Timeline for Action of Frobenius on the Étale Cohomology of the variety of Borel subgroups for an arbitrary local system
Current License: CC BY-SA 3.0
8 events
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| Oct 20, 2012 at 20:47 | comment | added | Jay Taylor | 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. | |
| Oct 20, 2012 at 20:40 | vote | accept | Jay Taylor | ||
| Oct 20, 2012 at 19:29 | comment | added | Geordie Williamson | 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 $Gal(Spec\mathbb{F}_p)$). | |
| Oct 20, 2012 at 17:57 | answer | added | Will Sawin | timeline score: 3 | |
| Oct 20, 2012 at 16:06 | history | edited | Jay Taylor | CC BY-SA 3.0 |
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| Oct 20, 2012 at 15:32 | comment | added | Jay Taylor |
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?
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| Oct 20, 2012 at 13:50 | comment | added | user27056 | 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. | |
| Oct 20, 2012 at 13:19 | history | asked | Jay Taylor | CC BY-SA 3.0 |