The Springer Correspondence, Part II: The Resolution

This is a continuation of The Springer Correspondence, Part I. Here we will work with unipotent matrices to construct the Springer resolution and the cohomology of its fibers.

Unipotent Matrices and Partitions

A unipotent element of a linear algebraic group $G$ is any element $u\in G$ such that $1-u$ is nilpotent. That is, $u=1+n$ where $n^k=0$ for some $k$.

To get a sense of what unipotent matrices look like, consider the type A situation in which $\DeclareMathOperator{\GL}{GL}\newcommand{\CC}{\mathbb{C}} G=\GL_n(\CC)$. Given a unipotent element $u$, we can conjugate it by some matrix to put it in Jordan normal form. It will look something like this: \[gug^{-1}=\left(\begin{array}{ccccccc} \lambda_1 & 1 & & & & & \\ & \lambda_1 & 1 & & & & \\ & & \lambda_1 & & & & \\ & & & \lambda_2 & 1 & & \\ & & & & \lambda_2 & & \\ & & & & & \ddots & \\ & & & & & & \lambda_k \end{array}\right)\]

The Springer Correspondence, Part I: The Flag Variety

In prior posts, we’ve seen that the irreducible representations of the symmetric group $S_n$ are in one-to-one correspondence with the partitions of $n$, and the Schur functions give an elegant encoding of their characters as symmetric polynomials. Now we can dive a bit deeper: a geometric construction known as the Springer resolution allows us to obtain all the irreducible representations of $S_n$ geometrically, and as a side bonus give natural graded representations that will allow us to define a $q$-analog of the Schur functions known as the Hall-Littlewood polynomials.

Quite a mouthful of terminology. Let’s start at the beginning.

Digging deeper: The isotypic components

In last week’s post, we made use of the coinvariant ring \[\mathbb{C}[x_1,\ldots,x_n]/I\] where $I=(p_1,\ldots,p_n)$ is the ideal generated by the positive-degree homogeneous $S_n$-invariants (symmetric polynomials). We saw that this was an $S_n$-module with Hilbert series $(n)_q!$, and claimed that it was the regular representation.

Let’s see why that is, and see if we can understand where the irreducible components occur.

More precisely, our goal is to understand the series \[\sum_{d} H_{\chi^\mu}(d)q^d\] where $H_{\chi^\mu}(d)$ is the number of copies of the $\mu$th irreducible representation of $S_n$ occurring in the $d$th degree component of $\mathbb{C}[x_1,\ldots,x_n]/I$. In Stanley’s paper on invariants of finite groups, he states without proof the answer as the following ``unpublished result of Lusztig’’:

Writing polynomials in terms of invariants and coinvariants

There is a fun little fact regarding polynomials in two variables $x$ and $y$:

Any two-variable polynomial $f(x,y)$ can be uniquely written as a sum of a symmetric polynomial and an antisymmetric polynomial.

(To be more precise, this is true for polynomials over any field of characteristic not equal to $2$. For simplicity, in what follows we will assume that our polynomials have coefficients in $\mathbb{C}$.)

Recall that a polynomial $g$ is symmetric if it does not change upon permuting its variables. In this case, with two variables, $g(x,y)=g(y,x)$. It is antisymmetric if swapping any two of the variables negates it, in this case $g(x,y)=-g(y,x)$.

It is not hard to prove the fact above. To show existence of the decomposition, set $g(x,y)=\frac{f(x,y)+f(y,x)}{2}$ and $h(x,y)=\frac{f(x,y)-f(y,x)}{2}$. Then \[f(x,y)=g(x,y)+h(x,y),\] and $g$ is symmetric while $h$ is antisymmetric. For instance, if $f(x,y)=x^2$, then we can write \[x^2=\frac{x^2+y^2}{2}+\frac{x^2-y^2}{2}.\]

Expii: Learning, connected

It’s been several months since I posted a gemstone, and the main reason is that much of my free-time mathematics energy recently became channeled into a new project: Expii.

Expii (currently beta) is a new online crowdsourced learning site that aims to fill the gaps in users’ understanding of topics, with the goal of making math, science, and other topics easy for everyone in the universe. Its motto? Learning, connected.

With an addictive, game-like format (hence the XP pun) in which users are awarded “fame points” for writing good explanations and “experience points” for successfully making it through tutorials, Expii is more interactive and community oriented than other online learning resources like Wikipedia. It is also more structured than question-and-answer sites like Quora or Stack Exchange, in that the primary “graph structure” for the topics is organized by our team, and users fill in the content in the nodes.