# Idea of using etale site

I have just read an article which mentions that, when Grothendieck considered using etale morphism, he did borrow the idea from Riemann that multivalued function on an open subset of complex plane should live in Riemann surface covering it instead of the open subset itself. The question is: how these things are related? Any detailed explanation is very welcome and appreciated. More generally, what are the good properties etale morphism has which make it essential in solving Weil conjecture.I understand that Grothendieck tried to search for a cohomology theory in algebraic geometry which is similar to classical cohomology theory on manifold theory but I dont know what obstructions are there preventing a cohomology theory in AG behaves similar to one in manifold for which etale site was introduced to overcome them.

• My outsider's impression is that the etale topology serves to make the Inverse Function ("A map is locally invertible if it is infinitesimall invertible") is true in algebraic geometry, by changing the meaning of the word "locally". – Tom Goodwillie Jul 2 '15 at 20:03
• Tom, I know that by using etale topology,we can have an analogue of your statement "A map is locally invertible if it is infinitesimall invertible".What I am not sure is that, is this an essential obstruction to have 'good' cohomology theory AND is it the only obstruction? – wkf Jul 2 '15 at 20:12
• For the origins of etale cohomology, see the article "The Riemann Hypothesis..." on Milne's website. – abz Jul 3 '15 at 1:20
• To supplement Tom's comment above, I will give an example showing why we need to change the meaning of the word "locally" (or really, the topology). Let $k$ be a field of characteristic not $2$, and define $X := \operatorname{Spec} k[u]_u$, $Y:= \operatorname{Spec} k[t]$. Let $f : X \to Y$ be the map induced from $t \mapsto u^2$. Since $\operatorname{char} k \neq 2$, it is easy to check that $f$ is \'{e}tale. I claim that there is no Zariski open subscheme $U \subseteq Y$ such that $f|_{f^{-1}(U)} : f^{-1}(U) \to U$ is an isomorphism of schemes. – Ben Lim Jul 3 '15 at 18:50
• (cont'd) Indeed, the existence of such a $U$ would imply that $f$ is birational, and so $k(u^2) \to k(u)$ is an isomorphism, a contradiction. – Ben Lim Jul 3 '15 at 18:51

I can tell you how they are related.

Before Riemann people would say, for example, the complex square root function (for $z\neq 0$) is two valued, but for any small region of (non-zero) complex numbers you can make it single valued by picking one branch. Riemann had a vastly better idea: there is a two-sheeted covering surface for the complex plane (ramified at 0) with square root a single-valued function on that cover.

Serre, who was well aware of the connection to Riemann, found a theory of 1-dimensional cohomology that worked correctly for the Weil conjectures, using not sheaves but fiber bundles, where a fiber bundle is considered locally trivial (and called "isotrivial"), not when it restricts to product bundles on small enough parts, but if it can be made into a product bundle by pulling it back along such a cover.

Well, Serre also saw how he could state the algebraic conditions needed to make this work, not only over the complex numbers, but over any field. Those conditions are now taken as the definition of a finite etale map. Grothendieck, with Artin and others, including Serre, made it work in all dimensions and for that purpose preferred to drop the requirement that the map be finite.

As to this works for the Weil Conjectures, let add a bit on why Serre first thought his "unramified maps" (which later gave way to the slightly different etale maps) were the way to such a cohomology, and why Grothendieck then decided this was exactly the way. You should combine this with Peter Dalakov's concise modern statement of the facts in his comment, and Will Sawin's beautiful account of what a cohomology theory for those conjectures would have to be like.

No one who was interested in the Weil Conjectures when they first appeared believed fields in finite characteristic would support any close analogue to the analytic topology on complex numbers. In hindsight people today pretty much agree with that, but at the time most considered this a decisive obstacle to any cohomological proof of the Weil Conjectures. And no one before Serre's FAC saw how to use Zariski topology to prove any very serious results. Serre's FAC immediately persuaded a lot of people that algebraic geometry over arbitrary fields could, and in fact must, use the Zariski topology.

But many structures which intuitively ought to be "locally trivial" are clearly not so if "locally" means "on small enough Zariski open sets." Zariski open sets just never are small -- they are dense on any connected component. Serre wrestled with precisely this problem for several years. And then in 1958, with Riemann's original works explicitly in mind, Serre said let us allow "local trivialization" of fiber bundles just the way Riemann "trivialized" multiple valued functions into single valued ones-- let us trivialize them by pullback along unramified Riemann surface covers -- except using a purely algebraic definition of "unramified" so it works over any field, and indeed for varieties of any dimension. A strikingly plausible idea once you think of it. But does it work?

By the kind of deep, detailed skill that Serre typically conjoins to his insights, he got it to work for dimension one cohomology (of varieties of any dimension). It works in the precise sense that it delivers the $H^1$ part of the long exact cohomology sequences you would want for the Weil Conjectures.

Serre knew well how hard he had to work to get these $H^1$s. So he was skeptical when Grothendieck first announced this had to work for cohomology in all dimensions. But Grothendieck had utter faith in his general theory of derived functor cohomology: once Serre identified the correct basics, they had to deliver the whole theory.

Well it turned out to take a lot more specific work, and there is the long and on-going story of the standard conjectures which were meant to make the cohomological proof much simpler than it yet is, but Grothendieck's faith was essentially justified.

As to the history I would slightly modify what Will Sawin says. He puts the key issues very well. But Weil did not believe there could be an actual cohomology theory for varieties in finite characteristic. I believe he believed there would be some more direct comparison theorem between varieties in finite characteristic, and their lifts to characteristic zero, which would make the conjectures follow from simplicial cohomology. And he did not especially believe that such a comparison would be the way to prove the conjectures. He probably leaned to the idea that the relation to simplicial cohomology of complex manifolds would be an enlightening corollary to some other kind of proof.

• Colin, do you have any reference of the idea of Serre you mentioned? – wkf Jul 3 '15 at 17:01
• @wkf: See e.g. Serre, "Espaces Fibres Algebriques", Sem.Bourbaki, 1951-54, exp.82, p.305-311 or "Espaces fibres algebriques", Sem. C.Chevalley, tome 3 (1958), exp.1, p.1-37. Both are on Numdam. As a different way of phrasing what Colin is saying: think of $\mathbb{C}^\times \to \mathbb{C}^\times$, $z\mapsto z^2$. This is a holomorphic Z/2 principal bundle: you can trivialise it on any simply-connected open (e.g. remove two rays). But it is not Zariski locally trivial. But you can trivialise it by pulling back by squaring - an etale map. – Peter Dalakov Jul 3 '15 at 20:11
• @wkf Reference? Did you check the reference in my comment yesterday? – abz Jul 3 '15 at 20:37
• @wkf Yes Dalakov gives the original references. And they are very worth reading, as so often with Serre. I have not seem the Milne reference but of course I've read other things by him and I will look at that. – Colin McLarty Jul 3 '15 at 23:23
• @abz I cant find the article you mentioned. I didnt see Milne has any article on his site of title starting with 'the Riemann hypothesis'. – wkf Jul 4 '15 at 16:29

With regards to the Weil conjectures:

I believe it was known to Weil that, to prove most of the Weil conjectures, a cohomology theory for varieties with coefficients in a field of characteristic zero would just have to satisfy a few properties that are familiar from singular cohomology:

1. The Lefschetz fixed point formula, applied to the endomorphism Frob_p, and finiteness of Betti numbers immediately give rationality of the zeta function.

2. Poincare duality, suitably twisted to account for the fact that Frob_p should act on $H^{2n} (X)$ for $X$ an $n$-dimensional smooth projective variety by multiplication by $p^n$, gives the functional equation.

3. The Riemann hypothesis is the only one I'm not sure about - Deligne's proof of this involves many clever steps that had no antecedent in singular cohomology.

4. The fact that Betti numbers are constant in smooth proper fibrations suffices for the Betti number conjecture, and a comparison theorem to singular cohomology over the complex numbers, (although Weil may not have had the full picture of an arithmetic family of varieties as being the same as a geometric family as clearly as Grothendieck did)

So why does etale cohomology have these features? I think experience shows that any cohomology theory we define has these features as long as it satisfies some very basic tests - off the top of my head, I can't think of any cohomology theory that has the right Betti numbers for a curve of genus $g$ that doesn't have all these properties.

$\ell$-adic etale cohomology avoids the trivial failure of all the cohomology theories you can construct with just the Zariski topology - either cohomology of the constant sheaf $\mathbb Z$ in the Zariski topology, or cohomology of some coherent sheaf, probably powers of the cotangent bundle to mimic Dolbeaut cohomology in characteristic zero. The first one has the wrong Betti numbers, and the second one has coefficient field the base field, so has coefficients of characteristic $p$ in characteristic $p$, and so cannot prove the Weil conjectures.

• "a cohomology theory for varieties with coefficients in a field of characteristic zero" - Do you mean to say in a field of positive characteristic? – Sam Hopkins Jul 4 '15 at 16:04
• Will, could you tell what are the 'basic tests'?sorry for my ignorance. – wkf Jul 4 '15 at 16:25
• @SamHopkins No, I mean to say that the coefficients of the cohomology theory are in a field of characteristic zero. The varieties, I should have said, are characteristic $p$. – Will Sawin Jul 4 '15 at 16:37
• @wkf I don't think there's a formal statement. The first basic test I would try is whether $H^1$ of a smooth projective curve of genus $g$ is $2g$-dimensional. Grothendieck said that if you have a good theory of $H^1$ you should be able to find a good theory of all the other cohomology groups, so that might be enough. – Will Sawin Jul 4 '15 at 16:39
• @Wkf After that I would try showing that, more generally, $H^1$ of a variety is twice the dimension of the Albanese, and then to show that $H^2$ of a curve is one-dimensional. I would consider that enough basic tests and then try to prove some theorems about a cohomology theory that passed the tests. – Will Sawin Jul 4 '15 at 16:40