# Branching rule of $S_n$ and Springer theory

Let $$u\in\mathrm{GL}_n$$ be a unipotent element, let $$\mathcal{B}_u$$ be the variety of Borel subgroups containing $$u$$, and let $$d=\dim \mathcal{B}_u$$. Then Springer theory tells us that $$H^{2d}(\mathcal{B}_u,\overline{\mathbb{Q}}_{\ell})$$ is an irreducible representation of $$S_n$$ in a natural way, and that all irreducible representations of $$S_n$$ come this way.

Now, combining with the branching rule of symmetric groups, viewing as representations of $$S_{n-1}$$ one has that $$$$H^{2d}(\mathcal{B}_u,\overline{\mathbb{Q}}_{\ell})\cong\bigoplus_{u'}H^{2d'}(\mathcal{B}_{u'},\overline{\mathbb{Q}}_{\ell}),$$$$ where $$u'$$ runs over some unipotent elements of $$\mathrm{GL}_{n-1}$$ whoese corresponding Young diagrams are obtained from that of $$u$$ by removing a corner box.

Question. Is there a purely geometric proof of the above displayed formula?

This is a nice question. I have never seen this before.

Let us write $$\mathcal B_u = \{ V_0 \subset V_1 \subset \cdots \subset V_{n-1} \subset V_n = \mathbb C^n : u V_i \subset V_i \}.$$ Let $$\lambda$$ be the Jordan type of $$u$$. Then we can partition $$B_u$$ into locally closed subsets depending on the Jordan type of $$u \rvert_{V_{n-1}}$$; such a Jordan type $$\mu$$ must necessarily be obtained by removing one box from $$\lambda$$. So we have $$\mathcal B_u = \bigsqcup_\mu B_u^\mu \quad B_u^\mu = \{ V_\bullet : u\rvert_{V_{n-1}} \text{ has Jordan type \mu } \}.$$ Let $$G(n-1, n)_u$$ denote the set of all $$n-1$$-dimensional $$u$$-invariant subspaces of $$\mathbb C^n$$ and partition $$G(n-1,n)_u$$ into locally closed subsets $$G(n-1,n)_u^\mu$$ according to the Jordan type of the restriction of $$u$$ to the subspace.

We have $$\mathcal B_u \rightarrow G(n-1,n)_u$$ and $$\mathcal B^\mu_u$$ is the preimage of $$G(n-1,n)^\mu_u$$. Note that the fibre of $$\mathcal B_u \rightarrow G(n-1,n)_u$$ is $$\mathcal B_{u\rvert_{ V_{n-1}}}$$.

Now, here are two facts that I don't see right away, but which must be true:

1. Each piece $$\mathcal B_u^\mu$$ has the same dimension.
2. $$G(n-1,n)^\mu_u$$ is irreducible and simply-connected (or at least that the bundle $$\mathcal B^\mu_u \rightarrow G(n-1,n)\mu_u$$ is topologically trivial on components).

Edit: I think these facts should be contained in the classic paper by Spaltenstein https://www.sciencedirect.com/science/article/pii/S138572587680008X

Assuming these facts, we get $$H(\mathcal B_u) = \bigoplus_\mu H(\mathcal B_u^\mu) = \bigoplus_\mu H(\mathcal B_{u(\mu)})$$ where again the direct sum ranges over all partitions made by deleting one box from $$\lambda$$ and where $$u(\mu)$$ denotes a unipotent element of Jordan type $$\mu$$, and $$H(X)$$ denotes top (co)homology. This gives the desired decomposition.