ct.category theory - When do two objects become isomorphic in the stable category?

Suppose $X$ and $Y$ are stably isomorphic, so that there exist a morphism $f:Xto Y$ whose image $underline f:Xto Y$ in the stable category is an isomorphism. Then $underline f$ has an inverse: there exists $g:Yto X$ such that $underline gcircunderline f=1_X$ and $underline fcircunderline g=1_Y$, and this means in particular that there is a projective-injective $P$ and maps $r:Xto P$ and $s:Pto X$ such that $gcirc f-1_X=scirc r$.

This gives us maps $F=left(begin{smallmatrix}f\aend{smallmatrix}right):Xto Yoplus P$ and $G=left(begin{smallmatrix}g&-bend{smallmatrix}right):Yoplus Pto X$ such that $Gcirc F=1_X$. If now we assume that $mathcal A$ has all its idempotents split, then we can conclude that there is a $Q$ such that there is an isomorphism $H:Xoplus Qxrightarrow{cong} Yoplus P$ such that $F=Hcirciota:Xto Yoplus P$, with $iota:Xto Xoplus Q$ the canonical map.

Notice that $underline F$ and $underline H$ are isomorphisms in $underline{mathcal A}$, so that also $underlineiota$ is an isomorphism there. If $p:Xoplus Qto X$ is the projection, then $underline pcircunderlineiota=1_X$ in $underline{mathcal A}$, so in fact $(underlineiota)^{-1}=underline p$, and in consequence the composition $iotacirc p:Xoplus Qto Xoplus Q$ is the identity of $Xoplus Q$ in $underline{mathcal A}$. In other words, there exists a projective-injective $R$ and morphisms $u:Xoplus Qto R$ and $v:Rto Xoplus Q$ such that $pcirciota-1_{Xoplus Q}=vcirc u$.

If now $j:Qto Xoplus Q$ and $q:Xoplus Qto Q$ are the canonical maps, we have $qcirc vcirc ucirc j=-1_Q$, so that the morphism $underline{1_Q}:Qto Q$ is zero. This implies that in fact $Qcong 0$ in $underline{mathcal A}$. By what you showed in your question, this implies that $Q$ is a projective-injective in $mathcal A$.

All in all, we have shown that there exists projective-injectives $P$ and $Q$ such that $Xoplus Qcong Yoplus P$ in $mathcal A$, as you wanted.

(I do not think your question will have a positive answer when $mathcal A$ does not have all its idempotents split... I do not have a counterexample, though)

soft question - Interesting complexity classes $PR subsetneq c subsetneq R$

First of all, it’s certainly possible to obtain some intermediate class by taking a language that only computes PR functions (say, an imperative programming language using only for loops) and adding any total computable but non PR function (e.g., Ackermann’s function). The resulting language L is non-universal, because it only computes total functions: you can still construct a computable but non-L-computable function by diagonalisation. However, L is clearly more powerful than the original language.

As for “interesting”, I guess it really depends on what you mean by that.

If “interesting” means “of practical use”, then one could answer that all computable functions of practical use are PR, since a non-PR function requires an amount of time to compute that is not, in turn, PR. Considering that time bounds such as 2ⁿ, 2²ⁿ, 2²²ⁿ, …, are all PR, you see that there isn’t much hope to compute non-PR functions for large values of n.

If “interesting” means “logically interesting”, then I think the answer is “yes”. I’m somewhat familiar with Girard’s System F (also called “second order λ-calculus” or “polymorphic λ-calculus”), described for instance in Girard’s Proofs and Types (freely available here). The functions that can be computed in F are “exactly those which are provably total in [second order Peano arithmetic]” (page 123), and among these we have Ackermann’s function. There is an explicit λ-term for it on these slides (page 20).

If I recall correctly, the standard calculus of constructions includes System F and only computes total functions, so it also provides an example.

books - What are some good beginner graph theory texts?

The thing about graph theory (and combinatorics more generally, although it's especially true for graph theory) is that the basic definitions are very simple, and there is a lot of interesting math you can do without using anything but the basics.

As such, there's really only a limited benefit to just reading a text -- if you want to learn the subject, you have to have to have to do problems. Of the textbooks mentioned, I personally own Diestel (free online edition) and Bollobas; of these two, Bollobas has more and better exercises (although Diestel's a wonderful reference, and has the advantage of including hints.)

Honestly, I think the really important thing when teaching/learning graph theory is for the lecturer to know what he or she is doing. Obviously some books are better than others, but none of them are very good if they're not being used correctly.

lo.logic - Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)

I have solved the problem affirmatively at least for sets of reals.

Theorem. Every set of reals admits a rigid binary relation (with no use of Axiom of Choice). Equivalently, every set of reals is the vertex set of a rigid directed graph.

Proof. Suppose that A is a set of reals. We may freely regard A as a subset of Cantor space 2^omega. Let us break into several cases.

Case 1. A is countable. This is the easy case, since we may simply impose a rigid structure on it by making it a linear order isomorphic to omega (or a finite linear order if A is finite).

Case 2. A is uncountable, but A has a countably infinite subset. Fix such a subset Z={z_0, z_1, ...} and fix a point z* in A-Z. For each finite binary sequence s, let U_s be the neighborhood in 2^omega of all sequences extending s, so that U_s(x) iff x extends s. Clearly, the structure (A,U_s)_s is rigid, since if you move any point x in A to another point, you will move it out of some neighborhood U_s that it was formerly in. We now reduce this structure to a binary relation. Let R be a relation on A that places all the z_n below z*, ordered like omega, and makes R(z*,z*) true. Next, enumerate the finite binary sequences as s_0, s_1, etc., (this does not require AC). We define R(x,y) iff x=z_n for some n, y is not z_m for any m, y is not z*, and U_{s_n}(y). That is, the first coordinate gives you some z_n, and hence some s_n, and then you use this to determine which neighborhood predicate to apply to y, but we only do this for y outside of Z union {z*}. I claim that the structure (A,R) is rigid. The reason is that z* and the reals z_n are definable in the structure (A,R), and so they are fixed by all automorphisms. (The real z* is the only one such that R(z*,z*), and the z_n are the only predecessors of z* wrt R.) Since every z_n is fixed, it follows that every automorphism must respect the neighborhood U_{s_n} intersect A, and hence fix all reals. So there are no nontrivial automorphisms.

Case 3. Weird A. The only remaining case occurs when A is uncountable, but has no countably infinite subset. (It follows that A will be Dedekind finite, but not finite.) In this case, every permutation of A will consist of disjoint orbits of finite length, since if there were an infinite orbit, then we could build a countably infinite subset of A by iterating it. But if every permutation of A is like that, then A has no permutations that respect the usual linear order < of the reals. Thus, (A,<) is rigid. QED

In particular, it is not true that the usual counterexample to AC in the symmetric forcing models is a counterexample to this rigidity question. Those sets are sets of reals, and this argument shows that they have rigid binary relations, without being well-orderable.

I'm not sure how far one can extend this idea. How about subsets of 2^kappa for any cardinal kappa? I think, however, that even this still won't give a full positive answer for all sets.

arithmetic geometry - Lifting the p-torsion of a supersingular elliptic curve.

La réponse est oui. La raison est la suivante. Si $S$ est un schéma sur p est localement nilpotent, par définition (cf. thèse de Messing, chapitre I), un groupe de Barsotti-Tate tronqué d'échelon $1$ sur $S$ est un schéma en groupes fini localement libre sur $S$ annulé par $p$ tel que si $G_0$ désigne la réduction de $G$ modulo $p$ on ait
$$
Im (F_{G_0}) = ker (V_{G_0})
$$
comme faisceaux fppf sur $S_0$, $S_0$ désignant la réduction modulo $p$ de $S$. Cependant, il résulte du critère de platitude fibre à fibre de EGA IV que le morphisme de $S_0$-schémas en groupes de présentation finie
$$
F:G_0 rightarrow ker (V_{G_0})
$$
est fidèlement plat si et seulement si il en est de même fibre à fibre sur $S_0$. De cela on déduit que dans la définition d'un $BT_1$ on peut remplacer $S_0$ par $S_{red}$ !

Dans les considérations précédentes j'ai pris un schéma $S$ sur lequel $p$ est localement nilpotent, mais il en résulte que l'on a le même type de définition-résultat sur une base qui est un schéma formel $p$-adique.

Revenons maintenant à nos moutons, c'est à dire la question de Kevin. On a donc $G$ un schéma en groupes fini et plat sur l'anneau des entiers de $K$
dont la fibre spéciale sur le corps résiduel de $K$ est un $BT_1$. Le schéme en groupes $G$ est donc un $BT_1$.

Nous allons maintenant utiliser le théorème suivant (cf. l'article d'Illusie aux journées arithmétiques de Rennes: "Déformations de groupes de Barsotti-Tate (d'après A. Grothendieck)", Astérisque 127). Si $mathcal{BT}_1$, resp. $mathcal{BT}$, désigne le champ des groupes de Barsotti-Tate tronqués d'échelon $1$, resp. des groupes de Barsotti-Tate, sur des bases qui sont des schémas sur lesquels $p$ est localement nilpotent, alors le morphisme "points de $p$-torsion"

$$
mathcal{BT} longrightarrow mathcal{BT}_1
$$

est formellement lisse. De cela on déduit la chose suivante. Soit $k$ un corps parfait de caractéristique $p$ et $H$ un groupe $p$-divisible sur $k$. Soit $mathfrak{X}$ l'espace des déformations de $H$, un $spf (W(k))$-schéma formel non-canoniquement isomorphe à
$spf big (W(k)[[x_1,dots,x_{d(h-d)}]]big )$ où $h$ désigne la hauteur de $H$ et $d$ sa dimension.
Soit $mathcal{H}$ la déformation universelle sur $mathfrak{X}$. Alors, $mathcal{H}[p]$ est une déformation verselle de $H[p]$.

Pour conclure et répondre à la question de Kevin il suffit maintenant d'invoquer le théorème de Serre-Tate qui montre que si $H=E[p^infty]$ où $E$ est une courbe elliptique supersingulière sur $k$ alors $mathfrak{X}$ est également l'espace des déformation de $E$. Appliquant le théorème d'algébrisation de Grothendieck (GAGF) on en déduit que si $mathfrak{X}= spf (R)$, la déformation universelle de la courbe elliptique $E$ sur $mathfrak{X}$ provient en fait d'une courbe elliptique sur $spec (R)$. Le résultat s'en déduit par spécialisation sur $mathcal{O}_K$.

P.S.: I read the rules for this forum: it is nowhere written the questions and answers should be written in english !!!!!!!!

ag.algebraic geometry - Principal Bundles over Complex Projective Varieties

As I interpret your question, you're looking for principal bundles where the base is projective, and both the fiber and total space are both compact groups.

There's an obvious class of these, which is taking any compact group $P$ and modding out by a Levi subgroup $G$ (I'm trying to match the notation in your question. I apologize to the Lie theorists in the audience for using the most confusing notation ever) [EDIT: in the original, this had read "containing the maximal torus" since I'd forgotten that there are non-Levi subgroups which contain the maximal torus]. This quotient will always be a projective variety.

I believe that one can prove that this is the only way of getting a quotient which is projective (note that the complexification of $P$ acts on the quotient, since it is compact, and the complexification of the Lie algebra acts; now use the Borel fixed point theorem). [EDIT: it seems that this isn't true. For example, elliptic curves exist.]

It's possible that there's some strange way of making $G$ act on $P$ that's not a subgroup, but I think the above are the right class of examples generalizing the Grassmannian. [EDIT: this paragraph at least is vague enough to not be false, but as pointed out in comments, there are other examples]

mathematics education - Why is a topology made up of 'open' sets? Part II

Because the display was getting quite cluttered, I thought I'd post a second part to this question separately I'll go now into some details with which I wished intially to avoid prejudicing the replies.

Here are three natural areas of progression in the study of topology:

(1) point-set topology;

(3) sheaves and their cohomology;

and a very important

(2) middle ground

that I won't give a name to, involving matters like the classification of two-manifolds. I don't quite agree that the metric space intuition is sufficient even for a first course, simply because we always look ahead to the way the material will help or hinder students' future understanding.

Now, it is in (3) that a topology in the abstract sense plays a very active role, as the relation to global invariants emerges prominently. At this point, the definition needs no other motivation. However, in an introductory course on (1), I have yet to incorporate successfully interesting material from (3). When dealing with (2), curiously enough, the definition of a topology plays a very passive role. Many arguments are strongly intuitive. However, it is clear that one needs to have already absorbed (1) for the material in (2) to feel really comfortable. Of course there are exceptionallly intuitive people who can work fluently on (2) without a firm foundation in point-set topology. But for most ordinary folks (like me), the rigor of (1) is needed if only as a psychological crutch. Note here that most of the natural spaces that come up in (2) are in fact metrizable. However, getting bogged down with worrying about the metric would be a definite hindrance in working through the operations that come up constantly: stretching, bending, and perhaps most conspicuously, gluing. A quotient metric is a rather tricky notion, while a quotient topology is obvious.

With this future work in mind, how best then should one get this background in (1)? When I was an undergraduate, in fact, there was plenty of motivation in course (1). We used the first part of Munkres' book, and had exercises dug out of 'Counterexamples in Topology,' involving many strange spaces that have one property but not another. It was great fun. You may wonder then, what exactly I'm worried about. It's that I felt later that this kind of motivation had not been quite right. It wasn't entirely wrong either. Certainly we became very confident in working with the definitions, and that was good. However, and this is a big 'however,' when I moved on to (2), I had a distinct sense that the motivational material and exercises used in (1) were actually preventing me from learning the new notions efficiently. It took me quite a long time to make the transition, bugged by a persistent longing for the axioms and the exotic examples. A number of conversations I've had over the years indicate that my experience was far from unique. At any rate, this experience makes the issue for me quite different from a course on, say, linear algebra. No doubt the motivational material for diagonalization used in most courses is hardly convincing. Substantive examples come later, sometimes much later. But most of the operators and eigenvectors in standard textbooks are good toy models for a wide range of serious objects.

One obvious remedy would be to incorporate material from (2) into (1). I found this very hard, mostly because of the point already made above, that the role of (1) in (2) is quite implicit. When I first posed the question, I was hoping someone had figured out a good way of doing something like this. Incidentally, sheaves came around much later than the definition of a topology, so the historical question remains as well. How were the standard properties decided upon?

Let me make clear that I am not arguing that everyone has to go through the progression just outlined. Obviously, some people will take (1) elsewhere, in a way that the issues become quite different. Perhaps many people will never need more than metric spaces (or normed spaces, for that matter). But (1)-->(2)-->(3) is certainly common enough (perhaps especially for arithmetic geometers) to call for some reasonable methodology.

Meanwhile, I also appreciate Andrew Stacey's point (possibly even more than he does!). The long paragraphs above notwithstanding, the pedagogical question isn't something I lose sleep over. But it would be nice to have a few concrete and systematic ideas to use. They would certainly help me to understand the subject better!

Added:

Perhaps I should rephrase the question: How should we teach point set topology so as to facilitate the transition to the topology of natural spaces?

Somehow, this way of putting it seems much vaguer to me. 'How to teach X?' is such a broad question I would never be able to answer it myself in a finite amount of time. It seems to invite a diffuse discussion of everythng under the sun. That's why I had preferred to focus on one rather precise mathematical facet of that question. But I don't feel too strongly about it either way.

gr.group theory - Is Thompson's group F residually finite?

The answer is "no" -- the quickest way to see this appeals to the following nontrivial fact: the commutator subgroup of $F$, denoted by $F'$ as usual, is infinite and simple.

Armed with this, we argue as follows. Let $N$ be a normal subgroup in $F$ of finite-index; then $Ncap F'$ is going to be normal in $F'$ and of finite index in $F'$. Hence $Ncap F'=F'$, that is, $N$ contains $F'$. So the intersection of all finite-index normal subgroups of $F$ must contain $F'$. But if $F$ were residually finite then this intersection would only contain the identity element, and the result follows.

If you don't mind me asking: is this a question out of curiosity, or one that you've run into during your studies or research?

set theory - Axiom of Choice and Order Types

Order equivalence is an equivalence relations on ordered sets, not on sets. It is just the isomorphism relation on ordered structures. An ordered structure is a set, together with an order.

The Axiom of Choice says that every set has a well-order. Since the order-types of well-orders are well-ordered (given any two, one of them is uniquely isomorphic to a unique initial segment of the other), it follows under AC that for every set, we can associate to it the smallest order-type of a well-order on that set. This is called the cardinality of the set.

There is another more general concept of cardinality, which does not rely on AC or on orderings at all, and it is just the equinumerosity class of the set.

pr.probability - Does central limit theorem hold for general weakly dependent variables?

Your double subscripts are extraneous. Let's consider a simpler situation, where we have a single family of random variables ${X_i}$.

As Yuri Bakhtin says above, your condition is not sufficient for a CLT to hold. Here is a simpler situation, however: suppose that $X_i$ and $X_j$ satisfy finite-range dependence. That is, there exists a positive integer $R$ such that if $|i-j| ge R$, then $X_i$ and $X_j$ are independent. We will prove a law of large numbers for ${X_i}$. If you're interested, you can push it farther to prove a central limit theorem. Suppose that $X_i$ has mean $mu$ for each $i$.

Let $S_N = tfrac{1}{N} sum_{i=1}^N X_i$ as usual. Without loss of generality, we may consider indices only divisible by $R$: $S_{RN} = tfrac{1}{RN} sum_{i=1}^{RN} X_i$. Let $$S_{RN}^{(k)} = tfrac{1}{N} sum_{j=0}^{N-1} X_{Rj+k}$$ for $k= 1, dots, R$, so that $$S_{RN} = tfrac{1}{R} left( S_{RN}^{(1)} + dots + S_{RN}^{(R)} right).$$Each sum $S_{RN}^{(k)}$ is comprised of independent random variables, so the classical law of large numbers applies and $S_{RN}^{(k)} to mu$ both in probability and almost surely. Consequently, $S_{RN} to mu$.

Obviously, this argument breaks down when $R = infty$. In that case, the problem is no longer trivial and you will have to be more cautious with your assumptions.

textbook recommendation - Suggestions for good books on class field theory

When you are first learning class field theory, it helps to start by getting some idea of what the fuss is about. I am not sure if you have already gotten past this stage, but if not, I recommend B. F. Wyman's article "What is a Reciprocity Law?" in the American Mathematical Monthly, Vol. 79, No. 6 (Jun. - Jul., 1972), pp. 571-586. I also highly recommend David Cox's book Primes of the Form $x^2 + ny^2$ (mentioned by Daniel Larsson). Cox's book will show you what class field theory is good for and will get you to the statements of the main theorems quickly in a very accessible way. (You can safely skim through most the earlier sections of the book if your goal is to get to the class field theory section quickly.) As a bonus, the book will also give you an introduction to complex multiplication on elliptic curves.

However, Cox's book does not prove the main theorems of class field theory. You will need to look elsewhere for the proofs. There are several different approaches and someone else's favorite book may be unappealing to you and vice versa. You will have to dip into several different books and see which approach appeals to you. One book that has not been mentioned yet is Serge Lang's Algebraic Number Theory. Even if you ultimately choose not to use Lang's book as your main text, there is a short essay by Lang in that book, summarizing the different approaches to class field theory, that is worth its weight in gold.

big list - Which popular games are the most mathematical?

I consider a game to be mathematical if there is interesting mathematics (to a mathematician) involved in

• the game's structure,


• optimal strategies,


• practical strategies,


• analysis of the game results/performance.

Which popular games are particularly mathematical by this definition?

---

Motivation: I got into backgammon a bit over 10 years ago after overhearing Rob Kirby say to another mathematician at MSRI that he thought backgammon was a game worth studying. Since then, I have written over 100 articles on the mathematics of backgammon as a columnist for a backgammon magazine. My target audience is backgammon players, not mathematicians, so much of the material I cover is not mathematically interesting to a mathematician. However, I have been able to include topics such as martingale decomposition, deconvolution, divergent series, first passage times, stable distributions, stochastic differential equations, the reflection principle in combinatorics, asymptotic behavior of recurrences, $chi^2$ statistical analysis, variance reduction in Monte Carlo simulations, etc. I have also made a few videos for a poker instruction site, and I am collaborating on a book on practical applications of mathematics to poker aimed at poker players. I would like to know which other games can be used similarly as a way to popularize mathematics, and which games I am likely to appreciate more as a mathematician than the general population will.

Other examples:

• go


• bridge


• Set.

Non-example: I do not believe chess is mathematical, despite the popular conception that chess and mathematics are related. Game theory says almost nothing about chess. The rules seem mathematically arbitrary. Most of the analysis in chess is mathematically meaningless, since positions are won, drawn, or tied (some minor complications can occur with the 50 move rule), and yet chess players distinguish strong moves from even stronger moves, and usually can't determine the true value of a position.

To me, the most mathematical aspect of chess is that the linear evaluation of piece strength is highly correlated which side can win in the end game. Second, there is a logarithmic rating system in which all chess players say they are underrated by 150 points. (Not all games have good rating systems.) However, these are not enough for me to consider chess to be mathematical. I can't imagine writing many columns on the mathematics of chess aimed at chess players.

Non-example: I would exclude Nim. Nim has a highly mathematical structure and optimal strategy, but I do not consider it a popular game since I don't know people who actually play Nim for fun.

---

To clarify, I want the games as played to be mathematical. It does not count if there are mathematical puzzles you can describe on the same board. Does being a mathematician help you to learn the game faster, to play the game better, or to analyze the game more accurately? (As opposed to a smart philosopher or engineer...) If mathematics helps significantly in a game people actually play, particularly if interesting mathematics is involved in a surprising way, then it qualifies to be in this collection.

If my criteria seem horribly arbitrary as some have commented, so be it, but this seems in line with questions like Real world applications of math, by arxive subject area? or Cocktail party math. I'm asking for applications of real mathematics to games people actually play. If someone is unconvinced that mathematics says anything they care about, and you find out he plays go, then you can describe something he might appreciate if you understand combinatorial game theory and how this arises naturally in go endgames.

nt.number theory - The inverse Galois problem, what is it good for?

The previous answers are all on point; let me just say a little more.

First, the IGP as a problem is a sink, not a source (or something with both inward and outward flow!): I know of no nontrivial consequences of assuming that every finite group over Q (or even over every Hilbertian field) is a Galois group. This does not mean it's a bad problem: the same holds for Fermat's Last Theorem.

As with FLT, if IGP were easy to prove, then it would be of little interest. (As a good example, if you know Dirichlet's theorem on primes in arithmetic progressions, it's easy to prove that every finite abelian group occurs as a Galois group of Q. What does this tell you about the maximal abelian extension of Q? Not much -- the Kronecker-Weber theorem is an order of magnitude deeper.) But as with FLT, the special cases of IGP that have been established use a wide array of fascinating techniques and provide an important border-crossing between algebra and geometry.

Arguably more interesting than IGP itself is the Regular Inverse Galois Problem: for any field K and any finite group G, there exists a regular function field K(C)/K(t) with Galois group isomorphic to G. (If K is Hilbertian -- e.g. a global field -- then RIGP for K implies IGP for K.) Now RIGP is of great interest in arithmetic geometry: given any finite group G there are infinitely many moduli spaces (Hurwitz spaces) attached to the problem of realizing G regularly over K (because we have discrete invariants which can take infinitely many possible values, like the number of branch points). If even one of these Hurwitz schemes has a K-rational point, then G occurs regularly over K. In general, the prevailing wisdom about varieties over fields like Q is that they should have very few rational points other than the ones that stare you in the face. (Yes, it is difficult or impossible to formalize this precisely.) So it is somewhat reasonable to say that the chance that a given Hurwitz space -- say of general type -- has a Q-rational point is zero, but what about the chance that at least one of infinitely many Hurwitz spaces, related to each other by various functorialities, has a Q-rational point? To me that is one of mathematics' most fascinating questions: to learn the answer either way would be tremendously exciting.

The core question of topology

As others have noted, it's hopeless to try to answer this question for general topological spaces. However, there are a few positive results if you assume, say, that X and Y are both simply connected closed manifolds of a given dimension. For example, Freedman showed that if X and Y are oriented and have dimension four, then to check whether they're homeomorphic you just need to compute (i) the bilinear "intersection" forms on H^2(X;Z) and H^2(Y;Z) induced by the cup product; and (ii) a Z/2-valued invariant called the Kirby-Siebenmann invariant. The invariant in (ii) obstructs the existence of a smooth structure, so if you happened to know that both X and Y were smooth manifolds (hence that their Kirby-Siebenmann invariants vanished) you'd just have to look at their intersection forms to determine whether they're homeomorphic (however a great many examples show that this wouldn't suffice to show that they're diffeomorphic).

In higher dimensions, Smale's h-cobordism theorem shows that two simply connected smooth manifolds are diffeomorphic as soon as there is a cobordism between them for which the inclusion of both manifolds is a homotopy equivalence. Checking this criterion can still be subtle, but work of Wall and Barden shows that in the simply-connected 5-dimensional case it suffices to check that there's an isomorphism on second homology H2 which preserves both (i) the second Stiefel-Whitney classes, and (ii) a certain "linking form" on the torsion subgroup of H2.

If you drop the simply-connected assumption, things get rather harder--indeed if n>3 then any finitely presented group is the fundamental group of a closed n-manifold (which can be constructed in a canonical way given a presentation), and Markov (son of the probabilist) showed that the impossibility of algorithmically distinguishing whether two presentations yield the same group translates to the impossibility of algorithmically classifying manifolds. Even assuming you already knew the fundamental groups were isomorphic, there are still complications beyond what happens in the simply-connected case, but these can sometimes be overcome with the s-cobordism theorem.

In a somewhat different direction, in dimension 3 one can represent manifolds by link diagrams, and Kirby showed that two such manifolds are diffeomorphic (which in dimension 3 is equivalent to homeomorphic) iff you can get from one diagram to the other by a sequence of moves of a certain kind. (see Kirby calculus in Wikipedia; similar statements exist in dimension 4). I suppose that one could argue that this isn't an example of what you were looking for, since if one felt like it one could extract diffeomorphisms from the moves in a fairly explicit way, and one can't (AFAIK) just directly extract some invariants from the diagrams which completely determine whether the moves exist.

gn.general topology - A question about unbounded connected subsets of the plane.

The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is closed. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though.

Let $Gamma$ be a closed unbounded connected subset of the plane. Let $xinGamma$ and let $B:=B(x,r)$ be an open ball around $x$. I claim that the connected component of $x$ in $Gammacap bar{B}$ meets $partial B$, which shows that $Gamma$ does contain non-trivial bounded connected subsets.

For any $epsilon>0$, consider the $epsilon$-neighborhood of $Gamma,$ that is $Gamma_epsilon:=cup_{yinGamma}B(y,epsilon).$ It is an open unbounded connected subset of the plane.
Let $U_epsilon$ be the connected component of $x$ in $Gamma_epsiloncap B$. Since the latter is locally connected, $U_epsilon$ is both an open and closed subset of it in the relative topology. It is therefore an open subset of $Gamma_epsilon$; however it is not closed in it, because $Gamma_epsilon$ is connected. Therefore $bar U_epsilon$ is a closed connected set that meets $partial B, $ and of course contains $x$. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as $epsilonto0$ we get a bounded connected subset of $Gamma$ connecting $x$ with $partial B$ (this also passes to the limit).

Rmk One could state the above in terms of the one-point compactification of $Gamma$, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines $X_epsilon$ as an $epsilon$ nbd of $X$ in the embedding).

PS: Of course the same affirmative conclusion holds, even more directely, if $Gamma$ is assumed to be open, which is another case included in the original assumption of completely metrizable.

gr.group theory - non-abelian groups of prescribed order

I am also unsure of what "nontriviality" conditions you want to impose. Without any further conditions, the following answers your question:

Call a positive integer $n$ nilpotent if every group of order $n$ is nilpotent.

Call a positive integer $n$ abelian if every group of order $n$ is abelian.

Suppose that the prime factorization of $n$ is $p_1^{a_1} cdots p_r^{a_r}$. Then:

1) $n$ is nilpotent iff for all $i,j,k$ with $1 leq k leq a_i$, $p_i^k not equiv 1 pmod{p_j}$.

2) $n$ is abelian iff it is nilpotent and $a_i leq 2$ for all $i$.

These results are proved in

---

Pakianathan, Jonathan(1-WI); Shankar, Krishnan(1-MI)
Nilpotent numbers.
Amer. Math. Monthly 107 (2000), no. 7, 631--634.

---

The proofs are constructive: for any $n$ which is not nilpotent (resp. abelian), they give an explicit group of that order which is not nilpotent (resp. abelian).

The paper is available at

http://www.math.uga.edu/~pete/nilpotentnumbers.pdf

at.algebraic topology - Involutions of $S^2$

are there some complete results on the involutions of 2 sphere?

at least I have three involutions:
(let $mathbb{Z}_2={1,g}$,and $S^2={(x,y,z)inmathbb{R}^3|x^2+y^2+z^2=1}$)

1.$g(x,y,z)=(-x,-y,-z)$(antipodal map) with null fixed point set,and orbit space $mathbb{R}P^2$
actully,for free involution on $S^n$ with $nleq3$,the orbit space is homeomorphic to real projective space (Livesay 1960)

2.$g(x,y,z)=(-x,-y,z)$ (rotation $pi$ rad around $z$ axis) with fixed point set $S^0$(the north pole and south pole) and orbit space $S^2$.

3.$g(x,y,z)=(x,y,-z)$(reflection along $z=0$) with fixed point set $S^1$ (the equator)and orbit space $D^2$

i want to know if there are some other involutions over 2-sphere.
here we take two involutions as equivalent if there are conjugate in the homeomorphism group of $S^2$

ac.commutative algebra - Atiyah-MacDonald, exercise 2.11

I posted this question on a different site a couple of years ago. Eventually I found that a book of T.Y. Lam has a very nice treatment. Here is the writeup I posted on the other site:

---

After paging through several algebra books, I found that T.Y. Lam's GTM Lectures on
Rings and Modules has a beautiful treatment of this question.

The above property of a (possibly noncommutative) ring is called the "strong rank
condition." It is indeed stronger than the corresponding statement for
surjections ("the rank condition") which is stronger than the isomorphism version
"Invariant basis number property". However, in fact it is the case that all
commutative rings satisfy the strong rank condition. Lam gives two proofs
[pp. 12-16], and I will now sketch both of them.

First proof:

Step 1: The result holds for (left-) Noetherian rings. For this we establish:

Lemma: Let $M$ and $N$ be (left-) $A$-modules, with $N$ nonzero. If the direct sum
$M oplus N$ can be embedded in $M$, then $M$ is not a Noetherian $A$-module.

Proof: By hypothesis $M$ has a submodule
$M_1 oplus N_1$, with $M_1 cong M$ and $N_1 cong N$. But we can also
embed $M oplus N$ in $M_1$, meaning that $M_1$ contains a submodule $M_2 oplus N_2$ with $M_2 cong M$ and $N_2 cong N$. Continuing in this way we construct an ascending
chain of submodules $N_1$, $N_1 oplus N_2$,..., contradiction.

So if A is (left-) Noetherian, apply the Lemma with $M = A^n$ and $N = A^{m-n}$.
$M$ is a Noetherian $A$-module, and we conclude that $A^m$ cannot be embedded in $A^n$.

Step 2: We do the case of a commutative, but not necessarily Noetherian, ring.
First observe that, defining linear independent subsets in the usual way, the
strong rank condition precisely asserts that any set of more than $n$ elements in $A^n$
is linearly dependent. Thus a ring $A$ satisfies the strong rank condition iff: for
all $m > n$, any homogeneous linear system of $n$ linear equations and $m$ unknowns has a
nonzero solution in $A$.

So, let $MX = 0$ be any homogeneous linear system with coefficient matrix $M = (m_{ij}),
1 leq i leq n, 1 leq j leq m$. We want to show that it has a nonzero solution in $A$. But the subring $A' = mathbb{Z}[a_{ij}]$, being a quotient of a polynomial ring in finitely many variables over a Noetherian ring, is Noetherian (by the Hilbert basis theorem), so by Step 1 there is (even) a nonzero solution $(x_1,...,x_m) in (A')^m$.

This makes one wonder if it is necessary to consider the Noetherian case separately,
and it is not. Lam's second proof comes from Bourbaki's Algebra, Chapter III, §7.9, Prop. 12, page 519. [Thanks to Georges Elencwajg for tracking down the reference.] It uses the following elegant characterization of linear independence in free modules:

Theorem: A subset ${u_1,...,u_m}$ in $M = A^n$ is linearly independent iff: if $a in A$ is such that $a cdot (u_1 wedge ldots wedge u_m) = 0$, then $a = 0$.

Here $u_1wedge ldots wedge u_m$ is an element of the exterior power $Lambda^m(M)$.

(I will omit the proof here; the relevant passage is reproduced on Google books.)

This gives the result right away: if $m > n$, $Lambda^m(A^n) = 0$.

ac.commutative algebra - What is the insight of Quillen's proof that all projective modules over a polynomial ring are free?

Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 here.

First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$-module. When is $M$ extended from $A$, meaning there is $A$-module $N$ such that $M = A[T]otimes_AN$?

The proof can be broken down to 2 punches:

Theorem 1 (Horrocks) If $A$ is local and there is a monic $f in A[T]$ such that $M_f$ is free over $A_f$, then $M$ is $A$-free

(this statement is much more elementary than what was stated in Quillen's paper).

Theorem 2 (Quillen) If for each maximal ideal $m subset A$, $M_m$ is extended from $A_m[T]$, then $M$ is extended from $A$

(on $A$, locally extended implies globally extended).

So the proof of Serre's conjecture goes as follows: Let $A=k[x_1,cdots,x_{n-1}]$, $T=x_n$, $M$ projective over $A[T]$. Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ + Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$, so by induction must be free.

Eisenbud note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric).

As Lieven wrote, the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$, namely $N=M/TM$, so let $M'=A[T]otimes_AN$ and build an isomorphism $M to M'$ from the known isomorphism locally.

It is hard to answer your question: what did Serre miss (-:? I don't know what he tried? Anyone knows?

ag.algebraic geometry - Intersection of open affines is affine

I would like to see a clear, rigorous and elementary proof of the following statement:

Let X be a (not necessary quasi-projective, separated) algebraic variety over the complex numbers, and let U,V be two affine open subsets of X. Then the intersection of U and V is affine.

Does the proof change if one substitutes "scheme" for "variety"?

nt.number theory - Slick proof related to choosing points from an interval in order

Choose a point anywhere in the unit interval $[0, 1]$. Now choose a second point from the same interval so that there is one point in each half, $[0, frac12]$ and $[frac12, 1]$. Now choose a third point so that there is one point in each third of the interval, and so on. How long can you choose points like this?

In other words, what is the largest $n$ so that there exists a (finite) sequence $(a_i)_{i=1}^n$ so that for all $k$, among the first $k$ points $(a_i)_{i=1}^k$, there is at least one in each interval $[0, frac1k], [frac1k, frac2k], ldots, [frac{k-1}k,1]$?

My question

Is there a slick proof that $n$ is bounded? (Note that by compactness, the existence of such a sequence for all $n$ is equivalent to the existence of an infinite sequence with the same property.)

Backstory

I was recently re-reading old columns of Martin Gardner's, and he describes this problem in "Bulgarian solitaire and other seemingly endless tasks". (Original column 1983, available in his collections "The last recreations: ..." and "The colossal book of mathematics".) He says that the problem first appeared in "One hundred problems in elementary mathematics" by Hugo Steinhaus. There, he shows that there is a sequence of length 14, but there is none of length 75. The first published proof of the longest sequence, which has length 17, was by Elwyn Berlekamp and Ron Graham in their paper "Irregularities in the distributions of finite sequences" in the Journal of Number Theory. This was followed by a shorter proof by Mieczysƚaw Warmus in "A supplementary note on the irregularities of distributions" in the same journal.

Now, these proofs mostly use case analysis to some degree or other. They are of varying complexities, oddly with Warmus's proof of the optimal $n$ being the shortest. It's also not too hard to write a computer check oneself that finds the optimal $n$. However, I feel that because of the elegant nature of the problem, there should be some "nice" proof -- not of optimality, but simply that such a sequence can't continue forever. Can anyone find one?

Technical note: The problem is usually stated with half-open intervals. I made them closed so that compactness worked. (Edit: The possibility that this changes the answer did occur to me. I assume it doesn't, and I'll check by computer soon. I am fine with answers for any kind of intervals -- open, closed, half-open.)

dg.differential geometry - Is the volume form on an oriented Riemannian manifold parallel?

These are important questions.
When one starts to define connections, one usually imposes the Leibnitz condition on it. But I am proceeding too fast. First, let's recall what a connection is.
Here we come to one very confusing point, at least, it was for me, when I started learning differential geometry: There are various ways of defining connections.
Just to name some of them:

1. Define the Levi-Civita connection on a (Riemannian) manifold.
   The Levi-Civita connection is an affine connection that preserves the metric structure, i.e.,
   $nabla g = 0$ and that is torsion-free, i.e., $nabla_{X}Y-nabla{Y}X = [Y, X]$, where $[. , .]$ denotes the Lie-bracket of two elements of $Gamma(TM)$. Then one usually starts by showing that, by this definition, the Levi-Civita connection (hereafter called LC connection) is uniquely determined.




2. However, one can even generalize this approach by at first considering linear connections, then metric ones and then affine ones. After all that one can turn to the derivation of the Levi-Civita connection. Personally, I like this approach pretty much because it shows very well why the LC connection is so important. Namely, it eases computations for the Lie-derivative which normally does not require the notion of the LC connection at all. Also there are many fascinating mathematical objects, such as Killing fields which can be calculated (or at least defined) effectively, if one introduces the notion of the LC connection.




3. One can even start more generally, by considering fiber bundles. This approach was taken by the French mathematician Charles Ehresmann (1905-1979). This approach has one fundamental advantage: Generality, however, it lacks the easy-to-grasp-effect (I like to assign a definition this value if it has a very intuitive meaning). Interestingly, this approach does yield the LC connection as a special case when one reduces fiber bundles to vector bundles and considers the bundles $pi : TM longrightarrow M$ as a special case and requires the base manifold $M$ to be equipped with a symmetric, positive definite 2-times covariant tensor field, i.e., the metric tensor $(g_{ij})_{i,j}$.

Why am I writing all this? Simply to sum up what we already know and how beautifully, at least, in my opinion, all these definitions harmonize.

To come to your question (I apologize for relying so much on your patience):

If one has finally obtained a manifold that is equipped with a Riemannian metric, then one goes further and asks what the term orientation means for general Riemannian manifolds. Although for submanifolds of $mathbb{R}^{n}$ one has a very intuitive picture of what orientability means. We could expand this picture by making use of the Nash Embedding theorem, but it turns that this is not necessary.

Example: Consider for instance the so-called Möbius strip. It is a typical example of a non-orientable two-dimensional submanifold of the Euclidean 3-space.

Then, one can ask further what it means for a manifold to be orientable. Consider now the tangent and the cotangent bundle. It is now possible to associate to each vector space a certain orientation if one introduces one starting point for that. Consider a basis that is called "unit basis". Define the orientation of this basis to be positive. Then, one imposes further conditions on the unit basis, namely orthonormality w.r.t the Riemannian metric. This way, one can define an orientation on the cotangent bundle $TM^{*}$ as well (Duality principle).

Now we simply ask what the Riemannian volume form is: It is certainly $SO(n)$-invariant, because we have chosen an orientation and require the 1-forms that constitute a basis of $TM^{*}$ to be positively oriented (c.f. what I said above).

OK, now to question one:
Via the Leibnitz rule and the duality principle it is possible to define an induced connection on the cotangent bundle and, by that way, further expanding this definition to the k-th exterior power of the cotangent bundle.

Namely, from the tensor product spaces can be obtained the k-th exterior cotangent space. I would like to refer you to the book Introduction to Kähler Manifolds which contains and excellnt treatment of exactly this. Via the equivalence relation one then breaks down the Leibnitz rule to the k-h exterior power cotangent bundle. The rules is exactly what Matt has written above.

Let's come to question two: We know that the volume form is $SO(n)$ invariant, where $n=dim(M)$ Since we define the volume form $omega := dx^{1} wedge dots wedge dx^{n}$, with the $dx^{i}$ for $i=1, dots, n$ to be the product of the basis of the cotangent bundle, we can make the following thought.

We know how to "differentiate" the basis w.r.t to the LC connection. When we calculate all that, we finally obtain that the volume form is parallel. This was how I learned it. Of course one can use representation theory here which is, in my eyes, very interesting and also more elegant.

If there are questions, don't hesitate to contact me.

Prime decomposition for knots in manifolds

This was addressed in Miyazaki, Conjugation and the prime decomposition of knots in closed, oriented 3-manifolds, using the definition of connected sum suggested by Ryan Budney in the comments.

The main results are that a prime decomposition of K exists iff a meridian of K is not null-homotopic in the complement of K, and if a prime decomposition of K does not contain a particular knot R in S¹×S² then it is the unique decomposition of K, whereas knots with this summand R can admit several prime decompositions.

pr.probability - probability mass function fitting

I have a probability mass function of some experimental data who's log looks like the following: (please ignore the fact that it is not normalized)
![alt text][1]

[image shack image removed]

(meaning if p(x) is the pmf, this is log(p(x)) )
Does anyone know what parametric family it might belong to? (note that this is a discrete distribution)

ag.algebraic geometry - Cartier divisor on an open subscheme whose complement is of codim 2

1 - This is true for locally factorial schemes; in this case all Weil divisors are Cartier, and extending Weil divisors is not a problem.

2 - This is not true in general (for example, if you glue together two points of the affine plane over a field k, the Picard group of the resulting scheme is $k^*$, while the Picard group of the complement is trivial). It is true for schemes satisfying Serre's S2 condition.

ag.algebraic geometry - Rosenlicht differentials for possibly non-reduced curves

Let $X$ be proper Cohen-Macaulay scheme of pure dimension 1 over an algebraically closed field $k$. When $X$ is moreover reduced, Rosenlicht's theory of regular differential forms gives a beautiful explicit description of Grothendieck duality for $X$. I am wondering if there is an analogue of Rosenlicht's theory in the general context of proper CM curves.

More precisely, in the reduced case Rosenlicht defines a sheaf of $cal{O}_X$-modules $omega$

whose sections over an open $U$ in $X$ are those meromorphic differentials
$eta$ on the inverse image of $U$ in the normalization $X'$ of $X$ with the property that for each closed point $xin U$,
the sum of the residues of $feta$ over all points $y$ of $X'$ mapping to $x$ is zero for all
$fin mathcal{O}_{X,x}$. He also defines a trace morphism $Tr:H^1(X,omega)rightarrow k$
in terms of "sums of residues" with the property that the pair $(omega,Tr)$ is canonically isomorphic to the relative dualizing sheaf with its (Grothendieck) trace mapping (one proves this by showing that Roesnlicht's construction satisfies the right universal property).

Is there a similarly explicit (i.e. in terms of certain kinds of differential forms and residues) description of the sections of the dualizing sheaf and of the trace map in the general proper CM setting?

Things to note: A random normal proper and flat curve (=scheme of pure relative dimension 1) $cal{X}$ over $W(k)$ will always have CM special fiber. However, this special fiber is "very often" not reduced, so there are many examples of non-reduced CM curves.

I asked a previous question on MO hinting at this one: Adjunction for underlying reduced subschemes

I'm happy to assume that $X$ is Gorenstein, if that is at all useful.

Eigenvalues of an element in a Weyl algebra

I have an operator acting on the polynomial algebra $mathbb{C}[x,y,z]$ that I would like to find the eigenvalues/eigenvectors of. More specifically, let $P(x_1, ldots, x_6)$ be a homogeneous polynomial, my operator has the form $P(x,y,z, frac{partial}{partial x}, frac{partial}{partial y}, frac{partial}{partial z})$. Are there any general strategies that could help me? For instance, say my operator were: $$z^2yfrac{partial^3}{partial x^2 partial y} + y^3zfrac{partial^4}{partial y^2 partial z^2} + xz^2frac{partial^3}{partial x partial z^2}.$$ My actual operator is degree 6 and more complicated, but other than that the same type of object. Any thoughts or references on how to attack this type of problem will be very welcome. Thanks a lot!

EDIT: My first example-operator was very poorly chosen, since every monomial would automatically be an eigenvector. I have now altered it a little to avoid this. Keep in mind this was only an example to show what type of object I am considering, and I am looking for general strategies. None the less, thanks for the quick response.

EDIT 2: I had misread the degree of my actual operator - it is 6.

hopf algebras - Action of Co-quasi-triangular Universal r-form on $a otimes 1$

Like David said, the proof is almost identical to the earlier one for $R$-matrices:

$r(xotimes 1) = rcirc (idotimesmu)(xotimes 1otimes 1) = (r_{13}ast r_{12})(xotimes 1otimes1)= sum r(x'otimes 1)r(x''otimes 1) = (r ast r)(xotimes 1).$

Since $r$ is invertible, $r(xotimes 1)=epsilon(x)$.

rt.representation theory - Is the classification of supercuspidal representations a wild problem?

I cannot directly answer your question because I am not a specialist of wild classification problems. However I want to make the following remark that could help.

I was told that the problem of classifying the irreducible (smooth) complex representations of ${rm GL}_n ({mathfrak o}_F )$, where ${mathfrak o}_F$ is the ring of integers of a local field $F$, is wild (at least for $n$ large).

Moreover to construct and classify supercuspidal representations, Bushnell and kutzko obtain them as compactly induced representations from irreducible smooth representations of compact mod center subgroups. In particular we obtain a large class of supercuspidals by inducing certain irreducible complex representations of $F^{times}{rm GL}_n ({mathfrak o}_F)$.

However in Bushnell and Kutzko's theory not all representations of ${rm GL}_n ({mathfrak o}_F )$ are needed, but only very particular ones and one knows how to construct them effectively. Indeed the whole Bushnell and Kutzko's construction is based on the theory of 'simple characters' which is in principal entirely effective.

pr.probability - Porbability of selecting balls from boxes

Well, this looks more like someone trying to get their homework done, but for the first part:

$ p = 0.2 * frac{3}{3+7} + 0.2 * frac{5}{5+5} + 0.6 * frac{2}{2+8}$

$ p = 0.06 + 0.10 + 0.12 $

$ p = 0.28 $

Showed the work for you too.

So if the probability that a chosen ball is red is 28%, then the probability that a chosen ball is green is 72%.

So what is {probability that chosen ball came from B3 | chosen ball is green}? Look up conditional probability, look up bayesian, etc.

$ p_g = 0.2 * frac{7}{3+7} + 0.2 * frac{5}{5+5} + 0.6 * frac{8}{2+8}$

$ p_g = 0.14 + 0.10 + 0.48 $

$ p_g = 0.72 $

{ $p_g3$ | green ball} = (picked from box 3 and green) / (picked green)

= 0.48 / 0.72

= 2 /3

Please do your homework yourself.
Showed the work for you too.

teaching - Blackboard rendering of math fonts

Suggestion number one:

Learn Calligraphy! It's a lot of fun and does mean that you can write the fonts in genuinely nice ways. Books on calligraphy tend to have detailed instructions on how to do at least the basic alphabets: explaining which stroke to do first, and how to hold the pen. Although not all of it transfers to the blackboard, it helps a lot. For example, once you seen how the different "g"s are written, you'll know how to write the Lie algebra symbol correctly. However, I do find that a script S ($mathcal{S}$ is not even close) can take me a couple of goes to make it look right - it shouldn't be pointy at the top but should sweap backwards.

(Note: to anyone reading the comments, I originally had this answer together with my other answer in the same post. Given the comments, I decided to split them.)

Following up on aleksander's comment to the original question, I had a go at doing a video of how to draw a fraktur g (𝔤) (well, actually it's gothic but if you know the difference you don't need this video, and the gothic g is probably more distinguishable from a normal g than a fraktur one on a blackboard). It's not very polished, but you can see what it looks like here. It was quite fun to do so if this would be helpful, I can easily do more.

ct.category theory - Is there a category of non-well-founded sets?

Starting from any model of a membership-based set theory, be it well-founded or ill-founded, you can construct a category of sets and functions. The basic properties of this category (e.g. it is a well-pointed topos) don't depend on whether the model you started from was well- or ill-founded. In fact, in the presence of the axiom of choice, every ill-founded set is still well-orderable, hence bijective to a well-founded set (a von Neumman ordinal) -- thus the category of sets obtained from a model of ill-founded set theory + choice is equivalent to the subcategory obtained from its submodel of well-founded sets.

You can then ask whether you can reconstruct a model of membership-based set theory from its category of sets; here is where the graphs come in (to model hereditary membership relations), as at nlab:pure set. You can choose to use well-founded graphs or ill-founded ones. If the one you choose matches the type of set theory you started from, then (as long as your set theory is strong enough otherwise) you'll reconstruct the same model. If you choose well-founded graphs starting from an ill-founded set theory, then you'll reconstruct the submodel of well-founded sets, reproducing the proof of relative consistency of the axiom of foundation. And if you choose ill-founded graphs starting from a well-founded set theory, you'll reproduce Aczel's original proof of the relative consistency of the anti-foundation axiom.

dg.differential geometry - Normal coordinates for a manifold with volume form

I'm hoping that the following are true. In fact, they are probably easy, but I'm not seeing the answers immediately.

Let $M$ be a smooth $m$-dimensional manifold with chosen positive smooth density $mu$, i.e. a chosen (adjectives) volume form. (A density on $M$ is a section of a certain trivial line bundle. In local coordinates, the line bundle is given by the transition maps $tildemu = left| det frac{partial tilde x}{partial x} right| mu$. When $M$ is oriented, this bundle can be identified with the top exterior power of the cotangent bundle.) Hope 1: Near each point in $M$ there exist local coordinates $x: U to mathbb R^m$ so that $mu$ pushes forward to the canonical volume form $dx$ on $mathbb R^m$.

Hope 1 is certainly true for volume forms that arise as top powers of symplectic forms, for example, by always working in Darboux coordinates. If Hope 1 is true, then $M$ has an atlas in which all transition maps are volume-preserving. My second Hope tries to describe these coordinate-changes more carefully.

Let $U$ be a domain in $mathbb R^m$. Recall that a change-of-coordinates $tilde x(x): U to mathbb R^m$ is oriented-volume-preserving iff $frac{partial tilde x}{partial x}$ is a section of a trivial ${rm SL}(n)$ bundle on $U$. An infinitesimal change-of-coordinates is a vector field $v$ on $U$, thought of as the map $x mapsto x + epsilon v(x)$. An infinitesimal change-of-coordinates is necessarily orientation-preserving; it is volume preserving iff $frac{partial v}{partial x}(x)$ is a section of a trivial $mathfrak{sl}(n)$ bundle on $U$. Hope 2: The space of oriented-volume-preserving changes-of-coordinates is generated by the infinitesimal volume-preserving changes-of-coordinates, analogous to the way a finite-dimensional connected Lie group is generated by its Lie algebra.

Hope 2 is not particularly well-written, so Hope 2.1 is that someone will clarify the statement. Presumably the most precise statement uses infinite-dimensional Lie groupoids. The point is to show that a certain a priori coordinate-dependent construction in fact depends only on the volume form by showing that the infinitesimal changes of coordinates preserve the construction.

Edit: I have preciseified Hope 2 as this question.

mathematics education - Journals for undergraduates

``Plus"- an online-only British journal (similar to Math Horizons):

http://plus.maths.org/content/

From their home page: ``Plus provides articles and podcasts on any aspect of mathematics, covering topics as diverse as art, medicine, cosmology and sport, a news section, showing how recent news stories were often based on some underlying piece of maths that never made it to the newspapers, reviews of popular maths books, and puzzles for you to sharpen your wits. We have a regular interview with someone in a maths-related career, showing the wide range of uses maths gets put to in the real world."

The intended audience is not limited to math undergraduates; science undergrads would find something interesting there, too. Some articles can be accessible even to high-school students.

And for those who read Polish, there is ``Delta" (which I used to read regularly in the paper version as an undergrad):

http://www.deltami.edu.pl/

The 01/2013 issue contains e.g. an article on William Thurston and geometrization conjecture (by Zdzislaw Pogoda).

ca.analysis and odes - Estimate for tail of power series of exponential function?

The exponential inequality yields explicit bounds as follows. For every $Bge A$, consider the series
$$
S(A,B)=sum_{k=B}^{+infty}frac{A^k}{k!}.
$$
Then, as Michael Lugo noticed, $mathrm{e}^{-A}S(A,B)=P(N_Age B)$, where $N_A$ is a Poisson random variable with parameter $A$.

For every positive $t$, $N_Age B$ if and only if $mathrm{e}^{tN_A-tB}ge1$, and for every nonnegative random variable $X$, $P(Xge1)le E(X)$. Using this for $X=mathrm{e}^{tN_A-tB}$, one gets
$$
mathrm{e}^{-A}S(A,B)le E(X)=E(mathrm{e}^{tN_A}) mathrm{e}^{-tB}.
$$
To go further, one uses the fact that the Laplace transform $E(mathrm{e}^{tN_A})$ of a Poisson distribution is $mathrm{e}^{A(mathrm{e}^t-1)}$ and one optimizes the upper bound with respect to $tge0$. That is, one plugs in the inequality the value of $t$ such that $mathrm{e}^t=B/A$, which yields
$$
S(A,B)lemathrm{e}^{B-Blog(B/A)}.
$$
Note that this upper bound is not asymptotic (which is nice) but that, in general, it is not optimal in the regime of the central limit theorem you are interested in (which is not so nice).

Turning to the case where $Atoinfty$ and $B=A+Csqrt{A}$ with $C>0$ fixed, the expansion of $log(B/A)$ up to the order $o(1/A)$ yields
$$
S(A,A+Csqrt{A})lemathrm{e}^{A-C^2/2+o(1)}=mathrm{e}^{-C^2/2}mathrm{e}^A(1+o(1)).
$$
Which is a (very odd) way to check that $Phi(C)lemathrm{e}^{-C^2/2}$...

Edit : Non asymptotic upper bound :
$$
S(A,A+Csqrt{A})lemathrm{e}^{-C^2/2}mathrm{e}^{A}exp(C^3/(2sqrt{A})).
$$

nt.number theory - Strict Class Numbers of Totally Real Fields

Maybe it's worth a word about why Cohen-Lenstra predicts this behavior. Suppose K is a field with r archimedean places. Then Spec O_K can be thought of as analogous to a curve over a finite field k with r punctures, which is an affine scheme Spec R. Write C for the (unpunctured) curve. Then the class group of R is the quotient of Pic(C)(k) by the subgroup generated by the classes of the punctures -- or, what is the same, the quotient of Jac(C)(k) by the subgroup generated by degree-0 divisors supported on the punctures. (This last subgroup is just the image of a natural homomorphim from Z^{r-1} to Jac(C)(k).)

The Cohen-Lenstra philosophy is that these groups and the puncture data are "random" -- that is, you should expect that the p-part of the class group of R looks just like what you would get if you chose a random finite abelian p-group (where a group A is weighted by 1/|Aut(A)|) and mod out by the image of a random homomorphism from Z^{r-1}. (There are various ways in which this description is slightly off the mark but this gives the general point.)

It turns out that when r > 1 the chance is quite good that a random homomorphism from Z^{r-1} to A is surjective. In fact, the probability is close enough to 1 that when you take a product over all p you still get a positive number. In other words, when r > 1 Cohen-Lenstra predicts a positive probability that the class group will have trivial p-part for all p; in other words, it is trivial. (In fact, it predicts a precise probability, which fits experimental data quite well.)

When r = 1, on the other hand, the class group is just A itself, and the probability its p-part is trivial is on order 1-1/p. Now the product over all p is 0, so one does NOT expect to see a positive proportion of trivial class groups. And in fact, when there is just one archimedean place -- i.e. when K is imaginary quadratic -- this is just what happens!

soft question - Too old for advanced mathematics?

With all this unanimous enthusiasm, I can't help but add a cautionary note. I will say, however, that what I'm about to say applies to anyone of any age trying to get a Ph.D. and pursue a career as an academic mathematician.

If you think you want a Ph.D. in mathematics, you should first try your best to talk yourself out of it. It's a little like aspiring to be a pro athlete. Even under the best of circumstances, the chances are too high that you'll end up in a not-very-well-paying job in a not-very-attractive geographic location. Or, if you insist on living in, say, New York, you may end up teaching as an adjunct at several different places.

Someone with your mathematical talents and skills can often find much more rewarding careers elsewhere.

You should pursue the Ph.D. only if you love learning, doing, and teaching mathematics so much that you can't bear the thought of doing anything else, so you're willing to live with the consequences of trying to make a living with one. Or you have an exit strategy should things not work out.

Having said all that, I have a story. When I was at Rice in the mid 80's, a guy in his 40's or 50's came to the math department and told us he really wanted to become a college math teacher. He had always loved math but went into sales(!) and had a very successful career. With enough money stashed away, he wanted to switch to a career in math. To put it mildly, we were really skeptical, mostly because he had the overly cheery outgoing personality of a salesman and therefore was completely unlike anyone else in the math department. It was unthinkable that someone like that could be serious about math. Anyway, we warned him that his goal was probably unrealistic but he was welcome to try taking our undergraduate math curriculum to prepare. Not surprisingly, he found this rather painful, but eventually to our amazement he started to do well in our courses, including all the proofs in analysis. By the end, we told the guy that we thought he really had a shot at getting a Ph.D. and have a modest career as a college math teacher. He thanked us but told us that he had changed his mind. As much as he loved doing the math, it was a solitary struggle and took too much of his time away from his family and friends. In the end, he chose them over a career in math (which of course was a rather shocking choice to us).

So if you really want to do math and can afford to live with the consequences, by all means go for it.

nt.number theory - What do you call this ring?

As David Speyer says, the most common ways of referring to $prod_p mathbb{Z}_p$ are "$mathbb{Z}$-hat" or "the profinite completion of $mathbb{Z}$''.

However, I have also heard it called "the Prufer ring", see e.g.

http://mathworld.wolfram.com/PrueferRing.html

I do not endorse this usage, since nowadays "Prufer ring" is a certain kind of commutative ring. An integral domain is a Prufer ring if every finitely generated ideal is invertible. Of course $widehat{mathbb{Z}}$ is not a domain. More recently there have been definitions of Prufer ring for non-domains; unfortunately not all of the characterizations of Prufer domains carry over to rings with zero divisors and it is not completely clear to me that there is one standard definition of a Prufer ring. For instance, I have seen that a Prufer ring is a ring in which every finitely generated regular (i.e., containing a non zero-divisor) ideal is invertible, and also that a Prufer ring is a ring in which each finitely generated ideal is flat. (In fact I do not know off-hand whether these two conditions are equivalent! In any case, there are many others...)

For every definition of Prufer ring I have seen, it is at least true that $widehat{mathbb{Z}}$ is a Prufer ring.

Also, in Fried and Jarden's authoritative text Field Arithmetic, they refer to
$widehat{mathbb{Z}}$ as the Prufer group (p. 14 of the third edition). Again, I have heard other things referred to as Prufer groups in the literature...

In summary, "Z-hat" is probably your best bet.

Complex powers in finite fields

There is no standard way to define this idea, and I doubt there is a useful one.

If I were trying to invent a definition, I'd say the following:

We should definitely have $(ab)^i = a^i b^i$ and $(a^i)^i=a^{-1}$. The first says that taking the $i$-th power is a homomorphism from the multiplicative group of the field to itself. Since the multiplicative group of a finite field is cyclic, the only homomorphisms from it to itself are of the form $a mapsto a^k$ for some integer $k$.

Which integers $k$ should we accept? The second property imposes $a^{k^2} = a^{-1}$ modulo $p$. This implies that $k^2 = -1 mod (p-1)$.

For most primes $p$, there are no solutions to $k^2 = -1 mod (p-1)$. In the cases that there are, you could define raising to the $i$ power to means raising to the $k$ power for such a $k$. For example, when $p=11$, you could define $a^i$ to mean $a^3$, and you would get the two properties above.

I wouldn't do this though. At least in the areas of math where I work, it is considered a bad idea to define an operation on $F_p$ that doesn't extend nicely to the fields $F_{p^j}$. And there will never be a good definition of raising to the $i$ power in $F_{p^2}$. In order to get one, you'd need a solution to $k^2 = -1 mod p^2-1$. But, for any prime other than $3$, $p^2-1$ is divisible by $3$, and $k^2+1$ is never divisible by $3$.

rt.representation theory - Are there positive formulae for the inner product between elements of a Lie algebra representation in the Shapovalov form?

This may not be exactly what you want, but I'd suggest you look at Kashiwara's paper ``On crystal bass of the q-analogue of the universal enveloping algebra" (see MR1115118 ).

In section 2.5 Kashiwara discusses the quantum version of the Shapovalov form. More relevant to what I want to say is Proposition 3.4.4, which defines/proves existence of a modification of the Shapovalov form defined on $U_q^-$. Roughly what he is doing is the following: Consider the pairing of, for example, $F_1F_2F_1v_lambda$ and $F_1^2 F_2v_lambda$,
using the Shapovalov form on $M(lambda)$, and allow $lambda$ to vary. What you get is $$P(lambda)/(q-q^{-1})^3$$ where $P$ is some Laurent polynomial in the $K_i$ and $q$, and evaluation at $lambda$ means
setting $K_i$ equal to $q^{(alpha_i, lambda)}.$ $P$ has a well defined highest order term in the $K_i$, and the coefficient of this term is a Laurent polynomial in $q$. Let $(F_1F_2F_1, F_1^2 F_2)$ be that leading coefficient. This will be Kashiwara's inner product on $U_q^-$, up to a power of $q$. In general, when pairing two monomials in the $F_i$ applied to $v_lambda$, the denominator in the above equation has a factor of $q_i-q_i^{-1}$ for each $F_i$ in the first monomial.

The way Kashiwara sets things up, it is clear that the inner product of two monomials in the $F_i$ is a Laurent polynomial in $q$ with positive coefficients. For monomials $m_1$ and $m_2$, the inner product will be zero unless $m_1$ and $m_2$ have the same weight (i.e. they are both the products of the same number of each $F_i$, but possibly in a different order). Furthermore, the sum of the coefficients is the number of ways of matching each $F_i$ in $m_1$ with an $F_i$ in $m_2$ for all $i$. This all follows from Equations (3.3.1) and (3.4.6) in Kashiwara's paper. I believe you can find the power of $q$ associated to a given matching by arranging the monomials correctly, drawing a line between each matching pair, and counting a contribution for each crossing in the resulting picture. So there should be a completely combinatorial formula.

As I said, this may well not be what you want. For instance, this construction does not depend on $lambda$. But maybe it is related.

In case you are wondering about the connection with crystal bases (i.e. the title of Kashiwara's paper), Kashiwara shows that the inner product of any two elements in the crystal lattice $L(infty)$ is regular at $q=0$, and a crystal basis is an orthonormal basis for the evaluation $(cdot, cdot)_0$ of the inner product at $q=0$. Of course for this to be true you need to get the powers of $q$ right, which I have not done here.