Hard Lefschetz Theorem for the Flag Manifolds

Question

In the case of a generalized flag manifold $G/P$, we have an explicit description of their cohomology groups due to Borel.(See herehere for a description.) I would like to know what the hard Lefschetz theorem looks like in this presentation.

Answer 1

I'll spell out Hard Lefschetz as an explicit combinatorial statement about the Grassmannian $G(d,n)$. I can definitely give you a version of this for the full flag manifold if you want it and probably for any $G/P$. I have no idea about a combinatorial proof.

$H^{\ast}(G(d,n))$ is entirely in even degrees, and has a standard basis given by the Schubert classes $X_{\lambda}$. The Schubert classes are indexed by partitions $(\lambda_1, \ldots, \lambda_d)$: Sequences of integers with $n-d \geq \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_d \geq 0$. The class $X_{\lambda}$ lives in cohomological degree $2 \sum \lambda_i$. We set $\sum \lambda_i = |\lambda|$.

I'll take $G(2,5)$ as my running example. There are $10$ total partitions with $5-2 \geq \lambda_1 \geq \lambda_2 \geq 0$: namely $(0,0)$, $(1,0)$, $(2,0)$, $(1,1)$, $(3,0)$, $(2,1)$, $(3,1)$, $(2,2)$, $(3,2)$ and $(3,3)$. The groups $H^{2 \cdot 2}(G(2,5))$ and $H^{2 \cdot 4}(G(2,5))$ both have dimension $2$, with bases $([X_{2,0}], [X_{1,1}])$ and $([X_{3,1}], [X_{2,2}])$ respectively.

$H^2(G(d,n))$ is always one dimensional, with generator $\zeta = [X_{1000\ldots0}]$, and positive multiples of $\zeta$ are ample. Multiplication is given by the Pieri rule: $$\zeta \cdot [X_{\lambda}] = \sum_{|\mu| = |\lambda|+1,\ \mu_j \geq \lambda_j} [X_{\mu}].$$ For example, $\zeta \cdot X_{2,0} = X_{3,0} + X_{2,1}$. So the coefficient of $X_{\mu}$ in $\zeta^{|\mu| - |\lambda|} X_{\lambda}$ is the number of chains of partitions $(\lambda = \alpha^0, \alpha^1, \ldots, \alpha^N = \mu)$ with $|\alpha^{i+1}| = |\alpha^i|$ and $\alpha^{i+1}_j \geq \alpha^i_j$. In other words, the number of standard Young tableaux of shape $\mu \setminus \lambda$.

Taking our running example of the map $\zeta^2 : H^{2 \cdot 2}(G(2,5)) \to H^{2 \cdot 4}(G(2,5))$, we see that $$\zeta^2 [X_{2,0}] = 2 [X_{3,1}] + [X_{2,2}]$$ $$\zeta^2 [X_{1,1}] = [X_{3,1}] + [X_{2,2}].$$

So Hard Lefschetz in this case is the statement that $$\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$$ is invertible.

More generally, fix $(d,n,r)$. Let $A(d,n,r)$ be the set of partitions $n-d \geq \lambda_1 \geq \cdots \geq \lambda_d \geq 0$ with $\sum \lambda_i=r$. Hard Lefschetz says that, if I take a matrix whose rows are indexed by $A(d,n,r)$, and whose columns are indexed by $A(d,n,d(n-d)-r)$, and where the entry $L_{\lambda \mu}$ is the number of SYT of shape $\mu / \lambda$, then this matrix is invertible.

I think Harry Tamvakis has thought about whether this can be proved directly from symmetric function combinatorics.

Is this the sort of thing you are looking for? Is it worth it to you for me to give you the analogous versions for other $G/P$?

Answer 2

To add to David's answer a bit:

The Borel description of $H(G/P)$ is the $W_P$-invariant subring of the $W$ coinvariants, $S^{W_P}_W$, where $W$ is the associated Weyl group and $W_P\subseteq W$ is the parabolic subgroup associated to $P$. As David indicates above, we can choose a Schubert basis for $S^{W_P}_W$, and compute the matrices for the Lefschetz maps in that basis. The Pieri rule allows us to interpret these Lefschetz matrices as so-called ''weighted path matrices'' with respect to the Bruhat order restricted to the set of maximal coset representatives of $W/W_P$. A well known ''lemma'' of Gessel and Viennot is then helpful in computing these determinants (the original paper is available here). These ideas are used in this paper (now published in Journal of Algebra) to give a purely algebraic proof of the hard Lefschetz theorem for $S_W$ where $W$ is any finite Coxeter group, as well as for $S^{W_P}_W$ for certain maximal parabolics $W_P\subseteq W$. One could probably also use these ideas to prove that $S^{W_P}_W$ is hard Lefschetz for any parabolic subgroup, although the Bruhat order for $W/W_P$ is more complicated.