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.

  • $\begingroup$ Even in degree $0$, this is unclear: How does $f$ induce a natural map $\Gamma(X,F) \to \Gamma(X,F)$? $\endgroup$ 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


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).

  • $\begingroup$ 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. $\endgroup$ Commented Nov 1, 2011 at 12:00
  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.