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In the search for a Weil cohomology theory $H$ over a field $K$ (with $\text{char}(K)=0$) for varieties in characteristic $p$, a classical argument by Serre shows that the coefficient field cannot be a subfield of $\mathbb{R}$ or of $\mathbb{Q}_p$ ; an obvious choice is to take $\mathbb{Q}_\ell$ for a prime $\ell \neq p$.

Now, we can try to make a Weil cohomology theory by taking the sheaf cohomology with constant sheaves with the Zariski topology, but this does not work as all cohomology vanishes.
Grothendieck's insight was that we can find a different topology, for example the étale topology. Then we can build a Weil cohomology theory with coefficients in $\mathbb{Q}_\ell$ by taking cohomology with coefficients in the constant sheaves $\mathbb{Z}/ \ell^n\mathbb{Z}$ and then taking the inverse limit with respect to $n$ and tensor with $\mathbb{Q}_\ell$ : this gives $\ell$-adic cohomology.

But it is not so clear to me why the étale topology is best suited at this task. What happens if we repeat the above procedure on other sites? Does the cohomology theory we get fail to be a Weil cohomology theory?

P.S.: Information for fields other than $\mathbb{Q}_\ell$ would also be nice!

edited Feb 21 2010 at 19:03

asked Feb 21 2010 at 18:17

Sam Derbyshire
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Also, does anyone know of a reference for Serre's argument? – Sam Derbyshire Feb 21 2010 at 18:19

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The point is that for a supersingular curve E over a finite field, $End(E)\otimes Q$ is a quaternion algebra (skew field of degree 4 over Q). It can't act on a two-dimensional space over a field F unless F splits the algebra (i.e., tensoring with F gives a matrix algebra). Since $End(E)\otimes Q$ isn't split by R or Q_p, there can't be a Weil cohomology with coefficients in those fields. This is always credited to Serre because he put it in print, but was surely common knowledge, e.g., Weil surely knew it. – JS Milne Feb 21 2010 at 18:29

Thanks for the comment. I was aware of the argument, but I don't know where Serre wrote it down. It's nice to know that it was known before. – Sam Derbyshire Feb 21 2010 at 18:41

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I'm a bit confused by the statement that "there can't be a Weil cohomology with coefficients in those fields." Perhaps I am mis-remembering what the axiomatic definition of a Weil cohomology is, but isn't rigid cohomology supposed to be a Weil cohomology? – B. Cais Feb 21 2010 at 19:05

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@B. Cais, Sam D: The argument of Serre shows that a you can't have a Weil cohomology over say $\mathbf{F}_{p^2}$ which takes values in $\mathbf{Z}_p$. The argument breaks down for a supersingular curve over $\mathbf{F}_p$ because not all of its endomorphism ring is defined over $\mathbf{F}_p$. So for instance crystalline (resp. rigid) cohomology takes values in W(F) (resp. W(f)[1/p]) – David Zureick-Brown♦ Feb 21 2010 at 19:34

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JS Milne · Accepted Answer · 2010-02-21 18:38:43Z UTC

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As observed, the Zariski topology has too few open subsets to compute cohomology with constant coefficients. It had already been observed by Serre that etale covers were enough to trivialize principal bundles for many algebraic groups. That suggested using etale covers. The etale "topology" is the coarsest for which the inverse function theorem holds. The flat topology also gives Weil cohomologies (the same ones as the etale topology), but why use flat covers when etale covers are enough.

answered Feb 21 2010 at 18:38

JS Milne
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As I understand it, we want the inverse function theorem to hold so that the étale cohomology compares favourably to singular cohomology in the complex analytic case; in particular they are isomorphic for finite coefficient abelian groups, which is what we need to set up $\ell$-adic cohomology. Will this necessarily fail if we have a topology in which the inverse function theorem fails? – Sam Derbyshire Feb 21 2010 at 18:57

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The group $H^1(X,A)$ with A a finite commutative group, should classify the A-torsors over X. In order for it to do this, the topology must split the torsors. This means you must allow etale covers, and the miracle is once you do that you get the "correct" cohomology groups in all degrees. – JS Milne Feb 21 2010 at 19:26

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I think the best explanation for this miracle is homotopy-theoretic. The insufficiency of a topology for calculating the homotopy type of a space loosely amounts to the lack of "good covers", that is covers by contractible open sets (given good covers, one recovers the homotopy type from the combinatorics of a suitable hypercover by good opens). Now, in this sense, the Zariski topology is not too far from giving the right homotopy type: [cont'd] – Dustin Clausen Feb 21 2010 at 20:15

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as Artin shows in SGA, every variety (over C say) has a Zariski cover by varieties which iteratively fiber down to the point in hyperbolic curves, and thus, while not contractible, have vanishing higher homotopy groups: they are K(pi,1)'s. Then when we allow ourselves etale covers as well, we can take any finite covering space of these K(pi,1)'s (Riemann existence). Thus, though we can't make etale covers completely contractible, we can make them contractible enough for the purposes of finite coefficient systems. – Dustin Clausen Feb 21 2010 at 20:15

Motivation for the étale topology over other possibilities

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