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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 $F^*$.

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I think you mean to consider an absolutely simple adjoint semisimple group $G$ over a finite field $k$ (say of size $q$) and $u \in G(k)$, so the scheme $\mathfrak{B}_u$ of Borels containing $u$ is defined over $k$. For a lisse $\overline{\mathbf{Q}}_{\ell}$-sheaf $\mathcal{F}$ on the $k$-scheme $\mathfrak{B}_u$, let $\mathcal{F}'$ be its pullback to $\mathfrak{B}'_u = (\mathfrak{B}_u)_{\overline{k}}$. There is a natural $q$-Frobenius endomorphism $F^{\ast}$ of $H^i_c(\mathfrak{B}'_u,\mathcal{F}')$. That being said, the answer is "no" (as in the topology version) and paragraph 2 won't help. –  user27056 Oct 20 '12 at 13:50
    
I don't see the problem with the situation as given above. The Frobenius endomorphsim $F$ is a finite morphism of $\mathfrak{B}_u$ 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? –  Jay Taylor Oct 20 '12 at 15:32
    
Hi Jay! I thought typically your $\BC_u$ are (geometrically) simply connected. Hence their fundamental groups over $\mathbb{F}_p$ will simply be equal to the fundamental group of $Spec\mathbb{F}_p$ and any local system will be pulled back from a local system on $Spec\mathbb{F})_p$. In this case I would guess that $H^*_c(\BC_u,\pi^*V)$ would just be $H^*_c(\BC_u)\otimes V$ (where $V$ is a local system on $Spec\mathbb{F}_p$ aka continuous rep of $Gal(Spec\mathbb{F}_p)$). –  Geordie Williamson Oct 20 '12 at 19:29
    
Hi Geordie! That I am not sure about, although it sounds plausible. Probably the person to ask about that would be Daniel. This would be useful to know if it were the case. –  Jay Taylor Oct 20 '12 at 20:47

1 Answer 1

up vote 1 down vote accepted

First, note that you only get an induced $F^* $ map on cohomology if you fix a map $F^* \mathcal F \to \mathcal F$. Note that you get this for free if $\mathcal F$ is defined over $\mathbb F_p$.

The Frobenius action on the cohomology group of a sheaf is the main object of study in etale cohomology over a finite field. Entire books have been written about the action of Frobenius on the etale cohomology of certain sorts of sheaves. Thus, you shouldn't expect to easily escape this difficulty.

The reason that the induced map $\mathcal F \to \bar{\mathbb Q}_l$ doesn't work is that the map fits into a short exact sequence, giving a long exact sequence on cohomology and unless you have a good understanding of the kernel of that sequence there is no hope of computing the cohomology groups of $\mathcal F$. If $\mathcal F$ is very close to $\bar{\mathbb Q}_l$, and say the kernel is supported on a subvariety of smaller dimension, this is the right approach.

The reason that the other thing doesn't work is that it's not in general true. To see this, suppose $\mathcal F$ is something nice like the extension by $0$ of a lisse sheaf $\mathcal G$ on an open set $U$. Then the compact cohomology of $\mathcal F$ is the compact cohomology of $\mathcal G$, which is Poincare dual to the regular cohomology of $\mathcal G^\vee$. So $H^{2d}_c(\mathcal F) = H^0(\mathcal G^\vee)^{\vee}(-d)$.

Since it is often easy to compute the action of $F$ on the global sections of $\mathcal G^v$, this is usually the best approach, as long as the sheaf is actually the extension by zero of a lisse sheaf. But note that by choosing $\mathcal G$s dual to sheaves with different groups of global sections, you can get many different sorts of cohomology groups with different $Frob_p$ actions which are not the same as the $Frob_p$ action on $H^{2d}_c(\bar{\mathbb Q}_l)$.

In fact, the "typical" thing, in a lot of situations, is for $\mathcal G^\vee$ to have no global sections at all. This gives you a nice answer, but perhaps not the one that you were looking for. To say more I'd have to know more about your sheaf.

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Thanks for your answer Will! Indeed, I had been assuming that I had indeed even an isomorphism $F^*\mathcal{F} \to \mathcal{F}$, which I should have put in the post. It is good to know that this situation is more complicated for an arbitrary local system! –  Jay Taylor Oct 20 '12 at 20:42

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