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https://mathoverflow.net/questions/5209 | 5 | There are a few standard notions of matrix derivatives, e.g.
* If *f* is a function defined on the entries of a matrix *A*, then one can talk about the matrix of partial derivatives of *f*.
* If the entries of a matrix are all functions of a scalar *x*, then it makes sense to talk about the derivative of the matrix as the matrix of derivatives of the entries.
In the second case, it makes sense to talk about higher order derivatives, but in the first example the derivative provides a matrix from a scalar function, so you have to massage it a little bit to define a higher order derivative (e.g. take the trace of the resulting matrix).
I was wondering what other notions of matrix differentiation might exist out there, particularly any notions that allow for higher-order differentiation. I am also interested in any connections between various forms of matrix derivatives. As this is related to an undergraduate research project, I am mostly looking for answers that include a minimum of advanced terminology, but a discussion of how more general concepts (e.g. differential forms, matrix exponentials, etc.) relate to matrix derivatives would also be helpful.
| https://mathoverflow.net/users/498 | Notions of Matrix Differentiation | There is another interpretation of Elisha's question that I think has not yet been addressed: How, and to what extent, can you do differential calculus with functional expressions of square matrices? For instance, how do you differentiate $\exp(A)$, which is defined for all square matrices $A$?
There is a good answer to this question for polynomials and an even better answer for the trace of a univariate polynomial. Both of these good answers extend automatically to analytic functions $f(z)$ evaluated at matrices by means of their Taylor series. (Which of course includes exponentiation.) I like to write the answer in terms of differentials. The differential of matrices $A$, $B$, etc., is $dA$, $dB$, etc., which you can take to mean that the matrices are unspecified functions of some dummy parameter $t$, and $dA$ then represents the formal numerator of the matrix-valued derivative $dA/dt$.
In this formalism, the Leibniz rule still holds, as long as you remember that matrix multiplication is non-commutative: $d(AB) = (dA)B + A(dB)$. Indeed, the matrices don't actually need to be square. The sum rule holds trivially. The negative power rule has a creative correct answer: $d(A^{-1}) = A^{-1}(dA)A^{-1}$. You can then differentiate exponentiation:
$$d\exp(A) = dA + \frac{(dA)A + A(dA)}2 + \frac{(dA)A^2 + A(dA)A+A^2(dA)}6 + \cdots.$$
As you can see from this example, you can differentiate a power series, but nothing all that great happens because $dA$ might not commute with $A$. However, if you're computing the differential of $\mathrm{Tr}(f(A))$, then something very nice happens: You can cyclically permute each term of the differentiated power series to put $dA$ at the end. Thus, you get the very nice formula
$$d\mathrm{Tr}(f(A)) = \mathrm{Tr}(f'(A)dA).$$
I have only derived this formula when $A$ is within the radius of convergence of the Taylor series of $f$. However, by continuity it applies generally when $f(A)$ and $f'(A)$ are well-defined. (For instance, if $A$ and $dA$ are both real symmetric or Hermitian, then it is enough for $f$ to exist as a real function and have one continuous derivative near the eigenvalues of $A$.)
Higher derivatives work basically the same way as first derivatives. Again, you should take $A$ to be a function of a dummy parameter $t$. To get the simplest expressions for higher derivatives, you should assume that $A$ is a linear function of $t$ (including a constant). Then for example:
$$d^2\exp(A) = (dA)^2 + \frac{(dA)^2A + (dA)A(dA) + A(dA)^2}3 + \cdots.$$
The trace of this looks good at first, but the quadratic term in $A$ has both $\mathrm{Tr}((dA)^2A^2)$ and $\mathrm{Tr}((A(dA))^2)$, and I don't think that the rest of the traced Taylor series simplifies the way that it did for the first derivative.
A final remark: All of this works the best for functions $f(x)$ that are entire (in the sense of complex analysis, i.e., an infinite radius of convergence). One of the definitions of an entire function is one whose Taylor series decays superexponentially, and this is also a good condition for a non-commutative multivariate Taylor series, in the sense that it will converge for any matrices that you plug in.
| 9 | https://mathoverflow.net/users/1450 | 5318 | 3,579 |
https://mathoverflow.net/questions/5321 | 25 | In order get a minimal model for a given a variety $X$, we can carry out a sequence of contractions $X\rightarrow X\_1\ldots \rightarrow X\_n$ in such a way that that every map contracts some curves on which the canonical divisor $K\_{X\_j}$ is negative.
Here we have, at least, the following technical problem: in contracting curves, the resulting variety $X\_j$ might have become singular. In order to fix this issue, people consider a ***flip***.
Here are my questions. What is the intuition to understand such a flip?
Are there examples of such things in other contexts of math or is it an *ad hoc* construction?
| https://mathoverflow.net/users/1547 | Flips in the Minimal Model Program | I am also just learning this stuff, and I'm partly writing this out for my own benefit. Experts, please correct and up/down vote as appropriate!
The goal of the minimal model program is to give a standard, nonsingular, representative for each birational class of algebraic variety. As stated, this goal is too ambitious, but it will help us to understand the minimal model program if we think of it as a partially successful attempt at this goal.
Let $X$ be a compact, smooth algebraic variety of dimension $n$. Let $\omega$ be the top wedge power of the holomorphic cotangent bundle. Then the vector space, $V:=H^0(X, \omega)$, of holomorphic $n$-forms on $X$ is a birational invariant of $X$. This means that we should be able to see $V$ from just the field of meromorphic functions on $X$; here is a [sketch](https://mathoverflow.net/questions/152/how-do-you-see-the-genus-of-a-curve-just-looking-at-its-function-field/345#345) of how to do that. So we get a rational map $X \to \mathbb{P}(V^{\\*})$ by the standard recipe. More generally, we can replace $\mathbb{P}(V)$ with Proj of the ring $\bigoplus H^0(X, \omega^{\otimes n})$. This is called the canonical ring; you may have heard of the recent breakthrough in proving that the canonical ring is finitely generated. We can map $X$ rationally to this Proj; the image is called the log model. This is a partial success: it is a canonical, birational construction, but it may not be birational to $X$ and may not be smooth.
There are certain well understood rules of thumb for how various subobjects of $X$ behave in the log model. For example, if $X$ is a surface and $C$ a curve with negative self intersection, then $C$ will be blown down in the log model.
Here is a more complicated example, which is relevant to your question. Let $Y$ be some variety that locally looks like the cone on the Segre embedding of $\mathbb{P}^1 \times \mathbb{P}^1$. So $Y$ is a $3$-fold with an isolated singularity. If you are familiar with the toric1 picture, it looks like the tip of a square pyramid. Inside $Y$, let $Z$ be the cone on one of the $\mathbb{P}^1$'s. This is a surface, but not a Cartier divisor. Let $X$ be $Y$ blown up along $Z$; so that the isolated singularity becomes a line. In the toric picture, the point of the pyramid has lengthened into a line segment, and two of the faces which used to touch at the point now border along an entire edge. In the log model, the line will blow back down to become a point. So the log model can turn a smooth variety, like $X$, into a singular one like $Y$.
Now, birational geometers did not rest on their laurels when they had constructed the log model. They made other constructions, which are smoother but less canonical. Many of these constructions can be thought of as taking the log model and modifying it in some way. If the log model looks like the example of the previous paragraph, they want to take the singular point of $Y$ and replace it by a line, to look like $X$. But they have two ways they can do this; they can blow up one $\mathbb{P}^1$ or the other; giving either $X$ or $X'$. Often, replacing $X$ by $X'$ is crucial in order to improve the model somewhere else. The relationship between $X$ and $X'$ is called a flip, because we take the line inside $X$ and flip it around to point in a different direction.
1 Cautionary note: although the toric picture is excellent for visualizing what is going on locally, you shouldn't take $X$ itself to be a toric variety. There are no global sections of $\omega$ on a toric variety, so the log model is empty. You want $X$ to locally look like a toric variety, but have global geometry which is nontoric in a way that creates lots of sections of $\omega$.
| 16 | https://mathoverflow.net/users/297 | 5341 | 3,594 |
https://mathoverflow.net/questions/5342 | 3 | I wanted to learn prediction, forecasting etc. I also have time series data on millions of online videos. I would like to test out prediction algorithms etc on this data set, for eg. Linear Prediction, Kalman filter.
Are there any good resources out there to get me started on those?
Edit: Let me rephrase the question. If you were given such a time series data set with a million videos with some number of attributes to each, what steps/path path would you take to come up with a decent view prediction scheme? For example, factor analysis or PCA etc first.
| https://mathoverflow.net/users/1761 | Where can i find 'getting started' resources for statistical prediction | The following might be of some help
[linear regression](http://en.wikipedia.org/wiki/Linear_regression)
[Bayesian inference](http://en.wikipedia.org/wiki/Bayesian_inference)
(possibly).
| 2 | https://mathoverflow.net/users/1536 | 5350 | 3,601 |
https://mathoverflow.net/questions/5329 | 9 | I think I'm a bit confused about the relationship between some concepts in mathematical logic, namely constructions that require the axiom of choice and "explicit" results.
For example, let's take the existence of well-orderings on $\mathbb{R}$. As we all know after reading [this answer by Ori Gurel-Gurevich](https://mathoverflow.net/questions/5116/is-the-existence-of-a-well-ordering-on-r-independent-of-zf), this is independent of ZF, so it "requires the axiom of choice." However, the proof of the well-ordering theorem that I (and probably others) have seen using the axiom of choice is nonconstructive: it doesn't produce an *explicit* well-ordering. By an *explicit* well-ordering, I simply mean a formal predicate $P(x,y)$ with domain $\mathbb{R}\times\mathbb{R}$ (i. e., a subset of the domain defined by an explicit set-theoretic formula) along with a proof (in ZFC, say, or some natural extension) of the formal sentence "$P$ defines a well-ordering." Does there exist such a $P$, and does that answer relate to the independence result mentioned above?
More generally, we can consider an existential set-theoretic statement $\exists P: F(P)$ where $F$ is some set-theoretic formula. Looking to the previous example, $F(P)$ could be the formal version of "$P$ defines a well-ordering on $\mathbb{R}$." (We would probably begin by rewording that as something like "for all $z\in P$, $z$ is an ordered pair of real numbers, and for all real numbers $x$ and $y$ with $x\neq y$, $((x,y)\in P \vee (y,x)\in P) \wedge \lnot ((x,y)\in P \wedge (y,x)\in P)$, etc.) On the one hand, such a statement may be a theorem of ZF, or it may be independent of ZF but a theorem of ZFC. On the other hand, we can ask whether there is an explicit set-theoretic formula defining a set $P^\\*$ and a proof that $F(P^\\*)$ holds. How are these concepts related:
* the theoremhood of "$\exists P: F(P)$" in ZF, or its independence from ZF and theoremhood in ZFC;
* the existence of an explicit $P^\\*$ (defined by a formula) with $F(P^\*)$ being provable.
Are they related at all?
| https://mathoverflow.net/users/302 | "Requires axiom of choice" vs. "explicitly constructible" | In Goedel's proof of consistency of AC, we in fact get much more. There is an explicit relation defined, which is (provably in ZF) a well-ordering of a certain subset of the reals. It is consistent (and follows from the axiom V=L) that the subset is all of the reals.
| 11 | https://mathoverflow.net/users/454 | 5356 | 3,604 |
https://mathoverflow.net/questions/5323 | 29 | At the time of writing, question 5191 is closed with the accusation of homework. But I don't have a clue about what is going on in that question (other than part 3) [Edit: Anton's comments at 5191 clarify at least some of the things going on and are well worth reading] [Edit: FC's excellent answers shows that my lack of clueness is merely due to ignorance on my own part] so I'll ask a related one.
My impression is that it's generally believed that there are infinitely many Mersenne primes, that is, primes of the form $2^n-1$. My impression is also that it's suspected that there are only finitely many Fermat primes, that is, primes of the form $2^n+1$ (a heuristic argument is on the wikipedia page for Fermat primes). [EDIT: on the Wikipedia page there is also a heuristic argument that there are infinitely many Fermat primes!]
So I'm going to basically re-ask some parts of Q5191, because I don't know how to ask that a question be re-opened in any other way, plus some generalisations.
1) For which odd integers $c$ is it generally conjectured that there are infinitely many primes of the form $2^n+c$? For which $c$ is it generally conjectured that there are only finitely many? For which $c$ don't we have a clue what to conjecture? [Edit: FC has shown us that there will be loads of $c$'s for which $2^n+c$ is (provably) prime only finitely often. Do we still only have one $c$ (namely $c=-1$) for which it's generally believed that $2^n+c$ is prime infinitely often?]
2) Are there any odd $c$ for which it is a sensible conjecture that there are infinitely many $n$ such that $2^n+c$ and $2^{n+1}+c$ are simultaneously prime? Same question for "finitely many $n$".
3) Are there any pairs $c,d$ of odd integers for which it's a sensible conjecture that $2^n+c$ and $2^n+d$ are simultaneously prime infinitely often? Same for "finitely often".
| https://mathoverflow.net/users/1384 | Infinitely many primes of the form $2^n+c$ as $n$ varies? | Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime.
[**edit** Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.]
Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences).
I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance:
www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf
Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to [question 2339](https://mathoverflow.net/questions/2339/).
| 34 | https://mathoverflow.net/users/nan | 5367 | 3,612 |
https://mathoverflow.net/questions/5344 | 25 | Given a compact smooth manifold $M$, it's relatively well known that $C^\infty(M)$ determines $M$ up to diffeomorphism. That is, if $M$ and $N$ are two smooth manifolds and there is an $\mathbb{R}$-algebra isomorphism between $C^\infty(M)$ and $C^\infty(N)$, then $M$ and $N$ are diffeomorphic.
(See, for example, Milnor and Stasheff's "Characteristic Classes" book where an exercise walks one through the proof of this fact).
Thus, in some sense, all the information about the manifold is contained in $C^\infty(M)$.
Further, the tools of logic/set theory/model theory etc. have clearly been applied with greater success to purely algebraic structures than to, say, differential or Riemannian geometry. This is partly do the fact that many interesting algebraic structures can be defined via first-order formulas, whereas in the geometric setting, one often uses (needs?) second-order formulas.
So, my question is two-fold:
First, is there a known characterization of when a given (commutive, unital) $\mathbb{R}$-algebra is isomorphic to $C^\infty(M)$ for some compact smooth manifold $M$? I imagine the answer is either known, very difficult, or both.
Second, has anyone applied the machinery of logic/etc to, say, prove an independence result in differential or Riemannian geometry? What are the references?
| https://mathoverflow.net/users/1708 | Algebraic description of compact smooth manifolds? | There is a very cool answer to your question, and it goes by the name [well-adapted models](http://ncatlab.org/nlab/show/Models+for+Smooth+Infinitesimal+Analysis) for [synthetic differential geometry](http://ncatlab.org/nlab/show/synthetic+differential+geometry). Andrew Stacey already indicated it in his reply, but maybe I can expand a bit more on this.
Synthetic differential geometry is an axiom system that characterizes those categories whose objects may sensibly be regarded as spaces on which differential calculus makes sense. These categories are called [smooth toposes](http://ncatlab.org/nlab/show/smooth+topos).
A *model* for this is a particular such category with these properties. A *well-adapted model* is one which has a [full and faithful embedding](http://ncatlab.org/nlab/show/full+and+faithful+functor) of the category of smooth manifolds. (This is "well adapted" from the point of view of ordinary differential geoemtry: ordinary differential geometry embeds into these more powerful theories of smooth structures).
The striking insight is that this perspective in particular usefully unifies the ideas of algebraic geometry with that of differential geoemtry to a grander whole.
Indeed, the category of presheaves on the opposite of (finitely generated) commutative rings is a model for the axioms, and of course this is the context in which algebraic geometry takes place.
But we are entitled to take probe categories considerably richer than just that of duals of commutative rings. In particular, we may consider a category of commutative rings that have a notion of being "smooth" the way a ring of smooth functions is "smooth". These are the C-infinity rings or [generalized smooth algebra](http://ncatlab.org/nlab/show/generalized+smooth+algebra)s. Every ring of smooth functions on a smooth manifold is an example, but there are more.
The formal dual of these rings are spaces called [smooth loci](http://ncatlab.org/nlab/show/smooth+loci). This is a smooth analog of the notion of affine scheme. (Notice that the notion of "smooth" as used here is that of differential geometry, not quite that of algebraic geometry, which is more like "singularity free". But they are not unrelated).
The main theorem going in the direction of an answer to your question is that the category of manifolds embeds fully and faithfully into that of smooth loci. See at the link [smooth loci](http://ncatlab.org/nlab/show/smooth+loci) for the details.
But inside the category SmoothLoci, manifolds are characterized as the formal dual to their smooth rings of functions, so that's one way to answer your question.
There is a grand story developing from this point on, but for the moment this much is maybe sufficient as a reply.
| 25 | https://mathoverflow.net/users/381 | 5368 | 3,613 |
https://mathoverflow.net/questions/5364 | 39 | I have several naive and possibly stupid questions about deformations of categories. I hope that someone can at least point me to some appropriate references.
What is a deformation of a (linear, dg, A-infinity) category? Is it a "bundle of categories" over a scheme? How can you make such a notion rigorous? Maybe via stacks? Suppose we take some nice scheme $X$ and we consider $D^b\text{Coh}(X)$; if we deform $X$, then do we also get a corresponding deformation of the derived category? What does this corresponding deformation "look like"? Are we deforming the morphisms? The objects? Both?
Similarly, what about in the situation where we have a category of modules over an algebra $A$? If we deform the algebra, then do we also get a corresponding deformation of the category? Again, what does it "look like"?
And finally, in either of the above cases, are there deformations of the respective categories that *don't* correspond to deformations of $X$ or of $A$ respectively? I expect the answer to be "yes"; then my next question is: Are there any nice examples of such deformations that can still be described in an explicit or geometric way?
I am most of all interested in concrete examples, and less interested in general theory.
---
Edit 1: I probably should have mentioned this when I first posted this (almost 3 months ago now!), but somehow I forgot. Kontsevich has been at least implicitly talking about deformations of categories since at least 1994, in the original paper introducing homological mirror symmetry. The idea (or "philosophy") seems to be that the deformation theory of a category should have something to do with its Hochschild (co)homology. But I still do not understand this connection, at least in any sort of generality. Perhaps this is explained in some of the papers already listed in the answers below --- what I'd most like to see is how to relate "deformation of a (linear/dg/A-infinity) category", however one defines it, to Hochschild (co)homology.
Perhaps it's somehow obvious... but I'm pretty dense and would like to see it spelled out...
So I hope someone can explain this to me, or point me to a spot in a paper where it is explained.
I'm adding a bounty to this question just for the heck of it.
Edit 2: See my answer below.
| https://mathoverflow.net/users/83 | What is a deformation of a category? | Certainly not my field, but you might want to check the paper by Lowen and Van den Bergh [Deformation theory of abelian categories](https://arxiv.org/abs/math/0405226). I think that's where the first notion of deformation of a category appeared.
| 17 | https://mathoverflow.net/users/914 | 5371 | 3,616 |
https://mathoverflow.net/questions/2339 | 18 | Suppose that $N$ is prime, and consider the normalized cuspidal Hecke eigenforms of weight 2 and level $\Gamma\_0(N)$.
For such an eigenform $f$, the coefficients generate (an order in) the ring of integers of a totally real field $E$. The corresponding simple abelian variety quotient $A\_f$ of $J\_0(N)$ has dimension $[E:\mathbb{Q}]$. Let $C(N)$ denote the maximal degree $[E:\mathbb{Q}]$ amongst all such eigenforms. It is not too hard to prove that $C(N)$ is unbounded, and not much harder to show that the $\lim \inf$ of $C(N)$ as $N\to \infty$ is infinite. (I don't want to mention the argument here because I don't think it will apply to my question below.)
Is the same result true over $\mathbb{Q}\_2$? Namely, fix an embedding of $\bar{\mathbb{Q}}$ into $\bar{\mathbb{Q}}\_2$. Then, for any $d$, is it true that for all sufficiently large prime $N$ there exists a cuspidal eigenform of weight 2 and level $\Gamma\_0(N)$ whose coefficients generate some field $E/\mathbb{Q}\_2$ with $[E:\mathbb{Q}\_2] > d$?
This would be interesting to know even for $d = 1$.
| https://mathoverflow.net/users/nan | 2-adic Coefficients of Modular Hecke Eigenforms | A natural way to approach this question is to ask for an eigenform $f$ with coefficients in a local field $K$ with residue field degree $\ge 2$. Or, asking for slightly more, with coefficient ring $\mathcal{O}$ admitting a surjective map to a field $\mathbf{F}$ of order divisible by $4$ (this is slighly more since the rings $\mathcal{O}$ can typically be non-trivial orders in $\mathcal{O}\_K$). By Serre's conjecture, this is equivalent to asking for the existence of an irreducible Galois representation
$$\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}\_2(\mathbf{F})$$
with Serre weight and conductor $(2,p)$ (note, no oddness condition!). It's hard to construct such representations, but one natural way is to use induced ($=$ projectively dihedral) representations.
If $E$ is the corresponding quadratic extension, and $F/E$ the cyclic extension, then the Serre weight and level will be $(2,p)$ providing that $F/E$ is totally unramified and $E = \mathbf{Q}(\sqrt{-1})$, $\mathbf{Q}(\sqrt{p})$, or $\mathbf{Q}(\sqrt{-p})$.
(EDIT: In a previous version of this post, I only thought about extensions that were totally unramified at $2$ for some unknown reason.)
One is in good shape providing that the odd part of the class group of one of the fields $\mathbf{Q}(\sqrt{\pm p})$ is not $(\mathbf{Z}/3 \mathbf{Z})^n$. Since one can't really say much about real quadratic fields, let's think about the imaginary quadratic fields.
It's a theorem, proved by (amongst other people) Lillian Pierce, and Jordan Ellenberg and Akshay Venketesh, that
the 3-torsion part of the class group is a negligible part of the class group. Thus we are done if $p$ is big *providing* that the $2$-part of the class group is not so big. The $2$-part is trivial if $p$ is $3$ mod $4$, and has order $2$ if $p$ is $5$ mod $8$. However, it is quite possible that the class group of $\mathbf{Q}(\sqrt{-p})$ is $\mathbf{Z}/2^n \mathbf{Z}$, and very rough heuristics suggest that this might happen infinitely often.
[EDIT: LaTeX issues now mostly sorted in the sequel]
If $\mathbf{T}$ denotes the Hecke algebra tensored with $\mathbf{Z}\_2$, then
$\mathbf{T}$ is a free $\mathbf{Z}\_2$-module of rank $n = \mathrm{dim}(S\_2(\Gamma\_0(p)))$.
It is also a semi-local ring. For each maximal ideal $m$ of $\mathbf{T}$, we are asking that
$$\mathbf{T}\_{m} \otimes \mathbf{Q}$$
is not a number of copies of the 2-adic numbers. If $m$ is the $2$-adic Eisenstein ideal, then in a paper with Matthew Emerton, I prove that
$$\mathbf{T}\_{m}/2$$
has rank $2^{e-1}-1$, where the $2$-part of the class group of
$\mathbf{Q}(\sqrt{-p})$ (we may assume that $p$ is $1$ modulo $8$) has order $2^{e}$.
If $p$ is $9$ modulo $16$ then $\mathbf{T}$ localized at $m$ is also a domain, and so we are done in this case if $2^{e-1} -1 \ge d$. Even if $p$ is $1$ modulo $16$, this shows that the bulk of $\mathbf{T}/2$ must come from non-Eisenstein primes $m$, since
$2^e \le h \ll p^{1/2 + \epsilon} \ll p/12 \simeq n$.
So, in the case $d = 1$, one may assume that the odd part of the class group is
$3$-torsion, and then hope that for the corresponding $m$, one can prove that
the corresponding rings $\mathbf{T}\_{m}/2$ can't be too large.
It turns out that the relevant question becomes: Given a $\overline{\rho}$ corresponding to an $S\_3$ extension, and given a finite flat deformation of
$\overline{\rho}$ to $\mathrm{SL}\_2(A)$ for a local ring $A$ killed by $2$, can one bound the rank of $A$ as an $\mathbf{F}\_2$-module? For example, can one prove that $A$ has rank $p^{1 - \epsilon}$? The example of Eisenstein $m$ suggests that one can not neccesarily do better than $p^{1/2 + \epsilon}$.
---
Here is an answer that shows works for d = 1, and works for general d assuming GRH.
Step I. Let $C$ be the class group of $\mathbf{Q}(\sqrt{-p})$. The group $C$ decomposes as
$C\_{odd} \oplus C\_{even}$.
Step II. Suppose that $C\_{odd}$ is not annihilated by $2^{2d} - 1$. Then there exists a dihedral representation induced from a character of $C\_{odd}$ which is finite flat at $2$, ordinary at $p$, and has which has image in
$$\mathrm{SL}\_2(\mathbf{F})$$
for a field $\mathbf{F}$ of degree $>d$.
Thus we are done unless $C\_{odd} = C\_{odd}[2^{2d} - 1]$.
Step III. If $d = 1$, then then the $2^{2d} - 1 = 3$-torsion of $C\_{odd}$ has small order, namely, of order at most
$$p^{1/2 - \delta}$$
for some explicit $\delta > 0$. For general $d$, the $2^{2d} -1$-torsion has order at most $p^{\epsilon}$, assuming GRH.
Step IV: If $p$ is $-1$ modulo $4$ or $5$ modulo $8$, then $C\_{even}$ has order $\le 2$.
This implies by Steps II and III that $C$ has order $\le p^{1/2 - \delta}$, which contradicts the estimate $h \gg p^{1/2 - \epsilon}$.
Step V: We may assume that $p$ is $1$ modulo $8$. Let $|C\_{even}| = 2m$. Using the estimate $h \gg p^{1/2 - \epsilon}$, we deduce that
$m \gg p^{\delta - \epsilon}$ for some explicit $\delta > 0$.
Step VI: Let $\mathbf{T}$ denote the localization of the Hecke algebra at the Eisenstein prime of residual characteristic $2$. It is a consequence of one of the main theorems of FC-ME that one has $\mathbf{T} = \mathbf{Z}\_2[x]/f(x)$, where:
$$f(x) \equiv x^{m - 1} \ \mathrm{mod} \ 2,$$
$$\mathbf{Z}\_2/f(0) \mathbf{Z}\_2 \simeq \mathbf{Z}\_2/n \mathbf{Z}\_2,$$
and $n$ is the numerator of $(p-1)/12$. (Compare the discussion in Mazur's Eisenstein ideal paper, section 19, page 140.)
Step VII: If the coefficients of all forms of level $\Gamma\_0(p)$ define extensions of degree $\le d$, then the normalized $2$-adic valuation of every root of $f(x)$ is at *least* $1/d$. We deduce that
$$\frac{m - 1}{d} \le v\_2(n),$$
Since $v\_2(n) \le \log\_2(n) < \log\_2(p)$, we deduce that
$$ m \le d \log\_2(p).$$
This contradicts the previous estimate $m \gg p^{\delta - \epsilon}$ for sufficiently large $p$.
Remark: For $d = 1$, it should be easy to use the unconditional results to give an explicit lower bound on possible $p$.
---
To summarize, for any $d$, and sufficiently large $p$, there exists either:
(i) A modular form $f$ of level $\Gamma\_0(p)$ with coefficients in an extension of $\mathbf{Q}\_2$ which contains an unramified extension of degree $> d$, and whose residual representation is dihedral,
(ii) A modular form $f$ of level $\Gamma\_0(p)$ which coefficients in an extension of $\mathbf{Q}\_2$ which contains a ramified extension of degree $> d$, and whose residual representation is Eisenstein.
---
Extra: For *any* $p$, the coefficients of an $f$ of level sufficiently divisible by $p$ contains the totally real subfield of the $p^n$th roots of unity.
For any $p$, and sufficiently large $n$, this contains a large extension of the $2$-adic numbers.
Reason why this argument still sucks: Doesn't work at all for primes $> 2$, and uses GRH for $d > 1$.
| 12 | https://mathoverflow.net/users/nan | 5407 | 3,646 |
https://mathoverflow.net/questions/4665 | 36 | Fix an integer n. Can you find two finite CW-complexes X and Y which
\* are both n connected,
\* are not homotopy equivalent, yet
\* $\pi\_q X \approx \pi\_q Y$ for all $q$.
In [Are there two non-homotopy equivalent spaces with equal homotopy groups?](https://mathoverflow.net/questions/3540/) some solutions are given with n=0 or 1. Along the same lines, you can get an example with n=3, as follows. If $F\to E\to B$ is a fiber bundle of connected spaces such that the inclusion $F\to E$ is null homotopic, then there is a weak equivalence $\Omega B\approx F\times \Omega E$. Thus two such fibrations with the same $F$ and $E$ have base spaces with isomorphic homotopy groups.
Let $E=S^{4m-1}\times S^{4n-1}$. Think of the spheres as unit spheres in the quaternionic vector spaces $\mathbb{H}^m, \mathbb{H}^n$, so that the group of unit quarternions $S^3\subset \mathbb{H}$ acts freely on both. Quotienting out by the action on one factor or another, we get fibrations
$$ S^3 \to E \to \mathbb{HP}^{m-1} \times S^{4n-1},\qquad S^3\to E\to S^{4m-1}\times \mathbb{HP}^{n-1}.$$
The inclusions of the fibers are null homotopic if $m,n>1$, so the base spaces have the same homotopy groups and are 3-connected, but aren't homotopy equivalent if $m\neq n$.
There aren't any n-connected lie groups (or even finite loop spaces) for $n\geq 3$, so you can't push this trick any further.
Is there any way to approach this problem? Or reduce it to some well-known hard problem?
(Note: the finiteness condition is crucial; without it, you can easily build examples using fibrations of Eilenberg-MacLane spaces, for instance.)
| https://mathoverflow.net/users/437 | Are there pairs of highly connected finite CW-complexes with the same homotopy groups? | Here is a method for constructing examples. If a fiber bundle $F \to E \to B$ has a section, the associated long exact sequence of homotopy groups splits, so the homotopy groups of $E$ are the same as for the product $F \times B$, at least when these spaces are simply-connected so the homotopy groups are all abelian. To apply this idea we need an example using highly-connected finite complexes $F$ and $B$ where $E$ is not homotopy equivalent to $F \times B$ and the bundle has a section.
The simplest thing to try would be to take $F$ and $B$ to be high-dimensional spheres. A sphere bundle with a section can be constructed by taking the fiberwise one-point compactification $E^\bullet$ of a vector bundle $E$. (This is equivalent to taking the unit sphere bundle in the direct sum of $E$ with a trivial line bundle.) The set of points added in the compactification then gives a section at infinity. For the case that the base is a sphere $S^k$, take a vector bundle $E\_f$ with clutching function $f : S^{k-1} \to SO(n)$ so the fibers are $n$-dimensional. Suppose that the compactification $E\_f^\bullet$ is homotopy equivalent to the product $S^n \times S^k$. If $n > k$, such a homotopy equivalence can be deformed to a homotopy equivalence between the pairs $(E\_f^\bullet,S^k)$ and $(S^n \times S^k,S^k)$ where $S^k$ is identified with the image of a section in both cases. The quotients with the images of the section collapsed to a point are then homotopy equivalent. Taking the section at infinity in $E\_f^\bullet$, this says that the Thom space $T(E\_f)$ is homotopy equivalent to the wedge $S^n \vee S^{n+k}$. It is a classical elementary fact that $T(E\_f)$ is the mapping cone of the image $Jf$ of $f$ under the $J$ homomorphism $\pi\_{k-1}SO(n) \to \pi\_{n+k-1}S^n$. The $J$ homomorphism is known to be nontrivial when $k = 4i$ and $n\gg k>0$. Choosing $f$ so that $Jf$ is nontrivial, it follows that the mapping cone of $Jf$ is not homotopy equivalent to the wedge $S^n \vee S^{n+k}$, using the fact that a complex obtained by attaching an $(n+k)$-cell to $S^n$ is homotopy equivalent to $S^n \vee S^{n+k}$ only when the attaching map is nullhomotopic (an exercise). Thus we have a contradiction to the assumption that $E\_f^\bullet$ was homotopy equivalent to the product $S^n \times S^k$. This gives the desired examples since $k$ can be arbitrarily large.
These examples use Bott periodicity and nontriviality of the $J$ homomorphism, so they are not as elementary as one might wish. (One can use complex vector bundles and then complex Bott periodicity suffices, which is easier than in the real case.) Perhaps there are simpler examples. It might be interesting to find examples where just homology suffices to distinguish the two spaces.
| 34 | https://mathoverflow.net/users/23571 | 5431 | 3,665 |
https://mathoverflow.net/questions/5340 | 11 | Given that $Q\_d$ is the hypercube graph of dimension $d$ then it is a known fact (not so trivial to prove though) that given a perfect matching $M$ of $Q\_d$ ($d\geq 2$) it is possible to find another perfect matching $N$ of $Q\_d$ such that $M \cup N$ is a Hamiltonian cycle in $Q\_d$.
The question now is - given a (non necessarily perfect) matching $M$ of $Q\_d$ ($d\geq 2$) is it possible to find a set of edges $N$ such that $M \cup N$ is a Hamiltonian cycle in $Q\_d$.
The statement is proven to be true for $d \in\{2,3,4\}$.
| https://mathoverflow.net/users/1737 | Is every matching of the hypercube graph extensible to a Hamiltonian cycle | This is a known open problem. See "[Matchings extend to Hamiltonian cycles in hypercubes](http://garden.irmacs.sfu.ca/?q=op/matchings_extends_to_hamilton_cycles_in_hypercubes)" over at the Open Problem Garden.
| 15 | https://mathoverflow.net/users/1079 | 5433 | 3,667 |
https://mathoverflow.net/questions/5427 | 9 | I am wondering if there are analytic tools to find asymptotic formulae for the coefficients of a complex power series of a function with branch singularities. For example, it is possible to show using elementary means that, for $q>1$, the coefficients of $\frac{1}{1-z} log(\frac{1}{1-qz})$ are asymptotic to $\frac{1}{1-q^{-1}} \frac{q^n}{n}$, but I would like to know if it is possible to see this from analytic properties of the function itself.
Motivation: There are nice asymptotic formulae for the coefficients of power series of meromorphic functions. As a simple example, the coefficients of an entire function must be $O(\epsilon^n)$ for any $\epsilon$. In general, the sum of the principal parts of the poles of smallest modulus will provide a very good first approximation, and contour integration can be applied to get more delicate bounds, see e.g. Wilf's Generatingfunctionology.
| https://mathoverflow.net/users/1770 | Asymptotics of Power Series With Branch Singularities | The answer is "Quite often yes, but the error terms are seldom as good as in the meromorphic case". The reason the asymptotics for the meromorphic functions works is that we know the exact coefficients for $c\_k(z-z\_0)^{-k}$ and can always remove the singularity by subtracting a few terms of this kind thus increasing the radius of convergence and boosting the decay of the coefficients, so that the error is exponentially small compared to the main terms.
When you have some branching singularity $S(z)$ (like $\log(z-z\_ 0)$ or $\sqrt[]{z-z\_ 0}$) for which you know the coefficients when it stands alone, you can still play the trick to get the asymptotics for a function $F(z)S(z)$ where $F$ is alnalytic in some larger disk by writing the sum as $F(z\_ 0)S(z)+(F(z)-F(z\_ 0))S(z)$ and noting that the second term is smoother than the first one on the boundary circle, so its Fourier coefficients decay a bit faster. If you take more terms in the Taylor series for $F$ at $z\_ 0$, you can get any degree of smoothness in the remainder you wish, so the error terms can be made of size $n^{-k}$ times the leading terms with any given $k$, but that's about as far as you can go unless you want integral expressions instead of algebraic quantities.
In the latter case, you should make a radial cut starting at singularity and ending at some bigger circle and take the contour integral over that larger circle and 2 sides of the cut in the Cauchy formula. The integral over the larger circle will decay exponentially faster than the typical coefficient, so all asymptotics will come from the cut. The point is that the integral over the cut is not that of something oscillating but that of a fast decaying function, so you can use real variable methods to find its asymptotics in nice algebraic terms (or you can just leave it as is).
| 8 | https://mathoverflow.net/users/1131 | 5434 | 3,668 |
https://mathoverflow.net/questions/5436 | 3 | Let X be a (finite dimensional) manifold. Consider smooth mapping space $$PX = C^\infty(I, X)$$ where I = [0,1] is the closed interval. Is this space paracompact? What if we fix a point x in X and consider the *pointed* path space, is this space paracompact?
| https://mathoverflow.net/users/184 | Are mapping spaces paracompact? | What topology do you want on $C^\infty(I,X)$? One natural topology, the one of uniform convergence of all derivatives, is metrizable, so paracompact.
| 5 | https://mathoverflow.net/users/1409 | 5438 | 3,671 |
https://mathoverflow.net/questions/5457 | 10 | I've been reading through Lurie's book on higher topos theory, where he develops the theory of $(\infty,1)$-toposes, which leads me to the following question: Is there any sort of higher topos theory on the more general $\omega$-categories, where we don't require all higher morphisms to be invertible?
| https://mathoverflow.net/users/1353 | $\omega$-topos theory? | The short answer is no. Even 2-toposes are poorly understood -- we don't know what the right definition is. For higher dimensions, including $\infty$, we *definitely* don't have the answers.
Just as the primordial example of a (1-)topos is $\mathbf{Set}$ (the 1-category of sets and functions), the primordial example of a 2-topos should be $\mathbf{Cat}$ (the 2-category of categories, functors and natural transformations).
Mark Weber has done some work on 2-toposes, building on earlier ideas of Ross Street. But I think Mark is quite open about the tentative nature of this so far.
There was a [good discussion](http://golem.ph.utexas.edu/category/2008/01/2toposes.html) of the current state of 2-toposes (and more generally n-toposes) at the $n$-Category Café last year:
| 9 | https://mathoverflow.net/users/586 | 5461 | 3,688 |
https://mathoverflow.net/questions/5378 | 47 | This question is being asked on behalf of a colleague of mine.
Let $X$ be a topological space. It is well known that the abelian category of sheaves on $X$ has enough injectives: that is, every sheaf can be monomorphically mapped to an injective sheaf. The proof is similarly well known: one uses the concept of "generators" of an abelian category.
It is also a standard remark in texts on the subject that on a general topological space $X$, the category of sheaves need *not* have enough projectives: there may exist sheaves which cannot be epimorphically mapped to by a projective sheaf. (Dangerous bend: this means projective in the categorical sense, not a locally free sheaf of modules.) For instance, Wikipedia remarks that projective space with Zariski topology does not have enough projectives, but that on any spectral space, a space homeomorphic to $\operatorname{Spec}R$, there are enough projective sheaves.
Two questions:
1. Who knows an actual proof that there are not enough projectives on, say, $\boldsymbol{P}^1$ over the complex numbers with the Zariski topology? What about the analytic topology, i.e. $\boldsymbol{S}^2$?
2. Is there a known necessary and sufficient condition on a topological space $X$ for there to be enough projectives?
---
EDIT: I meant the question to be purely for sheaves of abelian groups. Eric Wofsey points out that the results alluded to above on Wikipedia are not consistent when interpreted in this way, since $\boldsymbol{A}^1$ and $\boldsymbol{P}^1$, over an algebraically closed field $k$, with the Zariski topology are homeomorphic spaces: both have the cofinite topology.
I am pretty sure that when my colleague asked the question, he meant it in the topological category, so I won't try to change that. But the other case is interesting too. What if $(X,\mathcal{O}\_X)$ is a locally ringed space. Does the abelian category of sheaves of $\mathcal{O}\_X$-modules have enough projectives? I think my earlier warning still applies. Since for a scheme $X$ not every $\mathcal{O}\_X$-module is coherent, it is not clear that "projective sheaves" means "locally free sheaves" here, even if we made finiteness and Noetherianity assumptions to get locally free and projective to coincide.
| https://mathoverflow.net/users/1149 | When are there enough projective sheaves on a space X? | About [Jon Woolf's answer](https://mathoverflow.net/a/5401/64073), it seems to me that the condition that "$x$ is a closed point" was implicitly used: the extension by zero $Z\_A$ is only defined for a locally closed subset $A$ (see e.g. Tennison "*Sheaf theory,*" 3.8.6). So $X-x$ must be locally closed. How about the following trivial modification: instead of $Z\_X$, consider the sheaf $i\_\ast Z$, where $i$ is the inclusion of a point $x$ into $X$.
Suppose that $x$ does not have the smallest open neighborhood and $x$ has a basis of connected neighborhoods. Then $i\_\ast Z$ is not a quotient of a projective sheaf $P$. Suppose otherwise. Then for any connected open neighborhood $U$ of $x$, the homomorphism $P(U) \to i\_\ast Z(U)$ is zero. This implies that the homomorphism $P \to i\_\ast Z$ is zero since it is equivalent to a homomorphism from the stalk $P\_x$ to $Z$. Indeed, pick a neighborhood $V$ which is smaller than $U$. We have a surjection $Z\_V \to i\_\ast Z$. The homomorphism $P \to i\_\ast Z$ must factor through $Z\_V$, so $P(U) \to i\_\ast Z(U)$ must factor through $Z\_V(U)$. But $Z\_V(U)=0$. This equality may fail if $U$ is not connected however.
So to summarize Jon Woolf and David Treumann, the category of sheaves of abelian groups on a locally connected topological space $X$ has enough projectives iff $X$ is an [Alexandrov space](https://en.wikipedia.org/wiki/Alexandrov_topology).
---
Surely this must appear in some standard text. Anybody knows a reference? And what about non-locally connected spaces?
For ringed spaces $(X,\mathcal{O}\_X)$ one direction is still clear: $X$ being an Alexandrov space implies you'll have enough projectives. But on reflection the other direction, $X$ being a locally connected space without minimal open neighborhoods implies you don't have enough projectives, appears to be rather tricky. One can think of some weird structure sheaves for which the above argument does not go through, in particular $\mathcal{O}\_V(U)\ne 0$. So I still wonder what the answer is for ringed spaces.
| 20 | https://mathoverflow.net/users/1784 | 5470 | 3,695 |
https://mathoverflow.net/questions/3344 | 6 | The other day I came across the statement that in the triangulated category $\mathfrak{KK}$ (of C\*-algebras with KK-groups as morphism sets) "there are many other sources of exact triangles besides extensions". Except for mapping cone triangles I don't know what is meant. What can you come up with?
I would also appreciate answers focussing on triangulated categories in general.
| https://mathoverflow.net/users/1291 | Sources for exact triangles in triangulated categories. | This question has kind of been bothering me since I started thinking about it - I am far from an expert on KK-theory but I thought I'd throw something out there and maybe someone else will see it and come along and agree with me or correct me.
I think this statement is a way of thinking rather than something precise. Indeed by definition every distinguished triangle in KK is an isomorph of an extension triangle (although in the equivariant case I believe life is not so simple). Alternatively one can define the triangulation by taking distinguished triangles to be those candidates triangle (i.e. $X\to Y\to Z \to \Sigma X$ where each pair of composites vanishes) which are isomorphic to mapping cone triangles. So this is really all one has. The same story is true in the derived category of an abelian category for instance.
But (here is the punchline) one does not generally build the triangles one needs to prove things by considering short exact sequences of chain complexes! Indeed, one of the virtues of triangulated categories is that one has the ability to produce lots of new objects and triangles starting with very little. Often it is hard/impossible to do this in any explicit way - in fact one generally just knows that some collection of triangles doing the job exists and has no idea what they look like. So even though every triangle one might construct is (up to KK equivalence) an extension there is a very good chance that one didn't obtain it by writing down an explicit extension.
I guess there is also the fact that a triangle which is just isomorphic in KK to an extension triangle is not itself literally an extension of $C^\*$-algebras. I don't know the stuff well enough to know whether or not one can produce interesting triangles via other constructions where there is a guarantee that some extension exists to make it distinguished. This is entirely possible (and in my opinion viewing such a construction as a different source of triangles is a worthwhile psychological distinction).
| 4 | https://mathoverflow.net/users/310 | 5472 | 3,697 |
https://mathoverflow.net/questions/5485 | 86 | Although we are not so numerous as other respected professionals, like for example lawyers, I wonder if we could come up with a reasonable estimate of our population.
Needless to say, the question more or less amounts to the definition of"mathematician".
Since I should like to count only research mathematicians (and not, say, high-school teachers) some criterion of publishing should be applied. But it should not be too strict in order not to exclude Grothendieck, for example, who has not published any mathematics for a long time.
An excuse for asking a question so soft as to verge on the flabby is that it might be considered an exercise in Fermi-type order of magnitude estimation.
| https://mathoverflow.net/users/450 | How many mathematicians are there? | Current count of [Mathematics Genealogy Project](http://genealogy.math.ndsu.nodak.edu/) is 137672 (I am assuming that the PhD students that graduated are ranked as "research mathematicians"). But the problem is.. Mathematics Genealogy is mostly for universities of developed countries. There could be some really good university in Russia, China or Korea out there that doesn't give us the correct statistics. Another problem is.. Mathematics Genealogy Project counts even the dead mathematicians (like Hilbert, Hasse, Kepler and so on).. and I am assuming you want a report of living mathematicians.. but hey, I'm quite surprised by the number even 200k is pretty low for the living!
| 25 | https://mathoverflow.net/users/1245 | 5487 | 3,709 |
https://mathoverflow.net/questions/5518 | 26 | The standard approach to proving that $H^n(X; G)$ is represented by $K(G, n)$ seems to be to prove that $\text{Hom}(X, K(G, n))$ defines a cohomology theory and then use Eilenberg-Steenrod uniqueness. This is utterly spiffing, but as far as I can see gives little geometric intuition. In his treatment, Hatcher mentions that there is a more direct cell-by-cell proof, albeit a somewhat messy and tedious one. I haven't been able to find any such proof, but I'd really like to see one; I think it would help me solidify my mental picture of Eilenberg-MacLane spaces. Does anyone have a reference?
| https://mathoverflow.net/users/1202 | "Dirty" proof that Eilenberg-MacLane spaces represent cohomology? | I'd suggest looking up some basic material on obstruction theory. There, you generally find classification of maps $X \to Y$ with domain a CW-complex in terms of cohomology groups $H^s(X;\pi\_t(Y))$. The proofs are often very cellular indeed.
In the case where the range is an Eilenberg-Maclane space (for an abelian group), the dirty proof is something like:
* Any map from $X$ is homotopic to one where the (n-1)-skeleton $X^{(n-1)}$ maps to the basepoint of $Y$.
* A map on the n-skeleton $X^{(n)}$ sending the (n-1)-skeleton to the basepoint is determined, up to homotopy, by a choice of element of $G$ for each n-cell of $X$, essentially by definition of homotopy. This is an element in the n'th CW-chain group $C^n\_{CW}(X;G)$.
* Such a map extends to all higher skeleta if and only if the attaching maps for all the (n+1)-cells become nullhomotopic in $Y$. Thus the map extends if and only if it's represented by a cocycle, i.e. an element of $Z^n\_{CW}(X;G)$.
* This is a complete invariant, up to homotopy, of maps that are trivial on the (n-1)-skeleton. (Higher cells have basically unique maps up to homotopy.)
* Any homotopy between two such maps can be pushed to a homotopy that's trivial on the (n-2)-skeleton of $X$.
* Such a homotopy is determined, up to a "track" (a homotopy between homotopies), by a choice of element of $G$ for each (n-1)-cell of $X$.
* Such a homotopy alters the map on the n-skeleton (as an element of $C^n\_{CW}(X;G)$) by adding a coboundary element, something in $B^n\_{CW}(X;G)$.
* Therefore, the full mapping space mod homotopy is $H^n\_{CW}(X;G)$.
This is a little messy. Often it's nice to use the filtration of $X$ by subcomplexes $X^{(n)}$ and use that each inclusion in the filtration induces a fibration of mapping spaces
$$F(X^{(n)}/X^{(n-1)},Y) \to F(X^{(n)},Y) \to F(X^{(n-1)},Y)$$
to clean this homotopical analysis up a little into something slightly more systematic. This leads to a spectral sequence for the homotopy groups of the mapping spaces in terms of the cohomology of $X$ with coefficients in the homotopy groups of $Y$, but you have to be a little careful because there is a "fringe" that exhibits some non-abelian-group-like behavior.
| 31 | https://mathoverflow.net/users/360 | 5521 | 3,739 |
https://mathoverflow.net/questions/5532 | 17 | It seems as though many consider ZF to be the foundational set of axioms for all of mathematics (or at least, a crucial part of the foundations); when a theorem is found to be independent of ZF, it's generally accepted that there will never be a proof of the theorem one way or another. My question is, why is this? It seems as though ZF is flawed in a number of ways, since propositions like the axiom of choice and the continuum hypothesis are independent of it. Shouldn't our axioms of set theory be able to give us firm answers to questions like these? Yes, the incompleteness theorems say that we'll never develop a perfect set of axioms, and many of the theorems independent of our axioms will probably be quite interesting, but is ZF really the best we can do? Is there hard evidence that ZF is the "best" set theory we can come up with, or is it merely a philosophical argument that ZF is what set theory "should" look like?
| https://mathoverflow.net/users/1455 | The Importance of ZF | Very few mathematicions these days wish to base their mathematics on ZF without the axiom of choice, as your question seems to imply. Yes, there is an intuitionist school about which I don't know a whole lot, but they seem to be quite the minority. So let's consider ZFC. I think the main philosophical argument for it is that the axioms seem obviuosly true, if you are willing to believe that such things as infinite sets exist in some fashion, and that nobody has been able to come up with an equally obviously “true” statement about sets that is not a consequence of these axioms. The independence of the continuum hypothesis doesn't seem to bother people much, since to most of us it doesn't seem either obviously true or obviously false. Though some people are working on settling it by finding other axioms that will at least be considered likely true or at least useful. There was a recent article in the *Notices* about these efforts, in fact. But is there “hard evidence” that ZFC is the best we can come up with? Not by most mathematicians' standards I think, but there seems to be plenty of soft evidence.
I might add that many working mathematicians (I am talking here mostly about analysts, since they are the people I know best) don't care one whit about these questions, but happily go about their business using Zorn's lemma whenever it seems necessary and never let it bother them. And there are some who would rather base all mathematics on category theory rather than ZFC set theory, but that is not my cup of tea, so I will leave it unstirred.
| 11 | https://mathoverflow.net/users/802 | 5540 | 3,750 |
https://mathoverflow.net/questions/5522 | 10 | I recall the following question from Ulam's book "Unsolved math problems": show that the ring of Dirichlet series with integer coefficients is a factorial ring. I believe that soon after Ulam wrote his book, this problem was solved; essentially, this ring is isomorphic to the ring of formal power series in infinitely many variables (one for each prime) with integer coefficients, and once it is reformulated this way it looks much less mysterious. However, my actual question is not about the proof, - I was always curious if a result like that (in its original formulation) can be really applied or interpreted from the Dirichlet series point of view. Has anyone heard/thought of any reasonable number-theoretic interpretation of this fact, or does the interpretation as a power series ring kill any hope of that sort?
| https://mathoverflow.net/users/1306 | Dirichlet series with integer coefficients as a UFD | First, I want to nail down a reference to the solution to the problem alluded to above: the Dirichlet ring of functions f: Z+ -> Z with pointwise addition and convolution product is a UFD. This was proved by L. Durst in his 1961 master's thesis at Rice University and independently by Cashwell and Everett. References:
Deckard, Don; Durst, L. K.
Unique factorization in power series rings and semigroups.
Pacific J. Math. 16 1966 239--242.
Cashwell, E. D.; Everett, C. J.
Formal power series.
Pacific J. Math. 13 1963 45--64.
Note that the latter is the second paper of Cashwell and Everett on the subject of factorization in Dirichlet rings. They also had a 1959 paper:
Cashwell, E. D.; Everett, C. J.
The ring of number-theoretic functions.
Pacific J. Math. 9 1959 975--985.
Ulam makes reference to the 1959 paper in his statement of the problem in the first (1960) edition of his book. In the 1964 paperback edition of his book, he announces the problem as solved by 1961 papers of Buchsbaum and Samuel.
I want to make the remark that since the original 1959 paper of Cashwell and Everett certainly makes the connection to formal power series rings, there is more to the solution of the problem than this. (Indeed, the 1961 result of Samuel, that if R is a regular UFD, then R[[t]] is again a UFD, seems to be the key.)
There is a 2001 arxiv preprint of Durst which claims a more elementary proof of the result:
<http://arxiv.org/PS_cache/math/pdf/0105/0105219v1.pdf>
As to whether the result has number-theoretic significance: not as far as I know. I should say that I first learned of the Cashwell-Everett theorem by reading an analytic number theory text -- Tenenbaum, *Introduction to Analytic and Probabilistic Number Theory*, p. 26 -- but the author seems to mention it just for culture.
Finally, remember that this is a result about factorization in the ring of **formal** Dirichlet series, an inherently algebraic beast. So perhaps the following result is more relevant:
Bayart, Frédéric; Mouze, Augustin Factorialité de l'anneau des séries de Dirichlet analytiques. (French) [The ring of analytic Dirichlet series is factorial] C. R. Math. Acad. Sci. Paris 336 (2003), no. 3, 213--218
| 4 | https://mathoverflow.net/users/1149 | 5549 | 3,756 |
https://mathoverflow.net/questions/5547 | 18 | I work in planar algebras and subfactors, where the idea of path algebras on a graph (alternately known as graph algebras, graph planar algebras, etc.) is quite useful. The particular result I'm thinking of is a forthcoming result of Jones and some others; it says that any subfactor planar algebra can be found inside the planar algebra of its principal graph. If you're not into subfactors/planar algebras, the importance of this result is that it says you know a concrete place to begin looking for a particular abstract object.
At Birge Huisgen-Zimmermann's talk on quivers at the AMS meeting at Riverside last weekend, I encountered what seemed to be a similar result: Gabriel's theorem, which says that any finite-dimensional algebra is equivalent (Morita equivalent I think?) to a path algebra modulo some relations. (As far as I can tell, "quiver" is a fancy word for a directed finite graph). I also know, though I don't know why, that path algebras are used in particular constructions in C\*-algebras.
This got me thinking:
1. What are some other places that path algebras appear, and what are they used for?
2. Why is this idea so useful in these different fields? Is it simply that path algebras are a convenient place to do calculations? Or is there some philosophical reason path algebras are important?
| https://mathoverflow.net/users/699 | ubiquity, importance of path algebras | *As far as I can tell, "quiver" is a fancy word for a directed finite graph*
Yes. It doesn't even have to be finite.
*Or is there some philosophical reason path algebras are important?*
A huge application of path algebras lately is the path algebra of a quiver of Dynkin type. Following the ideas of Lusztig and Ringel, the representation varieties of these quivers are a main method to categorify quantum groups. A big share of the interest in quivers is either this specific purpose, or generalizations of features of this application. Lusztig's papers on this and his book are a big revelation.
| 10 | https://mathoverflow.net/users/1450 | 5556 | 3,762 |
https://mathoverflow.net/questions/5558 | 7 | What is the impact of Mathematics in social science today?. That is to say, what are the mathematics that a social scientist is using and, from the point of view of a mathematician, what are the mathematics they should use more or start out using.
Let me give a particular example to start with: Social Inequality. Every day I can see governments trying to eradicate the social inequality in terms of wealth. Assuming wealth can be measure, Is there a theoretical result saying that it is possible (or impossible) to work out such a thing exactly (mathematically speaking)?
| https://mathoverflow.net/users/1547 | Math Vs Social Science | Right now, the idea of actually applying the results of Social Choice theory (aka voting theory) is gaining steam. There are plenty of mathematical results that show that plurality voting method (what US has) is one of the least effective ways to measure the will of the electorate, and something like Borda count or Approval voting would result in all kinds of benefits for promoting meaningful democracy.
| 17 | https://mathoverflow.net/users/619 | 5575 | 3,778 |
https://mathoverflow.net/questions/5568 | 3 | I've heard about this construction on the lecture about **higher representation theory**:
>
> Given a Lie algebra $g$, one constructs $\mathcal A$, a category whose $K\_0$ is the universal enveloping algebra of $g$. Conjecture: any $\mathcal A$-acted triangulated category $\mathcal V$ (with its $K$ locally finite) decomposes to $\oplus \mathcal V\_\lambda$ with braid action; and there is bijection between $g$-representations and minimal such categories.
>
>
>
Is there a good — if possible, non-$sl\_2$ — example of such a category $\mathcal A$, minimal categories $V\_\lambda$ and braid action which explains why one would have such a construction?
**Update:** Found the [notes of the talk](http://www.math.utexas.edu/users/benzvi/GRASP/lectures/IAS/rouquierhigher.pdf) that has two $sl\_n$ examples, one from quivers, another from sheaves on the grassmannian, $\mathcal V :=\oplus^n\_i D^b\mathop{\rm constr}/\mathop{\rm Gr}(i,n)$.
A more accessible text for either example would be welcome! Because if the best way to understand these is to "get" quantum groups, that's quite a big topic. My idea was more like "maybe this is a good place to start".
| https://mathoverflow.net/users/65 | Categorifying the group representations | The $sl\_n$ version of this shouldn't so bad. I think it's just self-dual objects in parabolic category O and shuffling functors, though I'll admit, I haven't checked this myself, and doubt it's written properly somewhere. Probably the best reference is the papers of Brundan and Kleshchev (for example "Schur-Weyl duality for higher levels").
I suspect the inspiration for such a conjecture isn't really particular examples so much a philosophy about what sort of structures on a quantum group should be categorifiable.
By the way, I think your conjecture might be a bit too strong (at least as I interpret it). The 2-representations of a 2-Kac-Moody algebra aren't semi-simple (I've got a huge supply of non-semisimple examples categorifying tensor products).
| 4 | https://mathoverflow.net/users/66 | 5592 | 3,793 |
https://mathoverflow.net/questions/5305 | 14 | If you have two Möbius transformations represented as:
$f(z) = \frac{az + b}{cz + d}$
$g(z) = \frac{pz + q}{rz + s}$
where $a, b, c, d, p, q, r, s, z \in \mathbb{C}$
Is it possible to derive a third function $h(z, t)$, where $t \in \mathbb{R}$ and $0 \leq t \leq 1$, which "smoothly" interpolates between the transformations represented by $f(z)$ and $g(z)$?
Let me try to clarify the meaning of "smoothly". I'm working in a the Poincaré Disc model of Hyperbolic geometry. The functions $f$ & $g$ represent transformations which preserve congruence within the model, i.e. a Poincaré line segment $PQ$ will have the same Poincaré distance as $P'Q'$, where $P' = f(P)$ & $Q' = f(Q)$. I would like the interpolated transform to preserve this property as well.
**Clarification**:
The answers involving diagonalization of the matrix "almost" work. Let me be clearer about how these mobius transformations are used, so that I can give some more context.
$f$ represents the position and orientation of a tile on the Poincare disc. I use this to transform a tile centered at the origin. $g$ represents a neighboring tile. I'd like to "animate" the $f$ tile moving it smoothly on top of the $g$ tile. While animating, the center of $f$ should travel along the poincare line connecting the two tiles at rest position.
The problem with the diagonalization approach is that the centers of the two tiles will travel about a corner of one of the tiles, sometimes the "long" way around.
Of course it could just be a bug in my code...
P.S. This is for an iphone game I'm developing called [Circull](http://www.circull.com)
<http://www.youtube.com/watch?v=DiWijYb-xus>
| https://mathoverflow.net/users/1747 | How to smootly interpolate between möbius transformations? | I'd like to address two issues about how to implement these interpolations and what they mean. The group that you ask about is actually equivalent to the group of real M\"obius transformations. If you were in the upper half plane model, the coefficients would be real numbers. You actually don't need to explicitly use the upper half plane model, but you do need the fact that comes from it that all of the motions are elliptic, hyperbolic, or parabolic. The group is called $\mathrm{PSL}(2,\mathbb{R})$.
It is widely understood that a good way to make smooth motions in a Lie group is with geodesics. This is an important principle in ordinary 3D computer graphics, where the Lie group is instead $\mathrm{SO}(3)$. A standard method to find the geodesics in this comparison case is with quaternions. As it happens, this introduces a wrinkle with doubled angles that also appears in your case.
I assume that you are using $f$ and $g$ as follows: $z$ is a point in the standard bit map of your bird, and then $f(z)$ or $g(z)$ is the mapped position of the bird in your Poincare disk model of the hyperbolic plane. If this is what you are doing, then a correct derivation of the geodesic is $h\_t(f(z))$, where $h\_t(z)$ is an exponential path from the identity to $g(f^{-1}(z))$. I suppose that the swapped formula that other people have used, $f^{-1} \circ g$, is equivalent.
You can find the exponential path by diagonalizing the matrix of $h\_1 = g \circ f^{-1}$. However, there is the important wrinkle that the group of matrices is $\mathrm{SL}(2,\mathbb{R})$, which is twice as big as $\mathrm{PSL}(2,\mathbb{R})$. Let $M$ be the matrix of $h\_1$. If $M$ is hyperbolic, then it has real eigenvalues. In this case, you should first switch to $-M$ if the eigenvalues are negative. In the basis in which $M$ is diagonal, you should then specifically use
$$M\_t = \begin{bmatrix} \exp(ct) & 0 \\\\ 0 & \exp(-ct)\end{bmatrix}.$$
If you use some other choice, you will not follow the geodesic at a constant rate. If $M$ is elliptic, then you should use
$$M\_t = \begin{bmatrix} \exp(i\theta t) & 0 \\\\ 0 & \exp(-i\theta t)\end{bmatrix},$$
again in a (complex) basis in which $M$ is diagonal. But here, how do you choose between $M$ and $-M$? The total geometric rotation is $2\theta$, not $\theta$. You should negate $M$ if after you diagonalized, $\theta$ is more than $\pi/2$; if you don't do that you might rotate by more than 180 degrees. In the parabolic case (which you might not see because it requires a numerical coincidence),
$$M\_t = \begin{bmatrix} 1 & t \\\\ 0 & 1\end{bmatrix}.$$
Finally, in computer graphics you might want a smooth trajectory accelerate from 0 at the beginning and decelerate to 0 at the end. With the confidence that $t$ in the above formulas follows the geodesic at a uniform rate, you can do something like $t = \sin(s)^2$ and use $s$ as the time parameter instead.
| 11 | https://mathoverflow.net/users/1450 | 5608 | 3,807 |
https://mathoverflow.net/questions/5597 | 6 | ... are all isomorphic to $l^1$ on some other index set. At least, that much I "know" from 2nd-hand sources, since the original proof is apparently in a paper of Köthe from the 1930s 1960s (in German) that I can't get hold of have had trouble digesting. Since there are some Banach space specialists reading MO, I wondered if someone could sketch how the proof differs from the countable case, or point to a more recent text, preferably in English or French, that gives the proof? I hope this is a well-defined question for MO, since I'm not baiting with something where I know the answer.
(Some background for other readers: the analogous result when $X$ is countably infinite follows from combining two steps: one first uses a block basis argument to show that a closed, complemented subspace $V$ inside $l^1(\bf N)$ must either be finite-dimensional, or contain an infinite-dimensional, closed complemented subspace $W$ that is isomorphic to $\ell^1({\bf N})$. In the former case, $V$ is then obviously isomorphic to some finite-dimensional $\ell^1$. In the latter case, one applies Pelczynski decomposition. My impression is that it's the *first step* which might prove problematic if attempted for $\ell^1(X)$ when $X$ is uncountably infinite, but I could well be wrong and would welcome corrections.)
| https://mathoverflow.net/users/763 | Closed, complemented subspaces of $l^1(X)$ when $X$ is uncountable | A proof in English (modulo some details involving the Pelczynski decomposition method) can be found in the article 'On relatively disjoint families of measures' (Studia Math, 37, p.28-29) by Haskell Rosenthal.
Regarding the analogous result for $\ell\_p (X)$ ($p\in (1,\infty)$) and $c\_0 (X)$, I seem to recall reading somewhere that it was solved by Joram Lindenstrauss, but now I can't seem to find any reference to it. I seem to think that I saw something about it in the Appendix to the English translation of Banach's book on linear operations (the appendix is by Bessaga and Pelczynski), but it would take me a while to sift through it to find it, and family dinner is being dished up very shortly. I wonder anyhow how much can be gleaned from Matthew Daws' classification of the closed, two-sided ideals in $\mathcal{B}(\ell\_p (X))$, the Banach algebra of all bounded linear operators on $\ell\_p (X)$? The relevant paper can be downloaded at <http://www.amsta.leeds.ac.uk/~mdaws/pubs/ideals.pdf> . The paper 'The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space' by Laustsen, Schlumprecht and Zsak illustrates how classification of the complemented subspaces of a Banach space $E$ can follow from the classification of closed, two-sided ideals in $\mathcal{B}(E)$ if all the closed, two-sided ideals are generated by projections onto complemented subspaces having certain nice properties. How much of this can be done using Matt's results I haven't checked, but I think that at the very least some partial results could be obtained. Matt might comment of this if he passes by, or if no one else does I might try to look into it in the next day or so and edit this answer accordingly.
The analogous result for $\ell\_\infty (X)$ does not hold for uncountable $X$ in general. Indeed, every $\ell\_\infty (X)$ is the dual of $\ell\_1 (X)$, every Banach space embeds isomorphically into some $\ell\_\infty (X)$, but there are injective Banach spaces that are not isomorphic to any dual Banach space; the first such example seems to have been found by Haskell Rosenthal in his paper 'On injective Banach spaces and the spaces $L^\infty (\mu)$ for finite measure $\mu$' (Acta Mathematica, 124, Corollary 4.4), and the existence of such a space provides the desired counterexample.
| 7 | https://mathoverflow.net/users/848 | 5610 | 3,809 |
https://mathoverflow.net/questions/5553 | 16 | This is a follow-up to [a previous question](https://mathoverflow.net/questions/4912/ "Graphs with incidence matrices whose pseudoinverses are proportional to their transposes"). What graphs have incidence matrices of full rank?
Obvious members of the class: complete graphs.
Obvious counterexamples: Graph with more than two vertices but only one edge.
I'm tempted to guess that the answer is graphs that contain spanning trees as subgraphs. However, I haven't put much thought into this.
| https://mathoverflow.net/users/1674 | Which graphs have incidence matrices of full rank? | The first answer identifies "incidence matrix" with "adjacency matrix". The latter is the
vertices-by-vertices matrix that Sciriha writes about. But the original question
appears to concern the incidence matrix, which is vertices-by-edges. The precise
answer is as follows.
Theorem: The rows of the incidence matrix of a graph are linearly independent over the reals if and only if no connected component is bipartite.
Proof. Some steps are left for the reader :-)
Note first that the sum of rows indexed by the vertices
in one color class of a bipartite component is equal to the sum of the rows indexed by
the other color class. Hence if some component is bipartite, the rows of the incidence matrix are linearly dependent.
For the converse, we have to show that the incidence matrix of a connected non-bipartite
graph has full rank. Select a spanning tree $T$ of our graph $G$. Since $G$ is not bipartite, there is an edge $e$ of $G$ such that the subgraph $H$ formed by $T$ and $e$ is not bipartite.
The trick is to show that the columns of the incidence matrix indexed by the edges of H
form an invertible matrix. We see that $H$ is built by "planting trees on an odd cycle".
We complete the proof by induction on the number of edges not in the cycle.
The base case is when $H$ is an odd cycle. It is easy to show that its incidence matrix is invertible. Otherwise there is a vertex of valency one, $x$ say, such that $H \setminus x$ is connected and not bipartite. Then the incidence matrix of $H \setminus x$ is invertible and again it is easy to see this implies that the incidence matrix of $H$ is invertible.
Remark: I do not know who first wrote this result down. It is old, and is rediscovered
at regular intervals.
| 27 | https://mathoverflow.net/users/1266 | 5621 | 3,814 |
https://mathoverflow.net/questions/5618 | 5 | This question has been inspired by [covering 3-torus post](https://mathoverflow.net/questions/5546/ramified-covers-of-3-torus).
>
> Is it true that any good (smooth, compact, oriented) $n$-manifold can be mapped to $S^n$ in such a way that the map is true covering away from codimension 2?
>
>
>
| https://mathoverflow.net/users/65 | Ramified covers of S^n | Yes. See Feighn's short [note](http://www.raco.cat/index.php/CollectaneaMathematica/article/view/56965) "Branched covers according to J.W. Alexander".
| 11 | https://mathoverflow.net/users/1650 | 5626 | 3,819 |
https://mathoverflow.net/questions/5179 | 12 | I'm reading the classical Brown-Gersten's paper "Algebraic K-theory as generalized sheaf cohomology" and I'm stuck with their choose of global fibrations. Namely, a morphism of simplicial sheaves $p : E \longrightarrow B$ is a global fibration if for every inclusion of open sets $U\subset V$ the natural map $E(V) \longrightarrow B(V) \times\_{B(U)} E(U)$ is a (Kan) fibration of simplicial sets.
My problem is: why these fibrations? As far as I can see, when they make use of this definition in constructing the factorizations of the model category structure, they could have chosen the fibrations to be defined open-wise: $p : E \longrightarrow B$ is a fibration if $p(V) : E(V) \longrightarrow B(V)$ is a (Kan) fibration of simplicial sets for every open set $V$ and apply as well the small object argument they use at this point.
In other contexts I understand this kind of fibrations. For instance, for the model structure of the category of diagrams $C^I$ of a model category $C$ when $I$ is a 'very small' category (Dwyer-Spalinski, "Homotopy theories"), or a Reedy category. In these cases, this kind of fibrations ensures that you can extend your liftings by induction. But I don't see if this is their role with a category of sheaves, since no induction seems to be at hand.
A colleague of mine has said to me thas this choice of fibrations is the consequence of choosing the cofibrations to be the monomorphism, following Joyal's "Letter to Grothendieck"; that is, these are precisely the fibrations if you choose monomorphisms as cofibrations and ask fibrations to have the RLP with respect to trivial cofibrations. But I couldn't find anywhere this famous Joyal's letter, so I would also be glad if someone could tell me where I can find it.
Thanks in advance for any hints.
| https://mathoverflow.net/users/1246 | global fibrations of simplicial sheaves | For model structures on simplicial sheaves, there is a difference between the Joyal-Jardine approach and the Brown-Gersten approach. This is well explained in Voevodsky's preprint: *Homotopy theory of simplicial presheaves in completely decomposable topologies*, available [here](https://arxiv.org/abs/0805.4578). Briefly, the Brown-Gersten approach does not work for arbitrary sites, but it works for a class of sites defined in Voevodsky's paper - this class includes Noetherian finite-dimensional spaces. When the B-G approach works, the resulting model structure has better finiteness properties than the Joyal-Jardine model structure, which on the other hand can be defined for simplicial (pre)sheaves on any site.
| 5 | https://mathoverflow.net/users/349 | 5637 | 3,827 |
https://mathoverflow.net/questions/5635 | 170 | Purely for fun, I was playing around with iteratively applying $\DeclareMathOperator{\Aut}{Aut}\Aut$ to a group $G$; that is, studying groups of the form
$$ {\Aut}^n(G):= \Aut(\Aut(\dots\Aut(G)\dots)). $$
Some quick results:
* For finitely-generated abelian groups, it isn't hard to see that this sequence eventually arrives at trivial group.
* For $S\_n$, $n\neq 2,6$, the group has no center and no outer automorphisms, and so the conjugation action provides an isomorphism $G\simeq \Aut(G)$.
* Furthermore, I *believe* that if $G$ is a non-abelian finite simple group, then $\Aut(\Aut(G))\simeq \Aut(G)$, though this is based on hearsay.
* If one considers topological automorphisms of topological groups, then $\Aut(\mathbb{R})\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{R}$, and so $$\Aut(\Aut(\mathbb{R}))\simeq \Aut(\mathbb{Z}/2\mathbb{Z}\times \mathbb{R})\simeq\mathbb{Z}/2\mathbb{Z}\times \mathbb{R}.$$
When the sequence $\Aut^n(G)$ is constant for sufficiently large $n$, we will say the sequence **stabilizes**. Despite my best efforts, I have been unable to find a group $G$ such that the sequence $\Aut^n(G)$ is provably non-stabilizing. Is this possible?
A slightly deeper question is whether there are groups $G$ such that the sequence becomes periodic after some amount of time. That is, $\Aut^n(G)\simeq \Aut^{n+p}(G)$ for some $p$ and for $n$ large enough, but $\Aut^n(G)\not\simeq \Aut^{n+m}(G)$ for $m$ between 0 and $p$. A simple way to produce such an example would be to give two groups $G \not\simeq H$ such that $\Aut(G)\simeq H$ and $\Aut(H)\simeq G$. Does anyone know an example of such a pair?
| https://mathoverflow.net/users/750 | Does $\DeclareMathOperator\Aut{Aut}\Aut(\Aut(\dots\Aut(G)\dots))$ stabilize? | I remember that my old grad classmate from Berkeley, Joel David Hamkins, worked on the transfinite version of this problem. [The Automorphism Tower Problem](http://www.math.rutgers.edu/%7Esthomas/book.ps), by Simon Thomas, is an entire book on this subject. The beginning of the book gives the example of the infinite dihedral group $D\_\infty$, in the sense of $\mathbb{Z}/2 \ltimes \mathbb{Z}$. It says that the automorphism tower of this group has height $\omega+1$. It also treats Joel's theorem, which says that every automorphism tower does stabilize, transfinitely. A [Proceedings paper](http://www.jstor.org/pss/2044505) with the same author and title says that Wielandt showed that every finite centerless group has a finite automorphism tower.
---
An improved answer: Simon's book later shows that the automorphism tower of the finite group $D\_8$ has height $\omega+1$, and that for general finite groups no one even knows a good transfinite bound. (The $8$ may look like a typo for $\infty$, but it's not :-).) Apparently the centerless condition is essential in Wielandt's condition.
Also, to clarify what these references mean by the automorphism tower, they specifically use the direct limit of the conjugation homomorphisms $G \to \mbox{Aut}(G)$. $D\_8$ is abstractly isomorphic to its automorphism group. This is a different version of the question that I suppose does not have a transfinite extension. Section 5 of Thomas' book implies that it's an open problem whether the tower terminates in this weaker sense, for finite groups.
Finally an arXiv link to Joel Hamkins' charming paper, [Every group has a terminating transfinite automorphism tower](http://arxiv.org/abs/math/9808014).
---
As other people in this thread have pointed out, it's unsatisfying to make an automorphism tower that only stabilizes transfinitely as a direct limit, when all of the finite terms of the tower are abstractly isomorphic to the base group $G$. I Googled around a bit more and came back to the same two sources, Thomas' book, and this time a joint result of Hamkins and Thomas which is in chapter 8 of the book.
If an automorphism tower stabilizes after exactly $n \in \mathbb{N}$ steps in the direct limit sense, then it also stabilizes after exactly $n$ steps in the weaker abstract isomorphism sense. (Otherwise the direct limit "wouldn't know to stop".) Hamkins and Thomas do better than that. For any *two* ordinals $\alpha$ and $\beta$, which may or may not be finite numbers, they find *one* group $G$ whose automorphism tower has height $\alpha$ and $\beta$ in two different models of ZFC set theory. (Whether it's really the "same" group in different worlds is unclear to me, but their models are built to argue that it is so.) I would suppose that it is possible to make a tower without isomorphic terms by taking a product of these groups, even without the two-for-one property.
Other than [one paper on the Grigorchuk group](http://arxiv.org/abs/math.GR/0308127) by Bartholdi and Sidki, I haven't found anything on automorphism towers of finitely generated groups. The Grigorchuk group has a countably infinite tower, but I'd have to learn more to know whether the terms are abstractly isomorphic.
| 134 | https://mathoverflow.net/users/1450 | 5638 | 3,828 |
https://mathoverflow.net/questions/5546 | 10 | It is known that every orientable 3-manfiold can be obtained as a ramified cover of S3 with a ramification (of some order) at a link in S3. I am curious if there is a reasonable characterization of 3-manifolds that cover 3-torus?
Added. Notice that such a manifold is enlargeble, so it does not admit a metric of positive scalar curvature, so for example a connected sum of n copies of S2 x S1 does not admit a ramified cover of T3 (as far as I understand).
| https://mathoverflow.net/users/943 | Ramified covers of 3-torus | Note that a branched covering induces an injection of rational cohomology rings, by transfer considerations. Therefore the cohomology of a manifold that is a branched covering of $T^3$ must contain three classes of degree 1 whose triple cup product is nontrivial.
This condition on a manifold $M^3$ implies the existence of a map $M^3\to T^3$ of nonzero degree. Passing to a covering space of $T^3$ if necessary we obtain such a map that is also surjective on $\pi\_1$. Assuming the resulting map is of degree $\ge 3$, the main result of [Edmonds, Allan L.Deformation of maps to branched coverings in dimension three. Math. Ann. 245 (1979), no. 3, 273--279.] implies that this map is homotopic to a branched covering.
| 13 | https://mathoverflow.net/users/1822 | 5644 | 3,834 |
https://mathoverflow.net/questions/5650 | 2 | The [splitting lemma](http://en.wikipedia.org/wiki/Splitting_lemma) says:
>
> Given a short exact sequence with maps $q$ and $r$:
>
>
> $0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$
>
>
> then the following are equivalent:
>
>
> 1. ...
> 2. there exists a map $u : C \rightarrow B$ such that $r \circ u = \mathrm{id}\_C$
> 3. $B \cong A \oplus C$
>
>
>
Now I figured, since $r(B) = \ker 0$, $r$ is surjective. Hence, for every $c \in C$ we have some $b \in B$ such that $r(b) = c$. Simply set $u(c) = b$ for one of these $b$s and you have your map $u$.
However, it turns out that this construction assumes the axiom of choice; it chooses one element from each of an infinite number of sets. So my question is: assuming the axiom of choice, is condition (2) always satisfied? Because this would imply that (3) holds for any such short exact sequence. Or am I making some mistake here?
| https://mathoverflow.net/users/1825 | Splitting lemma under assumption of the axiom of choice | I assume you are working in some fixed abelian category $\mathcal{A}$.
It is not true in general that every short exact sequence in $\mathcal{A}$ will split. The problem is that although you can pick a preimage for every 'element' $c\in C$ there is no guarantee that you can assemble this into a morphism in $\mathcal{A}$. It is true in the category of sets that every surjection splits if one assumes the axiom of choice but this is only a set map.
For instance in the category of abelian groups
$$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{2}{\to} \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$
is exact but $\mathbb{Z}/4\mathbb{Z}$ is indecomposable.
Also if you are not in an abelian category it is not necessarily true that exact sequences display this symmetry. See for example [this](https://mathoverflow.net/questions/3757/when-does-splits-imply-cosplits) question where the notion is considered in the category of groups.
| 7 | https://mathoverflow.net/users/310 | 5653 | 3,841 |
https://mathoverflow.net/questions/5611 | 24 | If K is a number field then the Dedekind zeta function Zeta\_K(s) can be written as a sum over ideal classes A of Zeta\_K(s, A) = sum over ideals I in A of 1/N(I)^s. The class number formula follows from calculation of the residue of the (simple) pole of Zeta\_K(s, A) at s = 1 (which turns out to be independent of A).
Let E/Q be an elliptic curve. One might try to prove the (strong) Birch and Swinnerton-Dyer conjecture for E/Q in an analogous way: by trying to define L-functions L(E/Q, A, s) for each A in the Tate-Shafarevich group, writing L(E/Q, s) as a sum of zeta functions L(E/Q, A, s) where A ranges over the elements of Sha, then trying to compute the first nonvanishing Taylor coefficient of L(E/Q, A, s) at s = 1.
Has there been work in the direction of defining such zeta functions L(E/Q, A, s)? If so, what are some references and/or what are such zeta functions called?
Also, taking K to be quadratic, there is not only a formula for the first nonvanishing Laurent coefficient of Zeta\_K(s) (the class number formula), but there is a formula for the second nonvanishing Laurent coefficient of Zeta\_K(s) (coming from a determination of the second nonvanishing Laurent coefficient of Zeta\_K(s, A) - something not independent of K - this is the Kronecker limit formula). Does the Kronecker limit formula have a conjectural analog for the L-function attached to an elliptic curve over Q?
A less sharp question : are there any ideas whatsoever as to whether any of the Taylor coefficients beyond the first for L(E/Q, s) expanded about s = 1 have systematic arithmetic significance?
| https://mathoverflow.net/users/683 | The class number formula, the BSD conjecture, and the Kronecker limit formula | I apologize for in advance for making just a few superificial remarks. These are:
1. The question is not uninteresting. Just because Sha doesn't appear in the definition of L easily, there's no reason one shouldn't ask about manifestations more fundamental than the usual one.
2. An approach might be to think about the p-adic L-function rather than the complex one. I'm far from an expert on this subject, but the algebraic L-function is supposed to be a characteristic element of a dual Selmer group over some large extension of the ground field. The Selmer group (over the ground field) of course does break up into cosets indexed by Sha. Perhaps one could examine carefully the papers of Rubin, where various versions of the Iwasawa main conjectures are proved for CM elliptic curves.]
Added, 8 July:
This old question came back to me today and I realized that I had forgotten to make one rather obvious remark. However, I still won't answer the original question.
You see, instead of the $L$-function of an elliptic curve $E$, we can consider the zeta function $\zeta({\bf E},s)$ of a regular minimal model ${\bf E}$ of $E$, which, in any case, is the better analogue of the Dedekind zeta function. One definition of this zeta function is given the product
$$\zeta({\bf E},s)=\prod\_{x\in {\bf E}\_0} (1-N(x)^{-s})^{-1},$$
where ${\bf E}\_0$ denotes the set of closed points of ${\bf E}$ and $N(x)$ counts the number of elements in the residue field at $x$. It is not hard to check the expression
$$\zeta({\bf E},s)=L(E,s)/\zeta(s)\zeta(s-1)$$
in terms of the usual $L$-function and the Riemann zeta function.
The product expansion, which converges on a half-plane, can also be written as a Dirichlet series
$$\zeta({\bf E},s)=\sum\_{D}N(D)^{-s},$$
where $D$ now runs over the *effective zero cycles* on ${\bf E}$. This way, you see the decomposition
$$ \zeta({\bf E},s)=\sum\_{c\in CH\_0({\bf E})}\zeta\_c({\bf E},s), $$
in a manner entirely analogous to the Dedekind zeta. Here, $CH\_0({\bf E})$ denotes the rational equivalence classes of zero cycles, and we now have the partial zetas
$$\zeta\_c({\bf E},s)=\sum\_{D\in c}N(D)^{-s}.$$
It is a fact that $CH\_0({\bf E})$ is finite. I forget alas to whom this is due, although the extension to arbitrary schemes of finite type over $\mathbb{Z}$ can be found in the papers of Kato and Saito.
It's not entirely unreasonable to ask at this point if the group $CH\_0({\bf E})$ is related to $Sha (E)$. At least, this formulation seems to give the original question some additional structure.
Added, 31, July, 2010:
This question came back yet again when I realized two errors, which I'll correct explicitly since such things can be really confusing to students. The expression for the zeta function in terms of $L$-functions above should be inverted:
$$\zeta({\bf E},s)=\zeta(s)\zeta(s-1)/L(E,s).$$
The second error is slightly more subtle and likely to cause even more confusion if left uncorrected. For this precise equality, ${\bf E}$ needs to be the Weierstrass minimal model, rather than the regular minimal model. I hope I've got it right now.
| 12 | https://mathoverflow.net/users/1826 | 5655 | 3,843 |
https://mathoverflow.net/questions/5666 | 2 | For the question, everything is over an algebraically closed field and by a scheme we mean a scheme of finite type. The theorem of the cube is the following:
Let $X$ be a complete variety, $Y$ a scheme, and $L$ a line bundle on $X\times Y$. Then there is a unique closed subscheme $Y\_1$ of $Y$ satisfying the following properties: (a) If $L\_1$ is the restriction of $L$ to $X\times Y\_1$, there is a line bundle $M\_1$ on $Y\_1$ and an isomorphism $p\_2^\*M\_1\rightarrow L\_1$ on $X\times Y\_1$. (b), since (b) has little to do with the question, let us focus on (a). I am puzzled by the following thing: by the Künneth formula, if $p\_2^\*M\_1\rightarrow L\_1$ is an isomorphism (and $M\_1$ is a line bundle on $Y\_1$), then $M\_1 \rightarrow {p\_2}\_\*(L\_1)$ is an isomorphism. Does anyone knows what is the content of the Künneth formula mentioned here and more importantly, why the statement is true? Thanks!
| https://mathoverflow.net/users/1468 | An application of the Künneth formula in the proof of the theorem of the cube | One uses the following trick.
By the projection formula we have
$${p\_2}\_\*(\mathcal{O}\_{X\times Y\_1}) \cong {p\_2}\_\*(p\_2^\*M\_1^{-1} \otimes L\_1) \cong M\_1^{-1} \otimes {p\_2}\_\*(L\_1)$$
and since $X$ is complete (and $k$ is algebraically closed) it follows from the Künneth formula that
$${p\_2}\_\*\mathcal{O}\_{X\times Y\_1} \cong \mathcal{O}\_{Y\_1}$$
which gives the desired isomorphism on $Y\_1$ by uniqueness of inverses up to isomorphism in the Picard group.
In case I misunderstood and you were asking what the relevant version of the Künneth formula is in algebraic geometry there is an explanation [here](http://eom.springer.de/k/k056010.htm) at the Encyclopedia of Mathematics (search for algebraic geometry to get to the relevant part).
To address Wayne's comment (where by the way it should be $M\_1$ not its inverse): I think it is not necessarily true (well at least it isn't clear to me why it should be) that the adjunct of your original isomorphism will still be an isomorphism. The point (at least in the proof I know from Mumford's Abelian Varieties) is that the existence of such an isomorphism allows one to reduce to checking that ${p\_2}\_\*L\_1$ is a line bundle and that the counit of the adjunction is an isomorphism (which is good because this is a natural map which we get for free).
| 3 | https://mathoverflow.net/users/310 | 5669 | 3,853 |
https://mathoverflow.net/questions/5299 | 5 | How do you define unbounded measurable functions for a general von Neumann algebra?
For the commutative algebra $L^\infty(X,\mu)$, we can consider the space of all measurable functions that are almost everywhere finite. This set has certain nice properties: it is closed under multiplication and there is the notion of convergence almost everywhere.
For the the noncommutative algebra of bounded operators on a Hilbert space, we can consider the set of all closed unbounded operators with dense domain. These operators are quite important in PDE, since differential operators are always unbounded. It is not obvious to me, however, why the product of two unbounded operators will be again a nice operator or how to generalize convergence almost everywhere to this setting.
Is there any construction like the above ones for an arbitrary von Neumann algebra? Can one get any standard properties of measurable functions from this construction?
If the closure of the spectrum of an unbounded operator $T$ is not the whole $\mathbb C$, and $z$ does not lie in this set, then we can consider instead of $T$ the resolvent $(z-T)^{-1}$, which will be a bounded operator. However, it is not clear to me how to proceed when the spectrum is the whole $\mathbb C$ or whether there is a more conceptual way to define these objects.
Update: I must have phrased the question inaccurately; the answer to the original question would be given by affiliated operators, a construction beautiful and useful, but not quite what I had in mind. I am sorry for the confusion caused (by the way, does anyone know how I can mark two answers as accepted, one for the original question and one for the rephrased one?) The rephrased question is:
For each von Neumann algebra, define canonically a set $S$ such that:
\* For the algebra $L^\infty(X)$, $S$ is the set of all measurable functions on $X$.
\* For the algebra of bounded operators on a Hilbert space, $S$ is the set of all unbounded operators on the same Hilbert space.
Alternatively, explain why it impossible (or unreasonable to try) to define such a set for all von Neumann algebras. So, I wish to somehow see the set of objects that will somehow remind of unbounded operators, not to study some specific unbounded operators on given spaces. These objects need not be actual operators in any sense for an arbitrary algebra.
| https://mathoverflow.net/users/1704 | Measurable functions and unbounded operators in von Neumann algebras | I think your question should be as follows:
Given a von Neumann algebra $M$, can we define a canonical set $S$ such that
* if $M=L^\infty(X)$ acting on $L^2(X)$, then $S$ is (isomorphic to) the set of a.e. defined measurable functions, and
* if $M=B(H)$ acting on $H$, then $S$ is the **densely defined, closed** unbounded operators?
If we take this slight alteration of the question, then I believe the answer is the closed affiliated operators. Here's a sketch of a proof which I think should work, but you should check the details just to make sure.
Let $A=L^\infty(X)$. If $f$ is an a.e. defined measurable function, define as in my other answer
$$
D(M\_f)=\{ \xi\in L^2(X) | f\xi\in L^2(X)\}.
$$
Then $M\_f\colon D(M\_f)\to L^2(X)$ is closed and affiliated with $A$.
Now suppose $T$ is a closed, densely defined operator affiliated to $A$ acting in $L^2(X)$ (abbreviated $T\eta A$). Then we can do polar decomposition to get $T=U|T|$ where $U\in A$ and $|T|\eta A$. Hence we have reduced to the case where $T$ is positive and self adjoint. Since $T\eta A$, we must have that $f(T)\in A$ for all bounded Borel functions $f$. In particular, for $n\geq 1$, $T\_n=\chi\_{[0,n]}(T)\in A$, and $T\_n$ increases to $T$. It should be clear how to proceed now to get that $T$ is multiplication by an a.e. defined measurable function.
| 6 | https://mathoverflow.net/users/351 | 5679 | 3,857 |
https://mathoverflow.net/questions/5658 | 3 | $l\_1$ minimization / compressed sensing comes to mind. Does anyone have any concrete examples? Or is such a construct completely useless?
| https://mathoverflow.net/users/1745 | Is there a use for a Hilbert space that uses a different norm than the one induced by the inner product? | One place where this is used is with Hilbert bundles. Taking $L^2$-functions on a space generally works very badly, so one instead takes $L^{2,1}$-functions - functions which are differentiable (almost everywhere) with square-integrable first derivative. However, the transition functions aren't isometries with respect to this norm so one does tend to use the $L^2$-norm on this space, remembering that the fibres aren't complete with respect to the norm.
Another use is in Wiener integration. Depending on one's point of view, one either has a Hilbert space with a weaker norm, or a Banach space with a dense subspace equipped with a Hilbert norm. Essentially, having the two norms means that one can "tame" stuff with respect to one norm by using the stronger one.
This extends further to the notion of "rigged Hilbert spaces".
Generally, the idea is that you want to work in one space but you don't have enough control over convergence, so you introduce a stronger norm and then ordinary convergence with respect to the stronger norm implies fantastic convergence with respect to the weaker one.
(I apologise for using vague language, but the question doesn't give much indication of how deep an answer to give. For those who know a little about ideals of operators, one generally wants the inclusion from the strong norm to the weak to be at least trace class, and often Hilbert-Schmidt.)
| 7 | https://mathoverflow.net/users/45 | 5683 | 3,860 |
https://mathoverflow.net/questions/5684 | 11 | If I have the periods $$\pi\_1(\lambda)=\int\_0^1\frac{dx}{\sqrt{x(x-1)(x-\lambda)}}$$ and $\pi\_2(\lambda)$ similarly defined of the cubic curve $$y^2z=z(x-z)(x-\lambda z)$$ Such functions will be holomorphic on $\lambda \in \mathbb{C}-\{0,1\}$. Then the sum $\pi(\lambda)=\pi\_1(\lambda)+\pi\_2(\lambda)$, as well as each $\pi\_i$, satisfy the differential equation called **Picard-Fuchs** equation $$0=\tfrac{1}{4}\pi+(2\lambda-1)\pi'+\lambda(\lambda-1)\pi''$$ derivation with respect to $\lambda$.
My questions are the following. Where does such a differential equation come from?
why is it important for?
has it something to do with Gromov-Witten Theory?
I'll appreciate any kind of comment related with such an equation.
| https://mathoverflow.net/users/1547 | Picard-Fuchs equations | Here is a very rough outline:
Take your family $E$ of elliptic curves over $B := \mathbb{C} - \{0,1\}$. Then take the associated "(co)homology bundle" over $B$, whose fibre over $\lambda$ is the (singular) (co)homology of the elliptic curve $E\_\lambda$. To be rigorous, the $i$-th cohomology bundle is $R^i \pi\_\ast\mathbb{C}$, where $\pi$ is the map $E \to B$ and $\mathbb{C}$ is the constant sheaf (let us work in the analytic topology). Actually to be precise I should say that $R^i\pi\_\ast\mathbb{C}$ is a (locally constant) sheaf of $\mathbb{C}$ vector spaces, and the corresponding vector bundle is $R^i\pi\_\ast\mathbb{C} \otimes\_\mathbb{C} \mathcal{O}\\_B$. It is a fact that these cohomology bundles come with flat (Gauss-Manin) connections $\nabla$. One way to see that the vector bundles are flat is to observe that there are integral lattices $R^i\pi\_\ast \mathbb{Z} \subset R^i\pi\_\ast\mathbb{C}$.
Let $\omega$ be a 1-form on the family $E$. Note that the 1st cohomology of an elliptic curve is rank 2, so the cohomology bundle $R^1\pi\_\ast \mathbb{C}$ is rank 2, thus if we have 3 sections, then they will be (fiber-wise) linearly dependent. So here are 3 sections: $\omega, \nabla\_{d/d\lambda}\omega, (\nabla\_{d/d\lambda})^2 \omega$. The Picard-Fuchs equation is essentially just the equation which expresses that these sections are linearly dependent. Your equation involving "$\pi$" (not the same as what I am calling "$\pi$") and its derivatives comes from taking this linear dependence equation and "plugging in" (i.e. integrating along) homology classes extended by parallel transport.
The story that I've described above generalizes to arbitrary families of smooth compact varieties.
Thomas Riepe's answer explains some of the more classical reasons why we might be interested in period integrals and Picard-Fuchs equations, so let me say a few words about the relation to Gromov-Witten theory.
The relation to GW theory arises from mirror symmetry, which is a duality between type IIA and type IIB string theories. One of the reasons why mathematicians first became interested in mirror symmetry was because of the prediction of the physicists Candelas-de la Ossa-Green-Parkes in the early 90s that the genus 0 GW invariants of a quintic threefold (type IIA theory) could be computed via an analysis of period integrals and Picard-Fuchs equations coming from a "mirror" family of Calabi-Yau manifolds (type IIB theory). This is a general principle of mirror symmetry: that GW invariants of certain manifolds can be computed via completely different methods on the "mirror manifold". Usually, studying the mirror manifold is "easier" than trying to study the GW theory of the original manifold directly; although by now our knowledge of GW theory has grown considerably, so this is less true than it used to be.
A very nice introductory paper on this material is "Picard-Fuchs equations and mirror maps for hypersurfaces" by David Morrison: <http://arxiv.org/abs/alg-geom/9202026>
If you're interested in reading further, you should check out the book "Mirror symmetry and algebraic geometry" by Cox-Katz, which covers all of this material in detail and explains the proofs (due to Givental and Lian-Liu-Yau) of the Candelas-et. al. prediction.
| 17 | https://mathoverflow.net/users/83 | 5695 | 3,868 |
https://mathoverflow.net/questions/5691 | 5 | The exercise is the following:
>
> Let $A$ be a valuation ring, $K$ its field of fractions. Show that every subring of $K$ which contains $A$ is a local ring of $A$.
>
>
>
Does anyone know what is meant by "to be a local ring of a valuation ring"?
| https://mathoverflow.net/users/1841 | Atiyah-MacDonald: exercise 5.29 - "local ring of a valuation ring" | Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ring of A.
First B is a local ring [by Atiyah-MacDonald Prop 5.18 i)] ; let $M\_B$ be its maximal ideal.
Similarly let $M\_A$ be the maximal ideal of A.
Define $P=A\cap M\_B$. Then of course $P\subset M\_A$ is prime and if we localize A at P, I claim that we get B:
a) If $\pi \in A-P$ then $\pi \notin M\_B$ and so $\pi$ is invertible in B. Hence for all $ a \in A$ we have $a/ \pi \in B$.
This shows $A\_P \subset B$.
b) Now we see that B dominates $A\_P$: this means we have an inclusion of local rings $A\_P \subset B$ and of their maximal ideals: $PA\_P \subset M\_B$. But $A\_P$ is a valuation ring of K because A is (This is the preceding exercise 28 : these guys know where they are heading!)
and valuations rings are maximal for the relation of domination (Exercise 27: ditto !).
Hence we have the claimed equality $B=A\_P$.
| 4 | https://mathoverflow.net/users/450 | 5703 | 3,875 |
https://mathoverflow.net/questions/5085 | 3 | I have a 2D rectangular domain. The governing equation on this domain is Laplace equation:
$\nabla^2 f = 0$
In the left edge there is [Neumann boundary conditon](http://en.wikipedia.org/wiki/Neumann_boundary_condition) :
$\frac{\partial f}{\partial n} = -a$
n is the normal vector to the domain's boundary(here on the left edge it's equal to the negative direction of x axis) and 'a' is a given date and it's a constant.
There is a Dirichlet boundary condition at the bottom edge and there is no boundary condition on right and top edge.
My problem is how to apply that Neumann boundary condition. I'm using finite element method (with first order triangulation)
As you may know, in finite element method first we make stiffness matrix (or global coefficient matrix from local coefficient matrix). Then we apply our governing equation(here the Laplace equation).
| https://mathoverflow.net/users/1591 | How to apply Neuman boundary condition to Finite-Element-Method problems? | You need to modify right-hand vector b of an equation Kx = b, where K is your stiffness matrix.
Here's how to do it, depending on which edge is on von Neumann boundary:
* edge 1-2 (i.e. connecting local nodes 1 and 2),
```
l = sqrt(x21*x21 + y21*y21);
b[node1] += a * l / 2.0f;
b[node2] += a * l / 2.0f;
b[node3] += 0;
```
* edge 1-3,
```
l = sqrt(x13*x13 + y13*y13);
b[node1] += a * l / 2.0f;
b[node2] += 0;
b[node3] += a * l / 2.0f;
```
* edge 2-3,
```
l = sqrt(x32*x32 + y32*y32);
b[node1] += 0;
b[node2] += a * l / 2.0f;
b[node3] += a * l / 2.0f;
```
| 1 | https://mathoverflow.net/users/1846 | 5718 | 3,885 |
https://mathoverflow.net/questions/5723 | 3 | I'm trying to understand the Hodge-Index Theorem at the moment. What does it say explicitly for the case of $\mathbb{CP}^2$?
| https://mathoverflow.net/users/1095 | Hodge-Index Theorem for $\mathbb{C}P^2$ | That the inner product on $H^2(\mathbb{CP}^2)$ has signature $(1,0)$.
| 4 | https://mathoverflow.net/users/297 | 5725 | 3,891 |
https://mathoverflow.net/questions/5711 | 5 | The Fell-Doran problem is a problem in functional analysis. It goes as follows: Let $A$ be a complex unital algebra, $X$ a locally convex space, and $L(X)$ the algebra of all continuous endomorphisms of $X$. Suppose that we have a representation of $A$ on $X$, by which we simply mean an algebra homomorphism
$$
T : A \rightarrow L(X)
$$
which is irreducible (no proper closed invariant subspace) and has trivial commutant (any bounded operator commuting with all the $T\_a$ must be a multiple of the identity). The Fell-Doran problem is: Is $T(A)$ dense in $L(X)$ in the strong operator topology?
My question is: Is this a problem having to do with the fact that we didn't require a topology on our algebra? In other words, what can be said about the case when $A$ is actually a 'topological algebra' and the map $T$ is required to be continuous in some sense? Does that make the problem trivial, i.e. is the answer then automatically yes?
By the way, I have heard that so far there is almost no progress on the Fell-Doran problem in general; not even for Hilbert spaces! The only thing that is known is that there exists a certain concrete space where the answer is affirmative.
| https://mathoverflow.net/users/401 | Is the Fell-Doran problem trivial in a topological setting? | I don't agree with Andrew: more specifically, if $A$ is the algebra of compact operators on a Hilbert space $H$, then let $T\in L(H)$. If, say, $H$ is separable, then let $(e\_n)$ be an orthonormal sequence, and let $P\_n$ be the orthogonal projection on the span of $e\_1,\cdots,e\_n$. Then $P\_n$ is compact, and $P\_n(x)\rightarrow x$ in norm for each $x\in H$. In particular, $P\_nT\in A$ for each $A$ and $P\_nT\rightarrow T$ is the strong operator topology.
More generally, the Kaplansky Density Theorem (see your favourite book on Operator Algebras, or Wikipedia) shows that if $A\subseteq L(H)$ is a $\*$-closed algebra, then the unit ball of $A$ is strong operator dense in the unit ball of the von Neumann algebra which $A$ generates. So if $A$ has trivial commutant, then the von Neumann bicommutant theorem shows that $A$ generates all of $L(H)$. In particular, $A$ is strong operator dense in $L(H)$.
Of course, the question is: what happens if $A$ isn't $\*$-closed...
| 2 | https://mathoverflow.net/users/406 | 5729 | 3,894 |
https://mathoverflow.net/questions/5733 | 9 | Let $X$ be a random variable with mean $0$ and variance $1$, and let $X\_1, X\_2, X\_3, \dots$ be iid copies of $X$. Under what conditions can we say that the density of $\frac{X\_1+\dots+X\_n}{\sqrt{n}}$ converges pointwise to $N(0,1)$?
In particular, when can I say that for any sequence $\epsilon\_n \rightarrow 0$ we have
$$\frac{P(|\frac{X\_1+\dots+X\_n}{\sqrt{n}}|<\epsilon\_n)-P(|N(0,1)|<\epsilon\_n)}{\epsilon\_n} \rightarrow 0?$$
In flavor this is somewhat similar to what I've seen termed as "local limit theorems", except a little bit stronger; for example if $X$ is a Bernoulli variable the above would not hold (take $\epsilon\_n=2^{-n}$). My guess would be that a sufficient condition would be for the usual CLT to hold and $X$ to have bounded density functions, though I haven't seen this cited anywhere.
| https://mathoverflow.net/users/405 | When does a pointwise CLT hold? | Bounded density will suffice, I think. Basically what one needs is for the Fourier transforms (aka characteristic functions) of the $X\_1 + \ldots + X\_n / \sqrt{n}$ to converge pointwise to the Fourier transform of normal distribution while being dominated by something integrable plus something whose L^1 norm goes to zero, so that the (noisy) Lebesgue dominated convergence theorem applies and will give uniform convergence of the density function. Pointwise convergence is not a problem, because the finite second moment will make the characteristic function of X in the class $C^2$ (twice continuously differentiable). This function cannot equal 1 except at the origin (because X is not discrete), so by continuity and Riemann-Lebesgue it is bounded by $1-\epsilon$ outside of a small neighbourhood of the origin; this together with Plancherel (here we use the bounded density - actually square integrable density will suffice) is enough to get the required domination.
| 15 | https://mathoverflow.net/users/766 | 5738 | 3,901 |
https://mathoverflow.net/questions/5740 | 9 | If $d(x,y)$ and $e(x,y)$ are metrics then $d(x,y)+e(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$ are metrics.
If $d\_i(x,y)$ for $i=1,\dots,n$ are metrics then so is $\sqrt{\sum\_{i=1}^n{d\_i^2(x,y)}}$
Are there other interesting ways of constructing new metrics from old metrics?
| https://mathoverflow.net/users/812 | What are some interesting ways of making new metrics out of old metrics? | Personally, I prefer $\min(d(x,y),\epsilon)$ over the standard $d(x,y)/(1+d(x,y))$ trick if the goal is to turn a metric into an equivalent metric – it's a lot easier to prove the metric axioms for this one.
From that follows one way to turn a countably infinite number of metrics into a new metric: $d(x,y)=\sum\\_{n=1}^\infty \min(d\\_n(x,y),2^{-n})$. The point being that convergence in the new metric is the same as convergence in all of the original ones. So this can be used to prove stuff like $C(\mathbb{R})$ with the topology of uniform convergence on compact sets is a metrizable space, and similarly for the space of test functions used in distribution theory. (The individual $d\\_n$ could even be pseudometrics rather than metrics; it doesn't matter, so long as $x\ne y$ implies $d\\_n(x,y)\ne0$ for *some* $n$.)
| 8 | https://mathoverflow.net/users/802 | 5748 | 3,907 |
https://mathoverflow.net/questions/5745 | 5 |
>
> For a fixed $d$, is there a relationship between the homotopy groups of smooth $d$-manifolds?
>
>
>
The $d=1$ case is trivial, but I already don't know how to approach $d=3$ (I should have said that the case of $d=2$ is simple as well, since there is only a sphere to consider, but I don't know how to formulate the property of "having the same homotopy groups as $S^2$" in a simpler way).
Note about the discussion on the comments: it's unreasonable to expect an easy *complete* characterization of homotopy groups of $S^2$, even less for other manifolds. But I think one could try some *partial* relations. An interesting relationship would be: for some $d$, the groups $\pi\_n$ can be determined from groups $\pi\_m$ for $m<N<n$ (this is unlikely to be true though).
| https://mathoverflow.net/users/65 | Homotopy groups of smooth manifolds? | For $d=3$ the homotopy groups can be pretty elaborate. Consider the connect-sum of some lens spaces. The universal cover embeds in $S^3$ as the complement of a cantor set (except for a few degenerate cases where you have $\mathbb RP^3$ summands). So the homotopy-groups are pretty complicated ($\pi\_2$ is finitely generated over $\pi\_1$). You could probably make an argument that this is about the worst thing that can happen for the homotopy-groups of 3-manifolds.
You might want to phrase your question as a question about the Postnikov towers of manifolds. Eilenberg-Maclane spaces are rarely compact boundaryless manifolds.
edit: I guess another spin on your question could go like this. We know the fundamental groups of compact manifolds are all possible finitely presentable groups provided $n \geq 4$. So is there a sense in which the homotopy-algebras of manifolds can be anything finitely presentable? Say, for example, $\pi\_2$. As a module over the group-ring of $\pi\_1$, are there any restrictions beyond being finitely generated? I suppose you could construct a compact $6$-manifold with $\pi\_2$ (almost) any finitely-presented thing over any finitely-presented $\pi\_1$ pretty much the exact same way $4$-manifolds with any finitely presented $\pi\_1$ are constructed. I think if $H\_2(\pi\_1)$ is non-trivial you might run into problems following the analogous construction, in that $\pi\_2$ might strictly contain the $\pi\_2$ you're trying to create.
2nd edit: So regarding 3-manifolds I think your question has something of an answer now, right? $\pi\_n M$ is $\pi\_n$ of the universal cover provided $n > 1$. The universal cover of a geometric 3-manfold is homeomorphic to $\mathbb R^3$ or $S^3$. So by climbing up the JSJ and connect sum decomposition of a 3-manifold, the universal cover is diffeomorphic to a punctured $S^3$ -- the number of punctures is either $0$, $1$, $2$ or a Cantor set's worth of punctures. In the Cantor set case we're giving this complement the compactly generated topology induced from the Cantor set complement's subspace topology. So among other things, $\pi\_2 M$ is a direct sum of copies of $\mathbb Z$, similarly $\pi\_3 M$, torsion first appears in $\pi\_4 M$. The complement you think of as a directed system of wedges of $S^2$'s so the Hilton-Milnor theorem tells you the homotopy groups.
| 9 | https://mathoverflow.net/users/1465 | 5749 | 3,908 |
https://mathoverflow.net/questions/5196 | 0 | Let $L:K$ be a field extension. Let $A$ be a set of elements in $L$, all of which are algebraic over $K$. Construct the field extension $M=K(A)$. I have two questions:
[1] Is $M:K$ an algebraic field extension?
[2] Take $\beta\in M$ where $\beta$ is algebraic over $K$. Then $K(\beta):K$ is a finite extension. Can I assert that $\beta$ lies in a finite extension $K(a1,..., an)$ where $a1,..., an\in A$?
Both questions are trivial when $A$ is finite. So assume that $A$ is infinite; indeed assume that $[M:K]$ is infinite. Now what?
| https://mathoverflow.net/users/801 | On algebraic field extensions | The question was a bit confusing because it was not clear what was meant as the definition of "the field extension $K(A)$". Nick says that "the smallest field containing" is the definition that he wanted, to make the question non-trivial. (But still routine, in my opinion.)
Let's agree then that both (1) and (2) are answered by this summary: The smallest field that contains $K$ and $A$ is the set of rational expressions in $A$ with coefficients in $K$. An element $\beta$ of this field $M$ can only use finitely many elements of $A$, so $\beta$ lies in the subfield generated by them. Moreover, $\beta$ is algebraic by the lemma that the sum or product of two algebraic numbers is algebraic, and the reciprocal of an algebraic number is algebraic.
I'm putting this summary here to hopefully take the question off of the unanswered stack.
| 3 | https://mathoverflow.net/users/1450 | 5763 | 3,917 |
https://mathoverflow.net/questions/5760 | 27 | In particular, suppose I have an affine algebraic variety over $\mathbb{R}^n$ described by generators of a radical ideal $I$ and I want to find (perhaps not all of the) points in the variety. Several important questions come up in practice:
1. are there versions of Buchberger's algorithm that work with inexact data? For instance, suppose that the coefficients of the polynomials generating $I$ are known only to floating point precision. Some CAS will try to find solutions assuming that these coefficients are *exact*. Are there CAS that do something more intelligent (e.g., make certain guarantees given that the numerical coefficients are the truncation of exact coefficients)?
2. does a sparse system of polynomial equations yield a Gröbner basis with sparse elements? In other words, if each polynomial in the original system has a small number of non-zero coefficients relative to $n$, do the basis elements also have this property?
3. what bounds are known for the size of a Gröbner basis in terms of size and sparsity of the original system?
4. are there more appropriate algorithms (than Buchberger's) if we just want to find a *single* point in the variety? (Suppose that any such point is sufficient.) More generally, which algorithms are better suited to address the kinds of issues mentioned above?
| https://mathoverflow.net/users/1557 | Can Gröbner bases be used to compute solutions to large, real-world problems? | First, the Gröbner basis is not sparse. I am speaking a little off-the-cuff, but empirically when I ask SAGE for the Gröbner basis of $(y^n-1,xy+x+1)$ in the ring $\mathbb{Q}[x,y]$, it gets worse and worse as $n$ increases. Any bound would have to be in terms of the degrees of the original generators as well as their sparseness, and I suspect that the overall picture is bad.
Overall your questions play to the weaknesses of Gröbner bases. You would need new ideas to make not just the computations of the bases, but also the actual answer numerically stable. You would also need new ideas to make Gröbner bases sparse.
You are probably better off with three standard ideas from numerical analysis: Divide and conquer, chasing zeroes with an ODE, and Newton's method. If you have the generators for the variety in an explicit polynomial form, then you are actually much better off than many uses for these methods that involve messy transcendental functions. Because you can use standard analysis bounds, specifically bounding the norms of derivatives, to rigorously establish a scale to switch between divide-and-conquer and Newton's method, for instance. Moreover you can subdivide space adaptively; the derivative norms might let you stop much faster when you are far away from the variety.
---
To explain what I mean by derivative bounds, imagine for simplicity finding a zero of one polynomial in the unit interval $[0,1]$. If the polynomial is $100-x-x^5$, then a simple derivative bound shows that it has no zeroes. If the polynomial is $40-100x+x^5$, then a simple derivative bound shows that it has a unique zero and Newton's method must converge everywhere within the interval. If the polynomial is something much more complicated like $1-x+x^5$, then you can subdivide the interval and eventually the derivative bounds become true. Also, with polynomials, you can make bounds to know that there are no zeroes in an infinite interval under suitable conditions.
You can do something similar in higher dimensions. You can divide space into rectangular boxes, and just median subdivisions. It's not very elegant, but it works well enough in low dimensions. In high dimensions, the whole problem can be intractable; you need to say something about why you think that the solution locus is well-behaved to know what algorithm is suitable.
| 23 | https://mathoverflow.net/users/1450 | 5767 | 3,921 |
https://mathoverflow.net/questions/5772 | 22 | What is the exact relationship between principal bundles, representations, and vector bundles?
| https://mathoverflow.net/users/1648 | Principal bundles, representations, and vector bundles | Let G be an algebraic group (or, since the question was tagged as differential geometry, a Lie group). Then if we're given a principal G-bundle $E\_G$ and a representation V of G, we get a vector bundle out of this through the associated bundle construction: $(E\_G \times V)/G$ is a vector bundle with generic fiber V. Here, G acts on $E\_G \times V$ as $g(x,v) = (xg^{-1},gv).$
This shows that fixing a G-bundle determines an exact tensor functor from the category of representations of G to the category of vector bundles. There's a converse to this which says that giving an exact tensor functor from representations of G to vector bundles is equivalent to a G-bundle.
| 28 | https://mathoverflow.net/users/916 | 5776 | 3,927 |
https://mathoverflow.net/questions/5769 | 8 | I'm trying to understand the Hodge-Index Theorem at the moment. What does it say explicitly for the case of the complex Grassmannian Gr($n,k$), and can this be established without recourse to the theorem?
| https://mathoverflow.net/users/1095 | Hodge Index Theorem for Gr(n,k) | The answer is a happy surprise for me: The Hodge index theorem for a Grassmannian matches a special case of John Stembridge's $q=-1$ phenomenon, that I also studied in an old paper.
First, some generalities about what is going on, and about when you do or don't "need" the Hodge Index Theorem. I am following the Hodge Index theorem described in Claire Voisin's book, that says that the signature of a complex $n$-manifold is a certain alternating sum of Hodge numbers. A Grassmannian is a type of partial flag variety, and flag varieties have Schubert cell decompositions. That makes it easy to compute both the Hodge numbers, and the signature of the manifold separately so that the Hodge Index Theorem becomes an identity between two calculations.
More precisely, the Schubert cells are a basis for the cohomology: they make a CW complex with even-dimensional cells, which forces the CW differential to vanish. For complex geometry reasons, the Schubert cells are all in $H^{k,k}$ for some $k$, i.e., all on the diagonal of the Hodge diamond. The Hodge Index theorem then says that the signature is the alternating sum of these numbers. At the same time, the intersection matrix of such a manifold is just a symmetric permutation matrix connecting dual Schubert cells. So the signature is also the number of self-dual Schubert cells. The Hodge Index theorem thus equates the number of self-dual cells with signed sum of all of the cells.
In the case of a Grassmannian $\mathrm{Gr}(a,a+b)$, the Schubert cells are labelled by partitions in an $a \times b$ rectangle. The Hodge index formula is the $q$-enumeration of these partitions with $q = -1$. On the other hand, the direct calculation is the number of self-complementary partitions, where you replace a partition with its complement in the rectangle and turn the rectangle upside-down. It is easy to calculate the self-complementary partitions in the case. The lattice path of the partition is symmetric about the middle of the rectangle, so these are just partitions in an $\lfloor a/2 \rfloor \times \lfloor b/2 \rfloor$ rectangle. Or there are none if $a$ and $b$ are both odd: in this case the signature is trivial a priori because the middle dimension has odd degree. This combinatorial equality is exactly the $q=-1$ phenomenon.
What I wonder now is whether every case of the $q=-1$ phenomenon, for the $q$-dimension of an irreducible representation of a complex simple or compact simple Lie group, can be explained by a Hodge Index theorem with coefficients, combined with some geometric model of the representation. This would be sophisticated (to the point of overkill, if you just want the combinatorics), because the involution in general is something called "evacuation", which John related to canonical bases.
| 14 | https://mathoverflow.net/users/1450 | 5780 | 3,929 |
https://mathoverflow.net/questions/5786 | 89 | Because adjoint functors are just cool, and knowing that a pair of functors is an adjoint pair gives you a bunch of information from generalized abstract nonsense, I often find myself saying, "Hey, cool functor. Wonder if it has an adjoint?" The problem is, I don't know enough category theory to be able to check this for myself, which means I can either run and ask MO or someone else who might know, or give up.
I know a couple of necessary conditions for a functor to have a left/right adjoint. If it doesn't preserve either limits or colimits, for example, I know I can give up. Is there an easy-to-check sufficient condition to know when a functor's half of an adjoint pair? Are there at least heuristics, other than "this looks like it might work?"
| https://mathoverflow.net/users/382 | How do I check if a functor has a (left/right) adjoint? | The adjoint functor theorem [as stated here](http://en.wikipedia.org/wiki/Adjoint_functor_theorem#Existence) and the special adjoint functor theorem (which can also both be found in Mac Lane) are both very handy for showing the existence of adjoint functors.
First here is the statement of the special adjoint functor theorem:
**Theorem** Let $G\colon D\to C$ be a functor and suppose that the following conditions are satisfied:
(i) $D$ and $C$ have small hom-sets
(ii) $D$ has small limits
(iii) $D$ is well-powered i.e., every object has a set of subobjects (where by a subobject we mean an equivalence class of monics)
(iv) $D$ has a small cogenerating set $S$
(v) $G$ preserves limits
Then $G$ has a left adjoint.
**Example** I think this is a pretty standard example. Consider the inclusion **CHaus**$\to$**Top** of the category of compact Hausdorff spaces into the category of all topological spaces. Both of these categories have small hom-sets, it follows from Tychonoff's Theorem that **CHaus** has all small products and it is not so hard to check it has equalizers so it has all small limits and the the inclusion preserves these. **CHaus** is well-powered since monics are just injective continuous maps and there are only a small collection of topologies making any subspace compact and Hausdorff. Finally, one can check that $[0,1]$ is a cogenerator for **CHaus**. So $G$ has a left adjoint $F$ and we have just proved that the Stone-Čech compactification exists.
If you have a candidate for an adjoint (say the pair $(F,G)$) and you want to check directly it is often easiest to try and cook up a unit and/or a counit and verify that there is an adjunction that way - either by using them to give an explicit bijection of hom-sets or by checking that the composites
$$G \stackrel{\eta G}{\to} GFG \stackrel{G \epsilon}{\to} G$$
and
$$F \stackrel{F \eta}{\to} FGF \stackrel{\epsilon F}{\to} F$$
are identities of $G$ and $F$ respectively.
I thought (although I am at the risk of this getting excessively long) that I would add another approach. One can often use existing formalism to produce adjoints (although this is secretly using one of the adjoint functor theorems in most cases so in some sense is only psychologically different). For instance as in Reid Barton's [nice answer](https://mathoverflow.net/questions/1620/is-there-a-free-digraph-associated-to-a-graph/1703#1703) if one can interpret the situation in terms of categories of presheaves or sheaves it is immediate that certain pairs of adjoints exist. Andrew's great answer gives another large class of examples where the content of the special adjoint functor theorem is working behind the scenes to make verifying the existence of adjoints very easy. Another class of examples is given by torsion theories where one can produce adjoints to the inclusions of certain subcategories of abelian (more generally pre-triangulated) categories by checking that certain orthogonality/decomposition properties hold.
I can't help remarking that one instance where it is very easy to produce adjoints is in the setting of compactly generated (and well generated) triangulated categories. In the land of compactly generated triangulated categories one can wave the magic wand of Brown representability and (provided the target has small hom-sets) the only obstruction for a triangulated functor to have a right/left adjoint is preserving coproducts/products (and the adjoint is automatically triangulated).
| 79 | https://mathoverflow.net/users/310 | 5788 | 3,935 |
https://mathoverflow.net/questions/5800 | 14 | The classic puzzle goes something like this: "You are standing in front of a lake with a 3 gallon bucket and a 5 gallon bucket, how can you get 4 gallons of water?"
Is there an easy way to generate the triple (A,B,C) where you can get C gallons of water using buckets of size A and B?
| https://mathoverflow.net/users/1857 | Generalization of the two bucket puzzle | Simon's answer points out that the Euclidean algorithm shows that gcd(A,B) divides C is necessary, but the lack of large container makes the problem more difficult, because obviously you can't get $C$ if $C \gt A+B$. However, the following modification of the algorithm seems to work.
Let's assume $A \lt B$ and gcd$(A, B)=1$ for simplicity. By pouring from B into A and dumping A, you can get any positive integer $B-nA$ left in B. Do this until the answer is less than $A$. Then by transferring the contents to bucket A and filling it from B into it, we get $2B-(n+1)A$. Then subtract $A$ again until you have $0 \lt 2B-kA \lt A$, and we can iterate this process to get $3B-(k+1)A$, etc. This gives any integer linear combination $rB-sA$ up to the size of $A+B$ because once you get the right multiple of $B$ into the combination, you can always add bucketsful of $A$.
You just need to get $rB\equiv C$ (mod A) in order to find a combination for $C$, which happens if gcd$(A, B)=1$. With this algorithm run in the general case you can get any multiple of gcd($A, B$) up to $A+B$ (this holds trivially when $A=B$).
(Sorry, I had to edit this answer several times because parts disappeared, until I realized that my inequality signs were being parsed as HTML tag starts even after dollar signs.)
| 23 | https://mathoverflow.net/users/1198 | 5807 | 3,949 |
https://mathoverflow.net/questions/5808 | 5 | It is well known that left [Bousfield localizations](http://ncatlab.org/nlab/show/Bousfield+localization) of the [global functor model category](http://ncatlab.org/nlab/show/global+model+structure+on+functors) $Func(C^{op}, SSet\_{standard})$ of functors with values in simplicial sets equipped with the [standard model structure on simplicial sets](http://ncatlab.org/nlab/show/model+structure+on+simplicial+sets) yields the local [model structure on simplicial presheaves](http://ncatlab.org/nlab/show/model+structure+on+simplicial+presheaves) that models [hypercomplete oo-stacks](http://ncatlab.org/nlab/show/hypercomplete+(infinity%2C1)-topos).
What is known, though, about left Bousfield localizations of $Func(C^{op}, SSet\_{Joyal})$, with simplicial sets equipped with the [Joyal model structure](http://ncatlab.org/nlab/show/model+structure+on+simplicial+sets#joyals_model_structure_9)?
If it exists, this should model $(\infty,2)$-sheaves on $C$. But possibly it doesn't exist.
| https://mathoverflow.net/users/381 | Local Joyal-simplicial presheaves? | If $V$ is a reasonnable model category (i.e. combinatorial, etc), and $C$ a small category endowed with a Grothendieck topology $\tau$, there are $\tau$-local model structures on the category $Fun(C^{op},V)$ (a projective version as well as an injective version). You may have a look at this [paper](http://www.intlpress.com/HHA/v12/n2/a9/v12n2a9.pdf) of Clark Barwick (see also section 4.4 of Joseph Ayoub's [book](http://user.math.uzh.ch/ayoub/PDF-Files/THESE.PDF) for the hypercomplete versions). In the case $V$ is the standard model structure on simplicial sets, we get the usual homotopy theory of stacks in $\infty$-groupoids, while, if $V$ is the Joyal model structure, you get the homotopy theory of stacks in $(\infty,1)$-categories. If you consider the hypercomplete version, then these model structures will behave in the usual way; for instance, if moreover the topos of sheaves on $C$ has enough points, a morphism of simplicial presheaves on $C$ will be a weak equivalence for the hypercomplete $\tau$-local Joyal model structure if and only if its stalkwise is a weak equivalence for the Joyal model structure.
You may as well consider the case where $V$ is the [Rezk model structure](http://arxiv.org/abs/0901.3602) for $(\infty,n)$-categories to get the homotopy theory of stacks in $(\infty,n)$-categories (for $0\leq n\leq \infty$) to obtain the $(\infty,n+1)$-topos of stacks in $(\infty,n)$-categories (whatever this means). Alternatively, you may as well consider the case where $V$ is the model category of Segal $n$-categories, and get something rather close to [Simpson and Hirschowitz](http://arxiv.org/abs/math/9807049) theory of higher stacks. It is also possible to consider the case where $C$ is a simplicial category and $\tau$ is a Grothendieck topology on $C$ (in the sense of Toën-Vezzosi-Lurie, see [HAG I](http://www.math.univ-toulouse.fr/~toen/hagI.pdf) and [Lurie's book](http://www.math.harvard.edu/~lurie/papers/highertopoi.pdf)), and obtain stacks of $(\infty,n)$-categories over any $(\infty,1)$-topos.
| 11 | https://mathoverflow.net/users/1017 | 5818 | 3,957 |
https://mathoverflow.net/questions/5823 | 2 | I'm looking for a list of examples of random variables to use in teaching a measure-theoretic probability course. For example, the Rademacher functions are an explicit construction of independent Bernoulli random variables.
If you were teaching such a course and had a list of canonical examples to illustrate definitions and theorems, what would be on the list?
| https://mathoverflow.net/users/136 | Examples of random variables | That will be not quite an answer for your question. Anyway it may be helpful.
1. If you have a sequence of independent Bernoulli r.v. $(B\_i)$ then you can define a uniform variable by $U = \sum 2^{-i} B\_i$ and further you can obtain an infinite sequence of independent uniform r.v $U\_i$. (just by splitting $B\_i$ into infinitely many subsequences). Finally from this sequence you may get a sequence of independet variables of any continuous distribution $F^{-1}(U\_i)$ where $F^{-1}$ is the generalised inverse of the cdf of the distribution we want to have.
2. There is a construction of the Wiener process using Haar functions. I guess it is may be to difficult for your students. But I can look for references in English (in a moment I have only a Polish book).
| 5 | https://mathoverflow.net/users/1302 | 5829 | 3,965 |
https://mathoverflow.net/questions/5761 | 6 | Let $\psi(x):=\sum\_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$?
Thanks in advance,
efq
| https://mathoverflow.net/users/1849 | Change rates of the 2nd Chebyshev function | There is the asymptotic estimate $\psi(x+h) - \psi(x) \sim h$ for $x^{7/12 + \epsilon} \leq h \leq x$, valid for any $\epsilon > 0$. This is due to M. N. Huxley, and dates to 1972. I am not aware of any better range for $h$ if you want asymptotic equality. But if you are satisfied with an order of magnitude result, you can have $c\_1h \leq \psi(x+h) - \psi(x) \leq c\_2h$ with $c\_1$ and $c\_2$ positive constants and $x^{\theta} \leq h \leq x$ for some $\theta$ slightly larger than $0.5$. I can't give references offhand, but you should be able to find such papers by searching on R. C. Baker, G. Harman, J. Pintz in Mathscinet.
| 7 | https://mathoverflow.net/users/3304 | 5836 | 3,970 |
https://mathoverflow.net/questions/5833 | 7 | Let V and W be irreducible representations of $S\_n$ and $S\_m$ over a field of characteristic 0. Then the Littlewood-Richardson coefficients allow us to compute the isomorphism type of the induced $S\_{n+m}$-module V⊗W↑. This induction comes from the inclusion
$S\_n\times S\_m \rightarrow S\_{n+m}$.
Now suppose V=W. Then V⊗V↑ is a $S\_{2n}$-module. But actually there's a symmetry coming from the symmetric monoidal category structure, so there is another induction up to an $S\_{2n}$-module structure:
Extend the action of $S\_n\times S\_n$ on V⊗V by including the symmetry $c\_V$, this naturally extends the group to the wreath product $S\_n\sim S\_2$. Induction along
$S\_n\sim S\_2 \rightarrow S\_{2n}$
gives the representation that I want:
$(V\otimes V)\_{S\_n\sim S\_2}\uparrow^{S\_{2n}} \hookrightarrow V\otimes V\uparrow^{S\_{2n}}$.
Using the Littlewood-Richardson rules we know the structure of the last term in terms of semi-standard skew tableau. My question is, how do we characterise the inclusion?
| https://mathoverflow.net/users/109 | Symmetric Powers, Tableau and Wreath Products | You want to read [Splitting the square of a Schur function into its symmetric and antisymmetric parts](http://www.ams.org/mathscinet-getitem?mr=1331743), where this question is answered in terms of domino tableaux. I am also a big fan of [Domino tableaux, Schützenberger involution, and the symmetric group action](http://www.ams.org/mathscinet-getitem?mr=1798322) which, to my mind, gives the "right" formulation of the domino tableaux rule.
| 7 | https://mathoverflow.net/users/297 | 5837 | 3,971 |
https://mathoverflow.net/questions/5845 | 2 | Can anyone give an explicit basis of the universal (noncommutative) differential calculus over an algebra $A$ with basis ${e\_i}$. (The universal calculus over $A$ is the kernel of the multiplication map $m:A \otimes A \to A$.)
| https://mathoverflow.net/users/1867 | Basis for Universal Calculus | You can describe $\Omega\_A=\ker(m:A\otimes A\to A)$ as the quotient of the free $A$-bimodule generated by symbols $d(a)$, one for each element $a\in A$, by the sub-bimodule generated by the elements of the form $$d(ab)-d(a)\\,b-a\\,d(b), \qquad a,b\in A,$$ together with the elements of the form $$d(\lambda 1), \qquad \lambda\in k$$ with $k$ being the base field.
The elements $\{a d(b):a,b\in A\}$, when seen in $\Omega\_A$, span $\Omega\_A$ over $K$ but are not linearly independent over $k$.
To extract from this a $k$-*basis* of $\Omega\_A$ you need to know more than a basis of $A$. For example, if you know a presentation of $A$ given by generators and relations, you can obtain a basis using essentially Groebner bases.
| 3 | https://mathoverflow.net/users/1409 | 5849 | 3,980 |
https://mathoverflow.net/questions/5841 | 2 | This question is motivated from my [last question](https://mathoverflow.net/questions/4317/integrally-closed-factor-rings-and-projective-modules) here. I wonder if one has a ring A and an over-ring of this ring say B, and if we know that B is a projective A-module can we have a particular idea of how Spec B would look like if we know how Spec A would look like?
This question does make sense to me. Because for instance given a local ring A, then B's are some form of copies of A. If A were zero dimensional reduced and commutative then Spec B would look like copies of clopen subsets of Spec A (because projective modules over von Neumann regular rings are isomorphic to direct sum of principal ideals). So what else do we know?
| https://mathoverflow.net/users/1245 | Spectra of rings that are projective module over a subring | So, for example, A could be a field k, and B could be any k-algebra whatsoever. Basically you would be trying to recover all of classical algebraic geometry from Spec k. It does not seem likely.
| 6 | https://mathoverflow.net/users/1698 | 5858 | 3,987 |
https://mathoverflow.net/questions/5865 | 2 | As is well known, every differential calculus $(\Omega,d)$ over an algebra $A$ is a quotient of the universal calculus $(\Omega\_A,d)$, by some ideal $I$. In the classical case, when $A$ is the coordinate ring of a variety $V(J)$ (for some ideal of polynomials $J$), and $(\Omega,d)$ is its ordinary calculus, how is $I$ related to $J$?
| https://mathoverflow.net/users/1867 | Classical Calculi as Universal Quotients | In the classical case, if $\Omega(A)$ is the kernel of the multiplication map $m:A\otimes A\to A$, then—since $A$ is commutative, so that $m$ is not only a map of $A$-bimodules but also a morphism of $k$-algebras,—it turns out that $\Omega(A)$ is an *ideal* of $A\otimes A$, not only a sub-$A$-bimodule. In particular, you can compute its square $(\Omega(A))^2$. Then the classical module of Kähler differentials $\Omega^1\_{A/k}$ is the quotient $\Omega(A)/\Omega(A)^2$.
(This is the construction used by Grothendieck in EGA IV, for example)
| 4 | https://mathoverflow.net/users/1409 | 5870 | 3,995 |
https://mathoverflow.net/questions/5857 | 5 | In a [previous question](https://mathoverflow.net/questions/3887/quantum-frobenius), I asked how Lusztig's quantum Frobenius generalizes the classical Frobenius map on a variety over a finite field. I was directed to a very interesting [paper](http://arxiv.org/abs/math/0005246) by Kumar and Littlemann in which a quantized analog of the multicone over a flag varietywas constructed. The [response](https://mathoverflow.net/questions/3887/quantum-frobenius) claimed that "Upon specialization and base change this morphism becomes the standard Frobenius morphism on the flag variety." I am finding the paper pretty tough going. Could someone explain, for the simple case of $\mathbb{CP}^1$, what exactly its quantum analogue is, whether it has anything to do with the standard Podles' q-sphere, and how the quantum Frobenius is defined upon it.
| https://mathoverflow.net/users/1095 | Quantum Frobenius II | In general, the idea of the Kumar-Littelmann paper is the following: For a semisimple group G, set $V := \displaystyle \bigoplus\_{n \geq 0} H^0(\lambda)$, where $\lambda$ is a fixed regular dominant weight for G. Then $V$ is the projective coordinate ring of $G/B$ under the embedding $G/B \hookrightarrow \mathbb P( V( \lambda ) )$, where $V$ is the Weyl module for $G$ of highest weight $\lambda$. In particular, one can obtain the sheaf of regular functions on $G/B$ in the natural way from $V$.
Now, the (absolute) Frobenius morphism on the flag variety $G/B$ induces an automorphism of $V$ as an $\mathbb F\_p$-vector space, and in fact the converse holds: the appropriate $p^{th}$-power morphism $V \to V$ (which is just the morphism of taking $p^{th}$ powers of sections) induces the Frobenius morphism on $G/B$ (this is the process called "sheafification" in their paper, cf section 6). The point of the paper is now that one can define a module (let's call it $V'$) for the quantum group associated to $G$ such that upon base change, $V'$ becomes $V$. Furthermore, Lusztig's Frobenius morphism induces a morphism $V \to V'$ (which they call $Fr^\*$) which, upon base change, becomes exactly the desired $p^{th}$-power morphism $V \to V$.
Let me give an explicit example for $\mathbb P^1$. In this case, $\mathbb P^1$ is the flag variety of $G = SL\_2$. Since the weights of $SL\_2$ are parametrized by integers, I'll write $H^0(n)$ for the global sections of the corresponding line bundle on $G/B$ (which is just a complicated way of saying that $H^0(n) = H^0( \mathbb P^1, O(n) )$, where that $O$ should be a \mathscr O but that doesn't seem to work). Then in this case, we can take $V = \displaystyle \bigoplus\_{n \geq 0} H^0(n)$, and $V$ is just $k[x, y]$. The scheme-theoretic Frobenius morphism on $G/B$ is induced by the natural $p^{th}$-power morphism $V \to V$, $\; s \mapsto s^{ \otimes p }$ (which is just the natural $p^{th}$-power morphism on the ring $k[x, y]$). We now quantize this picture: Set $$V' := \bigoplus\_{n \geq 0} H^0( X, \chi\_{n}^\xi ) ,$$ where here I'm using their notation from the paper (note that the "X" should be a mathfrak X as in the paper, but somehow I can't do mathfrak here). That is, $H^0( X, \chi\_{n}^\xi )$ is the induction functor from $U\_q(b)$ modules to $U\_q(g)$ modules, applied to the 1-dimensional $U\_q(b)$-module $\chi\_{n}^\xi$ (cf section 2 of the paper). The point is that $V'$ is a quantized version of $V$, and Lusztig's Frobenius morphism induces a morphism $Fr^\* : V \to V'$ that, upon base change, becomes the $p^{th}$-power morphism $V \to V$.
(As for the Podles' q-sphere, I don't know what that is, so I can't speak to that part of your question).
(Edit: I realized that there is a slight white lie in what I wrote above, namely that the morphism Fr\* initially isn't quite a morphism from $V$ to $V'$, but from a characteristic-0 version of $V$ to $V'$; one only gets $V$ after base change to positive characteristic. Kumar and Littelmann first construct Fr\* in characteristic 0. Morally, though, one can ignore this issue on a first pass; it's a bit confusing because Fr\* appears in various incarnations, both before and after base change).
| 9 | https://mathoverflow.net/users/1528 | 5883 | 4,008 |
https://mathoverflow.net/questions/5867 | 3 | I started writing [nLab:Theta space](http://ncatlab.org/nlab/show/Theta+space). Not done yet, but while I am working on it:
is there a good proposal for what the "$(n+1,r+1)$-$\Theta$-space of all $(n,r)$-$\Theta$-spaces" would be?
| https://mathoverflow.net/users/381 | (n+1,r+1)-Theta space of (n,r)-Theta spaces? | Let me assume $n=\infty$, to make things easier to write, so "$(\infty,r)$-$\Theta$-space" equals "$r$-$\Theta$-space".
The totality of $r$-$\Theta$-spaces forms a (large) category enriched over $r$-$\Theta$-spaces, which I'll call $C$. Given this, you can form a presheaf of spaces $X$ on the category $\Theta\_{r+1}$, where
$$X[0] = \text{class of objects of $C\_r$},$$
and
$$X(\[m\](\theta\_1,\dots,\theta\_m)) = \coprod\_{a\_0,\dots,a\_m} C(a\_0,a\_1)(\theta\_1)\times \cdots \times C(a\_{m-1},a\_m)(\theta\_m).$$
Here "$\[m\](\theta\_1,\dots,\theta\_m)$" represents a typical object in $\Theta\_{r+1}$ (so each $\theta\_i\in \Theta\_r$). The coproduct is over tuples of objects of $C$.
The structure maps in the presheaf use the fact that $C$ is a category object. (It's like the way you get a Segal category from a category enriched over spaces.)
The gadget $X$ is *almost* an $(r+1)$-$\Theta$-space. It satisfies all the "Segal" conditions, and also all the completeness conditions *except* for the one in bottom dimension. You get an honest $(r+1)$-$\Theta$-space $X'$ from $X$ by applying a suitable localization.
The gadget $X'$ should be the thing you want. (None of the proofs involved here have been written up, or at least not by me, though we're working on it.)
| 5 | https://mathoverflow.net/users/437 | 5886 | 4,010 |
https://mathoverflow.net/questions/5885 | 4 | In classical geometry the calculation of the Chern classes of a vector bundle using a connection is independent of the choice of connection. Does any such result hold for projective modules in non-commutative geometry?
| https://mathoverflow.net/users/1095 | Does the non-commutative Chern class depend on the choice of connection? | Yes.
You can see the construction in detail, for example, in Max Karoubi's ‘Homologie cyclique et $K$-theorie’ (Asterisque 149, SMF; you can get this from his web page), where he constructs the Chern classes $K\_0(A)\to H(A)$ using connections much à Chern-Weyl (Here $H(A)$ is the non-commutative de Rham theory, or one of the various cyclic homologies of $A$) He also constructs higher Chern classes on the higher algebraic $K$-theory by a similar procedure. The book by Loday on cyclic homology also covers this.
| 7 | https://mathoverflow.net/users/1409 | 5887 | 4,011 |
https://mathoverflow.net/questions/5847 | 10 | Faa di Bruno's formula ([MathWorld](http://mathworld.wolfram.com/FaadiBrunosFormula.html), [Wikipedia](http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula)) gives the kth derivative of f(g(t)) as a sum over set partitions. (I'm not sure how well-known the combinatorial interpretation is; for example see [a 2006 paper of Michael Hardy](http://www.combinatorics.org/Volume_13/Abstracts/v13i1r1.html)).
Is there a similar formula for the kth derivative of f(g(h(t))), i. e. of a threefold composition? One has, I think,
${d^2 \over dt^2} f(g(h(t))) = f^{\prime\prime}(g(h(t))) g^\prime(h(t))^2 (h^\prime(t))^2 + f^\prime(g(h(t))) g^{\prime\prime}(h(t)) h^{\prime}(t)^2$
$ + f^\prime(g(h(t))) g^\prime(h(t)) h^{\prime\prime}(t)$
but I am not even sure this is right (computation is tricky!) and I do not know how to generalize this.
| https://mathoverflow.net/users/143 | Is there a Faa di Bruno-like formula for composition of three functions? | The Faa di Bruno formula can be well reinterpreted in terms of trees, and then the generalization is obvious. Here's how:
First of all, the $k$th derivative of $f\circ g \circ h$ depends only on the first $k$ derivatives of $f,g,h$, evaluated at the appropriate spots. So we can without loss of generality identify smooth functions with their power series --- these should be appropriately centered, so we might as well assume that $f(0) = g(0) = h(0) = 0$. So the data of the "function" $f$ is the sequence $f^{(1)},f^{(2)},f^{(3)},\dots$ of derivatives at $0$. (Any combinatorial fact about derivatives can be checked by composing polynomials.)
Second, rather than trying to guess formulas by thinking only in terms of functions of a single variable, moving to vector land. So $f,g,h$ are functions (rather, formal power series) between different vector spaces. For the Faa di Bruno formula in this generality, see M. Hardy, "Combinatorics of Partial Derivatives", the electronic journal of combinatorics, vol. 13 (R1), 2006.
Third, as soon as you're working in vectors, you realize that if you thought $f$ was a smooth vector-valued function $V \to W$, then $f^{(n)}$ is a linear function $V^{\otimes n} \to W$. And then it's natural to draw it in the Penrose notation (R. Penrose, "Applications of negative dimensional tensors". In D.J.A. Welsh, editor, *Combinatorial mathematics and its applications*, pages 221–244, London, 1971. Mathematical Institute, Oxford, Academic Press) as a directed tree. You should label the $n$ incoming edges with the vector space $V$, the one outgoing edge with $W$, and the vertex with $f$ --- you don't need to label it with $f^{(n)}$ because the $n$ is counted by the number of incoming edges.
Fourth, work a la Feynman. Declare that a vertex with $n$ incoming and one outgoing edge *is* $f^{(n)}$. But when you evaluate it, you should divide by a symmetry factor, which counts the number of automorphisms of the diagram ($n!$). Put an $x$ label on each incoming edge. Then:
$$ f(x) = \sum\_\Gamma \frac{\Gamma}{|{\rm Aut} \Gamma|} $$
where the sum ranges over all diagrams $\Gamma$ that can be drawn with a single vertex, which is labeled by $f$:
```
x x x x x x
| | | | | |
| | | \ | /
1 | 1 \ / 1 \|/
f(x) = - * f + - * f + - * f + ...
1 | 2 | 6 |
| | |
| | |
```
Ok, then the Faa di Bruno formula says the following:
$$ f(g(x)) = \sum\_\Gamma \frac{\Gamma}{|{\rm Aut} \Gamma|} $$
where the sum ranges over all diagrams with two "rows": an $f$ on the bottom, and some $g$s on the top:
```
x x x x x
| | | | |
| \ / | |
1 * g 1 * g 1 * *
f(g(x)) = - | + - | + - | | +
1 | 2 | 2 \ /
* f * f * f
| | |
x x x x x x x x x
| | | | | | | | |
\|/ \ / / | | |
1 * g 1 *g *g 1 *g *g *g
+ - | + - | | + - \ | / +
6 | 2 \ / 6 \|/
* f *f *f
| | |
x x x x x x x x x x x x x x x x x x x x
\ | | / | | | | | | | | | | | | | | | |
\\ // | | | | | | | | | | | | | | | |
\V/ \|/ / \| |/ \| | | | | | |
1 *g 1 *g *g 1 *g *g 1 *g *g *g 1 *g *g *g *g
+ -- | + - | | + - | | + - \ | / + -- \ \ / / +
24 | 6 \ / 8 \ / 4 \|/ 24 \ V /
*f *f *f *f *f
| | | | |
+ ...
```
Here I've sorted the diagrams into the degree in $x$. The numerical factor counts the number of combinatorial automorphisms of each diagram (see e.g. the 1/8 and the 1/4), and I list every diagram once up to isomorphism. Which is to say that I'm really considering the whole groupoid of diagrams, and computing a "groupoid integral" in the sense of Baez and Dolan.
Anyway, the composition of three functions can again be understood as counting labeled graphs with symmetry.
```
x x x x x x x
| \ / | | | |
*h *h *h *h *h *h
1 | 1 | 1 \ / 1 | |
f(g(h(x))) = - *g + - *g + - *g + - *g *g + ...
1 | 2 | 2 | 2 \ /
*f *f *f *f
| | | |
```
I should mention one more thing. The above sums of diagrams compute the whole power series, and so they're computing the $n$th derivatives of the composition divided by $n!$. For example,
$$ \frac1{4!} (f\circ g)^{(4)} = \frac1{24} f^{(1)}g^{(4)} + \frac16 f^{(2)} g^{(3)} g^{(1)} + \frac18 f^{(2)} (g^{(2)})^2 + \frac14 f^{(3)} g^{(2)} (f^{(1)})^2 + \frac1{24} f^{(4)} (g^{(1)})^4 $$
If you want to evaluate the honest derivative, that's fine too: label the $x$s so that you can distinguish between them, and demand that isomorphisms of diagrams preserve the labellings. Then no diagrams have nontrivial automorphisms (the automorphism group of any diagram is a subgroup of $S\_n$ acting on the $n$ $x$s), and it's still true that you should think in terms of the sum over all isomoprhism classes of diagrams.
| 22 | https://mathoverflow.net/users/78 | 5888 | 4,012 |
https://mathoverflow.net/questions/5889 | 2 | What's the correct mathematical name for the partial ordering on vectors based on what is sometimes called "Pareto Dominance"?
Does Pareto Dominance have an alternative name in fields other than economics?
For two vectors of the same dimension, one Pareto Dominates the other if all its elements are greater than the corresponding element in the other vector.
| https://mathoverflow.net/users/1875 | Is there an agreed name for partial ordering based on Pareto Dominance relation? | Depending on if you allow "greater" to mean "greater or equal" or to mean "strictly greater" we have two answers coming from Order Theory, respectively:
* The product order: <http://en.wikipedia.org/wiki/Product_order>
* The reflexive closure of the direct product of two strict total orders
Those two, along with the lexicographic order (which I searched to find this answer), are the three best-known choices among the possible orders on the Cartesian product of two totally ordered sets, as is stated in
<http://en.wikipedia.org/wiki/Total_order#Orders_on_the_Cartesian_product_of_totally_ordered_sets>
| 1 | https://mathoverflow.net/users/1234 | 5898 | 4,017 |
https://mathoverflow.net/questions/5897 | 6 | In Hartshorne p. 109 he defines a sheaf $\mathcal{F}$ of $O\_X$-modules to be locally free if there is an open cover of $X$, s.t. on each $U$, $\mathcal{F}|\_U$ is a free $O\_X|\_U$ module of rank $I$. Then if $X$ is connected, rank $I$ is globally well-defined. Here $(X,O\_X)$ is any ringed topological space (e.g. not necessarily the structure sheaf of a ring). A similar definition is here: [Locally free sheaf - Encyclopedia of Mathematics](https://encyclopediaofmath.org/wiki/Locally_free_sheaf).
However, it didn't seem obvious to me that if $V$ is a smaller open set included in $U$ (say $U$ connected), then the number of copies $J$ of $(O\_X|\_V)^J=\mathcal{F}|\_V$ would remain the same, because in general the restriction map of the sheaf $O\_X$ or $\mathcal{F}$ from $U$ to $V$ need be neither subjective nor injective, why would the index $J$ stay the same as $I$?
| https://mathoverflow.net/users/1877 | Why is the rank of a locally free sheaf well-defined? | Actually, there are two different restriction maps:
1. The first one (the one you correctly say is neither surjective nor injective in general) is that on **sections**: for $\mathcal{F}$ a sheaf on a scheme $X$ and two open subsets $V \subseteq U$, there is a map $\mathcal{F}(U) \to \mathcal{F}(V)$.
2. On the other hand, the inclusion map $i\_{VU}: V \to U$ induces a functor $i\_{VU}^{-1}$ from the category of sheaves on $U$ to the category of sheaves on $V$. This is called restriction of **sheaves**. By functoriality, $W \subseteq V \subseteq U$ yields $i\_{WV}^{-1}\circ i\_{VU}^{-1} = i\_{WU}^{-1}$, so the notation is usually shortened to just $ -|\_{W}$.
The second one is the one that you want to look at: the statement is then that $\mathcal{F}|\_{U} \cong \mathcal{O}\_X^{\oplus I}|\_{U}$ implies $\mathcal{F}|\_{V} \cong \mathcal{O}\_X^{\oplus I}|\_{V}$ for $V \subseteq U$.
| 19 | https://mathoverflow.net/users/1797 | 5905 | 4,020 |
https://mathoverflow.net/questions/5895 | 12 | * What are tame and wild hereditary algebras?
* Are they related to hereditary rings? (Those are rings for which every left (resp. right) ideal is projective, equivalently, for which every left (resp. right) submodule of a projective module is again projective).
* Googling them I can see they seem related to the "tame representation type", but this concept is also new to me.
* I would also like to know what is their relation to path algebras, since sometimes they appear mentioned together.
Do you know any good (newbie) references for all this? (Or can you elaborate in any of the questions?)
| https://mathoverflow.net/users/1234 | What are tame and wild hereditary algebras? | An $k$-algebra $A$ is *tame* (or, equivalently, it has tame representation type) if, for every dimension $d\geq0$, you can parametrize all isoclasses of indecomposable $A$-modules of dimension $d$, apart from a finite number of them, by a finite number of $1$-parameter families. On the other hand, a finite dimensional $k$-algebra $A$ is *wild* (or, equivalently, it has wild representation type) if in the category $\mathrm{mod}\_A$ of finite dimensional modules contains a copy of the category $\mathrm{mod}\_{k\langle x,y\rangle}$ of modules over the free $k$-algebra on two generators. It is an amazing theorem of Drozd that a finite dimensional algebra is either tame or wild; this is the so called dichotomy theory. One of the reasons that make this theorem so amazing is that one can show that if $A$ is wild then $\mathrm{mod}\_A$ contains copies of the module categories of *all* finite dimensional algebras; in other words, wild algebras are really wild...
In particular, this concepts of tame and wild apply to hereditary algebras, which are those of global dimension $1$.
Now, a finite dimensional hereditary algebra is Morita equivalent to the path algebra $kQ$ on a quiver without oriented cycles. A well-known theorem of Gabriel and others tells us that such a path algebra $kQ$ is tame iff the quiver $Q$ is, when you forget the orientation of the arrows, an Dynkin or an extended-Dynkin diagram. In all other cases the parth algebra is wild.
Two great references on all of this are the (first volume of the) book by Assem, Skowroński and Simson, or the book by Auslander, Reiten and Smalø.
| 32 | https://mathoverflow.net/users/1409 | 5906 | 4,021 |
https://mathoverflow.net/questions/5907 | 4 | Let $X$ be a set and let $f: X\longrightarrow X$ be a function on $X$. Introduce a topology on $X$ by the following basis of open sets: for any subset $S$ of $X$, let $B\_S$ be the set of forward images of $S$ under $f$; i.e. $$B\_S = \{f^n(s): s\in S, n\in \textbf{Z}^+\}.$$
My question is, is this topology well-known and well-understood? Is there a theory which relates properties of $f$ to the resulting topology?
| https://mathoverflow.net/users/960 | Is there a name for this topology? | This topology is rather combinatorial in nature. You certainly could call it well-understood, but I would not expect it to have a name in the context of topology. It is essentially a special case of a concept in combinatorics known as an [order ideal](http://en.wikipedia.org/wiki/Ideal_%28order_theory%29), or if you like the set of all order ideals. The set of all order ideals of a partially ordered set is indeed a topology, but not one that looks very topological. On the one hand, it does not satisfy any of the separation axioms other than $T\_0$. On the other hand, it satisfies an even stronger axiom than finite intersection: The intersection of any collection of open sets is open. The set of order ideals is better understood as a distributive lattice than as a topology, even though it is both.
We can clean things up a little as follows. First, if we switch to the image of $f$, we can let $B\_S$ instead be the union of images of $S$ including $S$ itself. Second, in each periodic orbit of $f$, any open set that contains one point contains the other one, so we might as well collapse that orbit to a point. After that, we can say that $x \prec y$ when $x = f^n(y)$. This defines a partial ordering on $X$ with respect to which the topology is just the set of order ideals.
| 13 | https://mathoverflow.net/users/1450 | 5910 | 4,024 |
https://mathoverflow.net/questions/5892 | 190 | If random variable $X$ has a probability distribution of $f(x)$ and random variable $Y$ has a probability distribution $g(x)$ then $(f\*g)(x)$, the convolution of $f$ and $g$, is the probability distribution of $X+Y$. This is the only intuition I have for what convolution means.
Are there any other intuitive models for the process of convolution?
| https://mathoverflow.net/users/812 | What is convolution intuitively? | I remember as a graduate student that Ingrid Daubechies frequently referred to convolution by a bump function as "blurring" - its effect on images is similar to what a short-sighted person experiences when taking off his or her glasses (and, indeed, if one works through the geometric optics, convolution is not a bad first approximation for this effect). I found this to be very helpful, not just for understanding convolution per se, but as a lesson that one should try to use physical intuition to model mathematical concepts whenever one can.
More generally, if one thinks of functions as fuzzy versions of points, then convolution is the fuzzy version of addition (or sometimes multiplication, depending on the context). The probabilistic interpretation is one example of this (where the fuzz is a a probability distribution), but one can also have signed, complex-valued, or vector-valued fuzz, of course.
| 228 | https://mathoverflow.net/users/766 | 5916 | 4,029 |
https://mathoverflow.net/questions/5249 | 16 | Everybody knows that there are $D\_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ [derangements](http://en.wikipedia.org/wiki/Derangement) of $\{1,2,\dots,n\}$ and that there are $D\_n(q)=(n)\_q! \left( 1-\frac{1}{(1)\_q!}+\frac1{(2)\_q!}-\frac1{(3)\_q!}+\cdots+(-1)^{n}\frac1{(n)\_q!} \right)$ elements in $\mathrm{GL}(q,n)$ which do not have $1$ as an eigenvalue; here $q$ is a prime-power, $(k)\_q!=(1)\_q(2)\_q\cdots(k)\_q$ are the $q$-factorials, and $(k)\_q=1+q+q^2+\cdots+q^{k-1}$ are the $q$-numbers.
Now, there are $D\_n^+=\tfrac12\bigl(|D\_n|-(-1)^n(n-1)\bigr)$ and $D\_n^-=\tfrac12\bigl(|D\_n|+(-1)^n(n-1))\bigr)$ even and odd derangements of $\{1,2,\dots,n\}$, as one can see, for example, by computing the determinant $\left| \begin{array}{cccccc} 0 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 0 & 1 & \cdots & 1 & 1 \\ 1 & 1 & 0 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 0 \end{array} \right|$ and looking at the result.
How should one define $D\_n^+(q)$ and $D\_n^-(q)$?
| https://mathoverflow.net/users/1409 | Derangements and q-variants | Both of Reid's suggestions sort-of work and lead to the same formula. However, it's easier to first change the question to a $q$-analogue of the difference
$$D^{\pm}\_n = D^+\_n - D^-\_n = (-1)^n(n-1).$$
(Also, the formula for $D\_n(q)$ is missing a factor of $q^{n(n-1)/2}$.)
You can find $D^{\pm}\_n$ by the inclusion-exclusion formula just as you can find $D\_n$. In fact, it's even easier since only the last two terms of inclusion-exclusion survive. This proof has a straightforward $q$-generalization, although the final formula isn't the same. But I don't know how likely the latter is in the $q$-linear algebra context. The determinant of a matrix seems like the reasonable $q$-analogue of the sign of a permutation, yet it is not a perfect analogue.
Let $\chi$ be a complex character of the multiplicative group of $\mathbb{F}\_q$. Let $D^\chi(q)$ be the corresponding sum of $\chi(\det M)$, summed over deranged matrices $M$. Then the $q$-analogue is that $D^\chi(q) = (-1)^nq^{n(n-1)/2}$ for all non-constant $\chi$.
The proof uses the Möbius function of the lattice of subspaces, just like for the ordinary enumeration of deranged matrices. This time only the last term of the Möbius inversion survives, the term for the identity matrix.
| 5 | https://mathoverflow.net/users/1450 | 5929 | 4,038 |
https://mathoverflow.net/questions/5893 | 10 | I have just created a presentation using beamer, and I want the
"one" command at the top of the file that creates a printable version. It is true that I can recompile having searched for all the \pause commands and percent them out, but I remember there is an elegant way of doing this.
For the record, I am using the Singapore style, and the talk will not be of much interest to mathematicians.
---
(*This has now been closed as "off-topic" with one of the reasons being the existence of [TeX-SX](http://tex.stackexchange.com). People interested in this question should therefore consider the question [Is there a nice way to compile a beamer presentation without the pauses?](https://tex.stackexchange.com/q/1423/86) on the TeX-SX site.*)
| https://mathoverflow.net/users/36108 | Beamer printout | The first answer is the basic correct answer, but there are variants.
1. Don't change the background colour (waste of ink), rather use pgfpages to put a border around each frame (this isn't one of the standard page-type declarations, but it isn't hard and I can make mine available if anyone wants it).
2. It's possible to change the type of the output (between beamer, handout, trans, or article) without modifying the file. The trick is to put the main document in one file, say `geometry.tex` but *without* the documentclass declaration. Then you create a new file for each type with just the documentclass declaration. For example, `geometry.beamer.tex` could contain:
```
\documentclass[12pt,t,xcolor=dvipsnames,ignorenonframetext]{beamer}
\input{geometry.tex}
```
whilst `geometry.handout.tex` might contain
```
\documentclass[12pt,xcolor=dvipsname,ignorenonframetext,handout,%
notes=only%
]{beamer}
\input{geometry.tex}
```
and `geometry.article.tex` might be
```
\documentclass[a4paper,10pt]{article}
\usepackage[envcountsect]{beamerarticle}
\setjobnamebeamerversion{geometry.beamer}
\input{geometry.tex}
```
Not only does this make sure that you are always compiling the correct version of the document, it also means that if you use a version control system then it doesn't keep complaining about you modifying the file just because you change the output type.
3. If you are strong in the ways of beamer and TeX, you can go one step further. I use beamer for lectures which means that one single file contains the beamer versions and the handout versions of nearly 30 lectures. To produce a given version of a given lecture, I need to have a way of telling TeX what I want. I could have 60 separate files all with variations on the above, but I've found a simpler way is to have TeX examine the jobname to determine this. Then I just have to have 60 symlinks to the main file (and I can create all 60 symlinks with a single `zsh` command). That is, `lecture.beamer.2009-11-19.tex` is a symlink to `lectures.tex` and when I run LaTeX on it then I get tomorrow's lecture in beamer format (well, I would if I'd written it yet). Again, I'd be happy to share the code for this if anyone's interested.
---
*Addendum 2012-11-13* Unsurprisingly, an early question on [TeX-SX](http://tex.stackexchange.com) was essentially exactly this one. Also unsurprisingly, I answered there as well and as that place is more suited to TeX answers than this one I incorporated this answer there. That answer can be found at <https://tex.stackexchange.com/a/1426/86> which also contains links to my code for the above.
| 10 | https://mathoverflow.net/users/45 | 5933 | 4,042 |
https://mathoverflow.net/questions/5954 | 49 | According to the [Norwegian meterological institute](http://www.yr.no/nyheter/1.6870525), the answer is that it is best to run. According to Mythbusters (quoted in the comments to that article), the answer is that it is best to walk.
My guess would be that this is something that can be properly modelled mathematically and solved. I suspect that I'm not alone in making a start on this, but as it's usually during a rainstorm that I think of it, I don't get far before deciding that the best strategy is to call a taxi!
I would also guess that someone has already done this and figured out the answer. If I'm right about that, I would like to read about it and learn what the real answer is. Thus my question is most likely: can someone provide a reference for this?
(If I'm wrong that it's already been solved, then some pointers as to how to solve it or even an explanation of what the main issues are, would be most interesting.)
As well as simply satisfying my curiosity, this is also one of those questions that non-mathematicians often ask (or at least, that non-mathematicians can easily understand the motivation for the question) so if there's a reasonable answer then it would be a good way to show off some mathematics.
---
**Update** This question has a long and curious history. There's a discussion about it at the meta site: <http://mathoverflow.tqft.net/discussion/90/question-about-walking-in-the-rain>
| https://mathoverflow.net/users/45 | Is it best to run or walk in the rain? | try this, the latest in a long line of recreational mathematics on the topic
"Keeping Dry: The Mathematics of Running in the Rain"
Dan Kalman and Bruce Torrence,
Mathematics Magazine, Volume 82, Number 4 (2009) 266-277
| 40 | https://mathoverflow.net/users/454 | 5962 | 4,062 |
https://mathoverflow.net/questions/5955 | 14 | May be it's not the right place for this, but I don't know the right definition of a strange attractor. Wikipedia states that "An attractor is informally described as strange if it has non-integer dimension or if the dynamics on it are chaotic." In the Tucker's paper about Lorenz system is written that "an attractor is called strange if for almost all pairs of different points in B( Λf ), their forward orbits eventually separate by at least a constant δ (depending only on Λf )." I feel that this definitions are not equivalent. I also would welcome links to useful literature.
| https://mathoverflow.net/users/1888 | Definition of a strange attractor. | This is a good question. For some reason, terminology in dynamical systems is not standardized at all--and it's interesting to disentangle various definitions. A good book to look at is [Differential equations, dynamical systems, and an introduction to chaos](http://books.google.com/books?id=GNOmchErrMgC). The authors (Hirsch, Smale, Devaney) are at the center of the field, and they point out there's no standard definition of even "attractor"! (Let alone "strange attractor," which they only use once, informally.)
In my view, definitions based on the shape of an attractor (like the first part of Wikipedia's) are a little odd, since you can have chaotic dynamics on geometrically simple attractors. Think of a map with a circle as an attractor on which the dynamics are like $\theta \mapsto 2\theta$.
The second Wikipedia definition, that the dynamics are chaotic, might imply the Tucker definition--but that in turn depends on your definition of chaos. There's interesting work on how various criteria for "chaos" relate. A good entry point might be ["On Devaney's Definition of Chaos"](http://www.jstor.org/pss/2324899) (Banks et al, American Math. Monthly, 1992).
| 15 | https://mathoverflow.net/users/1227 | 5972 | 4,070 |
https://mathoverflow.net/questions/5975 | 1 | Given two positive integers `a,b` what is the minimal integer `n`, so that there exist two positive integers `u,v` for which `n=au=av`?
It is easy to verify that `n=ab/gcd(a,b)`.
But what happens if instead of requiring `au=bv`, or `|au-bv|`≤`0`, we require that `|au-bv|`≤`k` for some number k?
That is, given two positive integers `a,b`, what are the minimal integers `u,v` for which `|au-bv|`≤`k`, for some `k`? If there's no direct formula, is there an easy way to find `u,v`?
| https://mathoverflow.net/users/55 | For which integers u,v does au=bv *approximately*? | A good heuristic is to compute the [continued fraction](http://en.wikipedia.org/wiki/Continued_fraction) of $a/b$ and drop the last few terms. The continued fraction will equal $u/v$ with $a/b-u/v$ very small. Since $a/b-u/v=(av-bu)/bv$, we will have $au-bv$ small.
| 2 | https://mathoverflow.net/users/297 | 5978 | 4,074 |
https://mathoverflow.net/questions/5915 | 8 | Here is a topological question which seems quite elementary. The answer to this question may be useful e.g. in estimating the orders of the automorphism groups of some algebraic varieties and in computing the cohomology of some moduli spaces.
Let $K$ be a compact Lie group, $T$ a maximal torus of $K$ and $V$ a complex vector space of dimension $d$. (If one wishes, instead of $K$ and $T$ one can consider a complex reductive group and its Borel subgroup.) From a representation $R:T\to GL(V)$ we can construct a homogeneous vector bundle with total space
$$E=K\times\_T V.$$
(I.e., we identify $(kt,v)$ with $(k,R(t)v$ for all $k\in K, t\in T, v\in V$.)
Set $E\_0$ to be $E$ minus the zero section. This space is fibered over $K/T$ with fiber $V$ minus the origin.
Suppose the Euler (=top Chern) class of the vector bundle is zero. Then $H^{2d-1}(E\_0,\mathbf{Z})\cong \mathbf{Z}$, because $K/T$ has only even-dimensional cells. Take a generator $a$ of $H^{2d-1}(E\_0,\mathbf{Z})$. We can identify the integral cohomology of $E\_0$ with $H^\*(K/T,\mathbf{Z})\otimes\Lambda(a)$ where $\Lambda$ stands for exterior $\mathbf{Z}$-algebra.
We have the natural group action map $K\times E\_0\to E\_0$. The cohomology of $E\_0$ has no torsion, so we can identify $H^\*(K\times E\_0,\mathbf{Z})$ with
$H^\*(K,\mathbf{Z})\otimes H^\*(E\_0,\mathbf{Z})$ using the projections. The latter can be written as $$H^\*(K,\mathbf{Z})\otimes H^\*(K/T,\mathbf{Z})\otimes\Lambda(a).$$
The pullback of $a$ under the action map is $1\otimes a+x\otimes 1$ for some $x\in H^{\*}(K,\mathbf{Z})\otimes H^{\*}(K/T,\mathbf{Z})$. Question: find $x$. The weights of the representation are assumed to be known; for simplicity one can assume that $K=U(n)$ or $SU(n)$.
For real cohomology I know how to reduce the problem to linear algebra using Lie algebra cohomology. But the result I am able to get in this way is not very illuminating. And moreover, I have no idea how to extract the integral structure out of it.
For projectivized (and not spherized) bundles, the question becomes trivial. But this does not seem to help.
| https://mathoverflow.net/users/2349 | Cohomology map induced by the group actions on homogeneous vector bundles | I'll answer in the case of K = U(n) because I'm less familiar with the cohomology of the other compact Lie groups. Let BK and BT be the classifying spaces.
The representation of T on V gives rise to a vector bundle on BT with total space F, and complement of the zero section F0. There is an Euler class c in H`*`(BT). As H`*`(BT) is a polynomial algebra in these cases, there are basically two cases: either the Euler class is zero, or it is not a zero divisor. There is a map from K/T to BT which is, up to homotopy, the "quotient" by the action of K.
In the first case, you will actually find that your element 'a' in H`*`(E0) is pulled back from H`*`(F0), where the action of K is trivial.
In the second case, you have an exact sequence (the Gysin sequence)
$$
0 \to H^\\*(BT) {\mathop\to^c} H^\\*(BT) \to H^\\*(F\_0) \to 0
$$
of modules over H\*(BK), which is actually a free resolution.
I find it easier to describe the "coaction" you want in terms of an action on homology. The space E0 is the pullback of F0 along the map from K/T to BT. An Eilenberg-Moore argument finds that there are isomorphisms
$$
H\\_\\*(K) \cong Ext\_{H^\\*(BK)} (\mathbb Z, \mathbb Z)
$$
and
$$
H\\_\*(E\\_0) \cong Ext\\_{H^\\*(BK)} (H^\\*(F\\_0), \mathbb Z).
$$
(I have to be a little careful about the gradings; these Ext-groups are bigraded according to a grading on the ring and a grading on the Ext-group, and the elements in Ext^s actually have their grading shifted down by s.)
The coaction on homology you describe is actually just the Yoneda pairing on Ext. For a specific choice of Euler class this is something that you can work out by writing down a resolution of **Z** over H`*`(BK) and using that to find generators for your Ext-algebra.
I don't know if an example would be more helpful.
| 5 | https://mathoverflow.net/users/360 | 5983 | 4,078 |
https://mathoverflow.net/questions/5790 | 9 | I seem to recall that there is a straightforward subfactor construction that yields fusion categories given by G-graded vector spaces and representations of G, for finite groups G. Is there an analogous construction for 2-groups?
Some background: A 2-group is a monoidal groupoid, for which the isomorphism classes of objects form a group. Sinh showed that up to monoidal equivalence, these are classified by a group G (isom. classes of objects), a G-module H (automorphisms of identity), and an element of H3(G,H). In the context of this discussion, we can limit our attention to G finite, H=Cx. One notable feature is that when the action of G on H is trivial, the three-cocycle twists the associator in the G-graded vector space category.
I'm mostly curious about how to tell when two elements of H3(G,H) yield Morita-equivalent fusion categories, and am wondering if subfactors or planar algebras make it easy to detect this.
| https://mathoverflow.net/users/121 | Is there a subfactor construction involving 2-groups? | This is a standard construction in Subfactor theory see the intro of <http://arxiv.org/abs/0811.1084v2> for details. The construction goes back a long long way (if I remember correctly both Vaughan Jones and Adrian Ocneanu's theses were related to this question, but I could be wrong there).
From a category theory perspective recall that a subfactor (N < M) is a unitary tensor category C (the N-N bimodules) together with a Frobenius algebra object A in C (M as an N-N bimodule with conditional expectation as trace). In this case the tensor category C is the twisted category of G-graded vector spaces (where you use the 3-cocycle to change the associator), and the algebra object is a twisted version of the group algebra (or maybe just the group algebra? I'm getting confused, shouldn't group algebras be twisted by 2-cocycles?).
| 6 | https://mathoverflow.net/users/22 | 5984 | 4,079 |
https://mathoverflow.net/questions/5198 | 2 | Is there a specific name for matrices with nonsingular principal submatrices?
| https://mathoverflow.net/users/1172 | Is there a specific name for matrices with nonsingular principal submatrices? | I've heard them called "strongly nonsingular matrices" in numerical linear algebra. Google that and you'll find some literature.
| 4 | https://mathoverflow.net/users/1898 | 5988 | 4,081 |
https://mathoverflow.net/questions/5986 | 14 | I want a good introduction to localization in equivariant $K$-theory. This introduction can be simple in several ways:
1. I only care about torus actions.
2. I only care about $K^0$.
3. I only care about very nice spaces. I would be fine if the only spaces considered were $G/P$'s and smooth projective toric varieties.
However, I want this exposition to include the following:
1. How to compute a $K$-class from a Hilbert series.
2. Given $X \to Y$ nice spaces, how describe push back and pull forward in terms of the map $X^T \to Y^T$.
3. Ideally, this reference would also give the generators and relations presentation of $K$-theory for the sort of examples mentioned above. I mean formulas like
$$K^0(\mathbb{P}^{n-1}) = K^0(\mathrm{pt})[t, t^{-1}]/(t-\chi\\_1)(t-\chi\\_2) ... (t - \chi\_n)$$
where $\chi\_i$ are the characters of the torus action. But I can find other references for this sort of thing.
The best reference I currently know is the appendix to [Knutson-Rosu](http://www.ams.org/mathscinet-getitem?mr=1970011).
| https://mathoverflow.net/users/297 | References for equivariant K-theory | I like the book by Chriss and Ginzburg (Representation Theory and Complex Geometry, <https://doi.org/10.1007/978-0-8176-4938-8>) very much, and I think it fits many of your requirements.
| 8 | https://mathoverflow.net/users/582 | 5991 | 4,083 |
https://mathoverflow.net/questions/5993 | 26 | What is known about the classification of n-transitive group actions for n large without using the classification of finite simple groups? With the classification of finite simple groups a complete list of all 2-transitive group actions is known, in particular there are no 6-transitive groups other than the symmetric groups and the alternating groups. I want something like "there are no interesting n-transitive group actions for n sufficiently large" but without the classification theorem (however, I'd be happy if the n in that statement was obscenely large). Even any partial (but unconditional) results would interest me (like any n-transitive group for n sufficiently large needs to have properties X, Y, and Z).
| https://mathoverflow.net/users/22 | Highly transitive groups (without assuming the classification of finite simple groups) | Marshall Hall's *The Theory of finite groups* only cites an asymptotic bound: a permutation group of degree n that isn't Sn or An can be at most t-transitive for t less than 3 log n. I suppose that was the state of the art at the time (late 1960s). There is an earlier paper by G.A. Miller on JSTOR that you can find by searching for "multiply transitive group".
There is a classical theorem of Jordan that classified sharply quadruply transitive permutation groups, i.e., those for which only the identity stabilizes a given set of 4 elements (from Wikipedia).
| 18 | https://mathoverflow.net/users/121 | 5996 | 4,085 |
https://mathoverflow.net/questions/5997 | 21 | One time I heard a talk about "the" random tree. This tree has one vertex for each natural number, and the edges are constructed probabilistically. Connect vertex $2$ to vertex $1$. Connect vertex $3$ to vertex $1$ or $2$ with probability $\frac{1}{2}$. Connect vertex $n+1$ to exactly one of vertices $1,\dots, n$ with equal probability $\frac{1}{n}$. This procedure will construct an infinite tree. The theorem is that with probability $1$, any tree constructed this way will be the same (up to permutation of the vertices).
My question is, does anyone know of a reference for this result? What is the automorphism group of this tree? Can anyone draw a picture of it?
I don't have any reason for knowing about this, just curiosity, and I wasn't able to turn up anything with a (not too extensive) internet/mathscinet search.
| https://mathoverflow.net/users/1345 | "The" random tree | I learned about [The random graph](http://en.wikipedia.org/wiki/Rado_graph) a week ago from [the blog post on n-cafe](http://golem.ph.utexas.edu/category/2009/11/fraisse%5Flimits.html) (thanks to Andrew and sdcvvc for the links!).
Your construction, while slightly different, can be examined in the same way as the Rado graph. Note, by the way, that expected number of edges meeting at each point is infinity, as is with Rado's graph.
**The lemma** you want to prove is, I think, this:
>
> Fix an isomorphism between two finite sets $A\to B$ and number $a\notin A$. Take random trees $X$, $Y$ having property that $X|\_A = Y|\_B$. Then with probability 1 there exists $b\notin B$ such that the isomorphism can be extended to $A\cup\{a\}$ and $B\cup\{b\}$.
>
>
>
A repeated application of this (or very similar) lemma should yield your result: there is an isomorphism between two random trees with probability 1.
---
Your trees have actually quite simple structure: with probability 1 all vertices have infinite degree. Any two *connected* trees with this property can be drawn starting from one point and adding infinitely many edges to it, then to leaves, then adding infinitely edges to each leaves and so on.
Sorry, the quality of the picture isn't good:
```
+ --- + --- + ...
| |
| + --- + ...
|
+ --- + --- + ...
| |
| + --- + ...
|
+ --- + --- + ...
| |
| + --- + ...
```
Since the trees you descrived are always connected (see comments), this is "the canonical drawing".
In particular, the stabilizer of a point has a filtration by groups of the form $S\_\infty$ and $(S\_\infty)^\infty$ (interchanging the "outer" edges of a given leaf is $S\_\infty$).
| 11 | https://mathoverflow.net/users/65 | 5999 | 4,087 |
https://mathoverflow.net/questions/6009 | 8 | What is the best algorithm for computing covariance that would be accurate for a large number of values like 100,000 or more?
| https://mathoverflow.net/users/812 | What's an efficient way to calculate covariance for a large data set? | Check out [How to calculate correlation accurately](http://www.johndcook.com/blog/2008/11/05/how-to-calculate-pearson-correlation-accurately/). There are two common formulas that are algebraically equivalent but one has much better numerical properties than the other.
| 7 | https://mathoverflow.net/users/136 | 6013 | 4,097 |
https://mathoverflow.net/questions/6019 | 18 | How would you calculate the order of a list of reviews sorted by "Most Helpful" to "Least Helpful"?
Here's an example inspired by product reviews on Amazon:
Say a product has 8 total reviews and they are sorted by "Most Helpful" to "Least Helpful" based on the part that says "x of y people found this review helpful".
Here is how the reviews are sorted starting with "Most Helpful" and ending with "Least Helpful":
7 of 7
21 of 26
9 of 10
6 of 6
8 of 9
5 of 5
7 of 8
12 of 15
What equation do I need to use to calculate this sort order correctly? I thought I had it a few times but the "7 of 7" and "6 of 6" and "5 of 5" always throw me off. What am I missing?
| https://mathoverflow.net/users/1901 | Calculating the "Most Helpful" review | See [How not to sort by average ranking](http://www.evanmiller.org/how-not-to-sort-by-average-rating.html).
| 27 | https://mathoverflow.net/users/297 | 6021 | 4,101 |
https://mathoverflow.net/questions/5813 | 8 | It seems that there is no g.r.r for stack yet according to dejong. Does anyone know anything about it? But as you might know, there are some complex manifold which is not scheme having atiyah singer index theorem. So I was wondering if there exists some analogue of g.r.r for special stack.
| https://mathoverflow.net/users/1851 | Is there any Grothendieck Riemman Roch theorem for general stack ? | If you work with the naive Chow-groups and allow non-representable morphisms the GRR-Theorem does not hold! In the paper by Toen quoted above and in some of the papers by Joshua there are explicit counterexamples. They always involve non-representable morphisms.
There are two ways to get around this.
The first is to modify the definition of the Chow-groups. This is what Toen does. He takes Chow groups with coefficients in the characters of the stack, which is quite an involved definition. But it leads to a GRR-theorem for DM-stacks.
The second approach by Joshua is modify the topology to keep track of the stabilizer groups. He introduces a topology which he calls the isovariant etale site, which is motivated by ideas of Thomason. This gives a different kind of Chow groups. For this recall that you can define the Chow groups as cohomology of some sheaves using higher K-theory. For stacks this was done by Gillet. You can then get new kinds of Chow groups by calculating the cohomology of these sheaves in the isovariant etale topology. In a series of papers Joshua proves GRR-theorems in this context.
| 12 | https://mathoverflow.net/users/473 | 6027 | 4,106 |
https://mathoverflow.net/questions/6026 | 7 | According to the Wikipedia page on [generalized continued fractions](http://en.wikipedia.org/wiki/Generalized_continued_fraction), $\pi$ can be given several GCF representations which have very regular structures; for example, one has the partial denominators as (1, 2, 2, 2, ...) and the partial numerators as (4, 12, 32, 52, 72, ...). My question is, can every real number be represented as a GCF that exhibits some sort of "structure"? I put that word in quotes because I'm not really sure how to define it concretely; ideally, the sequence of either partial numerators or partial denominators should be expressable using a predictable, explicit formula. Are there any theorems establishing results like this, or are there some numbers that have no GCF representation that is indistinguishable from random noise? If some numbers need to be expressed in other forms for a pattern to emerge, like Engel expansions, I'd be interested in knowing about that too.
| https://mathoverflow.net/users/1455 | Patterns in Generalized Continued Fractions | I think under any reasonable definition there will be only countably many explicit formulas or patterns. So in fact most reals can't be expressed this way. (See also the concept of ["computable number."](http://en.wikipedia.org/wiki/Computable_number))
| 15 | https://mathoverflow.net/users/1227 | 6031 | 4,109 |
https://mathoverflow.net/questions/6010 | -3 | Here is something that isn't yet very clear to me. Say, I've got a commutative ring A. I consider the affine scheme from A, so it's a sheaf of rings over Spec A.
**EDIT**: And additionally let's say Spec A is Hausdorff.
Now additionally let's say I know an A-module M and from that I can make a sheave of modules over O\_Spec(A), call it M~. All standard stuffs till now. But now I want to have one more information, namely I have an open subset of Spec(A), say U, that is dense in Spec(A). And I know additionally that the stalks M~\_x are isomorphic to O\_x for all x in U.. Can one conclude that M and A are A-module isomorphic? (if so can one follow the same argument for general schemes with modules over them?) What are the conditions by which one can conclude this?
| https://mathoverflow.net/users/1245 | Dense section of sheaves of modules | If U is an open set in X, but U isn't X, then there are non-zero sheaves on X whose support lies outside U. Now add O\_X to one of these to get a counterexample.
| 2 | https://mathoverflow.net/users/1384 | 6034 | 4,111 |
https://mathoverflow.net/questions/6033 | 6 | Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C\_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution. Suppose $e \in C\_c(G)$ is an idempotent, so that $H = eC\_c(G)e$ is an algebra with identity. Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well?
This may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$. *Now* is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well?
[I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$. So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction.]
| https://mathoverflow.net/users/379 | Inverses in convolution algebras | I don't have a solution, but here are some thoughts which might be of use or interest.
You may have seen this already, but if your group is *discrete* then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. I think this property is inherited by the algebra obtained when one compresses by an idempotent in $C\_c(G)$.
The earliest reference I know of is somewhere in Kaplansky's *Fields and Rings*; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in
Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc. 75 1969 539--540. MR0238967 (39 #327)
(The proof uses the existence of a faithful tracial state on $VN(G)$, plus the fact that every idempotent in a $C^\*$-algebra is similar in the algebra to a self-adjoint idempotent -- something which was not all that obvious to me the first time I saw this result.)
I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^\*$-algebras of *some* totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments.
Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind.
| 4 | https://mathoverflow.net/users/763 | 6044 | 4,115 |
https://mathoverflow.net/questions/6006 | 9 | Any map of finite graphs (1-dimensional CW-complexes) factors as a composition of
1. a finite sequence of folds;
2. an inclusion; and
3. a finite-to-one covering map.
There should be a corresponding result for handlebodies, which presumably should say that, after a homotopy, a continuous map of handlebodies factors as:
1. a compression (by which I mean a map of a handle into the complement of its interior);
2. an inclusion; and
3. a finite-to-one covering map.
Is my intuition correct, and does anyone have a reference? I'm specifically interested in how well-behaved the homotopy can be taken to be. For instance, can it be made to respect the boundary?
**Notes**
A *fold* is a map that identifies two edges with a common endpoint. Many folds don't change the homotopy type of a graph, and one would expect not to need these in the handlebody setting. The important folds are the ones that kill a loop. In handlebody terms, you can think of this as gluing in a two-handle, or as cutting a one-handle - hence my use of the word "compression". Is this word acceptable in this context?
The graph-theoretic result is due to [Stallings](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=RT&pg7=ALLF&pg8=ET&review_format=html&s4=stallings&s5=finite%20graphs&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=695906).
By an *inclusion* of handlebodies, I mean that the new one should be obtained from the old by attaching 1-handles.
**EDIT** (prompted by Sam's comments below) I'm not quite sure what "respect the boundary" should mean, at this point. Suggestions welcome!
| https://mathoverflow.net/users/1463 | Factoring maps of handlebodies | Suppose that we are given a PL map from a handlebody $W$ to a handlebody
$V$. Choose a spine for $W$. Homotope the map until the image is a
regular neighborhood of the image of the spine. By general position,
our map is now an embedding. Fix a pants decomposition of disks $D =
(D\_i)$, for $V$. Suppose that $P$ is a component of $V - D$ (so $P$ is a
three-ball with three distinguished disks on its boundary). Consider a
component $X$ of $f(W)
\cap P$. This is essentially a knotted graph. Via a homotopy (keeping
$X \cap D$ fixed) unknot $X$. If the rank of $X$ is positive then another
homotopy produces small lollipops which we shall compress a bit later.
If $X$ meets any disk $D\_i \subset \partial P$ more than once
then we may homotope a leg of
$X$ through $D\_i$. Let $Y$ be the resulting component of $f(W) \cap
P$. (Note that this reduces $f(W) \cap D$.)
Homotoping in this fashion we eventually arrive at an embedding of $W$
so that every component in every three-ball of $V - D$ is either a
tripod or an interval, possibly with lollipops attached. The feet of
the tripod/interval lie in distinct disks in the boundary of the
containing solid pants.
Now compress all of the lollipops to get $f'(W')$ (a new handlebody,
because we compressed and a new map because we have to extend it over
the two-handles we added).
EDIT:
This reproduces, in our context, part of Stallings paper (eg sliding
the leg is a fold, arriving at only tripods and intervals produces
an immersion.)
Since $f'$ is an immersion, it follows from Stallings paper that $f'$
is $\pi\_1$ injective and that $W'$ embeds into a finite cover of $V$.
| 4 | https://mathoverflow.net/users/1650 | 6047 | 4,116 |
https://mathoverflow.net/questions/6043 | -1 | I need a way to split output pdf-file (a book) into chapters on such a way that cross-references will survive.
A simple example with a **solution** (based on answers below) can be found [here](http://www.math.psu.edu/petrunin/papers/splitting-pdf-latex/)
| https://mathoverflow.net/users/1441 | Splitting book into chapters | If you are using a book documentclass, then you can create a template file, with \include{chapter3} or \includeonly{chapter2}. I don't know if this works particularly. A hacky way to do it is to save your .aux file as say a .auk file, process, and then write over the .aux file with the saved version.
If you are on a mac, and have leopard or snow leopard, you can open files in preview and drag sets of pages into different preview windows, then you can save those files while editing others. Again it's a hack, but it works.
| 6 | https://mathoverflow.net/users/36108 | 6049 | 4,118 |
https://mathoverflow.net/questions/6052 | 2 | I read the following fact: if $U$ is an open subset of $P\_k^1$ and $f: U \to U$ is an automorphism of schemes, then $f$ extends to an automorphism of $P\_k^1$. Thus I was curious: is there a general criteria for when a continuous map defined on an open subset $U \subset X$ extends to $X$ (especially in non-Hausdorff settings, preferably including the aforementioned case)?
| https://mathoverflow.net/users/344 | How do we know that a map $f: U \to Y$ extends to $\bar{U}$? | Your example is in the category of schemes, but you then ask about maps of topological spaces.
Assuming you care about schemes, the condition is that the source is one dimensional and regular, and the target is proper. That you can extend in this case is essentially the valuative criterion of properness; that you can't extend in any other case is basically the valuative criterion of properness plus Scott's argument.
If you care about topological spaces, I don't know that there is any good criterion. Having the source be a one dimensional manifold and the target compact is not enough: Consider the line segment $(0,1)$ mapping to a closed disc in $\mathbb{R}^2$ by $x \mapsto (x, \sin 1/x)$; you can't extend to $[0,1]$. (Take the closed disc is large enough to contain the image of the map.)
| 5 | https://mathoverflow.net/users/297 | 6056 | 4,123 |
https://mathoverflow.net/questions/6020 | 2 | I'm not the person to understand everything in *[Geometric Endoscopy and Mirror Symmetry](http://arxiv.org/abs/0710.5939)*, but some parts of it are reasonably clear to me.
In particular, one of the main objects, mathematically speaking, is the category of coherent sheaves on an orbifold point $\mathrm{pt}/\Gamma$, where $\Gamma$ is a finite group of automorphisms of some ${}^LG$-local system. This category is well-known in algebraic geometry to be just $\mathrm{Rep}(\Gamma)$.
The main point of the paper is that some other, less obvious, additive category happens to be isomorphic to this well-known $\mathrm{Rep}(\Gamma)$. This means, in particular, that its objects are actually sums of things like $R\otimes V\_R$ where $R$ goes over irreps of $\Gamma$.
But (9.5), (9.8) (numbers from the [version 3](http://arxiv.org/pdf/0710.5939v3)) are different:
$${\mathcal F}\_{\mathrm{Reg}(\Gamma)} = \bigoplus\_{R \in \mathrm{Irrep}({}^LG)} R^\* \otimes
{\mathcal F}\_R$$
Note the sum goes by $\mathrm{Irrep}({}^LG)$ where I thought $\mathrm{Irrep}(\Gamma)$ is appropriate. The source of the chain of equations seems to be on page 112, where, quote, the regular representation
$$\mathrm{Reg}(\Gamma) = \bigoplus\_{R \in \mathrm{Irrep}({}^LG)} R^\* \otimes R,$$
unquote. Thus I've decided it's a typo in the formula for the regular representation carried over for a next several pages. Yet I feel out of place until I'm completely sure there is no other explanation — theoretically there *could* be some relationship between the representations of $\Gamma$ and those of ${}^LG$, after all.
>
> **Question:** do these two formulas have a typo or is there a meaning I miss?
>
>
>
| https://mathoverflow.net/users/65 | Understanding formula in Frenkel-Witten | It's a typo. R always denotes representations of Gamma, and representations of LG are named with a V.
| 2 | https://mathoverflow.net/users/121 | 6059 | 4,125 |
https://mathoverflow.net/questions/6057 | 25 | [*Edit (June 20, 2010):* I posted [an answer](https://mathoverflow.net/a/28891/14094) to this which summarizes one that I received verbally a few weeks after posting this question. I hope it is useful to someone.]
I am presently seeking references which introduce "formal geometry". So far as I can tell, this idea was presented by I.M. Gel'fand at the ICM in Nice in 1970. There is his lecture, a paper by him and Fuks, and also a paper by Bernshtein and Rozenfeld with some applications that I don't understand too well. What I am unable to find is a thorough exposition of the foundations. It seems like a canonical enough construction that it should have been included in some later textbook, (though apparently not called "formal geometry" since that is not turning up anything useful).
Below is what I understand, which is several main ideas, but missing many details; this will most certainly be riddled with errors, because I am only able to give what I have roughly figured out from reading incomplete (though well-written and interesting!) sources, and asking questions. I am including it in the hopes it will be familiar to some kind reader.
>
> I'm sorry to not ask a specific question. Hopefully some answers will help me edit the below description to remove inaccuracies, and some others will suggest references. Both would be very helpful.
>
>
>
Let $X$ be a smooth complex algebraic variety of dimension $n$ (could just as well be a complex analytic or smooth real manifold so far as I understand; probably can be algebraic over any field, at least for awhile). There is a completely general torsor over $X$: its fiber over a point $x$ is the set of all coordinate charts on the formal neighborhood of $x$ in $X$. This is a torsor over the infinite dimensional group G of algebra endomorphisms of $\mathbb{C}[\![x\_1,...,x\_n]\!]$ which preserve the augmentation ideal and are invertible modulo quadratic terms (and hence invertible over power series of endomorphisms). It's a torsor because any two coordinate systems are related by such an endomorphism, but there isn't a canonical choice of coordinate system along the variety.
I think one can rephrase the conditions of the previous paragraph more precisely by first noting that an endomorphism of $\mathbb{C}[\![V]\!]$ preserving augmentation ideal (where we use notation $V=\operatorname{span}\_\mathbb{C}(\{x\_1,...,x\_n\})$ is given by a linear map $V\to V\ast \mathbb{C}[\![V]\!]$, which then uniquely extends to an algebra map. Then the condition of the last paragraph is that
$$V \to V\ast \mathbb{C}[\![V]\!]\to V\ast \mathbb{C}[\![V]\!] / V\ast V\ast \mathbb{C}[\![V]\!] = V$$ is invertible.
It's not hard to see that these in fact form a group, and that this group acts simply transitively on the set of coordinate systems.
The Lie algebra $\mathfrak{g}$ of $G$ (once one makes sense of this) is a subalgebra $W^0$ (described below) of the Lie algebra $W\_n$ of derivations of $\mathbb{C}[\![x\_1,...,x\_n]\!]$. $W\_n$ is the free $\mathbb{C}[\![x\_1,...,x\_n]\!]$-module generated by $\partial\_1,\dots,\partial\_n$, with the usual bracket.
$W=W\_n$ has a subalgebra $W^0$ of vector fields which vanish at the origin (i.e. constant term in coefficients of ∂i are all zero), and another $W^{00}$ of vector fields which vanish to second order (so constant and linear terms vanish). It's fairly clear that $W^0/W^{00}$ is isomorphic to $\mathfrak{gl}\_n$. One now considers W\_n modules which are locally finite for the induced $\mathfrak{gl}\_n$-action. It turns out that these can be "integrated" to the group $G$, because $G$ is built out of $\operatorname{GL}\_n$ and a unipotent part consisting of those endomorphisms which are the identity modulo $V\ast V$. So the integrability of the $\mathfrak{gl}\_n$-action is all one needs to integrate to all of $G$.
Now one performs the "associated bundle construction" in this context, to produce a sheaf of vector spaces out of a W\_n module of the sort above. One could instead start with a f.d. module $V$ over $\mathfrak{gl}\_n$, and there's a canonical way to turn it into a $W\_n$-module (in coordinates you tensor it with $\mathbb{C}[\![x\_1,...,x\_n]\!]$ and take a diagonal action: W\_n acts through $\mathfrak{gl}\_n$ on the module $V$ and by derivations on $\mathbb{C}[\![x\_1,...,x\_n]\!]$). The sheaves you get aren't a priori quasi-coherent; some can be given a quasi-coherent structure (i.e. an action of the structure sheaf on X) and some can't. However, the sheaves you get are very interesting. By taking the trivial $\mathfrak{gl}\_n$-bundle you get the sheaf of smooth functions on the manifold (this was heuristically explained to me as saying that to give a smooth function on a manifold is to give its Taylor series at every point, together with some compatibilities under change of coordinates, which are given by the $W\_n$-action). By taking exterior powers of $\mathbb{C}^n$ you recover the sheaves of differential forms of each degree (these examples can be made into quasi-coherent sheaves in a natural way). The $W\_n$-modules associated to the exterior powers are not irreducible; they have submodules, which yield the subsheaves of closed forms (these give an example of a sheaf built this way which isn't quasi-coherent: function times closed form isn't necessarily closed).
Finally, one is supposed to see that the existence of the extra operators $\partial\_i$ of W which aren't in $W^0$ further induce a flat connection on your associated bundle. I don't yet understand the underpinnings of that, but it's very important for what I am trying to do.
>
> Is this familiar to any readers? Is there a good exposition, or a textbook which discusses the foundations? Can anyone explain the last paragraph to me?
>
>
>
| https://mathoverflow.net/users/1040 | Formal geometry | This was briefly discussed [previously on this site](https://mathoverflow.net/questions/518/what-is-the-connection-between-d-modules-and-coordinate-bundles/523). You can find discussions of it in:
1. Chapters 6 and 17 of Frenkel and Ben-Zvi: *Vertex algebras and algebraic curves.*
2. Section 2.6.5 of Beilinson and Drinfeld's *Quantization of Hitchin's Integrable system* (available online)
3. Section 2.9.9 of Beilinson and Drinfeld's *Chiral Algebras*.
| 8 | https://mathoverflow.net/users/121 | 6060 | 4,126 |
https://mathoverflow.net/questions/6061 | 9 | I am currently trying to apply some results from Choquet theory - i.e., the generalisation of results by Minkowski and Krein-Milman for representing points in a compact, convex set C by probability measures over its extreme points, ext C = { x ∈ C : C - { x } is convex }.
My main problem is with explicitly describing the set of extreme points for a particular convex set, namely the set C of concave functions over the k-simplex that vanish at the vertices of the simplex and have sup-norm at most 1. I've convinced myself that this set of functions in compact and convex and so the Choquet's theorem applies. However, apart from the case of the 1-simplex I am struggling to say anything about what the extreme functions might be.
In the case of the 1-simplex, the functions in ext C are "tents" with height 1, that is, functions f that are zero on the boundaries and rise linearly to a single point x where f(x)=1. I suspect that in the case of the 2-simplex the extreme functions are also piece-wise linear concave functions with height 1. I have considered a number of candidates (the functions formed by the taking the minimum of 3 affine functions, each zero on a different vertex) but am having trouble showing that the candidates are actually extreme.
Does anyone know of any techniques for identifying extreme points of convex sets?
Pointers to applications of Choquet's theorem that explicitly construct ext C and the probability measure for a given point in C would also be much appreciated. My reading in this area has only got me as far as Phelps' monograph "Lectures on Choquet Theory" and a survey article by Nina Roy titled "Extreme Points of Convex Sets in Infinite Dimensional Spaces".
| https://mathoverflow.net/users/1915 | Explicitly describing extreme points of infinite dimensional convex sets | There are more candidates for the extreme points. Take any compact convex subset of the simplex, then take the infimum of all affine functions that are ≥ the characteristic function of the subset. You could start with more arbitrary subsets, but the result is the same.
As far as general techniques go, I don't have any, except when the set is given by inequalities, try to make as many of them equalities as you can. In this case, make your functions equal to 1 or to 0 as much as possible, and as affine as you can make them.
By the way, don't you have a problem with compactness? Take tents on points $x\_n$ and let $x\_n$ converge to a vertex of the simplex.
**Edit:** I realized that you wanted a proof. Let *S* be the simplex, let *K* be a closed convex subset (not containing a corner of *S*), and let *f* be the inf of all affine functions ≥ 0 on *S* and ≥ 1 on *K*.
Assume $2f=g+h$. Then *g* and *h* are both equal to 1 on *K*, and since they are concave they are infimums of affine functions, and so both are $\ge f$. But then they must both be equal to *f*.
| 7 | https://mathoverflow.net/users/802 | 6063 | 4,128 |
https://mathoverflow.net/questions/5739 | 8 | Hi,
I have recently got interested in multi-index (multi-dimensional) Dirichlet series, i.e. series of the form $F(s\_1,...,s\_k)=\sum\_{(n\_1,...,n\_k)\in\mathbb{N}^k}\frac{a\_{n\_1,...,n\_k}}{n\_1^{s\_1}...n\_k^{s\_k}}$. I found some papers suggesting that multi-index Dirichlet series are in fact a distinct subfield for itself within analytic number theory. So, I´m now looking for some 'basic' learning materials/books or similar on this subject.
Any suggestions are greatly appreciated!
efq
PS: I believe I have already checked most books on multi-dimensional complex analysis/several complex variables.
| https://mathoverflow.net/users/1849 | multi-index Dirichlet series | De la Breteche proved recently a Tauberian theorem for multiple Dirichlet series (MR1858338 (2002j:11106)). This is useful stuff in applications. It fails shortly of proving the main result in Balazard, et. al recent paper: <http://iml.univ-mrs.fr/~balazard/pdfdjvu/19.pdf> (but does so assuming the Riemann Hypothesis). Finally Daniel Bump (look up his homepage on google) did a lot of work on multiple Dirichlet series - unfortunately I am not familiar with any of it - it also seems to have a more algebraic flavor to it.
P.S: It is remarkable that De La Breteche avoids using several complex variables.
| 3 | https://mathoverflow.net/users/1919 | 6065 | 4,130 |
https://mathoverflow.net/questions/6070 | 84 | I know (at least I think I know) that some of the main motivating problems in the development of etale cohomology were the Weil conjectures. I'd like to know what other problems one can solve using the machinery of etale cohomology. I know a little bit about how etale cohomology groups appear in algebraic number theory but I'd like to know about ways that these things show up in other mathematical subjects as well. Is there anything that an algebraic topologist should really know about etale cohomology? What about a differential geometer?
| https://mathoverflow.net/users/493 | Etale cohomology -- Why study it? | $\DeclareMathOperator{\gal}{Gal}$
Here's a comment which one can make to differential geometers which at least explains what etale cohomology "does". Given an algebraic variety over the reals, say a smooth one, its complex points are a complex manifold but with a little extra structure: the complex points admit an automorphism coming from complex conjugation. Hence the singular cohomology groups inherit an induced automorphism, which is extra information that is sometimes worth carrying around. In short: the cohomology of an algebraic variety defined over the reals inherits an action of $\gal(\mathbb{C}/\mathbb{R})$.
The great thing about etale cohomology is that a number theorist can now do the same trick with algebraic varieties defined over $\mathbb{Q}$. The etale cohomology groups of this variety will have the same dimension as the singular cohomology groups (and are indeed isomorphic to them via a comparison theorem, once the coefficient ring is big enough) but the advantage is that that they inherit a structure of the amazingly rich and complicated group $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$. I've often found that this comment sees off differential geometers, with the thought "well at least I sort-of know the point of it now". A differential geometer probably doesn't want to study $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$ though.
If I put my Langlands-philosophy hat on though, I can see a huge motivation for etale cohomology: Langlands says that automorphic forms should give rise to representations of Galois groups, and etale cohomology is a very powerful machine for constructing representations of Galois groups, so that's why I might be interested in it even if I'm not an algebraic geometer.
Finally, I guess a much simpler motivating good reason for etale cohomology is that geometry is definitely facilitated when you have cohomology theories around. That much is clear. But if you're doing algebraic geometry over a field that isn't $\mathbb C$ or $\mathbb R$ then classical cohomology theories aren't going to cut it, and the Zariski topology is so awful that you can't use it alone to do geometry---you're going to need some help. Hence etale cohomology, which gives the right answers: e.g. a smooth projective curve over *any* field has a genus, and etale cohomology is a theory which assigns to it an $H^1$ of dimension $2g$ (<pedant> at least if you use $\ell$-adic cohomology for $\ell$ not zero in the field <\pedant>).
| 120 | https://mathoverflow.net/users/1384 | 6076 | 4,135 |
https://mathoverflow.net/questions/6082 | 2 | We have a probability game, where we have $N$ number of events, each of which outcome can be $A,B$ or $C$. We do/will NOT know real probabilities afterwards: only the discrete outcome ($A, B$ or $C$) of each event.
Player 1 forecasts these events with certain probabilities (not only guess what is the outcome, but gives probabilities for each outcome option), and Player 2 as well with own probability estimates.
How we can know how accurate Player 1 and Player 2's predictions were (relation to reality) and how to measure the accuracy?
I have heard that one can use Akaike's information criterion to solve the problem. I was wondering another way but I need expert's opinion if this works:
I heard that one can start solving the problem by modeling the process by multinomic distribution and then take its Dirichlet's distribution. But how this leads to a solution?
Okay, I agree that one can write the solution like "Take some Dirichlet's distribution. Now use Akaike's information criterion". But I would like to know if this problem can be solved by using Dirichlet's distribution in some relative reasonable way so that you can't remove the distribution argument and the solution is still valid.
| https://mathoverflow.net/users/1928 | Which fortuneteller is better | I could be missing something here but I'd compute
P[These events occur|Fortune teller 1 is telling the truth] and P[These events occur|Fortune teller 2 is telling the truth].
More explicitly
Let N\_a, N\_b and N\_c be the number of times outcomes A,B and C occured. Then
P(This happened given fortune teller 1 told the truth)= (N choose a,b,c)[(P\_1,a)^N\_a][(P\_1,b)^N\_b][(P\_1,c)^N\_c]
Where P\_1,a is the probabiliy fortune teller one assigns to event a. (similiar for b and c).
Do the same computation for fortuneteller 2 and compare these 2. This tells us who assigned a higher probability to the outcome that came out. Suppose we might have some other criteria for "better" though (who is more often "close" for example but doubt it'll change the answer for "reasonable" definitions of "close").
| 1 | https://mathoverflow.net/users/1000 | 6087 | 4,141 |
https://mathoverflow.net/questions/6079 | 45 | I would like to study/understand the (complete) classification of compact lie groups. I know there are a lot of books on this subject, but I'd like to hear what's the best route I can follow (in your opinion, obviously), since there are a lot of different ideas involved.
| https://mathoverflow.net/users/1049 | Classification of (compact) Lie groups | First, here's a rough outline of how the classification works:
1. Prove that if G and H are simply connected and have the same Lie algebra, then G and H are isomorphic as Lie groups.
2. Prove that if G is any Lie group, its universal cover $\tilde{G}$ inherits a natural Lie group structure for which G = $\tilde{G}/Z$ where $Z\subseteq Z(\tilde{G})$.
This reduces classification to a) understanding the Lie algebras and b) understanding the centers of simply connected Lie groups.
`3.` Classify (simple) Lie algebras. This is done via root diagrams (Dynkin diagrams).
`4.` For each simply connected compact Lie group, compute its center.
For references, I'd check out Fulton and Harris' book "Representation Theory". I'm not sure if it actually does 4., but that's a fairly easy exercise afterwards (except for perhaps the exceptional groups).
| 43 | https://mathoverflow.net/users/1708 | 6095 | 4,145 |
https://mathoverflow.net/questions/6089 | 5 | Propositional dynamic logic (PDL) is an example of a (multi)modal logic with a structure on the set of modalities. In particular, the set of its modalities is indexed by "programs" and one can use program constructors such as composition, choice and iteration to make new programs out of old (of course, there is a set of basic programs to begin with). So if π and ρ are programs, so are π;ρ, π∪ρ and ρ\*. The intuitive interpretations are, respectively, "run π, then run ρ", "nondeterministically run either π or ρ", and "run ρ finite number of times".
The formulas in this language are formed from propositional letters using boolean connectives. Also, if π is a program and φ a formula, then <π>φ is a formula with intuitive interpretation "there is a state of computation accessible by program π in which φ is satisfied". Modality <π> is read "diamond π".
Not to dwell any deeper right now, I refer you to a very well written [article about PDL](http://plato.stanford.edu/entries/logic-dynamic/ "Propositional dynamic logic").
Now to get to my question. PDL was created in a line of formal systems that were ment to be used to "talk about" programs, prove program correctness etc. Its predecessors were Hoare logic (HL, also known as Floyd-Hoare logic) and its modal version, the dynamic logic (DL). But, since PDL is propositional, it lacks the expressiveness given by first-order constructions found in HL and DL. So, somehow I haven't really been able to find any real example of its original use.
Does anyone know an example of a program and its specifications that I can express and verify for correctness in PDL? I would be happy with any academic or real-world example, or any reference of such an example.
I would even be happy with an example of use of PDL in other areas. Most books that I've seen mention its use in philosophy, artificial intelligence, linguistics etc., but never give an example of this use.
| https://mathoverflow.net/users/1716 | Applications of propositional dynamic logic | The practical applications might be more obvious once you observe that these propositional "programs" are *regular expressions* -- which is to say, state machines. So you can expect it to have applications in the study of things [like program analyses](http://www.cs.cornell.edu/~kozen/papers/opti.pdf) and verifying concurrent protocols.
[Dexter Kozen](http://www.cs.cornell.edu/~kozen/papers/papers_expanded.htm) at Cornell has done a great deal of work in this area. In fact, he's mostly focused on a subsystem of PDL, called ["Kleene algebra with tests"](http://www.cs.cornell.edu/~kozen/papers/kat.ps), which has an easier decision problem (PSPACE rather than EXPTIME) and tends to have nicer equational proofs.
| 5 | https://mathoverflow.net/users/1610 | 6098 | 4,148 |
https://mathoverflow.net/questions/6093 | 5 | Does it help to learn statistical mechanics or thermodynamics (as in physics or mathematical physics) in order to learn thermodynamic formalism: the study of equilibrium states, Gibbs measure, maximal measures mostly on shift spaces or ℤn?
I've not taken a course on thermodynamics, but so far my learning of the concept of Gibbs states, pressures, equilibrium states on shift spaces seems to be going fine. But then maybe I'm missing physical intuitions that may be necessary later.
| https://mathoverflow.net/users/1354 | Does it help to learn statistical mechanics in order to learn thermodynamic formalism? | If you're learning thermodynamic formalism in order to apply it to physics then it's fairly clear that you ought to learn the physical context, so I infer from this that you are more interested in using it either to understand dynamical systems, or for the sake of its applications to other areas of mathematics entirely (such as those explored by Pollicott, Baladi, Lalley, Urbanski, Sharp, and so on). Of the twenty or so people I know who are active in thermodynamic formalism, only two or three have any background whatsoever in mathematical physics - so I think it's safe to say that understanding the physical background isn't necessary. For example, I've learned enough thermodynamic formalism to write a paper on it, and I know no statistical mechanics whatsoever.
| 5 | https://mathoverflow.net/users/1840 | 6109 | 4,155 |
https://mathoverflow.net/questions/6088 | 6 | Hey
A friend and I are thinking of having an algebraic statistics seminar next semester. Does anyone know of a good book to try learn it out of?
| https://mathoverflow.net/users/1000 | Algebraic Statistics textbook | I've heard that [the book of Sturmfels and Pachter](http://bio.math.berkeley.edu/ascb/) is supposed to be good. But it's a bit slanted towards biological applications, which may or may not be what you're into.
| 3 | https://mathoverflow.net/users/143 | 6119 | 4,161 |
https://mathoverflow.net/questions/6111 | 10 | Does anyone know of a global proof (involving no local argument) of Serre duality at the level of varieties or manifolds (as opposed to schemes).
| https://mathoverflow.net/users/1095 | Global proof of Serre duality | You might like the proof in section 5.3 of Voisin's book *Hodge theory and complex algebraic geometry*.
| 9 | https://mathoverflow.net/users/297 | 6120 | 4,162 |
https://mathoverflow.net/questions/6108 | 9 | Let $S$ the blow up of $P^2$ in nine points. Why is the anticanonical divisor $-K\_S$ not semiample?
| https://mathoverflow.net/users/1937 | Anticanonical divisor of the blow up of P^2 in 9 points | Your nine points must be [**EDIT: very**] general [see MP's answer], otherwise it *can* be semiample.
The only effective anticanonical divisor is then (the strict transform of) the cubic C through the nine points. Since there is no other cubic curve cutting C in your nine points, the restriction of -K\_S to C is a noneffective divisor of degree 0 (C has genus 1). So the restriction of -mK\_S is also noneffective for all m (the points are [**very**] general in C! [I guess] Torsion points can make a difference) which means the only effective divisor in -mK\_S is mC. Thus -mK\_S is never base point free.
| 10 | https://mathoverflow.net/users/1939 | 6121 | 4,163 |
https://mathoverflow.net/questions/6074 | 71 | What's the relationship between Kahler differentials and ordinary differential forms?
| https://mathoverflow.net/users/1867 | Kahler differentials and Ordinary Differentials | Let $M$ be a differentiable manifold, $A=C^\infty (M)$ its ring of global differentiable functions and $\Omega^1 (M)$ the A-module of global differential forms of class $C^\infty$.
The A-module of Kähler differentials $\Omega\_k(A)$ is the free A-module over the symbols $db$ ($b \in A$) divided out by the relations
$d(a+b)=da+db,\quad d(ab)= adb+bda,\quad d\lambda=0 \quad(a,b\in A, \quad \lambda \in k)$
There is an obvious surjective map $\quad \Omega\_k(A) \to \Omega^1 (M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule).
However, I do not believe at all that it is injective.
For example, if $\: M=\mathbb R$, I see absolutely no reason why $\mathrm{d}\sin(x)=\cos(x)\mathrm{d}x$ should be true
in $\Omega\_k(A) $ (beware the sirens of calculus). Things would be worse if we considered $C^\infty$
functions which, unlike the sine, are not analytic.
The same sort of reasoning applies to holomorphic manifolds and also to local rings of differentiable or holomorphic functions on manifolds.
To sum up: the differentials used in differentiable or holomorphic manifold theory are a quotient of the corresponding Kähler differentials but are not isomorphic to them.
(And I think David's claim that they are isomorphic is mistaken.)
| 64 | https://mathoverflow.net/users/450 | 6138 | 4,178 |
https://mathoverflow.net/questions/6132 | 24 | Is it true that *any finitely presented group can be realized as fundamental group of compact 3-manifold **with boundary***?
| https://mathoverflow.net/users/1441 | Fundamental group of 3-manifold with boundary | A couple of extra points.
Any compact 3-manifold with boundary $M$ can be doubled to give a closed 3-manifold $D$. As $M$ is a retract of $D$, it follows that $\pi\_1(M)$ injects into $\pi\_1(D)$. Therefore, any "poison subgroup" (such as the Baumslag--Solitar groups that Autumn mentions above) applies just as well to compact 3-manifolds as closed 3-manifolds.
Other classes of poison subgroups can be constructed from cohomological conditions. The Kneser--Milnor Theorem implies that any closed, irreducible 3-manifold with infinite fundamental group is aspherical. It follows that any freely indecomposable infinite group with cohomologial dimension greater than 3 cannot be a subgroup of a closed 3-manifold (and hence of a compact 3-manifold, by the previous paragraph).
EDIT:
Oh, and yet *another* source of poison subgroups comes from Scott's theorem that 3-manifold groups are *coherent*, meaning that every finitely generated subgroup is finitely presented. This rules out subgroups like $F\times F$ (where $F$ is a free group), which is not coherent.
| 24 | https://mathoverflow.net/users/1463 | 6143 | 4,181 |
https://mathoverflow.net/questions/6142 | 12 | Does anybody know if orientable, closed $3$-manifolds that are circle bundles over $RP^2$ have been classified?
One can determine the isomorphism classes of bundles using obstruction theory, but I am interested in what total spaces can appear.
I am not assuming the bundle is principal.
Thank you.
| https://mathoverflow.net/users/1944 | Circle bundles over $RP^2$ | Such manifolds are examples of Seifert fibered spaces, which have, indeed, been classified. A good reference is Montesinos "Classical Tessellations and Three-Manifolds". Basically, such manifolds (over any nonorientable surface base) are classified by their Euler class, which measures the obstruction to the existence of a section.
| 17 | https://mathoverflow.net/users/1672 | 6145 | 4,182 |
https://mathoverflow.net/questions/1959 | 22 | There's an important piece of geometric knowledge usually quoted as Beilinson-Bernstein-Deligne.
Here's a refresher: by $IC$ one means the intersection complex, which is just $\mathbb Q$ for a smooth scheme but more complicated for others, and by $IC\_i$ one denotes the complex constructed from a pair ($Y\_i$, $\mathcal L\_i$) of subvariety together with the local system as $IC\_i := j\_{!\*}\mathcal L\_i$.
Now
for a projective morphism $f: X\to Y$ turns out you can decompose in the derived category
$$f\_\*IC = \oplus IC\_i[n\_i].$$
The special beauty of this decomposition theorem is in its examples. Here are some I think I know:
* For a **free action** of a group G on some X, you get the decomposition by representation of G.
* For a **resolution of singularities**, you get $f\_\*\mathbb Q = IC\_Y \oplus F$ (and $F$ should have support on the exceptional divisor.)
* For a **smooth algebraic bundle** $f\_\*\mathbb Q = \oplus\\, \mathbb Q[-]$ (spectral sequence degenerates)
There are many known applications of the theorem, described, e.g. in the review
>
> [The Decomposition Theorem and the topology of algebraic maps\* by de Cataldo and Migliorini](http://arxiv.org/abs/0712.0349),
>
>
>
but I wonder if there are more examples that would *continue the list above*, that is, "corner cases" which highlight particularly specific aspects of the decomposition theorem?
>
> **Question:** What are other examples, especially the "corner" cases?
>
>
>
| https://mathoverflow.net/users/65 | Examples for Decomposition Theorem | I can think of several additions to your list which don't seem to be represented yet.
1. Semismall resolutions
------------------------
This first example is rather general, but afterward I will discuss how it is used in Springer theory.
First, suppose that $f:X \to Y$ is a proper map of stratified irreducible complex algebraic varieties with $X$ rationally smooth such that, if $Y = \cup Y\_n$ is the stratification of $Y,$ the restriction of $f$ to $f^{-1}(Y\_n) \to Y\_n$ is topologically locally trivial (there's a theorem (not sure who it's by) that says we can always find a stratification such that this condition holds). Furthermore, we say that $f$ is **semi-small** if for each stratum $Y\_n,$the dimension of the fiber of $f^{-1}(Y\_n) \to Y\_n$ is less than or equal to the half of codimension of $Y\_n$ inside $Y.$ This condition is important largely because of the following theorem:
>
> **Fact**. The pushforward of the constant perverse sheaf under a semismall map is still perverse.
>
>
>
Furthermore, we say that a stratum $Y\_n$ is *relevant* whenever equality holds above, i.e., twice the fiber dimension is equal to the codimension. These will be important soon, as they will be the subvarieties appearing in the decomposition theorem.
By the assumptions we made on $f:X \to Y,$ we have a monodromy action of $\pi\_1(Y\_n)$ on the top dimensional cohomology group of the fiber of $f^{-1}(Y\_n) \to Y\_n.$ This corresponds to a local system $L\_{Y\_n},$ which we can decompose into irreducible components: $L\_{Y\_n} = \oplus L\_{\rho}^{d\_{\rho}}$ where $\rho$ runs over the set of irreducible representations of $\pi\_1(Y\_n)$ and $d\_{\rho}$ are non-negative integers. We then say that a pair $(Y\_n, \rho)$ is *relevant* iff $Y\_n$ is a relevant stratum and $d\_{\rho} \neq 0$ (i.e., $\rho$ appears in the decomposition of the representation of $\pi\_1(Y\_n)$).
Now we can finally state a theorem, which I believe is due to Borho and Macpherson, but perhaps others deserve credit as well. Keep the initial assumptions on $f:X \to Y,$ but now assume in addition that $X$ is smooth. Then a little work plus the decomposition theorem establish the following.
>
> **Theorem**. $f\_{\ast}IC\_X = \oplus IC\_{Z\_n}(L\_{\rho})^{d\_{\rho}}$ where $Z\_n$ is the closure of $Y\_n$ and the sum ranges over all relevant pairs $(Y\_n, \rho).$
>
>
>
This theorem is used in **Springer theory** (and perhaps other places as well). In this case, we want $f:X \to Y$ to be the Springer resolution. That is, $Y = \mathcal{N},$ the nilpotent cone of a Lie algebra $g$ associated to a reductive group $G$, and $Y = \widetilde{\mathcal{N}},$ the variety of pairs $(x,b)$ where $x \in \mathcal{N},$ $b$ is a Borel subalgebra, and $x \in b.$ If we stratify $\mathcal{N}$ using the $Ad(G)$-orbits (of which there are finitely many), then it turns out that the Springer resolution is semismall and every stratum is relevant.
It can furthermore be shown that the $L\_{\rho}$ appearing in the theorem above correspond to the irreducible components of the regular representation of the Weyl group of $G.$ This can be seen as follows. There's an analog of the Springer resolution $\pi:\widetilde{g} \to g$ defined as above but with g in place of $\mathcal{N}.$ By proper base change, the pushforward of the constant sheaf on $\widetilde{\mathcal{N}}$ coincides with the pull-back (under the inclusion $\mathcal{N} \to g$) of the pushforward of the constant sheaf on $\widetilde{g}.$ Finally, since $\pi$ is what's known as a small map, the pushforward of the constant sheaf on $\widetilde{g}$ is equal to $IC\_g(L)$ where $L$ is the local system on the dense open subset $g^{rs}$ of regular semisimple elements obtained from the $W$-torsor $\widetilde{g^{rs}} \to g^{rs}.$ From all this we obtain that the top-dimensional cohomology groups of Springer fibers produce all irreducible representations of $W.$
2. Geometric Satake
-------------------
In a different direction, let me mention how the decomposition theorem is used in the geometric Satake correspondence (see the [Mirkovic-Vilonen paper](http://arxiv.org/abs/math/0401222) or the Ginzburg paper on this topic).
**Geometric Satake** is concerned with proving a tensor equivalence between the category of spherical perverse sheaves on the affine Grassmannian (i.e., perverse sheaves which are direct sums of IC sheaves) associated to a reductive group $G$ and the category of representations of the Langlands dual of $G.$ This is done through the Tannakian formalism, which in particular requires a tensor structure on spherical perverse sheaves. This tensor structure comes from a convolution product on perverse sheaves, meaning that it comes from a pull-back followed by a tensor product followed by a pushforward. In order to ensure that this operation takes spherical perverse sheaves to spherical perverse sheaves, we need the decomposition theorem.
Edit: According to the comments below, the decomposition theorem isn't actually needed to define the convolution product.
Comment on Kazhdan-Lusztig
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I'm going to assume that Gil Kalai is referring to the work of Lusztig on [Kazhdan-Lusztig polynomials](http://en.wikipedia.org/wiki/Kazhdan%E2%80%93Lusztig_polynomial) and the Kazhdan-Lusztig conjecture (mentioned in his answer). In particular, they have a paper,
* [KL] *Schubert varieties and Poincaré duality*, D. Kazhdan, G. Lusztig, Proc. Symp. Pure Math, 1980
in which the coefficients of the Kazhdan-Lusztig polynomials are related to the dimensions of the intersection cohomology of Schubert varieties (which are not generally smooth, hence the appearance of intersection cohomology). At this point, the Decomposition Theorem had not been proved and was not used in [KL]. However, the proof of the Decomposition Theorem heavily uses Deligne's Purity Theorem, which also had not been proved at the time of [KL]. Kazhdan and Lusztig ended up giving a proof of the Purity Theorem in the special case they were considering (i.e., a proof for Schubert varieties). Given this, it's not too surprising that a few years later Macpherson and Gelfand gave a proof of the aforementioned result of [KL] using the decomposition theorem and the result explained at the beginning of this answer.
It's my understanding that Lusztig has another paper from the mid-eighties on finite Chevalley groups which uses the Kazhdan-Lusztig conjecture (proved in 1981) and the full machinery of perverse sheaves and the Decomposition Theorem (I've never looked at it though). Additionally, Lusztig's work in the late seventies and early eighties on Springer theory certainly hints at the Decomposition Theory methods eventually used by Borho and Macpherson (some of his conjectures are proved by Borho and Macpherson, for example).
A wonderful history and reference guide to much of this can be found in this [article](http://arxiv.org/abs/math/0701462) by Steve Kleiman.
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