nt.number theory - How much complex geometry does the zeta-function of a variety know

Although the question is phrased a bit sloppily, there is a standard interpretation: Given a smooth complex proper variety $X$, choose a smooth proper model over a finitely generated ring $R$. Then one can reduce modulo maximal ideals of $R$ to get a variety $X_m$ over a finite field, and ask what information about $X$ it retains. As has been remarked, the zeta function of $X_m$ gives back the Betti numbers of $X$.

I believe Batyrev shows in this paper that the zeta function of a Calabi-Yau is a birational invariant and deduces from this the birational invariance of Betti numbers for Calabi-Yau's. And then, Tetsushi Ito showed here that knowledge of the zeta function at all but finitely many primes contains info about the Hodge numbers. (He did this for smooth proper varieties over a number field, but a formulation in the 'general' situation should be possible.)

For an algebraic surface, once you have the Hodge numbers, you can get the Chern numbers back by combining the fact that

$c_2=chi_{top},$

the topological Euler characteristic, and Noether's formula:

$chi(O_X)=(c_1^2+c_2)/12.$

I guess this formula also shows that if you know a priori that $m$ is a maximal ideal of ordinary reduction for both $H^1$ and $H^2$ of the surface, then you can recover the Chern numbers from the zeta functions, since $H^1(O_X)$ and $H^2(O_X)$ can then be read off from the number of Frobenius eigenvalues of slope 0 and of weights one and two.

You might be amused to know that the homeomorphism class of a simply-connected smooth projective surface can be recovered from the isomorphism class of $X_m$. (One needs to formulate this statement also a bit more carefully, but in an obvious way.) However, not from the zeta function. If you compare $P^1times P^1$ and $P^2$ blown up at one point, the zeta functions are the same but even the rational homotopy types are different, as can be seen from the cup product in rational cohomology. See this paper.

Added: Although people can see from the paper, I should have mentioned that Ito even deduces the birational invariance of the Hodge (and hence Betti) numbers for smooth minimal projective varieties, that is, varieties whose canonical classes are nef. Regarding the last example, I might also point out that this is a situation where the real homotopy types are the same.

Added again: I'm sorry to return repeatedly to this question, but someone reminded me that Ito in fact does not need the zeta function at 'all but finitely many primes.' He only needs, in fact, the number of points in the residue field itself, not in any extension.

computer science - Counting subgraphs of bipartite graphs

Unfortunately, I don't have an answer. The following observation lets me think that a
clever brute force solution is near at hand, and may even find its way to an enumeration.

If m were n-1, the problem would be easy: determine the parity of the number of edges
of the graph G, and determine the parity of the degree of each edge. Then the number
of subgraphs on n-1 vertices with an odd number of edges is the same as the number of vertices whose degree has parity opposite that of parity of G.

Now if I were doing a brute force algorithm, and m satisfied 2*m > n, then I might
use the above observation to look at the induced subgraphs on (m+1) vertices and
then do the calculations over a covering set of the (m+1) subgraphs. Or there might
be a 2-step version where I could use pairs of vertices in an (m+2) subgraph. It is tempting to think there is also an (n-m)-step version using generating functions or Polya enumeration, but I don't know enough to tell you about those.

When I am able to comment, then I might turn this musing from an answer into a comment.

Gerhard "Ask Me About System Design" Paseman, 2010.02.04

ag.algebraic geometry - Projective closure of affine curve

When the curve is a plane curve of degree $d$, the formula is simple (and in fact, this works for any hypersurface) you take $f(x,y)=0$ and replace it by $z^d f(x/z,y/z)=0$. This will be homogeneous of degree $d$, and when $zneq 0$, you recover your curve.

Now, if you have a more general affine variety, given by $langle f_1,ldots,f_krangle$ in $k[x_1,ldots,x_n]$, then you first compute a Groebner Basis, which you can learn how to do using Cox, Little and O'Shea's "Ideals, Varieties and Algorithms" (assumes no background whatsoever) and then you do this same trick with each polynomial in the Groebner basis.

Why can't you just use any basis? Look at the twisted cubic. $tmapsto (t,t^2,t^3)$. This is cut out, in affine space, by $y-x^2$ and $z-x^3$, though $y-x^2$ and $xy-z$ are better to use. But still, $yw-x^2$ and $xy-zw$ don't give the twisted cubic! Bezout tells us that they give something of degree four containing the cubic, and it's not a hypersurface, so you get the twisted cubic plus a line. To get the cubic itself, you have to use a third quadratic polynomial that you get in the ideal, $y^2-xz$, which is not in the ideal given by the homogenized generators. These three, however, form a Groebner basis for the ideal, and then homogenization gives the homogeneous ideal of the projective twisted cubic.

Good Algebraic Number Theory Books

In particular, in view of the focus of your studies, I suggest the following additional book; where additional is meant that I would not suggest it as the only book (see below for explanation).

There is a fairly recent book (in two volumes) by Henri Cohen entitled "Number Theory" (Graduate Texts in Mathematics, Volumes 239 and 240, Springer).
[To avoid any risk of confusion: these are not the two GTM-books by the same author on computational number theory.]

It contains material related to Diophantine equations and the tools used to study them, in particular, but not only, those from Algebraic Number Theory.
Yet, this is not really an introduction to Algebraic Number Theory; while the book contains a chapter Basic Algebraic Number Theory, covering the 'standard results', it does not contain all proofs and the author explictly refers to other books (including several of those already mentioned).

However, I could imagine that a rich exposition of how the theory you are learning can be applied to various Diophantine problems could be valuable.

Final note: the book is in two volumes, the second one is mainlyon analytic tools, linear forms in logarithms and modular forms applied to Diophantine equations; for the present context (or at least initially), the first volume is the relevant one.

mp.mathematical physics - Do there exist small neighborhoods in a classical mechanical system without pairs of focal points?

The question I will ask makes sense in much more generality, but I will leave the translation to the experts, since I'm only looking for a special case (and it would not surprise me if the answer does not generalize). I will give some background, and then ask my question as a conjecture, set apart from the main text.

Let $mathbb R^n$ have its usual metric, and pick a differential one-form (= vector field) $B$ (the "magnetic potential") and a differential zero-form (= function) $C$ (the "electric potential"). Then consider the following second-order ODE for parameterized paths $gamma: [0,T] to mathbb R^n$:

$$0 = ddot gamma + dB cdot dotgamma + dC quadquad text{(EOM)}$$
I'll let you pick the signs for how the two-form $dB$ eats the vector $dotgamma$; just be consistent.

Then (EOM) is nondegenerate, and so a solution is determined by its initial conditions $(dotgamma(0),gamma(0))$. For each $T in mathbb R$, let $phi_T: mathbb R^{2n} to mathbb R^n$ be the "flow by time $T$" (actually, it is defined only on an open subset of $mathbb R^{2n}$, given by $phi_T(v,q) = gamma(T)$ for the solution $gamma\,$ to (EOM) with initial conditions $(dotgamma(0),gamma(0)) = (v,q)$. Then $phi_T$ is smooth; in fact, it is smooth in the $T$ variable as well. This follows from some standard fundamental result in ODEs, for which I don't have a good reference.

A path $gamma: [0,T] to mathbb R^n$ is classical if it satisfies (EOM); its duration is the number $T$. We can also consider paths with negative duration by flowing backwards, although we will not need to do so.

Definition: A point $(v,q) in mathbb R^{2n}$ is focal for duration $T$ iff ($phi_T(v,q)$ is defined and) $det(partial phi_T(v,q)/partial v) = 0$; i.e. fix the $q$, think of $phi_T(-,q)$ as a function of $v$ only, and ask that its differential is degenerate. By identifying $(v,q)$ with its classical path, we will talk about "focal (classical) paths" for given durations.

It is a standard results (see e.g. Milnor's Morse Theory) that for a given point $(v,q) in mathbb R^{2n}$, the durations $Tin mathbb R$ for which it is focal are discretely separated. Note that every $(v,q)$ is focal for duration $T=0$.

Proposition: Let $gamma$ be a classical path of duration $T$. Then it is non-focal if and only if it extends to a family of classical paths smoothly parametrized by the boundary positions $(gamma(0),gamma(T))$.

Sketch of Proof: Being focal for duration $T$ is a closed condition on $mathbb R^{2n}$, so we can vary $gamma(0) = q$ while remaining non-focal. But for non-focal paths we can vary $gamma(T)$ via the inverse function theorem.

Anyway, pick $q in mathbb R^n$, and $v = B(q)$ (or $-B(q)$ depending on your sign convention: for experts, I want the momentum to vanish). Then for some $epsilon>0$, for all $Tin (0,epsilon)$, $(v,q)$ is non-focal for duration $T$. Thus, for each $T in (0,epsilon)$, I can find an open neighborhood $q in mathcal O_0 subseteq mathbb R^n$ and another open neighborhood $mathcal O_1 subseteq mathbb R^n$ so that for $(q_0,q_1) in mathcal O_0 times mathcal O_1$, there is a non-focal classical path $gamma$ of duration $T$ with $gamma(0) = q_0$, $gamma(T) = q_1$, depending smoothly on the boundary conditions, and such that the classical path of duration $T$ and initial conditions $(dotgamma(0),gamma(0)) = (v,q)$ is contained within this family.

Note that as $T to 0$, the classical path with initial conditions $(dotgamma(0),gamma(0)) = (v,q)$ ends at a point very close to $q$. I don't know if I can take $mathcal O_1$ to actually contain $q$.

I would like to reverse the direction of choices: I'd like to pick $mathcal O_0,mathcal O_1$ first.

Question/Conjecture: Let $q in mathbb R^n$. Then there exist open neighborhood $mathcal O_0,mathcal O_1 subseteq mathbb R^n$, with $q in mathcal O_0,mathcal O_1$, and $epsilon>0$ such that:

1. There exists a family of classical paths $gamma$ with boundary values varying in $mathcal O_0,mathcal O_1$ and with duration varying in $(0,epsilon)$. I.e. let $Delta = { (T,t) in mathbb R^2 : T in (0,epsilon), tin [0,T] }$; then there is a smooth function $gamma: mathcal O_0 times mathcal O_1 times Delta to mathbb R^n$ with: (a) $gamma(q_0,q_1,T,-)$ is classical for each $(q_0,q_1,T) in mathcal O_0 times mathcal O_1 times (0,epsilon)$, and (b) $gamma(q_0,q_1,T,0) = q_0$ and $gamma(q_0,q_1,T,T) = q_1$.


2. For each $T in (0,epsilon)$, the classical path of duration $T$ with initial conditions $(B(q),q)$ appears as some $gamma(q,q_1,T,-)$.

For comparison, the corresponding theorem about geodesics on a Riemannian manifold is standard: around any point you can find a small neighborhood such that any two points in the neighborhood can be connected by a unique geodesic that does not leave the neighborhood. In fact, it follows from the proposition and the observation that changing the duration of a geodesic for fixed boundary conditions amounts just to a linear reparameterization.

pr.probability - Notions of "independent" and "uncorrelated" for subsets of the natural numbers

There are some ways to assign probability measures to the set of antural numbers. Consider the probability measure $P_s$ on the positive integers which assigns "probability" $n^{-s}/zeta(s)$ to the integer $n$. ($s$ is a constant real number greater than $1$.)

Then under this measure being a multiple of $r$ and a multiple of $s$ are independent events, in the probabilistic sense, if $r$ and $s$ don't have a common multiple. You can show this starting form the fact that the measure assigned to the set of multiples of $k$, for some positive integer $k$, is

$${1 over zeta(s)} sum_{n=1}^infty {1 over (kn)^s} = {1 over zeta(s)} {1 over k^s} zeta(s) = {1 over k^s}.$$
That is, the probability that a random positive integers is divisible by $k$ is $k^{-s}$. Of course you really want all integers to be equally likely, which should correspond to $s = 1$.

(I learned this from Gian-Carlo Rota, Combinatorial Snapshots. Link goes to SpringerLink; sorry if you don't have access.)

Under "suitable conditions", which I don't know what they are because Rota doesn't say, the density of any set of natural numbers $A$ is the limit $lim_{s to 1^+} P_s(A)$.

In particular it might be reasonable to define correlation between sets of natural numbers in the same way. Let $A$ and $B$ be two sets of natural numbers. Let $X$ and $Y$ be the indicator random variables of the sets $A$ and $B$ in the measure $P_s$. The Pearson correlation coefficient between $X$ and $Y$ is

$${(E(XY) - E(X) E(Y)) over sigma_X sigma_Y }$$
where $E$ is expectation and $sigma$ is standard deviation. Of course this can be simplified in the case where $X$ and $Y$ are indicators (and thus only take the values $0$ or $1$) -- in particular it simplifies to
$${P_s(A cap B) - P_s(A) P_s(B) over sqrt{P_s(A) P_s(B) (1-P_s(A)) (1-P_s(B))}}$$
We could then deifne the correlation between $A$ and $B$ to be the limit of this as $s to 1+$.

In the case where $A$ is the event divisible by 2'', for example, and $B$ is the eventdivisible by 3'', then $A cap B$ is the event ``divisible by 6''. So $P_s(A cap B) = 6^{-s}$, $P_s(A) = 2^{-s}$, and $P_s(A) = 3^{-s}$, so the numerator here is 0 and so the correlation is zero.

But in the case where $A$ is the event divisible by 4'' and $B$ is the eventdivisible by 6'', then $A cap B$ is the event ``divisible by 12''. So the correlation with respect to $P_s$ is

$${12^{-s} - 24^{-s} over sqrt{4^{-s} 6^{-s} (1-4^{-s}) (1-6^{-s})}}$$
which has the limit $1/sqrt{15}$ as $s to 1^+$; more generally the correlation between being divisible by $a$ and being divisible by $b$ is
$${ab - lcm(a,b) over lcm(a,b) sqrt{(a-1)(b-1)}}$$
and this may or may not be what you want.

ca.analysis and odes - cauchy product for general case

It's not a problem to multiply the series: the product is $sum_{(t,k)inmathbb Z^2} a_tb_k$. The question is how to sum the double series that we have.

For series with nonnegative terms summation is not a problem either: we take the supremum of all finite sums. And since any finite sum is contained in a sufficiently large square, it follows that $sum_{(t,k)inmathbb Z^2} |a_tb_k|$ is finite whenever $sum_{tinmathbb Z} |a_t|$ and $sum_{kinmathbb Z} |b_k|$ are.

In general, $sum_{(t,k)inmathbb Z^2} a_tb_k=S$ if for any $epsilon>0$ there is a finite subset $Asubset mathbb Z^2$ such that $|sum_{(t,k)in B}a_tb_k - S|<epsilon$ whenever $B$ is finite and $Bsupset A$. Now if both given series converge absolutely, then the contribution from outside of a large square is small, and it follows that $S$ is the product of two sums.

homotopy theory - Natural examples of finite dimensional spaces with interesting 2-type

In low dimensional topology there are host of examples along the lines of the curve complex of a surface. Someone like Andy Putman will be able to provide a more detailed explanation, but here is a summary. You build a simplicial complex in which vertices are isotopy classes of closed curves in a surface, and n-simplices are disjoint (n+1)-tuples of curves. If you require that no curve bounds a disc (or an annulus if your surface has boundary) then it is fairly easy to see that the complex is finite dimensional. The mapping class group of the underlying surface acts on this complex.

Now, the homotopy types of this complex and its variations, and of the quotient by the mapping class group, are very interesting objects. In particular, studying the low dimensional homotopy groups allow you to do things like construct presentations for the mapping class group.

This sort of construction is really a small industry in low dim topology/geometric group theory. Braid groups, automorphism groups of free groups, surface mapping class groups, 3-manifold mapping class groups, and various subgroups of these, can all be studied via complexes of this type, and often we only understand the complex as far as its 2-type.

knot theory - What are the points of Spec(Vassiliev Invariants)?

Background

Recall that a (oriented) knot is a smoothly embedded circle $S^1$ in $mathbb R^3$, up to some natural equivalence relation (which is not quite trivial to write down). The collection of oriented knots has a binary operation called connected sum: if $K_1,K_2$ are knots, then $K_1 # K_2$ is formed by spatially separating the knots, then connecting them by a very thin rectangle, which is glued on so that all the orientations are correct. Connect sum is commutative and associative, making the space of knots into a commutative monoid. In fact, by a theorem of Schubert, this is the free commutative monoid on countably many generators. A ($mathbb C$-valued) knot invariant is a $mathbb C$-valued function on this monoid; under "pointwise" multiplication, the space of knot invariants is a commutative algebra $I$, and $#$ makes $I$ into a cocommutative bialgebra. I.e. $I$ is a commutative monoid object in $(text{CAlg})^{rm{op}}$, where $text{CAlg}$ is the category of commutative algebras.

Warm-up question: Any knot $K$ defines an algebra morphism $I to mathbb C$, i.e. a global point of $I in (text{CAlg})^{rm{op}}$. Are there any other global points?

Edit: In response to Ilya N's comment below, I've made this into its own question.

Finite type invariants

Recall that a singular knot is a smooth map $S^1 to mathbb R^3$ with finitely many transverse self-intersections (and otherwise it is an embedding), again up to a natural equivalence. Any knot invariant extends to an invariant of singular knots, as follows: in a singular knot $K_0$, there are two ways to blow up any singularity, and the orientation determines one as the "right-handed" blow-up $K_+$ and the other as the "left-handed" blow-up $K_0$. Evaluate your knot invariant $i$ on each blow-up, and then define $i(K_0) = i(K_+) - i(K_-)$. Note that although the connect-sum of singular knots is not well-defined as a singular knot, if $iin I$ is a knot-invariant, then it cannot distinguish different connect-sums of singular knots. Note also that the product of knot invariants (i.e. the product in the algebra $I$) is not the point-wise product on singular knots.

A Vassiliev (or finite type) invariant of type $leq n$ is any knot invariant that vanishes on singular knots with $> n$ self-intersections. The space of all Vassiliev invariants is a filtered bialgebra $V$ (filtered by type). The corresponding associated-graded bialgebra $W$ (of "weight systems") has been well-studied (some names: Kontsevich, Bar-Natan, Vaintrob, and I'm sure there are others I haven't read yet) and in fact is more-or-less completely understood (e.g. Hinich and Vaintrob, 2002, "Cyclic operads and algebra of chord diagrams", MR1913297, where its graded dual $A$ of "chord diagrams" is described as a sort of universal enveloping algebra). In fact, this algebra $W$ is Hopf. I learned from this question that this implies that the bialgebra $V$ of Vassiliev invariants is also Hopf. Thus it is a Hopf sub-bialgebra of the algebra $I$ of knot invariants.

I believe that it is an open question whether Vassiliev invariants separate knots (i.e. whether two knots all of whose Vassiliev invariants agree are necessarily the same). But perhaps this has been answered — I feel reasonably caught-up with the state of knowledge in the mid- to late-90s, but I don't know the literature from the 00s.

Geometrically, then, $V in (text{CAlg})^{rm{op}}$ is a commutative group object, and is a quotient (or something) of the monoid-object $I in (text{CAlg})^{rm{op}}$ of knot invariants. The global points of $V$ (i.e. the algebra maps $V to mathbb C$ in $text{CAlg}$) are a group.

Main Questions

Supposing that Vassiliev invariants separate knots, there must be global points of $V$ that do not correspond to knots, as by Mazur's swindle there are no "negative knots" among the monoid $I$. Thus my question.

Main question. What do the global points of $V$ look like?

If Vassiliev invariants do separate knots, are there still more global points of $V$ than just the free abelian group on countably many generators (i.e. the group generated by the free monoid of knots)? Yes: the singular knots. (Edit: The rule for being a global point is that you can evaluate any knot invariant at it, and that the value of the invariant given by pointwise multiplication on knots is the multiplication of the values at the global point. Let $K_0$ be a singular knot with one crossing and with non-singular blow-ups $K_+$ and $K_-$, and let $f,g$ be two knot invariants. Then

$$begin{aligned} (fcdot g)(K_0) & = (fcdot g)(K_+) - (fcdot g)(K_-) = f(K_+)cdot g(K_+) - f(K_-)cdot g(K_-) neq \\ f(K_0) cdot g(K_0) & = f(K_+)cdot g(K_+) - f(K_+)cdot g(K_-) - f(K_-)cdot g(K_+) + f(K_-)cdot g(K_-)end{aligned}$$ .) What else is there?

What can be said without knowing whether Vassiliev invariants separate knots?

lie groups - A terminology issue with the Killing Form

Let me add a few comments to the answers by Mariano and Theo.

There is a one-to-one correspondence between bi-invariant metrics (of any signature) in a Lie group and ad-invariant nondegenerate symmetric bilinear forms on its Lie algebra.

In a simple Lie algebra every nondegenerate symmetric bilinear form is proportional to the Killing form which you wrote down in your question, hence a simple Lie group has precisely one conformal class of bi-invariant metrics. If (and only if) the group is compact, are these metrics positive-definite. (Some people call this riemannian, reserving the word pseudo-riemannian (or sometimes also semi-riemannian) for indefinite signature metrics. Personally I prefer to use riemannian for general metrics.)

One can ask the question: which Lie groups admit bi-invariant metrics of any signature? which is the same thing as asking which Lie algebras admint ad-invariant non-degenerate symmetric bilinear forms. Such Lie algebras are called metric (or also sometimes quadratic, orthogonal,...) and although there is no classification except in small index (index 0 = positive-definite, index 1 = lorentzian, etc...) there is a structure theorem proved by Alberto Medina and Philippe Revoy in this paper (in French). Their theorem says that the class of such Lie algebras is generated by the simple and the one-dimensional Lie algebras under two operations: orthogonal direct sum and double extension, a construction explained in that paper.
Double extension always results in indefinite signature, so if you are only interested in the positive-definite case, you get back to the well-known result that every positive-definite metric Lie algebra $mathfrak{g}$ is isomorphic to the orthogonal direct sum of a compact semisimple Lie algebra and an abelian Lie algebra, or in other words,

$$mathfrak{g} cong mathfrak{s}_1 oplus cdots oplus mathfrak{s}_N oplus mathfrak{a}$$
where the $mathfrak{s}_i$ are the simple factors and $mathfrak{a}$ is abelian.
Up to automorphisms, the most general positive-definite inner product on such a Lie algebra is given by choosing for each simple factor $mathfrak{s}_i$ a positive multiple $lambda_i > 0$ of the Killing form.

These Lie algebras are precisely the Lie algebras of compact Lie groups. Their metricity can also be understood as follows: take any positive-definite inner product on $mathfrak{g}$ and averageng it over the adjoint representation.

So in summary, although there are metric Lie algebras which are not semisimple (or even reductive), their inner product is always an additional structure, unlike the Killing form which comes for free with the Lie algebra.

As for question concerning the difference between Killing form and Cartan(-Killing) metric it depends on who says this. In much of the Physics literature people refer to an inner product on a vector space as a "metric". But assuming that this is not the case, then the Killing form is a bilinear form on the Lie algebra, whereas the metric is a metric (in the sense of riemannian geometry) on the Lie group. If $G$ is a Lie group whose Lie algebra is semisimple, then the Killing form on its Lie algebra defines a bi-invariant metric on the Lie group, which I suppose you could call the Cartan-Killing metric.

computational complexity - Practical use of probability amplification for randomized algorithms

There seems to be some confusion in your notation. When you says error $epsilon$, do you mean the probability of failure is $epsilon$, which means that the algorithm outputs the wrong answer with probability $epsilon$. In this case, if $epsilon$ is polynomially small, that's great. If you run it a few times you should keep getting good answers.

But then you talk about the case where the error is $1/2 - epsilon$ (in the comments), where $epsilon$ is inverse polynomial. This situation is bad, because your algorithm will seem to output random bits. Thus you should error amplify to get constant error (say 1% error). The number of steps taken for this is logarithmic in $1/epsilon$.

So in conclusion, if the error is like $1/2 - 1/poly$, you can amplify this to 1/3 in polynomial time. Once your error is a constant, you can also amplify it to inverse exponential, i.e., $epsilon = 1/exp$ in polynomial time.

sp.spectral theory - An inverse eigenvalue problem on Jacobi matrices

I am interested in trying to design a Hermitian Jacobi (tridiagonal) matrix $H$ that has specific properties. The basic property, which is simple enough to construct, is that for an $Ntimes N$ matrix with basis elements $|1rangle$ to $|Nrangle$, it should be capable of generating the evolution $|langle N|e^{-iHt}|1rangle|^2=1$ for some time $t$. However, I'm also interested in trying to make the system have the property $|langle 1|e^{-iHt'}|1rangle|^2=0$ at regular intervals in the intervening times (and specifically, how small that interval can be).

I've reduced this problem to a question about the eigenvalues $lambda_n$ of the matrix, for $n=1,dots,N$, such that $lambda_N > lambda_{N-1} > dots > lambda_1 =0$. If $n$ is even $lambda_n$ is an even integer and if $n$ is odd, $lambda_n$ is an odd integer. We define

$R_k=sum_{lambda_n equiv k rm{ mod } M}frac{(-1)^n}{B'(lambda_n)}$

for $k=0,dots,M-1$ for some integer $M$, and where $B'(lambda)$ is the derivative of the function
$B(lambda)=prod_{n=1}^N(lambda-lambda_n)$ with respect to $lambda$.

I need to determine what the maximum value of $M$ is such that all the $R_k$ can be equal, and would like an example of the set of eigenvalues that achieves this. If this is not possible, are there any examples of sets of eigenvalues such that the $R_k$ are all equal for $M>2$?

In this context, $langle a|H|brangle$ denotes the matrix element $H_{ab}$.

oc.optimization control - Maximizing Sparsity in l1 Minimization?

Consider the optimization problem

$$min_x ||Ax||_1 + lambda||x-b||^2,$$

where $A in mathbb{R}^{n times n}$, $x,b in mathbb{R}^n$ and $lambda$ is strictly greater than 0. (This problem is closely related to the "lasso" problem in basis pursuit.) Can anything be said about the value of $lambda$ for which $Ax^*$ is sparsest? Clearly some values are bad: for instance, if $lambda$ is huge and $b$ is dense then it is unlikely that $Ax^star$ will be very sparse.

In other words: among all $lambda > 0$ there is at least one value $lambda^star$ such that $||Ax^star(lambda)||_0$ is minimized. Are there, say, bounds on $lambda^star$ in terms of $A$ and $b$? I'd also be interested in results pertaining to basis pursuit or other similar problems.

Edit: I'm primarily interested in problems where ideal sparsity cannot be achieved, i.e., $||Ax^star(lambda^star)||_0 > 0.$ (Assume that $A$ is square w/ full rank and $b ne 0$.)

ag.algebraic geometry - Is anything known about the enumeration of degree d, genus g curves in CP^2 where g >1 ?

There is a combinatorial formula for the number of geometric genus $g$ curves of degree $d$ (possibly reducible) passing through $3d-1+g$ generic points of $mathbb{P}^2$ (or a more general toric surface) derived by tropical techniques, see

Grigory Mikhalkin, Enumerative tropical algebraic geometry in $mathbb{R}^2,$
J. Amer. Math. Soc. 18 (2005), no. 2, 313–377 MR

ct.category theory - "Philosophical" meaning of the Yoneda Lemma

Here is an example on representable functors. Yoneda's lemma gives down-to-earth, morpshim oriented interpretation of representable functors, and vice versa.

I will explain this with an example.

In a category $mathscr{C}$, the product of $A$ and $B$ is the pair of object $Atimes B$ in $mathscr{C}$ and a fixed natural isomorphism

$$sigma colon mathrm{Hom}(-,Atimes B)to mathrm{Hom}(-,A)times mathrm{Hom}(-,B).$$

This definition of products only uses terminology of functors. By applying Yoneda's lemma, we arrive at a morphism oriented definiton of products. Yoneda's lemma says that there is a bijection

$$Psi colon mathrm{Hom}left( mathrm{Hom}(-,Atimes B),mathrm{Hom}(-,A)times mathrm{Hom}(-,B)right) to mathrm{Hom}(Atimes B,A)times mathrm{Hom}(Atimes B,B).$$
In particular, we apply this to $sigma$ and denote
$$Psi(sigma)=sigma(Atimes B)(mathrm{id}_{Atimes B})=(pi^{A}colon Atimes Bto A,pi^{B}colon Atimes Bto B).$$
Next, by applying the inverse of $Psi$, we compute
$$sigma(X)=Psi^{-1}left( Psi(sigma)right)(X):mathrm{Hom}(X,Atimes B)to mathrm{Hom}(X,A)times mathrm{Hom}(X,B)$$
$$fcolon Xto Atimes Bmapsto (pi^{A}circ f,pi^{B}circ f).$$
Since $sigma$ is a natural isomorphism, $sigma(X)$ is a bijection. This bijectivity is the usual definition of product based on morphisms (universality):

For any pair of morphisms $f^{A}colon Xto A$ and $f^{B}colon Xto B$, there exists a unique morphism $fcolon Xto Atimes B$ with $pi^{A}circ f=f^{A}$ and $pi^{B}circ f=f^{B}$.

I think the Philosophy behind Yoneda's lemma is that, it connects the world of functors (and natural transformations) $mathfrak{Set}^{mathscr{C}^{mathrm{op}}}$ and the world of morphisms $mathscr{C}$.

nt.number theory - Gaps in nx (mod 1)

This reminds me of the following cute statement (which I believe has been recently generalized to arbitrary manifolds, but I couldn't find the paper): Given the sequence of points $kalpha (mod 1)$, there are at most 3 different distances between nearest neighbors.

Actually, I first heard it stated as $alpha^k$ on the unit circle. To prove it, notice that the first cycle of points around the circle all have the same length between them, and then there is the remainder at the end.

When the wrap occurs, this is the remainder just shifted (because the distance between two consecutive points is always the same), and it keeps appearing between previous points in the same way at each step (and then there are the remaining points it has yet to appear between). This still leaves only 3 lengths (the two from the "cut" it produced in the previous length, and the previous lengths it has yet to cut).

I hope this helps.

ag.algebraic geometry - When can we cancel vector bundles from tensor products?

The answer to your question is yes, if $F$ is a direct sum of line bundles. So, let us assume that and its rank is $m$. And as you observed, we will assume that we are working over $mathbb{P}^2$. Let $0to F_0to F_1to Eto 0$ be the minimal resolution of $E$, where $F_i$ are direct sum of line bundles with rank of $F_0=n$ so that rank of $F_1=n+m$. Tensoring with $G$, we get $0to Gotimes F_0to Gotimes F_1to Gotimes Fto 0$, the last by the assumption. Taking cohomologies, letting $H_*^0(G)=M,H_*^1(G)=N$, we get an exact sequence $N^nto N^{n+m}to N^m$, which for length considerations (length of $N$ is finite) can easily seen to be exact on the left (and right). In particular, we have surjectivity of global sections and thus an exact sequence $0to M^nto M^{n+m}to M^mto 0$. This splits (I have forgotten whose theorem it is, but I think Ihave read it recently in some comment by Graham Leuschke) which is impossible since the first map has all entries in the maximal ideal unless $n=0$.

ag.algebraic geometry - If we can define a topology on the set of all the ideals of a commutative ring?

Obviously, it is possible to define some topology. However, the obvious generalization of the Zariski topology, to take a set to be closed if and only if it is of the form $V(I) = {J : I subset J}$ for some fixed ideal $I$, does not work since this collection of sets is not generally closed under finite unions. Consider the set $A$ of all ideals of $mathbb{Z}$ containing (2) or (3). If $A$ were of the form $V(n)$ for any $n in mathbb{Z}$, then $(n) in A$; thus, would contain (2) or (3), i.e., n would divide 2 or 3, which is absurd. However, the set of all prime ideals containing (2) or (3) is equal to the set of all prime ideals containing (6).

Nevertheless, using schemes, one can give geometric meaning to the set of all ideals of a ring. Specifically, the ideals of $R$ correspond naturally to the set of closed subschemes of Spec $R$.

rt.representation theory - An identity for sheaf cohomology of flag varieties

Let $G$ be a connected complex semisimple Lie-group, $T$ a maximal torus and $B$ a Borel subgroup containing it. Let $phi:Grightarrow G/B$ denote the projection.
Given a representation ($theta,V$) of $B$, we can define a $G$-equivariant holomorphic vector bundle over the flag variety $X:=G/B$ by

$$Gtimes_B V :=(Gtimes V)/{(g,v)sim(gb^{-1},theta(b)v),forall bin B}.$$
Its sheaf of sections $mathcal{I}(theta)$ may be described as the holomorphic functions
$$mathcal{I}(theta)(U)={f:phi^{-1}(U)rightarrow V mid f(gb^{-1}) = theta(b)f(g)}.$$
$G$ acts on a section by $(gf)(x)=f(g^{-1}x)$.

An integral weight $lambda$ of $T$ gives a character $chi_lambda$ of $B$. Let $thetaotimeschi_lambda$ denote the tensor product of the representations $theta$ and $chi_lambda$.
Suppose ($theta,V$) is the restriction of a representation ($pi,V$) of $G$, then the associated ($G$-equivariant) vector bundle is trivial (i.e. isomorphic to $Xtimes V$ with $(g,(x,v))mapsto (gx,pi(g)v)$). Is the identity

$$mathrm{H}^i(X,mathcal{I}(thetaotimeschi_lambda))simeq mathrm{H}^i(X,mathcal{I}(chi_lambda)) otimes V$$
as $G$ or $mathfrak{g}$-modules correct?

Background/Motivation

I have been reading about the Borel-Weil theorem lately, and

Edit:
This question arose from an attempt to fix a mistake in a book.
The identity is indeed correct and I believe I have found the error elsewhere.
Thanks Chuck and Jim!

at.algebraic topology - Applications of the Brown Representability Theorem

I think it might be pretty hard to get an answer that says nothing about cohomology. I interpret/think of Brown Representability as saying that if you want to think about these types of invariants you should really look at spectra. That somehow cohomology is really about spectra. Once you see this you then interpret the axioms for a cohomology theory or rather what you know about singular cohomology and try to think of the structure you have on the representing object that comes from this and you get ring spectra. I dont know if non-topologists really think about these things or if i just think that they should. They bring up interesting questions about operads... that may have come up independently.

You also get cohomology operations for any cohomology theory from this, but only as a theoretical tool, that is i am not completely sure how to use Brown Representability to compute Adams operations, in fact i dont think that we know all of $K^*K$.

This is a pretty biased answer though, but it is a unifying idea of algebraic topology that we should look at ring spectra, or rather the representing objects of all "nice" cohomology theories.

anyone should feel free to appropriately fix this, especially if they have better historical or computational information

ct.category theory - Iterated adjoint functors

A simple example where there are adjoint strings of arbitrary length is given by the simplex category, or rather the simplex 2-category, the sub-2-category of Cat whose objects are finite ordinals (so the 1-cells or functors are order-preserving maps, and the 2-cells or transformations are instances of the order relation f ≤ g). Notice that the functor 0: [1] --> [2] = {0, 1} is left adjoint to the unique functor !: [2] --> [1] which is left adjoint to the functor 1: [1] --> [2] = {0, 1}.

Using this and the monoidal structure, you can generate adjoint strings of arbitrary length which zig-zag between the cofaces i_k: [n] --> [n+1] and codegeneracies p_k: [n+1] --> [n]. Specifically, if i_0 < i_1 < ... < i_n name the n+1 injections [n] --> [n+1] and p_1 < ... < p_n name the n surjections [n+1] --> [n], then there is an adjoint string of the form

$i_0 dashv p_1 dashv i_1 dashv ldots dashv p_n dashv i_n$

and clearly there is no periodicity here.

measure theory - What is the "continuity" in "absolute continuity", in general?

The wikipedia article on absolute continuity gives a delta-epsilon definition for a measure $mu$ defined on the Borel $sigma$-algebra on the real line, with respect to the Lebesgue measure $lambda$:

$mu<<lambda$ if and only if for every $epsilon>0$ and for every bounded real interval $I$ there is a $delta>0$ such that for every (finite or infinite) sequence of pairwise disjoint sub-intervals [$x_i,y_i$] of $I$ with

$sum_{i} |y_i - x_i| < delta$

it follows that

$sum_{i} |mu((-infty, y_i])-mu((-infty, x_i])| < epsilon$.

My questions are: Does this connection generalise? What would be a topological reformulation of $mu<<nu$? If the notion does not generalise for arbitrary measures, does it generalise for the Lebesgue measure on $R^n$? How would a sketch of the proof look like?

ag.algebraic geometry - Why can't subvarieties separate?

I'm posting my answer to this question as its own question:

Let $V$ be an irreducible projective variety over $mathbb{C}$. Let $U$ be a Zariski open set in $V$. I'll use $V(mathbb{C})$ and $U(mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.

What is the easiest proof that $U(mathbb{C})$ is connected?

Here's the proof I know: Suppose that $U(mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$. Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $overline{A}$ and $overline{B}$ of $A$ and $B$ in $V(mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $overline{A}$ and $overline{B}$ are subvarieties of $V$. Since they're proper, $V$ is not irreducible, and we have a contradiction.

I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively. A reference would be fine.

(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine. It is just not obvious to me that it is irreducible (without appealing to the above argument).)

ag.algebraic geometry - Does this condition reduce to the correct notion of irreducibility on schemes?

Consider the category of sheaves (of sets) on the affine étale site. It's a well known fact that a morphism of schemes is a Zariski-open immersion if and only if it is an étale monomorphism, so we extend this idea to all sheaves as follows:

We say that a sheaf $S$ satisfies the condition if given any two étale monomorphisms $Ato S$ and $Bto S$, then $Atimes_S B$ is representable by $Spec(0)$ (the image of the 0 ring in the opposite category of commutative rings) if and only if either $A$ or $B$ is representable by $Spec(0)$.

Motivation: It would be nice if we could define the notion of irreducibility only in terms of functors of points. The condition that we are trying to simulate is the intersection of two open subsets of an irreducible topological space being empty if and only if one of the open subsets is empty. The problem is that the fiber-product of schemes does not necessarily coincide with the fiber product of the underlying topological space.

Questions:

Are there any cases of schemes where this condition and irreducibility are not equivalent?

If this definition does work for schemes, does it work for algebraic spaces (perhaps with some tweaking)?

Edit: Recall that a morphism of sheaves $Fto G$ is an étale monomorphism if it is a monomorphism and the pullback (fiber product) by any morphism from an affine scheme $Xto G$ is an algebraic space with an atlas of affine schemes given by ${U_ito F times_G X}$ such that the composition $U_ito F times_G Xto X$ with the projection is an étale morphism of affine schemes (maps corresponding to étale maps of rings).

tag removed - Sum Equals Product

Let the positive, unequal integers be $a_1 < a_2 < cdots < a_k$ with $a_1+a_2+cdots+a_k = a_1a_2cdots a_k = n$. Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$. Note that $a_k ge k$. Then we have $a_1 + a_2 + cdots + a_k < ka_k$, while $a_1a_2cdots a_k ge (k-1)!a_k$. So we must have $ka_k > (k-1)!a_k$, which means that $k > (k-1)!$. This means $k le 3$. If $k = 2$, then we have $a_1 + a_2 = a_1a_2$, or $frac{1}{a_1}+frac{1}{a_2} = 1$. This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a_i$ be unequal.

The case $k = 3$ is then the only tricky case. If $a_1 > 1$, then we have $a_1+a_2+a_3 < 3a_3$ and $a_1a_2a_3 ge 6a_3$. So $a_1 = 1$. We then must find solutions to $1 + a_2 + a_3 = a_2a_3$. This can be rewritten as $(a_2-1)(a_3-1) = 2$, which as $a_2 < a_3$ are integers immediately gives $a_2 = 2, a_3 = 3$.

In summary, all possible solutions are ${n}$ for all positive integers $n$ and ${1,2,3}$.

History: this is definitely a classic problem; see for example 2006 USA Mathematical Olympiad problem #4 (pdf, problem is on second page), which is your problem with several restrictions removed.

ct.category theory - Completion of the rationals to the reals as an inverse limit construction?

Although your question is not at all vague, there are a few completely different ways to interpret what a good answer would be. Knowing a bit about your personal preferences, I suspect the following is not at all what you wanted, but it is still of interest to the community.

The fact that $mathbb{Q}_p$, for example, can be seen as an inverse limit boils down to the fact that the series representations

$$sum_{n in mathbb{Z}} a_n p^{-n}$$
give a homeomorphism between $mathbb{Q}_p$ and the subspace of ${0,1,dots,p-1}^{mathbb{Z}}$ of eventually zero sequences (on the positive side). At first glance, it would seem that the same works for $mathbb{R}$ since every element of $[0,infty)$ has a very similar binary expansion
$$sum_{n in mathbb{Z}} a_n 2^n$$
where $(a_n)_{n in mathbb{Z}}$ is again in the space of all eventually zero sequences in ${0,1}^{mathbb{Z}}$. Unfortunately this is not a bijection. In fact, there is no way to select such a sequence $(a_n)_{ninmathbb{Z}}$ continuously — binary rationals are always points of discontinuity no matter how the selection is made.

There is a funny way to remedy this which is commonly used in Computable Analysis and Reverse Mathematics. The idea is to represent real numbers again with series of the form

$$sum_{n in mathbb{Z}} a_n 2^n$$
but where $(a_n)_{n in mathbb{Z}}$ is now an eventually zero sequence from the extended set ${-1,0,1}^{mathbb{Z}}$. The obvious cost is that no number has a unique representation in this form, but the side benefit is that the binary rationals are no longer 'special' in this way. In the end, this representation is extremely well behaved compared to ordinary binary expansions. This can be seen from the fact that any invariant continuous function between representations (see note) gives rise to a continuous function $mathbb{R}tomathbb{R}$, and every continuous function $mathbb{R}tomathbb{R}$ admits such a lifting.

The conclusion to draw from this is that, up to some very nice blurring, $mathbb{R}$ is indeed an inverse limit just like $mathbb{Q}_p$.

Note: The topology on eventually zero sequences is not exactly the product topology, it is given by the metric

$$d(vec{x},vec{y}) = inf{2^n : forall m geq n,({x_m = y_m})}.$$ The invariance of a function $f$ between eventually zero sequences from ${-1,0,1}^{mathbb{Z}}$ means that if $vec{x}$ and $vec{y}$ represent the same real number then so do $f(vec{x})$ and $f(vec{y})$.

ag.algebraic geometry - Moduli space of flat bundles

Is there a good place to learn about the structure of moduli stack of flat $G$-bundles on an algebraic curve?

Of course, we're just studying representations of a group $pi_1(X)to G$ modulo some conjugation (that's why it should be a stack). Since this is very similar to Galois representations in number theory, I wonder if there's a reference that also explains the similarities and differences between the two cases.

Tropical mathematics and enriched category theory

Is there a connection between tropical mathematics and the Lawvere enriched category theory approach to metric spaces? I guess I will give a partial answer to this below, but I mean can they be formally be put on the same level in some sense?

In the Lawverian point of view one does category theory with the extended non-negative real numbers, [0,∞] or R_≥0∪∞, equipped with + as the 'tensor' product and max as the 'categorical' product or sum. In tropical mathematics you work (it seems) with the the extended reals R∪∞ equipped with the 'product' + and the 'sum' max (or min depending on your point of view I think).

In the enriched category theory approach to metric spaces, one has the notion of a kernel (or bimodule or profunctor depending on your point of view) between two metric spaces X and Y which is just a distance non-increasing function K:X×Y->[0,∞]. The correct notion of function on a metric space here is a distance non-increasing function φ:X->[0,∞]. Then the transform of a function φ by a kernel K is a function on Y defined by

K(φ)(y):= inf_xεX ( φ(x) + K(x,y) ).

There is similarly a dual notion which takes functions on Y to functions on X.

K^{^}(ψ)(x):= sup_yεY ( ψ(y) - K(x,y) ).

This is explained in a bit more detail in a post in at the n-Category Café.

It was pointed out to me that these look similar to the Legendre transform. And looking on the internet I found that tropical mathematics is one way to interpret the Legendre transform as an 'integral transform'.

So has anyone ever considered any formal connections between these two points of view?

ag.algebraic geometry - In what degrees does Ext(S/(f),S) vanish?

Let $S=k[x_0,...,x_n]$ be the polynomial ring over a field $k$ and $fin S$ non-zero and homogeneous. Is it true that $Ext^m(S/(f),S)$ is zero?

This would help me to show that $Ext^m(S/fI,S)cong Ext^m(S/I,S)(deg f)$ for $mgeq 2$ and a homogeneous ideal $I$ of codim $geq 2$. I tried the following approach:
Applying the long exact sequence of $Ext$ to the exact sequence of graded $S$-modules

$$0to S/Ixrightarrow{cdot f} S/fI(deg f)xrightarrow{tau}S/(f)(deg f)to 0,$$
where $tau$ is the canonical morphism $s+fImapsto s+(f)$,
brings
$$ldotsto Ext^m(S/(f)(deg f),S)to Ext^m(S/fI(deg f),S)to Ext^m(S/I,S)to$$
$$Ext^{m+1}(S/(f)(deg f),S)to Ext^{m+1}(S/fI(deg f),S)to Ext^{m+1}(S/I,S)toldots$$
and two zeros on the left would suffice.

geometry - Systems of conics

As long as you can get an elementary proof of the fact that one of the families consists of ellipses with foci $A=(-c,0)$ and $B=(c,0)$ and the other consists of hyperbolas with the same foci, you can say that for any intersection point $P$ the angle between the lines $PA$ and $PB$ is dissected by the tangent to either curve - otherwise moving on the tangent would cause a first-order error in the sum $|PA|+|PB|$ or the difference $|PA|-|PB|$, respectively. Hence the two tangents are just the two angle bisectors of a pair of lines, and are thus orthogonal.

linear algebra - Is the direction of the longest line of a polytope unique?

I think the answer to your question is yes. edit: NO

First I'll set some notation. Assume that $X$ is rank $m$. I'll denote by $L$ the m-dimensional plane defined by $X u = y$. Subscripts will denote components of vectors. Instead of $u_{max}$ I'll use $v$. I'll denote by $A_{v}$ the hypercube $0leq u_ileq v$ for $1leq ileq p$. The problem as stated is about the intersection between $A_v$ and $L$, which is a polytope I'll call $P_v$.

We can rescale the coordinates by taking $urightarrow vu'$ so that $A_v$ has side length 1 in the $u'$ coordinates. Under this transformation, $L$ keeps its orientation but is shifted. In particular, $L$ is now defined by $X (vu')=y$, or $X u'=y'$ where $y' = 1/v*y$. As $v$ gets larger and larger, $y'$ gets closer and closer to the origin. Note that if $y$ were the zero vector, your problem is scale invariant and hence has a positive answer.

If $y$ was not the zero vector, then to understand what $P_v$ looks like for large $v$, we need to understand how a slice through the hypercube behaves very close to one of its vertices. Is there a result (for convex polytopes in general?) that tells us that the "shape" of a slice is stable to small translations of the slicing plane when we're close to a vertex? I haven't found any counter-examples in the low-dimensional cases I've (unsystematically) tried.

edit: I spoke way to soon. Consider a plane slicing through the 3-dimensional cube such that the plane makes right angles with the top and bottom faces of the cube. In general the intersection will be a rectangle whose aspect ratio changes and becomes skinnier and skinnier as the plane gets closer to a vertex. The direction of the longest segment in this rectangle (either of the diagonals) obviously does not stabilize. But is there a positive result lurking here for suitably "generic" planes?

gr.group theory - Extension of induced reps over Z: is it a sum of induced reps?

Let $G$ be a finite group. If $L$ is a finite free $mathbf{Z}$-module with an action of $G$, say $L$ is induced if it's isomorphic as a $G$-module to $Ind_H^G(mathbf{Z})$ with $H$ a subgroup of $G$ and $H$ acting trivially on $mathbf{Z}$. And, for want of better terminology, let's say $L$ is a sum-of-induceds if it's isomorphic to a direct sum of induced modules in the sense above. [EDIT: Ben Webster points out that "permutation representation" is a rather better name for this notion! It's just the $mathbf{Z}$-module coming from the action of $G$ on a finite set.]

The question: Is a finite free $mathbf{Z}$-module which is an extension of one sum-of-induceds by another, also a sum-of-induceds?

Over $mathbf{Q}$ this is trivial because every short exact sequence splits. But this is not true over $mathbf{Z}$. For example there are two actions of $mathbf{Z}/2mathbf{Z}$ on $mathbf{Z}^2$ which become the group ring over $mathbf{Q}$: one has the non-trivial element acting as $(1,0;0,-1)$ and the other has it acting as $(0,1;1,0)$. The latter is a non-split extension of the trivial 1-d representation by the non-trivial one (and also a non-split extension of the non-trivial one by the trivial one). Note that this non-split extension is induced.

There are mod $p$ extensions that don't split either. For example over $mathbf{Z}/pmathbf{Z}$ there is a non-trivial extension of the trivial representation by itself. But this extension does not lift to an extension of the trivial $mathbf{Z}$-module by itself.

Why am I interested? For those that know what a $z$-extension of a connected reductive group over a number field is, my "real" question is: is a $z$-extension of a $z$-extension still a $z$-extension? I've checked the geometric issues here but the arithmetic one above is the one I haven't resolved. $G$ is a Galois group and the $mathbf{Z}$-modules are the character groups of the central tori in question. If I've understood things correctly, a $z$-extension of a $z$-extension is a $z$-extension iff the question I ask above has a positive answer.

Note finally that applying the long exact sequence of cohomology, and using the fact that induced representations have no cohomology by Shapiro's Lemma, we see that the extension I'm interested in also has no cohomology (and furthermore its restriction to any subgroup has no cohomology either). Is this enough to show it's induced?

dg.differential geometry - Does every smooth manifold of infinite topological type admit a complete Riemannian metric?

To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.

I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.

lie algebras - Figure out the roots from the Dynkin diagram

This question is answered, probably in many textbooks on Lie algebras, Chevalley groups, representation theory, etc.. I always think that Bourbaki's treatment of Lie groups and algebras is a great place to look (but I don't have it with me at the moment). I also tend to look things up in Humphreys, and Knapp's "Lie Groups, Beyond an Introduction"

Here's a method that's very good for pen-and-paper computations, and suffices for many examples. I believe that it will algorithmically answer your question in general as well.

The key is the "rank two case", and the key result is the following fact about root strings. I'll assume that we are working with the root system associated to a semisimple complex Lie algebra. Let $Phi$ be the set of roots (zero is not a root, for me and most authors).

Definition:
If $alpha, beta$ are roots, the $beta$-string through $alpha$ is the set of elements of $Lambda cup { 0 }$ of the form $alpha + n beta$ for some integer $n$.

Theorem: There exist integers $p leq 0$ and $q geq 0$, such that the $beta$-string through $alpha$ has the form:

$${ alpha + n beta : p leq n leq q }.$$
Furthermore, the length of this string can be determined from the inner products:
$$p+q = frac{ -2 langle alpha, beta rangle }{langle beta, beta rangle}.$$

Example: Let $alpha$ be the short root and $beta$ be the long root, corresponding to the two vertices of the Dynkin diagram of type $G_2$. The $alpha$-string through $beta$ consists of:

$$beta, beta+alpha, beta + 2 alpha, beta + 3 alpha,$$
as remarked in the question.

Here $p = 0$ and $q = 3$. Indeed, we find that

$$3 = 0+3 = frac{ -2 langle beta, alpha rangle }{langle alpha,alpha rangle} = frac{(-2)(-3)}{2},$$
using the Cartan matrix (which is easily encoded in the Dynkin diagram).

Perhaps you already knew this, since you found those roots "clear". But from here, you can continue again with $alpha$ and $beta$ root strings through these roots.

For example, let's consider the $beta$-string through $beta + 3 alpha$. These are the roots of the form $beta + 3 alpha + n beta$, for integers $p leq n leq q$. We know that $p = 0$, since for $n = -1$ we find that $3 alpha$ is not a root. The length of the root string is:

$$q = frac{ -2 langle beta+3alpha, beta rangle}{langle beta, beta rangle} = frac{ -2(2 - 3)}{2} = frac{2}{2} = 1.$$
From this we find that (corresponding to $n = q = 1$), there is another root $2 beta + 3 alpha$, and furthermore, $m beta + 3 alpha$ is not a root for $m geq 3$.

By using root strings, together with bounds on how long roots can be, one can find all of the roots without taking too much time. It should also be mentioned that, for a simple root system, there is a unique "highest root", in which the simple roots occur with maximal multiplicity. The multiplicities of simple roots in the highest root can be looked up in any decent table, and computed quickly by hand for type A-D-E (using a trick from the McKay correspondence). This is useful for a bound, so you don't mess around with root strings for longer than necessary.

gn.general topology - A question about homeomorphic subsets of Hilbert space

Yes. All we need is to construct a continuous on $H$ function $f$ that is unbounded on $C$. After that, ${x,f(x)}subset Htimes mathbb C$ is closed, unbounded, and homeomorhic to $C$ (in the obvious way) and $Htimes mathbb C$ is isometric to $H$. Being closed and non-compact in an arbitrary separable metric space $X$ implies the existence of such a function. The simplest construction is to take a countable open cover $U_j$ of $X$ that contains no finite subcover of $C$ and to put $f(x)=min f_j(x)$ where $f_j(x)=frac j{min(1,operatorname{dist}(x,Xsetminus U_j))}$. Separability has actually nothing to do with it but in the non-separable case things become a bit more complicated.

ac.commutative algebra - Class groups of normal domains over finite fields

As requested in the comments, here's an example of a local, normal $2$-dimensional domain R in positive characteristic such that $mathrm{Cl}(R)$ is not torsion: choose an elliptic curve $E subset mathbf{P}^2$ over a field $k$ such that $E(k)$ is not torsion, and take R to be the local ring at the origin of the affine cone on $E$ (i.e., $R = k[x,y,z]/(f)_{(x,y,z)}$ where $f$ is a homoegenous cubic defining $E$). This can be done over $k = overline{mathbf{F}_p(t)}$.

Proof: The normality follows from the fact that R is a hypersurface singularity (hence even Gorenstein) and isolated and $2$-dimensional (hence regular in codim 1). Blowing up at the origin defines a map $f:X to mathrm{Spec}(R)$. One can then show the following: $X$ is smooth, and $X$ can be identified with the Zariski localisation along the zero section of the total space of the line bundle $L = mathcal{O}_{mathbf{P^2}}(-1)|_E$ (these are general facts about cones). By Lipman's theorem, it suffices to show that $mathrm{Pic}^0(X)$ contains non-torsion elements. As $X$ is fibered over $E$ with a section, the pullback $mathrm{Pic}^0(E) to mathrm{Pic}^0(X)$ is a direct summand. As $mathrm{Pic}^0(E) simeq E(k)$ has non-torsion elements by assumption, so does $mathrm{Pic}^0(X)$.

Also, an additional comment: In general, Lipman's theorem tells you that $mathrm{Cl}(R)$ is torsion if and only if $mathrm{Pic}^0(X)$ is torsion. Now $mathrm{Pic}(X) simeq lim_n mathrm{Pic}(X_n)$ where $X_n$ is the $n$-th order thickening of the exceptional fibre $E$. Because we are blowing up a point, the sheaf of ideals $I$ defining $E$ is ample on $E$. The kernel and cokernel of $mathrm{Pic}(X_n) to mathrm{Pic}(X_{n-1})$ are identified with $H^1(E,I|_E^{otimes n+1})$ and $H^2(E,I|_E^{otimes n+1})$. As $I|_E$ is ample, it follows that the system "$lim_n mathrm{Pic}(X_n)$" is eventually stable. Thus, $mathrm{Pic}(X) simeq mathrm{Pic}(X_n)$ for $n$ sufficiently big. As $X_n$ is a proper variety, it follows that if we are working over a finite field (resp. an algebraic closure of a finite field), then $mathrm{Pic}^0(X)$ is finite (resp. ind-finite).

Solution for a system of PDEs

I recently came across a system of PDEs
$frac{partial S}{partial z}= f_1(x,y,z,w,t)$,
$frac{partial S}{partial w}= f_2(x,y,z,w,t)$,
$frac{partial S}{partial t}= f_3(x,y,z,w,t)$,
$S(x,y,1,1,1)=f_4(x,y)$,
where $S$ is an unknown function of five variables $x,y,z,w,t$ and $f_i$ are known.
The question is how to obtain a general solution for $S$?

soft question - What are some slogans that express mathematical tricks?

"It is easy to prove existence when there is only one, or when there are many"

explanation:

If there is only one object with a certain property, you can sometimes use it to define it. For example, in geometric situations, you can sometimes define it locally and glue the patches since uniqueness guarantees compatibility on overlaps. It suggests that you should try proving uniqueness before proving existence and if uniqueness fails, maybe you should add constraints (thus, paradoxically, adding constrains can help in proving existence). On the other hand, sometimes it is easier to prove that there are many than to point out one specific example (transcendental numbers, continues nowhere differentiable functions,...). Therefore, you may want to seek for the right notion of "many" in your universe (cardinality, measure, "topological bigness" like the baire property,...) and try to prove that actually there are "few" objects that don't have the required property.

comment: This relates to the answer saying that when you can't avoid making a choice, make all of them simultaneously. This happens when there are more than one, but not many...

abstract algebra - Best way to teach concept of real numbers using a hands-on activity?

There is an argument to be made that the real numbers, by which I mean the completed reals, does not belong in an algebra course. In the answer above, all of the motivation comes from the real-algebraic-closure of $mathbb{Q}$ (the largest algebraic field extension not containing $sqrt{-1}$).

The reason one might want to introduce the whole continuum comes from numbers like $e$ and $pi$, which are transcendental (the fact that these numbers are transcendental is not immediate and requires a proof that I would consider past middle-school level). If you're willing to state those facts without proof, you can give a moral argument for $e$ by showing that it is the limit of the sequence $((1+1/n)^n)_{nin mathbb{N}}$, which is Cauchy, and its inclusion in the real numbers follows from the completeness of $mathbb{R}$. However, this argument may still be somewhat sophisticated for a middle-school algebra course.

Edit: On Prof. Clark's suggestion, I've copied my comments into the body text (with an additional section as well):

The notion of a sequence converging to a limiting value has a very intuitive geometric interpretation, so it wouldn't be hard to give a geometric argument (say on a graph, for example) that $e$ is a real number, since after relatively few iterations, the graph does level out. Showing that it is not the solution to a polynomial is effectively proving that it is transcendental, and I can't think of an informal argument showing this, but since this answer is community wiki, if someone has an idea, this would give a "moral" argument for the study of the "whole" continuum.

I find this approach useful because it can be introduced using the compound interest formula, which is often taught in an introductory algebra class.

We can see this as follows

$$A=P(1+r/n)^{nt}.$$

Let $N:=n/r$. Then we have

$$A=P(1+1/N)^{Nrt},$$
which we can rewrite as $$A=P((1+1/N)^N)^{rt}.$$

If we increase $N$ and leave $r$ fixed, this is equivalent to increasing $n$ (this is obvious because $r>0$), which amounts to increasing the number of compounding periods per unit time. Taking the limit (in some informal geometric sense), we can see that as we increase the number of compounding periods, we approach the continuously compounded interest formula

$$A=Pe^{rt}.$$

oc.optimization control - Navigation solution for frictionless vehicles.

Looked around a bit and couldn't seem to find a similar question. (either that or it was worded with vocabulary above the multivariable calculus I've taken. :))

Roughly worded: I would like to develop an algorithm (either in the form of "action to take each discrete time step" or "do these actions at exactly these times") for navigating a rigid body vehicle time-optimally in a frictionless environment from point A to point B.

This specific instance happens to be a Spaceship, whom I want to get from point A to point B utilizing Forward thrusters, reverse thrusters (for braking), torque thrusters, and an omnidirectional thruster for minor position/velocity corrections. The environment is 2 dimensional, though if someone knows preexisting work in 3 dimensions I can extrapolate a simpler solution from that. I've looked around the web a bit, and was unable to find anything other than some work on steering behaviors, which always assume point particles and thus do not factor torque into the equations.

I've worked for a couple of days on this problem using standard Newtonian equations, (p(t) = at^2/2 + vt + p(0)), but the entire problem is polluted by the torque calculations, such that torque is applied to turn towards and then slow down and stop at an angle that changes based on the objects velocity and the time it would take to turn to that angle. :-S 9 pages of scribbling and several frustrated nights later, a friend Reccomend I ask here.

A generic solution (which incorporates initial velocity and angle) with a high degree of accuracy would be preferred, though in the end if I have to just "fake it" to look nice (space based RTS), that would be fine too.

nt.number theory - Existence of multi-variable p-adic L-functions

What's the "state of the art" in constructing multi-variable p-adic L-functions for number fields?

More precisely: if K is a number field, and $K_{infty} / K$ is an infinite Galois extension, unramified outside a finite set S of primes of K, containing the cyclotomic $mathbb{Z}_p$ - extension K_cyc of K and whose Galois group G is abelian with an open subgroup isomorphic to ${mathbb{Z}_p}^d$ for some $d$, then does there exist an element of the Iwasawa algebra $mathbb{Z}_p[[G]]$ (or some localisation of it) whose values at finite order characters χ of $G$ are the special values of the Hecke L-functions $L_S(chi, 1)$ (with the Euler factors at primes in S removed)?

When K is totally real Leopoldt's conjecture forces $d = 1$; but I'm interested in other cases. I know there is a 2-variable Iwasawa main conjecture for imaginary quadratic fields, which I understand has been proved by Rubin, but I'm just asking about the existence of the L-function (not about any connection to annihilators of class groups). What is known in this direction for more general $K$?

Suggestions for a good Measure Theory book

Well, I personally HATE Halmos' Measure Theory, even though an entire generation grew up on it. My favorite book on measure and integration is available in Dover paperback and is one of my all time favorite analysis texts: Angus Taylor's General Theory Of Functions And Integration. Lots of wonderful examples and GREAT exercises along with discussions of point set topology, measure theory both on $mathbb{R}$ and in abstract spaces and the Daneill approach. And all written by a master analyst with lots of references for further reading. It's one of my all time favorites and I heartily recommend it.
Folland's Real Analysis is a fine book, but it's much harder and it's really more of a general first year graduate analysis course. On the plus side, it does have many applications, including probability and harmonic analysis. It's definitely worth having, but it's going to take a lot more effort then Taylor. The IDEAL thing to do would be to work through both books simultaneously for a fantastic course in first year graduate analysis.
And please don't torture yourself with Rudin's Real And Complex Analysis. It's
sole purpose seems to be to see how much analysis can be crammed incomprehensibly into a single text. Folland is the same level and is much more accessible.
That should get you started. Good luck!

reference request - A Riemannian metric on S^2 times S^2 of nonnegative curvature that is not a product

One has to be careful with perturbations of metrics of nonnegative curvature, because that may introduce negative curvature.

Here's another approach which gives you a nonnegatively curved metric.

Start with $S^3times S^2$ with the product of round metrics. (Note that the round metric on S^3 is its biinvariant metric).

Consider the $S^1$ action on $S^3 times S^2$ where it acts as the Hopf action on $S^3$ and simultaneously rotates the $S^2$ factor $2k$ times around for some integer $k$.

To make it explicit, thinking of $S^2$ as the unit sphere in the imaginary quaternions, the action can be described as $z*(p,q) = (zp, z^k q overline{z}^k)$.

The action is clearly free and the quotient is diffeomorphic to $S^2 times S^2$. Since the circle is acting isometrically, there is an induced submersion metric on $S^2times S^2$. By the O'Neill formulas for a submersion, this metric has nonnegative curvature. When k = 0, one gets the usual product of round metrics, but when $kneq 0$ the metric is, in general, not a product.

Edit I'm now no longer certain that for $kneq 0$, the metric is not a product. I am confident that for $kneq 0$, the metric is not a product of round metrics, but I don't see any reason they can't be a product of two nonnegatively curved metrics.

However, here is an example (sorry for doubling the length of my post!): Let $G = S^3times S^3$. Let $g_0$ denote a biinvariant metric on $G$. Writing $mathfrak{g}$ for the Lie algebra of $G$, set $mathfrak{p}$ to be the Lie algebra of the diagonal $S^3$ and choose $mathfrak{q}$ to be $g_0$-orthogonal to $mathfrak{p}$.

Fix a positive real number $t$ and define a new inner product $g_1 = g_0|_{mathfrak{q}} + frac{t}{t+1}g_0|_{mathfrak{p}}$ and left translate it around $G$ to give a left invariant, right $Delta S^3$ invariant metric. Such a metric is called a Cheeger deformation of $g_0$ and it is known that $g_1$ has nonnegative sectional curvature.

Give $Gtimes G$ the product metric $g_0+g_1$ and consider the space $Delta S^3 backslash Gtimes G/ T^2$ where the $T^2$ acts on $Gtimes G$ as $(z,w)*(p,q,r,s) = (pz^{-1}, q, rw^{-1},sw^{-1})$.

(The map $Gtimes Grightarrow G$ sending $(p,q,r,s)$ to $(r^{-1}p, s^{-1}q)$, or something like it if I've made a mistake, induces a diffeomorphism between $Delta S^3backslash Gtimes G/T^2$ and $G/T^2 = S^2times S^2$, where the $G/T^2$ is referring to the action of $T^2$ on $G$ spelled out before the edit with k=1).

As above, there is an induced submersion metric of nonnegative sectional curvature by the O'Neill formulas. Finally, to prove that this is NOT a product metric, one observes that at generic points, there is a unique plane with 0 sectional curvature, while for a product metric, there should be infinitely many planes of 0 curvature.

The observation comes from

P.Müter, Krümmungserhöhende Deformationen mittels Gruppenaktionen, Ph.D. thesis, University of Münster, 1987.

modular tensor categories - Do all 3D TQFTs come from Reshetikhin-Turaev?

If you have a 3d TQFT, with no anomaly, and which goes down to points, and where things are sufficiently finite and semisimple, then I think you can show that it comes from a Turaev-Viro type construction on the 2-category Z(pt).

If you have a 3d TQFT, possibly with anomaly, which goes down to circles, and where things are sufficiently finite and semisimple, then I agree with Noah: Z(S^1) is a MTC and the RT construction on the MTC reproduces the TQFT.

Relating these two statements, TV(C) = RT(double(C)), where C is a 2-cat, double(C) is the Drinfeld double (or maybe center), TV is the Turaev-Viro construction, and RT is the Reshetikhin-Turaev construction.

it.information theory - Question regarding divergence

The inequality $D(P'|Q') ge D(P^star| Q^star)$ does not need to hold.

Here is an example.

Let $A$ be the set ${1,2,3,...,n}$. Let $E$ be the set of measures $P$ on $A$ such that $P({1}) = 0$. Projecting a measure $P$ on $E$ using $D$ is equivalent to conditioning $P$ on $A- {1}$. Choose $P'$ and $Q'$ such that they both put equal and nonzero mass on ${1}$. By direct computation one sees: $D(P^star| Q^star) = frac{1}{1-P'({1})} D(P'|Q') > D(P' | Q')$.

The details of the above computation are as follows.

For ease of notation set $n=3$. Let $E$ be the set of measures $P$ with $P({1}) =epsilon$; to obtain the example above, one sets $epsilon = 0$. Let us parametrize the measures on ${1,2,3}$ as follows: $P({1}) = p_1$, $P({2}) =p_2$ and $P({3}) = 1-p_1 -p_2$. Our problem is:

$$inf_{ Q in E}left[ p_1 log frac{p_1}{q_1} + p_2 log frac{p_2}{q_2} + (1-p_1 -p_2) logfrac{ 1- p_1 - p_2}{ 1- q_1 - q_2 } right].$$
Let $F$ denote the expression after the $inf$.
$F$ is strictly convex in $Q$ and therefore will have a unique optimizer. In the above coordinates, the normal to $E$ is the vector $(1,0)$. Then
$$frac{partial F} {partial q_1} = -frac{p_1}{q_1} + frac{1-p_1-p_2}{1-q_1-q_2} = lambda$$
and
$$frac{partial F} {partial q_2} = -frac{p_2}{q_2} + frac{1-p_1-p_2}{1-q_1-q_2} = 0.$$
We have the constraint that $Qin E$, i.e., $q_1 =epsilon$.
From the last two equalities one infers:
$$q_2 = frac{(1-epsilon) p_2}{ 1-p_1}.$$

Going back to the coordinates $(p_1,p_2,p_3)$ to denote a measure on ${1,2,3}$,
projecting a measure on $E$ using $D$ corresponds to the following map:

$$(p_1,p_2,p_3) rightarrow left(epsilon, (1-epsilon)frac{p_2}{p_2+p_3}, (1-epsilon)frac{p_3}{p_2 + p_3}right).$$
For $epsilon =0$, this is the same as conditioning $P$ on ${2,3}$.

One obtains the expression for the relative entropy given above by directly computing it using this formula for the projections.