# Étale homotopy type of non-archimedean analytic spaces

The following is likely all obvious to the experts. But since the field looks tricky to an outsider, maybe I may be excused for asking anyway.

I am wondering about basic facts of what would naturally be called the étale homotopy type of good versions of non-archimedean smooth rigid analytic spaces. I suspect what I am after works best for smooth Berkovich analytic spaces, but my main interest is not in studying one fixed model for analytic spaces, but in knowing those models which do have a good étale homotopy theory. Therefore in the following I'll say "analytic space" as shorthand for "non-archimedean rigid or Berkovich-style or otherwise analytic space, whichever works best".

By the étale homotopy type of such an analytic space I want to mean the hopefully obvious definition directly analogous to the familiar definition in algebraic geometry.

Now, at least in Berkovich's theory there is already a topological space underlying an analytic space by way of the Berkovich analytic spectrum. My first concrete question is:

Is the étale homotopy type of a sufficiently well-behaved analytic space equivalent to that of its underlying topological space?

Berkovich showed that the topological space underlying a smooth Berkovich-analytic space is locally contractible (see here). So the above question has the following sub-question:

Are polydiscs and/or analytic domains étale contractible?

Are analytic spaces locally étale contractible?

(The last one is really the main point that I am after, since it would imply that the hypercomplete $\infty$-topos over analytic spaces is cohesive, by a similar argument as for complex-analytic geometry. This is something I had been wondering about here on MO a good while back but it really boils down to knowing that smooth analytic spaces are locally étale contractible.)

I imagine that eventually this kind of questions should be particularly interesting when combined with a non-archimedean analytification map from some kind of smooth arithmetic schemes. Then one would naturally wonder if there is a non-archimedean analog of theorem 5.2 in

• Daniel Dugger, Daniel Isaksen, Hypercovers in topology, 2005 (K-theory archive)

which says, in particular, that complex analytification sends hypercovers of smooth schemes over $k \hookrightarrow \mathbb{C}$ to hypercovers of topological spaces/simplicial sets. It seems natural to wonder whether this kind of theorem has a non-archimedean analog. What is known?

• One of the major problems: the underlying space of a Berkovich space associated to the projective line (or a Grassmannian or more generally things which have smooth formal models) is contractible. I do not think that the etale homotopy type of the above things should be contractible - therefore I expect the answer to your first concrete question to be no. – Matthias Wendt Jun 11 '14 at 13:11
• In the same direction as Matthias, the underlying topological space of an annulus is contractible too. – Jérôme Poineau Jun 11 '14 at 13:36
• @UrsSchreiber: no, that is not right. An analogue of $\mathbb{A}^1$-homotopy theory has been defined by Joseph Ayoub. It procedes via simplicial sheaves on a Nisnevich site of rigid varieties and then forces the unit disc $\mathbb{B}^1$ to be contractible. As in $\mathbb{A}^1$-homotopy theory in characteristic $p$, there is no étale realization functor because of the existence of non-trivial étale covers of the affine line (as in Jérome's answer). Working away from the residue characteristic allows to define a realization functor as in the Dugger-Isaksen paper you mentioned. – Matthias Wendt Jun 11 '14 at 16:24
• @UrsSchreiber: Before addressing a non-archimedean version of Theorem 5.2, I again raise the question: is there any serious content in the complex-analytic version? It seems on the surface to be nothing more than the observation that the analytification of an etale morphism is a local analytic isomorphism (due to the Zariski-local structure theorem for etale morphisms in EGA IV$_4$, something underlying all work with etale maps). The same works in the non-archimedean setting. So what problem is there in just applying the complex-analytic argument verbatim in the non-archimedean case? – user27920 Jun 11 '14 at 18:30
• @user52824: on the one hand your are right, getting a realization functor on the model category of simplicial sheaves to analytic spaces is nothing more than the observation you mentioned. But getting this realization functor to descend to the $A^1$-local model structure requires the contractibility of $\mathbb{C}$. This is clear, but the non-archimedean analogue (in mixed characteristic) is not true, due to the etale covers mentioned in the comments and Jérome's answer. – Matthias Wendt Jun 11 '14 at 20:40

## 3 Answers

This might help to clarify your first question. The underlying topological space of the Berkovich analytification of $X$ encodes the weight zero part of the cohomology of $X$ - i.e. the singular cohomology of $X^{an}$ is the weight zero cohomology of $X$. See for instance,

V. Berkovich, A non-Archimedean interpretation of the weight zero subspaces of limit mixed Hodge structures

In this paper,

On the comparison theorem for étale cohomology of non-Archimedean analytic spaces Israel J. Math. 92 (1995), 45-60.

Berkovich proves a comparison theorem that says roughly, algebraic and analytic etale cohomology agree under some mild and reasonable hypotheses. I believe this should imply that the analytic etale homotopy type agrees with the algebraic etale homotopy type.

• Thanks! Yes, I was reminded of that in the comments above, by Antoine Chambert-Loir and others. This is certainly good to keep in mind, so thanks for the concrete pointers to references! For later use, I have briefly recorded them here: ncatlab.org/nlab/show/Berkovich+space#Cohomology – Urs Schreiber Jun 12 '14 at 10:17

EDIT: I undestand this topic a lot better after a year of thinking about it. I will also shamelessly shill my new preprint: https://arxiv.org/abs/1708.03657

After a longer time thinking about this I can pretty confidently answer the following.

The exact model of non-archimedean geometry you work with does not matter so much. In the situation where the suitable categories of Berkovich spaces, rigid spaces, and adic spaces are all defined and equivalent, they actually have equivalent étale homotopy types. There are actually subtle distinctions between the étale theory of Berkovich spaces and the other two, for example the étale topos of the Berkovich space is smaller than the étale topos of the corresponding adic space, however they do not matter cohomologically.

Is the étale homotopy type of a sufficiently well-behaved analytic space equivalent to that of its underlying topological space?

No. Even $\mathbb{P}^1$ can fail to be étale simply connected, even though the underlying space is simply connected. It is true in a wide generality that rational étale cohomology agrees with rational singular cohomology of the underlying space. Presumably one can turn this into an equivalence of "rational étale homotopy types", except that the fundamental pro-groups will frequently not be trivial.

Are polydiscs and/or analytic domains étale contractible?

No. The fundamental pro-group(oid) of the shape of an $\infty$-topos classifies local systems on the shape. Local systems on the shape are in a mildly-functorial bijection with locally constant sheaves internal to the topos. Locally constant sheaves are representable, and thus the fundamental pro-group(oid) will agree with the one defined by A.J. de Jong. In particular Jérôme's answer shows that polydisks may already be non-simply connected.

Are analytic spaces locally étale contractible?

In mixed or positive characteristic, I'd expect not. Even in pure characteristic zero, if you're not over an algebraically closed non-archimedean field the answer must still be no since the Galois cohomology of the field will be seen by the analytic space. A good place to start looking might be analytic spaces over the Puiseaux series ring $\mathbb{C}((\cup_n t^{1/n}))$, where it might be possible.

I cannot say that I am familiar with étale homotopy types, but I hope that the following remark is relevant: over a field $k$ of mixed characteristic $(0,p)$, with $p>0$, a closed disc will have non-trivial étale covers of degree $p$. Consider for instance the cover defined by $Y^p = 1+X$ over a closed disc (with coordinate $X$) of center 0 and radius $r<1$ close enough to 1 (so that the cover is not split).

• Thanks for your reply. Let's see, so the étale homotopy type of the disc would be obtained by looking at hypercovers over the disc: covers of the disc itself which are furthermore equipped with covers of their double intersections, and those equipped in turn with covers of their triple intersections -- and so ever on. Such constructions result in a simplicial space and contracting each connected component in each simplicial degree to a point yields a simplicial set, hence a homotopy type. The étale homotopy type is the colimit over that under refinement of hypercovers. – Urs Schreiber Jun 11 '14 at 12:38
• To clarify my above comment: is it clear what the existence of the étale covers that you consider implies for the étale homotopy type of the disc? – Urs Schreiber Jun 12 '14 at 10:03