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.

The Classical Flag Variety

When you think of a flag, you might imagine something like this:

Roughly speaking, a flag on a flagpole consists of:

• a point (the bulbous part at the top of the pole),
• a line passing through that point (the pole),
• a plane passing through that line (the plane containing the flag), and
• space to put it in.

Mathematically, this is the data of a complete flag in three dimensions. However, higher-dimensional beings would require more complicated flags. So in general, a complete flag in $n$-dimensional space $\mathbb{C}^n$ is a chain of vector spaces of each dimension from $0$ to $n$, each containing the previous:

$$0=V_0\subset V_1 \subset V_2 \subset \cdots \subset V_n=\mathbb{C}^n$$

with $\dim V_i=i$ for all $i$.

(Our higher-dimensional flag-waving imaginary friends are living in a world of complex numbers because $\mathbb{C}$ is algebraically closed and therefore easier to work with. However, one could define the flag variety similarly over any field $k$.)

Variety Structure

Now that we’ve defined our flags, let’s see what happens when we wave them around continuously in space. It turns out we get a smooth algebraic variety!

Indeed, the set of all possible flags in $\mathbb{C}^n$ forms an algebraic variety of dimension $n(n-1)$ (over $\mathbb{R}$), covered by open sets similar to the Schubert cells of the Grassmannian. In particular, given a flag $\{V_i\}_{i=1}^n$, we can choose $n$ vectors $v_1,\ldots,v_n$ such that the span of $v_1,\ldots,v_i$ is $V_i$ for each $i$, and list the vectors $v_i$ as row vectors of an $n\times n$ matrix. We can then perform certain row reduction operations to form a different basis $v_1^\prime,\ldots,v_n^\prime$ that also span the subspaces of the flag, but whose matrix is in the following canonical form: it has $1$’s in a permutation matrix shape, $0$’s to the left and below each $1$, and arbitrary complex numbers in all other entries.

For instance, say we start with the flag in three dimensions generated by the vectors $\langle 0,2,3\rangle$, $\langle 1, 1, 4\rangle$, and $\langle 1, 2, -3\rangle$. The corresponding matrix is $$\left(\begin{array}{ccc} 0 & 2 & 3 \\ 1 & 1 & 4 \\ 1 & 2 & -3\end{array}\right).$$ We start by finding the leftmost nonzero element in the first row and scale that row so that this element is $\newcommand{\PP}{\mathbb{P}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\ZZ}{\mathbb{Z}}
\DeclareMathOperator{\Gr}{Gr}
\DeclareMathOperator{\Fl}{F\ell}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\inv}{inv}1$. Then subtract multiples of this row from the rows below it so that all the entries below that $1$ are $0$. Continue the process on all further rows:

$$\left(\begin{array}{ccc} 0 & 2 & 3 \\ 1 & 1 & 4 \\ 1 & 2 & -3\end{array}\right) \to \left(\begin{array}{ccc} 0 & 1 & 1.5 \\ 1 & 0 & 2.5 \\ 1 & 0 & -6\end{array}\right)\to \left(\begin{array}{ccc} 0 & 1 & 1.5 \\ 1 & 0 & 2.5 \\ 0 & 0 & 1\end{array}\right)$$

It is easy to see that this process does not change the flag formed by the partial row spans, and that any two matrices in canonical form define different flags. So, the flag variety is covered by $n!$ open sets given by choosing a permutation and forming the corresponding canonical form. For instance, one such open set in the $5$-dimensional flag variety is the open set given by all matrices of the form
$$\left(\begin{array}{ccccc}
0 & 1 & \ast & \ast & \ast \\
1 & 0 & \ast & \ast & \ast \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & \ast & 0 \\
0 & 0 & 0 & 1 & 0 \end{array}\right)$$

We call this open set $X_{45132}$ because it corresponds to the permutation matrix formed by placing a $1$ in the $4$th column from the right in the first row, in the $5$th from the right in the second row, and so on. The maximum number of $\ast$’s we can have in such a matrix is when the permutation is $w_0=n(n-1)\cdots 3 2 1$, in which case the dimension of the open set $X_{12\cdots n}$ is $n(n-1)/2$ over $\CC$ — or $n(n-1)$ over $\RR$, since $\CC$ is two-dimensional over $\RR$. In general, the number of $\ast$’s in the set $X_w$ is the inversion number $\inv(w)$, the number of pairs of entries of $w$ which are out of order.

Finally, in order to paste these disjoint open sets together to form a smooth manifold, we consider the closures of the sets $X_w$ as a disjoint union of other $X_w$’s. The partial ordering in which $\overline{X_w}=\sqcup_{v\le w} X_v$ is called the Bruhat order, a famous partial ordering on permutations. (For a nice introduction to Bruhat order, one place to start is Yufei Zhao’s expository paper on the subject.)

Intersection Cohomology

Now suppose we wish to answer incidence questions about our flags: which flags satisfy certain given constraints? As in the case of the Grassmannians, this boils down to understanding how the Schubert cells $X_w$ intersect. This question is equaivalent to studying the cohomology ring of the flag variety $\Fl_n$ over $\mathbb{Z}$, where we consider the Schubert cells as forming a cell complex structure on the flag variety.

The cohomology ring $H^\ast(\Fl_n)$, as it turns out, is the coinvariant ring that we discussed in the last post! For full details I will refer the interested reader to Fulton’s book on Young tableaux. To give the reader the general idea here, the Schubert cell $X_w$ can be thought of as a cohomology class in $H^{2i}(\Fl_n)$ where $i=\inv(w)$. We call this cohomology class $\sigma_w$, and note that for the transpositions $s_i$ formed by swapping $i$ and $i+1$, we have $\sigma_{s_i}\in H^2(\Fl_n)$. It turns out that setting $x_i=\sigma_i-\sigma_{i+1}$ for $i\le n-1$ and $x_n=-\sigma_{s_{n-1}}$ gives a set of generators for the cohomology ring, and in fact $$H^\ast(\Fl_n)=\mathbb{Z}[x_1,\ldots,x_n]/(e_1,\ldots,e_n)$$ where $e_1,\ldots,e_n$ are the elementary symmetric polynomials in $x_1,\ldots,x_n$.