st.statistics - Using chi-squared?

(This question relates to a problem I had at work a while ago, doing a little data mining at a car rental company. Names changed, of course.)

There was a flight of steps out the front of our building. It had a dodgy step on it, on which people often stub their toes.

I had records for everyone who works in the building, detailing how many times they climbed these steps and how many of these times they stubbed their toes on the dodgy step. There's a total of 3000 stair-climbing incidents and 1000 toe-stubbing incidents.

Jack climbed the steps 15 times and stubbed his toes 7 times, which is 2 more than you'd expect. What's the probability that this is just random, vs the probability that Joe is actually clumsy?

I'm pretty sure from half-remembered statistics 1 that its something to do with chi-squared, but beats me where to go from there.

...

Of course, we actually had several flights of steps, each with different rates of toe stubbing and instep bashing. How would I combine the stats from those to get a more accurate better likelihood of Joe being clumsy? We can assume that there's no systematic bias in respect of more clumsy people being inclined to use certain flights of steps.

ag.algebraic geometry - Why is Proj of any graded ring isomorphic to Proj of a graded ring generated in degree one?

I have seen it stated that Proj of any graded ring $A$, finitely generated as an $A_0$-algebra, is isomorphic to Proj of a graded ring $B$ such that $B_0 = A_0$ and $B$ is generated as a $B_0$-algebra by $B_1$.

Could someone either supply a reference for or a sketch a proof of this statement?

Note: An obvious approach to this question is to make $B$ a Veronese subring of $A$. However, when I try this approach, I end up getting a terrible combinatorics problem that I do not know how to approach.

set theory - a question about diagonal prikry forcing

Suppose <kappa_n|n<omega> is a strictly increasing sequence of measurable cardinals,

kappa is the limit of this sequence. For each n<omega, U_n is a normal measure on

kappa_n. P is the diagonal Prikry forcing corresponding to kappa_n's and U_n's.
Suppose g is P-generic sequence over V. We have known that for each strictly increasing

sequence x of length omega such that each x(i)<kappa_i and xin{V}, x is eventually

dominated by g. In V[g], suppose A is a subset of kappa, A is not in V. Is there a strictly

increasing sequence y of length omega such that each y(i)<kappa_i and yin{V[A]}, y is not

eventually dominated by g?

(g can eventually dominate all such sequences in V, V[A] is greater than V, I feel g can not

eventually dominate all such sequences in V[A].)

random matrices - Probability on the distance

Let $A$ be an $ntimes n$ gaussian matrix whose entries are i.i.d.
copies of a gaussian variable, and $left{ a_{j}right} _{j=1}^{n}$
be the column vectors of $A$. How to show that the probability
$mathbb{P}left(dgeq tright)leq Ce^{-ct}$
for some $c,C>0$ and every $t>0$, where $d$ is the distance between
$a_{1}$ and the $n-1$-dimensional subspace spanned by $a_{2},cdots,a_{n}$.

Thanks a lot!

sg.symplectic geometry - Manifolds distinguished by Gromov-Witten invariants?

Here is an answer to the REFINED question given to me by Richard Thomas.
In this refined version we want an example such that the cohomology
classes of two symplectic forms coincide.

In a later paper 1996, Duke Vol. 83
TOPOLOGICAL SIGMA MODEL AND DONALDSON TYPE
INVARIANTS IN GROMOV THEORY, Ruan proved that such refined examples exist.
He admitted in this paper that for the $Vtimes S^2$
examples from the paper in JDG 1994
(cited by Mike Usher) he does not know if the classes of
constructed symplectic forms can coincide too. In fact
this does not seem very plausible.

These refined examples are two $3$-dimensional Calabi-Yau manifolds,
constructed by Mark Gross. The construction is described in the paper
of Mark Gross (1997): "The deformation space of Calabi-Yau $n$-folds with canonical singularities can be obstructed". One $3$-dimensional Calabi-Yau
is a smooth anti-canonical section of $P^1times P^3$ and the over is
a smooth anti-canonical section of the projectivsation of the bundle
$O(-1)+O+O+O(1)$ over $P^1$.

The construction of Gross is recalled on the pages 47-48 of
http://xxx.soton.ac.uk/PS_cache/math/pdf/9806/9806111v4.pdf

Using Wall's theorem Ruan proves that these two Calabi-Yau
manifolds are differomorphic. Then he studies the quantum cohomology
rings of these Calabi-Yaus and proves that there they are different.

cherednik algebra - Quantum equivariant $K$-theory and DAHA.

My best guess is that either

1. this is true for $mathbb{CP}^1$, and it's pretty easy to generalize that given what's already in that paper, or


2. this is false for $mathbb{CP}^1$, and you're hosed.

The bit one has to understand is the map from the 2 point genus 0 moduli space to the Steinberg variety. BMO get away with just noting that the two spaces have the same dimension, so the pushforward of the fundamental class of the moduli space has to be a sum of fundamental classes of components of Steinberg, whose coefficients they work out by deforming to an almost generic situation and doing the calculation for $mathbb{CP}^1$.

I think by looking at the pushforward of the structure sheaf on the 2-point moduli space, you'll find that quantum correction is some K-class on the Steinberg variety and thus something in the affine Hecke algebra, and I think it should be the sum of SL(2) contributions for each root by the same deformation arguments that BMO use.

---

I just spoke to Davesh Maulik about this, and it seems my intuition has failed me: he claims it is just hard, and the techniques of that paper will not work.

nt.number theory - Modular forms with prime Fourier coefficients zero

Write $f=sum c_i f_i$ as a sum over new eigenforms. Your condition is thus equivalent to $sum c_i lambda_i(p)=0$ for all $p$. Taking the absolute value squared of this and summing over $pleq X$ gives

$0=sum_{i,j}c_i overline{c_j} sum_{pleq X} lambda_i(p)overline{lambda_j(p)}$.

By the pnt for Rankin-Selberg L-functions, the inner sum over primes is $sim X (log{X})^{-1}$ if $i=j$, and is $o(X (log{X})^{-1})$ otherwise. Taking $X$ very large we obtain $0=cX(log{X})^{-1}+o(X(log{X})^{-1})$, so contradiction.

reference request - Differential equation for a ratio of consecutive Bessel functions

I do not know how helpful this would be to you however it was very helpful to understand the physics and numerics of Bessel.
if you are studying elastic wave propagation. The solution of the differential equations of potential wave is the cylindrical Bessel:

$<math>r^2 frac{d^2 R}{dr^2} + r frac{dR}{dr} + (r^2 - alpha^2)R = f(r)</math>$

for an arbitrary real integer number α (the '''order''' of the Bessel function). In solving problems in cylindrical coordinate systems, Bessel functions are of integer order (α = ''n''). Since this is a second-order differential equation, there must be two [[linearly independent]] solutions. My solutions use Bessel J(n,.) and Hankel H(n,.) (as previously mentioned)

The potential is assumed for each media to be:

$<math>phi=left(a_{1}J_{n}(K*r)+a_{2}*H_{n}(K*r)right)*e^{intheta} ,</math>$

$<math>psi(t) = left(a_{3}J_{n}(k*r)+a_{4}*H_{n}(k*r)right)*e^{intheta} ,</math>$

However for numerical stability:
They (ref. 1) normalize the potential for each layer and for each nth iteration The potential will have Hankel function equal to 1 at the inner radius, while Bessel J will be multiplied by Hankel at outer radius.

$<math>phi=left(a_{1}J_{n}(K*r)*H_{n}(K*r_{out})+a_{2}*frac{H_{n}(K*r)}{H_{n}(K*r_{in})}right)*e^{intheta},</math>$

$<math>psi(t)=left(a_{3}J_{n}(k*r)*H_{n}(k*r_{out})+a_{4}*frac{H_{n}(k*r)}{H_{n}(K*r_{in})}right)*e^{intheta},</math>$

For more details:
Bessel functions in wave propagation and scattering
Reference:
David C. Ricks and Henrik Schmidt, "A numerically stable global matrix method for cylindrically layered shells excited by ring forces" 1994

fa.functional analysis - Show a linear operator is not compact

Anton already gave a very clean answer. Another way to see it is to work backwards: start from a sequence of functions $F_j$ in $L^2$ that does is non-compact, and define $f_j(x) = frac{d}{dx} (x F_j(x) )$.

For example, let $phi(x)$ be an arbitrary smooth bump function supported in $[-1/4,1/4]$, then the sequence of functions $F_j(x) = 2^j phi( 4^j x - 1)$ all have disjoint support, but all have the same $L^2$ norm, so obviously does not have a converging subsequence in $L^2$.

Now set $f_j = (xF_j)' = F_j(x) + 8^j x phi' (4^j x - 1)$. Since $phi'$ has support only in $[-1/4,1/4]$, on the support of $f_j$ we can bound $4^j x$ absolutely by, say, 2. So we have that $f_j$ is a bounded sequence in $L^2$, whose corresponding $F_j = Tf_j$ cannot have a Cauchy subsequence.

Edit: I should also provide some motivation: observe that the scaling argument also works the other way (replace $j$ by $-j$, so that you can dilate). The Hardy-type inequality that you are using is a scaling invariant inequality: you estimate $f/x$ in $L^2$ by its derivative $f'$. If we treat $x$ as having units of distance, then the two objects have the same units regardless of what units $f$ has. This gives scaling invariance of the estimate. In other words, the estimate is invariant under the natural scaling action of $mathbb{R}_{+}$ on $L^2(mathbb{R}_+)$, where the group operation for $mathbb{R}_{+}$ is multiplication.

Observe that $(mathbb{R}_+, times)$ is a non-compact Lie group. Generally, if you have an inequality/operator that is invariant under the action of a non-compact Lie group, the inequality/operator cannot be compact. You just need to start with some test function and act on it by the Lie group action to generate a bounded sequence that runs off non-compactly in the "infinity dimension" direction. Terry summarised it in his Buzz http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness a short while back.

This is, of course, closely related to the notion of concentration compactness.

ag.algebraic geometry - Morphism between polarized abelian varieties

That should be true, yes.

A polarization of $A$ is given by a bilinear form on $H_1(A, Z)$; this is equivalent to a map $H_1(A,Z) to H_1(A,Z)^vee$, which is an isomorphism if the polarization is principal.

A map between two abelian varieties is given by a corresponding linear map $H_1(A_1, Z) to H_1(A_2, Z)$. The map between the varieties is an isomorphism if the map on $H_1$ is.

The map induced on the bilinear form then is the composition

$$H_1(A_1,Z)to H_1(A_2,Z) to H_1(A_2,Z)^vee to H_1(A_1,Z)^vee$$
If the form is respected by this map, then this is an isomorphism. Consequently, the left-hand map must be as well, as claimed.

cv.complex variables - Simultaneous convergence of powers of unit complex numbers

Yes.

A standard lemma is that some element of ${alpha, 2alpha, ..., malpha}$ is within $1/(m+1)$ of 0 mod 1, since otherwise, there would have to be two multiples of $alpha$ between some $k/(m+1)$ and $(k+1)/(m+1)$ which means their difference would be close to 1.

The same pigeonhole argument works on $(S^1)^n$. Consider the first $m^n+1$ multiples of $(alpha_1,...alpha_n)$. Two must be in the same $n$-dimensional box $[k_1/m,(k_1+1)/m]times...[k_n/m,(k_n+1)/m]$ which means their difference is a multiple $t_mtimes(alpha_1,..,alpha_n)$ within $1/m$ of 0 on each coordinate.

If you want more details and better approximations, then there are some multidimensional versions of simple continued fractions which might work, but this suffices to show that a sequence of integers ${t_m}$ exists so that $z_i^{t_m}$ is within $2pi/m$ of 1.

linear algebra - Is there a version of inclusion/exclusion for vector spaces?

One way to look at this question is via quiver representations. Two subspaces of a vector space form a representation of the quiver $A_3$ with orientations $bullet rightarrow bullet leftarrow bullet$ with the additional condition that both maps are injective (that's tautology). Now, every representation of $A_3$ is a sum of indecomposables, whose dimension vectors are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (0,1,1), (1,1,1), where for the first and the third one the maps are not injective, and for the remaining four the maps are injective. Thus, the dimension vector for a generic representation with injective maps is a(0,1,0)+b(1,1,0)+c(0,1,1)+d(1,1,1)=(b+d,a+b+c+d,c+d). Clearly, the dimension of the sum of the two subspaces is b+c+d (the complement is represented by the first summand a(0,1,0)), which is (b+d)+(c+d)-d, and d is the dimension of the intersection.

Now, for the three subspaces we deal with representations of the quiver $D_4$ with injective maps. (I am too lazy to draw $D_4$ on MO, sorry!). Indecomposable representations have dimension vectors $(d_1,d_2,d_3,d)$ (note different ordering of dimensions - the largest one is the last one) being (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1), (1,0,0,1), (0,1,0,1), (0,0,1,1), (1,1,0,1), (1,0,1,1), (0,1,1,1), (1,1,1,1), (1,1,1,2) - altogether 12 vectors. Among them, the first three have non-injective maps, and the fourth one captures the complement of the sum of our three subspaces. Thus, there are 8 numbers through which the dimension can be expressed (not 7, as in the inclusion-exclusion formula), and what remains is to choose the 8th number, in addition to the dimensions of all possible intersections, reasonably for your needs.

For $k>3$ subspaces the classification problem stops being of finite type so it becomes a bit more nasty...

pr.probability - sum of order statistics

The inequality $E(S_K) geq E(S_i)$ holds.

To avoid any doubt, let me be more specific. Let $Y_1, Y_2, ..., Y_N$ be a collection of random variables, and write $X_1 geq X_2 geq ... geq X_N$ for their reordering in non-increasing order.

Suppose $K < N$ is fixed and let $S_K$ be the sum of the $K$ largest of the random variables, that is $S_K=X_1+...+X_K$.

Let $R$ be a random variable taking values in $0,1,...,N$ which is independent of the random variables $Y_i$. The independence from the $Y_i$ is important of course (this is how I interpret your "based on some criteria". If the $R$ is allowed to depend on the realisation of the $Y_i$ then all sorts of different behaviours are possible).

Now let $S_R$ be the sum of the $R$ largest of the random variables, that is $S_R=X_1+...+X_R$. (In your notation this is $S_i$).

Suppose that $ER=K$. Then I claim that $E S_R leq E S_K$, with equality iff $R=K$ with probability 1. (Unless the $Y_i$ are somehow degenerate, in which case equality can occur in other cases as well).

Proof: Write $p_k=P(Rgeq k)$ for $k=1,2,...,N$. We have $sum p_k=ER=K$.

Also

$S_R=sum_{k=1}^N X_k I(Rgeq k)$

so

$ES_R=sum_{k=1}^N P(Rgeq k) E X_k = sum_{k=1}^N p_k E X_k$.

(Here we used the independence of $R$ from the $X_i$).

Consider maximising this sum subject to the constraints that $sum p_k=K$ and that
$1geq p_1 geq p_2 geq p_3geq ...$.

Since the terms $E X_k$ are decreasing in $k$,
the maximum is achieved when $p_k=1$ for $kleq K$ and $p_k=0$ for $k>K$.
(Provided the $Y_i$ are not degenerate, the terms $E X_k$ are strictly decreasing,
and this is the only way to achieve the maximum. If not, the maximum may be achieved in some other cases too).

That is, the maximum value of $ES_R$ occurs precisely if $R$ is equal to $K$ with probability 1.

It doesn't matter whether the $Y_i$ are identically distributed, and also they don't need to be independent. However, it is important that $R$ is independent of the $Y_i$.

pr.probability - Integrating a simple exponential over the space of matrices that define a metric

I want to interpret an $ntimes n$ matrix $D$ as a set of pairwise distances, and assume that $D$ obeys metric properties. Namely, $D_{ii} = 0$, $D_{ij} geq 0$, $D_{ij} = D_{ji}$ and $D_{ij} leq D_{ik} + D_{kj}$ for all $1 leq i,j,k leq n$. For convenience, let $bigtriangleup_n$ denote the set of such matrices.

Now, I need to integrate some "simple" functions over this set. The simplest would be an exponential. Namely, I want to compute something like $int_{bigtriangleup_n} expleft[-lambda sum_{i=1}^n sum_{j=i+1}^n D_{ij}right] d D$.

I've been able to work this out for the simplest nontrivial case: namely $n=3$. But for higher $n$, my brute force way of calculation gets really ugly. The approach I've been taking is to basically first integrate over $D_{11}, D_{12}, D_{13}, ..., D_{1n}$, all of which have no constraints... then integrate over $D_{23}$ which is just a definite integral from $|D_{12} - D_{23}|$ to $D_{12}+D_{23}$ of $exp[-lambda D_{13}]$ and then, in the general case, integrating over $D_ij$ becomes the definite integral from $max_{k neq i,j} |D_{ik}-D_{jk}|$ to $min_{kneq i,j} D_{ik}+D_{jk}$, but this is the point at which I get stuck, because these things becomes nasty quite quickly (even for just $n=4$).

At the end of the day, I've love to be able to integrate more complex functions, like a chi-square type function rather than an exponential type function, but the exponential is the most trivial case that is interesting...

To be precise, I'm looking for a closed form evaluation of the integral above, preferably with some derivation that will help me work out more complex examples.

dg.differential geometry - How do I make the conceptual transition from multivariable calculus to differential forms?

I have struggled with this question myself, and I couldn't find a perfectly satisfactory answer. In the end, I decided that the definition of a differential form is a rather strange compromise between geometric intuition and algebraic simplicity, and that it cannot be motivated by either of these by itself. Here, by geometric intuition I mean the idea that "differential forms are things that can be integrated" (as in Bachmann's notes), and by algebraic simplicity I mean the idea that they are linear.

The two parts of the definition that make perfect geometric sense are the d operator and the wedge product. The operator d is simply that operator for which Stokes' theorem holds, namely if you integrate d of a n-form over an n+1-dimensional manifold, you get the same thing as if you integrated the form over the n-dimensional boundary.

The wedge product is a bit harder to see geometrically, but it is in fact the proper analogy to the product measure. Here's how it works for one-forms. Suppose you have two one-forms a and b (on a vector space, for simplicity). Think of them as a way of measuring lengths, and suppose you want to measure area. Here's how you do it: pick a vector $vec v$ such that $a(vec v) neq 0$ but $b(vec v) = 0$ and a vector $vec w$ s.t. $a(vec w) = 0$ but $b(vec w) neq 0$. Declare the area of the parallelogram determined by $vec v$ and $vec w$ to be $a(vec v) cdot b(vec w)$. By linearity, this will determine area of any parallelogram. So, we get a two-form, which is in fact precisely $a wedge b$.

---

Now, the part that makes no sense to me geometrically is why the hell differential forms have to be linear. This implies all kinds of things that seem counter-intuitive to me; for example there is always a direction in which a one-form is zero, and so for any one-form you can draw a curve whose "length" with respect to the form is zero. More generally, when I was learning about forms, I was used to measures as those things which we integrate, and I still see no geometric reason as to why measures (and, in particular, areas) are not forms.

However, this does make perfect sense algebraically: we like linear forms, they are simple. For example (according to Bachmann), their linearity is the thing that allows the differential operator d to be defined in such a way that Stokes' theorem holds. Ultimately, however, I think the justification for this are all the short and sweet formulas (e.g. Cartan's formula) that make all kinds of calculations easier, and all depend on this linearity. Also, the crucial magical fact that d-s, wedges, and inner products of differential forms all remain differential forms needs this linearity.

Of course, if we want them to be linear, they will be also signed, and so measures will not be differential forms. To me, this seems as a small sacrifice of geometry for the sake of algebra. Still, I don't believe it's possible to motivate differential forms by algebra alone. In particular, the only way I could explain to myself why take the "Alt" of a product of forms in the definition of the wedge product is the geometric explanation above.

So, I think the motivation and power behind differential forms is that, without wholly belonging to either the algebraic or geometric worlds, they serve as a nice bridge in between. One thing that made me happier about all this is that, once you accept their definition as a given and get used to it, most of the proofs (again, I'm thinking of Cartan's formula) can be understood with the geometric intuition.

Needless to say, if anybody can improve on any of the above, I'll be very grateful to them.

P.S. For the sake of completeness: I think that "inner products" make perfect algebraic sense, but are easy to see geometrically as well.

How do you find the potential function V of the gradient system?

I'm going to assume everything is happening in $mathbb{R}^n$, which I think is what you intended.

Start by defining V(0) = 0.

Now, for each $xin mathbb{R}^n$, let $gamma_x:[0,b]rightarrow mathbb{R}^n$ be a (piecewise) smooth curve with $gamma_x(0) = 0$ and $gamma_x(b) = x$, i.e., $gamma_x$ is any continuous curve joining $0$ to $x$. Define $V(x) = int_0^b W cdot dgamma_x(t) = int_0^b W(gamma_x(t))cdot gamma_x'dt$, that is, $V(x)$ is the result obtained by integrating $W$ along $gamma_x$.

First note that by the fundamental theorem for line integrals, $V(x)$ is independent of the choice of curve. Thus, to actually compute $V(x)$, one may as well take $gamma_x$ to be a straight line joining $0$ and $x$ (or, if some other path gives a nicer integral, use that).

However, to actually prove that this $V(x)$ satisfies $nabla V = W$, it'll help to be able to pick the curves however we want.

So, why does this $V(x)$ work? Well, suppose one wants to compute $frac{d}{dx_1} V(x)$. Formally, this is lim$_{hrightarrow 0} frac{V(x+he_1) - V(x)}{h}$, where $e_1$ is a unit vector in the direction of $x_1$.

To actually evaluate this, make life as easy as possible by picking $gamma_{x+he_1}$ and $gamma_x$ nicely. So, pick $gamma(t)$ to be a smooth curve which starts at 0, and when , near $x$, looks like a straight line pointing in the direction of $e_1$, with $gamma(1) = x$. Thus, $gamma(t)$ is both $gamma_x$ and $gamma_{x+he_1}$, if you travel along it long enough.

Then $V(x+he_1) - V(x) = int_1^{1+h} Wcdot e_1 dt$.

But then $frac{d}{dx_1} V(x) =$lim$_{hrightarrow 0}frac{1}{h} int_1^{1+h} Wcdot e_1 dt$. But then, by the fundamental theorem of calculus (the usual one variable version), this is exactly $Wcdot e_1$, i.e., it's the first component of $W$. Of course, the other components work analogously.

(Incidentally, using Petya's isomorphism between vector fields and one forms, and using Stokes' theorem in place of the fundamental theorem of line integrals, this proves that $H^1_{text{de Rham}}(mathbb{R}^n) = 0$)

Fixed Point Property in Algebraic Geometry

I am wondering about the following problem: for which (say smooth, complex, connected) algebraic varieties $X$ does the statement any regular map $Xto X$ has a fixed point hold?
MathSciNet search does not reveal anything in this topic.

This is true for $mathbb{P}^n$ (because its cohomology is $mathbb{Z}$ in even dimensions
and $0$ otherwise, and the pullback of an effective cycle is effective, so all summands
in the Lefschetz fixed point formula are nonnegative, and the 0-th is positive -- is this a correct argument?). Is it true for varieties with cohomology generated by algebraic cycles (i.e. $h^{p,q}(X)=0$ unless $p=q$ and satisfying Hodge conjecture), for example for Grassmannians, toric varieties, etc.? This is not at all clear that the traces of $f$ on cohomology will be nonnegative.

Probably you have lots of counterexamples. What about positive results?

Suggest effective heuristic (not precise) graph colouring algorithm

There are a number of heuristics that work fairly well. They all work by prescribing some kind of ordering on the vertices, and then coloring the vertices one by one, using the least unused color to color the next one.

• First Fit does precisely the above, with an arbitrary initial ordering. It's fast, but needless to say performs rather poorly.


• LDO orders the vertices in decreasing order of degree, the idea being that the large degree vertices can be colored more easily.


• SDO (saturation degree ordering) is a variant on LDO where the vertices are ordered in decreasing order by "saturation degree", defined as the number of distinct colors in the vertex neighborhood.


• IDO (incidence degree ordering) is a variant of SDO where the "degree" of a vertex is defined as the number of colored vertices in its neighborhood.

The latter two heuristics require the order to be rebuilt after each step, and so are more expensive, but there's empirical evidence suggesting that they do reasonably well, especially in parallel.

None of these algorithms come with any kind of formal guarantees, so be warned.

ca.analysis and odes - Asymptotics of iterated polynomials

Let the sequence $u_1, u_2, ldots$ satisfy $u_{n+1} = u_n - u_n^2 + O(u_n^3)$. Then it can be shown that if $u_n to 0$ as $n to infty$, then $u_n = n^{-1} + O(n^{-2} log n)$. (See N. G. de Bruijn, Asymptotic methods in analysis, Section 8.5.)

This can be used to obtain asymptotics for $v_{n+1} = Av_n - Bv_n^2 + O(v_n^3)$, where $A$ and $B$ are constants. Let $w_n = A^{-n} v_n$; this gives

$$A^{n+1} w_{n+1} = A^{n+1} w_n - B A^n w_n^2 + O(A^n w_n^3)$$
and so
$$w_{n+1} = w_n - BA^{-1} w_n^2 + O(w_n^3).$$
Then let $w_n = Ax_n/B$ to get
$$Ax_{n+1}/B = Ax_n/B - B/A cdot (Ax_n/B)^2 + O(x_n^3)$$
and after simplifying
$x_{n+1} = x_n - x_n^2 + O(x_n^3)$. This satisfies the initial requirements for $u_n$ (with some checking of the side condition); then substitute back.

But say I actually know that $u_{n+1} = P(u_n)$ for some polynomial $P$, with $P(z) = z - z^2 + a_3 z^3 + cdots + a_d z^d.$ In this case it seems like it should be possible to get more explicit information about $u_n$. Is there a known algorithm for computing an asymptotic series for $u_n$ as $n to infty$?

nt.number theory - Irreducibility of polynomials in two variables

What follows is more a series of considerations than a practical algorithm, but could still be of interest. The main idea is that it's easier to work with one-variable polynoms, so we trade a bad problem in two variables for several bad problems in one variable.

The key point is the following lemma (assume $k$ is of characteristic zero!): if $P(X,Y)$ is a two-variable polynom and there are enough distinct values of $a$ such that $P(X,a)$ is a constant polynomial, then $P(X,Y)$ is a polynom in $Y$ only. The proof is by arguing that $P(X,a)$ being a constant polynom means that $a$ is a root for the polynom (in $Y$) coefficient of $X^n$ for all $n>0$. And that can't happen too often in characteristic zero, unless those coefficients are zero polynoms, hence $P(X,Y)$ is reduced to its constant term as a polynom in $X$, hence is only a polynom in $Y$, as was to be proved.

Now, for your question : if you suppose a $P(X,Y)$ isn't irreducible, say factors as $Q(X,Y)R(X,Y)$, but many $P(X,a)$ are irreductible, then that means for each such $a$ either $Q(X,a)$ or $R(X,a)$ is a constant, hence given enough of those $a$, one of $Q$ or $R$ at least is only a polynom in $Y$ by the lemma, say $R$.

Then if you manage to fully factor $P(a,Y)=Q(a,Y)R(Y)$ for some $a$ (again dropping to a one-variable polynom), you get a list with $R$ (and divisors of $R$): check each element for divisibility of $P(X,Y)$. If none is good, $P$ is irreducible.

EDIT:

1. In fact, after you have found $R$ is a polynom in $Y$, just consider the gcd of the coefficients of the $X^n$ -- if you get $1$, $P$ is irreducible.


2. The previous considerations mostly prove that a cheating polynom (ie: not irreducible although it appears to be when evaluated along a variable) necessarily has a very precise form, which makes it susceptible to easy factorisation.

lo.logic - Does the axiom of specification prevent writing any proof?

I think with such problems it helps to make it clear what the syntactical rules actually "mean". The rule you mention sais that - basically - if you dont have any further assumptions about a term $t$ in your proof, then you could do the same proof for any term, thus, if you can derive $Sigma vdash theta(t)$, and no further assumption is made about $t$, then the prove is independent of the actual choice of $t$, and thus, $Sigma forall x theta(x)$.

However, in general, you dont have to allow general terms $t$ - you can require $t$ to be a variable, thus having a proper notion of "free" and "bounded". This makes discussing a lot easier (and doesnt change your possibilities of deriving). If $t$ doesnt occur free anywhere in $Sigma$, then from $theta(t)$ follows $forall x theta(x)$, because $t$ is arbitrary. If $t$ occurs somewhere, but it (or - if you want - a subterm of $t$) is bound there, it also makes no problems, since then its not an assumption about $t$. For example, $exists n forall m (m notin n)$ proclaims the existence of an empty set, but makes no proposition about $n$ itself.

Now, since $theta[x_2]$ has $x_2$ as the only free variable - as you said - your $t$ cannot occur freely in $theta[x_2]$, except for $t=x_2$. But if $t=x_2$, you still have no problem, since $x_2$ is bound in the axiom. Thus, you have no problems. Hope this could help you.

complex geometry - Newlander-Nirenberg for surfaces

Quite a long ago, I tried to work out explicitly the content of the Newlander-Nirenberg theorem. My aim was trying to understand wether a direct proof could work in the simplest possible case, namely that of surfaces. The result is that the most explicit statement I could get is a PDE I was not able to solve.

Assume a quasi-complex structure $J$ is given on the surface $S$; we want to prove that this is induced by a complex one (in this case there are no compatibility conditions). This can be easily transformed in the problem of local existence for a second order PDE, as follows.

We look for local charts on $S$ which are holomorphic (with respect to the quasi-complex structure on $S$). Two such charts are then automatically compatible. So the problem is local.

Fix a small open set $U subset S$ and identify it with a neighboorhood of $0 in mathbb{R}^2$ via a differentiable chart. Locally we can write $J = left(begin{matrix}[a | b] \ [c | d ]end{matrix}right)$ for some functions $a, cdots, d$ (pretend it is a two by two matrix).

A chart is given by a complex valued function $f = u + iv$. The condition that the differential is $mathbb{C}$-linear can be verified on a basis of the tangent space; moreover if it is true for a vector v, it remains true for Jv, which is linearly independent. Here we have used that $J^2 = -1$.

So we need only to check it for the vector $partial_x$. Since $J partial_x = a partial_x + c partial_y$, the condition says

$-v_x = a u_x + c u_y$

$u_x = a v_x + c v_y$

Hence we need to solve this system, with $f = u + i v$ non singular ($f$ will be then locally invertible). Since $a$ and $c$ do not vanish simultaneously, we can assume $c(0) neq 0$, hence $c neq 0$ on $U$ provided $U$ is small.

We can then solve for $u_y$ and get the equivalent system

$u_x = a v_x + c v_y$

$-u_y = frac{1 + a^2}{c}v_x +a v_y$

Moreover the Jacobian $J_f = u_x v_y + u_y v_x = frac{1}{c}(v_x^2 + (a v_x + c v_y)^2)$, so $f$ is nonsigular if $v$ is. By Poincaré's lemma, the system admits a local solution if and only if

$frac{partial}{partial_y} left( a v_x + c v_y right) - frac{partial}{partial_x} left( frac{1 + a^2}{c}v_x +a v_y right) = 0$.

Hence we are looking for a local solution of the last equation with $(v_x(0), v_y(0)) neq (0, 0)$.

So my question is:

Is there a simple way to prove local existence for a nonsingular solution of the last displayed equation?

I should make clear that I'm not looking for a proof of Newlander-Nirenberg; of this there are plenty. I am more interested in seeing what Newlander-Nirenberg becomes in terms of PDE in the simplest possible case, and then see that the PDE thus obtained is solvable. According to the answer of Andy, the equation which comes out is the Beltrami equation, so I will have a look at it. Still, I'm curious if any standard PDE technique can solve the equation I derived in the most stupid way above.

gt.geometric topology - Minimal volume of 4-manifolds

This question came up in a talk of Dieter Kotschick yesterday. The minimal volume of a manifold is the infimum of volumes of Riemannian metrics on the manifold with sectional curvatures bounded in absolute value by 1. Kotschick proved
that there are distinct smooth structures on $k(S^2 times S^2) sharp (1 + k)(S^1 times S^3)$, $k$ sufficiently large,
for which in the standard smooth structure, the minimal volume $=0$ (by finding a fixed-point free circle action), and another smooth structure for which the minimal volume is bounded away from 0.

My question is whether the converse is true: if there is a metric in which the minimal volume $=0$, must the smooth structure be standard? The existence of a polarized F-structure in this case may be relevant.

cv.complex variables - Ways to prove the fundamental theorem of algebra

Here is a translation into English of a second "real" proof from the journal Ilya mentioned in his answer. This proof is due to Petya Pushkar', it is found at

http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=7&option_lang=eng

and it is based on the notion of the degree of a map. Recall that for a smooth proper mapping of oriented manifolds, its degree is defined by picking a regular value and adding up the signs of the determinants of the differential of the mapping at the points in the inverse image. That the degree is well-defined is rather complicated to prove, but it explains the following topological fact.

Fact: Let $M^n$ and $N^n$ be smooth connected oriented manifolds and
$f colon M^n rightarrow N^n$ be a smooth proper mapping of degree
not equal to zero. Then $f$ is surjective.

To prove the Fundamental Theorem of Algebra in a "real" version, we will focus on polynomials of even degree (any of odd degree have a real root). We will show that
any real polynomial of degree $2n$ can be factored into a product of $n$ polynomials
of the second degree.

We identify each monic polynomial $x^d + a_{d-1}x^{d-1} + cdots + a_1x + a_0$ with the point $(a_{d-1},dots,a_0)$ in $mathbf R^d$. We will be particularly interested in the space of monic quadratic polynomials $x^2 + ax + b$, which are identified with the plane $mathbf R^2$. Consider the multiplication mapping

$$u colon (mathbf R^2)^n rightarrow mathbf R^{2n} text{ where} (f_1,f_2,dots,f_n) mapsto f_1f_2cdots f_n.$$
Proving the Fundamental Theorem of Algebra amounts to showing that $u$ is surjective.

First we show $u$ is proper.

For any $d geq 1$, identify the nonzero polynomials of degree at most $d$, considered up to scaling by nonzero real numbers, with $mathbf P^d(mathbf R)$ by $[a_dx^d + a_{d-1}x^{d-1} +cdots + a_0] mapsto [a_d,a_{d-1},dots,a_0]$. (The polynomials of exact degree $d$, after being scaled to be monic, are a copy of $mathbf R^d$ in $mathbf P^d(mathbf R)$.) Consider the multiplication mapping

$$widehat{u} colon (mathbf P^2(mathbf R))^n rightarrow mathbf P^{2n}(mathbf R) text{ where} ([f_1],[f_2],dots,[f_n]) mapsto [f_1f_2cdots f_n].$$
The mapping $widehat{u}$ is proper since it is defined on a compact manifold and is continuous.

The mapping $widehat{u}$ is a natural "compactification" of the mapping $u$. The space
$(mathbf P^2(mathbf R))^n$ can be written as the union of $(mathbf R^{2})^n$ and an "infinitely distant part" $B_1$ ($n$-tuples of polynomials of degree at most 2 where at least one polynomial has degree less than 2), while $mathbf P^{2n}(mathbf R)$ can be written as the union of $mathbf R^{2n}$ and an "infinitely distant part" $B_2$ (polynomials of degree less than $2n$). From this point of view, $widehat{u}$ on $(mathbf R^2)^n$ agrees with $u$ and, clearly, $widehat{u}^{-1}(B_2) = B_1$. Therefore the map $u$ is proper.

Next we show the degree of $u$ is equal to $n!$.
Orient the space of monic polynomials of degree 2 (we denote this space as $mathbf R^2$) arbitrarily and give $(mathbf R^2)^n$ the product orientation (as a product of oriented manifolds). As an exercise, show
the polynomial $p(x) = prod_{i=1}^n (x^2+i)$ is a regular value of the mapping $u$. (Hint:
This polynomial is a product of distinct monic irreducibles. Now use the description of the regular values of the multiplication mappings $mu_k$ in Pukhlikov's proof of the Fundamental Theorem of Algebra, which is written in a separate answer on this page.)

The polynomial $p(x)$ has $n!$ inverse images under $u$: all ordered $n$-tuples
with coordinates $x^2+i$ for $i = 1,dots,n$. Let's prove that these points
all contribute the same sign to the degree.

The mapping $u$ is invariant under permutations of its arguments, and
any such permutation preserves orientation (exercise). Therefore the
sign of the determinant of the differential at all the inverse images is the same, which shows $u$ has degree $n!$. By the topological fact at the start, $u$ is surjective, so all monic real polynomials of degree $2n$ are a product of monic quadratic real polynomials.

soft question - Can Wikipedia be a reliable (and sustainable) resource for advanced mathematics?

References to Wikipedia articles are fairly common and
elsewhere, but I'm one of those people who wonder whether the Wikipedia
framework will evolve toward more rather than less useful information at
relatively advanced levels of mathematics. There are at the moment
approximately 23581 mathematics entries, which of course I haven't read
systematically. But my smaller sample has usually left me with some
doubts about balance, reliability, thoroughness of entries. Often the
coverage is spotty, while the references and links are erratic. Outright
falsehoods seem rare compared with skewed or outdated versions of what is
known. Obviously it takes considerable effort by individuals to make
Wikipedia entries complete, accurate, up-to-date. And will the site
itself be sustainable over decades to come?

Over the centuries print encyclopedias of all sizes and shapes have
existed, some more useful and reliable than others but all impossible
to update continuously. Advanced mathematics has benefited relatively
little from these volumes. Commercially published mathematical book series
called "encyclopedias" tend to be uneven at best. One series collects
monographs on special topics, of varying quality and coverage. So the
Internet might promise better things. But many general-reader Wikipedia entries such as biographical sketches are disappointing. Editing is
possible, but sometimes the site is the target of those wanting to rewrite history. (I've made only one foray into editing, to correct the
common misspelling of our family name in a biographical sketch of my
oldest brother's thesis advisor. But I could see other fuzzy parts of
that sketch that would be complicated to rewrite in detail.)

A problem on Algebraic Number Theory, Norm of Ideals

Here is a proof that the ideal norm as defined in the books by Serre and Lang is equal to the ideal norm as defined in Swinnerton-Dyer's book. We will start from the definition given by Serre and Lang, state some of its properties, and use those to derive the formula as given by Swinnerton-Dyer.

Background: Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension, and $B$ be the integral closure of $A$ in $L$. For any prime $mathfrak P$ in $B$ we define ${rm N}_{B/A}({mathfrak P}) = mathfrak p^f$, where $f = f({mathfrak P}|{mathfrak p})$ is the residue field degree of $mathfrak P$ over $mathfrak p$, and this norm function is extended to all nonzero ideals of $B$ by multiplicativity from its definition on (nonzero) primes in $B$.

Properties.

1) The map ${rm N}_{B/A}$ is multiplcative (immediate from its definition).

2) Good behavior under localization: for any (nonzero) prime ${mathfrak p}$ in $A$, ${rm N}_{B/A}({mathfrak b})A_{mathfrak p} = {rm N}_{B_{mathfrak p}/A_{mathfrak p}}({mathfrak b}B_{mathfrak p})$. Note that $A_{mathfrak p}$ is a PID and $B_{mathfrak p}$ is its integral closure in $L$; the ideal norm on the right side is defined by the definition above for Dedekind domains, but it's more easily computable because $B_{mathfrak p}$ is a finite free $A_{mathfrak p}$-module on account of $A_{mathfrak p}$ being a PID and $L/K$ being separable. The proof of this good behavior under localization is omitted, but you should find it in books like those by Serre or Lang.

3) For nonzero $beta$ in $B$, ${rm N}_{B/A}(beta{B}) = {rm N}_{L/K}(beta)A$, where the norm of $beta$ on the right is the field-theoretic norm (determinant of multiplication by $beta$ as a $K$-linear map on $L$). To prove this formula, it is enough to check both sides localize the same way for all (nonzero) primes $mathfrak p$: ${rm N}_{B_{mathfrak p}/A_{mathfrak p}}(beta{B}_{mathfrak p}) = N_{L/K}(beta)A_{mathfrak p}$ for all $mathfrak p$. If you know how to prove over the integers that $[{mathcal O}_F:alpha{mathcal O}_F] = |{rm N}_{F/{mathbf Q}}(alpha)|$ for any number field $F$ then I hope the method you know can be adapted to the case of $B_{mathfrak p}/A_{mathfrak p}$, replacing ${mathbf Z}$ with the PID $A_{mathfrak p}$. That is all I have time to say now about explaining the equality after localizing.

Now we are ready to show ${rm N}_{B/A}({mathfrak b})$ equals the ideal in $A$ generated by all numbers ${rm N}_{E/F}(beta)$ as $beta$ runs over $mathfrak b$.

For any $beta in mathfrak b$, we have $beta{B} subset mathfrak b$, so ${mathfrak b}|beta{B}$. Since ${rm N}_{B/A}$ is multiplicative, ${rm N}_{B/A}({mathfrak b})|{rm N}_{E/F}(beta)A$ as ideals in $A$. In particular, ${rm N}_{E/F}(beta) in {rm N}_{B/A}({mathfrak b})$. Let $mathfrak a$ be the ideal in $A$ generated by all numbers ${rm N}_{E/F}(beta)$, so we have shown $mathfrak a subset {rm N}_{B/A}(mathfrak b)$, or equivalently ${rm N}_{B/A}(mathfrak b)|mathfrak a$. To prove this divisibility is an equality, pick any prime power ${mathfrak p}^k$ dividing $mathfrak a$. We will show ${mathfrak p}^k$ divides ${rm N}_{B/A}(mathfrak b)$.

To prove ${mathfrak p}^k$ divides ${rm N}_{B/A}(mathfrak b)$ when ${mathfrak p}^k$ divides $mathfrak a$, it suffices to look in the localization of $A$ at $mathfrak p$ and prove ${mathfrak p}^kA_{mathfrak p}$ divides ${rm N}_{B/A}(mathfrak b)A_{mathfrak p}$, which by the 2nd property of ideal norms is equal to ${rm N}_{B_{mathfrak p}/A_{mathfrak p}}(mathfrak b{B_{mathfrak p}})$. Since $B_{mathfrak p}$ is a PID, the ideal ${mathfrak b}B_{mathfrak p}$ is principal: let $x$ be a generator, and we can choose $x$ to come from $mathfrak b$ itself. By the 3rd property of ideal norms, ${rm N}_{B_{mathfrak p}/A_{mathfrak p}}(xB_{mathfrak p}) = {rm N}_{E/F}(x)A_{mathfrak p}$.
Showing ${mathfrak p}^kA_{mathfrak p}$ divides ${rm N}_{E/F}(x)A_{mathfrak p}$ is the same as showing ${rm N}_{E/F}(x) in {mathfrak p}^kA_{mathfrak p}$. Since $x$ is in in $mathfrak b$, ${rm N}_{E/F}(x) in mathfrak a subset {mathfrak p}^k$, so ${rm N}_{E/F}(x) in {mathfrak p}^kA_{mathfrak p}$. QED

gr.group theory - Unipotent linear algebraic groups

Assume characteristic 0. I do not know how much of this extends to finite characteristic.

Let $mathbf u$ be the Lie algebra of the unipotent subgroup $U$, and $mathbf t$ that of the torus (1- dimensional or not, it doesn't matter).

Define $Delta(mathbf g,mathbf t)$ as the sets of roots of $mathbf g$ w.r.t. $mathbf t$ (the usual definition is fine, even if $mathbf t$ is not maximal, however the root spaces will in general not be 1-dimensional). Let $C$ denote the centralizer.

Then you have $mathbf u=C_{mathbf u}(mathbf t)oplussum mathbf u_alpha$ for $alphainDelta(mathbf g,mathbf t)$. Here $mathbf u_alpha=mathbf ucapmathbf g_alpha$ or equivalently the set {$Xinmathbf umid [H,X]=alpha(H)X forall Hinmathbf t$}.

Let now $mathbf t_{max}$ be a maximal torus containing $mathbf t$, and $Delta(mathbf g,mathbf t_{max})$ the corresponding root system (this is the "usual" root system).
An element $T$ of $mathbf t_{max}$ is called regular if
$alpha(T)neqbeta(T)$ and $alpha(T)neq 0$ for all roots $alphaneqbetainDelta(mathbf g,mathbf t_{max})$.

If the torus $mathbf t$ contains a regular element $T$, the roots w.r.t. $mathbf t$ are in bijection with those w.r.t. $mathbf t_{max}$, and in particular the root spaces are 1-dimensional. It follows that if $mathbf u_alphaneq 0$ then $mathbf u_alpha=mathbf g_alpha$, and $mathbf u$ is a sum of root spaces.

Most important domains, extension theorems, and functions in several complex variables

Here are a few points to guide you into the beautiful subject you had the good taste to choose.

1) Hartogs extension phenomenon :given two concentric balls in $mathbb C^n$, any holomorphic function $B(0;M) setminus B(0;m) to mathbb C$ extends to a holomorphic function $B(0;M) to mathbb C$.This really launched the subject and showed that function theory in several variables is not just an extension of the theory in one variable.You should study a few such classes of examples. Key words: Hartogs figures, Reinhardt domains.

2) Domains for which such extensions do not exist are called holomorphy domains: balls are holomorphy domains but as we just saw "shells" $B(0;M) setminus B(0;m)$ are not.
If a region is not a domain of holomorphy, it has a holomorphic hull, but this is no longer included in $mathbb C^n$ : you get étalé spaces ( Yes, you algebraic geometers out there, this is where they were introduced ! ). This is an important subject and you can test whether a domain is a holomorphy domain at its boundary. Key-words: Levi problem, plurisubharmonic functions.

3)Holomorphic manifolds and Stein manifolds: these are abstractions of domains of $mathbb C^n$ and holomorphic domains respectively . In retrospect they were inevitable because of the nature of holomorphy hulls (cf. 2). Stein manifolds are highly analogous to the affine varieties of algebraic geometers.

4) Sheaf theory, cohomology: these are all powerful techniques that you MUST master if you want to read anything at all in the subject. In particular you must understand coherent sheaves, which have a flavour of Noetherianness in them, but are a more subtle notion.
The most important result here is Cartan's theorem B : coherent sheaves have no cohomology in positive dimension.

To help you learn all this , I would recommend:

B.Kaup, L. Kaup: Holomorphic Functions of several Variables (de Gruyter). [Quite friendly]

H.Grauert R.Remmert: Theory of Stein Spaces (Springer). [The ultimate source by the Masters]

All my wishes for success in your study of complex geometry.

ag.algebraic geometry - Torsion line bundles with non-vanishing cohomology on smooth ACM surfaces

I am looking for an example of a smooth surface $X$ with a fixed very ample $mathcal O_X(1)$ such that $H^1(mathcal O(k))=0$ for all $k$
(such thing is called an ACM surface, I think) and a globally generated line bundle $L$ such that $L$ is torsion in $Pic(X)$ and $H^1(L) neq 0$.

Does such surface exist? How can I construct one if it does exist? What if one ask for even nicer surface, such as arithmetically Gorenstein? If not, then I am willing to drop smooth or globally generated, but would like to keep the torsion condition.

More motivations(thanks Andrew): Such a line bundle would give a cyclic cover of $X$ which is not ACM, which would be of interest to me. I suppose one can think of this as a special counter example to a weaker (CM) version of purity of branch locus.

To the best of my knowledge this is not a homework question (: But I do not know much geometry, so may be some one can tell me where to find an answer. Thanks.

EDIT: Removed the global generation condition, by Dmitri's answer. I realized I did not really need it that much.

at.algebraic topology - Poincaré-Hopf and Mathai-Quillen for Chern classes?

One. The Poincaré-Hopf theorem is usually stated as a formula for the Euler characteristic of the tangent bundle TM. Is there a version for Euler classes, of oriented real vector bundles?

It seems like one should be able to use the section to lift the map $M to BO(n)$ to a map $M to mathcal V$, where $mathcal V$ is the universal bundle, and pull back a Thom form from there. I'd much rather reference this than work it out.

Two. Is there a version of it for Chern classes, not just the Euler class ( = the top Chern class)?

Here I guess one would probably use several sections to lift the map $M to BU(n)$.

Mathai and Quillen (Superconnections, Thom classes, and equivariant differential forms,
Topology 25 (1986), no. 1, 85--110) interpolate between the Gauss-Bonnet theorem, which computes an Euler class using a connection on a vector bundle, and the Poincaré-Hopf theorem, which computes an Euler class using a section. Mathai and Quillen make a form using both a section and a connection. Scaling the section to 0 gives Gauss-Bonnet, scaling to $infty$ gives Poincaré-Hopf.

Three. Is there a Mathai-Quillen theorem for Chern classes, interpolating between Chern-Weil and Q#2 above?

co.combinatorics - alternating sums of terms of the Vandermonde identity

So, you are interested in $f(n,k)=sum_{i=0}^k (-1)^ibinom{k}{i}binom{2n-k}{n-i}$.
Simple manipulations show $f(n,k)=frac{k!(2n-k)!}{(n!)^2}left[sum_{i=0}^n (-1)^i binom{n}{i}binom{n}{k-i}right]$
Now the second factor counts the coefficient of $x^k$ in $(1-x^2)^n$ and therefore if $k$ is odd $f=0$ otherwise $f=(-1)^{frac{k}{2}}frac{k!(2n-k)!}{n!(k/2)!(n-k/2)!}$ which is far from zero...

EDIT: On a different note I see the result is a signed generalized Catalan number of degree 2 (I was not aware they satisfied such simple identities). Since usually providing combinatorial interpretations for generalized Catalan numbers is not easy, may I ask in what combinatorial context did you face the above calculation?

dg.differential geometry - Frobenius Theorem for subbundle of low regularity?

Let me conisder the case when the distribution of planes is of codimension 1 and explain why in this case it is enough to have $C^1$ smoothness in order to ensure the existence of the folitation.

In the case when the distribution is of codimension 1, you can formulate Frobenius Theorem in terms of 1-forms. Namely you can define a non-zero 1-form $A$, whose kernel is the distribution. The smoothness of this 1-form will be the same as the smoothness of the distribution. Now, you can say that the distribution is integrable if $Awedge dA=0$. This quantity is well defined is A is $C^1$. Let me give a sketch of the proof that $Awedge dA=0$ garanties existence of the foliation is A is $C^1$.

The proof is by induction

1) Consider the case $n=2$. In this case it is a standard fact of ODE, that for a $C^1$ smooth distribution of directions on the plane the integral lines are uniquelly defined.

2) Conisder the case $n=3$. We will show that the foliation exists locally near any point, say the origin $O$ of $R^3$. The 1-form A, that defines the distribution is non vanishing on one of the coordinate planes, say $(x,y)$ plane in the neighborhood of $O$. Take a $C^1$
smooth vector field in the neigborhood of $O$ that is transversal to planes $z=const$
and satisfies $A(v)=0$. Take the flow correponding to this vector field. The flow is $C^1$ smooth and moreover it preserves the distribution of planes $A=0$. Indeed, dA vanishes on the planes A=0 (by the condition of integrability), and we can apply the formula for Lie derivative $L_v(A)=d(i_v(A))+i_v(dA)=i_v(dA)$. Finally, we take the integral curve of the restriction of $A=0$ to the plane $(x,y)$ and for evey curve conisder the surface it covers unders the flow of $v$. This gives the foliation.

This reasoning can be repeated by induction.

A good refference is Arnold, Geometric methods of ordinary differential equations. I don't know if this book was transalted to English

intuition - What is torsion in differential geometry intuitively?

Torsion is easy to understand but this knowledge seems to be lost. I had to go back to Elie Cartan's articles to find an intuitive explanation (for example, chapter 2 of http://www.numdam.org/numdam-bin/fitem?id=ASENS_1923_3_40__325_0).

Let $M$ be a manifold with a connection on its tangent bundle.
The basic idea is that any path $gamma$ in $M$ starting at $xin M$ can be lifted as a path $tildegamma$ in $T_xM$, but is the $gamma$ is a loop $tilde gamma$ need not be a loop. The resulting translation of the end point is the torsion (or its macroscopic version).

The situation is easy in a Lie group $G$ (which I imagine Cartan had in mind).
$G$ has a canonical flat connection for which the parallel vectors fields are left invariant vectors fields. For this connection the parallel transport is simply the left translation. The Maurer-Cartan form $alpha$ is then the parallel transport to the tangent space $T_1G$ at the identity $1in G$.

If $gamma:[0,1]to G$ is a path in $G$ starting at $1$. $gamma'$ is a path in $TG$ and $alpha(gamma')$ is a path in $T_1M$. $alpha(gamma')$ can be integrated to another path $tilde gamma$ in $T_1M$. Let $gamma_{leq x}$ be the path $gamma:[0,x]to G$, then we define

$$tilde gamma(x) = int_0^xalpha(gamma'(t))dt = int_{gamma_{leq x}}alpha.$$
In the sense given by the connection, $gamma$ and $tildegamma$ have the same speed and the same starting point, so they are the same path (but in different spaces).

If $gamma$ is a loop and $D$ a disk bounding $gamma$,
$tildegamma$ is a loop iff $tildegamma(1)=0in T_1G$.
We have

$$tildegamma(1) = int_gammaalpha = int_Ddalpha.$$
$tildegamma$ is a loop iff this integral is zero.

Now, $alpha$ can be viewed as the solder form for $TG$, so the torsion is the covariant differential $T=d^nablaalpha$. As the connection is flat $T$ reduces to $T=dalpha$.
The Maurer-Cartan equation gives an explicit formula: $T=dalpha = -frac{1}{2}[alpha,alpha]$.
The previous integral is then the integral of the torsion

$$tildegamma(1) = int_Ddalpha = -frac{1}{2}int_D[alpha,alpha]$$
and may not be zero.

The situation is the same for a general manifold, but the parallel transport is not explicit and formulas are harder.

The notion behing this is that of affine connection. As I understand it, an affine connection is a data that authorize to picture the geometry of $M$ inside the tangent space $T_xM$ of some point $x$. If I move away from $x$ in $M$, there will be a corresponding movement away from the origin in $T_xM$ (this is the above lifting of path). If I transport in parallel a frame with me, the frame will move in $T_xM$. Globally the movement of my point and frame is encoded by a family of affine transformations in $T_xM$.

Of course this picture of the geometry of $M$ in $T_xM$ is not faithful.
Because of the torsion, if I have two paths in $G$ starting at $x$ and ending at the same point, they may not end at the same point in $T_xM$.
Because of curvature, even if my two lifts end at the same point, my two frames may not be parallel.
The picture is faithful if $M$ is an affine space iff both torsion and curvature vanish (Cartan's structural equations for affine space).

I think torsion is beautiful :)

rt.representation theory - Apocryphal Maschke theorem?

The result about bimodules is true, and standard. Here is one way to see it.

By Frobenius reciprocity, $Hom_G(V,k[G]) = Hom_k(V,k)$, since $k[G]$ is the induction (or coinduction, depending on your terminology) of the trivial representation of the trivial subgroup of G to G.$

Since Frobenius reciprocity is functorial, one easily sees that this canonical isomorphism
is an isomorphism of right $G$-representations, where the source has a right $G$-action coming from the right $G$-action on $k[G]$, and the target has a right $G$-action coming as the transpose of the left $G$-action on $V$.

Now if by Maschke's semisimplicity theorem, we know that
$k[G] = bigoplus_V V otimes_k Hom_G(V,k[G])$, where the sum is over all irreducible left $G$-representations. (Indeed, Mashke shows that this is true
for any left $G$-module in place of $k[G]$.) Again, this is a natural isomorphism, and so respects the right $G$-actions on source and target.

Combined with the preceding computation, we find that
$k[G] = bigoplus_V Votimes_k Hom_k(V,k) = bigoplus_V End(V),$ as both left and right $G$-modules,
as required.

[Edit:] Leonid's remark about $k$ being needing to be big enough in his answer below is correct. Each simple $V$ comes equipped with an associated division algebra of $G$-endomorphisms
$A_V := End_G(V)$. The representation $V$ is absolutely irreducible (i.e stays irred. after passing to any extension field) if and only if $A_V = k$. When we consider $Hom_k(V,W)$ for another left $G$-module $W$, this is naturally an $A_V$-module, and Maschke's theorem
will say that $W = bigoplus_V Hom_k(V,W)otimes_{A_V} V$. (I have written the factors in the tensor product in this order because $V$ is naturally a left $A_V$-module (if we think of endomorphisms acting on the left), and then $Hom_k(V,W)$ becomes a right $A_V$-module.)

So in the case of $W$ being the group algebra, we have

$$k[G] = bigoplus_V Hom_k(V,k)otimes_{A_V} V$$
(an isomorphism of $G$-bimodules).

If all the $V$ are absolutely irreducible, e.g. if $k$ is algebraically closed,
then all the $A_V$ just equal $k$, and the preceding direct sum reduces to what I wrote above, and what was written in the question.

gr.group theory - does every right-angled coxeter group have a right-angled artin group as a subgroup of finite index?

As James points out, the paper of Davis and Januskiewicz proves the inverse. To see that the answer to your question is 'no', consider the right-angled Coxeter group whose nerve graph is a pentagon. That is, it's the group with presentation
$langle a_1,ldots, a_5 mid a_i^2=1, [a_i,a_{i+1}]=1rangle$
where the indices are considered mod 5.

This group acts properly discontinuously and cocompactly on the hyperbolic plane, and it's not hard to see that it has a finite-index subgroup which is the fundamental group of a closed hyperbolic surface. Every finite-index subgroup of a right-angled Artin group is either free or contains a copy of $mathbb{Z}^2$, but the fundamental group of a closed hyperbolic surface has no finite-index subgroups of this form.

nt.number theory - elliptic curve with j-invariant T

The idea for (2) is the following: the modular curve $Y(ell^n)$ classifying elliptic curves
over ${mathbb C}$ together with an isomorphism $({mathbb Z}/ell^n)^2 cong E[ell^n]$
identifying the standard symplectic pairing on the left (i.e. $langle (a_1,a_2),(b_1 ,b_2)rangle = e^{2pi i (a_1b_2-a_2b_1)/ell^n}$) with the Weil pairing on the right,
is irreducible. (It is isomorphic to $mathcal H/Gamma(ell^n)$, where
$mathcal H$ is the complex upper half-plane and $Gamma(ell^n)$ is the congruence
subgroup of full level $ell^n$.)

(3) follows from (2) and the irreducibility of cyclotomic polynomials over ${mathbb Q}$.