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Set-up for classical Springer Correspondence:

$G$ = reductive group over $\mathbb{C}$, with Borel subgroup and maximal torus $B \supset T$, Weyl group $W=N_G(T)/T$.

Fix a unipotent $u \in G$ with component group $A(u) = C_G(u)/C_G(u)^\circ$.

$\mathcal{B} = G/B$ (flag variety), containing Springer fiber $\mathcal{B}_u$ = fixed points under $u$, $d=\dim \mathcal{B}_u$ (= half codimension of class of $u$ in unipotent variety)

Then $W \times A(u)$ acts on cohomology ($\ell$-adic or classical) $H^*(\mathcal{B}_u)$ with top cohomology in degree $2d$.

(Springer) Each irreducible representation of $W$ occurs, for some pair $(u,\phi)$ with $u$ unipotent and$\phi$ an irreducible character of $A(u)$, as an isotypic component of the $W \times A(u)$ representation on the top cohomology. Here all pairs $(u,1)$ occur.

Assume $A(u) \neq 1$ (possible except in type A).

(1) Must some pair $(u,\phi)$ with $\phi \neq 1$ occur?

(2) Is the representation of $A(u)$ on $H^*(\mathcal{B}_u)$ always a permutation permutation?

The answers to both questions seem to be yes, but I don't know any uniform approach using Springer theory. For example, (1) can be checked using case-by-case study of simple types, but is there a general reason for it? For (2) there is a sophisticated indirect argument using work of Bezrukavnikov, Mirkovic, Rumynin. All of this ties in naturally with some unsolved problems about representations of related Lie algebras in characteristic $p$.

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General facts about semi-small maps show that the action of $A_G(u)$ on $H^{2d}(B_u)$ is a permutation rep, but I guess what you are asking is much stronger! –  Geordie Williamson Mar 14 '10 at 18:38
    
Also, I thought not all pairs occur, but would have to hunt around a bit before justifying this comment! –  Geordie Williamson Mar 14 '10 at 18:52
    
Not all pairs $(u,\phi)$ occur that's true, hence the generalized Springer correspondence! The pair $(u,triv)$ however always occurs which is what I think is meant no? That follows from just from the fact that $H^{2d}(B_u)$ is a permutation rep. –  Kevin McGerty Mar 14 '10 at 19:20
    
Yes, the component group permutes irreducible components of the Springer fiber, inducing a permutation rep on top cohomology. (This plays a nice role for subregular unipotents.) But the action in lower degrees is harder to study directly. And yes, some pairs do fail to occur. These are accounted for in Lusztig's generalized Springer correspondence. But it would be undesirable (for me) if all nontrivial characters of the component group failed to occur. –  Jim Humphreys Mar 14 '10 at 19:22
    
Sorry ... I misunderstood what you are asking. –  Geordie Williamson Mar 14 '10 at 22:05
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