# In what sense is the étale topology equivalent to the Euclidean topology?

I have heard it said more than once—on Wikipedia, for example—that the étale topology on the category of, say, smooth varieties over $\mathbb{C}$, is equivalent to the Euclidean topology. I have not seen a good explanation for this statement, however.

If we consider the relatively simple example of $\mathbb{P}^1_\mathbb{C}$, it seems to me that an étale map is just a branched cover by a Riemann surface, together with a Zariski open subset of $\mathbb{P}^1_\mathbb{C}$ that is disjoint from the ramification locus. (If there is a misconception there, small or large, please let me know) The connection to the Euclidean topology on $\mathbb{P}^1_\mathbb{C}$, however, is not obvious to me.

What is the correct formulation of the statement that the two topologies are equivalent, or what is a good way to compare them?

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It has to do with the fact that the étale cohomology groups are closely related (they might be isomorphic, but I don't remember) to the singular cohomology groups in the case of smooth complex varieties. –  Harry Gindi Apr 5 '11 at 4:01
Vague answer: I think there are two parts. First, by some GAGA principle, algebraic etale covers = Euclidean etale covers, say when they are both finite. The Euclidean etale topology is equivalent just to the Euclidean topology, since every covering map is a local homeomorphism. To see that the GAGA correspondence really does capture the fineness of the Euclidean topology, you need to check that you can make enough refinements; the "refinement" of a covering is measured in some sense by its Cech cohomology, and as Harry says, etale and singular cohomology are isomorphic, morally. –  Ryan Reich Apr 5 '11 at 4:46
On page 9 of Bhargav Bhatt's lovely notes on the étale topology there is the following statement: "For an analytic space $X$, the étale topos $Et(X)$ is equivalent, as a category, to the standard topological topos $Top(X)$ whose cohomology is, by definition, singular cohomology." The étale site of an analytic space $X$ is defined just like in the algebraic case, only "with (analytic) local isomorphisms replacing the étale morphisms". –  Ingo Blechschmidt Jul 3 at 13:11

Saying that the étale topology is equivalent to the euclidean topology is vastly overstating the case. For example, if you compute the cohomology of a complex algebraic variety with coefficients in $\mathbb Q$ in the étale topology, typically you get 0. On the other hand, it is a deep result that the étale cohomology of such a variety with coefficients in a finite abelian group coincides with its cohomology in the euclidean topology.

Similarly, you can't capture the whole fundamental group with the étale topology, but only its finite quotients (and the fact that you can indeed describe the finite quotients of the fundamental group via étale covers is, again, a deep result).

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This might be a stupid question, but is there any Grothendieck topology that you can put on an arbitrary scheme (say, fppf, fpqc, nisnevich, etc) that recovers the euclidean cohomology groups when restricted to complex algebraic varieties? If not, is there a proof that no such topology could exist? –  Harry Gindi Apr 5 '11 at 8:32
It is not stupid at all. An indication of the fact that this should not be possible is the fact that $\ell$-adic cohomology of elliptic curves over an algebraically closed field of positive characteristic cannot be defined over $\mathbb Z$. At least, I believe this is known and due Serre, but I can't recall the precise argument (I would guess it has to do with the existence of elliptic curves with exotic complex multiplication, but I am not sure). –  Angelo Apr 5 '11 at 10:04
Harry, there's no purely algebraic definition of the $\mathbb{Q}$ cohomology of an algebraic variety $X$ over $\mathbb{C}$. The choice of a model of $X$ over a subfield $K$ of $\mathbb{C}$ will define an action of $Aut(C/K)$ on $X$ and hence on the cohomology. But we can choose $K$ so that such that $\mathbb{C}/K$ is an infinite Galois extension. Then $Gal(C/K)$ is uncountable, and should typically act through an uncountable quotient on the $\mathbb{Q}$ cohomology (because it does on the etale cohomology). But the $\mathbb{Q}$ cohomology is countable. –  mephisto Apr 5 '11 at 11:34
Thanks, guys! –  Harry Gindi Apr 5 '11 at 15:18
I think that the argument that Angelo is referring to is the second explanation of why there is no universal cohomology theory over $\mathbf Q$ given in Milne's notes on motives: jmilne.org/math/xnotes/mot.html (The first explanation in Milne's notes is the one given above by mephisto.) –  Dan Petersen Apr 5 '11 at 15:26

Your final question, i.e. in what way does one compare the classical and the étale topology of a scheme, is answered for instance in section 2 of Mumford's classical "Picard Groups of Moduli Problems". There are also some nicely written parts in Vistoli's notes on descent on how to compare Grothendieck topologies in general.

As you say there is no direct way to compare the two. What you do is introduce an auxilliary topology which refines both of them. If you take a sep. finite type scheme X over $\mathbf C$, then you can put a topology on complex analytic spaces over X by taking for open subsets those maps $U \to X(\mathbf C)$ that form a covering space over an open subset of $X(\mathbf C)$. Coverings are jointly surjective. We call this site $X_{cx}^\ast$. Then every every open set in both the étale and classical topologies are also open sets of this site, so there are maps $\alpha : X_{cx}^\ast \to X_{cx}$ and $\beta : X_{cx}^\ast \to X_{ét}$. Moreover, $\alpha$ is an equivalence of topologies. (I see now that most of what I said here was already stated in the comment by Ryan Reich.)

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Dear Andrew, here is another take on your question, in the direction of (ramified) covers.
1. Riemann There is an analytification functor $X\mapsto X^\{an}$ from the category of $\mathbb C$-schemes locally of finite type to that of (non-reduced) complex analytic spaces. Its introduction is due principally to Riemann, Chow, Serre and Grothendieck. It has all the desirable properties and of course the set of points of $X^\{an}$ is $X(\mathbb C)$. Riemann's existence theorem states that this functor induces an equivalence of categories between the finite étale covers of $X$ and the finite étale analytic covers of $X^\{an}$.
This is the deep result (alluded to in Angelo's fine answer) which, in particular, yields the identification of the topological fundamental group of $X^\{an}$ with a completion of the scheme-theoretic fundamental group.

2. Grauert-Remmert In a sense the classification of algebraic covers has been reduced to that of analytic ones. We can then apply the following result, due to Grauert and Remmert :
Let $X$ be a normal analytic space and $U\subset X$ an open subset with analytic complement. Then any (ramified) finite normal cover of $U$ uniquely extends to a normal finite cover of $X$.
(Grothendieck, in SGA 1, gave a slick proof of this theorem by invoking Hironaka's resolution of singularities)

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A good start is to read Chapter 21 of Milne's textbook on etale cohomology. I suspect the results there are not stated with maximal power, but it at least gives you a feeling for what's true.

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