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?



Branching rule of $S_n$ and Springer theory
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.


