Timeline for Frobenius actions on de Rham cohomology, clarify questions on a paper of Kedlaya
Current License: CC BY-SA 4.0
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| Jun 16 at 13:03 | comment | added | user267839 | ps: I see the problem with the point I tried to pose in previous comments as I misread that the argument you gave was more "Frobenius specific" (...including more data then just from endomorphism inducing universal homeom on topol side) then I initially wrongly assumed. So it's fine, the question in my last comments don't make sense | |
| Jun 16 at 11:36 | comment | added | user267839 | yes, the question deviated a bit from initially posed one. I tried to phrase my concern here, could you by occasion skim briefly through? (so far I understood your argument correctly, it is nothing abolute Frobenius specific, but depends only on requirement $F$ to be universally homeo (what for instance abs Frob is) | |
| Jun 16 at 1:58 | comment | added | R. van Dobben de Bruyn | I think this is not really the place for extended discussion. Maybe you could ask a new question on MO or MSE if there is something else you want to know. | |
| Jun 16 at 1:09 | comment | added | user267839 | pose instead as "compensation for this weakening" an identical question about how it would act on Zariski, instead of etale cohomology. Motivation: To which extent here "universality" (of the homeom) goes into? Is it only due to that etale sheaves are much finer than Zariski, or would the identical question but with Zariski, instead of of etale cohomology require the same assumption that the homeomorphism $F$ has still to be assumed to be universal? | |
| Jun 16 at 1:07 | comment | added | user267839 | so this argumentation goes through for any universal homeo $F:X \to X$ to assure that the adjoint functors are inverse to each other. But what one should expect if we drop the "universal" assumption, ie we have only a homeomorphism $H:X\to H$ and asking what can we say how does $H^∗$ act on (etale, or if that's too strong, then say Zariski) cohomology? My concern is that so far I know a homeo between schemes induces equivalence of their Zariski sheaves, and so I'm wondering if the argumentation you gave would go through if we weaken the assumption for $F$ to be only a homeomorphism, but | |
| Jun 15 at 22:26 | comment | added | R. van Dobben de Bruyn | While you normally get pullback maps $H^i(X,-)\to H^i(X,F^*(-))$, composing with $F^*\stackrel\sim\to\operatorname{id}$ defines maps $H^i(X,-)\to H^i(X,-)$. For sheaves of sets, the functor $H^0(X,-)$ is represented by the terminal object $X \in \mathbf{Sh}(X_{\text{ét}})$, so the Yoneda lemma shows that the only natural transformation $H^0(X,-) \to H^0(X,-)$ is the identity. This implies that $H^0(X,-) \to H^0(X,-)$ is also the identity for abelian sheaves, and then use injective resolutions to get the result for all $i$. | |
| Jun 15 at 22:26 | comment | added | R. van Dobben de Bruyn | You can find this for instance in SGA 5, Exp. XV, §2, Prop. 2(c). The proof basically goes as follows: because the absolute Frobenius $F\colon X\to X$ is a universal homeomorphism, the adjoint functors $F_*\!:\mathbf{Sh}(X_{\text{ét}})\leftrightarrows\mathbf{Sh}(X_{\text{ét}}):\!F^*$ are inverses. The relative Frobenius $F_{U/X}\colon U\to U^{(p)}=F^*U$ is an isomorphism for $U\to X$ étale, defining an isomorphism $F_*\stackrel\sim\to\operatorname{id}$. The adjoint of its inverse gives an isomorphism $F^*\stackrel\sim\to\operatorname{id}$. | |
| Jun 15 at 19:06 | comment | added | user267839 | Re on question $2$: is there a standard argument why the the absolute Frobenius here acts trivially on the etale cohomology? | |
| May 6, 2018 at 19:38 | vote | accept | Wenzhe | ||
| May 6, 2018 at 19:35 | vote | accept | Wenzhe | ||
| May 6, 2018 at 19:38 | |||||
| May 6, 2018 at 19:29 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |