# 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?

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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
As is the underlying topological space of an elliptic curve with good reduction. In the case of degenerating families over a disk, a theorem of Berkovich states that the topology of Berkovich spaces (only) explains for the weight-0 part of the limit Hodge structure. – ACL Jun 11 '14 at 14:00
@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

## 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.

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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

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).

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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

This is something I've been thinking about for some time, I think I can at least say a little bit. There is a general theory of "[BLANK] homotopy type" for topoi and/or sites, which is provided by Lurie's shape functor from HTT on the level of $\infty$-topoi, but goes back to a paper by Toën-Vaquié on Segal Topoi.

There is a general result which --as far as I know-- was first published at about the same time by Hoyois in Higher Galois Theory and Barnea-Harpaz-Horel in Pro-Categories in Homotopy Theory which states that for reasonable sites that the shape of the hypercomplete $\infty$-topos is co-represented by the diagram of hypercovers of the terminal object of the site. This has the property that the category of local systems on the shape is categorically equivalent to the category of locally constant abelian sheaves on the site, and under the equivalence the singular cohomology agrees with the sheaf cohomology. It also works in such a way that the $p$-cover Jérôme constructs becomes a surjection $\pi_1^{\acute{e}t}(B(0;1-\epsilon)) \rightarrow \mathbb{Z}/p$, essentially witnessing that the ball is not contractible.

One can then make the definition of the ("integral") etale homotopy type of a Berkovich space $X$ as the shape of the associated hypercomplete etale $\infty$-topos. Since --again, as far as I know-- the theory for integral etale cohomology of Berkovich spaces is relatively unexplored, we can also do a profinite completion operation. Call these two pro-objects by $\operatorname{b\acute{e}t}(X)$ and $\widehat{\operatorname{b\acute{e}t}}(X)$. We can also complete at a prime $l$ away from the characteristic of the residue field and/or the fraction field, which we'll denote by $\widehat{\operatorname{b\acute{e}t}_l}(X)$

What is true is that if $Y$ is a projective variety over the non-archimedean field $K$, then one has a weak equivalence

$\widehat{\operatorname{\acute{e}t}}(Y) \sim \widehat{\operatorname{b\acute{e}t}}(Y^{an})$

From the classical etale homotopy type of $Y$ to the Berkovich etale homotopy type of the analytification $Y^{an}$. There is a general result that etale cohomology with constructible coefficients agrees between the two. From there we need to compare the fundamental groups. This is essentially due to GAGA: locally constant sheaves of finite groups are representable in Berkovich spaces. Consider this as a commutative algebra object in the coherent module category, then use GAGA to algebraify such a thing.

For non-proper algebraic varieties in the mixed characteristic case I am unsure what to expect to be true on the $p$-part of the homotopy types. I certainly expect the pro-$l$ homotopy types to agree. The result in the second paper linked by Jeffrey shows that the abelian cohomologies agree even if the groups are $p$ groups. However, there is an example from de Jong of an infinite abelian $p$-primary Galois cover by the $p$-adic logarithm for $\mathbb{A}^1_{\mathbb{C}_p}$. I do not understand this example very well, and it may be that this Galois group 'disappears' when one takes pro-finite completion. I have a very crude argument for why the profinite etale homotopy type of an algebraic curve should agree with the profinite Berkovich etale homotopy type of the analytification, but I haven't really sat down and hammered it out. That relies on the Riemann existence theorem to reduce to the compact situation alowing ramification at the added points, and the fact that one can extend birational maps for curves. Neither of which are true in higher dimension.

Now as to your direct questions:

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

I'd expect only in the case both are contractible, given that for smooth projective algebraic Berkovich spaces of dimension $n$ they should have cohomology in degree $2n$. I believe in general smooth proper things are contractible in the Hausdorff topology. Additionally, the map $X \rightarrow \operatorname{sp} K$ should induce a fibration over $B\operatorname{Gal}(K)$ which is split by rational points adding an arithmetic component to the picture.

Are polydiscs and/or analytic domains étale contractible?

Not if you include the $p$-part for polydiscs, and possibly if you work away from $p$. As far as analytic domains we can again compare in the case where they are algebraic, in which they can have non-trivial cohomology.

Are analytic spaces locally étale contractible?

Given the above example it seems unlikely. Perhaps a smaller disk could be taken on which that cover splits, but I'd presume one can then just take a different cover.

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