User nicholas proudfoot - MathOverflow

Fundamental groups of symplectic leaves

2013-02-23T01:26:27Z

Let $X = \operatorname{Spec}(R)$ be a conical Poisson variety over $\mathbb{C}$. This means that $R$ is a non-negatively graded Poisson algebra over $\mathbb{C}$ with $R_0 = \mathbb{C}$ and that the Poisson bracket is homogeneous of negative degree. Assume that $X$ admits a conical symplectic resolution. Examples to keep in mind are:

the nilpotent cone in a simple Lie algebra
a Slodowy slice to a nilpotent orbit inside the nilpotent cone
the symmetric scheme of $n$ points on a Kleinian singularity
a hypertoric variety
a quiver variety.

Let $S$ be a symplectic leaf of $X$.

Q: Is it possible for the fundamental group of $S$ to be infinite?

In the first four examples listed above, the fundamental group of every symplectic leaf is finite. I don't know whether or not this property holds for quiver varieties, or for other examples.

Stable isomorphisms of groups

2012-04-19T20:43:43Z

Possible Duplicate:
Stably isomorphic groups

If $G$ and $H$ are two groups (finitely presented, if you wish) with the property that $G\times\mathbb{Z}$ is isomorphic to $H\times\mathbb{Z}$, does that imply that $G$ is isomorphic to $H$?

Answer by Nicholas Proudfoot for Homotopy type of Hilbert schemes of points of $\mathbb C^2$

2011-04-22T16:24:52Z

Here's how I understand the argument in Slodowy's book. Sticking with Tamas' notation, let $X$ denote the Hilbert scheme and $D$ the punctual Hilbert scheme. You'd like to define a deformation retraction from $X$ to $D$ by sending every point to its limit under the dilation action. Unfortunately, as Ben points out, $D$ is not fixed by the dilation action, and this function isn't even continuous! So that doesn't work.

Instead, you choose a closed tubular neighborhood $A$ of $D$ such that $D$ is a deformation retract of $A$; this is possible by a basic result in algebraic topology (Slodowy cites Spanier's book). You then use the dilation action to define a deformation retraction from $X$ to $A$. More precisely, send every element $x\in X$ to the first element of $A$ that it hits when you dilate it inward. This is a perfectly well-behaved map, and it does the trick.

The scheme-theoretic flow-in locus

2011-02-16T00:45:11Z

Let $R$ be a ring with an $\mathbb{N}\times\mathbb{Z}$-grading. The $\mathbb{N}$-grading allows you to construct the scheme $X = \operatorname{Proj} R$, and the $\mathbb{Z}$-grading defines an action of $T = \mathbb{G}_m$ on $X$.

Let $S$ be an $\mathbb{N}\times\mathbb{Z}$-graded quotient of $R$ such that every element of $R$ whose $\mathbb{Z}$-degree is nonzero vanishes in $S$, and let $Y = \operatorname{Proj} S$. Then we have an inclusion from $Y$ into the fixed locus $X^T$. $T$ acts trivially on the tautological bundle of $Y$, which is equal to the restriction of the tautological bundle of $X$. Assume that $X$ and $Y$ are smooth and that $Y$ is a connected component of $X^T$.

Let $R^+ \subset R$ be the subring generated by elements of non-negative $\mathbb{Z}$-degree, and let $V = \operatorname{Proj} \left(R\;\otimes_{R^+}S\right)\subset X$.

Morally, $V$ should be thought of as the locus of points $x$ with the property that $\lim_{t\to 0}t\cdot x \in Y$, though in fact the support of $V$ may be strictly bigger than this set. For example, let $X= T^*\mathbb{P^1}$, where the action of $T$ on $X$ is induced from the standard action on $\mathbb{P^1}$. Then $X$ has two fixed points which I'll call $N$ and $S$ (for the North and South poles). If we take $Y$ to be $N$, then $V$ will be the cotangent fiber over $N$. If we take $Y$ to be $S$, then $V$ will be the union of the cotangent fiber over $N$ with the zero section (even though the set-theoretic flow-in locus consists only of the zero-section minus $N$). In particular, $V$ need not be irreducible.

More generally, if $X = T^{*}(G/B)$ and $T$ acts on $X$ via a generic cocharacter of $G$, $V$ will be a union of conormal varieties to Schubert strata that appear with various multiplicities. In particular, $V$ need not be reduced.

The example that most interests me is when $X$ is the family over $\mathbb{A}^1$ with special fiber $T^{*}(G/B)$ and general fiber $G/H$, where $H$ is a maximal torus of $G$. This family is $U(1)$-equivariantly diffeomorphic (but not isomorphic) to $T^{*}(G/B)\times \mathbb{A}^1$, where $U(1)\subset T$ acts trivially on $\mathbb{A}^1$. Thus the components of $X^T$ are images of sections of the map from $X$ to $\mathbb{A}^1$. In this example $V$ will always be irreducible and reduced, even though its intersection with $T^{*}(G/B)$ is neither irreducible nor reduced (see the previous paragraph).

What I would like to know is that if I take $X$ to be a variety like this (a twistor families for a "nice" symplectic variety on which $T$ acts hamiltonianly with isolated fixed points), then $V$ will be reduced. In an attempt to get this, I'll try posing the following general question:

Suppose that $V$ is irreducible and generically reduced. Does it follow that $V$ is reduced?

Answer by Nicholas Proudfoot for Cohomology of a configuration space

2010-11-21T02:37:13Z

This isn't exactly an answer to your question, but here's how I like to think about the fact that you quoted.

Let's assume for a minute that $n=2$, so that we can think of $X$ as the complement of the braid arrangement in $\mathbb{C}^k$. Let $G = \mathbb{Z}/2\mathbb{Z}$, which acts on $X$ by complex conjugation.

Replace $\mathbb{Q}$ with a field $F$ of characteristic $2$. The $G$-equivariant cohomology ring $H^*_G(X; F)$ is a free module over $H^*_G(pt; F) \cong F[x]$ with the property that specializing at $x=0$ gives $H^* (X; F)$ and specializing at $x=1$ gives $H^*(X^G;F)$.

Thus we have a family of $\Sigma_k$ representations over the $F$-affine line interpolating between $H^* (X; F)$ and $H^*(X^G; F)$. Since the category of $\Sigma_k$ representations is semisimple, these two representations have to be isomorphic. The fact that $H^*(X^G; F)$ is the regular representation is obvious.

This is a good way to see that $H^*(X; F)$ is isomorphic to the regular representation of $\Sigma_k$. I'm not sure how to modify this argument to get $H^* (X; \mathbb{Q})$. I'm also not sure if this will help you find a cyclic vector, since it does not give you an explicit isomorphism between $H^* (X; F)$ and $H^* (X^G; F)$.

By the way, for $n>2$ you can do something similar, where $G$ acts on $\mathbb{R}^n$ by negating the last $n-1$ coordinates.

Counting solutions to x^{p+1}=y^4 in a finite field

2009-11-19T22:25:16Z

I need to compute the number of solutions to the equation $x^{p+1} = y^4$ in the field with $p^2$ elements (for p sufficiently large). The form of the equation suggests to me that the solution would depend on the congruence class of p mod 4, but I have reason to believe that the answer is a single polynomial in p.

I feel as if this should be easy, and I'm missing an obvious approach. Can anyone help me out?

Comment by Nicholas Proudfoot

2013-04-14T18:18:41Z

In this paper, Bezrukavnikov cites a paper of Kohno and Oda (1987) in which they prove (among other things) an LCS formula for the Poincare polynomial in question. Assuming that one can compute the ranks of the subquotients in the lower central series of the fundamental group, this completely answers Christin's question. However, Roman points out on page 133 of his paper that there are some incorrect results in the Kohno-Oda paper. Do you know if the Kohno-Oda LCS formula is correct as stated?

Comment by Nicholas Proudfoot

2013-04-14T15:59:20Z

@Dan Peterson: Totaro's paper does not give a formula for the Poincare polynomial. There's a big gap between being able to write down a DGA in terms of generators and relations and actually having a formula for its Poincare polynomial.

Comment by Nicholas Proudfoot

2013-03-18T05:31:36Z

Thanks, your reference to Namikawa's paper is very helpful! In fact, finiteness of the algebraic fundamental group is sufficient for the application that I have in mind, so this is perfect.

Comment by Nicholas Proudfoot

2013-02-26T04:58:34Z

Do there exist any examples where the fundamental groups of these strata are infinite? (See also mathoverflow.net/questions/122683/… for the same question.)

Comment by Nicholas Proudfoot

2012-11-20T20:18:51Z

David, do you know of any way to understand the non-coherent refinements of cone(S)? Put differently (if I understand correctly), let X be the affine toric variety associated to cone(S). The cones of the secondary fan classify projective toric maps Y->X that are bijective on orbits of codimension 0 or 1. The non-coherent refinements correspond to such maps that are proper but not projective. Can these be nicely classified?

Comment by Nicholas Proudfoot

2012-04-19T20:40:16Z

If $V = \mathbb{R}^3$ and you consider not the full group $SO(3)$ but rather the circle subgroup consisting of rotations about some axis in $V$, a presentation is given by Daniel Moseley in Theorem 3.4 of arxiv.org/abs/1110.5369. Whatever the answer to your question is, it should admit lots of natural maps to Moseley's ring (one for every choice of oriented line in $V$).

Comment by Nicholas Proudfoot

2011-02-16T17:23:19Z

@Dave continued again: If you choose the $\mathbb{Z}$-degrees of $x$ and $y$ to be 1 and $-1$ as you say, then you get the square of the standard action. Furthermore, you get it with a $T$-structure on the tautological line bundle such that $T$ acts nontrivially on the fibers at both fixed points. Instead, you should assign degrees $(2,0)$ or $(0,-2)$, depending on which of the two fixed points you want to look at. Finally, you're right, it's no coincidence that Ben and I are asking similar questions!

Comment by Nicholas Proudfoot

2011-02-16T17:20:09Z

@Dave continued: If you want to get the inclusion of the other fixed point, you should let $x$ have degree 0 and $y$ have degree $-1$, and then define your map from $R$ to $S$ by sending $x$ to $t$ and $y$ to $0$. These two choices of $\mathbb{Z}$-grading on $R$ induce the same $T$-action, but with a different equivariant structure on the tautological line bundle. With my conventions, the restriction of the tautological line bundle from $X$ to $Y$ will have the trivial $T$-action on every fiber.

Comment by Nicholas Proudfoot

2011-02-16T17:16:29Z

@Dave: As you say, let $R = k[x,y]$, where $x$ and $y$ both have $\mathbb{N}$-degree 1, so that $X = \operatorname{Proj} R = \mathbb{P}^1$. To define a $T$-action on $X$, we need to choose the $\mathbb{Z}$-degrees of $x$ and $y$. If you want to get the standard action, you should let $x$ have degree 1 and $y$ have degree $0$. Then you can let $S = k[t]$, where $t$ has $\mathbb{N}$-degree 1 and $\mathbb{Z}$-degree 0, and the map from $R$ to $S$ sending $x$ to $0$ and $y$ to $t$ induces the inclusion of one of the two fixed points.

Comment by Nicholas Proudfoot

2010-11-27T14:48:56Z

To Semen: When $k=2$, both the regular representation and $H^*(X; F)$ are trivial. (Recall that in my argument $F$ is supposed to have characteristic 2.)

Comment by Nicholas Proudfoot

2009-11-20T00:23:35Z

This is a small piece of a larger calculation in which I'm trying to understand the action of a particular finite group on the intersection cohomology of a particular complex variety. The intersection cohomology is pure and concentrated in even degrees, so this computation is amenable to finite field techniques. The fact that the answer to my question (as Sivek points out) DOES depend on the residue of p mod 4 means either that this dependence will be magically cancelled out by another part of the calculation, or that I've made a mistake earlier on.