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
\begin{equation}
H^{2d}(\mathcal{B}_u,\overline{\mathbb{Q}}_{\ell})\cong\bigoplus_{u'}H^{2d'}(\mathcal{B}_{u'},\overline{\mathbb{Q}}_{\ell}),
\end{equation}
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?
 A: 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:

*

*Each piece $ \mathcal B_u^\mu $ has the same dimension.

*$ 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.
