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Let $\pi:\tilde{\mathfrak{g}}\rightarrow \mathfrak{g}$, be the Grothendieck-Springer resolution, where $\mathfrak{g}$ is a semisimple Lie algebra over $\mathbb{C}$.

We know that $\pi$ is small thus $\pi_{*}\mathbb{Q}_{\ell}[\mathrm{dim}\,\mathfrak{g}]$ is an irreducible perverse sheaf on $\mathfrak{g}$.

Is this sheaf locally constant on the stratas where the dimension of Springer fibers is constant?

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  • $\begingroup$ Don't you need to shift by $-\dim\mathfrak g$ to make it perverse? $\endgroup$
    – SashaP
    Commented Sep 17, 2016 at 15:56
  • $\begingroup$ What do you call the Springer resolution of $\mathfrak{g}$? $\mathfrak{g}$ is smooth. $\endgroup$
    – abx
    Commented Sep 17, 2016 at 16:02
  • $\begingroup$ Probably $\mathfrak{g}$ should be replaced by the nilpotent cone of it. $\endgroup$
    – SashaP
    Commented Sep 17, 2016 at 16:49
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    $\begingroup$ Why do you think so? The stalks of this sheaf should reflect the topology of the fibers which is more rich than just their dimension... $\endgroup$
    – Sasha
    Commented Sep 17, 2016 at 19:51
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    $\begingroup$ The sentence starting "We know that..." is false: small tells us that it is the IC extension of the local system on the regular semi-simple locus. This representation is the regular representation and is not irreducible. Also, the answer to "Is this sheaf..." is no: over the regular (not necessarily semi-simple) elements the dimension is constantly zero, however the number of points in the fibre can change, varying between order of Weyl group and 1, thus the sheaf is not locally constant over this locus. $\endgroup$
    – Geordie Williamson
    Commented Sep 18, 2016 at 1:08

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