Assume $\mathbf{G}$ is a simple adjoint algebraic group over an algebraic closure $\overline{\mathbb{F}_p}$ of the finite field of characteristic $p > 0$ and $u \in \mathbf{G}$ is a unipotent element. Denote by $\mathfrak{B}_u$ the variety of all Borel subgroups of $\mathbf{G}$ containing $u$. Assume now that $\mathcal{F}$ is an arbitrary $\overline{\mathbb{Q}}_{\ell}$-local system on $\mathfrak{B}_u$, (i.e. a locally constant $\overline{\mathbb{Q}}_{\ell}$-constructible sheaf with finite dimensional stalks), and denote by $H^i_c(\mathfrak{B}_u,\mathcal{F})$ the compactly supported $\ell$-adic cohomology group with coefficients in $\mathcal{F}$. Now assume that $F : \mathbf{G} \to \mathbf{G}$ is a Frobenius endomorphism and $u$ is fixed by $F$ then we have an induced map $F : \mathfrak{B}_u^{\mathbf{G}} \to \mathfrak{B}_u^{\mathbf{G}}$ and in turn an induced map $F^* : H^i(\mathfrak{B}_u,\mathcal{F}) \to H^i_c(\mathfrak{B}_u,\mathcal{F})$ in cohomology. The question is can I determine the action of $F^*$ on $H^i_c(\mathfrak{B}_u,\mathcal{F})$ directly from its action on $H^i_c(\mathfrak{B}_u,\overline{\mathbb{Q}}_{\ell})$?

If we had a morphism of sheaves $\mathcal{F} \to \overline{\mathbb{Q}}_{\ell}$ then we would get an induced map $H^i_c(\mathfrak{B}_u,\mathcal{F}) \to H^i_c(\mathfrak{B}_u,\overline{\mathbb{Q}}_{\ell})$ in cohomology. However would this necessarily respect the induced actions of $F$?

EDIT: I should mention that I am really only interested in the action of $F$ on the top non-vanishing cohomology group, so this is the one of degree $2d_u$ where $d_u = \dim\mathfrak{B}_u$. As the above idea won't work, maybe this is a different approach. In their paper on the Green functions of exceptional groups Beynon--Spaltenstein, (Journal of Algebra - 1984), state the following: "As the irreducible components of $\mathfrak{B}_u$ all have the same dimension they form a basis for the cohomology group $H^{2d_u}_c(\mathfrak{B}_u,\overline{\mathbb{Q}}_{\ell})$. The action of $F^*$ is multiplication by $q^{d_u}$ followed by the permutation of the irreducible components induced by $F$". I'm not sure why this statement is true but is it possible that the same statement is also true for $H^{2d_u}_c(\mathfrak{B}_u,\mathcal{F})$? This would then give the action of $F^*$.

`$\mathfrak{B}_u$`

to itself, hence it induces a map in $\ell$-adic cohomology (by the functoriality of such cohomology). Could you explain why paragraph 2 won't help? $\endgroup$ – Jay Taylor Oct 20 '12 at 15:32