for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity (though please let me know if I am wrong). 
So my question is: is it possible for a non-strictly hyperbolic system (but one still hyperbolic) where some of the sheets have higher multiplicity to not come from "the square of a quadratic sheet" but from a genuinely quartic or higher polynomial? 
Postscript
Please do let me know if you need any clarification on my question. Thanks.
Update
I struck out question 1 for the following reason: in view of my motivation from hyperbolic polynomials arising from second order PDEs, the answer is negative. The argument is thus: for a hyperbolic system of PDEs, the time-like direction $\xi_4$ should have its corresponding $h^{44}_{ij}$ negative definite, whereas the space-like directions $\xi_1,\xi_2,\xi_3$ should have their corresponding $h^{aa}_{ij}$ positive definite. A simple computation shows that the coefficient to the $\xi_a^4$ term in $\det H(\xi)$ must be $\det h^{aa}_{ij}$. If the target space is two dimensional, both positive definite and negative definite matrices have positive determinants. So for any hyperbolic polynomial arising from a second order system of PDEs, the coefficients for $\xi_a^4$ must be positive.

REPLY [2 votes]: There is a natural example where an irreducible component of the symbol, a hyperbolic polynomial, is of degree higher than $2$. It occurs in compressible magnetohydrodynamics (MHD). It is a coupling of the Euler system of compressible, inviscid gas dynamics, with Maxwell's equations, when the magnetic part dominates the EM field. It consists of $8$ conservation laws, governing mass density ($1$), momentum ($3$), energy ($1$) and magnetic field ($3$). When linearizing around a uniform state, we obtain a linear, first-order hyperbolic system. The determinant of the symbol splits into the square of a linear factor (pure transport), a quadratic factor (Alfvén mode) and an irreducible factor of degree $4$. The roots of the latter are the velocities in direction $\xi$ of the fast/slow backward/forward waves.
You can also get a complex structure in nonlinear models of EM fields. See here.<|endoftext|>
TITLE: The Fundamental Theorem of Calculus in Lebesgue Theory
QUESTION [13 upvotes]: I am interested to what extent the famous identity
$$
\int_a^b f'(x) \ dx=f(b)-f(a)
$$
is true for a function $f:[a,b]\to \mathbb C$ continuous on $[a,b]$ and differentiable on $(a,b)$. One famous easy case of this problem is where $f'$ is continuous. In the above identity, the integral is with respect to Lebesgue measure on $\mathbb R$.
I have proven so far that $f'$ is always measurable on $(a,b)$ and that if $f'$ is bounded on $(a,b)$ then the result holds. The proof was reasonably elementary, making heavy use of the mean value theorem and the so-called bounded convergence theorem.
I felt that my condition was an artifact of the proof, as the bounded convergence theorem is considerably weaker than the dominated convergence theorem and its strengthened forms.
So does anyone know of a strengthened version of this result, or perhaps even a full description of all differentiable functions such that the above identity holds?
Thank you for your time and effort.

REPLY [3 votes]: N.L. Carothers's Real Analysis has a fairly good bit devoted to this in Chapter 20 "Differentiation" but unfortunately the relevant part of the book is not on Google books.
Carothers's exposition focuses entirely on the real line allowing for more focus on what can be proven in this single case while forgoing questions about abstract measures.<|endoftext|>
TITLE: Decomposing a large colimit as a pushout of smaller colimits
QUESTION [10 upvotes]: I would like to find a reference in the literature for the following result.  I have it on high authority that it isn't in 'Categories for the Working Mathematician' and I can't find it in Borceux's handbook.  It's a result that I'm confident is true (at least when stated correctly) and is probably second nature to category theorists.  I however am writing for group theorists and so want to reference results thoroughly.
I have a functor $F:\mathcal{C} \rightarrow \mathcal{D}$.  The target category $\mathcal{D}$ is cocomplete.  The source category $\mathcal{C}$ is finite and can be decomposed as the pushout of smaller categories $\mathcal{C}_1\leftarrow\mathcal{C}_0\rightarrow\mathcal{C}_2$.
The functors from these into $\mathcal{D}$ are denoted $F_1,F_0$ and $F_2$ respectively.
I need to take the colimit of $F$ and I think that it can be taken to be the pushout of
$\text{colim}F_1\leftarrow\text{colim}F_0\rightarrow\text{colim}F_2$.
Obviously if $\mathcal{C}$ were constructed from a different colimit rather than a pushout one might expect an analogous result.
Giving a proof is an option, but would be out of context with the rest of the paper and probably consigned to an unread appendix.  Or I could just quote it without proof.  Help, or just opinions would be very welcome.

REPLY [7 votes]: I'd agree that the result is true, and “well-known” to category-theorists!  Unfortunately I don't know a specific reference, but my best guess would be something like Kelly’s “Elements of Enriched CT”, which proves lots of useful things about (co)ends and weighted (co)limits, which specialise to lots of useful things about limits.  It's also hard not to wonder about the legendary treatise of Chevalley on “all possible properties of limits” that got lost in the mail...
If you can't find a reference, though, the proof can certainly be made pretty short — I used 6 well-spaced or 2 cramped lines, and it can probably be compressed further...
[this was meant to be just a comment, but it got a bit too long]
$\newcommand{\C}{\mathbf{C}} \newcommand{\D}{\mathbf{D}} \DeclareMathOperator{\colim}{colim}$
Edit: For clarification, the precise statement I had in mind is that
$$\colim_{I}\ (\colim_{\C_i}\ F_i)\ \cong\ \colim_{\left( \colim_{I} \C_i \right)} [F_i]_{i \in I}$$
where $I$ is a small cat, $\C_i$ is an $I$-indexed diagram of small cats, $F_i : \C_i \to \D$ is a co-cone of functors, $[F\_i]\_{i \in I}$ denotes the induced cotuple functor $\colim_{I} \C\_i \to \D$, and all the relevant colimits exist in $ \D$.<|endoftext|>
TITLE: Representations of finite Coxeter groups
QUESTION [5 upvotes]: What is reference for complex irreducible representations of Hecke algebra of finite Coxeter groups (say generic case q =1)? I am interested in knowing its Wedderburn decomposition. So want explicit information regarding the number of irreducible representations (and parameterization, if any) with their multiplicities. I looked at Kazhdan-Lusztig's paper but not able to get required information. 
For example: In case of symmetric group $S_n$, these representations are parametrized by all partitions of $n$ with multiplicities equal to number of standard tableaux.

REPLY [9 votes]: There are many relevant papers, but the most convenient book to consult is:
MR1778802 (2002k:20017) 20C15 (20C08 20F55),
Geck, Meinolf (F-LYON-GD); Pfeiffer,G¨otz (IRL-GLWY)
Characters of finite Coxeter groups and Iwahori-Hecke algebras.
London Mathematical Society Monographs. New Series, 21.
The Clarendon Press, Oxford University Press, New York, 2000. xvi+446 pp.   
See also their earlier paper:   MR1250466 (94m:20018) 20C15,
Geck, Meinolf (D-AACH-DM); Pfeiffer,G¨otz (D-AACH-DM),
On the irreducible characters of Hecke algebras.
Adv. Math. 102 (1993), no. 1, 79–94.
In the last chapter of my book Reflection Groups and Coxeter Groups (Cambridge, 1990) there is a brief summary of earlier work done on irreducible 
representations or character tables of finite Coxeter groups and their Iwahori-Hecke algebras.    The 1979 Kazhdan-Lusztig paper was partly motivated by earlier papers of people like Iwahori and Curtis, but especially by Springer's theory of Weyl group representations on cohomology of flag varieties.   The powerful "cell" construction by Kazhdan-Lusztig does not in general give explicitly the irreducible representations, however.   Carter's 1985 book on characters of finite groups of Lie type discusses how all of this feeds into that kind of representation theory.   Much of the progress has been due to Lusztig.
It's important to distinguish between finite crystallographic Coxeter groups (Weyl groups) and the remaining dihedral groups along with exceptions $H_3, H_4$.  Even in the latter cases, much of the Weyl group theory has good analogues in spite of being outside the classical framework of groups of Lie type.    In any case, whether the results in the literature are explicit enough for some purposes may be an open question.    Certainly the case of symmetric groups and their Iwahori-Hecke algebras has been developed most concretely.

REPLY [5 votes]: More context is needed for this question. I am going to address the case of generic representations in the spherical case.
It is an old theorem of Lusztig that over the ring $\mathbb{C}[v,v^{-1}]$, the generic Hecke algebra $H_v$ with generators $T_i$ and relations
$$T_i T_j \ldots = T_j T_i \ldots \ (m_{ij} \text{ factors}), \qquad (T_i+v^2)(T_i-v^2)=0$$
is isomorphic to the group ring of the corresponding finite Coxeter group $W.$ Thus every representation of $W$ can be canonically deformed to a representation of $H_v.$ In the course of developing representation theory of reductive groups over a finite field, Lusztig developed quite a bit of machinery describing these representations (fake degrees, etc). This is described in his book
G. Lusztig, Characters of reductive groups over a finite field. Annals of Mathematics Studies, 107. Princeton University Press, Princeton, NJ, 1984
A more recent source, in a more general situation and with improved proofs, is
G. Lusztig, Hecke algebras with unequal parameters. CRM Monograph Series, 18. American Mathematical Society, Providence, RI, 2003 
For specific non-zero $v$ that are not roots of unity, the same result holds. The case of roots of unity and of more general Hecke algebras (in particular, for Coxeter system of affine type) has also been studied.<|endoftext|>
TITLE: Growth rate of number of loops in a graph
QUESTION [17 upvotes]: Let G be a directed graph with a countable number of vertices, and suppose G is strongly connected (given any two vertices v and w, there exists a path from v to w).  Fix a base vertex v0∈G, and let Ln denote the number of loops of length n based at v0; that is, the number of sequences of vertices v0, v1, ..., vn such that vn = v0 and there is an edge from vi to vi+1 for every 0 ≤ i < n.  We allow the loops to self-intersect, repeat segments, etc.
Let h be the exponential growth rate of the number of such loops: $h=\lim_{n\to\infty} \frac 1n \log L_n$.  The value of h may be either finite or infinite, and I am interested in finding conditions on the graph that help determine which of these is the case.
Question:  Is there any characterisation of the set of graphs for which h < ∞?  A necessary and sufficient condition would be ideal, but anything that is known would be appreciated.
Partial answer so far:  If G is uniformly locally finite -- that is, if there exists C < ∞ such that every vertex of G has incoming degree ≤ C or every vertex has outgoing degree ≤ C -- then Ln ≤ Cn, and so h ≤ log(C) < ∞.  However, it is not difficult to construct locally finite graphs with unbounded degree, or even graphs that are not locally finite, for which h < ∞, so this condition is not necessary.  If G is undirected, or equivalently, if v → w implies w → v, then one can show that this condition is both necessary and sufficient.  However, the directed case is more subtle.
Motivation:  One can define a topological Markov chain as the space of all infinite paths through the graph G together with a shift that maps v0v1v2... to v1v2v3....  The value h defined above is the Gurevich entropy of this dynamical system, and it is of interest to know when the Gurevich entropy is finite.

REPLY [5 votes]: This is sort of an anti-answer/extended comment. First let me comment that this is the same as the problem of determining which graphs have finite spectral radius. If your graph has adjacency matrix $A=(a_{ij})$, then the number of paths of length $n$, $v_i \to v_j$ is given by $(A^n)_{ij}$, and the spectral radius is given by $\limsup _{n\to \infty} (A^n) _{i,j}^{1/n}$. This value doesn't depend on the values of $i,j$ obviously, since we are in the connected (resp. strongly connected) case.
There is a not-so-recent survey on spectra of infinite graphs by B. Mohar and W. Woess (who also has a book on the related subject of random walks on infinite graphs and groups) but it doesn't go beyond the locally finite case and it doesn't treat directed graphs. They give many references there, including works of A. Torgasev, who studied spectral properties of graphs that aren't necessarily locally-finite.
Now spectra of infinite graphs can be approximated  by the spectra of finite graphs which converge to our graph, and the reason why undirected graphs with finite spectral radius are easy to describe (locally finite and moreover bounded degree) is because one can check this property locally. That is I can look at the neighborhood of vertices with arbitrarily large degree and loops based on them to conclude that there is no upper bound on the spectral radius. However this property is completely lost in the case of directed graphs, given an outgoing edge or an incoming edge, I have no way of telling if they are part of a loop of some bounded length. So a characterization in terms of the degrees seems out of question. A geometric criterion is an other possibility, but as suggested from the related problem of determining transitivity/recurrence of graphs, maybe it is only possible if we are dealing with a Cayley graph.
So, I guess what I'm trying to say that a universal characterization of when an infinite digraph has finite spectral radius seems very hard, however one can deal with the cases which can be approximated through finite digraphs which have well understood spectral properties (for example, when you have a sparsity condition). I may be wrong, but it seems that spectral theory of directed graphs is a subject that hasn't been studied a lot, at least compared to the undirected case.<|endoftext|>
TITLE: What is the history of the Y-combinator?
QUESTION [17 upvotes]: Inspired by the comments to  this question, I wonder if someone can explain the history of the fixed point combinator (often called the Y combinator) in lambda calculus.  
Where did it first appear?  Was it directly inspired by the Arithmetic Fixed Point Theorem?  The two are very similar in spirit.  
Based on the dates of Church's introduction of lambda calculus and Goedel's incompleteness theorem, it seems to me the Arithmetic Fixed Point Theorem must have come first.

REPLY [12 votes]: Although I don't have historical information, let me comment on the suggestion that the $Y$ combinator may have been inspired by the arithmetical fixed-point theorem.  I think a more likely inspiration would be Russell's paradox.  If you think of application (of functions to arguments) as analogous to membership in sets (i.e., think if $x(y)$ as analogous to $y\in x$), then the $Y$ combinator builds a fixed point $Yf$ for a function $f$ in the same way that Russell built a fixed-point for negation.<|endoftext|>
TITLE: Closed 3-manifolds with free abelian fundamental groups
QUESTION [9 upvotes]: Which free abelian groups can be realized as the fundamental group of a closed 3-manifold?  The only one I can come up with is $\mathbb{Z}$, which is the fundamental group of $S^1 \times S^2$.  For the application I have in mind, the key case is $\mathbb{Z}^2$.  Here it is easy if you allow boundary (just take $T^2 \times [0,1]$), but I don't see how to do it without the boundary.

REPLY [15 votes]: (I assume all occuring 3-manifolds to be orientable and closed)
A manifold with a free abelian fundamental group cannot be a connected sum of non-trivial 3-manifolds since its fundamental group is not a free product. A prime manifold is either $S^1\times S^2$ or irreducible (Hatcher's notes on 3-manifolds, 1.4). By 3.9 in the same source, an irreducible 3-manifold $M$ with infinite fundamental group is a $K(\pi,1)$. If $\pi = \mathbb{Z}^n$, then $M$ needs therefore to be homotopy equivalent to $(S^1)^n$. This can only be if $n=3$.

REPLY [14 votes]: Only $\mathbb Z$ and $\mathbb Z^3$ (for $T^3$) are free abelian groups that appear as fundamental groups of $3$-manifolds. Hopefully the following is an approximative proof.
The manifold must be prime (otherwise the $\pi_1$ is not ableian), hence it is $K(\pi,1)$.
Hence  its cohomology are just cohomology of the group $\mathbb Z^n$. So the can not get 
$\mathbb Z^2$ since $H^3(\mathbb Z^2)=0$, and we can not get $\mathbb Z^n$ with $n>3$ since $H^n(\mathbb Z^n)=\mathbb Z$.

REPLY [6 votes]: John Hempel, in his book $3$-manifolds, shows that if $G$ is a finitely generated abelian group which is a subgroup of the fundamental group of a closed $3$-manifold, then $G$ is one of $\mathbb Z$,  $\mathbb Z\oplus\mathbb Z$,   $\mathbb Z\oplus\mathbb Z \oplus\mathbb Z$, $\mathbb Z_p$ or $\mathbb Z\oplus\mathbb Z_2$. This is theorem 9.13 in the book.
He also proves, in theorem 9.14, that an abelian group which is not finitely generated and a subgroup of the fundamental group of a $3$-manifold, then it is isomorphic to a subgroup of $\mathbb Q$ (and proposes, as an exercise, to show that all such groups in fact occur)<|endoftext|>
TITLE: Some basic questions about Chern-Simons theory
QUESTION [13 upvotes]: Let the Chern-Simons lagrangian for a group $G$ be,
$$L= k \epsilon^{\mu \nu \rho} Tr[A_\mu \partial _ \nu A_\rho + \frac{2}{3} A_\mu A_\nu A_\rho]$$
Then it is claimed that on "infinitesimal" variation of the gauge field ("connection") the lagrangian changes by,
$$\delta L = k \epsilon^{\mu \nu \rho} Tr[\delta A_\mu F_{\nu \rho}]$$
where the "curvature" $F_{\mu \nu}$ is given as $\partial _ \mu A_\nu - \partial _ \nu A_\mu +[A_\mu,A_\nu]$
Under gauge transformations on $A$ by $g \in G$ it changes to say $A'$ whose $\mu$ component is given as $g^{-1}A_\mu g+g^{-1}\partial _ \mu g$  (This makes sense once a representation of $G$ has been fixed after which $A$ and $g$ are both represented as matrices on the same vector space)
Say the Lagrangian under the above gauge transformations change to $L'$ and then one has the relation,
$$L'-L=-k \epsilon^{\mu \nu \rho}\partial_\mu Tr[\partial_\nu g g^{-1}A_\rho]-\frac{k}{3}\epsilon^{\mu \nu \rho} Tr[g^{-1}\partial_\mu gg^{-1}\partial_\nu gg^{-1}\partial_\rho g]$$
The second term of the above expression is what is proportional to the "winding number density"
 of the Chern-Simons lagrangian and thats what eventually gets quantized. 
I would like to know the following things,

Is there a neat coordinate free way of proving the above two variation change equations? Doing this in the above coordinate way is turning out to be quite intractable!
Since the Lagrangian is just a complex number one can talk of the "real" and the "imaginary" part of it. But I get the feeling that at times a split of this kind is done at the level of the gauge field itself. Is this true and if yes the how is it defined? (Definitely there is lot of interest in doing analytic continuation of the "level" $k$)
The Euler-Lagrange equations of this action give us only the "flat" connections and in that sense it is a topological theory since only boundary conditions seem to matter. Still all the flat connection configurations are not equivalent but are labelled by homomorphisms from the first fundamental group of the 3-manifold on which the theory is defined to $G$. How to see this? What is the background theory from which this comes? And why is this called "holonomy"? (I am familiar with "holonomy" as in the context of taking a vector and parallel transporting it around a loop etc)

REPLY [5 votes]: There is a nice geometrical treatment of the Chern-Simons action functional for a general Lie group $G$ - not necessarily simply connected - that relates it to "holonomy". It also clarifies why Chern-Simons theories are classified by $H^4(BG,\mathbb{Z})$.
This point of view is described in a paper by Carey et. al. (http://arxiv.org/abs/math/0410013). 
The group $H^4(BG,\mathbb{Z})$ classifies geometrical objects over $G$ called "multiplicative bundle gerbes". Given a multiplicative bundle gerbe $\mathcal{G}$ and a closed oriented 3-manifold $M$ with a principal $G$-bundle $P$ with connection over it, Carey et. al. describe a beautiful construction of a bundle 2-gerbe $CS(\mathcal{G},P)$ over $M$, the Chern-Simons 2-gerbe. They prove that this 2-gerbe carries a connection such that its holonomy around $M$, 
$$
Hol_{CS(\mathcal{G},P)}(M) \in S^1
$$
is the (exponentiated) Chern-Simons action. This works without any assumptions on the manifold $M$ or on the Lie group $G$. 
For a little background, recall that connections on $n$-gerbes have holonomies around closed oriented $(n+1)$-manifolds. Ordinary bundles are 0-gerbes. Part of a connection on an $n$-gerbe is a curving: a $(n+1)$-form on some covering space. In case of the 2-gerbe $CS(\mathcal{G},P)$, the covering space is the total space of the given bundle $P$, and the curving 3-form is the Chern-Simons 3-form $CS_A$ on $P$.<|endoftext|>
TITLE: Curves of constant curvature on S^2
QUESTION [22 upvotes]: Most probably this is a well known question.
Consider $S^2$ with a Riemannian metric. I would like to ask what is known about the structure of the set of simple (without self-intersections) closed curves on it of constant  geodesic curvature. 
Here is a series of questions.
1) Is this  true that through each point of $S^2$ passes a simple closed curve of constant curvature? If not, can one estimate from below the proportion of the area of $S^2$ covered by such curves?
2) Is it true that for each value of curvature there are at least $2$ simple closed curves on $S^2$ of this curvature? Or maybe even more than $2$?
3) What can be said about the global structure of these curves on a generic $S^2$?
Taking the union of all such closed curves we could try to cook up from them a surface
(that maps naturally to $S^2$). Is something known about the topology of this surface?
Comments
1)  The theorem of Birkhoff states that each Riemannian $S^2$ contains at least three simple closed geodesics, as Joseph remarks below.
2) For a generic metric on $S^2$ the set of such curves  this set should be one dimensional. Indeed for each fixed value of curvature you can consider an analogue of the geodesic flow on the space of unite tangent vectors to $S^2$ and you expect that closed orbits will be isolated. 
ADDED. Is seems indeed that these are open (and I guess hard) questions. Macbeth gave a very nice reference, that tells in particular that similar questions were raised previously by Arnold, I copy the Macbeth's reference here, so it is visible to everyone: http://count.ucsc.edu/~ginzburg/ARNOLD/mag-post.pdf
Update. The following reference : http://arxiv.org/abs/0903.1128 gives a positive answer to question 2) for spheres of non-negative Gaussian curvature provided we consider not only simple curves on $S^2$ but also curves that bound immersed disks.
One more update. There is a new nice article  http://arxiv.org/abs/1105.1609 that provides some further results concerning question 2) for $S^2$ of positive curvature. This article also gives all necessary references from which one can conclude that 
question 2) was considered by Poincare in 1905, as it is written in the article of S.P. Novikov  http://iopscience.iop.org/0036-0279/37/5/R01/pdf/0036-0279_37_5_R01.pdf

REPLY [3 votes]: This is an update on new results (and old conjectures) on closed curves of constant curvature. I've just spotted a new paper on arxiv on immersed curves of constant curvature. The authors prove the existence of immersed closed curves of constant geodesic curvature
in an arbitrary Riemannian $2$-sphere for almost every prescribed curvature.
https://arxiv.org/pdf/2106.12374.pdf
There is a nice short survey in this paper about the history of the topic, on page 2. In particular, there is a reference to the following paper of Ketover and Liokumovich
https://arxiv.org/pdf/1810.09308.pdf
which attributes the following conjecture to Novikov:
Conjecture (Novikov). Every Riemannian two-sphere contains a smoothly embedded curve of curvature c for any $0 < c < \infty$.
The authors say that this conjecture is from Section 5 of the following Novikov's paper:
http://www.mi-ras.ru/~snovikov/74.pdf
I was not able to spot this conjecture in Novikov's paper immediately, but hopefully it's there. So it looks like part 2 of the original question is indeed an old open problem.<|endoftext|>
TITLE: A probability question related to extremal combinatorics
QUESTION [20 upvotes]: $k$ people play the following game: person $i$ independently picks a subset $S_i$ of $\{ 1,2,\ldots,n \}$ according to some distribution $p$ on the $2^n$ subsets; each person uses the same distribution $p$. If some $S_i$ is contained in $\cup_{j \neq i} S_j$, they all lose; else, they all win. What distribution $p$ maximizes the probability of winning? 
I am actually only interested in the case where $n/k$ is an integer, in which case I would conjecture that the optimal distribution is for each person to pick a random subset with $n/k$ elements. I can prove this only for $k=2$, in which case it follows straightforwardly from Sperner's theorem. 
Edit: JBL points out in the comments that its also easy to confirm the $n=k$ case of the
conjecture in the previous paragraph.

REPLY [4 votes]: Marek Chrobak, Leszek Gasieniec, Dariusz Kowalski, and I looked at the following variant of this problem in section $8$ of this paper (published version):

Let $k$ be large.  For what $n$, as a function of $k$, is it possible
for the players to win with high probability?

Our original motivation here was a problem in communication: If you have a large number of identical sensors, each transmitting randomly to the same receiver, how long do you need to have them transmitting to guarantee that (with probability approaching $1$) each one transmits at least once without interference from the other sensors?  The $S_j$ from your problem correspond to the set of times at which each sensor transmits.
It turns out that the sharp threshold here is $n=\frac{k \ln k}{(\ln 2)^2}$, in the following sense:

If $n=ck \ln k$ and $c < \frac{1}{(\ln 2)^2}$, then the players lose with probability $1-o(1)$ as $k$ tends to infinity, regardless of what $p$ is.  Here the $o(1)$ error term is a function depending  on $c$ and $k$ (but not on $p$) that tends to $0$ as $k$ tends to infinity for any fixed $c$.

If $n=c k \ln k$ and $c > \frac{1}{(\ln 2)^2}$, then the players can win with probability $1-o(1)$.
The distribution $p$ we use for the second part is to divide the interval into subintervals of length $\frac{k}{\ln 2}$, and have players choose exactly one number from each subinterval.  This should be pretty much equivalent to choosing uniform subsets of size $\frac{n}{k/\ln 2}$ -- the main reason we used subintervals was to make the calculation cleaner.


So in a sense, choosing uniform subsets of a given size is, asymptotically, optimal in this regime of $n$ and $k$.  The size just ends up being slightly smaller than $n/k$.<|endoftext|>
TITLE: "Spec" of graded rings?
QUESTION [14 upvotes]: From the discussion here, it seems that general Hochschild cohomology classes correspond to deformations where the deformation parameter can have nonzero degree.
So I have some naive and maybe stupid questions:
How can I interpret this geometrically? What is the "base space" of the deformation? What kind of object is it?
In other words, what is the "Spec" of a graded ring or a graded algebra (e.g. $k[t]$ or $k[[t]]$ or $k[t]/(t^n)$ with the variable $t$ having some nonzero degree)?
(..... maybe what I'm really asking is: Is there a theory of "schemes" where the "affine schemes" correspond to graded commutative rings rather than commutative rings? .....)

REPLY [2 votes]: maybe these notes by Vezzosi can be helpful for some
http://www.dma.unifi.it/~vezzosi/papers/derivedintctgtcplx.pdf<|endoftext|>
TITLE: Hessian as a tensor, multi-dimensional taylor series, and generalizations
QUESTION [16 upvotes]: The Hessian matrix $\{\partial_i \partial_j f \}$ of a function $f:\mathbb{R}^n \to \mathbb{R}$ depends on the coordinate system you choose. If $x_1,\cdots,x_n$ and $y_1,\cdots,y_n$ are two sets of coordinates (say, in some open neighborhood of a manifold), then $\frac{\partial f(y(x))}{\partial x_i} = \sum_{k} \frac{\partial f}{\partial y_k} \frac{\partial y_k}{\partial x_i}$. Differentiating again, this time with respect to $x_j$, we get $\frac{\partial^2 f(y(x))}{\partial x_i \partial x_j} = \sum_{k} \sum_{l} \frac{\partial^2 f}{\partial y_k \partial y_l} \frac{\partial y_l}{\partial x_j} \frac{\partial y_k}{\partial x_i}+\frac{\partial f(y(x))}{\partial y_k}\frac{\partial^2y}{\partial x_i \partial x_j}$. At a critical point, the second term goes away, so we will consider such a case.
In other words, if the derivative is a differential $1$-form, i.e. $\sum_{i} \frac{\partial f}{\partial x_i} dx_i$, a section of the cotangent bundle, then the second derivative should be $\sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} dy_k \otimes dy_l$. This makes sense since $dy_k=\sum_{i} \frac{\partial y_k}{\partial x_i} dx_i$, and $dy_l=\sum_{j} \frac{\partial y_l}{\partial x_j} dx_i$, meaning that $\sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} dy_k \otimes dy_l = \sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} (\sum_{i} \frac{\partial y_k}{\partial x_i} dx_i) \otimes (\sum_{j} \frac{\partial y_l}{\partial x_j} dx_j) = \sum_{i,j,k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} \frac{\partial y_k}{\partial x_i} \frac{\partial y_l}{\partial x_j} dx_i dx_j = \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j}$, making it coordinate independent. Note that I did not use exterior powers, I used tensor powers, since I wanted to actually find a way to make sense of second derivatives, rather than having $d^2=0$. This means the Hessian should be a rank $2$ tensor ((2,0) or (0,2), I can't remember which, but definitely not (1,1)).
Does this make sense? Can we then express the third, etc, derivative as a tensor? More interestingly, how can this help us make sense of Taylor's formula? Can we come up with a coordinate-free Taylor series of a function at a point on a manifold?
EDIT: An in general, if the first $n$ derivatives vanish, then the $n+1$ derivative should be a rank $n+1$ tensor, right?

REPLY [3 votes]: More generally, for any vector bundle $V \to M$, the vanishing of a section at a point to a given order $k$ is independent of local coordinates and choice of basis of local sections, and when this happens at some point $m \in M$, the $(k+1)$-jet of the local section at that point is a well defined element of $S^kT^*_mM \otimes V_m$.<|endoftext|>
TITLE: Choosing lines and points in D^2
QUESTION [8 upvotes]: I recently heard of a game between two players "Line" and "Point" and wanted to look for more information on it. However, without knowing the name of it (if it has one) finding more information is hard, has anyone heard of it? Is there a winning strategy for one of the players?
The game is as follows, it is played on the unit disk $D^2$ in $\mathbb{R}^2$ with the point $p_0 = (0,0)$ marked to begin with. Play alternates between L and P (starting with L) and on turn $n$ they do the following:
L chooses a new line $l_n$ through point $p_{n-1}$ and then P chooses a new point $p_n$ on line $l_n$ inside $D^2$.
This forms a sequence of points $(p_n)_{n = 1}^\infty$ in $D^2$. L wins if this sequence converges to a point in $D^2$, P wins if it does not.
As far as I can tell P has a winning strategy, but I my formal proof for this is a sketch at best.

REPLY [5 votes]: Line actually has a winning strategy: it can force a convergent sequence. The problem was posed and solved in the following paper:
J. Maly and M. Zeleny (2006), A note on Buczolich's solution of the Weil gradient problem: a construction based on an infinite game, Acta Mathematica Hungarica, Vol. 113, pp. 145-158.<|endoftext|>
TITLE: Neusis constructions
QUESTION [14 upvotes]: Is there some simple description of which complex numbers are "constructible" with straightedge and compass and neusis?
See http://en.wikipedia.org/wiki/Constructible_number and http://en.wikipedia.org/wiki/Neusis.

REPLY [9 votes]: There are various flavors of neusis construction.  In the weakest flavor, in addition to having the marked straightedge pass through the pole point, the two marks on it are required to lie, one each, on two specified lines; we might call that tool a line-line neusis.  For a line-circle neusis, one mark must lie on a specified line while the other must lie on a specified circle.  For a circle-circle neusis, the two marks must lie, one each, on two specified circles.  (We view a line as a special case of a circle; so each of these tools is at least as powerful as its predecessors.)
If we allow ourselves a straightedge, a compass, and a line-line neusis, then, as Stillwell tells us, we get precisely the closure of the rationals under complex square roots and cube roots.  The Alperin paper that Stillwell mentions is a high-level reference.  Here are some more details.
In one direction, consider the line through the pole point that has slope $s$.  We can intersect that line with the two specified lines.  The distance between the two resulting intersections equals the fixed distance between the two marks on the straightedge just when a certain quartic equation in $s$ holds.  And, of course, any quartic can be solved using complex square roots and cube roots.
In the other direction, the compass allows us to bisect any angle and to extract any real square root; so we can take complex square roots.  To show that the line-line neusis can take complex cube roots, we need to show two things: that it can trisect any angle and that it can extract any real cube root.  
Trisecting first:  There is a well-known neusis angle-trisection credited to Archimedes; but that construction uses a line-circle neusis, and hence doesn't help us here.  But the Greeks also knew of a trisection using a line-line neusis.  Alperin credits that construction to Apollonius, but gives no details.  For the details, see either A History of Greek Mathematics, Volume 1: From Thales to Euclid, by Sir Thomas Heath, reprinted by Dover in 1981, pages 236-238.  Or see Exercise 10 on page 245 of Michael O'Leary's Revolutions in Geometry, published by Wiley in 2010.  (Note that, in this construction, of the four slopes for the neusis straightedge that satisfy the distance requirement, all four are real; one is trivial and should be ignored, while the other three are the three trisectors.)
Now for real cube roots:  The Greeks also knew a line-line neusis construction, credited to Nicomedes, for extracting real cube roots.  One source for that construction, pointed out by Gerry Myerson, is the article "Constructions using a compass and twice-notched sraightedge", by Arthur Baragar, pages 151-164 in volume 109, number 2 of the American Mathematical Monthly.  (In this construction, of the four slopes mentioned above, one is trivial and should be ignored, a second gives the required real cube root, and the remaining two are complex.)
One side remark:  The Nicomedes cube-root construction is a bit subtle.  But Conway and Guy give a dead-simple line-line neusis construction for the special case of the cube root of 2, on page 195 of The Book of Numbers, published by Springer in 1996.
So the line-line neusis gives us precisely the power to solve quadratic and cubic (and hence also quartic) equations, which makes it equivalent to "conic constructability" or, as the Greeks called it, "solid constructability".
What about the other flavors of neusis construction?  Baragar shows that, for either a line-circle neusis or a circle-circle neusis, there are in general six slopes for the line through the pole that have the proper distance relationship -- six, rather than the four of the line-line case.  He then gives an explicit example of a line-circle neusis construction in which one of these slopes is real and trivial, three others are real, and the final two are complex.  Furthermore, the five nontrivial slopes are the roots of a irreducible quintic equation whose Galois group is all of $S_5$, and which hence cannot be solved with radicals.  Thus, the line-circle neusis is a strictly more powerful tool than the line-line neusis.
As an upper bound on the power of these more general neusis constructions, Baragar shows that any point generated by either the line-circle neusis or the circle-circle neusis lies in an extension field of the rationals that can be reached by a tower of fields in which each adjacent pair has index either 2, 3, 5, or 6.  So the jump in power over the line-line neusis, where the adjacent-pair indices are either 2 or 3, is not too great.  
Baragar's paper closes with some interesting open problems.  One problem that he doesn't mention is this:  Is the circle-circle neusis strictly more powerful than the line-circle neusis?<|endoftext|>
TITLE: Are Finsler manifolds semi-Riemannian? 
QUESTION [8 upvotes]: A Finsler manifold is defined as a differentiable manifold with a metric defined on so that any well-defined curves of finite arc length is given by a generalized arc length integral of an asymmetric norm over each tangent space defined at a point. This generalizes the Riemannian manifold structure since the norm is no longer required to be induced by an inner product and therefore the Finsler manifold is not necessarily Euclidean in the tangent space structure. 
A collegue of mine and I recently got into an arguement over whether or not Finsler manifolds are semi- or psuedo-Reimannian. I say no-by definition, A semi-Reimannian manifold-like an Loretzian manifold in relativity theory-is still required to have a metric tensor as it's normed structure,which is clearly an inner product. We simply weaken the condition of positive definiteness(i.e. the associated quadratic form of the norm is real valued) to nondegeneracy (i.e. the tangent space is isomorphic with its dual space). Both conditions require the distance structure to be induced by an inner product. 
My colleague's argument is that the key property of semi-Riemannian manifolds is that they admit local signed coordinate structures that allow the distinction of different kinds of tangent spaces on the manifold. Local isomorphisms can be defined-especially in infinite-dimensional extensions of classical relativistic spaces-that make certain Fisler manifolds eqivilent to relativistic models of space-time. 
I honestly don't know enough about the research that's been done on this. Is he right? This seems very bizarre to me, but it may indeed be possible to use specially constructed mappings to convert Finsler spaces to semi-Riemannian ones and vice-versa. I seriously doubt it could be done globally without running into serious topological barriers. 
I'd like the geometers to chime in on this,particularly ones who are well-versed in relativistic geometry: Am I right? Can Finsler manifolds be defined in such a manner as to be true semi-Reimannian manifolds? Can local isomorphisms or differentiomorphisms be defined to interconvert them?

REPLY [9 votes]: No, a Finsler metric is in general not semiriemannian. As you and José indicate, a semiriemannian metric is always given by a nondegenerate quadratic form on tangent vectors at each point in the manifold. In other words, if you fix a basis of the tangent space for a given point, then the norm of that vector is given by a homogeneous quadratic polynomial in the coefficients of the vector with respect to the basis.
On the other hand, a Finsler metric is given by a norm function on the tangent space of each point in the manifold. This norm function must be convex, and additional regularity and convexity assumptions are often made. However, there is no requirement that the norm function be given by a quadratic form. It could be given by a higher even degree polynomial in the coefficients of a vector with respect to a basis. But it could be an arbitrary sufficiently smooth sufficiently convex function, too.
One way to think about this is to consider the standard flat models. The standard flat semiriemannian model is just $R^n$ with the metric given by a non-degenerate quadratic form. The standard flat Finsler model is $R^n$ with a (sufficiently smooth and convex) Banach norm, i.e. a finite dimensional Banach space. There are obviously a lot more of the latter than the former.

REPLY [4 votes]: The answer is no. Any smooth manifold admits a Riemannian metric using paracompactness and partitions of unity: in short, a convex sum of positive definite symmetric matrices is positive definite symmetric. So any manifold has such a structure. But there are topological obstructions to the existence of global pseudo-Riemmannian metrics of other prescribed signatures.
http://en.wikipedia.org/wiki/Semi-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds 
EDIT, 16 July:  I was looking for an example of said topological obstructions, and with an assist fom Willie Wong it has worked out: the ordinary sphere $$ \mathbb S^2 \subseteq \mathbb R^3 $$ does not possess a signature $(+,-) $ metric which I suppose ought to be called Lorentzian for this dimension. The topological obstruction is that $ \mathbb S^2$  cannot have a smooth (tangent) "line field," just as it cannot have a smooth nonzero tangent vector field by Brouwer. Now, a Lorentzian metric would give (pointwise)  null cones, in this case a pair of distinct but intersecting lines in each tangent plane. As we are using $\mathbb R^3, $ we can cheat and define a line field from the angle bisector of the $+$ part of the cone.<|endoftext|>
TITLE: Weil conjecture for algebraic surfaces
QUESTION [17 upvotes]: Deligne's proof of the Weil conjecture is difficult.
On the other hand, there are some "simpler" proofs of the Weil conjecture in the case of algebraic curves.
For instance, in GTM52, one see it eventually reduced to the Hodge index theorem, which is the geometric input.
And there even exists some elementary proofs, by Bombieri or Stephanov.
So what I am asking is, will there be some "simple" proofs of the Weil conjecture for algebraic surfaces, at least for some special classes of them? And in that case, what can be the geometric inputs, without using the Standard Conjecture?

REPLY [2 votes]: I proved it for Zariski Surfaces in 1970. See chapter 1 of my book with Jeff Lang. Zariski Lives!!! 

Blass, Piotr; Lang, Jeffrey (1987), Zariski surfaces and differential equations in characteristic p>0, Monographs and Textbooks in Pure and Applied Mathematics 106, New York: Marcel Dekker Inc., ISBN 978-0-8247-7637-4, MR 879599<|endoftext|>
TITLE: Finite dimensional Feynman integrals
QUESTION [14 upvotes]: In a sense this is a follow up question to The mathematical theory of Feynman integrals although by all rights it should precede that question.
Let $S$ be a polynomial with real coefficients in $n$ variables. Is there a criterion which would say when the integral $$\int_{\mathbf{R}^n}e^{iS(x)}dx$$ converges? Here $dx$ is the Lebesgue measure on $\mathbf{R}^n$ and the integral is understood as the limit of the integrals over the balls of radius $r$ centered at the origin (with respect to the standard metric) as $r\to\infty$.
Some obvious remarks:

If $\deg S=2$, the integral converges if and only if the quadratic part of $S$ is nondegenerate.
If $n=1$, the integral converges if and only if $\deg S>1$.
The answer to the above question is probably classical (but it is unknown to me).

upd: Conjecture (inspired by Jeff's answer below). A sufficient condition for the integral to converge is as follows: let $S_i$ be the degree $i$ part of $S$ and let $V_i,i=1,\ldots,d=\deg S$ be the subvariety of the real projective space $\mathbf{P}^{n-1}(\mathbf{R})$ given by $S_i=0$. The integral converges if $V_2\cap\cdots\cap V_d={\emptyset}$. Here is how one can try to prove this: the above condition is equivalent to saying that the integral along any line converges, so one can try to first integrate along all half-lines emanating from the origin, get a continuous function on the sphere (hopefully) and then integrate it along the sphere. As remark 1. above shows, this condition may be sufficient but it is not necessary.

REPLY [6 votes]: Let $d=\deg S$. If $d\ge 2$ and $S$ is non-degenerate near infinity, in the sense that $|\nabla S(x)|^2 \ge c|x|^{2d-2}$ for $|x|\ge R$ for some positive $R$ and $c$ then the integral converges in a sense close to what you propose.  Namely, 
$$\lim_{r\rightarrow \infty} \int_{\mathbb R^n} \phi(|x|/r) e^{iS(x)} dx$$ exists where $\phi$ is any sufficiently smooth function with $\phi(t)\equiv 1$ for $0\le t\le 1$ and $\phi (t)\equiv 0$ for $t\ge 2$. 
To prove this use integration by parts.  Fix $r < r'$ and let
$$I_{r,r'} = \int_{\mathbb R^n} \phi_{r,r'}(x) e^{i S(x)}dx,$$where $\phi_{r,r'}(x)=\phi(|x|/r)-\phi(|x|/r')$.  Assume $r >R$. Multiply and divide by $|\nabla S(x)|^2$, noting that 
$$|\nabla S(x)|^2e^{i S(x)}= -i\nabla S(x)\cdot \nabla e^{i S(x)}.$$
So we have, after IBP,
$$I_{r,r'} = -i \int_{\mathbb R^n} \left [ \nabla \cdot \left ( \frac{\phi_{r,r'}(x)}{|\nabla S(x)|^2} \nabla S(x) \right ) \right ] e^{i S(x)} d x.$$
The integrand is supported in the region $r \le |x| \le 2 r'$, and you can check that it is bounded in magnitude by some constant times $|x|^{1-d}$.  (It is useful to note that an $m$-th derivative of $S$ is bounded from above by $|x|^{d-m}$ since it is a polynomial of degree $d-2$.)  If $d$ is large enough this may be enough to control the integral.  If not repeat the procedure as many times as necessary to produce a factor that is integrable and you can use to show that $I_{r,r'}$ is small for $r,r'$ sufficiently large.  Basically, each time you integrate by parts after multiplying and dividing by $|\nabla S(x)|^2$ you produce a an extra factor of size $|x|^{1-d}$.  
All of the above can be extended to integrals with $S$ not necessarily a polynomial, but sufficiently smooth in a neighborhood of infinity and with derivatives that satisfy suitable estimates.  The real difficulty with such integrals is not proving that they exist, but estimating their size.  Here stationary phase is useful, when applicable, but I don't know of much else.<|endoftext|>
TITLE: Todd class and Baker-Campbell-Hausdorff, or the curious number $12$
QUESTION [6 upvotes]: The number $12$ (or, probably we shall say Bernoulli numbers in general) appears in many places in Mathematics, sometimes leading to unexpected connections between different topics.
For instance, some time ago there was a very interesting explanation for
1) its occurrence in the Todd class
and
2) its occurrence in the Euler-Maclaurin formula
in terms of Riemann-Roch for toric varieties, as explained in:
Euler-Maclaurin formula and Riemann-Roch
===========================================================
My question is, will there be some relation between 1) and 
3) its occurrence in the Baker-Campbell-Hausdorff formula.
I guess this might be related to some explicit local expressions in some method of proving the index theorem on Lie groups, or even the Duflo map (which I don't really understand).
Thank you very much.

REPLY [15 votes]: The answer to your question is the following: given two non-commutative variables $x$ and $y$ one has 
$$
log(e^xe^y)=x+e^{ad_x}\frac{ad_x}{e^{ad_x}-1}(y)+O(y^2)
$$
It is not the appearance of $12$ that is intriguing, but the appearance of the Todd series in algebraic geometry. It suggests that there is a group hidden somewhere... and this is indeed the case. This group is the derived loop space of your favorite algebraic variety $X$, and its tangent Lie algebra is the shifted tangent sheaf $T_X[-1]$, with Lie bracket given by the Atiyah class (the fact that the Atiyah class gives rize to a Lie structure was discovered by Kapranov). 
The universal enveloping algebra of this Lie algebra is the Hochschild complex of $X$. One then gets a nice dictionnary between the Lie side and the algebraic geometry side. E.g.: 

any object in the derived category of $X$ turns out to be a representation of this Lie algebra. 
Poincare-Birkhoff-Witt is Hochschild-Kostant-Rosenberg. 
the Duflo isomorphism is the Kontsevich-Caldararu isomorphism between the Harmonic and Hochschild structures. 
there is also an relation between closed embeddings in algebraic geometry and inclusions of Lie algebras. 
...<|endoftext|>
TITLE: "Simple" Kahler manifolds
QUESTION [13 upvotes]: I have some lecture notes from Demailly on Kahler geometry where he talks about "variétés Kahleriennes simples", which are defined as Kahler manifolds $X$ such that for very generic points $x_0$ in $X$ there exists no irreducible submanifold $Y$ of $X$ of dimension $0 < \dim Y < \dim X$ which contains $x_0$.
Examples of these kinds of manifolds are very general complex tori and quotients thereof, and they're interesting because they give counterexamples to the Hodge conjecture in the analytic category.
I thought I'd take a look at these things, but I can't find any mention of "simple Kahler manifolds" either here or on google. Did I get the name wrong? Do any of you know what I'm talking about and know of some references?

REPLY [4 votes]: These manifolds have actually been studied to some extent by Campana and Peternell in their series of papers "Towards a Mori theory on compact Kähler threefolds I, II, III". In those papers they mention a folklore conjecture (which can be found for example here), that says that every simple Kähler manifold of odd dimension $n>1$ must be Kummer (i.e. bimeromorphic to the quotient of a complex torus by a finite group).
In the paper number II Peternell shows that this conjecture in dimension $3$ follows from MMP plus abundance for Kähler threefolds. On the other hand in paper number III he shows that abundance does hold for Kähler threefolds with the possible exception of simple non-Kummer manifolds (which should not exist).
On the other hand more recent developments using model theory seem to suggest that in the even-dimensional case apart from Kummer manifolds the only other simple Kähler manifolds are in generically finite-to-finite correspondence with an irreducible hyperkähler manifold, like in Misha's answer.
Anyway, even after all this work it's unclear whether simple non-Kummer odd-dimensional manifolds exist or not. It's certainly an interesting problem, but probably a very hard one.<|endoftext|>
TITLE: Continuous functions remaining constant
QUESTION [5 upvotes]: I solved a problem in analysis and i was thinking of generalizing this question which i couldn't succeed. 
If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function which satisfies $f(x)=f(2x+1)$, for all $x \in \mathbb{R}$ then prove that $f$ is constant. I was able to prove it considering $g(x)=f(x-1)$ and showing that $g(x) \to g(0)$.
Now my question is suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function and satisfies $f(p(x))=f(x)$ for every polynomial $p(x) \in \mathbb{R}$, then what should be the condition on $p(x)$ such that $f$ remains constant.

REPLY [3 votes]: I'll assume, per Pete's comment above, that you are looking for polynomials $p(x)$ with real coefficients such that $f(p(x)) = f(x)$ implies $f$ is constant. 
It suffices (though I am not sure if this is necessary) then that $p(x)$ has a fixed point that is strictly unstable or strictly stable in the strict sense. In other words, there exists $x_0$ such that $p(x_0) = x_0$ and such that $|p(x - x_0)| > (1+\epsilon) |x-x_0|$ for some $\epsilon > 0$ or $|p(x - x_0)| < (1 - \epsilon)|x - x_0|$. Basically the exact same argument you used to show the property for $2x+1$ can be used for this case.<|endoftext|>
TITLE: Conjugacy problem in a conjugacy separable group
QUESTION [20 upvotes]: Here is a question that has been bothering me for some time:
Let G be a finitely generated conjugacy separable group with solvable word problem. Does it follow that the conjugacy problem in G is solvable?
Background. A group G is said to be conjugacy separable is for any two non-conjugate elements x,y in G there is a homomorphism from G to a finite group F such that the images of x and y are not conjugate in F. Equivalently, G is conjugacy separable if each conjugacy class is closed in the profinite topology on G.
A well-known theorem of Mal'cev states that a finitely presented conjugacy separable group has solvable conjugacy problem (in this case it is possible to recursively enumerate all the finite quotients, simultaneously checking conjugacy of the images of two given elements in each of them).
On the first page of the paper 'Conjugacy separability of certain torsion groups.'
(Arch. Math. (Basel) 68 (1997), no. 6, 441--449.) Wilson and Zalesskii claim that the conjugacy problem is solvable in finitely generated recursively presented conjugacy separable groups (which, of course, implies a positive answer to my question), and refer to a work of J. McKinsey, 'The decision problem for some classes of sentences without quantifiers' (J. Symbolic Logic 8, 61 – 76 (1943)). 
However, I could not find anything in the latter paper that would allow to deal with infinite recursive presentations. Moreover, the corresponding property for residually finite groups simply fails. More precisely, there exist finitely generated residually finite recursively (infinitely!) presented groups with unsolvable word problem (cf. 'A Finitely Generated Residually Finite Group with an Unsolvable Word Problem' by S. Meskin, Proceedings of the American Mathematical Society, Vol. 43, No. 1 (Mar., 1974), pp. 8-10).

REPLY [3 votes]: This is another incomplete answer, but which hopefully can still be useful.
Say that a finitely generated group G has computable finite quotients if there is an algorithm that, on input a finite group F given by a finite presentation and a function that maps the generators of G to those of F, decides whether or not this function could be extended as a homomorphism.
McKinsey’s argument then clearly yields: in a finitely generated, conjugacy separable group, with computable finite quotients, there is an algorithm that decides when two elements are not conjugated.
And thus if that group is recursively presented, it must have solvable conjugacy problem.
I studied this property in my article: « Computability of finite quotients of finitely generated groups », where I prove that this property is independent from the solvability of the word problem: a residually finite group can have computable finite quotients while having unsolvable word problem, and there exist finitely generated residually finite groups with solvable word problem but without computable finite quotients. (This answers negatively Ian Agol’s question.)
Because of this, we have no reasons to believe that for finitely generated groups with solvable word problem, being conjugacy separable is a sufficient condition to have solvable conjugacy problem, since this hypothesis cannot be used without the ability to detect finite quotients.
However, no matter how much I would like to plainly answer « no» to your question, I don’t know how to change my construction to build a conjugacy separable group with solvable word problem and without computable finite quotients, less again one with unsolvable conjugacy problem.
Note finally that it follows from an article of René Hartung (Coset enumeration for certain infinitely presented groups, International Journal of Algebra and Computation) that the first Grigorchuk group has computable finite quotients, and thus the article of Wilson and Zalesskii which you mention, and which proves that this group is conjugacy separable, does help proving that it has solvable conjugacy problem.<|endoftext|>
TITLE: Direct proof of irrationality?
QUESTION [35 upvotes]: There are plenty of simple proofs out there that $\sqrt{2}$ is irrational. But does there exist a proof which is not a proof by contradiction? I.e. which is not of the form: 
Suppose $a/b=\sqrt{2}$ for integers $a,b$. 
[deduce a contradiction here] 
$\rightarrow\leftarrow$, QED
Is it impossible (or at least difficult) to find a direct proof because ir-rational is a negative definition, so "not-ness" is inherent to the question? I have a hard time even thinking how to begin a direct proof, or what it would look like. How about: 

$\forall a,b\in\cal I \;\exists\; \epsilon$ such that $\mid a^2/b^2 - 2\mid > \epsilon$.

REPLY [3 votes]: Most common axiom systems I've seen are a list of $\forall$ and $\exists$ axioms. If you look at a minimal underlying logic, most of the common rules for transforming these axioms shouldn't change the $\forall$ or $\exists$ into a $\neg \exists$. So you could fix a logic system, and argue that the only method that results in a $\neg \exists$ statement is the equivalent of proof by contradiction.
You haven't fixed a logic system in your original question, but the "proof by direct substitution" method won't be sufficient.<|endoftext|>
TITLE: Finding recurrence relation for a sequence of polynomials
QUESTION [9 upvotes]: The sequence
A059710
starts 1,0,1,1,4,10,35,... 
This satisfies the polynomial recurrence relation
$$ (n+5)(n+6)a(n)=2(n-1)(2n+5)a(n-1)+(n-1)(19n+18)a(n-2)+14(n-1)(n-2)a(n-3) $$
I have a $q$-analogue of this sequence. The first few terms are:
$$1$$
$$0$$
$$1$$
$$q^{3}$$
$$q^{6} + q^{4} + q^{2} + 1$$
$$q^{9} + q^{8} + 2 q^{7} + 2 q^{6} + 2 q^{5} + q^{4} + q^{3}$$
$$q^{14} + q^{13} + 4 q^{12} + 2 q^{11} + 5 q^{10} + 4 q^{9} + 5 q^{8} + 
2q^{7} + 5 q^{6} + q^{5} + 2 q^{4} + q^{3} + q^{2} + 1$$
$$q^{21} + q^{19} + 2 q^{18} + 4 q^{17} + 5 q^{16} + 9 q^{15} + 10 q^{14}
+ 13 q^{13} + 13 q^{12} + 14 q^{11} + 12 q^{10} + 12 q^{9} + 8 q^{8} + 7
q^{7} + 4 q^{6} + 3 q^{5} + q^{4} + q^{3}$$
These are $q$-analogues since if you put $q=1$ you get the original sequence.  
Would anyone like to suggest a $q$-analogue of the polynomial recurrence relation?
I have asked a closely related question in 
17610
I can calculate a few more terms than I have posted here.  
Since you asked, the polynomial is constructed as follows: take $V$ to be the seven dimensional representation of $G_2$; take the invariant tensors in $\otimes^nV$;
take the Frobenius character of this representation of $S(n)$; take the fake degree polynomial of this symmetric function (almost the principal specialisation).
Further information In response to Will's comment:
Evaluating at $q=-1$ gives
$$ 1,0,1,-1,4,-2,13,-10,55,-40,241,-190,\ldots $$
Reducing modulo $1+q+q^2$ gives
$$1,0,1,1,1,1,5,3,5,19,15,19,\ldots$$
Reducing modulo $1+q^2$ gives 
$$1,0,1,-q,0,0,q,q-1,3,0,2q+3,-q-1,\ldots$$
Reducing modulo $(1-q^5)/(1-q)$ gives 
$$1,0,1,q^3,-q^3,0,0,0,0,-q^3-q-1,3,0,\ldots$$
Reducing modulo $1-q+q^2$ gives 
$$1,0,1,-1,1,1,1,-1,1,-1,1,-1,\ldots$$
As requested by Jacques, I have put the first fifteen polynomials in a file which you
should be able to access here
G2 polynomials
I have put the first forty polynomials of a second example in a file which you
should be able to access here
A1 polynomials
These are $q$-analogues of the 
Riordan numbers
The linear recurrence relation is given there as
$$ (n+1)*a[n] = (n - 1)*(2*a[n - 1] + 3*a[n - 2]) $$

REPLY [5 votes]: Using FriCAS, one can indeed guess a q-recurrence, given the first 50 terms or so. It is not nice, though.  The command issued is
guessHolo(q)(cons(1, [qRiordan n for n in 1..60]), debug==true, safety==10)
for the q-differential equation (a linear combination with polynomial coefficients of $f(x), f(qx),\dots,f(q^5 x)$, degree in $x$ is 6), or
guessPRec(q)(cons(1, [qRiordan n for n in 1..48]), debug==true, safety==2)
for the q-recurrence.<|endoftext|>
TITLE: Classification of t-structures in derived category of R-mod?
QUESTION [7 upvotes]: I am looking for a reference talking about the complete(or not)description of t-structures in bounded derived category of $R-mod$, i.e. $D^b(R-mod)$.where $R$ is commutative ring, in particular, polynomial ring, say $C[x]$.
Thanks!

REPLY [5 votes]: Provided $R$ has a dualizing complex an answer is given in this (very nice) preprint of Alonso, Jeremias, and, Saorin in terms of certain filtrations on the spectrum of $R$. Corollary 6.11 is the result you are after.<|endoftext|>
TITLE: On a theorem of Jacobson
QUESTION [20 upvotes]: In a comment to an answer to a MO question, in which Bill Dubuque mentioned Jacobson's theorem stating that a ring in which $X^n=X$ is an identity is commutative (theorem which has shown up on MO quite a bit recently, e.g. here), Pierre-Yves Gaillard observed that there is a more general theorem in which $n$ is allowed to be different for each element of the ring, 
so that in fact we can rephrase the theorem as saying that the set $S=\{X^n-X:n>1\}\subset\mathbb Z[X]$ has the following property:

If $A$ is a ring such that for every $a\in A$ there is an $f\in S$ such that $f(a)=0$, then $A$ is commutative. 

Of course, $S\cup (-S)$ also has this property, and even if we construct $S'$ from $S\cup(-S)$ by closing it under the operation of taking divisors in $\mathbb Z[X]$, it also has the same property. Pierre-Yves then asked:

Is $S'$ maximal for this property?

So, is it?

REPLY [6 votes]: Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).
EDIT.  Herstein's set may be maximal.  The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained by the condition that if there is $p$-torsion then no $p^2$-torsion.<|endoftext|>
TITLE: Points of continuity of Baire class one functions
QUESTION [10 upvotes]: This is an idle question motivated by two comments I made to a previous MO question (which I just searched for, unsuccessfully).  That question asked if the characteristic function of the rationals is a pointwise limit of continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$.  
My first reply was that the answer is no: a function is said to be of Baire class one if it is a pointwise limit of continuous functions, and (so I remembered) every Baire class one function has a dense set of points of continuity, whereas of course $\bf{1}_{\mathbb{Q}}$ is discontinuous everywhere.
Then I checked up on this result (I am not an analyst, if that is not already clear) and realized that I had misremembered it slightly: the Baire Characterization Theorem says that a real function is of Baire class one iff for every nonempty closed subset $S \subset \mathbb{R}$, the restriction of $f$ to $S$ has at least one point of continuity.  (Note that in my comment I required that $S$ be perfect, i.e., without isolated points, because some sources state it this way.  This hypothesis is however superfluous because an isolated point of a topological space is a point of continuity for every real-valued function on that space.)  Taking $S = [a,b]$, we find that either (i) our BC1 function restricted to $S$ is continuous at $c \in (a,b)$ -- meaning that $f$ itself is continuous at $c$, since $S$ is a neighborhood of $c$ -- or (ii) $f|_S$ is continuous at $a$ -- meaning that $f$ itself is right-continuous at $a$, or finally (iii) $f|_S$ is continuous at $b$ -- so $f$ itself is left-continuous at $b$.  So we find that any BC1 function has a dense set of points at which it is either left-continuous or right-continuous (or both).  This is good enough to answer the question because $\bf{1}_{\mathbb{Q}}$ has no points of left- or right-continuity.  
But later I wondered whether my original claim was actually incorrect:

Do there exist BC1 functions for which the set of points of continuity is not dense?  Or even empty?  

I did of course try looking on the internet, but I didn't find a reference which gave a solid treatment of Baire classes of real functions: could someone supply one?  Moreover, I do see that by taking intervals of the form $(-\infty,b]$ and $[a,\infty)$ a BC1 function must have infinitely many points at which it is left-continuous and infinitely many points at which it is right-continuous.  And then I got stumped.  Probably this is in the category of things that were much better understood almost one hundred years ago than they are now...

REPLY [13 votes]: This is impossible. Baire proved that if a function defined on $\mathbb R$ is of Baire class 1, then it is continuous everywhere except, possibly, for a meagre set. And by another Baire's theorem a complement of a meagre set in $\mathbb R$ is dense.
An elementary exposition of this and related results can be found in the nice little book by Oxtoby. 
Edit. Another good reference which covers Baire's theorems and provides some historical background is "The calculus gallery" by W. Dunham.<|endoftext|>
TITLE: Are nets and filters useful in geometry and topology?
QUESTION [12 upvotes]: Many results in topology can be restated using the concepts of nets and ultrafilters. This seems to be of interest for set theorists, maybe even logicians. But for geometers and topologists, those who use point-set topology only as a tool in proving theorems about manifolds, varieties, schemes, homology groups, etc, can this reformulation be useful? If it is, please give examples of how it might be used.
In the case of Tychonoff's Theorem, it may provide an interesting way to prove a result in point-set topology which is useful for geometers, but even in this case, its use does not seem to shed much insight once one has obtained the technical point-set result.

REPLY [2 votes]: In formalised mathematics, Lean's mathlib library uses filters extensively when developing the basic theory of topological spaces.  I do not know exactly why that decision was taken.  However, when a non-traditional formulation turns out to be more convenient for formalisation, it is often worth asking whether that formulation has advantages in other contexts as well.<|endoftext|>
TITLE: Class number of non-maximal order in imaginary quadratic function field?
QUESTION [10 upvotes]: It is well known that for $K=\mathbb{Q}(\sqrt{D})$, $D < 0$, the non-maximal order of squarefree conductor $f$, relatively prime to $D$, has class number $$h_K \prod_{p|f} (p-(\frac{D}{p}))$$
What is the class number of a non-maximal order in an imaginary quadratic extension of $\mathbb{F}_p[t]$? Is it proven using ideas involving zeta function, as appears in many books for the case above?

REPLY [19 votes]: There is a formula that works in all degrees, not just imaginary quadratic.  In a global field $K$, let $O$ be integral over ${\mathbf Z}$ or ${\mathbf F}[t]$ (${\mathbf F}$ a finite field) and be "big", i.e., it has fraction field $K$.  Let $\mathfrak c$ be the conductor ideal of $O$ in its integral closure $R$.  Then 
$$h(O) = \frac{h(R)}{[R^\times:O^\times]}\frac{\varphi_{R}({\mathfrak c})}{\varphi_O(\mathfrak c)},$$
where $\varphi_O(\mathfrak c)$ is the number of units in $O/\mathfrak c$ and $\varphi_R(\mathfrak c)$ is the number of units in $R/\mathfrak c$.  This is derived in Neukirch's alg. number theory book in the number field case, but it goes through to any one-dimensional Noetherian domain with a finite residue rings and a finite class group.
In the imag. quadratic case the unit index $[R^\times:O^\times]$ is 1 most of the time so you don't notice it.  
Both $\varphi_R(\mathfrak c)$ and $\varphi_O(\mathfrak c)$ can be written in the form ${\text N}(\mathfrak c)\prod_{\mathfrak p \supset \mathfrak c}(1 - 1/{\text{N}(\mathfrak p)})$, where the ideal norm $\text N$ means the index in $R$ or $O$ and $\mathfrak p$ runs over primes in $R$ or $O$ for the two cases.<|endoftext|>
TITLE: Are there homotopy equivalences that are not weak homotopy equivalences?
QUESTION [8 upvotes]: I can imagine a map $f: X\to Y$ which is a homotopy equivalence of unpointed spaces, but which is not a homotopy equivalence of pointed spaces, no matter what basepoint is chosen.  That being the case, I don't see why $f$ would have to be a weak homotopy equivalence.
More detail:  by choosing  $x\in  X$, and its image $y\in Y$ as basepoints, we get a pointed map and an induced map on homotopy groups.  To be a weak homotopy equivalence, this needs to be an isomorphism (and one point is as good as any point if $X$ is path connected).  But this (hypothetical) pointed map is not invertible in the homotopy category of pointed spaces, so why should it induce an isomorphism?

REPLY [15 votes]: f is always a weak homotopy equivalence. This is related to the following assertion: Let $g\colon X\to X$ be homotopic to the identity. Then for any $x_0\in X$. $g$ induces an isomorphism $g_*\colon\pi_*(X,x_0)\to \pi_*(X, g(x_0))$. This is so because $g_*$ is the same as conjugation by the path $H(x_0\times I)$, where $H:X\times I \to X$ is the homotopy of $g$ with the identity. Conjugation by a path induces an isomorphism.
Now let $h$ be a homotopy inverse of $f$ and consider the homomorphisms
$$\pi_*(X, x_0)\to \pi_*(Y, f(x_0))\to \pi_*(X, hf(x_0)) \to \pi_*(Y, fhf(x_0))$$
the first and the third being induced by $f$ and the second by $h$.
The composition of the first two is an isomorphism and so is the composition of the last two.
It follows that all three homomorphisms are isomorphisms.<|endoftext|>
TITLE: When can we factor out the time dimension?
QUESTION [5 upvotes]: First of all, my knowledge of both GR and differential geometry is quite weak, so forgive me if the physics here doesn't make much sense.
Let $(M, g)$ be a smooth, connected Lorentzian manifold of dimension $n$.  Let $f: \mathbb{R}\to M$ be a smooth curve such that the pullback of $g$ through $f$ is everywhere negative (where we've chosen an orientation on $\mathbb{R}$); we say that $f$ is time-like.  Say that we can "factor" $f$ out of $M$ if there exists a manifold $S$ of dimension $n-1$ and an isomorphism $M\simeq S\times \mathbb{R}$ so that the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to S$ is constant and the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to \mathbb{R}$ is the identity.  Intuitively, this factorization exhibits $f$ as "time" in some reference frame, and $S$ as space.  My question is:

For which $(M, g)$ can every time-like path be factored out?

Minkowski space seems like an obvious example unless I'm missing something; it seems one can take a tangent vector to $f$ at any point and consider a perpendicular subspace to that vector as $S$.  I'd accept as an answer a characterization of all such $(M, g)$ in dimension $4$, or some nice sufficient condition on $M$ for factorization to always work.
If the motivation isn't obvious already, this is supposed to codify the intuition that in my reference frame, I seem to be standing still -- and that the same is true for everyone else, even if they seem to me to be moving.  My apologies if I've overloaded terms, or used them incorrectly.

Added:  Note that this condition is much stronger than stable causality; indeed, it certainly implies stable causality, as choosing any timelike path $f$ and then considering the given projection to $\mathbb{R}$ gives a global time function.  However, I am asking for (1) a product structure on $M$ for each path $f$ and (2) in order to formalize the notion that I seem to be standing still (to myself), the projection of $f$ to $S$ must be constant.  

Added:  I don't think global hyperbolicity suffices either.  The theorem of Geroch (it and other splitting theorems are discussed here, for example) does indeed give a decomposition of $M$ as $\mathbb{R}\times S$.  But I don't think this is enough.  In particular, I am asking for the following---for every timelike path $f: \mathbb{R}\to M$, there is a product structure $M\simeq \mathbb{R}\times S$ such that the projection to $\mathbb{R}$ is a section of $f$, and that $f$ is constant upon projection to $S$.  This is much stronger than Geroch's splitting theorem, as far as I can tell.

Added:  As the accepted answerer rightly points out in the comments to his question, I was wrong to claim that my condition is stronger than global hyperbolicity.  They are in fact equivalent.

REPLY [5 votes]: This sounds to me like you're asking that your spacetime admit a family of Cauchy surfaces (modulo annoyances like having $f$ be closed and acausal).   There's a theorem of Geroch which guarantees that this is equivalent to global hyperbolicity.  
I don't think that globally hyperbolic and stably causal are equivalent conditions, but I'm not an expert on causality conditions in GR, so take this claim with a grain of salt.<|endoftext|>
TITLE: What is Out(G-mod) for a finite group G?
QUESTION [6 upvotes]: Following the notation of Etingof-Nikshych-Ostrik what is Out(G-mod) for a finite group G?
That is what are all bimodule cateogries over the fusion category G-mod of complex G-modules which have the property that they're just G-mod as a left (resp. right) G-module up to equivalence of bimodules?  I think this is the same as the group of tensor autoequivalences of G-mod which do not come from conjugating by 1-dimensional objects, but ENO only prove that in the case where there are no 1-dimensional objects.
If you have an automorphism of the group G then you get an automorphism of G-mod.  I'm not totally sure though if outer automorphisms of G necessarily give outer automorphisms (for example could you have an outer automorphism that fixed all conjugacy classes?) and I also don't know whether all outer automorphisms come up this way.
The reason that I'm asking is that I want to understand the relationship between Out(C) and Out(D) where C and D are Morita equivalent fusion categories.  However, I realized I don't have enough examples where I understand what Out(C) actually is.

REPLY [4 votes]: The group $\operatorname{Out}(G\operatorname{-Mod})$ is equivalent to the group $\operatorname{Aut}_{\textbf{Tw}}(k[G])$ of gauge equivalence classes of twisted automorphisms of the Hopf algebra $k[G]$ defined by Davydov in Twisted automorphisms of Hopf algebras. In Davydov's other paper, Twisted automorphisms of group algebras, he describes this group in the case that $\lvert G \rvert$ is relatively prime to 6. (The proposed description I gave in the earlier version of this answer fails in general because $\operatorname{Out}(G)$ is not in general a normal subgroup of $\operatorname{Aut}_{\textbf{Tw}}(k[G])$.)
First, we note that since $G\operatorname{-Mod}$ is symmetric, any inner tensor autoequivalence is automatically tensor equivalent to the identity, so it suffices to describe the group of isomorphism classes of tensor autoequivalences. Let $\omega: G\operatorname{-Mod} \to \operatorname{Vect}$ be the usual fiber functor. Any tensor autoequivalence $F: G\operatorname{-Mod} \stackrel{\cong}{\to} G\operatorname{-Mod}$ gives a new (tensor) fiber functor $\omega \circ F: G\operatorname{-Mod} \to \operatorname{Vect}$. This gives $G\operatorname{-Mod}$ the structure of the category of representations of some Hopf algebra; it is shown in the paper On families of triangular Hopf algebras by Etingof and Gelaki that this Hopf algebra will be a twist of the original one, and thus $F$ induces a twisted automorphism of $H$. Conversely, any twisted automorphism of $F$ gives rise to an autoequivalence of $G\operatorname{-Mod}$. Natural transformations of monoidal functors correspond to gauge equivalences of twisted automorphisms. Thus, we can identify isomorphism classes of monoidal autoequivalences of $G\operatorname{-Mod}$ with gauge equivalence classes of twisted automorphisms of $k[G]$.<|endoftext|>
TITLE: Deformations of Nakajima quiver varieties
QUESTION [22 upvotes]: Are deformations of Nakajima quiver varieties also Nakajima quiver varieties ?
In case the answer to this is (don't k)no(w), here are some simpler things to ask for.

(If you're a differential geometer) Is any hyperkahler rotation / twistor deformation of a Nakajima quiver variety also a Nakajima quiver variety ?
(An example, for algebraic geometers) Consider the Hilbert scheme $H$ of $k$ points on the minimal resolution of $\mathbb C^2/(\mathbb Z/n)$. Or restrict to $n=2=k$ and consider $Hilb^2T^*\mathbb P^1$. Its exceptional divisor over the symmetric product defines a class in $H^1(\Omega_H)$ (despite the noncompactness). Using the holomorphic symplectic form, we get a deformation class in $H^1(T_H)$. (The corresponding deformation is not so far from the twistor deformation, and can be realised as a composition of a deformation of the ALE space followed by a twistor deformation.) Is there a Nakajima quiver variety in the direction of this deformation ?

For instance if I take the quiver $\bullet^{\ \rightrightarrows}_{\ \leftleftarrows}\bullet$, dimension vector (1,1), and an appropriate stability condition (or value of the real moment map) then I get $T^*\mathbb P^1$ as the moduli space over the value $0$ of the complex moment map, and the smoothing of the surface ordinary double point over nonzero values.
Now if I take dimension vector (2,2) I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. However, as I vary the value of the complex moment map I simply vary the surface that I take $Hilb^2$ of, rather than getting the deformation I'm after. (The hyperkahler rotation is not a $Hilb^2$, since the exceptional divisor disappears in this deformation.)
But is there another quivery way of producing this deformation ?

REPLY [4 votes]: One more example:
Take the quiver of type $A_1$, and vector spaces $V$, $W$ with $\dim V > \dim W$. Then the quiver variety is empty regardless of (generic) stability parematers nor complex moment map parameters. But we have one dimensional deformation space for the moment map equation.
Does someone know the deformation theory of the empty set ?<|endoftext|>
TITLE: Is the dual notion of a presheaf useful?
QUESTION [16 upvotes]: It seems that there is a common theme in mathematics where, if we want to find out about a category C, then we look at $\hat{C}$ (the category of contravariant functors from $C$ to $Set$). There are all sorts of good reasons for this (Yoneda's lemma being a big one, and the fact that this is a topos). There are other versions, sometimes you look at contravariant functors into some other category, like groups (e.g. for algebraic groups). But I have never seen the dual notion: covariant functors from some other category INTO $C$. 
Is this notion as useful as the notion of a presheaf? I guess it's like looking at a category "over" $C$ as opposed to a category "under" it. I guess, hidden in this question is the question: Can one learn anything about a category $C$ by looking at presheaves of $C$? For example, does the difference between presheaves of sets and presheaves of ableian groups tell us anything about the differences between the category of Groups and the category of Sets? 
EDIT: There seems to be a bit of confusion of what I mean by "dual of a presheaf." I don't mean "copresheaves" (which I didn't know existed before I asked this), I mean what you get by reversing the arrow of the functor not taking the opposite category. So I'm looking at functors into the category of interest, as opposed to out of them. I can see how this is confusing because usually "dual" doesn't turn around functors, just arrows inside categories. So I guess I mean presheaves in $C^{op}$... (but covariant instead of contravariant?)... who knows.

REPLY [14 votes]: Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be much longer than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc.
To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then:
(1) Functors into $\mathcal{A}$ are interesting.
(2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$.
(3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof).
(4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality.
The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples.
To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of generalized elements.
Definition: Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A generalized element of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in_{d} a$.
We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows out of $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised:

The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma.
Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them).

For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples.

Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a very large class of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting.
If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes.

Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category.
For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples.

In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$.
Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$.

Hope it helps, regards,
G. Rodrigues<|endoftext|>
TITLE: Contracting a geodesic on a space of curvature less than 1
QUESTION [6 upvotes]: I would like to ask for a reference to the following statement (hopefully correct):
Let $M$ be a manifold of sectional curvature at most $1$ and let $\gamma$ be a closed geodesic.
Suppose that $\gamma$ is contractible. Then for any contraction of this geodesic at some point its length will be equal to $2\pi$. 
It would be even better if there is a reference for the case when $M$ a locally $CAT(1)$ space (not necessarily manifold)

REPLY [8 votes]: There is an article by B. Bowditch, Notes on locally CAT(1) spaces, in Geometric Group Theory,  R.Charney, M.Davis, and M.Shapiro eds., de Gruyter (1995) that will likely be of help.  The article is about curve shortening in locally CAT(1) spaces.<|endoftext|>
TITLE: For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular
QUESTION [5 upvotes]: For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.
If the answer is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?

REPLY [14 votes]: Over a field of characteristic zero, your result is true. This is Corollary III.10.7 in Hartshorne.  

In characteristic $p$ no. The simplest example is to take $k$ an algebraically closed field and map $\mathbb{A}^1_k$ to itself by $x \mapsto x^p$. For every $t \in k$, the fiber above $t$ is $\mathrm{Spec}\ k[x]/(x^p-t) \cong \mathrm{Spec} \ k[y]/y^p$ where the isomorphism is $x-t^{1/p} = y$. So every fiber is singular.
There is a more interesting counter-example due to Serre: Let $k$ be an algebraically closed field of characteristic $2$. Consider the planar cubic
$$ F_{a,b,c}(x,y,z) :=a  (y^2 z + y z^2) + b (x^2 z + x z^2) + c  (x^2 y + x y^2) \quad (*)$$
We leave it to the reader to check that $F=0$ is singular at $(\sqrt{a}:\sqrt{b}:\sqrt{c})$. Generically, the singularity is a node cusp. Choose $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ in $k^3$ and try to map $\mathbb{P}^2_k \to \mathbb{P}^1_k$ by
$$(x:y:z) \mapsto (F_{(a_1,\ b_1,\ c_1)}(x,y,z) : F_{(a_2,\ b_2,\ c_2)}(x,y,z))$$
Then the fiber over $(t_1:t_2)$ is $F_{(t_1 a_1+t_2 a_2,\ t_1 b_1+t_2 b_2,\ t_1 c_1+t_2 c_2)}=0$ which, as we just computed, is singular. More precisely, the above map is undefined at the $9$ points where $F_{(a_1,\ b_1,\ c_1)} = F_{(a_2,\ b_2,\ c_2)} =0$. But, if we take $X$ to be $\mathbb{P}^2$ blown up at those $9$ points, then we get a map from the regular $X$ to the regular $S$, where every fiber is a nodal cuspidal cubic or worse.
Remark: Both of these counter-examples are still counter-examples if you replace "algebraically closed" by "perfect", but it would make my exposition messier.<|endoftext|>
TITLE: Are there natural examples of mathematical statements which follow from consistency statements?
QUESTION [13 upvotes]: Motivation
One of the methods for strictly extending a theory $T$ (which is axiomatizable and consistent, and includes enough arithmetic) is adding the sentence expressing the consistency of $T$ ( $Con(T)$ ) to $T$. But this extension ( $T+Con(T)$ ) looks very artificial from the mathematical viewpoint, i.e. does not seem to have any mathematically interesting new consequences, and therefore is probably of no interest to a typical mathematician.

I would like to know if there is a natural theory (like PA, ZFC, ... ) which by adding the consistency statement we can prove new mathematically interesting statements.
I don't have a definition for what is a natural theory or a mathematically interesting statement, but a theory artificially build for the sole purpose of this question would not be natural, and a purely metamathematical statement (like consistency of $T$, or a statement depending on the encoding of $T$ or its language, or ...) would not count as a mathematically interesting statement.
Questions:


Is there a natural theory $T$ and an mathematically interesting statement $\varphi$, such that it is not known that $T \vdash \varphi$, but $T + Con(T) \vdash \varphi$?

Is there a natural theory $T$ and an interesting mathematical statement $\varphi$, such that $T \nvdash \varphi$ but $T + Con(T) \vdash \varphi$?

REPLY [17 votes]: Vitali famously constructed a set of reals that is not Lebesgue measurable by using the Axiom of Choice. Most people expect that it is not possible to carry out such a construction without the Axiom of Choice.
Solovay and Shelah, however, proved that this expectation is exactly equiconsistent with the existence of an inaccessible cardinal over ZFC. Thus, the consistency statement Con(ZFC + inaccessible) is exactly equivalent to our inability to carry out a Vitali construction without appealing to AC (beyond Dependent Choice). 
Thus, if $T$ is the theory $ZFC+$inaccessible, then T+Con(T) can prove "You will not be able to perform a Vitali construction without AC", but $T$, if consistent, does not prove this.
I find both this theory and the statement to be natural (even though the statement can also be expressed itself as a consistency statement). Most mathematicians simply believe the statement to be true, and are often surprised to learn that it has large cardinal strength.

There is another general observation to be made. For any consistent theory $T$ whose axioms can be computably enumerated, and this likely includes most or all of the natural theories you might have in mind, there is a polynomial $p(\vec x)$ over the integers such that $T$ does not prove that $p(\vec x)=0$ has no solutions in the integers, but $T+Con(T)$ does prove this. So if you regard the question of whether these diophantine equations have solutions as natural, then they would be examples of the kind you seek. And the argument shows that every computable theory has such examples. 
The proof of this fact is to use the MRDP solution of Hilbert's 10th problem. Namely, Con(T) is the assertion that there is no proof of a contradiction from $T$, and the MRDP methods show that such computable properties can be coded into diophantine equations. Basically, the polynomial $p(\vec x)$ has a solution exactly at a Goedel code of a proof of a contradiction from $T$, so the existence of a solution to $p(\vec x)=0$ is equivalent to $Con(T)$. If $T$ is consistent, then it will not prove $Con(T)$, and so will not prove there are no integer solutions, but $T+Con(T)$ does prove that there are no integer solutions. 
By the way, it is not true in general that if $T$ is consistent, then so is $T+Con(T)$. Although it might be surprising, some consistent theories actually prove their own inconsistency! For example, if PA is consistent, then so is the theory $T=PA+\neg Con(PA)$, but this theory $T$ proves $\neg Con(T)$. Thus, there are interesting consistent theories $T$, such as the one I just gave, such that $T+Con(T)$ proves any statement at all!<|endoftext|>
TITLE: Positivity of sequences via generating series
QUESTION [12 upvotes]: There are different ways of showing that a given sequence $a_0,a_1,a_2,\dots$
of integers, say, is nonnegative. For example, one can show that $a_n$ count
something, or express $a_n$ as a (multiple) sum of obviously positive numbers.
Another recipe is manipulating with the corresponding generating series
$A(x)=\sum_{n=0}^\infty a_nx^n$ and showing that $A(x)\ge0$ (this is the
notation for a series which has all coefficients nonnegative, and this extends
to formal power series in as many variables as needed).
An example of criterion in this direction is
$$
(*) \qquad
A(xy)\ge0 \iff \frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)\ge0
$$
(the multiple $1/(1-x)(1-y)$ is introduced for cosmetic purposes only and, of course, both $A(x)\ge0$ and $A(xy)\ge0$ are by definition equivalent to the nonnegativity of the sequence $a_n$).
The latter can be verified by comparing the corresponding coefficients
in the power series expansion
$$
\frac1{(1-x)(1-y)}A\biggl(\frac{xy}{(1-x)(1-y)}\biggr)
=\sum_{n=0}^\infty a_n\sum_{k,m=0}^\infty\binom{n+k}n\binom{n+m}nx^{n+k}y^{n+m}.
$$
On the other hand, the one-dimensional version of $( * )$,
$$
A(x)\ge0 \iff \frac1{1-x}A\biggl(\frac{x}{1-x}\biggr)\ge0,
$$
is simply false.
My question is whether it is possible to find two nontrivial rational functions
$p(x)\in\mathbb Q[[x]]$ and $r(x)\in x\mathbb Q[[x]]$ in one variable $x$
such that
$$
A(x)\ge0 \iff p(x)A\bigl(r(x)\bigr)\ge0.
$$
Although I am not supposed to put several problems in one question, I
would also ask about a more direct proof of $( * )$ and about general ways
of constructing such $p$ and $r$ in more than one variable.
Motivation. Basically I am interested in proving nonnegativity of certain
$q$-series sequences $a_0(q),a_1(q),a_2(q),\dots$ by manipulating with the
corresponding generating series $A_q(x)=\sum_{n=0}^\infty a_n(q)x^n$. Some
of them can be "guessed" from non-$q$-versions, for example there is a neat
$q$-analogue of the criterion $( * )$.

REPLY [3 votes]: It is impossible, and not just for rational functions.  To see this, let's consider the coefficients $b_n$ of $p(x) A(r(x))$ as functions of the $a_n$, the coefficients of $A(x)$.  Since $r(0) = 0$ (as it must be) we see that:

Each $b_n$ is a linear combination of $a_0, \dots, a_n$; i.e. we have an upper-triangular infinite matrix $F$, not depending on $a_n$, such that (writing $\vec{a} = (a_n), \vec{b} = (b_n)$) $\vec{b} = F \vec{a}$.  Suppose for now that $p(0), r'(0) \neq 0$; then $F$ has a nonzero diagonal, and so is invertible.
If $\vec{b} \geq 0$ for all $\vec{a} \geq 0$, then in particular this is true of the columns of $F$, taking $\vec{a}$ to be infinite "basis" vectors.  Conversely, we have equivalently that $\vec{a} = F^{-1} \vec{b}$, so if $\vec{a} \geq 0$ for all $\vec{b} \geq 0$ this must be true of the columns of $F^{-1}$.  We conclude that both $F$ and $F^{-1}$ have nonnegative entries.
Lemma in linear algebra: if $F$ is upper-triangular and both it and $F^{-1}$ have nonnegative real entries, then $F$ is diagonal.  Proof by induction: true for $1 \times 1$ matrices vacuously.  In general, by induction we may assume that the upper-left and lower-right $(n - 1) \times (n - 1)$ blocks of $F$ are diagonal, so only the $(1,n)$ entry of $F$ is nonzero off the diagonal.  Then we have $(F^{-1})_{1n} = -F_{1n} F_{nn}/F_{11}$.  Since $F_{11}$ and $F_{nn}$ are both positive, $F_{1n}, (F^{-1})_{1n} \geq 0$ implies $F_{1n} = 0$.
This is also true of infinite matrices, since we can compute the finite upper-left blocks independently of the rest of the matrix.
However, if $F$ is diagonal then we see that $p(x)A(r(x)) = \sum a_n p(x) r(x)^n = \sum F_{nn} a_n x^n$ for all choices of $a_n$, so (for example, taking $a_n = t^n$ for a new variable $t$ and rewriting both sides as power series in $t$) we have $p(x) r(x)^n = F_{nn} x^n$ for all $n$ (and some $F_{nn} > 0$).  That is, $(r(x)/x)^n = F_{nn} p(x)^{-1}$ for all $n$, so in fact $r(x)/x = F_{11}/F_{00}$ is constant, and finally, $p(x)$ is constant as well.
Now we remove the assumptions that $p(0), r'(0) \neq 0$.  If $x^k$ divides $p(x)$, then replacing $p(x)$ by $p(x)/x^k$ does not change positivity of the coefficients.  Now suppose the bottom exponent of $r(x)$ is $x^m$ with positive coefficient; then in $\mathbb{R}[[x]]$ we can write $r(x) = s(x)^m$, and if we denote $A_m(x) = A(x^m)$, we have $A(r(x)) = A_m(s(x))$.  Clearly, $A_m$ has nonnegative coefficients if and only if $A$ does, and $s'(0) \neq 0$, so the previous proof applies and $s(x)$ is a multiple of $x$, i.e. $r(x)$ is a multiple of $x^m$, and $p(x)$ is a multiple of $x^k$.<|endoftext|>
TITLE: What is Lagrange Inversion good for?
QUESTION [71 upvotes]: I am planning an introductory combinatorics course (mixed grad-undergrad) and am trying to decide whether it is worth budgeting a day for Lagrange inversion. The reason I hesitate is that I know of very few applications for it -- basically just enumeration of trees and some slight variants on this. I checked van Lint and Wilson, Enumerative Combinatorics II (but not the exercises) and Concrete Mathematics, and they all only present this application.
So, besides counting trees, where can we use Lagrange inversion?

REPLY [4 votes]: An elementary proof of Rodrigues' formula for Legendre polynomials (which is usually done via the orthogonality properties of the polynomials). If we define the polynomials by the classical generating function,
$$ \frac{1}{\sqrt{1-2z_{0}w+w^{2}}}=\sum_{n=0}^{\infty}P_{n}\left(z_{0}\right)w^{n} $$
and consider the solution $z(w)$ of $$w=f(z)=2\frac{z-z_{0}}{z^{2}-1}$$ fixed by $z(0)=z_0$, then the direct application of the Lagrange's formula (formulated for the solution $z(w)$ of the equation $w=f(z)$ with $w_0=f(z_0)$)
$$\frac{\Phi\left(z\left(w\right)\right)}{1-\left(w-w_{0}\right)\left.\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{z-z_{0}}{f(z)-w_{0}}\right)\right|_{z=z(w)}}=\sum_{n=0}^{\infty}c_{n}\left(w-w_{0}\right)^{n},\quad c_{n}=\frac{1}{n!}\left.\frac{\mathrm{d}^{n}}{\mathrm{d}z^{n}}\left[\Phi\left(z\right)\left(\frac{z-z_{0}}{f(z)-w_{0}}\right)^{n}\right]\right|_{z=z_{0}}$$
to $f(z)$, $w_0=f(z_0)=0$ and $\Phi(z)\equiv 1$ gives
$$P_{n}(z)=\frac{1}{2^{n}n!}\frac{\mathrm{d}^{n}}{\mathrm{d}z^{n}}\left[\left(z^{2}-1\right)^{n}\right].$$<|endoftext|>
TITLE: Set of real numbers with positive measure containing no midpoints
QUESTION [13 upvotes]: Does there exists a subset E of R with positive measure and without containing any midpoints (i.e. x,y distinct in E, (x+y)/2 not in E)?

REPLY [9 votes]: As already  said by Gerry, the answer to your question is negative. However, it becomes positive if you only ask your set to have Hausdorff dimension 1 instead of positive Lebesgue measure, see 
Salem, R.; Spencer, D. C. On sets which do not contain a given number of terms in arithmetical progression.  Nieuw Arch. Wiskunde (2)  23,  (1950). 133--143.
For a more general recent result see
Tamás Keleti Construction of one-dimensional subsets of the reals   not containing similar copies of given patterns, Analysis and PDE Vol. 1 (2008), No. 1, 29-33 
(if you do not know this journal, you should have a look at it and more generally to the web site of the Mathematical Science publishers, by the way.)

REPLY [6 votes]: No, such a set cannot exist and one can prove this using Lebesgue Density Theorem and a simple pegionhole argument. Infact all points $x$ which are density points of $E$ will
be a midpoint for some $y,z \in E$ i.e., $x=\frac{y+z}{2}$. 
Let $F \subseteq E$ be the set of density points of E, and $x \in F$.
Then there exists a $\epsilon > 0$ such that $m( B_{\epsilon}(x)\cap F) > \epsilon$. Now if $x$ is not a midpoint of $E$ then $\forall d \in (0,\epsilon)$, atleast one of $x-d$ or $x+d$ does not belong to $F$. 
But then $m( B_{\epsilon}(x)\cap F)= \int_0^{\epsilon} |F\cap \lbrace x-t,x+t\rbrace| dt < \epsilon$, a contradiction !!
A set $A$ of real number is called Universal if every measurable set of positive measure necessarily contains an affine image of $A$. A simple variation of the above argument will give that all finite set $A$ are infact Universal. However, no example of an infinite Universal set is knwon and its a conjecture of Erdos that no infinite universal sets exists.
This paper has a nice discussion and references to this problem 
M. Kolountzakis: Infinite Patterns That Can Be Avoided by Measure,
Bull. London Math. Soc. 29 (1997), 4, 415-424.  http://fourier.math.uoc.gr/~mk/ps/universal.pdf
As Gerry and Benoît Kloeckner has mentioned the problem becomes interesting when one considers Hausdroff measure instead of Lebesgue measure. 
Recently I. Laba and M. Pramanik proved existence of 3 term arithmetic progression even in closed sets which has Hausdroff dimension close to 1, `under the condition that E supports a probability measure obeying appropriate dimensionality and Fourier decay conditions'
I. Laba and M. Pramanik: "Arithmetic progressions in sets of fractional dimension",, Geom. Funct. Anal. 19 (2009), 429-456. http://www.math.ubc.ca/~ilaba/preprints/progressions-may15.pdf<|endoftext|>
TITLE: Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
QUESTION [28 upvotes]: Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\operatorname{det}(A-xI) = p_0 + p_1x + \dots + p_n x^n$.
I am looking for a proof that
$-\operatorname{adj}(A) = p_1 I + p_2 A + \dots + p_n A^{n-1}$.
In the case where $\operatorname{det}(A)$ is a unit, $A$ is invertible, and the proof follows from the Cayley–Hamilton theorem. But what about the case where $A$ is not invertible?

REPLY [6 votes]: This formula can be obtained during a proof of the Cayley-Hamilton theorem, as is indicated on its Wikipedia article. The essence of the argument is that Euclidean division by a monic polynomial (on the left, say), can be performed in the polynomial ring over any (unitary) ring, not necessarily commutative; this follows directly from consideration of what Euclidean division does, or by a simple inductive argument.
Since I care about polynomials being monic, I'l define the characteristic polynomial of a matrix $A$ to be $\chi_A=\det(I_nX-A)=\sum_{i=0}^nc_iX^i$ where $c_n=1$ (and $c_0=\det(-A)$), so the result to prove becomes $\mathrm{adj}(-A)=c_1I_n+c_2A+\cdots+c_{n-1}A^{n-2}+A^{n-1}=\sum_{i=1}^nc_iA^{i-1}$
Consider the noncommutative ring  $M=\mathrm{Mat}_n(R)$, and using Euclidean division in $M[X]$ (in which $R[X]$ is embedded by mapping $r$ to $rI_n$) divide $\chi_A$ on the left by $X-A$. Since we know that $(X-A)\mathrm{adj}(X-A)=\det(X-A)=\chi_A$, uniqueness of quotient and remainder in Euclidean division implies they will have to be $\mathrm{adj}(X-A)$ and $0$, respectively. Writing the quotient $\mathrm{adj}(X-A)=\sum_{i=0}^{n-1}B_iX^i$, its coefficients $B_i\in M$ are determined in the division successively as $B_{n-1}=c_n=1$ and $B_{i-1}=c_i+AB_i$ for $i=n-1,\ldots,1$ (these are just the intermediate values while computing the evaluation of $\chi_A$ at$~X=A$ using the Horner scheme), which expands to $B_{i-1}=c_iA^0+c_{i+1}A^1+\cdots+c_nA^{n-i}$. In particular the constant coefficient of the quotient $\mathrm{adj}(X-A)$ equals $B_0=\sum_{i=1}^nc_iA^{i-1}$, but this is also $\mathrm{adj}(-A)$ (by substituting $X=0$).
To retrieve the Cayley-Hamilton theorem from the formula found, multiply on the left or right by $A$ and move the left hand side to the right.<|endoftext|>
TITLE: The associated prime ideals of $Ext^i_R(M,N)$
QUESTION [9 upvotes]: If R is a commutative noetherian ring, M and N are modules with M finite. It is well known in commutative algebra that $AssHom_R(M,N)=Supp(M)\cap Ass(N)$. But I want to know whether there is a formular for $AssExt_R^i(M,N) ?$ Thanks!

REPLY [5 votes]: The short answer is no, as hinted at in the comments by Karl and Graham. I would argue that even the question of understanding the minimal primes of $\text{Ext}^i(M,N)$ (which is the minimal set of the associated primes, hence an easier question) is intractable. Let's assume that $R$ is Noetherian and $M,N$ are finitely generated. 
Since $\text{Ext}$ localizes, one essentially needs to understand when $\text{Ext}^i(M,N)$ vanishes over a local ring $R$. There is no good answer in general. Even in the very nice case when $N$ is the canonical module (assuming it exists), these $\text{Ext}$ are Matlis dual to the local cohomology modules of $M$, and while there are good bounds on the indices when they vanish (below the depth and beyond the dimension), there are no general result which can tell you exactly when.  
Here is one more way to see the complexity of the problem even when $i=1$. Take a free cover of $M$ to get an exact sequence $0 \to M_1 \to F \to M \to 0$. Applying $\text{Hom}(-,N)$ to get: 
$$0 \to \text{Hom}(M,N) \to \text{Hom}(F,N) \to \text{Hom}(M_1,N) \to  \text{Ext}^1(M,N) \to 0$$
Assuming some mild condition on $M,N$ (reflexive for example), this shows that the set of associated prime of $\text{Ext}^1(M,N)$ is a subset of the set of primes $p$ such that $\text{depth}(\text{Hom}(M,N)_p) = 2$. Again, this set is not very well understood in general.<|endoftext|>
TITLE: Examples of groups without the n-positive approximation property
QUESTION [10 upvotes]: Let $G$ be a locally compact group and let $A(G)$ be the http://eom.springer.de/f/f120080.htm>Fourier algebra of $G$, which we view as the predual of the group von Neumann algebra $\mathcal M(G)$.  Let $MA(G)$ be the space of multipliers of $A(G)$, i.e., $\varphi \in MA(G)$ if and only if $\varphi \psi \in A(G)$ for all $\psi \in A(G)$.  Then $\varphi \in MA(G)$ induces a bounded operator $m_\varphi: A(G) \rightarrow A(G)$, and hence also a bounded operator $M_\varphi = m_\varphi^*$ on $\mathcal M(G)$.  
$M_\varphi$ is completely bounded if $\| M_\varphi \|_{CB} = \sup_n \| M_\varphi \otimes {\rm id}_n \| < \infty$, where ${\rm id}_n$ is the identity operator on the $n \times n$ matrices $\mathbb M_n(\mathbb C)$.  $M_\varphi$ is $n$-positive if $M_\varphi \otimes {\rm id}_n$ takes the positive cone $\mathcal M(G)_+$ into itself, or equivalently $\| M_\varphi \otimes {\rm id}_n \| = \varphi(e)$.  $M_\varphi$ is completely positive if it is $n$-positive for every $n \in \mathcal N$.
A well known result is that $G$ is amenable if and only if $A(G)$ has an approximate unit $\{ \varphi_k \}_k$ such that $M_{\varphi_k}$ is completely positive for all $k$.  Haagerup showed that $SL_2(\mathbb R)$, and all of its lattices have the completely bounded approximation property:  For these groups, $A(G)$ has an approximate unit $\{ \varphi_k \}_k$ such that $\sup_k \| M_{\varphi_k} \|_{CB} < \infty$, (in fact he showed that this supremum can be 1 for $SL_2(\mathbb R)$, and all of its lattices).  To contrast, he also showed that $SL_m(\mathbb R)$, and all of its lattices do not have the completely bounded approximation property whenever $m \geq 3$.  De Canniere and Haagerup have also shown that free groups have the $n$-positive approximation property for every $n \in \mathbb N$:  For these groups, given any $n \in \mathbb N$, $A(G)$ has an approximate unit $\{ \varphi_k \}$ of compactly supported functions such that $\varphi_k$ is $n$-positive.
Recently, I was at a conference and Mikael de la Salle asked me if I knew of any examples of groups which do not have the $n$-positive approximation property.  I do not and so I thought I would ask here.
1)  What is an example of a group which for some $n$ does not have the $n$-positive approximation property?
2)  What is an example of a group which for any $n$ does not have the $n$-positive approximation property?
3)  Are there groups which have the $n$-positive approximation property for some $n$, but not for $n + 1$?

REPLY [13 votes]: Haagerup actually proved (lattices of) higher rank Lie groups do not have $1$-positive approximation property, nor bounded approximation property, i.e., there is no uniformly bounded sequence of compactly supported multipliers that converges pointwise to $1$. (It's still open whether the reduced group C$^*$-algebra of such a group also fails bounded approximation property.) Also, lamplighter groups on non-amenable groups do not have $1$-positive approximation property (bounded approximation property with constant $1$). The reason is that the proofs of no $1$-positive approximation property boil down to non-existence of certain types of multipliers on an amenable group; and for multipliers $M_{\varphi}$ on an amenable group, cb-norm coincides with norm. (See Cowling et al., Duke Math. J. 127 (2005), 429--486 for a survey.) Problem 3 seems inaccessible.<|endoftext|>
TITLE: Koszul duality and modules over the Chevalley complex
QUESTION [8 upvotes]: Let $g$ be a Lie algebra over $\mathbb{C}$. Then the equivalence between the derived category of modules over $U(g)$ and the coderived category of co-modules over it's Chevalley complex $C_*(g)$ in which $M\rightarrow C_*(g,M)$ is a classical example of Koszul duality. In the case when $g$ is a finite dimensional Lie algebra, it is easier for most humans to dualize everything and discuss modules over the Chevalley cochain complex $C^*(g)$. Most of the time, people are most interested in various subcategories of modules over a given Lie-algebra. For example the category of semi-simple modules or finite dimensional modules over $g$. I would like to know if for any Lie-algebra $g$ if one can understand what happens to the subcategories of the derived category generated by these objects under the Koszul duality. Are the corresponding subcategories equally natural to describe? My intuition is that the semisimple modules are sent to something like (co) perfect modules based upon the observation that $\mathbb{C}$ is sent to $C_*(g) $. But I don't think this is exactly right and I wonder if someone could help me out.

REPLY [11 votes]: The derived category of finite-dimensional $g$-modules is not a full subcategory of the derived category of arbitrary $g$-modules for a finite-dimensional Lie algebra $g$, in general (e.g., for a semi-simple Lie algebra $g$).  However, one can consider the full subcategory of the derived category of $g$-modules consisting of complexes of $g$-modules with bounded and finite-dimensional cohomology.  This triangulated category is equivalent to the homotopy category of DG-modules over $C^\ast(g)$ that are free/projective and finitely generated as graded modules.
The proof (based on the general Koszul duality theorem mentioned in the question) proceeds as follows.  The category of DG-modules over $C^\ast(g)$ that are injective/projective as graded modules is a full subcategory of the co/contraderived category of DG-modules.  The category of complexes of $g$-modules with bounded and finite-dimensional cohomology is generated by finite-dimensional $g$-modules, and these are transformed by the Koszul duality functor into DG-modules that are free and finitely generated as graded modules.  This provides a fully faithful functor in one direction.
To prove that it is essentially surjective, consider the functor assigning to a complex of $g$-modules the underlying complex of vector spaces.  Let us presume that our Koszul duality functor is $M\longmapsto C^\ast(g,M)$.  Then this Koszul duality functor transforms our functor of forgetting the action of $g$ into the functor assigning to a DG-module over $C^\ast(g)$, projective as a graded module, its quotient complex by the action of the augmentation ideal of $C^\ast(g)$.  Thus the complex of $g$-modules corresponding to a DG-module that is projective and finitely generated as a graded module has bounded and finite-dimensional cohomology.<|endoftext|>
TITLE: Rolling a convex body: Geodesics vs. rolling curves
QUESTION [7 upvotes]: What are the curves of contact on a convex body $B$ rolling down an inclined plane?
Assume $B$ is smooth, and there is sufficient friction to prevent slippage.
Certainly, one can develop a geodesic to a straight line on a plane by rolling $B$ so that the geodesic
is the point of contact, but it doesn't appear that this in general would be the point of contact
in the physical situation of free rolling under gravity.  It is for a sphere, which will roll along
great circles.  And an ellipsoid should roll along its three simple closed geodesics (although there
are significant instabilities with all but the shortest closed geodesic—I'd prefer to ignore
stability issues). But for other shapes,
I imagine that an off-center center of gravity (and perhaps rotational momentum?) will cause the rolling to deviate from a
geodesic.  (But I am uncertain of this.  Please correct me if I'm wrong!)
Assuming this is correct (that rolling curves are not always geodesics),
what are the conditions that determine if a curve $\gamma$ is a rolling curve:
$\gamma$ is the trace of the point of contact between $B$ and an inclined plane as it rolls,
from some initial position?
Perhaps: If $p \in \gamma$ is a point of contact, then (a) the normal vector $N$ of $\gamma$ at $p$ must be
perpendicular to the plane, and (b) the center of gravity of $B$
must lie in the osculating plane of $\gamma$ at $p$ (the plane containing $N$ and
the tangent $T$ vector at $p$)?
Have what I christened rolling curves been studied in the literature?  If so, under what name?
My searches have been unsuccessful.  Can you think of shapes outside of {sphere, ellipsoid, cylinder} where the rolling curves can be determined? Thanks for any thoughts or pointers!

REPLY [5 votes]: The rolling motion of a convex symmetric body on a horizontal plane is a classical problem. In the symmetric case, Chaplygin was the first who showed that the full equations of motion can be reduced to a linear integrable system of two ODEs. A modern exposition of Chaplygin's results can be found in the very recent book by Cushman, Śniatycki and Duistermaat.
The problem of rolling motion on an inclined plane is, in general, nonintegrable (this problem was studied, in particular, by V.V. Kozlov in 1990s).
As for the tracing trajectories of the point of contact,  you might be interested in this article (also available on arXiv) and the references therein. The authors discuss the case of a disc (i.e. a convex body of revolution) rolling on a horizontal plane.
From the Introduction:

It appears that the point of contact performs the composite bounded motion: it periodically traces some closed curve which rotates as a rigid body with some constant angular velocity about the fixed point...<|endoftext|>
TITLE: Literature for gauge field theory on the lattice in geometrical formulation
QUESTION [9 upvotes]: I have found an article by Huebschmann, Rudolph and Schmidt: http://www.springerlink.com/content/b8v216v0m8h16264/  about "A Gauge Model for Quantum Mechanics on a Stratified Space" and I am very interested in this subject, but I don't have any background in gauge-field theory or something like that. 
So my question is, if there are any good introductory books or overview articles which cover Hamiltonian (quantum) gauge field theory on the lattice in a geometrical mathematical language like the article mentioned above? Especially I am interested in references which deal with more regular cases rather than the singular cases discussed in the article above. Finally it is important that it covers this topic in a way that one can gain a bit deeper physical understanding (without giving up a clear, rigorous and geometrical mathematical language). 
Added
Thanks for your answers. Gauge field theory seems to be a very wide field, so I should perhaps mention why I want to learn some basics about gauge field theory beside of very strong intrinsic interest. 
I am studying phase-space reduction in the context of deformation quantization of systems with finite degrees of freedom. Now I want to know if it is possible to reinterpret this situation (at least as a  toy model) in some way in terms of gauge field theories. So my aim is to learn at least as much of gauge field theory that I can understand if and why such an reinterpretation is possible or how far one can go.
The idea to look for lattice gauge theories was the following quote from the paper by Huebschmann et. al.
"Gauge theory in the Hamiltonian approach, phrased on a finite spatial lattice, leads
to tractable finite-dimensional models for which one can analyze the role of singularities
explicitly. Under such circumstances, after a choice of tree gauge has been made, the
unreduced classical phase space amounts to the total space $T^* (K \times \dots \times K)$ of the cotangent bundle on a product of finitely many copies of the manifold underlying the
structure group $K$. Gauge transformations are then given by the lift of the action of K
on $K \times \dots \times K$ by diagonal conjugation. This leads to a finite-dimensional Hamiltonian system with symmetries."
Why a asked for a "geometric language" is clear because the strength of deformation quantization is in fact in the area of the quantization of systems with a more complex phase-space geometry.
Having this additional information it is perhaps easier for you to give me some hints where to start in literature.

REPLY [5 votes]: Rudolph&Schmidt just released their second volume Differential Geometry and Mathematical Physics II: Fiber bundles, Topology and Gauge fields, and it covers exactly this stuff.
These two volumes instantly became my favorite intro to geometric methods in theoretical physics. I'm sure in a few years they'll replace the likes of Nakahara due to their incredible mathematical completeness and physical insight.<|endoftext|>
TITLE: Why are they called isothermal coordinates?
QUESTION [20 upvotes]: On a Riemannian manifold, a coordinate system is called "isothermal" if the Riemannian metric in those coordinates is conformal to the Euclidean metric:
$$g_{ij} = e^{f} \delta_{ij}$$
My question is: Why are such coordinate systems called "isothermal"? It must have something to do with classical thermal physics. I tried looking for a reason online, with no success.
It is well known that when the dimension $n=2$, there always exist isothermal coordinates, and this is probably where they were first introduced. So maybe the nomenclature has something to do with heat diffusion in the plane?
(The reason I ask is because I am planning to give a seminar talk next week giving a proof that such coordinates exist when $n=2$, and thought it would be nice to explain to the students where the name comes from...)

REPLY [22 votes]: Isothermal coordinates are harmonic. In other words, it solves $\triangle_g u = 0$. So locally it is a stationary solution of the heat equation. In physics, for a steady state distribution of temperatures, each level set is called an isotherm.

REPLY [15 votes]: According to Gray, Abbena and Salamon, that's the name given to such coordinate systems by Gabriel Lamé in his 1833 study of heat transfer. The reason is, if you've got a thermally isolated surface of constant heat conduction, the constant coordinate lines are isotherms iff the coordinates are isothermal.<|endoftext|>
TITLE: Infinite games: are they well defined?
QUESTION [8 upvotes]: It is just my curiosity about this question where we have an infinite game and (according to the answers) winning strategies for both players. I am familiar with terminating games only, and I am pretty sure that nonterminating ones can cause many paradoxes. As this is a community wiki mode, I would be happy to see formal definitions, existence of wining strategies as well as possible collisions.
Thanks in advance for keeping my mind finite!

REPLY [2 votes]: In recent years, a relatively abstract and general formulation of extensive form games has been developed by Carlos Alos-Ferrer and Klaus Ritzberger. The formulation is general enough to discuss when some useful properties hold- and when they fail. In particular, in Trees and Extensive Forms (WP version can be found here here), they give necessary and sufficient conditions for when every pure strategy in an extensive game induces an outcome and when this outcome is unique.<|endoftext|>
TITLE: Common quotients of direct products
QUESTION [5 upvotes]: In a paper I read recently, the authors use the fact that if two groups G and H have no nontrivial common quotient, then neither do GxG and HxH. It's unclear from the context whether this is true for all groups, or just groups of the type that are important for this paper, and they don't prove the claim.
I've been trying to prove the statement for general groups without any success. Is it true?
I've also tried restricting to the case I really care about, as follows. Let K, K', M, and M' be normal subgroups of U, with $U/K \simeq U/K'$ and $U/M \simeq U/M'$. Suppose that KK' = KM = KM' = K'M = K'M' = MM' = U. Is it also the case that $(K \cap K')(M \cap M') = U$? The reason this is connected to the above is that $U/(K \cap K')(M \cap M')$ is a common quotient of $U/(K \cap K') \simeq U/K \times U/K' \simeq U/K \times U/K$ and $U/(M \cap M') \simeq U/M \times U/M' \simeq U/M \times U/M$. So if U/K and U/M have no nontrivial common quotients, is the same true of $U/(K \cap K')$ and $U/(M \cap M')$?
Finally, if the result is not true in general for either of these cases, what about if we restrict to finite groups G and H?

REPLY [13 votes]: If $Q$ is a quotient of $G\times G$, then it has a normal subgroup $A$ (the image of $G\times 1$) such that both $A$ and $Q/A$ are quotients of $G$. If it is also a quotient of $H\times H$, then it has a normal subgroup $B$ such that both $B$ and $Q/B$ are quotients of $H$. Now assume that every common quotient of $G$ and $H$ is trivial. Then $Q/AB$ is trivial, so $AB=Q$. So $Q/A=AB/A=B/(A\cap B)$ is trivial, so $A=Q$. Likewise $B=Q$. So $Q$ is trivial.<|endoftext|>
TITLE: Why did people originally like Frobenius algebras?
QUESTION [16 upvotes]: These days, lots of people are excited by Frobenius algebras because commutative Frobenius algebras are the same thing as 2D topological quantum field theories.
...but this seems like teaching an old dog new tricks. Can anyone sum up (using only diet representation theory :-P), why Frobenius algebras were invented and what was so good about them?
Also, any nice texts and/or papers along this line would be much appreciated. I'm working through the old Nakayama papers now, but perhaps this material exists in a friendlier and more modernised form somewhere?

REPLY [9 votes]: Frobenius's original turn-of-the-century perspective was the nonvanishing of a determinant. Brauer–Nesbitt–Nakayama studied some equivalent definitions in the late 30s and early 40s.  For instance, an equivalence between the left and right regular representations is a rare and beautiful thing; this gives an equivalence between projectivity and injectivity that is explained in modern language in Lam's Lectures on Modules and Rings.  This also gives a "perfect duality" studied by Dieudonné in the late 50s.  I added the missing early sources to the wikipedia article, including the Brauer–Nesbitt announcement in PNAS which is pretty easy to read.<|endoftext|>
TITLE: Model of a scheme regular over the generic point
QUESTION [5 upvotes]: Let all schemes below be excellent.
Let $X_0$ be a regular (not necessarily smooth, projective) non-empty scheme of finite type over the generic point $\eta$ of a regular connected scheme $S$. As the answers to my question 
For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular
show, there does not have to exist a dense open $U\subset S$ such that $X_0$ possesses a fibrewise regular model over $U$. Yet, should there always exist a pseudo-finite dominant morphism  $j:U\to S$ and some model $X$ of $X_0$ over $S$ such that $U$ is regular and the reduced scheme associated to $X_U$ is smooth over $U$? Can we assume that the morphism $U\to S$ is radiciel?
Upd.: it is probably sufficient for my purposes to find such a $U$ for a variety $X_0$ that is smooth over the perfect closure of $\eta$. It seems that (the first)  BCnrd comment solves the problem; thanks!!

REPLY [6 votes]: To synthersize a bit: let $S$ be an excellent (quasi-excellent is enough) integral scheme, let $X_0$ be a regular scheme of finite type over $K=K(S)$. I will suppose $X_0$ separated to avoid possible pathologies. 

There exists a separated integral noetherian scheme $X$ over $S$ with generic fiber isomorphic to $X_0$. 
The singular locus of $X$ is closed (excellence), its projection in $S$ is constructible (Chevalley) and does not contain the generic point of $S$, therefore is contained in a proper closed subset $F$ of $S$. The scheme $X_U$, where $U=S \setminus F$,  is a regular model of $X_0$ over $U$ (or $S$). 
If $(\overline{X_0})_{\rm red}$ is smooth, then, as explained by BCnrd, there exits a radicial finite flat (hence dominant) morphism $U\to S$ with $U$ integral such that $(X_U)_{\rm red}\to U$ is smooth. 
[Edit] The  assumption of smoothness is also necessary if $(X_U)_{\rm red}\to U$ is smooth for some dominant morphism $U\to S$ (just consider the generic fiber). 
The above assumption of smoothness is essentially sharp [Edit] if we want to have a quasi-finite (i.e. finite type with finite fibers) and dominant base change $U\to S$ such that $(X_U)_{\rm red}\to U$ is fiberwise regular. Consider the example $S={\rm Spec} (k[t])$ with $k$ perfect of characteristic $p>2$, and let $X_0$ be the regular affine curve over $K=k(t)$ defined by the equation $y^2=x^p-t$. Then $X_0$ is geometrically integral but not smooth over $K$. Let $X\to S$ be a model of $X_0$ (i.e. finite type separated morphism with generic fiber isomorphic to $X_0$). Shrinking $S$ if necessary, we can suppose that $X\to S$ is flat with geometrically integral fibers ([EGA], IV.9 or directly compare with the obvious projective model associated to the equation $y^2=x^p-t$ over $S$) [Edit] and that the Zariski closure in $X$ of the non-smooth point $(y, x^p-t)\in X_0$ meets every fiber. Let $U\to S$ be any quasi-finite dominant morphism $U\to S$ (with $U$ integral). Then $X_U$ is integral. But none of the closed fibers of $X_U\to U$ is regular because it would be smooth (any finite extension of $k$ is perfect) and then $X_0$ would be smooth at $(y, x^p-t)$, contradiction.<|endoftext|>
TITLE: Can any properties of a ring other than being a field be captured by the geometry of its 2-dimensional free module?
QUESTION [7 upvotes]: Can any properties of a ring other than being a field be captured by the geometry of its 2-dimensional free module?
Background:
In his wonderful, wonderful book Geometric Algebra, Emil Artin describes the following way of putting coordinates on affine geometries:
Take a 2-dimensional Affine geometry A, that is, a set of lines and a set of points, together with the following axioms:


Any two points determine a line
Through any point not on a given line, there passes exactly one line parallel to the given one.
There exist 3 non-collinear points.


Define a dilatation to be a map of points that sends any line to a line parallel to it (modulo some technical details). Because of the parallel postulate, dilatations are uniquely defined by the images of 2 points.
Then translations are dilatations without any fixed points (or the degenerate translation given by the identity map that has 2 fixed points and hence fixes every point). The direction of a translation τ is defined as the pencil (equivalence class of parallel lines) of the line joining the point P and τ(P). (the pencil does not depend on the choice of point P).
Translations form a group T (the identity mapping being the identity), and if there exist translations in two different directions, the group is commutative.
In the case where T is commutative, the endomorphisms of T that preserve directions form a (not necessarily commutative) ring with identity R.
Another two additional axioms, guarantee a) that R is a division ring and b) that R acts regularly on translations of the same direction, i.e. that if τ1 and τ2 have the same direction, then τ1 is sent to τ2 by some element of R.
These axioms are:

4a. For any two points P and Q, there exists a translation mapping P to Q.
4b. For any three points P, Q and R, there exists a dilatation that fixes P and sends Q to R.

Note that 4a. and 4b. are respectively equivalent to the affine formulations of Desargues' Theorem when the three lines are parallel and when the three lines intersect at a single point.
With these four axioms, it follows that choosing a point O and translations τ1 and τ2 in different directions, we can make A into an affine space over R with (0,0) corresponding corresponding to O, (1,0) to τ1(O) and (0,1) corresponding to τ2(O).
Note that R is commutative (and hence a field) if and only if A satisfies Pappus' Theorem.
The Question
The above construction is also reversible and establishes a correspondence between affine geometries satisfying Desargues' theorem and division rings.
It seems to me that we can associate a 'geometry' to any ring via the 2d free module.

Are there any rings (or classes of rings) other than division rings and fields whose 'geometry' can be axiomatized similarly to affine geometries? Are any of them also uniquely determined by that `geometry'?

For example, if the ring is ℤ the geometry consists of a lattice of points in the plane, and seems to me is 'hyperbolic' in the sense that through any point not on a given line there are infinitely many parallel lines (join the given point to any point with non-integer coordinates on the line in question).

REPLY [2 votes]: I'm not sure that this is the answer you are looking for, but you might be interested in Mnev's universality theorem. Unwinding the language of semialgebraic sets, this says the following: Consider any finite set of polynomial equations and inequalities with integer coefficients, such as
$$\zeta^3=1 \quad \zeta \neq 1.$$
Then there is an arrangement $A$ of points and lines such that $A$ can be realized over a commutative ring $R$ if and only if $R$ contains a solution to those equalities and inequalities. 
For example, the Fano plane is representable only if $2=0$ in $R$. The statement "you can find $9$ points $x(i,j)$ labeled by $\mathbb{F}_3^2$ such that $x(i_1, j_1)$, $x(i_2, j_2)$ and $x(i_3, j_3)$ are collinear if and only if $\sum (i_s, j_s)=0$" is equivalent to the statement "$R$ contains a nontrivial cube root of unity." 
But all of this is for $R$ commutative; I don't know if there is a noncommutative version of Mnev.<|endoftext|>
TITLE: Techniques to bound products of upper triangular matrices and their inverses
QUESTION [8 upvotes]: Let $A_0, \dots, A_{n-1}$ be upper triangular matrices with ones on the diagonal.  Let $B_{n-1}, \dots, B_0$ be of the same form.
I am interested in bounding
$$|| A_0 \dots A_{n-1} B_{n-1}^{-1} \dots B_0^{-1}||$$
and in particular showing that this product is close to the identity when $A_i$ and $B_i$ are close (specifically, $||A_i - B_i|| \leq \delta \lambda^{- \min(i, n-i)}$ where $\lambda>1$).
In general, this will probably not be possible, but there may be conditions when there are interesting bounds.
The problem is, I don't really know anything about techniques for bounding products of matrices.  There is the obvious multiplicativity of the matrix (i.e., operator) norm, but it is far from best possible in this case.  For instance, it is easy to check that the norm can grow at most polynomially in $n$ if the norms of the $A_i, B_i$ are bounded above by some constant.
Hence:
Question: What techniques/tricks are there for bounding these kinds of products of matrices?
Unfortunately I don't know exactly which techniques would be useful, so I'd appreciate any pointers to relevant papers or books. 
I can at least explain how the problem is motivated.  Namely, we have a Holder-continuous matrix-valued function $A: X \to GL_m(\mathbb{R})$ for $X$ a compact metric space.
Then the $A_i$ will be the values $A(T^ix)$ and the $B_i$ will be $A(T^ip)$ for $p$ close to $x$.  The multiplicative ergodic theorem states that the norm products $||A_0 \dots A_{n-1}||$ grow like $e^{\lambda n}$ almost everywhere (in $x$) with respect to any invariant measure (where $\lambda$ is the maximal Lyapunov exponent), but in the case of upper triangular matrices, this is actually immediate---we even have polynomial growth if ones are on the diagonal (a corollary of the nilpotence of strictly upper triangular matrices).  The multiplicative ergodic theorem and its variants were used in a paper that motivated the problem I'm working on, but it doesn't quite seem to help here (because we make slightly different assumptions), which is why I was curious about other techniques.

REPLY [2 votes]: A variant of the previous answer which uses the particular form of the matrices. 
I shall just take four matrices $A_{1}, A_{2}, B_{2}, B_{1}$, upper triangular with 1's on the diagonal, because the general case reduces to this one. 
Consider the group of $n \times n$  real upper triangular matrices with 1's on the diagonal. That is a Carnot group! Identify it with its Lie algebra because the exponential is bijective. Let $Q= \sum_{i=1}^{n-1} (n-i) i$, that is called the homogeneous dimension. 
Then you have two norms on this group: the first one is a Carnot-Caratheodory 
norm, call it '$\| A\|_{CC}$' , the other one is just an euclidean norm, call it $\|A \|$ . 
These norms  have the properties: 

$\| AB \|_{CC} \leq \| A \|_{CC} + \|B\|_{CC}$, 
$ \|A\|_{CC}$ grows like (is comparable with) $\|A \|^{1/Q}$. 

Now, suppose we want to estimate $\|A_{1} A_{2} B_{2}^{-1} B_{1}^{-1}\|_{CC}$, then we do like this: 
$\|A_{1} A_{2} B_{2}^{-1} B_{1}^{-1}\|_{CC} = \|A_{1} A_{2} B_{2}^{-1} A_{1}^{-1} A_{1}B_{1}^{-1}\|_{CC} \leq \|A_{1} A_{2} B_{2}^{-1} A_{1}^{-1} \|_{CC} + \|A_{1}B_{1}^{-1}\|_{CC} $
You only need to estimate now the first term from the right. Remember that because the group is nilpotent the Baker-Campbell-Hausdorff formula (i e the multiplication law when you identify the Lie algebra with the Lie group) is polynomial. Then you get, by simply taking 
the Euclidean norm of the multiplication, that 
$\|A_{1} A_{2} B_{2}^{-1} A_{1}^{-1} A_{1}B_{1}^{-1}\| \leq \mathcal{O}(\|A_{1}\|) 
 \|B_{2}^{-1} A_{1}^{-1}\|_{CC}^{Q}$ 
And that is it, you may iterate the procedure if you have more matrices!<|endoftext|>
TITLE: The total curvature of very knotty knots
QUESTION [9 upvotes]: One of my favorite theorems is that of Fáry-Milnor, stating that the total curvature of a knot in $\mathbb R^3$ which is not an unknot (an ununknot) is at least $4\pi$. 
Can one quantify the way in which knottedness forces an increase in total curvature?

REPLY [10 votes]: Fields medalist Michael Freedman got involved with knots using a very simple technique, assign a sort of energy integral that becomes infinite if there is a genuine self-crossing, that is if you try to force the curve to change isotopy class. The results   relate, at least, to Ryan's comment "the figure 8 knot is twice as knotted as the trefoil". 
http://www.jstor.org/pss/2946626 
Excerpts from a column by Ian Stewart...the quoting process does not seem to have rendered the mathematical symbols very well, and I do not know what the letter p means, but here is the link:
http://www.fortunecity.com/emachines/e11/86/knotprob.html 

But it now looks as if the most
  interesting "energy" concept for knots
  is not elastic, but electrostatic, as
  suggested in 1987 by S. Fukuhara of
  Tsuda College, Tokyo. Imagine the knot
  to be a flexible wire of fixed length,
  which can pass through itself if
  necessary and which has a uniform
  electrostatic charge along its length.
  Because like charges repel each other,
  a knot that is free to move will
  arrange itself so as to keep
  neighbouring strands as far apart as
  possible in order to minimise its
  electrostatic energy. This minimum
  energy value is the invariant.
But is it a useful one? Does it have
  simple, natural properties that
  mathematicians can exploit? In 1991,
  Jun O'Hara of Tokyo Metropolitan
  University proved that the minimum
  energy of a knot really does increase
  as the knot becomes more complicated.
  Only a finite number of topologically
  different knots exist with energy less
  than or equal to any chosen value.
  This means that topological types of
  knots can be "ordered" in terms of
  their energy. There is a natural
  numerical scale of complexity for
  knots, ranging from simple knots at
  the low energy end to more complicated
  ones higher up. 
What are the simplest knots? In the
  most recent issue of the Bulletin of
  the American Mathematical Society, a
  team of four topologists - Steve
  Bryson of NASA's Ames Research Center
  in California, Michael Freedman and
  Zhenghan Wang of the University of
  California at San Diego, and Zheng-Xu
  He of Princeton University- prove that
  the simplest knots are exactly what
  you would expect. They are "round
  circles"- that is, circles in the
  everyday sense. Topologists, whose
  "circles" are usually bent and
  twisted, have to append an adjective
  to remind themselves when, as in this
  case, they are not.
The energy of a round circle is 4, and
  all other closed loops have higher
  energy. Any loop with energy less than
  $6 \pi + 4$ is topologically unknotted - it
  is a bent circle. More generally, a
  knot with $c$ crossings in some
  two-dimensional picture has energy at
  least $2 \pi c + 4,$ though this bound is
  probably not the best possible, as the
  lowest known energy for an overhand
  knot is about 74. The number of
  topologically distinct knots of energy
  $E$ is at most $(0.264) x (1.658)^E .$<|endoftext|>
TITLE: Extraordinary cohomology as a derived functor?
QUESTION [8 upvotes]: The purpose of this question is to find out whether one can view the Atiyah-Hirzebruch spectral sequence as a particular case of the "composition of derived functors" spectral sequence.
The Leray spectral sequence of a continuous map $f:X\to Y$ of topological spaces can be constructed as follows. Let $a_X:X\to pt$ and $a_Y:Y\to pt$ be the maps from $X$ and $Y$ respectively to the one point space $pt$; we obviously have $a_X=a_Y\circ f$. So for any sheaf $L$ on $X$ we have $(a_X)_\ast L=(a_Y)_*f_\ast L$. But $(a_X)_\ast$ is just the functor of the global sections and so is $(a_Y)_*$. Recall that if $A,B$ and $C$ are abelian categories and $F:A\to B,G:B\to C$ are left exact functors, then (under mild hypotheses) $R_\ast(G\circ F)$ is isomorphic to $R_\ast\circ G_\ast$ and for any object $X$ of $A$ there exists a spectral sequence abutting $R^\ast (G\circ F) X$ with the $E_2$ sheet given by $E_2^{pq}=R^pG(R^q F(X))$. See e.g Gelfand-Manin, Methods of homological algebra, 3.7.
Applying this to the case when $A$, $B$ and $C$ are the categories of sheaves of $X$, $Y$ and the point respectively we get a spectral sequence $(E^{pq}_r,d_d)$ abutting to $H^*(X,F)$ with the $E_2$ term given by $$E_2^{pq}=H^p(Y,R^q f_\ast L).$$
If $X$ and $Y$ are sufficiently nice (say finite CW complexes), $F$ is constant and $f$ is a locally trivial fibration with fiber $F$, then we get (assuming for simplicity that $Y$ is simply-connected) $$E_2^{pq}=H^q(Y,H^q(F)).$$
Now, if we have an extraordinary cohomology theory $h^\ast$, we can construct (under the hypotheses of the previous paragraph) the Atiyah-Hirzebruch spectral sequence: the $E_2$ sheet is given by $$E_2^{pq}=H^q(Y,h^q(F))$$ and the spectral sequence abuts to $h^*(X)$. This looks pretty similar to the Leray spectral sequence, so it seems natural to ask whether it can be obtained in a way similar to the one described above.
Namely, given an extraordinary cohomology theory $h^\ast$ and a continuous map $f:X\to Y$ of topological spaces, is there an "extraordinary direct image" functor $f^{ex}_*$ from sheaves on $X$ to sheaves on $Y$ which would be "functorial in $f$" and which would give $h^{\ast}(X)$ after deriving when $Y$ is a point?
If not, is there still a way to view the Atiyah-Hirzebruch spectral sequence as (a version of) the spectral sequence of the composition of two derived functors? (It may happen that one has to consider something other than the categories of sheaves, but I have no idea what this could be.)

REPLY [9 votes]: I'd like to propose that the answer is "No, but..".  The viewpoint I'd like to suggest is that thinking of "derived functors" is probably insufficient here (because we're secretly interested in homotopical categories that are not derived categories of abelian categories), but that we are relying essentially on the observation that $R\Gamma(X, -) = R\Gamma(Y, Rf_* -)$.  So what follows is a sketch-construction that gives a positive answer to a related question you could've been asking: "We can get the Serre spectral sequence from the Leray spectral sequence.  Can we get the Atiyah-Hirzebruch spectral sequence in a similarly sheaf-theoretic way?"

We get the Leray spectral sequence by studying sheaves of complexes of $\mathbb{Z}$-modules.  The "derived" category (of sheaves of complexes of $\mathbb{Z}$-modules) is the derived category of its heart (sheaves of $\mathbb{Z}$-modules) w.r.t. the usual $t$-structure, so it's reasonable that we get lots of mileage from looking at derived functors, composites of derived functors, etc.
Analogously, we can think of more general Atiyah-Hirzebruch type spectral sequences as arising from studying sheaves of spectra.  This is not the derived category of its heart w.r.t. the usual $t$-structure (this heart is again sheaves of $\mathbb{Z}$-modules).  But, this does give a way of thinking about Atiyah-Hirzebruch type spectral sequences:
We start with (say) the "constant sheaf of spectra" $\mathbf{E}$ on $X$, and we're interested in computing the (homotopy groups of the spectrum) $$R\Gamma(X, \mathbf{E}) = R\Gamma(Y, Rf_* \mathbf{E})$$  The spectral sequence of interests arises by filtering $Rf_* \mathbf{E}$ using the $t$-structure (Postnikov sections "on values"), $$\cdots \to \tau_{\geq k} Rf_* \mathbf{E} \to \tau_{\geq (k-1)} Rf_* \mathbf{E}\to \cdots \to \mathbf{E}$$  The $k$-th "associated graded piece" of this filtration is $\pi_k Rf_* \mathbf{E}$ (which, recall, is an object in the heart --- i.e., a sheaf of $\mathbb{Z}$-modules on $Y$). This gives rise to a spectral sequence (excuse the funny indexing!, and no comment on convergence)
$$ E^2_{p,q} = \pi_{-p} R\Gamma\left(Y, \pi_{-q} Rf_* \mathbf{E}\right) \Rightarrow \pi_{-p-q} R\Gamma\left(Y, Rf_*\mathbf{E}\right) = \pi_{-p-q} R\Gamma(X,\mathbf{E})$$  The funny indexing was picked so that I can rewrite it as
$$ H^p(Y, \pi_{-q} Rf_* \mathbf{E}) \Rightarrow E^{p+q}(X) $$
To recover the usual form of AH-SS we need the following observation (analogous to what we'd need to get the Serre spectral sequence via sheaf theory): If $f$ is nice (i.e., a fibration between reasonable spaces), then $\pi_{-q} Rf_* E$ will be the locally constant sheaf associated to $E^{q}(F)$ with its monodromy action.<|endoftext|>
TITLE: Intersection product in a manifold, taking values in one factor
QUESTION [7 upvotes]: In a joint paper that I am working on, we are interested in taking the intersection product $[X] \cap [Y]$ of the fundamental classes of two compact, oriented pseudomanifolds $X$ and $Y$ in a compact, oriented manifold $M$.  Now, the usual intersection product takes values in $H_*(M)$, but I need an intersection product that takes values in $H_*(Y)$.  I think that if $X$ and $Y$ are favorably stratified, whatever that means, then such an intersection product should exist.  My question is, what is a good approach for such a construction?  What is a good proof that it's well-defined?  What is a good reference?
The intuitive idea is to wiggle $X$ generically to make it transverse to $Y$, and then take transverse intersections along strata.  But there are various different rigorous frameworks that you could use to prove that this is well-defined.
For example, if $X$, $Y$, and $M$ are all PL, then one way to define the right thing is to restrict the Poincaré dual $[X]^*$ to $H^*(M,M\setminus N) \cong H^*(N,\partial N)$, where $N$ is a regular neighborhood of $Y$, then take the Poincaré dual of that to obtain an element of $H_*(N) \cong H_*(Y)$.  This way to define the intersection is based on very standard ideas.  You can prove that it works is to use the fact that $N$ is unique up to a relevant isotopy.
In the case that I/we need, $M$ is a smooth complex projective variety and $X$ and $Y$ are singular complex subvarieties.  Now, it's easy to suppose that the definition and proof from the PL case still work using the fact that $X$ and $Y$ are Whitney-stratified.  However, I no longer know how standard the argument is.  For all I know, a regular neighborhood argument is not as standard as a direct argument with transverse position.  Or, for all I know, $X$ and $Y$ don't have to be Whitney-stratified; maybe they can be much more general.  Certainly it seems like window dressing to demand that they be complex projective.

REPLY [14 votes]: I don't think you have to get involved with strata or transversality at all. Poincare duality (Edit: and cap products) will do all the work. Let's assume:
$M$ is a compact oriented $n$-manifold. 
$\xi$ is a $p$-dimensional homology class of $M$. (We don't care if it comes as the fundamental class $[X]$ of some pseudomanifold in $M$.)
$Y$ is a subset of $M$ and $[Y]$ is an element of $H_q(Y)$. (We don't care if $Y$ is a pseudomanifold, or $q$-dimensional, or anything.) 
Then duality in $M$ yields an element $\xi^*\in H^{n-p}(M)$. Restrict it to get an element of $H^{n-p}(Y)$. Cap with $[Y]$ to get an element of $H_{p+q-n}(Y)$.
(Final paragraph now simplified after comment from Greg.)<|endoftext|>
TITLE: Recent work on hypergeometric functions
QUESTION [9 upvotes]: Does anyone know of a monograph/survey on the modern history of (basic or elliptic) hypergeometric functions and their applications?
I haven't had much time to search the literature, and because it is summer it is hard to reach professors or specialists, which is why I am asking the question here. It is also likely that there are obvious choices out there that I am unaware of because of my ignorance in the field. I would appreciate it a lot if along with the suggestions you could give a quick description of what the book/article treats.

REPLY [5 votes]: Hyper Geometric Functions, My Love: Modular Interpretations of Configuration Spaces (Aspects of Mathematics) by Masaaki Yoshida 
Discriminants, Resultants, and Multidimensional Determinants 
Israel M. Gelfand, Mikhail Kapranov, Andrei Zelevinsky 
I heard a talk by Rivoal a decade back when he had just had his breakthrough concerning irrationality of zeta values.  He quoted some very classical but not well-known results and got asked where he learned them.  He recommended this book:
Confluent Hypergeometric Functions  L. J. Slater<|endoftext|>
TITLE: Inequality involving probability measures
QUESTION [5 upvotes]: I have been working on a problem(alternate minimization) where I want to establish an inequality in which I am stuck.
An $\alpha$- parameterized version of the divergence(Kullback-Leibler) takes the following form:
$$I_{\alpha}(P,Q)=\frac{1}{\rho}\log\left[\frac{h(Q)^{1-\alpha}}{h(P)}\sum_{a\in A}P(a)Q(a)^{\alpha-1}\right] $$
where $\alpha=\frac{1}{1+\rho}, \rho>0$
 and I let $h(P)=\left(\sum P(a)^{\alpha}\right)^{\frac{1}{\alpha}}$, $A$ is a finite set and $P, Q$ are distributions on $A$.
By Holder's inequality I have shown that $\sum_{a\in A}P(a)Q(a)^{\alpha-1}\ge h(Q)^{\alpha-1}h(P)$. So $I_{\alpha}(P,Q)\ge 0$ and is $0$ iff $P=Q$.
The problem is the following:
Let $P$ be a distribution and $E$ be a closed and convex set. Let $I_{\alpha}(P,Q^*)=\min_{Q\in E}I_{\alpha}(P,Q)$.
I want to establish $I_{\alpha}(P',P)+I_{\alpha}(P',Q')\ge I_{\alpha}(P',Q^*)$ for any $P'$ and any $Q'\in E$.
The equivalent form of the above ineq. is
$$\frac{h(P)^{1-\alpha}}{h(P')}\sum_{a\in A}P'(a)P(a)^{\alpha-1}\cdot \frac{h(Q')^{1-\alpha}}{h(P')}\sum_{a\in A}P'(a)Q'(a)^{\alpha-1}\ge \frac{h(Q^*)^{1-\alpha}}{h(P')}\sum_{a\in A}P'(a)Q^*(a)^{\alpha-1}$$
Imre Csiszar has proved the same inequality for the Kullback Leibler divergence where he employs a derivative argument. When I do the similar thing I get the following:
Let $Q_t=(1-t)Q^*+tQ'\in E, 0\le t\le 1$. Then 
$$0\le \frac{1}{t}\left[I_{\alpha}(P,Q_t)-I_{\alpha}(P,Q^*)\right]=\left[\frac{d}{dt}I_{\alpha}(P,Q_t)\right]_{t=\tilde{t}}, \quad 0<\tilde{t}\le t$$
As $t\to 0$, I get the following:
$$\frac{1}{h(Q^*)^{\alpha}}\sum_{a\in A}P(a)Q^*(a)^{\alpha-1}. \sum_{a\in A}Q'(a)Q^*(a)^{\alpha-1}\ge \sum_{a\in A}P(a)Q'(a)Q^*(a)^{\alpha-2}$$
I don't know how to use this to prove the desired inequality. Can anyone help?
You may refer Csiszar's book on Information theory and statistics- a tutorial page no 445-446, available here http://www.nowpublishers.com/product.aspx?product=CIT&doi=0100000004
Update 14, September.
If I denote $\frac{P'(a)}{h(P')}$'s by $a_i$, $\frac{P(a)}{h(P)}$'s by $b_i$, $\frac{Q'(a)}{h(Q')}$'s by $c_i$, and $\frac{Q^{\star}(a)}{h(Q^{\star})}$'s by $d_i$, then a, b, c, d are vectors whose $\alpha$th norm is $1$.
Problem would be solved if we show the following.
$$\min_{\|a\|_{\alpha}=1}\sum_{i=1}^n a_i b_i^{\alpha-1}.\sum_{i=1}^n a_i c_i^{\alpha-1}-\sum_{i=1}^n a_i d_i^{\alpha-1}\ge \sum_{i=1}^n b_i d_i^{\alpha-1}. \sum_{i=1}^n c_i d_i^{\alpha-1}-\sum_{i=1}^n b_i c_i d_i^{\alpha-2}$$ as the right hand side is $\ge 0$.

REPLY [3 votes]: If you can show that your $\alpha$-divergence above is a Bregman divergence that is convex in both its arguments, then the following paper points out a result that implies the abovementioned four-points property that you are trying to prove.
Link: http://math.haifa.ac.il/yair/AORfulltext.pdf<|endoftext|>
TITLE: Inverse of Baker-Campbell-Hausdorff
QUESTION [10 upvotes]: This should be quite simple to answer. I have a situation in which I must have an explicit expression for the inverse of Baker-Campbell-Hausdorff. More precisely:
I have two power series $P_1(X,Y)$, $P_2(X,Y)$ in noncommuting variables $X,Y$. I want an expression for 
$$exp\left(log(1+P_1) + \ log(1 +P_2) \right) .$$
I searched some books and google; but beyond the fact that it exists, I am unable to find an explicit expression. If it already exists in print, it would save me some work of trying to derive it myself.
Thank you very much.

REPLY [4 votes]: The following recursion formula for the homogeneous factors of $A(P,Q)$ exists:
Define:
$A(tP,tQ) = 1 + t \frac{B_1}{1!}+t^2 \frac{B_2}{2!}+t^3 \frac{B_3}{3!}+...$
$C_n  = P^n+Q^n$
$\Delta C_n = -n C_{n+1}$
Then:
$B_1 = C_1$
$n! B_n  = [(\Delta+C_1) B_{n-1}] _{symm}$
Where the symmetrization is of all words of the type $C_{i1} C_{i2} C_{i3} ... $
This formula can be obtained by differentiation with respect to t. I compared the results up to
the cubic term to Theo's expression with a full agreement.
The coefficient of the string $\[(C_{i1} C_{i2} C_{i3} ...]_{symm}$ in $B_n$ is
equal to the coefficient of $tr(M^{i1}) tr(M^{i2}) tr(M^{i3})...$ in the coefficient of $t^n$
in $det(I-tM)$. These coefficients are the charcters of $\{{{1^n\}}}$ of $S_n$.<|endoftext|>
TITLE: Representing cohomology of a sheaf à la Eilenberg-Maclane
QUESTION [13 upvotes]: Suppose that we are given a nice space $X$ and a sheaf of abelian groups $F$ on $X$. Fix an integer $n$. Then We have a contravariant functor from nice spaces over $X$ to abelian groups; Namely, to a space $f: Y \to X$ we associate the abelian group $H^n (Y, f^{*}F)$ (Sheaf cohomology).
If $X$ is a point, Then this functor is represented, in the homotopy category, by the Eilenberg-Maclane space $K(F,n)$.
My questions are:
1) Can one formalize what will it mean for our functor to be representable, "homotopically"? I am not very sure, but I suspect that the most naive definition of homotopy category of spaces over $X$, and the requirement that the functor is representable in this category, are not right (and I did not check that the functor actually factors to this "homotopy" category).
2) Is it representable?

REPLY [8 votes]: Sheaf cohomology over $X$ is representable in the homotopy category of oo-stacks over X / spaces over X, yes. 
Details, links and references are at  http://ncatlab.org/nlab/show/cohomology<|endoftext|>
TITLE: Why pi-systems and Dynkin/lambda systems? On the relative merits of approaches in measure theory.
QUESTION [21 upvotes]: What is the point of $\pi$-systems and
  $\mathcal{D}$ / Dynkin /
  $\lambda$-systems? 

I am an analyst in the process of consolidating my measure theory knowledge before moving on to harder/newer things, having been first introduced to measure theory in a course with a probability as opposed to an analysis viewpoint. So far, everything that I've needed from elementary measure theory for analysis can be done (and is done in all of my analysis textbooks) without mention of the $\pi$-systems and $\mathcal{D}$-systems which were used in my first course. Do these set systems belong strictly to probability and not analysis? Heuristically, are they useful or important in any way? Why?

REPLY [4 votes]: I think there's indeed a lot of confusion here, in that in many situations the $\pi-\lambda$ is applied where a weaker result would suffice. What one uses most of the time is not the $\pi-\lambda$ theorem itself, but the following corollary: 

If $\mu_1$ and $\mu_2$ are two probability  measures that agree on a
  $\pi$-system, then they agree on the $\sigma$-algebra generated by
  that $\pi$-system.

The example about independent events given above by Peter Luthy falls into this category.
In many textbook situations the $\pi$-system in question is actually a semi-ring. For ($\sigma$-)finite premeasures, Carathéodory extension is unique for very simple reasons: any extension of the measure is bounded from above by the outer measure, and by passing to the complement it is also bounded from below. Thus, for semi-rings the above corollary is trivial.
A typical example of a $\pi$-system that is not a semi-ring are closed sets in a topological space. But here, the fact that a Borel measure is determined by its values on closed sets follows from regularity, which does not require $\pi-\lambda$ theorem either.
One place where you need $\pi-\lambda$ theorem (or a similar result) in an essential way is Fubini. There, you have a $\pi$-system (which is also a semi-ring) of sets of the form $E_1\times E_2$, and two ways to extend the pre-measure: by product-measure (= Carathéodory extension), and by integrating the measures of the slices. What is tricky to show is that the collection of sets for which the latter is well defined is a $\sigma$-algebra, in particular, that it is closed under pairwise intersections. Indeed, it's not clear under such operation, the function under the integral remains measurable. The $\pi-\lambda$ theorem allows one to bypass this difficulty in a very neat way.
That said, once you have the $\pi-\lambda$ theorem at hand, it is often just shorter to write down the proof that something is a $\lambda$-system than that it is a $\sigma$-algebra. No wonder the authors of textbooks use it systematically even when it's an overkill.<|endoftext|>
TITLE: What are some of the big open problems in 3-manifold theory?
QUESTION [60 upvotes]: From what I understand, the geometrization theorem and its proof helped to settle a lot of outstanding questions about the geometry and topology of 3-manifolds, but there still seems to be quite a lot of activity.  I am not prepared to make a full push to familiarize myself with the literature in the near future, but I am still curious to know what people are working on, what techniques are being developed, etc.
So I was hoping people could briefly explain some of the main open questions and programs that are motivating current research on 3-manifolds.  I'll let the community decide if this undertaking is too broad, but I'm hoping it is possible to give a rough impression of what is going on.  References to survey articles are appreciated, especially if they are accessible to non-experts like myself.
It seems like the community wiki designation is appropriate for this question, and the usual rules ought apply.

REPLY [7 votes]: Thurston's remaining challenge:
Show that volumes of hyperbolic 3-manifolds are not all rationally related.
THURSTON'S VISION AND THE VIRTUAL FIBERING THEOREM
Is there an algorithm which determines whether two funda-
mental groups of 3-manifolds are isomorphic?
DECISION PROBLEMS FOR 3-MANIFOLDS AND THEIR
FUNDAMENTAL GROUPS<|endoftext|>
TITLE: Left and right eigenvalues
QUESTION [13 upvotes]: A quaternionic matrix $A$ gives rise to a 
function $\mathbb{H}^n \to \mathbb{H}^n$
given by $x \mapsto A \cdot x$.  This is real linear, 
but not complex- or quaternionic-linear
(in general) if we consider $\mathbb{H}^n$ as 
a left $\mathbb{C}$ or $\mathbb{H}$ module, 
but is pretty good if we use right actions.
A right eigvenvalue of $A$ is a 
quaternion $q$ such that $A\cdot x = x \cdot q$
for some $x\in \mathbb{H}^n$; 
a left eigenvalue is quaterion $q$ 
such that $A \cdot x = q\cdot x$ for some  $x\in \mathbb{H}^n$.
The algebra of right eigenvalues is pretty good, 
but the algebra of left eigenvalues is quite
interesting.  For example, it is not hard to 
see that there are matrices $A$ with infinitely 
many left eigenvalues, even for $2$-by-$2$ matrices.
Now let's assume that $A\in Sp(n)$, so that the left eigenvalues are all contained in 
$S^3\subseteq \mathbb{H}$.  What sort of geometric properties must the set $L(A)$ of left
eigenvalues have?   
EDIT:  An example is 
$$
\begin{pmatrix} 
0 & 1 
\cr
-1 & 0
\end{pmatrix}   \cdot \begin{pmatrix} 
1 
\cr
q  
\end{pmatrix}
   = 
q \cdot
{\begin{pmatrix}
1
\cr
q  
\end{pmatrix}}$$
for any $q\in S^3 \subseteq \mathbb{H}$ with zero real part, since then $q^2 = -1$.
EDIT 2:  Examples like this show that for some symplectic matrices, the set of left 
eigenvalues is a union of copies of $S^2$.

REPLY [3 votes]: The symplectic 2x2 quaternionic matrices with infinite left eigenvalues are completely characterized [Macias-Pereira, Elect. J. Lin. Alg 2009]. There is a fundamental difference between left and right eigenvalues: the associate eigenvectors form a vector space (left eigenvalues) or not (right eigenvalues). In the latter case the quaternions similar to a rigth eigenvalue are eigenvalues too, but I should not call that to have infinite eigenvalues.<|endoftext|>
TITLE: Periodicity theorems in (generalized) cohomology theories
QUESTION [18 upvotes]: It is well-known that topological K-theory is blessed with the Bott periodicity theorem, which specifies an isomorphism between $K^2(X)$ and $K^0(X)$ (where $K^n$ is defined from $K^0$ by taking suspensions).  I am wondering if other generalized cohomology theories have their own periodicity theorems, and if there is a general framework for conceptualizing them.  I am interested in any substantive answer to this question, but there are two specific avenues for generalization that I am particularly curious about.
The first avenue begins with the Clifford algebra approach to Bott periodicity.  This approach relates periodicity in K-theory to a certain natural periodicity present in the theory of complex Clifford algebras, and it generalizes the 8-fold periodicity of real K-theory (corresponding to an 8-fold periodicity in real Clifford algebras).  Can one fruitfully generalize the notion of a Clifford algebra, associate to it a generalized cohomology theory, and analogously produce a periodicity theorem?
The second avenue involves Cuntz's proof of Bott periodicity for C*-algebras (which in particular implies topological Bott periodicity by specializing to commutative C*-algebras).  Cuntz proves Bott periodicity for any functor from the category of C*-algebras to the category of Abelian groups which is stable (i.e. insensitive to tensoring with the C*-algebra of compact operators on Hilbert space), half exact, and homotopy invariant.  The proof uses topological properties of Toeplitz algebras in an essential way.  Because of the generality of his approach, I am left wondering if the essential features of his argument can be translated into more general contexts.
Any ideas are welcome!

REPLY [4 votes]: In this video of a lecture given at Atiyah's 80th Birthday Conference, Mike Hopkins gives a description of the solution to the Kervaire Invariant Problem which relies heavily on some periodicity in a cohomology theory created for this problem. He mentions Clifford algebras and Bott Periodicity at one point towards the end of the talk... I don't know a whole lot about this topic so I didn't understand too much, but maybe it will be helpful for you!<|endoftext|>
TITLE: Explicit integral representation theory
QUESTION [15 upvotes]: The representation theory of finite groups over the complex numbers is classical und it is usually quite easy to compute the set of isomorphism classes of irreducible representations, at least for small examples. Now, sometimes one is not content with the representation theory over the complex numbers or even over any field, but one wants to consider representations over $\mathbb{Z}$ or $\mathbb{Z}_{(p)}$. At least for the non-expert it is not so easy to obtain a complete list of isomorphism classes of indecomposable representations in these cases even for small examples. So, what I want to ask is the following:
1) Can one formulate a "guide" how to obtain such a list?
2) Is there a place in the literature where a list of indecomposables/irreducibles is given for some small examples as $S_3$ (the symmetric group on three elements)? In the last example I am especially interested.

REPLY [25 votes]: You might be asking about four separate types of modules: 

irreducible Z[G] modules,
Z-forms of irreducible Q[G] modules,
indecomposable Z[G] modules, or
indecomposable Z[G] modules that are finitely generated and free as Z-modules.

I'll assume the last is the main concern.
The irreducible modules of ZS3 are all finite and have an elementary abelian p-group as their additive group.  For p=2,3 there are 2 each, and for p>3, there are 3 each.
The irreducible CS3 modules are all realizable over Q.  Every such module may be realized over Z, but the two-dimensional representation has two distinct Z-forms, giving four total "irreducible" Z-free ZS3 modules, that is, four total Z-forms of irreducible QS3 modules.
Indecomposable ZS3 modules up to isomorphism are more complicated than the human mind can possibly comprehend.  Indeed, even those in which S_3 acts as the identity are much too complex.  Luckily they divide up into several types: annihilated by a prime p (then classified by modular representation theory), torsion (more complicated, but basically now p-adic integral reps), Gorenstein projective (Z-free, so covered in the next bullet point), or madness (that is, the rest).
The indecomposable ZS3 modules that are free as Z-modules are classified in:
Lee, Myrna Pike.  "Integral representations of dihedral groups of order 2p."
Trans. Amer. Math. Soc. 110 (1964) 213–231.
MR 156896
doi:10.2307/1993702
There are 10 of them, and the Krull-Schmidt theorem fails for them.  Not only are indecomposables not completely reducible, the decomposition of a finitely generated Z-free module into indecomposable summands is not unique.  In other words, integral representations of even very small groups are quite complicated.<|endoftext|>
TITLE: Has the Lie group E8 really been detected experimentally?
QUESTION [103 upvotes]: A few months ago there were several math talks about how the Lie group E8 had been detected in some physics experiment. I recently looked up the original paper where this was announced,
"Quantum Criticality in an Ising Chain: Experimental Evidence for Emergent E8 Symmetry", Science 327 (5962): 177–180, doi:10.1126/science.1180085
and was less than convinced by it. The evidence for the detection of E8 appears to be that they found a couple of peaks in some experiment, at points whose ratio is close to the golden ratio, which is apparently a prediction of some paper that I have not yet tracked down. The peaks are quite fuzzy and all one can really say is that their ratio is somewhere around 1.6. This seems to me to be a rather weak reason for claiming detection of a 248 dimensional Lie group; I would guess that a significant percentage of all experimental physics papers have a pair of peaks looking somewhat like this.
Does anyone know enough about the physics to comment on whether the claim is plausible? Or has anyone heard anything more about this from a reliable source? (Most of what I found with google consisted of uninformed blogs and journalists quoting each other.)
Update added later: I had a look at the paper mentioned by  Willie Wong below, where Zamoldchikov predicts the expected masses. In fact he predicts there should be 8 peaks, and while the experimental results are consistent with the first 2 peaks, there are no signs of any of the other peaks. My feeling is that the interpretation of the experimental results as confirmation of an E8 symmetry is somewhat overenthusiastic.

REPLY [41 votes]: This is a great question, but I don't think a reasonable answer can be given in this short space.  So I wrote an expository note jointly with a colleague who was trained as a physicist.  You can read it by following the link above -- comments are welcome.
But here are a couple of highlights:

 It's not true that they measured this one number, and so claimed to have detected E8.*  There is a bit more data than that.  And there is a lot more history!  Back around 1990, there was a series of theoretical "deductions" (in the weak sense of physics) investigating what the appropriate theoretical model should be for the situation in the magnet experiment.  This led to a unique candidate for a model, one built out of E8.  I would say that the experiment corroborated the series of deductions, with the sensational bonus that the deductions led to E8.

 Which E8 appears in the theoretical model?  The obvious answer is that it is the compact real E8 and not just the root system or root lattice.  For example, even though the masses of the 8 particles are given as entries in an eigenvector for the Cartan matrix (which makes it sound like it's just the root system), the proof of this statement is a calculation within the compact Lie algebra.

One can argue about both of these points, of course.  But these seem to be what the physicists claim and what they use in their papers.

To address Wadim's concerns about fringe science: Whether or not you find the E8 angle interesting or plausible, it seems that the experiment is interesting for entirely different reasons.  The experimenters themselves claim that their main achievement is realizing this 1-dimensional quantum Ising model in the laboratory in a situation where the external field can be tuned to be above, below, or at the critical point.  The Physics Today article on the subject paraphrases Subir Sachdev: 

only recently could researchers reach
  the high fields and low temperatures
  needed to access the critical point
  and have high enough instrumental
  resolution to resolve the masses of at
  least some of the quasiparticles
  they excited. (Temperatures have to be
  low enough to suppress any impact of
  thermal fluctuations.)



Footnote: To be precise: Also, Coldea, the author of this particular article in Science, uses the more-cautious "detected evidence of E8 symmetry" as opposed to the stronger "detected E8".<|endoftext|>
TITLE: Most general formulation of Gödel's incompleteness theorems
QUESTION [11 upvotes]: Modern statements of Gödel's incompleteness theorems are usually in terms of first-order predicate logic. However, I've often read the claim that they extend to arbitrary formal systems that can prove basic propositions about numbers. Indeed, according to Wikipedia, the original theorems referred to the type theory of Principia Mathematica, which is apparently not based on predicate logic.
My two questions are:

What is the most general concept/definition of a formal system for which Gödel's theorems have been stated?
How does their proof differ from the predicate logic variant?

Regarding 1., I could imagine several equally general definitions, for example based on either strings of symbols or abstract syntax trees. Being a little biased, I actually think of formal systems as data structures of more or less arbitrary computer programs, so maybe there is a definition based on Turing machines... In any case, it would need to specify what a "proof" of a "theorem" is, but I would like to do without the concept of "axioms." (See also Derivation rules and Godel theorem.)
Regarding 2., I'm specifically thinking about the diagonal lemma (or arithmetic fixed-point theorem, or whatever it is really called). The version I know refers to a "formula with one free variable," but that presupposes such concepts as "formula" and "free variable" in the formal system, and I'm wondering how to generalize that to arbitrary formal systems. I know there are proofs of the first incompleteness theorem which take an entirely different route, but AFAIK they don't carry over to the second incompleteness theorem.
I would like to add that I don't doubt the generality of Gödel's incompleteness theorems in any way. I just feel there is a gap between their claimed general nature and the way they are usually presented. A year ago, I devised a nonstandard formal system for a proof assistant. Although it could easily formalize its own concepts and express its own consistency, a translation of Gödel's incompleteness theorems from predicate logic turned out to be surprisingly nontrivial.

REPLY [14 votes]: Raymond Smullyan gave a very general formulation in terms of representation systems.  They appear in his "Theory of Formal Systems", and in the first and last chapters of "Godel's Incompleteness Theorems".  They generalise first- and higher-order systems of logic, type theories, and Post production systems.
A representation system consists of:

A countably infinite set $E$ of expressions.
A subset $S \subseteq E$, the set of sentences.
A subset $T \subseteq S$, the set of provable sentences.
A subset $R \subseteq S$, the set of refutable sentences.
A subset $P \subseteq E$, the set of (unary) predicates.
A function $\Phi : E \times \mathbb{N} \rightarrow E$ such that, whenever $H$ is a predicate, then $\Phi(H,n)$ is a sentence.

The system is complete iff every sentence is either provable or refutable.  It is inconsistent iff some sentence is both provable and refutable.
We say a predicate $H$ represents the set $A \subseteq \mathbb{N}$ iff $A = \{ n : \Phi(H,n) \in T \}$.
Let $g$ be a bijection from $E$ to $\mathbb{N}$.  We call $g(X)$ the Godel number of $X$.
We write $E_n$ for the expression with Godel number $n$.
Let $\overline{A} = \mathbb{N} \setminus A$ and $Q^* = \{ n : \Phi(E_n,n) \in Q \}$.
We have:

(Generalised Tarski Theorem) The set $\overline{T^*}$ is not representable.
(Generalised Godel Theorem) If $R^*$ is representable, then the system is either inconsistent or incomplete.
(Generalised Rosser Theorem) If some superset of $R^* $ disjoint from $T^*$ is representable, then the system is incomplete.

In case it's not clear: in a first-order system, we can take $P$ to be the set of formulas whose only free variable is $x_1$, and $\Phi(H,n) = [\overline{n}/x_1]H$.<|endoftext|>
TITLE: What is the relationship between "translation" and time complexity?
QUESTION [8 upvotes]: Consider the problem of deciding a language $L$; for concreteness, say that this is the graph isomorphism problem.  That is, $L$ consists of pairs of graphs $(G, H)$ such that $G\simeq H$.  Now the time complexity of deciding this problem as stated depends on how the graphs are encoded.  For example, if one were to have a "canonical" encoding of graphs (such that encoding strings are in bijective correspondence with isomorphism classes of graphs) the problem would be $O(n)$, as we could decide whether $G\simeq H$ simply by comparing the string representing $G$ to the string representing $H$.
On the other hand, if we represent a graph via its adjacency matrix, the best known algorithm (according to Wikipedia) gives only a subfactorial bound.  Now consider the time complexity of converting from one language to another.  If we let $T_1, T_2$ be the time complexity of deciding languages $L_1$ and $L_2$ respectively, and $T_{ij}$ be the time it takes a Turing machine to take a string $S$ and output another string $S'$ which is in language $j$ if and only if $S$ is in language $i$.  We have
$$T_1\leq T_2+T_{12}$$
$$T_2\leq T_1+ T_{21}$$
as given a string that we want to test for its belonging to $L_i$, we may run it through the translation $L_i \to L_j$ and then decide language $j$.  Indeed, this is a special case of a trivial "triangle inequality" for translation; the time it takes to translate from $L_1$ to $L_2$ plus the time it takes to translate from $L_2$ to $L_3$ is greater than or equal to the time it takes to translate from $L_1$ to $L_3$.   (I say it is a special case because a decision problem is the same as converting a language $L$ to the language $\{ 1 \}$.)
What I want to know is:

Can  we better quantify the relationship between the time complexity of a decision problem and the nature of the encoding?

So that this question is not prohibitively vague, let us say that I am looking for (1) related references, and (2) a measure of the complexity of an encoding which more tightly relates to time complexity of the "underlying" decision problem.

Added (7/19/2010):  The answers below, particularly Ryan Williams' excellent survey of the dependence of the time complexity of various problems on their encoding, get at the motivation to my question but not at my question itself.  In particular, it's clear that every problem may be re-encoded to allow (say) $O(\log n)$ time complexity, by padding.  My question is whether there's a reasonable way to measure this dependence.
For example, say the decision problem for $L_1$ is reducible to the decision problem for $L_2$, and vice versa, so that $L_1$ and $L_2$ in some sense represent the same problem.  Is there a way to formalize this last statement (about "representing the same problem")?  I am imagining, for example, a measure $C_i$ of the complexity of a language so that if $T_i$ is the time complexity of the language, and $L_1$ and $L_2$ are, say, easily reducible to one another, then $T_1/C_1\sim T_2/C_2$.  (Of course $C_i=T_i$ works, but ideally $C_i$ would be somehow a property of the language, rather than the decision problem.)  This is unfortunately becoming quite speculative, so again, related references would be a great answer.

REPLY [3 votes]: The general abstract setting for the issue driving your
question is the notion of reduction of equivalence
relations. The idea of this is that one equivalence
relation $E$ reduces to another $F$ with respect to some
complexity concept if there is a function $f$ in this class
such that

$x\, E\, y$ if and only if $f(x)\, F\, f(y)$

You can imagine that $E$ is the equivalence relation
arising from one way of representing mathematical objects
(graphs, algebraic structures, whatever) and $F$ is the
relation corresponding to an alternative method. The
reduction is saying that equivalence with respect to the
$E$ way of representing the objects is no more difficult
than equivalence with respect to the $F$ way of
representing them. 
I claim that understanding this reducibility relation amounts to
understanding exactly what your question is aimed at, the
question of how one manner of representing the same objects
can be simpler than another. More generally, this reducibility relation provides a very precise way to understand what it means to say that one classification problem is strictly harder than another, even when the objects in the two cases seem totally unrelated at first.
In the case you seem most interested, you could regard $E$
and $F$ as NP equivalence relations and insist that $f$ is polynomial time
computable. This is a case that has been recently
investigated by Sy Friedman, and this MO
question
arose out of a talk he gave on this topic here in New York,
and discusses as motivation some of the relevant general theory.
This appears to be a completely new research area, ripe for
progress. I would encourage anyone to enter into it.
Much of that theory is inspired by the enormous successes
of the much more developed instance of this concept,
occurring when $E$ and $F$ are Borel relations on the reals
and $f$ is a Borel function. This case is the
emerging-but-possibly-now-mature field of Borel equivalence
relation theory (see Greg Hjorth's survey
article
and Simon Thomas'
notes).
The theory of Borel equivalence relation theory has to deal
explicitly with the Borel analogues of the precise issues
you mention in your question, and has made huge
illuminating progress in understanding the structure of
Borel equivalence relations under Borel reducibility. I
mention some of the basic results in this MO
answer.
In general, for each notion of complexity, the goal is to
study the whole hierarchy of equivalence relations, to
discover its features and general structural results. The
Borel case is quite well developed by now, exhibiting many
fascinating features, but the NP case is much less well
developed. I do know personally, however, that other
researchers are working on several other natural contexts
of this idea.<|endoftext|>
TITLE: What is known about the transcendence of zeroes of Riemann zeta?
QUESTION [24 upvotes]: I was wondering if there are any well-known results or hunches about whether the non-trivial zeroes of Riemann-zeta (or zeta/L-functions in general) are algebraic or not.

REPLY [21 votes]: There is a paper by A. E. Ingham, "On two conjectures in the theory of numbers", Amer. J. Math. 64 (1942), 313-319, where he shows that if the ordinates of the non-trivial zeros of the Riemann zeta-function are linearly independent over $\mathbb{Q}$ then Merten's conjecture is false. This is, of course, weaker than the Rubinstein-Sarnak conjecture, but related and much earlier.<|endoftext|>
TITLE: Doubly-transitive groups
QUESTION [8 upvotes]: I want to know what all doubly-transitive groups look like. Do you know some good reference where I can read about it?

REPLY [9 votes]: In Section 7.7 "The Finite 2-transitive Groups" of the book Permutation groups by John D. Dixon and Brian Mortimer, the authors describe the complete list of finite 2-transitive groups without proofs but with references.
They list eight infinite families: the alternating, symmetric, affine and projective groups in their natural actions, as well as the less known groups of Lie type: the symplectic groups, the Suzuki groups, the unitary groups and the Ree groups. The symplectic groups have two distinct 2-transitive actions, the last three classes are 2-transitive on the sets of points in their action on appropriate Steiner systems. Additional there are 10 sporadic examples of 2-transitive groups.<|endoftext|>
TITLE: Pythagorean theorem for right-corner hyperbolic simplices?
QUESTION [14 upvotes]: My answer to the "Favorite equations" question was the Pythagorean theorem for right-corner tetrahedra:

Euclidean: $A^2+B^2+C^2=D^2$
  Hyperbolic: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}−\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$
  Spherical: $\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos\frac{D}{2}$

where $A$, $B$, $C$ are the areas of the "leg-faces" and $D$ is the area of the "hypotenuse-face".
For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex.
I mentioned not knowing the non-Euclidean counterparts of the generalization. After a couple of days of toying, I find these counterparts elusive. (It would probably help if I were better-versed in differential geometry.) So, I turn to MO to ask:

What are the non-Euclidean analogues in higher dimensions? 
  (I'm particularly interested in the case relating volumes in a hyperbolic 4-simplex.)

While the Pythagorean Theorems for Euclidean Simplices progress in a straightforward manner (just add another leg-content-square), the Pythagorean Theorem for Hyperbolic Tetrahedra already diverges somewhat dramatically from its 2-dimensional counterpart, $\cosh a \cosh b = \cosh c$. So, it's not even clear to me what form the relations would take in general. 
Edit to add a couple of references
"The laws of cosines for non-Euclidean tetrahedra" (pdf file) (by me) derives a Law of Cosines for which each non-Euclidean Pythagorean Theorem is a special case. I do not know if these results exist elsewhere in the literature. (BTW: There's a lot of unnecessary equation manipulation shown; I was using this as an opportunity to learn LaTeX. :)
A passage in above jumps into a discussion of "pseudofaces" by making passing reference to the Euclidean case. You can read about Euclidean pseudofaces and how I find them useful here:
"Heron-like results for [Euclidean] tetrahedral volume" (by me).
Edit2: A Special Case
(Edit 3: A strikingly-similar pair of volume formulas.)
Consider the special case of a right-corner hyperbolic $4$-simplex whose "legs" are congruent right-corner tetrahedra with isosceles right-triangle faces; the "hypotenuse" is a regular tetrahedron with equilateral faces. Each face of the simplex's hypotenuse is an hypotenuse-face of one of the simplex's legs.
Let the volume of each leg-tetrahedron be $L$ and the volume of the hypotenuse-tetrahedron be $H$. Then we have these formulas:
$$L=3 \int_{\rm{acos}\sqrt{x}}^{\rm{acos}\sqrt{\frac{1}{3}}} \rm{atanh}\sqrt{\frac{3\cos^2t-1}{1-\cos^2t}} \, \mathrm{d}t$$
$$H=6\int_{\rm{acos}x}^{\rm{acos}\frac{1}{3}} \rm{atanh}\sqrt{\frac{3\cos t-1}{1-\cos t}} \, \mathrm{d}t$$
where $\frac{1}{3} \le x = \cos^2{\theta_L} = \cos{\theta_H} \le \frac{1}{2}$, for $\theta_L$ the acute dihedral angle in the leg-tetrahedron and $\theta_H$ the dihedral angle in the hypotenuse-tetrahedron.
The similarity in form is rather intriguing, though not necessarily encouraging: swapping "$\cos^2 t$" for "$\cos t$" in an integrand, or "$\sqrt{x}$" for "$x$" in a limit, can completely change the nature of an integral, so there's no reason to expect a straightforward relationship between these formulas. And yet, Pythagoras beckons: there must be some connection here!
Now, it's possible to write series for $H$ and $L$, then invert the second series, then substitute back in to the first series to arrive at a series for $H$ in terms of $L$.
$$H = \frac{4}{2!\;3!} M^3 
+ \frac{18}{3!\;5!}M^5
- \frac{ 918 }{4!\;7!}M^7
+ \frac{24786}{5!\;9!}M^9
- \frac{ 6018759 }{8 \cdot 6!\;11!} M^{11}
- \frac{ 8233607961 }{80\cdot 7!\;13!} M^{13} 
- \cdots
$$
where $M := (6L)^{1/3}$. This isn't the relation I'm seeking, but observe that, for infinitesimal $L$, we have $H \approx 2 L$; that is, $H^2 \approx 4L^2 = L^2 + L^2 + L^2 + L^2$, which is the corresponding Pythagorean relation for Euclidean $4$-simplices. Even so, despite my best efforts of playing with these series, I have yet to get any insights into the nature of a non-infinitesimal connection.
I'll close here by mentioning a special-special case: at the extreme, a quadrupally-asymptotic right-corner $4$-simplex has legs that are triply-asymptotic right-corner tetrahedra with doubly-asymptotic right-triangle leg-faces; the simplex's hypotenuse is a quadrupally-asymptotic regular tetrahedron with triply-asymptotic equilateral faces. The volumes of the components (with parameter $x = 1/2$) attain significant values that happen to have interesting series representations of their own:
$$L^\star := \frac{1}{2} \; \Im\left(Li_2\left(\exp\frac{i\pi}{2}\right)\right) = \frac{1}{2} \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{2}} = 0.45798\dots$$
$$H^\star := \Im\left(Li_2\left(\exp\frac{i\pi}{3}\right)\right) = \; \sum_{k=1}^{\infty} \frac{1}{k^2}\sin{\frac{\pi k}{3}} = 1.01494\dots$$
... where $Li_2$ is the dilogarithm.
Here, $L^\star$ is half of Catalan's constant, and $H^\star$ is also known in the literature. (It is, for instance, the maximum of the Clausen function $\mathrm{Cl}_2$.)
If there's going to be a Pythagorean theorem for hyperbolic simplices, then it must apply to this case, ideally relating these values in the non-Euclidean Pythagorean tradition:
$$\text{function}(H^\star) = \text{combination of related functions}(L_1=L^\star;L_2=L^\star;L_3=L^\star;L_4=L^\star)$$
where the right-hand side is symmetric in the four parameters $L_i$ (representing the volumes of the four legs of the simplex), which are all set equal to $L^\star$. (When the formula is populated with infinitesimal quantities, it should collapse to the Euclidean sum-of-squares relation.) However, while the dilogarithm has many interesting properties, the connection between $H^\star$ and $L^\star$ is not obvious (to me). In a separate MO post, I note a hypergeometric series "similar" to a series for $H^\star$ (what I call "$T(1/2)$" there) that has a direct relation to Catalan's constant (and therefore $L^\star$), but this hasn't yet provided appropriate insights into relating $H^\star$ to $L^\star$ directly.
Have I perhaps lost the forest amid a bunch of trees?

REPLY [4 votes]: I think that the referenced paper has all you might want (it is obviously related to the Derevnyn-Mednykh paper, but is considerably more extensive).
They do not specifically talk about orthoschemes, but their results specialize nicely.
A. D. Mednykh and  M. G. Pashkevich,
"Elementary formulas for a hyperbolic tetrahedron",
Sib. Math. J.
47
(2006),
687-695.<|endoftext|>
TITLE: Computer power in plane geometry
QUESTION [6 upvotes]: I often hear that modern computer programs "may prove any theorem in elementary Eucledian geometry". Of course, as stated it is false - say, they can not prove theorems about $n$-gons for arbitrary or large enough $n$, and so on. But I wonder, how really powerful are they in problems with not so many points, lines and circles. Morley's theorem is the often example of hard theorem, which may be easily proved by computer, but this does not impress me much, since Morley theorem has two free parameters, and some equality has to be proved. The question is how number of parameters, conditions on them and inequalities make the process of automatic proof slower. To make it more precise, let consider specific example. 
Can any computer program prove Erdos-Mordell inequality?

REPLY [11 votes]: First off, I would be skeptical of the claim that computer programs "may prove any theorem in elementary Euclidean geometry", simply because it is so wide and general that is prone to be false. Secondly, I am not directly an expert in this field myself, but I hope my references are not too much off.
However, modern automatic geometric theorem proving is definitely capable of dealing with a large number of geometric problems, including those which involve geometric inequalities. The older methods (going back to Wu), translate a geometric statement is translated into an implication of the form
$$ \bigwedge_{i=1}^n f_i(x_1,\ldots,x_m) \implies f_0(x_1,\ldots,x_m)$$
where the $f_i$ are polynomials. From this, with various methods one then obtains a prove of the statement, or a counter example. I am suppressing here that often you need to specify further side conditions for a proof to be possible, e.g. that a triangle is non-degenerate; in fact, Wu's approach and the Gröbner basis even allow deducing sufficient conditions to make a theorem true in retrospect. The Wikipedia page for Wu's method gives some more details and a few references to relevant papers; you can easily google more.
Anyway, this allows encoding things like multiple points being collinear, points being contained on a circle, intersection of lines, perpendicularity of lines, and so on. However, this does indeed not allow encoding inequalities effectively; e.g. just specifying that a point is 'inside' a triangle, or that one value is less than another, in general is not possible.
But since Wu's original work, there have been many advances. If one looks a bit closer, then one notices that the above techniques actually prove theorems about complex geometry, as we are arguing about zeros of polynomials, and this all happens over an algebraically closed field. But we are usually interested in real geometry only. There are surprisingly many classical theorems from "real" geometry which remain true in the complex case, and this somewhat surprising (and as far as I know rather mysterious) fact ensures that nevertheless Wu's method and its relatives are quite successful. Still, people have worked on overcoming this limitation, as well as that of inequalities.
One approach is described "A New Approach for Automatic Theorem Proving in Real Geometry", by Dolzmann, Sturm & Weispfenning, who use quantifier elimination (from logic) to prove theorems in real geometry (as the title suggests), using their Reduce package Redlog. They use that, for example, to prove Pedoe's inequality. I am, however, not sure if this can be used to prove the Erdős–Mordell inequality; one could ask them. I think Sturm wrote his PhD thesis on the subject.
Another paper to look at is "A Practical Program of Automated Proving for a Class of Geometric Inequalities" by Lu Yang and Ju Zhang. There they describe a Maple package "Bottema" (unfortunately, this does not seem to be available on the net, at least I couldn't find it). They use it to prove a load of inequalities, and give many examples involving inequalities. To give you a flavor, here is an example (which they proved using their package):

Example 4. By $m_a$, $m_b$, $m_c$ and $2 s$
  denote the three medians and perimeter
  of a triangle, show that
  $$\frac{1}{m_a}+\frac{1}{m_b}+\frac{1}{m_c}\geq
> \frac{5}{2}.$$

And here is another excerpt (I included it as it also goes back to Erdős).

A. Oppenheim studied the following inequality in order
  to answer a problem proposed by P. Erdös.
Example 9. Let $a$, $b$, $c$ and $m_a$, $m_b$, $m_c$ denote the side
  lengths and medians of a triangle,
  respectively. If $c = \min\{a, b, c\}$, then
  $$2m_a+2m_b+2m_c \leq 2a+2b+(3\sqrt{3}-4)c.$$

This all does not answer your question about the Erdős–Mordell inequality. And I am afraid I don't know the answer! But I hope my above explanations at least made it plausible that the answer could be yes, however vague that statement is.<|endoftext|>
TITLE: Computation of homology groups of $M_{g,n}$
QUESTION [6 upvotes]: First some definitions: $\bar{M_{g,n}}$ is Deligne-Mumford space, i.e., the moduli space of stable nodal complex projective curves of genus $g$ with $n$ marked points. It is a complex orbifold, $\partial \bar{M_{g,n}}$ is the locus in $\bar{M_{g,n}}$ corresponding to nodal curves (with singularity). Do we have any relationship between the homology groups $H_{*}(M_{g,n},Q)$  and the homology groups of $\bar{M_{g,n}}$, $\partial \bar{M_{g,n}}$, the pair $(\bar{M_{g,n}},\partial \bar{M_{g,n}})$ (relative homology); here $M_{g,n}$ is $\bar{M_{g,n}}\setminus \partial \bar{M_{g,n}}$ the locus of smooth curves?
The point is for any pair of compact oriented manifolds $(X, Y), Y\subset X$, can we calculate the homology groups of $X\setminus Y$ from those of $X, Y$ and the relative homology groups $(X,Y)$ (it is not an excision case)?
This is a problem I find on page 23 of the paper: Costello, "Gromov-Witten potential associated to a TCFT", (although there it is $\bar{M_{g,n}}/S_n$, modulo the action of permutation of marked points, but it is not a big deal).
One more question is: is $\bar{M_{g,n}}/S_n$ orbifold?

REPLY [5 votes]: There is some general machinery that is perfectly suited to this question.
Consider the following setup:  let $U$ be the complement of a normal crossing divisor $D$ in a compact complex manifold (or orbifold) $X$.  (In the special case at hand, $U = \mathcal{M}_{g,n}$, and $X$ is the Deligne-Mumford compactification.)   With a bit of work one can see that the Leray spectral sequence for the inclusion $U\hookrightarrow X$ has $E_2$ page given by
\[
E^{p,q}_2 = \oplus_S H^{q-2p}(D_S;\mathbb{C})
\]
where the sum runs over the closed boundary strata of codimension $p$ and the differential on this page is given by $da = \Sigma_T \pm (i_{S,T})_! a$  where $a\in H^{q-2p}(D_S;\mathbb{C})$ and $T$ runs over codimension $p-1$ boundary strata that contain $D_S$, and $(i_{S,T})_!$ is the pushforward along the inclusion $D_S \hookrightarrow D_T$.  One can also get this spectral sequence from the weight filtration on the complex of forms with logarithmic poles along $D$.
Deligne proved that this spectral sequence degenerates at the $E_2$ page. So the cohomology of this page is the associated graded for $H^*(U;\mathbb{C})$.
In the case of the moduli space of curves and its Deligne-Mumford-Knudsen compactification, the $E_2$ page is described in terms of the cohomology of the various strata of the boundary, which are isomorphic to smaller compactified moduli spaces.  There have been quite a few papers that used this spectral sequence to prove interesting things.  For instance, there is an old paper of Voronov (alg-geom/9708019) (from just a couple of years before the proof of the Madsen-Weiss theorem) in which he analyzes this SS to show that the rational homotopy type of $\mathcal{M}_{g,n}$ is stable in the Harer-Ivanov stable range and moreover, it is formal in this range.  
As an aside, very shortly after Voronov's paper, there was Tillmann's paper in which she showed that $\mathcal{M}_{g,n}$ has the homology of an infinite loop space in the stable range, which also implies rational homotopy formality in the stable range.<|endoftext|>
TITLE: 'Eigenvectors' of evolute operation
QUESTION [13 upvotes]: The evolute of a curve is the locus of its centers of curvature.
The evolute of some plane curves is a scaled, or scaled and
reflected/rotated, version of that curve.
For example, the evolute of a cardiod is a reflected cardiod at one-third
the scale:
     (source: MathWorld)
Loosely speaking, we might say that a cardiod is an eigenvector
of the evolute operation with eigenvalue $\frac{1}{3}$.
(I am not sure if this eigenvector language can be made
technically accurate.)
There are many classical curves that are evolute eigenvectors
in this sense.  Here is a partial list
(based on this table), where I indicate the scale but not reflection: 

 Evolute( cardiod ) $\mapsto$ $\frac{1}{3}$ cardiod
 Evolute( nephroid ) $\mapsto$ $\frac{1}{2}$ nephroid
 Evolute( astroid ) $\mapsto$ 2 astroid
 Evolute( cycloid ) $\mapsto$ 1 cycloid
 Evolute( logarithm spiral ) $\mapsto$ 1 logarithm spiral
 Evolute( deltoid ) $\mapsto$ 3 deltoid
 Evolute( epicycloid ) $\mapsto$ $\lambda$ epicycloid
 Evolute( hypocycloid ) $\mapsto$ $\lambda$ hypocycloid

My question is: Has the complete class of plane
curves whose evolutes are scaled versions of themselves been
studied?
The same question can be asked for evolute surfaces, and in 
higher dimensions.
This question is (very tenuously!) related to some research into
cut loci, but at this point I'm primarily curious and interested
in learning.
Edit.  If I may extend my question slightly: Can indeed "this eigenvector language" be made
"technically accurate"?  Is there a formalization where the plane curves are vectors (perhaps in Hilbert space?) and
evolution is an operator whose eigenvectors correspond to the class of curves I seek?
I am woefully ignorant in this area and formalization is beyond my expertise.  Thanks for any hints!

REPLY [6 votes]: All these curves are "multihedgehogs" : for clear definitions and neat examples see
Y. Martinez-Maure, A Sturm-type comparison theorem by a geometric study of plane multihedgehogs, Illinois Journal of Mathematics 52 (2008), 981-993
or
Y. Martinez-Maure, Les multihérissons et le théorème de Sturm-Hurwitz, Archiv der Mathematik 80, 2003, p. 79-86.
Such a curve can be defined by  a 2Nπ-periodic function of class $C^2$ on $R$, (the number $N$ is merely the number of full rotations of the coorienting normal vector $u (t) = (cos t, sin t)$. Its evolute as support function $h'(t-π/2)$. When you say that the curve is an "eigenvector", that means in fact that its support function is a spherical harmonic that is, an eigenfunction of the circular Laplacian.
There are of course extensions of these notions in higher dimensions.<|endoftext|>
TITLE: Good reduction and blow-ups
QUESTION [7 upvotes]: Let $X$ be a projective variety over $\mathbb{Z}$, and suppose that $X$ has everywhere good reduction. Let $Y$ be the blow-up of $X$ at an integral point.
Then is it the case that $Y$ also has everywhere good reduction?
The example situation that I have in mind is the following (my main motivation is del Pezzo surfaces).
Take $X= \mathbb{P}^2$ over $\mathbb{Z}$. This clearly has good reduction everywhere. Next let $Y$ be the blow-up of $\mathbb{P}^2$ at the integral point $(0:0:1)$. This can be realised as the subvariety of $\mathbb{P}^2 \times \mathbb{P}^1$ (with variables $x_0,x_1,x_2$ and $y_1,y_2$) given by the equation $x_1 y_2 = x_2 y_1$. Then $Y$ has everywhere good reduction (at least if my caluclations are correct). 
I am curious to know if this happens after successively blowing up more integral points to obtain other del Pezzo surfaces. Note however that I am not claiming that all del Pezzo surfaces have everywhere good reduction!
Thanks in advance!

REPLY [9 votes]: More generally, if $X\to S$ is flat and finitely presented, and if $T$ is a closed subscheme of $X$ which is a relative local complete intersection over $S$, then the blow-up $Y$ of $T$ in $X$ is flat over $S$ and commutes with every base change $S'\to S$. This is just because the powers of the defining ideal $I$ of $T$ are all flat and commute with base change, and $Y$ is by definition Proj($\bigoplus_{n\geq0}I^n$).
This applies in particular if $X$ and $T$ are both smooth over $S$. In this case, compatibility with base change implies that the geometric fibers of $Y\to S$ are blow-ups of smooth subvarieties in smooth varieties, hence smooth. Since $Y$ is flat over $S$, it is smooth.<|endoftext|>
TITLE: Vector spaces without natural bases
QUESTION [15 upvotes]: Does anyone know any nice examples of vector spaces without a basis that is in some sense "natural".
To clarify what I mean, suppose we look at $\mathbb{R}^2$. We define $\mathbb{R}^2$ as pairs of real numbers. In some sense, what we are doing is expressing vectors in terms of a natural basis : (1,0) and (0,1). This is not what I want. 
An example that I thought of is a tangent space to a manifold. When one picks a tangent space to a manifold, there is no natural basis that one can pick. 
Are there other nice examples?

REPLY [6 votes]: One example I like to use is the $1$-dimensional vector space of multiples of some physical unit (length, time, mass): for example, the meter is a basis of the $1$-dimensional vector space of lengths, and the light-year is also a basis of it, but there is no natural basis of this vector space.
This example can also be used to illustrate multilinear algebra constructs on $1$-dimensional vector spaces: the space of speeds is the (still $1$-dimensional) space of linear maps between the $1$-dimensional vector space of time spans and the $1$-dimensional vector space of lengths (it turns out that, in special relativity, but not in classical mechanics, there is a canonical isomorphism between these spaces, i.e., a canonical basis for the space of speeds).  The space of areas it the tensor square of the space of lengths, and the space of volumes is its tensor cube.  And so on.
This kind of example makes it clear why for $1$-dimensional vector spaces $V$, the tensor product of $V$ with its dual is canonically isomorphic to the base field, so such spaces can be called "invertible" (as in "invertible sheaf").<|endoftext|>
TITLE: Can the image of a Schur functor always be made an irreducible representation?
QUESTION [6 upvotes]: For a partition $\lambda$ let $S^{\lambda}$ be the corresponding Schur functor.  Is it true that for every $\lambda$ there exists an irreducible representation $V$ of a finite nonabelian group $G$ such that $S^{\lambda}(V)$ is still irreducible?
This is not obvious to me even for the symmetric and exterior powers (although maybe I'm not thinking hard enough), so any partial results would be appreciated.

REPLY [3 votes]: Since the Guralnick and Tiep paper is very long, I thought I would summarize my understanding of it. Note that I only learned about this result from moonface last night, so I am hardly an expert. This answer is community wiki, in case anyone can improve on my summary.
We want to establish the following result: Let $G$ be a finite noncommutative subgroup of $GL(V)$, with $V$ a $\mathbb{C}$ vector space. Let $k \geq 6$. Then $\mathrm{Sym}^k(V)$ is reducible. (If $G$ is commutative and $V$ is one dimensional then, of course, $\mathrm{Sym}^k(V)$ is always one dimensional and hence irreducible.)
Our proof is by induction on $|G|$. Our base case will be when $G$ is a central extension of a simple group.
Choose a nontrivial normal subgroup $H$ of $G$ such that $G/H$ is simple. If we can't do this then $G$ is simple and we are in the base case. Let $V \cong \bigoplus U_i$ be the decomposition of $V$ into $H$-isotypic components. If there is more than one summand, then $\bigoplus \mathrm{Sym}^k(U_i)$ is a nontrivial $G$-subrep of $\mathrm{Sym}^k(V)$. So we may assume that $V \cong W \otimes X$, where $W$ is an $H$-irrep and $H$ acts trivially on $X$. Then one can show (proof of lemma 2.5) then $G$ is contained in $GL(W) \times GL(X)$. (This is nontrivial but not deep; you could give it as a problem in a graduate-level representation theory course.)
Then $\mathrm{Sym}^k(W) \otimes \mathrm{Sym}^k(X)$ is a subrepresentation of $\mathrm{Sym}^k(V)$. This subrep is proper unless $\dim X=1$ or $\dim W=1$. If $\dim X=1$, then $V$ is an irrep of $H$ and $\mathrm{Sym}^k(V)$ is irreducible as an $H$-rep, so we are done by induction.
If $W$ is one dimensional, then $H$ acts on $V$ by scalars, so $H$ is central. So we are in our base case: a central extension of a simple group. This case is done by group cohomology and the classification of simple groups. See section 4 for groups of Lie type, section 6 for alternating groups and section 7 for sporadic groups.<|endoftext|>
TITLE: Question on consecutive integers with similar prime factorizations
QUESTION [21 upvotes]: Suppose that $n=\prod_{i=1}^{k} p_i^{e_i}$ and $m=\prod_{i=1}^{l} q_i^{f_i}$ are prime factorizations of two positive integers $n$ and $m$, with the primes permuted so that $e_1 \le e_2 \cdots \le e_k$, and $f_1 \le f_2 \le \cdots \le f_l$. Then if $k=l$ and $e_i=f_i$ for all $i$, we say that $n$ and $m$ are factorially equivalent. In other words, two integers are factorially equivalent if their prime signatures are identical. In particular, $d(n)=d(m)$ if the two are factorially equivalent.
There's a question I've had for a long time, which is: Are there infinitely many integers $n$ such that $n$ is factorially equivalent to $n+1$? There are numerous curious pairs of consecutive integers for which this holds: $(2,3)$, $(14,15)$, $(21,22)$, $(33,34)$, $(34,35)$, $(38,39)$, $(44,45)$, as well as $(98,99)$, and many more. As you can see, many of them are almost-primes, but the last two pairs are quite striking. Although there are so many of them, a proof that there are infinitely many such pairs seems elusive. Has anyone made any progress on (or even asked) such a question? Does anyone here have a solution or progress for this?
Edit: As an added bonus, the $k$th such $n$, as a function of $k$, seems almost linear. It would be interesting to express and prove an asymptotic formula for this. Can anyone guess heuristically what the slope of this line is?
What I'll add, though I'd like to keep my question focused on the above, is that there are many other questions you can ask: How many integers $n$ are there such that $n$ is factorially equivalent to $n^2+1$, or $n^4+5n+3$, or $2^n+1$? You can generate an almost unending list of seemingly uncrackable number-theoretic conjectures this way.
Many of these questions seem to relate to other well-known number theoretic conjectures. The Twin Prime Conjecture would imply that there are infinitely many $n$ such that $n$ is factorially equivalent to $n+2$. The truth of my question above would imply that there are infinitely many $n$ such that $d(n)=d(n+1)$, a result which has actually been proven, so my conjecture is a strengthening of it. Furthermore, the proof of the infinitude of Mersenne primes would prove the infinitude of $n$ factorially equivalent to $2^n-1$. But beyond all these connections to well-known conjectures, I think the question about and its generalizations are aesthetically interesting.

REPLY [8 votes]: I'm coming into this late, but am wondering why no one seems to have mentioned the results of Goldston, Graham, Pintz and Yildirim:
http://arxiv.org/pdf/0803.2636.pdf
In particular, their Theorem 4 answers the OPs first question in the affirmative.<|endoftext|>
TITLE: Maximal ideal that annihilates entire ring
QUESTION [12 upvotes]: Does there exist a ring $R$ with a nonzero maximal ideal $M$ such that $R^2=R$ and $MR = RM = 0$?
Here $R$ is associative but does not have an identity (obviously).  It seems a simple enough question but I'm not even sure of the answer if I insist that $M$ be the only proper ideal.  I know $R$ can't be commutative and that the conditions can't be relaxed (much); for example, there is a commutative ring satisfying these conditions if $M$ need not be maximal.

REPLY [2 votes]: Ok, I think I worked out the bugs in my previous answer.
Let $F$ be the field with two elements (for simplicity).  Let $T=F< a_{i}, x_{k,i},y_{k,i}:i\in \mathbb{N}, k\in K>$ be the free algebra over $F$, generated by the non-commuting variables $a_{i}$, $x_{k,i}$, and $y_{k,i}$ (where $k$ runs over an indexing set $K$ we will describe later, and $i$ runs over the non-negative integers), with no constant term.  Let $J$ be the ideal generated by the relations $a_{i}=a_{2i+1}a_{0}a_{2i+2}$, $x_{k,i}=x_{k,2i+1}a_{0}x_{2i+2}$, $y_{k,i}=y_{k,2i+1}a_{0}y_{k,2i+2}$ and $a_{0}^{2}a_{i}=a_{0}^{2}x_{k,i}=a_{0}^{2}y_{k,i}=a_{i}a_{0}^{2}=x_{k,i}a_{0}^{2}=y_{k,i}a_{0}^{2}=0$ for each $i$.  Let $S=T/J$ and identify each variable with its image in the quotient (and continue this practice below).  Notice that $S$ is generated by $a_{0}$ and $S^{2}=S$.  Also note that $a_{0}^{2}$ is a universal zero-divisor.  The ring $R$ we want will be a quotient of $S$, so will still be generated by $a_{0}$, we will still have $R^{2}=R$, and $a_{0}^{2}$ will still be a universal zero-divisor.  We will construct $R$ so that $a_{0}^{2}$ remains nonzero and generates a maximal ideal.
Let $I'$ be the set of words $w\in T$, $w\notin a_{0}^{2}+J$,  such that there is a word $w_{1}\in T$ of length 1 (i.e. a variable) with $w_{1}w-a_{0}^{2}\in J$ or $ww_{1}-a_{0}^{2}\in J$ but no words $w_{2},w_{3}\in T$ with $w_{2}w-a_{0}\in J$ or $ww_{3}-a_{0}\in J$ or $w_{2}ww_{3}-a_{0}\in J$. [For example, $a_{0}a_{1}a_{0}a_{5}a_{0}$ is such a word.  Multiplying on the right by $a_{6}$ it reduces to $a_{0}^{2}$, but it does not equal $a_{0}^{2}$ modulo $J$, and can never be multiplied to $a_{0}$.]  If we make words in $I'$ zero (or even zero divisors) that will possibly make $a_{0}^{2}$ zero.  So, let $I$ be the ideal generated by the following relations: $x_{w,0}wy_{w,0}=a_{0}$ if $w\in I''$, and take $S_{1}=S/I$.
At this point, we repeat the argument in the previous paragraph on $S_{1}$ (our new set $I_{1}'$ will have new words in it).  We do this countably many times, the resulting ring is $S_{\infty}$.
Next, mod out by the ideal generated by words $w\neq a_{0}^{2}$ such that there exist no words $w_{1}, w_{2}$ so that $w_{1}w=a_{0}^{2}$ or $ww_{2}=a_{0}^{2}$ or $w_{1}ww_{2}=a_{0}^{2}$ (in $S_{\infty}$).
The resulting quotient ring should be the structure you are looking for.  (The index set $K$ could initially be all words in $T$, then cut down to words that appeared in any of the $I'$'s above.)
This is complicated enough that I might have made a mistake somewhere--this is just a sketch of my ideals.  One of the main points that should be noted is that (besides $a_{0}$) each variable appears in exactly two relations (except those when a word is made equal to 0): one where it is used to reduce the size of a word, and one where the size expands.<|endoftext|>
TITLE: Grothendieck's Galois Theory today
QUESTION [19 upvotes]: I have recently become aware of, and started to study in my free time (abundant in these summer months) Grothendieck's Galois Theory (GGT), as formulated in SGA 1 and later by Grothendieck's contemporaries. I understand there to have been a number of unresolved and open questions relating to GGT upon its formulation, some of which seem to persist. This is a truly gorgeous subject, and I wonder whether it is still studied rigorously or researched at all today. Where/Who produces interesting results regarding things such as Galois and Atomic topoi, applications of the Grothendieck fundamental group, etc., today?
Is there a great deal of utility to GGT beyond the foundational work in algebraic geometry later formulated by Grothendieck?

REPLY [10 votes]: The basic Grothendieck's assumptions means we are dealing with an connected atomic site $\mathcal{C}$ with a point, whose inverse image is the fiber functor $F: \mathcal{C} \to \mathcal{S}et$:
(i) Every arrow $X \to Y$ in $\mathcal{C}$ is an strict epimorphism.
(ii) For every $X \in \mathcal{C}$ $F(X) \neq \emptyset$.
(iii) $F$ preseves strict epimorphisms.
(iv) The diagram of $F$, $\Gamma_F$ is a cofiltered category.
Let $G = Aut(F)$ be the localic group of automorphisms of $F$.
Let $F: \widetilde{\mathcal{C}} \to \mathcal{S}et$ the pointed atomic topos of sheaves for the canonical topology on $\mathcal{C}$. We can assume that $\mathcal{C}$ are the connected objects of $\widetilde{\mathcal{C}}$.
(i) means that the objects are connected, (ii) means that the topos is connected, (iii) that $F$ is continous, and (iv) that it is flat. 
By considering stonger finite limit preserving conditions (iv) on $F$ (corresponding to stronger cofiltering conditions on $\Gamma_F$)  we obtain different Grothendieck-Galois situations (for details and full proofs see [1]):
S1) F preserves all inverse limits in $\widetilde{\mathcal{C}}$ of objets in $\mathcal{C}$, that is $F$ is essential. In this case $\Gamma_F$ has an initial object $(a,A)$ (we have a "universal covering"), $F$ is representable, $a: [A, -] \cong F$, and $G = Aut(A)^{op}$ is a discrete group. 
S2) F preserves arbritrary products in $\widetilde{\mathcal{C}}$ of a same $X \in \mathcal{C}$ (we introduce the name "proessential for such a point [1]). In this case there exists galois closures (which is a cofiltering-type  property of $\Gamma_F)$, and $G$ is a prodiscrete localic group, inverse limit in the category of localic groups of the discrete groups $Aut(A)^{op}$, $A$ running over all the galois objects in $\mathcal{C}$.
S2-finite) F takes values on finite sets. This is the original situation in SGA1. In this case the condition "F preserves finite products in $\widetilde{\mathcal{C}}$ of a same $X \in \mathcal{C}$ holds automatically by condition (iv) ($F$ preserves finite limits), thus there exists galois closures, the groups 
$Aut(A)^{op}$ are finite, and $G$ is a profinite group, inverse limit in the category of topological groups of the finite groups $Aut(A)^{op}$.
NOTE. The projections of a inverse limit of finite groups are surjective. This is a key property. The projection of a inverse limit of groups are not necessarily surjective, but if the limit is taken in the category of localic groups, they are indeed surjective (proved by Joyal-Tierney). This is the reason we have to take a localic group in 2). Grothendieck follows an equivalent approach in SGA4 by taking the limit in the category of Pro-groups.
S3) No condition on $F$ other than preservation of finite limits (iv).  This  is the case of a general pointed atomic topos. The development of this case we call "Localic galois theory" see [2], its fundamental theorem first proved by Joyal-Tierney.
[1] "On the representation theory of Galois and Atomic topoi", JPAA 186 (2004)
[2] "Localic galois theory", Advances in mathematics", 175/1 (2003).<|endoftext|>
TITLE: Vector spaces with natural bases
QUESTION [9 upvotes]: Sergeib's question asks about vector spaces without a natural basis.
Actually, I would claim (apparently in accord with many comments and answers to Sergeib's question ) that this is the default situation and that it is a rare event if a naturally occurring vector space does have a natural basis or even if it has any basis that can be described explicitly: the notorious $\mathbb Q \; -$ vector space $\mathbb R$ being the foremost example of this impossibility of exhibiting an explicit basis . Of course  tautological examples like polynomial rings don't contradict this thesis, since they are defined as free vector spaces on some set!
In fact, the only example I can see without thinking twice of a natural vector space  with a big natural non-tautological basis is the rational function field $k(X)$ seen as a $k$ - vector space .
Namely, consider a field $k$ and the set of monic polynomials  $f_i(X) \; (i \in I)$ irreducible over $k$. Then  $k(X)$ has the following $k$ - basis:
the mononomials $X^k (k\in \mathbb N) $ and the rational fractions $\frac{X^m}{f_i(X)^s}$
(with $i\in I,\; s>0$ and $ m< deg f_i $ )
In particular this natural basis is non denumerable if $k$ is non denumerable..
Question  Which other vector spaces do you know for which some (preferably big) explicit basis can be given, but which are not clearly constructed as free vector spaces over a set?

REPLY [2 votes]: I somewhat disagree with the premise of the question: I want to point out that after imposing more structure, this also becomes a default situation. The general setting I have in mind is commutative algebra and there are various techniques to construct a "preferred" SAGBI basis in the coordinate ring or homogeneous coordinate ring of a variety. Particularly useful in representation theory are standard monomial bases in $K[G/U],$ where $G$ is a semisimple algebraic group and $U$ is the unipotent radical of a parabolic subgroup. 
Gröbner bases of various kinds are of similar ilk. Finally, one should mention orthogonal polynomials: starting with an "obvious" basis, perhaps a naive monomial basis in some polynomial algebra, produce a different standard basis.<|endoftext|>
TITLE: Left U_n-invariants of SL_n - an exercise in Kraft-Procesi
QUESTION [5 upvotes]: I am sorry for spamming MO with questions I have not thought about for more than 3 hours, but currently I am quite busy with preparing a talk on representations of $S_n$, and I don't want these to get lost. I hope this one is not quite as vague as the last one.
This here is an attempt to generalize Exercise I.21 in Kraft-Procesi, Classical Invariant Theory.
Let $K$ be a field - say, infinite, since we are going to do classical invariant theory. Let $K\left[\mathrm{SL}_n K\right]$ denote the $K$-algebra of polynomial functions on $\mathrm{SL}_n K$, which I define either as
$\left\lbrace f\mid_{\mathrm{SL}_n K} \ \mid \ f\in K\left[\mathrm{M}_n K\right]\right\rbrace$
or as $K\left[\mathrm{M}_n K\right]\diagup \left(\det-1\right)$ (proving the equivalence of these two definitions is not the matter, it's rather easy - even easier than Kraft and Procesi try to make one believe).
Now, the group $\mathrm{U}_n K$ of unipotent upper triangular matrices acts on $\mathrm{SL}_n K$ from the left. What is the invariant ring? It is easily seen that
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j,i+1,i+2,...,n\right)$
is an invariant for any $i\geq j$. These generate the fraction field of the invariants, but do they also generate the ring of the invariants itself?
(The above-mentioned exercise is the above for $n=2$.)
Arguments using Victorian age methods (as opposed to Zariski-topological or other algebro-geometrical) would be particularly preferred.
EDIT: As Allen Knutson has pointed out, my question has a negative answer. However, the (larger) collection of determinants of the form
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j_1, j_2, ..., j_{n-i+1}\right)$
for $1 < i \leq n$ and $1 \leq j_1 < j_2 < ... < j_{n-i+1} \leq n$ does generate the ring of invariants. When $K$ has characteristic $0$, this can be proven using the standard theory of highest-weight modules and multiplicity-free algebras explained in Kraft-Procesi (see my errata, "Page 9, Exercise 21" for a proof). I am still wondering whether it holds for arbitrary $K$ and has a more elementary or combinatorial proof.

REPLY [2 votes]: Your question (in char 0) has already been answered by Knutson and Victor Protsak. I just wanted to say that this holds in greater generality (char 0) but the method is not Victorian. 
So, consider the algebraically closed field $K$ of char 0, and $G=SL_n$. Given an irreducible $V_{\lambda}$ with highest weight $\lambda$ (relative to  the standard upper triangular subgroup $B=TU$where $T$ is the group of diagonals), we have the decomposition for the action of $G\times G$ on the coordinate ring $k[G]$
$$k[G]=\bigoplus V_{\lambda }^* \otimes V_{\lambda}.$$ Taking $U$ invariants on the left, we have 
$$k[U\backslash G]=\bigoplus V_{\lambda},$$ i.e. every irreducible representation $V_{\lambda}$  occurs exactly once and is generated as a $G$ module by the vector $v_{\lambda}$ which is invariant under the group $V$ of $lower \quad triangular$ unipotent matrices. If $\lambda_1, \cdots, \lambda_{n-1}$ are the highest weights of the fundamental representations, and $\lambda =\sum a_i\lambda _i$ with $a_i\geq 0$ then clearly (by multiplicity one) $v_{\lambda}=\prod v_i^{a_i}$ is a product of powers of $v_{\lambda_1}, \cdots ,v_{\lambda _{n-1}}$. Hence the ring of invariants is generated by the functions spanning $V_{\lambda_1},\cdots  V_{\lambda _{n-1}}$. Thus the ring of invariants is finitely generated.
The above proof works for any connected reductive group.<|endoftext|>
TITLE: Does the quantum subgroup of quantum su_2 called E_8 have anything at all to do with the Lie algebra E_8?
QUESTION [26 upvotes]: The ordinary McKay correspondence relates the subgroups of SU(2) to the affine ADE Dynkin diagrams.  The correspondence is that the vertices correspond to irreducible representations of the subgroup, and edges correspond to tensor product with the 2-dimensional representation (which is given by the inclusion of G in SU(2)).
The quantum McKay correspondence does the same thing for "quantum subgroups" of the quantum group U_q(su(2)) for q a root of unity and the ordinary ADE Dynkin diagrams.  One way to make this precise is that "quantum subgroups" are module categories over semisimplified categories of modules for U_q(su(2)), so the simple objects in the module category are the vertices, and the edges come from the tensor product with the defining rep of U_q(su(2)).  In this language the result is due to Kirillov-Ostrik, but it goes back further in both the subfactor literature (due to Ocneanu and others) and the conformal field theory literature (which I don't know as well).  These quantum subgroups come in two sorts, type 1 and type 2, and the type 1 ones can also be realized as fusion categories where the module structure comes from "restrict and tensor."  The A_n, the D_2n, E_6 and E_8 are Type 1.
Ok, so now for the question, does the E_8 "quantum subgroup of quantum su(2)" have anything whatesoever to do with Lie algebra E_8?
I think the answer should be "no."  The main reason I'm guessing this is that although the D_2n fusion categories are related to certain SO quantum groups under level-rank duality, the level-rank dual fusion category is not related to the quantum group for the Lie algebra D_2n it's instead related to the Lie algebra D_n-1 (see section 4.3 of my paper with Scott and Emily and the references therein).
The reason that I ask is in reference to Borcherds's question about detecting E_8 experimentally.  I don't totally understand Will Orrick's answer but it sure sounds to me like he's saying that they've detected the quantum subgroup E_8, not the Lie algebra E_8.  In particular its the quantum subgroup E_8 that has 8 objects ("particle types" in physics speak) one of which (all the way at the far end) has dimension the golden ratio (and so the golden ratio should come up in comparing it with the trivial object in some experiment).

REPLY [4 votes]: This is not really an answer. It's more like a stub.
I hope that by making it wiki, I'll encourage others to contribute.

So there are plenty of things classified by A-D-E (or variants thereof, such as A-B-C-D-E-F-G, or A-D-E-T, or A-Deven-Eeven). In particular, there are plently of things called "the E8 ...".
Instead of making a complete graph, and showing that for any X and Y, "the E8 X" is directly related to "the E8 Y", one should maybe be less ambitious, and only construct a connected graph.
So, our task is to connect "the E8 Lie group" to "the E8 quantum subgroup of SU(2)".
I suggest the following chain:

E8 Lie group --- E8 Lie algebra --- E8 surface singularity --- E8 subgroup of SU(2) --- E8 quantum subgroup of SU(2).

(1) E8 Lie group --- E8 Lie algebra
No comment.
(2) E8 Lie algebra --- E8 surface singularity
Look at the nilpotent cone $C$ inside $\mathfrak g_{E_8}$. That's a singular algebraic variety with a singular stratum in codimension 2. The transverse geometry of that singular stratum yields a surface singularity.
(3) E8 surface singularity --- E8 subgroup of SU(2)
Given a finite subgroup $\Gamma\subset SU(2)$, the surface singularity is $X:=\mathbb C^1/\Gamma$. Conversely, $\Gamma$ is the fundamental group of $X\setminus \{0\}$.
(4) E8 subgroup of SU(2) --- E8 quantum subgroup of SU(2)
Is this quantization?!?
So let's recall what one really means when one talks about the "E8 quantum subgroup of SU(2)". We start with the fusion category Rep(SU(2))28, which one can realize either using quantum groups, or loop groups, or vertex algebras. That category is a truncated version of Rep(SU(2)): whereas Rep(SU(2)) has infinitely many simple objects, Rep(SU(2))28 has only finitely many, 29 to be precise. 
Now this is what the "E8 quantum subgroup of SU(2)" really is: it's a module category for Rep(SU(2))28. In other words, it's category M equipped with a functor Rep(SU(2))28 × M → M, etc. etc.
That's where one sees that "quantum subgroup of SU(2)" is really a big  abuse of language.
...so I don't know how to relate subgroups of SU(2) with the corresponding "quantum subgroups". Can anybody help?<|endoftext|>
TITLE: When do cofinal chains of universal codings of the integers exist?
QUESTION [5 upvotes]: Universal codings of integers
A (binary) coding of the integers is a prefix-free code of the natural numbers, whose codewords are non-decreasing in size. A coding is universal if it is short enough (log n + o(log n)), but that's not important.
Some examples:

The unary coding 0, 10, 110, ...; code length is n
Code first the length of the number in unary, then the number itself in binary; code length is about 2log n
Code first the length of the number using the previous coding, then the number itself in binary; code length is about log n + 2log log n
...
Diagonalize the construction to get code length of log n + log log n + ... + 2log* n
Continue this way through the constructible ordinals

The diagonalized code, known as the $\omega$-code, is due to Peter Elias.
A partial ordering of codes
The sequence of codes above are progressively better, in the following sense:

A coding a is better than a coding b if |b(n)| - |a(n)| tends to infinity.

There are some natural questions to ask:

Is there a best code?
If not, is there an optimal sequence of codes?

As it turns out, not only is there no best code, but given any sequence of codes, we can always find a code which is better than all of them; the proof from one of Hausdorff's papers (Untersuchungen über Ordnungtypen V from 1907) can be adapted to our setting.
Scales
The best thing that can be hoped for is a chain of codes which is cofinal for the poset of codes, i.e. a set of mutually comparable codings, such that for each arbitrary coding, our scale contains a superior one (such a beast Hausdorff called a Pantachie).
The problem of scales is well-known, and it is easy to show the existence of a scale given CH (following Hausdorff's steps). In other settings (and possibly this one), existence already follows from MA. However, most of the literature deals with somewhat different posets, and it is not clear that their results apply in this case.
Here are some pointers:

Hausdorff Gaps and Limits by Frankiewicz and Zbierski, which deals with the ordering f > g if f(n) > g(n) infinitely often.
Gaps in $\omega^\omega$ by Marion Scheepers, which deals with the ordering f > g if f(n) - g(n) tends to infinity.

In their settings, Hechler forcing can be used to produce worlds in which there is no scale.

Is the existence of scale (in the context of monotone codings of integers) independent of set theory?

Codings and series
Some easy reductions connect our problem with problems involving convergent series and divergent series satisfying some extra conditions, which stem from our monotonicity requirements; the key is Kraft's inequality, stating that a code with codeword lengths wi exists iff the sum $\sum 2^{-w_i}$ converges.
The reductions are most easily stated if we extend our posets with some equivalence relation. We then say that two posets are interlacing if there are two order-preserving mappings (between the two posets in both directions) which are pseudo-inverses, i.e. their composition sends a point to an equivalent one. Given two interlacing posets, one has a scale iff the other one has a scale.
The following posets are interlacing:

Arbitrary (non-monotone) codes, with a < b if b is better than a, and a ~ b if |a(n)-b(n)| is bounded.
Convergent positive series, with a < b if b(n) = o(a(n)), and a ~ b if a(n) = O(b(n)) and b(n) = O(a(n)).
Divergent positive series (reverse definition of <).

Monotonicity complicates the picture (the corresponding series are no longer arbitrary) but seems necessary, since one can give a non-monotone code with the property that no monotone code is better than it.
Effective and efficient codings
The motivation behind the question is the actual usage of universal codings by computer engineers. New, impractical methods of codings are suggested all the time, but no one seems to have tackled the fundamental question.
This prompts us to ask similar questions for effective (computable) codings.

Can classical recursion theory hierarchies be adapted to the setting of codes?

It would be nice to get an analog of the fast-growing hierarchy, for example.
We could further wonder what happens if we ask our coding procedure to be efficient, for example linear-time computable.

REPLY [4 votes]: I believe I can answer your first question.  But the answer involves forcing, which I cannot explain here (see Kunen's Set Theory. An introduction to independence proofs).
Assuming CH, there is a scale of codes.  Why? Enumerates all codes as
$(c_\alpha)_{\alpha<\omega_1}$.   Construct a sequence $(a_\alpha)_{\alpha<\omega_1}$
such that for each $\alpha$, the code $a_\alpha$ is better than $c_\alpha$ and all
$a_{\beta}$, $\beta<\alpha$.  This is possible since for every countable set of codes
there is one code that is better than all of them, if I understand you correctly.
Now $(a_\alpha)_{\alpha<\omega_1}$ is a scale of codes.
For the other consistency result, namely ZFC is consistent with the non-existence of scales, consider the partial order of finite initial segments of monotone prefix-free codes, where a finite initial segment $c$ is stronger than $d$ ($c\leq d$) if $c$ extends $d$.
This partial order is countable and every element has two extensions that don't have a common
extension.  This implies that the partial order is forcing equivalent to the so called
Cohen forcing.
Whenever $c$ is a finite initial segment of a code, $n$ is a natural number and $a$ is a code, then $c$ can be extended to a finite initial segment $d$ of a code such that
for some $m>n$, $d$ already contains the code word of $m$ and this code word for $m$ is longer than the code word for $m$ that $a$ has.
Similarly, $c$ can be extended to a finite $d$ that for some $m>n$ has a shorter code word for $m$ than $a$ has.  
This shows that forcing with this partial order adds a code that is incomparable with all
codes in the ground model.  
Now we start from a model of set theory that satisfies CH and force over it with a finite support product of $\aleph_2$ copies of the countable partial order defined above.
This forcing adds a family $(c_\alpha)_{\alpha<\omega_2}$ generic codes.
(CH fails in this generic extension).
I claim that no subfamily of size $\aleph_1$ of this family of codes has an upper bound.
Why?  Let $A\subseteq\omega_2$ be of size $\aleph_1$.  By the properties of Cohen forcing
(c.c.c. in particular) we may assume, after enlarging $A$ if necessary, that
$A$ is already in the ground model.
Now, whenever $a$ is a code in the generic extension, then $a$ has a name that only depends 
on countably many of the $c_\alpha$'s.
Take $\beta\in A$ outside this countable set of indices.
Then, by the argument above, $c_\beta$ and $a$ are incomparable as codes 
(this is because $c_\beta$ is generic over a model containing $a$) and hence
$a$ is not an upper bound of the $c_\alpha$, $\alpha\in A$.
A similar argument shows that in the model of set theory that we have constructed (which is in fact Cohen's original model that refutes CH) no set of codes of size $<\aleph_2$
is cofinal.  (Why?  If $C$ is a set of $\aleph_1$ codes, then there still is some 
$\alpha<\omega_2$ such that $c_\alpha$ is generic over a model that contains $C$
and now no element of $C$ is an upper bound for $c_\alpha$.)
We now have a model of set theory in which there is an unbounded set of codes of size $\aleph_1$ but no cofinal set of size $<\aleph_2$.  This implies that there is no scale.  
I hope it is possible to get something out of this answer.
I am fully aware that I am using forcing jargon here, but to really give a complete proof
would take a lot of space and time.  I would guess that the code problem can actually be reduced to some partial order which has been studied in the literature.<|endoftext|>
TITLE: Recognizing regular neighborhoods
QUESTION [6 upvotes]: In a Riemannian manifold consider two compact smooth submanifolds $S$, $S^\prime$ that intersect transversely. It seems intuitively obvious that for a sufficiently small number  $r$, the union of $r$-neighborhoods of $S$ and $S^\prime$ is a regular neighborhood of $S\cup S^\prime$ in some triangulation of the ambient manifold. Is this written somewhere?
By the way, a standard reference for regular neighborhoods is  this paper  by Marshall Cohen, published in Trans. Amer. Math. Soc.  136, 1969 189--229.
EDIT: it may be worth discussing why any small $r$-neighborhood $N_r(S)$ of a
smooth compact submanifold $S$ is a regular neighborhood. This is written on page 527 of [M. Hirsch, Smooth regular neighborhoods., Ann. of Math. (2) 76 1962 524--530]. First, one uses results of J. Whitehead to show that $N_r(S)$ is a subcomplex in some smooth triangulation. By choosing $r$ small enough we arrange that $N_r(S)$ is inside the second derived neighborhood $R$ of $S$, i.e. the star of $S$ in the second barycentric subdivision.  The second derived neighborhood is always a regular neighborhood. Furthermore, Hirsch says $R$ is the mapping cylinder of a retraction $\partial R\to S$, and each radial segment in the cylinder is transverse both to $S$ and to $\partial R$. This transversality implies that for small $r$ removing the interior of $N_r(S)$ from $R$ gives $\partial R\times I$, so that $N_r$ is also a regular neighborhood of $S$.
The same argument may work for $S\cup S^\prime$. In fact it is believable that 
the second derived neighborhood $R$ of any subcomplex $K$ can be written as a mapping cylinder whose radial segments are PL-transverse to $\partial R$ and $K$. If true, we just need to see that the segments are also transverse to the boundary of the union of $r$-neighborhoods of $S$ and $S^\prime$.

REPLY [2 votes]: You're not really forced to leave the smooth category at an earlier stage than before.  There is a good pseudogroup for the category of "smooth manifolds with transverse submanifolds".  If $r$ is small enough, then $S$, $S'$, $\partial N_r(S)$, $\partial N_r(S')$ are all transverse, so what you have exists in that category.  Of course there is more work to do than just this remark.
For simplicity, suppose that the normal exponential maps of $S$ and $S'$ commute up to some radius $r$.  Then the local structure works out exactly as you expect, and it should be routine to generalize Moe Hirsch's argument.  Suppose instead that the normal exponential maps do not commute.  Then using the implicit function theorem, you can bend the normal exponential map of $S'$ out to a radius of $r' < r$ to make it commute with normal exponential map of $S$.  You can do this in such a way that $\partial N_{r'}(S')$ is not moved.  (Well, I think so.  I admit that I'm handwaving some.)  Then it should work as in the simpler case.<|endoftext|>
TITLE: Chevalley–Shephard–Todd theorem
QUESTION [15 upvotes]: The wikipedia article claims that the theorem "was first proved by G. C. Shephard and J. A. Todd (1954) who gave a case-by-case proof. Claude Chevalley (1955) soon afterwards gave a uniform proof". I read the paper by Chevalley and it seems that he only proves the implication: "If the group is generated by pseudo-reflections, then the ring of invariants is polynomial". I wonder whether there is a uniform proof of the inverse implication? Where is it written?

REPLY [10 votes]: Torsten's argument is of course beautiful, but it might be worth recording that there is also a slick combinatorial argument, in case you need to teach this to students without algebraic geometry. (After rereading, it looks like Josh Swanson is linking to the same argument, so upvote his answer if you like this.) Let $G$ be any finite subgroup of $GL(V)$ for $V$ a vector space of characteristic $0$; let $R = \mathrm{Sym}(V)$ and $S = R^G$. 
For any finite dimensional vector space $U$ with an action of $G$, we have 
$$\dim U^G = \frac{1}{|G|} \sum_{g \in G} \mathrm{Tr}(g: U \to U).$$
So, working degree by degree,
$$\sum_{k=0}^{\infty} (\dim S_k) t^k = \frac{1}{|G|} \sum_{g\in G} \sum_{k=0}^{\infty} t^k Tr(g: Sym^k V \to Sym^k V).$$
If the eigenvalues of $g$ on $U$ are $\lambda_1(g)$, ..., $\lambda_n(g)$, we deduce
$$\sum_{k=0}^{\infty} (\dim S_k) t^k= \frac{1}{|G|} \sum_{g\in G} \prod_{j=1}^n \frac{1}{1-\lambda_j(g) t} \quad (\ast). $$
This is called Molien's formula. 
In particular, if $S$ is a polynomial ring with generators in degrees $e_1$, ..., $e_n$, then
$$\prod_{i=1}^n \frac{1}{1-t^{e_i}} = \frac{1}{|G|} \sum_{g\in G} \prod_{j=1}^n \frac{1}{1-\lambda_j(g) t} \quad (\dagger).$$
Let $T$ be the set of pseudoreflections. Expanding both sides in Laurent series around $t=1$, we have
$$\prod_{i=1}^n \left( \frac{1}{e_i (1-t)} + \frac{e_i-1}{2 e_i} + O(1-t) \right)$$
$$= \frac{1}{|G|} \left( \frac{1}{(1-t)^n} + \sum_{g \in T} \frac{1}{(1-t)^{n-1} (1-\lambda_n(g))} + O((1-t)^{-n+2}) \right)$$
 where $\lambda_n(g)$, for $g \in T$, is taken to be the eigenvalue which is not $1$. Grouping together elements of $T$ which generate the same subgroup and matching coefficients, one concludes:
$$|G| = \prod e_i \ \mbox{and}\ |T| = \sum (e_i-1). \quad (\S)$$
As in Torsten's proof, let $H$ be the subgroup of $G$ generated by $T$. By the reverse direction of CST, $R^H$ is a polynomial ring, say with generators in degrees $d_1$, ... $d_n$. Since the first generator of $R^G$ must involve at least one generator of $R^H$, the first two generators of $R^G$ must involve at least two generators of $R^H$ etc we see that, after reordering $d_1 \leq d_2 \leq  \cdots \leq d_n$ and $e_1 \leq e_2 \leq \cdots \leq e_n$, we have $e_i \geq d_i$.
But $\sum (e_i -1) = \sum (d_i-1) = |T|$. So $e_i = d_i$. Then $|G| = \prod e_i = \prod d_i = |H|$, so $G=H$.<|endoftext|>
TITLE: Smallest area shape that covers all unit length curve
QUESTION [16 upvotes]: On a euclidean plane, what is the minimal area shape S, such that for every unit length curve, a translation and a rotation of S can cover the curve.
What are the bounds of the shape's area if this is a open problem?
When I asked this problem few years ago, someone told me it's open. I don't know if this is still open and I can't find any reference on it. 
I don't even know what branch of mathematics it falls under. so I can't even tag this question.

REPLY [10 votes]: The result of mine alluded to in A B's answer, giving an improved lower bound, is now on arxiv: http://arxiv.org/abs/1101.5638<|endoftext|>
TITLE: A nice explanation of what is a smooth (l-adic) sheaf?
QUESTION [13 upvotes]: I would like to understand this concept. It seems to be important (for the theory of perverse sheaves), yet I don't know any nice exposition of the properties of smooth sheaves.

REPLY [26 votes]: I'll give an answer, only because I'm interested in chasing down these references
myself. But all I'm doing is assembling references. I  assume that
BCnrd will keep me honest. [July 21: I've added some remarks about constructibility, to
makes this more useful (at least to me).]
Since I'm a complex geometer rather an arithmetic one, let me start with the first case for intuition. If $X_{an}$ is a (connected) complex variety endowed with the classical topology then one knows that representations of the usual $\pi_1(X_{an},x)$ correspond to locally constant sheaves on $X_{an}$. This is classical. A good source of examples are as follows:
if $f:Y\to X$ is a surjective smooth proper map, then it is topologically
a fibre bundle (Ereshmann). Therefore $R^if_*\mathbb{Z}$ is locally constant. The 
corresponding $\pi_1(X)$-module is the monodromy representation. The most
general statement one can make, without making any assumptions
on $f$, is that the proper direct image $R^if_!\mathbb{Z}$ is constructible.
Note that constructibility can mean different things in the topological world.
The best notion (from my point of view) is what is sometimes called algebraic constructibility: there exists a partition of the base into Zariski locally closed strata
such that the restrictions of the sheaf are locally constant. The only reference that I know which takes this viewpoint is Verdier, Classe d'homologie associée à un cycle. If people are aware of other sources, please let me know.
Remarkably, the analogous results hold in the $\ell$-adic case, although for different reasons. Let $X$ be variety over some  field.
A lisse (resp. constructible) $\ell$-adic sheaf is now a prosheaf 
$$\ldots \mathcal{F}_n\to \mathcal{F}_{n-1}\ldots $$
on the etale site $X_{et}$ such that  each item above is a locally constant (resp. constructible) $\mathbb{Z}/\ell^n$-module etc. (see Freitag-Kiehl, pp 118-131, for the precise conditions). For lisse sheaves, each $\mathcal{F}_n$ gives a representation of the etale fundamental group 
$$\pi_1^{et}(X,x)\to GL_N(\mathbb{Z}/\ell^n)$$
($x$ a geom. pt.). So passing to the limit, we get a continuous representation
$$\pi_1^{et}(X,x)\to GL_N(\mathbb{Z}_\ell)$$
This constuction is an equivalence [FK,p 286].
The corresponding result that $R^if_*\mathbb{Z}_\ell$
is lisse, when $f$ is smooth, proper and surjective, should follow from Theorem 20.2 of Milne "Lectures on etale cohomology" from his website. The contrucibility of $R^if_!\mathbb{Z}_\ell$ would follow from SGA4 exp XIV 1.1 (It ought to be in [FK,M], but I probably didn't look hard enough.)
When $X$ is defined over $\mathbb{C}$, one can compare cohomology for the classical
and etale topologies with general coefficients by applying SGA4 exp XVI 4.1 and taking inverse limits. A more
general comparison result for the "6 operations" is given in [Beilinson-Bernstein-Deligne
p 150], but the proof seems a bit sketchy.
Remark added July 22: Unfortunately, this part of the story appears to be inadequately 
addressed in the literature. See BCnrd's comment below.<|endoftext|>
TITLE: What are some mathematical sculptures?
QUESTION [54 upvotes]: Either intentionally or unintentionally.
Include location and sculptor, if known.

REPLY [3 votes]: Credit to the photographer for this shot of Charles O. Perry's 'Solstice' in my downtown Tampa (a '2/3-twist triangular torus Mobius strip' according to him in this article about his work)<|endoftext|>
TITLE: How much can small modifications change the nef cone? 
QUESTION [9 upvotes]: First let me give a precise formulation of the question; I'll give some background/motivation at the end.
If X is a projective variety which is Q-factorial (meaning X is normal, and some sufficiently high multiple of every Weil divisor is Cartier), then a small Q-factorial modification or SQM of X means a birational map φ: X --> Y (where Y is another Q-factorial projective variety) which is an isomorphism in codimension 1. (Examples: flips and flops.) The question is then this:

Is there an example of a Q-factorial projective variety X and φ: X --> Y an SQM of X such that the nef cone Nef(X) is rational polyhedral but Nef(Y) is not rational polyhedral? Or (highly unlikely I think) can one prove that this situation is impossible?

And to push my luck:

Give sufficient conditions to ensure that in the above situation, Nef(X) rational polyhedral implies Nef(Y) rational polyhedral.

Really I think only the first question has a hope of being answered positively, but it would be nice to know if I was wrong.  
Background: The question was prompted by a theorem of Hu and Keel ("Mori dream spaces and GIT", Michigan Math. J., 2000) which gives a characterisation of so-called Mori dream spaces --- certain varieties which behave very well with respect to the operations of the minimal model program. In particular, if X is a Mori dream space, then Nef(X) is rational polyhedral and so too is Nef(Y) for any SQM Y of X. It then seems natural to ask for an example where the first condition here holds but the second fails.
Hu--Keel's theorem gives one answer to my second question above, since Birkar--Cascini--Hacon--McKernan proved that any Fano variety (or more generally, any variety of Fano type) is a Mori dream space. But it would be great to know of any sufficient condition that applies outside the Fano domain. 
Edit (July 23): Balazs' answer settles the first question in the affirmative: flops can change the nef cone from finite to infinite. But it seems to remain open (and interesting to me) whether there are examples $X$ of this phenomenon with Kodaira dimension $kd(X)=-\infty$.

REPLY [2 votes]: Let me start by a disclaimer: I misread the question and wrote what's below thinking that it was relevant. Then, just before posting it I looked at the question one more time and realized the mistake. 
As it happens, the statement and proof below is mathematically correct (I think), except that it does not answer the question, or one could perhaps even say that it does not have much to do with the question and probably everyone before me who have looked at this question knows what I wrote down here. 
At the same time, I would hate to just erase what I typed up and perhaps someone finds this useful, so I will just post it. If some MO users think this is inappropriate, I will be happy to erase it, just let me know. 
So, here come some mathematical ramblings loosely related to the question above.
(The point is that there does not exist a morphism, but the question was about rational maps).

There does not exist a morphism that is an SQM. This is because if $f:X\to Y$ is a small morphism, then $Y$ is not $\mathbb Q$-factorial. In fact a little more seems to be true:
Proposition. Let $f:X\to Y$ be a proper birational morphism and assume that $X$ is quasi-projective and $Y$ is $\mathbb Q$-factorial. Then the exceptional set of $f$ is of pure codimension $1$.
Proof. Let $H\subset X$ be a very ample Cartier divisor on $X$ and consider $f_*H$ the cycle theoretic image of $H$. This is a priori only a Weil divisor, but since $Y$ is $\mathbb Q$-factorial, it is $\mathbb Q$-factorial and replacing $H$ with an appropriate multiple we may assume that it is in fact Cartier.
Now consider $F:=f^*f_*H-H$ and notice that a) $F$ is effective b) $-F$ is $f$-ample. These show that $F$ must be exceptional. This already proves that $f$ cannot be small, but we can prove the stronger statement stated above.
${\rm Supp}\, F$ is obviously of pure codimension $1$, so if we prove that the exceptional set of $f$ is equal to this, then we are done. We have already seen that ${\rm Supp}\, F$ is contained in the exceptional set, so we only need to demonstrate that $f$ is an isomorphism on the complement of $F$.  
Now, observe that by the definition of $F$, we have that 
$$ 
\mathcal O_X(-F)\simeq \mathcal O_X(H)\otimes\mathcal O_X(-f^*f_*H)
$$
and
$$
\begin{multline*}
f^*f_* \mathcal O_X(-F)\simeq f^*f_*\mathcal O_X(H-f^*f_*H)\simeq \\ \simeq  f^*\left(f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f_*H)\right)\simeq f^*f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f^*f_*H) 
\end{multline*}
$$
Furthermore, since $H$ is very ample, it is generated by global sections, so 
$$
f^*f_*\mathcal O_X(H) \to \mathcal O_X(H)
$$
is surjective. Then it follows from the above displayed lines that 
$$
f^*f_*\mathcal O_X(-F) \to \mathcal O_X(-F)
$$
is also surjective. This implies that $f^{-1}(f({\rm Supp}\, F)) = {\rm Supp}\, F$. It follows then that $H$ and $f^*f_*H$ agree on $X\setminus {\rm Supp}\, F$. In particular, $f^*({\rm some divisor})$ is ample on this $X\setminus {\rm Supp}\, F$. Finally this implies that then $f$ has to be an isomorphism on $X\setminus {\rm Supp}\, F$. 
Q.E.D.<|endoftext|>
TITLE: *-homomorphisms between matrix algebras
QUESTION [8 upvotes]: Edited question:
Are there any other non-trivial *-homomorphisms between matrix algebras apart from the unitary homomorphisms? 
Original question:
Does there exist a surjective (but not bijective) *-homomorphism between matrix algebras over the complex numbers? If so, are there any nice examples?
(I had not realized the matrix algebras were simple but since they are, the answer to the original question is indeed obvious)

REPLY [2 votes]: I do understand this is an old question but, considering that:
1. you might still be interested in a simpler (and surely less elegant) proof;
2. I had the same problem, so this might be helpful for others in the future
I'll tell you how I solved this problem
Take your *-homomorphism $\lambda:M_n\to B$, where $B$ is any other C*-algebra. Set
$$F_{ij}:=\lambda(E_{ij}),\qquad\forall i,j=1,\ldots,n,$$
where the $E_{ij}$ is the canonical basis of the underlying vector space of $M_n$ (then the $F_{ij}$ satisfy the very same algebra, i.e. $F_{ij}F_{mn}=\delta_{jm}F_{in}$), and suppose that the kernel of $\lambda$ is non-trivial. Then there exists $a\in M_n\smallsetminus \{0\}$ s.t. $\lambda(a)=0$. This is a constraint between all the $F_{ij}$, namely there are coefficients $\alpha_{ij}$ such that
$$\sum \alpha_{ij}F_{ij}=0.$$
Note that if just one of the $F_{ij}$ for some $(i,j)$ is 0, then $\lambda$ is 0, because $F_{kk}$ are all Murray-von Neumann equivalent projections, and the "off-diagonal" elements $F_{ij}$, $i\neq j$ are partial isometries linking them. Therefore sandwich the above linear combination between $F_{kk}$ and $F_{mm}$ to obtain
$$0=\sum_{ij}\alpha_{ij}F_{kk}F_{ij}F_{mm} = \alpha_{km}F_{km},$$
which implies
$$\alpha_{km}=0$$
for arbitrary $(k,m)$. Hence all the $F_{ij}$ are linearly independent, meaning that, as a vector space, $B$ must have enough space to accommodate at least a copy of $M_n$. But taking into account once again that all the projections $F_{kk}$ are Murray-von Neumann equivalent, we can conclude that there must be a subalgebra $C$ of $B$ such that $M_n\otimes C\subset B$ and a projection $P\in C$ such that, up to unitary equivalence
$$\lambda(E_{ij})=E_{ij}\otimes P\in M_n\otimes C.$$
If the rank of $P$ is $k$, then $B$ should be large enough to accommodate $k$ copies of $M_n$, i.e. the underlying vector space of B must have dimension greater than $kn$ (if it is not exactly $kn$ the you'll have some 0-padding).
Hope this helped like it did for me!<|endoftext|>
TITLE: Complex subvarieties of hermitian symmetric spaces
QUESTION [5 upvotes]: Assume that M=G/K is a non-compact Hermitian symmetric space, for G the real points of  a semisimple (or even simple) algerbaic group and K a maximal compact subgroup. M admits the structure of a complex manifold. Now, if X is a complex subvariety of M then its pre-image in G is a real analytic subvariety of G. My question is: how does one recognize, among all real analytic subvarieties of G, those  which project onto complex subvarieties in M? 
More concretely, if H is a real Lie subgroup of G (even real algebraic), when is it the case that its image under the projection map (into M=G/K) gives a complex subvariety of M?

REPLY [2 votes]: I will explain how to check it in any particular case, but I am not sure whether there is a good classification available (perhaps, it is even easy and I did not think enough about it). Helgason's book and papers of Joe Wolf and Alan Huckleberry may contain some relevant information.
Since the complex structure on $M=G/K$ is homogeneous, the image of $H$ is a complex subvariety if and only if its tangent subspace at $e$ is a complex subspace of $T_e M,$ i.e. invariant under the complex structure operator $J: T_e M\to T_e M.$ Let $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ be the Cartan decomposition, then $T_e (G/K)$ may be identified with $\mathfrak{p}$  via the projection onto the second summand and $J=\text{ad}(X_0)$ for a certain element $X_0$ in the center of $\mathfrak{k}$ (if $G$ is simple then this center is one-dimensional). Thus the answer is "yes" if and only if
$$T_e(H/K\cap H)=\mathfrak{h}/\mathfrak{k}\cap\mathfrak{h} \subset 
\mathfrak{p} \text{ is } \text{ad}(X_0)\text{-invariant.}$$ 
Subalgebras $\mathfrak{h}$ with this property would need to be classified modulo conjugation by $K$.
One can proceed a bit further by complexifying $\mathfrak{g}$, so that $\mathfrak{p}_\mathbb{C}=\mathfrak{p}_{+}\oplus\mathfrak{p}_{-}$ is the eigenspace decomposition for $J_{\mathbb{C}}$ (the eigenvalues are $\pm i$ and $\mathfrak{p}_{\pm}$ are abelian subalgebras of $\mathfrak{g}_{\mathbb{C}}$), keeping in mind that a complex Lie subalgebra of $\mathfrak{g}_\mathbb{C}=\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complexification of a Lie subalgebra of $\mathfrak{g}$ if and only if it is invariant under the complex conjugation. 
If $H$ is a connected semisimple group of Hermitian type and $f:H\to G$ is an embedding, then after a suitable conjugation, the image of a maximal compact subgroup $L<H$ is contained in $K,$ $f(H)=(f(H)\cap K)(f(H)\cap P)$ and $f(H)$ mod $K$ is isomorphic to a Hermitian symmetric space $H/L,$ which is thus realized in $G/K.$ This allows one to construct many examples by starting with (1) a group $H$ as above and (2) a faithful complex representation of $H$ that preserves an indefinite unitary form, so that $G=SU(p,q)$ if the signature is $(p,q).$  
A final elementary observation is that there always exist nontrivial subgroups $H$ that do not arise from the previous construction, for example, $NA$ from  the Iwasawa decomposition $G=NAK$ projects onto $G/K.$<|endoftext|>
TITLE: What is the product in the 2-category of spans?
QUESTION [9 upvotes]: Let $\mathcal S$ be a category with finite limits.  The 2-category $\operatorname{Span}(\mathcal S)$ has the same objects as are in $\mathcal S$.  For objects $X,Y$, the hom category in $\operatorname{Span}(\mathcal S)$ between $X$ and $Y$ is the category of diagrams in $\mathcal S$ of the form $X \leftarrow \bullet \rightarrow Y$, and morphisms are natural transformations of such diagrams that restrict to $\operatorname{id}_X,\operatorname{id}_Y\,\,$ at the endpoints.  The 1-composition of 1-morphisms is given by the obvious pull-back diagrams: $$\{X\leftarrow A \rightarrow Y\} \circ \{Y\leftarrow B \rightarrow Z\} = \{X\leftarrow A\underset Y \times B \rightarrow Z\}$$
Thinking of $\mathcal S$ as a 2-category with only identity morphisms, the "spanishization" functor (does this functor have another name?) $\mathcal S \to \operatorname{Span}(S)$ is the identity on objects and takes $\{X \overset f \to Y\}$ to $\{X = X \overset f \to Y\}$.  There is also an obvious isomorphism $\mathcal S \cong \operatorname{Hom}_{\operatorname{Span}}(1,1)$, where $1 \in \mathcal S$ is the terminal object.
I believe that the correct weakened notion of "cartesian product" in a 2-category is that $X\times Y$ is determined up to equivalence (not isomorphism) as the representing object for the 2-functor $Z \mapsto \operatorname{Hom}(Z,X) \times \operatorname{Hom}(Z,Y)$, where on the right-hand side is the usual product of categories.  (Incidentally, what's a good reference for $n$-Yoneda's Lemma?)  Even if $\mathcal S$ has finite limits, or even all small limits, then I'm not sure whether $\operatorname{Span}(\mathcal S)$ has finite products.  But for good enough categories $\mathcal S$, I feel like $\operatorname{Span}(\mathcal S)$ should also be good.
However, I believe that the product in $\operatorname{Span}(\mathcal S)$ is not the product in $\mathcal S$, i.e. spanishization does not respect limits.  Provided all my beliefs are correct:

Is there an easy description of the product in $\operatorname{Span}(\mathcal S)$ in terms of $\mathcal S$?  Do any products at all exist in $\operatorname{Span}(\mathcal S)$?

REPLY [7 votes]: If $\mathcal S = Set$, it looks to me like the product in $Span(\mathcal S)$ is given by disjoint union.
Define functors $F_Z: Span(Z,X\sqcup Y) \to Span(Z,X)\times Span(Z,Y)$
$$
F_Z(Z \leftarrow A \rightarrow X \sqcup Y) = ((Z \leftarrow A \times _{X\sqcup Y}X \rightarrow X), (Z \leftarrow A \times _{X\sqcup Y}Y \rightarrow Y))
$$
and $ G_Z: Span(Z,X)\times Span(Z,Y) \to Span(Z,X\sqcup Y)$
$$
G_Z((Z \leftarrow B \rightarrow X), (Z \leftarrow C \rightarrow Y)) = (Z \leftarrow A\sqcup B \rightarrow X\sqcup Y).
$$
Basically, if you have a span between $Z$ and $X\sqcup Y$, take the preimage of each component to get a pair of spans. I think it should be clear what happens on morphisms. Now, in Set (or maybe we just require that products distribute over coproducts), I think these form an equivalence. This should also behave well under maps $Z \to Z^\prime$.
This seems pretty weird to me, and it may not be correct (my category theory is pretty rudimentary). I am interested to find out more though, as it seems related to a question I was once asked about limits and colimits in cobordism categories (thinking of cobordisms as certain cospans of manifolds).<|endoftext|>
TITLE: Flat regions on surfaces of genus greater than 1
QUESTION [5 upvotes]: Here is a polygonal disk + gluing scheme model of a surface we are attempting to construct.  We want the regions of the surface bounded by the two vertical, dotted lines $\alpha,\beta$ to have zero curvature throughout, while the rest of the surface should be "smoothed out" with some kind of metric which gives it negative curvature and thus allows the entire shape to satisfy the requirement that a two-holed torus have negative total curvature.  
Assuming that we can adjust the circled (and, separately, squared) angles so that they add up to $2\pi$ and the relevant identifications (side $a$ with side $a$, $b$ with $b$) can be made smoothly while maintaining the zero curvature of the inner region, the question is the following.  Will the smoothing out of edges $1$ through $4$ necessarily need to be continued into the region inside of which we want to have zero curvature, thus giving it non-zero curvature instead?  In other words, we have to smooth out the triangled vertices to some extent because their corresponding angles add up to more than $2\pi$.  But, thinking geometrically (we have a limited knowledge of the underlying Riemannian geometry here), it seems that we would need to smooth out neighborhoods of these vertices and of the edges $1$ through $4$.  Will we be able to stop before reaching the region inside of $\alpha,\beta$?
Thank you.  Any references that might be able to help us are welcome.  We're pretty new to this but if the answer to these questions is that we can maintain zero curvature, we've made progress on our problem!

REPLY [4 votes]: Let $S$ be the underlying surface and call the image of the dotted lines $\gamma$.  If we cut $S$ along $\gamma$ then $S$ falls apart into two components $X$ and $Y$ (both tori with a single boundary component).  Let $X$ be the component that contains the edges $a$ and $b$.  If I read your question correctly, you want the metric inside of $X$ to be flat and are willing to accept any smooth, negatively curved metric on $Y$.  
I'll assume that you also want the surface $S$ to be a Riemannian manifold at the end of the day.  I'll also assume that you are insisting that the two edges of $\gamma$ (on its $X$ side) be straight (otherwise you can proceed as Daniel Mehkeri suggests -ie in this situation you allow $\gamma$ to "bow-in" to the $X$ side.  Eg, you could take any flat torus and cut out a round disk to get $X$.)  
If my assumptions are correct then the answer to your question is negative.  This is because in a Riemannian manifold the exponential map is well defined -- however in your surface, the dotted edges are limits of geodesic segments and hence geodesic.  Thus at the square vertex the upward pointing tangent vector has two geodesic continuations, a contradiction.<|endoftext|>
TITLE: Finite axiom of choice: how do you prove it from just ZF?
QUESTION [24 upvotes]: The axiom of choice asserts the existence of a choice function for any family of sets F. Suppose, however, that F is finite, or even that F just has one set. Then how do we prove the existence of a choice function? 
The usual answer is that we just go from set to set, picking an element from each set. Since F is finite, this process will terminate. What I'm really wondering is how we can always choose from a single set. The informal answer seems to be just that it's possible... but this isn't an axiom, so it must be justified some other way. 
So: how do you prove from the axioms of just ZF without choice, that for any nonempty x there exists a function f:{x}->x?

REPLY [8 votes]: The Axiom of Choice does not allow to "choose" a choice function, it only says that a choice function exists. To show that a choice function for a single nonempty set exists, you do not need to "choose" an element in the set, it is enough to show that at least one element exists (i.e. the set is nonempty). Every element in the set will give a different choice function.
What did you mean by "choosing" an element from a set? ;)<|endoftext|>
TITLE: Are there more Nullstellensätze?
QUESTION [36 upvotes]: Over which fields $k$ is there a reasonable analogue of Hilbert's Nullstellensatz?
Here is a more precise formulation: let $k$ be an arbitrary field, $n$ a positive integer, and $R = k[t_1,..,t_n]$.  There is a natural relation between $k^n$ and $R$: for $x \in k^n$ and $f \in R$, $(x,f)$ lies in the relation if $f(x) = 0$.
This relation induces a Galois connection between the power set of $k^n$ and the set of all ideals of $R$ (both partially ordered by inclusion).  In more standard algebraic-geometric language, if $S$ is a subset of $k^n$ and $J$ is an ideal of $R$, put 
$I(S) = \{f \in R \ | \ \forall x \in S, \ f(x) = 0\}$
and
$V(J) = \{x \in k^n \ | \ \forall f \in J, \ f(x) = 0\}$.
There are induced closure operators: for a subset $S$, $\overline{S} := V(I(S))$ and for an ideal $J$, $\overline{J} := I(V(S))$.  
The closure operator on subsets is compatible with finite unions so is the closure operator for a topology on $k^n$, the Zariski topology.
The question is: what is the closure operator $I \mapsto \overline{I}$ on ideals of $R$?  By a Nullstellensatz, I mean a nice description of this closure operator.
Some remarks and examples:
1) Over any field $k$, one sees that $\overline{I}$ is a radical ideal hence contains $\operatorname{rad}(I) = \{x \in R \ | \ \exists n \in \mathbb{Z}^+ \ | \ x^n \in I\}$.
If $k$ is algebraically closed, then Hilbert's Nullstellensatz says that $\overline{I} = \operatorname{rad}(I)$.
It is easy to see that if $\overline{I} = \operatorname{rad}(I)$ for all maximal ideals of $k[t]$, then $k$ is algebraically closed.
2) If $k$ is formally real, then for any ideal $I$ of $R$, $\overline{I}$ is a real ideal, i.e., $x_1,\ldots,x_n \in R, \ x_1^2 + \ldots + x_n^2 \in I \implies x_1,\ldots,x_n \in I$.  Moreover, for any ideal $I$ in a commutative ring, there is a unique minimal real ideal containing $I$, its real radical $\mathbb{R}ad(I)$, which is the intersection of all real prime ideals $\mathfrak{p}$ containing $I$.
If $k$ is real-closed, then for any ideal $I$ in $k[t_1,\ldots,t_n]$, $\overline{I} = \mathbb{R}ad(I)$: this is Risler's Nullstellensatz.
3) There is also a Nullstellensatz for p-adically closed fields (in particular, for $p$-adic fields) due to Jarden and Roquette: see 
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jmsj/1240234705

Are there further Nullstellensätze (say, for non-Henselian fields to rule out variations on 3)?  

Although I haven't been precise on what a description of $\overline{I}$ means (I don't know how), it seems reasonable to guess that there is no good Nullstellensatz over a field like $\mathbb{Q}$ for which it is believed that Hilbert's 10th problem has a negative answer.  Briefly: if you had a system of polynomial equations $P_1,\ldots,P_m$ with $\mathbb{Q}$-coefficients, then they have a simultaneous solution over $\mathbb{Q}$ iff 
the closure of $\langle P_1,\ldots,P_m \rangle$ is a proper ideal, so if you had a sufficiently nice description of the closure operation, you could use it to answer H10 over $\mathbb{Q}$ affirmatively.
A case of persistent interest to me over recent years is that of a finite field.  In some sense this is the worst case, since it is not hard to show that the zero ideal in $k[t_1,\ldots,t_n]$ is closed iff $k$ is infinite.  Nevertheless, I vaguely feel like there should be something to say here, possibly something having to do with reduced polynomials -- i.e., for which each exponent of each variable is at most $\# k - 1$ -- as in one of the proofs of the Chevalley-Warning theorem.  
P.S.: I am aware of other algebraic results about $k[x_1,\ldots,x_n]$ over a general field $k$ which, when $k$ is algebraically closed, imply Hilbert's Nullstellensatz, e.g. that a finitely generated $k$-algebra which is a field is finite-dimensional over $k$, or that every prime ideal in $k[t_1,\ldots,t_n]$ is an intersection of maximal ideals.  These are interesting and useful, but here I am really interested in $I \mapsto \overline{I}$.

REPLY [11 votes]: If $k$ is a finite field with $q$ elements and $I$ an ideal of $k[x_1,\dots,x_n]$, then $\overline I=I+I_0$, where $I_0=(x_1^q-x_1,\dots,x_n^q-x_n)$. This follows immediately from Hilbert’s Nullstellensatz applied to the algebraic closure of $k$, and the observation that any ideal extending $I_0$ is a radical ideal (as it contains all polynomials of the form $f^q-f$).
On an unrelated note, a more explicit description for the case of $k$ real-closed follows from Stengle’s (Positiv- and) Nullstellensatz: $f\in\overline I$ iff $-f^{2n}\in I+\Sigma$ for some $n\in\mathbb N$, where $\Sigma$ is the set of all sums of squares of polynomials.<|endoftext|>
TITLE: Variety defined by a recursively enumerable set of polynomials
QUESTION [9 upvotes]: Suppose we have a recursively enumerable set of polynomials $\mathcal{P}=\{ p_1({\bf x}), p_2({\bf x}), \ldots\}, p_i \in \mathbb{Z}[{\bf x}], {\bf x} = (x_1, \ldots, x_n)$. Let $V(\mathcal{P})$ denote the affine variety in $\mathbb{C}^n$ defined by $\mathcal{P}$. Is there an algorithm to compute $V(\mathcal{P})$? By the Nullstellensatz, we know that we need only use finitely many of the polynomials $p_i$ to cut out $V(\mathcal{P})$. We can recursively compute varieties cut out by $\{p_1, \ldots, p_k\}$, for example by computing a Grobner basis for the radical ideal of $(p_1,\ldots,p_k)$. But is there a way to compute $k$ such that $V(\mathcal{P})=V(p_1,\ldots,p_k)$? 
Please let me know if this question needs clarification or if I'm not using the correct notation. 
Addendum: This problem was motivated by this MO question. It would follow from:
If one has a finitely generated group $G$ with solvable word problem, for any $n$ can one compute the representation variety $G\to SL_n(\mathbb{C})$? 
I view $G$ as being given as the homomorphic image of a free group $\\langle g_1,\ldots,g_k\\rangle$. Moreover, there is a Turing machine which takes as input any element $h\in \\langle g_1,\ldots,g_k\\rangle$ and tells if $h$ is trivial in $G$. The space of representations  $\rho:G\to SL_n(\mathbb{C})$ is an affine variety, with $kn^2$ variables given by the entries of the matrices of $\rho(g_i)$. One can recursively generate polynomials 
which are the entries of the matrices $\rho(h)-I$ which cut out the representation variety (together with $det(\rho(g_i))-1$). So the algorithm should depend on how these polynomials are generated, if one wants to be able to compute the representation variety for each $n$. 
I suspect that the answer is no, although I'm not sure how to generalize Borcherds or Groves' answers to this context.  
If one could compute the representation variety, then one could determine if $G$ has a homomorphism to a finite group.

REPLY [2 votes]: Agol's other question (If one has a finitely generated group G  with solvable word problem, for any n  can one compute the representation variety G -> SLn(C)) also has a negative answer. In fact one cannot even compute the number of homomorphisms to the group of order 2 when G is generated by one element g. 
Take the relations to be g^(2k+1)=1 if  some given Turing machine halts for the first time after k steps. A little thought shows that this group has an effectively solvable word problem and an effectively recursive set of relations, but one cannot decide if it has a homomorphism to the group of order 2 without solving the halting problem for the given Turing machine.<|endoftext|>
TITLE: Compact surfaces of negative curvature
QUESTION [29 upvotes]: John Hubbard recently told me that he has been asking people if there are compact surfaces of negative curvature in $\mathbb{R}^4$ without getting any definite answers.  I had assumed it was possible, but couldn't come up with an easy example off the top of my head.
In $\mathbb{R}^3$ it is easy to show that surfaces of negative curvature can't be compact:  throw planes at your surface from very far away.  At the point of first contact, your plane and the surface are tangent.  But the surface is everywhere saddle-shaped, so it cannot be tangent to your plane without actually piercing it, contradicting first contact.  
This easy argument fails in $\mathbb{R}^4$.  Can the failure of the easy argument be used to construct an example?  Is there a simple source of compact negative curvature surfaces in $\mathbb{R}^4$?

REPLY [3 votes]: In "Y. Martinez-Maure, A counter-example to a conjectured characterization of the sphere. (Contre-exemple à une caractérisation conjecturée de la sphère.) (French), C. R. Acad. Sci., Paris, Sér. I, Math. 332, 41-44 (2001), the author disproves an old characterization of the 2-sphere by giving an exemple of a "hyperbolic hedgehog" of R^3 (a sphere-homeomorphic envelope parametrized by its Gauss map whose Gaussian curvature K is everywhere negative excepted at four singular points where K is infinite). 
By projective duality, this implies the existence of a 2-sphere C^2 embedded in the 3-sphere with a nonpositive extrinsic curvature but not totally geodesic.<|endoftext|>
TITLE: Reciprocity law for number fields defined by torsion points of modular elliptic curves
QUESTION [6 upvotes]: Shimura (Crelle 221, 1966) considers the elliptic curve $E:y^2+y=x^3-x^2$ (although he doesn't use this equation) of conductor $11$ whose associated modular form is
$$
q\prod_{k=1}^{+\infty}(1-q^k)^2(1-q^{11k})^2=\sum_{n=1}^{+\infty}c_nq^n
$$
where $q=e^{2i\pi\tau}$ and $\tau$ is in the upper half of $\bf C$.  For a prime $l$, he denotes by $K_l$ the extension of $\bf Q$ obtained by adjoining the $l$-torsion points of $E$ and shows that if $l\in[7,97]$, then ${\rm Gal}(K_l|{\bf Q})$ is isomorphic to ${\rm GL}_2({\bf F}_l)$. 
Question. Is ${\rm Gal}(K_l|{\bf Q})$ now known to be isomorphic to ${\rm GL}_2({\bf F}_l)$ even for $l>97$ ?
Even if the faithful representation ${\rm Gal}(K_l|{\bf Q})\rightarrow{\rm GL}_2({\bf F}_l)$ fails to be surjective for a few $l>97$, does the recent proof of Serre's modularity conjecture not imply the
Statement.  For every prime $l>5$ and every prime $p\neq11,l$, the characteristic polynomial of ${\rm Frob}_p$ (thought of as an element of ${\rm GL}_2({\bf F}_l)$) is
$\equiv X^2-c_pX+p \pmod l$ ?
Shimura shows this only for $l\in[7,97]$. 
Addendum.  (2010/07/24)  Looking at Shimura's paper beyond the first page shows that  he actually proves (Section 3) that the characteristic polynomial for the action of ${\rm Frob}_p$ on the $l$-adic Tate module $T_l(E)$ is 
$X^2-a_pX+p\in{\bf  Z}_l[X]$ 
for all primes $l$ and $p\neq11, l$ and (Section 6) that $a_p=c_p$ for all $p\neq11$.  And yes, he does use the Eichler-Shimura relation.

REPLY [5 votes]: The first question is answered in Serre's 
[Propriétés galoisiennes des points d'ordre fini des courbes elliptiques,
Invent. Math. 15:4 (1972) 259--331], on page 304, section 5.2, exactly for this curve. In general this paper give a good way to determine for which $\ell$ the mod-$\ell$ representation is not surjective. Sage can do that efficiently for a given curve.
For every prime $p$ different from $\ell$ and $11$, the characteristic polynomial of $\rm{Frob}_p$ is indeed $T^2 - c_p T +p$ in   $ \mathbb F_{\ell}[T]$. The isomorphism
$\rm{Gal}(K _{\ell}/\mathbb Q) \to \rm{Aut}(E[\ell])=\rm{GL} _2(\mathbb F _{\ell})$ sends $\rm{Frob}_p$ to the Frobenius endomorphism $\phi:E[\ell] \to E[\ell]$ on $E/\mathbb{F}_p$.
Your $c_p$ is the trace of $ \phi$ and $p$ is the determinant of it since the Eichler--Shimura relation shows that $c_p$ is the Fourier coefficent of the associated modular form. See this answer for why it is so.<|endoftext|>
TITLE: Applications of classifying thick subcategories
QUESTION [10 upvotes]: So, relatively recently, Balmer introduced this notion of a spectrum for a tensor triangulated category and used it to prove a generalization of a classification theorem done in several areas of mathematics. Of course, the precursor to this was the work done by Devinatz, Hopkins, and Smith in classifying the thick subcategories of the stable hootopy category of (finite) spectra. They famously used this classification to prove the periodicity theorem, and I can see why it is helpful: the classification theorem reduces the periodicity theorem down to the (still nontrivial!) task of finding a single type $n$ complex with a periodic self-map. I'm sure similar uses have been found for the other classification theorems, but I am left to wonder, more generally:
What kinds of problems are made simpler with a classification theorem? What questions does it answer? 
I'm looking for some general heuristics here. They should satisfy the following conditions:

They should apply to theorems already proven by classification theorems; examples and references would be lovely here!
These heuristics should come with some sense of why one would think to use the classification theorem in this way. For example, I can see how the classification theorem makes the periodicity theorem manageable to prove, but why would one think to use it in the first place?
This last one is more of a throw away or a bonus, but it's worth a shot: If there are any areas of mathematics, or open problems that you think are begging for a classification theorem type application then please share! It would be a useful test of the proposed heuristics if they are able to predict the solution of a problem that has not been solved... 

Basically I'm looking for some intuition here. My logic being: if we know more about what kinds of questions a classification can answer then we will know more about the information contained in a classification. This, in turn, may provide clues for how to compute or construct such a classification (which is, of course, the next step in Balmer's program).
(P.S. I've tagged the areas that I know of with classification theorems. If I'm forgetting some, do remind me in the comments Looks like there's a limit on tags :).)

REPLY [4 votes]: Let me go through a possible answer to my own question- just do some thinking out loud here. It seems to me that a classification of thick subcategories of a (tensor-) triangulated category is a good thing to have when you have some, relatively generic, thick subcategory, $S$, at hand and you want to prove something about it. Then you can say, "Well, $S$ can only be one of a couple subcategories and I have them all listed here. It can't be any of these because __, and if it's any of these other ones, then I can easily prove my claim.
Here's an example that I thought of, but I'm not sure if this has been done. Suppose you have some functor $F: \mathcal{T} \rightarrow \mathcal{A}$ from a triangulated category to an abelian category. If, for some reason, it turned out that the kernel of this functor was a thick subcategory then we could apply the thick subcategory theorem to ask the question "When is $Fa = 0$?" In the case that the functor at hand satisfies $Fa = 0$ only when $a = 0$, then perhaps you could run through your classification and find a single object in each nonzero thick subcategory that maps to something nontrivial in $\mathcal{A}$. This would then prove that the kernel has to be whatever thing you had left.
Actually, what's good about this use of the classification theorem is that it can be used very easily to get partial results. Maybe the thick subcategories of your triangulated category come in multiple flavours and it is easy to show that one flavour is not the kernel of the given functor, but you don't know for the other flavours. Maybe that's all you need to prove something useful regarding the functor at hand.
I'm not sure how many functors from triangulated categories to abelian categories happen to satisfy the property that their kernel is a thick subcategory... On the other hand, any (triangulated) functor between triangulated categories does satisfy this property. So maybe you could apply this technique in the following way: You have a map between two objects $A \rightarrow B$ in some category (like noetherian schemes, or finite group schemes, or topological spectra). This should give rise to (hopefully triangulated) functors betweensome triangulated categories related to $A$ and $B$ (like derived category of perfect complexes on a scheme, or stable module category, etc.) I'm pretty sure that, at least in the usual cases, the original map will be trivial if and only if this induced map is. To find out if the induced map is trivial, we may pull the same trick as before and check to see if the thick subcategory corresponding to the kernel must be zero.
That seems like a high-powered way to check if a map between to objects is trivial! But maybe it's useful? If anyone has seen anything like what I've described then please post a reference! I would love to know if this application has been done before (I'm sure it has, it was the first thing that came to mind...)<|endoftext|>
TITLE: Lehmer's conjecture for Ramanujan's tau function
QUESTION [12 upvotes]: Lehmer's conjecture for Ramanujan's tau function,
$$
\Delta(q)=q\prod_{n=1}^\infty(1-q^n)^{24}=\sum_{m=1}^\infty\tau(m)q^m,
$$
asserts that $\tau(m)$ never vanishes for $m=1,2,\dots$.
In the recent question
it was asked why it is important to have the nonvanishing.
I am wondering whether there are upper bounds,
unconditional or conditional (modulo some other
known conjectures), in terms of $x\in\mathbb R_+$ for the number
of integers $m\le x$ satisfying $\tau(m)=0$ (maybe better,
for the number of primes $p\le x$ satisfying $\tau(p)=0$)?
It looks like the series $\Delta(q)$ is very far from being "lacunary".
But besides Deligne's upper bound $|\tau(m)|\le d(m)m^{11/2}$
(where $d(\ )$ counts the number of divisors) and the lower bound
$$
\operatorname{card}\lbrace\tau(n):n\le x\rbrace\ge \operatorname{const}\cdot x^{1/2}e^{-4\log x/\log\log x}
$$
from
[M.Z. Garaev, V.C. Garcia, and S.V. Konyagin,
A note on the Ramanujan $\tau$-function, Arch. Math. (Basel) 89:5 (2007) 411--418]
for the distribution of tau values, I cannot find any quantitative progress
towards Lehmer's original question.

REPLY [3 votes]: Lehmer's conjecture has an equivalent result in the theory of Harmonic Maass forms.  The non-vanishing of the tau function is equivalent to the irrationality of the coefficients of Harmonic Maass forms.  
Specifically there is a correspondence between the spaces $ \zeta_{2-k} : H_{2-k}(N, \chi) \rightarrow S_k(N, \chi) $. where 

$ \zeta_{2-k}$ is a differential operator
H = Harmonic Maass forms
S = cusp forms (referred to as the shadow of the Maass form)

The discriminant function $\Delta(z)$ is the shadow of the Harmonic Maass form $\frac{1}{11!} Q^+(-1, 12, 1; z) $  
See Theorem 12.5 in the paper Unearthing the visions of a master: harmonic Maass forms and number theory by Ken Ono.  Also see Algebraicity of Harmonic Maass forms<|endoftext|>
TITLE: Special cases of Dirichlet's theorem
QUESTION [19 upvotes]: Dirichlet's theorem states that for any coprime $k$ and $m$ there exists infinitely many primes $p$ such that $p \equiv k \pmod m$.
Some special cases of this theorem are easy to prove without any analytic methods. Those cases include, for example, $m=4, k=1$ and $m=4, k=3$.
Both cases could be proved by considering first $t$ prime numbers $p_i \equiv k \pmod m$ and constructing a new number which is proved to have prime divisor $p \equiv k \pmod m$ that is not equal to any $p_i$.
For case $m=4, k=1$ we can consider number $(p_1 p_2 \cdots p_t)^2 + 1$. And for case $m=4, k=3$ number $4p_1 p_2 \cdots p_t + 3$.
Those constructions could also be applied to some other special cases as well.
Are there any other special cases for which there exists a simple non-analytic proof which don't use any of those two constructions?

REPLY [3 votes]: I found this generalization of the "$3 \pmod{4}$" version while teaching number theory a few years ago.
Let $G$ be a proper subgroup of $(\mathbb{Z}/n)^\times$.  Then there are infinitely many primes $p$ such that $[p]\in (\mathbb{Z}/n)^\times$ and $[p]\not\in G$.
Proof:  Suppose as usual that there are finitely many, $p_1, p_2, \ldots, p_r$, and find a number $g$ such that $(p_i,g) = 1$ for all $i$ and $[g]\not\in G$.  Then the number $N = np_1 p_2 \cdots p_r + g$ has a prime factorization $N = q_1q_2 \cdots q_s$ satisfying

$q_i \neq p_j$ for all $i$ and $j$ and
since $[N]=[g]\not\in G$, $[q_i]\not\in G$ for at least on $i$.

EDIT 8-29-20
Here is a detailed proof.  There's a bit of fun messing around to find the right number $N$.
Theorem. Let $m\in \mathbb{N}$ and let  $G\subseteq (\mathbb{Z}/m)^\times$  be a proper subgroup.  Then  %for each  %$\alpha\in (\mathbb{Z}/m)^\times - G$,  there are infinitely many primes $p$ such that  $[p] \in (\mathbb{Z}/m)^\times - G$.
Proof.
Assume to the contrary that there are only finitely
such primes,
$$
\mathcal{P}
=
\{
\mbox{all primes $p$ such that 
$[p] \in (\mathbb{Z}/m)^\times -G$}
\}
=
\{ p_1, p_2, \ldots, p_r\}.  
$$
Since each
$[p_i]\in (\mathbb{Z}/m)^\times$
we have $(p_i,m) = 1$ for $i= 1, 2, \ldots, r$.
Since $G$ is a proper subgroup of $(\mathbb{Z}/m)^\times$,
we can find an integer $a$ such that
$[a]\in (\mathbb{Z}/m)^\times - G$;
again $(a, m) = 1$.
Now we inductively define a sequence of integers
$N_k$ for $k = 0, 1, 2, \ldots, r$ with the properties

$N_k \equiv a$ mod $m$
$(p_i, N_k) = 1$ for $i=1, 2, \ldots, k$.

The construction begins with
$
N_0 = mp_1p_2 \cdots p_r + a  .
$
Once we have $N_k$, we define
$$
N_{k+1} = 
\left\{ 
\begin{array}{ll}
N_k  & \mbox{if $(p_{k+1}, N_k) = 1$}
\\
\\
N_k + m p_1p_2\cdots p_k & \mbox{if $p_{k+1} | N_k$.}
\end{array}
\right.
$$
Obviously
$N_{k+1} \equiv N_k\equiv a$ mod $m$,
and $N_{k+1} \equiv N_k$ mod $p_i$ for each
$i = 1, 2, \ldots, k$, so that
$$
(N_{k+1}, p_i) = (N_k , p_i) = 1
$$
for $i = 1, 2, \ldots, k$.  Furthermore,
if $p_{k+1}| N_k$, then $p_{k+1}$ cannot
divide $N_{k+1}$, lest $p_{k+1}$ divide
$m p_1p_2\cdots p_k$, which is impossible.  Therefore
the construction continues, and ultimately
obtain the integer $N_r$.
Now consider  its prime factorization
$
 N_r = q_1q_2 \cdots q_s
 $.
We can say from what we have done that

$q_j \neq p_i$ for any $i$ and $j$, so
$[q_j]  \in G$ for all $j =1, 2, \ldots, s$, and so
$[N_r] = [q_1]\cdot [q_2] \cdots [q_s] \in G$,  but
$[N_r] = [ a] \not\in G$.

This contradiction of the last two lines
shows that our assumption that
there are only finitely many such primes $p$
such that $[p]\in (\mathbb{Z}/m)^\times - G$
must be wrong,
and this completes the proof.<|endoftext|>
TITLE: How to shuffle a deck by parts?
QUESTION [15 upvotes]: This question is mainly a curiosity, but comes from a practical experience (all players of Race for the galaxy, for example, must have ask themselves the question).
Assume I have a deck of cards that I would like to shuffle. Unfortunately, the deck is so big that I cannot hold it entirely in my hands. Let's say that the deck contains $kn$ cards, and that the operation I can perform are: 1. cut a deck into any number of sub-decks, without looking at the cards but remembering for all $i$ where the $i$-th card from top of the original deck has been put; 2. gather several decks into one deck in any order (but assume that we do not intertwin the various decks, nor change the order inside any of them); 3. shuffle any deck of at most $n$ cards. Assume moreover that such a shuffle consist in applying an unknown random permutation drawn uniformly.
Here is the question: is it possible to design a finite number of such operations so that the resulting deck has uniform law among all possible permutations of the original deck? If yes, how many shuffles are necessary, or sufficient, to achieve that ?
The case $k=2$ seems already interesting.

REPLY [2 votes]: Assuming you want a practical answer to "I have too many cards to hold in my hands at once; how do I shuffle them reasonably well in a relatively short amount of time?",
you might want to consider a "parallel shuffle", distributing the work over several players in hopes that we can get an adequately shuffled deck in less wall-clock minutes than a single-person shuffle, even if it requires more total operations and player-minutes than a single-person shuffle.
I am reminded of the "FFT butterfly diagram" used in digital signal processing and the "Omega Network" used in some computer clusters, based on the "perfect shuffle interconnection".
http://www.ece.ucsb.edu/~kastner/ece15b/project1/fft_description_files/image032.jpg
http://github.com/vijendra/Omega-network/raw/master/16X16.png
Parallel shuffle-deal-shuffle algorithm: (for $k \le n$)

somehow give k players n cards each (either grab a block of n cards off the top for each player, or evenly deal the cards to the k players)
shuffle: each of the k players uniformly shuffles their sub-deck of n cards
deal: each of the k players evenly deals -- face down -- her sub-deck to the k other players (including herself). Equivalently, each player breaks her sub-deck into k equal sub-sub-decks, and distributes one sub-sub-deck to each player (including herself). After all the players have dealt, each player gathers her cards (a few from each player, including herself) into one sub-deck of n cards.
shuffle: (as above)

By this stage (1 round), we have done the equivalent to randomizing each row of a matrix, then each column.
Any particular single card could be anywhere after one round of shuffle-deal-shuffle, with equal probability.
Alas, at this stage, there are still a few permutations that have probability zero.
For example, the possible permutations equivalent to a rotation by shear (RBS) ("how do I rotate a bitmap?") require 3 shears. The closest that a single round of shuffle-deal-shuffle can produce is 2 shears, which is not enough to produce those permutations.
So we continue with the second round:

deal: (as above)
shuffle: (as above)
gather all the sub-decks into one large full deck

The full 2-round shuffle-deal-shuffle-deal-shuffle algorithm can produce any possible permutation, but each permutation does not have exactly the same probability.
Each of the two "deal" steps mixes at least as well as a single riffle shuffle of the entire kn cards. The paper -- by Dave Bayer and Persi Diaconis -- that David Speyer mentioned proves that $m = \frac{3}{2} \log_2 (kn) + \theta$ riffle shuffles are sufficient.<|endoftext|>
TITLE: When are modules and representations not the same thing?
QUESTION [27 upvotes]: I've been trying for a while to get a real concrete handle on the relationship between representations and modules. To frame the question, I'll put here the standard situation I have in mind:
A ring $R$ lives in the category Ab of Abelian groups as an internal monoid $(\mu_R, \eta_R)$. A module is then just an Abelian group $A$ and a map $m : R \otimes A \rightarrow A$ that commutes with the monoid structure in the way you'd expect.
Alternatively, take an Abelian group $A$ and look at its group of endomorphisms $[A,A]$. This has an internal monoid $(\mu_A, \eta_A)$ just taking composition and identity. Then a representation is just a monoid homomorphism $(R, \mu_R, \eta_R) \rightarrow (A, \mu_A, \eta_A)$ in Ab. I.e. a ring homomorphism.
But then, Ab is monoidal closed, so these are the same concept under the iso
$$\hom(R\otimes A, A) \cong \hom(R, [A,A])$$
This idea seems to work for any closed category where one wants to relate a multiplication to composition. So, my question is, since these things are isomorphic in such a general context, why are they taught as two separate concepts? Is it merely pedagogical, or are there useful examples where modules and representations are distinct?

REPLY [14 votes]: Here is my representation theorist's perspective: the key difference between representations and modules is that representations are "non-linear", whereas modules are "linear". I'll concentrate on the case of groups as the most familiar, but this applies more generally. 
As Greg has already mentioned, in the most general sense, a representation is a homomorphism $f:G\to H,$ and usually there is no linear (or additive) structure on $H$, i.e. the set $f(g)$ need not be closed under sums; in fact, if $H$ is a non-abelian group, e.g. the symmetric group, the notion of sum doesn't even make sense (if $H=GL(V)$ then we may view its elements as endomorphisms of $V$ and add them, but this is unnatural since, by definition, $f$ is compatible with multiplicative structure). By contrast, a module involves a linear action $G\times V\to V,$ which is then "completed" by allowing arbitrary linear combinations, leading to certain technical advantages.
Here is an example of a construction that is very useful and makes perfect sense module-theoretically, but not representation-theoretically: change of scalars. Given a module $M$ over a group ring $R[G]$ and a commutative ring homomorphism $R\to S,$ one gets a module $S\otimes_R M$ over the group ring $S[G]$. Common examples involve extensions of scalars (e.g. from $\mathbb{R}$ to $\mathbb{C}$, from a field $K$ of definition to the splitting field, from $\mathbb{Z}$ to $\mathbb{Z}_p$) and, more to the point, reductions (e.g. from $\mathbb{Z}$ or $\mathbb{Z}_p$ to $\mathbb{Z}/p\mathbb{Z}$). The module language is, predictably, also very useful in providing categorical descriptions of various operations on representations, such as functors of induction and restriction, 
$$Ind_H^G: H\text{-mod}\to G\text{-mod}\ \text{ and }\ Res_H^G: G\text{-mod}\to H\text{-mod},$$ 
where $H$ is a subgroup of $G,$ or the monoidal structure on $G$-mod.
Finally, here are two illustrations of the complementary nature of the two approaches besides the group case, in linear algebra. A single linear transformation $T:V\to V$ on a finite-dimensional vector space $V$ over $K$ is most naturally viewed as a representation (no additive structure); in this case, it's a representation of the quiver with a single vertex and a single loop. From this point of view, classification up to isomorphism is a problem about conjugacy classes of linear transformations, 
$$T\to gTg^{-1},\ g\in GL(V).$$ 
By contrast, in the module style description we associate with $T$ a module over the ring $K[x]$ of polynomials in one variable over $K$ and classification problem reduces to the structure of modules over $K[x]$, which is a PID, with all the usual consequences. (Here the module picture is more illuminating.) If we consider a linear operator $S:V\to W$ between two different vector spaces, 
$$S\to hSg^{-1},\ g\in GL(V),\ h\in GL(W),$$
and a classification up to isomorphism is accomplished by row and column reduction. The corresponding quiver $\circ\to\circ$ is a single arrow connecting two distinct vertices, but its path algebra is less familiar. (Here the representation theory picture is more illuminating.)<|endoftext|>
TITLE: Status of Hilbert-Smith conjecture and H-S conjecture for Hölder actions
QUESTION [8 upvotes]: The Hilbert-Smith conjecture states that 

If $G$ is a locally compact group which acts effectively on a connected manifold as a
  topological transformation group then is $G$ a Lie group.

It was established for actions by diffeomorphisms by Bochner and Montgomery. Later on it was also established for (compact?) actions by Lipschitz homeomorphisms (Repovs and  Shchepin) and Hölder actions with very large exponent (>dim M/ dim M+2). 
I am interested if the conjecture holds for Hölder actions (with small exponents). Is it plausible these arguments can be pushed to get the conjecture for Hölder actions? Or there is a fundamental obstruction?
Also, there is a 2001 preprint "A Proof of the Hilbert-Smith Conjecture" on arxiv that claims the full conjecture. I assume it's wrong as it wasn't published, but a comment from an expert would be highly appreciated.

REPLY [10 votes]: If you search mathscinet for Hilbert-Smith conjecture you shall find some results including something in the Hölder case [Maleshich, The Hilbert-Smith conjecture for Hölder actions.  Uspekhi Mat. Nauk  52  (1997),  no. 2(314), 173--174;  translation in Russian Math. Surveys 52 (1997), no. 2, 407--408]. 
I am no expert in the subject, but I have heard that McAuley's preprint "A Proof of the Hilbert-Smith Conjecture" is incorrect. A while ago there was a conference in Istanbul, where McAuley worked at the time. The conference was devoted to studying this preprint, and in particular, Shepin, who earlier proved with Repov the HS-conjecture for Lipschitz actions, was there. If memory serves me, Shepin found that McAuley's argument actually proves something stronger to which there was a counterexample known.<|endoftext|>
TITLE: How should a homotopy theorist think about sheaf cohomology?
QUESTION [32 upvotes]: As a student of homotopy theory or algebraic topology, I have a certain outlook as to how one ought to think of a cohomology theory. There are axioms that help us with rudimentary computations, there are some spectral sequences and then there is Brown representability. This is far away from the starting point of looking at the singular (co)chain complex or a simplicial complex and trying to compute its homology, it seems a bit more refined. Even the ring structure is a bit clearer, it comes from the fact that we are mapping into a ring object.
There are more and more instances where i feel like i would benefit from understanding a bit more of sheaf cohomology than just "it's the derived functor of the global sections functor of a sheaf." This is a tidge helpful, but it does not really help too much with computations from my point of view. It feels like resolutions of sheaves are large hard objects mostly because sheaves contain so much data.
My question is essentially the following:

Are there homotopy theorists out there who have over come these feelings? what advice do you have? In fact any advice that someone might have that understands the uses of sheaf theory in homotopy theory would be helpful.
Are there things resembling the Eilenberg-Steenrod axioms for sheaf cohomology? not directly due to their classification theorem, but things that help you to compute the Sheaf cohomology like a MVS sequence or what have you. I mostly would like help in doing computations in the way that the E-S axioms do? so things like the Grothendieck-Riemann-Roch Theorem, which i am told can be used in such a way.
Is there a book that goes through explicit toy computations of sheaf cohomology? Are there toy examples you would suggest for getting to be more comfortable with these things? Are there examples that live on simpler spaces than schemes? these might help a bit more than others

Some addendums : I think i need to make a few comments. I am not well versed in algebraic geometry. I find this to be a fault of mine and this is an attempt to help bridge the gap. Suggestions for references are appreciated. I really appreciate all of the excellent answers so far!
thanks for your time
EDIT MAIN QUESTION: I think what i am really asking about is the six functor formalism, but i don't really know since i don't know what that is. A friend started explaining it to me and it seemed like what i am looking for, but he said he did not feel confident in writing an answer explaining them. Hopefully someone will see this edit and give a fun answer.

REPLY [5 votes]: This is mostly an answer to 3: 
While I mostly use sheaves arising in topology (and since your question explicitly involves homotopy theory, that might also be your point of view), I think it's a useful learning tool to do some reading about how sheaves come up and are used in complex analysis and Hodge theory. The big benefit there is that most of the sheaves there are already nice (fine, soft,...), so one can do a lot without having to take resolutions by sheaves one has less control over. Wells's book Differential Analysis on Complex Manifolds might be a good starting point, or even Voisin's book on Hodge Theory (which is more advanced). 
In the topological world, I second Bredon and Gelfand/Manin (the second edition of Methods of Homological Algebra is in better shape than the first regarding typos). Also Dimca has a book Sheaves in Topology, which is a nice introduction to the derived category point of view that includes some useful examples and exercises. 
One last note: in doing actual computations, the exact sequences coming from adjunction triangles (see e.g. Section 2.4 of Dimca) are often your best friends :-)<|endoftext|>
TITLE: Planar sets where any line through the center of mass divides the set into two regions of equal area.
QUESTION [17 upvotes]: This question is influenced by the following riddle:

You are given a rectangular set in the plane with a rectangular hole cut out (in any orientation). How do you cut the region into two sets of equal area?

SPOILER ALERT!! - The answer is that you can cut through the center of both rectangles, and because any line through the center of a rectangle divides it into two pieces of equal area, this cut works. 

I have been wondering about the following question - What sort of conditions on a set guarantee that it has this property, that any line through the center of mass divides it into two regions of equal area?
The only thing that I have been able to think of is $\pi$ rotational symmetry around the center of mass. For example the rectangle has this symmetry. This symmetry means that in fact the two regions cut by any line through the center of mass are congruent, and not just equal area. 
Thus, my question is:

Suppose that we have a planar (measurable) set $A \subset \mathbb{R}^2$ (with positive measure). If there is a point $a\in \mathbb{R}^2$ such that: for any line $\ell \subset \mathbb{R}^2$ through $a$, denoting the regions of $A$ on either side of the line $B$ and $C$ then we have $|B| = |C|$ (Lebesgue measure), then is it necessarily true that
  (1) $a$ is the center of mass of $A$ and (2) that $A$ has $\pi$ rotational symmetry around $A$ in the a.e. sense, i.e. if $\tilde A$ is $A$ rotated by $\pi$ around $a$ then the symmetric difference between $A$ and $\tilde A$ has measure zero, i.e. $ |A \  \Delta \ \tilde A| = 0$.

I feel like (1) should be true, but I'm not so sure about (2). If the answer to (2) is no, then what sort of sufficient conditions are there? I'm mostly just curious about the answer, so by all means feel free to strengthen the assumptions on $A$, like requiring it to be a region bounded by a smooth boundary, etc. Thanks!

REPLY [7 votes]: This is to answer a natural and interesting question raised 
by Joseph O'Rourke's in a comment above. Indeed we have:

Any open subset $A$ of $\mathbb{R}^n$, star-shaped wrto a point $P$ and which is  partitioned into two pieces of equal measure by each hyperplane through $P$, is center-symmetric wrto $P$.

Here below I'm describing (with some freedom) the main argument of the proof, that I extracted by this short paper by K.J.Falconer, for those who have no access to Jstor. The key-point is a consequence of the Funk-Hecke theorem: the integral mean of a spherical harmonic $S$ over the hemi-sphere centered at $\theta\in \mathbb{S}^m$, as a function of $\theta$, is a non-zero scalar multiple of $S$ (the F-H theorem says much more; so I think hopefully there is also a  short proof of this fact).
Assume $P$ is the origin, and let $f:\mathbb{S}^{n-1}\to \mathbb{R}_+$ describe the boundary of $A$ in polar coordinates (that is, for $x=r\theta \in \mathbb{R}^n$ with $r\ge 0$ and $\theta\in \mathbb{S}^{n-1}:=\partial B _ {\mathbb{R}^n }(0,1 )$, then $x\in A$ if and only if $r< f(\theta)\, $ ).
By integrating in polar coordinates, the condition on $A$ writes:
$$\int_{(\psi\cdot \theta)\ge0} f(\theta)^n d\theta=\int_{(\psi\cdot \theta)\ge0} f(-\theta)^n d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1}   $$
and we are to show that this implies that $f^n$, thus $f$ itself, is an even function (the other implication is of course quite obvious, and reflects the fact that a center-symmetric $A$ is equi-partitioned by any hyperplane through the origin). To this end, consider the transformation 
$u\in L^2(\mathbb{S}^{n-1})\mapsto \tilde u\in L^2(\mathbb{S}^{n-1})$ defined by 
$$\tilde u (\psi):=\int_{(\psi\cdot \theta)\ge0} u(\theta)d\theta=\int_\mathbb{{S}^{n-1}} \chi_ { \mathbb{R}_+}(\psi\cdot \theta) u(\theta) d\theta\, , \quad\forall\psi\in\mathbb{S}^{n-1}  \, . $$
Due to the symmetry of the integral kernel $\chi_ { \mathbb{R}_+}(\psi\cdot \theta)$ we have 
$(\tilde u\cdot v) _ {L^2}=(u\cdot \tilde v) _ {L^2}$; moreover, as recalled, spherical harmonics are eigenfunctions of this transformation, with non-zero eigenvalues. Therefore, if $\tilde u$ is even then for any odd spherical harmonic $S$ we have $(u\cdot \tilde S) _ {L^2}=(\tilde u\cdot S) _ {L^2}=0$, thus also $(u\cdot S) _ {L^2}=0$, and $u$ is an even function too, ending the proof.
Moreover, just differentiating with respect to $\psi$ it is easy to see that for a star-shaped open subset $A$ the above property of equipartition by all hyperplanes is equivalent to: any section of $A$ by a hyperplane through $P$ has $n-1$ dimensional barycenter located in $P$ (note that in dimension 2 this immediately implies that $A$ is center-symmetric).<|endoftext|>
TITLE: From function field to curve: non-algebraically closed ground field and functor of points
QUESTION [8 upvotes]: This question concerns the (re)construction of a smooth projective curve $C$ over a field $k$, using the function field $K=K(C)$ of $C$. When $k$ is algebraically closed, this is described for instance in Hartshorne, I.6.
My questions are the following:

At a few points in the construction, Hartshorne uses that $k$ is algebraically closed, at other points at least that $k$ is infinite. Can one get around this, and use non-algebraically closed or even finite ground fields for reconstructing a curve from its function field?
In constructing a smooth projective curve from a finitely generated field $K$ of transcendence degree $1$ over $k$, one takes as the underlying points of the curve the discrete valuations on $K$, defines a topology on this set by declaring closed sets to be fintie or the whole set, and then building an appropriate sheaf of rings. Then one checks that the result is a smooth projective curve. Is there a slick way to describe the functor of points for this curve, instead of the associated locally ringed space?

REPLY [8 votes]: The construction holds for any base field $k$. But if $k$ is not perfect, you get a projective curve which is normal but not necessarily smooth. For example, if $k$ has characteristic $p>2$ and $t\in k$ is a not a $p$-th power in $k$, consider the function field $K=k(x,y)$ defined by the relation $y^2=x^P-t$. The curve you get is not smooth, and there is no projective smooth curve over $k$ whose function field is $K$. 
Note that smooth curves are always normal, and the converse is true if $k$ is perfect. 
Pick any transcendental element $x\in K$. Then $K$ is finite over $k(x)$. Let $A$ be the integral closure of $k[x]$ in $K$ and let $B$ be the integral closure of $k[1/x]$ in $K$. Then the localizations $A_x$ and $B_{1/x}$ are both equal to the integral closure of $k[x, 1/x]$ in $K$. Therefore we can glue the affine curves ${\rm Spec} A$ and ${\rm Spec} B$ along ${\rm Spec} A_x$ and get a curve $C$. By constuction $C$ is normal and integral, with field of functions $K$,  and there is  finite morphism $C\to \mathbb P^1={\rm Spec} k[x] \cup {\rm Spec} k[1/x]$ (obtained by glueing ${\rm Spec} A\to {\rm Spec} k[x]$ and ${\rm Spec} B\to {\rm Spec} (k[1/x])$). Hence  $C$ is its projective, and it is the projective normal curve associated to $K$. 
As a bonus, one also has a nice correspondance betweeen finite morphisms of curves and extensions of function fields of one variable. If $f : C\to D$ is a finite morphism of projective normal integral curves over $k$, then it induces a finite extension $k(D)\to k(C)$. One can show  that this establises a anti-equivalence from the category of integral normal projective curves over $k$ (where morphisms are finite morphisms of $k$-curves)  to the category of function fields of one variable over $k$ (where morphisms are morphisms of $k$-extensions).<|endoftext|>
TITLE: Irrational Numbers and the Riemann Surface of a Multi-Valued Function
QUESTION [17 upvotes]: Suppose a meromorphic function $f(z)$ has two poles, with residues $1$ and $\gamma$, respectively. Then the topology of the Riemann surface of the anti-derivative of $f(z)$ depends on whether or not $\gamma$ is irrational. More generally, the topology of a meromorphic function with $n$ poles with residues $\gamma_1,\gamma_2,\cdots,\gamma_n$ depends on the linearity of $\gamma_1,\cdots,\gamma_n$ over $\mathbb{Q}$. Has anyone considered this phenomenon? Are there relationships between the (co)homology groups of the covering and the residues? Could one attempt to prove the irrationality (or rationality) of a given complex number by considering the residues of the poles of a meromorphic function in this way? (In such a case, one would need other ways of extracting topological information about the given meromorphic function.)

REPLY [7 votes]: EDIT 1/5/2010:  I was a little dissatisfied with the informality of what I wrote below, so here is a somewhat more formal writeup.  The statement is somehow geometrically obvious, but the proof is still a bit nice.
EDIT:  I initially claimed the sequence below is short exact, which is false; it is right exact.  It is fixed below, with some explanation, and a bit of a geometric explanation of what's going on.
I know little about the theory of linear independence over $\mathbb{Q}$, but I'll attempt an answer to this part of the question:

Are there relationships between the (co)homology groups of the covering and the residues?

The answer is "yes."  Let $f$ be a meromorphic function on $\mathbb{C}$, and for convenience let's assume that it has poles $z_1, ..., z_n$ of order $1$, and no other singularities.  Let $g$ be an antiderivative of $f$.  Then the Riemann surface of $f$ is $M_f=\mathbb{C}-\{z_1, ..., z_n\}$ and the Riemann surface of $g$, which we will denote by $M_g$, is a covering space of $M_f$ with covering map $\pi: M_g\to M_f$.  Let $V\subset \mathbb{C}$ be the $\mathbb{Q}$-vector space spanned by $\operatorname{Res}_{z_1}(f), ..., \operatorname{Res}_{z_n}(f)$.  Then there is a short exact sequence $$H_1(M_g, \mathbb{Q})\overset{H_1(\pi)}{\longrightarrow} H_1(M_f, \mathbb{Q})\overset{\int}{\longrightarrow} V\to 0,$$
where the map $\int$ is given as follows.  Namely, $\int: H_1(M_f, \mathbb{Q})\to V$ is given by $[\gamma]\mapsto \frac{1}{2\pi i}\int_\gamma f~\operatorname{dz}$.
Let's elucidate the connection to the linear independence of $\operatorname{Res}_{z_1}(f), ..., \operatorname{Res}_{z_n}(f)$ over $\mathbb{Q}$.  $H_1(M_f, \mathbb{Q})$ is a $\mathbb{Q}$-vector space with a basis of cycles $[\lambda_1], ..., [\lambda_n]$ corresponding to the punctures $z_1, ..., z_n$.  Then the map $\int$ sends $[\lambda_i]$ to $\operatorname{Res}_{z_i}(f)$.  So the image of $H_1(M_g, \mathbb{Q})$ in $H_1(M_f, \mathbb{Q})$ is precisely the vector space of relations between the residues of $f$.
Added:  We can extend this right exact sequence into a longer sequence.  In particular, by covering space theory we have that $\pi_1(M_g)\to \pi_1(M_f)$ is an injection.  It is easy to see that the commutator subgroup of $\pi_1(M_f)$ is contained in the image of $\pi_1(M_g)$.  By the Hurewicz theorem $$H_1(M_g, \mathbb{Q})\simeq \pi_1(M_g)^{Ab}\underset{\mathbb{Z}}{\otimes} \mathbb{Q}.$$
So the kernel of the map $H_1(M_g, \mathbb{Q})\to H_1(M_f, \mathbb{Q})$ is given by the image of $[\pi_1(M_f), \pi_1(M_f)]$ (which is contained in $\pi_1(M_g)$) in $H_1(M_g, \mathbb{Q})$.  One can extend the exact sequence further back by looking at quotients of commutators in this manner.
This first extension has a geometrical interpretation.  Namely, let $h$ be a meromorphic function whose poles have the same locations as those of $f$, but whose residues are linearly independent over $\mathbb{Q}$.  Then the antiderivative of $h$, denoted $s$ has Riemann surface $M_s$, which is a covering space over $M_f$, with covering map $\pi': M_s\to M_f$.  By the properties of covering spaces, $\pi'$ factors through $\pi$, and it is not hard to see that $\pi_1(M_s)$ is exactly the commutator subgroup of $\pi_1(M_f)$.  Then the sequence $$H_1(M_s, \mathbb{Q})\overset{H_1(\pi')}{\longrightarrow} H_1(M_g, \mathbb{Q})\overset{H_1(\pi)}{\longrightarrow} H_1(M_f, \mathbb{Q})\overset{\int}{\longrightarrow} V\to 0,$$
is exact, and coincides with the sequence described above.
I don't know if the continuing left extensions of this sequence have similar geometric interpretations.  Also, it would be nice to have a naturally arising description of this sequence, rather than the somewhat ad hoc one I've given.<|endoftext|>
TITLE: Dependence of trace norm on matrix size for smooth vs. random matrices.
QUESTION [5 upvotes]: Problem
Consider two d x d complex matrices, R and S, whose entries lie in the unit disk:
$\quad |R_{i,j}|<1   \quad$ and $\quad |S_{i,j}|<1 $.
Say that R is constructed by randomly choosing complex numbers from the unit disk, but S is constructed as
$\quad S_{i,j} = f(i/d,j/d)$
where $f(x,y)$ is a smooth function for $x,y \in [0,1]$, with $|f(x,y)|<1$.  In other words, the entries of S are smooth functions of the indices (in the limit of large d), but those of R are not.
Question
How do the trace norms 
$\quad ||R||=Tr[\sqrt{R^\dagger R}] \quad$   and   $\quad ||S||=Tr[\sqrt{S^\dagger S}]$
of these matrices behave as $d \to \infty$ ?
Numerical Evidence
A few lines of Mathematica strongly suggest that 
$\quad ||R|| \propto d^{3/2}$
but
$\quad ||S|| \propto d$ 
for large d.  (The proportionality constants depend on the probability distribution used to pick numbers from the unit disk for R and the function $f(x,y)$ used to pick entries for S, respectively.)
What explains this behavior?
Addendum
After Willie's excellent answer below, I thought I'd mention that it's really fast to see the scaling behavior once you discretize the function.  Let $F$ be some matrix of discrete values for the function, and let $J_n$ be the $n \times n$ matrix with all elements equal to unity.
$||F \otimes J_n|| = \mathrm{Tr} \sqrt {(F^\dagger \otimes J_n)( F \otimes J_n)} = \mathrm{Tr} \sqrt {(F^\dagger F) \otimes (J_n J_n)} = ||F|| \cdot ||J_n|| = n ||F||$
Basically, the idea is that once the dimension of $F$ is large enough to capture the important detail in the function, increasing the dimension is really just increasing the dimension of $J$.

REPLY [5 votes]: To flesh out Helge's answer a bit before I go to bed:
Assume that $f(x,y)$ is a smooth function on the unit square. 
Define the functions $f_n(x,y) = f(\frac{\lfloor nx \rfloor}{n}, \frac{\lfloor ny\rfloor}{n})$. This is a piecewise step function. Observe that the operator $S$ of dimension $d$ is the same if you define it relative to $f$ or $f_d$. It is elementary to show that $f_n\to f$ uniformly (as functions).
Define an action of $f_n$ on $L^2[0,1]$ by 
$$ g(x) \mapsto \int_0^1 f_n(x,y)g(y) dy $$
and note that for a $n$-vector $v = (v_1,\ldots, v_n)$ we can associate 
$$ g_v(x) = \sum v_i \chi_{[\frac{i-1}{n},\frac{i}{n})} $$ 
we observe that 
$$ f_n(g_v(x)) = \frac{1}{n} g_{Sv}(x)$$
the $1/n$ factor coming from the fact that the length of the segment $[(i-1) / n, i/n)$ is $1/n$. 
Now consider $V^n$ as the subspace of $L^2[0,1]$ spanned by $\chi_{[\frac{i-1}{n},\frac{i}{n})}$. By definition $f_n$ annihilates its orthogonal complement, and $f_n$ restricted to $V^n$ is equivalent to a rescaled version of $S$, so in particular you have that the trace norms of $f_n$ (acting on $L^2$) is the same as $1/n$ times the trace norm of $S$ (acting on $\mathbb{R}^n$). 
To finish you just need to note that via some functional analysis voodoo the corresponding operators $f_n\to f$, and so the trace norms of $f_n$ converges. Therefore you have that $1/n$ times the trace norm of $S$ converges to a constant (which may be zero). Note that the regularity for $f$ is only needed in this last step, and you probably just need  uniform continuity to assure that the operators converge in a strong enough sense.
[Addendum: all the functional analysis voodoo you need (which is not very much for this problem) can be found in Reed & Simon, volume 1, chapter 6.]<|endoftext|>
TITLE: What is the standard notation for a multiplicative integral?
QUESTION [34 upvotes]: If $f: [a,b] \to V$ is a (nice) function taking values in a vector space, one can define the definite integral $\int_a^b f(t)\ dt \in V$ as the limit of Riemann sums $\sum_{i=1}^n f(t_i^*) dt_i$, or as the final value $F(b)$ of the solution $F: [a,b] \to V$ to the ODE problem $F'(t) = f(t); F(a) = 0$.
In a similar spirit, given a (nice) function $f: [a,b] \to {\mathfrak g}$ taking values in a Lie algebra $\mathfrak g$ of a Lie group $G$, one can define the multiplicative definite integral, which for sake of discussion I will denote $\Pi_a^b \exp(f(t)\ dt) \in G$, either as the limit of Riemann products $\prod_{i=1}^n \exp(f(t_i^*) dt_i)$ (with the product read from left to right), or as the final value $F(b)$ of the solution $F: [a,b] \to G$ of the ODE $F'(t) = F(t) f(t); F(a) = 1$.
Thus, for instance, when the Lie algebra is abelian, the multiplicative integral is just the exponential of the ordinary integral,
$$\Pi_a^b \exp(f(t)\ dt) = \exp( \int_a^b f(t)\ dt)$$
but in general the two are a little bit different, though still quite analogous.
This notion arises implicitly in many places (solving ODE, integrating connections along curves, dynamics and random walks on Lie groups (e.g. in the work of Terry Lyons), the "noncommutative Fourier transform" from scattering theory, etc.), but I am sure that it must be studied explicitly in some body of literature (and even vaguely recall seeing such at some point in the past).  But I am having difficulty locating this literature because I am not sure I have the correct terminology for this concept.  So my questions are:

What is the accepted name and notation for this concept in the literature? (Perhaps there is more than one such notation, coming from separate bodies of literature.)
What are the references for the theory of this concept?

REPLY [8 votes]: In the physics world, the notation is $P\exp(\int_a^b f(t)\,dt)$ or $T\exp(\int_a^b f(t)\,dt)$, where the "$P$" and "$T$" stand for "path ordered" and "time ordered".  The idea of time-ordered arithmetic I think is originally due to Feynman:

R.P. Feynman (1951). An operator calculus having applications in quantum electrodynamics. Physical Review. vol. 84 (1) pp. 108-128.

In the UC Berkeley 2008 course on Lie theory by Mark Haiman (my edited lecture notes are available as a PDF), we called it just $\int$, which was a bit of an abuse of notation.  Or rather, for any ODE, we referred to the corresponding "flow" as $\int$: $\int_p(\vec x)(t)$ was the point that you get to by starting at a point $p$ and flowing via the vector field $\vec x$ by $t$ seconds.  I'm not a fan of this notation, myself, whereas I'm reasonably happy with "$T\exp$".<|endoftext|>
TITLE: What is the right way to define the nerve of an unbiased monoidal category?
QUESTION [6 upvotes]: I've been toying around with unbiased composition in higher categorical structures on and off for a while now. In particular, I've been playing around with unbiased monoidal 2-categories. One motivation for this, as I discussed in my last question on the matter, is that unbiased tensor products and compositions often seem to be better descriptions of what goes on "in nature" than biased ones.
Another motivation was the hope that such gadgets would provide a cleaner notion of nerve than what one gets in the biased setting, where higher associators are floating around everywhere. However, directly transcribing the ordinary notion of nerve seems to work poorly, even for unbiased monoidal categories, for two reasons. First, in each dimension, one is forced to consider products of fixed numbers of objects, which is antithetical to the unbiased philosophy. Secondly, degeneracies are difficult to write down because one has, in place of a unit object, a zero-fold tensor product, which requires a bit of care to handle. A more natural "nerve" for an unbiased monoidal categories might involve having simplicies of dimension $n$ correspond to nested tensor products of depth $n$. I can't quite get such a definition to work, although I'm pretty sure that something like it should be possible.
Is there a construction of the nerve of an unbiased monoidal category that is natural to write down? (The definition of unbiased monoidal category can be found in section 3.1 of Leinster's Higher Operads, Higher Categories.) It strikes me that the problem might be simplicial sets themselves; are there some more exotic combinatorial objects that are better suited to capturing unbiased compositions? I'm aware of the existence of things like opetopes, but I have no idea if they're relevant to this particular issue.

EDIT:
I'd like to clarify why I'm interested in nerves (and consequently, why I'd really prefer that my nerve be a simplicial set instead of something more exotic, unless I can be convinced that more exotic objects can be easily adapted to my needs).
My poking around in all of this was inspired by the preprint by Etingof, Nikshych, and Ostrik, "Fusion categories and homotopy theory." The main results of this paper are proved by formulating their questions in terms of classical obstruction theory on the nerves of certain 3-groupoids. The obstruction theory itself can be justified using elementary fiddling with simplicial sets, as the reference Gregory Arone provided to my earlier question on obstrucion theory reveals. However, I wanted to understand the category theory side of the equation better, which led me to try to formulate things in terms of unbiased monoidal 2-categories.
So ultimately, the goal is to have a definition of the nerve to which I can apply my favorite classical obstruction theory techniques. While some people appear to have studied obstruction theory in more general settings (eg Obstruction Theory in Model Categories), it's not clear to me how to squeeze out the appropriate concrete computational gadgets (e.g., the cohomology groups $H^n(X; \pi_{n - 1}(Y))$) from the relevant abstract nonsense. Of course, if somebody could elucidate how that works, that would be wonderful, although perhaps that should be the subject of another question...

REPLY [6 votes]: Monoidal categories are a special kind of (coloured) operad. It is this viewpoint in which the unbiased version is most natural. In light of this, it would be most natural to consider the nerve of the associated operad. So, you're on to something in your suspecting that simplicial sets are not the most natural choice for capturing the nerve of a monoidal category. If your monoidal category is symmetric, then its nerve can be taken as a dendroidal set (otherwise, you it should be taken as a planar dendroidal set). Dendroidal sets were invented (discovered?) by Ieke Moerdijk and Ittay Weiss. Roughly speaking, dendroidal sets are to operads as simplicial sets are to categories. Like simplicial sets, they are presheaves on a certain test-category. Simplicial sets are $Set^{\Delta^{op}}$ where $\Delta$ consists of finite linear orders. Dendroidal sets are $Set^{\Omega^{op}}$, where $\Omega$ consists of finite rooted trees. One can even extend Joyal's model structure for quasicategories to dendroidal sets, and start talking about infinity-operads.
In fact, there is some very general machinery developed by Mark Weber which automatically produces the "correct" combinatorial objects to use to take a nerve of a certain type of algebraic object (given as an algebra for a monad). See: http://golem.ph.utexas.edu/category/2008/01/mark_weber_on_nerves_of_catego.html . Basically, given a monad $T$, this machinery produces a category $\Theta(T)$, consisting of certain "linear-like" free $T$-algebras, which has $Set^{\Theta(T)}$ some how the canonical choice for taking nerves of $T$-algebras. So, you COULD plug in the free-unbiased-monoidal category monad into this machinery and see what you get out, and use THIS to take the nerve. However, I think this is a bit over-kill. Using dendroidal sets should do just fine. However, it is worth mentioning that Cisinski has extended Joyal's model structure in this setting as well.<|endoftext|>
TITLE: What is the Cheeger constant of a cubical subset of the cubic lattice?
QUESTION [6 upvotes]: The Cheeger constant of a finite graph measures the "bottleneckedness" of the graph, and is defined as:
$$h(G) := \min\Bigg\lbrace\frac{|\partial A|}{|A|} \Bigg| A\subset V, 0<|A|\leq \frac{|V|}{2} \Bigg\rbrace$$
Here $V$ is the vertex set of $G$ and $\partial A$ denotes the collection of all edges going from a vertex in $A$ to a vertex in $V\setminus A$. The idea is that $h(G)$ is small if there is a bottleneck somewhere in $G$.
Now let $G$ have vertices $\lbrace 1,2,\ldots,n\rbrace^3\subset\mathbb{Z}^3$, and with an edge between two vertices if the distance between them is 1. Suppose that $n$ is even. Then it seems intuitively obvious that the minimum should be achieved with an "orthogonal half", that is $A= \lbrace 1,2,\ldots,n/2\rbrace\times\lbrace 1,2,\ldots,n\rbrace\times\lbrace 1,2,\ldots,n\rbrace$, and so $h(G)$ would be $n^2/(n^3/2) = 2/n$. Is this in fact the minimum, and how could one prove such a thing?

REPLY [2 votes]: The result (for 3 dimensions and I think easily generalises to any dimension) follows from Theorem 3 of the Bollobás and Leader paper. The theorem (in 3 dimensions) states that for any subset $A$ of the vertices $V$ of a cubical grid of side length $N$ with $|A|\leq\frac{N^3}{2}$ that $$|\partial A| \geq \min_{r=1,2,3}\left\lbrace|A|^{1-1/r}rN^{(3/r)-1}\right\rbrace$$
So:
$$\min_{r=1,2,3}\left\lbrace\left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N}\right\rbrace \leq \frac{|\partial A|}{|A|}$$
Now $|A|  \leq  \frac{N^3}{2}$, so $2  \leq  \frac{N^3}{|A|}$, so $\frac{r2^{1/r}}{N}  \leq  \left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N}$.
We can check for $r=1,2,3$ that $2\leq r2^{1/r}$
 so we get that 
$$ \frac{2}{N}  \leq  \left(\frac{N^3}{|A|}\right)^{1/r}r\frac{1}{N} $$
for each $r$, and therefore for the minimum, and so the ``orthogonal half'' subset of the cube, 
$(1,2,\ldots,N/2)\times (1,2,\ldots,N)\times (1,2,\ldots,N)$ which gives $\frac{|\partial A|}{|A|}=\frac{N^2}{N^3/2} = \frac{2}{N}$, is best possible.<|endoftext|>
TITLE: Non-Borel sets without axiom of choice
QUESTION [67 upvotes]: This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel  sets would be a Hamel basis, which needs axiom of choice, and examples of non Lebesgue measurable set would be Vitali sets, which seems to be unprovable without axiom of choice. Then I saw the answer of François G. Dorais. His construction of an uncountable $\mathbb{Q}$-independent set in $\mathbb R$ does not require axiom of choice. Which leaves a faint hope for the following:

Is it possible to construct without using the axiom of choice examples of non Borel sets?

There is a classic example of an analytic but non-Borel set due to Lusin, described by Gerald Edgar here, but it is not clear to me whether it needs axiom of choice since it seems to be putting restrictions on higher and higher terms in the continued fraction expansion. Since I do not know logic and set theory, I hoped of asking the experts.

REPLY [20 votes]: There are some very nice examples of non-Borel sets. Two that I particularly like are the differentiable functions (as a subset of the space of continuous functions on [0,1], say) and the set of all infinite graphs that contain an infinite clique (as a subset of the set of all graphs with vertex set $\mathbb{N}$ with the product topology). In general, a continuous image of a Borel set need not be Borel, and many natural non-Borel sets arise this way.<|endoftext|>
TITLE: website for jobs in applied or industrial mathematics (or financial math)
QUESTION [9 upvotes]: What are the websites for general position in applied or industrial mathematics(or financial mathematics) related jobs (that is if we have to find a non academic job temporarily) ?Thanks!

REPLY [6 votes]: For finance, I recommend wilmott.com and gloriamundi.org. There's also www.mathjobs.org.<|endoftext|>
TITLE: Jordan Curve Homotopy
QUESTION [7 upvotes]: Does there exist a notion of Jordan curve homotopy? 
In particular, suppose we have two Jordan curves $C_0 : S^1 \rightarrow \mathbb{R}^2$ and $C_1 : S^1 \rightarrow \mathbb{R}^2$. When does there exist a continuous function $f: S^1 \times [0,1] \rightarrow \mathbb{R}^2$ such that:
$f(x,0) = C_0(x)$, $f(x,1) = C_1(x)$, and for all $t \in [0,1]$, the function $C_t: S^1 \rightarrow \mathbb{R}^2$ defined by $C_t(x) = f(x,t)$ is a Jordan curve. 
My intuition tells me that such a function always exists, but I'm unsure about how to go about proving this. Also, if this is a known result, are there similar results for manifolds other than $\mathbb{R}^2$?

REPLY [10 votes]: Homotopies through embeddings are usually called isotopies. 
There is a subtlety called local flatness that comes up in higher dimensions. Let $E$ be any embedding of $\mathbb R$ in $\mathbb R^3$ such that $E(s)=(s,0,0)$ when $s<-1$ or $s>1$. Define a homotopy $H_t$ with $E_0(s)=(s,0,0)$ for all $s\in \mathbb R$ and $E_1=E$, as follows: $E_t(s)=tE(s/t)$ if $-t\le s\le t$ and otherwise $E_t(s)=(s,0,0)$. This is a homotopy through embeddings, but it (un)ties the knot. This is easily adapted to apply to examples of embeddings of $S^1$ in $\mathbb R^3$, for example. 
The way to fix this problem is to only consider embeddings that are locally flat and isotopies that are locally flat. An embedding $E:M\to N$ is (topologically) locally flat if for every point $p\in M$ there exist charts around $p$ and $E(p)$ such that $E$ looks like $(x_1,\dots,x_m)\mapsto (x_1,\dots,x_m,0,\dots,0)$. An isotopy $E_t$ is locally flat if for each point $p\in M$ and time $\tau\in I$ there are charts around $(p,\tau)\in M\times I$ and around $(E_\tau(p),\tau)\in N\times I$, both of them using projection to $I$ as last coordinate, such that locally $(x,t)\mapsto (E_t(x),t)$ looks like $(x_1,\dots,x_m,t)\mapsto (x_1,\dots,x_m,0,\dots,0,t)$. Local flatness is automatic when $m=1$ and $n=2$. The example I gave (with $m=1$ and $n=3$) was such that the isotopy was not locally flat although if the original embedding $E_1$  was locally flat then for every $t$ the embedding $E_t$ was, too. 
The Alexander horned sphere ($m=2$, $n=3$, not locally flat) can be smoothed out by such a procedure, too. 
Another way of limiting oneself to the right kind of isotopies is to use ambient isotopies: to require $E_t$ to be $H_t\circ E_0$ where $H_t$ is a homeomorphism $N\to N$ depending on $t$. (Local flatness in the case $m=n$ follows from invariance of domain.)
Another way is to limit oneself to smooth embeddings (meaning, as usual, smooth maps that are topological embeddings and that are one to one on the tangent-space level, or equivalently locally flat in the smooth category) and smooth isotopies.<|endoftext|>
TITLE: Is there any generalization of the Dold-Kan correspondence?
QUESTION [13 upvotes]: The Dold-Kan correspondence gives an equivalence between simplicial abelian groups and chain complexes of abelian groups supported on negative degrees. It actually works for any abelian category. I'm wondering if there is any generalization of this corrensponces, more precisely, for other category of algebraic objects, like groups, rings, monoids, A-algebras (where A is a ring), should simplicial objects in them correspond to some more familiar objects? (hopefully some graded thing and maybe we can still take cohomologies of them?)

REPLY [5 votes]: For simplicial groups, there is the generalised Dold-Kan-Puppe theorem by Carrasco-Cegarra: "Group-theoretic algebraic models for homotopy types", J. Pure Appl. Alg. 75 (1991), no. 3, 195--235 (see also MathSciNet MR1137837). (The additive notation in the article is to be read non-commutatively.)<|endoftext|>
TITLE: Tractability of forcing-invariant statements under large cardinals
QUESTION [7 upvotes]: It is usual to mention theorems of the kind:
Th. Assume there is a proper class of Woodin cardinals, $\mathbb{P} $ is a partial order and $G \subseteq \mathbb{P}$ is V-generic, then $V \models \phi \iff V[G] \models \phi$ 
where $\phi$ is some set theoretic statement (like "the Strong Omega Conjecture holds"), as some sort of evidence that $\phi$ is a statement less intractable than other statements like CH which are not focing-invariant.
My question is, in wich sense are these statements more tractable? What kind of "empirical evidence" gives support to the hope that they can be decided by large cardinal axioms?

REPLY [13 votes]: Typically, generic absoluteness is a consequence of a stronger property, that in many cases is really the goal one is after. To explain this stronger property, let me begin by reviewing two important absoluteness results.
1) The first is Mostowski's Absoluteness. Suppose $\phi$ is $\Sigma^1_1(a)$, where $a\in\omega^\omega$. This means that $\phi(x)$ for $x\in\omega^\omega$, asserts

$∃y\in\omega^\omega\forall n\in \omega R(x\upharpoonright n, y\upharpoonright n,a\upharpoonright n)$,

where $R$ is a predicate recursive in $a$. These statements are very absolute: Suppose that $M$ is a well-founded model of a decent fragment of ZF, and that $a,x\in M$. Then $\phi(x)$ holds iff $M\models \phi(x)$. 
In particular, whether or not $\phi(x)$ holds cannot be changed by forcing, or by passing to an inner or outer model. Note that $M$ could be countable. In fact, only needs to be an $\omega$-model; well-foundedness is not necessary.
This is how the result is usually stated. What is going on is the following:
Suppose that $T$ is a tree (in the descriptive set theoretic sense) and that $T\in M$. Then $T$ is ill-founded iff $M\models T$ is ill-founded. 
In particular, $T$ could be the tree associated to $\phi$. This is the tree of all $(s,t)$ such that $s,t$ are finite sequences of the same length $l$, and $\forall n\le l$ $ R(s\upharpoonright l,t\upharpoonright n,a\upharpoonright n)$. So $T$ is the tree of attempts to verify $\phi$: $\phi(x)$ holds iff (there is a $y$ such that for all $n$, $(x\upharpoonright n,y\upharpoonright n)\in T$) iff the tree $T_x$ is ill-founded. Recall that $T_x$ consists of all $t$ such that, if $l$ is the length of $t$, then $(x\upharpoonright n,t\upharpoonright n)\in T$ for all $n\le l$.
The point is that $T$ is a very simple object. As soon as $T,x$ are present, $T_x$ can be built, and the result of the construction of $T_x$ is the same whether it is performed in $V$ or in $M$. Since well-foundedness is absolute, whether or not $T_x$ is ill-founded is the same whether we check this in $M$ or in $V$. Of course, $T_x$ is ill-founded iff $\phi(x)$ holds. 
The moral is that the truth of $\Sigma^1_1$ statements is certified by trees. And I think that this is saying that in a strong sense, $\Sigma^1_1$ statements are very tractable. All we need to check their validity is a very easy to build tree and, once we have it, the tree is our certificate of truth or falsehood, this cannot be altered. 
Recall that proofs in first-order logic can be described by means of certain finite trees. If something is provable, the tree is a very robust certificate. This is a natural weakening of that idea.
Of course one could argue that, if a $\Sigma^1_1$ statement is not provable, then in fact it may be very hard to establish its truth value, so tractability is not clear. Sure. But, independently of whether or not one can prove something or other, the certificate establishing this truth value is present from the beginning. One does not need to worry that this truth value may change by changing the model one is investigating.
2) The second, and best known, absoluteness result, is Shoenfield's absoluteness. Suppose $\phi$ is $\Sigma^1_2(a)$. This means that $\phi(x)$ holds iff 

$\exists y\forall z\exists n$ $R(y\upharpoonright n,z\upharpoonright n,x\upharpoonright n,a\upharpoonright n)$,

where $R$ is recursive in $a$. Let $M$ be any transitive model of a decent fragment of ZFC, and suppose that $\omega_1\subset M$ and $a,x\in M$. Then $\phi(x)$ holds iff $M\models\phi(x)$.
This is again a very strong absoluteness statement. Again, if one manages to show the consistency of $\phi(x)$ by, for example, passing to an inner model or a forcing extension, then in fact one has proved its truth. 
Again, one could say that if $\phi$ is not provable, then it is in fact not very tractable at all. But the point is that to investigate $\phi$, one can use any tools whatsoever. One only needs to worry about its consistency, for example, and one can make use of any combinatorial statements that one could add to the universe by forcing. 
Just as in the previous case, $\Sigma^1_2$ statements can be certified by trees. The tree associated to $\phi$ is more complicated than in the previous case, and it is now a tree of $\omega\times\omega_1$. (Jech's and Kanamori's books explain carefully its construction.) Again, the tree is very absolute: As soon as we have $a$ and all the countable ordinals at our disposal, the tree can be constructed. (One comparing two models $M\subset V$, we mean all the countable ordinals of $V$, even if $M$'s version of $\omega_1$ is smaller.) 
3) Generic absoluteness of a statement $\phi$ is typically a consequence of the existence of absolutely complemented trees associated to $\phi$. In fact, all generic absoluteness results I'm aware of are established by proving that there are such trees ("conditional" generic absoluteness results, such as only for proper forcings, or only in the presence of additional axioms, are somewhat different). This is a direct generalization of the situations above.
To define the key concept, recall that if $A$ is a tree of $\omega\times X$, then the projection $p[A]$ of $A$ is the set of all $x\in\omega^\omega$ such that $\exists f\in X^\omega\forall n\in\omega\,(x\upharpoonright n,f\upharpoonright n)\in A$.
Two (proper class) trees $A$ and $B$ on $\omega\times ORD$ are absolutely complemented iff:

$p[A]\cap p[B]=\emptyset$ and $p[A]\cup p[B]=\omega^\omega$.
Item 1 holds in all generic extensions.

A statement $\phi$ admits such a pair iff, in addition,

In any forcing extension, $\phi(x)$ iff $x\in p[A]$.

The idea is that this is a precise, formal, definable approximation to the intuitive statement one would actually like, namely, that there are such trees describing $\phi$ that have this ``complementary'' behavior in all outer models. First-order logic limits ourselves to considering forcing extensions.
Let me point out that $\Sigma^1_1$ and $\Sigma^1_2$ statements admit absolutely complemented pairs, so the existence of such a pair is a natural (far reaching) generalization of the two cases above. 
Once we accept large cardinals, we can show that much larger classes than $\Sigma^1_2$ admit absolutely complemented trees. For example, any projective statement does. Once again, the point is that as soon as we have the large cardinals and real parameters in our universe, we have the trees, and the trees certify in unambiguous forcing-unchangeable terms, whether the statements hold at any given real. It is in this sense that we consider these statements more tractable.
Here is a rough sketch of an example I particularly like, due to Martin-Solovay (for measurables) and Woodin (in its optimal form). For details, see my paper with Ralf Schindler, ``projective well-orderings of the reals,'' Arch. Math. Logic (2006) 45:783–793:

$V$ is closed under sharps iff $V$ is $\Sigma^1_3$-absolute. $(*)$

The right hand side means that for any $\Sigma^1_3$ statement $\phi$ (so now we have three alternations of quantifiers) and any two step forcing ${\mathbb P}∗\dot{\mathbb Q}$, for any $G$, ${\mathbb P}$-generic over $V$, any $H$, ${\mathbb Q}$-generic over $V[G]$, and for any real $x$ in $V[G]$, we have 
 $$ V[G]\models\phi(x)\Longleftrightarrow V[G][H]\models\phi(x). $$
The left hand side of $(*)$ is a weakening of "there is a proper class of measurable cardinals", which is how the statement is usually presented. 
The proof of the implication from left to right in $(*)$ goes by building a tree of attempts to find a witness to the negation of a $\Sigma^1_2$ statement. The goal is that if such a witness can be added by forcing, then in fact we can find one in the ground model. If there is a forcing adding a witness, there is a countable transitive model where this is the case. Essentially, the tree gives the construction of such a model, bit by bit, and if we have a branch, then we have such a model. 
So: If there is a witness added in a forcing extension, the tree will have there a branch. So it is illfounded. By absoluteness of well-foundedness, the tree has a branch in $V$. The sharps are used to ``stretch'' the model so that we can use Shoenfield absoluteness, and conclude that there must be a witness in $V$. 
4) Projective absoluteness, a consequence of large cardinals, is established by showing the existence of absolutely complemented trees for any projective statement. The theory of universally Baire sets originates with this concept, and the closely related notion of homogeneously Suslin trees. All of this is also connected to determinacy. Once again, to drive the point home: Generic absoluteness is not the goal. The goal is the existence of the pair of trees. Once we have them, we have a strong certificate of truth or falsehood of a statement. I do not know if one is to accept the search for such trees as a more tractable problem than the original statement whose pair of trees we are now searching for. But it certainly says that consistency of the statement, using large cardinals or any combinatorial tools whatsoever, is enough to have a proof of the statement. This seems much more hopeful and generous an approach than if only proofs in the usual sense are allowed. The existence of these trees for projective statements is what I meant in a comment by ``large cardinals settle second order arithmetic.'' Put yet another way: If you show, for example, that a projective statement is not 'decidable' (in the presence of large cardinals), meaning that it is consistent and so is its negation, then you have either actually showed that certain large cardinals are inconsistent, or you have found a way of changing the truth of arithmetic statements, and both of these options are much more significant events than the proof of whatever the projective statement you were interested in was. More likely than not, the truth value of the statement will be uncovered at some point, and you know there is no ambiguity as of what it would be, since the witnessing trees are already present in the universe.  
(In spite of its length, I am not completely happy with this answer, but I would need to get much more technical to expand on the many interesting points that your question raises. Hopefully there is some food for thought here. For nice references to some of the issues I mention here, Woodin's article in the Notices is a good place to start, and Steel's paper on the derived model theorem has much of the details.)<|endoftext|>
TITLE: What's after natural transformations?
QUESTION [14 upvotes]: If functors are morphisms between categories, and natural transformations are morphisms between functors, what's a morphism between natural transformations? Is there ever a need for such a notion?

REPLY [7 votes]: This is not really a sophisticated answer as the other ones, but maybe it makes visible why higher structures are needed to get an interesting notion of morphism between natural transformations.
Assume $F,G : C \to D$ are functors and $\alpha, \beta$ are natural transformations $F \to G$. What could be a morphism $\alpha \to \beta$? Since $\alpha$ and $\beta$ consist of their components $\alpha(c) : F(c) \to G(c), \beta(c) : F(c) \to G(c)$, the only reasonable way of "connecting" these data in our category $D$ is by means of two morphisms $\gamma(c) : F(c) \to F(c), \delta(c) : G(c) \to G(c)$, so that the resulting diagram becomes commutative. Furthermore, $\gamma$ and $\delta$ should become natural transformations $F \to F, G \to G$.
But this comes from a more general concept, namely the arrow category: If $C$ is a category (in the above case, this is a functor category), its arrow category has as objects the morphisms of $C$ and a morphism between two morphisms $x \to y, x' \to y'$ is a pair of morphisms $x \to x', y \to y'$, making the obvious diagram commutative. This may be also described as the functor category $C^{\textbf{2}}$.<|endoftext|>
TITLE: Torsion-freeness of Picard group
QUESTION [23 upvotes]: Let $X$ be a complex normal projective variety.
Is there any sufficient condition to guarantee the torsion-freeness of Picard group of $X$?
One technique I sometimes use is following:
If $X$ can be represented by GIT quotient $Y//G$ for some projective variety $Y$ with well-known Picard group, then by using Kempf's descent lemma, we can attack the computation of integral Picard group. 
Of course, if we can make a sequence of smooth blow-ups/downs between $X$ and $X'$ with well known Picard group, then we can get the information of $\mathrm{Pic}(X)$ from $\mathrm{Pic}(X')$.
Is there any way to attack this problem?

REPLY [9 votes]: You have my sympathy, I have struggled with torsion-freeness of Picard and class groups too. 
As Donu hinted in his comment on Damiano's answer, a torsion element of the Picard group gives you a cyclic cover of $X$, so triviality would follow from certain purity results for etale covers. A fact that is interesting (at least for me) is that one could use the purity of the local ring at the origin of the affine cone over $X$, see Lemma 9 of this paper. 
(purity for the local ring here is in the sense of Grothendieck, that is to say, the restriction of etale covers over $\text{Spec} A$ to the punctured spectrum is an equivalence) .
This shows, for example, Grothendieck's classical result that if $X$ is a complete intersection of dimension at least $2$, then $\text{Pic}(X)$ is torsion-free. But Cutkosky's paper above gives quite more general results, see Theorem 19, the Cor after Theorem 26 and the Examples after Theorem 22 of his paper. Basically, his results say that if $X$ is locally a complete  intersection in high enough codimension relative to the deviation of the local ring at the vertex of the cone, then $\text{Pic} (X)$ is torsion-free. These apply to a few Grasmanians and  Pfaffians, and hopefully they can be helpful to what you want to do.  
Also note that various results which give vanishing of $H_1$ or $\pi_1$ when $X$ is relatively small codimension subvariety of the projective space can be found in the paper of Lyubeznik and the references therein. 
ADDED: I had a chance to look at Lyubeznik's paper and actually he has Theorem 11.2, which says that if  $Y\subset \mathbb P^n_{\mathbb C}$ is an irreducible algebraic set  of codimension $c< n/2$, then $\text{Pic}(Y)$ is torsion-free.<|endoftext|>
TITLE: Does there exist a pure recurrence formula with polynomial coefficients for Fibonacci(2^n)?
QUESTION [9 upvotes]: Let F(n) be the Fibonacci sequence as defined by F(1)=1, F(2)=1, F(n)=F(n-1)+F(n-2) for n>=3.  I'm looking for a pure recurrence formula for the function X(i)=F(2i) whose coefficients may be polynomials in i.  This is Sloane's A058635.  I also would like it to be "pure" in the sense that there is no auxiliary function involved.  Is such a formula known?
I attempted using Sister Celine's technique (as described in A=B) with the data up to 221 without success.
My motivation is that I have a fairly complicated recurrence formula for another sequence, but I am only interested in the terms whose indices are of the form 2i-3.  The existence (or non-existence) of a recursion for X(i) would be a kind of "proof of concept" as to whether or not I should explore the possibility of finding such a recursion for my sequence.

REPLY [2 votes]: Interestingly, the recursion $u_{n+1} = (u_n + 5/u_n)/2$, with $u_0=1$, gives the fractions $Lucas(2^n)/Fibonacci(2^n)$.<|endoftext|>
TITLE: Word problem for fundamental group of submanifolds of the 4-sphere
QUESTION [22 upvotes]: Given any finitely-presented group $G$, there are a few equivalent techniques for constructing smooth/PL 4-manifolds $M$ such that $\pi_1 M$ is isomorphic to $G$.  For most constructions of these 4-manifolds, they embed naturally in $S^5$ (as the boundary of regular neighbourhoods of $2$-complexes in $S^5$.)  
Question: Are there are any smooth/PL 4-dimensional submanifolds $M$ of $S^4$ such that $\pi_1 M$ has an unsolvable word problem?  $M$ would of course have to be a smooth $4$-manifold with non-empty boundary.
I'm aware there are several constructions and obstructions to $2$-complexes embedding in $S^4$.  Moreover, I've heard some of the construction techniques fall into the tame topological world and may not be smoothable.  The condition given by Kranjc (that $H^2$ of the 2-complex is cyclic) is generally a non-computable condition for a group with non-solvable word problem. Although, perhaps there are many groups with non-solvable word problem and $H^2$ trivial.  The closest to references on the subject that I know:
M. Kranjc, "Embedding a 2-complex K in R^4 when H^2(K) is a cyclic group," Pac. J. Math. 150 (1991), 329-339. 
A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.
edit:  Thanks for the comments people.  Now that I'm back in Canada with proper internet (+MathSciNet) access, I did a little digging and came across this:
A. Dranisnikov, D. Repovs, "Embeddings up to homotopy type in Euclidean Space" Bull. Austral. Math. Soc (1993). 
They show that any finitely-presented group is the fundamental group of a 2-dimensional polyhedron in $\mathbb R^4$. This was apparently a question of Matthias Kreck's.  
And yes, Sam Nead, this question was in part motivated by the concern that 2-knots could have undecidable word problems for the fundamental groups of their complements.  I've been thinking about the fundamental groups of 2-knot complements recently, and this is a concern.

REPLY [15 votes]: Update:
My memory was quite blurry about this when I originally answered.  
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.
They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.
Original answer:
You most likely would like a finitely presented group, but this might be of interest anyway:
Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group 
$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$
This has unsolvable word problem.  See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)
This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$.  You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.
I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.<|endoftext|>
TITLE: Illuminating piecewise-flat manifolds with geodesics
QUESTION [11 upvotes]: If one issues a geodesic in every direction from a point $p$ on a piecewise-flat 2-manifold, will
it necessarily illuminate the entire surface?  I know the answer is 'No,' but I would like to explore
the question further.
I am using the term piecewise-flat manifold in the sense that David Glickenstein uses it,
e.g., in "Introduction to piecewise flat manifolds:" a gluing of Euclidean triangles edge-to-edge.
This is also called a polyhedral manifold.  For the purposes of this question, whether it is
embedded in $\mathbb{R}^3$ is not relevant.  More generally, these manifolds are formed by edge-to-edge gluings of planar polygons (each of which could be triangulated).  The manifold is flat everywhere but at a finite number of vertices (or cone points) at which the surrounding angle differs from $2 \pi$.
Because geodesics do not pass through vertices (or, more accurately, I stipulate they cannot), it is conceivable that there is some $p$ from
which geodesics shot in every direction fail to reach every point of the manifold.
This was established in a rather different context in the paper by
George Tokarsky, "Polygonal Rooms Not Illuminable from Every Point"
[Amer. Math. Monthly, 102:867-879 (1995)].  Mathworld has a nice description, including this figure:



If you glue two copies of either of these polygons back-to-back, it forms a polyhedral 2-manifold
with the property that geodesics (light rays) from one red point cannot reach the other red point.
One can ask many questions here, but these three interest me:

Tokarsky's example is a doubly covered polygon.  If one generalizes instead to arbitrary
polyhedral manifolds, are there other, perhaps more straightforward examples where from some 
$p$ not all the manifold is covered its geodesics?
I conjectured long ago that the measure of the "dark points" is zero.
Is there an example (of a polyhedral manifold) where more than isolated points are unilluminated?
Could a segment be unilluminated?  A region of positive area?

Are there examples of these same phenomena in piecewise-flat 3-manifolds
(gluings of Euclidean tetrahedra)?

Edit. In response to Henrik's example below, I should have said that ideally two further conditions should be satisfied: (a) $p$ is not at a vertex (so it is surrounded by $2\pi$ of surface); and (b) the manifold should be closed, without boundary.  This is not to say that $p$ at a vertex and a manifold with boundary are not of interest!
Addendum: Thanks for the interest and help!  I have much to learn on the topic of translation surfaces!

REPLY [2 votes]: The illumination problem for translation surfaces was solved in this paper by Lelievre, Monteil and Weiss here (https://arxiv.org/pdf/1407.2975.pdf). In particular, they answer the "Conjecture 1" in your comment to Alex's answer above.<|endoftext|>
TITLE: If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*?
QUESTION [7 upvotes]: Let X be a normed space and denote by X* the space of all bounded linear functionals on X. Take a linear subspace G ≤ X* which separates the elements of X, i.e., for each x ∈ X, there is an f ∈ G with f(x) ≠ 0. Denote by B the closed unit ball in X. Now consider a linear subspace Y ≤ X. The question is:
If Y is dense in X in the weak topology induced by G, is Y ∩ B necessarily dense in X ∩ B in that topology?
REMARKS, BACKGROUND AND MOTIVATION
Without the assumption that G separates points, there exists a trivial counter-example.
Take X := ℝ2 with the supremum norm, i.e., ∥(x, y)∥ :=
max{|x|, |y|}. For G, take the linear span of the linear functional f(x, y) := x + y. Finally, take Y := {(x, 0) ; x ∈ ℝ}. Then Y is G-dense in X because (x, y) and (x + y, 0) are indistinguishable in the G-topology. However, the element (1, 1) ∈ B is not in the closure of Y ∩ B because f(1, 1) = 2
and f(x, 0) ≤ 1 for each x with (x, 0) ∈ B.
An interesting example is to take the space G := L∞(S), the space of all bounded measurable functions on a measurable space S, equipped with the supremum norm. Take X := G*, with the corresponding dual norm. The space G can be naturally considered as a subspace of X*. Clearly, it separates the points in X, and the G-topology is exactly the weak *-topology.
An important subspace of X is Y := M(S), the space of all real measures on S (with finite total variation). If S is large enough, let's say, ℕ, then Y is a proper subspace of X. It is well-known that Y is weakly *-dense in X, but it is also interesting that Y is weakly *-complete by sequences (see Diestel: Sequences and Series in Banach spaces, Springer-Verlag, 1984).
By the Banach-Alaoglu theorem, B is weakly *-compact. One may wonder whether X ∩ B is also weakly *-compact. The answer is no. However, the argument that Y is weakly *-dense in X is insufficient; a sufficient argument is that that Y ∩ B is weakly *-dense in X ∩ B. Though this is not difficult to prove in our particular case, it might by a non-trivial issue in more general cases. If the answer to my initial question is yes, it will be sufficient to only prove that Y is dense in X.
Many thanks in advance for any answer, reference or comment!

REPLY [3 votes]: You can derive a complete answer to your question and more from W. Davis, J. Lindenstrauss ``On total nonnorming subspaces", PAMS 31 no. 1 109-111 (1972), although their theorem is stated in the dual space.

REPLY [2 votes]: If I understand the question correctly, then maybe you are after special cases, as well as a general comment.  So, as one of your examples suggests, one special case is to let G be a Banach space, considered as sitting inside its own bidual, and let $X=G^*$.  Thus G induces the usual weak*-topology on X.
So an example of a positive answer is furnished by the Kaplansky Density Theorem: here G would be the predual of a von Neumann algebra, X would be a von Neumann algebra, and we let Y be any self-adjoint subalgebra which is weak*-dense.  Then Kaplansky Density tells us that indeed the unit ball of Y is weak*-dense in the unit ball of X.  This is an incredibly useful tool in Operator Algebra theory.
This then suggests that the result is unlikely to be true in general.  Indeed, I think rpotrie's counter-example works!  But here's an easier variant.  Let $G=c_0$ and $X=\ell^1$, and let Y be the span of vectors $e_n+ne_{n+1}$.  To see that this is weak*-dense, suppose that $\sum_k a_k e_k^* \in c_0$ annihilates all of Y.  Then $a_n + na_{n+1}=0$ for all $n$, so $a_1 + a_2=0$ and $0 = a_2+2a_3 = 2a_3 - a_1$ and $0=a_3+3a_4 = (1/2)a_1+3a_4$, so $a_1 = -a_2 = a_3/2 = -a_4/3$ and an easy induction shows $a_1 = (-1)^{n-1}a_n/n$.  Thus $|a_n| = n|a_1|$ for all $n$, but as $|a_n|\rightarrow 0$, it
follows that $a_1=0$, and so actually $a_n=0$ for all $n$.  Hence Y is weak*-dense.  However, $e_1$ is in the closed unit ball of X, but it's pretty clear that we can't approximate it by norm one elements in Y (to do this without a tedious calculation defeats me right now).<|endoftext|>
TITLE: Spherical Harmonics - a bunch of questions about them
QUESTION [5 upvotes]: Hi there,
Please tell me if I should divide these into individual questions next time.
Short intro:
Spherical Harmonics are a nice collection of functions. They are orthogonal and allow you to take a function $f$ and break it into a linear combination of spherical harmonic functions $Y_n^m$. This is very similar to how the Fourier Transform allows you to break a function into a linear combination, which is the Fourier Series.
Questions about Spherical Harmonics:

What kind of functions can be broken down using this linear combination?
I read in Wikipedia here about having the Laplacian of $f$ equal to zero. How is that connected to this linear combination (or is that an unrelated and has more to do with physics)?
For the following linear combination of $f$: $f=\displaystyle\sum_n^\infty\displaystyle\sum_{m=-n}^n\alpha_{nm}Y_{nm}(\theta,\phi)$ it seems as though $f$ takes only two parameters $\theta$ and $\phi$ but that doesn't make a lot of sense because (I think) there should be a radial parameter $r$ somewhere. Where am I wrong?

Thank you very much, help saves lives :)
Ofer

REPLY [18 votes]: If you are only interested in expanding polynomials then you can forgo Hilbert spaces and get an exact expansion into a finite linear combination.

 Spherically-invariant case  Let $f$ be a polynomial in $x_1,\ldots, x_n$ (the coefficients may be real, complex, or even a field of characteristic zero). If $f$ is invariant under isometries of the $n$-dimensional space, i.e. $f(Ax)=f(x)$ for any $n\times n$ orthogonal matrix $A,$ then $f=g(r^2),$ where $g$ is a polynomial in one variable of degree $\deg(f)/2$ and 
$$r^2=x_1^2+x_2^2+\ldots+x_n^2.$$ 
This is intuitively clear: a rotationally invariant function depends only on the distance to the center and $g(t)=f(t,0,\ldots,0)$ is a polynomial. (For $n\geq 2$ it's sufficient that $f$ be invariant under rotations, i.e. the special orthogonal group $SO(n)$.) 

The theory of spherical harmonics, which is closely related to the representation theory of the orthogonal group $O(n),$ generalizes this observations to polynomials (and even more general functions) that are not necessarily spherically invariant. 

General case  The first key theorem states that a polynomial $f$ can be expanded as 
$$ f(x)=\sum_{k=0}^d r^{2k}h_k(x) \qquad (*)$$
where $d\leq\deg(f)/2$ and $h_k$ is a harmonic polynomial of degree at most $\deg(f)-2k.$ Recall that a polynomial is $h$ is harmonic if $\Delta h=0,$ where $\Delta$ is the Laplace operator,
$$\Delta h=\frac{\partial^2 h}{\partial x_1^2}+\frac{\partial^2 h}{\partial x_2^2}+\ldots+\frac{\partial^2 h}{\partial x_n^2}.$$
Moreover, the harmonic polynomials $h_k$ in $(*)$ are uniquely determined. If $f$ is spherically invariant then $h_k$ is simply the coefficient of $t^k$ in the polynomial $g(t)$ from above, viewed as a constant (i.e. degree $0$) polynomial. 
So now every polynomial can be expanded in terms of the powers of $r^2$ and harmonic polynomials. What can we say about the latter? Let $\mathcal{H_\ell}$ (respectively, $\mathcal{H}$) denote the space of homogeneous harmonic polynomials of degree $\ell$ (respectively, all harmonic polynomials). Every harmonic polynomial is uniquely decomposed into the sum of homogeneous polynomials of various degrees, and these components are themselves harmonic:
$$\mathcal{H}=\bigoplus_{\ell\geq 0}\mathcal{H}_\ell\qquad (**)$$
The second key theorem states that $\mathcal{H}_\ell$ is an irreducible representation of the orhtogonal group $O(n)$ of dimension $\binom{n+\ell-1}{n-1}-\binom{n+\ell-3}{n-1}$ (for $\ell=0$ or $1$, the second term is zero). For $n\geq 3,$ this representations remains irreducible upon restriction to the special orthogonal group $SO(n)$ (for $n=3$, this is the rotation group).
For small $n$, namely $n=2,3,4$, there are standard bases in the spaces $\mathcal{H}_\ell$ that have been tabulated and extensively studied. In particular, for $n=3,$ $\mathcal{H}_\ell$ is a $(2\ell+1)$-dimensional representation of $SO(3)$, or spin $\ell$ representation in physics language, which has a basis indexed by an integer $m,\ -\ell\leq m\leq \ell$ constructed using associate Legendre polynomials $P_\ell^m.$ Depending on your goals, you may want a suitable explicit description of these functions in rectangular or spherical coordinates (the Wikipedia article Spherical harmonics, including the references, is a good starting point). 
Let me also mention a common point of confusion related to your question 3: in view of $(*),$ for any polynomial $f$ as above, there exists a unique harmonic polynomial $h$ such that the restrictions of $f$ and $h$ to the unit sphere 
$$S^{n-1}: x_1^2+x_2^2+\ldots+x_n^2=1$$ coincide, $f|_{S^{n-1}}=h|_{S^{n-1}}.$ In other words, the space $\mathcal{H}$ of harmonic polynomials can be naturally identified with the space of polynomial functions on the unit sphere $S^{n-1}$ and the decomposition $(**)$ becomes the decomposition of the polynomial functions on the sphere into irreducible representations of $O(n),$ each of which occurs with multiplicity one. These functions are frequently called spherical harmonics.

Addendum Here is a high level view of polynomial spherical harmonics from a representation theory vantage point. The decomposition $(*)$ is related to the representation theory of the Lie algebra $\mathfrak{sl}_2.$ The operators 
$$E=r^2/2,\ F=-\Delta/2,\ H=\deg+n/2$$ 
on the vector space of polynomials in $n$ variables commute with orthogonal transformations and form a representation of $\mathfrak{sl}_2.$ (I've described the skew-symmetric analogue in this answer.) Homogeneous harmonic polynomials are precisely lowest weight vectors for $\mathfrak{sl}_2,$ and the second key theorem amounts to saying that the lowest weight spaces are irreducible representations of $O(n).$ One consequence of this description is that the coefficients $h_k(x)$ in $(*)$ can be found inductively starting with $h_d(x)$ using repeated applications of the Laplace operator (the precise statement is omitted due to bulky notation). Furthermore, the first and second key theorems can be combined into the statement that the space $\mathcal{P}$ of polynomials in $n$ variables has the following decomposition into irreducible components under the joint $O(n)$ and $\mathfrak{sl}_2$ actions: 
$$\mathcal{P}=\bigoplus_{\ell\geq 0} \mathcal{H}_\ell \otimes V_\ell, $$
where $V_\ell$ is the lowest weight $\mathfrak{sl}_2$-module with lowest weight $\ell+n/2.$
This is one of the starting points of Roger Howe's theory of reductive dual pairs. In the present case, the reductive dual pair is $(O(n),SL(2,\mathbb{R}))$ over the real numbers. 
Remark The group $SL(2)$ is secretely a symplectic group, in fact, the isometry group of a 2-dimensional symplectic vector space, and the theory extends to reductive dual pairs consisting of an orthogonal group and a symplectic group of any rank. The same theory over $\mathbb{Q}$ and its adeles is behind some classical results in the theory of theta functions due to Siegel and Weil. In particular, it explains the importance of theta functions with harmonic coefficients.<|endoftext|>
TITLE: How large (small) can be the measure of a set where a polynomial takes small values ?
QUESTION [7 upvotes]: A $n$-th degree polynomial has precisely $n$ roots. So it is natural to ask the question
how large ( and small) can be the measure of a set where a polynomial takes small values ?
This, and other interesting variation of this must have been studied in depth. 
I would really appreciate any reference to the relevant literature. 
Also, if there are some interesting variation of this problem I would like to know. 
Thank you.

REPLY [2 votes]: Dear Vagabond,
I think a useful reference containing a lemma in the direction of your question is:
http://www.ams.org/mathscinet-getitem?mr=1652916 (see in particular proposition 3.2)
In this paper, Kleinbock and Margulis show the following result:
$\lambda(\{x\in I: |f(x)|<\varepsilon\})\leq 2k (k+1)^{1/k} (\varepsilon/\|f\|_I)^{1/k} \lambda(I)$
Here $\lambda$ is the Lebesgue measure, $f$ is a polynomial of degree $k$ and $\|f\|_I$ is the supremum of $f$ on $I$.
Best,
Matheus<|endoftext|>
TITLE: Structure theorems for Turing-decidable languages?
QUESTION [16 upvotes]: Languages decidable by weak models of computation often have certain necessary characteristics, e.g. the pumping lemma for regular languages or the pumping lemma for context-free languages.  Such characteristics are useful for determining when a language cannot be decided by that weak model of computation.

Do Turing-decidable languages have any such necessary characteristics?  If not, why not?

Such a characteristic should avoid mention of Turing machines.
Motivation:  The only way I know to show that a language is not Turing-decidable is diagonalization, or reducing to a language where diagonalization has already been applied.  It would be nice to have another method.
I suspect that there aren't any nice characteristics of this type known, or they'd probably be very well-known; but an argument for why they can't exist could be illuminating.
I'd also be interested in the same question with the phrase "Turing-decidable" replaced with any complexity class, e.g. NP.  This question seems more tractable (as there are fewer languages under examination).
Edit:  Given the comments, I'll be a bit more precise about what I mean by "characteristics."  I would be thrilled with either of the following two possibilities:

1) A property that allowed one to show a language was not in the relevant class (Turing-decidable, NP, whatever) without using diagonalization.

Or

2)  A property that can be stated in a language too weak to define Turing machines.

REPLY [5 votes]: It should be noted that the idea of finding a "property" that can be used to show non-membership in a certain class of languages was a driving force in computability theory in the 1950's or so.  Specifically, Emil Post posed the question: are there computably enumerable (aka "c.e.", also known as Turing-recognizable (but not necessarily decidable)) sets that are undecidable and strictly Turing-below the Halting problem (note that any c.e. set is Turing-below the halting problem).  That is, we want a c.e. set A such that $\emptyset <_T A <_T H$, where $H$ is the halting problem.  This became known as Post's Problem.  Further, he proposed Post's Program, which was to find a property of c.e. sets that implied the required conditions, and then to find an A with this property.  Examples of such candidate properties are being "simple" (its a technical term).  
Post's problem was solved in the positive by Friedberg and Muchnik, but by using a clever sort of delayed diagonalization, a sort of "injury argument".  However, this did not show that Post's program could be solved in the positive, but indeed Harrington and Soare showed in 1991 that there is such a property satisfying Post's conditions (and a corresponding set A satisfying this property).
See the wikipedia article: http://en.wikipedia.org/wiki/Turing_degree#Post.27s_problem as well as the Harrington-Soare article explaining some of the background (its open-access): http://www.ncbi.nlm.nih.gov/pmc/articles/PMC52904/<|endoftext|>
TITLE: Replacing Spectrum with Valuations of a Field - An Alternative to Schemes?
QUESTION [12 upvotes]: A scheme is defined to be a sheaf which is locally isomorphic to the spectrum of a ring. The idea behind this is that given an affine coordinate ring of a variety over an algebraically closed field, we can recover the variety, i.e. the geometric object, by looking at the maximal ideals of this affine coordinate ring. Including the prime ideals (which add in the irreducible subvarieties), we get the notion of a scheme, which is something which is gotten essentially from the spectrum of a ring.
Another way to recover a variety from the algebra associated with it is to consider the valuations of its function field. Specifically, the points of a non-singular complete variety correspond to the valuations on the function field of the variety. We can actually define the variety as the set of valuations. If $K$ is the field and $v(K)$ denotes the set of valuations on $K$, then we declare $\{v \in v(K) \mid v(x) > 0\}$ for each $x \in K$ to be closed, giving a topology on the set of valuations. Finally, we can define the local ring at each point to be the valuation ring for that valuation. My question is, what if, instead of looking at spectra of rings, we defined a new object, which is locally the set of valuations of a field? For Dedekind rings, these seems to give something similar to the spectrum of the given Dedekind domain. Is this interesting in other contexts? Can one gain something by looking at it from this perspective?
Edit: Although valuations do not give varieties up to isomorphism, our new object could still be something along the lines of "variety up to birational equivalence."

REPLY [13 votes]: This is an old approach to finding models for varieties, introduced by Zariski in  1944 in his work on resolution of singularities. (See "The compactness of the Riemann manifold of an abstract field of algebraic functions", Bulletin of the American Mathematical Society 50: 683–691, doi:10.1090/S0002-9904-1944-08206-2, MR0011573)
He defined a Zariski topology on a space of valuations, which seems to have inspired Grothendieck's definition of Zariski topology on a scheme. Much of Zariski's work on these spaces is rather similar to Grothedieck's work on the foundations of schemes. 
Zariski called the space of valuations the "Riemann manifold" of a variety, though it is now called the Zariski-Riemann space. 
Volume 2, chapter VI section 17 of Zariski and Samuel's book on commutative algebra gives more details.

REPLY [7 votes]: This "space of valuations" to me sounds like the Riemann-Zariski space. Generally this should be pretty nasty, for surfaces eg it is some kind of limit of the system of all possible blow-ups. It is an old idea.
For a different direction towards compactifying $Spec(\mathbb{Z})$, you can read about Arakelov theory and look at the recent papers of Connes and Consani on the arXiv about geometry over the "field" with one element.<|endoftext|>
TITLE: Is there a non-projective flat module over a local ring?
QUESTION [5 upvotes]: Is there a non-projective flat module over a local ring?
Here I assume the ring is commutative with unit.

REPLY [6 votes]: It is related to Bass' theorem. Flat modules are projective iff the ring is perfect. $p$-adic integers or formal power series are examples of local rings which are not perfect and have nonprojective flat modules.<|endoftext|>
TITLE: Looking for references talking about category of topological vector spaces
QUESTION [6 upvotes]: It's known that category of topological vector spaces is not abelian but quasi-abelian or exact category. I am looking for the references playing with this category(category theory). All the related references are welcome which means
The abstract work on quasi abelian category is ok, although I am more interested in some concrete examples, some special topological vector spaces. 
I realized that it seems that p-adic analysis and p-adic representation theory has something to do with this kind of stuff, however, I do not know anything about this area. 
I am not sure whether Grothendieck's work on topological vector spaces writing about something on this category. Somebody told me that Grothendieck formulated the algebraic inductive limits and many other categorical constructions in his book. I have not checked out this book to see.
My motivation for asking this question is that I am considering unitary representation of compact group which are Hilbert spaces. I want to consider category of unitary representations of this group and formulate some categorical construction, such as limits, colimits and so on so forth which might be helpful to study representation theory. 
I noted that there are some work by Fabienne Prosmans on derived category and functional analysis. 
Thanks!

REPLY [3 votes]: Some of the crucial ideas and results on quasi-abelian categories in the context of topological vector spaces are already contained in the work of Lucien Waelbroeck (see here) which predates the work of Schneiders and Prosmans considerably. Waelbroeck's book contains also a lot of examples.<|endoftext|>
TITLE: Lifting units from modulus n to modulus mn.
QUESTION [9 upvotes]: Background
In his beautifully short answer to a previous question of mine, Robin Chapman asserted the following.

Let $m,n,r$ be natural numbers with $r$ coprime to $n$.  Then there is $r' \equiv r \mod n$ which is coprime to 
  $mn$.

Letting $C_n$ denote the cyclic group of order $n$, the above statement is equivalent to this:

Every automorphism of $C_n$ lifts to an automorphism of $C_{nm}$ for all $m$.

Since that is the context of the question I asked, I thought that this fact ought to have an elementary group-theoretical derivation, but alas I have been unable to find one.  I asked a number theorist colleague of mine and he gave me this "sledgehammer proof" (his words):
Since $r$ is coprime to $n$, the arithmetic progression $r + kn$ for $k=1,2,\ldots$ contains an infinite number of primes (by a theorem of Dirichlet's).  Since only a finite number of those primes can divide $m$, there is some $k$ for which $r'= r+kn$ is a prime which does not divide $m$, and hence neither does it divide $nm$.
Question
Is there an elementary (and preferably group-theoretical) proof of this result?

REPLY [5 votes]: There is an explicit formula for $r'$, and no need to invoke the Chinese Remainder Theorem; $$r'=r+kn{\rm\ where\ }k=\prod_{p\mid m,p\nmid r}p.$$ We need to show that $r'$ is relatively prime to $mn$. It suffices to show that $r'$ is relatively prime to $m$. Let $p$ be a prime dividing $m$. If $p$ divides $r$, then it doesn't divide $n$ (since $r$ and $n$ are relatively prime), and it doesn't divide $k$ (by construction of $k$), so it doesn't divide $kn$, so it doesn't divide $r'$. If $p$ doesn't divide $r$, then, by construction, it divides $k$, so it divides $kn$, so it doesn't divide $r'$. 
Schinzel showed me this construction 35 years ago.<|endoftext|>
TITLE: Selberg's advisor?
QUESTION [11 upvotes]: Does anybody know who was Atle Selberg's advisor?
I find it interesting to know the advisor's impact on his students.
Unfortunately, in Selberg case, this information (even his advisor's name) seems to be nowhere to be found.

REPLY [26 votes]: I'm a student at the university of Oslo, so I thought I'd have a go at this. I just talked to Erling Størmer (Carl's grandson) who is a professor emeritus here. He said that in practice Atle had no advisor. Of course someone must have signed the papers but he doesn't know who (I don't really see what difference it makes anyway). Erling told me that according to Atle the reason why so many Norwegian mathematicians at the time worked in number theory is that they were all self-taught, and number theory is more accessible to the autodidact.<|endoftext|>
TITLE: A few questions about Kontsevich formality
QUESTION [21 upvotes]: [K] refers to Kontsevich's paper "Deformation quantization of Poisson manifolds, I".
Background
Let $X$ be a smooth affine variety (over $\mathbb{C}$ or maybe a field of characteristic zero) or resp. a smooth (compact?) real manifold. Let $A = \Gamma(X; \mathcal{O}_X)$ or resp. $C^\infty(X)$.
Denote the dg Lie algebra of polyvector fields on $X$ (with Schouten-Nijenhuis bracket and zero differential) by $T$. Denote the dg Lie algebra of the shifted Hochschild cochain complex of $A$ (with Gerstenhaber bracket and Hochschild differential) by $D$.
Then the Hochschild-Konstant-Rosenberg theorem states that there is a quasi-isomorphism of dg vector spaces from $T$ to $D$. However, the HKR map is not a map of dg Lie algebras. It is not a map of dg algebras, either (where the multiplication on $T$ is given by the wedge product and the multiplication on $D$ is given by the cup product of Hochschild cochains).
I believe "Kontsevich formality" refers to the statement that, while the HKR map is not a quasi-isomorphism --- or even a morphism --- of dg Lie algebras, there is an $L_\infty$ quasi-isomorphism $U$ from $T$ to $D$, and therefore $D$ is in fact formal as a dg Lie algebra.
The first "Taylor coefficient" of the $L_\infty$ morphism $U$ is precisely the HKR map (see section 4.6.2 of [K]).
Moreover, this quasi-isomorphism $U$ is compatible with the dg algebra structures on $T$ and $D$ (see section 8.2 of [K]), and it yields a "corrected HKR map" which is a dg algebra quasi-isomorphism. The "correction" comes from the square root of the $\hat{A}$ class of $X$. See this previous MO question.
Questions
(0) Are all of my statements above correct?
(1) In what way is the $L_\infty$ morphism $U$ compatible with the dg algebra structures? I don't understand what this means.
(2) When $X$ is a smooth (compact?) real manifold, I think that all of the statements above are proved in [K]. When $X$ is a smooth affine variety, I think that the statements should all still be true. Where can I find proofs?
(3) Moreover, the last section of [K] suggests that the statements are all still true when $X$ is a smooth possibly non-affine variety. For a general smooth variety, though, instead of taking the Hochschild cochain complex of $A = \Gamma(X;\mathcal{O}_X)$, presumably we should take the Hochschild cochain complex of the (dg?) derived category of $X$. Is this correct? If so, where can I find proofs?
In the second-to-last sentence of [K], Kontsevich seems to claim that the statements for varieties are corollaries of the statements for real manifolds, but I don't see how this can possibly be true. In the last sentence of the paper, he says that he will prove these statements "in the next paper", but I'm not sure which paper "the next paper" is, nor am I even sure that it exists, since "Deformation quantization of Poisson manifolds, II" doesn't exist.
P.S. I am not sure how to tag this question. Feel free to tag it as you wish.

REPLY [5 votes]: Here are answers to questions 2, 3 and 1 (in this order). 
2) Let $X = \operatorname{Spec} A$ be a smooth affine variety over a field $K$ of characteristic $0$. Then there is a canonical bijection between Poisson deformations of $A$ and associative deformations of $A$ (all up to gauge equivalence). This was first proved (in slightly greater generality, yet assuming $K$ contains the real numbers) in my paper

Deformation Quantization in Algebraic Geometry, Amnon Yekutieli, 
  Advances in Mathematics 198 (2005), 383-432. Erratum: Advances in Mathematics  217 (2008), 2897-2906.

The condition about real numbers is not essential for the proof; all that is needed is a Formality Theorem for the power series ring 
$K[[t_1, \ldots, t_n]]$ satifying the invariance properties of Kontsevich. 
It should be noted that the proof above is of a global nature. A local argument can only be made on an affine scheme $X$ that admits etale coordinates (namely an etale map to $\mathbf{A}^n_K$). Bigger affine schemes (that do not admit such coordinates) require a gluing argument, based on formal geometry, and the vanishing of some cohomological obstructions.  
There are other later proofs of this same result (listed in another answer). I am only familiar with the proof by Van den Bergh (which is a variation on my proof).   
3) If $X$ is an arbitrary smooth scheme (over a field $K$ of characteristic $0$), then quantization is possible, but only in a stacky sense. This was suggested by Kontsevich in his paper 

M. Kontsevich, Deformation quantization of algebraic varieties, Lett. Math. Phys. 56 (2001), no. 3, 271-294.

The precise statement, and its proof, are in the paper (still not published!) 

[TDQ] Amnon Yekutieli, Twisted Deformation Quantization of Algebraic Varieties, http://arxiv.org/abs/0905.0488 .

In this more general situation one has to consider twisted (i.e. stacky) Poisson deformations of the sheaf $\mathcal{O}_X$, and twisted associative deformations of the sheaf $\mathcal{O}_X$. 
A twisted associative deformation $\mathcal{A}$ is a refined sort of stack of algebroids. The category of coherent left $\mathcal{A}$-modules is a deformation (as a stack of abelian categories, in the sense studied by Lowen and Van den Bergh) of the category of coherent $\mathcal{O}_X$-modules. This also goes in reverse: $\mathcal{A}$ can be recovered from its module category (basically by Morita theory). 
A twisted Poisson deformation is a new kind of algebro-geometric object. There is no Morita theory connected to it. 
Theorem 0.10.1 of [TDQ] states that there is a canonical bijection, called the twisted quantization map, between twisted Poisson deformations of $\mathcal{O}_X$  and twisted associative deformations of $\mathcal{O}_X$ (up to twisted gauge equivalence). 
There are lecture notes on this work: 
http://www.math.bgu.ac.il/~amyekut/lectures/twisted-defs-2013/notes.pdf 
and a survey article: 
http://www.math.bgu.ac.il/~amyekut/publications/tw-defs-surv/tw-defs-surv.html
The main paper [TDQ] relies on several auxiliary papers, including:

Central Extensions of Gerbes, Amnon Yekutieli, 
  Advances in Mathematics 225, Issue 1 (2010), 445-486.
MC Elements in Pronilpotent DG Lie Algebras,
  Amnon Yekutieli, J. Pure Appl. Algebra 216 (2012), 2338–2360
Deformations of Affine Varieties and the Deligne Crossed Groupoid,
  Amnon Yekutieli, Journal of Algebra 382 (2013), 115–143.
Combinatorial Descent Data for Gerbes,
  Amnon Yekutieli, to appear in J. Noncommutative Geometry. Eprint arXiv:1109.1919 at http://arxiv.org. 

There are several intriguing open questions in this subject. For instance Question 0.10.2 of [TDQ] regarding twisted deformation quantization of Calabi-Yau surfaces. 
1) The compatibility of the Kontsevich formality morphism with the associative DG algebra structures on polyvector fields and polydifferential operators is made precise and proved in the paper 

Hochschild cohomology and Atiyah classes,
  Damien Calaque and Michel Van den Bergh,
  Advances in Mathematics 224 (2010) 1839–1889. 

As I recall, they prove that the universal formality morphism is (in additions to being L_infty) an A_2 morphism (namely A_infty truncated at level 2). This is enough to deduce the formula for twisting the HKR morphism (by the square root of the Todd class) in order to respect the multiplications.<|endoftext|>
TITLE: Does anyone know a polynomial whose lack of roots can't be proved?
QUESTION [61 upvotes]: In Ebbinghaus-Flum-Thomas's Introduction to Mathematical Logic, the following  assertion is made:
If ZFC is consistent, then one can obtain a  polynomial $P(x_1, ..., x_n)$ which has no roots in the integers. However, this cannot be proved (within ZFC).
So if $P$ has no roots, then mathematics (=ZFC, for now) cannot prove it.
The justification is that Matiyasevich's solution to Hilbert's tenth problem allows one to turn statements about provable truths in a formal system to the existence of integer roots to polynomial equations.  The statement is "ZFC is consistent," which cannot be proved within ZFC thanks to Gödel's theorem.
Question: Has such a polynomial ever been computed?
(This arose in a comment thread on the beta site math.SE.)

REPLY [32 votes]: For every consistent recursively axiomatizable theory $T$ there exists (and can be effectively computed from the axioms of $T$) an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this fact cannot be proved in $T$:
\begin{align}&(elg^2 + \alpha - bq^2)^2 + (q - b^{5^{60}})^2 + (\lambda + q^4 - 1 - \lambda b^5)^2 + \\
&(\theta + 2z - b^5)^2 + (u + t \theta - l)^2 + (y + m \theta - e)^2 + (n - q^{16})^2 + \\
&[(g + eq^3 + lq^5 + (2(e - z \lambda)(1 + g)^4 + \lambda b^5 + \lambda b^5 q^4)q^4)(n^2 - n) + \\
&\quad\quad(q^3 - bl + l + \theta \lambda q^3 + (b^5-2)q^5)(n^2 - 1) - r]^2 + \\
&(p - 2w s^2 r^2 n^2)^2 + (p^2 k^2 - k^2 + 1 - \tau^2)^2 + \\
&(4(c - ksn^2)^2 + \eta - k^2)^2 + (r + 1 + hp - h - k)^2 + \\
&(a - (wn^2 + 1)rsn^2)^2 + (2r + 1 + \phi - c)^2 + \\
&(bw + ca - 2c + 4\alpha \gamma - 5\gamma - d)^2 + \\
&((a^2 - 1)c^2 + 1 - d^2)^2 + ((a^2 - 1)i^2c^4 + 1 - f^2)^2 + \\
&(((a + f^2(d^2 - a))^2 - 1) (2r + 1 + jc)^2 + 1 - (d + of)^2)^2 + \\
&(((z+u+y)^2+u)^2 + y-K)^2 = 0.
\end{align}
Moreover, for every such theory, the set of numbers with this property is infinite and not recursively enumerable.
The equation is derived from Undecidable Diophantine Equations, James P. Jones, Bull. Amer. Math. Soc. (N.S.), Vol. 3(2), Sep. 1980, pp. 859–862, DOI: 10.1090/S0273-0979-1980-14832-6.<|endoftext|>
TITLE: Cycles of length 1(mod 3) in regular graphs
QUESTION [10 upvotes]: Does every 4-regular graph contain a cycle of length (number of edges in the cycle) $1 (\mod3)$? Are there only finitely many exceptions?
I suspect such cycles exist for most 3-regular graphs but 4-regularity is enough for what I'm investigating.

REPLY [7 votes]: According to this paper, N. Dean et al. have shown that if a simple graph G

is 2-connected,
has minimum degree at least 3, and
is not isomorphic to the Petersen graph,

then G contains a cycle of length 1 mod 3.
Now consider any 4-regular graph H. If I'm not mistaken, it's fairly easy to show that H contains a subgraph G such that two nodes of G have degree 3, all other nodes of G have degree 4, and G is 2-connected. Clearly G can't be the Petersen graph, and thus the above theorem implies that G (and therefore also H) contains a cycle of length 1 mod 3.
Edit: Here is how would construct G given H. Recall that H has a 2-factorisation, i.e., it consists of cycles such that each node of H is "covered" by exactly 2 cycles. If there are no articulation points in H, the claim is clear. Otherwise consider the block tree of H and pick a biconnected component K that contains only one articulation point; let x be the articulation point. Consider the subgraph K' = K − x; this graph consists of one path P and a set of cycles, and each node of K' is covered by either P and one cycle or exactly 2 cycles. It may be the case that K' contains articulation points. However, each articulation point is an endpoint of a bridge, and each bridge is an edge of P. Remove all bridges and let G be one of the connected components in the resulting bridgeless graph.<|endoftext|>
TITLE: Three questions on large simple groups and model theory
QUESTION [21 upvotes]: Yesterday, in the short course on model theory I am currently teaching, I gave the following nice application of downward Lowenheim-Skolem which I found in W. Hodges A Shorter Model Theory:
Thm: Let $G$ be an infinite simple group, and let $\kappa$ be an infinite cardinal with $\kappa \leq |G|$.  Then there exists a simple subgroup $H \subset G$ with $|H| = \kappa$. 
(The proof, which is short but rather clever, is reproduced on p. 10 of http://www.math.uga.edu/~pete/modeltheory2010Chapter2.pdf.)
This example led both the students and I (and, course mechanics aside, I am certainly still a student of model theory) to ask some questions:
$1$. The theorem is certainly striking, but to guarantee content we need to see an uncountable simple group without, say, an obvious countable simple subgroup.  I don't know that many uncountable simple groups.  The most familiar examples are linear algebraic groups like $\operatorname{PSL}_n(F)$ for $F$ an uncountable field like $\mathbb{R}$ or $\mathbb{C}$.  But this doesn't help, an infinite field has infinite subfields of all infinite cardinalities -- as one does not need Lowenheim-Skolem to see!  (I also mentioned the case of a simple Lie group with trivial center, although how different this is from the previous example I'm not sure.)  The one good example I know is supplied by the Schreier-Ulam-Baer theorem: let $X$ be an infinite set.  Then the quotient of $\operatorname{Sym}(X)$ by the normal subgroup of all permutations moving less than $|X|$ elements is a simple group of cardinality $2^{|X|}$.  (Hmm -- at least it is when $X$ is countably infinite.  I'm getting a little nervous about the cardinality of the normal subgroup in the general case.  Maybe I want an inaccessible cardinal or somesuch, but I'm getting a little out of my depth.)  So:

Are there there other nice examples of uncountable simple groups?

$2$. At the beginning of the proof of the theorem, I remarked that straightforward application of Lowenheim-Skolem to produce a subgroup $H$ of cardinality $\kappa$ which is elementarily embedded in $G$ is not enough, because it is not clear whether the class of simple groups, or its negation, is elementary.  Afterwards I wrote this on a sideboard as a question:

Is the class of simple groups (or the class of nonsimple groups) an elementary class?

Someone asked me what techniques one could apply to try to answer a problem like this.  Good question!
$3$. The way I stated Hodges' result above is the way it is in my lecture notes.  But when I wrote it on the board, for no particular reason I decided to write $\kappa < |G|$ instead of $\kappa \leq |G|$.  I got asked about this, and was ready with my defense: $G$ itself is a simple subgroup of $G$ of cardinality $|G|$.  But then we mutually remarked that in the case of $\kappa = |G|$ we could ask for a proper simple subgroup $H$ of $G$ of cardinality $|G|$.  My response was: well, let's see whether the proof gives us this stronger result.  It doesn't.  Thus:

Let $G$ be an infinite simple group.  Must there exist a proper simple subgroup $H$ of $G$ with $|H| = |G|$?

Wait, I just remembered about the existence of Tarski monsters.  So the answer is no.  But what if we require $G$ to be uncountable?

REPLY [9 votes]: Another rather nice example: the group of automorphisms of $[0,1]$ endowed with the Lebesgue measure (i.e, bi-measurable maps preserving the measure, identified when they coincide outside a set of Lebesgue measure zero). In a related vein, the full group of any countable, ergodic, probability measure preserving Borel equivalence relation on $[0,1]$ is simple.

REPLY [3 votes]: More uncountable simple groups:  The automorphism group of the random graph is of size $2^{\aleph_0}$ and simple (Truss), the same holds for the automorphism group of the unique countable atomless Boolean algebra, i.e., the group of auto-homeomorphisms of the Cantor space $\{0,1\}^{\mathbb N}$.<|endoftext|>
TITLE: Alexander polynomial or Reidemeister torsion for knotted surfaces?
QUESTION [10 upvotes]: An important invariant of a knot in $S^3$ is its Alexander polynomial, related also to Reidemeister torsion. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?

REPLY [7 votes]: Yes for $S^2$, and more generally, depending on what you are after.
Given any torsion module $M$ over the PID $Q[t,t^{-1}]$, the order of $M$ is well defined in 
$Q[t, t^{-1}]$ up to units, in the usual way.   Moreover $M\otimes Q(t)=0$. These two facts are at the heart of why the Alexander polynomial is related to Reidemeister torsion.  The way it works is that if $C$ is a f.g. chain complex over $Q[t, t^{-1}]$ its homology is torsion iff  $C\otimes Q(t)$ is acyclic. Then the two notions of torsion are related by 
$$\prod_k order(H_k(C))^{(-1)^k}=\tau(C)$$
If  $S^n\subset S^{n+2}$ and $X$ its complement, then $X$ has a $Z$ cover $\tilde{X}$ and hence an exact sequence of chain complexes  $0\to C(\tilde{X})\to C(\tilde{X})\to C(X)\to 0.$ The corresponding long exact sequence shows that $H_k(\tilde{X})$ is torsion for all $k$, since $H_k(X)=H_k(S^1)$.  
Setting  $\Delta_k$ to be the order of $H_k(\tilde{X})$ gives the $k$-th Alexander polynomial. You get Reidemeister torsion equals the multiplicative Euler characteristic: 
$$\tau= \prod \Delta_k^{(-1)^k}$$
In the case of $S^1\subset S^3$ Poincare duality relates $\Delta_2$ and $\Delta_1$, and $\Delta_0$ is independent of the knot, so you recover Alexander poly "=" Reidemeister torsion. But in general $\tau$  combines all the $\Delta_k$.
All this (including how to pick the basis of the acyclic complex) is explained in Milnor's article "infinite cyclic covers."
As far as what happens more generally, all this extends, but you have to be careful when the homology is not torsion, which can happen for surfaces in $S^4$. Then you have to view Reidemeister torsion as a function of the homology , etc. Milnor's other Torsion articles are worthwhile reading for these topics.<|endoftext|>
TITLE: Where is it shown how to construct a decomposition tree for a series-parallel graph in linear time?
QUESTION [6 upvotes]: There are two common ways to define a series-parallel graph (or 2-terminal series-parallel graph).
Definition 1

start with K_2 marking both vertices as terminals
repeatedly join two smaller 2-terminal s/p graphs either in series or in parallel 

Definition 2

start with K_2
repeatedly replace a single edge by two in series or two in parallel

A "decomposition tree" for an s/p graph shows how it is constructed according to Definition 1; each node of the tree is a s/p graph and the children of each node are the components from which that graph was built by series or parallel composition.
It is well known that a series-parallel graph can be recognized in linear time; the usual reference to this is Valdes, Tarjan and Lawler (Valdes, Jacobo; Tarjan, Robert E.; Lawler, Eugene L. The recognition of series parallel digraphs.  SIAM J. Comput.  11  (1982), no. 2, 298--313.)
It is also frequently stated in the literature that the decomposition tree can be found in linear time, either just as an assertion or with a reference to the same Valdes/Tarjan/Lawler paper.
However, when you actually read Valdes, Tarjan and Lawler, they do not construct the decomposition tree in linear time, but rather they run "Definition 2" in reverse and work on reducing the graph to a single edge by series and parallel reductions. So they recognize that the graph is s/p but they do not actually give the decomposition tree. 
Does anyone know if there is an explicit reference in the literature to actually constructing the sp-tree for a series-parallel graph in linear time?

REPLY [2 votes]: The answer can be found in Bodlaender, Fluiter 1996 "Parallel algorithms for series parallel graphs". They construct an sp-tree from an input graph if it is series-parallel.
[I know the question is three years old, but if that answer had been posted on this page, this would have saved me a lot of work. That's why I found it reasonable to answer it here, even now.]<|endoftext|>
TITLE: A q-analogue of Ramanujan's tau function
QUESTION [8 upvotes]: There have been a couple of questions on Ramanujan's $\tau$ function.
Lehmer's conjecture for Ramanujan's tau function
The Vanishing of Ramanujan's Function tau(n)
A $q$-analogue is given by
$$ \tau_q(n) = \sum_{|\lambda|=n}\prod_{(i,j)}
\frac{[5-h(i,j)][5+h(i,j)]}{[h(i,j)]^2} $$
Here $[k]=\frac{q^k-q^{-k}}{q-q^{-1}}$. The sum is over all partitions of $n$.
The product is over all boxes in the diagram of $\lambda$ and $h(i,j)$ is the hook length of the box $(i,j)$.
This is a $q$-analogue because (despite appearances) it is a Laurent polynomial in $q$ and substituting $q=1$ gives $\tau(n)$.
It would be ridiculous to ask if this is $q$-holonomic.
Is this $q$-analogue known? and does it have any significance in number theory?

REPLY [4 votes]: I am not very well familiar with the notion "$q$-holonomic" but learned it quite recently from this question. If the sequence $\tau_q(n)$ were $q$-holonomic, then $\tau(n)$  would be holonomic. The latter means that $$\Delta(x)=x\prod_{n=1}^\infty(1-x^n)^{24}=\sum_{k=1}^\infty\tau(k)x^k$$ satisfies a linear differential equation with rational function coefficients. However, this is known to be false: the function $\Delta(x)$ has singularities at all roots of unity.
Note that the function $\Delta(x)$ satisfies an algebraic differential equation with constant coefficients; the result due to Halphen (and independently, although later, by Ramanujan).<|endoftext|>
TITLE: How do proof verifiers work?
QUESTION [18 upvotes]: I'm currently trying to understand the concepts and theory behind some of the common proof verifiers out there, but am not quite sure on the exact nature and construction of the sort of systems/proof calculi they use. Are they essentially based on higher-order logics that use Henkin semantics, or is there something more to it? As I understand, extending Henkin semantics to higher-order logic does not render the formal system any less sound, though I am not too clear on that.
Though I'm mainly looking for a general answer with useful examples, here are a few specific questions:

What exactly is the role of type theory in creating higher-order logics? Same goes with category theory/model theory, which I believe is an alternative.
Is extending a) natural deduction, b) sequent calculus, or c) some other formal system the best way to go for creating higher order logics?
Where does typed lambda calculus come into proof verification?
Are there any other approaches than higher order logic to proof verification?
What are the limitations/shortcomings of existing proof verification systems (see below)?

The Wikipedia pages on proof verification programs such as HOL Light Coq, and Metamath give some idea, but these pages contain limited/unclear information, and there are rather few specific high-level resources elsewhere. There are so many variations on formal logics/systems used in proof theory that I'm not sure quite what the base ideas of these systems are - what is required or optimal and what is open to experimentation.
Perhaps a good way of answering this, certainly one I would appreciate, would be a brief guide (albeit with some technical detail/specifics) on how one might go about generating a complete proof calculus (proof verification system) from scratch? Any other information in the form of explanations and examples would be great too, however.

REPLY [4 votes]: You ask how proof verifiers work, but it seems you are asking a somewhat more precise question: how do proof verifiers based on various higher-order logics work?  If you are interested only in classical first-order logic (or less, e.g., propositional logic), then you don't need to worry about higher-order issues. (There are important caveats.  For example, the expressiveness of plain first-order logic has limits that you might consider unwelcome for various purposes, such as its inability to capture, e.g., reachability in a graph.)  The Mizar system, for example, is based on classical first-order logic plus (a rather strong) set theory.<|endoftext|>
TITLE: Higher dimensional berkovich spaces
QUESTION [22 upvotes]: I am looking for a geometric and topological way to make a visualization of higher dimensional berkovich spaces, starting with the berkovich plane. Of course, this is just a collection of bounded semi-norms, but the question remains: 

Is there a visualization possible for $\mathbb A^2_{\text{Berk}}$ like the infinite branched tree for $\mathbb A^1_{\text{Berk}}$? 

(For $\mathbb A^1_{\text{Berk}}$ see for example Baker and Rumely's Potential Theory and Dynamics on the Berkovich projective line, Chapters 1 - 2.)
I think you get a simplicial complex, but I don't know exactly how. On the one hand (reading Favre and Johnsson's The valuative tree), you have this list of valuations (thus, also of seminorms, allthough this book discusses seminorms on $\mathbb{C}^2$). On the other hand we have Berkovich's theory of Type I - Type IV points. I guess there just more Type I - Type IV points in a plane (i.e. more seminorms that occur as it were type I points), and some can be only represented by faces (two dimensional simplices), only I don't know how.
Are there any references on the visualization part?

REPLY [2 votes]: Since I don't seem to be able to edit my own question. I found s link which might be useful here:

http://users.math.yale.edu/~sp547/pdf/Anayltification-tropicalizations.pdf, in which a homeomorphism is constructed between $X^{an}$ (or $X_{Berk}$, if that is you cup of coffee) and an inverse limit of tropicalizations of embeddings of (toric) subvarieties of $X(K)$, where $K$ is the basefiel. Payne first takes the field $\mathbb{C}((t^{\mathbb{R}}))$ (in which, for example in $\mathbb{P}^1_{Berk,K}$ there are only Type 1 and Type 2 points), then switches to fields with trivial norms.<|endoftext|>
TITLE: Quaternary quadratic forms and Elliptic curves via Langlands?
QUESTION [15 upvotes]: The content of this note was the topic of a lecture by Günter Harder at the School on Automorphic Forms, Trieste 2000. The actual problem comes from the article 

A little bit of number theory by Langlands.
The problem is about a connection between two quite different objects. The first object
is the following pair of positive definite quadratic forms:
$$ P(x,y,u,v)  =  x^2 + xy + 3y^2 + u^2 + uv + 3v^2 $$
$$ Q(x,y,u,v)  =  2(x^2 + y^2 + u^2 + v^2) + 2xu + xv + yu - 2yv $$
The second object is the elliptic curve
$$ E: y^2 + y = x^3 - x^2 - 10x - 20. $$
To each of our objects we now associate a series of integers. 
For each integer $k \ge 0$ define
$$ n(P,k)  =  | \{(a,b,c,d) \in {\mathbb Z}^4: P(a,b,c,d) = k\} |, $$
$$ n(Q,k)  =  | \{(a,b,c,d) \in {\mathbb Z}^4: Q(a,b,c,d) = k\} |. $$
As a matter of fact, these integers are divisible by $4$ for any
$k \ge 1$ because of the transformations $(a,b,c,d) \to (-a,-b,-c,-d)$
and $(a,b,c,d) \to (c,d,a,b)$.
For any prime $p \ne 11$ we now put
$$ a_p =  |E({\mathbb F}_p)| - (p+1),$$
where $E({\mathbb F}_p)$ is the elliptic curve over ${\mathbb F}_p$ defined above. 
Then Langlands claims
For any prime $p \ne 11$, we have 
$ 4a_p = n(P,p) - n(Q,p).$
The "classical" explanation proceeds as follows: Given the series of integers $n(P,p)$ and $n(Q,p)$, we form the generating series
$$ \Theta_P(q)  = \sum \limits_{k=0}^\infty n(P,k) q^k 
         = 1 + 4q + 4q^2 + 8q^3 + \ldots, $$
$$ \Theta_Q(q)  = \sum \limits_{k=0}^\infty n(Q,k) q^k
         = 1 + 12q^2 + 12q^3 + \ldots. $$
If we put $q = e^{2\pi i z}$ for $z$ in the upper half plane, then $\Theta_P$ and $\Theta_Q$ become ${\mathbb Z}$-periodic holomorphic functions on the upper half plane. As a matter of fact, the classical theory of modular forms shows that the function 
$$ f(z)  =  \frac14 (\Theta_P(q) - \Theta_Q(q))  
         =  q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 - 2q^9 - 2q^{10}
  + q^{11} - 2q^{12} \ldots $$
is a modular form (in fact a cusp form since it vanishes at $\infty$) of weight $2$ for 
$\Gamma_0(11)$. More precisely, we have $ f(z) = \eta(z)^2 \eta(11z)^2,$ where $\eta(z)$ is Dedekind's eta function, a modular form of weight $\frac12$. 
Now we have connected the quadratic forms to a cusp form for $\Gamma_0(11)$. This group has two orbits on the projective line over the rationals, which means that the associated Riemann surface can be compactified by adding twocusps: the result is a compact Riemann surface $X_0(11)$ of genus $1$. Already Fricke has given a model for this Riemann surface: he found that $X_0(11) \simeq E$ for the elliptic curve defined above.
Now consider the space of cusp forms for $\Gamma_0(11)$. There are Hecke operators $T_p$ acting on it, and since it has dimension $1$, we must have $T_p f = \lambda_p f$ for
certain eigenvalues $\lambda_p \in {\mathbb Z}$. A classical result due to Hecke then predicts that the eigenvalue $\lambda_p$ is the $p$-th coefficient in the $q$-expansion of $f(z)$. Eichler-Shimura finally tells us that $\lambda_p = a_p$. Putting everything together gives Langlands' claim.
Way back then I asked Harder how all this follows from the general Langlands conjecture, and he replied that he did not know. Langlands himself said his examples came "from 16 of  Jacquet-Langlands". So here's my question:
 Does anyone here know how to dream up concrete results like the one above from Langlands' conjectures, or from "16 of Jacquet-Langlands"?

REPLY [11 votes]: I would like to add to Emerton's answer that Eichler gave a proof for square free level in Modular functions of one variable I, Springer Lecture Notes 320.
Further classical work on the basis problem is in:
MR0960090 (90d:11056) Hijikata, Hiroaki; Pizer, Arnold K.; Shemanske, Thomas R. The basis problem for modular forms on $\Gamma_0(N)$. Mem. Amer. Math. Soc. 82 (1989), no. 418
Yet another reference:
MR0333081 (48 #11406) Shimizu, Hideo Theta series and automorphic forms on ${\rm GL}_{2}$. J. Math. Soc. Japan 24 (1972), 638--683<|endoftext|>
TITLE: Have any long-suspected irrational numbers turned out to be rational?
QUESTION [219 upvotes]: The history of proving numbers irrational is full of interesting stories, from the ancient proofs for $\sqrt{2}$, to Lambert's irrationality proof for $\pi$, to Roger Apéry's surprise demonstration that $\zeta(3)$ is irrational in 1979.
There are many numbers that seem to be waiting in the wings to have their irrationality status resolved. Famous examples are $\pi+e$, $2^e$, $\pi^{\sqrt 2}$, and the Euler–Mascheroni constant $\gamma$. Correct me if I'm wrong, but wouldn't most mathematicians find it a great deal more surprising if any of these numbers turned out to be rational rather than irrational?
Are there examples of numbers that, while their status was unknown, were "assumed" to be irrational, but eventually shown to be rational?

REPLY [2 votes]: Another constant that has "no business being rational" I think, although a bit elementary:  Choose a point at random in the unit disk $D=\left\{x^2+y^2\leq 1\right\}$.  Then the expected value $E$ of its distance to the origin is a rational number!  (click below for solution).

 $E=\frac{2}{3}$<|endoftext|>
TITLE: Slick ways to make annoying verifications
QUESTION [39 upvotes]: There are many times in mathematics that one needs to make verifications that are annoying and distract from the main point of the argument. Often, there are lemmas that can make this much easier, at least in many important cases. 
For instance, in topology, it can be quite annoying to verify directly from the definition that a particular quotient space is what you think it is, and not something else with the same underlying set. (In fact, I suspect that many topologists habitually skip this verification.) However, the following lemma can in many cases make this verification much simpler:

Lemma: If $X$ is compact, $Y$ is Hausdorff, and $f \colon X \to Y$ is surjective, then $f$ is a quotient map.

This lemma can be made more powerful using the fact that it suffices to show a map is locally a quotient map.
Another such difficulty is to verify that a category is abelian; if you go directly from the definition, there is an annoyingly long list of things to verify.  However, unless I am mistaken, once you have an abelian category $\mathcal{A}$, there are a number of other categories that are guaranteed to give other abelian categories.  These (I think) include the category of functors into $\mathcal{A}$ from a fixed other category, the category of sheaves in $\mathcal{A}$ on a (topological space? other category?) (assuming $\mathcal{A}$ is nice enough for this to make sense), and any full subcategory of $\mathcal{A}$ that is closed under 0, $\oplus$, kernels, and cokernels.  Using these in combination, together with the fact that $R$-mod is an abelian category for every ring $R$, I believe one can get to every abelian category used in Hartshorne.  (Note: I am not too confident in this example, so if someone wanted to elaborate this in an answer, it would be appreciated.)
EDIT: As is pointed out in the comments below, the category of $\mathcal{O}_X$-modules is not of this form.  (I came up with this example while writing the question, and did not think it through too carefully.)  Thus, I would doubly appreciate a good answer specifically addressing, "How do you show a category is abelian?"

Question: What are some more of these useful lemmas / collections of lemmas, and how are they used?

REPLY [14 votes]: To show that an object S has property P, first show that the collection of all objects satisfying P is closed under a bunch of operations, prove that certain very simple objects have property P, and show that S can be "decomposed" or "filtered" or somehow unscrewed into these simple objects using the operations mentioned above.
This is a sort of induction, and it is used all the time to turn annoying verifications into verifying that something is true for like... a point. Maybe a shorthand for this "slick method" would be "think like Grothendieck."
For lots of examples of this see any proof in Higher Topos Theory or Higher Algebra by Lurie.<|endoftext|>
TITLE: Can a reductive group act non-linearly on a vector group?
QUESTION [14 upvotes]: Let $k$ be a field; I'm going to discuss linear algebraic groups over $k$. The question I'll pose is only interesting when the characteristic is $p>0$.
1. Some motivation
A vector group is an algebraic group isomorphic (over $k$) to a product of (finitely many) copies of the additive group $\mathbf{G}_a$.  Let $U$ be any connected unipotent group over $k$. If $k$ is perfect, or if $U$ is $k$-split, then $U$ has a filtration 
$1 = U_0 \subset U_1 \subset \cdots \subset U_n = U$
where each $U_i$ is normal in $U$ and each $U_i/U_{i-1}$ is a vector group.  If $U$ is a normal subgroup of a linear group $G$, you can
arrange that each $U_i$ is invariant under conjugation by $G$.  This suggests that to study the group extension $$(*) \quad 1 \to U \to G \to G/U \to 1,$$ one might profitably study first the case where $U$ is a vector group.
Let $G$ be a linear group and suppose that the unipotent radical $R$
of $G$ is defined over $k$ and is $k$-split (each of these conditions
can fail in general; they always hold when $k$ is perfect). Then the
question of whether $(*)$ splits when $U=R$ is precisely the question
of whether $G$ has a Levi factor; cf. this question of Jim Humphreys.
2. Action on a vector group
Let $U$ be a vector group and suppose that the linear group $G$ acts
on $U$ by algebraic group automorphisms. 
The action of $G$ on $U$ determines an action of $G$ on
$\mathfrak{u}=\operatorname{Lie}(U)$.

Question (first approximation): Is there a $G$-equivariant
  isomorphism $\mathfrak{u} \to U$ (where the vector space
  $\mathfrak{u}$ is viewed as a vector group in the obvious fashion)?

I'll say that the action of $G$ on $U$ is linearizable if
there is such an equivariant isomorphism.  
Some remarks: If the action of $G$ on $U$ is linearizable, then $G$
centralizes the action of the multiplicative group $\mathbf{G}_m$ on
$U$ obtained by transport of structure from scalar multiplication on
$\mathfrak{u}$.  This $\mathbf{G}_m$-action determines a grading on
the algebra $k[U]$ of regular functions on $U$ which is stable
for the action of $G$ on $k[U]$.
3. Partial answers
Postive: If the characteristic of
$k$ is $0$, the above question has always an affirmative answer. (Use the exponential mapping $\mathfrak{u} \to U$; this is $G$ equivariant e.g. because there is a faithful linear representation of the semidirect product $G \ltimes U$).
Negative: If $G = \mathbf{G}_a$, the
question has a negative answer in char. $p>0$. Consider the action of
$G$ on $V = {\mathbf{G}_a}^2$ given by the rule $$x.(a,b) = (a +
xb^p,b).$$ 
The action of $G$ on $\operatorname{Lie}(V)$ is trivial, so there is no
$G$-equivariant isomorphism $\operatorname{Lie}(V) \to V$.
4. What I meant to ask.

Refined question: If $G$ is reductive, is the action of $G$ on a vector group $U$
  always linearizable?

An affirmative answer would mean that when $G/U$ is reductive, one can study the group extension $(*)$ using cohomology of linear representations of $G/U$. 
Note that are known examples even in char. 0 (Schwarz, Kraft,...) of 
actions of reductive $G$ on
affine space $A=\mathbf{A}^n$ which are not linear, but, at least in
char. 0, these actions don't respect a vector group structure on $A$.

REPLY [4 votes]: I somehow just realized that although I've now know the answer for a while, I had failed to give the answer to this question...
In fact, for (any) field of characteristic $p>0$, there are (plenty of) examples of reductive groups $G$ which act by group automorphisms on a vector group $U$ for which the action is not linear -- i.e. there is no $G$-equivariant isomorphism $\operatorname{Lie}(U) 
\simeq U$.
For an example, see < my preprint @ gmcninch.math.tufts.edu > (especially section 5). Given a non-split
extension
$$(\flat) \quad 0 \to W \to E \to V \to 0$$
of finite dimensional $G$-modules, one can view the extension as being defined by a suitable sort of cocycle. One can then "twist" this cocycle by Frobenius,
yielding a vector group $\tilde E$ on which $G$ acts by group automorphisms.
As a $G$-module, the Lie algebra $\operatorname{Lie}(\tilde E)$ is isomorphic
to $V \oplus W^{(1)}$, where the exponent on $W$ denotes the first Frobenius twist.
By construction,
there is a $G$-equivariant surjection $\pi:\tilde E \to V$, and the kernel
of $\pi$ identifies with $W^{(1)}$. 
But there is no $G$-equivariant homomorphism of algebraic groups $V \to \tilde E$ which is a section
to $\pi$, hence there can be no $G$-equivariant isomorphism between $\tilde E$ and
$\operatorname{Lie}(\tilde E)$.
Since there are plenty of non-split extensions $(\flat)$, there are plenty 
of examples...
On a more positive note, let $V$ be a vector group on which $G$ acts.
The main result of the preprint just mentioned shows that if $G$ is connected,
if the unipotent radical of $G$ is defined over the ground field, and if 
$\operatorname{Lie}(V)$ is a simple $G$-module, then the action of $G$ on $V$ is linear.
In particular, it follows (for $G$ as above) that if $G$ acts on a split unipotent
group $U$, there is a filtration of $U$ by closed $G$-stable subgroups such that
the quotients are vector groups on which $G$ acts linearly.<|endoftext|>
TITLE: How fast are a ruler and compass?
QUESTION [27 upvotes]: This may be more of a recreational mathematics question than a research question, but I have wondered about it for a while.  I hope it is not inappropriate for MO.
Consider the standard assumptions for ruler and compass constructions: We have an infinitely large sheet of paper, which we associate with the complex plane, that is initially blank aside from the points 0 and 1 being marked.  In addition we have an infinite ruler and a compass that can be stretched to an arbitrary length.
Let us define a move to be one of the two actions normally associated with a ruler and compass:

Use the ruler to draw the line defined by any two distinct points already marked on the paper.
Stretch the compass from any one marked point to another and draw the resulting circle.

Assume that all intersection points among lines and circles drawn by these operations are automatically marked on the paper.  
Now define $D(n)$ to be the maximum distance between any two marked points that can be constructed in this way with $n$ moves.
Questions:

Is anyone aware of results about the function $D(n)$ or something equivalent?
It is not difficult to prove $D(n) > 2^{2^{cn}}$ for some positive constant $c$ for sufficiently large $n$.  Can one do better?  If so, can one prove an upper bound on $D(n)$?

REPLY [4 votes]: Here is a construction showing that $D(n) \ge 2^{2^{n-O(1)}}$.  This shows that any constant $c<1$ in the original question can be achieved for sufficiently large $n$.  The construction assumes some fixed (but arbitrary) positive integer $m$.

Using a constant number of moves, draw both the real and imaginary axes, and mark the points $A = -i$ and $B = -1$.  Let $z_0 = 1$.
For $k = 1,2,3,\ldots,m$, do

a)  If $k$ is odd, then draw a circle centered at $A$ and passing through $z_{k-1}$.  Let $z_k$ be the intersection of this circle with the positive imaginary axis.
b)  If $k$ is even, then draw a circle centered at $B$ and passing through $z_{k-1}$.  Let $z_k$ be the intersection of this circle with the positive real axis.

Using a constant number of moves, construct the reciprocal of $|z_m|$.

Some things to observe:

Step 2 uses exactly $m$ moves, so the total number of moves is $n = m + O(1)$.
$z_k > 0$ for all even $k$, and $-iz_k > 0$ for all odd $k$.
For all $k\in\{1,\ldots,m\}$, we have $|z_k| = \sqrt{1+|z_{k-1}|^2}-1 \le |z_{k-1}|^2/2$.
The previous fact combined with induction on $k$ gives $|z_k| \le 2^{-(2^k-1)}$.
Thus the point constructed in Step 3 has norm at least $2^{2^m-1} \ge 2^{2^{n-O(1)}}$.<|endoftext|>
TITLE: Density and sums of reciprocals
QUESTION [7 upvotes]: Is there a notion of density for a (strictly increasing) sequence of natural numbers that decides whether the sum of the reciprocals of that sequence converges?

REPLY [5 votes]: The notion of natural density gives a suficient condition. Namely, one can prove that if $A \subseteq \mathbb{N}$ has positive upper natural density then $\sum_{a \in A} \frac{1}{a}$ diverges. This condition is not a necessary one, though. A well-known result in number theory ascertains that the series of the reciprocals of the prime numbers diverges whereas $\pi(n) = o(n)$.
In the following list you are to encounter some previous discussions on MO that are closely related to the current inquiry of yours:
[1] Erdos Conjecture on arithmetic progressions
[2] On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...
References
I. Erdös's brilliant proof of the divergence of $\sum_{p} \frac{1}{p}$:
http://www.renyi.hu/~p_erdos/1938-13.pdf
II. A. Rice, Density and substance: an investigation into the size of integer subsets.<|endoftext|>
TITLE: Association scheme on injective functions
QUESTION [7 upvotes]: This problem arises while studying the complexity of algorithms and I am quite unfamiliar with the subject.
Consider the set F of injective functions from {1..N} to {1..M}
we can define an association scheme on F x F by
(f,f') and (g,g') are in the same class if there is a permutation $\pi\in S_M$ and a permutation $\tau \in S_N$ such that $g = \pi \circ f \circ \tau$ and $g' = \pi \circ f' \circ \tau$.
I checked that this really defines an association scheme. In a way it is an "ordered" version of the Johnson scheme. It seems to me that it is a natural extension of the Johnson scheme, but I did not find any reference about it.

Q1: Has this association scheme ever been studied? What is its name?
Q2: Can this scheme be obtained by a combination (tensor product? suprema?) of the Johnson scheme and another quantity?

More precisely, I am interested in the "Bose-Mesner Algebra" point of view on this scheme. It is known that all the matrices in the algebra defined by this association scheme diagonalize in the same basis.

Q3: How can we construct/characterize these eigenspaces? 

--
Some background on Association Schemes.
An association scheme is a set of symmetric boolean matrices $A_1, \dots , A_S$ such that
1) $\sum_{i=1}^s A_i =J$ the all-one-matrix
2) $A_1 = I$ the identity matrix 
3) $\forall i,j \; A_iA_j \in {\rm span} ( A_i )$
The matrices $A_i$ can be seen as adjency matrix for some graph (but I don't think it might help here)
The span{$A_i$} defines an algebra called the Bose-Mesner Algebra. Condition (3) implies that all matrices commute so they diagonalize in the same basis.
--
In the case I'm considering here, the dimension of the $A_i$ is ${M \choose N}N!\times {M \choose N}N!$. The $A_i$ are not explicitly defined but we know that $[A_i]_{fg}=[A_i]_{f'g'}$ if there is a permutation $\pi\in S_M$ and a permutation $\tau \in S_N$ such that $g = \pi \circ f \circ \tau$ and $g' = \pi \circ f' \circ \tau$.
--
About the Johnson scheme: The $A_i$ have size ${M \choose N}$. The rows and the columns of the matrices are labeled by subsets of size $N$ of {$1,\dots,M$}. (in my case, the labels are injective functions, ie. ordered sets of subsets of size $N$ of  {$1,\dots,M$}.
$[A_i]_{ab}=[A_i]_{a'b'}$ if there is a permutation $\pi\in S_M$ such that $\pi(a) = a'$ and $\pi(b) = b'$. (where $\pi(a)$ denotes the subset of {$1,\dots,M$} obtained by applying the permutation $\pi$ to the elements of the sets $a$.

REPLY [5 votes]: I don't think this scheme has a particular name, and am not aware of any study of it. Its Bose-Mesner algebra is the commutant of a multiplicity-free representation of the wreath product of $S_m$ by $S_n$. To get the eigenspaces you need to find the decomposition of the representation into irreducibles.
The most useful reference I know is "Harmonic Analysis on Finite Groups: Representation Theory, Gelfand Pairs and Markov Chains" by Ceccherini-Silberstein, Scarabotti and Tolli.
for what it's worth, I think there's a chance that getting the decomposition is actually
doable, but it would not be quick :-(<|endoftext|>
TITLE: distance formula in Farey graph?
QUESTION [11 upvotes]: Consider the usual "Farey graph", i.e. the 1-skeleton of the (essentially unique) triangulation of the disk by ideal triangles.  If one insists that 1/0, 0/1, and 1/1 are vertices of a triangle then the vertices of this graph are naturally in bijective correspondence with the rationals, together with a single point 1/0.  Another name for this is the "Hatcher-Thurston complex", or "pants complex" or "curve complex" associated to simple, closed, unoriented curves on a torus (or punctured torus, or four-punctured sphere, with appropriate modifications to the definitions).
Is there a reference for a formula for the (combinatorial) distance function d(p/q, r/s) in this graph?  There seem to be plenty of references to connections to continued fractions and intersection numbers, but I've been unable to locate an explicit statement with a formula.

REPLY [10 votes]: Given $p/q, r/s \in \mathbb{Q}$, where $\gcd(p,q)=\gcd(r,s)=1$, choose the triangles in the Farey tessellation whose interior intersects the geodesic connecting $p/q$ and $r/s$ in $\mathbb{H}^2$: 

If $ps-qr=1$, then $d(p/q,r/s)=1$ and there is a geodesic of the Farey tessellation
connecting the two points. Otherwise, there will be a non-trivial triangle intersecting the geodesic. The triangles may be grouped together into maximal collections of triangles sharing a common vertex (a "pivot"). Let $p_i$ be the number of triangles in the $i$th group.
Associated to this sequence of grouped triangles is a canonical sequence $[p_1,p_2, \ldots, p_k]$,
where $p_1>1, p_k>1$. Define $l([p_1,p_2,\ldots,p_k])=d(p/q,r/s)$. 

Note that the sequence of triangles $[1,p_2, \ldots, p_k]$ is the same as the
sequence $[1+p_2, \ldots, p_k]$, since the lone triangle shares a vertex with the
pivot of the adjacent group, which is why we may assume that $p_1>1$,
and similarly $p_k>1$. 
The formula for the the Farey distance is computed inductively by
$l([p_1,p_2,\ldots,p_k]) = 1+ l([p_2,\ldots,p_k])$, and for $k=1$, 
$l([p_1])= 2$ (we are assuming that $p_1>1$ when $k=1$, since otherwise $d(p/q,r/s)=1$).
Here, if $p_2=1$, then $l([p_2,\ldots,p_k])=l([1+p_3,\ldots,p_k])$. 
To prove this formula, one
can see that a shortest path from $p/q$ to $r/s$ must go through
the adjacent pivot to $p/q$. If not, a case-by-case analysis shows that one
can find a shorter path, unless $p_1=2$ and the path goes through
the lower two edges of the first group of triangles. But then one can take an equal length path going
through the first pivot. 
To relate this to the continued fraction expansion, use an element of $A\in PGL_2(\mathbb{Z})$ such that $A(p/q)=\infty$, and $0\leq A(r/s) \leq 1/2$. Then $d(p/q,r/s)=d(A(p/q),A(r/s))=d(\infty,A(r/s))$. To compute the triangle
sequence in this case, we may use the continued fraction expansion for $A(r/s)$ to get $[p_1,\ldots,p_k]$ via $A(r/s)=1/(p_1+1/(p_2+1/(\cdots +1/p_k)\cdots )))$: 

In this example, $p/q=\infty=1/0$, $r/s=2/5$, and the continued fraction is $[2,2]$, since $2/5=1/(2+1/2)$.<|endoftext|>
TITLE: Existence of triply orthogonal coordinates
QUESTION [8 upvotes]: Definition
Consider a domain $\Omega \subseteq \mathbb{R}^3$ and three differentiable maps $f_i:\Omega \rightarrow \mathbb{R}$, $i = 1, 2, 3$.  If at every point $x \in \Omega$, $\nabla f_i(x) \cdot \nabla f_j(x) = 0$ whenever $i \ne j$, then the $f_i$ are triply orthogonal coordinates on $\Omega$.
Context
Dupin's theorem states that orthogonal coordinate surfaces (i.e., level sets of orthogonal coordinate functions) intersect along lines of principal curvature.  A classic example is Monge's ellipsoid -- see Jorge Sotomayor, "Historical Comments on Monge's Ellipsoid and the Configuration of Lines of Curvature on Surfaces Immersed in \mathbb{R}^3."
Question
Suppose we start with a compact, connected, orientable smooth surface $\mathcal{M}$ embedded in $\mathbb{R}^3$, and let $\Omega$ be a band of uniform thickness around $\mathcal{M}$ (i.e., a union of $\epsilon$-balls centered at each point on $\mathcal{M}$) small enough so that at each point $x \in \Omega$ there is a unique closest point on $\mathcal{M}$.  Further, suppose $f_1:\Omega \rightarrow \mathbb{R}$ gives the signed distance to $\mathcal{M}$, so that $f^{-1}(0) = \mathcal{M}$.
Can we always find (nontrivial) functions $f_2, f_3$ such that $(f_1,f_2,f_3)$ are triply orthogonal coordinates?
Or less formally, can we always find orthogonal coordinates "around" a given surface?

REPLY [12 votes]: I give an explicit counterexample below, so I have modified my answer: No, these sorts of orthogonal coordinates on 'shells' of parallel compact surfaces don't always (or even usually) exist.   The reason has to do with whether there are nonconstant functions on the given surface $\mathcal{M}$ that are constant on one of the families of principal curves. 
Such functions don't always exist: Suppose you take an oriented surface $\mathcal{M}\subset\mathbb{E}^3$, maybe, say, an embedded torus such that it has no umbilics and such that at least one family of its principal curves (i.e., the lines of curvature) wind densely around the torus.  It then turns out that, for $|t|<\epsilon$, the oriented-distance $t$ parallel surface $\mathcal{M}_t$ (where $\mathcal{M} = \mathcal{M}_0$) also has these properties.  An example is constructed below.  [Note that the standard ellipsoid is very different:  The lines of curvature on an ellipsoid are all closed.  However, this is very unusual behavior.  Most differential geometry books give the ellipsoid example because it's pretty and computable, and this tends to make students think that lines of curvature are 'usually' closed, but they are not.]
I claim that, for a surface as above, the desired functions $f_2$ and $f_3$ will not exist on the $\epsilon$-shell $\mathcal{S}$ around $\mathcal{M}$.  The reason is that each noncritical level set of $f_2=c$ (for example) will be closed in $\mathcal{S}$ and will have to intersect each $\mathcal{M}_t$ in a line of curvature, which, by supposition, is a dense curve in $\mathcal{M}_t$.  Thus, this (closed) level set will have to include $\mathcal{M}_t$.  Hence, this level set must be a disjoint union of the parallel surfaces $\mathcal{M}_t$.  It follows that $\nabla f_1$ and $\nabla f_2$ must be parallel along any noncritical level set of $f_2$.  However, this is absurd, because $\nabla f_1$ and $\nabla f_2$ must be perpendicular wherever $\nabla f_2$ is nonvanishing and $\nabla f_2$ does not vanish along a noncritical level set (by definition).  It follows, then that $f_2$ cannot have any noncritical level sets, i.e., $\nabla f_2$ must vanish identically.  Since you don't consider this to be a nontrivial solution, there are no nontrivial solutions. 
Added comment (an explicit example):  It turns out that there's an easy explicit example of a torus for which there is no nontrivial solution.   Let $C\subset\mathbb{E}^3$ be a closed, embedded, regular space curve, parametrized by arclength $ds$.  Thus, associated to the inclusion mapping $X:C\to\mathbb{E}^3$, there will be its Frenet apparatus, i.e., an orthonormal frame field $T,N,B:C\to S^2$ and functions $\kappa,\tau:C\to\mathbb{R}$ with $\kappa >0$ such that 
$$
dX = T\ ds,\quad dT = \kappa N\ ds,\quad dN = (\tau B -\kappa T)\ ds,\quad
dB = -\tau N\ ds.
$$
It is easy to see that there are such curves with $\int_C \tau\ ds$ being an irrational multiple of $\pi$, so let's assume this.  Now, for sufficiently small $t>0$, consider the tube of radius $t$ about $C$, parametrized by $Y_t:C\times S^1\to\mathbb{E}^3$ where 
$$
Y_t = X + t\cos\theta\ N + t\sin\theta\ B.
$$
(We require only that $t$ be so small that $Y_t$ be a smooth embedding of $C\times S^1$. In particular, we want $t$ to be less than $1/\kappa_{max}$.)  This is a parallel family of surfaces.
It is easy to compute that the principal curves on $Y_t$ are the leaves of the two foliations $ds=0$ and $d\theta + \tau(s)\ ds = 0$.  Because I required that $\int_C\tau\ ds$ be an irrational multiple of $\pi$, it follows that each of the leaves of $d\theta + \tau(s)\ ds = 0$ is dense in $C\times S^1$.  Consequently, the above argument applies to show that, even though you can take $f_1$ to be $t$ for this family, and, say $f_3$ to be some function of $s$, any function $f_2$ on a shell constructed by taking $t$ in some small interval in $(0,1/\kappa_{max})$ that satisfies $d f_2 \wedge(d\theta + \tau(s)\ ds) =0$ will have to be constant on the level sets of $f_1$.
One further remark: It turns out that a necessary and sufficient condition for the existence of a non-trival solution $(f_2,f_3)$ for the given $f_1$ is simply this:  A nontrivial (in the OP's sense) solution exists if and only if there exist functions $g_2$ and $g_3$ on $\mathcal{M}$ with the properties that $dg_2\wedge dg_3$ is nonvanishing on a dense open set in $\mathcal{M}$ and that each $g_i$ is constant on a family of principal curves on $\mathcal{M}$.  Sufficiency follows since one can simply take $f_i = \pi^*(g_i)$ for $i=2$ or $3$ where $\pi:\mathcal{S}\to\mathcal{M}$ is the retraction of the shell $\mathcal{S}$ to $\mathcal{M}$ long the normal lines from $\mathcal{M}$.  Necessity follows since, if $f_2$ and $f_3$ exist, then one can simply define $g_i$ to be the restriction of $f_i$ to $\mathcal{M}$.  Thus, one can see that the behavior of the principal curves on $\mathcal{M}$ completely determines whether or not there are solutions.<|endoftext|>
TITLE: Birational invariants and fundamental groups
QUESTION [28 upvotes]: In pondering this MO question and people's efforts to answer it, and recalling also something that I learned in my youth about using Morse theory ideas to prove some results of Lefschetz in the complex case, I seem to have learned two things -- things that I suppose are absorbed in the cradle by those who study algebraic geometry as opposed to learning it by osmosis -- or maybe I haven't got them quite right. Anyway:
(1) Blowing up a surface at a smooth point does not change the fundamental group, and (therefore ?) the fundamental group of a smooth projective surface is a birational invariant.
(2) Surfaces in $\mathbb P^3$ are simply connected, and more generally for a set $X\subset\mathbb P^n$ defined by a single homogeneous equation of degree $>0$ the pair $(\mathbb P^n,X)$ is at least $(n-1)$-connected. That is, the relative homotopy groups and therefore the relative homology groups vanish up through dimension $n-1$.)
(This is all over the complex numbers.) 
Is this correct? And, taking off from (1), what are some other simple statements about invariants from homotopy theory that are birational invariants? And what are the first things to know about birational invariants that do not come from topology?
EDIT  I wish I could accept more than one answer.

REPLY [6 votes]: The paper of Keum and Zhang 
"Fundamental groups of open K3 surfaces, Enriques surfaces and Fano 3-folds", 
Journal of Pure and Applied Algebra vol. 170
provides some answers to Tom's question "What happens to the fundamental group of a singular variety when removing the singular points?"
It appears that the fundamental group of the smooth part can be quite complicate also in simple situations. For instance, one of the results in the paper is the following:
"THEOREM. Let X be a $K3$ surface with at worst Du Val singularities (then X is still simply connected), and let $X^0$ be its smooth part. The number $c$=#(Sing $X$) is bounded by $16$, and if $c=16$ then $\pi_1(X^0)$ is infinite".
So, given for instance a quartic surface $X \subset \mathbb{P}^3$ with $16$ nodes (a Kummer surface) we have 
$\pi_1(X)=\{1\}$, but $\pi_1(X^0)$ is infinite! 
This follows from the fact that $X^0$ has an étale $\mathbb{Z}_2$-cover $Y^0 \to X^0$, where $Y^0$ is an Abelian surface minus 16 points.
For smaller values of $c$, the group $\pi_1(X^0)$ is finite, but not trivial in general. 
In higher dimension, there is the following
"CONJECTURE. Let $V$ be a $\mathbb{Q}$-Fano $n$-fold. Then the topological fundamental group $\pi_1(V^0)$ of the smooth part $V^0$ of $V$ is finite". 
(a normal variety $V$ with at worst log terminal singularities is $\mathbb{Q}$-Fano if, by definition, the anti-canonical divisor $−K_V$ is $\mathbb{Q}$-Cartier and ample).<|endoftext|>
TITLE: Can iterating countable unions give every set? (ZF)
QUESTION [7 upvotes]: Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under at-most-countable unions?

REPLY [2 votes]: If we assume ZF plus the assertion that $\omega_1$ is regular (which is provable from countable choice), or that indeed there is any uncountable regular cardinal $\delta$, 
then such a set $S$ exists. (Note that François provided a model having no uncountable regular cardinal, where there is no well-orderable counterexample.)
Let $S$ simply contain elements of Levy rank unbounded in $\delta$. To be specific, you could let $S$ be $\delta$ itself, or $\omega_1$ if this is regular as we usually expect. If a set $X$ has Levy rank $\alpha$, so that $X\in V_{\alpha+1}$, then the union set $\bigcup X$ is also in $V_{\alpha+1}$. In particular, $V_\delta$ is closed under arbitrary unions of its elements. Also, it contains every element of $S$ and also {s} for $s\in S$. 
Furthermore, since $\delta$ has uncountable cofinality, $V_\delta$ contains as elements all of its countable subsets, since any such subset would have rank bounded below $\delta$. Thus, the clsoure of your set is contained within $V_\delta$, but $S$ is not in $V_\delta$.

A simpler instance of the idea: 
The union of any set of ordinals is still an ordinal. Thus, if $\delta$ is an ordinal with uncountable cofinality, then it is already closed under the process of taking countable unions of its elements, but doesn't contain $\delta$ itself as a member. 
More specifically, if we take $S=\omega_1$, provided this is regular, then if we start with {s} for all $s\in S$ and close under the process of taking countable unions, we simply get all countable ordinals, and do not generate $\omega_1$ itself this way.<|endoftext|>
TITLE: How much reading do you do before you attack a problem?
QUESTION [47 upvotes]: When going off on a tangent from your regular area, where, presumably, you have such mastery of all cutting-edge research from your routine reading that you hardly need to do any extra (if this is false, please correct me), how much do you try to familiarize yourself with that area before beginning to directly attack your problem? Do you read just a few canonical papers and surveys, look thoroughly over a dozen and glance at a couple dozen more, or do enough to write a whole survey article of your own?

REPLY [37 votes]: When I attacked the problem of finding a quantum factoring algorithm, I had read four or five papers on the subject, which constituted nearly all the literature on quantum algorithms at the time. However, there were lots of other relevant papers that I didn't even know about in the field which would later be called quantum information theory, and I didn't feel compelled to do a literature search to find them.
This is one extreme. If you try to go this route, you may very likely miss some important techniques that are commonly used in the new field, so I would actually recommend substantial reading in the new field. I had to do that when I started working on quantum information theory.
It would make a lot of sense if you worked on your problem while you did this substantial reading (even though you're liable to go in the wrong direction), because that will be a good guide for choosing which papers you should read. It also helps if you have a colleague in the new area you can talk to or collaborate with. So, as Deane says in the comments, "It depends."<|endoftext|>
TITLE: Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?
QUESTION [7 upvotes]: Let $S$ = { $a^2b^3$ : $a, b \in \mathbb{Z}_{>1}$ }.
Does there exist $n$ such that $n$, $n+1 \in S$?
Motivation: I was thinking about Question on consecutive integers with similar prime factorizations, wondering whether any such pair had to have prime signatures with at least one 1.  This would follow if the answer to the above question is negative.  (This would also follow from weaker versions of the above question too, such as taking out perfect $n$th powers from $S$.)
Please note that $a$ and $b$ in the set definition are not allowed to be equal to 1.  Otherwise, there'd be solutions like 8, 9 or 465124, 465125. (465124 = $(2\cdot 11 \cdot 31)^2$ and 465125 = $61^25^3$.)

REPLY [2 votes]: See also http://oeis.org/classic/A076445 and this thread on the search for consecutive odd powerful numbers: http://www.mersenneforum.org/showthread.php?t=3474
Similar technique can be used for search for just consecutive powerful numbers (i.e., without the oddness restriction).
P.S. And of course, http://oeis.org/classic/A060355 is relevant.<|endoftext|>
TITLE: Countable Hom/Ext implies finitely generated
QUESTION [15 upvotes]: Today I learned this interesting fact from Jerry Kaminker: If $A$ is an abelian group such that $\mathrm{Hom}(A,\mathbb{Z})$ and $\mathrm{Ext}(A,\mathbb{Z})$ are both countably generated, then in fact $A$ is finitely generated.  This is known in the literature, in some old papers by Nunke-Rotman, Chase, and Mitchell.  It makes me interested in possible generalizations.
Suppose that $M$ is a left module over a ring $R$ and that $\mathrm{Ext}^k(M,R)$ is countably generated for all $k$.  For which $R$ can you conclude that $M$ is finitely generated or, better, finitely resolved?  Any commutative Noetherian ring with finite projective dimension?  Is there a countability restriction missing from this proposed generalization?  What about non-commutative rings?
The result has been stated for any countable PID rather than just for $\mathbb{Z}$.  In fact Mitchell says that if $R$ is a countable PID and $M$ is infinitely generated, then
$$|\mathrm{Hom}(M,R)|\cdot|\mathrm{Ext}(M,R)| = 2^{|M|}.$$

Victor in the comments asks for an application for this result for abelian groups.  The original motivation was to extract information about what is possible for the cohomology of a topological space.  The homology can be anything, provided that $H_0(X)$ is free and non-trivial.  There are various obvious and not-so-obvious impossible choices for cohomology.  For instance, this result implies that $H^*(X)$ cannot be countably infinitely generated as an abelian group.  (And if it is so as a ring, then the degrees of the generators have to go to $\infty$.)  I suspect that Jerry needs it for a similar reason.
I found the result interesting because it gives an "external" criterion for whether a countable abelian group is finitely generated.  Even though the forgetful functor to Set does not distinguish countable groups, it does sometimes distinguish their dual or derived dual groups.  One of the things that it can determine is whether the original group was finitely generated.  Also, assuming that the result holds for all PIDs, it is a natural and non-trivial generalization of the fact that the algebraic dual of an infinite-dimensional vector space $V$ satisfies
$$\dim V^* = 2^{\dim V}.$$
Everyone learns this for vector spaces, so it's cool to have a module version.

REPLY [3 votes]: This is an answer to one of the questions asked:
It is not sufficient for $R$ to be a commutative noetherian ring with finite global dimension.
Let $R=\mathbb{Z}_p$, the $p$-adic integers, which is a principal ideal domain, and so has global dimension $1$. It is well known that $\operatorname{Ext}^1_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)=0$, and clearly $\operatorname{Hom}_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)=0$.
So for $M=\mathbb{Q}_p$, $\operatorname{Ext}_R^k(M,R)=0$ (so certainly countably generated!) for all $k\geq0$, but $M$ is not finitely generated as an $R$-module.<|endoftext|>
TITLE: Eta-products and modular elliptic curves
QUESTION [17 upvotes]: Recently the elliptic curve $E:y^2+y=x^3-x^2$ of conductor $11$ (which appears in my answer) became my favourite elliptic over $\bf Q$ because the associated modular form
$$
F=q\prod_{n>0}(1-q^n)^2(1-q^{11n})^2
$$
is such a nice "$\eta$-product".  (This modular form is also associated to the isogenous elliptic curve 
$y^2+y=x^3-x^2-10x-20$ which appears in Franz's question.)
Question. Are there other elliptic curves over $\bf Q$ which have a simple minimal equation and whose associated modular form is a nice $\eta$-product or even a nice $\eta$-quotient?
I know two references which might have a bearing on the question 
--- Koike's article on McKay's conjecture
and 
--- p.18 of Ono's Web of modularity on $\eta$-quotients.
Can someone provide a partial or exhaustive list of such nice pairs $(E,F)$ ?

REPLY [2 votes]: Today I came across the following theorem of Mersmann: There are precisely 14 primitive eta-products which are holomorphic modular forms of weight $\frac{1}{2}$, namely...
He also proves a conjecture of Zagier to the effect that there are essentially only finitely many such products of any given weight.
I learnt these facts from Zagier's contribution to the The 1-2-3 of modular forms (which happens to have been metioned by Wadim Zudilin).
Addendum (2011/02/10)  Came across an advertisement for the book Eta products and theta series identities by Günter Köhler.<|endoftext|>
TITLE: Moduli space of semistable bundles
QUESTION [14 upvotes]: It is well-known that the space of $S$-equivalence classes of rank 2 semistable holomorphic vector bundles with trivial determinant on a genus 2 Riemann surface $M$ is $CP^3$ (more concretely $PH^0(Jac(M),L(2\theta)$). Especially, the points corresponding to semistable (and not stable) bundles are smooth points. On the open dense subspace consisting of points corresponding to stable bundles, there is a natural symplectic structure, compatible with the natural Riemannian metric. It defines a Kaehler structure. It should be true that this Kaehler structure extends to the semistable points (and therefore $CP^3$ is equipped with its natural (only) Kaehler structure). Does anyone know a reference, or a proof of this?
Thank you.

REPLY [5 votes]: Disclaimer:  This answer is rewritten in response to Chris Woodward's insightful comments.
The moduli space of rank 2 semistable holomorphic vector bundles with trivial determinant on a genus 2 surface $\Sigma$ is homeomorphic to the character variety $$\mathfrak{X}_\Sigma(SU(2)):=\mathrm{Hom}(\pi_1(\Sigma),SU(2))/SU(2).$$  This homeomorphism restricts to a diffeomorphism between stable bundles and irreducible representations.
In the case of fixed determinant Higgs bundles on $\Sigma$ there is a similar homeomorphic correspondence with $$\mathfrak{X}_\Sigma(SL(2,\mathbb{C})):=\mathrm{Hom}(\pi_1(\Sigma),SL(2,\mathbb{C}))/\!/SL(2,
\mathbb{C}).$$  Carlos Simpson has shown under this latter correspondence all points, even singular points, have isomorphic corresponding étale neighborhoods  (Isosingularity Theorem).  In short, they are locally isomorphic.  However, it is known these moduli spaces are not even biholomorphic let alone biregular, since the complex structure of the Higgs moduli space depends on the complex structure of $\Sigma$ as a Riemann surface whereas the complex structure on the character variety only depends on the group $SL(2,\mathbb{C})$.
The moduli space of holomorphic vector bundles naturally embeds into the moduli space of Higgs bundles as those with trivial Higgs field.  Likewise, $\mathfrak{X}_\Sigma(SU(2))$ embeds into $\mathfrak{X}_\Sigma(SL(2,\mathbb{C}))$, and the homeomorphisms above respect these embeddings.
So it is natural to think that the stratified analytic smooth structures on the moduli space of vector bundles and that of the semi-algebraic space $\mathfrak{X}_\Sigma(SU(2))$ also correspond; as they do generically and also do on their "complexifications".  However, as pointed out by Chris Woodward, Johannes Huebschmann proves in Smooth Structures on Certain Moduli Spaces for Bundles on a Surface that this is not the case.  In particular, in Section 8 he explicitly addresses the case of a genus 2 surface where the moduli spaces in question are homeomorphic to $\mathbb{C}P^3$, showing explicitly that the natural smooth structure on $\mathfrak{X}_\Sigma(SU(2))$ differs from that of $\mathbb{C}P^3$. 
That is sufficient to answer the original question as: NO, if one interprets the question as "Is $\mathfrak{X}_\Sigma(SU(2))$ Kähler isomorphic to $\mathbb{C}P^3$?"
Remark: With regard to the (singular) symplectic structure on $\mathfrak{X}_\Sigma(SU(2))$.   The Goldman Poisson structure, a Lie algebra structure and derivation, is defined globally on the coordiante ring of the character variety and makes sense even in a singular setting. In general, such a Poisson structure gives a foliation of the smooth locus by symplectic sub-manifolds.  In this case there is one leaf (since the surface in question is closed).
Remark: For a direct proof, not using the theory of holomorphic bundles, that the character variety is $\mathbb{C}P^3$ see here.  We can see the singularities directly in Choi's construction.<|endoftext|>
TITLE: Functorial point of view for formal schemes
QUESTION [22 upvotes]: Giving a scheme is the same as giving the corresponding functor from the category of rings to the category of set, and there are characterization of what functors arise in this way. This is explained in the book by Demazure and Gabriel.
This "functorial point of view" is sometimes very useful, so I was wondering whether there is something similar for formal schemes rather that algebraic schemes. I searched for this but I wasn't able to find anything, it seems that Demazure and Gabriel don't speak about formal schemes at all.
Ricky

REPLY [3 votes]: Although this doesn't cover every example, formal completions $\hat{X}_Y$  of a scheme $X$ along a closed subscheme $Y$ have a particularly nice functorial presentation. We have
$$\hat{X}_Y \simeq Y_{dR} \times_{X_{dR}} X.$$
Here the subscript $dR$ means we are considering the de Rham stack of a space $X$; i.e. the stack whose $S$-points are $\operatorname{Hom}(S^{red},X)$.
In general, a formal scheme is a ind-scheme $X$ that has a presentation
$$ X = \operatorname{colim}_{i \in I} X_i, $$
where each $X_i$ is a finite scheme and the transition maps $X_i \to X_j$ are closed embeddings. By definition, this colimit takes place in some category of sheaves (e.g. fppf) on the category of affine schemes, and therefore automatically has a functor of points interpretation.<|endoftext|>
TITLE: Are there good product rules on the $k$-sphere?
QUESTION [7 upvotes]: I have heard sometimes that the only dimensions $k$ for which there exists a "good" smooth product $P:S^k\times S^k\to S^k$ are $k = 0,1,3,7$ (the above products corresponding to $\mathbb Z_2, U(1)\subset\mathbb C$, the product of unit quaternions and of unit Cayley numbers).
I would like to ask for references about such a result. 
More precisely, I am interested in finding out how is it that one can define "good" so that the above is true (with proofs, preferably).

REPLY [12 votes]: $S^0$, $S^1$, and $S^3$ have well-known smooth group structures. These can be obtained from well-known bilinear multiplications in $\mathbb R$, $\mathbb R^2$, and $\mathbb R^4$. $S^7$ does not have a smooth group structure; this fact can be obtained from the theory of Lie groups and their Lie algebras. Stronger statement: $S^7$ does not have a continuous group structure; this can be deduced from the previous fact using the deep solution of a Hilbert problem. But $S^7$ does have an $H$-space structure, by which I mean a continuous multiplication law with a two-sided identity. This can be obtained from a (non-associative) bilinear multiplication on $\mathbb R^8$, so it can even be chosen in such a way that multiplication is smooth and multiplication by a fixed element on either side has a smooth inverse.
Outside these dimensions, $S^{n-1}$ has no $H$-space structure. For $n>1$ odd, this is easy using cup products in $S^{n-1}\times S^{n-1}$. For $n$ not a power of $2$, it can be done using Steenrod operations in mod $2$ cohomology. The general case uses $K$-theory. 
A corollary is that real division algebras (even in a nonassociative sense) are impossible in dimensions other than $1$, $2$, $4$, and $8$. Another corollary is that no spheres other than those are parallelizable.
The question of how many linearly independent tangent vector fields the $(n-1)$-sphere admits was also settled (by Adams) using more subtle $K$-theory arguments.<|endoftext|>
TITLE: Entropy of a general prob. measure
QUESTION [6 upvotes]: How is entropy of a general probability measure defined?

REPLY [7 votes]: It is not.  If a probability measure on $\mathbb{R}$ is absolutely continuous and has density $f$, then "entropy" usually refers to the differential entropy, defined in the Wikipedia page falagar linked to.  If the probability measure has discrete support, entropy is defined by an analogous formula, given in this Wikipedia page.  In the most classical treatments, these are the only situations covered at all.
However, both of these are special cases of the more general notion of relative entropy that Helge and robin girard pointed out: in the continuous case the reference measure ($\nu$ in Helge's notation, $\mu$ in robin's notation) is Lebesgue measure, and in the discrete case the reference measure is counting measure on the support.<|endoftext|>
TITLE: How/where are semi-log resolutions used?
QUESTION [5 upvotes]: In the paper, by János Kollár there is problem 19 (page 8). 
It is one more strict resolution. A resolution that leaves untouched the semi-simple-normal-crossings singularities of pairs.
My question is: How/where is that kind of resolution used/needed?
Quick definitions: 
Pair: $(X,D)$ with $X$ algebraic variety and $D$ a Weil divisor on it.
Semi-simple-normal-crossings: A point in $X$ where $X$ is (locally) a union of coordinates hyperplanes and $D$ is given by intersecting $X$ with some of the other coordinate hyperplanes not contained in $X$.

REPLY [19 votes]: Perhaps a little more explanation would be this:
One of the first things we learn in algebraic geometry is normalization and we are told that it is "harmless" to assume that something is normal since the normalization exists and canonical and all that jazz. This is fine as long as one studies a stand alone object, but once they come in a family, it is no longer true. 
Consider a family of curves. Recall that for curves normal=smooth so if not all members of the family are smooth, which is the likely scenario, then they are also not all normal. Now the problem is, there is no way to normalize the family members that they stay in a family. For instance, a family of smooth cubic plane curves degenerates to a singular one, but the normalization of the singular cubic has genus $0$, while the cubic curves have genus $1$ so they can't be members of the same family. Also, if you try to resolve the singularities by blowing up you'll see that you can resolve all singularities to be normal crossings, but you cannot do better and you also add new irreducible components to the singular fibers. This leads one to do semi-stable reduction, which is actually another story, so I won't get into that.
Anyway, for curves, we can actually make do with handling only smooth and simple normal crossing points. In higher dimensions if one tries to do the same, then there are other singularities that one must allow and these are the semi-log canonical (a.k.a. slc) singularities Zsolt mentioned. 
OK, so maybe the above convinces you that if you want to do moduli theory and you want to study compact moduli spaces, that is, you actually would like to understand degenerations as well and not just the nice part, then you have to deal with non-normal, in particular with slc singularities. (Actually, "s-something" is usually the non-normal version of "something", accordingly slc is the non-normal version of lc).
Well, now how do you define a non-normal version of a singularity that is otherwise defined via some properties of exceptional divisors (or saying it in a more enlightened way: exceptional set)? You cannot take a full fledged resolution of singularities, because it will resolve the non-normality of the singularity as well. This would not be a huge problem from the point of view of making it simple, but it sabotages the entire operation. The issue is, that if you look at the definition of lc (and klt, dlt, etc) more closely, then it becomes clear that it kind of needs that the resolution used in the definition is an isomorphism in codimension $1$ on the target, that is, the singular guy. It is also important that the exceptional set is a divisor. These will fail for non-normal but $S_2$ singularities, for instance for slc but not lc singularities.
So, you need a partial resolution that resolves the singularities to something that is close to being smooth but has the above properties. The "close to being smooth" is called "semi-smooth", these are double normal crossings and pinch points, exactly the singularities that cannot be made better by only changing something in codimension $2$. (This last statement is left to the reader. If you have difficulty with it, ask).
OK, I better wrap this up. So the point of a semi-resolution is that it has those properties that make it possible to define discrepancies but it does not go "too far". However, it produces varieties with very mild singularities that are almost as good as smooth, at least from the point of view of this definition.<|endoftext|>
TITLE: Reading materials for mathematical logic
QUESTION [28 upvotes]: Hi everyone, the summer break is coming and I am thinking of reading something about mathematical logic. Could anyone please give me some reading materials on this subject?

REPLY [3 votes]: Joe Mileti wrote a really nice set of course notes on mathematical logic (approx 20 weeks of lectures). It's a draft for a book titled, I think, "Mathematical Logic for Mathematicians."  The course notes are beautifully written (and beautifully delivered, if you've had the chance to see him lecture).  He's a very nice guy, and I would suggest contacting him about it.  
My memory is a bit hazy about the topics he covered, but we discussed propositional and first order logic, nonstandard analysis, and axiomatic set theory.  I also remember that some highlights included connections to graph theory and algebra (I guess this sort of touches upon his research themes).<|endoftext|>
TITLE: Twisted curves, admissible covers, and an algebraic analogue of a specific monodromy computation  
QUESTION [7 upvotes]: This problem arose when trying to understand the stack of twisted stable maps into a stack (specifically BG), as introduced by Dan Abramovich, Angelo Vistoli and several co-authors (Olsson, Graber, Corti,...). However, my specific question can be formulated without mentioning such beasts and I will do so. If anyone wants background I can give some.
Suppose that we have the following set-up. One is given:

A smooth projective curve C
and a finite abelian group G acting on C,
such that the quotient map $\pi : C \to C/G$ is an admissible cover.

Suppose moreover that $C/G \cong \mathbf{P}^1$. Over the complex numbers, one can make the following computation. For each ramification point of the covering, consider a small loop centered around that point, with orientation induced by the complex structure. Choose a lifting of that loop to a path on C. The path will start and end in the same fiber. Since the restriction of $\pi$ away from the ramification locus is a G-torsor, the difference between start- and endpoint will give us a well-defined element of G. (This uses that G is abelian -- in general, one would only get an element up to conjugation, since there are several choices of liftings of the loop.) Now consider the product of all these elements over all ramification points. By considering the fundamental group of $\mathbf{P}^1$ minus the ramification locus, it is clear that this product is the identity in G. 
I would like to express this computation algebraically, i.e. without recourse to any monodromy or the classical fundamental group, working over an arbitrary base where the order of G is invertible. Unless I'm mistaken, one can still define an evaluation map associating an element of G to each ramification point, since the following (to me rather mysterious) construction should work. Consider the stack quotient $[C/G]$. This is a twisted curve in the sense of Abramovich et al, which basically means that the ramification points on C/G have been cut out and replaced by cyclotomic gerbes ("stacky points"). Now this twisted curve has an actual G-torsor over it, so we get a map $[C/G] \to BG$. Restricting it to one of the stacky points, we obtain a cyclotomic gerbe over the base scheme with a G-torsor over it. But this is by definition an object of the rigidified inertia stack of BG, and the points of the (rigidified) inertia stack correspond to the elements of G.
Question 1: Is there a way of formulating this without using the language of twisted curves, i.e. associating an element of G to each ramification point only in terms of the admissible cover? There should be, since the moduli space of maps from twisted stable curves to BG is isomorphic to the moduli space of stable curves with an admissible cover which is  a G-torsor away from the branch locus, but I don't see any sensible way of expressing such a construction algebraically.
Question 2: Can one show that in the algebraic setting, the product of the elements over all ramification points is the identity?

REPLY [6 votes]: (written hurriedly, hope it is semi-helpful and -correct)
Let's say you're over some field k.  If it's not separably closed, basechange until it is.  The strictly completed local ring of a ramification point in your base P^1 is going to look like k[[t]].  Now remove that point to give yourself Spec k((t)).  The restriction of your cover to this punctured formal neighborhood is going to be a G-cover of Spec k((t)), which is to say a map from the absolute Galois group of k((t)) to G.  But since you've wisely demanded that the order of G is prime to char k, this map kills wild inertia, and factors through the tame inertia group, which is cyclic.  So the image of this map at least gives you a well-defined cyclic subgroup of G.  You have to be a bit more careful to get an element of G, and indeed there isn't really a canonical choice locally; there is a choice of generator of tame inertia involved.  Via class field theory you can make these choices "compatibly" at different ramification points (e.g. this will make the product vanish, as you say) but I don't think there's a "correct" such compatible choice.

REPLY [5 votes]: What naturally associates to each ramification point (or more generally, irreducible divisor in the ramification locus) is a pair $(H,\psi)$ where $H$ is a cyclic subgroup of $G$ (with $\ge 2$ elements) and $\psi$ is an injective character. If the ground field is the complex numbers, you can have a bijection between such pairs $(H,\psi)$ and the nonidentity elements of $G$ by mapping an element $g$ of order $n$ to the subgroup generated by $g$ and the character $\psi$ which has value $e^{2\pi i/n}$, as explained by Arend.
This is written in very classical, nonstacky terms in Pardini's Abelian Covers paper. On the other hand, it works in arbitrary dimensions, so the paper is a bit more complicated than it would be if only curves where involved.<|endoftext|>
TITLE: Estimate probability( 0 is in the convex hull of N random points ) ?
QUESTION [5 upvotes]: Can anyone estimate N such that Prob( 0 is in the convex hull of $N$ points ) >= .95
for points uniformly scatterered in $[-1,1]^d$, $d = 2, 3, 4, 10$ ?
The application is nearest-neghbour interpolation:
given values $z_j$ at sample points $X_j$, and a query point $P$,
one chooses the $N$ $X_j$ nearest to $P$ ($N$ fixed) and averages their $z_j$.
If $P$ is not in the convex hull of the $N$ $X_j$,
the interpolation will be one-sided, not so good.
I'd like to be able to say
"taking 6 neighbors in 2d, 10 in 3d, is seldom one-sided".
If anyone could point me to selfcontained pseudocode for the function Inhull( $N$ points )
(without calling full LP), that would be useful too.
(Please add tags interpolation convex-geometry ?)

REPLY [17 votes]: This is a classical and essentially geometric problem. In fact, the answer does not depend 
on the distribution of the points (as long as the distribution is centrally symmetric). 
The following result is due to Wendel (link).

Theorem. If $X_1$, ..., $X_N$ are i.i.d. random points in $R^d$ whose distribution
  is symmetric with respect to $0$ and assigns measure zero to every hyperplane
  through $0$, then
  $$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\frac{1}{2^{N-1}}\sum\limits_{k=0}^{d-1}{N-1 \choose  k}.$$

The proof is straightforward. Let $\mu$ be the distribution of $X_k$, and set
$$ f(x_1,\dots,x_N) = \begin{cases} 1, & \mbox{if } x_1,\dots,x_N\ \mbox{ lie in an open halfspace of $\mathbb R^d$ with $0$ in the boundary}, \newline 0, & \mbox{else.} \end{cases}$$
Then due to the invariance of $\mu$ under reflection in the origin, we have that
$$\mathbb P(0\notin \mbox{conv}\{X_1,\dots,X_N\})=\int_{\mathbb R^d}\dots \int_{\mathbb R^d} \frac{1}{2^N}\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)\ \mu(dx_1)\dots\mu(dx_N).$$
Now, the sum
$$C(N,d)=\sum\limits_{\varepsilon_i=\pm1}f(\varepsilon_1x_1,\dots,\varepsilon_Nx_N)$$
can be interpreted as the number of connected components of the set $\mathbb R^d\backslash (H_1\cup\dots\cup H_N)$  induced by the hyperplanes $H_1$, ..., $H_N$ through $0$ which are in general position. But there is a classical calculation going back to to Steiner and Schläfli, which shows that
$$C(N,d)= 2\sum\limits_{k=0}^{d-1}{N-1 \choose  k}.$$<|endoftext|>
TITLE: Question about von Neumann algebra generated by a complete algebra of projections
QUESTION [6 upvotes]: Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:
1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.
2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).
Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?
I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum_i \lambda_i p_i$, where $p_i \in P$ and $\lambda_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.

REPLY [5 votes]: The answer "yes" follows from Theorem 2.8 of Bade's "On Boolean algebras of projections and algebras of operators," 1955, which is in the more general context of algebras of operators on a Banach space. 
Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "Weak and strong limits of spectral operators," 1954.<|endoftext|>
TITLE: Probability of random permutation having certain cycles
QUESTION [5 upvotes]: Are there any good references (either books or on-line) on the subject of the distribution of various cycle properties amongst permutations, particularly ones containing exact, closed-forms?
For example, what is the probability that a random permutation of N objects contains some cycle having length between A and B? Or, what is the probability that all cycles in a random permutation of N objects have lengths between A and B?
I would be particularly interested in references that survey what is known in this field, preferably with a little detail.
Thanks.

REPLY [2 votes]: The following paper may be what you're looking for:
Schramm, Oded. Compositions of random transpositions. Israel J. Math. 147 (2005), 221--243. MR2166362 (2006h:60024)
http://www.springerlink.com/content/f572513876635mj5/
This paper is concerned with the distribution of a random permutation in $S_n$ generated by $c n$ random transpositions (where $c>1/2$). Schramm calculates the limiting distribution of the cycle lengths (ordered from largest to smallest). 
A recent paper of Nathanael Berestycki, Oded Schramm, and Ofer Zeitouni has extended the techniques of Schramm's earlier paper to the case where the random permutation is generated by random $k$-cylces. I don't know if this recent paper answers the same questions, but it might be relevant. 
http://arxiv.org/abs/1001.1894<|endoftext|>
TITLE: Is there a reference containing standard mathematical notations?
QUESTION [11 upvotes]: Suppose you are writing a mathematical text (say an article) and you want to call an object (for example, a set) by a letter. It would be cool then to have some reference (optimally available on the internet) where you could find some standard letters and notations of mathematical objects and pick one that you like. Does such a "notation dictionary" exist?
ADDED. Thanks everybody for interesting answers! Maybe it is worth to add that I had in mind rather basic things. The question was trigged by my attempt to find a good letter to denote a subset of the segment $[0,1]$. Finally I decided to call it $T$ (in the course of the proof it turns out that $T$ is equal $[0,1]$ :) ).

REPLY [2 votes]: The International Organization for Standardization (ISO) offers a standard for mathematical notation as part of the more general ISO 80000 standard for quantities and units. It is widely used in the physical sciences and also adopted by the National Institute of Standards and Technology (NIST) in the USA, for example. The full standard can be found around in the internet.
It is a self-consistent very well-thought standard. It generally agrees with common notation in maths literature. It covers areas from basic set-theory, first-order logic, tensors, functional analysis and special functions, and much more. The specialized notation of some areas such as differential geometry or category theory or higher-level formal logic is not quite covered. Notation for probability theory is given separately in the Guide to the Expression of Uncertainty in Measurement.<|endoftext|>
TITLE: Use of Conjectures to Prove a Theorem
QUESTION [11 upvotes]: Name a theorem T that has a proof based upon the truth of a conjecture C, and also has another proof based upon the falsehood of the same conjecture C, but for longtime has no known direct proof that is independent of C. For instance, it would be a claim that can be proved if P=NP, and can also be proved if P is different from NP.
I apologize if the question was previously asked (I also apologize for the title if it does not faithfully reflect the question).

REPLY [31 votes]: Another standard example is Littlewood's theorem that the number of primes less than x is sometimes greater than Li(x). His proof used different arguments depending on whether the Riemann hypothesis is true or false. See http://en.wikipedia.org/wiki/Skewes%27_number

REPLY [22 votes]: How about this basic one:
Theorem: There exists an irrational number $p$ and an irrational number $q$ such that $p^q$ is rational.
Proof: Let $q=\sqrt{2}$ and note that it is irrational.  Conjecture: $q^q$ is irrational.  If this conjecture is false, then we are done (the theorem is proved with $p=q$).  If the conjecture is true, then let $p=q^q$ and note that it is irrational.  The theorem is true in this case as well since $$p^q=(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=\sqrt{2}^2=2$$ is rational.<|endoftext|>
TITLE: How divisible is the average integer?
QUESTION [14 upvotes]: I don't know any number theory, so excuse me if the following notions have names that I'm not using.  
For a positive natural number $n\in{\mathbb N}_{\geq 1}$, define $Log(n)\in{\mathbb N}$ to be the ``total exponent" of $n$.  That is, in the prime factorization of $n$ it is the total number of primes being multiplied together (counted with multiplicity); for example $Log(20)=3.$  I'll define $log_2(n)\in{\mathbb R}$ to be the usual log-base-2 of $n$, so $log_2(20)\approx 4.32$.
One can think of $log_2(n)$ as "the most factors that $n$ could have" and think of $Log(n)$ as the number of factors it actually has.  Define $D(n)$ to be the ratio of those quantities $$D(n)=\frac{Log(n)}{log_2(n)}\in(0,1],$$ and call it the divisibility of $n$.  Hence, powers of 2 are maximally divisible, and large primes have divisibility close to 0.  Another example: $D(5040)=\frac{8}{12.3}\approx 0.65$, whereas  $D(5041)\approx\frac{2}{12.3}\approx 0.16$.
Question: What is the expected divisibility $D(n)$ for a positive integer $n$?  That is, if we define $$E(p):=\frac{\sum_{n=1}^p D(n)}{p},$$ the expected divisibility for integers between 1 and $p$, I want to know the value of $$E:=lim_{p\rightarrow\infty}E(p),$$ the expected divisibility for positive integers.
Hints: 

I once wrote and ran a program to determine $E(p)$ for input $p$.  My recollection is a bit faint, but I believe it calculated $E(10^9)$ to be about $0.19.$
A friend of mine who is a professor in number theory at a university once guessed that $E$ should be 0.  I never understood why that would be.

REPLY [8 votes]: Hardy and Wright define $$ \Omega (n) $$ to be the sum of the exponents. This is on page 354 in the fifth edition, section 22.10.  Then we have Theorem 430, the "average order" of $ \Omega (n) $ is $ \log \log n .$ Then Theorem 431, same answer for the "normal order."
$$ $$ I just saw Cam's answer, I think I will leave this anyway. The result on the average order answers your question. The book  is "An Introduction to the Theory of Numbers."<|endoftext|>
TITLE: (Good) effective version of Kronecker's theorem?
QUESTION [10 upvotes]: Thm (Kronecker).- If all conjugates of an algebraic integer lie on the unit circle, then the integer is a root of unity.
Question: Can one provide a good effective version of this? That is: given that we have an algebraic integer alpha of degree <=d, can we show that alpha has a conjugate that is at least epsilon away from the unit circle, where epsilon depends only on d? It actually isn't hard to do this (from the standard proof of Kronecker, viz.: alpha, alpha^2, alpha^3... are all algebraic integers, and their minimal polynomials would eventually repeat (being bounded) if all conjugates of alpha lied on the unit circle) with
epsilon exponential on d, i.e., epsilon of the form epsilon = 1/C^d; what we actually want is an epsilon of the form 1/d^C, say.
(Question really due to B. Bukh.)

REPLY [14 votes]: If $M(\alpha)$ is the Mahler measure of $\alpha$, then the largest conjugate of $\alpha$
has absolute value at least
$$1 + \frac{\log(M(\alpha))}{d}.$$
Lehmer's conjecture implies that this at least $O(d^{-1})$ away from one. 
Dobrowolski's lower bound for $M(\alpha)$ shows that there is a conjugate at least
$O(d^{-1-\epsilon})$ away from $1$ for any $\epsilon > 0$. Better bounds are available if one has more information, for example, the signature of $\mathbf{Q}(\alpha)$, or whether $\alpha$ is conjugate to $\alpha^{-1}$ or not.
For explicit references, see
http://www.maths.ed.ac.uk/~chris/Smyth240707.pdf<|endoftext|>
TITLE: A slick definition of the Kan extension?
QUESTION [7 upvotes]: Background:
The limit of a functor $F:D^{op}\to C$ is an object $\lim F$ representing the functor $$\ell F(x):=\operatorname{Psh}_D(\ast,C(x, F(\cdot))),$$ where $*$ denotes the terminal presheaf on $D$. (Notice that $C(x, F(\cdot)))$ is a presheaf on $D$). 
We can define the limit of a functor weighted by a presheaf in much the same way (by replacing $\ast$ with a fixed presheaf on $D$ called the weight).  
Why am I bringing this up?  It is a very slick definition.  Nowhere do we have anything like universal arrows popping up.  Adjunctions are out of sight and out of mind.  Indeed, this definition generalizes straightforwardly to S-enriched categories for S symmetric monoidal closed (and all of the other requirements you need for the S-enriched Yoneda lemma to work).  
The usual definition of the Kan extension is as a functor completing a certain commutative triangle such that it is universal in a specific sense in a certain functor category (intentionally vague...).  This definition is pretty annoying to work with and is avoided whenever possible by instead insisting that all Kan extensions be pointwise (for instance, Kelly does this his book on enriched categories).  
Question:
Does there exist a similar slick definition of the Kan extension (not necessarily pointwise)?  By slick here, we mean free of adjoint functors (and their less conspicuous cousins, universal arrows) and free of commutative diagrams (translating the content of a commutative diagram into prose does not count).

REPLY [10 votes]: Firstly, the $W$-weighted limit $\lim^W F$ is defined to be a representation of $\operatorname{Psh}_D(W,C(-,F-))$; the definition you've given isn't even well-typed.
There is no difference at all in $\mathrm{Set}$-enriched category theory between a representation and a universal arrow -- each determines the other, and when these exist for all suitable objects then adjoints are there whether you like it or not (this is all in Mac Lane).  So I'm not sure what's particularly slick about this approach.
Anyway, the non-pointwise right Kan extension is given, for $E \overset{K}{\leftarrow} C \overset{F}{\to} D $ by
$$[C,D] (G K, F) \cong [E,D] (G, \operatorname{Ran}_K F)$$
which is exactly the same as the usual definition in terms of universal arrows.  The pointwise extension is $(\operatorname{Ran}_K F)e = \lim^{E(e,K-)} F$.  The difference is discussed in Kelly chapter 4.3.  Incidentally, Kelly sticks to pointwise extensions because the non-pointwise ones aren't much use, not because he doesn't like the usual definition.
Is that the kind of thing you're looking for?<|endoftext|>
TITLE: Galoisian sets and the Langlands programme
QUESTION [14 upvotes]: Note: I've revised the question just a little bit in the hope of making it easier.

Given an algebraic number field $F$, which we may as well take to be Galois over $\mathbb{Q}$, we denote by $S_F$ the set of rational primes that split in $F$. Sets of the form $S_F$ are called Galoisian. At some point, there was a discussion
Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?
of the fact that the abelian Galoisian sets, that is, $S_F$ corresponding to $F$ abelian over $\mathbb{Q}$, are exactly the sets of primes defined by congruence conditions.
A while later, Matthew Emerton gave this nice answer
Galoisian sets of prime numbers
to a question of Chandan Singh Dalawat about non-abelian Galoisian sets. 
I made a comment there I thought I would upgrade to a question. As Matthew points out,  Neukirch's remark that the Langlands program  provides a characterization of all Galoisian sets is probably meant as a metaphor for some other process. However, I couldn't help but hope that the characterization could be taken literally, at least for some special families. For example, we will refer to a number field $F$ as being of $GL_2$ type if it is the fixed field of
$Ker(\rho)$, where
$$\rho: Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow GL_2(\mathbb{C})$$
is an irreducible two-dimensional Artin representation of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$*. 
 Now call a set of primes a $GL_2$ Galoisian set if it is of the form $S_F$ for some extension $F$ of $GL_2$-type.
The question then  is: can one use the Langlands program (or anything else) to give a sensible characterization of $GL_2$ Galoisian sets?
One could obviously change this question in any way that would make it more tractable. One could try to characterize, for example:
-Solvable $GL_2$ Galoisian sets, where the $GL_2$-field $F$ is further required to be solvable;
-Odd $GL_2$ Galoisian sets: $S_F$ where $F$ is the fixed field of a representation $\rho_f$ arising from a holomorphic modular form $f$ of weight one;
-Odd $GL_2$ Galoisan sets of conductor $N$, where we further require the form $f$ to have level $N$;
and so on. The last case probably admits a tautological answer of sorts, in that we can in principle list the finitely many forms (sorted by Dirichlet characters $\epsilon$), and then make some statement about the $p$'s where
$$X^2-a_pX+\epsilon(p)=(X-1)^2.$$
Is it entirely unreasonable to hope for something more compact?

*The idea that we should simply organize fields in this manner corresponding to representations is perhaps a valuable perspective coming out of the Langlands program.

Added, 25 July:
Having thought about it a bit more, it occurs to me that this is yet another situation where Langlands urges us to go beyond a classical framework in seeking answers to non-abelian questions. For example, when we associate to an odd Artin represention 
$$\rho:Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow Aut(V)$$
of dimension two an Artin $L$-function 
$$
L(\rho,s)=\sum a_n/n^s,$$
we can perfectly sensibly assert that the $a_n$'s follows a pattern. When asked what that pattern is, the answer, satisfying to some and mysterious to others, is that
$$\sum a_nq^n$$
is a modular form. This is the kind of thing that comes out of Langlands. 
Now, if we want to 'characterize,' say, odd $GL_2$ Galoisian sets, we can say the following: Enumerate the normalized holomorphic Hecke (new) eigenforms $f$ of weight one sorted by level $N$ and character $\epsilon$.
For each such form, run over the prime numbers $p$ not dividing $N$, and take the number $a_p$ defined by the equation
$$T_pf=a_pf.$$
for the $p$-th Hecke operator $T_p$. Now look at the set
$S_f$ of primes  $p$ such that 
$$(p,N)=1, \epsilon (p)=1, a_p=2.$$
These $S_f$'s are exactly the odd $GL_2$ Galoisian sets.
Perhaps it's unreasonable to want more from the Langlands' programme. Whether or not this is the final word on all such questions, well, that's a different matter.

REPLY [4 votes]: Here is an extremely naive answer to a case of my own question, which is surely obvious to experts. For me, even coming up with this silly version required quite a bit of conversation with Sugwoo Shin (who is of course blameless of any errors).
We give a description of the odd $GL_2$-Galoisian sets of level $N$ in terms of 'higher congruence conditions.' The point is to consider the $\mathbb{Q}$-Hecke algebra $H(N)$ determined by the Hecke operators acting on modular forms of weight 1, level $N$.
The maximal ideals in $H(N)$ are in correspondence with Galois conjugacy classes of normalized new weight one eigenforms of level $N$. There is also a map $$p\mapsto T_p$$
from primes not dividing $N$ to $H(N)$.
Any given maximal ideal $m$ determines a Dirichlet character $\epsilon_m$, and one considers the set of primes $S(m)$ defined by the 'congruence conditions'
$$(p,N)=1, \ \epsilon_m(p)=1, \ T_p\equiv 2 \ \ \mod \ m$$
These $S_m$ are exactly the odd $GL_2$ Galoisian sets of level $N$. 
With a bit more care, one should be able to give a similar description that doesn't refer to the level beforehand.
Of course this is no different from what I wrote before, but focussing on the Hecke algebra seems to allow a formulation that's rather analogous to the classical one. That is, one can forget about modular forms for a moment and examine sets of primes determined by congruences in the Hecke algebra.<|endoftext|>
TITLE: Elliptic curves — general structure of the group
QUESTION [7 upvotes]: Let $K$ be a field and $E$ be an elliptic curve defined over $K$. It well understood the $K$-points on $E$ forms an abelian group. What is the structure of this group?(Depending on char($K$)?) Is it a direct sum of some well known abelian groups such as $\mathbb{Z}/m\mathbb{Z}$?

REPLY [7 votes]: If you don't specify more about the structure of the field $K$, then we can't say much about the structure of the group $E(K)$.  There are special cases (described in the Wikipedia article):

If $K$ is a number field, then the Mordell–Weil theorem implies the group is finitely generated (and this has been generalized, as Anweshi mentioned).  In fact, for each number field, there is a global bound on the size of the torsion of any elliptic curve over that field.  In particular, if $K = \mathbf{Q}$, then it is a direct sum of a free abelian group of finite rank with a torsion group that is one of 15 types.
If $K$ is finite of order $q$, then (by a theorem of Hasse) the group is finite of order about $q+1$ with error bounded by $2\sqrt{q}$.  It is a sum of two cyclic groups.

If $K$ is larger than that, then $E(K)$ can be quite large.  For example, if $K$ is separably closed, then $E(K)$ is divisible.  In this case, if $K$ has characteristic zero, then $E(K) \cong (\mathbf{Q}/\mathbf{Z})^2 \oplus \bigoplus \mathbf{Q}$.  If $K$ has characteristic $p>0$, then $E(K) \cong \bigoplus_{\ell \neq p} (\mathbf{Q}_\ell/\mathbf{Z}_\ell)^2 \oplus (\mathbf{Q}_p/\mathbf{Z}_p)^h \oplus \bigoplus \mathbf{Q}$.  Here, $h$ is zero or one depending on whether the curve is supersingular or ordinary, and the $\bigoplus \mathbf{Q}$ is a vector space whose dimension is:

zero if K is an algebraic closure of a finite field.
countably infinite if $K$ is countable and not an algebraic closure of a finite field.
equal to the cardinality of $K$ otherwise.

Away from the separably closed case, you get a subgroup of one of these groups, but you can have very complicated subgroups of $\mathbf{Q}$ as summands, and very complicated torsion subgroups.<|endoftext|>
TITLE: When is the independence number of a graph equal to its clique cover number?
QUESTION [8 upvotes]: Let G be a connected graph without loops. 
The (vertex) independence number is the maximum size of a set of vertices such that no two vertices in the set are connected by an edge. 
The (vertex) clique cover number is the minimum number of cliques in G which cover all the vertices of G (not necessarily all the edges of G).
My question is - under what conditions are those two numbers equal?

REPLY [7 votes]: Denote the independence number of $G$ by $\alpha(G)$, and the clique cover number by $\overline{\chi}(G)$.  It is obvious that $\overline{\chi}(G) \ge \alpha(G)$.  You are asking when $\alpha(G) = \overline{\chi}(G)$.
This seems to be the question that drove the definition and investigation of perfect graphs.  See Berge's historical overview.  The definition of perfect graphs requires the equality between independence number and clique cover number for every induced subgraph.  This is Berge's "Beautiful Property", defining what he calls "class 2".  Rephrasing a comment by Lovász, the induced subgraph condition is needed to remove the artificial examples where a large set $X$ of new vertices is added to a graph such that each vertex in $X$ is adjacent to each vertex of the original graph.  Applying this construction to any graph yields a new graph that satisfies your condition.
Berge also defined "class 3", those graphs for which the chromatic number equals the clique number for every induced subgraph, and "class 4", the graphs which contain no induced odd hole or induced odd antihole.  In the paper mentioned above, Lovász proved the perfect graph theorem, that the complement of a perfect graph is perfect.  This shows that "class 2" and "class 3" are the same.  Perfect graphs are now usually defined as "class 3", and the strong perfect graph theorem shows that "class 4", of Berge graphs, is the same as "class 3".  Berge claimed he was originally more interested in whether "class 2" and "class 3" coincided.
One final remark: if $\alpha(G) = \overline{\chi}(G)$, then the Shannon capacity of $G$ is $\log \alpha(G)$.  This is Berge's "class 1" (note that the "log" is missing in the paper, clearly a typo).  This is really saying that in the limit, when $\alpha(G) = \overline{\chi}(G)$ then the independence number dominates the channel performance in the long run.  From this applied point of view, it makes sense to look for the kinds of graphs which have this property inherently, and not just by a large independent set being glued on.
However, there may be other applications for graphs for which $\alpha(G) = \overline{\chi}(G)$ holds, but where this relationship fails for some induced subgraphs.  This would then make it interesting to find the graphs that have this property but not via a glued-on independent set.  To end this answer on a question: do you have such an application in mind?<|endoftext|>
TITLE: On $\gamma$-graded pieces of the localization sequence for G-theory (i.e. for K'-theory)
QUESTION [5 upvotes]: There is a well-known Quillen's localization sequence for (algebraic) K-theory: $\dots\to K_p^Y(X)\to K_p(X)\to K_p(X-Y)\to \dots$, where $Y\to X$ is a closed embedding of schemes. 
Now suppose that $X$ is regular (and excellent of finite dimension, if needed).
Another well-known fact is that (in this case) the relative K-theory group 
$K_p^Y(X)$ is isomorphic to $K'_p(Y)$ (some authors denote this by $G_p(Y)$; note that $Y$ is not necessarily regular!).
Now, I tensor this long exact sequence by $\mathbb{Q}$. Can I consider the $i$-th graded piece of the $\gamma$-filtration for this long exact sequence? Certainly, $K^{}(X)\otimes \mathbb{Q}$  and $K^*(X-Y)\otimes \mathbb{Q}$ are endowed with $\gamma$-filtration, but I am not quite sure about $K'_p(Y)\otimes \mathbb{Q}$ (one of my problems here is that I am interested in quite a general situation). Also, could I say that the $i$-th level of the $\gamma$-filtration for $K'_p(Y)$ is some (which one??) level of its niveau filtration? 
Which references are most appropriate for these matters? I believe that for rational coefficients these things are easier than for integral ones.

REPLY [6 votes]: The basic reference is Soule's paper "Operations en K-theorie algebrique" (Can. J. Math. 37 (1985) 488-550). Essentially, there is a grading on $K'$-theory (with rational coefficients) enabling one to interpret the localisation sequence as the long exact sequence of motivic homology. The filtration comes from the $\gamma$-filtration on $K_*^Y(X)$, the $K$-theory with support. 
Tamme's article in the Beilinson conjectures book (http://wwwmath.uni-muenster.de/math/u/schneider/publ/beilinson-volume/index.html) is a good survey of this, but he sticks to the case of schemes over a field.
I am not sure I understand your final question, since even for $K$-theory of nice schemes (e.g. spectra of fields) the $\gamma$- and niveau filtrations don't agree except for $K_0$.<|endoftext|>
TITLE: Small neighborhoods of singularities on varieties
QUESTION [8 upvotes]: In Singular points of complex hypersurfaces, John Milnor proves the following theorem:
Let $x \in V$ be a point on a variety $V$ in $\mathbb{R}^n$ or $\mathbb{C}^n$. Assume $x$ is either a smooth point or an isolated singularity. Let $D_{\epsilon}$ be the closed $\epsilon$-ball about $x$, $S_{\epsilon}$ its boundary (the sphere about $x$ of radius $\epsilon$), and $K = V \cap S_{\epsilon}$.  Then for $\epsilon$ sufficiently small, the pair $(D_{\epsilon}, V \cap D_{\epsilon})$ is homeomorphic to the pair $(CS_{\epsilon}, CK)$, where $C$ denotes taking the cone. (Theorem 2.10)
In Remark 2.11, Milnor observes that this theorem "likely" holds even if $x$ is a non-isolated singularity; in particular, it is known even in this case that "a suitably chosen neighborhood of any point is homeomorphic to the cone over something."
This book was written in 1968. What is the current status of this problem?

REPLY [4 votes]: Indeed the following theorem to me seems exactly you were looking for (see J. Bochnak, M. Coste, M-F. Roy, "Real algebraic geometry", Theorem 9.3.6 [Local conic structure]):
Let $E$ be a semialgebraic susbet of $\mathbb{R}^n$ and $x$ be a nonisolated point of $E.$ Let also $D_\epsilon$ be the closed $\epsilon$-ball around $x$ and $S_\epsilon$ its boundary. Set $K=S_\epsilon \cap E$. Then there for $\epsilon>0$ small enough the pair $(D_\epsilon,E∩D_\epsilon)$ is semialgebraically homeomorphic to the pair $(CS_\epsilon,CK)$, where $C$ denotes taking the cone.
Moreover the semialgebraic homeomorphism can be chosen as to preserve the distance from $x.$
Two words of remarks on the previous statement: 

Every real or complex algebraic set in $\mathbb{R}^n$ or in $\mathbb{C}^n\simeq \mathbb{R}^{2n}$ is a semialgebraic set. 
The point $x$ is any nonisolated point of $E$ (no matter singular - in whatever meaning this word has for a general semialgebraic set - or regular).<|endoftext|>
TITLE: Tor and projective dimension
QUESTION [6 upvotes]: Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$?
What I do know is that if $(A,\mathfrak{m})$ is Noetherian local and $M$ is finitely generated over $A$ then $\mbox{Tor }^{r+1}(M,A/\mathfrak{m})=0 $ if and only if $\mbox{proj. dim }M\leq r$.
Generally speaking, is $\mbox{Tor }$ functor as good a tool to measure projective dimension as $\mbox{Ext }$ even when the ring/module is not Noetherian or local?
I suspect we can use $\mbox{Tor }$ to measure projective dimension when ring is Neotherian local and module is finitely generated because flatness and projectivity coincide in such case.

REPLY [11 votes]: Since your question is really about projective dimension of flat modules, it is worth noting the following result  (see Raynaud-Gruson MR0308104, Cor 3.3.2 or Jensen MR0407091, Thm 5.8) which complements Richard's example:
The projective dimension of a flat module over a commutative Noetherian ring $R$ is bounded by $n+1$ if  the cardinality of $R$ is at most $\aleph_n$. 
EDIT: It looks like commutativity is not needed!<|endoftext|>
TITLE: A question about definability in first order theories based upon classical logic
QUESTION [5 upvotes]: Let T be a theory formalized in the classical first order predicate calculus with equality. If P(x) is a formula of T in which one and only one variable of T--here denoted by 'x'--occurs free, and if the following pair of statements are provable in T, then P(x) is said 
to be a defining formula of T. 
(1) "There exists an x such that P(x)" and 
(2) "For any y and any z, P(y) and P(z) imply y=z"-where "y" and "z" denote variables of T.

Now if T were ZFC, for example, consider the formula of T which states "The continuum 
hypothesis implies that x is the set of all positive integers and the negation of the
continuum hypothesis implies that x is the set of all negative integers". This is a defining formula of T and Sierpinski often used formulae of this type to define sets which were "non-effective" in some way. But many might ask whether such definitions are really "legitimate". They arise when the formula P(x) contains sub-formulae which are sentences
of T (i.e. closed formulae in which no variables occur free). My question is, to what
extent (if any) would such first order theories as Peano's Arithmetic or ZFC be weakened,
if their defining formulae were not allowed to contain sub-formulae that are closed?

REPLY [4 votes]: Not weakened at all!
In PA and ZFC (and a wide class of other f-o theories), every defining formula is equivalent to one with no closed subformula.  Let's start with ZFC.  Suppose $\varphi(x)$ is a defining formula, with $x$ free; now let $z$ be any variable not appearing anywhere in $\varphi$, and for each subformula $\psi$ of $\varphi$, define a new formula $\psi^z(z)$ (with all the same free variables as $\psi$, plus $z$) by:

if $\psi$ is an atomic formula, take $\psi^z\ :=\ \psi \land (z=z)$;
if $\psi = \top$, take $\psi^z\ :=\ (z=z)$
if $\psi = \bot$, take $\psi^z\ :=\ \forall z'.\ z \in z'$
if $\psi = \psi_1 \land \psi_2$, then take $\psi^z\ :=\ \psi_1^z \land \psi_2^z$

…and so on: all the remaining cases (non-nullary connectives and quantifiers) just commute with $(-)^z$, same as $\land$.  Anyway, we've defined these new versions of all subformulas; by induction, they all have $z$ free, have no closed subformulas, and are equivalent to the original versions; so up at the top we apply it to our original formula, and have $\varphi^z(z,x)$; now $\forall z.\ \varphi^z (z,x)$ is equivalent to $\varphi(x)$ and has no closed subformula.
Now, this relied on the fact that ZFC has (in most presentations) no closed terms (indeed, no terms except variables), for the atomic formula case to work: in PA, for instance, $0=0$ gets bumped up to $0=0 \land z=z$, which still has a closed subformula.  So for eg PA, we have to work a bit harder at defining that case:

for $\psi$ an atomic formula $R(t_1,\ldots,t_n)$, take $\psi^z\ :=\ \exists w.\ [w = t_1\ \land\ R(w,\ldots,t_n)\ \land\ z=z]$.

This now makes it all work again!  But, this relied on all basic relations $R$ taking at least one argument.  In any language with this property, we're good.  The one thing we can't generally deal with is theories with nullary relation symbols, aka propositional constants — although we can sometimes still handle them like we handled $\top$ and $\bot$ in $ZFC$.
(Actually, I'm not quite sure what you mean by “how would ZFC and PA be weakened if we changed the definition of a defining formula” – since I don't know axiomatisations of those theories that involve this term.  But the standard axiomatisation of ZFC does involve functions (in the replacement axiom), which are just defining formulas $f(x,y)$ with an extra free variable $y$, so I guess what you have in mind might be something like strengthening the definition of function allowed in there?  But I think the above construction should answer the question, in any case!)<|endoftext|>
TITLE: Simple proof that these graphs are perfect
QUESTION [7 upvotes]: I recently came across a family of infinite graphs (in the context of two-dimensional convexity) that don't have induced 4-paths (paths with 4 vertices). 
Note that the complement of a 4-path is again a 4-path.
Clearly, every induced $n+1$-cycle contains an induced $n$-path.
Hence, by the Strong Perfect Graph Theorem of Chudnowski, Robertson, Seymour, and Thomas,
graphs without induced 4-paths are perfect.  
Can anyone provide a simple proof of that fact?
Having no induced 4-paths seems like a very strong condition.

REPLY [11 votes]: These P4-free graphs are also known as cographs. A simple proof of the perfectness of such graphs was given by Seinsche, On a property of the class of n-colorable graphs, J. Comb. Th. Ser. B 16 (1974), 191–193. MR0337679 The key to the proof is the fact that these graphs are also characterized by the property that every subset of $V(G)$ with more than one element is either not $G$-connected or not $\overline{G}$-connected. It follows that every such graph can be obtained from a single vertex by repeatedly duplicating vertices with or without an edge between the two duplicates. Since these two duplication operations preserve perfectness, all such graphs are perfect. (This quick argument is due to Lovász.)<|endoftext|>
TITLE: Matrices whose exponential is stochastic
QUESTION [5 upvotes]: The complex matrix exponential of a Hermitian matrix is unitary: $e^{-iH} = U$.  Is there a name or a characterization for matrices Q whose real exponential is stochastic: $e^{-Q} = S$?

REPLY [6 votes]: A matrix $A$ such that $\exp(tA)$ is (right) stochastic for all $t > 0$ should be called a "generator of a semigroup of stochastic matrices" or an "infinitesimally stochastic matrix". Clearly, since $A=\lim_{t\to0} (\exp(tA)-I)/t$, (i) the sum of the elements in each row of $A$ has to be 0, and (ii) all non-diagonal elements must be non-negative. Conversely, a matrix $A$ satisfing (i) and (ii), for large enough $n$ produces a stochastic matrix $I+A/n$, hence $(I+A/n)^n$ and  $\exp(A)=\lim_{n\to\infty}(I+A/n)^n$ are also stochastic (and so is $\exp(tA)$). That said, I would have a look at the results of a Google search with "infinitesimally stochastic" (I can't do it now).
(Edit: as observed, the above is a stronger condition than the one you asked for; though it's a more close analog to your example.)<|endoftext|>
TITLE: Is there a common genesis for ADE classifications?
QUESTION [32 upvotes]: Recall that a certain type of object admits an ADE classification if there is a notion of equivalence relative to which equivalence classes of objects of the given type can be placed in one-to-one correspondence with the collection of simply laced Dynkin diagrams.  The simply laced Dynkin diagrams have themselves been classified into two infinite families (denoted $\mathrm{A}_n$ and $\mathrm{D}_n$) and three exceptional examples (denoted $\mathrm{E}_6, \mathrm{E}_7$, and $\mathrm{E}_8$).  See the Wikipedia page for more details.
There is a veritable laundry list of objects that admit an ADE classification.  Examples include (modulo a number of qualifiers that I don't want to get into):

Semisimple Lie algebras
Conformal field theories
Tame quivers
Platonic solids
Positive definite quadratic forms on graphs

The list goes on.  Accoding to the aforementioned wikipedia page, Vladmir Arnold asked in 1976 if there is a connection between these different kinds of objects which really explains why they all admit a common classification.  The page also makes an offhand comment about how such a connection might be suggested by string theory.  
I am hoping that somebody can explain some of the progress that has been made (if any) on Arnold's question.  A good answer to this question is not one which explains the proofs that various different types of objects admit ADE classifications, nor one which aimlessly extends the list above.  Rather, I would like to see someone take a collection of objects which on the surface are unrelated but which all have ADE classifications and then outline a deeper connection between them which at least suggests that they might all have a common classification.  Bonus points if anyone can justify wikipedia's invocation of string theory in this context.

REPLY [7 votes]: I will first address the string theory part of the question. 
String theory provides examples of physical systems admitting several descriptions that provide natural bridges between  Kleinian singularities (and therefore Platonic solids), ALE spaces, quiver diagrams, ADE diagrams and two dimensional Conformal Field Theories. 
The scene is given by compactifications of string theory on Kleinian orbifolds  $M_\Gamma=\mathbb{C}^2/\Gamma$ where $\Gamma$ is a discrete subgroup of $SU(2)$. The space $M_\Gamma$ admits a Kleinian singularity at the origin. After studying this physical system, one is less surprised to see that Kleinian singularities, quiver diagrams, ALE spaces, ADE diagrams and 2 dimensional  Conformal Field Theories all admit the same ADE classifications since they provide different descriptions of the same underlying physical system. 
Michael Douglas and Gregory Moore have studied the compactification of string theory on Kleinian orbifold $M_\Gamma$ using D-branes as probes of the geometry. 
D-branes are extended objects on which strings can end. 
 D-branes provide  a physical description of the geometry  in terms of  supersymmetric gauge theories. Such supersymmetric gauge theories are efficiently summarized by a  quiver diagram  with a very natural physical interpretation: the nodes correspond to D-branes with specific gauge groups on them and the links between the nodes are open strings ending on the branes. 
The minimal energy configurations (the vacua) of these supersymmetric gauge theories are obtained finding the extrema of a potential whose construction  is equivalent to the hyperkhäler quotient construction of   Asymptotic Locally Euclidian Spaces (ALE spaces) first obtained by Kronheimer. ALE spaces are HyperKähler four dimensional real manifolds  whose anti-self-dual  metrics are asymptotic  to a Kleinian orbifold $M_\Gamma=\mathbb{C}^4/ \Gamma$. Physically ALE spaces described gravitational instantons. 
 ALE spaces provide small resolutions of the Kleinian singularities where the singular point is replaced by a system of spheres whose intersection matrix is equivalent to the Cartan matrix of an  ADE Dynkin diagram. One can also consider Yang-Mills instantons on such spaces.  The gauge group associated with the Yang-Mills instantons  is given by the type of ADE diagram obtained by the resolution of the singularity. This was analyzed in the math literature by Kronheimer and Nakajima. Physically the ALE instantons  moduli space is equivalent to the vacua of the gauge theory description of D-branes located at the singularities. 
The link between D-branes on ALE spaces (or equivalently Kleinian singularities) and the ADE classification of  two dimensional Conformal Field Theories (CFT) was studied by Lershe, Lutken and Schweigert. Although the geometry is singular, the CFT description is smooth. The 2 dimensional CFT is coming directly from the string description: as a string evolves it described a 2 dimensional surface called the string worldsheet. D-branes enter the CFT as boundary states.  In the description of the CFT, one recovers Arnold's ADE list of simple isolated singularities. 
Updates
I would like to comment on the non-stringy part of the question. This is motivated by the comments of Victor Protsak.
If one removes all the string theory interpretation in the discussion above. What is left is 
Kronheimer's description of ALE spaces.  Kronheimer's construction provides a beautiful realization of McKay's correspondence between Kleinian singularities, their crepant resolutions and ADE diagrams. This is reviewed in chapter 7 of Dominic Joyce's book "Compact Manifolds  with Special Holonomy". From that perspective, the string theory description provides a physical interpretation of Kronheimer's construction and adds  a natural link with quiver diagrams and 2 dimensional Conformal Field Theories.<|endoftext|>
TITLE: Continuous Linear Programming: Estimating a Solution
QUESTION [5 upvotes]: I have a "continuous" linear programming problem that involves maximizing a linear function over a curved convex space. In typical LP problems, the convex space is a polytope, but in this case the convex space is piecewise curved -- that is, it has faces, edges, and vertices, but the edges aren't straight and the faces aren't flat. Instead of being specified by a finite number of linear inequalities, I have a continuously infinite number. I'm interested in estimating solutions numerically, and my current method is to approximate the surface by a polytope, which means discretizing the continuously infinite number of constraints into a very large finite number of constraints.  Unfortunately, typical linear programming algorithms run in something like cubic-time in the number of constraints, so I'm getting a huge performance hit as I make the discretization finer.  Firstly, I'm interested to know if this kind of problem has been studied before, and what's been done.  Secondly, I'm looking for good strategies for approaching my problem numerically (good LP packages, suggested algorithms, optimizations, etc.). 
For concreteness, here is a simplified version of the problem I'm trying to solve:
I have $N$ fixed functions $f_i:[0,\infty]\to \mathbb{R}$. I want to find $x_i$ $(i=1,\dots,N)$ that minimize $\sum_{i=1}^N x_i f_i(0)$, subject to the constraints:
$\sum_{i=1}^N x_i f_i(1) = 1$, and
$\sum_{i=1}^N x_i f_i(y) \geq 0$ for all $y>2$
More succinctly, if we define the function $F(y)=\sum_{i=1}^N x_i f_i(y)$, then I want to minimize $F(0)$ subject to the condition that $F(1)=1$, and $F(y)$ is positive on the entire interval $[2,\infty)$. Note that this latter positivity condition is really an infinite number of linear constraints on the $x_i$'s, one for each $y$.  A specific $y_0$ restricts me to the half-space $F(y_0) \geq 0$ in the space of $x_i$'s. As I vary $y_0$ between 2 and infinity, these half-spaces change continuously, carving out a curved convex shape. The geometry of this shape depends implicitly (and in a complicated way) on the functions $f_i$.
The reason I suspect there should be an approach that's better than just discretizing the number of constraints is that continuity of the $f_i$'s implies a kind of local structure on the space of constraints that becomes invisible under discretization.  If we sit on the boundary of our convex space (so that at least N constraints are saturated, corresponding to some $y_k$), and we want to move along the boundary, then generically only those constraints corresponding to small neighborhoods of the $y_k$ are important.  Sometimes when the function $F(y)$ develops a new zero, new $y$ can become important, but this is nongeneric.
NOTE: I asked this question first on stackoverflow.net, and was told it was a nonstandard enough CS problem that I should ask about it here.

REPLY [6 votes]: Even though the $f_i$ are not polynomials I'll give the answer in that case because it is very nice and it seems like there is some interest.  I have to stress in advance though that the answer exploits the resulting algebraic structure in a fundamental way, and so is unlikely to extend to the case when the $f_i$ are not polynomials.
First of all, a semidefinite program (SDP) is an optimization problem with matrix variables, linear objective, and positive semidefiniteness constraints on symmetric (real) matrices in addition to the standard linear (in)equalities allowed in linear programming.  They are a generalization of linear programs (LP) and are vastly more expressive.  LPs are the case when the matrices are constrained to be diagonal.  SDP can also be viewed as a noncommutative version of LP.
The relationship with semi-infinite programming suggested by Gilead is I think the fact that one can view the constraint "A is positive semidefinite" as $x^TAx\geq 0$ for all $x$, which is an infinite number of constraints.  On the other hand, one can view any convex constraint in this way, because any closed convex set can be described by (infinitely  many) linear inequalities.
Theoretically SDPs are important both because many problems can be written as SDPs and because they can be solved using interior point methods in polynomial time, almost as efficiently as LPs in theory.  In practice, the technology is much newer than that for LPs so one cannot solve SDPs which are nearly as big using off-the-shelf software, but those days seem to be getting closer.
To see how to turn your problem into an SDP if the $f_i$ were polynomials, let $x_i$ be your decision variables.  Note that $f_i(1)$ is just a constant, so your first constraint is just a linear equation on the $x_i$ and that is no problem in an SDP.  For the others, we need to do a little work.
Let $H$ denote the operator which sends a symmetric matrix to its sums along antidiagonals, so
$H: \begin{bmatrix}a & b \\\\ b & c \end{bmatrix}\mapsto\begin{bmatrix}a & 2b & c\end{bmatrix}$, and so on for bigger matrices.  If we identify polynomials with their sequences of coefficients, then $p = q^2$ as polynomials if and only if $p = H(qq^T)$ as vectors.  Therefore $p$ is a sum of squares (SOS) if and only if $p = H(Q)$ for a positive semidefinite $Q$ (any such $Q$ is the sum of matrices of the form $qq^T$).  Now, if a polynomial $p$ is SOS then it is automatically nonnegative everywhere.  Conversely, one can show than any univariate nonnegative polynomial is a sum of squares.  This gives us an exact characterization of nonnegative polynomials in terms of positive semidefinite matrices.  Thinking of the matrix $Q$ as a new decision variable, we can write the constraint "$p$ is nonnegative" in a semidefinite program; the solver will find a $Q$ which certifies this.
Similarly, a polynomial $p$ is nonnegative on an interval $[w,\infty)$ if and only if it is of the form $p(x) = SOS_1(x) + (x-w)\cdot SOS_2(x)$ for some SOS polynomials $SOS_i$.  Again, one direction is obvious and the other requires a little effort (write $p$ in factored form and group the factors cleverly).  Therefore we can also write nonnegativity on an interval in terms of positive semidefinite matrices, and hence use it as a constraint in an SDP.
Now note that when we use $H$ to define the constraint for a polynomial $p$ to be SOS, we are writing linear equality constraints between the coefficients of $p$ and some linear functionals of the matrix inside $H$.  Similarly for when we write the constraint that $p$ is nonnegative on an interval: it is a linear equality between some decision variables, plus the constraint that certain matrices (the ones defining the SOS polynomials) are positive semidefinite.
Until now we've been thinking of $p$ as a constant polynomial.  But because arbitrary linear equalities between decision variables are allowed in an SDP, we can just as easily write the constraint "$\sum_{i=1}^N x_i f_i(y)$ is nonnegative for $y\in[2,\infty)$" using this method, because our polynomial in question has coefficients which are linear in the decision variables.
Putting this all together gives a semidefinite program which would express exactly what you want in the case that the $f_i$ are polynomials.  One could then either find a feasible $x_i$, prove that none exists, or optimize a linear functional of the $x_i$ all in polynomial time.  Unfortunately because of the way we have used the algebraic structure of the problem, this is unlikely to extend to non-polynomial $f_i$.
Finally, I should note that if you're interested in this sort of thing I highly recommend checking out my advisor Pablo Parrilo's course notes on MIT OpenCourseWare.  You can find the link on his website.<|endoftext|>
TITLE: Heuristic reason for Polya's conjecture
QUESTION [15 upvotes]: Let $\lambda(n)$ be Liouville's function, so that for each positive integer $n = p_1^{m_1}\cdots p_r^{m_r}$, we have that $\lambda(n) = (-1)^{\sum^{r}_{k=1}{m_k}}$. In 1919, Polya conjectured that $L(x) = \sum_{n \leq x}{\lambda(n)} \leq 0$ for all $x \geq 2$; his reasoning was based on some limited numerical evidence (up to $x = 1500$, I believe), its connection to the Riemann Hypothesis (it implies RH and the simplicity of the zeroes of $\zeta(s)$), and Polya showed that for $p \equiv 3 \pmod{4}$ with class number $h(-p) = 1$, $L(p) = 0$. Unfortunately, Polya's conjecture is false; it is known that the first counterexample occurs at $x = 906150257$ (so one can't really blame Polya for trying), and that there exist infinitely many positive integers $n$ such that $L(n) \geq 0.061867 \ldots$.
Nevertheless, Polya's conjecture does seem to be usually true, in that $L(x) \leq 0$ "most" of the time. There are a couple of different arguments that give an indication of why one would expect $L(x)$ to often be negative. For example, standard methods show (under RH, of course) that
$${\sum_{n \leq x}}'{\lambda(n)} = \frac{\sqrt{x}}{\zeta(1/2)} + \sum_{\rho}{\frac{\zeta(2\rho)}{\zeta'(\rho)}\frac{x^{\rho}}{\rho}} - 1 + O\left(\frac{1}{\sqrt{x}}\right),$$
and one expects the terms in the sum over the zeroes to generally be very small, whereas $1/\zeta(1/2) = -0.684765\ldots$, so it would be expected that $L(x)$ is "usually" negative. Another method is via Lambert series; I mentioned here that one can show that
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x},$$
where $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}} = O(e^{-\pi x})$; this Lambert series is in some sense a smoothed version of $L(x)$. Again, the leading term is negative, suggesting that $L(x) \leq 0$ often.
My question is: what other methods (elementary, analytic, or probabilistic) can be used to show why we would expect $L(x)$ to usually be negative?

REPLY [7 votes]: I'm not answering your actual question about other methods, but I can provide some information about the analytic approach.
Suppose that the Riemann hypothesis is true and that all the imaginary parts of the zeros are linearly independent over the rational numbers. Then one can show, using the methods of Rubinstein and Sarnak ("Chebyshev's bias", Experiment. Math., 1994), that the function $L(x)/\sqrt x$ has the same limiting (logarithmic) distribution function as the random variable
$$
\frac1{\zeta(1/2)} + \sum_{\rho\colon \Im\rho>0} 2\bigg| \frac{\zeta(2\rho)}{\rho\zeta'(\rho)} \bigg| X_\rho,
$$
where the $X_\rho$ are independent random variables each taking values in $[-1,1]$ according to the sine distribution (that is, each $X_\rho$ is the real part of a random variable uniformly distributed on the unit circle in the complex plane).
The distribution of the sum is symmetric around 0, which explains why the distribution itself is predominantly negative (since $\zeta(1/2) \lt 0$).<|endoftext|>
TITLE: Strongest known version of Baker's theorem
QUESTION [5 upvotes]: The article I have checked for Baker's theorem is Waldschmidt's. But the article and the citations therein are from the time of '88. Question:

What is the the strongest known lower bound for Baker's theorem on linear forms on logarithms?

Similarly for p-adic Baker.

REPLY [7 votes]: There is a big difference between linear forms in many logarithms and in two (or three) logarithms. The first case is covered in the archimedean case by the work of E. Matveev; Matveev's original works are hard even to specialists but there is a very nice survey [Yu. Nesterenko, Linear forms in logarithms of rational numbers, in Diophantine approximation (Cetraro, 2000),  53--106, Lecture Notes in Math., 1819, Springer, Berlin, 2003. MR2009829  (2004i:11082)]. The $p$-adic case was mostly done by Kunrui Yu. The best estimate for the case of two logarithms, which is of importance because of Tijdeman's application to Catalan's equation, is given in [M. Laurent, M. Mignotte et Y. Nesterenko, Formes linéaires en deux logarithmes et déterminants d'interpolation, J. Number Theory 55 (1995), no. 2, 285--321. MR1366574 (96h:11073)]. The latest news in the last direction (also in relation to Catalan's) are reviewed in [M. Mignotte, Linear forms in two and three logarithms and interpolation determinants.  Diophantine equations, 151--166, Tata Inst. Fund. Res. Stud. Math., 20, Tata Inst. Fund. Res., Mumbai, 2008. MR1500224  (2010h:11119)]<|endoftext|>
TITLE: How do we know that Fermat wrote his famous note in 1637?
QUESTION [55 upvotes]: It is widely stated that Fermat wrote his famous note on sums of powers ("Fermat's last theorem") in, or around, 1637. How do we know the date, if the note was only discovered after his death, in 1665?
My interest in this stems from the fact that if this is true, we can be absolutely certain that whatever proof Fermat in mind was wrong, and he must have noticed (or he would have mentioned it to his correspondents in later years). On the other hand, if the note had been written much later the reasoning would fail. I have used this argument previously in talks for the general public, rather acritically, and I would like to make sure it is sound.

REPLY [5 votes]: There is a relevant piece of information that does not seem to have been mentioned in the other answers so here it is.
Fermat became acquainted with the book by Diophantus by studying in the library of d'Espagnet (father Jean and son Etienne) in Bordeaux around 1629.  This was a period of great creative activity for Fermat, including the invention of the method of adequality (incidentally the term derives from Bachet's latin term adaequalitat which itself derives from Diophantus's Greek parisotes). 
So Fermat may well have written the comment in the  margin of the book Arithmetic by Diophantus as early as 1629.<|endoftext|>
TITLE: Fontaine-Mazur for GL_1
QUESTION [33 upvotes]: For any number field $K$, the Fontaine-Mazur conjecture predicts that any potentially semistable $p$-adic representation of the absolute Galois group $G_K$ of $K$ that is almost everywhere unramified comes from algebraic geometry (i.e., is a subquotient of the etale cohomology of some variety over $K$, up to Tate twist). As far as I can see, the only cases where any progress has been made concerns the case that $K$ is totally real or CM.
This made me wonder: Is the Fontaine-Mazur conjecture known to be true for $1$-dimensional representations for any number field $K$? For CM fields, the theory of CM abelian varieties gives varieties whose cohomology realizes nontrivial characters (and I guess that easy variations should produce all characters). What are the geometric objects appearing for other fields?
[edit: The word 'geometric' is avoided now, see the comments.]

REPLY [20 votes]: Let $\chi$ be a one-dimensional geometric (in the sense of FM) $p$-adic Galois representation of $G_K$ and let $\psi$ be the Hecke character of $K$ associated to $\chi$ by class field theory. The fact that $\chi$ is de Rham (=pst) at all primes above $p$ imples that $\psi$ is an algebraic Hecke character. Generally, the only algebraic Hecke characters of $K$ are of the form $(\text{finite order})\cdot\mathcal{N}^n$ where $\mathcal{N}$ is the norm character. Under class field theory, $\mathcal{N}$ corresponds to the cyclotomic character, so it comes from geometry; additionally, any finite order character comes from geometry (it arises as the subquotient of the $H^0$ of a zero-dimensional variety). The only time there are more algebraic Hecke characters is when $K$ contains a CM field. Denoting $L$ the maximal CM field in $K$, every algebraic Hecke character of $K$ is of the form $(\text{finite order})\cdot(\psi_L\circ\mathcal{N}_{K/L})$ where $\psi_L$ is an algebraic Hecke character of $L$ and $\mathcal{N}_{K/L}$ is the norm from $K$ to $L$. Again, finite order characters come from geometry, so this case is reduced to the CM case. As you've mentioned the CM case has been dealt with, so Fontaine–Mazur is true for $\mathrm{GL}(1)$.<|endoftext|>
TITLE: Combinations of multisets with finite multiplicities
QUESTION [6 upvotes]: The question may be of little interest to most people here on MathOverflow, but after browsing a pile of books in combinatorics, I had to ask it somewhere:
What are the most efficient formulae for calculating the number of $k$-combinations (and $k$-permutations) of multisets with finite multiplicities (i.e. combinations and permutations with repetition, but with restrictions on the number of repetition)?
I know that generating functions are often used for solving this kind of problems, but there has been a number of formulae used for such counting, such as Percy MacMahon's one ($m_i$ denotes multiplicities of $n$ different elements in the multiset):
$$C(k;m_{1},m_{2},\ldots,m_{n})=\sum_{p=0}^{n}(-1)^{p}\sum_{1\le i_{1}\le i_{2}\le\cdots\le i_{p}\le n}{n+k-m_{i_{1}}-m_{i_{2}}-\ldots-m_{i_{p}}-p-1 \choose n-1}$$
Are you aware of other formulae for it, or useful references in literature? 
EDIT: Clearing up the statement: a $k$-combination means simply picking $k$ elements from the multiset (order not important). $k$-permutation is basically the same, but order is important. In the example above, the multiset is $\{ m_1\cdot a_1,m_2\cdot a_2,\ldots m_n\cdot a_n\}$, $a_i$  being the elements, $m_i$  being the multiplicities.

REPLY [2 votes]: In addition to the OP's 2011 paper with Ž. Jurić:
A New Formula for the Number of Combinations of Permutations of Multisets
Applied Mathematical Sciences, Vol. 5, 2011, no. 18, 875-881
there is a paper by Thomas Wieder, also in 2011:
Generation of All Possible Multiselections from a Multiset
CSCanada Progress in Applied Mathematics, Vol. 2, No. 1, 2011, pp. 61-66
which approaches counting the $k$-combinations of a multiset (sub-multisets of a multiset) in terms of "selection matrices" (similar to contingency tables).
In addition to the references cited in these two papers, Frank Ruskey's 2003 work-in-progress Combinatorial Generation has Sec. 4.5.1 with algorithms based on representing $k$-combinations as weak compositions with restricted parts.<|endoftext|>
TITLE: Can we have A={A} ? 
QUESTION [10 upvotes]: Does there exist  a set $A$ such that  $A=\{A\}$ ? 
Edit(Peter LL): Such sets are called Quine atoms.
Naive set theory  By Paul Richard Halmos On page three, the same question is asked.
Using the usual set notation, I tried to construct such a set: First 
with finite number of brackets and it turns out that after deleting 
those finite number of pairs of brackets, we circle back to the original
 question.
For instance, assuming $A=\{B\}$, we proceed as follows: 
$\{B\}=\{A\}\Leftrightarrow
B=A  \Leftrightarrow B=\{B\}$  which is equivalent to the original 
equation.
So the only remaining possibility is to have infinitely many pairs of 
brackets, but I can't make sense of such set. (Literally, such a set is 
both a subset and an element of itself. Further more, It can be shown 
that it is singleton.)
For some time, I thought this set is unique and corresponded to $\infty$
 in some set-theoretic construction of naturals. 
To recap, my question is whether this set exists and if so what 
"concrete" examples there are. (Maybe this set is axiomatically 
prevented from existing.)

REPLY [4 votes]: (This is intended as a comment on Thomas Forster’s point above, “i think there is a literature about them that goes back earlier than Quine,” but I’m over the character limit.)
    Much earlier than Quine; Frege uses the trick in §10 of the Grundgesetze, setting (loosely, in modern terms) the True and False to their own singletons.  He considers using the trick in greater generality, à la Quine, in footnote 17: “A natural suggestion is to generalize our stipulation so that every object is regarded as a course-of-values, viz., as the extension of a concept under which it and it alone falls.”  I would be stunned if this did not influence Quine’s use, either directly or indirectly, subconsciously or consciously.  
    Richard Heck discusses the stipulation is great detail in Part I of Reading Frege’s Grundgesetze, as part of his treatment of The Julius Caesar Problem, i.e., how can we be sure that the extension of some concept is not equal to Julius Caesar?
    Personally, I feel that it is still too soon for Frege to worry about The Julius Caesar Problem. As he pointed out elsewhere, discussing a language that hasn’t yet been fully formalized is fraught with peril, and “Julius Caesar” is not yet a precisely-defined term in a formal language.  (Perhaps Caesar will end up being formalized as the extension of a concept after all—I don’t want to pre-judge the formalization of zoology.)  And obviously his proof of referentiality in §10 has to break down somewhere, or it would have been a consistency proof for naïve set theory.<|endoftext|>
TITLE: Implications of the abc conjecture in Arakelov theory
QUESTION [11 upvotes]: It is apparent that the abc conjecture is deeply related to Arakelov theory. In one direction, it is shown in S. Lang, "Introduction to Arakelov Theory", that a certain height inequality in Arakelov theory implies the abc conjecture. I am wondering about the other direction and precise implications. Are there such results? 
More importantly, if there exist such results, what are some Diophantine implications? That is, what are the Diophantine implications of abc conjecture, that factor through Arakelov Theory/Arithmetic Geometry? 
(I once read that Joseph Oesterle came up with abc conjecture while trying to do computations towards the Taniyama-Shimura conjecture. But that story does not make clear the connection with Arakelov theory.)

REPLY [8 votes]: ABC is equivalent to the conjectured height inequality that Lang (or more precisely Vojta, in an appendix to Lang's book, following the ABC appendix he wrote for his own book) uses.  This is shown in several papers by van Frankenhuijsen.  
http://research.uvu.edu/machiel/papers/abcrvhi.pdf
http://research.uvu.edu/machiel/papers/ABCRothMord.pdf
http://research.uvu.edu/machiel/bibliography.html
So ABC is equivalent to some of Vojta's conjectures in arithmetic geometry.
Also, Elkies' "ABC implies effective Mordell" is a sort of counter-application in that it shows that, given the ABC conjecture, one does not need Arakelov methods to prove the Mordell conjecture.
http://imrn.oxfordjournals.org/cgi/pdf_extract/1991/7/99
Lang wrote an article on Diophantine inequalities related to ABC that outlines some of the relationships between conjectures that were known 20 years ago.
http://www.ams.org/bull/1990-23-01/S0273-0979-1990-15899-9/S0273-0979-1990-15899-9.pdf<|endoftext|>
TITLE: Cardinal of maximal linearly independent subsets of a free module
QUESTION [16 upvotes]: Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero commutative ring. I know that it's true if $A$ is Noetherian or integral domain. I thought it was not true in general but I came up with something that looks like a proof and I can't figure out where it went wrong.

REPLY [19 votes]: I think I have a counter-example. Let $A$ be the ring of functions $f$ from $\mathbb{C}^2 \setminus (0,0) \to \mathbb{C}$ such there is a polynomial $\widetilde{f} \in \mathbb{C}[x,y]$ such that $\widetilde{f}(x,y)=f(x,y)$ for all but finitely many $(x,y)$ in $\mathbb{C}^2$.
Map $A$ into $A^2$ by $f \mapsto (fx, fy)$. We check that this is injective: If $fx=0$ then $f$ is zero off of the $x$-axis. Similarly, if $fy=0$, then $f$ is zero off of the $y$-axis. So $(fx, fy) = (0,0)$ implies that $f$ is zero everywhere on $\mathbb{C}^2 \setminus (0,0)$.
We now claim that there do not exist $(u,v)$ in $A^2$ such that $(f,g) \mapsto (fx+gu, \ fy+gv)$ is injective. Suppose such a $(u,v)$ exists. Let $\widetilde{u}$ and $\widetilde{v}$ be the polynomials in $\mathbb{C}[x,y]$ which coincide with $u$ and $v$ at all but finitely many points. Let $\Delta=\widetilde{u} y - \widetilde{v} x$. Since $\Delta$ is a polynomial which vanishes at $(0,0)$, it is not a non-zero constant. Thus, $\Delta$ vanishes on an entire infinite subset of $\mathbb{C}^2$. Let $(p,q)$ be a point in $\mathbb{C}^2 \setminus (0,0)$ such that $\Delta(p,q)=0$, $\widetilde{u}(p,q)= u(p,q)$ and $\widetilde{v}(p,q)=v(p,q)$. 
So $q u(p,q) - p v(p,q) =0$. Since $(p,q) \neq (0,0)$, there is some $k \in \mathbb{C}$ such that $(u(p,q), v(p,q)) = (kp, kq)$. Take $f$ to be $-k$ at $(p,q)$ and $0$ elsewhere; let $g$ be $1$ at $(p,q)$ and $0$ elsewhere. So $(fx+gu, fy+gv)=0$, and the map $(f,g) \mapsto (fx+gu, \ fy+gv)$ is not injective.<|endoftext|>
TITLE: What braking strategy is most fuel-efficient?
QUESTION [16 upvotes]: You notice a stop-light ahead of you and it is currently red.  You can't run the red light, so you will have to brake, but braking wastes energy and you want to be as fuel efficient as possible.  What braking strategy maximizes efficiency?
Let's set down some notation and move slowly toward a well-defined question. Suppose you are currently a distance $d$ from the stop-light, and suppose that the stop-light is on a timer whereby it switches from red to green after $T$ seconds.  You know the value of $T$, but you don't know how far in the cycle the stop-light is right now -- perhaps it will turn green in 1 second or perhaps in $T$ seconds.  So if $t$ is the amount of time until it actually turns green, then $t$ is a random variable uniformly distributed on $[0,T]$.  
Your initial speed is $v$, so that if you don't slow down you'll be at the light in $\frac{d}{v}$ seconds.  If $t<\frac{d}{v}$ then ``you win" by not slowing down, because the light will turn green before you get to it and you will have lost no energy to heat.  So if $T\leq\frac{d}{v}$ then clearly the best strategy is not to slow down.  Thus we may suppose $T>\frac{d}{v}$.  We assume no friction.
I'm looking for a strategy for applying the brake minimally, not knowing the status of the stop-light's cycle.  Perhaps we apply the brakes uniformly to end up stopped at the light, or perhaps we do not apply the brakes at all until we are almost to the stop-light, or perhaps we apply the brakes at the very beginning and coast at that reduced speed until we get very close to the light.  What strategy minimizes the expected brake usage? 
More precisely, I'm looking for a non-increasing differentiable function (for the car's velocity in terms of its distance to the stop-light) $$f\colon[0,d]\to{\mathbb R}_{\geq 0}$$ such that $f(d)=v$, $f(0)=0$, and such that if you solve the differential equation to find velocity in terms of time, then the expected value of that velocity at time $t$ is maximized.

REPLY [5 votes]: I think the problem would have been more naturally stated in the context of bicycles.  In any case, the answer is as follows:
You are looking for an optimal velocity function $v: [0, T] \to \mathbb{R}_{\geq 0}$ satisfying some conditions.  Each such function represents the strategy, "if the light is still red at time $t$, travel at speed $v(t)$; when the light turns green, coast."  One of the conditions on $v$ is that you may not run the red light.  In terms of the function $v$, this condition may be written as $\int_0^T v(t) \, dt \leq d$.
The quantity you wish to compute is the expected speed at which you will pass through the light after it turns green.  By the givens (uniform distribution, the nature of our strategy), this expected speed is precisely the average value of $v(t)$, i.e., it is $\frac{1}{T} \int_0^T v(t) \, dt$.
Putting the last two paragraphs together, we see that the optimal expected speed is $\frac{d}{T}$.  Moreover, this expected speed is achieved for any choice $v(t)$ with the property $\int_0^T v(t) \, dt = d$, i.e., for any strategy that will get you to the stoplight within time $T$.
Added in edit:
I agree with Willie Wong that maximizing the expected kinetic energy with which you pass through the light should be more physically relevant to, say, a bicyclist coasting on a shallow down-hill.<|endoftext|>
TITLE: lists of computed cohomologies?
QUESTION [7 upvotes]: Is there any comprehensive list of examples for computed
1) de-Rham cohomology-groups
2) Lie-algebra-cohomology groups $H^i(\mathfrak{g},\mathbb{R})$
3) equivariant de-Rham cohomology groups ?
Especially I am interested in a comprehensive list of examples (or classes of examples) of
a) manifolds with vanishing first and second de-Rham-cohomology
b) lie-algebras with vanishing first and second lie-algebra-cohomology

REPLY [11 votes]: Chevalley and Borel calculated the cohomology of many Lie groups and Landweber collected those results here in just two pages: it describes the cohomology of $U(n)$, $SU(n)$, $SO(n)$, $G_2$, $F_4$, $E_6$, $E_7$ and $E_8$. 
You might also enjoy Neil Strickland's bestiary of topological spaces and spectra, which lists the (co)homology of many topological spaces.<|endoftext|>
TITLE: Extra assumption in Hodges' lemma on the resultant of a first-order formula?
QUESTION [10 upvotes]: Background
I am working through a particular result in a paper of Cherlin, Shelah, and Shi, and am satisfied that it follows from basic model theory material - but I'm stuck on one point in the background material.  
In Hodges' "A Shorter Model Theory", Lemma 7.2.5 on page 191 seems to have an unused assumption.  He
considers, for a $\forall_2$ $L$-theory $T$, an existential formula $\phi$ and the set of all universal formulas implied (under $T$) by $\phi$.  He calls this set the resultant of $\phi$:   
$Res_\phi(x) =$ {universal $\psi(x)$ | $T \vdash \forall x (\phi(x) \rightarrow \psi(x))$}
Then he shows that an arbitrary $L$-structure $A$
and element $a$ satisfy $Res_\phi(a)$ iff there is some extension $B$
of $A$ with $B \models T$ and $B \models \phi(a)$.  
Key Issue
I can't tell where Hodges is using the assumption that $T$ is $\forall_2$.  The given proof (below) looks like it goes through without that assumption:
$\Rightarrow$ Suppose some extension $B$ of $A$ satisfies $\phi(a)$ and $T$.  Then $B$ satisfies
$Res_\phi(a)$.  Any universal formula satisfied in $B$ must be satisfied in the substructure $A$, also.  Since $Res_\phi(x)$ contains only universal formulas, $B \models Res_\phi(a) \implies A \models Res_\phi(a)$.
$\Leftarrow$:  Suppose $A$ satisfies $Res_\phi(a)$.  If there is no extension
$B$ of $A$ that satisfies $\phi(a)$ and $T$, then add to the language a constant $c$ representing $a$, and consider (in the language $L_c$) $Diag(A) \cup \{\phi(c)\} \cup T$.  By assumption, this set of sentences has no model, so by compactness
$$
T \vdash \phi(c) \rightarrow \neg \sigma(c,d^A)
$$ 
for some $d^A$ in $A$ and some
quantifier-free formula $\sigma$.  Since $c$ and $d^A$ do not appear in $T$, we can replace
them with universally quantified variables:
$$
T \vdash \forall x [ \phi(x) \rightarrow \forall y \neg \sigma(x,y)  ]
$$
Thus, $\forall y(\neg \sigma(x,y)) \in Res_\phi(x)$ so $A \models \forall y (\neg \sigma(a,y))$.  But we got $\sigma$ by considering the $L_c$-diagram of $A$, so 
$A \models \exists y \sigma(a,y)$.  Contradiction.
I don't see anywhere that the $\forall_2$ assumption on T comes in.
Perhaps this assumption is just for context in the rest of the neighboring material?

REPLY [8 votes]: The result doesn't need the assumption that $T$ is an $\forall_2$-theory, nor does it need that the formula $\phi(x)$ is existential.  The relevant general result is that a structure $A$ has an extension satisfying a theory $T$ (in the same vocabulary) if and only if $A$ satisfies all the universal sentences that are provable from $T$.  To get the result quoted in the question, minus the unnecessary hypotheses, apply this general result not to the given $T$ but to $T$ plus $\phi(c)$, where $c$ is a new constant symbol interpreted as $a$.  (Actually, there is an even more general version of the result; see the "Model Extension Theorem" in Section 5.2 of Shoenfield's "Mathematical Logic" or Lemma 3.5.10(i) of Hinman's "Fundamentals of Mathematical Logic.")<|endoftext|>
TITLE: Question about projections on a Hilbert space
QUESTION [7 upvotes]: Sorry for the vague title, I can't think of a better one that isn't overly long.
Suppose that $S$ is a commuting set of projection operators on a Hilbert space. I'll introduce the following notation: if $p \in S$, let $p^+ \equiv p$ and $p^- \equiv 1 - p$. Let $I \equiv  ${$+, -$}. The projections are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$; this makes the set of all projections into a complete lattice. Is the following identity true?
$\sup_{f \in I^S} \inf_{p \in S} p^{f(p)} = 1$
In the case where $S$ is finite with elements $p_1, p_2, \ldots p_n$, the left hand side of this equation is simply the product over $i$ of $p_i + (1 - p_i)$, so I'm interested in whether this can be generalised to the infinite case. It's easy to see that the following two statements are equivalent to the above:
If $\inf_p p^{f(p)} x = 0$ for all $f$, then $x = 0$
If $\sup_p p^{f(p)} x = x$ for all $f$, then $x = 0$
but I have no idea how to prove either of these.
My reason for asking is that I'm trying to show that, if $\mathcal{H}_1$ and $\mathcal{H}_2$ are Hilbert spaces, then if a projection on $\mathcal{H}_1 \otimes \mathcal{H}_2$ is of the form $\sup_i p_i \otimes q_i$, with $p_i$ and $q_i$ drawn from some complete Boolean algebras of projections on $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively, then the $q_i$ may be chosen to satisfy $q_i q_j = 0$ when ever $i \neq j$. So if anybody knows of an alternative way to prove that, or knows that it's false, then by all means say so.

REPLY [5 votes]: Here's a counterexample.  For each natural number $n$, let $D_n$ be the set of those $x\in[0,1]$ whose binary expansion has a 1 in the $n$-th place.  Then the operation of multiplication by the characteristic function of $D_n$ is a projection operator $p_n$ on $L^2[0,1]$.  Let $S$ be the set of these operators $p_n$; they commute.  For any $f:S\to I$, the infimum of the operators $p^{f(p)}$ is zero.  For example, if $f$ were identically $+$, then the range of $p_n$ consists of functions supported on $D_n$ (up to measure 0, of course), and so the infimum would project to functions supported on the intersection of the $D_n$, i.e., a one-point set (up to measure 0 again); such functions are 0 in $L^2$.  For the general case, where some of the values of $f$ are $-$, you can use the same argument with some of the $D_n$ replaced by their complements.<|endoftext|>
TITLE: theorem of Borel and Tits
QUESTION [7 upvotes]: Is there anywhere where I can read a complete proof in English of this theorem by Borel and Tits:

Suppose that $G$ is a simple algebraic group over an infinite field $k$, and that $H$ is a subgroup of $G(k)$ containing the subgroup of $G(k)$ generated by the rational points of the unipotent radicals of the $k$-parabolic subgroups, and that $\alpha \colon H \to G'(k')$ is a homomorphism, where $G'$ is a simple algebraic group over an infinite field $k'$, such that $\alpha(G'')$ is Zariski dense in $G'$. Then there exists a homomorphism $\phi\colon k \to k'$, a $k'$-isogeny $\beta\colon G^\phi\to G'$ with $d \beta \ne 0$, and a homomorphism $\gamma\colon H \to Z_{G'(k')}$ to the centre, all three unique, such that $\alpha(h)=\gamma(h)\beta(\phi^0(h))$ for all $h \in H$.

There is a proof in French in Borel and Tits, Homomorphismes 'abstraits' de groupes algebriques simples, Annals of Mathematics, Second Series, Vol. 97, No. 3 (May, 1973), pp. 499-571.

REPLY [3 votes]: The paper Abstract homomorphisms of simple algebraic groups by Robert Steinberg (Semináire N. Bourbaki, 1972-1973), is written in English and includes a proof of the Borel-Tits theorem.<|endoftext|>
TITLE: Why is proving P != NP so hard?
QUESTION [28 upvotes]: Does anyone have any insight into why it is so hard to prove that P != NP conjecture? There seems to be so much evidence in its favor, and so many problems and techniques with which to attack it, that I don't get why it has remained unproven for so long.

REPLY [32 votes]: On the contrary, there are two major results in complexity theory that rule out a wide class of methods to show that $P \ne NP$.  The first is the theorem of Baker, Gill, and Solovay, that a proof that $P \ne NP$ (or a proof that they are equal) cannot relativize.   In other words, they showed that there exists an oracle relative to which they are equal, and an oracle relative to which they are different.  The second result is the theorem of Razborov and Rudich, that if a widely accepted refinement of the $P \ne NP$ conjecture is true, then there does not exist a "natural proof" that they are different.  By a natural proof, they mean a proof from a large class of combinatorial constructions.  In light of those two theorems, there actually aren't very many known promising techniques left, even though there is a lot of evidence by example that the conjecture seems to be true.  As Razborov and Rudich explain, these two results rule out candidate approaches to P vs NP for sort-of opposite reasons.
There is a CS professor named Ketan Mulmuley who has expressed some optimism that P vs NP can be solved with "geometric complexity theory".  I can believe that Mulmuley is doing interesting work of some kind (which seems to involve quantum algebra and representation theory), but I haven't heard of many complexity theorists who are optimistic along with him that he can really solve P vs NP.  (But hey, Perelman surprised everyone with a proof of the Poincare conjecture, so who knows.)

Some additional remarks.  First, there are plenty of conjectures have ample evidence yet are difficult for no obvious reason.  The P vs NP problem has an unusual status in that people have thought of rigorous reasons that it's hard.
Second, when people prove a "barrier result" (meaning, a negative result about how not to prove a conjecture), obviously the community will take it as a challenge to find new ideas that circumvent the barrier.  As mentioned in the comments, there was even a conference last year on doing exactly that!  Baker, Gill, and Solovay was published in 1975, and it took about 15 years to find convincing exceptions to their point about relativization.  (Unconvincing exceptions that can be explained as unfairly restricted oracle access came more quickly.)  Nonetheless, when I did a computer-assisted survey of binary relations between complexity classes a few years ago, it was clear that the vast majority of these proven relations still relativize.  It is true that by now the research focus is on non-relativizing results, with the exception of quantum complexity classes, where relativizing results are still popular.
Third, the newer Razborov-Rudich theorem made people start all over again to look for barrier loopholes.  Moreover the Baker-Gill-Solovay theorem, as an obstruction to P vs NP, was sharpened somewhat by Aaronson and Wigderson in their paper on "algebrization" of complexity class relations.  My third point is that I can't speak with any authority on efforts to overcome the current set of barriers.<|endoftext|>
TITLE: Partial sums of the Chu--Vandermonde identity
QUESTION [9 upvotes]: I am interested in finding a lower bound of the sum:
$$\sum_{i=0}^d \left(\genfrac{}{}{0pt}{}{n}{i}\right)
\left(\genfrac{}{}{0pt}{}{m}{k-i}\right)$$
when $d < k$ (and assuming both $n\geq k$, $m\geq k$). When $d=k$ this sum  is equal to
$$\left(\genfrac{}{}{0pt}{}{n+m}{k}\right) $$ (the Chu--Vandermonde identity). What I would like to know if there is some good and standard lower bound for the first $d$ terms of this sum in terms of the relation between $d$ and $k$. Is there some standard reference book where I can hope to find such a bound?

REPLY [6 votes]: Carla, I don't think it is a good idea to give a detailed solution here.
My point is that this problem is pretty standard: if you look inside
N.G. de Bruijn's Asymptotic Methods in Analysis,
especially in the 3rd chapter, you will find that your sum (generically) falls into the
category c (p. 54, a comparatively small number of terms somewhere in the middle);
the method is discussed in full details (Section 3.4) on the example of a very similar binomial sum.<|endoftext|>
TITLE: Preservation of limits
QUESTION [7 upvotes]: Is there a functor that preserves all small limits but not a large one?

REPLY [18 votes]: Let's try this. I'll use colimits, so take the opposite. 
The class of all ordinals is ordered. Add one more element $\infty$ at the end, bigger than all of them. View this "large ordered set" as a large category $\cal C$. A small diagram in $\cal C$ has colimit $\infty$ if $\infty$ occurs in the diagram, and otherwise it has a colimit less than $\infty$. But the large diagram consisting of everything except $\infty$ has colimit $\infty$. The functor to the ordered set $\lbrace 0<1\rbrace$ that sends $\infty$ to $1$ and everything else to $0$ preserves small colimits but not all colimits.<|endoftext|>
TITLE: existence of Morse functions satisfying the Palais-Smale condition
QUESTION [6 upvotes]: Given an arbitrary Hilbert manifold, can on find a complete Riemannian metric and a Morse function satisfying the Palais-Smale condition?

REPLY [8 votes]: I recently learned that the answer to the question is YES, answered in the ETH preprint "H-cobordism for Hilbert Manifolds" by Dan Burghelea. I found the reference in the article "On the differential topology of Hilbert manifolds" of Eells and Elworthy.<|endoftext|>
TITLE: Why are two notions of Gaussian curvature are the same - what is the simplest & most didactic proof?
QUESTION [23 upvotes]: This question is still wide open - all of the answers so far rely on magical calculations. I've only accepted an answer because, by bounty rules, otherwise one would be accepted automatically. I can't change the accepted answer, but it would be amazing to have more discussion on this question.
I'd like a nice proof (or a convincing demonstration), for a surface in $\mathbb R^3$, that explains why the following notions are equivalent:
1) Curvature, as defined by the area of the sphere that Gauss map traces out on a region.
1.5) The integral of the product of principal curvatures.
2) The angle defect of parallel transport about a geodesic triangle.
(This equivalence may be considered as either a part of the Theorema Egregium or a part of Gauss-Bonnet. Proving that numbers 1 and 1.5 are the same is pretty easy).
Motivation: I'm teaching a five-day class for very bright high-school students. The idea is to give them an impression of what geometry is about. However, when I looked at Spivak's proof of this, it was much more of a messy calculation than I expected. I'd like, if at all possible, something more conceptual, ideally with a nice picture attached to it.
Since this doesn't have to be a perfectly complete class, I'll be perfectly happy with a good illustration of why this is true instead of a rigorous proof, if a conceptual and rigorous proof is completely out of the question. 
One idea I had is to show the example of a sphere and the hyperbolic plane, and then explain that on very small scales the curvature is constant. However, then I would need a nice proof that the embeddings of the hyperbolic plane in $\mathbb R^3$ have curvature -1.
Thank you very much!
P.S. This question is related, but not quite the same (I hope), to this question:
Equivalent definitions of Gaussian curvature
P.P.S. Thank you to whomever recommended to Berger's "Panoramic View of Riemannian Geometry". it was quite useful to me. I do not know why you deleted your answer. 
That books claims there is no conceptual proof. However, I'd still be very happy with a nice illustration of why one should believe this, especially for negative curvature.

REPLY [4 votes]: There is a physical "proof" of this fact which I learned from Mark Levy;
it is in his book "THE MATHEMATICAL MECHANIC: Using Physical Reasoning to Solve Problems".
Imagine that you keep the axis of a bicycle wheel
and move it in such a way that the bicycle wheel lies in
the tangent plane to the surface.
In the initial position the wheel stays still; 
you go along the loop in the surface and stop. After that your wheel rotates by some angle $\alpha$.
If your loop was triangular this angle is its defect; this is (2) in your list. 
The parallel motion does not rotates the wheel, 
so the same result will be the same if you only rotate the axis without moving the center of the wheel.
This tells you that $\alpha$ depends only on the spherical image of the loop
and from here it is easy to see that it is proportional to the algebraic area of the domain bounded by the spherical image of the loop.
I.e., the area of the sphere that Gauss map traces out on a region;
this is (1) in your list.<|endoftext|>
TITLE: Simply-connected domain around a curve
QUESTION [6 upvotes]: In a current project with a colleague, we have come across the following reasonably classical-sounding geometric question. While not vital to our work, it would be interesting if anyone has seen this type of issue discussed before, and if there are any references.
To keep things simple, I will formulate a special case, which appears to retain all issues present in the general case.

Question
Let $\gamma:[0,\infty)\to\mathbb{C}\setminus\{0\}$ be a continuous and injective curve in the punctured complex number plane, with $\gamma(t)\to\infty$ as $t\to\infty$.
Is there a simply-connected domain $V\subset\mathbb{C}\setminus\{0\}$ with $\gamma\subset V$ such that $\gamma$ tends to the boundary of $V$ horocyclically in $V$?

(The latter condition means the following: There is a conformal isomorphism $\phi$ that maps $V$ to the right half plane in such a way that $\operatorname{Re}(\phi(\gamma(t)))\to+\infty$ as $t\to\infty$.)

Remarks

We may additionally suppose that a function $\delta(t)$ is given, and require that $V\subset\bigcup_{t\geq 0}B(\gamma(t),\delta(t))$, where $B(z,\delta)$ is the open disk of radius $\delta$ around $z$. This makes the statement of the question slightly more complicated, but makes the discussion of examples less cumbersome. (It is also part of the more general setup I mentioned above, which additionally replaces $\mathbb{C}\setminus\{0\}$ by an arbitrary open Riemann surface.)
If the curve $\gamma$ is $C^1$, then it is easy to construct such a domain, by taking $V$ to be a (shrinking) "tubular" neighborhood of $\gamma$. In this case, it is even possible to ensure that the convergence is non-tangential, meaning that the argument of $\phi(\gamma(t))$ stays bounded as $t\to\infty$. 
It is not too difficult to see that non-tangential convergence cannot be obtained without some regularity assumption. However, perhaps a suitable uniform Lipschitz condition is sufficient.
However, for horocyclic convergence, one can easily make do with much weaker conditions. For example, it seems enough to ask that the curve is $C^1$ on some sequence of intervals tending to infinity, while the behaviour in between can be as bad as you like.
I tend to think that one can probably construct a counterexample for horocyclic convergence, but I may be wrong and in any case it is likely to involve some thought and effort. So I am hoping that someone can tell me that this or a similar question has been discussed in the literature.

REPLY [3 votes]: First of all, notice that if $D$ is a compact and $\Gamma$ is a compact Jordan curve that starts on the boundary of $D$ but otherwise lies in the infinite component of the open complement of $D$, then you can find a polynomial that is approximately $0$ on $D$ and approximately $f$ on $\Gamma$ where $f$ is any continuous function on $\Gamma$ that is almost $0$ at the starting point of $\Gamma$. That is just Mergelyan (we can close all holes in $D$ losing nothing).
Using this, we can easily construct an entire function that approximates any continuous function $f$ we want on $\gamma$ with any precision. Indeed, choose $R_1>0$ and approximate $f$ on the piece $\Gamma_1$ of $\gamma$ that goes from the starting point until the last point of $\gamma$ on the circle of radius $R_1$ by $p_1$. Next choose $R_2$ and find a polynomial that approximates $0$ on $\Gamma_1\cup D(0,R_1)$ and $f-p_1$ on the part $\Gamma_2$ of $\gamma$ that runs from the end of $\Gamma_1$ to the last point on the circle $D(0,R_2)$, and so on. The series $\sum_j p_j$ will converge on the entire plane and be almost equal to $f$ on $\gamma$.
Now we conclude that we may have a harmonic on the entire plane function $u$ that is positive and tends to plus infinity along $\gamma$. It remains to take $V$ to be the connected component of the set $u>0$ containing $\gamma$ (then $u$ has to be the real part of the conformal mapping in question up to a positive constant factor just because there are very few positive harmonic functions in the right half plane that are $0$ on the boundary).
I leave the (harder) task to find this argument in the literature to somebody else :).<|endoftext|>
TITLE: Seeing stacks in the Calculus of Functors
QUESTION [27 upvotes]: Recently I was told (by an algebraic geometer) that when algebraic geometers look at the Calculus of Functors, they think of stacks.
When I look at the Calculus of Functors, I see a categorification of polynomial approximation. While I am at best a beginner at algebraic geometry, I would like to understand why he is saying this. 
My motivation is twofold. First, I want to know why he is saying this, and second, because I am beginning to learn about stacks, and I want to come at it with some intuition. I have pursued the obvious routes of reading about them in general (such as Tolland's Blog Post).
Specifically, my question is 

How does one see Calculus of Functors as stacks?

A secondary question, 

Is there some highly degenerate way to look at stacks to see polynomial approximation?

Thanks!

REPLY [4 votes]: EDIT:

Regarding the original question: I think that ultimately the sense in which functor calculus is like stacks is that we are peforming localizations of a functor category. Moreover, these localizations are left exact. This leads to similarities in how the localizations are described (via descent with respect to certain "covering" diagrams, which can be described as (hyper)covers if you like) and how the localizations are constructed. More on this below.

Regarding the analogy between manifold calculus and Goodwillie calculus: In both cases, we have a tower of localizations of a functor category $Fun(C,D)$. In both cases, we have natural descriptions of the localizations in terms of descent with respect to certain "covering" diagrams, and in both cases the tower of localizations corresponds to a filtration of the collection of "covering" diagrams for the strongest localization.
But I don't think these filtrations arise in analogous ways. In the manifold calculus, the strongest localization comes from the usual Grothendieck topology on the site of manifolds: "covers" are Cech covers in the usual sense; the $n$th localization is at the subcollection of $n$-Weiss covers as described by Dmitri Pavlov. In the homotopy calculus, we start with a "cover" being any strongly cartesian cube in $Spaces^{op}$; the $n$th localization is obtained by restricting to cubes of dimension $\geq n+1$. There's no Weiss anything in sight.



Here is a nuts-and-bolts comparison, which I think lurks behind the Rezk / Lurie approach to Goodwillie calculus in Higher Algebra, 6.1.1. I'm comparing to 6.2.2 in Higher Topos Theory for the sheaf case.

Recall that a functor $F: \mathcal C \to \mathcal D$ is $n$-excisive if it takes strongly cocartesian $n$-cubes to cartesian $n$-cubes. So what we have are a bunch of "covering" diagrams on $\mathcal C^{op}$ (given by the $n$-cubes which are strongly cocartesian in $\mathcal C$) and we are localizing $Fun(\mathcal (C^{op})^{op}, \mathcal D)$ with respect to these "covers", just like when we localize a category of presheaves at the "covers" of a Grothendieck topology to get sheaves.

So much could be said of any localization, but the analogy extends further, to the way that the localization is computed. For sheafification, we have the Grothendieck plus construction $F^+(C) = \varinjlim_U \varprojlim_{C' \in U} F(C')$ where the colimit is over covers $U$ of $C$. The sheafification of $F$ is computed by iterating this construction [1]. In Goodwillie calculus we do exactly the same thing, but it's simpler because there's always a unique finest cover of any object $C \in \mathcal C$, namely the unique strongly cocartesian cube with $C$ at the initial vertex and the terminal object at all of the vertices adjacent to the terminal vertex. This construction is called $T_n(F)(C) = \varprojlim_{C'} F(C')$ where the limit is over the aforementioned finest cover (the colimit was computed by evaluating at the finest cover). We iterate this construction to produce the $n$th polynomial approximation $P_n(F)$.

If you look at Lurie's treatment, the analogy extends further to the reasoning why this construction correctly computes the localization of $F$. In each case, one shows that the resulting functor is local by inputting a cover $U$ of an object $C$, and then factoring the natural transformation $FU \Rightarrow F^+U$ (resp. $FU \Rightarrow T_n(F)U$) through a limit diagram; applying this factorization to each $F^{+\alpha}U \Rightarrow F^{+\alpha^+}U$ (resp. $T_n^\alpha(F)U \Rightarrow T_n^{\alpha^+}(F)U$) produces a cofinal chain of limit diagrams [2]. Universality comes because the map $F \to F^+$ (resp. $F \to T_n(F)$) is local by construction.

Then the verification that this localization is left exact proceeds in the same way in both cases: the construction $F \mapsto F^+$ (resp. $F \mapsto T_n(F)$) preserves finite limits because (finite) limits commute with limits [3]. Then the localization functor commutes is a filtered colimit of these functors, so it also commutes with finite limits, because finite limits commute with filtered colimits.



[1] In the case of ordinary sheaves, the iteration actually stabilizes after two steps, but in homotopical contexts it may need to be iterated transfinitely.
[2] In the sheaf case, we need to iterate long enough so that the colimit over the $F^{+\alpha}$'s commutes with th limit; in the Goodwillie case, we usually just assume we're in a situation where finite limits commute with filtered colimits, and then since our "covers" are finite, this suffices.
[3] In the sheaf case, we need to know that so do filtered colimits.<|endoftext|>
TITLE: Geometric flavored textbook on algebra
QUESTION [14 upvotes]: I am interested in topology, while I am not so comfortable with some 
algebraic flavored textbook on algebra. Actually, it was not until I learned some topology that I began to understand some abstract algebra. I believe that behind every algebraic theorem, there is a geometric analogue, and this is what I am interested in.
I want to find algebraic textbooks that are of geometric flavor. Say, Armstrong's book, Groups and Symmetry is a lovely book on group theory that is of this type. Are there some more such textbooks (on ring theory and homological algebra)?
Thank you!

REPLY [2 votes]: manin's linear algebra and geometry<|endoftext|>
TITLE: Poll about your proof of resolution of singularities and a request for advice
QUESTION [5 upvotes]: The questions first: What is the proof of resolution of singularities that you know? 
Why am I asking?: There are a number of proofs of resolution of singularities of varieties over a field of characteristic zero, all with more or less similar flavor but different in technical details and in choices that the resolution algorithm allows us to make. When writing a proof that uses specific features of some of those details I can't stop being uneasy about assuming the reader read about the specific constructions elsewhere. 
I would like to know from MOers what proof you have seen and if you have a reason for the choice, if it was a choice, I would like to hear it too. 
Maybe asking about what you know is too invasive. I am just asking for the proof that you happened to find in your way, even you have only read a few lines of it. 
The purpose of the question: The conspicuous one. To get a sense, by a rough approximation and a small sample, of what proofs are more culturally known. Have a concrete feeling when sending a reader to find the details in other paper, either of feeling OK with it or of guilt. 
It is a question about fashion, which also has its role in mathematics... and knowing what the fashion is is useful. 
What details?: Although I had in mind a specific detail of the proofs I didn't mention it because it is not the only one that changes from proof to proof and because the result of the poll  gives information about all of them. Examples are: the resolution invariant, the ways of making the local descriptions of the centers of blowings-up match to form a globally defined smooth subvariety, the ways of getting functoriality and the different meanings that functoriality can have... 
(edit) Forgot the "request for advise": If you have would like to give advise about how you have dealt with similar situations and describe your example that is welcomed.  
It is a community wiki question, so feel free to change what is said here if needed or if you want the poll to also give information about other questions that you would like to be answered. (or for correcting the English!)

REPLY [5 votes]: As a bit of an update on the subject in characteristic 0, there have been some more recent papers which try to vary the desingularization algorithm slightly to achieve a better resolution of singularities, for example algorithms which avoid simple normal crossings or semi-simple normal crossings or even "minimal" singularities. These proofs shed light on the original algorithm while showing how far we can push this approach. If you read through some of these articles, you will not only get a glimpse of the original algorithm, but you will see how the desingularization invariant can be used as a computational tool. Here are the more recent articles on the topic:

Bierstone/Da Silva/Milman/Pacheco, Desingularization by blowings-up avoiding simple normal crossings

Bierstone/Pacheco, Resolution of singularities of pairs preserving semi-simple normal crossings

Bierstone/Pacheco, Desingularization preserving stable simple normal crossings

Bierstone/Milman, Resolution except for minimal singularities 1 and Bierstone/Lairez/Milman, [...] 2 (the case of four variables)


I found the following two articles very enlightening when I was learning about the topic:

Bierstone/Milman, Resolution of singularities

Bierstone/Milman, Functoriality in resolution of singularities


The first goes into great detail about some of the subtle points of the algorithm, while the second gives a rather concise proof of resolution of singularities.
Finally, the article below is quite distinct in that it is the first time (as far as I know) that the complexity of the desingularization algorithm has been studied - that is, how many steps in the algorithm occur before it terminates? Can we find an upper bound? You can read more about that here:

Bierstone/Grigoriev/Milman/Wlodarczyk, Effective Hironaka resolution and its complexity (with appendix on applications in positive characteristic)<|endoftext|>
TITLE: Measurable cardinals under Axiom of Determinacy
QUESTION [9 upvotes]: I seem to remember reading somewhere that ZF+AD proves that $\omega_1$ and $\omega_2$ are measurable cardinals.
Is that right?
If so, can someone [point me to or give here] a [sketch or proof] of these results?

REPLY [2 votes]: For a different proof of the measurability of $\omega_1$ under AD, you could look at Eugene Kleinberg's book Infinitary Combinatorics and the Axiom of Determinateness. He shows it using the strong partition relation  on $\omega_1$(that any function colouring the $\omega_1$-cardinality subsets of $\omega_1$ into 2 colours has a homogeneous subset of size $\omega_1$), showing that the filter of $\omega$-closed unbounded sets of $\omega_1$ is actually a normal ultrafilter. The only issue is that proving the strong partition relation is slightly involved. 
P.S. Could someone please tell me how to get the arrow notation for representing partition relations right? I have absolutely no idea. Also, please correct me if I'm wrong above; it's my first post on Math Overflow!<|endoftext|>
TITLE: Is tensor product of local algebras local?
QUESTION [12 upvotes]: In general, the tensor product of two local rings is not local. For example, $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}\ $ is not a local ring. 
Let $\mathbb{F}_{p}$ denote the finite field with $p$ elements. Let $A,B$ be two complete local noetherian $\mathbb{Z}_p$-algebras with residue field $\mathbb{F}_p$. Let $m_A, m_B$ denote the maximal ideals of $A,B$, respectively.
Question:
Is it true that $A \otimes_{\mathbb{Z}_p} B\ $ is a local ring?
Clearly, the ideal $m_A \otimes B + A \otimes m_B$ is a maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $ with the residue field $F_p$. Is it the only maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $?

REPLY [6 votes]: Let $A=\mathbb F_p[[t]],B=\mathbb F_p[[u]]$. Then, $1\otimes1-t\otimes u$ is neither in $\mathfrak m_A\otimes B+A\otimes\mathfrak m_B$ nor a unit, so it is contained in some other maximal ideal of $A\otimes B$. (Proof that $1\otimes1-t\otimes u$ is not a unit: An element of $\mathbb F_p[[t]][[u]]$ coming from $A\otimes B$ has the property that its coefficients with respect to $u$ span a finite-dimensional subspace of $\mathbb F_p[[t]]$, but this fails for the coefficients $t^k$ of $(1-tu)^{-1}=\sum t^ku^k$.)<|endoftext|>
TITLE: Did Durov's work give an example of noncommutative schemes?
QUESTION [6 upvotes]: I just took a look at the nlab entry: Nikolai Durov. It seems that Skoda never mentioned that what Durov introduced was a special case of generalized scheme theory. I did not read his dissertation carefully or completely. I wonder whether his "generalized scheme" is a special case of noncommutative scheme in the sense of A. Rosenberg. 
I will give a talk on noncommutative schemes in a few days. Now I am collecting interesting examples of noncommutative schemes (quasi schemes). Examples that I already know are:
commutative schemes; D-modules; quantum D-modules; almost schemes by Gabber; general Grothendieck category (abelian category); Artin-Zhang noncommutative projective schemes; quantum flag variety in the sense of Rosenberg and Lunts; holonomic D-modules. 
Thanks!

REPLY [6 votes]: No, there is no need for the Rosenberg noncommutative scheme that the categories be abelian in general; whatever being said in his 1998 paper he does not mean so. He defines a relative scheme over a base category given by exactness properties of direct and inverse image functors describing covers used for gluing and the affinity property. In general one has to be careful, with formulating correctly the exactness properties for nonabelian context. 
Of course, to justify this one needs to say that the Durov's schemes are determined by the categories of quasicoherent modules. I do not know if the reconstruction theorem a la Gabriel-Rosenberg holds for the generalizes schemes of Durov. Durov glues schems along categorical localizations which he calls for some reason "pseudolocalizations". They just have the correct exactness properties. On the other hand, for commutative monads, Durov has two nice versions of prime spectra; now one should compare those with a version of Rosenberg's spectrum for nonabelian context. Now one version if the spectrum for right exact categories of Rosenberg. What is a right exact structure for the case of categories like the ones in Durov's work ? Well Durov has spent some time to develop a theory of vectoids which generalize topoi, but also the categories of modules over finitary monads in Set and, more generally, the categories of quasicoherent sheaves of $\mathcal{O}$-modules over generalized schemes. Durov wrote a draft text in Russian about vectoids (which I have seen but is not released yet), and a video of a talk at Steklov. Is there a canonical right exact structure on a vectoid for which the spectrum of the right exact category in the sense of Rosenberg gives a sensible reconstruction theorem, it would be very interesting to investigate. 
Remark: I believe that that the quantum flag variety of Lunts-Rosenberg is isomorphic in the case of $SL_n$ to the flag variety studied in my thesis from the dual point of view and with explicit Ore localizations. I never had time to check and publish all the details.<|endoftext|>
TITLE: Geometric interpretation of characteristic polynomial
QUESTION [63 upvotes]: The coefficients of lowest and next-highest degree of a linear operator's characteristic polynomial are its determinant and trace. These have well-known geometric interpretations. But what about its intermediate coefficients?
For a linear operator $f : V \to V$, we have the beautiful formula
$$\chi(f) = det(f - t) = \sum_{i=0}^n (-1)^i\  tr(\wedge^{n-i}(f))\ t^i,$$
where $\wedge^{p}(f)$ is the map induced by $f$ on grade $p$ of $V$'s exterior algebra.
While this formula is rarely mentioned (at least I haven't seen it in any of the standard textbooks), it is not too surprising if you have a good grasp of exterior algebra. It presents $\chi(f)$ as a generating function for the exterior traces of $f$.
My question is whether these traces have a simple geometric interpretation on par with $tr$ and $det$.

REPLY [10 votes]: I am reluctant to answer a question this old that already has a very nice answer, however, looking at the title the first thing that comes to my mind is something quite different from the existing answer (and maybe it will be useful to someone who comes across this question).
When $V$ is a finite dimensional vector space over $\mathbb{C}$, then the coefficients of the characteristic polynomial are global coordinates on certain moduli spaces.  I think of global coordinates as geometric since they generally give an embedding of the space in question, and sometimes they also give an explicit geometric description (as with the examples below).
First, the coefficients give a coordiante system on the moduli space of representations of $\mathbb{Z}$ into $GL(V)$.  To see this first identify $Hom(\mathbb{Z}, GL(V))$ with $GL(V)$, and then the coefficients of the characteristic polynomial, call them $c_1,....,c_n$ where $n=\dim(V)$, are conjugation invariant functions $GL(V)\to \mathbb{C}$.  As they are symmetric polynomials in the eigenvalues (on the open dense subset of diagonalizable matrices) they are algebraically independent.  As the dimension of the moduli space $Hom(\mathbb{Z}, GL(V))//GL(V)$ is $n$, we have a global coordinate system.  The moduli space is then seen to be $\mathbb{C}^{n-1}\times \mathbb{C}^*$.
Now, if one wants to allow general linear operators (still assuming $V$ is finite dimensional over $\mathbb{C}$), then similar ideas give that the coefficients are global coordinates on $End(V)//GL(V)$ which is seen to be isomorphic to $\mathbb{C}^n.$  From this point-of-view, the "geometric interpretation of the characteristic polynomial" is that its coefficients give the global geometry of the moduli space of operators themselves.<|endoftext|>
TITLE: 4-manifolds in the 4-sphere such that it, *and* its complement have unsolvable word problem
QUESTION [11 upvotes]: In an earlier thread I had asked whether or not one can find a smooth 4-dimensional submanifold of $S^4$ whose fundamental group has an unsolvable word problem.  The answer is yes, and the reference is in that thread. 
see: Word problem for fundamental group of submanifolds of the 4-sphere
I'm curious if one can go a step further.  Can one find a smooth $4$-dimensional submanifold $M$ of $S^4$ such that both $\pi_1 M$ and $\pi_1 (S^4 \setminus M)$ have unsolvable word problems? 
My motivation for this question has to do with the decidability of a classical 3-manifold theory problem.  Given a smooth $3$-manifold, determine whether or not it admits a smooth embedding into $S^4$.  
It's possible to algorithmically generate all smooth embeddings of $3$-manifolds in $S^4$ using normal surface theory for triangulations of the $4$-sphere ((a) start with a given triangulation of $S^4$, (b) enumerate vertex-normal $3$-manifolds, (c) barycentrically subdivide, goto (b)).  This is an unfortunately horribly memory-intensive algorithm, but it's exponential run-time so it could be a lot worse.
Whether or not the embedding problem is algorithmically decidable boils down to whether or not there is a computable function $\beta : \mathbb N \to \mathbb N$ such that if a $3$-manifold $M$ has a triangulation with $n$ tetrahedra, you want to know that you only need to barycentrically subdivide a triangulation of $S^4$ at most $\beta(n)$ times to find $M$ as a vertex-normal 3-manifold in the triangulated $4$-sphere. 
Another way to look at this is that perhaps some $3$-manifolds have to embed in $S^4$ in extremely "twisted" ways.   If $M$ embeds in $S^4$ such that both $4$-manifolds that it separates $S^4$ into have unsolvable word problem, that qualifies for "really twisted" in my opinion.   
This also gets to the issue of invariants.  If you want to say that a $3$-manifold does not embed in $S^4$, perhaps invariants sensitive to the fundamental group should be important.  But any such invariant would have to be relatable to $\pi_1$ of the two $4$-manifolds that $M$ separates $S^4$ into.  So issues of computability also come up here.  At present I know of no obstruction to a $3$-manifold embedding in $S^4$ that sees really "deeply" into $\pi_1 M$ but it seems like such obstructions are a realistic possibility.  Alexander modules are known to play a role as an obstruction. Twisted Alexander modules should play a role as well, for example, although the literature on this is currently contained in the (closely related) knot-concordance world. 
Presumably this is an open problem (my apologies) but I'm curious if there are any insights out there. 
An offshoot of this question would be whether or not every $3$-manifold that admits an embedding in $S^4$ has some embedding in $S^4$ where the two $4$-manifolds it splits $S^4$ into have solvable fundamental group word problems.

REPLY [6 votes]: The answer to the  question in the beginning should be  YES and it follows from the answer to your previous question. We just need to use the fact that if $G$ has unsolvable word problem then $G*F$ too, where $F$ is a group and $G*F$ is the free product.
To construct the example take the solution to the previous question, namely a $4$-manifold  $M^4 $embeddable  in $S^4$  with unsolvable word problem. Now take an open ball $B^4$ in $M^4$ and cut from it a small copy of $M^4$, that we call $N^4$. Finally dig a wormhole that connects $N^4$ with $S^4\setminus M^4$. This divides the sphere into two connected 4-manifolds each of which is homotopic to a connected sum of two manifolds, one of which has same fundamental group as $M^4$.<|endoftext|>
TITLE: Justification of the term "invertible sheaf"
QUESTION [12 upvotes]: Let $X$ be a locally ringed space (or a scheme) and $M,N$ two $\mathcal{O}_X$-modules such that $M \otimes N \cong \mathcal{O}_X$. Does it follow that $M$ is invertible in the usual sense, namely that $M$ is locally free of rank $1$?
It is true if $M$ is locally of finite type (which is, of course, also necessary). 
Proof: Let $x \in X$. Then $M_x \otimes N_x \cong \mathcal{O}_{X,x}$. Now tensor with the residue field of $\mathcal{O}_{X,x}$ and use linear algebra to conclude that $M_x / \mathfrak{m}_x M_x$ is $1$-dimensional. Since $M_x$ is of finite type over $\mathcal{O}_{X,x}$, Nakayama shows that $M_x$ is generated by just one element. Since $M$ is of finite type in a neighborhood of $x$, it follows that the generator at $x$ is also a generator in a neighborhood of $x$. Also $N$ has one generator, and their tensor product is a generator of $M \otimes N \cong \mathcal{O}_X$, which must be free. Thus also the generators of $M$ and $N$ are free.
But I don't know what happens in the general case. Here are some intermediate questions:

Does it follow that $M$ is flat?
Is the resulting morphism $M \to Hom(N,\mathcal{O}_X)$ an isomorphism?
Is the claim true for $X$ a point, i.e. a local ring?
Is the claim true if $X$ is an affine scheme and $M,N$ are quasi-coherent? (Thus in the question, replace $\mathcal{O}_X$ by a usual ring.)

REPLY [12 votes]: Yes, for quasi-coherent sheaves on a scheme it is true that if $M \otimes N \cong \mathcal{O}_X$, then 
$M$ is locally free of rank one. It is enough to prove the 
REDUCTION  Let $M,N$ be $A$- modules such that $M \otimes_A N\cong A$. Then $M$ is projective of finite type.
Proof : We are given an isomorphism $f:M \otimes_A N\cong A$ . Say $\; f( \Sigma m_i\otimes n_i)= 1$ (FINITE index set!). The composition of the isomorphisms
$g_M:M \to M\otimes (N\otimes M):m\mapsto \Sigma m\otimes (n_i\otimes m_i)$
$assoc: M\otimes (N\otimes M) \to (M\otimes N) \otimes M:
m\otimes (n \otimes m') \mapsto (m\otimes n)\otimes m'$ 
$f_M:(M\otimes N) \otimes M \to M: (m\otimes n) \otimes m_1\mapsto f (m \otimes n).m_1 $
is the isomorphism $j:M\to M: m\mapsto \Sigma  f(m\otimes n_i).m_i$
By introducing the linear forms $\nu_i:M\to A: m\mapsto f(m\otimes n_i)$ we see that we have an isomorphism
$j:M\to M: m\mapsto \Sigma \nu _i(m).m_i$ and we deduce that for all $m\in M$ we can write  $m=\Sigma \nu _i(m).j^{-1}(m_i)$.
It is well known that the existence of such a dual basis $(j^{-1 }m_i, \nu_i)$ proves that $M$ is a finitely generated projective module.<|endoftext|>
TITLE: Degree sequences of multigraphs with bounded multiplicity
QUESTION [5 upvotes]: I got to thinking about this problem while sifting through the math puzzles for dinner thread.  There's a fun puzzle by rgrig which asks the guests to prove that when they came to dinner two of them shook hands the same amount of times.  The solution is to make the handshake graph, and apply the pigeonhole principle on the vertex degrees.  The key observation is that if there are $n$ guests, then the degrees 0 and $n-1$ cannot both occur.
Yaakov Baruch made the astute comment that the result is false if the forgetful mathematicians shake hands twice with each other.  However, note that mathematicians are probably not so forgetful to shake hands three times with someone.  This leads us to the following questions:
Question 1.  Is it true that for large $n$, every loopless multigraph on $n$ vertices with at most two parallel edges between any two vertices has two vertices of the same degree?  This is false for $n=3,4,5,6,7$.
Question 2. Is there a nice characterization of such degree sequences?
I'll finish with what is known.  If we allow loops, then the problem is trivial, because all sequences whose sum is even are degree sequences.  To see this, put a matching on the odd degree vertices, and then add loops.  For simple graphs, there are many nice characterizations.  For example, the Erdos-Gallai theorem says that a decreasing sequence $(d_1, \dots, d_n)$ is a degree sequence of a simple graph if and only if $\sum_i d_i$ is even and for all $k \in [n]$
\[
\sum_{i=1}^k d_i \leq k(k-1)+ \sum_{i=k+1}^n \min (k, d_i). 
\]
This sort of answers Question 2, since a degree sequence of a multigraph with multiplicity at most 2 is the sum of two degree sequences of simple graphs.  However, this is a rather convoluted characterization, and it is unclear how to answer Question 1 from it.  
I'll end by mentioning that if we do not bound the multiplicity of edges, then there is  a nice characterization of degree sequences of multigraphs by Hakimi (1962).

REPLY [2 votes]: Here is the answer of Question 2: (Fulkerson,Hoffman,McAndrew[1965]) 
A decreasing sequence $(d_1,d_2,...,d_n)$ of non-negative integers is a degree sequence of a loopless multiple graph with each edge capacity atmost 2  if and only if $ \sum_{i=1}^{n} d_i$ is even and for each subset A of [n]
 $$ \sum_{i=1}^{n} min(d_i,2|A-\{i\}|) \geq \sum_{j \in A} d_j$$<|endoftext|>
TITLE: Denominators in the solution to Hilbert's XVII
QUESTION [6 upvotes]: Hilbert's seventeenth problem asks to prove that every positive semidefinite form can be written as the sum of squares of rational functions. Currently we don't seem to have a good understanding of how many such squares are sufficient, or how ugly the denominators will eventually get.
One theorem that bears some psychological value in this direction is the following result of Pólya (discussed somewhere in "Inequalities" by Hardy, Littlewood and Pólya):

If a homogeneous polynomial $p$ in $n$ variables $x_1,x_2,\dots,x_n$ is positive in the standard simplex then there exists an $N=N(p)\in \mathbb N$ so that $(x_1+x_2+\cdots+x_n)^N p$ has all coefficients positive.

This means that every even positive definite form can be multiplied by a high enough power of  $x_1^2+\cdots+x_n^2$ to have all coefficients positive and thus be a sum of squares of polynomials. In fact it is true that after multiplying any positive definite form with a high enough power of $x_1^2+\cdots+x_n^2$ one gets a sum of squares of polynomials. 
Now, let $S_{n,d}$ be the set of positive semidefinite forms in $n$ variables of degree $d$. Let $\Sigma$ be the set of all forms that are a sum of squares of polynomials. 

Question: Can we always choose a countable collection of forms $Q=\{q_1,q_2\dots\}$ so that for each  $p\in S_{n,d}$ we have $q_i^2p\in \Sigma$ for some $i\in \mathbb N$?

I have proven that a finite $Q$ does not suffice, however I can't seem to prove that one can get a way with countably many denominators. It would be nice if this was true and moreover one could find a $Q$ with some simple structure like in Polya's theorem. On the other hand, if it wasn't true it would be interesting to know if one can put any condition at all.

REPLY [2 votes]: I passed this question on to Bruce Reznick, who points to http://www.math.uiuc.edu/~reznick/aud.pdf, 
MR2159759 (2006d:12001)
Reznick, Bruce(1-IL)
On the absence of uniform denominators in Hilbert's 17th problem. (English summary)
Proc. Amer. Math. Soc. 133 (2005), no. 10, 2829--2834 (electronic).
Bruce adds, "I already proved that finitely many won't work."<|endoftext|>
TITLE: Flatness and local freeness
QUESTION [35 upvotes]: The following statement is well-known:
Let $A$ be a commutative Noetherian ring and $M$ a finitely generated $A$-module. Then $M$ is flat if and only if $M_{\mathfrak{p}}$ is a free $A_{\mathfrak{p}}$-module for all $\mathfrak{p}$.
My question is: do we need the assumption that $A$ is Noetherian? I have a proof (from Matsumura) that doesn't require that assumption, but the fact that other references (e.g. Atiyah, Wikipedia) are including this assumption makes me rather uneasy.

REPLY [5 votes]: This is too long for a comment, so I'm forced to include it as an answer, even though it isn't one.  In defense of ashpool's question:  until reading Brian Conrad's excellent answer, I'd assumed that finite presentation (not just finite type) hypotheses were necessary (for the statement that flat modules over a local ring are free).  Reason 1:  there is a (slightly) easier proof in the finite presentation case, and when I proved it myself (something my advisor wisely told me to always do, rather than looking things up), I was led to the finite presentation argument (see 25.4.2 on the June 27 2011 version of the notes here --- it will be fixed in later versions), and didn't think to go further.  Reason 2:  When sources such as Mumford (see the 2nd edition of the Red Book, p. 215) state the result with finite presentation hypotheses, my prejudices are confirmed.  Then when I read Matsumura, I see what I expect to see, and not what is on the page.
Also, I liked Matt's interesting point:  you only need Noetherian hypotheses to get from "free stalks" to "locally free sheaf".<|endoftext|>
TITLE: Deformations and the dual numbers
QUESTION [10 upvotes]: The question I have is pretty straightforward, and its answer could very well be contained in some more complicated question(s) asked previously.  Here it is:
Why are deformations typically considered as structures over the ring of dual numbers?
My motivation is rather general: I just want to know more about deformation theory.  And this question seems like a place to start.  The problem for me is that all resources I have read just jump into the fray with "Consider the ring of dual numbers..." and define first-order deformations straight away.
I am sure that I am just missing something obvious; any nudge in the right direction will be greatly appreciated.

REPLY [6 votes]: Here are two main ideas which I found helpful when I was learning this stuff (very basic and handwavy, but good starting points to see where more precise and sophisticated statements are going):

see how, for instance, an algebra defined over $k[x]$ can be seen as a continuously-(in some sense)-varying $k$-indexed family of $k$-algebras.  In algebraical terms, you recover the elements of the family by localising at the ideal $x=t$ for each specific $t \in k$.  Geometrically, think of the Grothendieck-style picture where rings really represent some kind of nice spaces, but by duality, so the maps go the “wrong” way.  So a $k[x]$-algebra $R$ corresponds to a ring map $k[x] \to R$, and hence (since $k[x]$ represents the affine line $\mathbb{A}^1(k)$) to a map of spaces $X \to \mathbb{A}^1(k)$.  But then we can think of $X$ as the bundle of all its fibers, a family varying over $\mathbb{A}^1(k)$.
More generally, a widget defined over $k[x]$ can be thought of a a continuous (in some sense) map from $\mathbb{A}^1(k)$ into the space of widgets. 
now, see how just as $k[x]$ represented the affine line, $k[x]/(x^2)$ represents an “infinitessimally short line segment”, or a “walking tangent vector” (other answers include good sketches of what's meant by this and why it's a powerful viewpoint).  In particular, if $R$ correpsonds to a space $X$, a map $R \to k[x]/(x^2)$ corresponds to a map from the walking tangent vector into the space $X$, or in other words a point of $X$ and a tangent vector at that point.  Algebraically, we get a $k$-point of $R$ (by setting $x=0$), and (under good circumstances) a point of the Zariski tangent space.

So then putting these two together, one can see why a widget defined over the dual numbers should behave like some kind of an infinitessimal deformation: a widget, together with a tangent vector at that point in the space of widgets.  (This may be literally true if there's a good moduli space of widgets!)<|endoftext|>
TITLE: What is the ring of integers of the Pythagorean field?
QUESTION [23 upvotes]: Following Hilbert, we call the complex numbers constructible via
compass and straight-edge the field of Euclidean numbers, and
the totally real such numbers the field of Pythagorean numbers.  (Among other possible definitions, an algebraic number is totally real if its minimal polynomial has all real roots).  For a reference, Richard Alperin gives a description of these and related fields from a constructibility viewpoint in his paper "Trisections and Totally Real Origami."
There is a remarkably nice characterization of the Pythagorean numbers
-- the Pythagorean field is the smallest field containing the
rationals and closed under the operation $x\rightarrow \sqrt{1+x^2}$.  Or, from an only slightly different viewpoint, it is the Pythagorean closure of $\mathbb{Q}$, in the sense of
Link
Because it's a nice "hands-on" intro to this field, let me include in the question Daniel Litt's comment below that since $\sqrt{2}=\sqrt{1+1^2}$, and $\sqrt{3}=\sqrt{1+\sqrt{2}^2}$, and so on, the Pythagorean field contains $\sqrt{n}$ for all $n\geq 0$, and hence contains the compositum of all real quadratic fields.
My Question:

What is the ring of integers of the Pythagorean field?

Note that the most naive guess of it being the smallest subring of algebraic integers closed under the operation $x\rightarrow \sqrt{1+x^2}$ is incorrect -- this ring does not include $\frac{1+\sqrt{5}}{2}$, which is certainly a totally real Euclidean algebraic integer.  I suspect/hope (though this may just be the second most naive guess) that there's some description of the form "smallest subring of the algebraic integers closed under $x\rightarrow \sqrt{1+x^2}$ and division by 2 when certain conditions are met."  I've done a little bit of a literature search on rings of integers of totally real multiquadratic extensions of $\mathbb{Q}$, but haven't found anything even remotely inspiring something of this form.
I don't have much to offer in terms of motivation, except that I have come across a variety of rings of integers in my research, and I'm trying to decide if any are exactly the ring of Pythagorean integers.  It would be nice to be able to compare them to the Pythagorean integers just by seeing whether or not one of these rings satisfies certain closure operations.

REPLY [4 votes]: The first idea I tried to look at was replacing the iteration by
$$ x \mapsto  x' = \frac{1 + \sqrt{1+4x^2}}{2}, $$
which eliminates the problem with $\frac{1+\sqrt{5}}{2}$; on the other hand I cannot see that e.g. the integral element $\frac{1 + \sqrt{3}}{\sqrt{2}}$ lies in the ring generated by this iteration.
By modifying this approach we can eliminate all obvious counterexamples. Define $v_0 = 2$ and 
$$  v_1 = \sqrt{2}, v_2 = \sqrt{2+\sqrt{2}}, ...., v_{n+1} = \sqrt{2+v_n}, ... . $$
Given an algebraic integer $x$ in the Pythagorean ring, let $v(x)$ denote the "maximal" product of elements $v_i$ dividing both $2$ and $x$ (I guess that a proof of the existence of $v(x)$ is not too hard; Krull, Herbrand and Scholz studied the arithmetic in these "infinite number fields" in the 1930s, and their methods should suffice). Then the iteration
$$ x \mapsto x' = \frac{1+x + \sqrt{1+x^2}}{v(x)} $$
sends algebraic integers to algebraic integers, and this map has at least a chance of
generating the ring of all integers in the Pythagorean field.<|endoftext|>
TITLE: Finitely generated subgroups with infinite cyclic quotient
QUESTION [12 upvotes]: Suppose that $G$ is a finitely presented group and $H$ is a finitely generated normal subgroup such that $G/H$ is infinite cyclic. Is it true that $H$ is finitely presented?

REPLY [5 votes]: It worth noticing that the Sigma-invariants (also known as BNS- or BNSR-invariants) provide a strategy to answer such a question. 
Given a group $G$ of type $F_n$ and an integer $k \leq n$, the $k$-th Sigma-invariant $\Sigma^k(G)$ is a (complicated) subset of the character sphere
$$S(G):= \left( \mathrm{Hom}(G,\mathbb{R}) \backslash \{ 0 \} \right) / \text{positive scaling}$$
of $G$. Notice that the class of every non trivial morphism $G \to \mathbb{Z}$ is an element of $S(G)$.
Theorem: The kernel of $\chi : G \twoheadrightarrow \mathbb{Z}$ is of type $F_k$ if and only if $[\chi], [-\chi] \in \Sigma^k(G)$.
So the case $k=2$ corresponds to $\mathrm{ker}(\chi)$ finitely presented. Unfortunately, these invariants are usually quite difficult to compute. Nevertheless, there exists a useful method, based on Bestvina and Brady's Morse theory, extending their work on right-angled Artin groups (mentioned in Henry's answer). 
For instance, all the Sigma-invariants are completely known for right-angled Artin groups and some Thompson-like groups. An application I really like is:
Theorem: Any finitely presented normal subgroup of Thompson's group $F$ is of type $F_{\infty}$.
A few references on the subjet:

Strebel, Notes on the Sigma-invariants.
Bux & Gonzales, The Bestvina-Brady construction revisited - Geometric computation of $\Sigma$-invariants for right-angled Artin groups.
Witzel & Zaremsky, The $\Sigma$-invariants of Thompson's group $F$ via Morse theory.
Zaremsky, On the $\Sigma$-invariants of generalized Thompson groups and Houghton groups.
Zaremsky, Symmetric automorphisms of free groups, BNSR-invariants, and finiteness properties.<|endoftext|>
TITLE: Associated graded and flatness
QUESTION [9 upvotes]: Let $M$ be a filtered module over a filtered algebra $A$, and suppose $gr(M)$ is flat over $gr(A)$, where $gr$ means the associated graded module and algebra, respectively.
What can one say in general about the flatness of $M$ over $A$, or with relevant assumptions (for instance in the above, we should assume both filtrations are complete to avoid dumb counterexamples)?  Are there good references for this sort of question?  I have played with the various definitions of flatness trying to find an obvious relationship, but I find flatness proofs confusing.
The particular examples I have in mind are comparing $U(\mathfrak{g})$-modules to $S(\mathfrak{g})$-modules, and $D(X)$-modules to $O(T^*X)$-modules for affine varieties $X$, if it helps.  I suspect the answer doesn't depend on any of the details though.

REPLY [8 votes]: Let me suppose, as in your examples, that we have a base field $k$.
It is well known that to check that a right $A$-module $M$ is flat it is enough to show that whenever $I\leq_\ell A$ is a left ideal, the map $M\otimes_AI\to M\otimes_A A$ induced by the inclusion $I\to A$ is injective. This condition can be rewritten: $M$ is flat iff for each left ideal $I\leq_\ell A$ we have $\mathrm{Tor}^A_1(M,A/I)=0$.
So now suppose $A$ and $M$ are (exhaustively, separatedly, increasingly from zero) filtered in such a way that $\mathrm{gr}M$ is a flat $\mathrm{gr}A$-module. 
Pick a left ideal $I\leq_\ell A$; notice that the filtration on $A$ induces a filtration on the quotient $A/I$. We can compute $\mathrm{Tor}^A_\bullet(M,A/I)$ as the homology of the homologically graded complex $$\cdots\to M\otimes_kA^{\otimes_kp}\otimes_kA/I\to M\otimes_kA^{\otimes_k(p-1)}\otimes_kA/I\to\cdots$$ with certain differentials whose formula does not fit in this margin, coming from the bar resolution. Now the filtrations on $M$, on $A$ and on $A/I$ all collaborate to provide a filtration of our complex. We've gotten ourselves a positively homologicaly graded with a canonically bounded below, increasing, exhaustive and separated filtration. The corresponding spectral sequence then converges, and its limit is $\mathrm{Tor}^A_\bullet(M,A/I)$. Its $E^0$ term is the complex
$$\cdots\to\mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_kp}\otimes_k\mathrm{gr}(A/I)\to \mathrm{gr}M\otimes_k\mathrm{gr}A^{\otimes_k(p-1)}\otimes_k\mathrm{gr}(A/I)\to\cdots$$
with, again, the bar differential, and its homology, which is the $E^1$ page of the spectral sequence, is then precisely $\mathrm{Tor}^{\mathrm{gr}A}_\bullet(\mathrm{gr}M,\mathrm{gr}(A/I))$. Since we are assuming that $\mathrm{gr}M$ is $\mathrm{gr}A$-flat, this last $\mathrm{Tor}$ vanishes in positive degrees, so the limit of the spectral sequence also vanishes in positive degrees. In particular, $\mathrm{Tor}^A_1(M,A/I)=0$.
NB: As Victor observed above in a comment, Bjork's Rings of differential operators proves in its Proposition 3.12 that $\mathrm{w.dim}_AM\leq\mathrm{w.dim}_{\mathrm{gr}A}\mathrm{gr}M$ (here $\mathrm{w.dim}$ is the flat dimension) from which it follows at once that $M$ is flat as soon as $\mathrm{gr}M$ is; the argument given is essentialy the same one as mine. I am very suprised about not having found this result in McConnell and Robson's!<|endoftext|>
TITLE: Striking applications of Baker's theorem
QUESTION [19 upvotes]: I saw that there are many "applications" questions in Mathoverflow; so hopefully this is an appropriate question. I was rather surprised that there were only five questions at Mathoverflow so far with the tag diophantine-approximation, while there are almost 900 questions on number theory overall. It is my intention to promote the important subject a little bit by asking one more question.
Question: 

What are some striking applications of Baker's theorem on lower bounds for linear forms on logarithms of algebraic numbers?

If, for example, I were in a discussion with a person who has no experience with diophantic approximation, to impress upon the person the importance of Baker's theorem I would cite the following two examples:

Giving effective bounds for solutions of (most of the time exponential) diophantine equations under favorable condition. For example, Tijdeman's work on the Catalan conjecture, or giving effective bounds for Siegel's theorem, Fermat's last theorem, Falting's theorem, etc., in certain cases.
Transcendence results which are significant improvements over Gelfond-Schneider. In particular, the theorem that if $\alpha_1, \ldots, \alpha_n$ are $\mathbb{Q}$-linearly independent, then their exponentials are algebraically independent over $\mathbb Q$. I would cite the expose of Waldschmidt for more details. 

These are, to me, quite compelling reasons to study Baker's theorem. But as I do not know much more on the subject, I would run out of arguments after these two. I would appreciate any more striking examples of the power of Bakers' theorem. This is 1. for my own enlightenment, 2., for future use if such an argument as I hypothesized above actually happens, 3. To promote the subject of diophantine approximation in this forum, especially in the form of Baker's theorem.

REPLY [2 votes]: This recent result Root separation for trinomials by Koiran also uses Baker's Theorem. In my PhD thesis, we use this result to demonstrate a polynomial time algorithm to isolate real roots of integer trinomials.<|endoftext|>
TITLE: Do homotopy colimits always commute with homotopy colimits?
QUESTION [8 upvotes]: Do homotopy colimits commute with homotopy colimits? The setting I am thinking of is that of a triangulated category with a model, but it would be interesting to have more general answers as well. A good reference would also be appreciated.

REPLY [8 votes]: Yes.  See theorem 24.9 of this paper.  Here's a somewhat less abstract reference.<|endoftext|>
TITLE: Intuition for Model Theoretic Proof of the Nullstellensatz
QUESTION [13 upvotes]: I recently read the model-theoretic proof of the Nullstellensatz using quantifier elimination (see www.msri.org/publications/books/Book39/files/marker.pdf). I'm convinced that the Nullstellensatz is true, i.e. by showing that $\exists y_1 \cdots \exists y_n (\bigwedge_{i} (f_i(y) = 0)$ is equivalent to a quantifier free formula, hence if it is true in an algebraically closed extension, it is true in the original algebraically closed field. However, I don't have any intuition for how this formal sequence of steps leads to the truth of the Nullstellensatz. In particular, what I don't see is what the quantifier-free formulas are that these are equivalent to. For example, what is a quantifier-free formula in $a,b,c,d,e,f,g,h,i,j,k,l$ which is equivalent to the system $ax^2+bxy+cy^2+dx+ey+f=gx^2+hxy+iy^2+jx+ky+l=0$ having a non-trivial solution? And generalizations?
Can one simplify the model-theoretic proof by showing more directly that the existence of solutions to systems of equations (or non-equations) in algebraically closed fields are equivalent to polynomial conditions on the coefficients? I.e. apply essentially the same argument but restrict to the case of algebraically closed fields and avoid general results like Godel's Completeness Theorem to make the argument more clear. My expectation is that this might involve proving an algorithm for determining the quantifier-free formula from the formula with quantifiers.

REPLY [3 votes]: After some more searching, I found the notes written by Ford for an REU. These notes prove the Nullstellensatz using model theory, though the presentation is a little bit more concrete than that in Marker's notes.
On page 7, in Theorem 4.3, the notes prove quantifier elimination for ACF without reference to any preceding lemmas. The basic idea is this: First note that we only need consider formulas of the form $\exists x \phi(\bar{v},x)$. In other words, if our system of equations has $n$ variables, we need to first find quantifier-free conditions on $x_1,\cdots,x_{n-1}$ which will give the system of equations a solution in $x_n$. (In general, we will be dealing with a system of equations and inequations joined by and/or, but I ignore that for simplicity). Next, we find conditions which will make those conditions have a solution in $x_{n-1}$, etc. Repeating this process, we have a series of algebraic equations and inequations in the coefficients of the polynomials joined by and/or which gives a necessary and sufficient condition for the equation to have solutions.
Thus the problem is to find an algebraic condition (i.e. a sequence of expressions of the form $p(x)=0$ and $p(x) \neq 0$ joined by and/or operators) on the coefficients $\bar{v}$ of a system of $m$ equations in one variable $x$ which will ensure that the equations have a solution over an algebraically closed field.
To give a rough sketch, assume our system contains at least two non-constant polynomials in $x$, with degrees $1 \le d_1 \le d_2$ and leading coefficients $a_i(\bar{v})$. Then our new condition is $$(a_1(\bar{v})=0 \wedge p_1(x)-a_1(\bar{v})x^{d_1}=0 \wedge p_2(x)=0 \wedge\cdots \wedge p_m(x)=0) \vee (a_1(\bar{v}) \neq 0 \wedge p_1(x)=0 \wedge a_1(\bar{v})p_2(x)-a_2(\bar{v}) x^{d_2-d_1} p_1(x) = 0 \wedge p_3(x)=0 \wedge \cdots \wedge p_m(x)=0)$$ The sums of the degrees of the polynomials is smaller than before, and we can repeat this process until we don't have at least two non-constant polynomials. Then we use the Euclidean algorithm to deal with the $\neq$, and we are done.
What's nice is that this actually gives an elementary and algorithmic proof of the Nullstellensatz: first reduce the existence of a solution over an algebraically closed field to a system of equations and inequations in the coefficients. Next, note that there is a solution over some algebraically closed extension. Therefore, the coefficients satisfy the necessary conditions, so there is a solution over our original algebraically closed field, and we are done. I believe this is correct and much simpler than the one which Terry Tao provides.
Now we tackle the example which I give (again, a relatively rough sketch). We want to find conditions on $a,b,c,d,e,f,g,h,i,j,k,l,x$ which ensure that the equations have a solution in $y$.
Let the first equation be $p_1$, the second $p_2$. We thus reduce it to either $c=0$ and $(e+bx)y+ax^2+dx+f=iy^2+(hx+k)y+gx^2+jx+l=0$ (Case 1) or $c \neq 0$ and $cy^2+(e+bx)y+ax^2+dx+f= (chx+ck-ie-ibx)y+cgx^2+cjx+cl-iax^2-idx-if=0$ (Case 2).
In Case 1, the first polynomial is $p_1$, the second is $p_2$, and if $e+bx=0$, the equation having a solution is equivalent to $ax^2+dx+f=0$ and either $i \neq 0$ or $hx+k \neq 0$ or $gx^2+jx+l=0$. If $e+bx \neq 0$, we subtract $iy$ times the first polynomial from $e+bx$ times the first to get $
(e+bx)(iy^2+(hx+k)y+gx^2+jx+l)-iy((e+bx)y+ax^2+dx+f)=$
$bx gx^2 + e gx^2 + bx hx + e hx - ax^2 iy - bx iy - dx iy - e iy - 
 f iy + bx iy^2 + e iy^2 + bx jx + e jx + bx k + e k + bx l + e l$ We now have a system of simultaneous linear equations, so to check whether there's a solution, we only need to check whether each has a solution and, degenerate cases aside, whether a certain determinant is $0$, i.e. an algebraic condition.
In Case 2, the first polynomial is $p_2$, the second is $p_1$. If $chx+ck-ie-ibx=0$, then the existence of a solution is equivalent to $cgx^2+cjx+cl-iax^2-idx-if=0$ and either $c \neq 0$, $e+bx \neq 0$, or $ax^2+dx+f=0$. If $chx+ck-ie-ibx \neq 0$, then we compute $(chx+ck-ie-ibx)(cy^2+(e+bx)y+ax^2+dx+f)-cy((chx+ck-ie-ibx)y+cgx^2+cjx+cl-iax^2-idx-if)=$
$ax^2 chx + ax^2 ck - cgx^2 cy - cjx cy - cl cy + chx cy^2 + ck cy^2 + 
 chx dx + ck dx + chx f + ck f + cy iax^2 - ax^2 ibx - cy^2 ibx - 
 dx ibx - f ibx + cy idx - ax^2 ie - cy^2 ie - dx ie - f ie + cy if$. We then proceed as in Case 1.<|endoftext|>
TITLE: Monic polynomial from absolute value information
QUESTION [8 upvotes]: I'm trying to find the minimal (monic) polynomial $M(x)$ (over the rationals) for an algebraic number. I know the degree of the polynomial (call it $d$) and I have $d+1$ data points of the form $(x_i, |M(x_i)|)$. The $x_i$ are all rational numbers, so the $|\cdot|$ is just regular absolute value.
If it wasn't for the absolute value sign, it'd be a straight-forward polynomial fit. However, to only way I've been able to solve for $M(x)$ is fitting a polynomial for each of the $2^d$ choices of plus/minus on the second coordinate, and then finding one that's monic.
Luckily, so far this has always produced one and only one such polynomial. Anybody know a more feasible/elegant way to do this?
(For more motivation than you actually care for, see http://course1.winona.edu/eerrthum/Papers/MAANCS081018.pdf where on slide 8 (page 34) I mention the brute force method.The last slide contains an example calculation.)
(Feel free to re-tag as appropriate.)

REPLY [2 votes]: Here's a solution using lattice reduction:
1) Find degree $d$ polynomials $p_i(x)$ such that $p_i(x_j) = |M(x_i)| \delta_{i j}$.
2) Let $c_i$ be the coefficient of $x^d$ in $p_i(x)$, and $c$ the $d+1$ long column vector whose coordinates are $c_i$.
3) Find a matrix $U \in SL_{d+1}(\mathbb{Z})$ such that $U c = e$, where $e$ is the $d+1$ long column vector with a 1 in the first coordinate and zeros elsewhere.  Added later:
not quite.  Let $D$ be a common denominator of all the elements of $c$, and form a $d+1 \times d+1$ matrix, $A$, whose first column is $cD$ and the lower $d \times d$ block is the identity matrix (with the rest of the top row 0).  Find the Hermite normal form: $U \in SL_{d+1}(\mathbb{Z})$, $U A = H$.  The first column of $H$ will be zeros below the first entry, which will be a positive integer $r$.  In order for there to be a solution it is necessary that $r | D$.  Form a new matrix $U'$ by multiplying the top row of $U$ by $D/r$.
4) The answer (see below) is a vector in the $\mathbb{Z}$-lattice generated by the bottom $d$ rows of $U'$ which is close to the top row.
Namely, form the matrix $W$ by adjoining a $d+1 \times d+1$ identity matrix to the right of $c$.  Since only the coefficient of $x^d$ matters in the answer, we can see that an answer will be given by some integer linear combination of the rows of $W$ which has $\pm 1$ in the last $d$ coordinates.  The squared Euclidean length of that vector will be $d+1$, which is quite short.  There are a number of algorithms for finding a closest vector (in theory for general lattices it's a hard problem, but in practice in a lattice like this it's not too hard).  For a nice account of how to do it look in http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.81.8089 starting around page 14.
The idea here is to prepend a column to $U'$ which is a unit vector with 1 in the first position and 0's everywhere else.  Now multiply the rest of the matrix (all columns but the first) by a large scaling factor, $s$.  This will make sure that the first row will show up in the linear combination forming the shortest basis vector.
Lattice reduction will supply a short vector in the lattice which we know that our answer is.  We then read off the coefficients of the $p_i$ in the last $d+1$ coordinates.
I've programmed this, and tested it on random polynomials of degree 20, and it successfully finds the $\pm 1$ combination leading to a monic polynomial.
In Tony Scholl's example of three different polynomials having the same data, the lattice generated has a lot of short vectors, so in that case one needs to enumerate short vectors to pick out the answers.<|endoftext|>
TITLE: What does this naive attempt at $S^1$-equivariant homology describe?
QUESTION [11 upvotes]: After reading Cohen and Voronov's notes on string topology, one can find the following construction: Suppose we have a topological space $X$ with continuous action of $S^1$. This means we have a map $\rho: S^1 \times X \to X$. If we choose a fundamental class $[S^1]$ of the circle, we can form the operator $\Delta: H_\ast(X) \to H_{\ast+1}(X)$ by setting $\Delta(a) = \rho_\ast([S^1] \times a)$. For dimensional reasons, $\Delta$ squares to zero and therefore turns $H_\ast(X)$ into a cochain complex. Let's denote the cohomology of this complex by $H_\ast^\Delta(X)$. Because in this cohomology we have as representatives homology classes in $X$ which are annihilated by the action of $S^1$ "in a homological sense", one would think this is relevant to the $S^1$-equivariant homology of $X$.
However, a few simple examples show that $\Delta$-cohomology and equivariant homology are certainly not equal. For example, take the point with trivial action, then $H^\Delta_\ast(pt)$ is $\mathbb{Z}$ in degree 0 and zero in all other degrees, but $H^{S^1}_\ast(pt) = \mathbb{Z}[a_2]$ where $|a_2| = 2$. Another example is $LS^1$.
So my question is: is there a different (hopefully geometric) description of $\Delta$-cohomology? Is it related to $S^1$-equivariant homology in any way? 
If this is not possible in the general case, is it at least possible for $X$ of the form $LM$, with $M$ a manifold? My main motivation for considering $\Delta$-cohomology is string topology, where $\Delta$ is also known as the BV-operator.

REPLY [14 votes]: By $G$-equivariant homology of $X$ you mean the homology of the Borel construction or homotopy orbit space $EG\times_GX$. (I only mention this because the same phrase can refer to other kinds of homology for $G$-spaces, satisfying a weaker kind of homotopy axiom. This "Borel homology" gives you an isomorphism for every $G$-equivariant map $X\to Y$ that is nonequivariantly a homotopy equivalence.)
Yes, the operator that you call $\Delta$ is related to $S^1$-homology. There is a first-quadrant spectral sequence for the $G$-homology of $X$, coming from the fact that $EG\times _GX$ is a bundle over $BG$ with fiber $X$. The $E^2$ is $H_i(BG;H_j(X)$. It is a spectral sequence of modules for the ring $H^{-*}BG$. When $G=S^1$, the groups are $E^2_{2k,j}=H_j(X)$ if $k\ge 0$ and otherwise $0$, and the differential $d^2_{2k,j}\to E^2_{2k-2,j+1}$ is independent of $k$. This first differential $d^2$ is the operator $\Delta$. The $2$-periodicity continues in $E^r$ until the differentials start crossing the vertical axis. So basically your $\Delta$-homology is $E^3=E^4$ and then there is an operator $d^4$ going from $ker(\Delta)/im(\Delta)$ to $ker(\Delta)/im(\Delta)$ and raising $j$ by $3$; and so on. 
This is related to cyclic homology.<|endoftext|>
TITLE: Making the branching rule for the symmetric group concrete
QUESTION [16 upvotes]: This question concerns the characteristic $0$ representation theory of the symmetric group $S_n$.  I'm a topologist, not a representation theorist, so I apologize if I state it in an odd way.
First, a bit of background.  The finite-dimensional irreducible representations of $S_n$ are given by the Specht modules $S^{\mu}$.  Here $\mu$ is a partition of $n$, which is best visualized as a Young diagram.  There are classical rules for restricting $S^{\mu}$ to $S_{n-1}$ and inducing $S^{\mu}$ to $S_{n+1}$ (these are known as branching rules).  Namely, we have the following.

The restriction of $S^{\mu}$ to $S_{n-1}$ is isomorphic to the direct sum of the $S_{n-1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of removing a box from the Young diagram for $\mu$ (while staying in the world of Young diagrams).

The induction of $S^{\mu}$ to $S_{n+1}$ is isomorphic to the direct sum of the $S_{n+1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of adding a box to the Young diagram for $\mu$ (again while staying in the world of Young diagrams).


These two rules are equivalent by Frobenius reciprocity.
There is a nice concrete proof of the restriction rule (I believe that it is due to Peel, though I first learned about it from James's book "The representation theory of the symmetric groups").  Assume that the rows of the Young diagram for $\mu$ from which we can remove a box are $r_1 < \ldots < r_k$, and denote by $\mu_i$ the partition of $n-1$ obtained by removing a box from the $r_i^{\text{th}}$ row of $\mu$.  There is then a sequence
$$0=V_0 \subset V_1 \subset \cdots \subset V_k = S^{\mu}$$
of $S_{n-1}$-modules such that $V_i/V_{i-1} \cong S^{\mu_i}$.  In fact, recalling that $S^{\mu}$ is generated by elements corresponding to Young tableaux of shape $\mu$ (known as polytabloids), the module $V_i$ is the subspace spanned by the polytabloids in which $n$ appears in a row between $1$ and $i$.
Question : Is there a similarly concrete proof of the induction rule (in particular, a proof which does not appeal to Frobenius reciprocity and the restriction rule)?

REPLY [5 votes]: I guess this is an old thread now, but I always thought that James' proof was a little indirect and not very explicit. A much nicer proof was given by Steen Ryom-Hansen in his paper “Grading the translation functors in type A.” Journal of Algebra 274, no. 1 (2004): 138–63; see arXiv:math/0301285.
To blow my own trumpet, a "cellular algebra" proof in the cyclotomic case (which includes the symmetric groups) is given in arXiv:0903.4493. For the corresponding result in the graded setting, following Ryom-Hansen, see arXiv:1008.1462.
All of these arguments work over arbitrary rings.<|endoftext|>
TITLE: dg objects: Z-graded vs. Z/2Z-graded
QUESTION [7 upvotes]: I am wondering: Are there any general theorems or principles relating the theory of Z-graded dg objects and the theory of Z/2Z-graded dg objects? I am mainly interested in dg algebras, dg Lie algebras, and dg categories over fields of characteristic zero.

REPLY [2 votes]: To me, the main difference between the Z-graded and Z/2-graded cases is that the former allows certain simplifying boundedness restrictions, which in the latter do not seem to make sense.  Typically, one considers a nonpositive (in the cohomological grading) Z-graded dg-algebra and dg-modules bounded above or below over it, as appropriate.  The case of connected, simply connected (in the sense of cochains, not just cohomology) nonnegative dg-algebra is similar.
The typical simplification achieved under such restrictions is that a dg-module whose underlying graded module is projective is always homotopy projective.  Also, the two ways of defining the differential derived functors (by taking infinite direct sums or products along the diagonals) become equivalent, since the sums/products are actually finite.
When one has to consider dg-algebras that do not satisfy the above kind of restrictions and/or unbounded dg-modules, the Z-graded situation is not any simpler than, and not much different from, the Z/2-graded situation.
References: 1. Keller "Deriving DG-categories"; 2. Husemoller, Moore, Stasheff "Differential homological algebra and homogeneous spaces".<|endoftext|>
TITLE: Does elliptic regularity guarantee analytic solutions?
QUESTION [22 upvotes]: Let $D$ be an elliptic operator on $\mathbb{R}^n$ with real analytic coefficients.  Must its solutions also be real analytic?  If not, are there any helpful supplementary assumptions?  Standard Sobolev methods seem useless here, and I can't find any mention of this question in my PDE books.
I began thinking about this because I overheard someone using elliptic regularity to explain why holomorphic functions are smooth.  Aside from the fact that I find that explanation to be in poor mathematical taste (I regard the beautiful regularity properties of holmorphic functions as fundamentally topological phenomena), it occurred to me that standard elliptic theory falls short of exhibiting a holomorphic function as the limit of its Taylor series.  So I'm left wondering if this is an actual limitation of elliptic regularity which could vindicate and entrench my topological bias.
In the unfortunate event of an affirmative answer to my question, I would be greatly interested in geometric applications (if any).

REPLY [7 votes]: Also: there is a classical result due to Charles Morrey, "Analyticity of the solutions of analytic non-linear elliptic systems of partial differential equations", that says that if $F(x,u,\nabla u,\nabla^2 u,...)$ is analytic in its arguments and elliptic then the solution of $F(x,u,\nabla u, \nabla^2 u,...)=0$ will be as well. (It actually goes one step further to deal with systems, but the notion of ellipticity is complicated to explain.) This result generalizes work done since the early 1900's; references can be found in Fritz John's (and two other author's I can't recall) pde book.<|endoftext|>
TITLE: How to construct a vector fields with isolated zeros?
QUESTION [7 upvotes]: The Poincare-Hopf theorem tell us that the sum of the indices of a vector field at isolated zeros on a compact, oriented manifold is the same as the Euler characteristic of the manifold. But how to construct a vector fiedls with isolated zeros?

REPLY [5 votes]: If one takes the differential of a Morse function, one gets
a differential form (a cotangent field) with isolated zeros.
If one has a Riemannian metric on the manifold one can
convert between covector fields and vector fields. So,
from a Riemannian metric and a Morse function you can write
down a vector field with isolated zeros.

REPLY [4 votes]: Your question isn't very well defined.  A manifold on its own is not an object where constructions come by easily.  But there is a generic way to construct vector fields with isolated zeros.  Any vector field can be approximated by one with isolated zeros.  This is a consequence of Sard's theorem.  So start off with the zero vector field and choose any small random perturbation of that, and there you go. 
If you want a more constructive answer you'll have to assume a more constructive situation.  Like say if your manifold is triangulated, or has a handle decomposition, or a morse function.  
Chapman describes the Morse situation so I'll give the triangulation situation. 
The vector field has these properties: 
There is a critical point at the barycentre of every cell in the triangulation.  The vertices are repellors.  The barycentres of the top-dimensional simplices are the attractors.  A 1-simplex is a (1,n-1)-index critical point -- meaning there's two orbits approaching (along the 1-simplex) and an n-2-dimensional family of reverse orbits attracting.  Etc.  A j-simplex barycentre has a j-1-dimensional family of attracting orbits, and an n-j-1-dimensional family of reverse orbits attracting. 
That isn't quite explicit as one needs an explicit smoothing of the triangulation to put this all together.  But it gives you the idea.

REPLY [3 votes]: Just use the transversally theorem, an application of Sard's theorem: the generic vector field intersects the zero-section of the tangent bundle transverse, therefore the zeros are isolated.<|endoftext|>
TITLE: Cubical cohomology and de Rham cohomology
QUESTION [11 upvotes]: Qiaochu's question on a discrete analogue of harmonic function theory reminded me of some thoughts I had a long time ago about the relationship between cubical cohomology and de Rham cohomology.
The main reason de Rham cohomology is often used as a first introduction to algebraic topology is due to the few technical prerequisites. You can set up the differential graded ring structure with very little work. One of the important results is the Stokes theorem,
$$\int_C d\omega = \int_{\partial C} \omega,$$
where $\omega$ is a $p$-form and $C$ is a $(p+1)$-chain. Conceptually, it says the differential operator $d$ is adjoint to the boundary operator $\partial$ with respect to the non-degenerate bilinear pairing $\int$.
It is sometimes remarked that you could turn around and use this result to justify defining $d$ as the adjoint of $\partial$. Indeed, that is how simplicial cohomology is usually defined in terms of simplicial homology. But it seems clear that the better analogy is with cubical cohomology:
Establishing homotopy invariance and the cup product for singular simplicial cohomology is a bit technical. In contrast, it's easy with singular cubical cohomology: $I^m\ I^n$ = $I^{m+n}$, so cubical prisms and products are manifestly cubical, no subdivision needed. Demonstrating these properties for de Rham comology can be done with similar ease. The advantages and disadvantages of cubes over simplices have been discussed here before.
The analogy goes further. If you write out products and exterior derivatives of differential forms in coordinates and compare them with the corresponding formulas in cubical cohomology in a lattice-like neighborhood, they are exactly analogous in the same way that freshman calculus and the calculus of finite differences are analogous. If you have trouble wrapping your head around the geometrical interpretation of concepts in de Rham cohomology like products and differentials and especially more advanced concepts in Hodge theory, the analogy with cubical cohomology is very enlightening (or at least it was for me).
My question is whether anyone has worked out this relationship in detail. Maybe in a context where infinitesimals have a concrete existence as in algebraic geometry and synthetic differential geometry.

REPLY [7 votes]: M. Carmen Minguez in this article constructs a homomorphism between de Rham and Cubical Singular Cohomology without showing that is an isomorphism. This is done in the context of Synthetic Differential Geometry.
In general Synthetic Differential geometers seem to be quite aware of the cubical setup, probably because differential forms naturally get a cubical structure if defined via infinitesimals. This is very visible in chapter IV, section 1 of Moerdijk/Reyes' book on synthetic differential geometry. In remark 1.8 (p. 145) they mention a bridge to cubical (co)homology - then, however, they choose to establish the isomorphism between de Rham and simplicial singular cohomology.<|endoftext|>
TITLE: What's the simplest rational not expressible as a sum of a given number of unit fractions?
QUESTION [20 upvotes]: This is essentially the same as the closed question Representation of rational numbers as the sum of 1/k but I hope I can make a case for it as an MO-worthy question. 
Ed Pegg, Jr., in his Math Games column for 19 July 2004 at the MAA website, http://www.maa.org/editorial/mathgames/mathgames_07_19_04.html writes, "Here is an interesting sequence of fractions that would likely would [sic] have fascinated Ahmes: $$1/2, 2/3, 4/5, 8/11, 14/17, 19/23, 24/29, 49/59, 65/71, 76/83, 61/157, 183/191, 260/269, 289/299.$$ $8/11 = 1/2 + 1/6 + 1/21 + 1/77$. This is the simplest Egyptian fraction that requires 4 parts. 
$14/17 = 1/2 + 1/4 + 1/20 + 1/55 + 1/187$ requires 5 parts. 289/299 is the simplest fraction that requires 14 parts. One might think that this sort of thing was well known, but it isn't.... What is the simplest fraction that requires 15 parts, 16 parts, and beyond?" 
Pegg never defines "simplest," but presumably it means smallest (positive) denominator and, among fractions with the same denominator, smallest (positive) numerator. So the general question would be, given $s$, what's the simplest rational that can be expressed as a sum of $s$ unit fractions, but not fewer? 
In this form, it's probably an open, and maybe impossible, problem (that is, I don't think anyone will find a simple formula for the rational as a function of $s$), so let me ask a bit less. Has there been any advance beyond 14 since 2004? Are there any bounds in the literature (that is, bounds on the "complexity" of the rational as a function of $s$)? 
I note that Pegg gives no source for his list of 14. The Online Encyclopedia of Integer Sequences does not recognize the sequence of numerators, nor the sequence of denominators. Before anyone suggests typing "Egyptian fractions" into Google, or looking at the Wikipedia article on that subject, I hope he or she will verify that the particular question I'm asking is in fact answerable by such means. 
EDIT: As per the comments, it appears that only the first four terms in Pegg's list are correct, and that the current state of knowledge is $${1\over2},{2\over3},{4\over5},{8\over11},{16\over17},{77\over79},{732\over733}.$$ 
Also as per the comments, if we are after
$f(s)=\min\lbrace b:N(a,b)=s{\rm\ for\ some\ }a,1\le a\lt b\rbrace$ then 
$f(s)\ge e^{Cn^2}$ for some $C>0$, and, conjecturally, $f(s)\ge e^{e^{Cn}}$ for some $C>0$. 
At this point I will gladly settle for a calculation of $s(8)$.

REPLY [16 votes]: $s(8) = \frac{27538}{27539}$.
I have made the code I used available at http://crypt.org/hv/maths/least_eg-0.01.tar.gz, with a README file at http://crypt.org/hv/maths/least_eg-0.01-README.
Update: those links no longer available, code is available via github at https://github.com/hvds/seq/tree/master/least_eg.
The package includes both PARI/GP code and C code using the GNU GMP library to calculate the results, as well as a synopsis of the results for each denominator from 2 to 27539 which may be of use for further analysis.
I estimate the PARI code would have taken about a CPU-year to find the result; the C code runs over 20 times faster on my machine, and I don't understand why the difference is so great. (I'd appreciate email if someone can explain.)<|endoftext|>
TITLE: Is the subobject functor really a presheaf?
QUESTION [5 upvotes]: I refer to "Sheaves in Geometry and Logic", by S. MacLane.
Let C be a category. Dealing with a subobject of an object $D \in \text{Ob}_{\mathbf C}$, one defines an equivalence relation between morphisms towards D:

Two monomorphisms $f:A\to D$, $g:B\to D$ with a common codomain D are called equivalent if there exists an isomorphism $h\colon A\to B$ such that gh= f.
  A sbobject of D is an equivalence class of monos towards D. The collection SubC(D) of subobject of D carries a natural partial order [...].
  Then SubC(D) is the set of all subobjects of D in the category C.

I can't figure out why SubC(D) is a set, rather than a proper class! Indeed, we are considering something like an qeuivalence relation on
$\displaystyle \coprod_{A\in \text{Ob}} \text{Hom}_{\bf C}(A,D)$
which is not a set, as soon as C isn't small.
So, how can I avoid the problem?

REPLY [10 votes]: For a general category the subobjects do indeed not have to form a set. 
In the context of MacLane/Moerdijk you only look at toposes and there one has a natural isomorphism $Sub_{\mathbf{C}}(D) \cong Hom_{\mathbf{C}}(D,\Omega)$, where $\Omega$ is the subobject classifier. 
So it follows from the axioms of a topos, (edit, thanks Mike:) if it is locally small, that $Sub_{\mathbf{C}}(D)$ is a set. When you prove that the basic examples, sheaves, finite sets, products of those, etc. are toposes you exhibit an object $\Omega$ and establish the above bijection. Before that point the left hand side could a priori be a proper class but the right hand side is a set, since you know that your category is locally small, and your bijection then shows that subobjects form a set.
Knowing this you can conclude that objects in full subcategories (edit, thanks again:) whose embedding preserves monos (e.g. if they are reflective) of toposes also have a set of subobjects, e.g. in all locally presentable categories...

REPLY [3 votes]: A reasonable reformulation of the question is, if there exists a set of representatives for subobjects; i.e. if there is a set of monomorphisms into our object $D$, such that every other monomorphism into $D$ is isomorphic to one of them.
This is, of course, false. Take a preorder $P$, which is a proper class, and has a maximal element $\infty$ (for example, the ordinals plus a maximal element). Then $\infty$ has no set of representatives for its subobjects.
However, it happens very often that there is a set of representatives. The category of sets or topological spaces are examples. If $C$ is a category which has the property, then the same is true for every algebraic finitary category over $C$. Thus, for example, the category of (topological) groups has the property.<|endoftext|>
TITLE: Are there "motivic" proofs of Weil conjectures in special cases?
QUESTION [23 upvotes]: This is a question meant as a first step to get into reading more on Weil conjectures and standard conjectures. It is known that the standard conjectures on vanishing of cycles would imply the Weil conjectures. So, are there proofs of Weil conjectures in special cases using partial results on the standard conjectures? If so, which cases, and what are the references?
Background: Borcherds mentions here that Manin proved a few special cases in higher dimensions using motives.

REPLY [19 votes]: Of course, there's Serre's Analogues kählériens de certaines conjectures de Weil, Annals 1960, where he deduces an analogue of the Weil-Riemann hypothesis  over $\mathbb{C}$ using standard facts from Hodge theory.
This is technically not an answer at all, but I thought I'd mention it since I had the
(perhaps mistaken) impression that this was partly the inspiration for the standard conjectures.
The other more relevant comment is that one can give an elementary proof of the Weil conjecture for any smooth variety whose Grothendieck motive lies in the tensor
category generated by curves. I should explain,
especially in light of Minhyong's comments, that this could be understood as shorthand
for saying the variety can built up from curves by taking products, taking images, blow ups along centres which  of the same type, and so on. Actually, for such varieties, the
Frobenius can be seen to act semisimply. I think this open in general. So perhaps there's
some value in this.<|endoftext|>
TITLE: Is there a quantum Hermite reciprocity?
QUESTION [8 upvotes]: It is well known that there is an isomorphism of $SL_2=SL(V)$ representations 
$$
Sym^n(Sym^m(V))\simeq Sym^m(Sym^n(V))
$$
called Hermite reciprocity (discovered in 1854). 
My question is: Is there anything like this isomorphism
for $U_q(sl_2)$, at least for generic $q$?

REPLY [7 votes]: There is in fact a reasonable way to define quantum analogues of symmetric and exterior powers of a finite-dimensional representation of $U_q(\mathfrak{g})$.  Let $V$ be such a representation, and let $\hat{R} : V \otimes V \to V \otimes V$ be the braiding of $V$ coming from the universal R-matrix.
It is a fact (see, for instance, Proposition 22 and Corollary 23 in Chapter 8 of the book Quantum Groups and Their Representations, by Klimyk and Schmudgen) that the eigenvalues of $\hat{R}$ are all of the form $\pm q^{t_i}$, where $t_i \in \mathbb{Q}$.  Call the eigenvalues of the form $+q^{t_i}$ positive, and those of the form $-q^{t_i}$ negative (this notion is well-defined if $q$ is not a root of unity).  Then call eigenvectors for $\hat{R}$ positive or negative, respectively, if their eigenvalues are positive or negative.
The idea is that positive eigenvectors are $q$-symmetric, while negative eigenvectors are $q$-antisymmetric.
Then define
$$
S_q^2 V = \mathrm{span} \{ \text{positive eigenvectors} \}
$$
and 
$$
\Lambda_q^2 V = \mathrm{span} \{ \text{negative eigenvectors} \}.
$$
Since $\hat{R}$ is diagonalizable, $V \otimes V = S_q^2V \oplus \Lambda^2_q V$.
For example, when $V$ is the 2-dimensional representation of $U_q(\mathfrak{sl}_2)$ with weight basis $x,y$, where $x$ is the highest weight vector, we have
$$
S_q^2 V = \mathrm{span} \{ x \otimes x, y \otimes x - q x \otimes y, y \otimes y  \},
$$
and 
$$
\Lambda_q^2 V = \mathrm{span} \{ y \otimes x + q^{-1} x \otimes y  \}.
$$
Finally, you can define higher quantum symmetric powers $S^n_qV$ to be the submodules of $V^{\otimes n}$ created by intersecting the submodules of tensors that are $q$-symmetric in all $n-1$ consecutive pairs of entries:
$$
S^n_q V = (S^2_q V \otimes V^{\otimes n -2}) \cap  \dots \cap (V^{\otimes n -2} \otimes S_q^2 V).
$$
There is also a closely related notion of quantum symmetric algebra, which is a graded $U_q(\mathfrak{g})$-module algebra whose homogeneous components are isomorphic to the quantum symmetric algebra defined above.
Anyway, that's the good news; there is a not-too-bad definition of quantum symmetric powers.  The bad news is that it doesn't always give you the classical result.  The quantum symmetric powers of a module are no larger than their classical counterparts, and the module is called flat (in a different sense than the usual homological one) if all of its q-symmetric powers (or equivalently, just the q-symmetric cube) are the right size.
The flat simple modules $V_\lambda$ have been classified by Sebastian Zwicknagl in his paper R-Matrix Poisson Algebras and their Deformations.
For each semisimple Lie algebra there are only finitely many flat simple modules.
In the paper Braided Symmetric Algebras of $U_q(\mathfrak{sl_2})$-modules and Their Geometry, he computes all of the quantum symmetric powers of simple $U_q(\mathfrak{sl_2})$-modules.
It turns out that if $V$ is the 2-dimensional simple module, then its symmetric powers are the right size, i.e.
$$
S^n_q V \cong V_n,
$$
where $V_n$ is the $(n+1)$-dimensional simple module.
So your question about Hermite Reciprocity boils down to the question: are $S^m_q V_n$ and $S^n_q V_m$ isomorphic for any $m$ and $n$?
The answer is that they are not.  The first example is $m=3$ and $n = 4$, which follows from the computation in Theorem 3.1 of the second paper I referenced.  The decompositions into simple modules are:
$$ S^4_q(V_3) \cong V_{12} \oplus V_8,
$$
while
$$ S^3_q(V_4) \cong V_{12} \oplus V_{8} \oplus V_{4} \oplus V_{0}.
$$
Of course, this doesn't rule out the possibility of a better definition which does satisfy Hermite Reciprocity, but nobody has come up with one yet.  And if you want everything to be $U_q(\mathfrak{g})$-equivariant, then your choices are pretty rigid.  But perhaps if you let go of that requirement then something more is possible.<|endoftext|>
TITLE: How do I visualize finite covers of curves over non-algebraically closed fields?
QUESTION [8 upvotes]: If $L$ is algebraically closed, fields of transcendence degree one over $L$ correspond to algebraic curves over $L$ up to birational equivalence, and finite extensions correspond to finite Galois coverings. A common example is $L(\sqrt{t})/L(t)$, which defines a two-sheeted covering ramified at one point. Furthermore, if $K$ is a subfield of $L$, then fields of transcendence degree one over $K$ whose intersection with $\bar{K}$ is equal to $K$ correspond to curves defined over $K$. If we consider a covering of such curves, we are considering an extension which does not add anything algebraic over $K$. Such a covering is what I mean by 'geometric' in the title. The fact that $L(\sqrt{t})/L(t)$ is a lift of the extension $K(\sqrt{t})/K(t)$ corresponds to the fact that we can define this covering morphism over $K$.
Now, my question is, if we also include field extensions which add elements algebraic over $K$ ('constant' ones), what is the geometric picture? For example, if $F/K$ is a finite non-trivial (Galois) extension, what is the geometry behind the extension $F(t)/K(t)$? (If you'd like, take more concrete example, such as $\mathbb{Q}(i)(t)/\mathbb{Q}(t)$.) It should be something like, the covering of the projective line over $K$ by the projective line over $F$. Or, what happens if we act both geometrically and arithmetically at the same time, and consider an extension like $\mathbb{Q}(\sqrt[3]{2},\sqrt{-3})(\sqrt{t})/\mathbb{Q}(t)$? Or, to be a little more geometric (i.e. a case which also counts as a manifold), an extension like $\mathbb{C}(\sqrt{t})/\mathbb{R}(t)$ or even $\mathbb{C}(t)[x]/(x^2-t^4-t)/\mathbb{C}(t)$? Furthermore, the Galois group of $F/K$ is the Galois group of $F(t)/K(t)$, so can we view the Galois group as a set of geometric transformations in the same way that we view $\mathrm{Gal}(L(t)/L(t^2))$ as the monodromy group? And in the case where we have both 'geometric' and 'arithmetic' components in our extension, how do we interpret the Galois group then?
What's interesting is that if we view $K[t]$ as the affine coordinate ring of the variety associated to $K(t)$, and similarly $F[t]$, then $F/K$ is separable iff our extension $F[t]/K[t]$ is unramified, and separability might be viewed in light of ramification theory. In addition, might we combine class field theory for number fields and function fields by considering abelian extensions of fields like $K(t)[x]/(f(x))$, where $K$ is a number field?
I believe the better way to view this is to consider not just the locus of points corresponding to the projective line over $K$ (i.e. $K \cup \{\infty\}$), but the affine scheme $Spec(K[x])$ (or its projective completion?). Then we are, in either case, looking at a covering of one scheme by another. (Since there is not in fact a nice map $\mathbb{P}^1(\mathbb{C}) \to \mathbb{P}^1(\mathbb{R})$ corresponding to the extension $\mathbb{C}(t)/\mathbb{R}(t)$, for example! But there is if we consider schemes...)
I'm aware that we can draw a geometric picture for an extension $E/F$ of number fields by looking at Spec of their integer rings. This might be sufficient when looking at affine lines, but I'm also interested in the more general case, with other curves and even possibly higher-dimensional varieties.
Edit: I changed 'arithmetic' to 'constant' to reflect more standard terminology.

REPLY [4 votes]: I think it's possible that non-geometric extensions are indeed not as directly visualizable as geometric ones.
Some terminology: let $k$ be a field, and either assume $k$ has characteristic $0$ or beware that some separability issues are being omitted in what follows.  A (one variable) function field over $k$ is a finitely generated field extension $K/k$ of transcendendence degree 
one.  This already allows for the possibility of a nontrivial constant extension, which is often excluded in geometric endeavors: for instance, according to this definiton, $\mathbb{C}(t)$ is a function field over $\mathbb{R}$, but a sort of weird[1] one: e.g. it has no $\mathbb{R}$-points.  
One says a function field $K/k$ is regular if $k$ is algebraically closed in $K$; i.e., any element of $K$ which is algebraic over $k$ already lies in $k$ [plus separability stuff in positive characteristic].  Any function field can be made regular just by enlarging the constant field to be the algebraic closure of $k$ in $K$; e.g., the previous example is a regular function field over $\mathbb{C}$.  
Regularity is what one needs to think about function fields as geometric objects: namely, there is a bijective correspondence between regular function fields $K/k$ and complete, nonsingular algebraic curves $X_{/k}$.  
Now, on to covers.  Let $L/K$ be a finite degree extension of function fields over $k$.  One says (often; this is slightly less standard terminology) that the exension $L/K$ is geometric over $k$ if both $L$ and $K$ are regular function fields.  And again, there is a bijective correspondence between geometric extensions of function fields and finite $k$-rational morphisms of algebraic curves $Y \rightarrow X$.  
Assuming that the bottom function field $K$ is regular, every extension $L/K$ may be decomposed into a tower of a constant extension $lK/K$ followed by a geometric extension $L/lK$.  Constant extensions have a role to play in the theory -- see for instance the chapter on constant extensions in Rosen's Number theory in function fields, but I think it is fair to describe their role as algebraic rather than geometric: at least that's the standard view.
In fact, the issue that not all extensions of regular function fields are geometric is an important technical one in the subject, because sometimes natural algebraic constructions do not preserve the class of geometric extensions.  
Here is an example very close to my own heart: let $p$ be an odd prime.  The elliptic modular curves $X(1)$ and $X_0(p)$ have canonical models over $\mathbb{Q}$ and there is a natural "forgetful modular" covering $X_0(p) \rightarrow X(1)$.  This corresponds to a geometric extension of function fields $\mathbb{Q}(X_0(p)) / \mathbb{Q}(X(1))$.  This is not a Galois extension: what is the Galois closure and what is its Galois group?  If -- as was classically the case -- our constant field were $\mathbb{C}$ -- then the Galois closure is the function field of the modular curve $X(p)$ and the Galois group of the covering $X(p)/X(1)$ is 
$\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$.  However, over $\mathbb{Q}$ the Galois closure also contains the quadratic field $\mathbb{Q}\left(\sqrt{(-1)^{\frac{p-1}{2}} p}\right)$ so is an extension of a cyclic group of order $2$ by  $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ (in fact it is $\operatorname{PGL}_2(\mathbb{Z}/p\mathbb{Z})$).  Thus the extension is not geometric.  This is unfortunate, because Hilbert's Irreducibility Theorem says that if one has a geometric Galois extension $L/k(t)$ with $k$ a number field, then one can realize $\operatorname{Aut}(L/k(t))$ as a Galois group over $k$.  So in this case, this obtains $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ as a Galois group over not $\mathbb{Q}$ but over the variable quadratic field given above.  K.-y. Shih found a brilliant way to "tweak" this construction to realize $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ over $\mathbb{Q}$ in certain (infinitely many) cases, and other mathematicians -- e.g. Serre, myself, my graduate student Jim Stankewicz -- have put a lot of thought into extending Shih's work, but with only very limited success.  
Added: Brian's example in the comments is very nice.  Maybe another remark to make is that in the arithmetic theory of coverings of curves (an active branch of arithmetic geometry) the distinction between a Galois extension and a geometrically Galois extension of fields (i.e., one which becomes Galois after base change to $\overline{k}$) is a key one: it's certainly something that many arithmetic geometer think a lot about.  It just doesn't come with an obvious "visualization", at least not to me.  Not everything in algebraic or arithmetic geometry can be visualized, or at least not visualized in a way common to different workers in the field.  For instance, an inseparable field extension $l/k$ is by definition ramified, but I have never seen anyone describe this visually.  (There are things you can say to justify that this is not a "covering map", e.g. by pointing to the nonreducedness of $l \otimes_k l$, but I don't think this is direct visualization either.  Maybe some would disagree?)  What you do is think of the case of a ramified cover of Riemann surfaces, and take away the (key) piece of intuition that an inseparable field extension -- which is, visually speaking, just one closed point mapping to another -- behaves like a ramified cover of Riemann surfaces in many ways.  So, as Brian says, in this subject a lot of geometric reasoning proceeds by analogy.  Unlike in, say, certain branches of low-dimensional topology, one does not prove a theorem by referring to (allegedly) visually apparent features of one's constructions.  
[1]: Those who know me well know that I certainly don't think that a curve is weird just because it has no degree one closed points.  More accurate is to say that this curve doesn't have any degree one closed points for a "weird reason".<|endoftext|>
TITLE: Ideals in the ring of single-variable Laurent polynomials with integer coefficients
QUESTION [7 upvotes]: I'm writing some software to automatically compute things like Alexander modules, Alexander ideals, Milnor signatures and Farber-Levine pairings for 1 and 2-knot complements.   So among other things I'd like to have a quick way of comparing ideals in the Laurent polynomial ring $\mathbb Z[t^{\pm 1}]$.  
So my question:
Is there an efficient way of determining whether or not two  ideals in $\mathbb Z[t^{\pm 1}]$ are equal?  What's the most pleasant procedure you can think of? 
My ideals are given by a finite number of generators. 
The knot theory literature tends to stop at "$\mathbb Z[t^{\pm 1}]$ isn't a PID" but I'm hopeful there's some reasonable tools out there.  Understanding ideals in $\mathbb Z[t^{\pm 1}]$ is essentially the same thing as understanding cyclic group actions on finitely generated abelian groups so I could imagine answers might be more complicated than I like.  But I hope to be pleasantly surprised by some sort of "division algorithm + details" type algorithm. 
For example, is there a better way to go about this problem than:
Given an ideal $I \subset \mathbb Z[t^{\pm 1}]$, consider the finitely-generated abelian group $G = \mathbb Z[t^{\pm 1}]/I$ together with the action of $\mathbb Z$ on $G$ given by multiplication by $t$.  We want to determine $G$ as a $\mathbb Z$-module.  So presumably you would do this on the $\mathbb Z$-torsion subgroup of $G$ (call it $\tau G$), then work out the conjugacy class of the action of $\mathbb Z$ on $G/\tau G$ (which amounts to the conjugacy problem in $GL_n(\mathbb Z)$), then there would be an extension problem to deal with.  
I'm hoping there's a simpler way to deal with this problem -- simpler in the sense of implementation.

REPLY [5 votes]: It is fairly straightforward to adapt standard Gröbner basis techniques to such algebras, e.g. see the paper [1]. See also the paper [0] which applies such algorithms to the problem at hand.
[0] Jesus Gago-Vargas; Isabel Hartillo-Hermoso; Jose Marya Ucha-Enryquez
Algorithmic Invariants for Alexander Modules. LNCS 4194, 149-154
http://www.springerlink.com/content/m704326653727425/fulltext.pdf
Abstract. Let G be a group given by generators and relations. It is possible
to compute a presentation matrix of a module over a ring through
Fox's differential calculus. We show how to use Grobner bases as an algorithmic
tool to compare the chains of elementary ideals defined by the
matrix. We apply this technique to classical examples of groups and to
compute the elementary ideals of Alexander matrix of knots up to 11
crossings with the same Alexander polynomial.
[1] Franz Pauer, Andreas Unterkircher.
Gröbner Bases for Ideals in Laurent Polynomial Rings and their Application to Systems of Difference Equations.
AAECC 9, 271-291 (1999)
http://www.springerlink.com/content/qgbwymag351atn71/fulltext.pdf
Abstract. We develop a basic theory of Grobner bases for ideals in the algebra
of Laurent polynomials (and, more generally, in its monomial subalgebras). For
this we have to generalize the notion of term order. The theory is applied to
systems of linear partial difference equations (with constant coefficients) on
${\mathbb Z}^n$. Furthermore, we present a method to compute the intersection of an ideal
in the algebra of Laurent polynomials with the subalgebra of all polynomials.<|endoftext|>
TITLE: Assistance with understanding parent/child relationships in Pythagorean Triples
QUESTION [17 upvotes]: I want to start by apologising for what is probably a weak attempt at a question on a site like this, but I'm having trouble understand a concept that doesn't seem to be properly explained elsewhere - perhaps someone can put it in layman's terms.
I've been trying to understand the idea on Wikipedia that discusses Pythagorean Triple
- namely the section entitled Parent/child relationships. It talks about a Swedish man called Berggren who devised a set of equations that would allow you to determine the children of this parental triple. Each parent created 3, which in turn created 3 and so on.
When I started running the code, I couldn't pick up a certain triple - (200, 375, 425) 
Basically, I wondered if someone could provide a little clarification. Is it either that...

My code is wrong, it's definitely
possible to get to that triple from
a starting point of (3,4,5). 
I haven't understood what Berggren
used these equations for, and I need
to back and read it properly.

Any clarification would be superb,
Thanks
P.S - Could someone also tag this appropriately? I have no idea which subject it comes under.

REPLY [4 votes]: The Classic Tree shown above was probably first found by B. Berggren in 1933. Recently, an entirely New Tree was found. [see: H. Lee Price (2008).The Pythagorean tree: A new species. ArXiv e-prints, arXiv:0809.4324,  pp 14.]<|endoftext|>
TITLE: Comodule exercises desired
QUESTION [15 upvotes]: This Question is inspired by a Quote of Moore's 
"There are two ‘evil’ influences at work here:
1. we are toilet trained with algebras not coalgebras
2. some of us are addicted to manifolds and so think of differential forms as given by God and all the rest are the works of man."
I understand what a coalgebra is and what a comodule over a coalgebra is. I am curious if anyone knows of some lecture notes, a text or fun little toy examples that i could work on. Ideally I would like to better understand what a cotensor product is and maybe feel comfortable computing some Cotor groups. I am currently looking at such objects in the category of chain complexes, so more concrete suggestions will be appreciated.
So if you have:
1. the name of a text that has lots of examples that it works through, essentially something that tries to correct for Moore's first point, or
2. a list of examples to work through, like "hey, look at this algebra and this module etc...", or
3. illuminating thoughts or mantras that I can recite while working with such gadgets
I would really appreciate hearing from you.
Thanks,
Sean

REPLY [17 votes]: My favorite exercise is: prove that a comodule (over a coalgebra over a field) is coflat if and only if it is injective.  This presumes that you already know that any coalgebra is the union of its finite-dimensional subcoalgebras, and any comodule is the union of its finite-dimensional subcomodules (if you don't know that, prove that first).
Also: construct a counterexample showing that infinite products in the category of comodules are not exact in general.  Describe as explicitly as possible the cofree coalgebra with one cogenerator, i.e. the coalgebra $F$ such that coalgebra morphisms from any coalgebra $C$ into $F$ correspond bijectively to linear functions on $C$.  Prove existence of the cofree coalgebra with any vector space of cogenerators.
Define a cosimple coalgebra as a coalgebra having no nonzero proper subcoalgebras, and a cosemisimple coalgebra as a direct sum of cosimple coalgebras.  Prove that a coalgebra is cosemisimple if and only if the abelian category of comodules over it is semisimple.  Prove that any coalgebra $C$ contains a maximal cosemisimple subcoalgebra which contains any other cosemisimple subcoalgebra of this coalgebra.  Consider the quotient coalgebra of $C$ by this maximal cosemisimple subcoalgebra; it will be a coalgebra without counit.  Prove that any element of this quotient coalgebra $D$ is annihilated by the iterated comultiplication map $D\to D^{\otimes n}$ for a large enough $n$ (depending on the element).
Define Cotor between an unbounded complex of right comodules and an unbounded complex of left comodules as the cohomology of the total complex of the cobar complex of the coalgebra with coefficients in these two complexes of comodules, the total complex being constructed by taking infinite direct sums along the diagonal planes.  Construct a counterexample showing that the Cotor between two acyclic complexes can be nonzero.<|endoftext|>
TITLE: If p is a prime congruent to 9 mod 16, can 4 divide the class number of Q(p^(1/4))?
QUESTION [17 upvotes]: When $p$ is a prime $\equiv9\bmod16$, the class number, $h$, of $\mathbb Q(p^{1/4})$ is known to be even. In 
[Charles J. Parry,
A genus theory for quartic fields.
Crelle's Journal 314 (1980), 40--71]
it is shown that $h/2$ is odd when 2 is not a fourth power in $\mathbb Z/p\mathbb Z$. Does this still hold
when 2 is a fourth power?
Some years ago I gave an (unpublished) proof that this is true
provided the elliptic curve $y^2=x^3-px$ has positive rank, and in particular that it is
true on the B. Sw.-D. hypothesis. It's known that the above curve has positive rank for
primes that $\equiv5$ or $7\bmod16$, but to my knowledge $p\equiv9\bmod16$ remains untouched. But perhaps
there's an elliptic-curve free approach to my question?

REPLY [16 votes]: Let $p \equiv 1 \bmod 8$ be a prime number, let $K = {\mathbb Q}(\sqrt[4]{p})$, and let $F$ be the quartic subfield of the field of $p$-th roots of unity. An easy exercise involving Abhyankar's Lemma shows that $FK/K$ is an unramified quadratic extension, hence the class number of $K$ is always even.
The field $KF$ has the quartic subfield $L = {\mathbb Q}(\sqrt{u})$, where $u$ is the fundamental unit of $k = {\mathbb Q}(\sqrt{p})$. An routine application of the ambiguous class number formula to $L/k$ shows that $L$ has odd class number (there are two ramified primes, one infinite and the other one above $2$; clearly $-1$ is not a norm residue at the infinite prime). 
Now I claim that if $p \equiv 9 \bmod 16$, the class number of $KF$ is odd. By class field theory, this implies that the $2$-class number of $K$ must be $2$. An application of the ambiguous class number formula to $KF/L$ shows that the $2$-part of the ambiguous class group has order
$$ h = \frac{2}{(E:H)}, $$
where  $E$ is the unit group of $L$ and $H$ its subgroup of units that are norms from all completions of $KF$: in fact, only the two prime ideals above $p$ are ramified in $KF/L$. Thus it is sufficient to show that $E \ne H$. I will show that $\sqrt{u}$ is a quadratic nonresidue modulo the primes $\mathfrak p$ above $p$. But if $u = T + U \sqrt{p}$ (replace $u$ by $u^3$ in order to guarantee that $T$ and $U$ are integers), then $(\sqrt{u}/{\mathfrak p})_2 = (u/\mathfrak p)_4 = (T/p)_4 = (T^2/p)_8 = (-1/p)_8 = -1$
because $p \equiv 9 \bmod 16$; here we have used the congruence $T^2 \equiv -1 \bmod p$.
The reason why the case $(2/p)_4 = -1$ is easier is because in this case, the ideal above
$2$ ramified in $K$ generates a class of order $2$ in the $2$-class group, whereas this prime generates a class with odd order if $(2/p)_4 = +1$, which means that there is no strongly ambiguous ideal class in this case.
Edit. Paul Monsky has kindly written up this argument, filled in all the details, and made it available here. Thanks!<|endoftext|>
TITLE: Is there an analogue of the hive model for Littlewood-Richardson coefficients of types $B$, $C$ and $D$?
QUESTION [19 upvotes]: If $V_\lambda$, $V_\mu$ and $V_\nu$ are irreducible representations of $\operatorname{GL}_n$, the Littlewood-Richardson coefficient $c_{\lambda\mu}^\nu$ denotes the multiplicity of  $V_\nu$ in the direct sum decomposition of the tensor product of $V_\lambda$ and $V_\nu$.
Knutson and Tao (JAMS 1999, link at AMS site) proposed a  "hive model" for Littlewood-Richardson coefficients.
Is there an analogous model for such tensor product multiplicities for Lie groups of types B, C or D?

REPLY [6 votes]: There is some recent progress on this question.  JiaRui Fei https://arxiv.org/pdf/1603.02521.pdf has given a generalization of the hive model for all semisimple Lie algebras, based on the theory of cluster algebras.  In type A, his model specializes to the Knutson-Tao hive model.<|endoftext|>
TITLE: Existence conditions for twisted cohomological equations?
QUESTION [5 upvotes]: Let $T: X \to X$ be an Anosov diffeomorphism.  Suppose $f: X \to \mathbb{R}$ is Holder continuous (say with exponent $\alpha$).  The question arises as to when $f$ can be written as $g \circ T - g $ for some $\alpha$-Hölder $g: X \to \mathbb{R}$.  It is easily checked that a necessary condition is that the periodic data vanishes, i.e. the sum of $f$ over every periodic orbit is zero.  It is also a (nontrivial) theorem of Livsic that the converse is true, namely vanishing periodic data implies that this cohomological equation is solvable.
I'm interested in the following variant of the cohomological equation. Let $A:  \mathbb{R}^k \to \mathbb{R}^k$ be an isomorphism.  Suppose $f: X \to \mathbb{R}^k$ is $\alpha$-Hölder. When is the equation $f = A( g\circ T) - g$ solvable in $g$?
There seem to be sufficient conditions known for the case of compact groups. Are there conditions known in the noncompact but abelian case?

REPLY [7 votes]: Part I:
The answer is yes under additional conditions:

Periodic data conditions are satisfied. That is, for any periodic point $p$
$$
\sum_{x\in O(p)}f(x)=0.
$$
Exponent $\alpha$ is sufficiently close to 1.
Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exists an $\alpha$-Holder solution $g$. The result is more general: the target group is any Lie group with a bi-invariant metric and the equation is the cohomological equation for two cocycles rather than just coboundary equation.
Part II:
Notice however that if $A\neq Id$ then the periodic conditions may be no longer necessary. Let's restrict to the case when $k=1$ then our equation takes form
$$
f=\lambda g\circ T-g,
$$
where $\lambda<1$. Direct computation shows that 
$$
g=-\sum_{i\ge 0} \lambda^if\circ T^i
$$
is a solution. It is also clear that $g$ is Holder continuous. Moreover, in this case uniqueness is clear as well since the above formula for $g$ is obtained recurrently from
$$
g=-f+\lambda g\circ T.
$$
It seems that this generalizes rather straightforwardly to the case when $A$ is hyperbolic. And I think it's worthwhile to see if anything interesting happens in the case then $A$ has some eigenvalues on the unit circle and some off.<|endoftext|>
TITLE: Cox rings of toric varieties over arbitrary fields
QUESTION [10 upvotes]: The Cox ring of a toric variety X can be viewed as a generalisation of the homogeneous coordinate ring of projective n-space.  Over the complex numbers, the theory is outlined in The Homogeneous Coordinate Ring of a Toric Variety.
We associate to X a graded polynomial ring S and an ideal B.  The space $V=Spec(S)$ is isomorphic to $\mathbb{C}^n$ and has a subvariety Z cut out by the ideal B.  There is an action of a torus G on V such that V - Z is G-invariant with quotient isomorphic to X.  Furthermore, when X is smooth and complete, the automorphism group Aut(X) of the toric variety has a cover $\widetilde{Aut}(X)$ fitting into the exact sequence
$1 \to G \to \widetilde{Aut}(X) \to Aut(X) \to 1$.
The construction can be generalised to other varieties and base fields, often under the name of "universal torsors."  Unfortunately, this added generality lacks the wonderful combinatorial description in Cox's paper (for example, the Demazure root system).
Toric varieties over general base fields can be studied by considering the actions of Galois groups on the fan.  It should be fairly straghtforward to extend these ideas to the Cox ring, as well.  Rather then reinvent the wheel, I'd like to see if this has already been done. 

Does there exist a reference?  A combinatorial description of Cox rings of toric varieties over arbitrary base fields, especially descriptions of analogs of $\widetilde{Aut}(X)$.

REPLY [2 votes]: Parametrizations of toric varieties over any field: also available via google<|endoftext|>
TITLE: literature on geometrical viewpoint on calculus of variations for physics
QUESTION [5 upvotes]: What is a good reference for a geometrical viewpoint on the calculus of variations for physics, using differential forms etc. to derive Yang-Mills equations and other topics of the standard model?
Thanks for ideas.

REPLY [4 votes]: I think that the book 

David Bleecker: Gauge Theory and Variational Principles, Addison-Wesley, 1981

contains exactly what you are looking for.<|endoftext|>
TITLE: What makes the stable module category stable?
QUESTION [12 upvotes]: When geometrically flavoured words  like "mapping cone" or "chain homotopy" crop up in homological algebra, there's usually a good reason. (In this case, looking at the chain complexes associated with each geometric construction gives the algebraic construction).
But, for the life of me, I cannot find a reason why the "stable module category" deserves to be called "stable." It doesn't seem, superficially at least, to be related to the stable homotopy category. So:

Are these categories related in any nontrivial manner that gives merit to the terminology?
If not, is there some other reason why the stable module category deserves to be called stable? (Mind you, this sounds like it might have a higher-categorical answer, and my knowledge of higher category theory is very sketchy... so do be gentle if that's the direction your answer is taking!)

EDIT: The stable module category for a ring $R$ has $R$-modules for objects and maps "modulo projectives" for morphisms. That is, we put an equivalence relation on $Hom_{R}(M,N)$ by declaring $f \sim 0$ if $f$ factors through a projective module. For certain rings (e.g. group rings), this category is tensor triangulated (which is why I'm interested).

REPLY [9 votes]: The answer to the first question is yes, the two categories are nontrivially related: Both the traditional stable homotopy category and the stable module category are examples of stable homotopy categories in the sense of Hovey-Palmieri-Strickland, Axiomatic Stable Homotopy Theory. See the first few pages of Schwede's Stable model categories are categories of modules for a convenient overview. Of special interest are Definition 2.1.1 and the subsequent paragraph on page 107, and Example 2.4.(v) on page 111. You may also find it useful to consult Hovey's Model Categories, especially Chapters 2 and 7, as well as Chapter I.2 in Quillen's Homotopical Algebra.
As for the second question, observe the following extract from Example 2.4.(v):


Fortunately, the two different meanings of ‘stable’ fit together nicely; the stable module category is the homotopy category associated to an underlying stable model category structure [21, Section 2].


so it sounds like Schwede would agree with Mariano and Tom that, at least originally, the selection of the word 'stable' in stable module category likely had nothing to do with stable homotopy and all that.
Corrections are welcome.<|endoftext|>
TITLE: Is this a well-known bound for the derived length of a finite group?
QUESTION [13 upvotes]: Let $cd(G)$ be the set of degrees of the irreducible complex characters of the finite group $G$.
It is conjectured that if $G$ is solvable then $dl(G)\leq |cd(G)|$ and it is a result by Gluck that $dl(G) \leq 2|cd(G)|$.
I have managed to show that $dl(G)\leq 2|cd(G)|-3$ and I was wondering if this is a well-known bound and whether even better bounds are known.
Edit Forgot to mention that this is for $|cd(G)|\geq 3$.

REPLY [8 votes]: Taketa proved that $dl(G) \le |cd(G)|$ for M-groups, so this is called the "Taketa inequality", and it is conjectured that it holds for all solvable groups. I think that perhaps G. Seitz first posed that conjecture, but it is sometimes attributed to me. I was the first person to prove any bound at all, and then Gluck squeezed a better bound out of my method. (My bound was something like $3|cd(G)|$.)
Although no one has succeeded in proving the "Taketa inequality" for all solvable groups, examples suggest that much better bounds might exist. Perhaps some sort of logarithmic bound is the correct one. The situation seems to be wide open even for $p$-groups, where the Taketa inequality definitely does hold because $p$-groups are M-groups. I think that it is unknown, for example, if a $p$-group $G$ with $|cd G| = 4$, can have derived length as large as $4$. (If this is known for $4$, I am sure it is not known for $5$.)<|endoftext|>
TITLE: Existence of zero cycles of degree one vs existence of rational points
QUESTION [26 upvotes]: Let $k$ be a field (I'm mainly interested in the case where $k$ is a number field, however results for other fields would be interesting), and $X$ a smooth projective variety over $k$. 
By a zero cycle on $X$ over $k$ I mean a formal sum of finitely many (geometric) points on $X$, which is fixed under the action of the absolute Galois group of $k$. We can define the degree of a zero cycle to be the sum of the multiplicities of the points. 
Now, if $X$ contains a $k$-rational point then it is clear that $X$ contains a zero cycle of degree one over $k$. 
What is known in general about the converse? That is, which classes of varieties are known to satisfy the property that the existence of a zero cycle of degree one over $k$ implies the existence of a $k$-rational point? For example what about rational varieties and abelian varieties?
As motivation I shall briefly mention that the case of curves is easy. Since here zero cycles are the same as divisors we can use Riemann-Roch to show that the converse result holds if the genus of the curve is zero or one, and there are plently of counter-examples for curves of higher genus. However in higher dimensions this kind of cohomological argument seems to fail as we don't (to my knowledge) have such tools available to us.

REPLY [7 votes]: On the positive side (with no restriction on the field), I don't think anyone has mentioned quadrics. For a smooth quadric, the existence of a 0-cycle of degree one is equivalent to the existence of a point with odd degree. This implies the existence of a rational point, by a theorem of Springer.<|endoftext|>
TITLE: Does the ideal class of the different have a functorial square root?
QUESTION [11 upvotes]: This question is inspired by Emerton's question whether the ideal class of the different has a canonical square root. 
Consider the diagram (of elements; the groups these lie in are the ideal class group of $L$ in the top row and that of $K$ in the bottom row)
 $$ \matrix{
       ?  & \to & [diff(L/K)] \cr
      \downarrow & & \downarrow \cr
     [St(L/K)]  & \to  & [disc(L/K)]}  $$
where the horizontal maps are squaring and the vertical maps are taking norms from $L$ down to $K$. Here $diff(L/K)$ is the different of an extension of number fields, and $disc(L/K)$ its discriminant. The element $[St(L/K)]$ of the class group of the base field $K$ is the Steinitz class (see KConrad's comment to Emerton's question). 
My question is whether there exists an element in the class group $Cl(L)$ that is at home in the left upper corner of this diagram. The most simple question would be whether the Steinitz class is always a norm; and if it is, whether it is the norm of a class whose square is the different class. The last question would be whether this element "?" is unique up to elements that lie in the intersection of  the kernel of the norm and that of squaring.

REPLY [12 votes]: Taking $L$ to be the Hilbert class field of $K$, such a construction
would imply that the Steinitz class of $L/K$ is always trivial. 
Yet this is false - take $K = \mathbb{Q}(\sqrt{-15})$ for example.
(EDIT): If $L = K(\sqrt{\alpha})$ is a tamely ramified extension of $K$, then
the Steinitz class is represented by an ideal $I$ such that $I^2 (\alpha) = \Delta_{L/K}$.
In particular, if $L/K$ is unramified, so $(\alpha) = \mathfrak{n}^2$, then the Steinitz class is trivial if and only if $\mathfrak{n}$ is principal, i.e., if and only if we may take
$\alpha$ to be a unit in $K$. Clearly this is not the case for $L/K$ above.
In fact, I just found a source that works out this example in explicit detail - see Theorems 2.2 and 3.1 of the following:
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf<|endoftext|>
TITLE: Local characterization of semistability
QUESTION [5 upvotes]: It is known that a morphism of schemes $f\colon X \to S$ is smooth at a point $x \in X$ if and only if there is an open neighborhood $U$ of $x$ and an étale map $g \colon U \to \mathbb A^n_S$ such that $g \circ p=f_{|U}$, where $p \colon \mathbb A^n_S \to S$ is the natural projection.
I'm looking for a similar characterization for semistable curves $f \colon X \to S$. I'm interested in the case $S=Spec(k)$, with $k$ a field, and in the case $S=Spec(V)$, with $V$ a discrete valuation ring, where now $X$ is generically smooth.
In particular my question is: in the second case it is true that we can find $\lbrace Spec(R_i)\rbrace _{i \in I}$, an affine open covering of $X$, such that for each $i$, there is an étale map $V[x,y]/(xy-\pi) \to R_i$, where $\pi$ is a uniformizer of $V$?
Thanks.
Ricky

REPLY [2 votes]: If you take the irreducible nodal cubic $y^2=x^2+x^3 \subset \mathbb{C}^2$, which is stable,  no Zariski open set of it can be isomorphic to the reducible curve $xy=0$. For this reason, it seems to me  that the answer to your question should be "no".<|endoftext|>
TITLE: Can a the q-expansion of a p-adic modular form be a non-constant polynomial?
QUESTION [7 upvotes]: Let $E_k$ be the normalized Eisenstein series of weight k and let p be an odd prime. Then 
$$
E_{p^m(p-1)} = 1 \mod p^{m+1},
$$
and so the p-adic limit $\lim E_{p^m(p-1)} = 1$ is a p-adic modular form of weight 0. (It is even overconvergent.)

Question: Suppose f is a p-adic modular form whose q-expansion is a polynomial. Is the q-expansion of f a constant?

REPLY [13 votes]: It is. I want to argue the following way: if the polynomial is non-constant then after scaling it has integral coefficients and so the reduction of the p-adic form mod p^n will be a classical form whose q-expansion is a non-constant polynomial. But I think Katz proved in his Antwerp paper that the only modular forms which are polynomials in q are the constants, which is a contradiction. One needs to dot some i's and cross some t's here, but I am optimistic.<|endoftext|>
TITLE: Why is the x->x^1/3 atlas on R diffeomorphic with the x->x atlas on R?
QUESTION [5 upvotes]: The fact that the atlas using $\phi: x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $\psi: x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases.  Apologies in advance for clustering several questions, but I'm not sure how to disentangle them.
First of all, which is the more relevant/correct statement here:  that the atlases are diffeomorphic, that the manifolds are diffeomorphic ($M \mapsto N$), that the manifold-atlas pairs are diffeomorphic ($(M,\psi) \mapsto (N,\phi)$), or that the manifolds and their differential structures are diffeomorphic ($(M,\mathcal{A}) \mapsto (N,\mathcal{B})$)? 
Is the relevant diffeomorphism patently obvious here given that the domain of $x \mapsto x^{1/3}$ is the same as that of $x \mapsto x$?
If we had two homeomorphic manifolds with atlases that were not diffeomorphic, would it be obvious?

REPLY [21 votes]: The fact that the atlas using $x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases.

Other differential topologists should weigh in on this to confirm or deny, but as a differential topologist I have never come across the notion of a diffeomorphism between two atlases, or even two smooth structures.  Moreover, what you have here is not two atlases but two charts.  These may seem like picky points, but if you find yourself getting confused then one good technique to learn is to sharpen your definitions.  By that, I mean be a bit more careful about distinguishing between things that although often used synonymously are actually distinct.

So you have two charts, $x \mapsto x^{1/3}$ and $x \mapsto x$.  As both of these have image $\mathbb{R}$, they each define an atlas: $\{x \mapsto x^{1/3}\}$ and $\{x \mapsto x\}$.  Each of these atlases then defines a smooth structure on $\mathbb{R}$.  Each of these smooth structures defines a smooth manifold with underlying topological space $\mathbb{R}$.  Although each of these constructions follows in a unique way from the previous step, technically each is a different thing.

Back to the confusion about diffeomorphisms.  We talk of two atlases being equivalent if they generate the same smooth structure, or if they define the same smooth manifold.  In concrete terms, we can test this by looking to see if the identity map is smooth in both directions (using the atlases to test smoothness).
But I can have inequivalent atlases that nonetheless define diffeomorphic manifolds.  This is because the condition of being equivalent is stronger than that of defining diffeomorphic manifolds.  Equivalence rests on the smoothness of the identity map (in both directions), the manifolds being diffeomorphic rests on the smoothness of some map (and its inverse).  So although the two atlases given are inequivalent, they define diffeomorphic manifolds because I'm free to take the map $x \mapsto x^{1/3}$ and its inverse to define the diffeomorphism.
It's a good exercise to help with sorting out the definitions to check that you really understand why these two manifold structures on $\mathbb{R}$ are diffeomorphic.  That is, write down the map and write out its compositions with the transition maps and see that it works.<|endoftext|>
TITLE: Angle Maximizing the Distance of a Projectile
QUESTION [22 upvotes]: It is well-known that to maximize the horizontal distance traveled by a projectile fired from the ground at a given speed, one should fire it at a $45^\circ$ angle. What's less-known, though not too difficult to prove with some manipulation of derivatives, is that if one is on an incline of angle $\phi$ below the horizontal, then to maximize horizontal distance traveled, one should fire the projectile at an angle of $\frac{\pi}{4}-\frac{\phi}{2}$ above the horizontal.
To see this, suppose we fire it at an angle of $\theta$ at speed $v_0$ with gravitational constant $g$. Then the equation of the path is $(v_0 t \cos{\theta}, -gt^2+v_0 t \sin{\theta})$, so we are looking for the $t$ at which $v_0 t \cos{\theta} \sin{\phi}-gt^2 \cos{\phi}+v_0 t \sin{\theta} \cos{\phi} = 0$, or $t = \frac{v_0(\cos{\theta}\sin{\phi}+\sin{\theta}\cos{\phi})}{g} = \frac{v_0}{g} \sin(\theta+\phi)$. Then we wish to maximize $\cos{\theta} t$ at this point, or equivalently $\cos{\theta} \sin{(\theta+\phi)}$. Taking the derivative with respect to $\theta$, we find $-\sin{\theta} \sin{\theta+\phi}+\cos{\theta} \cos(\theta+\phi) = \cos{(2\theta+\phi)} = 0$. Thus $2\theta+\phi = \frac{\pi}{2}$, giving us our answer.
Since this formula looks so nice, I'm wondering whether there is a nicer proof of this fact, possibly using symmetry and/or with more physical intuition. In particular, it might involve some kind of rotation (maybe by $\frac{\phi}{2}$?).

REPLY [5 votes]: I've given two answers here earlier, but thought I'd add a third.  This one focuses on the OP's original request for an argument that utilizes symmetry.  It also avoids doing any explicit algebraic calculations for projectile motion, using instead just general principles and physical intuition.
The key lies in the fact that the answer to the problem is the angle bisector between the vertical (i.e., the direction of gravity) and the slope of the ground.  This suggests that one should be able to use the angle bisector as an axis of mirror-image symmetry.  But there's an obvious problem:  Gravity and ground are not physically symmetric.  The angle bisector is not an axis of symmetry for the parabola of projectile motion.
The solution to this is equally obvious (at least in hindsight):  Pretend that gravity and ground are symmetric. That is, go ahead and take the asymmetric picture, symmetrize it with a reflection across the angle bisector, and then try to attach some meaning to things.
It's helpful to start with the classic flat-ground case.
Draw any (downward-pointing) parabola that leaves from the origin (with positive slope) and passes through the $x$-axis again at some point, say $(D,0)$.  Go ahead and extend the drawing past $D$ (we'll see why in a couple of paragraphs).  Note that physics (or the geometry) tells you that the slope of the parabola at $D$ is the negative of its slope at the origin.
Now reflect this drawing through the line $y=x$ (i.e., the angle bisector between the horizontal and the vertical), so that you have a second parabola, this one passing through the origin and the point $(0,D)$ on the $y$-axis.  Picture a projectile $P$ moving along the first parabola and its mirror image $Q$ moving along the second.  We can think of $Q$ as a mass that is impervious to the normal form of gravity, but instead is subject to an identically strong "horizontal gravity" that pulls it toward the $y$-axis.  The physics is perfectly symmetric across the angle bisector of the $x$- and $y$-axes (i.e., the line $y=x$).
Because of the symmetry, the midpoint, call it $M$, between $P$ and $Q$ starts at the origin, travels out along the angle bisector to some maximum distance, and then heads back toward the origin, passing through it if you allow $P$ and $Q$ to continue past where they hit their respective axes.  It's easy to see (by drawing a picture, connecting the points $(D,0)$ and $(0,D)$ with a line of slope $-1$, and comparing that slope to the slopes of the parabolas at those points) that if the initial slope of the $P$-parabola is less than 1 (i.e., if you fire $P$ at an angle below 45 degrees), the maximum distance $M$ reaches occurs after the parabolas for $P$ and $Q$ pass through the axes (which is why I recommended continuing the $P$-parabola past the point of impact at $(D,0)$), that if the initial slope of the $P$-parabola is greater than 1, $M$'s maximum occurs before $P$ and $Q$ hit their axes, and that if the initial slope is exactly 1, $M$ reaches its maximum exactly when the projectiles $P$ and $Q$ hit their respective "grounds."
What's crucial here is that the distance $D$ that projectile $P$ travels before it hits the ground is never greater than what's allowed by the maximum distance that $M$ gets from the origin, with equality being achieved when $P$ is fired at 45 degrees.
But what is $M$?  Because it stays on the line $y=x$, it can be interpreted as a projectile fired straight out along that line subject to the vector sum of the two gravities, which is a force that points straight back along the same line.  Consequently, the maximum "height" (above the "$y=-x$" axis) achievable by $M$ is determined by its initial outward velocity, which is the component of the projectiles' ($P$ and $Q$) initial velocity along the line $y=x$.  And that component is obviously maximized when the entire velocity points in that direction.
So that takes care of the flat-ground case.  To go to the general case, it's helpful to rotate the flat-ground picture by 45 degrees, so that the axis of symmetry (i.e., the angle bisector) is now the $y$-axis.  The ground is now along the line $y=x$ and (normal) gravity points along the line $y=-x$.  The projectiles $P$ and $Q$ still travel along parabolas, but the line connecting them is now horiontal, and the midpoint projectile $M$ goes up and down the $y$-axis.  The maximum corresponds to a picture in which the two parabolas are tangent to the $y$-axis at the origin and to the horiontal line connecting them at the point of impact.
Now apply a linear transformation that takes $(x,y)$ to $(mx,y)$.  If $m\gt1$, this stretches out the angle between the direction of (normal) gravity and the direction of the ground, corresponding (if you re-rotate things) to sloping ground, but it leaves tangencies intact.  It has no effect on the $y$-axis (i.e., the angle bisector), and it stretches out the line connecting $P$ and $Q$ but keeps it horizontal.  Moreover, the parabolas are still parabolas -- you're really just using a different coordinate system to describe the same physics.  Case closed.
Footnote:  For lovers of conic sections, I heartily recommend the 1937 short Parabola, by experimental film maker Mary Ellen Bute, which highlights (literally!) sculptures by the artist Rutherford Boyd, whose work appeared regularly in the journal Scripta Mathematica.  It can be viewed in thumbnail size online  but for anyone interested it's available on disc 1 of Unseen Cinema: Early American Avant-Garde Film.<|endoftext|>
TITLE: Lower bounds for chromatic number of a graph
QUESTION [16 upvotes]: I am trying to find a good lower bound for chromatic number of one family of graphs. I'm curious what are the known lower bounds for chromatic number. There are two obvious: $\chi(G) \geq \omega(G)$ and $\chi(G) \geq n / \alpha(G)$. One can also employ fancy Lovasz theta-function.
Chromatic number of Kneser graph can be obtained by the means of topological combinatorics (particularly, using Borsuk-Ulam theorem). But it is not clear if this method is general enough.
So, what are the more or less general techniques for lowerbounding chromatic number?

REPLY [3 votes]: Knowing more about your graph class would be very helpful.  I know Bill Cook and Stephan Held are doing some work on lower-bounding $\chi$ using LP duality and branch-and-bound.  Basically they look for a lower bound on the fractional chromatic number by finding a reasonably good feasible solution for the dual LP, i.e. a fractional clique.
A fractional clique is just a non-negative vertex weighting on the graph so that no stable set has weight more than 1.  The total weight of a fractional clique is a lower bound for the fractional chromatic number, and so is in turn a lower bound for the chromatic number.
Of course, even with only 560 vertices this will not necessarily get you very far in a short amount of time.  You can do tricks to help yourself with the time cost.  Obviously you can start by throwing away vertices with degree lower than the bound you're hoping for.  You can also partition the vertices of the graph into dense subgraphs, then try to bound the chromatic number of these subgraphs individually.  Doing this using my RNSC (restricted neighbourhood search clustering) algorithm, which was originally used to find clusters in biological networks, helped a little bit with what Cook and Held were doing, but not too much.
I'm copy-pasting an abstract from a talk that Bill Cook gave this winter, which includes a link to their colouring page:
DATE: Tuesday, February 9
SPEAKER: Bill Cook (Georgia Tech)
TITLE: Computing the chromatic number of graphs
ABSTRACT: It can be very difficult in practice to optimally color a graph. For 
example, a set of randomly-generated test instances introduced by David Johnson 
in 1989 remain unsolved, the smallest example having only 125 vertices.  We 
discuss the use of linear-programming methods to compute safe lower bounds on 
the chromatic number.  Our methods do not depend on the floating-point accuracy 
of linear- programming software.  This talk is based on joint work with Stephan 
Held (University of Bonn). Computational results and computer codes are freely 
available at site: http://code.google.com/p/exactcolors/<|endoftext|>
TITLE: Picard groups of non-projective varieties
QUESTION [12 upvotes]: As far as I know, the main representability result for the relative Picard functor $Pic_{X/k}$, for a noeth. sep. scheme of finite type over a field $k$ is:
If $X$ is proper then $Pic_{X/k}$ is representable by a $k$-scheme loc. of finite type.
(This is attributed to Murre and Oort in Bosch-Lüttkebohmert-Raynaud)

I am interested in what can be said once the requirement of properness is dropped, e.g. what can be said for quasi-projective varieties?
Representability is probably to much to ask for (even as an algebraic space), but do you have references or know of examples where the Picard functor of a non-projective quasi-projective variety is representable?
Is there a weaker sense of representability in which sense the "open" Picard functor is representable?
Is the group somehow controlled by (the group of $k$-points of) representable objects. (I have the naive impression that if $X$ is my quasi-projective variety, then a proper hypercovering of $X$ should be able to compute $H^1(X,\mathcal{O}_X^*)$, and that then one might be able to use representability theorems for proper/projective maps, but I know nearly nothing about the involved technical requirements.)
Edit: I should have added that I do not want to assume resolution of singularities.

REPLY [6 votes]: The first thing to consider is the case of affine curves : let $k$ be an algebraically closed field, $C/k$ a smooth affine curve, $\bar{C}/k$ its smooth projective compactification, $\bar{C}=C\cup{p_0,p_1,...,p_n}$, $J=J(\bar{C})$ the jacobian, $\theta:C\rightarrow J$ the map induced by the choice of the base point $p_0$. Then $Pic^0(C)$  is identified with the quotient $J/\langle\theta(p_1),\ldots,\theta(p_n))\rangle$. This is always divisible but depends somehow on what this subgroup of the groups of rational points of an abelian variety look like (does it land in the torsion, etc.). 
Let's think about it over $\mathbb{C}$ : there you have the quotient of a complex torus by a finitely generated subgroup : when this subgroup is not discrete the quotient does look like it is not representable as the $\mathbb{C}-$points of a scheme.
*Edit : * As Emerton pointed out in the comments, in this case the correct "geometric" object is the 1-motive associated to C. But there is a general construction of Picard 1-motives associated to varieties over a field of characteristic 0 due to Barbieri-Viale and Srinivas, which encode the $Pic^0$ geometrically :
Albanese and Picard 1-motives
Luca Barbieri-Viale - Vasudevan Srinivas
Mémoires de la SMF 87 (2001), vi+104 pages 
http://arxiv.org/abs/math/9906165<|endoftext|>
TITLE: Work on independence of pi and e
QUESTION [31 upvotes]: It is an open problem to prove that $\pi$ and $e$ are algebraically independent over $\mathbb{Q}$.

What are some of the important results leading toward proving this?
What are the most promising theories and approaches for this problem?

REPLY [34 votes]: People in model theory are currently studying the complex numbers with exponentiation.
Z'ilber has an axiomatisation of an exponential field (field with exponential function) that looks like the complex numbers with exp. but satisfies Schanuel's conjecture.
He proved that there is exactly one such field of the size of $\mathbb C$.
I would find it odd if Z'ilber's field turned out to be different from the complex numbers.
By results of Wilkie, the reals with exponentiation are well understood, and the complex numbers with exponentiation is in some way the next step up.  The model theoretic frame work
(o-minimality) that works for the reals with exp. fails for the complex numbers, but there might be a similar theory that works for the complex field with exponentiation.

REPLY [15 votes]: There is a proof of the algebraic independence of $\pi$ and $e^\pi$ in Introduction to Algebraic Independence Theory and a detailed exposition of methods created in last the 25 years although I have not read it.<|endoftext|>
TITLE: Is every graded manifold affine, and is this definition of graded manifold the right one?
QUESTION [9 upvotes]: The following definition is from:

Dmitry Roytenberg, "AKSZ-BV formalism and Courant algebroid-induced topological field theories", Letters in Mathematical Physics, 2007 vol. 79 (2) pp. 143-159, MR2301393.


A graded manifold $M$ over base $M_0$ is a sheaf of $\mathbb Z$-graded commutative algebras ${\rm C}(M)$ over a smooth manifold $M_0$ locally isomorphic to an algebra of the form ${\rm C}^\infty(U) \otimes {\rm S}(V)$ where $U \subseteq M$ is an open set, $V$ is a graded vector space whose degree-zero component $V_0$ vanishes, and ${\rm S}(V)$ is the free graded-commutative algebra on $V$.

Presumably there is an equivalent definition that builds in the axioms for $M_0$ to be a smooth classical manifold — we could start with the topological space with a sheaf of Frechet algebras locally isomorphic to something.
Recall, from, for example the MO question Algebraic description of compact smooth manifolds?, that all classical manifolds are affine in the sense that, although presented as sheaves, a complete invariant is the algebra of global functions.

Question: Is it true that the algebra of global sections of the sheaf ${\rm C}(M)$ is a complete invariant?  I.e. can I recover a graded manifold from its algebra of global smooth functions?

The follow-up question would be to describe, a la the above-linked MO question, which algebras are algebras of smooth functions on a graded manifold.  The orthogonal follow-up question is the same one for "dg manifolds", which is a graded manifold with a square-zero degree $-1$ vector field, thought of as a derivation of the sheaf algebras of functions: does knowing the derivation on global sections determine it on all local sections?

Since no one has yet posted an answer, let me generalize the question to include:

Question: Is the above definition the right one?

Now let me motivate this generalization.  First, the answer to my original question above is trivially "yes" provided that the underlying manifold $M_0$ can be read from the algebra, as then I would have access to the right partitions of unity to really get my hands on the whole sheaf.  This certainly happens when the "fiber" (whose linear functions are) $V$ has only positive gradings: then the manifold is recovered from the degree-zero subalgebra.  But if $V$ has both positive and negative degrees, then the degree-zero subalgebra is too big.  Instead, the base $M_0$ should be recovered as the maximal degree-zero quotient algebra, and I have less intuition for whether such a thing should exist.
But then it's clear that the sheaf in the above definition is not, generally, a sheaf of Frechet algebras.  For example, consider the vector space $\mathbb R^{1t^2 + 1t^{-2}}$ with one dimension in degree $2$ and one in degree $-2$, and call the coordinate functions $x$ and $y$.  Then there is a quadratic map $\mathbb R^{1t^2 + 1t^{-2}} \to \mathbb R^1$ corresponding to the subalgebra of polynomials in $xy$, and this degree-zero subalgebra is not Frechet-complete.  Rather, to exhibit the map as a map of smooth rather than algebraic spaces requires that ${\rm C^\infty}(\mathbb R^{1t^2 + 1t^{-2}})$ include the Frechet algebra ${\rm C^\infty}(\mathbb R^1)$ as a subalgebra.
So what's going on?  If I add the words "Frechet completion" in the correct spot in the above definition, is it the answer to the first question "yes"?

There is another, older meaning of "affine", namely "properly embeds into (finite-dimensional) affine space".  Before adding this extra content, the title of this question included a parenthetical "and in what sense".  So at the risk of making this question long enough that the correct answer is a good reference:

Question: For the correct definition of "graded manifold", is it true that every graded manifold embeds smoothly into some finite-dimensional graded vector space"?

REPLY [6 votes]: First of all, you're completely right in that I didn't need to stipulate that $M_0$ be a manifold in the definition: it follows from a graded manifold being a locally ringed space with a particular local model (it was a physically motivated survey article, so I wasn't much concerned with foundational issues). It also follows that a map of graded manifolds must be polynomial in coordinates of nonzero degree, so things like $e^{xy}$ on $\mathbb{R}[-2]\times\mathbb{R}[2]$ are not allowed. Furthermore, the body of $\mathbb{R}[-2]\times\mathbb{R}[2]$ is a single point, so there are no non-constant maps from $\mathbb{R}[-2]\times\mathbb{R}[2]$ to $\mathbb{R}$; in particular, your example is not a map of graded manifolds. In general, the body $M_0$ of a graded manifold $M$ can be recovered as the spec of $C^0(M)/(I\cap C^0(M))$, where $I$ is the ideal generated by coordinates of non-zero degree -- much like it is done for supermanifolds. Of course, in the non-negatively graded case, $M_0$ can also be recovered as $0\cdot M$.
Now, manifolds as well as supermanifolds and graded (super)manifolds are "affine" in the sense you asked. More precisely, taking global sections of the structure sheaf defines a fully faithful (contravariant) functor to $\mathbb{R}$-algebras (notice that you don't need a compactness assumption on the body, nor any topology on the algebra, which is a blessing for those who, like me, are put off by functional analysis). For manifolds this is classical (see eg. Cor 35.10, p. 301 of Kolar-Michor-Slovak
http://www.emis.de/monographs/KSM/kmsbookh.pdf ), the super/graded case follows easily. The differentials are okay as well, since they're nothing but vector fields. 
Finally, whether my definition is the "right" one depends on the application one has in mind, I guess. For me, coordinates of positive degree generate the symmetries, while those of negative degrees serve to cut out the zero locus of the homological vector field (the solution set of Maurer-Cartan or Euler-Lagrange equations) and resolve its singularities. The relevant completions are then pro/ind completions, not something as awful as Frechet. Of course, that doesn't rule out applications for which the latter would be relevant.
ADDENDUM. Oh yes, graded manifolds are affine in the other sense as well, at least the non-negatively graded ones are (I haven't thought through the general case). It follows from the fact that they are isomorphic to total spaces of graded vector bundles over manifolds, by a result analogous to Batchelor's theorem for supermanifolds.<|endoftext|>
TITLE: Monochromatic triangles in every two-coloring of the plane?
QUESTION [23 upvotes]: An old problem (possibly due to Erdős and Graham?): given a triangle $T$ and a two-coloring of the plane, does there necessary exist a monochromatic congruent copy of $T$?  Here "monochromatic" means that all three vertices receive the same color.  It is known that the answer depends on $T$.
There are several instances known where the answer is "yes".  For example, an amusing exercise is to show that this holds if $T$ is a triangle with side lengths $1, \sqrt{3}, 2$.  I believe Erdős and Graham gave infinite families of $T$ for which the answer is yes.
On the other hand, one can give a two-coloring of the plane so that there is no monochromatic triangle with side lengths $1, 1, 1$.
If I remember correctly, Erdős conjectured that there is always a monochromatic copy of $T$, except for the equilateral triangle which is the only exception.
(1)  Does anyone know if there has been any recent progress on this conjecture?
(2)  What I'd really like to know:  what about the (degenerate) special case of a $1, 1, 2$ triangle?  This question can be seen as a hypergraph analogue of the Hadwiger-Nelson problem, and suggests an interesting intersection of Euclidean Ramsey theory and additive combinatorics.

REPLY [18 votes]: One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.
In http://nucularpower.com/papers/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle; I have been unable to find a 2-coloring that does so. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.
The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.
Edit, December 2016: I have updated the link to my note. In case the link becomes outdated again, the 3-coloring that avoids a degenerate triangle is just the coloring shown below, where the upper hexagon illustrates how boundaries are colored.<|endoftext|>
TITLE: What is a principal refinement of a Postnikov system?
QUESTION [8 upvotes]: I've been reading the book of Hilton, Mislin, and Roitberg on Localization of Nilpotent Groups and Spaces. In Section II.2 they define a principal refinement at stage $n$ of a Postnikov system $$\cdots \to X_n\overset{p_n}{\to} X_{n-1}\to \cdots$$
to be a factorization of $p_n$ into a finite sequence of fibrations 
$$X_n=Y_c \overset{q_c}{\to}\cdots\to Y_1\overset{q_1}{\to}Y_0=X_{n-1}$$
whose fibers are Eilenberg-MacLane spaces $K(G_i,n)$.
But isn't the point of a Postnikov system that $p_n$ is already a fibration whose fibers are Eilenberg-MacLane spaces? So I don't understand why the condition of the definition isn't satisfied trivially at every stage. Perhaps there's some subtlety involving the condition also given that each $q_i$ be induced by a map $g_i: Y_{i-1}\to K(G_i, n+1)$, but aren't all fibrations with fiber $K(G_i, n)$ induced this way since $K(G_i, n+1)$ is the base of a path-space fibration with fiber $K(G_i, n)$?

REPLY [7 votes]: A fibration with $K(G,n)$ fibers need not be principal. The base of the universal fibration with fiber $K(G,n)$ is not a $K(G,n+1)$, even if $G$ is abelian. For example, in the case $G=\mathbb Z$ and $n=1$ it is $BO(2)$, with $\pi_1=\mathbb Z/2$ and $\pi_2=\mathbb Z$. In fact, in general it has homotopy groups $\pi_1=Aut(G)$, $\pi_{n+1}=G$, all the rest trivial. That's if $n>1$; in the case $n=1$, $\pi_1$ is the outer automorphisms of $G$ (the automorphisms modulo inner automorphisms) and $\pi_2$ is the center of $G$.
You can see this by viewing that space (base of universal fibration with fiber $F$) as a delooping of the space of homotopy equivalences of $F$. The latter fibers over $F$ with fiber the pointed self-equivalences; in the case $F=K(G,n)$ the space of pointed self-equivalences is equivalent to the discrete space $Aut(G)$.<|endoftext|>
TITLE: Geometric models for classifying spaces of $GLn(Fq)$.
QUESTION [15 upvotes]: The title pretty much says it.  In a follow-up to my question about alternating groups, does anyone know of a "geometric" model for $BGL_n(F_q)$?  By "geometric" I mean "a space you would have heard about even if you aren't studying classifying spaces"; I mean to exclude general constructions such as standard nerve constructions, infinite joins, intractable quotients of frame bundles, etc.

REPLY [3 votes]: Quillens' paper on the Adams conjecture (doi:10.1016/0040-9383(71)90018-8) almost gives an answer. He maps a limit of spaces BGL_n(F_q) to BU and shows that it is not far from an isomorphism. This is related to the plus construction, but cant remember the details offhand. The space BU in turn can be described in terms of Grassmannians.<|endoftext|>
TITLE: Is it true that exotic smooth R^4 cannot be diffeomorphic to RxN, where N is a 3-manifold?
QUESTION [11 upvotes]: Since $\mathbb{R}$ and any 3-manifold $N$ must be non-exotic, their product $\mathbb{R}\times N$ cannot possibly be diffeomorphic to exotic $\mathbb{R}^4$, correct?
Update: Andy Putman already answered this question in a different thread, as pointed out by Steven Sivek below.    The answer is yes, but not for the reasoning I implied above, because, I gather, the product could in principle be taken in a nontrivial way that alters the differentiable structure.
The proof outlined by Andy relies on $\mathbb{R}\times N$ being piecewise linearly isomorphic to $\mathbb{R}^4$, which is said to be proved in "Cartesian products of contractible open manifolds" by McMillan, which happens to be freely available here:
http://www.ams.org/journals/bull/1961-67-05/S0002-9904-1961-10662-9/S0002-9904-1961-10662-9.pdf .
The relevant part of that paper is as follows:

"A recent result of M. Brown asserts that a space is topologically $E^n$ if it is the sum of an ascending sequence of open subsets each homeomorphic to $E^n$. 
THEOREM 2. Let $U$ be a $W$-space.  Then $U\times E^1$ is topologically $E^4$
Proof.  Let $U=\sum_{i=1}^{\infty}H_i$ where $H_i$ is a cube with handles and $H_i\subseteq \text{Int } H_{i+1}$.  By the above result of Brown, it suffices to show that  if $i$ is a positive integer and $[a,b]$ an interval of real numbers ($a\lt b$), then there is a 4-cell $C$ such that
$H_i\times[a,b]\subseteq\text{Int }C\subseteq C\subseteq U\times E^1$.
There is a finite graph $G$ in $(\text{Int }H_i)\times\{(a+b)/2\}$ such that if $V$ is an open set in $U\times E^1$ containing $G$ then there is a homeomorphism $h$ of $U\times E^1$ onto itself such that $h(H_i\times[a,b])\subseteq V$.  But $G$ is contractible to a point in $U\times E^1$.  Hence, by Lemma 8 of [Bull. Amer. Math. Soc. 66, 485 (1960)], a 4-cell in $U\times E^1$ contains $G$, and the result follows."

A $W$-space was earlier defined as a contractible open 3-manifold, each compact subset of which is embeddable in a 3-sphere.
I'm not sure what it means for a simply connected manifold such as $\mathbb{R}^3$ to be equal to an infinite sum of cubes with handles, but given that, can we say that the above machination qualifies as a piecewise linear isomorphism because each $H_i\times[a,b]$ can be covered with a chart, and each $C$ can be covered with a chart, such that there is a linear mapping between the two?

REPLY [2 votes]: To say that a 3-manifold $W$ is an infinite sum of cubes with handles means that there is an exhausting sequence $W_0 \subset W_1 \subset W_2 \subset \dots$  such that $W = \cup W_i$ and each $W_i$ is a compact handlebody (i.e. homeomorphic to a closed regular neighborhood of a finite graph in $\mathbb{R}^3$). It is a theorem that any contractible 3-manifold has such an exhaustion. In fact, if $W$ is open, irreducible and contains no closed essential surface then $W$ has such an exhaustion. The proof is not too difficult and can be found in several places including Theorem 2 of Freedman and Freedman's article "Kneser-Haken finiteness for bounded 3-manifolds, locally free groups, and cyclic covers" (Topology Vol 37 No 1).<|endoftext|>
TITLE: Suzuki and Ree groups, from the algebraic group standpoint
QUESTION [22 upvotes]: The Suzuki and Ree groups are usually treated at the level of points.  For example, if $F$ is a perfect field of characteristic $3$, then the Chevalley group $G_2(F)$ has an unusual automorphism of order $2$, which switches long root subgroups with short root subgroups.  The fixed points of this automorphism, form a subgroup of $G_2(F)$, which I think is called a Ree group.
A similar construction is possible, when $F$ is a perfect field of characteristic $2$, using Chevalley groups of type $B$, $C$, and $F$, leading to Suzuki groups.  I apologize if my naming is not quite on-target.  I'm not sure which groups are attributable to Suzuki, to Ree, to Tits, etc..
Unfortunately (for me), most treatments of these Suzuki-Ree groups use abstract group theory (generators and relations).  Is there a treatment of these groups, as algebraic groups over a base field?  Or am I being dumb and these are not obtainable as $F$-points of algebraic groups.
I'm trying to wrap my head around the following two ideas:  first, that there might be algebraic groups obtained as fixed points of an algebraic automorphism that swaps long and short root spaces.  Second, that the outer automorphism group of a simple simply-connected split group like $G_2$ is trivial (automorphisms of Dynkin diagrams mean automorphisms that preserve root lengths).  
So I guess that these Suzuki-Ree groups are inner forms... so there must be some unusual Cayley algebra popping up in characteristic 3 to explain an unusual form of $G_2$.  Or maybe these groups don't arise from algebraic groups at all.
Can someone identify and settle my confusion?
Lastly, can someone identify precisely which fields of characteristic $3$ or $2$ are required for the constructions of Suzuki-Ree groups to work?

REPLY [7 votes]: The second part of Karsten Naert's thesis constructs Suzuki—Ree groups as groups of points of "algebraic groups over $\mathbb{F}_{\!\sqrt{p}}$". Such fields being nonexistent, these "algebraic groups" are understood not as schemes, but as the so-called "twisted schemes". They live in the category of pairs $(X, \Phi_X)$, where $X$ is a scheme and $\Phi_X\in \operatorname{End}(X)$ satisfies $\Phi_X\circ\Phi_X=F_X$, the Frobenius morphism.
Another way to describe Suzuki—Ree groups as something "defined by equations" (not algebraic, but difference-algebraic) is to compare two fundamental representations of the ambient Chevalley group in characteristic 2 or 3, whose highest weights $\lambda$ and $\mu$ are interchanged by the exceptional symmetry. These representations happen to have the same dimension, so one considers the group of all $g$ such that $\tau g_
\lambda=g_\mu\tau$, where $\tau^2$ is the Frobenius morphism. This produces the Suzuki—Ree groups, requires no generators and relations and works over any ring of characteristic $p$ admitting such $\tau$ (in particular, the field of definition need not be perfect).<|endoftext|>
TITLE: integer solutions to quadratic forms
QUESTION [6 upvotes]: How does one solve the diophantine equation $x^2 + y^2 = z^2 + w^2 $?  Can solutions be parameterized in three variables analogously to the Pythagorean triples case?

REPLY [14 votes]: Here is the standard geometric argument: after extracting common factors, you are asking for rational points on the quadric $Q\colon x^2-w^2=z^2-y^2$ in $\mathbb{P}^3$, which is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ (as Matt Young's comment explains). Projection from a point on $Q$ gives a birational map $p:Q\to \mathbb{P}^2$ hence a rational parameterisation of $Q$. Explicitly, projecting from $P_0=(0:1:0:1)$ to the $(w=0)$-plane gives the parameterisation 
$$
p : (a:b:c)\in\mathbb{P}^2 \mapsto (2ab:c^2-a^2+b^2:2bc:c^2-a^2-b^2) \in Q
$$
of all the points except for the two lines on $Q$ passing through $P_0$ - i.e. the intersection of $Q$ with the tangent plane at $P_0$, which are $z=w$, $x=\pm y$. Notice also that the points $\{b=0,a\ne\pm c\}$ all go to $P_0$, for the same reason.
The advantage of this method is that there is nothing particularly special about the quadratic form $x^2+y^2-z^2-w^2$ here; the same argument parameterises the rational zeros of any nonsingular quadratic form in 3 or more variables, as soon as it has one non-trivial zero.<|endoftext|>
TITLE: Reconstructing a polynomial from resultants
QUESTION [9 upvotes]: I am trying to compute a monic polynomial $f(x)$ with integer coefficients and known degree $d$.  I am given $n$ pairwise coprime polynomials $g_1(x),\ldots,g_n(x)$, also with integer coefficients, each monic of degree at most $e < d$.  I am also given the values of the $n$ resultants $\mathrm{Res}_x(f(x),g_i(x))$ for $i = 1,\ldots,n$.

The question is: find an algorithm that recovers $f(x)$ from these inputs, for some value of $n$ (depending on $d$ and $e$).

If I take $n = (d+1)$ choose $e$, then an algorithm is as follows: write the coefficients of $f$ as indeterminates, and write out each resultant in terms of these variables.  Then I get $n$ polynomial equations in $d$ variables of degree at most $e$.  I linearize the system, and if $n$ is as above then I have enough equations to find a solution.
I suspect there is some algorithm involving Gröbner bases, but I doubt it is any faster than the above.
Ideally I would like an algorithm that is polynomial-time in $d$ and $e$.  (In my application I have $e = O(\sqrt{d})$.)  I have no idea if such an algorithm is reasonable to expect.  Even something better than $O(d^e)$ would be nice.

[EDIT] What gives me hope are these two papers:
C. Hillar, Cyclic Resultants, Journal of Symbolic Computation, 39 (2005), 653-669.
C. Hillar and L. Levine, Polynomial recurrences and cyclic resultants, Proceedings of the American Mathematical Society, 135 (2007), 1607-1618.
They show that if $g_i(x) = x^i - 1$ then there is an algorithm.  The algorithm requires exponentially many resultants, but they conjecture that it is possible with polynomially many.  I was hoping that if we allow a larger set of $g_i$ but strongly constrain the degrees than we can still recover something.

REPLY [5 votes]: I really like this question, but unfortunately I think that it doesn't have any algorithm in the general case, as I'll explain below.  Thus, to reconstruct your desired $f$, either you need to use $g_m$'s that are special, or assume that $f$ has some special structure (or both).
My claim goes against your stated algorithm (which is a shame, as it is a nice idea), so I'll first describe why I don't think that the algorithm works.  I thought I had a formal proof that the algorithm was faulty (aside from the below counter-example), but it didn't go through.  The problems I see are two fold:

Even if one solves the system, I feel like it may not always be possible to get back to the actual coefficients of $f$ (I wouldn't be surprised if it was possible, though).
You claim that once we have enough equations then we should be able to solve the system of equations, but I don't currently see a proof that these equations are linearly independent (I feel like this is really where the algorithm breaks down).

So now I'll give my counter-example: in a nutshell I'll construct an infinity family of monic, integer, coprime polynomials $g_m$ such that there are (at least) two monic integer polynomials $f_{-1}$ and $f_1$ such that $Res_x(f_1(x),g_m(x))=Res_x(f_{-1}(x),g_m(x))$ for all $m$. Thus, the resultants do not contain enough information to reconstruct $f$, no matter how long the algorithm takes.
Now for the details.  First I'll assume that both $d$ and $e$ are even (I don't feel like this is a big restriction, but who knows?) [but I don't actually need $d>e$].  Define $g_m(x):=x^e-m^e$, for $m\in\{2,3,\ldots\}$. Clearly they are monic, integer, and coprime. Define $f_1=(x-1)^d$ and $f_{-1}=(x+1)^d$.  Recall the formula for the resultant of monic polynomials (over the closure of whatever field we are working over):

$Res_x(P,Q)=\Pi_{(a,b):P(a)=Q(b)=0}(a-b)$

where the product is over roots, taken with multiplicity.  So then

$Res_x(f_1,g_m)=\Pi_{j=1}^e (1-m\omega^j)^d$

where $\omega$ is a primitive $e$-root of unity (again, over the closure of the field).  But as $d$ is even, this means that $-1$ is a $d$-root of unity so,

$=\Pi_{j=1}^e (1-m\omega^j)^d=\Pi_{j=1}^e (-(1-m\omega^j))^d=\Pi_{j=1}^e (-1+m\omega^j)^d$

and using $e$ is even, and thus that $-\omega$ is an $e$-th root iff $\omega$ is, we see that via reindexing

$=\Pi_{j=1}^e (-1-m\omega^j)^d=Res_x(f_{-1},g_m)$

so I've established the equality of the resultants, for any $m$.  So clearly any algorithm that attempts reconstruction will fail, as it cannot distinguish between $f_{-1}$ and $f_{1}$.
Clearly, this result relies crucially on the fact that both $e$ and $d$ are even, but otherwise has no restriction.  I feel like something could be done for cases when a small prime divides both $d$ and $e$, but at the moment this result seems sufficient for your purposes.
I suppose in a sense this counter-example "shows" that the "linearized" system you proposed cannot be invertible in general.  It could be an interesting to ask for conditions for when the $g_m$ do form such an invertible system. However, as you say above, your $g_m$ are given you, so I'm not sure of a quick way to avoid this counter-example. 
I hope that something is still recoverable for your application.<|endoftext|>
TITLE: Large cardinals
QUESTION [10 upvotes]: Looking at the chart of cardinals in Kanamori's book, one realizes that all large cardinals are implied by stronger ones and imply weaker ones. For instance measurable implies Jonsson which implies zero sharp which implies weakly compact which implies Mahlo which implies inaccessible. So it seems as if all these large cardinal assumptions are linearly ordered by consistency strength. Is there a some assumption above ZFC that is not implied by and does not imply any of the linearly ordered large cardinals?

REPLY [12 votes]: Woodin's Ω-conjecture implies that all large cardinal axioms are well ordered under the relation ``its consistency implies the consistency of''. See his paper in the Notices 2001/7. For this, of course, he does define what a large cardinal is.<|endoftext|>
TITLE: Indeterminate "$x$" in algebra/ring Theory
QUESTION [6 upvotes]: How do you interpret the indeterminate "$x$" in ring theory from the set theory viewpoint? How do you write down $R[x]$ as a set? Is it appropriate/correct to just say that
$$R[x] = \{ f: R \to R \mid \exists n \in \mathbb{N}, \mbox{ and } c_0, \dots, c_n \in R \mbox{ s.t. } f(x) = c_0 + c_1 x + \dots + c_n x^n \}$$
This appears to be a very analytic definition. Is there a better definition that highlights the algebraic aspect of the set of polynomials? $$ $$
EDIT: There is a thread on this at http://mathoverflow.tqft.net/discussion/568/inconsistent-and-closedminded-question-closing/#Item_0 
There is a single user who is composing a lengthy response to this question. Anyway, take a look.

REPLY [3 votes]: This began as a comment on Anweshi's answer, but it got too long.  
I consider the functorial definition (as in Anweshi's answer but slightly modified --- see below) to be the best one, because it tells us what is really important about $R[x]$.  Other definitions (using sequences, modules with products, symbols, etc.) are better viewed as constructions rather than definitions; they provide proofs that the functorial definition can be satisfied.  The situation is analogous to the case of the real numbers, whose important property is (and whose definition therefore should be) that they form a complete Archimedean ordered field; Cauchy sequences, Dedekind cuts, etc. provide constructions.  
The slight modification that I'd make in Anweshi's formulation of the functorial definition is that I regard this definition not as simply defining $R[x]$ but as defining the pair $(R[x],x)$.  The most obvious reason for the change is that $x$ is explicitly used in the statement of the universal property.  In fact, the definition simply says that the pair $(R[x],x)$ is the universal example of a ring with a homomorphism from R and a specified element.  (I've been assuming that we're dealing with commutative rings; otherwise, I should either say "a specified central element" or work with fancier "polynomials" where the coefficients and $x$ can be interleaved in monomials, rather than just having a coefficient times a power of $x$.)  A side benefit of this form of the definition is that it gives an immediate answer to the question "what is $x$": it is the specified element in the universal example.<|endoftext|>
TITLE: Hochschild (co)homology of A and of Mod_A
QUESTION [10 upvotes]: Let A be an algebra (or dg algebra). Where can I find a proof of HH_*(A) = HH_*(Mod_A) and HH^*(A) = HH^*(Mod_A)? (And does this hold for any A?) Here Mod_A is, e.g., the category of left A-modules.
One reason why this is interesting/important/useful is because many categories which arise "in nature" are of the form Mod_A. For example, there is a theorem of Bondal and van den Bergh which states that derived categories of a large class of varieties (I forget their exact hypotheses) are equivalent to Mod_A for some A. Dyckerhoff also proved that categories of matrix factorizations are of this form. By mirror symmetry, Fukaya-type categories should be of this form as well...
Anyway, so to compute HH of such a category, it suffices to find this A and then compute HH(A). I think that it generally(?) should be easier to compute HH of an algebra than HH of a category. (Of course finding this A can be a very nontrivial task.)

REPLY [8 votes]: Basically this follows from the fact that the derived category of bimodules over two algebras is equivalent to the (suitably defined) functor category between the derived category of modules of each algebra. Say, Toen's paper on derived Morita equivalence. Then, the identity functor is given by the algebra itself interpreted as a bimodule, so the Hochschild cohomology is $\mathrm{Ext}^i_{A-A}(A,A)$. You can compute this using the bar resolution and a quick calculation gives you the usual definition of Hochschild cohomology.<|endoftext|>
TITLE: How do you exponentiate a section of the adjoint bundle to get a gauge transformation?
QUESTION [5 upvotes]: Suppose $E$ is a vector bundle with structure group $G$ and let $P = F(E)$ be the frame bundle. Let $\mathfrak{g}_P$ denote the associated bundle to the adjoint representation of $G$ on its Lie algebra (i.e. $\mathfrak{g}_P = P \times_G \mathfrak{g}$ where $\mathfrak{g}$ is the Lie algebra of $G$). Given a section of $\mathfrak{g}_P$, I should be able to "exponentiate" it pointwise to get a gauge transformation. How is this defined? I couldn't come up with anything well-defined.
For some context, I was reading Chapter 2 of Donaldson and Kronheimer's "The Geometry of 4-Manifolds," and they mention this pointwise exponential in passing on p. 33. I'm guessing they assume the reader is familiar with it from a more elementary text, but I looked in a few other books and couldn't find it.

REPLY [7 votes]: I'm not sure what it is that you tried, but the exponential map should work.  First of all, let $U_i$ be a trivialising cover for $P$ and its associated bundles.  A section through $\mathfrak{g}_P$ is given by functions $\omega_i: U_i \to \mathfrak{g}$ which, on overlaps, transform according to
$$\omega_i(p) = \operatorname{Ad}_{g_{ij}(p)} \omega_j(p) \qquad \forall p \in U_i \cap U_j,$$
where I use $\operatorname{Ad}$ to mean the adjoint representation of $G$ on its Lie algebra $\mathfrak{g}$.
Now simply compose $\omega_i$ with the exponential map $\exp: \mathfrak{g} \to G$, resulting in functions $\exp\omega_i : U_i \to G$ which, on overlaps, transform according to
$$\exp\omega_i(p) = g_{ij}(p) \exp\omega_j(p) g_{ij}(p)^{-1} \qquad \forall p \in U_i \cap U_j.$$
But this is just a section through the associated fibre bundle usually denoted $\operatorname{Ad} P$, and that is the same thing as a gauge transformation.<|endoftext|>
TITLE: Proving equality of varieties by dimension counting
QUESTION [6 upvotes]: For finite sets $A$ and $B$, it is clear that $A \subseteq B$ and $|A| \geq |B|$ implies $A = B$. While an obvious fact, it can sometimes be a nice shortcut in proofs.
Analogously, if $V$ and $W$ are finite-dimensional vector spaces such that $V \subseteq W$ and $dim\ V \geq dim\ W$ then $V = W$. This is an especially useful tool when $V$ is defined parametrically and $W$ is defined implicitly. Then you can easily prove $V \subseteq W$ by plugging the parametric expression for $V$ into the equations for $W$. Counting the dimensions can take more work, but you sometimes get lucky.
For a while I've been wondering how this extends to algebraic varieties.
Here's an attempted application to proving the spectral theorem. Fix the dimension $n$; all matrices will be $n \times n$. The theorem says that for every symmetric matrix $S$ there exists an orthogonal matrix $Q$ and a diagonal matrix $D$ such that
$$Q^T\ D\ Q = S.$$
Let $A$ and $B$ respectively denote the matrices of the form on the left-hand and right-hand side. $A$ is defined parametrically by a function $f$ from $D$ and $Q$, and $B$ is defined implicitly by the symmetry condition. We want to prove $A = B$. It is clear that $A \subseteq B$:
$$(Q^T\ D\ Q)^T = Q^T\ D^T\ (Q^T)^T = Q^T\ D\ Q,$$
so $Q^T\ D\ Q$ is symmetric.
The domain of $f$ has dimension $dim\ D + dim\ Q$ where $dim\ D = n$ and $dim\ Q = (n-1) + \cdots + 1$, while $S$'s space has dimension $n + (n-1) + \cdots + 1$, so the dimensions seem to match.
But what about $f$'s degree of injectivity? It isn't perfectly injective: if $D$ and $D'$ equal the identity matrix then $Q^T\ D\ Q = Q'^T\ D'\ Q'$  for any independent combination of $Q$ and $Q'$.
My question is thus twofold:

What is the right generalization of the theorem for vector spaces to algebraic varieties?

and

Can the attempted proof of the spectral theorem be salvaged with a genericity argument?

I'm happy with the extant proofs of the spectral theorem, so this is more curiosity than anything else.

REPLY [2 votes]: Another generalization of the theorem for vector spaces could be the following.
Assume that $X \subset Y$ is a closed subvariety, with $Y$ connected. If 
$\dim X = \dim Y $ and 
$\textrm{mult}_xX = \textrm{mult}_x Y$ for all $x \in X$,
then $X=Y$. 
In fact, assume $X \neq Y$. Since $Y$ is connected there exists $x \in X  \cap \overline{Y \setminus X}$, and for such a $x$ we have 
$\textrm{mult}_xY \geq \textrm{mult}_xX + \textrm{mult}_x \overline{Y \setminus X}   \geq \textrm{mult}_xX+1$,
contradiction. This can be useful, since it does not require the irreducibility of $X$ and $Y$. However, the assumption about the connectedness of $Y$ cannot be dropped: consider the case where $Y$ is the disjoint union of two copies of $X$.
EDIT. In a first version of this answer I required the condition $\dim T_xX = \dim T_xY$ instead of $\textrm{mult}_xX = \textrm{mult}_x Y$. This actually did not work, see damiano's comment below.<|endoftext|>
TITLE: Representations of products of groups (and monoids)
QUESTION [5 upvotes]: I have very little knowledge of representation theory, but the following has come up in my summer undergrad research project (relates to conformal field theory and geometric function theory).
Suppose we have a group $G$ and subgroups $A$ and $B$ such that $A \cap B = \{1\}$ and for every $g \in G$, there exists $a \in A, b \in B$ such that $g = ab$.  Then we can express $G = A \bowtie B$ as a Zappa–Szép product.  This of course reduces to the semidirect or direct product in the nice cases.
Then suppose, we have sufficiently nice representations of $A$ on an $F$-vector space V, and $B$ on an $F$-vector space W, then can we find a representation of $G$ which in some sense preserves the representations of $A$ and $B$?
I've been told that the solution for semidirect products uses something called Clifford Theory, but we don't have a semidirect product here.
Our problem involves a monoid, not a group, but the Zappa-Szep product is constructed the same way there.

REPLY [5 votes]: It is hard to prove that there is no "nice" relationship between the representations of $A$ and $B$ on one hand and $G$ on the other, but experience with representation theory suggests that no nice relationship exists. For example, $A$ and $B$ can have nontrivial one-dimensional representations while $G$ has no such representation. 
Even in the case where $A$ is normal, so we have a semidirect product, there is not really a clear relationship. In that case, each irreducible of $G$ is associated with a conjugacy class of representations of $A$, but there is no connection in general with representations of $B$, but only with appropriate "projective representations" of certain subgroups of $B$.<|endoftext|>
TITLE: How are these two ways of thinking about the cross product related?
QUESTION [25 upvotes]: I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner.  I now know, not one, but two ways of doing this, and I can't quite see how they're related:

The cross product is the Lie bracket in the Lie algebra of $\text{SO}(3)$.
The cross product is the Hodge star map $\Lambda^2(V) \to V$ where $V$ is an oriented $3$-dimensional real inner product space.

Okay, so there's one obvious relation here: $V$ has automorphism group $\text{SO}(3)$.  But for some reason I can't figure out where to go from here.  A good starting point would be to exhibit a canonical isomorphism between an oriented $3$-dimensional inner product space $V$ and the Lie algebra of $\text{Aut}(V)$.  Maybe this is obvious.  In any case, I would appreciate some clarification.

REPLY [3 votes]: Geometric (Clifford) algebra provides yet another way to understand the cross product.  If a vector represents a 1D subspace containing the origin, the wedge (outer) product is a bilinear form on 2 vectors ($a \wedge b$) that represents the subspace containing both vectors.  The vector has basis $e_1, e_2, ..., e_n$ ($n$ elements), so the wedge product has basis $e_1\wedge e_2, e_1\wedge e_3, \ldots, e_{n-1}\wedge e_n$ ($\binom{n}{2}$ elements).  The cross product is a way to represent this subspace (using its dual, the normal vector).  This works only for 3D, where the basis sizes match, using $e_1 e_2 \to e_3,\, e_2 e_3 \to e_1,\, e_3 e_1 \to e_2$.<|endoftext|>
TITLE: Quantum cohomology of projective bundles
QUESTION [5 upvotes]: There is classical description of cohomology ring of projective bundle. Is there an analog in quantum cohomology?

REPLY [4 votes]: There is a quantum Leray-Hirsch Theorem, do to Maulik and Pandharipande here.  It relates the Gromov-Witten theory of the projective line bundle to that of the base.<|endoftext|>
TITLE: Question regarding divergence
QUESTION [5 upvotes]: Let $E$ be a closed and convex set of distributions on a finite set $A$. Let $P',Q'\notin E$ and let $P^{\star},Q^{\star}$ be their respective estimates in $E$ with respect to  the KL-divergence, i.e., $D(P'\|P^{\star})=\min_{P\in E}D(P'\|P)$ and similarly for $Q^{\star}$. I am wondering whether $D(P'\|Q')\ge D(P^{\star}\|Q^{\star})$.

REPLY [3 votes]: The inequality $D(P'|Q')  \ge D(P^\star| Q^\star)$ does not need to hold.
Here is an example.
Let $A$ be the set $\{1,2,3,...,n\}$. Let $E$ be the set of measures $P$ on $A$ such that $P(\{1\}) = 0$. Projecting a measure $P$ on $E$ using $D$ is equivalent to conditioning $P$ on $ A- \{1\}$. Choose $P'$ and $Q'$ such that they both put equal and nonzero mass on $\{1\}$. By direct computation one sees: $D(P^\star| Q^\star) = \frac{1}{1-P'(\{1\})} D(P'|Q') > D(P' | Q')$.
The details of the above computation are as follows.
For ease of notation set $n=3$. Let $E$ be the set of measures $P$ with $P(\{1\}) =\epsilon$; to obtain the example above, one sets $\epsilon = 0$. Let us parametrize the measures on $\{1,2,3\}$ as follows: $P(\{1\}) = p_1$, $P(\{2\}) =p_2$ and $P(\{3\}) = 1-p_1 -p_2$. Our problem is:
$$
\inf_{ Q \in E}\left[ p_1 \log \frac{p_1}{q_1} + p_2 \log \frac{p_2}{q_2} + (1-p_1 -p_2) \log\frac{ 1- p_1 - p_2}{ 1- q_1 - q_2 } \right].
$$
Let $F$ denote the expression after the $\inf$.
$F$ is strictly convex in $Q$ and therefore will have a unique optimizer. In the above coordinates, the normal to $E$ is the vector $(1,0)$. Then
$$
\frac{\partial F} {\partial q_1} = -\frac{p_1}{q_1} + \frac{1-p_1-p_2}{1-q_1-q_2} = \lambda 
$$
and
$$
\frac{\partial F} {\partial q_2} = -\frac{p_2}{q_2} + \frac{1-p_1-p_2}{1-q_1-q_2} = 0. 
$$
We have the constraint that $Q\in E$, i.e., $q_1 =\epsilon$.
From the last two equalities one infers:
$$
q_2 = \frac{(1-\epsilon) p_2}{ 1-p_1}.
$$
Going back to the coordinates $(p_1,p_2,p_3)$ to denote a measure on $\{1,2,3\}$,
projecting a measure on $E$ using $D$ corresponds to the following map:
$$
(p_1,p_2,p_3) \rightarrow \left(\epsilon, (1-\epsilon)\frac{p_2}{p_2+p_3}, (1-\epsilon)\frac{p_3}{p_2 + p_3}\right).
$$
For $\epsilon =0$, this is the same as conditioning $P$ on $\{2,3\}$.
One obtains the expression for the relative entropy given above by directly computing it using this formula for the projections.<|endoftext|>
TITLE: Possible singularities of the base of a Mori fiber space
QUESTION [5 upvotes]: Suppose X is a normal projective complex variety, (X, $\Delta$) is a klt pair and f : X $\to$ Z is a Mori fiber space given by a contraction of an extremal ray for this pair. Here I mean that the relative dimension is at least 1. Do we know anything about the singularities of Z? Z is normal more or less by construction and we do know that if X was Q-factorial then so is Z. Can we say anything more? Eg: is there a $\Delta'$ on Z with (Z, $\Delta'$) klt? Does Z have rational singularities? What if $\Delta$ = 0, is Z terminal? Canonical?

REPLY [6 votes]: Under the assumption that $X$ is $\mathbb{Q}$-factorial, section 5 of the paper http://arxiv.org/pdf/math/0606666 addressed this issue, which was also proved earlier in Ambro's paper. Basically, if you assume the pair $(X,\Delta)$ is klt, so is the base $(Z,\Delta_Z)$ for some $\Delta_Z$. As the example of Prokhorov shows,  this is optimal, namely, the base may not be terminal (canonical) even you assume the $(X,\Delta)$ is. For lc case, I think dlt modification+perturbation reduce the question to the klt case.<|endoftext|>
TITLE: Reference Request for Drinfeld and Laumon Compactifications
QUESTION [12 upvotes]: Background
Let $X$ denote a smooth projective curve over $\mathbb{C}$ and let $G$ denote a semi-simple simply connected algebraic group over $\mathbb{C},$ which has associated flag variety $G/B.$
Then we can consider the variety $Maps^d(X, G/B)$ of maps from $X$ to $G/B$ of fixed degree $d$ where $d$ is an $\mathbb{N}$-linear combination of coroots of $G.$  See the top of page 2 of this paper by Alexander Kuznetsov Kuznetsov for the definition of degree.  The Plucker embedding of the flag variety into projective space gives an alternative formulation of $Maps^d(X, G/B)$ which can be found in section 1.2 of Kuznetsov or in this survey article of Alexander Braverman Braverman.
In general, $Maps^d(X, G/B)$ is not compact, but there is a compactification due to Drinfeld, which is referred to as the variety of quasi-maps and denoted $QMaps^d(X, G/B).$  See Kuznetsov or Braverman.
On the other hand, when $G = SL_n,$ there is a second compactfication due to Laumon.  This is because when $G = SL_n,$ we have both the Plucker embedding description of the flag variety, but also the description of the flag variety as flags of vector spaces.  This latter description gives another formulation of $Maps^d(X, G/B)$ but leads to a compactification known as quasi-flags.  Once again, see Kuznetsov.  When $n>2,$ varieties of quasi-maps and of quasi-flags are different.  It turns out that quasi-flags are always smooth, while quasi-maps have singularities.
Broadening our focus somewhat, we could instead consider the representable map of stacks $Bun_B(X) \to Bun_G(X),$ and note that the fiber over the trivial $G$-bundle is the union of all the $Maps^d(X, G/B)$ for all possible degrees (note that the degree just tells us which connected component of $Bun_B$ we live in).
Just as the variety of maps above was not compact, the map $Bun_B \to Bun_G$ is not proper.  But there exists a relative compactification of $Bun_B,$ also referred to as the Drinfeld compactification, which I will denote $Bun_B^D.$  This compactification still maps to $Bun_G,$ but the map is now proper.  The fiber over the trivial bundle of this map coincides with the union of all $QMaps^d(X, G/B).$
As before, when $G = SL_n,$ there is a second compactification of $Bun_B$ which I will denote $Bun_B^L$ whose fiber over the trivial bundle coincides with the union of all the quasi-flags varieties.  See this paper by Braverman and Gaitsgory BG or this follow-up paper by Braverman, Gaitsgorgy, Finkelberg, and Mirkovic BGFK for more details.
Question
In Kuznetsov, Kuznetsov proves that when $X = \mathbb{P}^1$ and $G = SL_n,$ there is a map from the space of quasi-flags of degree $d$ to the space of quasi-maps of degree $d$ which is a small resolution of singularities.
Later, in BG, it is asserted that Kuznetsov proved that $Bun_B^L(X)$ is a small resolution of singularities of $Bun_B^D(X)$ for any smooth projective curve $X.$
It seems to me that there are two discrepancies here.  One has to do with an arbitrary smooth projective curve versus $\mathbb{P}^1.$  The second has to do with moving from the varieties of quasi-maps and quasi-flags to the stacks $Bun_B^D$ and $Bun_B^L.$
Does anyone know a reference which explains the bridge between Kuznetsov and the assertions of BG?  Or perhaps this was just something clear to the experts which never warranted an explanation?

REPLY [10 votes]: I guess that formally this is not written anywhere, but it is indeed easy to deduce the general case from Kuznetsov's result. The point is that both Drinfeld and Laumon compactifications consist of G-bundles with some kind of degenerate B-structure, where
the degeneration occurs at finitely many points of the curve. It is easy to see that the 
fiber of of the map $Bun_B^L\to Bun_B^D$ depends only on the  behaviour of everything
in the formal neighbourhood of those points (using this observation you can formally reduce
your question to Kuznetsov's result).<|endoftext|>
TITLE: Do the base 3 digits of $2^n$ avoid the digit 2 infinitely often -- what is the status of this problem?
QUESTION [12 upvotes]: I believe this question is due to Erdős and Graham, and I think it is still open: does the base 3 expansion of $2^n$ avoid the digit 2 for infinitely many $n$?
If we concatenate the digits of $2^i$, $i \geq 0$, we produce the number $0.110100100010000...$. This number is not simply normal in base 2, so it is not normal. Is it simply normal in base 3? I think even that result would not imply that for sufficiently large $n$, 2 doesn't appear in the base 3 expansion of $2^n$.
The number 20 here is not special:
$2^{20} = 1222021101011_3, \;\;\;\; 2^{21} = 10221112202022_3, \;\;\;
2^{22} = 21220002111121_3$
Statistically, we seem to be flipping a fair 3-sided coin, and statistical analysis for larger $n$ bears this out (in the past, I did a p-test on the digits, but don't have the data available here). If we actually produced these digits by flipping this 3-sided coin, for fixed $n$ we would have probability about
$$(2/3)^{n\ln2/\ln3}$$
of having no 2s in the base-3 digit expansion.
What is the state of the art for this problem? Is there a good number-theoretic reason why this problem should be very difficult (e.g. an analogy with other supposed-hard problems)? Are there related problems that have been solved?

REPLY [6 votes]: As of a few months ago, the status of the problem was: still unsolved. See the slides Jeff Lagarias put up from a talk he gave in September 2009: http://www.math.lsa.umich.edu/~lagarias/TALK-SLIDES/ternary-fields-2009sep.pdf
An older reference is 
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.65.6976&rep=rep1&type=pdf (Brian Hayes, Third Base, American Scientist) which says the problem was still open in late 2001; also that Ilan Vardi searched up to $2^{6973568802}$ without finding any 2-less powers of 2 (other than $2^2$ and $2^8$).<|endoftext|>
TITLE: What is the spectrum of the ring of entire functions?
QUESTION [17 upvotes]: Let $\mathcal{O}(\mathbb{C})$ be the ring of entire functions, that is, those functions $f : \mathbb{C} \to \mathbb{C}$ which are holomorphic for all $z \in \mathbb{C}.$ For each $z_0 \in \mathbb{C}$.
Are there any other maximal ideals in $\mathcal{O}(\mathbb{C})$ besides these obvious ones?
If anyone can give a concise description of $\text{Spec }\mathcal{O}(\mathbb{C})$, that would be extremely helpful. I'm trying to understand wether or not knowing the closed subset $V(f)$ of $\text{Spec }\mathcal{O}(\mathbb{C})$ of ideals containing $f$ gives you more information about $f$ than simply knowing the vanishing set of $f$ in the classical sense.

REPLY [9 votes]: An addendum to Richard Borcherds's answer: Of course, instead of the integers, you could use any closed, discrete subset $A$ of $\mathbb C$.  By taking different such $A$'s and different ultrafilters $U$ on them, the ideals $I$ that you get are all of the maximal ideals in the ring of entire functions.  
More generally, with any $A$ and $U$ as above, suppose $S$ is a nonempty, proper initial segment of the ultrapower $\mathbb N^A/U$ that is closed under addition.  Let $P$ be the collection those entire functions $f$ such that either $f$  is identically 0 or the function $v_f:A\to \mathbb N$ sending $a$ to the order-of-vanishing of $f$ at $a$ represents, in the ultrapower, an element not in $S$.  Then $P$ is a prime ideal in the ring of entire functions, and every prime ideal arises in this way.  (The maximal ideals correspond to the case where $S=\{0\}$.)<|endoftext|>
TITLE: Topological spaces that resemble the space of irrationals
QUESTION [11 upvotes]: (This question actually arose in some research on number theory.)
I once learned that any countable dense subspace of any Euclidean space $\mathbb R^n$ is homeomorphic to the rationals $\mathbb Q$.
Now I wonder if something similar is true for the irrationals $J \mathrel{:=} \mathbb R \setminus \mathbb Q$ (with the subspace topology from $\mathbb R$).
Let $\mathfrak c$ denote the cardinality of the continuum.

I.    Is each cartesian power $J^n$ homeomorphic to $J$?

Also, how far can this be pushed?

II.   Let $X$ be a dense totally disconnected subspace of $\mathfrak R$ such that every neighborhood of each point of $X$ contains $\mathfrak c$ points.  Is $X$ homeomorphic to $J$?

What about for such subspaces of fairly nice subspaces of $\mathbb R^n$?

IIa.  Let $X$ be any subspace of $\mathbb R^n$ as described in II., and let $B$ denote any subspace of $\mathbb R^n$ homeomorphic to [the open unit ball in $\mathbb R^n$ $\cup$ any subset of its boundary].   Then is $X \cap B$ homeomorphic to $J$?

And what about greater generality?

III.  Is there a simple set of conditions that describe exactly all spaces (or subspaces of $\mathbb R^n$) that are homeomorphic to $J$?  What about $J^n$? (Perhaps the word homogeneous or metric needs to be included.)

(I found nothing relevant via Google, in MathSciNet, or here on MathOverflow.)

REPLY [2 votes]: Several answers point out the following:

The space of irrationals is homeomorphic to the Baire space $\mathbb N^{\mathbb N}$ of functions from $\mathbb N$ to $\mathbb N$.

No one gave an explicit homeomorphism. [PS: It's now pointed out that the answer by Richard Borcherds did so, if very tersely. I skimmed too fast.]
Let $a_1,a_2,a_3,\ldots$ be a sequence of positive integers. Then
$$
a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {a_4 +\ddots}}} \in \mathbb R \smallsetminus\mathbb Q.
$$
This number cannot be rational since an expansion of a rational number in this way must terminate because of well-ordering of $\mathbb N.$
That gives you the positive irrationals. To see that that is homeomorphic to the space of all irrationals, recall a fact proved by Georg Cantor: any two countable densely linearly ordered sets without endpoints are order-isomorphic to each other. ("Densely" means only that between any two elements there is another (so no knowledge of topology is needed to understand that word).) And an order isomorphism of those two sets of rationals will give you an order isomorphism of those two sets of irrationals.<|endoftext|>
TITLE: Generalization of Schur's lemma (Update)
QUESTION [6 upvotes]: I am not a mathematician nor physicist. I just know the basics of the representation theory. In my research, I realized that there is an orthogonality relation between the unitary group matrix elements as follows:
$$I_1 = \int \mathrm{D}\mathbf{U} \; U_{i j}^{(\mathbf{r})} U_{k l }^{*(\mathbf{r}^{\prime})} = \frac{1}{ d_{ \mathbf{r} } } \delta_{\mathbf{r} \mathbf{r}^{\prime} }  \delta_{i k} \delta_{j l}
$$
where $\mathbf{U} \in \mathcal{U}(N)$, $\mathrm{D}\mathbf{U}$ is the standard Haar measure, $U_{ij}^{(\mathbf{r})}$ denotes the $(i,j)$-th element of the representation matrix of $\mathbf{U}$, and $d_{ \mathbf{r} }$ is the dimension of the irreducible representation $\mathbf{r}$. 
Now, I need to know the answer for this integral:
$$I_2 = \int \mathrm{D} \mathbf{U} \; U_{i_1 j_1}^{(\mathbf{r})} U_{ k_1 l_1 }^{ * ( \mathbf{r} ) } U_{ i_2 j_2 }^{(\mathbf{r}^{\prime})} U_{ k_2 l_2 }^{* ( \mathbf{r}^{ \prime \prime } ) }
$$
I appreciate any help.
p.s. Here is my conjecture for the answer:
$$
I_2 = \delta_{ \mathbf{r}^{\prime} \mathbf{r}^{\prime \prime} } \times \left\{
 \eqalign{
\frac{1}{ d_{ \mathbf{r} } d_{ \mathbf{r}^{ \prime } } -1 } \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } ( 1- \delta_{ \mathbf{r} \mathbf{r}^{\prime} } ) \\
 + \delta_{ \mathbf{r} \mathbf{r}^{\prime} } \left[
     \eqalign{
      \frac{ 1 }{ d_{ \mathbf{r} }^2 -1 }
 ( \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 }
 + \delta_{ i_1 k_2 } \delta_{ j_1 l_2 } \delta_{ i_2 k_1 } \delta_{ j_2 l_1 } ) \\
 - \frac{ 1 }{ d_{ \mathbf{r} } ( d_{ \mathbf{r} }^2 -1 ) } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_2 } \delta_{ i_2 k_2 } \delta_{ j_2 l_1 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_1 }  \delta_{ i_2 k_1 } \delta_{ j_2 l_2 } )  }
    \right] 
} \right\} $$
UPDATE:
I have been advised that it might be helpful if I can find the tensor product of two irreducible representations, $ \mathbf{s} = \mathbf{r} \otimes \mathbf{r}^{\prime}$, which most likely leads to a reducible representation, and then I need to decompose $\mathbf{s}$ into its irreducible components (by using the Clebsch–Gordan coefficients, according to wikipedia), to be able to use the Schur's lemma to get the answer!!!
However, it is hard for me to do this, and needs awful background.

REPLY [2 votes]: I'm afraid I do not know the answer to your problem, but here's a counterexample to your conjecture that the result is zero if $r'\ne r''$.
take $r=(2,1)$ (i.e., the Young diagram with two boxes one the first row and one on the second). then in the decomposition into irreducible representations of $r=(2,1)$ with its dual $r^\star=(...,-1,-2)$ one finds $(2,...,-1,-1)$ and $(1,1,...,-2)$. now these occur naturally as the tensor product of $r'=(2)$ and the dual of $r''=(1,1)$, or vice versa. ergo, $r\otimes r^\star\otimes r'\otimes r''^\star$ contains the trivial representation so that the integral of some of its matrix elements will be non-zero.<|endoftext|>
TITLE: Question on the decimal expansion of algebraic numbers
QUESTION [5 upvotes]: Is the countability of the set of irrational algebraic numbers somehow reflected in a characteristic property of their decimal expansions?

REPLY [2 votes]: Well, something may be said for sure: decimal expression of irrational algebraic number can not have too many small periodic blocks in a row (like 0.1239239239239...), since it would imply the existence of too good rational approximations and so contradict the Roth's theorem.<|endoftext|>
TITLE: How many people fully understand the proof of Fermat's Last Theorem?
QUESTION [16 upvotes]: What is a rough order of magnitude estimate? $$ $$ There is a thread on Meta about this question, http://mathoverflow.tqft.net/discussion/567/rapid-closing-of-questions/#Item_0

REPLY [53 votes]: Dear Michael, 
The methods introduced by Wiles, and by Taylor and Wiles, in the two papers that proved FLT, as well as the methods introduced by Ribet in his earlier paper reducing FLT to Shimura--Taniyama, are at the heart of much modern work in algebraic number theory and automorphic forms, such as (in addition to the proofs of Shimura--Taniyama and FLT) the proofs of Serre's conjectures and the Sato--Tate conjecture.  
Conferences/workshops in these fields typically attract on the order of magnitude of 100 or so particants, which gives you some sense of the number of students/researchers thinking about these questions: its in the tens or hundreds, but probably not in the thousands. Of course, not all these people know all the details, but the people at the top of the field surely do. (Of course, there is a question of what "understand" means exactly. I don't know how many people have both carefully studied  all the details of the trace formula arguments that underly Jacquet--Langlands, Langlands--Tunnell, and base-change,  and also carefully studied the details of the p-adic Hodge theory and other arithmetic geometry that is used in the arguments. But certainly the top people do understand the significance of these techinques, and are fluent in their use and application, and understand both the overall structure and strategy, as well the technical details, of the proof of FLT itself (and of various more recent related results).  
Finally, let me note that the best evidence for the final claim of the previous paragraph is that this is currently an extremely vibrant area of research, which has progressed at a rapid clip over the last ten years or so.  (The reason for this being that people have not only assimilated the arguments of Wiles/Taylor--Wiles but have improved upon them and pushed them further.)<|endoftext|>
TITLE: Poincaré disk model: is this locus a known curve?
QUESTION [7 upvotes]: Please, consider a line segment $AB$ in the Poincaré disk model. Now, consider the set $S$ of all point $P$ in the disk such that the angle $\angle APB$ is constant. 
Question: is $S$ a known curve?
Thanks!

REPLY [4 votes]: In the Klein model, one may see that this is also a circle. Consider a line segment with one point on the center of the disk. One side of the triangle goes through the center. Then orthogonal lines to a line through the center are also orthogonal in the hyperbolic metric, e.g. since they are preserved by reflection. So one sees that a circle is traced out which goes through the origin. If you'd rather center the curve at the origin, then it will be an ellipse, since hyperbolic isometries of the Klein model are projective transformations. 
To convert to the Poincare model, take a hemisphere sitting over the disk, and project vertically. The projection of the circle is given by the intersection of a cylinder over the circle with the upper hemisphere. This upper hemisphere is conformally equivalent to the Poincare model, e.g. by inversion through a sphere centered at the south pole of the lower hemisphere. I haven't computed the curve this traces out though.<|endoftext|>
TITLE: Is {Ø,{Ø},{Ø,{Ø}}, ... } the only known universe?
QUESTION [17 upvotes]: In the first pages of SGA4 I read

[...] Cependant le seul univers connu est l'ensemble des symboles du type {Ø,{Ø},{Ø,{Ø}}, ... } etc. (tous les éléments de cet univers sont des ensembles finis et cet univers est dénombrable). En particulier, on ne connaît pas d'univers qui contienne un élément de cardinal infini. [...]

(the sole known universe is like {Ø,{Ø},{Ø,{Ø}}, ... }, and we don't know any universe with a infinite cardinal).
Mais, c'est vrai? I wonder if during all these years somebody discovered a universe "bigger" than that exhibited by Grothendieck.

REPLY [17 votes]: Let me rephrase part of what Joel David Hamkins and Anon already said, but without mentioning inaccessible cardinals: 
A Grothendieck universe strictly bigger than the one in the question would be a model of ZFC.  (More precisely, it would become a model once we interpret the membership symbol of ZFC as actual membership.)  So the existence of such a Grothendieck universe would imply the consistency of ZFC.  G"oedel's second incompleteness theorem implies that ZFC cannot prove the consistency of ZFC.  Therefore ZFC cannot prove the existence of a Grothendieck universe.<|endoftext|>
TITLE: Exceptional Lie algebras 
QUESTION [8 upvotes]: I have some questions regarding the exceptional Lie algebras e(n), n=6,7,8.
Can anybody explain to me what prevents us from constructing e(9) from e(8)? One can use the e(8) lattice vectors and try to construct an e(9) vector; one could go even further and try e(10) etc. I know that the Cartan Matrix becomes zero (or negative for 10, ...) which is forbidden, but what does that mean if one would try to write down the generators for e(9)? What's wrong with them as Lie algebra generators? Where does the re-construction of the Lie algebra from the Dynkin diagram / Cartan matrix fail?
Another question I have is related to E(n) as symmetry groups. For the A, B, C and D series one can understand the (fundamental or defining representation of) Lie groups acting on a certain vector space and leaving a certain scalar product invariant. For SO(n) it's (x,y) with x,y living in a real vector space, for SU(n) it's (x*, y) with x,y living in a complex vector space. What about E(n)? Is there a similar scalar product which is invariant? What is the corresponding vector space? Are there other invariants?
(my background is theoretical physics, particle physics, gauge symmetries etc.)
Thanks
Tom

REPLY [3 votes]: The appearance of exceptional Lie algebra, Kac-Moody algebras and Borcherds algebra in gravitational theories is a very elegant  and exciting corner of current research in supergravity and string theory. I would like to discuss how exceptional Lie algebras naturally appear in the context of maximal supergravity theories and how this is connected to $E_{11}$ and Borcherds superalgebras.  I will also discuss how Kac-Moody algebra  occurs naturally in the context of gravitational singularities. 
11 dimensional supergravity and exceptional Lie algebra
11 dimensional supergravity was constructed in 1978 by  Cremmer, Julia and Scherk. 
Nowadays it is considered as the low energy limit of M-theory. 
The  field content of 11-dimensional supergravity  is simply given by a metric $g$ and a 3-form $A_{(3)}$. 
 We can construct all the (massless)
 maximal supergravity theory in
 $D$-dimension with $2 < D <11$, by
 considering  (Kaluza-Klein) reduction
 of 11 dimensional supergravity on a
 $(11-D)$-torus.  This process produces a lot of additional fields (2-forms, 1-forms and 0-forms) coming from the reduction of the metric and the 3-form. 
 In general for a compactification of 11 dimensional supergravity on
 a torus $T^{11-D}$ to a D-dimensional
 spacetime, we produce a scalar
 manifold   $$
 \frac{E_{11-D}}{K(E_{11-D})},$$ 
where $K(G)$ is the maximal 
 compact subalgebra of $G$.  In particular, we have in dimension 5,
 4 and 3
$$ 5D\rightarrow \frac{E_6}{USp(8)},
 \quad 4D\rightarrow \frac{E_7}{SU(8)},
 \quad 3D\rightarrow
 \frac{E_8}{SO(16)}.  $$
$E_{11}$ conjecture and Borcherds algebras
We recall that $E_9=E_8^+$ is understood as the extended Dynkin diagram of $E_8$. In the same way $E_{10}=E_8^{++}$ and $E_{11}=E_8^{+++}$ are the over-extended  and the very-extended Dynkin diagram of $E_8$.
 There is a conjecture introduced by
 Peter West in 2001 and supported by
 several facts that the Kac-Moody
 algebra $E_{11}$ is related to a 
 non-linear realization of M-theory and
 that $E_{11}$ can provide an 11
 dimensional origin not only of all
 massless maximal supergravity theories
 (including type IIB) but also of the
  massive ones.
A beautiful duality was discovered by Iqbal, Neitzke and Vafa between compactifications of M-theory on tori and  the second cohomology of some associated del Pezzo surfaces.  Now the full cohomology of theses surfaces spans the root lattice of a Borcherds superalgebra.
 Henry-Labordere Julia and Paulot have
 shown that some truncations of these
 Borchers algebras provide a
 classification of $p$-forms coming from
 tori reduction of (massive) maximal
 supergravity.  This classification matches the one of the $E_{11}$ conjecture of Peter West. The Borcherds description was recently proven to be systematically derived from the split real form of $E_{11}$ by Henneaux, Julia and Levie. 
Space-time singularities, Kac-Moody algebra and Cosmic billiards
A fascinating and non-speculative occurrence of $E_9$ and $E_{10}$ in a theory of gravity occurs when studying the behavior of gravity near a spacetime singularity. 
Belinskii, Khalatnikov and Lifchitz (BKL) have studied in details the general solution of Einstein equations near a spacetime singularity. As one reaches the singularity, the Einstein equations admit a chaotic behavior in time. Chitre and Misner has reformulated the BKL analysis in terms of a billiard motion in a 2 dimensional hyperbolic space. 
In higher dimension, the chaotic behavior disappear in spacetime dimensions greater than 10. 
In particular, in 11 dimensions, there is no chaos at all. But if one add a 3-form (like the one of 11 dimensional supergravity), chaos comes back. In higher dimension one can also describe the chaotic behavior by a billiard in a higher dimensional hyperboloic space. 
When a theory admits a
 compactification to three dimensions
 on a higher dimensional torus such
 that in the reduced 3 dimensional
 theory, the Lagrangian is given by
 Einstein-Hilbert action and a sigma
 model with target space a $G/H$  such
 that $G$ is a simple Lie group and $H$
 its maximal compact subgroup, the
 billiard table is a Coxeter polyhedron
 and the billiard group is a Coxeter
 group. The table billiard can be
 described by the over-extended
 Kac-Moody algebra $G^{++}$ associated
 with the group $G$.
 In particular $$ \text{The billiard
 associated with eleven supergravity is
 } E_8^{++}=E_{10}.$$
One can formulate the billiard dynamics as a motion in the Cartan subalgebra of the Kac-Moody algebra.<|endoftext|>
TITLE: Demystifying the Caratheodory Approach to Measurability
QUESTION [79 upvotes]: Nowadays, the usual way to extend a measure on an algebra of sets to a measure on a $\sigma$-algebra, the Caratheodory approach, is by using the outer measure $m^* $ and then taking the family of all sets $A$ satisfying $m^* (S)=m^* (S\cap A)+m^* (S\cap A^c)$ for every set $S$ to be the family of measurable sets. It can then be shown that this family forms a $\sigma$-algebra and $m^*$ restricted to this family is a complete measure. The approach is elegant, short, uses only elementary methods and is quite powerful. It is also, almost universally, seen as completely unintuitive (just google "Caratheodory unintuitve" ). 
Given that the problem of extending measures is fundamental to all of measure theory, I would like to know if anyone can provide a perspective that renders the Caratheodory approach natural and intuitive.
I'm familiar with the fact that there is a topological approach to the extension problem (see here or link text) for the $\sigma$-finite case due to M.H. Stone (Maharam has actually shown how to extend it to the general case), but it doesn't give much of an insight into why the Caratheodoy approach works and that is what I`m interested in here.

REPLY [2 votes]: The notion of exterior measure seems quite natural to me after having seen the definition of the Lebesgue exterior measure as the infimum of the measures associated to countable covers of the sets by intervals or rectangles.
Let us first assume that we are on a space $X$ of finite measure, say the unit cube. Let us try to single out a class of sets for which $\sigma$-additivity may holds. Let $A$ be a set in such a class. If we want to measure $A$, we also want to measure its complement in $X$ and the two sets are disjoints so that at least the following equality should be satisfied. 
$$\mu^*(A)+\mu^*(X\backslash A) = \mu^*(X).$$
This property can be shown to be equivalent to Caratheodory definition of measurability if $X$ is of finite measure and the measure is regular (for all subsets $A$ of $X$, there exists a measurable set $B$ containing $A$ with the same outer measure, a fact that can be guaranted by replacing $\mu^*(A)$ by $inf\{\mu^*(B) \mid A \subset B, \ B \ \mu^*-\hbox{measurable}\}$), and it looks quite natural to me.
If $X$ is not of finite measure, this definition is not very restrictive because $everything + \infty = \infty$. In particular, it is satisfied for all bounded sets in ${\bf R}^d$ wrt to Lebesgue exterior measure. So we need to be slightly more restrictive and asks that the previous property  holds in restriction, say, to balls $B_N$ of radius $N$, $N>0$ if they are of finite measure.
$$ \mu^*(B_N) = \mu^*(B_N \cap A) + \mu^*(B_N \cap A^c), \quad \forall N.$$
And here again, this can be shown to be equivalent to Caratheodory's definition.
If we are not on ${\bf R}^d$ but on some general abstract set $X$ with infinite measure, the only choice here is to restrict to all sets $E$ of finite measure.
$$ \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c), \quad \forall E \subset X \hbox{ such that } \mu^*(E)<\infty.$$
Note that this definition is obviously satisfied if $E$ is of infinite measure and we get the general definition of Caratheodory.<|endoftext|>
TITLE: Explicit elements of $K((x))((y)) \setminus K((x,y))$
QUESTION [40 upvotes]: In an answer to the popular question on common false beliefs in mathematics
Examples of common false beliefs in mathematics
I mentioned that many people conflate the two different kinds of formal Laurent series field in two variables.  Let $K$ be any field.  Then we have the joint Laurent series field, the field of fractions
$K((x,y)) = \operatorname{Frac}(K[[x,y]])$ 
and also the iterated Laurent series field 
$K((x))((y)) = \left(K((x))\right)((y))$.  
These fields are not the same, roughly because when we write out an arbitrary element of the iterated field as a formal Laurent series in $y$, for each (say non-negative) $n$, the coefficient of $y^n$ is allowed to be an arbitrary formal Laurent series in $x$.  In particular, as $n$ varies, arbitrarily large negative powers of $x$ may appear.  
However, this is rather far from a convincing argument.  Indeed, I gave the following explicit (and fallacious!) example: $\sum_{n=0}^{\infty} x^{-n} y^n$.  But in a comment to my answer, user AS points out that this element is equal (in the iterated field, say) to $\frac{1}{1-\frac{y}{x}}$ and therefore it must lie in the fraction field of $K[[x,y]]$.  Evidently the fallacy here is that the fraction field of $K[[x,y]]$ is the field of all formal Laurent series which are finite-tailed in both $x$ and $y$.  But as this example shows, the latter isn't even a field, unlike the one-variable case.
[At least when $K = \mathbb{C}$, by less explicit means one can see that these two fields are very different: e.g., the joint field is Hilbertian so has nonabelian Galois group, whereas the iterated field has Galois group $\widehat{\mathbb{Z}}^2$.]
AS offered to write down, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$, so I decided to post a question asking for such a guy.  Of course, there is more than one such element -- or better put, more than one type of construction of such elements -- so I would be interested to see multiple answers to:

Please exhibit, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$.

REPLY [6 votes]: Addendum: actually there is a chain of subfields between $K((x,y))$ and $K((x))((y))$ with cardinality c.
For instance, for any $\lambda>1$ we may consider the subset $R_\lambda$ of $K((x))((y))$ of all Laurent series $\sum_k c_k(x)y^k$ with $c_k\in K((x))$ satisfying $$\inf_k \lambda ^{-k} \mathrm{ord}(c_k)   > -\infty.$$
It's easy to check that it's a subfield of $K((x))((y))$, containing $K[[x,y]]$.  
Moreover, since in place of $\lambda^k$ we can use functions $\mathbb{N}\to\mathbb{N}$ with arbitrarily large growth, one can also show that the subfields between $K((x,y))$ and $K((x))((y))$ have uncountable cofinality also.<|endoftext|>
TITLE: How do branched coverings of complex surfaces "fit" with branched coverings of curves?
QUESTION [8 upvotes]: Since I'm used to working with algebraic $\pi_1$'s, which don't work well with surfaces, I find myself lacking geometric intuition when I attempt to do these types of purely geometric arguments. I'm hoping someone can point out a fact or two that will help me gain some insight.
Let's say we have a complex surface, $X$, and we are given a map $X \rightarrow \mathbb{A}^1_{\mathbb{C}}$ such that the fibers are all $\mathbb{P}^1_{\mathbb{C}}$'s. Let's say we are given a prescribed branch divisor, $B$, made up of the prime divisors $B_1, ..., B_r$ (I want them to be horizontal - meaning that each $B_i$ considered over any $a$ in $\mathbb{A}^1_{\mathbb{C}}$ is just a point). Assume for the sake of simplicity that $B_1$ meets $B_2$ once and no two other branch points meet.
Let's fix a complex $t$ (over which $B_1$ and $B_2$ don't meet), and pick our basepoint to be some point over $t$ that doesn't meet the branch points. A $G$-Galois cover $Y \rightarrow X$ branched at most at $B$ gives a $G$-Galois cover of $t \times_{\mathbb{A}^1_{\mathbb{C}}} X \cong \mathbb{P}^1_{\mathbb{C}}$. This cover can now be described as $\gamma_i \mapsto g_i$ where $\gamma_i$ is a loop around $B_i \times_{\mathbb{A}^1_{\mathbb{C}}} t$ (meaning the fiber of $B_i$ over $t$). This would imply in the cover of the curve $\langle g_1 \rangle$ is the inertia of some point over $B_1 \times_{\mathbb{A}^1_{\mathbb{C}}} t$ and $\langle g_2 \rangle$ is the inertia of some point over $B_2 \times_{\mathbb{A}^1_{\mathbb{C}}} t$. In fact, we know what point it is, if you fix a basepoint above. Fix such a basepoint.
It seems that this data alone determines the cover over the entire surface. Is that right? By this I mean: take $\pi_1(X \setminus B, basept)$ and map $\gamma_i$, which is a loop in our curve which lies in our surface, $X$, to $g_i$. And if it is true -- can we somehow use this to know what the inertia groups are for the prime divisors over $B_1$ and $B_2$ that meet? (after all, they don't all meet. all we know is there is some prime divisor over $B_1$ that meets some prime divisor over $B_2$.) Finding the inertia groups of those divisors in terms of whatever I can get my hands on through curves (which means including the cover over $t$) is really my ultimate goal.

REPLY [4 votes]: My answer does not intend to completely solve your problem, but I wish to write down a couple of examples that (I hope) could help you gain some geometrical insight.
I will take $G$ abelian, since in this case the map $\pi_1(X \setminus B) \to G$    factors through $H_1(X \setminus B, \mathbb{Z}) \to G$. Moreover I will take 
$X=\mathbb{P}^2$ (if you want a map with fibre $\mathbb{P}^1$, just blow-up a point).
EXAMPLE 1. 
$G$=$\mathbb{Z}_2=\langle g | g^2=1 \rangle$, 
$B_1$ and $B_2$ two distinct lines intersecting in a point $p$, 
$B=B_1 + B_2$. 
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the two loops $\gamma_1$ and $\gamma_2$ (around $B_1$ and $B_2$, respectively) with the relation $\gamma_1+\gamma_2=0$, hence  it is isomorphic to $\mathbb{Z}$. The unique $\mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_1$, and consequently $ \gamma_2$, to the generator $g$. The inertia group (stabilizer) over $B_1 + B_2$ is of course isomorphic to $\mathbb{Z}_2$, but notice that the preimage of $p$ in $Y$ is $singular$. In fact, $Y$ is isomorphic to a quadric cone in $\mathbb{P}^3$.
EXAMPLE 2.
$G$=$\mathbb{Z}_2 \times \mathbb{Z}_2=\langle g_1, g_2, g_3 | g_i^2=1, g_1g_2g_3=1, [g_i, g_j]=1 \rangle$, 
$B_1$, $B_2$, $B_3$ three distinct lines intersecting pairwise in three distinct points; 
set $p_{ij}:=B_i \cap B_j$. 
$B=B_1+B_2+B_3$.
Then $H_1(X \setminus B, \mathbb{Z})$ is generated by the three loops $\gamma_1$,  $\gamma_2$ and $\gamma_3$ (around $B_1$, $B_2$ and $B_3$, respectively) with the  relation $\gamma_1+\gamma_2+\gamma_3=0$, hence  it is isomorphic to $\mathbb{Z}^2$. Up to permutation of the indices, the unique $\mathbb{Z}_2 \times \mathbb{Z}_2$-cover $Y \to X$ branched on $B$ is obtained by sending $\gamma_i$ to $g_i$, for all $i=1,2,3$. 
The inertia group of a point over $B_1$, distinct from $p_{12}$ and $p_{13}$, is isomorphic to $\langle g_1 \rangle$ and similarly for the other two lines. But the inertia group over the three points $p_{ij}$ is the whole group $\mathbb{Z}_2 \times \mathbb{Z}_2$.  
In this case $Y$ is smooth, in fact it is isomorphic to $\mathbb{P}^2$. However, there are three "intermediate covers" $Y_1$, $Y_2$, $Y_3$, corresponding to the three non-trivial subgroups of $G$; each $Y_i$ is then isomorphic to a quadric cone.
A good reference for these topics is Pardini's paper "Abelian covers of algebraic varieties". If you like a more topological approach, you can read Catanese's article "On the moduli space of surfaces of general type", where the case $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ is developed in full details.<|endoftext|>
TITLE: Group cannot be the union of conjugates
QUESTION [6 upvotes]: I have seen that if $G$ is a finite group and $H$ is a proper subgroup of $G$ with finite index then $ G \neq \bigcup\limits_{g \in G} gHg^{-1}$. Does this remain true for the infinite case also?

REPLY [10 votes]: In your original question you require $H$ to have finite index in $G$. Most of the other answers are treating this as unintended. If you actually did want to require this, then the result is true for infinite groups as well. 
Proof: Let $[G:H]=n$. The action of $G$ on $G/H$ gives a map $G \to S_n$; let $K$ be the kernel of this map. Then $H/K$ and $G/K$ are finite groups, so we know $G/K \neq \bigcup g (H/K) g^{-1}$. In particular, there is some coset $fK$ which is not in any $g H g^{-1}$. So $f$ is not in any $g H g^{-1}$.<|endoftext|>
TITLE: Function satisfying $f^{-1} =f'$
QUESTION [49 upvotes]: How many functions are there which are differentiable on $(0,\infty)$ and that satisfy the relation $f^{-1}=f'$?

REPLY [7 votes]: This question is more complicated than it seems
If $f^{-1}(x)=f'(x)$ then $f(f^{-1}(x))=f(f'(x))$ or $x=f(f'(x))$
First derivative is giving
$\displaystyle 1=f'(f'(x))f''(x)$
Now replace $y=f'(x)$
$\displaystyle y(y(x))y'(x)=1$
Instead of the above we will solve a more general:
$\displaystyle g(y(x))y'(x)=1$
This one has a solution expressed as
$\displaystyle \int_{1}^{y(x)}g(r) \mathrm{d} r = x+c$
Now since we want $y(x)=g(x)$ we have
$\displaystyle \int_{1}^{y(x)}y(r) \mathrm{d} r = x+c$
$\int_{1}^{y(x)}y(r) \mathrm{d} r$ is actually a curious operator which we will represent as $\mathfrak{R}(y(x))_{1}$
Now this operator has its inverse and this is what we are looking for. For a function $h(x)$ we want to know $s(x)$ so that
$\displaystyle \mathfrak{R}(s(x))=h(x)$
We will call $\mathfrak{R}^{-1}()$ the integral root.
As a new operator we need to examine it a little bit. We can create a table for some known functions first
$\begin{matrix} h(x) & \mathfrak{R}^{-1}(h(x))_{1} \\  & \\ \frac{1}{2}(x^2-1) & x\\ \frac{1}{3}(x^6-1) & x^2\\ 1 - \ln(x)(1 - \ln(\ln(x))) & \ln(x)\\ e^{e^x}-e & e^x\\ -\ln(x) & \frac{1}{x}\\ \cos(1)-\cos(\sin(x)) & \sin(x)\\ \frac{2}{3}(x^{\frac{3}{4}}-1) & \sqrt{x} \end{matrix}$
We do not ignore constant term since integral root is very sensitive operator and we define in general integral root with base $b$ as:
$\mathfrak{R}(h(x))_{b}=\int_{b}^{f(x)}f(r) \mathrm{d} r$
We will return to the importance of base later but for now we say that the function does not have to be defined everywhere and base can help about it. So, all we need to find the integral root of $h(x)=x$.
The operator resembles a normal derivative/integration except that it is extending it all, it is giving sort of a faster result.
It is a deep question if this operator has unique values. (We will hint something into that direction but overall this is a very good subject for some semester work.)
Notice that the operator is very sensitive to the constant value and this one cannot be ignored.
Since powers of $x$ appears from the powers of $x$, we could try the solution in the most general form as $g(x)=ax^b+f(x)$ however this will convince us very quickly that for any solution defined everywhere $f(x)=0$, since such solutions are very restrictive and it is not possible to drive constants the way we would like to, even adding $ax^b+c$ would create a solution that is useless. The way of dealing with constant term is by changing the base in general.
So, if we restrict ourselves to global world we have
$\displaystyle \mathfrak{R}(ax^b)_{1}=\frac{a^{b+2}x^{b(b+1)}}{b+1}-\frac{a}{b+1}$
Since we want
$\displaystyle \mathfrak{R}(ax^b)_{1}= x+c$
that requires $b(b+1)=1$ making the result
$ba^{b+2}x-ab$
This requires $ba^{b+2}=1$ as well which makes $a=(b+1)^{\frac{1}{b+2}}$
(Notice that these are not two different integral roots of the same function.)
Now our solution is $y(x)=f'(x)$ so we have it back as
$\displaystyle f(x)=\frac{a}{b+1}x^{b+1}=(b+1)^{-\frac{1}{b+1}}x^{b+1}$
where as we have mentioned $b(b+1)=1$
The reason we opt for this operator is that the function in question has derivative thus it is assumed that it is a nicely behaving function.
Let us prove that these two are the solutions
Derivative
$\displaystyle f'(x)=(b+1)^{-\frac{1}{b+1}}(b+1)x^{b}$
$\displaystyle f'(x)=(b+1)^{-\frac{1}{b+1}+1}x^{b}$
$\displaystyle f'(x)=(b+1)^{\frac{b+1-1}{b+1}}x^{b}$
$\displaystyle f'(x)=(b+1)^{\frac{b}{b+1}}x^{b}$
Inverse
$\displaystyle x=(b+1)^{-\frac{1}{b+1}}h(x)^{b+1}$
$\displaystyle (b+1)^{\frac{1}{b+1}}x=h(x)^{b+1}$
$\displaystyle (b+1)^{\frac{1}{(b+1)(b+1)}}x^{\frac{1}{b+1}}=h(x)$
$\displaystyle f^{-1}=h(x)=(b+1)^{\frac{b}{b+1}}x^{b}$
Finally let us use golden ratio to find what the solutions we are talking about
$\displaystyle b_{1}=-\phi$,$ b_{2}=\phi-1$
making the first solution actually complex and the second real:
$\displaystyle (-\phi+1)^{-\frac{1}{-\phi+1}}x^{-\phi+1}=(-\phi)^{-\phi}x^{-\frac{1}{\phi}}$
$\displaystyle \phi^{-\frac{1}{\phi}}x^{\phi}$
The two solutions we have are driven by differing a constant only, otherwise the integral root has the unique solution just like normal root has. It is possible that there are solutions that are not analytical or defined everywhere, otherwise the two given solutions are the only nice global solutions.
Apart from these global solutions that do not change if we change the integral root base, there is an option of having a local solution. For this we extend the function analytically at some point $x_{0}$ and ask that the function behaves as $x$ regarding the integral root at that point.
Then near $x_{0}$ that function will behave as required.
$\displaystyle f(x)=\sum\limits_{n=0}^{+\infty}c_{n}(x-x_{0})^n $
Basically we are asking
$\displaystyle \mathfrak{R}(\sum\limits_{n=0}^{+\infty}c_{n}(x-x_{0})^n)_{x_{0}} = x+c$
Notice that the change of base is very important, albeit purely technical for this particular task, as we have assumed that we know this function only around $x_{0}$
Then we need to have:
$\displaystyle \sum\limits_{n=0}^{+\infty}\frac{c_{n}}{n+1}(\sum\limits_{m=0}^{+\infty}c_{m}(x-x_{0})^m-x_{0})^{n+1}=x+f(x_{0})+c$
Obviously we can match all coefficients since the final coefficient by $x$ should be $1$ and all higher powers $0$. Solving all coefficient for $c_{k}$ will give local solutions for each $x_{0}$. We already have one such non trivial solution, but there could be more than that global one.
However, notice that the integral root base does not affect the answer, since the base deals with constant term only, and constant term is arbitrary. Instead of $1$ for our function we could take any constant.
This is to say that our local solution again has the same resolution which we have found when we started from $1$.
Unless there is some curious and undiscovered way the integral root can have more than two given solutions, one real and another complex, each local solution is equal to either of the two found global solutions.<|endoftext|>
TITLE: Transcendence of PI
QUESTION [7 upvotes]: Can anyone suggest me an ingenious proof of the transcendence of $\pi$. I have seen Lindemann's proof but it appears intricate.

REPLY [7 votes]: An unfortunate thing is that proving the transcendence of $\pi$ has the same complexity as the Lindemann--Weierstrass theorem. (Proving the transcendence of $e$ is much easier.) I am quite surprised to see that the book 
[A.I. Galochkin, Yu.N. Nesterenko, and A.B. Shidlovskiĭ,
Введение в теорию чисел [Introduction to number theory],
2nd edition, Moscow State Univ., 1995. 160 pp. ISBN: 5-211-03075-3. MR1367734 (96i:11001)] is not translated into English (I remember that there was an attempt to negotiate the translation about 10 years ago). I put here an extract from Yu. Nesterenko's lectures (in Russian!) on the Lindemann--Weierstrass theorem which are "cleaned", so I believe this is the simplest version of proof.
There are more exotic ways to state the transcendence of $\pi$, of course, these proofs are more involved. One example in this direction is [V.N. Sorokin, On the measure of transcendency of the number $\pi^2$, Sb. Math. 187  (1996),  no. 12, 1819--1852] where the author shows the transcendence of $\pi^2$ (rather than $\pi$ itself) by constructing linear forms involving the numbers
$$
\sum_{n_1\ge n_2\ge\dots\ge n_r\ge1}\frac1{n_1^2n_2^2\cdots n_r^2}
$$
which are (now) known to be simple rational multiple of $\pi^{2r}$, $r=1,2,\dots$.<|endoftext|>
TITLE: If $f$ is infinitely differentiable then $f$ coincides with a polynomial
QUESTION [222 upvotes]: Let $f$ be an infinitely differentiable function on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how.
I thought of using Weierstrass approximation theorem, but couldn't succeed.

REPLY [24 votes]: Let me add one more solution. It is not really different from the accepted one, but it includes all details. The problem is that a student without sufficient experience will not even see necessity to fill details.  

Theorem.
If $f\in C^\infty(\mathbb{R})$ and for every $x\in\mathbb{R}$ there is a nonnegative integer $n$ such that $f^{(n)}(x)=0$, then $f$ is a
  polynomial.

The following exercise shows that the result cannot be to easy.

Exercise. Prove that there is a function $f\in C^{1000}(\mathbb{R})$ which is not a polynomial, but has the property described in the above
  theorem.

Proof of the theorem.
Let $\Omega\subset\mathbb{R}$ be the union of all open intervals $(a,b)\subset\mathbb{R}$
such that $f|_{(a,b)}$ is a polynomial. The set $\Omega$ is open, so
$$
\Omega=\bigcup_{i=1}^N (a_i, b_i)\, ,
\qquad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   (1)
$$
where $a_i<b_i$ and $(a_i, b_i)\cap (a_j, b_j) = \emptyset$ for $i\neq j$, $1\leq N\leq\infty$.
Observe that $f|_{(a_i, b_i)}$ is a polynomial
(Why?)*.
We want to prove that $\Omega=\mathbb{R}$. First we will prove that
$\overline{\Omega}=\mathbb{R}$. To this end it suffices to prove that for any
interval $[a,b]$, $a<b$ we have $[a,b]\cap\Omega\neq\emptyset$.
Let
$$
E_n=\{x\in\mathbb{R}:\, f^{(n)}(x)=0\}\, .
$$
The sets $E_n\cap [a,b]$ are closed and
$$
[a,b]=\bigcup_{n=0}^\infty E_n\cap [a,b]\, .
$$
Since $[a,b]$ is complete, it follows from the Baire theorem that for some $n$
the set $E_n\cap [a,b]$ has nonempty interior (in the topology of $[a,b]$),
so there is $(c,d)\subset E_n\cap [a,b]$ such that $f^{(n)}=0$ on $(c,d)$.
Accordingly $f$ is a polynomial on $(c,d)$ and hence
$$
(c,d)\subset\Omega\cap [a,b]\neq\emptyset.
$$
The set $X=\mathbb{R}\setminus\Omega$ is closed and hence complete.
It remains to prove that $X=\emptyset$. Suppose not.
Observe that every point $x\in X$ is an accumulation point of the set,
i.e. there is a sequence $x_i\in X$, $x_i\neq x$, $x_i\to x$. Indeed,
otherwise $x$ would be an isolated point, i.e. there would be two intervals
$$
(a,x),\, (x,b)\subset\Omega,\ x\not\in\Omega\, . 
\qquad \ \ \ \ \ \ \ \ \ \ \ \ (2)
$$
The function $f$ restricted to each of the two intervals is a polynomial, say
of degrees $n_1$ and $n_2$. If $n>\max\{ n_1, n_2\}$, then
$f^{(n)}=0$ on $(a,x)\cup (x,b)$. Since
$f^{(n)}$ is continuous on $(a,b)$, it must be zero on the entire interval
and hence $f$ is a polynomal of degree $\leq n-1$ on $(a,b)$, so
$(a,b)\subset\Omega$ which contradicts (2).
The space $X=\mathbb{R}\setminus\Omega$ is complete.
Since
$$
X=\bigcup_{n=1}^\infty X\cap E_n\, ,
$$
the second application of the Baire theorem
gives that $X\cap E_n$ has a nonempty interior in the topology of $X$, i.e.
there is an interval $(a,b)$ such that
$$
X\cap (a,b)\subset X\cap E_n\neq\emptyset\, .
\qquad \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
$$
Accordingly $f^{(n)}(x)=0$ for all $x\in X\cap (a,b)$.
Since for every $x\in X\cap (a,b)$ there is a sequence $x_i\to x$, $x_i\neq x$
such that $f^{(n)}(x_i)=0$
it follows from the definition of the derivative that
$f^{(n+1)}(x)=0$ for every $x\in X\cap (a,b)$, and by induction
$f^{(m)}(x)=0$ for all $m\geq n$ and all $x\in X\cap (a,b)$.
We will prove that $f^{(n)}=0$ on $(a,b)$. This will imply that
$(a,b)\subset\Omega$ which is a contradiction with (3).
Since $f^{(n)}=0$ on $X\cap (a,b)=(a,b)\setminus\Omega$ it remains to prove that
$f^{(n)}=0$ on $(a,b)\cap\Omega$.
To this end it suffices to prove that for any interval
$(a_i, b_i)$ that appears in (1) such that
$(a_i, b_i)\cap (a,b)\neq\emptyset$, $f^{(n)}=0$ on $(a_i, b_i)$. Since
$(a,b)$ is not contained in $(a_i, b_i)$ one of the endpoints belongs
to $(a,b)$, say $a_i\in (a,b)$. Clearly
$a_i\in X\cap (a,b)$ and hence $f^{(m)}(a_i)=0$ for all $m\geq n$.
If $f$ is a polynomial of degree $k$ on $(a_i, b_i)$, then
$f^{(k)}$ is a nonzero constant on $(a_i, b_i)$, so $f^{(k)}(a_i)\neq 0$ by
continuity of the derivative. Thus $k<n$ and hence
$f^{(n)}=0$ on $(a_i,b_i)$.
$\Box$

Exercise.
As the previous exercise shows the theorem is not true if we only assume that $f\in C^{1000}$. Where did we use in the proof the
  assumption $f\in C^\infty(\mathbb{R})$?


*It suffices to prove that $f$ is a polynomial on every compact
subinterval $[c,d]\subset (a_i, b_i)$. This subinterval has a finite covering by
open intervals
on which $f$ is a polynomial. Taking an integer $n$ larger than the maximum
of the degrees of these polynomials, we see that $f^{(n)}=0$ on $[c,d]$
and hence $f$ is a polynomial of degree $<n$ on $[c,d]$.<|endoftext|>
TITLE: When does graph minor containment imply subgraph containment?
QUESTION [15 upvotes]: Consider a path of length 3.  Any graph G which contains this graph as a minor must also contain it as a subgraph. For paths of any length this is easy to prove.
In general this happens for any graph in which each connected component is either a path or a subdivision of the claw grah. (The claw graph is the star graph on 4 vertices, or the complete bipartite graph $K_{1,3}$.) 
Does anyone know where I can find a proof of this fact? I know how to prove it, but if it has already appeared in some book/paper, it's easier for me to cite the result instead.
I am also interested in the following more general question, whose answer I do not know: How does one characterize sets of graphs H={H1,H2,...,Hk}, such that containing any of the Hi as a minor is equivalent to containing some graph from a finite set G={G1,G2,...,Gm} as a subgraph. In other words, for which sets H is H-minor containment equivalent to G-subgraph containment for some finite set G. (Note that this is trivial if we allow G to be infinite.)
For example, containing any graph from the set {path of length 3, claw} as a minor is equivalent to containing any graph from that set as a subgraph. As a non-trivial example, containing any from from {path of length 4, cycle of length 3} as a minor is equivalent to containing one of {path of length 4, cycle of length 3, cycle of length 4} as a subgraph.
Edit: For the single graph problem, I think I have stated a complete characterization of such graphs. I only wish to know if this appears in the literature somewhere. For the second problem I do not know a characterization (other than some special cases), and would welcome any information about the problem.

REPLY [4 votes]: The case when H has more than one forbidden minor is messy, but here's a partial analysis.
If H contains a linear forest (disjoint union of paths) then some path P_k is forbidden and the H-minor-free graphs are obviously describable by a finite set of forbidden subgraphs (P_k and the expansions of H that do not have a P_k subgraph).
If H does not contain a linear forest or disjoint union of subdivided claws, then the H-minor-free graphs include all subdivided claws. In this case the H-minor-free graphs are obviously not describable by a finite set of forbidden subgraphs -- any minimal forbidden minor of H can have its vertices blown up to degree-three trees and then its edges blown up to long paths so that the local neighborhood of any vertex looks the same as a subdivided claw, producing a graph that is not H-minor-free but looks locally like an H-minor-free graph.
If H does contain a cycle, and doesn't contain a linear forest, then the H-minor-free graphs are again obviously not describable by a finite set of forbidden subgraphs -- long cycles are not H-minor-free but look locally like paths, which are H-minor-free.
If H doesn't contain a linear forest or a cycle, but does contain a linear forest + claw in which only one of the claw edges is subdivided, then the H-minor-free graphs are describable by a finite set of forbidden subgraphs (the subdivided claw and the expansions of H that do not contain it -- because H doesn't contain a path or a cycle, each edge in each forbidden minor of H can only be expanded to a length shorter than the subdivided edge of the forbidden claw).
The remaining case is when H contains linear forest + claw subgraphs, but only those in which two or three of the claw edges have been subdivided, or when it contains graphs with more than one claw component.<|endoftext|>
TITLE: Irreducible components of quotients of Cohen-Macaulay rings of the "correct" dimension
QUESTION [5 upvotes]: Suppose $R$ is a Cohen-Macaulay ring.  It is well known that if $I$ is an ideal of $R$ generated by $n$ elements, and $I$ has codimension $n$, then $R/I$ is also Cohen-Macaulay.
Now suppose that $I$ does not have codimension $n$, but (the scheme defined by) $R/I$ has several irreducible components, one of which has codimension $n$.  Is (the coordinate ring of) that component necessarily Cohen-Macaulay?
Because being Cohen-Macaulay is a local condition, it is clear that the component is generically (in fact everywhere it does not intersect the other components) Cohen-Macaulay, but there is no reason obvious to me why this would extend to the whole component.

REPLY [6 votes]: Hi Alex, I think this fails for $n=2$. Start with  a polynomial ring $S$ and  height $2$ prime $P$ such that $S/P$ is not Cohen-Macaulay (for example let $P$ be the kernel of the map $S=k[a,b,c,d] \to k[x^4,x^3y,xy^3,y^4]$). Let $a,b$ be a regular sequence in $P$. Let $R=S[t]$ and $I=(ta,tb)$. Then $I$ has height $1$ and is $2$-generated but the components of $I$ is $(t)$ and the components of $(a,b)$, which include $P$. But $R/P$ is not Cohen-Macaulay because $S/P$ isn't.

REPLY [4 votes]: To complement Hailong's answer, here is another way of seeing that an irreducible component of a Cohen-Macaulay scheme need not have special properties. In the example below, the whole scheme is Cohen-Macaulay, so all of its components have the same dimension, unlike in Hailong's example.
Let $X$ be a reduced and irreducible subscheme of projective space $\mathbb{P}^n$ of codimension $c$.  Choose $c$ elements $f_1,\ldots,f_c$ in the ideal of $X$ in $\mathbb{P}^n$ with the property that the vanishing set $X'$ of $f_1,\ldots,f_c$ is reduced and has codimension $c$.  Thus the scheme $X'$ contains $X$ as a component, it is a complete intersection, and hence it is Cohen-Macaulay.  On the other hand, $X$ was essentially arbitrary, so you could have chosen it to be not Cohen-Macaulay!
For a more explicit example, let $X$ be a surface in $\mathbb{P}^4$ with a point that analytically locally looks like the union of two planes at a single point. (This is one of the standard example of a scheme that is not Cohen-Macaulay.)  Choose two "general" elements of the ideal of $X$, and let $X'$ be the scheme defined by those two elements.  Thus, $X'$ is Cohen-Macaulay and $X$ is a component of $X'$, but the surface $X$ it is not Cohen-Macaulay.
EDIT As Hailong pointed out, I answered a question that is different than what was asked.  To answer the initial question, it suffices to argue as above, choosing also a hypersurface $H$ in $\mathbb{P}^n$ not containing $X$.  Denote by $h$ an equation for $H$ and replace $f_1 , \ldots , f_c$ by $h f_1 , \ldots , h f_c$.  The vanishing set $\overline{X}$ of these equations consists of the union of $H$ and the scheme $X'$ we had before.  Thus $X$ is still a component of $\overline{X}$, of codimension $c$ in $\mathbb{P}^n$, the ideal of $\overline{X}$ is generated by $c$ equations, but the component $X$ is (essentially) arbitrary, in particular it need not be Cohen-Macaulay.  This should now answer the question that was asked!<|endoftext|>
TITLE: Motivation of Moment Generating Functions
QUESTION [8 upvotes]: What is the motivation of defining the moment generating function of a random variable $X$ as: $E[e^{tX}]$? I know that one can obtain the mean, second moment, etc.. after computing it. But what was the intuition in using $e^{x}$? Is it because its one-to-one and always increasing?

REPLY [6 votes]: The goal is to to put all the moments in one package. Since
$$
 e^{tx} = \sum \frac{x^n}{n!} t^n
$$
the coefficients of $t^n$ in $E(e^{tx})$ are (scaled) moments. In other contexts we can use
$$
(1-xt)^{-1} = \sum x^n t^n
$$
in place of $e^{tx}$. This gives more or less what engineers call the "z-transform"
and in combinatorics it is known as "ordinary generating function". Using the exponential
has the happy advantage that convolution of random variables translates to product
of moment generating functions.<|endoftext|>
TITLE: What are the obstructions for a Henstock-Kurzweil integral in more than one dimension?
QUESTION [29 upvotes]: I have recently come across the book The Kurzweil-Henstock Integral and its Differentials by Solomon Leader, in which, if I understand correctly, the HK integration process is modified in a way that makes it also work for dimensions higher than 1 (there's a proof of Green's theorem at the end). It has always been my impression that HK-integration doesn't extend to n dimensions, but truth be told, I don't actually know why. 
So my question(s) is (are):


In what sense can the Henstock-Kurzweil integral not be extended to more than one dimension?
Leader's construction via summants below seems very reminiscent of Jenny Harrison's work on chainlets. Are the two related?
Does the relationship to measures from the one-dimensional case go both ways, i.e. every measure is a differential? Would this relationship be preserved in higher dimensions?


Below I've summarized the key features of Leader's construction.

A cell is a closed interval [a,b] in $[-\infty, \infty]$. A figure is a finite union of cells. A tagged cell in a figure K is a pair (I,t) where I is an interval contained in K, and t is an endpoint of I (according to Leader, the restriction of tags to be endpoints is key).
A gauge is a function $\delta:[-\infty,\infty]\to (0,\infty)$. Every gauge associated to every point t a neighborhood '$N_\delta(t)$ which is $(t-\delta(t),t+\delta(t))$ for finite t, $[-\infty,-\frac1{\delta(-\infty)}]$ and $[\frac1{\delta(-\infty)},\infty]$ for the infinite points. This ensures that $N_\alpha(t) \subset N_\beta(t)$ if $\alpha(t)\leq\beta(t)$, and then we can define division of a figure K into tagged cells to be $\delta$-fine if for each tagged cell $(I,t)$ we have $I\subset N_\delta(t)$.
Then where I understand Leader's theory to take a departure from the normal development, he defines a summant S to be a function on tagged cells, and then he constructs $\int_K S$ of a summant S over a figure K as the directed limit of $\sum_{(I,t)\in\mathcal{K}} S(I,t)$ over gauges $\delta$, where $\mathcal{K}$ are $\delta$-fine divisions of K (he actually defines a limit supremum and a limit infimum and works with those).
Some summants are for example $\Delta([a,b])=b-a$ and $|\Delta|([a,b]=|b-a|$. Any summant S can be multiplied by a function by way of $(fS)(I,t)=f(t)S(I,t)$, and any function can be canonically extended to a summant $f\Delta$.
Where his theory really gets interesting is that he defines differentials as equivalence classes of summants under the equivalence relation $S~T$ if $\int_K|S-T|=0$. From there he defines the differential of any function g by $dg=[g\Delta]$, where $[S]$ is the equivalence class of the summant S.
From this he calls a differential integrable if its representative summants S are integrable, and show that every integrable summant is of the form df where f is a function. Then absolutely integrable summants (ones such that $|df|$ is integrable) give rise to measures. For example, the differential dx, where x is the identity function, corresponds to standard Lebesgue measure.
A final point of interest is that the fundamental theorem of calculus can be formulated as $f\Delta=f'df$ where $f'$ is the usual derivative of f (which I actually think is a great way to motivate the definition of pointwise derivative in the first place).
On to n-dimensional theory, though. An n-cell [a,b] consists of the parallelepiped with opposite vertices (a,a,a,...,a) and (b,b,b,b,...,b). A tagged n-cell (I,t) has tag one of the vertices, it is $\delta$-fine if it's diameter is less than $\delta(t)$. The $\Delta^{(n)}g$ is given by the alternating sum $\sum_{t\in V_I} (-1)^{\mathcal{N}_I(v)}g(t)$, where $V_I$ are the vertices of I, and $N_I(v)$ are the number of coordinates of v that are the same as those of the tag t.
Allegedly (Solomon doesn't give the details in his book), integration goes through, though some results regarding fundamental theorem and such allegedly do not. I am unclear on what happens to the relationship with measures.

REPLY [5 votes]: A reasonable method of defining an integral that includes the HK integral is to say a Schwartz distribution $f$ is integrable if it is the distributional derivative of a continuous function $F$.  Then the integral $\int_a^b f =F(b) - F(a)$.  The resulting space of integrable distributions is a Banach space that includes the space of HK integrable functions and is isometrically isomorphic (with Alexiewicz norm) to the continuous functions vanishing at a (with uniform norm).  In Euclidean spaces one does much the same but integrable distributions are those that are the distributional derivative taken once in each Cartesian variable.  Again, we get a Banach space of integrable distributions.  A Fubini theorem holds but the only transformations that can be done are those that map Cartesian intervals to Cartesian intervals.  This rules out rotations.  For details, see D.D. Ang and L.K. Vy,
On the Denjoy--Perron--Henstock--Kurzweil integral,
Vietnam J. Math. 31 (2003), 381--389.<|endoftext|>
TITLE: Order matters when choosing sets
QUESTION [16 upvotes]: Warren Moors and Julia Novak in a paper entitled "Order matters when choosing sets" proved that if 1 < w < t < v are integers then
$${{{v}\choose {w}}\choose {t}} >  {{{v}\choose {t}}\choose {w}}.$$
In words: the number of t-subsets from the family of w-subsets of [v]={1,2,...,v} is larger than the number of w-subsets from the family of t-subsets of [v].
The question
(proposed by Steve Wilson in a problem session of this conference)

Find a combinatorial explanation for Moors and Novak's result.

Remark
Moors and Warren give an elementary proof an even stronger inequality.
$$(w!)^{t-1}{{{v}\choose {w}}\choose {t}} > (t!)^{w-1} {{{v}\choose {t}}\choose {w}}.$$

REPLY [4 votes]: This is not a full answer but I'm throwing some observations out. Note that the even stronger inequality also has a clear combinatorial interpretation. Namely, if one denotes by inj(A,B) the set of injections from A to B, then the inequality says that 
|inj(T,inj(W,V))| > |inj(W,inj(T,V))|.
To see this, multiply both sides of the stronger inequality by t!w!. Of course, here t = |T|, w = |W| and v = |V|.
In this form, it is easy to believe the inequality. Consider a function $f : T \times  W \to V$, thought of as a $t \times w$ matrix with entries in V. It defines an element of the left hand side if there is no repeated entry in any row, and there are no repeated entire rows. The right hand side counts the same thing with columns. We have assumed that the columns are longer than the rows. Then if the entries of the matrix are picked uniformly at random, it should fail to define an element of the right hand side with greater probability, since most functions that fail should do so because of repeated entries in a row/column rather than an entire repeated row/column, and it is more likely to be a repeated entry in a column than in a row.
However, I don't see any nice way of producing a proof out of the above heuristic. Bijective proofs seem to get really complicated...<|endoftext|>
TITLE: A geometric interpretation of the Levi-Civita connection?
QUESTION [42 upvotes]: Let $M$ be a Riemannian manifold. There exists a unique torsion-free connection in the (co)tangent bundle of $M$ such that the metric of $M$ is covariantly constant. This connection is called the Levi-Civita connection and its existence and uniqueness are usually proven by a direct calculation in coordinates. See e.g. Milnor, Morse theory, chapter 2, \S 8. This is short and easy but not very illuminating.
According to C. Ehresmann, a connection in a fiber bundle $p:E\to B$ (where $E$ and $B$ are smooth manifolds and $p$ is a smooth fibration) is just a complementary subbundle of the vertical bundle $\ker dp$ in $T^*E$. If $G$ is the structure group of the bundle and $P\to B$ is the corresponding $G$-principal bundle, then to give a connection whose holonomy takes values in $G$ is the same as to give a $G$-equivariant connection on $P$.
If $p:E\to B$ is a rank $r$ vector bundle with a metric, then one can assume that the structure group is $O(r)$; the corresponding principal bundle $P\to B$ will in fact be the bundle of all orthogonal $r$-frames in $E$. One can then construct an $O(r)$-equivariant connection by taking any metric on $P$, averaging so as to get an $O(r)$-equivariant metric and then taking the orthogonal complement of the vertical bundle.
Notice that in general one can have several $O(r)$-equivariant connections: take $P$ to be the total space constant $U(1)$-bundle on the circle; $P$ is a 2-torus and every rational foliation of $P$ that is non-constant in the "circle" direction gives a $U(1)$-equivariant connection. (All these connections are gauge equivalent but different.)
So I would like to ask: given a Riemannian manifold $M$, is there a way to interpret the Levi-Civita connection as a subbundle of the frame bundle of the tangent bundle of $M$ so that its existence and uniqueness become clear without any calculations in coordinates?

REPLY [5 votes]: I think that a good way to understand the Levi-Civita connection is to say that is is the Ehresmann connection in $TTM$ obtained from the linearization of the geodesic flow by a natural geometric construction.
I described this construction in my answer to this MO question, but I'll do so again with some improvements.
Dynamic construction.
Let $c(t)$ be an orbit of the geodesic flow in $TM$, consider the vertical subspaces $V(t)$ in $TTM$ along $c(t)$ and bring them back to the tangent space of the cotangent bundle over the point $c(0)$ by using the differential of the flow. You get a family of (Lagrangian) subspaces $l(t) := D\phi_{-t}(V(t))$ in the symplectic vector space $T_{c(0)}TM$. 
Now forget you ever had a geodesic flow: all that you need is the curve of subspaces. A bit of differential projective geometry---described below---shows that you also get a second curve $h(t)$ of (Lagrangian subspaces) in $T_{c(0)}(T^*M)$ that is transversal to $l(t)$. The subspace $h(0)$ is the horizontal subspace of the connection and  $T_{c(0)}(T^*M) = l(0) \oplus h(0)$ is the decomposition into vertical and horizontal subspaces.
Projective construction.
Now I'll describe as succintly as possible the projective-geometric construction
that underlies both the Levi-Civita connection and the Schwartzian derivative.
For the detais of what follows see this paper What's new in the description here is that I explicitly use the Springer resolution (Duran and I used implicitly in the paper).
First we need two remarks on the geometry of the Grassmannian $G_n(\mathbb{R}^{2n})$ of $n$-dimensional subspaces in $\mathbb{R}^{2n}$ 
1. The tangent  space  of $G_n(\mathbb{R}^{2n})$ at a subspace $\ell$
is canonically identified with the space of linear maps from $\ell$ to $\mathbb{R}^{2n}/\ell$ or, equivalently, with the space $(\mathbb{R}^{2n}/\ell) \otimes \ell^*$. Since $\mathbb{R}^{2n}/\ell$ and $\ell$ have the same dimension, we may distinguish a class of differentiable curves $\gamma$ on the Grassmannian  by requiring that at each instant $t$ their velocities are invertible linear maps from $\gamma(t)$ to  $\mathbb{R}^{2n}/\gamma(t)$. These curves are called fanning or regular.
Using that the cotangent space of $G_n(\mathbb{R}^{2n})$ at a subspace $\ell$
is canonically isomorphic to $\ell \otimes (\mathbb{R}^{2n}/\ell)^*$, we can lift
every fanning curve $\gamma(t)$ to  a  curve on the cotangent bundle of the Grassmannian by $t \mapsto (\dot{\gamma}(t))^{-1}$. 
2. Consider the action of the linear group $GL(2n;\mathbb{R})$ on the Grassmannian $G_n(\mathbb{R}^{2n})$ and lift it to an action on its cotangent bundle. The moment map of this action takes values on the set of nilpotent matrices. 
Now consider a fanning curve $\gamma(t)$ on the Grassmannian $G_n(\mathbb{R}^{2n})$ and lift it to  the curve $(\dot{\gamma}(t))^{-1}$ on its cotangent bundle. Use the moment map to obtain a curve $F(t)$ of nilpotent matrices. Note that everything we have done is $GL(2n,\mathbb{R})$-equivariant. 
Finally we come to the little miracle: the time derivative of $F(t)$ is a curve of reflections $\dot{F}(t)$ (i.e., $\dot{F}(t)^2 = I$) whose -1 eigenspace is the curve of subspaces $\gamma(t)$ and whose $1$-eigenspace defines a "horizontal curve" $h(t)$ equivariantly attached to $\gamma(t)$. This is the construction that yields the Levi-Civita connection (and what is behind the formalisms of Grifone and Foulon for connections of second order ODE's on manifolds).
Differentiate $F(t)$ a second time to find the Schwartzian derivative. Geometrically, it just describes how the curve $h(t)$ moves with respect to $\gamma(t)$. For comparison, recall that the curvature of a connection is obtained by differentiating (i.e., bracketing) horizonal vector fields and projecting onto the vertical bundle.<|endoftext|>
TITLE: Interpolation on real Riemann surfaces
QUESTION [6 upvotes]: Background:
Generalizing the notion of upper half plane to compact Riemann surfaces:
Suppose $p(x,y) \in \mathbb{R}[x,y]$ is a polynomial in 2 variables with real coefficients, defining a smooth complex plane algebraic curve $C_0 = \{(x,y) \in \mathbb{C}^2:p(x,y)=0\}$. Let $C$ be the projective closure of $C_0$ in $P^2\mathbb{C}$, and assume that $C$ is also smooth. Since $C$ is defined over the real numbers, it comes equipped with an involution $\sigma:C\rightarrow C$, $\sigma(x,y) = (\overline{x},\overline{y})$. Denote by $X$ the compact Riemann surface associated to $C$, and let $X_\mathbb{R}$ be the set of fixed points of $\sigma$.
If the space $X - X_\mathbb{R}$ has exactly two connected components, then $X$ is called a real compact Riemann surface of dividing type, and the two connected components are denoted by $X_+$ and $X_{-}$ (the decision between "the positive half plane" and  "the negative half plane" arbitrarily).
And finally, to the question:
I am given a real compact Riemann surface of dividing type $X$, and interested in interpolation problems of meromorphic functions with conditions such as "all the poles of $f$ lie in the upper half plane". Does anybody knows of any previous work in the area? Any known techniques to relate these topological and algebraic constructions?

REPLY [2 votes]: Let me give a version of the question in the comment:
Let $X$ be a curve of genus $g$ with a real separting involution, and conisder the map 
$Sym^n(X_+)\to Jac^n(X)$.
For wich $n$ is this map surjective? Or, in other words, what is the minimal number of poles of a meromorphic function with poles in $X_+$ that garanties 
that zeros can happen at any collection of points?
This sounlds like a very nice question. In the case $g=1$ you can always take $n=2$.
Also for any $g$ you sould take $n>g$ because $Sym^g(X)$ maps to $Jac^g(X)$ with degree $1$.    
Added. The notation $Sym^n(X)$ means the symmetric power of $X$. Let me explain also why what is above is a reformulation of the original question. Indeed, a divisor $\sum_i x_i-\sum_i y_i$ on $X$ is a divisor of a meromorfic function iff it represent zero in $Jac^0(X)$. So if we want to chose arbitraly zeros $x_i$ of a meromorphic function $f$ keeping the poles $y_i$ in $X_+$ it is enouth to know that $\sum_i y_i$ can take any value in $Jac^n(X)$ (to cancel the point $\sum_i x_i$). This is eactly the condition that $Sym^n(X_+)\to Jac^n(X)$  is surjective.<|endoftext|>
TITLE: Algebraic geometry examples
QUESTION [77 upvotes]: What are some surprising or memorable examples in algebraic geometry, suitable for a course I'll be teaching on chapters 1-2 of Hartshorne (varieties, introductory schemes)? 
I'd prefer examples that are unusual or nonstandard, as I already know many of the standard ones (27 lines on a cubic surface, etc). 
To illustrate the sorts of things I am looking for, here are some examples that would have been useful answers if I had not already thought of them: 
The  proof of the Kakeya conjecture for finite fields.
Free sheaves need not be projective
The Hilbert scheme of m points on an n dimensional variety can have dimension larger than mn.
The scheme of nilpotent matrices is not reduced. 
The 1-line proof of Pascal's theorem from Bezout's theorem.
Spec Z[x]
Resolution of the Whitney umbrella
The related threads
What should be learned in a first serious schemes course? and Interesting results in algebraic geometry accessible to 3rd year undergraduates also have some good examples, such as Grassmannians.
Added later: Thanks for all the examples; I'm embarrassed to admit that some of them are counterexamples to statements I would have guessed were true. Over the next few weeks I will  gradually add answers below (with credit to those who suggested them) to my  draft course notes (These notes still need a lot of corrections/expansion/rewriting; they should have reached a more stable state by Dec 2010)

REPLY [4 votes]: Given in an order from less challenging to do in such a course to more challenging.

Conics (to "connect" with high-school algebraic geometry)
All smooth conics in $\mathbb{P}^2_{k}$ that have a rational point are isomorphic to $\mathbb{P}^1$. In particular, ellipse, hyperbola and parabola are the same; one may also learn how to distinguish them in the affine plane over $\mathbb{R}$.
Steiner's construction of a conic through 5 points, no three of which are collinear is an example of how (synthetic) projective geometry can dispense with algebra! If you are adventurous, you can also give Steiner's construction of the rational normal curve.

Flatness (example learnt from S. Ramanan)
The morphism from $V((x)\cap(x-t))$ in $\mathbb{A}^2$ to $\mathbb{A}^1$ given by $(x,t)\mapsto t$ is flat.
However, the morphism from $V((x,y)\cap(x-t,y-s))$ in $\mathbb{A}^4$ to $\mathbb{A}^2$ given by $(x,y,t,s)\mapsto (s,t)$ is not flat.
This is because the "degenerate doubled point" in the plane must keep track of the "direction" in which it was doubled.
The "real" flat family of "pairs of points" in the plane is parametrized by the blow-up of $\mathbb{A}^2$ at a point.
The insight that "flatness" is an essential condition for extending the notion of families to singular situations was a "game-changer" for algebraic geometry. Flatness is a definition that seems to require algebra and yet is important for geometry.

Tangent bundle.
The tangent bundle of a smooth variety $X$ over $k$ is the scheme of $k[\epsilon]$ points of $X$. (This is an extension of the above example, in a way.)
This example clearly shows how infinitesimals have a meaning in algebraic geometry.

Hirzebruch surfaces (learnt from Beauville's book on surfaces).
Given the rational normal curve $C_k$ in $\mathbb{P}^k$ and the rational normal curve $C_m$ in $\mathbb{P}^m$. Choose an isomorphism $g$ between $C_k$ and $C_m$, and choose disjoint linear embeddings of $\mathbb{P}^k$ and $\mathbb{P}^m$ in $\mathbb{P}^n$ (so $n\geq k+m+1$). Let $J$ be the union of the lines joining $g$-pairs of points in $C_k$ and $C_m$.
Then $J$ is the Hirzebruch surface $F_{|m-k|}$. Moreover, if $m<k$ then $C_m$ is the curve on $J$ with negative self-intersection.
Bonus points for understanding "elementary transformations" $F_{a}\to F_{a\pm 1}$ in terms of projection of this surface from a point on $C_k$ or $C_m$!

Quartic curve.
Redo the analysis of a quartic equation from Chapter 12 (section 95+, page 351+) of Weber's book on "Algebra" in modern language. He demonstrates the existence of 28 bitangents. It is completely "high-school" algebra and geometry argument which does not involve and Abelian varieties and theta characteristics! Essentially he re-write the equation in the form $\sqrt{a_1a_2}+\sqrt{b_1b_2}+\sqrt{c_1c_2}$ where $a_i, b_i, c_i$ are linear forms.
In my humble opinion, this is more challenging and interesting than 27 lines on a cubic surface. In fact, the latter follows from the above analysis by doubling the plane along the quartic curve.<|endoftext|>
TITLE: Sylow's theorem 3rd Proof  Page 96 I.N.Herstein
QUESTION [5 upvotes]: I was just going through the 3rd Proof of Sylow's theorem given in the "Topics In Algebra" Book by I.N. Herstein. It looked very interesting and i really liked its Philosophy. My question what is its significance, and how can it be applied to problems, or something else.

REPLY [14 votes]: This is the proof that uses the lemma that if a finite group $G$ has
a Sylow $p$-subgroup then so does each subgroup of $G$. To complete
the proof of existence of Sylow $p$-subgroups, it suffices to show
one can embed each group in a group with a Sylow $p$-subgroup.
By Cayley's theorem each finite $G$ embeds in $S_n$ with $n=|G|$
and $S_n$ embeds in $S_{p^k}$ where $p^k\ge n$. One then writes
down a Sylow $p$-subgroup of $S_{p^k}$ (essentially an iterated
wreath product of $C_p$s).
But a slicker conclusion is to embed $S_n$ in $GL_n(p)$
(via permutation matrices),
as one sees with little effort that the upper triangular matrices
with $1$s on the diagonal form a Sylow $p$-subgroup of $GL_n(p)$.<|endoftext|>
TITLE: Is Galois theory necessary (in a basic graduate algebra course)?
QUESTION [50 upvotes]: By definition, a basic graduate algebra course in a U.S. (or similar) university with
a Ph.D. program in mathematics lasts part or all of an academic year and is taken 
by first (sometimes second) year graduate students who are usually attracted 
primarily to "pure" mathematics and hope for a career combining some mixture
of teaching, problem-solving, basic research.  This definition already covers a lot of
possibilities, especially if broadened to include those institutions offering only a 
master's degree.   Most users of MO have probably had (or avoided) such
an algebra course along the way, beyond an undergraduate introduction.
There is a long "abstract" or "modern" algebra  tradition  going back to E. Noether 
and B. van der Waerden, but the steady growth of mathematics has added a huge
amount of material to textbooks and has also created too much competition for the
time of beginning graduate students.    In practice many students can and do bypass
algebra at this level.    My own sporadic teaching of algebra took place in three
quite different departments (Oregon, Courant, UMass) with varying research
emphasis on algebraic number theory: the most likely place where mathematicians
will really need a lot of Galois theory.    
Galois theory has an illustrious history and (to quote Lang) "gives very quickly an
impression of depth".   It exposes students to real mathematics, combining the 
study of polynomial rings, fields, and groups in unexpected ways.   But it also
takes quite a bit of time to develop properly, together with supporting material.
And people no longer care as much about solving polynomial equations exactly
as about using sophisticated computational methods to estimate roots.   In real
life the eigenvalues of a big matrix are not estimated by factoring the characteristic
polynomial.
Especially in a first semester algebra course taken by a wider range of students,
I've found it more rewarding to spend time developing the parallel between 
finite abelian groups and finitely generated torsion modules over $F[x]$ (unified
in the theory of finitely generated modules over PIDs).   This is challenging 
material but gets at some of the canonical form theory for operators in the way 
most mathematicians should understand it for theory and applications.  The minimal polynomial comes into its own here.
Even in a second semester course, where tradition at UMass and many other
departments has favored Galois theory, there may be a stronger case to make for 
teaching  basic character theory of finite groups.   This too is a meeting ground
for many subjects and has even broader applicability than Galois theory when 
developed into full scale representation theory.   (For number theorists, there is 
the neat proof that degrees of irreducible characters divide the group order.)
Working in algebraic Lie theory and representation theory, subjects unseen by most
Ph.D. students, I am especially conscious of choices about which subjects students
get exposed to formally.  Algebraic and differential geometry often have their
own standard (but not first year) courses in departments like UMass, but most 
people with a Ph.D. in mathematics get by without even those subjects in their
background.     "What should every mathematician know?" seems more elusive 
than ever.

Is Galois theory necessary (in a basic graduate algebra course)?

POSTSCRIPT: I appreciate the fact that so many people have actually given the whole issue careful thought, since it bothered me all through my own teaching years.   With so little time and so much to learn, choices are inevitable.   And it's always easiest to follow the existing course tradition and textbooks.   My definition above of "basic graduate course" doesn't fit everywhere, to be sure, but U.S. students usually don't learn much mathematics before that level no matter what their potential is.   So the issue won't go away in most U.S. universities that offer advanced work.   (It will also continue to be true that most people with a Ph.D. here in "mathematical sciences" will never encounter rigorous Galois theory in courses or in real life.)

REPLY [3 votes]: Double majoring in physics and math, I was undecided about entering graduate school, and if I did, I was anticipating going further into physics instead of math, probably. My junior year abstract algebra course and ESPECIALLY GALOIS THEORY is the reason I chose to apply to math Ph. D. programs. I'm glad I did. 'Nuf said!<|endoftext|>
TITLE: Simple/efficient representation of Stirling numbers of the first kind
QUESTION [13 upvotes]: Stirling numbers of the second kind can be expressed by means of a simple hypergeometric (considering $n$ fixed) sum
$$S_2(n,k) = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j} j^n. \qquad (1)$$
This can be used for direct calculation of $S_2(n,k)$, without the need to compute any preceding values. But for Stirling numbers of the first kind, one seems to need a nested sum or a recurrence over preceding values, the most direct known representation perhaps being
$$S_1(n,k) = \sum_{j=0}^{n-k} (-1)^j {n+j-1\choose n-k+j} {2n-k \choose n-k-j} S_2(n-k+j,j). \qquad (2)$$
Is there a reason to believe that no formula similar to (1) exists for Stirling numbers of the first kind? Does a formula better than (2)+(1) for calculations exist (assume that I have no interest in generating a table of all preceding values)?

REPLY [4 votes]: $S_1(n,0)=S_1(0,n)= \delta_n \; \;$ and, for $n > 0$, 
$$S_1(n,k)= \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^j \binom{k}{j} \; \frac{(-j \; y)!}{(-j \; y-n)!} \;$$
$$ = \lim_{y \to 0} \;  \frac{y^{-k}}{k!} \; \sum_{j=1}^k (-1)^{n-j} \binom{k}{j} \; \frac{(j \; y - 1 + n)!}{(j \; y-1)!} \; $$
$$= \sum_{j=k}^n \; S_1(n,j)\; (-y)^{j-k}\;S_2(j,k) \; |_{y=0} \; . $$
For a derivation, see A class of differential operators and the Stirling numbers. Note that with $y$ small enough taking the nearest integer generates $S_1$.<|endoftext|>
TITLE: Fast matrix multiplication
QUESTION [36 upvotes]: Suppose we have two $n$ by $n$ matrices over particular ring. We want to multiply them as fast as possible. According to wikipedia there is an algorithm of Coppersmith and Winograd that can do it in $O(n^{2.376})$ time. I have tried to look at the original paper and it scares me. It seems that it is impossible to understand current state of the art.
So, the question is the following. Is there any 'gentle' introduction or survey for beginners in this particular field? I took only introductory course in algebra, so it would be nice to know what parts of algebra do these techniques rely on.

REPLY [3 votes]: A recent article by Kevin Hartnett in Quanta magazine (Matrix Multiplication Inches Closer to Mythic Goal) is targeted at (math-inclined) laypeople and discusses the recent result of Josh Alman and Virginia Vassilevska Williams (A Refined Laser Method and Faster Matrix Multiplication) where they obtain the bound 2.37286.<|endoftext|>
TITLE: Converse to basic result on prime decomposition
QUESTION [8 upvotes]: We know that in a normal extension of number fields $L/K$, for any prime $P$ of $K$ and any primes $Q_1,Q_2$ of $L$ lying over $P$, the ramification indices and inertial degrees are the same,
$$e(Q_1|P)=e(Q_2|P),\quad f(Q_1|P)=f(Q_2|P),$$
and that this is not necessarily the case if $L/K$ is not normal. Is it never the case when $L/K$ is not normal? That is, must there be at least one $P$ and $Q_1,Q_2$ for which $e(Q_1|P)\neq e(Q_2|P)$ or $f(Q_1|P)\neq f(Q_2|P)$?

REPLY [14 votes]: Forget the ramification indices: they're nearly always 1.  Let's focus on the residue field degrees. Let $E$ be the Galois closure of $L$ over $K$. For a prime $P$ in $K$, write $f_P(E/K)$ for the common residue field degree of all primes in $E$ over $P$. Since $E$ is determined (up to isom. as an extension of $K$) by $L$, one can imagine there might be a formula for $f_P(E/K)$ in terms of the $f(Q_i|P)$'s as $Q_i$ runs over the primes in $K$ that lie over $P$.  There is such a formula if $P$ is unramified in $E$:  $f_P(E/K) = \text{lcm } f(Q_i|P)$. (This is basically because a composite of finite fields over a particular finite field has degree equal to the lcm of the degrees of the extensions.)
Now imagine that for all $P$ the numbers $f(Q_i|P)$ are equal.  Then their lcm is that common value, so when $P$ is unramified in $E$ we have $f_P(E/K) = f(Q_i|P)$ for all $Q_i$ over $P$ in $L$. Then $f_{Q_i}(E/L) = 1$ for all $Q_i$, so all but finitely many primes in $L$ are split completely in $E$.  By Chebotarev's density theorem that implies $E = L$, so $L/K$ is Galois. (Strictly speaking we are using a result that is logically much weaker than Chebotarev: that the primes which split completely determine a Galois extension is due to Bauer from 1916 or so, before Chebotarev proved his general theorem.)
Since Chebotarev works with a density 0 set of primes taken out, instead of assuming for all $P$ that the numbers $f(Q_i|P)$ (when $Q_i|P$) are equal you only need to assume that for all but finitely many $P$ or even for all but a density 0 set of $P$.<|endoftext|>
TITLE: Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions?
QUESTION [36 upvotes]: This question is related to this recent but currently
unanswered MO
question
of Ricky Demer, where it arose as a comment.
Consider the structure $R^n$ consisting of $n\times n$
matrices over the reals $\mathbb{R}$, $n$-dimensional row
vectors, column vectors and real scalars, with the ordered
field structure on the scalars. Thus, we can add and
multiply matrices; we can multiply anything by a scalar; we
can multiply matrices by vectors (on the suitable side);
and we can add and multiply vectors of the suitable shape.
The corresponding matrix algebra language has four
variable sorts - scalars, matrices, row vectors and column
vectors - together with the rules for forming terms so that
these expressions make sense in any $R^n$. In this
language, you can quantify over matrices, vectors and
scalars, form equations (and inequalities with the
scalars), but you cannot quantify over the dimension. The
idea is that an assertion in this language can be
interpreted in any dimension, one $R^n$ at a time. You have to make assertions that do not refer to the dimension; the
language is making assertions about matrices and vectors in
some fixed but unspecified dimension.
My question is whether truth in this real matrix algebra
obeys a 0-1 law as the dimension increases, that is:

Question. Is every statement in the matrix algebra
  language either eventually true or eventually false in
  $R^n$ for all sufficiently large dimensions $n$?

To give some trivial examples:

the statement asserting matrix commutativity
$\forall A,B\, AB=BA$ is true in dimension $1$ but false in all
higher dimensions.
the statement that the dimension is at least 17, or at
most 25, or an odd number less than 1000, are all
expressible, since you can quantify over enough vectors
and the assertions that they are independent or that they
span are expressible. The truth values of these
statements all stabilize in sufficiently high dimension.
the assertion that a particular real number is an
eigenvalue for a matrix is expressible.

But it isn't clear to me how one could express, for
example, that the dimension is even. (Edit: Gerry and Ryan below have explained how this is easily done.)
In the previous question, Ricky inquired
whether there is a decision procedure to determine which
assertions of matrix algebra are true for all $n$. For any
particular $n$, then Tarski's theorem on the decidability
of real-closed fields shows that the theory of the
structure $R^n$ is decidable: when $n$ is fixed, we may
translate any statement about matrices and vectors into
statements about real numbers by talking about the
components. (We may also add to the language the functions
that map a matrix or vector to the value of any particular
entry, as well as $det(A)$ etc.)
If my question here has a positive answer, and the
stabilizing bound is computable from the formula, then this
would provide an affirmative answer to Ricky's question,
since we could just determine truth in a large enough
$R^n$.
Lastly, I don't think it will fundamentally change the
problem to work in the complex field, since the
corresponding structure $C^n$ with complex matrices and
vectors is interpretable in $R^n$. For example, I think we
could freely refer to complex eigenvalues.

Edit. The real case was quickly dispatched by Gerry and Ryan, below. Let us therefore consider the complex case. So we have for each dimension $n$ the structure $C^n$ with $n\times n$ matrices, row vectors, column vectors and complex scalars. The question is: Does the truth of every statement of matrix algebra stabilize in $C^n$ for sufficiently large $n$?
Ricky proposed that we add Hermitian transpose (conjugation on scalars) to the language. This would would also allow us to refer to the real scalars. If we expand the language so that we are able to define the class of real matrices and vectors, however, then we can still express Gerry's and Ryan's solutions for a negative answer here.

Edit 2. As in the comments, let us say that the truth set of a formula $\phi$ in the language is the set of $n$ for which $\phi$ is true in dimension $n$. These truth sets form a Boolean algebra, closed under finite differences. Which sets of natural numbers are realizable as truth sets? (Note that there are only countably many truth sets.) And how does it depend on the field?

REPLY [2 votes]: What about the following easy-looking statement: 

for every $n\times n$ matrix, one has $\det(A^T-A)=0$, 

which is true if $n$ is odd, but false if $n$ is even ?<|endoftext|>
TITLE: $n$th root of $(a,b) \mapsto (\operatorname{gm}, \operatorname{am})$
QUESTION [7 upvotes]: Suppose $0 < a < b$, and let GM and AM be respectively the geometric and arithmetic means of $a$ and $b$.  Does the mapping $(a,b) \mapsto (\operatorname{GM}, \operatorname{AM})$ have a well-behaved compositional $n$th root?  In what senses might such an $n$th root be unique?

REPLY [2 votes]: A simple scheme (a parallel of which can be used to define the Gamma function from the factorial) is as follows: call $T$ your function, then define the functional square root as
$$T^{\frac{1}{2}}=\lim_{n\rightarrow\infty}T^{n+1}ST^{-n}$$
where $S$ is the sensible approximation of $T^{-\frac{1}{2}}$ for $b\gg a$, given by
$$S(a,b)=\Big(\frac{a^\sqrt{2}}{2^{2\sqrt{2}-5/2}b^{\sqrt{2}-1}}, \sqrt{2}b\Big)$$
Simple numerical testing shows $T^{n+1}ST^{-n}$ becoming rapidly more accurate (as $n$ grows) than $TS$ as a functional square root of $T$, and the limit above most likely converges to the only reasonable definition of $T^{\frac{1}{2}}$.
A (tedious) generalization to all fractional functional powers should be straightforward.<|endoftext|>
TITLE: Are all primes in a PAP-3?
QUESTION [22 upvotes]: Van der Corput [1] proved that there are infinitely many arithmetic progressions of primes of length 3 (PAP-3).  (Green & Tao [2] famously extended this theorem to length $k$.)
But taking this in a different direction, are all odd primes in a PAP-3?  That is, for every prime $p>2$, is there a $k$ such that $p+k$ and $p+2k$ are prime?
Unsurprisingly, the first 100,000 primes have this property; the largest value of $k$ needed is just 1584 (see [4] and also [5] where this is greatly extended).  Heuristically, you'd expect a given prime to be in
$$\int_2^\infty\frac{a\ dx}{\log(x\log x)}=+\infty$$
different PAP-3s, and there are no small prime obstructions, so the conclusion seems reasonable.  On the other hand, it seems to involve Goldbach-like (or better, Sophie Germaine-like) additive patterns in the primes: in essence, we're looking for prime $q$, $2q-n$ for a fixed odd $n$, so I don't imagine this has been resolved.
Basically, I'm just looking for more information on this problem.  Surely it's been posed before, but does it have a common name and/or citation?  Have any partial results been proved?  Perhaps this is a consequence of a well-known conjecture?
[1] A. G. van der Corput (1939). "Über Summen von Primzahlen und Primzahlquadraten", Mathematische Annalen 116, pp. 1-50.
[2] Ben Green and Terence Tao (2008). "The primes contain arbitrarily long arithmetic progressions", Annals of Mathematics 167, pp. 481–547. http://arxiv.org/abs/math/0404188
[3] Amarnath Murthy, http://oeis.org/A084704
[4] Giovanni Teofilatto, http://oeis.org/A120627
[5] Charles R Greathouse IV, https://oeis.org/A190423

REPLY [25 votes]: This question is extremely close to this one
Covering the primes by 3-term APs ?
though not exactly the same.
For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.
To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.
For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.
I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.<|endoftext|>
TITLE: Bijective proof of weak form of Stirling's approximation
QUESTION [6 upvotes]: There are short and sweet proofs of various forms of Stirling's approximation. But even the sweetest among them don't instill the same conviction in the reader as a direct bijective proof.
Computer scientists can often get away with a very weak form of Stirling's approximation:
$$(n/2)^{n/2} \leq n! \leq n^n$$
From this follows
$$(n/2)\ log(n) - (n/2)\ log(2) \leq log(n!) \leq n\ log(n)$$
and therefore $log(n!) = \Theta(n\ log(n))$. This is sufficient to establish the lower bound on comparison-based sorting algorithms and many other asymptotic bounds.

Does anyone know a natural bijective proof of $(n/2)^{n/2} \leq n! \leq n^n$?

REPLY [6 votes]: Think of all maps from the first $n/2$ elements of {$1,...,n$} to the last $n/2$.  Say, let $a_1< \ldots < a_k \to z$.  Make a cycle $a_1 \to a_2 \to \ldots \to a_k \to z \to a_1$.  Do this for all $z$.  The details are straightforward.  This proves the lower bound.<|endoftext|>
TITLE: How do professional mathematicians learn new things?
QUESTION [24 upvotes]: How do professional mathematicians learn new things? How do they expand their comfort zone? By talking to colleagues?

REPLY [12 votes]: Teaching a course in something is the only way that I can really learn something new.<|endoftext|>
TITLE: Does every nontrivial sheaf of rings have a maximal ideal?
QUESTION [7 upvotes]: Let $R$ be a sheaf of rings on a topological space $X$. Assume $R \neq 0$. Does then $R$ have a maximal ideal? So this is a spacified analogon of the theorem, that every nontrivial ring has a maximal ideal. Currently I try to develope this sort of spacified commutative algebra and algebraic geometry. If anyone knows some literature about it, please let me know.
So let's try to imitate the known proof for rings and use Zorn's Lemma. For that, we need that for every linear ordered set $(J_k)_{k \in K}$ of proper ideals in $R$, their sum $\sum_{k \in K} J_k$ is also a proper ideal. Note that if we replace $R \neq 0$ by $R_x \neq 0$ for all $x \in X$ and the notion proper by "stalkwise proper", then everything works out fine since stalks and sum commute. However, global sections do not commute with (infinite) sums. Anyway, let's try to continue:
Assume $\sum_{k \in K} J_k = R$, that is, $1$ is a global section of the sum. Then there is an open covering $X = \cup_{i \in I} U_i$, such that $1 \in \sum_{k \in K} J_k(U_i) = \cup_{k \in K} J_k(U_i)$. Thus we get a function $I \to K, i \mapsto k_i$, such that $1 \in J_{k_i}(U_i)$. If this function has an upper bound, say $k$, then we get a contradiction $J_k=R$. Thus the function is unbounded. And now?
I think that this already indicates that there will be counterexamples, but I'm not sure. Also note that everything is fine when $X$ is quasi compact.

REPLY [8 votes]: One could have guessed that the answer is "no" through the following reasoning:
A sheaf of rings on a space $X$ is a ring object in the topos of sheaves on $X$, and your question is about ring theory in that topos. But such a topos (unless $X$ is very close to being discrete, see e.g. comments here) can be seen as an intuitionistic universe of sets, where the axiom of choice and Zorn's lemma are not valid.
This can be made precise; you can interpret formal languages in a topos and you have certain proof systems obeying intuitionistic rules which are sound and complete with respect to the topos interpretation. The soundness part tells you that whenever you manage to prove a theorem about rings according to the intuitionistic rules it will also be valid for sheaves of rings. Here you have to make a distinction between first order and higher order languages - both are interpretable in toposes. Your question about ideals is a higher order question since it talks about subsets of a ring. Anyway both justify the point of view that you are doing ring theory in an intuitionistic set universe.
A nice and very friendly written example of such reasoning is this article of Mulvey, "Intuitionistic Algebra and Representations of rings", in which he gives an intuitionistic proof of a theorem of Kaplansky to conclude that it holds for sheaves of rings (and then draws nice consequences). If you are interested in this way of reasoning about sheaves of rings it is definitely the article for you - and you don't have to read Johnstone first, he does it all from scratch and talks about both, 1st and higher order, in a colloquial way.
The clearest written precise accounts of this are IMHO section D1 of Johnstone's Elephant for 1st order and D4 for higher order, but they are a bit more general than what you need, see the last paragraph, the treatment of Mulvey, is closer to your setting. 
A beautiful feature of the 1st order version (e.g. Johnstone, section D3) is that you have a universal ring object living in a certain topos (the "classifying topos" for rings) which satisfies exactly those 1st order statements which are provable in any topos. So whenever you can prove something about this particular ring object you can be sure it holds for all sheaves of rings on a space; see MacLane/Moerdijk's Sheaves in geometry and Logic, section VIII.5 for this universal ring object.
The existence of a classifying topos also leads to the following excellent news: For formulas of a certain syntactic form, called "geometric formulas", there are the theorems of Deligne and Barr (see Johnstone) which tell you that whenever you can prove such a formula using classical reasoning there also exists an intuitionistic proof - so you don't have to bother restricting your logic in these cases!
Since you said that you are also interested in "spacified" algebraic geometry: There is an article by Anders Kock, "Universal Projective Geometry via Topos Theory", J. Pure Appl. Algebra 9 (1976), 1-24, where he does projective geometry over this universal ring, presumably again obtaining statements which hold over any sheaf of rings (but I only skimmed that one long ago).
To sum up: If you can prove something intuitionistically (in a non-defined loose sense) you have good chances to be able to translate your proof into the formal systems from above and then you know that your result is true about rings in any topos. If you manage to express a 1st order statement in a certain syntactic form (as a "geometric formula") and can prove it with classical logic you can also conclude that it holds for any ring in a topos. 
Important note: "Ring in a topos" is a more general class of objects than you seem to be interested in, you want to know about "rings in a spatial topos" (i.e. topos of sheaves on a topological space), and there more will be true. So even if a statement does e.g. not hold for the universal ring (which lives in a non-spatial topos) it might still be true for all rings in spatial toposes.<|endoftext|>
TITLE: Compact cover of a Hausdorff compact space
QUESTION [6 upvotes]: In his book "Riemannian Geometry" do Carmo cites the Hopf-Rinow theorem in chapter 7. (theorem 2.8). One of the equivalences there deals with the cover of the manifold using nested sequence of compact subsets. This made me wonder whether the following lemma holds:
Lemma: Let $M$ be a compact Hausdorff space, and let $K_i \subset M$ be a sequence of compact subsets such that: $K_i\subset K_{i+1}$ and $\cup_{i=1}^{\infty} K_i = M$. Then there exists an index $i_0$ such that $K_i = M$ for all $i \geq i_0$.
Here is my proof to this:
Proof: Assume that $K_i \neq M$ for all $i$, that is all $K_i$'s are proper subsets of $M$. With out loss of generality we can then assume that $K_i\subsetneq K_{i+1}$. This implies that $\forall i$ there exists $x_i$ such that $x_i\notin K_i$ but $x_i\in K_{i+1}$. Since $K_i$ is compact subset of a Hausdorff space, there exist open $U_i$ and $V_i$, such that $K_i \subset U_i$, $x_i\in V_i$ and $U_i \cap V_i = \emptyset$.
Now, if $x\in \cup_{i=1}^{\infty} U_i$ then clearly $x\in M$ since $U_i \subset M$. On the otehr hand, if $x\in M$, then $x\in K_{i_0}$ for some $i_0$, and thus it is also in $U_{i_0}$. This yields that $M= \cup_{i=0}^{\infty}U_i$. 
Let us now assume that $\cup_{j=1}^n U_{i_j}=M$ is a finite cover. If $n_0 = \max_{j=1,\ldots,n}\{i_j\}$ then $x_{n_0+1}\notin \cup_{j=1}^nU_{i_j}$. This in turn means, that we cannot find a finite sub-cover of $M$ using the open cover $\cup_{i=1}^{\infty}U_i$. But this contradicts the compactness of $M$. This completes the proof. $\square$
Finally, here's my question. Is this lemma correct? Is my proof correct?
Thanks in advance and all the best!
Dror, Edit: As I verified with the author, he meant that the last equivalence is valid when the manifold is not compact. Thus, my false lemma, is irrelevant from the first place.

REPLY [12 votes]: Note: whilst typing this, Martin posted his answer.  As I come to a completely different conclusion, I'd be very interested in knowing who's right!

False.  Let $M = [0,1]$ and $K_i = \{0\} \cup [\frac{1}{i},1]$.
The flaw in the proof is the assumption that the $U_j$ are increasing; ie that $U_j \subseteq U_{j+1}$.  Thus the sentence "If $n_0=\max_{j=1, \dots, n} i_j$ then $x_{n_0+1} \notin \bigcup_{j=1}^n U_{i_j}$." is not true.

REPLY [5 votes]: [I edited this after Andrew's comment!]
$M$ has to carry the colimit topology for the lemma to become true. Besides it is well-known and frequently used in, say, homotopy theory. Indeed, it implies for example, that homotopy groups commute with nice filtrations.
Here is a generalization to non-hausdorff spaces (which was useful in my research on algebraic geometry):
Let $X_1 \subseteq X_2 \subseteq X_3 \subseteq ...$ a sequence of closed subspaces of a topological space $X$ and let $X_\infty$ their colimit (i.e. the union with the colimit topology). Assume that $X_\infty$ is quasicompact and that for every $i < j$, such that $X_i$ is a proper subset of $X_j$, $X_j \setminus X_i$ contains a closed point of $X_j$. Then there is some $i$ such that $X_\infty = X_i$.
Proof: Assume $X_i \neq X_\infty$ for all $i$. Then you can construct $i_0 < i_1 < ...$ and $x_{i_k} \in X_{i_k} \setminus X_{i_{k+1}}$, closed in $x_{i_k}$ and thus in $X_\infty$. Now consider $D = \{a_{i_k} : k \in \mathbb{N}\}$. Then $D \cap X_i$ is a finite set of closed points of $X_\infty$ resp. $X_i$, thus closed. Therefore $D \subseteq X_\infty$ is closed. In the same way, we see that $D \setminus \{d\}$ is closed in $X_\infty$ for every $d \in D$. But then it is also closed in $D$. Hence $D$ is discrete. Since $D$ is, as a closed subset of $X_\infty$, quasicompact, we get that $D$ is finite, which is a contradiction.
EDIT: I wonder if Andrew's counterexample is so popular compared to the correct statement, which is, as I said, used frequently. :-)<|endoftext|>
TITLE: Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$
QUESTION [7 upvotes]: Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be an injection for $n>m$. Can $f$ be continuous? Why?
I got this question in mind when I was trying to find a continuous map from $\mathbb{R}^{2}$ to $\mathbb{R}$.

REPLY [2 votes]: I taught an introductory topology course this last autumn where I covered this theorem from an elementary point of view. The argument just uses the Brouwer fixed point theorem (which itself has a proof via Stokes' theorem which is readily accessible to students with several variable calculus) plus elementary point set topology. In particular no homology or Jordan--Brouwer separation theorems are used. The treatment was based upon that of Hurewicz & Wallman but was also inspired by Larry Guth's ICM-2010 presentation in Hyderabad.
If you're interested the notes are available on 
http://www.maths.ed.ac.uk/uploads/assets/32_section6.pdf<|endoftext|>
TITLE: When are two symplectic forms "isotopic"?
QUESTION [28 upvotes]: I've been skulking around MathOverflow for about a month, reading questions and answers and comments, and I guess it's about time I asked a question myself, so here is one has interested me for a long time.
Suppose $M$ is a compact even dimensional smooth manifold with two symplectic forms $\omega_0$ and $\omega_1$ When are they "isotopic", i.e., when does there exist a 1-parameter family of diffeos $\phi_t$ of $M$, starting from the identity, such that $\phi_1^*(\omega_0) = \omega_1$? Of course a necessary condition is that $\omega_0$ and $\omega_1$ should define the same 2-dimensional cohomology classes. Is this also sufficient? One can ask the same question for volume forms.  I asked Juergen Moser about this twenty-five years ago, and he came back with an elegant proof of sufficiency for the volume element case a few months later in a well-known paper in TAMS. He remarks in that paper as follows:
"The statement concerning 2-forms was also suggested by R. Palais. Unfortunately 
it seems very difficult to decide when two 2-forms which are closed, 
belong to the same cohomology class and are nondegenerate can be deformed 
homotopically into each other within the class of these differential forms."
So my question is, what if any progress has been made on this question. Poking around here and in Google hasn't turned up anything. Does anyone know if there are any progress?

REPLY [2 votes]: Just a reformulation of Dick's question: How to describe the orbits of the identity component of the group of diffeomorphisms of a compact manifold, acting naturally on the subspace of cohomology classes of its symplectic forms?<|endoftext|>
TITLE: Are computable models sufficient?
QUESTION [7 upvotes]: What I mean is this. By downward Lowenheim-Skolem theorem, first-order formula Q is a always true iff it is true in every countable structure. But is there some first-order formula Q which is true in every computable structure and false in some non-computable structure? My feeling is that of course the answer should be "yes", but I can't construct an example. I feel also that the questions of the sort has been widely studied. (For example, maybe some study of conditions under which first-order theory has computable model or hasn't). Do you know anything about that?
Thanks in advance.

REPLY [5 votes]: Carl's answer is correct. 
There is also a to more direct way to achieve the same thing. The Tennenbaum's theorem holds for much weaker theories, e.g. $I\Delta_0$ (even for far weaker theories like $IOpen$ plus some number theoretic principles, but not for $IOpen$, a result due to Shepherdson). 
$I\Delta_0+exp$ is finitely axiomatizable, see Haim Gaifman and Constantine Dimitracopoulos, "Fragments of Peano's Arithmetic and the MRDP Theorem". It is also a sub-theory of $PA$.
For more on Tennenbaum's theorem and weak arithmetics, have a look at this paper: 
Shahram Mohsenipour, "Hierarchies of Subsystems of Weak Arithmetic", to appear 
in "Set theory, Arithmetic, Philosophy: Essays in Memory of Stanley Tennenbaum" (edited by J. Kennedy and R. Kossak), Cambridge University Press.<|endoftext|>
TITLE: Laplacian operator and relation to the Laplace Transform
QUESTION [5 upvotes]: I'm trying to understand why the Laplacian operator is used in blob detection in image analysis. I must admit that in trying to figure out why the Laplacian is useful in this application, I've really confused myself with the different uses of the word 'Laplace.' For instance, Wikipedia has many articles on this, and the ones I'm having trouble unifying conceptually are the Laplace Transform and the Laplace Operator.
From co-workers and some reading on the internet, I have come to very shallowly think of my Laplacian convolutions on images as performing something similar to the second derivative, where the most quickly changing areas on the image are what become highlighted in the new, convolved image. From the page on the Laplace Operator this makes a lot of sense. This doesn't make sense to me from the page on the Laplace Transform. My question then, I think, is how are the Laplace Operator and the Laplace Transform related? If I can see, from the definition, that the Laplace Operator is basically doing the second derivative, I would think I should be able to see something similar from the Laplace Transform. But I don't. Am I mistaken in thinking that the Laplace Transform and the Laplace operator are the same thing? How are they related?

REPLY [5 votes]: They are certainly not the same thing.
You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.
Here's some intuition you might find helpful.
The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simple:
In an area of constant color the Laplacian is zero. Indeed, even if you have linear variation it remains zero, e.g. in the neighborhood [1 2 3] the Laplacian's value at the center point is
$$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0.$$
But quadratic and higher-order variation excites the Laplacian and results in non-zero values. Thus it's especially useful for detecting 'jumps'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to pre-filter out any small-scale features or noise that might cause spurious edges to be detected.
I should mention that the Laplacian in two dimensions and higher is significantly richer than the one-dimensional case might suggest. For one, not all two-dimensional images with a uniformly zero Laplacian are linear. But qualitatively a lot of the same intuition holds true as to how the Laplacian reacts to variation.

REPLY [2 votes]: There is no relation on the basic level between the Laplace operator and Laplace transform. From the point of view of learning about them, put the coincidence of names out of your mind.<|endoftext|>
TITLE: Eigenvalues of A+B where A is symmetric positive definite and B is diagonal
QUESTION [8 upvotes]: If I have a symmetric positive definite matrix A and a diagonal matrix B, and I know the eigenvalues of both A and B (by iterative numerical computation in A's case and trivially for B), is there any way I can rapidly find the eigenvalues of the matrix M=A+B?
(I would be surprised if it helps, but I actually have the stronger condition that A is Laplacian.
Unfortunately, the entries of the matrix B are large, and so B cannot be considered a small "perturbation" to A. Finally, I only really have the extremal eigenvalues of A, though I am only hoping to find the extremal eigenvalues of A+B.)

REPLY [8 votes]: The ambiguity in your question is the word 'rapidly'.
If you want to have an information on the eigenvalues of $A+B$, without any extra information besides those given in the question, then this is the problem raised By H. Weyl in 1912. The answer was conjectured in 1962 by A. Horn, and this conjecture was proved by A. Knutson and T. Tao in 1999. It is one of the works for which Tao received a Fields medal. So, the answer is that the spectrum may be any vector in a polytope in ${\mathbb R}^n$ whose definition is given recursively in terms of the size $n$ of the matrix. A nice expository paper is R. Bhatia, Linear algebra to quantum cohomology: the story of A. Horn’s inequalities. Amer. Math. Monthly, 108 (2001), pp 289–318.
Historically, the interest in this question came from Quantum Mechanics.<|endoftext|>
TITLE: Examples of badly behaved derivatives 
QUESTION [10 upvotes]: Consider a real valued function g on an open interval $(a,b)$ which is the derivative of a function continuous on $[a,b]$ at each point of $(a,b)$. The function $g$ has the intermediate value property, so a monotone $g$ will have to be continuous, a general $g$ cannot have simple discontinuities, etc. With such constraints how badly can a derivative behave in terms of continuity, can it get much worse than the derivative of say, $x^2 \sin(1/x)
$?

REPLY [5 votes]: In Y. Katznelson and K. Stromberg’s paper “Everywhere differentiable, nowhere
monotone, functions,” Amer. Math. Monthly 81 (1974), no. 4, 349–354 (doi: 10.1080/00029890.1974.11993558, JSTOR), there is a construction, based on somewhat similar ideas to those described in Majer’s above mentioned article, whereby for any two disjoint countable sets $A, B \subset \mathbb{R}$ there exists a differentiable function whose derivative equals $1$ on $A$ and is less than $1$ on $B$.   So if $A$ and $B$ are both dense in $\mathbb{R}$ then this derivative must be discontinuous at every point of $B$.<|endoftext|>
TITLE: CM rational points on modular curves
QUESTION [9 upvotes]: Dear MO Community, 
I am trying to understand Mazur's 1976 notes "Rational points on Modular Curves" (which can be found in Springer Lecture Notes in Mathematics 601).
Let N be a prime number, and let $X_0(N)$ be the usual modular curve over $\mathbb{Q}$. Say that a point on it is 'CM' if the elliptic curve corresponding to the point has Complex Multiplication. I am interested in the following sorts of questions (which are suggested by what happens over $\mathbb{Q}$):

Let $K$ be a number field that is not $\mathbb{Q}$. For which N can I construct CM points on $X_0(N)(K)$ that are not defined over $\mathbb{Q}$? Are there finitely many such N?
For such N, are there finitely many such CM points?

For example, let $K = \mathbb{Q}(\sqrt{-6})$, and let $R = O_K = \mathbb{Z}[\sqrt{-6}]$. Set $E = \mathbb{C}/R$. Choose $N$ such that $R/NR$ has nontrivial radical, or equivalently such that $R/NR = \mathbb{F}_N[\epsilon]$ for $\epsilon$ some nontrivial element in the radical. Then (E,ker $\epsilon$) determines a point $a_E(N)$ on $X_0(N)$ which "is defined over a subfield of index 2 in the ray class field of $R \otimes \mathbb{Q}$, with conductor equal to the conductor of $R$ (which in my example is 1). Is $a_E(N)$ defined over $K$? What is this index-2 subfield to which Mazur refers?
I think my second question is equivalent to asking: For such N, are there finitely many imaginary quadratic orders $R$ such that the aforementioned index 2 subfield is K, and such that R/NR has nontrivial radical?

REPLY [6 votes]: Dear Barinder,
This is the subject of the arithmetic theory of complex multiplication.  Given any CM 
elliptic curve $E$, and any kind of level structure on $E$, this theory determines the precise
field of definition of $E$ with its given level structure.   You can find a discussion of this
in Silverman II ("Advanced topics") and in Shimura's book.
In your example: the $j$-invariant of $\mathbb C/R$ lives in the totally real subfield of
the Hilbert class field of $K$ (which is $\mathbb Q(\sqrt{2})$, and is also the ray class field of conductor 1 that you mention).  The $\Gamma_0(N)$ level structure ker $\epsilon$
is defined over this same field.  So this is the field of definition of $a_E(N)$.<|endoftext|>
TITLE: What are some interesting corollaries of the classification of finite simple groups?
QUESTION [39 upvotes]: The classification of finite simple groups, whether it be viewed as finished, or as a work in progress, is (or will be) without doubt an enormous achievement. It clearly sheds a great deal of light on the structure of finite groups. However, as with the classification of simple Lie algebras, one might expect this to have a significant impact outside of the immediate subject.  So what are some of the known, or expected, applications to the classification outside of finite group theory?

NB: 

this question was significantly edited by other users soon after being posted (Aug 2 '10). The few critical comments below were actually addressed before the question was edited and improved.

To anyone who has doubts about the proof, please see Aschbacher's Notices paper "The status of the classification of the finite simple groups"
and please rather don't debate that particular matter here.

REPLY [5 votes]: I have a few comments, but will make this an answer due to length, and I just got a brainstorm of a real answer anyhow. I've now rewritten this to more answer the question: how does the Classification attach itself to other mathematical areas.

Group actions are quite common in mathematics. Showing that only finitely many types of group actions occur in a problem is a typical idea. The theorem of Fried involving indecomposable polynomials fits into this, as there is an action on branched covers.

There are a number of corollaries of CFSG, which are essentially classifications in their own right. The above Fried result depends on a type of doubly transitive action being classified. Another example, rooted in Dunfield/Thurston (Finite covers of random 3-manifolds, page 45), they note that for the orbit in question in their application, a result of Gilman (Finite quotients of the automorphism group of a free group) suffices (they use CFSG to assert that a finite 6-transitive group action contains $A_n$). For some of these, asking whether all of CFSG is needed could be apropos.

Another type of corollary is that some bounds are lowered, due to the fact that we now know (for instance) that all groups have (say) a representation satisfying a certain bound. The existence of a qualitatively different bound under CFSG (say polynomial opposed to exponential) has more interest than just making numbers smaller. Looking at the Babai and Codenotti paper Isomorphism of hypergraphs of low rank in moderately exponential time for graph isomorphism, they even note (Theorem 3.1) that a easier weaker result suffices.

The work of Aschbacher, followed by the book of Kleidman and Liebeck (The subgroup structure of the finite classical groups), on maximal subgroups of finite classical groups is another source. Here $\operatorname{SL}$, $\operatorname{SO}$, $\operatorname{Sp}$ and $\operatorname{SU}$ are all involved. Aschbacher's theorem (King has a survey The subgroup structure of finite classical groups in terms of geometric configurations) says that there are 8 types of subgroups (stablizers of: subspaces, direct sums, spreads, forms, extension fields, tensor products, subfields; extraspecial normalizers, plus the exotic ninth class). Another survey (precursor to their book) is by Kleidman and Liebeck (A survey of the maximal subgroups of the finite simple groups. Geometries and groups). Once the degree is above 14, I think, the 8 classes become uniform in description (though the exotics persist). This relates directly to group theory of course, but many math branches use these constructs.


My specific example was a paper of Bachoc and Nebe (Extremal lattices of minimum $8$ related to the Mathieu group $M_{22}$) that showed that an 80-dimensional lattice with a large minimal norm (of 8) had a known automorphism group (related to $M_{22}$) that was maximal finite in $\operatorname{GL}_{80}(Z)$. They then used to show that their lattice was not isometric to a different one they constructed. More generally, if there is no possible common finite supergroup of the known automorphisms of two lattices, they are not isometric. To prove this can require CFSG in one form or another.
I agree with what was stated in a comment by Jim Humphries above: "But probably the more interesting question is where the classification has impact on mathematics outside group theory."<|endoftext|>
TITLE: Mathematical software for computing in integral group rings of discrete groups?
QUESTION [9 upvotes]: I'm doing computations in the integral group ring of a discrete group,
in particular the discrete Heisenberg group. In this case elements
are integral combinations of monomials $x^k y^m z^n$, where the
generators $x$, $y$, and $z$ satisfy $xz=zx$, $yz=zy$, and $yx=xyz$.
Is there mathematical software for doing calculations with such 
objects, for example to compute powers of an element? 
I've not been able to do this inside standard mathematical software like
Mathematica.

REPLY [13 votes]: You can do this with GAP. The example below assumes that you have the polycyclic package installed.
First, you tell GAP which group you want to work with. Luckily, Heisenberg groups are polycyclic, and the polycyclic package provides a command to obtain them:
gap> G:=HeisenbergPcpGroup(1);
Pcp-group with orders [ 0, 0, 0 ]

Note that we could have also defined it by some other means if polycyclic was not available (e.g. as a matrix group), but this way is the most convenient. Now let's form the integral group ring:
gap> ZG:=GroupRing(Integers,G);
<free left module over Integers, and ring-with-one, with 6 generators>

The extra three generators come from the inverses of x, y and z (note that internally it calls them g1,g2,g3; it would be possible to change that with some effort, but that's beyond the scope here). Let's assign the corresponding generators of the group ring to variables x, y, z, and verify the relations you have given:
gap> x:=ZG.1;; y:=ZG.2;; z:=ZG.3;;
gap> x*z=z*x and y*z=z*y and y*x=x*y*z;
true

Here is an example of powering a group element (this works with more complicated ones, too, but I picked a small one to keep the output readable).
gap> (x+7*y)^2;
(1)*g1^2+(7)*g1*g2*g3+(7)*g1*g2+(49)*g2^2

I hope this helps.<|endoftext|>
TITLE: When is an algebra of commuting matrices (contained in one) generated by a single matrix?
QUESTION [14 upvotes]: Let C be an nxn  matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first question (adding several conditions) is: Suppose A and B are symmetric integer matrices with AB=BA, must there be a matrix C such that both A and B are polynomials in C?  This would imply that the algebra which A and B generate has dimension n or less. In general: Let A be an algebra of pairwise commuting nxn matrices over an integral domain D, Under what conditions (if any) on D and A can we conclude that there is a matrix C so that A is contained in the D-algebra generated by C? I am thinking of conditions such as D is an algebraically closed field, D has characteristic zero, the matrices are symmetric. etc.

REPLY [5 votes]: Thanks for the answers. Just to wrap up a bit, here are a few examples.

Sometimes an ACM (algebra of commuting matrices) is sure to be generated by one of its members

Other times it has dimension too large to possibly be (embedded in) an ACM with a single generator.

An ACM might be generated by $2$ matrices, not generated by any of its members, but embed in a larger ACM which does have a single generator.

An ACM might be generated by $2$ matrices, not generated by any of its members, but not embed in a larger ACM which does have a single generator (even in the $3\times 3$ case).

If the matrices are all normal then they can be simultaneously diagonalized. This reduces the problem to an algebra of diagonal matrices, which is easy to understand. Such an algebra is actually generated by one of its members.

The $5$ dimensional algebra ${\cal{A}}_5$ mentioned by Mariano ($4\times 4$ matrices with $2 \times 2$ blocks $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ has dimension too large to be generated by a single matrix. Furthermore, each member M  generates only the $2$ dimensional algebra of matrices $jI+kM$ so no subalgebra of dimension $3$ or $4$ has a single generator.

Consider the subalgebra ${\cal{A}}_3$ of  ${\cal{A}}_5$ generated by $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&B\\0&0\end{smallmatrix}\right)$  with  $A$ and $B$   both $2 \times 2$ invertible matrices (neither a scalar multiple of the other). As mentioned, we can't embed ${\cal{A}}_3$ in a singly generated $4$ dimensional subalgebra of ${\cal{A}}_5$
However it also embeds in other $4$ dimensional algebras.


For example a $4 \times 4$ matrix $\left(\begin{smallmatrix}BA^{-1}&C\\0&A^{-1} B\end{smallmatrix}\right)$ will generate an ACM which commutes with everything in ${\cal{A}}_3$. I guess in this case it would automatically contain ${\cal{A}}_3$. I certainly verified that randomly filling in the C does this in several cases. In many cases I tested one can get away with one or both of $A$ and $B$ having rank $1$... but not always. The two $4 \times 4$ matrices made from matrices with $A=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ give an example of that. One can shrink this to $3 \times 3$ (in the case that the underlying ring is $\mathbb{Z}_2$ as noted by Martin and missed by me), so I will:

The two $3 \times 3$  matrices $\left(\begin{smallmatrix}0&1&0\\0&0&0\\0&0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0&1\\0&0&0\\0&0&0\end{smallmatrix}\right)$ generate an algebra A with eight members $\left(\begin{smallmatrix}a&b&c\\0&a&0\\0&0&a\end{smallmatrix}\right)$.
A is maximal but not generated by any of its members as each member generates a $2$ dimensional (or smaller) subalgebra.<|endoftext|>
TITLE: The only great book that Bourbaki ever wrote?
QUESTION [18 upvotes]: OK, the title is opinionated and contentious, but I have a definite
question. I know that the title refers to the Bourbaki volume
Groupes et Algèbres de Lie (Chapters 4-6), published in 1968, but

who said that it is the only great book that Bourbaki ever wrote?

The only reference I can find is the 2009 Prize Booklet for the AMS-MAA
Joint Meetings, where no source is given, but I'm sure I've seen the
claim somewhere else.
Edit. I have rolled back the title of this question to almost its original
form, because putting the title in quotes misled some people into thinking
I sought a source for the exact phrase "the only great book that Bourbaki
ever wrote." Rather, I wanted a source (not necessarily unique) for the idea
that Chapters 4-6 of Groupes et Algèbres de Lie  is Bourbaki's one great
book. Gerald's answer and Jim's comment together are exactly what I wanted.

REPLY [4 votes]: In a similar vein, Godement wrote in (1982, p. 6.28; translation):

... the previous lemma, which we have taken from N. Bourbaki, Lie Groups and Lie Algebras, chap. III, (the most unreadable presentation of the theory of Lie groups ever published since Sophus Lie, but fortunately the chapters on semisimple Lie groups and algebras make up for this)

Also Borel (1998):

A good example is provided by Chapters 4, 5, and 6 on reflection groups and root systems.
It started with a draft of about 70 pages on root systems. The author was almost apologetic in presenting to Bourbaki such a technical and special topic, but asserted this would be justified later by many applications. When the next draft, of some 130 pages, was submitted, one member remarked that it was all right, but really Bourbaki was spending too much time on such a minor topic, and others acquiesced. Well, the final outcome is well known: 288 pages, one of the most successful books by Bourbaki. It is a truly collective work, involving very actively about seven of us, none of whom could have written it by himself.<|endoftext|>
TITLE: How to tell whether a compact manifold can be realized as a nontrivial fiber bundle?
QUESTION [8 upvotes]: Let $M$ be a compact closed path-connected manifold.
There is the Postnikov tower as in
http://en.wikipedia.org/wiki/Postnikov_system
which tells us $M$ can be realized as an inverse limit of a tower of fibrations, up to homotopy, and this tower is often infinite, if $\pi_n(M)$ does not vanish when $n$ is large enough.
So I wonder, what if we work in the homeomorphic category instead of the homotopy category, and try to realize $M$ as a fiber bundle 
$F \rightarrow M \rightarrow B$
instead of a fibration (which is more general). Are there certain obstructions to this, such that one can tell whether this is possible by looking at the algebraic topology of $M$?
As an example of what I have in mind, note we have Hopf fibrations for $S^3, S^7, S^{15}$, and these are the only possibilities if the fiber $F$ and the base space $B$ are required to be spheres as well. I wonder what other spheres can be realized as non-trivial fiber bundles, if we don't put any restrictions on $F$ and $B$.
Thank you very much.

REPLY [5 votes]: For manifolds fibering over $S^1$, there are certain conditions. The 2-dim case is trivial. If $dim(M)=3$, then Stallings proved that $M$ fibers over $S^1$ if and only if there  is an epimorphism $\varphi:\pi_1(M)\to \mathbb{Z}$ such $ker(\varphi)$ is finitely generated. In fact, using the Thurston norm and Gabai's theory of sutured manifold hierarchies, one may algorithmically determine if a 3-manifold fibers over $S^1$.
Farrell gave algebraic conditions for fibering over $S^1$ when $dim(M)>5$. 
Another low-dimensional case may be determined. If $dim(M)=3$, and one wants $M$ to fiber over a 2-dimensional manifold, then $M$ must be a Seifert-fibered space with no exceptional fibers. After possibly passing to a 2-fold cover, the group $\pi_1(M)/Z(\pi_1(M))$ should be the fundamental group of a surface (where $Z(\pi_1(M))=\mathbb{Z}$ is the fundamental group of the fiber). Again, one may check this algorithmically.<|endoftext|>
TITLE: Consequences of not requiring ring homomorphisms to be unital?
QUESTION [31 upvotes]: As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1_R)=1_S$. But there has been a minority who do not require this, one prominent example being Herstein in Topics in Algebra.
What are some of the most striking consequences of not requiring ring homomorphisms to be unital? For example, what aspects of algebraic geometry would need to be reworked if we no longer required it? What interesting theorems or techniques arise in the not-necessarily-unital theory which do not apply (or are degenerate) for unital homomorphisms?

REPLY [2 votes]: If we permit rings to have 1 = 0, then the trivial ring {0} becomes the zero object of the category.
For comparison, if we required morphisms to keep 1 (but still permitted for 1 = 0), then {0} would be the terminal object only – any object has (unique) morphism to {0}, and 1 maps to 1 because 1 = 0 in the trivial ring. But there would be no morphisms from {0} to anything else since 1 ≠ 0 in any non-trivial ring.<|endoftext|>
TITLE: How well can we localize the "exoticness" in exotic R^4?
QUESTION [17 upvotes]: My question concerns whether there is a contradiction between two particular papers on exotic smoothness, Exotic Structures on smooth 4-manifolds by Selman Akbulut and Localized Exotic Smoothness by Carl H. Brans.  The former asserts:
"Let $M$ be a smooth closed simply connected $4$-manifold, and  $M'$ be an exotic copy of $M$ (a smooth manifold homeomorphic but not diffeomorphic to $M$). Then we can find a  compact contractible  codimension zero submanifold $W\subset M$ with complement $N$, and an involution $f:\partial W\to \partial W$ giving a decomposition:
$M=N\cup_{id}W$, $M'=N\cup_{f}W$."
The latter states:
"Gompf's end-sum techniques are used to establish the existence of an infinity of non-diffeomorphic manifolds, all having the same trivial ${\bf R^4}$ topology, but for which the exotic differentiable structure is confined to a region which is spatially limited. Thus, the smoothness is standard outside of a region which is topologically (but not smoothly) ${\bf B^3}\times {\bf R^1}$, where ${\bf B^3}$ is the compact three ball. The exterior of this region is diffeomorphic to standard ${\bf R^1}\times {\bf S^2}\times{\bf R^1}$.  In a space-time diagram, the
confined exoticness sweeps out a world tube..."
and further:
"The smoothness properties of the ${\bf R^4_\Theta}$... can be
summarized by saying the global $C^0$ coordinates, $(t,x,y,z)$, are
smooth in the exterior region $[a,\infty){\bf\times S^2\times R^1}$ given
by $x^2+y^2+z^2>a^2$ for some positive constant $a$, while the closure of
the complement of this is clearly an exotic ${\bf B^3\times_\Theta
R^1}$. (Here the 'exotic' can be understood as referring to the
product which is continuous but cannot be smooth...)"
The theorem from the first paper applies to closed manifolds.  Is it generalizable to open manifolds (such as $\mathbb{R}^4$)?  If so, then its confinement of "exoticness" to a compact submanifold seems inconsistent with the world tube construction implied in the statements from the second paper.

REPLY [4 votes]: The compactness of the 4-manifold is really necessary for the existence of the Akbulut cork. The non-compact case cannot be done with this method. If there is a compact subset in the exotic $\mathbb{R}^4$ determing the exoticness, then one can make a one-point compactifications of the $\mathbb{R}^4$ getting an exotic 4-sphere $S^4$ (constructing a  counterexample to the smooth Poince conjecture in dimension 4 (SPC4)).
More exact: Let $X$ be an exotic $\mathbb{R}^4$ and $R$ the standard $\mathbb{R}^4$. Let $C\subset X, C'\subset R$ be compact, contractable subsets. Now we assume that the subsets $C,C'$ are like Akbulut corks, i.e. $C$ and $C'$ are homeomorphic but non-diffeomorphic and $X-C$ is diffeomorphic to $R-C'$. The compactification of $X$ and $R$ result in a homotopy $S^4$ homeomorphic to the $S^4$. Then $X-C$ and $R-C'$ are changed to $\hat{X}-C$ and $\hat{R}-C'$ where $\hat{X},\hat{R}$ are homeomorphic to the $S^4$. Then $\hat{X}-C$ and $\hat{R}-C'$ are diffeomorphic (the assumption above) but $\hat{X},\hat{R}$ are not diffeomorphic. Thus we produce a counterexample to SPC4.
But there is not such a compact subset (someday I heart the formulation: "the exoticness is located at infinity").
Therefore the approach in the 94'paper is correct, the world tube is a non-compact area. No contradiction.<|endoftext|>
TITLE: Reconstruction from category of D-modules on variety
QUESTION [6 upvotes]: Arinkin has a theorem which says that an abelian variety can be reconstructed from its derived category of coherent D-modules. 
D.Orlov conjectured that this theorem is true for any variety. 
My question is:
Is this conjecture proved or disproved? I wonder know the related work, examples and any other related observations, comments. 
Thanks

REPLY [6 votes]: As far as I understand from your statement of the conjecture, the conjecture is false, although there are similar statements that are true. If I understand correctly, a weaker question (more likely to have the answer yes) would be "can one recover a variety from its category of D-modules."
For a non-example of the weaker question, if $X = Spec(\mathbb{C}[x])$ and $Y = Spec(\mathbb{C}[x^2,x^3])$, then D(X) and D(Y) are Morita equivalent. If X is a smooth curve and Y is another curve, then D(X) is Morita equivalent to D(Y) iff X and Y are homeomorphic (in the example above, the normalization map gives a homeomorphism $X \to Y$). If $X = Spec(\mathbb{C}[x])$, then the natural numbers parameterize isomorphism classes of curves Y with D(X) Morita equivalent to D(Y). 
A similar-sounding statement which is true is "If X and Y are smooth curves, they are isomorphic iff D(X) and D(Y) are isomorphic (as algebras)." A paper with these and many more facts can be found here http://arxiv.org/abs/math/0304320<|endoftext|>
TITLE: Cliquewidth of Cographs + kv
QUESTION [8 upvotes]: I have a question about graph width measures of undirected simple graphs. It is well-known that cographs (graphs which can be built by the operations of disjoint union and complementation, starting from isolated vertices) have cliquewidth at most 2. (Courcelle et al, Upper bounds to the clique width of graphs). Now consider some fixed non-negative integer k, and consider the class of graphs $\mathcal{G} _k$ of graphs such that for every $G = (V,E) \in \mathcal{G} _k$ there is a set $S$ of at most k vertices such that $G[V - S]$ is a cograph. Since the graph class $\mathcal{G} _k$ can also be seen as the class of graphs that can be built out of cographs by adding at most $k$ vertices, this class has also been called cographs + $kv$.
My question is: what is a tight bound on the cliquewidth of graphs in $\mathcal{G}_k$? 
It is known that if a graph $G$ is obtained from $H$ by deleting $k$ vertices then $cw(H) \leq 2^k (cw(G) + 1)$. This shows that if a cograph $G$ can be obtained from a graph $H$ by deleting $k$ vertices, then $cw(H) \leq 2^k (3 + 1)$, and hence the cliquewidth of a graph in $\mathcal{G}_k$ is at most $2^k 4$. I am unsure whether this exponential dependency on $k$ is necessary. In this context I would also be interested in the maximum decrease in the cliquewidth by deleting a vertex; i.e. if we delete a single vertex from a graph, how much can the cliquewidth decrease?

REPLY [4 votes]: I think the exponential bound is necessary.  Here is why.
Consider the disconnected graph $G$ on $n$ vertices, labelled $1$ to $n$.  Denote the set of vertices that new vertex $i$ is connected to by $f(i)$, a subset of $\lbrace 1,2,\ldots,n \rbrace$.
The question then becomes: is it possible to find such a function $f$ so that the set $\lbrace f(i) \mid 1 \le i \le k \rbrace$ of subsets of $V(G)$ generates a set of subsets of size $2^k$?
With "generates" I mean the closure under the operation of taking pairwise subsets.
The idea here is that to distinguish between any subsets one needs to label them differently.  However, due to the way the cliquewidth operations are restricted, this can only happen if their intersection is constructed first, and then the remaining vertices.
As far as I can tell, this scenario is possible, but clearly the above is quite far from a rigorous construction of an example.<|endoftext|>
TITLE: On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational.
QUESTION [9 upvotes]: Hi,
I'm interested in the behaviour of the sequence $(\sin(n!\pi x))$, when $x$ is irrational, as $n$ tends to infinity.
1) Is the sequence dense in $(-1,1)$?
or
2) Is it possible that for some irrational $x$, $\sin(n!\pi x)$ tends to $0$ as $n$ tends to infinity?
Any reference would be appreciated,
Thank you

REPLY [7 votes]: Denote by G the set of all x for which $sin(n! \pi x)$ approaches 0 as n approaches infinity. Every real number $0 < x < 1$ can be uniquely written as $ \sum_{n \ge 2} \frac{x(n)}{n!}$ where $0< x(n) < n$. From this it can be immediately seen that $x \in G$ iff $\frac{x(n)}{n}$ approaches 0 or 1. This immediately shows that there are continuum many points of G in every interval.
It can also be shown that G is an F-sigma-delta (countable union of countable intersections of open sets) additive subgroup of reals but it's neither G-delta nor F-sigma. One may iterate this construction to obtain additive groups of reals at arbitrarily high finite levels of Borel hierarchy in the following way: Let $G_0 = G$. Let $G_{n+1}$ be the set of all x such that the fractional part of $(n!x)$ converges to a point in $G_n$. Then $G_n$'s form an increasing chain of additive subgroups of reals and their union is a Borel additive subgroup of reals which is not at any finite level of Borel hierarchy.
I wrote a note on this here.<|endoftext|>
TITLE: Any finite dimensional admissible(smooth) irreducible representation of GL(2,Q_p) is 1-dim
QUESTION [5 upvotes]: Just like the title. I want a simple proof of the statment in the title.
$\mathbb{Q}_p$ is the p-adic field.
I wonder which module (or vector space) will be chosen as the space for the representation. Is this statement true for arbitarily module/vector space?
Thanks!

REPLY [14 votes]: What does "admissible" mean for you? Does it imply smoothness (stabilisers are open)? If not then I think the statement might be false (choose some hopelessly discontinuous injection from $\mathbf{Q}_p$ into $\mathbf{C}$ and then consider the induced "natural" representation of $GL(2,\mathbf{Q}_p)$ on $\mathbf{C}^2$; that's definitely irreducible, and the space of vectors fixed by some compact open will definitely be finite-dimensional).
But assuming smoothness too, it is true that any finite-dimensional smooth irreducible representation of $GL(2,\mathbf{Q}_p)$ is 1-dimensional. The proof is: if $V$ is such
a thing, then choose a basis for $V$ and for each basis vector choose a compact
open subgroup stabilising it. The intersection of these guys is still compact and
open, and fixes everything. So the kernel of the representation contains a compact
open subgroup. But this is a bit worrying because the kernel is normal. Now use
the fact that the normal subgroup generated by matrices $(1,e;0,1)$ and $(1,0;e,1)$
for $e$ small is still the whole of $SL(2,\mathbf{Q}_p)$ to deduce that $SL(2,\mathbf{Q}_p)$
is in the kernel, and now the action has to factor through $GL(2,\mathbf{Q}_p)/SL(2,\mathbf{Q}_p)=\mathbf{Q}_p^\times$. But now Schur's Lemma, which works for smooth irreducible
representations, says $V$ is 1-dimensional.<|endoftext|>
TITLE: Order of finite unitary group
QUESTION [13 upvotes]: This may be an easy exercise but I am not getting it. Let $\mathbf F_q$ be a finite field with $q$ elements and $\mathbf F_{q^2}$ be its degree two extension. Define an automorphism $\sigma$ of $\mathbf F_{q^2}$ by $\sigma (x) = x^q$. For any matrix $A = (a_{ij}) \in M_n(\mathbf F_{q^2})$, let $A^{\star} = (a_{ji}^{\sigma})$ (i.e., $A^{\star} = A^{\sigma \mathrm{t}}$). Then finite  unitary group $U_n(q)$ is a given by
$ U_n(q) = \left\lbrace A \in GL_n(\mathbf F_{q^2}) | A A^{*} = I_{n} \right\rbrace $
In a paper of Wall (page 33), it is mentioned that the order of this group is $q^{(n^2-n)/2} \prod\limits_{i=1}^{n} (q^i - (-1)^i)$. 

How to prove this? 

Any help will be appreciated.

REPLY [6 votes]: This question from 2010 was just listed as "active", apparently because someone (not myself) downvoted it today.  Anyway, rather than prolong the previous list of comments, I'll offer an explicit answer by pointing to an online resource that slightly predates Steinberg's 1968 AMS Memoir and is still a useful way to get into the details about Chevalley groups and their twisted analogues.   In 1967-68, Steinberg gave a course at Yale which was written up by listeners and published in mimeographed form by the math department there.   These typed notes are not as readable as typeset material and lack an index, but have been scanned in PDF format and placed on Steinberg's homepage at UCLA http://www.math.ucla.edu/~rst/
In particular, Section 11 deals with twisted groups, their Bruhat decompositions, and their orders.    One advantage of this broader approach via Chevalley groups is that it makes transparent the close analogy between the order formulas for finite unitary groups and for special linear groups of similar size.   In any case, as previous comments already indicate, the suggestion in the question that "This may be an easy exercise ..." is misguided.    But the order formula itself is standard and well studied from a variety of directions in the literature cited.<|endoftext|>
TITLE: On  proving that a certain set is not empty by proving that it is actually large
QUESTION [12 upvotes]: It happens occasionally that one can prove that a given set is not empty by proving that it is actually large. The word "large" here may refer to different properties.
For example, one can prove that a certain set is not empty by proving that its cardinality is big, as in the proof that there exist transcendental numbers : The set of algebraic numbers is countable, but the set of real numbers is uncountable, so there is uncountably many transcendental numbers.
One could also prove that a certain set is not empty by proving, for example, that it has positive measure, that it is dense, etc.
What are some good examples of such proofs?

REPLY [3 votes]: The theme is prevalent in combinatorial number theory and ergodic theory. Consider the ergodic Szemerédi theorem, for instance. It says that if $(X,\mathcal{B},\mu,T)$ is a measure-preserving system and $A \in \mathcal{B}$ has $\mu (A) >0$, then $\forall k \in \mathbb{N}$ $\exists n \in \mathbb{N}$ so that $\mu (A \cap T^{-n}A \cap T^{-2n}A \cap \cdots \cap T^{-kn}A)>0$. It turns out that it's easier to prove that measure of the intersection is positive for infinitely many $n$. In fact, Furstenberg's original proof showed that $\liminf_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} \mu (A \cap T^{-n}A \cap T^{-2n}A \cap \cdots \cap T^{-kn}A) > 0$.<|endoftext|>
TITLE: G-spaces and manifolds
QUESTION [6 upvotes]: In his book "The geometry of geodesics" H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:

The space is metric
The space is finitely compact, i.e., a bounded infinite set has at least one accumulation point
[metric convexity] For every $x\neq z$ there exists a third point $y$ different from $x$ and $z$ such that $d(x,y)+d(y,z)=d(x,z)$
[local prolongation] To every point $p$ there corresponds $\rho_p>0$ such that for every two point $x,y\in S(p,\rho_p)$ there exists a point $z$ such that $d(x,y)+d(y,z)=d(x,z)$
[uniqueness of prolongation] If $d(x,y)+d(y,z_1)=d(x,z_1)$ and $d(x,y)+d(y,z_2)=d(x,z_2)$ and $d(y,z_1)=d(y,z_2)$ then $z_1=z_2$.

Busemann conjectured that every $G$-space is a topological manifold. My question is does every topological/smooth/Riemannian manifold is also a $G$-space?
As for connected complete Riemannian manifold, I figured out that 1 holds since by the metric. 3 holds since every two points can be joined by a minimal geodesic, and then we can pick $y$ to be a point on it. 4 holds since it is a manifold and locally it is homeomorphic to some Euclidean space. Unfortunately, even in this case, I couldn't figure out 5 and 2.

REPLY [3 votes]: If I am not wrong, I believe that every topological manifold admits a complete metric. With this metric, it is possible to give a Path Metric Space structure (in the sense of Gromov, see the book Metric Structures for Riemannian and non-Riemannian spaces). I guess that this structure allows to make the same arguments as for Riemannian manifolds for topological manifolds. 
Anyway, I recomend this survey about Busemann conjecture which also discusses a stronger conjecture (the Bing-Borsuk conjecture).<|endoftext|>
TITLE: Complete extensions of first order logic (or language)
QUESTION [10 upvotes]: Lindstrom's theorem states that any extension of first order logic (FOL) more expressible than FOL fails to have either compactness or Lowenheim-Skolem. When I first read Lindstrom's theorem my first reaction was: "Does it mean incompleteness of any more expressible extension of FOL? And why this obviously important question isn't discussed in standard logic texts?" Standard extensions (such as second order logic) in fact are incomplete. After some attempts of proving incompleteness I found reference to Vaught's paper in which he proves completeness of extension of FOL by adding quantifier Qx = "there are uncountably many x such that..." It would be very interesting (at least for me) to understand complete extensions in general. Such an extensions may be of great importance, for instance, for computer science, because FOL isn't enough expressive for its purposes.
So, my questions are:
1) Do you know some other examples of complete FOL extensions?
2) Are there some results concerning characterization of complete FOL extensions?
3) Do you know any people, papers, books... studying general properties of FOL extensions? 
Thanks in advance.

REPLY [4 votes]: Bumping an old question, I think it's relevant that Lindstrom actually had a second theorem:

No proper strengthening of FOL has the downward Lowenheim-Skolem property and a r.e. set of validities. 

(This latter condition is what the OP calls "complete" - note that we're not assuming the existence of any proof system, it's purely a hypothesis on the "end-of-the-day" complexity.)
Put another way, downward Lowenheim-Skolem is a fundamental barrier to completeness beyond FOL. We can replace dLS here with a more technical property designed to do exactly what we need from dLS, but I'm not sure how valuable that is in this context and I don't know of any fundamentally different properties we can use in its place, so I'll provocatively claim (until my silliness is revealed) that dLS is the fundamental barrier to completeness beyond FOL.
The proof of both Lindstrom theorems can be found in Ebbinghaus-Flum-Thomas; they're very similar, but the second also brings in Trakhtenbrot's theorem. Incidentally, this paper by Barwise is a wonderful summary of various Lindstrom-style theorems. It's been a while since I read it and at a glance it doesn't discuss completeness very directly, but I can't help but mention it here.<|endoftext|>
TITLE: The upper semi-continuous rank of a module sheaf 
QUESTION [5 upvotes]: The article "Intuitionistic algebra and representations of rings" is fantastic; it develops the language, logic etc. for the topos $Sh(X)$, where $X$ is a fixed topological space, from scratch and thereby illuminating most of the known (and more) notions of sheaf theory in the sense that they become more natural. Thanks again Peter Arndt for this reference.
On page 37, the rank of a module $M$ is introduced as a upper Dedekind cut in the sheaf $\mathbb{N}$, or equivalently as a upper semi-continuous function $r : X \to \mathbb{N} \cup \{\infty\}$. Now it is claimed that it is given explicitely by mapping $x$ to the minimal number of generators needed for the stalks $M_x$. However, I don't see why this should be upper semi-continuous. I can show that only in the case that $M$ is locally of finite type. In fact, when I track back the constructions I get, that $r(x)$ is the infimum of numbers of sections which generate $M|_U$, where $U$ varies over the open neighborhoods of $x$.
Is this the correct definition of the rank?
Basically the same question applies to the definition of the independence $i(x)$, which is claimed to be the maximal number of linearly independent elements in $M_x$. Again, I don't see why this should be lower semi-continuous.

REPLY [4 votes]: This ought to be a comment, but I don't have the necessary reputation. I believe the claim is indeed false unless $M$ is locally of finite type. Here's a bit of context:

It's always the case that an internal upper Dedekind cut $A$ in $\mathrm{Sh}(X)$ induces an upper semi-continuous function $X \to \mathbb{N} \cup \{ \infty \}$, by mapping $x \in X$ to $\inf\{ n \in \mathbb{N} | \exists \text{open} V \subseteq X, x \in V: V \models (n \in A) \}$. This function takes values in $\mathbb{N}$ iff the cut is inhabited (in the internal sense).
The internally defined cut $A = \{ n \in \mathbb{N} | \text{there exists a generating family for $M$ consisting of $n$ elements} \}$ is inhabited iff $M$ is in the internal sense a finitely generated module, i.e. in the usual external sense locally of finite type.
Let $r$ be the function induced by this cut. It's easy to prove that the stalk $M_x$ can be generated by $r(x)$ elements. The reverse implication, that if $M_x$ can be generated by $n$ elements then $r(x) \leq n$, can be proven if $M$ locally of finite type.

So $r(x)$ coincides with the minimal number of generators needed for the stalk $M_x$ if $M$ is locally of finite type (at least in an open neighbourhood of $x$). If $M$ is not locally of finite type, $r$ will still be upper semi-continuous, but might not coincide with the minimal number of generators.
I don't know the "correct" definition of the rank if the module is not locally of finite type.<|endoftext|>
TITLE: Categories whose objects are CW-complexes
QUESTION [5 upvotes]: For a category whose objects are CW-complexes (with a chosen cell structure), what is the most natural notion of morphism? Are there choices, and if so, what are the pros and cons?
Types of maps that immediate come to mind would be continuous maps where the direct image a cell in X is a cell Y, or where the inverse image of a cell in Y is a cell in X. There are probably lots of elegant ways to get to maps like this, perhaps involving conditions on chain maps. It seems such an obvious thing to do that surely there are good references on this.

REPLY [4 votes]: My work on Higher Order Seifert-van Kampen Theorems has led to the conclusion that a very useful category is that of filtered spaces. Of course a CW-complex gives rise to a filtered space with the skeletal filtration. 
The reason behind this utility is that we can define (non trivially) a functor  $\rho$ from filtered spaces to a form of strict cubical  $\omega$-groupoid nicely related to classical invariants (i.e. relative homotopy groups) and which satisfies such a Seifert-van Kampen theorem; this has as a Corollary the Relative Hurewicz Theorem, as well as allowing the computation of the  homotopy 2-types of certain pushouts, in nonabelian terms. This then allows some computations of, for example,  second homotopy groups, as modules over the fundamental group. The functor $\rho$, being cubical,  is also very useful for consideration of tensor products.  
For more information, see 
http://groupoids.org.uk/nonab-a-t.html
and also this 2016 preprint on Modelling and Computing Homotopy Types: I. 
Notice also that Grothendieck in Section 5 of `Esquisse d'un programme' makes some interesting comments on the lack of relevance to geometry of the standard notion of topological space, and feels much more structure is needed. This comment needs further discussion in this blog!<|endoftext|>
TITLE: Relation between Hausdorff dimension and Bowen's equation
QUESTION [6 upvotes]: I am reading the paper  Hausdorff dimension for Horseshoes, by McCluskey and Manning. In the following theorem

 Theorem: 
Let $\Lambda$ be a basic set for a $C^1$ axiom A diffeomorphism $f:M^2\to M^2$ with $(1,1)$ splitting
$$
T_{\Lambda}M=E^s\oplus E^u.
$$
Define $\phi:W^u(\Lambda)\to\mathbb R$ by
$$
\phi(x)=-\log\|Df_x|_{E^u_x}\|.
$$
Then the Hausdorff dimension of $W^u(x)\cap\Lambda$ is given by the unique $\delta$ for which
$$
P_{f|_{\Lambda}}(\delta\phi)=0.     \qquad \qquad (1)
$$
The authors compute the Hausdorff dimension of $\Lambda\cap W^u_x$ by using the Bowen's equation (1). I read the proof but I was not able to figure out the intuition behind the Bowen's equation in this theorem. Could you give me explanation (do not need to be rigorous) about that or point out a reference ?

REPLY [3 votes]: In my opinion, Vaughn Climenhaga gave a nice answer and one of his papers named "Bowen's equation in the non-uniform setting" is a good one . You can also refer to a note given by Godofredo Iommi, i.e., ``The Bowen Formula: Dimension Theory and Thermodynamic Formalism '' which is available at:http://www.mat.puc.cl/~giommi/<|endoftext|>
TITLE: Maurer-Cartan form
QUESTION [8 upvotes]: I suppose given a Lie Group ($G$) and its corresponding Lie Algebra ($\mathfrak{g}$) every element in its dual defines a Maurer-Cartan form on the whole Lie Group? 
Let $\omega \in \mathfrak{g}^*$ be a Maurer-Cartan form and let $X$ and $Y$ be two elements of $g$ then in what sense are $\omega(X)$ and $\omega(Y)$ "constant functions" on $G$ ? (such that we can write $X\omega(Y)=Y\omega(X) =0$)
Assuming the above one can immediately write the Maurer-Cartan equation, $d\omega(X,Y)= -\frac{1}{2}\omega([X,Y])$ 
Thinking of $\omega$ as Lie Algebra valued 1-form on the Lie Group and using the fact from linear algebra that $V^* \otimes W = Hom(V,W)$ one can write them as $\omega = \sum _i \omega_i \otimes B_i$ where $\omega_i$ are ordinary 1-forms on $G$ and $B_i$ are a basis on $\mathfrak{g}$. (should there be some arbitrary coefficients in front of every term in the above sum?) 
Say $c^i_{jk}$ are the structure constants of the Lie Algebra then I do not understand how the Maurer-Cartan equations can be recast as,
$$d\omega_i = -\frac{1}{2}\sum_{j,k} c^i_{jk}  \omega_j \wedge \omega_k$$
which apparently can be further recast as the equation,
$$d\omega = -\frac{1}{2} [\omega,\omega]$$
I would be happy if someone can explain how the above two forms of the Maurer-Cartan equation can be obtained knowing the first form which is more familiar form to me. 
Also finding the structure constants of a Lie Algebra is not so hard for at least the common ones. Knowing that one fully "knows" the Maurer-Cartan Equation.  Now is there any sense in which one can "solve" this to find out the Maurer-Cartan forms? (I would guess a basis might be obtainable)

REPLY [3 votes]: The Maurer-Cartan form for matrix groups is very well explained in Santalo's book about Integral Geometry.<|endoftext|>
TITLE: Which continuous functions are polynomials?
QUESTION [17 upvotes]: I posted this on the new math.SE website but didn't get much of a response, so I am reposting it here.
Suppose $f$ is a continuous $\mathbb{R}$-valued function on $\mathbb{R}^n$.  What type of conditions on $f$ guarantee it is a polynomial up to homeomorphism. That is, when can I find a homeomorphism $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $\phi^* f = f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?
Some related questions:

A necessary condition in the case of $n = 1$ is that point inverse images of $f$ must be finite (since a polynomial has only finitely many roots).  Is this sufficient?
What if we replace $\mathbb{R}$ by $\mathbb{C}$?  
What if we look at smooth functions and diffeomorphism instead?  (I tried playing around with the implicit function theorem but didn't get anywhere).
What about the complex analytic case?

I'm not quite sure how to tag this, so feel free to edit them.

REPLY [4 votes]: I'm going to go out on a limb and make a partial conjecture based on Tom Goodwillie's comment.  A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p(x)$ only if it is topologically conjugate to a continuous function $q(x0$ of finite type which is nowhere constant.  By "finite type" I mean that there is a tiling of $\mathbb{R}^n$ by finitely many convex regions, such that the restriction of $q(x)$ to each region is non-constant and either linear or of the form $1/\phi(x)+c$, where $\phi$ is linear and $c$ is constant.
In this version of the answer, I'm going out on a limb for a second time.  I first conjectured that $q$ should simply be linear on each convex piece, and Richard Borcherds quickly found a counterexample to that in two variables.
I don't mean this to be a sufficient condition, since clearly it is not sufficient when $n=1$.  A "finite type" function in the above sense can be bounded, while a polynomial cannot be bounded.  Maybe it is a sufficient condition as a topological characterization of rational functions with no poles.  For polynomials specifically, there are strong restrictions on the behavior at infinity, but per Richard's example, they are somewhat looser than I first thought.
There is a relevant pair of results due to Whitney and Goresky.  Whitney proved that every analytic variety in $\mathbb{R}^n$ is a Whitney stratified space.  Goresky proved that every Whitney stratified space in $\mathbb{R}^n$ is supported on a piecewise smooth triangulation (but not necessarily one which is finite type).  It is easy to ride roughshod over subtleties as I already did, but these results seem like a good way to get started with the problem.
The smooth and complex cases of the problem seem more complicated for various reasons.<|endoftext|>
TITLE: Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?
QUESTION [27 upvotes]: This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that:

L4(2) and L3(4) both have order 20160
O2n+1(q) and S2n(q) have the same order for q odd, n > 2

I think this means that for each integer g, there are 0, 1 or 2 simple groups of order g.
Do we need the full strength of the Classification of Finite Simple Groups to prove this, or is there a simpler way of proving it?
(Originally asked at math.stackexchange.com).

REPLY [16 votes]: Emil Artin proved in 1955 in two papers that the above mentioned examples are the only instances of non-isomorphic finite simple groups having the same order. He proved the result only for the groups that were known till then. As new groups were being discovered Jacques Tits took the responsibility of checking that there were no such further cases. For an exposition of this, one may look in `Kimmerle and others, Proc. London Math. Soc. 60(3) (1990) 89–122'.
So, indeed the classification is used to some extent.<|endoftext|>
TITLE: An elementary question about adjunctions between presheaf categories preserving pullbacks. 
QUESTION [12 upvotes]: A functor $C \to D$ between categories induces a morphism of presheaf categories $Pre(D) \to Pre(C)$. This functor has a left adjoint given by left Kan extension and I am interested in knowing when this left adjoint preserves pull-back squares. 
I'm interested in any conditions that make this happen, but I am particularly interested in a special case. Let me say a little more about the context I am working in, and why I am interested. In the situation that this came up $C$ is a "lluf" subcategory of $D$, that is $C$ has the same objects as $D$ and the functor $C \to D$ is faithful. In that case it is good to call the functor $U:Pre(D) \to Pre(C)$ the forgetful functor. It automatically preserves limits and colimits. Let L be its left-adjoint.
Since C has the same objects as D, this forgetful functor is also conservative, meaning that it reflects isomorphisms. So by general non-sense (specifically Beck's monadicity theorem) this is a monadic adjunction. This means that $Pre(D) = T-alg$ is the category of T-algebras in $Pre(C)$ where T is the monad $T= UL$. 
I'm trying to understand conditions under which this monad is cartesian in the sense described at the n-lab. This means, among other things, that the monad T is supposed to send fiber products to fiber products. This is equivalent to having L send fiber products to fiber products. I want to understand when this happens. Does it always happen? Are there reasonable conditions on C or D that ensure that this happens?
Notice that I am not asking for L to be "left-exact", i.e. to commute with all finite limits. This property is generally much too strong. In particular L will not usually preserve terminal objects. This means it doesn't preserve products, but should instead send products to fiber products over $L(1)$. 
Here is an example. Let $C = pt$ be the singleton category and $D = G$ be the one object category with morphisms a group G. There is a unique inclusion $C \to D$ which is obviously faithful. The forgetful functor 
 $$U:Pre(D) \to Pre(C)$$ 
sends a G-set to its underlying set. The left adjoint L sends a set S to the free G-set $S \times G$. This doesn't preserve terminal objects, but it does commute with fiber products. What is more, the monad $T=UL$ is a classic example of a cartesian monad in the n-lab sense.  
I've played around with this, but can't seem to get it to work. I feel like this is going to be a well known result or there is going to be a counter example which sheds light on the situation. 

Question: In the context I described above (where $C \to D$ is lluf), does the left adjoint $$L: Pre(C) \to Pre(D)$$ always commute with fiber products? If not what is a counter example, and are there conditions one can place on C and D to ensure that L does commute with fiber products?

REPLY [5 votes]: Here's another way of getting to the same answer as Peter's.  A functor $F\colon A\to B$ preserves pullbacks if and only if the induced functor $F/1 \colon A \to B/F1$ preserves all finite limits, where 1 is the terminal object of A.  When F is left Kan extension $L\colon Psh(C) \to Psh(D)$ along a functor $f\colon C\to D$, it's not hard to check that Psh(D)/L1 is equivalent to presheaves on the opposite of the category el(L1) of elements of L1 (this is true with any presheaf replacing L1), and that L/1 is left Kan extension along the induced functor $f'\colon C\to el(L1)^{op}$.  Thus, we need to know when that functor is flat.
Now an object of $el(L1)^{op}$ is an object $d\in D$ together with a connected component, call it X, of the comma category $(d\downarrow f)$.  And a morphism in $el(L1)^{op}$ is a morphism $d_1\to d_2$ such that the induced functor $(d_2\downarrow f) \to (d_1\downarrow f)$ maps $X_2$ to $X_1$.  You can then check that the comma category $((d,X)\downarrow f')$ is precisely the connected component X of the comma category $(d\downarrow f)$.  Therefore, since $f'$ is flat just when all categories $((d,X)\downarrow f')$ are cofiltered, we conclude that left Kan extension along f preserves pullbacks iff all connected components of all comma categories $(d\downarrow f)$ are cofiltered, i.e. if all $(d\downarrow f)$ are "semi-filtered" in Peter's terminology.
Edit: you also asked for a specific counterexample when $f\colon C\to D$ is the inclusion of a lluf subcategory.  Let D be the walking commutative square generated by arrows $a\to b$, $a\to c$, $b\to d$, and $c\to d$, and let C be its lluf subcategory containing the identities and the arrows $b\to d$ and $c\to d$.  Then the comma category $(a\downarrow f)$ has two connected components, one of which is not semi-cofiltered.<|endoftext|>
TITLE: Automatic proving some expression is positive
QUESTION [5 upvotes]: Is there any automated (i.e., some algorithm) to prove that a certain algebraic expression is always non-negative in some range ? If so, is there any implementation you would suggest? My concrete problem is that I want to prove that for $f \in [0,1], 1 \leq a \leq L-2$ the following is true:
$$2^{(-a - L)} f^{-a} (1 + f)^{(-1 - a)} \left\{2^{(1 + a)} f^a (1 + f)^L (1 + 2 f) \left(-(1 + f)^{(1 + a)} + 2^a (1 + f^{(1 + a)})\right)\right.$$
$$ + 2^L (1 + f)^a \left[-2^a (1 + f) \left(-f^{(1 + 2 a)} + f^L + 3 f^{(a + L)} + 
3 f^{(1 + a + L)}\right) \right.$$
$$ \left.\left.+ (1 + f)^a \left((-3 + f) f^{(1 + a)} - a (-1 + f) (1 + 2 f) (f^a - f^L) + f^L (2 + 3 f (3 + f))\right)\right]\right\} >=0$$

REPLY [2 votes]: Here is a possible approach, more ad hoc than those previously suggested.  Let $E=E(f,L,a)$ be the expression without the "manifestly positive" factor that Willie Wong noticed is irrelevant.  Hope that establishing it for integers $a$ will lead to settling it for real $a$ (that's just a hope).
So focus on integral $a$.  Because $a=1$ is a bit different, separate that case off.  So now explore
$E(f,L,a)$ for $1 < a \le L-2$, where both $a$ and $L$ are integers.
For $L$ even,
$$E = -2^a \; f^{a+1} \; (1+f)^{a+1} \; \mathrm{poly}(f^{L+1}),$$
where $\mathrm{poly}(f^{L+1})$ is a polynomial in $f$ of degree $L+1$.
For $L$ odd,
$$E = -2^a \; f^a \; (1+f)^{a+2} \; \mathrm{poly}(f^L).$$
Examples, $L$ even:
$$L=6,a=2: \quad E = -8 f^2 (f+1)^3 \left(34 f^7-31
   f^6-56 f^5+59 f^4+10 f^3+23
   f^2-20 f-19\right).$$
$$L=6,a=3: \quad E = -16 f^3 (f+1)^4 \left(46
   f^7-27 f^6-76 f^5+19
   f^4+110 f^3-5 f^2-48
   f-19\right).$$
$$L=6,a=4: \quad E = -32 f^4 (f+1)^5 \left(44
   f^7-15 f^6-90 f^5+55 f^4+80
   f^3+15 f^2-66 f-23\right).$$
$L$ odd:
$$L=7,a=2: \quad E =
-8 f^2 (f+1)^4 \left(74
   f^7-127 f^6+91 f^4-46
   f^3+39 f^2+4 f-35\right).$$
$$L=7,a=3: \quad E =
-16 f^3 (f+1)^5 \left(106
   f^7-147 f^6-36 f^5+55
   f^4+74 f^3+27 f^2-48
   f-31\right).$$
$$L=7,a=4: \quad E =
-32 f^4 (f+1)^6 \left(118
   f^7-143 f^6-56 f^5+15
   f^4+174 f^3-f^2-76
   f-31\right).$$
$$L=7,a=5: \quad E =
-64 f^5 (f+1)^7 \left(104
   f^7-105 f^6-106 f^5+137
   f^4+28 f^3+73 f^2-90
   f-41\right).$$
Now the task is prove that $\mathrm{poly}(\;)$ is negative for $f$ in $[0,1]$.
As observed previously, $f=1$ is a root, so $(f-1)$ is a factor.
Just taking the last polynomial above as an example, it has a root at $f=-0.346213$
and is negative between there and $f=1$.  It seems feasible to analyze the structure of
$\mathrm{poly}(\;)$ and prove that it has no roots in $[0,1]$, which would settle it
for integers $a>1$.
Of course I am aware that I am leaving much to hope and further work.<|endoftext|>
TITLE: Symmetric powers and duals of vector bundles in char p
QUESTION [32 upvotes]: Suppose that $X$ is a smooth projective variety (eg $P^n$) and $E$ is a vector bundle (eg the tangent bundle). If the characteristic is zero, then taking symmetric powers "commutes" with taking duals:
$
Sym_m(E)^* 
$
and 
$Sym_m(E^*)$ are canonically isomorphic.
This is not true in characteristic $p>0$ (one has a canonical isomorphism with a divided power, instead.) But even in characteristic $p$, if the bundle is trivial, then there are non-canonical isomorphisms, and it is not hard to show that the Chern classes are the same.
Question: Is $Sym_m(E)^*$ isomorphic to $Sym_m(E^*)$ (non-canonically) in general?

REPLY [13 votes]: Here's another look at the case $m=2$. Maybe it can be used to shed more light on Torsten's nice example. $S^2E$ is part of an exact sequence $0\to \Lambda^2E\to E\otimes E\to S^2E\to 0$. $\Gamma^2E$ is part of an exact sequence $0\to \Gamma^2E\to E\otimes E\to \Lambda^2E\to 0$. The composition of $E\otimes E\to \Lambda^2E \to E\otimes E$ is $1-T$ where $T$ is the involution $x\otimes y\mapsto y\otimes x$. Its kernel $\Gamma^2E$ may be considered as the symmetric bilinear forms on $E^*$, while its cokernel $S^2E$ is the quadratic forms. If $2=0$ then the equation $(1-T)(1+T)=0$ says that the image of $1-T$ is contained in the kernel of $1-T$; we have $\Lambda^2E$ injecting into $\Gamma^2E$. We have in fact an exact sequence $0\to \Lambda^2E\to\Gamma^2E\to E'\to 0$, and another one $0\to E'\to S^2E\to \Lambda^2E\to 0$, where I am writing $E'$ for "$E$ twisted by Frobenius". 
If you're not in characteristic $2$ then there's no reason for $S^2E$ (or $\Gamma^2E$) to have a proper nontrivial subbundle.
Added: In the case when $E$ is the tangent bundle of $P^2$, or alternatively the rank $2$ quotient bundle of a trivial rank $3$ bundle which, as Torsten mentions, is the tangent bundle twisted by a line bundle, I believe it is not hard to work out by hand that the only global maps $E\otimes E\to E\otimes E$ are the linear combinations of the identity and the involution $v\otimes w\mapsto w\otimes v$. Of these, the only ones that kill the image of $\Lambda^2 E$ and so give a map $S^2E=coker(\Lambda^2E\to E\otimes E)\to E\otimes E$ are the multiples of $v\otimes w\mapsto v\otimes w+w\otimes v$, so that the only maps $S^2E\to \Gamma^2E=ker(E\otimes E\to\Lambda E)$ are the multiples of the usual one. This argument works over any ground ring, and shows that the two bundles are isomorphic only if $2$ is invertible.<|endoftext|>
TITLE: Complete problems for randomized complexity classes
QUESTION [7 upvotes]: It is believed that $BPP$ has no complete problems. Even for $BPP^O$ for a suitable oracle $O$ it is believed not to have complete problems, unless P=BPP. I wonder if the class MA (the randomized version of NP) has complete problems. For example, IP has complete problems (given that it is equal to PSPACE). There is a similar post asking about complexity classes with no complete problems. Here, I'm interested specifically on complete problems for MA. If the answer is positive, can you give some examples? I've tried google and the complexity zoo, but with no success.

REPLY [5 votes]: My own favorite complete problems are promise problems for the promise problem
version of SZK (Statistical Zero-Knowledge). These complete problems have played
a major role in the study of SZK, and the relations among them are fascinating per se.
See "On the complexity of computational problems regarding distributions (a survey)" 
at http://eccc.hpi-web.de/report/2011/004.
Oded<|endoftext|>
TITLE: Is there a procedure for obtaining all knots in S^3?
QUESTION [12 upvotes]: (Just to be precise, in this question, the word "knot" means "ambient isotopy class of a (EDIT: smooth) knot in $S^3$.") A knot in $S^3$ is called prime if it is not the connected sum of two other non-trivial knots in $S^3$. Clearly any knot is a sum of prime knots, and it is a theorem that this decomposition is unique. One downside of this is that there are infinitely many prime knots (for example, all non-trivial torus knots are prime). Here's a vague version of the question - Is there a way to trade off the uniqueness result (and add finitely many operations) in exchange for starting with a finite list of knots?
To try to make my question a little more precise, I'll define an "operation" as a function that takes a list of knots as input and outputs a finite list of knots. I'm not sure how to say this well, but I'd like to avoid very dumb operations like "fix an ordered list of all knots $K_1,K_2,\ldots$ and if $K_i$ is input, output $K_{i+1}$." However, I am interested in dumb answers, just not as dumb as that :-)

Is there finite list of knots $L$ and a finite list of operations $O$ on knots with the property that if $S$ is the smallest set of knots which contains $L$ and is closed under the operations $O$, then $S$ contains all knots?

For example, the first paragraph, phrased in this language, is 
L = all prime knots
O = input two knots and output their connected sum
S = all knots

REPLY [9 votes]: If you allow yourself to consider knotted trivalent graphs, instead of just knots, then you can start with just the tetrahedron and Mobius bands, and use the operations unzip, bubbling, and connect sum to get all knotted trivalent graphs.  This result is due to Dylan Thurston and Dror Bar-Natan and is written up by Dylan in a GT monograph.  Their motivation was the lack of a good candidate for an answer to your question which stayed entirely in the language of knots.  However for many situations KTGs are just as good as knots (e.g. click on a random link at Dror's wiki).<|endoftext|>
TITLE: Hilbert schemes of points and exceptional collections
QUESTION [7 upvotes]: Does there exist an exceptional collection of coherent sheaves on the Hilbert scheme of points on projective plane? If so, Could it be a strong full collection?

REPLY [9 votes]: It's more than 7 years after the question, but better late than never. The result of @Sasha holds for any number of points, and for surfaces more general than $\mathbb{P}^2$: once $\mathbf{D}^{\mathrm{b}}(S)$ has a full exceptional collection, then so does $\mathbf{D}^{\mathrm{b}}(\mathop{\mathrm{Hilb}}^nS)$. This is proposition 1.3 of
Krug, Andreas; Sosna, Pawel, On the derived category of the Hilbert scheme of points on an Enriques surface, Sel. Math., New Ser. 21, No. 4, 1339-1360 (2015). ZBL1331.18015.<|endoftext|>
TITLE: What are the most important results (and papers) in complexity theory that every one should know?
QUESTION [13 upvotes]: A few years ago Lance Fortnow listed his favorite theorems in complexity theory:
(1965-1974)
(1975-1984)
(1985-1994)
(1995-2004)
But he restricted himself (check the third one) and his last post is now 6 years old. An updated and more comprehensive list can be helpful.

What are the most important results (and papers) in complexity theory that every one should know? What are your favorites?

REPLY [4 votes]: There's the Bazzi/Razborov/Braverman sequence on fooling AC0 circuits.<|endoftext|>
TITLE: Zeroes of Maass forms
QUESTION [14 upvotes]: By a Maass form I just mean--maybe a bit loosely--any real analytic $\Bbb C$-valued function $f$ on the upper halfplane $\cal{H}$ which is automorphic of weight $k\in\Bbb Z$ with respect to a discrete subgroup $\Gamma<{\rm SL}_2(\Bbb R)$ such that $\Gamma\backslash\cal H$ has finite volume, and an eigenfunction for the Laplacian operator corresponding to the Casimir element in the universal enveloping algebra of the complexified $\rm{sl}_2$.
Is it true that the zeroes of these forms are isolated?
The answer is obviously affirmative in the case of holomorphic modular forms.
$\textbf{Edit}$: Scott's comment and Matt's answer below show that the answer is generally negative when the weight is $0$. Then, one can construct real valued Maass forms which have nodal curves.
Thus, let's make the assumption that $k\neq0$ and in particular that the Maass form $f$ belongs to a discrete series representation space (generated by a holomorphic modular form).
To make things as explicit as possible assume also that $\Gamma$ is either a congruence subgroup of ${\rm SL}_2(\Bbb Z)$ or the group of norm 1 elements in an Eichler order of an indefinite quaternion algebra over $\Bbb Q$

REPLY [14 votes]: Maass forms will generally have nodal lines rather than zeros.  Let's take the case where $\Gamma = SL_2(\mathbb{Z})$ and the weight $k=0$.
First recall that a Maass form has a Fourier expansion of the form $$f(x+iy) = \sum_{n \neq 0} a_n \sqrt{y} K_{\nu}(2 \pi |n| y) e^{2 \pi i n x},$$ where $K_{\nu}$ is the usual $K$-Bessel function.  Here $\nu$ is purely imaginary.  It is not difficult to show that $K_{it}(v)$ is real for $t, v$ real.  If $f$ is an eigenfunction of all the Hecke operators then the $a_n$'s are real also, which is a consequence of the fact that the Hecke operators are self-adjoint with respect to the Petersson inner-product.  Furthermore, by considering the reflection operator $T_{-1}f(x+iy) = f(-x+iy)$ which commutes with all the Hecke operators as well as the Laplacian, we should suppose $f$ is an eigenfunction of $T_{-1}$ also.  The only possible eigenvalues of $T_{-1}$ are $\pm 1$ so in the above Fourier expansion we can replace $e^{2 \pi i n x}$ by $\cos(2 \pi n x)$ or $\sin(2 \pi n x)$.  Then in either case $f$ is patently real-valued.
One can find some nice pictures of Maass forms at Fredrik Strömberg's homepage
Edit: I should also point out the simple fact that if $f$ is odd (i.e. $f(-x+iy) = -f(x+iy)$) then $f(iy) = 0$ for all $y$.  This can be clearly seen in Strömberg's pictures.<|endoftext|>
TITLE: Banach and Knaster-Tarski fixed point theorems -- are they related?
QUESTION [16 upvotes]: It there any known way of obtaining the Banach fixed-point theorem from the Tarski fixed-point theorem or vice-versa?

REPLY [4 votes]: As suggested by Jacques, I turn my comment into an answer.
This is not exactly what you ask for, but it is related. Efe Ok in Section 3.4 in Chapter 6 of his yet-to-be-written book on ordered sets gives a proof of the Banach fixed point theorem using the Kantorovitch-Tarski fixed point theorem: files.nyu.edu/eo1/public/Book-PDF/CHAPTER%205.pdf https://sites.google.com/a/nyu.edu/efeok/books/CHAPTER%205.pdf<|endoftext|>
TITLE: Analogues of Luzin's theorem
QUESTION [25 upvotes]: If $X$ is a compact metric space and $\mu$ is a Borel probability measure on $X$, then the space $C(X)$ of continuous real-valued functions on $X$ is a closed nowhere dense subset of $L^\infty(X,\mu)$, and hence bounded measurable functions are generically discontinuous.  Nevertheless, Luzin's theorem says that every measurable function is in fact continuous on a set of arbitrarily large measure.  This allows us to gain continuity from measurability at the cost of ignoring a small portion of $X$.
Question: Are there any analogues of Luzin's theorem that allow us to go from continuity to Hölder continuity?
A direct analogue would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X_\epsilon \subset X$ such that $\mu(X_\epsilon) > 1-\epsilon$ and the restriction of $f$ to $X_\epsilon$ is Hölder continuous.  (For my purposes, it would be all right if the Hölder exponent and coefficient become arbitrarily bad as $\epsilon\to 0$.)
Another possible analogue, and one that I am actually more interested in, would be a statement that given a continuous function $f\in C(X)$ and an arbitrary $\epsilon>0$, there exists a set $X_\epsilon \subset X$ such that the restriction of $f$ to $X_\epsilon$ is Hölder continuous (again with arbitrarily bad exponent and coefficient) and instead of an estimate on the measure of $X_\epsilon$, we have
$$
\dim_H(X_\epsilon) > \dim_H(X) - \epsilon,
$$
where $\dim_H$ is Hausdorff dimension.
Full motivation:  Ideally I would like to consider the setting where $T\colon X\to X$ is a continuous map, and obtain a similar statement about the restriction of a continuous potential $f\in C(X)$ to a set of large topological pressure,
$$
P_{X_\epsilon}(f) > P_X(f) - \epsilon,
$$
such that $f$ restricted to $X_\epsilon$ has the Walters property, which deals with variation on Bowen balls rather than on metric balls.  But the purely static version stated above for Hausdorff dimension seems like a good place to start.  Does anybody know of any results in this direction?  Or counterexamples showing that such a theorem can't be true in full generality?
Edit: I've accepted Anonymous's answer, which shows quite nicely that the direct analogue (using measures) fails.  However, I remain very interested in the indirect analogue (using dimensions), which seems to still have a chance of holding, so any information in that direction would be welcomed.

REPLY [2 votes]: I'm posting some references given to me by Jon Chaika as a CW answer, which addresses the question of whether we can get a set $X_\alpha$ such that the restriction to $X_\alpha$ is Hölder continuous.  It seems this question was posed by Márton Elekes in a 2004/5 paper, where a partial answer was given: for every $\alpha\in (0,1]$, a typical continuous function $[0,1]\to\mathbb{R}$ is not $\alpha$-Hölder on any set of Hausdorff dimension greater than $1-\alpha$.  The converse direction was provided in the 2009 PhD thesis of András Máthé, and appeared in a 2013 paper: for every $\alpha$ and every Borel measurable $f\colon [0,1]\to \mathbb{R}$, there is a set $X_\alpha\subset [0,1]$ with Hausdorff dimension $1-\alpha$ such that $f|_{X_\alpha}$ is $\alpha$-Hölder.<|endoftext|>
TITLE: Proof of the result that the Galois group of a specialization is a subgroup of the original group?
QUESTION [17 upvotes]: I have been using the following result:
Given a polynomial $f(x,t)$ of degree $n$ in $\mathbb{Q}[x,t]$, if a rational specialization of $t$ results in a separable polynomial $g(x)$ of the same degree, then the Galois group of $g$ over $\mathbb{Q}$ is a subgroup of that of $f$ over $\mathbb{Q}(t)$.
However, I have been unable to prove this for myself, and cannot seem to find a proof of it anywhere.  Is there an elementary proof?  And if not, can anyone direct me to a source containing one, or at least explain the general principle?
My need to understand the result arose from considering the following:
If I specialize $t$ such that $g$ factorizes as $x^k.h(x)$, where $h$ is an irreducible polynomial of degree $n-k$, is it legitimate to surmise that the Galois group of $h$ over $\mathbb{Q}$ is a subgroup of the original?
If nothing else, I'd be very grateful for an answer to this!

REPLY [18 votes]: Here is a broader setup for your question.  Let $A$ be a Dedekind domain with fraction field $F$, $E/F$ be a finite Galois extension, and $B$ be the integral closure of $A$ in $E$. 
Pick a prime $\mathfrak p$ in $A$ and a prime $\mathfrak P$ in $B$ lying over $\mathfrak p$.
The decomposition group $D(\mathfrak P|\mathfrak p)$ naturally maps by reduction mod $\mathfrak P$ to the automorphism group $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$ and Frobenius showed this is surjective.  The kernel is the inertia group, so if $\mathfrak p$ is unramified in $B$ then we get an isomorphism from $D(\mathfrak P|\mathfrak p)$ to $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$, whose inverse is an embedding 
of the automorphism group of the residue field extension into $\text{Gal}(E/F)$.
If we take $A = {\mathbf Z}$ then we're in the number field situation and this is where Frobenius elements in Galois groups come from. 
In your case you want to take $A = {\mathbf Q}[t]$, so $F = {\mathbf Q}(t)$. 
You did not give any assumptions about $f(x,t)$ as a polynomial in ${\mathbf Q}[x,t]$. (Stylistic quibble: I think it is better to write the polynomial as $f(t,x)$, specializing the first variable, but I'll use your notation.) Let's assume $f(x,t)$ is absolutely irreducible, so the ring $A' = {\mathbf Q}[x,t]/(f)$ is integrally closed. [EDIT:  I should have included the assumption that $f$ is smooth, as otherwise $A'$ will not be integrally closed, but this "global" int. closed business is actually not so important.  See comments below.] Write $F'$ for the fraction field of $A'$. After a linear change of variables we can assume $f(x,t)$ has a constant coefficient for the highest power of $x$, so $A'$ is the integral closure of $A$ in $F'$. 
Saying for some rational $t_0$ that the specialization $g(x) = f(x,t_0)$ is separable in ${\mathbf Q}[x] = (A/(t-t_0))[x]$ implies the prime $(t-t_0)$ is unramified in $A'$.  Let $E$ be the Galois closure of $F'/F$ and $B$ be the integral closure of $A$ in $E$. A prime ideal that is unramified in a finite extension is unramified in the Galois closure, so $(t-t_0)$ is unramified in $B$. For any prime $\mathfrak P$ in $B$ that lies over $(t-t_0)$, the 
residue field $B/\mathfrak P$ is a finite extension of $A/(t-t_0) = \mathbf Q$ and since 
$E/F$ is Galois the field $B/\mathfrak P$ is normal over $A/(t-t_0)$. These residue fields have characteristic 0, so they're separable: $B/\mathfrak P$ is a finite Galois extension of $\mathbf Q$.  I leave it to you to check that $B/\mathfrak P$ is the Galois closure of $g(x) = f(t_0,x)$ over $\mathbf Q$.  Then the isomorphism of $D(\mathfrak P|(t-t_0))$ with $\text{Aut}((B/\mathfrak P)/\mathbf Q) = \text{Gal}((B/\mathfrak P)/\mathbf Q)$ provides (by looking at the inverse map) an embedding of the Galois group of $g$ over $\mathbf Q$ into the Galois group of $f(x,t)$ over $F = {\mathbf Q}(t)$.
I agree with Damiano that there are problems when the specialization is not separable. In that case what happens is that the Galois group of the residue field extension is identified not with the decomposition group (a subgroup of the Galois group of $E/F$) but with the quotient group $D/I$ where $I = I(\mathfrak P|\mathfrak p)$ is the inertia group, and you don't generally expect a proper quotient group of a subgroup to naturally embed into the original group.<|endoftext|>
TITLE: When and how is it appropriate for an undergraduate to email a professor out of the blue?
QUESTION [118 upvotes]: This may not be appropriate for MathOverflow, as I haven't seen precedent for this type of question. But the answer is certainly of interest to me, and (I think) would be of interest to many other undergraduates. 
Often, while seeing some lecture online, or reading lecture notes from the internet, or hearing about someone who is an expert in a topic I'm interested in, I feel as though I want to email the professor in charge of whatever I'm looking at or hearing about. Usually this is about some question; these questions can range from the very technical to the very speculative. In the past, when I have mustered up enough guts to actually email the question, I've had good results (question answered, or reference given, etc.) However, every time I send a question I get the same sort of anxiety of "I have no idea what I'm doing!"
So here are my questions:

When is it appropriate to email a professor (that you don't know) a question about their work? (Remember; I'm an undergraduate, so these questions have a high chance of being inane in one way or another).
How should this email be formatted? Do I give my name in the first line? Or do I give a bit of background, ask the question, and sign my name. (The latter is what I have been doing).
How specific should the subject line be in order to increase the chances that the email gets read?
Should I mention anywhere in the email that I am an undergraduate?
This one is kind of silly, but I always have trouble deciding what to do: Suppose that I've emailed a professor, and they email me back with some answer... is it appropriate to email back with just a "Thank you"? I always feel like I'm wasting their time with such a contentless email, but at the same time I do want to thank them... Actually, in general I feel as though my questions are a waste of the professor's time (probably what feeds the email anxiety)... which brings me to:
If you happen to be a professor who has received such emails in the past; are they a waste of your time? Please be honest!

EDIT: Small new question...

If you receive a response and the professor signs with their first name, are you supposed to refer to them by first name the next time you email them?

Once again, if you think this is an inappropriate question, I totally understand. But please start a thread on meta to discuss it, because I think this might be borderline, at least.

REPLY [9 votes]: Let me also try to answer the questions:
So here are my questions:

1.When is it appropriate to email a professor (that you don't know) a
  question about their work? (Remember;
  I'm an undergraduate, so these
  questions have a high chance of being
  inane in one way or another).

If you have a research level question about a professor's work or related to his or her field of interest it is appropriate to email.

2.How should this email be formatted? Do I give my name in the first line?
  Or do I give a bit of background, ask
  the question, and sign my name. (The
  latter is what I have been doing).

You can first introduce yourself. You may give the background to the question. Try to be clear and rather brief.

3.How specific should the subject line be in order to increase the chances
  that the email gets read?

As informative as possible, 
"A question about convex bodies", "Are there only finitely many smooth toric varieties" 
"An idea related to the Erdos-Faber-Lovasz conjecture".

4.Should I mention anywhere in the email that I am an undergraduate?

Yes. If this is a good research level question it will will be impressing and if not it will not harm.  

5.This one is kind of silly, but I always have trouble deciding what to
  do: Suppose that I've emailed a
  professor, and they email me back with
  some answer... is it appropriate to
  email back with just a "Thank you"? I
  always feel like I'm wasting their
  time with such a contentless email,
  but at the same time I do want to
  thank them... Actually, in general I
  feel as though my questions are a
  waste of the professor's time
  (probably what feeds the email
  anxiety)...

Yes. email back with "thank you".

which brings me to:
6.If you happen to be a professor who has received such emails in the past;
  are they a waste of your time? Please
  be honest! EDIT: Small new question...

I have received quite a few of emails from undergraduates. Usually it is not a waste of time.

7.If you receive a response and the professor signs with their first name,
  are you supposed to refer to them by
  first name the next time you email
  them?

No.<|endoftext|>
TITLE: Realization space of matroids
QUESTION [8 upvotes]: Let $M$ be a matroid admitting a coordinatization over a complex vector space.   If we know that the complex coordinatization space for $M$ is connected, then may we conclude that the matroid admits a coordinatization over the real numbers?
The only examples that I am able to construct which do not have real coordinatizations have disconnected coordinatization spaces. 
Note: Some texts refere to 'coordinatization over a vector space' as 'realizability over a vector space.'

REPLY [9 votes]: Mnev wrote two papers in English about his theorem, as far as I know. (I don't read Russian.) A two-page paper (in Doklady I think) states the result for arbitrary "partially-oriented" matroids (including unoriented matroids), citing his dissertation, but with no proof. The paper in LNM that is usually cited has a sketch of the proof, but treats only the oriented case. Vakil's paper doesn't have a proof, but refers to the beginning of Lafforgue's book, where there is an algebraic argument (at the level of schemes) for the unoriented case, using the same method as Sturmfels used to prove a birational version of the universality theorem in the unoriented case, independently and simultaneously to Mnev. (Bernd doesn't get enough credit, I think.)
While that answers the question in principle, modeling the equation $x^2+y^2 = -1$ with a matroid requires a very large number of points. I once built the matroid for $x^2-1=0$, based on Sturmfels' construction, and needed 17 points (if I recall correctly), while there is a nine-point matroid with real, disconnected realization space.
So, consider the nine-point matroid of the line arrangement consisting of the irreducible components of $(x^3-y^3)(y^3-z^3)(z^3-x^3)=0$. (This is the matroid AG(2,3).) The matroid is (obviously) realizable over $\mathbb C$, but is projectively unique. (I'm pretty sure it is, but I don't have a reference.) So the realization space is a single point (connected), which, when put in a normal form, is not real. One can then argue that there is no linear change of variables that carries this matrix to a real matrix (up to scaling the rows). (What this means is that the conjugate of any realization is linearly equivalent to the original.)<|endoftext|>
TITLE: Decidable but nonrecursive sets
QUESTION [7 upvotes]: Until recently, I believed that recursive=decidable,
subscribing to this Wikipedia quote:
"In computability theory, a set is decidable, computable, or recursive if there 
is an algorithm that terminates after a finite amount of time and correctly decides 
whether or not a given object belongs to the set."
But my confidence in equating 'recursive' and 'decidable'
has been undermined by encountering intriguing work of Benjamin Wells that seems to
point to decidable but nonrecursive theories and sets:
[1] B. Wells, 
"Is There a Nonrecursive Decidable Equational Theory?"
J. Minds Machines, Volume 12, Number 2, 2002, 301-324.
[2] B. Wells,
"Hypercomputation by definition," 
Theor. Comput. Sci., 317, 1-3 (Jun. 2004), 191-207.
He has constructed "finitely based pseudorecursive" equational
theories which Tarski (his advisor) believed are decidable.
My (shakey) understanding based on these two papers is as follows.
$T$ is the equational theory, and $T_n$ is the subset of $T$ consisting
of the equations in which no more than $n$ distinct variables occur.
Each $T_n$ is recursive, but $T$ is not recursive.
For each $T_n$, there is a procedure for deciding whether an
arbitrary equation is in $T_n$;
so there is a "catalog" of these procedures indexed by $n$.
The $T_n$ are individually recursive but they are not
"uniformly recursive" in $n$,
apparently because the catalog is too chaotically arranged for indexing.
Despite reading Wells' description[1] of the sense in which 
$T$ might or should be considered decidable, I do not understand it.
(I am well beyond my expertise here, 
pushed into this unfamiliar territory by work with a student.)
I have two concrete questions.
First, is there later work on nonrecursive but decidable theories?
Has Wells' challenge to resolve "the current impasse" been
addressed by others?
Is there acceptance that there is indeed an impasse?
Second and relatedly, I am especially interested if
other models of computation ("hypercomputation"?)
have been suggested to capture the sense in which
$T$ is decidable by extensions of Turing machines.
Wells' 2004 summary[2] is negative:
"So far, there has emerged no concrete extension of computing
models corresponding to the extension of decidability
to finitely based pseudorecursive theories."

REPLY [3 votes]: In some situations there is a distinction between decidability and computability, but it works in the other direction, with decidability being stronger than computability.
The example is this:  A (possibly partial) function $f$ from $\mathbb N$ to $\mathbb N$ is decidable if its graph is decidable.  It is computable (or recursive) if there is an algorithm that for each $n\in\mathbb N$ computes $f(n)$ as long as $n$ is in the domain of $f$ and does not terminate on input $n$, otherwise.  
Now, if $f$ is computable (or even decidable), then its domain is c.e. (recursively enumerable).  Assume that $f$ has a domain that is c.e. but not computable.  Assume further that $f$ has only finitely many values (for instance only the single value $0$).
Such computable functions exist.  An example is the function that returns $0$ on every $n$ in the halting problem and is otherwise undefined.  In this case $f$ is not decidable.<|endoftext|>
TITLE: Semisimple quantum cohomology
QUESTION [7 upvotes]: Dubrovin's conjecture (or Bayer's modified version, if prefer) establishes a condition for the semisimplicity of the quantum cohomology of a manifold X, but, Why is important to know that the quantum cohomology of some manifold X is semisimple?

REPLY [4 votes]: One of the first interesting questions in GW theory asks how much information is needed to determine all GW invariants. The germ of a semisimple Frobenius is determined by a finite amount of data - this follows either from Dubrovin's or Manin's classification data via their structure connections. This implies that all genus-zero GW invariants are determined by this data; combined with Givental's conjecture/Teleman's theorem, this shows that all GW invariants are determined by the same finite amount of data. It also means that you could prove a mirror theorem (an isomorphism of Frobenius manifolds, in this case) by comparing finite amount of data on each side.
The mirror partner of $P^n$ (or any toric variety) is, as Kevin mentioned, constructed from the versal deformation of a function with isolated singularities (essentially this was proved by Givental as the first step of his mirror theorem). The Frobenius manifold associated to such a versal deformation is always generically semisimple. It seems reasonable to ask which other manifolds could have a function with isolated singularities as mirror partner. Clearly, having semisimple quantum cohomology is necessary (and one could motivate Dubrovin's conjecture by suggesting that it is a sufficient condition as well).

While I post here I should as well point out that "Bayer's modified version" was a little to optimistic. A counterexample is given by minimal surfaces of general type which have an exceptional vector bundle. At least one set of examples are the surfaces constructed by Yongnam Lee, Jongil Park in arXiv:math/0609072, on which a construction by Paul Hacking (which is included in arXiv:0808.1550) applies to produce exceptional vector bundles. On the other hand, all genus zero invariants vanish on minimal surfaces of general type by dimension reasons.<|endoftext|>
TITLE: Eisenstein series as sections of line bundles on moduli spaces
QUESTION [16 upvotes]: It is well known that a modular form of weight k and level \Gamma is a global section of k-power of a Hodge line bundle over some modular curve. e.g. H^0(X,E^k). 
My question is 
How to characterize Eisenstein series among such sections using geometric datas? 
For example, we know cusp forms are just sections of H^0(X,E^k(-cusps)).But how about Eisenstein series?
Actually in his Introduction to "Abelian Varieties" 1970, Mumford writes:
"It is interesting to ask whether further ties between the analytic and algebraic theories exist: e.g. an algebraic defintion of the Eisenstein series as a section of a line bundle on the moduli space. ..."
Could somebody explain the analytic-algebraic-representation aspects of Eisenstein series in some detail?  
Thank you!

REPLY [12 votes]: Here is one construction:
We have the exact sequence
$$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k})
\to H^0(\text{cusps}, \omega^{\otimes k}_{| \text{cusps}}).$$
(Here I am using $\omega$ for what you called $E$; this is the traditional notation
for modular forms people.) It is easy to define a Hecke action on the third $H^0$ so that this
exact sequence is Hecke equivariant.  
The right hand map is surjective if $k > 2$, and its image has codimension one when
$k = 2$.  In any event, write $\mathcal I$ to denote the image, so that
$$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k})
\to \mathcal I \to 0$$
is short exact.  One then shows that this short exact sequence has a unique Hecke
equivariant splitting; i.e. there is a uniquely determined Hecke equivariant subspace
$\mathcal E \subset H^0(\omega^{\otimes k})$ such that $\mathcal E$ projects isomorphically
onto $\mathcal I$.  This space $\mathcal E$ is the space of weight $k$ Eisenstein series (for whatever level we are working at).<|endoftext|>
TITLE: Polar body of a convex body that avoids a lattice
QUESTION [32 upvotes]: Let $K \subset {\bf R}^d$ be a symmetric convex body (an open bounded convex neighbourhood of the origin with $K = -K$) with the property that $K + {\bf Z}^d \neq {\bf R}^d$, i.e. the projection of $K$ to the standard torus ${\bf R}^d/{\bf Z}^d$ is not surjective, or equivalently $K$ is disjoint from some coset $x + {\bf Z}^d$ of the standard lattice.  My question is: what does this say about the polar body
$$K^* := \{ \xi \in {\bf R}^d: \xi \cdot x < 1 \hbox{ for all } x \in K \}?$$
Intuitively, the property $K + {\bf Z}^d \neq {\bf R}^d$ is a "smallness" condition on K, and is thus a "largeness" condition on $K^*$.
If $K^*$ contains a non-trivial element $n$ of $2 {\bf Z}^d$, then $K$ is contained in the strip $\{ x: |n \cdot x| < 1/2 \}$, and will thus avoid the coset $x+{\bf Z}^d$ whenever $x \cdot n = 1/2$.  So this is a sufficient condition for $K + {\bf Z}^d \neq {\bf R}^d$, but it is not necessary.  Indeed, if one takes $K$ to be the octahedron 
$$K := \{ (x_1,\ldots,x_d) \in {\bf R}^d: |x_1|+\ldots+|x_d| < d/2 \}$$
then $K$ avoids $(1/2,\ldots,1/2)+{\bf Z}^d$, but the dual body
$$ K^* = \{ (\xi_1,\ldots,\xi_d) \in {\bf R}^d: |\xi_1|,\ldots,|\xi_d| < 2/d \}$$
is quite far from reaching a non-trivial element of $2 {\bf Z}^d$.
On the other hand, by using the theory of Mahler bases or Fourier analysis one can show that if $K + {\bf Z}^d \neq {\bf R}^d$, then $K^*$ must contain a non-trivial element of $\varepsilon_d {\bf Z}^d$ for some $\varepsilon_d > 0$ depending only on $d$.  However the bounds I can get here are exponentially poor in $d$.
Based on the octahedron example (which intuitively seems to be the "biggest" convex set that still avoids a coset of ${\bf Z}^d$), one might tentatively conjecture that if $K + {\bf Z}^d \neq {\bf R}^d$, then the closure of $K^*$ contains a non-trivial element of $\frac{2}{d} {\bf Z}^d$.  I do not know how to prove or disprove this conjecture (though I think the $d=2$ case might be worked out by ad hoc methods, and the $d=1$ case is trivial), so I am posing it here as a question.

REPLY [3 votes]: This is not really an answer, but a comment on a natural variation of the original problem: assume a convex body in $\mathbb{R}^n$ that is not necessarily centrally symmetric contains only one lattice point as an interior point. What can be said about the dual of $K$ defined with respect to the unique lattice point in its interior?
In general, you can say that the volume of the dual body cannot be too small. In this paper Balacheff, Tzanev, and I conjecture that the volume of the dual cannot be less than $(n+1)/n!$. We can prove this in dimension two under the following equivalent guise:
Theorem. The area of a convex body in the plane that intersects every
integer line $mx + ny = 1$ is at least $3/2$. Moreover, equality holds only
for the triangle with vertices $(1, 0)$, $(0, 1)$, $(−1,−1)$ and its images under
(integer) unimodular transformations.
In higher dimensions, we rely (heavily!) on Greg Kuperberg's lower bound for the volume product to prove.
Theorem. The volume of a convex body in $\mathbb{R}^n$ that intersects every
integer hyperplane $m_1x_1 + · · · + m_nx_n = 1$ is at least $(\pi/8)^n(n + 1)/n!$.
But the fun part of the paper is that is also shows that these inequalities are (in the case of the first theorem) or should be extended to inequalities about Finsler metrics or even contact manifolds.<|endoftext|>
TITLE: Compactification of a manifold
QUESTION [14 upvotes]: This is just a curiosity and the question is really foggy. I'm wondering if there can exist a notion of "minimal smooth compactification" (when I say minimal I think something like adding a finite number of points or at least cells of dimension less than than the dimension of manifold) for a smooth non-compact manifold, in this sense: if the one point compactification of the manifold is smooth and the embedding is smooth, we are done; but what if the one point compactification is singular? Can I embed the manifold in a "minimal" compact manifold of the same dimension?

REPLY [17 votes]: A "surface of infinite genus" $S$ is an example of a manifold that is not an open subset of a compact manifold.
The reason $S$ cannot be embedded in a compact manifold is straightforward: we can find simple closed curves $a_1 , b_1 , \ldots , a_n , b_n , \ldots $ on $S$ such that, for each positive integer $g$, the curves $a_1 , b_1 , \ldots , a_g , b_g$ form a standard basis for a surface of genus $g$.  Thus, considering the product in homology of the classes of these curves, we deduce that they are independent.  If $S$ were an open subset of a compact manifold $M$, the same argument would imply that the images of the curves constructed above would also be independent in the homology of $M$. This contradicts the fact that the homology of the compact manifold $M$ is finite dimensional.
Observe that this example is not particularly different from the example of the complement in the complex plane of the integers.  Indeed, $S$ can be realized as the double cover of $\mathbb{C}$ branched along the integers.<|endoftext|>
TITLE: When is an irreducible scheme quasi-compact?
QUESTION [22 upvotes]: The standard examples of schemes that are not quasi-compact are either non-noetherian or have an infinite number of irreducible components. It is also easy to find non-separated irreducible examples. But are there other examples?
Question: Let $X$ be a locally noetherian scheme and assume that $X$ is irreducible (or has a finite number of irreducible components) and separated. Is $X$ quasi-compact (i.e., noetherian)?
If the answer is no in general, what conditions on $X$ are sufficient? Locally of finite type over a noetherian base scheme $S$? Fraction field finitely generated over a base? What if $X$ is regular? In general, the question is easily reduced to the case where $X$ is normal and integral.
It certainly feels like the answer is yes when $X$ is locally of finite type over $S$. Idea of proof: Choose an open dense affine $U\subseteq X$, choose a compactification $\overline{U}$ and modify $X$ and $\overline{U}$ such that the gluing $Y=X\cup_U \overline{U}$ is separated. Then, $Y=\overline{U}$ (by density and separatedness) is proper and hence quasi-compact.
Remark 1: If $X\to S$ is a proper morphism, then the irreducible components of the Hilbert scheme Hilb(X/S) are proper. The subtle point (in the non-projective case) is the quasi-compactness of the components (which can be proven by a similar trick as outlined above).
Remark 2: If $X\to S$ is universally closed, then $X\to S$ is quasi-compact. This is question 23337.

REPLY [8 votes]: Actually, Nick Proudfoot and I have been talking for years about the irreducible smooth surface constructed from countably many copies of ${\mathbb A}^2$ by gluing $(p,q)$ in the $n$th copy to $(p^2q,p^{-1})$ in the $n+1$th copy.  This even has a ${\mathbb G}_m$ action $\lambda \cdot (p,q) = (\lambda p, \lambda^{-1} q)$, a symplectic form $dp \wedge dq$, and a moment map $(p,q) \mapsto pq$ whose zero fiber is an infinite chain of projective lines.   This too can be regarded as the toric variety associated to a fan of infinite type in the plane.  It appears to be another way of describing Ekedahl's example.<|endoftext|>
TITLE: Finite subgroups of unitary groups
QUESTION [25 upvotes]: Let $n$ be an integer. Camille Jordan showed that there exists some $m \in {\mathbb N}$ (depending on $n$), such that for any pair of $n \times n$-unitaries $u,v \in U(n)$ which generate a finite group, one has $[u^m,v^m] = 1_n$. (In fact, he showed that any finite subgroup of $GL(n)$ has a abelian normal subgroup of finite index, bounded independently of the subgroup.)
In particular, the set of pairs that generate a finite subgroup cannot be dense in $U(n) \times U(n)$ for $n \geq 2$. Indeed, a generic pair of unitaries in $U(2)$ generates a free group. I am asking whether a stable version could still be true. 

Question  Let $n$ be an integer, let $u,v$ be a pair of unitaries in $U(n)$ and let $\varepsilon>0$. Is there some integer $k \in {\mathbb N}$ and unitaries $u',v' \in U(nk)$, such that  $\|u \otimes 1_k-u'\| \leq \varepsilon$ and $\|v \otimes 1_k - v'\| \leq \varepsilon$ (with respect to the standard embedding $U(n) \subset U(nk)$, $u \mapsto u \otimes 1_k$), and
  the unitaries $u',v'$ generate a finite group?

The question does only make sense once one has fixed a norm on $M_n {\mathbb C}$ for all $n \in {\mathbb N}$. I would either take the normalized Hilbert-Schmidt norm, i.e. $\|x\| = \frac1n Tr(x^*x)^{\frac12}$ for $x \in M_n \mathbb C$, or the operator norm.
(More globally seen, the questions are equivalent to the questions whether for the unitary group of certain UHF-algebras or the hyperfinite $II_1$-factor, pairs of unitaries generating a finite subgroup are dense in the natural topologies.)

REPLY [4 votes]: In the meantime I could answer the question myself; at least if one considers the metric induced by the norm. This will be contained in a forthcoming joint preprint with Ken Dykema. We can show two theorems:

Theorem 1: For every $\varepsilon>0$, there exists $n \in \mathbb N$ and unitaries $a,t \in U(n)$ such that $\|a-1\|\geq 1$ and
  $$\|1-tat^{-1}ata^{-1}t^{-1}a^{-2}\| < \varepsilon.$$

and

Theorem 2: There exists $\varepsilon_0>0$ such that for all $n \in \mathbb N$, and any two unitaries $a,t \in U(n)$ which generate a finite group: 
  $$\|1-tat^{-1}ata^{-1}t^{-1}a^{-2}\| < \varepsilon \leq \varepsilon_0$$
  implies
  $$\|1-a\|< 1/4-\sqrt{1/16-\varepsilon}.$$

This together implies that the answer to my question above is negative. The details are not complicated but a bit too long for putting them in this answer. I will post a link to the preprint as soon as it is ready.<|endoftext|>
TITLE: Dimensions of Jordan blocks associated to representations
QUESTION [5 upvotes]: Given a linear representation $\rho$ of $SL_n(\mathbb C)$ of finite dimension $m$,
the image $\rho(U)$ of a maximal unipotent Jordan block $U\in SL_n$ decomposes
into generally several Jordan blocks of size $m_1,\dots,m_k$.
Is it possible to describe the partition $m=m_1+m_2+\dots+m_k$, say in terms
of the highest weight vector associated to an irreducible representation $\rho$?

REPLY [3 votes]: As Ben has mentioned, this question can be best formulated in terms of restrictions
of finite-dimensional simple highest weight modules over a complex semisimple Lie algebra to a principal $sl_2$-subalgebra.
A classical reference:

Kostant, Bertram, The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group. Amer. J. Math. 81, 1959, 973-1032, doi:10.2307/2372999 (collected papers version: https://doi.org/10.1007/b94535_11)

More refined information about these restrictions is contained in the $q$-analogues of weight multiplicities introduced by Lusztig,

Lusztig, George, Singularities, character formulas, and a q-analog of weight multiplicities. Asterisque 101-102, 208-229, Numdam

This is explained in Sec 4 of Ginzburg's paper,

Victor Ginzburg, Perverse sheaves on a Loop group and Langlands' duality, arXiv:alg-geom/9511007<|endoftext|>
TITLE: Structure on $X(k)$ for separated finite type alg. space $X$, for complete valued $k$.
QUESTION [29 upvotes]: Let $k$ be a field complete with respect to a non-archimedean absolute value, and $X$ a separated algebraic space of finite type over $k$.  
If $X$ is a scheme then $X(k)$ inherits a natural (Hausdorff) topology from $k$ (uniquely determined by functoriality, compatibility with open immersion, closed immersions, fiber products, and the special case of the affine line), locally compact
when $k$ is locally compact. There is even a natural $k$-analytic manifold structure if $X$ is smooth.  This can be seen in (at least) two ways: algebraically by using Zariski-open covers by affines, and for smooth case take them to be "standard etale" over affine spaces (and then the $k$-analytic inverse function theorem can be applied), or analytically by identifying $X(k)$ with $X^{\rm{an}}(k)$ for the analytification $X^{\rm{an}}$ in the sense of rigid-analytic spaces (so then affinoid open covering of $X^{\rm{an}}$ does the job). 
Now suppose $X$ is not a scheme (as often happens with moduli spaces which only exist as algebraic spaces; e.g., Rapoport's thesis, etc.). The 2nd method above can be used, but  constructing $X^{\rm{an}}$ as a rigid-analytic space is really hard (the only method I know is to take a very long detour through Berkovich spaces to make the required rigid-analytic quotients from an etale scheme chart for $X$). 
My question is this: is there a known simple procedure (much as the method in the scheme case is simple), bypassing the use of $X^{\rm{an}}$, to make a functorial Hausdorff topology on $X(k)$ (and $k$-analytic manifold structure when $X$ is smooth) which is locally compact
when $k$ is and recovers the usual topology when $X$ is a scheme and shares the same basic properties (good behavior for open immersions, closed immersions, and fiber products; carries etale maps to local homeomorphisms as well as local $k$-analytic isomorphisms in case $X$ is smooth; and $X(k) \rightarrow X(k')$ is a closed embedding any extension
$k'/k$ of such fields)? 
A natural idea for the topology aspect is to choose an etale scheme cover $U \rightarrow X$ and identify $X(k_s)$ with a quotient of $U(k_s)$ as a set. Give it the quotient topology of the natural topology on $U(k_s)$ arising from the topology on $k_s$, and give $X(k)$ the subspace topology from $X(k_s)$. This does give the right answer (same as via the rigid-analytic method), but the only way I see this is locally compact when $k$ is
(noting that $X(k_s)$ is essentially never locally compact) and functorial is to invoke comparison with the rigid-analytic method.  (Otherwise I get bogged down in rising chains of finite Galois extensions and it feels like it becomes a mess. Doing the $k$-analytic manifold structure for smooth $X$ in this way also seems to become quite unpleasant. Note, by the way, that $k_s$ is not complete, so can't speak of $k_s$-analytic manifolds.) So this idea doesn't seem to provide an answer. Or maybe I am missing a simple trick? 
I am happy to restrict to the case of locally compact $k$ (though allowing completed algebraic closures is perfectly interesting), but do not want to restrict to characteristic 0 (though ideas in that case are appreciated too).

REPLY [21 votes]: Assume $k$ is a field and $F$ is a contravariant functor from $k$-schemes of finite type to sets. For each $X/k$ and $x\in F(X)$ we get a "characteristic map" $X(k)\to F(k)$ by pulling back $x$. Now if $k$ is a topological field, we can define a topology on $F(k)$, which is the finest making all these maps (for all pairs $(X,x)$) continuous. Some facts are easy to check, for instance:

if $F$ is representable we get the "usual" topology on $F(k)$,
every morphism of functors $F\to G$ gives rise to a continuous map $F(k)\to G(k)$,
restricting $X$ to affine $k$-schemes does not change anything.

With this generality, of course, one doesn't get very far without making some assumptions on $F$ and/or $k$. For instance, compatibility with products is not a formal consequence of the definition and therefore, probably, does not hold in general (example, please?). The nasty thing about this topology is that it is a quotient topology ($F(k)$ is a quotient space of the disjoint sum of all $X(k)$ indexed by pairs $(X,x)$) and general quotient topologies don't behave nicely.
Assume now that $k$ satisfies the following "weak henselian" property: 
(WH) Every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$.
This is satisfied by local fields and more generally by henselian valued fields. It is of course particularly relevant if $F$ is an algebraic space of finite type over $k$. In this case, it is easy to check that if $U\to F$ is an étale presentation, then, assuming (WH), $U(k)\to F(k)$ is an open map. This (together with the existence of pointed étale neighborhoods) is enough to imply, for instance, that the topology on $F(k)$ is compatible with products of algebraic spaces. (The point is that we now have a quotient by an open equivalence relation, and these quotients do behave nicely). I have not checked the other conditions in detail but I am confident that it works.
Some remarks:

I have used this trick in my paper "Problèmes de Skolem sur les champs algébriques" (Compositio 125 (2001), 1-30) to define open subcategories of $X(k)$ when $X$ is an algebraic stack of finite type over a local field $k$.
I am currently working with P. Gille on the case where $F(X)=H^1_{fppf}(X, G)$, where $G$ is an affine $k$-group scheme of finite type. 
Condition (WH) seems to become even nicer when, in addition, every finite morphism gives rise to a closed map. This holds for local fields, and more generally for any field with a (rank 1?) valuation, which is algebraically closed in its completion, such as the quotient field of an excellent henselian DVR.<|endoftext|>
TITLE: Local vs. infinitesimal rigidity
QUESTION [11 upvotes]: Can someone please explain the difference between local rigidity and infinitesimal rigidity? Does either version of rigidity imply the other?
In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension $\geq 3$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction.

REPLY [13 votes]: Infinitesimal rigidity implies local rigidity, but not conversely.  Local rigidity means a representation has no deformations, whereas infinitesimal rigidity means the natural tangent space to the character variety is 0-dimensional (this tangent space is a certain cohomology group with twisted coefficients).  Weil proved both in the context you mention.   See e.d. David Fisher's survey paper, where Weil's theorem is Theorem 3.2.  
See also Section 5 of this paper.

REPLY [12 votes]: Local rigidity means that the structure in question is an isolated point in its deformation space (which is typically an algebraic set).  Infinitesimal rigidity means that there are no first-order deformations of the structure in question.  A first-order deformation is a nonzero element of a certain cohomology group.
Because you can take the derivative of a path of structures and get a first-order deformation, infinitesimal rigidity implies local rigidity.
Because a first-order deformation may or may not correspond to an actual path (due to higher-order obstructions), local rigidity does NOT necessarily imply infinitesimal rigidity.<|endoftext|>
TITLE: Is the set of exponentials open?
QUESTION [8 upvotes]: Let $A$ be a $C^*$-algebra or some norm-closed algebra of operators on a Hilbert space. 
In the old paper 
Hille, E. On Roots and Logarithms of Elements of a Complex Banach Algebra, Math. Annalen, Bd. 136, S. 46--.57 (1958) 
the question is studied, which elements $x \in A$ have the property that the exponential function is open at $x$ (i.e. every neighborhood of $x$ maps to a neighborhood of $\exp(x)$). Under some conditions on the spectrum of $x$, Hille shows that the answer is positive.

Question: Is there an example of a unital $C^*$-algebra (or operator algebra) with the property that the exponential function is not open?

More specifically: Is the exponential function open for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$?
Is there an example of a Banach algebra where the exponential function is not open? 
EDIT: Jonas Meyer has clarified that $B(H)$ is a counterexample. It remains unclear what happens in general, or whether there is any description of the class of Banach algebras, where the exponential map is open. In particular, it remains unclear for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$.

REPLY [5 votes]: The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space.  A 1987 article by Conway and Morrel shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity.  On the other hand, a bilateral shift U has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for U under the exponential map.  Hence U is in the image but not in the interior of the image.  
I learned about Conway and Morrel's article from this answer by David Speyer.<|endoftext|>
TITLE: Permute Wada Lakes keeping the coastline intact? (still open in dim >2)
QUESTION [11 upvotes]: Wada Lakes are three disjoint open subsets of $\mathbb R^2$ with common boundary. Originally they were constructed by hand, but they also arise naturally in the real life, that is, theory of dynamical systems. (See, for example, this http://www.math.cornell.edu/~hubbard/pendulum.pdf paper of John Hubbard.)
From dynamical construction it is clear that one can have a homeomorphism that permutes the lakes. For example, this expository paper http://www.ams.org/notices/200601/fea-coudene.pdf gives such a homeomorphism (in fact, a diffeomorphism) for lakes on the sphere $S^2$. The dynamics on the boundary of the lakes is chaotic. 
==
Question 1: Let $U_1$, $U_2$ and $U_3$ be open subset of $\mathbb R^d$ with common boundary $C$. Assume that each $U_i$ is an image of an injective continuous map $\mathbb R^d\to\mathbb R^d$. Is there a homeomorphism of $\mathbb R^d$ that permutes open sets $U_i$ and is identity on $C$?
Question 2: Let $U_i\subset \mathbb R^d$, $i\in \mathbb Z$, be a disjoint collection of open sets each of which is homeomorphic to $\mathbb R^d$. Is there a homeomorphism of $\mathbb R^d$ that satisfies the following

$h$ maps $U_i$ onto $U_{i+1}$
every point in $\partial U_i$ is periodic under $h$

==
Edit: Below André Henriques has produced two examples when such a homeomorphism does not exist. Both examples make use of some local structure that cannot be permuted. It is still not clear what happens for "abstact Wada Lakes": are these examples exceptions or arbitrary Wada Lakes cannot be permuted as well? 
Edit2 (Jan. 9th 2012) 

In is clear now that there is no such homeomorphism if the dimension $d=2$. (See the comments below).
The answer to Q1 is also negative in higher dimension as explained in the reference given by Andres Koropecki.
For $d\ge 3$ there are examples by André Henriques of open sets for which one cannot find such a homeomorphism. In general, Question 2 is still open.

REPLY [6 votes]: I'm late by a year, but just in case, the result from this one-page paper seems to answer your question (but it has nothing to do with the Wada property):
M. Brown and J. M. Kister, Invariance of complementary domains of a fixed point set, Proc. Amer. Math. Soc. 91 (1984), no. 3, 503–504. MR 744656
The theorem is as follows:
Let $f$ be a homeomorphism of a connected topological manifold $M$ with fixed point set $F$. Then either 
(1) each connected component of $M-F$ is invariant, or 
(2) there are exactly two components and $f$ interchanges them.<|endoftext|>
TITLE: Is there a Whitney Embedding Theorem for non-smooth manifolds?
QUESTION [52 upvotes]: For smooth $n$-manifolds, we know that they can always be embedded in $\mathbb R^{2n}$ via a differentiable map.  However, is there any corresponding theorem for the topological category?  (i.e. Can every topological manifold embed continuously into some $\mathbb R^N$, and do we get the same bound for $N$?)

REPLY [4 votes]: My contribution here is very limited:
Every topological 4-manifold admits an embedding in $R^8$.
Proof. It suffices to consider the case of connected manifolds. If a connected 4-manifold is noncompact then it admits a smooth structure (equivalently, a triangulation). A proof can be found in
Quinn, Frank, Ends of maps. III: Dimensions 4 and 5, J. Differ. Geom. 17, 503-521 (1982). ZBL0533.57009.
Hence, such manifold admits an embedding in $R^8$ by Whitney's theorem.
It is a general result that every compact $n$-dimensional manifold admits an embedding in $R^{2n}$. This observation is due to Sergey Melikhov (in comments to Andy Putman's answer). All the available references prove a stronger result, namely, an embedding theorem for compact generalized manifolds:
Bryant, J. L.; Mio, W., Embeddings in generalized manifolds, Trans. Am. Math. Soc. 352, No. 3, 1131-1147 (2000). ZBL0934.57023.
and for certain classes of compact ANRs (which include all compact manifolds):
S. Melikhov, E.Shchepin,
The telescope approach to embeddability of compacta, unpublished, 2006.
Combining the two, one gets the embedding theorem for 4-manifolds. qed

My sense is that similar arguments to the ones by Bryant-Mio and Melikhov-Shchepin will go through in the noncompact case (I looked only at the paper by Melikhov and Shchepin) and can be used to prove an embedding theorem for noncompact manifolds but somebody would have to check the details...<|endoftext|>
TITLE: Kunneth formula for sheaf cohomology of varieties
QUESTION [46 upvotes]: What is a good reference for the following fact (the hypotheses may not be quite right):

Let $X$ and $Y$ be projective varieties over a field $k$. Let $\mathcal{F}$ and $\mathcal{G}$ be coherent sheaves on $X$ and $Y$, respectively. Let $\mathcal{F} \boxtimes \mathcal{G}$ denote $p_1^*(\mathcal{F}) \otimes_{\mathcal{O}_{X \times Y}} p_2^* \mathcal{G}$. Then
  $$H^m(X \times Y, \mathcal{F} \boxtimes \mathcal{G}) \cong \bigoplus_{p+q=m} H^p(X,\mathcal{F}) \otimes_k H^q(Y, \mathcal{G}).$$

Note: Wikipedia leads me to believe that this may be related to Theorem 6.7.3 in EGA III2, but I find this theorem quite intimidating. Although I would be willing to study this if there is no more basic reference, I would at least like some confirmation that I am studying the right thing.

REPLY [18 votes]: One can find this in section 9.2 of Kempf's book "Algebraic Varieties". 
The slightly more general case where $X, Y$ are over an affine scheme $\operatorname{Spec} R$ and $\mathcal{F}, \mathcal{G}$ are quasi-coherent sheaves flat over $\operatorname{Spec} R$ can be found in 
Kempf: "Some elementary proofs of basic theorems in the cohomology of quasi-coherent sheaves"
Rocky Mountain J. Math. Volume 10, Number 3 (1980), 637-646. link<|endoftext|>
TITLE: Example of a CW complex not homeomorphic to the realization of a simplicial set?
QUESTION [11 upvotes]: I've often heard that we can give examples of CW complexes that aren't homeomorphic to the realization of any simplicial set (although I haven't heard that there exist Kan complexes that aren't isomorphic to the total singular complex of a CGWH space. Are there?)  Would someone mind providing an example of one (and an example for the opposite statement as well, if it is true)?

REPLY [10 votes]: The geometric realization of a simplicial set is always triangulable. See Corollary 4.6.12 in Cellular Structures in Topology by Fritsch and Piccinini. They also give an explicit example (in section 3.4) of a non-triangulable CW-complex (which uses, I think, essentially the same idea as Tom Goodwillie's suggestion). This paper contains another example and shows, on the other hand, that every CW-complex with cells in at most two dimensions is triangulable.<|endoftext|>
TITLE: Counting/constructing Toric Varieties
QUESTION [10 upvotes]: Given a torus $T$ is there way to classify all the toric varieties it gives rise to?  That is, classify all toric varieties $X$ whose torus is isomorphic to $T$.  Is there a way to construct these toric varieties (i.e. give equations for them)?
Remark: as has been explained in the comments and answers, this question is I'll posed. But it seems to have generated good discussion so I'm leaving the phrasing as is.

REPLY [12 votes]: As far as my understanding goes the answer is no, and I will try to explain why and clarify the list of comments (I have little reputation so I cannot comment there) and give you a partial answer. I hope I do not patronise you, since you may now already part of it.
First of all, as Torsten said, it depends what you understand for classification. In this context a torus $T$ of dimension $r$ is always an algebraic variety isomorphic to $(\mathbb{C}^*)^r$ as a group. A complex algebraic variety $X$ of finite type is toric if there exists an embedding $\iota: (\mathbb{C}^\ast)^r \hookrightarrow X$, such that the image of $\iota$ is an open set whose Zariski closure is $X$ itself and the usual multiplication in $T=\iota((\mathbb{C}^\ast)^r)$ extends to $X$ (i.e. $T$ acts on $X$).
Think about all toric varieties. It is hard to find a complete classification, i.e. being able to give the coordinates ring for each affine patch and the morphisms among them for all toric varieties.
However, when the toric varieties we consider are normal there is a structure called the fan $\Sigma$ made out of cones. All cones live in $N_\mathbb{R}\cong N\otimes \mathbb{R}$ where $N\cong \mathbb{Z}$ is a lattice. A cone is generated by several vectors of the lattices (like a high school cone, really) and a fan is a union of cones which mainly have to satisfy that they do not overlap unless the overlap is a face of the cone (another cone of smaller dimension). There is a concept of morphism of fans and hence we can speak of fans 'up to isomorphism' (elements of $\mathbf{SL}(n,\mathbb{Z})$). Given a lattice N, there is an associated torus $T_N=N\otimes (\mathbb{C}^*)$, isomorphic to the standard torus.
Then we have a 1:1 correspondence between separated normal toric varieties $X$ (which contain the torus $T_N$ as a subset) up to isomorphism and fans in $N_\mathbb{R}$ up to isomorphism. There are algorithms to compute the fan from the variety and the variety from the fan and they are not difficult at all. You can easily learn them in chapter seven of the Mirror Symmetry book, available for free. Given any toric variety (even non-normal ones) we can compute its fan, but computing back the variety of this fan many not give us the original variety unless the original is normal. You can check this easily by computing the fan of a $\mathbf{V}(x^2-y^3)$ (torus embedding $(t^3,t^2)$) which is the same as $\mathbb{C}^1$ but obviously they are not isomorphic (the former has a singularity at (0,0)). In fact, since there are only two non-isomorphic fans of dimension 1 (the one generated by $1\in \mathbb{Z}$ and the one generated by 1 and -1) we see that there are only three normal toric varieties of dimension 1, the projective line and the affine line, and the standard torus.
The proof of this statement is not easy and to be honest I have never seen it written down complete (and I would appreciate any reference if someone saw it) but I know more or less the reason, as it is explained in the book about to be published by Cox, Little and Schenck (partly available) This theorem is part of my first year report which is due by the end of September, so if you want me to send you a copy when it is finished send me an e-mail.
So, yes, in the case of normal varieties there is some 'classification' via combinatorics, but in the case of non-normal I doubt there is (I never worked with them anyways).
Become a toric fan!.<|endoftext|>
TITLE: Is a pseudonatural transformation of strict 2-functors to Cat isomorphic to a 2-natural transformation?
QUESTION [7 upvotes]: Let $\mathcal{C}$ be a strict 2-category. A corollary of the bicategorical Yoneda lemma says that any pseudofunctor $\mathcal{C} \to \operatorname{Cat}$ is pseudonaturally equivalent to a strict 2-functor. I would like to know if the "next level" of strictification is true; namely, is it true that any pseudonatural transformation of strict 2-functors $\mathcal{C} \to \operatorname{Cat}$ is isomorphic (via an invertible modification) to a 2-natural transformation?
The specific case of this that I am interested is when $\mathcal{C}$ is the delooping of a monoidal category: the first statement says that any left $\mathcal{C}$-module can be made into a "strict" left $\mathcal{C}$-module, while the second would imply that any morphism of left $\mathcal{C}$-modules can be strictified.

REPLY [13 votes]: No, it is not true.  For a counterexample, let C be the delooping of the group Z/2 (regarded as a discrete monoidal category).  A strict C-module is then a category equipped with an involution, a strict C-module morphism is a functor preserving the involution strictly, and a pseudo C-module morphism preserves the involution up to coherent isomorphism.  In particular, a strict C-module morphism must map an object fixed by the involution to another fixed object, but a pseudo C-module morphism can map a fixed object to one which is only fixed up to isomorphism.  Thus, if A is a C-module with fixed objects and B is a C-module with no fixed objects, there can be pseudo C-module morphisms from A to B, but there cannot be any strict C-module morphisms from A to B.
It is, however, true that any strict (or even pseudo) 2-functor $F:C\to Cat$ is pseudonaturally equivalent to a strict 2-functor $F':C\to Cat$ which has the property that any pseudonatural transformation $F'\to G$ is isomorphic to a strict 2-natural transformation.  An $F'$ with this property is called flexible (if you're a 2-category theorist) or cofibrant (if you're a homotopy theorist).  The possibility of flexible replacement follows from generalities about 2-monads: there is a strict 2-monad T on the 2-category $Cat^{ob(C)}$ for which strict T-algebras are strict 2-functors $C\to Cat$, strict T-morphisms are strict 2-natural transformations, and pseudo T-morphisms are pseudonatural transformations.  The general coherence theorems of Power and Lack (see the papers "A general coherence result" and "Codescent objects and coherence") apply to this 2-monad and specialize to the statement I quoted above.  The homotopy theorist can instead construct a model structure on the 2-category of strict 2-functors and strict 2-natural transformations in which the cofibrant objects are cofibrant/flexible; see Lack's paper "Homotopy-theoretic aspects of 2-monads."<|endoftext|>
TITLE: Approaches to Riemann hypothesis using methods outside number theory
QUESTION [42 upvotes]: Background: Once an analytic number theorist remarked to me that all attempts to prove the Riemann hypothesis using number theoretic methods have failed. Since then that remark stuck in my mind.
The veracity of the above alluded number theorist's opinion does not really matter for the question to make sense; I just included it for background.
Question:

What are some promising methods from outside number theory to approach Riemann hypothesis?

I know two:

The geometric approach of Artin, Hasse, Weil and Deligne, the most important result being the proof of the Weil Conjectures.
The Bost-Connes approach. This is outlined by Lieven Le Bruyn for instance and has a hint of thermodynamics . 

I imagine that both of the above are cited by some people as the basis for the hopes that the theory of the field with one element will prove the Riemann hypothesis. Again, this question formally has no need to be connected the theory of field with one element to make sense. Other than just mentioning the above, let us not get into that aspect.
I am interested in other possible and promising methods. I am not interested in an equivalent formulation of Riemann hypothesis which is no better than the original. Both the above are very promising in terms of undiscovered things and might give a much better "big picture".
An approach I am ambivalent about, is that of Baez-Duarte. Though it does provide some evidence. I do not know whether it is any easier to prove Riemann hypothesis that way, rather than the original statement.
Please give me examples of any other methods; preferably very "promising" ones.
Edit 1: The meaning of "methods outside number theory" is the following: Nothing in the book of Ivic or Titchmarsch and Heath-Brown. More precisely, methods outside the traditional sybjects of elementary number theory and analytic number theory. I have given two examples above. One with algebraic geometry and one with thermodynamics.

REPLY [10 votes]: Hey,
I remember that in an interview, Selberg said that what Conne did was essentially getting a "different access" to the explicit formula, but that his proof did not yield new "hard facts" about the riemann zeta function. However take me with a grain of salt: I am not an expert in algebraic fields.
At the same time analytic methods give us information that are of a statistical nature. So for example they are quite well suited to prove that there is a positive proportion of the zeros on the half-line (40%, which is impressive enough I think). They can also show that almost all zeroes are concentrated in a box of height T and width $10^{10}/\log T$ around the line $\sigma = 1/2$. So those results are very useful in their own rights: For example they allow us to prove that the prime number theorem holds in intervals of length $x^{1/2+1/10}$ something you would expect to know only assuming the RH or a quasi-RH. In fact for many arithmetic applications the results we already know allow us (often/sometimes?) to by-pass the Riemann Hypothesis. 
So I wouldn't say that all approaches failed hopelessly. We are armed to deal with the problem, and paraphrasing Montgomery: "Sometimes I have the impression that we are missing just a fundamental insight to prove RH" .<|endoftext|>
TITLE: Is the topological concept of collapsible useful?
QUESTION [9 upvotes]: I ask this question because in the process of reviewing for my topology comp, I began rereading Alg Topology by Hatcher.  In the introduction is the famous Bing's House of Two Rooms.  I thought this was an interesting example and began reading about it on the web (procrastinating).  Several sites note that Bing's house is contractable (as described in Hatcher) but not collapsible.  The definition of collapsible does not appear in any of my topology or alg topology books (Munkres, Hatcher, Spanier) and the only definition I have found is on wikipedia.  So this brings me to my question, is collapsible a useful topological concept? And can anyone show me why Bing's house is not collapsible (I guess I probably do not fully comprehend the definition)?

REPLY [4 votes]: A small addendum: Collapses seem also useful in some applications, namely practical homology computations, as preprocessing steps for algorithms that compute homology. (Concretely - say you have to compute the homology of a simplicial complex with lots of facets; before starting to write down adjacency matrices etc. you might want to check if you can first "simplify" the complex with collapsing steps.) Sometimes this approach fails, as you see if you start with the Bing house. Yet sometimes one sequence of collapses works much better than another sequence, so it's also a matter of luck. 
However, an advantage of collapsibility over contractibility is that the former is an algorithmically-decidable property. (A non-efficient algorithm to decide collapsibility consists in trying all possible collapsing sequences, and see if one leaves us with a point.) In contrast, contractibility is in some cases algorithmically undecidable (by reduction to the word problem). For more info, cf. the recent preprint by Martin Tancer.<|endoftext|>
TITLE: Projective & injective dimensions
QUESTION [9 upvotes]: $A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. I'm not quite sure about the relation between finiteness of projective and injective dimensions of $M$. Does the finiteness (or infiniteness) of one necessarily imply the finiteness (or infiniteness) of another?

REPLY [11 votes]: To complement Mariano's answer: If finite projective dimension implies finite injective dimension for any module $M$, then $R$ better have finite injective dimension (the converse is also quite easy).
The local rings $R$ which have finite inj. dim. over themselves are also known as Gorenstein rings. In fact, a theorem by Foxby says that $R$ possesses a module of both finite proj. and inj. dim. if and only if $R$ is Gorenstein.<|endoftext|>
TITLE: Succinctly naming big numbers: ZFC versus Busy-Beaver
QUESTION [74 upvotes]: Years ago, I wrote an essay called Who Can Name the Bigger Number?, which posed the following challenge:

You have fifteen seconds. Using standard math notation, English words, or both, name a single whole number---not an infinity---on a blank index card. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature.

The essay went on to discuss systems for naming increasingly huge numbers concisely---including the Ackermann function, the Busy Beaver function, and super-recursive generalizations of Busy Beaver.
Recently (via Eliezer Yudkowsky), the claim has come to my attention that there are ways to concisely define vastly bigger numbers than even the super-recursive Busy Beaver numbers, using set theory.  (See for example this page by Agustín Rayo, which proposes a definition based on second-order set theory.)  However, whether these definitions work or not seems to hinge on some very delicate issues about definability in set theory.
So, I have a specific question about fast-growing integer sequences that are "well-defined," as I understand the term.  But first, let me be clear about some ground rules: I'm certainly fine with integer sequences whose values are unprovable from (say) the axioms of ZFC, as sufficiently large Busy Beaver numbers are.  Crucially, though, the values of the sequence must not depend on any controversial beliefs about transfinite sets.  So for example, the "definition"

n := 1 if CH is true, 2 if CH is false

makes sense in the language of ZFC, but it wouldn't be acceptable for my purposes.  Even a formalist---someone who sees CH, AC, large-cardinal axioms, etc. as having no definite truth-values---should be able to agree that we've picked out a specific positive integer.

Let me now describe the biggest numbers I know how to name, consistent with the above rules, and then maybe you can blow my numbers out of the water.
Given a Turing machine M, let S(M) be the number of steps made by M on an initially blank tape, or 0 if M runs forever.  Then recall that BB(n), or the nth Busy Beaver number, is defined as the maximum of S(M) over all n-state Turing machines M.  BB(n) is easily seen to grow faster than any computable function.  But for our purposes, BB(n) is puny!  So let's define $BB_1(n)$ to be the analogue of BB(n), for Turing machines equipped with an oracle that computes $BB_0(n):=BB(n)$.  Likewise, for all positive integers k, let $BB_k$ be the Busy Beaver function for Turing machines that are equipped with an oracle for $BB_{k-1}$.  It's not hard to see that $BB_k$ grows faster than any function computable with a $BB_{k-1}$ oracle. 
But we can go further: let $BB_{\omega}$ be the Busy Beaver function for Turing machines equipped with an oracle that computes $BB_k(n)$ given (k,n) as input.  Then let $BB_{\omega+1}$ be the Busy Beaver function for Turing machines with an oracle for $BB_{\omega}$, and so on.  It's clear that we can continue in this way through all the computable ordinals --- i.e. those countable ordinals $\alpha$ for which there exists a way to describe any $\beta < \alpha$ using a finite number of symbols, together with a Turing machine that decides whether $\beta < \beta'$ given the descriptions of each.
Next, let $\alpha(n)$ be the largest computable ordinal that can defined (in the sense above) by a Turing machine with at most n states.  Then we can define
$f(n) := BB_{\alpha(n)}(n),$
which grows faster than $BB_{\alpha}$ for any fixed computable ordinal $\alpha$.

A different way to define huge numbers is the following.  Given a predicate $\phi$ in the language of ZFC, with one free variable, say that $\phi$ "defines" a positive integer m if m is the unique positive integer that satisfies $\phi$, and the value of $m$ is the same in all models of ZFC.
Then let z(n) be the largest number defined by any predicate with n symbols or fewer.
One question that immediately arises is the relationship between f(n) and z(n).  I don't think it's hard to show that there exists a constant c such that $f(n) < z(n+c)$ for all n (please correct me if I'm wrong!)  But what about the other direction?  Does z(n) grow faster than any function definable in terms of Turing machines, or can we find a function similar to f(n) that dominates z(n)?  And are there other ways of specifying big numbers that dominate them both?

Update (8/5): After reading the first few comments, it occurred to me that the motivation for this question might not make sense to you, if you don't recognize a distinction between those mathematical questions that are "ultimately about finite processes" (for example: whether a given Turing machine halts or doesn't halt; the values of the super-recursive Busy Beaver numbers; most other mathematical questions), and those that aren't (for example: CH, AC, the existence of large cardinals).  The former I regard as having a definite answer, independently of the answer's provability in any formal system such as ZFC.  (If you doubt that there's a fact of the matter about whether a given Turing machine halts or runs forever, then you might as well also doubt that there's a fact of the matter about whether a given statement is or isn't provable in ZFC!)  For questions like CH and AC, by contrast, one can debate whether it even means anything to discuss their truth independently of their provability in some formal system.
In this question, I'm asking about integer sequences that are "ultimately definable in terms of finite processes," and which one can therefore regard as taking definite values, independently of one's beliefs about set-theoretic questions.  Of course, "ultimately definable in terms of finite processes" is a vague term.  But one can list many statements that certainly satisfy the criterion (for example: anything expressible in terms of Turing machines and whether they halt), and others that certainly don't (for example: CH and AC).  A large part of what I'm asking here is just how far the boundaries of the "definable in terms of finite processes" extend!
Yes, it's possible that my question could degenerate into philosophical disagreements.  But a priori, it's also possible that someone can give a sequence that everyone agrees is "definable in terms of finite processes," and that blows my f(n) and z(n) out of the water.  The latter would constitute a completely satisfying answer to the question.

Update (8/6): It's now been demonstrated to my satisfaction that z (as I defined it) is blown out of the water by f.  The reason is that z is defined by quantifying over all models of ZFC.  But by the Completeness Theorem, this means that z can also be defined "syntactically," in terms of provability in ZFC.  In particular, we can compute z using an oracle for the $BB_1$ function (or possibly even the BB function?), by defining a Turing machine that enumerates all positive integers m as well as all ZFC-proofs that the predicate $\phi$ picks out m.
So thanks -- I didn't want to prejudice things, but this is actually the answer I was hoping for!  If it wasn't clear already, I'm interested in big numbers not so much for their own sake, but as a concrete way of comparing the expressive power of different notational systems.  And I have a strong intuition that Turing machines are a "maximally expressive" notational system, at least for those numbers that meet my criterion of being "ultimately defined in terms of finite processes" (so in particular, independent of the truth or falsehood of statements like CH).  If one could use ZFC to define integer sequences that blew my sequence f(n) out of the water (and that did so in a model-independent way), that would be a serious challenge to my intuition.
So let me refocus the question: is my intuition correct, or is there some more clever way to use ZFC to define an integer sequence that blows f(n) out of the water?
Actually, a proposal for using ZFC to at least match the growth rate of f now occurs to me.  Recall that we defined the sequence z by maximizing over all models M of ZFC.  However, this definition ran into problems, related to the "self-hating models" that contain nonstandard integers encoding proofs of Not(Con(ZFC)).  So instead, given a model M of ZFC and a positive integer k, let's call M "k-true" if every $\Pi_k$ arithmetical sentence S is true in M if and only if S is semantically true (i.e., true for the standard integers).  (Here a $\Pi_k$ arithmetical sentence means a sentence with k alternating quantifiers, all of which range only over integers.)
Now, let's define the function
$z_k(n)$
exactly the same way as z(n), except that now we only take the maximum over those models M of ZFC that are k-true.
This remains to be proved, but my guess is that $z_k(n)$ should grow more-or-less like $BB_{k+c}(n)$, for some constant c.  Then, to get faster-growing sequences, one could strengthen the k-truth requirement, to require the models of ZFC being maximized over to agree with what's semantically true, even for sentences about integers that are defined using various computable ordinals.  But by these sorts of devices, it seems clear that one can match f but not blow it out of the water---and indeed, it seems simpler just to forget ZFC and talk directly about Turing machines.

REPLY [9 votes]: It seems that the philosophical position you're taking for granted is close to the position called "predicativism" advocated by mathematicians such as Poincarè and Weyl, and whose boundaries have been delineated more precisely by Solomon Feferman. Roughly, Feferman's position is that the notion of an "arbitrary set" is not well-defined enough for statements like AC and CH to take on definite values. So instead we should restrict our attention to the natural numbers and any questions which are "implicit" in our concept of the natural numbers; for example, talking within Peano Arithmetic (PA) makes sense, but so does talking about the truth of PA-strings, and talking about the truth of strings which talk about the truth of PA-strings, and so on. In his system, questions about the Busy Beaver functions $BB_\alpha$ have definite values whenever $\alpha$ is an ordinal less than $\Gamma_0$, where $\Gamma_0$ is the Feferman-Schutte ordinal.
It might be possible to extend predicativity to some ordinals greater than $\Gamma_0$ without losing "truth-definiteness"; for example, this is argued by Nik Weaver. But I think extending to the first non-recursive ordinal (i.e. the Church-Kleene ordinal), which is implicit in the function $f(n) = BB_{\alpha(n)}(n)$ defined in the OP, is problematic. This is because the notion of "ordinalhood" is not a notion which is "ultimately definable in terms of finite processes", in your terms. To be more specific, let's recall that a computable ordinal is a computable ordering of $\mathbb N$ with the property that every nonempty subset of $\mathbb N$ has a least element with respect to this ordering. The notion of a computable ordering is certainly well-defined, but the ordinalhood property depends on universal quantification over the realm of subsets of the integers. According to Feferman, such quantification is ill-defined because "the concept of the totality of arbitrary subsets of $\mathbb N$ is essentially underdetermined or vague" -- because sets are introduced by definitions, and no formal system for describing how we define things can capture all of the different ways in which we can define things.
Now, you may disagree with Feferman on this point, perhaps being skeptical only of definite totalities consisting of all subsets of $\mathbb R$ or higher types. In which case I think that would be a new philosophical position, and it would be worth expanding on what exactly the boundaries of this system are, and what the motivation for those boundaries are.
On the other hand, without accepting the notion of a totality of subsets of $\mathbb N$, one can philosophically accept the notion of ordinalhood as somehow describing the "well-definedness" of concepts -- that is, an ordering of the integers is said to be an ordinal if any recursive definition (in a particular formal language) of a function $f:\mathbb N\to\mathbb N$ of the form
$$
f(n) = F(f\upharpoonleft \{m : m < n\})
$$
determines a well-defined total function $f:\mathbb N\to\mathbb N$. But this raises issues of whether the concept of "well-definedness" is really a well-defined concept. Certainly, a diagonal argument may cause problems if we intend to mean the same thing by the two uses of the phrase "well-defined" in the previous sentence. Also, the dependence of this concept on the formal language used for the universal quantification over $F$ may also be problematic. (You can't quantify over all formal languages for the same reason you can't quantify over all sets.) But perhaps these philosophical issues are not too troublesome, in which case the question of whether or not a given ordering of the natural numbers is a well-ordering always has a definite truth value.
But in any case, it seems to me that if one accepts the notion of "well-definedness" described above, then one is no longer talking about those things which are "ultimately definable in terms of finite processes" -- instead, one has introduced new notions and is now talking about them.
So in conclusion, I don't think that the entry $f(n) = BB_{\alpha(n)}(n)$ given in the original post satisfies the rules for the contest laid down in the original post.
........
I've also spent some time thinking about the main question of this post, that is, what is the best strategy for winning a largest-number contest? After looking at a number of case studies (which I hope to write up at some point), I concluded that there are only three types of largest-number contests:

Those in which whoever has the (slightly) bigger paper to write on wins.
Those in which whoever has the most faith in (true) large cardinal axioms wins.
Those which are not well-defined as mathematical contests, but instead ask philosophical questions.

The first kind of contest can usually be judged on a reasonable timescale, while the second kind of contest can often be judged on an unreasonable timescale. The third kind can't really be judged at all.
The simplest example is the Busy Beaver Contest: Given a (reasonably large) $n$, the challenge is to write a Turing machine on $n$ symbols, and the winner is whoever has the longest runtime (and is thus the best approximation to $BB(n)$). The optimal strategy is this:

Pick a large cardinal axiom $\kappa$ that you think is $\omega$-consistent.
Enter the following program into the contest:

For every $\sigma\leq 3\uparrow\uparrow\uparrow 3$(

For every Turing machine $m$ of Gödel number $\leq 3\uparrow\uparrow\uparrow 3$(

If $\sigma$ is the Gödel number of a proof in $ZFC+\kappa$ that $m$ halts(

Run $m$)))

        Essentially, what this strategy does is "run all of the strategies that your opponent might be using that you can prove halt". If your opponent is honest enough that he won't submit any entry that he doesn't believe will halt, and he doesn't believe any mathematical statement that can't be proven in a formal system of consistency strength $\leq ZFC+\kappa$, then you will win. So this is a contest of type 2.




An obvious modification to this contest would be to only allow entries which provably halt within some formal system, e.g. $\Sigma = PA$. (The proof of halting must be submitted with the entry, as otherwise the requirement is trivial since every program which halts, halts provably in PA.) Then the best strategy is to fix a large $n$ consider the formal system $\Sigma(n)$ which is the same as PA except that only statements of with $\leq n$ nested quantifiers are allowed in the proofs of theorems. Then $\Sigma(n)$ is provably $\omega$-consistent in PA, although the length of the proof of $\omega$-consistency increases as $n\to\infty$. So then you just replace $ZFC+\kappa$ by $\Sigma(n)$ in the previous strategy; now that strategy provably halts within PA. So as long as your $n$ is greater than your opponent's $n$, then you win. So this is a contest of type 1.
The original posting seems to be a largest-number contest of type 3...<|endoftext|>
TITLE: Analytic tools in algebraic geometry 
QUESTION [150 upvotes]: This is not a very precise question, but I hope it will get some good answers.
As someone with a background in smooth manifold theory, I have experienced algebraic geometry as a beautiful but foreign territory. The strangeness has a lot to do with the lack of an Inverse Function Theorem. One day, after many years of thinking I knew what the word "etale" was all about, it dawned on me that what the etale site really is is a place where people like me don't have to feel so homesick: Grothendieck provided a way for the statement "infinitesimally invertible implies locally invertible" to be true in algebraic geometry, by the astounding device of changing the meaning of the word "locally". 
Bott might have called this an instance of "the old French trick of turning a theorem into a definition". He referred in that way to Schwarz's derivative of a distribution (theorem: integration by parts) and Serre's definition of fibration (theorem: homotopy lifting in fiber bundles). (Yes, I know Grothendieck is not French.)
My question is, if we list some other facts from calculus or analysis that are everyday tools in smooth manifold theory or analytic geometry, do some of them also become available in algebraic geometry when the right topology is chosen? I suspect that the word "crystalline" will come in here somewhere. For example:
Existence and uniqueness of solutions of ODEs (with dependence on initial data).
Sard's Theorem (Is there some topology that is good to invoke when proving "moving lemmas"?)
Various forms of the Fundamental Theorem of Calculus, such as: Stokes's Theorem, Poincare Lemma, or just existence of antiderivatives of one-variable functions.
Added: In characteristic zero algebraic geometry, of course de Rham cohomology has the familiar property that $X\times \mathbb A^1$ looks like $X$, and (therefore) that $\mathbb A^n$ looks like a point. But is the de Rham complex a resolution of the constant sheaf in any sense? I mean, this is not true in the etale topology, even though in some sense all smooth $n$-dimensional things are etale locally the same, right?

REPLY [4 votes]: Let me try to give a few answers to the first question, on ODEs.
The first answer is that you can solve ODEs in characteristic zero in the ring of formal power series. You also have the appropriate existence and uniqueness theorems for PDEs and systems of PDEs. Passing to a complete local ring is a completely valid means of localization in algebraic geometry (and also number theory, where one almost always takes a local ring to be complete). However, it has some weaknesses. In particular, when you localize at a closed point in this way, you lose track of all the other closed points.
A second approach would be to try to use a topology finer than the etale topology, like the smooth topology. (You could guess that the smooth topology works because it's known how to write the complete local ring as a limit of smooth morphisms) To pull back your ODE to some smooth covering, you need to lift whatever vector field you defined it to the covering.  So we can just define a topoology on $X$ with a fixed vector field $v$ where an open set is a smooth map $f: Y \to X$ and a vector field $w$ on $Y$ such that $df(w(y)) = v(f(y))$. In this topoology, the existence of local solutions is trivial. Simply take $Y$ to be the space of pairs of a point $x \in X$ and a function on $X$ modulo the $n$th power of the maximal ideal at $x$, $n$ the order of the ODE (i.e. an $n-1$-jet on $X$). To define a connection on this bundle, and hence a lift of the vector field, observe that we have exactly enough information to extend our function to a power series that solves the ODE, and use the natural connection on power series. By construction, we have a function that satisfies this differential equation, which sends a pair $(x,f)$ to $f(x)$. This again requires characteristic zero to work properly.
A third approach would be the closest to the "old French trick", and I think requires category theory. We can simply define an open set on $X$ to be a usual open set $Y$, or perhaps an etale open set $Y$, plus a fiber functor of the Tannakian category of vector bundles with a flat connection on $Y$. We can define a point of this open set to be a point $y\in Y$ plus an isomorphism between this fiber functor and the functor taking a vector bundle with flat connection to the sections at the point $y$. We define a local section of the ODE to be a vector in the evaluation of the fiber functor at the corresponding vector bundle with flat connection (which is the vector bundle of functions mod the appropriate power of the maximal ideal, as in the previous paragraph). We can dutifully check that a local section, as we have defined it, can actually be evaluated at each point, as we have defined it (by applying the isomorphism to the chosen vector).
Of course this construction is all a bit silly, because in most cases you just want to work with the Tannakian category of vector bundles with flat connection. The previous constructions I described are not, as far as I know, practically used in algebraic geometry, but this Tannakian category is - most often in a slightly extended form as the category of $D$-modules.
A final approach (which I will only sketch) is to say perhaps the etale topology was all you needed all along. To do this, you take whatever differential equation you care about, find the corresponding vector bundle with a flat connection, then apply the Riemann-Hilbert correspondence to obtain a corresponding representation of the fundamental group of $X$. You then choose a map from the defining ring of that representation to a ring of $\ell$-adic integers for some prime $\ell$, and then extend the representation, using the fact that $GL_n (\mathbb Z_\ell)$ is profinite, to a representation of the profinite completion of the fundamental group of $X$, which is the etale fundamental group of $X$. Finally you observe that modulo any finite power of $\ell$, this representation can be trivialized etale-locally. Almost anything you'd want to do with ODEs in a cohomological direction can be transferred, eventually, into the framework of lisse $\ell$-adic sheaves by this strategy. This theory, too, is used in practical algebraic geometry.<|endoftext|>
TITLE: What is the best known upper bound for the number of twin primes?
QUESTION [11 upvotes]: A quantitative form of the twin prime conjecture asserts that the the number of twin primes less than $n$ is asymptotically equal to $2\, C\, n/ \ln^2(n)$ where $C$ is the so-called twin prime constant. A variety of sieve methods (originating with Brun) can be used show that the number of twin primes less than $n$ is at most $A\, n/ \ln^2 (n) $ for some constant $A>2C$. My question is: What is the smallest known value of $A$? I'd also be interested in learning what the best known constants are for the prime k-tuple conjecture?

REPLY [18 votes]: J Wu, Chen's double sieve, Goldbach's conjecture, and the twin prime problem, Acta Arith 114 (2004) 215-273, MR 2005e:11128, bounds the number of twin primes above by $2aCx/\log^2x$, with  $C=\prod p(p-2)/(p-1)^2$, and $a=3.3996$; I don't know whether there have been any improvements.<|endoftext|>
TITLE: Counting certain arrangements of n triangles. Does the count grow superexponentially?
QUESTION [16 upvotes]: Overview
For integers n ≥ 1, let T(n) = {0,1,...,n}n and B(n)= {0,1}n. Note that |T(n)|=(n+1)n and |B(n)| = 2n.
A certain set S(n) ⊂ T(n), defined below, contains B(n). The question is about the growth rate of |S(n)|. Does it grow exponentially, like |B(n)|, so that |S(n)| ~ cn for some c, or does it grow superexponentially, so that cn/|S(n)| approaches 0 for all c> 0?
Definition
The set S(n) is defined as follows: an n-tuple t = (t1,t2,...,tn) ∈ T(n) is in S(n) if and only if ti+j ≤ j whenever 1 ≤ j< t i. For example, if t ∈ T(10) with t 4=5, t 5 can be at most 1, t 6 can be at most 2, , t 7 can be at most 3, and t 8 at most 4, but there is no restriction (at least not due to the value of t 4) on t 9 or t 10; t 9 and t 10 can have any values in {1,...,10}.
Alternate formulation (counting triangles)
The elements of S(n) can be put into one-to-one correspondence with certain configurations of n right isosceles triangles, so that |S(n)| counts the number of such configurations. 
For integers k>0 (size) and v≥0 (vertical position), let Δ k,v be the triangle with vertices (0,v), (k,k+v), and (k,v). (Δ0,v is the degenerate triangle with all three vertices at (0,v).)
Now associate with an n-tuple t = (t1,t2,...,tn) ∈ T(n) the set Dt = $\lbrace\Delta_{t_k,k}:1\le k \le n\rbrace$. (That's "\lbrace\Delta_{t_k,k}:1\le k \le n\rbrace," if you can't read it.) The set D t contains n isosceles right triangles that extend to the right of the y-axis, one triangle at each of the points (0,k) for 1 ≤ k ≤ n.
The tuple t is in S(n) if and only if the triangles in D t have disjoint interiors. (This isn't hard to show, and if it is, I've probably made a mistake in my definitions, so let me know.)  Thus |S(n)| counts the number of ways one can arrange n isosceles right triangles of various sizes (between size zero and size n) at n consecutive integer points on the y-axis so the triangle can extend to the right and up without overlapping. Triangles of the same size are indistiguishable for the purpose of counting the number of arrangements. (It may help to think of right isosceles pennants attached at an acute-angle corner to a flagpole in a stiff wind.)
Question
Does |S(n)| grow exponentially with n, or faster?
Calculations
If I’ve counted correctly, the first few terms of the sequence {|S(n)|} beginning with n=1 are 2, 8, 38, 184, 904, and 4384. This sequence (and some sequences resulting from minor variations of the problem) fails to match anything in the Online Encyclopedia of Integer Sequence.
Links to similar counting problems mentioned or solved in the literature would help. 
Thanks!

REPLY [6 votes]: Your sequence is bounded by $(125+\epsilon)^n$. Obviously, this isn't close to a good bound, but it answers the question.
We start by bounding a different question: Let $\Gamma_n$ be the convex hull of $(0,0)$, $(0,n)$ and $(n,n)$. (So $\Gamma$ is rotated $180^{\circ}$ with respect to your $\Delta$.) Let $q_n$ be the number of ways to pack non-overlapping triangles into $\Gamma_n$. 
Given any packing of triangles in $\Gamma_n$, which uses at least one triangle, let the largest triangle be of size $k$ and have a vertex at $(0,r)$. (If there is more than one largest triangle, make an arbitrary choice; this will just lead to a larger bound in the end.) So all the other triangles must fit into one of two trapezoids: one with base $r$ and height $k$ and the other with base $n-r$ and height $k$. In any case, these two trapezoids fit into translations of $\Gamma_r$ and $\Gamma_{n-r}$. So we obtain the inequality
$$q_n \leq \sum_{r=1}^{n-1} q_r q_{n-r} + 1,$$
where the $+1$ is because we have to remember the possibility that there might be no triangles in the packing. If we take $q_0=0$ for convenience, we get that $\sum q_n z^n$ is term by term dominated by the solution of 
$$Q(z) = Q(z)^2 + \frac{z}{1-z}.$$
Solving the quadratic,
$$Q(z) = \frac{1}{2} \left( 1 - \sqrt{1-\frac{4z}{1-z}} \right).$$
Notice that $Q(z)$ has radius of convergence $1/5$ so $q_n \leq (5+\epsilon)^n$.
I previously had an argument here that didn't work, so here is something even more sloppy. All the triangles you are considering fit inside $\Gamma_{3n}$. So your quantity is bounded by $q_{3n}$, and hence by $(125+\epsilon)^n$.
I suspect that $5^n$ may be pretty close to the right rate of growth, especially given Roland Bacher's computation.<|endoftext|>
TITLE: Ado's theorem for metric Lie algebras?
QUESTION [12 upvotes]: Background
Ado's Theorem states that every finite-dimensional Lie algebra over a field of zero characteristic admits a faithful representation.
More precisely, if $\mathfrak{g}$ is a finite-dimensional Lie algebra over a field $K$ of zero characteristic, then there is a Lie algebra monomorphism $\rho: \mathfrak{g} \to \operatorname{End}(K^N)$ for some $N$, where $\operatorname{End}(K^N)$ is the Lie algebra of endomorphisms of $K^N$ relative to the commutator.
Now let $\mathfrak{g}$ be a finite-dimensional Lie algebra over $K$ and let $\langle-,-\rangle$ be an ad-invariant symmetric inner product; that is, a nondegenerate symmetric $K$-bilinear form such that for all $x,y,z \in \mathfrak{g}$,
$$\langle [x,y], z\rangle = \langle x,[y,z]\rangle.$$
We call such a Lie algebra a metric Lie algebra.
For example, $\operatorname{End}(K^N)$ itself is a metric Lie algebra, relative to the inner product
$$\langle X,Y \rangle := \operatorname{Tr}(XY),$$
for endomorphisms $X,Y$.
Semisimple and, more generally, reductive Lie algebras are metric, but there are others.  The relevant structure theorem is due to Medina and Revoy (MathSciNet link).
A final definition, added after Victor's comment below, is the following.   Given two metric Lie algebras, the orthogonal direct sum of the underlying vector spaces can again be given the structure of a metric Lie algebra, in which the original Lie algebras sit as orthogonal ideals.  This gives rise to the notion of an indecomposable metric Lie algebra as one which is not isomorphic (as metric Lie algebra) to the direct product of orthogonal (nonzero) ideals.
The following question is motivated by trying to construct Chern-Simons forms for Lie groups admitting a bi-invariant metric.  But this motivation aside, I think the question is natural.
Question

Does every (finite-dimensional) indecomposable metric
  Lie algebra admit a faithful representation
  $$\rho: \mathfrak{g} \to \operatorname{End}(K^N),$$
  for some $N$, such that for all $x,y\in\mathfrak{g}$,
  $$\langle x,y \rangle = c \operatorname{Tr}(\rho(x)\rho(y)),$$
  for some nonzero $c \in K$?

REPLY [9 votes]: The answer is negative if $\mathfrak g$ is solvable and non commutative. It follows from the "Critère de Cartan" (Bourbaki, algèbres de Lie, chapitre 1, par. 5).<|endoftext|>
TITLE: Are the homology and cohomology Serre spectral sequences dual to each other?
QUESTION [17 upvotes]: If we use homology and cohomology over a field $k$, if a space has homology and cohomology groups of finite type in each degree, then $H_\ast(X;k)$ is dual to $H^\ast(X;k)$ using the universal coefficient theorem for cohomology.
Now, suppose I have a fibration $F \to E \to B$ such that $F$ and $B$ have homology and cohomology over $k$ of finite type in each degree and $\pi_1(B) = 0$ for simplicity. Certainly, the $E_2$-page of the cohomology Serre spectral sequence will be dual to the $E^2$-page homology Serre spectral sequence. My first question is: Is it also true that the differentials for the cohomology spectral sequence are dual as a linear map to the differentials of the homology spectral sequence, and vice versa?
Secondly, the cohomology Serre spectral sequence is a multiplicative one. Is the homology one comultiplicative? If so, is the product for cohomology dual to the coproduct for homology?
Finally, if all of this holds, to which extend can it generalized?

REPLY [4 votes]: There is, what my advisor calls, a Atiyah-Hirzebruch-Serre SS that works in generalized homology. You filter the total space just as Tyler says, except instead of applying cellular chains with coefficients in $Z=\pi\_* (HZ)$ we apply cellular chains with coefficients in $E\_* =\pi\_* (E)$ to the filtration. This will work if the usual AH SS collapses/converges for the theory $E$.
As to your question about whether or not the homology SS is comultiplicative, it probably should be. What is true though, is that if X is an H-space, then the differential in the homology Serre SS is a derivation like it is in cohomology.<|endoftext|>
TITLE: What is the Hopf algebra structures in the homology of the based loop spaces of $E_7$ and $E_8$?
QUESTION [17 upvotes]: Since $\Omega X$ is a $H$-space, if it has homology of finite type, the homology acquires the structure of a Hopf algebra. Bott has shown that for $X=G$ a Lie group, in fact $H_*(\Omega X)$ is free and concentrated in even degree. 
Bott has also, using generating varieties, calculated this Hopf algebra structure for $G = SO(n)$, $Spin(n)$, $SU(n)$, $G_2$ in The space of loops on a Lie group. Michigan Math. J., 5:35-61, 1958. The same technique was used by Watanabe in The homology of the loop space of the exceptional group F4. Osaka J. Math., 15:463-474, 1978 for $F_4$ and by Nakagawa in The space of loops on the exceptional Lie group E6. Osaka J. Math., 40:429-448, 2003 for $E_6$. Using Bott periodicity, Kono and Kozima calculated this Hopf algebra structure on the homology for $Sp(n)$ in The space of loops on a symplectic group. Japanese J. Math., 4:461-480, 1978.
However, I haven't been able to locate similar calculations for the exceptional Lie groups $E_7$ and $E_8$. Does anyone know where these can be found? Seeing that it is already quite involved to do these calculations for $F_4$ and $E_6$, I'd rather not try to them myself. However, if some simple method exists, answers explaining it are also welcome.

Also note that Bott's result for $G_2$ in The space of loops on a Lie group seems to be incorrect. Watanabe writes in The homology of the loop space of the exceptional group F4:

There is a misprint in Bott's result on $H_\ast(\Omega G_2)$ [5;p. 60]. The
  coproduct formula for $w \in H_{10}(ΩG_2)$ is an error. It is corrected by exchanging
  2 for 3.

This can also be found in Clarke's On the K -theory of the loop space of a Lie group, Proc. Camb. Phil. Soc. (1974), 76, 1.

REPLY [2 votes]: This is more of a suggestion than an answer. Let $K$ be a compact Lie group and $G$ its complexification. It is known (see Pressley-Segal) that the based loop space of $K$ and the affine Grassmannian $Gr_G$ of $G$ have the same (co)homology groups. It is a theorem of Ginzburg that the the cohomology ring of $Gr_G$ may be (canonically) identified with the enveloping algebra of the centraliser of a regular nilpotent element $e \in \mathfrak{g}^{\vee}$ (here $\mathfrak{g}^{\vee}$ denotes the Lie algebra of the group $G^{\vee}$ Langlands dual to $G$).
One can dualise this to get that the homology of $Gr_G$ can be identified with the functions on $B^{\vee}_e$, where $B^{\vee}$ denotes a Borel subgroup of $G^{\vee}$ and $e$ is a regular nilpotent element in the Lie algebra of $B^{\vee}$. This is explained in a very nice paper by Yun and Zhu called "Integral homology of loop groups via Langlands dual groups". They prove that this induces an isomorphism of the corresponding group schemes (with the Hopf algebra structure on the homology of $Gr_G$ inducing the group scheme structure on one side).
So this doesn't provide explicit formulas, but it does give you a way to work out the Hopf algebra structure (even over the integers in the simply laced case -- again see Yun-Zhu for this statement) in terms of combinatorics of root systems.
If you are interested in torsion phenomena the paper "Some arithmetical results on semi-simple Lie algebras" by Springer might be useful.<|endoftext|>
TITLE: Continuous + holomorphic on a dense open => holomorphic?
QUESTION [26 upvotes]: Let D ⊂ ℂ be the closed unit disc in the complex plane, and let C be a continuously embedded path in D between the points -1 and 1. The curve C splits D into two halfs $D_1$ and $D_2$.
Let f : D→ℂ be a continuous function that is holomorphic on the interiors of $D_1$ and $D_2$. Is f then necessarily holomorphic?

PS: If he path C is sufficiently smooth (so that ∫C f(z) dz makes sense), then f is necessarily holomorphic, as it is then given by Cauchy's formula $f(w)=1/(2\pi i)\int _ {\partial D} f(z) /(z-w) dz$.

REPLY [18 votes]: Denjoy makes a detailed study of this question, and in particular constructs counterexamples where the curve C is the graph of a continuous function. Apparently, the construction works for curves which are 'very' non rectifiable, i.e., the local variation is infinite of a suitably high order at each point.<|endoftext|>
TITLE: Proving Independence of Axioms by Exhibiting Models Which Don't Satisfy Our Intuition
QUESTION [12 upvotes]: I recently saw the proof of the independence of ZF (with allowance for multiple empty sets) and AC. The proof constructed the model based on a set theory generated by infinitely many empty sets and then restricted this model to those with a hereditary finite basis (HFB). However, this model did not seem to conform to my intuition about set theory. Rather, it seemed like an odd construction which is not what I think of as the theory of sets, yet which did in fact formally satisfy all the axioms of ZF as well as negation of AC. Although we may have proved that the ZF axioms do not imply choice, I do not feel at all convinced that AC does not have to be true in set theory. Rather, although what I'm about to say is imprecise, I feel that in any model of set theory which is actually like our intuitive notion of set theory, AC should be true.
Similarly, at this thead, one constructed a model of arithmetic (without induction) in which $\pi$ is rational. However, I know the integers very well, and even though this model satisfied the axioms of the integers, these were intuitively clearly not the integers.
My question is, I feel that with my of these independence proofs, if you precisely identify the notion we're talking about (like the integers, set theory), then these pathological models don't exist. Maybe it means these axioms aren't sufficient - are there any better sets of axioms? Or maybe it means that we should be focusing on particular models rather than theories in general (as in, a different philosophy of doing mathematical logic)? I'm trying to understand whether there's a way to precise-ify things so that any independence proof we do really shows that something is independent of the actual thing we're considering (not some set of axioms which happen to conform to that thing). This is guided in part by the intuition that if we really know which mathematical object (or collection of objects) we are talking about, then in some sense, any statement should simply either be true or false.

REPLY [25 votes]: I'm inclined to agree that "if you precisely identify the notion we're talking about (like the integers, set theory), then these pathological models don't exist."  The problem is that it's not so easy to precisely identify such structures.  
The usual approach to precisely identifying a structure is to write down its essential properties, the axioms governing it.  To make use of such axioms, we need to derive consequences from them, and here we find ourselves in a dilemma.  On the one hand, there is a perfectly clear notion of logical deduction in the context of first-order logic.  On the other hand, the L"owenheim-Skolem-Tarski theorem guarantees that, in the context of first-order logic, there will be unintended models of the axioms (as long as the intended model was infinite).  So first-order logic does not accomplish what is wanted.
So let's use second- (or higher-)order logic insterad.   (Here you can quantify over subsets of the structure, and those quantified variables are assumed to really range over all subsets, not just, say, definable ones.)  Now structures like the integers and the reals can be uniquely specified.  But there is no complete deductive system for second-order logic.  (More precisely, the set of valid second-order sentences is not recursively enumerable.)  Furthermore, the intended meaning of second-order quantifiers depends on the general notion of "set," which is one of the concepts that we were hoping to precisely specify.
So the bottom line, in my opinion, is that, while we might want to build mathematics on the basis of unique specifications of the relevant structures, it simply can't be done, at least not if we want the specifications to contain actual information about the structures (as opposed to just saying "I mean the genuine integers, you know") and to be able to deduce the logical consequences of that information.

REPLY [19 votes]: You have run into one of the main themes of contemporary logic: the difference between "truth in the standard model" and "provability". This is an extremely deep issue, so I'm sure other people will also have something to say about it. 
The difficulty with focusing only on standard models rather than on theories in general is that essentially the only way to convince someone else that the standard model has some property is to prove it, and then you're back to the problem of choosing axioms. For example, the reason you know π is transcendental is because you recognize certain axioms that are true in the standard model and which allow you to prove π is transcendental.  If someone else did not already believe π is transcendental, you would try to convince them by getting them to accept the axioms you used to prove it. 
In some cases, we can make a set of axioms that completely describes a standard model. For example, there are complete axiomatizations of Euclidean geometry, which allow you to prove any statement in the language of geometry that is true about the standard Euclidean plane model and disprove any statement false on it.
But for other models, like the standard model of the natural numbers, there are theorems that show we can never find an effective, complete, consistent axiomatization. The axiom systems we use to study these models are called "essentially incomplete". For these models, it isn't clear in what sense you could make the underlying concept (e.g. "natural number") precise enough to eliminate independence results. 
That being said, one nice property of models in classical logic is that any sentence in the language of the model is either true in the model or false in the model. So there is no concept of independence from a model. The downside is that for models like the standard natural numbers, it takes stronger and stronger axiom systems to determine more and more true statements about the model.<|endoftext|>
TITLE: Are finite colimits of topological spaces stable under pull-back? 
QUESTION [6 upvotes]: The category of topological spaces has a forgetful functor to set which commutes with both small limits and colimits (it has both a left and a right adjoint). Moreover Set is a Grothendieck topos and in any Grothendieck topos colimits are stable under pull-backs in the sense that
$$ colim_\alpha (X_\alpha \times_Y Z) \cong (colim_\alpha X_\alpha) \times_Y Z.$$
I'm not an expert, but I don't think the topological spaces form a Grothendieck topos. I'd be happy to learn the contrary. But this raises the question: Does this formula hold in topological spaces? Notice that because this equation holds in Set this is just a question about two topologies on the same set. 
We could also ask some related questions:

If this doesn't hold generally, then does it hold if we assume the map $Z \to Y$ is open?
What if $Z \to Y$ and all the maps $X_\alpha \to Y$ are open?
What if we assume something nice about our spaces? For example suppose they are sober? or compactly generated? or Hausdorff?

REPLY [9 votes]: In compactly generated weak hausdorf spaces (CGWH), proclusions are compatible with base change.  (A proclusion $f:X\to Y$ is a map of topological spaces which is homeomorphic to a quotient map.)  That is, if $f:X\to Y$ is a proclusion between CGWH-spaces, and $B\to Y$ any map between CGWH-spaces, then $A=X\times_Y B\to B$ is also a proclusion, where the fiber product is taken in CGWH.  (Note that $A$ is set theoretically the fiber product, with the topology obtained by replacing the product topology with the closest compactly generated one.)
So in CGWH, quotients are always compatible with base change.
In Top, a colimit is always be computed as a quotient of a coproduct.  Unfortunately, this is not true in CGWH; the colimit in Top of a diagram of CGWH-spaces will be compactly generated, but not necessarily weak hausdorf.  
There are some results which will tell you that a certain colimit in Top of CGWH-spaces is also the CGWH-colimit.  For instance, given $f:X\to Y$ and $g:X\to Z$ maps of CGWH-spaces, if $f$ is the inclusion of a closed subspace, then the Top-pushout of $f$ along $g$ is actually in CGWH.  Since the pushout is a quotient of the coproduct of $Y$ and $Z$, you can see that this particular pushout is compatible with fiber products, by the proclusion result I mentioned above.
If you're looking for a counterexample in CGWH, I'd look for a Top-pushout of CGWH-spaces which is not weak hausdorf.  I don't have a particular one handy, though I'm sure there are many.
Most or all of the above comes from Gaunce Lewis's thesis, though I don't have it at hand to give precise references.<|endoftext|>
TITLE: Dirichlet's regulator vs Beilinson's regulator
QUESTION [14 upvotes]: Consider a number field $F$ with ring of integers $O_F$. The Beilinson regulator can in this particular setting be viewed as a map from $K_n(O_F)$ to a suitable real vector space. Here $n$ is any positive odd integer. For $n \geq 3$, there is also another regulator map, defined by Borel, and Burgos has proved that the Borel regulator (suitably normalized) is twice Beilinson's regulator. For $n=1$, the Borel regulator is not defined (as far as I understand, correct me if I'm wrong), but we do have the original and most basic example of a regulator, which is that of Dirichlet.
Question: Is there some form of comparison theorem between the Beilinson regulator and the Dirichlet regulator?

REPLY [2 votes]: As you comment, the Beilinson regulator is defined using Chern classes for Deligne cohomology.  In particular, for $K_1(\mathbb C)$, the first Chern class induces the identity from $K_1(\mathbb C)\simeq \mathbb C^* $ to $H^1_{Deligne}(pt, \mathbb Z(1))\simeq \mathbb C^*$.  One passes to  $H^1_{Deligne}(pt, \mathbb R(1))\simeq \mathbb R$ by taking the logarithm.  So the Belinson regulator for $K_1$ is induced by the map $\log |\; |:K_1(\mathbb C)\to \mathbb R$.<|endoftext|>
TITLE: What was the relative importance of FLT vs. higher reciprocity laws in Kummer's invention of algebraic number theory?
QUESTION [44 upvotes]: This question is inspired in part by this answer of Bill Dubuque, in which he remarks that the fairly common belief that Kummer was motivated by FLT to develop his theory of cyclotomic number fields is essentially unfounded, and that Kummer was instead movitated by the desire to formulate and prove general higher reciprocity laws.
My own (not particular well-informed) undestanding is that the problem of higher reciprocity laws was indeed one of Kummer's substantial motivations; after all, this problem is a direct outgrowth of the work of Gauss, Eisenstein, and Jacobi (and others?) in number theory.  However, Kummer did also work on FLT, so he must have regarded it to be of some importance (i.e. important enough to work on).  
Is there a consensus view on the role of FLT as motivating factor in Kummer's work?  Was his work on it an afterthought, something that he saw was possible using all the machinery he had developed to study higher reciprocity laws?  Or did he place more importance on it than that?
(Am I right in also thinking that there were prizes attached to its solution which could also
have played a role in directing his attention to it?  If so, did they actually play any such role?)

REPLY [12 votes]: I too defer to Franz's deep knowledge on such topics. Everything that I have read in both the primary and secondary literature completely agrees with what Franz has written here and elsewhere. Besides the link I gave to the discussion in his interesting paper on Jacobi and Kummer's ideal numbers, one may also find helpful the following passage from p. 15 of his beautiful book on reciprocity laws. It provides a concise summary of what we currently know about such matters. I quote it below since some readers may not have convenient access to the book.

The role of Fermat's Last Theorem in the development of algebraic
  number theory is often overrated, probably due to Hensel's (false)
  story claiming Kummer's first manuscript to have been an incorrect
  proof of this problem.  The crème de la crème of French mathematicians
  - Lame, Legendre, Liouville, and Cauchy - tried their luck but didn't
  really advance algebraic number theory during their work on FLT.
  Gauss did not value it very highly, but admits that it made him take
  up his investigations in number theory again: in a letter to Olbers
  from March 21, 1816, he writes

I admit, that Fermat's Theorem, as an isolated result, has little
    interest for me, since I can easily make a lot of such claims that
    can be neither proved nor disproved. Nevertheless it made me take up
    again some old ideas about a large extension of the higher arithmetic. 
    [...] Yet I am convinced, if luck should do more than I may expect
    and if I succeed in making some of the main steps in that theory,
    then Fermat's Theorem will appear as one of the less interesting
    corollaries.

Gauss's last remark clearly indicates that he was at least thinking
  about the arithmetic in cyclotomic number fields ${\mathbb Q}(\zeta_p)$, even
  when, in a letter to Bessel a few months later, he reveals that the
  investigations in question had to do with the part that he eventually
  would publish, namely the theory of biquadratic residues. Parts of
  his research were published in 1828 [272] and 1832 [273], and the
  last paper contains the statement that cubic reciprocity is best
  described in ${\mathbb Z}[\rho]$, where $\rho^2 + \rho + 1 = 0$, and that, more generally,
  the study of higher reciprocity laws should be done after adjoining
  higher roots of unity.
Even Kummer, who is responsible for the greatest advance towards of
  Fermat's Last Theorem before the recent developments, got the main
  motivation for studying cyclotomic fields from his desire to find a
  general reciprocity law (which he called his "main enemy" in [Ku, Feb
  25, 1848]). In almost every letter to Kronecker written between 1842
  and 1848, Kummer mentions results related to reciprocity; the Fermat
  equation is mentioned for the first time in [Ku, Apr. 02, 1847]. In
  [Ku, Sept. 17, 1849] he informs Kronecker about the prize of 3000
  Francs that the French Academy had offered to pay for a solution of
  Fermat's Last Theorem, and in [Ku, Jan. 14, 1850] he writes

Once I will have fathomed whether this is so or not, then I will drop
    the avaricious plans and work again only for the science, especially
    for the reciprocity laws for which I have already envisaged some ideas.

It is therefore safe to say that it was the quest for higher
  reciprocity laws that made Kummer study abelian extensions of $\mathbb Q$,
  Eisenstein those of ${\mathbb Q}(i)$ and Hilbert those of general number fields.
  Ironically, it was Hilbert himself who started the rumour that it was
  FLT that led Kummer to his ideal numbers: in his famous address at
  the ICM in Paris 1900, he wrote

stimulated by Fermat's problem, Kummer arrived at the introduction of
    his ideal numbers and discovered the theorem of unique factorization
    of the integers of cyclotomic fields into ideal prime factors.

Already in 1910, Hensel talked about "incontestable evidence" (actually it 
  was something that Gundelfinger had heard from H.G. Grassmann) for the 
  existence of a manuscript in which Kummer had claimed to have solved 
  Fermat's Last Theorem, a rumour eventually dismissed by Edwards [Ed1, Ed2] 
  and Neumann [Neu].<|endoftext|>
TITLE: "inversion" of a convolution
QUESTION [6 upvotes]: I have the following relation:
$$
    \sum_{d|n} (1+1/x)^{d-1} F_{n/d}(x^d)=L_n(x)
$$
where the right hand side is (for every $n$) a polynomial in $x$, which I have an expression for, but it's not extremely beautiful.  The family of polynomials $F_k(x)$ is unknown, and is what I'm looking for.
Since this is close to Dirichlet convolution, I have not quite given up hope that there is something similar to Möbius inversion, that would give me $F_k(x)$ explicitely.  Is this possible?  Related instances of such a problem may also be interesting.
A possibly weaker, but still sufficient solution would be an expression in terms of $L_k$ and $R_k$ of the expression
$$
    \sum_{d|n} R_d(x) F_{n/d}(x^d)
$$
where $R_k(x)$ is another family of polynomials, which is also unknown.

REPLY [2 votes]: At least computing $F_k(x)$ turned out not to be that hard after
all.  Slightly more generally, consider
$$
\sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x),
$$
with $Z_1(x)=1$.
Then we have
$$
F_n(x)=\sum_{1=d_0|d_1|\dots|d_k|n}L_{n/d_k}(x^{d_k}) (-1)^k\prod_{i=0}^{k-1} Z_{d_{i+1}/d_i}(x^{d_i}),
$$
where in the sum $d_0 < d_1 < \dots < d_k \leq n$.  In other words, we are summing over all chains starting at $1$, below $n$. The formula is easily shown by induction.<|endoftext|>
TITLE: A morphism from proper to affine is constant?
QUESTION [6 upvotes]: Let $S$ be a base scheme and $f \colon X \to S$ and $Y \to S$ finitely presented  morphisms. Suppose that $g$ is affine and $f$ is faithfully flat and separated with connected reduced geometric fibers. Also, suppose that $f$ is proper over a (topologically) dense subscheme $U$ of $S$. Is it true that every morphism $g \colon X \to Y$ over $S$ is constant (i.e., factors through a section $S \to Y$)?

REPLY [2 votes]: This is essentially the proof of the rigidity theorem. As Brian says, it is enough to show that $O_S \rightarrow f_ {\ast}O_X$ is an isomorphism. If $p \in S$, then $\mathrm{H}^0(X_p, \mathcal{O}_{X_p}) = k(p)$, because the fibers are geometrically connected and geometrically reduced. Then by Grothendieck's base change theorem, $O_S \rightarrow f_ {\ast}O_X$ is in fact an isomorphism.
Edit: This does not answer Behrang's question, in view of his first comment (which should be really incorporated in the question).<|endoftext|>
TITLE: Correct to characterise NP set as P-time image of P set?
QUESTION [14 upvotes]: [I'm not familiar with the terminology, so when I write P (resp. NP) set, I mean a subset of the integers whose membership function is a decision problem in P (resp. NP).]
Is it correct to say that a set is NP if and only if it is the image of a P set under a polynomial time function?
This seems fairly clear to me: the P set acts as the collection of verification certificates for the elements of the NP set.  I'm not particularly familiar with the field though, so I wanted to check.

REPLY [17 votes]: No. In particular, the images of sets in $P$ under polynomial time reductions contain many more sets than $NP$. Below I show that the halting problem is among them.
Let {$M_i$} be a list of all Turing machines. 
Let $f$ be a mapping from {$M_i$} $\times$ {0,1}*  $\times$ {0,1}* to {$M_i$} $\times$ {0,1}* , such that

If $H \in~$ {0,1}* is a valid computation history that witnesses $M_i$ on $x$ halt, then $f(M_i,x,H) = (M_i,x)$. More precisely, a configuration of $M$ on $x$ is a bit string encoding the entire status of the computation at some step of $M$ on $x$: it encodes the currrent head position, the current non-blank tape contents (finite), and the current state of $M$. $H$ encodes a chronological list of every configuration of $M$ on $x$, starting from its initial configuration to its halting configuration. ($H$ could be very, very long, but it is finite, provided that $M$ halts on $x$.)
Otherwise, $f(M_i,x,H) = (M,0)$ where $M$ is a fixed Turing machine that halts on input $0$. We can hard-code such an $M$ into a polynomial time computable function, as it is of constant length.

Observe that $f$ is polynomial time computable! If $H$ is actually valid, then we have all the information we need to determine if $M_i(x)$ halts: the length of $H$ is at least the number of steps needed for $M_i(x)$ to halt. If $H$ isn't valid, then we can determine this by finding two adjacent configurations on the list such that the "next" configuration doesn't follow from the "previous" configuration. Each of these have length at most $|H|$ and checking the property amounts to checking a constant number of bits in each configuration: since $M_i$ is a Turing machine, at most a constant number of bits of the tape (and its current state) are manipulated in every step.
However, it is clear that the image of this polynomial time mapping is precisely the halting problem: the list of all pairs $(M_i,x)$ such that $M_i$ halts on $x$.
The gap in Joel's proof in the converse direction: the problem is that $a$ may be much much smaller in length than $b$.<|endoftext|>
TITLE: What is Realistic Mathematics?
QUESTION [43 upvotes]: This post is partially about opinions and partially about more precise mathematical questions. Most of this post is not as formal as a precise mathematical question. However, I hope that most readers will understand this post and the nature of the question.
I will first try to explain what I would call Realistic Mathematics. Let us say that mathematics is about the formalization, organization and expression of thought.
At the same time one could have the feeling that thought is usually trying to capture some aspect of physical reality; and let that be anything from a feeling, an impression of something, to experimental data of an experiment, or the observed geometric properties of lines and points in two-dimensional space. Of course thought itself (as described above) can be about anything and hence anything axiomatizable could then be seen as mathematics (that is David Hilbert's point of view). At the other hand if thought is primarily about physical reality then the focus of mathematics should be more restrictive. (I remember that Arnold argued in favor of this view; also von Neumann.) Let us call this restricted part of mathematics for the moment Realistic Mathematics. I am not saying that such a restriction of focus would be good or necessary; and I do not want to start a discussion about this. I just want to find out and discuss, whether mathematicians could agree on what we could call Realistic Mathematics. Let us suppose for a moment that we made some sense of the concept of Realistic Mathematics, and observed that it is a science that is about some part or aspect of physical reality. My naive question is now: 

Question: What is Realistic Mathematics about from a mathematical (or model theoretic) perspective?

or

Question: Is there any mathematical structure which serves as the part of mathematics which is about physical reality?

Just to give some examples: I consider anything related to finite-dimensional geometry (manifolds, simplicial complexes, convex sets etc.), number theory, operators or algebras of operators on separable Hilbert spaces, differential equations, discrete geometry, combinatorics (under countability assumptions) etc. as being part of observed physical reality or potentially useful for the study of physical reality. At the other side, existence of large cardinals, non-measurable subsets of the reals, etc. are not (immediately) useful for such a study. In particular, my view is that the Axiom of Choice does not add anything to the understanding of physical reality. It produces highly counter-intuitive statements (not observed in nature) about subsets of finite-dimensional euclidean space and has its merits (in Realistic Mathematics) only through short proofs and the knowledge that many statements are provable in ZFC if and only they are provable with more realistic assumptions.) 
What about $L(R)$? (See Wikipedia for definitions.) That would be a concrete model and maybe a realistic mathematician just studies properties of $L(R)$? Maybe a realistic mathematician studies what can be proved using $ZF + DC$? Is there any other canonical candidate which arises? My question here is mainly about opinions or some sort of vision which explains why this or that model or object of study arises naturally.

Question: Is there any mathematical application to the study of physical reality which is not captured by the study of the model $L(R)$?

More specifically: What about concrete statements which are undecidable in $ZF$? Does the Continuum Hypothesis belong to the statements we want to be true in Realistic Mathematics? What about the Open Coloring Axiom? Here, I am also asking for opinions or some consistent perspective on the realistic part of mathematics which captures my imprecise way of describing it.

REPLY [2 votes]: Update: Having read below a second time, I think you should change your usage from 'realistic mathematics' to 'empirical mathematics'. I think this is much more apt, plus it has the advantage that everyone can give it their own interpretation, which I suspect is about as much as you can hope for.

You mention Von Neumann. Actually he wrote:
"…mathematical ideas originate in empirics, although the genealogy is sometimes long and obscure. But, once they are so conceived, the subject begins to live a peculiar life of its own and is better compared to a creative one, governed by almost entirely aesthetically motivations, than to anything else and, in particular, to an empirical science. There is, however, a further point which, I believe, needs stressing. As a mathematical discipline travels far from its empirical source, or still more, if it is a second and third generation only indirectly inspired by ideas coming from ‘reality’, it is beset with very grave dangers. It becomes more and more purely aestheticising, more and more purely l’art pour l’art. This need not be bad, if the field is surrounded by correlated subjects, which still have closer empirical connections, or if the discipline is under the influence of men with exceptionally well-developed taste. But there is a grave danger that the subject will develop along the line of least resistance, that the stream, so for from its source, will separate into a multitude of insignificant branches, and that the discipline will become a disorganised mass of details and complexities. In other words, at a great distance from its empirical source, or after much ‘abstract’ inbreeding, a mathematical subject is in danger of degeneration."
I not sure I agree with all of this (as much as I am a judge), especially the tone. But I thought it was relevant.<|endoftext|>
TITLE: Are topological manifolds homotopy equivalent to smooth manifolds?
QUESTION [23 upvotes]: There exist topological manifolds which don't admit a smooth structure in dimensions > 3, but I haven't seen much discussion on homotopy type.  It seems much more reasonable that we can find a smooth manifold (of the same dimension) homotopy equivalent to a given topological manifold.  Is this true, or is there a counterexample?

REPLY [23 votes]: For every $n\ge 4$ there exists a closed aspherical  topological $n$-manifold $N$ which is not homotopy-equivalent to a PL manifold. Furthermore, $\pi_1(N)$ is a CAT(0) group. This is a theorem of Davis and Januszkiewicz, see theorems 5a1, 5b1 in their "Hyperbolziation of polyhedra" paper http://intlpress.com/JDG/archive/1991/34-2-347.pdf The construction relies heavily on Freedman's result about E8-manifold. As Vitali said, this is an old post, but it is good to have an answer that works in all dimensions since dimension 4 is exceptional in many ways.<|endoftext|>
TITLE: sheaves of modules on an $\ell$-space
QUESTION [5 upvotes]: Let $X$ be a Hausdorff, locally compact, and totally disconnected topological space, which I call an $\ell$-space, and write $A = C^{\infty}_C(X)$ for the algebra of locally constant complex-valued functions on $X$ with compact support (under pointwise multiplication). The algebra $A$ is nonunital when $X$ is noncompact, so let's look at $A$-modules $M$ which are nondegenerate in the sense that $A \cdot M = M$. Now there is an equivalence between the category of nondegenerate $A$-modules and the category of sheaves of complex vector spaces on $M$, which brings me to my question: can anyone improve on the following construction of this equivalence? I know this is vague, but I would very much like to have a slightly more concrete or "geometrical" equivalence.
As I understand it, given a nondegenerate $A$-module $M$, we start by forming the constant sheaf $[M]$ of locally constant $M$-valued functions on $X$. Denote by $\mathcal{A}$ the sheaf of locally constant complex-valued functions on $X$, so that $\mathcal{A}$ has the structure of a sheaf of $[A]$-modules (if this is as confusing to you as it was to me, notice that a section of $[A]$ can be "evaluated twice" at a point in its domain to give a complex number). Finally we define a sheaf of vector spaces $\widetilde{M} = \mathcal{A} \otimes_{[A]} [M]$, at which point I claim $M \mapsto \widetilde{M}$ is the desired equivalence. Does this seem unnecessarily complicated? Can anyone think of a way to streamline this construction, or for that matter any alternative construction?
By way of motivation, maybe I should mention that this stuff is useful in the study of smooth (sometimes called algebraic) representations of $\ell$-groups (topological groups whose underlying space is an $\ell$-space), such as $p$-adic linear algebraic groups. If you have access to it, Rodier has a nice paper called "Decomposition Spectrale des Representations Lisses" which is unfortunately behind a paywall on Springerlink.

REPLY [2 votes]: You probably know about this already, but this is discussed in the great paper of Bernstein and Zelevinsky, "Representations of the group GL(n, F), where F is a non-Archimedean local field" to some extent.  They define an "l-sheaf", which is a sort of intermediate concept between sheaves (which make sense on any space) and nondegenerate $A$-modules.  It consists of the following data:

Stalks (vector spaces) $\mathcal{F}_x$ at every point $x$ and a vector space $\Gamma(X, \mathcal{F})$ of global sections, consisting of functions $s$ with $s(x) \in \mathcal{F}_x$, such that:
If you have a function which is not necessarily one of these distinguished cross-sections, but which is locally equal to cross sections everywhere on $X$, then it is in fact in $\Gamma(X, \mathcal{F})$;
If some $s \in \Gamma(X, \mathcal{F})$ vanishes at $x$, then it vanishes in a neighborhood of $x$;
The evaluation maps $\Gamma(X, \mathcal{F}) \to \mathcal{F}_x$ are surjective.

Since $X$ is an l-space, in fact this data is enough to reconstruct an entire sheaf as it is usually defined, so this is nothing really new.  Proposition 1.14 of that paper asserts that an l-sheaf is "the same" as a nondegenerate $A$-module, in that every such module is obtained as the space of compactly supported global sections of an l-sheaf.<|endoftext|>
TITLE: (How) is category theory actually useful in actual physics?
QUESTION [50 upvotes]: An answer to a recent question motivated the following question:

(how) is category theory actually
  useful in actual physics?

By "actual physics" I mean to refer to areas where the underlying theoretical principle has solid if not conclusive experimental justification, thus ruling out not only string theory (at least for the moment) but also everything I could notice on this nLab page (though it is possible that I missed something).
Note that I do not ask (e.g.) whether or not category theory has been used in connection with hypothetical models in physics. I've read Baez' blog from time to time over the decades and have already demonstrated knowledge of the existence of the nLab. I am dimly aware of stuff like (e.g.) the connection between between Hopf algebras and renormalization, but I have yet to encounter something that seems like it has a nontrivial category theoretic-component and cannot be expressed in some other more "traditional" language.
Note finally that I am ignorant of category theory beyond the words "morphism" and "functor" and (in my youth) "direct limit". So answers that take this into account are particularly welcome.

REPLY [4 votes]: Monoidal category theory (especially dagger-compact categories) and its associated string diagram calculi are a very useful language, especially for quantum mechanics, quantum computing, and QFT. See this nice article by John Baez & Mike Stay for some of the details. It seems that quite a good deal of basic principles in quantum mechanics, such as the no-cloning theorem are really just statements about monoidal categories. And Feynman diagrams are essentially string diagrams for monoidal categories of representations.
Topos-theoretic QFT is a thing as well, though I honestly don't know anything about this approach.<|endoftext|>
TITLE: Elementary problem about triangles inside a convex polygon
QUESTION [6 upvotes]: Let P be a convex polygon with area A(P), and to each side of P, attach the largest area triangle possible that lies entirely within P. Must the sum S(P) of the areas of these triangles always satisfy $S(P) \geq 2A(P)$? If P is a rectangle, then $S(P) = 2A(P)$.
Supposing this is true, is there a lower bound with some parameter that would detect, for instance, that for P a triangle, $S(P) = 3A(P)$?
A friend mentioned this question to me several months ago, and it has been bothering me (and several other people) since. I've heard claims of a Fourier-analytic proof, so I believe it isn't open. On the other hand, it sounds like a problem from an IMO or Putnam competition, and it feels like it ought to have a clean solution.

REPLY [5 votes]: (This should be a comment, but I can't comment yet)
Indeed, the problem is from an IMO. You can look at the solutions posted here<|endoftext|>
TITLE: Why are divisible abelian groups important?
QUESTION [8 upvotes]: I just quote wikipedia:
"Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups."
I am asking for detail explanation of this application (i.e. concrete theorems, methods and so forth), or/and sources on that issue. Are there some other applications of DAG besides abelian groups theory or even besides algebra? 
Thanks in advance.

REPLY [13 votes]: In my work I come upon divisible abelian groups all the time, especially those of the form $\mathbb Q_p/\mathbb Z_p$, or direct sums of such groups.  One frequently uses the fact that they are injective (to conclude that certain sequences of Hom groups, which a priori would only be left exact, are in fact exact on the right as well), and other properties as well; for example, groups of type mentioned above, related to $\mathbb Q_p/\mathbb Z_p$, have non-zero $p$-adic Tate modules, which can often be useful.  
This last property is related to the role that divisible groups play in duality theory (in the sense of Pontrjagin duality), and this is another reason that divisible groups are important.
I think I probably speak for a lot of algebraic number theorists when I say that a lot of what we do is a kind of applied algebra.  We don't care about algebraic properties of structures for their own sake, but in our work we encounter lots of different groups, rings, and modules, and we like to understand their properties as well as we can, so that we can use them to gain control of the number theoretic computations that we are trying to make.  Divisibility of an abelian group is one such useful property, which we are trained to recognize and exploit.<|endoftext|>
TITLE: A language complete for NP intersection co-NP
QUESTION [13 upvotes]: Hi,
Is there any language $L$ know to be complete for $NP \cap co-NP$, i.e. any language $L^{\prime} \in NP\cap co-NP$ reduces in polynomial-time to $L$ and it is known that $L\in NP\cap co-NP$?
Thanks

REPLY [21 votes]: $NP \cap coNP$ is not known to have complete languages, but I don't know of any consequences as strong as what Marcos claims. Juris Hartmanis and his students worked extensively on this problem in the early 80's. Two references I know are:

Michael Sipser: On Relativization and the Existence of Complete Sets. ICALP 1982: 523-531

This paper shows that there is an oracle relative to which $NP \cap coNP$ does not have complete sets. So proving that it does have complete sets would at least require non-relativizing techniques. Also, as mentioned by Peter in Marcos' comment, there's also Hartmanis and Immerman's work:

Juris Hartmanis, Neil Immerman: On Complete Problems for NP$\cap$CoNP. ICALP 1985: 250-259

They give several interesting structural results about the problem. For one, they complement Sipser's result, giving an oracle relative to which $NP \cap coNP$ has complete languages yet $P \neq NP \cap coNP \neq NP$. They also show that $NP \cap coNP$ has complete languages under many-one reductions iff it has complete languages under Turing reductions. This paper also cites the following neat characterization by Kowalczyk (1985): $NP \cap coNP$ has a complete language iff there is a recursively enumerable list of pairs of $NP$ machines {$(N_{i,1}, N_{i,2})$} such that $\overline{L(N_{i,1})} = L(N_{i,2})$ and $\bigcup_i L(N_i) = NP \cap coNP$.
But since then, there hasn't been much progress on the question, to my knowledge. I'd be very happy if I were corrected...<|endoftext|>
TITLE: Characterizations of complex Abelian varieties (especially 3-folds) among projective nonsingular varieties?
QUESTION [16 upvotes]: If $X$ is a complex Abelian variety of dimension $g$, then 

The canonical sheaf is trivial
$\dim {\rm H}^i(X; \mathcal{O}_X) = \binom{g}{i}$.

When $g =1,2$, then any connected, projective nonsingular $X$ satisfying the above two must be an Abelian variety. Is this true for higher $g$? If not, what other conditions can I add? Or is such a request unreasonable?
Disclaimer: I don't know very much about Abelian varieties, so apologies if this material is standard. A search in the literature turned up some papers about characterizations of Abelian varieties up to birational equivalence, but under weaker assumptions. I really want to know if I'm given a variety $X$, how to tell if it is Abelian or not via some sort of reasonably accessible sheaf-related conditions. I'm most interested in the case $g=3$, but results for other $g$ are welcome also.

REPLY [22 votes]: A result of Kawamata (Kawamata, Yujiro, Characterization of abelian varieties. Compositio Mathematica, 43 no. 2 (1981), p. 253-276) implies that, under your assumptions, $X$ is birational to an abelian variety (in fact you just need the Kodaira dimension of $X$ to be zero and the irregularity to be equal to the dimension of $X$).
Once you know that $X$ is birational to an abelian variety $A$, a Lemma of Deligne implies that if the canonical divisor on $X$ is trivial, then $X$ is in fact an abelian variety. This is not a particularly deep result.  First, the rational map $f \colon X \to A$ is actually a morphism (essentially because $A$ cannot contain rational curves). Second, the morphism $f$ induces a morphism $df$ between the cotangent bundles.  The determinant of $df$ is a morphism between the canonical bundles of $A$ and $X$, that are both trivial by assumption. Thus the determinant is either identically zero, or it is an isomorphism. Since the morphism $f$ is generically etale, the determinant is not identically zero. But then it is an isomorphism, so that the morphism $f$ is always etale, and we conclude that $X$ is an abelian variety.
EDIT: Ah, as Pete remarked below, I did not answer the question! The answer is "Yes"!  Even under weaker assumption: namely it suffices to know that the canonical bundle is trivial and that $\dim (X) = h^1(X,\mathcal{O}_X)$.<|endoftext|>
TITLE: Minimizing variance of distances between points when mean distance is fixed
QUESTION [5 upvotes]: In Rd, I have n > d+1 points. The mean distance between pairs of points is 1. How can I minimize the variance of the distances (equivalently, the mean squared distance)? I'm mainly interested in d ∈ {1,2,3} although I'd be curious if there were any patterns for larger values of d. As for values of n, I'm interested both in specific solutions for small values, and general patterns in larger values.

REPLY [2 votes]: For $d=1$, it is minimized by arithmetic progression.<|endoftext|>
TITLE: Conjugacy classes and reduced group $C^*$-algebra of an amenable group
QUESTION [9 upvotes]: The reduced $C^*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star_{red} G$. On the other side, for the group ${\mathbb Z}$, separation is clearly possible.

Question: Let $G$ be an amenable group. Does the reduced group $C^\star$-algebra of $G$ support sufficiently many traces to distinguish between conjugacy classes of group elements?

EDIT: The question seems already interesting for $S_{\infty} = \cup_n S_n$. Let's get explicit and pick $g \in S_n$ (for some $n$) and consider the canonical trace $\tau_{g,n}$ which sends every conjugate of $g$ to $1$ and all other elements to zero. (This can be done for any $n' \geq n$ since $S_n \subset S_{n'}$.) The function $\tau_{g,n} \colon {\mathbb C}S_n \to {\mathbb C}$ is a conjugation invariant function and hence, it must be a linear combination of the normalized traces of irreducible representations of $S_n$. 

Question: What is the sum of the absolute values of the coefficients that come up in this linear combination of traces? 

This is (as one can check) the norm of $\tau_{g,n}$, call it $c(g,n)$. So, we see that the compatible family of maps $\tau_{g,n}$ extends from ${\mathbb C}S_{\infty}$ to $C^* S_{\infty}$ if and only if $c(g,n)$ remains bounded.

REPLY [5 votes]: There is an easy example, namely $SL_3(F)$ where $F$ is the algebraic closure of some finite field. This group does not admit non-trivial characters (a result of Kirillov) and is locally finite, hence amenable. This gives a negative answer to the first question.
Sorry, I deleted my previous answer since it contained a mistake.<|endoftext|>
TITLE: Reconstruction Conjecture: Group theoretic formulation
QUESTION [6 upvotes]: As we read from wiki, informally, the reconstruction conjecture in graph theory says that graphs are determined uniquely by their subgraphs.

Is there a group-theoretic formulation of this conjecture?
Has an analogous conjecture been made in group theory(in any sensible way)?

References such as books will do, thanks.

REPLY [8 votes]: There are continuum 2-generated groups where all proper subgroups are cyclic of order $p$ (for the same prime $p\sim 10^{70}$), the Tarski monsters. All these groups have the same lists of proper subgroups, and the same lattice of subgroups. Also all cyclic groups of prime order have the same proper subgroups: the trivial group. So in general the answer is "no". But every finite non-cyclic abelian group is determined by the list of its proper subgroups, which follows easily from the description of finite abelian groups (note that I and, I think, the question, are not talking about the lattice of subgroups, but about the list of all proper subgroups). I suspect that at least for a large class of finite solvable groups, the list of proper subgroups determines the group. I would look first at A-groups, finite groups with all Sylow subgroups abelian (see the book of Huppert, "Endliche Gruppen") because the structure of subgroups of A-groups is more known, and one can use induction on the order of the group.<|endoftext|>
TITLE: Are the path components of a loop space homotopy equivalent?
QUESTION [12 upvotes]: If $X$ is a based space, then we have $\pi_1(X) \cong \pi_0(\Omega X)$.  This is to say we can identify elements in the fundamental group of $X$ with path components of the first loop space of $X$.  My question is this: do all the path components of $\Omega X$ necessarily have the same homotopy type?
I ask because when you iterate this idea by considering $\pi_2(X) \cong \pi_1(\Omega X) \cong \pi_0(\Omega^2 X)$, the space $\Omega^2 X$ only sees the component of $\Omega X$ containing the basepoint (in this case the constant map).  So when I imagine climbing up the tower of $X$'s homotopy groups, at each stage we apply the loop functor and discard all the information in the non-basepoint components.
If all the components of $\Omega X$ are homotopy equivalent, then this is clearly no loss of information, since we can identify the homotopy groups of $X$ with the (properly shifted) homotopy groups of any of the connected components.  Otherwise, don't we generate much more information about $X$ by looking at the homotopy types of all the path components of $\Omega X$, $\Omega^2 X$, etc.?
I keep thinking there might be some argument about a certain map being a fibration, and this giving an equivalence between the various path components of $\Omega X$, realized as fibers of the map, or something like this, but I haven't found it yet, and I can think of no canonical maps which might serve as the desired equivalences.

REPLY [14 votes]: Does the following work?
Let $A$ and $B$ be components of $\Omega X$ and assume that $A$ is
the component containing the trivial path based at $x_0$. Let $f$
be any element of $B$. Then $f$ is a path from $x_0$ to $x_0$.
There's a map $\phi:A\to B$ obtained by following a path in
$A$ by $f$.
Similarly there's $\psi:B\to A$ obtained by following a path by the
reverse of $f$.
Surely $\phi$ and $\psi$ are homotopy inverses of each other?
(Warning: I'm not a topologist by trade, so I fear that I may
have missed something subtle, or indeed something obvious, here).<|endoftext|>
TITLE: Logically independent but true sentences
QUESTION [20 upvotes]: My question is of a logical nature and concerns what I perceive to be two different types of mathematical independence.
Suppose we have a (sufficiently strong) axiomatic theory $T$. Gödel's Incompleteness Theorems state that:

$T$ is not a complete theory. That is, there is a sentence (expressible in the language of the theory) which is true, but not provable in the theory. In what follows, I will refer to such a sentence as a Gödel sentence and denote it by $G$.
$T$ cannot prove its own consistency. That is, assuming that $T$ is consistent, $\not\vdash_T\mathrm{Con}(T)$.

For my question to make sense, I must lay out the following principle, which I take to be "self-evident" (by which I mean that I believe most people would endorse it):
'If one is committed to a theory $T$, then one is also committed to $\mathrm{Con}(T)$.'
In other words, suppose that I accept the axioms of $PA$ (for instance). That means that I am committed to $PA$, in the sense that I believe it to be true, and therefore consistent. As such, it would be incoherent for me to disbelieve $\mathrm{Con}(PA)$.
This situation gives rise to the following state of affairs:

On the one hand, there are statements which are independent from a theory $T$, but whose truth is nevertheless implied by $T$, even though $T$ cannot prove them. This is the paradigm of the First Gödel Theorem (cited above) applied to arithmetic: in the context of $PA$, it says that there is a Gödel sentence $G$ which is not provable in $PA$, but that if $PA$ is consistent, then $G$ must nevertheless be true. Thus, if one is committed to $PA$, one is committed to $\mathrm{Con}(PA)$ (by the above principle) and therefore one is committed to the truth of $G$.
On the other hand, there are statements which are independent from a theory $T$, and in addition, no judgment regarding their truth value may be inferred from $T$. This is the paradigm of Set Theory ($T=ZFC$) and the Continuum Hypothesis ($CH$). One's commitment to $ZFC$ does not imply anything about the truth of $CH$, since both $ZFC+CH$ and $ZFC+\neg CH$ are consistent. Note that this is different from the first case, in which $PA+\neg G$ is inconsistent.

In essence, I see a dichotomy between statements which are independent from a theory $T$ and also from $\mathrm{Con}(T)$, and those which are independent from $T$ but nevertheless implied by $\mathrm{Con}(T)$. I am tempted to say that there are two types of logical independence; is such a division valid, or would anyone care to contest it?
In case this is a very well-known issue, are there any other examples (aside from Gödel sentences) of statements which are independent from a theory but provable if one assumes consistency? In particular, I am wondering if there are any "natural" such questions. (Of course, the statement $\mathrm{Con}(T)$ is itself an example, albeit a trivial one.)
Thank you!

REPLY [17 votes]: First, a minor correction to the end of the second bullet item in the question: $PA+\neg G$ is consistent (that's the same as saying that $PA$ doesn't prove $G$); what you presumably meant is that $PA+Con(PA)+\neg G$ is inconsistent.
Now concerning the actual question: What you've described is the beginning of a whole hierarchy of "commitments implicit in accepting a theory."  Your principle, that anyone accepting a theory $T$ must also accept $Con(T)$, can be iterated.  Having accepted PA, and therefore having accepted $Con(PA)$, I ought to also accept $Con(PA+Con(PA))$ and then $Con(PA+Con(PA+Con(PA)))$, etc.  In fact, this process can even be iterated transfinitely.  Ultimately, what I ought to accept (in this sense) will depend on my knowledge that certain orderings are well-orderings (so that iterating the "Con" process along those orderings is justified).  Such iterations have been studied by Turing and (later) Feferman.  If I remember correctly, if I'm willing to iterate along all computable well-orderings, this process will produce enough "Con" axioms to imply all true $\Pi^0_1$ statements about the natural numbers.
The philosophy behind your argument about being committed to $Con(T)$ seems to actually justify a stronger principle, namely that anyone who accepts $T$ should also accept the principle that anything provable in $T$ is true.  Because of the undefinability of truth, this stronger principle should be formulated as a schema, called the reflection schema: $Prov_T(\lceil\phi\rceil)\to\phi$, where $Prov_T(\lceil\phi\rceil)$ formalizes, in terms of the G"odel number \lceil\phi\rceil, the statement that $\phi$ is provable in $T$.  If I rememeber correctly, iteration of this reflection principle, along all computable well-orderings, suffices to prove all arithmetical truths.  (That doesn't contradict G"odel's incompleteness theorems, because the notion of "all computable well-orderings" isn't computable nor even arithmetically definable.)  Note, by the way, that $Con(T)$ is equivalent to the particular instance of the reflection principle where $\phi$ is a contradiction.
In light of all this, I'd say that your distinction between those statements that are independent of $T$ but provable in $T+Con(T)$ and those that are independent of $T+Con(T)$ is "valid" in the sense of being a well-defined mathematical distinction, but it is not "valid" in the sense of capturing the full import of your dictum that whoever accepts $T$ should accept $Con(T)$, because it does not take into account the possibility of iterating this dictum.  In addition, it seems to me that the reflection principle more completely captures the intuition behind your dictum.<|endoftext|>
TITLE: Modeling free Lie algebras with matrix algebras
QUESTION [5 upvotes]: I am approximating some algebraic expressions of operators from a free Lie algebra. It is possible but messy to collect all independent operator objects of a given degree (same as grading?) that appears in such expressions. These are collectively referred to as the "Hall basis". [See here ]
I would like to find a shortcut in terms of matrix algebras. In particular I was wondering if there is a matrix algebra of $n$-dimensional matrices in which any member can be written as a sum of commutators of up to degree $d$. I think $n$ will grow exponentially with $d$. 
I am trying to use such a method to streamline some calculations in the dynamics of open quantum systems were such expansions (e.g. Magnus expansion ) occur naturally but are hard to work with. I think the solution must have to do with triangular matrices but I might be missing something. Perhaps the same thing can be done with a little overhead by rephrasing in terms of products and nil-potency as opposed to commutators. Any pointers in helpful directions are much appreciated!

REPLY [2 votes]: This is not supposed to be an answer to your question, just a bunch of casual remarks.

Free Lie algebras indeed may be approximated (in some strict sense) by, say, Lie algebras $sl_n(K)$ (observing that there is no nontrivial identity satisfied by $sl_n(K)$'s for all $n$), but this appears to be not very constructive, at least I cannot see how it can buy something for kind of computations you are interested in.
In arXiv:1008.2380 , Bremner and Hu study (in a very computational-oriented way) invariants of actions of low-dimensional matrix Lie algebras on a free Lie algebra. May be this is related to what you are looking for.
There is a lot of other works dedicated to various computations in free Lie algebras, sometimes on the borderline with computer science. (Some of them are referenced in the paper by Munthe-Kaas and Owren you link to, so you are probably aware of that). Usually these computatuions are performed not in terms of matrix algebras, but, again, may be it could be useful for the sort of calculus you are interested in.
In: M. Grayson and R. Grossman, Models for free nilpotent Lie algebras, J. Algebra 35 (1990), 177-191 http://pubs.rgrossman.com/dl/journal-005.pdf , they tackle a related problem: computations in free nilpotent Lie algebras. The latters admit a represemtation by (upper-triangular) matrices - unlike in the case of free algebras this is not approximation, but a true embedding - but the authors are unsatisfied with it and pursue another one, by vector fields.<|endoftext|>
TITLE: When does iterating $z \mapsto z^2 + c$ have an exact solution?
QUESTION [44 upvotes]: If one iterates the map $z \mapsto z^2 + c$ there is obviously a simple formula for the sequence one gets if $c=0$. Less obviously, there is also a simple formula when $c = -2$ (use the identity $2 \cos(2x) = (2\cos(x))^2 - 2)$. Are there any other values of $c$ for which one can solve this recurrence explicitly? (For all initial values of course: there are many trivial explicit solutions for special initial values, such as fixed points.)
Related links:
http://en.wikipedia.org/wiki/Mandelbrot_set (the points $c$ where 0 remains bounded under iteration of this map: this strongly suggests that there is no simple exact solution for general $c$).
http://en.wikipedia.org/wiki/Logistic_map (gives the explicit solutions above, after a change of variable)
Motivation: I once used the map with $c=-2$ in a lecture to show that one could prove limits exist even without a formula for the exact solution. A first year calculus student  pointed out the non-obvious exact solution above, and I don't want to be caught out like this again.

REPLY [4 votes]: Actually, there are many other cases : if $f$ is a surjective map from $C$ to $C$ such that $f(az)=f(z)^2+c$, we get the simple formula $z_n=f(a^nZ0)$, where $f(Z_0)=z_0$. So  the question is : for which values of $a$ and $c$ does such an $f$ exists, with perhaps some more constraints  on $f$ such as smoothness. It turns out that for any $c$ large enough (at least of module greater than 1) there exists exactly one such entire function $f$, with $f'(0)=1$ and $a=2f(0)=1+\sqrt{1-4c}$ (this is shown by an elementary, but tedious calculation of the coefficients of the series by recurrence (done in French here, page 8)), followed by a simple argument of domination of those coefficients, proving convergence for $R>0$ ;  then, analytic continuation shows that the radius of convergence is infinite) ; $f$ is (almost) surjective by Picard theorem. Of course, those $f$ are never "usual" or elementary functions, except for  $c=0$ or $c=-2$, but so what?<|endoftext|>
TITLE: Realizing D_8 as a Galois group over C(x) with prescribed decomposition groups
QUESTION [12 upvotes]: Coming up with examples of $D_8$-covers of $\mathbb{C}(x)$ is easy. For example:
$Quot(\mathbb{C}(x)[y,z]/(y^2=x(x-7), z^4=(y+\sqrt{-6})^2(y-\sqrt{-6})^2(y+\sqrt{-10})(-y+\sqrt{-10})^3))$
defines a $D_8$ extension of $\mathbb{C}(x)$. If you view this field extension as an extension of the corresponding smooth projective models, it is branched at $x=0,1,2,7$.
I wish to control the inertia groups (=decomposition groups, in this case) associated to the ramification.
I'll use the presentation $D_8= \langle a,b | a^4, b^2, ab=ba^{-1} \rangle$.
Take $\mathbb{P}^1_{\mathbb{C}} \setminus 0,7,1,2$. Its fundamental group is $\langle \gamma_1, \gamma_2, \gamma_3, \gamma_4| \gamma_1\gamma_2\gamma_3\gamma_4=1 \rangle$ (where $(\gamma_1, \gamma_2, \gamma_3, \gamma_4)$ are respectively the loops from a fixed base point around the points $(0,7,1,2)$). The surjection: $(\gamma_1, \gamma_2, \gamma_3, \gamma_4)$ goes to $(b, ab, a^2, a^3)$, gives a $D_8$-Galois cover of $\mathbb{P}^1_{\mathbb{C}}$.
I wish to find the equations that define THAT $D_8$ cover. The equations I wrote in the beginning won't work because there the inertia groups of the ramification points above $x=0$ and above $x=7$ are the same groups; whereas if done through the description, since no conjugate of $b$ is equal to any conjugate of $ab$, no inertia group of any ramification point above $x=0$ will equal the inertia group of any ramification point above $x=7$.
For some reason I easily confuse myself in such computations.
If you mod out the description by $\langle a \rangle$ you see that the description of the $D_8 / \langle a \rangle$ sub-cover of the original cover, has description $(b\langle a \rangle, b\langle a \rangle, \langle a \rangle, \langle a \rangle)$, and therefore must be given by $y^2=x(x-7)$ (if it's, say, $y'^2=\frac{x}{x-7}$, then $y'=\frac{y}{x-7}$). But I can't find the right other equation that would make this work.
Note that this should be easy, because this $D_8/\langle a \rangle$-cover is a $\mathbb{P}^1_{\mathbb{C}}$ with parameter $y$, so there's no funny business. And yet I'm having some trouble finishing this example up.

REPLY [5 votes]: As you say, the fixed field by $\langle a\rangle$ is $\mathbb{C}(u)$ with $u^2=-5x/2(x-7)$, and the fixed field by $\langle a^2,b \rangle$ is similarly $\mathbb{C}(v)$ with $v^2=7(x-2)/2(x-7)$ (constants chosen so that $u^2+v^2=1$). So their compositum, the fixed field by $\langle a^2\rangle$, has genus zero and is $\mathbb{C}(t)$, where $u=2t/(t^2+1)$, $v=(t^2-1)/(t^2+1)$. This means that the last quadratic step of your tower is determined entirely by its ramification points, which are the 2 points above $x=2$ (which are the points $t=\pm 1$) and the 4 points above $x=1$, call them $t=c_1,\dots,c_4$. So your desired field is $\mathbb{C}(t,w)$ where $w^2=(t-c_1)(t-c_2)(t-c_3)/(t-c_4)(t^2-1)$.<|endoftext|>
TITLE: Point cloud that maximizes the minimum pairwise distance in Euclidean space
QUESTION [8 upvotes]: I am interested finding the collection of points in the Euclidean space that has the maximal minimal pairwise distance subject to an average norm constraint, that is, how to maximize
$min_{i \neq j} |x_i - x_j|$
subject to $\frac{1}{n} \sum_{i=1}^n |x_j|^2 \leq1$ where $\{x_1, \ldots, x_n\} \subset \mathbb{R}^d$.
I wonder if this problem has a name and what is known about it. Of course $d = 1$ is easy: just choose $n$ uniformly spaced points that satisfies the constraint with equality. I am especially interested in $d=2$. If little is known in the non-asymptotic case, maybe we know more when $n$ and/or $d$ is large? Is it related to sphere packing?
(BTW, I heard that the answer is given by vertices on the simplex when $n \leq d -1$ (or maybe the other way around?))

REPLY [2 votes]: Consider fixing $\min_{i \ne j}|x_i-x_j|=1$ and try to minimize $S=\sum_{1}^{n}|x_j|^2$. Then it is a sphere packing problem and the answer to your question would be $\frac{1}{\sqrt{S}}$<|endoftext|>
TITLE: Are there space filling curves for the Hilbert cube ?
QUESTION [12 upvotes]: There is a surjective continuous map $[0;1]\rightarrow [0;1]^2$ ("space filling curve"). Using such a map one can easily get space filling curves for all finite dimensional cubes. 
So my question is: Is there a space filling curve of the Hilbert cube 
$[0;1]\rightarrow [0;1]^\mathbb{N}$ ?

REPLY [10 votes]: Well there is indeed a "simple" construction of such a space filling curve. 
Let $\gamma:[0;1]\rightarrow [0;1]^2$ be a space filling curve. Then one can obtain a space filling curve for $[0;1]^3$ by postcomposing with $id_\mathbb{R}\times \gamma$. Then one can postcompose with $id_{\mathbb{R}^2}\times \gamma$ and so on. Note that the first coordinates didn't change in the last step. 
Putting all this together we get a map 
$f:[0;1]\rightarrow [0;1]^\mathbb{N} \qquad t\mapsto (pr_1\circ \gamma \circ (pr_2\circ \gamma)^{n-1}(t))_{n\in \mathbb{N}}, $
where $pr_i$ denote the obvious projections. This map can be seen as the infinite composition of the maps above. By the definition of the product topology this map is continuous. 
Especially if we postcompose $f$ with the projection on the first $n$ coordinates, we just get a space filling curve (see above). Let us show, that a  arbitrary element $x=(x_i)_{i\in \mathbb{N}}\in [0;1]^\mathbb{N}$ lies in the Image of $f$. We already know, that for each $n$ there is a element $y^n$ in Im$(f)$ agreeing with $x$ in the first $n$ coordinates.
As $[0;1]$ is compact, Im$(f)$ is compact and hence closed ($[0;1]^\mathbb{N}$ is Hausdorff). 
And $\lim_{n\to\infty}y^n=x$. Hence $x\in$ Im$(f)$. So $f$  is a continuous surjective map $[0;1]\rightarrow [0;1]^\mathbb{N}$.<|endoftext|>
TITLE: Maximum number of distinct diagonals generated by permutations
QUESTION [13 upvotes]: Given a matrix $A \in \{0,1\}^{n \times n}$, let $diag(A)$ be the set of vectors $D \in \{0,1\}^n$ that are the diagonal of one of the $n!$ matrices obtained from $A$ via row permutations. 

What is the maximum size of $|diag(A)|$ over all matrices $A \in \{0,1\}^{n \times n}$?

REPLY [10 votes]: The answer is $2^n -n$.
Let $I$ be the $n \times n$ identity matrix and let $D$ be a diagonal whose non-zero entries are indexed by $S$.  Further suppose that $D$ does not have exactly one zero entry. Let $\pi$ be a permutation of the rows of $I$ whose fixed points are exactly $S$.  Then the diagonal of $\pi(I)$ is $D$, and there are $2^n-n$ such diagonals.  
We finish by showing that $2^n-n$ is the best one can do.  
An $n \times n$ bipartite graph is a bipartite graph with bipartition $([n]_r, [n]_c)$, where $[n]_r$ and $[n]_c$ are both copies of $[n]$.  Let $G$ be an $n \times n$ bipartite graph.  Define $G'$ to be equivalent to $G$, if $G'$ can be obtained from $G$ by complementing the neighbourhoods of some vertices in $[n_r]$.  Note that the equivalence class of $G$, denoted $[G]$, has size $2^n$.  
It is easy to check that the following lemma proves the tightness of the bound.
Lemma. For any $n \times n$ bipartite graph $G$, at most $2^n-n$ members of $[G]$ have a perfect matching.  
Proof. For each $i \in [n]_c$ there is a graph $G^i \in [G]$ such that $i \in [n_c]$ has degree 0 in $G^i$.  Just pick the vertices in $[n]_r$ that are adjacent to $i$ in $G$ and complement their neighbourhoods.  If all $G^i$ are distinct, then the lemma clearly follows.   Otherwise, $G^i=G^j$ for some $i \neq j$.  Thus, both $i$ and $j$ have degree 0 in $G^i$.  But now, the $n$ graphs obtained from $G^i$ by performing a single complementation each do not have a perfect matching.<|endoftext|>
TITLE: Horn clauses and satisfiability
QUESTION [6 upvotes]: It is well known that satisfiability of Horn formulae can be checked in polynomial time using unit propagation.
But suppose we relax the condition for horn clauses from at most one un-negated literals to two un-negated literals. Then is it possible to prove that satisfiability of such a formula can be checked in time polynomial in the size of the formula?

REPLY [10 votes]: In the paper The complexity of satisfiability problems MR0521057, Tom Schaefer characterizes exactly which general classes of satisfiability problems are in P and which are NP-complete. Those problems which are in P fall into six cases:

Every relation in S is satisfied when all variables are 0.
Every relation in S is satisfied when all variables are 1.
Every relation in S is definable by a CNF formula in which each conjunct has at most one negated variable.
Every relation in S is definable by a CNF formula in which each conjunct has at most one unnegated variable.
Every relation in S is definable by a CNF formula having at most 2 literals in each conjunct.
Every relation in S is the set of solutions of a system of linear equation over the two-element field {0,1}.

Here, S is a set of boolean relations that one takes as primitives for the language; the associated satisfiability problem is then deciding the satisfiability of a finite conjunction of such primitives. Schaefer moreover shows that any set of relations which does not fall into one of the above has a NP-complete satisfiability problem. In your example, S would be a set of boolean relations definable by a CNF formulas in which each conjunct has at most two unnegated variables. This is not in the above list, so the corresponding satisfiability problem is NP-complete.<|endoftext|>
TITLE: Does EXP $\in$ P/poly imply NP=RP?
QUESTION [11 upvotes]: I guess the answer is that this unknown.
Maybe this implies some "lowness" result on NP relative to BPP?

REPLY [12 votes]: Note that it would suffice to prove $EXP \subseteq P/poly$ implies $NP = BPP$ (since the latter implies $NP = RP$).
I am pretty sure this is not known. $NP \neq RP$ does not seem to imply any circuit lower bounds for $EXP$.
Note that $EXP \subseteq P/poly$ does imply $P \neq NP$. If $EXP \subseteq P/poly$ then there is some fixed $k$ for which $TIME[2^n]$ has $n^k$-size circuits. Therefore $P \subseteq TIME[2^n]$ has $n^k$-size circuits. But $P = NP$ implies that $\Sigma_2 P = P$, and we know by Kannan's theorem that $\Sigma_2 P$ does not have $n^k$-size circuits for any fixed $k$ -- this is a contradiction. (This result is due to Meyer, cited in the famous Karp-Lipton paper.)
I would suspect that one can prove $EXP \subseteq P/poly$ implies $RP \neq NP$ or maybe $ZPP \neq NP$, but the above proof doesn't do the job. (Technically, since I believe $EXP \not\subseteq P/poly$, the negation should imply everything...)<|endoftext|>
TITLE: Collapsing of exptime and alternation bounded turing machine  
QUESTION [6 upvotes]: Hello
Let C be a set of function (let say time-computable increasing function to avoid pathological cases).
Let's call $\rm{ATIME}(C,j)$ the class of langages decided by a Turing Machine begining in existantial state, alternating at most $j-1$ times and whose time is bounded by a function $f\in C$.
Let suppose that $\rm{ATIME}(C,j)=\rm{ATIME}(C,j+1)$ then, can we prove that $\forall k>j \rm{ATIME}(C,j)=\rm{ATIME}(C,k)$ ? Or at least, what are the condition over $C$ ?
It is true if $C$ is the class of polynomials because $C$ is then closed under composition. I would have expected it to be true for every polynomially closed set (for example for the exponential hierarchy, a class like $C=\{2^{2^{\dots^{2^{n^{O(1)}}}}}\}$ for a bouned tower of 2.)
I guess and hope that some things are known about it, but I was not able to find any reference, neither my advisor could. 
An equivalent way to ask the question in a finite model theory setting, is:
Let HO$^{r,f}_j$ be the class of formula of order $r$ with free variable of order up to $f$ and with $j$ alternations of quantification of order $r$ beginning with an existantial quantification. 
If there exists $k>2$ such that HO$^{r,2}_j$=HO$^{r,f}_k$ then can we proove that  $\forall l>j$  HO$^{r,2}_j$=HO$^{r,2}_l$ ?

REPLY [6 votes]: Let suppose that $ATIME(C;j)=ATIME(C;j+1)$  then can we prove that $\forall~k>j$, $ATIME(C;j)=ATIME(C;k)$? Or at least, what are the condition over $C$?

Yes, when $C$ is the class of polynomial functions of $n$, or the class of linear functions of $n$. First, note that your $ATIME(C;j)$ is typically written as $\Sigma_j TIME[C(n)]$. I will use this notation as it is more standard. Moreover it distinguishes between the classes where the machine starts in a universal state instead of an existential one. The universal version is written as $\Pi_j TIME[C(n)]$.
Now, $\Sigma_{j+1} TIME[C(n)]=\Sigma_j TIME[C(n)]$ implies $\forall~k>j$, $\Sigma_k TIME[C(n)]=\Sigma_j TIME[C(n)]$, when $C(n)$ is the class of all polynomial functions of $n$, or the class of linear functions of $n$. In fact, you can get away with the weaker assumption $\Pi_j TIME[C(n)] = \Sigma_j TIME[C(n)]$ in place of $\Sigma_{j+1} TIME[C(n)]=\Sigma_j TIME[C(n)]$. This follows from standard stuff in the chapters on alternations and the polynomial hierarchy of any complexity theory book. (You probably are well aware of this, but other readers may not be. So please bear with me.) For example, one result you may see in a complexity course is $NP = coNP$ implies that the polynomial hierarchy collapses to $NP$. This is exactly the same as saying $\Pi_1 TIME[n^{O(1)}] = \Sigma_1 TIME[n^{O(1)}]$ implies $\bigcup_{k \geq 1} \Sigma_k TIME[n^{O(1)}] = \Sigma_1 TIME[n^{O(1)}]$, which is the case $j=1$ in your question.
For superpolynomial functions $C(n)$, one runs into trouble. Suppose $\Sigma_j TIME[2^{O(n)}] = \Sigma_{j+1} TIME[2^{O(n)}]$, and you want to show $\Sigma_{j+2}TIME[2^{O(n)}] \subseteq \Sigma_{j+1} TIME[2^{O(n)}]$. The usual way of doing this is to take a $\Sigma_{j+2}$ machine, and consider the language $L'$ of pairs ${(x,y)}$ with the property that, if I feed $x$ to the machine, and substitute the string $y$ in place of the guesses for the first existential mode, the remaining $\Pi_{j+1}$ computation accepts. $L'$ is in $\Pi_{j+1}$, and so you usually apply your assumption to conclude $L'$ is in $\Pi_j$, hence the whole computation is in $\Sigma_{j+1}$. But observe that this "remaining $\Pi_{j+1}$ computation" runs in polynomial time in the length of the input, since the string $y$ can be of length $2^{O(n)}$, and the remaining computation takes $2^{O(n)}$ time. So it appears you need an assumption about polynomial time alternating computation in order to get this collapse. If you apply $\Sigma_j TIME[2^{O(n)}] = \Sigma_{j+1} TIME[2^{O(n)}]$ you end up with a doubly-exponential time computation.
I don't think any alternative argument for this kind of collapse is known, which gives you what you want. If you find one, please tell me! It may have applications to separating complexity classes.<|endoftext|>
TITLE: Descriptive complexity theoretic-characterizations of P and NP
QUESTION [11 upvotes]: Prompted by Vinay Deolalikar's purported proof of P != NP, I've been reading up on Descriptive Complexity for some background material. 
The major successes of Descriptive Complexity include Fagin's result that $NP=SO\exists$ (that is, the class NP is equal to the class of models of a second-order existential query over some vocabulary), and also that $P = FO(LFP)$ (that the class P is equal to the class of models of first-order queries that might use a Least-Fixed-Point operator), and also $PH = SO$.
My understanding of mathematical logic is quite shaky, but from what I understand, second-order formulas are not expressible in first order logic - how does this fact stand in relation to the results I mentioned above? Why does it not separate NP from P, or PH from P?

REPLY [11 votes]: Remember that this is finite model theory and it is quite different from logic on infinite structures, e.g. satisfiability of first-order formulas on finite structures is $\Sigma_1$, whereas the same question is $\Pi_1$ for general structures (where the formula is satisfiable iff it does not lead to (i.e. prove) contradiction).
It is not known yet if the properties of the finite structures expressible in SO are different from those expressible in $FO(LFP)$.<|endoftext|>
TITLE: What is known about higher-categorical reconstruction theorems? (reference request)
QUESTION [10 upvotes]: The answer to my question is almost certainly "not much" — at least, I've asked a few people, and that was their answer.  But I'd like to refine this answer, and MathOverflow seems like the best place.
I learned from David Ben-Zvi in an answer to this question the following theorem:

Let $\mathcal C$ be a 1-category, and consider the category $\operatorname{Rep}(\mathcal C)$ of 1-functors $\mathcal C \to \operatorname{1Vect}$.  It is a (symmetric) monoidal category by "pointwise tensor product", i.e. pulling back along the diagonal map $\mathcal C \to \mathcal C^{\times 2}$.  Conversely, we can consider some sort of "spec" of $\operatorname{Rep}(\mathcal C)$, namely the category of monoidal functors (and monoidal natural transformations) $\operatorname{Rep}(\mathcal C) \to \operatorname{1Vect}$.  In fact, this "spec" is equivalent as a category to $\mathcal C$.

Given this, it is natural to ask the following three questions (or combinations thereof):

Recognition: which monoidal categories are of the form $\operatorname{Rep}(\mathcal C)$ for some $\mathcal C$?
Bump up $n$: modulo definitions, it is clear what the statement is with "$1$" replaced by "$n$".  For example, the "$0$" version of the above says that a set is recoverable up to isomorphism from its algebra of all functions (the 0-category $\operatorname{0Vect}$ is precisely the ground field).
Internalize: is there a similar statement for "topological categories" and "continuous functors", for example?  A version of in algebrogeometric land is in these questions (see also the answer here).

I'm not asking for definite answers to any of these directions, because I expect that telling the complete story is hard.  But I am hoping for references to the existing literature.  Hence: "What's already known (in the literature) about higher-categorical reconstruction theorems?"

REPLY [4 votes]: In his PhD, Giorgio Trentinaglia
(see http://arxiv.org/abs/0809.3394)
considers the reconstruction of an orbifold groupoid from its category of "representations".
At first, one might guess that "representation" means vector bundle.
But Trentinaglia works with a larger category of non-locally-trivial vector bundles: 
the dimension is allowed to jump.
Wether or not the same reconstruction result holds true with actual vector bundles depends on the global quotient conjecture that claims that every orbifold is a global quotient by a compact Lie group.<|endoftext|>
TITLE: Families of genus 2 curves with positive rank jacobians
QUESTION [11 upvotes]: It's fairly easy to write down families of elliptic curves over $\mathbb{Q}(t)$ such that almost every (i.e. when the "height" of $t$ is sufficiently large) curve in the family has positive rank over $\mathbb{Q}$.  One can do this by constructing the fibration so it has sections /$\mathbb{Q}$ a priori, or by fiddling around with Gauss sums; see papers of Fermigier, Mestre, Arms/Miller/Lozano-Robledo, etc.  Are there any known examples of families of genus two curves over $\mathbb{Q}(t)$ such that the Jacobian of almost every curve $C_t$ in the family has provably positive rank?  Can one do this while requiring that $\mathrm{Jac}(C_t)$ be $\overline{\mathbb{Q}}$-simple for almost all $t$?

REPLY [4 votes]: My guess for some examples is the family of  (genus 2) hyperelliptic curves y2=degree 6 poly in x passing through n "randomly chosen" rational points (for n=2,3,4,5, or 6). The family of such curves has dimension 7-n, and I would guess that if a hyperelliptic curve has n "random" rational points on it then the Q rank of its Jacobian is usually at least n-1. 
An obvious place to look for explicit examples is
MR1406090 Cassels, J. W. S.; Flynn, E. V. Prolegomena to a middlebrow arithmetic of curves of genus 2. London Mathematical Society Lecture Note Series, 230. Cambridge University Press, Cambridge, 1996.  ISBN: 0-521-48370-0<|endoftext|>
TITLE: When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?
QUESTION [26 upvotes]: David's question Families of genus 2 curves with positive rank jacobians reminded me of a question that once very much interested me: when is a product of elliptic curves isogenous to the jacobian of a hyperelliptic curve?
How is this related to David's question? Well, if we can multiply two elliptic curves over $\mathbb{Q}(t)$ with large rank, and the result is isogenous to the jacobian of a hyperelliptic curve, then this will probably produce record families answering David's question, i.e. genus two curves with very large rank. It is also interesting for all genera, so don't restrict answers to 2. On the other hand, answers containing arithmetic information, for example on elliptic curves over the rationals, are more than welcome.

REPLY [2 votes]: In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $A$ over $\mathbb{C}$ with the product of two $CM$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $A$ may be $E_1\times E_2$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.
By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.
Here is the statement for the $CM$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.:
Let $E_1\sim E_2$ be two isogenous $CM$ elliptic curves over an algebraically closed field $K$ with its characteristic is zero. Then there is no genus 2 curve on $E_1\times E_2$ if and only if the degree map on $Hom(E_1,E_2)$ is equivalent to one of the 15 forms $f(x,y)=ax^2+bxy+cy ^2$ whose coefficients $(a,b,c)$ are in the following list:
$\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$
The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.
Some notes: The assumption of the characteristic 0 can be removed. This was also done by Ibukiyama/Katsura/Oort.
The exact list is only known under the GRH in the nonCM case. But Kani's solution doesn't depend on the GRH.<|endoftext|>
TITLE: A Pachner complex for triangulated manifolds
QUESTION [31 upvotes]: A theorem of Pachner's states that if two triangulated PL-manifolds are PL-homeomorphic, the two triangulations are related via a finite sequence of moves, nowadays called "Pachner moves".  
A Pachner move has a very simple picture.  If $N$ is a triangulated $n$-manifold and $C \subset N$ is a co-dimension zero subcomplex of $N$ which is equivalent to a subcomplex $D'$ of $\partial \Delta_{n+1}$ where $\Delta_{n+1}$ is an $(n+1)$-simplex equipped with its standard triangulation, then you can replace $N'$ in $N$ by $(\partial \Delta_{n+1}) \setminus D'$ from $\partial \Delta_{n+1}$, the gluing maps being the only ones available to you. 
One way to say Pachner's theorem is that there is a ''graph'' of triangulations of $N$, and it is connected.  Specifically, the vertices of this graph consist of triangulations of $N$ taken up to the equivalence that two triangulations are equivalent if there is an automorphism of $N$ sending one triangulation to the other.  The edges of this graph are the Pachner moves.  
This formulation hints at an idea.  Is there a useful notion of "Pachner complex"?  
Of course this would lead to many further questions such as what geometric / topological properties does such a complex have, is it contractible for instance, are there "short" paths connecting any two points in the complex, and so on. 
I'm curious if people have a sense for what such a Pachner complex should be.   For example, some Pachner moves commute in the sense that the subcomplexes $C_1, C_2 \subset N$ are disjoint, so you can apply the Pachner moves independantly of each other.  Presumably this should give rise to a "square" in any reasonable Pachner complex.  
This feels related to the kind of complexes that Waldhausen and Hatcher used to use in the 70's but it's also a little different.  

Udo Pachner, P.L. homeomorphic manifolds are equivalent by elementary shellings, European J. Combin. 12 (1991), 129–145.

REPLY [2 votes]: I discussed this complex with a few folks, a while back (perhaps including Ben Burton and Jonathan Spreer?).  We decided that its "coarse geometry" is not very nice.
Suppose that $M$ is a surface.  Let $P(M)$ be its Pachner graph.  We perform Pachner moves to ensure that there are $k$ disjoint disks in the current triangulation.  We now subdivide independently inside of these disks.  We deduce that $P(M)$ contains "quasi-$k$-orthants": subgraphs quasi-isometric to $\mathbb{N}^k$.  This is true for all $k$.
If we fix the number of zero-cells in dimension two (that is, disallow (3,1) and (1,3) moves) then the "flip graph" is very interesting, and appears in Teichmuller theory.  Fixing the number of zero-cells does not save us in dimension three - the graph again has quasi-orthants of all dimensions.  This suggests the following question:

What if we restrict our attention to a compact, triangulated, four-manifold $X^4$, and only allow $(3, 3)$ moves?

Can we hope that this graph has some sensible connection to the mapping class group of $X^4$?<|endoftext|>
TITLE: Must the left and right unitors of a monoidal category coincide at the neutral object?
QUESTION [5 upvotes]: A monoidal category has (among other things) a pair of natural transformations called the left and right unitors
$$ \lambda_A:I\otimes A \cong A $$
  $$ \rho_A:A\otimes I \cong A $$
The categories I work with on a daily basis have $\lambda_I=\rho_I:I\otimes I \cong I$, but I think that is mostly a consequence of the fact that most of them have a terminal object as their $I$.  Is this true in general?
I'm sort of hoping this is true, because otherwise it seems that some of the definitions of enriched categories need to make an arbitrary choice of one unitor or the other.  For example, see display (1.10) of Max Kelly's introduction to enriched category theory on page 10.
(I'm actually interested in the even-more-general case of premonoidal categories, but mathematicians don't use those very often, so I've posed the question in terms of monoidal categories instead).

REPLY [6 votes]: In Categories for the Working Mathematician MacLane included $\lambda_I = \rho_I$ as one of three diagrams involving the associator and the two unitors that must commute (the other two being the usual pentagon and triangle diagrams) as the axioms defining monoidal categories.
It was however proven to follow from the other axioms in Max Kelly's 1964 paper On MacLane's Conditions for Coherence of Natural Associativities, Commutativities, etc. (Journal of Algebra 1, 397–402)
EDIT: For the case of premonoidal categories, I think $\lambda_I = \rho_I$ follows from the results of section 4 of John Power's Premonoidal categories as categories with algebraic structure (Theoretical Computer Science 278, 1-2, 303-321). In that paper, the unitors in a premonoidal category are defined to be central natural transformations; further down, the centre of a premonoidal category is defined to be the category with the same objects but with only the central morphisms of the original category, and this turns out to be a monoidal category, hence reducing to Kelly's proof.<|endoftext|>
TITLE: The topological analog of flatness?
QUESTION [61 upvotes]: Recall that a map $f:X\to Y$ of schemes is called flat iff for any $x\in X$ the ring $O_{X,x}$ is a flat $O_{Y,f(x)}$-module.
Briefly the question is: what is the topological analog of this?
Many notions and constructions in scheme theory have obvious topological counterparts (which probably were the inspiration at least in some cases, but I'm not a historian to tell this for sure). Gluing schemes is analogous to gluing smooth manifolds out of copies of Euclidean balls. Proper, \'etale and smooth morphisms all have obvious topological analogs: these are proper maps, local homeomorphisms and smooth maps of smooth manifolds such that the differential at each point is surjective (submersions). A separated scheme is the analog of a Hausdorff space. Moreover, in all these cases it seems clear that there is just one way of translating the corresponding topological notion in the language of schemes.
Flat morphisms seem trickier (to me). I'm aware of two interpretations. One is too vague ("a map such that the preimages of points don't vary too wildly"). The other ("a Serre fibration") is not completely satisfactory: all fibers of a Serre fibration are homotopy equivalent and are even homeomorphic if the fibration is locally trivial. However, there are plenty of flat maps which do not look like Serre fibrations at all: for example the projection of the union of the lines $x=\pm y$ in the plane onto the $x$-axis.
One way to make the above question a bit more precise is this: is there a way to define the notion of a "flat" map (of sufficiently nice topological spaces, say smooth manifolds or CW complexes or polyhedra) in terms of topology or differential geometry so that when $X(\mathbf{C})$ and $Y(\mathbf{C})$ are the sets of closed points of varieties (= reduced separated schemes of finite type) $X$ and $Y$ over $\mathbf{C}$ a morphism $X\to Y$ is flat if and only if induced map $X(\mathbf{C})\to Y(\mathbf{C})$ of topological spaces is "topologically" flat? Maybe this is too much to ask for, in which case I'd be interested to know if there is a variation of this which holds.
An obvious guess: one should take the notion of a submersion and relax it, but I'm not sure how.

REPLY [33 votes]: Here is a statement that goes into the direction you are looking for:

When $X$ and $Y$ are smooth varieties over $\mathbb C$, then $f$ is flat if and only if every fiber of $f$ has dimension $\dim X - \dim Y$.

Thiis is perhaps not quite as well-known as it should be; I learned it as a student in my algebraic geometry class taught by Jens Franke. I would also be glad if someone could tell me a reference, as Jens Franke doesn't seem to be planning to turn his lecture notes into a book...
(Here is an example of one of the several precise statements he proved:
Let $f \colon X \to Y$ be a morphism of finite type between locally Noetherian prescheme, such that $X$ is Cohen-Macaulay and $Y$ is regular. Then $f$ is flat if either of the following two conditions holds:

For every irreducible closed subset $Z \subset Y$ and every irreducible component $Z'$ of $f^{-1}(Z)$ we have $\mathrm{codim}(Z', X) = \mathrm{codim}(Z, Y)$
$Y$ is ``equicodimensional'', $f$ maps closed points to closed points, and every non-empty fiber of $f$ has dimension $\dim X - \dim Y$.

Here ``equicodimensional'' means that every closed point has the same codimension.)<|endoftext|>
TITLE: Is the tangent bundle of the long line $L$ homeomorphic to $L\times\mathbb R$?
QUESTION [14 upvotes]: Question: Is the tangent bundle of the long line $L$ homeomorphic to $L\times\mathbb R$? I'd guess that the answer doesn't depend on choice of differentiable structure, but maybe it does. 
Motivation: One night at dinner, someone brought up a puzzle involving infinitely many prisoners standing in a line, and someone asked if there was a physical reason that the collection of prisoners had to be countable. In other words, might (one of) the directions in the physical universe be modeled after the long line?
The answer to that question is no: the universe has a metric, but the long line has no Riemannian structure. The standard explanation for this is that a Riemannian manifold is metrizable, and a non-paracompact space isn't. Without using fancy theorems, one could instead suppose that $L$ was Riemannian and look at the exponential map starting at a point $x$ going in the increasing direction. This is an increasing function from $\mathbb R$ to $L$, so it converges to some point $y$. Then the exponential map from $y$ downwards reaches $x$ in finite time, contradiction.
In any case, the basic result is that $L$ is not Riemannian, so its tangent bundle must be nontrivial, but only in the differential sense. One could try to instead consider a continuous metric (if the tangent bundle were indeed topologically trivial), but this wouldn't give rise to an exponential map, nor, as far as I know, a metric space structure on $L$.

REPLY [14 votes]: Lots of information about this is available in a paper by Peter Nyikos:
Various smoothings of the long line and their tangent bundles. 
Adv. Math. 93 (1992), no. 2, 129--213.<|endoftext|>
TITLE: Uniqueness/motivation for the Suslin-Voevodsky theory of relative cycles.
QUESTION [7 upvotes]: In the Suslin-Voevodsky paper "Relative cycles and Chow sheaves" they define abelian groups $z(X/S, r)$ for schemes $X$ of finite type over Noetherian schemes $S$, and then they show that these groups satisfy a number of properties, for example 1) they are cdh-sheaves in $S$, and 2) the obvious maps $cycl: \mathbb{Z} Hilb(X / S, r) \to z(X/S, r)$ form a natural transformation of presheaves, etc
The only motivation I can think of for the definition of $z(X/S, r)$ is that it gives a "good theory". However, we could also consider, for example, the presheaf obtained as the image of cycl and we would also get a theory of relative cycles with well defined pullback although we would lose some other proprieties. Then there are also the presheaves $z_{equi}(X/S, r), c(X/S, r), c_{equi}(X/S, r)$.
So my question is: (1) is there a list of properties that determines the groups $z(X / S, r)$ uniquely, and (2) why do we want each of these properties.
For (1) the answer is of course yes and so the trick is to find a list of properties that we can justify wanting, apart from saying "they are nice".

REPLY [4 votes]: I will just sum up the situation as I see it (too big for the comment box).
One important goal is to set up a good intersection theory for cycles without quotienting by rational equivalence, and using it to get a composition product for finite correspondences, which are by definition elements of groups of the form $c_{equi}(X\times_S Y/X,0)$
It is true that the variety of definitions of cycle groups in the paper is somewhat confusing. There are 16 possible groups because starting from the "bare" notion of relative cycles (def. 3.1.3)  there are 4 binary conditions : being effective, being equidimensional, having compact support (c, PropCycl), and being "special", i.e satisfying the equivalent conditions of lemma 3.3.9 (everything except Cycl and PropCycl). So you have 
1)$z_{equi}(X/S,r)\subset z(X/S,r)\subset Cycl(X/S,r) \supset Cycl_{equi}(X/S,r)$
and their effective counter-parts.
2)$c_{equi}(X/S,r)\subset c(X/S,r)\subset PropCycl(X/S,r) \supset PropCycl_{equi}(X/S,r)$
and their effective counter-parts.
(1) is then a "subline" of 2))
In a sense, the most satisfying definition would be to use only cycles which are flat over $S$ (the $\mathbb{Z}Hilb$-groups, or the closely related $z_{equi}$) but pullbacks along arbitrary morphisms are not defined there in general.
With the groups Cycl, thanks to the relative cycle condition built in Cycl, you have pullbacks along arbitrary morphism, but only with rational coefficients (thm 3.3.1, the denominators of the multiplicities are divisible by residue characteristics)
The main interest of the "special" relative cycles $z(-,-)$ is in their definition : they admit integral pullbacks ! Then you have the small miracle that this condition is stable by those pullbacks and you get a subpresheaf. This means that using them you can set up intersection theory with integral coefficients even on singular car p schemes.
All this zoology simplifies when $S$ is nice : there are some results when $S$ is geometrically unibranch, but the nicest case is $S$ regular, in which the chains of inclusions I wrote down collapse, you are left with two distinctions which are reasonable from the point of view of classical intersection theory : effective/non-effective, general/with compact support. Furthermore, the intersection multiplicities are computed by the Tor multiplicity formula, so the Suslin-Voevodsky theory is really an extension of local intersection theory of regular rings as in Serre's book.<|endoftext|>
TITLE: A Riemannian metric on $S^2 \times S^2$ of nonnegative curvature that is not a product
QUESTION [11 upvotes]: Good afternoon,
There is an example of a Riemannian metric on $S^2 \times S^2$  of nonnegative sectional curvature that is not a product metric. I know there is one; however, I cannot find a specific reference. Any suggestions?

REPLY [11 votes]: A discussion of nonnegatively curved metrics on $S^2\times S^2$ can be found in the  survey by B. Wilking, see page 26 and last paragraph on page 25. In particular, there is a one parametric family of metrics of nonnegative curvature on $S^2\times S^2$ which are not moved by any diffeomorphism  to a product metric.<|endoftext|>
TITLE: Why does the algebraic condition of flatness on the structure sheaves  give a good definition of family?
QUESTION [18 upvotes]: Hartshorne remarks that is is something of a mystery as to why the algebraic condition of flatness on the structure sheaves gives a good definition of a family (see below). Are there any known enlightening explanations that help serve to unravel this mystery?  Below is Hartshorne's introductory motivation to flat families containing said remark:

For many reasons it is important to have a good notion of an algebraic 
  family of varieties or schemes. The most naive definition would be just to 
  take the fibres of a morphism. To get a good notion, however, we should 
  require that certain numerical invariants remain constant in a family, such 
  as the dimension of the fibres. It turns out that if we are dealing with non- 
  singular (or even normal) varieties over a field, then the naive definition is 
  already a good one. Evidence for this is the theorem (9.13) that in such a 
  family, the arithmetic genus is constant. 
On the other hand, if we deal with nonnormal varieties, or more general 
  schemes, the naive definition will not do. So we consider a flat family of 
  schemes, which means the fibres of a flat morphism, and this is a very good 
  notion. Why the algebraic condition of flatness on the structure sheaves 
  should give a good definition of a family is something of a mystery. But 
  at least we will justify this choice by showing that flat families have many 
  good properties, and by giving necessary and sufficient conditions for 
  flatness in some special cases. In particular, we will show that a family 
  of closed subschemes of projective space (over an integral scheme) is flat if 
  and only if the Hilbert polynomials of the fibres are the same. 
  -- Hartshorne, Algebraic Geometry, 1977, III.9.5, p. 256

REPLY [5 votes]: There is also the following (probably unhistorical) point of view (it is a version of Hailong Dao's answer). Namely, you don't have to work with flat families at all, so if you want, you can just declare all morphisms to be 'families'. The problem with this approach is that this is a family of `derived' objects. Here's an example:
Let $S$ be a scheme, and let $F$ be a coherent sheaf on S. When is it a 'family' of its fibers? If it is flat, it definitely deserves to be called a family of vector spaces (a vector bundle). But even if it is not flat, you can still view it as a family, but the family of what? The (derived) fibers of $F$ are no longer vector spaces, they are complexes of vector spaces (precisely because $F$ fails to be flat), so we can view $F$ as a nice family of complexes of vector spaces, even though $F$ itself is a sheaf, not a complex.
To summarize: by all means, let's forget about flatness and declare any morphism to be a family. . . of some kind of derived objects. If we now want members of the family to be actual objects (schemes, vector spaces, sheaves, or whatever it is we are trying to include in a family), flatness is forced on you more or less by definition.<|endoftext|>
TITLE: Fourier coefficients for elliptic curves on average
QUESTION [8 upvotes]: Fix a prime p, and look at elliptic curves in some family (e.g. all elliptic curves ordered by height). How often do the Fourier coefficients a_p occur? Are there any conjectures?

REPLY [6 votes]: As others have mentioned, if $p$ is fixed then you're really looking at elliptic curves over a fixed finite field.  
From some points of view an interesting variant would be to look at elliptic curves say $E_{a,b}:y^2 = x^3 + ax + b$ where $a$ and $b$ vary over integers in a box, say $|a| \leq A$ and $|b| \leq B$ and relatively small compared to $p$.  The one might try to find asymptotic results that hold as $p$, $A$, $B$ get large together.  If $A$ and $B$ aren't too big then this is giving more information about individual curves.  For example, in bounding the average analytic rank of elliptic curves it is important to get a good bound on $$\frac{1}{AB} \sum_{p < P} \sum_{|a| \leq A} \sum_{|b| \leq B} a_P(E_{a,b})$$ with $A$ and $B$ as small as possible.  For example, see A. Brumer, The average rank of elliptic curves. I, Invent. Math. 109(3), 445–472 (1992).
In a different but related direction, there is a paper of David and Pappalardi, Average Frobenius distributions of elliptic curves (it's the fourth from the bottom) on this subject.  They get a kind of Lang-Trotter on average, so they are varying both $p$ and the coefficients defining the elliptic curves.  Stephan Baier later made some improvements on this problem here.<|endoftext|>
TITLE: Interesting applications (in pure mathematics) of first-year calculus
QUESTION [35 upvotes]: What interesting applications are there for theorems or other results studied in first-year calculus courses?
A good example for such an application would be using a calculus theorem to prove a result in group theory. On the other hand, the importance of calculus in applied mathematics or in physics is well known, therefore is not a good example.

REPLY [5 votes]: In number theory, here are four applications of techniques or results in first-year calculus.
(1) Finding equations of tangent lines by first-semester calculus methods lets us add points on elliptic curves using the Weierstrass equation for the curve. This is more algebraic geometry than number theory, so I'll add that the methods show if the Weierstrass equation has rational coefficients then the sum of two rational points is again a rational point.
(2) The recursion in Newton's method from differential calculus is the basic idea behind Hensel's lemma in $p$-adic analysis (or, more simply, lifting solutions of congruences from modulus $p$ to modulus $p^k$ for all $k \geq 1$).
(3) The infinitude of the primes can be derived from the divergence of the harmonic series (the zeta-function at 1), which is based on a bound involving the definition of the natural logarithm as an integral.
(4) Unique factorization in the Gaussian integers can be derived from the Leibniz formula
$$
\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \sum_{n \geq 0} \frac{(-1)^n}{2n+1}
$$
by interpreting it as a case of Dirichlet's class number formula $2\pi h/(w\sqrt{|D|}) = L(1,\chi_D)$ for $\chi_D$ the primitive quadratic character associated to ${\mathbf Q}(\sqrt{D})$ where $D$ is a negative fundamental discriminant, $h$ is the class number of ${\mathbf Q}(\sqrt{D})$ and $w$ is the number of roots of unity in ${\mathbf Q}(\sqrt{D})$. Taking $D = -4$ turns the left side into $2\pi h/(4\sqrt{4}) = (\pi/4)h$, so the Leibniz formula is equivalent to $h = 1$, which is another way of saying $\mathbf Z[i]$ is a PID or equivalently (for Dedekind domains) a UFD.
Here are two more applications, not in number theory directly.
(5) Gerry Edgar mentions in his answer Niven's proof of the irrationality of $\pi$, which is available in Spivak's calculus book. The same ideas imply irrationality of $e^a$ for every positive integer $a$, which in turns easily implies irrationality of $e^r$ for nonzero rational $r$ and thus also irrationality of $\log r$ for positive rational $r \not= 1$. The calculus fact in the proof of irrationality of the numbers $e^a$ is that for all positive integers $n$ the polynomial 
$$
\frac{x^n(1-x)^n}{n!}
$$
and all of its higher derivatives take integer values at $0$ and $1$. That implies a certain expression involving a definite integral is a positive integer, and then with the fundamental theorem of calculus that same expression turns out to be less than 1 for large $n$ (where "large" depends on the hypothetical denominator of a rational formula for $e^a$), and that is a contradiction.
(6) Prove that if $f$ is a smooth function (= infinitely differentiable) on the real line and $f(0) = 0$ then $f(x) = xg(x)$ where $g$ is a smooth function on the real line. There is no difficulty in defining what $g(x)$ has to be if it exists at all, namely 
$$
g(x) = \begin{cases}
f(x)/x, & \text{ if } x \not= 0, \\
f'(0), & \text{ if } x = 0.
\end{cases}
$$
And easily the function defined this way is continuous on the real line and satisfies $f(x) = xg(x)$. But why is this function smooth at $x = 0$ (smoothness away from $x = 0$ is easy)? You can try to do it using progressively messier formulas for higher derivatives of $g$ at 0 by taking limits, but a much slicker technique is to use the fundamental theorem of calculus to write
$$
f(x) = f(x) - f(0) = \int_0^x f'(t)\,dt = x\int_0^1 f'(xu)\,du,
$$
which leads to a different formula for $g(x)$ that doesn't involve cases:
$$
g(x) = \int_0^1 f'(xu)\,du.
$$
If you're willing to accept differentiation under the integral sign (maybe that's not in the first-year calculus curriculum, but we used first-year calculus to get the slick formula for $g(x)$) then the right side is easily checked to be a smooth function of $x$ from $f$ being smooth.<|endoftext|>
TITLE: Sum Equals Product
QUESTION [9 upvotes]: Sum Equal Product
There are many articles in re this issue on the Web, although many are restricted to special cases (e.g., John Cook’s sum of tangents = product of tangents), and even some involving mixed numbers. 
There is one excellent article in re: sum equal product written by two Polish mathematicians which is replete with wonderful proofs that I am not sophisticated enough to understand. (Kurlandchick & Nowicki, Nov `98)
Actually, my question is quite straightforward. Is there a formula or some kind of algorithm that can show which sums of positive, unequal integers equal their products?
Falco

REPLY [18 votes]: Let the positive, unequal integers be $a_1 < a_2 < \cdots < a_k$ with $a_1+a_2+\cdots+a_k = a_1a_2\cdots a_k = n$.  Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$.  Note that $a_k \ge k$.  Then we have $a_1 + a_2 + \cdots + a_k < ka_k$, while $a_1a_2\cdots a_k \ge (k-1)!a_k$.  So we must have $ka_k > (k-1)!a_k$, which means that $k > (k-1)!$.  This means $k \le 3$.  If $k = 2$, then we have $a_1 + a_2 = a_1a_2$, or $\frac{1}{a_1}+\frac{1}{a_2} = 1$.  This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a_i$ be unequal.
The case $k = 3$ is then the only tricky case.  If $a_1 > 1$, then we have $a_1+a_2+a_3 < 3a_3$ and $a_1a_2a_3 \ge 6a_3$.  So $a_1 = 1$.  We then must find solutions to $1 + a_2 + a_3 = a_2a_3$.  This can be rewritten as $(a_2-1)(a_3-1) = 2$, which as $a_2 < a_3$ are integers immediately gives $a_2 = 2, a_3 = 3$.
In summary, all possible solutions are $\{n\}$ for all positive integers $n$ and $\{1,2,3\}$.
History: this is definitely a classic problem; see for example 2006 USA Mathematical Olympiad problem #4 (pdf, problem is on second page), which is your problem with several restrictions removed.<|endoftext|>
TITLE: What impact would P!=NP  have on the characterization of BQP?
QUESTION [11 upvotes]: Many complexity theorists assume that $P\ne NP.$ If this is proved, how would it impact quantum computing and quantum algorithms? Would the proof immediately disallow quantum algorithms from ever solving NP-Complete problems in Quantum Polynomial time?
According to Wikipedia, quantum complexity classes BQP and QMA are the bounded-error quantum analogues of P and NP. Is it likely that a proof that $P\ne NP$ can be adapted to the quantum setting to show that $BQP \ne QMA?$

REPLY [2 votes]: To your second question. It is unlikely that the current under-review proof of P != NP will allow you to seperate BQP and QMA (or BQP and P, or BQP and NP, or even BPP and NP...). Deolalikar's proof uses descriptive complexity, in particular it uses a correspondence between statements expressible in certain logics and the complexity classes P and NP. As far as I know there is no such nice one-to-one correspondences between BPP, BQP, or QMA and nice logics.<|endoftext|>
TITLE: How do you define the Euler Characteristic of a scheme?
QUESTION [22 upvotes]: If X is a variety over the complex numbers, one reasonable thing to do is to consider the associated analytic space $X_{an}$  and to take the topological Euler characteristic of that.
Is there a purely algebraic way to obtain this number?
If X is non-singular then one might define it as the integral of the top Chern class of its tangent bundle.
The reason I ask is that I'm currently reading Joyce's survey on Donaldson-Thomas invariants and I wanted to know if by any chance he were using some more sophisticated notion.
On related note: if X is a non-proper scheme over C, why is its Euler characteristic well-defined?

REPLY [8 votes]: One comment on the "excision" property for Euler characteristics of complex algebraic varieties: $\chi(X) = \chi(Z) + \chi(X\backslash Z)$.  There is a short proof of this in Fulton's Introduction to Toric Varieties, p. 142, which avoids the sticky triangulation issue and doesn't seem to be so well-known.  In fact, it's equivalent to the statement that $\chi(X) = \chi_c(X)$ for any algebraic variety (the latter being the Euler characteristic for compactly supported cohomology). 
This doesn't exactly help with the first question (about an algebraic definition), but it does show why the Euler characteristic is well-defined for any variety, per Arend's comment.<|endoftext|>
TITLE: Topology of function spaces?
QUESTION [33 upvotes]: Let $X,Y$ be finite-dimensional differentiable manifolds, and let's assume that they are connected.  In fact, in applications I would like both $X$ and $Y$ to be riemannian manifolds.
Let $C^\infty(X,Y)$ denote the space of smooth maps $f: X \to Y$.  I'm interested in, say, the connected components, fundamental group,... of this space, but I'm really not sure where to start looking.  I realise that I first need to topologise this space.  My only experience in this realm is an introductory point-set topology course (based on Munkres) I took as a graduate student.  Munkres talks about the compact-open topology for the space $C^0(X,Y)$ of continuous maps between two topological spaces and shows, for instance, that if $X$ is locally compact Hausdorff then the evaluation map $X \times C^0(X,Y) \to Y$ is continuous.  Later in the book he also applies this to give a slick proof of the existence of covering spaces with prescribed covering group.
Back to the differentiable category, ideally I'd like to be able to do calculus on $C^\infty(X,Y)$, hence I'd like to think of $C^\infty(X,Y)$ as an infinite-dimensional differentiable manifold and possibly even riemannian whenever so are $X$ and $Y$.
In case it helps to focus the question, let me say a few words of (physical, I fear) motivation.
When $X,Y$ are riemannian, $C^\infty(X,Y)$ plays the rôle of the configuration space for a physical model known as the nonlinear sigma model, whose action functional, assigning to $\sigma: X \to Y$, the value of the integral (either take $X$ to be compact or else restrict the possible functions further to assure convergence)
$$ S[\sigma] = \int_X |d\sigma|^2 \operatorname{dvol}_X,$$
where I'd like to think of $d\sigma$ as a one-form on $X$ with values in the pullback $\sigma^*TY$ by $\sigma$ of the tangent bundle to $Y$, and $|d\sigma|^2$ involves the metric on the bundle $T^*X \otimes \sigma^*TY$ induced from the riemannian metrics on $X$ and on $Y$.  The extrema of $S$ are then the harmonic maps.
We are often interested in the quantum theory (un)defined formally by a path integral.  A mathematically conservative point of view is that the path integral simply gives a recipe for the perturbative treatment of the quantum theory, where we fix an extremum $\sigma_0$ of $S$ and quantise the fluctuations around $\sigma_0$.  By definition, fluctuations around $\sigma_0$ lie in the connected component of $\sigma_0$ and as a first approximation, the path integral becomes a sum over the connected components of the space of maps.  Hence the interest in determining the connected components of $C^\infty(X,Y)$, which in this context are often called superselection sectors.
So in summary, a possible question would be this:

What can be said about the topology (e.g., homotopy type) of $C^\infty(X,Y)$ in terms of $X$ and $Y$?

I'm not asking for a tutorial, just for some orientation to the available literature.
Thanks in advance.
Added
In response to the helpful answers I have received already, I'd like to point out that my interest is really on the homotopy type of the space of maps.  I only mentioned the analytic aspects in case that narrows down the topology one would put on the space.  As pointed out in the answers, most reasonable topologies are equivalent, so this is a relief.
As for concrete examples, I am particularly interested in the case where $X$ is a compact Riemann surface and $Y$ a compact Lie group.  I'm happy to put the constant curvature metric on $X$ and a bi-invariant metric on $Y$.

REPLY [4 votes]: The rational homotopy type of $C^\infty(X,Y)$ or $C^0(X,Y)$ has been determined by 

MR0667163 (84a:55010) 
Haefliger, André:
Rational homotopy of the space of sections of a nilpotent bundle. 
Trans. Amer. Math. Soc. 273 (1982), no. 2, 609–620. 

See also

MR0850369 (87h:55009)
Vigué-Poirrier, Micheline:
Sur l'homotopie rationnelle des espaces fonctionnels. (French. English summary) [The rational homotopy of function spaces] 
Manuscripta Math. 56 (1986), no. 2, 177–191. 

I remember vaguely that one needs a minimal model of $X$ and of $Y$; then a model of the function space is the tensor product of the minimal models.
There are some conditions, which are satisfied for finite dimensional smooth manifolds.<|endoftext|>
TITLE: A differentiable approximation to the minimum function
QUESTION [14 upvotes]: Suppose we have a function $f : \mathbb{R}^N \rightarrow \mathbb{R}$ which, given a vector, returns the value of its smallest element. How can I approximate $f$ with a differentiable function(s)?

REPLY [6 votes]: I would use $$f(x) = -\frac{1}{\rho}\log \frac{1}{N} \sum_{i=1}^N e^{-\rho x_i},$$ which approaches $f(x) \rightarrow \min_i |x_i|$ as $\rho \rightarrow +\infty$.
There has been some debate about normalization, so let's compare these four:
$$f_{A}(x) = -\frac{1}{\rho}\log \frac{1}{N}\sum_{i=1}^N e^{-\rho x_i} $$
$$f_{B}(x) = -\frac{1}{\rho}\log \sum_{i=1}^N e^{-\rho x_i} $$
$$f_C(x) = \left(\frac{1}{N}\sum x_i^{-\rho}\right)^{-1/\rho}$$
$$f_D(x) = \left(\sum x_i^{-\rho}\right)^{-1/\rho}$$
For $x$=[1,2,3,4,5] and $\rho=10$, $f_A=1.16$, $f_B=1.0$, $f_C=1.1745$, $f_D=1.0$.
For $x$=[1,2,3,4,5] and $\rho=100$, $f_A=1.016$, $f_B=1.0$, $f_C=1.0162$, $f_D=1.0$.
For $x$=[1,1,1,1,1] and $\rho=10$, $f_A=1$, $f_B=0.8391$, $f_C=1$, $f_D=0.8513$.
For $x$=[1,1,1,1,1] and $\rho=100$, $f_A=1$, $f_B=0.9839$, $f_C=1$, $f_D=0.9840$.
For $x$=[0,1,10,100,1000] and $\rho=10$, $f_A=0.1609$, $f_B=0$, $f_C=0$, $f_D=0$.
For $x$=[0,1,10,100,1000] and $\rho=100$, $f_A=0.0161$, $f_B=0$, $f_C=0$, $f_D=0$.
(For the last two cases, $f_C$ and $f_D$ can be evaluated by using $\log(0)=\infty$ and $\infty^{-1/\rho}=0$).
So as we would hope, when $\rho$ is large, all versions give reasonably good approximations to the minimum.
The question then is which is best when $\rho$ is not that big -- this is an important question since large $\rho$ can give practical difficulties with finite precision arithmetic.  Comparing the four versions above, we can see that the mean versions (A and C) overestimate the minimum when $x$ has a lot of values above the minimum.  Conversely, the sum versions (B and D) underestimate the minimum when $x$ has a lot of values that are all equal to the minimum.  Which is better is ultimately a question of your application.  But to me, the mean version gives an answer that makes much more sense.  The approximate minimum should be similar to the minimum but the approximation should get pulled in the direction of all of the individual values in $x$.  This is what happens in the mean version (A and C), which also keep the approximate minimum inside the range of the data, i.e.
$$\min(x) \le f_A(x) , f_C(x) \le \max(x)$$
On the other hand, the sum versions (B and D) can give a minimum value that is actually lower than any value contained in the data (see the third and fourth examples above).  In other words, it is possible that
$$f_B(x), f_D(x) < \min(x),$$
which is a property I want to avoid in an approximate minimum since it makes it very hard to interpret that approximate-minimum value.  So I find A,C to be more useful than B,D.
The last question is A vs C.  Version A is fairly stable as $x_i \rightarrow 0$, but version C however runs out of precision. For example $x^{-100}$ overflows double precision below $x \approx 8.27e^{-4}$ and overflows single precision below $x \approx 0.412$.
Therefore, at least in my own uses, $f_A$ gives the most useful approximation and is the most numerically stable.
Edit:  For practical implementation, note that instead of directly aggregating the raw values $x$, you can also instead do the aggregation relative to some arbitrary value $\mu$, without analytically changing the result:
$$\mu + f_A\left(x-\mu\right) = f_A$$
$$\mu + f_B\left(x-\mu\right) = f_B$$
$$\mu f_C\left(x / \mu\right) = f_C$$
$$\mu f_D\left(x / \mu\right) = f_D$$
The end result is unchanged (aside from numerical issues), but a clever choice of $\mu$ can help improve numerical stability.  Specifically, if possible I would recommend using the true minimum, $\mu = \min(x)$, which gives
$$\begin{aligned}
\hat{f}_A\left(x\right) &= f_A\left(x-\min(x)\right) + \min(x) \\
&= \min(x) -\frac{1}{\rho}\log \frac{1}{N}\sum_{i=1}^N e^{-\rho \left(x_i - \min(x)\right)} \end{aligned}$$
The other versions can also be modified similarly.  You can probably do similar tricks when calculating the derivative.<|endoftext|>
TITLE: Complex vector bundles with trivial Chern classes on k-tori
QUESTION [12 upvotes]: Let $E\to X$ be a principal $U(N)$-bundle over a (nice) topological space $X$. It is well known that vanishing of the Chern classes of $E$ is not a sufficient condition for $E$ to be trivial, the simplest example being probably the nontrivial $U(2)$-bundle over $S^5$. However one may wonder what happens for a specific base $X$, e.g, for $k$-tori. In this case it is completely trivial that any $U(N)$-bundle on $S^1$ is trivial and that an $U(N)$-bundle on $S^1\times S^1$ is trivial precisely when its first Chern class vanishes. A very little obstruction theory then shows that this holds true for $U(N)$-bundles over $S^1\times S^1\times S^1$. Such simple arguments, however do not work for $(S^1)^k$ for $k\geq 4$. Can anything general be said?

REPLY [15 votes]: As the cohomology of $(S^1)^n$ is torsion free every stable bundle on $(S^1)^n$ is
determined by Chern classes (this also follows from the $K$-theory Künneth
formula) so just as for the spheres it is an unstable problem. As for the
unstable problem unless I have miscalculated, if $(S^1)^5\rightarrow S^5$ is a
degree $1$ map, then the pullback of the non-trivial $U(2)$ bundle on $S^5$ with
trivial Chern class is non-trivial. (The proof uses that the $5$'th step in the
Postnikov tower of $\mathrm{BU}(2)$ is a principal fibration $K(\mathbb
Z/2,5)\rightarrow U\rightarrow K(\mathbb Z,4)\times K(\mathbb Z,2)$.)
Some more details of the calculation: The first and second Chern class gives a
map 
$$\mathrm{BU}(2)\rightarrow K((\mathbb Z,4)\times K(\mathbb Z,2)$$
which induces an isomorphism on homotopy groups in degrees up to $4$. As
$\pi_i(\mathrm{BU}(2))=\pi_{i-1}(\mathrm{SU}(2))$ for $i>2$ we get that
$\pi_5(\mathrm{BU}(2))=\pi_4(S^3)=\mathbb Z/2$. Hence, the next step $U$ in the
Postnikov tower of $\mathrm{BU}(2)$ is the pullback of the path space fibration
of a morphism $K(\mathbb Z,4)\times K(\mathbb Z,2)\rightarrow K(\mathbb Z/2,6)$.
In particular we have a principal fibration 
$$K(\mathbb Z/2,5)\rightarrow
U\rightarrow K(\mathbb Z,4)\times K(\mathbb Z,2).$$
This means that for any space
$X$, the image of $[X,K(\mathbb Z/2,5)]$ in $[X,U]$ is in bijection with the
cokernel of $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ obtained by applying $[X,-]$ to the looping of the structure map
$K(\mathbb Z,4)\times K(\mathbb Z,2)\rightarrow K(\mathbb Z/2,6)$. As
$H^4(K(\mathbb Z,3),\mathbb Z/2)=0$ the Künneth formula shows that any map
$K(\mathbb Z,3)\times K(\mathbb Z,1)\rightarrow K(\mathbb Z/2,5)$ factors
through the projection $K(\mathbb Z,3)\times K(\mathbb Z,1)\rightarrow K(\mathbb
Z,3)$ and $H^5(K(\mathbb Z,3),\mathbb Z/2)=\mathbb Z/2\mathrm{Sq}^2\rho\iota$ (where
$\iota$ is the canonical class, $\iota\in H^3(K(\mathbb Z,3),\mathbb Z)$ and
$\rho$ is induced by the reduction $\mathbb Z\rightarrow\mathbb Z/2$). Hence,
the map $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ is either the zero map or given by the composite of the projection to
$H^3(X,\mathbb Z)$, the reduction to $\mathbb Z/2$ coefficients and
$\mathrm{Sq}^2$ (I actually think it is non-zero as otherwise the cohomology of
$H^\ast(\mathrm{BU}(2),\mathbb Z)$ would have $2$-torsion). If we apply it to
$X=(S^1)^5$ we get that $[X,K(\mathbb Z,3)\times K(\mathbb Z,1)]\rightarrow[X,K(\mathbb
Z/2,5)]$ is zero provided that 
$$\mathrm{Sq}^2\colon H^3((S^1)^5,\mathbb
Z/2)\rightarrow H^5((S^1)^5,\mathbb
Z/2)$$
 is zero. However, all Steenrod squares are zero on all of
$H^*((S^1)^n,\mathbb Z/2)$. Indeed, the Künneth and Cartan formulas reduce this to
$n=1$ where it is obvious.<|endoftext|>
TITLE: Smooth homotopy theory
QUESTION [33 upvotes]: Let us define the nth smooth homotopy group of a smooth manifold $M$ to be the group $\pi_n^\infty(S^k)$ of smooth maps $S^n \to S^k$ modulo smooth homotopy.  Of course, some care must be taken to define the product, but I don't think this is a serious issue.  The key is to construct a smooth map $S^n \to S^n \lor S^n$ (regarded as subspaces of $\mathbb{R}^{n+1}$) which collapses the equator to a point; we then define the product of two (pointed) maps $f, g: S^n \to S^k$ to be the map $S^n \lor S^n \to S^k$ which restricts to $f$ on the left half and $g$ on the right half.  To accomplish this, use bump functions to bend $S^n$ into a smooth "dumbell" shape consisting of a cylinder $S^{n-1} \times [0,1]$ with two large orbs attached to the ends, and retract $S^{n-1}$ to a point while preserving smoothness at the ends.  Then retract $[0,1]$ to a point, and we're done.

Question: is the natural "forget smoothness" homomorphism $\phi: \pi_n^\infty(S^k) \to \pi_n(S^k)$ an isomorphism?  If not, what is known about $\pi_n^\infty(S^k)$ and what tools are used?

In chapter 6 of "From Calculus to Cohomology", Madsen and Tornehave prove that every continuous map between open subsets of Euclidean spaces is homotopic to a smooth map.  Thus every continuous map $f$ between smooth manifolds is "locally smooth up to homotopy", meaning that every point in the source has a neighborhood $U$ such that $f|_U$ is homotopic to a smooth map.  However it is not clear to me that the local homotopies can be chosen in such a way that they glue together to form a global homotopy between $f$ and a smooth map.  This suggests that $\phi$ need not be surjective.
In the same reference as above, it is shown that given any two smooth maps between open subsets of Euclidean spaces which are continuously homotopic, there is a smooth homotopy between them.  As above this says that two smooth, continuously homotopic maps between smooth manifolds are locally smoothly homotopic, but I again see no reason why the local smooth homotopies should necessarily glue to form a global smooth homotopy.  This suggests that $\phi$ need not be injective.
I am certainly no expert on homotopy theory, but I have read enough to be surprised that this sort of question doesn't seem to be commonly addressed in the basic literature.  This leads me to worry that my question is either fatally flawed, trivial, useless, or hopeless.  Still, I'm retaining some hope that something interesting can be said.

REPLY [15 votes]: Dear Paul, as Ryan says the smooth and continuous homotopy groups of a manifold coincide.
This is stated as Corollary 17.8.1 in Bott and Tu's book Differential Forms in Algebraic Topology (Springer Graduate Texts in Mathematics, #82).The Corollary is to the preceding Proposition 17.8, which says that a continuous map is homotopic to a differentiable one.This is easy but relies on Whitney's embedding theorem for which the authors refer to De Rham's book Variétés différentiables ; you might prefer Hirsch's book in the same Springer series, GTM #33, which is more modern and in English.
As an application Bott and Tu calculate $\pi_q S^n$ for $q\leq n$ by differential methods.<|endoftext|>
TITLE: Ackermann function in the Primitive recursive arithmetic 
QUESTION [7 upvotes]: Hello.
I study primitive recursive arithmetic and have the following questions.
1) Is it possible to express in the PRA that Ackermann function is total?
2) If yes, is such expression decidable in the PRA ?
Can u suggest some literature on this topic?
Thank you.

REPLY [15 votes]: You can express the totality of any computable function in PRA, using Kleene's T predicate, which is primitive recursive. So if you pick any index $e$ for the Ackermann function, the formula $(\forall n)(\exists t) T(\underline{e}, n, t)$ is already in the language of PRA. 
However, you cannot prove the totality of the Ackermann function in PRA. One way to see this is to note that PRA is a subtheory of $\text{I-}\Sigma^0_1$, modulo an interpretation of the language of PRA into $\text{I-}\Sigma^0_1$. The provably total functions of $\text{I-}\Sigma^0_1$ are well-known to be exactly the primitive recursive functions. 
There is a lot of proof theory literature on provably total functions, which are also called provably recursive functions. But I don't know how much of it focuses specifically on primitive recursive arithmetic. One place to look might be Hájek and Pudlák, Metamthematics of First-Order Arithmetic.<|endoftext|>
TITLE: Finding a subnormal series with specified quotients and end group of specific depth (defect)
QUESTION [7 upvotes]: Suppose $n \ge 3$ and $A_1, A_2, \dots, A_n$ are nontrivial groups. Under what conditions can we find a group G and a subgroup H with a chain:
$$H = H_0 \le H_1 \le H_2 \le \dots \le H_n = G$$
such that each $H_{i-1}$ is normal in $H_i$ and $H_i/H_{i-1} \cong A_i$ and such that the subnormal depth (also called the subnormal defect) of H in G is exactly n. In other words, there is no shorter subnormal series for H in G.
Incidentally, this is always possible for $n = 2$. I have a proof here. The key idea is to take the wreath product of the two groups and then take H as "all but one coordinate" in the base normal subgroup of the corresponding semidirect product.
Aside: The wreath product construction probably does not work for $n \ge 3$ because iterating wreath products does not get us beyond a subnormal depth of 2. See this, for instance.

REPLY [4 votes]: This can always be done:
Given nontrivial groups $A_i$ for $0 \le i \le n$, there exists a group $G$ and a subnormal series $H = H_0 < \cdots < H_n = G$ such that $H_i/H_{i-1} \cong A_i$ for $0 \le i < n$ and such that no shorter subnormal series from $H$ to $G$ exists.
Here is my proof:
We can assume $n > 1$, and we induct on $n$. By the inductive hypothesis, let $W$ be a group with subnormal series $V = V_1 < \cdots < V_n$, such that $V_i/V_{i-1} \cong A_i$ for $1 \le i < n$, and such that there exists no shorter subnormal series for $V$ in $W$. Write $A = A_0$ and let $G$ be the wreath product of $A$ with $W$ corresponding to the action of $W$ on the right cosets in $V$. In other words, $G = BW$ is a semidirect product, where $B \triangleleft G$ and $B$ is the direct product of $|W:V|$ copies of $A$. Also, $W$ acts to permute these direct factors of $B$, and this action is permutation isomorphic to the action of $W$ on the cosets of $V$ in $W$. (In fact, we assume that we are given a specific bijection from the set of cosets of $V$ onto the set of direct factors of $B$.)
Now let $C$ be the product of all of the direct factors of $B$ that correspond to nontrivial cosets of $V$, and note that ${\bf N}_W(C) = V$. Let $H = H_0$ be the group $CV$, and for $i > 0$, let $H_i = BV_i$. It is easy to see that  $H_0 < H_1 < \cdots < H_n = G$ is a subnormal series with factors $A_i$ as wanted. We must show that no shorter subnormal series for $H$ exists. Note that the subnormal depth of $H_1$ is exactly $n - 1$. (This can be seen by intersecting a subnornal series for $H_1$ in $G$ with $W$. This yields a subnormal series for $V$ in $W$.)
Suppose $H \triangleleft K$. We argue that $BK = BV$. Otherwise,
$BK > BV$, so $BK \cap W > V$. But $BK$ normalizes $C$ since $C = B \cap H$, and this contradicts the fact that $V$ is the full normalizer of $C$ in $W$. Now if $H = K_0 < K_1 < \cdots < K_m = G$ is a subnormal series for $H$, then  $H_1 = BV = BK_1 \subseteq \cdots \subseteq BK_m = G$ is a subnormal series for $H_1$ with length at most $m-1$, and thus $m \ge n$, as wanted.<|endoftext|>
TITLE: Negativity of contraction
QUESTION [8 upvotes]: Let $f:X\to Y$ be a birational morphism, $X, Y$ projective, $X$ smooth (threefold if this helps). Let $Exc(f)\subseteq X$ be the exceptional locus of $f$ and let $E\subseteq Exc(f)$ be an irreducible divisor. Is it true that for any curve $C\subseteq E$ contracted by $f$ one has $C\cdot E<0$? I can see this is true if $C$ is not contained in any other divisor sitting in $Exc(f)$, but what if it is?

REPLY [5 votes]: Dear Carlos, the statement is false in general. For example let $Y$ be $\mathbb{C}^3$, let $f_1 : X_1 \rightarrow Y$ be the blowup of a point on $Y$, and $f_2 : X \rightarrow X_1$ the blowup of a point on the exceptional divisor of $f_1$. Let $f : X \rightarrow Y$ be the composition. The exceptional locus of $f$ has two components: a copy $F$ of $\mathbb{P}^2$ (the exceptional divisor of $f_2$) and a copy $E$ of the blowup of $\mathbb{P}^2$ in one point (the strict transform of the exceptional divisor of $f_1$). The intersection $C = E \cap F$ is a copy of $\mathbb{P}^1$, it is a line on $F$ and the $f_2$-exceptional curve on $E$. Now $E \cdot C$ equals $C^2$ computed on $F$ (because $C=E \cap F$),
so $E \cdot C = +1$.<|endoftext|>
TITLE: Is there a syntactic characterization for BPP, BQP, or QMA?
QUESTION [26 upvotes]: Background
The complexity classes BPP, BQP, and QMA are defined semantically. Let me try to explain a little bit what is the difference between a semantic definition and a syntactic one. The complexity class P is usually defined as the class of languages accepted in polynomial time by a deterministic Turing machine. Although it seems to be a semantic definition at first, $P$ has an easy syntactic characterization, i.e. deterministic Turing machines with a clock counting the steps up to a fixed polynomial (take a deterministic Turing machine, add a polynomial clock to it such that the new machine will calculate the length of the input $n$, then the value of the polynomial $p(n)$, and simulate the original machine for $p(n)$ steps. The languages accepted by these machines will be in $P$ and there is at least one such machine for each set in $P$). There are also other syntactic characterizations for $P$ in descriptive complexity like $FO(LFP)$, first-order logic with the least fixed point operator. 
The situation is similar for PP. 
Having a syntactic characterization is useful, for example a syntactic characterization would allow us to enumerate the sets in the class effectively, and if the enumeration is efficient enough, we can diagonalize against the class to obtain a separation result like  time and space hierarchy theorems.

My main question is:

Is there a syntactic characterization for BPP, BQP, or QMA?

I would also like to know about any time or space hierarchy theorem for semantic classes mentioned above.

The motivation for this question came from here.
I used Google Scholar, the only result that seemed to be relevant was a citation to a master's thesis titled "A logical characterization of the computational complexity class BPP and a quantum algorithm for concentrating entanglement", but I was not able find an online version of it.

REPLY [15 votes]: No, I don't think any syntactic characterization is known for BPP, BQP or QMA. (BPP might turn out to be P, and then we'd have such a characterization of course.)
In particular we don't know any languages that are complete for either of these classes. A lot of people believe that classes like QMA do not even have complete languages. (See John Watrous' survey, where he says that "indeed it would be surprising if QMA were shown to have a complete problem having a vacuous promise.")
There are hierarchy theorems for BPP with 1 bit of advice, but I don't think we have any for BPP, BQP or QMA. For the advice-based results, see Hierarchy Theorems for Probabilistic Polynomial Time.<|endoftext|>
TITLE: The definition of homotopy in algebraic topology
QUESTION [18 upvotes]: In this post, let $I=[0,1]$.
Something about the definition of homotopy in algebraic topology (and in particular in the study of the fundamental group) always puzzled me. Most books on the fundamental group often begin with the basic notion of a homotopy of curves (or more generally, continuous functions between topological spaces) and describe it intuitively as "a continuous deformation of one curve into another". They often supplement this statement with some nice picture, like this one in Wikipedia. When I was taught algebraic topology, I too had heard a motivating explanation as above and was shown a picture of this sort. From this I could already guess what a (supposedly) natural formal definition would be. I expected it to look something like this:

Let $X$ be a topological space and let
  $f,g : I \to X$ be two curves in $X$.
  Then a homotopy between $f$ and  $g$
  is a family of curves $h_t: I \to X$
  indexed by $t \in I$ (the "time"
  parameter) such that $h_0 = f$, $h_1 =
> g$ and the function $t \mapsto f_t$ is
  continuous from $I$ to $C(I,X)$ (the
  space of curves in $X$ with domain
  $I$, equipped with some suitable
  topology).

However, the definition given (which is used in every book on algebraic topology which I sampled) is similar, but not quite what I thought. It is defined as a continuous function $H: I \times I \to X$ such that $H(s,0)=f(s)$ and $H(s,1)=g(s)$ for all $s \in I$.
This actually quite surprised me, for several reasons. First, the intuitive definition of a homotopy as a "continuous deformation" contains no mention of points in the space $X$ - it gives the feeling that it is the paths that matter, not the points of the underlying space (though obviously one needs the space in question to define the space of paths $C(I,X)$). However, the above definition, while formally almost equivalent to the definition I thought of (up to a definition of a "good" topology on $C(I,X)$), makes the underlying space $X$ quite explicit, it appearing explicitly in the range of the homotopy.
Moreover, many of the properties related to homotopies, the fundamental group and covering spaces can be expressed using the vocabulary of category theory, using universal properties. Now, from a categorical-theoretic point of view, wouldn't one want to suppress the role of the underlying space as much as one can (in favor of its maps and morphisms)?
Additionally, the definition of homotopy (as used) seems notationally inconvenient to me, in that it is less clear which of the two variables is the time parameter (each mathematician has his own preference, it seems). Also, the definition of many specific homotopies looks needlessly complicated in this notation, IMO. For instance, if $f,g$ are two curves in $\mathbb{R}^n$ then they are homotopic, and one can write the obvious homotopy either as $H(s,t)=tf(s)+(1-t)g(s)$ or as $h_t = tf+(1-t)g$. Maybe that's just me, but the second notation seems much more natural and easier to understand than the first one. Formulae of this sort appear frequently in the study of the fundamental group of various spaces (and in the verification that the fundamental group is indeed a group), and using the $H(s,t)$ notation makes these formulae much more cumbersome, in my opinion.
So, to sum up, I have two questions:

1) For a topological space $X$, can
  $C(I,X)$ be (naturally) topologized so
  that "my" definition of homotopy (see above) and
  the usual definition coincide (by
  setting $h_t (x) = H(x,t)$)?
2) If so, why isn't such a definition
  preferred? See my arguments above.

REPLY [7 votes]: In regards to your question 2, I think there are at least two reasons people may like the standard, general definition of homotopy between maps.  First, it lets you get right to the point without defining an extra space and a somewhat complicated topology on that space.  Of course at some point, one doesn't mind these complications, but for a beginning student (e.g. the reader of a basic topology book) thinking of homotopies as maps $X\times I \to Y$ is probably simpler.
The second thing is that I think it's often easier to check continuity of a specific homotopy by viewing it as a map $X\times I \to Y$.  For example, there are lots of places in topology where one uses what are essentially piecewise linear homotopies $I\times I \to Y$.  These are formed by cutting up the square into closed sets in some nice way, such that on each piece the homotopy is a linear function in two variables.  Then by the "gluing lemma," to check continuity you need only check that the maps agree on the overlaps.  I think you'll find examples of this if you work through the details of basic arguments with the fundmental group, for instance.  Some wonderful examples show up in May's Geometry of Iterated Loop Spaces.
All that being said, mapping spaces with the compact-open topology are really important, and it's a shame that they're not emphasized in many topology text books.  Ioan James' books are a good source, though (he has two books about fiber-wise topology that say everything I've ever needed to know) and Munkres has a good discussion too.<|endoftext|>
TITLE: Do free higher homotopy classes of compact Riemannian manifolds have preferred representatives?
QUESTION [25 upvotes]: A well known theorem of Cartan states that every free homotopy class of closed paths in a compact Riemannian manifold is represented by a closed geodesic (theorem 2.2 of Do Carmo, chapter 12, for example).  This means that every closed path is homotopic to a closed geodesic (through a homotopy that need not fix base points).
I am wondering if there is a higher dimensional generalization of this result.  Let us define a free n-homotopy class of $M$ to be a set $L_n$ of continuous maps $S^n \to M$ such that if $f \in L_n$ and $g: S^n \to M$ is any continuous map which is homotopic to $f$ (no base points required) then $g \in L_n$.

Question 1: Can one use the geometry of $M$ to intelligently single out a preferred class in any free n-homotopy class?

The naive guess is that every free n-homotopy class has an area minimizing representative, but my intuition tells me that this is either not true or really hard to prove.  This is because the proof in the 1-dimensional case develops the argument from certain continuity properties of arclength that tend to either fail or require tremendous care when generalized to area.  But perhaps this idea abstracts the wrong property of geodesics; maybe one should instead look for representatives which extremize some intelligently chosen functional instead.  I'm hoping someone has already thought about this and come up with a good answer.

Question 2: Can anything more be said in the presence of negative curvature?

In the 1-dimensional case, I believe that the closed geodesic guaranteed by Cartan is in fact unique if $M$ is compact with strictly negative curvature.  If there is an affirmative answer to question 1, I would be curious to know if there is a corresponding uniqueness statement in negative curvature.  And if question 1 doesn't seem to have a nice answer in general, maybe negative curvature helps.
Thanks in advance!

REPLY [12 votes]: A theorem of Sacks and Uhlenbeck tells us
http://www.jstor.org/pss/1971131
that we are still in the good shape for free homotopy $2$-classes. Namely, any such class can be represented by a collection of minimal spheres joined by geodesic segments. I would guess that in the case when you consider $\pi_n(M^{n+1})$ sometimes there will be a minimal hypersurface that realises such a class, for example in the case $M^{n+1}=S^1\times S^n$.
But for a general situation I remember a "claim" of Gromov, that such an preferred class does not exist, in particular for the case of homotopy groups of spheres. Unfortunately, I don't remember a reference (in fact I might be wrong here, see the article of Gromov below). Though the complexity of the question can be traced in the article of Larry Guth, treating in particular the case of maps $S^3\to S^2$.
Isoperimetric inequalities and rational homotopy invariants
http://arxiv.org/abs/0802.3550
ADDED.  In his new article Gromov says a bit about "good representatives" of maps in fixed homothopy classes $S^N\to S^n$. But this ounds more like a collection of questions and possible directions rather than some definite theorems (I guess it reflect the current state of these questions). This is contained in section 2,
http://www.ihes.fr/~gromov/PDF/manifolds-Poincare.pdf<|endoftext|>
TITLE: $L$-functions for $\Theta$-lifts
QUESTION [9 upvotes]: Let $E/F$ be a quadratic extension of number fields.  Let $W$ be a hermitian space over $E$ of dimension $2,$ and let $V$ be a skew-hermitian space of dimension $3$ over $E.$  Consider the associated unitary groups $H:=U(W)$ and $G:=U(V).$  Let $\sigma$ be an irreducible, cuspidal, automorphic representation of $H(\mathbb{A}_F).$  Let $\pi=\Theta(\sigma,\psi,\gamma)$ be a theta lift of $\sigma$ to $G(\mathbb{A}_F)$.  ($\psi:\mathbb{A}_F/F\to \mathbb{C}^\times$ and $\gamma:\mathbb{A}_E^\times/E^\times\to\mathbb{C}^\times$ are the splitting data necessary to define the theta-lift for unitary groups.)
My question is, how do automorphic $L$-functions (standard, adjoint, etc.) for $\pi$ relate to those for $\sigma$?

REPLY [3 votes]: This question is answered in a paper of Gan, Gross, and D. Prasad.  Here's a link:
http://www.math.ucsd.edu/~wgan/ggp-evidence4-1.pdf
The relation between L-parameters of representations and their theta-lifts (at least locally) is discussed in section 7 of the paper.<|endoftext|>
TITLE: Category and homotopy theoretic methods in set theory
QUESTION [8 upvotes]: I am looking for applications of category theory and homotopy theory in set theory and particularly in cardinal arithmetics. "Applications" in the broad sense of the word --- this would include theorems, definitions, questions, points of view (and papers) in set theory that could be motivated or understood with help of category theory and homotopy theory. I am aware of some applications of set theory in category theory, e.g. large cardinal axioms (Vopenka principle) are used to construct localisations in homotopy theory, but this is not what I am asking for. However, I would be interested to hear if Vopenka principle is equivalent to a statement in category or homotopy theory.
The reason for the question is that I am trying to better understand this sketch of an attempt to understand an invariant in PCF theory in terms of homotopy theory. I am most interested in applications to cardinal arithmetic.

REPLY [5 votes]: Peter Freyd wrote a paper, "The Axiom of Choice," in which he used topos-theoretic methods to prove that the axiom of choice is independent of (classical) Zermelo-Fraenkel set theory.  Unlike earlier topos-versions of set-theoretic independence proofs, Freyd's construction does not merely provide a category-theoretic view of a model that had already been considered by set theorists.  His models can be obtained by set-theoretic forcing methods (by another result of Freyd, in "All topoi are localic"), but those particular forcing constructions had not been considered until Andre Scedrov and I analyzed them (in "Freyd's models for the independence of the axiom of choice").  So I think it's fair to say that these models were a contribution from category theory to set theory.<|endoftext|>
TITLE: Origins of names of algebraic structures
QUESTION [43 upvotes]: Consider the names of basic algebraic structures: 'group', 'ring', 'space', 'field', 'Körper', even the name 'structure' itself  - all of them time-honoured terms, deeply rooted in our history and culture. 
But what has an algebraic field to do with an acre? What has an algebraic group to do with a group of people?
Even when it's known who coined these names (of algebraic structures), it's not obvious why they were choosen and what the connection is between the named structures and what was named originally (or later on). Only those who coined the names could tell.

Are there etymological studies concerning
  these names - 'group', 'ring',
  'space', 'field',... - which
  elucidate this connection?

REPLY [15 votes]: For the origin of all names, see: http://jeff560.tripod.com/mathword.html
For the origin of all symbols, see: http://jeff560.tripod.com/mathsym.html<|endoftext|>
TITLE: Undergraduate roadmap to algebraic geometry?
QUESTION [41 upvotes]: Hello,
I'm sorry if this question isn't posted correctly. I hope that it is (since other questions regarding roadmaps have been allowed). Now to my question:
From what I've heard from professors and such, algebraic geometry seems like an interesting branch of mathematics. I'd like to learn some basic results and maybe do some kind of thesis in a few years on the subject. So, what I'm curious about is you have any tips on what books to read? Say that one has read Artin's Algebra and Herstein's Topics in Algebra, and also has the basic courses in real analysis and topology, complex variables etc. down, where should one go to learn? What books? I'm also curious if algebraic geometry (at an "easy level") requires deep knowledge about other fields of mathematics too, so that one might have to read books that at first seems to have no relevance to algebraic geometry?
Best regards.

REPLY [6 votes]: I think the best way is to read a book on commutative algebra (Atiyah & MacDonald) and then, you can start reading Hartshorne's. Chapter 1 will give you a fairly concrete idea of what classical alg. geo. is about. You should also try to do all the exercises even though that will be time consuming. I think this is the best way one can do it.<|endoftext|>
TITLE: P/poly algorithm for polynomial identity testing
QUESTION [8 upvotes]: By the Schwartz–Zippel lemma, "Is this arithmetic formula identically zero?" is in coRP $\subseteq$ BPP $\subset$ P/poly, with the second inclusion by Adleman's theorem.  By basically following the proof, but using the improved error bound that comes from the original algorithm only having one-sided error, one gets an algorithm that computes suitable advice. (equivalently, a suitable circuit)
Is there any known P/poly algorithm for this problem with advice that can be computed faster?
(I already know about www.cs.sfu.ca/~kabanets/Research/poly.html)

REPLY [10 votes]: The Schwartz-Zippel lemma is very fast, only one evaluation of the formula at one random point. There's nothing better known that minimizes time and error as well as Schwartz-Zippel. But Schwartz-Zippel requires a lot of randomness in each repetition: a fresh new point of n elements.
Have you tried some of the polynomial identity tests with better tradeoffs between randomness and error? Their running time (and the running time dependence on the error) is a bit worse than Schwartz-Zippel, but the number of random bits needed is much less than Schwartz-Zippel. So in the application of Adleman's theorem, the sizes of the witnesses you need to hard-code in the non-uniform circuit will shrink, but the time dependence on error increases, potentially making the number of necessary witnesses increase. Given these complex tradeoffs, I'm not sure which of them would work best for obtaining small circuits.
For a quick overview of these alternative identity tests and their tradeoffs, see the table on p.3 in Agrawal and Biswas: http://www.cse.iitk.ac.in/users/manindra/algebra/identity.pdf<|endoftext|>
TITLE: Perron-Frobenius "inverse eigenvalue problem"
QUESTION [29 upvotes]: The Perron-Frobenius theorem says that the largest eigenvalue of a positive real matrix (all entries positive) is real.  Moreover, that eigenvalue has a positive eigenvector, and it is the only eigenvalue having a positive eigenvector.
Now suppose we want to construct a positive rational matrix with a particular Perron-Frobenius eigenvalue.  Specifically, consider a positive real algebraic number $\lambda$ which is greater in absolute value than all of its Galois conjugates.  Does there exist a positive rational matrix $A$ with $\lambda$ as its Perron-Frobenius eigenvalue?

REPLY [4 votes]: Perhaps you're thinking of the strong form given in the Roy Adler conference
Symbolic Dynamics and Its Applications (Contemporary Mathematics) [1992].
One of the papers there by me was answering a question of Doug Lind, to show that if $\lambda $ exceeds the absolute value of its algebraic conjugates, then it appears as the Perron eigenvalue of a primitive integer matrix. In general, you can arrange it so that the nonzero spectrum consists of $\lambda$ and its conjugates (multiplicity one each of course) together with a usually large number of $1$s. The presence of many more eigenvalues (hence there is no control on the size of the realizing matrix) is necessary, since we must also have $\text{tr} A^n \geq 0$ for all $n$ (and other conditions) if $A$ is a primitive integer matrix.<|endoftext|>
TITLE: Is being torsion a local property of module elements?
QUESTION [16 upvotes]: Say $R$ is a ring, not necessarily a domain, and $M$ is an $R$-module. All rings are commutative with 1.  An element $m\in M$ is called torsion if $r.m=0$ for some regular element (non-zerodivisor) $r\in R$.  
(I learned this definition over non-domains from a lecture of Irena Swanson, and it's noted in the last paragraph of the definition on Wikipedia.)
It's easy to see that torsion elements localize to torsion elements (note $0$ is always torsion), because regular elements never disappear.  How about the converse?  If an element is locally torsion, is it torsion?  That is,

If $f_1,\ldots,f_n$ generate the unit ideal in $R$, and $m\in M_{f_i}$ is torsion over $R_{f_i}$ for each $i$, then is $m$ torsion over $R$?

I started trying to make a high-dimensional variety as a counterexample, but instead wound up proving that for $R$ Noetherian, torsionality is stalk-local 1, so in that case the answer is yes.  
So how about a proof or a counterexample for rings in general?  This will affect when and whether I think about sections of $O_X$-modules on a non-integral scheme $X$ as "torsion" or not... 

1 Proof of stalk-locality in Noetherian case: Say $m\in M$ is non-torsion, meaning $ann(m)$ is contained in the set of zero-divisors of $R$, which equals the union of the associated primes of $R$.  By prime avoidance, $ann(m)$ is contained in some associated prime of $R$, say $p=ann(x)$ for $x\in R$.  I claim $m$ localizes to a non-torsion element of $M_p$.  If $r.m=0$ in $M_p$ for $r\in R$ (WLOG), it means $rs.m=0$ in $M$ for some $s\notin p$.  Now, $rs\in ann(m)\subseteq p$, so $r\in p$ and $rx=0$ in $R$.  But $x\neq 0$ in $R_p$ since $ann(x)$ is (contained in) $p$, so $r$ is not regular in $R_p$, as required.  Hence an element of $M$ which is torsion in every $M_p$ must be torsion in $M$.

REPLY [6 votes]: No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, this should work now. Apologies for the length of this answer, but I don't see any quick constructions].
The idea is to construct a ring $R$ and an ideal $I$ contained in the zero divisors of $R$, and $f_1,f_2\in R$ satisfying $f_1+f_2=1$ such that $I_{f_i}$ contains a regular element of $R_{f_i}$ for each $i$.
This provides a counterexample to the question by taking the module $M=R/I$ and $m=I+1$. Then, ${\rm ann}(m)=I$ consists of zero divisors, so $m$ is not torsion. However, mapping $m$ into $M_{f_i}$ takes ${\rm ann}(m)$ to $I_{f_i}$, which contains regular elements of $R_{f_i}$. So, $m$ is torsion in each $M_{f_i}$.
This does get rather involved, so let's start simple and construct an example showing that being torsion is not a stalk-local property.
Choose a field $k$, set $A=k[X_0,X_1,X_1,X_2,\ldots]$ and let $J$ be the ideal generated by $X_iX_j$ for $i\not=j$ and $X_i(X_i-1)$ for $i\ge1$. Then, $R=A/J$ is the $k$-algebra generated by elements $x_0,x_1,\ldots$ satisfying the relations $x_ix_j=0$ for $i\not=j$ and $x_i(x_i-1)=0$ for $i\ge1$. Let $I\subseteq R$ be the ideal generated by $x_0,x_1,\ldots$.
We can see that $x_i\not=0$ by considering the $k$-morphism $A\to k$ taking $X_j$ to 1 (some fixed $j$) and $X_i$ to 0 for $i\not=j$. This takes $J$ to 0, so it defines a $k$-morphism $R\to k$ mapping $x_j$ to 1, so $x_j\not=0$. Then, every $a\in I$ satisfies $ax_j=0$ for large $j$, showing that it is a zero divisor. Also, the $k$-morphism $A\to k$ taking each $X_i$ to zero contains $J$ in its kernel, and defines a morphism $R\to\mathcal{k}$ with kernel $I$, showing that $R/I\cong k$. So $I$ is a maximal ideal. For any prime $\mathfrak{p}$ we either have $\mathfrak{p}\not=I$, in which case the non-empty set $I\setminus\mathfrak{p}$ maps to units (and hence, regular elements) in $R_{\mathfrak{p}}$. Or, we have $\mathfrak{p}=I$ in which case $x_i-1$ maps to a unit and $x_i$ goes to zero in $R_{\mathfrak{p}}$ ($i\ge1$). So, $R_{\mathfrak{p}}\cong k[X]$ with $x_0$ going to the regular element $X$. This shows that $I$ contains regular elements in the localization at any prime, giving the required counterexample for the stalk-local case.
Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, I will start with a simple example of a polynomial ring and then transform it in such a way as to give the properties we are looking for. I find it helpful to first fix the following notation: Start with the base (polynomial) ring $R=\mathbb{Z}[x,y,z]$. A (commutative, unitial) R-algebra is simply a ring with three distinguished elements $x,y,z$, and a morphism of R-algebras is just a ring homomorphism respecting these distinguished elements. For an R-algebra $A$, define $K(A)\subseteq A$ to be the smallest ideal such that, for all $a\in A$,
$$
\begin{align}
ax\in K(A)&\Rightarrow az\in K(A),\\\\
ay\in K(A)&\Rightarrow a(1-z)\in K(A).
\end{align}
$$
In particular, $K(A)=0$ implies that $x$ is a regular element in the localization $A_z$ and $y$ is a regular element in $A_{1-z}$. If we can construct such an example where the ideal $Ax+Ay$ consists purely of zero divisors, then that will give the counterexample needed. The idea is to start with $A=\mathbb{Z}[x,y,z]$ and transform it using the following steps.

Force the elements of $I=Ax+Ay$ to be zero divisors. So, for each $a\in I$, add an element $b$ to $A$ in as free a way as possible such that $ab=0$. Adding elements to $A$ also has the effect of adding elements to $I$. So, this step needs to be iterated to force these new elements of $I$ to also be zero divisors.
Replace $A$ by the quotient $A/K(A)$ to force the condition $K(A)=0$.

The first step above is easy enough. However, we do need to be careful to check that the second step does not undo the first. Suppose that $a\in A$ is a zero divisor, so that $ab=0$ for some non-zero $b$. It is possible that taking the quotient in the second step above takes $b$ to zero, so that $a$ becomes a regular element again. To get around this, we need some stronger condition on $b$ which implies $b\not=0$ and is also stable under taking the quotient. Note that $A(1-b)$ being a proper ideal or, equivalently, $A/(1-b)$ being nontrivial, will imply that $b\not=0$. In turn, this is implied by $K(A/(1-b))$ being a proper ideal. As it turns out, this property of $b$ does remain stable under each of the steps above, and can be used to show that this construction does give the counterexample required. However, note that if $ab=0$ and $K(A/(1-b))$ is proper, then $a=a(1-b)\in A(1-b)$, from which we can deduce that $K(A/(a))$ is a proper ideal. This necessary condition is unchanged by either of the steps above, so we had better check that elements $a\in Ax+Ay$ in our R-algebra do satisfy this from the outset. I'll make the following definition: $A$ satisfies property (P) if $K(A/(a))$ is proper for every $a\in Ax+Ay$. As it turns out, polynomial rings do satisfy this property and, consequently, the construction outlined above works fine.
Now on to the details of the argument.

(1) Let $f\colon A\to B$ be an R-morphism. Then $f(K(A))\subseteq K(B)$. Furthermore,

If $I\subseteq A$, $J\subseteq B$ are ideals with $f(I)\subseteq J$ and $K(B/J)$ is proper, then $K(A/I)$ is proper.
If $B$ satisfies (P) then so does $A$.


As $f^{-1}(K(B))$ satisfies the defining properties for $K(A)$ (other than minimality), it contains $K(A)$. In particular, if $K(B)$ is proper then $K(A)\subseteq f^{-1}(K(B))$ is proper. Next, if $f(I)\subseteq J$ are ideals, then $f$ induces an R-morphism $A/I\to B/J$ so, if $K(B/J)$ is proper then so is $K(A/I)$.
If $B$ satisfies property (P) and $a\in Ax+Ay$ then $f(a)\in Bx+By$ and $K(B/f(a))$ is proper. So, $K(A/(a))$ is proper and $A$ also satisfies property (P).

(2) If $A$ is a non-trivial ring then the polynomial ring $R\otimes A\cong A[x,y,z]$ satisfies (P).

As $A$ is non-trivial, it must have a maximal ideal $\mathfrak{m}$. Applying (1) to the R-morphism $A[x,y,z]\to(A/\mathfrak{m})[x,y,z]$ reduces to the case where $A=k$ is a field. Then, letting $\bar k$ be the algebraic closure, applying (1) to $k[x,y,z]\to\bar k[x,y,z]$ reduces to the case where $A=k$ is an algebraically closed field.
Now set $B=k[x,y,z]$ and choose $a\in Bx+By$. The idea is to look at the morphism $\theta\colon B/(a)\to k$ taking $x,y,z$ to some $x_0,y_0,z_0\in k$ with $a(x_0,y_0,z_0)=0$. As long as these satisfy $ux_0=0\Rightarrow uz_0=0$ and $uy_0=0\Rightarrow u(1-z_0)=0$ (all $u\in k$) then $K(B/(a))$ will be contained in the kernel of $\theta$, so will be proper. For this to be the case it is enough that both ($x_0\not=0$ or $z_0=0$) and ($y_0\not=0$ or $z_0=1$).
Case 1: We can find $a(x_0,y_0,z_0)=0$ such that $x_0y_0\not=0$. This satisfies the required condition.
Case 2: Whenever $a(x_0,y_0,z_0)=0$ then $x_0y_0=0$. This means that $xy$ is contained in the radical ideal generated by $a$, so $a$ divides $x^ry^r$ some $r\ge1$. Then $a$ is a multiple of $x$ or $y$ and one of $(x_0,y_0,z_0)=(0,1,0)$ or $(1,0,1)$ satisfies the required condition. So, $K(B/(a))$ is proper.
Next, we construct extensions of the R-algebra forcing elements of $Ax+Ay$ to be zero-divisors.

(3) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ with a left-inverse and such that, for every $a\in Ax+Ay$, there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

To construct the morphism, set $I=Ax+Ay$ and let $(X_a)_{a\in I}$ be indeterminates over $A$. Let $J$ be the ideal in $A[(X_a)_{a\in I}]$ generated by $(aX_a)_{a\in I}$. Then define $B=R[(X_a)_{a\in I}]/J$ and let $f$ be the canonical homomorphism. Its left inverse is the map taking $X_a$ to 0.
Now, for a fixed $a\in I$, set $b=J+X_a$, so $ab=0$. Consider the morphism $A[(X_c)_{c\in I}]\to A\to A/(a)$ taking each $X_c$ to 0 (for $c\not=a$) and $X_a$ to 1. As its kernel contains $J$, it defines a morphism $g\colon B\to A/(a)$, which takes $b$ to one. Therefore, the ideal $B(1-b)$ maps to 0 and, as $K(A/(a))$ is proper, (1) says that $K(B/(1-b))$ is proper.

(4) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ such that, for every $a\in Bx+By$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

Set $A_0=A$ and use (3) to construct a sequence of extensions $f_i\colon A_i\to A_{i+1}$ with left inverses such that, for every $a\in A_ix+A_iy$ there is a $b\in A_{i+1}$ with $ab=0$ and $K(A_{i+1}/(1-b))$ is proper. Note that, as each $f_i$ has a left inverse, (1) says that $A_{i+1}$ satisfies (P) whenever $A_i$ does. So, we can keep applying (3) to build up the entire sequence of extensions.
We now take the colimit $B={\rm colim}A_i$ and let $f$ be the induced morphism (i.e, if we consider $A_i\subseteq A_{i+1}$ then $B$ is the union and $f$ is inclusion). As each $A_i\to B$ has a left-inverse, (1) shows that the required properties for $B$ are inherited from the individual $A_i$.

(5) Suppose that $A$ satisfies the following: for every $a\in Ax+Ay$ there is a $b\in A$ with $ab=0$ and $K(A/(1-b))$ is proper. Then, the R-algebra $B=A/K(A)$ satisfies the same property, and also $K(B)=0$.

That $K(B)=0$ follows quickly from the definition of $K$. 
Suppose $a\in Ax+Ay,b\in A$ are such that $ab=0$ and $K(A/(1-b))$ is proper. Set $C=A/(1-b)$, so that $C/K(C)$ is not trivial. By (1), the canonical morphism $A\to C$ maps $K(A)$ into $K(C)$. So, it induces a morphism $B\to C/K(C)$. This takes $1-b$ to zero, so it induces an R-morphism $B/(1-b)\to C/K(C)$. As $K(C/K(C))=0$, (1) implies that  $K(B/(1-b))$ maps to zero, so is proper.

(6) If $B$ is the R-algebra constructed in (5), then $I=Bx+By$ contains only zero-divisors but $x,y\in I$ map to regular elements in $R_z$ and $R_{1-z}$ respectively.

For any $a\in I$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ proper. In particular, $(1-b)$ must be a proper ideal, so that $b\not=0$ and $a$ is a zero divisor.
Finally, the property $K(B)=0$ implies that $x$ is regular in $B_z$ and $y$ is regular in $B_{1-z}$.<|endoftext|>
TITLE: Does there exist a potential which realizes this strange quantum mechanical system?
QUESTION [7 upvotes]: I have done some courses on quantum mechanics and statistical mechanics in the past. Since I also do math, I wonder about converge issues which are usually not such a problem in physics. One of those questions is the following. I will describe the background, but in the end it boils down to a question about ordinary differential equations.
In quantum mechanics on the real line, we start with a potential $V: \mathbb{R} \to \mathbb{R}$ and try to solve the Schrödinger question $i\hbar \frac{\partial}{\partial t}\Psi(x,t) = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)$. In many cases this can be accomplished by seperating variables, in which case we obtain the equation $E\Psi(x,t) = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)$ which we try to solve for $E$ and $\Psi$ to obtain a basis for our space of states together with an associated energy spectrum. For example, if we have a harmonic oscillator, $V(x) = \frac{1}{2}m\omega^2x^2$ and we get $E_n = \hbar \omega (n+\frac{1}{2})$ and $\Psi_n$ a certain product of exponentials and Hermite polynomials. We assume that the energy in normalized such that the lowest energy state has energy $0$.
If the states of our system are non-degenerate, i.e. there is only one state for each energy level in the spectrum, then the partition function in statistical mechanics for this system is given by the sum $Z(\beta) = \sum_n \exp(-\beta E_n)$, where $\beta$ is the inverse temperature $\frac{1}{k_B T}$. It is clear that this sum can be divergent; in fact for a free particle ($V = 0$), it is not even well defined since spectrum is a continuum.
However, I was wondering about the following question: Is there a system such that $Z(\beta)$ diverges for $\beta < \alpha$ and converges for $\beta > \alpha$ for some $\alpha \in \mathbb{R}_{> 0}$? Am I correct in thinking that such a system is most likely an approximation of another system, which undergoes a phase transition at $\beta = \alpha$?
Anyway, an obvious candidate would be a potential $V$ such that the spectrum is given $E_n = C \log(n+1)$ for $n \geq 0$ and $C > 0$. This gets me to my main mathematical question: Does such a potential (or one with spectrum asymptotically similar) exist? If so, can you give it explicitly?
One the circle, the theory of Sturm-Liouville equations tells us that the eigenvalues must go asymptotically as $C n^2$, so in this case such problems can't occur. I don't know much about spectral theory for Sturm-Liouville equations on the real line though. The second question is therefore: What is known about the asymptotics of the spectrum of a Sturm-Liouville operator on the real line?

REPLY [5 votes]: If I understand your first question correctly, then the answer is yes. In fact, all physical matter exhibits this behavior. Allow me to answer in the following mathematically nonrigorous way:
Consider that even in a lone hydrogen atom, the Hamiltonian operator for the nonrelativistic electron
$H = - \frac 1 2 \nabla^2 + \frac{1}{r}$
has a discrete spectrum of bound states corresponding to the 1s, 2s, 2p, 3s, ... atomic orbitals and a continuous spectrum of unbound states corresponding to an electron that is unbound for all practical purposes. Thus at sufficiently high temperature (probably at $\beta^{-1}$ =  kT ~ 0.5) there will be significant population of the continuous spectrum and you would have to deal with counting the continuous spectrum in the partition function.
The same phenomenon exists for all atoms and collections of atoms, even when the nuclear and interactions terms are turned on.
I am not 100% confident that the same thing holds in the relativistic case too, but I would be surprised if it did not.

Regarding your discussion of the harmonic oscillator, and the comment that "such a system [exhibiting such divergence at a critical temperature] is most likely an approximation of another system", I would go so far as to say that it is the other way round, that almost all the time "nice" systems like the harmonic oscillator are in fact derived as asympotic approximations to messier Hamiltonians. For example, you could write down the molecular Hamiltonian
$H = \sum_i -\frac 1 2 \nabla_i^2 + \sum_{ij} \frac 1 {r_{ij}} - \sum_{Ki} \frac {Z_K} {r_{iK}} + \sum_K -\frac 1 2 \nabla_K^2 $
which as mentioned above has both a discrete part and a continuous part to its spectrum, and assume that we are interested only in the regime where we care about slow atomic nuclear motions, and that they move very little, and from there derive an effective lattice Hamiltonian of coupled harmonic oscillators. While the phase transition can be observed in the original molecular Hamiltonian, it would not be possible to see this occur in the simplified Hamiltonian since the the discrete spectrum of the harmonic oscillators would go on forever without becoming continuous.<|endoftext|>
TITLE: Integration and Stokes' theorem for vector bundle-valued differential forms?
QUESTION [24 upvotes]: Is there a version of Stokes' theorem for vector bundle-valued (or just vector-valued) differential forms?
Concretely: Let $E \rightarrow M$ be a smooth vector bundle over an $n$-manifold $M$ equipped with a connection. First of all, is there an $E$-valued integration defined on the space $\Omega^{n-1}(M,E)$ of smooth sections of $E\otimes \Lambda^{n-1}T^\ast M$ mimicking what you have for $\mathbb{R}$-valued forms? If so, does
$\int_{\partial M} \omega = \int_M d\omega$
hold (or even make sense) for $\omega\in\Omega^{n-1}(M,E)$ if $d$ is the exterior derivative coming from our connection on $E$?
In this setting, letting $E$ be the trivial bundle $M\times \mathbb{R}$ should give the ordinary integral and Stokes' theorem.
Sorry if the answer is too obvious; I just don't have any of my textbooks available at the moment, and have never thought about Stokes' theorem for anything other than scalar-valued forms before.

REPLY [6 votes]: After looking at this question for a few days in the context of the Riemann curvature tensor, holonomy for a given affine connection, and the (false) conjecture that the parallel transport around the boundary curve could equal the integral of the Riemann tensor within the span of the closed curve, I've concluded that the Stokes theorem cannot be applied to this conjecture except when the connection is flat.
The reason for the failure of the conjectured relation between curvature and parallel transport is that the Stokes theorem's integrals are themselves not really well defined. But it is not that simple. I'll explain...
When even constructing a simple Riemann integral from the fundamentals, one has to add vectors at different points inside the region. Even if you have a connection, you have to decide which paths to use to connect the points of the region. You can do a kind of a "raster scan" of the image of a rectangular region of $\mathbb{R}^2$, parallel transporting the vectors back to the left of the scan to add them on the left hand side, and then you can parallel transport all of these X-scans down the Y-axis by transporting them down to the bottom left of the rectangle. But then what do you have? It's clearly not geometrically meaningful. And then you have the same problem with the boundary integral of a vector function.
A second conclusion which I came to is that if you do apply the Stokes theorem to this scenario, you get a mathematically correct identity, which has a practical value as the first iteration of the Picard iteration procedure to compute the parallel transport around the curve. This, clearly, is not extremely useful. But in my opinion, the Stokes theorem is applicable to this situation. It just doesn't give anything geometrically meaningful for a non-flat geometry, and it has limited value for relating curvature to parallel transport and holonomy. On the other hand, it does get the right answer in the limit of a shrinking region to a point, which gives the correct answer for the "Cartan characterization of curvature".
This issue is related to questions 16850 and 50051.<|endoftext|>
TITLE: Polynomial with prime powers values
QUESTION [6 upvotes]: It's well known that there are no non-constant polynomials with integer coefficients whose values at integer points are primes. Could this result be generalized to the case of prime powers?
The question is whether there exists a polynomial $p(x) \in \mathbb{Z}[x]$ with degree at least one such that for all $x \in \mathbb{Z}$ $|p(x)|$ is prime power.

REPLY [5 votes]: Actually you can prove a lot more.
Theorem
For any non-constant polynomial $p(x)\in\mathbb{Z}[x]$ and any positive integer $k$ there is an integer $n$ such that $p(n)$ is divisible by at least $k$ distinct primes.
Proof
If we prove that there exist integers $n_1,\ldots,n_k$ and distinct primes $p_1,\ldots,p_k$  such that $p(n_i)\equiv 0 \bmod{p_i}$ then we are done, because there exists an $n$ such that $n\equiv n_i\bmod{p_i}$ by the Chinese Remainder Theorem, and any such $n$ satisfies $f(n)\equiv f(n_i)\equiv 0 \bmod{p_i}$, as desired.
Now by contradiction suppose that $p$ is only divisible by the primes $p_1,\ldots,p_l$, $l\le k-1$. Since we have that $p(0)\neq 0$, let $p(0)=\pm p_1^{\alpha_1}\ldots p_l^{\alpha_l}$, $x\equiv 0\bmod{p_1^{\alpha_1+1}\ldots p_l^{\alpha_l+1}}$.
Then $p(x)\equiv p(0)\bmod{p_i^{\alpha_i+1}}$, $1\leq l\leq k-1$, so that the greatest power of $p_i$ that divides $p(x)$ is $p_i^{\alpha_i}$. 
But by hypothesis $p(x)$ is only divisible by the $p_i$, so we conclude that $p(x)=\pm p(0)$. Using the pigeonhole principle and the fact that a non-constant polynomial can only assume a value a finite amount of times this is a contradiction, as desired.<|endoftext|>
TITLE: Are there any (interesting) consequences of the irrationality of π?
QUESTION [12 upvotes]: I am not sure  how appropriate this question is for MO. If it is not, I apologize in advance but I could not resist asking it and if by any chance I get some interesting answers, it will for sure be very useful to keep my students excited about mathematics and physics as September arrives. 
We all know very well that $\pi$ (the ratio of the circumference of a circle to its diameter in Euclidean space) is irrational and even  transcendental. These  are some of the famous results in all mathematics. 
So I was wondering what will go wrong if $\pi$  was just an integer number? 

Are
  there important theorems that are based on the fact that it is actually irrational
  and/or transcendental?

REPLY [12 votes]: The fact that π is irrational has few direct applications. However the techniques used to prove this, or rather used to prove the stronger statement that it is transcendental, have many applications. For example, Baker proved that 1 and the logs of algebraic numbers are linearly independent over algebraic numbers except in trivial cases. (This includes the fact that π is irrational as a special case because π = log(-1)/i.) Baker used his theorem to give effective bounds on the solutions of Diophantine equations and to solve Gauss's class number problem for imaginary quadratic fields, among other things. See Baker's book on transcendental number theory for more details.<|endoftext|>
TITLE: Ideal of "Compact Operators" in a W*-algebra which gives the sigma-strong-* topology.
QUESTION [5 upvotes]: In the case of bounded operators on a Hilbert space $\mathcal{H}$, $L(\mathcal{H})$, there are multiple descriptions of the $\sigma$-strong-* topology, namely:
1) As given by seminorms $p_{\phi},~p_{\phi}^{*}$, indexed by normal linear functionals on $L(\mathcal{H})$, and defined on operators $T\in L(\mathcal{H})$ by $p_{\phi}(T)=\phi(T^{*}T)^{1/2}$ and $p_{\phi}^{*}(T)=\phi(TT^{*})^{1/2}$.
2) As given by semi-norms $p_{\xi_{i}}(T),~p_{\xi_{i}}^{*}(T)$, indexed by sequences $\lbrace \xi_{i}\rbrace_{i\in\mathbb{N}}$ such that $\sum_{i} ||\xi_{i}||\leq \infty$, and defined on operators $T\in L(\mathcal{H})$ by $p_{\xi_{i}}(T)= \sum_{i\in\mathbb{N}}||T\xi_{i}||$ and $p_{\xi_{i}}^{*}(T)=p_{\xi_{i}}(T^{*})$.
3) The strict topology coming from the strict topology when $L(\mathcal{H})$ is identified with the multiplier algebra of the compact operators $C(\mathcal{H})$. More precissely given by the seminorms $p_{K},~p_{K}^{*}$, indexed by operators $K\in C(\mathcal{H})$, and defined on operators $T\in L(\mathcal{H})$ by $p_{K}(T)=||KT||$ and $p_{K}^{*}(T)=||TK||$.
The first description is the usual definition of the $\sigma$-strong-* topology on any W*-algebra, the second is the usual definition when the W*-algebra is faithfully represented on a Hilbert space.
The question I now like to ask, is: "Is there for any W*-algebra $A$ an ideal $I$ such that $M(I)=A$ and the corresponding strict topology coincides with the $\sigma$-strong-* topology?"
May be the standard form of a Von Neumann algebra can be of help, but I am not very into Tomita-Takesaki theory and I am hoping to avoid that.

REPLY [2 votes]: You might want to make clear what you mean by "ideal". If it is a von Neumann ideal, then the answer is simply no because in that case you would have $M(I)=I$. 
Now, in the motivating example the ideal is an essential C*-ideal. But even then the general answer to the question has to be no. If you consider a II$_1$-factor, then it is simple as a C*-algebra, so the only nonzero ideal is $A$ itself. Of course $M(A)=A$, but the strict topology cannot agree with the $\sigma^*$-strong: because the ideal contains the identity, the strict topology in this case agrees with the norm topology.
Regarding Dmitri's answer, in a II$_\infty$-factor $M$ the "compact" ideal $K$ (norm closure of the span of the finite projections) is an essential ideal, and so $M(K)=M$ by standard results. But I don't immediately see that the strict topology in this case agrees with the $\sigma^*$-strong on bounded sets (because the key of the argument for $K(H)\subset B(H)$ is that for a finite projection $p\in B(H)$, $\|xp\|$ can be calculated over a finite-dimensional subspace).<|endoftext|>
TITLE: Criterion for an abelian group to have a commutative endomorphism ring
QUESTION [8 upvotes]: Given an abelian group $G$, one can form the endomorphism ring $\mbox{End}(G)$ by letting $\alpha+\beta=\alpha(x)+\beta(x)$, and $\alpha\beta=\alpha(\beta(x))$, where $\alpha$ and $\beta$ are endomorphisms. Clearly, composition distributes over addition, and addition is commutative, so $\mbox{End}(G)$ is a ring. My question is: when is $\mbox{End}(G)$ commutative? Are there a nice set of criteria, or, if there is no such nice set of criteria, is there a nice class of abelian groups with commutative endomorphism rings.

REPLY [16 votes]: Amongst the finitely generated abelian groups, those with commutative endomorphism ring are exactly the cyclic groups.
Torsion abelian groups with commutative endomorphism rings are exactly the locally cyclic groups, that is, the subgroups of Q/Z.  They were classified in:

Szele, T.; Szendrei, J.
"On abelian groups with commutative endomorphism ring."
Acta Math. Acad. Sci. Hungar. 2, (1951). 309–324
MR51835
DOI:10.1007/BF02020735

This papers also gives more complicated examples of mixed groups with commutative endomorphism ring.
The mixed case was completed in:

Schultz, Ph.
"On a paper of Szele and Szendrei on groups with commutative endomorphism rings."
Acta Math. Acad. Sci. Hungar. 24 (1973), 59–63.
MR316598
DOI:10.1007/BF01894610

This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").<|endoftext|>
TITLE: Consecutive integers with many prime factors
QUESTION [7 upvotes]: This question is inspired by Project Euler's Problem 47.
Let m and n be positive integers. Consider the following four (related) statements:

There exists m consecutive integers, each of which has at least n distinct prime factors.
There exists m consecutive integers, each of which has precisely n distinct prime factors.
There exists m consecutive integers, each of which has at least n prime factors (counted with multiplicity)
There exists m consecutive integers, each of which has precisely n  prime factors (counted with multiplicity).

Question $i$: For which pairs $(m,n)$ is statement $i$ true?
For $n = 2$ and any $m$, statement 3 is obviously true (it is just the elementary fact that there are prime gaps of arbitrary length). I suppose it is not too hard to avoid prime powers, so that statement 1. is also true for $n=2$ and arbitrary $m$ (although at the moment I cannot see if this follows by an easy modification of the usual $(m+1)!+2, \ldots, (m+1)!+m+1$ argument). Are there other elementary cases? In particular, is there an easy proof that Problem 47 (which is to find the first instance of Statement 1 for $m=n=4$) is actually solvable? The actual answer is small enough that it is found rather quickly by a naive brute force approach, but it would be nice to know in advance that there is an answer (perhaps even having an upper bound, so that one also knows that one may find the answer before the heat death of the universe).
Statement $i$ can be strengthened by requiring the existence of infinitely many $m$-tuples; call this Statement $i'$.
Statements 2 and 4 are probably too hard to say anything about in general (but maybe not?), but for $m=2$, Statements $2'$ and $4'$ are both weaker versions of this question.

REPLY [5 votes]: Regarding questions 1 and 3: the counting function of the set of integers with at most n-1 prime factors is $O(x(\log\log x)^{n-2}/\log x)$, where the implicit O-constant depends on n. (This is true for either distinct prime factors or factors counted with multiplicity. See the references in this problem to Montgomery and Vaughan or to Tenenbaum.) In particular, the set of numbers with only a few prime factors has density zero, which implies the existence of arbitrarily large gaps, hence implies 1 and 3 are true.
Regarding questions 2 and 4, for suitable values of m and n (as noted in other comments, 2 and 4 aren't true for all m and n) they should follow from conjectures on prime k-tuples. For example, to get three consecutive integers each with exactly two prime factors (which will be different, so addressing both 2 and 4), one "only" needs to find integers n such that 10n+1, 15n+2, and 6n+1 are simultaneously prime, for then 30n+3, 30n+4, and 30n+5 will each have exactly two prime factors.<|endoftext|>
TITLE: Is there a notion of congruence relation for essentially algebraic structures?
QUESTION [5 upvotes]: In universal algebra there is the notion of congruence relation: Consider a (1-sorted) algebraic structure, i.e. a set $A$ with a bunch of finitary operations $f_i$ satisfying equations.
A congruence relation is an equivalence relation $\sim$ on $A$ such that the operations on $A$ produce well-defined operations on the set $A/\sim$ of equivalence classes by applying them to representatives. I.e. for any operation $f$ on $A$ the operation $\bar{f}$ on $A/\sim$ given by $\bar{f}([x_1],...,[x_n]):=[f(x_1,...x_n)]$ is well-defined, i.e. if $x_1 \sim y_1, ... , x_n \sim y_n$ then $f(x_1,...,x_n) \sim f(y_1,...,y_n)$ for all operations $f$ of the given structure.
Thus these relations are the right ones to form quotients inside the given category of algebraic structures.
An essentially algebraic structure is a (if 1-sorted) or several (if many-sorted) sets with partially defined operations satisfying equational laws, where the domain of any given operation is a subset defined by equations between previously defined operations (equivalenty: it is a $Set$-model of a finite limit sketch). The standard example are categories, where one has three global operations, identity, source and target, and a partial operation, composition, defined only for certain pairs of morphisms.
My question is: Is there a notion of congruence relation for these more general algebraic structures? E.g. one equivalence relation on each set satisfying the analogous properties to the above? If so have these "congruence relations" been studied, do they e.g. form lattices?
Motivation: Just curiosity really. I asked myself this question, after reading this MO-question of Colin Tan, which might be a special case. He asks whether there is a way to collapse two objects in a category. If there was a lattice of congruence relations on a category, there might be the congruence relation generated by the relation which identifies just the two objects (this would of course mean to treat categories in an "evil", non-two-categorical way, but that was what the question sounded like to me).
Googling did reveal nothing, so I ask you people...

REPLY [3 votes]: As Finn says, lfp categories have all coequalizers.  However, there is a fly in the ointment.  In set-models of algebraic theories, the underlying functor preserves the congruences and the quotients of the congruences.  This means that a congruence is an equivalence relation on the underlying set and the quotient alegbraic structure has underlying set the set-quotient of the congruence. This is true even in many sorted algebraic theories.  
But in models of finite limit sketches the quotient can blow up.  The quotient need not be a structure whose underlying set is the quotient of the equivalence relation on the underlying set.  An example of this happens in the category of small categories when you merge nonisomorphic objects.  All of a sudden arrows may compose that didn't meet each other before, creating new arrows.  (So the underlying set of arrows in the quotient is not the quotient of the congruence on the underlying set.)
This is spelled out and proved in Toposes, Triples and Theories, by Michael Barr and Charles Wells, in Theorem 4.1 of Chapter 8.  In that theorem, "LE" means "Finite Limits" and "EE" means "effective   equivalence relations whose quotients are preserved by the underlying set functor".  This book is available for free on the internet -- just google it.  Exercises EEPO and ORTHODOX give specific examples of that behavior.  I think there is a specific example for small categories somewhere but at the moment I can't find it.
ADDED: The specific example is in section 1.8, exercise CBB.  Section 1.8 talks about effective equivalence relations in general and calls them congruences.<|endoftext|>
TITLE: A chain homotopy that does not arise from a homotopy of spaces?
QUESTION [6 upvotes]: Algebraic topologists like to cook up algebraic invariants on topological spaces in order to answer questions, so they are often concerned with how strong those invariants are. Currently, I am concerned with just how much information is lost when moving from a space to `the' chain complex associated to that space.
Now, I should be a bit more specific here. There are many homology theories in which one takes a space, cooks up a chain complex, and takes its homology. I am mainly interested in just singular homology for this question, but if you can only think of an answer using sheaf cohomology or some other homology theory then that's alright. Now, the actual question is:

Do there exist two spaces that have chain homotopic associated chain complexes, but are not, themselves, homotopic?

I imagine that, unless the answer to the question is "no," it would be a bit difficult to show that the two spaces are not homotopic, since we have taken away the tool we usually use to prove such facts. My first thought was that one might be able to cook up a counterexample by looking at two spaces whose compactifications give different homologies... but I wasn't able to come up with anything immediately. (Another method may be looking at homotopy groups... but they're so hard to compute, I didn't even try this approach). I probably didn't give this as much thought as I should have, so if the answer to this question is somewhat trivial, then go ahead and scold me and I'll go put some more effort into thinking about it.

REPLY [11 votes]: A bounded below complex of free $R$-modules is acyclic if and only if it is contractible (in the sense that the identity map is chain homotopic to zero). Since the singular chain complex of a space is constructed out of free $\mathbb{Z}$-modules, any space with no homology would have to contract to a point, which is not the case (check out the wikipedia page for some examples/references).
Thus the chain homotopy exhibiting contractibility of the identity does not come from a topological homotopy.<|endoftext|>
TITLE: Can all terms of the Johnson filtration be hom-mapped onto the same nontrival group?
QUESTION [5 upvotes]: Let $F_n$ where $n \ge 3$ be a free group and let $(\mathcal A_n(k))$ where $k \ge 1$ be the
kernel of the homomorphism $Aut(F_n) \to Aut(F_n/\gamma_{k+1}(F))$
determined by the natural homomorphism $F_n \to F_n/\gamma_{k+1}(F).$
($(\mathcal A_n(k) : k \ge 1)$ is called the Johnson filtration of $Aut(F_n);$
$\gamma_k(G)$ denotes the $k$-th terms of the lower central series of a group $G,$
$\gamma_1(G)$ being equal to $G$).
I do not know an example of a group homomorphism
$f : Aut(F_n) \to G$ which
takes all terms of the Johnson filtration
$(\mathcal A_n(k))$ to the same nontrivial
subgroup:
$$
1 \ne f(\mathcal A_n(1))=f(\mathcal A_n(2)) = \ldots = f(\mathcal A_n(k)) = \ldots
$$
I would be very grateful for such an example, or for an argument
that homomorphisms like that do not exist.

REPLY [5 votes]: The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one (see his comment above). Let $K=\mathcal A(1).$ Then
$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (*)$
Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.
Then 
$$  
A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K).  
$$
It follows that 
$$  
\Delta( [K,K,\ldots,K])=A_q  
$$ 
and by $(*)$  $\Delta( \mathcal A(m))=A_q  $ for every $m \ge 1.$<|endoftext|>
TITLE: Infinite direct products and derived subgroups
QUESTION [5 upvotes]: Suppose $G_1, G_2, \dots, G_n, \dots$ are groups (I use countable sequences, though the question is also applicable for uncountable collections of groups). Suppose G is the unrestricted external direct product of the $G_i$s. What is the derived subgroup (commutator subgroup) of G?
I get two things:

The derived subgroup of G contains the restricted direct product of the derived subgroups of the $G_i$s, i.e., the set of elements with finitely many nonzero coordinates where each coordinate is in the derived subgroup of the corresponding group.
The derived subgroup of G is contained in the unrestricted direct product of the derived subgroup of the $G_i$s, i.e., the set of elements where all coordinates are in the derived subgroups of their respective groups.

A more precise characterization of the derived subgroup seems to be: the set of those elements for which there exists some finite number c (depending on the element) such that every coordinate can be expressed as a product of at most c commutators in its coordinate group $G_i$.
Is this characterization correct? Also, under what conditions does the derived subgroup become either of the two extremes (the restricted direct product, or the unrestricted direct product)?
Also, the above suggests that there could be examples where all the $G_i$s are perfect groups [ADDED: A perfect group is a group that equals its own derived subgroup; however, it is not necessary that every element be a commutator] but their unrestricted direct product is not a perfect group. Could somebody come up with an example of this sort?

REPLY [2 votes]: I believe your characterization is right. 
If you take all the $G_i$ to be finite nonabelian simple groups, then Ore's conjecture implies that every element in each $G_i$ is a commutators. The conjecture was completed recently by Liebeck, O'Brien, Shalev, and Tiep, see http://www.math.auckland.ac.nz/~obrien/research/ore.pdf. So in this case your $G$ is actually perfect. (Actually this follows from weaker results then the full conjecture, which were known before).<|endoftext|>
TITLE: Naturally occuring groups with cardinality greater than the reals.
QUESTION [9 upvotes]: In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, uncountably infinite simply means a cardinality of $\aleph_{1}$, the same as $\mathbb{R}$. My question is: is there anywhere that groups with cardinality strictly greater than $\aleph_{1}$ arise naturally? Of course, it is easy enough to construct groups with arbitrarily large cardinality, but I cannot recall ever seeing them used.

REPLY [5 votes]: Any product group like $\{0, 1\}^I$ for index sets $I$, using mod 2 addition coordinatewise. It's just (isomorphic to) the power set of $I$ using symmetric difference as the addition. It's of course also a ring (pointwise multiplication / intersection ). These Boolean groups often come up in general topology.<|endoftext|>
TITLE: Most 'obvious' open problems in complexity theory
QUESTION [14 upvotes]: What open problems in computational complexity theory have the most 'obvious' answers, regardless of whether that answer is true or false?  The problems I'm talking about certainly have more 'obvious' answers than P =?= NP.
I'll start this with
ZPP =?= EXPTIME

REPLY [2 votes]: A) Multiplication (of $n$--bit numbers) obviously takes longer than addition.
B) Sorting $n$ objects obviously takes longer than linear time.  (I suppose that this is obvious because it is true: there is a lower bound of $n \log n$ for comparison-based sorts.  However, radix sort of integers can be faster.  In fact, for any collection (of a particular type of object), there just might be a super clever way to sort it faster than $n \log n$....)<|endoftext|>
TITLE: Which algebraic varieties admit a morphism to a curve?
QUESTION [25 upvotes]: I'm somewhat surprised to figure out that P^n's does not admits any non-constant morphism to a curve: in fact this is not fancy, for P^1 it follows by Riemann-Hurwitz, and for P^n, note that it doesn't admit any non-constant morphism to P^1, while any morphism to a curve can be composed with a branched cover of P^1, one sees that any map from P^n to curves are constant.
So my question is, given an projective variety, when can one find a non-constant morphism to P^1? And when can one find a non-constant morphism to a curve of genus greater than 0?

REPLY [7 votes]: This is a substantial revision of my previous answer based on the comments and the other answers. I am making it community wiki for two reasons: it incorporates ideas from my answer (for which I already have gained reputation), as well as ideas form the other posts (for which I do not deserve reputation!); people can edit it to correct mistakes and typos that are certainly present.
In what follows I will frequently not specify that a morphism is assumed to be non-constant; I will leave it to the reader to figure out when the word "non-constant" can be added!  Moreover, I will only address the question of a variety of dimension at least two mapping to a curve: the case of curves can be made very difficult by specifying the question further, but I will not worry about this.
First, the question in the title asks about morphisms to a curve.  We shall see that there are differences based on the Kodaira dimension of the target curve, the higher the dimension, the more rigid the situation: we go from negative Kodaira dimension (target of genus zero) where the property is neither a birational, nor deformation invariant; to zero Kodaira dimension (target is a curve of genus one) where the property is a birational invariant, but not deformation invariant; to positive Kodaira dimension (target of genus at least two) where the property is both a birational invariant and a deformation invariant (and more!).  Let's start.
If a variety $X$ admits a non-constant morphism to a curve, then it admits a non-constant morphism to $\mathbb{P}^1$ (just compose the given morphism with a morphism of the curve to $\mathbb{P}^1$). Thus, if you only care about existence of a non-constant morphism to a curve, you may as well restrict your attention to the question of existence of a non-constant morphism to $\mathbb{P}^1$.  The property of admitting a morphism to $\mathbb{P}^1$ is clearly not a birationally invariant property, as any non-constant rational function determines a rational map to $\mathbb{P}^1$ (see Charles Matthews' answer and think of $\mathbb{P}^2$). It is also not a deformation invariant property, as the example of a family of quartic surfaces in $\mathbb{P}^3$ containing a quartic with Picard number one and a quartic with a line clearly shows.
For this general case there is an easy necessary condition.
Lemma 1.
Let $X$ be a smooth projective variety of dimension at least two and let $f \colon X \to \mathbb{P}^1$ be a non-constant morphism.  Then there are on $X$ two non-zero disjoint effective linearly equivalent divisors $D_0,D_1$.  In particular, there are on $X$ nef divisors that are not ample and hence the Picard number of $X$ is at least two.
Proof. Just let $D_0$ and $D_1$ be the fibers of the morphism $f$ above two distinct points. For the second statement, if the Picard number of $X$ were equal to one, then any non-zero effective divisor $H$ on $X$ would be ample, and in particular we would have $D_0 \cdot D_1 \equiv H^2 \neq 0$, contradicting the fact that $D_0$ and $D_1$ are disjoint. $\square$
As Donu Arapura remarks, a variation of Lemma 1 also works for non-constant morphisms to varieties $Y$ of dimension smaller than the dimension $X$.  In this case we loose the fact that the divisors are linearly equivalent, but we can still find $\dim(Y)+1 \leq \dim(X)$ (Weil) divisors on $Y$ whose intersection is empty. Their pull-backs to $X$ give us $\dim(Y)+1$ divisors with empty intersection; again it follows that the Picard number of $X$ is at least two (see Donu Arapura's answer for a statement with Chow groups and cohomology).
As jvp remarks, there is a refinement of Lemma 1 by Totaro to an equivalence (see jvp's answer for more details).
Let us move on to the case of morphisms to smooth non-rational curves. The existence of a morphism to a curve of genus at least one is a birational property: any such morphism factors through the morphism between the Albanese variety of $X$ and the Jacobian of the curve, and we are done, since the image of $X$ in its Albanese variety is a birational invariant and is all we need to decide the existence of a morphism.
Also in this case, though, the property is not deformation invariant: already in the case of jacobians of genus two curves we can find examples of simple abelian surfaces and of abelian surfaces that are products of elliptic curves (or also just isogenous to products; see, for instance, this MO question).
There is an easy necessary criterion that everyone knows, but no one explicitly mentioned, so I will mention it here: if a variety $X$ admits a (generically smooth) morphism to a curve of genus $g$, then the irregularity of $X$ is at least $g$ (this means that the space of holomorphic one-forms has dimension at least $g$).  The proof is obtained simply by pulling back forms from the curve to $X$.  This rules out several possibilities, but is quite far from being sufficient - it is an "abelian version" of something that we will see later.
In this case we can state a criterion that might give a feel for what are the difficulties involved.
Lemma 2.
Let $X$ be a smooth projective variety. The following are equivalent

$X$ admits a non-constant morphism to a curve of genus one; 
the Albanese variety of $X$ is isogenous to a product of an elliptic curve and an abelian variety; 
the Albanese variety contains an elliptic curve.


The equivalence of the statements above follows at once from the comment about the morphism factoring through the Albanese variety, as well as well-known statements about morphisms of abelian varieties.  We could replace the Albanese variety in Lemma 2 by the Picard group (or the connected component of the identity), using the duality between the Albanese variety and the Jacobian.  Thus deciding whether a variety admits a morphism to a curve of genus one reduces to the question of understanding the dimensions of the simple factors of its Albanese variety.  This is tricky, but at least it feels like we made some progress, compared to the case of morphisms to curves of genus zero!
On to the case of morphisms to curves of genus at least two. We have already seen that the property is in this case a birational invariant; in fact the property is a deformation invariant (and more!).  This follows from a beautiful characterization that only involves topological fundamental groups (over fields of characteristic zero). This characterization starts (in the case of surfaces) with the Theorem of Castelnuovo and de Franchis, later generalized (by Siu, Beauville and Catanese) to work for arbitrary smooth projective varieties.  For more details, see the comment by Angelo to my previous answer, as well as the answers of jvp and Donu Arapura. As Donu mentions, it would be interesting to generalize this to varieties over general fields.  For instance, what can be said of a variety whose étale fundamental group surjects onto the étale fundamental group of a curve of general type?
Finally, I will conclude this answer by analyzing the case minimal surfaces.
A surface of negative Kodaira dimension (i.e. a ruled surface or $\mathbb{P}^2$, here we do not need the surface to be minimal) admits a morphism to $\mathbb{P}^1$ if and only if it is not isomorphic to $\mathbb{P}^2$. Morphisms to curves of higher genus are also easy to analyze and quickly reduce to the case of morphisms between curves.
A surface of Kodaira dimension zero (i.e. a K3, Enriques, Abelian of bielliptic surface) not admitting a morphism to $\mathbb{P}^1$ is either a non-elliptic K3 surface or a simple abelian surface (i.e. an abelian surface that is not isogenous to a product of two elliptic curves). Morphisms to curves of genus at least one are ruled out for K3 surfaces by the irregularity; for abelian varieties every non-constant morphism to a curve factors through a morphism to an elliptic curve (and clearly there are no morphisms to curves of genus at least two).
Every surface of Kodaira dimension one (a properly elliptic surface) admits a morphism to $\mathbb{P}^1$; more precisely, the Stein factorization of the morphism determined by a sufficiently large multiple of the canonical divisor is a canonical morphism to a curve.
For surfaces of Kodaira dimension two, I do not know what else to say, besides what has been already remarked above!
Thus, in conclusion, for minimal surfaces of special type, the only surfaces not admitting a morphism to a curve are $\mathbb{P}^2$, non-elliptic K3 surfaces and simple abelian surfaces (which does not look so bad, after all!). Note that every irreducible component of the moduli space of polarized K3 surfaces contains elliptic surfaces and non-elliptic ones.<|endoftext|>
TITLE: The algebro-geometric counterpart of the Dijkgraaf-Witten model
QUESTION [9 upvotes]: Can the Dikgraaf-Witten model for a finite gauge group $G$ [Robbert Dijkgraaf and Edward Witten, Topological Gauge Theories and Group Cohomology, Commun. Math. Phys. 129 (1990), 393] be described in terms of the geometry of moduli spaces $\overline{\mathcal{M}}_{g,n,\beta}([*//G])$ of stable maps to the stack $[*//G]$? I strongly suspect the answer is yes, in view of the classical relation between 3-dimensional topological quantum field theories and complex analytic 2-dimensional modular functors, but I'm unaware of rigorous results in this direction.

REPLY [6 votes]: This has been done, in a variety of related ways.  A lot of the difficulty is in defining an appropriate notion of a "stable" map to [pt/G].
The earliest mathematical work I know of is Chen & Ruan's "orbifold cohomology", which is done in the symplectic category.  (Caveats:  Abramovich's lecture notes on orbifold GW theory quote a 1996 letter from Kontsevich, who outlines a lot of the basic ideas in 2 pages.  Also, string theorists were looking at non-topological sigma models to orbifolds at least as far back as Dixon, Harvey, Vafa, & Witten's 1985 papers.)  
In algebraic geometry, this stuff has been studied by Jarvis, Kaufmann, & Kimura, who focused on G-bundles, and by Abramovich, Graber, & Vistoli, who figured out how to deal with D-M stacks.
(You can also carry out these constructions in K-theory for finite-dimensional Lie groups.  See, for example, Frenkel, Teleman, & [cough].)<|endoftext|>
TITLE: Why isn't the theorem of approximation applicable in Banach spaces?
QUESTION [6 upvotes]: Let X be a Hilbert space, A a convex, closed subset of X. Then there exists for every x in X exactly one best approximation in A, that is there exists a y in A such that || x-y || = d(x,A) = inf { || x-z || : z in A}.
Why isn't that theorem true for Banach spaces?

REPLY [13 votes]: Happened to have this slide in one of my lectures.  The green circle and green dot correspond to the Euclidean norm, the purple square and purple line to the sup norm.  With the Euclidean norm, the green dot is the closest point in the square to the blue dot; with the sup norm, the purple line is the set of closest points in the square to the blue dot.<|endoftext|>
TITLE: Does the hypergraph structure of the set of subgroups of a finite group characterize isomorphism type?
QUESTION [33 upvotes]: Question 
Suppose there is a bijection between the underlying sets of two finite groups $G, H$, such that every subgroup of $G$ corresponds to a subgroup of $H$, and that every subgroup of $H$ corresponds to a subgroup of $G$. Does this imply that $G, H$ are isomorphic? Note that we do not require the bijection to actually be the isomorphism.
 Motivation 
The question is interesting to me because I am considering maps of groups which aren't homomorphisms but preserve the subgroup structure in some sense - given a group, we can forget the multiplication operation and look only at the closure operator that maps a subset of $G$ to the subgroup generated by it. If the question is resolved in the affirmative, then the forgetful functor from the usual category $Grp$ to this category won't create any new isomorphisms. (Note that I didn't precisely specify the morphisms this new category -- you could just use the usual definition of a homomorphism, and say that if the mapping commutes with the closure operator, then its a morphism. The definition I actually care about is, a morphism of this category is a mapping such that every closed set in the source object is the preimage of a closed set of the target object. It doesn't make much difference as far as this question is concerned, the isomorphisms of both categories are the same.)
I asked a friend at Mathcamp about this a few weeks ago, he said a bunch of people started thinking about it but got stumped after a while. The consensus seems to have been that it is probably false, but the only counter examples may be very large. I don't really have any good ideas / tools for how to prove it might be true, I mostly wanted to just ask if anyone knew offhand / had good intuition for how to find a finite counterexample. 

Edit (YCor): (a) the question has reappeared in the following formulation: does the hypergraph structure of the set of subgroups of a (finite) group determine its isomorphism type? A hypergraph is a set endowed with a set of subsets. The hypergraph of subgroups is the data of the set of subgroups, and therefore to say that groups $G,H$ have isomorphic hypergraphs of subgroups means that there's a bijection $f:G\to H$ such that for every subset $A\subset G$, $f(A)$ is a subgroup of $H$ if and only $A$ is a subgroup of $G$. Several answers, complementing the one given here, have been provided in this question.
(b) There a weaker well-studied notion for groups, namely to have isomorphic subgroup lattices. Having isomorphic hypergraphs of subgroups requires such an isomorphism to be implemented by a bijection (this is not always the case: take two groups of distinct prime order).

REPLY [48 votes]: The answer is no in general. I.e, there are finite non-isomorphic groups G and H such that there exists a bijection between their elements which also induces a bijection between their subgroups.
For this, I used two non-isomorphic groups which not only have the same subgroup lattice (which certainly is necessary), but also have the same conjugacy classes. There are two such groups of size 605, both a semidirect product $(C_{11}\times C_{11}) \rtimes C_5$ (see this site for details on the construction). In the small group library of GAP, these are the groups with id [ 605, 5 ] and. [ 605, 6 ]. These are provably non-isomorphic (you can construct the groups as described in the reference I gave, and then use GAPs IdSmallGroup command to verify that the groups described there are the same as the ones I am working with here). With a short computer program, one can now construct a suitable bijection.
First, let us take the two groups:
gap> G:=SmallGroup(605, 5);    
<pc group of size 605 with 3 generators>
gap> H:=SmallGroup(605, 6);
<pc group of size 605 with 3 generators>

The elements of these groups are of order 1, 5 or 11, and there are 1, 484 and 120 of each. We will sort them in a "nice" way (that is, we try to match each subgroup of order 5 to another one, element by element) and obtain a bijection from this. First, a helper function to give us all elements in "nice" order:
ElementsInNiceOrder := function (K)
    local elts, cc;
    elts := [ One(K) ];
    cc := ConjugacyClassSubgroups(K, Group(K.1));
    Append(elts, Concatenation(List(cc, g -> Filtered(g,h->Order(h)=5))));
    Append(elts, Filtered(Group(K.2, K.3), g -> Order(g)=11));
    return elts;
end;;

Now we can take the elements in the nice order and define the bijection $f$:
gap> Gelts := ElementsInNiceOrder(G);;
gap> Helts := ElementsInNiceOrder(H);;
gap> f := g -> Helts[Position(Gelts, g)];;

Finally, we compute the sets of all subgroups of $G$ resp. $H$, and verify that $f$ induces a bijection between them:
gap> Gsubs := Union(ConjugacyClassesSubgroups(G));;          
gap> Hsubs := Union(ConjugacyClassesSubgroups(H));;
gap> Set(Gsubs, g -> Group(List(g, f))) = Hsubs;
true

Thus we have established the claim with help of a computer algebra system. From this, one could now obtain a pen & paper proof for the claim, if one desires so. I have not done this in full detail, but here are some hints.
Say $G$ is generated by three generators $g_1,g_2,g_3$, where $g_1$ generates the $C_5$ factor and $g_2,g_3$ generate the characteristic subgroup $C_{11}\times C_{11}$. We choose a similar generating set $h_1,h_2,h_3$ for $H$. We now define $f$ in two steps: First, for $0\leq n,m <11$ it shall map $g_2^n g_3^m$ to $h_2^n h_3^m$. 
This covers all elements of order 1 or 11, so in step two we specify how to map the remaining elements, which all have order 5. These are split into four conjugacy classes: $g_1^G$, $(g_1^2)^G$, $(g_1^3)^G$ and $(g_1^4)^G$. We fix any bijection between $g_1^G$ and $h_1^H$ and extend that to a bijection on all elements of order 5 by the rule $f((g_1^g)^n)=f(g_1^g)^n$. With some effort, one can now verify that this is a well-defined bijection between $G$ and $H$ with the desired properties. You will need to determine the subgroup lattice in each case; linear algebra helps a bit, as well as the fact that all subgroups have order 1, 5, 11, 55, 121 (unique) or 605. I'll leave the details to the reader, as I myself am happy enough with the computer result. 
UPDATE: as pointed out in another answer below by @dvitek (explained by @Ian Agol in comments), there is actually a much simpler example, which I somehow overlooked when I did my computer search. Credit to them, but just in case people want to reproduce their example with GAP, here is an input session doing just that:
gap> G:=SmallGroup(16,5);; StructureDescription(G);
"C8 x C2"
gap> H:=SmallGroup(16,6);; StructureDescription(H);
"C8 : C2"
gap> Gelts := ListX([1..8],[1,2],{i,j}->G.1^i*G.2^j);;
gap> Helts := ListX([1..8],[1,2],{i,j}->H.1^i*H.2^j);;
gap> f := g -> Helts[Position(Gelts, g)];;
gap> Gsubs := Union(ConjugacyClassesSubgroups(G));;
gap> Hsubs := Union(ConjugacyClassesSubgroups(H));;
gap> Set(Gsubs, g -> Group(List(g, f))) = Hsubs;
true<|endoftext|>
TITLE: Interesting complexity classes $PR \subsetneq c \subsetneq R$
QUESTION [7 upvotes]: I'm working on a proof-checker that can verify termination proofs.  The fundamental method it provides for constructing such proofs is to translate the program into primitive recursion.  Basically, I provide a combinator $\rho$ typed as:
$\rho: \forall A,B:(A\rightarrow Nat \rightarrow A)\rightarrow (A \rightarrow B)\rightarrow A\rightarrow Nat \rightarrow B$
which, in the notation defined here, constructs $h$ given $f$ and $g$.
Although the term language contains a fixed-point combinator and is therefore Turing-complete, terms that use it have a "tentative" flag in their type that indicate this.  The $\rho$ combinator and the fixed-point combinator are the only two language primitives that allow for recursion or looping of any sort (i.e., without either of these two combinators, all you've got is a finite-state machine).  Therefore, all terms that are well-typed and non-tentatively typed are primitive recursive.
What I'm wondering is if there are any interesting complexity classes that you can build by starting with primitive-recursive constructions, and adding a finite number of other functions $Nat \rightarrow Nat$, each of which is in R but not in PR, and allowing composition with these functions.  It's easy to come up with non-interesting examples of such classes, e.g. "primitive recursion plus the Ackermann function", but I'm looking for any that have sufficiently interesting properties that it would be worth adding the functions which characterize them as admitted axioms in the proof system.

REPLY [7 votes]: First of all, it’s certainly possible to obtain some intermediate class by taking a language that only computes PR functions (say, an imperative programming language using only for loops) and adding any total computable but non PR function (e.g., Ackermann’s function). The resulting language L is non-universal, because it only computes total functions: you can still construct a computable but non-L-computable function by diagonalisation. However, L is clearly more powerful than the original language.
As for “interesting”, I guess it really depends on what you mean by that.
If “interesting” means “of practical use”, then one could answer that all computable functions of practical use are PR, since a non-PR function requires an amount of time to compute that is not, in turn, PR. Considering that time bounds such as 2n, 22n, 222n, …, are all PR, you see that there isn’t much hope to compute non-PR functions for large values of n.
If “interesting” means “logically interesting”, then I think the answer is “yes”. I’m somewhat familiar with Girard’s System F (also called “second order λ-calculus” or “polymorphic λ-calculus”), described for instance in Girard’s Proofs and Types (freely available here). The functions that can be computed in F are “exactly those which are provably total in [second order Peano arithmetic]” (page 123), and among these we have Ackermann’s function. There is an explicit λ-term for it on these slides (page 20).
If I recall correctly, the standard calculus of constructions includes System F and only computes total functions, so it also provides an example.<|endoftext|>
TITLE: Widely accepted mathematical results that were later shown to be wrong?
QUESTION [340 upvotes]: Are there any examples in the history of mathematics of a mathematical proof that was initially reviewed and widely accepted as valid, only to be disproved a significant amount of time later, possibly even after being used in proofs of other results?
(I realise it's a bit vague, but if there is significant doubt in the mathematical community then the alleged proof probably doesn't qualify. What I'm interested in is whether the human race as a whole is known to have ever made serious mathematical blunders.)

REPLY [3 votes]: R. V. Gurjar and M. Miyanishi [Affine surfaces with $\bar {\kappa}\leq 1$, Algebraic geometry and commutative algebra, in Honor of Masayoshi Nagata, Vol. I, 99-124 (1988) ZBL0707.14015] gave a classification of smooth affine surfaces of logarithmic Kodaira dimension zero, whose coordinate ring is factorial and has trivial units. G. Freudenburg et al. [``Smooth factorial affine surfaces of logarithmic Kodaira dimension zero with trivial units'', https://arxiv.org/abs/1910.03494] identified a gap in this classification, which was closed in Tomasz Pełka and Paweł Raźny, Classification of smooth factorial affine surfaces of Kodaira dimension zero with trivial units,  ZBL07383176 who classified an additional infinite series of such surfaces.<|endoftext|>
TITLE: Up to projectivities, which configurations of four lines in $\mathbb{P}^3$ can one distinguish?
QUESTION [6 upvotes]: Background
I am interested in the projective classification of reduced curves of degree four in $\mathbb{P}^3(\mathbb{R})$ (and more generally of degree $n+1$ in $\mathbb{P}^n(\mathbb{R})$). More precisely, I am looking at the case where the curve is a union of four distinct lines. I need this classification because I want to make sure that I consider all possible cases in a problem in interpolation theory.

For instance, there are two types of configurations of three lines in $\mathbb{P}^2$. Either three lines meet in a single point, or three lines meet in three distinct points. More generally, according to this integer sequence, there are 3 configurations of four lines in $\mathbb{P}^2$, 5 configurations of five lines in $\mathbb{P}^2$, and 18 configurations of 6 lines in $\mathbb{P}^2$. These configurations are shown in this figure (except for the configurations in which all lines are concurrent).
I believe there are six configurations of three lines in $\mathbb{P}^3$: Two configurations for which the three lines lie in a plane, three configurations for which precisely two of the three lines lie in a plane, and one configuration where none of the lines intersect.
My (related) questions are now as follows:


How many configurations are there of four lines in $\mathbb{P}^3$ (and more generally of $n+1$ lines in $\mathbb{P}^n$)?
Is there a convenient way to enumerate these?

REPLY [6 votes]: This is one of my favourite projective geometry examples; it is a case of
classification involving moduli and if done right most of the arguments can be
done with a combination of geometry and linear algebra with no explicit
calculation.
I assume that we are talking about an ordered quadruple of lines in $\mathbb
P^3(k)$ (I do it over any field). The trick is to not think of $\mathbb P^3$ as
the set of lines in $k^4$ but rather of lines in some $4$-dimensional vector
space V and then use the date to get closer to an adapted coordinatisation.
Assume first that no two of the lines are skew. Then a simple geometric argument
shows that either all lines lie in a plane or pass through a common point. Both
of those cases are reduced to the problem of four points in $\mathbb P^2$ which
I skip. We can then assume that the first two lines are skew and think of $V$ as
$V_1\bigoplus V_2$, where the projectivisations of $V_1$ and $V_2$ are the two
first lines. Assume then that the third line is skew with the first and
second line. This means that it is the projectivisation of the graph $V_3\subset
V_1\bigoplus V_2$ of an isomorphism. Hence, we may assume that $V_1=V_2$ and
$V_3$ is the diagonal in $V_1\bigoplus V_1$. If we also assume that the fourth
line is skew with the first two, then it is also the graph $V_4\subset
V_1\bigoplus V_1$ of an automorphism $V_1\rightarrow V_1$. Hence, four lines,
the first two of which are skew and the last two are skew with the first two up
to projective transformations correspond to isomorphism classes of pairs
$(V_1,\varphi)$ where $V_1$ is a two-dimensional vector space and $\varphi$ is
an automorphism of it distinct from the identity. This is the same thing as
conjugacy classes of $\mathrm{GL}_2(k)$ distinct from the identity element. The
condition that the last two lines be skew is exactly that $\varphi$ does not
have $1$ as an eigenvalue. Of course the characteristic polynomial distinguish
between conjugacy classes so there are continuous families of configuration
(i.e., it has non-trivial moduli).
The remaining case of two skew lines and two lines which are not both skew with
respect to both of the first lines is easy but a little bit tedious; given a
pair of non skew lines one looks at the plane spanned by them and the position
of the other lines with respect to it.
Addendum: I did not mean to suggest that it is the simplest problem with proper moduli. Of course the classification of four points in $\mathbb P^1$ has the cross ratio is its moduli. (The classification is proven almost word for word in the same way as the above case.)<|endoftext|>
TITLE: Values of the multiplicative group over a ring spectrum
QUESTION [5 upvotes]: In his notes on elliptic cohomology, Lurie defines the multiplicative group $\mathbb{G}_m$ over a ring spectrum $A$ as $\operatorname{Spec} A[\mathbb{Z}]$. What is the value $\mathbb{G}_m(B)$ of the represented functor at an $A$-algebra $B$? If this is too hard to say in general: Are there at least any specific examples, other than Eilenberg-MacLane spectra, where one does know the answer?

REPLY [7 votes]: It is slightly complicated.
One has a number of adjunctions:
$$
\begin{eqnarray*}
\mathbb{G}_m(B) &=& Hom_{A-alg}(A[\mathbb{Z}],B) \\
&\simeq& Hom_{E_\infty-rings}(\mathbb{S}[\mathbb{Z}],B)\\
&\simeq& Hom_{E_\infty-spaces}(\mathbb{Z},GL_1(B))\\
&\simeq& Hom_{spectra}(H\mathbb{Z},gl_1(B)).
\end{eqnarray*}
$$
(Note these adjunctions are weak equivalences of spaces, and the last two adjunctions require a fair amount of theory to make rigorous.)
The problem is that it is usually quite difficult to compute the maps out of the Eilenberg-Mac Lane spectrum $H\mathbb{Z}$ unless the target is also an Eilenberg-Mac Lane space.  In the case where the algebra $B$ comes from a simplicial commutative ring, this is true and so one at least knows that the set of homotopy classes of maps $[H\mathbb{Z}, gl_1(B)]$ surjects onto $\pi_0(B)^\times$.  Even for complex K-theory, the calculation is somewhat involved (but doable), but the only method that I can immediately think of involves the Bousfield-Kuhn functor.<|endoftext|>
TITLE: Minimally 2-vertex-connected graphs?
QUESTION [8 upvotes]: A class of "minimally 2-vertex-connected graphs" - that is, 2-vertex-connected graphs which have the property that removing any one vertex (and all incident edges) renders the graph no longer 2-connected - have come up in my research.
Dirac wrote a paper on "minimally 2-connected graphs" (G. A. Dirac, Minimally 2-connected graphs, J. Reine Angew. Math. 228 (1967),. 204-216), which gives quite a detailed description of the structure of such graphs.  However, in his sense, minimal 2-connectivity means that deleting any EDGE leaves a graph which is not 2-connected, which is not an equivalent property to the vertex-deletion one.  Does anyone know anything about graphs with the latter property?
In the hope of stimulating some discussion, here is a wildly speculative and vague conjecture:  The only graphs satisfying this property are simple cycles, and certain cycles with chords.

REPLY [2 votes]: Here is a more general family:
Draw your favorite tree in the plane, with circles for the nodes and "thick" lines for the edges.  Now turn every circle into a cycle, and every thick line into a pair of parallel paths $p_1, \ldots, p_m$ and $q_1, \ldots, q_n$ with various crossbraces.  The crossbraces just have to follow the rule that if $p_i$ is connected to $q_\ell$ and $p_j$ is connected to $q_k$, for $i&lt;j$ and $k&lt;\ell$, then $j =i+1$ or $\ell=k+1$.
This is probably still not close to a complete characterization, but at least shows that the class is a lot broader than the small class you posited to promote discussion.<|endoftext|>
TITLE: Grothendieck-Messing theory for finite flat group schemes
QUESTION [14 upvotes]: Classical Grothendieck-Messing theory relates deformations of $p$-divisible groups to lifts of the Hodge filtration (if the ideal defining the nilpotent immersion is equipped with a PD-structure). If I understand it correctly, Faltings in his article "Group schemes with strict $\mathcal{O}$-action" proves a version of this result for finite locally free group schemes $G$ (and even for his variant with $\mathcal{O}$-action, but that is not important for my question), with $M(G)$ now being a filtered perfect complex rather than a filtered locally free module. Is there another reference for this result?

REPLY [9 votes]: Dear Peter, I will answer to the question in the comment since it seems this is you main interest. For the algebraicity of the $p^n$-torsion points of the universal deformation, I gave a proof of this to Matthias Strauch a few years ago. He included it in his article "Deformation spaces of one-dimensional formal groups and their cohomology", this is theorem 2.3.1 (the proof as written in Matthias article is for Lubin-Tate spaces but works in general without changing anything), see his webpage. You don't need deformation theory for finite flat group schemes for this...look at the proof there's a trick (due to Artin).
For Brian, you say "Do you mean there's a BT-group over a finitely presented algebra whose pullback to completion at some point is the universal formal deformation? If so, then I find that hard to believe". But in fact this is conjecturally true ! This would follow from the non-emptiness of Newton strata in unitary PEL type Shimura varieties at a split prime $p$. For example this is known for the deformation space of a principally polarized BT group thanks to the non-emptiness of Newton strata of Siegel modular varieties (I mean you deform not the BT group but the BT group together with its principal polarization).<|endoftext|>
TITLE: How to study a math text
QUESTION [6 upvotes]: Hello,
recently I've been trying various attempts regarding how to approach a math book to learn in the best way. Should one memorize the theorems and proofs so that one can recite them? I tend to sometimes forget proofs and such after some time, so my question is the following.
When you're trying to learn from a book and it's a new subject, how do you usually do? Do you learn all theorems and proofs? Do you write it down? Or do you just read? This is a general question, I'm aware of that, I do however hope that this question will fit the community.

REPLY [2 votes]: I am a big believer in learning by exploration: It builds independence and confidence... something you need if you plan to pursue further your math journey. That means when you find an interesting result or theorem in a book, explore it's consequences, see the details of the proof, etc. It helps your understanding.
Most important when learning math: do a lot of exercises. If possible, do a lot of difficult ones. Really, do them, there is no learning if don't get your hands dirty! Don't expect to solve every single difficult problem you try, but seriously, trying them will help you tremendously. As a rule of thumb, I never spend more that half an hour on a problem, unless it's really something very motivating. 
And last: Life is sequence of choices, and exploration takes time, a lot of it!
Which means: you learn much faster by reading (and paying attention to what you read, they call it focus), but you don't master something unless you get your hands dirty, as I said.<|endoftext|>
TITLE: What is the commutative analogue of a C*-subalgebra?
QUESTION [12 upvotes]: Using the duality between locally compact Hausdorff spaces and commutative $C^*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions regarding its ring of functions $C_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences:
$$
\;\;\;\text{open subset of $X$}\quad \longleftrightarrow\quad\text{ideal in $C_0(X)$}
$$
$$
\;\;\;\;\;\quad\text{dense open subset of $X$}\quad \longleftrightarrow\quad\text{essential ideal in $C_0(X)$}
$$
$$
\;\;\;\quad\text{closed subset of $X$}\quad \longleftrightarrow\quad\text{quotient of $C_0(X)$}
$$
$$
\text{locally closed subset of $X$}\quad \longleftrightarrow\quad\text{subquotient of $C_0(X)$}
$$
$$
\;\;\;\quad\qquad\qquad\qquad\qquad\text{???}\qquad\qquad \longleftrightarrow\quad\text{$C^*$-subalgebra in $C_0(X)$}
$$
By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.
Well, I can't quite see how to reconvert a $C^*$-subalgebra in $C_0(X)$ into something topological involving only the space $X$ (and some data describing the subalgebra in topological terms). Can you come up with something handy?

Example: A simple example of a subalgebra of a commutative $C^*$-algebra not being an ideal is
$$
\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C.
$$

First attempts: Instead of talking about a subalgebra, we should probably talk about the injective $^* $-homomorphism given by the inclusion of this subalgebra. But is this inclusion proper (i.e., does it preserve approximate units) in general? Well, at least when we restrict to compact spaces. Then an injective $^* $-homomorphism $C(Y)\to C(X)$ will induce a surjective continuous map $X\to Y$. How to proceed?

Remark: Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.

Disclaimer: I posted this question yesterday on MSE. I also got an interesting answer. However, I'm not yet fully satisfied. If I violate any policy by reposting the question here, please tell me about it.

REPLY [13 votes]: Let $A$ be a commutative $C^\star$-algebra and $B$ a $C^\star$-subalgebra (in general, this will not preserve approximate units as the example ${\mathbb C} \oplus 0 \subset {\mathbb C} \oplus{\mathbb C}$ shows). This gives also rise to an inclusion of the unitalization $B^+$ into $A^+$. Now you are in the realm of unital $C^\star$-algebras and your own remark applies. If $A$ was isomorphic to the algebra of continuous functions on the locally compact space $X$, then $A^+$ is the algebra of continuous functions on the one-point compactification $X^+$ of $X$. Now, $B^+ \cong C(Y)$, where $Y$ is some quotient of the one-point compactification of $X$. And again, $Y$ is the one-point compactification of some space locally compact space $Z$ such that $B = C_0(Z)$. What is the relation between $Z$ and $X$? Well, the continuous quotient map $f \colon X^+ \to Z^+=Y$ gives rise to a canonical open subset $U \subset X$, which arises as the pre-image of $Z$. Moreover, one gets a quotient map $f|_U \colon U \to Z$ and $f|_U$ is easily seen to be proper. This is everything that can be said.
(Conversely, let $U \subset X$ be open and $g \colon U \to Z$ be a proper surjection. Then
$C_0(Z) \to C_0(U) \subset C_0(X)$ gives rise to a sub-algebra.)
This is explained in more detail in the book by Higson and Roe (see here).<|endoftext|>
TITLE: Pair of curves joining opposite corners of a square must intersect---proof?
QUESTION [23 upvotes]: Reposting something I posted a while back to Google Groups.
In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says
"... every pair of curves in the square joining different pairs of
opposite corners must intersect".
This is obvious geometrically but I was wondering how one could go
about proving this rigorously. I have thought of a proof using
Brouwer's Fixed Point Theorem which I describe below. I would greatly
appreciate the group's
comments on whether this proof is right and if a simpler proof is
possible.
We take a square with side of length 1. Let the two curves be
$(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ where the $x_i$ and $y_i$ are
continuous functions from $[0,1]$ to $[0,1]$. The condition that the
curves join different pairs of opposite corners implies,
$$(x_1(0),y_1(0)) = (0,0)$$
$$(x_2(0),y_2(0)) = (1,0)$$
$$(x_1(1),y_1(1)) = (1,1)$$
$$(x_2(1),y_2(1)) = (0,1)$$
The two curves will intersect if there are numbers $a$ and $b$ in $[0,1]$
such that
$$p(a,b) = x_2(b) - x_1(a) = 0$$
$$q(a,b) = y_1(a) - y_2(b) = 0$$
We define the two functions
$$f(a,b) = a + p(a,b)/2 + |p(a,b)| (1/2 - a)$$
$$g(a,b) = b + q(a,b)/2 + |q(a,b)| (1/2 - b)$$
Then $(f,g)$ is a continuous function from $[0,1]\times [0,1]$ into itself and
hence must have a fixed point by Brouwer's Fixed Point Theorem. But at
a fixed point of $(f,g)$ it must be the case that $p(a,b)=0$ and $q(a,b)=0$
so the two curves intersect.
Figuring out what $f$ and $g$ to use and checking the conditions in the
last para is a tedious. Can there be a simpler proof?

REPLY [4 votes]: How about the following, using the  Nested Intervals Theorem  (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:
We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$). 
If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.<|endoftext|>
TITLE: Upper bound for tetrahedron packing?
QUESTION [8 upvotes]: There have been several recent advances on packing regular tetrahedra in $\mathbb{R}^3$.  All the results I've seen have been lower bounds -- first John Conway and Sal Torquato showed that there exists an arrangement of tetraheda filling about 72% of space.  This has been improved in a series of papers, and the latest result of which I am aware is Elizabeth Chen's record of 85.63%.  (A NYTimes article summarizing the history of the problem can be found here.)
My question is does anyone know of any upper bounds, either published or unpublished?  I saw a colloquium by Jeff Lagarias, and he said someone was claiming that they had proved something like $1 - 10^{-26}$, but that it was still unpublished.
(A compactness argument gives that since regular tetrahedra don't tile space the maximum volume is strictly less than one, but this argument does not give a quantitative bound.)

REPLY [5 votes]: A paper just appeared on the arXiv that supports Jeff Lagarias's claim: "Upper bound on the packing density of regular tetrahedra and octahedra," by Simon Gravel, Veit Elser, and Yoav Kallus (medicine and physics researchers at Stanford and Cornell):

In this article, we obtain an explicit bound to the packing density of regular tetrahedra, namely
  $\phi < 1-\delta$ with $\delta = 2.6...\times 10^{-25}$. 
  [...]
  In order to obtain a bound to the packing density, we show the existence, in any tetrahedron packing, of a 
  set of disjoint balls whose intersection with the packing is particularly simple, and whose density can be 
  bounded below. The construction is such that the density of the packing within each of the balls can be 
  bounded away from one. The combination of these two bounds gives the main result. 

They modify their argument to obtain a $10^{-12}$ bound on regular octahedra.<|endoftext|>
TITLE: How fast can the base-bumping function in Goodstein's theorem grow?
QUESTION [14 upvotes]: In the usual presentation of Goodstein's theorem, the base is bumped up by the "add 1" function. Does the theorem still hold when we replace this function by a fast-growing one (e.g. Ackermann or busy beaver)? How far can we push this? For example, let's define $g_0(n)$ to be the number of Goodstein iterations needed to reach 0 when we start with base 2 and seed $n$ (so that $g_0(0)$ = 0). Then we can build a hierarchy of functions by defining $g_{k+1}(n)$ as the number of Goodstein iterations needed to reach 0 with seed $n$ and base-bumping function $g_k$ ($k$ = 0, 1, ...), continuing through the ordinals by diagonalization at each limit ordinal. Surely it's got to break down when we go past $\varepsilon_0$, if not long before that!

REPLY [9 votes]: Dear all,
let me give the following remark:
"Goodstein actually employed arbitrary increasing base-bumping functions. He showed that the convergence of all such is equivalent to transfinite induction below ϵ0."
This statement has to be taken with care when it comes to weakly increasing base bumping
functions. When we reach functions in the neighboorhood of log* then the Goodsteinprocess
becomes provable in PRA. 
But when we take a fixed iterate of log then of course termination of Goodstein
sequences is equivalent to the 1 consistency
of PA.
If the base bumping function growth faster than H_epsilon_0 then Goodstein
can of course yield more than the 1 consistency of PA.
Best,
Andreas<|endoftext|>
TITLE: First: upper-star, then: lower-star, finally: lower-shriek
QUESTION [15 upvotes]: For a category $\mathcal{C}$, let $\mathcal{C}-Set$ denote the category of functors $\mathcal{C}\to{\bf Set}$.  Recall that given a functor $F\colon\mathcal{B}\to\mathcal{C}$, the ``composition with $F$" functor is denoted $F^*\colon\mathcal{C}-Set\to\mathcal{B}-Set.$  It has a left and a right adjoint, $F_!$ and $F_*$.  I call these functors the pullback, the left pushforward, and the right pushforward.
Let $p\colon A\to B$ be a function of sets, thought of as a functor $P\colon[1]\to{\bf Set}$, where $[1]$ is the "free-arrow category," $[1]="\bullet\to\bullet$."  Suppose one wants to find the image of $p$, but he or she can only use pull-backs, left pushforwards, and right pushforwards to manufacture it.  In other words, suppose one wants to find a zigzag of functors $[1]=:C_0\leftarrow C_1\rightarrow C_2\leftarrow C_3\rightarrow\cdots\rightarrow C_n=[0]$ such that if we perform a pullback along all leftward functors and either a left pushforward or a right pushforward along rightward functors, then the end result will be the image set $im(p)$ of $p$ (considered as a functor $[0]\to{\bf Set}$).
This can be done.   To do it, I used a sequence of the form $$[1]\leftarrow C_1\rightarrow C_2\rightarrow [0].$$  If the functors are denoted (left to right) by $F,G,$ and $H$, I found that $H_! \circ G_*\circ F^\ast (P)=im(P)$.
I'm not going to bore you with the details of $C_1, C_2$ and $F,G,H$.  
Here's the question.  I've seen things like $H_! \circ G_*\circ F^\ast$ before in the context of polynomial functors.  Unfortunately, I don't know enough about them to know if there's a connection.  Is there?  
I also don't know if I can get the whole epi-mono factorization somehow.  I haven't worked that long at it, but suppose I want not to end up with the set $im(p)$ but instead the maps $A\to im(f)\to B$. Can I achieve that by use of pullbacks and pushforwards as above (with $C_n=[2]$ now)?  Is there any rhyme or reason to such constructions?
Thanks.

REPLY [7 votes]: First of all: yes, there's certainly a connection. See http://ncatlab.org/nlab/show/polynomial+functor. If the base category is $Set$, the composite 
$$Set/W \stackrel{f^\ast}{\to} Set/X \stackrel{g_\ast}{\to} Set/Y \stackrel{h_!}{\to} Set/Z$$ 
first takes a $W$-indexed set $S_w$ to an $X$-indexed set $T_x = S_{f(x)}$, then takes this to the $Y$-indexed set $U_y = \prod_{x: g(x) = y} T_x$, then takes this to the $Z$-indexed set $V_z = \sum_{y: h(y) = z} U_y$. Putting this together, the composite is a family of polynomials, each a sum of monomial terms 
$$P(\ldots, S_w, \ldots) = (z \mapsto \sum_{y \in h^{-1}(z)} \prod_{x \in g^{-1}(y)} S_{f(x)})$$
I'll give a quick example. Suppose we want to express the free monoid functor 
$$F(S) = \sum_{n \geq 0} S^n$$ 
in this form. Then we take $W = 1$, $X = \mathbb{N} \times \mathbb{N}$, $Y = \mathbb{N}$, $Z = 1$. There's only one choice for $f$ and $h$, and $g$ is rigged so that the fiber over $n \in \mathbb{N}$ is an $n$-element set: $g(m, n) = m + n + 1$. One can easily check this works. 
As for the other question: it would have been nice if you had "bored" us! Because I don't see how to reconstruct what you did. What I have to get the image is a zig-zag of length 4 
$$Set^{[1]} \stackrel{F^\ast}{\to} Set^{C_1} \stackrel{G_\ast}{\to} Set^{C_2} \stackrel{H^\ast}{\to} Set^{C_3} \stackrel{J_!}{\to} Set^{[0]}$$ 
where $C_1$ is the generic cospan $a \to c \leftarrow b$, $C_2$ is the generic commutative square, $C_3$ is the generic span $a \leftarrow d \to b$, and then $G$ and $H$ are the evident inclusion functors, and $F$ takes each arrow of the generic span to the arrow of $[1]$. Then $F^\ast$ takes $p: A \to B$ to the cospan consisting of two copies of $p$; hitting this with $G_\ast$ takes this cospan to the pullback square (pulling back $p$ against itself); hitting this with $H^*$ restricts the pullback square to the span consisting of the pullback projections; finally, hitting this with $J_!$ takes this span to its colimit = pushout, which is the same as the coequalizer of the pullback projections (because they have a common right inverse). (Based on his comment, I'm guessing that some guy on the street was doing more or less the same thing.) 
Could you tell us what you had in mind?<|endoftext|>
TITLE: Generalisation of Lebesgue decomposition theorem
QUESTION [14 upvotes]: Background.  The Lebesgue decomposition theorem states that if $(X,\Omega)$ is a measurable space and $\mu$ is a finite measure on $X$, then for every measure $\nu$, there is a unique decomposition $\nu = \nu_1 + \nu_2$ such that $\nu_1 \ll \mu$ and $\nu_2 \perp \mu$.
Let us denote the space of all finite measures on $(X,\Omega)$ by $\mathcal{M}$.  Then the above is equivalent to the statement that $\mathcal{M} = \mathcal{S} \oplus\mathcal{T}$, where $\mathcal{S}$ is the space of all measures that are absolutely continuous with respect to $\mu$, while $\mathcal{T}$ is the space of all measures that are singular with respect to $\mu$.
We can characterise $\mathcal{T}$ in terms of $\mathcal{S}$ as
$$
\mathcal{T} = \mathcal{S}^\perp = \lbrace \nu\in \mathcal{M} \mid \nu \perp m \text{ for all } m\in \mathcal{S} \rbrace.
$$
Let us say that a subspace $\mathcal{S} \subset \mathcal{M}$ has property D (for decomposition) if $\mathcal{M} = \mathcal{S} \oplus \mathcal{S}^\perp$.  Then the Lebesgue decomposition theorem says that $\lbrace \nu \mid \nu \ll \mu\rbrace$ has property D for any fixed $\mu$.
Question.  Which subspaces have property D?  Are there conditions on $\mathcal{S}$ that are equivalent to property D, or at least imply it?  Presumably $\mathcal{S}$ should have the property that if $\nu \ll \mu \in \mathcal{S}$, then $\nu\in \mathcal{S}$ as well; is this sufficient, or are there other requirements?

REPLY [3 votes]: A little further searching turned up a simple proof of Lebesgue's decomposition theorem in "The Lebesgue Decomposition Theorem for Measures", J. K. Brooks, The American Mathematical Monthly, 78 (1971), pp. 660-662.  Without much extra work, it admits the following generalisation.
Theorem.  Let $\mathcal{N} \subset \Omega$ be a collection of subsets such that

if $E\in \mathcal{N}$ and $F\in \Omega$, $F\subset E$, then $F\in \mathcal{N}$;
if $E_n \in \mathcal{N}$ is a countable collection, then $\bigcup_{n} E_n \in \mathcal{N}$ as well.

Consider the subspace $\mathcal{S} = \lbrace \mu \mid \mu(E) = 0 \text{ for all } E\in \mathcal{N} \rbrace$.  Then $\mathcal{M} = \mathcal{S} \oplus \mathcal{S}^\perp$.
Proof.  Fix $\nu\in \mathcal{M}$, and consider the following collection of subsets:
$$
\mathcal{R} = \lbrace E \in \mathcal{N} \mid \nu(E) > 0 \rbrace.
$$
Let $\alpha = \sup \lbrace \nu(E) \mid E\in \mathcal{R} \rbrace$, and let $E_n\in \mathcal{R}$ be a sequence of sets such that $\nu(E_n) \to \alpha$.  Let $A = \bigcup_n E_n$.  Then $\nu(A) = \alpha$ and $A \in \mathcal{N}$.
Furthermore, given any $E\in \mathcal{R}$, we have $\nu(E\setminus A) = 0$.  Indeed, if $\nu(E\setminus A)>0$, then $\nu(E) = \nu(A) + \nu(E\setminus A) > \alpha$, contradicting the definition of $\alpha$.  Similarly, $\nu(E\setminus A) = 0$ for every $E\in \mathcal{N}$.
Thus we may take $\nu_1 = \nu|_{X\setminus A}$ and $\nu_2 = \nu|_A$.  It follows that $\nu_2 \in \mathcal{S}^\perp$, since $\nu_2(A)=1$ and $A\in \mathcal{N}$, and $\nu_1\in \mathcal{S}$, since $\nu(E\setminus A) = 0$ for every $E\in \mathcal{N}$.
Finally, uniqueness follows since $\mathcal{S} \cap \mathcal{S}^\perp = \lbrace 0\rbrace$.<|endoftext|>
TITLE: Is anything known about the enumeration of degree d, genus g curves in CP^2 where g >1 ?
QUESTION [5 upvotes]: I wanted to know if there is something analogous to Kontsevich's recursion formula for 
enumeration of genus zero curves in $\mathbb{C}\mathbb{P}^2$, for higher genus curves. 
There is a 
similar formula for genus one curves. See the book "Mirror Symmetry and Algebraic Geometry"
by Katz, Page 211.
Any partial results known for g>1? That is, maybe its not known for all d, but for some small values of d?

REPLY [9 votes]: The formula is due to Caporaso and Harris, Counting plane curves of any genus, Invent. Math. 131 (1998), no. 2, 345-392, http://arxiv.org/abs/alg-geom/9608025<|endoftext|>
TITLE: How to construct a ring with global dimension m and weak dimension n?
QUESTION [8 upvotes]: Given two integers $m,n$ such that $n < m$, it is easy to construct a ring with global dimension $m$ or weak dimension $n$. But I wonder whether there exists a ring satisfying both the conditions?

REPLY [5 votes]: If $R$ is Noetherian then they are equal. 
For $n=0$ one can use the fact that any Boolean ring has weak dimension $0$ (any module is flat), but a free Boolean ring on $\aleph_n$ generators have global dimension $n+1$, see the last paragraph of this paper. 
For any given pair of $(m,n)$ one can perhaps use polynomial rings over the examples for $n=0$ case (The global dimensions do go up properly, but the behavior of weak dimensions seem to be trickier, may be someone who is a real expert can confirm this?)<|endoftext|>
TITLE: History of the Normal Basis Theorem
QUESTION [14 upvotes]: The Normal Basis Theorem: If $E/F$ is a finite Galois extension, then there exists $a \in E$ such that the orbit of $a$ under the action of $\mathrm{Gal}(E/F)$ is a basis for $E$ as a vector space over $F.$
Who discovered this?
I've looked through the collected works of Frobenius and Dedekind, which are the earliest works I've seen referring to it, but it looks like the theorem led Dedekind to what is called the group determinant, and he doesn't give a reference. (p. 433 of Dedekind's Gesammelte Werke, v. 2, via Curtis's Pioneers of Representation Theory. See KConrad's answer below.) Among others, I've also looked at some of the correspondence of Hasse and Noether. The works are in German, which is second language to me, so it's possible I missed something. Needless to say, I've searched using Google to no avail. If anyone knows something, I'd be very grateful.

REPLY [14 votes]: The cached page 
http://webcache.googleusercontent.com/search?q=cache:q5q43iNq1SQJ:siba2.unile.it/ese/issues/1/690/Notematv27n1p5.ps+normal+basis+theorem&cd=3&hl=en&ct=clnk&gl=us&client=safari
gives some information: Eisenstein conjectured it in 1850 for extensions of finite fields and  Hensel gave a proof for finite fields in 1888.  Dedekind used such bases in number fields in his work on the discriminant in 1880, but he had no general proof. (See the quote by Dedekind on the bottom of page 51 of Curtis's "Pioneers of Representation Theory: Frobenius, Burnside, Schur, and Brauer".) In 1932 Noether gave a proof for some infinite fields while Deuring gave a uniform proof for all fields (also in 1932).
In Narkiewicz's "Elementary and Analytic Theory of Algebraic Numbers" (3rd ed.) he writes on the bottom of p. 186 that the normal basis theorem is due to Noether, but as usual the history is slightly more complicated.<|endoftext|>
TITLE: Does Ricci flow with surgery come from sections of a smooth Riemannian manifold?
QUESTION [26 upvotes]: More precisely, is Ricci flow with surgery on a 3-dimensional Riemannian manifold M given by the "constant-time" sections of some canonical smooth 4-dimensional Riemannian manifold? 
There would be a discrete set of times  corresponding to the surgeries, but the 4-dimensional manifold might still be smooth at these points even though its sections would have singularities.  The existence of such a 4-manifold is well known if there are no singularities: the problem is whether one can still construct it in the presence of singularities. 
Background: Ricci flow on M  in general has finite time singularities. These are usually dealt with by a rather complicated procedure, where one stops the flow just before the singularities, then carefully cuts up M into smaller pieces and caps off the holes, and constructs a Riemannian metric on each of these pieces by modifying the metric on M, and then restarts the Ricci flow. This seems rather a mess: my impression is that  it involves making several choices so is not really canonical, and has a discontinuity in the metric and topology on M. If the flow were given by sections of a canonical smooth 4-manifold as in the question this would give a cleaner way to look at the surgeries of Ricci flow. 
(Presumably if the answer to the question is "yes" this is not easy to show, otherwise people would not spend so much time on the complicated surgery procedure. But maybe Ricci flow experts know some reason why this does not work.)

REPLY [9 votes]: To build a little on what Agol said:
For mean curvature flow of hypersurfaces, the analogous question is at least partially known to be true (sort of).  The advantage of the mean curvature flow over the Ricci flow in this case is that there are already are good notions of weak solution--that is solutions defined through singularities but that agree with classical solutions when the latter exist.  The two most important are the Brakke flow and the level set flow.  In principal the latter does exactly what you want, namely given a $\Sigma_0$ a closed hypersurface in $\mathbb{R}^{n+1}_x$ there is a ``hypersurface'' $\mathcal{M}$ in $\mathbb{R}^{n+1}\times \mathbb{R}^{\geq 0}$.  So that $\partial \mathcal{M}=\Sigma_0 \subset \mathbb{R}^{n+1} \times 0$ and each level set $\lbrace x_{n+2}=t \rbrace \cap \mathcal{M}$ can be interpreted as the flow of $\Sigma_0$ at time $t$ (indeed if $\Sigma_0$ is smooth these level sets agree with the usual flow up to the first singular time, thereafter they continue to exist while the classical flow ceases to make sense).  It should be pointed out that in principal $\mathcal{M}$ is quite singular and there are some subtleties about when the level sets have non-empty interior.   If you are willing to start with a mean convex $\Sigma_0$, White has shown that the latter never occurs and that (I believe) the levels are almost everywhere smooth.
On the other hand the surgery question for mean curvature flow is a bit trickier.  Huisken and Sinestrari give such a surgery when (if I recall correctly) $n\geq 3$ and the initial surface is 2-convex (i.e. the sum of the lowest two principal curvatures is positive...a stronger condition than mean convex).  Recently, there were a couple of papers on the arxiv where it was shown that if you took the surgery times (of the H-S surgery procedure) closer and closer to the singular time then you you would limit to the level set flow. 
This seems to be the flavor of what you are looking for, though it should be pointed out that the regularity of the higher dimensional guy $\mathcal{M}$ is somewhat unclear in general.<|endoftext|>
TITLE: Is #k-XORSAT #P-complete?
QUESTION [17 upvotes]: k-XORSAT is the problem of deciding whether a Boolean formula $$\bigwedge_{i \in I} \oplus_{j=1}^k l_{s_{ij}}$$ is satisfiable.  Here $\oplus$ denotes the binary XOR operation, $I$ is some index set, and each clause has $k$ distinct literals $l_{s_{ij}}$ each of which is either a variable $x_{s_{ij}}$ or its negation.
Equivalently, $k$-XORSAT requires deciding whether a set of linear equations, each of the form $\sum_{j=1}^k x_{s_{ij}}\equiv 1\; (\mod 2)$, has a solution over $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$.
Every decision problem Q has an associated counting problem #Q, which (informally speaking) requires establishing the number of distinct solutions.  Such counting problems form the complexity class #P.  The "hardest" problems in #P are #P-complete, as any problem in #P can be reduced to a #P-complete problem.
The counting problem associated with any NP-complete decision problem is #P-complete.  However, many "easy" decision problems (some even solvable in linear time) also lead to #P-complete counting problems.  For instance, Leslie Valiant's original 1979 paper The Complexity of Computing the Permanent shows that computing the permanent of a 0-1 matrix is #P-complete.  As a second example, the list of #P-complete problems in the companion paper The Complexity of Enumeration and Reliability Problems includes #MONOTONE 2-SAT; this problem requires counting the number of solutions to Boolean formulas in conjunctive normal form, where each clause has two variables and no negated variables are allowed.  (MONOTONE 2-SAT is of course rather trivial as a decision problem.)
Andrea Montanari has written about the partition function of $k$-XORSAT in some lecture notes, and his book with Marc Mézard apparently discusses this (unfortunately I do not have a copy available to hand, and the relevant Chapter 17 is not included in Montanari's online draft).
These considerations lead to the following question:

Is #$k$-XORSAT #P-complete?

Note that the formula in Montanari's notes does not obviously appear to answer this question.  Just because there is a nice closed form solution, doesn't mean we can evaluate it efficiently: consider the Tutte polynomial.
Some difficult counting problems in #P can still be approximated in a certain sense, by means of a scheme called an FPRAS.  Jerrum, Sinclair, and collaborators have linked the existence of an FPRAS for #P problems to the question of whether $NP = RP$.  I would therefore also be interested in the subsidiary question

Does #$k$-XORSAT have an FPRAS?

Edit: clarified second question as per comment by Tsuyoshi Ito.  Note that Peter Shor's answer renders this part of the question moot.

REPLY [3 votes]: FYI: In our book we explain that the number of solutions of $Ax=b$ is $2^{n-rank(A)}$ as mentioned by Peter.<|endoftext|>
TITLE: Simultaneous rational approximation of two reals using their continued fractions
QUESTION [10 upvotes]: Littlewood's well-known conjecture about simultaneous rational approximation is that for all $x, y \in \mathbb{R}$, $\liminf_{n \to \infty} n \Vert nx \Vert \Vert ny \Vert = 0$ (where $\Vert x \Vert$ denotes the distance from $x$ to the nearest integer).
A heuristic argument for this (mentioned in this survey article by Akshay Venkatesh) is as follows. Write $q_n(x)$ for the denominator of the $n$th convergent of the continued-fraction expansion of $x$. We know that $q_n(x) \Vert q_n(x) x \Vert$ is bounded, and it is reasonable to expect that $\Vert q_n(x) y \Vert$ will be small sometimes. Venkatesh points out that this argument doesn't work: in fact, given any sequence $q_n$ such that $\liminf q_{n+1} / q_n > 1$, we can find a $y \in \mathbb{R}$ such that $\Vert q_n y \Vert$ is bounded away from zero.
However, this doesn't quite rule out using the continued-fraction expansions of $x$ and $y$ to prove Littlewood's conjecture. The result would follow if it could be shown that for all $x, y \in \mathbb{R}$, either $\liminf_{n \to \infty} \Vert q_n(x) y \Vert = 0$ or $\liminf_{n \to \infty} \Vert q_n(y) x \Vert = 0$.
My question is whether this can be shown to be false. That is: do there exist $x$ and $y$ such that both $\Vert q_n(x) y \Vert$ and $\Vert q_n(y) x \Vert$ are bounded away from zero?
In light of gowers's answer below, let me add the condition that $x$ and $y$ are badly-approximable, i.e. that they have bounded partial quotients in their continued fractions. (It is still the case that a negative answer would imply Littlewood's conjecture.)
Here is an example from the realm of badly-approximable numbers. Let $x = \frac12 (1 + \sqrt{5})$ and $y = \frac12 (1 - 1/\sqrt{5})$. Then $q_n(x) = F_n$ (Fibonacci number), and $q_n(y) = 2F_n + F_{n+1}$ for $n \geq 1$. In this case, $\liminf \Vert q_n(x) y \Vert = \frac15 > 0$, but $\Vert q_n(y) x \Vert \to 0$.

REPLY [5 votes]: Yes, there are badly approximable numbers $x$ and $y$, in fact quadratic irrationals, such that $||q_n(x)y||$ and $||q_n(y)x||$ are bounded away from zero.  Specifically, we can take $x = \sqrt{2}/2$ and $y = \sqrt{2} + 1/2$.  It's straightforward to check that $q_n(\sqrt{2}/2)$ is always odd, from which it follows that $||q_n(x)y||$ tends to $1/2$ as $n \to \infty$.  Unfortunately, I don't know of a similarly easy way to analyze $||q_n(y)x||$, but one can do it as follows.  If we start numbering with $n=1$, then $q_n(y)$ is the nearest integer to $(1+\sqrt{2})^n$ times $(-2+3\sqrt{2})/8$, $\sqrt{2}/8$, $(2+3\sqrt{2})/8$, or $\sqrt{2}/4$, according to whether $n$ is $1$, $2$, $3$, or $4$ modulo $4$. (Specifically, $q_n(y)$ equals this number plus its conjugate over $\mathbb{Q}(\sqrt{2})$.)  If we multiply by $\sqrt{2}/2$, then we get an integer divided by $8$, $8$, $8$, or $4$.  One can check using the recurrence that this integer is always $2$ mod $4$, from which it follows that $||q_n(y)x||$ tends to $1/2$ when $n$ is $0$ mod $4$ and $1/4$ otherwise.<|endoftext|>
TITLE: What are your favorite puzzles/toys for introducing new mathematical concepts to students?
QUESTION [31 upvotes]: We all know that the Rubik's Cube provides a nice concrete introduction to group theory.  I'm wondering what other similar gadgets are out there that you've found useful for introducing new math to undergraduates and/or advanced high school students.

REPLY [2 votes]: Prof. Pedro Roitmann of University of Brasilia built these wonderful 'multiplying bottles' for teaching some properties of equilateral hyperbolas:<|endoftext|>
TITLE: some confusion about the explicit construction of irreducible representations of $S_n$
QUESTION [5 upvotes]: In this book chapter, the irreducible representations of the symmetric group $S_n$ is given in terms of polytabloids of a Ferrer's diagram $\lambda$, defined as 
$e_t = \sum_{\pi \in C_t} \text{sgn}(\pi) e_{\pi \lbrace t \rbrace}$.
 Here $t$ is a tableau of $\lambda$, $C_t$ is the column stablizing subgroup for $t$ in $S_n$. $\text{sgn}$ is the signature of the permutation $\pi$. Finally {t} is the equivalence class of tableau (called tabloid) represented by $t$, where two tableaux are considered equivalent if they have the same row entries. 
My question is, how is the definition of polytabloids above independent of the choice of $t$ in its equivalence class? For instance, if $t$ is the tableau {1,2},{3,4}, then it's equivalent to s={2,1},{3,4}, but $e_t \neq e_s$. So maybe it's not independent of representative. But then there seems to be too many polytabloids. I would also appreciate if someone could help me establish the connection with Fulton and Harris's book on representation theory problem 4.47. I am not sure what is meant by a standard tableau there.  Also in the second construction of the irreps of $S_n$ in the same problem, I am not sure how the action of $S_n$ on the polynomials is defined.

REPLY [6 votes]: Yes, equivalent tableaux $t$ may yield different $e_t$'s. However, equivalent tableaux $t$ yield the equivalent $\pi t$'s for any permutation $\pi$, so that the notation $\pi\left\lbrace t\right\rbrace$ on page 132 is justified. Nobody is claiming that $e_t$ depends on the tabloid $\left\lbrace t\right\rbrace$ only.
The Specht module $S^{\lambda}$ is defined as the vector space generated by $e_t$ for all Young tableaux $t$ corresponding to the partition $\lambda$. Now it turns out that there is a lot of redundancy in these $e_t$; that is, they are linearly dependent. One very nice basis of $S^{\lambda}$ is $\left\lbrace e_t \mid t\text{ is a standard tableau}\right\rbrace$, where a tableau is called standard if the rows are strictly monotonically increasing and the columns are strictly monotonically increasing (the word "strictly" is not of much importance here, because the numbers in our tableaus are pairwise distinct, but sometimes one also considers tableaux where the entries may be equal, and then it matters).
Concerning Fulton-Harris' problem 4.47, the first part (about the $E_T$) is exactly the definition of the Specht module that Diaconis gives. As for the second part (about the $F_T$), you have to show that there is an $S_d$-equivariant isomorphism $V_{\lambda}\to W_{\lambda}$, where $V_{\lambda}$ is the Specht module defined by means of the $E_T$'s, and $W_{\lambda}$ is the $k\left[S_d\right]$-submodule ($k\left[S_d\right]$ is what Fulton-Harris denotes by $\mathbb C\left[\mathfrak{S}_d\right]$) of $k\left[x_1,x_2,...,x_d\right]$ spanned by the polynomials $F_T=\prod\limits_{i < j;\ i\text{ and }j\text{ lie in the same column of }T}\left(x_i-x_j\right)$. To construct this isomorphism, let $\Psi$ be the vector space homomorphism $U_{\lambda}\to k\left[x_1,x_2,...,x_d\right]$ (where $U_{\lambda}$ is the representation of $S_d$ with basis the tabloids for the Young diagram $\lambda$) defined by $\Psi\left(\left\lbrace T\right\rbrace\right) = \prod\limits_{i=1}^{d}x_i^{\left(\text{number of the row in which }i\text{ lies in the tableau }T\right)-1}$ for every tableau $T$ (this is well-defined since the product on the right hand side depends only the equivalence class $\left\lbrace T\right\rbrace$ of $T$). Besides, $\Phi$ is easily seen to be $S_d$-equivariant and injective. Now, $\Psi\left(V_{\lambda}\right)=W_{\lambda}$, because every tableau $T$ satisfies $\Psi\left(E_T\right)=F_T$ or $\Psi\left(E_T\right)=-F_T$ (by Vandermonde's determinant, applied to the entries in every column of $T$), and thus the restriction of this homomorphism $\Psi$ to the subspace $V_{\lambda}$ of $U_{\lambda}$ is a $G$-equivariant bijective homomorphism $V_{\lambda}\to W_{\lambda}$. Thus, $V_{\lambda}\cong W_{\lambda}$ as representations of $S_d$.
My first actual source for the representation theory of $S_n$ were Etingof's lecture notes, but beware: they are very compressed and don't have much on $S_n$ (that's not the point of them either). Then, there is Fulton-Harris with a whole chapter on $S_n$ (but the proofs are mostly exiled into the exercises, which means that you often get hints rather than proofs). "The Representation Theory of the Symmetric Group" by James and Kerber looks very good as a comprehensive reference. There are also typewriter-style lecture notes by James (LNM 682: "The Representation Theory of the Symmetric Groups") which have the advantage of being just 136 pages long. I don't have any experience with them, however.<|endoftext|>
TITLE: Sanov's Theorem and Chernoff bound 
QUESTION [8 upvotes]: Sanov's Theorem (p.292, Thomas/Cover "Elements of Information Theory" (1991)) says that probability of a hypothesis $E$ according to distribution $Q$ is bounded above by 
$$(n+1)^k \exp (-n D(P^* \|Q) )$$ where $D$ is KL-divergence, $k$ is the size of sample space and $P^*$ is an element of $E$ closest to $Q$ according $D(\cdot\|Q)$.
This bound is quite loose for small $n$ because when $k=2$ and $E$ is simply connected, Chernoff's bound tells us that the probability is at most the following
$$\exp (-n D(P^* \|Q) )$$
Is there a natural restriction of $E$ that makes the bound above generalize to $k>2$? Alternatively, is there a natural restriction of $E$ and a modification of Sanov's theorem that gives it a similar level of tightness to Chernoff's bound for small n?

REPLY [4 votes]: For convex sets E of probability measures, the bound you want (i.e., probability of empirical measure being in E is smaller than $e^{-nD(E\|Q)}$) is true in great generality, not just for finite alphabet settings. This is contained, for instance, in the spectacular 1984 paper of Csiszar titled "Sanov Property, Generalized $I$-Projection and a Conditional Limit Theorem":
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aop/1176993227 ,
which also contains (when you reduce it to its essence) a beautifully short and elegant proof.<|endoftext|>
TITLE: How is representation theory used in modular/automorphic forms?
QUESTION [48 upvotes]: There is certainly an abundance of advanced books on Galois representations and automorphic forms. What I'm wondering is more simple: What is the basic connection between modular forms and representation theory?
I have a basic grounding in the complex analytic theory of modular forms (their dimension formulas, how they classify isomorphism classes of elliptic curves, some basic examples of level N modular forms and their relation to torsion points on elliptic curves, series expansions, theta functions, Hecke operators). This is all with an undergraduate background in complex analysis and algebra (Galois theory). I also know a little bit about the basics of algebraic number theory and algebraic geometry, if that helps. More importantly, I have a basic background in the representation theory of finite groups. My question is, then, could one example how modular forms and/or theta functions relate to representations of groups?
I'm asking this in part because I imagine a number of students with similar background as I have would have learned about modular forms and thus might be interested to understand how they relate to representation theory, despite not having an extensive background in more advanced results in algebraic geometry and commutative algebra needed for advanced study in the field.
Here are some ideas which might bear fruit: In analytic number theory, one often sees sums over characters - but characters are also very relevant in representation theory. In particular, Jacobi sums come up in both number theory and representation theory (and quadratic forms then relate to theta functions). Is there a connection here?
In addition, Hecke operators are symmetric-like sums over elements of groups, which would suggest a strong connection to representation theory.
Or is the connection to representations of $\mathrm{SL}_2(\mathbb{Z})$? Quotients of this group appear as Galois groups of extensions of spaces of modular forms, so they might be given representations by acting on these spaces?
The point of listing ideas is to show the kind of intuition I might be looking for. One of my ideas might be fruitful, or they all might have nothing to do with why representation theory connects to modular functions. The point is that I'm looking for basic ideas that someone with an elementary background might be able to understand.
I also added "reference request" because I imagine there might be a text which is at my level and discusses these ideas.
EDIT: The answer of paul garett here actually gives a nice history of how modular forms came to be viewed in terms of representations theory: What is the difference between an automorphic form and a modular form?

REPLY [2 votes]: Observation:
The modular forms/automorphic representations should be seen as generalization of Hecke quasi characters, which are simply characters $\chi: K^{\times }\backslash \mathbb{A}_K^{\times} \rightarrow \mathbb{C}^{\times}$. Tate's thesis gives a proof of the functional equation for the associated L-functions purely in terms of representation theory. He studies the right regular representation on the space of some functions $f:K^{\times }\backslash \mathbb{A} \rightarrow \mathbb{C}$. This is the so called $\mathrm{GL}_1$ case, since $\mathrm{GL}_1(K)=K^{\times}$. 
The next obvious choice is then to consider other reductive groups instead of $\mathrm{GL}_1$ e.g. $\mathrm{GL}_2$.
One remark how I think about Tate's thesis: The right regular representation on an (locally compact) abelian groups is in direct connection with its Fourier transform. The functional equation can be seen as an adelic version of the Poisson summation formula.
Observation:
The special linear group $\mathrm{SL}_2(\mathbb{R})$ acts on upper half plane $\mathbb{H}$ by Moebius transformations. Moreover this group is actually the group of all biholomorphic mappings $\mathbb{H} \rightarrow \mathbb{H}$. The isotropy group of $\mathrm{i}$ is $\mathrm{SO}_2$, i.e. the group of elements, which fix $\mathrm{i}$. Since the action is associative, i.e. for any $x, y \in \mathbb{H}$ there exists $g \in \mathrm{SL}_2(\mathbb{R})$ such that $gx = y$, we get an isomorphism of the orbit space with the space we act on. Hence, we actually have $\mathbb{H} \cong \mathrm{SL}_2(\mathbb{R}) /\mathrm{SO}_2$.
Observation
You can lift through weak approximation Dirichlet character $\chi : (\mathbb{Z}/n \mathbb{Z})^{\times} \rightarrow \mathbb{C}^{\times}$ to the adele space $K^{\times }\backslash \mathbb{A}_K^{\times}$. 
You can proceed similiar with strong approximation and obtain that the double coset space
$ SL_2(\mathbb{Q}) \backslash SL_2( \mathbb{A} ) / K $ is actually isomorphic to the orbit space $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}.$ Here $K$ means the product of all $\mathrm{SL}_2(\mathbb{Z}_p)$ for the finite places and $\mathrm{SO}_2$ for the archimedian place.<|endoftext|>
TITLE: Conjugate Gradient for a "slightly" singular system.
QUESTION [6 upvotes]: Suppose I have a symmetric $N \times N$ matrix A which has a one-dimensional Nullspace $N$.  A is positive definite on  $N^\bot$.  In my case $N$ is the space of constant vectors (i.e. generated by the all-one vector).
I have to solve the problem $Ax = b$, with $b \in R(A)$ which has infinitely many solutions.  I am looking for the minimum norm solution. The matrix $A$ is very large and sparse, direct methods aren't really an option.  The rank-deficient least squares algorithms I have seen also appear to be prohibitive.
I was solving a non-singular version of this problem with Conjugate Gradient.  Is there anyway I can modify the algorithm to solve this particular problem?
EDIT: Boiling the problem down to the bone the question is if $A$ is positive semi-definite with exactly one 0 eigenvalue, does CG work?
Thanks.

REPLY [4 votes]: Plain CG, without the regularization, will converge to the pseudo-inverse solution provided you start the iteration with $y=0$.  The regularized (preconditioned) solution will be different and the operation count will depend on the sparsity, so it's good to maintain this if you can. I guess partly it must depend on whether you can tolerate a solution that has a nonzero component in the null space.<|endoftext|>
TITLE: Extensions isomorphic as groups but not congruent or pseudo-congruent
QUESTION [10 upvotes]: I'm looking for an example of a finite abelian group A and a finite group G acting trivially on A such that there are two extensions $E_1$ and $E_2$ with base A and quotient G (i.e., they are both central extensions, and hence both give corresponding elements of $H^2(G,A)$) and:

$E_1$ and $E_2$ are isomorphic as abstract groups.
Under the natural action of $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ on $H^2(G,A)$ (by pre- and post-composition with 2-cocycles that then descends to action on cohomology classes), the cohomology classes corresponding to $E_1$ and $E_2$ are not in the same orbit.

Basically condition (2) states that $E_1$ and $E_2$ are not only not congruent extensions, they are not even congruent up to a relabeling of the subgroup A and the quotient G. Another way of putting this is that there is no isomorphism between $E_1$ and $E_2$ that sends the A inside $E_1$ to the A inside $E_2$.
The analogous statement with a nontrivial action of G on A is also of interest to me. In this latter case, though, the entire group $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ does not act.
I think that examples exist (because of my experience with finding examples for similar specifications) but there may well be a proof to the contrary.

REPLY [11 votes]: E = SmallGroup(32,28) is the first example.  It has two central subgroups A1 and A2 isomorphic to A ≅ 2 with quotient isomorphic to SmallGroup(16,11), but A1 and A2 are not conjugate in Aut(E).
Examples such as this are reasonably common in p-groups.
Edit: You can even have such an example with G abelian: G = 4×2, A = 4×2, E = SmallGroup(64,3) = 8⋉8, Z(E) = 4×4. E has two central copies of A=4×2 that are not conjugate in Aut(G), but the quotients are both abelian and isomorphic to G.
Edit: Vipul notes you can even have E abelian of order p7.<|endoftext|>
TITLE: Measure on real Grassmannians
QUESTION [6 upvotes]: OK, so I'm reading about this nice measure you can define on a (real) Grassmannian on Wikipedia. Basically, and to save you the trip through the link, consider the Haar measure $\theta$ on $O(n)$, fix a space V in your Grassmannian. Then for any subset A, the measure of A is $$\gamma(A)=\theta(g\in O(n) \mid gV \in A).$$
Fair enough. Two things Wikipedia does not really tell me, though:

Where does this construction originate? I would imagine something like this to be fairly folklore, but it would sure be nice if someone has a reference.
Does anyone know of especially interesting applications of this measure? Since the Wikipedia article cruelly lacks context, I would really like to see this idea in action.

Thanks in advance!

REPLY [5 votes]: First of all, a general fact. Any transitive homogeneous space $X$ of a
compact group $K$ has a unique $K$-invariant probability measure. Existence:
take the image $\nu$ of the normalized Haar measure $m$ on $K$ under the map
$g\mapsto gx_0$ (that's the construction you refer to). The measure $\nu$ is
well-defined, since the mass of $m$ is finite, it does not depend on $x_0\in
X$ by right invariance of $m$, and is $K$-invariant by left invariance of $m$.
Uniqueness: take an arbitrary $K$-invariant measure $\nu'$ on $X$, and
consider its convolution $m\ast\nu'$ with the measure $m$ (i.e., the image of
the product of $m$ and $\nu'$ under the map $(g,x)\mapsto gx$). Then, on one
hand $m\ast\nu'=\nu'$ by $K$-invariance of $\nu'$, on the other hand
$m\ast\nu'=\nu$ by the above construction.
Thus, since the Grassmannian in question has a transitive compact group of
automorphisms, it carries a "natural" invariant measure. So that
"platonically" it is always there - like, for instance, the Riemannian volume
on a Riemannian manifold (by the way, as mentioned before, any invariant
Riemannian metric on the Grassmannian produces the measure in question in this
way).
However, there is one subtlety here which has so far remained unnoticed. In
order to define the Grassmannian one needs a linear structure, whereas the
orthogonal group $O(n)$ is defined in terms of the Euclidean structure.
Therefore, the "canonical" measure we are talking about is only canonical with
respect to the given Euclidean structure on the linear space $V=R^n$. So, if
we look at the problem from this point of view, we obtain a map which assigns
to any Euclidean structure on $V$ a probability measure on the Grassmannian
$Gr_k(V)$. In fact, this measure depends only on the projective class of the
Euclidean structure (i.e., on the corresponding similarity structure), which
are parameterized by the Riemannian symmetric space $S=SL(n,R)/SO(n)$
(equivalently, one can say that we consider only the Euclidean structures on
$V$ with the same volume form, whence $SL$ instead of $GL$).
Thus, we have a map $x\mapsto\nu_x$ from $S$ to the space $P(Gr_k)$ of
probability measures on $Gr_k$. One can show that this map is an injection, so
that it can be used in order to compactify the symmetric space $S$ by taking
its closure in the weak$^*$ topology of $P(Gr_k)$. This is an example of a
so-called Satake-Furstenberg compactification, which can be defined for an
arbitrary non-compact Riemannian symmetric space. In the case of the space
$S=SL(n,R)/SO(n)$ all such compactifications are obtained by considering
rotation invariant measures on the flag space of $V$ and its equivariant
quotients (in particular, Grassmannians). In the general case the role of the
flag space is played by the so-called Furstenberg boundary, which is the
quotient of the semi-simple Lie group by its minimal parabolic subgroup. The
most recent reference for all this is the book by Borel and Ji.
The simplest non-compact symmetric space is the hyperbolic plane. In this case
the Furstenberg boundary (the associated "flag space") is just the boundary
circle in the disk model. Each point of the hyperbolic plane determines a
unique probability measure on the boundary circle invariant with respect to
the rotations around this point. These measures appear in the classical
Poisson formula for bounded harmonic functions in the unit disk (usually it is
written in terms of just a single measure corresponding to the Euclidean
center of the disk; the other measures appear in the guise of their
Radon-Nikodym derivatives with respect to this one, which is precisely the
Poisson kernel).<|endoftext|>
TITLE: Why relativization can't solve NP !=P?
QUESTION [21 upvotes]: If this problem is really stupid, please close it. But I really wanna get some answer for it. And I learnt computational complexity by reading books only.
When I learnt to the topic of relativization and oracle machines, I read the following theorem:

There exist oracles A, B such that $P^A = NP^A$ and $P^B \neq NP^B$.

And then the book said because of this, we can't solve the problem of NP = P by using relativization. But I think what it implies is that $NP \neq P$. The reasoning is like this:
First of all, it is quite easy to see that:
$$A = B \Leftrightarrow \forall \text{oracle O, }A^O = B^O$$
Though I think it is obvious, I still give a proof to it:
A simple proof of NP != P ?
And the negation of it is:
$$A \neq B \Leftrightarrow \exists \text{oracle O such that } A^O \neq B^O$$
Therefore since there is an oracle B such that:
$$ NP^B \neq P^B$$
we can conclude that $ NP \neq P $
What's the problem with the above reasoning?

REPLY [4 votes]: And of course, this is exactly what happens in the $\mathrm{IP}^A \ne \mathrm{PSPACE}^A$ proof -- IP is defined by its access to outside information, which can be considered akin to an oracle -- while PSPACE does not.  In the most cursory sense, both the IP-verifier and the IP-prover have access to an oracle of random bits.  PSPACE-bound machines in general do not.  A cursory way of looking at this, without even reference to the proof (which frankly I forget) is that granting IP an oracle of random bits would by definition do noting to it.  But granting PSPACE an oracle of random bits could improve its computational ability.<|endoftext|>
TITLE: subadditive implies concave
QUESTION [19 upvotes]: Let $f:R_+\to R_+$ be smooth on $(0,\infty)$, increasing, $f(0)=0$ and 
$\lim_{x\to\infty}=\infty$. Assume also that $f$ is subadditive:
$f(x+y)\le f(x)+f(y)$ for all $x,y\ge 0$. Must $f$ be concave? The converse is obvious.

REPLY [3 votes]: Subadditivity + $f(0)=0$ do not imply concavity, neither for regular functions.
Counterexamples are in Bruckner "Some relationships between locally superadditive functions and convex functions", Proc. Amer. Math. Soc., 15, 1964, 61-65, and in Bruckner-Ostrow "Some function classes related to the class of convex functions", Pacific J. Math. 12, 1962, 1203--1215.<|endoftext|>
TITLE: Complete knot invariant?
QUESTION [26 upvotes]: I've seen a couple papers (that I now can't find) that say that in his paper "On irreducible 3-manifolds which are sufficiently large" Waldhausen proved that the data $\pi_1(\partial (S^3\setminus K)) \to \pi_1(S^3\setminus K)$ is a complete knot invariant. However, the word "knot" doesn't appear in this paper (although the phrase "to avoid an orgy of notation" does :-). Is the claimed result a straightforward corollary of his main results? Or am I looking at the wrong paper?

REPLY [3 votes]: This topic is treated in G. Burde & H. Zieschang's book, Knots, 2nd edition, p. 40.<|endoftext|>
TITLE: Walking around Santa Cruz, track around the soccer field
QUESTION [11 upvotes]: I was recently walking around the track at UCSC, and I noticed that the track didn't always curve inward. Sometimes it curved the other way. Compare this (A convex track): http://www.irelandaerialphotography.com/dr_f2_3838.jpg to this (A concave track): http://www.motoyard.com/trackday/images/track-chuckwalla-l.gif . In the latter, there are portions of the track where the outer edge ( the edge bordering the infinite face, if this was a planar graph) is longer than the inner edge (the edge bordering the inner field). In a convex track, the outer edge is always longer than the inner edge. But is it possible to construct a (concave) track such that the inner edge is the same length as the outer edge? If not, how does one prove that it is impossible? The track must have a width greater than zero. (If the width is zero, the inner and outer edge will be one and the same, and the question is meaningless.) As an extension, is it possible to construct a track such that all paths along it are of the same length? A path is defined as the set of all points equidistant from the outside edge all the way around the track and equidistant from the inside edge all the way around the track, though these two distances do not have to be equal.

REPLY [3 votes]: Allow self-intersections.  Take the symmetric figure eight.
That does the trick. And it has the bonus that each half is convex. 
This is what Kloeckner might have been trying to say. 
(  As a kid, we used to drive model
slot cars on these. I imagine you can find such tracks raced by real cars. )<|endoftext|>
TITLE: Applications of fusion systems
QUESTION [10 upvotes]: What are the applications of theory of fusion systems to finite group theory
or representation theory of finite groups? More concretely, is there any important
result in finite group theory or representation theory of finite groups whose prove
uses fusion systems in the essential way?

REPLY [12 votes]: Lots of results in group cohomology only have topological proofs using the techniques of Bob Oliver and his (generalized) collaborators.  For instance, many results along the lines of "controls fusion iff controls cohomology" only have topological proofs using the same techniques that Bob Oliver called fusion systems (though I think some of the papers stick to the topological language).  "Controls fusion" is a very old term predating fusion systems by 50 years, so I think of this as a pretty pure finite group cohomology result.
I think it is common to call these topological results by the name "fusion" results, even if fusion systems are not used directly, but rather p-completed classifying spaces are used (also known as p-local groups and such), even if the fusion system used to define the topological space is not prominent.  Allowing such abuse of language, one of my favorite examples is the topological proof that the famous Z*-theorem holds for odd primes too: MR1125010.  It was proven later using more pure finite group theory, but I suspect lots of people prefer the topological proof.
In modular representation theory, the work of Puig on nilpotent blocks is (as far as I know) literally defined in terms of fusion systems and gets results on the representation theory using only conditions on the fusion systems.  One also has some nice descriptions of the representation type of blocks of dihedral defect in terms of the fusion system induced on the block (I believe Linckelmann's introduction gives this as an example).
More historically, the classification of finite groups with dihedral (and then for semi-dihedral/wreathed) begins by dividing clearly into cases based on which fusion system occurs.  This just organizes the (couple hundred page) papers, so I guess whether it is essential depends on how practical you are.
Perhaps the most "exciting" answer is not quite ready yet, but Michael Aschbacher and collaborators have begun to recast the classification of finite simple groups in terms of fusion systems.  It is quite surprising and comforting that several very important steps in the classification are radically easier in the fusion system case. The fusion system theorems would also have as side-effects corresponding theorems in modular representation theory and the theory of p-local groups.
Let me know if you want references for any of the wishy washy terms, but basically read Linckelmann's introduction, read a few papers of BLO (Broto–Levi–Oliver), and check out the slides from any recent finite group theory conference, and you should have some pretty convincing evidence that fusion systems are opening up a very bright future for finite group theory.<|endoftext|>
TITLE: Can any triangle be inscribed in any convex figure?
QUESTION [6 upvotes]: Can any triangle be inscribed in any convex figure?  i.e. given a convex figure and a triangle can we transpose and scale and rotate that triangle so that its vertices are on the boundary of the convex figure?

REPLY [12 votes]: A more general result is known: if $C$ is any Jordan curve and $T$ is a triangle then there exists a triangle similar to $T$ inscribed in $C.$ Moreover, the vertices of such triangles are dense in $C.$ See the references in the Wikipedia article on the Inscribed Square Problem.<|endoftext|>
TITLE: Abelianization of a semidirect product
QUESTION [17 upvotes]: I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})_{G}$.
(Here the subscript $G$ denotes taking the coinvariants with respect to $G$. That is, $(H^{ab})_{G}$ is a the quotient of $H^{ab}$ by the subgroup generated by elements of the form $h^g-h$ for $h$ in $H$ and $g$ in $G$, and where the superscript $g$ denotes the action of $G$ on $H^{ab}$ induced by the action of $G$ on $H$.)
Does anyone happen to know a good reference for this?

REPLY [3 votes]: If you have the semidirect product $H\rtimes G$ then you have the next group split extension
$1\rightarrow H\rightarrow H\rtimes G\rightarrow G\rightarrow 1$,
We have the Hochschild–Serre spectral sequence where $\mathbb{Z}$ is a trivial $H\rtimes G-$module
$E^{2}_{p,q}=H_{p}(G,H_{q}(H,\mathbb{Z}))\Rightarrow H_{p+q}(H\rtimes G,\mathbb{Z})$,
Since the the map $H\rtimes G\rightarrow G$ is a split surjection, it follows that the map (edge morphism)
$H_{n}(H\rtimes G,\mathbb{Z})\rightarrow H_{n}(G,\mathbb{Z})=E^{2}_{n,0}$ is a slit surjection and thus $E^{2}_{n,0}=E^{\infty}_{n,0}$
In particular we have that the diferenttial $d:E^{2}_{2,0}\rightarrow E^{2}_{0,1}$ is zero (since $E^{2}_{2,0}=E^{\infty}_{2,0}$). Therefore $E^{2}_{0,1}=E^{\infty}_{0,1}$.
It follows that there is a exact sequence
$0\rightarrow E^{\infty}_{0,1}\rightarrow H_{1}(H\rtimes G,\mathbb{Z})\rightarrow E^{\infty}_{1,0}\rightarrow 0$
which splits, in this case we have that
$H_{1}(H\rtimes G,\mathbb{Z})=E^{\infty}_{0,1}\times E^{\infty}_{1,0}$
Note that
$H_{1}(H\rtimes G,\mathbb{Z})=(H\rtimes G)^{Ab}$
$E^{\infty}_{1,0}=G^{Ab}$
and
$E^{\infty}_{0,1}=H_{0}(G, H^{Ab})=(H^{Ab})_{G}$
From this we have the result by using spectral sequences.<|endoftext|>
TITLE: Can any rectangle be inscribed in any convex figure?
QUESTION [8 upvotes]: Can any rectangle be inscribed in any convex figure?

REPLY [6 votes]: Yes, this follows from a more general result in 
Nielsen and Wright, Rectangles inscribed in symmetric continua. Geom. Dedicata 56 (1995), no. 3, 285–297 MR
(This is reference 4 in the Wikipedia article I quoted in my answer to your previous question.)
In their terminology, a simple closed curve $C$ is symmetric if there exists a point $P\notin C$ such that each straight line through $P$ intersects $C$ in exactly 2 points. This condition is trivially satisfied when $C$ is a boundary of a convex region.<|endoftext|>
TITLE: Community experiences writing Lamport's structured proofs
QUESTION [35 upvotes]: About two years ago, I came across this paper by Lamport
http://research.microsoft.com/en-us/um/people/lamport/pubs/lamport-how-to-write.pdf
on writing proofs hierarchically. It changed how I wrote all of my proofs for about six months and identified the gaps in my understanding and knowledge extremely well. These days, I won't use it for simpler proofs, but I find it indispensable when I want to thoroughly understand long and complex ones.
I think this is potentially a wonderful pedagogical tool, in that all steps and assumptions are organized and easily referred to, and is also useful for self-checking. Some discussion on the blog evaluating Deolalikar's claimed proof of $P\neq NP$, to give one example, Professor Tao's apposite remarks on the need for precision, (August 15, 2010, 3:05 PM - I hope he doesn't mind my quoting him)

One thing this illustrates is the
  importance of setting out precise
  definitions. I feel that if Deolalikar
  had written down a precise definition
  of what it meant for a solution space
  to be polylog parameterisable, the
  difficulties would have been found a
  lot sooner, and in particular probably
  by Deolalikar himself, well before he
  finished the preprint to share with
  others. 

reminded me of Lamport's essay. Lamport comments on something similar from his own experience:

The style was first applied to proofs
  of ordinary theorems in a paper I
  wrote with Martín Abadi. He had
  already written conventional
  proofs—proofs that were good enough to
  convince us and, presumably, the
  referees. Rewriting the proofs in a
  structured style, we discovered that
  almost every one had serious
  mistakes, though the theorems were
  correct. Any hope that incorrect
  proofs might not lead to incorrect
  theorems was destroyed in our next
  collaboration. Time and again,
  we would make a conjecture and write a
  proof sketch on the blackboard—a
  sketch that could easily have been
  turned into a convincing conventional
  proof—only to discover, by trying to
  write a structured proof, that the
  conjecture was false. Since then, I
  have never believed a result without a
  careful, structured proof. My
  skepticism has helped avoid numerous
  errors.

Has anyone had experience with this style of writing proofs?

REPLY [22 votes]: From a proof-theoretic point of view, Lamport essentially suggests is writing proofs in natural deduction style, along with a system of conventions to structure proofs by the relevant level of detail. (It would be very interesting to study how to formalize this kind of convention -- it's something common in mathematical practice missing from proof theory.) 
I have written proofs in this style, and once taught it to students. I find that this system -- or indeed any variant of natural deduction -- is extremely valuable for teaching proof to students, because it associates to each logical connective the exact mathematical language needed to use it and to construct it. This is particularly helpful when you are teaching students how to manipulate quantifiers, and how to use the axiom of induction. 
When doing proofs myself, I find that this kind of structured proof works fantastically well, except when working with quotients -- i.e., modulo an equivalence relation. The reason for this is that the natural deduction rules for quotient types are rather awkward. Introducing elements of a set modulo an equivalence relation is quite natural:
$$
\frac{\Gamma \vdash e \in A \qquad R \;\mathrm{equivalence\;relation}}
     {\Gamma \vdash [e]_R \in A/R}
$$
That is, we just need to produce an element of $A$, and then say we're talking about the equivalence class of which it is a member.
But using this fact is rather painful: 
$$
\frac{\Gamma \vdash e \in A/R \qquad \Gamma, x\in A \vdash t \in B \qquad \Gamma \vdash \forall y,z:A, (y,z) \in R.\;t[y/x] = t[z/x]}{\Gamma \vdash \mbox{choose}\;[x]_R\;\mbox{from}\;e\;\mbox{in}\;t \in B}
$$
This rule says that if you know that 

$e$ is an element of $A/R$, and 
$t$ is some element of $B$  with a free variable $x$ in set $A$, and
if you can show that for any $x$ and $y$ in $R$, that $t(y) = t(z)$ (ie, $t$ doesn't distinguish between elements of the same equivalence class)

Then you can form an element of $B$ by picking an element of the equivalence class and substituting it for $x$. (This works because $t$ doesn't care about the specific choice of representative.) 
What makes this rule so ungainly is the equality premise -- it requires proving something about the whole subderivation which uses the member of the quotient set. It's so painful that I tend to avoid structured proofs when working with quotients, even though this is when I need them the most (since it's so easy to forget to work mod the equivalence relation in one little corner of the proof). 
I would pay money for a better elimination rule for quotients, and I'm not sure I mean this as a figure of speech, either.<|endoftext|>
TITLE: The canonical line bundle of a normal variety
QUESTION [16 upvotes]: I have heard that the canonical divisor can be defined on a normal variety X since the smooth locus has codimension 2. Then, I have heard as well that for ANY algebraic variety such that the canonical bundle is defined:
$$\mathcal{K}=\mathcal{O}_{X,-\sum D_i}$$
where the $D_i$ are representatives of all divisors in the Class Group.
I want to prove that formula or I want to find a reference for that formula, or I want someone to rephrase it in a similar way if they heard about it.
Why do I want to prove it? Well, I use the definition that something is Calabi Yau if its canonical bundle is 0. In the case of toric varieties, $\sum D_i$~0 if all the primitive generators for the divisors lie on a hyperplane. Then the sum is 0 and therefore the toric variety is Calabi-Yau.
Can someone confirm or fix the above formula? I do not ask for a debate on when something is Calabi-Yau, I handle that OK, I just ask whether the above formula is correct. A reference would be enough. I have little access to references at the moment.

REPLY [47 votes]: Edit (11/12/12): I added an explanation of the phrase "this is essentially equivalent to $X$ being $S_2$" at the end to answer aglearner's question in the comments.
  [See also here and here]

Dear Jesus,
I think there are several problems with your question/desire to define a canonical divisor on any algebraic variety. 
First of all, what is any algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.
OK, let's assume that $X$ is such a variety. Then what is a divisor on $X$? Of course, you could just say it is a formal linear combination of prime divisors, where a prime divisor is just a codimension 1 irreducible subvariety. 
OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible. 
Still, if you want to talk about divisors, you would surely want to say when two divisors are linearly equivalent. OK, we know what that is, $D_1$ and $D_2$ are linearly equivalent iff $D_1-D_2$ is a principal divisor.
But, what is a principal divisor? Here it starts to become clear why one usually assumes that $X$ is normal even to just talk about divisors, let alone defining the canonical divisor. In order to define principal divisors, one would need to define something like the order of vanishing of a regular function along a prime divisor. It's not obvious how to define this unless the local ring of the general point of any prime divisor is a DVR. Well, then this leads to one to want to assume that $X$ is $R_1$, that is, regular in codimension $1$ which is equivalent to those local rings being DVRs.
OK, now once we have this we might also want another property: If $f$ is a regular function, we would expect, that the zero set of $f$ should be 1-codimensional in $X$. In other words, we would expect that if $Z\subset X$ is a closed subset of codimension at least $2$, then if  $f$ is nowhere zero on $X\setminus Z$, then it is nowhere zero on $X$. In (yet) other words, if $1/f$ is a regular function on $X\setminus Z$, then we expect that it is a regular function on $X$. This in the language of sheaves means that we expect that the push-forward of $\mathscr O_{X\setminus Z}$ to $X$ is isomorphic to $\mathscr O_X$. Now this is essentially equivalent to $X$ being $S_2$.
So we get that in order to define divisors as we are used to them, we would need that $X$ be $R_1$ and $S_2$, that is, normal. 
Now, actually, one can work with objects that behave very much like divisors even on non-normal varieties/schemes, but one has to be very careful what properties work for them.
As far as I can tell, the best way is to work with Weil divisorial sheaves which are really reflexive sheaves of rank $1$. On a normal variety, the sheaf associated to a Weil divisor $D$, usually denoted by $\mathcal O_X(D)$, is indeed a reflexive sheaf of rank $1$, and conversely every reflexive sheaf of rank $1$ on a normal variety is the sheaf associated to a Weil divisor (in particular a reflexive sheaf of rank $1$ on a regular variety is an invertible sheaf) so this is indeed a direct generalization. One word of caution here: $\mathcal O_X(D)$ may be defined for Weil divisors that are not Cartier, but then this is (obviously) not an invertible sheaf.
Finally, to answer your original question about canonical divisors. Indeed it is possible to define a canonical divisor (=Weil divisorial sheaf) for all quasi-projective varieties. If $X\subseteq \mathbb P^N$ and $\overline X$ denotes the closure of $X$ in $\mathbb P^N$, then the dualizing complex of $\overline X$ is 
$$
\omega_{\overline X}^\bullet=R{\mathscr H}om_{\mathbb P^N}(\mathscr O_{\overline X}, \omega_{\mathbb P^N}[N])
$$
and the canonical sheaf of $X$ is
$$
\omega_X=h^{-n}(\omega_{\overline X}^\bullet)|_X=\mathscr Ext^{N-n}_{\mathbb P^N}(\mathscr O_{\overline X},\omega_{\mathbb P^N})|_X
$$ 
where $n=\dim X$. (Notice that you may disregard the derived category stuff and the dualizing complex, and just make the definition using $\mathscr Ext$.)  Notice further, that 
if $X$ is normal, this is the same as the one you are used to and otherwise it is a reflexive sheaf of rank $1$.
As for your formula, I am not entirely sure what you mean by "where the $D_i$ are representatives of all divisors in the Class Group". For toric varieties this can be made sense as in Josh's answer, but otherwise I am not sure what you had in mind.

(Added on 11/12/12):

Lemma A scheme $X$ is $S_2$ if and only if for any $\iota:Z\to X$ closed subset of codimension at least $2$, the natural 
map $\mathscr O_X\to \iota_*\mathscr O_{X\setminus Z}$ is an isomorphism.
Proof
Since both statements are local we may assume that $X$ is affine.
Let $x\in X$ be a point and $Z\subseteq X$ its closure in $X$. If $x$ is a codimension at most $1$ point, there is nothing to prove, so we may assume that $Z$ is of codimension at least $2$.
Considering the exact sequence (recall that $X$ is affine):
$$
0\to H^0_Z(X,\mathscr O_X) \to H^0(X,\mathscr O_X) \to H^0(X\setminus Z,\mathscr O_X) \to H^1_Z(X,\mathscr O_X) \to 0
$$
shows that $\mathscr O_X\to \iota_*\mathscr O_{X\setminus Z}$ is an isomorphism
if and only if 
$H^0_Z(X,\mathscr O_X)=H^1_Z(X,\mathscr O_X)=0$ the latter condition is equivalent to 
$$
\mathrm{depth}\mathscr O_{X,x}\geq 2,
$$
which given the assumption on the codimension is exactly the condition that $X$ is $S_2$ at $x\in X$. $\qquad\square$<|endoftext|>
TITLE: Conjectures on iterated polynomial maps on finite fields
QUESTION [13 upvotes]: Let $p$ be a prime, and consider the sequence $x_0, x_1, \dots$ of elements of the finite field $\mathbf F_p$ given by $x_0 = 0$ and $x_{i+1} = x_i^2 + 1$ for all $i \ge 0$.  This sequence must eventually start repeating;  let's write $T(p)$ and $U(p)$ for the period and preperiod (resp.) of the sequence.
There's an informal idea, used for example as the basis of Pollard's Rho method for integer factorisation, that for a 'randomly chosen' prime $p$, $T(p)$ and $U(p)$ should behave like the period and preperiod of the sequence of iterates of a randomly chosen function $\mathbf F_p\rightarrow \mathbf F_p$.  For example, it's expected that the quantity $T(p) + U(p)$ (that is, the index of the first repeated value in the sequence) is comparable in magnitude to $\sqrt p$, exceeding $x\sqrt p$ with 'probability' $\exp(-x^2/2)$.
Question: Does anyone know of formal statements to this effect in the literature?  I'm looking for a conjecture that would imply the following (or something similar) as a special case:

Notation. For each positive integer $M \ge 2$, let $X_M$ be a discrete random variable that takes values in the set of primes not greater than $M$, with all primes in $[2, M]$ being equally likely to occur.  Let $X_\infty$ be a continuous random variable on $[0, \infty)$ that satisfies $\mathbf P(X_\infty > x) = e^{-x^2/2}$.
Conjecture. The sequence $$\frac{T(X_M) + U(X_M)}{\sqrt{X_M}}, \quad M \ge 2$$
of random variables converges (in distribution, say) to $X_\infty$.

Similarly, one might conjecture that $T(X_M) / (T(X_M) + U(X_M))$ is, in the limit, uniformly distributed on $(0, 1]$, and it should be possible to make (not prove!) some sort of statement about the independence of $T(X_M) / (T(X_M) + U(X_M))$ and $(T(X_M) + U(X_M)) / \sqrt{X_M}$.
I've so far failed to find any concrete statements of this form in the literature.  Any pointers?
[It's difficult to know how to tag this.  I guess it's really about discrete dynamical systems; please retag as appropriate!]

REPLY [14 votes]: "ds.Dynamical-Systems" and "nt.Number-Theory" are good tags. Another one you could add is "Arithmetic-Dynamics". You might look at the arithmetic dynamics bibliography that I've assembled at
http://www.math.brown.edu/~jhs/ADSBIB.pdf
and search for titles that include the words "finite field". (Sorry, it hasn't been updated in a while.) 
I'll also mention the following three articles. 

H. Niederreiter and I. E. Shparlinski. Dynamical systems generated by rational functions.
In Applied Algebra, Algebraic Algorithms and Error-Correcting Codes (Toulouse, 2003),
volume 2643 of Lecture Notes in Comput. Sci., pages 6–17. Springer, Berlin, 2003.

(The summary says " "We consider dynamical systems generated by iterations of rational functions over finite fields and residue class rings. We present a survey of recent developments and outline several open problems.")
Here's a paper of mine which might be relevant, although the estimates are pretty weak:

Variation of periods modulo p in arithmetic dynamics, New York J. Math. 14 (2008), 601-616.

And one more whose summary says "The orbits produced by the iterations of the mapping $x\mapsto x^2+c$, defined over $\mathbb{F}_q$, are studied. Several upper bounds for their periods are obtained, depending on the coefficient $c$ and the number of elements $q$."

A. Peinado, F. Montoya, J. Munoz, and A. J. Yuste. Maximal periods of $x^2 + c$ in $\mathbb{F}_q$. In Applied Algebra, Algebraic Algorithms and Error-Correcting Codes (Melbourne, 2001),
volume 2227 of Lecture Notes in Comput. Sci., pages 219–228. Springer, Berlin, 2001.

If you forward and backward reference these papers, you'll certainly find other papers dealing with iteration over finite fields.<|endoftext|>
TITLE: Inaccessible cardinals and Andrew Wiles's proof
QUESTION [74 upvotes]: In a recent issue of New Scientist (16 Aug 2010), I was surprised to read that a part of Wiles' proof of Taniyama-Shimura conjecture relies on inaccessible cardinals.
Here's the article

Richard Elwes, To infinity and beyond: The struggle to save arithmetic, New Scientist, August 2010.

Here's the relevant bit from the article:

"Large cardinals have been studied by logicians for a century, but their intangibility means they have seldom featured in mainstream mathematics. A notable exception is the most celebrated result of recent years, the proof of Fermat's last theorem by the British mathematician Andrew Wiles in 1994 [...]
To complete his proof, Wiles assumed the existence of a type of large cardinal known as an inaccessible cardinal, technically overstepping the bounds of conventional arithmetic"

Is this true ?
If so, could someone please outline how they are used ?

REPLY [14 votes]: At the suggestion of François G. Dorais I am moving this answer from Recent claim that inaccessibles are inconsistent with ZF to here. The text below was written before I read the above, very well-informed answers/comments, so there is considerable overlap.

This is just a quick answer regarding FLT, mentioned by the OP.
Colin McLarty is working on showing a small part of Friedman's grand conjecture, namely that the Fermat-Wiles(-Taylor?) theorem is provable in a weak system of arithmetic. Since originally the semi-stable case of Taniyama-Shimura-Weil that Wiles proved (not to mention the work by others such as Frey, Serre, Ribet to get to that point) required large parts of algebraic and arithmetic geometry developed by the Grothendieck school, which among other things uses sheaves, cohomology and so on, and so a priori requires some foundational care. Universes (~inaccessible cardinals) were introduced to take care of the problem of e.g. forming categories of sheaves on categories of sheaves on a site.
However, one can use a version of set theory much weaker than ZFC+Universe(s), indeed a fair bit weaker than ZFC, as McLarty has shown, and still get pretty much all of EGA/SGA (this is based on general arguments, he hasn't sat down and worked through it all). However, the arithmetic geometry needed for FLT [edit: and indeed a lot of number theory] really only needs to consider countable sites, rather than generic small sites, and so one really only needs to assume much weaker assumptions about infinite objects.
A more recent talk (July 2011 - slides not publicly available as far as I know) by McLarty had the statement that derived functor cohomology is a finitely axiomatised first-order theory, and so not really the complicated logical beast it appears to be.
EDIT: More work has been published in the meantime:

Colin McLarty, The large structures of Grothendieck founded on finite-order arithmetic, Rev. Symb. Log. 13 No. 2 (2020) pp. 296–325, doi:10.1017/S1755020319000340, arXiv:1102.1773

To quote from the abstract:

Such large-structure tools of cohomology as toposes and derived categories stay close to arithmetic in practice, yet existing foundations for them go beyond the strong set theory ZFC. We formalize the practical insight by founding the theorems of EGA and SGA, plus derived categories, at the level of finite order arithmetic.

Some partial results to getting to a specific finite order are in

Colin McLarty, Zariski cohomology in second order arithmetic, arXiv:1207.0276


The cohomology of coherent sheaves and sheaves of Abelian groups on Noetherian schemes are interpreted in second order arithmetic by means of a finiteness theorem

though the tools in that paper are insufficient to apply to étale cohomology, this means we are a long, long way from needing universes.<|endoftext|>
TITLE: Is completeness of a field an algebraic property?
QUESTION [12 upvotes]: Pretty straitforward:
If a field has a metric in which it is complete can it have a metric in which it is not complete?
By metric I mean field norm of course

REPLY [13 votes]: I claim that, for a field $K$, the following are equivalent:
(i) $K$ can be given a nontrivial norm -- i.e., there exists $x \in K$ with $|x| \neq 0,1$.
(ii) $K$ admits a nontrivial rank one valuation $v$.
(iii) $K$ admits infinitely many inequivalent rank one valuations $v$ such that $(K,v)$ is not complete.
(iv) $K$ is not an algebraic extension of a finite field.
Some of these facts are proved in 
http://math.uga.edu/~pete/8410Chapter2v2.pdf
(see e.g. Theorem 1).  
Let me prove here that (iv) $\implies$ (iii), which answers the OP's question in a rather definitive way.  
1) Suppose first that $K$ has characteristic $0$.  Then $K$ contains $\mathbb{Q}$, which admits the $p$-adic valuations $v_p$.  By Theorem 1 of loc. cit., each $v_p$ extends to a valuation on $K$.  
Now suppose that $K$ has characteristic $p$ and contains an element $t$ which is not algebraic over $\mathbb{F}_p$.  Thus $K$ contains the rational function field $\mathbb{F}_p(t)$, which carries infinitely many inequivalent nontrivial valuations $v_P$ corresponding to the irreducible polynomials $P \in \mathbb{F}_p[t]$ (and one more corresponding to the point at infinity on the projective line).
2) (F.K. Schmidt) If a field $K$ is complete with respect to two inequivalent rank one valuations, it is algebraically closed and uncountable.  See e.g. Theorem 24 of 
http://math.uga.edu/~pete/8410Chapter3.pdf
3) So we are reduced to the case in which $K$ is algebraically closed and uncountable.  Then $K$ is isomorphic to the algebraic closure of $K(t)$.  If we give $K$ the trivial valuation and $K(t)$ the Gauss norm $v$, then the algebraic closure of $K(t)$ has infinite degree over $K(t)$ so any extension of $v$ to the algebraic closure is not complete.  The image of the Gauss norm $v$ under the group $PGL_2(K)$ of linear fractional transformations gives us infinitely more pairwise inequivalent valuations.<|endoftext|>
TITLE: Decimating the infinite grid graph
QUESTION [8 upvotes]: Let $G$ be the graph whose nodes are the points of
$\mathbb{Z}^d$ in the nonnegative orthant (i.e., all
coordinates are $\ge 0$), with edges connecting each
pair of points separated by unit distance.
So the degree of each node not on the boundary is $2d$.
Now delete each node with probability $\delta$,
except always retain the origin $o=(0,0,\ldots,0)$.
Let $G_\delta$ be the component connected to $o$.

Q1. Does $G_\delta$ contain a simple path
  from the origin of infinite length?  

The length of a path is its number of edges.
A simple path does not cross itself.
For $d{=}1$, the answer is 'No' for any $\delta > 0$,
because eventually the run of nodes connected to $o$ will
be broken.  So, almost surely every path is of finite length.
The situation is less clear to me for $d \ge 2$.
Some experimentation tentatively suggests that for $d{=}2$ and $\delta=\frac{1}{2}$, the answer is again 'No.'
When the answer to Question 1 is 'No,'
let $r(d,\delta)$ be the radius of $G_\delta$, defined to be the expected length of the longest of the shortest paths 
from $o$ within $G_\delta$.

Q2. What is $r(d,\delta)$?

For $d{=}1$, I believe the radius is
$$\sum_{k=1}^{\infty} k (1-\delta)^k \delta = (1-\delta)/\delta \;.$$
For example, $r(1,\frac{1}{4})=3$.
The $d{=}2$ example below shows a shortest path of length 18 connecting
$(0,0)$ to $(10,4)$. (Yellow=deleted nodes, green=component connected to origin,
blue=undeleted but disconnected from origin.)
I produced this example with $\delta=0.55$.



I suspect these questions have been addressed in the literature
on random graphs, with which I am not so familiar.
Any references, reformulations, proof ideas, or partial solutions ($d{=}2$ and $d{=}3$
are of special interest to me), would be appreciated.
Thanks!
Edit.  Thanks for all the references and corrections.  From what I have learned so far, the model I defined is known as site percolation in the literature (in contrast to bond percolation).
My restriction to the positive orthant is not generally followed in the literature, but that
aside, there is much known, and much unknown.  In general there is a critical probability
$\delta_c$ for each dimension $d$ that answers my first question: for $\delta < \delta_c$,
the origin belongs to an infinite component with positive probability,
and for $\delta > \delta_c$, it belongs to a finite
component with probability $1$.
Remarkably, exact values for $\delta_c$ for site percolation on $\mathbb{Z}^d$ for $d \ge 2$ are not known.  For $d=2$, it is estimated via numerical simulations to be 0.59; for $d=3$, it is about 0.31.

REPLY [3 votes]: For an accessible introduction to percolation theory see the course handout at https://www.jyu.fi/science/muut_yksikot/summerschool/en/history/JSS19/courses/MA/main#ma2-percolation-theory by Jeffrey Steif.<|endoftext|>
TITLE: Maurer-Cartan and representable functors on differential graded commutative algebras
QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a differential graded Lie algebra on a charcteristic zero field, whose underlying chain complex is bounded below and degreewise finite dimensional. Then $\mathfrak{g}$ defines a Set-valued functor on differential graded commutative algebras (also, with some finiteness and boundedness assumption) mapping each cgda $\Omega$ to the set $MC(\Omega\otimes \mathfrak{g})$ of Maurer-Cartan elements of the dgla $\Omega\otimes \mathfrak{g}$ (with the natural dgla structure on the tensor product of a dgla with a cdga). It is well known that this functor is representable: $MC(\Omega\otimes \mathfrak{g})\cong Hom_{dgca}(CE(\mathfrak{g}),\Omega)$, where the Chevalley-Eilenberg dgca $CE(\mathfrak{g})$ is the free graded commutative algebra on the shifted linear dual of $\mathfrak{g}$ endowed with the differential induced by the dgla structure on $\mathfrak{g}$. 
If one looks at the category of commutative graded algebras instead (i.e., one forgets the differential), then $CE(\mathfrak{g})$ represents the functor $\Omega\mapsto (\Omega\otimes\mathfrak{g})^1$.
My question is: is this latter functor representable also in the category of differential graded commutative algebras? if yes, by which dgca?

REPLY [7 votes]: By the Weil algebra of $\mathfrak{g}$.<|endoftext|>
TITLE: Expected number of steps for a discrete random walk to visit every point on an N-dimensional rectangular lattice
QUESTION [8 upvotes]: Please imagine a discrete random walk on an N-dimensional rectangular lattice with dimensional lengths $(l_1, ..., l_N) \in L$ and total lattice points $P = \prod{l_i}$, for $i = 1, ..., N$.  At each time step, the walker will move to one of it's adjacent lattice points with equal probability.  The N-dimensional random walk is non-self-avoiding, the walker must move with each time step, and the boundaries of the lattice are reflecting.  However, jump probabilities must be adjusted at edges and corners due to a reduction in the number of adjacent nodes - i.e. jump probabilities will vary from $\frac{1}{2N}$ internal to the lattice to $\frac{1}{N}$ at the edges of the lattice.  
Provided the random walk specifications above, what might be the expected step-time distribution for the walker visiting every position in the N-dimensional rectangular lattice with dimensional lengths $L$?

REPLY [10 votes]: If you look for "cover time of graph" you will find a lot of references, cf. e.g. "Jonasson, Schramm, ON THE COVER TIME OF PLANAR GRAPHS, Elect. Comm. in Probab. 5 (2000) 85-90, http://www.emis.de/journals/EJP-ECP/_ejpecp/ECP/include/getdocbfb7.pdf. In this paper you find the following result by Zuckermann (for details pls. cf the paper):  
If $G$ is a finite portion of a $d$-dimensional lattice $\mathbb{Z}$ with $n$ vertices, the expected cover time (to visit all vertices) is $\Theta(n^2)$ for $d=1$, $\Theta(n (\log n)^2)$ for $d=2$, and $\Theta(n \log n)$ for $d \ge 3$.<|endoftext|>
TITLE: Hochschild and cyclic homology of smooth varieties 
QUESTION [6 upvotes]: Many of the standard sources which discuss the Hochschild Kostant Rosenberg theorem and cyclic homology for smooth varieties such as Loday and Weibel's paper "The Hodge Filtration and Cyclic Homology" ignore the positive characteristic case. Based upon these sources, I wasn't really sure if this was from a lack of knowledge or because the theorems are just not really that good in characteristic p. Here are a few rather simple(and hopefully correct!) observations about Hochschild homology and cyclic homology of smooth varieties over a field of characteristic p>0 which hopefully get the ball rolling. These are all trivial observations(as long as they are right) but they seem to suggest that there is some interesting math in the characteristic p>0 case, and I was wondering whether I had made a mistake or misunderstood something in the literature/what the opinion of experts was on these questions.
1) If $X$ is a smooth scheme over a field k of characteristic $p>0$, then we can prove the Hochschild Kostant Rosenberg
theorem as follows. The basic observation is that to compute $HH_*(X)=Tor_{O_{X\times X}}(\Delta_{*}O_X,\Delta_{*}OX)$ by the adjunction this is the same as $Tor_{O_X}(\Delta^*\Delta_*O_X,O_X)$ but the complex in the first argument is canonically
isomorphic to the tangent complex $\bigoplus\Lambda^iT(X)[-i]$ as proven on page 247 in Huybrecht's book on the Fourier Mukai
transform.
2)If $A$ is a smooth commutative ring, again over a field k of characteristic $p>d$, the Krull dimension of the ring, then
all the arguments that are given in Loday's book regarding the relationship between de Rham cohomology and cyclic homology seem to work exactly the same when the characteristic $p>d$. In particular, the spectral sequence converging to cyclic homology still degenerates on the second page.  This should lead to the following scheme theoretic theorem as well. The periodic cyclic homology is isomorphic to $\prod H^*_{dr}$ if the characteristic $p>d$.
3) The above theorems seems to suggest that maybe the above degenerates for smooth algebras over a field, independent of the
characteristic. Somewhat independent of that one could wonder if the de Rham cohomology and periodic cyclic homology always agree for smooth
varieties over a field. Does anyone know of any counterexamples to this? Again, in the affine case, this might for example follow from a sort of Cartier isomorphism, optimistically a quasi-iso from $(C^*(A,A)((u)),d+uB) \mapsto C^*(A,A)((u)),d)$. In the case of ordinary de Rham theory, for general schemes there are obstructions to realizing the Cartier isomorphism at the chain level like this--- but I think these obstructions all vanish for affine schemes, hence this guess. Anyways, I have the impression that Kaledin proved something like this, but I haven't had a chance to study it yet, so I thought I'd just ask the MO community.

REPLY [3 votes]: HKR is true in characteristic $p>0$ as soon as you assume that $p$ is greater than the dimension of $X$. 
You can find a very nice proof of that in a recent preprint of Arinkin-Caldararu: http://arxiv.org/abs/1007.1671
I am sure this is not new, but the reason I am quoting this paper is that you will find in it a very nice generalization of HKR for closed embeddings $X\subset Y$ (where the result works in positive characteristic whenever it is greater than the codimension). The Hochschild case is the diagonal inclusion of $X$ into $X\times X$. 
EDIT: I am not an expert in algebraic geometry, but I believe your question is related to Deligne-Illusie proof of Hodge-to-de Rham degeneration, which has the same kind of restriction (p greater than the dimension).<|endoftext|>
TITLE: What does the representation theory of the reduced C*-algebra correspond to?
QUESTION [7 upvotes]: Let $G$ be a locally compact group. The group C*-algebra $C^* (G)$ is designed to come with a natural bijection between its (nondegenerate) representations and the (strongly continuous, unitary) representations of $G$.
Question: Is there a similar statement for the reduced group C*-algebra $C^*_r (G)$?
If the answer is no, I'll probably end up asking for the actual purpose of defining $C^*_r (G)$. So far, I know that its isomorphic to $C^* (G)$ in important cases, and that its construction is in some sense simpler than the one of $C^* (G)$.
(The definitions and the claims used above can be found in Blackadar's Operator Algebras.)

REPLY [4 votes]: For suitable locally compact groups $G$ (separable, unimodular, type I), there is a measure $\mu$ on the dual $\hat{G}$ such that, for every function $f\in L^1(G)\cap L^2(G)$:
$$\int_G |f(g)|^2dg=\int_{\hat{G}}\|\pi(f)\|^2_{HS}d\mu(\pi)$$
where $\|.\|_{HS}$ denotes the Hilbert-Schmidt norm. The measure $\mu$ is the Plancherel measure of $G$ and its support is exactly the reduced dual, i.e. the dual of $C^*_r(G)$. For all this, see section 18.8 in J. Dixmier, $C^*$-algebras, North Holland, 1977.<|endoftext|>
TITLE: Profunctors corresponding to "partial functors"
QUESTION [6 upvotes]: Suppose we have a span of categories $C \overset{F}{\hookleftarrow} D \overset{G}{\rightarrow} E$, where $F$ is a subcategory embedding. We can lift these normal functors to profunctors $\hat F$ and $\hat G$ and compose the "formal" adjoint $\hat F^\dagger$ with $\hat G$ to obtain a profunctor $\hat G \hat F^\dagger : C \nrightarrow E$. To what extent could one think of this as a partial functor, and what nice behaviours could it inherit from $F$ and $G$? For instance, if $F$ reflects products and $G$ preserves them, does $\hat G \hat F^\dagger$ preserve them where ever it is defined?

REPLY [4 votes]: One thing along these lines that you can say is that if D has, and G preserves, finite limits (or more generally is flat), then so does $\hat{G}$ considered as a cocontinuous functor $[D^{op},Set] \to [E^{op},Set]$.  Since $\hat{F}^\dagger : [C^{op},Set] \to [D^{op},Set]$ is just precomposition with F, it preserves all limits and colimits; thus $\hat{G} \hat{F}^\dagger$ preserves finite limits as soon as G does, without any hypothesis on F.  I don't know whether this can be extended to other kinds of limits.
Regarding the more general question of whether $\hat{G} \hat{F}^\dagger$ could be considered a "partial functor," another way to describe it is as the left Kan extension of the composite $D \overset{G}{\to} E \hookrightarrow [E^{op},Set]$ along F.  If F is fully faithful, then such an extension is an honest extension, i.e. it restricts back along F to the original functor.  So one could think of it as obtained by extending G to objects not in D in the most universal way possible: it maps an object $c\in C$ to the formal colimit (viewing $[E^{op},Set]$ as the free cocompletion of E) over all approximations to c by objects of D.
On the other hand, every profunctor can be obtained as a composite $\hat{G} \hat{F}^\dagger$ for some functors G and F (not necessarily an embedding): let the intermediate category D be the two-sided discrete fibration corresponding to that profunctor.  An arbitrary profunctor can be thought of as a "generalized functor," but usually not specifically a "partial functor."  However, perhaps faithfulness, or full-and-faithfulness, of F implies some properties of the resulting profunctor which makes it seem more like a "partial functor."<|endoftext|>
TITLE: Extending vector bundles on a given open subscheme, reprise
QUESTION [20 upvotes]: In this question, Ariyan asks about the question of uniqueness of extensions of vector bundles when they exist.
Sasha's answer suggests that extensions of vector bundles don't always exist.
More precisely, if $F$ is a vector bundle on an open subscheme $U$, there does not always exist a vector bundle $F'$ on the ambient space $X$ such that $F'|_U \cong F$.
Can anyone give me a simple example of such an $F$?
I am mainly interested in the case when $X$ is a variety (over $\mathbb{C}$), and $U$ is an open subvariety. Probably I want $X$ to be smooth.

REPLY [3 votes]: Generalizing @Sasha's example.

Let $X$ be a regular projective scheme over a noetherian ring $A$, and let $Z$ be a closed subscheme of $X$ of codimension $\ge 3$.
  The ideal sheaf $I_Z$ is generated by sections after tensoring by $O_X(l)$ for some $l \gg 0$, it follows that we have an exact sequence
  $$
0 \to E \to O_X(l)^m \to O_X \to O_Z \to 0
$$
   for some $m$ large.
  If $U \subset X - Z$ is any open subset whose complement is of codimension $\ge 2$, then $E|U$ is locally free but does not extend to the whole of $X$, or to any open set containing an associated point of $Z$.


For any $x\in X$, the ring $O_{X,x}$ is regular and thus the module $E_x$ is of finite projective dimension. 
For module with finite projective dimension, being a second syzygy is equivalent to being reflexive. 
It follows that $E_x$ is reflexive for all $x\in X$ and thus $E$ is reflexive. 
Clearly $E$ is locally free on $X-Z$.
Let $z$ be an associated point of $Z$, then we claim $E_z$ is not free at $z$.
The module $O_{Z,z}$ has finite projective dimension pdim $O_{Z,z}$  = depth $O_{X,z}$ - depth $O_{Z,z}$ by the Auslander-Buchsbaum formula. Since $z$ is an associated point of $Z$, we have depth $O_{Z,z} = 0$. Since Z has codimension $\ge 3$, and $O_{X,z}$ is Cohen-Macaulay, we conclude that pdim $O_{Z,z}$ =  depth $O_{X,z}$ = dim $O_{X,z} \ge 3$. It follows that $E_z$ cannot be free, otherwise 
$$
0 \to  E_z \to O(l)^m_z \to O_{X,z} \to O_{Z,z} \to 0
$$ would be a free resolution of $O_{Z,z}$ of length 2.
On a scheme satisfying (G1) (Gorenstein in codimension one) and Serre's condition (S2), e.g. normal, Cohen-Macaulay, regular, etc., a reflexive sheaf admits a unique extension from an open set whose complement has codimension $\ge 2$ to the whole space.<|endoftext|>
TITLE: Highest weights of the restriction of an irreducible representation of a simple group to a Levi subgroup
QUESTION [11 upvotes]: Let $G$ be a simple Lie group over ${\mathbb C}$, $P \subset G$ a parabolic subgroup, and $L \subset P$ its Levi subgroup. Let $\lambda$ be a $G$-dominant weight and $V_G^\lambda$ an irreducible representation of $G$ with highest weight $\lambda$. I am interested in restrictions on highest weights of irreducible components of the restriction
$$
(V_G^\lambda)_{|L}.
$$
In the simplest case, when $P = B$ is the Borel subgroup, $L = T$ is the maximal torus, and there is a well-known restriction --- the weights of $V_G^\lambda$ all lie in
$$
Conv(\lbrace w\lambda \rbrace_{w \in W}),
$$
where $W$ is the Weyl group. I would be happy to know something of the same sort for arbitrary parabolic subgroup.

REPLY [2 votes]: This is not a complete answer, but maybe it will be of use.
You're asking which $L$-high weights $\mu$ occur in the $G$-irrep $V_\lambda$. Let me say that $\mu$ occurs classically if for some $N>0$, $N\mu$ occurs in $V_{N\lambda}$.

The set of such $\mu$ form a rational polytope lying inside $L$'s positive Weyl chamber.
The vertices of this polytope strictly inside $L$'s chamber ("regular vertices") are exactly those of the form $w\cdot \lambda$ that are lucky enough to be in there.
The vertices of this polytope lying on $L$'s Weyl walls are very likely to be very complicated. In particular they may not be integral weights of $L$. As I recall this already happens for $GL(3) \supset GL(2)\times GL(1)$.

Parts 1 & 3 apply to any branching problem (and much further). Part 2 is special to your case that $L$ has the same rank as $G$ (I'm not actually using that it's a Levi). 
If all you want is an upper bound, as your comment to Jim suggests, then that's easy: the $L$-high weights that can occur are a subset of the $T$-weights that occur, which you already described. Probably you want something better than that though. In principle it wouldn't be too hard to figure out the local structure of your polytope nearby the regular vertices, but I expect that not all facets contain regular vertices.
Littelmann describes (in the case of a Levi) the highest weights that occur and their multiplicities: one looks at all the Littelmann paths for the irrep $V_\lambda$ that lie entirely inside the closed $L$-chamber.<|endoftext|>
TITLE: Discontinuous convolutions
QUESTION [7 upvotes]: Is the following true?
The convolution of two infinitely differentiable as well as integrable real functions can be nowhere continuous.
A reference/proof idea would be very helpful.

REPLY [5 votes]: I believe the answer is yes. though I don't quite have a precise construction.  The basic idea is to find a series $f = \sum_n f_n$ of increasingly narrow bump functions $f_n$ whose $L^1$ norms decay very quickly (e.g. exponentially fast in $n$), but such that the series converges to a nowhere continuous function $f$; I think some sort of "typewriter function" construction will do this.  If the bump functions $f_n$ are reasonable, then they should be expressible as (or modifiable to) a convolution $f_n = g_n * h_n$, where the $g_n$ are also narrow (but perhaps tall) bump functions whose $L^1$ norms also decay very quickly.  If one then sets $g(x) := \sum_n g_n(x-x_n)$ and $h(x) := \sum_n h_n(x+x_n)$ where $x_n$ goes to infinity extremely quickly with $n$ (e.g. $x_n = 2^{2^n}$) then I think $g*h$ should differ from $f$ by a continuous function and thus also be nowhere continuous, despite $g, h$ being infinitely smooth and integrable.<|endoftext|>
TITLE: Square of an elliptic curve and projective plane
QUESTION [5 upvotes]: Let's assume one takes $E = \mathbb{C}^* / \langle p \rangle$ an elliptic (Tate) curve over the complex field ($p = e^{2 \pi i \tau}$ where $1, \tau$ are the 2 periods in additive notation; $\Im \tau > 0$). On this take points $u_1, u_2, u_3$ such that $u_1 u_2 u_3 = 1$ and then mod out by the action of the symmetric group $S_3$. So we essentially have a hypersurface in $E^3$ - a copy of $E^2$ with coordinates $(u_1, u_2)$ and we mod out by permuting $u_1, u_2$ and $1/u_1 u_2$ (the $u_i$'s are zeros and their reciprocals poles of an elliptic function - essentially the only one up to constant with these zeros and poles). 
The question: is this quotient space $\mathbb{P}^2$? I believe the answer is yes, but I can't see a way of using theta functions or other gadgets to explicitly give the isomorphism (whereby a theta function I mean $$\theta_p(x) = \prod_{l \ge 0}(1-p^l x)(1-p^{l+1}/x)$$ which reduces to the Jacobi theta via the triple product identity).
Finally, does this work over other fields (reals, finite fields, other reasonable fields)?

REPLY [3 votes]: Consider some $a,b,c\in E$. Then $a\oplus b\oplus c=0_E$ in the group $E$ iff $a+b+c = 3\cdot 0_E$ in $\mathrm{Pic}^3(E)$ iff $a,b,c$ are colinear in the complete linear system $|\mathcal{O}_E(3\cdot 0_E)|\cong\mathbb{P}^2$. I.e. you "unordered triplets" scheme is ${\mathbb{P}^2}^*$. This is true over any field.<|endoftext|>
TITLE: Bounds on remainder term of power series of elementary functions
QUESTION [7 upvotes]: This is mainly a question about the remainder term of power series for elementary functions.
I'm very interested in aspects of calculating or computing elementary operations and functions, by which I mean:

trigonometric: $\sin$, $\cos$, $\tan$
inverse trig.: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$
log and exponential: $\ln$, $\exp$
hyperbolic: $\sinh$, $\cosh$, $\tanh$
inverse hyp.: $\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$
powers, reciprocation, $\sqrt{\ \ \ }$

perhaps also:

gamma function: $\Gamma$
and a few other important functions

There are many contexts (of calculation). For example:

real versus complex arguments
known, fixed precision versus variable precision
numerical versus symbolic

There are many approaches and techniques available too. For example:

power series expansions and polynomial approximations
use of relationships between the functions
use of periodic or similar properties to shrink the domain
lookup tables and interpolation
CORDIC (used within some hand calculators I believe)
exact methods
interval or other error-tracking methods

Some good references to certain aspects include:

Digital Library of Mathematical Functions: Elementary Functions
Chee-Keng Yap, Fundamental problems of algorithmic algebra
Behrooz Parhami, Computer Arithmetic: Algorithms and Hardware Designs

The main gap in my knowledge is in finding bounds for the error or remainder term in partial power series expansions of certain of the above functions. Some are fairly simple to determine, whilst others seem to be awkward.
Any pointers on this matter would be much appreciated.
Likewise for any further references on any other aspects of or techniques for calculating elementary functions.

REPLY [5 votes]: finding bounds for the error or remainder term in partial power series expansions

I think you want the Euler-Maclaurin Summation formula.  That bounds the remainder terms, although it would require knowing the closed form of the integral representation of the function you are calculating.  
$ \sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right) $
The paper by Apostol "Elementary view of Euler-Maclaurin" AMM vol 106 (1999) pp. 409-418  is very accessible.  
The following papers/books may also be helpful 

R.P. Boas "Estimating
Remainders." Math. Mag. 51,
pp 83-89, (1978)
http://www.tricki.org/article/Estimating_sums
Bridger and Frampton Bounding Power
Series Remainders Math. Mag. 71 (1998), pp. 204-207
Sofo. Computational Techniques for
the Summation of Series
Ross.  Methods of Summation
Davis.  Summation of Series.<|endoftext|>
TITLE: Is this a $C^{\infty}$ function ?
QUESTION [11 upvotes]: Let be $(a_n)\in\ell^2(\mathbb N)$ and consider the mapping $f:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ given by
$$
f\Big((a_n)\Big)=(a_n^n).
$$
Question: Is $f$ a Fréchet $C^{\infty}$ function in whole $\ell^2(\mathbb N)$ ?
If the answer for the previous question is no. Is there a non-linear $C^{\infty}$ function, defined in some Banach space that maps a closed bounded set onto a non bounded set ?

REPLY [6 votes]: The behaviour of this function is rather more subtle (and interesting) than suggested by the answers so far.  Let me begin with the general remark that the theory of holomorphic functions on Banach spaces (even locally convex spaces) is very well developed---a readable introduction which contains enough material to answer the above questions is a survey article  by Nachbin in vol. 79 of the Bulletin of the American Mathematical Society 
(1973) but a great deal of work has been done since and several monographs have been written on the subject.  The function of the question is indeed holomorphic but only on the open unit ball.   The problem is that in larger balls there are elements with components larger than $1$ and the powers make them huge.  Interestingly, it is holomorphic in weaker senses
everywhere, e.g., it is finitely holomorphic, i.e., its restriction to finite dimensional subspaces is holomorphic (in view of Hartogs' theorem it suffices to check this for one-dimensional spaces).  It is also holomorphic as a function with values in $\omega$, the product of countable copies of the real line. Very mild smoothness properties combined with finite homomorphicity imply holomorphicity in the strong sense and so these cannot hold for the function in question outside of the unit ball.
With regard to the boundedness question, it is a central fact that holomorphic functions on Banach spaces do not necessarily preserve boundedness---in fact, this behaviour is the exception
in a certain sense.  For example, a result quoted in the above-mentioned article states that if a closed, bounded subset of a reflexive or separable space is such that each holomorphic mapping is bounded on it, then it is compact.<|endoftext|>
TITLE: How many pairs of edges can disconnect a biconnected graph?
QUESTION [8 upvotes]: Consider some biconnected graph $G$. Removing any single edge will not disconnect $G$. However, unless $G$ is triconnected, there is some pair of edges whose removal will disconnect $G$. For a cycle of length $l$, the removal of any pair will disconnect the graph (if the edges are adjacent, there will be an isolated vertex).

Conjecture: any biconnected graph on $l$ vertices has at most $\binom{l}{2}$ pairs of edges that disconnect it. Furthermore, the bound is tight only for cycles.

Is the conjecture true? I have a candidate proof but I suspect that there's a simpler one.

REPLY [15 votes]: The statement is true.  In fact, much more general statements are true.  If $G$ is a graph with $n$ vertices and $c$ is the cardinality of a minimum edge cut of $G$, then the number of edge cuts of cardinality $c$ is at most $\binom{n}{2}$, and for every half-integer $k \geq 1$, the number of edge cuts containing at most $kc$ edges is bounded above by $2^{2k-1} \binom{n}{2k}.$
The upper bound of $\binom{n}{2}$ on the number of minimum cuts is attributed to Bixby and Dinitz-Karzanov-Lomonosov.  The more general bound on the number of approximate minimum cuts is due to Karger (Global min-cuts in RNC, and other ramifications of a simple min-cut algorithm), who also re-proved the $\binom{n}{2}$ bound on minimum cuts.  His appealingly simple proof rests on the analysis of a simple "randomized contraction" algorithm.  Here we present the proof that the number of minimum cuts is at most $\binom{n}{2}$.
Suppose that $G$ is a multigraph with $n$ vertices, $c>0$ is the number of edges in a minimum cut of $G$, and $A$ is a specific set of $c$ edges whose removal disconnects $G$.  Repeatedly perform the following process to obtain a sequence of multigraphs $G = G_0, G_1, \ldots, G_{n-2}$: choose a uniformly random edge of $G_t$ and contract it to obtain $G_{t+1}$.  In other words, if $(u,v)$ is the edge chosen in step $t$, then we replace $u$ and $v$ with a single vertex $z$ in $G_{t+1}$, and we replace every edge of $G_t$ having exactly one endpoint in $\{u,v\}$ with a corresponding edge of $G_{t+1}$ with endpoint $z$.  (Edges from $u$ to $v$ in $G_t$ are deleted during this step.)  Note that $G_{n-2}$ has exactly two vertices $a,b$, these vertices correspond to a partition of $V(G)$ into two nonempty sets $A,B$ (those vertices that were merged together to form $a \in V(G_2)$, and those that were merged together to form $b$), and that the edges of $G_{n-2}$ are in one-to-one correspondence with the edges of the cut separating $A$ from $B$ in $G$.  Denote this random cut by $R$.
Now consider a specific cut $C$ of cardinality $c$.  We claim that the probability of the event $R=C$ is at least $1 \left/ \binom{n}{2} \right.$, from which it follows immediately that the number of distinct cuts of cardinality $c$ is at most $\binom{n}{2}$.  To prove the upper bound on the probability that $R=C$, observe that for all $t = 0,\ldots,n-2$, every vertex of $G_t$ has degree at least $c$.  (Otherwise, that vertex of $G_t$ corresponds to a set of vertices in $G$ having fewer than $c$ edges leaving it, contradicting our assumption about the edge connectivity of $G$.)  Consequently, the number of edges of $G_t$ is at least $(n-t)c/2$, and the probability that an edge of $C$ is contracted in step $t$, given that no edge of $C$ was previously contracted, is at most $c/|E(G_t)| \leq 2/(n-t)$.  Combining these bounds, we find that the probability that no edge of $C$ is ever contracted is bounded below by $\prod_{t=0}^{n-3} \left(1 - \frac{2}{n-t}\right) = \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \frac{1}{3} = \frac{2}{n(n-1)}.$<|endoftext|>
TITLE: monochromatic cycle-free colouring of the complete graph on R?
QUESTION [12 upvotes]: Hi 
So there is an edge-colouring of a complete graph on R (the reals), with countably many colours that as no monochromatic triangle. To find it map R to (0,1) write the numbers in binary and if 2 numbers differ 1st in the kth digit use colour k. 
Now this colouring has cycles of length 4. (1/4, 3/4, 1/3, 2/3 for example). You can get rid of cycles of length 4 by considering the 1st 2  binarary digits in which 2 numbers differ (and of course seperate colours if they only differ in 1 digit). Anyway my question is can we avoid cycles completely? i.e. does there exist a colouring of the complete graph on R such that there is no monochromatic cycle.

REPLY [4 votes]: Let me add to the above answer by noting that a stronger result holds and that, in the absence of the Continuum Hypothesis, the problems with the attempted construction of a 4-cycle-free coloring are unavoidable.
To fix notation, if $X$ is a set, $[X]^2$ is the set of 2-element subsets of $X$. We show that any coloring $c:[\omega_2]^2 \rightarrow \omega$ has a monochromatic 4-cycle. This will also give a shorter proof of the negative direction of the Erdos-Kakutani result mentioned above.
Fix $c:[\omega_2]^2 \rightarrow \omega$. For each $\omega_1 \leq \gamma < \omega_2$, find $n_\gamma < \omega$ such that $A^\gamma_{n_\gamma} := \{\alpha < \gamma \mid c(\{\alpha, \gamma\}) = n_\gamma\}$ is uncountable. Find a stationary $S \subseteq \omega_2$ and an $n^* < \omega$ such that, for all $\gamma \in S$, $n_\gamma = n^*$. For each $\gamma \in S$, let $\alpha_\gamma = \min(A^\gamma_{n^*})$, and let $\beta_\gamma = \min(A^\gamma_{n^*} \setminus (\alpha_\gamma + 1))$. By Fodor's Lemma, we can find $\alpha^* < \beta^* < \omega_2$ and a stationary $T \subseteq S$ such that, for all $\gamma \in T$, $(\alpha_\gamma, \beta_\gamma) = (\alpha^*, \beta^*)$. Fix $\gamma < \delta$, both in $T$. Then $c(\{\alpha^*, \gamma\}) = c(\{\gamma, \beta^*\}) = c(\{\beta^*, \delta\}) = c(\{\delta, \alpha^*\}) = n^*$, giving us a monochromatic 4-cycle.<|endoftext|>
TITLE: Indecomposable objects in a category
QUESTION [10 upvotes]: According to the Elephant, and these notes, an object X in a category C is indecomposable if given an isomorphism $X \cong \coprod_i U_i$ there is a unique $i$ such that $X \cong U_i$ and $U_j \cong 0$ for $j\neq i$ where 0 is the initial object.  If C is extensive, then X is indecomposable iff it is connected (proof and details here).
Lambek and Scott give a different definition: they say that X is indecomposable if given an epi $[k,l] \colon U + V \twoheadrightarrow X$, one of k or l is epi.  I suppose this can be generalised to say that X is indecomposable if any jointly epic family into X contains at least one epi.
Perhaps I'm missing something obvious, but I can't see that either definition implies the other.  So my question is

Are these definitions equivalent, or does one imply the other, in general or in some specific class of categories?  Do you know of a reference that compares or discusses the two?

REPLY [10 votes]: Briefly: there's a simple difference in how they treat 0.  That fixed, still neither implies the other in general.  In a regular extensive category, a slight modification of the LS definition implies the Elephant one.  I suspect they're not fully equivalent in anything short of a topos.  As Mike Shulman points out, even in a topos they are not equivalent.

The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty).  So, let's temporarily change one of the definitions to fix this.  I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition.  (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)
In eg Top, however, we can see that the Elephant def still doesn't imply the LS def.  $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi).  Even more, it’s decomposable by a regular epi (more on this distinction below).
Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but both of $f$ and $g$ are mono).
When do they imply each other?  If we upgrade the LS definition to involve regular epis, then in a regular lextensive category, it implies the Elephant definition, if I'm not mistaken.  For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$.  The coproduct inclusions are then jointly reg epi, so one of them is reg epi.  But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso.  There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.
Edit from Mike Shulman's comments: if moreover we're in a pretopos, all epis are regular, so there the original LS definition will imply the Elephant definition.  On the other hand, the Elephant definition doesn't imply the LS even in a topos: the terminal object of $\mathbf{Sh}([0,1])$ is a counterexample, essentially for the same reasons that $[0,1]$ was a counterexample in $\mathbf{Top}$.
However, the two definitions are equivalent for projective objects… and I guess that's how this situation has arisen, since a common use of indecomposable objects in topos theory is the theorem that the indecomposable projectives in a presheaf category are exactly the retracts of representables.  (This is useful because it lets us recover the idempotent-completion of  $\mathbf{C}$, which is very close to $\mathbf{C}$ itself, from $[\mathbf{C}^\mathrm{op},\mathbf{Set}]$.)<|endoftext|>
TITLE: Generating the derived category with line bundles
QUESTION [20 upvotes]: The following lemma is useful and well-known:
LEMMA If $L^{\pm 1}$ is ample on proper scheme over a field $k$, then some number of powers $\mathcal{O},L,...,L^{m}$ generate the unbounded derived category of quasi-coherent sheaves $D(X)$ (or split generate the subcategory of perfect complexes).
QUESTION: What about a converse? Suppose that I know some number of powers of $L$ generate $D(X)$. Then can I conclude that $L^{\pm 1}$ is ample?
The best I can do so far is see that the restriction of $L$ to any integral curve 
$C$ in $X$ has non-zero degree. (Since by adjunction $\mathcal{O},L,...,L^{m}$ generates $D(C)$, but if $L$ had degree $0$ on $C$, there would be something orthogonal $\mathcal{O},L,...,L^{m}$, for instance a generic line bundle of degree $g-1$ having no cohomology.) 
Something I don't know yet: does the degree of $L$ must have the same sign on all curves?
This would be useful for numerical tests of ampleness.  
Note: I think that one doesn't need properness in the above lemma, but I am willing to assume it to get a converse. It makes life easier when restricting to closed subschemes.
Note 2: When saying a collection of objects generates a triangulated category with all coproducts, like $D(X)$, one usually means that you take the smallest triangulated subcategory closed under all coproducts and containing the the collection. Once you have all coproducts, then idempotents automatically split, by a standard argument called, I think, the Eilenberg swindle. If you are working with a smaller triangulated category having only finite coproducts, like perfect complexes on a scheme, then the smallest triangulated subcategory containing a collection might not be 'thick', in the sense that some idempotents might not split, so in this case one usually adds in the missing summands. To emphasize this, some people speak of 'split generation'.

REPLY [2 votes]: I think you can construct a (many) counter-examples as follows. Let $X$ be the blow-up of $\mathbb{P}^2$ along a point $p_0 \in \mathbb{P}^2$. I denote by $E$ the exceptional divisor. Let $\pi : X \longrightarrow \mathbb{P}^2$ be the blow-up map and let $L = \pi^* \mathcal{O}_{\mathbb{P}^2}(1) \otimes \mathcal{O}_{X}(E)$.
The line bundle $L$ is not ample because its restriction to $E$ has negative degree. The dual of $L$ is not ample because its retsriction to any curve in $X$ disjoint from $E$ has negative degree.
Now, I claim that $L^{-2},L^{-1},\mathcal{O}_X$ generates $D^b(X)$. This can be proved as follows. I denote by $A$ the full sucategory of $D^b(X)$, closed under taking direct summands, which is generated by $L^{-2}, L^{-1}, \mathcal{O}_X$. Let $F \in D^b(X)$ such that $\mathrm{Ext}^k(a, F) = 0$ for all $a \in A$. I want to prove that $F = 0$ in $D^b(X)$.
Let $x \in X \backslash E$. There exists a line $l \subset \mathbb{P}^2$ through $\pi(x)$, disjoint from $p_0$ so that there is a section $s$ of $L$ whose vanishing locus is exactly $\pi^{-1}(l) \cup E$. Hence, we have an exact sequence:
$$ 0 \rightarrow L^{-1} \stackrel{s}\longrightarrow \mathcal{O}_X \rightarrow \mathcal{O}_{\pi^{-1}(l)} \oplus \mathcal{O}_E \rightarrow 0.$$
We deduce that $\mathcal{O}_E$ and $\mathcal{O}_{\pi^{-1}(l)}$ are in $A$. Twisting the above exact sequence by $L^{-1}$, we get that $\mathcal{O}_{\pi^{-1}(l)} \otimes \pi^* \mathcal{O}_{\mathbb{P}^2}(-1)$ and $\mathcal{O}_{E}(-E)$ are also in $A$.
Now, we have an exact sequence:
$$0 \rightarrow \mathcal{O}_{\pi^{-1}(l)} \otimes \pi^* \mathcal{O}_{\mathbb{P}^2}(-1) \rightarrow \mathcal{O}_{\pi^{-1}(l)} \rightarrow \mathcal{O}_x \rightarrow 0.$$
As a consequence, we have $\mathrm{Ext}^k(\mathcal{O}_x, F) = 0$ for all $k\in \mathbb{Z}$. In particular, the support of $F$ does not contain $x$.
This is true for all $x \in X \backslash E$ so that the reduced support of $F$ is  included in $E$. 
Let $y \in E$. We have an exact sequence on $E$:
$$0 \rightarrow \mathcal{O}_E(E) \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_y \rightarrow 0.$$
Twisting by $\mathcal{O}_{E}(-E)$, we get an exact sequence on $E$:
$$0 \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_E(-E) \rightarrow \mathcal{O}_y \rightarrow 0.$$
Pushibg forward by $\mathrm{R} i_*$, we get an exact squence on $X$:
$$0 \rightarrow \mathcal{O}_E \rightarrow \mathcal{O}_E(-E) \rightarrow \mathcal{O}_y \rightarrow 0.$$
Since we know that $\mathcal{O}_E$ and $\mathcal{O}_E(-E)$ are in $A$, we deduce that $\mathrm{Ext}^k(\mathcal{O}_y, F) = 0$ for all $k\in \mathbb{Z}$. But this is true for all $y \in E$. We conclude that the support of $F$ is zero and that $F \simeq 0$ in $D^b(X)$.
Of course it seems difficult to find any condition on $X$ to prevent such a counter-example to appear. Indeed, we'll get a counter-example similar to the one above a soon as we get a birational map $X \rightarrow Y$ which contracts a curve ($Y$ may well be singular!).<|endoftext|>
TITLE: Geometric meaning of fiber of modular parameterization over a point of an elliptic curve?
QUESTION [16 upvotes]: Given an elliptic curve $E/\mathbb{Q}$ of conductor $N$, parameterization $\psi : X_0(N) \rightarrow E$, and a point $P \in E$, take the fiber $\psi^{-1}(P)$. Its points, being on $X_0(N)$, correspond to equivalence classes of pairs $(E_x, C_x)$.
Is there a geometric meaning for these pairs in relation to the point $P$? Something about these elliptic curves that has something to do with $P$? (Other than the j-invariant solving some polynomial equation with coefficients depending only on $P$ (and $E$ of course))
Anything special about the $C_x$'s in relation to $P$?
Modular parameterization is fascinating, but I just don't understand where it comes from. $X_0$ is for a whole bunch of elliptic curves. $E$ is a single specific one.What's the connection between points on the modular curve, to a single specific point of $E$?

REPLY [6 votes]: If your elliptic curve $E$ has rank $1$ over $\mathbb{Q}$, then there is a point $P$, a Heegner point, in $E(\mathbb{Q})$ with a special point $x=(E_x,C_x)$ in the fibre of $\psi$, i.e. such that $E_x$ has extra endomorphisms (as described in the other answers).
If instead your point $P$ of infinite order is on a curve $E$ of rank $2$ over $\mathbb{Q}$, then I believe there is nothing known about the points in the fibre. The degree $d$ of $\psi$ is getting large quite quickly. Each of the $d$ points will be defined over a large number field $K_x$. Probably in most cases the points in the fibre will form one single Galois-orbit. I believe this was verified for the generators of the curve of conductor 389. And there did not seem anything special about the points in the fibre.
It is a strange coincidence that elliptic curves over $\mathbb{Q}$ admit a cover by a curve which parametrises elliptic curves with extra structure. 
Given that there is this coincidence, one can ask what happens if one takes the points $x=(E,C)$ for different $C$'s on $X_0(N)$ and map them to $E$. This point $P=\psi(x)$ will not be defined over $\mathbb{Q}$ unless $N=27$. (There are two cm curves of conductor 27 with a 27 isogeny between them; the point $P$ is torsion in this case.) Otherwise, if the $j$-invariant of $E$ is not in $\tfrac{1}{2}\mathbb{Z}$ then $P$  will be of infinite order, but it will be defined over a number field much larger than $\mathbb{Q}$.<|endoftext|>
TITLE: Spectrum of the sum of generators for irrational rotation algebra
QUESTION [7 upvotes]: Let $\theta \in \mathbb{R}\backslash\mathbb{Q}$. The irrational rotation C*-algebra $\mathcal{A}_{\theta}$ is the universal C*-algebra generated by unitary elements $u$ and $v$ with $vu=e^{2\pi i \theta}uv$.
What is the spectrum of u+v?
(Note: in the case where the unitary elements u and v are the standard generators of a free group factor, the spectrum of u+v is computable by work of Haagerup and Larsen.)

REPLY [3 votes]: The question is nicely resolved here: http://arxiv.org/abs/1210.4771<|endoftext|>
TITLE: How to escape the inclination to be a universalist or: How to learn to stop worrying and do some research.
QUESTION [165 upvotes]: As an undergraduate we are trained as mathematicians to be universalists. We are expected to embrace a wide spectrum of mathematics. Both algebra and analysis are presented on equal footing with geometry/topology coming in later, but given its fair share(save the inherent bias of professors). Number theory, and more applied fields like numerical analysis are often given less emphasis, but it is highly recommended that we at least dabble in these areas.
As a graduate student, we begin by satisfying the breadth requirement, and thus increasing these universalist tendencies. We are expected to have a strong background in all of undergraduate mathematics, and be comfortable working in most areas at a elementary level. For economic reasons, if our inclinations are for the more pure side, we are recommended to familiarize ourselves with the applied fields, in case we fall short of landing an academic position.
However, after passing preliminary exams, this perspective changes. Very suddenly we are expected to focus on research, and abandon these preinclinations of learning first, then doing something. Professors espouse the idea that working graduate student should stop studying theories, stop working through textbooks, and get to work on research.
I am finding it difficult to eschew my habits of long self-study to gain familiarity with a subject before working. Even during my REU and as an undergrad, I was provided with more time and expectation to study the background.
I am a third year graduate student who has picked an area of study and has a general thesis problem. My advisor is a well known mathematician, and I am very interested in this research area. However, my background in some of the related material is weak. My normal mode of behavior, would be to pick up a few textbooks and fix my weak background. Furthermore, to take many more graduate courses on these subjects. However, both of my major professors have made it clear that this is the wrong approach. Their suggestion is to learn the relevant material as I go, and that learning everything I will need up front would be impossible. They suggest begin to work and when I need something, pick up a book and check that particular detail.
So in short my question is: 

How can I get over this desire to take
  lots of time and learn this material
  from the bottom-up approach, and
  instead attack from above, learning
  the essentials necessary to move more
  quickly to making original
  contributions? Additionally, for those of you advising students, do you recommend them the same as my advisor is recommending me?

A relevant MO post to cite is How much reading do you do before attacking a problem. I found relevant advice there also.
As a secondary question, in relation to the question of universalist. I find it difficult to restrain myself to working on one problem at a time. My interests are broad, and have difficulty telling people no. So when asked if I am interested in taking part in other projects, I almost always say yes. While enjoyable(and on at least one occasion quite fruitful), this is also not conducive to finishing a Ph.D.(even keeping in mind the advice of Noah Snyder to do one early side project). With E.A. Abbot's claim that Poincaré was the last universalist, with an attempt at modesty I wonder

How to get over this bred desire to work on everything of interest, and instead focus on one area?

I ask this question knowing full well that some mathematicians referred to as modern universalists visit this site. (I withhold names for fear of leaving some out.)
Also, I apologize for the anonymity.
Thank you for your time!
EDIT: CW since I cannot imagine there is one "right answer". At best there is one right answer for me, but even that is not clear.

REPLY [8 votes]: One thing that might help is to think about this advice as saying you need to develop new skills rather than being about the one true way to do mathematics.  That is, you have a lot of experience and skill at learning math systematically from the ground up like you did in your classes, but that's only one skill out of many that are necessary to be an effective researcher.  You also need to hone your problem solving skills, you need to learn how to identify appropriate interesting problems and research programs, you need to learn how to efficiently wade through the literature to find what you need, etc.  You have less practice at these things, so naturally you're not as good at them and so it's going to feel uncomfortable, but that's normal when you're learning new skills.
In other words, you're focusing too much on being well-rounded in terms of subject matter, but you're missing out on being well-rounded in terms of your skills.
That said, I do think it's worth following your interests, provided you still have enough time for your main research program.  There's more time in a day than my brain can handle working hard on my main research program, and the more of the rest of the time that was spent learning other mathematics, filling in gaps in my background, or working on a side project, rather than say playing 2048, the better.  So keep learning broadly and deeply, but don't let it eat into the time you're getting your main work done.<|endoftext|>
TITLE: Random, Linear, Homogeneous Difference Equations and Time Integration Methods for ODEs
QUESTION [7 upvotes]: Most methods (that I know of) of numerically approximating the solution of ODEs are "general linear methods". For this type of method, the so-called 'linear stability' is examined by applying the method to the linear, constant-coefficient, (complex) scalar ODE
$\dot{y} = \lambda y $.
Because the same procedure is used to generate a new approximation at each time step, the methods result in linear, homogeneous, constant-coefficient difference equations for the approximate solution values at each time step. As a result, linear stability analysis for these methods amounts to analyzing the solutions of constant coefficient, homogeneous linear difference equations.
If you vary the method randomly at each time step, the coefficients of the difference equation are no longer constant. For example, second order explicit linear multistep methods can be written as a one-parameter family. If you choose this parameter randomly at each time step, the difference equation looks like this
$y_{n+1} + F(a(n))y_n + G(a(n))y_{n-1} = 0$
where $a(n)$ is a random variable (the parameter in the family of methods) and the functions $F$ and $G$ are known. My main question is whether there is any theory giving conditions on $F$, $G$, and the distribution of $a$ such that solutions remain bounded in the limit of large $n$. It would be nice if the theory generalized to higher order difference equations too.
If you were to randomly decide to use second-order multistep versus second-order Runge-Kutta methods, for example, then $F$ and $G$ would also depend on $n$. A theory to handle that case would be welcome too.
Since the numerical analyst is free to choose the distribution of $a$, and to some extent also $F$ and $G$, I'm wondering if it might be possible to design 'random' methods that have better linear stability properties than the usual ones which repeat the same process over and over. I'm posting here because I know almost nothing about stochastic/random processes.

REPLY [2 votes]: The usual way of analyzing three-term recurrences is to look at the asymptotic behavior of your $F(a(n))$ and $G(a(n))$ and then invoke the machinery of (Poincaré-)Perron. What often happens is that there is one solution that is decaying (the "minimal" solution) and one that is growing (the "dominant" solution).
Usually people are interested in the bounded or minimal solution, and attempting to propagate the recurrence forward (increasing n) in finite precision is courting certain numerical disaster. Due to rounding, the initial values of your recurrence can be expressed as a linear combination of the minimal and dominant solution; thus, if the recurrence is executed forward, there will come a point where the growth of the dominant solution will swamp the behavior of the bounded solution. (In numerical work, people usually recurse backwards or use special techniques like the Miller algorithm when they are interested in minimal solutions)
Related to this, it has long been known that certain multistep methods for solving ODEs (e.g. Milne's method) can exhibit the phenomenon of "parasitic solutions", where no matter how small a step size $h$ you take, a growing solution to the recurrence that alternates in sign with every step can easily contaminate your bounded solution, rendering your numbers meaningless.
At this point, I would like to direct you to this excellent survey article by Walter Gautschi and this old but nice book by Jet Wimp.<|endoftext|>
TITLE: When is a symplectic manifold equivalent to a cotangent bundle?
QUESTION [33 upvotes]: Let $X$ be a differentiable manifold. Its cotangent bundle $T^*X$ carries a canonical 1-form $
\alpha$ whose exterior differential $\omega = d\alpha$ endows $T^*X$ with the structure of a symplectic manifold.
But what about the converse question? Which symplectic manifolds are cotangent bundles? 
Clearly a necessary condition is that $\omega$ must be exact, so cohomological obstructions are relevant. Is that all? Compact symplectic manifolds have non-trivial de Rham cohomology in grade two, so the cohomological test passes muster for that important class of examples.
I'm also interested in examples of manifolds where different symplectic forms (modulo exact 2-forms) give qualitatively different dynamics with the same Hamiltonian. As symplectic structures on the same manifold are all locally equivalent by Darboux's theorem, one expects such phenomena would occur only on a global scale.

REPLY [5 votes]: I don't know if this question : "when a symplectic manifold is isomorphic to a cotangent bundle" has a complete and simple answer in the literature, in the way you want, but this is some comments that come in mind about this question. 
A cotangent bundle $T^*Q$ has two main characteristics: 

A Lagrangian foliation over its base $(q,p) \mapsto q$. 
A one parameter dilatation group $t\mapsto [(q,p) \mapsto (q,e^t
    p)]$.

On a symplectic manifold $(M,\omega)$, we usually call polarization a Lagrangian foliation when the space of leaves is a manifold. And we call Liouville 1-parameter group a 1-parameter group of diffeomorphisms $\varphi_t$ such that $\varphi_t^*(\omega) = e^t\omega$. The infinitesimal action of the Liouville group is a (complete) Liouville vector field, it satisfies ${\cal L}_\xi(\omega) = \omega$. Every Liouville vector field on a symplectic manifold gives a primitive of the symplectic form: $\alpha(\cdot) = \omega(\xi,\cdot)$, that is, $d\alpha = \omega$. We have then two necessary conditions for the symplectic manifold $(M,\omega)$ to be a cotangent space: 

There exists a polarization $\pi : x \mapsto q$ onto some manifold $Q$.
There exists a complete Liouville vector field $\xi$, tangent
to the polarization $\pi$.

Let's assume now that these two conditions are satisfied, we can define a natural map $\Phi : M \to T^*Q$ by 
$$
\Phi(x) = (q = \pi(x), p = [\delta q \mapsto \omega_x(\xi(x),\delta x)]) 
\quad \mbox{with} \quad \pi_*(\delta x) = \delta q.
$$
Here $\delta q \in T_qQ$, $\delta x \in T_xM$. Because the Liouville field is tangent to the polarization, $\omega_x(\xi(x),\delta x)$ depends only on $\delta q = \pi_*(\delta x)$, and therefore $p$ belongs to $T^*_qQ$. Now this map $\Phi$ satisfies:
$$
\Phi(\lambda) = \alpha \quad \mbox{and then} \quad \phi^*(d\lambda) = \omega,
$$
where $\lambda$ is the canonical Liouville 1-form $pdq$ on $T^*Q$. Now, since $d\lambda$ and $\omega$ are symplectic the tangent linear map $D(\Phi)_x$ is non degenerate, $\ker D(\Phi)_x = \{0\}$ for all $x$. Thus, $\Phi$ is an étale map, that is, a local diffeomorphism everywhere. Hence, $(M,\omega)$ is not far to be a cotangent bundle, we can already say that $\omega$ is the pullback of the standard symplectic form $d\lambda$ on a cotangent $T^*Q$ by an étale map which is already a bit of interesting information. It remains to give some conditions on the polarization to move from an étale map to a diffeomorphism. I don't know if it is exactly the sense you gave to your question but it may help to apprehend the situation.<|endoftext|>
TITLE: Does Weyl's Inequality prove equidistribution?
QUESTION [13 upvotes]: Let $f(n) = \theta n^d + a_{d-1} n^{d-1} + \cdots a_1 n + a_0$ be a polynomial with real coefficients, and $\theta$ irrational. Let $S_N = \sum_{n=1}^N e^{2 \pi i f(n)}$. Weyl's Equidistribution theorem for polynomials is equivalent to the claim that $S_N/N \to 0$ as $N \to \infty$. You can read a nice proof of this theorem on Terry Tao's blog (see Corollary's 5 and 6). I had thought that Weyl's Inequality was supposed to be a more precise version of this bound. However, I can't actually figure out how to get Weyl's Inequality to imply the required claim!
Specifically, let $p/q$ be a rational number in lowest terms with $|\theta - p/q| \leq 1/q^2$. Weyl's Inequality is the bound:
$$S_N/N \leq 100 \left( \log N \right)^{d/2^d} \left( \frac{1}{q} + \frac{1}{N} + \frac{q}{N^d} \right)^{1/(2^d-1)}$$
Here are I am quoting from Timothy Gowers' notes. (UPDATE: George Lowther, below, suggests that Gowers may have a typo.) Wikipedia has a softer version, with more freedom in choosing parameters; I think my question applies to both versions.
Now, suppose that the convergents $p_i/q_i$ of $\theta$ grow so fast that $q_{i+1} > e^{(d+1) q_i}$. 
And take $N \approx e^{q_i}$. I get that, for any choice of $q$ with $|\theta - p/q| < 1/q^2$, either $1/q > 1/\log N$ or $q/N^d > 1$. This gives infinitely many $N$'s for which the right hand bound is useless (greater than $1$). So it seems that Weyl's inequality does not prove $S_N/N \to 0$.
Am I missing something?
The motivation for this question was my attempt to answer this question over at math.SE. So any useful comments you have on that question would be appreciated as well.

REPLY [2 votes]: This response is in answer to David's further question about whether it is possible to bound the rate at which SN/N tends to zero, as he was wanting to use Weyl's inequality to do.
This is not possible, even in the case d=2 and f(n)=θn2. (for d=1 it is not hard to show that SN is bounded so $S_N/N=O(N^{-1})$).
Set
$$
S_N(\theta)=\sum_{n=1}^Ne^{2\pi i\theta n^2}
$$
in the following. Given any function h: ℕ → ℝ+ with liminfnh(n) = 0, I show that there are irrational θ with
$$
\begin{array}{}\displaystyle\sup_N\vert S_N(\theta)/(h(N)N)\vert=\infty.&&(*)\end{array}
$$
[Note: The following is a much simpler argument than the original version]. I'll use the Baire category theorem to find counterexamples

For any countable collection An of open dense subsets of ℝ, the intersection A = ∩nAn is dense in ℝ.

In particular, any such A is nonempty. We can say more than this; if S is a countable subset of the reals then $A\setminus S=\left(\bigcap_nA_n\right)\cap\left(\bigcap_{s\in S}\mathbb{R}\setminus\{s\}\right)$ is an intersection of dense open sets, so is dense. In particular, A will contain a dense set of irrational values.
To construct counterexamples then, it is only necessary to show that the set of all θ at which the sequence diverges to infinity is an intersection of countably many open sets, and show that it contains a dense set of rational numbers. The Baire category theorem implies that it will also diverge at a dense set of irrationals.
In fact, for any sequence xn(θ) depending continuously on a real parameter θ, the set of values of θ for which it diverges to infinity is an intersection of countably many open sets
$$
\{\theta\colon\sup_n\vert x_n(\theta)\vert=\infty\}=\bigcap_n\bigcup_m\{\theta\colon\vert x_m(\theta)\vert>n\\}.
$$
So, we only need to find a dense set of rational numbers at which (*) holds.

Let θ = a/b for integers a,b with b > 0. Setting $x=S_b(\theta)/b$ then $S_N(\theta)/N\to x$ as $N\to\infty$.

Proof:
If m ≡ n (mod b) then θm2 - θn2 is an integer, and $e^{2\pi i\theta m^2}=e^{2\pi i \theta n^2}$. So $n\mapsto e^{2\pi i\theta n^2}$ has period b, giving
$$
S_{bN}(\theta)=\sum_{j=0}^{N-1}\sum_{k=1}^{b}e^{2\pi i\theta(jb+k)^2}=N\sum_{k=1}^be^{2\pi i\theta k^2}.
$$
So, SbN(θ) = NSb(θ). Now, any N can be written as N = bM + R for some R < b. Then, $\vert S_N-MS_b\vert\le R$ and, dividing by N gives $\vert S_N/N-S_b/b\vert\to0$ as N goes to infinity.
As |SN(θ)/(h(N)N)| ∼ |x|/h(N) → ∞ whenever x is nonzero, the following shows that (*) holds whenever θ is of the form a/p for an odd prime p not dividing a. Such rationals are dense, so the existence of irrational θ for which (*) holds follows from the Baire category theorem.

Let θ = a/p for integers a,p with p an odd prime not dividing a. Then $x=S_p(\theta)/p$ is nonzero.

Proof:
Note that $u=e^{2\pi i a/p}$ is a primitive p'th root of unity with minimal polynomial $X^{p-1}+X^{p-2}+\cdots+X+1$ over the rationals. Then, all proper subsets of $\{1,u,u^2,\ldots,u^{p-1}\}$ are linearly independent over the rationals and
$$
S_p(\theta)=\sum_{k=1}^{p}u^{k^2}=1+2\sum_{k=1}^{(p-1)/2}u^{k^2}
$$
is nonzero.
In fact as pointed out by David below, Sp is a Gauss sum and has size √p.<|endoftext|>
TITLE: What is the topological/smooth mapping class group of an n-dimensional torus?
QUESTION [11 upvotes]: The page "mapping class groups" on wikipedia says the topological MCG of T^n is GL(n,Z), but does anyone know a reference? Also, is the smooth MCG of T^n known?

REPLY [13 votes]: Indeed, $MCG(\mathbb T^n)=GL(n,\mathbb Z)$ in dimension $n<4$, but it is not simple. In dimension 2 it was first proved by Earle and Eells using complex analysis.[Edit: As Allen Hatcher points out this was known for a long time, Earle and Eells prove much stronger statement: $\mathbb T^2$ is deformation retraction of $Diff_0(\mathbb T^2)$]
I am not sure what happens in dimension 4. 
This is not correct in dimension >4, the MCG is semidirect product of $GL(n,\mathbb Z)$ with another (non-finitely generated group). Let me just quote Hatcher:


If $n\ge 5$ then there are split exact sequences
  $$
0\to \mathbb Z_2^\infty\to\pi_0(Top(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
  $$
0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\to\pi_0(PL(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
  $$
0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\oplus\sum_{i=0}^n\binom n i\Gamma_{i+1}\to\pi_0(Diff(\mathbb T^n))\to GL(n,\mathbb Z)\to 0
$$
  where $\Gamma_i$ are Kervaire-Milnor finite abelian groups of homotopy spheres, $\mathbb Z_2$ is just the group of order 2 and $\mathbb Z_2^\infty$ is the group of finite strings.

The above quote is from Hatcher "Concordance spaces, higher simple homotopy theory and application."<|endoftext|>
TITLE: What is the probability a random Turing machine is isomorphic to a DFA?
QUESTION [6 upvotes]: This is a sort of Chaitin/Omega constant type question, and so I do not expect this probability to be computable to arbitrary precision.  However, it is also a very practical thing to know from the perspective of inductive learning.  The motivation to ask this question comes from reading some of Solomonoff's papers on algorithmic probability and resource bounded probability.
In Solomonoff's algorithmic probability, one studies induction / machine learning under the assumption that all observed data is the output of a universal Turing machine with random input.  With this prior, expectation maximization via Bayesian methods leads to the natural notion that the best model for a particular observed sequence of data is the one generated by the shortest program (aka minimum description length / Kolmogorov complexity).
One cute thing about this prior is that it beats the so-called "no free lunch theorems" as best as possible.  Because Kolmogorov complexity is unique up to an additive constant across Turing machines, as long as the underlying sequence is non-uniformly random it can at least theoretically be learned given a finite number of samples.
The bad thing of course is that Kolmogorov complexity is uncomputable, so this is really more of a theoretical curiosity than an actual tool for machine learning.  To get around this, Solomonoff proposed the notion of resource bounded probability, where the resources for the Turing machine are ultimately limited (ie bounded compute time/space).  In the case of a space bound, this limitation ultimately transforms the Turing machines into DFAs, which leads to my question:
Given the algorithmic prior (ie observed data is generated by some universal Turing machine fed uniform random inputs), what is the probability that the data observed is the output of a DFA?
Of course if it is a DFA, life is really great since there are a bunch of tricks out there for estimating such a DFA.  Also it might be interesting to maybe look at other classes of automata as means to get other variants of resource limited types of results.

REPLY [15 votes]: Let me focus on the question of your title, and mention
that there is another quite robust way to understand what
it means to say that a random Turing machine has
such-and-such property.
Specifically, we use the concept of asymptotic density as
the size of the program increases. For any natural number
$n$, there are only finitely many Turing machine program
using $n$ states. The asymptotic density or asymptotic probability
of a set $A$ of Turing machine programs is the limit (if it
exists)

$\lim_{n\to\infty} \frac{|A\cap P_n|}{|P_n|}$,

where $P_n$ is finite the set of Turing machine programs
with exactly $n$ states. Thus, the asymptotic probability
of a set $A$ of Turing machine programs is simply the limit
of the proportion of $n$-state programs in $A$. In
particular, if a set $A$ has asymptotic density $1$, then
it means that more than $99\%$, more than $99.9\%$, of
Turing machine programs are in $A$, as close to $1$ as
desired as the number of states increases. In this case, we
would seem to be justified in saying that almost every
Turing machine program is in $A$.
One can interpret your title question this way as
inquiring: what is the asymptotic density of the set of
Turing machine programs that decide sets that are
equivalent to a DFA?
To give an elementary sample calculation, a Turing machine
program $p$ in finite alphabet $\Sigma$ with states $S$
(not counting the halt state) is a function
$\Sigma\times S\to \Sigma\times
(S\cup\{halt\})\times\{L,R\}$. For example, if the alphabet
has $2$ symbols and there are $n$ states, then there are
$(4(n+1))^{2n}$ many programs. The number of programs that
never transition to the halt state, however, is
$(4n)^{2n}$, which has proportion $(\frac{n}{n+1})^{2n}$,
which goes to $\frac{1}{e^2}$ as $n\to\infty$. Thus, the
density of programs that never halt at all, because they
can never transition to the halt state, is $\frac{1}{e^2}$,
or about $13.5\%$. Of course, all such programs decide the
empty language, which is also decided by a DFA, so this is
a lower bound on the title question.
This way of thinking is the foundation of the topic of
generic case
complexity.
A central concern of this topic is the fact that many
undecidable or unfeasible decision problems admit a black
hole, a very small region where the problem is difficult,
outside of which it is easy. For example, it is not good to
base a financial encryption scheme on a problem whose
difficulty is confined to a black hole---a robber is after
all satisfied to rob the bank even only $90\%$ of the time,
or even only $1\%$ of the time. Alexei Miasnikov inquired
whether the halting problem itself admits a black hole, and
it turned out that for one of the standard models of
computability, the answer is yes:
Theorem.(Hamkins+Miasnikov) For the Turing machine
model with one-way infinite tapes, there is a set of Turing
machine programs $A$ such that

$A$ has asymptotic density $1$, so almost every program
is in $A$.
$A$ is polynomial time decidable.
The halting problem is polynomial time decidable for
programs in $A$.

Thus, for this model of computation, the halting problem is
decidable with probability $1$. The reason has to do with
the fact that for the one-way infinite tape Turing machine
model, it turns out that almost every Turing machine
program, like Polya's drunken man, falls off the tape
before repeating a state. And this is something that can be
detected in linear time. It follows that with asymptotic
probability one, a Turing machine program computes a finite
set. Since all finite sets are DFA computable, we conclude:
Corollary. For the standard one-way infinite tape
model of Turing machines, with asymptotic probability one,
a random Turing machine computes a set that is DFA
decidable.
See J. D. Hamkins, A. Miasnikov, The halting problem is
decidable on a set of asymptotic probability
one, Notre Dame Journal
of Formal Logic, Notre Dame J. Formal Logic 47 (2006),
515–524.
See also this MO
answer,
which mentions similar ideas.<|endoftext|>
TITLE: Natural examples of finite dimensional spaces with interesting 2-type
QUESTION [5 upvotes]: Riemann surfaces provide interesting examples of 1-types - interesting as they have roles in diverse areas. However, apart from 2-dimensional lens spaces, I can't readily bring to mind natural examples of spaces with non-trivial first two homotopy groups (non-trivial firt $k$-invariant optional, I suppose). Given a crossed module one can get an interesting 2-type, but this is via geometric realisation, so hardly finite-dimensional and not smooth in the usual sense (maybe in some exotic notion of smoothness).
Do natural examples of spaces with interesting 2-types turn up anywhere?
I tried to find out if I could construct one in a naive way at this question, but it fell over.

REPLY [4 votes]: Cappell-Shaneson knots are a special class of knots in homotopy 4-spheres.  In a sense they were designed as an example of knots with a well-known but prescribed 2-type.  
The definition is like this.  Look at bundles over $S^1$ whose fiber is a once-punctured $(S^1)^3$.   The monodromy is an element of $A \in GL_3(\mathbb Z)$.  A non-trivial (but fun) exercise is to check this manifold $M = \left((S^1)^3\setminus\{*\}\right) \rtimes_A S^1$ is the complement of a smoothly embedded $S^2$ in a homotopy $4$-sphere if and only if $det(A) = \pm 1$.  
$\pi_2 M$ has a single generator as a module over $\pi_1 M$, but it's far from a free module over $\pi_1 M$.  Perhaps the best way to describe $\pi_2 M$ is that it's a Laurent polynomial ring $\mathbb Z[x^\pm,y^\pm,z^\pm]$ where the $x,y,z$ correspond to the generators of $\pi_1 ((S^1)^3 \setminus\{*\})$  i.e. this is $H_2$ of the universal cover, which has a natural identification with $(\mathbb R^3 \setminus \mathbb Z^3) \times \mathbb R$.  The action of $\pi_1 M$ is just the action on this covering space, so you get not just multiplication by units $x^ay^bz^c$ but also the automorphisms of $\mathbb Z^3$ (coming from $A \in GL_3(\mathbb Z))$ acting on the exponents of the polynomials. 
The homotopy 4-spheres that contain Cappell-Shaneson knots were once considered possible counter-examples to the smooth 4-dimensional Poincare conjecture.  Recent work of Akbulut and Gompf seem to have largely removed this possibility.<|endoftext|>
TITLE: Asymptotic density of k-almost primes
QUESTION [26 upvotes]: Let $\pi_k(x)=|\{n\le x:n=p_1p_2\cdots p_k\}|$ be the counting function for the k-almost primes, generalizing $\pi(x)=\pi_1(x)$. A result of Landau is
$$\pi_k(x)\sim\frac{x(\log\log x)^{k-1}}{(k-1)!\log x}\qquad\qquad(1)$$
but this approximation is very poor for $k>1$.
For $\pi(x)$ much more is known. A (divergent) asymptotic series
$$\pi(x)=\frac{x}{\log x}\left(1+\frac{1}{\log x}+\frac{2}{\log^2x}+\frac{6}{\log^3x}\cdots\right)\qquad\qquad(2)$$
exists (see. e.g., the historical paper of Cipolla [1] who inverted this to produce a series for $p_n$). And of course it is well-known that
$$\pi(x)=\operatorname{Li}(x)+e(x)\qquad\qquad(3)$$
for an error term $e(x)$ (not sure what the best current result) that can be taken [4], on the RH, to be $O(\sqrt x\log x)$.  Even better, Schoenfeld [6] famously transformed this into an effective version with
$$|e(x)|<\sqrt x\log x/8\pi\qquad\qquad(4)$$
for $x\ge2657$. For the heretics rejecting the Riemann Hypothesis, Pierre Dusart has a preprint [2] which improves on the results in his thesis [3]; in particular, for $x\ge2953652302$,
$$\frac{x}{\log x}\left(1+\frac{1}{\log x}+\frac{2}{\log^2x}\right)\le\pi(x)\le\frac{x}{\log x}\left(1+\frac{1}{\log x}+\frac{2.334}{\log^2x}\right)\qquad\qquad(5)$$
But I know of no results even as weak as (2) for almost primes. Even if nothing effective like (5) exists, I would be happy for an estimate like (3).
Partial results
Montgomery & Vaughan [5] show that
$$\pi_k=G\left(\frac{k-1}{\log\log x}\right)\frac{x(\log\log x)^{k-1}}{(k-1)!\log x}\left(1+O\left(\frac{k}{(\log\log x)^2}\right)\right)$$
for any fixed k (and, indeed, uniformly for any $1\le k\le(2-\varepsilon)\log\log x$ though the O depends (exponentially?) on the $\varepsilon$), where
$$G(z)=F(1,z)/\Gamma(z+1)$$
and
$$F(s,z)=\prod_p\left(1-\frac{z}{p^s}\right)^{-1}\left(1-\frac{1}{p^s}\right)^z$$
though I'm not quite sure how to calculate $F$.
If this is the best result known (rather than simply the best result provable at textbook level) then this shows that far less is known about the distribution of, e.g., semiprimes than about primes.
References
[1] M. Cipolla, “La determinazione assintotica dell n$^\mathrm{imo}$ numero primo”, Matematiche Napoli 3 (1902), pp. 132-166.
[2] Pierre Dusart, "Estimates of Some Functions Over Primes without R.H." (2010) http://arxiv.org/abs/1002.0442
[3] Pierre Dusart, "Autour de la fonction qui compte le nombre de nombres premiers" (1998) http://www.unilim.fr/laco/theses/1998/T1998_01.html
[4] Helge von Koch, "Sur la distribution des nombres premiers". Acta Mathematica 24:1 (1901), pp. 159-182.
[5] Hugh Montgomery & Robert Vaughan, Multiplicative Number Theory I. Classical Theory. (2007). Cambridge University Press.
[6] Lowell Schoenfeld, "Sharper Bounds for the Chebyshev Functions θ(x) and ψ(x). II". Mathematics of Computation 30:134 (1976), pp. 337-360.
[7] Robert G. Wilson v, Number of semiprimes <= 2^n. In the On-Line Encyclopedia of Integer Sequences, A125527. http://oeis.org/A125527 ; c.f. http://oeis.org/A007053

EDIT, by Joël. I edit this old question to bump it up and observe that one aspect has not been answered. Namely, is there under the Riemann Hypothesis an asymptotic estimate for $\pi_k(x)$ analog to (3), (4) for $\pi(x)$ (that is $\pi(x) = Li(x) + O(\sqrt{x} \log x)$)? Or any estimate for $\pi_k(x)$, with a principal term given by some classical functions, and an error term in $O(x^\delta)$ with some $\delta<1$? Micah's answer gives a principal term which is a rational function of $x$, $\log x$, $\log \log x$, but with a much less good error term, which is not surprising since even for $\pi(x)$ it is well-known that the principal term must be written as $Li(x)$, not $x/\log(x)$, if we want to have some hope of and rarer term in $O(x^\delta)$, $\delta<1$ (let alone $O(\sqrt{x}\log x)$).

REPLY [2 votes]: As not necessarily proven results were asked for, I have found the following quite accurate:
$$N_k(x):=\ \mid\{n\leq x : \Omega(n)=k\}\mid \ \sim \Re\bigg(\frac{2^{1-k}\alpha e^{1+e}x\log(1+e+\log(2^{1-k}\alpha x))^{\beta}}{\beta!(1+e+\log(2^{1+e}\alpha x)}\bigg)
$$
for $1 \leq k\leq \lfloor \log_2 (x) \rfloor$, where $\log_2$ is $\log$ base $2$, $\gamma $ is Euler's constant, 
$\beta=1+e+ \log \alpha +(1+e+ \log \alpha) ^{1/\pi}$,  and$$
\alpha=\frac{1}{2}\ \rm{erfc}\bigg(-\frac{k-(2e^{\gamma}+\frac{1}{4})}{(2e^{\gamma}+\frac{1}{4})\sqrt{2}\ }\bigg)-2\rm{T}\bigg(\bigg(\frac{k}{(2e^{\gamma}+\frac{1}{4})}-1\bigg),e^{\gamma}-\frac{1}{4}\bigg)\\
$$
where $\rm{erfc}$ is the complementary error function and $\rm{T}$ is the Owen T-function. 
In integral form, 
$$\alpha=
\frac{1}{\pi}\int_{(-3+8e^\gamma)/(\sqrt{2}(1+8e^\gamma))}^\infty e^{-t^2}\rm{d} t +\int_0^{1/4\ -\ e^\gamma}\frac{e^{-(3\ -\ 8e^\gamma)^2(1+t^2)/(2(1+8e^\gamma)^2)}}{1+t^2}\rm{d} t.$$
As $k\rightarrow \infty$, $\alpha\rightarrow 1$, so
$$\lim_{k \rightarrow \infty}N_{k}(x \cdot 2^{k-1})\sim\frac { {e^{e+1}} x\log\log( {e^{e+1}} x)^{\beta}}{\log( {e^{e+1}} x)\beta!},
$$
where $\beta=\log(e^{e+1})+\log(e^{e+1})^{1/\pi}.$
For $k\leqslant 3$, improvements to the above can certainly be made, but as $k\rightarrow \infty$ (or more correctly, as $k\rightarrow \lfloor \log_2 (x) \rfloor$), the formulae above, as far as have been tested, seem to be fairly accurate.
For convenience, I include the following Mathematica code:
cdf[k_, x_] :=
Re[N[
(2^-k E^(1 + E) x Log[1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])]]^(1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] +4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/\[Pi])) (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma]))/((1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])] + (1 + E + Log[1/2 (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2]
(1 + 8 E^EulerGamma))] + 4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma),
1/4 - E^EulerGamma])])^(1/\[Pi]))!
(1 + E + Log[2^-k x (Erfc[(1 + 8 E^EulerGamma - 4 k)/(Sqrt[2] (1 + 8 E^EulerGamma))] +
4 OwenT[(1 + 8 E^EulerGamma - 4 k)/(1 + 8 E^EulerGamma), 1/4 - E^EulerGamma])]))]];

landau[k_, x_] := N[(x Log[Log[x]]^(-1 + k))/((-1 + k)! Log[x])];

actual[k_, x_] := N[Sum[1, ##] & @@ Transpose[{#, Prepend[Most[#], 1], PrimePi@
Prepend[ Prime[First[#]]^(1 - k) Rest@FoldList[Times, x, Prime@First[#]/Prime@Most[#]],
x^(1/k)]}] &@Table[Unique[], {k}]];

I warmly welcome any criticism or comments on the above, and apologise in advance if I have made any serious errors.
Note: Table code included as requested:
a = 7;
x = 10^a;
kk = 20;
TableForm[Transpose[{Table[x, {x, 1, kk}], Table[Round[landau[k, x]], {k, 1, kk}], 
Table[Round[cdf[k, x]], {k, 1, kk}], Table[actual[k, x], {k, 1, kk}]}], 
TableHeadings -> {None, {"k  ", "Landau", "CDF   ", "Actual"}}, 
TableSpacing -> {2, 3, 0}]<|endoftext|>
TITLE: How do you calculate the group scheme of E[p] for a an elliptic curve E in characteristic p?
QUESTION [15 upvotes]: I know that the answer is $\mu_p \times \mathbb{Z}/p\mathbb{Z}$ if $E$ is ordinary, and $\alpha_p$ if $E$ is supersingular, where $\mu_p$ and $\alpha_p$ are the kernels of Frobenius  on $\mathbb{G}_m$ and $\mathbb{G}_a$ respectively. But why is it this true? 
Suppose $E'$ is a lift of $E$ to characteristic 0. Then $E'[p] = (\mathbb{Z}/p\mathbb{Z})^2$. If $E$ is ordinary then we have $E[p] =\mathbb{Z}/p\mathbb{Z}$, and one way to reconcile these two facts is to have $E[p] =  \mu_p \times \mathbb{Z}/p\mathbb{Z}$, since the group of closed points of $\mu_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ in characteristic 0, whereas in characteristic p, it is just a single (nonreduced) point. I'm not sure why this is the only possibility though--I think it has to do with the height of the formal group, but I just can't nut out the details. In Katz-Mazur "Arithmeticic moduli of elliptic curves" (proof of theorem 2.9.3) they say that "any p-divisible group over an algebraically closed field is the product of a p-divisible commutative formal Lie group and  finite number of copies of $\mathbb{Q}_p/\mathbb{Z}_p$," but I don't see why this is true. 
For the supersingular case, I'm even hazier. Is $\alpha_p$ the unique one-paramater formal group of height 2, and if so, how can you see this? For an affine scheme Spec(R), we have $\alpha_p(R) = \mathrm{Spec}(R[x]/(x^p))$. Is it true that in characteristic 0 we have (for example) $\alpha_p(\mathbb{C}_p) = (\mathbb{Z}/p\mathbb{Z})^2$?
Sorry if this is a mess, i'm really confused, and I haven't been able to find a sufficiently dumbed down explanation of this stuff anywhere.

REPLY [31 votes]: Dear Maxmoo,
Just to offer a slightly different perspective than that given by Kevin and Brian:
While their advice is certainly correct, when I was learning this I also found it very
helpful to make a couple of "bare hands" computations, as a kind of reality check.
For this, begin with an elliptic curve in char. $2$, in fact with two, of the form:
$$y^2 + y = x^3$$
and
$$y^2 + x y = x^3 + x $$
One of these is supersingular, the other ordinary.  (I won't tell you which here!)
Now try computing the $2$-torsion concretely, using lines passing through three points
and so on.
Remember that in the end you are looking for a degree $4$ equation (you may need to change
variables to see the point at infinity; this won't show up in the affine equations I've
given you).  By general theory, you know this equation won't be separable: non-reduced
group scheme structure will show up concretely as inseparability in this polynomial.
In one case (the s.s. case) it will be entirely inseparable; in the other (ordinary) case
it will have inseparability degree $2$ (so "half" inseparable, "half" separable).
Once you've done the case of char. $2$, you might want to try char. $3$ as well (since
computing the equation for the 3-torsion is also just about in reach by hand).
The reason I suggest this is that I remember, when I was learning this stuff, that all
these group schemes (especially the non-reduced ones) seemed fairly ephemeral, but after
I had made these kind of explicit computations, I had a much more concrete mental model
for what the general theory was talking about, which gave me a lot more confidence in
reading and making arguments about these kinds of things.
Best wishes,
Matt<|endoftext|>
TITLE: How expensive is knowledge? Knots, Links, 3 and 4-manifold algorithms. 
QUESTION [24 upvotes]: With geometrization, Rubinstein's 3-sphere recognition algorithm and the Manning algorithm, 3-manifold theory has reached a certain maturity where many questions are "readily" answerable about 3-manifolds, knots and links.  I'd like to have a community wiki post where we collectively build up a knowledge of just how much is algorithmically known, and how "expensive" it is to know something, in the sense of run-time upper bounds, and memory-usage upper bounds (if available).  At worst I'd like to know "there's no known algorithm". 
I hope to shape the discussion in this way: This initial post will list the "things we want to know", and the responses will pick off one of these topics and provide details.  Whatever details you know, with references and examples of implementations (if available).  Preference for algorithms you imagine are genuinely implementable -- ones with reasonable run-time estimates, reasonable memory-consumption, and assume reasonable start-up datum, something that you could feed to a computer without too much pain.   As topics from the top post are answered, they will be erased from the top post to keep clutter low. 
So if there are things you want to know -- for example, is there an estimate on how long it would take to compute (?triangulations?) of all finite-volume complete hyperbolic 3-manifolds which have the n-th largest volume (among the volumes of all finite-volume manifolds), where n is the input ordinal? -- add your question to the top post. 
Of course, upper bounds on run-times is all I'm looking for. If you know the complexity class on the nose, great.  The bias is more towards actual run-times of algorithms you'd consider using. 
Let me get the ball rolling.  I know answers to a few of these, and I'll try to get around to a few of them in the coming days. 
 knots and links 

Given a planar diagram for a knot, how expensive is it to compute: the Alexander polynomial, the Jones polynomial, or the HOMFLYPT polynomial? To what extent do these benefit from a quantum computer? 
Given a planar diagram for a knot, how expensive is it to compute a presentation matrix for the Alexander module?  How about the Tristram-Levine or Milnor signatures? 
What are the best run-times for unknot recognition, such as (any modified version of) the Haken algorithm?  Dynnikov's work on unknot recognition would go here, as well.
Are there run-time estimates on how long it would take to determine if a knot is slice? ribbon?

 3-manifolds 
These questions all assume triangulations as input.

How expensive is 3-sphere recognition? The connect-sum decomposition? (Jaco, Rubinstein, Burton, etc)
How expensive is the compression-body decomposition, and the JSJ-decomposition? 
How expensive is hyperbolisation (for a triangulated, hyperbolisable 3-manifold) i.e. the closed+cusped Manning algorithm. (Manning, ?Tillman?, others?)
How expensive is geometrization? (?)
How expensive is it to compute the Alexander ideals of a triangulated 3-manifold?  
How expensive is it to produce a surgery presentation of a 3-manifold from a triangulation?  (D.Thurston and Costantino's work is the closest related to this that I know)
Given an ordinal $n$ representing the volume of a hyperbolic $3$-manifold of finite volume, I want to know the actual volume (as a real number).  How difficult is that to know? How about reconstructing the 3-manifold as well? 
Given a triangulated cusped hyperbolisable $3$-manifold, is there an efficient algorithm to construct the Epstein-Penner decomposition? 

 4-manifolds 

Given a triangulated rational homology 3-sphere, how expensive is it to compute the generalized Rochlin invariant?   (or the Rochlin invariant for a homology 3-sphere)
Same question, but given a surgery presentation for the rational homology 3-sphere. In this case there is the Kaplan algorithm. 
What computable invariants of Farber-Levine pairings are there, and how hard are they to compute from a surgery presentation of a triangulation of a 4-manifold? 
Is the Oszvath-Szabo ''d''-invariant of $spin^c$ rational homology spheres algorithmically computable now, given a surgery presentation?  How are run-times?

REPLY [4 votes]: The Alexander polynomial is calculatable in polynomial time in the number of crossings, as is implemented in the Knot Atlas. So is the Burau representation of links. 
I do not know whether the Gassner representation, or the multivariable Alexander polynomial, are calculable in polynomial time. The trick which allows for Gaussian elimination (and therefore polynomial-time calculation of determinants) over $\mathbb{Z}[t^{\pm 1}]$ does not work over $\mathbb{Z}[t_1^{\pm1},\ldots,t_\mu^{\pm 1}]$.

REPLY [3 votes]: Saul Schleimer proved that sphere recognition lies in NP.  See his paper here (n.b. : the version at the arXiv is out of date).<|endoftext|>
TITLE: Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice.
QUESTION [15 upvotes]: It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the group generated by the basis or the vector subspace generated by some proper uncountable set of the basis). 
However, the first step (constructing the basis) requires the axiom of choice.
So does anyone know of any proper uncountable subgroup of $\mathbb{R}$ that does not require choice to construct?
or is this not possible.
Meaning are there models not involving choice where every uncountable subgroup of $\mathbb{R}$ is equal to $\mathbb{R}$.

REPLY [6 votes]: For any subset $S$ of the positive integers, let $\alpha_S =\sum_{i\in S} 10^{-i!}$. Then it's not hard to show (if I haven't made a mistake) that the subgroup of $\mathbb{R}$ generated by the $\alpha_S$'s is uncountable and proper.<|endoftext|>
TITLE: Crossed module structure on homotopy groups
QUESTION [7 upvotes]: A crossed module is a pair of groups $C$ and $G$, an action of $G$ on $C$, and a homomorphism $\partial: C \to G$ that satisfy 

$\partial(g\cdot c)=g(\partial c)g^{-1}$, and 
$cc'c^{-1}=(\partial c)\cdot c'$

Let $(X,A)$ be a pointed pair of spaces. Whitehead proved that, in the homotopy long exact sequence of the pair, 
$$
\pi_2(X,A) \stackrel{\partial}{\rightarrow} \pi_1(A)
$$
is a crossed module.
Simply put, my question is: what does this give us, other than an extra bit of structure? Does knowing that this is true aid on calculation? Does it aid in distinguishing spaces? Does it give us something really cool that I haven't thought of? (Probably the answer to this one is "yes".)

REPLY [6 votes]: I can't resist referring to the paper 
R. Brown and P.J. Higgins, ``On the connection between the second relative homotopy groups of some related spaces'', {\em Proc. London Math.  Soc.} (3) 36 (1978) 193-212, 
which gives some explicit calculations with pushouts of crossed modules, but which is little referred to, and further examples are in 
(with C.D.WENSLEY), `Computation and homotopical applications of induced crossed modules', J. Symbolic Computation 35 (2003) 59-72. 
The key to the proof of the theorem is the notion of homotopy double groupoid of a based pair. 
The paper arXiv:0909.3387v2 gives applications of higher van Kampen theorems to homotopy groups of spheres. 
January 22, 2016  Tom refers to "an extra bit of structure". This structure is crucial for applications of  a colimit theorem (see the above 1978 paper, and the 2011 book partially titled 2011 book  Nonabelian Algebraic Topology, NAT,  pdf available);  altering the structure alters the colimits. Part I of this book gives lots of explicit calculations, for example showing how for $A$ connected, and $f:A \to X$ a pointed map, the crossed module $$\pi_2(X \cup _f CA,X,x) \to \pi_1(X,x)$$is determined by the morphism $f_*:\pi_1(A) \to \pi_1(X)$. This is a  generalisation of a theorem of JHC Whitehead in "Combinatorial Homotopy II" on free crossed modules, which is the case  $A$ is a wedge of circles. 
There is a general assumption that we most want to calculate the second homotopy group $\pi_2$; but this, even as a module over $\pi_1$ is but a pale shadow of the 2-type. 
One of the problems in homotopy theory is that identifications in low dimensions have influence on high dimensional invariants. So part of the aim is to study this influence using algebraic structures which have information in a range of dimensions, and to develop and apply nonabelian colimit theorems in higher homotopy. 
Tim Porter refers to the nonabelian tensor square, which generalises to tensor products. I have kept up a bibliography  on this topic; the idea arose from considering pushouts of crossed squares, and the bibliography currently has 144 items, dating back to 1952. 
January 23,2016 I thought of another point which maybe answers the question better, namely as to the import of these crossed module rules on this boundary map for second relative homotopy groups. The situation becomes clearer if you work with the homotopy double groupoid of a pointed pair of spaces, see the NAT book and also the presentation at Galway on my preprint page. In terms of double groupoids, the first rule is a boundary rule for a certain subdivided square, and the second rule is equivalent to the interchange law for the two compositions. So if you draw the right pictures, then these rules become necessities. 
The emphasis on homotopy groups is part of the "squashing" of 2-dimensional situation into a single dimension, on a line. Then some natural rules become obscure, and it becomes impossible to do the 2-dimensional compositions which are necessary for the proof of the Seifert-van Kampen Theorem for homotopy double groupoids and so for the homotopy  crossed module of the question.<|endoftext|>
TITLE: Model theory stressing order type of universe.
QUESTION [7 upvotes]: In Appendix B to their Model Theory, Chang and Keisler list some problems and conjectures that, at the time of publication, were unsolved. A few of them take imperative form, for instance:
"Develop a theory of models which stresses the order type of the model $\mathfrak{A} = \langle A, <, \dots\rangle$ rather than the cardinality of the set $A$."
Has anyone worked on this?

REPLY [7 votes]: I don't think many model theorists have worked on this.  Granted, I'm a little unclear what Chang and Keisler were asking here, but here's one possible precisification:
Question: Suppose we are given a (complete?) theory T in a language with a binary relation < such that T proves "< is a strict linear ordering."  Try to develop a theory of the "order-type spectrum" $I(\alpha, T)$, which is defined as the number of nonisomorphic models $M$ of $T$ such that $(M, <^M)$ has order type $\alpha$.
You could start by trying to think about what it means for $T$ to be "$\alpha$-categorical" for some order type $\alpha$, meaning, what are necessary and sufficient conditions for $I(\alpha, T)$ to equal $1$?  (I have no idea whether anybody has investigated this before.)  For example, if $T$ proves that the ordering is dense without endpoints and $\eta$ is the order type of the rationals, then $I(\eta, T) = 1$ if and only if $T$ is $\omega$-categorical, since the complete theory of $(\mathbb{Q}, <)$ itself is $\omega$-categorical.
An immediate complication I see to this project is that I don't know if there is any good analogue of the upward Lowenheim-Skolem theorem.  It seems like it would be difficult to answer the question: "Given a theory $T$, for which infinite order types $\alpha$ is $I(\alpha, T) \neq 0$?"  (The corresponding question for cardinalities of the universe for a $T$ with infinite models is trivial, by Lowenheim-Skolem.)  For example: your theory $T$ could force the order type of any model to not be Dedekind complete (e.g. take the complete theory of an densely-ordered ring with a unary predicate for a proper convex subring).
Are there Morley-like categoricity theorems?  If $T$ is countable, say, and $I(\alpha, T) = 1$ for some uncountable order type $\alpha$, can we conclude that $I(\beta, T) \leq 1$ for every uncountable $\beta$?  Possibly this could be an interesting question; offhand I have no idea what the answer is.
Illustrating the difficulties of this, here's a paper just on the possible order types of a particular theory, PA:
"Order-types of models of Peano arithmetic," by Andrey Bovykin and Richard Kaye, pp. 275-285 of Logic and Algebra, edited by Yi Zhang with a preface by Oleg Belegradek, Contemporary Mathematics 302, AMS.
A different interpretation of the original question would be: given a particular order type $\alpha$, investigate structures with order type $\alpha$.  Along these lines, many model theorists (such as Pillay, van den Dries, Wilkie, and others) have been studying expansions of the ordered field of the real numbers under the rubric of "o-minimal theories," though generally the interest has been in definable sets rather than models per se.  Chris Miller is an example of an o-minimalist who has done a significant amount of research just on structures expanding the field of reals; check out his webpage for some state-of-the-art papers in this area.<|endoftext|>
TITLE: Deformations of hypersurfaces
QUESTION [8 upvotes]: Suppose I have a smooth hypersurface $X$ in $\mathbb{P}^n$ which is invariant under a (say finite) group $G$ of projective transformations. What can be said about the action of $G$ on the deformation space $H^1(X,T_X)$? I could imagine that many examples (especially with $dim(X)=1$ or $dim(X)=2$) have been worked out, if this is of any interest at all.

REPLY [5 votes]: Suppose that the action of $G$ on the smooth hypersurface $X$ of degree $d$ in $\mathbb P^n$ comes from a linear action of $G$ on $k^{n+1}$, as in Francesco's example. Assume also that $n \geq 3$, and $d \geq 2$. From the conormal sequence
$$
0 \longrightarrow \mathrm T_X \longrightarrow \mathrm T_{\mathbb P^n}{\mid}_X \longrightarrow
\mathcal O_X(d) \longrightarrow 0
$$
we get an exact sequence
$$
\mathrm H^0(X, \mathrm T_{\mathbb P^n}{\mid}_X)\longrightarrow 
\mathrm H^0(X, \mathcal O_X(d)) \longrightarrow \mathrm H^1(X, \mathrm T_X)\ .
$$
One can show that the homomorphism $\mathrm H^0(X, \mathcal O_X(d)) \longrightarrow \mathrm H^1(X, \mathrm T_X)$ is surjective, except in the single case $n = 3$, $d = 4$, where there is a 1-dimensional cokernel. Let us exclude this particular case (which can also be treated).
For each $i \geq 0$, let $V_i$ be the space of forms of degree $i$ in $n+1$ variables; there is a natural action of $G$ on $V_i$. In representation theoretic terms, $V_i = \mathop{\rm Sym}^i(k^{n+1})^\vee$. Let $f \in V_d$ an equation for $X$ and $L$ the substspace generated by $f$. From the Euler sequence
$$
0 \longrightarrow \mathcal O_X \longrightarrow \mathcal O_X(1)^{n+1} \longrightarrow 
\mathrm T_{\mathbb P^n}{\mid}_X \longrightarrow 0
$$
we get a surjection $\mathrm H^0(X, \mathcal O_X(1))^{n+1} \to
\mathrm H^0(X, \mathrm T_{\mathbb P^n}{\mid}_X)$; thus $\mathrm H^1(X, T_X)$ can be interpreted as the cokernel of a map $\phi \colon \mathrm H^0(X, \mathcal O_X(1))^{n+1} \to \mathrm H^0(X, \mathcal O_X(d))$. We have $\mathrm H^0(X, \mathcal O_X(1)) = V_1$ and $\mathrm H^0(X, \mathcal O_X(d)) = V_d/L$; furthermore, by unwinding the definitions one can show that the map $\phi \colon V_1^{n+1} \to V_d/L$ sends $(\ell_0, \dots, \ell_n)$ into the class of $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$.
So, one can describe $\mathrm H^1(X, \mathrm T_X)$ as the quotient of $V_d$ modulo the subspace generated by $f$ and by the classes of the form $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$, where the $\ell_i$ are homogeneous of degree 1. If the characteristic of $k$ does not divide $d$, then from Euler's formula we see that $f$ is of the form $\ell_0f_{x_0} + \cdots + \ell_n f_{x_n}$, so we don't need to add it.
This gives a description of the action of $G$ on $H^1(X, \mathrm T_X)$, which allows to compute it, at least in simple cases (the calculations could become unwieldy, particularly in the non-abelian case).<|endoftext|>
TITLE: Measure 0 sets on the line with Hausdorff dimension 1
QUESTION [7 upvotes]: I use $\dim_H(E)$ to denote the Hausdorff dimension of a set $E \subseteq \mathbb{R}$ and $|E|$ to denote its Lebesgue measure. It is easy to see from the definition of Hausdorff dimension that if $\dim_H(E) < 1$, then $|E| = 0$. The converse is not true, and there are many cases where $\dim_H(E) = 1$ yet $|E| = 0$. So the question:
What was the first (or most elementary) example of this phenomenon?
After some looking around, I was able to prove that a central Cantor set $C$ with ratio of dissection $r_k = 1/(2+\frac{1}{k})$ satisfies the condition I want. It is easy to see that $|C| = 0$ since at step $n$ of the process to construct this Cantor set, it has measure $2^n(r_1 \cdots r_n)$ which in this case limits to 0, but for the Hausdorff dimension I required a non-trivial result from the paper Sums of Cantor sets (Cabrelli, Hare, Molter) that gave the formula 
$\dim_H(C) = \liminf_n \frac{n \ln 2}{\ln r_1 \cdots r_n}$.
This result is fairly recent and sophisticated, and I feel that there should be older and simpler examples.

REPLY [3 votes]: It is a very common phenomenon in ergodic theory when the set of points which do not satisfy the Birkhoff ergodic theorem (i.e., a set of zero measure) has full Hausdorff dimension.
See, for instance, http://www.math.psu.edu/pesin/papers_www/birk.pdf<|endoftext|>
TITLE: Why were matrix determinants once such a big deal?
QUESTION [107 upvotes]: I have been told that the study of matrix determinants once comprised the bulk of linear algebra. Today, few textbooks spend more than a few pages to define it and use it to compute a matrix inverse. I am curious about why determinants were such a big deal. So here's my question:
a) What are examples of cool tricks you can use matrix determinants for? (Cramer's rule comes to mind, but I can't come up with much more.) What kind of amazing properties do matrix determinants have that made them such popular objects of study?
b) Why did the use of matrix determinants fall out of favor? Some back history would be very welcome.
Update: From the responses below, it seems appropriate to turn this question into a community wiki. I think it would be useful to generalize the original series of questions with:
c) What significance do matrix determinants have for other branches of mathematics? (For example, the geometric significance of the determinant as the signed volume of a parallelepiped.) What developments in mathematics have been inspired by/aided by the theory of matrix determinants?
d) For computational and theoretical applications that matrix determinants are no longer widely used for today, what have supplanted them?

REPLY [6 votes]: The area of an arbitrary triangle is most easily computed from its vertex coordinates as a 3x3 determinant (no square roots, trig, etc.).  This generalizes to 3D (volume  of a tetrahedron) and higher dimensions.
The cross product of two vectors a,b (essential tool in mechanics) is easily understood and memorized once one observes that its coordinates are 2x2 minor determinants of a 2x3 array formed by a and b.  This too generalizes and provides a definition for cross product of n-1 vectors in n dimensions.
More generally, determinants combined with homogeneous coordinates are all one needs to derive elegant formulas for the basic operations of n-dimensional projective geometry, such as plane through 3 points, intersection of 3 planes.  They also provide a coordinate representation (Pluecker coordinates) for lines in 3-space, or more generally for k-dim subspaces of n-dim prijective space.
The sign of the determinant of a matrix tells whether the rows are a left- or right-handed frame, and whether the corresponding linear map preserves or reverses orientations.
Determinants are more efficient than Gaussian elimination to compute the inverse of a 2x2 or 3x3 matrix, perhaps even 4x4. Unlike standard G.E. these formulas do not require division and may require fewer bits per number in intermediate results.

REPLY [3 votes]: I quickly scanned the answers and didn't find the following:
Square matrices of order $n$ can be considered to be embedded in $\Bbb R^{n \times n}$. The determinant is a continuous function of the entries of the matrix, so the singular matrices are $\det^{-1}(0)$ and are therefore a closed set in $\Bbb R^{n \times n}$. Thus, there will be non-singular matrices arbitrarily close to a singular matrix in any convenient metric on $\Bbb R^{n \times n}$. Of course, they will have pretty ugly condition numbers.<|endoftext|>
TITLE: Applications of Banach-Tarski Paradox to Probability Theory?
QUESTION [10 upvotes]: I was just curious, since the B-T paradox is a measure theoretic result, if there are any consequences of this paradox in probability theory? Also, is there is a way of stating the B-T paradox in the language of probability theory?
I am ultimately interested in finding an application of the B-T paradox in physics which leads to an experimental prediction.

REPLY [7 votes]: I thought the whole point of having a $\sigma$-algebra for your probability space was to avoid non-measurable sets like the ones used in the proof of BT. Hence, it would seem that the BT paradox would be impossible to state in probability theory on account of the sets you need not being present in your algebra... but I might be mistaken, can someone else comment more?<|endoftext|>
TITLE: Ehrhart polynomial
QUESTION [6 upvotes]: What is the Ehrhart polynomial of the regular cross-polytope of dimension d? Are there published upper and lower estimates?

REPLY [11 votes]: If you mean the polytope with vertices $(0,\ldots,0,\pm1,0,\ldots,0)$
then it is easily seen to be
$$\sum_{k=0}^d 2^k{d\choose k}{x\choose k}.$$<|endoftext|>
TITLE: Understanding the etale space construction from a formal viewpoint
QUESTION [13 upvotes]: Suppose I have a topological space $X$. Let $\mathcal{O}(X)$ denote the poset of open subsets. There is a canonical functor $\mathcal{O}(X) \to Top/X$ which sends an open $U \in \mathcal{O}(X)$ to $U \hookrightarrow X$. By left-Kan extension, this produces an adjunction between $Set^{\mathcal{O}(X)^{op}}$ and $Top/X$. The left-adjoint, $L$, is precisely the etale space construction, whereas the right adjoint, $\Gamma$, is the "sheaf of sections" functor. $\Gamma \circ L$ is sheafification. By abstract nonsense, this adjunction restricts to an equivalence between, one one hand, the subcategory where the unit is an iso, and the other hand, the subcategory where the counit is an iso. The former is easily seen to be the category of sheaves over $X$. I know the later is the category of etale spaces over $X$, i.e. maps $Y \to X$ which are local homeomorphisms. This is traditionally proven usually an explicit description of the etale space of a sheaf, topologizing the germs of local sections etc. However, is there a way to see this at a higher level of abstraction, only appealing to the abstract definition of this induced adjunction and abstract properties of local homeomorphisms?

REPLY [5 votes]: Here is a sketch of why I think the condition that $Y$ is a local homeomorphism over $X$ should be sufficient for the counit to be a homeomorphism.  I haven't worked out the converse yet.
For a presheaf $F \in Set^{\mathcal{O}(X)^{op}}$, the formula for the left Kan extension should be $$L(F) = \mathrm{colim}_{y(U) \rightarrow F} U$$ where $y : \mathcal{O}(X) \rightarrow Set^{\mathcal{O}(X)^{op}}$ is the Yoneda embedding.  By Yoneda's lemma, the indexing category for the colimit is exactly the category of elements of $F$ which I will write as $\int F$.
Now, consider the case where $F = \Gamma_Y$ for some space $p: Y \rightarrow X$ and assume that $p$ is a local homeomorphism.  We have $\Gamma_Y(U) = \{\sigma : U \rightarrow Y | p \circ \sigma = \mathrm{id}_U \}$. So the objects in the category $\int \Gamma_Y$ are exactly the sections over the various open sets of $X$, and the morphisms are given by restriction of sections.  I'll write $d(\sigma)$ for the domain of a given section.
Our Kan extension formula becomes $$L(\Gamma_Y) = \mathrm{colim}_{\sigma \in \int \Gamma_Y} d(\sigma)$$ From here it's easy to see what the counit is: since our object is given by a colimit, it suffices to construct a map $d(\sigma) \rightarrow Y$ for each $\sigma \in \int \Gamma_Y$.  But clearly $\sigma$ itself qualifies.
Now choose an open covering $\{V_\alpha\}$ of $Y$ such that $p$ restricts to a homeomorphism on each $V_\alpha$.  We then have a collection $\{\sigma_\alpha : p(V_\alpha) \rightarrow V_\alpha\}$ of sections by choosing the inverse to each restriction.  My claim would be that this collection is cofinal (or final? I can never remember which) in the category $\int \Gamma_Y$ so that we can restrict our colimit to just this subcategory.  Notice that in this case, the components of the counit map above are homeomorphisms. 
Moreover, this subcategory should also be cofinal in $\mathcal{O}(Y)$ by associating $\sigma_\alpha$ with the open set $V_\alpha$.  Then the fact that the counit is a homeomorphism should be the statement that a topological space is the colimit of any of its open coverings.
Is this along the lines of what you were thinking?<|endoftext|>
TITLE: A question about the dispersion points of connected metric spaces
QUESTION [9 upvotes]: Let $C$ be an infinite, separable and connected metric space. If $C$ becomes totally disconnected when one of its points $p\in C$ is removed, does every closed ball of $C$ with
positive radius and center $p$ always contain an infinite connected subset?

REPLY [5 votes]: Here I construct an example which proves the answer is NO.
Take the KK fan:

Remove its dispersion point at the top. Now you have a Cantor set of lines of rationals/irrationals that cannot be separated horizontally. Stretch this into a "Cantor-like tube" and weave it closer and closer to a point $p$ in the plane while shrinking its diameter and making sure that every loop goes a distance of $7$ away from $p$.

Remove $p$ and you have a hereditarily disconnected space ($\simeq$ KK fan minus its vertex). If $A$ is nonempty and clopen in $X$ then $A$ must snake around the tube forever, so it limits to $p$ and thus $p\in A$. Therefore $X$ is connected.
The ball of radius $1$ around $p$ is hereditarily disconnected.
EDIT:  I am basically taking the space which consists of the curve below, and the origin $p=(0,0)$ (so $\{0\}\times (0,1]$ is not included). The difference is that instead of weaving a line, I am weaving this "Cantor tube" while shrinking its diameter. In the first case if I remove $p$ then I get something $\simeq [0,\infty)$, whereas in the second case I get something $\simeq$ my Cantor tube.<|endoftext|>
TITLE: Mod l local Galois representations (l different from p)
QUESTION [7 upvotes]: My question is referred to the statement and proof of Prop. 2.4 of Diamond's 
article "An extension of Wiles' Results", in Modular Forms and Fermat Last 
Theorem, page 479.
More precisely: fix $l$ and $p$ two distinct primes, with $l$ odd. Let $\sigma$ be an 
irreducible, continuous, degree 2 representation of the absolute Galois group 
$G_{p}$ of $Q_{p}$, with coefficients in $k$, an algebraic closure of the finite 
field with $l$ elements. 
Proposition 2.4 states that if the restriction of $\sigma$ to the inertia 
subgroup of $G_{p}$ is irreducible and $p$ is odd, then $\sigma$ is isomorphic to the 
representation induced from a character of the Galois group of a quadratic 
ramified extension $M$ of $Q_{p}$. 
The proof given works if the restriction of $\sigma$ to the wild inertia of $G_{p}$ is 
reducible (I think there's a typo in the first line of the proof). What if 
$\sigma$ is irreducible on wild inertia (and $p$ is always odd)? It seems to me that this case is not covered in the proof of the Proposition, but maybe I'm not seeing something obvious.. If such a representation exists, it cannot be induced from a quadratic extension $M$ as above, so how does it fit in the description given by the Proposition? Can one say something about such a representation (for example something about its projective image?).
Thanks

REPLY [8 votes]: The image of wild inertia is a finite $p$-group, and if $d$ is the degree of an irreducible representation of a $p$-group over an algebraically closed field of characteristic $\ne p$, then $d$ is a power of $p$. So for $p$ odd the image of wild inertia is always reducible.<|endoftext|>
TITLE: Modules and Square Zero Extensions
QUESTION [19 upvotes]: Let $R$ be a commutative ring, $RMod$ its category of modules and $CRing$ the category of commutative rings.  
There's an embedding $RMod \rightarrow CRing/R$ that sends an $R$-module $M$ to the ring $$R \oplus M$$ (the direct sum taken as modules) with multiplication $(r_0,m_0)(r_1,m_1) = (r_0 r_1, r_0 m_1 + r_1 m_0)$.  This functor restricts to an equivalence of categories between $RMod$ and $Ab(CRing/R)$, the category of abelian group objects in the slice category.
The projection $R \oplus M \rightarrow R$ makes this ring into a square-zero extension of $R$.  My understanding is that in algebraic geometry, one thinks of a square zero extension of a ring as a kind of infinitesimal extension of $Spec (R)$.  So the category of $R$-modules can be viewed geometrically as parameterizing a certain class of infinitesimal objects related to $R$.
On the other hand, of course, the category $RMod$ is equivalent to the category of quasicoherent sheaves on $Spec(R)$, which seems, to me at least, totally unrelated to my previous description.
So my question is: are these two views of the same category somehow related?  When I think of the sheaf associated to a module $M$, does it somehow contain information about the corresponding infinitesimal extension?  What about when I look at cohomology with coefficients in that sheaf?

REPLY [16 votes]: If $X$ is a scheme and $\mathcal M$ is a quasi-coherent sheaf on $X$ then we can
form a sheaf of rings $\mathcal A := \mathcal O_X \oplus \mathcal M$, on which multiplication
of sections is given just by the same formula as for $R \oplus M$.
The pair $(X,\mathcal A)$ is then a scheme which is an infinitesimal thickening of
$X$, and this is precisely how you pass from a quasi-coherent sheaf to the corresponding
thickening; it is just a sheafified version of the construction in your posting.
(Regarding cohomology, in your question you seemed most interested in the case when 
$X =$ Spec $R$ is affine, in which case quasi-coherent sheaves have vanishing higher cohomology, so I'm not sure there is much to say about this.)
Added in response to comment below:  To see how these come up geometrically,
consider for example a $k$-scheme $X$ embedded diagonally into $X \times X$.
(Here $k$ is a field, and everything is happening over Spec $k$.)
Let $\mathcal I_X$ be the ideal sheaf on $X \times X$ cutting out the diagonal,
and consider the square-zero thickening
$\mathcal O_{X\times X}/\mathcal I_X^2$ of $X$.
This sits in the short exact sequence 
$$0 \to \Omega^1_X = \mathcal I_X/\mathcal I_X^2
\to \mathcal O_{X\times X}/\mathcal I_X^2 \to \mathcal O_{X\times X}/I_X = \mathcal O_X
\to 0.$$  The projection $p_1:X\times X \to X$ gives a spliting of this short exact
sequence, and so we find that $\mathcal O_{X\times X}/\mathcal I_X^2 = \mathcal O_X \oplus
\Omega^1_X$.
Recapitulating, we see that in the special case $\mathcal M = \Omega^1_X$, then
$(X, \mathcal O_X \oplus \Omega^1_X)$ is equal to the first order infinitesimal neighbourhood of $X$ in $X\times X$.
Suppose for example that $X$ is a smooth curve, so that $\Omega^1_X$ is a line-bundle.
Then $(X,\mathcal O_X \oplus \Omega^1_X)$ is locally like the dual numbers (as you observe
in your comment) but is globally twisted (unless $X$ is an elliptic curve, i.e. the genus is 1, which is the one case when $\Omega^1_X$ is actually trivial).
This should give you some sense of how these kinds of objects arise geometrically (and 
why one would consider other examples rather than just the dual numbers).<|endoftext|>
TITLE: Embeddings and triangulations of real analytic varieties
QUESTION [8 upvotes]: This is a follow up question to my answer here How do you define the Euler Characteristic of a scheme?
A real analytic space is a ringed space locally isomorphic to $(X,O/I)$ where $X$ is the zero locus of some number of real analytic functions $f_1,\ldots, f_k$ on an open set $U$ of $\mathbf{R}^n$, $O$ is the sheaf of germs of real analytic functions on $U$ and $I$ is the ideal sheaf generated by $f_1,\ldots, f_k$ (see e.g. http://eom.springer.de/a/a012430.htm) I would like to ask if it is true that each real analytic space with a countable base can be embedded as a closed analytic subset of some Euclidean space.
The motivation behind this comes from the triangulation theorem for complex algebraic varieties: the only proof of that that I know of (Hironaka's 1974 notes) is based on triangulating analytic subvarieties of Euclidean spaces. So to apply this one must embed a complex algebraic variety as a real subvariety of a Euclidean space. This is easy for projective varieties and is probably possible in general, but I don't know a reference for the general case. (I'm mainly interested in the complex algebraic case, but I don't see why it should be any easier that embedding arbitrary real analytic spaces; however if it is easier, I'd be interested to know.)
A related question: is it possible to prove the triangulation theorem (for complex algebraic varieties or in general) without using embeddings in Euclidean spaces?

REPLY [5 votes]: If you just want a proper 1-1 real analytic map whose image is a real analytic variety then the result is theorem 2 page 593 of a paper of Tognoli and Tomassini in Ann.Scuola.Norm.Pisa
(3) vol 21 yr 1967 pages 575-598. This means there is no control over the differential of the
map.I am assuming that the real analytic space has finite dimension.If the dimension of the
Zariski tangent spaces of a connected reduced real analytic space is bounded then it can be
real analytically embedded in euclidean space, see paper by Aquistapace Broglia and Tognoli
Ann.Scuola.Norm.Pisa (4) vol 6 yr 1979 p 415-426.<|endoftext|>
TITLE: Getting a differential equation for a function from a functional equation of its Mellin transform
QUESTION [16 upvotes]: If $f$ is a locally integrable function then its Mellin transform
$\mathcal{M}[f]$ is defined by
$$ \mathcal{M}[f] (s) = \int_0^{\infty} x^{s - 1} f (x) dx . $$
This integral usually converges in a strip $\alpha < Re \; s < \beta$ and
defines an analytic function. For our purposes we can assume that
$\mathcal{M}[f]$ converges in the right half-plane.
Let us denote $F (s) =\mathcal{M}[f] (s)$. Provided that the corresponding
Mellin transforms exist, the basic general theory tells us that, for instance,
$$
    \mathcal{M} \left[ \frac{d}{d x} f (x) \right] = - (s - 1) F
    (s - 1),\quad
    \mathcal{M} \left[ x^{\mu} f (x) \right] = F (s + \mu) .
$$
This allows us to translate a differential equation for $f (x)$ into a
functional equation for its Mellin transform $F (s)$.
Example:
  For instance, the function $f (x) = e^{- x}$ satisfies the differential
  equation
  $$ f' (x) + f (x) = 0 $$
  which translates to the functional equation
  $$ - (s - 1) F (s - 1) + F (s) = 0 $$
  for its Mellin transform. Of course, the Mellin transform of $e^{- x}$ is
  nothing but the gamma function $\Gamma (s)$ which is well-known for
  satisfying exactly this functional equation.
Now, let us assume that we are given a function $f (x)$ and its Mellin
transform $F (s)$. Further, suppose that we know that $F (s)$, just as the
gamma function, can be analytically extended to the whole complex plane with
poles at certain nonpositive integers. We also know that $F (s)$ satisfies a
functional equation which we would like to translate back into a differential
equation for $f (x)$. Formally, we obtain, say, a third order differential
equation with polynomial coefficients. Can we conclude that $f (x)$ solves
this DE?
The issue is that in our case the derivatives of $f (x)$ develop singularities
in the domain and are no longer integrable. So Mellin transforms can't be
defined in the usual way for them (and so we can't just use Mellin inversion).
What I am looking for is conditions under which we can still conclude that the
functional equation for $F (s)$ translates into a differential equation for $f
(x)$. Preferably, these should be conditions on $F (s)$ and not on $f (x)$. If
it helps, we can assume $f (x)$ to be compactly supported.
Any help or references are greatly appreciated!

REPLY [3 votes]: By defining the Mellin transform for distributions as for instance done in Transform Analysis of Generalized Functions by O. Misra, J. Lavoine it follows that the functional equation for $F(s)$ translates into a differential equation of which $f(x)$ is a weak solution.<|endoftext|>
TITLE: Geometric interpretation of exceptional Symmetric spaces
QUESTION [21 upvotes]: Elie Cartan has classified all compact symmetric spaces admitting a  compact simple Lie group as their group of motion.There are 7  infinite series and 12 exceptional cases.  The exceptional cases are related to real forms of exceptional Lie algebra. 
Most of these symmetric spaces admit at least one  geometric interpretation usually  in terms of complex and real Grassmannians and their  generalizations  to  quaternions and octonions($\mathbb{H}$), octonions($\mathbb{\mathbb{O}}$), bioctonions ($\mathbb{C}\otimes \mathbb{O}$), quateroctonions ($\mathbb{H}\otimes \mathbb{O}$) and octooctonions ($\mathbb{O}\otimes \mathbb{O}$). See for example the Wikipedia's entry for  Symmetric Spaces.
Two of the exceptional symmetric spaces, don't seem to have such a geometric interpretation as far as I know. 
In Cartan notation, these two spaces are called   $EI$ and $EV$ and  correspond, respectively, to the exceptional symmetric spaces $\frac{E_6}{\mathrm{Sp}(4)/ \mathbb{Z}_2}$ and $\frac{E_7}{\mathrm{SU}(8)/\mathbb{Z}_2 }$ of respective ranks and dimensions $(6,42)$ and $(7,70)$.
Now that the stage is set, here is my question:

What is the geometric description of
  the symmetric spaces
  $\frac{E_{7}}{SU(8)/ \mathbb{Z}_2}$ and
  $\frac{E_6}{Sp(4)/ \mathbb{Z}_2}$?

References on the subject are also welcome. 
This question is motivated by an answer to 
 this  MO  question. 
In order to give a more precise idea of the kind of answer I expect, let me give some examples:  the symmetric space $\frac{F_4}{\mathrm{Spin}(9)}$ is geometrically described as the Cayley projective plane $\mathbb{O}P^2$, the space $\frac{E_6}{\mathrm{SO}(10) \mathrm{SO}(2)}$ is geometrically the Caylay bioctonion plane $(\mathbb{C}\otimes \mathbb{O}) \mathbb{P}^2$  and the symmetric space 
$\frac{E_6}{F_4}$ is the space of isometrically equivalent collineations of the Cayley plane $\mathbb{O}\mathbb{P}^2$.  
NB:These two spaces also show up as scalar manifolds in maximal supergravity theories, this is for example review in   this article of Boya.  But for this question, I won't consider supergravity  as a geometric interpretation.
Updates
Richard Borcherds  has provided an answer thanks to a reference to the book "Einstein manifolds", but the book gives the answer without any proofs or explanation. So we now have an answer but we don't understand it. So if anyone could help with explaining how "antichains" enter the story, it will be highly appreciated. I have put some extra information in the comments.

REPLY [2 votes]: Eschenburg present $E_I$ as $\{G_2\mathbb H^4/\mathbb Z_2 \subset \mathbb C \otimes \mathbb OP^2\}$ and $E_V$ as $\{G_4\mathbb C^8/\mathbb Z_2 \subset \mathbb C \otimes \mathbb OP^2\}$.
In my notes from 2008 I see interpretation of $E_I=E_6/Sp_4$ as $\{\mathbb HP^2 \times\mathbb HP^2 \subset \mathbb C \otimes \mathbb OP^2\}$ and and $E_V=E_7/SU_8$ as $\{G_2\mathbb C^6 \times G_2\mathbb C^6 \subset \mathbb C \otimes \mathbb OP^2\}$. Note that $G_2\mathbb C^6=\mathbb C \otimes \mathbb HP^2$. 
Without seeing the proof we cannot tell which one is true. There are some equalities in grassmanians e.g. I believe that $G_{2,2}\mathbb C$=$G_{2,4}^+$. It is possible that for example $G_2\mathbb H^4$ is equal to $\mathbb HP^2 \times \mathbb HP^2$ or rather $S^8 \times \mathbb HP^2$ - see below. 
I remember following argument for $G_{2,2}^+$. Let $1$ be fixed vector in $\mathbb R^4$. Complex structure is defined by unit vector $\color{red}v$ perpendicular to $1$. Complex lines for selected complex structure are forming $S^2=\mathbb CP^1$. Each element of $G_{2,2}^+$ belongs to exactly one complex structure, so we obtain $S^2\times S^2$.
Analogous arguments for two-dimensional grassmanian in complex and quaternion four dimensional spaces would give $\mathbb HP^1\times \mathbb CP^2$ and $\mathbb OP^1\times \mathbb HP^2$.<|endoftext|>
TITLE: Minimal Hausdorff
QUESTION [11 upvotes]: A Hausdorff space $(X,\tau)$ is said to be minimal Hausdorff if for each topology $\tau' \subseteq \tau$ with $\tau' \neq \tau$ the space $(X,\tau')$ is not Hausdorff.
Every compact Hausdorff space is minimal Hausdorff.
I would like to know:
1) Is every minimal Hausdorff space compact?
2) Does every Hausdorff topology contain a minimal Hausdorff topology?
Many thanks!

REPLY [12 votes]: The answer to both questions is no - see 7.5 in Porter and Woods book, "Extensions and Absolutes of Hausdorff Spaces", Springer-Verlag, 1988.  The space of rational numbers with the usual topology has no coarser minimal Hausdorff topology.  Every Hausdorff space can be embedded in a minimal Hausdorff space; in particular, if you start with a Hausdorff space X that is not Tychonoff and embed it in a minimal Hausdorff space Y, then Y can not be compact Hausdorff.<|endoftext|>
TITLE: Fields of definition for p-adic overconvergent modular eigenforms
QUESTION [8 upvotes]: If we consider the action of the $U_p$ operator on overconvergent $p$-adic modular forms, then we can get some information about the field over which the eigenforms are defined by looking at the slopes. For instance, my paper in Math Research Letters (MR2106238) proves that the slopes of $U_2$ acting on 2-adic overconvergent modular forms of level 4 with primitive Dirichlet character are distinct, so the field of definition has to be $\mathbf{Q}_2$. However, there are cases when the slopes fail to be distinct; for instance, in Emerton's thesis it is proved that the lowest slopes of T_2 acting on level 1 forms of weight congruent to 14 modulo 16 are 6 and 6.
For classical modular forms of level 1, we have Maeda's Conjecture which says that the field of definition is essentially as large as it can be; the Hecke polynomial is irreducible with Galois group $S_n$ where $n$ is the dimension. However, there is no reason that this should be true for overconvergent modular forms, and in fact it isn't. Discussions with Robert Coleman led me to the concrete example of 2-adic overconvergent modular forms of tame level 1 and weight 142, where there are two eigenforms of slope 6 which are both defined over the ground field $\mathbf{Q}_2$.
The question is, what should one expect here? Can one tell any more about the field of definition from the slopes than the absolute minimum?

REPLY [2 votes]: Professor Buzzard raises the a question of whether every normalized eigenform of level $1$ is defined over a quadratic extension of $\mathbb{Q}_2$. 
(This is Question 4.3 of http://www2.imperial.ac.uk/~buzzard/maths/research/papers/conjs.pdf)
In contrast, the multiplicity of the valuation of the set of $2$-adic slopes at level $1$ can be arbitrarily large,
as can be observed as follows: 
Consider the space
$S_k:=S^{new}_k(\Gamma_0(2))$ of newforms of weight $k$. Every newform has slope $(k-2)/2$. Thus, by
work of Coleman, the number of slopes of valuation $(k-2)/2$ at level $1$ and weight $k + 2^n$ for sufficiently large $n$ will be  at least $\mathrm{dim}(S_k)$, which is unbounded as $k$ increases.
EDIT: The point of the last example is that the answer to the original question is "not much", i.e., there can be many forms of the same slope, but all the forms are defined over a small (or even trivial) extension of $\mathbb{Q}_2$.<|endoftext|>
TITLE: nonasymptotic complexity results
QUESTION [8 upvotes]: I recall hearing about a result, or maybe a cluster of results, in some area of complexity theory, probably algebraic, to the effect that there are known, specific, short formulas whose minimal derivation is known to be exceedingly long. Or perhaps it is a specific function that requires an exceedingly deep curcuit. The "philosophy" seemed to be that such examples would threaten to make the older asymptotic question obsolete: "Who cares about asymptotics if the constants are huge." Can anyone, given these hints, describe ( and direct me to) the result I overheard?

REPLY [5 votes]: The general fact surrounding some of the other answers is
the following:
Every computably axiomatizable consistent theory $T$
containing trivial arithmetic admits very short theorems
requiring extremely long proofs.
The basic fact is that there can be no total computable
bound on the length of the proof required. To see this,
suppose that there is a computable total function $f$ such
that whenever statement $\psi$ is a statement of size at
most $n$ provable from $T$, then there is a proof of size
at most $f(n)$. In this case, the question of whether $T$
proves $\psi$ will be decidable, since we can simply
inspect all proofs of length $f(n)$, where $n=|\psi|$ and
check if any of them are proofs of $\psi$. But it is
impossible that $T$ proves $\psi$ is decidable, since we
could then produce a consistent completion of $T$, by the
usual process of completing a theory, which is effective
for decidable theories. This would produce a computable
complete consistent theory of arithmetic, in contradiction
to the incompleteness theorem.
One can make the conclusion fairly concrete via the halting
problem. If $T$ is true and contains some trivial
arithmetic, then $T$ will prove all true instances of the
assertion program $p$ halts on input $m$, and prove no
false instances. If there were a total computable function
$f(p,m)$ such that whenever $p$ halted on input $m$, then
there was a proof of this of length at most $f(p,m)$, then
we could decide the halting problem: on input $(p,m)$,
compute $f(p,m)$ and then look at all proofs of that length
and determine if there is a proof of halting or not.
In conclusion:
Theorem. For any computably axiomatizable true theory $T$ and for any total computable
function $f$, there is a program $p$ and input $m$ such
that $T$ proves that $p$ halts on input $m$, but there is
no proof of this in fewer than $f(p,m)$ steps.
Since as we all know, the computable functions can grow
quite outrageously, this means that for any theory there
will be short theorems that require extremely long proofs.<|endoftext|>
TITLE: Rankin-Selberg convolutions of motivic L-series
QUESTION [6 upvotes]: Background:
Let $M_{f_i}, i=1,2$ be two modular motives associated to cusp forms
  $f_i \in S_{w_i}(\Gamma_0(N_i))$ of weight $w_i$ and level $N_i$ respectively.
  The Rankin-Selberg convolution associates an L-series
  $L(f_1\otimes f_2,s)$ to this pair of modular forms.
  In the framework of automorphic motives a natural question is 
  whether the L-series $L(M,s)$ of a motive $M$ can be represented in
  terms of modular submotives $M_{f_i}$ as $L(M,s) \stackrel{?}{=}
  L(f_1\otimes f_2,s)$.
Question:
Is there a (practical) test that can be applied to a given
(motivic) L-series as to whether it admits a Rankin-Selberg
product representation?

REPLY [4 votes]: If $L(s,M)$ is an irreducible degree 4 motivic L-function, and $L(s,\mathrm{sym}^2(M))$ has a pole, then either $M=f_1 \otimes f_2$ for a pair of distinct classical modular forms $f_1, f_2$, or $M=\mathrm{Asai}(f)$ for $f$ cuspidal on $GL_2(K)$, with $K/\mathbb{Q}$ quadratic. You can rule out the Asai case if $L(s,\mathrm{sym}^2(M)\otimes \chi)$ is entire for any nontrivial quadratic character (in particular, entire for $\chi$ the character of $K$).  If by "practical" you mean "local", then I think the answer is no.<|endoftext|>
TITLE: On the size of balls in Cayley graphs
QUESTION [26 upvotes]: Next semester I will be teaching an introductory course on geometric group theory and there is a basic question that I do not know the answer to. Let $G$ be a finitely generated group with finite symmetric generating set $S$ and let $\Gamma$ be the corresponding Cayley graph. For each $n \geq 1$, let $B_{n}$ be the closed ball of radius $n$ in $\Gamma$ about the unit element $e$ and let
 $b_{n} = |B_{n}|$. Then it is known that $\lim b_{n}^{1/n}$ always exists. (For example, see de la Harpe's book.) My question is whether $\lim b_{n+1}/b_{n}$ always exists?

REPLY [31 votes]: In the article 
R. Grigorchuk and P. De La Harpe, On problems related to growth, entropy, and spectrum in group theory,  Journal of Dynamical and Control Systems, Volume 3, Number 1, 51-89

on the lower part of page 58 the authors mention the manuscript
A. Machi, Growth functions and growth matrices for finitely generated groups. Unpublished manuscript, Univ. di Roma La Sapienza, 1986.
and explain an example due to Machi. Machi showed that the convergence of $b_{n+1}/b_n$ can fail for one generating set of ${\mathbb Z}_2 \star {\mathbb Z}_3$ and hold for another. In particular, the limit does not always exist. The two generating sets are $\lbrace s,t\rbrace$ and $\lbrace s,st\rbrace$, where $s$ and $t$ are the natural generators with $s^2=t^3=e$.<|endoftext|>
TITLE: Computational Question about finite local rings: 
QUESTION [13 upvotes]: Let $(A,\mathfrak{m})$ be a local Artinian ring with
finite residue field, which I'm happy to assume is $\mathbf{F}_3$.
(In particular, $A$ has finitely many elements.)
I would like to do some computations of the following kind, as $I$ ranges over
all of the ideals of $A$.
(0) A way to enumerate all the ideals of $A$.
(1) For an ideal $I$ of $A$, compute the length of $I/I^2$.
(2) For an ideal $I$ of $A$, compute the ideal $J = \mathrm{Ann}(I)$.
(3) For an ideal $I$ of $A$, decide if $I$ is principal. (By computing the length of
$I/\mathfrak{m} I$ or otherwise.)
The ring $A$ itself will be given explicitly as a quotient of a power series
ring over $W(\mathbf{F}_3) = \mathbf{Z}_3$. For example, $A$ might be
given as $\mathbf{Z}_3[[x]]/(27,9x,x^3)$ or $\mathbf{Z}_3[[x]]/(9,x^2)$.
My question: What is the computer algebra package that is best suited to carry
out these computations? (I would like something that can be semi-automated for various possible $A$.) I would be interested in even a very simple one like $\mathbf{Z}_3[[x]]/(9,x^2)$
EDIT 2: There seems to be a consensus in the comments that this problem is significantly more manageable if $A$ is actually an algebra over its residue field. For example, in MAGMA, it is only possible to create ideals and quotient rings in univariate polynomial rings over fields. Other computer algebra packages have similar issues when the coefficient ring is not a field, although SINGULAR (for example) has some functionality with polynomials in several variables. As it happens, the problem I was interested in studying is still of interest for such fields.

REPLY [5 votes]: This is fleshed out from comments of Sam Lichtenstein; Some (or all) of what I have written is surely not the most elegant programming (at best), feel free to improve. One can do the following with MACAULAY2 (if you copy and paste this into the MACAULAY2 prompt it should work:) 
R = GF(2)[x,y]/(x^3,y^3);
m = ideal(x,y);
ModSquare  = ideal ->  length(ideal/(ideal*ideal));
Generators = ideal ->  length(ideal/(ideal*m));
I := ideal(x^2 + y^2);
J := ann(I);
[Generators(I),Generators(J),ModSquare(I),ModSquare(J)] 
The function ModSquare applied to an ideal $I$ computes the length of $I/I^2$, and
the function Generators computes the minimal number of generators of $I$. Similarly, ann computes the annihilator of $I$. From this we may compute that $I$ is principal, $J$ has two generators, and both $I/I^2$ and $J/J^2$ have length $4$. These functions only work for homogenous ideals. Following James Parson's suggestion, I also considered MAGMA (again with the restriction to affine algebras), and the following works for arbitrary ideals:
A:=AffineAlgebra<GF(2),x,y|x^3,y^3>;  x:=A.1;  y:=A.2;  AssignNames(~A,["x","y"]);
m:=ideal<A|x,y>;
Minus        := func<ideal | Dimension(quo<A|ideal>)>;
Generators   := func<ideal | Minus(ideal*m) - Minus(ideal)>;
ModSquare    := func<ideal | Minus(ideal*ideal) - Minus(ideal)>;
ann          := func<ideal | Annihilator(ideal)>;
I:=ideal<A|x^2+y^2>;
J:=ann(I);
[Generators(I),Generators(J),ModSquare(I),ModSquare(J)];<|endoftext|>
TITLE: Isogenies between Tate curves
QUESTION [9 upvotes]: Let $q$ and $q'$ be complex numbers with $0<|q|,|q'|<1$, and let $m$ and $n$ be positive integers.
Suppose that $q^m={q'}^n$.  Then the map
$$
f:\mathbb{C}^\times/q^{\mathbb{Z}} \to \mathbb{C}^\times/{q'}^{\mathbb{Z}}\qquad \text{defined by}\qquad u\mapsto u^m
$$
gives an isogeny of (analytic) elliptic curves over $\mathbb{C}$.  
The Tate curve $\mathrm{Tate}(q)$ is an (algebraic) elliptic curve over the Laurent series ring $\mathbb{Z}((q))$ which can be used to give a uniformization of the curve $\mathbb{C}^\times/q^\mathbb{Z}$ by means of certain well known explicit formulae.  
My question is: 

Does there exist an isogeny $\mathrm{Tate}(q)\to \mathrm{Tate}(q')$ of elliptic curves defined over $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ which "lifts" the map $f$ above, and if so, how do you prove it exists?

It should suffice to construct such an isogeny for $(m,n)=(m,1)$, and use dual isogenies and composition to get the general case.
(I'm being vague about "lifts", because one has to worry about convergence somewhere.  Probably you want to say that the isogeny is defined over some subring of $\mathbb{Z}((q,q'))/(q^m-{q'}^n)$ of power series which are analytically convergent near $q=0$, or something like that.)
I presume (though I probably can't prove) that the existence of the analytic isogenies means that such a map of schemes is defined over $\mathbb{C}((q,q'))/(q^m-{q'}^n)$, so that this is just a question about integrality.
This is very closely related to exercise 5.10 in Silverman, Advanced Topics in the Arithmetic of Elliptic Curves.  There, he apparently asks us to show that for a $p$-adic field $K$, if $q,q'\in K$, $0<|q|,|q'|<1$, and  $q^m={q'}^n$, then the function $\overline{K}^\times/q^\mathbb{Z}\to \overline{K}^\times/{q'}^{\mathbb{Z}}$ defined by $u\mapsto u^m$ lifts to an isogeny $E_q\to E_{q'}$ of elliptic curves over $K$, where $E_q$ and $E_{q'}$ are defined by the Tate curve equations.  (An answer to my question solves his exercise, right?)
Unfortunately, I have no idea how to do Silverman's exercise either (he marks it as difficult).  Any hints?

REPLY [7 votes]: No matter how you define Tate(q), it should have the following properties:
(a) for any $n$ it contains a subgroup $M_n$ canonically isomorphic to $\mu_n$ (which corresponds tho $\mu_n\subset\mathbb{C}^\times$ in the complex model),
(b) the (co)tangent space along the unit section is canonically trivialized (by $d\log u$ in the complex model).
Let me first treat the case $n=1$, as Charles suggests. The sought-for isogeny Tate($q$)$\to$Tate($q^m$) is characterized by two conditions:
(a') its kernel is $M_m$ (i.e. it induces an isomorphism Tate($q$)$/M_m\to$Tate($q^m$)),
(b') it induces multiplication by $m$ on the tangent space, modulo the identification (b).
Consider the scheme $X\to S:=\mathrm{Spec}\:\mathbb{Z}((q))$ parametrizing isomorphisms Tate($q$)$/M_m\to$Tate($q^m$). This is an unramified $S$-scheme, and in fact it is finite because Tate($q$) has no complex multiplication in any fiber over $S$ (I guess this has to be checked). Since it has a section over $\mathrm{Spec}\:\mathbb{C}((q))$ it is dominant, hence surjective over $S$. Since $S$ is normal it suffices to find a section at the generic point. But by flat descent, condition (b') guarantees that the above section over $\mathrm{Spec}\:\mathbb{C}((q))$ descends to the fraction field of $\mathbb{Z}((q))$. QED.
Remark: I am  a bit uncomfortable about the "no CM" stuff, but we can probably avoid it by noting that $X\to S$ satisfies the valuative criterion of properness, even when it is not of finite type. This (together with unramifiedness) is enough to imply that a section at the generic point extends over a normal base.
For arbitrary $n$, observe that we have just constructed $\alpha_m:\text{Tate}(q)\to \text{Tate}(q^m)$ with kernel killed by $m$, so multiplication by $m$ factors as $\beta_m\circ\alpha_m$ for some $\beta_m:\text{Tate}(q^m)\to \text{Tate}(q)$. You can now treat the general case by taking the composition
$$\text{Tate}(q)\to \text{Tate}(q^m)=\text{Tate}(q'^n)\to \text{Tate}(q')$$
where the two maps are $\alpha_m$ and $\beta_n$ respectively.<|endoftext|>
TITLE: Using Fisher Information to bound KL divergence
QUESTION [18 upvotes]: Is it possible to use Fisher Information at p to get a useful upper bound on KL(q,p)?
KL(q,p) is known as Kullback-Liebler divergence and is defined for discrete distributions over k outcomes as follows:
$$KL(q,p)=\sum_i^k q_i \log \frac{q_i}{p_i}$$
The most obvious approach is to use the fact that 1/2 x' I x is the second order Taylor expansion of KL(p+x,p) where I is Fisher Information Matrix evaluated at p and try to use that as an upper bound (derivation of expansion from Kullback's book, pages 26, 27, 28).
If $p(x,t)$ gives probability of observation $x$ in a discrete distribution parameterized by parameter vector $t$, Fisher Information Matrix is defined as follows
$$I_{ij}(t)=\sum_x p(x,t) (\frac{\partial}{\partial t_i} \log p(x,t)) ( \frac{\partial}{\partial t_j} \log p(x,t)) $$
where sum is taken over all possible observations.
Below is a visualization of sets of $k=3$ multinomial distributions for some random $p$'s (marked as black dots) where this bound holds. From plots it seems that this bound works for sets of distributions that are "between" $p$ and the "furthermost" 0 entropy distribution.
 (source)
Motivation: Sanov's theorem bounds probability of some event in terms of KL-divergence of the most likely outcome...but KL-divergence is unwieldy and it would be nicer to have a simpler bound, especially if it can be easily expressed in terms of parameters of the distribution we are working with

REPLY [4 votes]: I'm not sure if this is still of interest to you, but I think it is possible to get some reasonable bounds if you are okay with dropping the factor of $\frac{1}{2}$. Here's my work, which can be strengthened and refined.
We start by taking two probability mass functions $p$ and $q$ which we denote as $p_i$ and $q_i$. We define the function $f$ as $f_i= q_i-p_i$. Instead of doing anything fancy, we consider the line segment $p_i(t) = p_i +tf_i$. Since $f$ has total mass zero, the $p_i(t)$ are well defined probability distributions that form a straight line in the probability simplex. We can see that $p_i(0) = p_i$ and $p_i(1) = q_i$
Now we take the Taylor series for the Kullback-Liebler divergence, expanded at $t=0$. This will involve the Fisher metric, but we should expand it out further to get better results.
When we expand out $(p_i + t f_i)\log\left( \frac{p_i + t f_i}{p_i} \right)$, we get the following:
$$f_i t+\frac{f_i^2 t^2}{2 p_i}-\frac{f_i^3 t^3}{6 p_i^2}+\frac{f_i^4 t^4}{12 p_i^3}-\frac{f_i^5 t^5}{20 p_i^4}+\frac{f_i^6 t^6}{30 p_i^5}+O\left(t^7\right)$$
When we sum over $i$, the first term will vanish, and we can factor out a Fisher metric term from all of the others. I will use an integral sign to sum over $i$, as it is suggestive of what should happen in the continuous case.
$$\int f_i t+\frac{f_i^2 t^2}{2 p_i}-\frac{f_i^3 t^3}{6 p_i^2}+\frac{f_i^4 t^4}{12 p_i^3} \ldots \,di = \int \frac{f_i^2 t^2}{ p_i} \left( \frac{1}{2} - \frac{f_i t}{6 p_i} + \frac{f_i^2 t^2}{12 p_i^2}  \ldots \right) di $$
We find that the terms in the parenthesis on the right hand side can be simplified. We set $x_i = \frac{f_i t}{p_i}$ and can derive the following:
$$\left( \frac{1}{2} - \frac{x_i}{6} + \frac{x_i}{12}  \ldots \right) = \sum_{k=0}^\infty \frac{(-1)^k x_i^k}{(k+1)(k+2)} = \frac{(x_i +1)\log(x_i+1)-x_i}{x_i^2}$$
This should not be surprising; it's very closely related to the original formula for the Kullbeck-Liebler divergence. In fact, we didn't need Taylor series except to know to subtract off the pesky $t f_i$ term. Therefore, we don't need to worry about the convergence, this manipulation is valid without the series. Therefore,
$$KL(p(t), p) = \int \frac{f_i^2 t^2}{ p_i} \left(  \frac{( x_i +1)\log(x_i+1)-x_i}{x_i^2} \right) di $$
In order for this to make sense, we need to make sure that $x_i=  \frac{f_i t}{p_i}  \geq -1$. However, $\frac{f_i}{p_i} = \frac{q_i}{p_i} - 1  \geq -1$. Even better, it turns out that $ \frac{( x_i +1)\log(x_i+1)-x_i}{x_i^2}  \leq 1$ on its domain. With this, we are done, because this implies
$$KL(q,p)< I_p(f,f).$$
This implies that we can bound the Kullback-Liebler divergence by the Fisher information metric evaluated on a particular vector $f$. Since the KL-divergence can blow up, it's worthwhile to see what happens in this case. Whenever this happens, the tangent vector $f$ at $p$ is large in the Fisher metric.<|endoftext|>
TITLE: to test equivalence of representations under automorphism
QUESTION [6 upvotes]: Hi all,
I would like to propose the following problem:
Given two representations $\rho$ and $\tau$ of a group $G$ over complex number, we would like to know if there exists an automorphism $\phi$, such that $\rho\circ\phi$ and $\tau$ are equivalent. 
Is there any mathematical results concerning this problem? It seems that to understand the action of automorphisms on the set of irreducible representations is crucial.
Thank you!
Youming

REPLY [3 votes]: Victor's answer shows that it is important to understand the action of $Out(G)$ on the conjugacy classes of $G$. This can be interesting even in the abelian case, where the problem amounts to calculating the orbits of $Aut(G)$ on $G$. Even though the answer was well-known, we found a combinatorial approach to the question quite interesting.
The idea (which can be formulated for any group) is to define a pre-order on elements: $x\leq y$ if an endomorphism maps $y$ to $x$. The equivalence classes associated to this pre-order ($x\sim y$ if $x\leq y$ and $y\leq x$) are unions of outer equivalence classes. In the finite abelian case, these classes are precisely the $Aut(G)$-orbits.
The question of what other classes of groups this method can be applied to remains open.<|endoftext|>
TITLE: Is there any criteria for whether the automorphism group of G is homomorphic to G itself?
QUESTION [14 upvotes]: In the elementary group theory we know that for the symmetric groups $S_n$, except $S_6$, we have $Aut(S_n) \cong S_n$. Then the following question is natural:
What is the necessary and sufficient condition for $G$ such that $Aut(G) \cong G$?

REPLY [10 votes]: There does not exist a reasonable necessary and sufficient condition for an infinite centerless group to be complete. More precisely, letting $V$ be the set-theoretic universe, there exists an infinite complete group $G \in V$ and a $c.c.c$ notion of forcing $\mathbb{P}$ such that $G$ has an outer automorphism in the generic extension $V^{\mathbb{P}}$. An example can be found in :
S. Thomas, The automorphism tower problem II, Israel J. 
Math. 103 (1998), 93-109.
In fact, it can be shown that the group $G$ in this paper satisfies the stronger property that $G \not \cong Aut(G)$ as abstract groups in $V^{\mathbb{P}}$. In other words, there does not even exist a ''non-canonical isomorphism''.
For more on the ``nonabsoluteness'' of the height of automorphism towers, see:
J. Hamkins and S. Thomas, 
Changing the heights of automorphism towers, Annals Pure Appl. Logic 102 (2000), 139-157.<|endoftext|>
TITLE: What is the trace in the Chern-Simons action?
QUESTION [12 upvotes]: Warning: This is a very stupid question regarding a basic misunderstanding that I have. I realize that the question is very elementary, but I guess asking stupid questions is better than remaining ignorant.
To be explicit, consider the $\mathrm{SU}(2)$ Chern-Simons action on some very nice $3$-manifold $M$, i.e. the number
$S(A) = \frac{k}{4\pi}\int_M \mathrm{tr} \left(A\wedge\mathrm{d}A + \frac{2}{3}A\wedge A\wedge A\right)$,
where $A$ is an $\mathfrak{su}(2)$-valued $1$-form on $M$. 
What I simply cannot wrap my head around, and what is obviously a very silly basic question, is: What trace is this?! As I understand it, there is a certain abuse of notation in $\wedge$ on vector bundle-valued forms (namely, with $E$ the bundle, the wedge of two $E$-valued forms is an $E\otimes E$-valued one), but in the case of the Chern-Simons action this answer suggests that the $E\otimes E=\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathfrak{g}$ is supposed to be the Lie bracket on $\mathfrak{g}$. Anyway, that leads me to think that the trace is the trace on $\mathfrak{g}=\mathfrak{su}(2)$, which of course vanishes everywhere.
What am I misunderstanding here?

REPLY [6 votes]: I will give the physicist's answer.
I hope you are familiar with this notation.
$A=A_\mu dx^\mu$ and in this notation the action looks like
$S=\frac{k}{4\pi} \int_{\mathcal{M}} d^3x\ \epsilon^{\mu\nu\alpha}\ Tr \left( A_\mu \partial_\nu A_\alpha +\frac{2}{3}  A_\mu A_\nu A_\alpha   \right)$
where $\epsilon^{\mu\nu\alpha}$ is the Levi-Civita tensor in three dimensions. 
$A_\mu=A_\mu^at^a$ where $A_\mu^a$ are real functions, $t^a$ are the matrix representatives of generators of Lie algebra $[t^a,t^b]=if^{abc}t^c$. Here $a,b,c=1,2,...,dim G$ where G is the gauge group and $f^{abc}$ are structure constants.
So, that trace in the action is the trace of products of these $t^a$ matrices.
For example,
for $G=SU(N)$ and in the fundamental representation $Tr(t^at^b)=\frac{1}{2}\delta^{ab}$
$(a,b=1,2,...N^2-1)$<|endoftext|>
TITLE: Linear/Non-linear sigma model
QUESTION [13 upvotes]: This is slightly an open-ended invitation to discuss references and reasons for excitement about the linear and non-linear sigma model. 
I gauge from some other interactions that it has considerable interest even in mathematics (geometry?) apart from QFT. 
I would like if people can type in what are the canonical explanatory/expository references about the linear and non-linear sigma model from both the QFT and mathematical points of view. It would be illuminating if experts can shed light on why is this model exciting again from both the QFT and mathematics point of view. 
{As an example of the kind of expository reference I am looking for I would like to mention this one as an example for another interesting model in QFT}

REPLY [3 votes]: It is strange to distinguish the QFT point of view from the mathematics point of view. That's comparing apples to oranges.
One can distinguish the physics (non-mathematically rigorous) point of view from that of rigorous mathematics. In this case QFT has overlaps with both. I think the distinction here is between classical sigma models, an example of geometric PDE, and quantum sigma models which involves probability on a space of fields. As Willie Wong covered classical sigma models from the mathematics point of view and José covered quantum sigma models from the physics point of view; let me add some pointers to the literature regarding quantum sigma models from the mathematics point of view.
For rigorous work on perturbative (in the sense of formal power series) renormalization of the quantum sigma model, see:

P. K. Mitter and T. R. Ramadas,
"The two-dimensional $O(N)$ nonlinear $\sigma$-model: renormalisation and effective actions", Comm. Math. Phys. 122 (1989), no. 4, 575–596. 
T. Nguyen,
"Quantization of the nonlinear sigma model revisited",
J. Math. Phys. 57 (2016), no. 8, 082301, 40 pp. 

For rigorous nonperturbative results see:

C. Kopper,
"Mass generation in the large N-nonlinear σ-model",
Comm. Math. Phys. 202 (1999), no. 1, 89–126. This proves a mass gap but in the presence of an ultraviolet cutoff. 
K. Gawędzki and A. Kupiainen,
"Continuum limit of the hierarchical $O(N)$ nonlinear $\sigma$-model".
Comm. Math. Phys. 106 (1986), no. 4, 533–550. This removes the ultraviolet cutoff in a finite volume but concerns a hierarchical toy model.

Also, another account from the physics point of view and related to José's answer, the Ricci flow and all that, is the article:

M. Carfora, "Renormalization group and the Ricci flow",
Milan J. Math. 78 (2010), no. 1, 319–353.<|endoftext|>
TITLE: Finite dimensionality of certain $C^{\star}$-algebras
QUESTION [8 upvotes]: In the discussion about the question Finite-dimensional subalgebras of $C^{\star}$-algebras the following separate question came up:
Let $H$ be a Hilbert space and $a_1, \dots, a_n \in B(H)$ be self-adjoint operators. Consider the operators $x_1a_1+x_2a_2+\dots + x_n a_n$ , where the $x_i$'s  are complex variables and assume that there is a polynomial $p(z,x_1,\dots,x_n) \in \mathbb C[z,x_1,\dots,x_n]$ such that $z$ is in the spectrum of $x_1a_1+x_2a_2+\dots + x_n a_n$ if and only if $p(z,x_1,\dots,x_n)=0$. 

Question: Is the subalgebra of $B(H)$ which is generated by the operators $a_1 , \dots, a_n$ finite dimensional?

REPLY [2 votes]: This is an interesting question. The set is call multiparameter spectrum of the tuple g, it also called projective spectrum in my paper 
"http://www.worldscinet.com/jta/01/0103/S1793525309000126.html".
The paper didn't address this particular question, but said something in general.
Ron<|endoftext|>
TITLE: Finite group scheme acting on a scheme such that there is an orbit NOT contained in an open affine.
QUESTION [5 upvotes]: In Mumfords book on abelian varieties there is a theorem (on page 111) whose hypothesis is "Let G be a finite group scheme acting on a scheme X such that the orbit of any point is contained in an affine open subset of X".
Is there a well known example for when this condition is not fulfullied and in general, where should I look for the existance of quotients of schemes via actions of finite groups?

REPLY [5 votes]: The standard example is as follows: Take a $3$-fold $X$; for example let $X=\mathbb{P}^3$. Let $\sigma$ be an automorphism of $X$ of order $n$; for example, $(x_0:x_1:x_2:x_3) \mapsto (x_1:x_2:x_3:x_0)$. Let $C_1$, $C_2$, ..., $C_n$ be an $n$-gon of genus $0$ curves, with $C_i$ meeting $C_{i-1}$ and $C_{i+1}$ transversely and disjoint from the other $C_j$'s, with $\sigma(C_i)=C_{i+1}$. For example, $C_1 = \{(*:*:0:0) \}$, $C_2 = \{(0:*:*:0) \}$, $C_3 = \{(0:0:*:*) \}$ and $C_4 = \{(*:0:0:*) \}$. Let $p_i=C_{i-1} \cap C_{i}$.
Using this input data, we make our example. Take $X \setminus \{ p_1, p_2, \ldots, p_n \}$ and blow up the $C_i$. Call the result $X'$. Also, take a neighborhood $U_i$ of $p_i$, small enough to not contain any other $p_j$. Blow up $C_i \cap U_i$, then blow up the proper transform of $C_{i-1} \cap U_i$. Call the result $U'_i$. Glue together $X'$ and the $U'_i$'s to make a space $Y$. Clearly, $\sigma$ lifts to an action on $Y$.
There is a map $f: Y \to X$. The preimage $f^{-1}(p_i)$ consists of two genus zero curves, $A_i$ and $B_i$, meeting at a node $q_i$. The $q_i$ form an orbit for $\sigma$. We claim that there is no affine open $W$ containing the $q_i$. 
Suppose otherwise. The complement of $W$ must be a hypersurface, call it $K$. Since $K$ does not contain $q_i$, it must meet $A_i$ and $B_i$ in finitely many points. Since $W$ is affine, $W$ cannot contain the whole of $A_i$ or the whole of $B_i$, so $K$ meets $A_i$ and $B_i$. This means that the intersection numbers $K \cdot A_i$ and $K \cdot B_i$ are all positive. We will show that there is no hypersurface $K$ with this property.
Proof: Let $x$ be a point on $C_i$, not equal to $p_i$ or $p_{i+1}$. Then $f^{-1}(x)$ is a curve in $Y$. As we slide $x$ towards $p_i$, that curve splits into $A_i \cup B_i$. As we slide $x$ towards $p_{i+1}$, that curves becomes $A_{i+1}$. (Assuming that you chose the same convention I did for which one to call $A$ and which to call $B$.) Thus, $[A_i] + [B_i]$ is homologous to $A_{i+1}$. So $\sum [A_i] = \sum [A_i] + [B_i]$, 
$\sum [B_i] = 0$, and $\sum B_i \cdot K = 0$. Thus, it is impossible that all the $B_i \cdot K$ are positive. QED
I think this example is discussed in one of the appendices to Hartshorne.

REPLY [4 votes]: Oh, I guess we can just take the affine line with a double origin....:S<|endoftext|>
TITLE: In what cases does a Yoneda-like embedding preserve monoidal structure?
QUESTION [7 upvotes]: What kinds of Yoneda-like situations induce an embedding that preserves the tensor product for some arbitrary monoidal category?
The cases where the monoidal product is given by a limit or colimit give this immediately for the usual Yoneda embedding, but this breaks down for "real" monoidal categories like $(Vect, \otimes)$.
Are there $V$-enriched cases where the generalised embedding
$$ Y : C \to V^{C^{op}} $$
does preserve the tensor product for interesting monoidal categories $C$?

REPLY [5 votes]: Day showed that, for suitable V, any monoidal structure on a (V-)functor category $[C^{\mathrm{op}}, V]$ is essentially determined by its restriction to the representables as
$$ F \otimes G = \int^{A,B} F A \otimes G B \otimes P(A,B,-) $$
where $P(A,B,-) = C(-, A) \otimes C(-, B)$ is a profunctor $C \otimes C \otimes C^{\mathrm{op}} \to V$.  P (together with a unit and the usual structural isos) is said to endow C with a promonoidal structure.
If C is already a monoidal V-category, then there is a canonical promonoidal structure on it given by
$$ C(-, A) \otimes C(-, B) = C(-, A \otimes B) $$
In that case, the Yoneda embedding is strong monoidal by definition.  In fact it is the unit for the monoidal cocompletion of C.<|endoftext|>
TITLE: When completion of locally compact length space is locally compact?
QUESTION [9 upvotes]: As far as I know the answer to the question:
"Is it true that a completion of a locally compact length space is locally compact?" - Negative.
Does anybody know some metric and/or topological conditions for locally compact length space $(X,d)$ such that its completion $\bar{X}$ is locally compact?

REPLY [2 votes]: A necessary and sufficient condition (but I do not feel satisfied with that) for the locally compact length space $X$ to have a locally compact completion is that there exists some $r>0$ such that each ball of radius $r$ in $X$ is totally bounded.
In fact, if the condition holds closed balls of radius $r/2$ in $\overline{X}$ are compact.
On the other hand, suppose that $\overline{X}$ is locally compact. Then, as it is a complete length space, it is proper (this is called the Hopf-Rinow Theorem in the book by Bridson and Haefliger). This should imply that balls of any radius in $X$ are totally bounded.
The main reason why I am not satisfied with it is that the proof that the condition is sufficient does not use that $X$ is a length space, so this is not really the answer to what you asked. I thought it might be relevant, anyway...<|endoftext|>
TITLE: Light rays bouncing around inside a sphere in d-dimensions
QUESTION [6 upvotes]: Suppose $S=\mathbb{S}^d$ is a unit sphere in $(d-1)$ dimensional space, with $d=3$ of special interest.
The surface of $S$ is a perfect (internal) mirror.
You stand at point $x$ (not the sphere center $c$) inside $S$ and emit a single laser light ray in direction $u$.
What happens?  I believe that the light ray will remain within the plane containing the three points
$\{ x, x+u, c \}$.
Now suppose instead that from $x$ you shine a flashlight, a cone with angular extent $\pm \epsilon$.
Does this fill the sphere with constant-density energy for any $\epsilon > 0$?  Are there are no dark points within $S$?
A somewhat related question is: What would the flashlight-holder see from $x$?
What would the visual image be, say in a graphics ray-tracing system (in $d$ dimensions!)?  
I've asked enough questions for one MO posting, but ellipsoids in $\mathbb{R}^d$ are the
obvious extension.  Are they integrable or chaotic?

REPLY [2 votes]: Billiards inside an ellipsis all are integrable in dimension $2$, as is beautifully shown in the small book by Tabachnikov. I do not know if this property extends to higher dimensions, but the proof I know certainly doesn't (it relies to the metric bifocal caracterisation of ellipses).
Let me give a more precise statement and mention an important conjecture in this domain. Given an elliptic domain, there is an open neighborhood of its boundary that is foliated by curves (which are in fact confocal ellipses) in such a way that any billiard trajectory starting close enough to the boundary (in position and direction; here the proximity condition is in fact that the first segment does not go between the focal points), has each of its segments tangent to one and the same of these curves. The conjecture is that ellipses are the only convex 2-dimensional billiards having this property.<|endoftext|>
TITLE: What are natural questions to ask about an operad?
QUESTION [32 upvotes]: I just discovered that something I've been working with has the structure of an operad. So I'm wondering what natural basic questions does one ask about operads? For example, if I knew I had the structure of a group, I'd ask if it is abelian or has torsion, etc. So what are these questions for operads?

REPLY [12 votes]: Hi Connie. Let me use your question as an excuse for an extended answer.
A pair of brief papers "Definitions: operads, algebras and modules" and
"Operads, algebras and modules", which are available at 
http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf
and http://www.math.uchicago.edu/~may/PAPERS/handout.pdf (# 84,85 on my website)
give several variants and reformulations of the original definition together with
some history of antecedents, a variety of algebraic and topological examples,
and the crucial relationship with monads that led me to coin the word "operad".
There is also a discussion of the relationship to homological algebra, showing
how the homological theory simplifies if you work over a field of characteristic
zero and, in contrast, how operads encode homology operations (Steenrod operations
and Dyer-Lashof operation) if you work over a field of finite characteristic.  Notes
for a talk, http://www.math.uchicago.edu/~may/TALKS/SwitzerlandTalk.pdf, expand on
the last point.  
The distinction of characteristic illustrates a general point. Operads are defined
in any symmetric monoidal category, and the right questions to ask depend in large 
part on what category you are working in.  It may make no sense at all to ask 
algebraic questions of a topological operad or topological questions of an algebraic
operad.   There is also a distinction to be made about questions to ask about operads
and questions to ask about their algebras. Incidentally, groups are by design not 
examples of algebras over an operad: to define inverses, you need diagonals, and 
operads are not intended, or rather intended not, to incorporate such structure.
The questions to ask also depend on what role your observation plays.  Operads 
allow a taxonomy of certain types of algebraic structures, so the question may just 
be "what kind of structure am I looking at".
But you might also want to ask whether the algebras you are looking at give simpler
"approximations'' of more complicated or less accessible structures that occur "in
nature".  For example, spaces $\Omega^n\Sigma^n X$ occur in nature, but they can 
very usefully be approximated by the monads $C_nX$ associated to appropriate operads.
You might also want to ask if operads can be used to define rigorously new structures 
that you want to study.  A very recent example arose in work of Bertrand Guillou and myself
in equivariant infinite loop space theory: there is an intuition of what a genuine
strict symmetric monoidal $G$-category should be, one that gives rise to a genuine
$G$-spectrum; the best definition we know is that such a category is an algebra 
over a particular operad in $Cat$ (see http://front.math.ucdavis.edu/1207.3459).
Quite a few recent variants of the definition of an operad arose analogously.  
In algebra, very simple operads prescribe very natural and previously unstudied
kinds of algebras. Loday and some of his students (I'm blanking on names) gave a number of examples. 
While one can ask questions about the homotopy theory of operads in general,
using model category theory, that is perhaps my least favorite question to
ask: it rarely cuts to the heart of the applications, excepting those in higher
category theory, or so it seems to me.  Model categories of algebras
over particular operas do play a major role in many applications, albeit
sometimes only implicitly.
I'll stop here, since I could go on forever.
One comment. While the Martin-Shnider-Stasheff book is a useful compendium, its treatments 
of different topics are not all at the same level, and you might well prefer less 
comprehensive treatments that better address your directions of interest. And people 
should be warned that the definition of an operad in that book is actually incorrect: it omits a 
crucial equivariance property that is of real importance in applications. For example, it plays a 
key role in the proof of the Adem relations for the Steenrod and Dyer-Lashof operations.
Benoit Fresse's book "Modules over operads and functors" gives a quite different take
on operads, with a focus on modules over algebras over operads.<|endoftext|>
TITLE: Values of the j-function
QUESTION [7 upvotes]: In general, how do you compute the algebraic values of the modular j-function at quadratic imaginary points? (In other words, how do you compute the algebraic values of singular moduli?)
For instance, the Mathematica website (http://mathworld.wolfram.com/j-Function.html) gives the standard nine integral examples that result when the class number $h_k=1$, but then it gives 18 examples for when the class number is 2 without any specific references. How does one compute these? More importantly, can you also do it for higher degree cases? Or even just find the defining degree-$h_k$ polynomial?

REPLY [3 votes]: In case the Gross-Zagier paper doesn't meet your needs, you can also refer to the following 

Harold Baier Efficient computation
of singular moduli 
Noriko Yui 
    On The Singular Values Of Weber
    Modular Functions

and of course David Cox's book Primes of the form $x^2 + Ny^2$, Section 3.12
The bad thing about the j-function is that it is a level 1 modular function so the coefficients of its defining polynomial are going to explode with increasing degree.  Its easier to compute the singular moduli using a modular function of some higher level (e.g. Weber func has level 48) as demonstrated in the papers mentioned above as well in Cox's book.<|endoftext|>
TITLE: What can I say about the permutation $\alpha\beta$ if I know the permutation $\beta\alpha$?
QUESTION [10 upvotes]: I'm looking into a secret sharing scheme that has a secret permutation $\theta$ which has the cycle structure (n/2)+(n/2) (i.e. two (n/2)-cycles).
The permutation $\theta$ is decomposed into two permutations $\alpha$ and $\beta$, where $\alpha$ is generated uniformly at random.  So with knowledge of both $\alpha$ and $\beta$, we can find $\theta$, while with knowledge of $\alpha$ xor $\beta$, we cannot find $\theta$ (although, we could guess).
At this point, I want to make public $\beta\alpha(L)$ (L is actually a Latin square, but this is not too relevant for the question I want to ask).  It is possible that an attacker could find $\beta\alpha$ from $\beta\alpha(L)$.  However, I worry that knowledge of $\beta\alpha$ might give information about $\theta$.

If I know $\theta=\alpha\beta$, and I'm given the permutation $\beta\alpha$, what can I say about $\theta$?  (without a priori knowledge of $\alpha$, $\beta$ or $\theta$)

REPLY [24 votes]: $\theta$ could be any permutation of the form $\alpha (\beta \alpha) \alpha^{-1}$; in other words, it could be any permutation conjugate to $\beta \alpha$, so knowing $\beta \alpha$ tells you only the cycle type of $\theta$, no more and no less.  Since you already specified the cycle type this means an attacker gains no information (assuming $\alpha$ and $\beta$ really are chosen randomly).<|endoftext|>
TITLE: confusion about forcing
QUESTION [6 upvotes]: I learned from Kunen's book, besides forcing over countable transitive model (c.t.m.), there is an another way to do forcing, called the "syntactic method", i.e. forcing over V. 
Fixed a partial order P in the ground model. When one does forcing over a c.t.m. M, it is easy that for each p in P, there is a filter G which is P-generic over M and contains p. when we use the "syntactic method", can we alway assume there is a filter G which is P-generic over V and contains arbitrary given p in P ?
Let M be a c.t.m. for ZFC, P in M, G generic over M. Let A,B in M, r be a P-name for a function in M[G] from A into B,  for each a in A, choose  p(a) in P and b(a) in B s.t.
p(a)||- r(aˇ)=b(a)ˇ, let f(a)=b(a), then can we deduce that f is a fuction in M from A into B ?

REPLY [9 votes]: The syntactic method is simply to work inside the forcing relation, by looking at what is forced by which conditions. That is, rather than actually building the forcing extension $V[G]$, you reason as though you were inside it, by means of the forcing relation. For example, if you have condition $p$ forcing various statements $\varphi_1$, ... $\varphi_n$, and from those statements you can deduce $\psi$, then you can conclude that $p$ also forces $\psi$. In particular, note that any condition $p$ forces that $\dot G\subset\check P$ is $\check V$-generic and contains $\check p$, so syntactically, you have the filter you want in the sense I described. The point is that you don't actually need to build an actual model using an actually generic filter to gain insight into what is true there, because you have the forcing relation telling you what is forced to be true there.
If you want to build an actual model by forcing over $V$, then the thing to do is to use Boolean-valued models $V^B$, where there is Boolean-valued truth. You can quotient this $B$-valued structrue by an ultrafilter $U$ (no need for any genericity) to arrive at an actual first order 2-valued structure $V^B/U$, which satisfies a version of Los' theorem: $V^B/U\models\phi$ if and only if $[\phi]\in U$. 
For question 2, if you make your choice of $b(a)$ $p(a)$ inside $M$, then of course $f$ is a function in $M$. But there is little reason to expect that this function $f$ has much relation to the function in $M[G]$ named by $r$, since the conditions $p(a)$ might be incompatible. If you can choose a single $p=p(a)$, then it follows that $p$ forces $r=\check f$.

REPLY [9 votes]: One can speak about the independence results without models as follows. 
Let $(P,\leq)$ be a poset, with 1 as the largest element. There is a relation, called forcing, which is defined by formulainduction for all formulas, and which has the property that all axioms of ZFC are forced by 1, if some formula $\phi$ follows from ZFC then it is forced by 1. Now, if, for a particular $\phi$  we find a poset $(P,\leq)$ such that 1 does not force $\phi$, then $\phi$ is independent from ZFC (or there is a contradiction in ZFC). This does not work with models (it actually does assume the existence of models of ZFC, as that is equivalent to ZFC being noncontradictory by Godel), the argument is proof theoretic, however, all steps are essenitally the same as in the correspodning model theoretic argument.<|endoftext|>
TITLE: When is sin(r \pi) expressible in radicals for r rational?
QUESTION [13 upvotes]: Perhaps this question will not be considered appropriate for MO - so be it.  But hear me out before you dismiss it as completely elementary.
As the question suggests, I would like to know when $\sin(p\pi/q)$ can be expressed in radicals (in the way that $\sin(\pi/4) = \sqrt{2}/2$ and $\sin(\pi/3) = \sqrt{3}/2$ can).  Let $\alpha = \sin(x)$, and consider the field extension $\mathbb{Q}[\alpha]$.  Using $(\cos(x) + i\sin(x))^k = \cos(kx) + i \sin(kx)$ together with the binomial formula and the Pythagorean identity relating sine and cosine, we can see that $\sin(kx)$ lies in a solvable extension of $\mathbb{Q}[\alpha]$.  Thus $\sin(p\pi/q )$ is expressible in radicals if $\sin(\pi/q)$ is.
To handle $\sin(\pi/q)$, we start by using the same trick (which most people also learn in an elementary trig class).  Write $-1 = (\cos(\pi/q) + i\sin(\pi/q))^q$, use the binomial theorem to expand, compare imaginary parts, and express the right-hand-side in terms of sine using the Pythagorean identity.  This gives an explicit equation for any $q$ one of whose solutions is $\sin(\pi/q)$.  This equation is not a polynomial in $\sin(\pi/q)$ since it involves terms of the form $\sqrt{1 - \sin^2(\pi/q)}$, but it is enough to prove that $\sin(\pi/q)$ is algebraic.
So I am curious about the number theoretic properties of this equation.  What can be said about the Galois group of its "splitting field" over $\mathbb{Q}$?  Can we at least determine when it is solvable?  Note that if the prime factors of $q$ are $p_1, \ldots p_k$ and we can express each $\sin(\pi/p_j)$ in radicals, then the same is true for $\sin(\pi/q)$.  So it suffices to consider the case where $q$ is prime.  That's about all the progress I have made.

REPLY [8 votes]: Your question has been thoroughly answered, so I don't really have much to add, except that there is a paper where you can see examples of these ideas in action with a very elementary presentation (i.e. suitable for undergraduates who have a basic knowledge of Galois theory).
Skip Garibaldi
Somewhat more than governors need to know about trigonometry
Mathematics Magazine 81 (2008) #3, 191-200.<|endoftext|>
TITLE: Abelian varieties over local fields
QUESTION [6 upvotes]: Let $K$ be a local field of characteristic zero, $k$ its residue field, $R$ its ring of integers and $p$ the characteristic of the residue field $k$. Let $G$ be the Galois group of $K$, $I\subset G$ the inertia group and $P$ the maximal pro-$p$ subgroup of $I$. Let $I_t:=I/P$. 
Let $A_0$ be an abelian scheme over $R$ with generic fibre $A$. Then $A[p]$ is an $I$-module.
Let $V$ be a Jordan-Hölder quotient of the $I$-module $A[p]$.
I am interested in the representation $I\to Aut(V)$. 
Question (*): Is it true that $P$ acts trivially on $V$?
(I have seen that there are results
of Raynaud and Serre on the "action of $I_t$ on $V$". I want to study these things, but I am already stuck
with Question (*) at the moment, i.e. with the question whether $I_t$ acts at all.)  
Maybe someone can help?

REPLY [7 votes]: $V$ is an irreducible $\mathbb{F}_p$-representation of $I$. As $P$ is a pro-$p$ group, $V^P\ne 0$, and $P$ is normal in $I$ so $V^P$ is stable under $I$. Therefore $V^P=V$. You might like to look at Tate's article on finite flat group schemes in "Modular Forms and Fermat's Last Theorem" for more information.<|endoftext|>
TITLE: Sequential topological vector spaces
QUESTION [6 upvotes]: Since I'm dealing with the distinction between sequential continuous and continuous maps at the moment I came to ask myself once again what can be said about spaces where these two notions agree (sequential spaces). Of course we all know that metric spaces and more generally first-countable spaces are sequential and in the literatur it seems that often metrizability or first-countability is only assumed in order to not need to distinguish between sequential continuity and continuity.
I'm mostly interested in spaces that arise naturally in functional analysis, i.e. subspaces of topological vector spaces. A well known theorem says that a hausdorff topological vector space is metrizable iff it is first-countable. I tried to find out what could be said about sequential t.v.s. Are sequential t.v.s. metrizable too? Are there any reasonable t.v.s. that are sequential but not metrizable?

REPLY [12 votes]: The space of tempered distributions is sequential (for its usual strong topology). See, e.g., Dudley, and the references therein.

REPLY [3 votes]: Some of the common spaces of distributions are sequential but not metrizable.<|endoftext|>
TITLE: Holomorphic vector bundles and Swan's theorem
QUESTION [5 upvotes]: Is every holomorphic vector bundle a direct summand of a trivial vector bundle on submanifolds of C^n? What about projective varities? I believe Swan's theorem says something about the first question. But I wanted to make sure.

REPLY [7 votes]: The statement for Stein manifolds follows indeed from the analogue of the Serre-Swan theorem for Stein manifolds, which was proven first in 1967 in "Zur Theorie der Steinschen Algebren un Moduln" by O. Forster. The situation is a bit more complicated than the affine scheme or manifold case, but the final result relevant for the purposes of the question is the same
The category of locally free sheaves of finite rank is the same as the category of finitely generated projective modules over the global sections $\Gamma(O_X)$. Then one notes that a f.g. projective module is always a direct summand of a finite free module.<|endoftext|>
TITLE: How does curvature change under perturbations of a Riemannian metric?
QUESTION [5 upvotes]: Let $M$ be a compact subset of $\mathbb R^2$ with smooth boundary, and let $g$ be a Riemannian metric on $M$.  If $g'$ is another Riemannian metric which is "close" to $g$, then they should have almost identical curvature profiles.  I would like to prove a concrete estimate on the total difference of their curvatures in terms of the distance of $g'$ to $g$.  Before I state the question precisely, I need to introduce some notation.
Write $\operatorname{Sym}$ for the space of symmetric $2\times 2$ real matrices, and let $\operatorname{SPD} \subseteq \operatorname{Sym}$ be those matrices which are also positive-definite.  Consider the function space $\Omega = C^2(M, \operatorname{SPD})$.  Denote partial derivatives of $g_{ij} \in \Omega$ by additional subscripts following a comma, so that $\tfrac{\partial}{\partial x^k} g_{ij} = g_{ij,k}$, et cetera.  Endow the space $\Omega$ with the norm $$\|g\| = \sup_{x \in M} \max_{i,j,k,l} \left\{|g_{ij}(x)|, |g_{ij,k}(x)|, |g_{ij,kl}(x)| \right\},$$ so that it has the structure of an open cone within the Banach space $C^2(M, \operatorname{Sym})$.
Each $g \in \Omega$ defines a Riemannian structure on $M$ via the inner product $\langle v, g(x) v' \rangle$ for $v, v' \in T_x M$.  Let $K(g,x)$ be the scalar curvature of the metric $g$ at the point $x \in M$.
What I want to prove:  For each $g \in \Omega$, there exist constants $C$ and $\epsilon$ so that if $g' \in \Omega$ with $\|g - g'\| < \epsilon$, then $$\sup_{x \in M} \left| K(g,x) - K(g',x) \right| \le C\|g- g'\|.$$
My current approach to this is quite clunky, and involves calculating everything directly from the Christoffel symbols of the metrics.  Is there a better, more geometric approach to this than brute force calculations?
I'm sure this type of lemma is well known to geometric analysts.  Is a proof of a similar result written down somewhere?

REPLY [3 votes]: Elaborating on Deane Yang's answer and Willie Wong's comment: Since
$M^{2}\subset\mathbb{R}^{2}$ is a $C^{\infty}$ submanifold with boundary, the
Euclidean coordinates are global. Generally, if $M^{n}$ is a compact manifold
with boundary, we can cover it by a finite number of charts $\{x^{i}\}$, where
for any $C^{\infty}$ metric $g$ the functions $g^{ij}$ and $\partial^{\alpha
}g_{ij}$ are bounded (depending on $g$ and $|\alpha|$) and where $\alpha$ is a
multi-index with $|\alpha|\geq0$.
The scalar curvature $R_{g}$ (twice the Gauss curvature $K$ if $n=2$) is
$$
R_{g}=g^{jk}(\partial_{\ell}\Gamma_{jk}^{\ell}-\partial_{j}\Gamma_{\ell
k}^{\ell}+\Gamma_{jk}^{p}\Gamma_{\ell p}^{\ell}-\Gamma_{\ell k}^{p}\Gamma
_{jp}^{\ell})=(g^{-1})^{2}\ast\partial^{2}g+(g^{-1})^{3}\ast(\partial g)^{2}
$$
since the Christoffel symbols have the form $\Gamma=g^{-1}\ast\partial g$,
where $\partial^{k}g$ denotes some $k$-th partial derivative of $g_{ij}$ and
where $\ast$ denotes a linear combination of products while summing over
repeated indices. From the formula for $R$ we have for metrics $g,g^{\prime}$,
\begin{align*}
& |R_{g}(x)-R_{g^{\prime}}(x)|\\
& \leq C(|g^{-1}|^{2}+|g^{\prime-1}|^{2})|\partial^{2}g-\partial^{2}g^{\prime
}|+C(|g^{-1}|^{4}+|g^{\prime-1}|^{4})(|\partial g|^{2}+|\partial g^{\prime
}|^{2})|g-g^{\prime}|\\
& +C(|g^{-1}|^{3}+|g^{\prime-1}|^{3})\{(|\partial^{2}g|+|\partial^{2}
g^{\prime}|)|g-g^{\prime}|+(|\partial g|+|\partial g^{\prime}|)\left\vert
\partial g-\partial g^{\prime}\right\vert \}
\end{align*}
since $|g^{-1}-g^{\prime-1}|\leq C(|g^{-1}|^{2}+|g^{\prime-1}|^{2}
)|g-g^{\prime}|$.
Let $\hat{\Omega}=C^{2}(M,\operatorname{Sym})$. Given $h\in\hat{\Omega}$,
define $||h||=\sup_{x\in M}\max_{i,j,k,\ell}\{|h_{ij}(x)|,|h_{ij,k}
(x)|,|h_{ij,k\ell}(x)|\}$. Then $|R_{g}(x)-R_{g^{\prime}}(x)|\leq
C||g-g^{\prime}||$, where $C$ depends on bounds on the inverses and the first
and second derivatives of $g$ and $g^{\prime}$.
Elaborating on Terence Tao's answer and Deane Yang's comment: One reason it is
convenient to compute in local coordinates $\{x^{i}\}$ is that $[\partial
_{i},\partial_{j}]=0$. So the expression for the Christoffel symbols has only
$3$ terms instead of the $6$ terms comprising the formula for $\nabla$:
$\Gamma_{ij}^{k}=\frac{1}{2}g^{k\ell}(\partial_{i}g_{j\ell}+\partial
_{j}g_{i\ell}-\partial_{\ell}g_{ij})$, which is symmetric in $i$ and $j$. With
$\frac{\partial}{\partial s}g_{ij}=v_{ij}$, the variation formula is easy to
compute: $\frac{\partial}{\partial s}\Gamma_{ij}^{k}=\frac{1}{2}g^{k\ell
}(\nabla_{i}v_{j\ell}+\nabla_{j}v_{i\ell}-\nabla_{\ell}v_{ij})$, since the
computation of this tensor formula at any point $p$ may be done in coordinates
where $\partial_{i}g_{jk}(p)=0$ (such as normal coordinates centered at $p$);
this enables us to convert $\partial_{i}$ to $\nabla_{i}$ and to ignore the
$\frac{\partial}{\partial s}g^{k\ell}$ term since it is multiplied by terms of
the form $\partial g$. Now the variation of the Riemann curvature tensor is
$\frac{\partial}{\partial s}R_{ijk}^{\ell}=\nabla_{i}(\dfrac{\partial
}{\partial s}\Gamma_{jk}^{\ell})-\nabla_{j}(\dfrac{\partial}{\partial s}
\Gamma_{ik}^{\ell})$ using the same trick of computing at the center $p$ of
normal coordinates and replacing $\partial$ by $\nabla$ (note that
$\frac{\partial}{\partial s}(\Gamma\ast\Gamma)=0$ at $p$ by the product rule);
the resulting formula is true in any coordinates since it is tensorial.
Generally, it is convenient to compute in local coordinates because it can be
done more or less mechanically. For example, if $\alpha$ is a $1$-form, then
$\nabla_{i}\nabla_{j}\alpha_{k}-\nabla_{j}\nabla_{i}\alpha_{k}=-R_{ijk}^{\ell
}\alpha_{\ell}$. One can remember this as the contraction of
$-\operatorname{Rm}$ and $\alpha$, where the lower indices $i,j,k$ on
$\operatorname{Rm}$ appear in the same order as the first term on the left.
Similarly, if $\beta$ is a $2$-tensor, then $\nabla_{i}\nabla_{j}\beta_{k\ell
}-\nabla_{j}\nabla_{i}\beta_{k\ell}=-R_{ijk}^{m}\beta_{m\ell}-R_{ij\ell}%
^{m}\beta_{km}$, where the the lower indices of $\operatorname{Rm}$ are $i,j$
and then either $k$ or $\ell$, with upper dummy index $m$ on
$\operatorname{Rm}$ also replacing either $k$ or $\ell$ on $\beta$.<|endoftext|>
TITLE: Cusp forms and L^2
QUESTION [20 upvotes]: I am confused about the "bigger picture" when one goes from classical modular forms on $SL_2(\mathbb{Z})$ and its subgroups to automorphic forms (possibly non-holomorphic). 
For classical modular forms we have cusp forms and Eisenstein series. There is the Peterson product defined when one of the forms is a cusp form (since cusp forms vanish at the cusps) and Eisenstein series are the orthogonal complement of cusp forms. For cusp forms there is Atkin-Lehner theory but not (as far as I can understand) for Eisenstein series.
Now from reading various sources, in order to generalise this to nonholomorphic forms, the correct thing is to look at $L^2(\Gamma\backslash SL_2(\mathbb{R}))$.  I have figured out how to make a modular form into a function on $SL_2(\mathbb{R})$, but unless it is a cusp form it doesn't look like it is in $L^2$ (the inner product is just the generalization of Peterson, isn't it?)
On the other hand there are series which look like Eisenstein series like 
$$
E_\phi(z) = \sum_{\gamma\in\Gamma_\infty\backslash\Gamma} \phi(\mathrm{Im}(\gamma(z)))
$$
where $\phi$ is a compactly supported $C^\infty$-function on the positive reals. These obviously vanish at the cusps (and are in $L^2$) but they are (apparently) not cusp forms. 
So my specific questions are: what is the right way to generalize modular forms so that holomorphic Eisenstein series still stay in the picture? Where do these $E_\phi$ series fit in? And what happens to Hecke theory in this setting? 
I would like to also understand the role of weight 2 "almost-holomorphic" Eisenstein series in a more conceptual way.
(I don't know much functional analysis, so Gelbart's book is proving rather difficult for me. Do I have to understand all the functional analysis to get a feeling for what is going on?)

REPLY [27 votes]: This is a long answer because the question asks quite a lot of things. I agree that Gelbart's book, although inspirational, is hard for someone without a strong analytic background. The Boulder and Corvallis proceedings are full of articles which are worth studying if you want to get an understanding of automorphic theory.  It is hard to summarise but here's a (necessarily oversimplified) sketch of the "big picture" you may be looking for.
First, two important things to get straight:
1) "Cuspidal" is not the same as "vanishing at the cusps". Sticking to the upper half-place for the moment, a function is cuspidal if at each cusp the 0-th Fourier coefficient vanishes. For a holomorphic function the Fourier coefficients are constant, so cuspidal and vanishing at cusps are the same thing. But for those functions $E_\phi$ the 0-th Fourier coefficient at $\infty$ is some function $c_0(y)$ vanishing in a neighbourhood of infinity, but certainly non-zero in general. In fact, $f\in L^2(\Gamma\backslash\mathfrak{H})$ is cuspidal iff it is orthogonal to all the $E_\phi$ (compute the Petersson product to see this), so $\{E_\phi\}$ generates a complement to the cusp forms in $L^2$. Arithmetically, though, they aren't interesting.
2) Automorphic forms and $L^2(\Gamma\backslash G)$ are different beasts entirely. An automorphic form is a smooth function on $\Gamma\backslash G$  satisfying various properties (moderate growth, K-finite, and killed by an ideal of finite codimension in the centre of the universal enveloping algebra).  It does not have to be in $L^2$. For example, holomorphic Eisenstein series are automorphic forms which are not in $L^2$.
Hecke theory: for $SL_2(\mathbb{Z})$ you can just mimic the classical definitions in this more general setting, but for groups other than $GL_2/\mathbb{Q}$ you really need to look at the adelic setting. Here an automorphic form (for a reductive group $G/\mathbb{Q}$) is a smooth function on $G(\mathbb{Q})\backslash G(\mathbb{A})$, satisfying a bunch of properties (see Borel and Jacquet's article in Corvallis for a precise definition, indeed for just about everything here). The finite adelic group $G(\mathbb{A}_f)$ acts by right translation on the space of automorphic forms, and this is the right generalisation of Hecke operators. The Lie group $G(\mathbb{R})$ does not act on the space of automorphic forms (K-finiteness is not preserved) but there is a suitable algebra of invariant differential operators (the Hecke algebra of $G(\mathbb{R})$) which does. 
Cuspidal automorphic forms - in the adelic setting - are those $F$ for which the function
$$
F_N(g) = \int_{N(\mathbb{Q})\backslash N(\mathbb{A})} F(hg)\ dh
$$
vanishes for suitable unipotent subgroups $N$. For $GL_2$ these are just the unipotent radicals of Borel subgroups (defined over $\mathbb{Q}$), and since these are all conjugate it's enough to verify the vanishing for the upper triangular unipotent subgroup. 
A classical cusp newform $f$ of weight $k$ then gives rise to a cuspidal automorphic form $F$ on $GL_2/\mathbb{Q}$. There are two different things one can now do with $F$:
1) The translates of $F$ under $GL_2(\mathbf{A}_f)$ generate an irreducible representation $V_f$, which encodes the action of the Hecke operators (and more). Elements of this representation are none other than oldforms associated to $f$. More precisely, $V_f$ is an infinite tensor product of representations $V_p$ of $GL_2(\mathbb{Q}_p)$. If $p$ doesn't divide the level of $f$, then $V_p$ tells you the Hecke $T_p$ and $R_p$ eigenvalues. At bad primes, it contains much more delicate information than classical Atkin-Lehner theory does (one reason to use the adelic approach even for $GL_2$).  
2) The translates of $F$ under the Hecke algebra at infinity, on the other hand, generate an irreducible representation $V_\infty$ of a particular type (discrete series with parameter given by $k$), inside which $F$ is characterised as the lowest weight vector (this is the group-theoretic interpretation of the holomorphy of $f$).
The space of all translates of $F$ (i.e. by the finite adelic group and the Hecke algebra at infinity) is just $V_\infty\otimes V_f$. It is an example of an automorphic representation. (In general, an automorphic representation is any irreducible subquotient of the spaces of automorphic forms under these actions.)
The map $F\mapsto F_N$ allows one to describe the quotient (automorphic forms)/(cusp forms) by induced representations from parabolic subgroups. Although explicit, this quotient is quite a complicated representation - in particular, it is very far from being semisimple, even for $GL_2$.
So, if you are looking at this from a number-theorist's perspective, why care about $L^2$ ? The reason is the trace formula, which needs to be formulated in the setting of Hilbert spaces. There is essentially no difference between an $L^2$-form which is cuspidal and an automorphic cusp form, so the trace formula, appropriately wielded, can tell you a lot about cusp forms. It is a powerful and indispensable tool. But to apply it you need also to look at the rest of the $L^2$, in which lives the continuous spectrum, accounted for by real-analytic Eisenstein series at $Re(s)=1/2$.   Here there is a lot of analysis but you can usually find a friendly expert to help you out.
PS: Some people like to define a cuspidal automorphic representation as an irreducible subspace of $L^2_{\mathrm{cusp}}$. This has some advantages: (a) it's concise, and (b) at infinity one is working with genuine (unitary) representations of the real Lie group, rather than $(\mathfrak{g},K)$-modules (equivalently, representations of the real Hecke algebra). But it only gives the correct answer for cuspidal representations.<|endoftext|>
TITLE: Extending group actions on varieties
QUESTION [8 upvotes]: Let $X$ be a (irreducible) variety (over $\mathbb{C}$ if necessary, smooth orbifold if necessary), and $U\subset X$ a nonempty open subset, and let $G$ be a finite group with an algebraic action on $U$.
When does the action extend to an action of $G$ on $X$? Are there any conditions that are reasonable to verify, as I'd prefer to avoid having to construct the action on $X\setminus U$ (I know the action exists for other reasons, but the method I'm using gives easily that I have an action on a nonempty open, rather than the whole thing)

REPLY [6 votes]: Well, not always, e.g. take $X=\mathbb P^2$ with homogeneous co-ordinates $x,y,z$, $U$ the locus defined by $xyz\ne 0$ and $\sigma$ the involution of $U$ defined by $(x,y,z)\mapsto (yz,xz,xy)$ (the standard quadratic Cremona transformation of $X$). Then $\sigma$ is of order $2$ and does not extend to $X$. On the other hand some sufficient conditions, in characteristic zero, are: there are no rational curves in $X\setminus U$; $X$ has ample canonical class and is smooth (this can be weakened to having canonical singularities).
[Update: Charles Siegel asked for references, and I have none to hand, although this is all well known.] 
For the Cremona example, recall that every automorphism of $\mathbb P^2$ is linear, and $\sigma$ clearly isn't. More geometrically, $\sigma$ blows up the vertices of the triangle $xyz =0$ and collapses its sides.
No rational curves in $X\setminus U$: fix $g\in G$ and think of it as a rational map $g:X\to X$. By Hironaka, there is a minimal composite $p:Z\to X$ of blow-ups such that $g\circ p$ is regular. If $X\setminus U$ has no rational curves, then the rational curves in the exceptional divisor of the last blow-up are contracted, so the last blow-up was redundant, contradicting minimality.
Ample canonical class $\omega_X$: then $X=Proj(R(X,\omega_X))$, with $R(X)= \oplus_{n\ge 0}H^0(X,\omega_X^{\otimes n})$, the canonical ring. This is a birational invariant of $X$, so $G$ acting on $U$ acts on $R(X)$, so on $X$.<|endoftext|>
TITLE: How to transition from pure math PhD to nonacademic career?
QUESTION [33 upvotes]: I'm finishing up a PhD in math and am thinking about options outside of academia. So far, I've really only focused on pure mathematics, but I have a year left of grad school. Suppose I am interested in looking for a job in software engineering, finance, or some other quantitative field. What should I be doing in the next year? What classes should I take?

REPLY [2 votes]: In case you consider moving to finance, I suggest reading Options, Futures, and Other Derivatives by John C. Hull . It is a standard work in the field.<|endoftext|>
TITLE: Bounding the product of lengths of basis vectors of a unimodular lattice
QUESTION [5 upvotes]: Suppose $\Lambda\subset\mathbb{R}^n$ is a unimodular (i.e. volume $1$) lattice in Euclidean space.
Let $v_1,\dots, v_n\in\Lambda$ be a basis of $\Lambda$ such that the product of lengths $A=|v_1|\cdots|v_n|$ is minimal. I'd like to bound this minimal product $A$ from above as $\Lambda$ varies (and $n$ is fixed). I can prove that such an upper bound exists - for instance for $n=2$, it's attained by the A2 ("hexagonal") lattice since any lattice has a basis such that the angle between the two basis vectors is between 60 and 120 degrees. I don't know what it is for general $n$. It seems like this should be known, but I can't find it. Does anyone know of a good bound?

REPLY [2 votes]: I don't know how good the bound is you can obtain from this, but what about taking a Korkine-Zolotarev reduced basis of $\Lambda$, say $(b_1, \dots, b_n)$: then, by this paper, $\|b_i\|_2^2 \le \frac{i + 3}{4} \lambda_i(\Lambda)^2$, where $\lambda_i(\Lambda)$ is the $i$-th successive minimum of $\Lambda$. By Minkowski, $\prod_{i=1}^n \lambda_i(\Lambda) \le \gamma_n^{n/2} \det \Lambda = \gamma_n^{n/2}$ (in your case), $\gamma_n$ being the $n$-th Hermite constant, whence you get $A \le \prod_{i=1}^n \|b_i\|_2 \le \frac{\gamma_n^{n/2}}{2^n} \prod_{i=1}^n \sqrt{i + 3}$.<|endoftext|>
TITLE: Construction of Morava E-theory
QUESTION [33 upvotes]: I'm wrapping up a summer project that involved a computation in Morava $E$-theory.  As background knowledge I had to look into how the Johnson-Wilson theories $E(n)$ and Morava $K$-theories were constructed.  This was manageable since I'd been part-way down that road already and there's lots of support in, for example, the form of Hopkins' course notes.  Then, in early May I spent some time digging around for a construction of Morava $E$-theory, which led me to some conclusions:

The words "Morava $E$-theory" don't determine what object you're talking about; there's a whole bunch of slightly different Morava $E$-theories.
People frequently conflate Morava $E$-theory with (completed) Johnson-Wilson theory.  One source even claimed (without citation) that one was a finite free module over the other, and that I therefore shouldn't worry about the difference.

In the end, I didn't need to really know much about $E_n$ beyond a couple formal properties to work out the broad strokes of my computation, so I let the whole thing slide and pretended there existed a spectrum that did what I'd hoped.
However, I'm now getting to a point where understanding what I'm actually doing would be valuable.  In decreasing order of importance, can someone provide a reference that...

... constructs a family of Morava $E$-theories.  Any family would be a start!  I am particularly interested in one with a coefficient ring of the form $\mathbb{Z}_p[\![v_1, \ldots, v_{n-1}]\!][v_n^{\pm 1}]$.
... illustrates that $\mathrm{spf}\,(E_n)^* \mathbb{C}P^\infty$ is the universal deformation of $\mathrm{spf}\,K(n)^* \mathbb{C}P^\infty$ to formal groups over formal spectra of complete, local rings with residue field $\mathbb{F}_p[v_n^{\pm 1}]$.  The remark above about the comparison between Johnson-Wilson theory and Morava $E$-theory made me particularly uncomfortable in this respect; it's not clear to me that the formal group associated to Johnson-Wilson theory should be thought of as the universal deformation of the Honda formal group.  Clearing that up would be nice too.
... also shows that the reduction of the universal deformation to the "mod $p$" case exists as a spectrum, and the reduction map exists as a map of spectra.  That is, there is a complex-oriented, structured ring spectrum $E_n/p$ with coefficient ring $\mathbb{F}_p[\![v_1, \ldots, v_{n-1}]\!][v_n^{\pm 1}]$ whose associated formal group is the universal deformation of the Honda formal group to complete, local rings of characteristic $p$.
... also shows that the reduction of the universal deformation modulo the $n$th power of its maximal ideal exists as a spectrum, and the reduction map exists as a map of spectra.
... demonstrates this fact about $E_n$ being a finite free module over $E(n)$, at least for an appropriate interpretation of the symbol "$E_n$".

It may not be the case that points 3 and 4 are even true, but I'm hopeful.  Still, surely this is all catalogued somewhere!

REPLY [30 votes]: So far as what Morava E-theory should be: Morava E-theory always implicitly comes with a choice of a perfect residue field of positive characteristic and a formal group law of finite height over this field.  Sometimes people take a very specific formal group law, but there is no reason to be restrictive.  The underlying homotopy type of spectrum may not change in a very interesting way, but the multiplicative structure does (just as many rings that are different in interesting ways may have the same underlying abelian groups).  This actually becomes a more serious issue with things like $BP\langle n\rangle$ and the Johnson-Wilson theories because there are multiple possible inequivalent orientations that look basically the same when you write down the rings of homotopy groups.
Your points in order:

Constructing a family of Morava E-theories.  (The coefficient ring you list is implicitly the coefficient ring of a completed Johnson-Wilson theory, and that's only if I interpret your power series brackets as "invert vn and then complete" due to the grading issue.)  Johnson-Wilson theories and completed Johnson-Wilson theories, with the names you've given to the generators, all satisfy Landweber's criterion and so you can produce them the easy way as spectra via the Landweber exact functor theorem.  Morava E-theories are slightly more difficult; they're still Landweber exact, but to prove that you need to know that the universal deformation ring of a formal group law of finite height has a particular structure relative to the coordinates of its p-series.  However, if you want to construct any of these as genuine commutative ring spectra then you've got to use the Goerss-Hopkins-Miller theorem on the Morava E-theories, use it with its functoriality properties for the completed Johnson-Wilson theories, and you're out of luck for the uncompleted Johnson-Wilson theories except in a handful of very specific cases (some of which may require you to be slightly flexible about what "Johnson-Wilson" means).  For references for the Hopkins-Miller theorem in the associative case, there are of course Charles Rezk's Notes on the Hopkins-Miller theorem, although you have a far better source on hand.  The Goerss-Hopkins paper Moduli problems for structured ring spectra covers the obstruction theory for making these objects commutative.
So far as illustration goes, the method I was proposing in the previous paragraph started with the universal deformation ring and produced the E-theory as its manifestation, meaning that by the time you get here you've already finished this step.  Note that you have to be a little careful about the grading issue.  The coefficient ring of a Morava E-theory in its full, graded glory is the universal deformation ring of a formal group law equipped with a choice of generator of the relative cotangent space (a trivial torsor for the multiplicative group over the universal deformation of the formal group law).
The mod-p reduction exists as a spectrum, but not as a commutative ring spectrum.  Any commutative ring spectrum with $p=0$ in its homotopy groups is a module over the Eilenberg-Mac Lane spectrum ${\rm H}\mathbb{F}_p$ (this is because this Eilenberg-Mac Lane spectrum is actually the free algebra over the little 2-disks operad with $p=0$!) and these residue fields don't qualify.  You can show that these residue objects exist using obstruction theory; there are a number of references but let me specifically plug Vigleik Angeltveit's Topological Hochschild homology and cohomology of A-infinity ring spectra.
Modulo the n'th power of the maximal ideal is a little trickier and I would have to go check to see if the obstruction theory necessarily worked out.  If you're willing to instead kill the n'th powers of the generators of the maximal ideal then the obstruction theory from the previous point still works, because you are killing a regular sequence of generators.  Again, this only produces associative objects rather than commutative ones.
The fact that $E_n$ is a finite free module over the completed $E(n)$ is a formal consequence of the computation of the homotopy groups of both involved.  The left-hand object has homotopy groups
$$
\mathbb{Z}_{p^n}[\![u_1,\ldots,u_{n-1}]\!][u^{\pm 1}]
$$
with $|u_i| = 0, |u| = 2$ and the right-hand object has homotopy groups as the subring
$$
\mathbb{Z}[v_1,\cdots,v_n,v_n^{-1}]^\wedge
$$
where $v_i = u^{p^i - 1} u_i$ for $1 \leq i < n$, $v_n = u^{p^n - 1}$, and the completion is taken with respect to the intersection of the maximal ideal with this subring.  Once you have figured out what this entails, you find that the homotopy groups are a finite free module and as a consequence the spectrum itself is a finite free module.<|endoftext|>
TITLE: infinitely many linear equations in infinitely many variables
QUESTION [7 upvotes]: Let $(a_{mn})_{m,n\in\mathbb{N}}$ and $(b_m)$ be sequences of complex numbers.We say that $(a_{mn})$ and $(b_m)$ constitute an infinite system of linear equations in infinitely many variables if we seek a sequence $(x_n)$ of complex numbers such that $\forall m\in\mathbb{N}:$ $\sum_{n=1}^{\infty}a_{mn}x_n=b_m$. Note that in general the order of summation matters. 
I am sort of a undergraduate student with focus on number theory and have some background in functional analysis (2 semesters functional analysis, 1 semester non-linear functional analysis, 1 semester operator algebras, 2 semesters PDEs), so I am sort of a becoming number-theorist with bias for functional analysis :-) That is also why I am fascinated by the above defined object as a sort of natural extension of a practical problem from linear algebra.
We have never dealt with this type of objects and I wasn´t able to find much on google that I could start something with, maybe partly because I have searched in the wrong way. That is why I have a request if you could recommend some introductory literature focused on such infinite systems of linear equations in infinitely many unknowns over $\mathbb{C}$.
Thanks in advance!

REPLY [2 votes]: I would recommend taking a look at Hardy's "Divergent Series" it has quite a lot of nice ideas, in particular, I recall seeing exactly that example of a system of infinite equations in infinite unknowns related to fourier series.<|endoftext|>
TITLE: How come nowhere dense subsets implies discrete?
QUESTION [5 upvotes]: Hi, I am reading an article and have encountered a remark in a proof which is not clear to me.
Maybe someone can help?
The proposition is:
Let X be a topological space without isolated points having countable $ \pi $-weight and such that every nowhere dense subset in it is closed. Then it is a Pytkeev space.
Here is the begining of the proof:
Let $ x \in  Cl(A) \setminus A$. Then $ x \in Cl(Int(Cl(A))) $, because every nowhere dense set is closed (and hence discrete)...
The thing which is not clear to me:
Why can one conclude that every nowhere dense closed set is discrete? Suppose I take the set $ \mathbb N$ with the cofinite topology. Then the finite sets are closed and nowhere dense. But as far as I undesrtand they are not discrete since every open set in the topology that contains a finit set also has to contain other points since it is infinite.
Can somone see what am I missing?
The definition of a Ptkeev space:
Let X be a topological space. 
A point x is called a Pytkeev point if whenever $ x \in \overline {A\setminus{x}}$, there exists a countable $ \pi $-net of infinite subsets of A.  If every point of a space is a Pytkeev point then the space is called a Pytkeev space.
Thanks!

REPLY [5 votes]: Because a nowhere dense set minus a point is nowhere dense, the hypothesis implies that points are relatively open in nowhere dense subsets. Perhaps more to the point, every subset of a nowhere dense set is nowhere dense, hence closed by hypothesis.
As for your question about cofinite topologies: Finite subspaces of cofinite topological spaces are discrete, because finite subsets are closed. (The same is true in any $T_1$ space.) This doesn't mean that subsets of finite sets are open, but rather that they are relatively open.<|endoftext|>
TITLE: Computing the Mertens function
QUESTION [11 upvotes]: I wonder if anybody can help me with this problem.
I'm trying to compute the Mertens function for large $n$. The most obvious algorithm is just to compute all primes up to $\sqrt{n}$ and then to sieve. That takes at least an order of $n\log n$ operations, and really even more. 
The most recent article that I could find that discusses methods to compute the function directly is dated 1994, and it proposes to do exactly that.
Are there any known algorithms that let you compute Mertens faster than by sieving? I know that $\pi(n)$ can be computed in $O(n^{2/3})$, I looked into that algorithm but it does not seem to be easily adaptable to my task. 
Alternatively, I could use an algorithm to compute $M(n+dn)-M(n)$ for $dn\ll n$ (say $dn\sim \sqrt{n}$ ) in $O(\sqrt{n})$ time or less.

REPLY [4 votes]: Not to toot my own horn (joint with L. Thompson), but in https://arxiv.org/pdf/2101.08773.pdf we give an elementary algorithm running in time $O_\epsilon(x^{3/5} (\log x)^{3/5+\epsilon})$.<|endoftext|>
TITLE: Behavior of the projective dimension of modules in a continuous chain of extensions
QUESTION [5 upvotes]: Let $R$ be an arbitrary ring. Let $D$ be the class of $R$-modules of projective dimension less than or equal to a natural number $n$. If $L$ is the direct union of a continuous chain of submodules ${L_{\alpha},\alpha < \lambda}$ for some ordinal number $\lambda$ (this means that $L=\bigcup_{\alpha}L_{\alpha},\ L_{\alpha}\subseteq L_{\alpha'}$ if $\alpha \leq \alpha' <\lambda$ and $\ L_{\beta}=\bigcup_{\alpha <\beta} L_\alpha$ when $\beta < \lambda $ is a limit ordinal) with $L_{0}\in D$ and $L_{\alpha +1}/L_{\alpha}\in D, \forall \alpha<\lambda,$ can one show that $L \in D$?
PS: We know that when $R$ is a perfect ring, then $D$ is closed under direct limits, then we can prove the above by transfinite induction. But if $R$ is not perfect, how can we show that?

REPLY [2 votes]: When $n=0$, the module $L$ is isomorphic to the direct sum of $L_{\alpha+1}/L_\alpha$, so it is projective.  For an arbitrary natural number $n$, argue by induction.  Let $P_\alpha$ denote the free $R$-module with generators indexed by the elements of $L_\alpha$ and $P$ denote the free $R$-module with generators indexed by the elements of $L$.  Then $P$ is the direct union of $P_\alpha$.  There is a natural morphism of $R$-modules $P\to L$ mapping $P_\alpha$ surjectively onto $L_\alpha$.  Let $M$ be the kernel of the morphism $P\to L$ and $M_\alpha$ be the kernel of the morphism $P_\alpha\to L_\alpha$.  Then $M$ is the direct union of $M_\alpha$ and the quotient modules $M_{\alpha+1}/M_\alpha$, being isomorphic to the kernels of the morphisms $P_{\alpha+1}/P_\alpha\to L_{\alpha+1}/L_\alpha$, have projective dimensions not exceeding $n-1$.<|endoftext|>
TITLE: Characteristic power series for maps of E_{\infty} ring spectra
QUESTION [20 upvotes]: Let me admit right at the outset that I have a very superficial outsider's knowledge of homotopy theory. Nevertheless, I was trying to gain some understanding of Hopkins' ICM lecture 'algebraic topology and modular forms.' 
In section 6, he mentions two constructions. To a map 
$$\phi: MSpin\rightarrow KO$$
of $E_{\infty}$ ring spectra, he associates a characteristic power series $$K_{\phi}(x)\in \mathbb{Q}[[x]].$$ Similarly, to an $E_{\infty}$-map 
$$\psi: MO\langle 8\rangle \rightarrow tmf,$$
he associates a power series $$K_{\psi}(x)\in MF_{\mathbb{Q}}[[x]],$$ where $tmf$ is the topological modular form spectrum and $MF_{\mathbb{Q}}=MF\otimes _{\mathbb Z}\mathbb Q$ is the ring of modular forms with rational coefficients.
I wonder if someone could give a brief outline of how these associations are carried out. I presume it is something elementary having to do with the homotopy groups of $MSpin$ and $MO\langle 8\rangle$, but I don't quite have the resources right now to track these down. 
As usual with questions of this sort, I'm sure my level of ignorance is incongruous with the words I am employing already, but thank you in advance for any tolerant answers or references.
Added:
Maybe I should  summarize the point of this question for fellow number-theorists who are too busy to look into the paper. In the notation above, one associates to $\phi$ a characteristic sequence 
$$b(\phi)=(b_2, b_4, b_6,\ldots)$$
via the formula
$$\log(K_{\phi}(x))=-2\sum_{n>0} b_n\frac{x^n}{n!}.$$
Incredibly, this procedure sets up a bijection:
homotopy classes of $E_{\infty}$ maps from $MSpin$ to $KO$ $\leftrightarrow$ the set of sequences of rational numbers $(b_i)$ as above that satisfy
(1) $b_n\equiv B_{n}/n \ \ \mod \mathbb{Z}$, where the $B_n$ are the Bernouilli numbers;
(2) for each odd prime $p$ and $p$-adic unit $c$,
$$m\equiv n \ \mod p^k(p-1) \Rightarrow (1-c^n)(1-p^{n-1})b_n \equiv (1-c^m)(1-p^{m-1})b_m \ \mod p^{k+1};$$
(3) for each $2$-adic unit $c$,
$$m\equiv n \ \mod 2^k \Rightarrow (1-c^n)(1-2^{n-1})b_n \equiv (1-c^m)(1-2^{m-1})b_m \ \mod 2^{k+2}.$$
In the case of the homotopy classes of maps from $MO\langle 8\rangle$ to $tmf$, one gets similar congruences involving  Eisenstein series instead of their constant terms. Incidentally, perhaps these congruences imply the ones above?

REPLY [16 votes]: In short, the series $K_\phi$ is the "Hirzebruch characteristic series" which arises in the construction/calculation of genera, and in Hirzebruch-Riemann-Roch.  The first few chapters of Manifolds and modular forms by Hirzebruch et al. describe the classical version of this pretty well.
If I have a one dimensional formal group law $F$ over a ring $A$, then over $A_{\mathbb{Q}}$ there is an isomorphism $\mathrm{exp}_F: G_a\to F$ with the additive formal group.  Let $K(x)=x/\mathrm{exp}_F(x)$.
Now suppose $R$ is a "complex orientable cohomology theory", which means we are given a suitable isomorphism of rings $R^*(CP^\infty)\approx \pi_*R[[x]]$.  Such a theory has an associated formal group law $F$ (induced by the map $CP^\infty\times CP^\infty\to CP^\infty$ which classifies tensor product of line bundles), and thus there is an assocated series $K(x)=x/\mathrm{exp}_F(x)$ in $\pi_*R_\mathbb{Q}[[x]]$.
It turns out that a map of ring spectra $\phi:MU\to R$ corresponds exactly to giving a complex orientation of $R$.  By Thom, elements of $\pi_*MU$ correspond to cobordism classes of stably-almost-complex manifolds, and there is a standard calculus due to Hirzebruch of calculating the effect of the map $\pi_*MU \to \pi_*R_{\mathbb{Q}}$ using $K(x)$, which I might as well call $K_\phi(x)$, since it depends on $\phi$.  The formula (if I remember correctly), is that, if $[M]\in \pi_*MU$ is the class corresponding to a manifold of dimension $2n$, then 
$$
\phi(M) = \langle K_\phi(x_1)\dots K_\phi(x_n), [M] \rangle,
$$
where the $x_i$ are the "chern roots" of the tangent bundle of $M$, and $[M]\in H_{2n}M$ is the fundamental class.
There is a "universal example" of a $K_\phi$, corresponding to the identity map $\phi\colon MU\to MU$.  It turns out that $\pi_*MU_{\mathbb{Q}}$ is a polynomial ring on the coefficients of $K_\phi$, so that $K_\phi(x)=\sum a_{i-1}x^i$ (with $a_0=1$) and $\pi_*MU_{\mathbb{Q}}=\mathbb{Q}[a_1,a_2,\dots]$.  (I'll need this later.)
In his talk, Mike isn't talking about complex orientations, but rather orientations with respect to $MSpin$ or $MO\langle 8\rangle$ (instead of $MO\langle 8\rangle$, we call it $MString$ these days, for some reason).
There is a map of ring spectrum $MU\to MSO$, induced by the apparent homomorphisms $U(n)\to SO(2n)$ of Lie groups.  There is also a map $MSpin\to MSO$, induced by the double cover of lie groups.  Although $MSpin\neq MSO$, we have that $\pi_*MSpin_{\mathbb{Q}}\to \pi_*MSO_{\mathbb{Q}}$ is an isomorphsism.  Thus, a map $\phi\colon MSpin\to R$ induces
$$ \pi_*MU_{\mathbb{Q}}\to \pi_*MSO_{\mathbb{Q}}\approx \pi_*MSpin_{\mathbb{Q}}\to \pi_*R_{\mathbb{Q}},$$
and we can get $K_\phi(x)$ from this.
The $MO\langle 8\rangle$ case is a little trickier.
There is a map $MU\langle 6\rangle \to MO\langle 8\rangle$, so a ring spectrum map $\phi\colon MO\langle 8\rangle\to R$ gives rise to a map
$$ MU\langle 6\rangle_{\mathbb{Q}} \to R_{\mathbb{Q}}.$$
On the other hand, the effect of the map $MU\langle 6\rangle\to MU$, on homotopy groups tensored with $\mathbb{Q}$, is
$$ \mathbb{Q}[a_3,a_4,\dots]\to \mathbb{Q}[a_1,a_2,a_3,\dots].$$
So a map $\phi\colon MO\langle 8\rangle\to R$ gives us elements $\phi(a_i)\in \pi_{2i}R$ for $i\geq3$, which we can use as the coefficients of a series $K_\phi(x)\in \pi_*R_{\mathbb{Q}}$.
I must point out: there is actually an error in the statement of (2) and (3) given in Mike's talk.  What he writes down are the "Kummer congruences"; but what one really needs to require are the "generalized Kummer congruences", which are basically the collection of all possible $p$-adic congruences involving Bernoulli numbers, not just the ones listed in (2) and (3).  This comes from the theory of the "Mazur measure": the generalized Kummer congruences imply that the sequence $b_n(1-p^{n-1})(1-c^n)$ can be interpolated to a function $f$, so that $f(n)$ for $n$ an integer is the moment of a measure on $\mathbb{Z}_p^\times$.  With (2) and (3) replaced by "interpolates to the moments of a measure on $\mathbb{Z}_p^\times$", the result is correct.
Finally: There is a writeup of this at http://www.math.uiuc.edu/~mando/papers/koandtmf.pdf, which may or may not be of any use to you!<|endoftext|>
TITLE: Bundle-to-function correspondence
QUESTION [9 upvotes]: To a morphism of sets $f\colon E\to B$ with finite fibers, one may assign a function $$|f^{-1}|\colon B\to{\mathbb N}$$ sending an element $b\in B$ to the cardinality of the fiber $f^{-1}(b)$.

To a proper morphism of manifolds imbedded in Euclidean space $f\colon E\to B$ one may assign a function $$|f^{-1}|\colon B\to{\mathbb R}$$ by sending an element $b\in B$ to the volume of the fiber $f^{-1}(b)$.  ...  Or we could assign the dimension of the fiber (valuing in ${\mathbb N}$), or we could assign the number of points in the fiber (valuing in ${\mathbb N}\cup${$\infty$}).

If $M$ is a commutative monoid (thought of as a set with addition operation), let $Set_{/M}$ denote the category of sets equipped with a map to $M$.  Then to a morphism $f\colon E\to B$ in $Set{/M}$, where $f$ has finite fibers, one may assign a function $$|f^{-1}|\colon B\to M$$ sending an element $b\in B$ to the sum of the elements in the fiber.  ...  Or we could use a possibly non-commutative monoid $M$ but replace Sets-over-$M$ with Sequences-over-$M$.

To a discrete op-fibration of categories $f\colon E\to B$ one can assign a functor $$|f^{-1}|\colon B\to Set$$ sending $b$ to its fiber.


Question: What do all these have in common?  More specifically, where can I find some category-theoretic way to understand situations of this type?  The "type" here seems to be something like: a "finite type" morphism in a concrete category with "valuation" can be converted into a "valuation" on the base.  
One might call it "integration along the fiber" or "gysin" or "sheaf-to-function correspondence."  What is the generalized setup?  What is the notion of "valuation" or "measure" supposed to be?

REPLY [5 votes]: This answer comes by private correspondence from Mathieu Anel.  I record it here, with some minor clean-up, because it's exactly what I was looking for. --David Spivak

Here are some thoughts about what I've understood of your question.
We suppose that $f:E\to B$ is a (kind of) fibration.
If there exists a moduli object $M$ for fibrations, $f$ is classified by a function $[f]:B\to M$ (I've changed your notation).
Now I assume you want the map $f\mapsto [f]$ compatible with addition and products of fibers like $[f\times_Bg]=[f][g]$.
Remark : $M$ has a monoidal (or even a rig) structure iff the fibrations have a monoidal (or rig) structure. 
for example, discrete and Grothendieck fibrations are stable by disjoint sums and pull-backs and their moduli objects ($Sets$ and $Cat$) are rigs. 
other example, the direct sum and tensor product of vector bundles. 
in each case $f\mapsto [f]$ is always compatible with the operations (it's tautological).
Now compose with any rig morphism $g:M\to R$ where $R$ is any rig (e.g. a ring) and $g[f]$ is a "measure."  And, $[f]$ is the "universal measure."
Examples:

$M$ = moduli of (finite) sets $M\to {\mathbb N}$ = cardinality
$M$ = moduli for (finite dimensional) vector bundles $M\to {\mathbb N}$ = dimension (this one is compatible only with tensor product)
$M$ = moduli of compact riemannian manifolds $M\to {\mathbb R}$ = volume
K-theory : $M$ is anything that is a rig, and $g:M\to \Pi_0(M)^+$ (where $^+$ is the additive group completion)
$M$ = moduli for l-adic sheaves, $M\to R$ = trace of the frobenius operator<|endoftext|>
TITLE: Is a topology determined by its convergent sequences?
QUESTION [24 upvotes]: Just a basic point-set topology question: clearly we can detect differences in topologies using convergent sequences, but is there an example of two distinct topologies on the same set which have the same convergent sequences?

REPLY [3 votes]: I like the following example. Let $D$ be an infinite discrete space and $\beta D$ its Stone-Čech compactification. Then a sequence in $\beta D$ converges if and only if it is eventually constant. Thus convergent sequences do not distinguish between the compact topology of $\beta D$ and the discrete topology on its underlying set. Obviously - the same holds for every refinement of $\beta D$.<|endoftext|>
TITLE: V. A. Gorbatov and 4-color problem
QUESTION [13 upvotes]: V. A. Gorbatov claims that he solved 4-color problem in  1964, and he repeats this claim in his book which appears in 2000. (The paper and the book are in Russian; the book is also translated to Spanish --- see refs below.)  Obviousely, I doubt that it is true, but I do NOT want to read the book...
Question: Did anyone checked this proof?
Why do I care. This claim appears in Russian wikipedia and I would be happy to have a good reason to remove it.
(At the moment ru.wikipedia states Gorbatov's claim and says that nobody checked his proof.)

Горбатов В. А. Фундаментальные основы дискретной математики.
Gorbatov, V. A. Fundamentos de la matemática discreta.  
Горбатов, В. А. О минимальной раскраске графа // Доклады НТК. МЭИ, Подсекция вычислительной техники. --М.: МЭИ, 1964. -С. 17.

Summary of answers: I have got few answers, roughly stating that there is no sign that Gorbatov is a serious mathematician. I agree with that, but it is not an answer to my question...

REPLY [5 votes]: Check out the page http://community.livejournal.com/science_freaks/1639427.html.
The whole thing seems bizarre. The MathSciNet review of the book you mention says nothing about a claimed proof of the 4-color theorem. The earliest MathSciNet reference to a paper by any Gorbatov is from 1970; nothing from 1964. Did he write a proof and stuff it in a drawer or something? He's listed at http://www.newportuniversity.eu/faculty.php as having "degrees of the Doctor and Grand Doctor of Philosophy and a Rank of the Full Professor of the World Information Distributed University." That seems to be some strange academic structure between Belgium and Russia. (The description of certification of diplomas at WIDU on their page http://www.webspawner.com/users/eduardevreinov/index.html is hilarious: 
"Organization of the preparation and defense of theses in WIDU shall correspond to the common rules applied in the most of famous universities of the world. The Harvard and Yell Universities can be named as examples of these universities... The important difference of WIDU is an opportunity to study and defend thesis in the native language... A more perfect credit system is available, compared to the famous universities.") 
Gorbatov's own wikipedia page (the fifth person listed at http://ru.wikipedia.org/wiki/Горбатов) says nothing about this 4-color "achievement" and it also shows you're not exactly going to be getting any complaints from him: he passed away in June. 
All in all, the evidence supporting his claim is very weak, so it seems fair to cut it.

REPLY [2 votes]: From Gorbatov's book this seems to be the original reference: 
Горбатов, В. А. О минимальной раскраске графа // Доклады НТК. МЭИ, 27 марта - 10 апреля 1964 г. Подсекция вычислительной техники. --М.: МЭИ, 1964. -С. 17.
I'm not familiar enough with Russian journals to make sense of it. What is "НТК. МЭИ"? And is this a peer-reviewed journal or is this more in the nature of a technical report or unrefereed conference proceedings?<|endoftext|>
TITLE: Is the complex moduli of Quintic Calabi-Yau toric?
QUESTION [8 upvotes]: Complex moduli space (or Teichmuller space) of a Quintic Calabi-Yau 3-fold is a 
101-dimensional complex orbifold. Does it have a toric structure?

REPLY [7 votes]: The complex moduli space does not admit a toric strucutre, since the orbifold fundamental group of a toric orbifold must be abelian. Indeed, $\pi_1(\mathbb C^*)^n$ surjects on the orbifold fundamental group. Also, the orbifold stabisier of each point on a toric orbifold 
is a finite abelian group. At the same time the stabiliser of the quintic $\sum_i z^5=0$
is a non-comutative group. Also I am sure that the orbifold fundamental group of the moduli space of quintics contains free (non-abelian) subgroups, but I don't know how to prove it. 
Also it should be true that the Tiechmuller space is not algebraic. It least this happen in lower dimensions for cubics in $\mathbb CP^2$ and for quartics in $\mathbb CP^3$.
In the first case the Theichmuiller space is a disk, and in the second it is
a hermitian domain of type IV. Moduli spaces of polarised K3 are discussed here
for example, here:
http://people.bath.ac.uk/masgks/Papers/k3moduli.pdf<|endoftext|>
TITLE: Are orbifold singularities canonical?
QUESTION [5 upvotes]: This is a direct consequence of my previous question: Extending group actions on varieties
In his answer, inkspot said that group actions can be extended if the variety has ample canonical class and is smooth, but mentions that canonical singularities can be allowed.  Now, my situation doesn't involve a smooth variety, but instead I have an orbifold.
However, I know that for surfaces, the canonical singularities are the duVal singularities, and are all orbifold points (they're all $\mathbb{C}^2$ modulo a finite subgroup of $\mathrm{Sl}_2$.)
Now, I've not studied general singular points of surfaces, so I could be wrong already with surfaces, but are orbifold singularities canonical?

REPLY [11 votes]: For quotient singularities there is the so-called Reid-Tai criterion to check whether the singularity is canonical or not.
Suppose $G$ is a finite subgroup of $GL_n(\mathbb{C})$ without quasi-reflections. Let $m=|G|$ and fix a primitive $m$-th root of unity $\zeta$. Let $g\in G$ and let $0\leq a_i < m$ be such that $\zeta^{a_1},\dots,\zeta^{a_n}$ are the eigenvalues of $g$. Then the Reid-Tai sum of $g$ is defined as $\Sigma(g):=1/m(\sum a_i)$. 
The Reid-Tai criterion states that $\mathbb{C}^n/G$ has a canonical singularity at 0 if and only if $\Sigma(g)\geq 1$ for all $g\in G, g\neq id$.
(Taking a quotient by quasi-reflections does not yield a singularity, the Reid-Tai sum depends on the choice of $\zeta$, the criterion does not.)
For more on this see e.g. M. Reid's Young person's guide to canonical singularities (Bowdoin 1985 proceedings).<|endoftext|>
TITLE: Length of I/I^2 versus Ann(I)/Ann(I)^2 in Artinian rings.
QUESTION [13 upvotes]: Suppose that $(A,\mathfrak{m})$ is a local Artinian ring.
If $A$ is Gorenstein, then $A$ admits a dualizing functor
on finite length modules defined by $D(M):= Hom_A(M,A)$ which preserves
lengths.  If $M$ is a finite length $A$ module, let $\mathrm{length}(M)$
denote the length of $M$.
If $I$ is an ideal of $A$ and $J = \mathrm{Ann}(I)$, and
$A$ is Gorenstein, it is not hard to show that
$I = \mathrm{Ann}(J)$.
Question: Suppose that $A$ is a local Gorenstein Artinian ring, $I$ is an ideal,
and $J = \mathrm{Ann}(I)$. If $I$ is principal, is it the
case that
$$\mathrm{length}(J/J^2) \ge^{?} \mathrm{length}(I/I^2)$$
If so, can one characterize when equality holds?
Remarks:
1. If $J$ is also principal, then (reversing $I$ and $J$) one
predicts there should be an equality. One can show that this is the case.
However, equality sometimes occurs without $J$ being principal, as can be seen from
the example:
$A = \mathbf{F}_2[x,y]/(x^3,y^3)$, $I = (x^2+y^2)$, $J = (x^2 + y^2,xy)$, with
$\mathrm{length}(I/I^2) = \mathrm{length}(J/J^2) = 4$.
2. My reasons for believing that the answer to the question is "yes" are somewhat obscure, and I would not be entirely surprised if it turns out to be false.
3. I do not know a counterexample to the claim that $\mathrm{length}(J/J^2) \ge
\mathrm{length}(I/I^2)$ even without the Gorenstein hypothesis. I would
be interested in seeing such a counterexample, if one exists (although I'm more interested in the Gorenstein case).
EDIT: As usual with bounty questions, this has been "answered" although not yet completely solved --- further comments still welcome.

REPLY [6 votes]: The purpose of this separate answer is to address the other part of the question, when does equality happen? (also, my first answer is getting too long).
I.  As FC pointed out, equality happens if $J$ is principal (this can be proved using Fact 1 in my other answer). Consider the following:

$J$ is principal.
$R/I$ is Gorenstein.
$R/I$ is a complete intersection.

Then $(3)\Rightarrow (2) \Leftrightarrow (1)$. If $R$ is a complete intersection, then $(2)\Rightarrow (3)$. 
Sketch of proof: equivalent of (1) and (2) can be proved using the Facts of my other answer. For example, since $J$ is isomorphic to the canonical module of $R/I$, $R/I$ is Gorenstein forces $J$ to be principal. The last statement follows from a cute result by Kunz that an almost complete intersection is never Gorenstein! (E. Kunz, Almost complete intersections are not Gorenstein, J. Alg. 28(1974), 111–-115)
II. By my 10/04/10 update in the answer above, if $R$ has SLP (for example when $R$ is a monomial complete intersection, and conjecturally for all complete intersections), $k$ has char. $0$ and the socle degree is odd, then equality  occurs for $I$ generated by a general linear form (it may or may not happen when the socle degree is even).
For example, equality happens  $I=(x+y+u+v)$ in $R=\mathbb Q[x,y,u,v]/(x^6,y^7,u^7,v^7)$ (socle degree $23$) but not for $R=\mathbb Q[x,y,u,v]/(x^7,y^7,u^7,v^7)$ (socle degree $24$). Note that for such $I$, $J$ will typically not be principal unless if the number of variables is $2$, see part (III) below.
III. Here are some examples in positive characteristic, with the help of Macaulay 2 (thanks to Branden Stone for helping me with programming). Numerical evidences suggested:

Conjecture: Let $R=\mathbb Z/(p)[x,y]/(x^N,y^N)$ and $I=(x^n+y^n)$ with $N\geq n$. Then equality happens if and only if 
  there is an integer $k$ such that $N/n=k$ or $kp-1\leq N/n \leq kp+1$. 

Remark: the first condition is not surprising. In $2$ variables, if $n$ divides $N$  then $R/I$ is a complete intersection, so by part (I), equality happens. I think this one can be proved but did not have enough motivation to go through the details. The point is that $R/J$ is Gorenstein (Fact 3 above). In $2$ variables, this is the same as complete intersection, so $R/J$ is $k[x,y]/(f,g)$ so everything can be written down explicitly. Note that the length of $R/J$ is the product of degrees of $f,g$ by Bezout theorem.
I did not see any clear pattern when you have more variables. For example, when $R=\mathbb Z/(3)[x,y,z]/(x^N,y^N,z^N)$ and $I=(x^2+y^2+z^2)$, the values of $N$ between $2$ and $100$ such that equality fails to occur is  $3,9,27,33,75,81,99$. If you set the field to be $\mathbb Z/(5)$, those values will be $3,5,15,17,23,25,75,77,83,85,95,97$!<|endoftext|>
TITLE: What is known about the Gaussian measure of the unit ball in a Hilbert Space?
QUESTION [11 upvotes]: Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$.  What do we know about $\mu(B(0,1))$, where $B(0,1)$ is the unit ball w.r.t the norm?
This seems to me like a fundamental question but I cannot seem to find anything.  Any information/references would be most appreciated.
EDIT: A related question which is of interest to me:  Do there exist asymptotically tight bounds to 
$\int_{||u||> K}||u||^2 \mu(du)$?

REPLY [3 votes]: I think you can see the articles
entitled "concentration of measure phenomenon ". The idea is as
follows: Let $(X,d,\mu)$ be a metric measure space, such as
$\mu(X)=1$. Let
$$\alpha(\epsilon) = \sup {\mu(X \backslash A_\epsilon) \  | \ 
\mu(A) = 1/2 }$$ where
$$A_\epsilon = { x \  | \  d(x, A) < \epsilon }$$ is
the $\epsilon$-extension of a set $A$. The function $\alpha(.)$ is
called the concentration rate of the space $E$. The following
equivalent definition has many applications:$$\alpha(\epsilon) =
\sup { \mu( { F >= M + \epsilon }) },$$ where the supremum is
over all $1$-Lipschitz functions $F: X \to \mathbb{R}.$ For example
the median (or Levy mean) $M = \mathop{Med}(F) $ is defined by the
inequalities $$\mu ( F \geq M ) >= 1/2, \  \mu ( F <= M ) \geq
1/2.$$ More precisely, the space $X$ exhibits a concentration
phenomenon if $\alpha(\epsilon)$ decays very fast as $\epsilon$
grows. More formally, a family of metric measure spaces
$(X_n,d_n,\mu_n)$ is called a Levy family if the corresponding
concentration rates $\alpha(\epsilon)$ satisfy
$$\forall \epsilon > 0 \ \  \alpha_n(\epsilon) \to 0,$$ and a normal
Levy family if $$ \forall \epsilon \to 0 \ \  \alpha_n(\epsilon) =
O(\exp(-C n \epsilon^2))$$ for $C$ some positive constant. the last
inequality is obtained bay applying the "Hoeffding inequality" and in
the  case of  Hilbert space with concentration in small balls we do :
$$\forall (x_1,x_2)\in X^2, \ \  d(x_1,x_2)=\|x_1,x_2\|< r.$$<|endoftext|>
TITLE: Complex and Elementary Proofs in Number Theory
QUESTION [19 upvotes]: The Prime Number Theorem was originally proved using methods in complex analysis.  Erdos and Selberg gave an elementary proof of the Prime Number Theorem.  Here, "elementary" means no use of complex function theory.
Is it possible that any theorem in number theory can be proved without use of the complex numbers?
On the one hand, it seems a lot of the theorems using in analytic number theory are about the distributions of primes.  Since the Prime Number Theorem has an elementary proof, this might suggest that elementary proofs exist in other cases.  
On the other hand, the distribution of primes is intimately related to the zeros of the Riemann Zeta function.  Perhaps the proofs of other statements in analytic number theory require more direct references to the Riemann Zeta function.
This topic is more of a fascination for me, as I am not a number theorist.  I would be interested if there are other examples of elementary proofs of theorems originally proved with complex analytic methods.

REPLY [3 votes]: Here is a recent paper by Dimitris Koukoulopoulos, which obtains the strongest known form of the Prime Number Theorem without heavy use of complex analysis. One can find the symbol $i$ in the paper, but it does not rely on the analytic continuation of the zeta function.
This is a direct extension of the "pretentious" methods of Granville and Soundararajan, mentioned in an answer by Thomas Riepe.<|endoftext|>
TITLE: Is the solution bounded Diophantine problem NP-complete? 
QUESTION [11 upvotes]: Let a problem instance be given as $(\phi(x_1,x_2,\dots, x_J),M)$ where $\phi$ is a diophantine equation, $J\leq 9$, and $M$ is a natural number. The decision problem is whether or not a given instance has a solution in natural numbers such that $\sum_{j=1}^J x_j \leq M$. With no upper bound M, the problem is undecidable (if I have the literature correct). With the bound, what is the computational complexity?  If the equation does have such a solution, then the solution itself serves as a polytime certificate, putting it in NP. What else can be said about the complexity of this problem?

REPLY [18 votes]: A particular quadratic Diophantine equation is NP-complete.
$R(a,b,c) \Leftrightarrow \exists X \exists Y :aX^2 + bY - c = 0$
is NP-complete.  ($a$, $b$, and $c$ are given in their binary representations.  $a$, $b$, $c$, $X$, and $Y$ are positive integers).
Note that there are trivial bounds on the sizes of $X$ and $Y$ in terms of $a$, $b$, and $c$.
Kenneth L. Manders, Leonard M. Adleman: NP-Complete Decision Problems for Quadratic Polynomials. STOC 1976: 23-29<|endoftext|>
TITLE: Definition of an E-infinity algebra
QUESTION [42 upvotes]: Can anyone give me a plain-and-simple
definition of an E-infinity algebra without using
the words "operad," "ring spectrum," or
"stable homotopy"?
Sorry, but I honestly couldn't find it using
all on-line resources at my disposal.
Thanks!

REPLY [8 votes]: A completely explicit definition that works over any ring is given as Proposition 18 in the following preprint of Malte Dehling and Bruno Vallette that was posted on the arXiv today. http://arxiv.org/abs/1503.02701<|endoftext|>
TITLE: Relation between characteristic variety and support of D-Module
QUESTION [8 upvotes]: Given a coherent $\mathcal{D}_X$-Module $M$, one can assign to it its characteristic variety 
$$ch(M)\subseteq T^*X$$
or one could look at its support 
$$supp(M)\subseteq X$$
as an $\mathcal {O}_X$ module. Is there a relation between these two spaces?
Edit: Is it true, that the characteristic variety of a holonomic $\mathcal{D}_X$-Module is the conormal space of its support?

REPLY [7 votes]: I'm not an expert either but, at least for holonomic D-modules, the relation you're looking for should be 
$$ 
  supp(M) = ch(M) \cap T^*_X X
$$ 
where $T^*_XX$ is the zero section of the cotangent bundle identified with $X$.
You can check it in the basic examples: $X = \mathbb{C}$, $M$ the trivial bundle or the $\delta$-module at $x\in X$. The corresponding $ch(M)$ is $T^*_XX$ or $T^*_xX$. The general case should be an easy exercise in linear algebra.<|endoftext|>
TITLE: Smooth algebraic varieties with smooth Kahler quotients.
QUESTION [6 upvotes]: Let $V$ be a smooth algebraic variety defined over complex numbers. Suppose that $G$ admits a free discrete  action on $V$ so that $V/G$ is compact and Kahler (or algebraic). Is it ture that $G$ is virtually abelian (i.e., contains an abelian subgroup of finite index)?
If we don't ask $V/G$ to be algebraic (Kahler) there are lots of counterexamples, we can take 
a complex Lie group and quotient it by a co-compact lattice. But if $V/G$ is algebraic
the only example that I can imagine is when $V\cong M\times \mathbb C^n$, where M is compact and $G$ preserves the product. Are there other examples?  This question is related to 
Is the complex moduli of Quintic Calabi-Yau toric?
ADDED. Note that a compact complex nilmanifold, i.e. a complex manifold with a homogenious action of a nilpotent group is Kahler iff it is a trous, http://www.jstor.org/pss/2047375 . So a quotient of a complex nilpotent group by a co-compact subgroup never provides a counter-example.

REPLY [5 votes]: The following article http://arxiv.org/abs/1102.2762 of BENOÎT CLAUDON, ANDREAS HÖRING, AND JÁNOS KOLLÁR answers positively the question, provided $V/G$ is projective, $V$ is quasiprojective and $\pi_1(V)=0$.  Moreover, assuming abundance conjecture, they prove under the above conditions that $V$ is biholomophric to the product of $\mathbb C^n$ with a simply-connected variety.
Added. Interestingly, if we ask that $V/G$ is merely Kahler, the question seems to be open even for $V=\mathbb C^n$ for $n\ge 4$. It is related to (and follows immediately from) "Iitaka's conjecture" predicting that any such compact Kahler quotient  $\mathbb C^n/G$ has a finite cover bi-holomorphic to a torus. Iitaka conjecture is discussed in the article  "UNIFORMISATION IN DIMENSION FOUR:
TOWARDS A CONJECTURE OF IITAKA" of Horing, Peternell, and Radloff :  http://arxiv.org/abs/1103.5392<|endoftext|>
TITLE: Is it useful to consider cohomology of group representations?
QUESTION [5 upvotes]: In group representation theory, one attempts to explain and classify (some of) the modules over the group ring $k[G]$, for some field $k$. In group cohomology, one develops the machinery of the cohomology of groups out of the category of modules over the ring $\mathbb{Z}[G]$. In both cases, one has tensor products, restriction maps, induced modules, etc, and all these constructions look very similar in both cases, with $k$ replaced by $\mathbb{Z}$ in group cohomology.
My question is, might one develop a theory of cohomology of $G$-representations over a field $k$ analogous to the theory of cohomology of $G$-modules? Essentially, one would take a representation $M$, take an injective resolution of $M$ in the category of $k[G]$-modules, then find the fixed elements to get the cohomology. I believe this would be $\mathrm{Ext}_{k[G]}(k,M)$, where $k$ is given the trivial $G$-action.
Given that we know a lot about the structure of finite-dimensional representations of finite groups (character tables), what would the character table tell us about the cohomology? What might the cohomology tell us about the character table? What happens if we consider representations of infinite groups, such as Lie groups, algebraic groups, or Galois groups?

REPLY [4 votes]: You have already used the following fact in your statement:
For every representation $\rho: G\to \mathrm{Aut}_R(A)$ ($A$ an $R$-module) we can give $A$ the structure of an $R[G]$-module by $g\cdot a = \rho_g(a)$. Conversely, any torsion free $R[G]$-module $A$ admits a representation $\rho: G \to \mathrm{ Aut}_R(A)$ given by $\rho_g(a) := g\cdot a$.
So studying $R[G]$-modules IS studying representations of $G$. My answer would be if you want to study representations study $R[G]$-modules. $R[G]$-modules aren't like a cure-all, because they don't give you an action to start with. Does anyone know how to classify actions?

Studying a fixed $R[G]$-module is the same as studying a fixed representation. So it won't tell you anything about the character tables which involves different representations.
You can still use $R[G]$-modules to study representations of infinite groups and galois groups. You can use it to derive properties about galois groups acting on roots of unity or torsion points of elliptic curves. http://en.wikipedia.org/wiki/Kummer_theory
The wikipedia article http://en.wikipedia.org/wiki/Group_cohomology tells you that the Ext construction from resolutions is the same as the "standard" construction of group cohomology by cochain complexes.

I'm not sure if my answer was helpful. Hopefully it saved you some time. I don't think I understood what you meant about "taking the fixed elements to get the cohomology".<|endoftext|>
TITLE: A heart for stable equivariant homotopy theory
QUESTION [9 upvotes]: Let $G$ be a finite group. I wonder whether the following statement is true, known and written down:
There is a t-structure on the stable $G$-equivariant homotopy category such that the associated heart is isomorphic to the category of Mackey functors (on $B_G$).
I feel like someone has told me so, but I can't find a reference. Thanks for your help!

REPLY [12 votes]: Since G is finite, there is no problem with just repeating the proof in the 
case $G=e$, using $Z$-graded homotopy group functors on the orbit category. Take
$D^{\leq n}$ to be the spectra whose homotopy groups $\pi_q(X^H)$ are zero for $q>n$, 
and dually for $D^{\geq n}$.  The intersection for $n\leq 0$ and $n\geq 0$ consists 
of the Eilenberg-MacLane $G$-spectra $K(M,0)$ for Mackey functors $M$.   
Peter May<|endoftext|>
TITLE: Do we have non-abelian sheaf cohomology?
QUESTION [16 upvotes]: Lets $X$ be a complex manifold (algebraic variety), $N$ an integer, and consider the sheaf $F$ defined by:
$F(U)$ ={ holomorphic maps $f: U\rightarrow GL(N,\mathbb{C})$ } with multiplicative structure.
Can we define $H^i(X,F)$ ? Note that for $N=1$, this would be just $H^i(X,O_X^*)$.
(Please give reference for your claims)

REPLY [42 votes]: The quick reply is: not really for $i \gt 2$, and not in the way you perhaps expect for $i=2$, see below.

EDIT (Feb 2017): Debremaeker's PhD thesis [0] has now been translated into English and placed on the arXiv: Cohomology with values in a sheaf of crossed groups over a site, arXiv:1702.02128

The comment on Charles' answer about 'teaching you never to ask that question again' is partly true, partly not. The lesson to learn from Giraud is that really one does not use groups for coefficients of higher cohomology. For a start, Giraud's $H^2(X,G)$ is not functorial with respect to group homomorphisms $G\to H$! One also does not get the exact sequences that one expects (this is due to the lack of functoriality). But this is not a problem with his definition of the cohomology set, but a problem with what category you believe the coefficients lie in. This is because the coefficient object of Giraud's cohomology is actually the crossed module $AUT(G) = (G \to Aut(G))$, and the assignment $G \mapsto AUT(G)$ is not functorial. (Aside: Giraud contains lots of other important things on stacks and gerbes and sites and so on, so the book is not a waste of time by any means)
But little-known work by Debremaeker[1-3] from the 1970s fixed this up and showed that really the Giraud cohomology was functorial with respect to morphisms of crossed modules. This has been recently extended by Aldrovandi and Noohi [4] by showing that it is functorial with respect to weak maps of crossed modules aka butterflies/papillion.
It was realised by John E. Roberts (no relation) and Ross Street that the most general nonabelian cohomology has as coefficient objects higher categories. In fact, we now know that the coefficients of $n^{th}$ degree cohomology is an $n$-category (usually an $n$-groupoid, though), even when we are talking about usual abelian cohomology.
Everything I've talked about is just for groups etc in Set, but it can all be done internal to a topos, i.e. for sheaves of groups, and more generally a Barr-exact category (and probably weaker, but Barr-exact means that the monadic description of cohomology therein due to Duskin (probably going back to Beck) works fine).

[0] R. Debremaeker, Cohomologie met waarden in een gekruiste groepenschoof op een situs, PhD thesis, 1976 (Katholieke Universiteit te Leuven). English translation: Cohomology with values in a sheaf of crossed groups over a site, arXiv:1702.02128
[1] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. I, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 758 -- 764.
[2] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. II, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 765 -- 772.
[3] R. Debremaeker, Non abelian cohomology, Bull. Soc. Math. Belg., 29, (1977), 57 -- 72. 
[4] E. Aldrovandi and B. Noohi, Butterflies I: Morphisms of 2-group stacks, Advances in Mathematics, 221, (2009), 687 -- 773.<|endoftext|>
TITLE: Facts from algebraic geometry that are useful to non-algebraic geometers
QUESTION [75 upvotes]: A professor of mine (a geometric topologist, I believe) once criticized the core graduate curriculum at my institution because it teaches all sorts of esoteric algebra, but does not include basic information about Galois theory and algebraic geometry, which, according to him, are important even for non-algebraists.

What are some useful facts from algebraic geometry that are useful for non-algebraic geometers?  Ideally, the statements at least should be accessible without knowing much algebraic geometry.  

Edit: Please do not post results that are only relevant to people who already know massive amounts of algebraic geometry anyway.  In particular: Be very cautious about posting statements whose only applications are in number theory.
Example: Here is a basic statement that I have seen applied outside algebraic geometry, if not necessarily outside of algebra:
Let $U \subset \mathbb{C}^n$.  If there is some nonzero polynomial satisfied by every point of $\mathbb{C}^n \smallsetminus U$, then $U$ is dense in $\mathbb{C}^n$ (with the usual topology), and in fact contains a dense open subset of $\mathbb{C}^n$.
[Sketch of proof: Given any point $p \in \mathbb{C}^n$, find a complex line $L$ passing through $p$ that intersects $U$. Then $L \cap (\mathbb{C}^n \smallsetminus U)$ is algebraic, hence contains only finitely many points of $L$, and so $p$ is a limit point of $U$.]

REPLY [3 votes]: Bounds for minimal sized frames which are injective for phase retrieval (a problem in signal processing) have been obtained using the fact that the variety of $m \times n$ matrices of rank at most $k$ is irreducible of codimension $(m - k)(n - k)$ in the projective space of nonzero $m \times n$ matrices (specifically in the case $n=m$ and $k=2$). See the paper of Balan, Casazza and Edidin or Conca, Edidin, Hering, and Vinzant for details.<|endoftext|>
TITLE: Exact sequence in étale cohomology related with proper birational morphism
QUESTION [6 upvotes]: Let f:X→Y be proper birational morphism between two quasi-projective varieties over an algebraically closed field $k$. I am particularly interested in the case where the characteristic of $k$ is positive; Y is singular, X is smooth and both X and Y are not projective.
Let D be a closed subset of Y, let E=f-1(D) and assume that f|X\E:X\E→Y\D is an isomorphism. 
Based on results in characteristic 0 (and on results on rigid cohomology) I expect that there is a long exact sequence of étale cohomology groups
Hi(Y,ℚl)→
Hi(X,ℚl)⊕
Hi(D,ℚl)→
Hi(E,ℚl)→
Hi+1(Y,ℚl).
I hoped this exact sequence is well-known and would appear in a standard text, but I had trouble identifying such a text. I can think of a prove using Cox's étale version of tubular neighbourhoods for E in X and D in Y and you might be able to compare them using the Mayer-Vietoris sequences in étale cohomology, but such a proof seems quite involved (you need to define the image of an étale tubular nhd under a proper morphism (which seems non-trivially to me) and check that for certain exact sequences taking direct or projective limits turns out to be an exact functor.)
Before working out the details I would like to ask whether anyone knows a reference for the above sequence or knows a simpler/nicer proof.

REPLY [8 votes]: Let $V := X\setminus E$ and $U := Y\setminus D$ and $j\colon U \rightarrow Y$ and
$k\colon V \rightarrow X$ the inclusions. We have exact sequences
$\cdots\rightarrow H^\ast(X,k_!\mathbb Z_\ell)\rightarrow H^\ast(E,\mathbb
Z_\ell)\rightarrow H^\ast(X,\mathbb Z_\ell)\rightarrow\cdots$ and a similar one for
$Y$, $D$ and $j$. Furthermore, $f$ induces a map of these long exact
sequences. A standard diagram chasing (also used for instance to the
Mayer-Vietoris sequence from excision) can be used to show that to get the
desired exact sequence it is enough to show that $f$ induces an isomorphism
$H^\ast(Y,j_!\mathbb Z_\ell)\rightarrow H^\ast(X,k_!\mathbb Z_\ell)$. We have that
$H^\ast(X,k_!\mathbb Z_\ell)=H^\ast(Y,Rf_\ast k_!\mathbb Z_\ell)$ but $f$ is proper so
that $Rf_\ast k_!\mathbb Z_\ell=f_!k_!\mathbb Z_\ell$. However,
$f_!k_!=(fk)_!=(jf')_!=j_!f'_!$, where $f'=f_{|V}$, and as $f'$ is an isomorphism
we have $f'_!\mathbb Z_\ell=\mathbb Z_\ell$.<|endoftext|>
TITLE: Standard reduction to the artinian local case?
QUESTION [20 upvotes]: Where can I find a clear exposé of the so called "standard reduction to the local artinian (with algebraically closed residue field", a sentence I read everywhere but that is never completely unfold?

EDIT: Here, was a badly posed question.

REPLY [7 votes]: Since I was somewhat critical of the question, I would offer here some reasons why the phrase: "reduction to the artinian local case" may mean quite differently to an algebraist (I was also vaguely aware of the geometric use BCnrd explained, that's  why I asked the OP to clarify). 
To prove some statements about a commutative ring $R$, a very often used technique is to pass to the quotient of $R$ by a non-zerodivisor. This is a local analogue of "taking the hyperplane section" in geometry. For examples of  this technique, see the proof of the Auslander-Buchsbaum formula in Bruns-Herzog, or this answer. 
Now, if the ring you started with is local and Cohen-Macaulay, then after killing enough regular elements you ended up in a local Artinian ring $(A,m,k)$. There are a lot of nice things about such rings: all f.g. modules have finite lengths, the Grothendieck group is $\mathbb Z$, local duality is simple as $A$ is already complete, the representation theory is well-studied, blah blah...
Sometimes you also wish that the residue field $k$ is algebraically closed. This can be obtained by making a faithfully flat extension by the EGA result BCnrd mentioned. Here are a few examples where such extension is useful:
1) The Auslander-Reiten quiver is easier to write down (all arrows has trivial valuations).
2) If $A$ is a complete intersection, then one can attach to each pair of $A$-modules $M,N$ a $k$-variety, whose dimension equals to the rate of (polynomial) growth of the lengths of $\text{Ext}^i(M,N)$. This has a non-trivial consequence that $\text{Ext}^i(M,N)$  and $\text{Ext}^i(N,M)$ grow at the same rate! (see this paper).
3) To prove that  there are only finitely many semi-dualizing modules over Cohen-Macaulay algebras, the only known proof is to pass to the Artinian case with alg. closed field and then use relevant results from representation theory. 
My hope is although this post did not answer what the OP wanted to ask, perhaps it can be of help to someone who googles "reduction to artinian case" with different motivation and find this MO question.<|endoftext|>
TITLE: Is there a simple inductive procedure for generating labeled trees uniformly at random, without direct recourse to Prüfer sequences?
QUESTION [12 upvotes]: Suppose you have a labeled tree $T$ on vertices $V=\lbrace 1,\ldots,n\rbrace$ that is drawn uniformly at random from the set of all $n^{n-2}$ such trees. I am seeking an $f$ satisfying the following desiderata:
D1.  $f(T)$ is a (random) tree on the vertex set $V'=\lbrace 1,\ldots,n+1\rbrace$;
D2. the distribution of $f(T)$ is uniform over the set of all $(n+1)^{n-1}$ labeled trees on $V'$;
D3. $f$ is a simple graph-theoretic recipe.
This $f$ can be a random function. That is, it can flip coins in deciding what to do with $T$: for instance, removing an edge uniformly at random.
One obvious way to satisfy D1 and D2 is: 

convert $T$ to its Prüfer sequence $s$;
independently, with probability $1/(n+1)$, change each entry of $s$ to $n+1$ (otherwise leave it fixed); 
then add a random number to the end of $s$, drawn uniformly from $V'$;
let $f(T)$ be the tree corresponding to the new sequence.

But what this procedure "really does" to $T$ seems not so easy to describe in graph-theoretic terms. I am looking for a recipe that satisfies D1-3 by manipulating the graph "directly" (i.e., adding and removing edges) without opaque steps like (1) and (4) in the above procedure.

REPLY [4 votes]: Yes, there is a simple method if you allow some non-determinism.  Given a labelled tree $T$ on $n$ vertices, uniformly randomly choose a vertex $v$ and a number $k\in[1,n+1]$. Attach a new vertex via an edge to $v$, add 1 to every label that is $k$ or more, and label the new vertex with $k$. This gives a tree $T'$ with $n+1$ vertices. However, the probability of each labelled tree on $n+1$ being generated is proportional to its number of leaves, since each leaf represents a way this tree could have been generated. We can correct this with a probabilistic filter: if $\ell$ is the number of leaves of $T'$, reject $T'$ and start over with probability $1-2/\ell$. Keep doing this until a tree is accepted. The expected number of trials before a tree is accepted is at most $n/2$.  Probably it is close to $n/(2e)$ since the average number of leaves is $n/e$.<|endoftext|>
TITLE: quiver mutation
QUESTION [8 upvotes]: Hello to all,
The phrase "quiver mutation has been invented by Fomin and Zelevinsky and has found numerous applications throughout mathematics and physics" is one that some of us encountered on a near-daily basis. I personally know only of a few applications. I thus propose the following thread: how have you used quiver mutation in your area of expertise ?

REPLY [5 votes]: To the best of my understanding an exceptional collection of elements in $\mathcal{D}(X)$ determines a quiver - and quiver mutations come to represent  corresponding mutations among the elements of the collection, as developed by Rudakov et. al. (vertices of the quiver). In this context, there is a beautiful construction due to P. Seidel - on the mirror analog of such mutations in the Fukaya category<|endoftext|>
TITLE: Simultaneously orthogonally transform two SPD matrices to tridiagonal form?
QUESTION [6 upvotes]: Supposing you have two SPD matrices $A,B\in\mathbb{R}^{n\times n}$ are there any known results on the existence or non-existence of a unitary matrix $Q$ such that $Q^\top A Q=T_A$ and $Q^\top B Q=T_B$ are both tridiagonal. If such a transformation exists in general, it is not required for my purposes that it be computable in finitely many steps.
I am aware of non-orthogonal congruence transformations which tridiagonalize two matrices. 
Thanks!

Edit:
Thanks for the response. I am familiar with the papers of Tisseur and Garvey et. al, but they are using non-orthogonal transformations. In one paper they use alternating 1D Householder reflectors and matrices of the form $L=I+xy^\top$ to force portions of the leading columns to be in the same space. 
I tried finding a counter-example from the 3x3 case, but it looks like I have plenty of degrees of freedom to play with and higher dimensions become treacherously difficult to manage individual elements. 
Maybe this question is equivalent to finding a $Q$ such that for an arbitrary matrix $V$ that $Q^\top V$ is bidiagonal, which certainly looks hopeless to me.

REPLY [3 votes]: To elaborate on what I said in the comments... the reason I feel there might not be a way to do this is as follows: the problem of simultaneously tridiagonalizing two SPD matrices $A$ and $B$ is equivalent to tridiagonalizing either of $AB^{-1}$ or $A^{-1}B$; unfortunately both of these can be unsymmetric (the product of two symmetric matrices need not be symmetric), and orthogonal (unitary) similarity transformations of an unsymmetric matrix only manage to reduce to Hessenberg form.<|endoftext|>
TITLE: Subtle counterexample to $m\neq n$ but $R^m=R^n$ for some ring $R$ ?
QUESTION [6 upvotes]: To do Algebraic K-theory, we need a technical condition that a ring $R$ satisfies $R^m=R^n$ if and only if $m=n$. I know some counterexamples for a ring $R$ satisfies $R=R^2$. 
Are there any some example that $R\neq R^3$ but $R^2 = R^4$ or something like that?
(c.f. if $R^2=R^4$, then we need that $R^3=R^5=\ldots =R^{2n+1}$ for any $n>1$)

REPLY [4 votes]: Rings that satisfy the condition 
$R^n \cong R^m \iff n=m$ are said to have invariant basis number or the invariant basis property. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$.  
Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that 
$$
R^m\cong R^{m'} \iff m=m'\ \mbox{or}\ m,m'\ge h \ \mbox{and} \ m\equiv m'\ (\mod k)
$$
A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$. 
There are earlier constructions of rings of type $(h,k)$ by Leavitt [W. G. Leavitt: 
Modules without invariant basis number, Proc. Am. Math. Sot. 8 (1957), 322-328], [W. G. Leavitt: 
The module type of a ring, Trans. Am. Math. Sot. 103 (1962), 113-130], but they are far more complicated. 
Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K_0(R)\cong {\bf Z} \oplus \tilde{K}_0(R)$.<|endoftext|>
TITLE: Generators of the Principal Unit Group in Local Fields of Characteristic 0 
QUESTION [5 upvotes]: Let $p$ be a prime, $L$ be a finite, totally ramified-extension of $Q_p,$ and $U$ be the group of principal units in the ring of integers of $L.$ Then $U$ is a finitely generated $\mathbb{Z}_p$-module under exponentiation. I'm looking for an explicit set of generators for $U$ under this action; is there a resource where I can find this information?
Thanks Robin but am I afraid I need more control on the set of generators. In particular, I'm interested if G is a set of generators of the value m = max {$v_L(g-1):g\in G$}

REPLY [2 votes]: Well you can lift any set of generators for the finite group
$U/U^p$ and they will be a set of generators of $U$. If $n$ is
the degree of $L$ over $\mathbb{Q}_p$ then $U/U^p$ will have order
$p^{n+1}$ or $p^n$ according to whether or not $L$ has the $p$-th
roots of unity. Certainly there's a system of generators of
the form $1+\pi^j$ for a certain set of positive integers $j$
where $\pi$ is a uniformizer. You choose each $j$ to be the
next one such that $1+\pi^j$ is not in the group generated by
$U^p$ and the previous generators.<|endoftext|>
TITLE: Is all ordinary mathematics  contained in high school mathematics?
QUESTION [55 upvotes]: By high school mathematics I mean Elementary Function Arithmetic (EFA), where one is allowed +,×, xy, and a weak form of induction for formulas with bounded quantifiers. This is much weaker than primitive recursive arithmetic, which is in turn much weaker than Peano arithmetic, which is in turn much weaker than ZFC that we normally work in. 
However there seem to be very few theorems (about integers) that are known to require anything more than this incredibly weak system to prove them. The few theorems  that I know need more than this include:
*Consistency results for various stronger systems (following Godel). This includes results such as the Paris Harrington theorem and Goodstein sequences that are cleverly disguised forms of consistency results.
*Some results in Ramsey theory, saying that anything  possible will happen in a sufficiently large set.  Typical examples: Gowers proved a very large lower bound for Szemeredi's lemma showing that it cannot be proved in elementary function arithmetic, and the Robertson-Seymour graph minor theorem is known to require such large functions that it is unprovable in Peano arithmetic. 
I can think of no results at all (about integers) outside these areas (mathematical logic, variations of Ramsey theory) that are known to require anything more than EFA to prove. 
A good rule of thumb is that anything involving unbounded towers of exponentials is probably not provable in EFA, and conversely if there is no function this large then one might suspect the result is provable in EFA. 
So my question is : does anyone know of natural results  in "ordinary" mathematics (number theory, algebraic geometry, Lie groups, operator algebras, differential geometry, combinatorics, etc...) in which functions larger than a finite tower of exponentials occur in a serious way?  In practice this is probably more or less equivalent to asking for theorems about integers unprovable in EFA.
Related links:
http://en.wikipedia.org/wiki/Grand_conjecture about Friedman asking a similar question.
By the way, encoding deep results as Diophantine equations and so on is cheating. And please do not make remarks suggesting that Fermat's last theorem needs inaccessible cardinals unless you understand Wiles's proof.

REPLY [2 votes]: I think we're heading towards an era where "ordinary" mathematics includes logic, at least the parts of it that can be described as applied mathematics (used in practical problems of the real world).  For example, the ML programming language is based on polymorphic lambda calculus.  I'm no expert but I have the impression that the proof that polymorphic lambda calculus is strongly normalizing is equivalent to second order arithmetic.  There are fancier languages based in even more powerful (?) theories, like Coq implements Martin-Lof type theory more or less directly.  I'm just a programmer trying to learn these languages but people disigning them and writing compilers for them (e.g. implementing type inference) seem to me to often be up to their elbows in proof theory. I saw a thesis by someone about ordinal analysis of programs (after all, by the Curry-Howard correspondence, programs are proofs..).  I wondered if software engineers (at least those in high-assurance programming) will someday use proof-theoretic ordinals in their daily work just like electrical engineers now use complex numbers.  
In complexity theory, there are some proofs that P vs NP is independent of some sizeable fragments of PA, but maybe those fragments are weaker than EFA.<|endoftext|>
TITLE: range projection of an unbounded idempotent affiliated to a finite von Neumann algebra
QUESTION [5 upvotes]: To be slightly more precise: let $M\subset B(H)$ be a finite von Neumann algebra equipped with a faithful normal trace $\tau$, and let $L^0(M,\tau)$ be the completion of $M$ in the measure topology; this is an algebra, whose elements can be identified with those densely-defined and closed operators on $H$ that are affiliated with $M$. (See e.g. E. Nelson, Notes on noncommutative integration, JFA 1974). Let $e$ be an idempotent in $L^0(M,\tau)$, not necessarily self-adjoint; then it is not hard to show that $R=\{ \xi\in H : e\xi=\xi\}$ is a closed subspace of $H$.
Question: is the orthogonal projection onto $R$ affiliated with $M$?
I suspect the answer is yes (and would like it to be, for some calculations I'm doing at the moment) but am having difficulties nailing the argument down. Given that this should, if true, be a pretty basic bit of operator algebra theory, and standard knowledge, I'd be grateful if someone could point me to a reference. (I currently have somewhat limited library access, but this might well be covered in Kadison & Ringrose for instance.)

Edit/update: both Martin Argerami and Jonas Meyer have given straightforward proofs of the desired result, and a quick check in Kadison & Ringrose vol. 1 has not turned up any explicit statement (probably because the result turns out to be so basic). Since I can't accept both their answers, I'm accepting Martin's on grounds of personal preference.

REPLY [4 votes]: Disclaimer: I fear I may be missing some subtlety here, as is often the case when I think about unbounded operators.  This is an attempt to generalize the result.
A closed densely defined operator $T$ on $H$ has a unique polar decomposition $T=V|T|$ with $|T|=\sqrt{T^*T}$ and $V$ a partial isometry whose initial space is the closure of the range of $|T|$ and whose final space is the closure of the range of $T$.  If $T$ is affiliated with a von Neumann algebra $M$, then $V$ is in $M$ (as stated e.g. in Nelson's paper on the bottom of page 111).  Thus $VV^*$, the projection onto the closure of the range of $T$, is in $M$.<|endoftext|>
TITLE: Refereeing a Paper
QUESTION [219 upvotes]: I've refereed at least a dozen papers in my (short) career so far and I still find the process completely baffling.  I'm wondering what is actually expected and what people tend to do...
Some things I've wondered about:
1) Should you summarize the main results and or the argument?  If so, how much is a good amount. What purpose does this serve?
2) What do you do when the journal requests you to evaluate the quality of the result (or its appropriateness for the journal)?  Should one always make a recommendation regarding publication? 
3) What to do about papers with major grammatical or syntactical errors, but that are otherwise correct?  Does it matter if the author is clearly not a native English speaker?
4) On this note, at point does one correct such mistakes?  Ever?  If there are fewer than a dozen? Should one be proof-reading the paper?
5) What do you do when you do not understand an argument?  Does it matter if it "feels" correct.  How long should one spend trying to understand an argument?
6) What to do about papers that have no errors but whose exposition is hard to follow?

REPLY [25 votes]: The referee must check if the result is correct! That is, the referee should check the proofs. This point seems to be so under-estimated, but in my opinion it is the most important part of the referee's job. The referee is the only one who will do this, except the author. There does need to be a second check of correctness after the author. There are many flawed papers out there. This has cost many other researchers time and health.
If the referee is unsure, he should ask the author for help in understanding the proofs (and the author should be helpful). It can be a change in the paper or a note for the referee.
As for whether or not the paper is "interesting": I decline to referee a paper if it is not interesting (in a broad sense) enough for me to put the effort in checking whether or not it is correct. Thus if all referees were correctness minded, then "interesting" papers would get their advantage.<|endoftext|>
TITLE: How constructive is Doyle-Conway's 'Division by three'?
QUESTION [19 upvotes]: In the (whimsically written) article Division by three, Doyle and Conway describe a proof, (apparently) not using Choice, that an isomorphism $A \times 3 \simeq B\times 3$ of sets (where $3$ is a given three-element set) gives an isomorphism $A \simeq B$. The result is easy for well-ordered $A$ and $B$, but clearly assuming this isn't constructive. However, the authors seem (to me) to use excluded middle. Also, I don't know if I'm entirely convinced they avoid all forms of choice - it may be they are relying on some weak form of choice (aside from excluded middle).
Has anyone given this any thought? The article is from the mid-90s despite its ArXiV date, and purports to have discovered a lost proof of Lindenbaum from the 1920s which predates a different proof given by Tarski in 1949.

REPLY [12 votes]: The construction in the paper seems to rely on two non-constructive assumptions:


We can decide whether two elements in a set (involved in the division by 3) are equal.
A countable subset of $\mathbb{N}$ is infinite or not infinite.


(By "infinite" I mean "contains an infinite sequence of pairwise distinct elements".)
In computability theory the second assumption corresponds to Turing degree $0''$.
To see where the second assumption is used, consider the proof of Lemma 2 on page 26. There one is given an infinite sequence of 0's, 1's and 2's and one must decide whether there are infinitely many 0's in the sequence (and if not, whether there are infinitely many 1's). Let us show that such a decision procedure is equivalent to the second assumption. In one direction, a ternary sequence contains infinitely many 0's iff the set of indices at which the 0's appear is infinite. Conversely, given a countable subset $A \subseteq \mathbb{N}$ enumerated by $e$, consider the sequence $a_0, a_1, \ldots$ where

$a_k = 0$ if $e$ enumerates a new element at step $k$, and 
  $a_k = 1$ otherwise.

The sequence has infinitely many 0's iff $A$ is infinite.
Another kind of a typical construction in the paper requires one to determine whether the orbit of an element $x$ under a bijection is finite or infinite. We can do this using our assumptions as follows. Given $x \in A$ and a bijection $f : A \to A$, construct a sequence $a_0, a_1, \ldots$ as follows:

$a_k = 0$ if there is $m \leq k$ such that $x = f^m(x)$, and
  $a_k = 1$ otherwise.

Here we assumed that $A$ has decidable equality. The sequence $a_k$ contains a zero iff it contains infinitely many zeroes (thus we only require a $0'$ oracle to perform the following steps). If it contains a zero, say $ak = 0$ then the orbit $\lbrace x, f(x), f^2(x), \ldots \rbrace$ is finite with at most $k$ elements. If it does not contain a zero then the orbit is infinite because $f^k(x) \neq f^j(x)$ for all $k \neq j$, as otherwise we would have $f^{|k-j|+1}(x) = x$.
Let me just remark that one might be tempted to consider a non-constructive assumption such as:

A group generated by a single element is either infinite or finite.

(This would certainly allow us to determine whether orbits of elements under bijections are finite or infinite.) In fact division by 3 follows, but for a rather strange reason, namely that such a principle implies the law of excluded middle. Suppose $p$ is an arbitrary thruth value, and consider the Abelian group $G$ freely generated by the generators $p$ and $\top$ (true). Consider the subgroup of $G$ generated by the element $\top - p$. If it is infinite then there is $n \in \mathbb{N}$ such that $n \top \neq n p$, hence $\top \neq p$ and so $\lnot p$ holds. If the subgroup is finite then there is $n > 0$ such that $n \top = n p$, hence $\top = p$ and $p$ holds. (I hope I got this right, we have to be careful because finite sets need not have definite sizes.)
This hopefully gives us some idea about how non-constructive are the techniques employed in the proof of division by three (essentially it looks like a $0''$ oracle). It seems hard to quantify how non-constructive the theorem itself is. Knowing that I can divide by 3 does not obviously allow me to derive non-constructive consequences. For all I know, some suitably constructivized version of division by 3 might actually be constructively valid.
Can I have a green flag now, please?<|endoftext|>
TITLE: Is a valuation domain PID when its maximal ideal is principal?
QUESTION [5 upvotes]: It is known that a valuation domain is a principal ideal ring if and only if its prime ideals are principal. Is it a principal ideal ring when its unique maximal ideal is principal?

REPLY [9 votes]: I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x \in K^{\times}$, at least one of $x,x^{-1}$ lies in $R$.
In this case, I believe the answer is no.  Let $R$ be any valuation domain whose value group $K^{\times}/R^{\times}$ is isomorphic, as a totally ordered abelian group, to $\mathbb{Z} \times \mathbb{Z}$ with the lexicographic ordering.  (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.)  In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal.  
For some information on valuation rings, see e.g. Section 17 of 
http://math.uga.edu/~pete/integral.pdf<|endoftext|>
TITLE: obstruction to taking the square root of a Galois character
QUESTION [6 upvotes]: Suppose that $K$ is a number field, and (writing $G_K=\mathrm{Gal}(\overline{K}/K)$), suppose that $\phi:G_K\to \overline{\mathbb{Q}}$ is a finite order character of $G_K$. I believe that the obstruction to taking a square root of $\phi$ (that is, the obstruction to finding some finite order $\chi:G_K\to \overline{\mathbb{Q}}$ with $\chi^2=\phi$) can be identified with a 2-torsion element of the Brauer group of $K$, and I believe that the proof just involves taking a long exact sequence from a cunningly-chosen short exact sequence; but I cannot now reconstruct it. Can anyone remind me?

REPLY [11 votes]: With luck, this will be blunder-free (and if it's not, please tell me!):
Consider the short exact sequence 
$$1 \to \mu_2 \to \overline{\mathbb Q}^{\times} \to \overline{\mathbb Q}^{\times}
\to 1,$$
with the third arrow being squaring, and with trivial $G_k$-action.  Passing to cohomology,
the sequence of $H^0$s is just this same sequence, and the sequence of higher cohomology
becomes
$$0 \to Hom(G_K,\mu_2) \to Hom(G_K,\overline{\mathbb Q}^{\times}) \to
Hom(G_K,\overline{\mathbb Q}^{\times}) \to H^2(G_K,\mu_2),$$
with the last arrow being the obstruction you asked about. 
Added: See Brian Conrad's comment below for a cleaner point of view, showing
that $H^2(G_K,\mu_2)$ is the precise obstruction space.<|endoftext|>
TITLE: Effective algorithm to test positivity
QUESTION [12 upvotes]: Let $f(x_1,\ldots, x_n)$ be a real polynomial in several variables. Is there an effective algorithm to test whether $f$ is positive (or nonnegative) on the whole of ${\mathbb{R}}^n$?

REPLY [2 votes]: If you could find an upper bound for degrees of polynomials used in Positivstellensatz I guess you can create an algorithm for this.
You would like to take a look at Positivstellensatz, for instance in the book of Bochnak-Coste-Roy "Real Algebraic Geometry" under Corollary 4.4.3 (this corollary combines Positivstellnsatz, Negativstellensatz and Nullstellensatz for real closed fields). A more abstract form can be found in a German book "Einführung in die Reelle Algebra" by Knebusch and Scheiderer (the book is now freely available for download from the university of Regensburg library: here). You would also like to see the works by Schmüdgen and Putinar on various forms of Positivstellensätze.<|endoftext|>
TITLE: Additive combinatorics and large Fourier coefficients
QUESTION [6 upvotes]: Elon Lindenstrauss explains in his talk at the MSRI in Fall 2008 (the relevant comment is at minute 41 of the video) that the set of large Fourier coefficients of a probability measure $\mu$ on the torus ${\mathbb T}^n$ respects the additive structure. More precisely, he defines
$$A_{\delta} := \lbrace b \in {\mathbb Z}^n \mid |\hat \mu(b)| \geq \delta \rbrace$$
and says that it is "morally" true that $A_{\delta} - A_{\delta} \subset A_{\delta^2}$. (Here, the difference of two subsets is defined to be the set of all possible differences of elements in the respective subsets.) The precise statement (according to Lindenstrauss) is a consequence of the Balog-Szemeredi-Gowers Lemma. 
Can someone provide the precise statement or give some hint how the lemma can be used to obtain bounds on Fourier coefficients?

REPLY [2 votes]: I think I figured it out myself. What was meant is that for every finite subset $S$ of $A_{\delta}$ one has
$$| \lbrace (n,m) \in S \times S \mid n-m \in A_{\delta^2/2} \rbrace | \geq \frac{\delta^2 |S|^2}{2}.$$
This follows from the proof of the second part of Lemma 4.37 in Tao and Vu, Additive Combinatorics. (There, the inequality is stated incorrectly as $\leq$.)<|endoftext|>
TITLE: Interdependence between A^1 homotopy theory and algebraic cobordism
QUESTION [10 upvotes]: I would like to learn something about $\mathbb{A}^1$-homotopy theory. I know about standard references on the subject, but before dwelling into studying them I have a doubt which some expert could clear.
Having a look at Levine's survey it seems that algebraic cobordism has been very succesful at solving open problems in $\mathbb{A}^1$-homotopy theory.
How much the two topics are independent of each other? Is it possible to understand the full implications of $\mathbb{A}^1$-homotopy theory without knowing about algebraic cobordism? Has the latter theory in some sense superseded the former?

REPLY [15 votes]: The two topics are logically, if not morally, independent of one another.  $\mathbb{A}^1$-homotopy encodes objects like motivic cohomology & it's relatives which are of interest regardless of the framework.  There's no way for algebraic cobordism to supersede that -- algebraic cobordism is more directly comparable to Chow theory and $K^0$ than to motivic cohomology.
Conversely, algebraic cobordism provides a more geometric viewpoint on (a piece of) $MGL$ -- surely a valuable thing to have around as well as being of independent geometric interest.  That said, if your interests are more motivic than geometric, you could get by without knowing the details of algebraic cobordism provided that you know all the classical statements in complex cobordism that inspired it.
Motivic vs algebraic cobordism.
The $\mathbb{P}^1$-spectrum $MGL$, or "motivic cobordism," enjoys a privileged role in the world of $\mathbb{A}^1$-homotopy similar to that of $MU$ in classical homotopy.  There is a relationship between this "motivic cobordism" and "algebraic cobordism." The former is a bigraded theory, and Levine showed that $MGL^{2n,n}(X) = \omega^n(X)$.  (This bigrading issue is analogous to how Chow theory occurs as the $(2n,n)$-graded piece of motivic cohomology, and explains why you don't get a long exact sequence in algebraic cobordism, etc...)
So one can view Morel-Levine's (or Levine-Pandharipande's) algebraic cobordism as giving an axiomatic (or geometric) viewpoint on the motivic theory, like we had for $MU$.  Unlike the case of complex cobordism, where one can directly compare it to $MU$-cohomology using transversality results, here the comparison is much more difficult and computational.  The proof of this comparison relies on a (currently unpublished) spectral sequence due to Hopkins-Morel.  It should be noted that constructing this spectral sequence is hard, and by the time you've constructed it you've had to independently check lots of things that you might've wanted to deduce from the comparison with algebraic cobordism (for instance you pretty much end up computing $MGL^{2*,*}(Spec k)$, you can see the comparison of cobordism to Chow theory, etc.).
Degree formula (or, application to B-K)
The reference to open problems likely refers to the use of cobordism and Rost's degree formula in the final steps of proving Bloch-Kato for $\ell \neq 2$.  Cobordism is a tool in the proof, but introducing algebraic cobordism is not strictly necessary.  (One can get by with explicit computations with the characteristic numbers of interest.  It'd certainly be fair to like the Levine-Morel proof of the degree formula, though.)
The Bloch-Kato conjecture is concerned with the "Galois symbol" map
$$ K^n_M(k)/\ell = H^n(k, \mathbb{Z}/\ell(n)) \to H_{et}^n(k, \mathbb{Z}/\ell(n)) = H_{et}^n(k, \mu_\ell^{\otimes n}) $$
No cobordism in sight yet.  Suslin-Merkurjev's proof for $n=2$ and Voevodsky's proof for $\ell=2$ made use of "splitting varieties" that one could write down pretty much explicitly and then proceed to study: Brauer-Severi varieties and Pfister quadrics, respectively.  This doesn't seem to work for the general case, and instead one writes down a minimalist wishlist for splitting varieties and then has to show that they exist --- it is in this step where cobordism (or really, characteristic numbers) play a role.
A "splitting variety" for a non-zero symbol $0 \neq u = u_1 \otimes \cdots \otimes u_n \in K^n_M(k)/\ell$ should be a smooth variety $X/k$ such that $u$ pulls back to zero in $H^n(k(X), \mathbb{Z}/\ell(n))$ ("$X$ splits $u$"), with $\dim X = \ell^{n-1}-1$, and some more technical conditions, including a partial "universality" for this property:  $X'$ splits $u$ iff there is a rational map from (a degree prime-to-$\ell$ cover of) $X'$ to $X$; it follows that $X$ must have no degree prime-to-$\ell$ zero-cycles (or else $Spec k$ splits $u$, i.e., $u = 0$).  
Cobordism (of whatever flavor you like: complex cobordism suffices) enters when relating this to characteristic numbers: namely, to the property of being a $v_n$-variety (=representing a $v_n$ class in complex or algebraic cobordism, up to decomposables).  Here, one needs something like "Rost's degree formula", which implies for instance that the property of being a $v_n$-variety with no prime-to-$\ell$ zero-cycles is invariant under prime-to-$\ell$ degree covers.<|endoftext|>
TITLE: Distance of a barycentric coordinate from a triangle vertex
QUESTION [6 upvotes]: I have a triangle $ABC$ with side lengths $a,b,c$ (edges $BC, CA, AB$ respectively).
I have a point $p$ with barycentric coordinates $u:v:w$.
These are normalised: $u+v+w=1$.
$1:0:0$ corresponds to point $A$, $0:1:0$ is $B$ etc.
Is there a simple expression for the distance $d$ of the point $p$ from $A$ ?
(My initial naive guess based on $d(1:0:0)=0, d(0:1:0)=b, d(0:0:1)=c$ was that $d$ was linear $d(u,v,w)=v*b+w*c$ but this is clearly wrong as in the case of an equilateral triangle $a=b=c=1$ it returns $d=2/3$ for the centroid ($u:v:w = 1/3:1/3:1/3$), when the correct answer should be $1/\sqrt 3$ (the radius of the circumscribed circle)).

REPLY [2 votes]: There actually is a delightful formula -- see page 11 at 
http://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf
Your displacement vector is $(u-1,v,w)$, giving $d^2=-(a^2vw+b^2w(u-1)+c^2v(u-1))$<|endoftext|>
TITLE: Group Completions and Infinite-Loop Spaces
QUESTION [16 upvotes]: Let $X$ be a commutative H-space. A group completion is an H-map $X\to Y$, where $Y$ is another H-space, such that

$\pi_0(Y)$ is a group
The Pontrjagin ring $H(Y; R)$ is the localization of the Pontrjagin ring $H_*(X; R)$ at the multiplicative submonoid $\pi_0(X)$ for every coefficient ring $R$. 

Perhaps most interesting is the case where $X$ is a commutative monoid or more generally an $E_\infty$-space. In this case, May gives a functorial group completion $B_0$ via the two-sided bar construction which is an infinite-loop space. If I understand it right, it is defined as $colim_j \Omega^j|\Sigma^j(C_j\times C')^\bullet X|$, where $C_j\times C'$ denotes the monad associated to the product of the little $j$-cube operad and an $E_\infty$-operad and $\bullet$ the simplicial variable. I have some question concerning group completions:
1) Are all group completions equivalent? That is, does there always exist a homotopy equivalence of H-spaces between them?
2) Does the group completion preserve homotopy limits? For example, does the May functor preserve homotopy limits of $E_\infty$-spaces?
3) Suppose one knows that all loop spaces of $X$ are infinite loop spaces. Is there a simple relationship between the infinite-loop space $B_0 X$ and the loop spaces of $X$? Especially, I am interested in the homotopy groups of $B_0 X$. 
An answer to any of these question would be helpful to me.

REPLY [11 votes]: A well-written discussion of the group completion can be found on pp. 89--95 of
J.F. Adam: Infinite loop spaces, Ann. of Math. studies 90 (even though he only
discusses a particular group completion of a monoid). In particular you
assumption of commutativity comes in under the assumption that $\pi_0(M)$ is
commutative which makes localisation with respect to it well-behaved
(commutativity is not the most general condition what is needed is some kind of
Øre condition).
In any case if you really want conclusions on the homotopy equivalence level I
think you need to put yourself in some nice situation for instance requiring
that all spaces be homotopy equivalent to CW-spaces. If you don't want that you
should replace homotopy equivalences by weak equivalences, if not you will
probably find yourself in a lot of trouble. In any case I will assume that we
are dealing with spaces homotopy equivalent to CW-complexes.
Starting with 1) a first note is that your conditions does not have to involve
an arbitrary ring $R$. It is enough to have $R=\mathbb Z$ and one should
interpret the localisation in the way (for instance) Adams does:
$H_\ast(X,\mathbb Z)=\bigoplus_\alpha H_\ast(X_\alpha,\mathbb Z)$, where
$\alpha$ runs over $\pi_0(X)$, and a $\beta$ maps $H_\ast(X_\alpha,\mathbb Z)$
to $H_\ast(X_{\alpha\beta},\mathbb Z)$. Then your group completion condition is
that the natural map $\mathbb Z[\pi_0(Y)]\bigotimes_{\mathbb Z[\pi_0(X)]}
H_\ast(X,\mathbb Z)\rightarrow H_\ast(Y,\mathbb Z)$ should be an isomorphism.
This then implies the same for any coefficient group (and when the coefficient
group is a ring $R$ you get your condition). (Note that for this formula to even make sense we need at least associativity for the action of $\pi_0(X)$ on the homology. This is implied  by the associativity of the Pontryagin product of $H_*(X,\mathbb Z)$ which in turn is implied by the homotopy associativity of the H-space structure.)
Turning now to 1) it follows from standard obstruction theory. In fact maps into
simple (hope I got this terminology right!) homotopy types, i.e., spaces for
which the action of the fundamental groups on the homotopy groups is trivial (in
particular the fundamental group itself is commutative). The reason is that the
Postnikov tower of such a space consists of principal fibrations and the lifting
problem for maps into principal fibrations is controlled by cohomology groups
with ordinary coefficients. Hence no local systems are needed (they would be if
non-simple spaces were involved). The point now is that H-spaces are simple so
we get a homotopy equivalence between any two group completions and as
everything behaves well with respect to products these equivalences are H-maps.
Addendum:
As for 2) it seems to me that this question for homotopy limits can only be solved under supplementary conditions. The reason is that under some conditions we have the Bousfield-Kan spectral sequence (see Bousfield, Kan: Homotopy limits, completions and localizations, SLN 304) which shows that $\varprojlim^s(\pi_s X_i)$ for all $s$ will in general contribute to $\pi_0$ of the homotopy limit. As the higher homotopy groups can change rather drastically on group completion it seems difficult to say anything in general (the restriction to cosimplicial spaces which the OP makes in comments doesn't help as all homotopy limits can be given as homotopy limits over $\Delta$. Incidentally, for homotopy colimits you should be in better
shape. There is however an initial problem (which also exists in the homotopy limit case): If you do not assume that the
particular group completions you choose have any functorial properties it is not
clear that a diagram over a category will give you a diagram when you group
complete. This can be solved by either assuming that in your particular
situation you have enough functoriality to get that (which seems to be the case
for for instance May's setup) or accepting "homotopy everything" commutative
diagrams which you should get by the obstruction theory above. If this problem
is somehow solved you should be able to conclude by the Bousfield-Kan spectral
sequence $\injlim^\ast H_*(X_i,\mathbb Z)\implies
H_*(\mathrm{hocolim}X_i,\mathbb Z)$. We have that localisation is exact and
commutes with the higher derived colimits so that we get upon localisation a
spectral sequence that maps to the Bousfield-Kan spectral sequence for $\{Y_i\}$
and is an isomorphism on the $E_2$-term and hence is so also at the convergent.
As for 3) I don't altogether understand it. Possibly the following gives some
kind of answer. For the H-space $\coprod_n\mathrm{B}\Sigma_n$ which is the
disjoing union of classifying spaces of the symmetric groups its group
completion has homotopy groups equal to the stable homotopy groups of spheres
which shows that quite dramatic things can happen to the homotopy groups upon
group completion (all homotopy groups from degree $2$ on of the original space
are trivial).<|endoftext|>
TITLE: Algebrization of second-order logic
QUESTION [18 upvotes]: Is there an algebrization of second-order logic, analogous to Boolean algebras for propositional logic and cylindric and polyadic algebras for first-order logic?

REPLY [3 votes]: Did programme of predicate calculus algebraization succeed? In his essay An autobiography of polyadic algebras Halmos outlined why he is not satisfied with his baby. I suggest that Cylindric algebras are not genuine algebras either; what other algebraic structures have operators parametrized by something (cylindrification)? The situation is strikingly different from either boolean or relation algebra, each having a set of intuitive binary, unary, [and 0-ary] operations. 
I suggest that the culprit is positional perspective onto relation attributes. Positional perspective is ubiquitous in math (sequences, function arguments, etc), so is easy to see why it penetrated into the world of relations. Positional perspective makes perfect sense for binary relations, this is why nobody challenged its adequacy for n-ary ones. However, it is easy to see that attribute positions are not essential to the ability to match values of the two different attributes of two different relations. Named attribute perspective is widely used in database theory and practice, which implies yet another algebraization of predicate logic.<|endoftext|>
TITLE: Is the space of solutions to the Queens Domination Problem connected?
QUESTION [11 upvotes]: A configuration of queens on an 8 by 8 chessboard (or n by n if you like) is a queen domination if every square on the board lies in the same row, column, or diagonal as at least one of the queens. The Queens Domination Problem is to find the minimum number of queens necessary for a queen domination. A solution to the Queens Domination Problem is a queen domination using the minimum number of queens.
For the 8 by 8 chessboard, brute force has shown that 5 queens is the minimum number. More discussion about that here. 
The adjacency relation on the set of solutions to the Queens Domination Problem is defined as follows: we say solutions are adjacent if they differ by only the placement of one queen. For example: C3, E4, D5, B6, F4 is adjacent to C3, E4, D5, B6, F2.

How many equivalance classes are there in the equivalance relation generated by adjacency?

In other words, starting with one solution, can we reach any other solution by moving one queen at a time, such that the result of each move is itself a solution?

REPLY [11 votes]: Revised and correct:
589 equivalence classes.
http://www.math.ucsd.edu/~etressle/classes.txt
It seems that the pairs (#vertices,#components) are (1,388) (2,100) (3,40) (4,34) (5,20) (18,4) (20,2) (3804,1) – damiano 15 hours ago
1: Class I.D. 2, 4, 5, 6, 7,..., 589.
2: Class I.D. 22, 29, 35, 47, 48,..., 579.
3: Class I.D. 28, 33, 38, 49, 58,..., 574.
4: Class I.D. 13, 90, 95, 126, 141,..., 576. 
5: Class I.D. 32, 37, 62, 89, 97,..., 395. 
18: Class I.D. 17, 31, 185, 303.
20: Class I.D. 3, 75.
3804: Class I.D. 1.<|endoftext|>
TITLE: Nelson's program to show inconsistency of ZF
QUESTION [67 upvotes]: At the end of the paper Division by three by Peter G. Doyle and John H. Conway, the authors say:
Not that we believe there really are any such things as infinite sets, or that the Zermelo-Fraenkel axioms for set theory are necessarily even consistent. Indeed, we’re somewhat doubtful whether large natural numbers (like $80^{5000}$, or even $2^{200}$) exist in any very real sense, and we’re secretly hoping that Nelson will succeed in his program for proving that the usual axioms of arithmetic—and hence also of set theory—are inconsistent. (See Nelson [E. Nelson. Predicative Arithmetic. Princeton University Press, Princeton, 1986.].) All the more reason, then, for us to stick with methods which, because of their concrete, combinatorial nature, are likely to survive the possible collapse of set theory as we know it today.
Here are my questions:
What is the status of Nelson's program? Are there any obstruction to finding a relatively easy proof of the inconsistency of ZF? Is there anybody seriously working on this?

REPLY [29 votes]: Edward Nelson passed away at age 82 on September 10, 2014. You can read a tribute to Nelson's illustrious career from Princeton University.
Although he worked on the inconsistency of PA until the end, there were no standing claims to a proof of the inconsistency of PA at the time of Nelson's unfortunate passing.

On September 30, 2015, Nelson's unfinished manuscripts titled Inconsistency of Primitive Recursive Arithmetic and Elements have been posted on the arXiv, with a foreword by Sarah Jones Nelson and an afterword by Sam Buss and Terry Tao.

arxiv.org/abs/1509.09209
arxiv.org/abs/1510.00369<|endoftext|>
TITLE: Ring of closed manifolds modulo fiber bundles
QUESTION [26 upvotes]: Let $R$ be the ring which is generated by homeomorphism classes $[M]$ of compact closed manifolds (of arbitrary dimension) subject to the relations that
$$[F]\cdot [B] = [E]$$
if there exists a fibre bundle $F \to E \to B$, and
$$[M] + [N] = [M \cup N]$$
if $M$ and $N$ are of the same dimension. Clearly, $[pt]$ behaves as a unit  and we can write $[pt]=1$. Moreover, since $[F] \cdot [B] = [F \times B] = [B \times F] = [B] \cdot [F]$, we see that $R$ is a commutative ring.
It is clear that the Euler characteristic defines a homomorphism $\chi : R \to {\mathbb Z}$. What else can we say about the ring $R$ ? What can we say if everything is required to be oriented and/or smooth etc.? Is the ring $R$ finitely generated?
Example: Since $S^1$ is a double cover of itself, we get
$[S^0] \cdot [S^1] = [S^1]$, but $[S^0] = 2$ and hence $[S^1]=0$. In particular, the classes of all mapping tori of homeomorphisms vanish in $R$ since they are fiber bundles over $S^1$.

REPLY [6 votes]: Consider the variation where we ask for smooth manifolds and smooth fiber bundles. Then I claim that $R$ is not finitely generated. 
The starting observation is that if $F \to E \xrightarrow{p} B$ is a smooth fiber bundle then 
$$0 \to \text{ker}(p) \to T(E) \to p^{\ast}(T(B)) \to 0$$
gives a splitting of the tangent bundle of $E$. There are cohomological obstructions to such splittings existing in general, which we can compute. The upshot is that if $E$ is a simply connected (so it has no nontrivial covers) closed smooth manifold whose tangent bundle has no nontrivial subbundles, then $[E]$ does not participate in any of the interesting relations defining $R$, and in particular cannot lie in the subring of $R$ generated by manifolds of dimension smaller than $\dim E$. Hence to show that $R$ is not finitely generated it suffices to write down a sequence of such $E$ of arbitrarily large dimension.
But this is standard: we can take the even-dimensional spheres $S^{2n}$. First, observe that because $S^{2n}$ is simply connected, it has no nontrivial covering spaces, and in addition every real vector bundle over $S^{2n}$ is orientable, hence has well-defined Euler classes (after picking an orientation). Second, the Euler class $e(T)$ of the tangent bundle is $2$ times a generator of $H^{2n}(S^{2n})$, and in particular does not vanish. Since the Euler class is multiplicative with respect to direct sum, if $T = T_1 \oplus T_2$ is a nontrivial splitting of the tangent bundle then $e(T) = e(T_1) e(T_2)$. But the cohomology groups that $e(T_1)$ and $e(T_2)$ live in both vanish for $S^{2n}$; contradiction. Hence the tangent bundle of $S^{2n}$ admits no nontrivial splittings, and so $S^{2n}$ is not the total space of any nontrivial smooth fiber bundle of closed smooth manifolds. 
(Maybe this argument can be rescued in the topological setting using tangent microbundles?)
(Strictly speaking this argument's not quite complete: we also need to show that there aren't any interesting bundles with total space the disjoint union of $S^{2n}$ with something else. But fiber bundle maps $p : E \to B$ are open, so the image of $S^{2n}$ under such a map is a connected component of the base, and we can restrict our attention to this connected component without loss of generality. Then $E$ breaks up, as a fiber bundle, as a disjoint union of $S^{2n}$ and whatever else, and we can restrict our attention to $S^{2n}$ again without loss of generality. In other words, in the defining relations we can assume that $E$ and $B$ are both connected without loss of generality.)<|endoftext|>
TITLE: The space of compact subspaces of $R^\infty$ homotopy equivalent to a given finite complex.
QUESTION [11 upvotes]: Let $X$ be a finite (CW or simplicial - doesn't matter) complex and consider the space of all compact subspaces of $R^\infty$ which are homotopy equivalent to $X$, topologized say as a subspace of the space of all compact subspaces of $R^\infty$ with the Hausdorff metric.  What is known about this space?  I vaguely recall that it is related to Waldhausen's $A$-theory - are there any references for that?
Motivation: if one instead looks at the space of all subspaces homeomorphic to or respectively diffeomorphic (if $X$ is a manifold) to $X$ one has a model for $BHomeo(X)$ or $BDiff(X)$ (since the corresponding space of embeddings of $X$ in $R^\infty$ is weakly contractible and this space is the quotient of the space of embeddings by the homeo/diffeo group).  I have been having fun lately thinking about these models when $X$ is a finite set of points or a surface.  This space of subspaces homotopy equivalent to $X$ is different - in particular I do not see it (directly anyways) as a model for $BHomotopySelfEquiv(X)$ - so I'd like to know more what is known about it.

REPLY [4 votes]: To me, Hausdorff metric is an unaccustomed way of making such a space of spaces. I think I don't trust it because fixing a homotopy type gives you a set that is neither closed nor open in general. 
But yes I believe the picture is that some kind of "space of spaces of homotopy type $X$" is closely related to $A(X)$.
Let's start with smooth manifolds, but of codimension zero. For a fixed $n$ and a finite complex $K\subset \mathbb R^n$, let $M_n(K)$ be the space of smooth compact $n$-manifolds $N\subset \mathbb R^n$ containing $K$ in the interior as a deformation retract. (Let's say, the simplicial set where a $p$-simplex is a suitable thing in $\Delta^p\times \mathbb R^n$ such that the projection to $\Delta^p$ is a smooth fiber bundle.) You can map $M_n(K)\to M_{n+1}(K)$ by crossing with $[-1,1]$ (and doing something about corners), and you can consider the (homotopy) colimit over $n$. Using the classification of $h$-cobordisms you can work out that the set of components is the Whitehead group of $K$. The loopspace of one component is the smooth stable pseudoisotopy space of $K$. To get the idea, think of the case when $K$ is a point: the space $M_n(K)$ is then, after you discard extraneous components corresponding to cases where the boundary is not simply connected -- which were going to go away anyway upon stabilizing over $n$ -- your quotient of {embeddings $D^n\to\mathbb R^n$}~$O(n)$ by {diffeomorphisms $D^n\to D^n$}. It's also a kind of "space of all $h$-cobordisms on $S^{n-1}$, and thus a delooping of the (unstable) pseudoisotopy space of $S^{n-1}$.
When $K$ is more complicated than a point, it's important to distinguish between the space of all blah blah blah containing $K$ as a deformation retract and the space of all blah blah homotopy equivalent to $K$; they differ by the space of homotopy equivalences $K\to K$. 
There is a similar story for the piecewise linear or topological case.
The piecewise linear manifold version of this construction can, I believe, be shown to be equivalent to a non-manifold construction more like what you asked about: some kind of "space of compact PL spaces in $\mathbb R^n$ containing a fixed $K$ as deformation retract". 
Waldhausen tells us that the stable smooth construction above and the stable PL construction above are respectively (the underlying spaces of spectra which are) the fiber of a map from the suspension spectrum of $K\cup{point}$ to $A(K)$, and the fiber of a map from $A(*)\wedge (K\cup{point})$ to $A(K)$.<|endoftext|>
TITLE: Conventions for definitions of the cap product
QUESTION [12 upvotes]: In singular (co)homology, if $\alpha\in C^*(X)$ and $x\in C_*(X)$, then the cap product $\alpha \cap x$ is generally defined by the following process:

Apply to $x$ the diagonal map $C_*(X)\to C_*(X\times X)$ followed by some choice of Alexander-Whitney chain equivalence $ C_*(X\times X)\to C_*(X)\otimes C_*(X)$ to obtain an element $\sum y_i\otimes z_i$. 
Apply $\alpha$ to $\sum y_i\otimes z_i$ by the slant product, or, in other words and roughly speaking, apply $\alpha$ to ``half of the factors''. 
Depending on your favorite conventions (and what you're trying to accomplish), there may also be a sign (which might also be part of your definition of the slant product - but let's ignore this).

The reason I'm being so cagey with wording in part 2 is directly related to my question: In almost all major textbook sources I have consulted, step 2 is performed by forming $\sum y_i \alpha(z_i)$, which strikes me as somewhat unnatural, forcing the $\alpha$, which starts off on the left to jump all the way over the $y_i$ terms to get to the $z_i$ terms on the right. Is there a good mathematical reason for this convention? Why not define the cap product to be $\sum \alpha(y_i) z_i$?
The one major exception to this convention seems to be Hatcher. He does form $\sum \alpha(y_i) z_i$, but he also writes cap products as $x\cap \alpha$, so his cochain also has to jump, but it jumps over the $z$s instead!
(For the record, I'm not asking this question out of idle pickiness. Jim McClure and I have been doing a lot of work with cap products recently, and we're trying to be consistent amongst various conventions for various issues, but preferably with good reasons thrown in!)

REPLY [8 votes]: This is just an expanded version of Tyler's comment, I think.
Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$. Let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.
I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$. 
If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y_i\otimes z_i]$ where the coproduct of $x$ is $\Sigma y_i\otimes z_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get
$[a,bx]=[ab,x]=[a\otimes b,\Sigma y_i\otimes z_i]=\Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,\Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.
The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.
It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.
Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.<|endoftext|>
TITLE: Looking for references for an implicit differentiation formula
QUESTION [7 upvotes]: In a paper which I submitted to a peer-reviewed math journal in April 2010, I proved a formula for the n-th derivative $\frac{d^n z}{dw^n}$ in terms of (as a polynomial over the integers) of the partial derivatives of a given implicit function, $G(z,w)=0$, with respect to $z$ and $w$ (and negative integer powers of the "separant", $G_z$, the first partial derivative of $G$ with respect to $z$).
This is classic first-semester calculus homework exercise: to compute $\frac{d^n z}{dw^n}$ for n=1 and 2, namely, 
$\frac{dz}{dw} = - \frac{G_w}{G_z}$
$\frac{d^2 z}{dw^2}  = - \frac{G_{zz}G_{ww}}{G_{zzz}} + 2\frac{G_{zw}G_w}{G_{zz}} - \frac{G_{ww}}{G_z}$ 
I did so not knowing whether any one had proved the general formula first, because I am busy building on, generalizing, and using this result for other things, including chemical processing.
I have since proved the partial differential generalization of this implicit differentiation formula: i.e. given $G(z,w_1,...,w_N)=0$, compute
$\frac{d^{({u_1}+...+u_N})}{{dw_1}^{u_1}\cdots {dw_N}^{u_N}}z$ as a Laurent polynomial over the integers of the partial 
derivatives of $G$ with respect to $w_1,...,w_N$, and $z$  
I do not have access to most peer-reviewed journals. I have had to make do with Google searches, Wolfram Research's MathWorld, online searches through my county library, and the help from one mathematician friend who has sent me related papers. 
Most of the papers my friend sent me concern the Faa da Bruno formula (FdBF) and its generalizations, and the Lagrange Inversion Formula (LIF). Both the FdBF and LIF are very closely related to what I am doing, but they can not be trivially applied to get (my) general formula.  (I tried... for about 8 months.)  I have studied G.P. Egorychev's book: "Integral Representations of Combinatorial Sums" intensely, especially the back, with the multivariable generalizations of the LIF.
No, I am not in school. This is not a homework problem. This is "free-lance" research.
I am not asking for a solution to the problem (as I already solved it independently).
I simply want to know yes or no whether someone has already done this.
And, if so, where.
Thank you for the responses, in particular to go to arxiv.org, which I forgot, since I had submitted 2 papers there myself.

REPLY [6 votes]: Zentralblatt has a sample service where you get 3 responses. I think you might want to look at the first item:
Zbl 1186.92069 Pongor, Gábor; Eöri, János; Rohonczy, János; Kolos, Zsuzsanna
Direct inversion in the spectral subspace: a novel method for quantitative and qualitative analysis of chemical mixtures. (English)
J. Math. Chem. 47, No. 3, 1085-1105 (2010). MSC2000: *92E99 65F15 92E10
Zbl pre05669072 Wang, Weiping
Generalized higher order Bernoulli number pairs and generalized Stirling number pairs. (English)
J. Math. Anal. Appl. 364, No. 1, 255-274 (2010). MSC2000: *11-99 05-99 
Zbl 1183.74370 Wang, Meng-Fu; Au, F.T.K.
Precise integration methods based on Lagrange piecewise interpolation polynomials. (English)
Int. J. Numer. Methods Eng. 77, No. 7, 998-1014 (2009). MSC2000: *74S30 65D30
$$ $$
$$ $$
Zbl 1186.92069
Pongor, Gábor; Eöri, János; Rohonczy, János; Kolos, Zsuzsanna
Direct inversion in the spectral subspace: a novel method for quantitative and qualitative analysis of chemical mixtures. (English)
[J] J. Math. Chem. 47, No. 3, 1085-1105 (2010). ISSN 0259-9791; ISSN 1572-8897
Summary: A novel method, called Direct Inversion in the Spectral Subspace (DISS), has been developed for the quantitative (and partly qualitative) analysis of chemical mixtures. The method belongs to the broad group of supervised classification'' methods: its use necessitates the components'pure'' spectra, either experimental or computed. On the basis of three simple conditions, an elegant, linearized system of equations has been deduced, taking into account a sole restriction via the Lagrange multiplier method. This restriction is seemingly redundant but it has been shown that with its use the unknown normalization constant of the components' descriptive weighted average (CDWA) spectrum can be taken into consideration. The system of linearized equations can be solved repeatedly until convergence. Any kind of spectra can be used; the method does not require the non-negativity of spectral data points. Two versions of the new method have been developed: the normalized and the non-normalized versions regarding the components' spectra. In ideal cases, the non-normalized version of the DISS method provides a mixture's accurate composition due to the iteration for getting the correct norm of the CDWA spectrum. Realistically, the normalized version of the DISS method identifies a mixture's composition within a few molar percentage points accuracy, according to the test results in IR and $^{1}$H-NMR spectroscopy. The normalized method functions without any calibration measurements and needs only a control of accuracy; it is hoped that it will be a useful tool for chemical and biochemical analysis as well as for spectral databases. The DISS method is also useful for qualitative analyses in a limited sense: in the case of computed spectra of the components the set of the de facto components determined could be somewhat wider than those existing in the real system.
MSC 2000:
    *92E99 Appl. of mathematics to chemistry
    65F15 Eigenvalues (numerical linear algebra)
    92E10 Molecular structures
Keywords: decomposition of molecular spectra; Lagrange multiplier; quantitative analysis; qualitative analysis; IR; NMR; EPR; UV/Vis; Raman; CD; VCD; hexa-chloro-buta-1,3-diene; dioxane; D-Camphor; L-Menthol; supervised classification; spectral databases<|endoftext|>
TITLE: Do you recognise this variant of the cubes operad?
QUESTION [6 upvotes]: In the subject of operads one comes across many quirky variants of more or less the same operad. The cubes operad has many incarnations with various interesting properties.  In a recent paper I came across one such variant.  It's so simple I presume other people have come across it before, so it would be nice to have a reference if that's the case.  I'll give a sketch of this rather simple construction, below. 
Let me describe the operad, what I think it's good for, and how it relates to other operads. First, I'll set up some notation convention with the cubes operad.
Def'n: (cubes) An increasing affine-linear function $[-1,1] \to [-1,1]$ is a little interval.  A product of little intervals $[-1,1]^n \to [-1,1]^n$ is a little $n$-cube.  The space $\mathcal C_n(j)$ is the collection of $j$-tuples of
little $n$-cubes whose images are required to have disjoint interiors, $\mathcal C_n(0)=\{*\}$ is the empty cube.  The collection $\mathcal C_n = \sqcup_{j=0}^\infty \mathcal C_n(j)$ is the operad of little $n$-cubes, it is a $\Sigma$-operad with 
structure maps 
$$\mathcal C_n(k) \times \left( \mathcal C_n(j_1) \times \cdots \times \mathcal C_n(j_k) \right) \to \mathcal C_n(j_1+\cdots+j_k)$$ 
defined by 
$$(L,J_1,\cdots,J_k) \longmapsto (L_1 \circ J_1, \cdots, L_k \circ J_k)$$ 
and $\mathcal C_n(j) \times \Sigma_j \to \mathcal C_n(j)$ given by $(L, \sigma) \longmapsto L\circ \sigma$.
Def'n: (overlapping cubes) A collection of $j$ overlapping $n$-cubes is an equivalence class of pairs $(L, \sigma)$ where $L=(L_1,\cdots,L_j)$, each $L_i$ is a little $n$-cube and $\sigma \in \Sigma_j$.  Two collections of 
$j$ overlapping $n$-cubes  $(L,\sigma)$ and $(L',\sigma')$ are taken to be equivalent provided $L = L'$ and 
whenever the interiors of $L_i$ and $L_k$ intersect $\sigma^{-1}(i) < \sigma^{-1}(k) \Longleftrightarrow 
\sigma'^{-1}(i) < \sigma'^{-1}(k)$.  Given $j$ overlapping $n$-cubes $(L_1,\cdots,L_j,\sigma)$ say the $i$-th cube $L_i$ is at height $\sigma^{-1}(i)$. $\sigma(1)$ is the index of the bottom cube, and $\sigma(j)$ is the index of the  top cube.  Let $\mathcal C_n'(j)$ be the space of all $j$ overlapping $n$-cubes, with the quotient topology induced by the equivalence relation. 
The structure map 
$$\mathcal C_n'(k) \times \left( \mathcal C_n'(j_1) \times \cdots \times \mathcal C_n'(j_k) \right) \to \mathcal C_n'(j_1 + \cdots + j_k)$$
is defined by
$$\left((L,\sigma), (J_1,\alpha_1), \cdots, (J_k, \alpha_k)\right) \longmapsto ((L_1\circ J_1, \cdots, L_k\circ J_k), \beta)$$
the permutation $\beta$ is given for $1 \leq a \leq k$, $1 \leq b \leq j_a$ by
$$\beta^{-1}\left(\sum_{i<a} j_i + b\right) = \left( \sum_{i<\sigma^{-1}(a)} j_{\sigma(i)}\right) +\alpha^{-1}_a(b)$$
This permutation is obtained by taking the lexicographical order on the set $\{(a,b) : a \in \{1,\cdots,k\}, b \in \{1,\cdots,j_a\}\}$ and then identifying with $\{1, 2, \cdots,j_1+\cdots+j_k\}$ in the order-preserving way. 
== The point ==
So there is a map of operads $\mathcal C_{n+1} \to \mathcal C'_n$ given by sending $(L_1, \cdots, L_j)$ to $(L_1^\pi, \cdots, L_j^\pi, \sigma)$ where we write $L_i = L_i^\pi \times L_i^\nu$ where $L_i^\pi$ is an $n$-cube and $L_i^\nu$ a $1$-cube.  The permutation $\sigma$ is any element $\sigma \in \Sigma_j$ such that $L_{\sigma(j)}^\nu(1) \geq L_{\sigma(j-1)}^\nu(1) \geq \cdots \geq L_{\sigma(1)}^\nu(1)$.  
Some of the nice things about this operad are:

(1) it's a multiplicative operad, the inclusion of the associative operad is given by the elements $(Id_{\mathcal [-1,1]^n}, \cdots, Id_{\mathcal [-1,1]^n}, Id_{\{1,\cdots,j\}}) \in \mathcal C'_n(j)$. 
(2) The map above $\mathcal C_{n+1} \to \mathcal C'_n$ is an equivalence of operads. 
(3) While $\mathcal C_{n+1}$ acts on spaces such as $\Omega^{n+1} X$, $\mathcal C'_n$ does not.  $\mathcal C'_n$ acts on spaces of the form $\Omega^n M$ where $M$ is a topological monoid. 


The operad of overlapping intervals $\mathcal C'_1$ has a certain affinity to the cactus operad. For example, imagine $[-1/2,1/2]$ as an element of $\mathcal C'_1(1)$ as being represented by $[-1,1]$ with a $1$-cell attached at the points $-1/2$ and $1/2$. 
And there are all kinds of variants of this idea -- overlapping discs, or overlapping framed discs, etc. So you can get cyclic multiplicative operads out of these types of constructions.
The criterion for getting the answer "right" is either showing me an occurance of this operad in the literature, or coming up with some convincing argument it's a new construction.

REPLY [5 votes]: I just now noticed this question.
Your operad is a suboperad of the "surgery cylinder" operad described in arxiv 1009.5025 (more recent version available here).  See Section 8 and figures therein.  In the notation of that paper, your operad corresponds to case where all the manifolds $M_i$ and $N_i$ are $n$-cubes and all the homeomorphisms $f_i$ are the identity.
The surgery cylinder operad can be thought of as describing a sequence of (generalized) surgeries on an initial manifold $M_0$, yielding a final manifold $N_0$.  At the $i$-th stage ($1\le i \le k$) we remove a codimension-0 submanifold $M_i$ replace it with $N_i$, where $m_i$ and $N_i$ have the same boundary.  To get your overlapping $n$-cubes operad, let $M_0$ be the "big" $n$-cube and $M_i = N_i$ be the $i$-th little $n$-cube.  In other words we have a sequence of pointless surgeries in which little $n$-cubes are removed and replaced with identical copies of themselves.<|endoftext|>
TITLE: Why does finitely presented imply quasi-separated ?
QUESTION [15 upvotes]: By the EGA definition, a morphism of schemes of finite presentation is required to be quasi-separated. As far as I can see, removing this requirement does not prevent from proving the basic properties such as stability of the notion under composition, products, etc. So my question is :

where exactly in proving important theorems involving morphisms of finite presentation is the quasi-separated assumption crucial ?

Note that a morphism of finite type is not required to be quasi-separated.
All kinds of examples and counter-examples will be appreciated.

REPLY [28 votes]: One of the main interests in finitely presented morphisms comes from the various theorems in EGA IV,8. They show that for many questions about morphisms of schemes and sheaves on them, the condition of finite presentation allows one to reduce to a noetherian situation. For these theorems the assumption of quasi-separatedness is crucial. 
Let me quickly try to explain why. The heart of the reduction to the noetherian case are theorems like the following: Let X over Spec A be a finitely presented scheme. Then there is a subring $A_0$ of $A$ which is a finitely generated $\mathbb{Z}$-algebra (and in particular noetherian) and an $A_0$-scheme $X_0$ of finite presentation such that $X$ arises from $X_0$ via the base-change $A_0\to A$. If $X$ is affine, this is pretty clear, as $X$ is definied by finitely many equations in an affine space over $A$. In order to pass from the affine case to the general case, it does NOT suffice to know that we can cover $X$ by finitely many affine pieces (which would be the assumption of quasi-compactness), but we also need that the glueing data for the affine pieces are somehow described by a finite number of equations. This is ensured by the assumption that the intersection of two affine pieces is quasi-compact which corresponds precisely to the assumption that $X$ is quasi-separated over A.
I guess that these theorems were the reason for Grothendieck to include this condition in the definition of finitely presented.<|endoftext|>
TITLE: Detecting almost-primes quickly
QUESTION [12 upvotes]: There are many fast algorithms (deterministic and probabilistic) for detecting primality. Are there any fast algorithms (probabilistic ones allowed) known for detecting whether a number is the product of at most two primes?
In this context, can anybody think of a property fulfilled by all or most products of two primes, but by few other numbers? (You can't explicitly use divisor sums or the Moebius function, obviously, as then the game becomes (a) trivial (b) of doubtful computational utility.)

REPLY [2 votes]: There is a survey of open problems related to factoring algorithms, Open problems in number theoretic complexity, II, by Adleman and McCurley.  It lists so many questions related to distinguishing almost primes from other composite numbers that it seems safe to infer that it is an open problem to do it quickly, and that you would either use a factoring algorithm or a related method.  For instance, computing the Mobius function as you say is a known open problem, and detecting an almost prime can be viewed as a special case of computing the Mobius function.  Because, of course, it is easy to distinguish a prime from an almost prime.
As for relevant properties, the most important relevant property of a prime $p$ is that its multiplicative group $(\mathbb{Z}/p)^\times$ is cyclic of order $p-1$.  The Miller-Rabin primality test is based on that fact.  Now, if $n = pq$ is an odd almost prime, then the multiplicative group $(\mathbb{Z}/n)^\times$ is a product of two cyclic groups. Conversely, if $(\mathbb{Z}/n)^\times$ is a product of two cyclic groups and $n$ is odd, then $n$ is a product of two prime powers.  But, there is too little control over the cardinality of this group.  In the almost prime case, it has $(p-1)(q-1)$ elements.  Even if you found the 2-torsion elements, in other words all solutions to $x^2 \equiv 1 \bmod n$, you would be able to factor $n$.
The set of almost prime numbers is an important statistical ersatz for the set of prime numbers in analytic number theory.  However, in computational number theory, no particular reason leaps out that identifying almost primes is easier than other factoring-related questions that are accepted as hard.<|endoftext|>
TITLE: Difference between a 'calculus' and an 'algebra'
QUESTION [23 upvotes]: What is really the conceptual difference between a calculus and an algebra. 
Eg. Is SKI combinator calculus really a calculus?
A friend claims that free variables are fundamental for a calculus, and as such that SKI is not a calculus, but an algebra.

REPLY [2 votes]: a calculus is a symbolic system for computation where computation can most generally be seen to be a spatial reorganization of symbols. any kind of numerical computation can be described in terms of recursion which can fundamentally be seen as a symbolic manipulation process. logic calculi and the calculus that set theory uses is also describable as a symbolic computation system.
an algebra is a mathematical structure in the informal sense which turns out to be a vector space with the added ability to multiply the actual vectors together. the complex numbers with addition and multiplication is an algebra over the complex numbers; http://en.wikipedia.org/wiki/Algebra_over_a_field<|endoftext|>
TITLE: Finite dimensional spherical representation of $SO(n,1)(\mathbb{R})$
QUESTION [5 upvotes]: I'm looking for an explicit description of all the finite dimensional irreducible representation of the Lie group $SO(n,1)(\mathbb{R})$. Can you tell me, where I can find this description ? Thank you.

REPLY [6 votes]: Bonsoir Ludo! I am puzzled by the fact that your title asks something more restrictive than the OP, since the latter does not contain the word "spherical". Let me answer the latter first. Any finite-dimensional representation of $SO(n,1)(\mathbb{R})$ extends to a representation of the complexification, which is $SO_{n+1}(\mathbb{C})$. By Weyl's unitary trick, those are in 1-1 correspondence with unitary finite-dimensional representation of the maximal compact subgroup of the complexification, here $SO_{n+1}$. The finite-dimensional, unitary, irreducible representations of such a group are parametrized by their highest weight, and can be described via Verma modules, see Chapter IV in Knapp's "Representation theory of semi-simple groups" (Princeton UP, 1986). 
Now, if you need only spherical irrep's, this amounts to consider irreducible $SO_{n+1}$-representations having non-zero $SO_n$-invariant vectors; or, equivalently (by an easy case of Frobenius reciprocity), irreducible $SO_{n+1}$-sub-representations of $L^2(S^n)$ (where $S^n=SO_{n+1}/SO_n$ is the $n$-sphere). These correspond to homogeneous harmonic polynomials in $n+1$ variables.<|endoftext|>
TITLE: Emergence of English as the dominant mathematical language
QUESTION [31 upvotes]: My impression is that most math papers (and almost all of the most important ones) are now published in English.  Not long ago (historically) publishing in French, German, Russian, etc. were more common.  I'm curious when and how this transition occurred, and what it looked like while it was happening.
Of course English is the dominant language for many other areas such as publications in the sciences in general, international business, the internet, etc.  Nonetheless I have the impression that the transition occurred later and more slowly in math than in some of these other areas.  This may be false; I don't know why I think this.
Perhaps I have this idea because math graduate programs are unique among technical fields in having a language requirement, though that could also be explained by our tendency to read more old papers than say, biologists.  But it seems plausible that the transition would occur later in math because math can be more easily undertaken as a solitary activity than the sciences.
My main question is: where would I find data on the representation of different languages in papers in top math journals as, say, a fraction of total papers?  Of course I could try to determine what the top journals were at various times and go through their tables of contents for each issue and compile such data myself.  But it seems plausible someone may have already done a study of this and in much more depth than I would have time to do by myself.
Secondarily, perhaps there are published anecdotal accounts about how and when this transition occurred?  I can imagine there may have been resistance by some.  It may have happened at different times in different subfields for different reasons.  Perhaps some landmark papers in English paved the way.
I would be grateful for pointers to literature examining any of these issues.  Comparisons with other fields such as physics would also be interesting, but I'm primarily interested in math.

REPLY [4 votes]: In Physics and in Chemistry, particularly in Organic Chemistry, German was the lingua franca, the bridging language which it was necessary to know how to read and how to publish in, as the leading journals and the leading scientists communicated in German, but that was in the 19th century and the early parts of the 20th century.  The mass exodus of scientists from Germany starting immediately prior to the Second World War seems to be a big part of the change, at least in the Physical Sciences.<|endoftext|>
TITLE: Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology
QUESTION [12 upvotes]: It occurred to me that a subgroup of the modular group $\Gamma$ is a congruence subgroup iff it contains a subgroup of the form $\Gamma(N)$, while a subgroup of a general topological group is open iff it contains an open subgroup. This suggests making a topology on the modular group $\Gamma$ with the subgroups $\Gamma(N)$ as a basis of open neighborhoods of the origin so that $\Gamma$ becomes a topological group. It would then follow that a subgroup of $\Gamma$ is a congruence subgroup iff it is open.
Furthermore, for any $\gamma \in \Gamma$ not equal to the identity, there exists $N$ such that $\gamma \notin \Gamma(N)$, so this topology is Hausdorff, even totally disconnected.
I was inspired in part by this thread and looked at this paper but could not find anything about this idea.
Has anyone considered this topology? Does it provide insight into the problem of determining whether a group is a congruence subgroup?

REPLY [4 votes]: It's worth mentioning that there is a general definition of the "congruence subgroup topology" on the automorphism group of a profinite group $\widehat{A}$: namely, the open subgroups of $\text{Aut}(\widehat{A})$ under this topology are generated (as a topology) by the subgroups
$$\Gamma[K]: = \ker(\text{Aut}(\widehat{A})\rightarrow\text{Aut}(\widehat{A}/K))$$
as $K$ varies over finite index characteristic subgroups of $\widehat{A}$. You can find this for example in Ribes/Zalesskii's book Profinite Groups $(\S4.4)$. In the same way we also obtain a "congruence subgroup topology" on $\text{Out}(\widehat{A})$.
For a general abstract discrete group $G$, for every abstract group $A$ and a homomorphism $\phi : G\rightarrow\text{Aut}(A)$ (or $\text{Out}(A)$), one obtains the notion of "congruence subgroups" on $G$ (relative to $\phi$), which are by definition those subgroups which contain a preimage of some $\Gamma[K]$ under the map
$$G\stackrel{\phi}{\longrightarrow}\text{Aut}(A)\subset\text{Aut}(\widehat{A}),$$
where the hat denotes profinite completion.
For the modular group $G := \text{SL}_2(\mathbb{Z})$, one can naturally take $A = \mathbb{Z}^2$, and consider the natural homomorphism 
$$\phi : \text{SL}_2(\mathbb{Z})\rightarrow\text{Aut}(\mathbb{Z}^2)$$
 Then, $\text{Aut}(\widehat{\mathbb{Z}^2}) = \text{GL}_2(\widehat{\mathbb{Z}})$, the finite index characteristic subgroups of $\widehat{\mathbb{Z}}^2$ are $K_n := n\widehat{\mathbb{Z}}\times n\widehat{\mathbb{Z}}$, and the preimage of $\Gamma(K_n)$ in $\text{SL}_2(\mathbb{Z})$ is just the classical principal congruence subgroup $\Gamma(n)$.
However, for $G = \text{SL}_2(\mathbb{Z})$, there is another natural choice for $\phi$, given by noting that if $F_2$ denotes the free group of rank 2, then $\text{Out}(F_2)\cong\text{GL}_2(\mathbb{Z})$. From this, one obtains a natural map
$$\text{SL}_2(\mathbb{Z})\stackrel{\phi'}{\longrightarrow}\text{Out}(F_2)\subset\text{Out}(\widehat{F_2})$$
The preimages of $\Gamma[K]$ as $K$ varies over finite index characteristic subgroups of $\widehat{F_2}$ then generate the congruence subgroups (relative to $\phi'$) in the sense described above.
The cool thing is that while $\text{SL}_2(\mathbb{Z})$ doesn't have the congruence subgroup property relative to $\phi$, it does relative to $\phi'$. In particular, there are finite index subgroups of $\text{SL}_2(\mathbb{Z})$ which are congruence relative to $\phi'$, but not congruence in the usual sense (ie relative to $\phi$). There is a paper of Bux-Ershov-Rapinchuk which has a nice description of this phenomenon, though the result they prove is originally due to Asada.
In fact, the notion of congruence relative to $\phi$ gives rise to the usual moduli interpretations of "congruence" modular curves, whereas the more general notion of congruence relative to $\phi'$ gives rise to the moduli interpretations for "noncongruence" modular curves.<|endoftext|>
TITLE: Are there alternative proofs for existence/uniqueness of ODE solutions?
QUESTION [12 upvotes]: Consider the differential equation $\dot x = f(x)$. The standard proofs are 

The Picard iteration based proof of existence/uniqueness for Lipschitz $f$. 
The Peano existence theorem for continuous $f$.
The Caratheodory existence theorem for measurable $f$.

My question is as follows. Assuming a Lipschitz $f(x)$, are there any other proofs out there for existence of solutions (in some reasonable sense) for ODEs?

REPLY [4 votes]: The simplest proof (though 1 variable is probably not what you want)
is in Peter Lax's Calculus book, Springer, 1976, page 442. It uses
only the fundamental theorem of calculus. Existence, uniqueness, and 
asymptotics are proved for the case there.<|endoftext|>
TITLE: Is the 4x5 chessboard complex a link complement?
QUESTION [18 upvotes]: The 2x3 and 3x4 chessboard complexes (form a square grid of vertices and make a simplex for any set of vertices no two of which are in the same row or column) are a 6-cycle and a triangulated torus with 24 triangles, respectively. The 4x5 chessboard complex is only a pseudomanifold — each vertex has the 3x4 torus as its link, rather than a spherical link that a proper manifold would have — but if you delete its 20 vertices you get a bona-fide cusped hyperbolic 3-manifold, triangulated by 120 regular ideal tetrahedra. It sort of looks like the kind of manifold you might get as the complement of a 20-component link in Euclidean space. Is it a link complement? And if so, which link is it the complement of?
Edit: here are a couple of general references on chessboard complexes.
Ziegler, G. M. (1994). Shellability of chessboard complexes. Israel J. Math. 87: 97–110.
Björner, A.; Lovász, L.; Vrecica, S. T.; Zivaljevic, R. T. (1994). Chessboard complexes and matching complexes. J. London Math. Soc. 49: 25–39.
They also have important applications to the proof of colored Tverberg theorems in discrete geometry: see, e.g.
Pavle V. M. Blagojević, Benjamin Matschke, Günter M. Ziegler (2009). Optimal bounds for the colored Tverberg problem. arXiv:0910.4987.

REPLY [16 votes]: Figure 1.27 of 
http://math.berkeley.edu/~matthias/research/matthias_goerner_thesis_print.pdf
shows the link.<|endoftext|>
TITLE: Local linearization of ODE at singular point
QUESTION [15 upvotes]: I would like the simplest example of the failure of an ODE to be locally diffeomorphic to its linearization, despite being locally homeomorphic to it.  More precisely,  consider x' = f(x) with f(0) = 0 in R^n.  Let A = f'(0) so that the local linearization is x' = Ax.  Suppose the eigenvalues of A all have nonzero real part (i.e., 0 is a hyperbolic critical point).
The Hartman-Grobman theorem tells us that there is a homeomorphism of a neighborhood of 0 which conjugates the system x'=f(x) to its linearization x'=Ax.  If one reads the elementary `differential equations from the dynamical systems point of view' literature, however, you will gain the false impression that there is a diffeomorphism h : U --> U of a nhood of 0 which does this, and that further, one can even do this with h'(0) = I, the identity matrix.  The point of this is to ensure that the trajectories of the nonlinear system are tangent to the trajectories of the linearization:  if $h'(0)\neq I$ then this may be false.
Smale's stable manifold theorem gives partial information in this direction, saying that the stable manifolds of the system and its linearization are tangent, and similarly for the unstable manifolds.  In 2D, at a saddle, this is sufficient to imply that the separatrices of the original system are tangent to those of the linearization.  For a node in 2D, or in higher dimensions, I am under the impression this need not hold. I even think I had worked out an example many years ago, which I no longer recall.
Any enlightenment on this issue would be much appreciated.  I am not at all expert in these matters, so welcome any corrections, if I have distorted the facts.  I am hoping for a 2 dimensional example.
 Added later:
Yuri:  resonances and normal forms are definitely relevant. When I get time I will look into the references you suggest.
Here's an example of what I am trying to avoid.  Consider a flow on the unit disk with trajectories the radial lines y = mx.  Conjugate by $(r,\theta) \mapsto (r,f(r,\theta))$ where $f(0,\theta)$ is constant on, say, $[-\pi/2,\pi/2]$, e.g. $f(r,\theta) = r\theta$ on $[-\pi/2,\pi/2]$ and $(2\theta-\pi) + r(\pi-\theta)$ in the left half plane.  Now all the trajectories leaving the unit circle in the right half plane approach 0 along the positive x-axis.  So, conjugating with such a homeomorphism has replaced a single trajectory with horizontal tangent by an entire interval of such.
I strongly suspect that this sort of pathology doesn't happen with polynomial flows.  Perhaps the normal forms will show this.  What I would hope is that for each slope, there is a 1-1 correspondence between the trajectories in the original flow and its linearization approaching or leaving the singular point at that slope.  In particular, that you can't have a single trajectory in the linearization but a whole interval of them in the nonlinear flow.  A counterexample to this hope would be disappointing, but would settle the matter.

REPLY [5 votes]: In three dimensions, Hartman gave the example $dx/dt=ax$, $dy/dt=(a-b)y+cxz$, $dz/dt=-bz$ where $a>b>0$ and $c \neq 0$. On the other hand, any $C^2$ planar flow is $C^1$ linearizable (another result by Hartman), so you will not find any polynomial examples in the plane. See Linearization via the Lie Derivative  by Chicone & Swanson for references and more details.
(Strengthening the smoothness assumption will not give any more than $C^1$ linearizability even in the planar case, as shown by the $C^{\infty}$ example $dx/dt=-x$, $dy/dt=-2y+x^2$, which is easily solved explicitly: $x(t)=x_0 e^{-t}$, $y(t)=(y_0+x_0^2 t) e^{-2t}$. The solution curves are of the form $y=Ax^2-B x^2 \ln|x|$, and are therefore only $C^1$ at the origin, whereas the linearized system of course has smooth solution curves.)<|endoftext|>
TITLE: Applications of Math: Theory vs. Practice
QUESTION [9 upvotes]: I have a problem: I learned about a lot of the applications of mathematics from academics. Neither they nor I have had much contact with the "real world" to go and see for ourselves how mathematics are used today (rather than, say, in the pre-computer age).
So if there are non-academics out there reading MO, I would very much like to hear from them about how their use of mathematical tools may or may not differ from the academic training they received. I don't expect we'll get a very representative sample of math-users on MO, but that's quite all right, anecdotal evidence is all I'm after.
Precision: I want to read about the tools being used rather than, say, an underlying mathematical motivation (which is another legitimate role of mathematics, but not really part of my question). So for instance, you may say that Google's PageRank algorithm was motivated by the theory of Markov chains, but (from what I can tell), I would not say it uses Markov chains.

REPLY [6 votes]: My work draws on various bodies of mathematics. Here's a brief description (by no means exhaustive) of how I use math in my work:

Mathematical programming/optimization: Optimization is used for anything from reconciling actual data to a model's (regression), estimating unknown parameters in a system, to finding the best inputs that will extremize some functional in a dynamic system. The applications are endless. When people think mathematical programming, they think Linear Programming. But convex nonlinear programming is actually very well-established. In fact, large problems in nonconvex optimization are routinely solved (although modeling a nonconvex system can be quite an art). 
Real/functional analysis: useful for understanding optimization algorithms. An understanding of convex functions and sets is crucial -- they lead to global solutions (with guarantees) without solving an NP-hard problem, so we exploit convexity properties whenever possible. (Lipschitz) continuity is another important idea, subgradients etc. are important concepts in nonsmooth optimization. Real analysis is not applied directly, but a good understanding of it is required for reading convergence proofs or descriptions of optimization algorithms.
Computational Geometry: ideas like convex hulls, Voronoi diagrams, etc. are useful in optimization. I use them to partition a problem space into convex regions, or to parametrize a space. The region bounded by convex polytopes can be represented by a set of inequality constraints that can be enforced in an optimization problem. Discrete optimization is used to optimally switch between these regions. 
ODE/DAE theory: used for modeling dynamic systems. In particular, understanding the notion of index in DAEs can help one develop models that are amenable to reliable numerical solution. 
Calculus. Differential calculus is used everywhere (e.g. model sensitivity analysis, automatic differentiation, postoptimality analysis)
Statistics: projection methods like the Karhunen-Loewe transform (related to SVD) are used to reduce the dimensionality of large models constructed from data. They're also the only way to handled correlated/collinear data (in practice, most large datasets in the real world are correlated. The assumption of factor independence built into standard regression techniques often does not hold, so methods like multiple linear regression often have to be modified for instance into principal components regression in order for them to be usable on real world large datasets). Also, tools like time-series analysis are used to construct time series models from data. 
Linear algebra: used almost everywhere. They're the basic building blocks for working with nonlinear systems. In particular, efficient numerical solution of sparse structured matrices is crucial to the efficiency of large-scale nonlinear optimization algorithms (the bottleneck is often in the linear algebra solvers, not in the optimization algorithm itself). Tools like SVD are frequently used. 
Numerical methods: used everywhere. Understanding concepts like numerical conditioning is crucial; when modeling, one wants to end up with a system with a Jacobian that is well-conditioned with respect to inversion.
Misc: Diophantine equations are used to derive certain control laws. Laplace transforms are used for modeling linear-time-invariant systems because they allow differential equations to be manipulated as algebraic ones. Algebraic Riccati equations are solved in the derivation of the Kalman gain. Fixed-point iteration is used to converge decomposed models.<|endoftext|>
TITLE: Trost's Discriminant Trick
QUESTION [31 upvotes]: The following little trick was introduced by E. Trost 
(Eine Bemerkung zur Diophantischen Analysis, 
 Elem. Math. 26 (1971), 60-61). For showing that a diophantine equation
such as $x^4 - 2y^2 = 1$ has only the trivial solution, assume that 
$a^4 - 2b^2 = 1$; then the quadratic equation $a^4 t^2 - 2b^2 t - 1 = 0$
has the rational solution $t = 1$, hence its discriminant $4b^4 + 4a^4$ 
must be a square. Thus $a = 0$ or $b = 0$ by Fermat's Last Theorem for 
exponent $4$.
Trost gave a few other nice applications of this trick, but I have not seen
it in any textbook on number theory. My questions:

Was Trost's trick noticed by anyone (before or after Trost)?
Are there any other cute applications?

REPLY [5 votes]: As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.
The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.
$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$
where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$<|endoftext|>
TITLE: Fermat's Bachet-Mordell Equation 
QUESTION [16 upvotes]: Fermat once claimed that the only integral solutions to $y^2 = x^3 - 2$ are $(3, \pm 5)$.
Fermat knew Bachet's duplication formulas (more precisely, Bachet had a formula for computing what we call $-2P$), which for $y^2 = x^3 + ax + b$ says
$$ x_{2P} = \frac{x^4-8bx}{4x^3+4b} 
                = \frac x4 \cdot \frac{x^3 - 8b}{x^3+b}.$$
Using this formula it is easy to prove the following:
 Consider the point $P = (3,5)$ on the elliptic curve $y^2 = x^3 - 2$.
The $x$-coordinate $x_n$ of $[-2]^nP$ has a denominator divisible by 
$4^n$; in particular, $[-2]^nP$ has integral coordinates only if $n = 0$.
In fact, writing $x_n = p_n/q_n$ for coprime integers $p_n$, $q_n$, we find
$$ x_{n+1} = \frac{x_n}4 \cdot  \frac{x_n^3 + 16}{x_n^3 - 2} 
           = \frac{p_n}{4q_n} \cdot 
             \frac{p_n^3 + 16q_n^3}{p_n^3 - 2q_n^3}. $$
Since $p_n$ is odd for $n \ge 1$ and $q_n = 4^nu$ for some odd number $u$ (use induction), we deduce that the power of $2$ dividing $q_{n+1}$ is $4$ times that dividing $q_n$.
My question is whether the general result that $kP$ has integral affine coordinates if and only if $k = \pm 1$ can be proved along similarly simple lines. The modern proofs based on the group law, if I recall it correctly, use Baker's theorem on linear forms in logarithms.

REPLY [7 votes]: The above result for $p=2$ can be strengthen: the denominators of $x(kP)$ and $y(kP)$ are even if and only $k$ is even. In non-elementary language, this follows from the fact that the Tamagawa number at $2$ is 1 (or that $P$ has good reduction at $2$) and that the reduction is additive; so the kernel of reduction has index $2$. It also follows easily from the duplication formula and the fact that $2$ can not divide $x$.
Similarly, for any prime $p$, one could find a congruence for $k$ such that the denominator of $x(kP)$ is divisible by $p$. If $p$ has good reduction, i.e. $p>3$ then there is a number $M_p$ dividing $N_p=\vert\tilde E(\mathbb{F}_p)\vert$ such that $x(kP)$ is not $p$-integral if and only if $k$ is a multiple of $M_p$. 
So the answer to your question is, I guess, a "No". Just from looking at the group law, i.e. the addition and duplication formula, without using something further, one can not be certain that for any $k$ there is a $p$ such that $M_p$ divides $k$. E.g. the denominator of $x(5P)$ is the square of $29 \cdot 211\cdot 2069$.
As Kevin comments, there are of course other elementary ways, not along these lines.<|endoftext|>
TITLE: What does the computer suggest about the parity of p(n), for n in a fixed arithmetic progression?
QUESTION [11 upvotes]: Let p(n) be the number of partitions of n. A famous theorem of Euler allows one to compute
the parity of p(n) quickly for quite large n. In:
On the distribution of parity in the partition function, Math. Comp. 21 (1967) 466-480
Parkin and Shanks, on the basis of such computations, conjecture that for each rho> 1/2,
the number of n that are smaller than x and have p(n) even is (x/2)+ O(x^rho).
Nothing remotely like this has been proved. But I wonder:
Does the computer suggest that the analogous result holds when we restrict n to lie in a fixed
arithmetic progression? Or does it suggest that there are there arithmetic progressions in which the n with p(n) even predominate? (And are there MO viewers interested in making such
calculations, if they haven't yet been done?)

REPLY [5 votes]: The answer to your very last question is yes. As for the rest:
Computing the parity up to N is, in theory, quasi linear by using Pentagonal Number Theorem: compute the inverse of the pentagonal number power series mod $x^N$. This is of course faster than the quadratic-complexity "en masse" method of using simply the recurrence relation. I am not sure how inverse_mod is actually computed in sage, but it ran fine for the following code:
def get_parity(n):
    x = GF(2)['x'].0
    f = 1
    for k in xrange(1,sqrt(2*n/3)+10):
        f += x^((k*(3*k-1))/2) + x^((-k*(-3*k-1))/2)
    return f.inverse_mod(x^n).coeffs()

def get_best(l):
    n = len(l)
    highest, lowest = [0,0,0], [0,0,1]
    for a in [1..sqrt(n)]:
        for b in [0..a-1]:
            score = (l[b::a].count(1) + 0.0)*a/(n-b)
            if score > highest[2]:
                highest = [a, b, score]
            elif score < lowest[2]:
                lowest = [a, b, score]
    return highest, lowest

l = get_parity(2^22)
best = get_best(l[:2^21])
for x in best:
    print x[2], (l[x[1]::x[0]].count(1)+0.0)*x[0]/(len(l)-x[1])

What this does is calculate for every a.p. $an+b$ the density of odd values, and finds the two a.p.'s with most and least odd values.
get_parity took more than 20 minutes (not sure exactly since I was watching TV), and get_best took almost an hour (again, I think). I'm running a macbook pro with 4gb ram.
The results were:
(1442, 766) 0.557846694263366 0.531611381990907
(1389, 357) 0.440522320970815 0.468967411597574

This is to say that the most special a.p.'s up to $2^{21}$ become a bit less special when going up to $2^{22}$, and quite close to equidistribution. Hence, I would say that yes, the computer does suggest that the analogous result holds when we restrict n to lie in a fixed arithmetic progression.
Edit 1: If we change the "score" of an a.p. to:
$$\frac{\\\#\{odd\\ in\\ a.p.\} - \\\#\{even\\ in\\ a.p.\}}{\sqrt{length}}$$
Then up to $2^{21}$ the largest values are:
(712, 254) 4.84633129231862
(1389, 357) -4.60795692951670<|endoftext|>
TITLE: Explicit equations for Schubert varieties
QUESTION [8 upvotes]: How can one compute the Schubert variety (by compute I mean having actual polynomials that define it) for SL(n)? If this is well known forgive my ignorance and just point me to the right book/paper.
EDIT: Sorry I did not return here for quite some time. It is kind of amusing that the way I learned about Schubert varieties is not even mentioned.  Here is how I learned it:


$G$ an algebraic group with Lie algebra $\mathbf{g}$.
$L(\Lambda)$ is an integrable highest weight module for $\mathbf{g}$.
For $w$, an element of the Weyl group, consider the 1-dimensional root space $L(\Lambda )_{w \cdot \Lambda}$.
Denote the vector space $U(\mathbf{b}) \bullet L(\Lambda)_{w \cdot \Lambda}$ by $E_w(\Lambda)$ (you take the 1-dimensional root space and act on it by all the raising operators plus the Cartan). Then $E_w(\Lambda) \subset L(\Lambda)$.

Now we are ready: since $L(\Lambda )_{w \cdot \Lambda}$ is 1-dimensional it becomes a single point in $\mathbf{P} \left (E_w(\Lambda) \right)$. We look at the orbit $B \bullet L(\Lambda)_{w \cdot \Lambda} \subset \mathbf{P} \left (E_w(\Lambda) \right)$. We call its closure the Schubert variety associated to $w$ and $\Lambda$ and denote it by $S_{w, \Lambda}$.

I don't know if this is a good way of computing things but in principle it should give you any Schubert variety you need.

REPLY [6 votes]: Dave Anderson mentions the following basic fact, on which I want to elaborate: Let $V$ be an irreducible representation of the reductive group $G$, and let $G/P$ be the orbit of the high weight vector in $\mathbb{P}(V)$. Let $X=BwP/P$ be any Schubert variety in $G/P$. (For example, $V=\bigwedge^k \mathbb{C}^n$, $G/P$ is the Grassmannian $G(k,n)$ and $X$ is a Schubert variety in the Grassmannian.)
Then the homogenous ideal of $X$ is generated by the homogenous ideal of $G/P$ plus some linear forms. In other words, all you have to understand is (1) the equations of $G/P$ and (2) which elements of $V^*$ vanish on $X$. This result is due to Ramanathan, "Equations defining Schubert varieties and Frobenius splitting of diagonals", Pub. IHES 65 (1987) 61-90.
The equations defining $G/P$ are quadratic. In the case of Grassmannians in the smallest (Plucker) embedding, they are very classical and you can read about them in many places; I like Miller and Sturmfels book Combinatorial Commutative Algebra.
For Grassmanians in the Plucker embedding, it is also easy to describe the linear forms that vanish on a given Schubert variety; they will be the Plucker variables indexed by partitions which do not contain your partition. The general case isn't much worse, but I can't give an answer without first taking time to set up a notations for bases of $V$.<|endoftext|>
TITLE: Matrix factorization categories beyond the isolated singularity case
QUESTION [8 upvotes]: In his really nice thesis, Tobias Dyckerhoff proved the following theorems about matrix factorizations(of possibly infinite rank) over a regular local k-algebra R with a function w and residue field k such that the Tyurina algebra, T= $R/(w,dw)$ is finite dimensional. This last condition says that w has an isolated singularity. For further reference, let S denote the ring R/(w). 
1) The homotopy category of matrix factorizations has a compact generator as a triangulated category, which he denotes as $k^{stab}$.
2) As a consequence of 1), he derives that there is a natural complex which represents the identity functor thought of as an element of MF($R\otimes R,1\otimes w-w\otimes 1)$ which he denotes as the stabilization of the diagonal $R^{stab}$.
3) MF(R,w) is a Calabi Yau dg-category. 
Now my question is how much of the above remains true for when the singularity is non-isolated? In some writings, Kontsevich, while not explicitly saying so, writes as if the homotopy category always has a compact generator and that the category is there by "dg-affine", e.g. equivalent to D(A), the derived category of modules over a dg-algebra. Is this indeed known to be true or false? If not, is there a way to prove 2) without making reference to 1)? I'm asking because I haven't found anything about this stuff in the literature, but a lot of things in this field are not written or written in physics literature that I'm not familiar with. 
I've checked a few examples with non-isolated singularities and it appears that for example in the category of factorizations $(k[[x,y]], xy^2)$, that while $k^{stab}$ doesn't generate as Dyckerhoff proves, one has $(k\oplus k[[x]])^{stab}$, which I think does generate. The way I want to argue this is Dyckerhoff's theorem 3.6 that it is enough to show that $Tor_S(k\oplus k[[x]], M)$ implies that $Tor_S(N,M)=0$, where N is a finitely generated T module and M is any S module.Then one does an analysis of finitely generated modules over T(I didn't think about the characteristic 2 case) and does some devissage with short exact sequences. Please let me know if this sounds off. I also think that with a bit more calculation one can prove similarly that in (k[[x,y,z]], xyz) the module $(k[[x]]\oplus k[[y]]\oplus k[[z]]\oplus k)^{stab}$ is a compact generator.
Added: I think the right generalization of the above two examples is the following, in  http://websupport1.citytech.cuny.edu/faculty/hschoutens/pdf/finiteprojdim.pdf, the author introduces the notion of a "net". The above method should give a compact generator, whenever the net of finitely generated modules over T is generated as a net by finitely many modules. This happens for example when T has finitely prime ideals. The modules A/p, where p is a prime, generate the net of finitely generated projective modules over T, which is enough to prove the vanishing above. In particular, this should take care of the case when T has dimension 1. A question is what are some conditions on (R,w) which lead to the net of finitely generated modules over T being generated by finitely many objects?
Assuming that this right, I think that to derive 2 and 3 for these examples becomes a formality in view of Dyckerhoff's section 5. One just replaces his compact generator with the new one.

REPLY [9 votes]: The answer to (1) is yes for any local abstract hypersurface $S$ whose singular locus is closed (which is barely a hypothesis, and free in the case of interest). Let us write $\mathrm{Sing} \;S$ for the singular locus, which we can write as a union finitely many irreducible components corresponding to primes $\mathfrak{p}_i$ for $i=1,\ldots,n$. Then the image of the object
$$ \bigoplus_{i=1}^n S/\mathfrak{p}_i  $$
in the category of matrix factorizations is a compact generator. The point is that there is a notion of support for objects of the category of all matrix factorizations. For a hypersurface this support gives a classification of the localizing subcategories in terms of subsets of the singular locus. The image of the given object is supported everywhere so must generate.
The statement about the classification of localizing subcategories is not actually published anywhere (Iyengar has announced this as part of a more general result on complete intersections and I have an (independent and different) proof, also for complete intersections, which also works for certain singular schemes - but Iyengar has not yet released a preprint and the results of mine are in my thesis which isn't publicly available yet). Although there is  this paper by Takahashi, the main result of which is sufficient to give the statement about the generator above.
The generator above actually works for any noetherian ring which is locally a hypersurface if one works in one's favourite infinite completion of $D_{\mathrm{Sg}}(S)$.
 Edit:  As promised in the comments (well sort of, it didn't turn out to be soon)  here is the link to the preprint concerning subcategories of singularity categories from which one can deduce the existence of compact generators in certain cases as above.<|endoftext|>
TITLE: Applications of Measure, Integration and Banach Spaces to Combinatorics
QUESTION [17 upvotes]: I'm going to be teaching a Master's level analysis course(measure theory, Lebesgue integration, Banach and Hilbert spaces, and if there's time, some spectral or PDE stuff) in the fall. My problem is that while roughly half of the students will actually be using analysis in their further work, the rest of them are going to specialize in combinatorics, and while I want to convince them that they should know this stuff as part of their general mathematical culture, I'd also like to try to connect it to what they'll be working on. So far, I've found survey articles on applications of Ramsey theory to Banach spaces, and applications of harmonic analysis to additive number theory, but I was wondering whether anyone had some suggestions of references for applications of classical analysis to old-fashioned, classical combinatorics. (I realise that this is a pretty tall order, as on many levels, these two fields are at antipodes.) 
PS: I'm planning on talking about probability measures on discrete spaces, but I don't think that will convince the combinatorics people that hacking through the construction of the Lebesgue integral could have a practical payoff someday for them.

REPLY [3 votes]: As warm-up exercises using Cauchy-Schwarz and Holder's inequality, you could mention restricted subgraph bounds and incidence bounds. When you rearrange these bounds to bound the number of "rich/popular" points or vertices you get (weak) $L^p$ bounds for the degree of each vertex (viewed a function on the vertex set). In this setting it's easy to see how interpolation works, and it gives you a feel for how $L^p$ spaces work in a probability setting (i.e. on a compact domain). 
Here's a reference for the restricted subgraph bounds: http://murphmath.wordpress.com/2012/06/19/restricted-subgraph-bounds/. Instead of the convexity of ${x\choose s}$ in $x$, you can use Holder's inequality. The bounds can be rearranged using Chebyshev's inequality to state that (in the notation of that blog post):
$$|\{b\in B\colon \mathrm{deg}(b)\geq t\}|\leq C \frac{|A|^s}{t^s}$$
where $C$ is a constant depending on $s$ and $t$. Then using the "layer-cake" theorem (which is a nice exercise using Fubini's theorem!) you get that
$$||\mathrm{deg}(b)||_{L^s(B)}^s \leq C'|A|^s\log|A|.$$
I know this is pretty trivial, but this is actually what's at stake in the Kakeya maximal conjecture. Here one has a collection of tubes $T_1,\ldots,T_n$ and the goal is to prove $L^p$ bounds for the function
$$ f(x)=\sum_{i=1}^n \chi_{T_i}(x),$$
where $\chi_{T_i}$ is the indicator function on the tube $T_i$. Thus we are seeking $L^p$ bounds for the function that counts how many tubes are incident to to a point $x$! The proof of the two dimensional Kakeya conjecture is essentially the same as the proof restricted subgraph bound for $s=2$ in that blog post---the incidence graph for points and lines in a plane contains no $K_{2,2}$'s, and modulo some technicalities in the Kakeya case, the main tool is Cauchy-Schwarz. In fact, Wolff's paper on the Kakeya problem contains a finite field analog where the proof is exactly by Cauchy-Schwarz.<|endoftext|>
TITLE: Is there a tricategory of bicategories and biprofunctors?
QUESTION [15 upvotes]: Background
There is a bicategory where the objects are categories, the 1-morphisms are profunctors, and the 2-morphisms are morphisms of profunctors. The non-obvious part of this assertion is that profunctors admit a "good" (i.e., coherently associative) composition. The way one sees this is to use the fact that the Yoneda embedding is the free cocompletion, from which we may identify profunctors from $\mathcal{C}$ to $\mathcal{D}$ with cocontinuous functors from $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$ to $\operatorname{Set}^{\mathcal{D}^{\operatorname{op}}}$. Over here, there is a strictly associative composition, so life is good.
All of this extends easily to the enriched setting. In particular, if we enrich over $\mathcal{V} = \operatorname{Cat}$, we can talk about strict 2-profunctors between strict 2-categories, and we get a nice bicategory of 2-categories, strict 2-profunctors, and morphisms between these.
Question
I would like to know if there is an analogous way to obtain a tricategory of bicategories and biprofunctors. You can "follow your nose" and write down what the composition should be, but checking all the axioms of a tricategory does not seem like the simplest approach. Of course, we can strictify to get equivalent strict 2-categories and 2-profunctors, but I doubt the composition will be the same; I don't think the "strict" colimits will in general agree with the "weak" colimits. Also, using the standard enriched theory only gives a bicategory; the natural transformations are lost. It seems that an appropriate statement of the form "the bicategorical Yoneda embedding is a free cocompletion" will handle things just as it does in the setting of ordinary profunctors, but I am not sufficiently comfortable with limits and colimits in bicategories to try to figure out if such a statement makes sense. Has anybody worked all of this out?
(The closest thing I have seen to such a result is Justin Greenough’s paper Monoidal 2-structure of Bimodule Categories, which more or less establishes this result in the case of sufficiently nice $k$-linear monoidal categories. But his proof seems to be very specific to that setting, whereas one would hope that a proof could be established along the same lines as the result for usual profunctors.)

REPLY [10 votes]: For those coming across this question more recently, there is now an answer to the original question. In fact, the tricategory of pseudoprofunctors has been defined twice, independently, via different approaches.

Lawler defines the tricategory $2\text{-}\mathbb P\mathrm{rof}$ of bicategories and pseudoprofunctors (there called "2-profunctors") in Definition 4.1.1 of his 2014 thesis Fibrations of predicates and bicategories of relations. The hom-bicategories are given by the cocontinuous pseudofunctors between cocomplete bicategories, which is justified as corresponding to pseudoprofunctors by the results of Section 3.2.3. Defining the hom-bicategories in this way guarantees the coherence conditions hold immediately.
Chikhladze defines the tricategory $2\text{-}\mathcal P\mathrm{rof}$ of bicategories and pseudoprofunctors (there called "biprofunctors") in Section 8 of his 2014 paper Lax formal theory of monads, monoidal approach to bicategorical structures and generalized operads. Here, the approach taken is to define a bicategorical $\mathrm{Mat}$ construction, along with a bicategorical $\mathrm{Mod}$ construction, so that $2\text{-}\mathcal P\mathrm{rof} = \mathrm{Mod}(\mathrm{Mat}(\mathrm{Cat}))$. That $2\text{-}\mathcal P\mathrm{rof}$ is a tricategory then follows from the coherence of simpler constructions, which are shown to coincide with pseudoprofunctors in the same way as in the 1-categorical setting.<|endoftext|>
TITLE: Lie algebra automorphisms and detecting knot orientation by Vassiliev invariants
QUESTION [19 upvotes]: Recall that there are knots in $\mathbf{R}^3$ that are not invertible, i.e. not isotopic to themselves with the orientation reversed. However, it is not easy to tell whether or not a given knot is invertible; I believe the easiest known example involves 8 crossings. In particular, all knot invariants that are (more or less) easy to compute (e.g. the Jones or Homfly polynomials) fail to detect knot orientation -- there is a simple Lie algebra trick, the Cartan involution, that prohibits that. But this trick works for complex semi-simple Lie algebras and the question I'd like to ask is: can one perhaps circumvent this by using other kind of Lie algebras?
Here are some more details. A long standing problem is whether or not knot orientation can be detected by finite type (aka Vassiliev) invariants. Recall that these are the elements of the dual of a certain graded vector space $\mathcal{A}=\bigoplus_{i\geq 0} \mathcal{A}^i$; the vector space itself is infinite dimensional, but each graded piece $\mathcal{A}^i$ is finite dimensional and can be identified with the space spanned by all chord diagram with a given number of chords modulo the 1-term and 4-term relations (the 4-term relation is shown e.g. on figure 1, Bar-Natan, On Vassiliev knot invariants, Topology 34, and the 1-term relation says that the chord diagram containing an isolated chord is zero).
The above question on whether or not finite type invariants detect the orientation is equivalent to asking whether there are chord diagrams that are not equal to themselves with the orientation of the circle reversed modulo the 1-term and 4-term relations. (There are several ways to rephrase this using other kinds of diagrams, see e.g. Bar-Natan, ibid.)
However, although $\mathcal{A}^i$'s are finite-dimensional, their dimensions grow very fast as $i\to \infty$ (conjecturally faster that the exponential, I believe). So if we are given two diagrams with 20 or so chords, checking whether or not they are the same modulo the 1-term and 4-term relations by brute force is completely hopeless. Fortunately, there is a way to construct a linear function on $\mathcal{A}$ starting from a representation of a quadratic Lie algebra (i.e. a Lie algebra equipped with an ad-invariant quadratic form); these linear functions can be explicitly evaluated on each diagram and are zero on the relations. So sometimes one can tell whether two diagrams are equivalent using weight functions. But unfortunately, a weight function that comes from a representation of a complex semi-simple Lie algebra always takes the same value on a chord diagram  and the same diagram with the orientation reversed.
As explained in Bar-Natan, ibid, hint 7.9, the reason for that is that each complex semi-simple Lie algebra $g$ admits an automorphism $\tau:g\to g$ that interchanges a representation and its dual. (This means that if $\rho:g\to gl_n$ is a representation, then $\rho\tau$ is isomorphic to the dual representation.) Given a system of simple roots and the corresponding Weyl chamber $C$, $\tau$ acts as minus the element of the Weyl group that takes $C$ to the opposite chamber. On the level of the Dynkin diagrams $\tau$ gives the only non-trivial automorphism of the diagram (for $so_{4n+2}, n\geq 1,sl_n,n\geq 3$ and $E_6$) and the identity for other simple algebras. (Recall that the automorphism group of the diagram is the outer automorphism group of the Lie algebra.)
However, so far as I understand, the existence of such an automorphism $\tau$ that interchanges representations with their duals is somewhat of an accident. (I would be interested to know if it isn't.) So I'd like to ask: is there a quadratic Lie algebra $g$ in positive characteristic (or a non semi-simple algebra in characteristic 0) and a $g$-module $V$ such that there is no automorphism of $g$ taking $V$ to its dual?
(More precisely, if $\rho:g\to gl(V)$ is a representation, we require that there is no automorphism $\tau:g\to g$ such that if we equip $V$ with a $g$-module structure via $\rho\tau$, we get a $g$-module isomorphic to $V^{\ast}$.)

REPLY [9 votes]: Although, as Jose pointed out above, there are a lot of interesting non-reductive metric Lie algebras out there (see Medina and Revoy - Algèbres de Lie et produit scalaire invariant, Figueroa-O'Farrill and Stanciu - On the structure of symmetric self-dual Lie algebras, and A terminology issue with the Killing form), from the point of view of finite type knot invariants you might as well stick with the reductive ones.
Here's why.  Given a nice monoidal category (spherical, trivial object is simple) there is always a maximal planar ideal called the ideal of negligible morphisms.  This consists of all morphisms which always give you 0 when you close them off in any way to get an endomorphism of the trivial.  Equivalently, they are the kernel of the radical of the inner product given by the trace on the spherical category.  From the point of view of knot theory, it's always safe to kill off all negligible morphisms because knots are closed diagrams.
But, if you start with an abelian category (like a nice category of representations of a metric Lie algebra), then the quotient by the negligibles is always semisimple (see Are Abelian, non-degenerate tensor categories semisimple?, Deligne - La cat\'egorie des repr\'esentations du groupe sym\'etrique $S_t$, lorsque $t$ n'est pas un entier naturel, and Barrett and Westbury - Spherical categories).  Furthermore by Deligne's reconstruction theorem, once you know that all objects have integer dimensions your category must be the category of representations of a Lie superalgebra.
Combining the above, replace your category of $g$-modules with its quotient by the negligibles.  Realize this category as the category of representations of a different metric Lie algebra which is reductive but gives the same knot invariants.
There should be some way to make this construction concrete, i.e., any metric Lie algebra should have some sort of reductive "core" which gives the same knot invariants.  But my understanding of double extensions is not good enough to understand what's going on.  It would be interesting to see this worked out.<|endoftext|>
TITLE: Why is the largest signed 32 bit integer prime?
QUESTION [5 upvotes]: This may be subjective, but does anyone have any insight into why this is the case? This struck me while considering that it's also the eigth Mersenne prime (2^31-1=2147483647). 

I'm now wondering why this might be the case.

UPDATE:
It's been pointed out that the relationship doesn't necessarily hold for larger storage classes, e.g., 2^63 - 1 is not prime.

REPLY [7 votes]: Why is $3$ prime?  I don't really know that there are meaningful answers to these kinds of questions.  The best I can think of is some reasons it is not obviously composite, e.g. since $5$ is prime $2^5 - 1 = 31$ is not obviously composite (and it turns out to be prime) hence $2^{31} - 1$ is not obviously composite.  This is two applications of the "lemma" that if $p$ is prime then $2^p - 1$ is not obviously composite.
Note that any prime factor of $2^p - 1$ has to be congruent to $1 \bmod p$ by Fermat's little theorem, so it is "easier" for such numbers to be prime.<|endoftext|>
TITLE: Are non-PL manifolds CW-complexes?
QUESTION [26 upvotes]: Can every topological (not necessarily smooth or PL) manifold be given the structure of a CW complex?
I'm pretty sure that the answer is yes.  However, I have not managed to find a reference for this.

REPLY [15 votes]: Kirby and Siebenmann's paper "On the triangulation of manifolds and the Hauptvermutung" Bull AMS 75 (1969) is the standard reference for this, I believe. 
The result is that compact topological manifolds have the homotopy-type of CW-complexes, to be precise.

REPLY [7 votes]: See http://arxiv.org/abs/math/0609665<|endoftext|>
TITLE: What category without initial object do you care about?
QUESTION [7 upvotes]: Recently I have been listening to some constructions that have been designed to accommodate categories without an initial object. The speaker has given some idea of a category or two that he cares about, and thus why he was thinking in this direction, but I am now wondering;

As working mathematicians, what category are you concerned with that does not have an initial object?

I am sorry if this question is slightly strange, I have made it CW because it seems appropriate.
Thanks!

REPLY [4 votes]: In universal algebra and model theory, one usually requires that the underlying set be non-empty.  If the signature has constants in it, then this is no restriction, but otherwise, there is no initial object.  For example, under this definition there is no initial object in the category of semigroups. The reason for this restriction is that bad things happen to first-order logic when the underying carrier set is empty. Many standard theorems break down when you allow the empty domain.  For example, the following theorem of first-order logic
$$
\forall x P(x) \rightarrow \exists x P(x)
$$
becomess false.<|endoftext|>
TITLE: Books about the spectra of non-compact Riemann surfaces
QUESTION [5 upvotes]: Hello,
Thanks for reading my question ! Could anybody give me some references ( books, papers containing elementary results etc ) on the eigen values and eigenspectra of NON-compact Riemann surfaces. I studied the compact cases briefly and want to know the analogues or further results for the non-compact ones.
Also, I need some references on probabilistic results on compact or non-compact Riemann surfaces, like random walks and Brownian motion on them etc.
Thank you !

REPLY [4 votes]: If your French is alright, Nicolas Bergeron has written a beautiful book, Le spectre des surfaces hyperboliques, available here, which covers many of these topics in detail.<|endoftext|>
TITLE: Numerical evidence of Beilinson's conjecture in local fields and function fields
QUESTION [9 upvotes]: The famous Beilinson's conjecture predicts a relationship between the regulator map in $K$-theory and special value of $L$-function generalizing the Dirichlet's theorem in number theory. Please see this post for details. 
In this paper by Tim Dokchitser, Rob de Jeu, Don Zagier, the authors construct families of hyperelliptic curves over $\mathbf{Q}$ of arbitrary genus g with (at least) g integral elements in $K_2$. With these elements they compute the regulator numerical and compare the regulator with special value of $L$-function of these curves which give numerical evidence of Beilinson's conjecture for $K_2$ of curves.   
Here are my questions:
Is there explicitly formulated conjectures in p-adic fields and function fields as in number fields? Does one can do similar computations for curves over these fields? If not, what are the main obstacles?

REPLY [2 votes]: You may be interested in Colmez' Bourbaki survey "Fonctions L p-adiques", available here:
http://www.math.jussieu.fr/~colmez/bric-a-brac.html 
Section 2.8 is called "p-adic Beilinson conjectures". 
See also his other survey (on the same page) called "La conjecture de Birch et Swinnerton-Dyer p-adique".<|endoftext|>
TITLE: Conditions on a metric space so that boundedness implies total boundedness
QUESTION [5 upvotes]: I'm interested in conditions on a metric space $X$ which imply that boundedness is equivalent to total boundedness (or, assuming that $X$ is complete, that compactness is equivalent to precompactness). If $X$ is a normed space, then we know that this is true if and only if $X$ is finite-dimensional. Is there some suitable concept of dimension for general metric spaces such that boundedness implies total boundedness if and only if the space has finite dimension in this sense? You may assume any reasonable hypotheses on $X$ (e.g. that $X$ is a Polish metric space).

REPLY [6 votes]: Doubling metric spaces, which are studied a lot, have this property.  A metric space is doubling provided there exists $n$ so that for every $r>0$, a ball of radius $2r$ is covered by $n$ balls of radius $r$. Google "doubling metric spaces"; some of the half million hits should be useful.

REPLY [5 votes]: If $X$ is locally compact, then it has this Heine-Borel-property. For topological vector spaces locally compactness is equivalent to finite dimension if I remember correctly.
But there are other examples even vector spaces that have the Heine-Borel-property without being locally compact. The space $H(U)$ of holomorphic functions on an open set $U\subseteq\mathbb{C}$ with the topology of locally uniform convergence of all derivatives. This is Montel's theorem and therefore such spaces are called Montel spaces (well a certain additional condition is needed, but that's not the point) Another example is the Schwartz-Space $\mathcal{S}(\mathbb{R}^n)$ of rapidly decreasing functions. Because being Montel is stable under taking strong duals, the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$ has the Heine-Borel-property too (but is not metrizable).<|endoftext|>
TITLE: Looking for applications of a nice result in linear algebra
QUESTION [7 upvotes]: Hello everybody
There is a nice classical result in linear algebra: if $A, B$ are two matrices in $M_n(k),$ where $k$ is a field, and $B$ commutes with every element of $M_n(k)$ which commutes with $A$, then $B = f(A)$ for some polynomial $f(x)$ in $k[x].$
I was wondering if anybody knows any (important) theorem which is proved using this result. Thank you.

REPLY [4 votes]: Tate's famous "Endomorphisms of Abelian Varieties over Finite Fields," which proves the Tate conjecture in the finite field case, uses the full force of the theorem of bicommutation in a reduction lemma. As KConrad mentions in the comments, the result you've cited is the special case of this theorem where one works with the subalgebra generated by one element.<|endoftext|>
TITLE: Probabilistic interpretation of prime number theorem
QUESTION [10 upvotes]: Suppose there is a function $f(x)$ which is the "probability" that the integer $x$ is prime. The integer $x$ is prime with probability $f(x)$, and then divides the larger integers with probability $1/x$; so as $x$ changes from $x$ to $x+1$, $f(x)$ changes to (roughly)
$$f(x)\left(1-f(x)/{x} \right).$$
How do I show that? I can go on to show
$$\frac{df}{dx}  + \frac{f^2}{x}=0$$
and thus $\frac{1}{\log{x}} + c$ is solution but I can't show that step on how $f(x)$ changes. please advise

REPLY [2 votes]: Another way to obtain OP's differential equation is as follows:
$$f(x) \approx \prod_{p < x} \left( 1-1/p \right) \approx \exp( - \sum_{p<x} 1/p) \approx \exp( -\int_{1}^{x} \frac{f(t) dt}{t}).$$
Taking logarithms and differentiating w.r.t $x$, we obtain the desired ODE (of course, the argument is completely informal).<|endoftext|>
TITLE: Modular congruences related to sums of Catalan numbers
QUESTION [8 upvotes]: I am curious if somebody can be helpful concerning the following
experimental observation: 
There exist two rational sequences $\alpha_0,\alpha_1,\dots$ and
$\beta_0,\beta_1,\dots$, both with values in $\mathbb Z[1/3]$
such that 
$$\sum_{k=0}^{p-1}{2k\choose k}\frac{k^j}{k+1}\equiv 
\alpha_j+p\beta_j\pmod{p^2}$$
for every prime number $p\equiv 1\pmod 6$ and
$$\sum_{k=0}^{p-1}{2k\choose k}\frac{k^j}{k+1}\equiv 
-(-1)^j-\alpha_j+p\beta_j\pmod{p^2}$$
for every prime number $p\equiv 5\pmod 6$.
(More precisely, the sequences $3^n\alpha_n$ and $3^n\beta_n$ are seemingly integral.)
The sequence $\alpha_0,\alpha_1,\dots$ starts as
$$1, 0, -2/3, 4/3, -22/9, 140/27, -14, 1316/27, -17078/81, 87860/81, -1562042/243, 31323292/729, \dots$$
and the first terms $\beta_0,\beta_1,\dots$ are
$$0,0,2/3,-2,14/3,-34/3,98/3,-350/3,1526/3,-2622,46634/3,-311734/3,2316158/3,
-18920018/3,\dots$$
Let me end by remarking that one has as a special case a similar result when replacing Catalan numbers by central binomial coefficients.
Update: The existence of the sequence $\alpha_n$ is explained by the Zhi-Wei Sun paper,
see the answer by dke below.
Experimentally, the quotient sequence $\frac{\beta_n}{\alpha_n}$ (defined for $n\geq 2$)
seems to converge very quickly towards $-\frac{4\sqrt{3}\pi}{9}=-2.4183991523\dots$ (the error is
smaller than $10^{-78}$ for $n=120$).
The sequence $\frac{\alpha_{n+1}}{\alpha_n}-\frac{\alpha_n}{\alpha_{n-1}}$
converges perhaps (fairly slowly) towards something like $-.72\dots$.

REPLY [4 votes]: I wonder about the reason to consider
$$
A_j=\sum_{k=0}^{p-1}\frac{k^j}{k+1}\binom{2k}{k}
$$
instead of
$$
\tilde A_j = \sum_{k=0}^{p-1} k^j \binom{2k}{k},
$$
as $A_j\pm A_0$ ($+$ for odd $j$ and $-$ for even $j$) are reduced to linear combinations of the latter. Let me introduce the functions
$$
F_j(x)=\left(x\frac{d}{dx}\right)^j\sum_{k=0}^{p-1}\binom{2k}{k}x^k,
$$
so that $\tilde A_j=F_j(1)$. Note that
$$
F_0(x)=F(x)=\sum_{k=0}^{p-1}\binom{2k}{k}x^k
$$
satisfies the differential equation
$$
F(x)-\frac12(1-4x)F'(x)=\frac{(2p-1)!}{(p-1)!^2}x^{p-1}.
$$
Here the coefficient on the right is
$$
\frac{(2p-1)!}{(p-1)!^2}\equiv p\pmod{p^4}
$$
by Wolstenholme's theorem and we can write the equation in the form
$$
xF_0(x)-\frac12(1-4x)F_1(x)=\frac{(2p-1)!}{(p-1)!^2}x^p.
$$
Applying repeatedly the operator $x\dfrac{d}{dx}$ to both sides of the identity produces, for each $j$, a relation of the form
$$
P_{0,j}(x)F_0(x)+P_{1,j}(x)F_1(x)+\dots+P_{j-1,j}(x)F_{j-1}(x)-\frac12(1-4x)F_j(x)=p^{j-1}\frac{(2p-1)!}{(p-1)!^2}x^p,
$$
where all coefficients of the polynomials $P_{i,j}(x)$ are integral (and, of course, independent of $p$). This gives a way to express $\frac32F_j(1)=\frac32\tilde A_j$ as a $\mathbb Z$-linear combination of $\tilde A_0,\tilde A_1,\dots,\tilde A_{j-1}$ independently of $p$ modulo $p^j$. The remaining piece is to check that $\tilde A_0$ and $\tilde A_1$ as well as the starting $A_0$ do satisfy the "$p$-independence" property modulo $p^2$.
The construction above gives a recipe to construct recursively the sequence $\tilde A_j$ (hence $A_j$), although I do not try myself to figure out a closed-form evaluation of the auxilliary polynomials $P_{i,j}(x)$ and/or their values at $x=1$.<|endoftext|>
TITLE: Unusual Space-Filling Curve
QUESTION [7 upvotes]: Around 1998, I encountered a (forgotten) reference to a particularly strange space-filling curve.
Consider a foliation as a collection of continuous nonintersecting curves that start at (0,0) and end at (1,1) and collectively fill the unit square, such as the graphs of functions ft(x) = xt where t >=0.  Supposedly there exists a continuous curve G that starts at (1,0), ends at (0,1), fills the unit square, and crosses each ft curve only once.
This initially sounds even more impossible than the Cantor curve.  But intuitively a space-filling curve could trace back and forth over the ft curves and only cross at the corners (0,0) and (1,1).  Can someone please explain a construction of such a space-filling curve?

REPLY [5 votes]: André's answer is correct, but what you really are looking for (I think) is this example, due to Katok, but explained (beautifully) by Milnor.

There is a family of disjoint smooth real-analytic curves $\Gamma_{\beta}$ that fill the unit square $I^2$, and a subset $E \subset I^2$ of full measure, so that each curve $\Gamma_{\beta}$ intersects $E$ in at most one point.

This means that $E$ can be constructed by choosing one point for each parameter $\beta$, so in order not to contradict Fubini, the dependence of $\Gamma_{\beta}$ on $\beta$ cannot be continuous.<|endoftext|>
TITLE: The ring $C^{\infty}(M)$?
QUESTION [11 upvotes]: Let $M$ be a smooth paracompact manifold. I think that the ring $C^{\infty}(M)$ contains many (possibly almost all?) geometric or topological information about $M$. 
(e.g. Let $E$ be a vector bundle over $M$,$\Gamma(E)$ be a set of smooth section of $E$. Then, $\Gamma(E)$ is a $C^\infty(M)$-module. (Actually, I think $\Gamma(E)$ is projective $C^\infty(M)$-module because every a short exact sequence of vector bundle splits.)) 
But I have a feeling that $C^\infty(M)$ is too large to change the problem of Manifold theory into an algebraic problem or Ring theoretic problem. 
Are there any well-known concrete description about the ring $C^\infty(M)$ for some manifold $M$ with simple topology?

REPLY [6 votes]: Connes proved an analogue of the Hochschild-Kostant-Rosenberg theorem which asserts that for a compact manifold $M$, there is a canonical isomorphism between the continuous Hochschild cohomology groups of $C^\infty(M)$ and the spaces of de Rham currents on $M$, which are the dual to differential forms.  You can find this in Chapter 8 of the book Elements of Noncommutative Geometry, by Varilly, Gracia-Bondia, and Figueroa.
Also, in the paper On the Spectral Characterization of Manifolds, Connes shows how to reconstruct a manifold from a commutative spectral triple, i.e. take a commutative pre-$C^*$-algebra $A$ plus some extra data and build a manifold $M$ from it such that $A \simeq C^\infty(M)$.  This is really more than you were asking for but I thought it might be interesting nonetheless.<|endoftext|>
TITLE: What is known about the $k^{\mathrm{th}}$ powers of graphs of diameter $k+1$?
QUESTION [5 upvotes]: Let $G$ be a simple unweighted graph. The distance between two vertices $u,v$ in $G$ is the length of a shortest path in $G$ between $u$ and $v$. The diameter of $G$, denoted $diam(G)$, is the largest distance between two vertices in $G$. For a natural number $k$, The $k^{\mathrm{th}}$ power of $G$, denoted $G^{k}$, is the graph obtained from $G$ by adding edges between every two vertices $u,v$ where the distance between $u,v$ in $G$ is at most $k$. 

Let $\mathcal{F}_{k}$ be the family of all graphs that are the $k^{\mathrm{th}}$ powers of graphs of diameter $k+1$. That is,

$\mathcal{F}_{k} = \{G^{k}\mid diam(G)=k+1\}$


$\mathcal{F}_{1}$ is the set of all graphs of diameter $2$, and for all $k\ge2$, $\mathcal{F}_{k}\subseteq\mathcal{F}_{2}$. 
Is something more known about the structure of the graphs in $\mathcal{F}_{k}$ for $k\ge2$? In particular, are these containments strict? Do we get more structure the higher the value of $k$? 
Thanks for any ideas, pointers, and/or references on this.

REPLY [4 votes]: The containments must be strict for the trivial reason that each $\mathcal F_k$ contains a graph with $k+2$ nodes (the case $G$ = path) but no graph with $k+1$ nodes (if $G$ has diameter $k+1$, it must have at least $k+2$ nodes).
For further insight into the structure of $\mathcal F_k$, I'd study the structure of the complements of the graphs in $\mathcal F_k$. Each edge in the complement of $G^k$ corresponds to a diameter of $G$; that is, if you have an edge $\{u,v\}$ in the complement of $G^k$, the distance between $u$ and $v$ is exactly $k+1$ in $G$. Thus your question is really about the structure of the diameters of graphs.<|endoftext|>
TITLE: On the $\mathbb R$-algebra structure on $C^\infty(M)$.
QUESTION [9 upvotes]: As it is more or less well-know, and as it has come up on MO a couple of times, the $\mathbb R$-algebra $C^\infty(M)$ of smooth functions on a (say) compact manifold contains essentially everything there is to know about $M$ itself.

Does one really need to know the $\mathbb R$-algebra structure, though? Can we manage with $C^\infty(M)$ given as an abstract ring?

For comparison, if I recall correctly one can reconstruct (as an analytic manifold) an open subset $U\subseteq\mathbb C$ from the field of meromorphic functions on it, given as an abstract field. (It follows, of course, that the ring of holomorphic functions also can be used to reconstruct $U$.)

REPLY [16 votes]: You can determine the R-algebra structure of $C^\infty(M)$ purely from its ring structure. As Robin Chapman mentions, the constant function 1M is uniquely determined by the fact that it is the identity element, and multiplication by rationals is uniquely defined, so the functions equal to a constant rational value are uniquely determined.
Actually, the ring homomorphism $F\colon\mathbb{R}\to C^\infty(M)$ is unique, which also uniquely defines the R-algebra structure.
The positive elements $x\in\mathbb{R}$ are squares, so $F(x)$ must be a square in $C^\infty(M)$, hence nonnegative everywhere. Then, for any $x\in\mathbb{R}$ and rational numbers $a\le x\le b$ we have $F(x)-a1_M=F(x-a)\ge0$ and $b1_M-F(x)=F(b-x)\ge0$, so $F(x)\in C^\infty(M)$ takes values in the interval $[a,b]$. This shows that, in fact, $F(x)=x1_M$.
Thinking about it, this works because $\mathbb{R}/\mathbb{Q}$ has trivial Galois group. You can see this by asking if the C-algebra structure on $A\equiv C^{\infty}(M,\mathbb{C})$ is uniquely determined by its ring structure, for which the answer is no. For any $\sigma\in{\rm Gal}(\mathbb{C}/\mathbb{Q})$ it is not possible to distinguish a constant $f\in A$ from $\sigma(f)$ in terms of ring operations [edit: if $\sigma$ is continuous, that is. So, only considering the identity element and complex conjugation]. Instead, you could ask if it is possible to determine the C-algebra structure up to the action of the Galois group. If the manifold is connected then the answer to this is yes. The constant function taking the value $\pm i$ everywhere is given by $i_M^2+1_M=0$, and the constant functions $f\in A$ are those for which $f-\lambda1_M-\mu i_M$ are units for all but at most one choice of $\lambda,\mu\in\mathbb{Q}$. The constant functions are isomorphic to $\mathbb{C}$, which is determined up to the action of the Galois group. If it is not connected, then we can't even say that much. For any locally constant map $\sigma\colon M\to{\rm Gal}(\mathbb{C}/\mathbb{Q})$, it is not possible to distinguish $f\in A$ from $f_\sigma(P)\equiv\sigma(P)(f(P))$ using ring operations. The C-algebra structure is uniquely determined up to the action of such a locally constant $\sigma$ though, which should still be enough to tell you everything about the manifold. Working over the reals, none of this matters, because of the triviality of the Galois group.<|endoftext|>
TITLE: Some arithmetic terminology: "universal domain", "specialization", "Chow point"
QUESTION [9 upvotes]: As a non-connoisseur of arithmetic and arithmetic geometry, I would like to ask about some terminology, which meaning I haven't been able to find out on some books, nor on wikipedia, nor by google.
First,

What's a "universal domain" (of a given characteristic)?


What I knew is that, on a scheme, a (not necessarily closed) point x is called a specialization of a point y (which in turn is called a generization of x) if x lies on the topological closure of the sigleton {y}.

Does it make sense, in some context, to say that a given scheme (or subscheme of a fixed scheme) is the "specialization" of another one?

Suppose you are in the following context (that I will naively try to set). You are given a scheme $M$ over the integers, such that over points of $Spec \mathbb{Z}$ it has fibers that are algebraic varieties over the residue fields $\mathbb{F}_p$, $p\geq 0$, (or maybe over a "universal domain" of suitable characteristic?). Suppose also that it is kind of an "arithmetic moduli space" for e.g. curves of some genus so that closed points of its fiber "over p" parametrize curves of that genus in characteristic p. 

In the above  context, or in a similar one, does it make sense to say that "a curve $C'$ is a specialization of another curve $C$"? What about the assertion "the jacobian $J'$ is a specialization of the jacobian $J$"?

Also,

What's a "specialization over another specializaion"?



What's the "Chow method" to construct the jacobian of a nonsingular curve (of any characteristic)?



What's the "Chow point"? (I suppose it's a concept related to field extensions...)

REPLY [3 votes]: Igusa uses Weil's language, in a modified/enhanced version that deals
with reduction mod primes.  (My memory is that there is a paper of Shimura from the 50s that develops this language.)  It's not so easy to read it carefully, unfortunately.
Chow's method for constructing Jacobians (explained in his paper in the American Journal
from the 50s, again if memory serves) is, I think, as follows:
take $Sym^d C$ for $d > 2g - 2$. The fibres of the map $Sym^d C \to Pic^d(C)$
are then projective space of uniform dimension (by Riemann--Roch), and so it is not so hard to quotient out
by all of them to construct $Pic^d(C)$ (for $d > 2g - 2$), and hence to construct the Jacobian.  (I hope that I'm remembering correctly here; if not, hopefully someone will correct me.)
I think that this should be contrasted with the more traditional method of considering $Sym^g C$, which maps birationally to $Pic^g(C)$, i.e. with fibres that are generically points, but which has various exceptional fibres of varying dimensions, making it harder to form the quotient, thus inspiring in part Weil's "group chunk" method where he uses the group action
to form the quotient (in an indirect sort of way), and consequently loses some control of the
situation (e.g. he can't show that the Jacobian so constructed is projective).   I should
also say that it's been a long time since I looked at this old 1950s literature, and I'm not 
completely confident that I understand its thrust (i.e. I'm not sure what was considered easy and what was considered hard, and what was considered new and innovative in various papers as contrasted to what was considered routine), so take this as a very rough guide only.<|endoftext|>
TITLE: Intutive interpretation about Linking forms
QUESTION [8 upvotes]: Let $M^3$ be a rational homology 3-sphere. (i,e, $M^3$ is closed 3-manifold with 
$H_{*}(M;Q)=H_{*}(S^3;Q)$
As beautifully explained in Ranicki's Algebraic and Geometry surgery book and Davis-Kirk's Lecture notes in Algebraic toplogy book, we have a $Q/Z$ valued linking form, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ defined by the adjoint to following isomorphism. 
(Actually we can define linking form in more general setting, e.g.) odd dimensional manifold without the restriction such as rational homology sphere condition. Because I want to just intuitive idea about linking form, I restricted the case)
$H_1(M;Z)\cong H^2(M;Z)\cong H^1(M;Q/Z)=Hom(H_1(M;Z),Q/Z)$ 
First isomorphism : Poincare duality, 
Second isomorphism : Inverse of Bockstein homomorphism $\delta$ induced form $0\rightarrow Z \rightarrow Q\rightarrow Q/Z\rightarrow 0$. More precisely, induced long exact sequence is that $\ldots\rightarrow H^1(M;Q)\rightarrow H^1(M;Q/Z)\rightarrow H^2(M;Z)\rightarrow H^2(M;Q)$ and here $H^1(M;Q)$ and $H^2(M;Q)$ vanishes. Therefore, the long exact sequence shows that $H^1(M;Q/Z)$ and $H^2(M;Z)$ are naturally isomorphic (if $M$ is rational homology 3-sphere) and we call that homomorphism as Bockstein homomorphism.
Third isomorphism : Universal Coefficient theorem
In short, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ is defined by $\lambda(x,y)=<\tilde{x}\cup\delta^{-1}(\tilde{y}),[M]>$, where $\tilde{x},\tilde{y}$ are Poincare dual to $x,y$ and $\delta\colon H^1(M;Q/Z)\to H^2(M;Z)$ is a Bockstein homomorphism as defined above. Cup products are defined in $H^1(M;Z)\times H^2(M;Q/Z) \to H^3(M;Q/Z)$ induced from the Bilinear map $Z \times Q/Z \to Q/Z$.
I understand this linking form only algebraic viewpoint. Therefore, I can play with this form only by using dilluminating algebraic topology. 
I'm struggle to find geometric interpretation but I have no idea to express the Q/Z valued in terems of geometric language which seems to be highly algebraic. (Feeling like Injective, divisible, Ext or something linke that )
Are there any intuitve and geometric (clean) interpretation about this linking form? (e.g.)such as Alexander duality, like the argument that .........we can find a dual basis which represents meridian......)

REPLY [11 votes]: This may be a good place to explain the well-known principle 
$$\text{intersection in the interior = linking in the boundary}$$
in an oriented $m$-dimensional manifold with boundary $(M,\partial M)$. Let
$$f~:~(K,\partial K)\subset (M,\partial M)~,~g~:~(L,\partial L) \subset (M,\partial M)$$
be embeddings of oriented manifolds with boundary, such that
$${\rm dim}(K)~=~k~,~{\rm dim}(L)~=~\ell~,~k+\ell~=~m~,~
f(\partial K) \cap g(\partial L)~=~\emptyset \subset \partial M~.$$
Assume there exists an isotopy (= homotopy through embeddings) rel $\partial K$
$$f_t~:~K \to M~~(0 \leqslant t \leqslant 1)$$
such that $f_0=f$ and $f_1(K)\subset \partial M$, with each $f_t(K), g(L)$ intersecting transversely in $M$, so that $f_t(K) \cap g(L)\subset M$ is a finite set with an
intersection index $I(x)\in \{\pm 1\}$ at each point $x \in f_t(K) \cap g(L)$
according to the orientations. A continuity argument shows that the function
$$\lambda~:~[0,1] \to {\mathbb Z}~:~t\mapsto \lambda(t)=
\sum\limits_{x \in f_t(K) \cap g(L)}I(x)$$
is constant, so that 
$${\rm intersection}(f(K),g(L)\subset M)~=~\lambda(0)~=~\lambda(1)$$
$$=~{\rm linking}(f(\partial K),g(\partial L) \subset \partial M) \in {\mathbb Z}~.$$
This is best seen by drawing pictures for $(M,\partial M)=(D^2,S^1)$.
The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.<|endoftext|>
TITLE: Which (semi)regular polyhedra are combinations of two others?
QUESTION [10 upvotes]: The convex combination of convex polytopes is a convex polytope.
An example in $\mathbb{R}^2$ is that a regular octagon
can be obtained as $\frac{1}{2} S + \frac{1}{2} S'$,
where $S$ is a square and $S'$ is the same square rotated $45^\circ$—it is the mean of two squares:



Of special interest to me are (a) convex polyhedra in $\mathbb{R}^3$,
(b) combinations of just two polyhedra, and (c) where both are
regular or semi-regular.
For example,
the truncated cuboctahedron (a.k.a. the great rhombicuboctahedron)
"is the Minkowski sum of a cube and a truncated
octahedron" (quoting from a paper ["Zonohedra and Zonotopes"] by David Eppstein).
(It is also the sum of three cubes!)
My question is, essentially:

Which regular and semi-regular polyhedra can be obtained
as convex combinations of pairs of other regular and semi-regular polyhedra,
and which cannot be so represented—are 'prime' or unique in this respect?

I've searched around for a definitive tabulation of this information
without success.  It must be all known?
Particular subquestions and generalizations include:

 Can any of the Platonic solids be realized as sums
of two other Platonic solids? 
 Which of the semi-regular (Archimedean solids) can be realized as sums
of two Platonic solids? 
 I hesitate to delve into the [Johnson solids][14] ...
 Can any of the 4D regular polytopes be realized as sums of pairs
of other 4D regular polytopes?
 Can any of the three regular polytopes in dimension $d>4$ (the simplex, the hypercube, the orthoplex) be realized as sums of either of the other two?
 Is there some general underlying theorem lurking here?

I can make (informed?) guesses and proof sketches for several of these questions, but would prefer to defer to the experts.
Any observations on any aspect of these (sub)questions, references,
comments, would be appreciated.  Thanks!

REPLY [3 votes]: I think the whole list of Gjiergji (and lightly more, see below) follows from Wythoffian operations. For the cube-octahedron family, for example, consider the following diagram:
The polyhedron along an edge of the triangle is the Minkowski sum of the two vertices of that edge, which gives:

The truncated cube is the Minkowski sum of a cube and a cube octahedron
The truncated octahedron is the Minkowski sum of an octahedron and a cuboctahedron
The rhombicuboctahedron is the Minkowski sum of an octahedron and a cube

The polyhedron inside the triangle is the Minkowski sum of the three vertices or, equivalently, the Minkowski sum of a vertex and the opposite edge. This gives:

The truncated cuboctahedron is the Minkowski sum of a rhombicuboctahedron and a cuboctahedron

but also

The truncated cuboctahedron is the Minkowski sum of a cube and a truncated octahedron.
The truncated cuboctahedron is the Minkowski sum of an octahedron and a truncated cube.<|endoftext|>
TITLE: Asymptotic growth of a certain integer sequence
QUESTION [24 upvotes]: Some time ago, while putting my nose in the Sloane's Online Encyclopedia of Integer Sequences, I came over the sequence A019568 defined as follows:

$a(n):=$ the smallest positive integer $k$ such
  that the set $\{1^n, 2^n, 3^n,\dots k^n\}$ can be partitioned into two
  sets with equal sum.

In other words, $a(n)$ is the smallest $k$ such that there is a choice of signs + or - in the expression  $$1^n\pm2^n\pm\dots\pm k^n \qquad\qquad(1) $$ that makes it vanish. In order to show that this $a(n)$ is a well-defined integer (that is: that at least one such $k$ does exist), a simple observation gives in fact a bound $$a(n)<2^{n+1}.$$ Reason: $(1-x)^{n+1}$ divides the polynomial 
$$Q(x):=(1-x)(1-x^2)(1-x^4)\dots(1-x^{2^n})=+1-x-x^2+x^3-\dots +(-1)^n x^{2^{n+1}-1},$$
therefore, if $S$ is the shift operator on sequences, the operator $Q(S)$ has the $(n+1)$-th discrete difference $(I-S)^{n+1}$ as factor, hence annihilates any sequence that is polynomial of degree not greater than $n$. In particular, the algebraic sum (1) with the signs of the coefficients of $Q(x)$ vanishes (incidentally, the sequence of signs is the so called Thue-Morse sequence A106400, $+--+-++--++-+--+\dots$.
However, looking at the reported values of $a(n)$ for $n$ from $0$ to $12:$  
$$2,\ 3,\ 7,\ 12,\ 16,\ 24, \ 31,\ 39,\ 47,\ 44,\ 60,\ 71,\ 79,$$ 
it looks like the growth of $a(n)$ is much below $2^{n+1}$ (I have a weakness for sequences that grow slowly, here's possibly the main motivation of this question).

Question: Does anybody have a reference for the
  above sequence? Can you see how to
  prove an asymptotics, or a more
  realistic bound than $a(n)<2^{n+1}$?

REPLY [4 votes]: This is building off of the work of Richard Borherds' answer and the comment I made there.  We can provide a way to compute an upper bound which would be much better than the current exponential bound in $n$, but relies on enumerating solutions to certain Diophantine equations.
We are going to try and enumerate the distinct values of the expression $$e_11^n+e_22^n+\cdots+e_kk^n,$$ where $e_i = \pm 1$ for $i \in [1,n]$.  Let $S$ be the set of all values attained by this expression.  We have obviously $|S| \le 2^k$ as there are $2^k$ ways to choose the signs, but we know that two different choices for the $e_i$ may give the same numeric value.  An illustrative example occurs in case $(k,n) = (5,2)$, where we have $$+1^2+2^2+3^2+4^2-5^2 = +1^2+2^2-3^2-4^2+5^2 = 5.$$  If we have two choices for the $e_i$ - for now, call the two lists of coefficients $a_i$ and $b_i$ that coincide, we may form sets $A = \{i \in [1,n]: a_i = +1, b_i = -1\}$ and $B = \{i \in [1,n]: a_i = -1, b_i = +1 \}$.  Then we know that $A \cap B = \emptyset$, and more importantly, $\sum_{a\in A}a^n = \sum_{b \in B}b^n$.
Now define a function $s(k,n,a,b)$, with $a \ge b$ and $a+b \le k$, to count the number of pairs of sets $(A,B)$ satisfying the following conditions:

$A \cup B \subseteq [1,k]$;
$A \cap B = \emptyset$;
$|A| = a$ and $|B| = b$;
if $a = b$, then $\max{A} < \max{B}$;
$\sum_{a\in A}a^n = \sum_{b \in B}b^n$.

Then we may write $$|S| \ge 2^k - \sum_{m=3}^k2^{k-m-1}\sum_{1 \le j \le m/2}s(k,n,m-j,j).$$  Indeed, we see that if some set of choices of $e_i$ give the same numerical value, all but one will be removed by this inclusion-exclusion counting.  We may simplify this a bit further by writing $$f(k,n) = \sum_{m=3}^k2^{-m-1}\sum_{1 \le j \le m/2}s(k,n,m-j,j)$$ as our equation then becomes $|S| \ge 2^k(1-f(k,n))$.  Now once we have established suitable bounds on $f(k,n)$, we may proceed as Richard did, noting that we simply solve for the least valid value of $k$ in the inequality $$2^k(1-f(k,n)) \ge 2\sum_{i=1}^ki^n.$$
Where does Fermat's Last Theorem come in?  Well, in case $n \ge 3$, we need only sum from $m = 4$ to $k$ - there are no solutions for $m = 3$, as a solution for $m = 3$ takes the form $x^n + y^n = z^n$!  So we lose what is (possibly?) the largest contributing term to $f(k)$.
EDIT: I was off by a factor of two.  Darn.<|endoftext|>
TITLE: A good place to read about uniform spaces
QUESTION [19 upvotes]: I'd like to learn a bit about uniform spaces, why are they useful, how do they arise, what do they generalize, etc., without getting away from the context of general topology. I have to prepare an 1h30min talk on the subject, for an audience formed in standard general topology (i.e. Munkres), not so much in abstract algebra (so I'd like not to use topological groups).
The references I have are Kelley or Willard texts on Topology, Isbell's "Uniform Spaces" and James's "Topological and Uniform Spaces".
I discarded the last one because of its heavy use of filters from the beginning. I don't know about any other good references. What I'd like to see is the subject treated as Munkres does in his book: he gives good motivations, pictures, and is gentle to the reader.

REPLY [14 votes]: For a general audience it can be interesting to know that uniform spaces are just one of two opposite generalizations of metric spaces. Measuring distances with the help of metric, we can be interested in points lying on small distance (and exactly this aspect interests specialists in uniform spaces) but also is points lying on large distance (this is the subject of coarse geometry). So, uniform and coarse spaces are two opposite generalizations of metric spaces, corresponding to two ends of the real half-line $(0,\infty)$. 
Both of them allow us to compare distances between points without using real numbers.
Uniform spaces and coarse spaces have dual definitions. Both of them are pairs $(X,\mathcal U$) consisting of a set $X$ and a family $\mathcal U$ of subsets $U\subset X\times X$ containing the diagonal but uniform spaces obey the axioms:

(U1) for any $U\subset V\subset X\times X\quad U\in\mathcal U$ implies $V\in\mathcal U$;

(U2) each $U\in\mathcal U$ the entourage $-U:=\{(y,x):(x,y)\in U\}$ belongs to $\mathcal U$;

(U3) each $U\in\mathcal U$ contains $V+V:=\{(x,z):\exists y\in X\;(x,y),(y,z)\in V\}$ for some $V\in\mathcal U$.
Whereas coarse spaces obey the dual axioms:

(C1) for any $U\subset V\subset X\times X\quad V\in\mathcal U$ implies $U\in\mathcal U$;
(C2) each $U\in\mathcal U$ the entourage $-U:=\{(y,x):(x,y)\in U\}$ belongs to $\mathcal U$;
(C3) for each $U\in\mathcal U$ the set $U+U$ is contained in some $V\in\mathcal U$.
There is also a duality in morphisms. For the category of uniform spaces morphisms are uniformly continuous maps, i.e., functions $f:X\to Y$ between uniform spaces $(X,\mathcal U_X)$ and $(Y,\mathcal U_Y)$ such that for any $U\in\mathcal U_Y$ there exists $V\in\mathcal U_X$ such that $\{(f(x),f(y)):(x,y)\in V\}\subset U$.
For the category of coarse spaces morphisms are coarse maps, i.e., functions $f:X\to Y$ between coarse spaces $(X,\mathcal U_X)$ and $(Y,\mathcal U_Y)$ such that for any $V\in\mathcal U_X$ there exists $U\in\mathcal U_Y$ such that $\{(f(x),f(y)):(x,y)\in V\}\subset U$.
In fact, the structure of a uniform (coarse) space is the most general structure which allows us to speak about uniformly continuous (resp. coarse) maps whose $\varepsilon$-$\delta$-definitions were known in analysis long before the appearence of uniform (or coarse) spaces. And this was the reason for introducing uniform or coarse spaces.

Concerning the references, for me the basic reference in uniform spaces is Chapter 8 of Engelking's book "General topology" and for coarse spaces the book "Lectures on coarse geometry" of J.Roe.

For a bit deeper questions related to uniform or coarse spaces, which can be interesting for general audience, I would recommend the problem of dimension. Why does the space $\mathbb R^n$ have dimension $n$ and in which sense? In fact, there are many deep answers to this question in many various categories having $\mathbb R^n$ as an object (in particular, in the category of linear spaces, rings, metric spaces, topological spaces, etc). 
The categories of uniform and coarse spaces also have their own definitions of dimension, which are dual in some sense and yield the same dimension $n$ for the space $\mathbb R^n$.
The uniform dimension $dim(X,\mathcal U)$ of a uniform space $(X,\mathcal U)$ is the smallest number $n$ such that for any $U\in\mathcal U$ there exist $V\in\mathcal U$ and a cover $\mathcal C$ of $X$ by sets $C\subset X$ of diameter $<U$ (which means that $C\times C\subset U$) such that $\mathcal C$ can be written as the union $\mathcal C=\bigcup_{i=0}^n\mathcal C_i$ of $(n+1)$ subfamilies $\mathcal C_i$, which are $V$-separated in the sense that $(C\times D)\cap V=\emptyset$ for any distinct sets $C,D\in\mathcal C_i$.
The coarse (or else Gromov's asymptotic) dimension $Dim(X,\mathcal U)$ of a coarse space $(X,\mathcal U)$ is the smallest number $n$ such that for any $V\in\mathcal U$ there exist $U\in\mathcal U$ and a cover $\mathcal C$ of $X$ by sets $C\subset X$ of diameter $<U$ such that $\mathcal C$ can be written as the union $\mathcal C=\bigcup_{i=0}^n\mathcal C_i$ of $(n+1)$ subfamilies $\mathcal C_i$, which are $V$-separated.
These two definition have only one change in quantifiers but "look" in different directions -- uniform dimension takes small $U$, but the coarse direction takes large $V$!

The standard wall tiled with (small or large) bricks shows that the plane has both uniform and coarse dimension equal to 2. A non-trivial result of uniform and coarse dimension theories says that the uniform and coarse dimension of $\mathbb R^n$ is equal to $n$. By the way, the coarse dimension of $\mathbb Z^n$ also is $n$ whereas the uniform dimension of $\mathbb Z^n$ is zero. On the other hand, the uniform dimension of the cube $[0,1]^n$ is $n$ but the coarse dimension of $[0,1]^n$ is zero.

REPLY [3 votes]: About motivation. (Late answer, sorry) I don't know what you have chosen and how it went, but had I have found your question in time, I would have recommended pseudometrics. Your audience knows metric spaces ? they must feel very well pointwise convergence then (say for real-valued functions on a set $X$). But, if the base set $X$ is not denumerable it cannot be explained by a single metrics but by very simple pseudometrics (defined through the coordinate functions).
About place to learn uniform spaces. I would recommend Bourbaki, General Topology, because they give as well the equivalence with pseudometrics, the categorical (without saying) view of completion and the motivation by topological groups.<|endoftext|>
TITLE: Product Measure Only Possible Measure?
QUESTION [7 upvotes]: Let $X$ be a separable complete metric space and $Z$ be the set of all integers. Let $\nu$ be a Borel probability measure on $X^Z$ invariant under the shift function $S:X^Z \to X^Z$. Is it necessarily the case that $\nu = \mu^Z$ for some Borel probability measure $\mu$ on X?
Thanks

REPLY [7 votes]: Every such a $\nu$ is the law of some stationary process on $X^{\mathbb{Z}}$. Of course not every stationary process is i.i.d.<|endoftext|>
TITLE: Is Fourier analysis a special case of representation theory or an analogue?
QUESTION [137 upvotes]: I'm asking this question because I've been told by some people that Fourier analysis is "just representation theory of $S^1$."
I've been introduced to the idea that Fourier analysis is related to representation theory. Specifically, when considering the representations of a finite abelian group $A$, these representations are all $1$-dimensional, hence correspond to characters $A \to \mathbb{R}/\mathbb{Z} \cong S^1 \subseteq \mathbb{C}$. On the other side, finite Fourier analysis is, in a simplistic sense, the study of characters of finite abelian groups. Classical Fourier analysis is, then, the study of continuous characters of locally compact abelian groups like $\mathbb{R}$ (classical Fourier transform) or $S^1$ (Fourier series). However, in the case of Fourier analysis, we have something beyond characters/representations: We have the Fourier series / transform. In the finite case, this is a sum which looks like $\frac{1}{n} \sum_{0 \le r < n} \omega^r \rho(r)$ for some character $\rho$, and in the infinite case, we have the standard Fourier series and integrals (or, more generally, the abstract Fourier transform). So it seems like there is something more you're studying in Fourier analysis, beyond the representation theory of abelian groups. To phrase this as a question (or two):
(1) What is the general Fourier transform which applies to abelian and non-abelian groups?
(2) What is the category of group representations we consider (and attempt to classify) in Fourier analysis? That is, it seems like Fourier analysis is more than just the special case of representation theory for abelian groups. It seems like Fourier analysis is trying to do more than classify the category of representations of a locally compact abelian group $G$ on vector spaces over some fixed field. Am I right? Or can everything we do in Fourier analysis (including the Fourier transform) be seen as one piece in the general goal of classifying representations?
Let me illustrate this in another way. The basic result of Fourier series is that every function in $L^2(S^1)$ has a Fourier series, or in other words that $L^2$ decomposes as a (Hilbert space) direct sum of one dimensional subspaces corresponding to $e^{2 \pi i n x}$ for $n \in \mathbb{Z}$. If we encode this in a purely representation-theoretic fact, this says that $L^2(S^1)$ decomposes into a direct sum of the representations corresponding to the unitary characters of $S^1$ (which correspond to $\mathbb{Z}$). But this fact is not why Fourier analysis is interesting (at least in the sense of $L^2$-convergence; I'm not even worrying about pointwise convergence). Fourier analysis states furthermore an explicit formula for the function in $L^2$ giving this representation. Though I guess by knowing the character corresponding to the representation would tell you what the function is.
So is Fourier analysis merely similar to representation theory, or is it none other than the abelian case of representation theory?
(Aside: This leads into a more general question of mine about the use of representation theory  as a generalization of modular forms. My question is the following: I understand that a classical Hecke eigenform (of some level $N$) can be viewed as an element of $L^2(GL_2(\mathbb{Q})\ GL_2(\mathbb{A}_{\mathbb{Q}})$ which corresponds to a subrepresentation. But what I don't get is why the representation tells you everything you would have wanted to know about the classical modular form. A representation is nothing more than a vector space with an action of a group! So how does this encode the information about the modular form?)

REPLY [15 votes]: A long overdue comment. The paper:
Mackey, George W.
Harmonic analysis as the exploitation of symmetry—a historical survey.
Bull. Amer. Math. Soc. (N.S.) 3 (1980), no. 1, part 1, 543–698.
which is currently available here:
http://www.maths.ed.ac.uk/~jmf/Teaching/Projects/Poincare/S0273-0979-1980-14783-7.pdf
is wonderful (even a modern day classic?).
It answers the question very well, although it is quick to point out (on page 3) that "...many people use [harmonic analysis] to denote a different class of generalizations of the classical harmonic analysis of Fourier."<|endoftext|>
TITLE: Is non-connectedness of graphs first order axiomatizable?
QUESTION [13 upvotes]: A recent 
question
asked for graph properties that are first order axiomatizable but not finitely axiomatizable.
Connectedness was mentioned in the context.  Connectedness can be axiomatized in infinitary logic, but not in ordinary first order logic.  Just take an ultraproduct of the paths
$P_n$ of length $n$, $n\in\mathbb N$.  The paths are connected, but the ultraproduct has exactly two vertices of degree 1 and these two are not joined by a path of any finite length.
If connectedness was axiomatizable by a first order theory $\Phi$, then all the $P_n$ would satisfy $\Phi$ and hence the ultraproduct satisfies $\Phi$.  But the ultraproduct is not connected, a contradiction.  
I was wondering whether non-connectedness is first order axiomatizable.  I guess it is not,
but I don't have an argument for this right now.  
An attempted proof goes as follows:  Let $G_n$ be the disjoint union of two cycles of length $n$.  Take an ultraproduct $H$ of the $G_n$.  Now all vertices of $H$ have degree $2$ and 
there are no finite cycles.  In other words, $H$ is the disjoint union of a family of
infinite (in both directions) paths.
We were done if we could show that $H$ is elementary equivalent to the bi-infinite path
(is there a notation for this graph?).  I assume that this is the case, but I don't see why.
A different proof that non-connectedness is not first order axiomatizable would also be welcome (or an axiomatization).

REPLY [11 votes]: Stefan's original idea is realized in the following observation, which shows that one $\mathbb{Z}$-chain is elementary equivalent to two such chains.
Theorem. The theory of nontrivial cycle-free graphs where every vertex
has degree $2$ is complete.
Proof. All models of uncountable size $\kappa$ consist of
$\kappa$ many $\mathbb{Z}$ chains, and hence are isomorphic. Thus,
the theory is $\kappa$-categorical, and hence complete. QED
Thus, all cycle-free graphs with every vertex of degree
$2$ have the same first order theory. In particular, the graph consisting of one $\mathbb{Z}$-chain is
elementary equivalent to the graph consisting of any number
of such $\mathbb{Z}$ chains. Since the first graph is connected and
the latter are not, it follows that neither connectivity
nor disconnectivity are first-order expressible as theories
in the language of graph theory.

REPLY [10 votes]: The class of non-connected graphs is not axiomatizable. To see this, consider $\mathbb{Z}$ as a graph with $i$, $j$ connected by an edge if and only if $|i-j|=1$. Then a simple compactness argument yields a non-connected graph $\Gamma$ such that $\Gamma$ is elementarily equivalent to $\mathbb{Z}$; ie $\Gamma$ and $\mathbb{Z}$ satisfy precisely the same first order sentences. Since $\mathbb{Z}$ is connected and $\Gamma$ is non-connected, the result follows.<|endoftext|>
TITLE: Two interacting bodies in an external field
QUESTION [8 upvotes]: Hope, MO is the right place for this question (if not so: where would you pose it?).
Consider a two-body system in classical mechanics. As long as the interaction depends only on the distance of the two bodies, the two-body problem is integrable/solvable. Now consider the two bodies in a fixed external field. (This is only one step away from a three-body system that is known to be non-integrable in general, but obviously different from it.)

Question: Can the conditions on the combination
  of interaction and external field be
  explicitely given for the problem to
  be integrable/solvable? 

It might be the case that the problem is
always solvable. In this case the
following reference request becomes
predominant:

Reference request:
  Where can I find an explicit and
  elaborated treatment of this problem?

REPLY [8 votes]: I seriously doubt there is any general criteria.  However there are
more than one beautiful explicit examples of an external field which
lead to an integrable problem.  The simplest and probably best known is
that of a   constant field.
An absolutely beautiful description of this and its solution can be found in the book
`Essais sur le Mouvement des Corps Cosmiques' by V. Beletski, ch. 3.
(See eq. 3.2.1) It illustrates the plethora of qualitatively different phenomenon possible
within a single integrable system. 
As a wierd tangent, the `anisotropic Kepler problem': keep the same
potential but change the kinetic term to $a p_x ^2 + b p_y ^2$, $a \ne b$
is known to be non-integrable and Gutzwiller made an early career on this
problem.<|endoftext|>
TITLE: Cauchy-Davenport strengthening?
QUESTION [5 upvotes]: Is the following statement, refining classical Cauchy-Davenport Theorem (that states that for sets $A$, $B$ of residues modulo prime $p$, $|A+B|\geq |A|+|B|-1$ provided that RHS does not exceed $p$) true/known?
Let $A$, $B$ be two subsets of $\mathbb{F}_p$, $p$ being prime, and $|A|+|B|\leq p+1$. Then a complete bipartite graphs with parts $A$, $B$ (rigorously speaking, disjoint copies of $A$ and $B$, say $A\times \{0\}$ and $B\times\{1\}$) have a spanning tree such that all $|A|+|B|-1$ edgesums are different. (for edge $e=(a,b)$ its edgesum is defined as $\,a+b$).

REPLY [4 votes]: I believe that your statement follows from Cauchy-Davenport via matroid intersection theorem. (Matroid intersection theorem is stated in Chapter 41 of Alexander Schrijver's "Combinatorial optimization" book and can be also found here.) 
You want to find a "rainbow" spanning tree in a complete bipartite graph you define, where colors correspond to edgesums. "Rainbow" spanning trees, in fact, seem to be commonly used as an example of matroid intersection.
By matroid intersection it suffices to show that for any set of edges $U$ in your graph 
$r_1(U)+r_2(E \setminus U) \geq |A|+|B|-1,$
where: 
$E$ is the set of all $|A||B|$ edges,
$r_1(U)$ is the rank of $U$ in the cycle matroid, and is equal to $|A|+|B|-c(U)$ where $c(U)$ is the number of connected components in the graph induced by $U$, and 
$r_2(E \setminus U)$ is the number of edgesums obtained by the edges not in $U$.
If $c(U)=1$ then we are done.  Otherwise, let $A' \subseteq A$, $B' \subseteq B$ be obtained from $A$ and $B$ by choosing one element from each component of the graph induced by $U$, so that both are non-empty. Then the edges between $A'$ and $B'$ are not in $U$ and thus by Cauchy-Davenport
$r_2( E\setminus U) \geq c(U)-1$,
as desired.<|endoftext|>
TITLE: Determinant of a $3\times3$ magic square
QUESTION [10 upvotes]: This is my first question with mathOverflow so I hope my etiquette is up to par here.
My question is regarding a $3\times3$ magic square constructed using the la Loubère method (see la Loubère method)
Using the method, I have constructed a magic square and several semimagic squares (where one or both of the diagonals do not add up to a magic sum) with a program on written on my graphing calculator. After playing around with the program, I was shocked that the determinants of these $3\times3$ magic squares are all the same (specifically -360). Why is this so? (I am still an undergraduate so please go easy on the math :] )

REPLY [11 votes]: Kevin has done the magic squares; I'll do the semi-magics. So the rules are, we're using the 9 digits, once each, and all the row sums and all the column sums are equal. Note that this common sum must be 15. Now the only ways to use 1 and get 15 are $1+6+8$ and $1+5+9$. Since exchanging rows doesn't affect the (absolute value of) the determinant, and exchanging columns doesn't, either, and taking the transpose doesn't, and since further these operations don't affect semi-magicity, we may assume the matrix looks like this: $$\pmatrix{1&6&8\cr5&&\cr9&&\cr}$$ Now the only ways to use 9 are $1+5+9$, which we've already used, and $2+4+9$, so the matrix must look like $$\pmatrix{1&6&8\cr5&&\cr9&2&4\cr}\qquad{\rm or}\qquad\pmatrix{1&6&8\cr5&&\cr9&4&2\cr}$$ The first matrix fills out uniquely to $$\pmatrix{1&6&8\cr5&7&3\cr9&2&4\cr}$$ while the second one can't be completed. So, up to transpose, row swaps, and column swaps, there is only one order 3 semi-magic square, hence, only one determinant (up to sign).<|endoftext|>
TITLE: Normal Varieties
QUESTION [10 upvotes]: Let X be a complex normal variety and U a subvariety that is open in the analytic topology.  Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
edited to include complex

REPLY [4 votes]: I would like to risk an answer that does not use the language of algebraic geometry. For a pair (complex analytic variety $X$; closed analytic subvariety $Y$), $U=X\setminus Y$,
there exists a triangulation such that $Y$ is a subcomplex (see, for example Triangulations of algebraic sets - Hironaka 1974, can be found with google books). In other words $X$ is a simplicial complex, and $Y$ 
is a subcomplex. Now, if $X$ is normal its singularities are in real codimension at least $4$. I.e. $X$ is a $PL$ manifold in codimension $4$. 
In order to show that the fundamental group of $X\setminus Y$ surjects onto the fundamental group of $X$, it is sufficient to show that every loop in  $X$ can be homotoped into $X\setminus Y$. Since $Y$ it is contained in the simplicial subcomplex of codimension $2$ it is enough to show that any loop in $X$ can be homotoped so it does not touch any simplex of codim $2$, but this is true for every $PL$ space that is a manifold in codim $2$.<|endoftext|>
TITLE: What do mathematicians currently do in conformal field theory (or more general field theory)
QUESTION [21 upvotes]: I am wondering what currently our mathematicians do related to conformal field theory, (I know currently it is a central topic, but I have only a vague idea what mathematicians do in there), or more generally topological field theory, field theory... I am extremely appreicated if there is any survey paper.

REPLY [3 votes]: If you are interested in operator algebras, try this review by Kawahigashi From Operator Algebras to Superconformal Field Theory, where connections to subfactor theory, moonshine and noncommutative geometry are pointed out.<|endoftext|>
TITLE: Can knowing ahead the length of 3-SAT instance really help?
QUESTION [6 upvotes]: If I say I can solve 3-SAT ( known to be NP-complete) in polynomial time, yet with the following 'little' proviso:
Give me first $n$ the length of your 3-SAT formula, then give me some time on my own , then as soon as you give me your formula, I will answer in less that $n^k$.  
The $k$ will be constant independent of $n$ (this is not parametrized complexity)
Implicitly: after you give me $n$, I may pre-calculate as much as I want (say $n^n$ or even much more) and I may also store some results as much as I want.
Question : is this equivalent to 3-SAT?
Comment : I cannot find a polynomial solution like : calculate all solutions store them on a tree and then retrieve on question . So it seems to be as 'difficult' as 3-SAT.
Note : I took 3 SAT but any NP-complete problem Q will do : define generically the variation Q' with the length of the instance of the problem Q given ahead of the instance.

REPLY [17 votes]: If you're allowed to do an exponential amount of work, then you might just solve SAT on all formulae of that size. Then you create an exponential size lookup table. Now when you're given the instance, just look it up and you've solved the problem.
On the other hand, if you insist that the stored data should only be of polynomial size, then the question you're asking is whether 3SAT is in P/poly. The notion of unbounded pre-computation based only on the input length is nicely captured by the concept called advice in complexity theory. The "/poly" after P basically means that the polynomial-time algorithm has access to a trusted polynomial-size string which depends only on the input length and not the instance. The advice string itself may be very hard to compute. All we need is the existence of such a string, not necessarily computability.
It is not known if 3SAT is in P/poly, and it's not known if this implies P = NP. However, it does imply that the polynomial hierarchy collapses to the second level by the Karp-Lipton theorem.
Lastly, P/poly has a nice intuitive description. It is the set of problems that can be solved by a polynomial-size circuit family.

REPLY [2 votes]: It sounds like your're trying to place SAT in P/poly (or P/exp or P/unbounded) rather than P.<|endoftext|>
TITLE: Why can't proofs have infinitely many steps?
QUESTION [36 upvotes]: I recently saw the proof of the finite axiom of choice from the ZF axioms. The basic idea of the proof is as follows (I'll cover the case where we're choosing from three sets, but the general idea is obvious): Suppose we have $A,B,C$ non-empty, and we would like to show that the Cartesian product $A \times B \times C$ is non-empty. Then $\exists a \in A$, $\exists b \in B$, $\exists c \in C$, all because each set is non-empty. Then $a \times b \times c$ is a desired element of $A \times B \times C$, and we are done.
In the case where we have infinitely (in this case, countably) many sets, say $A_1 \times A_2 \times A_3 \times \cdots$, we can try the same proof. But in order to use only the ZF axioms, the proof requires the infinitely many steps $\exists a_1 \in A_1$, $\exists a_2 \in A_2$, $\exists a_3 \in A_3$, $\cdots$
My question is, why can't we do this? Or a better phrasing, since I know that mathematicians normally work in logical systems in which only finite proofs are allowed, is: Is there some sort of way of doing logic in which infinitely-long proofs like these are allowed?
One valid objection to such a system would be that it would allow us to prove Fermat's Last Theorem as follows: Consider each pair $(a,b,c,n)$ as a step in the proofs, and then we use countably many steps to show that the theorem is true.
I might argue that this really is a valid proof - it just isn't possible in our universe where we can only do finitely-many calculations. So we could suggest a system of logic in which a proof like this is valid.
On the other hand, I think the "proof" of Fermat's Last Theorem which uses infinitely many steps is very different from the "proof" of AC from ZF which uses infinitely many steps. In the proof of AC, we know how each step works, and we know that it will succeed, even without considering that step individually. In other words, we know what we mean by the concatenation of steps $(\exists a_i \in A_i)_{i \in \mathbb{N}}$. On the other hand, we can't, before doing all the infinitely many steps of the proof of FLT, know that each step is going to work out. What I'm suggesting in this paragraph is a system of logic in which the proof of AC above is an acceptable proof, whereas the proof of FLT outlined above is not acceptable.
So I'm wondering whether such a system of logic has been considered or whether any experts here have an idea for how it might work, if it has not been considered. And, of course, there might be a better term to use than "system of logic," and others can give suggestions for that.

REPLY [4 votes]: James Brotherston and Alex Simpson worked on non-well-founded proofs, see

J. Brotherston and A. Simpson: Sequent calculi for induction and infinite descent. J Logic Computation (2010)

as well as the talk "On Proof by Infinite Descent" by Alex Simpson at "Algebra & Coalgebra meet Proof Theory" in Bern, April 2012.<|endoftext|>
TITLE: What categorical mathematical structure(s) best describe the space of "localized events" in "relational quantum mechanics"?
QUESTION [5 upvotes]: In a recent (and to me, very beautiful) paper, entitled "Relational EPR",

Smerlak and Rovelli present a way of thinking about EPR which relies upon Rovelli's previously published work on relational quantum mechanics (see http://arxiv.org/abs/quant-ph/9609002 ). In relational quantum mechanics, there is no non-locality, but the definition of when an event occurs is weakened from Einstein's strict definition and instead is localized to each observer-measurement apparatus, including subsequent observers. There are (informal) coherence assumptions to ensure the consistency of reports from different subsequent observers (all possible friends of Wigner).
All of this seems very similar to various results in modern categorical mathematics. Is there a standard mathematical structure which well describes the structure of the space of localized measurements which Rovelli has envisioned? I know of Isham's work on topos theory and quantum mechanics, but I think he is aiming at something a little different.
PS I first asked this on mathunderflow, but was advised to repost here.

REPLY [2 votes]: I wouldn't say it is a standard structure (yet), but Samson Abramsky has recently given a relational/categorical account of a lot of properties like non-locality and no-signaling. Relations between these notions, including no-go theorems, that are surprisingly involved in other formulations, are quite simply derived, indicating that this could be a good way to state things. See arxiv:1007.2754.<|endoftext|>
TITLE: Appropriate journal to publish a determinantal inequality
QUESTION [7 upvotes]: I have recently made the following observation:
Let $v_i := (v_{i1}, v_{i2})$, $1 \leq i \leq k$, be  non-zero   positive  elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m_{ij} := \max${$v_{ik}/v_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$. 
The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)? 
It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated. 
 Edit:  Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.

REPLY [11 votes]: It seems to be true, but relatively simple (unless I made a mistake). Let's see:
First of all, scaling any pair $(v_{i,1},v_{i,2})$ by a constant $c$ does not change the determinant (one row of the matrix is multiplied by $c$, and one column is divided by $c$). We can therefore assume without losing generality that $(v_{i,1},v_{i,2})=(v_i,1)$.
Also, permutation of $v_i$'s does not change the determinant; therefore, we may assume that
$v_i$'s are strictly increasing. In this case, the matrix has the form
$$m_{ij}=\begin{cases} 1,& i\le j\cr v_i/v_j,& i>j\end{cases}.$$
Now subtract the top row from all others, and shift it to the bottom. You end up with a triangular matrix whose determinant is non-zero.<|endoftext|>
TITLE: Finite-dimensional version of the word problem for groups
QUESTION [10 upvotes]: The (uniform) word problem for groups can be stated in several equivalent ways:
Word Problem for Groups (WP)
Instance: A finite presentation of a group G and an element w of G as a product of generators and their inverses.
Question: Does every linear representation of G in a (not necessarily finite-dimensional) Hilbert space map w to the identity operator?
Equivalently: Does every unitary representation of G in a Hilbert space map w to the identity operator?
Equivalently: Does every normal subgroup of G contain w?
Equivalently: Is w=1 in G?
As is well-known, WP is undecidable; more specifically, it belongs to RE∖coRE, where RE is the class of recursively enumerable languages and coRE is the class of their complements.  (In fact, it is known that WP remains undecidable even if we fix the group G and its finite presentation suitably, but that is not the topic of this question.)
We can consider the finite-dimensional version of WP.
Finite-Dimensional Word Problem for Groups (FWP)
Instance: Same as WP.
Question: Does every matrix representation of G map w to the identity matrix?
Equivalently: Does every unitary matrix representation of G map w to the identity matrix?
Equivalently: Does every normal subgroup of G of finite index contain w?
(The equivalence is based on a result by Malcev; see this post by Greg Kuperberg and the comments attached to it.)
The two problems have different answers for some instances; that is, sometimes w≠1 in G but every matrix representation maps w to the identity matrix.  There even exists a finitely presented infinite group which does not have a nontrivial matrix representation.  See the answers to the question “Finitely presented sub-groups of GL(n,C)” by Dmitri.
In fact, FWP is in coRE unlike WP: if we also give the dimension as part of the input, the problem becomes decidable (because it is a system of algebraic equations over ℂ), and therefore FWP can be put in coRE by trying all dimensions.

Question: Is FWP decidable?

I do not even know whether FWP is NP-hard or not.  As far as I know, FWP can be in P or undecidable, or anywhere in between!
(Note: FWP can be viewed as a (very) special case of the problem discussed in the question “Decidability of matrix algebra” by Ricky Demer.)

REPLY [15 votes]: FWP is undecidable by a result of Slobodskoi. Slobodskoi shows that the "Universal theory" of finite groups is undecidable. What you are asking for is whether the "Q-theory" of the pseudovariety of finite groups is decidable. The universal theory and Q-theory are equivalent for the pseudovariety of finite groups (see Kharlampovich-Sapir, section 2.4 for a discussion), so the Q-theory is undecidable too.<|endoftext|>
TITLE: Contour integration problem from probability
QUESTION [7 upvotes]: Can integrals of the form
$$
\int_{-\infty}^{\infty}{\exp\left(-\left[x - c\right]^{2}\right) \over 1 + x^{2}}\, {\rm d}x
$$
be computed in closed form using contour integration (or any other technique)?  If
$c = 0$, the integral is $\pi{\rm e\ erfc}\left(1\right)$, but I'm interested in
$c$ real and non-zero.
( In probability terms, the integrand is a product of normal and Cauchy densities. )

REPLY [6 votes]: $$ J(c)=\int_{-\infty}^{\infty}\frac{\exp[-(x-c)^2]}{1+x^2}dx=e^{-c^2}\int_{-\infty}^{\infty}\frac{\exp[-x^2]}{1+x^2} e^{2cx}dx $$
The integral on the right can be treated as the Fourier transform $\mathcal{F}(\exp[-x^2]/(1+x^2))$, with the transform parameter equal to $\mbox{i}2c$. The function is actually symmetric wrt $x$, thus, it is the cosine Fourier transform we are talking about. The necessary transform is available in Vol. 1 of Bateman & Erdelyi's "Tables of Integral Transforms" (1954). I used a shortcut and computed the transform using Maple. The resulting expression is:
$$ J(c)=\frac{\pi\mbox{e}}{2}\left( \mbox{erfc}(1+\mbox{i}c)e^{\mbox{i}2c}+\mbox{erfc}(1-\mbox{i}c)e^{-\mbox{i}2c} \right) $$
It is easy to check that this answer satisfies the ODE obtained by fedja. Written as the sum of conjugate terms, the function $J(c)$ is clearly real-valued for real $c$. It remains an open question whether this is a "nicer" form compared to what you had originally!<|endoftext|>
TITLE: What is the smallest $C^*$-algebra containing the "standard" pseudodifferential operators?
QUESTION [8 upvotes]: Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying 
$
|\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|}
$
) a $C^*$-algebra? 
In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?

If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace? 

 Edit: Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described here, $\Psi^0(\mathbb{R})$ is a Fréchet $*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols:
$$
\Vert a \Vert_{\alpha,\beta} = \sup_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)|.
$$
So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:
Is there a reasonable description of the smallest $C^*$-algebra containing $\Psi^0$?

REPLY [4 votes]: A remark to add to Paul's answer: yes $\mathcal{B}(U)$ is precisely the $C^*$-algebra generated by $\Psi^0$, as follows.
Let $\mathcal{A}$ be the $C^\ast$-closure of $\Psi^0$. The image of the principal symbol map $\mathrm{Symb}\colon \Psi^0 \to C_0(S^\ast U)$ is dense, and so the image of $\mathcal{A}$ is all of $C_0(S^\ast U)$.  Thus, for any $T\in\mathcal{B}(U)$, we can find $T^\prime \in \mathcal{A}$ with the same symbol. Following Paul, $T-T^\prime$ is a compact operator. The compacts are in $\mathcal{A}$ (since the smoothing operators are dense therein). Thus, $T$ is in $\mathcal{A}$.
edit (AlexE): To phrase this result in another way, we have the so-called pseudodifferential operator extension (a short exact sequence) $$0 \to \mathcal{K} \to \mathcal{A} \stackrel{\mathrm{Symb}}\longrightarrow C_0(S^\ast U) \to 0,$$
where $\mathcal{K}$ are the compact operators.<|endoftext|>
TITLE: Ref request: A graph G contains H as a minor iff it contains one of finitely many graphs as a topological minor
QUESTION [5 upvotes]: For definitions of graph minors and topological minors, see wikipedia's article on graph minors.
Theorem: For every graph H, there is a finite set of graphs, say S(H), such that G contains H as a minor if and only if G contains some graph from S(H) as a topological minor.
Can anyone point me to a paper/book where this is proved? (I know how to prove it, I just want a reference to cite.)

REPLY [5 votes]: Hello !!! You will find this theorem as results 2.2 and 2.3 in Graph minors VIII : A Kuratowski theorem for general surfaces.
Nathann<|endoftext|>
TITLE: Stability of midpoints in CAT(0) spaces
QUESTION [5 upvotes]: Given a CAT(0) space $X$ and a compact, convex subset $A$ of $X$. One can define its midpoint $m(A)$ as the point, at which the following function attains its minimum.
$f:A\rightarrow \mathbb{R}\qquad x\mapsto \sup\{d(x,y)|y\in A\}$
One can show, that there is a unique such point. So my question is:
Given two compact,convex subset $A_1,A_2\subset X$. Is $d(m(A_1),m(A_2))$ less or equal to the Hausdorff distance between $A_1$ and $A_2$?

REPLY [7 votes]: No, even if $X=\mathbb R^2$.
Let $A_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $\sqrt 2$ centered at $(0,0)$. Shift $S_1$ a small distance $\varepsilon$ in the horizontal direction, denote the resulting circle by $S_2$. For each vertex of $A_1$, mark its nearest point on $S_2$. The marked points are vertices of a convex quadrangle $A_2$ inscribed in $S_2$ and containing its center $(\varepsilon,0)$. Hence $m(A_2)=(\varepsilon,0)$ but the Hausdorff distance between $A_1$ and $A_2$ is $\approx\varepsilon/\sqrt 2$.<|endoftext|>
TITLE: Classification of finite groups of isometries
QUESTION [24 upvotes]: Consider the problem of classifying the finite groups of isometries of $\mathbb{R}^n$.

For $n=2$ it is cyclic and dihedral groups.
For $n=3$ they are well known, probably from Kepler and are related to ade-classification.
For $n=4$ we can get them by taking the universal cover of $\mathrm{SO}(4)$ which is isomorphic to $\mathrm{SU}(2) \times \mathrm{SU}(2)$, though I do not know where the classification is available.

But my main question is for dimension $n\geq 5$. Does anybody knows the state of the art? A reference would be most helpful.
Note that the finite subgroups of $\mathrm{GL}_n(\mathbb{Z})$ are classified for $n\leq 10$.

REPLY [6 votes]: Surprisingly, I found explicit lists of discrete subgroups of the orthogonal group O(n) for up to n=8 dimensions on the wikipedia page for point groups, with rather unspecific references, however. Point groups is another name for discrete subgroups of O(n). [UPDATE+CORRECTION: For dimensions n=4 and larger, only the point groups which are generated by reflections (Coxeter groups) are listed. In particularly, subgroups of SO(n) (which include no matrix of determinant $-$1) are missing.]
There is an old sequence of two long papers by Threlfall and Seifert, part I Mathematische Annalen 1931, Volume 104, Issue 1, pp. 1-70, part II 1933, Volume 107, Issue 1, pp. 543-586, where they apparently do the classification of discrete subgroups of SO(4) by associating to each element of SO(4) a pair of rotations from SO(3). (Although my native language is German, I had a hard time reading (through) this, because I am not used to the terminology that was used at that time.) [Addition: These results are mentioned in the book by Conway and Smith on quaternions and octonions; Conway and Smith say that the list is complete, but contains duplicates.]
I have a rather wild conjecture (true up to three dimensions). [UPDATE 2: wrong in 4 dimensions]

Every discrete point group in n dimensions is the symmetry group of an n-dimensional polytope which is the Cartesian product of regular polytopes, or a subgroup thereof. 

[UPDATE 2: One counterexample in 4D is the group $\pm [I\times C_n]$ in Table 4.1 of Conway and SmithSloane's book, p. 44, for sufficiently high $n$. It is isomorphic to a subgroup of a direct product of lower-dimensional point groups, in the group-theoretic sense, but geometrically, it is not the symmetry group of a Cartesian product of two objects in orthogonal subspaces. An orbit $A$ of a point under this group can be constructed as follows. Consider the Hopf fibration $f\colon S^3\to S^2$, and take the twelve great circles which are the pre-images of the vertices of a regular icosahedron on $S^2$. The point set $A$ consists of 12 regular $n$-gons inscribed in these circles. Probably, the first group in the list, $\pm [I\times O]$, is also a counterexample, since it is not contained in an achiral group.]
[UPDATE 1:
Norman Johnson pointed out counterexamples: The symmetries of the root lattices E6, E7, E8 in 6, 7, and 8 dimensions. (I could not yet fully convinced myself that they are indeed counterexamples.) So dimensions 4 and 5 remain open. If I extend my conjecture to include the polytopes which have those E6, E7, or E8 symmetries, in addition to the regular polytopes, in which dimension would the next counterexamples be?]
For example, the symmetries of an $m$-gonal anti-prism in 3-space are
contained in the symmetries of the $2m$-sided prism,
which is the 1-simplex $\times$ the regular $2m$-gon.
Since the regular polytopes are known in all dimensions, this would give
an easy way to obtain all finite point groups.
(at least in principle).<|endoftext|>
TITLE: What are the big problems in probability theory?
QUESTION [108 upvotes]: Most branches of mathematics have big, sexy famous open problems. Number theory has the Riemann hypothesis and the Langlands program, among many others. Geometry had the Poincaré conjecture for a long time, and currently has the classification of 4-manifolds. PDE theory has the Navier-Stokes equation to deal with.
So what are the big problems in probability theory and stochastic analysis? 
I'm a grad student working in the field, but I can't name any major unsolved conjectures or open problems which are driving research. I've heard that stochastic Löwner evolutions are a big field of study these days, but I don't know what the conjectures or problems relating to them are. 
Does anyone have any suggestions?

REPLY [4 votes]: Forming tools for handling random surfaces and proving an universal central limit theorem for them given minimal conditions (think usual CLT):
a)The Gaussian Free Field (GFF) has shown up as the limiting universal object for many random surfaces (in KPZ 2+1, random tilings, random matrix theory Ginibre ensembles cf Borodin's and Kenyon's work)
b)Schramm-Loewner evolutions (SLE) have shown up as the limiting interfaces for families of statistical models.
c)Finally, merging the above two pictures since SLEs can be coupled to GFF (cf Sheffield).
Another powerful result would be showing the equivalence of Random planar maps and Liouville quantum gravity (LQG) (a promising approach by Miller and Sheffield). This is because it happens that statistical models become easier to handle over these random surfaces (Kazanov-Ising model, LERW-Duplantier).<|endoftext|>
TITLE: Random products of projections: bounds on convergence rate?
QUESTION [16 upvotes]: The von Neumann-Halperin [vN,H] theorem shows that iterating a fixed product of projection operators converges to the projector onto the intersection subspace of the individual projectors. A good bound on the rate of convergence using the concept of the Friedrichs number has recently been shown [BGM]. 
A generalization of this result due to Amemiya and Ando [AA] to the product of random sequences of projection operators drawn from a fixed set also shows convergence to the projector onto the intersection subspace.
My question is: are there any known bounds on the convergence rate for the latter problem analogous to the earlier one? In my application I'm only interested in the case of finite-dimensional Hilbert spaces.
[vN] J. von Neumann, Functional operators, Annals of Mathematics Studies No. 22, Princeton University Press (1950)
[H] I. Halperin, The product of projection operators, Acta. Sci. Math. (Szeged) 23 (1962), 96-99.
[BGM] C. Badea, S. Grivaux, and V. M¨uller. A generalization of the Friedrichs angle
and the method of alternating projections. Comptes Rendus Mathematique, 348(1–2):53–56, (2010).
[AA] I. Amemiya and T. Ando, Convergence of random products of contractions in Hilbert space, Acta. Sci. Math. (Szeged) 26 (1965), 239-244.

REPLY [2 votes]: Following Suvrit's post, you can also take a look at http://arxiv.org/abs/1205.5770 (Algorithm 3). It handles the case where the set of projectors have co-dimension one.
By the way, thanks for the links Martin.<|endoftext|>
TITLE: What are some open problems in algebraic geometry?
QUESTION [57 upvotes]: What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.

REPLY [8 votes]: The Maximal Rank Conjecture is a major outstanding problem in Brill-Noether theory, although recent advances in tropical techniques might point the way to a solution; see https://arxiv.org/abs/1505.05460.
EDIT: The Maximal Rank Conjecture was proved by Eric Larson in his PhD thesis; see: https://arxiv.org/abs/1711.04906

REPLY [6 votes]: The Tate conjecture: Let $k$ be a finitely generated field, $X/k$ a smooth projective geometrically integral variety and $\ell$ invertible in $k$.  Then the cycle class map $$\mathrm{CH}^r(X) \otimes_\mathbf{Z} \mathbf{Q}_\ell \to \mathrm{H}^{2r}(\bar{X},\mathbf{Q}_\ell(r))^{G_k}$$ is surjective.
It is e.g. proved for $r=1$ and Abelian varieties, a deep theorem. See http://www.math.harvard.edu/~chaoli/doc/TateConjecture.html.
This is analogous to the Hodge conjecture for complex varieties.<|endoftext|>
TITLE: Consistency results separating three cardinal characteristics simultaneously
QUESTION [13 upvotes]: (For information on cardinal characteristics of the continuum aka cardinal invariants see Joel David Hamkins' MO answer here; Andreas Blass's handbook article is an excellent reference.)
Problem 2.3 of  Shelah's  "On What I Do Not Understand (and Have Something to Say), Part I" (published in 2000 in Fundamenta Mathematicae) states, "Investigate cardinal invariants of the continuum showing $\geq 3$ may have prescribed order". One major barrier to such an investigation is the fact that countable support iteration of proper forcings yields models where the continuum is $\aleph_2$. In such models given any three cardinal characteristics at least two will have to be equal.
My question is the following. To what extent has such an investigation been pursued? In either the literature or folklore are there any results proving the consistency of inequalities $\mathfrak{c}_0<\mathfrak{c}_1<\mathfrak{c}_2$ where the $\mathfrak{c}_i$ are cardinal characteristics?

REPLY [18 votes]: Bumping an old question, because the situation has changed dramatically in the last year or so:
The most well-known cardinal characteristics of the continuum are those appearing in Cichoń's Diagram, which also presents all $ZFC$-provable relations between pairs of these characteristics: besides the dominating number $\mathfrak{d}$ and the bounding number $\mathfrak{b}$, these are all of the form $inv(\mathcal{I})$, for $\mathcal{I}$ the ideal of meager sets $(\mathcal{M})$ or null sets $(\mathcal{N})$, and $inv$ one of $add, cov, non, cof$:

$add$ of an ideal $\mathcal{I}$ is the least cardinality of a family of sets from $\mathcal{I}$ whose union is not in $\mathcal{I}$;

$cov$ of an ideal $\mathcal{I}$ is the least cardinality of a family of sets from $\mathcal{I}$ covering all of $\mathbb{R}$;

$non$ of an ideal $\mathcal{I}$ is the least cardinality of a set of reals not in $\mathcal{I}$; and

$cof$ of an ideal $\mathcal{I}$ is the cofinality of the partial order $\langle\mathcal{I}, \subseteq\rangle$.


As far as I know, the first results separating more than two cardinal characteristics from Cichon's diagram came out in the last two years:

In late 2012, Mejia constructed several examples of models separating multiple characteristics at once (see section 6).

In a paper arxived today(!), Fischer/Goldstern/Kellner/Shelah produced a model of $\aleph_1=\mathfrak{d}=cov(\mathcal{N})<non(\mathcal{M})<non(\mathcal{N})<cof(\mathcal{N})<2^{\aleph_0}$.


Now, I don't understand the proofs at all, but it seems the proofs by Mejia and by F/G/K/S are fundamentally different. The F/G/K/S paper has this to say about the two different approaches to separating multiple characteristics simultaneously:

"We cannot use the two best understood methods [to separate $\ge 3$ characteristics simultaneously], countable support iterations of proper forcings (as it forces $2^{\aleph_0}\le\aleph_2$) and, at least for the "right hand side" of the diagram, we cannot use finite support iterations of ccc forcings in the straightforward way (as it adds lots of Cohen reals, and thus increases $cov(\mathcal{M})$ to $2^{\aleph_0}$).
There are ways to overcome this obstacle. One way would be to first increase the continuum in a "long" finite support iteration, resulting in $cov(\mathcal{M})=2^{\aleph_0}$, and then "collapsing" $cov(\mathcal{M})$ in another, "short" finite support iteration. In a much more sophisticated version of this idea, Mejia [Mej13] recently constructed several models with many simultaneously different cardinal characteristics in Cichon's Diagram (building on work of Brendle [Bre91], Blass-Shelah [BS89] and Brendle-Fischer [BF11]).
We take a different approach, completely avoiding finite support, and use something in between a countable and finite support product (or: a form of iteration with very "restricted memory").
This construction avoids Cohen reals, in fact it is $\omega^\omega$-bounding, resulting in $\mathfrak{d}=\aleph_1$. This way we get an independence result "orthogonal" to the ccc/finite-support results of Mejia.
The fact that our construction is $\omega^\omega$-bounding is not incidental, but rather a necessary consequence of the two features which, in our construction, are needed to guarantee properness: a "compact" or "finite splitting" version of pure decision, and fusion . . ."
[Creature forcing and five cardinal characteristics in Cichon's diagram], pg. 2


I can't resist bumping this to add one fascinating result (from this morning!) of Goldstern, Kellner, and Shelah: that as many as possible of the cardinal characteristics in Cichon's diagram can be separated simultaneously! Now granted, this isn't a consistency result relative to ZFC - at present, it takes four strongly compact cardinals, which is no small assumption - but still, this is absolutely amazing.
EDIT: And as of a couple weeks ago, the other shoe has dropped (thanks to Jakob for bringing this to my attention below): Goldstern, Kellner, Mejia, and Shelah got it in ZFC.<|endoftext|>
TITLE: Different forms of compactness and their relation
QUESTION [10 upvotes]: Given a topological space X one can define several notion of compactness:
X is compact if every open cover has a finite subcover.
X is sequentially compact if every sequence has a convergent subsequence.
X is limit point compact (or Bolzano-Weierstrass) if every infinite set has an accumulation point.
X is countably compact if every countable open cover has a finite subcover.
X is σ-compact if it is the union of countably many compact subspaces.
X is pseudocompact if its if its image under any continuous function to $\mathbb{R}$ is 
bounded.
X is paracompact if every open cover admits an open locally finite refinement (i.e. every point of X has a neighborhood small enough to intersect only finitely many members of the cover).
X is metacompact if every open cover admits a point finite open refinement (i.e. if every point of X is in only finitely many members of the refinement). 
X is orthocompact if every open cover has an interior preserving open refinement (i.e. given an open cover there is a open subcover such that at any point, the intersection of all open sets in the subcover containing that point is also open).
X is mesocompact if every open cover has a compact-finite open refinement (i.e. given any open cover, we can find an open refinement such that every compact set is contained in finitely many members of the refinement).
So, there are quite a few notions of compactness (there are surely more than those I quoted up here). The question is: where are these definitions systematically studied? What I'm interested in particular is knowing when does one imply the other, when does it not (examples), &c.
I can fully answer the question for the first three notions:
Compact and first-countable --> Sequentially compact.
Sequentially compact and second-countable --> Compact.
Sequentially compact --> Limit-point compact.
Limit point compact, first-countable and $T_1$ --> Sequentially compact.
but I'm absolutely ignorant about the other cases. Has this been systematically studied somewhere? If so, where?

REPLY [7 votes]: I don't have Munkres' book.  So I don't know what is done there.  You should probably consult the "Counterexamples in Topology" as mentioned above.
My favourite book for questions of this type is "General Topology" by Ryszard Engelking.
It has a diagram in the back with interrelations between different properties of topological spaces.  It has all the notions you are interested in, except the last two.  Local compactness and various additional versions of paracompactness are also discussed in the book.
I assume, but haven't checked in every case, that the relations in the diagram are proved in the book.
The book also contains plenty of examples, showing that certain implications do not hold, sometimes in the form of exercises with hints.<|endoftext|>
TITLE: Derived algebraic geometry via dg rings?
QUESTION [13 upvotes]: Jacob Lurie's stuff seems to develop derived algebraic geometry via $E_\infty$ rings and/or maybe something like simplicial commutative rings. Ben Wieland's comment in this question indicates that Lurie never deals with commutative dg algebras. However, it is supposed to be true that all of these different things are the same (meaning more precisely that their model categories are Quillen equivalent) in characteristic zero.
So my question is:

Is the theory of derived algebraic geometry via dg rings or dg algebras in characteristic zero developed anywhere? If not, why not?

My motivations:

I feel like there must be a good reason why Lurie does not use dg rings/algebras, other than the fact that they apparently don't work well in positive characteristic. So I wonder what the reasons are.
I don't know very much about homotopy theory, so I find the $E_\infty$ rings approach to DAG a bit daunting. I am personally more comfortable with dg algebras.
I am personally more interested in things involving "sheaves of dg algebras" than things involving "sheaves of $E_\infty$ rings" (such as elliptic cohomology (and TMF), which I understand is one of Lurie's motivations).

REPLY [7 votes]: Dear Kevin,
This is more or less an amplification of Tyler's comment. You shouldn't take it too seriously, since I am certainly talking outside my area of expertise, but maybe it will be helpful.
My understanding is that homotopy theorists are extremely (perhaps primarily) interested in torsion phenomena.  (After all,
homotopy groups are often non-trivial but finite.)    TMF, for example, involves quite subtle torsion phenomena.  Coupled with Tyler's remark that homotopy theorists have no fear of $E_{\infty}$ rings, and so are (a) happy to identify them
with dg-algebras in char. zero, and (b) don't feel any psychological need to fall back on
the crutch of dg-algebras, this makes me suspect that your assumption (1) is likely to be wrong.  (I share your motivation (2), but this is a psychological weakness of algebraists that
homotopy theorists seem to have overcome!)
In particular, one of Lurie's achievements is (I believe) constructing equivariant versions of TMF,
which (as I understand it) involves (among other things) studying deformations of $p$-divisible groups of derived elliptic curves.  It seems hard to do this kind of thing
without having a theory that can cope with torsion phenomena.
Also, when Lurie thinks about elliptic cohomology, he surely includes under this umbrella TMF and its associated torsion phenomena.  (So your (3) may not include all the aspects
of elliptic cohomology that Lurie's theory is aimed at encompassing.)<|endoftext|>
TITLE: How to find the number of $k$-permutations of $n$ objects with $x$ types, and $r_1, r_2, r_3, \cdots , r_x$ = the number of each type of object?
QUESTION [10 upvotes]: I asked this on Math Stackexchange, but didn't really receive an answer - so I figured this might be the right place.

How can I find the number of k-permutations of n objects, where there are x types of objects, and r1, r2, r3 ... rx give the number of each type of object?
Here is an example with $n = 20, k = 15, x = 4,$ $ r_1 = 4 \quad r_2 = 5 \quad r_3 = 8 \quad r_4 = 3$.

I have 20 letters from the alphabet. There are some duplicates - 4 of them are a, 5 of them are b, 8 of them are c, and 3 are d. How many unique 15-letter permutations can I make?

Furthermore, if there isn't a straightforward solution: how efficiently can this problem be solved?

REPLY [2 votes]: Here's a sketch of an idea. Consider first the problem of computing the number of permutations of $k$ elements given the number of elements $n_i$ in each class. This is easy: it works out to $$\frac{k!}{\Pi_i n_i !}$$
because in any fixed string, you can permute the elements in a class without changing the string. 
Now consider the problem of writing down the different partitions of $k$ into the sum of $x$ labelled numbers. Merely finding the sum itself  can be done by determining the coefficient of $y^k$ in the polynomial 
$$ \Pi_i \frac{y^{r_i+1}-1}{y-1}$$
But this is not enough, since you need to weight each such partition by a different number. However, we know that the contribution of $y^i$ must be $1/i!$, so the overall number we are looking for is $k!$ times the coefficient of $y^k$ in the polynomial 
$$ \Pi_i \sum_{j=0}^{r_i} \frac{y^j}{j!}$$<|endoftext|>
TITLE: Why aren't there more classifying spaces in number theory?
QUESTION [29 upvotes]: Much of modern algebraic number theory can be phrased in the framework of group cohomology.  (Okay, this is a bit of a stretch -- much of the part of algebraic number theory that I'm interested in...).  As examples, Cornell and Rosen develop basically all of genus theory from cohomological point of view, a significant chunk of class field theory is encoded as a very elegant statement about a cup product in the Tate cohomology of the formation module, and Neukirch-Schmidt-Wingberg's fantastic tome "Cohomology of Number Fields" convincingly shows that cohomology is the principal beacon we have to shine light on prescribed-ramification Galois groups.  
Of course, we also know that group cohomology can be studied via topological methods via the (topological) group's classifying space.  My question is:

Question:  Why doesn't this actually happen?

More elaborately:  I'm fairly well-acquainted with the "Galois cohomology for number theory" literature, and not once have I come across an argument that passes to the classifying space to use a slick topological trick for a cohomological argument or computation (though I'd love to be enlightened).  On the other hand, for example, are things like Tyler's answer to my question 
Coboundary Representations for Trivial Cup Products
which strikes me as saying that there may be plenty of opportunities to carry over interesting constructions and/or lines of reasoning from the topological side to the number-theoretic one.  
Maybe the classifying spaces for gigantic profinite groups are too hideous to think about?  (Though there's plenty of interesting Galois cohomology going on for finite Galois groups...).  Or maybe I'm just ignorant to the history, and that indeed the topological viewpoint guided the development of group cohomology and was so fantastically successful at setting up a good theory (definition of differentials, cup/Massey products, spectral sequences, etc.) that the setup and proofs could be recast entirely without reference to the original topological arguments?  
(Edit:  This apparently is indeed the case.  In a comment, Richard Borcherds gives the link http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183537593 and JS Milne suggests MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280)., both of which look like good reads.)

REPLY [5 votes]: I suppose we should also mention algebraic k-theory. Quillen defined the k-groups as the homotopy groups of certain classifying spaces. For a unital, associative ring $R$, 
$$
K_n(R):=\pi_n(BGL(R)^+),
$$
where $GL(R)$ is the direct limit of the general linear groups and $^+$ is Quillen's plus-construction on spaces whose fundamental groups have perfect subgroups. Now I'm not sure how useful this has been for computation (these are homotopy groups, after all), but the classifying space is used. And there are number theory applications of algebraic k-theory.<|endoftext|>
TITLE:  $\pi_4$ of simply-connected 4-manifold  
QUESTION [15 upvotes]: In Baues "The homotopy category of simply conected 4-manifolds" there is some 
algebraic description of $\pi_4(M^4)$ where $M^4$ is simply-connected closed 4-manifold, but this description is pure algebraic.
And with it there is geometric description of $\pi_3(M^4)$:
$M^4 = \vee S^2 \cup e^4$, and attaching map of $e^4$ is element of $\pi_3(\vee S^2)$. So, $\pi_3(M^4)$ 
is a $\pi_3(\vee S^2)$ with additional relation.
Each element in $\pi_3(\vee S^2)$ has description in terms of linking number of point preimages (circles in $S^3$) of map $S^3\to \vee S^2$.
Does anyone knows geometric description (I think in terms of Whitehead product and typical preimages) of $\pi_4(M^4)$?

REPLY [4 votes]: I don't know exactly how to answer your question, but one observation is that if $M^4 \ncong S^4$, then for any map $\alpha : S^4\to M^4$, $H^4(\alpha):H^4(M^4)\to H^4(S^4)$ is the zero map. This follows because $[M]=a\cup b$ for $a,b\in H^2(M^4)$, by the Poincare conjecture and Poincare duality. Thus, one sees that the Hurewicz map is trivial.
If you have a geometric description of $M^4= \vee S^2 \cup e^4$, then for any point $x\in e^4$, the map $\alpha$ may be homotoped to miss $x$. To see this, make $\alpha$ transverse to $x$. Then the preimage of $x$ is finitely many points, whose orientations cancel since $H_4(\alpha)=0$. Choose arcs in $S^4$ pairing up the points in opposite signs, then since $M$ is simply connected, we may homotope neighborhoods of these arcs rel endpoints to be constant by homotopy extension, and then homotope off of $x$ (this is a version of the Whitney trick). Now by cellular approximation, the map may be homotoped onto $\vee S^2$. So the homotopy groups come from the homotopy groups of the 2-skeletion $\vee S^2$. I'm not certain though how to figure out what relations come from adding the 4-cell $e^4$ though.<|endoftext|>
TITLE: Ultraproducts of finite cyclic groups
QUESTION [19 upvotes]: Let G be the (non-principal) ultraproduct of all finite cyclic groups of orders n!, n=1,2,3,... . Is there a homomorphism from G onto the infinite cyclic group?

REPLY [2 votes]: I agree with the first answer. 
The ultraproduct in question is pure-injective (by a theorem of Jan Mycielski, Some compactifications of general agebras, Coll. Math., 13 (1964), 1-9). If an abelian group G contains a non-zero homomorphic image of a pure-injective group, then G is not slender. However, the infinite cyclic group is slender (by a theorem of Specker from 1950 (E. Specker, Additive Gruppen von folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131-140)), which is proved above as the Claim.
One might add: by a theorem of Sasiada (in L. Fuchs, Infinite Abelian Groups, vol.2, 1973), no reduced torsion-free abelian group of power less than continuum contains a non-trivial homomorphic image of the ultraproduct in question.
The results quoted above are also to be found in P.C. Eklof, A.H. Mekler, Almost Free Modules, rev. ed., 2002.<|endoftext|>
TITLE: Area of the boundary of the Mandelbrot set ?
QUESTION [16 upvotes]: My second question about Shishikura's result :
Shishikura (1991) proved that the Hausdorff Dimension of the boundary of the Mandelbrot set equals 2, in this paper 1. In a sense, could we consider it has an area ? If yes, has anybody measured or calculated its "size" (Hausdorff measure) ?
Thanks.

REPLY [11 votes]: (This used to be my research area, but I am not longer active in this topic, so I don't know all the latest references).  Nevertheless, last year X. Buff and A. Chéritat (see the X. Buff's preprint page) proved that there exist Julia sets with positive Lebesgue measure (a result which was presented at this years' ICM), which would lend credence to the conjecture that so does the Mandelbrot set.  But that, AFAIK, is still open.  Xavier and Arnaud would be the best people from whom to ask this question.<|endoftext|>
TITLE: Higher-dimensional braid group?
QUESTION [6 upvotes]: Let $\Delta$ be 2-disk. Let $C(\Delta;n)$ be a configuration space. 
i.e.) $C(\Delta;n)= \lbrace (z_1,\ldots,z_n)\in \Delta\times\ldots\Delta | z_i\neq z_j ~\textrm{if}~ i\neq j \rbrace $
Then, it is well known or direct to see that $\pi_1(C(\Delta;n))= PB_n$, where $PB_n$ is original pure braid group of n-strands.
I heard that by using this configuration space we can easily generalize Braid group in arbitrary topological space $X$.
i.e.) We can define $PB_n(X)$ (Pure Braid group of n-strands in $X$) by $PB_n(X)=\pi_1(C(X;n))$. 
Also, we can study some exact sequences analogous to original braid group situation if topology of $X$ is good.
Here, but I heard that if $X$ is manifold with dimension greater than 2. Then, the structure of $PB_n(X)$ is somewhat trivial because there are enough rooms to move 
I think that this phenomena occurs because we are observing too small objects in large space. Roughly speaking, this seems to be the same situation that Knot theory is trivial in codimension greater than 2.
Instead of this formulation, I hope that we could modify or generalize this configuration space set up by observing some space containg information like distribution or foliation or something like that?
Can I do this business? In short, Can we think nontrivial generalization of Braid group in higher dimensional manifold?

REPLY [5 votes]: There are natural generalizations of the braid group and pure braid group to higher dimensions, as follows: consider the space of arrangements of $n$ (labelled) affine hyperplanes in ${\mathbb C}^k$ in general position. If $k=1$ this is the set of (labelled) configurations of $n$ distinct points in the plane, with fundamental group the classical n-strand (pure) braid group. The fundamental group of this space, for any $k\geq 1$, is not trivial, and can be considered a higher braid group - in fact I think it's sometimes been called that in the literature. 
The space of labelled generic arrangements can be described as the space of $({\mathbb C}^*)^n$-orbits of $n \times (k+1)$ complex matrices with all $(k+1) \times (k+1)$ minors nonzero, and is closely related to the realization space of the uniform matroid $U_{n,k+1}$, and the "generic stratum of the Grassmannian." The symmetric group $S_n$ acts freely on the space of labelled generic arrangements; the fundamental group of the space of $S_n$-orbits is the full higher braid group. 
The pure higher braid group plays a role in the Aomoto-Gelfand theory of generalized hypergeometric functions. I think it's fair to say that almost nothing, certainly very little, is known about the groups themselves. Personally, I'd love to see a nice presentation of the fundamental group of the space of unlabelled generic arrangements of $n$ lines in ${\mathbb C}^2$. It might not be too hard for $n=4$.<|endoftext|>
TITLE: Are all sets totally ordered ?
QUESTION [32 upvotes]: The question is the title. 
Working in ZF, is it true that: for every nonempty set X, there exists a total order on X ? 
If it is false, do we have an example of a nonempty set that has no total order?
Thanks

REPLY [2 votes]: I'm just a student, so the deep logical principles behind François G. Dorais's answer are a drop over my head. However, I was able to find another Ultrafilter Lemma/Boolean Prime Ideal Theorem equivalent that seems a natural fit to the problem: the restricted Tukey-Teichmüller Theorem. Specifically, let $\mathscr A$ be the set of all relations on $X$ whose transitive closures are strict (partial) orderings. It's not hard to show that $\mathscr A$ satisfies the premises of rTT, where $(x,y)'$ is defined as $(y,x)$ for each $(x,y) \in S \times S$. Then rTT effectively shows that some element of $\mathscr A$ is connected, and thus its transitive closure is a strict total ordering of $S$ (in fact it is itself transitive, but this is not important). I've written up a more detailed outline on ProofWiki.<|endoftext|>
TITLE: Fairest way to choose gifts
QUESTION [21 upvotes]: Suppose that a parent brings home from a trip $2n$ gifts of roughly
equal value for his/her two children. The children get to choose one
at a time which gifts they want. What is the fairest way to do this?
For instance, if $n=1$ then clearly one child chooses first
(determined by a coin flip) and the other child chooses second. If we
denote the children by 0 and 1, then this method is described by the
choice sequence 01 (assuming, as I do from now on, that 0 choose
first). Now suppose $n=2$. The choice sequence 0101 is clearly biased
toward 0, since 0 has the first choice at the beginning and after both
have chosen one gift. The fairest sequence by any reasonable criterion
is 0110. What about general $n$?  If $n=2^k$, an argument can be made
that the fairest sequence is the first $n$ terms of the Thue-Morse
sequence
(http://mathworld.wolfram.com/Thue-MorseSequence.html). Another
argument can be made that the fairest sequence $a_1,\dots, a_n$ is one
that maximizes the value of $k$ for which the polynomial
$(1-2a_1)x^{n-1} + (1-2a_2)x^{n-2}+\cdots+(1-2a_n)$ and its first $k$
derivatives vanish at $x=1$. (The Thue-Morse sequence does not have
this property, though I cannot recall where I once saw this.)   
Has this problem received any attention? What is a reference for the
problem of maximizing $k$?

REPLY [5 votes]: The following paper apparently addresses exactly the question that you are interested in:
A General Elicitation-free Protocol for Allocating Indivisible Goods,  by: Sylvain Bouveret and Jérôme Lang, International Joint Conference on Artificial Intelligence (2011).<|endoftext|>
TITLE: (Stochastic) matrix for which a stochastic matrix logarithm exists?
QUESTION [5 upvotes]: I think this is basically the inverse question of Matrices whose exponential is stochastic. 
i.e. what are sufficient conditions on the matrix representation of an evolution operator of a (finite) discrete Markov chain for it to be embeddable in a continuous Markov chain?
I've found some old paper that may answer this (something about embeddability criteria) but I can't access it as it published in a closed-access journal.
I hope this is a sane question.

REPLY [5 votes]: Steve Hunstman's link above is good:
See the part leading up to Theorem 9 for something relevant to applications:

The main application of the following
  theorem may be to establish that
  certain Markov matrices arising in
  applications are not embeddable, and
  hence either that the entries are not
  numerically accurate or that the
  underlying process is not autonomous.
  The theorem is a quantitative
  strengthening of Lemma 8. It is of
  limited value except when n is fairly
  small, but this is often the case in
  applications.

Also the part on regularization for best compromises when matrices are not embeddable.<|endoftext|>
TITLE: Euler and the Four-Squares Theorem
QUESTION [39 upvotes]: There are several questions in the Euler-Goldbach correspondence that
I am unable to answer. Sometimes it does not take very much: in his 
letter to Goldbach dated June 9th, 1750, Euler conjectured 
that every odd number can be written as a sum of four squares
in such a way that $n = a^2 + b^2 + c^2 + d^2$ and $a+b+c+d = 1$.
I was just about to post this to MO when I saw that Euler's conjecture
can be reduced to the Three-Squares Theorem in one line (am I supposed to 
spoil this right away?). Here's another one where I haven't found a proof yet.
In his letter to Goldbach dated Apr.15, 1747, Euler wrote:
The theorem Any number can be split into four squares'' depends 
on this:Any number of the form $4m+2$ can always be split into 
two parts such as $4x+1$ and $4y+1$, none of which has any divisor 
of the form $4p-1$'' (which does not appear difficult, although I 
cannot yet prove it). 
Later, Euler attributed to Goldbach the much stronger claim that
the two summands can be chosen to be prime, which is a strong form
of the Goldbach conjecture.
Euler's intention was proving the Four-Squares Theorem (which he
almost 
did. Assuming this result, write 
$4m+2 = a^2+b^2+c^2+d^2$; then congruences modulo $8$ show that 
two numbers on the right hand side, say $a$ and $b$, are even, and 
the other two are odd. Now $a^2 + c^2 = 4x+1$ and $b^2 + d^2 = 4y+1$ 
satisfy Euler's conditions except when $a$ and $c$ (or $b$ and $d$) 
have a common prime factor of the form $4n-1$. 
Can this be excluded somehow?
Hermite [Oeuvres I, p. 259] considered a similar problem: 
Tout nombre impair est decomposable en quatre carres et,
parmi ces decompositions, il en existe toujours de telles
que la somme de deux carrees soit sans diviseurs communs
avec la somme de deux autres.
(Every odd number can be decomposed into four squares, and 
among these decompositions, there always exist some for which 
the sum of two squares is coprime to the sum of the other two.)
Hermite's proof contains a gap. Can Hermite's claim be proved somehow?

REPLY [9 votes]: The post below is jointly by Rainer Dietmann and Christian Elsholtz.
We had worked on the problem since a while and had an independent asymptotic solution
to Euler's problem.
Our argument is possibly easier, but in it's current form
it does does not achieve the correct order of magnitude of the number of solutions. This seems to be a very nice feature of Lucia's approach!
We had intended to make the argument entirely explicit
in order to prove the statement for all $n$, not only for sufficienly large $n$. (See also the comments
after the second argument below).
We also had intended to prepare these results for publication.
Lucia, we would appreciate if you could contact us by email, the email
adresses (RD in Royal Holloway and CE in Graz) are easy to find.
$\textbf{Theorem:}$
Let $n$ be a sufficiently large positive integer with
$n \equiv 2 \pmod 4$. Then $n$ can be written as the sum of two
positive integers, none of them having any prime factor
$p$ with $p \equiv 3 \pmod 4$.
This asymptotically answers a question of Euler.
Important partial results are due to R.D. James (TAMS 43 (1938), 296--302)
who proved the ternary case and an approximation to the binary case. Indeed the ternary case allows for an elementary proof,
based on Gaus' theorem on the sum of three triangular numbers: Any integer $k$
can be written as $k = \frac{x(x-1)}{2}+ \frac{y(y-1)}{2} + \frac{z(z-1)}{2}$ and therefore
$$ 4k + 3 = (x^2 + (x - 1)^2) + (y^2 + (y - 1)^2) + (z^2 + (z - 1)^2).$$
Observe that $ (x^2 + (x - 1)^2)$ is a sum of two adjacent squares, and thus cannot be divisible by any prime $ p = 3 \mod 4$. (Recall here and for later reference the following $\textbf{Fact:}$ if $p|n = s^2 +t^2$, with $p = 3 \mod 4$ prime, then $p|s$ and $p|t$.)
Using a well known result of the late George Greaves, one gets a short
proof of the Theorem.
$\textbf{Proof.}$
By a result of Greaves (Acta Arith 29 (1976), 257--274),
each sufficiently large positive integer $n$ with
$n \equiv 2 \pmod 4$ can be written in the form
$$  n = p^2+q^2+x^2+y^2
$$
for rational primes $p, q$ and integers $x, y$, and the number of such representations is at least of order of magnitude
$n (\log n)^{-5/2}$. We write  $a=p^2+x^2$ and $b=q^2+y^2$
and take multiplicities into account:
namely the number of representations $r_2(a)$ of
$a$ as a sum of two squares is $r_2(a) \leq d(a)\ll a^{\varepsilon}\ll n^{\varepsilon}$.
The same holds for $r_2(b)$.  We therefore find that there are
at least
$$
 n^{1-2\varepsilon}
$$
many tuples $(a, b)$ with positive
integers $a, b$, such that $n=a+b$ and both $a$ and $b$ are the sum
of the square of a prime and the square of an integer.
Now suppose that $w$ is a prime with $w \equiv 3 \pmod 4$ and
$w$ divides $a=p^2+x^2$, say. Then by the `fact' above and
as  $p$ is  prime this implies that $p=w$ and $x$ is
divisible by $w$. Therefore, at most $O(1+n^{1/2}/w)$ many
$a$ can be divisible by $w$, and for any such $a$ there will be only
one corresponding $b$ since $a+b=n$. The same argument applies if
$w$ divides $b$. Moreover, clearly $w$ can be at most $n^{1/2}$. Summing over all such $w$
we conclude that the number of tuples $(a,b)$ with $a+b=n$
and $a, b$ of the form above,
where one of $a$ and $b$ is divisible by any prime congruent
$3 \mod 4$, is at most $O(n^{1/2} \log \log n)$, which is of smaller order of magnitude than the expression $n^{1-2\varepsilon}$ above.
This finishes the proof.
$\textbf{Remark.}$
It seems likely that the number of representations $f(n)$ can be greatly improved by observing that one only needs $r_2(a)$ on average. This should produce  $f(n)$
within a logarithmic factor. Moreover it seems possible to adapt Greaves's argument by replacing $p^2$ and $q^2$ by squares of integers not containing a prime $3\bmod 4$, achieving a further logarithmic saving.
(Let us briefly reflect why the argument works: Greaves uses the fact
that Iwaniec's half dimensional sieve can also handle sums of two squares.
The contribution from the almost trivial 'fact' is also quite useful.)
Let us briefly sketch another possible approach, which could be more
suitable for getting a result for all positive integers:
Let $f(n)$ denote the number of representations as a sum of two integers, both not containing any prime factor $3 \bmod 4$.
Let $r_{(a,b,c,d)}(n)$ denote the number of representations as sum $ax^2+by^2+cz^2+dt^2$.
We intend to show that $$n^\varepsilon r(n) \gg r_{(1,1,1,1)}(n)- 2 \sum_{p=3 \bmod 4} r_{(1,1,p^2,p^2)}(n) \gg r_{(1,1,1,1)}(n).$$
Observe that $r_{(1,1,p^2,p^2)}(n)\approx  \frac{1}{p^2}r_{(1,1,1,1)} $ and that $\sum_{p =3\mod 4} \frac{1}{p^2}$ is a small and
finite number.
For a completely explicit result all we need is an explicit lower bound on
$r_{(1,1,1,1)}(n)$, which can be derived from Jacobi's formula,
and an explicit upper bound on expressions like $r_{(1,1,p^2,p^2)}(n)$,
which can be obtained either by the circle method using a Kloosterman
refinement, a modular forms approach, or via Dirichlet's hyperbola method. The big question then is if the
resulting numerical bounds allow the remaining finitely many cases
to be checked by a computer.<|endoftext|>
TITLE: Ends and coends as Kan extensions (without using the subdivision category of Mac Lane)?
QUESTION [10 upvotes]: Limits and colimits have very nice definitions in terms of Kan extensions, and therefore enjoy very nice adjointness properties.  Mac Lane's Categories for the Working Mathematician gives a construction called the subdivision category of a category $C$, which allows one to reduce the theory of ends and coends to the theory of limits and colimits (and therefore the theory of Kan extensions).  This construction feels a bit artificial and messy, although it is very useful for quick and dirty proofs of many of the details about ends and coends.  
Can we give a definition of the end and coend as some sort of Kan extension without invoking the subdivision category construction?

REPLY [7 votes]: I just happened across this old question and thought I would add another possible answer.  Instead of the subdivision category it is possible to use the twisted arrow category $\mathrm{tw}(C)$ in exactly the same way: there is a functor $\mathrm{tw}(C) \to C^{\mathrm{op}}\times C$, and the end of a functor $C^{\mathrm{op}}\times C\to D$ is obtained by restricting it to $\mathrm{tw}(C)$ and taking the limit.
This may not be as satisfying as the other answers, since it is very similar to Mac Lane's version, but at least the twisted arrow category is somewhat less ad hoc than the subdivision category.  It also has the advantage of being "homotopy correct", in that homotopy ends can be computed in the same way as homotopy limits over $\mathrm{tw}(C)$, which is not the case for the subdivision category (see e.g. this paper).<|endoftext|>
TITLE: Ramification in morphisms of surfaces
QUESTION [5 upvotes]: The question can be generalized, but we might as well restrict to this case.
Let $X \rightarrow Y$ be a morphism between nonsingular surfaces (say over $\mathbb{C}$). Let $R_1$ be an irreducible component of the ramification divisor (in $X$). Let $n$ be how much $R_1$ ramifies generically, and let $S$ be the finite set of points of $R_1$ that ramify to a degree which is not $n$. Is it true that the set $S$ is contained in the set of the points where $R_1$ intersects the other irreducible components of the ramification divisor (and maybe where $R_1$ is singular)?
The intuitive answer is yes, but I'm still somewhat skeptical.

REPLY [5 votes]: Let me expand jvp's answer, giving a picture of the situation in the case of a $general$ flat triple cover $f \colon X \to Y$.
Let  $R \subset Y$ be the ramification divisor and $B \subset Y$ the branch divisor, that is $B = f(R)$. Then $R$, $B$ are both reduced and irreducible, and $B$ has only a finite number of ordinary cusps $q_1, \ldots, q_t$ as singularities. These cusps are exactly the points over which $f$ is $totally$ $ramified$. Moreover $R$ is isomorphic to the normalization of $B$, in particular it is $smooth$.
One has the equality of divisors
$f^*(B)=2R + R'$,
where $R'$ is another irreducible curve, isomorphic to $R$, which meets $R$ in a finite number of points $p_1, \ldots, p_t$. Notice that $R'$ is $not$ a component of the ramification locus, since the latter consists of $R$ alone.
Moreover

$R$ and $R'$ are tangent at $p_1, \ldots, p_t$;
$p_1, \ldots ,p_t$ are the preimages of the cusps $q_1, \ldots, q_t$.

Summing up, in this case your $S$ is the set whose elements are the points $p_1, \ldots ,p_t$. They correspond to the points where the ramification divisor $R$ meets the curve $R'=f^*(B)   \setminus R$. In other words, they come from the singular points of the branch divisor $B$ (whereas the ramification divisor $R$ is smooth). 
This is easy to see; a good reference is Miranda's paper "Triple covers in algebraic geometry".
Anyway, the crucial fact here is  that a general triple cover is not a Galois cover, so over the branch locus $B$ there are both points where $f$ is ramified (the curve $R$) and points where it is not (the curve $R'$). 
If you consider instead any Galois cover, say with group $G$, then every preimage of a branch point is a ramification point (and the stabilizers of points lying on the same fibre are conjugated in $G$). In this case there are formulae relating the ramification number of a point on $X$ with the ramification numbers of the components of the ramification locus passing through it. 
See Pardini's paper "Abelian covers of algebraic varieties" for more details.<|endoftext|>
TITLE: Why did the word "exterior" get chosen for the idea of "exterior derivative"?
QUESTION [7 upvotes]: What are the intuitive and historical reasons for choosing the word "exterior" for the concept of an exterior derivative of a form? 
The reasoning I've heard about it is the following: let p(t) be a continuous parametric curve, then if you fix t_0, the tangent line to the curve p(t) at t_0 lies "exterior" of the curve p(t), since it is an approximation of p(t) itself.

REPLY [6 votes]: I) The term  exterior multiplication ("äussere Multiplication") is due to Grassmann, who introduced the term in his book (written in 1844)
Die Wissenschaft der extensiven Grösse oder die Ausdehnungslehre, eine neue Mathematische Disciplin"
As you can check in the table of contents of the book (on page 276), paragraphs §§34,35  are called  Grundgesetze der äusseren Multiplication (Basic laws of exterior multiplication).
Here is the scan of this  book  by Google .
II) The terminology exterior differential ("différentielle extérieure") was introduced in the 1930's by articles of Elie Cartan, inspired by Grassmann.
Here is a secondary reference from an Analysis course by Chatterji and another by Chern and Chevalley, in their analysis of Elie Cartan's mathematical contributions (cf. in particular pages 229 and 230 )<|endoftext|>
TITLE: Is a sub-stack of a scheme a scheme?
QUESTION [8 upvotes]: Let $S$ be a scheme. Let $\mathcal X$ be a algebraic $S$-stack and be $Y$ a $S$-scheme. Let $f:\mathcal X\longrightarrow Y$ be a $S$-morphism of algebraic stacks which is an open embedding (resp. a closed embedding). Is $\mathcal X$ automatically a open(resp. closed) subscheme of $Y$?

REPLY [2 votes]: Let $f:\mathcal{X}\rightarrow Y$ be a morphism from an Artin stack to a scheme such that $f$ is an immersion. Then $\mathcal{X}$ is automatically an algebraic space, so we're done by Knutson, Algebraic spaces, II.6.16.
Additions prompted by Brian's comment
Assume that $f:\mathcal{X}\rightarrow Y$ is a schematic map, and that $Y$ is a scheme; then $f$ is the pullback of $f$ over the map of schemes $\mathrm{id}_Y$, so $\mathcal{X}$ must be a scheme. Knutson needs lemma II.6.16 because he doesn't use the now-standard definition of schematic, but atlases instead.
When using immersion, I always mean $j\circ i$, where $i$ is a closed immersion and $j$ an open one, following EGA I. But I understand that this is not a better choice than the other way round, and that they are only equivalent when the morphism is quasicompact.<|endoftext|>
TITLE: Is there a known natural model of Peano Arithmetic where Goodstein's theorem fails?
QUESTION [23 upvotes]: (I've previously asked this question on the sister site here, but got no responses).
Goodstein's Theorem is the statement that every Goodstein sequence eventually hits 0.  It turns out that this theorem is unprovable in Peano Arithmetic ($PA$) but is provable in $ZFC$.
I'd like to "discuss" this (both the proof in $ZFC$ as well as the proof that it's impossible in $PA$) in an hour long lecture to a group of grad students (with no assumed background, handwaving is not only allowed, but encouraged).  Because of previous talks I've given, I think it will not take too long to cover/remind them of the basics of a semester course in first order logic (e.g., the compactness and completeness theorem, etc).
The problem is that the proofs of unprovability I've found (the same as those linked in the Wikipedia article) are rather too difficult for this setting.  In a nutshell, I'm looking for the easiest known proof.
For example, I would love a proof of unprovability which works by exhibiting a model of $PA$ in which Goodstein's theorem fails.  Such models neccesarily exist by the completeness theorem, since "$PA$ + Goodstein's theorem is false" is consistent.

Has anyone proven the independence of Goodstein's theorem by exhibiting a model of $PA$ where Goodstein's theorem has failed?

In the interest of getting as simple a proof as possible, I'd love to see a proof which uses the compactness and completeness ideas - something like showing there is a set $\Sigma = \{\phi_n\}$ of explicit first order sentences(in a slightly larger language, say) such that
1)  for any finite $\Sigma_0\subseteq PA\cup \Sigma$, the standard model $\mathbb{N}$ models $\Sigma_0$ and
2)  The theory $PA\cup \Sigma$ proves Goodstein's theorem is false.

Is such a proof known?  More generally, is there a known proof of the unprovability of Goodstein's theorem which is accessible to someone with only a semester or 2 of logic classes?

Thank you and please feel free to retag as necessary.

REPLY [20 votes]: Your proof strategy via (1) and (2) is impossible. If $PA\cup\Sigma$ proves that Goodstein's theorem is false, then the proof will have finite length, and so there will be some finite $\Sigma_0\subset\Sigma$ such that $PA\cup\Sigma_0$ proves that Goodstein's theorem is false. This would imply by (1) that Goodstein's theorem is false in the standard model. But Goodstein's theorem holds in the standard model, as Goodstein proved.
A second point is that you may find that there are no specific "natural" models of PA at all other than the standard model. For example, Tennenbaum proved that there are no computable nonstandard models of PA; that is, one cannot exhibit a nonstandard model of PA so explicitly that the addition and multiplication of the model are computable functions. (See this related MO question.) But I do not rule out that there could be natural nonstandard models in other senses.

REPLY [18 votes]: I'm not sure I understand what you mean by "natural." 
Anyway, the proof by Kirby-Paris is "model theoretic" or as "explicit" as you are likely to find. The reference is Accessible independence results for Peano arithmetic. Bulletin London Mathematical Society 14 (1982), 285–293. (I can email you the paper if you cannot find it easily. It is a nice read.)
The argument uses the method of indicators. The (very broad strokes) idea of the method is to start with a nonstandard model of PA (any would do, for some more specific results you may need to pay some attention to this step, but here any nonstandard model suffices), and use Goodstein's function $G$ (the map that assigns to $n$ the number of steps that the sequence takes to reach 0) to find a cut. This is an initial segment of your nonstandard model that is itself a model of PA, but the cut is built explicitly so that for some nonstandard $N$ in the cut the number of steps $G(N)$ is past the cut. This usually requires that you produce `indiscernibles' that are used to ensure the induction axioms hold in your cut. It is a very useful method, and the basis for most known independent results in PA. (The Paris-Harrington and Kanamori-McAloon theorems being other well-known examples).
I wrote a little paper showing that a well-known proof theoretic result about PA can be used to give the unprovability of Goodstein's theorem. The point of the paper is an explicit formula for $G$. You can find it in my page, or I can email it as well if you are interested.<|endoftext|>
TITLE: Coalescing random walks: a bound for the full coalescence time?
QUESTION [9 upvotes]: Start a random walk from each vertex of a graph $G$. Let the walkers evolve independently, except that when two of the walkers meet (ie. occupy the same vertex at the same time), they coalesce into one single walker. (Alternatively, one may label the walkers with numbers $1,\dots,n$ and say that walker $i$ is killed at the first time it meets a walker with smaller index.) 
It is not hard to show that, if $G$ is connected and finite, there will be a first time $C$ when all walks will have coalesced into one ($C$ is the first time when only one alive particle remains in the "killing description"). The following is Open Problem $13$ in Chapter $14$ of Aldous and Fill's manuscript:
http://www.stat.berkeley.edu/~aldous/RWG/book.html.
Problem: Prove that there exists a universal constant such that:
$$E(C)\leq K\max_{v,w\in V(G)}E_v(H_w)$$
where $V(G)$ is the vertex set of $G$ and $H_w$ is the hitting time of vertex $w$.
My question is: does anyone know of any work on this problem? I know there is a paper by Cox which studies the distributional limit of $C$ over $Z_n^d$ for $n\gg 1$ and $d$ fixed; a solution for the Problem follows from this in this specific family of graphs. What else is out there? Has the problem actually been solved by someone?

REPLY [2 votes]: fedja was right to guess that I had an idea to solve the problem. Reading that Yuval Peres is working on it scared me into writing something down as quickly as possible. Any comments on my manuscript (below) are very welcome.
http://arxiv.org/abs/1009.0664<|endoftext|>
TITLE: Why is the volume conjecture important?
QUESTION [26 upvotes]: The volume conjecture, a formula relating hyperbolic volume of a knot complement with the semiclassical limit of a family of coloured Jones polynomials, is widely considered the biggest open problem in quantum topology. It is one of a large family of conjectures and research programmes which have to do with  detecting classical geometry with semiclassical limits.
Embarassing as it is to say in public, I only very partially understand why people care so much about such conjectures.

What fantastic consequences would there be for low dimensional topology if the volume conjecture were proven tomorrow? What if all the related conjectures were proven too? How would it improve our understanding of classical topology? More broadly, how would it advance mathematics?

REPLY [16 votes]: I don't know any consequences for low-dimensional topology. My impression is that it would indicate how powerful TQFT invariants are at distinguishing 3-manifolds or knots. For example, the volume conjecture (or maybe variants by the Murakamis) would imply that the colored Jones polynomials distinguish knots from the unknot. This corollary is also claimed by Jorgen Andersen. There are only finitely many hyperbolic knots of a given volume, so the volume conjecture would imply that the colored Jones polynomials are close to distinguishing hyperbolic knots.<|endoftext|>
TITLE: Orders of products of permutations
QUESTION [40 upvotes]: Let $p$ be a prime, $n\gg p$ not divisible by $p$ (say, $n>2^{2^p}$). Are there two permutations $a, b$ of the set $\{1,...,n\}$ which together act transitively on $\{1,2,...,n\}$ and such that all products $w(a,b)=a^{k_1}b^{l_1}a^{k_2}...$ of length at most $n$ satisfy $w(a,b)^p=1$ (here $k_i,l_i\in {\mathbb Z}$)? 
Update: Following the discussion below (especially questions of Sergey Ivanov, here is a group theory problem closely related to the one before.
Is there a torsion residually finite infinite finitely generated group $G$ such that $G/FC(G)$ is bounded torsion? Here $FC(G)$ is the FC-radical of $G$, that is, the (normal) subgroup of $G$ which is the union of all finite conjugacy classes of $G$.
For explanations of relevance of this question see below (keep in mind that the direct product of finite groups coincides with its FC-radical).  Note that if we would ask $G$ to be bounded torsion itself, the question would be equivalent to the restricted Burnside problem and would have negative answer by Zelmanov. 
If the answer to any of the two questions above is negative for some $p>665$, then there exists a non-residually finite hyperbolic group.

REPLY [6 votes]: If $p=2$, then your condition  on words of length at most $2^p$ implies that $a$ and $b$ have order two and commute.  Therefore they generate a group of order at most four.  So, they together cannot act transitively on a large set.<|endoftext|>
TITLE: Alternating forms as skew-symmetric tensors: some inconsistency?
QUESTION [10 upvotes]: My trouble is best described by the following diagram:
$$ \begin{array}{ccccc} \mathrm{Alt}^k V &\stackrel{\sim}{\rightarrow}& (\Lambda^k V)^*  &\stackrel{\sim}{\rightarrow}& \Lambda^k V^* \cr
 i \downarrow &&&& \downarrow \mathrm{Sk}\cr
 \mathrm{Mult}^k V &\stackrel{\sim}{\leftarrow} & (\otimes^k V)^* & \stackrel{\sim}{\leftarrow} & \otimes^k V^*
  \end{array} $$ 
The problem is that this diagram is not commutative but let me explain the terminology first.
Here $\mathrm{Alt}^k V$ and $\mathrm{Mult}^k V$ are the spaces of alternating and multilinear $k$-forms, respectively, on the vector space $V$. All the horizontal isomorphisms are canonical. The left vertical arrow is the inclusion of the alternating forms in the multilinear ones. The only "questionable" arrow is the right-hand vertical one ($\mathrm{Sk}$, following the notation in Birkhoff-MacLane "Algebra", Section XVI.10). It is given as an extension of the following alternating map
$$ (*)\quad (v_1^*, \ldots, v_k^*) \mapsto {1\over k!} \sum_{\sigma\in S_k} (-1)^{\sigma} v_{\sigma(1)}^* \otimes \cdots \otimes v_{\sigma(k)}^*. $$
If the characteristic is zero (which I assume) then Sk is an inclusion. There are two good things about this inclusion: 
First, it is a linear inversion of the canonical projection modulo the graded ideal generated by squares of elements of grade 1, i.e. of $\otimes^k V^* \rightarrow \Lambda^k V^*$.
Second, if $\mathrm{Sk'}$ is a map $\otimes^k V^* \rightarrow \otimes^k V^* $ which is again obtained by extending a multilinear map (*), then we have
$$ Sk(a \wedge b) = Sk'(Sk(a)\otimes Sk(b)) $$
(i.e. to learn what $a\wedge b$ is you map both to tensors via $Sk$ and then antisymmetrize their tensor product in the tensor algebra).
So the above argument suggests that $Sk$ is somewhat canonical as well. However, here is a strange situation. Suppose that $e_1, \ldots, e_n$ are the basis of $V$. Then consider the alternating 2-form that operates on $V\times V$ as follows:
$$ f(v_1, v_2) = e_1^*(v_1) e_2^*(v_2) - e_1^*(v_2) e_2^*(v_1) $$
Its image in $\Lambda^k V^*$ is $e_1^* \wedge e_2^*$, and thus under $Sk$ it gets mapped to
$$ {1\over 2} (e_1^* \otimes e_2^* - e_2^* \otimes e_1^*) $$
Therefore applying the other two bottom isomorphisms we arrive at a multilinear form that operates on $V\times V$ as follows:
$$ g(v_1, v_2) = {1\over 2} (e_1^*(v_1) e_2^*(v_2) - e_1^*(v_2) e_2^*(v_1)) $$
Clearly $g\neq f$ and this is precisely the non-commutativity I was talking about.
Can somebody explain if I made a mistake somewhere? And if not, why then so many physicists happily use "skew-symmetric tensors" and refuse to use "differential forms" and still arrive to the very same answers never loosing coefficient like $1\over 2$? 
Thanks in advance! This looks really puzzling to me and I know this is too easy for MO, but I am in a situation much different from the rest of MO having zero mathematicians around to ask such silly questions to. Again, thanks for reading!
Added later: As Andrew and Georges point out it is easy to make the diagram commute by either redefining the $\mathrm{Sk}$ without the ${1\over k!}$ or by changing $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ from the $\mathrm{det}$-map to ${1\over k!}\mathrm{det}$. Let me explain why I think why either approach is confusing. 
First, redefining the $\mathrm{Sk}$ map as Georges suggests revokes its first property: namely it is no longer a right inverse of the projection $\otimes^k V \to \Lambda^k V$. On the other hand, map $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ determines what we call a wedge-product in the graded algebra $\mathrm{Alt}^* V$ (since the wedge-product in $\Lambda^* V^* $ canonically comes from $\otimes^* V^* $ via projection). Therefore, if we are to redefine the meaning of $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ as Andrew proposes, then we have to agree that now
$$ (* *)\quad (dx \wedge dy) (\partial_x, \partial_y) = {1\over 2},$$
which I think many will find somewhat weird. (Although, it seems things like Stokes theorem do not depend on the agreement (**), right?)
To sum up: if we agree what $\wedge$ means in $\mathrm{Alt}^k$ then this determines the definition of $\mathrm{Sk}$. And thus with the usual definition of $\wedge$ in $\mathrm{Alt}^k$ we unfortunatelly obtain the $\mathrm{Sk}$ which is not the right-inverse of the projection. Am I correct in this?

REPLY [7 votes]: Dear Paul, first of all let me congratulate you for the extremely clear formulation of your interesting question (which is not silly at all, contrary to what you say): +1.
The source of your trouble is the identification $Sk$: it is not the correct one.Why the cocksure statement? Because Laurent Schwartz wrote it! (In his book Les Tenseurs Hermann, 1975). Talking of (an analagon) of $Sk$, he writes on page 61:"il se trouve que cette identification n'est pas la bonne" (it happens that this identification is not the right one).
He recommends, for a vector space $E$,  the embedding $\Sigma k:\Lambda ^k E\to \otimes^k E$ where $\Sigma k=k! Sk$ (in your notation).The image of $\Sigma k$ is exactly the subspace $ A^k E \subset\otimes ^k E$ consisting of antisymmetric tensors This is valid in all characteristics $\neq 2$ and has  the crucial advantage that it preserves products: for $\alpha \in  \Lambda ^k E$ and
$\beta \in  \Lambda ^l E$, he proves $\Sigma (\alpha \wedge \beta)=\Sigma (\alpha)\otimes _a 
\Sigma (\beta)$, where $\otimes_a$ is the antisymmetric product defined on antisymmetric tensors and yielding antisymmetric tensors (This product is defined with shuffle permutations).The calculation is on page 104.
I find it very satisfying that Schwartz's point of view solves your trouble, without having to resort to ad hoc trickery. By the way, the book I mentioned is in French but many distinguished anglophones on this site have claimed energetically that mathematical French is no problem for an English-speaking person. I'll let you be the judge!
COMPLEMENT In case somebody is interested, here is the formula for the antisymmetric product I alluded to. Let $\alpha \in A^k E \subset \otimes ^k E$ and $\beta \in A^l E \subset \otimes ^l E$ be antisymmetric tensors. Their antisymmetric product is the antisymmetric tensor $\alpha \otimes_a \beta \in A^{k+l }E \subset \otimes ^{k+l} E$ defined by
$\alpha \otimes _a \beta=\sum sgn(s) s\bullet(\alpha \otimes \beta)$
(the sum is over shuffle permutations $s$,  i.e. permutations with   $s_1 < \ldots < s_k$ and $s_{k+1} < \ldots < s_{k+l}$ The bullet denotes the action of the symmetric group on tensors.) 
Of course Schwartz emphasizes that it is not a great idea to use antisymmetric tensors, which form a subspace of the tensor product: it is far better to use exterior products which are a quotient of said tensor product.<|endoftext|>
TITLE: Realizing groups as automorphism groups of graphs.
QUESTION [26 upvotes]: Frucht showed that every finite group is the automorphism group of a finite graph.  The paper is here.
The argument basically is that a group is the automorphism group of its (colored) Cayley graph
and that the colors of edge in the Cayley graph can be coded into an uncolored graph that has the same automorphism group.
This argument seems to carry over to the countably infinite case.
Does anybody know a reference for this?
In the uncountable, is it true that every group is the automorphism group of a graph?
(Reference?)
It seems like one has to code ordinals into rigid graphs in order to code the uncountably many colors of the Cayley graph.

REPLY [14 votes]: In the topological setting or if you want to relate the size of the graph to the size of the group, there are two relevant results:
(1) Any closed subgroup of $S_\infty$, i.e., of the group of all (not just finitary) permutations of $\mathbb N$, is topologically isomorphic to the automorphism group of a countable graph.
(2) The abstract group of  increasing homeomorphisms of $\mathbb R$, ${\rm Homeo}_+(\mathbb R)$, has no non-trivial actions on a set of size $<2^{\aleph_0}$. So in particular, it cannot be represented as the automorphism group of a graph with less than continuum many vertices.<|endoftext|>
TITLE: Approximating a probability distribution by a mixture
QUESTION [5 upvotes]: Let us consider a probability distribution $(g_n)_{n \in \mathbb{N}}$ which we want to approximate by a mixture of $(f_n(\lambda))_{n \in \mathbb{N}}$ where $\lambda \in \mathbb{R}$ is a parameter.
Are there known techniques that allow one to find the mixture minimizing the $L^1$ norm:
\begin{equation}
\min_{p} \sum_{n=0}^{\infty} \left|g_n - \int \rm{d} \lambda \;  p(\lambda) f_n(\lambda) \right|
\end{equation} 
where $p(\lambda)$ is a normalized probability distribution?
The motivation of this problem is linked to experimental physics: ideally one would like to generate an experimental process characterized by the probability distribution $g$ but this is really not practical. What is really easy, however, is to generate an experimental process with the distribution $f(\lambda)$ where $\lambda$ is a tunable parameter.
Therefore, the goal is to approximate $g$ as closely as possible with such a mixture of $f(\lambda)$, where the distance between the two distribution is computed with the $L^1$ norm, that is, I want to minimize the variation distance between the two distributions.
In the specific problem I consider, $f(\lambda)$ is a Poisson distribution with parameter $\lambda \geq 0$, but I really am interested in a general method to approach this problem-
Any pointer to the relevant literature would be greatly appreciated.
Thanks a lot!

REPLY [2 votes]: perhaps for starters you could take $p$ to be supported on a finite number of points. then the constraints on $p$ become simple linear inequalities and you have a [convex - perhaps even linear] programming problem. 
there has been attention in the statistical literature to fitting models using least absolute deviations, rather than least squares. [in the simplest case, the minimizer of
$$\sum_{i=1}^n |x_i - a|$$
is the sample median - rather than the sample mean one gets for $a$ using least squares.] 
you could see if references in the monograph by yadolah dodge [L$_1$ statistical procedures and related topics, ims lecture notes - monograph series vol 31 1997] give anything useful.<|endoftext|>
TITLE: What is a path in K-theory space?
QUESTION [19 upvotes]: In a comment on Tom Goodwillie's question about relating the Alexander polynomial and the Iwasawa polynomial, Minhyong Kim makes the cryptic but tantalizing statement:
In brief, the current view is that the Iwasawa polynomial=p-adic L-function should be viewed as a path in K-theory space.
As somebody who has been trying, for a long time, to understand the Alexander polynomial, and who believes there is something "more basic" underlying its appearances in mathematics in various guises (my particular interest has to do with why it turns up as the wheels part of the Kontsevich invariant, as per Melvin-Morton-Rozansky), I'd really like to understand this comment. 
Question: What is "K-theory space" supposed to be, and why should a path in it be a natural object to be examining? And why should we expect such a thing to give equivalent data to the Iwasawa polynomial/ Alexander polynomial?

REPLY [19 votes]: I know nothing about Alexander polynomials but let me try to answer the Iwasawa theory part. As is well known, in classical Iwasawa theory one considers cyclotomic $\mathbb{Z}_p$ extension $F{\infty}$ of $F$. We take the $p$-part of the ideal class group $A_n$ of the intermediate extension $F_n$ of $F$ of degree $p^n$. The inverse limit of $A_n$ with respect to norm maps, say $A$, has an action of $G= Gal(F_{\infty}/F)$. Since $A$ is pro-$p$, it becomes a $\mathbb{Z}_p[G]$-module. However, it is not finitely generated over this group ring (and for various other reasons) one considers the completion $\mathbb{Z}_p[[G]]$ of $\mathbb{Z}_p[G]$. Since $A$ is compact, it becomes a $\mathbb{Z}_p[[G]]$-module. As $G \cong \mathbb{Z}_p$, the ring $\mathbb{Z}_p[[G]] \cong \mathbb{Z}_p[[T]]$, the power series ring in variable $T$. There is a nice structure theory for finitely generated modules over $\mathbb{Z}_p[[T]]$. The module $A$ is a torsion $\mathbb{Z}_p[[G]]$-module (i.e.$Frac(\mathbb{Z}_p[[G]]) \otimes A = 0$). For such modules one can define the characteristic ideal using the structure theory. Iwasawa's main conjecture asserts that there is a canonical generator for this ideal called the $p$-adic $L$-function. 
In generalised Iwasawa theory (more precisely, to formulate the generalised main conjecture à la Kato), one wants to consider extensions whose Galois groups are not necessarily $\mathbb{Z}_p$ (but most formulations of the main conjecture still require that the cyclotomic $\mathbb{Z}_p$-extension of the base field be in the extension). For the completed $p$-adic groups rings of such Galois groups, the structure theory completely breaks down even if the Galois group is abelian. 
However, one can still show that $A$ is a torsion Iwasawa module (which again just means that $Frac(\mathbb{Z}_p[[G]]) \otimes A = 0$. Note that it is always possible to invert all non-zero divisors in a ring even in the non-commutative setting). Hence the class of $A$ in the group $K_0(\mathbb{Z}_p[[G]])$ is zero. Strictly speaking, here I must assume that $G$ has no $p$-torsion so that I can take a finite projective resolution of $A$, or I must work with complexes whose cohomologies are closely related to $A$. But I will sinfully ignore this technicality here. Now, since the class of $A$ in $K_0(\mathbb{Z}_p[[G]])$ is zero, there is a path from $A$ to the trivial module 0 in the $K$-theory space. In Iwasawa theory this is most commonly written as
There exists an isomorphism $Det_{\mathbb{Z}_p[[G]]}(A)$ $\to$ $Det(0)$. 
This isomorphism replaces the characteristic ideal used in the classical Iwasawa theory. The $p$-adic $L$-function then is a special isomorphism of this kind. (Well one has to be careful about the uniqueness statement in the noncommutative setting but it is a reasonably canonical isomorphism). Hence the main conjecture now just asserts existence of such a $p$-adic $L$-function. 
Thus the $p$-adic $L$-function may be thought of as a canonical path in the $K$-theory space joining the image of Selmer module (or better- a Selmer complex), such as the ideal class group in the above example, and the image of the trivial module.  
I hope this answer helps until Minhyong sheds more light on his remarks and relations between $p$-adic $L$-functions and the Alexander polynomials. 
[EDIT: Sep. 30th] To answer Daniel's questions below- 
1) Take the projective resolution of A to define its class in $K_0$. In any case the $K$-theory of the category of finitely generated modules is the same as the $K$-theory of finite generated projective modules. 2) I do not know if there are any sensible/canonical multiplicative sets at which to localise $\mathbb{Z}_p[G]$. The modules considered in Iwasawa theory are usually compact (or co-compact depending on whether you take inverse limit with respect to norms or direct limit with respect to inclusion of fields in a tower) and so the action of the ring $\mathbb{Z}_p[G]$ extends to an action of the completion $\mathbb{Z}_p[[G]]$ by which we mean $\varprojlim \mathbb{Z}_p[G/U]$, where $U$ runs through open normal subgroups of $G$. 3) We do not usually get a loop in the $K$-theory space of $\mathbb{Z}_p[[G]]$. However, modules which come up from arithmetic are usually torsion as $\mathbb{Z}_p[[G]]$-modules i.e. every element in the module is annihilated by a non-zero divisor or in other words if $X$ is the module then $Frac(\mathbb{Z}_p[[G]]) \otimes X=0$. (it is usually very hard to prove that the modules arising are in fact torsion). But once we know that they are torsion we get a loop in the $K$-theory space of $Frac(\mathbb{Z}_p[[G]])$. If we know that for some multiplicatively closed subset $S$ of $\mathbb{Z}_p[[T]]$ annihilates $X$ i.e. $(\mathbb{Z}_p[[G]]_S) \otimes X=0$, then we already get a loop in the $K$-theory space of $\mathbb{Z}_p[[G]]$. In noncommutative Iwasawa theory it is often necessary to work with a multiplicative set strictly smaller than all non-zero divisors.<|endoftext|>
TITLE: Statement of Millenium Problem: Yang-Mills Theory and Mass Gap
QUESTION [24 upvotes]: I'm wondering what the statement is that one has to prove for the Millenium Problem "Quantum Yang-Mills Theory".
According to the official article, it is required to show that for every simple Lie group G there exists a YM quantum field theory for G with a mass gap. Finding a quantum field theory amounts to finding a Hilbert space H, a representation of the restricted Lorentz group by unitary transformations of H and operator valued tempered distributions $\varphi_1,...,\varphi_m$ which satisfy density conditions, transformation properties under the Poincare-group, (anti-)commutativity of field operators for test functions of space-like separated supports, an asymptotic completeness property and existence of a unique vacuum state, the field operators acting on this vacuum state span (a dense subspace of) H.
A quantum field theory has a mass gap if the spectrum of the energy operator is contained in $\{0\} \cup [a,\infty)$ for $a>0$.
Now these are straightforward definitions, but what turns a quantum field theory into a YM theory for a group? I know of classical YM theory which gives a Lagrangian and thus equations of motion for the curvature components (fields). Does that mean that $\varphi_1,...,\varphi_n$ have to satisfy these equations? In what way? I could not find any reference for this. Is this millenium problem a mathematical statement at all?

REPLY [23 votes]: The term "Yang-Mills theory" in the mass gap problem refers to a particular QFT.  It is believed that this QFT (meaning its Hilbert space of states and its observable operators) should be defined in terms of a measure on the space of connections on $\mathbb{R}^4$; roughly speaking, the moments of this measure are the matrix elements of the operator-valued distributions.  (People also use the term "gauge theory" to refer to any QFT, like QCD, which has a Yang-Mills sub-theory.)
The mass gap problem really has two aspects:  First, one has to construct an appropriate  measure $d\mu$ on some space of connections.  Then, one has to work out which functions on the space of connections are integrable with respect to this measure, and show that the corresponding collection of operators includes an energy operator (i.e. a generator of time translations) which has a gap in its spectrum.
You'll have to read the literature to really learn anything about this stuff, but I can make a few points to help you on your way.  Be warned that what follows is a caricature.  (Hopefully, a helpful one for people trying to learn this stuff.)
About the Measure:
First, the measure isn't really defined on the space of connections.  Rather, it should be defined on space $\mathcal{F}$ of continuous linear functionals on the space $\mathcal{S}$ of smooth rapidly-vanishing $\mathfrak{g}$-valued vector fields on $\mathbb{R}^4$, where $\mathfrak{g}$ is the Lie algebra of the gauge group $G$.  The space ;$\mathcal{F}$ contains the space of connections, since any connection on $\mathbb{R}^4$ can be written as $d$ plus a $\mathfrak{g}$-valued $1$-form and paired with a vector field via the Killing form, but it also has lots of more "distributional" elements.
We're supposed to get $d\mu$ as the "infinite-volume/continuum limit" of a collection of regularized measures.  This means that we are going to write $\mathcal{S}$ as an increasing union of chosen finite-dimensional vector spaces $\mathcal{S}(V,\epsilon)$; these spaces are spanned by chosen vector fields which have support in some finite-volume $V \subset \mathbb{R}^4$ and which are slowly varying on distance scales smaller than $\epsilon$.  (You should imagine we're choosing a wavelet basis for $\mathcal{S}$.)  Then we're going to construct a measure $d\mu_\hbar(V,\epsilon)$ on the finite dimensional dual vector space $\mathcal{F}(V,\epsilon) = \mathcal{S}(V,\epsilon)' \subset \mathcal{F}$; these measures will have the form $\frac{1}{\mathcal{Z}(V,\epsilon,\hbar)} e^{-\frac{1}{\hbar}S_{V,\epsilon,\hbar}(A)}dA$.   Here, $dA$ is Lesbesgue-measure on $\mathcal{F}(V,\epsilon)$, and $S_{V,\epsilon,\hbar}$ is some discretization of the Yang-Mills action, which is adapted to the subspace $\mathcal{F}(V,\epsilon)$.  (Usually, people use some version of Wilson's lattice action.)  
Existence of the Yang-Mills measure means that one can choose $S_{V,\epsilon}$ as a function of $V$,$\epsilon$, and $\hbar$ so that the limit $d\mu_\hbar$ exists as a measure on $\mathcal{F}$ as $vol(V)/\epsilon \to \infty$.  We also demand that the $\hbar \to 0$ limit of $d\mu_\hbar$ is supported on the space of critical points of the classical Yang-Mills equations.  (We want to tune the discretized actions to fix the classical limit.)
About the Integrable Functions:
Generally speaking, the functions we'd like to integrate should be expressed in terms of the "coordinate functions" which map the $A$ to $A(f)$, where $f$ is one of the basis elements we used to define the subspaces $\mathcal{S}(V,\epsilon)$.  You should imagine that $f$ is a bump vector field, supported near $x \in \mathbb{R}^4$ so that these functions approximate the map sending a $\mathfrak{g}$-valued $1$-form to the value $A_{i,a}(x)$ of its $(i,a)$-th component.
There are three warnings to keep in mind:  
First, we only want to look at functions on $\mathcal{F}$ which are invariant under the group of gauge transformations.  So the coordinate functions themselves are not OK.  But gauge invariant combinations, like the trace of the curvature at a point, or the holonomy of a connection around a loop are OK.  
Second, when expressing observables in terms of the coordinate functions, you have to be careful, because the naive classical expressions don't always carry over.  The expectation value of the function $A \mapsto A_{i,a}(x)A_{j,b}(y)$ with respect to $d\mu_\hbar$ (for $\hbar \neq 0$) is going to be singular as $x \to y$.  This is OK, because we were expecting these moments to define the matrix elements of operator-valued distributions.  But it means we have to be careful when considering the expectation values of functions like $A \mapsto A(x)^2$.  Some modifications may be required to obtain well-defined quantities.  (The simplest example is normal-ordering, which you can see in many two-dimensional QFTs.)
Finally, the real problem.  Yang-Mills theory should confine.  This means, very very roughly, that there are some observables which make sense in the classical theory but which are not well-defined in the quantum theory; quantum mechanical effects prevent the phenomena that these observables describe.  In the measure theoretic formulation, you see this by watching the expection values of these suspect observables diverge (or otherwise fail to remain integrable) as you approach the infinite-volume limit.  
About the Operators:
In classical Yang-Mills theory, the coordinate observables $A \mapsto A_{i,a}(x)$ satisfy equations of motion, the Yang-Mills equations.  Moreover, in classical field theory, for pure states, the expectation value of a product of observables $\mathcal{O}_1\mathcal{O}_2...$ is the product of the individual expectation values.  In quantum YM, the situation is more complicated:  coordinate observables may not have be well-defined, thanks to confinement, and in any case, observables only satisfy equations of motion in a fairly weak sense:  If $\mathcal{O}_1$ is an expression which would vanish in the classical theory thanks to the equations of motion, then then the expectation value of $\mathcal{O}_1\mathcal{O}_2...$ is a distribution supported on the supports of $\mathcal{O}_2...$.<|endoftext|>
TITLE: Integral positive definite quadratic forms and graphs
QUESTION [12 upvotes]: Let me start with a question for which I know the answer. Consider a symmetric integral $n\times n$ matrix $A=(a_{ij})$ such that $a_{ii}=2$, and for $i\ne j$ one has $a_{ij}=0$ or $-1$. One can encode such a matrix by a graph $G$: it has $n$ vertices, and two vertices $i,j$ are connected by an edge iff $a_{ij}=-1$.
Question 1: Which graphs correspond to positive definite $A$?
Answer 1: (Classical, well-known, and easy) $G$ is a disjoint union of $A_n$, $D_n$, $E_n$. (https://en.wikipedia.org/wiki/Root_system)
Now let me put a little twist to this. Let us also allow $a_{ij}=+1$ for $i\ne j$, and encode this by putting a broken edge between $i$ and $j$ (or an edge of different color, if you prefer).
Real question: Which of these graphs correspond to positive definite $A$?

Let me add some partial considerations which do not quite go far enough. (Some of these were in the helpful Gjergji's answer.)
(1) Consider the set $R$ of shortest vectors in $\mathbb Z^n$; they have square 2. Reflections in elements $r\in R$ send $R$ to itself, and $R$ spans $\mathbb Z^n$ since it contains the standard basis vectors $e_i$. By the standard result about root lattices, $\mathbb Z^n$ is then a direct sum of the $A_n$, $D_n$, $E_n$ root lattices, and one can restrict to the case of a single direct summand.
Hence, the question equivalent to the following: what are the graphs corresponding to all possible bases of $\mathbb R^n$ in which the basis vectors are roots?
The case of $R=A_n$ is relatively easy. The roots are of the form $f_a-f_b$, with $a,b \in (1,\dots,n+1)$. Every collection $e_i$ corresponds to an oriented spanning tree on the set $(1,\dots,n+1)$. The 2-colored graph is computed from that tree. I don't see a clean description of the graphs obtained via this procedure, but it is something.
For $D_n$, similarly, a basis is described by an auxiliary connected graph $S$ on $n$ vertices with $n$ edges whose ends are labeled $+$ or $-$. The graph $G$ is computed from $S$.
And for $E_6,E_7,E_8$ there are of course only finitely many cases, but for me the emphasis is on MANY, very many.
So has anybody done this? Is there a table in some paper or book which contains all the 2-colored graphs obtained this way, or -- better still --  a clean characterization of such graphs?
(2) There is a notion of "weakly positive quadratic forms" used in the cluster algebra theory (see for example the first pages of LNM 1099 (Tame Algebras and Integral Quadratic Forms) by Ringel. And there is some kind of classification theory for them. Maybe I am mistaken, but this seems to be quite different: a quadratic form $q$ is "weakly positive" if $q\ge 0$ on the first quadrant $\mathbb Z_{\ge0}^n$. So there is no direct relation to my question, it seems.

REPLY [3 votes]: I have studied a closely related problem (but never published the results and my notes
are messy).
Consider a set $\mathcal S$ of roots in a simply laced root system. Associate to $\mathcal S$
the graph with edges given by pairs of distinct, non-opposite roots which are not orthogonal.
(In other terms, forget the colouring of the edges in your graph.)
Up to 8 vertices you get all graphs in this way. For more than 8 vertices, there are graphs which are not of this form. (There are some necessary conditions related to the Arf
invariant of an associated quadratic form over $\mathbb F_2$. One works of course
only with connected components.)
The trick, if I remember correctly, is to work modulo $2$ and to show that in certain 
situations, one can "lift" a solution into (projective) subsets of roots.
More precisely, one can define invertible combinatorial moves (chirurgies) on graphs which 
come from transformations on subsets contained in root systems (and which amount to convections in symplectic spaces over $\mathbb F_2$). 
For at most eight vertices, one can then look at the equivalence classes of such moves
and show that each equivalence class contains a member realizable by a suitable subset
of roots (in a uniquely defined minimal simply laced root-system). In particular, every
graph with at most $8$ vertices has a a certain "root-type".<|endoftext|>
TITLE: Groups as automorphism groups of small graphs and the number of rigid graphs of a given size
QUESTION [5 upvotes]: In a recent question of mine I asked whether every infinite group is (isomorphic to) the automorphism group of a graph.  The finite case was done by Frucht in 1939.  
The first answer to this question pointed out two papers answering my original 
question, one by Sabidussi
and one by de Groot.
Reading the 3-page paper by Sabidussi I thought "Wow, these graphs are huge":
Sabidussi realizes a group of size $\kappa$ as the automorphism group of a graph of size $\aleph_\kappa$. 
Indeed, de Groot in his paper notes that every countable group is the automorphism group
of a countable graph, every group of size $\leq 2^{\aleph_0}$ is the automorphism group of a graph of size $\leq 2^{\aleph_0}$, and every group of size $\kappa$ is the
the automorphism group of a graph of size $\leq 2^{\kappa}$.
But in general, he doesn't know how large a graph is needed to realize a given group.
Has this issue been resolved?  Is there a reason why for a given infinite group $G$ there shouldn't be a graph of size $|G|$ whose automorphism group is isomorphic to $G$?
As I said in my original question, by Frucht's construction (and the constructions of de Groot and Sabidussi) this is related to the question whether there are $\kappa$ many non-isomorphic rigid graphs of size $\kappa$, where a graph is rigid if the identity is the only automorphism.
Is this known?  I would guess that there are $2^\kappa$ pairwise non-isomorphic rigid graphs of infinite size $\kappa$, but maybe I am wrong.

REPLY [7 votes]: It is well-known that every infinite group $G$ can be realized as the automorphism group of a graph of size $|G|$. It is also well-known that for each infinite cardinal $\kappa$, there are $2^{\kappa}$ nonisomorphic rigid graphs of size $\kappa$. For example, both results are easily extracted from Section 4.2 of the following unpublished book:
http://www.math.rutgers.edu/~sthomas/book.ps<|endoftext|>
TITLE: Mirror of Flop?
QUESTION [5 upvotes]: If two Calabi-Yau 3-folds are bi-rational to each other via a Flop , then what is the relation between their mirrors ?

REPLY [7 votes]: I assume the question regards the coherent sheaves on these two CY's.
These CY's should be regarded as the "same" complex manifold with two
different choices of complexified symplectic forms ("Kahler form," in
physics terminology).
The mirrors are a "single" symplectic manifold with two different
complex structures on it.  There is a curve of complex structures
relating the two.
That's about it.  The tricky part is to "parallel transport" the
category of coherent sheaves along this curve, using
a "flat family of categories" defined by stability conditions.
Doing so should provide a preferred isomorphism of the categories.
Examples have been studied, but general statements (like the ones
I have glibly been making) are not proven.<|endoftext|>
TITLE: totally disconnected and zero-dimensional spaces
QUESTION [21 upvotes]: When do the notions of totally disconnected space and zero-dimensional space coincide? From what I gather, there are at least three common notions of topological dimension: covering dimension, small inductive dimension, and large inductive dimension. A secondary question, then, would be to what extent and under what assumptions the three different definitions of zero-dimensional coincide. For example, Wikipedia claims that a space has covering dimension zero if and only if it has large inductive dimension zero, and that a Hausdorff locally compact space is totally disconnected if and only if it is zero-dimensional, but I can't track down their source and would like to understand the proofs. I would appreciate any explanation, or a reference, since this is a pretty textbookish question.

REPLY [12 votes]: Following Victor Protsak's suggestion, I took the answer to this question and turned it into a paper found here Ultraparacompactness and Ultranormality, so it may be easier to read that paper than to read the answer here on MO.
The notions that you are looking for are ultraparacompactness (covering dimension depending on how you define it), ultranormality (large inductive dimension zero), and of course the notion of a zero-dimensional space (small inductive dimension zero). A Hausdorff space $X$ is said to be ultraparacompact if every open cover can be refined by a partition into clopen sets. It should be noted that there appears to be some disagreement on the definition of covering dimension since some people require your original cover to be finite and some people consider arbitrary covers, so the notion of ultraparacompactness may or may not coincide with the notion of covering dimension zero. It seems as if the standard  but not universal practice is to define the covering dimension in terms of finite covers. The spaces with large inductive dimension zero are known as ultranormal spaces. In other words, a Hausdorff space is ultranormal if and only if whenever $R,S$ are disjoint closed sets, there is a clopen set $C$ with $R\subseteq C,S\subseteq C^{c}$.  Ultraparacompactness, ultranormality, and zero-dimensionality are the zero-dimensional analogues of the notions of paracompactness, normality, and regularity. Many in the notions and results from general topology have analogous zero-dimensional notions and results. Clearly every ultraparacompact space is paracompact, and every ultranormal space is normal. It is easy to see that every ultraparacompact space is ultranormal and every ultranormal space is zero-dimensional. However, the converses fail to hold.
$\large\mathbf{Examples}$
Every compact totally disconnected space is ultraparacompact and hence ultranormal as well. 
The space $\omega_{1}$ of all countable ordinals with the order topology is ultranormal. If $R,S$ are two disjoint closed subsets of $\omega_{1}$, then either $R$ or $S$ is bounded, so say $R$ is bounded by an ordinal $\alpha$. Then since $[0,\alpha]$ is compact and zero-dimensional, the set $[0,\alpha]$ is ultranormal. Therefore there is a clopen subset $C\subseteq[0,\alpha]$ with $R\subseteq C$ and $S\cap C=\emptyset$. However, the set $C$ is clopen in $\omega_{1}$ as well. Therefore $\omega_{1}$ is ultranormal. To the contrary, the space $\omega_{1}$ is not even paracompact. The cover $\{[0,\alpha)|\alpha<\omega_{1}\}$ however does not have a locally finite open refinement since if $\mathcal{U}$ is an open refinement of $[0,\alpha)$, then for each $\alpha<\omega_{1}$ there is some $x_{\alpha}<\alpha$ where $(x_{\alpha},\alpha]\subseteq U$ for some $U\in\mathcal{U}$. However, since the mapping $\alpha\mapsto x_{\alpha}$ is regressive, everyone who knows anything about set theory can tell you that there is an ordinal $\beta$ where $x_{\alpha}=\beta$ for uncountably many $\alpha$. Since $\mathcal{U}$ refines $\{[0,\alpha)|\alpha<\omega_{1}\}$, each $U\in\mathcal{U}$ is bounded, so the ordinal $\beta$ must be contained in uncountably many $U\in\mathcal{U}$.
Now consider the space $\mathbb{R}$ with the lower limit topology. In other words, with this topology, $\mathbb{R}$ is generated by the basis $\{[a,b)|a<b\}$. Then $[0,\infty)$ is an ultraparacompact space with this topology. Let $\mathcal{U}$ be an open cover of $[0,\infty)$. Let $x_{0}=0$. If $\alpha$ is an ordinal and $\sup\{x_{\beta}|\beta<\alpha\}<\infty$, then let $x_{\alpha}$ be a real number such that $x_{\alpha}>\sup\{x_{\beta}|\beta<\alpha\}$ and $[\sup\{x_{\beta}|\beta<\alpha\},x_{\alpha})\subseteq U$ for some $U\in\mathcal{U}$. Then $\{[x_{\alpha},x_{\alpha+1})|\alpha\}$ is a partition of $[0,\infty)$ into clopen sets that refines $\mathcal{U}$. Thus, $[0,\infty)$ is ultraparacompact. The space $\mathbb{R}$ with the lower limit topology is ultraparacompact as well since $\mathbb{R}$ is isomorphic to the countable sum of spaces $[0,\infty)$ with the lower limit topology. However, it is a well known counterexample that the product $\mathbb{R}\times\mathbb{R}$ is not even normal. We conclude that $\mathbb{R}\times\mathbb{R}$ is a zero-dimensional space which is not ultranormal. 
In the paper Nonequality of Dimensions for Metric Spaces, Prabir Roy shows that certain space $\Delta$ is a complete metric space of cardinality continuum which is zero-dimensional, but not ultranormal, and hence not ultraparacompact. In fact, later in the paper Not every 0-dimensional realcompact space is $\mathbb{N}$-compact, Peter Nyikos shows that this space is not even $\mathbb{N}$-compact (a space is $\mathbb{N}$-compact if and only if it can be embedded as a closed subspace of a product $\mathbb{N}^{I}$ for some set $I$). This result strengthens Roy's result since every ultraparacompact space of non-measurable cardinality is $\mathbb{N}$-compact and the first measurable cardinal is terribly large if it even exists.
A metric $d$ on a set $X$ is said to be an ultrametric if $d$ satisfies the strong triangle inequality: $d(x,z)\leq Max(d(x,y),d(y,z))$, and a metric space $(X,d)$ is said to be an ultrametric space if $d$ is an ultrametric. Every ultrametric space is ultraparacompact.
There are zero-dimensional locally compact spaces that are not ultranormal. The Tychonoff plank $X:=((\omega_{1}+1)\times(\omega+1))\setminus\{(\omega_{1},\omega)\}$ is zero-dimensional (even strongly zero-dimensional; i.e. $\beta X$ is zero-dimensional) but not ultranormal.
$\large\mathbf{Results}$
A subset $Z$ of a space $X$ is said to be a zero set if there is a continuous function $f:X\rightarrow[0,1]$ such that $Z=f^{-1}[\{0\}]$. A completely regular space $X$ is said to be strongly zero-dimensional if whenever $Z_{1},Z_{2}$ are disjoint zero sets, there is a clopen set $C$ with $Z_{1}\subseteq C$ and $Z_{2}\subseteq C^{c}$. It is not too hard to show that a completely regular space $X$ is strongly zero-dimensional if and only if the Stone-Cech compactification $\beta X$ is zero-dimensional. We say that a cover $\mathcal{R} $ of a topological space $X$ is point-finite if $\{R\in\mathcal{R}|x\in R\}$ is finite for each $x\in X$, and we say that $\mathcal{R}$ is locally finite if each $x\in X$ has a neighborhood $U$ such that $\{R\in\mathcal{R}|U\cap R\neq\emptyset\}$ is finite. Recall that a Hausdorff space is paracompact iff every open cover has a locally finite open refinement.
A uniform space $(X,\mathcal{U})$ is said to be non-Archimedean if $\mathcal{U}$ is generated by equivalence relations. In other words, for each $R\in\mathcal{U}$ there is an equivalence relation $E\in\mathcal{U}$ with $E\subseteq R$. Clearly, every non-Archimedean uniform space is zero-dimensional. If $(X,\mathcal{U})$ is a uniform space, then let $H(X)$ denote the set of all closed subsets of $X$. If $E\in\mathcal{U}$, then let $\hat{E}$ be the relation on $H(X)$ where $(C,D)\in\hat{E}$ if and only if $C\subseteq E[D]=\{y\in X|(x,y)\in E\textrm{for some}x\in D\}$ and $D\subseteq E[C]$. Then the system $\{\hat{E}|E\in\mathcal{U}\}$ generates a uniformity $\hat{\mathcal{U}}$ on $H(X)$ called the hyperspace uniformity. A uniform space $(X,\mathcal{U})$ is said to be supercomplete if the hyperspace $H(X)$ is a complete uniform space.
$\mathbf{Proposition}$ Let $X$ be a locally compact zero-dimensional space. Then $X$ is ultraparacompact if and only if $X$ can be partitioned into a family of compact open sets.
$\mathbf{Proof}$ The direction $\leftarrow$ is fairly trivial. To prove $\rightarrow$ assume that $X$ is a locally compact zero-dimensional space. Then let $\mathcal{U}$ the collection of all open sets $U$ such that $\overline{U}$ is compact. Then since $X$ is locally compact, $\mathcal{U}$ is a cover for $X$. Therefore there is a partition $P$ of $X$ into clopen sets that refines $\mathcal{U}$. Clearly each $R\in P$ is a compact open subset of $X$. $\textrm{QED}$
$\textbf{Theorem}$ Let $X$ be a Hausdorff space. Then $X$ is normal if and only if whenever $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a point-finite open covering of $X$, there is an open covering $(V_{\alpha})_{\alpha\in\mathcal{A}}$ such that $\overline{V_{\alpha}}\subseteq U_{\alpha}$ for $\alpha\in\mathcal{A}$ and $V_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$.///
To prove the above result, one first well orders the set $\mathcal{A}$, then one shrinks the sets $U_{\alpha}$ to sets $V_{\alpha}$ in such a way that you still cover your space $X$ at every point in the induction process. See the book Topology by James Dugundji for a proof of the above result.
$\textbf{Theorem}$ Let $X$ be a Hausdorff space. The following are equivalent.

$X$ is ultranormal.
Whenever $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a point-finite open cover of $X$, there is a clopen cover $(V_{\alpha})_{\alpha\in\mathcal{A}}$ such that $V_{\alpha}\subseteq U_{\alpha}$ for each $\alpha$ and $V_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$.
If $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a locally-finite open cover of $X$, then there is system $(P_{\alpha})_{\alpha\in\mathcal{A}}$ of clopen sets such that $P_{\alpha}\subseteq U_{\alpha}$ for $\alpha\in\mathcal{A}$ and $P_{\alpha}\cap P_{\beta}=\emptyset$ whenever $\alpha,\beta\in\mathcal{A}$ and $\alpha\neq\beta$.
$X$ is normal and strongly zero-dimensional.

$\textbf{Proof}$
$1\rightarrow 2$. Since $X$ is normal, there is an open cover $(W_{\alpha})_{\alpha\in\mathcal{A}}$ such that $\overline{W_{\alpha}}\subseteq U_{\alpha}$ for each $\alpha\in\mathcal{A}$ and $W_{\alpha}\neq\emptyset$ whenever $U_{\alpha}\neq\emptyset$. Since $X$ is ultranormal, for each $\alpha\in\mathcal{A}$, there is some clopen set $V_{\alpha}$ with $\overline{W_{\alpha}}\subseteq V_{\alpha}\subseteq U_{\alpha}$.
$2\rightarrow 3$. Now assume that $(U_{\alpha})_{\alpha\in\mathcal{A}}$ is a locally-finite open over of $X$. Let $(V_{\alpha})_{\alpha\in\mathcal{A}}$ be a clopen cover of $X$ such that $V_{\alpha}\subseteq U_{\alpha}$ for each $\alpha\in\mathcal{A}$. Well order the set $\mathcal{A}$. The family $(V_{\alpha})_{\alpha\in\mathcal{A}}$ is locally-finite, so since each $V_{\alpha}$ is closed, for each $\alpha\in\mathcal{A}$, the union $\bigcup_{\beta<\alpha}V_{\beta}$ is closed. Clearly $\bigcup_{\beta<\alpha}V_{\beta}$ is  open as well, so $\bigcup_{\beta<\alpha}V_{\beta}$ is clopen. Let
$P_{\alpha}=V_{\alpha}\setminus(\bigcup_{\beta<\alpha}V_{\beta})$. Then $(P_{\alpha})_{\alpha\in\mathcal{A}}$ is the required partition of $X$ into clopen sets.
$3\rightarrow 1,1\rightarrow 4$. This is fairly obvious.
$4\rightarrow 1$. This is a trivial consequence of Urysohn's lemma.
$\textbf{QED}$
$\textbf{Theorem}$ Let $X$ be a Hausdorff space. The following are equivalent.

$X$ is ultraparacompact.
$X$ is ultranormal and paracompact.
$X$ is strongly zero-dimensional and paracompact.
Every open cover of $X$ has a locally finite clopen refinement.
$X$ is zero-dimensional and satisfies the following property: let $I$ be an ideal on the Boolean algebra $\mathfrak{B}(X)$ of clopen subsets of $X$ such that $\bigcup I=X$ and if $P$ is a partition of $X$ into clopen sets, then $\bigcup(P\cap I)\in I$. Then $I=\mathfrak{B}(X)$.
$X$ has a compatible supercomplete non-Archimedean uniformity. ///

The paper $\omega_{\mu}$-additive topological spaces by Giuliano Artico and Roberto Moresco gives several characterizations of when a $P_{\kappa}$-space (a space where the intersection of less than $\kappa$ many open sets is open) is ultraparacompact.
$\large\textbf{The Point-Free Context}$.
The following results on zero-dimensionality, ultranormality, and ultraparacompactness are part of my own research. If you think I have wrote too much already, you don't like me, or if you don't believe in pointless topology, then I suggest that you stop reading this answer here.
The notions of ultranormality, zero-dimensionality, and ultraparacompactness make sense in a point-free context, and by a generalization of Stone duality, the notions of ultranormality, zero-dimensionality, and ultraparacompactness translate nicely to certain kinds of Boolean algebras with extra structure.
A frame is a complete lattice that satisfies the following infinite distributivity law
$$x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i}).$$
If $X$ is a topological space, then the collection of all open subsets of $X$ forms a frame. Frames generalize the notion of a topological space, and frames are the central object of study in point-free topology. For Hausdorff spaces, no information is lost simply by considering the lattice of open sets of the topological space. More specifically, if $X,Y$ are Hausdorff spaces and the lattices of open sets of $X$ and $Y$ respectively are isomorphic, then $X$ and $Y$ are themselves isomorphic. Furthermore, many notions and theorems from general topology can be generalized to the point-free context. Moreover, the notion of a frame is very interesting from a purely lattice theoretic perspective without even looking at the topological perspective. The reader is referred to the excellent new book Frames and Locales: Topology Without Points by Picado and Pultr for more information on point-free topology.
For notation, if $X$ is a poset and $R,S\subseteq X$, then $R$ refines $S$ (written $R\preceq S$) if for each $r\in R$ there is some $s\in S$ with $r\leq s$. If $x\in X$, then define $\downarrow x:=\{y\in X|y\leq x\}$.
A Boolean admissibility system is a pair $(B,\mathcal{A})$ such that $B$ is a Boolean algebra and $\mathcal{A}$ is a collection of subsets of $B$ with least upper bounds such that
i. $\mathcal{A}$ contains all finite subsets of $B$,
ii. if $R\in\mathcal{A},S\subseteq\downarrow\bigvee R,R\preceq S$, then $S\in\mathcal{A}$ as well,
iii. if $R\in\mathcal{A}$ and $R_{r}\in\mathcal{A},\bigvee R_{r}=r$ for $r\in R$, then
$\bigcup_{r\in R}R_{r}\in\mathcal{A}$,
iv. if $R\in\mathcal{A}$, then $\{a\wedge r|r\in R\}\in\mathcal{A}$ as well.
Intuitively, a Boolean admissibility system is Boolean algebra along with a notion of which least upper bounds are important and which least upper bounds are not important. 
For example, if $B$ is a Boolean algebra, and $\mathcal{A}$ is the collection of all subsets of $B$ with least upper bounds, then $(B,\mathcal{A})$ is a Boolean admissibility system.
If $A$ is a Boolean subalgebra of $B$, and $\mathcal{A}$ is the collection of all subsets of $R$ where the least upper bound $\bigvee^{B}R$ exists in $B$ and $\bigvee^{B}R\in A$. Then $(A,\mathcal{A})$ is a Boolean admissibility system.
A Boolean admissibility system $(B,\mathcal{A})$ is said to be subcomplete if whenever $R,S\subseteq B$ and $R\cup S\in\mathcal{A}$ and $r\wedge s=\emptyset$ whenever $r\in R,s\in S$, then $R\in\mathcal{A}$ and $S\in\mathcal{A}$.
If $L$ is a frame, then an element $x\in L$ is said to be complemented if there is an element $y$ such that $x\wedge y=0$ and $x\vee y=1$. The element $y$ is said to be the complement of $y$ and one can easily show that the element $y$ is unique. The notion of a complemented element is the point-free generalization of the notion of a clopen set. Let $\mathfrak{B}(L)$ denote the set of complemented elements in $L$. Then $\mathfrak{B}(L)$ is a sublattice of $L$. In fact, $\mathfrak{B}(L)$ is a Boolean lattice. A frame $L$ is said to be zero-dimensional if $x=\bigvee\{y\in\mathfrak{B}(L)|y\leq x\}$ for each $x\in L$. 
A Boolean based frame is a pair $(L,B)$ where $L$ is a frame and $B$ is a Boolean subalgebra of $\mathfrak{B}(L)$ such that $x=\bigvee\{b\in\mathfrak{B}(L)|b\leq x\}$ for each $x\in L$. Clearly every Boolean based frame is zero-dimensional.
A zero-dimensional frame $L$ is said to be ultranormal if whenever $a\vee b=1$, there is a complemented element $c\in L$ such that $c\leq a$ and $c'\leq b$.
A cover of a frame $L$ is a subset $C\subseteq L$ with $\bigvee C=1$. A partition of a frame $L$ is a subset $p\subseteq L\setminus\{0\}$ with $\bigvee p=1$ and where $a\wedge b=0$ whenever $a,b\in p,a\neq b$.
A zero-dimensional frame $L$ is said to be ultraparacompact if whenever $C$ is a cover of $L$ there is a partition $p$ that refines $C$.
There is a duality between the category of Boolean admissibility systems and Boolean based frames. If $(B,\mathcal{A})$ is a Boolean admissibility system, then let $C_{\mathcal{A}}$ be the collection of all ideals $I\subseteq B$ such that if $R\in\mathcal{A}$ and $R\subseteq I$, then $\bigvee R\in I$ as well.
If $(B,\mathcal{A})$ is a Boolean admissibility system, then $\mathcal{V}(B,\mathcal{A}):=(C_{\mathcal{A}},\{\downarrow b|b\in B\})$ is a Boolean based frame. Similarly, if $(L,A)$ is a Boolean based frame, then $\mathcal{W}(L,A):=(A,\{R\subseteq A|\bigvee^{L}R\in A\})$ is a Boolean admissibility system. Furthermore, these correspondences give an equivalence between the category of Boolean admissibility systems and the category of Boolean based frames. Hence, we obtain a type of Stone-duality for zero-dimensional frames. If $(L,A)$ is a Boolean based frame, then $A=\mathfrak{B}(L)$ if and only if $\mathcal{W}(L,A)$ is subcomplete. Since $(L,\mathfrak{B}(L))$ is a Boolean based frame iff $L$ is a zero-dimensional frame, we conclude that the category of zero-dimensional frames is equivalent to the category of subcomplete Boolean admissibility systems.
If $(B,\mathcal{A})$ is a Boolean admissibility system, then $
\mathcal{V}(B,\mathcal{A})=(C_{\mathcal{A}},\{\downarrow b|b\in B\})$ is ultranormal with 
$\{\downarrow b|b\in B\}=\mathfrak{B}(C_{\mathcal{A}})$ if and only if whenever $I,J\in\mathcal{C}_{\mathcal{A}}$, then $\{a\vee b|a\in I,b\in J\}\in\mathcal{C}_{\mathcal{A}}$ as well. Hence, ultranormality simply means that the join of finitely many ideals in $\mathcal{C}_{\mathcal{A}}$ is an ideal in $\mathcal{C}_{\mathcal{A}}$.
If $(B,\mathcal{A})$ is a subcomplete Boolean admissibility system, then $\mathcal{V}(B,\mathcal{A})$ is ultraparacompact if and only if whenever $R\in\mathcal{A}$ there is some $S\in\mathcal{A}$ with $S\preceq R,\bigvee S=\bigvee R$ and $a\wedge b=0$ whenever $a,b\in S$ and $a\neq b$. Hence, the ultraparacompactness of frames translates into a version of ultraparacompactness in the dual subcomplete Boolean admissibility systems.<|endoftext|>
TITLE: Deformation quantization and quantum cohomology (or Fukaya category) -- are they related?
QUESTION [14 upvotes]: Good afternoon.
Let $M$ be, say, a compact symplectic manifold. Both deformation quantization (as in Kontsevich) and quantum cohomology yield "deformations" (in the appropriate respective senses) of "classical" data -- the Poisson algebra of functions $C^\infty(M)$ and the cohomology algebra (or rather, Frobenius algebra) $H^\ast(M; \mathbb{C})$ respectively.

Are these two things related somehow? 

I am interested in both mathematical and physical answers.
I apologize if this question is naive. I feel like, with a proper understanding of the physics, the answer to this question is probably obviously "yes" or obviously "no". Unfortunately, I don't have a good understanding of the physics.
Edit: From the looks of the discussion below, deformation quantization is perhaps more directly related to the Fukaya category. I welcome any additional remarks on the Fukaya category.

REPLY [5 votes]: I've scanned the paper I linked above for a few minutes(I am saying that so people judge what I say critically) and I'd like to expand on the comment that some category of modules over the deformation quantization is a sort of classical limit of the Fukaya category(e.g. a version without pseudo-holomorphic disks). I am sure there have been advances since this paper, since it goes back to 2002, and I'd be really interested to hear what they were. First of all, Bressler and Soibelman have the lemma that I was alluding to above that if $A_X$ is the quantized algebra of smooth functions, $HH^*(A_X-mod)= H^*(X)\otimes C((t))$(I guess Bressler et. al. work with complex valued smooth functions.) with the ordinary product. Meanwhile (assuming one has symplectic form that is integral??) the quantum cup product can be defined on the same vector space.The difference is that the product in the first vector space is the ordinary product, while the product in the second one is deformed by pseudo-holomorphic discs.
This can be some kind of closed string version of the classical limit analogy. Now, the observation that is really cool in the paper however, is that if one restricts to the category of what they call holonomic modules, one can get a much more precise version of this analogy(a categorical one). Let Hol(X), be the full subcategory of modules whose support(of M/tM) consists of modules are Lagrangian Submanifolds. Bressler and Soibelman claim that the data of a langrangian submanifold and a local system (L,p) can be used to produce a holonomic module in a canonical way. Furthermore, one has the following analogy with the Fukaya category... Let $M_{(L,p)}$ be the module produced above... One has $Ext^*(M_{(L,p)},M_{(L,p)})=H^*(L,p)\otimes C((t))$. The obvious proposition about two Lagrangian's with transversal support is correct too, see Prop 2. on page 12. So again, this category has the right Hom's as vector spaces, but not as an $A(\infty)$ category (the Hom's are again the Hom's in the absence of holomorphic disks). 
Bressler and Soibelman then continue with some speculation about algebraic ways to put in the discs, their basic idea being to find a functor $\phi : hol(X) \mapsto hol(X)$ such that  $Hom(A,\phi(B)$ agrees as an A-infinity category with the Fukaya category. By the dg category yoga, this should come in the form of a bi-module. They then explain roughly how to do it in the cotangent bundle case. Their vision seems to be that maybe one could have a purely algebraic approach to the Fukaya category via holonomic modules and thereby avoid some of the technical issues with moduli spaces of discs. This also gives an approach to the co-isotropic branes people have been talking about.<|endoftext|>
TITLE: Is there an easy way to describe the sheaf of smooth functions on a product manifold?
QUESTION [11 upvotes]: A smooth structure on a manifold $M$ can be given in the form of a sheaf of functions $\mathcal{F}$ such that there is an open cover $\mathcal{U}$ of $M$ with every $U\in \mathcal{U}$ isomorphic (along with $\mathcal{F}|_U$) to an open subset $V$ of $\mathbb{R}^n$ (along with $\mathcal{O}|_V$, where $\mathcal{O}$ is the sheaf of smooth functions on $\mathbb{R}^n$).  I think we might also need to say that this satisfies a smooth-coordinate-change axiom, although maybe that's already tied up in the definition of a sheaf.  In any case, here is my question:

Given two smooth manifolds
  $(M,\mathcal{F})$ and
  $(N,\mathcal{G})$, is there an easy
  way to write the sheaf of functions on
  $M\times N$ without reference to
  coordinate neighborhoods?

I'm wondering this because in one of my classes we defined smooth manifolds in this way (and we defined analytic and holomorphic manifolds similarly).  It seems like some people are very fond of this alternative definition because it doesn't refer to an atlas, which at first seems like it's an inherent part of the structure of the manifold.  So okay fine, everyone loves a canonical definition.  However, this is only going to be useful as long as we can tell our whole story in this canonical language.  In class, the only way the professor was able to give the sheaf on the product was by breaking down and using coordinates.  (Admittedly, he was on the spot and presumably unprepared for the question.)
This also suggests the broader, more open-ended question:

Are there longer-run advantages to the above definition (compared to the usual definition involving an atlas and perhaps a maximal atlas)?

REPLY [8 votes]: Smooth manifolds are affine, thus the sheaf of smooth functions is determined by its global sections.
Now C^∞(M×N)=C^∞(M)⊗C^∞(N).
The tensor product here is the projective tensor product
of complete locally convex Hausdorff topological algebras.<|endoftext|>
TITLE: Why are multinomial coefficients with same entropy equal? (usually)
QUESTION [5 upvotes]: Suppose $p_1,\ldots,p_d$ and $q_1,\ldots,q_d$ are positive real numbers such that 
$$p_1+\cdots+p_d=q_1+\cdots+q_d=n$$
and
$$p_1 \log p_1+\cdots+p_d\log p_d=q_1 \log q_1+\cdots+q_d \log q_d $$
Then the following seems to hold
$$\frac{n!}{p_1!\cdots p_d!}=\frac{n!}{q_1!\cdots q_d!}$$
why?
Edit: JBL correctly notices that it doesn't always hold. I just didn't go far enough.
Still, it's surprising to me that it holds so frequently.
If we put a black disk at x,y if equality seems to hold (in machine precision) for x=n,y=d, and positive integer coefficients, it'll look like this

(source)
Red circle is JBL's example. Blue circle is n=18,d=3 which fails for (12,3,3) and (9,8,1).

docheck[n_, d_] := (
   coefs = IntegerPartitions[n, {d}, Range[1, n]];
   entropy[x_] := N[Total[# Log[#] & /@ x]];
   groupedCoefs = GatherBy[coefs, entropy];
   allEqual[list_] := And @@ (First[list] == # & /@ list);
   multinomials = Apply[Multinomial, groupedCoefs, {2}];
   And @@ (allEqual /@ multinomials)
   );
vals = Table[docheck[#, d] & /@ Range[1, 30], {d, 1, 20}];
Graphics[Table[Disk[{n, d}, If[vals[[d, n]], .45, .1]], {d, 1, Length[vals]}, {n, 1, 30}]]

Edit:
Updated version that does exact checking and allows coefficients with 0 components. Still only one example of failure for d=3.

(source)

docheck[n_, d_] := (coefs = IntegerPartitions[n, {d}, Range[0, n]];
   entropy[x_] := Exp[Total[If[# == 0, 0, # Log[#]] & /@ x]];
   groupedCoefs = GatherBy[coefs, entropy];
   allEqual[list_] := And @@ (First[list] == # & /@ list);
   multinomials = Apply[Multinomial, groupedCoefs, {2}];
   And @@ (allEqual /@ multinomials));
maxn = 30; maxd = 20;
vals = Table[docheck[#, d] & /@ Range[1, maxn], {d, 1, maxd}];
Graphics[Table[Disk[{n, d}, If[vals[[d, n]], .45, .1]], {d, 1, Length[vals]}, {n, 1, maxn}]]

REPLY [2 votes]: EDIT 
You still have not given interesting examples of equal multinomal coefficients with equal entropy. I'm not saying there aren't any (or many) but it would help to see some. Or even interesting examples (especially for d=3) with at least the same prime divisors showing up.
Either of these tasks is pretty easy (I added the condition disjoint which is no loss) :

find n and d and two disjoint d-tuples of integers A and B both with sum n and $\prod_A a!=\prod_B b!$ (equal multinomials with the same n and d)
find n and d and two disjoint d-tuples of integers A and B with sum n 
$\prod_A a^a=\prod_B b^b$ (equal entropy)

I'd venture that most of the time a solution of one is not a solution of the other. 
For equal entropy: There is a 4 parameter family of solutions to the equal entropy problem (usually with d large) using 1,2,3,4,6,8,9,12 spread out between the sets A and B (Choose the number of 4s,8s,9s and 12s and where they go, there is a unique way to choose the 1s,2s,3s and 6s) . If one of the sets uses 12 then there will be a discrepancy in the multinomial coefficients related to the prime 11.
for equal multinomials: $<x,y,xy-1>$ and $<x-1,y-1,xy>$ give equal product of factorials so $<x,y,xy-1,u-1,v-1,uv>$ and $<x-1,y-1,xy,u,v,uv-1>$ gives a d=6 case of the multinomial problem. It is easy to arrange for cancelation. Set u=y and v=xy-1 to get a d=3 multinomial  $<x,xy-2,(xy-1)y>$ and $<x-1,xy,(xy-1)y-1>$. Here there will be primes present in one side but not the other so even the shifted entropy won't be exactly the same.
previous answer
- The missing black dot at n=19,d=4 is from  $<6,6,6,1>,<9,2,4,4>$ which have equal entropy yet the corresponding multinomial coefficients have ratio 25/28.
 - The missing black dot at n=20,d=5 is from  $<8,3,3,3,3>,<6,6,4,2,2>$ which have equal entropy yet the corresponding multinomial coefficients have ratio 21/20
.
 - What are some interesting examples of equal entropy and equal multinomial coefficients?
 - The ratio of similar binomial coeffcients will usually consist of powers of relatively small primes so coincidences are likely.
 - There are an enormous number of equal multinomial coefficients and they do not lead to equal entropy in most cases (not that you say they will).<|endoftext|>
TITLE: Best tablet computer for mathematics
QUESTION [42 upvotes]: I'm not sure if this is completely appropriate, but I thought I'd ask here.
I'm in the market for a tablet computer.  Unfortunately, my (mathematical) needs are very different from the needs of the sorts of people who usually review these things.  Namely, I want the tablet that is best for preparing lecture notes and taking notes in seminars.  It seems like most of the reviewers on "computer" websites are either looking for toys (like the iPad) or tools for creating art.  I just want the most efficient way to create a multi-page handwritten pdf.  Does anyone have any recommendations?

REPLY [5 votes]: To address part of your question, the iPad is far more than a toy for mathematicians, especially for reading mathematical articles.  Goodreader is the best means of reading and annotating PDFs I’ve ever seen; in particular, the usual size of LaTeX articles fits very well on the screen, with a little zooming, even in portrait mode.  I’ve virtually eliminated my use of my printer, both for reading others’ articles and correcting my own XeLaTeX output.  
Be very sure to test a PDF reader on real articles before it’s too late; even Adobe renders the Computer Modern font unacceptably on-screen, though with decidedly different results in different situations and platforms, and Okular on Linux is virtually unreadable at 100% zoom.
For note-taking, I’m not a big fan of raw handwritten text (disclaimer: I worked at Apple during the Newton days, and at Palm); I expected to have to buy a hardware keyboard for my iPad, but haven’t needed to yet.<|endoftext|>
TITLE: Pennies on a carpet problem
QUESTION [18 upvotes]: I recently read the following "open problem" titled "Pennies on a carpet" in   "An Introduction To Probability and Random Processes" by Baclawski and Rota (page viii of book, page 10 of following pdf), found here: http://www.ellerman.org/Davids-Stuff/Maths/Rota-Baclawski-Prob-Theory-79.pdf
"We have a rectangular carpet and an indefinite supply of perfect pennies. What is the probability that if we drop the pennies on the carpet at random no two of them will overlap? This problem is one of the most important problems in statistical mechanics. If we could answer it we would know, for example, why water boils at 100C, on the basis of purely atomic computations. Nothing is known about this problem.”
I was wondering if this problem goes by a more popular name and whether or not some form of progress has been made on it. In particular, references would be highly appreciated!

REPLY [10 votes]: This is the two-dimensional hard spheres model, sometimes called hard discs in a box.
See Section 4 of Persi Diaconis's recent survey article, The Markov Chain Monte Carlo Revolution.  The point here is that even though it very hard to sample a random configuration of nonoverlapping discs by dropping them on the carpet (because the probability of success is far too small for any reasonable number of discs), but it is nevertheless possible to sample a random configuration via Monte Carlo.<|endoftext|>
TITLE: covers of $Z^\infty$
QUESTION [11 upvotes]: Is it possible to cover $Z^\infty$ (the infinite direct sum of $Z$'s with the $l_1$-metric) by a finite set of collections of subsets $U^0,...,U^n$ such that each  collection $U^i$ consists of uniformly bounded sets $U_j^i$ that are 4-disjoint (the distance between any two subsets $U_j^i$, $U_k^i$ in each $U^i$ is at least 4)? The motivation is here: http://front.math.ucdavis.edu/1008.3868 .

REPLY [5 votes]: In fact there is also a negative answer (again with 3 instead of 4) in the case of the countably infinite sum of Z_2.  View the direct sum with the $\ell_1$-metric as being the collection of all finite subsets of N with the metric which counts the symmetric difference.  Let $[N]^k$ denote all of the k-element subsets of N.  Observe that it suffices to demonstrate that, for each k, if $[N]^k$ is partitioned into finitely many pieces, then there is a piece of the partition containing a sequence $a_i$ $(i < k)$ such that $a_0$ is disjoint from $a_{k-1}$ (thus $a_0$ and $a_{k-1}$ are 2k units apart) and 
$|a_i \Delta a_{i+1}| = 2$ of all $i < k-1$.  But this follows easily from Ramsey's theorem: given any such partition, there is a subset b of N of cardinality 2k, all of whose k element sets are in one piece of the partition (in fact we can find an infinite b if we like).  From this one can now easily construct the sequence of $a_i$ $(i < k)$:
if $b = \{m_i : i < 2k\}$, then set $a_i = \{m_j : i \leq j < i+k\}$.<|endoftext|>
TITLE: Maximum on unit ball (James' theorem).
QUESTION [5 upvotes]: James' theorem states that a Banach space $B$ is reflexive iff every bounded linear functional on $B$ attains its maximum on the closed unit ball in $B$.
Now I wonder if I can drop the constraint that it is a ball and replace it by "convex set". That is, I want to know if every bounded linear functional on a reflexive Banach space $B$ attains its maximum on a closed and bounded convex set in $B$.
By Pietro's answer this is known to be true. Is the maximum unique? In optimization by vector space methods it is know that this is true if the set is the closed ball. This was actually my biggest question since I want to show a optimization problem has a unique solution.

REPLY [9 votes]: A closed and bounded convex set of a reflexive Banach is w* compact, hence any bounded linear functional does attain its maximum and minimum there. On the other direction, the presence of a closed bounded convex set on which all bounded linear functionals have their maximum, of course, says nothing on the reflexivity of the space (the convex could be a single point).<|endoftext|>
TITLE: Random noncrossing chords of a circle
QUESTION [15 upvotes]: Suppose you generate random chords of a circle, with endpoints selected uniformly over the circumference, rejecting any chord that crosses a previously generated chord.
The disk is then partitioned into regions bounded by chords alternating with circular arcs.
For example, here are $n{=}100$ random noncrossing chords, with a region bounded by 5 chords highlighted
(in green).


I am interested in the statistics of the structure of the dual trees for these regions. 
Assign each region a node,
and connect two nodes by an edge if they share a chord.  In the example above, the highlighted
region's node has degree 5.
Example questions: What is the expected maximum degree of a node for $n$ chords?
Making a max-degree node the root, what is the expected height of the tree?
(In the example above, the height is 21.)
Etc.
Has anyone encountered this model before?
Or a model sufficiently analogous to help establish these statistics?
Thanks for any pointers!
Edit.  Many thanks for the wealth of information provided by the community!
So far I have not found the following specific question answered (although it is likely implied, perhaps
in the papers they cite):
What is the expected maximum degree of a node as $n \rightarrow \infty$? What brought me to this
topic in the first place is that I wondered if it might be near 3.

REPLY [8 votes]: Louigi is absolutely right. We control "typical" height not absolute one. Note that a similar discrete model has been investigated by physicists see http://www.phys.ens.fr/~wiese/pdf/hiraRNA.pdf
I think the maximal degree after n steps is logarithmic, but I don't have any exact expression for the expected value...<|endoftext|>
TITLE: Torsion submodule
QUESTION [8 upvotes]: $A$ a commutative Noetherian domain, $M$ a finitely generated $A$-module. How can I show that the kernel of the natural map $M\rightarrow M^{**}$, where $ M^{ * *}$ is the double dual (with respect to $A$), is the torsion submodule of $M$?
I do know that in this situation torsionlessness coincides with torsion-freeness. According to Auslander this result is ``well-know'' but I can't seem to prove it or find any reference on this.

REPLY [12 votes]: Let $K$ be the fraction field of $A$.  Then there is a natural isomorphism
$M^*\otimes_A K \cong (M\otimes_A K)^*$ (where the dual on the left is the $A$-dual,
and on the right is the $K$-dual).
Thus the double dual map $M \to M^{* *}$ becomes an isomorphism after tensoring with $K$
over $A$, and hence its kernel is contained in the kernel of the natural map $M \to
K\otimes_A M,$ which shows that its kernel is torsion.  On the other hand, clearly
the torsion submodule of $M$ is contained in this kernel, since $M^{* *}$ is torsion free.
This proves the result.<|endoftext|>
TITLE: Sequences of evenly-distributed points in a product of intervals
QUESTION [8 upvotes]: Let φ be the golden ratio, (1+√5)/2. Taking the fractional parts of its integer multiples, we obtain a sequence of values in (0,1) which are in some sense "evenly distributed" in a way which is due to the continued fraction form of φ, making the constant "as difficult as possible" to approximate using rational values (otherwise, the values in the sequence would cluster around multiples of such rational approximations). If one takes the first n values, especially if n is a Fibonacci number, they will be very evenly spaced; in fact, if n is a Fibonacci number, then the difference between two consecutive values (after ordering) is always one of two adjacent powers of φ, in correspondence with the fact that the Fibonacci numbers themselves are roughly of the form φk/√5.
Is there any related (or otherwise?) sequence of values in (0,1)d, where d > 1, which are similarly "evenly distributed"?
Edit: I've been a bit unclear about the way in which φ is "special", so I'll try to elucidate. My motivation was that, as drvitek says, φ has no "better-than-expected" rational convergents. So when nφ (mod 1) is plotted against n, not only is the entire set of residues uniformly distributed on (0,1) but also "locally" we have a roughly-uniform distribution on (0,1) × N. This property marks φ out as "special" compared with most irrational numbers. I'm afraid I'm not sure how to phrase it more precisely than that.

REPLY [3 votes]: Part 1. Equidistribution.
As mentioned in the comments, the equidistribution theorem states that any irrational value will produce an equidistributed sequence. That is, in the limit as $n \rightarrow \infty$, all finite subintervals of $(0,1)$ are equally likely. 
However, as you have mentioned, equidistribution is quite a weak criterion and does not necessarily imply that for finite $n$ the sequence will fill the interval 'evenly' without large gaps or clusters. 
Part 2. The Kronecker low discrepancy sequence in 1-dimension
Sequences that satisfy this stronger criterion of 'evenness' are called low discrepancy sequences. There are many ways to construct low discrepancy sequences, including the van der Corput / Halton sequences, Niederreiter and Sobol sequences to name a few. However, by far the most simple to construct are the Kronecker (sometimes called Weyl) additive recurrence sequences, which are defined exactly as you describe in your question. That is, 
$$ x_n =  \lfloor \alpha n\rfloor =  \alpha n \; (\textrm{mod} \; 1), \quad n=1,2,3,... $$
It turns out that only certain values of $\alpha$ produce low discrepancy sequences. For number theoretic reasons, these special values of $\alpha$ are also correspond to numbers that are badly approximable (by the rationals). Intuitively one can say that the more badly approximable a number is by the rationals, the 'more irrational' it is. This is why the golden ratio, which is the most badly approximable number, is often described as the most irrational number.  The more irrational the value of $\alpha$ is, the better ('more even') the low discrepancy sequence will be. Thus, the golden ratio $\varphi  = \frac{1+\sqrt{5}}{2} $ produces the optimal low discrepancy sequence.  
Two things worth mentioning.
First, is that any value of $\alpha$ related via the moebius transformation, 
$$ \alpha = \frac{a+b\varphi}{c+d\varphi} \quad \textrm{for integers} \;\;a,b,c,d \;\; \textrm{where} \;  |ad-bc|=1.$$
will also be optimal.
And secondly, $\alpha = 1+\sqrt{2} $ turns out to be the next most badly approximable number -- and therefore the next most optimal value for the construction of a low discrepancy sequence.  
Springborn as well as Spalding give number theoretic reasons why this is the second most badly approximable number, and also why $\alpha = \frac{1}{10} (9+\sqrt{221}$ is the third-most badly approximable number.
Interestingly the continued fraction for these three values are:
$$ \varphi  = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}} = [1,1,1,1,...]$$
$$ 1+\sqrt{2} = 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}} = [2,2,2,2,2,2] $$
$$ \frac{1}{10}(9+\sqrt{221}) = 2+\frac{1}{2+\frac{1}{1+\frac{1}{1+...}}} = [2,2,1,1,2,2,1,1,2,2,1,1,..] $$
Part 3. Kronecker low discrepancy sequences in higher dimensions
Regarding the main part of your question as to if there is an equivalent process for higher dimensions, the answer is a resounding 'Yes!'.
Some visualizations of various two dimensional low discrepancy sequences can be seen here.
The $d$-dimensional Kronecker sequence is a natural extension of the 1-dimensional case. That is, for a given constant $\pmb{\alpha} = (\alpha_1, \alpha_2,...,\alpha_d)$, the infinite sequence of $d$-dimensional vectors $\pmb{x}_n \in U[0,1]^d$ defined as follows:
$$ \pmb{x}_n =  n\pmb{\alpha}  \; (\textrm{mod} \; 1), \quad n=1,2,3,... $$
is equidistributed if $\alpha_i$ are all independently irrational numbers. Furthermore, in many cases if $\alpha_i$ are the square roots of prime numbers, than the sequence will be a low discrepancy sequence. Unfortunately, it is non-trivial to know in advance if a particular set of prime numbers will result in a low discrepancy sequence.
For example, $\pmb{\alpha} = (\sqrt{2}, \sqrt{3})$ is not great, but as seen in the diagram above  $\pmb{\alpha} = (\sqrt{3}, \sqrt{7})$ is quite good.
Thankfully, in most applications, for moderately large $d$, the construction of such a $d$-dimensional low discrepancy sequence that is based on the square roots of the first $d$ prime numbers, produces quite good results. 
Part 4. Generalizing the Golden ratio
My blog post "The unreasonable effectiveness of quasirandom sequences", describes another method that produces significantly better and more reliable results. In this case, the $d$-dimensional constant vector $\pmb{\alpha}$ is based on a special value $\phi_d$ that is a $d$-dimensional generalisation of the golden ratio, $\phi = \frac{1+\sqrt{5}}{2}$.
That is, 


$$ \pmb{t}_n =  n \pmb{\phi}  \; (\textrm{mod} \; 1),  \quad n=1,2,3,... $$
    where
    $$ \pmb{\phi} =(\frac{1}{\phi_d^1}, \frac{1}{\phi_d^2},\frac{1}{\phi_d^3},...\frac{1}{\phi_d^d}), $$
$$ \textrm{and} \; \phi_d\ \textrm{is the smallest value of } \; x>0 \; \textrm{such that} $$
$$ x^{d+1}\;=x+1. $$


For $d=1$,  $ \phi_1 = 1.618033989... $, which is the canonical golden ratio.
For $d=2$, $ \phi_2 = 1.3247179572... $, which  is often called the plastic constant or silver ratio, and has some beautiful properties. This value was conjectured to most likely be the optimal value for a related two-dimensional problem [Hensley, 2002]. Jacob Rus has posted a beautiful interactive/dynamic  visualization of this 2-dimensional low discrepancy sequence, which can be found here.
For $d=3$, $ \phi_3 = 1.2207440846... $
Summary
Here is some Python code to create the $d$-dimensional generalization of the golden ratio-based Kronecker sequence that you cite in your question.
# Use Newton-Rhapson-Method to calculate \phi_d
def gamma(d):
    x=1.0000
    for i in range(20):
        x = x-(pow(x,d+1)-x-1)/((d+1)*pow(x,d)-1)
    return x

d=3
n=100

g = gamma(d)
alpha = np.zeros(d)                 
for j in range(d):
    alpha[j] = pow(1/g,j+1) %1
z = np.zeros((n, d))    
seed = 0.5  #Optional seed value.
for i in range(n):
    z = (seed + alpha*(i+1)) %1

print(z)

Hope that helps!<|endoftext|>
TITLE: Hausdorff dimension of products of normal subgroups
QUESTION [7 upvotes]: Let $G$ be a metric group, and let $h$ be the associated Hausdorff dimension function on subsets of $G$.  (See for instance Barnea and Shalev, Hausdorff dimension, pro-p groups and Kac-Moody algebras, Trans. AMS 1997.)  When do we have $h(AB) = h(A) + h(B) - h(A \cap B)$ for normal subgroups of $G$?  If this property fails, is there still a general way to use $h$ (or some similar 'dimension' function) to construct a pseudometric on the lattice of normal subgroups of $G$?  What I am looking for here are some informative examples of bad behaviour, particularly for profinite groups.

REPLY [4 votes]: In the first conference I ever went to Slava Grigorchuk asked me a similar question and I didn’t have an answer. But when I have got back to Jerusalem I have talked with Elon Lindenstrauss about it and he suggested the following easy counterexample. Take $G=\mathbb{F}_p[[t]]$. Pick $S$ to be a subset of the integers with density one and with infinite complement $T$. Say $S=\left\{ n_i  \right\}$ and $T=\left\{ m_i  \right\}$. Take $A=\overline{< t^{n_i}>}$ and take $B=\overline{\left< t^{n_i} +t^{m_i} \right>}$. Cleary, $AB=G$, $h(A)=h(B)=1$, but $A \cap B=\emptyset$.
Now, $G$ is not finitely generated, if you would like to have a counterexample which is finitely generated, then you can take $G=SL_d(F_p[[t]])$ and construct in a similar way to the above $A$ which is made from upper triangular matrices and $B$ which is made from lower triangular matrices. However, $A$ and $B$ will not be normal any more.
I am not familiar with a counterexample in which $A$ and $B$ are normal and $G$ is finitely generated. I am also not familiar with a counterexample in which $A$ and $B$ are finitely generated. But as you can deduce from my story above this does not mean much.<|endoftext|>
TITLE: Zariski-style valuation theory
QUESTION [8 upvotes]: I've been trying to read some of Zariski's older works, and I'm having some trouble getting into his mindset. I'd appreciate some help with this.
To quote Zariski (in "normal varieties and birational correspondences"): "A zero-dimensional valuation, that is a place, with center at a point $P$, corresponds to a way of approaching P along some one-dimensional branch, which may be algebraic, analytic, or transcendental. Similarly an $s$-dimensional valuation with an $s$-dimensional center $W$ corresponds to a way of approaching $W$ along an $s+1$-dimensional branch through $W$."
This is leagues away from how I think of valuations. Can you help me fill in the gap?
Also, and more trivially, how do DVR's, the main valuative object that is normally discussed nowadays, fit in his theory?

REPLY [7 votes]: I'm not an expert here, but I'll try. Let $X$ be a proper irreducible variety and $K$ the field of meromorphic function on $X$. Let $v : K^* \to A$ be a valuation, where $A$ is a totally ordered abelian group. Recall that the valuation ring $R$ is $v^{-1}(A_{\geq 0})$, with maximal ideal $\mathfrak{m} = v^{-1}(A_{> 0})$. So $R$ is a local ring with fraction field $K$. By the valuative criterion of properness, we get a map $\mathrm{Spec} \ R \to X$; the image of the closed point is called the "center" of $v$. We denote the center by $Z$.
I'll start with the simplest cases, and move to the more general. The easiest case is that $X$ is defined over an algebraically closed field $k$, the point $Z$ is a closed point of $X$, and $A$ is $\mathbb{Z}$. Then the completion of $R$ at $\mathfrak{m}$ is isomorphic to $k[[t]]$; choose such an isomorphism. If $x_1$, $x_2$, ... $x_N$ are coordinates near $z$ then this isomorphism identifies $z_i$ with a power series $\sum x_i^j t^j$. 
Intuitively, we can think of $(x_1(t), x_2(t), \ldots, x_N(t))$ as giving an analytic map from a small disc $D$ into $X$, taking the origin to $Z$. (This interpretation makes rigorous sense if we are working over $\mathbb{C}$, and the power series $x_i$ converge somewhere.) The image of this map is the "branch" which Zariski speaks of. The valuation can be thought of as restricting a function to this branch and seeing to what order it vanishes at $Z$. Notice that, if there is a polynomial relation $f(x_1(t), \ldots, x_N(t))=0$ then we should have $v(f)=\infty$ and $v(1/f)= - \infty$. If you don't allow this then you need to require that the $x_i$ be algebraically independent; intuitively, this is the same as saying that the branch does not lie in any polynomial hypersurface. Since Zariski allows "algebraic" branches, I assume he IS permitting this and has some appropriate convention to deal with it.
Now, what if $A$ is a subgroup of $\mathbb{R}$, but is no longer discrete? Then $R$ is going to embed in some sort of Puiseux field. The details here can be subtle, but the intuition should be that the $x_i(t)$ can have real exponents. For example, if $K=k(x,y)$, and $v(f(x,y))$ is the order 
of vanishing of $f(t, t^{\sqrt{2}})$ at $t=0$, then we can think of $v$ as the order of vanishing along the branch $(t, t^{\sqrt{2}})$. Maybe that is what Zariski means by a transcendental branch?? 
Suppose now that $Z$ is the generic point of an $s$-dimensional variety. Let $L$ be the field of functions on $Z$ and suppose, for simplicity, that $L$ and $K$ have the same characteristic. If $A=\mathbb{Z}$ then $R$ embeds in $L[[t]]$. Again, we can take coordinates $x_1$, ..., $x_N$ and write them as power series. How to think of these power series? One way is to think of them as giving a map $U \times D$ into $X$, where $D$ is again a small disc and $U$ is a dense open subset of $X$. The image is the $(s+1)$-dimensional branch which Zariski discusses. The valuation is to restrict to this branch and work out the order of vanishing of this restriction along $U \times \{ 0 \}$.
Now, all of this is discussing the case where $A$ embeds (as an ordered group) in $\mathbb{R}$. In general, of course, there are more valuations. For example, let $A=\mathbb{Z} \times \mathbb{Z}$ ordered lexicographically. Let $v:k[x,y] \to A$ send a polynomial to its lowest degree monomial, and extend this to a valuation on $k(x,y)$. The way I would think of that is that we have a little disc near $(0,0)$, and a little curve passing through $(0,0)$ along the $x$-axis. So our valuation is to, first, restrict to the curve and the order of vanishing at the point and, second, to restrict to the surface and take the order of vanishing along the curve. So here we have a flag of branches, not just one. I'm not sure why your Zariski quote doesn't discuss this possibility.<|endoftext|>
TITLE: Why do statistical randomness tests seem so ad hoc?
QUESTION [26 upvotes]: Wikipedia describes Kendall and Smith's 1938 statistical randomness tests like this:


The frequency test, was very basic: checking to make sure that there were roughly the same number of 0s, 1s, 2s, 3s, etc.
The serial test, did the same thing but for sequences of two digits at a time (00, 01, 02, etc.), comparing their observed frequencies with their hypothetical predictions were they equally distributed.
The poker test, tested for certain sequences of five numbers at a time (aaaaa, aaaab, aaabb, etc.) based on hands in the game poker.
The gap test, looked at the distances between zeroes (00 would be a distance of 0, 030 would be a distance of 1, 02250 would be a distance of 3, etc.).


It is not obvious to me that these four particular tests were chosen with any deep understanding of how best to detect nonrandomness. Rather, it seems each one was probably chosen for simplicity.
Well, it's easy to see why early work in the field would be like that. But fast forward to 1995 and George Marsaglia's Diehard tests seem, on the surface, just as ad hoc:


Birthday spacings: Choose random points on a large interval. The spacings between the points should be asymptotically Poisson distributed. The name is based on the birthday paradox.
Overlapping permutations: Analyze sequences of five consecutive random numbers. The 120 possible orderings should occur with statistically equal probability.

(...and so on)

It is not obvious to me that these tests are really independent (i.e. that none is entirely redundant, rejecting only sequences also rejected by at least one of the other tests), or that there aren't obviously better generalizations of them, much less that they were chosen as a set to try to cover any particular space efficiently.
Ideally, a test suite would be designed to reject as many sequences of low Kolmogorov complexity as possible with minimal computation and false alarms. Is a more theoretical approach to this possible? Why hasn't it happened? —Or is there more to the state of the art than meets the eye?

REPLY [18 votes]: It's not clear that Marsaglia's tests are really good enough. See this Stack Overflow discussion.
Kolmogorov complexity is not the right criterion for statistical randomness tests, since any pseudorandom sequence has low Kolmogorov complexity. What you really want in a random number generator is for the sequence to be computationally pseudorandom; that is, without knowledge of the seed, no polynomial-time test can distinguish the sequence from a truly random sequence.  
In fact, there are a number of random number generators which are believed to be computationally pseudorandom. Nobody uses these, however, partly because they are computationally too expensive, and partly because of inertia and partly because the existing pseudorandom generators we have are good enough most of the time. A while ago, for one of the most common methods of pseudorandom number generation (linear congruential), I ran into cases where they unexpectedly produced the wrong answer.
Marsaglia's tests were developed over a number of years, and I believe each was designed to detect certain flaws that many pseudorandom number generators contained at the time. Once good-enough pseudorandom number generators were available, nobody bothered creating a more stringent series of tests.<|endoftext|>
TITLE: What is your favorite isomorphism?
QUESTION [19 upvotes]: The other day I was trying to figure out how to explain why isomorphisms are important. I pulled Boyer's A History of Mathematics off the bookshelf and was surprised to find that isomorphism isn't even listed in its index. The Wikipedia article on isomorphisms only gives two concrete examples.
There are many surprising, significant, classic isomorphisms. I'll refrain from giving examples. What are your favorites?
As usual, please limit yourself to one isomorphism per answer.
(Related: your favorite surprising connections in mathematics. But this question is looking for more concrete examples, particularly those that illustrate the power of the idea.)

REPLY [4 votes]: I like the isomorphism between a finite abelian group and its "Cartier" dual (not the bidual!) precisely because it's non-canonical. But I don't think it makes a good example for explaining isomorphism to non-mathematicians.<|endoftext|>
TITLE: Quotient of abelian variety by an abelian subvariety
QUESTION [12 upvotes]: Let $k$ be a field and $A$ an abelian variety over $k$.  Suppose that $B$ is an abelian subvariety of $A$.  Consider the following fact:
There exists an abelian variety $C$ over $k$ and a surjective morphism $A\twoheadrightarrow C$ with kernel exactly $B$.  
This is proved in section 9.5 of the book "Abelian varieties, theta functions and the Fourier transform" By Alexander Polishchuk on the way to proving Poincare reducibility.  The proof there seems (to me) to be a bit complicated, so I'm wondering if anyone knows of a "simple" proof. I think I could probably devise a proof of the above fact using Poincare reducibility (employing the proof of the latter result in Milne's chapter of Cornell-Silverman, Proposition 12.1 to avoid circular logic), but somehow I'm not so satisfied by this as it seems like it ought to be an "easy" fact.

REPLY [17 votes]: Let us work over $\mathbb{C}$.
The inclusion $u \colon B \to A$ induces a surjection $\hat{u} \colon A^{\vee} \to B^{\vee}$.
By general facts on Abelian varieties, the kernels of $u$ and $\hat{u}$ have the same number of connected components. Since $u$ is injective, its kernel is trivial, so it follows $\ker \hat{u}=(\ker \hat{u})_0$; in other words $\ker \hat{u}$ is an Abelian subvariety of $A^{\vee}$.
Therefore we have an exact sequence of Abelian varieties
$$0 \to \ker \hat{u} \to A^{\vee} \to B^{\vee}  \to 0.$$ By dualizing it, we obtain $$0 \to B \to A \to (\ker \hat{u})^{\vee} \to 0,$$
that is $C = (\ker \hat{u})^{\vee}$.<|endoftext|>
TITLE: Generalising the sphere-projective relationship to the flag manifold setting
QUESTION [5 upvotes]: As is well known, $CP^{N-1}$ is the base space of the principal bundle $SU(N)$ with fibre $U(N-1)$. Moreover, $S^{2N-1}$ is the base space of the principal bundle $SU(N)$ with fibre $SU(N-1)$. Finally, $CP^{N-1}$ is the base space of the principal bundle $S^{2N-1}$ with fibre $U(1)$. What is interesting is that all the line bundles of $CP^{N-1}$ arise as $U(1)$-associated bundles to $S^{2N-1}$. 
My question is, does this generalise to the setting of generalised flag manifolds $M=G/P$, where $P$ is a parabolic manifold and $G$ a semi-simple algebraic group? That is, if we take $P$ as generalising $U(N-1)$, $G$ as generalising $SU(N)$, and $M$ as generalising $CP^{N-1}$, what are the generalisations $X$ and $H$ of $S^{2N-1}$ and $U(1)$ so that all the line bundles of $M$ arise as $H$-associated bundles to $X$, or do such objects even exist?

REPLY [3 votes]: I would like to add a further analogy and an application for the case of the complex complete flag manifold $G/B$. In this case $ X = G = SU(N)$ is isomorphic (as a homogeneous space) to its principal homogeneous space: the Stiefel manifold $V_{n-1}(\mathbb{R}^n)$, as given in the following wikipedia page.
In both cases of the sphere and the Stiefel manifold, they can be given as (intersection of) quadrics. in $\mathbb{C}^n$ (the stiefel manifold can be identified with the space of full rank $n\times (n-1)$ matrices satisfying $\bar{M}^t M = 1$). Of course, the case of the standard Hopf fibration (X = S^3, M = S^2) is mutual to both cases.
This construction can be applied to perform integrations over the $SU(n)$ invariant measures
of the complex projective spaces and complete flag manifolds: Integration over spheres and Stiefel manifolds is relatively straightforward because they can be performed on quadric constraint surfaces in $\mathbb{C}^n$. The integrals can be converted to (a series of) Gaussian integrals by means of Fourier transforms.
Functions on the complex projective spaces and complete flag manifolds can be extended to functions on the total manifolds, the sphere and the Stiefel manifold by defining them to be constants along the torus fibers.
In the case of the complex projective spaces, the sphere is identified with the space of 
unit vectors in $\mathbb{C}^n$ and the complex projective space with the projectors onto these vectors. Thus any function on the sphere which depends solely on the projectors of the unit vector is an extension of a function on the complex projective space and its integral on the sphere is proportional to the integral of the original function on the complex projective space.
In the case of the complete flag manifold, the Stiefel manifold may be considered as the space of orthonormal frames in $C^n$, (the column vectors of $M$) and the flag manifold as the space of projectors onto these vectors. Again, functions on the Stiefel manifold depending solely on the projectors of the frame vectors are extensions of functions on the flag manifolds.<|endoftext|>
TITLE: Representability of polymatroids over $GF(2)$
QUESTION [8 upvotes]: A polymatroid is a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that
1) $d(\varnothing)=0$,
2) $A \subset B$ implies $d(A) \leq d(B)$, and
3) $d(A \cap B) + d(A \cup B) \leq d(A) + d(B)$ for all $A,B \in P(X)$.
A polymatroid is said to be representable over $GF(2)$ (the field with two elements), if there exists a collection of subvectorspaces $\lbrace V_x \mid x \in X \rbrace$ of $GF(2)^{\oplus d(X)}$, such that
$$d(A) = \dim_{GF(2)} \bigvee_{x \in A} V_x, \quad \forall A \in P(X).$$
(It is clear that any function $d$, which is defined this way is indeed a polymatroid.)
A polymatroid is called matroid if $d(\lbrace x\rbrace)=1$ for all $x \in X$.
Tutte proved that a matroid is representable over $GF(2)$ if and only if the matroid $U_{2,4}$ does not appear as a minor. Here, $U_{2,4}$ is the matroid formed by four points that lie on one line, i.e. the underlying set is $\lbrace x_1,\dots,x_4 \rbrace$ and $d(\lbrace x_i,x_j\rbrace) = 2$ for $i \neq j$ and $d(\lbrace x_1,\dots,x_4\rbrace)=2$. (The necessity of this condition is obvious, since $GF(2)^{\oplus 2}$ has only $3$ non-zero elements.)

Question: Is there any useful characterization of the representability of a polymatroid over $GF(2)$ ? What is known about this question ?

REPLY [5 votes]: Here's a construction due to Stefan van Zwam. 
Let $U_{2,4}$ be a 4-point line with ground set $[4]$, and rank function $r$.  For each $A \subset [4]$, let $\chi(A)$ be 1, if $1 \in A$, and 0 if $1 \notin A$.  For each $k \in \mathbb{N}$, let $S_k$ be the polymatroid with ground set $[4]$ and rank function $r+k\chi$.  
Lemma. $S_k$ is not representable over $\mathbb{F}_2$, for every $k$.
Proof.  Let $(V_i : i \in [4])$ be a representation of $S_k$ over $\mathbb{F}_2$.  Choose a basis $B_i$ for each $V_i$.  Since {1} has rank $k+1$ and {1,2,3,4} has rank $k+2$, we may assume that $B_1$ consists of the first $k+1$ standard basis vectors in $\mathbb{F}_2^{k+2}$.  Now, since {1,2}, {1,3}, and {1,4} all have rank $k+2$, it follows that $B_2, B_3$, and $B_4$ all must have last coordinate equal to 1.  In particular, $B_2 + B_3 +B_4 \neq 0$.   Also, since every two element subset of {2,3,4} has rank 2, it follows that $B_2, B_3$ and $B_4$ are distinct.  But now {2,3,4} must have rank 3 in $S_k$, a contradiction.  $\square$
Moreover, it is easy to check that every minor of $S_k$ is representable over $\mathbb{F}_2$, where minors of polymatroids are defined in the obvious way. Therefore, the set of binary polymatroids has an infinite set of excluded minors.<|endoftext|>
TITLE: A counter example to Hahn-Banach separation theorem of convex sets.
QUESTION [15 upvotes]: I'm trying to understand the necessity for the assumption in the Hahn-Banach theorem for one of the convex sets to have an interior point. The other way I've seen the theorem stated, one set is closed and the other one compact. My goal is to find a counter example when these hypotheses are not satisfied but the sets are still convex and disjoint. So here is my question:
Question: I would like a counter example to the Hahn-Banach separation theorem for convex sets when the two convex sets are disjoint but neither has an interior point. It is trivial to find a counter example for the strict separation but this is not what I want. I would like an example (in finite or infinite dimensions) such that we fail to have any separation of the two convex sets at all. 
In other words, we have $K_1$ and $K_2$ with $K_1 \cap K_2 = \emptyset$ with both $K_1$ and $K_2$ convex belonging to some normed linear space $X$. I would like an explicit example where there is no linear functional $l \in X^*$ such that $\sup_{x \in K_1} l(x) \leq \inf_{z \in K_2} l(z)$.
I'm quite sure that a counter example cannot arise in finite dimensions since I think you can get rid of these hypotheses in $\mathbb{R}^n$. I'm not positive though.

REPLY [2 votes]: A counterexample can be given even if both set are assumed to be closed, for any non-reflexive Banach space. See a proof of Klee's theorem 
http://www.johndcook.com/SeparationOfConvexSets.pdf 
Since in a reflexive space there exists a separation of two closed convex sets provided that one of them is bounded, I think above result is the best answer on your question.<|endoftext|>
TITLE: Other norms for Lattice reduction techniques (LLL, PSLQ)?
QUESTION [8 upvotes]: LLL and other lattice reduction techniques (such as PSLQ) try to find a short basis vector relative to the 2-norm, i.e. for a given basis that has $ \varepsilon $ as its shortest vector, $ \varepsilon \in \mathbb{Z}^n $, find a short vector s.t. $ b \in \mathbb{Z}^n, ||b||_2 < ||c^n \varepsilon||_2 $.
Has there been any work done to find short vectors based on other, potentially higher, norms?  Is this a meaningful question?

REPLY [5 votes]: There is an LLL analogue for arbitrary norms; the original paper by Lovász and Scarf can be found here. I recently found a bachelor thesis on lattice reduction in infinity norm, which contains several other references (for example, work by Kaib and Ritter).<|endoftext|>
TITLE: nth-order generalizations of the arithmetic-geometric mean
QUESTION [15 upvotes]: The arithmetic-geometric mean,
$a_{k+1}=\frac{a_k+b_k}{2} \quad b_{k+1}=\sqrt{a_k b_k}$
is one of the celebrated discoveries of Gauss, who found out that it is equivalent to computing a (complete) elliptic integral (which is a special case of the Gauss hypergeometric function ${}_2 F_1$).
I have been wondering if nth-order generalizations of the iteration,
$a_{k+1}=\frac{a_k+b_k}{n} \quad b_{k+1}=\sqrt[n]{a_k b_k}$
have ever been systematically studied. I've seen this paper by Borwein, but have had trouble searching for other papers. In particular, I'm interested if the coupled sequences also have a common limit, and if so, whether the limit is expressible as a hypergeometric function (or generalizations like those of Appell or Lauricella).
Another possible generalization I thought involves $n$ variables and makes use of the elementary symmetric polynomials. To use $n=4$ as an example:
$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$
$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{3}}$
$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{2}}$
$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$
Would these four sequences (and in general the $n$ sequences) tend to a common limit $F(a_0,b_0,c_0,d_0,\dots)$ like in the $n=2$ case, and if so, are they expressible in terms of known functions?

EDIT
Taking into account Darsh Ranjan's comments, I realized that what I should be looking at instead is the generalization whose denominators are binomial coefficients (thus, the general form $\sqrt[j]{\frac{e_j}{\binom{n}{j}}}$, for $j=1\dots n$ where $e_j$ is the jth elementary symmetric polynomial). The case $n=4$ now looks like
$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$
$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{6}}$
$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{4}}$
$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$
So, still the same question: is there a common limit, and if so, is the limit expressible in terms of known functions?

REPLY [12 votes]: Gauss's hypergeometric formula for the AGM can also be interpreted in terms of a complete elliptic integral
$\int_0^{\pi/2} \phantom. d\theta / \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta}$.  There's a remarkable generalization to complete hyperelliptic integrals in genus 2 arising from a four-variable AGM:
$$
(a,b,c,d) \mapsto \left(
  \frac14(a+b+c+d),
\frac12\bigl(\sqrt{ab}+\sqrt{cd}\phantom.\bigr),
\frac12\bigl(\sqrt{ac}+\sqrt{bd}\phantom.\bigr),
\frac12\bigl(\sqrt{ad}+\sqrt{bc}\phantom.\bigr)
\right)
$$
(whose specialization $a=b$, $c=d$ recovers the usual AGM).  One source available online is

Jarvis, Frazer: Higher genus arithmetic-geometric means, Ramanujan J. 17 (2008), 1–17.<|endoftext|>
TITLE: Convex hull of $k$ random points
QUESTION [12 upvotes]: Suppose we have $k$ realizations of a random variable uniformly distributed over the unit cube $[0,1]^n$.
What is the probability that their convex hull has all of the $k$ points as extreme points?
If it would be easier, "unit cube" can be replaced by "unit ball".

REPLY [2 votes]: The number of $k$-dimensional faces $f_k$ on a random polytope is well studied subject, and you are asking about the $k=0$ case. The distributions that have probably received the most attention are uniform distributions on convex bodies and the standard multivariate normal (Gaussian) distribution.  As Gjergji mentioned, Bárány has some of the strongest results in this area.  In particular Bárány and Vu proved central limit theorems for $f_k$.
This Bulletin survey article is a good place to start.
One amusing point worth noting:  if you look at uniform distributions on convex bodies, the answer will change drastically depending on the underlying body.  The convex hull of random points in a disk, for example, will have many more points than the convex hull of random points in a triangle.<|endoftext|>
TITLE: How big is the sum of smallest multinomial coefficients?
QUESTION [18 upvotes]: Given positive integers $n$ and $d$, let $S$ indicate the list of all $d$-tuples of non-negative integers $(c_1,\ldots,c_d)$ such that $c_1+\cdots+c_d=n$. Let $v_i$ be the value of the multinomial coefficient corresponding to $i$'th tuple in $S$, ie
$$v_i=\frac{n!}{c_1!\cdots c_d!}$$
What can we say about the sum of smallest coefficients, ie, the value of the following?
$$s(B)=\sum_{v_i < B} v_i$$
Motivation: upper bounds on multinomial tails would allow to give non-asymptotic error bounds for various learning algorithms
Update: 09/03
Here are all the relevant theoretical results I found so far. Let $B=\frac{n!}{c_1!\cdots c_d}$, $C=\max_i v_i$, $k=\min_i c_i$. Then for even n and $d=2$ the following are known to hold 
$$s(B)<\frac{B}{C} 2^n$$
Proof under Lemma 3.8.2 of Lovasz et al "Discrete Mathematics" (2003)
$$s(B)\le 2^n \exp(-\frac{(n/d-k)^2}{n-k})$$
Proof under Theorem 5.3.2 of Lovasz et al "Discrete Mathematics" (2003)
$$s(B)\le 2B(\frac{n-(k-1)}{n-(2k-1)}-1)$$
Michael Lugo gives outline of proof in another MO post 
$$s(B)<2(\exp(n \log n - \sum_i c_i \log c_i)-B)$$
Proof under Lemma 16.19 of Flum et al Parameterized Complexity Theory (2006)
To be practically useful for my application, these bounds need to be tight for tails, ie, for sums that are less than $d^n/10$. Here's a plot of logarithm of bound/exact ratio for such sums. X-axis is monotonically related to B.
 (source)
You can see that Michael Lugo's bound is by far the most accurate in that range.
Out of curiosity, I "plugged in" bounds above for sums of higher dimensional coefficients.

 (source)

 (source)

Mathematica notebook.

REPLY [2 votes]: If I'm not mistaken, you at least have the following bound:
$$
s(B) \ \; \leq\ \; d^n \,-\, \exp \left( -2 \left( \frac{\ln B}{\ln ( n! ) - d\ln ( (n/d)! )} \right)^2  \right) \,d^{n} .
$$
You may want to do a little fiddling around with Stirling's approximation to reduce the denominator in the fraction a bit further, but I didn't want to weaken the bound more than necessary.
I found this by upper-bounding the probability that a multinomial distribution with uniform bin probabilities yields an "atypical" observation, using Hoeffding's inequality. "Atypical" is here defined in the information-theoretic sense: An atypical observation is one whose logarithmic probability falls short of the expected logarithmic probability over the whole distribution (see Cover and Thomas' Elements of Information Theory, particularly chapter 11).
Some more details:
Let $p$ be point probabilities for a multinomial distribution with uniform bin probabilities:
$$
p(c) \ =\ \left( \frac{n!}{c_1!c_2!\cdots c_d!} \right) d^{-n}.
$$
Notice that
$$
\ln p(c) \,+\, n\ln d \,\ =\ \ln \frac{n!}{c_1!c_2!\cdots c_d!},
$$
and thus
$$
\ln p(c) \,+\, n\ln d \;\leq\; \ln B \;\quad \iff\quad  \frac{n!}{c_1!c_2!\cdots c_d!} \;\leq\; B.
$$
Further, the entropy of the flat multinomial distribution is less than the entropy of $n$ samples from the flat categorical distribution with $d$ bins: The categorical includes order information as well as frequency information, while the multinomial only includes frequency information.
We thus have the following bound on the expected value of $-\ln p(c)$:
$$
E \left[\, -\ln p(c)\, \right] \ \leq\ 
n \cdot H\left( \frac{1}{d}, \frac{1}{d}, \ldots, \frac{1}{d}  \right)
\ =\ n\ln d,
$$
or in other words, $-n\ln d \leq E[\ln p(c)]$.
Further, the minimum and maximum values of the random variable $\ln p(c)$ are
$$
a
\; =\; 
\min_c \ln p(c)
\; =\; 
\ln \frac{n!}{n!\, 0!\, 0!\, \cdots\, 0!} d^{-n}
\; =\;
-n\ln d;
$$
$$
b
\; =\; 
\max_c \ln p(c)
\; =\; 
\ln \frac{n!}{(n/d)!\, (n/d)!\, \cdots\, (n/d)!} d^{-n}
\; =\;
\ln ( n! ) - d\ln ( (n/d)! ) - n\ln d.
$$
The squared distance between these two extremes is consequently
$$
(a - b)^2 \ =\ \left( \ln ( n! ) - d\ln ( (n/d)! ) \right)^2.
$$
We can now make the following comparisons:
\begin{eqnarray}
\Pr\left\{ \frac{n!}{c_1!c_2!\cdots c_d!} < B \right\}
& = &
\Pr\left\{ \ln p(c) \,+\, n\ln d < \ln B \right\}
\\
& \leq &
\Pr\left\{ \ln p(c) \,-\, E[\ln p(c)] < \ln B \right\}
\\
& = &
1 \,-\, \Pr\left\{ \ln p(c) \,-\, E[\ln p(c)] \geq \ln B \right\}
\\
& \leq &
1 \,-\, \exp \left( -2 \left( \frac{\ln B}{\ln ( n! ) - d\ln ( (n/d)! )} \right)^2 \right).
\end{eqnarray}
The first inequality follows from the fact that a proposition always has a lower probability than its logical consequences; the second is the application of Heoffding's inequality. 
We thus have
$$
\Pr\left\{ \frac{n!}{c_1!c_2!\cdots c_d!} < B \right\}
\ \leq \ 
1\;-\;\exp \left( -2 \left( \frac{\ln B}{\ln ( n! ) - d\ln ( (n/d)! )} \right)^2 \right).
$$
By multiplying by $d^n$, we obtain the inequality stated above, since the probability of a sample from a flat multinomial distribution is equal to the corresponding multinomial coefficient divided by $d^n$.<|endoftext|>
TITLE: Easy special cases of the decomposition theorem?
QUESTION [5 upvotes]: The decomposition theorem states roughly, that the pushforward of an IC complex,
along a proper map decomposes into a direct sum of shifted IC complexes.
Are there special cases for the decomposition theorem, with "easy" proofs?
Are there heuristics, why the decomposition theorem should hold?

REPLY [5 votes]: Well, it depends on what you mean by "easy". A special case, which I find very instructive, is a theorem of Deligne from the late 1960's.

Theorem. $\mathbb{R} f_*\mathbb{Q}\cong \bigoplus_i R^if_*\mathbb{Q}[-i]$, when $f:X\to Y$  is a smooth projective morphism of varieties over $\mathbb{C}$. (This holds more generally with $\mathbb{Q}_\ell$-coefficients.)
Corollary. The Leray spectral sequence degenerates.

The result was deduced from the hard Lefschetz theorem.
An outline of a proof (of the corollary) can be found in Griffiths and Harris.
It is tricky but essentially elementary.
A much  less elementary, but more conceptual argument, uses
weights. Say $Y$ is smooth and projective, then
$E_2^{pq}=H^p(Y, R^qf_*\mathbb{Q})$ should be pure of weight $p+q$ (in the sense of Hodge
theory or $\ell$-adic cohomology). Since
$$d_2: E_2^{pq}\to E_2^{p+2,q-1}$$
maps a structure of one weight to another it must vanish. Similarly for higher differentials.
If $f$ is proper but not smooth, the decomposition theorem shows that $\mathbb{R} f_*\mathbb{Q}$ decomposes into sum of translates of intersection cohomology complexes.
This follows from more sophisticated purity arguments (either in the $\ell$-adic setting as in BBD, or the Hodge theoretic setting in Saito's work).
There is also a newer proof due to de Cataldo and Migliorini which seems a bit more geometric.
I have been working through some of this stuff slowly. So I may have more to say in a few months time. Rather than updating this post, it may be more efficient for the people
interested to check
 here  periodically.<|endoftext|>
TITLE: Gödel's Incompleteness Theorem and the complexity of arithmetic
QUESTION [13 upvotes]: In How complicated can structures be? Jouko Väänänen says:

The guiding result of mathematical logic is the Incompleteness Theorem of Gödel,
  which says that the logical structure of number theory is so complicated that it cannot be
  effectively axiomatized in its entirety. In other
  words, the theory is non-recursive, i.e. there
  is no Turing machine that could tell whether
  a sentence of number theory is true or not.

I've never seen Gödel's Incompleteness Theorem this way: that it's a matter of the overall complexity of the structure of the natural numbers that there are facts about them that cannot be proved. 
So I wonder whether I can take the quote above literally:

Can Gödel's Theorem be rigorously stated in terms
  of complexity? 

Somehow like this: "Every
system which exceeds complexity
threshold X is undecidable."
Or is it just a vague paraphrase, not to be taken too seriously?

REPLY [2 votes]: Hmm, nobody has mentioned Chaitin's incompleteness theorem.<|endoftext|>
TITLE: Roadmap to learning about Ricci Flow?
QUESTION [23 upvotes]: Hello,
I'm curious to what books etc. one could use to understand the basics of Ricci flow, what areas of math are needed and so? What areas should one specialize in? See it as a roadmap to understanding Ricci flow, or something.Say one wants to be able to read Perelman's proof of the Poincaré Conjecture.
Sorry if my english isn't that good and this seems a bit hurried, I'm on the run.

REPLY [8 votes]: Here is a list of literature which I compiled when I taught the course
on Ricci flow.
Basic differential geometry:
Einstein Manifolds (Besse). 
Riemannian geometry (Gallot S., Hulin D., Lafontaine J.)
Sign and geometric meaning of curvature (Gromov)
http://www.ihes.fr/~gromov/PDF/1%5B77%5D.pdf
Textbooks:
Lectures on the Ricci Flow (2006, 133 pp.) Topping P.
http://www.warwick.ac.uk/~maseq/RFnotes.html 
Hamilton's Ricci Flow  (Chow B., Lu P., Ni L.)
Standard texts:
http://en.wikipedia.org/wiki/Solution_of_the_Poincar%C3%A9_conjecture
(and links therein)
Perelman, Grisha (November 11, 2002). 
The entropy formula for the Ricci flow and its geometric applications.
 Perelman, Grisha (March 10, 2003). 
Ricci flow with surgery on three-manifolds. 
Perelman, Grisha (July 17, 2003). 
Finite extinction time for the solutions to the 
Ricci flow on certain three-manifolds.  
Bruce Kleiner, John Lott.  Notes on Perelman's papers 
Huai-Dong Cao, Xi-Ping Zhu.  Hamilton-Perelman's Proof of
the Poincaré  Conjecture and the Geometrization Conjecture. 
John W. Morgan, Gang Tian.  
Ricci Flow and the Poincaré Conjecture

It's very obsolete (2007), and does not contain much on short-term
existence of solutions of Ricci flow.<|endoftext|>
TITLE: Is there a group whose cardinality counts non-intersecting paths?
QUESTION [15 upvotes]: Introduction
Graphs are not only important combinatorial objects, but also related to many topological/algebraic structures. In this question I am going to talk about various group structures with combinatorial flavor that one can relate to a graph. In what follows all graphs are connected.
The first example that comes to mind is the fundamental group of a graph when viewed as a topological space. This is a free group and the information it carries is the Euler characteristic, or in simpler words the difference between the number of edges and vertices.
The second example is the critical group of a graph (also called the Sandpile group, Picard group and Jacobian group by different authors) which we can denote as $\mathcal{K}(G)$ for a graph $G$. If we let $L(G)$ be the Laplacian of $G=(V,E)$ this group is nothing more than just $$\mathcal{K}(G) := \mathbb{Z}^{V}/L(G)\mathbb{Z}^{V}$$
so maybe another name for it should be "Laplacian cokernel". By Kirchhoff's theorem the order of $\mathcal{K}(G)$ is precisely the number of spanning trees of $G$. A similar situtation occurs when we consider directed graphs instead. The nice thing about this group is that it behaves nicely under quotients. This means that if one has an automorphism $\phi$ of $G$ then $\mathcal{K}(G/\phi)$ is a subgroup of $\mathcal{K}(G)$. This is not hard to prove without using the critical group itself, see my answer here. The bad thing here is that it is impossible to put a "natural" bijection between $\mathcal{K}(G)$ and the set of spanning trees of $G$, however this is asking for too much.
A last example is that one can do something similar in the case of perfect matchings for planar bipartite graphs, and get Kasteleyn cokernels. Here one considers the Kasteleyn matrix instead of the Laplacian, and in some cases can give the group structure explicitly, for example G. Kuperberg in that paper shows that the Kasteleyn-Percus cokernel of the Aztec diamond is $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/2^{n}\mathbb{Z}$.
Question
Can we construct a group whose cardinality counts the number of non-intersecting paths in a directed graph which start and end in specified sources and sinks? Can we say anything about the structure of such a group and have they been studied before? I am particularly interested in subgraphs of $\mathbb{Z}^2$. Also any comments regarding the philosophy of counting objects by passing through a group structure first, are welcome.
Motivation
A positive answer to this question might help in giving a combinatorial proof to this question (see Qiaochu's comment for example). It would be interesting in general to study the subgroup structure of this (hypothetical) group in general.

REPLY [10 votes]: Hi Gjergji.  First, when there is a Gessel-Viennot matrix to count non-intersecting lattice paths, there is also a Gessel-Viennot cokernel.  This will happen when the graph is planar and when the sources are all "on the left" and the sinks are all "on the right".  Otherwise, if you can't find an integer matrix whose determinant counts the families of lattice paths or whatever, then there is no particular reason to believe in a cokernel.
Second, the Gessel-Viennot determinant is not essentially different from the Kasteleyn-Percus determinant.  Every planar non-intersecting lattice path problem can be expressed as a planar perfect matching problem for a modified graph, and the Kasteleyn-Percus matrix reduces to the Gessel-Viennot matrix in a predictable way.  In particular, the cokernels are the same.  This is explained in my paper that you cite.
Third, even the tree group of a planar graph is not all that different from either Gessel-Viennot cokernel or the Kasteleyn cokernel.  Again, you can modify a planar, bipartite graph to turn the matchings into trees.  The tree group is more general in the sense that it does not require planarity.

Actually, although the Gessel-Viennot method is a very nice and very important result, it is something of a social accident that it is (or has at times been) much more popular than the Kasteleyn method in enumerative combinatorics circles.  Kasteleyn published in mathematical physics journals, and his work was known but not widely known among combinatorialists until the 1990s.  Then, he published a more complicated Pfaffian expression that applies to non-bipartite graphs, and this was only simplified by Percus in the bipartite case.  (Although this simplification is easy.)  Then, although Kasteleyn-Percus determinants explain and generalize Gessel-Viennot determinants, the matrices are larger and the more condensed Gessel-Viennot form can look more convenient for explicit calculations.  Then too, unless you are studying cokernels, you might not need to know that Gessel-Viennot matrices can come from Kasteleyn-Percus matrices.
On the other hand, there are some things that Gessel-Viennot matrices do not easily show you.  Three examples:  (1) The same Kasteleyn-Percus matrix may have more than one Gessel-Viennot reduction, which will then have the same cokernel.  (2) The minors of a Kasteleyn-Percus matrix give you edge probabilities, a fact which has been used to great effect by Rick Kenyon.  (3) You may have to discard symmetry to write down the Gessel-Viennot matrix, which can make it more difficult to analyze matchings that are invariant under a group action.<|endoftext|>
TITLE: Demonstrating that rigour is important
QUESTION [204 upvotes]: Any pure mathematician will from time to time discuss, or think about, the question of why we care about proofs, or to put the question in a more precise form, why we seem to be so much happier with statements that have proofs than we are with statements that lack proofs but for which the evidence is so overwhelming that it is not reasonable to doubt them.
That is not the question I am asking here, though it is definitely relevant. What I am looking for is good examples where the difference between being pretty well certain that a result is true and actually having a proof turned out to be very important, and why. I am looking for reasons that go beyond replacing 99% certainty with 100% certainty. The reason I'm asking the question is that it occurred to me that I don't have a good stock of examples myself.
The best outcome I can think of for this question, though whether it will actually happen is another matter, is that in a few months' time if somebody suggests that proofs aren't all that important one can refer them to this page for lots of convincing examples that show that they are. 
Added after 13 answers: Interestingly, the focus so far has been almost entirely on the "You can't be sure if you don't have a proof" justification of proofs. But what if a physicist were to say, "OK I can't be 100% sure, and, yes, we sometimes get it wrong. But by and large our arguments get the right answer and that's good enough for me." To counter that, we would want to use one of the other reasons, such as the "Having a proof gives more insight into the problem" justification. It would be great to see some good examples of that. (There are one or two below, but it would be good to see more.)
Further addition: It occurs to me that my question as phrased is open to misinterpretation, so I would like to have another go at asking it. I think almost all people here would agree that proofs are important: they provide a level of certainty that we value, they often (but not always) tell us not just that a theorem is true but why it is true, they often lead us towards generalizations and related results that we would not have otherwise discovered, and so on and so forth. Now imagine a situation in which somebody says, "I can't understand why you pure mathematicians are so hung up on rigour. Surely if a statement is obviously true, that's good enough." One way of countering such an argument would be to give justifications such as the ones that I've just briefly sketched. But those are a bit abstract and will not be convincing if you can't back them up with some examples. So I'm looking for some good examples.
What I hadn't spotted was that an example of a statement that was widely believed to be true but turned out to be false is, indirectly, an example of the importance of proof, and so a legitimate answer to the question as I phrased it. But I was, and am, more interested in good examples of cases where a proof of a statement that was widely believed to be true and was true gave us much more than just a certificate of truth. There are a few below. The more the merrier.

REPLY [2 votes]: Gowers is particularly interested in cases in which a "proof of a statement that was widely believed to be true and was true gave us much more than just a certificate of truth." Reading through the answers, I am struck by a recurring theme.  There are many cases where

something seemed impossible;
it was impossible;
a lot of people didn't see the point of proving impossibility;
the search for an impossibility proof led to the discovery of structures that arguably would not have been discovered otherwise.

Just to list a few examples explicitly:

 It seemed impossible to prove the parallel postulate from the other axioms. The proof of impossibility led directly to non-Euclidean geometries.
 It seemed impossible to "square the circle."  The rigorous proof (Lindemann–Weierstrass) forms the foundation of modern transcendental number theory.
 It seemed impossible to solve all polynomial equations using radicals.  Rigorous investigation of this impossibility led to what we now call Galois theory.
 (Daniel Moscovich's answer) It seemed impossible to unknot a trefoil.  The search for a rigorous proof led to the discovery of all kinds of knot invariants.  More generally, topology and geometry are rife with examples of "obviously inequivalent" structures, and the search for rigorous proofs has uncovered all kinds of important invariants.
It seemed impossible to write down a procedure that would mechanically determine the truth or falsity of an arbitrary mathematical question. The rigorous proofs (incompleteness/undecidability) revealed fundamental limits to human knowledge.

In some cases, you could perhaps quibble with my claim that most if not all people thought these things were impossible, or that people didn't see the point of trying to prove impossibility.  In this regard, let me mention the article Why was Wantzel overlooked for a century? The changing importance of an impossibility result by Jesper Lützen. Lützen persuasively argues that Wantzel's proofs of the impossibility of trisecting the angle and duplicating the cube were all but ignored because people just weren't that interested in an impossibility proof of something that everybody already believed, or suspected, was impossible.  So even mathematicians are not immune to the tendency to undervalue negative results.<|endoftext|>
TITLE: Algorithms for Diophantine Systems
QUESTION [6 upvotes]: First off, I know that there's no general algorithm for determining if there's a solution to a general Diophantine equation, much less a system.
However, I'm wondering if there is an algorithm for solving a Diophantine system of linear and quadratic equations? In fact, I have a system which is "sparse" in some sense (the linear equations are all the sum of three variables equals a number, and all the same number, and the quadratics aren't much worse).
If so, then can it be extended to the case of countably many variables?

REPLY [14 votes]: No. Given any set of diophantine equations $f_1(z_1, \ldots, z_n) = \ldots = f_m(z_1, \ldots, z_n)=0$, we can rewrite in terms of linear equations and quadratics. Create a new variable $w_{k_1 \cdots k_n}$ for each monomial $z_1^{k_1} \cdots z_n^{k_n}$ which occurs in the $f$'s, or which divides any monomial which occurs in the $f$'s.  Turn each $f$ into a linear equation: For example, $x^3 y^2 + 7 x^2 y=5$ becomes $w_{32} + 7 w_{21} = 5$. Then create quadratic equations $z_i w_{k_1 \cdots k_i \cdots k_n} = w_{k_1 \cdots (k_i +1) \cdots k_n}$. For example, $x w_{22} = w_{32}.$  This shows that the solvability of Diophantine equations is equivalent to that of Diophantine equations of degree $\leq 2$.
I'll also mention a very concrete case. The intersection of two quadrics in $\mathbb{P}^3$ is a genus $1$ curve. To my knowledge, no algorithm is known to test for the existence of rational points even in this case. (But my knowledge is not very large.)<|endoftext|>
TITLE: Mostowski collapses and universal extensional relational classes
QUESTION [5 upvotes]: In the following, by a relational class I mean a pair$^1$ $(A,R)$, where $A$ is a class and $R \subseteq V \times V$ is a class relation, such that $R$ is well-founded and set-like on $A$ ($R$ is not assumed to be extensional). Homomorphisms of such classes are defined in the obvious way. Let $G : (A,R) \to (M,\in)$ be the Mostowski collapse of $(A,R)$, i.e. $G(x) = \{G(y) : y \in A , y R x\}$. Is this a functorial construction?
So let $f : (A,R) \to (A',R')$ a homomorphism, does this induce a homomorphism $(M,\in) \to (M',\in)$, such that the obvious diagram commutes? For this we have to check $G(x)=G(y) \Rightarrow G'(f(x))=G'(f(y))$ for all $x,y \in A$. Is this true?
If this works out fine and $R'$ is extensional, then $f$ extends to a homomorphism $(M,\in) \to (A',R')$. Is this extension unique? If yes, we have found a functor which is left-adjoint to the forgetful functor from relational classes to extensional classes.
If this does not work out, what about adding the assumption of transitivity? And if this also does not work, is there nevertheless some left-adjoint, which is then different from the Mostowski collapse?
$^1$ In $ZF$ we can't define pairs of classes, and perhaps the categories above are not well-defined. This does not affect the content of my question, which can be translated to a well-formed statement in $ZF$.

REPLY [2 votes]: The answer to your first question is no, unfortunately, it
is not functorial.
Here is a comparatively simple counterexample. The idea is
that two objects in $A$ can collapse to the same point, but
gain new elements in $A'$ that differentiate them. Let
$A=\{a,b\}$ have two objects, with $R$ the empty
relation. So the Mostowski collapse of $(A,R)$ maps both
$a$ and $b$ to the emptyset $\varnothing$. Let
$A'=\{a,b,c\}$, with $c\mathrel{R'}b$, but otherwise $R'$
is trivial. So the Mostowski collapse of $(A',R')$ maps
$a,c\mapsto\varnothing$ and $b\mapsto\{\varnothing\}$.
Let $f:A\to A'$ be the inclusion map, which is a
homomorphism, since $a$ and $b$ have no relation with
respect to $R$ or $R'$. Since we have $G(a)= G(b)$ but
$G'(f(a))\neq G'(f(b))$, there can be no commutative
diagram here.
But why not insistin on extensionality? In the category of extensional class relations, that is, if your structures $(A,R)$ and $(A',R')$ satisfy
extensionality, then the Mostowski collapse is functorial. This is because $G$
and $G'$ are isomorphisms, and it is easy to see that there
is the induced homomorphism $h:M\to M'$ defined by
$h(G(a))=G'(f(a))$. This is a homomorphism since $G(a)\in
G(b)\iff a\mathrel{R} b\iff f(a)\mathrel{R'}f(b)\iff
G'(f(a))\in G'(f(b))$, and the diagram commutes by
definition.
For the second question, in this extensional case, since
$(M',{\in})$ is isomorphic to $(A',R')$, then we get a map
$(M,{\in})\to (A',R')$, as you say. It is just the map
$h\circ G^{-1}$. In general, there is not a unique map from
$(M,{\in})$ to $(A',R')$, however, since for example, when
$(A,R)$ is a well-order, then the Mostowski collapse is an
ordinal, but if $(A',R')$ is a longer ordinal, there could
be many homomorphisms of $(M,{\in})$ to $(A',R')$, since
these are just the order-preserving maps. But this map will
be unique such that the diagram commutes, since it is
determined by those isomorphisms. (But I think you were not
actually asking this about the extensional case. I'm not
sure what you mean by "adding transitivity", since the
Mostowski collapse of any structure is a transitive set,
even when the relations are not extensional.)
If you really don't want extensionality, then it would also be sensible to restrict the homomorphism concept to functions $f:A\to A'$ which not only preserve the relation (in both directions), but which also respect equivalence of predecessors; this amounts to being well-defined on the Mostowski collapse. That is, in this category, we think of $\langle A,R\rangle$ as a code for its Mostowski collapse, and the notion of homomorphism should respect that. In this case, the Mostowski collapse will again be functorial. 
Lastly, let me mention that several other restrictions of your homomorphism concept are often studied. Namely, in set theory the concept of a $\Sigma_n$-elementary map is prominent for natural numbers $n$. Even $\Sigma_0$-elementary goes beyond the basic concept of homomorphism that I think you intend, but we are often interested in $\Sigma_1$-elementary embeddings or even fully elementary embeddings. Many large cardinal concepts, for example, are characterized by the existence of such embeddings defined on the entire universe into a transitive class.<|endoftext|>
TITLE: The Dold-Thom theorem for infinity categories?
QUESTION [24 upvotes]: Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces.  For a based space $X \in \mathcal T$, one has a canonical funtor $S_X : \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$.  The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism.
As is well know, the homotopy groups of $\mathrm{colim} S_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S_X = SP^\infty_h X$ given the stable homotopy of $X$.
Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit?
The motivation for this question comes from thinking about $\infty$-categories.  In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict categorical colimits.  So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology.  In short, is there any $\infty$-categorical analog of the Dold-Thom theorem?
Update: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations.  I'll just quickly sketch what I have so far:
Let $\mathcal O(\Sigma_n)$ denote the orbit category.  The objects are the homogeneous (discrete) spaces $\Sigma_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma_n$, and the morphisms are the $\Sigma_n$-equivariant maps.  There is a canonical functor $$\Sigma_n \rightarrow \mathcal O(\Sigma_n)^{op}$$ where we regard $\Sigma_n$ as a category with one object as usual.
Given a $\Sigma_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma_n/H) = X^H$$ It turns out that the above inclusion is final so that it induces an isomorphism of colimits.  Hence $\mathrm{colim}_{\mathcal O(\Sigma_n)} \tilde X \cong X_{\Sigma_n}$, i.e., the coinvariants.  It's also not hard to see that the undercategories are copies of $B\Sigma_n$, hence not contractible, so we don't expect an equivalence of homotopy colimits, which is good.
On the other hand, I can now show that when $X$ is discrete, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence.  My methods here do not generalize to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated.  (I think something like this must appear in May's book on equivariant homotopy theory if it's true, but I did not have it available this weekend.)
The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad.  (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \mathcal O(\Sigma_n)$.  Take the Grothendieck construction.  Some natural diagram on this category might give the right answer.)

REPLY [5 votes]: I just came across this old posting, and see that folks had correctly discovered presentations (dating back to the mid 1980's) by Emmanuel Dror about writing colimits as homotopy colimits.  
The special case in hand -- symmetric powers of spaces -- is particularly elegant, as the orbits which arise are very limited and special.  Kathryn Lesh and Greg Arone have a couple of lovely papers on this and related families of examples: Kathryn got this going with a 2000 paper in T.A.M.S., and then, with Greg, has a longer study in Crelle in 2007.  (They then use these ideas in a 2010 paper in Fund. Math.)  Kathryn's work was roughly contemporaneous with Emmanuel's - she has a 1997 Math. Zeit. paper where one can see the beginnings of the ideas.<|endoftext|>
TITLE: Riemannian surfaces with an explicit distance function?
QUESTION [72 upvotes]: I'm looking for explicit examples of Riemannian surfaces (two-dimensional Riemannian manifolds $(M,g)$) for which the distance function d(x,y) can be given explicitly in terms of local coordinates of x,y, assuming that x and y are sufficiently close.  By "explicit", I mean things like a closed form description in terms of special functions, by implicitly solving a transcendental equation or (at worst) by solving an ODE, as opposed to having to solve a variational problem or a PDE such as the eikonal equation, or an inverse problem for an ODE, or to sum an asymptotic series.
The only examples of this that I know of are the constant curvature surfaces, which can be locally modeled either by the Euclidean plane ${\bf R}^2$, the sphere ${\bf S}^2$, or the hyperbolic plane ${\bf H}^2$, for which we have classical formulae for the distance function.
But I don't know of any other examples.  For instance, the distance functions on the surface of the solid ellipsoid or solid torus in ${\bf R}^3$ look quite unpleasant already to write down explicitly.  Presumably Zoll surfaces would be the next thing to try, but I don't know of any tractable explicit examples of Zoll surfaces that are not already constant curvature.

REPLY [11 votes]: In the course of writing an answer to a related MO question, I realized that there is a surface with a complete Riemannian metric of non-constant negative curvature for which one can write down the distance function explicitly, so I thought I would record it here for those who might be interested.
Such metrics are quite rare; even when the geodesic flow is integrable (or even rotationally symmetric), one cannot generally compute the arc length along geodesics in a sufficiently explicit form that one can actually compute the geodesic distance between two given points in any explicit way.  This is the first complete example with nonconstant curvature that I have seen.  (There are many explicit but non-complete examples with non-constant curvature in the classical literature, c.f. Tome III of Darboux' monumental Leçons sur la théorie générale des surfaces et les applications géométriques du calcul infinitésimal.)
The surface is $\mathbb{R}^2$ and the metric in standard coordinates is
the rotationally symmetric metric
$$
g = (x^2+y^2+2)\,(\mathrm{d}x^2 + \mathrm{d}y^2).
$$
The Gauss curvature of $g$ is $K = -4/(x^2+y^2+2)^3<0$. It is  complete, since it dominates the standard flat metric.  It follows from general theory that any two points lie on a unique geodesic and each geodesic segment minimizes $g$-distance between its endpoints.
The geodesics of $g$ are easy to describe as curves: For every pair of numbers $(a,b)$ with $a^2+b^2\ge 1$, consider the equation
$$
(1+a)\,x^2 + 2b\,xy + (1-a)\,y^2 = a^2+b^2-1.
$$
When $a^2+b^2>1$, this is a hyperbola, and each of the branches is a geodesic.
When $a^2+b^2=1$, this is the equation of a line through the origin, which is also a geodesic.  Conversely, every geodesic of $g$ is either a line through the origin or a branch of one of the hyperbolae listed above.
The geodesic distance along a line through the origin is not hard to write down:
On the line $y=0$, the element of arc length is
$$
\mathrm{d}s = \sqrt{x^2+2}\,\mathrm{d}x = \mathrm{d}\left(\sinh^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{x\sqrt{x^2+2}}{2}\right).
$$
Set
$$
f(x) = \sinh^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{x\sqrt{x^2+2}}{2}
\approx \sqrt2\left(x + \frac{x^3}{12}-\frac{x^5}{160}+\cdots\right).
$$
I will now show that the $g$-distance between any two points $p,q\in\mathbb{R}^2$
is given by the formula
$$
\delta(p,q) = f\left(\frac{|p+q|+|p-q|}{2}\right)-f\left(\frac{|p+q|-|p-q|}{2}\right),
$$
where the norms are the Euclidean norms, i.e., taken with respect to the standard Euclidean inner product on $\mathbb{R}^2$.
Remark: In fact, the above formula also holds for the rotationally invariant metric $g = (2+x{\cdot}x)(\mathrm{d}x{\cdot}\mathrm{d}x)$ on $\mathbb{R}^n$ for $n\ge2$, since each geodesic for this metric lies in a (totally geodesic) $2$-plane through the origin $x=0$.
To prove the claim, first, note that, while the distance function $\delta:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ is not smooth along the diagonal, its square $\sigma = \delta^2$
is a smooth function on $\mathbb{R}^2\times\mathbb{R}^2$ that vanishes along the diagonal.  In fact, because $g$ is real-analytic, it follows that $\sigma$ is real-analytic.  Because $g$ is invariant under (Euclidean) rotation about the origin and reflection across lines through the origin, it follows that $\delta$ and $\sigma$ are also invariant under these rotations and reflections, now acting diagonally on $\mathbb{R}^2\times\mathbb{R}^2$.  Using this, one can show that $\sigma$ must be representable as
$$
\sigma(p,q) = C\bigl(|p|^2,\,p{\cdot}q,\,|q|^2\bigr)
\quad\text{for all}\ p,q\in\mathbb{R}^2,
$$
where $C(a,b,c)$ is a smooth function on the cone $\mathcal{C}_+$
defined by $a,c\ge 0$ and $ac-b^2\ge0$.
Now, for fixed $q\in\mathbb{R}^2$ the function $\delta_q:\mathbb{R}^2\to\mathbb{R}$, defined by $\delta_q(p) = \delta(p,q)$, vanishes at $q$ and satisfies $|\mathrm{d}(\delta_q)|^2_g = 1$ except at $q$ (where it is not differentiable).
This implies that the corresponding $\sigma_q = {\delta_q}^2$ attains its minimum value of $0$ at $q$ and satisfies the first-order PDE $|\mathrm{d}(\sigma_q)|^2_g = 4\sigma_q$.  Interpreting this in terms of the above representation of $\sigma$,
we find that $C$ must satisfy the first order PDE
$$
4aC_a^2 + 4bC_aC_b+cC_b^2 - 4(a+2)C^2 = 0.
$$
Similarly, using the fact that $C(a,b,c) = C(c,b,a)$ (since $\sigma(p,q) = \sigma(q,p)$), we find that
$$
4cC_c^2 + 4bC_cC_b+aC_b^2 - 4(c+2)C^2 = 0.
$$
This pair of first-order PDE for $C$ is singular at $(a,b,c) = (0,0,0)$, but, since $C$ must vanish when $a+c-2b = |p-q|^2 = 0$ but otherwise be positive in the cone $\mathcal{C}_+$, it is easy to show that $C$ has a Taylor expansion
$$
C\simeq (a{-}2b{+}c)\left(2 + \frac{(a{+}b{+}c)}{3}-\frac{(4a{+}7b{+}4c)(a{-}2b{+}c)}{360} + \cdots\right).
$$
In fact, examining the higher terms, it becomes apparent that $C$
should be a function of $u = a{+}c$ and $v = a{-}2b{+}c$.  Indeed, if
$$
C(a,b,c) = H(a{+}c,\,a{-}2b{+}c) = H(u,v)
$$
were to hold for some smooth function $H$ on the $uv$-domain defined by $0\le v\le 2u$, then one finds that $H$ would have to satisfy
$$
u\,{H_u}^2 + 2v\,(H_uH_v+{H_v}^2) - (u+4)\,H = 0.
$$
with $H \simeq v\,\bigr(2-\tfrac1{6}(v-3u)-\tfrac1{720}(15u-7v)v+\cdots\bigr)$.
Using the theory of singular analytic first-order PDE, it is not difficult to show that such an analytic solution $H(u,v)$ exists, is unique, and is a multiple of $v$.  (It's easy to show that there is a unique power series solution whose lowest term is $2v$, but one needs to show that this series converges.)
As a consequence, $C(a,b,c) = H(a{+}c,\,a{-}2b{+}c)$ satisfies the pair of first-order singular analytic PDE listed above.  Consequently,
$$
\sigma(p,q) = C\bigl(|p|^2,\,p{\cdot}q,\,|q|^2\bigr) 
= H\bigl(|p|^2{+}|q|^2,\,|p{-}q|^2\bigr),
$$
Since $H$ is a multiple of $v = |p{-}q|^2$, it follows that
$$
\delta(p,q) = |p{-}q|\,G\bigl(|p|^2{+}|q|^2,\,|p{-}q|^2\bigr)
$$
for some smooth positive function $G(u,v)$.  Meanwhile, for $b<a\in\mathbb{R}$,
taking $p = (a,0)$ and $q=(b,0)$, we have
$$
(a{-}b)\,G\bigl(a^2{+}b^2,\,(a{-}b)^2\bigr) = \delta(p,q) = f(a)-f(b).
$$
The function $G(u,v)$ is determined in the wedge $0\le v\le 2u$ by this equation
as $(a,b)$ vary over the half-plane $b<a$.  It follows from this that
$$
|p{-}q|\,G\bigl(|p|^2{+}|q|^2,\,|p{-}q|^2\bigr)
= f\left(\frac{|p+q|+|p-q|}{2}\right)-f\left(\frac{|p+q|-|p-q|}{2}\right),
$$
as desired.
Remark:  The reader might be startled (as I was initially) upon realizing that the above formula implies a seemingly strange identity
$$
f\left(\frac{|a+b|+|a-b|}{2}\right)-f\left(\frac{|a+b|-|a-b|}{2}\right)
= |f(a)-f(b)|
$$
for all real numbers $a$ and $b$, but, in fact, this identity holds for any increasing odd function $f$.
Added Remark (16 May 2020):  A similar analysis, yielding an explicit distance function, can be done for the incomplete metric
$$
g = (1-x^2-y^2)\bigl(\mathrm{d}x^2+\mathrm{d}y^2\bigr)
$$
on the interior of the unit disk $D$ defined by $x^2+y^2<1$.  This is a metric of positive curvature $K = 4/(1-x^2-y^2)^3$.  What one finds is that, setting
$$
s(x) = \tfrac12\arcsin(x) + \tfrac12x\sqrt{1-x^2}
\quad\text{for}\ |x|\le 1,
$$
the function
$$
\delta(p,q) = s\left(\frac{|p+q|+|p-q|}{2}\right)-s\left(\frac{|p+q|-|p-q|}{2}\right)
$$
gives the length of the shortest geodesic joining $p$ and $q$ when $|p+q|+|p-q|\le 2$.  (This inequality is also the condition for the existence of a geodesic joining $p$ and $q$ within the interior of $D$.)
Meanwhile, regarding the boundary circle $x^2+y^2=1$ as a single point $z$ whose distance from $p\in D$ is $s(1) - s(|p|) = \tfrac14\pi - s(|p|)$, we see that there is always a path from $p$ to $q$ (through z) of length $L(p,q) = \tfrac12\pi - s(|p|)-s(|q|)$.
It is now not difficult to show that the actual distance from $p$ to $q$ is $L(p,q)$ when $|p+q|+|p-q|\ge 2$ and is the minimum of $\delta(p,q)$ and $L(p,q)$ when $|p+q|+|p-q|\le 2$.<|endoftext|>
TITLE: Sum of reciprocals of primes modulo which a polynomial has a root
QUESTION [9 upvotes]: Dear all,
I am looking for a proof or a reference of the following statement:
Let $f$ be a non-constant polynomial with integer coefficients. Then the sum $\sum \{1/p \mid f \text{ has a root modulo } p\}$ diverges.
I am pretty sure that I saw it somewhere before but I cannot remember and I failed to find it in number theory books. A possible routes that has already been suggested to me is actually showing that the sums of reciprocals for which f even decomposes into linear factors has positive density which should stem from Galois theory. I am however not an expert in Galois theory so that I would prefer a direct proofs or a reference. 
Thanks in advance,
Alberto

REPLY [10 votes]: Charles is completely right that this follows from Frobenius' theorem. Since you don't like Galois theory, here is a proof which does not explicitly mention Galois theory. (But it is hiding just out of sight.)
We may assume that $f$ is irreducible as, if $g$ divides $f$, then the set of primes for which $f$ has a root contains the set for which $g$ does.
Let $K$ be the field $\mathbb{Q}[x]/f(x)$. Let $R$ be the ring of integers of $K$, and let $S=\mathbb{Z}[x]/f(x)$. Note that $S$ is a finite index sublattice of $R$ and, if $p$ is a prime which does not divide $|R/S|$, then $R/p \cong S/p$. Also, for any prime $p$ which does not divide the discriminant of $f$, the polynomial $f$ factors into distinct factors in $\mathbb{F}_p[x]$.
Thus, if $p$ is large enough to not divide either $|R/S|$ or the discriminant of $f$, then $R/p \cong S/p \cong \mathbb{F}_p[x]/f(x) \cong \bigoplus \mathbb{F}_p[x]/(f_i(x))$ where $f_i$ are the irreducible factors of $f$ mod $p$. So, for such a prime $p$, prime ideals of $R$ which contain $(p)$ are in bijection with irreducible factors of $f$ mod $p$, and the norm of such a prime is $p^{\deg f_i}$.
So, if $f$ has a root modulo $p$, then 
$$\frac{1}{p} \leq \sum_{\pi \supseteq (p), \ \pi \ \mbox{prime}} \frac{1}{N(\pi)} \leq \frac{\deg f}{p}$$
and, if $f$ does not have a root modulo $p$, then
$$\sum_{\pi \supseteq (p), \ \pi \ \mbox{prime}} \frac{1}{N(\pi)} \leq \frac{(\deg f)/2}{p^2}$$
We want to show that
$$\sum_{p: \exists \pi \ \mbox{a prime of} \ R \ \mbox{with} \ N(\pi)=p} \frac{1}{p}$$
diverges. By the above inequalities, it is equivalent to show that
$$\sum_{\pi \subset R, \ \pi \ \mbox{prime}} \frac{1}{N(\pi)}$$
diverges. (Note that the finitely many primes which divide $|R/S|$ or the discriminant of $f$ cannot change whether or not the sum converges.)
Now, we have unique factorization into prime ideals for $R$, so
$$\sum_{I \subseteq R} \frac{1}{N(I)^s} = \prod_{\pi \subset R, \ \pi \ \mbox{prime}} \left( 1 - \frac{1}{N(\pi)^s} \right)^{-1}.$$ 
The left hand side is the $\zeta$ function of $K$. By the class number formula (see most books on algebraic number theory), $\zeta_K(s) = C/(s-1) + O(1)$ for some positive constant $C$, as $s \to 1^{+}$. So
$$\log \zeta_K(s) = \log \frac{1}{s-1} + O(1) = \sum \log \left( \frac{1}{1-N(\pi)^{-s}} \right) = \sum \frac{1}{N(\pi)^s} + O(1/N(\pi)^{2s}).$$
We deduce that 
$$\sum \frac{1}{N(\pi)^s} = \log \frac{1}{s-1} + O(1)$$
so 
$$\sum \frac{1}{N(\pi)}$$
diverges.<|endoftext|>
TITLE: Degree Sequences and Graph Enumeration
QUESTION [9 upvotes]: I do recreational math from time to time, and I was wondering about a couple of graph enumeration issues.
First, is it possible to enumerate all simple graphs with a given degree sequence?
Second, is it possible to enumerate all valid degree sequences for simple graphs with a given number of vertices?
Based on my wikipedia surfing, we can use the Erdos-Gallai theorem to determine if a degree sequence is valid, but this doesn't really lend itself to enumerating valid degree sequences efficiently. Similarly, we can use the Havel-Hakimi algorithm to construct at least one graph for a given valid degree sequence, but this doesn't help to enumerate all graphs for that degree sequence. 
My (admittedly uneducated) guess is that it might be possible to work backwards using the Havel-Hakimi condition to construct graphs by building them up in different ways.  Any insight would be appreciated :D

REPLY [2 votes]: Hello !!!
I read your post several weeks ago as I also wanted to enumerate degree sequences but did not know how. I still do not know how to enumerate the graphs with a given degree sequence, but I did find a paper about the enumeration of degree sequences :
Alley CATs in search of good homes (1994)
by Frank Ruskey ,  Robert Cohen ,  Peter Eades ,  Aaron Scott
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.51.6493
The authors provide an algorithm in Pascal that I was not able to get to work, and I ended up re-implementing it myself in a different way, following the same idea of reversing the "Havel-Hakimi recognition algorithm"
I recently sent it as a Sage patch if you want to use it (even though my answer is almost one year late) :-)
http://trac.sagemath.org/sage_trac/ticket/11584
Nathann<|endoftext|>
TITLE: An Alternative to the Cook-Levin Theorem
QUESTION [6 upvotes]: In general to prove that a given problem is NP-complete we show that a known NP-complete problem is reducible to it. This process is possible since Cook and Levin used the logical structure of NP to prove that SAT, and as a corollary 3-SAT, are NP-complete. This makes SAT the "first" NP-complete problem and we reduce other canonical NP-complete problems (e.g. CLIQUE, HAM-PATH) from it.
My question is whether there is a way to prove directly from the definition/logical structure of NP that a different problem (i.e. not SAT) is NP-complete. A friend suggested that it would be possible to tailor the proof of the Cook-Levin Theorem to show that, for example, CLIQUE is NP-complete by introducing the reduction from SAT during the proof itself, but this is still pretty much the same thing.

REPLY [4 votes]: Yes, for example TMSAT (Turing Machine SAT):
TMSAT = $\{ \langle \alpha, x, 1^n, 1^t \rangle : \exists u \in \{0,1\}^n$ such that $M_\alpha$ outputs 1 on input $\langle x,u \rangle$ within $t$ steps.$\}$
(Theorem 2.9 of Arora-Barak. See Chapter 2 draft: http://www.cs.princeton.edu/theory/index.php/Compbook/Draft#np)
It is quite easy to show that this problem is NP-complete!<|endoftext|>
TITLE: Wasserstein distance in R^d from one dimensional marginals
QUESTION [11 upvotes]: This question occurred to me while I was reading Klartag's papers on central limit theorems for convex bodies. 
Given probability measures $\mu$, $\nu$ on (the Borel $\sigma$-field of) $R^d$ with finite first moments, their Wasserstein distance is given by:
$$W_{R^d}(\mu,\nu) = \sup \mbox{ of }\int_{R^d}f d\mu-\int_{R^d}f d\nu\mbox{ over all 1-Lipschitz }f:R^d\to R.$$
(NB: there was an error in this formula -- an inf in the place of the sup.)
Given a vector $v\in R^d$, let $\mu_v$ be the distribution of $X.v$, where $X$ has distribution $\mu$. Define $\nu_v$ analogously. Note that $\mu_v$ and $\nu_v$ are distributions over $R$.
Question: is there a constant $C_d>0$ depending on $d$ only such that: $$W_{R^d}(\mu,\nu)\leq C_d\sup_{v\in R^d, |v|=1}W_{R}(\mu_v,\nu_v)?$$
If so, how does $C_d$ grow with $d$?
An illustrative example: Assume $Z$ is uniform over a $D-1$ dimensional sphere $S^{D-1}$ in $R^D$. Any one-dimensional marginal of $\sqrt{D-1}Z$ is approximatelly Gaussian. Now let $\mu$ be the law of the first $d$ coordinates of $\sqrt{D-1}Z$, and $\nu$ be the standard Gaussian distribution on $R^d$. Can one deduce from the previous statement alone that $\mu$ and $\nu$ are close?
Another example: Let $Z$ be a random vector in $R^D$ with mean $0$ and covariance matrix $I_D$, $D\gg 1$. Old results of Sudakov (quoted here) show that "most" one-dimensional marginals of $Z$ are close to $|Z|N$ where $N$ is standard normal and independent from $Z$. A positive answer to the above question would lead to typical results for $d$-dimensional projections of $Z$.

REPLY [4 votes]: PS: I posted an answer in 2015. In late 2019, @Neyman identified a problem with my original post. 

Here is a non-constructive answer to the question. 
I don't know of any reference where you can find the solution. I started thinking about the problem and found a simple solution (using standard results).
I read in another question about Wasserstein distance the suggestion of using a finite cover of the unit ball in $Lip(\mathbb{R}^d) $ to reduce estimates of $ W_{\mathbb{R}^d} $ to a max over a finite set. I see that idea working for Lipschitz functions on a bounded set, but I don't see how to apply that idea when working in the whole $\mathbb{R}^d $. 
Here is my solution.
Let
$\quad Lip1(\mathbb{R}^d) = \{f:\mathbb{R}^d \longrightarrow\mathbb{R}: \forall x,y \in \mathbb{R} ^d \text{ with } x \neq y, \frac{|f(x)-f(y)|}{|x-y|} \leq 1)     \}$

be the set of 1-Lipschitz functions in $\mathbb{R}^d$.
Let $\mathcal{G}$ be the set of functions on $ \mathbb{R}^d  $ of the form:
$(1)\quad f(x) = \sum_{i \in F} a_if_i(x\cdot{v_i}) $
where
$\quad F$ is a finite set,
$\quad a_i \in \mathbb{R}, a_i \geq 0, \sum_{i \in F} a_i = 1,$
$\quad f_i \in Lip1(\mathbb{R}) ,$
$\quad v_i \in \mathbb{R}^d , |v_i| = 1 \text{  (Euclidean norm)}$
i.e.: $\mathcal{G}$ is the convex hull of $Lip1(\mathbb{R})$ functions composed with one dimensional projections. Clearly 
$\quad \mathcal{G} \subseteq \overline{\mathcal{G}} \subseteq Lip1(\mathbb{R}^d)$
where 
$\overline{\mathcal{G}}$ is the closure of 
$\mathcal{G}$ in any topology in which  $Lip1(\mathbb{R}^d)$ is closed. 
We will work with a weak* topology, namely the one in which the neighborhoods of 0 are generated by the sets
$(2)\quad \mathcal{N} = \{f \in Lip(\mathbb{R}^d): 
\arrowvert \int_{\mathbb{R}^d}  f(x) u(x) dx + 
\int_{\mathbb{R}^d}  \nabla f(x) \cdot{w(x)} dx\arrowvert < \epsilon \} $ 
for some $ u \in L^1(\mathbb{R}^d, (1+|x|)dx)$ with $\int u(x) dx = 0$, 
some $ w \in L^1(\mathbb{R}^d)^d $, and some $ \epsilon > 0 $.  
Here are some known facts:
1) If $f \in Lip(\mathbb{R}^d)$, then $ f$ is differentiable a.e and $ sup\{ \frac{|f(x)-f(y)|}{|x-y|}: x \neq y\} = \|\nabla f \|_\infty $ is finite, denoted $\|f\|_{Lip}$.  
2) If $f \in Lip(\mathbb{R}^d)$, then $ |f(x) - f(0)| \leq \|f\|_{Lip} |x| $, so the integral $\int_{\mathbb{R}^d}  f(x) u(x) dx $ is well defined for any $ u \in L^1(\mathbb{R}^d ,(1+|x|)dx)$. 
3) With this topology $ Lip(\mathbb{R}^d) $ is not separable. We will actually work in $ Lip(\mathbb{R}^d)/constants $, but we will not need be very explicit about it. 
4) With this topology, any linear continuous function is of the form 
$\quad \int_{\mathbb{R}^d}  f(x) u(x) dx + 
\int_{\mathbb{R}^d}  \nabla f(x) \cdot{w(x)} dx $ 
for some $ u \in L^1(\mathbb{R}^d, (1+|x|)dx)$ with $\int u(x) dx = 0$, 
and some $ w \in L^1(\mathbb{R}^d)^d $. The representation is not unique.
Claim 1: Let $\mathcal{S}$ be the space of functions like (1) with arbitrary $a_i$, i.e.: $ \mathcal{S}=\mathbb{R}\mathcal{G} $. The closure of $ \mathcal{S}$ in the topology defined in (2) is $Lip(\mathbb{R}^d)$.
Proof: If $ \overline{\mathcal{S}}  \neq Lip(\mathbb{R}^d)$ then, by Hahn-Banach, there would by a non-zero linear function $ L(f)=\int_{\mathbb{R}^d}  f(x) u(x) dx + 
\int_{\mathbb{R}^d}  \nabla f(x) \cdot{w(x)} dx$ such that $L(f) = 0$ for all $f \in \mathcal{S}$. 
Let $\rho$ be a smooth function with compact support. Since $\mathcal{S}$ is invariant under translations it follows that 
$(3) \quad 0 = \int u\ast{\rho}(x) f(x) + w\ast{\rho}(x)\cdot{\nabla f(x)} dx = $
$\quad  \int (u\ast{\rho}(x) - div(w\ast{\rho})(x))f(x) dx $   for all $f \in \mathcal{S}$.
The real and imaginary parts of functions of the form $ f(x)=e^{-2\pi x\cdot{\xi}}$ are in $\mathcal{S}$, therefore the Fourier transform of $(u\ast{\rho}(x) - div(w\ast{\rho}))$ is zero, so (3) holds for any $f \in Lip(\mathbb{R}^d)$. Letting $\rho \longrightarrow \delta $, we get that $L(f)=0$ for any $f \in Lip(\mathbb{R}^d)$, contradicting the fact that $L$ is not null.
Claim 2: $ Lip(\mathbb{R}^d) = \bigcup_{n \in \mathbb{N}} n\overline{\mathcal{G}}$. 
Proof: Let $f \in Lip(\mathbb{R}^d)$. From Claim 1, there is a net $\{f_\lambda\}_\lambda$ in $\mathcal{S}$ such that $f_\lambda \rightarrow f$. So the functionals defined as $L_\lambda(w)=\int_{\mathbb{R}^d} w(x)\cdot{\nabla f_\lambda(x)} dx $ for $w \in L^1(\mathbb{R}^d)^d$ are pointwise bounded. By the uniform boundness theorem, they are uniformly bounded. It follows that 
$ sup_\lambda \| \nabla f_\lambda \|_\infty $ 
is finite, since $ \| T_\lambda \| = \| \nabla f_\lambda \|_\infty $. 
Taking $ n \in \mathbb{N} $ sufficiently large we have $ sup_\lambda \| \nabla f_\lambda \|_\infty \leq n $, and so $ f \in n\overline{\mathcal{G}} $.
Claim 3: $ \overline{\mathcal{G}} $ is close in the strong topology (i.e.: the topology defined by the (semi) norm $\|f\|=_{def} \| \nabla f \|_\infty $).
Proof: Since $ L^1(\mathbb{R}^d) $ is separable, the weak* topology of $ Lip(\mathbb{R}^d) $ restricted to $Lip1(\mathbb{R}^d) $ is metrizable; let $d$ be a metric on $Lip1(\mathbb{R}^d) $ that defines the weak* topology. Let $ f $ be in the closure of $ \overline{\mathcal{G}} $ with the strong topology. Let $ \{f_n\}_n$  converge to $f $ in the strong topology, $f_n \in \overline{\mathcal{G}} $, i.e.: $ \|\nabla f_n - \nabla f \|_\infty \rightarrow 0 $; in particular, $ d(f_n,f) \rightarrow 0 $.  
For each $n$, since $f_n \in \overline{\mathcal{G}} $, there is $ h_n \in \mathcal{G} $ such that $ d(h_n,f_n) < \frac{1}{n} $. Then $ d(h_n,f) \leq d(h_n, f_n) + d(f_n,f) \rightarrow 0$, so $ f \in \overline{\mathcal{G}} $.  
Claim 4: There is a constant $C_d$ such that 
$ Lip1(\mathbb{R}^d) \subseteq C_d\overline{\mathcal{G}} $. 
Proof: From Claim 2, $ Lip(\mathbb{R}^d) = \bigcup_{n \in \mathbb{N}} n\overline{\mathcal{G}}$. From Claim 3, for each $ n \in \mathbb{N}, n\overline{\mathcal{G}}$ is closed in the strong topology. From Baire's theorem, at least one of the sets $n\overline{\mathcal{G}} \text{ } (n \in \mathbb{N}) $ has non-empty interior, i.e.: there is $n \in \mathbb{N}, \epsilon > 0, f \in Lip(\mathbb{R}^d)$ such that $f + \epsilon Lip1(\mathbb{R}^d) \subseteq n\overline{\mathcal{G}}$. From Claim 2, there is $a \in \mathbb{N}$ such that $ f \in a\overline{\mathcal{G}}$. Therefore, 
$ Lip1(\mathbb{R}^d) \subseteq C\overline{\mathcal{G}}$ with $ C = \frac{n+a}{\epsilon}$.
Finally, we can give a (non-constructive) answer the question in the affirmative. Since the definition of Wasserstein distance requires integration against functions in $Lip1(\mathbb{R}^d)$, we assume the measures involved have finite first moment. 
Claim 5: If $\mu, \nu $ are measures in $\mathbb{R}^d$ with finite first moment, then 
$ \quad W_{R^d}(\mu,\nu)\leq C_d\sup_{v\in R^d, |v|=1}W_{R}(\mu_v,\nu_v) $ 
where $C_d$ is the constant in Claim 4.
Proof: Let's assume first that $ \mu, \nu $ have densities $ u, w $ with respect to Lebesgue measure in $\mathbb{R}^d $. Then 
$ \quad W_{R^d}(\mu,\nu) = sup_{f \in Lip1(\mathbb{R}^d)} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx  \leq $ 
$\quad sup_{f \in C_d\overline{\mathcal{G}}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx $, by Claim 4 .
But $sup_{f \in C_d\overline{\mathcal{G}}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx = sup_{f \in \overline{\mathcal{G}}} \int_{\mathbb{R}^d} C_d f(x) (u(x)-w(x)) dx =$
$ C_d sup_{f \in \overline{\mathcal{G}}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx $. 
In turn, 
$ sup_{f \in \overline{\mathcal{G}}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx = 
sup_{f \in \mathcal{G}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx$, since 
$L(f)=\int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx$ is continuous in the weak* topology (2). 
By definition, $\mathcal{G}$ is the convex hull of functions on $\mathbb{R}^d$ of the form 
$f(x\cdot{v}) $ with $|v|=1$ and $f \in Lip1(\mathbb{R})$, so
$\quad  sup_{f \in \mathcal{G}} \int_{\mathbb{R}^d} f(x) (u(x)-w(x)) dx = $
$\quad sup_{f \in Lip1(\mathbb{R}), |v|=1} \int_{\mathbb{R}^d} f(x\cdot{v}) (u(x)-w(x)) dx = sup_{|v|=1} W_{\mathbb{R}}(\mu_v,\nu_v) $.
By continuity of the Wasserstein distance, we can pass from probability distributions with density to arbitrary distributions with finite first moments.$\square$<|endoftext|>
TITLE: Erik Westzynthius's cool upper bound argument: update?
QUESTION [20 upvotes]: Version 2 of this writeup is
  available, and includes a newer and simple upper bound thanks to
  MathOverflow 88777 as
  well as indirect references to future writeups.  Details of further work
  will be found in these writeups.  GRP 2014.06.04.


In a paper of Erik Westzynthius,

Ueber die Verteilung der Zahlen, die zu den n ersten Primzahlen teilerfremd sind,
  Comm. Phys. Math. Soc. Sci. Fenn., Helsingfors (5) 25 (1931), 1-37

I saw the following upper bound argument.  Having never
studied sieve theory, I was quite impressed by it.  The goal is to bound from above
the quantity max $(q_{i+1} - q_i)$, where the $q_i$ are the positive integers in
increasing order which are relatively prime to $P_n$, the product of the first
n primes.  Here is a sketch of the argument.
Let $a$ and $x$ be real parameters, with $x > 0$ .  Consider the
integers in the open interval $(a, a+x)$, and call this set $H$.  Let us look
at the subsets of $H$ consisting of those integers which are a multiple of the
positive integer $t$; call the size of each such subset $I_t$.  Step 1 is to use
inclusion-exclusion to 
estimate $I_0$, the number of integers in the interval $(a, a+x)$ which are relatively 
prime to $P_n$. (I.e. count integers, throw out multiples of 2, throw out multiples
of 3, add in multiples of (2*3) to compensate, etc.)  We get
$I_0 = \sum_{t \in R} [I_t * (-1)^{\mu(t)}] $.
Here $R$ is the set of positive integers which are of the form 
(warning: sloppy notation) $\prod_{J \subset n} p_j$,
that is all integers whose prime factorizations have only primes less than
or equal to the nth prime, and those occuring only to at most the first power.
(More succinctly, $R$ is also the set of positive divisors of $P_n$.)  For
such a number $t \in R$, $\mu(t)$ is precisely the number of prime factors in $t$,
and $\mu$ is chosen to suggest the Moebius function whose value at such $t$ is 
$(-1)^{\mu(t)}$.  The equality is exact.
Step 2 is to replace $I_t$ with a linearized approximation plus an error term which
I will call $E(t)$.    This substitution gives:
$I_0 = \sum_{t \in R} [ (E(t) + x/t) * (-1)^{\mu(t)} ]$  .
Since the number of multiples of $t$ in the interval $(a, a + x)$ is roughly $x/t$,
the error term $E(t)$ is bounded in absolute value by $1$.  Step 3 will rewrite
the RHS and estimate it pessimistically: $E(t) * (-1)^{\mu(t)}$ will be replaced
by $-1$, and the alternating sum of $x/t$ terms can be rewritten as a product
involving terms of the form $(1 - 1/p_i)$, where $p_i$ is the $i$th prime. There are 
$2^n$ terms of the form $E(t)$, so one gets:
$I_0 \geq [x * \prod_{1 <= i <= n} (1 - 1/p_i) ] - 2^n  = x/Q - 2^n$ .
Here $Q$ is an abbreviation for $1$ divided by the product of the n terms $(1 - 1/p_i)$.
It is roughly log n  for large n.  Here comes the kicker.  Step 4
notes that steps 1 through 3 are essentially independent of $a$, and if $x$ can be
chosen so that $x/Q - 2^n > 0$, then $I_0 > 0$ which means at least one of the $q_i$
is in the interval $(a, a+x)$ when $a > 0$, and such $x$ would be an upper bound for 
$q_{i+1} - q_i$ which is independent of $i$. So choose $x = Q * 2^n$ plus epsilon.
I thought it a neat enough argument (especially the kicker)
that I am sharing it here with other non-students of sieve theory.  Now to the questions.
1) Is there any work done which improves the upper bound for $q_{i+1} - q_i$?
The answer to this is yes, since in a footnote Westzynthius shows how to
improve the bound to $Q * 2^{n-1}$ by counting odd multiples.  So I really want
to know if there are even better bounds out there, done by additional researchers.
I would expect a provable bound to be $Q * 2^g$,  where $g$ is something like a 
polynomial in log(n), but even having $g$ be n to a fractional power would be something.
2) Is there work done which uses something like the Bonferroni inequalities to
improve the above argument?
3) Did Westzynthius publish any other work (possibly nonmathematical) besides the
paper that includes the argument above?
Motivation: I am considering improvements to this argument which do establish
better upper bounds, and am wondering how to push the exponent from n - o(1) 
 down to poly(log(n)).  Especially, I want to know if I am rediscovering how
to replace n by cn for some $c < 1$, as opposed to discovering how to do it.
Gerhard "Ask Me About System Design" Paseman, 2010.09.03

REPLY [2 votes]: After studying Kanold's 1967 paper on Jacobsthal's function, (and being inspired by a preprint http://arxiv.org/abs/1208.5342 that I discuss below,) I found an argument, mostly very simple, which gives some nice results for the effort given.  While Kanold deserves some of the credit for the argument, I have yet to see a statement by him or by anyone else that gives these results, so I present them here.  (Kanold wrote several articles on Jacobsthal's function, many of which I am tracking down, which might have this argument.  I am happy to accept help in obtaining electronic copies of them.)  This is the post I promised over a few months ago in a supplement to a question of Timothy Foo, 
Analogues of Jacobsthal's function  .
For maximum ooh-aah effect, I assume $n$ is squarefree and has $k \gt 2$ prime factors, one of which I call Peter, or $p$ for short.Now $1+tn$ is coprime to $n$ for any integer $t$.  So are most integers of the form $1 + tn/p$, the exceptions being those that are multiples of $p$, and those multiples do not occur as consecutive terms.  Thus, every interval of length $2n/p (=g(p)n/p)$ has at least one integer coprime to $n$ of the form $1+tn/p$.
Let's go further with this.  Let $d \gt 1$ and divide $n$, and let $f=n/d$.  (Here I use $n$ squarefree to get $f$ coprime to $d$.)  Then numbers of the shape $1+tf$ form an arithmetic progression, are coprime to $f$, and (as can be seen by multiplying by $f$'s inverse in the ring of integers mod $d$) you can't pick $g(d)$ consecutive members of this progression without hitting something coprime to $d$ also.  So $g(n) \leq g(d)f = g(d)n/d$ .
While I'm here, let me sharpen the inequality, assuming $f \gt 1$ and $d \gt 1$ are coprime:
there are $\phi(f)$ totients $c$ of $f$ in the interval $[0,f]$, so I can repeat the argument with $c+tf$ instead of $1+tf$.  In the worst case, using all $\phi(f)$ progressions, I get $g(fd) \leq g(d)f - f + g(f)$, which mildly improves upon Kanold's bound $g(d)f -\phi(f)+1$, and matches it when $f$ is prime.  (Of course, for $n=fd$ I really want $g(n)$ to be near $O(g(d)+g(f))$, but I don't yet know how to show that with grade school arithmetic.)
How to use this inequality? Pick the largest divisor $d$ for which one can comfortably compute (a subquadratic in $k$ upper bound for) $g(d)$; I pick $d$ to contain most of the large prime factors of $n$: find prime $q$ dividing $n$ so that $\sigma^{-1}(d)=\sum_{p \text{ prime,} p|n, p \geq q} 1/p$ is less than $1 + 1/2q$; a routine argument yields $g(d)$ is $O(qk)$.  The ugly part is to show that $q \lt k^{0.5}$ (or else $d=n$), that $n/d \lt 2^{3q/2}$ which for large $k$ approaches $2^{3(k^{\epsilon + 1/e})/2}$, and that asymptotically $g(n)$ is $O(e^{k^{1/e}+D\log(k)})$.  This isn't hard after using one of Mertens's theorems and a Chebyshev function; it just isn't pretty.  (Also for smaller $n$, $\epsilon + 1/e$ can be close to $1/2$, but with patience $\epsilon$ will tend to zero.)
This gives a bound that is asymptotically better than my first efforts at this, improves slightly ($k^{0.5}$ replaced by $Ck^{0.37}+ D\log(k)$ on Kanold's bound of $2^\sqrt{k}$ for $k$ not too large, and does not need Kanold's requirement that $k > e^{50}$.  Up until one chooses $d$ and crunches the formulae, it is also a very elementary argument; I suspect even Legendre knew about using the multiplicative inverse to transform a general arithmetic progression to a (effectively) consecutive sequence of integers and still preserve the property of interest here, being a unit in a certain ring (or missing it by that much).  
(One of the benefits of letting this sit for a few months before posting is that I can add cool observations like: If I could get the inequality down to $g(n) \leq g(d)g(n/d)$, I could iterate the
above simple estimate to get an explicit bound of $O(k^c)$, where $c$ is a positive number less than 3.  Or like: using more advanced work combined with the above, I can get $g(n) 
\leq e^{k^{e^{-a}}}Ck^{a}$ for some integers $a$, which seems better than $Ck^{4\log\log{k}}$ if you don't look too closely.)
Further, one can use a computer to refine the method slightly and get estimates which do quite well for small values of $n$, where small here means $k<100$.  Asymptotically though, Stevens's and my upper bounds eventually outperform this bound.
Also, there has been a nice result out of University College Dublin that I will briefly interpret.  Fintan Costello and Paul Watts find a way of presenting a related function recursively, then numerically compute a lower bound on this function which implies an upper bound on Jacobsthal's function computed on some particular values.  I thank them for reminding me about using a multiplicative inverse mod $d$ for $f$, so they deserve a "piece of the action".
These authors work in (and sometimes away from) the integer interval $BM = [b+1,b+2,\ldots,b+m]$.  Given squarefree $n$ and its distinct prime factors, listed in some order as $q_1$ to $q_k$, define $Q_i$ as  $\prod_{0 \lt j \leq i} q_j$. One approach to computing the size $\pi(b,m,n)$ of the set $CP(b,m,n)$ which has those integers in $BM$ coprime to $n$ is to do the standard inclusion-exclusion argument: if we represent by $F(b,m,d)$ the multiples of $d$ in $BM$, and say there are $f(b,m,d)$ many such multiples, and abuse some notation, I then write
$CP(b,m,n) = \sum_{d | n} sgn(d,F(b,m,d))$ .  Here $sgn$ is to suggest adding elements of the set $F(b,m,d)$ if $d$ has an even number of prime factors, and subtracting them instead when $d$ has an odd number of prime factors.
To set up for the recurrent expression, Costello and Watts use just some of the terms on the right hand side of the abused equation, and reorganize the rest of the terms.  In my interpretation of their work, they start with the multiset identity
$$CP(b,m,n) \cup \biguplus_{0 \lt i \leq k} F(b,m,q_i) =
BM \uplus  \biguplus_{0 \lt i \lt j \leq k} RCP(i,j)$$
where $RCP(i,j)$ is $F(b,m,q_iq_j) \cap CP(b,m,Q_{i-1})$, or the subset of $BM$ which has those multiples of $q_iq_j$ whose soonest prime factor in common with $n$ is $q_i$.  
One sees this identity holds by considering a member of $BM$ which has exactly $t$ distinct prime factors in common with $n$.
If $t$ is $0$, then the member occurs only once in $CP(b,m,n)$ and similarly only once in $BM$.  Otherwise, it occurs exactly $t$ times in the left hand side in $t$ distinct terms $F(b,m,q_i)$, and if $l$ is soonest such that $q_l$ is a prime factor of the member, the member occurs only once in each of $t-1$ sets
$RCP(l,j)$ (remember $l$ comes sooner than $j$) and only once in $BM$.
Now the term $RCP(i,j)$ is a subset of an arithmetic progression $A$ with common difference $q_iq_j$. By using the technique above of multiplying by a suitable inverse of $q_iq_j$ in the ring of integers mod $Q_{i-1}$, $A$ corresponds with an integer interval starting near some integer $c_{ijbm}$ of length $f(b,m,q_iq_j)$ which preserves the coprimality status with respect to $Q_{i-1}$: to wit, the size of $RCP(i,j)$ is $\pi(c_{ijbm},f(b,m,q_iq_j),Q_{i-1})$.  Using the $\pi$ term for the size of $CP$ and translating the other sets to numbers gives the numerical recurrent formula of Costello and Watts:
$$\pi(b,m,n) = m - \sum_{0 \lt i \leq k} f(b,m,q_i) 
+ \sum_{0 \lt i \lt j \leq k} \pi(c_{ijbm},f(b,m,q_iq_j),Q_{i-1})$$.
Following work of Hagedorn who computed $h(k)=g(P_k)$ for $k$ less than 50, where $P_k$ is the $k$th primorial, Costello and Watts use their formula and some analysis of coincidence of prime residues to compute an inequality for $\pi_{min}(m,n)$ which is the minimum over all integers $b$ of $\pi(b,m,n)$.  They underestimate $f(b,m,q_iq_j)$ by $\lfloor m/q_iq_j \rfloor$, ignore the $c$'s by using $\pi_min$, pull out the $i=1$ terms from the double sum and rewrite that portion to include a term $E$, depending only on $m$ and the $p_i$, which arises from looking at when estimates for the sizes of the $F(b,m,p_i)$  and $F(b,m,2p_i)$ sets can be improved, and come up with (a refined version, using $p$'s for $q$'s, of) the inequality 
$$m - \sum_{0 \lt i \leq k}  \lceil \frac{m}{p_i} \rceil + \sum_{1 \lt i \leq k} \lfloor \frac{m}{2p_i} \rfloor + E + \sum_{1 \lt i \lt j \leq k} \pi_{min}(\lfloor \frac{m}{p_ip_j} \rfloor,P_{i-1}) \leq \pi_{min}(m,P_k)$$.
With this inequality, Costello and Watts compute $\pi_{low}$, a lower bound approximation to $\pi_{min}$.  Since $h(k) \leq m$ iff $\pi_{min}(m,P_k) \gt 0$, computing $\pi_{low}(m,P_k)$ for various $m$ will give an upper bound on $h(k)$.  They say their computations for $k \leq 10000$ suggest $h(k) \leq Ck^2 \log k$, where $C$ is a constant less than $0.3$ .  Although this data is achieved using data from Hagedorn's work, even without that their algorithm yields values which are a vast improvement on known and easily computable bounds, even the ones listed above.
One item to explore is how an algorithm based on this approximation will perform given different orderings of the prime factors.  I suspect that letting the larger primes come first will give tighter results.  Another item to explore is to see if there is a better term $F$ that will supplant $E$ and some of the recurrent terms in the double sum.  The idea of rewriting the $\pi$ function recursively, while not new, is given new life in this double sum form, and suggests revisiting some old approaches with an eye toward computability.
Gerhard "Ask Me About Coprime Integers" Paseman, 2013.02.05<|endoftext|>
TITLE: Isomorphism between two universal p-typical formal group laws
QUESTION [16 upvotes]: EDIT: I've tried to alter the question so that its basic nature is clearer, as it's been unclear to a number of people now.
At any prime p, there is a graded polynomial ring $V \cong {\mathbb Z}_{(p)}[v_1, v_2, \ldots]$ carrying two formal group laws.  These formal group laws are of the form
$$
F(x,y) = \ell^{-1}(\ell(x) + \ell(y))
$$
for a logarithm $\ell(x) = \sum \ell_n x^{p^{n+1}} \in ({\mathbb Q} \otimes V)[\![x]\!]$ (where $\ell_0 = 1$ by convention).  Both of these formal group laws have the property that they are universal among so-called $p$-typical formal group laws, and see heavy computational use in stable homotopy theory.
These two are based on choices of recursive definition for the logarithm coefficients in terms of the generators $v_i$ of $V$.  The first definition (the Araki generators) satisfies:
$$
p \ell_n = \sum_{k=0}^n v_k^{p^{n-k}} \ell_k = v_n + \ell_1 v_{n-1}^p + \cdots + \ell_{n-1} v_1^{p^{n-1}} + \ell_n p^{p^n}
$$
The Hazewinkel generators are instead defined by:
$$
p \ell'_n = \sum_{k=1}^{n} v_k^{p^{n-k}} \ell'_k = v_n + \ell'_1 v_{n-1}^p + \cdots + \ell'_{n-1} v_1^{p^{n-1}}
$$
This gives the ring $V$ with two logarithms $\ell$ and $\ell'$, and two distinct universal formal group laws.
My question is: Are these two formal group laws isomorphic?  Strictly isomorphic?
ADDED: Since the ring is torsion free, any isomorphism between them is of the form $f(x) = (\ell')^{-1} (c \ell(x))$ for a unit $c \in \mathbb{Z}_{(p)}^\times = V^\times$.  They are therefore isomorphic if and only if they are strictly isomorphic.  Therefore, the question is equivalent to the following:
Does the power series $(\ell')^{-1} \circ \ell$ have coefficients in $V \subset V \otimes \mathbb{Q}$?
(The issue was brought up when thinking about truncated Brown-Peterson spectra ${\rm BP}\langle n\rangle$, whose rings of coefficients are $V/(v_{n+1},v_{n+2}, \cdots)$.  It then becomes a question as to whether these are equivalent as ring spectra depending on the choice of generators.  There are certainly different choices of generators for which they are inequivalent.)

REPLY [4 votes]: As a way of additional information - explicit expressions of the $l$s in terms of the generators look like this:  
for Hazewinkel generators,
$$
l_n=\sum_{\substack{n_1+...+n_k=n,\\1\leqslant k\leqslant n}}\frac1{p^k}v_{n_1}v_{n_2}^{p^{n_1}}v_{n_3}^{p^{n_1+n_2}}\cdots v_{n_k}^{p^{n_1+n_2+...+n_{k-1}}}
$$
while for the Araki ones,
$$
l_n=\frac1{p^{p^n-1}-1}\sum_{\substack{n_1+...+n_k=n,\\1\leqslant k\leqslant n}}-\frac{(-1)^k}{p^k}\frac{v_{n_1}v_{n_2}^{p^{n_1}}v_{n_3}^{p^{n_1+n_2}}\cdots v_{n_k}^{p^{n_1+n_2+...+n_{k-1}}}}{(p^{p^{n_1}-1}-1)(p^{p^{n_1+n_2}-1}-1)\cdots(p^{p^{n_1+n_2+...+n_{k-1}}-1}-1)}.
$$
I would not object if anybody finds this uselessly horrible --- such people may view this as a joke.
Then for the latter, as an additional joke, note that because of these expressions the following becomes tantalizingly close to being true:  
Araki's $v_n$ is ``almost'' equal to $-\frac1{\sqrt[p^n]{p^{p^n-1}-1}}$ times the Hazewinkel's $v_n$ :)<|endoftext|>
TITLE: Nash embedding theorem for 2D manifolds
QUESTION [22 upvotes]: The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$. What can we say in the special case of 2-manifolds? For example, can we always embed a 2-manifold in $R^3$?

REPLY [3 votes]: I had a conversation today with Christopher Croke of Penn., he pointed out that there was progress, extremely recent progress, on the local version, isometrically embedding a piece of surface in $\mathbf R^3.$ This is all in the $C^\infty$ category, it is always possible that more is known in $C^\omega$ and very likely less is known in, say $C^2.$
Anyway, if the curvature does not vanish in the neighborhood, then it can be isometrically embedded in  $\mathbf R^3.$ The same is true if the curvature vanishes at a point but some directional derivative does not. 
Well, from a new item on the arXiv by Qing Han and Marcus Khuri,

It is shown that if the Gaussian
  curvature vanishes to finite order and
  its zero set consists of two Lipschitz
  curves intersecting transversely at a
  point, then local sufficiently smooth
  isometric embeddings exist.

The title of the preprint is "On the Local Isometric Embedding in  $\mathbf R^3$ of Surfaces with Gaussian Curvature of Mixed Sign."
http://arxiv.org/abs/1009.6214 
Both Han and Khuri have published on this topic in the Journal of Differential Geometry, Han in 2003, Khuri in 2007.<|endoftext|>
TITLE: Which Riemannian manifolds admit a finite dimensional transitive Lie group action?
QUESTION [7 upvotes]: This is a basically an adjusted version of my earlier question about how to define a convolution algebra on a general Riemannian manifold.  The motivation for asking such a question of course comes from the observation that if G is a group and X is a manifold and the action of G on X is transitive, then the pullback from each point in X to its orbit is faithful.  This then cuts out an ideal in the convolution algebra on G which would (hopefully) correspond to some type of general convolution on X (and that would be pretty handy to have for many obvious reasons).
My intuition is that for 2D surfaces (which is the case I am most interested in right now), the group is going to be something like $PSL(2) / \pi(X)$ with the action obtained by the pushforward of the action of $PSL(2)$ on $RP^2$ by the universal covering of $RP^2 / \pi(X)$.  Of course, trying to work all of this out via quotient relations is an enormous pain in the neck, so it would be nice to maybe avoid some headaches and instead try to find maybe some standard references for this sort of thing (if they exist at all).

REPLY [7 votes]: I think that one of the nicest results in this direction may be the one in Ambrose and Singer's paper "On homogeneous Riemannian manifolds". They give a local NASC for a complete Riemannian manifold to be homogeneous. There is a follow-up book by Tricerri and van Hecke called Homogeneous structures on Riemannian manifolds.<|endoftext|>
TITLE: Is the Grothendieck ring of varieties reduced?
QUESTION [27 upvotes]: A neat construction of Bjorn Poonen shows that the Grothendieck ring of varieties (over a field of char. 0) is not a domain: http://arxiv.org/abs/math/0204306 
Is the Grothendieck ring of varieties reduced? (My guess: the answer is yes, the proof is easy enough that several people have observed this without writing it up anywhere. But I don't know how to show it.)

REPLY [15 votes]: Qing Liu's example probably works, only we don't know if an abelian variety in
positive characteric is determined by its class in
$K_0(\mathrm{Var}_k)$. However, we do know that in characteristic zero (this is
what Bjorn Poonen uses in his examples) and the non-cancellation is a purely
arithmetic phenomenon and hence can be realised in characteristic zero.
Hence, we let $\mathcal A$ be a maximal order in a definite (i.e., $\mathcal
A\otimes\mathbb R$ is non-split) quaternion algebra over $\mathbb Q$. There is
an abelian variety $A$ over some field $k$ of characteristic zero with $\mathcal
A=\mathrm{End}(A)$ (Bjorn works hard to get his example defined over $\mathbb
Q$, here I make no such claim). For any (right) f.g. projective (i.e., torsion
free) $\mathcal A$-module $M$ we may define an abelian variety
$M\bigotimes_{\mathcal A}A$ characterised by $\mathrm{Hom}(M\bigotimes_{\mathcal
  A}A,B)=\mathrm{Hom}_{\mathcal A}(M,\mathrm{Hom}(A,B))$ for all abelian
varieties (concretely it is constructed by realising $M$ is the kernel of an
idempotent of some $\mathcal A^n$ and then taking the kernel of the same
idempotent acting on $A^n$). In any case we see that $M$ and $N$ are isomorphic
precisely when $M\bigotimes_{\mathcal A}A$ is isomorphic to
$N\bigotimes_{\mathcal A}A$. 
Now (all the arithmetic results used below can be found in for instance Irving
Reiner: Maximal orders, Academic Press, London-New York), the class group of
$\mathcal A$ is equal to the ray class group of $\mathbb Q$ with respect to the
infinite prime, i.e., the group of fractional ideals of $\mathbb Q$ modulo
ideals with a strictly positive generators. As that is all ideals we find that
the class group is trivial. Furthermore, we have the Eichler stability theorem
which says that projective modules of rank $\geq2$ are determined by their rank and
image in the class group and hence are determined by their rank (the rank
condition comes in in that $\mathrm{M}_k(\mathcal A)$ is a central simple
algebra which is indefinite at the infinite prime). In particular if $M_1$ and
$M_2$ are two rank $1$ modules over $\mathcal A$ and $A_1$ and $A_2$ are the
corresponding abelian varieties we get that $A_1\bigoplus A_2\cong A\bigoplus A$
as the left (resp. right) hand side is associated to $M_1\bigoplus M_2$ (resp.
$\mathcal A^2$). Therefore, to get an example it is enough to give an example of
an $\mathcal A$ for which there exist $M_1\not\cong M_2$. The number (or more
easily the mass) of isomorphism classes of ideals can be computed using mass
formulas and tends to infinity with the discriminant of $\mathcal A$. It is
interesting to note that when the discriminant is a prime $p$ we can go
backwards using supersingular elliptic curves: The mass is equal to the mass of
supersingular elliptic curves in characteristic $p$ and the latter mass can be
computed geometrically to be equal to $(p-1)/24$.<|endoftext|>
TITLE: Projective dimension of zero module
QUESTION [13 upvotes]: Is there any consensus on what the projective dimension of the zero module should be? Here are three statements one commonly encounters in textbooks, sometimes with or without the condition $M\neq 0$:
(1) $\mbox{pd}(M)\leq n$ iff $\mbox{Ext}^{n+1}(M,-)=0$
(2) $\mbox{pd}(M)=0$ iff $M$ is projective
(3) $\mbox{grade}(M):=\infty$ if $M=0$
If one attempts to define $\mbox{pd}((0))$ by extending one of these results, (1), (2), (3) suggest $\mbox{pd}=-1, 0, \infty$, respectively.

REPLY [7 votes]: Let me explain a definition of projective dimension which gives the same result as the one given by Sándor Kovács, but without any restriction on the ring or the module we are talking about. This is, by the way, the one chosen by Bourbaki (A.X.8.1).
Let $A$ be a ring.
0) We write $\overline{\mathbb{Z}}=\mathbb{Z}\cup\{-\infty,\infty\}$ and furnish $\overline{\mathbb{Z}}$ with the ordering that extends the canonical ordering on $\mathbb{Z}$ and has $\infty$ as greatest and $-\infty$ as smallest element. We convene that suprema and infima of subsets of subsets of $\overline{\mathbb{Z}}$ are always understood to be taken in $\overline{\mathbb{Z}}$.
1) If $C$ is a complex of $A$-modules and $C_n$ denotes its component of degree $n\in\mathbb{Z}$, then we set $$b_d(C)=\inf\{n\in\mathbb{Z}\mid C_n\neq 0\}$$ and $$b_g(C)=\sup\{n\in\mathbb{Z}\mid C_n\neq 0\},$$ and we call $$l(C)=b_g(C)-b_d(C)$$ the length of $C$. Note that if $C$ is the zero complex then we have $b_d(C)=\infty$ and $b_g(C)=-\infty$, hence $l(C)=-\infty$.
2) If $M$ is an $A$-module and $(P,p)$ is a left resolution of $M$, then the length $l(P)$ of the complex $P$ is called the length of $(P,p)$. Note that if $P$ is the zero complex (which may be the case if and only if $M=0$) then the length of $(P,p)$ is $-\infty$.
3) If $M$ be an $A$-module, then the infimum of the lengths of all projective resolutions of $M$ is called the projective dimension of $M$. Hence, if $M=0$ then we have a projective resolution of length $-\infty$, and thus the projective dimension of $M$ is also $-\infty$. Conversely, if $M$ has projective dimension $-\infty$ then - since every $A$-module has a projective resolution - it necessarily has a projective resolution of length $-\infty$, and thus it follows $M=0$.
Note: This clearly makes sense in every abelian category with enough projectives, and there are obvious variants of the above that yield analogous definitions of injective or flat dimensions.<|endoftext|>
TITLE: Is the blowup of a normal scheme necessarily normal?
QUESTION [17 upvotes]: Is the blowup of an integral normal Noetherian scheme along a coherent sheaf of ideals necessarily normal?
I can show that there is an open cover of the blowup by schemes of the form $\text{Spec } C$, where $B \subset C \subset B_g$ for some integrally closed domain $B$ and some $g \in B$, but I don't see why this would imply that $C$ is integrally closed.  Intuitively, it seems reasonable that a blowup would be at least as "nice" as the original scheme, but that intuition may have more to do with how blowups are generally used than what they are capable of.

REPLY [23 votes]: Richard Borcherds sort of example will certainly work.  There's another type of example where the ambient space is smooth.
For example, blowing up $(x^2, y^2)$ in $\mathbb{A}^2$ will yield a pinch-point (aka Whitney's Umbrella) singularity.<|endoftext|>
TITLE: Which recursively-defined predicates can be expressed in Presburger Arithmetic?
QUESTION [14 upvotes]: In Presburger Arithmetic there is no predicate that can express divisibility, else Presburger Arithmetic would be as expressive as Peano Arithmetic.  Divisibility can be defined recursively, for example $D(a,c) \equiv \exists b \: M(a,b,c)$, $M(a,b,c) \equiv M(a-1,b,c-b)$, $M(1,b,c) \equiv (b=c)$.  But some predicates which can be expressed in Presburger Arithmetic also have recursive definitions, for example $P(x,y,z) \equiv (x+y=z)$ versus $P(x,y,z) \equiv P(x-1,y+1,z)$, $P(0,y,z) \equiv (y=z)$.
How to tell if a predicate, defined recursively without use of multiplication, has an equivalent non-recursive definition which can be expressed in Presburger Arithmetic?

REPLY [9 votes]: In addition to the answers above, it is worth mentioning that the existential fragment of Presburger arithmetic can actually be extended by a full divisibility predicate while retaining decidability [1]. Satisfiability in the resulting fragment is in NEXP, the lower bound being NP [2].
Also, it is possible to define a family of formulas $\Phi_n(x,m)$ such that for every natural number $m$ representable using $n$-bits, $\Phi_n(x,m)$ holds if and only if $x \equiv 0 \bmod m$ [3].

[1] L. Lipshitz, “The Diophantine problem for addition and divisibility,”
Transactions of the American Mathematical Society, vol. 235, pp. 271–
283, 1976.

[2] Lechner, A.; Ouaknine, J.; Worrell, J., "On the Complexity of Linear Arithmetic with Divisibility," in Proceedings of the 30th Annual ACM/IEEE Symposium on Logic in Computer Science (LICS), pp. 667-676, 2015.

[3] Haase, C., "Subclasses of Presburger arithmetic and the weak EXP hierarchy," in Proceedings of the Joint Meeting of the Twenty-Third EACSL Annual Conference on Computer Science Logic (CSL) and the Twenty-Ninth Annual ACM/IEEE Symposium on Logic in Computer Science (LICS) (CSL-LICS '14). ACM, New York, NY, USA, Article 47, 2014.<|endoftext|>
TITLE: Proving a hypergeometric  function identity
QUESTION [6 upvotes]: While playing around with the fractional calculus, I got stuck trying to show that two different ways of differintegrating the cosine give the same result. DLMF and the Wolfram Functions site don't seem to have this "identity" or something that can obviously be transformed into what I have, so I'm asking here.
The "identity" in question is
$(\alpha-1)\left({}_1 F_2 \left(1;\frac{1-\alpha}{2},\frac{2-\alpha}{2};-\frac{x^2}{4}\right)-{}_1 F_2 \left(-\frac{\alpha}{2};\frac12,\frac{2-\alpha}{2};-\frac{x^2}{4}\right)\cos(x)\right)\stackrel{?}{=}\alpha x \sin(x)\,{{}_1 F_2 \left(\frac{1-\alpha}{2};\frac32,\frac{3-\alpha}{2};-\frac{x^2}{4}\right)}$
Expanding the LHS minus the RHS in a Taylor series shows that the coefficients up to the 50th power are 0; trying out random complex values of $\alpha$ and $x$ seems to verify the identity. I would however like to see a way to confirm the identity analytically. How do I go about it?

REPLY [7 votes]: You can use the great HolonomicFunctions package by Christoph Koutschan to prove this identity in Mathematica. It automatically proves for you that both sides of your identity satisfy the sixth order differential equation
\begin{eqnarray}
   0=&&x^2 \left(2 a^2-11 a+18 x^2+14\right) D_x^6
-2 x \left(2 a^3-19 a^2+18 a x^2+58 a-54 x^2-56\right) D_x^5 \\\\
&&+\left(2 a^4-25 a^3+28 a^2 x^2+115 a^2-133 a x^2-230
   a+90 x^4+154 x^2+168\right) D_x^4 \\\\
&&-4 x
   \left(4 a^3-37 a^2+36 a x^2+115 a-99 x^2-114\right)
   D_x^3 \\\\
&&+4 \left(2 a^4-23 a^3+20 a^2
   x^2+96 a^2-71 a x^2-172 a+18 x^4+71 x^2+112\right) D_x^2 \\\\
&&+8 x \left(4 a^2-34
   a+36 x^2+43\right) D_x
+8 \left(2 a^2-17 a+18 x^2+35\right).
\end{eqnarray}
Together with your check that the first six Taylor coefficients (with respect to x) agree, this proves your identity.<|endoftext|>
TITLE: Why are there 1024 Hamiltonian cycles on an icosahedron?
QUESTION [52 upvotes]: Fix one edge $e$ of the graph (1-skeleton) of an icosahedron.
By a computer search, I found that there are 1024 Hamiltonian cycles that include $e$.
[But see edit below re directed vs. undirected!]
With the two endpoints of $e$ fixed, there are 10 "free" vertices in the cycle.
Because $1024=2^{10}$, it makes me wonder if there might be a combinatorial viewpoint that
makes it evident that there are 1024 cycles including a fixed edge.
It could just be a numerical coincidence, but if anyone
sees an idea for an argument, I'd appreciate hearing it. Thanks!



Incidentally, this MathWorld page says there are 2560 Hamiltonian cycles all together (without
the fixed edge condition). (Thanks to Kristal Cantwell for pointing me to this page.)
Edit. I apologize for misleading! :-/  When I looked at the full output of paths more carefully,
I realize I inadvertently computed directed cycles, so each is represented twice, i.e., both
$$ \lbrace 2, 7, 6, 11, 8, 9, 4, 10, 12, 5, 3, 1 \rbrace $$
$$ \lbrace 1, 3, 5, 12, 10, 4, 9, 8, 11, 6, 7, 2 \rbrace $$
are included, etc.  So there are 512 undirected cycles, 1024 directed cycles.
The paths are listed here: hpaths.html.

REPLY [17 votes]: I've fed these Hamiltonian cycles into Brendan McKay's NAUTY software.  They fall into the following isomorphic collections:

 6 asymmetric collections, with 48 examples of each
 3 collections with a rotational symmetry, with 24 examples of each
 5 collections with a reflectional symmetry, with 24 examples of each
 2 collections with two reflectional symmetries, with 12 examples of each
 1 collection with six-fold rotational symmetry, with 8 examples.

It makes some sense that an asymmetric cycle will have 48 examples: the specified link can be any of the 12 links, in either direction, and there is a two-fold symmetry in the graph once one directed link has been selected.  This gives the correct total (using undirected counts):
   6×48 + 3×24 + 5×24 + 2×12 + 1×8 = 512 
See below for a picture of the last type, by the way. This uses the same vertex numbers as in the original question.  From the picture, you can believe that there are two isomorphic classes of vertex, outer and inner -- different, for example, in that each outer vertex is connected to the next outer vertex in the cycle.  All links are between an outer and an inner vertex, but there are two isomorphic classes, and the cycle alternates between them, ABABABABABAB.  (You can see that they are different by comparing the result of ABA and BAB -- specifically, whether the end is connected to the start.)

I have
images
of the other symmetrical cycles.
  Anyway, the point is that only one of the types has 3-fold symmetry -- as Robin pointed out, there must be at least one for the total to be indivisible by 3.  But with only one, it will prove impossible to partition the collections into two groups of 256.  This looks like bad news for a base-2 explanation of 512.<|endoftext|>
TITLE: A possible generalization of the homotopy groups.
QUESTION [23 upvotes]: The homotopy groups $\pi_{n}(X)$ arise from considering equivalence classes of based maps from the $n$-sphere $S^{n}$ to the space $X$. As is well known, these maps can be composed, giving arise to a group operation. The resulting group contains a great deal of information about the given space. My question is: is there any extra information about a space that can be discovered by considering equivalence classes of based maps from the $n$-tori $T^{n}=S^{1}\times S^{1}\times \cdots \times S^{1}$. In the case of $T^{2}$, it would seem that since any path $S^{1}\to X$ can be "thickened" to create a path $T^{2}\to X$ if $X$ is three-dimensional, the group arising from based paths $T^{2}\to X$ would contain $\pi_{1}(X)$. Perhaps more generally, can useful information be gained by examining equivalence classes of based maps from some arbitrary space $Y$ to a given space $X$.

REPLY [8 votes]: I was told by Brian Griffiths that Fox was hoping to obtain a generalisation of the van Kampen theorem and so continue work of J.H.C Whitehead on adding relations to homotopy groups (see his 1941 paper with that title). 
However if one frees oneself from the base point fixation one might be led to consider Loday's cat$^n$-group of a based  $(n+1)$-ad, $X_*=(X;X_1, \ldots, X_n)$; let $\Phi X_*$  be the space of maps $I^n \to X$ which take the faces of the $n$-cube $I^n$ in direction $i$ into $X_i$ and the vertices to the base point. Then $\Phi$ has  compositions $+_i$ in direction $i$ which form a lax $n$-fold groupoid. However the group $\Pi X_*= \pi_1(\Phi, x)$, where $x$ is the constant map at the base point $x$, inherits these compositions to become a cat$^n$-group, i.e. a strict  $n$-fold groupoid internal to the category of groups (the proof is non trivial). 
There is a Higher Homotopy van Kampen Theorem for this functor $\Pi$ which enables some new nonabelian calculations in homotopy theory (see our paper in Topology 26 (1987) 311-334). 
So a key step is to move from spaces with base point to certain structured spaces. 
Comment Feb 16, 2013: The workers in algebraic topology near the beginning of the 20th century were looking for higher  dimensional versions of the fundamental group, since they knew that the nonabelian fundamental group was useful in problems of analysis and geometry. In 1932, Cech submitted a paper on Higher Homotopy Groups to the ICM at Zurich, but Alexandroff and Hopf quickly proved the groups were abelian for $n >1$ and on these grounds persuaded Cech to withdraw his paper, so that only a small paragraph appeared in the Proceedings. It is reported that Hurewicz attended that conference. In due course, the idea of higher versions of the fundamental group came to be seen as a mirage. 
One explanation of the abelian nature of the higher homotopy groups is that group objects in the category of groups are  abelian groups, as a result of the interchange law, also called the Eckmann-Hilton argument. However  group objects in the category of groupoids are equivalent to crossed modules, and so are in some sense "more nonabelian" than groups. Crossed modules were first defined by J.H.C. Whitehead, 1946, in relation to second relative homotopy groups. This leads to the possibility, now realised,   of "higher homotopy groupoids", Higher Homotopy Seifert-van Kampen Theorems,  and the notions of higher dimensional group theory. 
See this presentation for more background.<|endoftext|>
TITLE: How Do You Go About Learning Mathematics?
QUESTION [8 upvotes]: I really like mathematics, but I am not good at learning it. I find it takes me a long time to absorb new material by reading on my own and I haven't found a formula that works for me. I am hoping a few people out there will tell me how they go about learning math so I can try out their systems.
I need to know basic things. Should I use one book at a time or should I be reading many books on the same topic at once? Do you stop reading when you hit on a fact that you don't understand or do you keep reading?
Do you read all in one go or do you do a little bit and for how long (1 hr, 2 hr or more?)
Do you read all the chapters or do you do all the exercises before moving on from a chapter?
Do you adjust your technique in Calculus(calculation heavy) vs. Analysis (proof heavy)? If so, how?
When you make notes, what do you make notes about? Do you make notes while you read or after?
Is there some note writing system (eg. Cornell system) that you find superior for taking mathematics?
If you think these decisions all depend, can you say what they depend on?
I am really lost here. I would appreciate any input.
Full Disclosure: I have asked this question on Math Stack Exchange.  
I am looking for a diversity of approaches. I hope this question is on-topic here.

REPLY [6 votes]: "What is the most effective way to learn mathematics?"
I have been trying to answer this question for myself, and one measure I've taken towards this goal is to record all of my mathematical reading, work, and random thoughts in a journal.  I highly recommend the practice as it has been very illuminating to me since I started a few months ago.  Reviewing my previous readings allows me to ascertain how much math I actually end up retaining from my study sessions, and keeping all of my work in one place (as opposed to throwaway scrap paper) allows me to spot any particularly common mistakes.
So far, I've found that my memory is far more tenuous than I had previously assumed.  I'd look at last month's entries and realize that I'd only retained 20% of what I had learned; fine details being especially prone to slippage.  Yet from analyzing my mistakes, I've also found that those very details are much more crucial than I had thought.
The result of all of this is that I've started to shift my focus from "learning new math rapidly" (which has been my focus since I am still an undergraduate) to "winning the uphill battle against memory loss."  From this new perspective, the old adage: "the only way to learn mathematics is through doing" begins to make a lot more sense. While active learning is far from any cure to forgetfulness, given my own mnemonic capabilities I have come to see that it would probably be a better long-term investment to spend a month on fully working and understanding a chapter, than to spend the same time blazing through several chapters but skipping the exercises (having done both.)
I emphasize again that this is my own conclusion based on my own characteristics, and that is precisely why I recommend everyone to find their own answer to this question by keeping their own math notebook.<|endoftext|>
TITLE: A question about  homeomorphic subsets of  Hilbert  space
QUESTION [5 upvotes]: Let H be a an infinite dimensional and separable Hilbert space. Let C be a closed and
bounded subset of H that is not compact. Does there always exist a closed and unbounded
subset of H which is homeomorphic to C?

REPLY [5 votes]: Yes. All we need is to construct a continuous on $H$ function $f$ that is unbounded on $C$. After that, $\{x,f(x)\}\subset H\times \mathbb C$ is closed, unbounded, and homeomorhic to $C$ (in the obvious way) and $H\times \mathbb C$ is isometric to $H$. Being closed and non-compact in an arbitrary separable metric space $X$ implies the existence of such a function. The simplest construction is to take a countable open cover $U_j$ of $X$ that contains no finite subcover of $C$ and to put $f(x)=\min f_j(x)$ where $f_j(x)=\frac j{\min(1,\operatorname{dist}(x,X\setminus U_j))}$. Separability has actually nothing to do with it but in the non-separable case things become a bit more complicated.<|endoftext|>
TITLE: Long line fundamental groupoid
QUESTION [11 upvotes]: This question got me thinking about what makes the fundamental group (or groupoid) tick.  What is so special about the circle?  As another possible candidate for generalization, what about taking the one point compactification of the long closed ray R and thinking about homotopy theory with R in place of the interval?  Would a theory of "long homotopy" arise?  As a follow up, if this doesn't work, are there any other interesting instances of replacing the unit interval with another topological space and getting an interesting homotopy theory out of it?  If not is there some characterization of the interval as the unique space which induces a nice homotopy theory?

REPLY [2 votes]: Perhaps the article of J. Cannon and G. Conner, "The big fundamental group, big Hawaiian earrings, and the big free groups", will interest you. I believe they work with just what you said, the one-point compactification of the long closed ray, or something very similar.<|endoftext|>
TITLE: Does the Continuum Hypothesis hold for sets of a certain rank?
QUESTION [10 upvotes]: Hi
I hope this question is accurate. Gödel and Cohen could show that the Continuum Hypothesis (CH) is independent from ZFC using models in which CH holds, and fails respectively. 
My question now is: 
If we take a large enough part of the universe, say $V_{\omega_{\omega}}$ , does CH hold in it? 
And why can't we go on to argue like this:
If $V_{\omega_{\omega}} \models CH$ then there is a bijection between $\omega_1$ and $2^{\aleph_0}$, and as $V_{\omega_{\omega}}$ is transitive this is 'really' a bijection thus CH holds in ZFC.
If $V_{\omega_{\omega}} \models \lnot CH$ then, as any possible bijection between $\omega_1$ and $2^{\omega}$ is already an element of $V_{\omega_{\omega}}$, CH is inside ZFC refuteable?
This are maybe crude argumentations but I can't answer that by myself.

REPLY [11 votes]: The Continuum Hypothesis, viewed as the assertion that every subset of $P(\omega)$ is either countable or bijective with $P(\omega)$,  is expressible already in $V_{\omega+2}$, since that structure has the full $P(\omega)$ and all subsets of it, as well as all functions between such subsets (one should use a flat pairing function for ordered pairs, which doesn't require one to increase rank for pairs). Thus, CH holds if and only if it holds in any (or all) $V_\alpha$ for $\alpha\geq\omega+2$.

REPLY [7 votes]: Adding to the previous answers: Note that $V_{\aleph_\omega}$ is not a model of ZFC.
Hence, what holds or not in $V_{\aleph_\omega}$ does not formally tell you what follows from  ZFC and what doesn't.  This issue aside, as mentioned above, $V_{\aleph_\omega}$ knows whether or not CH holds in the real world.
What Gödel did to show the consistency of CH was to construct a "narrow" class $L$ inside a model of ZFC (resp. ZF if you also want to also prove the consistency of AC with ZF) such that 
$L$ satisfies ZFC+CH.  The point here is that even if CH fails in the large model of set theory, $L$ contains so few subsets of $\omega$ that CH actually holds (roughly).  
So, L is narrow, while $V_{\aleph_\omega}$ is short and wide (it contains all set up to a certain rank).  A short wide structure (as long it is long 
enough) is correct about CH, a narrow structure might not be.

REPLY [6 votes]: The outcome of CH is already known at very low rank, before rank $\omega+\omega$. Indeed, the set bijections between subsets of $\mathcal{P}(\omega)$ are already available in $V_{\omega+\omega}$, so the question whether there are sets of intermediate cardinality between $\aleph_0$ and $2^{\aleph_0}$ is already known there. (Also, note that there is a natural copy of $\omega_1$ inside $V_{\omega+\omega}$, namely the isomorphism classes of wellorderings of $\omega$.)<|endoftext|>
TITLE: Path connected coloured sets on the squared paper
QUESTION [5 upvotes]: Colour small squares on the standard squared paper in two colors A, B. Name two small squares with common side as "neighbor".
Let every colored set be "path connected": for any two small squares of the color A(resp. B) there is a sequence of color A(resp. B) neighbor squares from one to another. 
Could you help me to prove that must there exists square $3\times 3$ which has 6 squares of same color? 
(It is clear that there is infinite path of each color... I constructed a lot of finite examples without desired square $3\times 3$, but they haven't common structure...)
Question: does this $3\times 3$ square exists? I think, yes, but I can't prove it.
Added: There is a counterexample, two spirals without desired square $3\times 3$.

REPLY [2 votes]: I think the result is false.  Consider a sequence of drawings, one of which I will
represent here:

&&&&&&&&&&&&&&&
&  &  &  &  &
&  &  &  &  &
&& && && && &&
&  &  &  &  &
&  &  &  &  &
&& && && && &&
&& && && && &&

This is a coloring of a 9 x 15 region which satisfies the conditions and has no
3x3 square with six unit squares of the same color. (unfortunately, there are some
rendering problems as I am not seeing how to control the line spacing.)
 It should be clear how to extend 
this for mxn regions 
in which both m and n are arbitrarily large.  Now the idea is to develop a compactness
style argument which expresses connectedness of both regions, the lack of a 3x3
subregion with at least 6 squares of one color, and the arbitrary size of the diagram.
While I do not have the argument nailed down, I suspect one can use this to show an
infinite domain colored in such a way as to preserve all the properties.  This (plus
other poster's evidence to the contrary) is why I believe the poster's assertion that
such a 3x3 square exists that contains at least 6 squares of one color is false.
Gerhard "Ask Me About System Design" Paseman, 2010.09.05<|endoftext|>
TITLE: What are Jacob Lurie's key insights?
QUESTION [68 upvotes]: This question is inspired by this Tim Gowers blogpost.
I have some familiarity with the work of Jacob Lurie, which contains ideas which are simply astounding; but what I don't understand is which key insight allowed him to begin his programme and achieve things which nobody had been able to achieve before. People had looked at $\infty$-categories for years, and the idea of $(\infty,n)$-categories is not in itself new. What was the key new idea which started "Higher Topos Theory", the proof of the Baez-Dolan cobordism hypothesis, "Derived Algebraic Geometry", etc.?

REPLY [21 votes]: I think one of the insights leading to his successes, which is of course not unique to him, is that when doing higher category theory it is useful to add invertible higher cells going all the way up to the top before you try to add noninvertible ones anywhere.  This is along the same lines as Noah's answer: the cobordism hypothesis, which was originally stated in terms of n-categories, becomes easier once you generalize it to a stronger statement about (∞,n)-categories, since then induction becomes easier/possible.<|endoftext|>
TITLE: Closedness of finite-dimensional subspaces
QUESTION [5 upvotes]: Is the (algebraic) span a finite set of vectors in a Hausdorff topological vector space over a complete field always closed?
I suspect yes, but I can't come up with a proof, and it seems like locally convex might be needed to get this.

REPLY [7 votes]: This holds indeed for complete fields: see Theorem 2, Section I.2.3, of Bourbaki's "Espaces Vectoriels Topologiques".
Here is the argument.
Let $K$ be a (not necessarily commutative) field equipped with a complete nontrivial absolute value $x\mapsto|x|$, let $n$ be a positive integer, let $\tau$ be a Hausdorff vector space topology on $K^n$, and let $\pi$ be the product topology on $K^n$.
THEOREM $\tau=\pi$. 
REMINDER A topological group $G$ is Hausdorff iff {1} is closed. [Proof: {1} is closed $\Rightarrow$ the diagonal of $G\times G$ is closed (because it's the inverse image of {1} under $(x,y)\mapsto xy^{-1}$) $\Rightarrow$ $G$ is Hausdorff.]
LEMMA The Theorem holds for $n=1$.
The Lemma implies the Theorem. We argue by induction on $n$. The continuity of the identity from $K^n_\pi$ to $K^n_\tau$ (obvious notation) is clear (and doesn't use the Lemma). To prove the continuity of the identity from $K^n_\tau$ to $K^n_\pi$, it suffices to prove the continuity of an arbitrary nonzero linear form $f$ from $K^n_\tau$ to $K_\pi$. By induction hypothesis, the kernel of $f$ is closed, and the Theorem follows from the Reminder and the Lemma.
Proof of the Lemma. We'll use several times the fact that $K^\times$ contains elements of arbitrary large and arbitrary small absolute value. As already observed, we have $\tau\subset\pi$. If $x$ is in $K^\times$, write $B_x$ for the open ball of radius $|x|$ and center 0 in $K$. Let $a$ be in $K^\times$, and let $\tau_0$ be the set of those $U$ such that $0\in U\in\tau$. 
It suffices to check that $B_a$ contains some $U$ in $\tau_0$. 
We can find a $b$ in $K^\times$ and a $V$ in $\tau_0$ such that $a$ is not in $B_bV$, and then a $c$ in $K$ with $|c|>1$ and a $W$ in $\tau_0$ such that $a$ is not in $B_cW$. Then $U:=c^{-1}W$ does the job.<|endoftext|>
TITLE: To what extent can algorithms in undergraduate linear algebra be made continuous/polynomial/etc.?
QUESTION [6 upvotes]: I feel like many of the algorithms that I learned — indeed, that I have taught — in undergraduate linear algebra classes depend sensitively on whether certain numbers are $0$.  For example, many a linear algebra homework exercise consists of a matrix and a request that the student calculate a basis for the kernel or image.  The standard approach consists of row-reducing the matrix and reading off the answer.  The algorithm to row-reduce a matrix has many steps of the form "if $a \neq 0$, do something that involves a division by $a$, and if $a = 0$, do something else".  Such a step is unfortunate from many points of view.  In particular, it is not even continuous in $a$, so if you only have partial data about the value of $a$ (say, a truncated decimal expansion), then you cannot hope to apply this algorithm.
For the purpose of calculating kernel and image bases, perhaps this is not too surprising.  Indeed, the dimensions of the kernel and image of a basis do not depend polynomially on the coefficients — they don't even depend continuously, only semicontinuously — and so there is really no hope in writing an algorithm that computes the coefficients of a basis for either and that is algebraic in the matrix.
On the other hand, even the usual algorithm we teach to compute inverses to invertible matrices again uses row reduction.  The final answer is algebraic in the matrix coefficients, and there are algorithms that run algebraically (something to do with minors).  So my (slightly ambiguous and open-ended) questions are:

What problems in undergraduate linear algebra can be solved by algorithms that are totally polynomial/algebraic/continuous in the matrix coefficients?  In settings where there is not complete information about a matrix — e.g. when the coefficients are given by truncated decimal expansions — what types of algorithms do people actually use to "approximate" the answers to problems that cannot be solved continuously?  And how do people who do such problems measure/define how good their "approximations" are?

My motivation for asking this question was the (closed) question matrix that annihilates matrix, in which the asker asks for an algorithm that, when fed an $n\times m$ matrix $A$, computes an $m\times n$ matrix $B$ so that $\ker B = \operatorname{im} A$.  This is another example of a problem that is easy to do by row reduction, but it's not clear to me if there are solutions that run "more" continuously than that.  (There won't be a completely continuous algorithm.  Set $m=n$ and feed in an invertible matrix $A$.  Then $B = 0$.  But the invertibles are dense among all matrices.)

REPLY [2 votes]: If we talk about algorithms in a strict sense, then the data involved should be given constructively. This effectively limits us to rational numbers (represented as pairs of integers) or some 'easy' algebraic numbers (for which we can provide algorithms for basic arithmetic). And for these data types most linear algebra algorithms are polynomial in time and stable in the sense that they do not give any inaccuracy due to rounding/overflow/etc. This philosophy is implemented in the computer algebra system GAP.<|endoftext|>
TITLE: Is it a coincidence that the universal parabolic constant shows up in the solution to square point picking?
QUESTION [11 upvotes]: The expected distance $d$ of randomly selected points within a unit square to the square's center is
$d = \frac{1}{6} P$
where P is the universal parabolic constant
$P = \sqrt{2} + \ln{\left(1+\sqrt{2}\right)} = 2.2955871 \dots $
see

http://mathworld.wolfram.com/SquarePointPicking.html
http://mathworld.wolfram.com/UniversalParabolicConstant.html

Is this a mere coincidence or is there an (intuitive) reason why this constant shows up in the solution to this problem?

REPLY [2 votes]: The "reason" that the two given numbers are equal is "write up the integrals, they turn out to be the same integral".
An answer that might satisfy the "intuitive reason" criterion is that the sides of the unit square are given by axis-aligned straight lines; in polar coordinates $r=\frac{1}{cos(t)}$ or $r=\frac{1}{sin(t)}$; and $\frac{1}{cos(t)^2}=\frac{r^2}{r^2cos(t)^2}=\frac{\sqrt{x^2+y^2}}{x^2}$ is the integrand in the integral defining the arc length of the parabolic segment. Truth be told though, this isn't really much more than saying "the integrals turn out to be the same integral", so I'm not sure how much of an "intuitive" explanation this is.<|endoftext|>
TITLE: Kummer generator for the Ribet extension
QUESTION [17 upvotes]: Let $p$ be an odd prime and let $k\in[2,p-3]$ be an even integer such that $p$ divides (the numerator of) the Bernoulli number $B_k$ (the coefficient of $T^k/k!$ in the $T$-expansion of $T/(e^T-1)$).  This happens for example for $p=691$ and $k=12$.
Ribet (Inventiones, 1976) then provides an everywhere-unramified degree-$p$ cyclic extension $E$ of the cyclotomic field $K={\bf Q}(\zeta)$ (where $\zeta^p=1$, $\zeta\neq1$) which is galoisian over $\bf Q$ and such that the resulting conjugation action of $\Delta={\rm Gal}(K|{\bf Q})$ on the ${\bf F}_p$-line $H={\rm Gal}(E|K)$ is given by the character $\chi^{1-k}$, where $\chi:\Delta\to{\bf F}_p^\times$ is the ``mod-$p$ cyclotomic character''.
Kummer theory then tells us that there are units $u\in{\bf Z}[\zeta]^\times$ such that $E=K(\root p\of u)$.  Which units ?
More precisely, there is an ${\bf F}_p$-line $D\subset K^\times/K^{\times p}$ such that $E=K(\root p\of D)$.  Which line ?

REPLY [12 votes]: Let me first add that Herbrand wasn't the first to publish his result; it was obtained (but with a less clear exposition) by Pollaczek (Über die irregulären Kreiskörper der $\ell$-ten und $\ell^2$-ten Einheitswurzeln, Math. Z. 21 (1924), 1--38). 
Next the claim that the class field is generated by a unit is true if $p$ does not divide the class number of the real subfield, that is, if Vandiver's conjecture holds for the prime $p$.
Proof. (Takagi)
Let $K = {\mathbb Q}(\zeta_p)$, and assume that the class number of
its maximal real subfield $K^+$ is not divisible by $p$. Then any 
unramified cyclic extension $L/K$ of degree $p$ can be written in 
the form $L = K(\sqrt[p]{u})$ for some unit $u$ in $O_K^\times$.
In fact, we have $L = K(\sqrt[p]{\alpha})$ for some element 
$\alpha \in O_K$. By a result of Madden and Velez, $L/K^+$ is normal 
(this can easily be seen directly). If it were abelian, the subextension 
$F/K^+$ of degree $p$ inside $L/K^+$ would be an unramified cyclic 
extension of $K^+$, which contradicts our assumption that its class 
number $h^+$ is not divisible by $p$.
Thus $L/K^+$ is dihedral. Kummer theory demands that 
$\alpha /\alpha' = \beta^p$ for some $\beta \in K^+$, where
$\alpha'$ denotes the complex conjugate of $\alpha$.
Since $L/K$ is unramified, we must have $(\alpha) = {\mathfrak A}^p$. 
Thus $(\alpha \alpha') = {\mathfrak a}^p$, and since $p$ does not
divide $h^+$, we must have ${\mathfrak a} = (\gamma)$, hence
$\alpha \alpha' = u\gamma^p$ for some real unit $u$.
Putting everything together we get $\alpha^2 = u(\beta\gamma)^p$,
which implies $L = K(\sqrt[p]{u})$. 

If $p$ divides the plus class number $h^+$, I cannot exclude the possibility that the Kummer generator is an element that is a $p$-th ideal power, and I cannot see how this should follow from Kummer theory, with or without Herbrand-Ribet.
If $p$ satisfies the Vandiver conjecture, the unit in question can be given explicitly, and was given explicitly already by Kummer for $p = 37$ and by Herbrand for general irregular primes satisfying Vandiver: let $g$ denote a primitive root modulo $p$, and let $\sigma_a: \zeta \to \zeta^a$. Then 
$$ u = \eta_\nu  = \prod_{a=1}^{p-1} \bigg(\zeta^\frac{1-g}{2}\
  \frac{1-\zeta^g}{1-\zeta}\bigg)^{a^\nu \sigma_a^{-1}}, $$ 
where $\nu$ is determined by $p \mid B_{p-\nu}$. 
Here is a survey on class field towers based on my (unpublished) thesis on the explicit construction of Hilbert class fields that I have not really updated for quite some time. Section 2.6 contains the answer to your question for primes satisfying Vandiver.<|endoftext|>
TITLE: Restriction of Ext sheaves
QUESTION [7 upvotes]: Let $f \colon X \to Y$ be a map of schemes, $\mathcal{F}, \mathcal{G}$ two coherent sheaves on $Y$. I'm interested in conditions which guarantee an isomorphism
$$f^{*} \mathcal{E}xt^i(\mathcal{F}, \mathcal{G}) \cong \mathcal{E}xt^{i}(f^{*} \mathcal{F}, f^{*} \mathcal{G}).$$
I know this is true for $f$ flat by [EGA III.12.3.4]. Moreover it seems to me that this is true for $f$ arbitrary when $\mathcal{F}$ is locally free and $i = 0$, essentially by the argument in [EGA I.0.6.7.6]. I am also able to prove this in some other cases by ad hoc methods, but I could not find a general statement.

What are conditions which guarantee the existence of such an isomorphism? I am also interested in having some counterexamples at hand.

REPLY [6 votes]: In a pretty general situation there is an isomorphism of derived functors 
$$
Lf^* RHom(F,G) \cong RHom(Lf^* F,Lf^*G).
$$
Further you can write down the spectral sequences converging to each side,
(e.g. for the LHS it is quite simple --- $L_q f^* Ext^p(F,G)$, for the RHS
it is more complicated). Now, if $f$ is flat, the both spectral sequences
degenerate in the second term and give you the desired isomorphism. 
The same holds if $F$ is locally free. But the general statement is the above isomorphism of the derived functors.<|endoftext|>
TITLE: Walking to infinity on the primes: The prime-spiral moat problem
QUESTION [20 upvotes]: It is an unsolved problem to decide if it is possible to "walk to infinity" from the origin
with bounded-length steps, each touching a Gaussian prime as a stepping stone.
The paper by Ellen Gethner, Stan Wagon, and Brian Wick, 
"A Stroll through the Gaussian Primes"
(American Mathematical Monthly, 105: 327-337 (1998))
discusses this Gaussian moat problem and proves that steps of length $< \sqrt{26}$ are
insufficient.  Their result was improved to $\sqrt{36}$ in 2005.
My question is: 

Is the analogous question easier for the
  prime spiral (a.k.a. Ulam spiral)—Can one walk to infinity using bounded-length steps
  touching only the spiral coordinates of primes?  

What little I know of prime gaps
suggest that should be easier to walk to infinity.
For example, the first gap of 500 does not occur until about $10^{12}$
(more precisely, 499 and 303,371,455,241).
I ask this primarily out of curiosity, and have tagged it 'recreational.'
Edit1. In light of Gjergji's remarks below, I have tagged this as an open problem.
Edit2.
Just for fun, I computed which primes are reachable on a small portion of the spiral,
for step distances $d \le 3$ (left below) and $d \le 4$ (right below);
red=reached, blue=not reached.
The former does not reach 83, the 23rd prime blue dot barely discernable at spiral coordinates (5,-3); 
the latter does not reach 5087, the 680th prime blue dot at
spiral coordinates (36,10).

REPLY [8 votes]: Percolation theory suggests that the probability one can one can walk to infinity is 1 if the density of randomly chosen stepping stones is at least a certain critical number, and is 0 if the density is less than this number. Since the density of primes in the Ulam spiral and the density of Gaussian primes in the plane both tend to zero, the density of stepping stones is 0. This suggests that one cannot walk to infinity on either primes in the Ulam sprial or Gaussian primes, for any bounded size of step.<|endoftext|>
TITLE: Elliptic regularity for the Neumann problem
QUESTION [18 upvotes]: I'm trying to understand how to establish regularity for elliptic equations on bounded domains with Neumann data.
For simplicity, let's presume we are focusing on $-\Delta u = f$ in $\Omega$ and $\frac{\partial u}{\partial \nu} = 0$ on $\partial \Omega$. 
Interior regularity works the same as always.
When proving boundary regularity, for the dirichlet boundary case we first consider some ball $B(0,1) \cap \mathbb{R}_+^n$ and let $\xi = 1$ on $B(0,1/2)$, $\xi = 0$ on $\mathbb{R}^n - B(0,1)$ and then estimate all derivatives $\frac{\partial^2 u}{\partial x_i \partial x_j}$ except $\partial^2 u/\partial x_n^2$. Two main points are needed
1) $\xi$ vanishes on the curved part of $B(0,1) \cap \mathbb{R}_+^n$\
2) $u=0$ on $\{x_n=0\}$.
This allows us to let $-\partial_{x_i} (\xi \partial_{x_j} u)$ (with derivatvies replaced by difference quotients) be an admissible test function for our weak definitoin of a solution.
I presume the main difficulty in neumann boundary data is making your test function be admissible. In other words, we would need $\int v = 0$ since our existence was established on $H^1(\Omega)$ restricted to mean value zero functions. 
So in order to proceed, can we just subtract off a constant from our original $-\partial_{x_i}(\xi \partial_{x_j}u)$?
Is there some more natural way to establish regularity in this case? I do not want to take advantage of the fact that we have a green's function in this case however as I only chose the Laplace equation for simplicity.

REPLY [6 votes]: It's true that to prove existence for the Neumann problem you work in the space $X=\{v\in H^1(\Omega):\, \int_\Omega v\,dx=0\}$ and so a weak solution $u$ satisfies 
$$\int_\Omega\nabla u\cdot\nabla v\,dx=\int_\Omega fv\,dx$$
for all $v\in X$. However, if you take $w\in H^1(\Omega)$ and consider the function $v=w-\frac1{|\Omega|}\int_\Omega v\,dy$ you get 
$$\int_\Omega\nabla u\cdot\nabla w\,dx=\int_\Omega f\left(w-\frac1{|\Omega|}\int_\Omega v\,dy\right)\,dx=\int_\Omega fw\,dx-0$$
since $\int_\Omega f\,dx=0$. So You can actually work in the entire space $H^1(\Omega)$ rather than in the smaller space $X$.
Difference quotients work fine since the constraint is gone. If $\Omega=B^+(0,1)$ you can do difference quotients in the tangential directions $e_i$, $i=1,\cdots, n-1$ (so no need to assume $u=0$ on $\{x_n=0\}$) to prove that all second derivatives but $\partial^2 u/\partial x_n^2$ are in $L^2 (B^+(0,1/2))$ and then you use the equation (which is satisfied pointwise a.e. by interior regularity) to conclude that also $\partial^2 u/\partial x_n^2$ is  in $L^2 (B^+(0,1/2))$.<|endoftext|>
TITLE: What is a twisted D-Module intuitively?
QUESTION [17 upvotes]: When I think about $\mathcal{D}$-Modules, I find it very often useful to envison them as vectorbundles endowed with a rule that decides whether a given section is flat. Or alternatively a notion of parallel transport. 
Now my question is, what are good ways to think about modules over sheavers of twisted differential operators?

REPLY [10 votes]: Here's another perspective to complement the homogeneous approach given in Ben's and Scott's answers. One can look at twisted $D$-modules as connections with fixed scalar curvature. This is particularly powerful if you think complex-analytically: you can describe all possible twistings as follows:
Suppose $M$ is a complex manifold. In general, twistings of the sheaf of differential operators are parametrized by the hypercohomology of the truncated de Rham complex $$\Omega^1_{hol}\to\Omega^2_{hol}\to\dots.$$
(I use the lower index `hol' to distinguish the sheaf of holomorphic sections from the sheaf
of $C^\infty$-sections.)
If we use Dolbeault's complex to compute the hypercohomology, you see that twistings are represented by a closed 2-form $\omega$ whose $(0,2)$-part vanishes. $\omega$ matters only up to a differential of a $(1,0)$-form.
Let $\omega$ be such a closed 2-form. We can then consider vector bundles $F$ (or, more generally, quasicoherent sheaves) on $M$ equipped with a connection 
$$\nabla:F\to F\otimes\Omega$$ 
whose curvature is $\omega$.
More precisely, suppose $F$ is a holomorphic vector bundle. The sheaf of $C^\infty$-sections of $F$ carries an action of 
$$\overline\partial:F\to F\otimes\Omega^{0,1}.$$ 
An action of the TDO corresponding to $\omega$ on $F$ is the same as extension of $\overline\partial$ to $\nabla$ with prescribed curvature.
Remark. If one works algebraically (for instance, over fields other than ${\mathbb C}$), only some TDO's can be viewed in similar way; namely those whose class belongs to the image of the space $H^0(\Omega^2)$.<|endoftext|>
TITLE: What is the Implicit Function Theorem good for?
QUESTION [51 upvotes]: What are some applications of the  Implicit Function Theorem  in calculus?  The only applications I can think of are:

the result that the solution space of a non-degenerate system of equations naturally has the structure of a smooth manifold;
the Inverse Function Theorem.

I am looking for applications that would be interesting to an advanced US math undergraduate 
or 1-st year graduate student who is seeing the Implicit Function Theorem for the first or second time.  The context is I will be explaining this result as part of 
a review of manifold theory.

REPLY [7 votes]: I was just sitting down to brainstorm some applications, since we're going to finish the proof next week and I want to show some payoff. Here is what I came up with, not counting ideas already mentioned in other answers (many of which I like):

The orthogonal group is a manifold, of dimension $\binom{n}{2}$. It's not too hard to compute the derivative of the map $X \mapsto X X^T$ from $n \times n$ matrices to $n \times n$ symmetric matrices and show that it is rank $\binom{n+1}{2}$ when $X$ is invertible. So the preimage of $\mathrm{Id}$ is a manifold of dimension $n^2 - \binom{n+1}{2}$.
Suppose I have a smooth map $(x,y) \mapsto (f,g)$ from $\mathbb{R}^2 \to \mathbb{R}^2$, so that $\mathrm{rank} (Df)$ is everywhere $1$. Then one of $f$ and $g$ is locally a function of the other. (Proof: After some initial fiddling around, we can assume that $\partial f/\partial x \neq 0$. Then $f=\mathrm{constant}$ is a curve. Differentiating $g$ along this curve gives $0$.) I remember this trick being used to great effect in Kenyon-Okounkov to show that two functions obeyed a relation without actually constructing the relation; I'll try to find some simple examples where it is very nonobvious what the relation is.
This top voted proof of the fundamental theorem of algebra uses centrally that a submersion is an open map, which is the key lemma in all proofs of IFT that I know. 

I'll come back in a few weeks and report on how these went.<|endoftext|>
TITLE: Lecture notes by Thurston on tiling
QUESTION [47 upvotes]: I am looking for a copy of the following 
W. Thurston, Groups, tilings, and finite state automata, AMS Colloquium Lecture Notes.
I see that a lot of papers in the tiling literature refer to it but I doubt it was ever published. May be some notes are in circulation ?
Does anyone have access to it? I would be extremely grateful if you can send me a copy or tell me where can I find it.

REPLY [138 votes]: Unfortunately, the original to this is hard to locate. It was distributed by the AMS at the time of the colloqium lectures, but they apparently didn't keep the files they used. At one time it was distributed as a Geometry Center preprint, but the Geometry Center is now defunct. I've lost track of the source files through multiple moves, computer crashes, etc.
What I have is a scanned version of the Geometry Center version that is legible but not beautiful, which I can forward by email
Bill Thurston
ADDENDUM:
Renaud Dreyer told me of an online scanned version I wasn't aware of, which appears to be better quality than the one I have:
http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf<|endoftext|>
TITLE: Completeness and cocompleteness of the Kleisli category
QUESTION [10 upvotes]: If (T,η,μ) is a monad over a category C, which is complete and cocomplete, then what about the Kleisli category? And also, if C is cartesian closed, what about the Kleisli one?

REPLY [5 votes]: The article on limits and colimits in the Kleisli category shows that (with $C_T$ denoting the Kleisli category, and $C^T$ denoting the Eilenberg-Moore category):
1) If $C$ is complete, and the canonical functor $C_T\to C^T$ has a right adjoint, then $C_T$ is complete. 2) If $C$ has coproducts, then $C_T$ has coproducts (already mentioned in the comments to another answer to the question). 3) Cocompleteness is more subtle, and under some condition on partial algebras cocompleteness of $C_T$ is equivalent to the existence of a left adjoint of the canonical functor $C_T\to C^T$.<|endoftext|>
TITLE: When does a rational function have infinitely many integer values for integer inputs?
QUESTION [6 upvotes]: Consider rational functions $F(x)=P(x)/Q(x)$ with $P(x),Q(x) \in \mathbb{Z}[x]$. I'd like to know when I can expect $F(k) \in \mathbb{Z}$ for infinitely many positive integers $k$. Of course this doesn't always happen ($P(x)=1, Q(x)=x, F(x)=1/x$). I am particulary interested in answering this for the rational function $F(x)=\frac{x^{2}+3}{x-1}$.

REPLY [10 votes]: If $F=P/Q$ is integral infinitely often then $F$ is a polynomial.
Write $$P(x)=f(x)Q(x)+R(x)$$ for some polynomial $R$ of degree strictly less than the degree of $Q$. If you have infinitely many integral $x$ so that $P/Q$ is integral then you get infinitely many $x$ so that $NR/Q$ is integral, where $N$ is the product of all denominators of the coefficients in $f$. However $R/Q\to 0$ as $x\to \pm \infty$ so $R\equiv 0$ and so $Q(x)$ is a divisor of $P(x)$.
Now, as pointed out by Mark Sapir below, not all polynomials with rational coefficients take on integer values infinitely often (at integers), but you can check this  in all practical cases by seeing if $dF$ has a root $\pmod{d}$, where $d$ is the common denominator of the coefficients in $F$.

REPLY [7 votes]: $(x^2+3)/(x-1)=x+1+(4/(x-1))$ so this question, at least, is easy; you get an integer if and only if 4 is a multiple of $x-1$.<|endoftext|>
TITLE: Lebesgue dimension of closures satisfying the Z-set condition
QUESTION [5 upvotes]: Given any subspace $A\subset X$ of a topological space with Lebesgue dimension $\le N$. 
Let $\bar{A}$ denote the closure of $A$. Assume, that the pair $(\bar{A},A)$ satisfies the Z-set condition, i.e. there is a homotopy $H:\bar{A}\times [0;1]\rightarrow \bar{A}$, such that $H_1=id$ , Image$(H_t)\subset A$ for all $t<1$. 
Does this imply, that the Lebesgue dimension of $\bar{A}$ is at most the Lebesgue dimension of $A$?

REPLY [4 votes]: No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces).
Bad example: $X=\{a,b,c\}$ with open sets $\emptyset$, $X$, $\{a\}$, $\{a,b\}$, $\{a,c\}$. Let $A=\{a\}$, then $\bar A=X$. The homotopy is given by $H_1=id$, $H_t\equiv a$ for $t<1$. The dimension of $A$ is 0 but the dimension of $\bar A$ is 1.
On the positive side, let me begin with a quick and dirty proof in the case when $\bar A$ is a compact metric space. Let $\{U_i\}$ be an open covering of $\bar A$. We need to find a refined covering of multiplicity at most $N+1$ where $N=\dim A$. It suffices to find a continuous map $f:\bar A\to A$ and an open covering $\{V_j\}$ of $A$ such that $\{f^{-1}(V_j)\}$ is a refinement of $\{U_i\}$. Indeed, in this case we can find a refinement of $\{V_j\}$ of multiplicity at most $N+1$ and its $f$-preimage is the desired refinement of $\{U_i\}$.
In the compact case, let $\{V_j\}$ be the covering by $(\rho/3)$-balls where $\rho$ is the Lebesgue number of the covering $\{U_i\}$. Then, for some $t$ sufficiently close to 1, the map $f=H_t$ satisfies the desired property: the preimage of every $V_j$ has diameter less than $\rho$ and hence is contained in some of the sets $U_i$. Indeed, suppose the contrary. Then there is a sequence $t_k\to 1$ and sequences $x_k,y_k\in \bar A$ such that $|x_ky_k|\ge\rho$ but $|H_{t_k}(x_k)H_{t_k}(y_k)|<2\rho/3$. Due to compactness we may assume that $x_k$ and $y_k$ converge to some $x,y\in\bar A$. Then $|xy|\ge\rho$ but $|H_1(x)H_1(y)|\le 2\rho/3$, a contradiction.
In the general metric space case, let $\rho(x)$ denote the local Lebesgue number of $\{U_i\}$ at $x$, that is the supremum of $\rho$ such that the ball $B_\rho(x)$ is contained in one of the set $U_i$. Note that $x\mapsto\rho(x)$ is a positive 1-Lipschitz function on $\bar A$. It is easy to construct a continuous function $u:\bar A\to[0,1)$ such that the distance from $x$ to $H_t(x)$ is less that $\rho(x)/10$ for all $x\in\bar A$ and all $t>u(x)$. Then the map $f:\bar A\to A$ given by $f(x)=H_{u(x)}(x)$ and the covering of $A$ by the balls of the form $B_{\rho(x)/10}(x)$, $x\in A$, will do the job.<|endoftext|>
TITLE: Is there a "correct" general setting for the principle:  "tensoring any object with a projective object yields another projective"? 
QUESTION [10 upvotes]: Apparently this principle was first formulated for left modules over the group algebra $A=kG$ of a finite group, where $k$ is a field of characteristic $p>0$ dividing $|G|$.  (See Exercise 2 on p. 426 of Curtis & Reiner, Representation Theory of Finite Groups and Associative Algebras, 1962.)  Here the Hopf algebra structure of A yields a natural left module structure on the tensor product of two left modules over k.   
By the mid-1970s similar tensor product behavior was observed in other special cases for left A-modules and their tensor products, where A is a Hopf algebra over a commutative ring k:   (1)  the (finite dimensional) restricted enveloping algebra of a restricted Lie algebra $\mathfrak{g}$ over a field of prime characteristic; (2) more generally  the hyperalgebra of a higher Frobenius kernel  when $\mathfrak{g}$ is the Lie algebra of a reductive algebraic group;  (3) the universal enveloping algebra of a Kac-Moody algebra in characteristic 0;  (4) the full hyperalgebra of a reductive algebraic group in prime characteristic (with "projective" replaced by "injective" as in J.C. Jantzen's book Representations of Algebraic Groups, I.3).    Relevant references:
B. Pareigis, Kohomologie von p-Lie Algebren, Math. Z. 104 (1968); Lemma 2.5
J.E. Humphreys, Projective modules for SL(2,q), J. Algebra 25 (1973); Thms. 1, 2 (and note
added in proof referring to Pareigis)
J.E. Humphreys, Ordinary and modular representations of Chevalley groups, Springer
Lect. Notes in Math. 528 (1976); Appendix T (following Sweedler's suggestion)
H. Garland and J. Lepowsky, Lie algebra homology and the Macdonald-Kac formulas, Invent. Math. 34 (1976); 1.7 and Remark
J.E. Humphreys, On the hyperalgebra of a semisimple algebraic group, in Contributions to Algebra, Academic Press, 1977; 3.1
The arguments here typically involve special cases of a general theorem suggested by Sweedler (and closely related to the "tensor identity" discussed in a recent MO post 37709 ):   Let $A$ be a Hopf algebra (with antipode) over a commutative ring $k$, with Hopf subalgebra $B$ (possibly k).   Given an $A$-module  $M$ and a $B$-module $N$,  there is a natural  $A$-module isomorphism:
$$(A \otimes_B N) \otimes_k M \cong A \otimes_B (N \otimes_k M)$$  On the left side, A acts via comultiplication, while on the right it acts on the first factor. 

Is this the optimal generality, and if so is there a textbook reference?

REPLY [5 votes]: Dear Jim,
Perhaps this is the correct statement, which is proven in Etingof's lectures on tensor categories (though its proof follows from the yoga of tensor categories, as I'll explain):
http://www-math.mit.edu/~etingof/tenscat1.pdf
Proposition:
Let $P$ be a projective object in a multiring category C. If $X\in C$ has a right dual, then the object $P \otimes X$ is projective. Similarly, if $X \in C$ has a left dual, then the object $X \otimes P$ is projective. 
Multi-ring categories are not so common, so let me specialize a little:
Let $P$ be a projective object in a tensor category $C$.  Then $P\otimes X$ is projective, for any $X\in C$.
The other things I mentioned are automatic in a tensor category.
The proof is that being projective means that the functor $Hom(P, -)$ is exact.  Since we have $Hom(P\otimes X,-) \cong Hom(P,-\otimes ^*X)$ (right dual), it means that $P\otimes X$ is exact whenever $P$ is (tensoring with an object is always exact).<|endoftext|>
TITLE: The number of cusps of higher-dimensional hyperbolic manifolds
QUESTION [18 upvotes]: Suppose $n$ is an integer greater than 3. Sometimes ago I heard somewhere that it is still not known if there exist complete finite-volume hyperbolic $n$-manifolds having exactly one cusp.  
Could someone either confirm that the problem of finding such examples in every dimension is still open, or, preferably, give me a reference for examples of one-cusped hyperbolic manifolds in arbitrary dimension?

REPLY [6 votes]: Indeed, the general case is open. In the case of dimension four, we used the Coxeter $24$-cell, which is an ideal right-angled polytope (ideal means that all of its vertices are on $\partial \mathbb{H}^4$). It is know (a little result of mine) that there are now ideal right angled polytopes in dimensions greater than or equal to $7$ (however, I do not know anything about dimensions $5$ and $6$). Thus, there is no hope to use ideal right-angled polytopes in higher dimensions.<|endoftext|>
TITLE: Does this graph exist?
QUESTION [8 upvotes]: I have proved a certain result for all 2-connected graphs apart from those that fit the following criteria:

They are "minimally 2-connected", that is, deleting any vertex will produce a graph which is no longer 2-connected, and
They have circumference less than $\frac{n+2}{2}$, where $n$ is the number of vertices.

I have not been able to come up with an example of such a graph.  Can anyone help?
Of course the best possible outcome would be that they do not exist!

REPLY [9 votes]: There are lots of examples. For $t>5$, let $P_{1},..., P_{t}$ be internally disjoint paths with length $3$ such that each path has the vertices $x$ and $y$ as endpoints.

REPLY [7 votes]: What if you take $5$ paths with $k$ vertices each and glue them together at the endpoints? So $5(k-2)+2=5k-7$ vertices in all, and circumference $2(k-1)=2k-2$.<|endoftext|>
TITLE: Are root stacks characterized by their divisor multiplicities?
QUESTION [12 upvotes]: Definitions/Background
Suppose $S$ is a scheme and $D\subseteq S$ is an irreducible effective Cartier divisor. Then $D$ induces a morphism from $S$ to the stack $[\mathbb A^1/\mathbb G_m]$ (a morphism to this stack is the data of a line bundle and a global section of the line bundle, modulo scaling). For a positive integer $k$, the root stack $\sqrt[k]{D/S}$ is defined as the fiber product
$\begin{matrix}
 \sqrt[k]{D/S} & \longrightarrow & [\mathbb A^1/\mathbb G_m] \\
  p\downarrow & & \downarrow \wedge k \\
  S & \longrightarrow & [\mathbb A^1/\mathbb G_m]
\end{matrix}$
where the map $\wedge k: [\mathbb A^1/\mathbb G_m]\to [\mathbb A^1/\mathbb G_m]$ is induced by the maps $x\mapsto x^k$ (on $\mathbb A^1$) and $t\mapsto t^k$ (on $\mathbb G_m$). The morphism $p:\sqrt[k]{D/S}\to S$ is a coarse moduli space and is an isomorphism over $S\smallsetminus D$. Moreover, there is a divisor $D'$ on $\sqrt[k]{D/S}$ such that $p^*D$ is $kD'$.
The data of a morphism from $T$ to $\sqrt[k]{D/S}$ is equivalent to the data a morphism $f:T\to S$ and a divisor $E$ on $T$ such that $f^*D = kE$.
The question

Suppose $\mathcal X$ is a DM stack, that $f:\mathcal X\to S$ is a coarse moduli space, that $f$ is an isomorphism over $S\smallsetminus D$, and that $f^*D = kE$ for an irreducible Cartier divisor $E$ on $\mathcal X$. Is the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ an isomorphism?

I get the strong impression that the answer should be "yes", at least if additional conditions are placed on $\mathcal X$.
A counterexample
Here's a counterexample to show that some additional condition needs to be put on $\mathcal X$. Take $G$ to be $\mathbb A^1$ with a doubled origin, viewed as a group scheme over $\mathbb A^1$. Then $\mathcal X=[\mathbb A^1/G]\to \mathbb A^1$ is a coarse moduli space ("there's a $B(\mathbb Z/2)$ at the origin"). If we take $D\subseteq \mathbb A^1$ to be the origin, then the pullback to $\mathcal X$ is the closed $B(\mathbb Z/2)$ with multiplicity 1. Yet the induced morphism from $\mathcal X$ to $\sqrt[1]{D/\mathbb A^1}\cong \mathbb A^1$ is not an isomorphism.
In this case, $\mathcal X$ is a smooth DM stack, but has non-separated diagonal.

REPLY [10 votes]: What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^*D = kE$ for a reduced divisor $E$ on $\mathcal X$. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism.
I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame.<|endoftext|>
TITLE: The homotopy cofiber of the smash product of two maps of spectra
QUESTION [5 upvotes]: It is a standard fact that smashing with a fixed spectrum $Z$ preserves cofiber sequences.  So if I have a cofiber sequence $$X \xrightarrow{f} Y \rightarrow C_f$$ then there is also a cofiber sequence $$Z \wedge X \rightarrow Z \wedge Y \rightarrow Z \wedge C_f$$
If more generally I have a map $Z \xrightarrow{g} W$, is there any formula for the cofiber of the map $$Z \wedge X \xrightarrow{g \wedge f} W \wedge Y$$ in terms of $C_f$ and $C_g$?  (The above discussion corresponding to $g = \mathrm{id}_Z$).

REPLY [2 votes]: An axiomatization of exactly how smash products of cofiber sequences 
should behave is given in the context of triangulated categories in 
my paper
The additivity of traces in triangulated categories. 
Advances in Mathematics 163(2001), 34--73.
Of course, the axioms hold in the motivating example of the homotopy
category of spectra.<|endoftext|>
TITLE: Is every probability space a factor space of the Haar Measure on some group?
QUESTION [21 upvotes]: Let P be an arbitrary probability space.
I would like to find a compact topological group $G$ so that the Haar probability measure on $G$ admits a measurable map to the probability space $P$.
By a measurable map, I mean a function which lifts measurable sets to measurable sets of the same measure.  That is, $f : Q \longrightarrow P$ is measurable when for all measurable $E \subseteq P,$ the set $f^{-1}(E)$ is measurable and satisfies
$$ \mu_Q (f^{-1}(E)) = \mu_P (E). $$
I imagine that some enormous product space $(\mathbb{R} / \mathbb{Z})^{\kappa}$ will do.  Anyone see a nice way to make this work?
Edit:
My hunch that some power of the circle group will work is based on the game twenty questions.  One player thinks of an object (a trampoline), and the other players ask a sequence of at most twenty yes/no questions (Can it fly?  Is it legal to drive one on the highway?  Would it hurt to swallow?).  A strategy for twenty questions would consist of an enormous decision tree that says what question to ask next.
Here we're trying to guess a point of $P$ using at most $\kappa$ measurable yes/no questions.  The answers come from the coordinates of $[0,1]^{\kappa}$ which may be interpreted according to the probability of a yes.  The function $f$ could be built out of a suitable decision tree.
Perhaps a decision tree could be constructed out of a well-ordering of the sigma-algebra of measurable sets.
Update:
Thanks for all the help!
Here's the story:
What I really want is to perform convolution in $L^2(P)$.  I'll admit, $P$ is not a group, so I guess it's okay to switch to $L^2(G)$ provided that $G$ has $P$ as a factor space.  But it turns out even this was too greedy.
Let $S \subseteq G$ be a dense subset with a measure structure inherited from $G$.  Now we may perform convolution in $L^2(S)$, even though $S$ is not a group!  The continuous functions are dense in $L^2(S)$, so it will suffice to convolve two of them.  But any continuous function on $S$ extends uniquely to one on $G$.  So we convolve in $L^2(G)$ and then restrict the result to $L^2(S)$.
Because of this, George Lowther's weaker result will suffice for my purposes.  After all, a subset of full outer measure is certainly dense.  I will accept his answer unless a full answer to the original question materializes.

REPLY [3 votes]: I would consider Maharam's structure theorem...
But I guess that will get you only the measure algebra.  
So, consider a set $P \subset [0,1]$ that is badly nonmeasurable... but outer measure $1$.  Lebesgue measure induces a probability on its Borel sets, but is there a map $f$ as you require?<|endoftext|>
TITLE: Index of an Operator
QUESTION [7 upvotes]: I've been working through the heat-equation proof of the Atiyah-Singer index theorem.  My question is what is the motivation for the definition of the index of an operator?  I know there is the isomorphism between the homotopy classes of maps from a compact topological space to the space of Fredholm operators and the first K-group of the topological space given by the index map, but is there a simpler example of why we want to study the index?

REPLY [15 votes]: Usually one really wants to know the dimension of the kernel of an operator rather than its index. The problem is that the dimension of the kernel is not a continuous function of the operator, so is very hard to compute in terms of topological data. The key point about the index is that it is a continuous function of the operator, which makes it a lot easier to compute. And it's a modification of the kernel, so with some luck one can compute the kernel from it.

REPLY [5 votes]: Let me interpret the question as saying: "It's clear why one would be interested in (the dimension of) the kernel of a given differential operator, but why on earth would one be interested in the difference between the dimension of the kernel and the cokernel?"
I suppose one answer is that one can change the dimension of the (co)kernel of a Fredholm operator by adding a compact operator, whereas it is the difference (i.e., the index) which remains invariant.   This makes it easier to compute the index, since we can deform the operator without changing its index to another operator for which the computation is easier.  
Moreover, in many interesting cases one can show that cokernel vanishes, in which case the index agrees with the dimension of the kernel, which is what we were after all along.
One of my favourite examples of this is the computation of the dimension of the moduli space of instantons, as in the classic paper of Atiyah, Hitchin and Singer on Deformations of instantons.<|endoftext|>
TITLE: abelian categories enriched over sheaves
QUESTION [17 upvotes]: Let $X$ be a ringed space. The abelian category of (quasi-coherent) $\mathcal{O}_X$-modules does not behave as well as the category of $A$-modules for a commutative ring $A$. The reason is that there are not enough projective objects. In particular, $\mathcal{O}_X$ is not projective, since $Hom(\mathcal{O}_X,-)$ is the global section functor, which is not exact.
I want to fix this by enriching the abelian category over $Sh(X)$, the monoidal abelian category of abelian sheaves on $X$. Remark that $\underline{Hom}(\mathcal{O}_X,-)$ is just the identical functor, thus $\mathcal{O}_X$ should be projective from this point of view. Actually I want that $\mathcal{O}_X$-Mod has all the homological properties as $A$-Mod, except that we talk about them in another topos, namely $Sh(X)$ instead of $Set$.
In particular I want to generalize the following well-known theorem: Let $\mathcal{A}$ be an abelian category, which has a progenerator $P$ (i.e. a finite projective generator), whose coproducts exist. Then $\mathcal{A}$ is equivalent to $Mod_A$ for some (noncommutative) ring $A$, namely $A$ is the endomorphism ring of $P$.
So how do we define an abelian category $\mathcal{A}$ over $Sh(X)$ in reasonable way? I've already written down a possible definition, but I wonder if it is possible to avoid the category of elements of the enriched category, since this makes the whole story nonlocal. For example when you want to write down what a kernel of a morphism should be, or when you want to define the notions of generators or projective objects.
Or should we just say that everthing is a sheaf? A category is a pair $(O,M)$, consisting of a sheaf $O$ on $X$ (objects) and a sheaf $M$ on $X$ (morphisms) together with maps $M \to O \times O$, etc. ...
I'm pretty sure that somebody has worked this out in detail some decades ago, so a reference would be ok.
EDIT [12. 09]: I think the following definition works: A sheabelian ;) category is a pair $(X,A)$, consisting of a topological space $X$ and a presheaf $A$ on $X$, valued in the category of abelian categories, such that the following sheaf condition holds:
1) If $U = \cup_i U_i$ and $f,g : F \to G$ are morphisms in $A(U)$ and $\eta_i$ are natural transformations between $f|_{U_i}, g|_{U_i}$, which are compatible on the $U_i \cap U_j$, then they lift uniquely to a natural transformation between $f$ and $g$.
2) If $U = \cup_i U_i$ and $T_i \in A(U_i)$ are objects, which are isomorphic on $U_i \cap U_j$, such that the cocycle condition is satisfied, then there is an object $T \in A(U)$, which restricts to an object isomorphic to $T_i$, and the isomorphisms here are compatible with each other.
Basically this concept works for every $2$-category. Objects of $(X,A)$ are defined to be objects of $A(X)$, but properties and constructions of these objects are defined locally. For example, $Hom(P,-)$ for an object $P$ is a functor $A(X) \to Sh(X)$. Currently, I'm trying to work out the details for the generalization of the above theorem. Note that if $(X,\mathcal{O}_X)$ is a ringed space, then $O_X$ should be a progenerator of $Mod_{O_X}$.
EDIT [15.09] Ok I think I have just reinvented the notion of a stack. ;)
So my question is: Is there literature about a sort of homological algebra of stacks of abelian categories? Here I'm mainly interested in the site associated to a topological space, and some of the $2$-isomorphisms in the definition of a stack should become identities (since, for example, we have $(F|_V)|_W = F|_W$ for a sheaf $F$ on $X$).

REPLY [4 votes]: Yes, you do seem to be looking basically at stacks.  Note that stacks, and more generally prestacks and fibered categories, can be identified with categories enriched over the self-indexing of the topos of sheaves, or equivalently over the bicategory of spans in the topos of sheaves.  There is a bit of a dearth of good expositions of that point of view, but you can try B2.2 in Sketches of an Elephant and the paper "Variation through enrichment" (which is a bit dense and very category-theoretic).
However, it's not true that just working in another topos will resolve the issue of abelian sheaves not having enough projectives, because the proof in Set that module categories have enough projectives is not constructive, and so does not relativize to all topoi.  In particular, the fact that free modules are projective relies on the fact that all sets are projective in Set, which is equivalent to the axiom of choice.  You don't need the full strength of AC to show that there are enough projectives -- the presentation axiom suffices -- but it's still not true in most/all topoi.<|endoftext|>
TITLE: Zeros of Gradient of Positive Polynomials.
QUESTION [27 upvotes]: It was asked in the Putnam exam of 1969, to list all sets which can be the range of polynomials in two variables with real coefficients. Surprisingly, the set $(0,\infty )$ can be the range of such polynomials. These don't attain their global infimum although they are bounded below. But is it also possible that such polynomials with range $(0,\infty )$ also have a non zero gradient everywhere?

REPLY [52 votes]: $(1+x+x^2y)^2+x^2$<|endoftext|>
TITLE: Generating n-cycles
QUESTION [15 upvotes]: Let $G = S_n$ (the permutation group on $n$ elements).
Let $A\subset G$ such that $A$ generates $G$.
Is there an $n$-cycle $g$ in $G$ that can be expressed as
$g = a_1 a_2 ... a_k$
where $a_i\in A \cup A^{-1}$ 
and $k\leq c_1 n^{c_2}$, where $c_1$ and $c_2$ are constants?
What about $2$-cycles, or elements of any other particular form?

REPLY [7 votes]: To elaborate on Sergei Ivanov's point with some futher observations:
By an argument similar to his, if you can obtain some element of any conjugacy class of bounded support with polynomial wordlength, the entire symmetric group has polynomial wordlength. That's because there are only polynomially many elements to the conjugacy class, and members of a fixed bounded conjugacy class generates the symmetric group in at most quadratic$(n)$ word length. (Transpositions require quadratic word length, using bubble sort. You can obtain a transposition in bounded wordlength once you have all elements of some conjugacy class). Furthermore, for fixed $k$, any $k$-tuple can be taken to any other $k$-tuple in polynomial$(n)$ word length, so a word that has polynomial length in terms of elements of the conjugacy class can be rewritten to have polynomial length in terms of the generators.
One strategy for trying to obtain elements of small support, given a few miscellaneous permutations, is to first take powers that halt some of their cycles. Once the support is small enough, you can take commutators of pairs of words whose support intersect only modestly to get still smaller support.   These kinds of tricks make it hard to see how there could be counterexamples to the conjecture that the diameter of the group is a polynomial in n.
I think it would not be very hard to write a computer program that, given an arbitary generating set, would in practice produce a function $S_n \rightarrow $ polynomial-length word representative, because it shouldn't take many words to find permutations with reasonably arbitrary cycle shape.  But it would be hard to prove it worked reliably.
I'd like to mention that there are well-known fast algorithms for analyzing the group generated by a collection of permutations, but they use recursive words, that is, words in words in words ... in generators, which gets around the group efficiently and quickly.  Since permutation are easy to compose, the recursion is computationally cheap, and for many purposes, better than using words. (Actual experts should please elaborate or correct me if I'm mistaken).<|endoftext|>
TITLE: What is the "universal problem" that motivates the definition of homotopy limits/colimits (and more generally "derived" functors)?
QUESTION [12 upvotes]: The ordinary notions of limit and colimit are universal solutions to a problem, specifically, finding terminal/initial objects in slice/coslice categories.  In the context of homotopy right Kan extensions (it's not hard to show that the theory of homotopy limits reduces to this case (the same holds for left Kan extensions/colimits)), we say that given a functor $f:C\to C'$ between small categories, and a functor $F:C\to A$ where $A$ is a cofibrantly-generated model category, that a natural transformation of functors $\alpha: H\to Ran_f F$ exhibits $H$ as the homotopy right Kan extension of $F$ if there exists an injectively fibrant replacement $G$ of $F$ such that the composite $H\to Ran_fF \to Ran_fG$ is a weak equivalence.  
When $A$ is combinatorial, we can also simply define a homotopy right Kan extension functor along $f$ to be $Ran_f (Q (-))$, where $Q$ is a functorial injectively fibrant replacement functor.  
This is easy enough to define, but why is this the definition?  Why would we want to take fibrant/cofibrant replacements and consider their ordinary Kan extensions/limits/etc?  I suspect that it has to do with the fact that homotopy is an honest equivalence relation on arrows from a cofibrant object into a fibrant object, but I would appreciate an actual explanation.

REPLY [4 votes]: This could have been a comment to Peter's answer, except that it's going to be long. (Maybe I should be making it a new question.)
Suppose that $Ho(\mathcal C)$ is obtained from $\mathcal C$ by (universally) inverting a class $\mathcal W$ of morphisms. Let's call an object of $\mathcal C$ homotopically terminal if in $Ho(\mathcal C)$ it is terminal. 
I did not assume anything about $\mathcal W$ (except that this localization existed). In practice we usually have more to work with, like a model structure or at least the 2 out of 6 condition. Given enough structure or information, there are ways of defining a simplicial set for each pair of objects such that this serves as a "function space", in particular such that its set of components is the set of morphisms in $Ho(\mathcal C)$. I'm not a master of the known sufficient conditions for making these function spaces -- I suppose that there are several overlapping approaches. 
QUESTION: Can we say that (in any or all of these approaches) an object $X$ is homotopically terminal in the sense I gave above if and only if for every object $Y$ the function space $Hom(Y,X)$ is weakly equivalent to a point? I hope so.
Here is a general, relatively simple, approach to the idea of derived functor, really just a distillation of standard ideas as I understand them, but trying to steer away from extraneous optional technicalities:
Suppose that in $\mathcal C$ and in $\mathcal D$ there are classes of morphisms $\mathcal U$ and $\mathcal V$ respectively. Let $\mathcal F$ be the category of functors $\mathcal C\to \mathcal D$. In $\mathcal F$ let $h\mathcal F$ be the full subcategory of "homotopy-invariant" functors -- those which take $\mathcal U$ into $\mathcal V$. By a left derived functor of $F\in \mathcal F$ I mean the following: Form the category $hF$ of all homotopy-invariant functors over $F$. An object is a pair $(G,G\to F)$ where $G\in h\mathcal F$. A morphism is a map $G_1\to G_2$ compatible with the maps to $F$. Define the class of maps $\mathcal W$ in $hF$ by requiring that for every $X\in \mathcal C$ the map $G_1(X)\to G_2(X)$ belongs to $\mathcal V$. By a left derived funtor of $F$ let us mean a homotopically terminal object of the category $hF$ (with respect to this class $\mathcal W$).
The standard way to achieve such a thing is to use a "replacement functor" $Q:\mathcal C\to \mathcal C$. You need a map $Q\to 1$, and you need three things to be true: (1) $QX\to X$ always belongs to $\mathcal U$. (2) $Q$ takes $\mathcal U$ into $\mathcal U$. (3) As applied to maps between objects that are in the image of $Q$, $F$ takes $\mathcal U$ into $\mathcal V$. I believe that that's all you need to show that $F\circ Q$ is homotopically terminal in the sense I defined. The proof is basically contained in what Peter wrote, including the edits prompted by my comments.
You don't even need the 2 out of 3 condition for $\mathcal U$ or $\mathcal V$ to get this. 
Right?<|endoftext|>
TITLE: What, exactly, has Louis de Branges proved about the Riemann Hypothesis?
QUESTION [34 upvotes]: Hi,
I know this is a dangerous topic which could attract many cranks and nutters, but:
According to Wikipedia [and probably his own website, but I have a hard time seeing exactly what he's claiming] Louis de Branges has claimed, numerous times, to have proved the Riemann Hypothesis; but clearly few people believe him. His website is:
http://www.math.purdue.edu/~branges/site/Papers
but I find his papers difficult to follow. However, whether or not you believe him, his arguments presumably should prove something, even if not the full RH.
So, my question is:
Are there any theorems related to the Riemann Hypothesis and similar problems, arising from his work, which have been fully accepted by the mathematical community and published (or at least submitted)?

REPLY [45 votes]: The paper by Conrey and Li "A note on some positivity conditions related to zeta and L-functions"
http://arxiv.org/abs/math/9812166
discusses some of the problems with de Branges's argument. They describe a (correct) theorem about entire functions due to de Branges, which has a corollary that certain positivity conditions would imply the Riemann hypothesis. However Conrey and Li show that these positivity conditions are not satisfied in the case of the Riemann hypothesis. 
So the answer is that de Branges has proved theorems in this area that are accepted, and his work on the Riemann hypothesis has been checked and found to contain a serious gap. (At least the version of several years ago has a gap; I think he may have produced updated versions, but at some point people lose interest in checking every new version.)
Update: there is a more recent  paper by Lagarias discussing de Branges's work.<|endoftext|>
TITLE: A simple example where elliptic boundary regularity fails due to a kink in the domain
QUESTION [9 upvotes]: I'm seeking a simple example of where elliptic (preferably linear) boundary regularity fails due to a simple kink in the domain.
So far my gueses were to look at $-\Delta u = f$ on $[0,2\pi] \times [0,2\pi]$ with $0$ Dirichlet boundary conditions and choose an $f$ which was far from $0$. This hasn't seem to produce any results (I was checking regularity directly by the method of Fourier series).
So more precisely, I would like an example where 
1) $Lu = f$ in $\Omega \subset \mathbb{R}^n$ with $f$ smooth 
2) $L$ is elliptic and $u = 0$ on $\partial \Omega$ 
3) $\Omega$ is not smooth and consequently $u$ is not smooth up to the boundary.

REPLY [7 votes]: This is the same idea as timur's answer but with more details and less generality. A frequent test problem in numerical analysis is the Poisson equation $-\Delta u = 1$ on the L-shaped domain 
$\Omega = ([-1,1] \times [-1,1]) \setminus ([-1,0] \times [-1,0])$ 
with homogeneous Dirichlet boundary conditions: $u = 0$ on $\partial\Omega$. The solution has a singularity at the origin: it is continuous but not differentiable. More precisely, close to the origin we have  
$u(r,\theta) \approx r^{2/3} \sin \frac{2\theta+\pi}{3}$ 
in polar coordinates, according to equation (1.6) in http://eprints.ma.man.ac.uk/894/02/covered/MIMS_ep2007_156_Sample_Chapter.pdf (sample chapter from Elman, Silvester and Wathen, Finite Elements and Fast Iterative Solvers, Oxford University Press, 2005).
Added: I don't know the details and I don't have time to do the necessary computations, but I think that you can solve the PDE by converting the Laplacian to polar coordinates and applying separation of variables. I imagine that you get that
$u(r,\theta) = r^{2n/3} \sin \frac{2n}{3} (\theta + \frac{1}{2}\pi)$
with $n$ a positive integer satisfies the boundary conditions at $r=0$ and $\theta=-\pi/2$ and $\theta=\pi$ (as Dorian comments below, these are all harmonic functions, so there must be something else). Then take a linear combination of those to match the conditions on the rest of the boundary of the L-shaped domain. Close to the origin, the $n=1$ term dominates. Perhaps somebody else can confirm / amend?<|endoftext|>
TITLE: Idempotents in Rings of Differential Operators
QUESTION [26 upvotes]: Differential Operators on General Commutative Rings
Let k be an algebraically closed field of characteristic zero, and let R be a commutative k-algebra.  Then a (Grothendieck) differential operator on R is a k-linear endomorphism $\delta$ of R, with the property that there is some $n\in \mathbb{N}$ such that for any $r_0,r_1...r_n\in R$, the iterated commutator vanishes:
$$ [...[[\delta,r_0],r_1]...,r_n]=0$$
Let the smallest such $n$ be the order of $\delta$.
The set of all differential operators is then a subring of $End_k(R)$, which has an ascending filtration given by the order, and with $D_0(R)=R$. If $R=k[x_1,...x_r]$, then $D(R)$ will be polynomial differential operators (in the calculus sense) in r-variables.  More generally, if R is the ring of regular functions on a smooth affine variety, then $D(R)$ is the usual ring of differential operators generated by multiplication operators and directional derivatives.
However, if $Spec(R)$ is not smooth, then $D(R)$ does not have an obvious geometric interpretation.  For example, if $R=k[x]/x^n$, then all k-linear endomorphisms of R are differential operators, and so
$$D(k[x]/x^n)=Mat_n(k)$$
Idempotents
For both research reasons and curiosity, I am interested in idempotent elements in $D(R)$, for R a general commutative ring.  An idempotent is an element $\delta\in D(R)$ such that $\delta^2=\delta$.  Idempotents in a commutative ring $R$ correspond to projections onto disconnected components of $Spec(R)$, but $D(R)$ is not commutative.  If the base ring $R$ does have idempotents, then they will also be idempotents under the inclusion $R\subset D(R)$.
However, there can be idempotents of higher order.  Consider the example from before, of $R=k[x]/x^n$.  Here, $D(R)=Mat_n(k)$, and there are many idempotents in $Mat_n(k)$, even though $R$ here has none.  As an explicit example, take $k[x]/x^2$, and consider the endomorphism which sends 1 to 0 and x to itself.  This can be realized by the differential operator $x\partial_x$ (which has a well-defined action on $k[x]/x^2$), and it squares to itself.  In general, I believe that $R$ must have nilpotent elements if $D(R)$ will have idempotents of positive order (since the symbol needs to square to zero).
My general question is, what is known about general idempotent elements in $D(R)$?  Has anyone seriously looked at them?  Do they correspond to something geometric?  Is there a condition one can put on a subspace decomposition $V\oplus W=R$ such that the projection onto $V$ which kills $W$ is a differential operator for the algebra structure on $R$?

REPLY [5 votes]: Here's a cute partial result, classifying idempotents of order 1.
Let $\delta$ be an idempotent differential operator of order 1.  Then there is a unique decomposition $R\simeq A\oplus M$, with $A$ a subring and $M$ a square-zero ideal, and an element $m\in M$, such that 
$$ \delta = \epsilon + m - (1-2\epsilon) \pi_M $$
where $\epsilon$ is an idempotent in $R$ and $\pi_M$ is the projection onto $M$ with kernel $A$ (which is a derivation).  Note that if $R$ is an integral domain, then $\epsilon$ is $1$ or $0$.
As a consequence, when $R$ is an integral domain, the decomposition of $R$ corresponding to $\delta$ and $1-\delta$ is $R=A'\oplus M$, where $A'$ is a shear translation of $A$ given by $a\rightarrow a+m$ (for some fixed $m$).<|endoftext|>
TITLE: Notion of smoothness for set-valued functions
QUESTION [8 upvotes]: Is there a way of talking about continuity and smoothness for set valued functions? More precisely, consider $M$ and $N$ topological/smooth manifolds, and let $f$ a function that associates to each point $p\in M$ a subset $f(p) \subset N$ (I haven't made any assumptions on what target sets are allowed, but feel free to discuss cases where some restrictions are required). Is there a meaningful/canonical way of saying that $f$ is continuous or smooth? 
For my particular application in mind, $M$ is a smooth manifold, and $f$ associates to each $p\in M$ an open, convex cone inside $T_pM$. 
Edit: I should clarify that a convex cone $K$ in some real vector space $V$ is a subset such that
(a) conic: for any $v\in K$ and $r\in \mathbb{R}_+$ $\implies rv \in K$. 
(b) convex: for any $v,w\in K$ and $a,b \in \mathbb{R}_+$ $\implies av + bw \in K$
It is open if $K$ is an open subset of $V$, so in particular open cones do not contain the origin.

REPLY [3 votes]: Three outcomes of a short brainstorming (all inspired by Algebraic Geometry):

I second Todd Trimble's comment to your question (it deserves to appear in an answer, so I repeat it here): Synthetic differential geometry gives you a way to talk about "manifolds of subsets", and you have sort of an automatic smoothness built in. But getting into this probably takes you on a long detour...
The Algebraic Geometer's way of treating many-valued functions is, very sloppily: Identify functions $X \rightarrow Y$ with their graphs, i.e. subvarieties $\Gamma \subseteq X \times Y$ with $pr_X(\Gamma)=X$ (defined on all of X) and $|pr_X^{-1}(x) \cap \Gamma|\leq 1 \ \forall x \in X$ (single-valued), then drop the second requirement.
So maybe you can look at smooth submanifolds of $M \times TM$ subject to the conditions you want?
The most reasonable, I would say, and close to the previous: Go to differentiable stacks. For your particular situation you could look at the classifying stack of cones inside tangent bundles: This would be the category fibred in groupoids over the site of differentiable manifolds which has objects $(M,C)$ with $M$ a smooth manifold, $C \subseteq TM$ a submanifold (smooth, except for the tip of the cone) of the total space of the tangent bundle such that each fiber is a cone in $T_x(M)$. Morphisms $(M,C) \rightarrow (M',C')$ should probably be smooth morphisms $f:M \rightarrow M'$ such that $C$ is the pullpack of $C'$ along $Tf$, the differential of $f$. Now a map of differentiable stacks from a manifold $N$ into this stack is the same as a smoothly varying choice of cones in the fibers of the tangent bundle $TN$. The same technique should work with other set-valued maps and is very flexible if you want to modify conditions.<|endoftext|>
TITLE: Hypersurfaces orthogonal to a cone
QUESTION [7 upvotes]: This question is somewhat related to Differential inclusions for distributions but I am asking for something rather more specific, so I hope it is alright to leave this as a separate, new question. 
Let $M$ be a smooth manifold, then given a one-form $\omega$ on $M$, Frobenius' theorem gives a simple way to test whether the distribution defined by the kernel of $\omega$ in $TM$ is integrable. 
Now let $F\subset T^*M$ be a subset of the cotangent bundle such that the restriction of the canonical projection $\pi$ is onto, and such that $\pi^{-1}(p)$ is an open convex cone for every $p$ (I don't know whether this condition will affect the answer; I'm just including the information on what I know). I am interested in knowing conditions which will guarantee that there exists (locally) an integrable distribution in $F$. 
Trvially some restrictions must apply. One may imagine that $F$ is in some sense not continuous, such that for any $\omega_p\in F_p$, there does not exist any smooth extension $\omega$ to any neighborhood of $p$. An example would be taking $M$ to be $\mathbb{R}^2$, and $F_p$ to be the first quadrant for all $p\neq 0$, and the second quadrant for $p = 0$. 
So one specific question is: is this lack of freedom the only difficulty? Is the following statement true?

Suppose $F$ has the property that, for any $p$, and any $\omega_p \in F_p$, there exists some smooth one-form $\omega$ that is a section of $F$, such that $\omega |_p = \omega_p$, and $d\omega |_p = 0$, then for any $q\in M$, we can find an open neighborhood $U$ of $q$ and some one-form $\eta$ over $U$ such that $\eta$ is a section of $F|_U$ and $d\eta = 0$.

Feel free to ask for clarifications.

REPLY [11 votes]: There are a global obstrutions in some cases when local solutions exist, even in the case that the cone is an open subset of the cotangent bundle.  For example:  let $M^3$ be the tangent line bundle to any Riemannian surface $N^2$.  The Levi-Civita connection defines a 2-plane field, where curves tangent to these planes represent transportation of a tangent line that is parallel under the connection. This is not integrable except where the Gaussian curvature is 0. Let's relax the condition, allowing the open cone of motions of lines where they drift from parallelism by a rate say < .1 degree per unit of arc length.  By the h-principle, there are local solutions --- actually, you can get local solutions by parallel transport of lines along geodesics from a central point, up to a small radius.
However: note that for any global solution, the leaves of the foliation are covering spaces of $N^2$. If the surface is a sphere, this would mean they project homeomorphically, so each leaf would give a global line field --- but we all know you can't comb a billiard ball, so this is impossible.  It's also known that if the surface has negative Euler characteristic, it can't be done, by a theory developed by John Milnor and John Wood.  A circle bundle over a surface other than $S^2$ admits a foliation transverse to the fibers if and only if the Euler class of the bundle has absolute value that does not exceed -Euler characteristic of base. For the tangent sphere bundle, this just barely works:  some significant examples are the Anosov foliations of the geodesic flow for any metric of negative curvature.  For the tangent line bundle, it doesn't work, since its Euler class is doubled from the tangent sphere bundle.
Quite a bit more is known about existence and non-existence of foliations, but I think this answers your specific question.<|endoftext|>
TITLE: Which languages could appear on Weil's Rosetta Stone?
QUESTION [15 upvotes]: André Weil's likening his research to the quest to decipher the Rosetta Stone (see this letter to his sister) continues to inspire contemporary mathematicians, such as Edward Frenkel in Gauge Theory and Langlands Duality.
Remember that Weil's three 'languages' were: the 'Riemannian' theory of algebraic functions; the 'Galoisian' theory of algebraic functions over a Galois field; the 'arithmetic' theory of algebraic numbers. His rationale was the desire to bridge the gap between the arithmetic
and the Riemannian, using the 'Galoisian' curve-over-finite-field column as the best intermediary, so as to transfer constructions from one side to the other. (See also 'De la métaphysique aux mathématiques' 1960, in volume II of his Collected Works.)
That fitted rather neatly with demotic Egyptian mediating between priestly Egyptian
(hieroglyphs) and ordinary Greek on the real Rosetta Stone. But just as one might have expressed that text in
a range of other contemporary languages - Sanskrit, Aramaic, Old Latin, why should
there not be other columns in Weil's story? Frankel himself adds a fourth column (p. 11)
'Quantum Physics'.
So now the questions:


Are there other candidate languages for Weil's stone? Might there be a further language for which we would need intermediaries back to the arithmetic? Could there be a meta-viewpoint which determines all possible such languages.


Presumably the possession of a zeta function is too weak a condition as that would allow the language of
dynamical systems.

REPLY [11 votes]: All the 12 or more approaches to geometry over the field with one element are tentatives to create such intermediate languages. But you seemed to ask more about a pre-existing area of it's own which may serve as a bridge - in this direction there are
$\bullet$ Alexandru Buium's theory of Arithmetic Differential Equations which brings the theory of differential equations into play as an intermediate language.
$\bullet$ I remember Alexandru Buium saying that there is also a theory of difference equations, different from his, but don't know more about this.
$\bullet$ Shai Haran uses probability theory as an intermediate language in his book "Mysteries of the Real Prime". It connects to the quantum theory column, but I don't know whether it's in the same way that Frenkel suggested.
$\bullet$ Homotopy theory might be a future candidate for a column of its own, on the one hand via motivic homotopy theory as it is getting available over more and more general base schemes with more and more general coefficients and thus moving towards arithmetic. And on the other hand possibly via ring spectra which may serve as the base deeper than the integers which is hoped to trigger the translation process one day...
$\bullet$ You write that the possession of Zeta functions is too weak to make the theory of dynamical systems an intermediate language. But there is certainly much more connecting dynamical systems with Arithmetic, as shown in Deninger's work.
$\bullet$ The works of Bost, Connes, Marcolli, Meyer, Laca and others connect Arithmetic to the Theory of Operator Algebras, and via those again to dynamical systems, quantum physics and Thermodynamics<|endoftext|>
TITLE: What are examples of cogenerators in R-mod?
QUESTION [9 upvotes]: Fill in the blank, please :)

Let $\mathcal C$ be a complete and cocomplete abelian category.  A generator in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a colimit of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$.  When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a generator is $R$, thought of as an $R$-module.  A cogenerator in $\mathcal C$ is an object $X \in \mathcal C$ such that every object $Y \in \mathcal C$ is a limit of a (small) diagram made entirely of $X$s; in this way, $X$ knows everything there is to know about $\mathcal C$.  When $\mathcal C$ is the category of all modules of some ring $R$, then an example of a cogenerator is ________.

The only examples I know are: when $R = \mathbb Z$, an example of a cogenerator is the rational circle $\mathbb Q / \mathbb Z$; when $R = \mathbb K$ is a field, an example of a cogenerator is $\mathbb K$.  But by some version of the Law of Small Numbers, these examples are not enough for me to see how to (or, in fact, whether it is possible to) generalize.

REPLY [16 votes]: For any ring $R$, the $R$-module $Hom_{\mathbb Z}(R,\mathbb Q/\mathbb Z)$ is an injective cogenerator of the category of $R$-modules.  Here $\mathbb Q/\mathbb Z$ can be replaced with $\mathbb R/\mathbb Z$ or any other injective cogenerator of the category of abelian groups.  When $R$ is an algebra over a field $\mathbb K$, another example of an injective cogenerator of the category of $R$-modules is $Hom_{\mathbb K}(R,\mathbb K)$.

REPLY [7 votes]: (This is closely related to Hailong's comment above.)
You can say (albeit rather abstractly) what any cogenerator must look like.  The following can be found in T.Y. Lam's Lectures on Modules and Rings, Theorem 19.10.  Let $\{V_i\}$ be a complete set of simple right $R$-modules, with injective hulls $E(V_i)$.  Then $U_0 = \bigoplus E(V_i)$ is a cogenerator, called the canonical cogenerator of $\mathrm{Mod}_R$,  and any module $U_R$ is a cogenerator for $\mathrm{Mod}_R$ if and only if $U_0$ can be embedded in $U$.
Note that if $R$ is right noetherian, then the direct sum of injective right $R$-modules is again injective.  Hence $U_0$ is injective, and in this case $U_R$ is a cogenerator if and only if $U_0$ embeds in $U$, if and only if $U_0$ is a direct summand of $U$.
Referring to your examples of $R$ above:  If $R = \mathbb{Z}$ then $U_0 = \mathbb{Q}/\mathbb{Z}$.  If $R = \mathbb{K}$ then $U_0 = \mathbb{K}$.  Both of these rings are noetherian, so the previous paragraph applies.  (In particular, every nonzero $\mathbb{K}$-vector space is a cogenerator).<|endoftext|>
TITLE: Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V?
QUESTION [6 upvotes]: Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume $V$?  Or more generally a hyperbolic $n$-manifold of volume $V$?

REPLY [19 votes]: A couple of things are true:
1.  If you have any Riemannian manifold of bounded infinitesimal geometry (curvature pinched above and below), its thick part, where the injectivity radius $> \epsilon$, can be triangulated with a number of simplices bounded by a constant times volume, where the constant depends on the curvature bounds and the dimension.  I don't personally know the constant even for hyperbolic 3-manifolds,  but I think there are people who can produce explicit bounds. This is basically a consequence of the compactness of the set of manifolds of bounded infinitesimal geometry and injectivity radius bounded below, together with the fact that all smooth manifolds admit a smooth triangulation, and that any smooth triangulation of a closed subset can be extended.

For hyperbolic 3-manifolds, if you allow "spun triangulations" where some tetrahedra are allowed to have missing vertices that spiral infinitely around a short closed geodesic, then there is a similar bound, the number is less than some constant times volume.  To do it:  first triangulate the thick part leaving a boundary torus, then make cones on the boundary triangles that spiral around a short geodesic.

The answers are the same whether you're asking for a geodesic triangulation of a hyperbolic manifold, or any smooth triangulation.<|endoftext|>
TITLE: How do modular forms of half-integral weight relate to the (quasi-modular) Eisenstein series?
QUESTION [16 upvotes]: The Eisenstein series
$$
G_{2k} = \sum_{(m,n) \neq (0,0)} \frac{1}{(m + n\tau)^{2k}}
$$
are modular forms (if $k>1$) of weight $2k$ and quasi-modular if $k=1$. It is clear that given modular forms $f,g$ of weight $2k$ and $2\ell$ that $f\cdot g$ is a modular form of weight $2(k + \ell)$.
We can also define modular forms of half-integral weight if we are a little more careful. However, the functional equation
$$
f\Big(\frac{az + b}{cz+d}\Big) = (cz+d)^{2k}f(z)
$$
must be replaced with something more subtle.
In particular, the Dedekind $\eta$-function is a modular form of weight 1/2; it satisfies
$$
\eta(z + 1) = e^{\frac{\pi i}{12}}\eta(z) \qquad \eta\Big(-\frac{1}{z}\Big) = \sqrt{-iz}\ \eta(z)
$$
Now, it would be nice if $\eta^4$ were a modular form of weight 2; however, an easy check using the above relations shows that this is only the case up to roots of unity, and so $\eta^4$ is not a multiple of $G_2$.
My question is then the following: What is the relation between modular forms of half-integral weight and (quasi-)modular forms of even integer weight? I know that $\eta^{24}$ is an honest modular form of weight 12, so I'm more curious about the general setting, or even what can be said about things like $\eta^4$.
Edit: As was pointed out in the comments, $\eta^4$ is a modular form for a congruence subgroup of $SL_2(\mathbb{Z})$, but not for the full modular group. The space of modular forms for the full modular group is generated by $G_4$ and $G_6$ (with $G_2$ thrown in if we are looking at the space of quasi-modular forms); is there a corresponding statement for modular forms on congruence subgroups?

REPLY [7 votes]: To answer Scott; yes, every modular form of integer weight $k$ for a congruence group can
be expressed rationally in terms of $\eta(r\tau)$ for rational $r$. For a start,
$g=f/\eta^{2k}$ is a modular function for some $\Gamma(N)$, so it suffices
to consider these. Replacing $g$ by $g(N\tau)$ we can reduce to the case where
$g$ is a modular function for some $\Gamma_0(M)$. The field of modular functions
for $\Gamma_0(M)$ is generated by $j(\tau)$ and $j(M\tau)$. So all we need
is that $j(\tau)$ expressible in terms of the $\eta(r\tau)$. This is well-known;
one can express $j$ in terms of the Weber functions, and them in terms of
$\eta(\tau)$, $\eta(2\tau)$ and $\eta(\tau/2)$.
Note that this argument is essentially the same as the proof for a very similar theorem
in this paper by Kilford.<|endoftext|>
TITLE: Monoids with infinite products
QUESTION [12 upvotes]: Say a monoid $M$ has infinite products if, for any (possibly infinite) sequence $(m_1,m_2,\ldots)$ of elements of $M$, there exists an element $m_1m_2\cdots\in M$, satisfying some good properties.  First, if the sequence is finite, it should coincide with the usual product on $M$.  Second, concatenation of sequences results in multiplication of their products and is associative.  Third, identities can be "thrown out," as can consecutive inverses ($m_{i+1}=m_i^{-1}$).  (Other good properties to include?)
Another way to phrase this is: "$M$ is closed under small ordinal colimits."  That is, if $M$ is considered as a one-object category, then for any small ordinal $[\kappa]$ and functor $m\colon[\kappa]\to M$, the colimit of $m$ exists in $M$.
Example: let ${\mathbb N}^+$ denote the monoid with underlying set ${\mathbb N}\cup ${$\infty$} and whose operation on a sequence $m=(m_1,m_2,\ldots)$ is given by addition if $m$ has only finitely many non-zero elements, and by $\infty$ otherwise.
Now suppose that $M$ is any monoid and I want to replace it by a monoid that has infinite products.  I'm hoping there are two ways to do this.  One would be to add colimits freely, and the other would be to add a single "$\infty$" element that served as a catch-all (as in the example above).  
Q: Do these both exist (functorially in $M$)?  If so, can you describe them in elementary terms?  For example, I'm worried about sequences like $1-1+1-1+\cdots$.  So in a good answer I'd hope to see what happens with such infinite sums.

REPLY [3 votes]: Semigroups with an $\omega$-product play an important role in automata accepting infinite words and the second order monadic theory of the natural numbers with the successor operator.  See the book Infinite Words: Automata, Semigroups, Logic and Games by Perrin and Pin.<|endoftext|>
TITLE: Why not _co_free modules?
QUESTION [12 upvotes]: Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules.  There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and cocontinuous (in particular, exact).  Since $R\text{-Mod}$ is both complete and cocomplete, $\operatorname{Forget}$ has both a left adjoint $\operatorname{Free}: \text{AbGp} \to R\text{-Mod}$ and a right adjoint $\operatorname{Cofree}: \text{AbGp} \to R\text{-Mod}$.
You can see what these functors are explicitly.  Let me write $_R R_{\mathbb Z}$ for "$R$ as a left module" and $_{\mathbb Z} R _R$ for "$R$ as a right module".
The $\operatorname{Forget}$ functor is (isomorphic to) the functor $\operatorname{Hom}_R({_R R_{\mathbb Z}},-)$ — this description makes it clearly continuous, and its left adjoint is $\operatorname{Free} \cong {_R R_{\mathbb Z}} \otimes_\mathbb Z (-)$.  But we also have $\operatorname{Forget} \cong {_{\mathbb Z} R _R}\otimes_R (-)$, whence its right adjoint is $\operatorname{Cofree} \cong \operatorname{Hom}_{\mathbb Z}({_{\mathbb Z} R _R},-)$.
I feel like I have some positive amount of experience with free modules.  (I would say, given the above, that the correct definition of "free module" is "object in the essential image of $\operatorname{Free}$", although what's actually used is "object of the form $\operatorname{Free}(\mathbb Z^{\oplus \kappa})$ for some cardinal $\kappa$.)  But I hardly ever come across the essential image of $\operatorname{Cofree}$, or indeed the cofree functor at all.  (Again, maybe the "standard" definition of "cofree module" is "module isomorphic to $\operatorname{Cofree}((\mathbb Q/\mathbb Z)^{\times \kappa})$," or something.)  The functors are not the same: when $ R = \mathbb Z/2$, then $\operatorname{Free}(\mathbb Z) = \mathbb Z/2$, whereas $\operatorname{Cofree}(\mathbb Z) = 0$.  If you would rather replace $\mathbb Z$ by a field throughout, then they are still not the same when $R$ is infinite-dimensional (for example).
So: Do people use cofree modules?  If so, how?  If not, why not?  Are free modules just a lot nicer than cofree ones, and if so, how?

REPLY [8 votes]: This construction is used frequently (at least, I use it frequently in my work). 
For example, it appears in the usual proof that module categories have enough injectives.
(In this case one studies $Cofree(\mathbb Q/\mathbb Z)$, as you anticipated.)
If we generalize slightly, and replace $\mathbb Z$ by the group ring $k[H]$ and $R$ by
the group ring $k[G]$ (with $H$ being a subgroup of $G$), then 
$Hom_{k[H]}(k[G],\text{--})$ is precisely the functor of induction from $H$-representations to $G$-representations, and the adjointness you note is a form of Frobenius reciprocity.
If $R$ is a Hecke algebra (over $\mathbb Z$) on a space of weight $k$-cuspforms of some level, 
then $Cofree(\mathbb Z)$ is the space of modular forms of weight $k$ with coefficients in $\mathbb Z$.  (This technical relationship between Hecke operators and the space of modular
forms on which they operate is used frequently by number theorists working on the arithmetic of 
modular forms.)
There are lots of other contexts in which this functor (and its variants, replacing
$\mathbb Z$ by other rings) appear, but maybe I've said enough for now.<|endoftext|>
TITLE: A circle packing conjecture
QUESTION [19 upvotes]: Consider $n$ circles with variable radii $r_1,\ldots, r_n$ that pack inside a fixed circle of unit radius. In other words, all $n$ variable-radius circles are contained in the unit radius circle and their interiors have empty intersections. The tangency graph of a packing comprises $n+1$ vertices, one for each circle, and edges between vertices if the corresponding circles are tangent.

Conjecture: in a packing that
  maximizes $r_1+\cdots +r_n$,
  the corresponding tangency graph is
  planar and triangulated.

This conjecture looks like it might be related to the Koebe-Andreev-Thurston circle packing theorem. The latter states that for every planar triangulated graph there is a corresponding circle packing of the kind described and that this packing is unique up to conformal transformations. While it may turn out that the KAT theorem can provide some insights on proving the conjecture, I believe that something else is going on. For instance, the radius-sum objective function is not conformally invariant.
I have good numerical evidence in support of this conjecture. The optimum configurations I've found up to $n=20$ all have triangulated graphs. I'm posting this on MO because I also have something that looks like it may be "close" to a proof. Perhaps someone can close the gap or convince me that the gap is actually a bottomless chasm -- either would be helpful!
Here is my proof strategy:

Use convexity to show that an optimal configuration maximizes the number of edges in the tangency graph.
Use Euler's theorem to show that
a tangency graph that maximizes the
number of edges is triangulated.

This is a constrained optimization problem in $\mathbb{R}^{3n}$. Consider the constraint that applies to circles 1 and 2: $(x_1-x_2)^2+(y_1-y_2)^2 \ge (r_1+r_2)^2$. This type of constraint is called "reverse convex" (the feasible region is the complement of an open convex set). Feasible regions in reverse convex problems (intersection of open set complements) can be quite complex -- they may not even be connected. On the other hand, they have a very nice property when we are maximizing a convex function: an optimum can always be found at a "vertex" of the feasible region. In a reverse convex problem in $\mathbb{R}^{N}$, a vertex is a point of the feasible region where at least $N$ of the constraints are equalities. We can think of reverse convex problems as generalizing linear programming in a way that inherits all the nice local properties.
The existence of a global optimizer requires that the feasible region is non-empty and compact. This is not an issue for the circle packing problem since we can let the radii range over all the real numbers and add reverse convex constraints $r_1\ge 0,\ldots,r_n\ge 0$.
The alert reader will already have realized that not all of the constraints in the circle packing problem are reverse convex! The constraints that apply to the fixed unit circle have the wrong sense of the inequality, e.g. $x_1^2 + y_1^2 \le (1-r_1)^2$. One can try to fix this problem by replacing the fixed unit circle with a regular $M$-gon and taking the limit (in some sense) of large $M$. This has two nice consequences. First, the optimization is now truly reverse convex (half-plane constraint for every side of the polygon) and so there is an optimizer where exactly $3n$ constraints are active (at their equality value). To see the second nice feature we have to do some counting.
The tangency graph has one new feature when the fixed circle is replaced by a regular $M$-gon: it is no longer simple because it may have doubled edges between the variable-radius circles and the polygon (whenever a circle is tangent to adjacent polygon edges). Let the number of circles with double tangencies be $D$. If $E$ and $F$ are the number of edges and faces of the graph, and $\tilde{E}$ and $\tilde{F}$ are these quantities when the doubled edges are merged into single edges, then $E=\tilde{E}+D$ and $F=\tilde{F}+D$. Since our graph has $n+1$ vertices, and reverse convex programming tells us there is an optimum with $E=3n$ tangencies, Euler's theorem gives $n+1-3n+F=2$, or $F=2n+1$. We therefore have the following formulas for the "reduced graph" after merging doubled edges: $\tilde{E}=3n-D$, $\tilde{F}=2n+1-D$. The reduced graph is simple and planar and satisfies $2\tilde{E}\ge 3\tilde{F}$ where equality implies that the graph is triangulated. Using our formulas this inequality becomes $D\ge 3$.
The result of this analysis is that optimum configurations in the $M$-gon have at least 3 circles with double tangency, and that the reduced graph is triangulated when this minimum holds. The number 3 is interesting. I believe it corresponds to the fact that the conformal transformations are fixed by specifying 3 points on the boundary of the region (the $M$-gon) where the circles are mapped. Sacrificing the symmetry of the fixed circle paid off because it allowed the optimization problem to have discrete solutions (whose existence follow from reverse convex programming).
There are two gaps in the proof. How do we take results for the $M$-gon and by some limiting process prove a theorem about the circle? Second, how do we prove $D=3$? Optimal configurations with $D>3$ become more unlikely as $M$ becomes large because in that case more than the minimum number of active constraints arise from double tangencies. After all, the pair of constraints at a double tangency become degenerate at $M=\infty$.
I believe the conjecture is true for the class of objective functions $r_1^p+\cdots + r_n^p$ where $1\le p < 2$. The case $p\ge 2$ is uninteresting because the optima degenerate into a single circle that completely fills the fixed circle, the rest having zero radius.

REPLY [18 votes]: You have an elaborate set of ideas, and I haven't thought through all of what you outlined, but here's a suggestion:
Oded Schramm generalized the circle packing theory to include arbitrary convex shapes, and showed they work in much the same way.  (The famous case of packing squares is one instance included in this generalization).  His theory even allows the shape to be a function of position and size, but that generality  seems unnecessary here.  The suggestion:  consider a set of regular N-gons packed inside an N-gon. At corners, they will touch at more than one point, but it is a connected set, so there is a well-defined adjacency graph with out doubled edges, and no room for extra disks to try to hide in the corners. The reverse convex constraints become piecewise linear.   I think the limiting process should be straightforward to analyze.<|endoftext|>
TITLE: What's so special about the forgetful functor from G-rep to Vect?
QUESTION [19 upvotes]: The following is some version of Tannaka-Krein theory, and is reasonably well-known:

Let $G$ be a group (in Set is all I care about for now), and $G\text{-Rep}$ the category of all $G$-modules (over some field $\mathbb K$, say).  It is a fairly structured category (complete, cocomplete, abelian, $\mathbb K$-enriched, ...) and in particular carries a symmetric tensor product $\otimes$.  The forgetful functor $\operatorname{Forget}: G\text{-Rep} \to \text{Vect}$ respects all of this structure, and in particular is (symmetric) monoidal.  Let $\operatorname{End}_\otimes(\operatorname{Forget})$ denote the monoid of monoidal natural transformations of $\operatorname{Forget}$.  Then it is a group, and there is a canonical isomorphism $\operatorname{End}_\otimes(\operatorname{Forget}) \cong G$.

The following is probably also reasonably well-known, but I don't know it myself:

Let $G$, etc., be as above, but suppose that we have forgotten what $G$ the category $G\text{-Rep}$ came from, and in particular forgot, at least momentarily, the data of the forgetful functor.  We can nevertheless recover it, because in fact $\operatorname{Forget}$ is the unique-up-to-isomorphism ADJECTIVES functor $G\text{-Rep} \to \text{Vect}$.

My question is: what are the words that should go in place of "ADJECTIVES" above?  Certainly "linear, continuous, cocontinuous, monoidal" are all reasonable words, although my intuition has been that I can drop "cocontinuous" from the list.  But even with all these words, I don't see how to prove the uniqueness.  If I had to guess, I would guess that the latter claim is a result of Deligne's, although I don't read French well enough to skim a bunch of his papers and find it.  Any pointers to the literature?

REPLY [35 votes]: If $G$ is an affine algebraic group (for example a finite group), then the category of $k$-linear cocontinuous symmetric monoidal functors from $\mathsf{Rep}(G)$ to $\mathsf{Vect}_k$ is equivalent to the category of $G$-torsors over $k$. In particular, not every such functor needs to be isomorphic to the identity. For example, if $k'$ is finite Galois extension of k with Galois group $G$, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

REPLY [11 votes]: One needs to be careful.  One cannot recover the group $G$ from the tensor category alone, but only with the data of category, fiber functor.  There are examples of non-isomorphic (finite, even) groups with equivalent categories of representations.  For instance, see Pasquale Zito's answer to this question:
Finite groups with the same character table
However, as is discussed in the paper Zito links to, remembering the symmetry on the categories recovers the group, up to isomorphism.  I'm not sure who it's due to.<|endoftext|>
TITLE: an engineering Ph.D. teaching math in college
QUESTION [20 upvotes]: I have a friend who has been teaching college-level math (e.g., all levels of calculus) 
for about 4 years, although all of his education, including his Ph.D., was in engineering. 
Now he is considering applying for tenure track jobs at teaching schools [edit: in the USA] and he is bemoaning the fact that nearly all the ads he sees include "Ph.D. in mathematics", rather than "Ph.D. in mathematics or related area" as a qualification. This raised a question which I don't have the experience to answer for him: are these ads, for tenure-track positions at schools that focus more on teaching, generally so serious about the Ph.D. in math that they would dismiss an application from someone with a Ph.D. in engineering instead of math? [Edit: What I have in mind here about there being a problem are legal obligations that the hiring committee would have to stick exactly to the letter, not the general spirit, of the ad. 
This is not LawyerOverflow, so examples of lawsuits over job ad wording are not necessary.]
While there are other aspects of his application which are more important for a teaching job (he gets fantastic teaching evaluations and also quickly learns how to use new educational technology), the only thing I'm wondering about on this friend's behalf is the Ph.D.-not-in-math issue and whether it's a deal breaker. [Edit: If someone knows explicit examples where it was not an issue, that would be interesting to hear about, although you can keep information about the school and the hire anonymized.]
(Of course you may ask how he got his first job teaching math and why he can't do the same thing he did then. His first teaching job was short-term, not tenure track, and his hiring had some idiosyncratic features that are unlikely to repeat themselves for the next teaching job.)

REPLY [7 votes]: Having been in a similar situation, I feel qualified to add in my $0.02.  My BS and MA degrees are in Math, but my PhD is in Industrial Engineering.  The easiest thing you can do is simply write the search committee and ask if a PhD in a related field is acceptable.
There are multiple advantages to this approach.  The most obvious is that they will save you time putting together a packet if the answer is no.  Additionally, you get a chance to put a bug in their ear, so that your application will not be coming out of the blue.  Finally, you will be able to trump up the advantages of having a degree in a closely related field.  In my case, many of the schools I applied to had large teacher ed programs, and I was able to tout my rather large stockpile of practical applications of just about any mathematical topic.
That said, it still can be a tough sell.  Over the course of two years, I had four on-campus interviews (one was a visiting position) without an offer.  And I was coming in having successfully written a large NSF grant, a pair of journal articles in review, and glowing teaching evaluations.<|endoftext|>
TITLE: Is there an underlying explanation for the magical powers of the Schwarzian derivative?
QUESTION [129 upvotes]: Given a function $f(z)$ on the complex plane, define the Schwarzian derivative $S(f)$ to be the function
$S(f) = \frac{f'''}{f'} - \frac{3}{2} (\frac{f''}{f'})^2$
Here is a somewhat more conceptual definition, which justifies the terminology.  Define $[f, z, \epsilon]$ to be the cross ratio $[f(z), f(z + \epsilon); f(z + 2\epsilon), f(z + 3\epsilon)]$, and let $[z, \epsilon]$ denote the cross ratio $[z, z + \epsilon, z + 2\epsilon, z + 3\epsilon]$ (in fact this is just 4, but this notation makes my point clearer).  One can ask if $[f, z, \epsilon]$ is well approximated by $[z, \epsilon]$; indeed, it turns out that the error is $o(\epsilon)$.  So one pursues the second order error term and finds that $[f, z, \epsilon] = [z, \epsilon] - 2 S(f)(z) \epsilon^2 + o(\epsilon^2)$.  So the Schwarzian derivative measures the infinitesimal change in cross ratio caused by $f$.  In particular, $S(f)$ is identically zero precisely for Möbius transformations. 
That's all background.  From what I have said so far, the Schwarzian derivative is at best a curiosity.  What is not obvious at first glance is the fact that the Schwarzian derivative has magical powers.  Here are some examples:
First magical power: The Schwarzian derivative is deeply relevant to one dimensional dynamics, stemming from the fact that it behaves in a specific way under compositions.  For example, if $f$ is a smooth function from the unit interval to itself with negative Schwarzian derivative and $n$ critical points, then it has at most n+2 attracting periodic orbits.
Second magical power: It says something profound about the solutions to the Sturm-Liouville equation, $f''(z) + u(z) f(z) = 0$.  If $f_1$ and $f_2$ are two linearly independent solutions, then the ratio $g(z) = f_1(z)/f_2(z)$ satisfies $S(g) = 2u$.
Third magical power: The Schwarzian derivative is the unique projectively invariant 1-cocyle for the diffeomorphism group of $\mathbb{R}P^1$.  This is probably just a restatement of the conceptual definition I gave above, but I'm not sure; in any event, this gives the Schwarzian derivative a great deal of relevance to conformal field theory (or so I'm told).  
I'm sure there are more.  I'm wondering if all of these powers can be explained by some underlying geometric principle.  They all seem vaguely relevant to each other, but the first power in particular seems very hard to relate to the definition in any obvious way.  Does anybody have any insights?

REPLY [5 votes]: Let G(k,n) be the k'th order jet group in n variables which consists of the set of k-jets of local diffeomorphisms of R(n) fixing the origin under the operation of composition. In coordinates, this group operation can be written explicitly using the chain rule.
Now consider G(3,1) and G(2,1) and the obvious projection homomorphism from G(3,1) onto G(2,1) induced by jets. This projection splits, that is, there is an injective homomorphism which imbeds G(2,1) into G(3,1). Identifying the image of this injection with G(2,1), we can form the left (say) coset space G(3,1)/G(2,1). Now the expression for the Schwarzian derivative defines "coordinates" on G(3,1)/G(2,1). The details of these computations can be found on pages 152-153 of the book "An Alternative Approach to Lie Groups and Geometric Structures". This construction generalizes to arbitrary dimensions and is a very special case of defining geometric structures as explained on pages 174-182 of this book. 
To summarize here, these splittings arise from homogeneous spaces and define "geometric connections" which are not necessarily connections in the classical sense (keeping in mind that connections can be defined on general principal and vector bundles and are essentially topological objects). Furthermore, these splittings are built into the definition of a geometric structure and therefore there is no need to search for a "special connection" suitable for some geometric structure. Consequently, the magical powers of the Scwarzian derivative is a special case of the magical powers of "connections". 
As an interesting detail, the obvious left action of G(3,1) on G(3,1)/G(2,1) gives the transformation rule of the Schwarzian derivative found in classical textbooks. In affine case (the action of G(2,n) on G(2,n)/G(1,n)), for instance, we get the well known transformation rule of the "connection components" on the tangent bundle which is historically the starting point of the theory of connections on vector bundles! The action of G(1,n) on G(1,n)/G(0,n) = G(1,n) gives absolute parallelism studied in detail in this book.<|endoftext|>
TITLE: Why are spectra indexed over the natural numbers?
QUESTION [5 upvotes]: A spectrum is a sequence $X_0,X_1,...$ of spaces together with structure morphisms $\Sigma X_n\to X_{n+1}$. To get the usual model for the stable homotopy category based on the category of spectra, one "inverts" the suspension functor (or the shift functors) which is not an isomorphism in the category of spectra. It seems to me that this is a kind of allowing the objects to be indexed over the integers. So why does one not define a spectrum to be indexed over the integers or at least bounded below?

REPLY [2 votes]: The reason may be slightly historical, the first examples of spectra were most easily seen to be indexed over the natural numbers: $MU$, $HR$, and $\mathbb{S}$. Now though, we know we ought to be using finite dimensional inner product subspaces of $\mathbb{R}^\infty$! Also, we want there to be an underlying "point set" category of spectra that we do something to do in order to get the stable homotopy category. What we really care about is some spectrum, not the image of that spectrum in the stable homotopy category. I guess I am trying to say that even though we will end up inverting something later in order to work on it does not mean we should throw out the original and only remember the localization.<|endoftext|>
TITLE: Area of filled Julia sets
QUESTION [8 upvotes]: The recent question Area of the boundary of the Mandelbrot set ? prompted me to ask this question.
There has been some work on estimates for the area of the Mandelbrot set, e.g., a paper by John H. Ewing and Glenn Schober in Numerische Mathematik. Is there similar work for the estimating the area of quadratic filled Julia sets as a function of the parameter $c$? Perhaps the material in the book Computability of Julia Sets implies some estimates but I don't have the book handy and from what I recall of skimming it, there was none.

REPLY [3 votes]: This paper contains some information about the area of filled Julia sets, though not a formula:

Yang, Guoxiao, Some geometric properties of Julia sets and filled-in Julia sets of polynomials. Complex Var. Theory Appl. 47 (2002), no. 5, 383–391. MR1906990 (2003c:37067)

There is also this more promising paper by the same author, but it is in Chinese and I can't get a copy anyway:

Yang, Guo Xiao, The area and diameter of filled-in Julia sets and Mandelbrot sets.
  Acta Math. Sinica 38 (1995), no. 5, 607–613. MR1372560 (96m:30040)

If someone knows these papers, I'd be grateful for any insights.
Problem A-1 in Milnor's Dynamics in one complex variable contains a formula for the area expressed as a series based on Gronwall's area theorem:
$$
 \pi (1 - |a_2|^2 - 3|a_4|^2 - 5|a_6|^2 - \cdots)
$$
The series is said to converge slowly. The coefficients of the series can be easily computed recursively though by solving
$$
 \psi(w^2) = \psi(w)^2+c
$$
for
$$
\def\F#1{\frac{a_{#1}}{w^{#1}}}
 \psi(w) = w(1 + \F2 + \F4 + \F6 + \cdots)
$$<|endoftext|>
TITLE: Computation of vanishing cycles
QUESTION [9 upvotes]: Here's the problem I'm looking at: 
$F$ is a perverse sheaf (or a regular holonomic D-module, or even a mixed hodge module) on $\mathbb{C}^2$ stratified by $z_1 = 0$, $z_2=0$. It can be caracterized by a quiver
$$
  \begin{array}{ccc}
    \psi_{z_1}\psi_{z_2}(F) & \leftrightarrows & \psi_{z_1}\phi_{z_2}(F) \cr
    \uparrow \downarrow & & \uparrow \downarrow  \cr
    \phi_{z_1}\psi_{z_2}(F) & \leftrightarrows & \phi_{z_1}\phi_{z_2}(F)
  \end{array}
$$
where arrows are the canonical and variation maps.
Consider $f(z_1,z_2) = z_1+z_2$ (or any curve passing through $(0,0)$ transverse to the axises). 
How can we compute 
$$
  can: \psi_f(F) \leftrightarrows \phi_f(F) : var
$$
in terms of these data?

REPLY [4 votes]: Here's the bottom line. If $F$ is a perverse sheaf or a D-module, there is an isomorphism between $\psi_fF$ and the complex $[\psi \psi F \to \psi \phi F \oplus \phi \psi F]$. The reason why we can't get to it in a nice way is that it is really non canonical. 
To see it, consider a weight-filtrered perverse sheaf $(F,W)$ or an F-filtered D-module, the natual filtration on $\psi_f F$ does not correspond to the natural convolution filtration on $[\psi \psi F \to \psi \phi F \oplus \phi \psi F]$. Also the isomorphisms you get for perverse sheaves on one side and D-modules on the other side are not compatible with the Riemann-Hilbert correspondance. Non trival periods appear because integrals over the fibers $\int_{z_1+z_2 = 1} z_1^a z_2^b  = B(a+1,b+1)$ are related to Euler's Beta function.    
What is canonical (for $F$ bi-monodromic) is the isomorphism $\psi_f F = i_1^*Rf_*F$ between nearby cycles and the fiber over 1 of the direct image. The right way to approach the problem is to consider the additive convolution on the affine line as a new fundamental operation distinct from the tensor product.   
In this way one can prove a general Sebastiani-Thom theorem: in a neighborhood of $f(x) = g(y) = 0$, the vanishing cycles of $\phi_{f\oplus g}(F\boxtimes G)$ is the additive convolution of $\phi_f(F)$ and $\phi_g(G)$ (where vanishing cycles are interpreted as monodromic sheaves on the normal cones with 0 fiber on the zero sections). If $F$ and $G$ are perverse sheaves (resp. D-modules) then the additive convolution is non canonically isomorphic to the tensor product (this was known to Deligne). 
One has a similar result for nearby cycles and we can describe the canonical and variation morphism at $f(x) = g(y) = 0$. But I havent found how to describe the vertical monodromy of the nearby cycles yet. 
PS: I should mention that when I finally understood all this over a year ago. I sent a preprint to Claude Sabbah who informed me than M. Saito had proved a similar result more than 10 years ago but never published it. I still hope I will find the time to write things down properly and publish my version as I find the problem is both elementary, very deep and interesting.<|endoftext|>
TITLE: Where to publish a paper on the Mafia game?
QUESTION [16 upvotes]: I wrote a research paper "A mathematical model of the Mafia game" (arXiv:1009.1031 [math.PR]). However, I do not know where to publish it. As an undergraduate studying majorly physics, I have little knowledge of mathematical journals. Moreover, its not easy for me to classify its subject.
The paper itself is generally "using applied mathematics (and physicist's approach) to model a party game (psychology)". Just to give some of its features:

Uses pure death process and gives its closed-form solution
Is elementary and didactic (a tricky use of generating functions, Wallis formula, ...)
Somehow "cool" topic

I thought about The American Mathematical Monthly. Besides the obvious advantages I see two drawbacks:

In AMM there are usually an old results in a nice/didactic/cool form or short findings
It won't reach to psychologists who might investigate the Mafia game experimentally

Can you recommend me any proper journal? (popular mathematics, mathematics and psychology, ...?) 
Any other advice are appreciated as well.

REPLY [17 votes]: I have only glanced at your paper, but one possibility is to submit it to The Mathematical Intelligencer, in particular to Michael Kleber, who edits the "Mathematical Entertainments" column.  This is a great place for serious mathematical analysis of problems with a recreational flavor.  The Intelligencer has a wide readership, which is what you want for your article.
By the way, contrary to some other commentators, I don't see any reason why you necessarily need to find a mentor just because you're an undergraduate.  Your paper seems to be well-written enough.  Of course it doesn't hurt to find a mentor but what matters is the quality of your writing and not your employment status.
Also, regarding KConrad's comment that only existing players of the game will find it interesting, I don't believe that this is true.  First of all, as you note, Mafia is in fact a pretty well-known game.  Secondly, the subtlety of the game means that there is a lot of interesting mathematics buried in it.  If your article can draw more mathematicians into studying it, that would be a very good thing.<|endoftext|>
TITLE: Etale $\pi_1$ of Grassmannian
QUESTION [5 upvotes]: I am forced to know the etale fundamental group of the grassmannian over the rational field. I searched it but couldn't find any hint. I am wondering whether there are some positive results or recipe to compute it.
Thank you!

REPLY [15 votes]: The answer is that any Grassmannian is geometrically simply connected, so the etale fundamental group over $\mathbb{Q}$ is simply [edit: !!] the absolute Galois group $\operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$.
In more detail: let $X$ be a geometrically integral variety defined over $\mathbb{Q}$, let 
$\overline{X}$ be its basechange to an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $\mathfrak{g}_{\mathbb{Q}} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ be the absolute Galois group of $\mathbb{Q}$.  
1) Via a choice of geometric points (which we suppress), we get a short exact sequence of profinite groups
$1 \rightarrow \pi_1(\overline{X}) \rightarrow \pi_1(X) \rightarrow \mathfrak{g}_{\mathbb{Q}} \rightarrow 1$.  
2) Let $K$ be an algebraically closed field of characteristic $0$.  Suppose that either $X$ is complete or that $X$ is nonsingular [in our application, both hold].  Then the natural map
$\pi_1(\overline{X}) \rightarrow \pi_1(\overline{X} \otimes K)$ is an isomorphism.  
[Comment: if instead of $\mathbb{Q}$, our ground field was a field $k$ of positive characteristic and $K$ is an algebraically closed field containing $\overline{k}$, this map is still an isomorphism for complete varieties but not necessarily for all smooth varieties and indeed, not even for the affine line!]
3) Take $K = \mathbb{C}$.  Then the analytification functor induces an isomorphism from the 
profinite completion of the topological fundamental group of $X(\mathbb{C})$ (with the $\mathbb{C}$-analytic topology) to the etale fundamental group $\pi_1(X \otimes \mathbb{C})$.  
4) If $X/\mathbb{Q}$ is any Grassmannian, then it is nonsingular, complete and its analytification is the usual complex Grassmannian, which is simply connected: see e.g.
http://books.google.com/books?id=WHjO9K6xEm4C&pg=PA748&lpg=PA748&dq=simply+connected+Grassmannian&source=bl&ots=waYTv_whVx&sig=ErlPHYKL5FdQPUdBlIrIgYQGhbE&hl=en&ei=9FiITJuNNYjW9ASP_p3fDg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CC8Q6AEwBTgK#v=onepage&q&f=false
Therefore putting the previous parts together the geometric fundamental group $\pi_1(\overline{X})$ is trivial, so we naturally have $\pi_1(X) \cong \mathfrak{g}_{\mathbb{Q}}$.
References for these facts may be found, for instance, in Chapter 5 of Szamuely's Galois Groups and Fundamental Groups.<|endoftext|>
TITLE: Degree of $[K:K^p]$ and Completion
QUESTION [7 upvotes]: Suppose $K$ is a field endowed with a non-archimedian absolute value.  Assume $K$ has characteristic $p>0$ and that $[K:K^p] < \infty$.  Let $L$ be the completion of $K$ with respect to this absolute value.  Is it always true that $[L:L^p] = [K:K^p]$?

REPLY [8 votes]: In fact, BCnrd's comment says that for dvr's, only for non-excellent one's $[L:L^p]\ne [K:K^p]$ can happen. Actually, suppose $d=[K:K^p]$ is finite. Consider the canonical map $K\otimes_{K^p} L^p \to L$. The source is a $L^p$-vector space of dimension $d$, its image is therefore closed ($L^p$ is complete) and contains $K$, so the map is surjective. Therefore $[L:L^p]=d$ if and only if the above map is an isomorphism, or equivalently if $K$ and $L^p$ are linearly disjoint over $K^p$. Extracting $p$-th root in all these fields, this is also equivalent to $K^{1/p}$ and $L$ linearly disjoint over $K$. Which is also equivalent to $L$ separable (i.e. geometrically reduced) over $K$. This is the definition for the dvr of $K$ to be excellent. 
Example of non-excellent dvr with finite $[K : K^p]$: Let $s$  be an element of $\mathbb F_p((t))$   transcendent over $\mathbb F_p(t)$, let $K=\mathbb F_p(t,s)⊂\mathbb F_p((t))$ be endowed with the $t$-adic valuation. Then $[K:K^p]=p^2$  and $[L:L^p]=p$. If we take any number of algebraically independent elements $s_1,...,s_n$  (instead of just one $s$) over $\mathbb F_p(t)$, then we have $[K:K^p]=p^{n+1}$.<|endoftext|>
TITLE: Computable nonstandard models for weak systems of arithemtic
QUESTION [14 upvotes]: By Tennenbaum's theorem, PA itself does not have any computable nonstandard models.  The integer polynomials which are 0 or have a positive leading coefficient form a computable nonstandard model of Robinson arithmetic, which also happens to make the order relation total.  Since Presburger arithmetic is decidable, we can add axioms giving it a nonstandard number and work through Henkin's proof of the completeness theorem to get a computable nonstandard model of Presburger arithmetic.  (There's probably a simpler way to get one, though.)
Is any system strictly weaker than PA known to have no computable nonstandard models?
What other systems weaker than PA are known to have computable nonstandard models?
.
possible examples of either include:
I-Delta-0, I-Delta-0(exp), I-Sigma-1
Elementary Function Arithmetic
Elementary Recursive Arithmetic, Primitive Recursive Arithmetic
Robinson arithmetic + Euclidean division, Robinson arithmetic + Euclidean division + order relation is total

REPLY [6 votes]: $\mathsf{IE_1}$ doesn't have computable non-standard models. (George Wilmers, "Bounded existential induction", 1985) Any theory that contains it will not have computable non-standard models, e.g.: $\mathsf{I\Delta_0}$, $\mathsf{PRA}$, ...
On the other hand, $\mathsf{IOpen}$ does have computable non-standard models (J. C. Shepherdson, "A non-standard model for a free variable fragment of number theory", 1964). Any theory contained in it will also have computable non-standard models, e.g.: $\mathsf{Q}$.
The threshold(s) of having a computable non-standard model is somewhere between $\mathsf{IOpen}$ and $\mathsf{IE_1}$. There are various principles that one can add to $\mathsf{IOpen}$ and see if the resulting theory still has a computable non-standard model. 
It is known that $\mathsf{IOpen}$ plus cofinality of primes and Bezout axioms has computable non-standard models. Over $\mathsf{IOpen}$, Bezout implies normality and GCD axioms and is provable in $\mathsf{IE_1}$.<|endoftext|>
TITLE: Heuristic argument that finite simple groups _ought_ to be "classifiable"?
QUESTION [59 upvotes]: Obviously there exists a list of the finite simple groups, but why should it be a nice list, one that you can write down? 
Solomon's AMS article goes some way toward a historical / technical explanation of how work on the proof proceeded.  But, though I would like someday to attain some appreciation of the mathematics used in the proof, I'm hoping that there is some plausibility argument out there to convince the non-expert (like me!) that a classification ought to be feasible at all. A few possible lines of thought come to mind:

Groups have very simple axioms. So perhaps they should be easy to classify. This seems like not a very convincing argument, but perhaps there is some way to make it more convincing.
Lie groups have a nice classification, and many tools are available for their study and that of their finite analogues. And in fact, it turns out that almost all finite simple nonabelian groups fall under this heading.  Is it somehow clear a priori that these should be essentially all the examples? What sort of plausibility arguments might lead one to believe this?
If there are not currently any good heuristic arguments to convince a non-expert that a classification should be possible, then will this always be the case? Or will we someday understand things better...

There is probably a model-theoretic way to formalize this question. As a total guess, it might be something along the lines of "Do the finite simple groups have a finitely axiomatizable first-order theory?", except probably "finitely axiomatizable first-order theory" doesn't really capture the idea of a classification. If someone could point me towards how to formalize the idea of "classifiable", or "feasibly classifiable", I'd appreciate it.FSGs up to order SEFSGs up to order MO
EDIT:
To clarify, what I'd like is an argument that finite simple groups should be classifiable which does not boil down to an outline of the actual classification proof. Joseph O'Rourke asked on StackExchange Why are there only a finite number of sporadic simple groups?. There, Jack Schmidt pointed out the work of Michler towards a uniform construction of the sporadic groups, as reviewed here. Following the citation trail, one finds a 1976 lecture by Brauer in which he says that he's not sure whether there are finitely many or infinitely many sporadic groups, and which he concludes with some historical notes that describe a back-and-forth over the decades: at times it was believed there were infinitely many sporadic groups, and at times that there were only finitely many. So it appears that the answer to my question is no-- at least up to 1976, there was no evidence apart from the classification program as a whole to suggest that there should be only finitely many sporadic groups.
So I'd like to refocus my question: are such lines of argument developing today, or likely to develop in the (near? distant?) future? And has there been further clarification of what exactly is meant by a classification? (Is this too drastic a change? should I start a new thread?)

REPLY [8 votes]: There is an interesting result of U. Felgner (MR1107758, see also MR1477188): 
simplicity is an elementary statement in the class of finite non-abelian groups. 
E.i. he showed that there is a first order sentence $\sigma$ such that $G\models \sigma$, where $G$ is finite if and only if G is non-abelian and simple. However the proof uses the classification of finite simple groups.<|endoftext|>
TITLE: "Why" is every polynomial representation of SL(2) selfdual?
QUESTION [8 upvotes]: Given a field $K$ of characteristic $0$. It seems to me that every finite-dimensional polynomial representation of $\mathrm{SL}_2\left(K\right)$ is self-dual (i. e., isomorphic to its dual). In fact, every representation of $\mathrm{SL}_2\left(K\right)$ is a direct sum of irreducible representations (since $\mathrm{SL}_2\left(K\right)$ is semisimple), and the irreducible representations are the canonical representations on $K\left[x,y\right]_n$ which are known to be self-dual. But is there a simpler proof without subdividing into irreducibles?

REPLY [11 votes]: The question itself and some of the comments seem out of focus to me, so let me add to what Richard and George write the following summary version of an answer.  I'd stress that nothing here is really complicated or subtle to prove apart from the basic Cartan-Weyl classification and (in characteristic 0) complete reducibility for finite dimensional representations.   
First, the group itself is defined and split over the prime field (here $\mathbb{Q}$), hence over any larger field.   Chevalley's theory implies that the representations discussed here are absolutely irreducible over $K$.   (For a semisimple group defined but not split over a field, more analysis is needed of representations which require a field extension to become absolutely irreducdible.)
Anyway, for a connected semisimple group the "rational" and "polynomial" representations are the same, unlike the reductive group GL$(n,K)$.   The group also being simply connected in this case, the rational/polynomial representations are essentially those of the Lie algebra and are more easily classified by dominant integral highest weights in that setting.   So each irreducible representation or simple module in question has a unique highest weight $\lambda$.  The easy textbook criterion for such a module to be self-dual is just that $\lambda = -w_0 \lambda$ where $w_0$ is the longest element of the Weyl group.   As Richard Borcherds points out, this is -1 just for simple types listed, including type $A_1$.  
So far nothing really depends on characteristic 0.   But as George McNinch observes, there are plenty of cases where nonsimple modules in prime characteristic fail to be completely reducible and are typically not self-dual.   So you do need to invoke complete reducibility (and non-canonical direct sum decompositions) to dispose of the characteristic 0 question.   
P.S.  It's certainly possible to treat the rank 1 case here by direct ad hoc methods in characteristic 0, including the needed proof of complete reducibility (using the easily computed Casimir operator).   For irreducible representations, self-duality is a trivial consequence of the fact that these representations are uniquely classified (up to isomorphism) by their dimensions 1, 2, 3, ....   But such an ad hoc argument fails to provide much enlightenment.  And the general theory allows one to see that the group representations and Lie algebra representations are essentially the same, whether the groups are regarded as Lie groups or algebraic groups (or just as abstract groups).   Of course, finite dimensionality is a key point throughout, since the infinite dimensional representation theory involves harder questions.<|endoftext|>
TITLE: Is the category of commutative group schemes abelian?
QUESTION [23 upvotes]: I think, because in the category of schemes, all finite limits exist, the commutative group
objects with homomorphisms should form an abelian category. 
Is this true? And do you know anywhere to cite this?

REPLY [2 votes]: should be not true! What is the cokernel of the map $\mathbb{Z}\rightarrow A$ with $A$ an abelian variety and $\mathbb{Z}$ the constant group scheme represented by the ring of integers?  But if you consider only commutative algebraic group scheme over a field $k$(i.e. of finite type over $k$), then it should be abelian.<|endoftext|>
TITLE: Totally geodesic surfaces in fibered 3-manifolds
QUESTION [6 upvotes]: Is there an easy example of a (closed) hyperbolic 3-manifold that fibers over the circle but contains some totally geodesic surface?
(Of course such manifolds exist if the 'Virtually Fibered Conjecture' were correct, since a geodesic surface lifts to the fibered cover. But is there something more eplicit?)

REPLY [24 votes]: There are many specific known examples. Here is one construction:
Start with the 3-torus $T^3$, parametrize in the standard way as $R^3/Z^3$.  It fibers over the circle in many ways.
Let $a$, $b$ and $c$ be three disjoint circles, coming form lines parallel to the x, y and z axes.
For most fibrations, these three circles are transverse to the fibers.
Form a branched cover of the torus with two-fold branching over all preimages of these 3 circles. The resulting manifold has a hyperbolic structure that can be constructed from right-angled hyperbolic dodecahedra, and is commensurable with the 4-fold branched cover of $S^3$ over the Borromean rings. You can think of it this way:  you can take a unit cube as fundamental domain for the torus, and arrange that a, b and c lie on faces of the cube, each bisecting a pari of (glued together) opposite facce.  This induces a subdividision of the boundary of the cube into what look like rectangles, but are really pentagons. 
The map (x,y,z) -> x+y+z gives a fibration over the torus,  also works for any branched cover as described.  The preimage of any face of the cube is an extended face plane of a dodecahedron, and is always a totally geodesic immersed surface, but it splits into two embedded surfaces for suitable branched covers of $T^3$ (perhaps the one you first come up with.)  
The tiling of hyperbolic space by right-angled dodecahedra has a cameo appearance in the video "Not Knot" we made at the Geometry Center, available together with "Outside In" on DVD from AKPeters.  In the 1984 Scientific American Article The Mathematics of three-dimensional manifolds that Jeff Weeks and I wrote, a manifold in this family (constructed from right-angled hyperbolic dodecahedra and having the properties you asked for) was described as the configuration space of a mechanical linkage. I don't think these particular properties were pointed out in Scientific American.
This and other examples that are counterintuitive at first were a good part of my motivation when I raised the question whether all hyperbolic 3-manifolds virtually fiber over the circle, which at the time was a radical idea.<|endoftext|>
TITLE: Values of zeta at odd positive integers and Borel's computations
QUESTION [16 upvotes]: Someone recently quoted to me this recent article that claims to prove that $\zeta(2n+1) \notin (2\pi )^{2n+1} \mathbb{Q}$. 
I always assumed this was well known. More precisely I thought this result followed from the fact that the regulator
$$
  K_{2n-1}(\mathbb{Z})\otimes \mathbb{Q} = Ext^1_{MT(\mathbb{Z})}(\mathbb{Q}(0),\mathbb{Q}(n))    \longrightarrow Ext^1_{MHS}(\mathbb{Q}(0),\mathbb{Q}(n)) = \mathbb{C}/(2\pi i)^n\mathbb{Q}
$$
is injective (this is usually quoted as a consequence of Borel's computations of K-groups "Stable real cohomology of arithmetic groups", "Cohomologie de $SL_n$ et valeurs de fonctions zêta aux points entiers")
Am I mistaken? 
PS: corrected a typo thx to Pete L Clark

REPLY [16 votes]: It is not known (but conjectured) whether the numbers $\zeta(2n+1)/\pi^{2n+1}$ are irrational, $n=1,2,\dots$. It is not even known whether at least one of these numbers is irrational! In fact, the most general (folklore) conjecture states that $\pi$ and all odd zeta values are algebraically independent over $\mathbb Q$. There are natural links between this conjecture and the expected structure of the so-called multiple zeta values; the references I have in mind are papers by A. Goncharov and surveys/talks by M. Waldschmidt.<|endoftext|>
TITLE: Expectation of a simple function of multivariate gaussians iid rvs
QUESTION [7 upvotes]: I would like to compute analytically the following expected value:
$$ E\left( \frac{X_i^2}{\sum_j \lambda_j^2 X_j^2}\right) $$
where the $X_i \approx N(0,1)$ are iid.
It seems to be an elementary integral, but it is eluding me. Any pointer to a non-trivial solution technique, or the solution itself, of course, is highly appreciated.

REPLY [7 votes]: Here are some preliminary computations.
One wants to compute  $A_k^n=\lambda_k^2E\left(X_k^2S^{-1}\right)$, where $S=\sum\limits_{k=1}^n\lambda_k^2X_k^2$. Starting from the expression
$$
S^{-1}=\int_0^{+\infty}\mathrm{e}^{-tS}\mathrm{d}t,
$$
and using the independence property of the random variables $X_k$, one gets
$$
A_1^n=\int_0^{+\infty}\lambda_1^2E(X_1^2\mathrm{e}^{-tS})\mathrm{d}t=\int_0^{+\infty}\lambda_1^2E(X^2\mathrm{e}^{-t\lambda_1^2X^2})E(\mathrm{e}^{-t\lambda_2^2X^2})\cdots E(\mathrm{e}^{-t\lambda_n^2X^2})\mathrm{d}t,
$$
that is,
$$
A_1^n=-\lambda_1^2\int_0^{+\infty}u'(t\lambda_1^2)u(t\lambda_2^2)\cdots u(t\lambda_n^2)\mathrm{d}t,
\quad\text{where}\
u(t)=E(\mathrm{e}^{-tX^2}).
$$
By some simple computations,
$$
u(t)=(1+2t)^{-1/2},\quad u'(t)=-(1+2t)^{-3/2},
$$
hence
$$
A_1^n=\int_0^{+\infty}\frac{\lambda_1^2\mathrm{d}t}{(1+2\lambda_1^2t)\sqrt{(1+2\lambda_1^2t)(1+2\lambda_2^2t)\cdots(1+2\lambda_n^2t)}}.
$$
First example
When $\lambda_k^2=\lambda^2$ for every $k$, the change of variable $s=\sqrt{1+2\lambda^2t}$, $s\mathrm{d}s=\lambda^2\mathrm{d}t$,  yields
$$
A_1^n=\int_1^{+\infty}\frac{\mathrm{d}s}{s^{n+1}}=\frac1n,
$$
as was to be expected by symmetry. 
Second example
When $\lambda_1^2=\lambda^2$ and $\lambda_k^2=1$ for every $k\ge2$, the change of variable $s=\sqrt{1+2t}$, $s\mathrm{d}s=\mathrm{d}t$, with $1+2\lambda^2t=\lambda^2s^2+1-\lambda^2$ yields
$$
A_1^n=\int_1^{+\infty}\frac{\lambda^2\mathrm{d}s}{s^{n-2}(\lambda^2s^2+1-\lambda^2)^{3/2}}=1-(n-1)\int_1^{+\infty}\frac{\mathrm{d}s}{s^{n}(\lambda^2s^2+1-\lambda^2)^{1/2}}.
$$
When $n=2$ and $\lambda^2\ge1$, setting $\lambda^2=1/\cos^2 u$ yields $A_1^2=\displaystyle\frac1{1+\cos u}=\frac{\lambda}{1+\lambda}$. This last formula is also valid if $\lambda^2\le1$.
Further values for even integers $n$ are $$A_1^4=\dfrac{\lambda^2}{(1+\lambda)^2},\quad A_1^6=\dfrac{\lambda^2(1+3\lambda)}{3(1+\lambda)^3},\quad A_1^8=\dfrac{\lambda^2(1+4\lambda+5\lambda^2)}{5(1+\lambda)^4}.
$$
When $n=3$ and $\lambda^2\ge1$, setting $\lambda^2=1/\cos^2 u$ and some further computations yield
$$
A_1^3=\frac{1-u\cot u}{\sin^2u}.
$$
Likewise, if $\lambda^2\le1$, setting $\lambda^2=1/\cosh^2 u$ yields
$$
A_1^3=\frac{u\coth u-1}{\sinh^2 u}.
$$<|endoftext|>
TITLE: Tuples of sums of two squares
QUESTION [12 upvotes]: This is in response to a question of evarist that was closed. Evarist's question was this: Do there exist infinitely many triples of consecutive sums of two squares? The topic actually has an interesting history: In 1903 an anonymous reader submitted the following question to the section Mathematical Questions in the British journal The Educational Times - Find all consecutive triples of sums of two squares. Elementary though non-explicit solutions were submitted by A. J. Champneys Cunningham and two other British academics (The reference is Mathematical Questions 14955. Educ. Times (2) 3 (1903), 41-43.) The Cunningham in question is the same one as the Cunningham project is named after. He was an artillery officer in the Indian Army who turned to elementary number theory after he retired.
Later J. E. Littlewood posed the following research problem which generalizes Evarist's first question: Given distinct positive integers h and k, do there exist infinitely many triples n,n+h,n+k that are simultaneously sums of two squares? This was solved in 1973 by C. Hooley, On the Intervals between Numbers that are sums of Two Squares, II. J. Number Theory 5 (1973), 215-217. Hooley's proof that there are infinitely many such triples is not long, but uses the theory of ternary quadratic forms.
The following problem is the reason that I have information relevant to Evarist's question, and together with another Littlewood problem about sums of two squares, the source of my interest in this topic: 

For which choices of positive integers $h_1,\ldots,h_r$ are there infinitely many tuples $n + h_1,\ldots,n + h_r$ that are simultanously sums of two squares? 

This ties up with Evarist's second question. Clearly a necessary condition is that $h_1,\ldots,h_r$ do not cover all four residue classes modulo $4$, since integers congruent to 3 modulo 4 are not sums of two squares.
EDIT: Greg Martin (see below) has observed that there are other necessary congruence conditions 
than the one modulo 4. So the answer to the optimistic version of my question is NO, but the above highlighted question still stands. (The optimistic version was that the congruence condition modulo 4 would be sufficient).
(By the way, I have not been able to find Littlewood's statement of his research problem, and actually he may have considered the more general problem too. I could not find the n,n+h,n+k problem in the collection of research problems that he published, though his other and better known problem about sums of two squares is there).
I think the general case of the above problem is likely to be difficult. I had planned a project for a master's thesis to obtain more computational evidence, to establish special cases, and possibly to generalize the Cochrane-Dressler result discussed below, but no student with the requisite background was interested, so nothing came of it.
Actually, there is an unsolved problem directly connected to Evarist's first question too. In 1987 
T. Cochrane and R. E. Dressler used the Selberg sieve to obtain un upper bound 
$$
A(x) \ll \frac{x}{\log^{3/2}(x)}
$$
for the counting function $A(x)$ of the triples of consecutive sums of two squares. The reference is Consecutive triples of sums of two squares, Archiv der Math. 49 (1987), 301-304. One would expect a lower bound of the same order of growth, and actually an asymptotic estimate, but this is not known. 
(The distribution of consecutive pairs of sums of two squares has been investigated, and more is known in this case).
Cochrane and Dressler give a very simple argument for Evarist's first question. The integers 8,9,10 are sums of two squares. Suppose n+1,n,n-1 are sums of two squares. Then $n^2 + 1$, $n^2$ and $n^2 - 1 = (n - 1)(n +1)$ are also sums of two squares, the last by the Brahmagupta-Fibonacci identity 
$(x^2 + y^2)(u^2 + v^2) = (xu + yv)^2 + (xv - yu)^2$ that shows that products of sums of two squares are sums of two squares.

REPLY [2 votes]: As it stands, the problem is hopeless. The answer is yes for $h = (0,1,2)$ (by this I mean the vector $h_1 = 0$, $h_2 = 1$, $h_3 = 2$) by Cochrane and Dressler, and the same argument also works for $h = (0,1,2,5)$. Since $(0,1,2,3)$ is impossible, the "smallest" nontrivial problem is $(0,1,2,4)$. Is there a simple way of treating this one?
In general, the condition that the $h_i$ do not cover all residue classes modulo $4$ is much too weak. Consider e.g. $h = (0,2,4,6)$. If $n$ is odd, then $n$ or $n+2$ is $\equiv 3 \bmod 4$; if $n$ is even and $n+h$ is a sum of two squares, then so is $(n+h)/2$ because of
$$ a^2 + b^2 = \Big(\frac{a-b}2\Big)^2 + \Big(\frac{a+b}2\Big)^2. $$
Thus if the answer were positive for $h = (0,2,4,6)$, then it would also be positive for $h = (0,1,2,3)$, which it is not.<|endoftext|>
TITLE: Intuition on finite homotopy groups
QUESTION [16 upvotes]: As I have been studying algebraic topology, something that I found puzzling was the existence of finite homotopy groups. For instance, $\pi_{4}(S^{2})\cong\pi_{5}(S^{4})\cong\mathbb{Z}/2\mathbb{Z}$. I was wondering if there was any kind of intuitive reason for why this might be true, and if there are spaces $X$ such that $\pi_{1}(X)$ is finite. Speaking very roughly, it would seem that a finite, nontrivial fundamental group means that if you repeat a closed path enough times, it can be contracted to a point, something which I find rather hard to visualize. So the question is: Is there any intuitive reason for the existence of finite homotopy groups?

REPLY [3 votes]: Perhaps it isn't necessary to add another answer, but here is one more nevertheless.

Given any discrete group $G$, form the classifying space $BG$. A map of $S^1$ to $BG$
is equivalent to giving a principal $G$-bundle over $S^1$. Such a bundle can formed by
gluing the ends of the trivial bundle $[0,1]\times G$ with a twist given by an element
of $G$. So in this way, we get $\pi_1(BG)=G$.
Of course, invoking $BG$ is perhaps violating the spirit of the question. 
When $G$ is finite, we can approximate
this by something more concrete. Let $V$ be a faithful unitary representation of $G$ with no
trivial summands (e.g. the complement of $\mathbb{C}$ in the regular representation).
After replacing $V$ by $V\oplus V$ if necessary, we can assume that $\dim V>1$. 
Then the unit sphere $S\subset V$ will be simply connected, and $G$ will act on it without
fixed points. So the quotient $S/G$ will have fundamental group $G$.
This is the same construction used (implicitly) in many of the other answers.<|endoftext|>
TITLE: Homogeneous arithmetic progressions in difference sets
QUESTION [7 upvotes]: I have a nasty feeling that I ought to be able to answer this question, but I've got other things to think about right now and I'm interested in the answer just so that I can reply to a mathematical email I've received. (If anyone gives me substantial help I will of course acknowledge it when I reply.) 
This isn't precisely what was asked in the email, but it's closely related and would enable me to give a good answer. A result of Bourgain shows that if you take two dense subsets A and B of {1,2,...,n} then A+B must contain an arithmetic progression of length $\exp(c(\log n)^{1/3})$ or thereabouts. In particular this is true of A-A (since it contains arithmetic progressions of the same length as A-(n+1-A)). But what bounds can one get in the A-A case if one insists that the progression should be homogeneous? That is, suppose that A is a subset of {1,2,...,n} of density δ. How large an m can we guarantee to find such that there exists d such that all of -dm, -(d-1)m, ... , dm  are elements of A-A?
By Szemerédi's theorem applied to A, m at least tends to infinity with n and can be taken to be n logged a few times. But can we do a lot better than this? Another small observation is that if we apply Bourgain's theorem to A-A, we can obtain a quite long homogeneous arithmetic progression in A+A-A-A. 
It's been a little while since I looked at either Bourgain's proof or a subsequent improvement by Green to $\exp(c\sqrt{\log n})$, so I can't instantly say whether their arguments would give one a homogeneous progression in the case that B=-A. Based on my hazy memory, it feels as though it could go either way.
Although I think it is unlikely, there's just a small chance that this is an interesting question to which the answer is not known (or an easy consequence of known results or techniques).

REPLY [4 votes]: It's a bit late, but let me point out that there is a wonderfully short and elementary argument of Croot, Ruzsa and Schoen that gives a homogeneous arithmetic progression of length about $\log n$ in $A-A$: it can be found in a paper called Arithmetic progressions in sparse sumsets, available at http://people.math.gatech.edu/~ecroot/kterm.pdf.
It can also deal with much sparser sets.<|endoftext|>
TITLE: Symmetries of the Julia sets for $z^2+c$
QUESTION [6 upvotes]: The julia set seems to have symmetries roughly corresponding to translation, rotation and scaling.
In the following image

You can see the horizontal translation, which leaves the extremal left and right endpoints fixed is a symmetry.
The 21-fold rotational symmetry about any one of the 'whorls' is visible also - (I think that the number 21 corresponds to the denominator of the mandelbrot bulb the julia set comes from but I don't yet know how to compute this).
There is also a twofold rotation about the center (and any other part similar to it)
The scales are more difficult to describe and I don't think I have found them all so I will just avoid going into detail on this unless someone would like me to do so.


Do these have a mathematical interpretation, for example as automorphisms of the Julia set in some appropriately understood sense? Or is there some other way to describe these kinds of symmetries in mathematical terms?
Thank you.

REPLY [2 votes]: Yes, there is an explanation, but not in all cases.
Rotational symmetries of polynomial Julia sets are explained in the paper
MR0951972. For the case of rational functions, see also MR1092156.
And also MR1625697 about translational symmetries.
And several papers of Beardon, titled Symmetries of Julia sets.<|endoftext|>
TITLE: A principle of mathematical induction for partially ordered sets with infima?
QUESTION [43 upvotes]: Recently I learned that there is a useful analogue of mathematical induction over $\mathbb{R}$ (more precisely, over intervals of the form $[a,\infty)$ or $[a,b]$).  It turns out that this is an old idea: it goes back to Khinchin and Perron, but has for some reason never quite caught on and thus keeps getting rediscovered.  A nice recent paper is
http://alpha.math.uga.edu/~pete/Kalantari07.pdf
[As a side note, I gave a talk two days ago in the VIGRE Graduate Seminar at UGA -- essentially, a colloquium for graduate students, with talks which are supposed to be accessible to first and second year students -- about induction over the real numbers with applications to snappy proofs of essentially all the basic "hard theorems" of honors calculus / elementary real analysis.  It was a smashing success: the students both enjoyed stretching their minds around this new form of induction and appreciated the application to consolidated proofs of theorems which were, in their recent memeory, not so quick or easy to prove.]
The way I like to state the principle is a little different from Kalantari's approach.  [I actually got confused by Kalantari's "axioms for induction" when I first saw them and was under the impression that they were wrong.  In fact I was wrong, and he quickly responded to my email on the subject, setting me straight and, kindly, mentioning that others had made the same mistake.]  Here's my preferred version:
Let $(X,\leq)$ be a totally ordered set with a least element, which we may as well call $0$.  
Say that $X$ has infima if...every nonempty subset $S$ of $X$ has an infimum.  
Say that $S \subset X$ is an inductive subset if all of the following hold:
(POI1) $0 \in S$.
(POI2) For all $x \in S$, if there exists $z \in X$ such that $x < z$ -- in other words, if $x$ is not a maximum element of $X$ -- then there exists $y > x$ such that the interval $[x,y]$ is contained in $S$.
(POI3) For all $x \in X$, if $[0,x) \subset S$, then $x \in S$.
Finally, we say that $X$ satisfies the principle of ordered induction if the only inductive subset of $X$ is $X$ itself.  
Theorem: For a totally ordered set $X$ with a minimum element, TFAE:
(i) $X$ has infima.
(ii) $X$ satisfies the principle of ordered induction.
The proof is straightforward.  Applying this to half-closed intervals as above gives real induction.  Also, applying it to a well-ordered set recovers transfinite induction exactly as it is usually stated -- i.e., with an extra axiom for "limit elements", even though one could formally combine (POI2) and (POI3) into a single case.    
This issue came up on Tuesday on our sister site math.SE -- someone asked whether there was such a thing as real induction -- and I answered it, yes, as above.  Then someone commented on my answer: what about generalizations to partially ordered sets?
There is a well-known principle of induction on partially ordered sets satisfying the descending chain condition, or equivalently, in which every nonempty subset has a minimum.  This is called, by mathematicians of various stripes, well-founded induction or Noetherian induction.  (As far as I can see it should be called Artinian induction.  Anyone want to address that? Edit: I am satisfied by Dave Anderson's comment below.)  
Note that a partially ordered set with DCC need not have a minimum element, which is certainly necessary for the above setup in totally ordered sets.  But this isn't a big deal: if $(X,\leq)$ is a poset satisfying DCC, the poset $X_0$ obtained by adjoining a minimum element $0$ still satisfies DCC, and nothing is lost here.
[Edit: as Francois Dorais points out, adding a minimum is not enough; still a partially ordered set satisfying DCC need not have infima.  So what I am asking about really is different from Noetherian induction.]
After a little thought I was optimistic that there should be a version of induction partially ordered sets with a minimum element.  I even thought that the right definition of inductive subset should be essentially the one given above, with (POI2) modified slightly to
(POI2'): for every $x \in S$ and $z \in X$ with $x < z$, there exists $y \in (x,z]$ such that the entire interval $[x,y]$ is contained in $S$.
And then I tried to prove that any poset with a minimum and having infima satisfies the principle of ordered induction.   And I couldn't.  Eventually I found the following counterexample: let $A$ be an infinite set and $X = 2^A$ be its power set, partially ordered by inclusion.  Of course $X$ has infima: take the intersection.  Let $S$ be the collection of all finite subsets of $A$.  Then $S$ satisfies (POI1), (POI2') and (POI3) but is proper.  

Neither can I think of some small modification of (POI2') which evades this example.  I still think there should be some kind of principle of induction in partially ordered sets with infima, but I don't know what to do.  Can anyone state such a principle which recovers as special cases as a special case the principle of ordered induction and the principle of Noetherian induction?

REPLY [15 votes]: Something very close to François' conditions achieves the 
desired if-and-only-if version of the theorem for partial
orders, providing an induction-like characterization of the complete partial orders, just as Pete's theorem characterizes the complete total orders.
Suppose that $(P,\lt)$ is a partial order. We say that it
is complete if every subset $A$ has a least upper bound
$sup(A)$ and a greatest lower bound $inf(A)$. This implies
that $(P,\lt)$ has a least and greatest element, and in
this case, it is easy to see that completeness is equivalent to the assertion that
every set has a greatest lower bound (the least upper bound of a set with upper bounds is the greatest lower bound of its upper bounds). The complete partial
orders are exactly the complete
lattices.
For the purpose of this question, let us define that
$S\subset P$ is inductive if the following occur:

(1) $S$ is downward closed: $y\lt x\in S\to y\in S$;
(2) $S$ has no largest element, except possibly the largest element of $P$;
(3) If $d=sup(A)$ for some $A\subset S$, then $d\in S$. 

In (2), by a largest element, I mean an element that is
larger than all other elements (in distinction with
maximal element, a weaker concept). Conditions (2) and
(3) are both slightly weaker than François' conditions. The
difference in (2) is that we no longer assume that a
condition $x\in S$ can be extended in any particular
direction, and the difference in (3) is that we are not
assuming here that $P$ is complete, but only that $S$
contains the suprema of its subsets, when these suprema exist.
Lastly, define that a partial order $(P,\lt)$ has
partial-order induction if the only inductive set is all
of $P$.
Theorem. For any partial order $(P,\lt)$, the
following are equivalent:

$(P,\lt)$ is complete.
$(P,\lt)$ has least and greatest elements and satisfies partial-order induction.

Proof. For the forward implication, suppose $(P,\lt)$ is
complete. It follows that $(P,\lt)$ has least and greatest
elements, since these are the sup and inf of the emptyset.
Suppose that $S\subset P$ is inductive. By (3) we know
$sup(S)\in S$, which would make it the largest element of
$S$, contrary to (2), unless $sup(S)$ is largest in $P$, in which case
$S=P$ by (1). So $(P,\lt)$ has partial-order induction.
Conversely, suppose that $(P,\lt)$ has least and greatest
elements and satisfies partial-order induction. We want to
prove completeness. Suppose $B\subset P$ is nonempty and
has no greatest lower bound. Let $S$ be the set of lower
bounds of $B$. This is closed downwards, and so (1) holds.
Since $B$ has no greatest lower bound, it follows that $S$
has no largest element, and so (2) holds. Finally, if
$A\subset S$ and $sup(A)$ exists in $P$, then it remains a
lower bound of $B$, since every element of $B$ is an upper
bound of $A$ and $sup(A)$ is the least upper bound of $A$.
Thus, (3) holds and so $S$ is inductive, contrary to the
fact that it contains no elements of $B$. Thus, $B$ must
have a greatest lower bound after all, and so $(P,\lt)$ is
complete. QED
Note that when $(P,\lt)$ is a total order, then these
concepts reduce to the conditions Pete mentions in the
question (adding a least and greatest element if
necessary). Thus, this theorem seems to be the desired
generalization.
(I have deleted my other answer to the question, as it was
based on a misunderstanding of the question.)<|endoftext|>
TITLE: What is the easiest randomized algorithm to motivate to the layperson?
QUESTION [33 upvotes]: When trying to explain complexity theory to laypeople, I often mention randomized algorithms but seemingly lack good examples to motivate their usage.  I often want to mention primality testing but the standard randomized algorithms don't admit a simple description (or proof of correctness) in a lay-atmosphere.  I often resort to the saying that randomized algorithms allow "finding hay in a haystack", but that has little mathematical substance.  
The question: Is there a good example of a problem that:

is easily explained (and sufficiently interesting)
has a simple randomized algorithm
appears non-trivial to get an efficient deterministic algorithm

and ideally also satisfies:

the randomized algorithm has a somewhat understandable proof of correctness - so no Markov/Chernoff/random-walk-mixing-times

REPLY [3 votes]: Multiplication checking: is $ab$ equal to $c$ (or is $a(x)b(x)$ equal to $c(x)$)? Analysis of randomized solution is pretty easy; also it provides a huge gap in asymptotic compared to straightforward multiplication.<|endoftext|>
TITLE: Computable rings similar to Z
QUESTION [5 upvotes]: (This is related to my question at Computable nonstandard models for weak systems of arithemtic )
Is there a nontrivial computable discrete ordered ring with Euclidean division that is not isomorphic to Z?
If so, what other first-order properties could it share with Z?
.
Possible properties include:
all numbers with rational square roots are perfect squares
Lagrange 4-square theorem
prime elements are unbounded
.
Could it satisfy all Pi_1 properties satisfied by Z?
(bounded quantifiers referring to the absolute value being bounded)

REPLY [6 votes]: Berarducci and Otero in "A Recursive Nonstandard Model of Normal Open Induction" (Journal of Symbolic Logic v61, 1996)
give a discretely ordered ring $R$ with recursive operations having the following properties:

$R$ is integrally closed in its quotient field. (So elements with "rational" square roots are perfect squares.)
The prime elements of $R$ are cofinal
$R$ satisfies the induction axioms for quantifier-free formulas

In an earlier paper of Otero (Journal of Symbolic Logic, vol 55, 1990) Otero proves that every model of Open Induction extends to one in which Lagrange's Four-Square Theorem holds. Whether this can be done effectively I don't know.
As for the general question "What properties can a recursively presented discretely ordered ring share with $\mathbb{Z}$": Let's say that a discretely ordered ring $R$ is "diophantine correct" if it satisfies all universal sentences that hold in  $\mathbb{Z}$. Assuming that  the language of rings has signature $(+, -, \times , \le,0 ,1)$, diophantine correctness amounts to the requirement that any system of polynomial equations and inequalities  that are solvable in $R$ is  solvable in the standard integers. Incidentally, models of open induction satisfy a weaker property: Any system of equations solvable in some (at least one)  model of open induction has a p-adic solution.
The question whether a nonstandard diophantine correct model of open induction can be effectively constructed was raised by Adamowic and Morales-Luna in "A Recursive Model for Arithmetic with Weak Induction", (Journal of Symbolic Logic v50, 1985). I believe that this question is still open. I also believe that the models constructed in the Otero-Berarducci paper are in fact diophantine correct, but the proof of this seems to bump up against open problems in number theory. This is all discussed in an article on diophantine correct open induction by Sidney Raffer in "Set theory, Arithmetic, Philosophy: Essays in Memory of Stanley Tennenbaum (edited by J. Kennedy and R. Kossak), Cambridge University Press. (To appear.)
This is a response to a question from the comments: It is too long to fit there.
Proof of the Euclidean Division Theorem from the axioms of Open Induction:
The problem is to show that if $A$ is a model of open induction and if $x,y$ are elements of $A$ with $y>0$ and $x \ge 0 $ then there are unique elements $r, q$ of $A$ such that  $x=yq+r$ and $0 \le r < q$. (The statement actually holds for all $x$ and the proof for $x < 0 $ is similar.)

First show that for every $x\in A$ with $x\ge 0$, there is some $q\in A$ such that $yq\le x < y(q+1)$.  (Suppose, by way of by contradiction, that $x \ge 0$ and there is no such $q$. Let $S$ be the subset of $A$ defined by the quantifier-free formula $\sigma(q): q \ge 0 \wedge yq\le x$. Show that $S$ contains 0 and is closed under successor. By induction $S$ contains every positive element of $A$. But this is impossible because $x+1$ cannot be in $S$.)
Next, given $y > 0$ and $x \ge 0 $ choose $q$ (as in Part 1) such that $yq \le x < y(q+1)$. Put $r=x-yq$. Using that fact that $A$ is discretely ordered it is follows easily that $r$ and $q$ are the unique elements of $A$ satisfying $x=yq+r$ and $0 \le r < q$.<|endoftext|>
TITLE: What are the possible numbers of regions that 4 planes can divide space?
QUESTION [8 upvotes]: What are the possible numbers of regions that 4 planes can create? 
We know that the minimum number is 5 and the maximum number is 15.
(http://mathworld.wolfram.com/SpaceDivisionbyPlanes.html)
Is it possible to make a generalization based on the ways the planes 
could intersect?

REPLY [2 votes]: Computing the maximum number of pieces into which $r$ hyperplanes disconnect the space $\mathbb{R}^n$ is an old little problem, easily solved by induction, and the answer is
$${r \choose 0}+{r \choose 1}+\dots+{r \choose n},$$ and it is achieved if they are in generic position. So in dimension $n$ the number of pieces as $r$ increases is asymptotically $r^n/n!$. Note also that for $r\leq n$ it's $2^r$ and for $r=n+1$ it's $2^r-1$ (so in particular in dimension $3$ the number of pieces with $0,1,2,3,4$ generic planes is resp. $1,2,4,8$ and then $15$ (not $16$ as one would guess, so this may be used as an example of fallacious argument based on analogy).<|endoftext|>
TITLE: What are the shapes of rational functions?
QUESTION [127 upvotes]: I would like to understand and compute the shapes of rational functions, that is, holomorphic maps of the Riemann sphere to itself, or equivalently, ratios of two polynomials, up to Moebius transformations in both domain and range. For degree 1 and 2, there is only one equivalence class.  For degree 3, there is a well-understood one-complex-parameter family, so the real challenge is for higher degrees.

Given a set of points to be the critical values [in the range], along with a covering space of the complement homeomorphic to a punctured sphere, the uniformization theorem says this Riemann surface can be parametrized by $S^2$, thereby defining a rational function.  Is there a reasonable way to compute such a rational map?

I'm interested in ideas of good and bad ways to go about this.  Computer code would also be most welcome.

Given a set of $2d-2$ points on $CP^1$ to be critical points [in the domain], it has been known since Schubert that there are Catalan(d) rational functions with those critical points. Is there a conceptual way to describe and identify them?

In the case that all critical points are real, Eremenko and Gabrielov, Rational functions with real critical points and the B. and M. Shapiro conjecture in real enumerative geometry. Annals of Mathematics, v.155, p.105-129, 2002 gave a good description. They are determined by $f^{-1}(R)$, which is $R$ together with mirror-image subdivisions of the upper and lower half-plane by arcs. These correspond to the various standard things that are enumerated by Catalan numbers. Is there a global conceptual classification of this sort? And, is there a way to find a rational map with given critical points along with some kind of additional combinatorial data?
Note that for the case of polynomials, this is very trivial: the critical points are zeros of its derivative, so there is only one polynomial, which you get by integrating its the derivative.

Is there a complete characterization of the Schwarzian derivative for a rational map, starting with the generic case of $2d-2$ distinct critical points?

Cf. the recent question by Paul Siegel. The Schwarzian $q$ for a generic rational map has a double pole at each critical point.  As a quadratic differential, it defines a metric $|q|$ on the sphere - critical points which is isometric to an infinitely long cylinder of circumference $\sqrt 6 \pi$ near each. Negative real trajectories of the quadratic differential go from pole to pole, defining a planar graph. 

What planar graphs occur for Schwarzian derivatives of rational functions?  What convex (or other) inequalities do they satisfy?

The map from the configuration space of $(2d-2)$ points together with branching data to the configuration space of $2d-2$ points,  defined by mapping (configuration of critical values plus branched cover data) to (configuration of critical points) is a holomorphic map, which implies it is a contraction of the Teichmuller metric.

Is this map a contraction for other readily described metrics?

REPLY [10 votes]: The way that I view the shape of rational functions (and especially polynomials) is in terms of the configuration of their critical level curves.  Assume throughout that $r$ is a rational function with a simple pole at $\infty$ (which can always be achieved by pre and post composing with Moebius functions).  First lets classify the individual level curves of $r$.

THEOREM 1 Let $\Lambda$ be a level curve of $r$ (ie. a component of the set $\{z:|r(z)|=\epsilon\}$ for some $0<\epsilon<\infty$).  There are two possibilities.

If $\Lambda$ contains no critical points of $r$, then $\Lambda$ is a smooth Jordan curve in $\mathbb{C}$.
If $\Lambda$ contains critical points of $r$, then $\Lambda$ is a piecewise smooth connected finite graph having vertices at the critical points, which forms a "complicated figure eight".  That is, it satisfies the following properties:

There are evenly many (and more than two) edges of $\Lambda$ incident to each vertex of $\Lambda$.
Each edge of $\Lambda$ is incident to a bounded face of $\Lambda$ (if $r$ is a polynomial, now allowed to have any number of poles at $\infty$, there is the additional constraint that each edge is also incident to the unbounded face of $\Lambda$).



Call level curves of the second kind (ie. one containing critial points) critical level curves.  The two properties together imply that a critical level curve $\Lambda$ of $r$ is a sort of figure eight graph, which can be formed from a figure eight by iteratively joining circles to the graph at single points (with the circle in either a bounded face or an unbounded face).
Let us define a figure eight type graph to be one which has the properties described in Theorem 1, and a polynomial figure eight type graph to be one with the additional property described in Item 2 of Theorem 1.
We will now classify the way in which the critical level curves can lie amongst each other.

THEOREM 2 Let $\Lambda$ be a critical level curve (in fact any level curve) of $r$, and let $D$ denote one of its bounded faces.  One of the following obtain.

$r$ has a single distinct zero or pole in $D$.
There is some critical level curve $\Lambda_D$ which is "maximal" in $D$.  That is, maximal in the sense all zeros and poles of $r$ in $D$ are contained in the bounded faces of $\Lambda_D$, and every critical point of $r$ in $D$ is either in $\Lambda_D$, or contained in a bounded face of $\Lambda_D$.


The picture that arises from Theorem 2 is that, if $A$ is a set containing all the critical level curves of $r$, along with the zeros and poles of $r$, then every component of $A^c$ in $\mathbb{C}$ is a topological annulus.
More can be said in fact.

THEOREM 3 If $D$ is a component of $A^c$.  Then the following holds.

$D$ is topologically an annulus.

Let $D_-$ denote the bounded face of $D^c$.  Let $Z$ and $P$ denote the number of zeros and poles of $r$ in $D_-$ respectively.

$r^{\frac{1}{Z-P}}$ is a conformal map from $D$ to an annulus centered at the origin.

We could rephrase Item 2. as:

There is conformal map $\phi$ from $D$ to an annulus centered at the origin such that $r=\phi^{Z-P}$ on $D$.


We see now a clear idea of the shape of the function $r$ in terms of its critical level curves.  The critical level curves form a sort of skeleton, either one zero or pole, or one maximal critical level curve in each bounded face of any given level curve.  In between the critical level curves there is a smooth sheet of the function, conformally just a pure power.
Since the rational function is so simple between its critical level curves, it should come as no surprise that these critical level curves determine the rational function. That is, the geometric "skeleton" of the critical level curves is a strong conformal invariant of the rational function.
If some additional data is appended, (like the arguments of $r$ at the vertices, the net change in $\arg(r)$ along each edge of the graph, and the magnitude of $|r|$ on each graph), then the configuration of the critical level curves of a rational function is a strong conformal invariant.

THEOREM 4 If any two rational functions $r_1$ and $r_2$ have the same configuration of critical level curves (up to orientation preserving homeomorphism), then they are conformally equivalent.  That is, there is a Moebius map $M$ such that $$r_2=r_1\circ M$$ on $\overline{\mathbb{C}}$.

Finally, a natural question is to ask: Which such pictures arise from rational functions?  The answer is completely known for polynomials, and the answer is all of them.

THEOREM 5 Every configuration of finitely many polynomial figure eight type graphs arranged according to the conclusion of Theorem 2 is the critical level curve configuration for some polynomial.

The corresponding fact for rational functions (ie. all configurations correspond to some rational function, where the additional restriction from Theorem 1 Item 2 is dropped) seems to be almost certainly true, but at the moment defies proof.
Unfortunately I do not know of any good methods for computing the critical level curve configuration for a given rational function (other than approximating it with ContourPlot in Mathematica), or to find a rational function with a given critical level curve configuration.
All the above may be found in detail in papers 2 and 3 on the Arxiv here, or on my website here.  Background on level curve of analytic functions may be found in paper 4 on the Arxiv page linked to above.<|endoftext|>
TITLE: Computing the largest eigenvalue of a very large sparse matrix
QUESTION [5 upvotes]: I am trying to compute the asymptotic growth-rate in a specific combinatorial problem depending on a parameter $w$, using the Transfer-Matrix method. This amounts to computing the largest eigenvalue of the corresponding matrix.
For small values of $w$, the corresponding matrix is small and I can use the so-called power method — start with some vector, and multiply it by the matrix over and over, and under certain conditions you'll get the eigenvector corresponding to the largest eigenvalue. However, for the values of $w$ I am interested in, the matrix becomes too large, and thus the vector becomes too large, say, $n > 10^7$ entries or so, and it can't be contained in the computer's memory anymore and I need extra programming tricks or a very powerful computer.
As for the matrix itself, I don't need to store it in memory — I can access it as a black box, i.e., given $i, j$ I can return $A_{ij}$ via a simple computation. Also, the matrix has only $0$ and $1$ entries, and I believe it to be sparse (i.e., only around $\log n$ of the entries are $1$'s, $n$ being the number of rows/columns). However, the matrix is not symmetric.
Is there some method more space-effective for computation of eigenvalues for a case like this?

REPLY [4 votes]: You could use the Arnoldi Iteration algorithm. This algorithm only requires the matrix A for matrix-vector multiplication. I'm expecting that you will be able to black-box the function v→Av. What you generate is an upper Hessenberg matrix H whose eigenvalues whose can be computed cheaply (by a direct method or Rayleigh quotient iteration) and which approximate the eigenvalues of A. Arnoldi Iteration will give the best approximation to the dominant eigenvalue so I suspect you won't have to do many iterations before you have a good estimate.
An excellent introduction to this is: "Numerical Linear Algebra" by Trefethen and Bau. (p250)
The basic algorithm can be found here: http://en.wikipedia.org/wiki/Arnoldi_iteration
Now the only thing that is required to make this a fully functional algorithm is a termination condition. Since you don't seem to need the dominant eigenvalue to a high degree of accuracy I would not worry and just stop when the dominant eigenvalue estimate doesn't change too much.
If you have Matlab you can always use the built in function eigs(Afun,n,...) where Afun is the black-box function handle that computes Av.<|endoftext|>
TITLE: Virasoro constraints for the generating function of Hurwitz numbers.
QUESTION [8 upvotes]: Generating function of the simple Hurwitz numbers is known to be connected with Gromov-Witten potential of the point (Kontsevich $\tau$-function) (see e.g. Ian Goulden, David Jackson and Ravi Vakil). On the other side Virasoro constraints are known to play an imporatant role in the Gromov-Witten theory, in particular for the point. 
Is there known the set of Virasoro constraints for the generating function of simple (or double) Hurwitz numbers? Any ideas why it should (or should not) exist?

REPLY [8 votes]: This is something I've long been meaning to think about seriously, so maybe I can get back to you with something better later. In the meantime... 
I wasn't aware of anything explicitly in the literature about this until googling just a bit ago, when I found the paper "Virasoro constraints for Kontsevich-Hurwitz partition
function" by Mironov and Morozov. 
I haven't fully digested their paper yet, but they seem to be using the viewpoint of the Kazarian paper on Hodge integrals and KP hierarchy I mentioned in my answer to your other question.  
Briefly: the ELSV formula relates single Hurwitz numbers to Hodge integrals, essentially the GW theory of a point.  Kazarian shows how this transformation can be done explicitly by a certain operator M&M call $\hat{U}$ (the quantization of a quadratic function).  
M and M's point seems to be that since the generating functions are related by this operator, and one of them satisfies Virasoro, we can conjugate the Virasoro operators by $\hat{U}$ and get Virasoro operators for the other generating function.  The particular form they take seems to be a bit of a mess, and I worry about some details, but I've only skimmed that paper very quickly.
But philosophically, this seems to be going about things backwards: the Hurwitz side is really simpler, and the above setup is often used to show that the GW of a point satisfies Virasoro.  I feel we should be able to construct Virasoro operators for single Hurwitz numbers more easily with some kind of direct approach.
Hurwitz theory is all about statements about the symmetric group, and there are constructions (I read about it in a Frenkel-Wang paper) that build Virasoro actions out of the symmetric group.  
I'm not fully motivating this, but single and Hurwitz theory is very conveniently done on a certain Fock space.  Basically, you have the operator that multiplies by a transposition, (called $M_0$ by Kazarian, $\mathcal{F}_2$ by Okounkov-Pandharipande, physicists have some other notation for it...), and you have the operators $\alpha_n, n\in\mathbb{Z}$ that add or remove cycles of length $n$ from a conjugacy class, and that form a heisnberg algebra. 
These operators are exactly what you need to do Hurwitz theory.  But they're also what Frenkel and Wang use to construct a Virasoro algebra -- essentially, the commutators $L_n=[\alpha_n, M_0]$.  So we might hope that some similar construction would give us easier to understand Virasoro constraints for single Hurwitz numbers.  But I haven't spent the necessary time trying to nail it down.
As far as double Hurwitz numbers go, I'm a little less hopeful for the above vague ideas.   All I know is that Goulden, Jackson and Vakil have a few lines about trying and failing to construct Virasoro operators in their "Towards the Geometry of Double Hurwitz Numbers" paper.<|endoftext|>
TITLE: Gromov-Witten and integrability.
QUESTION [10 upvotes]: The generation function of the Gromow-Witten invariants (with descendants) of the point is known to be Kontsevich-Witten tau-function of KdV, partition functions of $P^1$ and equivariant $P^1$ are known to be tau-functions of extended Toda and 2-Toda respectively. Are there any other manifolds (except of orbifolds made of mentioned previously manifolds ) for which generation functions of GW invariants are identified with tau-functions of some integrable hierarchy?

REPLY [4 votes]: Let $X$ be a smooth projective variety. Actually, as long as the quantum cohomology of $X$ is semisimple, the partition function of the (descendent) GW-invariants of $X$ is always identified with a tau-function of the Dubrovin--Zhang integrable hierarchy associated to $X$. So for the case of $\mathbb{P}^N$, any $N$, there is a nice integrable hierarchy whose topological tau-function gives the partition function of GW-invariants of $\mathbb{P}^N$. The point is that for $N=0,1$, the hierarchy was known in the traditional literature of integrable systems; whereas for $N\geq 2$, the Dubrovin--Zhang hierarchy is new.<|endoftext|>
TITLE: How Symmetric is Diophantine Approximation using Fractions with Square Denominators?
QUESTION [7 upvotes]: Let $S$ be an infinite set of positive integers. 
Let us say that a  "best S-approximation" to a real irrational $r$ is a rational number
$p/q$, with $p$ and $q$ integers and $q \in S$, such that for any integer $m\in S$ with $1 \le m<q$, and any integer $n$, it holds that $|p/q-r|<|n/m-r|$. 
In the case that $S$ is the positive integers, it is well known that for any  $r$ there are infinitely many best $S$-approximations $p/q$ which are less than $r$ and infinitely many that are greater than $r$. This is proved for example in Khintchin's book "Continued Fractions." 
What if $S$ is the set of squares of integers? Is it known that for every irrational $r$ there are infinitely many best $S$-approximations less than $r$ and infinitely many greater than $r$? 
Experimentation with numerous commonplace irrationals gives the impression that the ratio of the number of under-estimates to the number of over-estimates tends to 1. Could this be true for every irrational $r$? Is this known?

REPLY [5 votes]: I believe that a natural version of the problem is finding how good are rational approximations of the form $p/q^k$ to a given irrational number $\theta$. This classical problem goes back to H. Heilbronn [Quart. J. Math. (Oxford) 19 (1948) 249--256] and I. Danicic [Mathematika 5 (1958) 30--37] and the best known results are due to C. Hooley [in: Analytic number theory, Vol. 2 (Allerton Park, IL, 1995) 471–-486,
Progr. Math. 139 Birkhäuser Boston, Boston, MA, 1996] and G. Harman [Glasgow Math. J. 38 (1996) 299–308]. They show that there are infinitely many coprime pairs $p,q$ with the property
$$
\left|\theta-\frac p{q^k}\right|<\frac1{q^{k+\rho_k-\varepsilon}}
$$
where $\rho_2=\frac25$ and $\rho_k=1/(3\cdot 2^{k-2}-1)$ if $k\geq 3$.<|endoftext|>
TITLE: Trapped rays bouncing between two convex bodies
QUESTION [23 upvotes]: At some point during my research I was confronted with this problem, but I did not dedicate serious time to it. Anyway it stayed in the back of my mind and I'm still interested in hints for it. Application: asymptotic properties of Schroedinger equations, scattering.
You have two convex (compact, smooth, everything) disjoint sets in the plane. Consider a ray starting in the complement of the two sets and bouncing on the boundary of the sets in the usual way, with the ingoing and outgoing rays forming equal angles with the normal to the boundary. Q.: does it always exist a trapped ray which never leaves a ball containing the two sets? This can happen of course if the ray keeps bouncing forever between the two bodies. A trivial example is obtained if the sets have two parallel sides, and the ray is chosen perpendicular to both. Less trivial examples (even strictly convex) can be constructed by choosing a trajectory first, and then joining the dots (i.e., the turning points of the trajectory) with convex curves; with some work and some adjustments in the trajectory, you can produce plenty of examples.
But, is this always the case? given two arbitrary bodies, does it always exist a trapped ray? 
EDIT (see Pietro's comment): I mean, another trapped ray besides the 'trivial' trapped ray bouncing between the closest points of the two sets (a general version of the trivial case mentioned above of two parallel sides).
EDIT 2 (quick summary of the discussion for the benefit of future readers): the answer is yes for smooth boundaries and large (in particular, with nonempty interior) sets of initial points. A continuity argument is enough to prove this. If the boundary is non smooth problems may arise. E.g. for two polygons with a couple of facing parallel sides, the only trapped ray is the periodic one.
PS in retrospect, the question was quite elementary, but I really enjoyed to discuss it here :)

REPLY [34 votes]: Yes, there is always a trapped ray. The simplest way to see it is to find the path between the two bodies that minimizes length.  It is necessarily perpendicular to both surfaces.
EDIT: I see the question was edited to ask for more than this trivial answer, so the new answer: there is a unique trapped ray from any starting point, but it is not trapped in backward time unless it is on the shortest path between the bodies. One can find it by minimizing distance of a zig-zag path alternately touching the two bodies a finite number of times, then passing to a limit.
Here is a generalization:  suppose you have a collection of smooth disjoint convex shapes $\{S_i\}$ in the plane arranged in a way that no straight line intersects more than two.  Then, for any doubly infinite sequence of indices $ \dots, i_{-1}, i_{0}, i_{1}, \dots $ such that $i_j \ne i_{j+1}$, there is a unique trajectory that intersects the shapes in that order, starting with $S_{i_1}$ in the positive direction and $S_{i_0}$ going backward.  If the sequence is periodic, you can find the trajectory just as for the case of two objects. For the infinite case, you can take limits. 
Even if the shapes are not convex, as long as they are smooth the trajectories still exist, but they are not necessarily unique.  If you want to say something about the case when the obstacles are not smooth, you can extend the rule to make it a non-deterministic dynamical system, where a ray hitting a corner has choices which way to go.
This kind of system is classical dynamical systems, which has been well-understood since early last century. Perhaps someone more knowledgable will supply appropriate references.
It is a limiting special case of the theory of the geodesic flow on surfaces of negative curvature.
In response to a comment, here is some more detail (that doesn't itself fit into a comment).
The question was about stability and how to prove convergence under the limiting process.
To prove existence, you don't need stability: just take a sequence of longer and longer rays, and choose a convergent subsequence. This exists because of compactness of the set of possible initial directions.
To prove uniqueness: this follows from the hyperbolicity of the flow. Think of the convex obstacles as trick mirrors that make you look skinny, cylinders with a convex cross-section. The convexity implies that reflected rays diverge at least as fast as they would from a flat mirror.  Successive reflected images of the two mirrors in each other get thinner and thinner, so they narrow down to a unique point. (In the three-dimensional picture, they're also narrowing vertically, just at the relatively slow rate at which images shrink with distance in Euclidean space rather than at the exponential rate resulting from mirrors that are convex to 2nd order).
One way to formalize the discussion above is by use of triangle comparison theorems. Double the complement of the convex bodies to make a surface. The surface can be smoothly approximated by a surface of nonpositive curvature if it's comforting, but that's not technically necessary; the (intuitively obvious) statement about image sizes above become cases of the  Toponogov comparison theorem.

REPLY [7 votes]: A reasonable conjecture (assuming smoothness and strict convexity of the two bodies) seems to me that any bounded bouncing ray necessarily lies completely on one side of the minimizing connecting line, and is asypmtotic to it as $t\to\infty.$ That there is at least one such ray on each side seems also true. Fix a parallel line $r$ to the minimizing line, still connecting the two bodies. Perturb the periodic trajectory of an angle $\epsilon$ from one of its two bouncing points, from the side where $r$ is. In a finite number of bouncings you will hit the line $r$ with a based vector $v_\epsilon$. Take a limit $ v_0$ of a subsequence of the $v_\epsilon$ as $\epsilon\to0$: I'd say that starting with initial point and velocity $-v_0$ you will get a ray trapped  between the two lines and the bodies (this seems easy to show) and having the minimizing segment as $\omega$-limit.
(As to the tagging issue, checking the tag list I think the most suitable are: dynamical-systems and convex-geometry. Actually, even more precise should be billiard (if not pinball); maybe we could create it).   
EDIT: actually, there is a whole open set of initial points that admit a ray asymptotic to the minimizing segment (let's call  $a$ and $b$ its endpoints, belonging resp. to the convex $A$ and $B$). 
Precisely, for $x\notin A\cup B$ let $x'\in A$ be the point of minimal distance to $x$. Assume that $x$ sees the whole arc $\Gamma$ of $\partial A$ connecting $a$ and $x'$ (here "sees $\Gamma$ " of course means that the convex hull of $x$ and $\Gamma$ does not meet $B$). It is easy to show that there exists a point $y$ on the arc  $\alpha$ such that a ray from $x$ to $y$ generates a ray asymptotic to the segment $[a,b]$. (I think it is also unique, and that this construction produces all the bounded rays.)<|endoftext|>
TITLE: fundamental domains for free fuchsian group.
QUESTION [10 upvotes]: I try to understand some of the topology of the space of pointed non-compact hyperbolic surfaces (with the pointed Gromov-Hausdorff topology). It is known that the fundamental
group of a non-compact surface is a free group, so I am interested in free Fuchsian groups
(discrete, free groups of direct isometries of the hyperbolic plane, not necessarily finitely generated).
Call ``ideal polygon'' any domain of the hyperbolic plane that is an intersection of half-planes limited by geodesics that are pairwise disjoint. 
Is it true that any free fuchsian group has a fundamental domain that is an ideal polygon?
I think that I can manage do do it by hand for the simplest examples (e.g. covering groups of an hyperbolic punctured tori, or a hyperbolic trice punctured sphere) and the result seems plausible, but I feel that either true or false it shall be well-known. Any reference on this, or more generally on uniformization of non-compact, possibly infinite genus hyperbolic surfaces would be welcome.

REPLY [12 votes]: Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to $R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want. 
Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$. 
If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal  points, or identifies two disks which are adjacent along a component of $I-X$ with a $\pi$ twist. The complement $U=S^2-(I\cup_i D_i)$ is again homemorphic to $R^2$. If we identify antipodal points of $D_j$, then this identifies two arcs in the boundary of $U$ to obtain an open Mobius strip. We add two arcs connecting antipodal points of $D_j$ to a point $x\in X$ at the end of the interval of $I-X$ which intersects $D_j$, which forms a single arc after identification of antipodal points of $D_j$, and cuts the Mobius strip back up into $R^2$. If adjacent disks $D_i, D_{i+1}$ are identified, then the complement $U$ gives a punctured torus. We add 4 arcs connecting these points to $x$ (again, $x\in X$ is at the end of the interval of $I-X$ containing $D_i$), cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.<|endoftext|>
TITLE: Did Pogorzelski claim to have a proof of Goldbach's Conjecture?
QUESTION [24 upvotes]: In 1977, Henry Pogorzelski published what some believed was a claimed proof of Goldbach's Conjecture in Crelle's Journal (292, 1977, 1-12). His argument has not been accepted as a proof of Goldbach's Conjecture, but as far as I know it has not been shown that his argument is incorrect.
Pogorzelski's argument is said to depend on the "Consistency Hypothesis," the "Extended Wittgenstein Thesis," and "Church's Thesis." Pogorzelski has a Ph.D. in mathematics (his advisor was Raymond Smullyan).
Daniel Shanks says in Solved and Unsolved Problems in Number Theory (fourth edition, 1993) that: "It seems unlikely that (most) number-theorists will accept this as a proof [of Goldbach's Conjecture] but perhaps we should wait for the dust to settle before we attempt a final assessment." (page 222)
Did Pogorzelski claim to present a proof of Goldbach's Conjecture? If so, and this claimed proof has not been disproven after 33 years, I am curious why this would be the case, given that Shanks considers it important enough to mention in his book.

REPLY [8 votes]: For what it's worth, the 12/12/2012 article at http://bostonglobe.com/news/science/2012/12/11/math-even-mathematicians-don-understand/e41V2lixnAVyB63X4NTPaO/story.html indicates that Pogorzelski believes that he has established a legitimate demonstration of the Goldbach conjecture (which he claims he obtained in the 1990s).  The article quotes from his manuscript's preface that his proof was submitted for publication in 2002, and he received no response.<|endoftext|>
TITLE: Class Numbers and 163
QUESTION [45 upvotes]: This is a bit fluffier of a question than I usually aim for, so apologies in advance if this doesn't pass the smell test for suitability.
Likely my favorite fun fact in all of number theory is the juxtaposition of two "extremal and opposite" properties about the prime 163 in relation to class numbers:


$p=163$ is the largest value of $p$ for which the quadratic imaginary number field $\mathbb{Q}(\sqrt{-p})$ has class number equal to one.  (Baker-Heegner-Stark) 



$p=163$ is the smallest value of $p$ for which the real cyclotomic field $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ has class number greater than one. (Schoof) 


Of the various ways I know of "pushing up" and "pushing down" class numbers (class field theory, Herglotz-type formulas, Scholz-type reflection theorems), none seem to give any indication that these two class numbers should be related, let alone inversely so.  Of course, since the smaller Heegner discriminants don't correspond to analogous real cyclotomic fields with positive class number, this is not surprising.
This leads me to wonder if there's an analytic link between these two quantities -- for example, by relating their zeta-functions and looking at the corresponding class number formulas.  My initial, admittedly naive, attempts to extract anything from the relationship between the zeta-functions for $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ have come up empty.  So my question is:

Are there fancier analytic (or other) techniques that might shed some light on the "miracle" above?

Of course, I'm also aware that this juxtaposition might be purely coincidental, a mildly large example of the law of small numbers at work.  I might even prefer it that way.
Edit to incorporate some computations and comments from below.
For primes congruent to 7 mod 12, the real cyclotomic field of conductor p contains a unique cyclic cubic subfield.  By class field theory, there is a surjection of class groups from the real cyclotomic to the cubic.  Since for 163, the cyclic cubic has class number 4, the non-triviality of the class number for the real cyclotomic can be said to "come from" the cubic.  In a sense, the coincidence thus reduces to the fact that the first cyclic cubic field ("first" with respect to ordering by conductor) of prime conductor 7 mod 12 with non-trivial class group is the one of conductor 163.  The fact that 163 is only the 11th prime in this congruence class may modify (in which direction I'm not sure) your opinion of whether or not this is a coincidence.
Barring any insight as to why the class number of this quadratic and cubic would be related, which may be unlikely given Franz Lemmermeyer's answer, it would be interesting to know if one could devise a clever probabilistic test for evaluating how surprised one should be to see ten class-number-one cubics in a row.  I imagine that it's not very unlikely -- I just ran a computation, and class number 1 seems to very prevalent for cyclic cubics of small conductor ($p<5000$), and some heuristic (sorry, Andrew) evidence in the literature seems to agree.

REPLY [3 votes]: Take the 53$^{rd}$ cyclotomic ring of integer and the homomorphism $\sigma: \zeta\to\zeta^2$ with the primitive root 2 modulo 53. Kummer's class number formula gives the prime 4889. Because every ideal is invariant under the homomorphism $\sigma$$^{52}\equiv id$ and the class group must be cyclic, the homomorphism $\sigma$ can only send every class $C$ to the class $C^\xi$ where $\xi$ must be a (not necessarily primitive) 52$^{nd}$ root modulo 4889 so that $\xi^{52}\equiv 1 \bmod{4889}$. There are 52 different prime ideals $\sigma^k\langle 107, \psi_{107}(\zeta)\rangle = \langle 107, \sigma^k\psi_{107}(\zeta)\rangle$. The second generator $\psi_{107}(\zeta)$ of the prime ideal can be taken from the first theorem in chapter 4.12 of Fermat's last theorem by Edwards. In Dedekind rings it is then easy to prove that the greatest common divisor of the ideals $\langle 107\rangle$ and $\langle\psi_{107}(\zeta)\rangle$ is the prime ideal $\langle 107, \psi_{107}(\zeta)\rangle$. A more detailed analysis of this ring then gives the class equivalence $\sigma\langle 107, \psi_{107}(\zeta)\rangle = \langle 107, \sigma\psi_{107}(\zeta)\rangle \sim \langle 107, \psi_{107}(\zeta)\rangle^{3637}$. The 'zip code' 3637 is of no further interest. However it is a primitive 52$^{nd}$ root modulo 4889 and we have a clear-cut relationship between the cyclic class group of this ring and the cyclic Galois group $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{\operatorname{id} = \sigma^0, \sigma^1, \dots, \sigma^{51}\}$.
Now take the 41$^{st}$ cyclotomic ring of integer, the homomorphism $\sigma: \zeta \to \zeta^6$ with the primitive root $6 \bmod{41}$ and the cyclotomic integer $g_1(\zeta) = \zeta^6 - \zeta^8 - \zeta^{20}$. The ideal $\langle g_1(\zeta)\rangle$ factorizes to $\langle g_1(\zeta)\rangle = \langle 83, \psi_{83}(\zeta)\rangle^2\cdot\langle 83, σ^8\psi_{83}(\zeta)\rangle$. The process of factorizing cyclotomic ideals is described in the chapters 4.11ff. of Edwards. I outlined this process in my answer to this question on math.stackexchange. The cyclotomic integer $g_2(\zeta) = - \zeta^4 + \zeta^{10} + \zeta^{21} + \zeta^{28} + \zeta^{39}$ factorizes to $\langle g_2(\zeta)\rangle = \langle 83, 
\sigma^8\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle$ so that we get the class equivalence $\langle 83, \psi_{83}(\zeta)\rangle^2 \sim \langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle$. Squaring this equivalence gives
$$\langle 83, \psi_{83}(\zeta)\rangle^4 \sim \langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle^2 \sim \sigma^{28}[\langle 83, \psi_{83}(\zeta)\rangle^2] \sim \sigma^{28}\langle 83, \sigma^{28}\psi_{83}(\zeta)\rangle = \langle 83, \sigma^{2·28}\psi_{83}(\zeta)\rangle.$$
Consecutive squaring gives $\langle 83, \psi_{83}(\zeta)\rangle^{2^k} \sim \langle 83, \sigma^{28k}\psi_{83}(\zeta)\rangle$ and we get the general relationship $\langle 83, \psi_{83}(\zeta)\rangle^n \sim \langle 83, \sigma^{4k}\psi_{83}(\zeta)\rangle$ for an integer $n$ because $\operatorname{gcd}(28, 40) = 4$. For $k=10$ we get $\langle 83, \psi_{83}(\zeta)\rangle^{1024} \sim \langle 83, \psi_{83}(\zeta)\rangle$ or the ideal $\langle 83, \psi_{83}(\rangle)\rangle^{1023}$ is principal. Kummer's class number formula gives $11^2$ so that the class order of the ideal $\langle 83, \psi_{83}(\zeta)\rangle$ is $11$ with $1023 = 3\cdot 11\cdot 31$. Then we have 4 class subgroups $\langle 83, \sigma^{m+4k}\psi_{83}(\zeta)\rangle, m \in \{0,1,2,3\}$. But the class number is $11^2$ so that the combination of the classes of 2 subgroups gives the classes of the remaining 2 class subgroups. For example the cyclotomic integer $s_2(\zeta) = \zeta^{28} + \zeta^{36} - \zeta^{39}$ factorizes to $\langle s_2(\zeta)\rangle = \langle 83, \sigma^{14}\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{24}\psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{33}\psi_{83}(\zeta)\rangle$. Therefore the homomorphism $\sigma$ sends the class of one class subgroup to the next class subgroup. And we only have a clear relationship between the homomorphism $\sigma^{4k}$ of the Galois group of this ring and the classes because the homomorphism $\sigma^4$ does not send the classes of one class subgroup to the next. However the class group $\mathbb{Z}/11\mathbb{Z} \oplus \mathbb{Z}/11\mathbb{Z}$ has been confirmed.
The matter gets worse with the 163$^{rd}$ cyclotomic ring of integers. We take the homomorphism $\sigma: \zeta \to \zeta^2$ with the primitive root $2 \bmod{163}$. I could only determine the class group of the prime ideals that factorize to maximal 27 conjugate prime ideals. The analysis gives a class subgroup of $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Take the four classes $c$, $d$, $e$, $f$, each the generator of one cyclic class subgroup $\mathbb{Z}/2\mathbb{Z}$ with class order 2. Then each ideal of order 2 can be assigned to a class $(c^r, d^s, e^t, f^u)$. We take the letter I for the principal divisor. The ideals $\langle q, \psi_{q}(\zeta)\rangle$ with $q = 61, 199, 347$ all have six conjugates (see chapter 4.9 of Edwards). The analysis similar to that above gives
\begin{align*}
\langle 61, \sigma^0\psi_{61}(\zeta)\rangle &\sim (c, I, I, I)\\
\langle 61, \sigma^1\psi_{61}(\zeta)\rangle &\sim (I, I, e, I)\\
\langle 61, \sigma^2\psi_{61}(\zeta)\rangle &\sim (I, d, I, I)\\
\langle 61, \sigma^3\psi_{61}(\zeta)\rangle &\sim (I, I, I, f)\\
\langle 61, \sigma^4\psi_{61}(\zeta)\rangle &\sim (c, d, I, I)\\
\langle 61, \sigma^5\psi_{61}(\zeta)\rangle &\sim (I, I, e, f)\\
\langle 199, \sigma^0\psi_{199}(\zeta)\rangle &\sim (I, d, e, I)\\
\langle 199, \sigma^1\psi_{199}(\zeta)\rangle &\sim (I, d, I, f)\\
\langle 199, \sigma^2\psi_{199}(\zeta)\rangle &\sim (c, d, I, f)\\
\langle 199, \sigma^3\psi_{199}(\zeta)\rangle &\sim (c, d, e, f)\\
\langle 199, \sigma^4\psi_{199}(\zeta)\rangle &\sim (c, I, e, f)\\
\langle 199, \sigma^5\psi_{199}(\zeta)\rangle &\sim (c, I, e, I)\\
\langle 347, \sigma^0\psi_{347}(\zeta)\rangle &\sim (c, I, I, f)\\
\langle 347, \sigma^1\psi_{347}(\zeta)\rangle &\sim (c, d, e, I)\\
\langle 347, \sigma^2\psi_{347}(\zeta)\rangle &\sim (I, d, e, f)
\end{align*}
and we have $\langle 347, \sigma^0\psi_{347}(\zeta)\rangle \sim \langle 347, \sigma^3\psi_{347}(\zeta)\rangle$. Here the action of the Galois group on the prime ideal $(61, \sigma^0\psi_{61}(\zeta))$ does not exhaust the classes of the class subgroup $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. It does not even form a class subgroup of $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Hence we have lost all relationship between the class group and the Galois group.
My experience with class groups of cyclotomic rings of integer taught me that the classes have a structure on their own though in general there does not seem to be any link between the classes and other algebraic structures of these rings.
You can get my determination of the class groups of the rings above and other cyclotomic rings here.<|endoftext|>
TITLE: Can there be two continuous real-valued functions such that at least one has rational values for all x?
QUESTION [11 upvotes]: Of course, no continuous real valued non-constant function can attain only rational or irrational values, but can there be a pair of nowhere-constant continuous functions f and g such that for all x, at least one of f(x) and g(x) is rational? Or maybe a countable collection of continuous functions, {f1, f2...} such that for all x there is n such that fn(x) is rational?
Thanks

REPLY [21 votes]: Since your functions are locally non-constant, the preimage of any (rational) point is nowhere dense (in $\mathbb{R}$; if a preimage of a point with respect to a continious function is dense on an interval, then the function is constant on this interval). Hence the union of all preimages of all rational points is of Baire category one (in $\mathbb{R}$); so it is not equal to the whole $\mathbb{R}$.

REPLY [16 votes]: If you allow the functions to be constant on some intervals, then there are some easy examples, and Ricky has provided one. 
But if you rule that out, then there can be no examples, even with countably many functions. To see this,
suppose that $f_n$ is a list of countably many continuous functions which are never constant on an interval. Enumerate the pairs $(r,n)$ of
rational numbers $r$ and natural numbers $n$ in a countable list
$\langle (r_0,n_0), (r_1,n_1),\ldots\rangle$. Let $C_0$ be any closed
interval. If the closed interval $C_i$ is defined, consider
the function $f_{n_i}$ and the rational value $r_i$. Since $f_{n_i}$ is not constant value $r_i$ on $C_i$, we may
shrink the interval to $C_{i+1}\subset C_i$ such that
$f_{n_i}$ on $C_{i+1}$ is bounded away from $r_i$. By
compactness, there is some $x\in C_i$ for all $i$. Thus,
$f_n(x)$ is not $r$ for any rational number $r$.

REPLY [7 votes]: f(x) := closest point in [-2,-1] to x

g(x) := closest point in [+1,+2] to x
If x≤0, then g(x) is rational.

If 0≤x, then f(x) is rational.
The question becomes more interesting if you demand that the functions be nowhere locally constant.<|endoftext|>
TITLE: The space of framed functions
QUESTION [13 upvotes]: Framed functions arose in the work of K. Igusa defining cohomology invariants for smooth manifold bundles (Igusa-Klein torsion). In the late 80's, he proved a strong connectivity result about the "space of framed functions" using Morse theory and conjectured that this space was, in fact, contractible. Jacob Lurie recently verified this conjecture using the language of higher category theory in the course of his proof of the Cobordism Hypothesis.
(1) Is there any progress on a more geometric proof of this? The relevant work of Igusa is very geometric and Lurie himself points out that a more direct proof should be found. Perhaps more realistically: is there a geometric heuristic for Lurie's result?
(2) Does anyone know of applications of framed functions beyond defining Igusa-Klein torsion and proving the Cobordism Hypothesis? Pointers into the literature on this topic would be much appreciated.
EDIT: A framed function on $M$ is a pair $(f,\xi)$ of a smooth function $f$ on $M$ having at most $A2$ singularities in the interior of $M$ and no singular points on the boundary, and an orthonormal framing $\xi$ of the negative eigenspace of the Hessian of $f$ at each critical point.

REPLY [12 votes]: Eliashberg once told me that the framed function theorem should be a consequence of his work on wrinkled maps. 
Igusa and I gave a fairly direct proof that the space of framed functions on the circle is contractible. The idea is that the assignment function $f \mapsto \Sigma(f)$ (= its singular set, suitably interpreted as a point of a configuration space) is is quasifibration with contractible fibers when $f$ varies through framed functions on $S^1$. The proof in this case appears in:
Igusa, Kiyoshi; Klein, John The Borel regulator map on pictures. II. An example from Morse theory. K-Theory 7 (1993), no. 3, 225–267.<|endoftext|>
TITLE: Are vertex and edge-transitive graphs determined by their spectrum?
QUESTION [8 upvotes]: A graph is called vertex and edge transitive if the automorphism group is transitive on both vertices and edges.
The spectrum of a graph is the collection (with multiplicities) of eigenvalues of the incidence matrix.
Supposedly, it is conjectured that almost all graphs have the property that they are the unique graph with their spectrum (at least, according to MathWorld).
If $\Gamma_1,\Gamma_2$ are two vertex and edge transitive graphs, with the same valence, which are isospectral (have the same spectrum) then does it follow that $\Gamma_1\cong \Gamma_2$?

REPLY [4 votes]: As Chris said, the answer is most probably no, and most likely there are known such examples. There are many examples and constructions of non isomorphic graphs with the same spectrum and some of these examples are Cayley graphs and other very symmetric graphs. Cayley graphs are always vertex transitive and quite often, for a suitable choise of generators, also edge transitive. Some of these examples are based on a famous paper 
of Sunada. Sunada's method was originally for creating isospectral manifolds but it can be applied (and is even easier) to create isospectral graphs. The following paper by Bob Brooks "Isospecrtal graphs and isospectral surfaces" is a good starting point.<|endoftext|>
TITLE: What heuristic evidence is there concerning the unboundedness or boundedness of Mordell-Weil ranks of elliptic curves over $\Bbb Q$?
QUESTION [31 upvotes]: Some experts have a hunch that for any nonnegative integer $r$ there are infinitely many elliptic curves over $\Bbb Q$ with Mordell-Weil rank at least $r$.
The best empirical evidence for this hunch can be found in Andrej Dujella's tables here and here, the strongest of this evidence being provided by Elkies using constructions involving K3 surfaces.
The only heuristic evidence I know of supporting this hunch is the results of Tate and Shafarevich (strengthened by Ulmer) that the Mordell-Weil ranks of elliptic curves over the a fixed function field $\Bbb F_q(t)$ can be arbitrarily large.

What other heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be unbounded?

Seeing as how some experts do no believe this conjecture, I'd also accept answer to the companion question:

What heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be uniformly bounded?

REPLY [11 votes]: An update from 2018: There has been some recent work on a heuristic suggesting that there are infinitely many elliptic curves of every rank $<21$ but only finitely many of rank $> 21$ (it's unclear to me what the model says about the case rank $=21.$)
I'm not sure what the full attribution should be, but one can read about these ideas in Bjorn Poonen's ICM article https://arxiv.org/abs/1711.10112v2 and the references within.<|endoftext|>
TITLE: stacks as Morita equivalence classes
QUESTION [21 upvotes]: I have often encountered definitions of the kind "stacks are equivalence classes of groupoids under Morita equivalence" in topological or differentiable context, with the notion of Morita equivalence appropriate for the context. 
I have two questions:
1) can one define Deligne-Mumford or Artin stacks this way, as Morita equivalence classes of algebraic groupoids?
2) if yes, what is the connection between the corresponding notion of Morita equivalence and the one for the rings (i.e. rings are called Morita equivalent if they have equivalent categories of modules over them)?

REPLY [8 votes]: Regarding your question relating Morita equivalence as defined for internal groupoids and as defined for rings:
Given an internal groupoid $G$ (say, Lie, topological or algebraic), it defines a presheaf of groupoids on the ambient category ($Diff$, $Top$ or $Sch$). The stackification of this presheaf is the category of (right, say) principal $G$-bundles. Thus two groupoids define equivalent stacks (isomorphic in the correct 2-categorical sense) whenever their categories of principal bundles are equivalent. This is the groupoid version of 'Morita equivalence' a la rings. But when these two groupoids present equivalent stacks, they are 'Morita equivalent' in the sense there is a span between them of 'essential equivalences' - fully faithful essentially surjective (-appropriately interpreted) functors. So the two notions coincide, at least when the ambient category has enough quotients of the right sort of reflexive coequalisers (as the examples I list do). In the more general case with no quotients, as I mention in my comment to the original question, I'm not sure what happens.

EDIT: It's been a while since I answered this, but now I know what happens in the case I mention in the final sentence above the line. One shouldn't take the category of principal bundles as being the stackification of the internal groupoid $G$, rather take the category of internal anafunctors with codomain $G$ and with domain a groupoid with no non-identity arrows. The arrows are a little tricky to describe, but one gets a stack over the base category with fibre over $X$ the hom-category $Hom(X,G)$ in the bicategory of internal groupoids, anafunctors and transformations. This will be covered in a forthcoming paper of mine.<|endoftext|>
TITLE: Restrictions of Diffeomorphisms
QUESTION [7 upvotes]: Notation: Let$M$ be a smooth, closed manifold, $S$ any submanifold of $M$, $Diff(M)$ the group of diffeomorphisms of $M$ and $Imb(S, M)$ the group of smooth imbeddings of $S$ into $M$.
A classical result of R. Palais from the 1960 paper Local triviality of the restriction map for embeddings says that the map $Diff(M)$  $\rightarrow$ $Imb(S, M)$ given by restriction is a fibration.
I feel like I've heard during numerous teas that there are various refinements and generalizations of this due to J. Cerf and (possibly) others.
(1) Can anyone summarize what else is known in this direction beyond the theorem of Palais? 
(2) Is there a way to see Palais' result easily? [added: from the responses it sounds like the original paper is still a great way to see this result. But see the answers of Randal-Williams and Palais for an alternate route.]

REPLY [3 votes]: This is closely related to (and can be proved by the same methods as) the fact that if we fix another manifold $N$ then the restriction map
$$\mathrm{Emb}(M, N) \to \mathrm{Emb}(S, N)$$
is a locally trivial fibration, which was proved in the same paper by Palais. However, I wanted to advertise a geodesic proof of this fact:
E. L. Lima, On the local triviality of the restriction map for embeddings,  Commentarii Mathematici Helvetici Volume 38, Number 1, pp 163-164.<|endoftext|>
TITLE: How can I visualize the nontrivial element of $\pi_4(S^3)$ and $\pi_5(S^3)$ ?
QUESTION [33 upvotes]: I've read in the textbooks that the non-trivial generator $\eta_n$ of $\pi_{n+1}(S^n)$ is the suspension of the Hopf map $S^3\to S^2$, and the generator $\chi$ of $\pi_5(S^3)$ is given by $\eta_3 \circ \eta_4$. Fine. 
My question is, how I can visualize them? Is there a nice explicit way to describe these maps $\eta_3$ and $\eta_3\circ \eta_4$ ? How about the generator of $\pi_6(S^3)$ ? 
(Other questions on MO look more serious. Hopefully this question is not out of place ...)
EDIT: anyone with rudimentary understanding of basic homotopy theory would say $\eta$ and $\eta\circ\eta$ are explicit enough, but I just can't visualize the suspension. I would be happy with a nice description of $SU(2)$ bundles over $S^n$, as my first exposure to homotopy is through quantum field theory...
Further edit: Thanks everyone for answers, I'm almost inclined to accept Per's answer, but I'm not still satisfied :p

REPLY [7 votes]: (This is a bit late, but I hope you find it interesting!)
Here's smooth representation of the generator of $\pi_4(Sp(1))$ (and so the same homotopy group of $S^3$ and $SU(2)$). Consider $S^4 = \mathbb{HP}^1$, and $Sp(1)$ the unit sphere in $\mathbb{H}$. Then the following function $t\colon \mathbb{HP}^4 \to Sp(1)$ represents the nontrivial homotopy class $S^4 \to S^3$:
$$
t[p;q] = \frac{2p\bar{q}i\bar{p}q - |p|^4 + |q|^4}{|p|^4 + |q|^4}
$$
where [p;q] are homogeneous coordinates on $\mathbb{HP}^4$. I don't know if this has appeared previously (I would love to know!), but I presented this as part of some slides at the Australian Mathematical Society's annual conference last year (see slide 6), and originally worked it out with a pointer from Michael Murray to the Hopf fibration described using quaternions (that is, $Sp(1) \to S(Im\mathbb{H})$, the unit sphere in the pure imaginaries). That this map is the generator (i.e. is not null-homotopic) I calculated following the answer at my question Detecting homotopy nontriviality of an element in a torsion homotopy group.
Note that this function followed by the inclusion $Sp(1) \hookrightarrow Sp(2)$ (as the top left entry) is the generator of $\pi_4(Sp(2))$ (by results of Mimura and Toda). And thus we also get a representative for the generator of $\pi_4$ of $Spin(5) = Sp(2)$.<|endoftext|>
TITLE: Exotic isomorphism of matrix rings
QUESTION [5 upvotes]: Let R and S be commutative rings with a 1 different from zero.  Let m and n be positive integers.  Assume the ring of m-by-m matrices over R is isomorphic to the ring of n-by-n matrices over S.  Does it follow that R is isomorphic to S?  Does it follow that m = n?  Does either of those follow from the other?  I'm interested in both where R,S are finite and where R,S are infinite.  (although the second question is trivial in the former case)

REPLY [8 votes]: Yes and yes. Let $T=M_m(R)=M_n(S)$. 
The center of $T$ is isomorphic to both $R$ and $S$.
The $1\times m$ matrices over $R$ form an $(R,T)$-bimodule and the $n\times 1$ matrices over $S$ form a $(T,S)$-bimodule. Tensor these over $T$ to get an $(R,S)$-bimodule. As an $S$-module the direct sum of $m$ copies of this is free of rank $n$. For a nonzero commutative ring this implies that $m$ divides $n$. (Tensor with a residue field to get a vector space of dimension $\frac{n}{m}$.) Likewise, looking at it as an $R$-module, $n$ divides $m$.<|endoftext|>
TITLE: Can we add two matrices by performing an operation on their eigenvalues & eigenvectors?
QUESTION [12 upvotes]: Can we find out what the eigenvalues of a sum of two initial matrices are, by performing an operation on the eigenvalues & eigenvectors of the initial matrices?

REPLY [35 votes]: I will interpret this to be the question `What do you need to know beyond the eigenvalues of $A$ and of $B$ to get the eigenvalues for $A+B$ from those of $A$ and $B$?'  It's a good question.   Here's one way to conceptualize it:
Let's modify it to talk of the roots of the characteristic polynomial, rather than just eigenvalues. Over the complexes, for most but not all matrices, these are the same.  Knowing the roots is equivalent to knowing the characteristic polynomials.
You can put the characteristic polynomial of $A$ into homogeneous form as the determinant of 
$\lambda_1 I + \lambda_2 A$ --- this translates easily back and forth to the usual form. With two matrices, there is a polynomial in one more variable: the determinant of $\lambda_1 I + \lambda_2 A + \lambda_3 B$.
The characteristic polynomial for $A+B$ is obtained by specializing this 3-variable polynomial. I believe this can be any homogeneous polynomial P( x,y,z) of degree n in 3 variables subject to the condition P(1,0,0)=(the determinant of the identity) = 1, but I don't know for sure: perhaps someone can elucidate this point. 
Update:  Dave Speyer elucidated (see his comment below the fold):

Yes, this is true. If the curve  is smooth, then the space of such representations is $J \setminus \Theta$, where $J$ is the Jacobian of the curve and $\Theta$ is the $\Theta$ divisor. See Vinnikov, "Complete description of determinantal representations of smooth irreducible curves", Linear Algebra Appl. 125 (1989), 103--140 – David Speyer

The question, what additional information do you need besides eigenvalues of $A$ and $B$ to get the eigenvalues of $A + B$, translates into a question about homogeneous polynomials.  There is a triangle's worth of coefficients. You are given the coefficients on two of the sides of the triangle.  What you want is the sum of coefficients along lines parallel to the third side.  There are many ways you might get this information, depending on context. For example, the characteristic polynomial of $A B^{-1}$ corresponds to the 3rd edge of the triangle: in dimension 2, that gives the missing coefficient (all you actually need is Trace$(AB^{-1})$), while in higher dimensions, it's not enough.
To illustrate, here's a plot for a pair of pseudo-random matrices of dimension 9 whose coefficients were chosen uniformly and independently in the interval $[0,1]$. This is the slice $\lambda_1 = 1$ of the homogeneous form of the characteristic polynomial, so on the two axes, its roots are the negative reciprocals of characteristic roots for $A$ and $B$, and on the diagonal, for $A+B$. The characteristic polynomial of $A B^{-1}$ determines the asymptotic behavior at infinity, in this slice. You can of course only see the real eigenvalues in this real picture, but you can see how they move around as you vary the slice through the origin.  Under varying circumstance, you will have more a priori information. If $n=2$, the curves are conics. If the matrices are symmetric, they will intersect each line through the origin $n$ in $n$ points --- etc.
 (source: Wayback Machine)
Addendum:  Out of curiosity, I drew some pictures in the symmetric case, again for dimension 9.  Here is a typical example, for a pair of pseudo-random 9-dimensional symmetric matrices.
Observe the eigenvalues meeting and crossing as lines through the origin sweep around. [The buttons are from TabView in Mathematica, and they don't work in this static image.]
 (source: Wayback Machine)<|endoftext|>
TITLE: What is the right citation for the power iteration method to find eigenvalues?
QUESTION [7 upvotes]: What is the right citation for the power iteration method to find eigenvalues, if I want to cite the method in a paper? I've seen some Google PageRank references in this context. But Brin and Page didn't invent the power iteration method, did they?

REPLY [3 votes]: In his Ph.D. thesis Erhard Schmidt proved the existence of an eigenfunction of a symmetric integral operator by using iterated kernels, the equivalent of matrix powers in a function space. In order to avoid convergence problems in the case of eigenvalues of equal modulus, he did not apply the integral operator directly to a function. 
https://gdz.sub.uni-goettingen.de/id/PPN235181684_0063?tify=%7B%22pages%22:[479]%7D
In his 1909 textbook "Determinantentheorie" Gerhard Kowalewski, a distant relative of Kovalevskaya's husband, gave a finite-dimensional version of Schmidt's existence proof for a symmetric matrix T, and then proposed "a simple infinite process to get to an invariant of T", the power method.<|endoftext|>
TITLE: Euler Characteristic of a manifold with non-vanishing vector field,
QUESTION [26 upvotes]: A friend of mine recently asked me if I knew any simple, conceptual argument (even one that is perhaps only heuristic) to show that if a triangulated manifold has a non-vanishing vector field, then Euler's formula (the alternating sum of the number of faces of given dimensions) vanishes. I didn't see how to get started, but it seems like a good MO question.

REPLY [3 votes]: This is basically a watered down version of Poincare-Hopf theorem. Assuming that your compact manifold $M$ has a tringulation. At each of the simplex put a vector field $V$ which is zero at the centre of its subsimplices and flows out from the centre of the higher simplex to its boundary simplices. This vector field has finitely many zeroes and the index of $V$ is precisely $\chi(M)$. Now notice that the index of a vector field doesn't depend on the vector field. This is because it's topologically the intersection number of the image of $V$ in $TM$ and $M$ sitting as the zero section and $V$ is homotopic to $M$. In other words, given your non-vanishing vector field $V'$ one can linealy homotope this to $V$ whose index calculates $\chi(M)$.<|endoftext|>
TITLE: What would an ideal mathematical note-taking/organizer/PIM software look like?
QUESTION [7 upvotes]: I'm struggling with this problem for a long time, and I'm sure a lot of you out there are having similar problems too: when studying from a e-textbook I read and annotate in one app and make flashcards in another, bookmark interesting sites in my browser, collaborate over e-mail or instant-messaging, jot random notes and ideas in yet another app, keep my bibliographies in yet another app, do calculations different software packages, etc. And over time I start losing track of things.
There are open-source packages for doing all these things, but there isn't a single one that glues them together. Another issue is that many note-taking apps are just too structured and unflexible. For example, you need to manually tag every note, make links between notes, etc. It would be a good thing if such an app had some kind of an underlying semantic engine, so that when you're working on something, the software shows you similar things that you've worked on in the past, or similar things your colleague is working on currently, etc.
I'm thinking of suggesting Computer Science Department at my faculty to start an open-source project for such an app, and I'd like to hear some feature brainstorming.

REPLY [5 votes]: I still find it most convenient to just use a pen and paper for taking notes. My mobile phone has a pen, and can even detect mathematical writing (which works surprisingly well), but the feeling of writing on a flat/glass screen is just not the same like writing on real paper.
When writing on the computer though, Emacs does almost everything I need (Flashcards, Email, Latex Editor, BibTex Manager, Pdf Viewer, Calendar, Instant Messaging, Interface to COQ, Interface to GAP, Sage, ...). More important though is having a good keyboard that is comfortable for typing and especially a US keyboard layout (typing latex code on a non-US keyboard is a pain, for example in my native language I have to either use both hands to type "\" or press 3 keys at the same time).<|endoftext|>
TITLE: Testing permutations to see if they generate $S_n$
QUESTION [6 upvotes]: Alright, so a similar question was recently asked about the theoretical bound for generating certain permutations in polynomial time.  I had been thinking about a related problem in algorithms (with applications to a specific problem in graph theory - namely, discrete moves of sets of points among the vertices of a graph) and H A Helfgott's question inspired me to ask here.
Suppose I have some "black box" that spits out permutations $\rho_i \in S_n$.  I know the following things about the permutations it spits out:

$\rho_i$ is of cycle type $(k_i,1,1,\cdots,1)$.
This "black box" is fast in $n$ (linear in $n$ or so, maybe plus a few log terms).
If I run this black box long enough, it will spit out all of the $k$-cycles in some subgroup $H \subseteq S_n$.  I don't know what $H$ is a priori, although I can tell you (based on other constraints of the general problem) if $H \subseteq A_n$.

Let $G \subseteq S_n$ be the group generated by the $\rho_i$.  (Note that $G$ may not in fact be either $H$ or $S_n$.)  
I'd like to test if $A_n \subseteq G$.

Is there a computationally efficient test to see if the $\rho_i$ act primitively on $[1,n]$?  I want to say that if they act transitively and if the $k_i$ do not all share some nontrivial factor, they act primitively, but I am not sure of this.
Assuming that the answer to (1) is yes, I can guarantee that the natural action of $G$ on $[1,n]$ is transitive and primitive.  Does this guarantee that $G = A_n$? If not, what computationally non-intensive criterion do I need to add to guarantee that $G = A_n$?

Note: right now my algorithm for solving this problem is somewhere in that scary, scary territory beyond $O(n!)$ (yeah, that's how I'm testing to see if the darn thing is the alternating group), so any polynomial-time algorithm here would be super-awesome.

REPLY [5 votes]: The answer to (1) is yes, primitivity can be checked in O(n^3) time and practical computer implementations have been widely available for decades.  See Butler's Fundamental Algorithms for Permutation Groups p.76 for this and various related algorithms (such as testing transitivity) explained in a friendly manner.  Holt et al.'s Handbook of Computational Group Theory also contains this material in textbook form.  GAP contains open-source implementations of most of the algorithms mentioned (for instance IsTransitive and IsPrimitive would be useful).
The answer to (2) is usually yes, since proper primitive groups do not tend to contain many k-cycles.  You are just looking for what are called "giant tests" that can be applied to your restricted setting.  Some old theorems of Jordan can be used for this in ways that are described in Seress's Permutation group algorithms, especially 10.2.1 and 10.2.2.  These are refined to give probabilistic runtime estimates, some of which you could probably use if your black box had a (vaguely) known probability distribution.  See also Holt's textbook and GAP's DoSnAnGiantTest.
Section 3.3 of Dixon–Mortimer's book also contains results which can be used (with other results) to rule out "small" k.  In some ways this is done in section 5.3 and 5.4: if k is smaller than √n, then G contains the alternating group.
Be careful about your guarantee in (2). In particular, can you tell if H is transitive and primitive from the graph?  If you are only gathering information from the group G, then be careful that G may not be transitive even if H is (where G is generated by all of the elements of H that happen to be cycles).<|endoftext|>
TITLE: Is there a reset sequence?
QUESTION [31 upvotes]: There is a question someone (I'm hazy as to who) told me years ago. I found it fascinating for a time, but then I forgot about it, and I'm out of touch with any subsequent developments. Can anyone better identify the problem or fill in the history, and say whether it's still unsolved? It's a challenging question if I've gotten it right. Here it is:
Suppose you have some kind of machine with two buttons, evidently designed by people with poor instinct for UI. The machine has many states in which the buttons do different things.
Here are the assumptions:

There is no periodic quotient of the state space: no way to label states by an n-cycle so that both buttons advance the label by 1 mod n.
It is not reversible: there are situations when two states merge into one.
It's ergodic: you can get from any state to any other state by some sequence of buttons.

Now suppose its dinky little LCD is faded or broken, so you can't actually tell what the state it's in.  Is there necessarily a universal reset code, a sequence that will get you to a known state no matter where you start?
(Formally, this is a finite state automaton, or an action of the free 2-generator semigroup on a finite set, and asks whether some element acts as a constant map).

REPLY [19 votes]: I've played with this problem in real life with a TiVo, wanting it to go to sleep (a low power consumption state) without having to turn on the monitor to watch as its states changed.  The TiVo, or any remote, uses an alphabet size of at least as many buttons as there are on the remote control.  However, a little hunting on wikipedia shows that "Synchronizing Word" is where "reset sequence" leads to.
For $n$-state DFAs over a $k$-letter input alphabet in which all state transitions preserve the cyclic order of the states, an algorithm by David Eppstein finds a synchronizing word in $O(n^3+kn^2)$ time and $O(n^2)$ space.  The name of that paper is "Reset Sequences for Monotonic Automata" .
Finding and estimating the length of the "reset sequence" for a Deterministic Finite Automaton has been studied since the 1960's.  The Černý conjecture posits $(n-1)^2$ as the upper bound of for the length of the shortest synchronizing word, for any $n$-state complete DFA (a DFA with complete state transition graph).
The way you've posed your question sets $k=2$, since the transitions can only be labeled by the two buttons as input, thus the Deterministic Finite Automaton underlying your question will have a directed graph with at most two outbound arcs at each state.

REPLY [17 votes]: I believe you are referring to the Road coloring theorem.  It was solved in this preprint.<|endoftext|>
TITLE: Integrals from a non-analytic point of view
QUESTION [23 upvotes]: I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.
This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?
I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).
If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.

REPLY [6 votes]: My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.
Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.
Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps.
proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).
The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)
$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$
Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.
Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.
To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:
$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$
Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the bounded finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:
Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.
Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).
$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$
It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.
Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is spectral or multiplicative, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.
The case of $\mathbf{L}_{\infty}(\Omega)$ does not need the introduction of measures but of course, this is not so with $\mathbf{L}_{1}$. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm
$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$
and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.
Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.
Before the conclusion let me address a few points.

Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).
Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.
I am not advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.

How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.
But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.<|endoftext|>
TITLE: Compact Hausdorff spaces without isolated points in ZF
QUESTION [8 upvotes]: S is uncountable :=  |$\mathbb{N}$| < |S|

S is noncountable :=  |S| $\not\leq |\mathbb{N}|$

(X,$T$) is a nice space :=  (X,$T$) is a compact Hausdorff space without isolated points
Does [ ZF / ZF + Countable Choice ] prove that every nice space is [ uncountable / noncountable ] ?

If not, is it known to prove that the statement implies some choice principle?

What if the spaces are additionally assumed to be metrizable?
Now, that's basically 12 questions, so I certainly don't anticipate answers for all of them.

If it matters, the one I'm most curious about is "Does ZF prove that every nice space is noncountable?".

REPLY [5 votes]: I'd like to extend on Joel's answer, and point that ZF itself cannot prove that every "nice space" is uncountable.
It is consistent that the axiom of choice fails and there exists a Dedekind-finite set $X$ which can be topologized as follows:

$X$ is Hausdorff compact;
$X$ is strongly connected (every real valued function is constant);
The topology is an order topology of a dense linear order with endpoints (if we remove the endpoints then this is a locally-compact space, and every closed interval is compact).

It follows that there are no isolated points, so this is a nice space. However the set itself is Dedekind-finite and so not uncountable (but it is still noncountable). Such topological space is called a Läuchli continuum.
For example if we consider Mostowski's ordered model, by adding two endpoints to the atoms the result is a Dedekind-finite $X$ with the above properties.
I elaborated on this in a recent math.SE answer which includes all the relevant references and more.

It should be pointed that in the great book of choice principles the existence of non-trivial Läuchli continua is the negation of Form 155; whereas countable choice is Form 8. I could not find much connection between the two forms in the site.<|endoftext|>
TITLE: Is the nearest walk to Brownian motion uniform?
QUESTION [13 upvotes]: Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$.
Let $F_n$ denote the collection of all the $2^n$ many piecewise linear continuous functions $f:[0,1]\rightarrow\mathbb R$ such that $f(0)=0$ and $f$ is linear with slope $\pm \sqrt{n}$ on the intervals $[\frac in,\frac{i+1}n]$ for $0\le i<n$.  
Let $\psi_n$ denote a uniformly randomly chosen element of $F_n$, i.e., $\mathbb P(\psi_n=f)=2^{-n}$ for each $f\in F_n$.
Let $\phi_n$ denote a uniformly randomly chosen element of
$$
\text{arg min}_{f\in F_n}\left(\sup_{0\le x\le 1}|W(x)-f(x)|\right).
$$
In other words, $\phi_n$ is an element of $F_n$ that minimizes the sup-norm distance to $W$. More simply, we can say that $\phi_n$ is a nearest walk to Brownian motion.
Question: Do $\phi_n$ and $\psi_n$ have the same distribution? In other words, is $\mathbb P(\phi_n=f)=2^{-n}$ for all $f\in F_n$? I am mostly interested in the case $n\rightarrow\infty$, but will accept a rigorous answer for $n=2$.
Motivation: Donsker's Theorem says that $\psi_n$ converges weakly to $W$, whereas it is clear that almost surely, $\phi_n$ converges to $W$ uniformly. 
EDIT: Here is a follow-up question.

REPLY [7 votes]: My understanding of the question is this:
There is a function $V_n$ that maps the $L^\infty([0,1])$ to the set $F_n$ of $2^n$ piecewise linear functions as defined. $V_n$ gives the Voronoi subdivision, taking each point of the big set to the nearest neighbor of the smaller set.
I believe you're asking whether $V_n$  pushes  Brownian measure on $L^\infty$ to uniform measure(?)
This doees not seem plausible, starting with $F_2$, because unlike random walks and Brownian motion, this process has a memory effect. 
Think of the provisional grouping of paths into an upper group and a lower group, based on behavior in the interval $[0,1/2]$. The two distributions of positions at time 1/2 are skewed, because of how Gaussian tails decay rapidly. The provisionally upper group has mean $\sqrt 2/2$, but more of them are less than $\sqrt 2/2$ than greater than $\sqrt 2/2$.  This means that those that start provisionally up have less than probability .5 to continue upward, more than probability .5 to go down or to switch to down-up, and low probability of switching to down-down.  In other words, I believe the Voronoi measure is biased toward paths that end at 0, rather than at $\pm \sqrt 2$.   
It should be possible for someone to compute the exact distribution in this case.<|endoftext|>
TITLE: What information does the completion of a stalk at its maximal ideal give us in the holomorphic, analytic, or algebraic cases?
QUESTION [9 upvotes]: In a bonus exercise last year, we were asked to compute the completion in general of such a stalk on a smooth manifold of dimension $n$ (it is isomorphic to the ring of formal power series over $\mathbb{R}$ in $n$ unknowns).  It's clear that this is a bad case to work with, since smooth manifolds admit bump functions (which allows us to prove that there exists a nonzero element in the intersection of all (finite) powers of the maximal ideal), and therefore the completion contains very little data.  
However, what kind of "stuff" does this technique allow one to do in the analytic/holomorphic cases?  Similarly, in the algebraic case, we can often look at the henselization for the same information, but I still am not really sure why one would want to do so in the first place. That is, what geometric idea corresponds to the idea of "completion" (including henselization and strict henselization depending on the context) in the same way that localization at a prime corresponds to taking stalks geometrically?

REPLY [14 votes]: First I think you are a little bit unfair when you say that the completion of
the ring of germs of $C^\infty$-functions "contains very little". Mapping a
function into that completion gives you the Taylor series of function which
contains a lot of data about the function even though you certainly miss a
significant amount of information...
Anyway, you are certainly right looking at the (strict) Henselisation of a local
ring is often better than looking at the completion. This is mainly because the
Henselisation is a direct limit of very well-behaved extension rings and hence
many things which are defined over the Henselisation will be defined over one of
these extension rings. However, if you look at the local rings of analytic
spaces they already are Henselian yet you still use the completion even in those
case. The reason for that is that constructing elements of the completion can be
done by a step by step by step procedure, constructing one term of a Taylor
expansion at the time. In classical complex analysis one then usually performs a
closer analysis and shows that the resulting power series is actually
convergent but the first step is still important. 
From that point of view the Artin approximation (and generalisations) gives a
very general criterion for when certain processes automatically give convergent
power series provided that it gives any power series at all. Note that there are
many more classical results which allow you to pass to the completion without
using something like the approximation theorem. One such is that the completion
(of a Noetherian local ring) gives a faithfully flat extensions which in
particular allows you to check many equalities by passing to the completion.
However, if we back down to an algebraic scheme over $\mathbb C$ then you get
several local rings; the local ring of a point, its Henselisation, the ring of
germs of analytic functions and the completion. The ring of germs of analytic
functions doesn't make algebraic sense so Henselisation and completion are
algebraic substitutes (from "both sides"). The Henselisation is closer to the
original ring which is an advantage but is often more difficulat to work with,
while the completion is easier to work with but it is more difficult to get back
to the original ring. The approximation theorems should be seen as a very
powerful way of getting back to the Henselisation and then one can work at
getting from the Henselisation to the original or one can stay at the
Henselisation (this is what the étale topology is all about). (This is the
optimistic view on the approximation theorems, rather than the pessimistic one
that they say that you never need to pass to the completion as everything
already lies in the Henselisation.)
Added: To add some specific examples. On is the proof that a regular local ring is a UFD (let us assume that it contains a copy of its residue field to simplify). This one shows by first showing that the ring is UFD it its completion is. For the completion which then is a power series ring over the residue field one can use the Weierstrass preparation theorem to show that one can reduce to the polynomial ring in the last variable over the power series ring in all but the last. This is a UFD by induction and the fact that the polynomial ring over a UFD is a UFD ("Gauss lemma"). Here the Henselisation does not appear at all.
Another (far more sophisticated) example is the one which I guess was one of Artin's motivation for the approximation theorem to begin with. Here one wants to show that some functor is representable by an algebraic space. This means constructing a universal element over some suitable base. The first step is to use deformation theory to show that the functor (for a fixed point over some field) is prorepresentable. This is done exactly by showing that it is representable over the category of local Artinian rings for which a fixed power of the maximal ideal is zero. This is done by induction over the fixed power (and hence can be said to do a Taylor expansion one going from one order to the next). The end result (if everything works!) is a formal deformation over some complete local ring. Then one uses further properties of the functor to show that this formal deformation is given by an actual element of the functor (this typically uses Grothendieck's GAGA-type results for formal schemes). Then one uses the approximation theorem to show that the element comes from the Henselisation of something of finite type (there is an extra trickyness in that the complete ring is not known beforehand to descend as some such Henselisation). One then stops there and uses the universality to get gluing data for an algebraic space.
Note that there are situations when this doesn't work. A particular class of
examples arise when one is dealing with differential equations: There are
differential equations (with coefficients in the local ring of a smooth variety)
which has no solution in the Henselisation but does have a solution in germs of
analytic functions and there are differential equations that have formal power
series solutions but no convergent ones.
As for this setup for a general locally ringed space I do not know of any
general results except for the situation where a problem can be reduced to a
commutative algebra type problem concerning the local rings of the space which
then can be solved by forgetting about the space altogether.<|endoftext|>
TITLE: When do two elliptic curves have equivalent small etale toposes?
QUESTION [21 upvotes]: Let $X$ and $Y$ be elliptic curves over an algebraically closed field $K$. If the characteristic of $K$ is nonzero, assume both curves are ordinary or both are supersingular. Does it follow that $X$ and $Y$ have equivalent small etale toposes?

REPLY [9 votes]: In characteristic zero, I believe the answer is "yes".  More generally, if $X$ and $Y$ are two smooth proper curves over algebraically closed fields $K$ and $L$ of characteristic zero, then $X$ and $Y$ have equivalent etale topoi if and only if they have the same genus and card($K$) = card($L$).  The "only if" claim follows from the fact that the etale topos has both the etale fundamental group and the Zariski site as invariants (the first being determined by the locally finite constant objects and the second being determined by the lattice of subobjects of the final object).  This is more of a remark; the question is about the "if" claim.
For that, choose a set S and bijections $f: S \to X(K)$ and $g: S \to Y(L)$.  Now, for every finite subset $T$ of $S$, let $$a_T: Spec(K) \to M_{g,T}$$ be the map classifying $f(T)$ inside $X$, and ditto $b_T$ for $g(T)$ inside $Y$.  Furthermore let $A_T$ be the Henselization of $a_T$, and ditto $B_T$ of $b_T$.  For $T$ inside $T'$ we have maps $A_{T'} \to A_T$ and $B_{T'} \to B_T$, mapping compatibly to $M_{g,T'} \to M_{g,T}$; we consider the inverse limit of the fiber products $$A_T \times_{M_{g,T}} B_T$$ along these maps.  Since it is (eventually) a filtered limit of nonempty schemes along affine dominant maps, it has a point over some algebraically closed field $F$.  Let $Z$ be the induced curve over $F$ and $h:S \to Z(F)$ be the induced injection.

Then I claim that the étale site of $X$ is equivalent to the subcategory of the étale site of $Z$ consisting of those $U \to Z$ with either $U$ empty or $Z-im(U)$ inside $h(S)$.  By symmetry this will prove our desired result.

Here is how that equivalence goes.  For $U \to X$ be etale (suppose nonempty), let $X'$ in $X$ be an open subset of $X$ for which the pullback $U' \to X'$ is finite (i.e. proper).  Take $T = f^{-1}(X-X')$, and recall we have maps $Spec(K) \to A_T \to M_{g,T}$ where the composite classifies $f(T)$ inside $X$, and similarly $Spec(F) \to A_T \to M_{g,T}$ classifying $h(T)$ inside $Z$.  Let $C'$ denote the pullback to $A_T$ of the universal $T$-punctured curve over $M_{g,T}$; it gives $X'$ over $K$ (special fiber) and $Z':=Z-h(T)$ over $F$ (generic fiber).  Since $C' \to A_T$ is a smooth map with a smooth proper relative normal crossings compactification, the maps $X' \to C' \leftarrow Z'$ induce equivalences on categories of finite étale covers; thus we can transport $U' \to X'$ uniquely over to a $V' \to Z'$, via some $W' \to C'$.
But in fact I claim that the same is true of $U \to X$.  Indeed, let $C$ be the completed curve of $C'$ (over $A_T$), and for $t$ in $T$ with corresponding section $c$ of $C-C'$ consider the completion of $C$ along $c$, then subtract the section $c$ to get $L \to A_T$.  On pullback to $K$ this becomes $L_x$, Spec of a Laurent field at $x=f(t)$ in $X$, and ditto for $F$ and $L_z$; furthermore once again the maps $L_x \to L \leftarrow L_z$ induce isomorphisms on categories of etale covers of $L$.  On the other hand, the Galois type of the pullback of $U' \to X'$ to $L_x$ completely determines the points at infinity of $U'$ lying above $x$ (and ditto on the $F$ side): they are the orbits of Galois (which is the profinite completion of $\mathbb Z$) over any geometric fiber.  Thus we can add in points at infinity on both sides in a consistent and canonical manner.
This finishes the proof, I suppose: functoriality of this one-to-one correspondence (and its independence of $X'$) comes from us having made our constructions compatible with changing $T$ and the fact that smooth proper compactifications of curves over fields are unique and functorial.

Some remarks:
1) This proof is close to the proof I gave in the comments for $K$ the complex numbers: the fact that all elliptic curves are homeomorphic is symptomatic of connected moduli, which was again a key fact in this proof (though in a hidden manner?)
2) However, the real surprise lies in the differences: to get an equivalence of &eactuetale topoi one needn't choose a homeomorphism -- in fact any bijection will do!  The field $F$ whose existence was ensured above guarantees some measure of continuity no matter what the bijection.  In fact I imagine that in higher dimensions the notions of being homeomorphic and having equivalent étale topoi diverges.  For instance, I'll bet that $E \times \mathbb A^1$ doesn't have the same étale topos as $\mathbb G_m \times \mathbb G_m$.  Similarly, I wonder about two varieties occurring in the same proper smooth family having inequivalent étale topoi.
3) The positive characteristic case is a bummer for me because of non-topological (i.e. "wild") phenomena and I don't have much to say.<|endoftext|>
TITLE: Additive, covariant functor preserve direct sum?
QUESTION [11 upvotes]: Does any additive, covariant functor preserve direct sum?

REPLY [18 votes]: I'm not sure what is being asked, but I'll try rephrasing what I suspect the question is: if $A$ and $B$ are enriched in abelian groups, and if $F: A \to B$ is a functor such that all the maps $\hom(a, a') \to \hom(F(a), F(a'))$ are homomorphisms, then is it true that $F$ takes finite sums (coproducts) in $A$ to finite sums in $B$? The answer to that question is 'yes'. José Figueroa-O'Farrill has already given some hints, but I think they deserve to be amplified. 
The result is standard in homological algebra textbooks (I believe you can find it in Hilton and Stammbach for instance). It helps to observe first that in $Ab$-enriched categories $A$, or even in categories enriched in commutative monoids, finite sums in $A$ are the same as biproducts, which roughly speaking are simultaneously products and coproducts (in a compatible way). 
Suppose $a + b$ is a coproduct in $A$. Then by definition there are coproduct inclusions $i_a: a \to a + b$, $i_b: b \to a + b$. Also by definition, a map $f: a + b \to c$ is uniquely determined by the pair of maps $f_a := f \circ i_a$, $f_b = f \circ i_b$. In particular, we have a map $p_a: a + b \to a$ determined by the pair $1_a: a \to a$, $0: b \to a$, where $0$ denotes the zero element in the abelian group $\hom(b, a)$. Similarly, we have a map $p_b: a + b \to b$. By definition, we have the equations 
$$p_a i_a = id_a \qquad p_a i_b = 0 \qquad p_b i_a = 0 \qquad p_b i_b = id_b$$ 
Note also that $i_a p_a + i_b p_b = 1_{a+b}$, because each side is a map $f$ satisfying the equations $f i_a = i_a$, $f i_b = i_b$, and there is only one such map (by the universal property of coproducts). 
Now, the crucial point is that whenever you have maps $i_a: a \to c$, $i_b: b \to c$, $p_a: c \to a$, $p_b: c \to b$ satisfying the four equations above and a fifth equation $i_a p_a + i_b p_b = 1_c$, then $c$ is simultaneously a product and coproduct of $a$ and $b$; in this situation we call $c$ a biproduct. I invite you to write out the proof. (Hint: given $g: d \to a$ and $h: d \to b$, define $\langle g, h \rangle: d \to c$ to be $i_a g + i_b h$. Show this is the unique map such that $p_a \langle g, h \rangle = g$ and $p_b \langle g, h \rangle = h$, so that $c$ satisfies the universal property of a product. By dualizing this argument, $c$ also satisfies the universal property of a coproduct.) 
But any $Ab$-enriched functor preserves these five equations since it preserves composition, identities, zero elements, and addition. Therefore $Ab$-enriched functors $F: A \to B$ take biproducts in $A$ to biproducts in $B$. That is the same as saying it takes coproducts in $A$ to coproducts in $B$. 
Just to round out the story: there are converses to these statements as follows. If a category $A$ has biproducts (see the nLab page cited above), then $A$ is automatically enriched in commutative monoids: the sum of two morphisms $f, g \in hom(a, b)$ is given by the composite 
$$a \stackrel{\Delta_a}{\to} a \oplus a \stackrel{f \oplus g}{\to} b \oplus b \stackrel{\nabla_b}{\to} b$$ 
where $\oplus$ denotes biproducts, and the diagonal and codiagonal are defined using the the biproduct structure. (The zero element in $\hom(a, b)$ is the composite $a \to 0 \to b$ factoring through the zero object.) And if $A$ and $B$ have biproducts and $F: A \to B$ preserves them, then $F$ is an enriched functor: preserves the addition and zero carried by the hom-sets.<|endoftext|>
TITLE: Obstruction Cocycles
QUESTION [5 upvotes]: Hey everyone, I was reading about obstruction theory, here's a bit of a summary. Take a cellular space $X$ and a fibre bundle $p:E \to X$ with fiber $F$; consider the problem of extending a section $s$, defined on the $(n-1)$-skeleton over to the $n$-skeleton. We work cell by cell pulling back the bundle via the characteristic map and the section via the restriction of the Char. Map to the boundary of our $n$-cell, since the cell is contractible, the bundle is isomorphic to $D^n \times F$ so the section defines a map from $S^{n-1} \to D^n \times F$ i.e. an element of $\pi_{n-1}(D^n \times F) \cong \pi_{n-1}(F)$. Define the obstruction cochain as the element in $C^n(X^n,\pi_{n-1}(F))$ taking each $n$-cell to the element in the $(n-1)$-homotopy group constructed before.
Here's what's bothering me, in Steenrod's book (The topology of Fibre Bundles) he proves that this cochain is actually a cocycle in a really weird way, it looks to me as if he makes no distinction between the homological boundary and the topological boundary of a cell. Roughly he writes the following composition:
$ C_{q+1}(X) \stackrel{\partial_*}\to Z_q(X) \stackrel{hurewicz}\to \pi_q(X^q) \stackrel{f=(p_2\circ \sigma)_*}\to \pi_q(F) $
And claims that this composition is the value of the obstruction cochain in an $n+1$ cell, how might one verify this?
Not being happy with this proof i went and looked at the one Kirk and Davis' book (Lecture notes on algebraic topology) and found it too complex (I know, I dont like anything sorry).
What I was wondering is if there was a way to prove this affirmation (the obstruction cochain is a cocycle) directly, i.e. denoting the cochain by $\Theta$ doing something like:
$\delta \Theta (e) = \Theta( \partial e) = \Theta (\sum [w_i;e]w_i) = \sum [w_i;e]\Theta(w_i) = \dots = 0$ 
(where $e$ is a $(n+1)$-cell, $w_i$ is a $n$-cell and $[w_i;e]$ is their incidence number, so that the third term is $\Theta$ evaluated on the cellular (homological) boundary).
Any help on the subject or a good refference is very very much appreciated!
Thanks and have a great week.

REPLY [2 votes]: How are you defining the cellular chain complex and  $[w_i;e]$? The usual way is to define $C_n(X):=H_n(X^n,X^{n-1})$ and the differential as the composite $H_n(X^n,X^{n-1})\to H_{n-1}(X^{n-1})\to H_{n-1}(X^{n-1},X^{n-2})$, where the first map is the connecting homomorphism for the pair. Steenrod's observation is then straightforward, and follows from the long exact sequence of the appropriate pair.  After the fact, you can show that $C_n(X)$ is free on the $n$ cells and that $[w_i;e]$ is the degree of the composite
of the attaching map $S^{n-1}\to X^{n-1}$ for $e$ composed with the projection $X^{n-1}\to X^{n-1}/X^{n-2}=\vee_i S^{n-1}\to w_i/\partial w_i\cong S^{n-1}$. That typically requires reconciling different notions of degree, e.g. homological and differential topological and perhaps this is where you are having difficulties.<|endoftext|>
TITLE: Gluing two diffeomorphisms together
QUESTION [7 upvotes]: A fundamental construction in a first course on manifolds is to build a smooth function $\psi\colon \mathbb{R} \to \mathbb{R}$ with the property that for some $0<\delta<\epsilon$ we have

$\psi(x)=1$ for $|x|\leq\delta$;
$\psi(x)=0$ for $|x|\geq\epsilon$.

Using this "bump function", one can do all sorts of "gluing" tricks:  for example, if $f\colon \mathbb{R}^n \to \mathbb{R}^n$ is any smooth map and $\epsilon>0$ is such that $|f(x)|<\epsilon$ for all $|x|\leq\delta$, then we can build a smooth map $F\colon \mathbb{R}^n\to \mathbb{R}^n$ such that

$F(x) = f(x)$ for $|x|\leq\delta$;
$F(x) = x$ for $|x|\geq\epsilon$.

However, the map we obtain in this manner does not necessarily preserve all the nice properties of $f$.  For example, if $f$ is a local diffeomorphism, it does not immediately follow from the above construction that $F$ can also be taken to be a local diffeomorphism.  
Intuitively, it seems clear that this is (usually) the case:

For linear maps with $n=1$, it's just a matter of drawing a smooth curve that starts on the line $y=\lambda x$ and goes to the line $y=x$ without ever having a horizontal tangent.  (Of course we must take $\lambda>0$ for this.)
For linear maps with $n=2$, the image of a small ball is an ellipse, so by smoothly deforming the ellipse into a ball, rotating so that the map is a multiple of the identity, and then using the trick from $n=1$ to make the eigenvalues equal to $1$, we can find a smooth homotopy $f_t$ such that $f_0=f$ and $f_1$ is the identity, and furthermore, setting $F(x) = f_t(x)$ for $|x| = \delta + t(\epsilon - \delta)$ makes $F$ a diffeomorphism.
Since linear maps approximate arbitrary maps, the above procedure ought to generalise.  (Modulo the restriction that $Df(0)$ should have positive determinant.)

I expect that there's a general result along these lines, and that it is quite standard and well-known.  But I don't know it (and I'd rather re-invent the wheel as few times as possible).  Can someone help me out with a statement of a general theorem, and ideally a reference?
Edit:  Since that was kind of rambling, here's the specific question.  Let $B(r)$ denote the (open) ball of radius $r$ in $\mathbb{R}^n$ centred at the origin.  Suppose $f\colon B(\delta) \to \mathbb{R}^n$ is a diffeomorphism onto its image, and suppose $\overline{f(B(\delta))} \subset B(\epsilon)$.  Let $\delta' < \delta$.  Does there necessarily exist a diffeomorphism $F\colon \mathbb{R}^n \to \mathbb{R}^n$ that agrees with $f$ on $B(\delta')$ and is the identity map outside of $B(\epsilon)$?
I think we need to require that $Df(0)$ have positive determinant.  Are there any other obstructions?  If there are, can they be removed (for a given $f$) by decreasing $\delta$ so that $f$ is close to being a linear map?  Is there a general theorem from which all this follows?

REPLY [11 votes]: The group $GL(n,R)$ has an affine structure, coming from the coefficients of the group.
It has an easily described and fairly large convex neighborhoods of the identity, satisfying that for no vector $V \ne 0$ is $A(V)$  negative real multiple of $V$ (i.e., no negative real eigenvalues).
The first derivative of a convex combinations of two diffeomorphisms, the first derivative is the convex combination of derivatives, plus a correction term of the difference of derivatives applied to the gradient of the convex coefficient.  Convex combinations of more terms are similar. In any case, if you bound the maximal gradient of a partition of unity, you can get an explicit bound for collections of first derivatives of local diffeomorphisms guaranteeing that there compostion is diffeomorphism.
I think a bigger issue to watch for is the difficulty in practice to give an actually  valid atlas for an abstract manifold that doesn't start life already embedded in $\mathbb R^m$. For two overlapping charts, an approximate diffeomorphism is fine. However, when three charts overlap, in principle the overlaps must satisfy a cocycle condition. In practice, there aren't many good choices of classes of gluing maps that are simply described and have reasonable group laws; note in particular that convex combinations disrupt relationships of compositions.
One solution is, to define how an intersection of $m$ charts patch together, work in the product of the collection of local coordinates, $\mathbb R^{nm}$. Build the graph of the relation among all the charts at once, rather than doing it pair by pair. Consistency is straightforward to arrange using partitions of unity on the graphs.  Even for an overlap of two charts, rather than specify a diffeomorphism which may be hard to invert explicitly, specify the graph of a diffeomorphism, which is trivial to invert.
For the specific question of extending a diffeomorphism defined on a small ball: there is a standard technique that works if you use one of Goodwillie's suggested modifications.
The set of differentiable embeddings of a ball in $\R^n$ retracts to the set of linear embeddings by conjugating with a family of contractions, $\phi_t(x) =  \phi((1-t) x)/(1-t)$.
The set of linear maps retracts to orthogonal maps, using a parametrized Gramm-Schmidt process. Every orhtogonal map is homotopic to either the identity, or some chosen reflection.   Then, apply the isotopy extension theorem to make all of $\R^n$ move along with the moving embedding.  You can compress everything that happens into the unit ball.
You could also do this in another way:  For any neighborhood $U$ of the identity in $GL(n,R)$, any diffeomorphism that is isotopic to the identity can be expressed as the composition of elements in the small ball (and conversely).  You can make a sequence of overlapping annular coordinate charts, where in the main part of each one you use an initial segment of the composition.  The overlap maps is now in a small neighborhood where the convex combination principle works.
One more remark:  in the particular case of dimension 2, if you define a lift of the derivatives to the universal cover of SL(2,R), then there is an averaging process for derviatives that always works, using complex logarithmic coordinates for the plane minus the origin.  In the particular case of dimension 3, the universal cover of $SO(3)$ is the group of unit quaternions, homeomorphic to $S^3$, so there is an averaging procedure in that group (and by quadratic form manipulation, in $SL(3,R)$ which works much better than averaging in the obvious coordinates.   Among other things, this is very useful for motion control. We used to get workable mathematical specifications for segments of the video Outside In. [``We'' primarily means the actual implementers---I especially recall discussions about the difficulties and solutions with my son Nathaniel Thurston].<|endoftext|>
TITLE: The fundamental lemma and the conjecture of Birch and Swinnerton-Dyer
QUESTION [25 upvotes]: Here is a rather pathetic question. In a comment on Tim Gower's weblog, I tentatively stated that the fundamental lemma was necessary for the work of Skinner and Urban relating ranks of Selmer groups of elliptic curves to the vanishing of their $p$-adic $L$-functions. Now, I believe it is correct that some endoscopic version of transfer from a unitary group to a general linear group is necessary for the construction of their $\Lambda$-adic representations. However, having a really poor understanding of the actual techniques, I don't know which version is crucial. That is to say, it's entirely likely that some earlier special case is sufficient for Skinner-Urban. Could I trouble some expert to give a brief outline of the situation? 
The pathetic part of this is that the journalist I mentioned in the comment will call in about 4 hours, so it would be nice to know before that. Of course I shouldn't have agreed to speak about something I know so little about, but it was hard to refuse under the circumstances. Oh, in case you're worried that I'm going to discuss Skinner-Urban with the fellow, don't. I just want to bone up on the background.

Added: 
For people who like the idea of linguistic diversity in mathematics, I am including a link to a report  written (with Sugwoo Shin) for the Korean Mathematical Society that expands on the comment to the journalist.

REPLY [20 votes]: Dear Minhyong,
My understanding, based on recalling talks of Skinner and also briefly looking over the ICM paper that you linked to is that, yes, they do rely on the fundamental lemma, namely, the fundamental lemma for unitary groups as proved by Laumon and Ngo.  Unfortunately, I'm not sufficiently educated in their work, or in the field of unitary Shimura varieties in general,
to be sure whether their Conjecture 4.1.1 is actually a theorem at this point or not.  Certainly related results have been proved by Sug Woo Shin and by Sophie Morel, both relying
on Laumon--Ngo.  But whether these results address the precise unitary groups considered by Skinner--Urban, I'm not sure.  (Just looking briefly at what Skinner and Urban write, it looks as if they are considering quasi-split groups, or at least groups that are not anisotropic,
so that their Shimura varieties are non-compact; thus the work of Morel (who considers the non-compact case) may be more directly relevant than that of Shin (who considers the compact case); but both Morel and Shin's work come with certain technical restrictions, so that the precise requirements of Conjecture 4.1.1 (in particular, precise local-global compatibility away from the residue characteristic) may not follow in the full generality considered by Skinner and Urban from the work of either.)
I'm sorry that I can't say more, but I do think that it's safe to say that the fundamental lemma is a crucial (albeit highly technical) ingredient in their program. 
At the risk of adding unsolicited and unwanted advice:
You could also mention the complete proof of Sato--Tate for elliptic curves over totally real fields (with no requirement that the $j$-invariant be non-integral, unlike in the original work of Clozel--Harris-Shepherd-Barron-Taylor), which follows from the work of CHSBT + that of Shin (the work on compact unitary Shimura varieties mentioned above, which as I already remarked requires Laumon--Ngo as an ingredient).
This is a pretty nice Diophantine statement (technical for a journalist obviously, but presumably one can convey the gist) with the fundamental lemma forming one of the pillars
that supports it.<|endoftext|>
TITLE: Spherical objects on Kummer surfaces
QUESTION [5 upvotes]: Spherical objects $E$ in the derived category of coherent sheaves over a K3 surface satisfy:

$\operatorname{Hom}(E,E)=\mathbb{C}$,
$\operatorname{Ext}^2(E,E)=\mathbb{C}$,
$\operatorname{Ext}^i(E,E)=0$ otherwise.

Are the structure sheaf $O_X$ and the sheaves associated with the exceptional divisors the only spherical objects on a Kummer surface?

REPLY [10 votes]: Certainly not. The group of autoequivalences acts on the set of all spherical objects, so a twist of a spherical object is spherical, hence any line bundle is spherical. Also, you can apply the reflection in one spherical object to another. E.g., applying the reflection in $O_X$ to $O_E(t)$ (where $E$ is a rational $(-2)$ curve and $t \ge 0$) one concludes that $Cone(O_X^{t+1} \to O_E(t))$ is spherical. You can continue by acting with another reflections to get a huge number of spherical objects.<|endoftext|>
TITLE: Ngo's support theorem & the Jacobian over the versal deformation of a planar curve
QUESTION [10 upvotes]: I am trying to understand what exactly Ngo's support theorem says about how the cohomology of compactified Jacobians varies over the versal deformation of a curve with plane singularities.  
That is, let $C$ be a compact complex curve with singularities of embedding dimension 2.  Let $\mathcal{C} \to \mathbf{V}$ be a versal deformation, i.e., there is a point $0 \in \mathbf{V}$ over which the fibre is $C$, and any deformation of $C$ pulls back from this family.  Let $\pi: \overline{\mathcal{J}} \to \mathbf{V}$ be the relative compactified Jacobian of the family; I want to understand the family of cohomologies $\mathrm{R}\pi_* \mathbb{Q}$.  As pointed out in the few references I've browsed, this is a situtation to which Ngo's support theorem applies: there is a relative action of the (uncompactified but generalized) Jacobian $\mathcal{J}$, and the locus $\mathcal{V}_h$ where the affine part of $\mathcal{J}$ is of dimension $h$ is of precisely codimension $h$ (by an old result of Teissier).  However, said references go on to treat a more famous example instead -- the Hitchin fibration. 
Unfortunately my competence with the machinery of perverse sheaves is so poor that I can barely understand the statement of the support theorem.  So, I apologize if the following questions are overly trivial or confused.
0.) Is the application of the support theorem to the situation I describe above treated explicitly in the literature?
1.)  Does the support theorem imply that $\mathrm{R} \pi_* \mathbb{Q}$ is the IC-sheaf on $\mathbf{V}$ determined by the local system of cohomologies of Jacobians on the locus $\mathbf{V}_{\mathrm{reg}} \subset \mathbf{V}$ of smooth curves?  If not, does it specify what other local systems on what other loci must be added?
2.) Does the knowledge of the Picard-Lefshetz monodromy around nodal degenerations determine (explicitly) the local system of cohomologies of Jacobians on $\mathbf{V}_{\mathrm{reg}}$, and, if so, does it allow an in principle calculation of the cohomology of the central fibre?  Can it be done in practice?
3.) Does it follow that the cohomology of the compactified Jacobian of the central fibre is determined by the topology of the singularity of the discriminant (the complement of the locus of smooth curves)?  Or -- perhaps it is necessary to know how monodromies around different nodes interact -- the stratification of the discriminant by the (closures of the) loci with a fixed number of nodes?
4.) Any other comments on how this (or anything else) would be used to compute the cohomology of the compactified Jacobian are welcome.

REPLY [5 votes]: Very partial answer - I don't think I can comment yet...
I found it helpful to rephrase the statement of the support theorem like this:
Let $R\pi_\ast \mathbb Q =  \bigoplus _i IC_{Z_i}(L_i)[d_i]$ be the decomposition of the pushforward sheaf. If Z is one of the $Z_i$ appearing in the sum above, then $Z$ is the support of a direct summand of $R^{2d}\pi_\ast \mathbb Q$ (i.e. there is extra stuff appearing in the top cohomology of the fibres). Such a summand only will occur when there are extra irreducible components in the fibres of $\pi$.
In particular, if the fibres of $\pi$ are irreducible, then the only possible support appearing in the direct sum is the whole of V. This means that the pushforward is the IC extension of the local system over the locus where $\pi$ is smooth.
I think this means that in your situation, the answer to (1) is yes (as compactified Jacobians are irreducible for such singularities). I would be interested to find out the answers to (2) and (3).<|endoftext|>
TITLE: Noether's bound for anticommutative invariant theory (diff. forms instead of polynomials)?
QUESTION [16 upvotes]: EDIT: Now with a concrete request to CAS experts (see the end of the post).
Let $G$ be a finite group, and $V$ a finite-dimensional representation of $G$. The classical invariant theory of $G$ and $V$ is a study of the $G$-invariants in the $K$-algebra $K\left[V\right]=\mathrm{S}\left(V^{\ast}\right)$, where $\mathrm{S}$ means the symmetric algebra. One has results like the Noether theorem, which says that over a field of characteristic $0$, the $G$-invariants in this $K$-algebra $\mathrm{S}\left(V^{\ast}\right)$ are generated (as a $K$-algebra) by the $G$-invariants in its submodule $\mathrm{S}^1\left(V^{\ast}\right)\oplus \mathrm{S}^2\left(V^{\ast}\right)\oplus ...\oplus \mathrm{S}^{\left|G\right|}\left(V^{\ast}\right)$.
Is there a similar theory, with similar bounds, for the $K$-algebra $\wedge \left(V^{\ast}\right)$ ? Actually this looks a bit simpler, because $\wedge^k \left(V^{\ast}\right)$ is nonzero only for $k\leq\dim V$; but $\dim V$ might still be greater than $\left|G\right|$. If there is an analogue of Weyl's theorem (Theorem A in 8.3 of Kraft-Procesi - see the Example in §8.5 for how I want to apply it), then we could limit ourselves to the case of $V$ being the regular representation, and then Noether's bound would follow. But there is no direct analogue of Weyl's theorem (in the sense of replacing $K\left[V^p\right]$ by $\wedge \left(V^{\ast p}\right)$ and $K\left[V^n\right]$ by $\wedge \left(V^{\ast n}\right)$).
The only thing from classical invariant theory which I can extend to this "anticommutative" (because $\wedge^k \left(V^{\ast}\right)$ can be seen as "anticommutative" polynomials on $V$) invariant theory is the Molien series, which takes the form
$\sum\limits_{d=0}^{\infty} \dim \left( \left(\wedge^d V\right)^G\right) T^d = \frac{1}{\left|G\right|}\sum\limits_{g\in G} \det\left(1+Tg\right)$
in our anticommutative setting, again only in the characteristic $0$ case. (Note that I have switched from $V$ to $V^{\ast}$, but this should not matter in the end since the the dimension of the invariants in some given representation always equals the dimension of the invariants in its dual - when we are in the characteristic $0$ case at least.)
Concrete request to anyone who has some experience with computer algebra systems:
Consider the exterior algebra freely generated by the differential forms $dx_1$, $dx_2$, ..., $dx_n$, $dy_1$, $dy_2$, ..., $dy_n$, $dz_1$, $dz_2$, ..., $dz_n$ over $\mathbb Q$. (This is the exterior algebra of a $3n$-dimensional vector space.)
Let the symmetric group $S_3$ act on this algebra by permuting $x$, $y$, $z$ (leaving the numeric indices unchanged).
For every $k\in\left\lbrace 1,2,...,n\right\rbrace $, the sum $dx_1\wedge dx_2\wedge ...\wedge dx_k + dy_1\wedge dy_2\wedge ...\wedge dy_k + dz_1\wedge dz_2\wedge ...\wedge dz_k$ is an invariant under this action. Does it lie in the subalgebra generated by invariants of smaller degree? (Note that it is enough to consider homogeneous invariants, so that there cannot be degree-reducing cancellations. Also note that it is easy to find a generating set of all invariants of given degree just by writing out all the monomials and averaging them over the action of $S_3$.) A negative answer for $n=k=7$ (or higher) would destroy the Noether bound in the anticommutative case.
I played with the idea of coding the exterior algebra as an algebraic type in Haskell, but finding out whether an element is generated by elements of smaller degree means solving a system of linear equations, and I don't have enough experience to code Gaussian elimination in Haskell. Other than that, there must surely be a CAS that knows how to work with anticommuting variables?

REPLY [2 votes]: PARTIAL answer: By a theorem of Solomon (Louis Solomon, Invariants of Finite Reflection Groups, Nagoya Math. J. Volume 22 (1963), pp. 57-64, also Julia Hartmann and Anne V. Shepler, Reflection Groups and Differential Forms), the Noether bound holds when the action of $G$ on $V$ is a reflection group. (Moreover, in this case, even the tensor product $S\left(V\right)\otimes \wedge\left(V\right)$ is generated in degree $\leq \left| G\right|$. I am wondering what can be said about $S\left(V\right)\otimes \wedge\left(W\right)$ for two different reflection representations $V$ and $W$ of $G$.)
This doesn't say anything about the $S_3$-module I am considering to be a potential counterexample, though ($S_3$ doesn't act as a reflection group on it). I would still be very indebted for some Sage code to work out this case.<|endoftext|>
TITLE: Is the Landau-Ramanujan constant irrational?
QUESTION [7 upvotes]: Hi, here, in wikipedia, the Landau-Ramanujan constant appears under a list of suspected transcendentals. I could not find anywhere a statement or a proof of it's irrationality. So, my question is, is the constant irrational( since it's under a list of suspected transcendentals, is it right to assume that it is irrational?) And if it is, is there a proof available? 
Thanks in advance.

REPLY [3 votes]: This is not a definitive answer, just a few links.
This link of "important irrational constants" lists it as irrational, without a citation.
The Encyclopedia of Integer Sequences has quite a few references at the page for Landau-Ramanujan, one or more of which may contain what you seek.<|endoftext|>
TITLE: Inclusions of constructible sets are *not* definable
QUESTION [7 upvotes]: Let $L$ be the constructible universe and $x,y \in L$ such that $x \subseteq y$. Is then $x \in D(y)$, i.e. $x$ a definable subset of $y$?
If this is not true, do we at least have the following: If $x,y \in L$, then $\rho(x) \leq \rho(y)$? Here $\rho$ denotes the $L$-rank.
And if this is also false, how do you prove, that for $x \in L$, we have $L \cap P(x) \in L$? So I need this to prove the power set axiom in $L$.

REPLY [7 votes]: If you just want to prove that $L$ satisfies the power set axiom, you don't need condensation; Kunen's comment is correct.  Given $x\in L$, to show that $L\cap P(x)\in L$, first consider an arbitrary $y\in P(x)\cap L$ and observe that, since $y\in L$, there is a first ordinal $\alpha(y)$ such that $y\in L_{\alpha(y)}$.  By replacement (in the real world), there is an ordinal $\beta$ that is larger than all these $\alpha(y)$ for all $y\in P(x)\cap L$.  So $P(x)\cap L\subseteq L_\beta$.  But then $P(x)\cap L$ is first-order definable over $L_\beta$, by the definition $\{y\in L_\beta:L_\beta\models y\subseteq x\}$, and is therefore an element of $L_{\beta+1}$.  
Note that this proof, unlike the one given by Francois, tells you nothing about how far you must go in the $L$ hierarchy before you find $P(x)\cap L$; it provides no information about the size of $\beta$.  To get such information, condensation is needed.
Concerning the first two parts of the original question, the answer is no for both in general.  Take $y$ to be $\omega$ and note that it has only countably many definable subsets and only countably many subsets of lower or equal $L$-rank, but it may have (and will have if, for example, $V=L$) uncountably many subsets in $L$.<|endoftext|>
TITLE: Colimits of manifolds
QUESTION [13 upvotes]: This question tells us that in general colimits do not exist in the category of manifolds.
However, this negative answer is not very satisfying. A manifold can be considered as a colimit of its altas. In this sense, it seems that constructing manifolds via atlas is a means of completing the model category of open subsets of $R^n$ (or more precisely a comma category based on this), so it should be reasonable that at least certain kinds of colimits should hold in the category of manifolds.
Considering the altas as a "good" system, do colimits of systems which are
1) of same-dimensional manifolds
2) have only open injections as morphisms,
3) is countable
and
4) (add your additional condition here),
exist?
PS the counter example in the above referred question fails conditions 1 and 2.
EDIT: After our good discussion below, perhaps let me restate my original question. I am motivated by the fact that a manifold is a colimit of its altas. What I'm looking for is a generalization of this fact. Can we extract out some general properties of an atlas that makes its colimit exist?
For instance do colimits of countable inverse limits of open smooth injections of (neccessarily) same-dimensional manifolds exist? (remove as many quantifiers as is uncessary).
I really do mean inverse system and not direct system. Observe that the altas (if we take all finite intersections) is filtered in the direction away from the colimit.

REPLY [6 votes]: This is a few years late, but here goes.
Associated to an open cover $\{U_i\}$ of a (Hausdorff paracompact) manifold $M$ there is a cover groupoid, whose space of objects is the disjoint union ${\mathcal U}:= \sqcup  U_i$ and whose space of arrows is the fiber product ${\mathcal U }\times  _M {\mathcal U}$.  The manifold $M$ is the quotient space of the cover groupoid, i.e., the colimit of ${\mathcal U }\times  _M {\mathcal U}\rightrightarrows {\mathcal U}$.  Note that the cover groupoid is proper so its quotient is Hausdorff.   The orbit spaces of two Morita equivalent groupoids are isomorphic.  Hence you could write a manifold as a colimit of a Lie groupoid which is Morita equivalent to a cover groupoid.  I believe the conditions for being equivalent to a cover groupoid are being proper and having all isotropy groups trivial ( i.e., $Hom (x,x)$ is a trivial group for any object $x$ of your Lie groupoid).<|endoftext|>
TITLE: Examples of certain locally compact totally disconnected groups
QUESTION [5 upvotes]: To find a counterexample disproving a generalization of a theorem in the theory of scale functions on locally compact totally disconnected groups, initiated by George Willis, I am looking for a group with certain properties. Alternatively I am looking for arguments why such a group cannot exist. The precise question is:

Is there an example of a totally disconnected locally compact topological group $P$ such that
  (1) $P$ is algebraically a (semi)direct product of subgroups $G$ and $H$
  (2) $G$ and $H$ are not closed in $P$, but locally compact in the topology inherited from $P$,
  (3) $K\cap G$ and $K\cap H$ are not compact for any compact open subgroup $K\subset P$,
  (4) $P$ is not compact itself?

Usual construction methods of totally disconnected locally compact groups such as direct limits of discrete groups seem to fail to produce such an example. In particular (3) seems hard to achieve. So maybe an example  which occurs naturally in some other context has a better chance to work than building one with bottom up methods.  
I hope I didn't overlook any obvious argument contradicting the existence of the desired group.

REPLY [2 votes]: This question just got bumped to the front page by the Mathoverflow bot. But the question was already answered in the comments by BCnrd. Following advice from this Meta article, I'm going to copy BCnrd's answer here so the OP can accept this answer and this question won't get bumped up again. My response is CW so I can't gain any reputation. 
The answer to the question is "no." Here's why:
First, note that every totally disconnected topological group is Hausdorff. Next any locally compact subspace of a Hausdorff space is locally closed, hence open in its closure (see e.g. Munkres Topology for a proof of this). If $P$ was a totally disconnected topological group and $G$ is a locally compact subgroup then $G$ is open in its closure $\overline{G} \subset P$, which in turn is a closed subgroup of P. But open subgroups of topological groups are closed, so $G$ is closed in $\overline{G}$ and hence $G=\overline{G}$. This proves (2) is impossible.<|endoftext|>
TITLE: Proof in the literature of an equality involving the prime counting function
QUESTION [5 upvotes]: Let 
\begin{equation}
R(x) = \sum_{k=1}^{\infty}\frac{\mu(k)}{k}li(x^{1/k})
\end{equation}
where $\mu$ is the Mobius function and 
\begin{equation}
li(x) = \int_0^x \frac{dt}{\log t}
\end{equation}
Is there a proof in the literature of
\begin{equation}
\pi(x)=R(x)-\sum_{\rho}R(x^{\rho})
\end{equation}
where $\pi$ is prime counting function and the sum is over all complex zeros of $\zeta(s)$. The literature seems to treat it as fact while stating no proof is available - a strange situation.
Thanks in advance.

REPLY [2 votes]: Stopple, A Primer of Analytic Number Theory, proves a theorem which looks something like the one under discussion. On page 248, he has $$\pi(x)=R(x)+\sum_{\rho}R(x^{\rho})+\sum_{n=1}^{\infty}{\mu(n)\over n}\int_{x^{1/n}}^{\infty}{dt\over t(t^2-1)\log t}$$
You say that the literature treats your formula as a fact, but you give no citation. Where in the literature do you find your formula?<|endoftext|>
TITLE: Smoothness in Mersenne numbers?
QUESTION [7 upvotes]: The $n$-th Mersenne number $M_n$ is defined as
$$M_n=2^n-1$$
A great deal of research focuses on Mersenne primes.  What is known in the opposite direction about Mersenne numbers with only small factors (i.e. smooth numbers)?  In particular, if we let $P_n$ denote the largest prime factor of $M_n$, are any results known of the form
$$\liminf_{n\rightarrow \infty}\frac{P_n}{f(n)}= 1$$
for some function $f$?
I've only come across two (fairly distant) bounds so far.  If we consider even-valued $n$, then $M_n=M_{n/2}(M_{n/2}+2)$, so:
$$\liminf_{n\rightarrow \infty}\frac{P_n}{2^{n/2}}\leq 1$$
In the other direction, [1] shows that $P_n\geq 2n+1$ for $n>12$, so
$$\liminf_{n\rightarrow \infty}\frac{P_n}{2n}\geq 1$$
[1] A. Schinzel, On primitive prime factors of $a^n-b^n$, Proc. Cambridge Philos. Soc. 58
(1962), 555-562.

REPLY [3 votes]: I can give you a slightly better upper bound.  Recall that $2^n - 1 = \prod_{d | n} \Phi_d(2)$ where $\Phi_d$ is a cyclotomic polynomial.  Now,
$$\Phi_d(2) = \prod_{(k, d) = 1} (2 - \zeta_d^k) \le 3^{\varphi(d)}$$
so that in particular the largest prime factor of $2^n - 1$ is at most $3^{\varphi(n)}$.  By taking $n$ to be a product of the first $k$ primes and letting $k$ tend to infinity we have $\liminf_{n \to \infty} \frac{\varphi(n)}{n} = 0$, hence
$$\liminf_{n \to \infty} \frac{P_n}{c^n} = 0$$
for any $c > 1$.  In fact if $n$ is the product of the first $k$ primes then we should expect something like $3^{\varphi(n)} \approx 3^{ \frac{n}{\log k} }$ but this doesn't seem like a big improvement to me.<|endoftext|>
TITLE: Are homological knot invariants of finite type?
QUESTION [13 upvotes]: It is well known that, after a change of variables, the quantum knot invariants (Jones, HOMFLY, Kauffman, etc.) can be written as power series whose coefficients are finite type (i.e., Vassiliev) invariants.  But what about their categorifications?  
Specifically, do the generating polynomials of dimensions of the Heegard-Floer, Khovanov/Khovanov-Rozansky, etc., homology theories admit a change of variable such that the coefficients of the resulting power series are finite type invariants?

REPLY [4 votes]: If the answer were "yes" by just a change of variable, I think that would imply that the generating polynomials satisfy some kind of skein relation.  That's surely false.<|endoftext|>
TITLE: Do the homological dimension and cohomological dimension for a group agree?
QUESTION [6 upvotes]: Or equivalently, if $G$ is a group, do the projective and injective dimension of $Z$ (viewed as a $ZG$-module) agree?
Thanks!

REPLY [10 votes]: The cohomological and homological dimension of a group do not agree in general. For example, the homological dimension of the group $Z[\frac{1}{2}]$ is one, while its cohomological dimension is 2.  However, if |G| is countable, then cohomological dimension is either equal to the homological dimension or one dimension greater, see Proposition 2.4 in the following paper:
R. Bieri, Normal subgroups in duality groups and in groups of cohomological dimension 2. J. Pure Appl. Algebra 7 (1976), no. 1, 35–51. 
Also , cohomological dimension of a group is defined by the highest dimension n such that $H^n(G,M)$ is nonzero, where M is a G-module. It is important to use a G-module instead of the usual cofficient. For example, the cohomological dimension for any nontrivial knot group is 2, while its cohomology with $Z$ cofficient, or any G-module with trivial group action is always the same as the cohomology of $S^1$.<|endoftext|>
TITLE: Elementary problem with  Fibonacci numbers
QUESTION [8 upvotes]: I need help in proving one elementary result with Fibonacci numbers.
Prove that for $n>2$, the product $F_1 \cdot F_2 \cdots F_n$ cannot be a perfect square,
where $F_1 = F_2 = 1, F_{n+1}=F_n + F_{n-1}$.

REPLY [11 votes]: Take the largest prime $p$ up to $n$ (which is greater than $n/2$ by Bertrand's postulate) and notice that since the Fibonacci sequence is a divisibility sequence: $gcd(F_p,F_k)=F_{gcd(p,k)}=1$ for $k\neq p$ then if $\prod F_k$ is a perfect square so is $F_p$, but according to this theorem the only non-trivial square in the Fibonacci sequence is 144 and so you get the result.
Remark: The paper "Diophantine equations with products of consecutive terms in Lucas sequences" by F. Luca, T.N. Shorey determines all products of consecutive Fibonacci numbers which are perfect powers.<|endoftext|>
TITLE: Two versions of hamiltonian reduction
QUESTION [6 upvotes]: Given a symplectic manifold $X$ with nice $G$ action, equivariant momentmap $\mu$ and 
 invariant $\chi \in \mathfrak{g}^*$ which is a regular value of $\mu$.
There are two ways to form the Hamiltonian reduction.
What one usually does is take the levelset $X_\chi=\mu^{-1}(\chi)$ and quotient out the group
$X_\chi/G$. 
However one could also quotient out $G$ first and then define $(X/G)_\chi$ to be the set of all points represented by elements of $\mu^{-1}(\chi)$.
I have sometimes heard that these two procedures are equal. 
Is this true, or more precisely in what situations is it true? Is this written down somewhere? Are there counterexamples one should have in mind?
I am also interested in settings where one replaces spaces by possibly noncommutative Poisson algebras.

REPLY [2 votes]: I seem to recall that, in the algebraic setting (so working with (affine, say) algebraic varieties and their algebraic functions, rather than manifolds and smooth functions), it is necessary to assume that G is reductive for the two processes you propose to yield the same result.  On the one hand, one has $(A/J)^\mathfrak{g}$, while on the other, one has $A^\mathfrak{g}/J^\mathfrak{g}$.  If it happens that $J$ has an complementary submodule $J^c$ in $A$, then one can write $A=J \oplus J^c$ as a $G$-module, so that $A^\mathfrak{g}=J^\mathfrak{g} \oplus (J^c)^\mathfrak{g}$, and one can finally conclude that $(A/J)^\mathfrak{g} \cong A^\mathfrak{g}/J^\mathfrak{g}$.  Otherwise, the definitions can disagree.  I learned this argument in a course which didn't produce notes, so I cannot provide a reference....
I didn't think about it, but I think the same sort of thing is going on for $C^\infty$ manifolds:  you want a reductive group.  I could be mistaken.<|endoftext|>
TITLE: Projective Plane of Order 12
QUESTION [39 upvotes]: I asked this question on the new Theoretical Computer Science "overflow" site, and commenters suggested I ask it here.  That question is here, and it contains additional links, which I doubt I can embed here because I don't have enough reputation.  Anyway, here goes:
Objective: Settle the conjecture that there is no projective plane of order 12.
In 1989, using computer search on a Cray, Lam proved that no projective plane of order 10 exists. Now that God's Number for Rubik's Cube has been determined after just a few weeks of massive brute force search (plus clever math of symmetry), it seems to me that this longstanding open problem might be within reach.  I'm hoping this question can serve as a sanity check.
The Cube was solved by reducing the total problem size to "only" 2,217,093,120 distinct tests, which could be run in parallel.
Questions:

There have been special cases of nonexistence shown (again, by computer search). Does anyone know, if we remove those and exhaustively (cleverly?) search the rest, if the problem size is on the order of the Cube search? (Maybe too much to hope for that someone knows this....)
Any partial information in this vein?

REPLY [65 votes]: I am actually not aware of many results on planes of order 12 in the vein of what Lam et. al. did (I list the few I know of below). There seems to be a plethora of papers proving restrictions on the collineation group of a hypothetical such plane, but I am not aware of how any of these could be used to settle the existence problem.
Moreover, I am quite skeptical that disproving the existence of planes of order 12 by a computer search would help for the general theory much. Though it certainly would be nice to know, and if one actually found a plane of order 12, that would be quite exciting; but it's hard to gain deep insights from these combinatorial brute force searches.
Extending the approach by Lam et. al. to planes of order 12 is in principle possible. But probably still not feasible with today's computers, as the search space is a lot bigger than for order 10. Anyway, here are some reasons why I think that, and at the same time a sketch of things that would have to be done. But my personal belief is that one will need some substantially new ideas to make progress on this. Then again, only by actually trying to do it can one be sure... :)
From here on, I'll assume you are familiar with Lam's "The Search for a Finite Projective Plane of Order 10" and the notation used within.
A crucial point was the reduction of the (non-)existence to the value of certain weight enumerator coefficients $w_0$ to $w_{n^2+n+1}$ (a good exposition can be found in "On the existence of a projective plane of order 10" by MacWilliams, Sloane and Thompson). But the real breakthrough was when Assmus and Mattson proved that one only needs to know $w_{12},w_{15},w_{16}$ to determine all others. I'll refer to these as essential weight enumerator coefficients. 
Some steps towards this for order 12 have been executed in "Ternary and binary codes for a plane of order $12$" by Hall and Wilkinson. Yet many nice properties and theorems will be hard to recover for order 12. E.g. for orders of the form $8m+2$, one knows the $\mathbb{F}_2$-rank of the incidence matrix. Not so for order 12, where working with a ternary code is in some ways more "natural." In particular, the $\mathbb{F}_3$-rank of the incidence matrix is known, but, alas, working with a ternary code means losing the natural identification of codewords with point sets, so tons of new machinery would be needed to exploit the ternary code.  Thus I'll focus on the binary code case here.
Anyway, let's assume we reduced the number of essential weight enumerator coefficients as much as we can (Hall and Wilkinson pushed it down to 16; remember, for $n=10$ we had only 3). We must compute the essential coefficients.
According to Lam, for $n=10$ and the case $w_{12}$, they estimated, using a Monte-Carlo method (before doing it) that $4\times 10^{11}$ configuration had to be checked. I don't have a good means to compute a good estimate for $n=12$, but for that there are 16 coefficients to determine, and I'd hazard to guess that some of them are much, much harder than the three cases for $n=10$ put together. Several orders of magnitude. However, this is just gut feeling.
So let's assume we had somehow managed to overcome this and had computed all essential weight enumerator coefficients. We then would have the full weight enumerator at hand (and no projective plane arose as a byproduct of our search). Now, the hard part starts (corresponding roughly to the second half of Lam's paper), the one that took them 2 years for $n=10$: We have to somehow derive a contradiction (or construct a plane). A lot of ground work needs to be done (extending stuff from $n=10$), before one can even start writing code...
Ah well. To anybody who wants to try out this strategy on $n=12$, I would recommend to first try reproducing the $n=10$ result -- with modern computers it should be possible to do this much, much quicker than it took Lam et. al. originally (this verification might already interest some people on its own). Actually, at the very start, try it with even smaller examples ($n=6,8$), then go up.<|endoftext|>
TITLE: A question about iterated forcing
QUESTION [7 upvotes]: I'm trying to get a better grasp of iterated forcing, and I ran across the following problem:
0) Let $P_\alpha$ be posets in a c.t.m. $M$, $\alpha<\beta$, and for each $\alpha$ let $G_\alpha$ be $P_\alpha$ generic. Let $P$ be the finite support iteration of the $P_\alpha$. Then is there necessarily a $G$ which is $P$-generic such that for all $\alpha$, $G_\alpha\in M[G]$?
My guess for how to do this would be transfinite induction, but I run into some problems when I try that. Specifically, the successor step makes sense, but the limit step and the base case $(\beta=2)$ don't.
[For what it's worth, here's the successor step. Suppose the answer to (0) is "yes" for all $\beta<\delta$, $\delta\ge 2$. Let $P_\alpha$ be posets for $\alpha<\delta+1$, let $G_\alpha$ be $P_\alpha$-generic, let $P$ be their finite support iteration, and let $P'$ be the finite support iteration of the $P_\alpha$ for $\alpha<\delta$. Then $P$ is equivalent (in fact, isomorphic) to $P'\times P_\delta$. By the induction hypothesis, we have some $G'$ which is $P'$-generic such that for all $\alpha<\delta$, $G_\alpha\in M[G']$. Applying the induction hypothesis again (since $\delta>1$) we have some $G$ which is $P'\times P_\delta$-generic such that $G', G_\delta\in M[G]$. But then $M[G']\subseteq M[G]$, so $G_\alpha\in M[G]$ for all $\alpha<\delta$.] 
Related questions:
1) What happens if we restrict to c.c.c. posets?
2) What happens if we specify $\beta=2$?
3) What happens if we try to generalize to other kinds of supports?
I have a feeling I'm missing something obvious, but I've thought about this for a while now with no success.

REPLY [10 votes]: I'm not sure I really understand the question, because of the mixture of iteration and product, but I believe the following negative answer is independent of such issues because it uses $\beta=2$ and takes both factors (or iterands) to be Cohen forcing (so the product forcing is equivalent to the iteration).  Recall that a real (regarded, as usual in such contexts, as an element of $2^\omega$) is Cohen-generic over $M$ if and only if it belongs to every dense $G_\delta$ set coded in $M$.  Since $M$ is countable, the set of Cohen-generic reals over $M$ is comeager.  Now choose some real $z$ that codes a well-ordering of $\omega$ longer than the height of $M$.  Since both $\{x\in 2^\omega:x\text{ Cohen over }M\}$ and its translate $\{x\in 2^\omega:x\oplus z\text{ Cohen over }M\}$ (where $\oplus$ is pointwise addition mod 2) are comeager, they have an element $x$ in common.  Take $G_0$ to be the generic filter coding $x$ and take $G_1$ to be the generic filter coding $x\oplus z$.  No generic extension of $M$ (by $P_0\times P_1$ or by any other forcing) can contain both of these $G_\alpha$'s.  [Proof: Such a model would contain both $x$ and $x\oplus z$, and would therefore contain $z$.  Being a model of ZF, it would have to contain the ordinal isomorphic to the ordering of $\omega$ coded by $z$.  But the ordinals of a forcing extension of $M$ are the same as those of $M$, so this contradicts the choice of $z$.]<|endoftext|>
TITLE: Online math history lectures
QUESTION [22 upvotes]: This question is somewhat similar to this: Best online mathematics videos?
I'm using the word "history" loosely here. What I'm looking for are those lectures that put various mathematical developments in perspective by explaining their origins. There's something very insightful about seeing someone talk about the origins of a concept, that makes things click. Especially if he or she partook in the inception of that development.
So: where can I find such lectures online?

REPLY [3 votes]: I just came across these BBC podcasts the other day.  These are almost certainly more populist than you had in mind, but given the title of your question I thought I would throw it up here anyway.  The are ten podcasts.  Some expire in a day or so, which made me think posting it for those interested was a decent idea.

Newton and Leibniz
Leonard Euler
Joseph Fourier
Evariste Galois
Carl Friedrich Gauss 
The Mathematicians Who Helped Einstein
Georg Cantor 
Henri Poincare 
Hardy and Ramanujan 
10.Nicolas Bourbaki

http://www.bbc.co.uk/programmes/b00srz5b

"Professor of Mathematics Marcus du
  Sautoy reveals the personalities
  behind the calculations and argues
  that mathematics is the driving force
  behind modern science"<|endoftext|>
TITLE: Minimal words of length n
QUESTION [7 upvotes]: Let W be a finite word on a two symbol alphabet {0,1}; let us say that W is maximal if it is the last item in the list of all its cyclic permutation (ordered lexicographically).
So, for instance:
{0,1}         are the maximal words of length 1;
{00, 10, 11}  are the maximal words of length 2;
{000, 100, 110, 111}    are the maximal words of length 3;
{0000, 1000, 1010, 1100, 1110, 1111} are the maximal words of length 4;
{00000, 10000, 10100, 11000, 11010, 11100, 11110, 11111} are the maximal words of length 5; ...
und so weiter.
Let 
k(n):= number of maximal words of length n
Is there some formula for k(n)?

REPLY [6 votes]: please see wikipedia :  http://en.wikipedia.org/wiki/Necklace_(combinatorics)
It represents a structure with n circularly connected beads of up to k different colors or numbers.<|endoftext|>
TITLE: Total energy of the universe
QUESTION [11 upvotes]: In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, "because the positive energy of matter is cancelled out by the negative energy of the gravitational field".
But I can't find anything concrete to substantiate this claim.
As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius R using Newton's Laws and threw in $E=m{c}^2$ for energy of the sphere, and it was by no means obvious that the answer is zero !
So, my questions:

What is the basis for the claim – does one require general relativity, or can one get it from Newtonian gravity ?

What conditions do you require in the model, in order for this to work ?

Could someone please refer me to a good paper about this ?

REPLY [6 votes]: When people claim that total energy is zero in a gravitational theory, they usually mean something kind of trivial. In an ordinary field theory, we can define the stress-energy tensor by varying the Lagrangian with respect to the metric. But in a gravitational theory, the metric is a dynamical field, so this variation just gives an equation of motion for the metric that should be zero on a classical solution.
What I've just said sounds fairly classical; the Wheeler-DeWitt equation is a version of this idea in the quantum context.
At low energies, where gravity is weak, you can of course just talk about the energy of a configuration of matter, independent of gravity, but this will not be conserved in general relativity. (A cosmological constant is a standard example where you can see that it's hard to make sense of energy conservation in the presence of gravity.)
As the others have said, the ADM energy is sometimes a more useful definition of energy in GR.<|endoftext|>
TITLE: Littlewood-Richardson rule and commutativity morphism
QUESTION [10 upvotes]: Background
Irreducible finite dimensional representations 
of the group $GL_n$ are parameterized by the highest weights,
that is by nonincreasing sequences of integers
$$
\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n. 
$$
Let us restrict to the case when $\lambda_n \ge 0$.
Then one can encode a highest weight by a Young diagram
with $\lambda_i$ boxes in the $i$-th row.
The Littlewood-Richardson rule describes the decomposition
of a tensor product $V^\lambda \otimes V^\mu$ into a direct
sum of irreducibles. It says that the multiplicity of $V^\nu$
in the tensor product $V^\lambda \otimes V^\mu$ is equal 
to the number of so-called Littlewood-Richardson tableux
in the skew-diagram $\nu\setminus\lambda$ of weight $\mu$.
See 
Littlewood-Richardson rule for precise definitions.
Note that the rule is not symmetric with respect to $\lambda$ and $\mu$!
On the other hand, the category of representations of $GL_n$ has 
a commutativity morphism: it is a bifunctorial isomorphism
$$
c_{V,W}:V\otimes W \to W\otimes V,
\qquad
v\otimes w \mapsto w\otimes v.
$$
Question

Is there a possibility to make Littlewood-Richardson rule 
  compatible with the commutativity morphism?

To be more precise, 
is there a way to associate with every Littlewood-Richardson tableau
in a skew diagram $\nu\setminus\lambda$ of weight $\mu$ an embedding
$V^\nu \to V^\lambda\otimes V^\mu$ such that the composition
$V^\nu \to V^\lambda\otimes V^\mu \stackrel{c}\to V^\mu\otimes V^\lambda$
is the embedding associated to an appropriate Littlewood-Richardson
tableau in a skew diagram $\nu\setminus\mu$ of weight $\lambda$?
Let me emphasize that I am asking about the $GL_n$ case, although 
this question has sense for any reductive group.

REPLY [8 votes]: Let $LR(\mu/\lambda;\nu)$ be the set of Littlewood-Richardson tableaux of shape $\mu/\lambda$ and weight $\nu$. Then there is a canonical bijection between $LR(\mu/\lambda;\nu)$ and $LR(\mu/\nu;\lambda)$, presented in a paper by Pak and Vallejo ("Fundamental Symmetry map"), in a paper by Danilov and Koshevoi ("Commutor"), and in a paper by Henriques and Kamnitz.
Is this what you want ?
In the paper by Pak and Vallejo actually two "Fundamental Symmetry maps" are presented. Danilov and Koshevoi show that they coincide, and that they coincide with their "commutor", and with the map defined by Henriques and Kamnitzer.
The references:
Igor Pak and Ernesto Vallejo.
Reductions of Young tableau bijections
SIAM J. Discrete Math.  24  (2010),  no. 1, 113--145.
doi: 10.1137/070689784
(Also http://arxiv.org/abs/math/0408171)
V.I. Danilov and G.A. Koshevoi
The Robinson-Schensted-Knuth correspondence and the bijections of commutativity and associativity.
2008 Izv. Math. 72 689
doi: 10.1070/IM2008v072n04ABEH002415
A. Henriques and J. Kamnitzer
The octahedron recurrence and $gl_n$-crystals
Adv. Math. 206:1 (2006), 211-249<|endoftext|>
TITLE: A sequence of generic filters that does not come from an iteration
QUESTION [7 upvotes]: Fix a countable transitive model $M$ of ZFC.
In my answer to this question I indicated that there are forcing iterations
$((Q_\alpha:\alpha\leq\omega),(\dot P_\alpha:\alpha<\omega))$ in $M$ and sequences
$(G_\alpha:\alpha<\omega)$ of filters such that the following happens:
Each $G_\alpha$ is a filter in the evaluation of $\dot P_\alpha$ with respect to the filter $G_0*\dots*G_{\alpha-1}$ and $G_\alpha$ is generic over $M[G_0,\dots,G_{\alpha-1}]$
(call such a sequence $(G_\alpha:\alpha<\omega)$ a sequence of generics), 
but there is no $Q_\omega$-generic filter over $M$ whose 
$\alpha$-th projection is $G_\alpha$ for all $\alpha<\omega$.  
An example can be obtained as follows:
Take the countable support iteration of Sacks forcing
(or any other nontrivial $\omega^\omega$-bounding proper forcing notion) of length $\omega$
(i.e., the supports are actually everything, but this doesn't matter).
This forcing adds no Cohen real.  
Compare this to the finite support iteration of the same forcing notions.
This iteration does add a Cohen real.  The Cohen real is coded by the sequence of generics
and hence this sequence of generics does not come from a generic filter for the countable support iteration mentioned before.  This sequence of generics is not even contained in
a forcing extension obtained using the countable support iteration.
Now here are two questions:
1) Is there an example of a sequence of generics (of length $\omega$) that cannot come from
any iteration of the $\dot P_\alpha$?  
I am asking here for iterations where the finite initial segments are as usual (just plain iteration) and we choose whatever ideal for the supports, including all finite subsets of the index set.
But I am open to more general forms of iteration.  For example take a large forcing notion
$Q$ along with commuting complete embeddings of the $Q_\alpha$, $\alpha<\omega$.
This would be an iteration of the $\dot P_\alpha$, too, the most general one that I can think of right now.
2) Is there an example of a sequence of generics over $M$ that is not contained in any countable transitive extension of $M$ with the same ordinals as $M$ that is a model of ZFC?
Obviously, a positive answer to 2) solves 1) as well.

REPLY [8 votes]: There are a number of interesting things to say. 
The answer to your first question is yes. Suppose that $M$ is a countable transitive model of set theory and we have a forcing iteration $P_\omega$ in $M$ of length $\omega$, forcing with, say, Cohen forcing $Q_n$ at stage $n$. Let $z$ be any real that cannot be added by forcing over $M$, such as a real that codes all the ordinals of $M$. This real cannot exist in any extension of $M$ to a model of ZFC with the same ordinals. Now, suppose that $G$ is any $M$-generic filter for the iteration, with $G_n$ being the stage $n$ generic filter. Let $H_n$ be the filter that results from $G_n$ by changing the first bit so as to agree with $z(n)$. That is, we change a single bit at each stage. The resulting sequence $\langle H_n | n\in\omega\rangle$ will be generic at every stage, since only finitely many bits are changed by a given stage, but the whole sequence computes $z$, which cannot be added by forcing. 
Second, a similar phenomenon occurs even just with 2-step product forcing:
Theorem. If $M$ is a countable transitive model of ZFC, then there are two $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common extension to model of ZFC with the same ordinals. 
The proof is to build $c$ and $d$ in stages. Fix a real $z$ which cannot exist in any extension of $M$ with the same ordinals, and enumerate the dense sets of $M$ by $D_0, D_1$ and so on. Build $c$ and $d$ in zig-zag fashion: first provide $c_0$ meeting $D_0$, and $d_0$ all $0$s of the same length as $c_0$, followed by the first digit of $z$. Now extend $d_0$ to $d_1$ meeting the dense set, adding all $0$s to $c_0$ making $c_1$ of the same length, and adding one more bit of $z$. And so on. The point is that we ensure that each of $c$ and $d$ is $M$-generic, but together, they reveal the coding points of $z$. So no model extending $M$ with the same ordinals can have both $c$ and $d$, for then it would have $z$.
Third, this is essentially the only kind of obstacle, for there is a positive result here. The following theorem is proved in a paper with G. Fuchs, myself and J. Reitz on the topic of set-theoretic geology:
Theorem. If $M$ is a countable transitive model of set theory, and $M[G_n]$ is a sequence of generic extensions of $M$ by forcing $G_n\subset P_n\in M$ of bounded size in $M$, such that the extensions are finitely amalgamble, in the sense that any finitely many of the $M[G_n]$ have a common forcing extension $M[H]$, then there is a single forcing extension $M[H]$ containing all $M[G_n]$. 
I'll try to post a proof sketch later, but the main idea is to perform a very large combination of forcing, and then perform surgey so as to replace certain coordinate generics with $G_n$, in such a way that the resulting extension can see only finite fragments of the sequence $\langle G_n | n\lt\omega\rangle$, without being able to construct the whole sequence. 
A special case of this theorem answers your second question, in a sense, for if we have a sequence of extensions $M\subset M[G_0]\subset M[G_1]\subset\cdots$, then these extensions are finitely amalgamable, and so there is indeed a common extension $M[H]$ containing every $M[G_n]$. This extension, however, is not an $\omega$-iteration of the forcing notions in your iteration, and in general we cannot expect that the sequence $\langle G_n | n\lt\omega\rangle$ is in $M[H]$, for the reasons described above.

REPLY [8 votes]: Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product.  So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$.  Fix some $z\in 2^\omega$ coding an ordinal larger than the height of $M$.  Define $y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$.  Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$.  The sequence $\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.<|endoftext|>
TITLE: Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
QUESTION [18 upvotes]: I'm fearful about putting this forward, because it seems the answer should be elementary.  Certainly, the Weak Approximation Theorem allows every system of simultaneous inequalities among archimedean absolute values to be satisfied.  But equality combined with inequality?

REPLY [5 votes]: One more comment on this frequently answered question: Let $a_0 x^{2n} + a_1 x^{2n-1} + \cdots + a_{n} x^{n} + \cdots a_1 x + a_0$ be a palindromic polynomial with real coefficients and $2k$ isolated roots on the unit circle. Then any sufficiently small perturbation of this polynomial to another real palindromic polynomial also has $2k$ roots on the unit circle.
Proof: Notice that $e^{\pm i \theta}$ is a root of this polynomial if and only if $a_0 \cos (n \theta) + a_1 \cos ((n-1) \theta) + \cdots + a_{n/2-1} \cos \theta + a_0/2=0$. Write $f(\theta)$ for the right hand side of this equation. Our hypothesis is that $f$ has $k$ isolated roots, $\theta_1$, $\theta_2$, ..., $\theta_k$. Then we can find $\epsilon>0$ and disjoint intervals $(a_i, b_i)$ around each $\theta_i$ such that (1) We have $f((a_i, b_i)) \supseteq (-\epsilon, \epsilon)$. (2) On $(a_i, b_i)$, we have $|f'|<\epsilon$ (3) Off of the $(a_i, b_i)$, we have $|f|>\epsilon/2$.
Then, if our perturbation is small enough that $f$ and $f'$ change by less than $\epsilon/4$ everywhere, then the perturbed $f$ still has one root in each $(a_i, b_i)$, and no roots elsewhere. QED
Why do I point this out? Take any of the above examples and perturb its coefficients slightly, while keeping them rational and palindromic. Then you get another example! This observation destroys most attempts to classify such polynomials by number theoretic criteria.<|endoftext|>
TITLE: Ramanujan's $\tau(n)$ and continued fractions
QUESTION [33 upvotes]: In D.H. Lehmer's paper "Ramanujan's function $\tau(n)$, (Duke J. Math v. 10 1943, pp. 483-492), Lehmer states the Ramanujan conjecture $|\tau( p )|< 2p^{11/2}$, so that $p^{-11/2}\tau( p )=2\cos(\theta_p)$ where $\theta_p$ is real.  He observes "It is interesting to note that $2\cos(\theta_{11})=1.000872909\ldots$" ie that $\theta_{11}$ is very nearly $\pi/3$.
In fact, Mathematica computes that 
$\theta_{11}/\pi=0.333172889904775\ldots$  There's nothing special about the expansion as a decimal, so I looked instead at the continued fraction expansion, which Mathematica can also do.  One finds it is
$
\{0, 3, 692, 5, 4, 1, 2, 3, 1, 2,\ldots\}
$
The $3$ is expected, as is the fact that the next term is large.  But that it's 692 is surprising to me, recalling that $\tau(n)\equiv \sigma_{11}(n) \bmod 691$ (which is due to Ramanujan himself).  
I've looked at other weights and levels with no insight.  I've also looked nontraditional continued fraction expansions with alternating $\pm$ signs, and numerators other than 1, to see if I could make 691 (v. 692) appear instead.  

The 'eigen-angles' $\theta_p$ clearly carry deep arithmetic information coming from Galois representations, as others on this site can explain much better than I.  But can something be proven about these continued fraction expansions?  Anything that's true is likely to be quite deep, so a more realistic (but vague) question is: 

Are there other examples of this kind of 'numerology'?

REPLY [13 votes]: Since the OP asked for other examples of this kind of numerology,I will give another one to support his observation
The function $\cos(\theta_{11})$  has the following closed form
$\cos(\theta_{11})=\frac{\sigma_{1}(11)}{22\sqrt{11}}-\frac{12\sum_{k=1}^{10} (2178k^2-572k^3+35k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag1$
and continued fraction
$\cos(\theta_{11})=\{0;1,1,572,3,2,1,2,1,2,2,4,3,1,6,\dots\}\tag2$
where we clearly see $572$ appearing both as a coefficient in the sum $(1)$ and largest partial quotient in the first few partial quotients of the continued fraction $(2)$
Is this a coincidence?
Edited:03 Sep 2017
And also 
$2\cos(\theta_{11})=\frac{\sigma_{1}(11)}{11\sqrt{11}}-\frac{12\sum_{k=1}^{10} (4356k^2-1144k^3+70k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag3$
with the following continued fraction
$2\cos(\theta_{11})=\{1;1145,1,1,2,6,2,2,1,1,1,1,1,2,3,\dots\}\tag4$
where $1145$ is the 0th partial qotient in the continued fraction $(4)$ and $1144$ appears as a coefficient in the formula $(3)$
Remark: Identity $(1)$ is a special case of the identity found in A000594 OEIS when $n=11$
$\tau(n)=n^4\sigma_{1}(n)-24\sum_{k=1}^{n-1} (18n^2k^2-52nk^3+35k^4)\sigma_{1}(n-k)\sigma_{1}(k)$<|endoftext|>
TITLE: What immersed closed curves on the double-torus are non-trivial when lifted to the unit tangent bundle?
QUESTION [5 upvotes]: Take an equator on the two sphere $S^2$ and parametrize it by arc-length obtaining a closed loop $\alpha: S^1 \to S^2$.  The curve $(\alpha,\alpha'):S^1 \to T^1S^2$ in the unit tangent bundle of $S^2$ is homotopically non-trivial.
However if you consider the concatenation $\beta$ of two copies of $\alpha$, you can take one copy and turn it about a diameter passing through two points of $\alpha$ so that it is now a copy of $\alpha$ but traversed in the opposite sense.  In other words, the curve $(\beta,\beta') \in T^1S^2$ is homotopically trivial.
Does this occur on other compact orientable surfaces?
On the torus the answer is no.
What about on the double-torus (which is a hyperbolic surface)?
Also, does anybody have references for the above statements about the sphere and the torus?  In particular, are they correct?
I've come across these problems while trying to picture the lift of a general geodesic flow to the universal covering space of $T^1M$ where $M$ is a compact orientable surface (in particular, what do homotopically trivial closed geodesics look like?).

REPLY [8 votes]: If $M$ is not $S^2$ or $RP^2$, then $\pi_1(T^1M)$ does not have elements of finite order
(in particular a double non-contractible loop is also non-contractible). Indeed, consider the long exact sequence of our fibration $E=T^1M\to M$:
$$
 \dots\to \pi_2(M)\to \pi_1(F)\to\pi_1(E)\to\pi_1(M)\to\dots
$$
where $F$ is a fiber (a circle).
Note that $\pi_2(M)=0$ (since the universal cover of $M$ is the plane), hence the arrow $\pi_1(F)\to\pi_1(E)$ is injective. Hence the kernel of the arrow $\pi_1(E)\to\pi_1(M)$ is isomorphic to $\pi_1(F)$ which is $\mathbb Z$. So this kernel does not contain elements of finite order. And if a non-kernel element has a finite order, then so does its image in $\pi_1(M)$. But $\pi_1(M)$ has no elements of finite order (e.g. due to existence of a nonpositively curved metric on $M$).<|endoftext|>
TITLE: Which group does not satisfy the Tits alternative?
QUESTION [24 upvotes]: A group is said to satisfy the Tits alternative if every finitely generated subgroup of $G$ is either virtually solvable or contains a nonabelian free subgroup.
Tits proved this for linear groups, and a MathSciNet search gives 38 papers with "Tits alternative" in the title (and 154 papers quoting Tits's original paper), so certainly a lot of groups do enjoy this property.
What then is an example of a group which does not satisfy the Tits alternative?

REPLY [12 votes]: $A_5\wr \mathbb{Z}$<|endoftext|>
TITLE: How to decompose a composition of representations?
QUESTION [9 upvotes]: Background
I would like to know if there is some slick machinery to solve the following representation-theoretic problem.
Let $\left(V,\langle-,-\rangle\right)$ be a finite-dimensional real inner product space and let $\mathfrak{so}(V)$ denote the Lie algebra of skewsymmetric endomorphisms.
Let $\mathfrak{g}$ be a Lie algebra and $\rho: \mathfrak{g} \to \mathfrak{so}(V)$ be a Lie algebra homomorphism.  This makes $V$ into an orthogonal $\mathfrak{g}$-module.
Let $\sigma: \mathfrak{so}(V) \to \operatorname{End}(\Delta)$ denote an irreducible spinor representation of $\mathfrak{so}(V)$.  Then the composition $\sigma \circ \rho : \mathfrak{g} \to \operatorname{End}(\Delta)$ of the representations turns $\Delta$ into a $\mathfrak{g}$-module.
Question

Is there a nice description of this module?  And in particular about its decomposition into irreducibles?

Contextualisation
This sort of problem arises often in my line of work; although usually with $\mathfrak{g}$ itself being an orthogonal Lie algebra and $V$ itself an irreducible spinor representation.  In other words, I end up considering spinors of spinors and this is useful in determining the structure of representations ("supermultiplets") of Lie superalgebras appearing in supersymmetric field theories.  I usually work this out "by hand" (or using LiE) in terms of roots and weights, but was wondering whether there was some more conceptual machinery available.

REPLY [4 votes]: I just came across this question, so apologies if it is no longer relevant to your research, but I figured I'd also point out my answer to this MO question which might be helpful: A Decomposition of the 'Spin' Representation of $\mathfrak{sl}_2$
There I describe a computational method for your question in the case of odd-dimensional irreducible representations of $\mathfrak{su}_2/\mathfrak{sl}_2$ using basic character theory. The first part of this method, wherein one lines up the weights, can be done for any orthogonal representation. The second part (using the special 'factorization' of the odd spinor representations) only works when $V$ is odd-dimensional; when $V$ is even-dimensional there is no analogous 'factorization' of the spinor irreps but nevertheless with a little work one can modify the odd-dimensional case slightly to handle this (by using the relationships between the spinor irreps of even and odd-dimensional orthogonal groups under restrictions).<|endoftext|>
TITLE: Topological degree of polynomial maps.
QUESTION [8 upvotes]: The $\mathbb{Z}_2$ topological degree of a (non-constant) polynomial in one variable, clearly, coincides with its degree as a polynomial, $\mod 2$.  
Consider further a polynomial self-mapping $F$ on $\mathbb{R}^2$, and assume it is a proper map (in case, even more generally a map in higher dimension) 

Is there a short way to decide what's
  the parity of the topological degree of $F$?

E.g. it's odd if $F$ is an odd map, or more generally, if $F$ can be transformed into an odd map by a proper homotopy. Actually: is there a short way to understand if a polynomial map is a proper map? What about the case of a gradient map (I mean, the gradient of a polynomial $f\in\mathbb{R}[x,y]$)? 
I'm somehow confident that there may be a simple criterion known, at least in $\mathbb{R}^2.$  After all, what is required is just a one-bit information (well this argument doesn't convince me either).

REPLY [3 votes]: Here is an idea, I'm not 100% confident that it makes sense in all cases, but I'll try it anyway. Assuming that $f(\mathbb{R}^2)$ is 2-dimensional, the degree mod 2 of your map is the cardinality of the preimage of a generic point. If your components have degree respectively d and e, then Bezout gives you a preimage size of de. 
That's projective solutions. In generic cases, you would expect no solutions at infinity. Unfortunately as David points out below, you may have an odd number of solutions at infinity.
So in the case when there's nothing at infinity, we have solutions over the complexes, but non-real solutions will come in complex conjugate pairs, so in that nice case the degree mod 2  is given by de.<|endoftext|>
TITLE: evil properties, higher category theory and well-chosen tensor products
QUESTION [12 upvotes]: Let's start with the following random example: If $F$ is a presheaf, then for every chain of open subsets $U \subseteq V \subseteq W$, the morphisms $F(W) \to F(V) \to F(U)$ and $F(W) \to F(U)$ coindice. But this may be an evil (nlab link) definition, especially when the values of $F$ are categories (which occurs in my current research). So we should just impose that these morphisms are equivalent. But then we arrive at compatibility conditions between these equivalences, which are, again, equalities, and may be evil. So what is the consequence of this: Should every mathematical theory take place in a $\infty$-category? Or is 'real' mathematics basically evil? I know this question is quite imprecise, but currently I just don't see where and why this process of increasing the "depth" of category theory should end. Anyway, I have to admit that my knowledge of higher category theory is very, very rudimental.
Here is a evil question, which is part of my confusion: Let $Ring$ denote the category of rings, ring homomorphisms and $AbCat$ denote the category of abelian categories, functors. Then $Ring \to AbCat, A \to Mod_A$ should be a functor, right? If $A \to B$ is a ring homomorphism, just tensor with $B$ over $A$. But then functoriality is only satisfied up to equivalence ($2$-isomorphism) in $AbCat$. To get a honest functor, we might mod out these equivalences to get a $1$-category $\tilde{AbCat}$. But I don't think that this is the most natural way to handle this. Or we may define quasi-functors (cf. the presheaf example). Anyway, may we think of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M \otimes_A B$ carefully? I think the latter is interesting, although I know that nobody really cares about such evil equalities like $(M \otimes_A B) \otimes_B C = M \otimes_A C$.

REPLY [13 votes]: There are a lot of questions here, but I'll try to answer them all.

Should every mathematical theory take place in a ∞-category? Or is 'real' mathematics basically evil?

I would say that all mathematics should take place in its natural context.  Sometimes you have things that are sets where equality makes sense, like an ordinary presheaf, and then you work in a 1-category.  Sometimes you have things where only isomorphism makes sense, like a presheaf of categories, and then you work in a 2-category.  Etc.
It is true that any n-category for finite n can be considered a special case of an ∞-category with only identity cells above n, so in this degenerate sense all n-categories are ∞-categories, and thus one might say that "all mathematics takes place in an ∞-category" — at least if one believes that all mathematics takes place in an n-category for some n!  But even that is not clear, e.g. some mathematics naturally takes place in other categorical structures, such as a double category or a proarrow equipment.  Some mathematics uses no category theory at all (at least as far as anyone has noticed so far), and so it would be a stretch to say that it takes place in any sort of category.

Anyway, may we think of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M\otimes_A B$ carefully?

I would say qualified yes, yes, and yes, respectively.  You can think of it as a usual functor as long as doing so doesn't cause you to think that it behaves in any way that a pseudofunctor doesn't!  Which is sort of a vacuous statement, but the point is that pseudofunctors really shouldn't be a very scary concept (as opposed to a technical definition, which might be a bit complicated, though cf. Harry's comment) — they really are just like ordinary functors, except that you're dealing with things (e.g. categories) for which it doesn't really make sense to ask morphisms to be equal, only isomorphic.
On the other hand, the "higher category theoretic" fact that pseudofunctors are not all strict functors is very important.  I believe that Benabou, the inventor of bicategories, once said that the important thing about bicategories is not that they themselves are "weak," but that the morphisms between them are weak.  In particular, although every bicategory is equivalent to a strict 2-category, not every pseudofunctor between bicategories is equivalent to a strict functor.
But on the third hard, it is true that any pseudofunctor with values in the 2-category Cat is equivalent to a strict functor.  In the language of fibrations, this says that any fibration is equivalent to a split one.  Tyler mentioned one construction of an equivalent strict functor in the case of modules and tensor products.  There is also a general construction which, applied to the case of modules, will replace $Mod_A$ by a category whose objects are pairs (M,φ) where M is an R-module and φ:R→A is a ring homomorphism.  We regard such a pair as a formal representative of $M\otimes_R A$ and define morphisms between them accordingly, to get a category eequivalent to $Mod_A$.  Now the extension-of-scalars functor $\psi_!:Mod_A \to Mod_B$ is represented by the functor taking a pair (M,φ) to (M,ψφ), which is strictly functorial since composition of ring homomorphisms is so.<|endoftext|>
TITLE: A question on the prime divisors of p-1
QUESTION [18 upvotes]: For each positive integer n we may define the convergent sum $$ s(n)=\sum_{p}\frac{(n,p-1)}{p^2} $$
where the summation is over primes p and $(a,b)$ denotes the greatest common divisor of a,b.
It is immediate to deduce that s(n) is bounded on average: 
Using $\sum \limits_{d|a, d|b}\phi(d)=(a,b)$ and inverting the order of summation we get
$s(n)=\sum_{d|n}\phi(d) a_d$ where $a_d=\sum_{p \equiv 1 (mod  \ d)}p^{-2}  $
Ignoring the fact that we sum over primes we get the bound $a_d \ll \frac{1}{d^2}$ which leads to $$\sum_{n \leq x}s(n) \ll x \sum_{d \geq 1}\frac{\phi(d)a_d}{d}=O(x) $$
and $$s(n) \ll  \exp(\sum_{p|n}\frac{1}{p}).$$ The last inequality  means that $s(n)$ stays bounded if $\omega(n)$ is 
bounded. Towards the other direction, it seems fair to expect 
that $s(n)$ grows to infinity if $\omega(n)$ is large in some quantitative sense, say 
$\omega(n) \geq (1+\epsilon) \log \log n$. Taking into account that the contribution to the sum $s(n)$ of the primes $p$ that satisfy $(p-1,n) \leq \frac{p}{\log p}$  is bounded, since $\sum_{p}\frac{1}{p \log p}$ converges, we see that $ s(n)=s'(n)+O(1)$ where $s'(n)=\sum_{ (p-1,n)>\frac{p}{\log p}} \frac{(n,p-1)}{p^2}$ We are therefore led  to the question as to whether a condition of the form $\frac{\omega(n)}{\log \log n}-1 \gg 1$ can guarantee that $s'(n) \to +\infty$ Are there any non-trivial techniques that can be used to answer this question ?

REPLY [9 votes]: Your guess that $s(n)$ gets large if $\omega(n)$ is large is not correct.
It is possible for $n$ to have many primes, and for $s(n)$ still to be small. 
This can be seen from some of the work in your question.  As you note $s(n) =\sum_{d|n} \phi(d) a_d$ where $a_d =\sum_{p\equiv 1\pmod d} p^{-2} \ll 1/d^2$.  Therefore 
$$ 
s(n) \ll \sum_{d|n} \frac{1}{d} \le \prod_{p|n} \Big(1-\frac 1p\Big)^{-1}. 
$$ 
If now every prime factor of $n$ exceeds $\log n$, then (since $\omega(n) \le \log n$ 
trivially) we have 
$$ 
s(n) \ll \Big(1-\frac{1}{\log n} \Big)^{-\log n} \ll 1.
$$ 
Thus $n$ can have about $\log n/\log \log n$ prime factors, all larger than $\log n$ 
and still $s(n)$ would be $\ll 1$.<|endoftext|>
TITLE: A Hölder continuous function which does not belong to any Sobolev space
QUESTION [19 upvotes]: I'm seeking a function which is Hölder continuous but does not belong to any Sobolev space.
Question: More precisely, I'm searching for a function $u$ which is in $C^{0,\gamma}(\Omega)$ for $\gamma \in (0,1)$ and $\Omega$ a bounded set such that $u \notin W_{loc}^{1,p}(\Omega)$ for any $1 \leq p \leq \infty$. Take $\Omega$ to be bounded, open.
My first guess is to do a construction with a Weierstrass function. I know this is differentiable 'nowhere' but that doesn't convince me it isn't weakly differentiable in some bizarre way. Hopefully someone knows of an explicit example.

REPLY [5 votes]: I believe that the Cantor ternary function (aka devil's staircase) is a simple example.<|endoftext|>
TITLE: Analysis from a categorical perspective
QUESTION [20 upvotes]: I have not studied category theory in extreme depth, so perhaps this question is a little naive, but I have always wondered if analysis could be taught naturally using categories. I ask this because it seems like a quite a lot of topological and group theoretic concepts can be defined most succinctly using categorical concepts, and the categorical definitions are more revealing. So my question is: (1) Is it possible/beneficial to teach analysis using category theory? and (2) Are there any good textbooks that use this method?

REPLY [27 votes]: I hesitate to let this out, but there's always this cute little note that I learned from another MO answer (I don't know which one): https://www.maths.ed.ac.uk/~tl/glasgowpssl/banach.pdf.  Maybe this will satisfy your curiosity, but I maintain that it takes a warped mind to identify such a categorical formulation of integration as the "right" way to think about integrals.

The advantage of categorical thinking in my view is that it helps to organize computations and arguments involving several different kinds of structures at the same time.  For instance, (co)homology is all about capturing useful invariants associated to a complicated structure (e.g. a geometric object) in a much simpler structure (e.g. an abelian group).  When we want to determine how the invariants behave under certain operations on the complicated structure (e.g. products, (co)limits) it helps to have a theory already set up to tell us what will happen to the simpler structure.  That's where category theory comes into its own, and instances of this paradigm are so ubiquitous in algebra and topology that category theory has taken on a life of its own.  It seems that people working in those areas have found it convenient to build categorical constructions into the foundations of their work in order to emphasize generality (one can treat algebraic varieties and solutions to diophantine equations on virtually the same footing), keep track of different notions of equivalence (e.g. homotopy versus homeomorphism), build new kinds of spaces (e.g. groupoids), and to achieve many other aims.
In many kinds of analysis, this kind of abstraction isn't necessary because there's often only one structure to keep track of: $\mathbb{R}$.  When you think about it, analysis is only possible because we are willing to seriously overburden $\mathbb{R}$.  Take, for example, the expression "$\frac{d}{dt}\int_X f_t(x) d\mu(x)$" and consider all of the different ways real numbers are being used.  It is used as a geometric object (odds are X is built out of some construction involving the real numbers or a subspace thereof), a way to give $X$ additional structure (it wouldn't hurt to guess that $\mu$ is a real valued measure), a parameter ($t$), and a reference system ($f$ probably takes values in $\mathbb{R}$ or something related to it).  In algebraic geometry, one would probably take each of these roles seriously and understand what kind of structure they are meant to bring to the problem.  But part of the power and flexibility of analysis is that we can sweep these considerations under the rug and ultimately reduce most complications to considerations involving the real numbers.
All that being said, the tools of category theory and homological algebra actually have started to make their way into analysis.  Because of the fact that analysts generally consider problems tied to certain very specific kinds of structure, they have historically focused on providing the sharpest and most detailed solutions to their problems rather than extracting the crude, qualitative invariants for which cohomological thinking is most appropriate.  However, as analysts have become more and more attuned to the deep relationships between functional analysis and geometry, they have turned to ideas from category theory to help keep things organized.  K-theory and K-homology have become indispensable tools in operator theory; there is even a bivariant functor $KK(-,-) $ from the category of C-algebras to the category of abelian groups relating the two constructions, and many deep theorems can be subsumed in the assertion that there is a category whose objects are C-algebras and whose morphism spaces are given by $KK(A,B)$.  Cyclic homology and cohomology has also become extremely relevant to the interface between analysis and topology.
So ultimately I think it all comes down to what kinds of subtleties are most relevant in a given problem.  There is just something fundamentally different about the kind of thinking required to estimate the propagation speed of the solution operator for a nonlinear PDE compared to the kind of thinking required to relate the fixed point theory in characteristic 0 of a linear group acting on a variety to that in characteristic p.<|endoftext|>
TITLE: Why is the symmetric monoidal structure on invertible modules strict?
QUESTION [14 upvotes]: Let $N$ be an object in a symmetric monoidal category.  Then the braid map $N\otimes N\to N\otimes N$ is almost never the identity, and this is the obstruction to making a symmetric monoidal category into a "strict" symmetric monoidal category, in which the functor $\otimes$ is commutative on the nose.  For example, when $\otimes$ is just the categorical product in the category of sets (or any other concrete category), this is the map $(x,y)\mapsto(y,x)$ which is almost never the identity.
However, one case in which the braid map is the identity is in the category of invertible modules over a commutative ring $R$, as a full subcategory of all modules.  Indeed, the braid map $R\otimes R\to R\otimes R$ is the identity, and any invertible module $I$ is locally isomorphic to $R$, so the braid map $I\otimes I\to I\otimes I$ is locally equal to the identity and hence equal to the identity.
What I'd like to have is a more conceptual explanation for why the braid map is the identity for invertible modules, which does not use the fact that they are locally free (indeed, I'm interested in this because I want to use this to prove invertible modules are locally free in a more general setting).  Unfortunately, the proof cannot just be abstract nonsense involving invertibility--for example, if we work with graded modules over a commutative ring instead of ordinary modules and use the usual sign conventions, then the braid map will be -1 rather than 1 on invertible modules concentrated in odd degrees.  Does anyone know of a better explanation, or know a reason I shouldn't expect there to be one?
EDIT: Inspired by Charles's answer, here's a closely related question.  I'm really interested in invertible objects in the derived category, and in the derived category dualizable objects can be represented by finite chain complexes of finite-rank projective modules.  Over a local ring, then, all dualizable objects have an Euler characteristic which is an integer (as opposed to an arbitrary element of $R$).  Since as Charles noted, the braid map of an invertible object can be identified with its Euler characteristic as a dualizable object, this implies that the braid map of any invertible object in the derived category of a local ring is $\pm 1$ (and so if you're willing to suspend your objects if necessary, you can assume it is 1).
Thus I would be satisfied with a conceptual answer to the following question: why is the Euler characteristic of a dualizable object in the derived category of a local ring always an integer?  (It may be more natural to not assume that the ring is local, in which case you should replace "integer" with "integral linear combination of idempotents".)

REPLY [7 votes]: Here's how I look at this. Let $(\mathcal C,\otimes,I, \sigma)$ be a symmetric monoidal category. Recall that the dimension of a dualizable object $C \in \mathcal C$ is the trace of the identity map on $C$; it lives in $End_{\mathcal C}(I)$. EDIT: This is what Charles referred to as the Euler characteristic.
Observations: 

Dimension is multiplicative: $\dim(C\otimes D) = \dim(C) \dim(D)$ for $C,D \in \mathcal C$ dualizable.
$\dim(C) = \dim(C^\vee)$.
If $L\in \mathcal C$ is $\otimes$-invertible, then $\dim(L)$ is invertible.

Corollary: Let $L \in \mathcal C$ be $\otimes$-invertible. Then

The symmetry on $L$ is given by $\sigma_{L,L} = \frac 1 {\dim L} 1_{L \otimes L}$.
$(\dim L)^2 = 1$. In particular, if $End_{\mathcal C}(I)$ is a domain, then $\dim L = \pm 1$.

Proof:

It suffices to verify this equation after tensoring with $L^{-1}$ on both sides and then capping off with some units and counits, where it becomes the true equation $\dim L = \frac 1 {\dim L} (\dim L)^2$.
This follows from the observations above since $L \otimes L^\vee = I$ and $\dim I = 1$.

So the fact that the braidings are trivial on $Pic(R)$ reduces to the fact that the dimension of an invertible module is $1$ rather than $-1$.<|endoftext|>
TITLE: Is $L^p(\mathbb{R})$ minus the zero function contractible?
QUESTION [22 upvotes]: Is $L^p(\mathbb{R}) \setminus 0$ contractible?  My intuition says that the answer is yes, but I'm afraid that this is based on thinking of this as somehow similar to a limit of $\mathbb{R}^n \setminus 0$ as n approaches $\infty$, which is of course nonsense.  In any case, every contraction I've tried ends up making some function pass through $0$.

REPLY [2 votes]: I got here via this mathoverflow question.  I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$ 
Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points.  Here is an easier proof that such a map exists.  Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$.  Note that $T$ is continuous and fixes the origin.  
Set $\One = (1,0,0,0,\ldots)$.  Working with the $L^2$ norm, define 
$$
f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x).
$$
Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere.  On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$.  So $f$ has no fixed points inside the ball.  We are done. 
Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.  
EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary.  The book goes on to say that Tychonoff found the required retraction.  See page 178 of "Spaces and fixed point theory" by Khamsi and Kirk for a brief discussion.<|endoftext|>
TITLE: Moser iteration for elliptic systems
QUESTION [8 upvotes]: I heard that De Giorgi-Nash-Moser type regularity arguments fail for elliptic systems, but do not know where to start looking for more substantial information. Why does the regularity fail? Is there some cases where the Moser iteration can be successfully applied to elliptic systems?

REPLY [5 votes]: Basically, the De-Giorgi - Moser - Nash regularity result fails for elliptic system. As Johannes pointed out, there is a counter-example in Giusti book.
For system, usually, one can get "partial regularity result". What I mean by that is: there exists $\Omega_0 \subset \Omega$ open such that $|\Omega\setminus \Omega_0|=0$ and $u \in C^{1,\alpha}(\Omega_0)$. Then with the smoothness of coefficients, you can have $u \in C^\infty(\Omega_0)$.
There are several references you can have a look at are:

M. GIAQUINTA, Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic 
Systems, Princeton U. Press, Princeton, 1983. 
M. GIAQUINTA • E. GIUSTI, On the regularity of the minima of variational inte- 
grals, Acta. Math. 148 (1982), 31-46.
Evans, Lawrence C. Quasiconvexity and partial regularity in the calculus of variations. Arch. Rational Mech. Anal. 95 (1986), no. 3, 227--252.

In the third one, Evans proved the "partial regularity result" for a minimizer of certain energy functional. The most important thing is that the Lagrangian only need to be uniformly stricly-quasiconvex instead of uniformly convex. And it has some very important applications in elasticity. You can search for some papers of John Ball for this issue.<|endoftext|>
TITLE: Dual Schroeder-Bernstein theorem
QUESTION [39 upvotes]: This question was motivated by the comments to Dual of Zorn's Lemma?
Let's denote by the Dual Schroeder-Bernstein theorem (DSB) the statement 

For any sets $A$ and $B$, if there are surjections from $A$ onto $B$ and from $B$ onto $A$, then there is a bijection between them.

In set theory without choice, assume that the Dual Schroeder-Bernstein theorem holds. Does it follow that choice must hold as well?
I strongly suspect this is open, though I would be glad to be proven wrong in this regard. In all models of ZF without choice that I have examined, DSB fails. This really does not say much, as there are plenty of models I have not looked at. In any case, I don't see how to even formlate an approach to show the consistency of DSB without AC.
The only reference I know for this is Bernhard Banaschewski, Gregory H. Moore, The dual Cantor-Bernstein theorem and the partition principle, Notre Dame J. Formal Logic 31 (3), (1990), 375–381. In this paper it is shown that a strengthening of DSB does imply AC, namely, that whenever there are surjections $f:A\to B$ and $g:B\to A$, then there is a bijection $h:A\to B$ contained in $f\cup g^{-1}$. (Note that the usual Schroeder-Bernstein theorem holds -without needing choice- in this fashion.)
The partition principle is the statement that whenever there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $A$. As far as I know, it is open whether this implies choice, or whether DSB implies the partition principle. Clearly, the reverse implications hold.
If you are interested in natural examples of failures of DSB in some of the usual models, Benjamin Miller wrote a nice note on this, available at his page.

Added Sep. 21. [Edited Aug. 14, 2012] It may be worthwhile to point out what is known, beyond the Banaschewski-Moore result mentioned above. 
Assume DSB, and suppose $x$ is equipotent with $x\sqcup x$. Then, if there is a surjection from $x$ onto a set $y$, we also have an injection from $y$ into $x$. (So we have a weak version of the partition principle.) This idemmultiple hypothesis that $x\sqcup x$ is equipotent to $x$, for all infinite sets $x$, is strictly weaker than choice, as shown in Gershon Sageev, An independence result concerning the axiom of choice, Ann. Math. Logic 8 (1975), 1–184, MR0366668 (51 #2915). 
Also, as indicated in Arturo Magidin's answer (and the links in the comments), H. Rubin proved that DSB implies that any infinite set contains a countable subset.

REPLY [7 votes]: This is only a partial answer because I'm having trouble reconstructing something I think I figured out seven years ago...
It would seem the Dual Cantor-Bernstein implies Countable Choice. In a post in sci.math in March 2003 discussing the dual of Cantor-Bernstein, Herman Rubin essentially points out that if the dual of Cantor-Bernstein holds, then every infinite set has a denumerable subset; this is equivalent, I believe, to Countable Choice.
Let $U$ be an infinite set. Let $A$ be the set of all $n$-tuples of elements of $U$ with $n\gt 0$ and even, and let $B$ be the set of all $n$-tuples of $U$ with $n$ odd. There are surjections from $A$ onto $B$ (delete the first element of the tuple) and from $B$ onto $A$ (for the $1$-tuples, map to a fixed element of $A$; for the rest, delete the first element of the tuple). If we assume the dual of Cantor-Bernstein holds, then there exists a one-to-one function from $f\colon B\to A$ (in fact, a bijection). Rubin writes that "a 1-1 mapping from $B$ to $A$ quickly gives a countable subset of $U$", but right now I'm not quite seeing it...<|endoftext|>
TITLE: Classifying functors of abelian categories
QUESTION [6 upvotes]: Let $AbCat$ denote the $2$-category of abelian categories with additive functors. Is the forgetful functor $AbCat \to Cat$ representable; i.e. is there an abelian category $T$ such that for every abelian category $A$, the category $Hom(T,A)$ is naturally isomorphic to (the category underlying) $A$?
This would be nice, because then it is possible to reconstruct an abelian category which has only a universal property.
I try to work out the structure of $T$: The isomorphism $Hom(T,T) \cong T$ maps $1_T$ to some object $x \in T$. If $A$ is arbitrary, then the isomorphism $Hom(T,A) \cong A$ is given by mapping a functor $F : T \to A$ to $F(x)$ and mapping a natural transformation $\eta : F \Rightarrow G$ to the morphism $\eta(x) : F(x) \to G(x)$.
Let $T'$ be the smallest full abelian subcategory of $T$, which contains $x$ (Existence: Construct inductively subcategories $T_{n+3k}$, which contain direct sums $(n=0)$, kernels $(n=1)$ and cokernels $(n=2)$ from $T_{n+3k-1}$.). Then $T'$ has the same universal property as $T$. Thus $T'=T$, and we see that $T$ is generated by $x$. Now we have to find such a $T$, which has no additional relations. In particular, we have to ensure that a natural transformation between additive functors on $T$ is determined by the morphism at $x$, although the functors don't have to be exact.

REPLY [3 votes]: If we require an isomorphism (as opposed to an equivalence) the answer is no, for reasons having little to do with the (interesting aspect of) the question.
Assume there exists an abelian category $T$ and an object $x$ of $T$ such that for every abelian category $A$ and object $a$ there is a unique functor $F_a:T\to A$ sending $x$ to $a$. Clearly, $x$ is nonzero, so the object $x^2$ has a nontrivial automorphism $\sigma$. Consider the functor $\Phi:T\to T$ defined as follows:
$\Phi$=identity on objects,
for a map $f:u\to v$, put:
$\Phi(f)=f$ if $u\neq x^2\neq v$ or $u=x^2=v$,
$\Phi(f)=\sigma f$ if $u\neq x^2$ and $v=x^2$,
$\Phi(f)=f\sigma^{-1}$ if $u=x^2$ and $v\neq x^2$.
Now $\Phi$ is indeed a functor (case-by-case inspection) sending $x$ to $x$. So it should be equal to the identity functor but isn't.<|endoftext|>
TITLE: Constructing prime numbers
QUESTION [26 upvotes]: The classical proof of the infiniteness of prime numbers is to take the $k$ first prime numbers $p_1,\ldots,p_k$, then to form
$$n_k:=1+p_1\cdots p_k.$$
Then $n_k$ has a prime factor, which is none of the $p_j$.
Notice that one could form instead the number
$$n_{k,I}:=\prod_{i\in I}p_i+\prod_{j\not\in I}p_j$$
where $I$ is any subset of $[1,\ldots,k]$ (we have $n_k=n_{k,\emptyset}$), or even
$$m_{k,I}:=\prod_{i\in I}p_i-\prod_{j\not\in I}p_j,$$
provided this number is not $\pm1$.
This suggests the following construction of prime numbers $q_k$ by induction. Start with $k=1$ and $q_1=2$. At step $k$, form the numbers
$$M_{k,I}:=\left|\prod_{i\in I}q_i-\prod_{j\not\in I}q_j\right|.$$
Then $q_{k+1}$ is the least of the prime factors of all these numbers (that is, of their product as $I$ runs over the subsets of $[1,\ldots,k]$). 

Is it true that $q_k=p_k$ for all $k$ ? If not, does the sequence covers all the prime numbers ? If not, how fast does this sequence increases as $k\rightarrow+\infty$ ?

One might enlarge the construction by taking in account the numbers $M_{k,I}$ and
$$N_{k,I}:=\prod_{i\in I}q_i+\prod_{j\not\in I}q_j,$$
but they are way larger.

REPLY [4 votes]: maybe you could take the primes on the products in any powers you want , it is a thought that i had before some years and it is also a natural question that one can make reading Euclid's proof since still the difference of the products is not divisible by any of these primes do you take in this way always the next prime?<|endoftext|>
TITLE: Proper subgroup of GL(n,Z) isomorphic to GL(n,Z)?
QUESTION [32 upvotes]: This is just a question originated from some random thoughts. I hope it's nevertheless fit for mo.
It's possible to find a proper subgroup of $GL(n,\mathbb{C})$ isomorphic to $GL(n,\mathbb{C})$ itself (simply as set-theoretical groups, not algebraic groups): just embed $\sigma:\mathbb{C}\hookrightarrow\mathbb{C}$ by a map which is identity on algebraic numbers and is "a shift" on a trascendence basis; then take invertible matrices with entries in $\sigma(\mathbb{C})$.
The ring $\mathbb{Z}$, instead, doesn't admit an injective non surjective morphism into itself, so the above trick does not apply to the following question:

Does $GL(n,\mathbb{Z})$ have any proper subgroup which is isomorphic to $GL(n,\mathbb{Z})$ itself?

REPLY [9 votes]: For $G=GL(2,\mathbb Z)$ there is no proper subgroup isomorphic to it. Consider the dihedral group $D$ of isometries of a regular 6-gon. There is only one conjugacy class in $G$ of subgroups isomorphic to $D$. Indeed given such a subgroup, simply call it $D$, one may adapt
the inner product to it, by averaging, so as to make $D$ consist of orthogonal matrices. Then take a basis of $\mathbb Z^2$ consisting of shortest vectors making an obtuse angle, say.
Let $s$ be the element that swaps the two basis vectors. We now look for an element $u$ of
order four with $susu=1$ so that $u^2$ commutes with the elements of $D$. There is
very little choice and we find that $u$ together with $D$ generates $G$.<|endoftext|>
TITLE: Tying knots with reflecting lightrays
QUESTION [60 upvotes]: Let a lightray bounce around inside a cube whose faces
are (internal) mirrors.
If its slopes are rational, it will eventually form a cycle.
For example, starting with a point $p_0$ in the interior
of the $-Y$ face of an axis-aligned cube, and initially heading in a direction 
$v_0=(1,1,1)$, the ray will rejoin $p_0$ after 5 reflections,
forming a hexagon.
The figure below shows a more complicated 16-cycle.

          

Assume that $p_0$ and $v_0$ are chosen so that 
(a) the ray never directly hits an edge or corner of the cube,
and (b) the ray path never self-intersects inside the cube.

Can every knot type be realized by a lightray reflecting inside in a cube?

The figure above is an unknot.
I believe (but am not certain) the 31-cycle below is knotted:

          

Any such knotted path is a stick representation of the knot,
but perhaps the many unsolved problems in stick representations
are not relevant to this situation.
My question is related to the probability of random knots
forming under various models, but usually those models are
aimed at polymers or DNA.  I have not seen this lightray model
explored, but would be interested to know of related models.
The choice of $(p_0,v_0)$ allows considerable freedom to "design"
a knot, but it seems difficult to control the structure of the path
to achieve a particular result.
I've explored tiling space by reflected cubes so
that the lightray may be viewed as a straight segment between
two images of $p_0$, but this viewpoint is not yielding
me insights.
If anyone has ideas, however partial, I would appreciate hearing them.
Thanks!
Edit1 (15Sep10).
I have not been able to yet access the
Jones-Przytycki paper that Pierre cites, but knowing the
keywords he kindly provided, I did
find related work by
Christoph Lamm 
("There are infinitely many Lissajous knots"
Manuscripta Mathematica
93(1): 29-37
(1997))
that provides useful information:

Theorem: Billiard knots in a cube are isotopic to Lissajous knots.
As Pierre said, many knots are unachievable in these models.  In particular,
algebraic knots cannot be achieved.
The technical result is this. 
Theorem: The Alexander polynomial of a billiard knot is a square
mod 2.
In 1997, there were several intriguing open problems, including
these two. (a) Is every knot a billiard knot in some convex polyhedron?
(b) Can the unknot be achieved in every convex polyhedron that supports periodic paths?

Edit2 (15Sep10). Here is a little more information on open problems ca. 2000,
found in a list by Jozef H. Przytycki in the book that resulted from 
Knots in Hellas '98:

Is there a manifold that supports every knot?
(By "supports every knot" he means there is a billiard path isotopic to every knot type.)
Is there a finite polyhedron that supports every knot?
There apparently is an "infinite polyhedron" that supports every knot.
More specifically, is there a convex polyhedron that supports every knot? (This is 3(a) above.) [See Bill Thurston's correction below!]
Even more specifically, is every knot supported by one of the Platonic solids?

I have not been successful in finding information on this topic later than 2000.
If anyone knows later status information, I would appreciate a pointer.  Thanks for the interest!
Edit3 (5Jul11). 
The question has been answered (affirmatively) in a paper by
Pierre-Vincent Koseleff and Daniel Pecker
recently (28Jun11) posted to the arXiv:
"Every knot is a billiard knot":

"We show that every knot can be realized as a billiard trajectory in a convex prism. ...
  Using a theorem of Manturov [M], we first prove that every knot has a diagram which is a star polygon. 
  [...Manturov’s theorem tells us that every knot (or link) is realized as the closure of a quasitoric braid...]
  Then, perturbing this polygon, we obtain an irregular diagram of the same knot. We deduce that it is possible to suppose that 1 and the arc lengths of the crossing points are linearly independent over $\mathbb{Q}$. Then, it is possible to use the classical Kronecker density theorem to prove our result." 

Edit4 (4Oct11).
A new paper was released by Daniel Pecker, "Poncelet's theorem and Billiard knots," arXiv:1110.0415v1. The context is that, earlier, "Lamm and Obermeyer proved that not all knots are billiard knots in a cylinder," and Lamm conjectured an elliptic cylinder would suffice.
Here is the abstract:

Let $D$ be any elliptic right cylinder. We prove that every type of knot can be realized as the trajectory of a ball in $D$. This proves a conjecture of Lamm and gives a new proof of a conjecture of Jones and Przytycki. We use Jacobi's proof of Poncelet's theorem by means of elliptic functions. 

Edit5 (13Nov12).
The Pecker paper cited above is now published:
Geometriae Dedicata,
December 2012, Volume 161, Issue 1, pp 323-333.

REPLY [7 votes]: Coming back to the sawteeth described by Bill: many (or all) crossing patterns can be generated if the parameters x of the crossings are rationally independent (or 'independent enough'). In particular, symmetries in the x-y-shadow must be avoided. 
A related article of mine on 'Fourier knots' discusses the question whether certain billiard curves in 2 dimensions allow the construction of all knots if the z-movement is arbitrary (I focus on the square and the circle). I will send this article to anyone who is interested (the 5 page article is from 1998 and part of my PhD thesis, that's the reason why it is not published elsewhere).
My favorite billiard room to generate all knots is the prism over any ellipse. In this case the x-y-paths are asymmetrical enough, but it would be necessary to prove that the coordinates of the crossings are, e.g., rationally independent. - Christoph Lamm<|endoftext|>
TITLE: What kind of Lagrangians can we have?
QUESTION [25 upvotes]: In any physics book I've read the Lagrangian is introuced as as a functional whose critical points govern the dynamics of the system. It is then usually shown that a finite collection of non-interacting particles has a Lagrangian $\frac{1}{2}(m_1\dot{x}_1^2 + \cdots + m_n \dot{x}_n^2)$. It is then generally argued that $L=T-U$. I feel like something is missing here. 
What exactly are the physical hypotheses that go into this? Can we have other forms of the Lagrangian? How do we know those are "right"? Do we always have to compare them to the form of the equations we derived previously? For example, the Lagrangian formalism seems to be justified usually in so far as it 'works' for a finite collection of particles. Then you can solve any dynamics problem involving a collection of particles.
I have been vague so let me try to be more precise in my question.
Is the principle of least action an experimental hypothesis? Is it always true that $L=T-U$? When we don't know what the Lagrangian is, do we have to just guess and hope it is compatible with the dynamical equations we had already? Or can we perhaps start with the ansatz of a Lagrangian in some cases?
I hope this is sufficiently precise.

REPLY [2 votes]: The Lagrangians in physics usually take the form $L=m\dot{x}^2/2+...$ (or $\mathcal{L}=(\partial \phi)^2/2+...$ in field theory), i.e. the velocity dependence is quadratic.
An important consequence of this form can be seen in the path integral formulation of quantum mechanics. The amplitude as a path integral in the phase space is $\int \mathcal{D}x \ \mathcal{D}p \ e^{i\int(p \ dx - H \ dt)}$ where the Hamlitonian $H$ is a function of $p, x$ and $t$. The thing is, this amplitude is in general NOT equal to the form we usually use $\int \mathcal{D}x \ e^{i\int L \ dt}$. The two forms are equal when $H$ depends on $p$ as $p^2$, i.e. when $L$ depends on $\dot{x}$ as $\dot{x}^2$. (the derivation can be found in Polchinski, Peskin etc., basically $e^{ip^2}$ will give a constant by Gaussian integration)
I am not sure if this is the reason that physical Lagrangian must take the form $L=m\dot{x}^2/2+...$, but at least I feel this shows some deep connection between the Lagrangian formulation and Hamiltonian formulation in quantum theory.<|endoftext|>
TITLE: Jokes in the sense of Littlewood: examples?
QUESTION [135 upvotes]: First, let me make it clear that I do not mean jokes of the
"abelian grape" variety. I take my cue from the following 
passage in A Mathematician's Miscellany by J.E. Littlewood
(Methuen 1953, p. 79):

I remembered the Euler formula $\sum n^{-s}=\prod (1-p^{-s})^{-1}$;
  it was introduced to us at school, as a joke (rightly enough, and
  in excellent taste).

Without trying to define Littlewood's concept of joke, I venture
to guess that another in the same category is the formula

$1+2+3+4+\cdots=-1/12$,

which Ramanujan sent to Hardy in 1913, admitting "If I tell you this 
you will at once point out to me the lunatic asylum as my goal."
Moving beyond zeta function jokes, I would suggest that the empty
set in ZF set theory is another joke in excellent taste. Not only
does ZF take the empty set seriously, it uses it to build the whole
universe of sets.
Is there an interesting concept lurking here -- a class of
mathematical ideas that look like jokes to the outsider, but which
turn out to be important? If so, let me know which ones appeal to
you.

REPLY [3 votes]: (An example from set theory, more specifically: forcing): 
The backslash $\backslash$ is used for set difference, the forward slash $/$ is often used for taking a quotient.   In my experience, students often have a hard time distinguishing between the two.  (Between the two symbols, not the two concepts.) 
But there is one case where the difference does not matter:  if $B$ is a Boolean algebra, and $I$ an ideal, then the sets $B\setminus I$ (= $I$-positive elements)  and $B/I$ (quotient Boolean algebra) are equivalent as forcing notions:  $B\setminus I  = B/I $. 
For example, random forcing can be seen as "Borel sets modulo measure zero sets", or equivalently as "Lebesgue-positive Borel sets (without any identification)".<|endoftext|>
TITLE: Re-seating a monad
QUESTION [9 upvotes]: Let $\mathcal C$ and $\mathcal D$ be categories with suitable limits and colimits for the following discussion. Is it possible to re-interpret, or "re-seat" a monad $T : \mathcal C \to \mathcal C$ as a monad over $\mathcal D$? When $T$ is finitary, I know at least one way to do this. Compute the Kleisli category $\mathcal C_T$ and consider $(\mathcal C_T)^{op}$ as a Lawvere theory. Then models of $(\mathcal C_T)^{op}$ on $\mathcal C$ are the same as $T$-algebras. If we then consider models of $(\mathcal C_T)^{op}$ on $\mathcal D$, we have a canonical, monadic forgetful functor $[(\mathcal C_T)^{op},\mathcal D] \to \mathcal D$ from which we can build a new finitary monad $T' : \mathcal D \to \mathcal D$. My question is as follows.

Is there an direct way to obtain $T'$ from $T$ (i.e. that doesn't go via the Lawvere theory)? If so, can it be extended to work for arbitrary monads?

REPLY [10 votes]: I like the question, Aleks - and I like the term "re-seating".
Yes, there is a direct way, at least in suitable circumstances.  First suppose that we're beginning with a finitary monad $T$ on $\text{Set}$.  For each set $X$, we have
$$
T(X) = \int^n T(n) \times X^n
$$
where the coend is over the category of finite sets.  (This is essentially the definition of finitariness, in disguise.  If you haven't seen this before, think of $T(n)$ as the set of $n$-ary operations in the theory $T$.)  The symbol $\times$ is of course product, but it could also be interpreted as copower: $T(n) \times X^n$ is a copower of the object $X^n$ of $\text{Set}$.  
Now take a category $\mathcal{D}$ with suitable colimits.  For $Y \in \mathcal{D}$, put
$$
T'(Y) = \int^n T(n) \times Y^n
$$
where again the coend is over the category of finite sets, and $\times$ is copower.  (We can no longer interpret $\times$ as product, because $T(n)$ and $Y^n$ belong to different categories.)  You can give $T'$ the structure of a monad, at least under mild assumptions on the behaviour of colimits in $\mathcal{D}$.  That's the same monad $T'$ you'd get if you followed the kind of procedure you describe.
Incidentally, I don't think your description of the procedure is quite right.  If, for example, $\mathcal{C} = \text{Set}$, then $(\mathcal{C}_T)^\text{op}$ is not the Lawvere theory of $T$.  The Lawvere theory is the full subcategory of $(\mathcal{C}_T)^\text{op}$ consisting of just the (free algebras on) finite sets.  
I think the coend formula for re-seating should work when $\mathcal{C}$ is an arbitrary locally finitely presentable category, not necessarily $\text{Set}$.  You'd have to replace the category of finite sets by the category $\mathcal{C}_{\text{fp}}$ of finitely presentable objects of $\mathcal{C}$.  And you'd need $\mathcal{C}_{\text{fp}}$ to act on $\mathcal{D}$, so that the copower defining $T'$ made sense.
Generalizing in a different direction, there's probably nothing special about finitariness: you could do it with any old rank.  But I suspect you do need the monad to have rank in order to re-seat a monad in the kind of way I'm suggesting.  The category of finite (or suitably small) sets provides a bridge between $\mathcal{C}$ and $\mathcal{D}$, and that's what enables the monad to be re-seated.<|endoftext|>
TITLE: Visualizing Orthogonal Polynomials
QUESTION [13 upvotes]: Recently I was introduced to the concept of Orthogonal Polynomials through the poly() function in the R programming language. These were introduced to me in the concept of polynomial transformations in order to do a linear regression. Bear in mind that I'm an economist and, as should be obvious, am not all that smart (choice of profession has an odd signaling characteristic). I'm really trying to wrap my head around what Orthogonal Polynomials are and how, if possible, to visualize them. Is there any way to visualize orthogonal polynomials vs. simple polynomials?

REPLY [23 votes]: Helge presented the continuous case in his answer; for the purposes of data fitting in statistics, one usually deals with discrete orthogonal polynomials. Associated with a set of abscissas $x_i$, $i=1\dots n$ is the discrete inner product
$$\langle f,g\rangle=\sum_{i=1}^n w(x_i)f(x_i)g(x_i)$$
where $w(x)$ is a weight function, a function that associates a "weight" or "importance" to each abscissa. A frequently occurring case is one where the $x_i$ are equispaced, $x_{i+1}-x_i=h$ where $h$ is a constant, and the weight function is $w(x)=1$; for this special case, special polynomials called Gram polynomials are used as the basis set for polynomial fitting. (I won't be dealing with the nonequispaced case in the rest of this answer, but I'll add a few words on it if asked).
Let's compare a plot of the regular monomials $x^k$ to a plot of the Gram polynomials:

On the left, you have the regular monomials. The "bad" thing about using them in data fitting is that for $k$ high enough, $x^k$ and $x^{k+1}$ are nigh-indistinguishable, and this spells trouble for data-fitting methods since the matrix associated with the linear system describing the fit is dangerously close to becoming singular.
On the right, you have the Gram polynomials. Each member of the family does not resemble its predecessor or successor, and thus the underlying matrix used for fitting is a lot less likely to be close to singularity.
This is the reason why discrete orthogonal polynomials are of interest in data fitting.<|endoftext|>
TITLE: Vector valued functions
QUESTION [5 upvotes]: Question : Does  there exist a surjective function $F$ that maps $\mathbb{R}^n_+$ to $\mathbb{R}^n$ (where $\mathbb{R}^n_+$ denotes the set of vectors of length $n$ with only positive entries). The answer is yes by considering the function $F$:
$$(x_1,\ldots,x_n)\to(\log x_1,\ldots,\log x_n)$$
It is easy to see that $F$ is surjective.
Now my real question is if there is such a function $F$ whose components are restricted to be polynomials of degree at most $n$ (and $F$ maps $\mathbb{R}^n_+$ to $\mathbb{R}^n$ and is surjective). For example:
$$(x_1,x_2,x_3,x_4)\to(1-x_1,x_1\cdot x_2+x_3,x_3^2+x_2-x_4,5x_4)$$
has polynomial components, but is not surjective as the restriction $x_i>0$ implies we cannot get a $1$ in the first coordinate.
Any help or advice is greatly appreciated. I think the answer is no but am not sure how to prove it.

REPLY [2 votes]: For n = 1, it is well-known and obvious that a polynomial $P(x)$ approaches either $+ \infty$ or else $-\infty$ as $x \to \infty$, so in particular it has all its values in either the positive half-axis or else in the negative half-axis for say $x > M$. But then it cannot cover the whole other half-axis on the compact interval $[0,M].<|endoftext|>
TITLE: Surface bundles over a surface
QUESTION [12 upvotes]: What can be used to distinguish two $\Sigma_g$-bundles over $\Sigma_h$ up to
(1) homotopy?
(2) homeomorphism?
(3) fiberwise homeomorphism?
(4) bundle isomorphism?
And can these always be computed given 2 specific surface bundles over $\Sigma_h$?

REPLY [5 votes]: I'll make some comments, but I don't know the complete answer to your questions. (I'll assume $g,h>0$ too.)
As Thurston says, the bundle is determined up to isomorphism by the image $\pi_1(\Sigma_h)\to Mod(g)$. I think this answers questions (3) and (4). If two fiber bundles are fiberwise homeomorphic, then the space of fibers determines the base space, and therefore the fibration, so I think (3) and (4) are equivalent. The algorithmic question is open, as far as I know (of determining when the image of two surface groups in $Mod(g)$ are conjugate). 
I think questions (1) and (2) are more difficult. One issue is that a 4-manifold that fibers over a surface might fiber many different ways. Because I'm assuming $g,h>0$, these manifolds are $K(\pi,1)$'s, and so the homotopy question reduces to determining if $\pi_1$ are isomorphic. The homeomorphism problem would follow from the the Borel conjecture, but this is wide open in the case of 4-manifold with fundamental group of exponential growth.
There is a well-known open question, whether there is a convex-cocompact map (in particular injective) $\pi_1(\Sigma_h)\to Mod(g)$. If this exists, the $\pi_1$ of the associated bundle is a word-hyperbolic group. In this case, there is a theorem of Sela which allows one to algorithmically distinguish the fundamental groups, and therefore determine the homotopy type of the associated manifolds. However, this could be a theory of the empty set, since no examples are known!
Addendum: There are some techniques for distinguishing the diffeomorphism types of these manifolds. If the fiber $\Sigma_g$ is homologically non-trivial in the manifold, then Thurston proved that $M$ admits a symplectic structure. Work of Taubes implies that the Seiberg-Witten invariants of $M$ may be computed from the Gromov invariants of $M$. This might give methods for detecting when two bundles are not diffeomorphic. I suppose that such manifolds might have many different smooth structures, but it's not clear to me that the natural smooth structures associated to different fiberings could be different.
One might be able to distinguish homotopy types of these manifolds using group invariants. For example, these groups are subgroups of mapping class groups, and are therefore residually finite. One could count homomorphisms from the fundamental group to a finite group to obtain invariants that may distinguish the bundles.<|endoftext|>
TITLE: Status of Quillen's conjecture on elementary abelian p-groups
QUESTION [15 upvotes]: These are questions on D. Quillen's 1978 paper Homotopy properties of the poset of nontrivial p-subgroups of a group.
Let $G$ be a finite group, $p$ a prime number, $\mathcal S(G)$ the poset of non-trivial $p$-subgroups of $G$, and $\mathcal A (G)$ the poset of non-trivial elementary Abelian $p$-subgroups of $G$, both ordered by inclusion.

Question One: Is $\mathcal S(G)$ homotopic or weakly homotopic to $\mathcal A (G)$? (Whichever is true is a theorem of Quillen.) If the latter, can someone give a specific example showing that the two posets are not homotopic?

Quillen also proved that for $G$ solvable, $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.
Conjecture (Quillen): $\mathcal A (G)$ is contractible if and only if $G$ has a non-trivial normal $p$-subgroup.

Question Two: Is this still open (I know it was a few years ago)? What are lines of attack on this problem? Have attempts to prove it led to any a priori unrelated work?

REPLY [16 votes]: The fact that $\mathcal{A}(G)\simeq\mathcal{S}(G)$ is normally proved using Quillen's Theorem A as explained by Dan, but here is another argument that I like better.  (I don't think it is in the literature.) For any poset $X$, let $sX$ be the poset of nonempty chains in $X$, ordered by inclusion.  There is a homeomorphism $m:|sX|\to |X|$ by barycentric subdivision.  There is also a poset map $\max:sX\to X$, and $|\max|$ is homotopic to $m$ (by a linear homotopy) so it is a homotopy equivalence.  Next, if $P$ is a nontrivial $p$-group then the group $TP=\{g\in ZP: g^p=1\}$ is nontrivial (by a standard lemma) and elementary abelian.  Unfortunately, the map $T:\mathcal{S}(G)\to\mathcal{A}(G)$ is not order-preserving.  However, suppose we have a chain $C=(P_0\leq\dotsb\leq P_r)\in s\mathcal{S}(G)$.  When $i\leq j$ we note that $TP_i\leq P_j$ and $TP_j$ is central in $P_j$ so $TP_i$ commutes with $TP_j$.  It follows that the product $UC=TP_0.TP_1.\dotsb.TP_r$ is again elementary abelian.  This defines a poset map $U:s\mathcal{S}(G)\to\mathcal{A}(G)$ and thus a map $|U|:|\mathcal{S}(G)|\simeq|s\mathcal{S}(G)|\to|\mathcal{A}(G)|$.  If we let $i:\mathcal{A}(G)\to\mathcal{S}(G)$ denote the inclusion then $i\circ U\leq\max$ so $|i|\circ|U|$ is homotopic to $|\max|$ and thus to $m$.   We also have $U\circ si=\max:s\mathcal{A}(G)\to\mathcal{A}(G)$.  It follows that $|i|$ is a homotopy equivalence as claimed.<|endoftext|>
TITLE: Is there a connected $k$-group scheme $G$ such that $G_{red}$ is not a subgroup?
QUESTION [18 upvotes]: I've been trying a learn a little more about group schemes by working through a set of exercises on Brian Conrad's website. Exercise 8.3 of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf reads:

If $k$ is a perfect field and $G$ is a locally finite type $k$-group prove that $G_{red}$ is a closed $k$-subgroup of $G$. Can you find a counterexample if $k$ is not perfect? (There's a third part to the question, but it's irrelevant here)

I'll try to not give away too much for the benefit of others who want to use these exercises, but the first part of the exercise is Lemma 7.10 of http://www.math.columbia.edu/algebraic_geometry/stacks-git/groupoids.pdf in DeJong's stacks project and the hint given for the omitted proof is that $G_{red}\times_k G_{red}$ is reduced if $k$ is perfect.
For the counterexample part, I was able to find a disconnected example over any imperfect field, but it really seemed to depend crucially on being disconnected.
This naturally suggests the question:

If $G_{red}$ is not a subgroup scheme of $G$, must $G$ be disconnected? 


edit: After almost a week out an answer in the zero-dimensional case was given and then retracted, but there has not been much action otherwise. Perhaps this is an open problem if $G$ is not finite.
For any finite group scheme $H$, $H_{red}$ is a subgroup of $H$ if and only if $H/H^0$ is a subgroup of $H$ (and in fact if $k$ is perfect, $H \cong H^0 \oplus H_{red}$, see Pink's notes Lecture 6- note this very explicitly depends on $H$ being finite)
For group schemes of locally finite type (as we assume here) there has been much done, for instance in SGA 3 where they prove among other things that if $G$ is connected, it's actually quasi-compact and thus of finite type (Proposition 2.4). Moreso, for a locally finite type group $H$ over a field, $(H^0)_{red}$ is a group in the category of reduced schemes. What's unclear is whether they believed it to be a group in the category of schemes but weren't able to prove that or if they knew of a counterexample but didn't include it.

REPLY [12 votes]: The new edition of SGA3 by Philippe Gille et Patrick Polo provides a connected example due to Raynaud. It is in SGA3 Exposé VIA Exemples 1.3.2 (2) and you may find it at http://www.math.jussieu.fr/~polo/SGA3/Exp6A-23mai11.pdf.
This example is $2$-dimensional, but it is easy to modify it to get a $1$-dimensional example if the characteristic of the base field is at least $3$. More precisely, let $G$  be defined by the equations $X^p-tY^p=Y^p-tZ^p=0$ in the additive group $\mathbb{G}_a^3$ over the field $\mathbb{F}_p(t)$, $p\neq 2$. Then $G$ is a connected group scheme of dimension $1$ whose reduction is not a subgroup scheme, as may be seen by following step by step the arguments of loc. cit.<|endoftext|>
TITLE: How can simple physical "proofs" of mathematical facts be made rigorous?
QUESTION [27 upvotes]: Mark Levi's The Mathematical Mechanic is a book of examples of how physical reasoning can be used to solve mathematical problems; another couple of examples is in this blog post at Concrete Nonsense.  Many of these proofs rely on conservation laws and other physical facts which, at least as I understand it, can be made rigorous by appealing to Noether's theorem.  So these proofs themselves ought to be presentable in a mathematically rigorous way.  But how?
There are many examples, so I will focus my question on two, one of which is in Levi's book and one of which I recently encountered on a friend's homework.

Problem: prove the Pythagorean theorem.  Solution: imagine a fish tank in the shape of a triangular prism.  The triangle is a right triangle with side lengths $a, b, c$; call the vertices of the two copies of the triangle $A, B, C$ and $A', B', C'$, where the right angle is at $C$ (resp. $C'$).  Drive a rod through $BB'$ and fill the tank with water.  This system is at equilibrium (by conservation of... angular momentum?), so the net torque through the rod is zero.  On the other hand the net torque through the rod is proportional to $c^2 - a^2 - b^2$.
Problem: on each face of a tetrahedron, not necessarily regular, place a vector pointing normal to and out from the face with magnitude equal to the area of the face.  Prove that the sum of these vectors is zero.  Solution: fill the tetrahedron with an ideal gas.  This system is at equilibrium (by conservation of... energy?), so the net force exerted by the gas on the tetrahedron is zero.  On the other hand the net force exerted on each face is proportional to its area and points normal to and out from each face.

As you can see, I'm not even totally clear what conservation law(s) I'm invoking at the crucial step of the physical arguments.  I would really appreciate mathematical insight into what's going on here.

REPLY [11 votes]: Both proofs in the question are of the following type. Suppose you want to prove a certain identity among numbers $X$, say $f(X)=0$. If you can find another function $g(X,Y)$ such that $f(X)=g(X,Y)$ for some $Y$. Then, proving that $g(X,Y)=0$ for this choice of $Y$, independent of $X$, automatically proves $f(X)=0$.
In the case of the Pythagorean theorem, $X=(a,b,c)$ are the dimensions of the triangular base of the water tank, $Y$ are all the other details of the physical setup and $g(X,Y)$ is the total torque about $BB'$ exerted on the tank. As long as $Y$ contains the fact that there are no torques acting on the tank other than the different parts of the tank acting on each other, physical law dictates that the total torque must be zero, hence $g(X,Y)=0$. The Kut-the-knot page linked from one of jc's comments shows why $g(X,Y)$ is proportional to $c^2-a^2-b^2$.
The case of the tetrahedron is similar, $X$ are the dimensions of the tetrahedron and $g(X,Y)$ is the total force acting on the gas container. The physical situation $Y$ is set up such that no forces act on the container external to the container itself, hence physical law dictates that the total force must be zero.
Now, what is it that allows us to conclude something about the value of $g(X,Y)$ in each case, given only $Y$ and independent of $X$? One way to think about it is to consider the physical world as a giant computer or oracle, $W$. If you set up a physical situation $Z$ and measure some observable quantity $G$, the world will give you a numerical answer $W(G,Z)$. The goal of physics, of course, is to build mathematical models of $W$, so that we don't have to consult the oracle every time we need an to know the value of $W(G,Z)$. But a mathematical model is not necessary if we are happy with just consulting the oracle when needed. Still, observation of the answers that we get from $W$ allows us to identify certain regularities in its output: these are the laws of physics.
Thus, supposing that we can find $G$ and $Z=(X,Y)$ such that $g(X,Y)=W(G,Z)$, the kind of of proof I outlined in the first paragraph can be carried out, appealing to the laws of physics as properties of the oracle $W$. To convert this kind of proof into a sequence of logical deductions, as a usual proof should be, requires a mathematical model for $W$, at least a restricted one that suffices for the purposes of making deductions about $g(X,Y)$. In both cases the model that does the job is plain old Newtonian mechanics of particles, solid bodies and fluids (if desired, solid bodies and fluids may be seen as a limiting cases of large assemblies of particles). The equality $g(X,Y)=f(X)$ follows from direct application of Newton's second law (an axiom of this mathematical model) and the identity $g(X,Y)=0$, independent of $X$, follows from Newton's third law (another axiom).
Personally, I think the appeal of such "physical" proofs is the fact that they rely only on some properties of $W$, and not on the details of a mathematical model for it. However, the validity of the proof only follows if one can show that a detailed mathematical model exists.
Hope this helps.<|endoftext|>
TITLE: A reference for Calculus of Functors for Model Categories
QUESTION [6 upvotes]: I am wondering where I might look to see what has been done in terms of Calculus of Functors for more general weak equivalences and Model Categories.
I am at least aware of some of the extended definitions of the main concepts in Calculus of Functors to weak equivalence such as homotopy limits, but I was wondering if a document existed that worked through the basic of COF in this setting. 
I am aware also of Lurie's work here.(Thanks Harry for pointing this out.)
I appreciate your help.

REPLY [2 votes]: You might look at http://arxiv.org/abs/1208.1919 "On calculus of functors in model categories, Stanculescu"<|endoftext|>
TITLE:  Graphs, K-theory and combinatorial balls: conjectures
QUESTION [22 upvotes]: The following conjectures from Kapranov and Saito's Hidden Stasheff polytopes in algebraic K-theory and in the space of Morse functions aren't as well-known as they aught to be, so I'd like to state them and then ask if any progress has been made since the paper was released in 1997.
I quote (more-or-less): 
Fix a ring $A$. By a hieroglyph we will mean an oriented graph $\Gamma$ without oriented loops, equipped with the following additional structure: 
(a) An assignment of a positive integer to each vertex of $\Gamma$ so that all these integers are distinct. 
(b) An assignment of a nonempty ordered sequence of elements of $A$ to each edge of $\Gamma$. 
The number of elements written on the edge of a hieroglyph is called the weight of the 
edge and the weight of the whole hieroglyph is by definition the sum of 
weights of all the edges. 

Conjecture For every hieroglyph $\Gamma$ there is a polyhedral ball $P (\Gamma)$ with the following properties: 

$(a)$ The dimension of $P (\Gamma)$ is equal to the weight of $\Gamma$. The combinatorial type of $P (\Gamma)$ depends only on the underlying graph of $\Gamma$ and on weights of the edges. 
$(b)$ If $\Gamma = \cup \Gamma_i$  is the irreducible decomposition of a hieroglyph $\Gamma$, then $P (\Gamma) = \Pi P (\Gamma_i)$.
$(c)$ The boundary of each $P (\Gamma)$ is composed of the balls $P (\Gamma)'$ for some hieroglyphs $\Gamma'$.
$(d)$ Let $B$ be the union of the polyhedral balls $P (\Gamma)$ for all the hieroglyphs $\Gamma$ according to the identifications of their boundaries given by part $(c)$ above. 
Then $B$ is the homotopy fiber of the natural map $BGL(A) \to BGL^+(A)$. Another model for this homotopy fiber is given by Volodin K-theory in terms of the classifying space of the groups of upper triangular matrices in $GL(A)$. So a proof here would give a "polyhedral model" for this K-theory space. 
$(e)$ For the hieroglyph of Dynkin type $A_n$, $P(A_n)$ is the associahedron (Stasheff polytope) $\cal K_{n+1}$....this should be doable, but I have not seen a proof in the literature.

This is wonderful stuff - does anyone know how to verify any of these claims? Or are they open?

The paper goes on to elucidate many connections of hieroglyphs to Morse theory and algebraic K-theory, so solutions should be very interesting. Is anyone working on this kind of thing? The constructions in this paper are very concrete, and I'd like to think that a lot more can be said.

REPLY [7 votes]: I have just stumbled on this, so sorry that I did not reply at the time of the question. 
Some years ago, Tony Bak, Gabriel Minian, Ronnie Brown and myself had serious discussions on this area. We sketched a proof of the connection with Volodin's theory and made some progress, but then Bangor was closed down and I have only recently looked back at this. Our proposed attack was to use ideas from Tony Bak's theory of homotopy for his global actions linked with some ideas from combinatorial group theory and Ronnie's theory of crossed complexes.  SOme of our preliminary observations were published in Global Actions, Groupoid Atlases and Applications, Journal of Homotopy and Related Structures, 1(1), 2006, pp.101 - 167.  
I should stress that our 'proof' was just a sketch and was never completed, nor checked. It also gave a different combinatorial model than their hieroglyphs, but the relations were clearly there beneath the surface. Thus the main part of the Kapranov Saito conjectures would seem still untouched. (I put in for a grant to pursue them but it was refused. I thought that the ideas that the investigations were using were neat and revealed new connections between areas, but apparently I was unable to convince others of this!)<|endoftext|>
TITLE: Generator of a Fukaya category with certain properties
QUESTION [5 upvotes]: There is an algebraic theory that I'm thinking of trying to develop and I wanted to know if it had any real world prevalence --- I'd like to know an example of a generator L of a Fukaya category on a compact symplectic manifold where this Lagrangian has the properties that :
1) Simply connected,but not a sphere or a product of spheres  2) The homology of the Floer complex HF*(L,L) is isomorphic as an algebra to H*(L). (Edit: as an ordinary algebra, the higher operations induced by the perturbation lemma can be different.) Extra points if the Floer theory can in a reasonable sense be defined over C. I'd also be really interested to know why this situation can't happen. 
Edit: I know it is kind of rare for a single Lagrangian(impossible?) to generate a Fukaya category. I'd be happy to have a bunch of such Lagrangians L_i where at least one of them was as above as long as Hom(L_i, L_j)=0 for i not equal to j...

REPLY [3 votes]: Even without the condition that $HF(L,L)$ is $A_\infty$-isomorphic to $H^\ast(L)$, your conditions are set up in such a way as to disable the standard tricks. This doesn't rule out the existence of examples, but it does make them hard to find.
One way to produce orthogonal objects in the Fukaya category is to take different spin structures on the same Lagrangian (this works for the Clifford torus in $\mathbb{CP}^{2n}$). Simple connectedness rules this out. So for orthogonality we need Lagrangians that are (at least Floer-theoretically) disjoint.
Free torus-orbits in toric symplectic manifolds give examples of interesting disjoint Lagrangians, but you've doubly disallowed tori! Thanks to Seidel, we know how to go about proving split generation by vanishing cycles of Lefschetz pencils, but these are spheres, and they tend to intersect one another.
We don't have any picture of the Fukaya category of a "generic" symplectic manifold, so it's hard to say how reasonable the conditions you impose really are. For all we know, it could be a common phenomenon that there are few unobstructed Lagrangians, or even none at all.<|endoftext|>
TITLE: Which elliptic curves over totally real fields are modular these days?
QUESTION [22 upvotes]: As the title says.  In particular, every elliptic curve over $\mathbb{Q}$ is modular; but what is the current state of the art for general totally real number fields?  I assume the answer is extractable from some papers of, say, Kisin, but I am not an expert in this material and hesitate to try that myself.

REPLY [9 votes]: Elliptic curves over real quadratic fields were proven to be modular very recently by Freitas, Le Hung and Siksek:

Nuno Freitas, Bao Le Hung, Samir Siksek, Elliptic Curves over Real Quadratic Fields are Modular (2015)

Some progress was alredy present in the thesis of Richard Taylor's student Le Hung.

Bao Le Hung, Modularity of some elliptic curves over totally real fields (2013)

Potentially modular is known unconditionally over arbitrary totally real fields, that is:

Theorem. Let $E/F$ be an elliptic curve over a totally real field. Then there is some totally real extension $F'/F$ such that $E/F'$ is modular.

As mentioned in the comments, for practical purposes this is almost as good as modularity.
The attribution of this theorem is complicated. There's some information about this in section 7 of our own Kevin Buzzard's very interesting survey on modularity:

Kevin Buzzard, Potential modularity—a survey (2011)

You can also consult:

Jean-Pierre Wintenberger, Appendix: Potential modularity of elliptic curves over totally real fields (2010)

For arbitrary totally real fields, modularity is still not known.<|endoftext|>
TITLE: Why is every quadratic subfield of a Galois extension of the rationals with the quaternions as Galois group real? 
QUESTION [6 upvotes]: Suppose that L is a field extension of the rationals with Galois group the quaternions Q={1,-1,i,-i,j,-j,k,-k}. Furthermore assume that L contains a quadratic subfield K. I have learned from this link 
http://www.math.princeton.edu/generals/bhargava_manjul 
that K is real. How can we prove this result? Of course one should use Galois theory There are three subgroups inside Q of order four. Each of them is cyclic and generated by i, j, or k, resp. But I don't know how to move on. Any hint is appreciated.

REPLY [2 votes]: A theorem of Witt asserts that a biquadratic extension $F(\sqrt{a},\sqrt{b}),\; a,b\in F^{\times}$ can be embedded in a quaternion Galois extension of $F$ iff the quadratic form $ax^2+by^2+abz^2$ is isomorphic over $F$ to $x^2+y^2+z^2$.  For $F=\mathbb{Q}$ this forces $a,b,ab>0$.<|endoftext|>
TITLE: An orthogonal companion matrix
QUESTION [6 upvotes]: Let $P\in{\mathbb R}[X]$ be a monic polynomial with roots on the unit circle. For the problem below, we may assume wlog that the roots are simple and distinct from $\pm1$. It can be shown that there exists a matrix $M\in{\bf SO}_n({\mathbb R})$, whose characteristic polynomial is $P$ (an orthogonal companion matrix of $P$, in short OCM). See for instance Exercise 99 on my list http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf .
Regretfully, this exercise uses the square root of Hermitian positive definite matrices, which cannot be computed in finitely many operations.
Does there exist a construction of an OCM that uses only finitely many elementary operations (including the square root of complex numbers) ?
Thanks to the reduction to Hessenberg form, which can be done in finite time and which preserves the orthogonal group, we may restrict our attention to a Hessenberg orthogonal matrix $M$. It writes
$$\left( \begin{array}{ccccc}
 c_1 & s_1c_2 & s_1s_2c_3 & s_1s_2s_3c_4 &  \ldots   \\\\
-s_1 & c_1c_2 & c_1s_2c_3 & c_1s_2s_3c_4 &  \ldots   \\\\ 
0 & -s_2 & c_2c_3 & s_2s_3c_4 &  \ldots     \\\\  
0 & 0 & -s_3 & c_3c_4 & \ldots     \\\\ 
0 & 0 & 0 & -s_4 & \ldots    
 \end{array} \right)$$
where $(c_j,s_j)$ are cosine/sine pairs.

REPLY [3 votes]: I'd do this in three steps:

Find any $2n \times 2n$ matrix $A$ whose eigenvalues are $e^{\pm i \theta}$.
Find a positive definite quadratic form preserved by $A$. In equations, we want $A P A^T = P$. 
Find an orthonormal basis for $P$, using the Gram-Schimdt algorithm. In equations, we want $S P S^T = \mathrm{Id}$. 

Then $S A S^{-1}$ is orthogonal and has the required eigenvalues.

I can think of two ways to do step 2. The first is more purely algebraic, the second I think would be much easier to implement.
Algebra: Let $f(x) = \prod_{j=1}^{n} (x-e^{i \theta_j}) (x - e^{-i \theta_j}) = \prod (x^2 - 2 \cos \theta_j + 1)$ be your characteristic polynomial. Let $V$ be the ring $\mathbb{R}[x]/f(x)$. Note $1$, $x$, ..., $x^{2n-1}$ is a basis for this ring, in which multiplication can be written down algebraically in terms of the coefficients of $f$. Also, multiplication by $x$ has the desired eigenvalues, so that accomplishes part 1.
For $y \in T$, let $T(y)$ be the trace of multiplication by $y$. Also, let $y \mapsto \overline{y}$ be the automorphism of $V$ induced by $x \mapsto x^{-1}$. Again, both of these can be written down, in the monomial basis, algebraically in terms of the coefficients of $f$.
Then $\langle y,z \rangle = T(y*\overline{z})$ is the desired positive definite quadratic form. Namely, observe that the ring $V$ is isomorphic to $\mathbb{C}^{\oplus n}$. In terms of this isomorphism, $\langle (z_1, \ldots, z_n), (z_1, \ldots, z_n) \rangle = 2 \sum |z_i|^2$.
In practice: The condition that $APA^T = P$ is a linear condition on $P$. Let $W$ be the subspace of the vector space of symmetric matrices where this condition is satisfied; finding $W$ is just algebra. Now, our goal is to find a positive definite element of $W$. For large $N$, $(1/N) \left( \mathrm{Id} + AA^T + A^2 (A^{T})^2 + \cdots + A^{N-1} (A^T)^{N-1} \right)$ is positive definite and is near $W$. I would guess that the orthogonal projection of this matrix onto $W$ would probably be positive definite for large $N$.<|endoftext|>
TITLE: Christoffel symbols on a Lie group in Riemann normal coordinates
QUESTION [11 upvotes]: Consider a coordinate patch around the identity element in a  Lie group given by the exponential mapping (Riemann normal coordinates).  We have  a Levi-Civita connection corresponding to the bi-invariant metric associated with the Killing form.
What are the Christoffel symbols for this metric in the Riemann normal coordinates?
Here is a refinement of this question:  The Christoffel symbols will have a Taylor expansion in the Riemann coordinates  with the coefficients being some tensors constructed out of the Lie algebra structure constants. The first nonvanishing tensor is a 4-tensor constructed out of two factors of the structure constants. That one is easy to compute – this is related to the curvature 4-tensor at the identity. The question is – are the other tensors known in any closed form?

REPLY [3 votes]: The Riemann normal coordinates on a group manifold are obtained by the application
of the exponential map to its Lie algebra.
This description (and notation) is based on the appendices of the following article by Marinov and terentyev.
http://physics.technion.ac.il/~site/wisw/memoriam/marinov/pub/Mar46.pdf
It is assumed that the group is semisimple.
I have written explicit expressions up to the third order. The Einstein summation convention is implemented.
The invariant metric on a semisimple Lie group manifold has the form:
$g_{ab}(\xi) = M_a^p(\xi) M_b^q(\xi) G_{pq}$
where: $G_{pq}$ is the Cartan-Killing form of the group Lie algebra.
The vielbeins $M_a^b(\xi)$ have the form:
$M_b^a(\xi) = \int_0^1 Ad(exp(\tau\xi))_b^a d\tau$
The Christoffel symbols are are given by:
$\Gamma_{ab}^c (\xi) = L_p^c (\xi)(\partial_a M_b^p +\partial_b M_a^p)$
where:  $L_a^b(\xi)$ are the inverse vielbeins:
$L_a^b(\xi)M_b^c(\xi) = \delta_a^c$
given by:
$L_b^a(\xi) = \frac{\partial(log(exp(\eta)exp(\xi))^a)}{\partial \eta^b}|_{\eta=0}$
Using the Baker Campbell Hausdorff formula, we obtain the following expression for the inversse vielbeins 
$L_b^a(\xi) = \delta_b^a +\frac{1}{2}C_{bc}^a\xi^c+\frac{1}{12}C_{bc}^pC_{pd}^a\xi^c\xi^d +\frac{1}{720}C_{bn}^m C_{mt}^r C_{rp}^q C_{qs}^a\xi^n \xi^t \xi^p \xi^s + ...$
Using the Hadamard formula, we obtain the following expression for the vielbeins 
$M_b^a(\xi) = \delta_b^a -\frac{1}{2}C_{bc}^a\xi^c +\frac{1}{6}C_{bc}^pC_{pd}^a\xi^c\xi^d -\frac{1}{24}C_{bn}^m C_{mt}^r C_{rp}^a \xi^n \xi^t \xi^p +\frac{1}{120}C_{bn}^m C_{mt}^r C_{rp}^q C_{qs}^a\xi^n \xi^t \xi^p \xi^s + ...$<|endoftext|>
TITLE: Banach-Mazur distance between $\ell^p$-norms
QUESTION [13 upvotes]: Let $E^n$ be the real or complex space of dimension $n$. If $N$ and $M$ are two norms over $E^n$, and if $A$ is an endomorphism, then 
$$\|A\|^M_N:=\sup_{x\ne0}\frac{M(Ax)}{N(x)}$$
is an operator norm of $A$.
The Banach-Mazur distance between $N$ and $M$ is 
$$d(N,M):=\log\inf_{A\in Aut(E)}\|A\|^M_N\|A^{-1}\|^N_M.$$
It is actually a semi-distance, with $d(N,M)=0$ iff one passes from $N$ to $M$ by a change of coordinates. Thus $d$ is a distance on some quotient space, in which there is only one Euclidian norm, for instance.
Let us define the usual $\ell^p$-norm by
$$\|x\|_p:=\left(\sum_j|x_j|^p\right)^{1/p}.$$
The $L^\infty$-norm is as usual.
It has been known for a long time that if $1\le p,q\le2$, or if $2\le p,q\le\infty$, then
$$d(\ell^p,\ell^q)=\left|\frac{1}{p}-\frac{1}{q}\right|\log n.$$
What is known if $1\le p\le2\le q\le+\infty$ ?

REPLY [13 votes]: I've just looked in the "Handbook of the Geometry of Banach Spaces", Chapter 1, Section 8, which gives that  $$d(\ell^p,\ell^q) = \log\max(n^{|1/p-1/2|}, n^{|1/q-1/2|})$$ up to a constant (which seems hard to find!)  The argument is not entirely elementary, using cotype estimates.  I'd also look in Tomczak-Jaegermann's book "Banach-Mazur distances and finite-dimensional operator ideals" but I don't have that on my shelf right now.<|endoftext|>
TITLE: What is sheaf cohomology intuitively?
QUESTION [85 upvotes]: What is sheaf cohomology intuitively? 
For local systems it is ordinary cohomology with twisted coefficients. But what 
if the sheaf in question is far from being constant? 
Can one still understand sheaf cohomology in some "geometric" way?
For example I would be very interested in the case of coherent $\mathcal{O}_X$-Modules. 
Or even just line bundles.

REPLY [12 votes]: This sounds like a cop-out, I know... what it is not very far from reality, at least in some contexts:
One is not really interested in the higher cohomology groups but only in the actual group of global sections, and higher cohomology groups' purpose in life is to allow us to do inductive proofs and other tricks like dimension shift.<|endoftext|>
TITLE: Betti numbers of moduli spaces of smooth Riemann surfaces
QUESTION [21 upvotes]: Where can I find a list of the known Betti numbers of the moduli spaces $\mathcal{M}_{g,n}$ of genus $g$ Riemann surfaces with $n$ marked points? I need it to cross check results by an implemented algorithm which should be producing them using Kontevich's graph complex. 
I am interested in the "open" moduli space consisting of smooth connected surfaces, not in its Deligne-Mumford compactification $\overline{\mathcal{M}}_{g,n}$. Also, I'm interested in the single Betti numbers and not in the Euler characteristic, which I know from e.g. Harar-Zagier and Bini-Gaiffi-Polito, and which I used to have a first check of the results of the algorithm. 
Thanks.
Edit: Riccardo Murri's paper with the algorithm and its implementation has now appeared on arXiv: http://arxiv.org/abs/1202.1820

REPLY [7 votes]: Complementing the earlier answers:
In genus zero, a very useful reference is Ezra Getzler, "Operads and moduli spaces of genus 0 Riemann surfaces". He determines generating series for the $\Sigma_n$-equivariant Poincaré polynomials of $\overline M_{0,n}$ and $M_{0,n}$, and proves the purity of the cohomology of $M_{0,n}$: the cohomology group $H^i(M_{0,n})$ carries a pure Hodge structure of weight $2i$ (of Tate type). 
Also two small remarks on algori's answer. (i) If I am not mistaken, the quotient map $F(\mathbf P^1,n) \to M_{0,n}$ is in fact a trivial $\mathrm{PGL}_2$-bundle, since it has a section given by putting the first three points at $0$, $1$ and $\infty$. So things are even simpler. (ii) If you don't care about the $\Sigma_n$-equivariance, then you can use exactly the same tool (the Cohen-Taylor-Totaro spectral sequence) but consider instead the configuration space $$ M_{0,n} \cong F(\mathbf P^1 \setminus \{0,1,\infty\},n-3). $$
Since $\mathbf P^1$ minus three points has $H^0$ of weight $0$ and $H^1$ of weight $2$, and the differentials are compatible with weights, it follows that the spectral sequence degenerates immediately so one can directly read off the Betti numbers and the purity.
In genus one, there is a recent preprint of Gorinov, available at http://www.liv.ac.uk/~gorinov/, that determines the cohomology of $M_{1,n}$ with its Hodge structure. I believe that what he does is the following: 
Consider the configuration space of $n$ points on an elliptic curve, $F(E,n)$, quotiented by the action of $E$ by translation. The cohomology of this space can be computed via the Cohen-Taylor-Totaro spectral sequence. If you are careful you can even determine how the cohomology of $F(E,n)/E$ is built out of (symmetric powers of) $H^1(E)$ and some Tate twists. 
Now consider the forgetful map $\pi \colon M_{1,n}\to M_{1,1}$. The results of the previous paragraph tell you how the local systems $R^i \pi_\ast \mathbf Q$ are built out of symmetric powers of the "standard local system", which is $R^1\pi_\ast \mathbf Q$ for $n=1$. Finally, the cohomology of these local systems is given by Eichler-Shimura theory; the cohomology groups are canonically given by spaces of modular forms for $\mathrm{SL}(2,\mathbf Z)$. As far as I can tell, the Eichler-Shimura theory is the only part which is specific to genus one: everything else would work without changes in higher genus as well. 
Maybe I should mention at this point that there is plenty to read about how to compute the cohomology of these local systems on $M_g$ when $g \geq 2$. For $g=2$ and $g=3$ you can read a sequence of papers of Faber, van der Geer and Bergström, who work on computing/conjecturing Euler characteristics of these local systems in the Grothendieck group of $\ell$-adic Galois representations by means of point counts. The Euler characteristic doesn't give you the cohomology, but it gives you partial information. In particular because you know that these local systems are pulled back from $A_g$, and on $A_g$, most cohomology groups of these local systems vanish. (There is a wealth of information in Faltings-Chai, chapter 6.) Then you can apply the Gysin exact sequence for the image of $M_g$ in $A_g$ under the Torelli map. 
Let me finally make a few remarks on genus two. If you only want the Betti numbers (i.e. you don't care about torsion), then there are easier ways than the papers linked by Oscar Randal-Williams. When $n=0$, there is an isomorphism $M_2 \cong M_{0,6}/\Sigma_6$ on coarse moduli spaces: every curve of genus two is hyperelliptic, and a hyperelliptic curve is uniquely determined by the $2g+2$ branch points on $\mathbf P^1$ under the hyperelliptic map. So rationally, you need only to take the $\Sigma_6$-invariants in Getzler's description of the cohomology of $M_{0,n}$, and you find that $M_2$ has the rational cohomology of a point.
When $n=1$, you apply the Leray spectral sequence for $\pi \colon M_{2,1} \to M_2$. The local systems $R^0\pi_\ast\mathbf Q$ and  $R^2\pi_\ast\mathbf Q$ are trivial, and $R^1\pi_\ast\mathbf Q$ has no cohomology since every point of $M_2$ has the hyperelliptic involution in its automorphism group, which acts as multiplication by $-1$ on the fibers of the latter local system (this also uses that you take rational coefficients). You conclude that $H^\ast(M_{2,1}) \cong H^\ast(\mathbf P^1)$. 
Note that this generalizes to hyperelliptic curves of any genus: you always have that $H_g$ has the rational cohomology of a point, and $H_{g,1}$ the cohomology of $\mathbf P^1$. 
For $n=2$ you need to compute the cohomology of the local systems $V_{2}$ and $V_{1,1}$ on $M_2$. Getzler manages to determine all but two Betti numbers without this information in "Topological recursion relations in genus two" and expresses the final Betti numbers in terms of their cohomology. The cohomology of these local systems can be found in the more recent paper by Hulek and Tommasi, "Cohomology of the second Voronoi compactification of $A_4$", Appendix A. (This appendix also contains some things useful for $M_3$.) Together these papers determine the Betti numbers of $M_{2,2}$. I would guess that you can compute also the Betti numbers of $M_{2,3}$ from this information, since the only "new" local systems that appear for $n=3$ have odd weight so their cohomology vanishes, but I have not sat down to compute the Leray spectral sequence here.
The Betti numbers for $M_{2,4}$ I suspect are unknown. Tom Church writes above that they can be found in Bergström-Tommasi, but I think this is a misreading of their paper. As Tom writes, much of the paper of Bergström and Tommasi summarizes their previous work, done separately and by different methods. Tommasi's work uses the Vassiliev-Gorinov method of computing the cohomology of complements of discriminants. This gives you the Poincaré-Serre polynomial, and in particular the Betti numbers. Bergström uses point counts over finite fields, which gives you the Euler characteristic in the category of $\ell$-adic Galois representations (or what they call the Hodge Euler characteristic), but NOT in general the Betti numbers. The results on $M_{2,n}$ are due to Bergström in "Equivariant counts of points of the moduli spaces of pointed hyperelliptic curves".<|endoftext|>
TITLE: Discriminant and Different
QUESTION [23 upvotes]: First some context. In most algebraic number theory textbooks, the notion of
discriminant and different of an extension of number fields $L/K$, or rather, of the corresponding extension $B/A$ of their rings of algebraic integers is defined.
The discriminant, an ideal of $A$, is the ideal generated by the discriminant of the quadratic form $\text{tr}(xy)$ on $B$. The different, an ideal of $B$, is the inverse of
the fractional ideal $c$ of $L$ defined by $c=\{x \in L, \text{tr}(xy) \in A \ \forall y \in B\}$. The norm of the different is the discriminant.
Now the discriminant makes sense in a much more general context, say for any extension of (commutative) rings $B/A$ that is finite projective, since the trace map $\operatorname{tr}$ makes sense in this context. My question is: is there a standard definition of the different 
in this context? if so, where can I find it in the literature, if possible with the basic results about it?
I am pretty sure the answer to the first question is yes, but I have not been able
to find a reference. The problem when I try to use google or MathSciNet seems to be that
"different" is not a very discriminant name: almost every paper in mathematics contains it.
Let me propose an answer to my own question: we could define the different of
$B/A$ by the Fitting ideal of the universal $B$-modules of differentials $\Omega_{B/A}$. The fact that it gives the correct definition in the number field cases is [Serre, Local Fields, chapter III, Prop. 14], and moreover it behaves well under base change. This definition may very possibly be a remembrance  of something I had heard in an earlier life. But even if it is the correct definition, I'd like to know a reference where it is stated.

REPLY [8 votes]: Tu peux regarder la section 1 de mon article "La filtration de Harder-Narasimhan des schémas  en groupes finis et plats":
https://www.imj-prg.fr/~laurent.fargues/HNgp.pdf
Tout ce dont tu rêves y est démontré.<|endoftext|>
TITLE: Random linear recurrence relations
QUESTION [8 upvotes]: Problem
I am interested in the random recurrence relation of the form $x_{n+1}=\alpha x_n \pm \beta x_{n-1}$ where $\alpha$, $\beta$ are known constants and the $\pm$ sign is chosen with equal probability. 
In particular I would like to investigate the limit of $|x_n|^{1/n}$ as $n\rightarrow\infty$. (i.e the exponential growth rate)
Motivation
I was considering the problem of finding $\lim_{n\rightarrow\infty}|x_n|^{\frac{1}{n}}$ where $x_n$ is the solution to $x_{n+1}=2x_n \pm x_{n-1}$, $x_0=x_1=1$. All numerical evidence suggests that $\lim_{n\rightarrow\infty}|x_n|^{\frac{1}{n}}\approx 1.91$ almost surely but I am having difficulty even proving that the sequence almost surely convergences.  
Survey
The solution to the recurrence relation of the from $x_{n+1}=x_n \pm \beta x_{n-1}$ has been show to have expontential growth almost surely. See http://en.wikipedia.org/wiki/Embree%E2%80%93Trefethen_constant
Also there has been work on the random fibonacci sequence $x_{n+1}=x_n \pm x_{n-1}$ for example Viswanath(2000), "Random Fibonacci sequences and the number 1.13198824"

REPLY [2 votes]: You question amounts to proving that the norms of products of i.i.d. random matrices  $({{\alpha \;\;\pm\beta}\atop{1\;\;\;\; 0}})$ have a non-zero exponential growth rate. This is a particular case of a famous theorem of Furstenberg ("Non-commuting random products", 1963) on positivity of the top Lyapunov exponent for products of random matrices. Existence of the limit a.e. is an earlier theorem of Furstenberg and Kesten, which is nowadays proved by referring to Kingman's subadditive ergodic theorem.<|endoftext|>
TITLE: Theorems first published in textbooks?
QUESTION [31 upvotes]: According to Wikipedia, the Bohr-Mollerup Theorem (discussed previously on MO here) was first published in a textbook. It says the authors did that instead of writing a paper because they didn't think the theorem was new. 
What other examples are there of significant theorems that first saw the light of day in a textbook? (I'm assuming Wikipedia is right about Bohr-Mollerup.) 
I recognize that the word "significant" is imprecise; I have in mind theorems that mathematicians have picked up on and used in their own work, but I'm open to other interpretations.

REPLY [4 votes]: Intersection theory has of course a long tradition, starting from the work of Schubert, Poincaré and many more, but the book Intersection theory by Fulton is by now considered the gold reference. Many results in the book were new at the time. Citing from the introduction

Several theorems are new or stronger than those which have appeared before, and some proofs are significantly simpler. Among the former are a new blow-up formula, a stronger residual intersection formula, and the removal of a projective hypothesis from intersection theory and Riemann-Roch theorems; the latter includes the proof of the Grothendieck-Riemann-Roch theorem. Some formulas from classical enumerative geometry receive a first modern or rigorous proof here.<|endoftext|>
TITLE: Example of what goes wrong with the functor category $D^C$ if $C$ is not small?
QUESTION [5 upvotes]: Typically, when defining the functor category $\mathcal{D}^\mathcal{C}$, where objects are functors $\mathcal{C}\rightarrow\mathcal{D}$ and the morphisms between such objects $F,G$ are the natural transformations $F\rightarrow G$ with the obvious composition and identities, one requires the category $\mathcal{C}$ to be small, i.e. that its objects form a set. $\mathcal{D}$ can be any category.
I've always thought that the reason for the smallness requirement on $\mathcal{C}$ is that "surely you'll get into trouble with the morphisms $F\rightarrow G$ in $\mathcal{D}^\mathcal{C}$ potentially not forming a set otherwise", and I've never given it much thought.
Now, does anyone know of an example which fixes a large category $\mathcal{C}$, a category $\mathcal{D}$ and two functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ such that that the collection of natural transformations $F\rightarrow G$ does not form a set? I've always thought in the back of my head that this should be easily doable with $\mathcal{C}=\mathcal{D}=\mathrm{Set}$ and some simple functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ by somehow cooking up at least one natural transformation $F\rightarrow G$ for each object of $\mathcal{C}$, i.e. for each set. Maybe my set theory fu is just weak.
I could of course just be missing something basic here. I'm asking because the books I've checked tend to just state $\mathcal{C}$ must be small. Apologies if this is all trivial!

REPLY [10 votes]: Hi,
let C be a discrete category (i.e. the only morphisms in C are the identity morphisms) and let D be the category consisting of two objects 0 and 1 and (apart from the identity morphisms) two parallel arrows $0\rightrightarrows 1$. Consider the constant functors F with value 0 and G with value 1. A natural transformation $F\to G$ corresponds to the choice of an element in a set of two elements for every object of C so that the conglomerate of all natural transformations $F\to G$ can be identified with the power conglomerate of Ob(C). This is not a set unless Ob(C) is a set.<|endoftext|>
TITLE: What is this operation on graphs called?
QUESTION [8 upvotes]: I am currently studying certain infinite graphs in terms of their finite induced subgraphs.
For the graphs that I am interested in the class of finite induced subgraphs is closed under the following operation:
Given two graphs $G=(V(G),E(G))$ and $H=(V(H),E(H))$ and a vertex $v$ of $G$,
let $G\otimes_v H$ (notation invented on the spot by myself) be the graph on the disjoint union of $V(G)\setminus\{v\}$ and $V(H)$ in which two vertices $x$ and $y$ are connected by an edge iff one of the following holds:
(1) $x,y\in V(G)\setminus\{v\}$ and $\{x,y\}\in E(G)$.
(2) Exactly one of $x$ and $y$ is in $V(H)$, say $y$ (wlog), and $\{x,v\}\in E(G)$.
(3) $\{x,y\}\in E(H)$.
In other words, the vertex $v$ of $G$ is replaced by a copy of $H$, and every
vertex $w$ of $G$ different from $v$ 
is connected to all vertices of the copy of $H$ if 
$w$ and $v$ are connected in $G$.  Otherwise $w$ is not connected to any of the vertices of the copy of $H$.
(Note that I am only doing this at a single vertex of $G$, not all of them. Otherwise I would get the wreath product or lexicographic product as mentioned in Nathann Cohen's answer below.)    
Since this is a natural operation between graphs (with a distinguished vertex of the first graph), I would guess this has a name.
If yes, how is this called?

REPLY [5 votes]: After looking through some of the literature, it seems that a common name for this operation
is substitution:  the graph $H$ is substituted for the vertex $v$ of the graph $G$.
This is what Lovasz calls the operation in his paper where he proves the perfect graph theorem
(PGT: complements of perfect graphs are perfect).
There is also the Lovasz substitution lemma which says that when a perfect graph is substituted for a vertex of a perfect graph, then the resulting graph is again perfect.<|endoftext|>
TITLE: How to distribute the source of programs used in a paper?
QUESTION [18 upvotes]: I have written a paper, which includes an appendix discussing how to obtain numerical evidence for the result of the paper. Now the computation essentially works as follows:

Create a large tridiagonal matrix.
Compute its eigenvalues.
Compute the difference of consecutive eigenvalues, and output it.

The implementation of such an algorithm is rather straightforward, but in order to look at large matrices, I started using algorithms from a package called LAPACK, which turned out to be faster then regular algorithms provided by Matlab. (I'm no specialist, so not exactly sure what happens).
I am curious if one should provide the source code for such a computation, and if yes in what form. I cam up with the following options:

Pseudocode (as  above)
Simplified matlab, that works with any installation of matlab, but is too slow to actually do the computations.
The real code, which most people will not be able to get to run without some effort.

I am also curious if one should include some sort of source code in the paper, and if yes, in what form? Or what people have done in such a case...
The simplified code is available at:
http://math.rice.edu/~hk7/ftp/matlab_code/SkewSpecDense.m
I have not put the real code online, because it requires external packages, and I am not sure how easy it is to install them...

REPLY [2 votes]: At least for numerical algorithms, submitting it and getting it accepted to ACM Transactions on Mathematical Software ensures that the code is mirrored on both the ACM site and the various mirrors of Netlib.

Added much later:
Nowadays, I would recommend setting up a GitHub or BitBucket account (or whatever is your preferred source management hosting site), put up your code there, and then include a link to it in your paper or preprint.<|endoftext|>
TITLE: The question about Kolmogorov tightness criterion
QUESTION [5 upvotes]: We know about Kolmogorov Criterion for the tightness of a stochastic process $X_n(t)$
1.The sequence $(X_{n}(0))_{n\geq0}$ is tight.
2.There exist constants $\gamma\geq0$,$\alpha>1$, $K>0$ and an integer $n_0$ such that
$$E(|X_{n}(t_{2})-X_{n}(t_{1})|^{\gamma})\leq K|t_{2}-t_{1}|^{\alpha}, \forall n \geq n_0$$
for all $t_{1},t_{2}$.
My first question: what should the $n_0$ depend? Could it depend on the $t_{1}$ and $t_{2}$?
My second question: Is there any other criterion for tightness with the parameter $\alpha=1$ for the version of the moment condition?

REPLY [2 votes]: $n_0$ must be independent of $t_1$ and $t_2$, of course. If it's not, the processes might be even discontinuous. For instance, $X_n$ is a Poisson process with parameter $1/n$. Then $$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$$ for all $n>|t_1-t_2|^{-1}$ (for all $n\ge 1$ if $t_1=t_2$). 
And the same answer works for the second question: when $\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower one polynomially, it may help (e.g. in the Gaussian case $\gamma=2$ and $\alpha=1$ is enough).<|endoftext|>
TITLE: Number of irreducible polynomials with some coefficients fixed over a finite field
QUESTION [8 upvotes]: I am interested in the following problem: I have a finite field $F_q$, two positive integers
$n>m$ and elements $a_1,...,a_m\in F_q$. How many of the polynomials
$x^n+a_1x^{n-1}+...+a_mx^{n-m}+c_{m+1}x^{n-m-1}+...+c_n,c_i\in F_q$ are irreducible?
What are the best known estimates, esp. for $q$ fixed and $m,n\to\infty$?

REPLY [10 votes]: This is similar to counting irreducibles in arithmetic progressions modulo $x^m$ (once you replace $x$ by $1/x$). You can turn the problem into counting rational points on a curve (coming from a "cyclotomic function field" in the sense of Carlitz) over $F_{q^n}$ and get an estimate $q^n/n + O(gq^{n/2})$, where $g$ is the genus of the curve. Unfortunately $g$ grows like $mq^m$ so you only get good estimates for $m$ small and nothing when $m$ gets close to $n/2$. There are plenty of papers on this (e.g. by S. Cohen). There is also some experimental work by Panario et al.. Mathscinet should help you locate these.<|endoftext|>
TITLE: Ideals in a noncommutative ring such that their product is their intersection?
QUESTION [6 upvotes]: If $R$ is a commutative ring and $I$ and $J$ are ideals in $R$ such that $I+J=R$ then $I \cap J=IJ$. This is not generally true in noncommutative rings, e.g. let $R$ be the lower triangular 2 x 2 matrices over $\mathbb{Z}$ and let $I$ be the ideal generated by $E_{1,1}$ and $J$ be the ideal generated by $E_{2,2}$. 
Is this ever possible in a noncommutative ring? If so, what are rings satisfying this property called? (Where can we find them in the literature?)

REPLY [3 votes]: I don't know non-commutative ring theory that well yet, but I ended up working this problem out anyway partially for my own edification.  Let us suppose that I, J are both left ideals in R.   Observe,
$(I \cap J) (I + J) = (I \cap J) I  + (I \cap J) J \subseteq IJ + JI$
If I + J = (1), then we get
$I \cap J \subseteq IJ + JI$
Now it can be shown that 
$I J \subseteq comm(I,J) \cap J$
where comm(I,J) denotes the set of all elements in I which commute with elements in J;
$comm(I, J) = ${$ x \in I : \forall y \in J; x y \in I $}
Therefore,
$I \cap J \subseteq IJ + JI \subseteq comm(I,J) \cap J + comm(J,I) \cap I$
If it is true that all of I commutes with J and vice-versa, then $comm(I,J) = I$, $comm(J,I) = J$ and we get:
$I \cap J = I J$
Now let us suppose that these bounds are not tight, or in other words there is some element $x \in I \setminus comm(I,J)$.  This implies that there exists some $y \in J$ such that $x y \not \in I$, and so $x J \not \subseteq I \cap J$.
Therefore if $I + J = (1)$, then $I \cap J = I J + J I$ if and only if $comm(I,J) = I$ and $comm(J,I) = J$.
(Of course there is probably an easier way to say this, so maybe it is also good to get an expert to weigh in on the issue.)<|endoftext|>
TITLE: Fibonacci sequence inversion
QUESTION [6 upvotes]: How do I get the index in the sequence from the Fibonacci number?
0 1 1 2 3 5 8 13 21 34...
For example
N(3) = 4 (starting from zero)
N(34) = 9 (starting from zero)
..
N(X) = ?
I've seen an equation in wikipedia

There are other ways to compute it?

REPLY [2 votes]: You can find the index using only integer arithmetic.
Since $F_n$ is monotone one can use Binary search
using the fact that $F_i$ can be computed in $O(\log(i))$.
This method works for all monotone sequences and can be used to check if a number is in the sequence.<|endoftext|>
TITLE: Is the sphere the only surface with circular projections? Or: Can we deduce a spherical Earth by observing that its shadows on the Moon are circular?
QUESTION [85 upvotes]: Several ancient arguments suggest a curved Earth, such as
the observation that ships disappear mast-last over the
horizon, and
Eratosthenes'
surprisingly accurate calculation of the size of the
Earth
by measuring a difference in shadow length between
Alexandria and Syene. These observations, however, suggest
merely a curved Earth rather than a spherical one.
Another ancient argument specifically suggesting a
spherical Earth is the fact that the shadow of the Earth on
the moon during a Lunar eclipse is circular.
My question is: is it true that the sphere is
the only surface all of whose projections are disks?
It surely seems to be true. The corresponding fact,
however, is not true in two dimensions. The Reuleaux
triangle
pictured below is a figure of constant
width,
meaning that every projection of it in the plane is a line
segment of the same length.

There are also surfaces of constant
width
in higher dimensions, meaning that any two parallel
bounding set of hyperplanes (touching the boundary) have
constant separation. But all of the non-spherical examples
of such surfaces I have seen have obviously non-circular
projections.
It also seems clear that finitely many circular projections
is insufficient, since intersecting finitely many cylinders
would produce a surface having corners and containing some
straight line segments.
The fact that you can spin such a surface with all circular
projections inside any bounding cylinder is suggestive, but it
is also true that you can spin the Reuleaux triangle
inside a square,
even though it isn't circular.

Further questions would include:

To what extent are other surfaces determined by their
projections? That is, which other shapes can we recognize
by the set of their shadows?
In particular, can we recognize the cube and other regular solids by their shadows?
Which sets of shadows are realizable as projections of a
surface? Is there some way to characterize these sets?
Clearly they must be continuously deformable to one
another and obey several other obvious conditions.

We had a great time discussing the question after our logic
seminar here in New York this week, when our speaker
Maryanthe Malliaris asked
the spherical Earth question.
December 20, 2010: In light (or dark, as it were) of the lunar eclipse tonight, I am bumping this question,  with the remark also that despite the truly outstanding answers we have received, several of the further questions stated above are not fully answered.

REPLY [6 votes]: The first two questions (and others) are discussed in the following two papers of Kuzminykh, A. V.: 

An isoprojection property of the sphere (1973) 
Reconstructibility of a convex body from the set of its projections (1984)<|endoftext|>
TITLE: Version of the Poincaré Inequality
QUESTION [8 upvotes]: Let $\Omega\subset \mathbb{R}^n$ open and bounded. The Poincaré inequality
$$\|u\|_p \le C \|\nabla u\|_p$$
($\|\cdot\|_p$ denotes the usual $L^p(\Omega)$-norm; the Lebesgue measure shall be used here)
is valid in the following cases:

$u$ is zero on $\partial\Omega$ (in the $W^{1,p}$-sense)
$u$ has an average value of zero (this implies an estimate of $\|u-\bar u\|_p$, but not of $\|u\|_p$ for general $u$)
$\mu := |\{u=0\}| > 0$ (Lebesgue measure), with a constant that blows up as $\mu\to 0$.

Of course the inequality cannot hold in general (take $u$ to be a constant function), but the only obstruction seems to be that $u$ can be far away from zero. Therefore it should be true that the inequality holds whenever $u$ is zero somewhere in $\Omega$ (in a suitable sense, for example, zero is contained in the essential range of $u$).
So the question is: Is there a version of the inequality for this case?

REPLY [13 votes]: The Poincaré inequality need not hold in this case.  The region where the function is near zero might be too small to force the integral of the gradient to be large enough to control the integral of the function.
For an explicit counterexample, let $$\Omega = \{(x,y) \in \mathbb{R}^2 : 0 < x < 1, 0 < y < x^2\}$$ be the region under the graph of a parabola, and take $p=2$.  For $\epsilon > 0$, let $$u_\epsilon(x,y) = \begin{cases} x/\epsilon, & x < \epsilon \\\ 1, & x \ge \epsilon \end{cases}.$$  $0$ is in the essential range of each $u_\epsilon$, but one can easily verify $||\nabla u_\epsilon||_2^2 = \epsilon/3 \to 0$ as $\epsilon \to 0$, whereas $||u_\epsilon||_2^2 \to |\Omega| = 1/3$.  
Edit: To address Piero's comment, the irregularity of the domain is not the main issue here.  For another counterexample, let $\Omega$ be the unit ball in $\mathbb{R}^d$, $d \ge 3$, and take
$$u_\epsilon(x) = \begin{cases} \frac{|x|^2}{\epsilon^2}, & |x| < \epsilon \\\ 1, & |x| \ge \epsilon \end{cases}.$$  Using polar coordinates, one easily computes $||\nabla u_\epsilon||_2^2 \sim \epsilon^{d-2}$ while again $||u_\epsilon||_2^2 \to |\Omega|$.
What's really the problem is that the set where $u$ vanishes has zero capacity.  If you can control from below the capacity of the set where $u$ vanishes, then you can get a Poincaré inequality.
Indeed, the following theorem may be what the asker wants: for "reasonable" $\Omega$ (Lipschitz suffices), if $\mathrm{Cap}(\{u = 0\}) \ge \delta$, then $||u||_2^2 \le \frac{C}{\delta} ||\nabla u||_2^2$.  The precise statement (for any value of $p$) can be found in section 4.5 of William P. Ziemer's Weakly Differentiable Functions (along with the definition of capacity).
Note in particular that $\mathrm{Cap}(E) \ge m(E)$, so it's enough to control the Lebesgue measure.  This is your bullet point 3.<|endoftext|>
TITLE: Efficiently computing a matrix's induced p-norm
QUESTION [12 upvotes]: Suppose $A$ is an $m\times n$ real matrix and we need to find $\left\|A\right\|_p$ for $p \notin \{ 1, 2, \infty \}$. What is the most efficient way to compute $\left\|A\right\|_p$? 
Here's one naive approach I can think of. Sample random points $\left\|x\right\|$ on the unit hypersphere  , computing $\left\|Ax\right\|_p$ for each such and take the maximum. What I would like to know is the runtime of this approach for the "average" A, and how we can optimize this for special classes of matrices (like Diagonal, Orthonormal, etc.)?

REPLY [7 votes]: S.W. Drury derives a method to find the operator norm of a general real matrix 
$$
A : \ell^p \longrightarrow \ell^q
$$
in a recent 
paper 
in Lin. Alg. Appl (and using it, refutes a long-standing conjecture of Matsaev).
In keeping with the answer of Alex Olshevsky, the algorithm seems have a complexity exponential in the number of columns of the matrix (but linear in the number of rows).
Drury's implementation for Visual C++ and Maple can be found here, and a C version targeted at Unix and with bindings for Matlab, Octave and Python can be found 
here.<|endoftext|>
TITLE: Isodiametric hull
QUESTION [7 upvotes]: Let A be a convex compact set in the plane (with a piecewise smooth boundary, say). We want to  `inflate' it in such a way that the diameter does not increase. 
More accurately, we are looking for all sets C such that 
a) A is a subset of C; 
b) diam(A)=diam(C)
Let now B is the largest possible set C which satisfies these two properties. 
By `largest' I mean either that it m(B) = max m(C), where m is the Lebesgue measure; or that B actually contains any C with these properties. Let us call B the isodiametric hull of A.
The simplest example of A is of course the square: here B is the superscribed disc, and it is the isodiametric hull of A in the strong sense.
Another example is the equilateral triangle, for which B is the Reuleaux triangle. Similarly, for any regular 2n-gon we have the disc, and for any regular (2n+1)-gon its isodiametric hull is a Reuleaux polygon. 
The first non-trivial example that comes to mind is an isosceles triangle that isn't equilateral. It is clear that the hull is always a set of constant diameter but how does one actually obtain it? It seems that its boundary - a curve of constant width - is not a finite union of circular arcs.
I wonder if all this is well known (being such a natural question!). In particular, does the isodiametric hull of a set always exists in the strong sense? 
Added: of course, if there is no IDH in the strong sense, B may not be unique. Its area is unique, though. How does one find it?

REPLY [5 votes]: For any shape that is not constant width, there are many different maximal elements of the same diameter containing it.

Constant width shapes are maximal. They are regions swept out by line segments of length $D$ swept by their midpoints moving perpendicularly along curves perpendicular to along smooth curves (with cusps) whose tangent line turns 180 degrees
Any shape that is not constant width has some projections that do not have maximal diameter.
For each projection that has maximal diameter, there is a diameter as above perpendicular to the line of projection.  
Any method of interpolating a positively turning curve between the existing diameters will create a maximal element of the set of shapes of the given diameter. There are always infinitely many ways to do this unless the set is already maximal --- any one solution for an interpolating curve can be perturbed anywhere it is not constrained.
This observation can be applied, for exmaple, to the case of squares mentioned in the question. For example, you can begin enlargement of a square by adding an arc of a circle centered at one corner and passing through an opposite corner, then take the convex hull.  More systematically: the square only defines two diameters.  There are many Legendrian sections of
tangent line bundle of plane $\to  \mathbb {RP|^2}$ (cf. Is the sphere the only surface all of whose projections are circles? Or: Can we deduce a spherical Earth by observing that its shadows on the Moon are always circular?) interpolating these two elements.
The total area is locally given by an easy formula, but globally it seems like a highly irregular function, since it depends on how the sweeping diagonal overlaps itself. I doubt if there is a good theory of a maximal area shape of the same diameter
 containing the given one.<|endoftext|>
TITLE: 12 balls weighing puzzle
QUESTION [8 upvotes]: In an article describing the twelve balls weighing problem, the author mentions a solution that involves the finite projective plane of order 3, discovered by Rick Wilson.  Does anyone know what this solution could have been?

REPLY [9 votes]: Will Orrick is right, the problem is solved by exhibiting a matrix $3\times 12$ with entries in $\{-1,0,1\}$ where all columns are pairwise independent and the row sums are zero, as mentioned in Wilson's book.
In general you can solve the $\frac{3^r-1}{2}-1$ coin problem using $r$ weighings. You need to use one of the generator matrices of the simplex code, so the columns are given by the points in the projective space $P(r-1,3)$ (you throw out the point at infinity) and by induction show that the choices can be made to arrange the zero sum rows. For the case of 12 coins it suffices to consider the projective plane, and you get a constant weight code and thus end up weighing groups of 4 coins each time.<|endoftext|>
TITLE: Are there examples of (successful) NSF mathematics proposals available anywhere online?
QUESTION [72 upvotes]: As anyone who has ever applied to the NSF for a grant knows, such a proposal is a slightly odd piece of writing, not quite like anything else mathematicians are called upon to write.  As such, it's a hard thing to learn to do well; of course, the basic requirements are set out in the grant proposal guide, and explained a bit more clearly in some other sources.  But there's no substitute for reading an actual specimen.  
Now, generally this is achieved by asking some older colleague to see an old proposal, an approach that works just fine in most cases.  On the other hand, if there are any publicly available old NSF proposals online, I think finding them would be a boon to all of us preparing proposals, if only to have more data points.  Does any one know of any (I didn't have much luck on Google).
I'll just note: I'm aware that the NSF has abstracts of all funded proposals on their website, which is useful, but doesn't give much insight into, say, what people write about broader impacts.

REPLY [3 votes]: The renewal of this 2010 question asks for more recent NSF math grant proposals.
A collection of proposals from 2000 through 2015 is at Christopher Bishop's home page.
This site maintained by Dan Margalit has one example of a successful NSF math proposal, and in addition contains resources on how to write such a proposal (notably the sections on intellectual merit and broader impact).

The renewed OP also notes broken links: some of these are on the Wayback Archive:

SCREMS:
The Computational Frontiers of Number Theory, Representation Theory,
and Mathematical Physics
James
Propp (1) and (2)<|endoftext|>
TITLE: Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n
QUESTION [7 upvotes]: Let $M_n$ be the set of $n$-by-$n$ matrices with complex entries, viewed as a variety over $k=\mathbb{C}$. Equip $M_n$ with the conjugation action of $\mathrm{GL}(n)=\mathrm{GL}(n,\mathbb{C})$. Consider $A:=\mathrm{Mor}^{\mathrm{GL}(n)}(M_n \oplus M_n, M_n)$, the set of $\mathrm{GL}(n)$-equivariant maps (of algebraic varieties) $M_n \oplus M_n \to M_n$, where $GL(n)$ acts diagonally on $M_n \oplus M_n$. Then the (standard) algebra structure of $M_n$ (with multiplication given by matrix multiplication) induces an algebra structure on $A$. 
The following maps belong to $A$:

$M_n \oplus M_n \to M_n \colon (A,B) \mapsto A$;
$M_n \oplus M_n \to M_n \colon (A,B) \mapsto B$;
$M_n \oplus M_n \to M_n \colon (A,B) \mapsto f(A,B)I_n$, where $I_n$ is the identity matrix and $f$ is an element of the ring of invariants $k[M_n \oplus M_n]^{\mathrm{GL}(n)}$. 

Do they generate $A$ as an algebra?

REPLY [5 votes]: The answer is affirmative not only in the case of 2 matrices, but also in the case of any number of matrices; in fact, an analogous statement is true for quiver representations (in characteristic 0).
The original question can be restated as follows.

Let $P$ be the space of polynomial functions of 2 $n\times n$ matrices, with the adjoint action of $GL_n$ and the ring of invariants $I.$ Consider the space $\text{Hom}_{GL_n}(M_n,P)$ as an $I$-module. Is it true that it is generated by the products of matrices?

For the case of any number of generic matrices $A_1,\ldots,A_k,$ Procesi proved that over a field $k$ of characteristic 0, $I$ is spanned by the traces of the products of matrices. Formally, consider words in the free monoid with $k$ generators, substitute the generic matrices, and take a trace. 
Procesi, C. The invariant theory of n×n matrices. Advances in Math. 19 (1976), no. 3, 306–381
The statement follows by adjoining an extra generic matrix $A_0$ and converting an $M_n$-space into a $GL_n$-invariant forming a product with $A_0$ and taking the trace, then undoing the trace of the term in the trace polynomial from Procesi's theorem containing $A_0.$ 

Here is a vast generalization due to Le Bruyn and Procesi. Given a finite quiver $Q$ and a dimension vector $\alpha,$ consider the corresponding representation space $R(Q,\alpha)$ with the action of the algebraic group $GL(\alpha)$ and the space $P$ of polynomial functions on $R.$ (If the quiver consists of a single vertex with $k$ loops and $\alpha=n$ then the representation space is given by $k$ generic $n\times n$ matrices with the simultaneous conjugation action by $GL_n.$) Then, over a field of characteristic 0, the algebra $I$ of polynomial invariants is spanned by the traces of matrix products over oriented cycles in $Q$ and for any pair of vertices $(i,j)$ of $Q,$ the space $\text{Hom}_{GL(\alpha)}(\text{Hom}_k(V_i,V_j),P)$ is generated as an $I$-module by the products over oriented paths connecting $i$ with $j.$
Lieven Le Bruyn, Claudio Procesi, Semisimple representations of quivers.
Trans. Amer. Math. Soc. 317 (1990), no. 2, 585–598<|endoftext|>
TITLE: Does second-order arithmetic prove every expressible instance of Dependent Choice?
QUESTION [5 upvotes]: Let bigset(X) and rel(X,Y) be otherwise arbitrary formulas in the language of second-order arithmetic with the indicated variables free, and thmemberof(Z,x,X) be the formula asserting that X is the xth member of the sequence of sets coded by Z.  Does it follow that second-order arithmetic proves
$((\exists X)(bigset(X)) \; \land \; (\forall X)(bigset(X) \implies (\exists Y)(bigset(Y) \; \land \; rel(X,Y)))) \implies$
$(\exists Z)(\forall x)(\exists X)(\exists Y)($
$bigset(X) \; \land \; bigset(Y) \; \land \; thmemberof(Z,x,X) \; \land \; thmemberof(Z,\operatorname{S}(x),Y) \; \land \; rel(X,Y))$
?

By second-order arithmetic, I mean Robinson arithmetic + full comprehension + the second-order induction axiom.

REPLY [9 votes]: Carl has pointed out that my previous answer missed a clause in the theorem I cited.
Simpson's book, Subsystems of Second Order Arithmetic, does address this in section VII.6.  He shows that dependent choice for $\Sigma^1_2$ formulas is equivalent to $\Delta^1_2$ comprehension plus $\Sigma^1_2$ induction (Theorem VII.6.9).
However even regular (non-dependent) $\Sigma^1_3$ choice is independent of full comprehension; he attributes this result to Feferman and Levy, and cites Theorem 8 of Levy's "Definability in axiomatic set theory, II".  
The result I mentioned before, that $\Sigma^1_k$ dependent choice is equivalent to $\Delta^1_k$ comprehension plus $\Sigma^1_k$ induction for $k\geq 2$, holds for $k\geq 3$ requires the additional assumption that the universe is constructible from from some set of integers.
Your statement of dependent choice is a bit more complicated than necessary; you can fold bigset into rel (take $rel'(X,Y)$ to hold if either $\neg bigset(X)$ and $bigset(Y)$, or if $bigset(X)$, $bigset(Y)$, and $rel(X,Y)$).  Conversely, it's a bit simpler than the version Simpson uses (which I believe is standard), in which $rel$ can depend on the parameter $x$ as well.<|endoftext|>
TITLE: Solve in positive integers: $n!=m(m+1)$
QUESTION [34 upvotes]: Does anybody know a solution to this problem? (Sorry, I've missed one summand in the previous post.)

REPLY [19 votes]: I thought I should share with you the results from my computations from the past two days.
All these computations could be carried out to significantly higher limits with better code and a bit more time. However, my motivation was just to play with them.
Please check the code before taking these computations as facts! I did not check the code too carefully.

Computation 1
The equation $$n! = m(m+1)$$ does not have integer solutions for $3 < n < 10^9$.
I used Mark Sapir's observation that if $n$ is a solution then $4n!+1$ is a square. I wrote a small program which calculates Legendre symbols $$\left(\frac{4n!+1}{p}\right),$$ with a set of primes $p>n$, to check if there might be a solution for $n$. Confirming that for $3 < n < 10^9$ there are no solutions took about one hour on my laptop using 40 primes. The code is available at http://users.jyu.fi/~tamaraja/temp/brogen.c for anyone interested to see it.

Computation 2
I wanted to test the slightly stronger guess in Mark Sapir's post that for $p>5$, $p^2$ does not divide any integer of the form $4n!+1$. This turned out not to be true.
I checked that for $5 < p <10^6$ there is one exception, namely $$761671^2 ~|~ 4\cdot446142!+1.$$ Here I used only brute force and the trivial observation that $$p ~|~ 4n!+1 \Rightarrow n < p.$$ Testing this range for $p$ took a couple of hours. The code is available at http://users.jyu.fi/~tamaraja/temp/sqrdiv.c

Computation 3
Byron Schmuland pointed out the similarity to Brocard's problem in his comment. Because the previous computations I found for the equation $$n! + 1 = m^2$$ were carried out over a decade ago, I decided to extend them a bit.
With the same approach which was used by Berndt and Galway in 2000 (and in Computation 1 above) I confirmed that the equation has no solutions for $7 < n < 10^{10}$. This took about one day. The code was the same as in Computation 1.

Computation 4
After the previous 3 computations I did a little bit of the "missing" fourth. The equation $$n!+1 \equiv 0 \pmod{p^2}$$ has no solutions $(p,n)$ with $p$ prime and $613 < p < 10^6$.<|endoftext|>
TITLE: Is there a natural random process that is rigorously known to produce Zipf's law?
QUESTION [86 upvotes]: Zipf's law is the empirical observation that in many real-life populations of n objects, the $k^{th}$ largest object has size proportional to $1/k$, at least for $k$ significantly smaller than $n$ (and one also sometimes needs to assume $k$ somewhat larger than 1).  It is a special case of a power law distribution (in which $1/k$ is replaced with $1/k^\alpha$ for some exponent $\alpha$), but the remarkable thing is that in many empirical cases (e.g. frequencies of words, or sizes of cities), the exponent is very close to 1.
My question is: is there a "natural" random process (e.g. a birth-death process) that one can rigorously demonstrate (or at least conjecture) to generate populations of n non-negative quantities $X_1,\ldots,X_n$ (with n large but possibly variable) that obey Zipf's law on average with exponent 1?  There are plenty of natural ways to generate processes that have power law tails (e.g. consider n positive quantities $X_1,\ldots,X_n$ evolving by iid copies of log-Brownian motion), but I don't see how to ensure the exponent is 1 without artificially setting the parameters to force this.   
Ideally, such processes should be at least somewhat plausible as models for an empirical situation in which Zipf's law is observed to hold, such as city sizes, but I would be happy with any non-artificial example of a process.
One obstruction here is the exponent one property is not invariant with respect to taking powers: if $X_1,\ldots,X_n$ obeys Zipf's law with exponent one, then for any fixed $\beta>0$, $X_1^\beta,\ldots,X_n^\beta$ obeys the power law with a different exponent $\beta$.  So whatever random process one would propose for Zipf's law must somehow be quite different from its powers.

REPLY [6 votes]: There's a recent paper on the arXiv by Schwab et al. that looks to contain a nice derivation of Zipf's law, which arises in random systems affected by a hidden variable (something like common inputs to a neural network): http://arxiv.org/abs/1310.0448<|endoftext|>
TITLE: Epsilon regularity: what does it say and where does it come from?
QUESTION [18 upvotes]: The $\varepsilon$-regularity phenomenon shows up in several different contexts. I try to describe it focussing on the harmonic map situation, but I really would like to understand the situation in general.
The following is the Schoen-Uhlenbeck $\varepsilon$-regularity lemma, extracted from
Tobias H. Colding, William P. Minicozzi II, An excursion into geometric analysis.
Let $N$ be a Riemannian manifold and $B_{r}$ be the ball of radius $r$ centred at the origin in $\mathbf{R}^k$. Then there exists $\varepsilon(k,N)$ such that if $u:B_{r}\in\mathbf{R}^k\rightarrow N$ is an energy minimizing map and
$$\frac{\int_{B_{r}}|\nabla u|^2}{r^{k-2}}<\varepsilon,$$
then $u$ is smooth in a neighborhood of $0$ and
$$|\nabla u|^2(0)\leq \frac{C}{r}.$$
Thus if a (conformally invariant) rescaling of the energy that $u$ minimizes is small (I suppose $u$ should be in a suitable Sobolev space), then $u$ is automatically smooth in some smaller ball. This rescaling is monotonically increasing thanks to a monotonicity lemma. I am not sure how to interpret the bound on the derivative at zero, though.
The $\varepsilon$-regularity lemma quickly implies that the singular set $S$ of $u$ has $(k-2)$-dimensional Hausdorff measure zero.
My questions are:

 What are the basic ingredients (I suppose I am talking about the properties of the energy functional here) that guarantee that such a lemma holds?
 What is the meaning of the supremum of the set of all $\varepsilon$ such that the energy bound holds, and how can it be computed?
 Is there a simple intuitive picture that I am missing that explains the situation?
 Is there an instance of this phenomenon that predates the Schoen-Uhlenbeck paper?

Many thanks.

REPLY [14 votes]: The way I think of it is to view semilinear PDEs, such as the harmonic map equation, as a contest between the linear portion of the equation ($\Delta u$ in this case) and the nonlinear portions (which, in the case of harmonic maps, are roughly of the shape $|\nabla u|^2$).  Intuitively, if the nonlinear part is small compared to the linear part then we expect the linear behaviour to dominate.  In the case of harmonic maps, this means that we expect the solutions to behave like solutions to Laplace's equation $\Delta u = 0$, which are known to be regular.
A bit of dimensional analysis then tells us that the condition $\frac{\int_{B_r} |\nabla u|^2}{r^{k-2}} < \varepsilon$ has the right scale-invariance properties to have a chance of making the nonlinear term smaller than the linear term.  (To make this rigorous, one of course needs to deploy various harmonic analysis estimates in well-chosen function space norms, such as Sobolev embedding.)
I discuss these heuristics (though more for dispersive equations than for elliptic ones) a bit at
http://terrytao.wordpress.com/2010/04/02/amplitude-frequency-dynamics-for-semilinear-dispersive-equations/
The question of what happens at the critical value of epsilon is an interesting one.  Often, the limiting non-regular solutions at that value of epsilon, after rescaling and taking limits, tend to be quite symmetric and smooth, away from a very simple singular set (e.g. a subspace).  I don't know the elliptic case too well, but one obvious candidate for such a solution would be a singular 2D harmonic map (such as the map from C -> S^1 given by x -> x/|x|) extended to k dimensions by adding k-2 dummy variables.  In the dispersive case, the analogous concept is that of the minimal energy blowup solutions, and these tend to be soliton solutions (so, typically, they obey a time translation invariance symmetry), associated to the ground state solution of the associated time-independent equation.<|endoftext|>
TITLE: What can be said about the Shadow hull and the Sight hull?
QUESTION [17 upvotes]: This is a question implicitly raised by Is the sphere the only surface all of whose projections are circles? Or: Can we deduce a spherical Earth by observing that its shadows on the Moon are always circular?.
Suppose $X$ is a closed subset of $\mathbb R^n$.
What can you say about 

The shadow hull of $X$: the intersection of complements of lines disjoint from $X$, and
The sight hull: the intersection of complements of rays disjoint from $X$?

Do these sets have standard mathematical or computer graphics / computational geometry names?
Are there other good characterizations?
How hard are they to compute (say for a polyhedron, so that complexity is fairly well-defined)?
Note: two sets with the same shadow hull  cast the same shadows (in light cast by any parallel beam).  Two sets with the same sight hull have the same visible surface.  Sight hull $\subset$ shadow hull $\subset$ convex hull.
Addendum  While we're at it: one could also define the "Knife Hull" to be the intersection of complements of half-planes disjoint from the set.  In other words, it's the best outer approximation of a small shape that you can carve using a long wide straight-edged knife.  This makes it sight hull $\subset$ shadow hull $\subset$ knife hull $\subset$ convex hull.
All of these are invariant under affine transformations.

REPLY [3 votes]: A very partial answer, just mentioning one reference, and one additional result (w/o a reference). 
Nina Amenta and Günter Ziegler computed the combinatorial complexity
of 2D shadows (and $k$-shadows) of convex polytopes in $\mathbb{R}^d$ 
in their now heavily cited 1999 paper
"Deformed Products and Maximal Shadows of Polytopes"
(Advances in Discrete and Computational Geometry 233:57-90 (1999)).
They showed that a polytope of $n$ facets can have $\Theta(n^{\lfloor d/2 \rfloor})$ vertices in
its 2D shadow, for fixed $d$—surprisingly large.  $k$-dimensional shadows have various applications to robotics and
computer vision.  For example, the set of stable 3-finger friction grasps of a polygon can be
represented as a 3-shadow of a 5-polytope.
I have seen your "shadow hull" called the line hull.
I cannot provide a reference, but I do know that Sergei Bespamyatnikh constructed
a (nonconvex) 3D polyhedron of $n$ vertices whose line hull has combinatorial complexity $\Theta(n^9)$.
Added. The shadow hull is called the visual hull in the computer vision literature.
A. Laurentini, "The Visual Hull Concept for Silhouette-Based Image Understanding"
IEEE Transactions on Pattern Analysis and Machine Intelligence
Volume 16,  Issue 2  (February 1994)
pp. 150 - 162.  From the Abstract: "it is the maximal object silhouette-equivalent to $S$, i.e., which can be substituted for $S$ without affecting any silhouette."<|endoftext|>
TITLE: Length of a module over different rings
QUESTION [5 upvotes]: Given a regular local ring $(R,m)$ and a finitely generated $R$-algebra $S$, which is free as an $R$-module. Let $M$ be a left $S$-module of finite length, $\ell_S(M)=r<\infty$.
Under what conditions is $\ell_R(M)<\infty$? If this is the case, can we compute $\ell_R(M)$ in terms of $\ell_S(M)$?
For example if $S=M_n(R)$, then i think we have $\ell_R(M)=n\ell_S(M)$.
If $S$ is commutative and local, with maximal ideal n, then according to Liu one has the following: $\ell_R(M)=[S/n:R/m]l_S(M)$.
Are there general formulas for length and "restriction of scalars"? I'm especially interested in the case when $S$ is a maximal $R$-order in a division algebra. Literature tips are also appreciated.

REPLY [5 votes]: Let $\{V_i\}$ be representatives from each of the isomorphism classes of simple left $S$-modules.  For any finite length module ${}_S M$, let $\ell_S(M; V_i)$ denote the number of times that $V_i$ occurs in a composition series for $M$.  Then the following formula holds (where almost all $\ell(M;V_i)$ are zero because $M$ has finite length, but any single $\ell_R(V_i)$ could be infinite, and $\infty \cdot 0 = 0$):$$\ell_R(M) = \sum \ell_R(V_i) \cdot \ell_S(M; V_i).$$
Thus, a finite length $S$-module $M$ has finite $R$-length if and only if every simple $S$-module that occurs in $M$ has finite $R$-length.  This makes it clear that every finite length left $S$-module has finite $R$-length if and only if all simple left $S$-modules have finite $R$-length.  The above discussion is true with no requirements on the ring $S$.  
Now let's assume that $S$ is finitely generated as an $R$-module.  (I'm not sure if this is what you meant by "finitely generated as an $R$-algebra," but it's probably true in the cases that you're studying if you're looking at maximal orders.)  I'll prove that $S$ has finitely many simple modules, each of which has finite $R$-length.  Let $J(A)$ denote the Jacobson radical of any ring $A$.  Then $\mathfrak{m} = J(R) \subseteq J(S)$ because $S$ is a module-finite $R$-algebra; for a proof of this fact, see Lam's A First Course in Noncommutative Rings, Proposition 5.7.  Now the simple $S$-modules are the same as the simple $S/J(S)$ modules (see Proposition 4.8 of the same text).  But $S/\mathfrak{m} S \twoheadrightarrow S/J(S)$.  Because $S$ is module-finite over $R$, $S/\mathfrak{m} S$ has finite $R$-length.  Thus $S/J(S)$ has finite $R$-length.  So $S/J(S)$ is artinian (the terminology here is that $S$ is a semilocal ring) and thus has finitely many simple modules, each of which has finite $R$-length.  The same must be true for the simple $S$-modules
The formula above also verifies your formula in the case of a matrix ring.  In case $S = \mathbb{M}_n(R)$, $S$ has a unique simple left module $V = (R/\mathfrak{m} \ \cdots \ R/\mathfrak{m})^T$, the set of all column vectors of length $n$ with entries in $R/\mathfrak{m}$.  (At least one way to see that this is the only simple $S$-module is through Morita theory: the Morita equivalence between $R$-$\operatorname{Mod}$ and $S$-$\operatorname{Mod}$ sends the unique simple $R$-module $R/\mathfrak{m}$ to $V$, so that $V$ must be the unique simple $S$-module.)  Since $\ell_R(V) = n$, the formula above reduces to $\ell_R(M) = n \cdot \ell_S(M)$.
(If something above doesn't make sense, you may want to review Jordan-Hoelder theory.)<|endoftext|>
TITLE: Broader impacts and synergistic activities
QUESTION [24 upvotes]: Since it was suggested on my last NSF proposal question, I thought I might also tap the collective brain on the question of broader impacts and synergistic activities.  These concepts are still slightly unclear to me, though I understand them a lot better than I did when I wrote this blog post (which actually resulted in some pretty good advice).  So let me ask:

What are some examples of synergistic activites and activities with "broader impacts" that many of us are doing, but haven't thought of to list in our NSF proposals?  What are some ones we probably should be doing, but maybe aren't?

I'm particularly interested in the answers for young mathematicians.  Most of the proposals I've read are from more senior people, who could point to their track record of mentoring graduate students and postdocs, or things like editing journals, in a way someone my age really can't.

REPLY [5 votes]: The Computer Science Directorate of NSF (CISE) held a
"Broader Impacts for Research and Discovery Summit" this past June. 
A summary of the meeting appeared in Computing Research News (p.6), including
this list of examples, only one of which is specific to computer science:

 Develop educational materials for elementary, high-school and undergraduate students.
 Involve high-school and undergraduate students in research where appropriate.
 Create or participate in existing effective mentoring programs. 
 Develop, maintain and operate a shared research infrastructure.
 Establish international, industrial or government collaborations.
 Form start-up companies. 
 Present research results to non-scientific audiences from policy-makers to average citizens. 
 Give presentations about the field to the public to foster life-long learning. 
 Develop exhibits in partnership with museums.<|endoftext|>
TITLE: (Finite) Classification Theory
QUESTION [13 upvotes]: In the context of asking about the classification of finite simple groups, the question arose: what exactly is meant by a "classification"?
Perhaps unsurprisingly, there is in fact a whole branch of model theory called classification theory, and an epynomous book by Shelah which apparently is unreadable. My impression is that the notion of "classifiable" in this subject is that of a stable theory, but probably there are other notions I'm not aware of.  Unfortunately, this notion seems best suited for talking about infinite models of a theory.  Moreover, it's not clear to me that this definition gives a complete answer to my question: 
What does it mean to say that the models of a theory (with some cardinality limit, particularly the case where the models are required to be finite) admit a classification?

REPLY [3 votes]: In addition to Andreas' remark on the number of non-isomorphic models, perhaps it's noteworthy to say that many non-structure theorems are aiming at the construction of many models which not only are non-isomorphic, but also are hard to tell apart. This may be achieved by requiring the models to satisfy the same set of sentences of some infinitary logic. By taking infinitary logics into consideration, we may consider a theory T to be classifiable if elementary equivalence in L_(infty, lambda) is a sufficient condition for an isomorphism between a given pair of models of cardinality lambda (this should be found in Shelah's book).<|endoftext|>
TITLE: Is the following map from Z(G) x H^3(G, C*) --> H^2(G, C*) ever nontrivial?
QUESTION [13 upvotes]: Suppose that G is a finite group, then we have the following map f which takes an element z in the center of G and a 3-cohomology class w and returns a 2-cohomology class f(z,w) (for concreteness let's take coefficients to be C* everywhere).
$f(z,w)(x,y) = \frac{w(z, x, y) w(x, y, z)}{w(x, z, y)}.$
Is this map ever nontrivial?  That is can you find a group G, a central element z, and an element of H^3 such that f(z,w) is not the trivial element of H^2?
The motivation for this question is that it should give an example where Z(G) did not lift to a subcategory of the Drinfel'd center Z(Vec(G,w)).

REPLY [3 votes]: Following up on what has already been said, this is an explanation taken from "Group cohomology and gauge equivalence of some twisted quantum doubles", by Geoffrey Mason and Siu-Hung Ng.
Let $G$ be a finite abelian group. We denote by $Z^3(G,\mathbb C^{\ast})$ the group of normalized $3$-cocycles. For any $\omega\in Z^3(G,\mathbb C^{\ast})$ and $g\in G$ we have the map
$$\omega_g (x,y)=\frac{\omega(g,x,y)\omega(x,y,g)}{\omega(x,g,y)}$$
Now let $Z^3(G,\mathbb C^{\ast})_{ab}$ denote the set of all normalized $3$-cocycles $\omega$ for which $\omega_g$ is a $2$-coboundary for all $g\in G$, and let $H^3(G,\mathbb C^{\ast}) _{ab}$ be the corresponding set of cohomology classes. It is not hard to check that $H^3(G,\mathbb C^{\ast})_{ab}$ is a subgroup of $H^3(G,\mathbb C^{\ast})$, and your question is asking for an example when it is a proper subgroup.
An easier way to do this is to look for a different description of $H^3(G,\mathbb C^{\ast})_{ab}$. It is not hard to check that $\omega_{g}(x,y)=\omega_g(y,x)$ for all $(x,y)\in G\times G$ iff $\omega_g$ is a $2$-coboundary. So if you define the map $\psi^{\ast}: H^3(G,\mathbb C^{\ast})\to Hom(\bigwedge^3 G,\mathbb C^{\ast})$ 
$$\psi^{\ast}([\omega])(x,y,z)=\frac{\omega(x,y,z)\omega(y,z,x)\omega(z,x,y)}{\omega(y,x,z)\omega(z,y,x)\omega(x,z,y)}=\frac{\omega_z(x,y)}{\omega_z(y,x)}$$
then $H^3(G,\mathbb C^{\ast})_{ab}$ is precisely the kernel of $\psi^{\ast}$ (Lemma 7.4 in the paper above). Now $\psi^{\ast}$ is surjective so the question becomes: when is $Hom(\bigwedge^3 G,\mathbb C^{\ast})$ non-trivial? This is the case whenever $G$ is the direct sum of at least $3$ cyclic factors, in particular any $(\mathbb Z/n\mathbb Z)^3$ works. 
An explicit example in this case is $\omega(x,y,z)=\mu^{x_1y_2z_3}$ where $x=(x_1,x_2,x_3)$ etc. and $\mu$ is a primitive $n$th root of unity. Then we have $\psi^{\ast}([\omega])(x,y,z)=\mu^{\det(x,y,z)}$ which is non-trivial. In particular $\omega_x$ is non-trivial in $H^2$.<|endoftext|>
TITLE: Showing the category of perfect complexes on a scheme is essentially small
QUESTION [8 upvotes]: So I've tried to set up a few meetings with professors to talk about this, but I think one of them forgot about it after a big conference, and the other is still on vacation... in the meantime...
At one point in Thomason's paper on the classification of thick subcategories of the derived category of perfect complexes over a (EDIT: quasi separated, quasi compact!) scheme, he needs to use the fact that this category is essentially small (that is: it has a SET of isomorphism classes). The reference for this is yet another paper by Thomason (and ghost buddy Trobaugh) that is HUGE (and very important) called "Higher algebraic K-theory of schemes and derived categories." The trouble is that the proof of essential smallness is tucked away in some appendix, paraphrased, and uses results that are scattered throughout this paper in a not-so-clear manner. 
So my question is: does anyone know of a proof, or an explanation of the proof, of the fact that the derived category of perfect complexes on a (quasi-separated, quasi-compact!) scheme is essentially small that doesn't require me to learn algebraic K-theory and read all of Thomason's big paper?
EDIT: A "strictly perfect complex" is a bounded complex of locally free $O_x$-modules of finite type. A "perfect complex" on $X$ is a complex $E$ of sheaves of $O_X$-modules such that there is an open cover of $X$ where $E$ restricted to each neighborhood in the cover is quasi-isomorphic to a strict perfect complex.  
Some background: be gentle, I know very little algebraic geometry... I learned everything backwards, so triangulated category talk is fine, but algebro-geometric talk should be dumbed down, and perhaps illustrated with some examples.

REPLY [7 votes]: I have just noticed this question. There is (reasonably) self-contained proof in section 3 of   arXiv:0905.2063v1, though it is phrased in the language of formal schemes. I think it works replacing every time that "noetherian formal scheme" occurs by "quasi-compact and quasi-separated scheme".
The affine case is easy enough, every perfect complex is quasi-isomorphic to a strict perfect complex which in turn is a bounded complex of finitely generated projective modules. Choosing a set of representatives of isomorphism classes of projective modules (as direct summands of free modules, for instance) you are done. 
The globalization is done by patching and I think is described with some detail in the paper.<|endoftext|>
TITLE: How to extract the diagonal from a bivariate generating function
QUESTION [10 upvotes]: Let $ F(s,t)= \sum_{i,j} f(i,j) s^i t^j$, which is a bivariate generating funcion of the number $f(i,j)$ for some enumeration problem. Sometimes we know about $F(s,t)$, but what we really need is the number $f(i,i)$ with the generating function $G(x)= \sum_i f(i,i) x^i$, called the diagonal of $F$.
The question is how to obtain the diagonal $G(x)$ from $F(s,t)$? Furthermore, how to get the asymtotic formula for $f(i,i)$ from $F(s,t)$?
One way to do this is shown as follows:
$G(x)$ is the constant term of $F(s,\frac{x}{s})$ regarded as a Laurent series in $s$ whose coefficients are power series in x. And we can use Cauchy's integral theorem and Residue Theorem to compute.
This method works when $F(s,t)$ is rational. But when $F(s,t)$ is more complicated, it seems not workable.
An example is 
$F(s,t)=\frac{4 s t}{\sqrt{1-4 s^2}\sqrt{1-4 t^2}(\sqrt{1-4 s^2}\sqrt{1-4 t^2}-4 s t)}$.

REPLY [3 votes]: The paper "A new method for computing asymptotics of diagonal coefficients of multivariate generating functions" by A. Raichev and M.C. Wilson (DMTCS 2007, arXiv, author homepages) explains more about this topic.
In particular, if you really want asymptotics, the explicit computation of the diagonal GF is unnecessary in most cases. There will be more on this topic in the forthcoming book by Pemantle and Wilson, "Analytic Combinatorics in Several Variables".
In the case of a 2-variable algebraic GF, I would see whether it arises naturally as a diagonal of a higher degree rational GF and start from there using methods as in the last paragraph.<|endoftext|>
TITLE: estimate the error term in CLT
QUESTION [9 upvotes]: Let $X_m = \frac{1}{\sqrt{m}}\sum_{k=1}^m Z_k$ where $Z_k$ are iid equally likely on $\{\pm 1\}$. Then $X_m$ convergens to $X \sim \mathcal{N}(0,1)$ in distribution by CLT.
Let $f$ be a smooth bounded function on $\mathbb{R}$. Then $\mathbb{E}[f(X_m)] \to \mathbb{E}[f(X)]$. I wonder if there is any general method to give sharp asymptotic estimate of the error term $\mathbb{E}[f(X_m)] - \mathbb{E}[f(X)]$, which I expect to be $\Theta(1/m)$. The scaling constant should depend on $f$ (as well as the distribution of $Z_k$ if they are not binary). 
For law of large number, this type of estimate can be done via the Delta method (e.g., to estimate $\mathbb{E}[f(\bar{Z})] - f(0)$). There must be a counterpart for CLT... I haven't found the Edgeworth expansion useful because it seems to work with distribution with densities.
Edited: To be clear, I am only interested in some specific nice function (e.g., $f(x) = x^2 e^{-x^2/4}$) and finding a sharp expansion for the error term of the form, say, $c/m + o(1/m)$, where $c$ will depend n $f$. As pointed by Mark, the worst-case rate of all bounded smooth function $f$ is $1/\sqrt{m}$, which agrees with the upper bound given by Stein's method.

REPLY [2 votes]: Your conjecture is correct. 
Suppose that (say) $f$ has a bounded $5$th derivative. Then
\begin{equation}
 Ef(X_m) - Ef(X)=-\frac{Ef''''(X)+o(1)}{12m}. \tag{1}
\end{equation}
Indeed, let 
\begin{equation}
 Z_{mj}:=\frac{Z_j}{\sqrt m},\quad Y_{mj}:=\frac{Y_j}{\sqrt m},\quad T_{mk}:=\sum_{j=1}^{k-1}Z_{mj}+\sum_{j=k+1}^m Y_{mj}, 
\end{equation}
where $Z_1,\dots,Z_m,Y_1,\dots,Y_m$ are independent random variables and $Y_j\sim N(0,1)$. 
Note that $EZ_{mj}^p=EY_{mj}^p$ for $p=0,\dots,3$, $EZ_{mj}^4=\tfrac1{m^2}$, $EY_{mj}^4=\tfrac3{m^2}$. 
Also, $Z_{mk}$ is independent of $T_{mk}$ for each $k$, and so is $Y_{mk}$. 
Moreover, $Ef''''(T_{mk})\to Ef''''(X)$ uniformly in $k=1,\dots,m$ (as $m\to\infty$); this should be as easy to show as the convergence $Ef(X_m) \to Ef(X)$, mentioned in the question. 
Now one has
\begin{equation}
 Ef(X_m) - Ef(X)=\sum_{k=1}^m D_k, \tag{2}
\end{equation}
where
\begin{equation}
 D_k:=E[f(T_{mk}+Z_{mk})-f(T_{mk}+Y_{mk})]
\end{equation}
\begin{equation}
 =\sum_{p=0}^4\frac1{p!}\,Ef^{(p)}(T_{mk})E(Z_{mk}^p-Y_{mk}^p)+O(E(|Z_{mk}|^5+|Y_{mk}|^5))
\end{equation}
\begin{equation}
 =\tfrac1{4!}\,(Ef''''(X)+o(1))(\tfrac1{m^2}-\tfrac3{m^2})+O(m^{-5/2}) 
\end{equation}
by Taylor's expansion, 
with $o(1)$ uniform in $k=1,\dots,m$. 
Now $(1)$ follows immediately from $(2)$.<|endoftext|>
TITLE: Non-trivial consequences of Baer's theorem and Lucchini's theorem in subnormality theory
QUESTION [9 upvotes]: There are a couple of beautiful results in finite group theory that look trivial, at least on a first glance, but require non-trivial facts to prove. I am basically interested in whether these results actually have relatively easier proofs to the ones I will outline below. More specifically, I am interested in whether these results may be proven without the aid of the powerful subnormality theory.
The first, and perhaps most beautiful result, is due to Horosevskii:

Let $\sigma\in Aut(G)$ where Aut($G$) denotes the automorphism group of a non-trivial finite group $G$. Then the order of $\sigma$ in Aut($G$) in less than $\left|G\right|$.

The second result, though perhaps more specialized, actually is an important ingredient in the purely group-theoretic proof of Burnside's $p^aq^b$-theorem:

Let $t$ be an involution in a finite group $G$, and assume that $t\not\in {\bf{O}}_2(G)$, where ${\bf{O}}_2(G)$ denotes the 2-core of $G$. Then there exists an element $x\in G$ of odd prime order such that $txt=x^{-1}$.

(I think that this result is due to Matsuyama but please do not take my word for this because I am not completely certain. For definiteness, I will refer to this result as the "result on involutions".)
Now the only proof I know of the second result (on involutions) relies on Baer's theorem in subnormality theory. For reference, Baer's theorem states that:

Let $H$ be a subgroup of a finite group $G$. Then $H\subseteq {\bf{F}}(G)$ if and only if $\left\langle H,H^x \right\rangle$ is nilpotent for all $x\in G$. 

(Here, ${\bf{F}}(G)$ denotes the Fitting subgroup of $G$ and $H^x$ denotes the conjugate of $H$ by $x$; that is, $H^x = \{x^{-1}hx|h\in H\}$.)
The proof (at least the one I know) of the result on involutions uses Baer's theorem, but actually, it really only uses a very special case of Baer's theorem: the subgroup $H$ in the statement of Baer's theorem is chosen to be $\{1,t\}$ in the proof, where $t$ is the involution quoted in the result. (The proof also uses a very easy fact about dihedral groups.) 
This leads me to wonder whether there exists a "Baer-free" proof of the result on involutions. More specifically, my question is:

Question 1: Does there exist a proof of the result on involutions independent of subnormality theory?

It would be interesting if such a proof existed since Baer's theorem relies on the non-trivial Wielandt "zipper lemma" (which I will quote at the end of my question) - and the zipper lemma really looks unrelated to the result on involutions.
Now Horosevskii's theorem relies on another non-trivial consequence of subnormality theory: Lucchini's theorem. For reference, Lucchini's theorem states that:

Let $A$ be a cyclic proper subgroup of a finite group $G$, and let $K=\mbox{core}_G(A)$. Then $\left|A:K\right|<\left|G:A\right|$, and in particular, if $\left|A\right|\geq \left|G:A\right|$, then $K>1$.

What is amazing, at least to me, is that a fact that one might consider fundamental about automorphisms (Horosevskii's theorem) relies on a result (Lucchini's theorem) that at least on a first glance seems very specialized. My second question is nearly identical to my first:

Question 2: Does there exist a proof of Horosevskii's theorem independent of subnormality theory?

(What follows is not really subsumed in my question, but it may be useful for answering my question.)
Succinctly, the proof of Baer's theorem (at least the proof I know) relies on the amazing Wielandt "zipper lemma" which states:

Suppose that $S\subseteq G$, where $G$ is a finite group, and assume that $S$ is subnormal in $H$ for every proper subgroup $H$ of $G$ that contains $S$. If $S$ is not subnormal in $G$, then there is a unique maximal subgroup of $G$ that conains $S$.

Likewise, the proof of Lucchini's theorem relies on Zenkov's theorem on intersections of abelian subgroups, the proof of which in turn relies on Baer's theorem. Zenkov's theorem states:

Let $A$ and $B$ be abelian subgroups of a finite group $G$ and let $M$ be a minimal member of the set $\{A\cap B^g|g\in G\}$. Then $M\subseteq {\bf{F}}(G)$.

(Note that "$B^g$" denotes the conjugate of $B$ by $g$; that is, $B^g=\{g^{-1}bg|b\in B\}$.)
(Note: I am well aware that there may be other theorems due to Baer, Lucchini and Horosevskii that are referred to as "Baer's theorem", "Lucchini's theorem" or "Horosevskii's theorem". However, I hope that by stating the results above, no confusion arises. For proofs of the theorems quoted, please see the book Finite Group Theory by I. Martin Isaacs. More specifically, please see Theorem 2.9., Chapter 2, Section 2A, page 50 for the statement and proof of the Wielandt "Zipper Lemma", Theorem 2.12., Chapter 2, Section 2B, page 55 for the statement and proof of Baer's theorem, Theorem 2.18., Chapter 2, Section 2D, page 61 for the statement and proof of Zenkov's theorem, and Theorem 2.20., Chapter 2, Section 2D, page 63 for the statement and proof of Lucchini's Theorem.)
Thanks!

REPLY [4 votes]: Baer–Suzuki: The subgroup generated by {x,x^g} is a p-group for all g in G if and only if x is contained in the p-core of G.
Baer's proof emphasized commutators, rather than subnormality.  In some sense these are the same thing, but perhaps it will feel different enough for you.  Baer's presentation is given in textbook form in Huppert's Endliche Gruppen as III.6.15, page 298.  See also IX.7.8 in Huppert–Blackburn, Finite Groups, Vol 2, p. 500.  Baer's original paper is:

Baer, Reinhold. "Engelsche elemente Noetherscher Gruppen."
Math. Ann. 133 (1957), 256–270.
MR86815
DOI:10.1007/BF02547953

Suzuki's proof is given in Gorenstein's Finite Groups, 3.8.2, p. 105.  It also avoids subnormality, rather using ideas about fusion of p-elements, and is probably how Bender thought of it.

Subnormality is a pretty critical idea, and many of Bender's insights use subnormality, so I would not suggest avoiding subnormality.  Other characterizations of the Fitting subgroup in terms of subnormality are given in Huppert's textbooks.  In particular, the Fitting subgroup as the elements that centralize chief factors is a very important viewpoint.  It generalizes to F-subnormality in the finite soluble world, and Bender's p*-nilpotency in the finite insoluble world.  Kegel and Carter have a number of nice papers that explore subnormality in ways that have heavily influenced both the soluble and the insoluble worlds.
Robinson's group theory textbook (and Lennox–Stonehewer MR902857) have a good description of subnormality in the infinite case.
Wielandt's collected works contains several good textbook style presentations of subnormality that are not properly contained in any other works that I have found.  They avoid assuming finiteness, and tend to have very interesting relationships between perfect subgroups and subnormality, that complement Bender's work.

One very nice thing about Isaacs's FGT is how it exposes you to important techniques.  It does not try to "optimize" the presentation either by using the bare minimum of tools or by using the most general tool here-to-fore created.  It just uses some nice results in a realistic way that more people should know.  Suzuki's textbooks also have this nice property, though they are not as easy to quote from.<|endoftext|>
TITLE: classification of small complete groups
QUESTION [7 upvotes]: This is "escalated" from stackexchange. 
So, $S_n$ for $n \ne 2,6$, $\text{Aut}(G)$ for $G$ non-cyclic simple, $\text{Hol}(C_p)$ for $p$ odd prime are well-known classes of complete group. 
What's the smallest complete group not in these classes? What's a possible way to generalise that group, and what's the smallest complete group not in that class either? And so on. I'd be interested in any partial result.

REPLY [4 votes]: Here's a very partial "experimental" result, using some GAP computations and the library of small groups in GAP. I am afraid this does not provide any deep theoretical insight, but it might at least give some ideas and starting points.
There is not really an efficient way to check whether a group is complete,  I think; however, one thing is sure: It can't be nilpotent (as nilpotent groups have non-trivial center). The small groups library knows for every group in it whether it is nilpotent, so we can filter those out. There are 1048 groups of order 1 to 100; of these 464 are not nilpotent, as the fol.owing code verifies:
gap> Sum(List([1..100],NumberSmallGroups));
1048
gap> grps:=AllSmallGroups([1..100], IsNilpotentGroup, false);;              
gap> Length(grps);
464

Computing the center of a group is also relatively efficient, so let's remove all with non-trivial center:
gap> grps:=Filtered(grps,g->IsTrivial(Center(g)));;
gap> Length(grps);
72

Finally, the condition on the automorphism group:
gap> grps:=Filtered(grps,g->Size(g)=Size(AutomorphismGroup(g)));;
gap> Length(grps);
5

So what are these groups? If we just print "grps", we don't see much, but we can ask GAP to try to come up with "nice" labels for the groups (they just give a rough idea of the group's structure; they are not enough to recover the group):
gap> List(grps, StructureDescription);
[ "S3", "C5 : C4", "S4", "(C7 : C3) : C2", "(C9 : C3) : C2" ]

Note that GAP denotes semidirect products by "N : H", where N is normal. Also note that in general there are many ways to write a group as, say, a semidirect product, and GAP may not pick the most "natural" one -- after all what is "natural" is highly subjective.
Anyway, so we get $S_3$, $S_4$ and the holomorphs of the cyclic groups of order 5, 7 and 9. The last one is not in your list (but almost)... Continuing with a slightly more elaborate program (looping over the orders instead of grabbing all groups of a lot of orders at once, as that will easily overflow memory), we find 27 complete groups up to and including order 500:

$S_3$
$Hol(C_5)$
$S_4 \cong Hol(C_2^2)$
$Hol(C_7)$
$Hol(C_9)$
$Hol(C_{11})$
$S_5$
$S_3 \times Hol(C_5)$
$((C_3^2) : C_8) : C_2$
$S_3 \times S_4$
$Hol(C_{13})$
$((C_2^3) : C_7) : C_3 < Hol(C_2^3)$
$(((C_2^2) : C_9) : C_3) : C_2$
$S_3 \times Hol(C_7)$
$Hol(C_{17})$
$((C_2^4) : C_5) : C_4$
$S_3 \times Hol(C_9)$
$PSL(3,2) : C_2$
$Hol(C_{19})$
$((((C_4^2) : C_3) : C_2) : C_2) : C_2$
$((((C_2^4) : C_3) : C_2) : C_2) : C_2$
$(((C_3^2) : C_3) : Q_8) : C_2$
$(((C_6^2) : C_3) : C_2) : C_2$
$(((C_3^2) : Q_8) : C_3) : C_2 \cong Hol(C_3^3)$
$Hol(C_5) \times S_4$
$Hol(C_{27})$
$Hol(C_{25})$ (order 500)

So, besides the classes you named, we also get holomorphs of cyclic groups of prime-power order; direct products of complete groups, plus some other groups which one should study a bit closer to understand.<|endoftext|>
TITLE: Unit fraction, equally spaced denominators not integer
QUESTION [15 upvotes]: I've been looking at unit fractions, and found a paper by Erdős "Some properties of partial sums of the harmonic series" that proves a few things, and gives a reference for the following theorem:
$$\sum_{k=0}^n \frac{1}{m+kd}$$ is not an integer.
The source is:

Cf. T. Nagell, Eine Eigenschaft gewissen Summen, Skrifter Oslo, no. 13 (1923) pp. 10-15.


 Question 
Although I would like to find this source, I've checked with my university library and it seems pretty out of reach.  What I'm really hoping for is a source that's more recent or even written in English.
Finding this specific source isn't everything, I'll be fine with pointers to places with similar results.

REPLY [21 votes]: I tracked down Nagell's paper in early 1992, since I had found a proof and wanted to see whether it was the same as his.  (It turned out to be essentially the same idea.)  Unfortunately, I've since lost my photocopy of his paper, but it came from UIUC, so that would be where I'd start looking.  If I remember right, the journal where it appeared is incredibly obscure, and UIUC was the only place in North America that had a copy.
Here's a proof copied from my very old TeX file.  A bit awkwardly written, but it explains how to do the case analysis, which is the part that makes this approach simpler than what Erdős did.
Suppose that $a$, $b$, and $n$ are positive integers and
$$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb} = c \in
\mathbb{Z}$$  We can take $\gcd(a,b) = 1$ and $a > 1$, and it is easy to
show that $n > 2$.
Suppose that $b$ is odd. In the arithmetic progression, there is then a
unique number $a + mb$ divisible by the highest possible power of $2$,
because for all $k$, the progression runs through a cycle modulo $2^k$ which
contains each value exactly once.  Then multiplication by $\ell =
\hbox{lcm}(a,a+b,\ldots,a+nb)$ gives
$$\frac{\ell}{a}+\frac{\ell}{a+b}+\frac{\ell}{a+2b}+\cdots+\frac{\ell}{a+nb}
= \ell c.$$  All of the terms here except $\frac{\ell}{a+mb}$ are even.
Thus, $b$ cannot be odd.
Now suppose that $b$ is even, and $b \le \frac{n-2}{3}$.  By Bertrand's
Postulate, there is a prime $p$ such that $\frac{n+1}{2} < p < n+1$.  Then
$p$ does not divide $b$, and $p$ is odd.  We must have $a \le n$, because
$$1 \le c = \frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{n+1}{a}.$$
Since $b$ generates the additive group modulo $p$ and $n+1 > p$, at least
one of the numbers $a, a+b, \dots, a+nb$ is divisible by $p$.  At most
two are, since $2p > n+1$.  Suppose that $p$ divides only the term $a+kb$.
Then
$$\frac{1}{a+kb} = c-\sum_{j \neq k}{\frac{1}{a+jb}}.$$
The denominator of the left side is divisible by $p$, but that is not true
of the right side.  Thus, $p$ must divide two terms.
Now suppose that $p$ divides $a+{\ell}b$ and $a+kb$, with $mp = a+{\ell}b <
a+kb = (m+b)p$.  Then
$$\frac{1}{p}\left(\frac{2m+b}{m(m+b)}\right) =
c - \sum_{j \neq \ell,k}{\frac{1}{a+jb}}.$$
This implies that $p \mid (2m+b)$.  However,
$$a+{\ell}b \le n + (n-p)\left(\frac{n-2}{3}\right) < n + \frac{n}{2}\left(\frac{n-2}{3}\right).$$
Therefore,
$$m < \frac{a+{\ell}b}{n/2} < \frac{n-2}{3} + 2.$$
It follows that $2m+b \le n+1$.  However, $n+1 < 2p$, so $2m+b = p$.  This
contradicts the fact that $b$ is even and $p$ is odd.
Finally, suppose that $b$ is even, and $b \ge \frac{n-1}{3}$.  Since $a > 1$
and $a$ is odd, we must have $a \ge 3$.  We must also have $b \ge 4$, since
if $b=2$, then Bertrand's Postulate and the fact that $n \ge a$ (as above)
imply that one of the terms is a prime, which does not divide any other
term.
Now, we show that $n \le 13$.  To do that, note that
$$\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{1}{3} + \frac{1}{7} + \frac{3}{n-1}\left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right).$$
A simple computation shows that for $n \ge 14$,
$$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} < \frac{11}{63}(n-1),$$
which implies that the sum is less than 1.  Hence, we must have $n \le 13$.
Thus, the sum is at most
$$\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \cdots + \frac{1}{47} + \frac{1}{51} + \frac{1}{55} < 1.$$
Therefore, the sum is never an integer, and the theorem holds.<|endoftext|>
TITLE: Is a quasi-iso in Lie algebra cohomology necessarily an iso?
QUESTION [7 upvotes]: Let $\mathfrak g$ be a Lie algebra (if it matters, right now I only care about finite-dimensional Lie algebras in characteristic $0$, although I'm never opposed to hearing about more general cases).  Recall that it determines a differential graded algebra ("the complex that computes Lie algebra cohomology"), with $k$th component $\wedge^k \mathfrak g^*$ and differential determined by the bracket.  It is a complex by the Jacobi identity.  Thinking in terms of supergeometry, I will call this dga $\mathcal C^\infty([-1]\mathfrak g)$.
Moreover, if $\mathfrak h \to \mathfrak g$ is a homomorphism of Lie algebras, then we get a homomorphism of dgas $\mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$, so that $\mathcal C^\infty\circ [-1]$ is a contravariant functor.  The dga is a complete invariant: the functor is full and faithful.

Suppose that $\mathfrak h,\mathfrak g$ are two Lie algebras and $f: \mathfrak h \to \mathfrak g$ a Lie algebra homomorphism.  Suppose furthermore that the corresponding map $f^*: \mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$ is a quasi-isomorphism, i.e. it induces an isomorphism on cohomology.  Does it follow that $f$ is an isomorphism?

When I ask it this way, it sounds strongly like the answer should be "no": almost never is cohomology a complete invariant.  For example, the cohomology in degree $1$ sees only the abelianizations of $\mathfrak g,\mathfrak h$.  But on the other hand, research I'm doing on Lie algebroids suggests that the similar statement with "algebra" replaced by "algebroid" throughout should be true.  I don't see a direct proof even in the "algebra" case, but I feel like there should be either a trivial counterexample or an easy argument in favor.  In either case, though, and maybe because it's late at night, I'm stuck.
Which is all to say that secretly I care about algebroids, so if any of y'all know a good reference for the problem at that generality, please send it my way.  But I will happily accept an answer just for algebras if one is provided.

REPLY [5 votes]: Here are some comments to your answer that I hope will be helpful (it's still sufficiently confusing that I might make some mistakes). Given an $L_{\infty}$ algebra, we can define the "Koszul dual" Chevalley $\it{chains}$ $C_*(L)$(for several reasons it's more natural to think of the Chevalley chains rather than cochains). The ordinary notion of a quasi-isomorphism of $L_{\infty}$ is a morphism of Chevalley complexes whose first Taylor coefficient induces a quasi-isomorphism of complexes. We could ask what happens if we declare two Lie algebras weakly equivalent if there is a quism of Chevalley complexes? We have seen that this notion can differ from the standard one. Another example is when $g_1=sl_2(\mathbb{C})$ and $g_2=\mathbb{C}[2]$ an abelian Lie algebra concentrated in degree 2. From this point of view, your notion is a very natural relaxing of the usual notion of quasi-isomorphism --- it's the notion of weak equivalence transported from Koszul duality. It's worth noting that the two notions will coincide if your dg-Lie algebra is concentrated in positive degree (or more generally with some nilpotency hypotheses on the action of the degree zero piece $g^0$ on your dg lie algebra.) This explains Trial's answer above. 
It's useful to have the rational homotopy picture in mind. Given a 1-connected CW complex Xwhich is finite in each degree, you can construct a Lie model L for X whose $H_{*}(L)=\pi_{*}(\Omega(M)) $ with Whitehead bracket. One gets chains on your space by taking $C_{*}(L)$. To go back, you have something known as the cobar construction which spits out $C_{*}(\Omega(M))$, which is U(L), the universal enveloping algebra of L. Again, things start to break down in the non-simply connected case if you don't assume some hypothesis about the action of $\pi_1$ on the higher homotopy groups. From this point of view, the comment at the end of the last paragraph is a reflection of the fact that in these instances, when you have an isomorphism on homology groups, you have an isomorphism on homotopy groups.
Finally when thinking about the idea that the more relaxed notion gives you a derived equivalence, you must define the derived category properly. Let's think about the abelian case. The most classical case of Koszul duality says that given a finite dimensional graded vector space, $D^{+}(SV) \cong D^{+}(\wedge(V^*))$. Here we are considering the derived category of modules which are bounded below. It's pretty easy to see that this cannot be extended an equivalence of unbounded derived categories. 
The right notion which makes the whole thing work for unbounded derived categories can be found in Positselski's works and is called the coderived category. Let g be a Lie algebra over $\mathbb{C}$. Then the equivalence between the derived category of modules over U(g) and the coderived category of co-modules over it's Chevalley complex $C_{∗}(g)$ in which $M \to C_{∗}(g,M)$. Under the assumption g is finite, we can consider these co-modules to be modules over C*(g)(now looking at cochains again) and look at the corresponding localization of the category of dg-modules over C*(g). What we see is that we are localizing at a smaller set of weak equivalences than quasi-isomorphisms.<|endoftext|>
TITLE: Finiteness and cardinality in abstract categories
QUESTION [5 upvotes]: My question is a very simple one.
What ways are there to generalize terms such as cardinality (or, more generally, the concept of finiteness) to abstract (and not concretizable) categories?
I have seen two ways to do so: One is to take a terminal object $\mathbf{T}$ and to count the numer of morphisms from $\mathbf{T}$ to an object $\mathbf{A}$ and define this number as the cardinality of $\mathbf{A}$. Clearly, this will coincide with the cardinality as defined in the category of sets as the number of morphisms from $\{1\}$ to another set is the cardinality of the letter. The downside of this approach is that it does not always work well: Even if a category has a terminal object, there is no garuantee that there are morphisms from this object to any other object and, thus, if we define cardinality in this way, it might give very misleading results: For instance, we can have a concrete category in which the number of morphisms from the terminal object does not reflect the finitness of an object that might be obtained by applying the forgetful functor to it.
Another way is to define an object as finite if it has a finite number of endomorphisms. Again, this coincides with the notional of finiteness in the category of sets. However, one can think of a category in which an object $\mathbf{A}$ only has the identity morphism as an endomorphism but an infinite number of morphisms exist between $\mathbf{A}$ and another object $\mathbf{B}$ (for which, again, only the identity morphism is an endomorphism). This creates a scenario in which it is hardly justified to think $\mathbf{A}$ and $\mathbf{B}$ as finite.
Are there other ways to define finitenes? In particular, is there a way to define finitness for a category such that, if this category is concretizable, it can be turned into a concrete category in which only the finite objects (abstractly defined) can have a finite image under the forgetful functor and this can be achieved for any abstractly defined finite object?

EDIT: In the comments, it was asked what categories I have in mind. Here is an answer to this:   
The categories I have in mind are only required to have the following: There must exist an object $\mathbf{A}$ such that all finite, non-empty powers of $\mathbf{A}$ are also in the category (the product is of course defined in the usual category-theoretic way and the point of all this is to work with a generalized definition of clones in these categories). Apart from this condition, however, the category can have any possible structure. Although it would be nice to have a general definition of cardinality, all I am really interested in is to distinguish between finite and non-finite objects. The definition should satisfy the three conditions below.

Of course, it should coincide with the usual definition of finiteness if my category is Set. 
If $\mathbf{A}$ and another object $\mathbf{B}$ is finite, then the number of morphisms from $\mathbf{B}$ to $\mathbf{A}$ should be finite (it would therefore be enough to have a notion of finiteness "with respect to $\mathbf{A}$", if such a thing makes sense to have). 
The definition should give me as many finite objects as the "best" concretezation. By this, I mean the following: If my category is concretizable, then you should not be able to get more finite objects by defining finiteness over the image under the forgetful functor.

REPLY [4 votes]: In Pareigis' Categories and functors an object $x$ of an abelian category $A$ is called finite if $Hom(x,-) : A \to Ab$ commutes with coproducts. For example, in $R$-Mod, every finitely generated free module is finite. There it is also proved that every cocomplete abelian category, which has a finite projective generator, is equivalent to some $R$-Mod.
There are various notions of finitely generated objects. See for example also Pareigis' book, this question or nlab.<|endoftext|>
TITLE: The reflex-free hull: Construction?
QUESTION [6 upvotes]: This is a followup to Bill Thurston's question
about different notions of hulls.
Here I want to raise a question about the
reflex-free hull, which is, intuitively, the smallest
enclosing shape to an object that cannot hold water in any
orientation.
Let $S$ be a closed solid object in $\mathbb{R}^3$, and $\partial S$ its surface.
Let $H$ be a closed hemispherical neighborhood, a ball intersected
with a closed halfspace through its center.
Define a reflex point $p$ on $\partial S$ to be one such
that it has a neighborhood $H$ such that
(a) $H \subset S$ and (b) $H \cap \partial S = p$.
An object is reflex-free if it has no reflex points.
Intuitively, a reflex point could hold a drop of water in its
exterior neighborhood in some orientation.
For example, this shape is reflex-free:

          


The reflex-free hull of an object $O$ is the intersection of all
reflex-free shapes that enclose $O$.
This notion was introduced in the interesting paper cited below.
It has application to manufacturing by molten-metal casting, 
and applications to architecture.They established a number of properties of the reflex-free hull,
but could not find
an algorithm to construct it.

Q1.
  Provide a finite algorithm to construct the reflex-free hull for
  a polyhedron.

They identified a number of difficulties that various ideas
for algorithms would encounter.
An algorithm that fills in cavities naively,
approaches, but never reaches, the reflex-free hull
of this example (their Fig. 7):

          



Q2.
  Is the reflex-free hull the same as Thurston's "knife hull"?
  (Answered by Bill Thurston below: No.)

Reference.
Hee-kap Ahn,  Siu-Wing Cheng,  Otfried Cheong,  Jack Snoeyink.
"The Reflex-Free Hull."
In Proc. 13th Canadian Conference on Computational Geometry, 2001, 
and in International Journal of Computational Geometry and Applications, 14(6):453-474, 2004. (CiteSeer link).

REPLY [5 votes]: If I understand correctly, reflex-free is equivalent to the property that the orthogonal projection to any line has no inside local minimum, that is, there is no point where the function is a local minimum on the boundary but not in the solid.  For a smooth surface, this means both principle directions can't be curving in the concave sense.
Consider any graph embedded in a ball, mostly trivalent but with some vertices of order 1 attached to faces of a ball, embedded in a way that each interior vertex is in the convex hull of its neighbors. Not all graphs admit such an embedding, but many do;  examples can be constructed,  using for example harmonic maps. For such a graph, you can hollow out of the ball a narrow tubular neighborhood whose complement is reflex-free  These give examples where the knife-hull is different from the reflex-hull --- just for a $Y$ graph, the knife hull could be convex.
Here's an alternative characterization of reflexes (a reflexive definition?):
a point $p$ is in a reflex for $S$ if there is a plane $P$ through $p$ and a region $R$
on the plane containing $p$ in its interior, such that the boundary of $R$ is contained in $\partial S$ and is null-homologous on  $ \partial S$ (bounds a region on $\partial S$). This is equivalent to the definition using the special case of half-balls, because if the shape is turned so that the plane is horizontal and the surface on $\partial S$ is below, we can look at a point where the height function has a local minimum.  If it's not a strict local minimum, turn by a slight random angle to make a strict minimum, in which case we can find a half-ball.
I wonder whether a variation of minimal surfaces would give a reasonable procedure to find the reflex-free hull. For a minimal surface, the two principal curvatures balance---you can think of it like rubber bands in perpendicular directions in a tug of war, trying to pull surface in opposite directions but coming to a draw. Now imagine a surface whose molecules align to pull very hard in the convex directions, and only weakly in the concave directions.   To those who have ... if you take the limit of inward-pulled surfaces, as insiders grow stronger and outsiders weaker, seems like a candidate reflex-free hull.
Minimal surfaces can be difficult to analyze because there are often local minima that are not global minima, and unstable critical points of the area function as well.  However, with inward-pulled surfaces, I have trouble thinking of examples where such a phenomena occurs.  If there is only one local minimum, then computation might be relatively efficient.<|endoftext|>
TITLE: Number of unique sortings of subset-sums
QUESTION [7 upvotes]: Take the set $A_n=\{a_1,...,a_n\}$. Let $S_n$ be the set of subset-sums of $A_n$. (The subset-sum of the empty set is assumed to be zero.) Assume that there are $2^n$ unique members of $S_n$. How many possible sortings are there of set $S_n$?
For instance, if $n=2$, we have $S_2=\{0,a_1,a_2,a_1+a_2\}$. The number of possible sortings of $S_2$ is 8: 
$$\left\{ \begin{matrix} 0<a_1<a_2<a_1+a_2,\\ 0<a_2<a_1<a_1+a_2,\\ a_2<0<a_1+a_2<a_1, \\ a_1<0<a_1+a_2<a_2, \\  a_2<a_1+a_2<0<a_1, \\ a_1<a_1+a_2<0<a_2, \\ a_1+a_2<a_1<a_2<0, \\ a_1+a_2<a_2<a_1<0 \end{matrix} \right\}.$$

REPLY [9 votes]: As David Speyer points out in a comment, this is equivalent to the number of regions resulting when $\mathbb{R}^n$ is divided by hyperplanes of the form $\sum_{i \in I}a_i = \sum_{i \in J}a_i$ for all disjoint pairs of subsets $I,J \subseteq [n]$.  Dual to this description, it is the number of ways that the $3^n-1$ nonzero vectors in $\{-1,0,1\}^n$ can be divided into "positive" and "negative" by a hyperplane (in general position) passing through the origin, which makes it easy to see that the answer is indeed $8$ for $n=2$.
As I mentioned in a comment, the total is always $2^n$ times what you get when making the assumption $a_i > 0$ for all $i$.  There is another symmetry that can also be exploited: making the assumption $a_1 < a_2 < \cdots < a_n$ reduces the total by a further factor of $n!$, which gives a known sequence:
http://www.oeis.org/A009997
http://arxiv.org/abs/math.CO/9809134
Various key words are "coherent boolean term order", "coherent generalized term order", and "additive antisymmetric comparative probability order".  It doesn't look like anyone knows the values beyond $n=7$.  You'll want to check Maclagan's reference to Fine and Gill 1976 to see if they give any asymptotics.
Including the $2^nn!$ symmetries gives these values:

 $2$ 
 $8$ 
 $96$ 
 $5$ $376$ 
 $1$ $981$ $440$ 
 $5$ $722$ $536$ $960$ 
 $138$ $430$ $238$ $607$ $360$<|endoftext|>
TITLE: Convex varieties that are not homogeneous
QUESTION [5 upvotes]: A projective variety $X$ is convex, if for any $f:\mathbb{P}^1 \to X$, the group $H^1(\mathbb{P}^1, f^*(T_X))$ vanishes. A big group of examples of convex varieties is made of homogeneous varieties. An homogeneous variety is the quotient variety $G/P$ of a Lie group $G$ by a parabolic subgroup $P$ of it.
My question is: is there an easy example of a projective variety, possibly smooth, that is convex but that does not admit such a description?
In the lectures: http://www.math.uic.edu/~coskun/utah-notes.pdf, the author asks whether every rationally connected smooth convex projective variety must also be homogeneous, thus suggesting that it should be rather easy to find such an example in the realm of non-rationally connected varieties.

REPLY [3 votes]: This is a variation of Angelo's first example. Consider a surface without rational curves and blow-up a finite number of distinct points. If $E$ is one of the exceptional divisors and $f \colon E \to X$ is the inclusion in the blow-up, since 
$H^0(E, N_{E|X})=H^1(E, N_{E|X})=0$,
we obtain $H^1(E, f^*T_X)=H^1(E, T_E)=0$, so $X$ is convex. On the other hand, $X$ is not homogeneous, since its automorphism group does not act transitively.
ADDED. As remarked by mdeland, this example does not really work since we can take 
finite covers of $E$ in order to obtain curves that violate the convexity condition. 
In order to avoid this problem, we must require that the splitting type 
of the tangent bundle of $X$ over any rational curve does not contain summands of negative degree. Let us give an example where this condition is satisfied.
Let $C$ be any curve of genus $g(C) \geq 1$ and let 
$\mathcal{E} = \mathcal{O}_C \oplus \mathcal{L}$, where
$\mathcal{L}$ is a line bundle of negative degree $-e$. 
Then $X = \mathbb{P}(\mathcal{E}) $ is a ruled surface over $C$ 
which contains a unique section $C_0$ such that 
$C_0^2 = -e$, in particular $X$ is not homogeneous.
Now let us show that $X$ is convex. Let $F \cong \mathbb{P}^1$
be any fibre of $p \colon X \to C$. We have a short exact sequence
$0 \to T_F \to (T_X)|_F \to N_F \to 0$.
Since $T_F=\mathcal{O}_{\mathbb{P^1}}(2)$ and $N_F=\mathcal{O}_{\mathbb{P}^1}$, 
it follows $\textrm{Ext}^1(N_F, T_F)=0$. Therefore the sequence above actually splits
and we obtain
$(T_X)|_F = \mathcal{O}(2) \oplus \mathcal{O}$.
On the other hand, since $g(C) \geq 1$ the unique rational curves on $X$ are the fibres of $p$,
so every non-constant holomorphic map $f \colon \mathbb{P}^1 \to X$ is given by the inclusion 
of a fibre composed with a finite cover. If $d$ is the degree of such a cover, we obtain 
$f^*T_X = \mathcal{O}(2d) \oplus \mathcal{O}$.
It follows $H^1(\mathbb{P}^1, f^*T_X)=0$, so $X$ is convex. Notice that $X$ is uniruled, 
but not rationally connected.<|endoftext|>
TITLE: connectivity of the group of orientation-preserving homeomorphisms of the sphere
QUESTION [26 upvotes]: In the paper "Local Contractions and a Theorem of Poincare" Sternberg has mentioned the following question which was open when the paper was written:
Is the group of orientation-preserving homeomorphisms of the $n$-sphere arc-wise connected?
According to Sternberg's paper Kneser has shown this to be true for $n=2$. Does anyone know the current status of the problem?

REPLY [30 votes]: This is known to be true for all $n$ as a consequence of the stable homeomorphism conjecture (SHC), itself closely related to the annulus conjecture. 
The SHC says that any orientation preserving homeomorphism of $\mathbb{R}^n$ is stable i.e. a (finite) product of homeomorphisms each of which is the identity on some non-empty open set. 
The same statement is then true for an homeomorphism $h$ of $\mathbb{S}^n$ : first, you can by composition with a (stable) homeomorphism of $\mathbb{S}^n$ assume that $h$ fixes the north pole $p$. Then by SHC $h$ restricted to $\mathbb{S}^n-p\simeq\mathbb{R}^n$ is a composition of homeomorphisms which are the identity on nonempty open sets. But an homeomorphism of $\mathbb{S}^n$ which is the identity on a non-empty open set is isotopic to the identity, by Alexander's trick, hence the conclusion.
PS : in fact, the detailed story is somewhat complicated. What was known relatively early (due to R. D. Anderson, G. M. Fisher around 1960) was that for any topological manifold $M$ (maybe non-compact, but paracompact), the group $H_c(M)$ of homeomorphisms generated by those compactly supported in domains of topological charts $\mathbb{R}^n\simeq U\subset M$ is the smallest nontrivial normal subgroup of $H(M)$ of all homeomorphisms, and that it is simple. In particular,  $H_c(M)$ is arcwise connected. The proof is ingenious, but not very difficult.
The (much harder) methods and results of geometric topology in dimension $3$ (Bing, Moise,...) then allowed Fisher to prove that for a closed $3$-manifold $M$, $H_c(M)$ is open in $H(M)$ and thus coincides with the identity component $H_0(M)$. In particular $H(M)$ is locally arcwise connected, a not at all obvious fact -- later generalized by Cernavskii and Edwards-Kirby, who proved the local contractibilty  (hence local arcwise connectedness) of $H(M)$ in any dimension.
Fisher also considered the group of stable homeomorphisms $H_s(M)$ (without the name) of a connected $M$ (otherwise the notion is empty). He managed to prove that it coincides with the group of orientation preserving ones $H_+(M)$ for closed oriented $3$-manifolds $M$ admitting an orientation reversing homeomorphism. For  $M=\mathbb{S}^3$, this implies that $H_+=H_0=H_c$ : any orientation preseving homeomorphism of $\mathbb{S}^3$ is isotopic to the identity (note that for all $n$, $H_s(\mathbb{S}^n)=H_c(\mathbb{S}^n)$). This is the $n=3$ case of your question.
Then M. Brown and H. Gluck named stable homeomorphisms, and studied stable structures on manifolds.
A puzzling aspect of the notion of stable homeomorphism is that it is very "contagious" : if $h\in H(M)$ coincides with an element $f$ of $H_s(M)$ on a nonempty open set $U$, then $h$ is in $H_s(M)$, since $h^{-1}f$ is the identity on $U$. So this is seen locally everywhere, like orientation preservation (to which it was eventually identified).
Brown and Gluck proved that SHC$_n$ (SHC in dimension $n$) implies the annulus conjecture in dimension $n$ (AC$_n$), and that AC$_k$ in all dimensions $k\leq n$ imply SHC$_n$. But this was still stuck at SHC$_3$.
After that came R. Kirby (and L. Siebenmann) in 1968, who proved (using the results of surgery by Wall et al), that SHC$_n$ (hence AC$_n$) is true in all dimensions $n> 4$.
But the remaining case SHC$_4$ was only solved by F. Quinn in 1982 (after work of A. Casson and M. Freedman),
who proved AC$_4$, hence SHC$_4$ since the $n\leq 3$ cases were known. See the survey by Edwards.<|endoftext|>
TITLE: "Are you taller than the average of those who are taller than the average?"
QUESTION [27 upvotes]: I've met tall people. That is: people taller than the average. Every now and then we encounter really tall people, even taller than the average of tall people i.e. taller than the average of those who are taller than the average. Meybe you've met someone who's even taller than the average of those who are taller than the average of those who are taller than the average... And so on.
So, take a quantity $X$ that we suppose normally distributed (caveat, I have no deep knowledge of probability theory), i.e. it's described by a gaussian distribution that we suppose standardized and call $f(x)$.
Now, define:
$M_0:= \int_{-\infty}^{\infty}f(x)dx=1$
$\mu_0:=\int_{-\infty}^{\infty}xf(x)dx=0$
and, inductively,
$M_{n+1}:= \int_{\mu_n}^{\infty}f(x)dx$
$\mu_{n+1}:=\frac{1}{M_n}\int_{\mu_n}^{\infty}xf(x)dx$
I think this describes the situation in which your $X$ (e.g. height) has the value $\mu_n$ precisely when you're as $X$ as the average of those who are more $X$ than the average of those who are more $X$ than...... (n times). If not, please explain why.
So my questions:

How does the sequence $\mu_n$ behave asymptotically? Does it converge? 
If yes, is there a nice expression for the limit? 
Is there even a reasonably explicit expression ("closed form") for $\mu_n$ as a function of $n$?

REPLY [26 votes]: As in Nate's answer, we are interested in iterating the function
$$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$
The numerator is $e^{-y^2}/2$ (elementary). The denominator is $e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$ (see Wikipedia). So $G(y) = y + (1/2) y^{-1} + O(y^{-3})$. 
Set $z_n = \mu_n^2$. Then 
$$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-2}) = z_n + 1 + O(z_n^{-1}).$$
So $z_n \approx n$ and we see that $\mu_n \to \infty$ like $\sqrt{n}$.
I haven't checked the details, but I think you should be able to get something like $\mu_n = n^{1/2} + O(1)$.<|endoftext|>
TITLE: On measurable functions of two variables
QUESTION [6 upvotes]: Suppose $f:[0,1]^2\to\mathbb{R}$, $(t,x)\mapsto f(t,x)$, is such that for each $t\in[0,1]$ $f(t,\cdot)$ is Lebesgue measurable on $[0,1]$, and for each $x\in[0,1]$ $f(\cdot,x)$ is continuous everywhere on $[0,1]\ni t$. 
1. Does this imply that $f(t,x)$ is measurable on $[0,1]^2$?
2. Does this imply that the function $g(x)=\min\limits_{t\in[0,1]}f(t,x)$ is measurable on $[0,1]$?

REPLY [7 votes]: Yes, by continuity in the $t$ variable $f(t,x)=\lim_n f(\lfloor n t\rfloor/n,x)$, which expresses $f$ as the pointwise limit of a sequence of measurable functions.
Yes, by continuity in the $t$ variable we have $\min_{t\in[0,1]} f(t,x)=\min_{t\in[0,1]\cap {\mathbb Q}} f(t,x)$, where $\mathbb Q$ means the rational numbers.<|endoftext|>
TITLE: Intuitive Proof of Cramer's Decomposition Theorem
QUESTION [20 upvotes]: Cramer's decomposition theorem states that if $X$ and $Y$ are independent real random variables and $X+Y$ has normal distribution, then both $X$ and $Y$ are normally distributed. I've seen a few proofs of this result which involve grunt work in the realm of complex analysis. I was wondering if there is an intuitive proof of this result. I would like to see a proof that exploits the unique nature of the normal distribution. For example, can one derive the result from the fact that on the real line, the normal distribution maximizes entropy for a given mean and standard deviation? 
Edit: The complex analysis proof I'm thinking of uses the fact that if $E[\exp(\alpha X^2)]<\infty$ for some $\alpha>0$ and the analytic continuation of the characteristic function of $X$ is nonzero, then $X$ is normally distributed.

REPLY [6 votes]: For example, can one derive the result from the fact that on the real line, the normal distribution maximizes entropy for a given mean and standard deviation?

That sounds like a workable approach. Here's an intuitive proof:
Visualize the convolution of two functions as the smoothing of one function by the other one's shape. Smoothing can only spread out probability mass and not concentrate it, so entropy cannot decrease in the process. If one function is kept fixed while the entropy of the other is increased, the entropy of the convolution also increases.
Now suppose $X$ and $Y$ are independent random variables such that $X + Y$ has maximum entropy among those random variables with its mean and variance, but $X$ does not have maximum entropy in its mean and variance class. Let $X'$ be a random variable with the same mean and variance as $X$ but greater entropy. Then $X' + Y$ has the same mean and variance as $X + Y$, and by the entropy convolution inequality in the previous paragraph $X' + Y$ has greater entropy than $X + Y$. But this is a contradiction, so $X$ and by symmetry $Y$ must have maximum entropy.<|endoftext|>
TITLE: Burnside's Lemma and Geometry
QUESTION [7 upvotes]: I think one of the most interesting results in Elementary Group Theory is the so-called "Burnside's Lemma", counting the numbers of orbits of a (finite) group action.
I wonder if there is any (interesting) application in Elementary Geometry (I mean Euclidean, hyperbolic or elliptic geometry).
Searching on Google, I've found the article "Applying Burnside’s lemma to a one-dimensional
Escher problem" by T. Pisanski, but it sounds to me rather a combinatorial result.

REPLY [7 votes]: Burnside Lemma can be used as a first step to classify all finite subgroups of $\mathrm{SO}(3)$: it gives you that there are at most $3$ orbits in the action of any finite group $G$ on the set of intersections between axes of elements of $G$ and the unit sphere.<|endoftext|>
TITLE: Do dualizable Hopf algebras in braided categories have invertible antipodes?
QUESTION [7 upvotes]: A classical result of Larson and Sweedler says that a finite dimensional Hopf algebra over a field has invertible antipode. Does this result extend to the setting of Hopf algebras in braided categories? In other words, if a Hopf algebra in a braided category is dualizable, is its antipode necessarily invertible? (Equivalently, if a dualizable bialgebra has invertible fusion operator, is its opfusion operator invertible?) If this result is not true, is there an "easy" example where it fails? Of course, if it fails, it should do so in the free braided category equipped with a Hopf algebra, but it seems difficult to see that things aren't invertible using string diagrams.

REPLY [6 votes]: Yes, they do. See Theorem 4.1 in "Finite Hopf algebra in braided tensor categories" M. Takeuchi, Journal Pure and Applid Algebra 138 (1999) 59-82<|endoftext|>
TITLE: How many non-equivalent sections of a regular 7-simplex?
QUESTION [10 upvotes]: Suppose we have a regular 7-simplex in $\mathbb{R}^8$ defined by vertices <1,0,0,...,0>, <0,1,0,..,0>,...,<0,...,0,1>. A section is a 3-dimensional linear subspace of $\mathbb{R}^8$ that contains simplex centroid and three other points, each of which is a centroid of a non-empty set of simplex vertices. Two sections are equivalent if they are identical spaces under permutation of coordinates. In other words, when some permutation of coordinates is the bijection between two spaces. How many non-equivalent sections are there? Is there an efficient way to enumerate them?
Motivation: visualizing symmetric priors over distributions over 8 outcomes
Update 12/09
I tried an automatic search and got 49 sections, same as Peter Shor below. Here they are. Note that grouping is a bit different since I group sections with or without unexpected centroids together.
No empty vertices:

One empty vertex:

Two empty vertices:

Three empty vertices:

Four empty vertices:

old stuff
Here's an illustration of solving this problem for 2-sections of a 3-simplex in 4 dimensions. There seems to be only 2 non-equivalent 2-sections (triangle and square). This solves the problem of visualizing entropy (contour lines) of distributions over 4 outcomes, and I'd like extend it to 8 outcomes.

Image sources: (1)
(2)
(3)
(4)
(5)
(6)

REPLY [10 votes]: Here's the answer. The main claims (Claims 1-4) I am fairly sure I got right, but I could easily have missed a case (or counted an extra case) in the later enumeration. If anybody finds a mistake, please comment. Let me remark that I find Claims 1-4 much more interesting than the subsequent enumeration based on them.
The simplex centroid $e=(1,1,1,1,1,1,1,1)$ is included in all our hyperplanes, but I'm generally not counting it as a centroid in the discussion below (so centroid means centroid of a $k$-dimensional face, with $k < 7$).
We'll divide the question into cases.
The first case we deal with is when we don't have any unexpected centroids. We start
with three centroids $a$, $b$, and $c$. These will automatically generate $\bar{a}$, $\bar{b}$, $\bar{c}$. We will call any centroid other than these six an unexpected centroid. We represent our centroids as subsets of
{$1,2,\ldots,8$}. If we have the centroid
{$1,2,3$}$=\langle 1,1,1,0,0,0,0,0\rangle $,
then we automatically have the centroid {$4,5,6,7,8$}$=\langle 0,0,0,1,1,1,1,1\rangle$ corresponding to
the complement of the set. So, to summarize, the first case consists of hyperplanes 
which pass through exactly six centroids: $a,b,c,\bar{a},\bar{b},\bar{c}$.
Now, let's represent this case by putting the numbers {$1,2,\ldots,8$} on
the vertices of a cube. The cube will have three faces
corresponding to $a,b,c$ and the three opposite faces will correspond to
$\bar{a},\bar{b},\bar{c}$. For example, if the sets were $a=${$1,2,3$},
$b=${ $1,5,6$} and $c=${$2,5,6,7$}, then the vertex $\bar{a}bc$ would contain
{$5,6$}, the vertex $\bar{a}\bar{b}\bar{c}$ would contain {$4,8$}, and the
vertex $\bar{a}b\bar{c}$ would be empty. It's not too hard to see that
whether there is an unexpected centroid only depends on the positions of the
empty vertices. It's also clear that rotations and reflections
of this cube give equivalent sections.
Claim 1: If two adjacent vertices are empty, there is an unexpected
centroid.
Proof: The two adjacent vertices form an edge. We might as well rotate
the cube so that the empty edge is the $ab$ edge. Then we have $a \cap b =
\emptyset$. This means that $a \cup b$ is an unexpected centroid
(here we have to use the fact $a \neq \bar{b}$). Example: if $a =${$1,2$}, $b =
${$3,4,5$}, then $a \cup b =${$1,2,3,4,5$} is in the linear span of $a$ and
$b$.
Claim 2: If two opposite vertices of the cube are empty, then there is an
unexpected centroid.
Proof: We can rotate the cube so the empty vertices correspond to
$\bar{a}\bar{b}\bar{c}$ and $abc$. Then, for example, if
$a=(1,1,1,0,0,0,0,0)$, $b=(0,0,1,1,1,1,0,0)$ and $c=(1,0,0,0,0,1,1,1)$,
we can take $a+b+c-e$ where $e$ is the all-ones vector, and get
$(1,0,1,0,0,1,0,0)$.
Claim 3: If the odd- or even-parity vertices of the cube are empty, then
there is an unexpected centroid.
Proof: Rotate the cube so that $\bar{a}\bar{b}\bar{c}$ is empty. Now, every
coordinate is in exactly 1 or 3 of $a,b,c$.  Thus, $\frac{1}{2}(a+b+c-e)$
is an unexpected centroid.  Example
$a=(1,1,1,0,0,0,1,1)$, $b=(0,0,0,1,1,0,1,1)$ and $c=(0,0,0,0,0,1,1,1)$, and
$\frac{1}{2}(a+b+c-e) = (0,0,0,0,0,0,1,1)$
Claim 4: If none of the situations in Claims 1,2,3 hold, then there is no
unexpected centroid.
Proof: We can rotate the cube so that the empty vertices are a subset of
$\bar{a}bc$, $a\bar{b}c$ and $ab\bar{c}$. For there to be an unexpected
centroid, you must be able to find $\alpha a + \beta b + \gamma c$ so that
the coordinates of this vector take on two values. One of these
coordinates is 0 (since $\bar{a}\bar{b}\bar{c}$ is not empty), meaning we
can assume wlog that $\alpha, \beta, \gamma$ are either 1 or 0. But
for $\alpha + \beta + \gamma$ to also be either 1 or 0, we need two
of $\alpha, \beta, \gamma$ to be 0, which means that we don't get an
unexpected centroid.
So now, we need to enumerate the number of ways of putting 8 elements
onto the vertices of a unit cube so that at least one element is on each of
the nonempty vertices. Since sections are equivalent under permutations of the coordinates, we should consider these to be  8 identical elements (so the only thing that matters is how many elements are on a vertex). There are four cases.
Case A: no empty vertices. There is just 1 way of doing this: putting
one element on each vertex.
Case B: one empty vertex. There are 3 ways of doing this. Exactly
one vertex will have two elements on it, and it can be either Hamming
distance one, two, or three from the empty vertex.
Case C: two empty vertices. In this case, these two vertices must have
Hamming distance 2. The two extra elements can either be on the same vertex
(3 ways) or two different vertices (7 ways).
Case D: three empty vertices. In this case, any pair of these three vertices
must have Hamming distance 2, so there's only one way of arranging them.
The three extra elements can either be all on the same vertex (3 ways),
divided two on one vertex and one on another (7 ways), or on three different
vertices (4 ways).
This gives 28 essentially different sections with no unexpected centroids.
We now must count the cases with unexpected centroids corresponding to Claims 1-3. 
We'll deal with the situation in the Claims 1,2,3 separately.
Case of Claim 3
Let's start
with the situation in Claim 3. First, we can assume that there are no
empty vertices of the cube other than the two opposite ones (if this happens,
we are in the Claim 1 situation, and we take care of it there).
We now can choose another centroid $d$ in the linear span of $a,b,c,e$
so that $a \cup b \cup c \cup d$ covers every coordinate exactly twice.
By the criterion that there are six non-empty vertices, none of the
six intersections $a \cap b$, $a \cap c$, etc. can be empty. We need to
put the eight elements into these six intersections. This corresponds
to putting 8 elements on the edges of a tetrahedron so that every edge
corresponds to at least one element. There are 3 ways to do this (two
extra on one edge, one extra on each of two opposite edges, and one
extra on each of two adjacent edges). Claim 3 thus gives 3 more
non-equivalent sections. Notice that if we had analysed Claim 3 by just
looking at the symmetries of the cube (as we did for the cases without
unexpected centroids), we would have obtained four non-equivalent sections.
Case of Claim 2
What I'd like to claim here is that this is really the situation in Claim 1
disguised. Maybe the best way to do this is by example. If we have
$a=${$1,2,7,8$}, $b=${$3,4,7,8$}, $c=${$5,6,7,8$}, then the centroid {$7,8$}
is in our hyperplane, and the hyperplane is thus generated by $a'=${$1,2$}, $b'=${$3,4$},
$c'=${$5,6$}, which is covered by Claim 1.
Case of Claim 1
Here, there are three possibilities. In the first one, there are four
centroids $a$, $b$, $c$, $d$, with pairwise empty intersections so that
$a \cup b \cup c \cup d =${$1,2,\ldots,8$}. The number of ways of doing this
is the number of partitions of 8 into four non-empty parts, which is 5:
{$(5,1,1,1), (4,2,1,1), (3,3,1,1),(3,2,2,1),(2,2,2,2)$}.
In the second possibility, we have three pairwise disjoint
centroids $a$, $b$, $c$, with $a \cup b \cup c = e$,
and also another centroid $d$ so that
both $d\cap x$ and $\bar{d} \cap x$ are non-empty for $x=a,b,c$.
The cardinalities of $a,b,c$ could be {$4,2,2$} or {$3,3,2$}. In either
case, we get two non-equivalent sections, giving 4 total non-equivalent
sections.
For the third possibility, we have three pairwise disjoint
centroids $a$, $b$, $c$, with $a \cup b \cup c = e$, and
we have two more pairwise disjoint centroids $f$ and $g$ so that $f \cup g = a \cup 
b$. In this case, the cardinality of $c$ can range from 1 to 4.
I'll just list representative vectors for these possibilities. The coordinates 
considered are those not in $c$.
$c=4$
$(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)$
$c=3$
$(1,1,1,0,0),(0,0,1,1,0),(0,0,0,1,1),(1,1,0,0,1)$
$c=2$
$(1,1,1,0,0,0),(0,0,1,1,1,0),(0,0,0,1,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,0,1,1,1,0),(0,0,0,0,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,1,1,1,1,0),(0,0,0,0,1,1),(1,0,0,0,0,1)$
$c=1$
$(1,1,1,1,1,0,0),(0,1,1,1,1,1,0),(0,0,0,0,0,1,1),(1,0,0,0,0,0,1)$
$(1,1,1,1,1,0,0),(0,0,1,1,1,1,0),(0,0,0,0,0,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,0,1,1,1,1,0),(0,0,0,0,1,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,1,1,1,1,0,0),(0,0,0,0,1,1,1),(1,0,0,0,0,1,1)$
This gives 9 more non-equivalent sections, making 49 altogether.<|endoftext|>
TITLE: Hopf algebras as cohomology of $\mathbb{CP}^\infty$, $\Omega S^3$ and related $H$-spaces
QUESTION [13 upvotes]: Let me begin by a couple of questions :

Consider a graded abelian group $V=\oplus_{i\geq 0} V_i$  such that $V_{2i}=\mathbb{Z}$ and $V_{\textrm{odd}}=0$. What are the possible Hopf algebra structures on it? 

One can ask a slightly stronger question :

When does a given Hopf algebra structure on $V$ (as above) arise as the integral cohomology of a $H$-space?

The motivation behind this question is purely my own curiosity. While discussing how to distinguish $\Omega S^3$ and $\mathbb{CP}^\infty$ rationally, we saw that the rational cohomology or the rational homotopy groups are unable to detect the difference. However, $H^\ast(\Omega S^3;\mathbb{Z})=\Gamma_{\mathbb{Z}}[\alpha]$, the divided polynomial algebra generated by $\alpha$ (of degree $2$) while $H^\ast(\mathbb{CP}^\infty;\mathbb{Z})=\mathbb{Z}[u]$ is the polynomial algebra generated by $u$ (of degree $2$). Moreover, a polynomial algebra such as $\mathbb{Z}[u]$ has a comultiplication map given by $u\stackrel{\Delta}{\longrightarrow}1\otimes u+u\otimes 1$ and extended naturally. One can check that the dual (as a Hopf algebra) of $\mathbb{Z}[u]$ is isomorphic to $\Gamma_{\mathbb{Z}}[u^\ast]$, where $u^\ast$ is the dual of $u$. 
From what I could conclude by playing around with coefficients is that for each prime $p$ and a positive integer $r$ one can cook up a Hopf algebra structure on $V$. I don't know if they come from a space. However, these structure constants must be compatible with the action of the Steenrod algebra (or the mod $p$ version) if $V=H^\ast(X;\mathbb{Z})$ for some $H$-space $X$. I vaguely remember that compatibility with the Steenrod algebra is not sufficient and Adams operations provide further obstructions (although I may be wrong on this point). This leads me to :

Is there a (list of) necessary and sufficient criteria (in general or at least in this case) which tells us when a given Hopf algebra structure on graded vector space arises as $H^\ast(X;\mathbb{Z})$ for some $H$-space?

This may be well known (and perhaps classical) to homotopy theorists and any reference to known results are good enough for me.

REPLY [9 votes]: Call a generating class in degree 1 'x'.  It is forced to be primitive.  The identity
$$
\Delta(x^n) = (1 \otimes x + x \otimes 1)^n \neq 0
$$
shows that, after tensoring with $\mathbb{Q}$, the resulting ring is a polynomial ring on your primitive class.  Thus, you find that your Hopf algebra is some sub-Hopf algebra of $\mathbb{Q}[x]$ and contains $\mathbb{Z}[x]$.
In particular, in each degree $n$ there is a unique positive integer $a_n$ such that $x^n/a_n$ is a generator of $V_n$.  We have $a_1 = 1$, and this subset being closed under multiplication is equivalent to $a_{n+m}/a_n a_m \in \mathbb{Z}$ for all $n$ and $m$.
Take duals.  The rationalization of the dual is also a polynomial algebra on $x^*$, and $V_n^*$ is generated by $${a_n} (x^n)^* = \frac{a_n}{n!} (x^*)^n.$$
This subset of $\mathbb{Q}[x^*]$ being closed under multiplication is equivalent to $\binom{n+m}{n} \cdot \frac{a_n a_m}{a_{n+m}} \in \mathbb{Z}$ for all n and m.
The possible Hopf algebra structures are therefore defined by all such sequences of positive integers $a_n$ so that $a_1 = 1$ and $a_n a_m$ divides $a_{n+m}$ divides  $\binom{n+m}{n} a_n a_m$.
The realizability problem is much harder and I do not have a concrete answer for you.<|endoftext|>
TITLE: Status of Harvey Friedman's grand conjecture?
QUESTION [34 upvotes]: Friedman [1] conjectured

Every theorem published in the Annals of Mathematics whose
  statement involves only finitary mathematical objects (i.e., what logicians
  call an arithmetical statement) can be proved in EFA. EFA is the weak
  fragment of Peano Arithmetic based on the usual quantifier free axioms for
  0,1,+,x,exp, together with the scheme of induction for all formulas in the
  language all of whose quantifiers are bounded. This has not even been
  carefully established for Peano Arithmetic. It is widely believed to be
  true for Peano Arithmetic, and I think that in every case where a logician
  has taken the time to learn the proofs, that logician also sees how to
  prove the theorem in Peano Arithmetic. However, there are some proofs which
  are very difficult to understand for all but a few people that have
  appeared in the Annals of Mathematics - e.g., Wiles' proof of FLT.

Have there been any serious challenges to this or the weaker conjecture with Peano arithmetic in place of exponential function arithmetic?
[1] http://cs.nyu.edu/pipermail/fom/1999-April/003014.html

REPLY [4 votes]: "Mathematische Annalen" and Annals of Mathematics aren't the same journal, but they have similar titles, so I'll try to slip this in.
Gentzen's proof of Con(PA) using transfinite induction on $\epsilon_0$ appeared in Mathematische Annalen in 1936 (cite pasted from Wikipedia):

G. Gentzen, 1936. 'Die Widerspruchfreiheit der reinen Zahlentheorie'. Mathematische Annalen, 112:493–565. Translated as 'The consistency of arithmetic', in (M. E. Szabo 1969). DOI: 10.1007/BF01565428, free eudml link.

Obviously this theorem can't be proved in PA itself, much less EFA.  I guess it doesn't really count but it seems sort of relevant.  It's surely a major theorem.<|endoftext|>
TITLE: Schroeder-Bernstein for Rings
QUESTION [8 upvotes]: Suppose $f: A \to B$ and $g: B \to A$ are injections of rings
(commutative with identity).  Must $A$ and $B$ be isomorphic as
rings?
According to this
question, this answer should be "no", but can someone give an example?
Thanks!

REPLY [2 votes]: Sam Lichtenstein poses the dual question in comments:  

What's a counterexample to "dual Schroeder-Bernstein" for rings? (That is, same question but with surjections rather than injections.) Is there one with A,B finite type over a field? 

That is,

Do there exist finite type $k$-algebras $A, B$ not isomorphic to each other, and surjections $A\to B, B\to A$? (*)

He gives an example in the non-Noetherian case; I claim the "dual Schroeder-Bernstein theorem" is true if $A$ and $B$ are Noetherian.  And in general, if two Noetherian schemes $X, Y$ admit maps $i: X\to Y, j: Y\to X$ exhibiting each as a closed subscheme of the other, then $i, j$ are isomorphisms.   So the answer to (*) is "no".
I'll prove the more general claim.  Assume to the contrary that one of $i,j$ is not an isomorphism.  Then $j\circ i: X\to X$ exhibits $X$ as a proper closed subscheme of itself, say $X_1$.  But then $X_1$ is isomorphic to some proper closed subscheme of itself, say $X_2$; continuing in this manner, we may construct a sequence $X_n$ where each $X_i$ is a proper closed subscheme of $X_{i-1}$.  Let $\mathcal{I}_n$ be the ideal sheaf of $X_i$ in $\mathcal{O}_X$.  By Noetherianness we must have that $\mathcal{I}_1\subset \mathcal{I}_2\subset \mathcal{I}_3\subset\cdots$ stabilizes, however, which contradicts the claim that each $X_i\subset X_{i-1}$ is a proper inclusion.  Here's a more formal write-up of the affine case.
This provides an example of a "surjunctive" category in the sense of John Goodnick's answer to this question.<|endoftext|>
TITLE: Gromov-Witten and integrability 2.
QUESTION [6 upvotes]: This is a followup of my previous question Gromov-Witten and integrability. As I have learned from the answer (but guessed before), GW potentials of the point and $P^1$ (with different modifications) are, more or less, the only examples of the GW generation functions with established integrable properties. So what about higher genera curves? Are they really so complicated to establish integrability, at least for stable sector? What is the main problem with them?

REPLY [7 votes]: Here's a sketch of my understanding of where the difficulty lies with higher genus curves.  It got kind of long and vague, at parts, but hopefully it explains a few problems.
In Gromov-Witten theory, I'm aware of two or three general approaches to integrability currently.  Certainly there's overlap among these approaches:

The quantumn cohomology of $X$ turns the cohomology of $X$ into a Frobenius manifold, and then Dubrovin has connections to various integrable hierarchies.
In some cases, the Gromov-Witten theory can be nicely expressed in terms of various Fock-spaces and the infinite wedges, and then you can get connections to the integrable hierarchies related to various infinite dimensional lie groups from the Kyoto school.
Matrix Models, and in particular the work of Eynard and Orantin.  This seems to have mostly implemented in terms of the Topological vertex, which is rephrasing things in terms of combinatorics.

For higher genus curves, approach 1 is completely a no-go: there are no positive degree maps from a sphere to a higher genus curve, so the quantum cohomology is just the usual cohomology.
The GW-theory of higher genus curves is computed by Okounkov and Pandharipande very much in the general method of 2. The GW/Hurwitz correspondence shows that what they call the "stationary sector" (descendents of point classes only -- not the identity or odd cohomology classes.  Is this the same as the stable sector?) is equivalent to Hurwitz theory, which is completely computable in terms of the symmetric group.  There are nice ways of computing these characters, and this is where the connections to the infinite dimensional lie algebras and come up.  However, this computation becomes much more complicated as we increase the genus.  I'd like to explain how it's an entirely different beast for genus 2 or bigger.
The GW/Hurwitz correspondence turns insertions of point classes into ramification data, and hence as far as counting goes, into multiplying elements in the symmetric group.  The ramification that shows up is nice and is easy to write in terms of the infinite wedge (free fermion) and I think can be written in terms of matrix model type stuff as well.  One particular nice bit is that any given insertion only produces permutations with bounded supports: even if we let the degree get big (so considering symmetric groups $S_d$ for $d$ large, we're only going to have a few nontrivial cycles in these permutations.  Most points in $S_d$ will be fixed.  
For genus 0, we're only multiplying these special elements in the symmetric group, and this is why everything is so beautiful here.
For higher genus, one way to keep everything in terms of just the symmetric group is degenerate the curve by pinching off $g$ cycles, so that we again have a genus zero curve, but now with $g$ pairs of points identified.  At each of these identified pairs of points, we're going to need to insert inverse permtuations, and we're going to have to sum over all permutations in $S_d$ this way for each pair of points.
This is where things get ugly -- in higher genus, we have to consider these arbitrary permutations.
In genus one, things aren't all that bad: we only have two arbitrary permutations.  So we can start with one of them, multiply in turn by the nice permutations that we know how to do, and then at end, instead of multiplying two arbitrary permutations, we only have to check that we have the same permutation, as the whole product has to be the identity.  Essentially, we're taking the trace of some nice operator on the infinite wedge, and this why the quasi-modular forms show up here -- work of Bloch-Okounkov shows these are quasimodular forms.
Once we get to genus two though, we loose this as well.  We really have to be able to multiply two arbitrary permutations of $S_d$.   I haven't read Mironov-Morozov's stuff on this closely at all, but seemed to recall them having some type of non-integrablity results for the general multiplication of three permutations, but couldn't find exactly this statement.  The start of section three of this paper might touch on this, though.
The representation theory viewpoint is probably better for integrable systems, but I think it's similarly more difficult.<|endoftext|>
TITLE: Are there motives which do not, or should not, show up in the cohomology of any Shimura variety?
QUESTION [42 upvotes]: Let $F$ be a real quadratic field and let $E/F$ be an elliptic curve with conductor 1 (i.e. with good reduction everywhere; these things can and do exist) (perhaps also I should assume E has no CM, even over F-bar, just to avoid some counterexamples to things I'll say later on). Let me assume that $E$ is modular. Then there will be some level 1 Hilbert modular form over $F$ corresponding to $E$. But my understanding is that the cohomology of $E$ will not show up in any of the "usual suspect" Shimura varieties associated to this situation (the level 1 Hilbert modular surface, or any Shimura curve [the reason it can't show up here is that a quaternion algebra ramified at precisely one infinite place must also ramify at one finite place]).
If you want a more concrete assertion, I am saying that the Tate module of $E$, or any twist of this, shouldn't show up as a subquotient of the etale cohomology of the Shimura varieties attached to $GL(2)$ or any of its inner forms over $F$ (my knowledge of the cohomology of Hilbert modular surfaces is poor though; I hope I have this right).
But here's the question. I have it in my head that someone once told me that $E$ (or perhaps more precisely the motive attached to $E$) should not show up in the cohomology of any Shimura variety. This is kind of interesting, because here is a programme for meromorphically continuing the L-function of an arbitrary smooth projective variety over a number field to the complex plane:
1) Observe that automorphic forms for GL_n have very well-behaved L-functions; prove that they extend to the whole complex plane. [standard stuff].
2) Prove the same for automorphic forms on any connected reductive algebraic group over a number field [i.e. prove Langlands functoriality]
3) Prove that the L-functions attached to the cohomology of Shimura varieties can be interpreted in terms of automorphic forms [i.e. prove conjectures of Langlands, known in many cases]
4) Prove that the cohomology of any algebraic variety at all (over a number field) shows up in the cohomology of a Shimura variety. [huge generalisation of Taniyama-Shimura-Weil modularity conjecture]
My understanding is that this programme, nice though it looks, is expected to fail because (4) is expected not to be true. And I believe I was once assured by an expert that the kind of variety for which problems might occur is the elliptic curve over $F$ mentioned above. At the time I did not understand the reasons given to me for why this should be the case, so of course now I can't reproduce them.
Have I got this right or have I got my wires crossed?
EDIT (more precisely, "addition"): Milne's comment below seems to indicate that I did misremember, and that in fact I was probably only told what Milne mentions below. So in fact I probably need to modify the question: the question I'd like to ask now is "is (4) a reasonable statement?".

REPLY [4 votes]: In their 2017 preprint
On subquotients of the étale cohomology of Shimura varieties
C.Johannsson and J.Thorne have made substantial progresses towards question 4. In particular, they show that under reasonable conjectures, the motive attached to an elliptic curve over an imaginary CM field does not appear as subquotient of the intersection cohomology of the minimal compactification of a Shimura variety, and hence should also not appear in the ordinary cohomology of open Shimura varieties.
Very roughly, they show that the conjectures of Arthur on $A$-packets and of Kottwitz on the intersection cohomology of Shimura varieties imply that any Galois representation of the absolute Galois group of a CM field which shows up as a subquotient of the intersection cohomology of Shimura varieties and which is strongly irreducible, that is to say which remains irreducible after restriction to the absolute Galois group of any finite extension, is conjugate self-dual up to a twist. As there are motivic strongly irreducible Galois representations which are not conjugate self-dual up to a twist, this gives a negative answer to statement 4.
It seems to me, by the way, that their work is a concrete implementation of the strategy suggested in the answer of Laurent F. (whoever that guy is).<|endoftext|>
TITLE: When is the K-theory presheaf a sheaf?
QUESTION [6 upvotes]: Let $F$ be a Deligne-Mumford stack that is of finite type, smooth and proper over $\mathrm{Spec~}k$ for a perfect field $k$. Consider $K_m$, the presheaf of $m$-th $K$-groups on $F_{et}$, the etale site of $F$:
$K_m : F_{et} \to Ab$
$(U \to F) \mapsto  K_m(U)$
$(f : U \to V) \mapsto (f^{*} : K_m(V) \to K_m(U))$
My question is, what are some simple cases when this is already a sheaf? For example, is it a sheaf when $F = BG$ for a finite group $G$?
Background
My question is aimed at a computation of motives of DM-stacks. The sheaffification $\mathcal{K}_m = K_m^{++}$ is one way to define the Chow groups of $F$:
$A^m(f) := H^m(F_{et}, \mathcal{K}_m \otimes {\bf Q})$
A twist on this definition leads to a well-behaved theory of motives for DM-stacks described by Toen
Etale site
Someone might be able to confirm that the cohomology can be computed using the etale site whose objects are etale morphisms from affine schemes, since Laumon and Moret-Bailly show it's equivalent (by the inclusion) to the larger site which contains all etale morphisms from algebraic spaces (Champs algebriques, p.102). This might simplify working with the $K$-groups.

REPLY [5 votes]: In general, these presheaves are not sheaves, even on the etale sites of fields. As an easy example, $K_2(\mathbb{C})$ is non-torsion divisible, but $K_2(\mathbb{R})$ has a $2$-torsion element given in symbols by $(-1,-1)$ in Milnor K-theory. But, $K_2(\mathbb{R})$, if $K_2$ were a sheaf, would be the $\mathbb{Z}/2$-fixed points of $K_2(\mathbb{C})$. This cannot happen in this example.
Using the fact that $K_{2i}$ of an algebraically closed field is a non-torsion uniquely divisible group, I imagine one can construct counter-examples for any even K-group.
I would imagine that odd K-groups are also not sheaves.
However, for finite fields, the situation might be different, by Quillen's computation. There, it looks as if the K-groups might be sheaves.
For details on $K_2$ and Milnor $K$-theory, look up Matsumoto's Theorem. For other K-groups of algebraically closed fields, see Suslin's paper On the K-theory of algebraically closed fields.
In general, the place to start thinking about the etale site and algebraic K-theory would be Thomason's paper Algebraic K-theory and etale cohomology.<|endoftext|>
TITLE: A down-to-earth introduction to the uses of derived categories
QUESTION [43 upvotes]: When I was learning about spectral sequences, one of the most helpful sources I found was Ravi Vakil's notes here.  These notes are very down-to-earth and give a kind of minimum knowledge needed about spectral sequences in order to use them.
Does anyone know of a similar source for derived categories?  Something that concentrates on showing how these things are used, without developing the entire theory, or necessarily even giving complete, rigorous definitions?

REPLY [2 votes]: "Derived Categories summer graduate school" is also a very good introduction.<|endoftext|>
TITLE: Must a ring which admits a Euclidean quadratic form be Euclidean?
QUESTION [22 upvotes]: The question is in the title, but employs some private terminology, so I had better explain.
Let $R$ be an integral domain with fraction field $K$, and write $R^{\bullet}$ for $R \setminus \{0\}$.  For my purposes here, a norm on $R$ will be a function $| \ |: R^{\bullet} \rightarrow \mathbb{Z}^+$ 
such that for all $x,y \in R^{\bullet}$, $|xy| = |x||y|$ and $|x| = 1$ iff $x \in R^{\times}$.  Also put $|0| = 0$.
[Edit: I forgot to mention that the norm map extends uniquely to a group homomorphism $K^{\bullet} \rightarrow \mathbb{Q}^{> 0}$.]
The norm is Euclidean if for all $x \in K \setminus R$, there exists $y \in R$ such that $|x-y| < 1$.
Now let $q(x) = q(x_1,\ldots,x_n)$ be a (nondegenerate) quadratic form over $R$, which I mean in the relatively naive sense of just an element $R[x_1,\ldots,x_n]$ which is homogeneous of degree $2$.  Supposing that a norm $| \ |$ on $R$ has been fixed, I say that the quadratic form $q$ is Euclidean if for all $x \in K^n \setminus R^n$, there exists $y \in R^n$ such that $0 < |q(x-y)| < 1$.  
(If $q$ is anistropic, then $x \in K^n \setminus R^n$, $y \in R^n$ implies that $x-y \neq 0$, so $|q(x-y)| \neq 0$, and the condition simplifies to $|q(x-y)| < 1$.)
As an example, the sum of $n$ squares form is Euclidean over $\mathbb{Z}$ iff $1 \leq n \leq 3$.
It is easy to see that the ring $R$ (with its fixed norm) is Euclidean iff the quadratic form $q(x) = x^2$ is Euclidean (iff the quadratic form $q(x,y) = xy$ is Euclidean), so the concept of a Euclidean quadratic form is indeed some kind of generalization of that of a Euclidean ring.  
Conversely, the following is an obvious question that I have not been able to answer.

Suppose that a normed ring $R$ admits some Euclidean quadratic form $q$.  Is $R$ then necessarily a Euclidean domain?  In other words, can there be any Euclidean quadratic forms if $q(x) = x^2$ is not Euclidean?

If the answer happens to be "no", I would of course like to know more: can this happen with an anisotropic form $q$?  What can one say about a ring $R$ which admits a Euclidean quadratic form?
I have some suspicions that a domain which admits a Euclidean quadratic form is at least a PID.  Indeed, let $q$ be a quadratic form over $R$.  By an $R$-linear change of variables we may write it as $a_1 x_1^2 + \sum_{j=2}^n a_{1j} x_1 x_j + \sum_{i=2}^n \sum_{j=1}^n a_{ij} x_i x_j$ with $a_1 \neq 0$.  Then if I take my vector $x$ to be $(x_1,0,\ldots,0)$ for $x_1 \in K \setminus R$, then the Euclidean condition implies the following: there exist $y_1,z \in R$ such that $|a(x_1-y_1)^2 - z| < 1$.  This is reminiscent of the Dedekind-Hasse property for a norm which is known to imply that $R$ is a PID.  In fact, it is much stronger in that the $a \in R$ is fixed (and the $z$ is not arbitrary either).  Unfortunately in place of an arbitrary element $x$ of $K$ we have a certain square, so this does not match up with the Dedekind-Hasse criterion...but it seems to me somewhat unlikely that a domain which is not a PID would satisfy it.
For more information on Euclidean forms, please feel free to consult
http://www.math.uga.edu/~pete/ADCforms.pdf 

Added: a new draft which takes into account comments of J. Hanke and F. Lemmermeyer is available at
http://www.math.uga.edu/~pete/ADCformsv2.pdf
I warn that the new Section 2.1 showing that (primitive) binary Euclidean forms correspond to Euclidean ideal classes in quadratic orders in the sense of Lenstra -- as pointed out by Franz Lemmermeyer in his answer -- is rather miserably written at the moment, but at least it's there.

REPLY [9 votes]: Following Pete's request, I give the following as a second answer. 
Take $R = {\mathbb Z}[\sqrt{34}]$ and $q(x,y) = x^2 - (3+\sqrt{34})xy+2y^2$; observe that the discriminant of $q$ is the fundamental unit $\varepsilon = 35 + 6 \sqrt{34}$ of $R$, and that its square root generates $L = K(\sqrt{2})$ since $2\varepsilon = (6+\sqrt{34})^2$. Then q is Euclidean over $R$ since the ring of integers in $L = K(\sqrt{2})$ is generated over $R$ by the roots of $q$, and since $L$ is Euclidean by results of J.-P. Cerri (see Simachew's 
 A Survey On Euclidean Number Fields). But $R$ is not principal ($L/K$ is an unramified quadratic extensions), so the answer to your question, if I am right, is negative.<|endoftext|>
TITLE: Effective contraction of a loop. Reference or a simple proof?
QUESTION [12 upvotes]: Let $M$ be a compact simply connected R. manifold. Let $x$ be a base point and let $\gamma$ be a smooth loop in $M$ starting and ending at $x$. 

Is there a base point preserving retraction of $\gamma$ to $x$  such that every point on $\gamma$ travels a distance at most, say, $2diam(M)$?

I think the answer is yes. The scheme I have in mind is the following
Consider the cut locus $CL(x)$ of $x$. By perturbing the metric and the curve we can assume that $CL(x)$ is triangulable and $\gamma$ intersects it at finitely many points. Then slide these points off the $CL(x)$. Then the new curve does not intersect $CL(x)$ and can be contracted to $x$ following the geodesics.
This is probably too complicated...

Is there a simple proof? A reference?

Remark: 1) I would be happy with any estimate (not necessarily $2diam(M)$) which is independent of $\gamma$.
2) If there is a good curve shortening procedure that is not base-point-preserving please share it as well.

REPLY [11 votes]: Yes there is a bound independent of $\gamma$. Fix a fine triangulation of $M$ (say, with simplices 10 times smaller than the injectivity radius of the metric). For each vertex $q$ of the triangulation, fix a shortest path $s_q$ connecting $q$ to the marked point $p$. It is easy to deform any loop $\gamma$, via a short homotopy, into a path in the 1-skeleton of the triangulation. Then, for every vertex occuring on this path, "pull" a small piece of the loop to $p$ along the corresponding $s_q$. Now the loop is a product of "elementary" loops of the form $s_q^{-1}[qq']s_{q'}$ where $[qq']$ is an edge of the triangulation. Contract these "elementary" loops one-by-one. Since there are only finitely many different elementary loops, and the same elementary loops work for all $\gamma$'s, we have a uniform upper bound (depending on $M$) for the length of the homotopy.
On the other hand, there is no bound for the homotopy length in terms of the diameter. In addition to the great Bill Thurston's answer, let me mention another construction. For every $\varepsilon>0$ there is a metric on $S^3$ with sectional curvature $\le 1$ and diameter $\le\varepsilon$ (it was first constructed in Gromov "Almost flat manifolds, see also Buser and Gromoll "On the almost negatively curved $3$-sphere"). In this metric, take a geodesic starting from $p$ of length $2\varepsilon$ and connect its endpoint back to $p$ by a shortest path. Since the curvature is bounded above by 1, any contraction of this curve will have a trajectory longer than $\pi$ (I think this fact is called Klingenberg's lemma but I am not sure).<|endoftext|>
TITLE: What is the naming reason of poles in complex analysis?
QUESTION [26 upvotes]: A function $f: \textbf{C} \to \textbf{C}$ has a pole of order $k$ if $f(z) = \frac{g(z)}{(z-z_0)^{k}}$ where $g(z)$ is a nonzero analytic function. Why do we call it poles?

REPLY [20 votes]: The answer suggested by @muad’s comment is supported by both P. Ullrich (1989, p. 163):

The expressions “pole” and “polar singularity” were apparently first used in the 1865 book “Vorlesungen über Riemann’s Theorie der Abel’schen Integrale”  of Carl Neumann (1832–1925): they are precisely the points that the function maps to the “pole” of the Riemann sphere. Neumann’s book, by the way, is also the first to develop the theory of the Riemann sphere, which Riemann himself had only presented in his lectures. (See Neuenschwander (1978) (...).)

(my emphasis) and A. Markushevich (1996, p. 191):

The classification of the singularities of an algebraic function as poles and critical points was carried out, for example, in the Théorie des fonctions elliptiques of Briot and Bouquet (1875). However, the term “pole,” as pointed out by E. Neuenschwander (1978) was first used in this sense by K. Neumann in his Vorlesungen über Riemann’s Theorie der Abelschen Integrale (1865) in connection with the fact that the point at infinity was depicted as the pole of the sphere in this book.

The quoted article of Neuenschwander notes that the histories by Brill-Noether (1894, p. 170) and Osgood in the Encyklopädie (1901, p. 18) both miscredit Briot and Bouquet. So does (as of now) the earliest uses site — quoting an edition (1859) where the word pôle does not even appear.
I have wondered if Neumann was perhaps inspired by prior uses of “pole” for homographies or inversions,
$$
f(z)=\frac{az+b}{cz+d}\qquad\text{or}\qquad f(z) = a+\frac k{\bar z-\bar a},
$$
(presumably then traceable to Stubbs, Chasles, Servois,...) but couldn't find solid evidence.<|endoftext|>
TITLE: Is there an elementary way to find the integer solutions to $x^2-y^3=1$?
QUESTION [26 upvotes]: I gave this problem to my undergraduate assistant, as I saw that Euler had originally solved it (although I am having trouble finding his proof).  After working on it for two weeks, we boiled the hard cases down to showing that (1) in $\mathbb{Z}[w]$ the fundamental unit is $1+w+w^2$ (where $w$ is the real cube-root of 2) [which I'm sorry to say, I'm not certain I know off the top of my head how to prove], and (2) using that fundamental unit, I found a crazy ad hoc computation to show that there are only the obvious solutions.
So I'm wondering if someone else out there is more clever, knows where I can find Euler's proof, or if there is another nice elementary proof in the literature.

REPLY [17 votes]: Only the second proof below is new. In 1., I explain where the units in the cubic field
are coming from, and  3. is a variation of Paul Monsky's proof.

We try working in ${\mathbb Z}$ for as long as possible.
Write $x^3 = y^2 - 1 = (y-1)(y+1)$; since the gcd of the 
factors on the right hand side divides $2$, there are
two possibilities:
a) $y$ is even; then $y+1 = \pm a^3$ and $y - 1 = \pm b^3$.
  Subtracting these equations gives 
  $2 = a^3 - b^3 = (a-b)(a^2+ab+b^2)$, hence $a-b$ divides
  $2$. Going through all cases gives $(x,y) = (-1,0)$.
b) $y$ is odd; then $y+1 = 2a^3$ and $y - 1 = 4b^3$. 
  Subtracting these equations gives $1 = a^3 - 2b^3$.
It remains to solve the last equation. One possibility is showing
that the unit $1 - \sqrt[3]{2}$ is fundamental in the cubic number
field ${\mathbb Q}(\sqrt[3]{2})$, and that its only powers
of the form $a+b\sqrt[3]{2}$ have exponent $0$ or $1$.
The other possibility is observing that this equation is a special
case of the twisted Fermat cubic $x^3 + y^3 = 2z^3$. But Paul
Monsky's proof shows directly that solving this equation is
all that is needed.
A more or less standard proof (this is basically a classical 
$2$-descent, but perhaps more direct than Euler's approach,
which avoids number fields altogether - it is well known that
on curves with a rational point of order $2$, a simple 2-decent
can be performed by working within the rationals)
proceeds as follows: write the equation in the form
$$ y^2 = (x+1)(x+\rho)(x+\rho^2), $$
where $\rho$ is a primitive cube root of unity. The gcd
of two factors divides $1-\rho$, hence there are two possibilities:
a) $x+1 = \pm a^2$, $x + \rho = (-\rho)^e (a+b\rho)^2$, 
  $x + \rho^2 =  (-\rho^2)^e (a+b\rho^2)^2$. Since
  $\rho = (1+\rho)^2$  is a square, we can subsume the powers
  of $\rho$ into the square and find
  $x + \rho = \pm (a+b\rho)^2$ and $x + \rho^2 = \pm (a+b\rho^2)^2$. 
  Subtracting these equations and dividing through by $\rho - \rho^2$
  gives $1 = \pm b(2a-b)$, leading to  $(x,y) = (0,\pm 1)$ and $(-1,0)$.
b) Here we find
   $$  x+1  = \ \pm 3a^2, \quad
        x + \rho    = \ \pm (1-\rho) (a+b\rho)^2, \quad
        x + \rho^2  = \ \pm (1-\rho^2) (a+b\rho^2)^2. $$
 Adding the last two equations gives $2x-1 = \pm 3(a^2 - b^2)$. 
 Eliminating $x$ from this and the first equation 
 (where we actually have $x = 3a^2$ since $x \ge -1$) yields 
 $(x,y) = (2,\pm 3)$.
Let me give here my rendition of Paul's beautiful proof:
Let $\alpha, \beta, \gamma \in {\mathbb Z}[\rho] \setminus \{0\}$. If
$\alpha + \beta + \gamma = 0$ and $\alpha\beta\gamma = 2\mu^3$, then
after a suitable permutation of the three numbers we have 
$\alpha = 0$ or $\beta = \gamma$.
Proof. Let $(\alpha,\beta,\gamma)$ be a counterexample. Then
$\alpha$, $\beta$ and $\gamma$ are pairwise coprime in ${\mathbb Z}[\rho]$, 
and after a suitable permutation we have 
$$ \alpha = 2 \rho^a A_1^3, \quad 
    \beta =   \rho^b B_1^3,  \quad
   \gamma =   \rho^c C_1^3. $$
Among all such counterexamples we now take one in which $N\alpha$
is minimal. Dividing all three numbers by $\rho^a$ we may assume
that $a = 0$.

Cubes in ${\mathbb Z}[\rho]$ are $\equiv 0, 1 \bmod 2$ by Fermat's Little Theorem. 
Thus $0 \equiv \alpha \equiv \beta + \gamma \equiv \rho^b + \rho^c \bmod 2$, which implies $b = c$. Since $\alpha\beta\gamma$ is a cube, we must have $a = b = c = 0$.
Thus 
$$ \alpha = 2 A_1^3, \quad \beta = B_1^3,  \quad \gamma = C_1^3. $$
Since $B_1^3 \equiv C_1^3 \equiv 1 \bmod 2$, we may assume that 
$B_1 \equiv C_1 \equiv 1 \bmod 2$ (after multiplying these numbers
through by a suitable power of $\rho$).
Now set $\alpha_1 = B_1 + C_1$, $\beta_1 = \rho B_1 + \rho^2 C_1$ 
and $\gamma_1 = \rho^2 B_1 + \rho C_1$. Then 

$\alpha_1 + \beta_1 + \gamma_1 = B_1(1+\rho+\rho^2) + C_1(1+\rho+\rho^2) = 0$.
$\alpha_1 \beta_1 \gamma_1  = B_1^3 + C_1^3 = \beta + \gamma 
             = -\alpha = 2(-A_1)^3$.
$\beta_1 + \gamma_1 = (B_1 + C_1)(\rho+\rho^2) = - (B_1 + C_1) \ne 0$
  since $\beta + \gamma = -\alpha \ne 0$.
$N(\alpha_1 \beta_1 \gamma_1) = N\alpha \mid N(\alpha\beta\gamma)$; 
if we had equality, it would follow that $N(\beta) = N(\gamma) = 1$,
hence $\beta, \gamma = \pm 1$. But then 
$\beta = 1$, $\gamma = -1$ and $\alpha = 0$ against our assumptions.

Now descent finishes the proof: $(\alpha_1, \beta_1, \gamma_1)$ is another
solution with $N(\alpha_1 \beta_1 \gamma_1) < N(\alpha\beta\gamma)$, 
As a final remark I would like to point out that the article by Cohn in Gjergji Zaimi's answer uses "well known results" such as the solution of $x^4 - 2y^2 = \pm 1$ and $x^4 - 3y^2 = 1$. I do not know offhand how elementary the corresponding proofs are.<|endoftext|>
TITLE: Publishing with Undergraduates
QUESTION [6 upvotes]: Is doing research with a student considered to be good for a dossier? Is it okay to have few research publications but a lot of student projects? I am finishing up a grad program and am looking at tenure track jobs at both big and small schools.

REPLY [4 votes]: One should also be aware (or beware!) that what a college's webpage or promotional materials advertise may be different than what a promotion and tenure committee values which may in turn be different from what an individual department values. Within a department faculty members may also have widely differing views on the value of undergraduate research (as UG research -- most people want good papers no matter who the authors are.) At some schools, like my own LA college, the research of a tenure candidate is evaluated mainly through letters written by people in related research areas. These people are likely to evaluate you solely on the basis of the (perceived) quality of your publications. Thus, if one has to make the decision between fewer or weaker publications with undergrads and more or stronger publications without undergrads, you really need to take institutional structure and personalities into account. It may be impossible to do this without reading the faculty handbook and without knowing the individuals in the department. I suspect most people find themselves constrained and guided by circumstances and don't actually choose one or the other.<|endoftext|>
TITLE: koszul duality and algebras over operads
QUESTION [12 upvotes]: Given a pair of Koszul dual algebras, say $S^*(V)$ and $\bigwedge^*(V^*)$ for some vector space $V$, one obtains a triangulated equivalence between their bounded derived categories of finitely-generated graded modules.  
Given a pair of Koszul dual operads, say the Lie and commutative operads, what is the precise analogue of a derived equivalence between their categories of algebras?

REPLY [12 votes]: The situation for graded modules over a pair of Koszul dual algebras is more complicated, actually.  What the question says is true for Koszul algebras $A$ and $A^!$ provided that $A$ is Noetherian and $A^!$ is finite-dimensional (including the case of the symmetric and exterior algebras) but not otherwise.  In general one can say that the unbounded derived categories of positively graded modules with finite-dimensional components over $A$ and $A^!$ are anti-equivalent.  The subcategories of complexes of positively graded modules bounded separately in every grading in these unbounded derived categories are also anti-equivalent.
One can replace the contravariant anti-equivalence with a covariant equivalence by considering positively graded modules over one of the algebras and negatively graded modules over the other one (both algebras being considered as positively graded).  In this case one does not have to require the components of the modules to be finite-dimensional.
With algebras over operads, the analogue of the equivalence for graded modules involves DG-algebras with an additional positive grading (there being only the ground field $k$ in the additional grading $0$ and nothing in the negative additional grading), with the additional grading preserved by the differential.  The Koszul duality is an anti-equivalence between the localizations of the categories of DG-algebras of this kind, with every component of fixed additional grading being a bounded complex of finite-dimensional vector spaces, by quasi-isomorphisms.  For some operads (e.g., for Lie and Com) one has to assume the field $k$ to have characteristic $0$, while for some others (e.g., Ass) one doesn't.
If one wishes to replace the contravariant anti-equivalence with a covariant equivalence in the case of algebras over operads, one has to consider algebras on one side of the equivalence and coalgebras on the other side.  Then the boundedness and finite-dimensionality requirements can be dropped.
What I've described above is the homogeneous Koszul duality; the nonhomogeneous case (with ungraded modules or algebras without the additional grading) is more complicated, though also possible.  See my answer to the question linked to from the question above.
References: 1. Beilinson, Ginzburg, Soergel "Koszul duality patterns in representation theory", 2. My preprint "Two kinds of derived categories, ...", arXiv:0905.2621, Appendix A.<|endoftext|>
TITLE: Missing document request
QUESTION [21 upvotes]: I received a request for another long-lost document:

I am wondering if there is any way I
  might obtain a copy of
The geometry of circles: Voronoi
  diagrams, Moebius transformations,
  convex hulls, Fortune's algorithm, the
  cut locus, and parametrization of
  shapes W.P. Thurston Technical Report,
  Princeton University, 1986. 
Is there a scanned version somewhere
  or might some library stock it?

Can anyone help out?
For context: I gave a series of presentations in a EECS course with Dave Dobkin, and I wrote these notes to go with it.  It was fun material at the time, but the notes were only moderately distributed.

REPLY [16 votes]: Here is the .tex not quite "as-is," but modified minimally so that it will compile: DTnotes.tex.
And here is .pdf produced by compiling that .tex:  DTnotes.pdf.<|endoftext|>
TITLE: Studying non-linear PDEs with manifolds
QUESTION [6 upvotes]: I'm sorry if this is an inappropriate forum to ask this question on, for I fear it is pretty undergraduate-level one :) I was contemplating on the study of non-linear PDEs. Is it possible to reduce a non-linear PDE on $\mathbb{R}^n$ to a distribution or a 'good' PDE on a smooth manifold? It seems to me like a natural step, but I don't know anything about it yet :(

REPLY [2 votes]: The question is vague (Denis is right) but it does make some sense. When looking at explicit solutions of nonlinear PDEs one frequently encounters singularities that remind of projections from some higher dimensional manifold (e.g. in Burger's equation). But I think the OP does not have a clear picture. Locally, working 'on' a manifold instead than on $R^n$ is essentially like changing coordinates in the equation. So usually you do not resolve nonlinearities, you just modify the coefficients of the equation. 
On the other hand, if you consider equations with values into a manifold, then sometimes a very difficult nonlinear term reveals to be a very simple geometric object, 'linear' in some sense. This does not happen by chance; your equation must have some geometric meaning. Best examples are harmonic maps (elliptic) and wave maps (hyperbolic). 
Some years ago, Serge Alinhac tried to develop a whole theory of 'geometric blow up': singularities in the solution which can be resolved by choosing an appropriate coordinate sistem in the target space. But I guess the results where not too satisfying since he abandoned the effort.<|endoftext|>
TITLE: Relations in symmetric group
QUESTION [6 upvotes]: It would be nice to find out what is known about the following problem.
First let us consider a free group $F$ with two generators $a$ and $b$. We are interested in its elements that are

not equal to identity,
of form $c_1 c_2 \ldots c_n d_1^{-1} d_2^{-1} \ldots d_n^{-1}$, where all $c_i$ and $d_i$ are either equal to $a$ or to $b$.

Let us denote all these elements by $W$.
Let us consider the group $S_k$ now. What is the shortest word from $W$ with the following property: whatever elements of $S_k$ we substitute for $a$ and $b$ we get identity (in $S_k$)?
The best bounds for the smallest $n$ I am aware of are $2^{O(k)}$ and $\Omega(k^2)$.

REPLY [7 votes]: You are asking for the shortest balanced semigroup identity in $S_k$. Some info can be found here: Pöschel, R.; Sapir, M. V.; Sauer, N. W.; Stone, M. G.; Volkov, M. V. Identities in full transformation semigroups.  Algebra Universalis  31  (1994),  no. 4, 580--588. But the bound there is exponential. I believe the lower bound should be exponential too, but I do not think there were any more recent papers on the subject. You may also try to read this paper. It is also relevant: Cherubini, Alessandra; Kisielewicz, Andrzej; Piochi, Brunetto, On the length of shortest 2-collapsing words. Discrete Math. Theor. Comput. Sci. 11 (2009), no. 1, 33--44.<|endoftext|>
TITLE: Timeline of cohomology (1935 to 1938)
QUESTION [35 upvotes]: There was a recent question on intuitions about sheaf cohomology, and I answered in part by suggesting the "genetic" approach (how did cohomology in general arise?). For historical material specific to sheaf cohomology, what Houzel writes in the Kashiwara-Schapira book Sheaves on Manifolds for sheaf theory 1945-1958 should be adequate.
The question really is about the earlier period 1935-1938. According to nLab, cohomology with local coefficients was proposed by Reidemeister in 1938 (http://ncatlab.org/nlab/show/history+of+cohomology+with+local+coefficients). The other bookend comes from Massey's article in History of Topology edited by Ioan James, suggesting that from 1895 and the inception of homology, it took four decades for "dual homology groups" to get onto the serious agenda of topologists. It happens that 1935 was also the date of a big international topology conference in Stalin's Moscow, organised by Alexandrov. This might be taken as the moment at which cohomology was "up in the air".
Now de Rham's theorem is definitely somewhat earlier. Duality on manifolds is quite a bit earlier in a homology formulation. 
It is apparently the case that At the Moscow conference of 1935 both Kolmogorov and Alexander announced the definition of cohomology, which they had discovered independently of one another. This is from http://www.math.purdue.edu/~gottlieb/Bibliography/53.pdf at p. 11, which then mentions the roles of Čech and Whitney in the next couple of years. This is fine as a narrative, as far as it goes. I have a few questions, though:
1) Is the axiomatic idea of cocycle as late as Eilenberg in the early 1940s?
2) What was the role of obstruction theory, which produces explicit cocycles?
Further, Weil has his own story. Present at the Moscow conference and in the USSR for a month or so after, his interest in cohomology was directed towards the integration of de Rham's approach into the theory. He comments in the notes to his works that he pretty much rebuffed Eilenberg's ideas. Bourbaki was going to write on "combinatorial topology" but the idea stalled (I suppose this is related). So I'd also like to understand better the following:
3) Should we be accepting the topologists' history of cohomology, if it means restricting attention to the "algebraic" theory, or should there be more differential topology as well as sheaf theory in the picture?
As said, restriction to a short period looks like a good idea to get some better grip on this chunk of history.

REPLY [3 votes]: As explained to us by Alan Mayer, sheaf cohomology is a generalization of cech cohmology.  I found this very helpful.
As to the question of how ordinary cohomology arose, Hermann Weyl implies in the revised version of his book Concept of a Riemann surface, that it is a generalization of the Weierstrass, Hensel,and Landsberg approach to Riemann surfaces, focusing first on the behavior of integrals, and passing from that to deductions about the paths of integration.
Bott also used to say that a cocyle was "something that hovers over a space and when it sees a cycle, pounces on it and spits out a number".  Such a thing he then observed was provided by an integral, and went on to introduce de Rham cohomology as the most natural type. So he too seemed to suggest that the fundamental example giving rise to cohomology was classical integration over cycles.<|endoftext|>
TITLE: Universe view vs. Multiverse view of Set Theory
QUESTION [18 upvotes]: Here I refer to Hamkins' slides:
http://lumiere.ens.fr/~dbonnay/files/talks/hamkins.pdf
particularly, to the "Universe view simulated inside Multiverse", p. 22.
My question is: is it very unsound to ask if the Multiverse view could be simulated (in a similar sense) inside Universe? 
If it is, why is it? If it is not, why should one prefer one view to the other?

REPLY [20 votes]: Update.(Sep 6, 2011) My paper on the multiverse is now available at the math arxiv at The set-theoretic multiverse, and gives a fuller account of the ideas in the slides mentioned in the question. The particular issue of the question arises in the discussion of the toy-model approach to formalization, discussed on page 23, and also at greater length in my paper Set-theoretic geology, G. Fuchs, J.D. Hamkins, J. Reitz.  

Thanks for the question; I'm glad you're interested in it.
The multiverse view in set theory is a philosophical position offered in contrast to the Universe view, an orthodox position, which asserts that there is a unique background set-theoretic context or universe in which all our mathematical activity takes place. On the Universe view, there are definitive final answers to the question of whether a given mathematical statement, such as the Continuum Hypothesis, is true or not, and we seek to find these answers. On the Universe view, the fact that such a statement is independent of ZFC or another weak theory is regarded as a distraction from the question of determining whether or not it is ultimately true. For example, many set theorists regard the accumulating regularity consequences of large cardinals for properties of sets of reals as indicating that the large cardinal hierarchy is on the right track towards the final set-theoretic truth. 
A paradox for the universe view, which I mention in the slides to which you link, is that the most powerful  set-theoretic tools that have informed a half-century of research in set theory are most naturally understood as methods of constructing alternative set-theoretic universes. That is, from a given set-theoretic universe we can construct others, by means of forcing, ultrapowers, inner models, definability, large cardinal embeddings and so on. Indeed, we can often construct models of set theory to exhibit exact precise properties, and forcing especially has led to a staggering diversity of models.
The multiverse view takes these diverse models seriously, holding that there are diverse incompatible concepts of set, each giving rise to a set-theoretic universe in which they are instantiated. The set-theoretic tools provide a means of modifying any given concept of set to a closely related concept of set, whose resulting universes can be fruitfully compared in a single mathematical context. For example, we can understand the relationship between a ground model concept of set and that of its forcing extensions. Although the multiverse includes all the familiar models of set theory that we have built by forcing and other methods, it likely also includes universes arising from other set concepts that we have not yet imagined. There seems to be little reason that any two given concepts of set can be compared together in one set-theoretic context.
Now, the Universe view seems simulable inside the multiverse view by the idea of picking a single universe $V$, call it the actual universe if you like, and then restricting attention only to the universes that are somehow describable from the perspective of $V$. 
But you ask about the other direction. There are a few ways to do it in a partial manner, but none of them seems fully satisfactory. 

First, one can mathematise the concept of multiverse, by just considering a multiverse as a collection of (set) models of set theory. For example, with Victoria Gitman (p. 44 of slides), we showed that if ZFC is consistent, then the set of all countable computably-saturated models of ZFC forms a multiverse satisfying all the multiverse axioms that I mention there (and others). This is just a straight theorem of ZFC. Since this multiverse does not include the set-theoretic background universe $V$ in which the collection was formed, however, we can recognize that it is not the full multiverse in which we are interested, and this is the sense in which we wouldn't really want to limit ourselves to that multiverse. 
Second, if one is interested only the generic multiverse---the part of the multiverse reachable by forcing, that is, by closing under forcing extensions and grounds in a zig-zag pattern---then one can formalize the whole set-up within ZFC. By describing exactly which forcing notions were used, one can index the models by their methods of construction. By this means, the concept of $\varphi$ is true throughout the generic multiverse of $V$'' is first-order expressible in $V$ in the language of set theory. This kind of analysis is a full answer to your question if you care only about the generic multiverse as opposed to the full multiverse. 

But other researchers have cared about the full multiverse, and in general there is no satisfactory way to simulate it from the universe perspective. For example, the Inner Model Hypothesis of Sy Friedman (see also Friedman, Welch, Woodin), is described as the assertion that if the universe $V$ has an outer model with an inner model having a certain property, then there is already an inner model of $V$ with the property. Such a statement is explicitly appealing to the multiverse concept, but one cannot seem easily to formalize it within set theory. Instead, the official account of IMH backs off to the case where $V$ is a countable model of set theory, which takes it somewhat away from its initial multiverse sense. 
So to finally come to an answer to your question, it doesn't seem possible to fully simulate the multiverse perspective from within the universe view, in a way that is fully satisfactory.
You ask, as a follow up, in this case why should we prefer one view to the other?
Well, these are philosophical positions on the fundamental nature of mathematical existence; this is a philosophical dispute rather than a mathematical one. Nevertheless, one's mathematical philosophy often suggests certain mathematical problems as interesting or solution methods as promising. The multiverse perspective naturally leads one to compare set-theoretic universes, and this led to the research on the modal logic of forcing and set-theoretic geology (mentioned in the slides). The Universe perpsective may lead elsewhere, perhaps towards an investigation of universes with highly structural features. 
Surely the mathematicians' measure of a mathematical philosophy is the value of the mathematics to which it leads...<|endoftext|>
TITLE: Is there a field which is the union of finitely many proper subfields?
QUESTION [6 upvotes]: Is there a field which is the union of finitely many proper subfields?

REPLY [4 votes]: See A Bialynicki-Birula, J Browkin, A Schinzel, On the representation of fields as finite unions of subfields, Colloq Math 7 (1959) 31-32, MR 22 #2601. At the risk of violating copyright, here's the review, by S W Golomb:
This article contains a short proof of the fact that no field can be represented as a finite union of proper subfields. A counterexample is exhibited to the corresponding assertion for integral domains.
EDIT: The paper may be available at http://matwbn.icm.edu.pl/ksiazki/cm/cm7/cm717.pdf<|endoftext|>
TITLE: Is there a reasonable way to refer to a 23 page article with 28 authors?
QUESTION [24 upvotes]: Most of us have spent time compiling reference lists for papers or longer documents, a task which used to be even more time-consuming before the Internet and TeX came along (all lists had to be typed and sometimes retyped).    With increased international communication as well as pressure by funding agencies to do collaborative work, more multi-author papers are apparently being written now.  For instance, recent VIGRE-supported algebra groups at the University of Georgia have been publishing papers with many authors.    This morning's automatic mailing from arXiv (in subject areas of special interest to me, mostly close to math.RT) brought a prize-winner: 1009.4134.  Are we looking at the future?
It's the result of an AIM conference, perhaps intended for formal publication but challenging in any event to those who might want to refer to it.    Page 23 of the paper itself consists mostly of an author listing.
Since the list of 28 authors goes from A to Z (Aguiar to Zabrocki), it would seem invidious to refer only to Aguiar et al.  Of course, if electronic-only publishing ever becomes the universal rule in mathematics, placing a link like the one I just posted in a numbered reference list might be enough.   (Provided the link is durable.)   

Is there a reasonable way to refer to a 23 page article with 28 authors?

P.S. I'm not planning to cite this particular paper, but am in the process of assembling a reference list for other purposes and might also need to cite Georgia VIGRE group papers at some point.   It's usually impossible in an alphabetical list of authors to identify the "leaders" or the people contributing the main ideas.   Theoretical progress does require ideas, whereas experimental work often depends more heavily on organization, teamwork, and of course funding.   (As an aside, if the current list of finite simple groups and the reasoning behind it are eventually accepted by all well-informed observers as correct, who will be cited for that theorem?)

REPLY [2 votes]: The question reminded me about one particular Ig Nobel Prize in Literature (1992):

Yuri Struchkov, unstoppable author from the Institute of Organoelement Compounds in Moscow, for the 948 scientific papers he published between the years 1981 and 1990, averaging more than one every 3.9 days.

This has been given to a physicist(!) and I wonder how many scientists coauthored the masterpieces. I also wonder whether the groups at the University of Georgia can be nominated in the nearest future...
Added. People outside mathematics would be hardly surprised by the 28/23 article. By mistake I came accross arXiv:1008.1753 which has
62($\pm$1) authors (there is even no room for the last 3 in the list!) and "11 pages (including Appendices), 6 figures".<|endoftext|>
TITLE: Bisymmetric Matrix, solving set of linear equations.
QUESTION [5 upvotes]: A bisymmetric matrix is a square matrix that is symmetric about both of its main diagonals.  
If $A$ is a bisymmetric matrix and I'm interested in solving $Ax=b$. 
Are there techniques used to exploit this structure when solving the system of linear equations?
Note: I'm looking for techniques which exploit more than just the fact that the matrix is symmetric.

REPLY [6 votes]: The condition of symmetry about the antidiagonal says that $A$ commutes with reversal of coordinates.  Call this operation $R$, so $R^2 = 1$ and $AR = RA$.
$R$ has a $+1$ eigenspace and a $-1$ eigenspace.  
For any solution, you can project both $x$ and $b$ to the two eigenspaces, by averaging them with either their reversals or $-$ the reversals. You can get the induced action of $A$ on these (roughly if in odd dimension) half-size eigenspaces similarly.  The two halves of $A$ are still symmetric, so you're left with the easier problem of solving two symmetric systems of equations in half the number of variables.<|endoftext|>
TITLE: Which groups have nice compactifications ?
QUESTION [8 upvotes]: Given a discrete group G. Is there a nice criterion to decide, whether there is a compact Hausdorff $G$- space X, that contains the discrete space $G$ as a subspace, such that the stabilizer of every point in $X$ is (virtually) cyclic ? 
For example the free group admits such a compactification (As well as any hyperbolic group I think). Is it possible to decide, whether $\mathbb{Z}^2$ admits such a compactification? .

REPLY [11 votes]: There are many compactifications of particular groups.
For your example of $\mathbb Z^2$: one construction for a compactification is to first embed it as a subgroup of
$S^1 = \mathbb R / \mathbb Z$ by picking two rationally independent numbers for the images of the generators.
Now compactify $\mathbb Z^2$ by making large elements connverge toward their image points in $S^1$.
The stabilizer of any point is trivial.
The same method works to get a compactification associated with any action of $G$ on a compact space $X$. Just pick a point $x \in X$, and adjoin the closure of the orbit of $X$ at infinity in $G$.  If the action has no fixed points in the cloure of the orbit, then stabilizers are trivial. It's easier to avoid all but cyclic stabilizers.  To make actions with small stabilizers, you can take products of examples; point stabilizes in the product become intersections of stabilizers in the factors.  There are many tricks, some of them useful, for making compactifications that are Hausdorff metric spaces.
There's an ultimate (but non-constructive and of large cardinality) compactification, the Stone-Cech compactification, which has trivial point stabilizers for any group,

REPLY [2 votes]: Regarding the question of `whether the CAT(0) boundary works, if the space doesn't contain $\mathbb{R}^2$ as a subspace' (see comments above), the Flat Plane Theorem asserts that any CAT(0) group that acts on a CAT(0) space without an isometrically embedded copy of $\mathbb{R}^2$ is word-hyperbolic.  So in this case you can use the usual hyperbolic boundary.  See Bridson & Haefliger for details.<|endoftext|>
TITLE: Is there a general setting for self-reference?
QUESTION [32 upvotes]: This is a question about self-reference: Has anyone established an abstract framework, maybe a certain kind of formal language with some extra structure, which makes it possible to define what is a self-referential statement?

REPLY [2 votes]: How relevant this is to "an abstract framework, maybe a certain kind of formal language with some extra structure" in the OP, I am not quite sure, and I am only beginning to read Halbach's works, yet I think they are general enough to properly belong to this thread, and they weren't yet mentioned. Very very roughly, what I gather from Halbach's work on this topic so far is that (my interpretation)

the 'say' in the old idea of formulae which 'say' about themselves they are not provable is not an idea which is synthetic a priori (like e.g. the idea of necessity), rather is an idea consisting of many sub-ideas. (For example, by filtering according to where the relevant formulae are placed in the arithmetical hierarchy.)

Philosophically, the approach to self-reference I am pointing to here could be called an 'analytic' approach to self-reference, as opposed to the 'synthetic' approach of Lawvere. (Both 'analytic' and 'synthetic' to be taken in their neutral technical sense, i.e. breaking a concept into parts in the former, combining the concept with other concepts in the latter.)
Some references on Halbach's work:

so new that it hasn't yet appeared: Volker Halbach has announced a book (with Graham Leigh) whose draft has the title Syntax and Circularity: A Study in Self-Reference and Paradox.

so new that it lies after the answers given so far:



Volker Halbach, Albert Visser: Self-reference in arithmetic I. Vol. 7(4), 2014 , pp. 671-691
Volker Halbach, Albert Visser: Self-reference in arithmetic II. Vol. 7(4), 2014 , pp. 692-712


a relevant lecture of Halbach's with a very general title, containing explanations on the work of Halbach and Visser:


V. Halbach: Self-reference. Talk at the Workshop in Mathematical Philosophy. Ludwig-Maximilians-Universität München. September 13, 2011<|endoftext|>
TITLE: Do Burnside Group Factors have Gamma?
QUESTION [12 upvotes]: The Free Burnside group $G=B(2,665)=\langle a,b|g^{665} \rangle$ is infinite, by the work of Adyan and Novikov. Furthermore, the centralizer of any nonidentity element in $G$ is finite cyclic, and so the group is an i.c.c. group and the associated left group von Neumann algebra $LG$ is a type $II_{1}$ factor. It is a fact, due to of Adyan, that this group is not amenable, so the group von Neumann algebra is not injective. 
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset $\{ x_{1}, x_{2},..., x_{n} \} \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.) 
I should mention that if a group is not inner amenable in the sense described in 
Is there an i.c.c. nonamenable simple group that is inner amenable? 
then its left group von Neumann algebra does not have property $\Gamma$. (There exist i.c.c. inner amenable groups whose group von Neumann algebras don't have $\Gamma$, as recently shown by Stefaan Vaes: http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.1485v1.pdf.)
My question is:

Does the group von Neumann algebra $LG$
  have Property
  $\Gamma$?

REPLY [5 votes]: The answer is no, the group von Neumann algebra of the Burnside group $B(2,665)$ does not have property Gamma. In fact, $B(2,665)$ is not inner amenable. 
To see this, first note that, as pointed out in the question, $B(2,665)$ is nonamenable, and the centralizer of every non-identity element of $B(2,665)$ is finite. However, every nonamenable inner amenable group $G$ contains a non-identity element whose centralizer is nonamenable. This is a consequence of applying the following well-known folklore lemma to the conjugation action of $G$ on $G-1$: 
Lemma: if an action of a nonamenable group $G$ on a set $X$ admits an invariant finitely additive probability measure $\mu$, then there is a point $x\in X$ whose stabilizer subgroup $G_x$ is nonamenable.
There are a variety of proofs of this Lemma. Here is a standard averaging argument: Suppose that every $G_x$ is amenable. Let $X_0\subset X$ contain one point from each $G$-orbit and for $x\in X_0$ let $\nu _x$ be a finitely additive probability measure on $G$ which is invariant under left translation by $G_x$. Then $x\mapsto \nu _x$ extends to a well-defined assignment on all of $X$ by taking $\nu _{g\cdot x} = g\cdot \nu _x$ for $x\in X_0$, $g\in G$. The barycenter $\nu = \int _X \nu _x \, d\mu (x)$ is then invariant under left translation by $G$, so $G$ is amenable, contrary to the hypothesis of the lemma.<|endoftext|>
TITLE: Applications of infinite graph theory
QUESTION [33 upvotes]: Finite graph theory abounds with applications inside mathematics itself, in computer science, and engineering. Therefore, I find it naturally to do research in graph theory and I also clearly see the necessity.
Now I'm wondering about infinite graph theory. Quite a bit of research seems to be done on it as well and of course they are a natural generalization of a useful concept. But I never saw an example where we actually need them.
I understand that they come up as infinite Cayley graphs in group theory, that the automorphism groups of infinite but locally finite graphs are topological groups, that they play some role in general topology, etc. But to me it seems they are "just there" and are not essential in the sense that a theorem about them proves something about groups or topology what we couldn't have done easily without using them.
Polemically phrased my question is

Why should we care about infinite graphs?

REPLY [2 votes]: In addition to the aforementioned many applications outside graph theory, another reason for caring about infinite graphs is that they are helpful for proving results about finite graphs. The papers below give some examples.
László Babai Vertex-transitive graphs and vertex-transitive maps
https://onlinelibrary.wiley.com/doi/10.1002/jgt.3190150605
Justin Salez Sparse expanders have negative curvature
https://arxiv.org/abs/2101.08242
Martin Grohe, Pascal Schweitzer, Daniel Wiebking,
Automorphism groups of graphs of bounded Hadwiger number
https://arxiv.org/abs/2012.14300
A short proof of Fleischner’s theorem
https://www.sciencedirect.com/science/article/pii/S0012365X09003355
was found via another excursion into the infinite.
According to this wikipedia article
https://en.wikipedia.org/wiki/Halin%27s_grid_theorem,
Halin's grid theorem
"is a precursor to the work of Robertson and Seymour linking treewidth to large grid minors", which if true provides perhaps the best example of how infinite graphs help you in the study of finite ones.<|endoftext|>
TITLE: Beyond an intro to topological graph theory...
QUESTION [8 upvotes]: I'm looking to find out what active areas of research there are in topological graph theory, particularly those that interface strongly with other areas of math (say, group theory, algebraic topology, Gromov-Witten theory, etc). 
Assuming a background given by, say, Topological Graph Theory by Gross and Tucker (a standard reference):   

(1) How does topological graph theory fit into and interface with 21st century mathematics, and where is the field going? 
(2) I'm particularly interested in applications to algebraic topology and algebraic geometry - can anyone summarize these or give a feel for what the most active areas are?

REPLY [16 votes]: I wouldn't describe it as an "area" of math, exactly, but there are certainly mathematicians who study embeddings of graphs into surfaces and related objects.  Let me recommend two sources to learn more about what people are doing.  I am not familiar with the book you listed, so I don't know if it sufficient preparation for reading them; however, I am a big fan of jumping in to an unknown field and then learning background information "as you go".
The first is the long book
MR2036721 (2005b:14068) 
Lando, Sergei K.(RS-IUM-M); Zvonkin, Alexander K.(F-BORD-LB)
Graphs on surfaces and their applications. 
With an appendix by Don B. Zagier. Encyclopaedia of Mathematical Sciences, 141. Low-Dimensional Topology, II. Springer-Verlag, Berlin, 2004. xvi+455 pp. ISBN: 3-540-00203-0 
This might be the closest to what you are looking for in that it focuses on connections with algebraic geometry and number theory, especially Grothendieck's theory of dessins d'enfants and questions related to the moduli space of curves.
The second is a very different direction, namely the study of "spatial graphs"; ie embeddings of graphs into 3-space and other 3-manifolds.  This is basically a generalization of knot theory.  One possible way of getting into this subject is to start with the survey
MR2179645 (2006e:57009) 
Ramírez Alfonsín, J. L.(F-PARIS6-CM)
Knots and links in spatial graphs: a survey. (English summary) 
Discrete Math. 302 (2005), no. 1-3, 225--242.<|endoftext|>
TITLE: A random walk on random lines
QUESTION [16 upvotes]: I am wondering if this random walk remains finite with positive probability.
Start with three lines $A,B,C$ that are extensions of an equilateral triangle.
Let $p_0$ be one corner. Generate a line $L_1$ through $p_0$ at a random orientation.
Now, from $p_0$ walk either right or left with equal probability along $L_1$ until
you hit the first intersection point $p_1$ of $L_1$ with $\{A,B,C\}$, or reach $\infty$.
Clearly there is a $\frac{1}{2}$ chance you stay finite.
Through $p_1$ generate a randomly oriented line $L_2$, and walk right or left to the
first intersection point $p_2$ of $L_2$ with $\{A,B,C,L_1\}$.  And so on.
On the one hand, there are always routes to $\infty$.  On the other hand,
as the line arrangement thickens, the number of steps to escape grows.
So if the process survives a few iterations, it becomes less likely it will escape.
Intuition is likely useless here, but it feels to me like this should remain finite. 
Can anyone see an argument?  
If it does remain finite, then other questions
suggest themselves, but perhaps I should start with the basic infinite/finite question.
Here is a hand-executed example (corrected from the original by Pablo):

     

Update. Here is a first attempt at an implementation:

     

I just stopped the iteration after $p_8$.
Addendum. Bill Thurston's convincing analysis coupled
with independent simulations suggest it remains
finite, perhaps with probability about $0.16$. Interesting questions
remain: What is the distribution of the number
of steps before shooting off to $\infty$?  When a path remains finite forever, does its total
length remain bounded?

REPLY [4 votes]: I don't know, how to find the probability analytically, but numerical simulation is actually simple. It suggests, that the process remains finite with probability $0.162\pm 0.003$. Here there is a simulation in Wolfram Mathematica 7.0.1. Typical examples of processes with 20+ steps are shown below. 20 steps are drawn. Initial triangle is always in the center, but on some pictures it is too small to see it.
, ,<|endoftext|>
TITLE: Random walk inside a random walk inside...
QUESTION [34 upvotes]: Let $G=(V,E)$ be a graph and consider a random walk on it. Let $G'=(V',E')$ be a subgraph consisting of the vertices and edges that are visited by the random walk.

Question 0: Is there a standard name for $G'$?

Intuitively $G'$ is a thin subgraph, so for instance, even when $G$ is transient, $G'$ can be recurrent.

Question 1: Is there a counterexample? So, Is there a transient graph $G$ so that $G'$ is transient with positive probability?

I'm also curious to know what happens when one iterates this procedure, $G,G',G'',\dots$. Does it eventually look like a path graph?

Question 2: What can one say about $G^{(n)}$ as $n\to \infty$?

REPLY [41 votes]: Question 0: $G'$ is known as the trace of the random walk. 
Question 1: $G'$ is always recurrent with probability one. This is a result of Benjamini, Gurel-Gurevich, and Lyons from 2007.
Question 2: Since $G'$ is recurrent, with probability one we have $G^{(n)}=G'$ for all $n \geq 1$.<|endoftext|>
TITLE: Morava on Shafarevich conjecture
QUESTION [37 upvotes]: $\DeclareMathOperator\Q{\mathbf{Q}}$Jack Morava has some interesting ideas stemming from stable homotopy theory and geometric topology on the Shafarevich Conjecture.
The Shafarevich Conjecture states: $\operatorname{Gal}(\bar \Q \,/ \,\Q_{cycl})$ is free. That is, the Galois group of the algebraic closure of the rationals over the cyclotomic closure of the rationals is a free  group (Added: or rather, a free profinite group).
References for Morava's thoughts are

The motivic Thom isomorphism 2003.

Toward a fundamental groupoid for the stable homotopy category Link is to the arxiv, last updated 2009. There is a journal version from 2007.

To the left of the Sphere Spectrum, from a talk given at Haynes Miller's 60th birthday conference in Bonn in 2008.

A theory of base motives 2009. A follow-up to the previous paper according to the introduction.


This is exciting material, but I'm having trouble coming up with a way to summarize the gist and have some questions.

(1)What exactly is Morava's definition of a mixed Tate motive?
(2) What exactly is the connection Morava is advocating between number theory and geometric topology by invoking the appearance of the Riemann zeta function in Waldhausen's A-theory/pseudo-isotopy?
(3) Morava states that the map from the K-theory of the integers to that of the sphere spectrum, $K(\mathbb {Z}) \to K(\mathbb {S})$, is a rational equivalence as a (partial) explanation of (2). How exactly does this work??
(4) Where does Shafarevich fit in here?

Down-to-earth answers to these would be much appreciated!!

REPLY [13 votes]: (3) The statement is that the map of ring spectra $S \to HZ$ induces a rational equivalence $K(S) \to K(Z)$.  A reference is Proposition 2.2 in:

Waldhausen, Friedhelm: Algebraic
  $K$-theory of topological spaces. I.
  Algebraic and geometric topology
  (Proc. Sympos. Pure Math., Stanford
  Univ., Stanford, Calif., 1976), Part
  1, pp. 35--60, Proc. Sympos. Pure
  Math., XXXII, Amer. Math. Soc.,
  Providence, R.I., 1978.

The proof uses the plus-construction definition of algebraic $K$-theory.
The map $BGL(S) \to BGL(Z)$ is a $\pi_1$-isomorphism and a rational equivalence,
since $\pi_{n+1} BGL(S) = \pi_n GL(S)$ is the group of infinite matrices over
$\pi_n(S)$ for $n\ge1$, which is torsion.
Hence $BGL(S)^+ \to BGL(Z)^+$ is also a rational equivalence.<|endoftext|>
TITLE: What is the strategy for "all words valid" scrabble?
QUESTION [10 upvotes]: The rules for "all words valid" scrabble are exactly the same as ordinary scrabble, except that every single combination of letters is in the dictionary. To make the game deterministic, we will also assume that every letter is worth the same amount of points (though it may be interesting to remove this restriction somehow)--so essentially it is just a game of placing tiles. Is there a winning strategy for either player?

REPLY [5 votes]: 'Pass' is a valid move in Scrabble. This means that a game can last for ever, so you need some criterion for ending the game after each player has passed. It also means that the game is either a win for the first player, or a draw (because if the starting position is losing, the first player can pass).
If you don't allow passes, then I can't see a way of calculating who wins. It looks more complicated than Draughts, which succumbed to rigorous evaluation only after years of dedicated effort -- see http://www.sciencemag.org/cgi/content/abstract/1144079v1.
I have some experience in these matters -- a Scrabble program I co-wrote won a Computer Olympiad gold medal many years ago.<|endoftext|>
TITLE: Which journals publish 1-page papers
QUESTION [20 upvotes]: Having read a thread on a similar question on expository papers I'm reminded of reason #99 to drop my math PhD thingy, late c20th: I just couldn't blow up this paper to 4 pages. (OK, one half-page calculation was left to the expert reader (and other experts could guess), but...)

REPLY [3 votes]: I just stumbled across a very recent example (2017) at the Archiv der Mathematik which I felt compelled to add to this list.<|endoftext|>
TITLE: A gamma function identity
QUESTION [21 upvotes]: In some of my previous work on mean values of Dirichlet L-functions, I came upon the following identity for the Gamma function:
\begin{equation} 
\frac{\Gamma(a) \Gamma(1-a-b)}{\Gamma(1-b)} 
+ \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} 
+ \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = 
\pi^{\frac12} 
\frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)}
\frac{\Gamma\left(\frac{a}{2}\right)}{\Gamma\left(\frac{1-a}{2}\right)}
\frac{\Gamma\left(\frac{b}{2}\right)}{\Gamma\left(\frac{1-b}{2}\right)}.
\end{equation}
As is often the case, once one knows such a formula should be true then it is easy to prove it.  I give my proof below.  My questions are 1) Has this formula been observed before?  I have no idea how to search the literature for such a thing.  2) Is there a better proof? (Of course this is totally subjective, but one thing that would please me would be to avoid trigonometric functions since they do not appear in the formula.)
Proof.
Using
\begin{equation}
\frac{\Gamma(\frac{s}{2})}{\Gamma(\frac{1-s}{2})} = \pi^{-\frac12} 2^{1-s} \cos({\textstyle \frac{\pi s}{2}}) \Gamma(s),
\end{equation}
the right hand side is
\begin{equation}
2 \frac{\cos(\frac{\pi a}{2}) \cos(\frac{\pi b}{2}) \Gamma(a) \Gamma(b)}{\cos(\frac{\pi (a + b)}{2}) \Gamma(a+b)}.
\end{equation}
On the other hand, the left hand side is
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(
 \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(b) \Gamma(1-b)} 
 + \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(a) \Gamma(1-a)}
 + 1
\right),
\end{equation}
which becomes after using $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}$,
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(\frac{\sin(\pi a) + \sin( \pi b) + \sin(\pi(a + b))}{\sin(\pi(a+b))} \right).
\end{equation}
Using trig formulas, we get that this is
\begin{equation}
 2 \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} \frac{\sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a-b)) + \sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a+b)) }{\sin(\pi(a+b))}
\end{equation}
I think I've run out of space?  The rest is easy trig.

REPLY [22 votes]: A proof of the statement has already been given, so I will just add a small historical remark.
The left hand side of your identity is the Veneziano amplitude in the case of four identical scalar particles. The right hand side corresponds to another crossing symmetric four-point amplitude for scalar particles suggested by Virasoro in "Alternative Constructions of Crossing-Symmetric Amplitudes with Regge Behavior". These are known to coincide for the choice of parameters $a+b+c=1$ giving your identity. I would recommend the article "The birth of string theory" for a quick survey of the Veneziano model and its generalization to N scalar particles and related mathematics.<|endoftext|>
TITLE: Non-vanishing of zeta(s), Re(s)=1, without complex analysis?
QUESTION [10 upvotes]: Say you are allowed to use Fourier analysis, complex variables, Euler-Maclaurin, etc., but no complex analysis - no holomorphic continuations, no definition of analytic function, and, in particular, no recourse to the concept that every analytic function $f(s)$ vanishing at $s_0$ behaves like $(s-s_0)$ near $s_0$. (You are still allowed to use that fact for a specific function $f$, if you can prove it for that function $f$.)
How would you prove that $lim_{\sigma->1^+} \zeta(\sigma+it) \ne 0$ ($t$ real and fixed)? All the proofs I know (with or without explicit recourse to $\zeta^3(\sigma) |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)|\geq 0$ or the like) use the fact that, if the limit were 0, then $\zeta(\sigma+it)\sim (\sigma+it-1)$ for $\sigma$ near 1.
(Motivation: of course, I am trying to present a proof of the prime number theorem with plenty of analytic ideas but no complex analysis.)

REPLY [3 votes]: You might want to look at the paper "Le théorème des nombres premiers et la transformation de Fourier" by Jean-Benoît Bost, available at http://www.math.polytechnique.fr/xups/xups02-01.pdf
It gives a proof of PNT freely using harmonic analysis and some basics of distributions, but as little complex analysis as possible. In particular, section 4 proves a statement that is equivalent to the non-vanishing of zeta(1+it) using only real analysis. Unfortunately, for my taste there still is too much complex reasoning in the preceding sections of the paper.
(I'm sure you know that the brevity of Don Zagier's proof of PNT can't be beaten, provided you accept complex analysis.)<|endoftext|>
TITLE: Duals of Abelian Categories
QUESTION [10 upvotes]: The dual of an abelian category is again abelian, since the axioms are all preserved by the reversing of arrows.  For example, the category of finite-dimensional vector spaces over a field is easily seen to be dual to itself, since we can just take linear duals of vector spaces.  However, this is the only example where I know a 'concrete' description of both an abelian category and its dual, where by concrete I mean describing the category as a variant of, say, modules over a ring or sheaves on some space.
What are other examples of dual pairs of abelian categories which can both be described 'nicely'? For example, is there a concrete description of the dual of the category of abelian groups?

REPLY [8 votes]: A few (mostly trivial) examples (may be related to references in Tim Porter's answer?):
First of all, the self-duality of the category of vector spaces can be enriched to other self-dualities:

The category of finite-dimensional representations of a group is self-dual.
The category of finite-dimensional representations of a Lie algebra is self-dual.

(Of course, now we can generalize this to Hopf algebras.)
Another simple class of examples comes from taking the dual of some object, and asking what additional structure is necessary to reconstruct the original object. For instance:

Let $V$ be a vector space. Consider $V^*$. It has a natural topology, in which open subspaces are orthogonal complements of finite-dimensional subspaces of $V$. (This is
essentially weak topology if $V$ is viewed as a discrete space.) This gives a duality
between vector spaces and certain class of topological vector spaces that can be easily described explicitly. (A fancier way to view this: vector spaces are 'ind-finite-dimensional', so their dual are 'pro-finite-dimensional'.)

This can be modified to duality on Tate vector spaces, if you happen to like this sort of things.
There are also some examples in (algebraic) geometry (that are probably not as straightforward as the others).

Category of perverse sheaves is self-dual.
Category of holonomic $D$-modules is self-dual (this is related to the previous comment
by the Riemann-Hilbert).

Finally, there is a (somewhat cheap) trick to start with a duality on the derived category, then take the abelian category sitting inside it, and realize that it is dual to its image. For instance:

Look at the derived category of constructible sheaves (say, on a scheme); it has Verdier's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of constructible sheaves is dual to perverse sheaves for a certain perversity.
Look at the derived category of coherent sheaves on a scheme; it has Serre's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of coherent sheaves is dual to perverse sheaves for a certain perversity.

REPLY [3 votes]: An $\epsilon$ generalization of your example: if $A$ is a $k$-algebra for a field $k$, then opping the category of finite-dimensional $A$-modules gives the category of $A^{\rm op}$ modules (the same algebra with the opposite multiplication).  I think you know already that this patently fails when you move to infinite dimensional things: the canonical map $\coprod^\infty M \to \prod^\infty M$ is a mono that is not an epi in $A$-mod, whereas it is an epi that is not a mono in $(A$-mod$)^{\rm op}$.
Letting $A = \mathbb Z$ and $k = \mathbb F_1$ (yeah, that sounds right), I think the analogy is that finite abelian groups are closed under Pontrjagin duality.
Hm, now I wonder what other "$\mathbb F_1$ algebras" have interesting finite representation theory.<|endoftext|>
TITLE: What do representations of infinite-dimensional Heisenberg groups look like?
QUESTION [5 upvotes]: I'm interested in an infinite dim'l Heisenberg group associated to the vector space $V = L\mathbb{C}/\mathbb{C}$ = {$f \colon S^1 \to \mathbb{C}$|$f$ smooth}/(const. maps). The group is $\mathbb{C}^\times \times V$ with group law  
$(z,f)(z',g) = (zz' e^{\pi i (f,g)}, f+g)$  
where $(f,g) = \int fdg$ is a symplectic form.
SOME BACKGROUND  
There's a pairing $e(f,g) = e^{2\pi i (f,g)}$.  The isotropic subspaces $W \subset V$ are the ones s.t. $e = 1$ on $W \times W$.  General theory says for such $W$ you can construct a representation $F(W)$.  Whenever you have a Lagrangian (= maximal isotropic) subspace $L \subset V$ you get up to equivalence a unique irreducible representation where $\mathbb{C}^\times$ acts by scalars.  
(actually I've heard this for groups that are extensions by $U(1)$ and then then there is a unique unitary representation; but I'm guessing it works with $\mathbb{C}^\times$ too by just removing unitary)  
One way to describe this representation is as continuous maps $\phi\colon V \to \mathbb{C}$ that satisfy $\phi(v + l) = e^{\pi i (v,\ l)}\phi(v)$ and $\int_{V/L} |\phi|^2 dk < \infty$ where $dk$ is a Haar measure on $V/L$.  
QUESTION  
For $V = L\mathbb{C}/\mathbb{C}$, $z = e^{i\theta}$ it seems that $z^k$ for $k \in \mathbb{Z} - 0$ forms a basis. Also $L^\pm =$ the vector spaces spanned by the positive/negative powers of $z$ are Lagrangian. `Span' here doesn't mean finite linear combinations but linear combinations where the coefficients form maybe absolutely convergent series?  
My question is what is an example of a $\phi \in F(L^+)$?  
I think all such $\phi$ should be described as follows.  Let $p_\pm \colon V \to L^\pm$ be the projections. Then $\phi(v) = e^{i\pi (p_-(v),\ p_+(v))}\overline{\phi}(p_i(V))$ where $\overline{\phi} \in L^2(L_-;dk)$  
Among my difficulties with answering this question is that $L_-$ is still a really big space and I don't know what a Haar measure would be on this a space.  
I should say, the answer to this question wont really help me in any research per se; I ask it because morally I feel better talking about $F(L)$ if I could write down at least one of its elements.

REPLY [5 votes]: I'll give a description on the level of the polynomial Lie algebra, and then wave my hands about integrating and completing.  As Victor Protsak mentioned in the comments, you can find a more precise treatment in section 9.5 of Pressley and Segal.  There, the unitary representation arises from a choice of complex structure.
The Lie algebra of the Heisenberg group is (topologically) spanned by operators $\{ x_k \}_{k \neq 0}$ and $c$, where $x_k$ describes the tangent vector corresponding to the basis vector $z^k$ in the group, and $c$ exponentiates to the central torus.  The element $c$ is central in the Lie algebra, and the other generators obey the commutation relation $[x_j, x_k] = j\delta_{j,-k} c$ (although you may need a factor of 1/2 with your choice of normalization).  The Lie algebra of $L^+$ is topologically spanned by $x_k$ for $k$ positive, and the corresponding statement holds for $L^-$ with $i$ negative.
The Fock representation is some completion of $\mathbb{C}[x_{-1},x_{-2},\dots,]$, i.e., ``finite energy'' elements are just finite sums of monomials in the generators of $L^-$.  The action is given by the following rules:

When $k$ is negative, $x_k$ acts by multiplication.
When $k$ is positive, $x_k$ acts by $k\frac{\partial}{\partial x_{-k}}$.
$c$ acts by the identity.

Exponentiating will give you a description of the action of group elements of the form $z^k$, and finite sums thereof, on finite energy elements of the representation.  Modulo normalization, this will yield essentially the formula you gave in the background section.  However, I don't think a Haar measure on $V/L$ exists.
Here is an explicit element: there is a distinguished vacuum vector $1$ (sometimes written $\Omega$ or $|0\rangle$), which is the unit element of the polynomial ring above.  It is annihilated by all $x_i$ for $i$ positive, so all $z^k$ act by identity on it when $k>0$.  The action of $z^k$ for $k<0$ is by exponentiating $x_k$ in the completion.  Elements of the central torus act by ordinary scaling.<|endoftext|>
TITLE: Hochschild homology of dga's
QUESTION [8 upvotes]: I'm sure the following statement is well-known to experts:
 Let $A$ be a dga. Let $perf(A)$ be the dg-category of perfect dg-modules over A. Then there is a quasi-isomorphism
$$C_\bullet(perf(A)) \to C_\bullet(A)$$
between their Hochschild chain-complexes.
Does anyone know a reference for it?
I'm aware of Keller's 2003 paper that I think gives the result for Hochschild cochain-complexes, but I need the chain version. Thanks!

REPLY [7 votes]: I don't know a reference, but the assertion is easy to prove.  The natural map goes in the opposite direction, $C_\bullet(A)\to C_\bullet(perf(A))$.  It is quite simply induced by the embedding of the DG-category with a single object associated with $A$ into the DG-category $perf(A)$, sending the only object to the DG-module $A$ over $A$.  To prove that it is a quasi-isomorphism, one can, e.g., interpret the Hochschild homology as the Tor of DG-bimodules, then use the fact that the derived categories of DG-bimodules over $A$ and $perf(A)$ are equivalent.
One can prove that what Caldararu-Tu call the Borel-Moore Hochschild homology are naturally isomorphic for a CDG-algebra $B$ and the DG-category $C$ of CDG-modules over $B$, projective and finitely generated as graded $B$-modules, in much the same way.  I am finishing writing (or rather, editing now) a paper about this.  It will be hopefully made public in a couple of weeks.<|endoftext|>
TITLE: In which commutative algebras does any derivation possess a flow?
QUESTION [9 upvotes]: Definitions
Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a derivation of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$. If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$. 
Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and
$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$
The question
1. I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?
2. Is there any simple condition for an algebra and it's derivation, from which it follows, that this derivation possesses a flow?
Examples
1. Algebra $C^\infty(M)$ of smooth functions on a closed manifold $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using Picard-Lindelof theorem.
1'. Algebra $C^\infty(M)$ of smooth functions on a non-compact manifold without boundary --- no (see example 2), but a derivation possesses a flow if it preserves some function $H\in A$, such that for any $c\in\mathbb{R}$ subspace $\{x\mid H(x) < c \}$ of the topological space $M$ is compact.
2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.
3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.
4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.
5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.
6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.
7.(from Greg Muller's answer) Algebra $A$ is finite-dimensional --- yes, every derivation possesses the flow.
This question was already posted here on math.stackexchange.com, but it has received no answers even with a bounty. Any help is appreciated, both in the theme of question and in improving its wording.

REPLY [3 votes]: If you want to work in the algebraic category you should think about an algebraic notion of flow. If $M$ is a smooth manifold then a flow is an $\mathbb{R}$-action $\mathbb{R} \times M \to M$; dualizing this map gives a coaction $C^{\infty}(M) \to C^{\infty}(\mathbb{R}) \otimes C^{\infty}(M)$ where by $\otimes$ I mean a suitably completed tensor product. (Note that the group structure on $\mathbb{R}$ gives $C^{\infty}(\mathbb{R})$ a Hopf algebra structure, at least with respect to a suitably completed tensor product as above, so we can talk about comodules over it.)
So the most algebraic notion of flow would be a "polynomial flow," namely a coaction $A \to k[t] \otimes_k A$, where $A$ is a $k$-algebra and we are thinking of $k[t]$ in its incarnation as the ring of functions on the additive group scheme $\mathbb{G}_a$ over $k$. In fact in this language a derivation is precisely a coaction $A \to k[t]/t^2 \otimes_k A$ and the problem of integrating this to a flow is a lifting problem. 
A sufficient condition for a polynomial flow to exist is that the original derivation $D$ be nilpotent, but you knew that already. A slightly more interesting sufficient condition is that it be locally nilpotent in the sense that for every $a \in A$ there is some $n$ such that $D^n a = 0$. For example this is true of the derivation $\frac{\partial}{\partial x}$ on $k[x]$. I think this condition is also necessary when $A$ is an integral domain.
There are variations on this theme, e.g. we can talk about formal flows $A \to k[[t]] \otimes_k A$ but these always exist in characteristic zero (formally exponentiate) so this is in some sense uninteresting. Replacing $k[[t]]$ with other variations of $k[t]$ give other notions of flow. 
For example, to talk about flows on smooth manifolds in this algebraic language we should really talk about smooth algebras. Smooth algebras admit a smooth tensor product (this is the suitably completed tensor product I wanted) with respect to which $C^{\infty}(M) \otimes C^{\infty}(N)$ really is just $C^{\infty}(M \times N)$, and in particular $C^{\infty}(\mathbb{R})$ really is a Hopf algebra in the category of smooth algebras.
Now a flow is a coaction $A \to C^{\infty}(\mathbb{R}) \otimes A$, and these exist for smooth compact manifolds but also for other smooth algebras, e.g. any embedding of a finite-dimensional real commutative algebra $A$ into a real matrix algebra gives it a smooth algebra structure and all derivations on such things exponentiate to smooth flows.<|endoftext|>
TITLE: Do there exist associative sudoku squares?
QUESTION [10 upvotes]: Last night I taught an algebra tutorial, and while writing out the multiplication table for the units of $\mathbb{Z}/5\mathbb{Z}$, a student remarked that it looked like a sudoku puzzle.  I noted that it was similar, as the rows and columns all satisfy the sudoku condition, however the 2 by 2 sub-squares do not. The reason I gave is that multiplication is commutative in this group, and so the squares along the diagonal are symmetric. Afterwards I realized that there is an even simpler reason which excludes the possibility of nonabelian groups having multiplication tables which are sudoku squares: the identity element always messes up one of the sub-squares along the diagonal. This leads to the natural question: do there exist associative sudoku squares of side length $n^2$? (Meaning interpret the sudoku square as defining a binary operation on the integers $1$ through $n^2$. Is this binary operation ever associative? Of course I care about when $n > 1$ as $n = 1$ is clear.)

REPLY [13 votes]: If we are not obligated to keep the same order in rows and columns of the table, the answer is yes, and the group can be chosen to be commutative. For example
the group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.
Otherwise, the answer is no for any $n>1$, because in any table of this type both $a\cdot 1$ and $1\cdot a$ are in the same $n\times n$ square for some $a\in G$.
Edit: If this operation doesn't need to define a group, then the answer is no too. Here is the proof.
Suppose $G$ is $\{1,\dots,n^2\}$ endowed with some associative operation and suppose, that the table of this operation is a sudoku table. Take any $x\in G$. In $x$-th line of the table we can find element $x$. Therefore $xy=x$ for some $y\in G$. Then take $a\in G$, such that $a\neq y$ and $ay$ and $ya$ are in the same square of the table. We have $(xy)a=x(ya)$ and, therefore $xa=x(ya)$. Using, that there is only one place in $x$-th line of the table, where we can find $xa$, we see that $a=ya$. From this we have $a(ya)=(ay)a$, $aa=(ay)a$ and $a=ay$. Therefore $ay=a=ya$ and the table is not a sudoku table.<|endoftext|>
TITLE: anti-measureable function
QUESTION [17 upvotes]: (ZFC)
Does there exist a function $f : \mathbb{R} \to \mathbb{R} \hspace{.1 in}$ such that for all $B$, if $B \subsetneq \mathbb{R}$ and $B$ is a nonempty Borel set, then $\lbrace x \in \mathbb{R} : f(x) \in B \rbrace$ is nonmeasurable?

REPLY [21 votes]: There are continuum many pairwise disjoint subsets of [0,1] each having Lebesgue outer measure 1. (By transfinite recursion of length continuum.) Assume that they are $\{A_x:0\leq x\leq 1\}$ and their union is [0,1]. Then set $f:[0,1]\to[0,1]$ so that $f(y)=x$ is $y\in A_x$. Now the inverse image of any $X\subseteq [0,1]$, $X\neq\emptyset, [0,1]$ has innermeasure 0, outer measure 1, hence $f^{-1}[X]$ is nonmeasurable.<|endoftext|>
TITLE: Fibered category with an adjoint inclusion
QUESTION [6 upvotes]: Suppose $X:D \to C$ is a fibered category (I do not assume the fibers to be groupoids). Suppose that $X$ is actually left adjoint to a fully faithful embedding $C \hookrightarrow D$. Is there a special name for such a fibration $X$?

REPLY [3 votes]: Isn't the answer simply that the fibration has fibred terminal objects?<|endoftext|>
TITLE: A question about saturation of quivers
QUESTION [6 upvotes]: Let $Q$ be an acylic quiver. Let $E$ and $F$ be finite dimensional representations, with $E$ indecomposable. Suppose that, for some positive integer $r$, the representation $F$ injects into $E^{\oplus r}$. Suppose also that, for every vertex $v$ of $Q$, we have $\dim F_v \leq \dim E_v$.
Does it follow that $F$ injects into $E$?
There are results like this in the work of Derksen, Schofield and Weyman, but I can't find this particular statement. Thanks!

REPLY [6 votes]: So I think you're asking if there is some kind of "saturation theorem" for generic rank.
The following is a counterexample. Let $Q$ be the ${\rm D}_4$ quiver with vertices 1,2,3,4 (4 is the center) where the orientation is $1 \to 4$, $2 \to 4$ and $4 \to 3$. Let $E$ be the unique indecomposable (up to isomorphism) of dimension $(1,1,1,2)$ and take $F$ to be the unique representation of dimension $(0,0,0,2)$. Now the simple $S_4$ injects into $E$ as the kernel of the map $E_4 \to E_3$, but $F$ does not inject into $E$. However, it does inject into $E \oplus E$.<|endoftext|>
TITLE: Number of ways to construct mathematical objects
QUESTION [6 upvotes]: This question stems from this other one mentioning 7 ways of constructing smooth manifolds. I quote:

At the 2010 Clay Research Conference, Gromov explained that we know of only 7 different methods for constructing smooth manifolds.[...]

 Algebraic geometry (affine and projective varieties, ...)
   Lie groups (homogeneous spaces, ...)
   General position arguments (Morse theory, Pontryagin-Thom construction, ...)
   Solutions to PDE (Moduli spaces in gauge theory, Floer theory, ...)
   Surgery (Cut and paste techniques, ...)
   Markov processes
   [and also bundles seems to be the consensus in the answers to the cited question]
  

Note that all of these methods are actually areas of mathematics in their own right (five of the six listed in that question involve trailing dots), so that got me thinking that 7 methods is actually a sign of a rich subject and a fairly ubiquitous concept. I have trouble comparing this to anything else: e.g., would you say that there are fewer ways to build a group? I don't know nearly enough about groups to answer that one. So here's my question:
Do you know of an abstract math construct that can be built in truly more than seven ways?
I realize that this is somewhat in the eye of the beholder (hence the soft question tag) since it may not be obvious where to draw the line between methods. But for the comparison to make sense, you need to consider broad categories like Gromov does.

REPLY [6 votes]: There are lots of different ways to construct a finitely generated (discrete) group. For example (in no particular order):

Fundamental groups of topological spaces;
Groups of symmetries of mathematical objects (including fields, manifolds, simplicial, real and $\Lambda$-trees,  etc.);
Groups given by presentations satisfying various small cancelation conditions (from Tartakovsky to Olshanskii to Gromov);
Lattices in Lie groups;
Wreath products of various kinds;
Direct limits of sequences of groups and their homomorphisms (including various "monsters", etc.);
Free constructions (HNN extensions and amalgamated products);
Groups simulating various computing devices (there are several different constructions here);
Groups acting on locally finite rooted trees (including Grigorchuk groups and iterated monodromy groups of Nekrashevych);

10 Automatic groups,
..................<|endoftext|>
TITLE: Explicit metrics
QUESTION [49 upvotes]: Every surface admits metrics of constant curvature, but there is usually a disconnect between
these metrics, the shapes of ordinary objects, and typical mathematical drawings of surfaces.

Can anyone give an explicit and intuitively meaningful formulas for negatively curved metrics that are related to an embedding of a surface in space?

There is an easy way to do this for an open subset of the plane.  If the metric of the plane is scaled by a function that is $\exp$ of a harmonic function, the scaling factor is at least locally the norm of the derivative of a complex analytic function, so the resulting metric is still flat; the converse is true as well.  Therefore, the sign of curvature of a conformally modified metric $\exp(g)$ depends only on the sign of the laplacian of the $g$. If the value of $g$ at a point is less than the average value in a disk centered at that point, then 
the metric $\exp(g) ds_E$ is negatively curved, where $ds_E = \sqrt(dx^2 + dy^2)$ is 
Euclidean arc length.
alt text http://dl.dropbox.com/u/5390048/NegativeMetrics.jpg
For example, in a region $R$, if we impose a limit that speed is not to exceed the distance to the complement of $R$, this defines a non-positively curved metric.  (The metric is 1/(distance to boundary)$ds_E$).
In this metric, geodesics bend around corners: it doesn't pay to cut too close, it's better to stay closer to the middle. If the domain is simply-connected, you see 
one and only one image of everything, no matter where you are.  
There are a number of other ways to write down explicit formulas for negatively curved or non-positively curved metrics for a subset of the plane, but that's not the question: what about for closed surfaces in space?  Any closed surface $M^2$ has at least a total of $4 \pi$ positive curvature, where the surface intersects its convex hull.  If $M$ is a double torus, how can this be modified to make it negative?  It would be interesting to see even one good example of a negatively curved metric defined in terms of  Euclidean geometry rather than an indirect construction. (In particular: it can be done by solving PDE's, but I wwant something more direct than that.)

REPLY [15 votes]: Here's an answer to an analogous question, not Bill's original question, but also a question about how to specify (in a simple way) a non-positively curved metric on a compact Riemann surface $C$ of genus $g>1$, in this instance, one that has been specified as an algebraic curve somewhere (as opposed to being given it as a surface in $3$-space).  
The construction is easy:  If the curve has been specified as an algebraic curve, then, more-or-less by algorithmic means, one can write down a basis for the holomorphic differentials on $C$ (which is a complex vector space of dimension $g$).  Now select two of these differentials, say, $\omega$ and $\eta$, that have no common zeroes on $C$.  (Again, this can be tested algebraically).  Now consider the metric $g = \omega\circ\bar\omega + \eta\circ\bar\eta$.  This $g$ will have non-positive curvature.  In fact, the curvature will vanish at only a finite number of points and will otherwise be strictly negative.   (Of course, you can add more terms.  If you take a basis $\omega_1,\ldots,\omega_g$ of the holomorphic differentials on $C$, then the metric $g = \omega_1\circ\bar{\omega_1} +\cdots + \omega_g\circ\bar{\omega_g}$ will have strictly negative curvature except when $C$ is hyperelliptic, in which case, the curvature will vanish at the Weierstrass points of $C$.)
For example, if you take a hyperelliptic curve, say $y^2 = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{2g+2})$ (with the $\lambda_i$ being distinct and, say, nonzero), then a basis for the holomorphic differentials will be given by $\omega_i = x^{i-1}dx/y$ for $i = 1,\ldots, g$.  Moreover, $\omega_1$ and $\omega_g$ (for example) have no common zeros.  Thus, the smooth metric $g = (1 + |x|^{2(g-1)})|dx|^2/|y|^2$ has negative curvature on this curve except at a finite number of points.<|endoftext|>
TITLE: The typical size of a random element in a Banach space
QUESTION [6 upvotes]: Let $X$ be a separable Banach space, and let $\mathbb P$ be a Radon probability measure on $X$ with zero mean and covariance operator $K : X^* \to X$.  Let $x$ be an $X$-valued random variable with distribution $\mathbb P$.  
I would like a simple upper bound on the size of $\mathbb E \|x\|^2$ in terms of the operator norm $\|K\|$.  In his textbook The Concentration of Measure Phenomenon, Ledoux proves that $$\mathbb E\|x\|^2 \le 4 \|K\|$$ as a consequence of a concentration inequality for a simpler example (where the vectors are sums of vectors with i.i.d. random coefficients $\eta_i$ such that $|\eta_i| \le 1$ a.s.).  Because of the boundedness assumption, the argument doesn't quite work in this setting, though I'm certain I could generalize it if I needed to.  
Nonetheless, I'm certain that the estimate I'm looking for is buried somewhere in the literature on concentration of measure (and in fact is probably due to Talagrand).  Could you please point me in the right direction?
Edit:  The inequality as I previously wrote it is incorrect.  The correct inequality should be $$\operatorname{Var}(\|x\|) \le 4\|K\|,$$
implying $$\mathbb E\|x\|^2 \le 4\|K\| + \left( \mathbb E\|x\| \right)^2.$$  That is, the size of typical random element could be quite large (i.e. $\mathbb E\|x\| \gg 1$, as in Mark Meckes's example in the comments), but the deviation is only of the order $\|K\|^{1/2}$.

REPLY [4 votes]: The inequality can't be true without additional assumptions.  To see this, let $X = \ell_2^n$ and let $x$ have a spherically symmetric distribution and let $R = \Vert x \Vert$.  Then $R$ is an essentially arbitrary nonnegative random variable; indeed we could start by picking $R$ and defining $x=R\Theta$, where $\Theta$ is a uniform random vector in $S^{n-1}$ independent of $R$. 
Now $\mathrm{Var}(\Vert x \Vert) = \mathrm{Var} (R)$, and $K = \frac{1}{n} \mathbb{E}(R^2) I_n$, so $\Vert K \Vert = \frac{1}{n} \mathbb{E} (R^2)$.  Thus for any fixed distribution of $R$, the inequality fails for sufficiently large $n$.<|endoftext|>
TITLE: Are the compact and Haagerup approximation properties equivalent?
QUESTION [7 upvotes]: The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102368918) and the Haagerup approximation property.
Let $M$ be a type $II_{1}$ factor with trace $\tau$. Let $\Omega$ denote the standard unit cyclic trace vector in $L^{2}(M)$ (associated to the element $1\in M$). If $\phi:M \rightarrow M$ is a normal completely positive map, we naturally associate an operator $T_{\phi}\in B(L^{2}(M))$ extending $T_{\phi}(x\Omega)=\phi(x)\Omega$.
If the map $x\mapsto \phi(x) \Omega$ is a compact linear map from $M$ with the operator norm into $L^{2}(M)$, is the operator $T_{\phi}$ compact?

REPLY [4 votes]: (I removed my first answer as it contained an egregious mistake, pointed out by Yemon; here's a second attempt)
I think that $T_\phi$ may fail to be compact.
Fix a sequence of projections $\{p_k\}$ in $M$, pairwise orthogonal, with $\tau(p_k)=2^{-k}$ ($\tau$ the trace in $M$) and sum 1. Now define 
$$
\phi:M\to M,\ \ \ \mbox{ given by } \phi(x)=\sum_k 2^k\tau(xp_k)p_k.
$$
(the series converges strongly because all of its terms are positive and any partial sum is bounded by $\|x\|$). This map is ucp (it is an infinite sum of cp), and since it commutes with $\tau$, it is normal. Let us also define the maps
$$
\phi_n=\sum_{k=1}^n 2^k\tau(xp_k)p_k.
$$
The maps $x\mapsto \phi_n(x)\Omega$ are all finite-rank.
If $\|x\|\leq1$, we have, using that the set $\{p_k\}$ is orthogonal in $L^2(M)$,
$$
\left\|\phi(x)-\phi_n(x)\right\|_2^2=\left\|\sum_{k>n}2^k\tau(xp_k)p_k\right\|_2^2
=\sum_{k>n}|\tau(xp_k)|^2\leq\sum_{k>n}\tau(p_k)^2\leq\frac1{3\times 4^n}
$$
This shows that $\|\phi-\phi_n\|<4^{-n}$ in $B(L^2(M))$: so the map $x\mapsto \phi(x)\Omega$ is compact. 
Now consider the orthonormal set $\{2^kp_k\}$ in the unit ball of $L^2(M)$. Since $\phi(p_k)=p_k$, we get that $T_\phi(2^kp_k)=2^kp_k$; so the range of $T_\phi$ contains an orthonormal set: $T_\phi$ is not compact.<|endoftext|>
TITLE: Pointer to literature on double enrichment and functors among enriching categories?
QUESTION [7 upvotes]: I'm currently working with the following two situations:

$\mathbb A$ is a monoidal category, $\mathbb B$ is an $\mathbb A$-enriched monoidal category, and $\mathbb C$ is a $\mathbb B$-enriched category (and ${\mathbb A}\ncong {\mathbb B}$, ${\mathbb B}\ncong {\mathbb C}$, ${\mathbb A}\ncong {\mathbb C}$).
$\mathbb V$ and $\mathbb W$ are monoidal categories, $v$ is a $\mathbb V$-enriched category, $w$ is a $\mathbb W$-enriched category, and there is a monoidal functor $F:{\mathbb V}\to{\mathbb W}$.

I haven't been able to find much in Kelly's Basic Concepts of Enriched Category Theory, but surely others have come across these two situations before; there are lots of easy results (like, the image of $v$ under $F$ is a $\mathbb W$-enriched category) which must have been known for ages.  Are there terms for the situations above?  That would help me search the literature.  If not, even just one paper on either of these cases with a bibliography would be a good foothold.
So far everything I've found on enriched category theory seems to focus on the case where there's only one enriching category in which all of the enriched categories are enriched (or the self-enrichment case where $\mathbb D$ is isomorphic to a $\mathbb D$-enriched category).
Thank you!

REPLY [2 votes]: The thesis of Dominic Verity treats base change in a very abstract 2-categorical setting - looks like a lot of specialization is necessary until you arrive at the sort of situation sketched in the question.<|endoftext|>
TITLE: The higher Van Kampen Theorems and computation of the unstable homotopy groups of spheres
QUESTION [13 upvotes]: Since Ronnie Brown and his collaborators have come up with a general proof of the higher Van Kampen theorems, what impediments are there to using these to compute the unstable homotopy groups of spheres?

REPLY [3 votes]: I think it is important to remember that the Brown-Loday theorem concerns colimits of cat-n groups obtained from an n stage filtration of the underlying space. Moreover, cat-n groups can only provide information on the n-type.
So, if you wanted to compute pi_200 of the two sphere, you would need a cat-200 group (at least). And if you wanted to apply the Brown_Loday VK theorem you would need a 200 stage topological filtration of the two sphere. Any thoughts ? or do I have this wrong?<|endoftext|>
TITLE: How do you decide whether a question in abstract algebra is worth studying?
QUESTION [112 upvotes]: Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.
So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:

Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.
The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.
Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.

The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.
Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.

How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?

Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?
Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.
I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is

How much would you subscribe to the statement that 
  EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?

Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.

Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.

Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".

REPLY [17 votes]: Dan Schechtman, winner of the 2011 Nobel Prize in Chemistry for the discovery of quasi crystals, said:  “The main lesson that I have learned over time is that a good scientist is a humble and listening scientist and not one that is sure 100 percent in what [they read] in the textbooks.”
My research on groupoids and higher groupoids was started in the 1960s by a dissatisfaction with a van Kampen theorem that did not compute the fundamental group of the circle, a basic example: but groupoids were at the time regarded as "rubbish" by many senior mathematicians, and the idea of higher van Kampen theorems using higher groupoids was described   by one such for 10 years as "ridiculous". (He gave in eventually!)
My worry is that people may be encouraged to follow high ups, rather than to analyse a programme on mathematical grounds,  and so to develop their own feeling for mathematical structures.
January, 2015: One needs a variety of strategies,  one of which is to look at what a theory does not do but somehow in principle should. This is the notion of anomaly. I have listed 5 anomalies in standard algebraic topology in this presentation Dec, 2014, Galway.
See also the advice given to me 1964 by S. Ulam, quoted in my web page discussing the issue of famous problems in category theory.
Alexander Grothendieck wrote to me that: "Throughout my whole life as a mathematician, the possibility of making explicit, elegant computations has always come out by itself, as a byproduct of a thorough conceptual understanding of what was going on. Thus I never bothered about whether what would come out would be suitable for this or that, but just tried to understand -- and it always turned out that understanding was all that mattered."
So I always advocate writing and rewriting to make things clear to yourself, testing that by explaining to other people. At Bangor we explained to research students that a thesis must have a "thesis". So having decided on the latter, the first thing for the student to do is write up the background to that "thesis", which  can always be expected to be a useful part of the final thesis. All sorts of things may turn up in that process.<|endoftext|>
TITLE: Why is the string group not a Lie group?
QUESTION [9 upvotes]: The string group $String(n)$ is by definition a 3-connected cover of $Spin(n)$. This definition determines the homotopy type of the string group.
[In a previous version of this question I screwed up the definition and caused some confusion, see the comments below.]
A common argument is saying that "the string group cannot be a Lie group because it has vanishing $\pi_3$". This is obviously not a complete argument because $(\mathbb{R},+)$ is a nice Lie group with vanishing $\pi_3$. 
What is the correct statement about Lie group structures on the string group, and how does one prove it?

REPLY [9 votes]: To follow up, there is now an infinite-dimensional Lie group model of String:

Thomas Nikolaus, Christoph Sachse, Christoph Wockel, A Smooth Model for the String Group, Int. Math. Res. Not. 16 (2013) 3678-3721, doi:10.1093/imrn/rns154, arXiv:1104.4288.<|endoftext|>
TITLE: Finite groups with elements of the same order
QUESTION [15 upvotes]: Given a finite group $G$, let $\{(1,1),(m_1,n_1),\ldots,(m_r,n_r)\}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n_1+\cdots+n_r$, and the pair $(1,1)$ accounts for the neutral element.
Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet not) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple. 
Edit. Nick's answer gives the correct terminology, of conformal groups. Ben's answer speaks of the refined notion of almost conjugate subgroups. Is there any other related notion ?

REPLY [15 votes]: This question was answered more than five years ago, but I am just now noticing it and wanted to point out that non-isomorphic groups with the same order portraits arise very naturally in spectral geometry and underlie Sunada's Method of constructing isospectral Riemannian manifolds.
Let $G$ be a finite group and $H_1, H_2$ be subgroups of $G$. We say that $H_1$ and $H_2$ are almost conjugate if every $G$-conjugacy class intersects both subgroups in the same number of elements. 
An easy argument shows that almost conjugate subgroups have the same order portraits. 
The connection of these groups to geometry stems from a theorem of Sunada that says that if $G$ acts by isometries on a Riemannian manifold $M$ and $H_1$ and $H_2$ are almost conjugate subgroups of $G$ then the quotient orbifolds $M/H_1$ and $M/H_2$ are isospectral with respect to the Laplace operator.
This paper by Brooks, Gornet and Gustafson gives explicit examples of arbitrarily large families of almost conjugate subgroups of Heisenberg groups defined over various commutative rings. The examples are then applied so as to construct families of isospectral non-isometric genus $g$ Riemann surfaces having cardinality $g^{c\log(g)}$ (here $c>0$ is a positive constant).
EDIT: In light of Nick's comment above I wanted to say a few words about Gassmann equivalence. To start with, by definition, two subgroups $H_1$ and $H_2$ of a finite group $G$ are Gassmann equivalent if and only if they are almost conjugate. The history of the term arises from a connection of these groups with number theory.
To an algebraic number field $k$ one can associate a Dedekind zeta function $\zeta_k(s)$. (If $k$ were the field of rational numbers then $\zeta_k(s)$ would be the Riemann zeta function.) This function determines all sorts of arithmetic properties of $k$: the discriminant, signature, splitting of primes, product of class number times regulator, etc. It is therefore natural to ask whether $k$ is determined up to isomorphism by $\zeta_k(s)$. It turns out that this is not in general true. We will therefore say that two number fields with the same Dedekind zeta function are arithmetically equivalent. The first examples of arithmetically equivalent number fields were discovered by Fritz Gassmann in 1925 and made use of the almost conjugacy condition defined above.
The term 'Gassmann equivalent' is due, I believe, to Robert Perlis. On page 344 of this paper Perlis defines subgroups $H_1$ and $H_2$ of $G$ to be Gassmann equivalent if every $G$-conjugacy class intersects $H_1$ the same number of times as $H_2$, which of course is precisely the same thing as saying that $H_1$ and $H_2$ are almost conjugate. Perlis then proves the following remarkable theorem.
Theorem (Perlis): Let $k$ be a Galois number field and $k_1, k_2$ be subfields of $k$. Then $\zeta_{k_1}(s)=\zeta_{k_2}(s)$ if and only if $\text{Gal}(k/k_1)$ and $\text{Gal}(k/k_2)$ are Gassmann equivalent subgroups of $\text{Gal}(k/\mathbb Q)$.

REPLY [10 votes]: The smallest counterexamples have order $16$.
Up to isomorphism, there are $14$ groups of order $16$;
these fall into $9$ distinct equivalence classes w.r.t. order portrait.
The $3$ equivalence classes containing more than one group can be found
with GAP as follows:
gap> OrderPortrait := G -> Collected(List(AsList(G),Order));;
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->List(cl,IdGroup)); # in terms of catalog id numbers of groups
[ [ [ 16, 5 ], [ 16, 6 ] ], [ [ 16, 2 ], [ 16, 4 ], [ 16, 12 ] ],
  [ [ 16, 3 ], [ 16, 10 ], [ 16, 13 ] ] ]
gap> List(last,Length); # two classes have length 3, and one has length 2
[ 2, 3, 3 ]
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->List(cl,StructureDescription)); # the structure of the groups
[ [ "C8 x C2", "C8 : C2" ], [ "C4 x C4", "C4 : C4", "C2 x Q8" ],
  [ "(C4 x C2) : C2", "C4 x C2 x C2", "(C4 x C2) : C2" ] ]
gap> List(Filtered(EquivalenceClasses(AllGroups(16),OrderPortrait),
>                  cl->Length(cl)>1),
>         cl->OrderPortrait(cl[1])); # the actual order portraits
[ [ [ 1, 1 ], [ 2, 3 ], [ 4, 4 ], [ 8, 8 ] ], 
  [ [ 1, 1 ], [ 2, 3 ], [ 4, 12 ] ], [ [ 1, 1 ], [ 2, 7 ], [ 4, 8 ] ] ]

REPLY [10 votes]: Let me join Stefan in resurrecting this question. I want to mention a fascinating observation of Thompson.
Apparently two groups $G$ and $G'$ that, in Denis' terms "have the same list" are, in the literature called conformal (see Mazurov-Shi in Groups St Andrews 1999).
J.G.~Thompson famously gave an example of two non-isomorphic conformal groups: $2^4:A_7$ and $L_3(4):2_2$. These two groups both appear as maximal subgroups of $M_{23}$, one of the sporadic Mathieu groups.
Along with this observation, Thompson posed the following question:

Suppose that $G$ and $G'$ are conformal, and that $G'$ is solvable. Is $G$ solvable?

As far as I know, this question remains open. However there has been a lot of progress in the study of conformality amongst non-solvable groups. I'm thinking of results of the kind: Suppose that $G$ and $G'$ are conformal, and that $G'$ is [insert name of some simple group], then $G\cong G'$. 
I refer those interested to the above-mentioned survey of Mazurov--Shi for several results of this kind (and many other interesting results).<|endoftext|>
TITLE: Finding solutions to $f'(x) = f(x + k)$
QUESTION [7 upvotes]: I'm trying to find non-trivial functions $f \colon \mathbb R \to \mathbb R $ that $f'(x) = f(x+k)$ with $k \in \mathbb R$.
For $k \le 0$, I've found functions based on $f(x)= e^x$, such as $f(x) = e^{x \frac{W(k)}{k}}$, where $W(k)$ is the Lambert W function.
However, for $k>0$, I can only find a solution if $k=2\pi n+\frac{\pi}{2} $ with $n\in\mathbb N$. The solution is $f(x) = \sin x$. Are there solutions for other values of $k$?
I was hoping that for $k>0$, a function whose graphic is similar to $f(x)=-e^x$ could exist, but it seems it doesn't exist. Or does it?
The method I used is basically trial and error. What other method could I have used?
Thanks.

REPLY [2 votes]: Let $a=\alpha+\beta i$, $\alpha,\beta\in\mathbb{R}$, be a complex number such that $a=e^{ak}$. In terms of $\alpha$ and $\beta$ this means that $\alpha=e^{\alpha k}\cos(\beta k)$ and $\beta=e^{\alpha k}\sin(\beta k)$. Then $f(x)=Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x)$ is a solution for any choice of $A,B\in \mathbb{R}$. Are there other solutions? Yes.
In the following I assume $k>0$. Let $\phi$ be a $C^\infty$ function with compact support in $[0,k]$. Let $n$ be a non negative integer. Define $g$ on $[nk,(n+1)k]$ by $g(x)=\phi^{(n)}(x-nk)$. Then $g$ verifies $g'(x)=g(x+k)$ on $[0,+\infty)$. We need to extend $g$ to $(-\infty,0]$. For this define indectively $g$ on $[-(n+1)k,-nk]$ by $g(x)=-\int_{x+k}^{-(n-1)k}g(s)ds$.
The general solution is of the form $f+g$ for some choice of $\alpha,\beta$ and $\phi$.<|endoftext|>
TITLE: Free commutative magma over a set
QUESTION [5 upvotes]: BOURBAKI, inside his book on ALGEBRA defines and provides explicit constructions concerning the concepts of free magma, free monoid (and implicitly free semi-group) and free group, and as well free commutative monoid (and implicitly free commutative semi-group) and free commutative group over a set X;
It seems clear that the concept of free commutative magma also makes sense, but does anyone know about an explicit construction for the free commutative magma over a set X ?

REPLY [12 votes]: The free magma on $X$ consists of finite sequences of length $1$ or $2$, which consist of finite sequences of length $1$ or $2$, etc., of elements of $X$; the magma operation is just concatenation, i.e. $mn = (m,n)$. For example, $(a,((b,c),d))$ is such a sequence, where $a,b,c,d$ are in $X$. Elements may be visualized as finite binary trees, whose leaves are labelled in $X$. The examples gives:

To give a formal construction, define by recursion the sets $X_n$ of elements of height $n$ by $X_0 = X$ and $X_{n+1} = X_n + X_n^2$ (disjoint union). Then the disjoint union of $X_n$ is the free magma on $X$.
Now the free commutative magma is obtained by taking the quotient with respect to the smallest congruence relation satisfying $(a,b) \sim (b,a)$. This can be visualized with trees: Every branch can be rotated freely. So for example, now we don't distinguish between the trees
.
Note, however, that this does not justify that we may replace every bracket $(...)$ with $\{...\}$. Namely, $(a,a)$, which has two leaves, has to be distinguished from $(a)$, which has only one.
Remark: These constructions are "abstract nonsense", the same procedure works for free algebraic structures of any type (free monoids, free groups, free Lie algebras, etc.). Usually a bit of work has to be done to find normal forms for the elements of free algebraic structures. For free magmas, every element is already in normal form. For the free commutative magma on $X$, choose a total order on $X$, and call an element in normal form if the occuring elements of $X$ (ignoring the brackets) are ordered from left to the right. In the picture above, when we order $a<b<c<d$, then the tree on the left is the normal form of the tree on the right.<|endoftext|>
TITLE: Are local, Noetherian rings with principal maximal ideal PIR?
QUESTION [9 upvotes]: A question asked by a friend. I believe it's false, but lack a decisive counterexample.
This question shows that it is true for valuation rings, but I know too little about them.
In the wider context, a solution to this problem would provide another proof that Artinian local rings whose maximal ideal is principal are principal ideal rings by shifting from Artinianness to Noetherianness instead of exploiting the nilpotence of the maximal ideal.
I'm tagging this commutative-rings because those are the only ones I really care about, but a noncommutative example would be just as decisive.

REPLY [2 votes]: Even in the noncommutative case, we can show that a local ring whose maximal ideal $\mathfrak{m}$ is left principal is a left PIR if and only if $\bigcap_{i \geq 0} \mathfrak{m}^i = 0$ (the conclusion of Krull's intersection theorem).
Recall that a noncommutative ring $R$ is local if the non-units form a left ideal $\mathfrak{m}$ (and in this case $\mathfrak{m}$ is in fact a two-sided ideal). Suppose this maximal ideal is left-principal, say $\mathfrak{m} = Rt$. 
The "if" direction follows the same proof as in the commutative case. 
For the "only if" direction, suppose that $R$ is a left PIR. Let $I = \bigcap_{i \geq 0} \mathfrak{m}^i$. Since $R$ is a PIR, we have $I = Rx$ for some $x$. Similar to the proof of Krull's intersection theorem in the commutative case, show that $tI =I$ (if $z \in I$, then $z = tz'$ for some $z'$; if $z' \notin I$, then $z' \notin \mathfrak{m}^j$ for some $j$, but then $z \notin \mathfrak{m}^{j+1}$, contradicting $z \in I$). Then there is some $y \in I$ such that $x = ty$. Since $y \in I=Rx$, we have $y=rx$ for some $r$, and thus $x = trx$, or $(1-tr)x=0$. Since $tr \in \mathfrak{m}$, $1-tr$ is a unit, and thus $x=0$ and $I = Rx = 0$.
(I don't even think you need Noetherian to prove this; in case $R$ ends up being a left PIR, Noetherian seems to follow.)<|endoftext|>
TITLE: Which knots' stick numbers are twice their crossing numbers?
QUESTION [7 upvotes]: Looking at a table of minimum stick numbers for knots (table here), 
it seems the known upper bound of $2 c(K)$ in terms of the knot crossing number $c(K)$
is realized by the trefoil $3_1$—it requires 6 sticks (see image below) and its crossing number is 3—but not by any other small knot, at least through cursory inspection.
Whence the question in the title: Are there other knots whose minimal stick number reaches the upper
bound of twice its crossing number? This is probably well-known (perhaps well-known to be unknown),
in which case a reference would suffice.  Thanks!



Addendum.
I found a 12-year old answer to my question
in a paper by 
Eric Furstenberg, Jie Lie, and Jodi Schneider [FLS]:

"Thus far, the trefoil is the only knot to realize Negami’s 
  upper bound of $2c[K]$ on the 
  stick number. 
  Do other such knots exist, and if so, what are their similarities to the 
  trefoil?"

If anyone knows of more recent information, I would appreciate
hearing of it.  Thanks!
[FLS]
"Stick Knots."
Eric Furstenberg, Jie Lie, and Jodi Schneider.
Chaos, Solitons & Fractals, Vol. 9, No. 4-5, pp. 561-568, 1998.
Elsevier link

REPLY [9 votes]: I just read the following paper, where an answer can be found:
Youngsik Huh, Seungsang Oh, An upper bound on stick number of knots, J. Knot Theory Ram. 20 (2011), no. 5, 741–747, doi: 10.1142/S0218216511008966, arXiv: 1512.03592. 
There it is shown that the trefoil is the only knot whose stick number equals twice its crossing number. 
This is a consequence of the authors' main result (Thm. 1.1), which states that any nontrivial knot $K$ satisfies $s(K)\leq \frac{3}{2}(c(K)+1)$ (thus improving Negami's upper bound).<|endoftext|>
TITLE: Is Dependent Choice equivalent to the statement that every PID is factorial?
QUESTION [18 upvotes]: In this question, it was asked if AC is needed in the proof of the well-known fact that every principal ideal domain is factorial. As KConrad and Joel David Hamkins have pointed out, only DC, the axiom of dependent choice, is needed, and in the comment box the question was raised if the statement is actually equivalent to DC. But this was not answered so far.
So I ask it here, is the statement equivalent to DC? It would not surprise me since there are already many algebraic equivalents of AC (existence of vector bases, existence of maximal ideals). Also note that the requirement of ideals being principal is somewhat analoguous to the assumption of entireness of the relation in DC, and the representation as a product of irreducibles is somewhat analoguous to the conclusion of DC.
If the statements turns out to be weaker, I would be also interested in a proof which is the set-theoretical equivalent (perhaps it's DC-fin as in Joel's answer?).
PS: Of course, this question takes place in ZF ;).

REPLY [2 votes]: Edit. This solution is incorrect. As François pointed out in the comments, the ring $B$ I construct is definitely not a domain; indeed, no nontrivial Boolean ring is a domain. But Martin asked me not to delete it, so I am leaving it here.
The idea was to take the tree $T$ through which a branch is desired and turn it into a ring, so that the lack of a branch would mean that every ideal is principal. One way to turn any partial order into a ring is to look at the regular open algebra, the completion as a Boolean algebra, and then look upon the Boolean algebra as a Boolean ring. 

I can prove a lower bound, showing that the statement implies at least a weakened version of DC. 
The principle of Dependent Choice is the assertion that if $R$ is a binary relation on a set $X$ and every $x\in X$ has some $y$ with $x\mathrel{R}y$, then there is an infinite sequence $x_0,x_1,\ldots $ with $x_n\mathrel{R} x_{n+1}$. Thus, DC asserts that one can make countably many choices in succession. 
Consider the weakened principle, call is DC-fin, where we make the conclusion of DC, but only for relations $R$ that are finitely-branching, so that every $x$ is $R$-related to at least one but at most finitely many $y$. This principle is closely related to Konig's Lemma, and is not provable in ZF, since it implies countable choice for finite sets, which is known not to be provable in ZF.
Theorem. If every PID is factorial, then DC-fin holds. 
Proof. Suppose that $R$ is a finitely branching relation having no infinite choice sequence. Let $T$ be the tree of all finite choice sequences. Thus, $T$ is a finitely branching tree, with no leaves, but $T$ has no infinite branch. (Weird, yes, but this is the world where choice fails.) Note that every node in $T$ has incompatible extensions, since otherwise there would be an isolated branch in $T$, which would give a choice sequence for $R$. 
There is a natural topology on the tree $T$, generated by the lower cones. That is, view $T$ as growing downwards, and use as the basic open sets the sets $U_t$ of all extensions of $t$ in $T$, for any $t\in T$. Let $B$ be the set of regular open subsets of $T$ (regular open = interior of its closure). This is a (complete) Boolean algebra, and is the completion of $T$ as a Boolean algebra (used commonly in forcing). 
In particular, $B$ is a Boolean ring. The Boolean operations can be used to define the ring operations by $a\cdot b=a\wedge b$ and $a+b=a\triangle b$, the symmetric difference, and the concepts of ideal and filter translate from ring theory to the Boolean algebra context this way. The only unit in $B$ is $1$, which corresponds to the root of $T$. There are no irreducible elements of $B$, since it is non-atomic: if $a\neq 1$, then $\neg a$ is a nonzero element, which has incompatible nonzero elements $p,q\lt\neg a$, and so $a=(a\vee p)\cdot(a\vee q)$ is a nontrivial factorization. 
But now, I argue under our assumption on $R$ that $B$ is a PID [Edit. This is wrong because it is not a domain]. Suppose that $I$ is an nonprincipal ideal in $B$. Let $F$ be the dual filter, which is therefore also not principal. So $1\in F$. Since the root $1$ is the join in $B$ of its immediate successors in $T$, it follows one and exactly one of them $x_0$ must be in $F$. Since $x_0$ is the join of its finitely many $R$-successors, again we get exactly one of these successors $x_1$ must be in $F$. And so on. The point is that the filter $F$ chooses exactly one successor at each level of the tree, and so defines a choice sequence through $R$. (Note, the filter cannot choose two different successors, since these are incompatible and hence meet to $0$ and so cannot both be in the filter, and it must choose something, or else $I$ is principal.) Since there is no such choice sequence, it must be that $F$ and hence $I$ is principal, and so $B$ is a PID that is not factorial. QED
The finite-branching assumption on $R$ was used in arguing that when an element of the tree is in $F$, then one of its successors is in $F$. I don't see how to make this part of the argument work if there are infinitely many $R$ successors. 
I believe that the principle DC-fin, which is a special case of Konig's lemma, is strictly weaker than DC.
It appears that the solutions in the other direction (at the other question) really use DC and not just DC-fin. (Although if one has factorizations for every element, then part of the argument involves choosing among the finitely many factors, and this would seems to just use DC-fin. But choice arises in the other part of the argument, where one must choose the factorizations in the first place.)
If one could somehow improve those arguments to need to make only choices from finite sets, then one would get that the statement is equivalent to DC-fin.<|endoftext|>
TITLE: Classification of $p$-groups of order $p^n$ with rank $n-1$
QUESTION [7 upvotes]: Hello,
i've been looking for a way to classify the non-trivial $p$-groups $G$ that live in an exact sequence of the form
    $ 0 \rightarrow \mathbb{Z}/p\mathbb{Z} \rightarrow G \rightarrow (\mathbb{Z}/p\mathbb{Z})^{n-1} \rightarrow 0 $. Was this question settled before? Or is there any explicit computation of $H^2((\mathbb{Z}/p\mathbb{Z})^{n-1}, \mathbb{Z}/p\mathbb{Z})$?
Thanks!

REPLY [4 votes]: Your group is such that $|G|=p^n$ and $|\Phi(G)|=p$.  Since $(C_p)^{n-1}$ is completely reducible, there is a subgroup $H$ of $G$ such that $G=HZ(G)$ and $H\cap Z(G)=\Phi(G)$. Thus $H$ is an extra-special group (possibly trivial), and we are taking the central product with the abelian group $Z(G)$, which is either of the form $(C_p)^m$ or $(C_{p^2})\times(C_p)^m$.  The first case is easy, since again, it is completely reducible, so we get a group of the form (extra-special) times (some copies of $C_p$).  The second case also gives (some group) times (some copies of $C_p$).  I believe the (some group) is uniquely determined by its order (that is the central product of either of the two non-abelian groups of order $p^3$ and $C_{p^2}$ are isomorphic), but I haven't checked any cases but $p=2$.
Steve<|endoftext|>
TITLE: Product of Borel sigma algebras
QUESTION [36 upvotes]: If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times  Y)$ of the product is the $\sigma$-algebra generated by $B(X)\times B(Y)$.  I am embarrassed to admit that I don't know the answers to: 
Question 1. What is a counterexample when $X$ and $Y$ are non-separable?
Question 2. If $X$ is an uncountable discrete metric space, does
$B(X) \times B(X)$ generate the Borel $\sigma$-algebra on $X \times X$?
Question  3. If $X$ and $Y$ are metric spaces, with $X$ separable, does
$B(X) \times B(Y)$ generate the Borel $\sigma$-algebra on $X \times Y$?

REPLY [9 votes]: This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.
Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.
Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.
See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.<|endoftext|>
TITLE: What spaces have well known horofunctions?
QUESTION [8 upvotes]: Following Gromov, take a metric space $(X,d)$ and consider $C(X)/\mathbb{R}$ the set of continuous functions to $\mathbb{R}$ with the topology of uniform convergence on compact sets after taking the quotient by constant functions (i.e. two functions are equivalent if the difference is a constant).   Embed $X$ into this space by means of the map:
$x \mapsto f(x) = [d(x,\cdot)]$
A horofunction is an element of the closure of $f(X)$ that is not in $f(X)$.
In $\mathbb{R}^d$ all horofunctions are given by inner product with a unit vector.
In the upper-half plane model of the hyperbolic disc the function $h(z) = \log(\text{Im}(z))$ is a horofunction and all others can be obtained by composing with hyperbolic isometries.
What other spaces have well known horofunctions?
For example, are horofunctions of the model geometries $\text{Nil}$, $\text{Sol}$, and $\widetilde{\text{Sl}(2,R)}$ in dimension $3$ known relatively explicitly?
I know there is a general relationship between horofunctions and geodesics (Busseman functions) in non-positively curved spaces.  However I'm interested in spaces where one can compute relatively explicitly (e.g. spaces where one knows what the distance function looks like).

REPLY [2 votes]: One possible place to look (other than symmetric spaces) is the warped product of hyperbolic metrics, a metric of the form $dr^2+e^{C_1r} dx^2 + e^{C_2r}dy^2$, $C_1, C_2 >0$. Then $r$ is a horofunction. If one could compute the horofunction for another point in the Gromov boundary, then since the isometry group acts transitively on these other points of the Gromov boundary, one would have a computation for all points. However, I imagine computing the other horofunctions could be tricky, since I don't know how to compute the geodesics explicitly.<|endoftext|>
TITLE: Set theory within the framework of category theory
QUESTION [6 upvotes]: I started studying the basics of category theory recently, and after seeing how a great deal of group theory could be described categorically, I began to wonder if it were possible to describe set theory, or set-theoretic concepts, without reference to elements, i.e., by only using sets and functions. For example, instead of saying $x\in S$, one could say that there is a map $f:0\to S$, where $0$ is a singleton. Similarly, one could describe the power set $\mathscr{P}(X)$ by saying that for any function $f:Y\to X$, there are functions $g:0\to \mathscr{P}(X)$ and $h:\mathscr{P}(X)\to X$ such that $f=hg$. Disjoint unions can be described as coproducts, and cartesian products can be described by means of a universal property. I was wondering if it were possible to describe all of naive set theory in this way, and if so, whether any attempts have been made to do so.

REPLY [7 votes]: Yes! Lawvere's Elementary Theory of the Category of Sets (ETCS) is probably just what you want. It is a characterisation of Set up to equivalence from category theoretic principles alone. Todd Trimble explains a lot of it in some blog posting that have now migrated to the nLab, see the links here. This is what can be called a structural approach to foundations: one cares less about membership as how sets related to each other by functions. 
It isn't the only one, Mike Shulman has SEAR (Sets Elements And Relations) which takes the named entities as primitive. It relies a little more on dependent type theory. This isn't too scary, it just means you keep track of what sort of set your elements live in (so questions like, 'is $\pi = (sin:\mathbb{R} \to [0,1]$?', don't come up) Note that ETCS a priori isn't about elements, but they are recovered as in your question: maps $1 \to S$, but it is by the time one uses the axiom 'Set is well pointed' that this comes about. SEAR on the other hand has elements are primitive, but then turn out to be what you say once functions are defined and subsets an so on.
Both of these can be taken to include Choice or not (ETCS has Choice by default, but it isn't necessary). SEAR-C is what we call SEAR with Choice, and it is equivalent to ZFC, and SEAR is equiconsistent with ZF (but not equivalent).
ETCS is slightly weaker than SEAR, but this hasn't been explored too much.
One can also consider the variant on SEAR called SEPS, which uses as primitives Sets, Elements, Pairing and Subsets (pairing is like canonical categorical product - stronger than the usual product in that it is specified. I wouldn't be surprised that if, after all's said and done, it's functorial). This is equivalent to SEAR<|endoftext|>
TITLE: The vector field of a given flow
QUESTION [14 upvotes]: Let $f:(0,1)\rightarrow(0,1)$ be a map with some regularity (${\mathcal C}^1$, ${\mathcal C}^2$, ${\mathcal C}^\infty$, analytic ?). We assume that $f(t)> t$ for every $t$, and that $f'> 0$.
Does there exist a vector field $X$ over $(0,1)$, whose flow at time $t=1$ is $f$ ?
If the answer is yes (as I bet), it will complete the existence proof in MO question link text

REPLY [18 votes]: I'll rephrase Pietro Majer's answer, and also give some more information.
For any $f: (0,1) \to (0,1)$ with $f(t) > t$, the quotient space $(0,1) / \left < f \right > $
is diffeomorphic to the circle.   Choose any diffeomorphism to $S^1 =\mathbb R /\mathbb Z$, and use the vector field $d/dt$ lifted back to $(0,1)$.
This construction is unlikely to be smooth at either endpoint, unless special care is taken.
In many cases, there are established techniques to embed the germ of a diffeomorphism in a flow near a fixed point. The most widely used case  in dynamics is that if a diffeomorphisms of $\mathbb R^n$ (say, $C^\infty$ or $C^\omega$) has a fixed point $p$ at which the first derivative has no zeros on the unit circle, then it is $C^\infty$ or $C^\omega$ conjugate to a linear diffeomorphism near that point. The conjugacy is unique up to a linear map. There are many other cases where the structure is known. I don't have a reference handy, but I think any text on dynamical systems will have this, and the proof is reasonably elementary.
In the analytic case, you can solve recursively for the Taylor series coefficients, and check local convergence. 
In the 1-dimensional case at hand, if $f'(0) > 1$, then a microscopic picture of $f$ near $0$ looks very close to the linear map $x \to  f'(0) x$.  We can conjugate a well-fitting microscopic coordinate system by a high power of $f$ to make a fixed-size coordinate in which $f$ is nearly linear: the formula for the coordinate is
$u_n(x) = (f'(0))^n f^{-n}(x)$.  These converge in the $C^\infty$ topology to a coordinate that $f$ maps linearly.  Another way to think of it is that to calculate the ratios of intervals in the new measure of length, send them both near $0$ by a high power of $f^{-1}$, and measure ratios there.  It is not hard to show that the limit $u(x))$ is smooth --- it's the limit of long compositions, but most of the composing takes place near 0 where $f$ is nearly linear.
Conjugating an orientation preserving map to a linear map gives an embedding in a flow near the fixed point, since orientation preserving linear maps are  embedded in linear flows ( i.e, in the image of the the exponential map in $GL(n,\mathbb R)$).
For a diffeomorphism of the closed interval $[0,1]$, if the derivative is not 1 at either endpoint, then their is a unique smooth conjugacy to a linear map near each endpoint (since the exponential map for $GL(1,\mathbb R)$ is univalent.)   An embedding in a flow near one endpoint extends uniquely over the entire open interval:  for any point, apply $f^{-1}$ enough times to get it to inside the last fundamental domain where the flow is defined.
The definitions from the two endpoints typically conflict.  Generic diffeomorphisms of a
closed interval are not embeddable in a smooth flow.   A diffeomorphism of an interval has a smooth invariant that measures the discrepancy between these: a pair of smooth probability measures on $S^1$  that coincide if and only if the overall diffeomorphism is embeddable in a flow.   In other words: the exponential map is not locally surjective in the group of diffeomorphisms of an interval.
Even when a diffeomorphism of a closed interval is not embeddable in a flow, it might still have a functional square root, a cube root, or whatever.  A similar analysis applies  --- near a hyperbolic fixed point, it is uniquely conjugate to a linear map, and it has a unique square root.
This determines a square root in the adjoining open intervals. You can compare  definitions from different fixed points and see if they match.
There is a lot more known about algebraic relationships that can and cannot occur in groups of diffeomorphisms of the interval, and in higher dimensional analogues as well. (The question concerns the simplest case, the groups $\mathbb Z$ and $\mathbb R$). This theory has had considerable attention motivated by trying to understand the structure of smooth foliations.   For example, any finitely generated group of diffeomorphisms of a closed interval must be residually torsion-free nilpotent (the generalized Reeb stability theorem).
From this it is possible to deduce that there are many groups of homeomorphisms of the interval that are not topologically equivalent to any smooth action.<|endoftext|>
TITLE: group of Yoneda extensions and the EXT groups defined via derived category
QUESTION [12 upvotes]: Given an abelian category C, we can form the Yoneda extensions $YExt^i(X,Y)$ to the equivalent classes of $i$-extensions of X by Y. 
Given any abelian category C, we can always formulate the derived category D(C), and define $Ext^i_C(X,Y)$ to be $Hom_{D(C)}(X,Y[i])$.
Now we can naturally associate a Yoneda $i$-extension
$$0\rightarrow Y\rightarrow Z_{i-1}\rightarrow \cdots\rightarrow Z_0\rightarrow X\rightarrow 0$$
 of X by Y the element $Y[i]\leftarrow[Y\rightarrow Z_{i-1}\rightarrow \cdots\rightarrow Z_0]\rightarrow X$ of $Hom_{D(C)}(X, Y[i])$.
I am wondering when this map is inj (resp. surjective, bijective)?
We know this is an isomorphism under the assumption that C admits enough injective objects or projective objects.

REPLY [11 votes]: It is true in general and Verdier did the computation in his thesis Des Catégories Dérivées des Catégories Abéliennes Chap III Sect 3 ... the thesis is on line!<|endoftext|>
TITLE: When does Lusztig's canonical basis have non-positive structure coefficients?
QUESTION [26 upvotes]: I've heard asserted in talks quite a few times that Lusztig's canonical basis for irreducible representations is known to not always have positive structure coefficents for the action of $E_i$ and $F_i$.  There are good geometric reasons the coefficents have to be positive in simply-laced situations, but no such arguments can work for non-simply laced examples.  However, this is quite a bit weaker than knowing the result is false.

Does anyone have a good example or reference for a situation where this positivity fails?

REPLY [23 votes]: Hi, 
The following formulas are examples of non-positive structure coefficients
for non-symmetric cases which are easily verified by the algorithm presented
in Leclerc's paper "Dual Canonical Bases, Quantum Shuffles, and q-characters" 
or quagroup package in GAP4.
Professor Masaki Kashiwara told me that he has known such non-positive 
structure coefficient for $G_2$ since Shigenori Yamane found it in 1994
as treated in his master thesis at Osaka University (written in Japanese).
You can see similar negative coefficients in at least case $A_{2n}^{(2)}, D_{n+1}^{(2)}$.
Anyway, conjecture 52 in Leclerc's paper is false 
(I already told Professor Leclerc about it).
Shunsuke Tsuchioka
Notation: $G(i_1,\cdots,i_n)$ stands for the canonical basis element
corresponds to a crystal element 
$b(i_1,\cdots,i_n)=\tilde{f}_{i_n}b(i_1,\cdots,i_{n-1})=\cdots$. 
$G_2$ (1 is the short root) : 
$f_2 G(121112211)
= G(1211122211)
  + [2]G(1111222211)
  + G(2111112221)$
$  + [2]G(1211112221)
  + G(1111122221)
  - G(1112211122)
  + [2]G(1122111122)$
$C_3$ (1,2 are short roots) :
$f_3 G(23122312)
= [2]G(222333121)
  + [2]G(312222331)
  + [2]G(231222331)$
$  + [2]G(122223331)
  + G(231223312)
  + [2]G(122233312)
  - G(223112233)
  + [2]G(231122233)$
$B_4$ (1,2,3 are long roots) :
$f_1G(4342341234)
= [2]G(43344423211)
  + [2]G(43423443211)
  - G(44233443211)$
$  + [2]G(43423344211)
  + [2]G(43423442311)
  + [2]G(34234442311)$
$  + [2]G(43422334411)
  + G(43423412341)$<|endoftext|>
TITLE: Does Qcoh(X) admit a generating set?
QUESTION [13 upvotes]: Let $X$ be a scheme (or more generally a ringed space, if it works). Does $Qcoh(X)$, the category of quasi-coherent sheaves on $X$, admit a generating set? This would be useful, because then every cocontinuous functor on $Qcoh(X)$ has a right adjoint (SAFT).
If $X$ is affine, then $\mathcal{O}_X$ is a generator. I doubt that this is true in general. If $X$ is quasi-separated, perhaps the direct images of the $\mathcal{O}_U$, $U$ affine, do the job, but the naive proof does not work. If $Qcoh(X)$ does not have a generating set in general, what conditions on $X$ guarantee this?
EDIT: It is true when $X$ is concentrated, i.e. quasi-compact and quasi-separated, in particular when $X$ is noetherian (see Philipp's comment). This is already satisfying. Anyway, are there other (counter)examples?
PS: Note that this question is somehow unnatural with the background of this question; $\underline{Qcoh}(X)$, considered as a stack of abelian categories, always has a "stack-generator", namely $U \mapsto \mathcal{O}_U$. Nevertheless, I think the question above is interesting.

REPLY [9 votes]: Gabber's argument also appears in print in 
Enochs and Estrada, "Relative homological algebra in the category of quasi-coherent sheaves," Adv. in Math. 194 (2005) 284--295.<|endoftext|>
TITLE: Collection of equivalent forms of Riemann Hypothesis
QUESTION [116 upvotes]: This forum brings together a broad enough base of mathematicians to collect a "big list" of equivalent forms of the Riemann Hypothesis...just for fun. Also, perhaps, this collection could include statements that imply RH or its negation.
Here is what I am suggesting we do:

Construct a more or less complete list of sufficiently diverse known reformulations of the Riemann Hypothesis and of statements that would resolve the Riemann Hypothesis.

Since it is in bad taste to directly attack RH, let me provide some rationale for suggesting this:
1) The resolution of RH is most likely to require a new point of view or a powerful new approach. It would serve us to collect existing attempts/perspectives in a single place in order to reveal new perspectives.
2) Perhaps the resolution of RH will need ideas from many areas of mathematics. One hopes that the solution of this problem will exemplify the unity of mathematics, and so it is of interest to see very diverse statements of RH in one place. Even in the event where no solution is near after this effort, the resulting compilation would itself help illustrate the depth of RH.
3) It would take very little effort for an expert in a given area to post a favorite known reformulation of RH whose statement is in the language of his area. Therefore, with very little effort, we could have access to many different points of view. This would be a case of many hands making light work. (OK, I guess not such light work!) 
Anyhow, in case this indeed turns out to be an appropriate forum for such a collection, you should try to include proper references for any reformulation you include.

REPLY [3 votes]: Matiyasevich has reformulated RH as a computer science problem : "a particular explicitly presented register machine with 29 registers and 130 instructions never halts", see this reference .<|endoftext|>
TITLE: Compatibility of braids as a simplicial set and as a braided monoidal category
QUESTION [6 upvotes]: We can form a braided monoidal category by taking the groupoid coproduct of the Artin braid groups $B_n$. We can also make the braids into a simiplical set where the ith face operation is removing the i-th strand and the ith degeneracy operation doubling the ith strand. (These operations are not group homomorphisms, so we don't get a simplicial group.)
Are these two notions compatible? For example, can we use the tensor product of the braided monodial category to describe these face and degeneracy operations? or vice versa? is there a 2-category or something that can combine these two notions?

REPLY [11 votes]: I think the best way to think about this is in terms of operads.
Braided monoidal categories are representations of an operad $\Pi$ in the category of (small) categories. 
The category $\Pi(n)$ has objects parenthesised permutations of $\{1,\ldots,n\}$ like $(4(23))1$. The morphisms $(\sigma) \to (\tau)$ are braids $\beta \in B_n$ so that $\beta$ maps to $\tau \sigma^{-1}$ under $B_n\to S_n$. 
The operad structure 
$$
  \Pi(n) \times  \Pi(k_1) \times  \ldots \times  \Pi(k_n) \to \Pi(k_1+\ldots+k_n)
$$
is given by replacing the $i$-th sting by the braid on $k_i$-strings. 
Now any representations $\Pi(n) \to \underline{Hom}(\mathcal{C}^n,\mathcal{C}) = End(\mathcal{C})(n)$ induces a braided monoïdal structure with  

$\otimes$ corresponding to the object $(12) \in \Pi(2)$
the associativity morphism corresponding to the trivial braid $(12)3 \to 1(23)$ in $\Pi(3)$
the braiding corresponding to the morphism $(12) \to (21)$ in $\Pi(2)$ induced by the generator of $B_2 = \mathbb{Z}$.

MacLane's coherence theorem tells you that this is an equivalence. 
All the operations you're looking at come from this operad structure. 
The $\Pi(n)$ have a nice geometric interpretation. They are the fundamental groupoids of the operad of little discs $C_2(n)$ (or equivalently of $F(\mathbb{C},n)$ the spaces of configurations of $n$ points in the plane) restricted to a suitable collection of basepoints. This generalizes the classical definition $P_n = \pi_1(F(\mathbb{C},n),p)$, $B_n = \pi_1(F(\mathbb{C},n)/S_n,\overline{p})$.
I think this makes the whole picture a lot clearer because we get all of these operations as part of the same structure and we get a universal characterisation of
that structure and a geometrical interpretation for it. 
This leads to other nice considerations. For example the Grothendieck-Teichmuller group $GT$ defined by Drinfeld is the automorphism group of the (prounipotent completion) operad $\Pi$. This explains why $GT$ is universal for quasi triangular quasi Hopf algebras as their representations form braided monoïdal categories and related to the Galois group of $\mathbb{Q}$ as $GT$ appears as an automorphism group of fundamental groupoïds of the algebraic varieties $F(\mathbb{A}^1_{\mathbb{Q}},n)$.<|endoftext|>
TITLE: Geodesic metrics that admit dilatation at each point
QUESTION [13 upvotes]: Consider the  class of geodesic  metrics $g$ on manifolds, that have the following
property:  for each point $x$ there exists a neighbourhood $U_x$ and 
a smooth vector field $v_x$ in $U_x$ that vanishes at  $x$ and whose flow (for small time) dilatates $g$ by a constant factor.
Let us call such metrics dilatatable.
An obvious example is provided by an Euclidean $\mathbb R^n$, the flow of the field 
$\sum_i x_i \frac{\partial}{\partial x_i}$ dilatates the Euclidean metric by 
a constant factor. More generally one can take any Banach space.
I would like to make a guess about the structure of such metrics in general.
Guess.  Suppose $g$ on $M^n$ is dilatatable. Then there exists a triangulation of $M^n$
such that the restriction of the metric $g$ to each simplex if flat with respect
to the flat structure on the simplex, and 
$g$ is flat on the complement to the union of all  co-dimension $2$ simplexes. 
The first question is the following: was such class of metrics considered somewhere and 
is this guess correct? Are there obvious counterexamples?
Second part of the question is about examples. It is not hard to construct an example 
of such a metric, if we don't require $M^n$ to be a smooth manifold. Namely, we can take any polyhedral metric
on $M^n$, i.e. glue $M^n$ from a union of Euclidean simplexes (glue the boundaries by isometries).
Then for each point there is a conical neighbourhood, and obviously we can always scale this neighbourhood 
by the radial field emanating from $x$. So now comes the
Second question. Take a topological manifold $M^n$ of dimension $n<7$ with such a polyhedral metric. 
It is known then that such a manifold has a smooth structure (because a PL structure in dimension
up to $6$ always defines a unique smooth structure).  Is it possible to chose this smooth structure 
in such a way, that the polyhedral metric is dilatatable for the smooth structure?
The answer to this question is positive for $n=2$, but I don't know already what happen for $n=3$.
At the same time, there are non-trivial examples in higher dimensions, coming from complex geometry.
For example one can quotient some complex tori $\mathbb T^n$ by a finite group of isometries to get 
$\mathbb CP^n$, the obtained polyheral metric on $\mathbb CP^n$ is dilatatable with respect 
to the canonical complex (and hence smooth) structure on $\mathbb CP^n$.

REPLY [2 votes]: Relative to comments by Sergei Ivanov and Bill  Thurston, maybe this line of research concerning "metric spaces with dilations" or "dilation structures" provides a precise 
answer, more general than Berestovskii result.  See this introduction and dig into the biblio. 
Concerning examples related to Carnot-Caratheodory geometry and nilpotent groups (precisely: "Carnot groups"), they appear naturally as models of the (metric) tangent space to a point in a space with dilations. 
If you stand to read a more algebraic account, see emergent algebras, where 
it is proven that this is not really a metric induced phenomenon.<|endoftext|>
TITLE: Minimal conditions for the exponential law for compact-open topologies
QUESTION [7 upvotes]: What are the minimal conditions on three topological spaces $X,Y$ and $Z$ such that with the compact-open topology the map
$${(X^Y)}^Z \to X^{Y \times Z}$$
given by taking adjoints is a homeomorpism. The map sends $f: Z \to X^Y$ to $g:Y \times Z \to X$ by the relation $g(y,z)=f(z)(y)$. 
This result is known for $Z$ Hausdorff and $Y$ locally compact. I'm interested in a proposition of the form the adjoint construction is a homeomorphism of mapping spaces if and only if some statement regarding the spaces $X$, $Y$ and $Z$. 
It would also be interesting to see some counterexamples, for example for $Z$ not Hausdorff, etc.

REPLY [13 votes]: A very closely related question (and maybe the one you meant to ask?) is: which spaces $Y$ in the category of topological spaces and continuous maps are exponentiable, i.e., for which $Y$ does the functor $- \times Y: Top \to Top$ have a right adjoint? A necessary and sufficient condition is that $Y$ is core-compact, as defined at the nLab. See also the references in that article. 
There are various ways of defining core-compactness; perhaps the fastest is that the topology is a continuous lattice. It is a slightly weaker condition than local compactness (if local compactness is defined as meaning that every point has a basis of compact neighborhoods), and coincides with local compactness if $Y$ is Hausdorff. 
If $Y$ and $Z$ are core-compact, then for every $X$ one can exhibit a canonical homeomorphism 
$$(X^Z)^Y \cong X^{Y \times Z}$$ 
by abstract nonsense (since any two right adjoints to $- \times (Y \times Z)$, in particular $((-)^Y)^Z$ and $(-)^{Y \times Z}$, are canonically naturally isomorphic). 
Your question is also interesting when interpreted for locales. See Johnstone's Stone Spaces, where it is shown that a locale is exponentiable if and only if it is locally compact. 
If $Y$ is not core-compact, then it is possible to show that there is no exponential $\mathbf{2}^Y$ where $\mathbf{2}$ is Sierpinski space (two points, one open, one closed). In other words, the functor $\hom_{Top}(- \times Y, \mathbf{2})$ is not representable. I once went through the detailed argument (in the case where $Y$ is the space of rational numbers, which I think is illustrative) over at the n-Category Café, see here and the ensuing discussion.<|endoftext|>
TITLE: Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]?
QUESTION [13 upvotes]: Given a subfactor $N\to M$ and a conditional expectation $E:M\to N$,
there is a numerical invariant $Ind(E)$ associated to to this situation, called the index of $E$.
The possible values of $Ind(E)$ are 
restricted to the set {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$.
The minimal conditional expectation is
the one that minimizes the value of $Ind(E)$.
The minimal index of the subfactor is then defined to be the index of its minimal conditional expectation.
Can the minimal index take all values in {$4\cdot \cos^2(\pi/n);n=3,4,5,...$} $\cup$ $[4,\infty]$? In other words, given a real number in the above set, is there a subfactor whose minimal index is that real number?

Remark: If the factors are of type $II_1$, there is another preferred conditional expectation: the one
that is compatible with the traces. The corresponding index is called the Jones index. This is not the index I care about.
Jones' index agrees with the minimal index in the case of irreducible subfactors,
but not in general.
Jones' index is known to take all the above values. But the subfactors used in the construction are not irreducible (and one can also check that their minimal index is different from their Jones index).

REPLY [6 votes]: There is an irreducible Temperley-Lieb subfactor at every allowed index. For $n\geq 3$, it has index $4\cos^2(\pi/n)$ and principal graph $A_{n-1}$ (in fact all subfactors of index less than $4$ are irreducible), and for every $r\geq 4$, it has index $r$ and principal graph $A_\infty$. Doesn't that do the job by your remark?<|endoftext|>
TITLE: Distinguishing pro-finite completions
QUESTION [12 upvotes]: Assume that we have two residually finite groups $G$ and $H$. Which properties of $G$ and $H$ could be used to show that their pro-finite (or pro-p) completions are different?
I asked a while ago in the group-pub mailing list whether finite presentability is such a property but Lubotzky pointed out that it is not the case. A finitely presented and an infinitely presented group can have isomorphic pro-finite completions.

REPLY [5 votes]: There's a theorem that two finitely generated residually finite groups have the same profinite completions if and only if they have the same finite quotients. A reference for the statement of this is Theorem 2 of this paper, but they cite Ribes and Zalesskii for the proof.<|endoftext|>
TITLE: Generalizing a problem to make it easier
QUESTION [153 upvotes]: One of the many articles on the Tricki that was planned but has never been written was about making it easier to solve a problem by generalizing it (which initially seems paradoxical because if you generalize something then you are trying to prove a stronger statement). I know that I've run into this phenomenon many times, and sometimes it has been extremely striking just how much simpler the generalized problem is. But now that I try to remember any of those examples I find that I can't. It has recently occurred to me that MO could be an ideal help to the Tricki: if you want to write a Tricki article but lack a supply of good examples, then you can ask for them on MO. 
I want to see whether this works by actually doing it, and this article is one that I'd particularly like to write. So if you have a good example up your sleeve (ideally, "good" means both that it illustrates the phenomenon well and that it is reasonably easy for others to understand) and are happy to share it, then I'd be grateful to hear it. I will then base an article on those examples, and I will also put a link from that article to this MO page so that if you think of a superb example then you will get the credit for it there as well as here.
Incidentally, here is the page on this idea as it is so far. It is divided into subpages, which may help you to think of examples.
Added later: In the light of Jonas's comment below (I looked, but not hard enough), perhaps the appropriate thing to do if you come up with a good example is to add it as an answer to the earlier question rather than this one. But I'd also like to leave this question here because I'm interested in the general idea of some kind of symbiosis between the Tricki and MO (even if it's mainly the Tricki benefiting from MO rather than the other way round).

REPLY [8 votes]: I remember there was an exercise in linear algebra:

Find the determinant of the following matrix:
\begin{pmatrix}
0 & a & a & \cdots & a\\
b & 0 & a & \cdots & a\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
b & b & \cdots & 0 & a\\
b & b & \cdots & b & 0
\end{pmatrix}

A simple solution is to generalize this to a function:
$$f(x) = \det\begin{pmatrix}
x & a + x & a + x & \cdots & a + x\\
b + x & x & a + x & \cdots & a + x\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
b + x & b + x & \cdots & x & a + x\\
b + x & b + x & \cdots & b + x & x\end{pmatrix}
$$
Namely, add $x$ to every entry of the matrix.
It is then obvious that:

$f$ is a linear function of $x$;
$f(-a)$ and $f(-b)$ are easily computed;
the original determinant is just $f(0)$.<|endoftext|>
TITLE: Finite dimensional automorphic representations of a definite quaternion with prime discriminant and Hecke action
QUESTION [9 upvotes]: Before stating the questions that I have, which are very specific and probably not so interesting to someone who has never thought about these things, I need to introduce some notation.
Let $p$ be any prime, and let $D$ be the quaternion algebra over $Q$ ramified precisely at $p$ and infinity. Choose a maximal order $R$ inside $D$. For any prime $\ell\neq p$, fix an isomorphism of $D_\ell:=D\otimes Q_\ell$ with the algebra $M_2(Q_\ell)$ in such a way that
the maximal compact subring $R_\ell:=R\otimes Z_\ell$ corresponds to $M_2(Z_\ell)$.
If $N$ is an integer $>0$ not divisible by $p$, let $U_\ell(N)$ be the subgroup of $R_\ell^\star\simeq GL_2(Z_\ell)$ given by those matrices whose bottom row is congruent to $(0$ $1)$ modulo $N$ (equivalently, modulo the highest power of $\ell$ dividing $N$).
The ring $R_p:=R\otimes Z_p$ has a unique maximal, two-sided principal ideal $(\pi)$ generated by any uniformizer $\pi$, the residue field $R/(\pi)$ is a finite field with $p^2$ elements. We let $R_p^\star(1)$ denote the subgroup of $R_p^\star$ given by the units that are congruent to $1$ modulo $(\pi)$.
Let now $D^\star$ be the multiplicative group of $D$, viewed as an algebraic group over $Q$. For any integer $N>0$ not divisible by $p$, we are going to define an open subgroup $U(1,N)$ of the group $D^\star_A$ of points of $D^\star$ valued in $A$, the adele ring of $Q$. Namely $U(1,N)$ is the product of all the $U_\ell(N)$, for $\ell\neq p$; of $R_p^\star(1)$;
and of the full (connected) component at infinity $D^\star_\infty$.
Consider the space $S(1,N)$ of complex valued functions on the double coset $D^\star\backslash D^\star_A/U(1,N)$, which is known to be finite. For any prime $\ell\neq p$ there is an Hecke operator $T_\ell$ acting on $S(1,N)$ that can be defined in terms of double cosets in the usual way.
(Let $\alpha_\ell$ be the matrix whose top row is $(\ell$ $0)$ and whose bottom is $(0$ $1)$; decompose $U_\ell(N)\alpha_\ell U_\ell(N)$ as a finite union of left cosets $\gamma_i U_\ell(N)$; for $f\in S(1,N)$ define $T_\ell(f)(x)=\sum f(x\gamma_i)$).
Let $V$ be the vector space of locally constant, complex valued functions on $D^\star_A$ that are left invariant by $D^\star$. Observe that $S(1,N)$ can be viewed as a finite dimensional subspace of $V$. Right translation defines an admissible representation of $D^\star_A$ on $V$ which is known to be completely decomposable into a discrete direct sum of irreducible admissible representations of $D^*_A$. If $f\in S(1,N)$, then denote by $V_f$ the smallest subspace of $V$ that is stable by $D^\star_A$.
Questions:
1) Let $f\in S(1,N)$, for some $N$. Is it true that the space $V_f$ is finite dimensional
if and only if $f:D^\star_A\rightarrow C$ factors through the reduced norm map $Nr:D^\star_A\rightarrow A^\star$?
2) Does the subspace of $S(1,N)$ given by those functions that factor through the reduced Norm admit an Hecke stable complement?
3) Is the action of $T_\ell$ on $S(1,N)$, for $\ell\nmid pN$, semisimple?
4) Assuming that 2) holds, and letting $S_0(1,N)$ be such complement, how do we relate the C-subalgebra $T_0(N)$ of End($S_0(1,N)$) generated by the Hecke operators $T_\ell$, with $\ell\nmid pN$, to a C-algebra of Hecke operators acting on weight 2 cusps forms of a certain level? What I mean is: out of the J-L correspondence, can we read off an isomorphism between $T_0(N)$ and some Hecke algebra coming from classical modular forms?
Thanks.
[EDIT: In the 2nd and 3rd lines above "Questions:" I should have probably have said "discrete Hilbert direct sum", the "direct sum" being only dense in V]

REPLY [6 votes]: 1) Yes, I think that's true. I guess it follows relatively easy from the statement that an automorphic representation of the algebraic group $D^\times$ is finite-dimensional iff it's 1-dimensional and factoring through the norm. This latter statement probably follows from strong approximation applied to (the adelic points of the algebraic group associated to) the norm 1 elements of $D^\times$, or you can apply Jacquet-Langlands and move to $GL(2)$ and basically use the standard classification of automorphic representations there.
2) Yes, I think it does. There's some analogue of the Petersson inner product here, right? It's just a combinatorial thing and you can use this. The space $S(1,n)$ is not a mysterious thing: multiplicity 1 etc is all true in this setting and the space is a sum of generalised eigenspaces just like classical spaces of cusp forms.
3) Yes. Use either Jacquet-Langlands and classical results, or mimic the standard proof for $GL(2)$ using the pairing on $S(1,N)$.
4) This question has a bit more meat to it. You seem to be asking what the Jacquet-Langlands correspondence is explicitly. I know something this but it's a bit messy. Let me write down some stuff; the answer isn't quite as "clean" as you'd like it to be though, perhaps.
In $S_0(N)$ you're seeing automorphic representations which have a fixed vector under $R_p^*(1)$. These come in two flavours. The first are those which actually have a fixed vector under all of $R_p^*$. These contribute one dimension, locally, to $S_1(N)$ and under J-L correspond to forms of level $Np$ which are new at $p$ with trivial character. The second sort are those without an $R_p^*$-fixed vector. These are 2-dimensional (I mean their $\pi_p$ is 2-dimensional) [Edit: as Tomasso pointed out, this isn't true. There are some more 1-dimensional ones factoring through the local norm map and coming from tamely ramified but not unramified quasicharacters; let's ignore them in what follows.] and all of $\pi_p$ is fixed by $R_p^*(1)$, so here you're getting 2-dimensional spaces in $S_0(N)$ on which Hecke is acting as scalars (so a super-naive version of multiplicity 1 is failing; what's really going on is that the classical theory of oldforms and newforms works in a completely different way to the behaviour of $S_0(N)$ at $p$). Applying JL, if memory serves, you see forms which are supercuspidal at $p$; their level will be something like $Np^2$, their character will be something like the norm of the character of $\mathbf{F}_{p^2}$ associated to the form on $S_0(N)$. So the answer to your question is "yes, you can write down a classical space of forms that corresponds to $S_0(N)$" but it's not pleasant to write it down---you have to understand the possibilities of what forms of level $Np^2$ look like, and only choose certain of them. It's much easier to understand from the point of view of representation theory, either of $GL(2,\mathbf{Q}_p)$ or the local Weil group. 
All this is very well-understood by lots of people here, so feel free to correct me or ask more.<|endoftext|>
TITLE: Half or more elements order two implies generalized dihedral?
QUESTION [19 upvotes]: The "generalized dihedral group" for an abelian group A is the semidirect product of A and a cyclic group of order two acting via the inverse map on A. A thus has index two in the whole group and all elements outside A have order two. Thus, at least half the elements of any generalized dihedral group have order two.
My question is the converse: if half or more the elements of a finite group G have order two, is it necessary that G is either an elementary abelian 2-group or a generalized dihedral group? [Note: Actually nontrivial elementary abelian 2-groups are also generalized dihedral, they're an extreme case. Also, note that the direct product of a generalized dihedral group with an elementary abelian 2-group is still generalized dihedral, because the elementary abelian 2-group can be pulled inside the abelian part.]
I have a proof when the order of G is twice an odd number m. In that case, there is a short an elegant elementary proof -- we consider the set of elements that can be written as a product of two elements of order two and show that this is a subgroup of order m, then show that any element of order two acts by the inverse map on it, and hence the subgroup is abelian. It can also be thought of as a toy example of Frobenius' theorem on Frobenius subgroups and complements (though we don't need Frobenius' theorem).
However, I am having some difficulty generalizing this, mainly because there are elements of order two that are inside the abelian group. 
Although I have a number of possible proof pathways, I'll refrain from mentioning them for now because the actual proof is likely to be much simpler and I don't want to bias others trying the problem.

REPLY [24 votes]: $D_8\times D_8$ is such a group (it has 35 involutions) that fails to be either elementary abelian or generalized dihedral.
There is an actual classification of groups with at least half the elements being involutions; it was first done by Miller.  A modern reference would be the paper by Wall, "On Groups Consisting Mostly of Involutions". A simpler treatment, though with a different goal in mind, is given in Liebeck and MacHale's paper "Groups with Automorphisms Inverting Most Elements".<|endoftext|>
TITLE: Complete metric on the space of Jordan curves?
QUESTION [18 upvotes]: I was interested in putting a complete metric on the space of Jordan curves. Say, just planar Jordan curves contained in $B(\bar{0}, 2) \backslash B(\bar{0}, 1)$ which separates $\bar{0}$ and infinity.
As we can see, the standard Hausdorff metric does not do the job, for example we can have a Cauchy sequence of Jordan curves with thinner and thinner 'fingers', the limit is a circle with a spike, hence not a Jordan curve. (see Figure below)

Hence in some sense, the metric needs to not only 'see' the curve but also what it encloses. I looked a bit into the literature and didn't find anything...Does anyone know such a metric? Any ideas?
An attempt of mine in constructing the metric: 
First we can consider the plane as in the Riemann sphere, so our space is merely the set of Jordan curves that separates the unit disk around the North pole and that around the South pole. For each curve we have (up to a Mobius automorphism of the disc) two Riemann maps from the standard unit disc to the North and South hemisphere separated by the curve.
Define the distance between two curves to be the minimum distance between the pair of 'North hemisphere Riemann maps', plus that of the South hemisphere. It's easy to check that this does give a metric. In our 'finger' case above, the $ \{ C_n \} $ is not Cauchy in our metric (because Riemann map will map less and less of the disc inside the finger so there is a subsequence of distance finger-length apart). 
I strongly believe this metric is indeed complete, but can't find a good argument.
What can be done is, given a Cauchy sequence of Jordan curves under the above metric, pick a pair of distance minimizing Riemann map for each curve (such thing always exist by a simple compactness argument) then apply Carathéodory, all Riemann maps extends to the homeo on the boundary, so we can take the pointwise limit of the boundary map. Do the same for both North and South hemisphere. All we need to show is this limit is still injective. (I think it should be because if not, either North or South will have a infinitely thin 'neck', so the Riemann maps on that half will not be Cauchy).
I ran into many topological difficulties when trying to make an argument for this (mainly because we know nothing about the limiting curve and it can be really complicated)...Any ideas on how one might go about that would be greatly appreciated. :-P

Figure for my comment on professor Thurston's response:

---A simple case to illustrate that the two Riemann maps does not match up on the equator. i.e. radial lines in the North hemisphere does not 'want' to pass through the thin tube and land on the tip, but radial lines in the South hemisphere does not care. Hence the pre-image of the black arc will be very short in the Northern disc but normal in the Southern disc. However, there of course exists unique homeo $h$ on the boundary to glue them up. I do not know how would that give a homeo from the sphere to itself, though.

REPLY [13 votes]: There is a very good theory of the boundary behavior for Riemann maps, the Caratheodory theory of ends.  Riemann maps for an open set extend continuously to the boundary of a disk provided the frontier is locally connected.
I assume your definition is in terms of the Riemann maps that fix the north or south pole and have derivative a positive real number at those points.  If so, then Riemann maps depend continuously in the $L^\infty$ metric on Jordan curves as you have described, so this gives a metric.
Is the metric complete?  A sequence of the pair (lower hemisphere map, upper hemisphere map) that is Cauchy in the uniform ($L^\infty$) topology converges to a pair of continuous maps that agree on the equator, so the glued maps give a continuous image of the sphere that is a homeomorphism at least in the complement of the equator.  If it is not injective, the preimages of points would need to be intervals, otherwise the topology would be destroyed.
But that's impossible.  If you push forward the measure $ds$ on the equator by a Riemann map, it never has atoms.  It is the same as hitting measure for Brownian motion in the image:  if you start at the north pole and follow a Brownian path, where does it first arrive at the boundary?  This is always a diffuse measure.
So, you're right:  it's a complete metric on the set of Jordan curves you described.
Note.  Any quasisymmetric map (I won't define it here, but Holder is sufficient) from the circle to the circle arises from a pair of Riemann maps for a Jordan curve, although not necessarily in the annulus you described, and the Jordan curve in this case is known as a quasi-circle.  However, there is no known good characterization of which gluing maps from the circle to the circle give Jordan curves in general.  Continuity is not sufficient:   there are counterexamples using things that are locally (for example) the graph of $\sin(1/x)$ plus the interval $[-1,1]$ on the $y$-axis, where the gluing map for the Riemann maps to the two sides extends continuously across the discontinuity of the graph.  Better characterizations of what gluing maps give what topology and geometry are very hard, but of great interest in complex dynamics and some other subjects.<|endoftext|>
TITLE: Roadmap for Mirror Symmetry
QUESTION [32 upvotes]: I am interested in learning Mirror Symmetry, both from the SYZ and Homological point of view. I am taking a reading course in Mirror Symmetry, which will focus on the SYZ side. 
I know basic Complex geometry, Kahler manifolds, Symplectic manifolds in the geometric side and also reading some material for my course on SYZ conjecture. My major concern is Homological side, about which I have little knowledge.
I am seeking a list of good references for SYZ conjecture, Homological Mirror Symmetry, physics of the theory, modern developments and on its relation to other areas of mathematics and some original papers (preferably in Chronological order).
What are your views about the Claire Voisin's book on Mirror Symmetry. 
And what is the present status of research in Mirror Symmetry, I mean what type of problems are people working on.

REPLY [14 votes]: Can't really say anything about the physics ... :-)
But for Kontsevich's HMS conjecture, my personal (very biased) all-time-favorite list is:

Two great survey papers to start with are: "A beginner's introduction to Fukaya categories" and "A symplectic prolegomenon". 
Following that, there are Denis's notes from a graduate class he taught two times. Notes can be found in: 18.969 (MIT)/Math 277 (Berkeley). 
Another great source is Nick Sheridan's lectures at IAS and Jussieu summer school. They are extremely clear and also include some overview of the operadic background.
James Pascaleff's class M392C (UT Austin) covers a lot of the basic aspects of Lagrangian Floer (co)homology (which is a neccesary component in defining the Fukaya category) beyond the usual 1-hour lecture. He also connects the actual theory with the standard heurestics ("Floer theory is Morse theory for the action functional!"). 
Sheel's graduate class Math 257B has a really nice summary of the analysis and algebra needed to define a Fukaya category with an eye toward LG-models
Seidel's notes from a topics class on equivariant mirror symmetry ("Lectures on Categorical Dynamics and Symplectic Topology") include a wealth of material on almost every subject you can imagine from Derived Picard groups, via Hochschild homology and cohomology, and to the actual setup of a Fukaya category for the surface case ... all in modular, concise (and dense) lectures.
Hiro Lee Tanaka taught a graduate class on Fukaya categories called: Fukaya Categories, Sheaves, and Cosheaves. It includes some nice examples of HMS, and also discusses the bigger abstract framework (like TFT's, $(\infty,1)$-categories, CY-categories,...)
For SYZ (besides Auroux's class), Siu Cheung Liu taught a couple of classes on mirror symmetry for an SYZ perspective a couple of years ago. The notes are on his website. 
A really nice book on derived categories is "Fourier-Mukai transforms in Algebraic Geometry". Sheel Gantara's thesis (more generally, the "Preliminaries" section in all the papers of Gantara-Perutz-Sheridan) provide a lot of background and references on $A_\infty$-algebra and noncommutative geometry with a mirror symmetry application in mind.

After that, one should probably start reading research papers (e.g. Paul's early papers like "A long exact sequence for symplectic Floer cohomology" and "Graded Lagrangian submanifolds" are a good source for the nuts and bolts of Floer theory, culminating in the definition of the Fukaya category in the exact setting). 
To understand the Fukaya category for a closed, non-monotone, symplectic manfold you will probably need virtual perturbation techniques. But this is a different story altogether ....<|endoftext|>
TITLE: Image of the cuspidal subgroup of J_0(N) in J_1(N)
QUESTION [8 upvotes]: Let N be a prime integer. We know that the element $c=(0)-(\infty)$ generates the torsion subgroup of $J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

REPLY [5 votes]: The fact you mention about $(0) - (\infty)$ generating the torsion subgroup of $J_0(N)$ is Theorem 1 (Ogg's conjecture) on the first page of Mazur's paper "The Eisenstein Ideal". I recommend you actually read this paper. If you get as far as page 2, you will find a "Theorem 2 (twisted Ogg's conjecture)" which concerns the Shimura subgroup $\Sigma$. The construction of this subgroup essentially identifies it with the kernel of $J_0(N) \rightarrow J_1(N)$, and Proposition 11.6 of ibid. shows that $\Sigma$ is of multiplicative type, so BCnrd's remarks apply to the general case.<|endoftext|>
TITLE: Prof. Murty's B. Sc. Thesis
QUESTION [7 upvotes]: Can any of you guys help me to find out if there is a retrodigitized copy of M. Ram Murty's 1976 thesis available on the online database of Carleton University Library?
I really hope this question is appropriate for MO. Thanks in advance for your replies.
Sources

Carleton University Library
M. Ram Murty. On the existence of euclidean proofs of Dirichlet's theorem on primes in arithmetic progressions. B. Sc. Thesis, 1976, Carleton University Library.

UPDATE (approx. one year later). I contacted the person in charge of the theses at the university library, she said there wasn't a copy of the said work in the central library and contacted the Math Dept. to see if they had a copy of it, but they answered in the negative. Is there a rationale for this situation?
ANOTHER UPDATE (Molley Malone Day 2017). Fellow MO-ers: take a look at what I have just come across while taking a look at the table of contents of the latest issue of the Nieuw Archief voor Wiskunde:
http://www.mast.queensu.ca/~murty/euclidean.pdf
I can't believe my eyes!

REPLY [10 votes]: If you are interested in the contents: the main part of Ram Murty's B.Sc. Thesis might be in his paper
Primes in certain arithmetic progressions, Journal of the Madras University, (1988) 161-169,  which is available online at
http://www.mast.queensu.ca/~murty/euclid.dvi
Also, another paper may perhaps be of interest (more abstract, an extension).
Primes in Certain Arithmetic Progressions,
Ram Murty and Nithum Thain.
Funct. Approx. Comment. Math. Volume 35, Number 1 (2006), 249-259.<|endoftext|>
TITLE: A sum involving irreducible characters of the symmetric group 
QUESTION [5 upvotes]: Recently, during the research, I came across a sum, denoted by $H(n,L)$, involving irreducible characters of the symmetric group,
\begin{equation}
H(n,L)\colon=\sum_{Y_{i,j,w}} \frac{\chi^{Y_{i,j,w}([2^n])} \chi^{Y_{i,j,w}}(\tau)}{\chi^{Y_{i,j,w}}([1^{2n}])} s_{Y_{i,j,w}}(1,\ldots,1).
\end{equation}
Notations: 
Let $S_{2n}$ denote the symmetric group of all permutations over $2n$ objects. Let $\chi^{Y}$ denote the irreducible character of $S_{2n}$ labeled by Ferrers diagram (or shape) $Y$. 
Let $Y_{i,j,w}$ denote a special type of Ferrers diagram having one row of $j+w+1$ boxes, one row of $j+1$ boxes, $i-1$ rows of two boxes and $w$ rows of one boxes, where  $i\geq 1$, $j\geq 1$ and $i+j+w=n$. 
Let $[2^n]$ and $[1^{2n}]$ denote the conjugacy classes of $S_{2n}$ having the cycle types $[2^n]$ and $[1^{2n}]$, respectively. Set $\tau = (1,\ldots,n) \circ (n+1,\ldots,2n)$, a permutation of cylce type $[n^2]$. 
Let $s_{Y_{i,j,w}}(x_1,\ldots,x_L)$ denote the Schur-polynomial of $Y_{i,j,w}$ over $L\geq 2n$ indeterminants.
The sum $H(n,L)$ runs over all the Ferrers diagrams $Y_{i,j,w}$ such that $i\geq 1$, $j\geq 1$ and $i+j+w=n$.
Using the Murnaghan-Nakayama Rule, we obtain $\chi^{Y_{i,j,w}}(\tau)=2 (-1)^w$. The Schur-polynomial can be experssed as $s_{Y_{i,j,w}}(1,\ldots,1)= \prod_{(p,q)\in Y_{i,j,w}}\frac{(L-p+q)}{h_{pq}} $, where $h_{pq}$ denote the hook length at position $(p,q)$ of $ Y_{i,j,w}$. Utilizing the hook length formula, we have $\chi^{Y_{i,j,w}}([1^{2n}]) =\frac{(2n)!}{\prod_{(p,q)\in Y_{i,j,w}}h_{pq} }$. Therefore, we can simplify the sum $H(n,L)$
\begin{equation}
H(n,L)=\frac{2}{(2n)!} \sum_{Y_{i,j,w}} (-1)^w \chi^{Y_{i,j,w}}([2^n])
 \prod_{(p,q)\in Y_{i,j,w}} (L-p+q)
\end{equation}
We can also use the Murnaghan-Nakayama Rule to compute $\chi^{Y_{i,j,w}}([2^n])$. But I can't find a formula! We can also show that the character values $\chi^{Y_{i,j,w}}([2^n])$ have the property: when $n$ is odd, $\chi^{Y_{i,j,w}}([2^n])=-\chi^{Y_{j,i,w}}([2^n])$, and when $n$ is even, $\chi^{Y_{i,j,w}}([2^n])=\chi^{Y_{j,i,w}}([2^n])$. For specific $n$, the  sum $H(n,L)$ is a polynomial of $L$. A few examples are listed below
\begin{equation}
H(2,L) = \frac{1}{6} L^2 (L^2-1) 
\end{equation}
\begin{equation}
H(3,L) = \frac{1}{20} L^3 -\frac{1}{20} L^5
\end{equation}
\begin{equation}
H(4,L) = \frac{1}{140} L^2 -\frac{131}{5040} L^4+\frac{47}{2520}L^6+\frac{1}{5040} L^8
\end{equation}
Questions:
I am wondering if there exists a formula for these character values $\chi^{Y_{i,j,w}}([2^n])$. Are there any other ways to compute this sum $H(n,L)$ smartly? Is there a formula for the sum $H(n,L)$? Any things about these character values $\chi^{Y_{i,j,w}}([2^n])$ and sums $H(n,L)$ would also be appreciated.

REPLY [10 votes]: For any irreducible character $\chi^\lambda$ of $S_{2n}$, the value
$\chi^\lambda([2^n])$ can be computed as follows. If $\lambda$ has a
nonempty 2-core, then $\chi^\lambda([2^n])=0$. Otherwise $\lambda$ has
a 2-quotient $(\mu,\nu)$, where $\mu$ and $\nu$ are partitions
satisfying $|\mu|+|\nu|=n$. Say $\mu$ is a partition of $k$, so $\nu$
is a partition of $n-k$. Then $\chi^\lambda([2^n])=\pm{n\choose k}f^\mu
f^\nu$, where $f^\rho$ denotes the dimension of the irrep of $S_n$
indexed by $\rho$ (given explicitly by the hook-length formula). The
sign is $(-1)^{m/2}$, where $\lambda$ has $m$ odd parts. This result
follow from the theory of cores and quotients, e.g., Section 2.7 of
James and Kerber, The Representation Theory of the Symmetric Group.<|endoftext|>
TITLE: Existence of finite index torsion free subgroups of hyperbolic groups 
QUESTION [11 upvotes]: Question. Is it true that each infinite hyperbolic group 
has a torsion free subgroup of finite index?
Are there counterexamples, or positive results for some large subclasses of hyperbolic groups? 
For example, is the answer positive for orbifold fundamental groups of negatively curved orbifolds? More precisely, I am interested the most in the case of orbifolds with locally $CAT(0)$ metric. I guess it will be hard to construct a counterexample in this category. 
Related question. Is it known that every hyperbolic group has a non-trivial subgroup of finite index? 
Just to recall, a definition of hyperbolic group is here  http://en.wikipedia.org/wiki/Hyperbolic_group .
Added. Note, that every hyperbolic group is finitely presented (thanks to Sam Nead).

REPLY [21 votes]: This is a well known open problem. The following properties are equivalent 
a) Every hyperbolic group is residualy finite 
b) Every hyperbolic group has a finite index torsion-free subgroup. 
The proof is either here: Olʹshanskiĭ, A. Yu.
On the Bass-Lubotzky question about quotients of hyperbolic groups. 
J. Algebra 226 (2000), no. 2, 807--817 or here: Kapovich, Ilya; Wise, Daniel T. The equivalence of some residual properties of word-hyperbolic groups.  J. Algebra  223  (2000),  no. 2, 562--583 or can be given by exactly the same methods as in these two papers (I do not remember exactly which of these three possibilities hold).<|endoftext|>
TITLE: Irrationality of $ \pi e, \pi^{\pi}$ and $e^{\pi^2}$
QUESTION [45 upvotes]: What is known about irrationality of $\pi e$, $\pi^\pi$ and $e^{\pi^2}$?

REPLY [47 votes]: I believe most such questions are still very far from being resolved. 
Apparently, it is not even known if $\pi^{\pi^{\pi^\pi}}$ is an integer (let alone irrational).<|endoftext|>
TITLE: Groups and rings which are not sets
QUESTION [10 upvotes]: An algebraic structure such as a group, ring, field, etc. is usually defined to be a set with some operations satisfying certain properties.  I am curious what, if anything, goes wrong when the underlying collection of elements is not a set.  I don't have any particular example in mind, but I know some exist (in particular the surreals).  I'm just asking out of curiosity and I'm not sure where one would look up such a thing -- I am not a set theorist or logician by any stretch.
For a concrete question: what are the most elementary algebra results (on the level of Dummit and Foote, say) which can fail for groups or rings with "too many elements"?  Perhaps things like the existence of maximal ideals in commutative rings?

REPLY [5 votes]: One issue that might come up is that some standard constructions won't make sense, typically those that involve looking at equivalence classes.  For example, if you have a set group and you take a quotient by a subgroup, you're resulting object is a collection of equivalence classes, in this case a set of sets.  But when trying to do the same thing with a class group, you might end up with a class of classes, or a set of classes, which is obviously no good.  Scott's trick is usually provides a suitable work-around for this problem: when some of your equivalence classes are proper classes, use the elements of minimal (von Neumann) rank in the equivalence class in place of the entire equivalence class.<|endoftext|>
TITLE: Why do we care about the Hilbert scheme of points?
QUESTION [32 upvotes]: If $X$ is a scheme, the Hilbert scheme of points $X^{[n]}$ parameterizes zero dimensional subschemes of $X$ of degree $n$.  


Why do we care about it?


Of course, there are lots of "in subject" reasons, which I summarize by saying that $X^{[n]}$ is maybe the simplest modern moduli space, and as such is an extremely fertile testing ground for ideas in moduli theory.  But it is not clear that this would be very convincing to someone who was not already interested in $X^{[n]}$.  
The question I am really asking is: 


Why would someone who does not study moduli care about $X^{[n]}$?


The main reason I ask is for the sake of having some relevant motivation sections in talks.  But an answer to the following version of the question would be extremely valuable as well: 


What can someone who knows  a lot  something* about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects?



*reworded in light of the answer of Nakajima

REPLY [8 votes]: One can use Hilbert schemes of points on a surface $S$ to detect sigularities of embedded curves $C \subset S$. For instance let $\xi \in S^{[3]}$ be a "fat point", i.e. a subscheme of $S$ isomorphic to $Spec(\mathbb{C}[x,y]/(x^2,xy,y^2))$, then $C$ contains $\xi$ if and only if $C$ has a singularity at $supp(\xi)=\{ p \}$.
In a recent preprint Thomas, Kool and Shende (http://front.math.ucdavis.edu/1010.3211) use this method to proof Goettsche's conjecture about counting singular members of linear systems on surfaces. The basic idea is to write the counting problem as an intersection product of class on  $S \times S^{[n]}$ and use results of Ellingsrud-Goettsche-Lehn (building on work of Nakajima) which describe the intersection theory on $S^{[n]}$.
REMARK: I just got aware of the fact, that V. Schende who coauthored the above paper is actually asking the question. So he probably knows about this application.<|endoftext|>
TITLE: Canonical geometric examples
QUESTION [10 upvotes]: The proofs without words post has some great entries.  I'm interested in a similar concept: examples where a problem in math or physics is accompanied by a geometric figure that illuminates some key relationship in a way that formulas might not.  These wouldn't be "proof by pictures," since they may not constitute a proof at all.
One example is the relationship between a quantity that varies inversely to the square of the distance to a source, and surface area of a sphere centered at the source.  Such an illustration gives insight into gravity, for example; that it behaves as light intensity does.  This is more significant than merely that "$g=mM/r^k$ for some $k$, and that $k$ happens to be $2$".
Another example is the integral over $\mathbb R$ of $1/(1+x^2)$ which is $\pi$.  My first experience with this calc problem was as a child, watching the landscape from the window of a moving car on the highway.  A tree, say 1 unit from the road, spins (relative to me, of course) at a rate of $1/(1+x^2)$ (where x is the position on the road), as can be seen from drawing a couple triangles.  The total spin is $\pi$.
I suppose there are many examples similar to this second one from calc books (and many from complex analysis), but this one stuck out in my mind as being extremely simple geometrically.  
I realize the question isn't precise, and most of math has some accompanying geometry.  But I'd love to see other typical examples where people have found a geometric "explanation" extremely simple or crucial to an otherwise opaque problem.

REPLY [3 votes]: The chapter of the Dimensions video series about the Hopf fibration is really nice. It visualizes, that the quotient $S^1\backslash S^3$ is  $S^2$.<|endoftext|>
TITLE: Does a compact negatively curved manfiold of  dimension 4 admit a cover of finite degree?
QUESTION [13 upvotes]: A $3$-dimensional compact manifold of negative sectional curvature admits (by geometrisation?) a metric of curvature $-1$, and so its fundamental group has subgroups of finite index. I wonder if an analagous question is open already in dimension $4$ -- i.e. it is not known that $M^4$ with negative sectional curvature always have a cover of finite degree? Are there any positive results in this direction?
This a folow up to question that turned up to be open
Existence of finite index torsion free subgroups of hyperbolic groups

REPLY [18 votes]: This is a well-known open problem. In fact, there are very few tools for studing general negatively curved manifolds. Even in dimension 3 it is unknown (I think) how to prove existence of proper finite index subgroups without using the geometrization. Geometrization implies residual finiteness of f.g. 3-manifold groups, and hence existence of proper finite index subgroups. As for positive results, lattices in semisimple Lie groups are residually finite. There is only one known method of constructing compact negatively curved manifolds that are not homotopy equivalent to locally symmetric ones, namely, branched covers (with examples given by Mostow-Siu, Gromov-Thurston, and Deraux). I do not know the answer to your question for these examples.<|endoftext|>
TITLE: Freshman's definition of sin(x)?
QUESTION [12 upvotes]: I would like to know how you would rigorously introduce the trigonometric functions ($\sin(x)$ and relatives) to first year calculus students. Suppose they have a reasonable definition of $\mathbb{R}$ (as Cauchy closure of the rationals, or as Dedekind cuts, or whatever), but otherwise require as few concepts as possible.
Some approaches I can think of are:

The "geometric way": $\sin(x)$ is the ordinate, on the usual unit-radius "trigonometric circle" in the $xy$-plane, of the end point of a circle arc of length equal to $x$.
The "power series way": define $\sin(x)$ as $\Sigma_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$.
The "complex exponential way": let $\exp:\mathbb{C}\rightarrow\mathbb{C}^{*}$ be the unique homomorphism of groups (blah blah), and define $\sin(x)$, $x$ real, to be the imaginary part of $\exp(ix)$.
The "differential equation way": $\sin(x)$ is the unique function $u(x)$ of class $\mathcal{C}^{\infty}$ such that $u''+u=0$, $u(0)=0$ and $u'(0)=1$.

Unfortunately, it seems to me that each of the above approaches has some drawbacks (developing the elementary properties of trigonometric funcions from some of these definitions may be not so straightforward), and need some "non elementary" notions to be introduced, where by "non elementary" here I mean notions involving e.g. the concept of limit or of derivative. One would like the standard functions like $\sin(x)$, $\cos(x)$ and $\exp(x)$ to be already available to the students before introducing limits, derivatives or integrals, let alone power series or differential equations.
Edit (example): when I was a first year student, the reals had been introduced axiomatically (in disguise) as an [but it was implicitely assumed that it was unique] ordered field with the "sup" property; but this is irrelevant: a lot of undergrads see the definition of $\mathbb{R}$ via Dedekind cuts [which is the definition is usually given in second or 3rd year of high school]. Then $\sin(x)$ was introduced as in 1 (geometric way). Then limits, continuity etc. were introduced (so, it made sense to ask "find the limit of $\sin(x)$ as $x\rightarrow0$"). My point is that the "geometric" definition 1 is actually cheating, as it already requires limits and differentiation: what is the "arc length" of the circle otherwise?
Edit: btw, I don't have to teach calculus to anybody now, I just asked myself this question by reading other m.o. questions related to teaching.

REPLY [12 votes]: I am fond of distinguishing between the "pre-rigorous", "rigorous", and "post-rigorous" phases of mathematical education, see
http://terrytao.wordpress.com/career-advice/theres-more-to-mathematics-than-rigour-and-proofs/
For the "pre-rigorous" stage (which, in the US, is basically everything up to undergraduate calculus), I don't see a pressing need for necessarily introducing and working with a concept (e.g. the sine function) before the rigorous foundations for that concept have been introduced; an informal appeal to Euclidean geometry should suffice at this stage.
Things do get more interesting at the "rigorous" stage (which, in the US, roughly starts at a good undergraduate real analysis class), when students already have plenty of pre-rigorous exposure to real numbers, limits, special functions, etc. but are now ready to revisit these concepts from a rigorous foundational point of view.  In my own textbook at this level, I proceed by this route:

Define rational numbers
Define Cauchy sequences of rational numbers, and equivalence of Cauchy sequences
Define reals as the space of Cauchy sequences of rationals modulo equivalence
Define limits (and other basic operations) in the reals
Cover a lot of foundational material including: complex numbers, power series, differentiation, and the complex exponential 
Eventually (Chapter 15!) define the trigonometric functions via the complex exponential.  Then show the equivalence to other definitions.

But certainly one can proceed in a different order to the above.
At the post-rigorous level, one can view of course trig functions as special cases of much more general operations, such as the exponential operation on a Lie algebra...<|endoftext|>
TITLE: The difference between Principal Components Analysis (PCA) and Factor Analysis (FA)
QUESTION [6 upvotes]: I am trying to understand the difference between PCA and FA. Through google research, I have come to understand that PCA accounts for all variance, while FA accounts for only common variance and ignores unique variance.
However, I am having a difficult time wrapping my head around how exactly this occurs. I know PCA rotates the axis used to describe the data in order to eliminate all covariance. Does this step still occur in FA? If not, what differentiates FA from PCA? Thanks in advance.

REPLY [9 votes]: The difference between PCA and FA can be thought of in terms of the underlying statistical models (regardless of estimation methods, although these will change depending on the model used).
Consider $n$ iid observations of a $p$ dimensional (column) vector $X$. Suppose that for each $X_i$, $i \in \lbrace 1, \dots, n\rbrace$, we also had a $k$ dimensional vector $f_i$, with $k \leq p$.  These are our "latent factors". A (linear) factor model assumes that $\mbox{E}(X_i \mid f_i) = Bf_i$, where $B$ is a $p \times k$ "factor loadings" matrix and $\mbox{Cov}(X_i \mid f_i) = \Psi$, a diagonal matrix.  If we further assume that $\mbox{V}(f_i) = \mbox{I}_k$ so that the factors are independent we see that the marginal covariance is $\Sigma \equiv \mbox{Cov}(X_i) = BB^t + \Psi$.  
Roughly, you can think of PCA as making the assumption that $\Psi$ is the zero matrix. In both cases the goal is to find/estimate rotations ($B$) that explain covariance patterns. 
If we remove the estimation part of the problem and assume we have $\Sigma$ in hand, the difference is between two ways of decomposing a covariance matrix. We either want a "factor decomposition" $\Sigma = BB^t + \Psi$ or a principle component decomposition $\Sigma = BB^t$.  

I think the key really is this: Any
  covariance matrix will admit either
  kind of decomposition, but often the
  rank of $B$ will be substantially
  smaller if we allow the diagonal
  elements of $\Psi$ to be non-zero as
  in the factor decomposition.

Incidentally, finding the factor decomposition for a given covariance that minimizes the rank of $B$ is known as the Frisch problem and is computationally demanding.
PS.  I hope this isn't merely a restatement of your remark that "PCA accounts for all variance, while FA accounts for only common variance and ignores unique variance".<|endoftext|>
TITLE: Verifying an example in the Geometry of Numbers and Quadratic Forms
QUESTION [7 upvotes]: In answer to Pete L. Clark's question Must a ring which admits a Euclidean quadratic form be Euclidean?  on Euclidean quadratic forms, I gave an example in seven variables, repeated below. Pete's Euclidean property is simply that for any point $\vec x \in \mathbf Q^7$ but $\vec x \notin \mathbf Z^7,$ we require that there be at least one $\vec y \in \mathbf Z^7$ such that $$ q(\vec x - \vec y) < 1. $$ 
[Edit: This is the definition for positive definite integral quadratic forms. --PLC]
I think my answer works (and the easier 6 variable one), based on extensive computer calculations, and Pete has been too polite to express much doubt.
Could someone please try to prove that this example works (and the 6 variable one)? It seems likely that this lies in the field http://en.wikipedia.org/wiki/Geometry_of_numbers but who can say? 
$$ q( \vec x) = x_1^2+ x_1 x_2 + x_2^2 + x_2 x_3 + x_3^2  + x_3 x_4 + x_4^2  + x_4 x_5 + x_5^2  + x_5 x_6 + x_6^2  + x_6 x_7 + x_7^2 + x_7 x_1. $$ This has the Euclidean property, its worst behavior is either when all $x_i = \frac{1}{4}$ or when all $x_i = \frac{3}{4},$ with ``Euclidean minimum'' equal to $\frac{7}{8}.$
Notice that with $\vec x = \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right),$ the integer lattice points $\vec y$ such that $ q( \vec x - \vec y)=\frac{7}{8} $ include $\vec y = \left( 0,0,0,0,0,0,0\right)$ and all seven cyclic permutations (including the identity) of
$\vec y = \left( 0,1,0,1,0,1,0\right),$ another seven for
$\vec y = \left( 1,0,0,0,0,0,0\right),$ another seven for
$\vec y = \left( 1,0,1,0,0,0,0\right),$ finally seven for
$\vec y = \left( 1,0,0,1,0,0,0\right),$ a total of 29 lattice points on the ellipsoid, of 128 in the standard unit 7-cube. The Gram matrix for the form is
$$  Q \; \; = \; \; 
\left(  \begin{array}{ccccccc}
  1 & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}\\\
  \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 \\\
  0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0  \\\
  0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0   \\\
  0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0    \\\
  0 &  0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2}     \\\
   \frac{1}{2} &   0 &  0 & 0 & 0 & \frac{1}{2} & 1
\end{array} 
  \right)  ,  $$
which has determinant $\frac{1}{32}$ and characteristic polynomial 
$$ \left( \frac{1}{64} \right) \left(x - 2 \right) \left(8 x^3 - 20 x^2 + 12 x - 1  \right)^2. $$ So the ellipsoids described are not oblate spheroids, there is less symmetry than that.
EDIT. I think it wise to describe what I am completely certain about and what is unclear. What I did is make a cubic grid, where each variable takes on values $\frac{i}{M}$ for $0 \leq i < M.$ So that makes a grid with $M^7$ points. For each point $\vec r$ in the grid, I find the 128 different values of $q(\vec r - \vec y)$ for
$\vec y \in \mathbf Z^7$ and all coordinates of $\vec y$ are either 0 or 1. For that point $\vec r,$ I take the smallest of the 128 values. Now, for every $M$ I have tried, and for every $\vec r$ in the grid, this best value out of 128 has never been larger than  $\frac{7}{8}.$ Now, using the fact that for any $\vec x \in \mathbf Q^7$ that is not in the grid, there is some point $\vec r$ such that $ | \vec r -\vec x | \leq \frac{1}{2 M \sqrt 7},$
I get that I can always find a $\vec y \in \mathbf Z^7$ such that
$$ g(x-y) \leq \frac{7}{8} + \frac{1}{7 M} + \frac{1}{14 M^2}, $$ using Cauchy-Schwarz and the maximum eigenvalue of $Q$ being 2. Anyway, this does not show that the Euclidean minimum is really  $\frac{7}{8},$ although I believe it is. What it does show is that the Euclidean minimum is less than 1, as soon as $M \geq 2.$

REPLY [6 votes]: The answer to the last question in Franz Lemmermeyer's answer (so maybe
this ought to be a comment?) is yes:
Since each $\delta_j^2\le\frac{1}{2}\delta_j$, you have
$-\delta_j+\delta_j^2\le-\frac{1}{2}\delta_j\le-\frac{1}{2}\delta_1$.
Then 
$\delta_1-\delta_2-\cdots-\delta_7+\delta_1^2+\cdots+\delta_7^2
 \le\delta_1+\delta_1^2-\frac{6}{2}\delta_1\le\frac{3-6}{2}\delta_1
 \le0$<|endoftext|>
TITLE: Reconstructing a model from its definable sets
QUESTION [6 upvotes]: Let $\mathcal{M}$ be an infinite model of a first-order language, and for each $n$, let $\mathcal{B}_n$ be the algebra of definable sets of $n$-tuples from $|\mathcal{M}|$.

Given $\{\mathcal{B}_n\mid _{n\in\mathbb{N}}\}$ (and, obviously, $|\mathcal{M}|$, is it possible to describe explicitly some $\mathcal{M}'$ whose definable sets are $\{\mathcal{B}_n\}$?  (I think the question makes the most sense assuming that the language itself is unknown, so I'm asking if there's a natural way to invent a language and a model which gives the chosen definable sets.  Obviously there is such a model, namely the original one, but I'm open minded about what it would mean to construct $\mathcal{M}'$ "explicitly".)
How well do the definable sets "pin down" the model?  Need two models with the same definable sets in the same language be elementarily equivalent?  Can anything be said about the relationship between two different models of different languages, but with the same definable sets?
There are some obvious restrictions on the $\mathcal{B}_n$: $\mathcal{B}_n\times\mathcal{B}_m\subseteq\mathcal{B}_{n+ m}$, each $\mathcal{B}_n$ is a Boolean algebra, $\mathcal{B}_m$ contains all projections from $\mathcal{B}_{m+n}$.  Are there any others?

REPLY [8 votes]: I like this question very much. 
For 1,  you can add a relation symbol for every set in the
family, and this will of course suffice to define every set
in the family, while creating no additional definable sets,
since giving a name to a definable set is clearly
conservative.
For 2, two models giving rise to the same family of definable
sets need not be elementarily equivalent. For example, you
can turn an order upside-down or use the complement of a
relation to get the same definable sets, but this can 
strongly affect the theory.
For 3, the properties you list do not suffice, since you need to be able to
permute variables within any one dimension, since definable
sets are closed under that operation. Also, you
will need to close under composition of relations and
substitution. For example, if you have the graph of a
function that is definable, then you will need to have all
substitution instances of this function into definable
sets. Finally, you will need to add the diagonal $\Delta=\{(x,x)\mid x\in M\}$, which is always a definable binary relation (if one assumes as usual that $=$ is included in all first order languages). This last property does not follow from your properties, since one could double every point while preserving your properties. Perhaps these additional requirements suffice to carry out Andreas' elimination of quantifiers idea...<|endoftext|>
TITLE: An alternative description of K^*/Nm(L^*)
QUESTION [7 upvotes]: Is there a nice explicit description for the group $K^*/Nm_{L/K}(L^*)$ for a finite field extension $L/K$? 
What if for example, $L$ is obtained from $K$ by ajoining an n-th root of some $\alpha \in K$
(and assuming that $K$ contains the n-th root of unity)?
I don't see a nice answer even for the case $n=2$.
Thank you. 
EDIT: Thanks for the answers! Is that correct that in the case of a cyclic extension this group is isomorphic to $Br(L/K)$, since both these groups are identified with $H^2(Gal(L/K), L^*)$?

REPLY [4 votes]: In addition to Alex's answer I would like to point out that studying the group  $K^\times/N(L^\times)$ is a perfectly fine goal (related of course to the validity of the Hasse norm principle in extensions of number fields), which should not simply be dismissed as a "wrong question". A good place to start is the work by Leonid Stern, the most recent article being
On the norm groups of Galois $2\frak n$-extensions of algebraic number fields,
J. Number Theory 129, No. 5, 1191-1204 (2009).<|endoftext|>
TITLE: Elementary reference for algebraic groups
QUESTION [19 upvotes]: I'm looking for a reference on algebraic groups which requires only knowledge of basic material on the theory of varieties which you could find in, for example, Basic Algebraic Geometry 1 by Shafarevich.
If possible, it would be nice to find such a book which also discusses representation theory, but that's not necessary.

REPLY [3 votes]: The following is an emended excerpt from my answer to a related question1 about books about Lie groups for someone with algebraic geometry background. I might add that Procesi's book ideally fits your goals, since you are also interested in representation theory.

For someone with algebraic geometry background, I would heartily recommend Procesi's Lie groups: An approach through invariants and representations. It is masterfully written, with a lot of explicit results, and covers a lot more ground than Fulton and Harris. If you like "theory through exercises" approach then Vinberg and Onishchik, Lie groups and algebraic groups is very good (the Russian title included the word "seminar" that disappeared in translation).
If you aren't put off by a bit archaic notation and language, vol 2 of Chevalley's Lie groups is still good.


1That question is exactly one year old and, according to Anton's MO birthday post on meta, was the second "real" question asked on Mathoverflow.<|endoftext|>
TITLE: geometric decomposition of J(11),
QUESTION [6 upvotes]: Let $N$ be a prime number. Let $J(N)$ be the jacobian of $X_\mu(N)$, the moduli space of elliptic curves with $E[N]$ symplectically isomorphic to $Z/NZ \times \mu_N$. Over complex numbers we get that J(N) is isogeneous to product of bunch of irreducible Abelian varieties. Is there a way of describing these Abelian varieties using $J_1(M)$ and $J_0(M)$? Specifically, what can we say about the decomposition of $J(11)$?
Note that $X_\mu(N)$ is birationally isomorphic as a curve to the fibre product $X_0(N^2) \times_{X_0(N)} X_1(N)$. (This is because $\Gamma(N)$ is conjugate to $\Gamma_0(N^2) \cap \Gamma_1(N)$, and the group generated by $\Gamma_0(N^2)$ and $\Gamma_1(N)$ is $\Gamma_0(N)$.)
Therefore, we have $J_1(N)$ and $J_0(N^2)$ are both some of the factors in $J(N)$. In fact, we know that $J(7)$ is three copies of $J_0(49)$. For N=11, the above fibre product to $X_0(121)$ is an unramified covering. If I was going to make a guess on what $J(11)$ going to decompose as, I would guess that it is five copies of $J_0^{new}(121)$ and six copies of $J_1(11)$. Is that reasonable? Is there a geometric way of arguing this?
Also, I'm guessing that the question about
 $SL_2(F_N)$ decompoposition of space of cusprforms  is related to this, and Jared Weienstein's thesis will come into play here, but I'm not sure how.

REPLY [8 votes]: The decomposition of $J(11)$ was known (at least over $\mathbf{C}$) to Hecke. It turns out that the Jacobian of the compactification of $\Gamma(11) \backslash \mathfrak{h}$ is isogenous to a product of 26 elliptic curves. All this is very well explained in the following article :
MR0463118 (57 #3079)  Ligozat, Gérard . Courbes modulaires de niveau $11$.
(French)  Modular functions of one variable, V (Proc. Second Internat. Conf.,
 Univ. Bonn, Bonn, 1976), pp. 149--237. Lecture Notes in Math., Vol. 601, Springer, Berlin,  1977.
http://www.springerlink.com/index/6722kj1764m8g50t.pdf
The idea is to look at the natural representation of the group $\mathrm{PSL}_2(\mathbf{F}_p)$ on the space of cusp forms $S_2(\Gamma(p))$. So, you're right that there is a geometric interpretation.
If I remember well, there are, among the factors of $J(11)$, elliptic curves of conductor $121$ which are $11$-isogenous to itself. These can be seen as rational points of the modular curve $X_0(11)$ which are not cusps (there are three such points).
EDIT : I remembered somewhat incorrectly. The three non-cuspidal points of $X_0(11)(\mathbf{Q})$ correspond to the elliptic curves 121B1, 121C1 and 121C2. The subgroups of order $11$ of these curves are described as follows : the elliptic curve 121B1 has CM by $\mathbf{Z}[\frac{1+i\sqrt{11}}{2}]$, so it is $11$-isogenous to itself, whereas 121C1 and 121C2 are $11$-isogenous to each other. Using the notations of Cremona's tables, the Jacobian of the compactification of $\Gamma(11)\backslash \mathfrak{h}$ is then isogenous to $(11A)^{11} \times (121B)^5 \times (121C)^{10}$.<|endoftext|>
TITLE: Are all homogeneous metric spaces bihomogeneous?
QUESTION [8 upvotes]: Let (X,d) be a metric space such that for all points p and q in X, there exists an isometry f such that f(p) = q.  Does it follow that for all points p and q in X, there exists an isometry f such that f(p) = q and f(q) = p?
This seems like an obvious enough question that I would be surprised if the answer isn't simply a reference, but I haven't found it mentioned anywhere.

REPLY [5 votes]: Here is a one-dimensional analogue of Richard's triangle
example, obtaining a counterexample in the set of
reals. Namely, replace every integer $n$ with two numbers at
fixed small distance $n\pm\epsilon$. One can suitably translate and reflect to realize
homogeneity, but there is no isometry swapping $\epsilon$ and $1+\epsilon$.<|endoftext|>
TITLE: Why is the Gaussian so pervasive in mathematics?
QUESTION [56 upvotes]: This is a heuristic question that I think was once asked by Serge Lang. The gaussian: $e^{-x^2}$ appears as the fixed point to the Fourier transform, in the punchline to the central limit theorem, as the solution to the heat equation, in a very nice proof of the Atiyah-Singer index theorem etc. Is this an artifact of the techniques (such as the Fourier Transform) that people like use to deal with certain problems or is this the tip of some deeper platonic iceberg?

REPLY [5 votes]: Edit 2/15/2022{
The utility of the Gaussian $e^{\frac{t^2}{2}}$--its numerous properties--derives from the nature of the coefficients of its Taylor series expansion, naturally. The coefficients, which are aerated OEIS A001147, the odd double factorials $n!!$, enumerate the number of perfect matchings of the vertices of the n-simplices / hypertetrahedra {and complete graphs}. In "Gaussian processes and Feynman diagrams", Faris succinctly sketches the relationship between this characterization and Feynman graphs. In "Graphs on surfaces and their applications", Lando and Zvonkin go into more detail on this and present further associations. The associated Appell sequence, one family of Hermite polynomials, $He_n(z)$ with the e.g.f. $e^{\frac{t^2}{2}} e^{zt}$ has two complementary generators: the raising op $z + \partial_z$ and the binomial transform $e^\frac{{\partial_z^2}}{2} \; z^n = He_n(z)$, from which the properties of the Gaussian, important in so many applications in analysis, algebra, probability, and physics, can be easily derived. When I see a family of Hermite polynomials, I think pair matchings, ribbon graphs, Heisenberg-Weyl ladder ops, orthogonality, heat/diffusion evolution equation, normal-ordering (A344678), quantum physics, and the Gaussian distribution, and, conversely.
(Revamped 2/25/21)
The Heisenberg-Weyl algebra associated with the Appell polynomial calculus of the Hermite polynomials provides a way to quickly derive and collate diverse properties of the Gaussian $e^{ \frac{t^2}{\sigma}}$ and connect these to important constructs in math and physics, such as those mentioned in the other responses. The quadratic argument suggests easy extensions to higher dimensions and metrics.
The Hermite polynomials and the Gaussian
The Gaussian is the exponential generating function (e.g.f) for the basic moment sequence for families of Hermite polynomials. One family, $H_n(x)= (H.(x))^n$, in umbral notation, can be characterized several equivalent ways. Three are by the
1) e.g.f.
$$e^{H.(x)t} = e^{t^2} \; e^{xt} = e^{h.t} \; e^{xt} = e^{(h.+x)t},$$
giving
$$H_n(x) = (h.+x)^n = (H.(0)+x)^n$$
with
$$h_n = H_n(0) = |\cos(\frac{\pi n}{2})| \; \frac{n!}{(\frac{n}{2})!} ,$$
2) binomial generating operator (BGO) and inverse (the heat operator in this case at unit time)
$$ e^{D^2} \; x^n = e^{h.D} \; x^n = (x+h.)^n = H_n(x)$$
with $h_n$ defined as above as the Taylor series coefficients of the Gaussian, and with the inverse
$$e^{-D^2} \; H_n(x) = x^n,$$
($D_{\omega} = \partial_{\omega} = \frac{\partial}{\partial \omega}$ can be considered a partial derivative throughout these notes with the variable often suppressed when obvious for readability),
3) raising (creation) operator
$$ R \; H_n(x) = H_{n+1}(x) $$
with
$$R = e^{D_x^2} \; x \; e^{-D_x^2} = x + D_{t=D_x} \; \ln[e^{t^2}] = x + 2 D_x.$$
The first equality in this last string is easily demonstrated using the action of the BGO and its inverse.
$$ R \; H_n(x) =  e^{D^2} \; x \; e^{-D^2} \; H_n(x) = e^{D^2} \; x \; x^n = e^{D^2}  \; x^{n+1} = H_{n+1}(x).$$
To relate a central concept in the Sheffer polynomial calculus, let $e^{-D^2} x^n = \hat{H}_n(x)$, the umbral compositional inverse (UCI) to the family $H_n(x)$. Then this last equation reads umbrally as
$$R \; H_n(x) =  e^{D^2} \; x \; e^{-D^2} \; H_n(x) = e^{D^2} \; x  \; H_n(\hat{H}.(x)) = e^{D^2} \; x \; x^n $$
$$ = e^{D^2}  \; x^{n+1} = H_{n+1}(x).$$
(The relation to the log can be shown via the Graves-Pincherle commutator/derivative as in this MO-Q.) Note this is an operator conjugation of the basic raising op for the fundamental Appell polynomial sequence $p_n(x) = x^n$.
4) lowering (annihilation / destruction) operator
The lowering op for all Appell sequences is the derivative since they all have the same binomial expansion as the Hermite polynomials, just with different moments,
$$D_x \; H_n(x) = D_x \; (h.+x)^n = n \; (h.+x)^{n-1} = n \; H_{n-1}(x).$$
In fact, any polynomial sequence with $p_0(x)=1$ such that $D \; p_n(x) = n \; p_{n-1}(x)$ is an Appell sequence.
Already, we have, in 1), a form, the e.g.f., related to  the integrand for characteristic functions for the Gaussian, used in dealing with the moments and cumulants of random variables; in 2), the heat operator $e^{-D_x^2}$ and its inverse, evaluated at unit time; and, in 3) and 4), forms of the ladder ops of quantum mechanics--the ops of the Graves-Lie-Heisenberg-Weyl group/algebra.
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A) Integral transforms of the Gaussian
These characterizations can be used to show that the Gaussian gives a Gaussian under a Fourier transformation (FT) and a two-sided Laplace transform (LPT). The solution to the heat equation (or Cauchy problem) can be derived using the FT, and the orthogonality of Hermite polynomials derived from the LPT.
The raising op gives
$$e^{tR} \; 1 = e^{H.(x)t}= e^{t^2} \; e^{xt},$$
so
$$D_t \; e^{t^2} \; e^{xt} = D_t \; e^{tR} \; 1 = R \; e^{t^2} \; e^{xt}.$$
Changing the variable $t \to it$ gives
$$ -iD_t  \;  e^{-t^2} \; e^{ixt} = R  \; e^{-t^2} \; e^{ixt} \;,  $$
and integrating gives
$$ i \; \int_{-\infty}^{\infty} \; D_t \; e^{-t^2} \; e^{ixt} \; dt = R \; \int_{-\infty}^{\infty}  \; e^{-t^2} \; e^{ixt} \; dt \;, $$
so
$$ 0 = (x + 2D_x) \; FT^{-1}[e^{-t^2}].$$
But also
$$ (x + 2D_x) \; e^{-{(\frac{x}{2})}^2} = 0,$$
so the complex conjugate gives the standard Fourier transform
$$ FT[e^{-t^2}] = C \; e^{-{(\frac{x}{2})}^2},$$
where $C=\sqrt{\pi}$. (The normalization factor can be read off from the Mellin transform of the Gaussian as shown below, or it can be determined the usual way by squaring the integral with $x=0$ and converting to polar coordinates.)
Substituting $-ix$ for $x$ in the FT or $ix$ in the inverse FT gives
$$ i \; \int_{-\infty}^{\infty} \; D_t \; e^{-t^2} \; e^{-xt} \; dt = i \;(x - 2D_x) \; \int_{-\infty}^{\infty}  \; e^{-t^2} \; e^{-xt} \; dt ,$$
(note $\hat{R} = x - 2D_x$ is the raising op for $\hat{H}(n(x)$),
and since
$-t^2 - xt = -(t + \frac{x}{2})^2 + (\frac{x}{2})^2,$
$$0 = (x - 2D_x) \; \int_{-\infty}^{\infty}  \; e^{-t^2} \; e^{-xt} \; dt .$$
Also
$$(x- 2D_x) \; e^{(\frac{x}{2})^2} =0,$$
so the two-sided LPT of the Gaussian is
$$LPT[e^{-t^2}] = \sqrt{\pi} \; e^{(\frac{x}{2})^2}.$$
Because the Gaussian is essentially self-reciprocal under the Fourier transform, the equality in the quantum mechanical uncertainty relation is achieved only for freely moving particles whose probability amplitudes are characterized as Gaussian wave packets.
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B) The heat equation (Cauchy problem)
Now with $t$ time and $x$ spatial displacement in this section,
$$ D_t \; f(t,x) = D_x^2 \; f(t,x)$$
is satisfied by
$$ e^{tD_x^2} \; f(0,x) = f(t,x).$$
The Green's function solution satisfies
$$ G(0^{+},x) = \delta(x) = \int_{-\infty}^{\infty} e^{i2\pi \omega x} \; d\omega$$
and
$$G(1,x) = e^{D^2} \; \delta(x) = \int_{-\infty}^{\infty} e^{i2\pi \omega H.(x) } \; d\omega  = \int_{-\infty}^{\infty} e^{-(2\pi \omega)^2} e^{i2\pi x \omega} \; d\omega = \frac{1}{2 \sqrt{\pi}} \; e^{-(\frac{x}{2})^2}  .$$
With $x = \sqrt{t} \; u$, then
$$G(t,x) = e^{tD_x^2} \; \delta(x) = e^{D_u^2} \; \delta(\sqrt{t} \; u) = \frac{1}{\sqrt{t}} e^{D_u^2} \; \delta(u)$$
$$ = \frac{1}{ \sqrt{t}} \frac{1}{2 \sqrt{\pi}} \; e^{-(\frac{u}{2})^2} = \frac{1}{\sqrt{4\pi t}} \; e^{-\frac{x^2}{4t}} ,$$
and the general solution is the convolution
$$f(t,x) = e^{tD^2_x} \; f(0,x) = \int_{-\infty}^{\infty} f(0,u) \; G(t,u-x) \; du.$$
This integral is a Segal-Bargmann transform discussed by Cartier in "Mathemagics" and by Hall in "Holomorphic methods in analysis and mathematical physics."
This is consistent with the BGO when $f(0,x) = x^n$, for which $f(1,x) = e^{D^2} \; x^n = H_n(x).$ And, that any Appell sequence/coefficient matrix can be transformed into any other Appell sequence/coefficient matrix suggests more general 'heat equations' associated to these Appell sequences can be transformed into the Hermite heat equation.
The time can be rescaled to include a diffusion constant (or adjustable variance in the probabilist's jargon).
For some of the physical import of the heat equation, see "Random Walk and the Theory of Brownian Motion" by Mark Kac and "Brownian motion and potential theory" by Hersh and Griego.
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C) The quantum harmonic oscillator
To dovetail with Cartier's arguments, best to jump to another family of Hermite polynomials, that of the probabilists or statisticians $SH_n(x)$, with the
1) e.g.f.
$$ e^{SH.(x)t} = e^{-\frac{t^2}{2}} \; e^{xt} \; ,$$
2) BGO and the inverse (the B-transform in Cartier)
$$ e^{-\frac{D^2}{2}} \; x^n = SH_n(x)  $$
and
$$ e^{\frac{D^2}{2}} \; SH_n(x) = x^n \; ,$$
3) raising op
$$R = e^{-\frac{D^2}{2}} \; x \; e^{\frac{D^2}{2}} = x - D \; .$$
Conjugations transform the between the spaces described and listed in the table on page 46 by Cartier. From the Fock $\mathfrak{F(c)}$ to $\mathfrak{L^2(d\gamma)}$ space, for the raising ops,
$$ e^{-\frac{D_z^2}{2}} \; z^n \; e^{\frac{D_z^2}{2}} = z - D_z,$$
the lowering op under this conjugation remains invariant as the derivative $D_z$, and for the bases
$$ e^{-\frac{D_z^2}{2}} \; z^n = SH_n(z).$$
From $\mathfrak{L^2(d\gamma)}$ to $\mathfrak{L^2(R)}$, for the raising and lowering ops,
$$ e^{-\frac{x^2}{4}} \; (x -D_x) \;  e^{\frac{x^2}{4}} = \frac{x}{2} - D_x,$$
$$ e^{-\frac{x^2}{4}} \; D_x \;  e^{\frac{x^2}{4}} = \frac{x}{2} + D_x,$$
and, for the bases,
$$ e^{-\frac{x^2}{4}} \; SH_n(x) = QH_n(x).$$
The number / state / Euler op for a basis set $p_n(x)$ characterized by lowering, $L$, and raising, $R$, ops is $E = RL$. For example, for $p_n(x) = x^n$, the Euler op is $xD_x$ and $xD_x \; p_n(x) = n \; p_n(x)$. For $\mathfrak{L^2(R)}$ space, action of the Euler op on the basis set gives essentially the Hamiltonian for the quantum harmonic oscillator (mod scaling factors)
$$E = (\frac{x}{2} - D_x) \;(\frac{x}{2} + D_x) = -D_x^2 + \frac{x^2}{4} - \frac{1}{2} ,$$
so
$$ (-D_x^2 + \frac{x^2}{4}) \; QH_n(x) =  (n + \frac{1}{2}) \; QH_n(x).$$
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D) The Gaussian and orthogonality of Hermite polynomials
As noted above the sequence of Hermite polynomials $\hat{H}_n(x)$ that is the UCI to the family $H_n(x)$ has the e.g.f.
1) e.g.f.
$$e^{\hat{H}.(x)t} = e^{-t^2} e^{xt}$$
with
$$\hat{h}_n = \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!},$$
2) BGO and inverse
$$e^{-D_x^2}  \; x^n = \hat{H}_n(x),$$
with inverse
$$e^{D_x^2}   \hat{H}_n(x) = \; x^n ,$$
3) raising op
$$\hat{R} = e^{-D^2} \; x \; e^{D^2}= x- 2 \: D_x.$$
From the e.g.f.,
$$   D_{t=0}^n \; e^{-t^2} \; e^{-2xt} \;  D_{s=0}^m  \; e^{-s^2} \; e^{-2xs} \;  = \hat{H}_n(-2x) \; \hat{H}_m(-2x) $$
$$=   D_{t=0}^n  \;  D_{s=0}^m  \; e^{-(t^2+s^2)} \;e^{-2x(t+s)}  ,$$
so
$$  D_{t=0}^n  \;  D_{s=0}^m  \; e^{-(t^2+s^2)} \; \int_{-\infty}^{\infty} \; e^{-x^2} \; e^{-2x(t+s)} \; dx = \int_{-\infty}^{\infty} \;  \hat{H}_n(-2x) \; \hat{H}_m(-2x) \; e^{-x^2} \; dx$$
$$ =  \; D_{t=0}^n  \;  D_{s=0}^m  \; \sqrt{\pi} \; e^{-(t^2+s^2) + (s+t)^2} = \; D_{t=0}^n  \;  D_{s=0}^m  \; \sqrt{\pi} \; e^{2st} = \sqrt{\pi} \; 2^n \; n! \; \delta_{n,m} , $$
or
$$\int_{-\infty}^{\infty} \;  \hat{H}_n(2x) \; \hat{H}_m(2x) \; e^{-x^2} \; dx = \sqrt{\pi} \; 2^n \; n! \; \delta_{n,m},$$
where $\delta_{n,m}$ is the Kronecker delta, which is unity for $n=m$ and vanishes otherwise.
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E) A Gaussian integral transform of Hermite polynomials
$$  D_{t=0}^n  \; e^{t^2} \; e^{xt}  = H_n(x)  $$
A Wick rotation gives
$$  \;(-i)^n D_{t=0}^n  \; e^{-t^2} \; e^{ixt}  = H_n(x)  .$$
Then
$$(-i)^n \; D_{t=0}^n  \; e^{-t^2} \; \int_{-\infty}^{\infty} \; e^{-x^2} \; e^{ixt} \; dx = \int_{-\infty}^{\infty} \;  H_n(x) \; e^{-x^2} \; dx$$
$$ =(-i)^n \; D_{t=0}^n  \sqrt{\pi} \; e^{-(1+\frac{1}{4})t^2}  =  \sqrt{\pi} \; (-\frac{5}{4})^{\frac{n}{2}} \cos(\pi\frac{n}{2}) \frac{n!}{(\frac{n}{2})!},$$
and
$$\int_{-\infty}^{\infty} \;  H_n(x) \; e^{-x^2} \; dx = \sqrt{\pi} \; (-\frac{5}{4})^{\frac{n}{2}}\; \hat{h}_n  = \sqrt{\pi} \; (-\frac{5}{4})^{\frac{n}{2}} \; \hat{H}_n(0) .$$
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F) Mellin transform of the Gaussian, Mellin transform interpolation of the coefficients of the Gaussian
The Mellin transform of the Gaussian is central to many discussions in analytic number theory and modular forms, being related to the Jacobi theta functions (e.g., see this MO-Q).
The modified Mellin transform can be read off from the coefficients of the e.g.f. of the Gaussian per Ramanujan's master formula as shown in MO-Q1, MO-Q2, and MSE-Q).
$\; \; \; \; \;g(t) = e^{-t^2} = e^{\hat{h}. t}= \sum_{n \ge 0} \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!} \;  \frac{t^n}{n!}, $
$\; \; \; \; \;f(t) = g(-t) =  e^{-t^2} =  g(t)= e^{-\hat{h}. t},$
and, for $ Re(s) > 0$, the modified Mellin transform gives
$$\hat{h}_{-s} = (\hat{h}.)^{-s} = \int_0^{\infty} e^{-\hat{h}. t} \; \frac{t^{s-1}}{(s-1)!} \; dt$$
$$ =  \int_0^{\infty} e^{-t^2} \; \frac{t^{s-1}}{(s-1)!} \; dt   =  \cos(\pi\frac{ s}{2}) \; \frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} .$$
The Mellin transform gives the general formula for the one-sided Gaussian integral transform of the monomials $x^n$ as
$$n! \; \hat{h}_{-n-1} = \int_0^{\infty} e^{-x^2} \; x^n \; dx = \frac{1}{2} \; (\frac{n-1}{2})! \; . $$
In particular for $s = 1$, or $n =0$,
$$ \hat{h}_{-1} = \int_0^{\infty} e^{-x^2} \;  dx   = \frac{1}{2}\; (-\frac{1}{2})! = \frac{\sqrt{\pi}}{2}.$$
For $n=1,3,5,7,9,...$, the sequence $n! \; \hat{h}_{-n-1}$ is $(\frac{1}{2}, \;\frac{1}{2}, \;1, \;3, \;12, \;60, ...)$ with the numerators OEIS A001710.
For $n=0,2,4,6,..$, the sequence $n! \; \hat{h}_{-n-1}$ is $\sqrt{\pi} \cdot (\frac{1}{2}, \; \frac{1}{2^2}, \;\frac{3}{2^3}, \;\frac{15}{2^4}, \;\frac{105}{2^5}, ...)$, a sequence with the double factorials of OEIS  A001147 (see also A094638) in the numerators.
Both OEIS entries have extensive references.
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G) Cumulants
The coefficients of the differential component of the raising op for an Appell sequence, determined from the natural log of the e.g.f. of the Appell moments, are proportional to the cumulants associated to the moment e.g.f. via the cumulant expansion formula of OEIS A127671. In particular, the raising ops of the Hermite polynomial families are $R = x + D_{t=D_x} \ln[e^{-\alpha \; t^2}]  = x - \alpha \;  D_x$, where $\alpha$ is proportional to the variance, or second cumulant, of the Gaussian probability density function (pdf).
A characteristic function of the normalized Gaussian pdf with $t$ treated as a random variable is, from the two-sided LPT above,
$$<e^{ xt}> \;  = \int_{-\infty}^{\infty} \frac{e^{-t^2}}{\sqrt{\pi}} \; e^{ xt} \; dt = \int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}} e^{\hat{H}.(x)t} \; dt = e^{\frac{x^2}{4}} = e^{\frac{h.}{2}x}  $$
with the moments given by
$$<t^n > \;  =  \; D_{x=0}^n \; <e^{xt}> \;  =  D_{x=0}^n e^{\frac{x^2}{4}} = \frac{h_n}{2^n} = \frac{H_n(0)}{2^n} =  |\cos(\frac{\pi n}{2})| \; \frac{n!}{(\frac{n}{2})!} \frac{1}{2^{n}},$$
agreeing with the Mellin transform result.
The formal cumulants in terms of formal moments (convergence of the e.g.f.s not required) are given by the cumulant expansion formula (see the associated partition polynomials in the Lang link in A127671)
$$\ln[<e^{xt}>] = \ln[\; e.g.f \; of \; moments \; ] = \; e.g.f. \; of \; cumulants, $$
and, in our case for the Gaussian, reduces to
$$  \ln[e^{\frac{x^2}{4}}] = \frac{1}{2} \frac{x^2}{2} = (<t^2> - <t>^2) \; \frac{x^2}{2}  = (<t-<t>>^2) \; \frac{x^2}{2}, $$
so the cumulants vanish except for the second cumulant, the variance, which is 1/2. The cumulant partition polynomials are invertible (a graded involution), so the formal cumulants completely define the formal moment e.g.f., and, if this is given by a characteristic function obtained by an invertible transform of a pdf, they define the pdf as well. See "Three lectures on free probability" by Novak and LaCroix for the significance of cumulants in combinatorics and statistics and for an elementary proof of the central limit theorem via cumulants. See also the stat mech refs in OEIS A036040. See the Wick's and Isserlis' theorems for generalizations.
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For more connections of the Gausssian, the Hermite polynomials, and the Graves-Lie-Heisenberg-Weyl group/algebra to other areas of math and physics, see "On the role of the Heisenberg group in harmonic analysis" by Roger Howe, "Heisenberg groups, theta functions, and Weil representation" by Yang, "An Exercise(?) in Fourier Analysis on the Heisenberg Group" by Bump, Diaconis, Hicks, and Widom, "Theta and Riemann xi function representations from harmonic oscillator eigensolutions" by Coffey, and "Quantum vs. classical integrability in Calogero-Moser systems" by Corrigan and Sasaki.

(Edit 5/14/21)
Generalization: Gaussian functions, moments, and the Heisenberg-Weyl algebra
Many important polynomial sequences $p_n(x)$ (in particular, all Sheffer polynomial sequences, related to invertible lower triangular matrices, including the Bernoulli and families of Laguerre and Hermite polynomials) have ladder ops--lowering and raising ops $L$ and $R$--such that
$ L \; p_n(x) = n \; p_{n-1}(x)$ and $R \; p_n(x) = p_{n+1}(x)$
with $L \; p_0(x) = 0.$
The commutator of the ops gives $[L,R] = 1$; i.e., $[L,R] \; p_n(x) = (LR -RL) \; p_n(x) = p_n(x).$
For operators that satisfy the commutator relations $[X,[X,Y]] = [Y,[Y,X]]= 0,$ the BCHD expansion reduces to
$$ e^{tX} e^{tY} = e^{t(X+Y) + \frac{t^2}{2}[X,Y]},$$
so the operator disentangling relationship
$$ e^{tR} e^{tL} = e^{t(R+L) + \frac{t^2}{2}[R,L]} = e^{\frac{-t^2}{2}} \; e^{t(R+L)}$$
holds, in which a Gaussian function appears.
For $p_0(x) =1$ with the umbral notation and maneuver $p.(x)^n = p_n(x)$,
$$ e^{tR} e^{tL}\; 1 =  e^{t R} \; 1= e^{t \; p.(x)} = e^{\frac{-t^2}{2}} \; e^{t(R+L)} \; 1  $$
so, with the moments $\bar{h}_n$ of the Gaussian function defined by  $e^{ \;\frac{t^2}{2}} = e^{\bar{h}.t}$,
$$e^{\frac{t^2}{2}} \; e^{t \; p.(x)} =  e^{t\; \bar{h}.} \; e^{t \; p.(x)} = e^{t\; (\bar{h}.+p.(x))} = e^{t \; \bar{H}.(p.(x))}  = e^{t(R+L)} \; 1,$$
with
$$\bar{H}_n(x) = (\bar{h}.+x)^n = \sum_{k=0}^n \; \binom{n}{k} \; \bar{h}_k \; x^{n-k},$$
and we can identify
$$(L+R)^n \; 1= (\bar{h}. + p.(x))^n = \bar{H}_n(p.(x));$$
that is,
$L+R$ is the raising op for the sequence $P_n(x) = \bar{H}_n(p.(x))$.
Since the lowering and raising ops for the monomial sequence $p_n(x) = x^n$ are $x$ and $D$, the raising op for the family of modified Hermite polynomials $\bar{H}.(x)$ above (OEIS A099174) is $x+D_x$, and the form of the polynomials gives $D_x$ as the lowering op. In the earlier answer above, you can see that diverse properties of the Gaussian function/distribution follow from this operator relation. Many can be generalized. For example, the raising op formula
$$(x+D) \; \bar{H}_n(x) = \bar{H}_{n+1}(x)$$
gives an instance
$$\bar{H}_{n+1}(x) = x  \; \bar{H}_n(x) - n \; \bar{H}_{n-1}(x)$$
of the general recurrence relation that is a sufficient and necessary condition for a sequence to be a set of orthonormal polynomials on the real line w.r.t. a weight function/distribution (see the discussion surrounding Favard's theorem in Operator Theory: A Comprehensive Course in Analysis, Part 4 by Barry Simon).
Another example is the evolution equation
$$ D_t \; f(x,t)  = (R+L) \; f(x,t),$$
which follows from the above exponential rep with the solution
$$f(x,t)  = e^{t \; P.(x)} = e^{t \; \bar{H}_n(p.(x))} = e^{t \; (\bar{h}.+p.(x))}$$
$$ =  e^{t \bar{h}.} \; e^{t \; p.(x)} = e^{ \frac{t^2}{2}} \; A(t) \; e^{x \; B(t)} , $$
an e.g.f for a Sheffer sequence, for which $A(t)$ and $B(t)$ are analytic about the origin and $A(0) =1$,  $B(0)=0$, and $B'(0) \neq 0$.
From the calculus of Appell Sheffer sequences, a solution of the deformed heat/diffusion equation
$$D_t \; \tilde{g}(x,t) = [ \; \frac{D_x^2}{2} + \sum_{n \geq 2}  \; c_n \; \frac{D_x^n}{n} \;] \; \tilde{g}(x,t) $$
is
$$\tilde{g}(x,t) = e^{t\frac{D_x^2}{2}} \; \exp[\; t \; \sum_{n \geq 2}  \; c_n \; \frac{D_x^n}{n} \;] \; g(x)$$
$$ =  g[ H.[S.(y,\gamma_2,\gamma_3,...)] \sqrt{t}] = g[ S.(H.(y),\gamma_2,\gamma_3,...) \sqrt{t}] =  g[ S.(x,t\bar{c}_2,tc_3,...)]$$
with
$ \gamma_n = \; t^{1-\frac{n}{2}} \; c_n$ for $n > 1$,
$y =x/ \sqrt{t} $,
$\bar{c}_2 = 1 + c_2$, and
$S_n(c_1,c_2,...,c_n) $ are the Stirling partition polynomials of the first kind (OEIS A036039), a.k.a. the cycle index polynomials of the symmetric groups, defined by
$e^{t \; S.(c_1,...)} =  \exp[-\ln(1-c.t)] = \exp[\sum_{k > 0} c_k \; \frac{t^k}{k}].$
Since these umbral compositions can be represented by multiplication of invertible lower triangular matrices, conjugations (or 'axes rotations') can be used to map between the nondeformed and deformed reps.
_____________________________________
Edit (5/30/21):
A Combinatorial Perspective
Underlying all of the relationships above is the fact $e^{t^2/2} = e^{\bar{h}. t}$ is the e.g.f. for the aerated odd double factorials ($\bar{h}_0=1,\;\bar{h}_1=0,\;\bar{h}_2=1,\;\bar{h}_3=0, \;\bar{h_4}=3,...$), which enumerate the number of perfect matchings of the n vertices of the regular (n-1)-dimensional simplices (hypertriangles, or hypertetrahedrons).
$(x+D_x)^{n}$ as a normal-ordered operator, i.e., expanded and expressed with all derivatives to the right of the variable $x$, has the same coefficients as the polynomial $\bar{H}_{n}(x+y) = (\bar{h}.+x+y)^{n}$ in the commutative variables $x$ and $y$ with the multinomial enumerating permutations of three families of objects--vertices labelled with either $x$ or $y$, or perfect (pair) matchings of the unlabeled vertices for the n vertices of the (n-1)-dimensional simplex. For example, $(x+D)^2 = xx +xD+Dx + D^2 = x^2 + 2xD + D^2+1$ corresponds to $H_n(x+y) = (h.+x+y)^2 = h_0 \dot (x+y)^2 + 2 h_1 \cdot (x+y) + h_2 =  x^2 +2xy +y^2+1$ which, in turn, corresponds to a line segment, the 1-D simplex, with both vertices labeled with $x$'s; or one with an $x$, the other a $y$; or both vertices labeled with $y$'s; or one unlabeled matched pair.<|endoftext|>
TITLE: Differential graded structures on free resolution?
QUESTION [11 upvotes]: Hello!
In "Homological Algebra on a Complete Intersection", Eisenbud proves the following: 
Let $A$ be a commutative ring, $M$ be an $A$-module and $F^{\ast}\to M$ an $A$-free resolution. Further, assume that $M$ is annihilated by $I := (x_1,...,x_n)$, and that $I$ contains a non zero divisor of $A$. Then there exist maps $s_{\alpha}: F^{\ast}\to F^{\ast}$, indexed by multiindices $\alpha$ of length $n$, of degree $2|\alpha|-1$, with the following properties:

$s_0$ is the differential of $F^{\ast}$.
$s_j := s_{(0,0,...,0,1,0,...,0)}$ is a nullhomotopy for the multiplication by $x_j$.
For any $\gamma$ with $|\gamma|\geq 2$ we have 
$\sum\limits_{\alpha+\beta=\gamma} s_{\alpha} s_{\beta} = 0$.

Now he asks the following: Assuming $(x_1,...,x_n)$ is regular and $F^{\ast}$ is the minimal free resolution of $M$, is is possible to choose the nullhomotopies $s_j$ in such a way that $s_j^2=0$ and $s_i s_j = -s_j s_i$? I.e. can we make $F^{\ast}$ into a differential graded module over the Koszul-Algebra of $(x_1,...,x_n)$? 
I'm interested in this question and would like to know if progress has been made to answer it. Is it possible to choose the $s_i$ as above, and, if not, how can the obstruction be described?
Edit:
Is it possible to handle all the ways the $s_j$ can be constructed? Does it have something to do with the $A_{\infty}$-stuff? 
Thank you!
Hanno

REPLY [8 votes]: Despite the existence of obstructions, there has also been progress in the other direction.  Namely, there are lots of classes of free resolutions where the question is known to have an affirmative answer.  (When $F^* $ can be given the structure of a dg module, it is sometimes said that $F^*$ has a "multiplicative structure".)
For instance, I believe that Srinivasan first showed that the Eagon-Northcott complex has a multiplicative structure.  This was then generalized by Pellikaan who showed that determinantal ideals also have multiplicative structures.  There is a reasonably large literature along these lines, and many of the papers have the phrase "multiplicative structure" in the title.<|endoftext|>
TITLE: Little disks operad and $Gal (\bar {Q}/Q)$ 
QUESTION [22 upvotes]: My question is simple: 

How do the little disks operad and $Gal (\bar {Q}/Q)$ relate?

I realize that a huge amount of heavy-machinery can be brought into an answer to this, but I'm struggling with the basics. All papers I've found just seem to jump into the deep-end or involve musings that are more inspirational than precise; so I'm eager to read what people here say.

REPLY [26 votes]: A short answer would be: $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts faithfully on the profinite fundamental groupoïd of the operad of little discs.
If $X$ is an algebraic variety over $\mathbb{Q}$ we have an exact sequence 
$$
  1 \to\pi_1(X\otimes \overline{\mathbb{Q}},p) \to 
  \pi_1(X,p) \to Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to 1
$$
Here $\pi_1(X\otimes \overline{\mathbb{Q}};p)$ is canonically identified with the profinite completion of the usual topological fundamental group $\pi_1(X(\mathbb{C}),p)$. 
If the basepoint is defined over $\mathbb{Q}$, this split and we have an action 
$$
  Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to Aut(\widehat{\pi}_1(X(\mathbb{C}),p)). 
$$  
The (profinite completion of the) fundamental groupoïds of the $C_2(n)$ inherit the operad structure. The trick is that all of it can be defined over $\mathbb{Q}$ as $C_2(n)$ is homotopy equivalent to the configuration space of points on the affine line $F(\mathbb{A}^1_{\mathbb{Q}},n)(\mathbb{C})$. One has to define rational "tangential base points" and check that the operad structure on the fundamental groupoïds is also defined over $\mathbb{Q}$. The resulting operad is described here. One can explicitly compute its automorphism group. This is the Grothendieck-Teichmuller group $\widehat{GT}$. 
As everything is defined over $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ operates on the whole operad. So we have a morphism 
$$
  Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to \widehat{GT}
$$ 
It follows from a theorem of Belyi that it is injective.<|endoftext|>
TITLE: Potts model simulation
QUESTION [8 upvotes]: I was wondering what were the state-of-the-art methods to simulate low temperature configurations of Potts-like models that exhibit a discontinuous phase transition. For models with a continuous phase transition, annealing methods (parallel tempering, simulated tempering and their friends) work pretty well: nevertheless, for discontinuous phase transition, these methods are typically not satisfying because a huge number of intermediate temperatures are needed.
Is there any other approach that has proven efficient?

REPLY [3 votes]: Hi Alekk,
There is a paper from several years ago by Bhatnagar and Randall on torpid mixing of simulated tempering on the Potts model. In it, they mention that they also have a proof of rapid mixing for a related tempering approach on the same model (I don't remember if this is actually in the paper; they haven't yet published everything that they know on the subject).
I've talked to a few people who were experts on this sort of thing in the intervening time, and they have mostly claimed that there has been very little progress on proving efficiency for anything that isn't a mean-field model. Of course, the people who I've been talking to tend towards proving 'good' (i.e. small-order polynomial in size of the lattice) for discrete models, and in L^{1} distance; its possible that other areas of the literature have analyses in other metrics or for other models.
Also, I don't know the etiquette here, but I've been thinking about a few simple related models for a little while now. I'd certainly be interested in talking & sharing results/simulations/etc if you are in fact working in the area (my guess from your blog is that you're a graduate student as well?). My (slightly embarrassing and very old) email is cauchie.p@gmail.com.
Cheers,<|endoftext|>
TITLE: Bernstein's presentation for the Hecke algebra
QUESTION [6 upvotes]: Any one know of any good references for reading about Bernstein's presentation of the Iwahori Hecke algebra? I need some notes which has an example or two. It would really help.

REPLY [3 votes]: In book form, the equivalence of the Coxeter and Bernstein presentations of the affine Hecke algebra appears in the first 4 chapters of Macdonald's book "Affine Hecke algebras and orthogonal polynomials".  It is very carefully written, but the notation can get a bit heavy.  When first reading it, I suggest you always assume that you are in case (1.4.1) in Macdonald's notation.
Chapter 6 does rank one examples.<|endoftext|>
TITLE: In Diff, are the surjective submersions precisely the local-section-admitting maps?
QUESTION [12 upvotes]: Question as in title (Diff = category of smooth manifolds and smooth maps)
I thought I'd convinced myself this is true, so this is just a sanity check.
Also, what about for settings other than smooth manifolds? (like analytic manifolds, complex manifolds, or less differentiable - say, $C^2$ manifolds)

REPLY [3 votes]: In case any one in interested, the following result is given in Lee's Introduction to smooth manifolds. It is theorem $4.26$ named Local section theorem. 

Suppose $M$ and $N$ are smooth manifolds and $\pi:M\rightarrow N$ is a smooth map. Then $\pi:M\rightarrow N$ is a smooth submersion if and only if every point in $M$ is in the image of a smooth local section of $\pi$.<|endoftext|>
TITLE: How to define the quotient of a measure which is invariant under group action?
QUESTION [5 upvotes]: I am a physicist, and I have the following problem. Consider a locally compact group G acting over a measure space $(X, {\cal B}, \mu)$, and assume that $\mu$ is G-invariant. My problem is how to "quotient" the measure $\mu$ for obtaining a measure $\mu/G$ on the quotient space $X/G$, i.e., the space wose elements are the orbits of G. The simple answer consisiting of defining $\mu/G(\Delta):=\mu(\Delta)$ for $\Delta \in X/G$ is not appropriate, as one can see from the following example.
Let $X:=\mathbb{R}^2$ be the configuration space of two one-dimensional particles, let $\mu$ be the Lebesgue measure, and let $G:=\mathbb{R}$ be the group of translations: $a(x, y)=(x + a, y + a)$ for $a \in G$ and $(x, y) \in X$. The orbit of a point $(x, y)$ is the line at $45^°$ passing for $(x, y)$. It is easy to see that the Lebesgue measure of any set of orbits is either $0$ or $\infty$ .
My tentative answer is the following. Let S be a section of the partition of the orbits, i.e., a subset of X composed by an element for every orbit. Let the map $h_S:G \times X/G \to X$ be defined as follows: $h_S(g, \xi)=g s_\xi$, where $s_\xi$ is the element of the section $S$ belonging to the orbit $\xi$. It is easy to see that $h_S$ is a bijection between $G \times X/G$ and $X$. It is also easy to see that a measure $\nu_S$ on $X/G$ exists such that $\mu=h_S(\alpha \times \nu_S)$, where $\alpha$ is the Haar measure on G. The measure $\nu_S$ is the measure I am looking for. The problem is to prove rigorously that it does not depend on the chosen section S.
I guess that this problem has already been addressed. Can anybody give me some reference?

REPLY [7 votes]: The references given in the comments above and in the answer below are, in my opinion, somewhat misleading because all of them discuss the case of a transitive action (but maybe this is what you meant). In the general situation the problem is precisely that the space of orbits may be quite nasty. Let us look at the irrational circle rotation, i.e., the action of $\mathbb Z$ defined as
$$
T^n x = x + n\alpha \;\; (\text{mod}\; 1)\;,
$$
where $\alpha$ is an irrational number. Then the set obtained by taking one point from each orbit is a classical example of a non-measurable set, and there is no way one can define a non-trivial measure on it. Another example like this can be obtained by a little modification of your action: this is the action of $\mathbb R$ on $\mathbb R^2/\mathbb Z^2$ defined by projecting the action
$$
T^t(x,y) = (x+t,y+\alpha t) \;,
$$
onto the torus (here $\alpha$ is again an irrational number). 
The reason why the orbit spaces are "bad" in these examples is that there are no non-trivial measurable sets (non-trivial in the sense that both the set and its complement are non-negligible with respect to the considered measure) which would be unions of orbits. Actions with this property are called $ergodic$. For an ergodic action the only reasonable quotient space is the one-point space.
For actions which are not ergodic there is a reasonable non-trivial quotient measure space, which is called the $space\; of\; ergodic\; components$. Its construction, roughly speaking, consists in first taking the $\sigma$-algebra of all measurable sets which are (up to subsets of measure 0) unions of orbits, and then using the fact that in all reasonable probability spaces any such $\sigma$-algebra can be realized as the preimage $\sigma$-algebra of a uniquely defined measurable quotient map. 
You can find details, for instance, in the article
MR1784210 (2001i:28021) 
Greschonig, Gernot(A-WIEN); Schmidt, Klaus(A-WIEN)
Ergodic decomposition of quasi-invariant probability measures. (English summary) 
Dedicated to the memory of Anzelm Iwanik. 
Colloq. Math. 84/85 (2000), , part 2, 495--514.

REPLY [3 votes]: There is a fair amount of work on this.  Since measures are roughly the same as cohomology, the standard approach in quantum field theory (one situation where such integrals are needed) roughly boils down to computing equivariant cohomology.  The physics buzzword for this is "BRST".
For the case of finite-dimensional manifolds, I think the best mathematical analysis is:

A. Weinstein, 2009.  The volume of a differentiable stack.  Lett. Math. Phys. 90(1-3) pp.353--371.

In particular, he explains what are the correct type of smooth measures when a Lie group acts on a manifold with compact stabilizers (the group itself need not be compact).
It is important to keep in mind that the search for invariant measures is simply the wrong way to go about defining measures on the quotient in general.  In particular, if you consider a group $G$ acting on a space $M$, then if $G$ is unimodular (left Haar measure equals right Haar measure), then the smooth measures on the stacky quotient $M/G$ are in bijection with the invariant measures on $M$ (a choice of bijection corresponds to a choice of Haar measure), but if $G$ is not unimodular (e.g. the two-dimensional nonabelian Lie group) then the correct thing to do is pick a measure on $M$ that fails to be invariant in precisely a way that cancels the failure of unimodularity.
More generally, the Weinstein point of view is that you should not think about a space and a group separately — that's too much data — but just think about the groupoid that encodes the action.  Then there is a well-developed theory of "equivalence" of groupoids, and Weinstein measures push and pull well across equivalences.
A particular example of an equivalence is given by a "full open neighborhood": let $U \subseteq M$ be any open set that intersects every orbit of $M/G$, and restrict your groupoid to it.  (Note: $G$ does not act on $U$, but we define a "partial action" in which two points in $U$ are equivalent iff they're equivalent in $M$, and with the same stabilizers.)  Then the restricted groupoid is equivalent to the original one.
So, let's take $M = \mathbb R^2$ and $G = \mathbb R$ as in your example.  Then we can restrict to the full subset $U = \{(x,y) \text{ s.t. } |x+y|<1$.
Now it is clear that we can pick an invariant Lebesgue measure with lots of finite-sized orbits.  We should divide the volume of any region — this is part of the definition, and in general see [ibid.] — by the length of the orbits: each orbit pulls back to some open subset of $G = \mathbb R$, and measure its length there against Haar measure.  You do get well-defined groupoid measure doing this.  At the end of the day, it's what you want it to be: the measure of a cylinder over a subset of the line $\{x+y = 0\}$ is just the length of the subset (up to your choice of normalizations).<|endoftext|>
TITLE: Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$
QUESTION [12 upvotes]: Recently, while playing around with infinite-divisibility, i arrived at the following metric:
$$d(x,y) := \sqrt{\log\left(\frac{x+y}{2\sqrt{xy}}\right)},$$
defined for positive reals $x$ and $y$. Proving that $d$ is a metric is trivial, except for the triangle-inequality. However, we can bypass a direct proof by appealing to Schoenberg's theorem (I. J. Schoenberg. Metric spaces and positive-definite functions, TAMS, 1938), from which the metricity follows easily because $-\log(x+y)$ is a conditionally positive-definite kernel.
However, i have been searching for following:


Applications / situations where this metric shows up? 
An elementary proof of $d(x,y)$ being a metric.


Remarks
a. A google search on "ratio arithmetic geometric mean" yields some applications of the ratio alone;
b. An elementary proof should exist, but my initial attempts have not been that successful, especially as i stubbornly did not want to use differential calculus.
c. Notice that while proving 
$$d(x,y) \le d(x,z) + d(y,z),$$ we may assume wlog $x < 1$ and $y > 1$ and $z=1$, as proving the other cases ranges from very-trivial to trivial.

REPLY [14 votes]: This is just the $L^2(\frac{dt}{t})$ distance between $e^{-xt}$ and $e^{-yt}$ (Frullani integrals) up to some positive factor. I'm not sure whether you'll call this "elementary" though.<|endoftext|>
TITLE: Ingenuity in mathematics
QUESTION [40 upvotes]: [This is just the kind of vague community-wiki question that I would almost certainly turn my nose up at if it were asked by someone else, so I apologise in advance, but these sorts of questions do come up on MO with some regularity now so I thought I'd try my luck]
I have just been asked by the Royal Society of Arts to come along to a lunchtime seminar on "ingenuity". As you can probably guess from the location, this is not a mathematical event. In the email to me with the invitation, it says they're inviting me "...as I suppose that some
mathematical proofs exhibit ingenuity in their methods." :-)
The email actually defines ingenuity for me: it says it's "ideas that solve a
problem in an unusually neat, clever, or surprising way.". My instinct now would usually be to collect a bunch of cute low-level mathematical results with snappy neat clever and/or surprising proofs, e.g. by scouring my memory for such things, over the next few weeks, and then to casually drop some of them into the conversation.
My instinct now, however, is to ask here first, and go back to the old method if this one fails.
Question: What are some mathematical results with surprising and/or unusually neat proofs?
Now let's see whether this question (a) bombs, (b) gets closed, (c) gets filled with rubbish,  (d) gets filled with mostly rubbish but a couple of gems, which I can use to amuse, amaze and impress my lunchtime arty companions and get all the credit myself.
This is Community Wiki of course, and I won't be offended if the general consensus is that these adjectives apply to the vast majority of results and the question gets closed. I'm not so sure they do though---sometimes the proof is "grind it out". Although I don't think I'll be telling the Royal Society of Arts people this, I always felt that Mazur's descent to prove his finiteness results for modular curves was pretty surprising (in that he had enough data to pull the descent off). But I'm sure there are some really neat low-level answers to this.

REPLY [9 votes]: Theorem [Inaba]. Any configuration of 10 points in
  the plane can be covered by disjoint unit disks.

Proof. Put a dense packing of unit disks on the plane at random.

Then the average number of covered points is $10\cdot \frac{\pi}{2\sqrt3}=9.069\ldots>9.$
See Covering Points with Disjoint Unit Disks for the history.<|endoftext|>
TITLE: Proving that a poset is a lattice
QUESTION [16 upvotes]: I discovered experimentally that a certain finite poset (sorry, I cannot give its definition here) seems to be in fact a (non-distributive, non-graded) lattice.  The covering relations are reasonably simple, but it seems not so easy to find out whether one element is smaller than the other, let alone find the meet (or join) of two elements.  However, it is (relatively) easy to see that the poset has a minimum and a maximum.
I wonder whether there are standard techniques for proving that a poset is a lattice, that do not need knowledge about how the meet of two elements looks like.  (In fact, any example would be very helpful.)
Some more hints:

I don't see a way to embed the poset in a larger lattice...

the poset is (in general) not self-dual, but the dual poset is itself a member of the set of posets I am looking at.

to get an idea, here (Wayback Machine) is a picture of one example (produced by sage-combinat and dot2tex)

REPLY [3 votes]: Since you seem at heart to be asking a question about how
to compute things with your partial order, let me offer
several observations from the perspective of computability
theory and computable model theory. I recognize, however,
that you may find this perspective unhelpful, and in this
case I apologize.

First, the order relation in any lattice is computable
from either the join or meet operations, since $x\leq y$ if
and only if $x\wedge y = x$ if and only if $x\vee y = y$.
Thus, the order relation can be no harder to compute than
the lub and glb's.
Second, the converse is not generally true in infinite lattices,
and one cannot generally expect to compute the meet and
join operations nor the cover relation from the order
relation itself. For example, there is a lattice order on
the natural numbers whose order relation is a computable
relation, but whose meet function and cover relation is not
computable. To construct this example, one arranges a
lattice of height $4$ in which a Turing machine program $e$
has meet $0$ with a fixed element $1$ if program $e$ does
not halt on trivial input, but otherwise the meet is a code
for the halting computation. In this way, the order
relation is computable in the sense that given $x$ and $y$
one can compute whether $x\leq y$, but there is in general
no way to compute $x\wedge y$, since from this function one
could solve the halting problem. Similarly, program $e$
covers $0$ if and only if $e$ does not halt, so the
covering relation is also not decidable. Such a kind of
model arises commonly in the subject known as computable
model theory.
Third, you mentioned that your order is finite, but any
finite structure is of course computable. That is, whatever
your relation and operations are, they are definitely
computable functions. From this perspective, the exact
content of your question becomes somewhat murky, and one
would request a greater clarity about what you are asking.
It sounds like you might have a uniform presentation of
infinitely many different partial orders, and you want to
know whether you can compute whether the $n^{\rm th}$ such
order is a lattice, or how to uniformly compute the
relations or the meets and joins. In this case, we would
need more details about what the orders are. For example,
one can easily construct examples of an infinite sequence
of partial orders, whose order relations are uniformly
computable, but the question of whether the $n^{\rm th}$
order is a lattice is undecidable. In contrast, if the
$n^{\rm th}$ order is necessarily a finite order of
computable size, then this question is always decidable by
brute force searching.
Fourth, you mention that the covering relation is easy.
In this case, the order relation should be low degree
polynomial time computable (in the size of the order),
since one can first compute the covering relation of all
pairs, and then successively compute the relation as the
transitive closure of this relation.
Finally, fifth, in response to item 1,
every partial $(P,\leq)$ embeds order-preservingly into a
lattice, since one may consider the set of downward closed
subsets of $P$. This collection is closed under unions and
intersections, and if one identifies a point in $P$ with
its lower cone, then one obtains an order-preserving map
into a lattice. This embedding preserves meets when they
exist, since the intersection of the cone below $x$ and the
cone below $y$ is the cone below $x\wedge y$, when this
meet exists. But unions will not generally preserve $\vee$
for this map. There are other possible completions that are
appropriate when the lattice exhibits certain other nice
properties. For example, when the lattice minus its minimal
element is separative, then it has a completion as a
complete Boolean algebra, the regular open algebra.<|endoftext|>
TITLE: edges minus vertices
QUESTION [6 upvotes]: Is there a more interesting name for this graph invariant: edges minus vertices?  It seems to have been called 'complexity' in

Remco van der Hofstad, Joel Spencer, Counting Connected Graphs Asymptotically, European Journal of Combinatorics 27 Issue 8 (2006) 1294–1320, doi:10.1016/j.ejc.2006.05.006, arXiv:math/0502579

and in

Joel Spencer, Probabilistic Methods in Combinatorics In: Chatterji S.D. (eds) Proceedings of the International Congress of Mathematicians,   Birkhäuser, Basel (1995), doi:10.1007/978-3-0348-9078-6_132, Wayback Machine pdf of article, big IMU pdf of whole volume

The motivation is that we want to talk about a quantity that is preserved under the graph transformation of collapsing two distinct vertices connected by an edge to a single vertex (thereby removing one edge and one vertex, preserving 'edges minus vertices').  So for example if the quantity 'edges minus vertices plus one' is more natural for some reason and has a name, then this would also be helpful.  The concept should not be restricted to e.g. planar graphs.

REPLY [13 votes]: Whether you're considering a multigraph (which may have multiple edges and/or loops) or a  simple graph, both are CW complexes.  For any finite CW complex $G$, the Euler characteristic $\chi(G)$ is defined as the alternating sum (#0-cells)-(#1-cells)+(#2-cells)-...  (see Wikipedia). Thus for a finite graph, the Euler characteristic is $|V|-|E|$.  It's a homotopy invariant, and the operation of collapsing one edge and its vertices to a single vertex is a homotopy equivalence, so any function of $|V|-|E|$ is invariant under this operation.
When the graph is connected, the quantity $|E|-|V|+1$ ($=1-\chi(G)$) is the smallest number of edges that must be removed to yield a graph with no cycles, called the cyclomatic number or the circuit rank (see Mathworld).  But if the graph is not connected, then "$+1$" must be replaced by "$+k$," where $k$ is the number of components.<|endoftext|>
TITLE: Verifying my other example in the Geometry of Numbers and Quadratic Forms
QUESTION [7 upvotes]: In answer to Pete L. Clark's question Must a ring which admits a Euclidean quadratic form be Euclidean? on Euclidean quadratic forms, I also gave an example in six or fewer variables, repeated below. Pete's Euclidean property (in the case of  positive definite integral quadratic forms) is simply that for any point $\vec x \in \mathbf Q^n$ but $\vec x \notin \mathbf Z^n,$ we require that there be at least one $\vec y \in \mathbf Z^n$ such that $$ q(\vec x - \vec y) < 1. $$
The question on the example with 7 variables was a big success, see Verifying an example in the Geometry of Numbers and Quadratic Forms
Could some kind soul please verify the example(s) below. Note how very symmetric this one is, I have little doubt that the "worst" point(s) must occur on the main diagonal $x_1 = x_2 = \cdots x_n.$ Indeed, I think that for any point the orthogonal projection onto the main diagonal has worse "Euclidean" value.
For six or fewer variables we can use one of the easiest constructions, include all mixed terms so that the Gram matrix becomes
$$  P_6 \; \; = \; \; 
\left(  \begin{array}{cccccc}
  1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}  & \frac{1}{2}\\\
  \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}  \\\
  \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}  \\\
  \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2}    \\\
  \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2}    \\\
  \frac{1}{2} &  \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1      
\end{array} 
  \right) .  $$
Then the worst $\vec x$ is either
$$ \vec x = \left( \frac{3}{7}, \frac{3}{7}, \frac{3}{7} ,  \frac{3}{7}, \frac{3}{7}, \frac{3}{7}   \right)  $$
or 
$$ \vec x = \left( \frac{4}{7}, \frac{4}{7}, \frac{4}{7} ,  \frac{4}{7}, \frac{4}{7}, \frac{4}{7}   \right)  $$
with ``Euclidean minimum''  $\frac{6}{7}.$
This construction is much easier to figure out. In dimension $ n$ we have determinant  $\frac{n +1}{2^n}$ and characteristic polynomial
 $$ \left( \frac{1}{2^n} \right) \left(2 x - (n+1) \right) \left(2 x - 1  \right)^{n-1}. $$ For even $n $ the worst $\vec x$ has either all entries $\frac{n}{2(n + 1)}$ or   $\frac{n + 2}{2(n + 1)}$  with a Euclidean minimum of  $\frac{n^2 + 2 n}{8 (n+1)}.$ 
For odd $n $ the worst $\vec x$ has all entries $\frac{1}{2}$  with a Euclidean minimum of  $\frac{n+1}{8}.$ 
The formulas $\frac{n^2 + 2 n}{8 (n+1)}$ and $\frac{n+1}{8}$ show that this recipe fails (just barely) for $n=7$ and more obviously for larger $n.$ I don't believe there are any examples with $n \geq 9$ and I have my doubts that there can be any for $n=8.$ I did try half of Gosset's root lattice for $E_8,$ see 
http://en.wikipedia.org/wiki/E8_lattice 
  but it does not seem to work to have any of the squared terms with a coefficient other than $1,$ in all likelihood as soon as $n \geq 4.$

REPLY [4 votes]: It's strange that I didn't see this question before, since Will and I have started thinking about these issues off-site.  Anyway, recently I found (in the sense of located, not discovered) the answer to this and a bit more.
First, the MAGMA computational system has a built in command to compute the Euclidean minimum of (the lattice associated to) a positive definite integral quadratic form, namely ${ \tt CoveringRadius }$.  In the examples I tried, this computation was almost instantaneous up until about $6$ variables, relatively quick for $7$ and $8$ variables, and didn't terminate when run on my local server starting in $9$ variables.  Here is how it goes for Will's six dimensional lattice:

$>$ F61 := MatrixRing(IntegerRing(), 6) ! [2,1,1,1,1,1, 1,2,1,1,1,1, 1,1,2,1,1,1\
  , 1,1,1,2,1,1, 1,1,1,1,2,1, 1,1,1,1,1,2];
  $>$ L61 := LatticeWithGram(F61);
  $>$ L61;
  Standard Lattice of rank 6 and degree 6
  Determinant: 7
  Factored Determinant: 7
  Inner Product Matrix:
  [2 1 1 1 1 1]
  [1 2 1 1 1 1]
  [1 1 2 1 1 1]
  [1 1 1 2 1 1]
  [1 1 1 1 2 1]
  [1 1 1 1 1 2]
  $>$ (1/2)*CoveringRadius(L61);
  6/7
  $>$ #GenusRepresentatives(L61);
  1
  $>$ IsIsomorphic(L61,Lattice("A",6));
  true  

Some comments: you can see that the Gram matrix is twice the one Will gives.  This is because MAGMA's lattice package likes matrices with integral entries.  Thus we compute half the covering radius rather than the covering radius to take care of this, and the answer is indeed what Will said it would be: $\frac{6}{7}$.  (In particular, this is a Euclidean quadratic form.)
The second calculation shows that this lattice has class number one in the sense of quadratic forms, i.e., its genus has only one class.  Will, Jon Hanke and I believe that any Euclidean quadratic form should have class number one, but we haven't (yet!) been able to prove this, so this was an interesting data point for that.  (I computed nearly $60$ other examples late last week.)  
The last calculation shows that Will's lattice (when rescaled by multiplication by $2$) is isomorphic to 
the $A_6$ root lattice.  The covering radii for all the root lattices are well known to the experts (though not to me): for instance, the covering radius of $A_n$ is given on p. 109 of Conway and Sloane's Sphere Packings, Lattices and Groups and the formula agrees with the one given in the question above: in particular it is asymptotic to $\frac{n}{8}$.  (Of course, in their book they explain what is going on geometrically, which is a much more desirable answer than just running a software package.)<|endoftext|>
TITLE: number of partial orders modulo a fixed number
QUESTION [22 upvotes]: Let p(n) be the number of partial orders on the set {1,...,n}.  From the Online Encyclopedia of Integer Sequences, we find that the known values of p(n) are 
{1,1,3,19,219,4231,130023,6129859,431723379,44511042511,6611065248783,1396281677105899,414864951055853499,171850728381587059351,98484324257128207032183,77567171020440688353049939,83480529785490157813844256579,122152541250295322862941281269151,241939392597201176602897820148085023}.
We see that the units digits of these numbers appear to cycle with a period of length four: 1, 3, 9, 9.
Experiments with other moduli indicate that given a prime modulus m, the sequence cycles with a period of length m-1.  If the modulus m is a prime power, then the period appears to be of length phi(m), where phi is Euler's phi-function.  For any modulus m, the period appears to be of length the least common multiple (LCM) of the constituent period lengths.  For example, if m=12, the period appears to be of length LCM(phi(4),phi(3))=LCM(2,2)=2.
I don't know how to prove this conjecture and I don't see any reference to it.  If proved, perhaps this result together with an asymptotic estimate for p(n) could be used to find higher values of p(n).

REPLY [17 votes]: For q prime, enlarge  $\{ 1,\cdots,m \}$ to a set of size $n=m+(q-1)$ by replacing $m$ by $q$ clones $m_1 , m_2 , \cdots , m_q$ and consider the $q$-cycle $\sigma=(m_1\ m_2\ \cdots \ m_q)$. It acts on the set of partial orders of the $n$-set and each of its orbits has size 1 or size q. Each orbit of size 1 arises from a unique partial order of the $m$-set by having all $p$ clones behave identically to the original. This proves that $p(m+(q-1)) \equiv p(m) \mod q $  I think I see how to generalize to $q^k$ but I'll have to think about it. The same idea should apply to a wider variety of structures, but which ones?
later The argument seems as if it should work for bipartite graphs on n labelled vertices and also connected bipartite graphs except for powers of 2 The data at OEIS supports this as far as it goes, ignoring the numbers for less than 3 vertices. 
http://www.oeis.org/A047864 http://www.oeis.org/A001832
It also works for appropriate restricted classes such as series parallel networks with n labelled vertices and parallel edges allowed. http://www.oeis.org/A053554 
Here is my argument for why $p(n+\phi(q^2)) \equiv p(n) \mod q^2$. I think it generalizes to $q^k$: Further enlarge the $n$ set above to one of size $m+q^2-1=n+\phi(q^2)=N$ by replacing each clone $m_i$ by $q$ clones $m_{i1}, m_{i2}, \cdots ,m_{iq}$ and consider the $q^2$ cycle $$\tau=(m_{11}m_{21}\cdots m_{q1}m_{12}m_{22}\cdots m_{q,q})$$ It acts on partial orders of the $N$-set and the action has orbits of size 1, $q$ and $q^2$. The orbits of size less than $q^2$ are in bijective correspondence with the orbits of the same size for the action of $\sigma$ on partial orders of the $n$-set.<|endoftext|>
TITLE: What part of the fundamental group is captured by the second homology group?
QUESTION [43 upvotes]: Let $X$ be a connected CW complex. One can ask to what extent $H_\ast(X)$ determines $\pi_1(X)$. For example, it determines its abelianization, because the Hurewicz Theorem implies that $H_1(X)$ is isomorphic to the abelianization of $\pi_1(X)$.
I'm thinking about invariants of 2-knots which can be extracted from have to do with the second homology of (covers of) their complements, and I'm therefore very much interested in the answer to the following question:
What part of the fundamental group is detected by $H_2(X)$?
In particular, is there an obvious map from $H_2(X)$ (or from part of it) into $\pi_1(X)$? Where in the derived series of $\pi_1(X)$ would the image of $H_2(X)$ live?

REPLY [6 votes]: Maybe it is worth to point out something that is at the origin of the results mentioned in several answers to the question. Hopf in one of his famous papers (Fundamentalgruppe und zweite Bettische Gruppe, Comment. Math. Helv. 14 (1940), 257-309) gives a characterization of the cokernel of the Hurewicz map in degree 2. His result states that for an arcwise-connected locally finite simplicial complex $X$, the Hurewicz map $h_2:\pi_2(X)\to H_2(X)$ has cokernel:
$$ H_2(X)/h_2(\pi_2(X))\simeq (R\cap [F,F])/[F,R], $$
where $\pi_1(X)\simeq F/R$ is any presentation of the fundamental group. 
In the same paper, he observes first that for any finitely generated group $G$ with a presentation $F/R$ the abelian group:
$$ G^*_1=(R\cap [F,F])/[F,R] $$
only depends on $G$, and not on the particular presentation $F/R$. 
In our modern day language, we would write Hopf's result as stating that:
$$ H_2(X)/h_2(\pi_2(X))\simeq H_2(\pi_1(X)), $$
which is the form that appears in some of the answers above.<|endoftext|>
TITLE: Difficult Infinite Sum
QUESTION [5 upvotes]: Does anyone know of a way to simplify this sum?
$$S(n)=\sum_{j=1}^{\rho(n)}\sum_{k=1}^\infty\frac{\sin[2\pi k n 2^{-j}]-\sin[2\pi k (n-1) 2^{-j}]}{k}$$
where $\rho(n)=[\log_2(n)]$ (and $[x]$ denotes the greatest integer less than $x$).
Note: This question is a follow-up to a previous question I asked:
Greatest power of two dividing an integer
EDIT: After following all the given suggestions, I found that for integer $n$,
$$\frac{S(n)}{\pi}=2^{-\rho(n)}-1+\frac{1}{1+(-1)^n}\sum_{j=1}^{\rho(n)}\left[\frac{n}{2^j}\right]-\left[\frac{n-1}{2^j}\right].$$
This is pretty much what I started with in my previous post, so if anyone knows of a way to take this sum, please let me know. Anyway, I will leave this result here in case anyone ever comes across $S(n)$ in some other context. Thanks to everyone who helped.

REPLY [10 votes]: The summand is even in k, and $\sin(x)/x$ has a nice Fourier transform, so it is very tempting to symmetrise in k (and throw in k=0, adopting the usual convention that the sinc function $\sin(x)/x$ equals 1 at x=0) and then apply Poisson summation.  This should lead to a nice expression for the inner sum that can then be summed in j.  (Admittedly there is the issue that the sinc function is not absolutely integrable, but one can still proceed formally for the purposes of getting the right answer, and then go back and make things more rigorous, e.g. by using the theory of distributions, if this becomes necessary.  I'm guessing that there is some cancellation between the two terms in the summand that will assist in this task.)
The logarithmic expressions of the form $\log(1 - e^{ix})$ that appeared in the other comments are not as scary as they seem.  Observe that the final sum is real, so one only needs the real parts of things like $i\log(1 - e^{ix})$, i.e. the phase of $\log(1-e^{ix})$, but this is basically something like $x/2 \pm \pi/2$ (depending on branch cuts).  So there is probably going to be quite a bit of simplification.  (But going via the Poisson summation route rather than the logarithmic power series route may be a bit more direct, even if both methods ought to give the same answer at the end.)<|endoftext|>
TITLE: Identifying the stacks in Devinatz-Hopkins-Smith
QUESTION [28 upvotes]: I read the Devinatz-Hopkins-Smith proof of the nilpotence conjectures last year, and while I followed along sentence to sentence I don't think I understood much of the motivating reasons for why what they did was a sensible mode of proof.  I intend to go back and figure some of these conceptual pieces out; this question is a step in that direction.
The proof of smash nilpotence is, on the face of it, accomplished by a sequence of interpolations.  The original statement is:

(Smash nilpotence:) If a map off a finite spectrum induces the zero map on $MU$-homology, then the map is in fact smash nilpotent.

Without any hassle, they reduce this to the following statement:

(Special case of Hurewicz nilpotence:) Suppose $\alpha$ is an element of the homotopy $\pi_* R$ of an associative ring spectrum $R$ of finite type.  If $\alpha$ is in the kernel of the map $\pi_* R \to MU_* R$ induced by smashing $R$ with the unit map $S \to MU$, then $\alpha$ is nilpotent.

To address this question, they interpolate between the sphere spectrum and $MU$ in two ways.  First, they produce a sequence of spectra $X(n)$, each given by the Thom spectrum associated to the composite $\Omega SU(n) \to \Omega SU \to BU$, sort of a restriction of the Bott map.  The colimit of the $X(n)$ is $MU$, and hence $X(N)_* \alpha$ is zero for some sufficiently large $N$, where $\alpha$ is considered as a map $S^t \to R$.  Since $X(1)$ is the sphere spectrum, the new goal is to show that nilpotence in $X(n+1)$-homology forces nilpotence in $X(n)$-homology for any $n$, then to walk down from $X(N)$ to the sphere spectrum.  To move between $X(n+1)$ and $X(n)$, they interpolate between these spectra by pulling $X(n+1)$ apart using the filtered James construction; this results in an increasing sequence of $X(n)$-module spectra $F_{n, k}$ satisfying $F_{n, 0} \simeq X(n)$ and converging to $X(n+1)$ in the limit.
The rest of the argument falls into two pieces:

1) If $\alpha: S^t \to R$ is nilpotent in $X(n+1)$-homology then it induces the zero map in $F_{n,p^k-1}$-homology for some large $k$ --- that is, our argument continues somewhere in the approximating tower between $X(n+1)$ and $X(n)$.
2) If it induces the zero map in $F_{n, p^k-1}$-homology it also induces the zero map in $F_{n, p^{k-1}-1}$-homology.

To prove part 1, they investigate the $X(n+1)$-based Adams spectral sequence
$$\mathrm{Ext}^{*, *}_{X(n+1)_* X(n+1)}(X(n+1)_*, X(n+1)_* F_{n, p^k-1} \wedge R) \Rightarrow (F_{n, p^k-1})_* R.$$
The key is the existence of vanishing lines in these spectral sequences, where the slope of the vanishing line can be made small by making $k$ large.  In order to establish these vanishing lines, they perform a sequence of approximations, finishing with spectral sequences with the following $E_2$ terms:
$$\mathrm{Ext}^{*, *}_{\mathbb{F}_p[b_n]}(\mathbb{F}_p, \mathbb{F}_p\{1, \ldots, b_n^{p^k-1}\}),$$
$$\mathrm{Ext}^{*, *}_{\mathbb{F}_p[b_n^{p^k}]}(\mathbb{F}_p, \mathbb{F}_p).$$

My question is: Is there a geometric interpretation for the stacks associated to the Hopf algebroids above? ( --or any of the other Hopf algebroids involved which I haven't listed.)  Or: what's the geometric content of Part II of D-H-S?

For instance, it's well-known that there's a spectral sequence computing the homotopy groups of real K-theory whose input corresponds to the moduli stack of quadratics and translations.  This stack is supposed to parametrize the available multiplicative groups over some non-algebraically closed field, which provides some geometric insight into the problem.  I'd like to know if there's some kind of geometry that corresponds to the controlling stacks that sit at the bottom of the D-H-S argument.
Note: Of course, a positive answer to this question as phrased might not mean much.  The process that D-H-S uses to reduce to this much smaller Ext calculation is an extremely lossy one with the very clear intention of just getting at the existence of a vanishing line.  The geometry of these bottom stacks may have very little to say about the geometry of the stacks we started with.

REPLY [12 votes]: The best I can do at this point is give you some interpretation of the Hopf algebroids associated to $X(n)$.
We can't really give a complete description of the stacks associated to these Hopf algebroids for the simple reason that their homotopy groups are pretty uncomputable.  However, we can say a little about the role these play in homotopy theory.
First, no matter what $R$ is, the Hopf algebroid in spectra $(R, R \wedge R)$ attempts to recover the sphere from its cobar construction, and if $R_*R$ is a flat $R_*$-module we get a Hopf algebroid $(R_*, R_* R)$ attempting to do the same.  In the cases of $X(n)$ these spectral sequences converge, so in some sense we should think of $(X(n)_*, X(n)_* X(n))$ as just another presentation of the moduli of formal groups.
In a little more detail, let's consider $MU_* X(n)$.  This is a subring $MU_*[b_1, b_2, \ldots, b_n] \subset MU_* MU$.  The latter ring parametrizes pairs of a formal group law together with a strict isomorphism $f(x) = \sum b_i x^{i+1}$ out to another formal group law; the former ring is the subring parametrizing a formal group law together with the truncation with a strict isomorphism out determined only up to the first $n$ coefficients.  This is a natural description in terms of the $MU_* MU$-coaction, and so you might think of $X(n)$ as being associated to a moduli $\mathcal{M}_{fg}(n)$ of "formal groups equipped with a coordinate, determined up to the n'th stage".
Even further, you can consider the map of Hopf algebroids
$$(MU_*, MU_* MU) \to (MU_*, \pi_* (MU \wedge_{X(n)} MU)).$$
The latter computes the homotopy groups of $X(n)$ and is definitely the Hopf algebroid associated to the cover $Spec(MU) \to \mathcal{M}_{fg}(n)$.
The $X(n)$-based Adams spectral sequence is then something you might think of as being associated to a Cartan-Eilenberg type spectral sequence associated to a composite sequence of covering stacks 
$$
Spec(MU) \to \mathcal{M}_{fg}(n) \to \mathcal{M}_{fg}.
$$<|endoftext|>
TITLE: Who first came up with the idea of essential/Morita equivalence of internal groupoids/categories?
QUESTION [8 upvotes]: The idea that stacks can be identified with groupoids internal to the base site $S$ up to what is variously called essential/Morita equivalence is well known. The basic idea is that one takes the 2-category of internal groupoids and localises with respect to internal functors $f:X\to Y$ such that
$$X_1 \simeq X_0^2\times_{f_0^2,Y_0^2,(s,t)} Y_1$$
("$f$ is fully faithful") and the composite map
$$X_0\times_{f_0,Y_0,s}Iso(Y_1) \stackrel{pr_2}{\to} Iso(Y_1) \stackrel{t}{\to} Y_0 \qquad (1)$$
is some sort of "surjective" map. What this means depends on the site one works with. For example, one could be working with Lie groupoids, then (1) is a surjective submersion. Or topological groupoids and (1) admits local sections. The oldest reference I know to this sort of functor is in

M. Bunge and R. Paré, Stacks and equivalence of indexed categories, Cahiers Topologie
Geom. Differentielle 20 (1979), no. 4, 373–399.

where they take $S$ to be finitely complete and regular, and (1) to be a regular epimorphism. Bunge and Paré call the externalisation of such an internal functor a weak equivalence.
Is this the earliest reference to this idea of weak equivalences of internal categories/groupoids?

REPLY [11 votes]: The stacks studied in the reference [1] are fibrations over a base topos S not internal to a site S. The notion of an intrinsic stack originates with Lawvere, where one replace a site S with the site consisting of the class epimorphisms in the topos S. In this sense, the reference given is the first one in the literature. It is not needed that
S be a topos as mentioned above by David Roberts.
[1] Marta Bunge and Robert Pare, Stacks and Equivalence of Indexed Categories, Cahiers de Topologie et Geometrie Differentielle Categoriques XX-4 (1979) 373-399<|endoftext|>
TITLE: Are there unique geodesics in the NIL and SOL geometry?
QUESTION [11 upvotes]: Is there a unique geodesics between any two points in the NIL (resp. SOL) geometry?
If so, is there a nice way of parametrizing them? For example geodesics in $S^3$ can be parametrized using the embedding in $\mathbb{R}^4$ and $\sin , \cos$ functions. Geodesics in hyperbolic space can be parametrized using the hyperboloid model and the functions sinh,cosh.

REPLY [3 votes]: The geodesics in SOL that are not contained in a vertical (hyperbolic) foliation tend to come in families meeting periodically and are thus highly non unique. They are precisely described in my paper L’Horizon de SOL.  Expos. Maths 1998.
In this paper, the SOL geodesics are explicitly parametrized using elliptic functions.<|endoftext|>
TITLE: Is a rhombus rigid on a sphere or torus? And generalizations
QUESTION [19 upvotes]: If a rectangle is formed from rigid bars for edges and joints
at vertices, then it is flexible in the plane: it can flex
to a parallelogram.
On any smooth surface with a metric, one can define a linkage
(e.g., a rectangle) whose edges are geodesics of fixed length,
and whose vertices are joints, and again ask if it is rigid
or flexible on the surface.
This leads to my first, specific question:

Q1.
  Is a rhombus, or a rectangle,
  always flexible on a sphere?


It seems the answer should be Yes but I am a bit uncertain
if there must be a restriction on the edge lengths.
(In the above figure, the four arcs are each $49^\circ$ in length, comfortably short.)

Q2. 
  The same question for other surfaces: Arbitrary convex surfaces? A torus?

I am especially interested to learn if there are situations where a linkage that is flexible
in the plane is rendered rigid when embedded on some surface.
It seems this should be possible...?

Q3.
  More generally, 
  Laman's theorem
  provides a combinatorial characterization of the rigid linkages in the plane.
  The $n{=}4$ rectangle is not rigid because it has fewer than $2n-3 = 5$ bars:
  it needs a 5th diagonal bar to rigidify.
  Has Laman's theorem been extended to arbitary (closed, smooth) surfaces
  embedded in $\mathbb{R}^3$?
  Perhaps at least to spheres, or to all convex surfaces?

Thanks for any ideas or pointers to relevant literature!
Addendum.
I found one paper related to my question:
"Rigidity of Frameworks Supported on Surfaces"
by A. Nixon, J.C. Owen, S.C. Power.
arXiv:1009.3772v1 math.CO
In it they prove an analog of Laman's theorem for
the circular cylinder in $\mathbb{R}^3$.
If one phrases Laman's theorem as requiring for
rigidity that the number of edges $E \ge 2 V - 3$
in both the graph and in all its subgraphs,
then their result (Thm. 5.3) is that, on the cylinder, rigidity requires
$E \ge 2 V -2$ in the graph and in all its subgraphs.
This is not the precise statement of their theorem.
They must also insist that the graph be regular
in a sense that depends on the rigidity matrix achieving maximal rank
(Def. 3.3).
They give as examples of irregular linkages on a sphere
one that
contains an edge with antipodal endpoints, or one that includes
a triangle all three of whose vertices lie on a great circle.
But modulo excluding irregular graphs and other minor technical
details, they essentially replace the constant 3 in Laman's
theorem for the plane with 2 for the cylinder.
Theirs is a very recent paper but contains few citations to related
work on surfaces, suggesting that perhaps the area
of linkages embedded on surfaces is
not yet well explored.
In light of this apparent paucity of information, it seems appropriate that I 'accept'
one of the excellent answers received.  Thanks!
Addendum [31Jan11].
I just learned of a 2010 paper by Justin Malestein and Louis Theran,
"Generic combinatorial rigidity of periodic frameworks"
arXiv:1008.1837v2 (math.CO),
which pretty much completely solves the problem of linkages on a flat 2-torus,
generalizing to flat orbifolds.  They obtain a combinatorial characterization for generic minimal
rigidity for "planar periodic frameworks," which encompass these surfaces.

REPLY [3 votes]: The first paragraph of Some notes on the equivalence of first-order rigidity in various geometries by Franco V. Saliola and Walter Whiteley states:

In this paper, we explore the connections among the theories of first-order rigidity of bar and joint frameworks (and associated structures) in various metric geometries extracted from the underlying projective space of dimension n, or $\mathbb{R}^{n+1}$. The standard examples include Euclidean space, elliptical (or spherical) space, hyperbolic space, and a metric on the exterior of hyperbolic space.

Section 4 of the notes proves that a framework is first-order rigid in the upper hemisphere $S^n_+$ iff a corresponding framework in $\mathbb{R}^n$ is first-order rigid.  
Section 5 extends this to geometries on the space $X^n_{c,k}=\{x\in\mathbb{R}^{n+1}|\langle x,x\rangle_k=c,x_{n+1}>0\}$ where $\langle x,x\rangle_k=x_1y_1+\cdots+x_{n-k+1}y_{n-k+1}-x_{n-k+2}y_{n-k+2}-\cdots-x_{n+1}y_{n+1}$.  Note that $k=1,c=-1$ is hyperbolic space $\mathbb{H}^n$, the case $k=0,c=1$ is $S^n_+$.<|endoftext|>
TITLE: Effective theta characteristics
QUESTION [8 upvotes]: Let $C$ be a complex smooth projective curve of genus $g$ and let $N$ be the number of effective theta-characteristics of $C$, or equivalently, the number of points of order two on the theta divisor of $J(C)$.
It is known that, if $C$ is generic, $N$ is the number of the odd theta characteristics. Mumford proves that on a principally polarized abelian variety the theta divisor cannot contain all the points of order two. It follows that $N<2^{2g}$.
Given an arbitrary curve $C$, is it known a upper bound for $N$ depending on $g$?

REPLY [4 votes]: Since the odd theta characteristics are always effective, one might equivalently ask how many even theta characteristics are effective. They are called vanishing theta characteristics.
If $C$ is a hyperelliptic curve of genus $g$, then there are 
$ c_g = \frac{1}{2} \left( \begin{array}{c} 2g+2 \\ g+1  \end{array} \right) $ even theta characteristics that do not vanish$^1$. Which means, the number $N$, defined in your question, for hyperelliptic curves of genus $g$ is $2^{2g} - c_g$. 
This at least provides a lower bound for an upper bound for $N$. 
Arnaud Beauville in Vanishing thetanulls on curves with involution looks further. However, he concludes his paper by saying that what you are asking is open for non-hyperelliptic curves even in $g \ge 6$. 
It could be possible that the maximum $N$ is attained on curves with involution, or even on hyperelliptic curves. If you could show that, you would be able to answer your question using these results.
[1]
See the proof of Lemma 5.2.2 in Dolgachev's Classical Algebraic Geometry.<|endoftext|>
TITLE: Orthogonal similarity of matrices
QUESTION [5 upvotes]: Given $M\in M_n({\mathbb R})$ 
and $\ell\in{0,\ldots,n-1}$, we define
$$d_\ell(M)=\sum_{j=1}^nm_{j,j+\ell},$$
where the indices are understood mod $n$. In particular, $d_0$ is the trace operator.
Let $A\in M_n({\mathbb R})$ be given. We define a map $\Delta: O_n({\mathbb R})\rightarrow{\mathbb R}^{n-1}$ by $$\Delta(R)=(d_1(R^TAR),\ldots,d_{n-1}(R^TAR)).$$ Mind that we omit $d_0(R^TAR)$, because we know in advance that it equals the trace of $A$.
Question. Does it exist an orthogonal matrix $R$ such that $\Delta(R)=(0,\ldots,0)$ ?
The requested property ressembles one for which the answer is known to be positive: find $R$ orthogonal such that the diagonal $R^TAR$ is constant (thus equal to $\frac{1}{n}{\rm Tr}A$). Both properties consist of $n-1$ linear constraints, and both are consistent with the fact that the mean value of $R^TAR$ over $SO_n$ is $(\frac{1}{n}{\rm Tr}A) I_n$. Thus the answer would certainly be positive if the stronger following statement is true.
Statement. The image of $SO_n$ under $\Delta$ is convex. True or False ?
This statement looks ambitious, since $\Delta$ is not linear, and $SO_n$ is not a convex set. But an optimistic mathematicien will say that it ressembles the Toeplitz-Hausdorff theorem about the convexity of the image of the complex unit sphere under the quadratic map $x\mapsto x^*Mx$. Note that the T-H thm is used to find an $R^TAR$ with constant diagonal.

REPLY [2 votes]: Believe me, I didn't know the answer when I asked the question. But now I do. It is No. Here is a counterexample, a $3\times3$ matrix $A$ for which $\Delta_A$ does not vanish over the orthogonal group.
The matrix is that of a rotation of angle $2\pi/3$ around some axis. For instance $A$ can be taken as the matrix of the permutation $[1,2,3]$. Its orbit under orthogonal conjugation is the set of all rotations of angle $2\pi/3$. So let $B$ be such a rotation, and let $v=(a,b,c)$ be the unitary vector about which the rotation takes place. Then $d_1(B)$ and $d_2(B)$ are
$$\frac{3}{2}(ab+bc+ca)\pm\frac{\sqrt3}{2}(a+b+c).$$
Thus $d_1=d_2=0$ means $ab+bc+ca=0$ and $a+b+c=0$, which are incompatible with $a^2+b^2+c^2=1$.<|endoftext|>
TITLE: Mori cone of homogeneous varieties
QUESTION [5 upvotes]: Let $X$ be an homogeneous projective variety, written as the quotient $G/P$, where $G$ is a Lie group and $P$ is a parabolic subgroup of it. It seems it is well-known that the monoid of effective curves on $X$, as a submonoid of $H_2(X, \mathbb{Z})$, is generated by finitely many curves $\beta_1, \ldots, \beta_p$. Moreover the $\beta_i$ are embedded $\mathbb{P}^1$'s in $X$. For instance in the expository notes http://arxiv.org/abs/alg-geom/9608011, the authors use this fact (e.g. at the beginning of p. 37).
The question is simply: how does one prove this fact? Is there a reference for a precise proof of it (perhaps in a slightly different generality, or proving the same statement for the effective monoid inside the Chow group, instead of homology)?

REPLY [3 votes]: One place this is treated is in Brion's notes, http://arxiv.org/abs/math/0410240, particularly Section 1.4, and the references in the notes at the end of that section.
A little more precisely, one shows that a divisor (line bundle) is nef iff it is globally generated, and this cone is generated by the Schubert divisors.  Using the fact that the Schubert classes form a self-dual basis for $H^*(X,{\Bbb Z})$ (as well as $A^*X$) under the intersection pairing, it follows that the "dual classes" to the Schubert divisors generate the cone of curves.  (As with many such things in $G/P$ world, the arguments for cohomology and Chow groups are exactly the same.)
By the way, it's an easy consequence of the existence of a dense open $B$-orbit that the pseudo-effective cone is generated by the complement of the open orbit --- that is, again by the Schubert divisors.  So for $G/P$, the nef and pseudo-effective cones are the same.<|endoftext|>
TITLE: Comparing cohomology over ${\mathbb C}$ and over ${\mathbb F}_q$
QUESTION [22 upvotes]: I have the following (probably well-known) question: let $X$ be a regular scheme over 
$\mathbb Z$. Let $p$ be a prime and  Let us denote the reduction of $X$ mod $p$ by $X_p$.
Let also $X_{\mathbb C}$ be the corresponding complex variety. Suppose that I know everything
about cohomology of $X_{\mathbb C}$ (i.e. I know the mixed Hodge structure). Is there a situation where I can use that to describe the
$\ell$-adic cohomology of $X_p$ with the action of Frobenius? More precisely, are there
conditions that would guarantee that all the eigen-values of Frobenius are powers of $q$
and that those powers are exactly those predicted by the Hodge structure on the cohomology
of $X_{\mathbb C}$?
UPDATE: I would like to emphasize that the scheme $X$ I am talking about is not proper
and its cohomology is not pure.
However, it is not very bad either: roughly
speaking nothing worse than ${\mathbb G}_m$ can appear there. In my specific situation conjecturally
(I don't know how to prove this) there is a ${\mathbb G}_m$-action on $X$ which
contracts it to a subscheme $Y$ (i.e. there exists a morphism ${\mathbb A}^1\times X\to X$ extending the ${\mathbb G}_m$-action such that
$\{ 0 \} \times X$ goes to $Y$) where $Y$ is a disjoint union of locally closed subschemes of the 
form ${\mathbb A}^k\times {\mathbb G}_m^l$. Note that in this case 
the cohomology of $X$ is the same as cohomology of $Y$, so this should give you an idea
how complicated the cohomology of $X$ is.

REPLY [13 votes]: You have to be a little careful because of the following sort of phenomenon: the pair of conjugate points $x^2 + 1 = 0$ has Frobenius eigenvalues equal to $1$ if $p$ is 1 mod 4,
but equal to $1$ and $-1$ if $p$ is $-1$ mod 4.  But its Hodge structure is no different
to the pair of rational points $x^2 - 1$, for which the Frobenius eigenvalues are always 1.  (If you want an example where the scheme is a variety, rather than geometrically disconnected, instead consider the plane blown up at
these two points.)
The condition you want is that the cohomology over $\mathbb C$ be generated by Hodge classes
(i.e. that the cohomology be all in even degree, and that in degree $2i$ it all be of type
$(i,i)$).  This will give that the Frobenius eigenvalues are powers of $q$ for almost all $p$,
up to multiplication by a root of unity.  (This is the Artin--Tate statement of Bhargav's answer.)  The above example shows that these roots of unity can appear.
A paper which discusses this kind of question pretty comprehensively, and where the preceding statement is proved, is the
linked paper of Kisin and Lehrer.
(See Corollary 2.3.)  See also this paper of Kisin and Wortmann, which provides key results for the argument of Kisin and Lehrer.  
The basic technique that you can use to connect the Hodge theory at infinity to the Frobenius elements mod primes is $p$-adic Hodge theory, and this is what Kisin--Lehrer
and Kisin--Wortmann use.<|endoftext|>
TITLE: Secondary fans and Stanley Reisner ideals
QUESTION [6 upvotes]: Consider a collection of points $S \subset \mathbb R^d$.  I would like to understand all possible fans $\Sigma$ whose support is the cone over $S$: $|\Sigma| = cone(S)$.  
I have heard that the secondary fan does this for me, but I am having trouble parsing the relevant sections of GKZ.  I would like to understand this completely, but to begin I would really like to know 

How to construct the resulting set of Stanley-Reisner ideals for each possible $\Sigma$.

Can someone explain this and/or give a readable explanation of

How the secondary fan is built and how each cone gives a refinement of $cone(S)$?

REPLY [5 votes]: Consider any function $f$ from $S \to \mathbb{R}$. Define a function $\tilde{f}: \mathrm{cone}(S) \to \mathbb{R}$ by 
$$\tilde{f}(w) = \max \{ \sum a_i f(s_i) : \ \sum a_i s_i = w,\ a_i \geq 0 \}.$$
That is, for every way of writing $w$ as a positive linear combination of the $a$'s, try extending $f$ linearly, and find the largest possible extension.
Then $\tilde{f}$ is a piecewise linear convex function. The domains of linearity for $\tilde{f}$ form a fan, each of whose cones is spanned by a subset of $S$. 
Subdivide $\mathbb{R}^S$ into cones, where functions $f$ and $g$ are in the same cone if $\tilde{f}$ and $\tilde{g}$ have the same domains of linearity. This is the secondary fan.
Thus, the cones of the secondary fan are in bijection with fans $F$ whose support is $\mathrm{cone}(S)$, where every face of $F$ is of the form $\mathrm{cone}(T)$ for some $T \subseteq S$, and where there is a convex piecewise linear function whose domains of linearity are the faces of $F$. 

Several warnings:
(0) You didn't say that you want every cone of your fan to be of the form $\mathrm{cone}(T)$, for $T \subseteq S$, but I assume that you meant to. If not, there will be infinitely many fans meetng your conditions, because you can always insert new vertices.
(1) There can be fans which do not support any convex piecewise linear function. Such fans are called non-coherent or non-regular. They do not correspond to cones of the secondary fan.
(2) As I have described it, the secondary fan lives in $\mathbb{R}^S$. Note that, if $\lambda$ is a linear function on $\mathbb{R}^d$, then $f$ and $f+\lambda$ always lie in the same cone of the secondary fan, so the secondary fan is invariant under this action of $(\mathbb{R}^d)^{\vee}$. Many references quotient by this action. 
(3) Many references specialize to the case that $S$ lies in an affine hyperplane, and then draw their pictures in $\mathbb{R}^{d-1}$. I think GKZ does this, which might be part of your difficulty.
(4) You mention Stanley-Reisner ideals. You don't say what your motivation is, but it sounds like you might be computing Grobner degenerations of toric varieties. If so, remember that these can be nonreduced, and the secondary fan will only see the reduced structure. See Sturmfels paper Gröbner bases of toric varieties for details.

Have you tried reading the Billera, Filliman and Sturmfels paper, Constructions and complexity of secondary polytopes? I seem to remember it is very clear, although it also has weakness (3).<|endoftext|>
TITLE: Extensions of an infinite product of copies of Z by Z
QUESTION [8 upvotes]: The question is simple:
Let $P$ be an infinite direct product of copies of $\mathbb Z$. Do there exist any nontrivial extensions
$$0 \to \mathbb Z \to E \to P \to 0$$
in the category of commutative groups?
In other words, I am asking whether the group $\mathrm{Ext}^1(P,\mathbb Z)$ is trivial. The problem here is of course that the group $P$ is not a free group.
Already a funny thing happens with $\mathrm{Hom}(P,\mathbb Z)$. For any finite or infinite index set $I$, the canonical evaluation map
$$\bigoplus_{i\in I}\mathbb Z \to \mathrm{Hom}\Big(\mathrm{Hom}\Big(\bigoplus_{i\in I}\mathbb Z,\:\mathbb Z \Big),\:\mathbb Z \Big) \cong \mathrm{Hom}\Big(\prod_{i\in I}\mathbb Z,\:\mathbb Z \Big)$$
is an isomorphism! That is a nontrivial statement (due to??), whose proof is not a formality at all. Replacing $\mathbb Z$ by, say, $\mathbb Z/p\mathbb Z$, the corresponding statement is wrong for infinite $I$.

REPLY [3 votes]: Here is a complete answer; I think it is more or less what Steve wrote in his comment, except I don't understand the appearance of $\mathbb{R}$ there. If $I$ is the infinite index set, let $L=\mathbb{Z}^{(I)}\subset P$ be the obvious free submodule. Then $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$. 
EDIT: the last formula is wrong, see Martin's and Steve's comments below.
Now $P/L$ has a big divisible subgroup $D$, whose inverse image in $P$ consists of maps $I\to\mathbb{Z}$ converging to zero in $\widehat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$). (For instance, if $I=\mathbb{N}$ take the sequence $n\mapsto n!$). Since $P/L$ is torsion-free (imediate), $D$ is a nonzero $\mathbb{Q}$-vector space. Since $D$ is divisible it is a direct summand of $P/L$; hence, $P/L$ admits $\mathbb{Q}$ as a direct summand. But it is well known (and easy to see) that $\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\cong\widehat{\mathbb{Z}}$, hence $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$.<|endoftext|>
TITLE: Distinct well-orderings of the same set
QUESTION [36 upvotes]: An easy consequence of the Erdős-Dushnik-Miller theorem $\kappa\to(\kappa,\omega)^2$ is the following, that  will denote $(*)$ (it appears as an exercise in Kunen's book, it was probably mentioned explicitly earlier by Dushnik-Miller or Erdős, but I haven't found a reference):

If $X$ is infinite and $<_1$ and $<_2$
  are two well-orderings of $X$, then
  there is a subset $Y$ of $X$ with
  $|Y|=|X|$ where $<_1$ and $<_2$
  coincide. 

The proof uses choice, but it follows that it holds in ZF: We may as well assume $X$ is an ordinal $\kappa$, so we can code $<_1$ and $<_2$ with a set $A\subseteq \kappa^2$, and argue in $L[A]$, which is a model of choice.
This proof has always looked bizarre to me. I remember a few years ago discussing this with Clinton Conley, and I believe we found a combinatorial argument, but I don't recall how it went, so I am not sure. 

What I am asking then is for a
  combinatorial proof of $(*)$ in ZF.

(``Combinatorial'' may signify several things, what I mostly mean is that no metamathematics appear in the proof.)
Perhaps there is some kind of inductive argument (I vaguely recall that's what we had), but I don't think it can be completely straightforward. For example, $X=\omega_1$ could have cofinality $\omega$, so the obvious approach may run out of room before $\omega_1$ steps.

REPLY [4 votes]: Just a little update on this question.  The argument in Andres' answer actually gives the following (in principle stronger but who knows) fact in ZF: any ordinal-valued function on a wellordered set $X$ is nondecreasing on a subset equinumerous with $X$.  This doesn't seem particularly revolutionary.  I don't see an adaptation of the argument that gives the full E-D-M theorem in the choiceless setting.<|endoftext|>
TITLE: Why is Maps(X,Y) an open subfunctor of Hilb(X x Y)?
QUESTION [10 upvotes]: Let $X$ and $Y$ be projective schemes. Then we can define the mapping scheme between them, $\rm{Maps}(X,Y)$ as follows:
To any map $f:X\rightarrow Y$ we consider the graph $\Gamma_f$ as a closed subscheme of $X \times Y$. So $\rm{Maps}(X,Y)$ is the set of all subschemes of $X \times Y$ that are graphs of morphisms. (Concretely, a subscheme $Z \subset X \times Y$ is the graph of a morphism iff the projection to $X$ is an isomorphism) Of course this all makes sense in families, so $\rm{Maps}(X,Y)$ is a subfunctor of the Hilbert scheme $\rm{Hilb}(X \times Y)$.
Now at this point, I have seen a number of sources casually claim that $\rm{Maps}(X,Y)$ is actually an $\it{open}$ subfunctor and is hence representable. None of these sources even remark on why this is true? So my question is: why is this true?

REPLY [10 votes]: See Grothendieck, seminaire bourbaki 221 "les schemas de Hilbert", bottom of page 221-19.<|endoftext|>
TITLE: subsets of products of trees
QUESTION [7 upvotes]: A subset of a geodesic metric space is called convex if for every two points in the subset one of the geodesics connecting these points lies in the subset. Is it true that every convex subset of a product of two trees with $l_1$-metric is a median space, that is for every three points A,B,C in the subset there exists a point D in the subset such that D is on (some) geodesics connecting A and B, B and C, A and C?

REPLY [5 votes]: I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take $\mathbb{R}^3$ with the $l^1$ metric and inside it the plane $x+y+z=1$. It is convex (because Euclidean segments are geodesics also for $l^1$), not strongly convex, and not median: it contains the points $(1,0,0), (0,1,0), (0,0,1)$ but not their median point $(0,0,0)$.<|endoftext|>
TITLE: Spencer-Brown's claimed proof of the four color theorem
QUESTION [5 upvotes]: I read on Wikipedia that G. Spencer-Brown gave a non-computer based proof of the four color theorem. As I'm not an expert in the subject I'm unable to verify that claim. Does any one have an idea about the proof or a reference to a serious discussion of the subject?

REPLY [10 votes]: I spent some time with this a couple of years ago out of curiosity, but did not make it any farther than you're like to be able to find online.  The water is definitely murky.
Here is a discussion of the work which makes it clear that there are substantial ideas in Spencer-Brown's work, though Kauffman makes it clear that he is not evaluating the work per se.  Kauffman reports that Spencer-Brown definitely gives (with proof) a reformulation of the 4-Color Theorem into something called the Primacy Principle, that "A minimal planar (non-empty) uncolorable trail is prime."  (You'll have to see the references for the terminology).  Kauffman points out that Spencer-Brown has set up his logical foundations to have basically unintentionally axiomatized this Principle (hence the line on Wikipedia that Spencer-Brown's work "straddles the boundary between mathematics and of philosophy"), which is the source of Spencer-Brown's claimed proof.  This seems to be far as Kauffman is willing to go in giving caution as to the validity of the proof.  
The Wikipedia article on Laws of Form is decidedly less charitable.<|endoftext|>
TITLE: Books for hyperbolic geometry ( surfaces ) with exercises?
QUESTION [9 upvotes]: what are good books on hyperbolic geometry/hyperbolic surfaces that have good number of exercises, just to get a good understanding of the literature . I know John Ratcliffe's book will be one of them, but it is kind of encyclopedic. I particularly want to solve some exercises where they will be using free homotopy, isotopy, Arzela-Ascoli arguments, covering spaces, lifting of curves and homotopies, and will be actually using hyperbolicity of the manifold.
Thanks !

REPLY [3 votes]: John Stillwell's book The Geometry of Surfaces is best if you are an undergraduate; if you are a more advanced than that, or an advanced undergraduate, Thurston's book Three Dimensional Geometry and Topology is the best. Either Katok's Fuchsian groups or Hubbard's new Teichmueller Theory Vol. 1, would be reasonable alternatives to Thurston; or, even better, supplemental reading to Thurston on weekend nights with some wine.<|endoftext|>
TITLE: Which sets of Stiefel-Whitney characteristic numbers can be realized as coming from a manifold?
QUESTION [10 upvotes]: EDIT: The original question was answered very quickly (and very nicely!) but the answer leads to a pretty obvious subsequent question, which I will now ask. The original question is maintained for motivational purposes below.
I now know that not every sequence of zeros and ones can be realized as the Stiefel-Whitney numbers for some manifold- as I'm sure many of you all already knew. What I don't know, and what I suspect is a more delicate question, is: Which ones are? Is there a relatively easy necessary condition? Any sufficient conditions? 
Along similar lines: are there estimates as to the number of cobordism classes in any dimension that are tighter than the number of "possible" Stiefel-Whitney numbers? A tighter bound, as it were.
Thanks!

(original question)
Well I just learned a very cool fact over tea: apparently there are finitely many (unoriented) cobordism classes of compact manifolds in any given dimension! The cobordism class is completely determined by the Stiefel-Whitney characteristic numbers (which were explained to me as "the various numbers one gets by cupping characteristic classes of the tangent bundle together and applying them to the fundamental class, all mod 2")... so that's pretty awesome.
While I get over this initial shock, I was wondering if anyone knew the answer to the following: we have an upper bound on the number of cobordism classes by looking at the number of possible Stiefel-Whitney numbers. But is this upper bound realized?
In other words, given a sequence of zeros and ones (the right number of them), can I always construct a manifold that has precisely that sequence of zeros and ones as its Stiefel-Whitney numbers?

REPLY [2 votes]: Here are the data from MathScinet of Dold's paper proving that precisely the so called Wu relations give a complete set of restrictions on the Stiefel Whitney numbers of manifolds:
MR0079765 (18,143a) Reviewed
Dold, Albrecht
Vollständigkeit der Wuschen Relationen zwischen den Stiefel-Whitneyschen Zahlen differenzierbarer Mannigfaltigkeiten. (German)
Math. Z. 65 (1956), 200–206.<|endoftext|>
TITLE: Calculations of Pic^0, Pic, NS of surfaces
QUESTION [9 upvotes]: I'm looking for an example in the literature where $\mbox{Pic}^0(X)$, $\mbox{Pic}(X)$, and $NS(X)$ of a projective surface $X$ over a field are calculated. I want them for an example I'm trying to work out, so ideally $X$ would be relatively simple, perhaps a cubic hypersurface in $\mathbb{P}^3$, or something along those lines. I know it's out there, but googling and browsing arXiv and MathSciNet haven't quite panned out.

REPLY [2 votes]: For Reid's list of 95 K3 surfaces Picard lattices have been computed by Belcastro. Her paper can be downloaded from the arXiv at http://arxiv.org/PS_cache/math/pdf/9809/9809008v2.pdf . She has also made her thesis available, which can be downloaded at 
http://www.toroidalsnark.net/sm_thesis.pdf .<|endoftext|>
TITLE: Proving the impossibility of an embedding of categories
QUESTION [11 upvotes]: A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is the category of all groups and homomorphisms. For examples, consider the homology groups $H_{n}(X)$ or the homotopy groups $\pi_{n}(X)$. Of course, a problem with these invariants is that they are not fully faithful functors, i.e., $H_{n}(X)\cong H_{n}(Y)$ does not imply that $X$ and $Y$ are homeomorphic. The existence of a fully faithful functor $F:\mathscr{T}\to\mathscr{G}$ would imply that $\mathscr{G}$ has a subcategory $F\mathscr{T}$ equivalent to $\mathscr{T}$. This would be both rather disturbing and extremely interesting. First, it would mean that in a sense, all of topology is just a subset of group theory, which would be rather disturbing to topologists, but it would also reveal a fundamental connection between two seemingly disparate disciplines. My question is: is it possible to prove that no such functor exists? In other words, could one exhibit some categorical property that $\mathscr{T}$ posesses that $\mathscr{G}$ does not. This question can naturally be extended to other important categories, like $\mathscr{M}$, the category of all modules, or $\mathscr{R}$, the category of all rings. So in general, given arbitrary categories $\mathscr{C}$, $\mathscr{D}$, is there any natural way of showing that no fully faithful functor $F:\mathscr{C}\to\mathscr{D}$ exists, i.e., are there any nice "categorical invariants?"
EDIT: A couple people pointed out that I really ought to be discussing fully faithful functors, rather than just faithful functors. Also, I have changed the title in accordance with Martin Brandenburg's recommendation.

REPLY [2 votes]: I think No (for reasons of category theory): 
Let $1\in Set$ the final object (the one-point set) considered as discrete topological space , then $\mathcal{T}(1, X)\cong |X|$ (where $|X|$ is the support set of $X$), then  $Gr(F(1), F(1)\cong \mathcal{T}(1, 1)$ has only one element, then $1_{F(1)}$ is the trivial morphism (constant on the identity) then $F(1)$ is the trivial (only-identity) group. Then  for any group $G$ the set $Gr(F(1), G)$ has only one element, then for any topological space $X$ follow that $|X|\cong \mathcal{T}(1, X)\cong Gr(F(1), F(X)$ has only one element (absurd).
Please, forgive me for my bad english.<|endoftext|>
TITLE: When are two ideals in a regular local ring generated by a regular sequence?
QUESTION [6 upvotes]: Hello!
Let $R$ be a regular local ring, and let $I,J\subset R$ be ideals. I'd like to understand the "meaning" of the existence of a regular sequence $(x_1,...,x_n)$ in $R$ such that $I$ is generated by $x_1,...,x_k$ and $J$ is generated by $(x_{k+1},...,x_n)$ for some $1\leq k\leq n$.
For example, the existence of such a sequence implies that $\text{Tor}^R_k(R/I,R/J)=0$ for all $k>0$. 
Is it possible to give an equivalent description of the above property in terms of the vanishing of certain Tor and/or Ext terms?
Thank you!
Hanno

REPLY [3 votes]: This question is a bit vague, but I will try my best. Equivalent conditions involving only vanishing of Tor or Ext over $R$ are unlikely to exist, as they tend to be able to detect only projective dimensions, or depth over $R$. 
Here is an example straight from yours: if $R/(I+J)$ has finite length, then $\text{Tor}_i^R(R/I,R/J)=0$ for all $i>0$ if and only if both $R/I, R/J$ are Cohen-Macaulay. This very neat result can be found in Serre's "Local Algebra".
If you are willing to look at finer data, then it is  not hard to detect complete intersection. For example, $I$ is generated by $R$-sequence iff the conormal module $I/I^2$ is free over $R/I$. This is equivalent to  the vanishing of $\text{Tor}_1^{R/I}
(I/I^2,R/m)$ alone ($m$ is the maximal ideal). 
The last one while great in theory, is not very helpful in practice, since you need to present $I/I^2$ as a $R/I$-module. It is simpler to compute $\text{Tor}_1^{R}(R/I,R/m)$ and compare with the codimension of $I$. If they are equal, $I$ must be complete intersection.<|endoftext|>
TITLE: Quasi-coherent envelope of a module
QUESTION [11 upvotes]: Let $X$ be a scheme. It is known that $Qcoh(X)$ is cocomplete, co-wellpowered and has a generating set. The special adjoint functor theorem tells us that then every(!) cocontinuous functor $Qcoh(X) \to A$ has a right-adjoint. Here $A$ is an arbitrary category (which I always assume to be locally small).
a) Is there a nice description of the right-adjoint to the forgetful functor $Qcoh(X) \to Mod(X)$? Here, you may impose finiteness conditions on $X$. This functor may be called a quasi-coherator.
b) Let $f : X \to Y$ be a morphism of schemes. Then $f^* : Qcoh(Y)  \to Qcoh(X)$ is cocontinuous, since $f^* : Mod(Y) \to Mod(X)$ is cocontinuous and the forgetful functor preserves and reflects colimits. In particular, there is a right-adjoint $f_+ : Qcoh(X) \to Qcoh(Y)$. If $f$ is quasi-separated ans quasi-compact, then this is the direct image functor $f_*$. Is there a nice description in general? Note that $f_+$ is the composition $Qcoh(X) \to Mod(X) \to Mod(Y) \to Qcoh(Y)$, where the latter is the quasi-coherator. This is only nice if we have answered a).
c) Since $Mod(X)$ is complete and $Qcoh(X) \to Mod(X)$ has a right adjoint, $Qcoh(X)$ is also complete. Is there a nice description for the products? They are given by taking the quasi-coherator of the product, can we simplify this? I mean, perhaps they turn out to be exact although the products in $Mod(X)$ are not exact?
Answer (after reading the article Leo Alonso has cited)
We have the following description of the quasi-coherator: Let $X$ be a concentrated scheme, i.e. quasi-compact and quasi-separated. If $X$ is separated, say $X = \cup U_i$ with finitely many affines $U_i$ such that the intersections $U_i \cap U_j$ are affine, then the quasi-coherator of a module $M$ on $X$ is the kernel of the obvious map
$\prod_i (u_i)_* \tilde{M(U_i)} \to \prod_{i,j} (u_{i,j})_*  \tilde{M(U_i \cap U_j)}$,
where $u_i : U_i \to X$ and $u_{ij} : U_i \cap U_j \to X$ are the inclusions. If $X$ is just quasi-separated, there is a similar description using the separated case.
The idea is quite simple and can be generalized to every flat ring representation of a finite partial order, which has suprema (for example the dual of the affine subsets of a quasi-compact separated scheme). On an affine part, the quasi-coherator consists of sections of all other affine parts over it, which are compatible in the obvious sense.
If we have no finiteness conditions, the description is basically also valid, but you have to take the quasi-coherators of the products or the direct images, since they don't have to be quasi-coherent. In general there is no nice description. Also in nice special cases, b) and c) have no nice answers (and infinite products are not exact, even in the category of quasi-coherent modules on nice schemes).

REPLY [2 votes]: Let us do the case of an affine scheme $X$ first. This is easy. If $M$ is an $O_X$-module, we define $\tilde{M}$ as the quasicoherent $O_X$-module defined by the global sections $M(X)$. Notice that the restriction gives the canonical map $\tilde{M}->M$.
As step 2, we extend to the case of quasiaffine $X$. Quasiaffinity means that $X-> Spec O_X(X)$ is an open embedding. Equivalently, all its quasicoherent modules are generated by global section. Hence, the coherator functor $M |-> \tilde{M}$ is defined in the same way. BTW, it is clear how the coherator works on maps too.
Finally, we have everything ready as a general scheme admits an affine open cover $X=\cup_i U_i$. The quasiaffine case is useful as double and triple intersection $U_{i,j}$, $U_{i,j,k}$ are all quasiaffine. Given an $O_X$-module $M$, we use its open pieces $M_i=M|_{U_i}$ and gluing maps $\phi_{i,j}$ from ${M_i}|_{U(i,j)}$ to ${M_j}|_{U(i,j)}$, 
where $U(i,j)=U_{i,j}$ (because tex translator is finding it difficult to comprehend it too without each formula starting in a new line), that satisfy the cocycle conditions on triple intersection.
Now the coherator is glued from open pieces $\tilde{M}_i$  
using isomorphisms $\tilde{\phi}_{i,j}$ which inherit the cocycle conditions. And That's All Folks!<|endoftext|>
TITLE: Hausdorff dimension of the boundary of an open set in the Euclidean space - lower bound
QUESTION [10 upvotes]: I consider a bounded open set $A$ in ${\mathbb R}^d$. Is the Hausdorff dimension of the boundary of $A$ at least $d-1$ ? I thought I would have found a result on this problem in any textbook about Hausdorff dimension but I failed. As you may guess I have never work with Hausdorff dimension.

REPLY [21 votes]: Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d-1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The latter is open in $\mathbb R^{d-1}$ and hence has Hausdorff dimension $d-1$. On the other hand, the projection map does not increase Hausdorff dimension since it does not increase distances.