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This is probably totally obvious but I have no clue how this is done: Say you have an endomorphism $f:X \rightarrow X$ of schemes. Why (if true, perhaps some additional assumptions are necessary!) do you get for a Zariski/étale/l-adic sheaf $\mathcal{F}$ on $X$ an induced endomorphism on the corresponding cohomology? How is this constructed? Are there conditions, when the induced morphism is an isomorphism (I'm having a Frobenius in mind)?

Perhaps the above is too general, so my real question is: Why does a group/monoid action on the Deligne-Lusztig variety induce a group/monoid action on the l-adic cohomology (with compact support) of this variety? In every book I looked at this is just mentioned but not explained.

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  • $\begingroup$ Even in degree $0$, this is unclear: How does $f$ induce a natural map $\Gamma(X,F) \to \Gamma(X,F)$? $\endgroup$
    – Martin Brandenburg
    Commented Nov 1, 2011 at 10:29
  • $\begingroup$ I just try. - I tend to think as cohomology as Cech cohomology (which in most cases is possible); take a covering $\mathcal{U}=${$U_i$} of $X$ such that also $f^{-1}(U_i)\in\mathcal{U}$. Then I think $f^{-1}$ preserves $\mathcal{U}\times_{X}\mathcal{U}$ and $\mathcal{U}\times_{X}\mathcal{U}\times_{X}\mathcal{U}$ etc. And $f$ acts on $\Gamma(U_1\cap \ldots\cap U_k,F)$ by pullback of sections. This should induce an action on Cech's $H^k(\mathcal{U},F)$ and, passing to the limit, on $H^k(X,F)$. $\endgroup$
    – Qfwfq
    Commented Nov 1, 2011 at 10:55
  • $\begingroup$ (by $\mathcal{U}\times_{X}\mathcal{U}$ I mean the set of opens of the form $U\cap V$ where $U,V\in\mathcal{U}$, and so on) $\endgroup$
    – Qfwfq
    Commented Nov 1, 2011 at 10:57

1 Answer 1

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If a (say constant) group $G$ acts on a scheme $X$, you may want to consider the notion of a $G$-sheaf : a sheaf $\mathcal F$ endowed with isomorphisms $\lambda_g: g^* \mathcal F\simeq \mathcal F$, for $g\in G$ satisfying the usual cocycle conditions. Then by functoriality of cohomology for $g:X\to X$ you get an isomorphism $H^i(X, \mathcal F) \to H^i(X, g^*\mathcal F)$ that you can compose with the morphism induced by $\lambda_g$, that is $H^i(X, g^* \mathcal F)\simeq H^i(X,\mathcal F)$. Thus you get for each $g\in G$ an automorphism of $H^i(X, \mathcal F)$ and it is easy to check that this gives an action of $G$ on this cohomology group.

A probably better way to see this is to use functoriality of $G$-sheaves : the global section functor goes from $G$-sheaves of abelian groups to abelian groups endowed with an action of $G$. Since the abelian category of $G$-sheaves has enough injectives (a classical fact, you can find it in Grothendieck's famous Tohoku paper) you can derive it. You get cohomology groups naturally endowed with an action of $G$. Once you apply the forgetful functor, you recover the usual cohomology groups (this is easy to see directly, or you can use Grothendieck's theorem on derivation of a composition of functors, the only point is that the forgetful functor is acyclic).

There is a natural generalization to action of non constant groups, and also to action of monoids.

I think you can apply this in your situation, since the $l$-adic sheaf defining $l$-adic cohomology is naturally endowed with the structure of a $G$-sheaf (the sheaf $\mathbb Z/l^k\mathbb Z$, as any constant sheaf, has a canonical structure of $G$-sheaf).

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  • $\begingroup$ Very good, thanks, I'll think about this! Two more questions: 1. What is a constant group? 2. Do you have a good reference for what you just explained? Perhaps something post-SGA to make it shorter :) $\endgroup$
    – George P.
    Commented Nov 1, 2011 at 11:48
  • $\begingroup$ The question of George P. was about general endomorphisms, not just constant group schemes. The functoriality of the cohomology is more difficult to construct for general endomorphisms. $\endgroup$
    – Damian Rössler
    Commented Nov 1, 2011 at 12:00
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    $\begingroup$ Sheaf cohomology is only functorial on the pairs $(X, F)$ consisting of a scheme $X$ and a sheaf $F$, where a map $(X, F)\to (Y, G)$ is a pair of maps $(f,\phi)$, $f : X \to Y$ and $\phi : f^*G \to F$. Any such map of pairs induces a map on the cohomology $H^i(Y,G)\to H^i(X,F)$ as explained by Niels. $\endgroup$
    – Nicolás
    Commented Nov 1, 2011 at 12:14
  • $\begingroup$ @ George P. 1 You can associate to each abstract group $G$ a "constant" group scheme $G_X$ on $X$, that represents the functor $Hom_X(\cdot,G\times X)$. In other words the sections of $G_X$ above the open $U\to X$ are $G^{\pi_O(U)}$, where $\pi_O(U)$ is the set of connected components of $U$. 2 The standard reference is Sur quelques points d'algèbre homologique, I Alexander Grothendieck Source: Tohoku Math. J. (2) Volume 9, Number 2 (1957), 119-221. projecteuclid.org/… $\endgroup$
    – Niels
    Commented Nov 1, 2011 at 13:12
  • $\begingroup$ @Damian Rössler: you are right, for endomorphisms one needs to be somewhat more careful and work with monoids, but I don't think it is fundamentally different. For actions of non-constant schemes, I am not longer sure what the question means (probably one then considers action of the global sections of the group scheme on the cohomology). $\endgroup$
    – Niels
    Commented Nov 1, 2011 at 13:15

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