ca.analysis and odes - Geometric group theory and analysis

Analytic ideas enter into several parts of geometric group theory. Tom already mentioned amenability, so I'll skip that.

1) Complex analysis and the theory of quasiconformal mappings plays an important role in understanding the mapping class group, which is one of the most important groups studied by geometric group theorists. For information on this, see Farb and Margalit's forthcoming book "A primer on mapping class groups", available on either of their webpages.

2) The study of groups with property (T) ends up using quite a bit of analysis. See the book "Kazhdan's Property (T)" by de la Harpe, Bekka, and Valette for information on this.

3) Analysis (together with ideas from ergodic theory, which is of course quite analytic) plays an important role in proving various rigidity theorems. The most famous is the Mostow Rigidity Theorem, whose original proof uses lots of analysis : quasiconformal mapping in high dimensions, the fact that Lipschitz functions are differentiable almost everywhere, ergodic theory, etc.

modular forms - Dedekind eta function history

After some searching I found some relevant references:

H. Rademacher and A. Whiteman, “Theorems on Dedekind sums,” Amer. J. Math. 63 (1941), 377–407.

Arias-de-Reyna, J. Riemann's fragment on limit values of elliptic modular functions.
Ramanujan J. 8 (2004), no. 1, 57--123.

Dedekind's paper is titled "Erlauterungen zu den vorstehenden Fragmenten" and was published published in Riemann's collected works. In this paper Dedekind attempts to elucidate Riemann's “Fragmente uber die Grenzfalle der Elliptischen Modulfunctionen,”

It looks like Dedekind's introduction of the eta function was inspired by these fragmentary notes of Riemann. It would take me some time to parse the relevant issues (e.g. to what extent did Riemann know the Dedekind eta function, how might things have seemed to Dedekind, etc.) and I will not do so at the moment, but am happy to have found these for future reference.

asymptotics - Approximating a multiple sum with an integral

Hi,

I want to approximate a multiple sum of the form
$$sum_{x_1+x_2+cdots+x_m leq n}e^{g(x_1,x_2,ldots,x_m)},$$
where each $x_i$ is an integer between $0$ and $n$,
by an integral
$$int_{x_1+x_2+cdots+x_m leq n}e^{g(x_1,x_2,ldots,x_m)}dx_1dx_2cdots dx_m,.$$
I know that the Euler-Maclaurin formula can be used to derive the error term when $m=1$ but often see sums of this form with $m > 1$ approximated by integrals, though with little justification. I do not have much of a background in mathematical analysis so am not sure where to look for a reference for this.

Any help will be much appreciated.

fa.functional analysis - When is a locally convex topological vector space normal or paracompact?

Thanks to your other question, I was on a LCTVS kick. I did find one general criterion that implies that a locally convex space is paracompact. According to the Encyclopedia of Mathematics, if it is Montel (which means that it is barrelled and the Heine-Borel theorem holds true for it), then it is paracompact. Although this criterion is important, it is of no use to your specific question.

I thought I had a proof for half of your question, which I wrote up as the first version of this answer, but I made a mistake and proved something different. My thinking is based on the fact that the normality axiom for a topological space is equivalent to the Tietze extension theorem. (Tietze extension follows from normality. In the other direction, if $A$ and $B$ are the two closed sets, you obtain disjoint open neighborhoods from a continuous function that is 0 on $A$ and 1 on $B$.) However, in my argument I conflated the locally convex direct sum of spaces with the topological direct sum. For a countable direct sum of copies of $mathbb{R}$, they are the same topology, and they agree with the box topology. But Waelbroeck, LNM 230 points out that they are different in the uncountable case.

Let $alpha$ be an ordinal, for instance an ordinal of cardinality $2^{aleph_0}$. Then $mathbb{R}^alpha$ in the topological direct sum topology satisfies Tietze extension. Let $A subset mathbb{R}^alpha$ be a closed set and let $f:A to mathbb{R}$ be a continuous function. For $beta < alpha$, let $A_beta$ be the intersection of $A$ and with $mathbb{R}^beta$. Suppose that $alpha = beta+1$ is a successor ordinal. If $alpha$ is finite, then the conclusion is standard. Otherwise, by induction, there is an extension $f_beta$ of $f$ to $mathbb{R}^beta$. Moreover, by induction in a different sense, we have already proved that $mathbb{R}^{beta+1}$ is normal, since $beta$ and $beta+1$ have the same cardinality. So there exists an extension $f_alpha$ to $mathbb{R}^alpha$. If instead $alpha$ is a limit ordinal, then the extensions all the way up to $alpha$ work just because they work; that's the behavior of topological direct limits.

Having failed to normality for the locally convex direct sum, I can't say much about paracompactness either. :-) However, there is an interesting result called the Michael selection theorem which seems to do for paracompactness what the Tietze theorem does for normality. If the Tietze theorem is useful for your spaces, then maybe the Michael selection theorem is too.

lo.logic - Models within a model of set theory

There is an interesting observation surrounding the latter
part of your post, recently discussed in an article of
Brice Halimi (Université Paris Ouest), that many set
theorists find surprising. It is the following:

Theorem. Every model of ZFC contains an element that
is a model of all the ZFC axioms.

To be more precise, if $langle M,{in}^Mrangle$ is a
model of of ZFC, then there is an object $n$ in $M$ which
$M$ thinks to be a model in the language of set theory and
furthermore, every individual axiom of ZFC is true in $n$.

The proof is relatively simple (and it seems that each part
of it was known classically, although Brice has pulled them
nicely together). Suppose that $langle M,{in}^Mrangle$
is a model of ZFC. This implies, of course, that ZFC is
consistent. If it happens that $M$ is $omega$-standard,
meaning that it has no nonstandard natural numbers, then
$M$ will have the same proofs that we do outside of $M$,
and so $M$ will agree that ZFC is consistent. Hence, by the
Completeness theorem, $M$ will be able to build a model of
ZFC, such as by the Henkin method. Alternatively, suppose
that $M$ is $omega$-nonstandard. We know that for any
natural number $m$, the Reflection theorem ensures that
some $V_alpha$ is a model of the $Sigma_m$-fragment of
ZFC. Thus, inside $M$, every standard $m$ has the property
that there is some $(V_alpha)^M$ satisfying the $Sigma_m$
fragment of the (nonstandard) version of ZFC inside $M$.
Since the standard cut of the natural numbers is not
amenable to $M$, it must be that there is some nonstandard
natural number $k$ such that $M$ believes that there is
some $(V_alpha)^M$ satisfying what $M$ thinks is the
$Sigma_k$ fragment of ZFC. Since this includes all the
standard part of ZFC, we find in this case that there is a
model of ZFC inside $M$, as desired. QED

The reason set-theorists often find the result surprising
is that on its face, it seems to conflict with the
consequence of the Incompleteness theorem, that if ZFC is
consistent, then there are models of
$ZFC+negtext{Con}(ZFC)$. A model of this theory can have
no element that it thinks is a model of ZFC. The difference
between this result and the result above is exactly the
difference that your question is about: the model $n$
inside $M$ is not a model of the full nonstandard version
of ZFC inside $M$, but it is a model of all the standard
ZFC axioms.

lo.logic - What is induction up to epsilon_0?

This is a question asked out of curiosity, and because I can't understand the wikipedia page.

I have often been told that PA cannot prove the validity of induction up to $epsilon_0$, which has been expressed to me roughly as the claim that $epsilon_0$ is well-ordered. I understand what ordinals are, and what $epsilon_0$ is. I also understand first order logic and axiom schemes, so I understand how the induction axiom scheme formalizes the notion that $omega$ is well-ordered.

What I don't understand is how one could formulate the statement that $epsilon_0$ is well-ordered as a first order sentence in arithmetic. Would someone mind spelling this out for me?

soft question - A single paper everyone should read?

Different people like different things in math, but sometimes you stand in awe before a beautiful and simple, but not universally known, result that you want to share with any of your colleagues.

Do you have such an example?

Let's try to go in the direction of papers that can actually be read online or accessible with little effort, e.g. in major libraries, so that people could actually follow your advice and read about it immediately.

And as usual let's do one per post and vote freely, vote a lot.

gn.general topology - Is a inverse limit of compact spaces again compact ?

Then one can construct a model for the inverse limit by taking all the compatible sequences.
This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace.

However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology.

But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.

ag.algebraic geometry - Is every smooth affine curve isomorphic to a smooth affine plane curve?

As suggested by Poonen in a comment to an answer of his question about the birationality of any curve with a smooth affine plane curve we ask the following questions:

Q) Is it true that every smooth affine curve is isomorphic to a smooth affine plane curve?

(a) In particular, given a smooth affine plane curve $X$ with an arbitrary Zariski open set $U$ in it, can one give a closed embedding of $U$ in the plane again?

(b) An extremely interesting special case of (Q) above: Suppose $X$ is a singular plane algebraic curve with $X_{sm}$ the smooth locus. Can one give a closed embedding of $X_{sm}$ in the plane?

All varieties in question are over $mathbb{C}$.

---

UPDATE:
Bloch, Murthy and Szpiro have already proven in their paper "Zero cycles and the number of generators of an ideal" , a much more general result (see Theorem 5.7, op.cit), that every reduced and irreducible prjective variety has an affine open set which is a hypersurface. This settles the above question birationally, in particular.
The authors give a very short and beautiful alternate proof of their result by M.V. Nori which I include here for its brevity and for anyone who may not have access to the paper:
Proof: Suppose $X$ is an integral projective variety of dimension $d$. By a generic projection, easily reduce to the case of a (possibly singular) integral hypersurface $X$ of $mathbb{A}^{d+1}$. Suppose the coordinate ring of $X$ is $A=mathbb{C}[x_1,dots,x_{d+1}]$ and its defining equation is $F=Sigma_0^m{f_i}x_{d+1}^{i}=0$ with $f_0neq{0}$. For some element $h$ in $Jcapmathbb{C}[x_1,dots,x_d]$, where $J$ defines the singular locus of $X$, put $x_{d+1}'=x_{d+1}/(hf_0^2)$ in $F=0$ to observe that $1/(hf_0)inmathbb{C}[x_1,dots,x_{d+1}']$ and $A_{hf_0}=mathbb{C}[x_1,dots,x_{d+1}']$. Clearly $rm{Spec} {A_{hf_0}}$ admits a closed immersion in $mathbb{A}^{d+1}$.

---

However, the above authors also prove in their Theorem 5.8 that there exist affine varieties of any dimension, which are not hypersurfaces. This answers our question in negative. This was also known to Sathaye for curves, see On planar curves. He gives a nice example of a double cover of a punctured elliptic curve, ramified at 9 points and also at the point at infinity. This curve cannot be embedded in $mathbb{A}^2$. Sathaye uses the value semigroup at the only point at infinity to prove this. His example has trivial canonical divisor. So it answers Poonen's question in the comments below, negatively.
In short, $K=0$ for an affine curve is necessary but not sufficient for the curve to be planar, however one should note that $K=0$ is necessary and sufficient for an affine curve to be a complete intersection.

co.combinatorics - Looking for references for an implicit differentiation formula

In a paper which I submitted to a peer-reviewed math journal in April 2010, I proved a formula for the n-th derivative $frac{d^n z}{dw^n}$ in terms of (as a polynomial over the integers) of the partial derivatives of a given implicit function, $G(z,w)=0$, with respect to $z$ and $w$ (and negative integer powers of the "separant", $G_z$, the first partial derivative of $G$ with respect to $z$).

This is classic first-semester calculus homework exercise: to compute $frac{d^n z}{dw^n}$ for n=1 and 2, namely,

$frac{dz}{dw} = - frac{G_w}{G_z}$

$frac{d^2 z}{dw^2} = - frac{G_{zz}G_{ww}}{G_{zzz}} + 2frac{G_{zw}G_w}{G_{zz}} - frac{G_{ww}}{G_z}$

I did so not knowing whether any one had proved the general formula first, because I am busy building on, generalizing, and using this result for other things, including chemical processing.

I have since proved the partial differential generalization of this implicit differentiation formula: i.e. given $G(z,w_1,...,w_N)=0$, compute

$frac{d^{({u_1}+...+u_N})}{{dw_1}^{u_1}cdots {dw_N}^{u_N}}z$ as a Laurent polynomial over the integers of the partial
derivatives of $G$ with respect to $w_1,...,w_N$, and $z$

I do not have access to most peer-reviewed journals. I have had to make do with Google searches, Wolfram Research's MathWorld, online searches through my county library, and the help from one mathematician friend who has sent me related papers.

Most of the papers my friend sent me concern the Faa da Bruno formula (FdBF) and its generalizations, and the Lagrange Inversion Formula (LIF). Both the FdBF and LIF are very closely related to what I am doing, but they can not be trivially applied to get (my) general formula. (I tried... for about 8 months.) I have studied G.P. Egorychev's book: "Integral Representations of Combinatorial Sums" intensely, especially the back, with the multivariable generalizations of the LIF.

No, I am not in school. This is not a homework problem. This is "free-lance" research.
I am not asking for a solution to the problem (as I already solved it independently).

I simply want to know yes or no whether someone has already done this.
And, if so, where.

Thank you for the responses, in particular to go to arxiv.org, which I forgot, since I had submitted 2 papers there myself.

linear algebra - Fast Fourier Transform for Graph Laplacian?

In the case of a regularly-sampled scalar-valued signal $f$ on the real line, we can construct a discrete linear operator $A$ such that $A(f)$ approximates $partial^2 f / partial x^2$. One way to interpret this operator is via spectral decomposition of the corresponding matrix:

$$A = UVU^T.$$

If our operator $A$ has spectral accuracy, then $U^T$ is precisely the discrete Fourier transform matrix. Hence, we could compute the Fourier transform of $x$ by computing all the eigenvectors of $A$ and sticking them in the columns of $U$. Of course, we all know there's a quicker way to do it: use the fast Fourier transform (FFT).

What about in a more general setting? In particular, consider the graph Laplacian $L=UVU^T$ which for a weighted, undirected graph on $n$ vertices is an $n times n$ matrix with the weights of incident vertices on the off-diagonal and (the additive inverse of) total incident weight on the diagonal.

Question:

Can we transform a signal on vertices into frequency space without computing the entire spectrum of $L$ (using something like the FFT)?

In particular I'm interested in the case where $L$ approximates the Laplace-Beltrami operator on some manifold $M$ -- here eigenvectors of $L$ approximate an orthogonal eigenbasis for square-integrable functions on $M$. However, pointers to nearby results (e.g., FFT for the combinatorial graph Laplacian) are appreciated.

ag.algebraic geometry - A hypersurface with many points

Here is a complete solution to the main question when $n$ and $q$ are both odd, and a partial solution for the other parities. The partial solution includes a reduction to the case $n=2$.

Let $text{Tr}_k$ denote the trace map from $mathbb{F}_{q^n}$ to $mathbb{F}_{q^k}$, assuming that $k|n$.

The equation is
$$z+z^q + z^{q^2} = x^{q^2+q+1} - xy^q - x^{q^2}y.$$
First, make the change of variables $y leftarrow -yx^{q^+1},$ so that the equation becomes
$$z+z^q+z^{q^2} = x^{q^2+q+1}(y^q+y+1).$$
Second, when $q$ is odd, we can clarify matters a little with the change of variables $y leftarrow y - frac12$ to get rid of the constant. The image of the $q$-linear map $z mapsto z + z^q + z^{q^2}$, acting on $mathbb{F}_{q^{3n}}$, consists of those elements $Z'$ such that $text{Tr}_3(z') = text{Tr}_1(z')$. Whenever such a $z'$ is reached by the right side, there are $q^2$ solutions for $Z$. On the right side, $y mapsto y' = y^q+y$ is an $q$-linear isomorphism when $n$ is odd, while when $n$ is even its image is the locus of $text{Tr}_2(y') = text{Tr}_1(y')$. Meanwhile $y' mapsto x^{q^2+q+1}y'$ is a $q$-linear isomorphism unless $x=0$.

Thus when $n$ and $q$ are both odd, there are $(q^{3n}-q^2)(q^{3n}-1)$ solutions with $x,y ne 0$. There are $q^2(2q^{3n}-1)$ more solutions when one of them is zero.

When $q$ is odd and $n$ is even, then in principle different $x' = x^{q^2 + q + 1}$ could behave differently in the equation $z' = x'y'$. For a fixed $x'$, the set of possible $x'y'$ is a certain $q$-linear hyperplane with codimension $1$, while the set of possible $z'$ is a certain $q$-linear hyperplane with codimension $2$. In the special case that ${x'y'}$ contains ${z'}$, then there are $q^{3n+1}$ solutions for that value of $x'$. For generic non-zero values of $x'$, there are $q^{3n}$ solutions. Dualize the hyperplanes with respect to $text{Tr}_1$. The dual of ${y'}$ is the line $L$ of trace 0 elements in $mathbb{F}_{q^2}$, while the dual of ${z'}$ is the plane $P$ of trace 0 elements in $mathbb{F}_{q^3}$. Meanwhile ${x'}$ is the subgroup $G$ of $mathbb{F}_{q^{3n}}$ of index $q^2+q+1$. A special value of $x'$ in this subgroup is one such that $x'L subset P$. A priori I am not sure that it never happens. What I can say is that if $x'$ is special, then it must lie in $mathbb{F}_{q^6}$ because both $L$ and $P$ do. So you can reduce the counting problem to the case that $3n = 6$ or $n = 2$.

If $q$ is even, then the equation is
$$z' = x'(y'+1),$$
where as before $z' = z+z^q+z^{q^2}$ and $y' = y + y^q$. In this case $y'$ is any element with zero trace, and the dual line $L$ is just $mathbb{F}_q$ itself. I have not worked out exactly how it looks, but I suppose that it reduces to the case $n=1$ for similar reasons as above.

Afterthought: I don't feel like changing all of the equations, but I'm wondering now whether there a dual change of variables to put the right side in the form $y''(x^{q^2}+x)$. I think that the map $x mapsto x^{q^2}+x$ is always non-singular when $q$ is odd.

---

A remark about where the trace conditions come from. If $a$ is an irreducible element of $mathbb{F}_{q^n}$, then the map $x mapsto x^q$ is a cyclic permutation matrix in the basis of conjugates of $a$. A map such as $z mapsto z+z^q+z^{q^2}$ is then a sum of disjoint permutation matrices and it easy to compute its image and cokernel.

---

Some remarks about Jared's second, more general question: C.f. the answer to this other mathoverflow question about counting points on varieties. For fixed $q$, the equation of a hypersurface is equivalent to a general Boolean expression, and there may not be much that you can do other than count one by one. There are several strategies that work in the presence of special structure: You can use zeta function information, if you have it, to extrapolate to large values of $q$. You can count the points on a variety if you happen to know that it's linear, or maybe quadratic, or the coset space of a group. And you can use standard combinatorial counting tricks, which in algebraic geometry form amount to looking at fibrations, blowups, inclusion-exclusion for constructible sets, etc.

This particular variety decomposes a lot because it can be made jointly linear in $Y$ and $Z$, and $X$ only enters in a multiplicative form.

co.combinatorics - What does the generating function $x/(1 - e^{-x})$ count?

Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series
$$ A(x) = frac{x}{1 - e^{-x}} = sum_{m=0}^infty left( -sum_{n=1}^infty frac{(-x)^n}{(n+1)!} right)^m = 1 + frac12 x + frac1{12}x^2 + 0x^3 - frac1{720}x^4 + dots $$
where I might have made an arithmetic error in expanding it out.

1. Are all the coefficients egyptian, in the sense that they are given by $A^{(n)}(0)/n! = 1/N$ for $N$ an integer? The answer is no, unless I made an error, e.g. the third coefficient. But maybe every non-zero coefficient is egyptian?




2. If all the coefficients were positive eqyptian, then the sequence of denominators might count something — one hopes that the $n$th element of any sequence of nonnegative integers counts the number of ways of putting some type of structure on an $n$-element set.

Of course, generating functions really come in two types: ordinary and exponential. The difference is whether you think of the coefficients as $sum a_n x^n$ or as $sum A^{(n)} x^n/n!$. If it makes more sense as an exponential generating function, that's cool too.

So my question really is: is there a way of computing the $n$th coefficient of $A(x)$, or equivalently of computing $A^{(n)}(0)/n!$, without expanding products of power series the long way?

Where you might have seen this series

Let $xi,psi$ be non-commuting variables over a field of characteristic $0$, and let $B(xi,psi) = log(exp xi exp psi)$ be the Baker-Campbell-Hausdorff series. Fixing $xi$ and thinking of this as a power series in $psi$, it is given by
$$B(xi,psi) = xi + A(text{ad }xi)(psi) + O(psi^2)$$
where $A$ is the series above, and $text{ad }xi$ is the linear operator given by the commutator: $(text{ad }xi)(psi) = [xi,psi] = xipsi - psixi$.

More generally, $B$ can be written entirely in terms of the commutator, and so makes sense as a $mathfrak g$-valued power series on $mathfrak g$ for any Lie algebra $mathfrak g$. It converges in a neighborhood of $0$ when $mathfrak g$ is finite-dimensional over $mathbb R$, in which case $mathfrak g$ is a (generally noncommutative) "partial group".

(More generally, you can consider the "formal group" of $mathfrak g$. Namely, take the commutative ring $mathcal P(mathfrak g)$ of formal power series on $mathfrak g$; then $B$ defines a non-cocommutative comultiplication, making $mathcal P = mathcal P(mathfrak g)$ into a Hopf algebra. Or rather, $B(mathcal P)$ does not land in the algebraic tensor product $mathcal P otimes mathcal P$. Instead, $mathcal P$ is cofiltered, in the sense that it is a limit $dots to mathcal P_2 to mathcal P_1 to mathcal P_0 = 0$, where (over characteristic 0, anyway) $mathcal P_n = text{Poly}(mathfrak g)/(mathfrak g text{Poly}(mathfrak g))^n$, where $text{Poly}(mathfrak g)$ is the ring of polynomial functions on $mathfrak g$, and $mathfrak g text{Poly}(mathfrak g)$ is the ideal of functions vanishing at $0$. Then $B$ lands in the cofiltered tensor product, which is just what it sounds like. (In arbitrary characteristic, $mathcal P$ is the cofiltered dual of the filtered Hopf algebra $mathcal S mathfrak g$, the symmetric algebra of $mathfrak g$, filtered by degree.))

Why I care

When $mathfrak g$ is finite-dimensional over $mathbb R$, and $U$ is the open neighborhood of $0$ in which $B$ converges, then $mathfrak g$ acts as left-invariant derivations on $U$, where by left-invariant I mean under the multiplication $B$. Hence there is a canonical identification of the universal enveloping algebra $mathcal Umathfrak g$ with the algebra of left-invariant differential operators on $U$. Since $mathfrak g$ is in particular a vector space, the "symbol" map gives a canonical identification between the algebra of differential operators on $U$ and the algebra of functions on the cotangent bundle $T^{ast} U$ that are polynomial (of uniformly bounded degree) in the cotangent directions. Left-invariance then means that the operators are uniquely determined by their restrictions to the fiber $T^{ast}_0mathfrak g = mathfrak g^{ast}$, and the space of polynomials on $mathfrak g^{ast}$ is canonically the symmetric algebra $mathcal S mathfrak g$. This gives a canonical PBW map $mathcal U mathfrak g to mathcal S mathfrak g$, a fact I learned from J. Baez and J. Dolan.

(In the formal group language, the noncocommutative cofiltered Hopf algebra $mathcal P(mathfrak g)$ is precisely the cofiltered dual to the filtered algebra $mathcal Umathfrak g$, whereas with its cocommutative Hopf structure $mathcal P(mathfrak g)$ is dual to $mathcal S mathfrak g$. But as algebras these are the same, and unpacking the dualizations gives the PBW map $mathcal Umathfrak g cong mathcal S mathfrak g$, and explains why it is actually an isomorphism of coalgebras.)

Anyway, in one direction, the isomorphism $mathcal Umathfrak g cong mathcal S mathfrak g$ is easy. Namely, the map $mathcal S mathfrak g to mathcal U mathfrak g$ is given on monomials by the "symmetrization map" $xi_1cdots xi_n mapsto frac1{n!} sum_{sigma in S_n} prod_{k=1}^n xi_{sigma(k)}$, where $S_n$ is the symmetric group on $n$ letters, and the product is ordered. (In this direction, the isomorphism of coalgebras is obvious. In fact, the corresponding symmetrization map into the full tensor algebra is a coalgebra homomorphism.)

In the reverse direction, I can explain the map $mathcal U mathfrak g to mathcal S mathfrak g$ as follows. On a monomial $xi_1cdots xi_n$, it acts as follows. Draw $n$ dots on a line, and label them $xi_1,dots,xi_n$. Draw arrows between the dots so that each arrow goes to the right (from a lower index to a higher index), and each dot has either 0 or 1 arrow out of it. At each dot, totally order the incoming arrows. Then for each such diagram, evaluate it as follows. What you want to do is collapse each arrow $psito phi$ into a dot labeled by $[psi,phi]$ at the spot that was $phi$, but never collapse $psito phi$ unless $psi$ has no incoming arrows, and if $phi$ has multiple incoming arrows, collapse them following your chosen total ordering. So at the end of the day, you'll have some dots with no arrows left, each labeled by an element of $mathfrak g$; multiply these elements together in $mathcal Smathfrak g$. Also, multiply each such element by a numerical coefficient as follows: for each dot in your original diagram, let $m$ be the number of incoming arrows, and multiply the final product by the $m$th coefficient of the power series $A(x)$. Sum over all diagrams.

Anyway, the previous paragraph is all well and cool, but it would be better if the numerical coefficient could be read more directly off the diagram somehow, without having to really think about the function $A(x)$.

ag.algebraic geometry - Schemy question..

Here is where I got lost .. I have a scheme Y over k (an algebraically
closed field), in it I have an irreducible closed subscheme X of
finite type (do I need finite type?). I also know that X is
universally closed (over k) and separated (do I need separated?) ..
then why is it that X should be either a point or Y itself?

how do you evaluate the p-adic modular form E_p-1 in the region |j|

One has $j = E_4^3/Delta$. In the region $|j|leq 1$, one is parameterizing elliptic curves with good reduction, and so $Delta$ is a unit. Thus $|j| = |E_4|^3$. This will help you
when $p = 5$.

When $p = 7,$ one can write $j = 1278 + E_6^2/Delta,$ hence $|E_6|^2 = | j - 1728|$ on
the region $|j| leq 1$.

For $p = 11$, these sort of explicit computations are harder (but maybe not much; see
the added material below), because there are two supersingular $j$-invariants. But the $p = 5$ and 7 cases will already be illustrative.

In the case when $p = 2$, I wrote something about this once, which appeared in an
appendix an article by Fernando Gouvea in a Park City proceedings volume. A slightly
butchered version (missing figures, among other things) can be found on my
web-page (near
the bottom). You might also look at the papers of Buzzard--Calegari
for related computations, as well as my thesis (available on my web-page) and later
work by Kilford and Buzzard--Kilford. (There is, or at least once was, a cottage industry
based on combining these sorts of explicit computations with some more theoretical
estimates, to compute information about slopes of overconvergent $p$-adic modular forms
for various small primes $p$.)

Added in response to the comment below: For $p = 11$, one has
$E_{10} = E_4 E_6,$
so $E_{10}^6 = j^2(j-1728)^3 Delta^5,$ and so when $|j| leq 1,$ one has
$|E_{10}|^6 = |j|^2|j-1728|^3.$ Perhaps this will help?

na.numerical analysis - Why does randomness work in numerical algorithms?

If one had infinite computing capacity, then you would be right that randomness would not be useful (unless an adversary also had access to randomness - see below), and it would always be better to maintain full control of one's parameters. ("God does not need to play dice".) However, our computing capacity is limited, and randomness offers a way to stretch that capacity further (though it is an important open problem to quantify exactly how much randomness actually stretches our capacity; see the discussion on P=BPP in other comments).

Let's give a simple example. Suppose one is playing the following game with a computer. This computer uses some (deterministic) algorithm to generate 1000 "bad" numbers from 1 to 1 million. Meanwhile, you pick your own number from 1 to 1 million. If you pick one of the bad numbers, you lose; otherwise, you win. You get to see the computer's algorithm in advance.

With infinite computing capacity, you have a strategy which is 100% guaranteed to succeed: you run the computer's algorithm, find all the bad numbers, and pick a number not chosen by the computer.

But there is a much lazier strategy that requires no computing power: just pick a number randomly, and one has a 99.9% chance of winning.

[Note also that if the computer used a randomised algorithm instead of a deterministic one, then there is no longer any 100% winning strategy for you, even with infinite computing power, though the random strategy works just fine. This is another indication of the power of randomness.]

Many randomised algorithms work, roughly speaking, by reducing the given problem to some version of the above game. For instance, Monte Carlo integration using a set of points to sample the function might give satisfactory levels of accuracy for all but a tiny fraction of the possible choices of sample points. With infinite computing capacity, one could locate all the bad configurations of sample points and then find a configuration which is guaranteed to give a good level of accuracy, but this is a lot of work - more work, in fact, than just doing classical numerical integration. Instead, one takes the lazy way out and picks the sample points randomly. (Well, in practice, one picks the points pseudorandomly instead of randomly, because true randomness is very hard to capture by a computer; this is a subtle issue - again related to P=BPP - that I don't want to get into here.)

How much of differential geometry can be developed entirely without atlases?

This is a comment, not an answer, but is too long to fit in the comment box: having read the question, answers, and comments, I don't quite follow the intent of this question:

We can define a manifold to be a locally ringed space in which each point has a neighbourhood
isomorphic to an open subset of ${mathbb R}^n$ (or even just ${mathbb R}^n$ itself) with its sheaf of smooth functions (plus second countability and Hausdorffness, if you like). As was remarked by Dmitri, the collection of all such will then form an atlas, but one doesn't need to say this.

As Pete Clark says, what I've said so far is evident.

But it seems that another aspect of the question is whether one can always avoid working in coordinates. This seems to have nothing to do with atlases.

E.g. in arguments in the locally ringed space set-up, one will certainly in many instances verify that a property can be checked locally, and then verify it on Euclidean space with
its natural smooth structure. (Just as in the theory of schemes, one often shows that a property is local, and then checks it in the affine case.)

Now one can ask: can one avoid the latter kinds of arguments? This seems unlikely: manifolds are defined to be locally Euclidean (no matter which of the possible formalisms one is using), and so if one is proving theorems about manifolds, one will
have to use this somewhere. For example, one can surely define the tangent sheaf in
a coordiante free way, but to prove that it is locally free of rank equal to the dimension
of the manifold, one is going to reduce to a local computation and then appeal to calculus
on Euclidean spaces; there is no other way!

[EDIT: The last sentence may be too categorical of a declaration; see Dmitri Pavlov's answer for a suggestion of a more substantially algebraic reformulation of the
notion of manifold.]

nt.number theory - What is the etymology for the term conductor?

It is a translation from the German Führer (which also is the reason that
in older literature, as well as a fair bit of current literature, the conductor
is denoted as f in various fonts). Originally the term conductor appeared in
complex multiplication and class field theory: the conductor of an abelian extension is a certain ideal that controls the situation. Then it drifted off into other
areas of number theory to describe parameters that control other situations.

Of course in English we tend not to think of conductor as a leader
in the strong sense of Führer, but more in a musical sense, so it seems like a weird translation. But back in the
1930s the English translation was leader rather than conductor, at least once: see the review of Fueter’s book on complex multiplication in the 1931 Bulletin of the Amer. Math.
Society, page 655. The reviewer writes in the second paragraph "First there is a careful treatment of those ray class fields whose leaders are multiples of the ideal..."
You can find the review yourself at
http://www.ams.org/bull/1931-37-09/S0002-9904-1931-05214-9/S0002-9904-1931-05214-9.pdf.

I stumbled onto that reference quite by chance (a couple of years ago). If anyone knows other places in older papers in English where conductors were called leaders, please post them as comments below. Thanks!

Concerning Artin's conductor, he was generalizing to non-abelian Galois extensions the parameter already defined for abelian extensions and called the conductor. So it was natural to use the same name for it in the general case.

Edit: I just did a google search on "leader conductor abelian" and the first hit
is this answer. Incredible: it was posted less than 15 minutes ago!

geometric rep theory - Can the Quantum Torus be realized as a Hall Algebra?

Background

The Quantum Torus

Let $q$ be an arbitrary complex number, and define (the algebra of) the quantum torus to be
$$T_q:=mathbb{C}langle x^{pm 1},y^{pm 1}rangle/xy-qyx$$
For $q=1$, this is the commutative ring of functions on the torus $mathbb{C}^timestimes mathbb{C}^times$; hence, for general $q$, this is regarded as a quantization of the torus.

Hall Algebras

Consider a small abelian category $A$, with the property that $Hom_A(M,N)$ and $Ext^i_A(M,N)$ are always finite sets for any $M,Nin A$ and $iin mathbb{Z}$. Let $overline{A}$ denote the set of isomorphism classes in $A$, and let
$$H(A)=oplus_{[M]in overline{A}}mathbb{C}[M]$$
denote the complex vector space spanned by $overline{A}$. Endow $H(A)$ with a multiplication by the formula
$$ [M]cdot [N]=sqrt{langle [M],[N]rangle)}sum_{[R]in overline{A}}frac{a_{MN}^R}{|Aut(M)||Aut(N)|}[R]$$
where $a_{MN}^R$ is the number of short exact sequences
$$0rightarrow Nrightarrow Rrightarrow Mrightarrow 0$$
and
$$langle [M],[N]rangle = sum (-1)^i |Ext^i_A(M,N)|$$
is the Euler form. This multiplication makes $H(A)$ into an associate algebra called the Hall algebra of $A$; the proof can be found e.g. here.

Finite Fields and Quantization

The categories $A$ appearing in the construction of a Hall algebra are usually linear over some finite field $mathbb{F}_q$. Often, it is possible to simultaneously define a category $A_q$ for each finite field $mathbb{F}_q$; usually by considering modules on the $mathbb{F}_q$-points of some scheme over $mathbb{Z}$. The corresponding Hall algebras $H(A_q)$ will then usually be closely related, and can often be defined by relations that are functions in $q$.

The Question

I know that there are cases where an algebra is deformed by a parameter $q$, and then the resulting family of algebras `magically' coincides with a family of Hall algebras $H(A_q)$ in the special cases when $q$ is a prime power. I think this happens in the case of the Hecke algebra (discussed here), and the case of quantum universal enveloping algebras (discussed here). I somewhat understand that this is a symptom of a related convolution algebra on the scheme used to define $A_q$.

Is there a family of categories $A_q$ such that the corresponding Hall algebras $H(A_q)$ are isomorphic to the Quantum Torus $T_q$ for all $q$ a prime power? If so, is there a convolution algebra realization of the Quantum Torus?

lo.logic - Are all mathematical theorems necessarily true?

Charles wrote:

Robert Hanna (Kant's Theory of Judgment, SEP 2009, sect. 2.2.2) interprets Kant as saying that "logically possible worlds are nothing but maximal logically consistent sets of concepts". If we apply this by saying that the logically possible worlds that can contain the concepts as expressed by an axiomatisation are the models of that theory, then it follows that the theorems of an axiomatisation are necessarily true.

The logical quantifiers interact in odd ways with this idea, and make "maximality" kind of a problematic notion. This is a familiar phenomenon for mathematicians and logicians, of course, but may be worth pointing out explicitly. So, if we have some set of atomic concepts or assertions about the world, then a model of a possible world is a Boolean algebra on this set. Then the basic logical connectives (conjunction, disjunction, negation) can be modelled by intersection, union and complement.

So far, so good. Now, if the set of atomic assertions is finite, then the Boolean algebra is finite, and so it is also a complete lattice, and so quantified statements also have interpretations in the model. But if the set of atomic assertions is not finite, then a Boolean algebra on this set doesn't have to be complete, and so quantified statements might not have interpretations! If we do demand that the Boolean algebra is complete, then what we consider to be "logically possible" depends upon what kinds of logical connectives we wish to consider. (For example, in probability theory the difference between the Kolmogorov and Bayesian axioms of probability is that the Bayesians don't demand countable additivity, which means that an existentially-quantified propositions might not have a Bayesian interpretation.)

computability theory - What are the most attractive Turing undecidable problems in mathematics?

What are the most attractive Turing undecidable problems in mathematics?

There are thousands of examples, so please post here only the most attractive, best examples. Some examples already appear on the Wikipedia page.

Standard community wiki rules. One example per post please. I will accept the answer I find to be the most attractive, according to the following criteria:

• Examples must be undecidable in the sense of Turing computability. (Please not that this is not the same as the sense of logical independence; think of word problem, not Continuum Hypothesis.)




• The best examples will arise from natural mathematical questions.




• The best examples will be easy to describe, and understandable by most or all mathematicians.




• (Challenge) The very best examples, if any, will in addition have intermediate Turing degree, strictly below the halting problem. That is, they will be undecidable, but not because the halting problem reduces to them.

Edit: This question is a version of a previous question by Qiaochu Yuan, inquiring which problems in mathematics are able to simulate Turing machines, with the example of the MRDP theorem on diophantine equations, as well as the simulation of Turing machines via PDEs. He has now graciously merged his question here.

commutative monoids have binary products?

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ag.algebraic geometry - Embeddings and triangulations of real analytic varieties

I do not know an answer to the embedability question, but the triangulation can be deduced from the existence of a (Thom-Mather) stratification by [Johnson "On the triangulation of stratified sets and singular varieties", Trans. Amer. Math. Soc. 275 (1983), no. 1, p. 333–343] or [Goresky "Triangulation of stratified objects", Proc. Amer. Math.
Soc. 72 (1978), no. 1, p. 193–200]; moreover it is known that analytic subvarieties of Euclidean space are Whitney stratified [Whitney "Local properties of analytic varieties", Differential and Combinatorial Topology (A Symposium in Honor of Marston Morse), Princeton Univ. Press, Princeton, N. J., 1965, p. 205–244 & "Tangents to an analytic variety", Ann. of Math. (2) 81 (1965), p. 496–549].

Moreover in [Mather "Notes on topological stability", 1970, Harvard University & "Stratifications and mappings", Dynamical systems (Proc. Sympos., Univ. Bahia, Salvador, 1971), Academic Press, New York, 1973, p. 195–232] you can find the result that Whitney stratified sets are Mather stratified. Finally, Mather stratification are given by local conditions (there might be an issue gluing the local strata but I do not think so), so these result should imply that analytic varieties are triangulable. I do not think it is the most efficient to do so.

ac.commutative algebra - Length of I/I^2 versus Ann(I)/Ann(I)^2 in Artinian rings.

UPDATE 10/04/10:

Here are some partial results in the graded case. Let $R=oplus_0^s R_i$ be a graded Gorenstein algebra over $k=R_0$ ($s$ is the socle degree). $R$ is said to have strong Lefschetz property (SLP) if for a general linear form $l$ in $R$, the multiplication map $times l^a: R_i to R_{i+a}$ has maximal rank for all $igeq 0, a geq 1$ (so it is either injective or surjective). Such $l$ is called a Lefschetz element.

By a result of Stanley, monomial complete intersections of characteristic $0$, e.g. $R=k[x_1cdots,x_d]/(x_1^{a_1},cdots, x_d^{a_d})$ have SLP (apply the Hard Lefschetz Theorem!). Many more classes of rings with SLP are known, and I think it is conjectured to hold for all complete intersections, at least in char. $0$ (the keywords are Weak Lefschetz and Strong Lefschetz Property, the literature is quite big, see for example this paper).

Now, suppose $R$ is Gorenstein with SLP, and $x$ be a Lefschetz element. I claim that such an element would satisfy your inequality. Let $L$ be the annihilator of $x^2$. By Fact 1 we have that $text{length} (x/x^2) = text{length}(L) - text{length}(J)$.

Let $h_i = text{dim}_k R_i$. Then since $R$ is Gorenstein with SLP, the sequence $h_0,cdots,h_s$ is unimodal and symmetric. Now since the maps $times x$ and $times x^2$ are always injective or surjective, it is not hard to compute that $text{length}(L)=h_n+h_{n+1}$ and $text{length}(J)=h_n$ with $n=lfloor s/2rfloor$. So $text{length} (x/x^2) =h_{n+1}$.

If $s$ is odd, then $h_n=h_{n+1}$ and $J$ must live in degrees at least $n+1$, thus $J^2=0$ and equality actually holds.

If $s$ is even, then $text{length} (x/x^2)=h_{n+1}$. Now if $h_n= h_{n+1}$, then as above $J^2=0$ and equality holds. If $h_n > h_{n+1}$, then $J$ lives only in degree $2n=s$, but $h_s=1$ (remember $R$ is Gorenstein) so $text{length} (J^2)$ is at most $1$. So $text{length}(J/J^2) geq h_n-1geq h_{n+1}$.

I think one can push this argument to show the inequality for $x=l^a$ for any Lefschetz element $l$ and $a>0$. Because the SLP is an open condition, this would imply the inequality for general forms of degree $a$.

I also have some examples of equality in positive characteristic with pretty interesting patterns, but I need more time to think about them.

End of UPDATE

(Some history: Yesterday when I saw this question I posted a simple solution, which I immediately realized is wrong. After a few email exchanges with FC we were both convinced that my first attempt would not work.)

I have thought about the question a bit today but could not quite prove it. Since I may not have time to work on it more in the next few days, I will put some of my thoughts here in case they help anyone.

Let $I=(x)$ and $J$ be the annihilator of $(x)$. Also, I will use $R$ instead of $A$.

Fact 1: $(x) cong R/Jcong D(R/J)$

Proof: For the first isomorphism, just look at the map $Rto R$ by multiplying with $x$. Now
$D(R/J) = text{Hom}(R/J,R)$ is isomorphic to the annihilator of $J$, which is $(x)$ again.

Fact 2: $x/x^2 cong R/(J+x)$.

Proof: $x/x^2 = (x)otimes R/(x) =R/Jotimes R/(x)$

Fact 3: $R/J$ is itself Gorenstein.

Proof: The canonical module of $R/J$ is $D(R/J)$ which is isomorphic to $R/J$ by Fact 1.

Fact 4: Let $(S,m,k)$ be a Gorenstein, artinian local ring. An $S$-module map $Sto M$ is injective iff the image of the generator of the socle of $S$ (which is $text{Hom}(k,S)$ and is 1-dimensional as $S$ is Gorenstein) is non-zero.

Proof: If the kernel is some non-zero ideal $K$, then some element in $K$ would have $m$ as the annihilator. But then such element has to be inside the socle of $S$, which is a 1-dim vector space.

Fact 5: For any ideal $L$, $L/L^2 = text{Tor}_1(R/L,R/L)$

Proof: tensor $0to L to Rto R/L$ with $R/L$.

Here are a couple of approaches I tried with some comments:

Random thoughts A:
By Fact 2 we need to prove $$text{length}(J) -text{length}(J^2)geq text{length}(R)- text{length}(J+x) $$
Rearranging, one needs to prove $$text{length}(R/J) leq text{length}((J+x)/J^2)$$
Let $S=R/J$ and $M=(J+x)/J^2$. One obvious thing to try is to show that $S$ (which is Gorenstein by Fact 3) can embed in $M$. Let $sin R$ be a lift of the socle generator of $S$. By Fact 4, for a counter example one would need
$$(J+x)s subseteq J^2 $$

This rules out many potential counter examples because of degree reasons, or if $J^2$ is too small. Note that $xs$ represents the socle generator of $R$.

Random thoughts B: Here is a formulation that only involves $x$. By Fact 5 we need to prove $$text{length}(text{Tor}_1(R/(x),R/(x)) leq text{length}(text{Tor}_1(R/J,R/J)$$
Which, by Fact 1 is really:
$$text{length}(text{Tor}_1(R/(x),R/(x)) leq text{length}(text{Tor}_3(R/(x),R/(x))$$

This equivalent statement actually makes me a little doubtful. It might be true, and sort of make sense, because the free resolution of $R/(x)$ will typically gets bigger and bigger, but many similar homological statements about Gorenstein rings turn out to be false (although it is often not easy to cook up examples).

So I would say that the statements is likely to be true for small (in terms of lengths or degrees) rings, because of Thoughts A, but might be false in general. Of course, I would be very happy to be wrong, and may be I was missing something really simple.

PS: thanks for asking a nice question in commutative algebra (-:

co.combinatorics - What (if anything) happened to Viennot's theory of Heaps of pieces?

Ok, this I know. Viennot basically invents lots of great stuff but rarely publishes his work. About "heaps of pieces" - this is a pretty little theory with very few original consequences. It is really equivalent to Cartier-Foata partially commutative monoid (available here). See Christian Krattenthaler's article for the connection and details. See there also some references to other recent papers.

Now, for many of Viennot's unpublished results, see his "Orthogonal polynomials..." book, which was unavailable for years, but is now on his web page. See there also several of his video lectures (mostly in French), where he outlined some interesting bijections based on the heaps (some related to various lattice animals were new to me, even if he may have come up with them some years ago - take a look). The reciprocal of R-R identities is an elegant single observation which he published separately: MR0989236. It really does not reprove the R-R identities, just gives a new combinatorial interpretation for one side using heaps of dimers (various related results were obtained by Andrews-Baxter a bit earlier).

About some recent applications outside of enumerative combinatorics. Philippe Marchal describes here that heaps of pieces easily imply David Wilson's theorem on Loop-erased random walks giving random spanning trees. Ellenberg and Tymoczko give a beautiful application to the diameter bound of certain Cayley graphs. Finally, in my paper with Matjaz Konvalinka, we use a heaps-of-pieces style bijection to give the "book proof" of the (non- and q-commutative) MacMahon Master theorem.

On your followup question regarding Kasteleyn's theorem and the Aztec diamond theorem - no, these are results of different kind, heaps don't really apply, at least as far as I know.

gr.group theory - Searching the symmetric group

You want to design a set of yes/no questions for quickly searching the symmetric group. The questions have to be of the form "Does your permutation move $a_1$ to $b_1$ or $a_2$ to $b_2$ or ... or $a_k$ to $b_k$?" Given a random permutation, you will ask all of your questions about that permutation, and your goal is to know what the permutation is once you know the answers to your questions.

In particular, note that you aren't allowed to have later questions depend on the answers to earlier questions. You always ask the same questions of whichever random permutation you get.

Here's a slightly different way to phrase the type of question you're allowed to ask. If we represent the elements of the symmetric group by permutation matrices, then the questions you can ask involve picking a (0,1)-matrix and asking if the random permutation matrix has any 1 where your chosen matrix has a 1 (i.e., if I apply the componentwise AND function to the random matrix and your chosen matrix, do I get 0 or 1?)

Two questions:

1. How many such questions are needed in order to search $S_n$? How much bigger than $log_2 n!$ is this?


2. Are there "good" strategies for designing sets of questions that can be used for any n? Can we achieve the minimum number of questions for each n with a single strategy?

ct.category theory - What are the products in the category of normed vector spaces with linear contractions?

Two words: sup norm.

I.e., the product of a family is the uniformly bounded subset of the cartesian product of the family, and the norm is the smallest uniform bound.

Explicitly, if ${X_i}_{iin I}$ is a family of normed vector spaces with all norms ambiguously denoted $|cdot|$, then the product is $X={{a_i}inprod X_i:sup|a_i|<infty}$, and for ${a_i}in X$, $|{a_i}|=sup|a_i|$.

(My intuition came from products of C*-algebras, where the $*$-homomorphisms are automatically contractive and products are defined in this way. So I had a good guess and it is easy to check that it works.)

motivic cohomology - Correspondences in Topology

For simplicity and definiteness, let's assume that $X$ and $Y$ are smooth and compact (and orientable, which will always be the case if they are complex varieties),
and let $n$ be the dimension of $Y$.
First of all, it might help to note that $H^n(Xotimes Y) cong H^*(X)otimes H^{n-*}(Y) cong
Hom(H^*(Y),H^*(X))$, where for the final assertion I am using that $Y$ is smooth and compact, so that its cohomology satisfies Poincare duality. Thus if $Z$ is a cycle in $Xotimes Y$,
of dimension equal to that of $X$ (and so of comdimension $n$) it induces a cycle class in $H^n(Xtimes Y)$, which in turn induces a map from cohomology of $X$ to that of $Y$.
If $f:Xto Y$ and $Z = Gamma_f$ is the graph of $f$ then this map is just the pull-back of cohomology classes by $f$.

So correspondences in the sense of physical cycles on $Xtimes Y$ induce correspondences in the sense of cohomology classes on $Xtimes Y$, which in turn
induce morphisms on cohomology. If you like, you can strengthen the analogy with the $Gamma_f$ case by thinking of a correspondence as a multi-valued function. Functions induce morphisms on cohomology;
but since cohomology is linear (you can add cohomology classes), correspondences also induce
morphisms on cohomology (you can simply add up the multiples values!). This gives the same construction as
the more formal one given above.

Ben Webster notes in his answer that geometric representation theory provides a ready supply of correspondences. So does the theory of arithmetic lattices in Lie groups and the associated symmetric space. (I am thinking of Hecke correspondences and the resulting action of Hecke operators on cohomology.)

A very general framework, which I think covers both contexts, is as follows: suppose that
a group $G$ acts on a space $X$, and that $H subset Aut(X)$ is another subgroup commensurable with $G$, i.e. such that $G cap H$ has finite index in each of $G$ and $H$.

Then (perhaps under some mild assumptions) $(Gcap H)backslash X hookrightarrow (Gbackslash X times Hbackslash X)
$ is a correspondence (in the physical, geometric sense) which will give a correspondence
in cohomology via its cycle class. The resulting maps on cohomology are (a very general form of) Hecke operators.

ca.analysis and odes - Distributions as presheaves?

While maybe not exactly what you were after, here is something that you might enjoy looking into, which relates presheaves and distributions.

There exists a category of sheaves on certain test objects, such that

• this category is a smooth topos into which the category of smooth manifolds embeds full and faithfully.




• in this topos, there exists not only a notion of infinitesimals, as in every smooth topos, but also of invertible infinitesimals, in fact, this topos provides a model for nonstandard analysis.




• Accordingly, in this topos distributions on manifolds are given by actual functions - internally in the topos.

So in a way, this topos makes precise and manifest the intuition that distributions are "generalized functions". They are functions in this topos.

The topos that I am talking about is described in great detail in section VI of the textbook Models for Smooth Infinitesimal Analysis. The test objects in this case, i.e. the objects in the site that the topos is a category of sheaves over, are smooth loci.

Distributions are discussed in section VII,3

pr.probability - Disintegrations are measurable measures - when are they continuous?

_{This is a sequel to another question I have asked.}

The notion of disintegration is a refinement of conditional probability to spaces which have more structure than abstract probability spaces; sometimes this is called regular conditional probability. Let $Y$ and $X$ be two nice metric spaces, let $mathbb P$ be a probability measure on $Y$, and let $pi : Y to X$ be a measurable function. Let $mathbb P_X(B) = mathbb P(pi^{-1} B)$ denote the push-forward measure of $mathbb P$ on $X$. The disintegration theorem says that for $mathbb P_X$-almost every $x in X$, there exists a nice measure $mathbb P^x$ on $Y$ such that $mathbb P$ "disintegrates":$$int_Y f(y) ~dmathbb P(y) = int_X int_{pi^{-1}(x)} f(y) ~dmathbb P^x(y) dmathbb P_X(x)$$
for every measurable $f$ on $Y$.

This is a beautiful theorem, but it's not strong enough for my needs. Fix a Borel set $B subseteq X$, and let $p(x) = mathbb P^x(B)$. Part of the theorem is that $p$ is a measurable function of $x$. Suppose that the map $pi : Y to X$ is continuous instead of simply measurable. My question: What is a general sufficient condition for $p(x)$ to be continuous?

To me, this is an obvious question to ask, since if $x$ and $x'$ are two close realizations of a random $x in X$, then the measures $mathbb P^x$ and $mathbb P^{x'}$ should be close too, at least in many natural situations. However, in my combing through the literature, I haven't been able to find an answer to this question. My guess is that most people are content to integrate over $x$ when they use the theorem. For my purposes, I need some estimates which I get by continuity.

At this point, I've managed to prove and write down a pretty good sufficient condition for the case I care about (Banach spaces), using an abstract Wiener space-type construction. However, I am hoping that an expert can point me toward a good reference that does this in wider generality.

books - Learning LaTeX properly

I've found l2tabu to be a very useful guide to avoiding bad habits. Yes, it's occasionally over-dramatic - one of its sections is entitled "Deadly Sins"! - but it does provide good reasons for all the taboos it lists.

But it does only cover sins of LaTeX coding, and not sins of bad typography. I tend to work on the principle that TeX and LaTeX were designed by people who know much more about typography than I ever will, so there's probably a good reason for the default settings. Thus, if in doubt, leave it alone.

That advice probably applies to the question of whether or not to create style files: unless you're creating a whole load of documents which you want to look identical in style, creating a style file only reduces the portability of your document, since you have to remember to include the extra file when you transfer it, etc., etc..

But, if you are writing style files, or class files (which are rather harder to get right!), Appendix A of the LaTeX Companion is exactly what you need, though if it gets complicated you might want to refer to the TeXbook when you need native TeX commands.

Finally, if you come to make pretty pictures and don't want to have to include EPS files and cart them around with the document, the LaTeX Graphics Companion tells you a million and one ways to get lovely diagrams with just a few lines of code; xy is particularly useful for commutative diagrams, though I mainly use pstricks for basically every kind of diagram anyone ever needs.

ct.category theory - On limits and Colimits

In general, your category has to admit small limits for that to even start to begin to make any sense at all. Also, as I noted in my question, I'm fairly sure that you've got it backwards. It should be:

Hom(colim(F(-)),X) is isomorpic to lim Hom(F(-),X), and Hom(Y,lim(F(-))) is isomorphic to lim Hom(Y,F(-)), where we're limiting and colimiting over the domain of F, where F is a functor into our category from some other category (Diagrams for example.)

I don't know if this is what you actually wanted, but if I read your question the way you typed it out, the first one doesn't make sense, since covariant hom is covariant. The second one might be true provided that the limit has certain restrictions on it or if covariant hom has an appropriate adjoint. There might be other cases, but it's not true in general. If you're just looking for the existence of a map in the second one, then it's trivial.

ac.commutative algebra - Why is an elliptic curve a group?

Short answer: because it's a complex torus. Explanation below would take as through many topics.

Topological covers

The curve should be considered over complex numbers, where it can be seen as a Riemann surface, therefore a two-dimensional oriented closed variety. How to find out whether this particular one is a sphere, torus or something else? Just consider a two-fold covering onto $x$-axis and count the Euler characteristics as $-2 cdot 2 + 4 = 0$ (don't forget the point at infinity.)

Complex tori

So this is a torus; now a torus with complex structure can be always defined as a quotient $mathbb C/Lambda$, where $Lambda$ is the lattice of periods. It can be written as integrals $int_gamma omega$ of any differential form $omega$ over all elements $gamma in pi_1$. The choice of differential form is unique up to $lambda in mathbb C$.

Algebraic addition

A complex map of a torus into itself that leaves lattice $Lambda$ fixed can be only given by a shift. Once you select a base point, these shifts are in one-to-one correspondence with points of $E$. We have unique distinguished point — infinity — so let's choose it as the base point. It follows that we now have an addition map $(u, v) to uoplus v$, though defined purely algebraically so far.

Geometric meaning

Now let's stop and ask ourselves: how to see this addition geometrically? For a start, consider map that sends $u$ to the third point of intersection with the line containing both $u$ and 0 (the infinity point). It's not hard to see that we fix 0 but change every class $gamma$ in a fundamental group into $-gamma$, so we must have the map $umapsto -u$ here.

Group theory laws

What would happen if you took a line through $u$ and $v$? By temporarily changing coordinates so that $u$ becomes the infinity point, one writes down that map as $(u, v) mapsto -(u+v)$.
Now if you took three points, there would be two different ways to add them; those would lead to $(u+v)+w$ and $u+(v+w)$ as complex numbers, which we know to be associative.

Logically proven

In the above, we worked over complex numbers, but we proved associativity which is a formal theorem about substitution of some rational expressions into others. Since it works over complex fields, it is required to work over all fields.

(In any case, the big discovery of mid-20th century was that you actually can take all of the intuition described above and apply it to the case of elliptic curves over arbitrary field)

Analytic computations (bonus)

Consider a line that passes through points $u$, $0$ and $-u$. This line is actually vertical, and $y$ is a well-defined function there which has two zeroes and one double pole at infinity. After a shift and multiplication of several such functions we'll be getting a meromorphic function on a complex torus with poles $p_i$ and zeroes $z_i$ having the property $sum p_i = sum z_i$. This method can give all such functions and only them; it's not hard to see that only meromorphic functions with this property are allowed on elliptic curve.

For example, $wp'$-functions are the ones that have triple pole at 0 and single zeroes at points $frac12w_1, frac12w_2, frac12(w_1+ w_2)$ where $w_1, w_2$ are generators of $Lambda$.

Jacobian of a curve (bonus 2)

The formula above describes what types of functions are allowed on our curve. It is a good idea to organize this information into a curve: in this case, the information is that a single expression $p_1 + p_2 + cdots + p_n - z_1 - cdots - z_n$, considered a point of the curve, must vanish. For curves of higher genus, more relations are necessary; for $mathbb Cmathbb P^1$, no relations beyond number of poles = number of zeroes are necessary. Those are relations in the group of classes of divisors (= Jacobian of a curve) mentioned in other answers.

In particular, elliptic curves coincide with their Jacobian and that's another explanation for the additive law.

quantum field theory - reference for wick product

The following article by E. R. Negrin provides the required formula for the antisymmetric Fock space
in the corollary on page 3644.

I want to point out that the Wick products (for the antisymmetric Fock space) can be constructed from a Gaussian generating function
which is Gaussian in (real) Grassmann variables, which is given for the case presented in the question by:

$G(mathbf{xi}) = exp((Sigma_{i=0}^{2k} xi_i f_i, Sigma_{j=0}^{2k} xi_j f_j))$

where $( , )$ denotes the Hilbert sapce $H_mathbb{C}$ inner product.

The required Wick product is obtained as the coefficient of $xi_1 xi_2 . . .xi_{2k}$.

at.algebraic topology - The space of Lie group homomorphisms

The general criterion is:

Given connected compact Lie groups $G,H$, $mathrm{Hom}(G,H)$ has finitely many components if and only if $H$ is semisimple or $G=1$.

Proof.
If $G=1$ there is nothing to prove.

Assume that H is not semisimple and $Gneq 1$. Then there exists a 1-dimensional torus $T$ in $H$ and a closed normal subgroup $N$ of codimension 1 such that $TN=H$. Define $T=Tcap N$, so that $Z$ is finite and define $T'=T/Z$. Then $mathrm{Hom}(T',G)$ can be identified to the set of elements in $mathfrak{g}$ whose exponential is 1; we have already seen that it has infinitely many components.

We have a continuous restriction map $mathrm{Hom}(H,G)to mathrm{Hom}(T,G)$. Its image contains $mathrm{Hom}(T',G)$ (identified to those homomorphisms trivial on $N$), and $mathrm{Hom}(T',G)$ is clopen in $mathrm{Hom}(T,G)$ (indeed, let $Z$ have exponent $n$; then 1 is an isolated point in the set $K$ of elements $gin G$ such that $g^n=1$ and $mathrm{Hom}(T',G)$ is the fiber of $(1,dots,1)$ for the mapping $mathrm{Hom}(T,G)to K^Z$ mapping $f$ to $zmapsto f(z)$). Thus $mathrm{Hom}(H,G)$ has infinitely many components (the proof even shows that it has a continuous map onto an infinite discrete set).

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Conversely, suppose $H$ semisimple.

Consider the map $L:mathrm{Hom}(H,G)tomathcal{L}(mathfrak{h},mathfrak{g})$ mapping $f$ to its tangent map between Lie algebras ($mathcal{L}(mathfrak{h},mathfrak{g})$ denoting the whole space of linear maps). Then $L$ is injective. Since $L(f)$ is locally conjugate to $f$ by the exponential map, $L$ is continuous, and actually is a homeomorphism to its image (because if $L(f_n)$ tends to $L(f)$, then in a given compact neighborhood of 1 we have $f_n$ tending to $f$ uniformly, and this implies that $f_n$ tends to $f$ uniformly on all of $H$).

If $H$ is simply connected, then the image of $f$ is equal to the set of Lie algebra homomorphisms $mathrm{Hom}(mathfrak{h},mathfrak{g})$, which is Zariski closed, and hence has finitely many components in the ordinary topology. In general (still assuming $H$ semisimple), let $tilde{H}$ be its universal covering and $Z$ the (finite) kernel of $tilde{H}to H$. Recalling that the exponential is surjective, write $Z=exp(Z')$ for some finite subset $Z'$ of $mathfrak{h}$; then the image of $L$ is the set of elements $u$ in $mathrm{Hom}(mathfrak{h},mathfrak{g})$ such that for all $zin Z'$ we have $u(exp(z))=1$. Let $mathfrak{g}_1$ be the set of elements $x$ in $mathfrak{g}$ such that $exp(x)=1$: then we have the restatement: the image of $L$ is
$${fin mathrm{Hom}(mathfrak{h},mathfrak{g}): f(Z')subset mathfrak{g}_1}.$$

Fix a faithful continuous linear representation of $G$ (in $mathrm{SO}(n)$ for some $n$), so that we have a faithful representation of $mathfrak{g}$ (by antisymmetric matrices): then $mathfrak{g}_1$ is the set of elements in $mathfrak{g}$ whose eigenvalues are in $2ipimathbf{Z}$. Note that as soon as $Gneq 1$, it has infinitely many connected components (since all these eigenvalues can be achieved, using a 1-parameter subgroup whose image in $G$ is a 1-dimensional torus).

Nevertheless, observe that (still assuming $H$ compact semisimple) $mathrm{Hom}(mathfrak{h},mathfrak{g})$ is a compact subset of the vector space $mathcal{L}(mathfrak{h},mathfrak{g})$. Indeed, $mathfrak{h}$ admits a basis $(e_j)$ such that each $e_j$ belongs to some subalgebra $mathfrak{h}_j$ isomorphic to $mathfrak{so}(3)$, such that some 2-dimensional complex representation of $mathfrak{h}_i$ maps $e_j$ to the diagonal matrix $(i,-i)$ (where $i^2=-1$). It follows from representation theory of $mathfrak{sl}_2$ that for every $n$-dimensional complex representation $rho$ of $mathfrak{h}_j$ (and hence of $mathfrak{h}$) maps $e_j$ to an element with eigenvalues in ${-i(n-1),dots,i(n-1)}$. Thus the trace of $-rho(e_j)^2$ is $le (n-1)^2n$. Since $-mathrm{trace}(XY)$ is a definite positive symmetric bilinear form on antisymmetric matrices, this shows that $mathrm{Hom}(mathfrak{h},mathfrak{g})$ is bounded, hence compact.

In particular, there exists $n_0$ such that eigenvalues of $f(z)$ for all $zin Z'$ and $finmathrm{Hom}(mathfrak{h},mathfrak{g})$ are in the interval $[-2ipi n_0,2ipi n_0]subset imathbf{R}$. Thus the image by $L$ of $mathrm{Hom}(H,G)$ is the set of $finmathrm{Hom}(mathfrak{h},mathfrak{g})$ such that for every $zin Z'$ we have $prod_{k=-n_0}^{n_0} (f(z)-2ipi k)=0$. This is a Zariski-closed subset, and hence has finitely many connected components in the ordinary topology, and hence so does the homeomorphic space $mathrm{Hom}(H,G)$.