Can you <i>prove</i> it?
As this ingenious post over at Math With Bad Drawings explained so clearly recently, there is a big difference between finding the answer to a math problem and being able to explain, beyond all reasonable doubt, why your answer is correct. It is the difference between solving it and proving it!
While mathematical proof is a huge part of math and science, it is unfortunately somewhat overlooked in the standard US curriculum. Partly for this reason, my family and I started a new math summer camp called Prove it! Math Academy. This year’s program will be a two-week crash course in proofs, using challenging problems and advanced mathematical concepts as examples.
Consider the following diagram which appears on our program website:
No explanation, no words, no proof. Just the picture. Enticing!
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 \[x^2=\frac{x^2+y^2}{2}+\frac{x^2-y^2}{2}.\]