*This is the third post in a series on the Springer correspondence. See Part I and Part II for background.*

In this post, we’ll restrict ourselves to the type A setting, in which $\DeclareMathOperator{\GL}{GL}\DeclareMathOperator{\inv}{inv} G=\GL_n(\mathbb{C})$, the Borel $B$ is the subgroup of invertible upper triangular matrices, and $U\subset G$ is the unipotent subvariety. In this setting, the flag variety is isomorphic to $G/B$ or $\mathcal{B}$ where $\mathcal{B}$ is the set of all subgroups conjugate to $B$.

# The Hall-Littlewood polynomials

For a given partition $\mu$, the Springer fiber $\mathcal{B}_\mu$ can be thought of as the set of all flags $F$ which are fixed by left multiplication by a unipotent element $u$ of Jordan type $\mu$. In other words, it is the set of complete flags $$F:0=F_0\subset F_1 \subset F_2 \subset \cdots \subset F_n=\mathbb{C}^n$$ where $\dim F_i=i$ and $uF_i=F_i$ for all $i$.

In the last post we saw that there is an action of the Weyl group, in this case the symmetric group $S_n$, on the cohomology rings $H^\ast(\mathcal{B}_\mu)$ of the Springer fibers. We let $R_\mu=H^\ast(\mathcal{B}_\mu)$ denote this ring, and we note that its graded Frobenius characteristic $$\DeclareMathOperator{\Frob}{Frob}\widetilde{H}_\mu(X;t):=\Frob_t(H^\ast(\mathcal{B}_\mu))=\sum_{d\ge 0}t^d \Frob(H^{2d}(\mathcal{B}_\mu))$$ encodes all of the data determining this ring as a graded $S_n$-module. The symmetric functions $\widetilde{H}_\mu(X,t)\in \Lambda_{\mathbb{Q}(t)}(x_1,x_2,\ldots)$ are called the **Hall-Littlewood polynomials**.

The first thing we might ask about a Hall-Littlewood polynomial $H_\mu$ is: what is its degree as a polynomial in $t$? In other words…

## What is the dimension of $\mathcal{B}_\mu$?

The dimension of $\mathcal{B}_\mu$ will tell us the highest possible degree of its cohomology ring, giving us at least an upper bound on the degree of $H_\mu$. To compute the dimension, we will decompose $\mathcal{B}_\mu$ into a disjoint union of subvarieties whose dimensions are easier to compute.

Let’s start with a simple example. If $\mu=(1,1,1,\ldots,1)$ is a single-column shape of size $n$, then $\mathcal{B}_\mu$ is the full flag variety $\mathcal{B}$, since here the unipotent element $1$ is in the conjugacy class of shape $\mu$, and we can interpret $\mathcal{B}_\mu$ as the set of flags fixed by the identity matrix (all flags). As described in Part I, the flag variety can be decomposed into Schubert cells $X_w$ where $w$ ranges over all permutations in $S_n$ and $\dim(X_w)=\inv(w)$. For instance, $X_{45132}$ is the set of flags defined by the initial row spans of a matrix 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)$$

because this matrix has its leftmost $1$’s in positions $4,5,1,3,2$ from the right in that order.

Thus the dimension of the flag variety is the maximum of the dimensions of these cells. The number of inversions in $w$ is maximized when $w=w_0=n(n-1)\cdots 2 1$, and so $$\dim(\mathcal{B})=\inv(w_0)=\binom{n}{2}.$$

We claim that in general, $\dim(\mathcal{B}_\mu)=n(\mu)$ where if $\mu^\ast$ denotes the conjugate partition, $n(\mu)=\sum \binom{\mu^\ast_i}{2}$. Another way of defining $n(\mu)$ is as the sum of the entries of the superstandard tableau formed by filling the bottom row of $\mu$ with $0$’s, the next row with $1$’s, and so on:

To show this, notice that since $\mathcal{B}_\mu$ is a subvariety of the full flag variety $\mathcal{B}$, and so $$\mathcal{B}_\mu=\mathcal{B}_\mu\cap \mathcal{B}=\bigsqcup \mathcal{B}_\mu\cap X_{w}.$$ It thus suffices to find the largest possible dimension of the varieties $\mathcal{B}_\mu\cap X_{w}$.

Let $u$ be the standard unipotent element of Jordan type $\mu$. For instance, the matrix below is the standard unipotent matrix of shape $(3,2,2)$.

Then the set $\mathcal{B}_\mu\cap X_{w}$ can be defined as the subset of $n\times n$ matrices defining flags in $X_w$ whose partial row spans are fixed by the action of $u$. Note that since the first vector is fixed by $u$, it must be equal to a linear combination of the unit vectors $e_{\mu_1}, e_{\mu_1+\mu_2},\ldots$. So we instantly see that the dimension of $\mathcal{B}_\mu\cap X_w$ is in general less than that of $X_w$.

Now, consider the permutation $$\hat{w}=n,n-\mu_1,n-\mu_1-\mu_2,\ldots,n-1,n-1-\mu_1,n-1-\mu_1-\mu_2,\ldots,\ldots.$$ Then it is not hard to see that the matrices in $X_{\hat{w}}$ whose flags are fixed by $u$ are those with $1$’s in positions according to $\hat{w}$, and with $0$’s in all other positions besides those in row $i$ from the top and column $k-\mu_1-\cdots-\mu_j$ from the right for some $i,j,k$ satisfying $i\le j$ and $\mu^\ast_1+\cdots+\mu^\ast_k< i \le \mu^\ast_1+\cdots+\mu^\ast_{k+1}$. This is a mouthful which is probably better expressed via an example. If $\mu=(3,2,2)$ as above, then $\hat{w}=7426315$, and $\mathcal{B}_\mu\cap X_{7426315}$ is the set of flags determined by the rows of matrices of the form $$\left(\begin{array}{ccccccc} 1 & 0 & 0 & \ast & 0 & \ast & 0 \\ 0 & 0 & 0 & 1 & 0 & \ast & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & \ast & 0 & \ast \\ 0 & 0 & 0 & 0 & 1 & 0 & \ast \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \end{array}\right)$$ The first three rows above correspond to the first column of $\mu$, the next three rows to the second column, and the final row to the last column. Notice that the stars in each such block of rows form a triangular pattern similar to that for $X_{w_0}$, and therefore there are $n(\mu)=\binom{\mu^\ast_1}{2}+\binom{\mu^\ast_2}{2}+\cdots$ stars in the diagram. Thus $\mathcal{B}_\mu\cap X_{\hat{w}}$ an open affine set of dimension $n(\mu)$, and so $\mathcal{B}_\mu$ has dimension at least $n(\mu)$. A bit more fiddling with linear algebra and multiplication by $u$ (try it!) shows that, in fact, for any permutation $w$, the row with a $1$ in the $i$th position contributes at most as many stars in $\mathcal{B}_\mu\cap X_{w}$ as it does in $\mathcal{B}_\mu\cap X_{\hat{w}}$. In other words, all other components $\mathcal{B}_\mu\cap X_{w}$ have dimension at most $n(\mu)$, and so $$\dim\mathcal{B}_\mu=n(\mu).$$

# The orthogonality relations

In Lusztig’s survey on character sheaves, he shows that the Hall-Littlewood polynomials (and similar functions for other Lie types) satisfy certain orthogonality and triangularity conditions that determine them completely. To state them in the type A case, we first define $\widetilde{H}_\mu[(1-t)X;t]$ to be the result of plugging in the monomials $x_1,-tx_1,x_2,-tx_2,\ldots$ for $x_1,x_2,\ldots$ in the Hall-Littlewood polynomials. (This is a special kind of plethystic substitution.) Then Lusztig’s work shows that:

- $\left\langle \widetilde{H}_\mu(X;t),s_\lambda\right\rangle=0$ for any $\lambda<\mu$ in dominance order, and $\langle\widetilde{H}_\mu,s_\mu\rangle=1$
- $\left\langle \widetilde{H}_\mu[(1-t)X;t],s_\lambda\right\rangle=0$ for any $\lambda>\mu$ in dominance order
- $\left\langle \widetilde{H}_\mu(X;t),\widetilde{H}_{\lambda}[(1-t)X;t]\right\rangle=0$ whenever $\lambda\neq \mu$.

In all three of the above, the inner product $\langle,\rangle$ is the *Hall inner product*, which can be defined as the unique inner product for which $$\langle s_\lambda,s_\mu\rangle = \delta_{\lambda\mu}$$ for all $\lambda$ and $\mu$.

Since the Schur functions $s_\lambda$ correspond to the irreducible representations $V_\lambda$ of $S_n$, we can therefore interpret these orthogonality conditions in a representation theoretic manner. The inner product $\left \langle \widetilde{H}_\mu(X;t),s_\lambda \right\rangle$ is the coefficient of $s_\lambda$ in the Schur expansion of $\widetilde{H}_\mu$, and is therefore the Hilbert series of the isotypic component of type $V_\lambda$ in the cohomology ring $R_\mu=H^\ast(\mathcal{B}_\mu)$. Moreover, the seemingly arbitrary substitution $X\mapsto (1-t)X$ actually corresponds to taking tensor products with the exterior powers of the permutation representation $V$ of $S_n$. To be precise: $$\widetilde{H}_\mu[(1-t)X;t]=\sum_{i\ge 0} (-1)^i t^i \Frob_t(R_\mu\otimes \Lambda^i(V)).$$

It turns out that any two of the three conditions above uniquely determine the Hall-Littlewood polynomials, and in fact can be used to calculate them explicitly. On the next page, we will work out an example using the first and third conditions above.

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