Lack of "good covers" in the étale topology

Question

Disclaimer: This question might be terribly naive and almost certainly reflects my own ignorance.

If $X$ is a topological space admitting a finite triangulation, then it admits a "good covering," i.e. an open covering by contractible sets $U_i$ such that any finite intersection of the $U_i$'s is also contractible. For such a covering, the Cech cohomology of the covering with coefficients in any constant sheaf coincides with the singular cohomology of $X$ with coefficients in the corresponding abelian group (I think I have sufficient hypotheses for this to work, and I think one can get by with less than a finite triangulation, but let me stick to that).

Now let $\mathscr{X}$ be a proper smooth $\mathbf{C}$-scheme (this is again likely overkill, but it ensures that $X=\mathscr{X}(\mathbf{C})$ is a topological space satisfying the conditions in the first paragraph). For a finite abelian group $A$, there is then a canonical isomorphism $H^i(\mathscr{X}_{et},\underline{A})\simeq H^i(X,A)$, where $\underline{A}$ is the constant étale sheaf on $\mathscr{X}$ associated to $A$.

But, as far as I know, there are in general no analogues of "good coverings" for the étale topology. That is, it's not usually possible to find a single étale covering of $X$ whose Cech cohomology with coefficients in $\underline{A}$ computes the étale cohomology of $\mathscr{X}$ with coefficients in $\underline{A}$. One instead needs to take the colimit over a cofinal set of étale coverings to the first cohomology groups to match up. There is a spectral sequence for any covering converging to the actual étale cohomology groups, but the problem seems to be that there aren't acyclic étale coverings for (finite) constant sheaves (I'd be happy to be informed that I'm mistaken about this).

I should be getting to a question about now. Is there a conceptual reason that these two cohomology groups, one defined algebraically and the other topologically, which coincide, should not be amenable to the same kind of Cech (or simplicial) computation, or that we shouldn't expect good (acyclic) coverings in the étale topology for (finite) constant sheaves? For example, one could prove something about the étale cohomology by working with a good covering of $X$, and there wouldn't necessarily be a "purely algebraic" proof just working on the étale side of things (I have some particular results in mind but it would require a large digression to describe them so for now I'll keep them to myself). I find this psychologically disconcerting.

I should mention there is the theory of hypercoverings, which I don't really understand, but which might provide a computational analogue for étale cohomology of the kind I'm asking about. There is also Bhatt-Scholze's pro-étale topology, a theory in which there are "enough contractible objects," but the affine contractible objects in this topology are (ostensibly) spectra of very unfamiliar (to me!) rings, and I don't think (though again, I'd be happy to be set straight on a misconception here) that this theory provides a computational technique analogous to what is available for singular cohomology on the complex analytic side.

Answer 1

Let $X$ be a compact Riemann surface / smooth projective complex algebraic curve of genus $g \ge 2$. Topologically it is an Eilenberg-MacLane space $B \pi_1(X)$, so its cohomology is the group cohomology of its fundamental group $\pi_1(X)$ (and its étale cohomology is the profinite group cohomology of the profinite completion $\widehat{\pi}_1(X)$). If you wanted to compute its cohomology using a Cech cover you'd just cover it with some disks or whatever. But of course you can't cover a Riemann surface with disks in algebraic geometry, even in the étale topology.

In algebraic geometry, and in particular in the étale topology, what you instead do is look at the finite covering spaces of $X$, since you're allowed to talk about those algebraically due to the Riemann existence theorem. Any particular finite cover, or even finite collection of finite covers, only sees a finite quotient of $\pi_1(X)$, so it's natural that you'd have to look at all of them in order to see all of $\pi_1(X)$ (resp. $\widehat{\pi}_1(X)$).

There is an analogous but somewhat better thing you can do in algebraic topology, which is to look at the universal cover of $X$. That sees all of $\pi_1(X)$ at once. But the universal cover also doesn't exist in algebraic geometry, even in the étale topology, which is of course why we take that profinite completion.

Answer 2

Disclaimer: I'm far from being an expert on this. The moral of the story of Čech cohomology is (as far as I understand it) as follows. Let $X$ be a "space" and $F$ a sheaf on $X$. The functor "take global sections" $(X,F)\mapsto \Gamma(X,F)$ is not exact, so we should derive it. You should think of $\Gamma(-,-)$ as a functor in two variables, because then it seems clear that there are two ways of deriving it.

1. We can find a resolution of the sheaf, $F\to I^\bullet$, and compute $\Gamma(X,I^\bullet)$.
2. We can find a resolution of the space, $U_\bullet \to X$, and compute $\Gamma(U_\bullet,F)$.

Of course, neither of these are very precise at the moment. By the first I mean that we can thicken the category $\mathrm{Sh}(X)$ by embedding it into the model category $\mathrm{K}(X)$ (complexes modulo homotopy), resolve $F$ in the category $\mathrm{K}(X)$, and think of $\Gamma(X,I^\bullet)$ as an object in $\mathrm{D}(\mathrm{Ab})$, the derived category of abelian groups.

What does it mean to resolve the space $X$? Suppose we have an open cover $\mathcal U=\{U_i\to X\}$. In Čech theory, one looks at the collections $$ \mathcal U_n = \left\{U_{i_1}\times_X \cdots \times_X U_{i_n}\right\}_{i_1,\dots,i_n} . $$ These form a simplicial space with augmentation $\epsilon:\mathcal U_\bullet \to X$. Thus $\Gamma(\mathcal U_\bullet,\epsilon^\ast F)$ is a simplicial abelian group. Now, if we are working with honest topological spaces, there is a good model structure (which I don't very well understand) in which it makes sense to call $\mathcal U_\bullet \to X$ a resolution of $X$ (this happens exactly when each $n$-fold intersection $U_{i_1}\cap \cdots \cap U_{i_n}$ is contractible). The category $\mathrm{sAb}$ of simplicial abelian groups has a model structure, and so we could define $\mathrm R \Gamma(X,F) = \Gamma(\mathcal U_\bullet,F)$; this lives in $\mathrm{Ho}(\mathrm{sAb})\simeq \mathrm{D}(\mathrm{Ab})$.

If we're working with schemes instead of topological spaces, I understand the situation less well. Roughly, for each étale (or fppf, or fpqc, ...) cover $\mathcal U=\{U_i\to X\}$, the corresponding simplicial cover $\mathcal U_\bullet \to X$ is a sort of "approximation to a resolution of $X$." Thus is makes sense that $\varinjlim \Gamma(\mathcal U_\bullet,F)$ should approximate (and, in nice settings, compute directly) the cohomology $\mathrm R\Gamma(X,F)$.

So to answer your question, I think what is going on is the following:

You can try to compute $\mathrm R\Gamma(X,F)$ either by resolving $X$ or resolving $F$. If $x$ is an honest topological space, there is a good theory of resolving $X$, and everything works out. But if $X$ is a variety and we're working in the étale topology, the notion of "resolving $X$" doesn't work very well, so it only gives us an approximation of $\mathrm R\Gamma(X,F)$.

Comments / corrections from people who actually understand what's going on here are greatly appreciated!

Edit: In case the idea of deriving a two-variable functor in one variable at a time seems weird, just think of, for any abelian category, the functor $\hom(-,-)\colon \mathcal{A}\times \mathcal{A}^\circ\to \mathrm{Ab}$. It's derived functor, $\mathrm{Rhom}(-,-)$ can be computed in two ways. Take an injective resolution $A\to I^\bullet$ and think of $\hom(I^\bullet,B)$ as living in the derived category, or take a projective resolution $P_\bullet \to B$ and consider $\hom(A,P_\bullet)$ in the derived category. One has to prove that these give equivalent definitions of $\mathrm{Rhom}(A,B)$.