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Suppose I have a sheaf $\mathcal{F}$ on the (small) étale site over $X$. By restriction, $\mathcal{F}$ is also a sheaf on $X$ (with the Zariski topology). When is it that the sheaf cohomologies (i.e. derived functors of the global section functors) agree in these two sites?

For example, in SGA 4 (Chapter VII, p355), Grothendieck proves that the above cohomologies agree for quasicoherent sheaves. I do not really understand the proof however, and would appreciate any elaboration on this. It uses the Leray spectral sequence, and it seems that the key ingredient is that the higher direct images $R^q f_*(\mathcal{F})$ are $0$, where $f_*$ is the direct image functor induced by the inclusion of the Zariski site in the étale one).
Question: Are there any precise conditions on $\mathcal{F}$ that guarantee that the two cohomologies agree? Is the above condition on $R^q f_*(\mathcal{F})$ necessary/sufficient, and when does it hold?

I should give an example of failure: constant sheaves seem like the obvious choice, and here the cohomologies don't agree even in the case of a point (Zariski cohomology is zero, étale cohomology reduces to Galois cohomology).

Maybe an easier question would be obtained by instead starting with a sheaf $\mathcal{G}$ on $X$ (with the Zariski topology).
Subquestion: Under what conditions can you extend $\mathcal{G}$ to a sheaf on the (small) étale site? (I'm thinking about extending by inverse image, i.e. for an étale $f \colon U \to X$, setting $\mathcal{G}(U) =f^*(\mathcal{G})(U)$, but maybe there are other possibilities)
(Possibly) Easier question: Supposing that $\mathcal{G}$ extends, are there any precise conditions on $\mathcal{G}$ that guarantee that the Zariski cohomology and the étale cohomology of $\mathcal{G}$ agree?

Here again I have in mind the case of quasicoherent sheaves, where everything works out nicely provided that you extend in the right way: starting with a quasicoherent sheaf on the Zariski topology, you need to make sure your sheaf ends up being quasicoherent for the étale topology. You need to take some tensor products to ensure this, as Scott points out in the comments.

(Of course, I have specialised this question to a particular example of sites/toposes, ie Zariski and étale. If you know how this works in more general cases, please share that too!)

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Regarding your subquestion, the restriction functor $\epsilon: X_{\'et} \to X_{Zar}$ has a right adjoint $\epsilon^{-1}$, and it returns the sheaf associated to the presheaf $(U,f) \mapsto \mathcal{G}(f(U))$. You have to take a tensor product to get the $\mathcal{O}$-module functor $\epsilon^*$, and this turns out to be an equivalence on quasicoherent modules. –  S. Carnahan Apr 3 '10 at 18:50
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A concrete way to see the vanishing for the qcoh case is Cartan's criterion for vanishing of higher sheaf cohomology (in terms of Cech-like cohomologies, applied in the etale topology) and the use of a cofinal system of affine etale covers of an affine scheme, coupled with the exactness of the long Cech-like complex built for a module and a faithfully flat ring extension (which in turn rests on the essential content of Grothendieck's brilliant trick with a section to prove fpqc descent for qcoh sheaves). –  BCnrd Apr 3 '10 at 23:42
    
Regarding the vanishing of higher direct images, it tells you that the Leray spectral sequence degenerates into a family of isomorphisms $H^p(X_{Zar},f_*(\mathcal{F}))\rightarrow H^p(X_{et},F)$. –  Keenan Kidwell Mar 26 '12 at 23:03
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1 Answer

up vote 6 down vote accepted

As you pointed out, your question (for general sheaves of abelian groups) follows from the vanishing of the derived functors $R^if_*{\mathcal F}$ for $i>0$, and that's all one can say.

A good example is the Beilinson-Lichtenbaum conjectures, which states that for a certain complex of sheaves $\mathbb Z(n)$, the etale and Zariski cohomology groups agree up to degree $n+1$, and then differ. The same example also show that many deep questions in arithmetic geometry can be formulated as a comparison result between the two cohomology groups, which you could view as a meta-proof that there is no good answer.

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