Let $k$ be a field complete with respect to a non-archimedean absolute value, and $X$ a separated algebraic space of finite type over $k$.

If $X$ is a scheme then $X(k)$ inherits a natural (Hausdorff) topology from $k$ (uniquely determined by functoriality, compatibility with open immersion, closed immersions, fiber products, and the special case of the affine line), locally compact when $k$ is locally compact. There is even a natural $k$-analytic manifold structure if $X$ is smooth. This can be seen in (at least) two ways: algebraically by using Zariski-open covers by affines, and for smooth case take them to be "standard etale" over affine spaces (and then the $k$-analytic inverse function theorem can be applied), or analytically by identifying $X(k)$ with $X^{\rm{an}}(k)$ for the analytification $X^{\rm{an}}$ in the sense of rigid-analytic spaces (so then affinoid open covering of $X^{\rm{an}}$ does the job).

Now suppose $X$ is not a scheme (as often happens with moduli spaces which only exist as algebraic spaces; e.g., Rapoport's thesis, etc.). The 2nd method above can be used, but constructing $X^{\rm{an}}$ as a rigid-analytic space is really hard (the only method I know is to take a very long detour through Berkovich spaces to make the required rigid-analytic quotients from an etale scheme chart for $X$).

My question is this: is there a known simple procedure (much as the method in the scheme case is simple), bypassing the use of $X^{\rm{an}}$, to make a functorial Hausdorff topology on $X(k)$ (and $k$-analytic manifold structure when $X$ is smooth) which is locally compact when $k$ is and recovers the usual topology when $X$ is a scheme and shares the same basic properties (good behavior for open immersions, closed immersions, and fiber products; carries etale maps to local homeomorphisms as well as local $k$-analytic isomorphisms in case $X$ is smooth; and $X(k) \rightarrow X(k')$ is a closed embedding any extension $k'/k$ of such fields)?

A natural idea for the topology aspect is to choose an etale scheme cover $U \rightarrow X$ and identify $X(k_s)$ with a quotient of $U(k_s)$ as a set. Give it the quotient topology of the natural topology on $U(k_s)$ arising from the topology on $k_s$, and give $X(k)$ the subspace topology from $X(k_s)$. This does give the right answer (same as via the rigid-analytic method), but the only way I see this is locally compact when $k$ is (noting that $X(k_s)$ is essentially never locally compact) and functorial is to invoke comparison with the rigid-analytic method. (Otherwise I get bogged down in rising chains of finite Galois extensions and it feels like it becomes a mess. Doing the $k$-analytic manifold structure for smooth $X$ in this way also seems to become quite unpleasant. Note, by the way, that $k_s$ is not complete, so can't speak of $k_s$-analytic manifolds.) So this idea doesn't seem to provide an answer. Or maybe I am missing a simple trick?

I am happy to restrict to the case of locally compact $k$ (though allowing completed algebraic closures is perfectly interesting), but do not want to restrict to characteristic 0 (though ideas in that case are appreciated too).

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    $\begingroup$ I started reading this question and said to myself, "Oh, BCnrd will answer this." Then I got to the end and said "Oh, dear." $\endgroup$
    – JSE
    Aug 5, 2010 at 16:07
  • $\begingroup$ But as for actual content; do you have to go all the way up to k_s? Each point in U(k_s) is defined over an extension k' of bounded degree; I guess I have some vague idea that it would help to work over the union of all the U(k'), which together cover X(k)? But maybe this is what you already did that led you to the increasing towers. $\endgroup$
    – JSE
    Aug 5, 2010 at 16:16
  • $\begingroup$ Dear JSE: Yes, your "vague idea" (as you call it) is what led to unpleasantly increasing towers. Maybe it can be made to work, but I don't see it. $\endgroup$
    – BCnrd
    Aug 5, 2010 at 16:56
  • $\begingroup$ Pursuing that idea, it appears to me that in the characteristic zero case there is a single finite extension $k'$ of $k$ such that every point of $U(k_s)$ which maps to a point of $X(k)$ is defined over $k'$. Because $k$ has only finitely many Galois extensions with a given Galois group. Is that right? $\endgroup$ Aug 8, 2010 at 5:30
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    $\begingroup$ Tom, I'm pessimistic that this finite group case (even if it works) could be adapted to handle the local compactness aspect of more general equivalence relations, but would be happy to be proved wrong. Anyway, I agree that (in char. 0) the space of interest is a union of images of finitely many closed subsets of $U(k_s^H)$. Then to control local compactness of an image, does it help to make a direct proof that the quotient topological map (hmm, which one?) is an open mapping (using openness of maps on rational points induced by \'etale maps between schemes of finite type)? $\endgroup$
    – BCnrd
    Aug 8, 2010 at 19:04

1 Answer 1


Assume $k$ is a field and $F$ is a contravariant functor from $k$-schemes of finite type to sets. For each $X/k$ and $x\in F(X)$ we get a "characteristic map" $X(k)\to F(k)$ by pulling back $x$. Now if $k$ is a topological field, we can define a topology on $F(k)$, which is the finest making all these maps (for all pairs $(X,x)$) continuous. Some facts are easy to check, for instance:

  • if $F$ is representable we get the "usual" topology on $F(k)$,
  • every morphism of functors $F\to G$ gives rise to a continuous map $F(k)\to G(k)$,
  • restricting $X$ to affine $k$-schemes does not change anything.

With this generality, of course, one doesn't get very far without making some assumptions on $F$ and/or $k$. For instance, compatibility with products is not a formal consequence of the definition and therefore, probably, does not hold in general (example, please?). The nasty thing about this topology is that it is a quotient topology ($F(k)$ is a quotient space of the disjoint sum of all $X(k)$ indexed by pairs $(X,x)$) and general quotient topologies don't behave nicely.

Assume now that $k$ satisfies the following "weak henselian" property:

(WH) Every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$.

This is satisfied by local fields and more generally by henselian valued fields. It is of course particularly relevant if $F$ is an algebraic space of finite type over $k$. In this case, it is easy to check that if $U\to F$ is an étale presentation, then, assuming (WH), $U(k)\to F(k)$ is an open map. This (together with the existence of pointed étale neighborhoods) is enough to imply, for instance, that the topology on $F(k)$ is compatible with products of algebraic spaces. (The point is that we now have a quotient by an open equivalence relation, and these quotients do behave nicely). I have not checked the other conditions in detail but I am confident that it works.

Some remarks:

  • I have used this trick in my paper "Problèmes de Skolem sur les champs algébriques" (Compositio 125 (2001), 1-30) to define open subcategories of $X(k)$ when $X$ is an algebraic stack of finite type over a local field $k$.
  • I am currently working with P. Gille on the case where $F(X)=H^1_{fppf}(X, G)$, where $G$ is an affine $k$-group scheme of finite type.
  • Condition (WH) seems to become even nicer when, in addition, every finite morphism gives rise to a closed map. This holds for local fields, and more generally for any field with a (rank 1?) valuation, which is algebraically closed in its completion, such as the quotient field of an excellent henselian DVR.
  • $\begingroup$ Thanks. I suppose "pointed etale neighborhoods" means every $x \in X(k)$ admits an etale affine neighborhood with a $k$-point over $X$. I thought long before that this fact wouldn't avoid the interference of rising towers of Galois extensions. Not sure why I thought that. I'll check the rest...except I wonder about the property that $X(k)$ is closed in $X(k')$ for an ext'n of complete fields $k'/k$; clear via the rigid-analytic method, less so otherwise. Also, seems sep'td needed for $U(k) \rightarrow X(k)$ to be "loc. injective" for affine etale chart $U \rightarrow X$, which is useful. $\endgroup$
    – BCnrd
    Aug 9, 2010 at 15:11
  • $\begingroup$ Dear Laurent: OK, I have checked this and it all works out nicely (assuming separatedness of $X$, which I don't know if you need for your more general functor stuff). The only "loose end" is the issue of behavior with respect to extension $k'/k$ of complete fields, though this can be seen via the rigid-analytic method so I am fine with that (though would be neat if it could be extracted by the kind of approach you have suggested). $\endgroup$
    – BCnrd
    Aug 9, 2010 at 21:05
  • $\begingroup$ sorry for digging out this ancient thread, but I would very glad if you could loose some words on explanation in which way your condition (WH) can be understood as "weak" henselian property. my main reference on characterizations of equivlent formulations of "classical" henselian property is Milne's book "Etale Cohomology" Thm 4.2 (on page 32). how are these related to (WH) and why we understand (WH) as a "weaker" form of henselian property? $\endgroup$
    – user267839
    Nov 12, 2019 at 18:25
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    $\begingroup$ @KarlPeter: Every henselian valued field satisfies (WH): see Proposition 3.1.4 in content.algebraicgeometry.nl/2014-5/2014-5-025.pdf. You can also find a summary of this kind of condition in math.u-psud.fr/~cesnavicius/topology-torsors.pdf. $\endgroup$ Nov 13, 2019 at 8:42

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