Let $X$ be a rigid analytic space over a non-Archimedean field $k$. If $U_1,\ldots,U_n\subseteq X$ are affinoid opens, then it's usually not clear whether or not the admissible open $U=U_1\cup\cdots\cup U_n$ is affinoid. But we have (due to the sheaf axioms, and the fact that the $U_i$ constitute an admissible covering of $U$) the canonical injection (arising from the restrictions $\mathscr{O}_X(U)\to\mathscr{O}_X(U_i)$)

$$\mathscr{O}_X(U)\hookrightarrow\prod_{i=1}^n\mathscr{O}_X(U_i)\quad (*)$$

The target is a product of affinoid $k$-algebras, and in particular is a Banach algebra over $k$. One can endow $\mathscr{O}_X(U)$ (or $\mathscr{O}_X(V)$ for any admissible open $V$ of $X$) with a locally convex topology via the $k$-algebra isomorphism (again due to the sheaf axioms) $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$, where $W$ runs over the affinoid opens of $X$ contained in $U$, by pulling back the projective limit topology on the target (using the canonical $k$-Banach topologies on each $\mathscr{O}_X(W)$). For this topology, the map (*) is certainly continuous. My question is as follows:

Is the map (*) a topological embedding?

This would follow if the sheaf of rings of a rigid space were in fact a sheaf of topological rings, but I don't think this is required in the definition of a $G$-ringed space in e.g. BGR like it is for adic spaces and formal schemes. I feel like this is the case for $X$ affinoid though...at least, I think this is true if it is the case that $k$-algebra maps between $k$-affinoid algebras, which are always continuous, are in fact strict (I don't know if this is true, though it's true for maps of finite modules over Noetherian $k$-Banach algebras, which affinoid algebras are by definition, but the base Tate algebra depends on the given affinoid algebra).

A positive answer to my question would imply that the locally convex topology of $\mathscr{O}_X(U)$, while possibly not that of an affinoid $k$-algebra, is at least defined by a norm (the one got by restricting the max norm for a choice of norms on each $\mathscr{O}_X(U_i)$ along (*)). This consequence is what I really want. (I guess if we really have a sheaf of topological rings, then the map would be a closed topological embedding, since it's the first arrow in an equalizer sequence of Hausdorff topological rings, but I don't actually need this.)

EDIT: As user grghxy points out, I should assume that $X$ is quasi-separated to ensure that my set $U$ is in fact admissible open with admissible covering given by the $U_i$.

  • 2
    $\begingroup$ Why is $U$ admissible (if using rigid-analytic spaces, not adic spaces)? Assume $X$ is quasi-separated, so (exercise!) $U$ is admissible with $\{U_i\}$ an admissible covering, so $U$ is qcqs. Hence, in such cases $O(U)$ is the equalizer of the natural map $\prod O(U_i) \rightarrow \prod O(V_{ijh})$ where $\{V_{ijh}\}$ is a finite (admissible) affinoid open cover of the quasi-compact $U_i \cap U_j$. You can check via the Banach open mapping theorem that the natural Banach space structure on the equalizer (a closed subspace of $\prod O(U_i)$) is the intrinsic topology you are considering. $\endgroup$
    – grghxy
    Jun 28, 2015 at 18:54
  • $\begingroup$ Dear @grghxy, Thanks for the response and for pointing out my oversight. I was thinking of affinoid spaces. $\endgroup$ Jun 28, 2015 at 19:28
  • $\begingroup$ Dear @grghxy, Perhaps I'm missing the point, and this is overkill, but the most general version of the open mapping theorem I know of has the source an LF-space (locally convex inductive limit of a sequence of Fréchet spaces), but I don't see why $\mathscr{O}(U)$ has this structure without further hypotheses. I'm assuming you're suggesting applying the open mapping theorem to the continuous bijection $\mathscr{O}(U)\to E$, where $E$ is the equalizer equipped with its Banach structure and deducing that this map is open. $\endgroup$ Jun 28, 2015 at 23:37
  • $\begingroup$ We define the topology on $O(U)$ to be that transported from the equalizer, and directly prove that this is independent of all choices. One has no need for any kind of open mapping theorem beyond the setting of Banach spaces. $\endgroup$
    – grghxy
    Jun 29, 2015 at 0:07
  • $\begingroup$ Okay, I expected the basic version was adequate. I think I understand. Thank you. $\endgroup$ Jun 29, 2015 at 0:09

1 Answer 1


So that this question doesn't remain unanswered, I will provide an elaboration of grghxy's answer in the comments. The quasi-separatedness of $X$ ensures that $U$ is admissible open with $U=\bigcup_{i=1}^n U_i$ an admissible covering (and in fact that any finite covering of $U$ by affinoid opens of $X$ is admissible). In my original formulation of the question I (again thinking about affinoid spaces) implicitly assumed each $U_i\cap U_j$ was affinoid, which would follow from separatedness of $X$, but is not needed. Quasi-separatedness ensures that each intersection $U_i\cap U_j$ is a finite union of admissible affinoid opens $V_{ijk}$, and we get an exact sequence


where the second and third terms are products of finitely many affinoid $k$-algebras, hence are themselves naturally Banach algebras. Since the second arrow is continuous (it comes from maps between $k$-affinoid algebras in the factors, which are always continuous), the kernel of this map is closed, so by transport of structure we get a Banach topology on $\mathscr{O}_X(U)$ for which the injection $\mathscr{O}_X(U)\to\prod_{i=1}^n\mathscr{O}_X(U_i)$ is a closed topological embedding. If we use another finite covering of $X$ by affinoid opens (necessarily admissible), then the same reasoning shows that we get another Banach topology on $\mathscr{O}_X(U)$ as a closed subspace of the product of the affinoid algebras of the members of the covering. By taking a common refinement if necessary, in comparing the Banach topologies, we may assume one covering refines the other. Then the Banach topologies are comparable, and so coincide by the open mapping theorem. So there is an canonical Banach space topology on $\mathscr{O}_X(U)$ which arises in the manner above from any finite covering by affinoid opens.

We can go slightly further using the above reasoning. If $W$ is any affinoid open contained in $U$, then the (redundant) covering $U=U_1\cup\cdots\cup U_n\cup W$ is admissible by quasi-separatedness, so, by the independence of the canonical Banach topology of the choice of finite affinoid covering, the map $\mathscr{O}_X(U)\to\mathscr{O}_X(W)\times\prod_{i=1}^n\mathscr{O}_X(U_i)$ is a closed topological embedding, and in particular, the restriction map $\mathscr{O}_X(U)\to\mathscr{O}_X(W)$ is continuous for the Banach topologies on source and target.

It follows that the $k$-linear bijection $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$ is continuous for the Banach topology on the source and the projective limit of Banach topologies on the target. Using the covering $U=\bigcup_{i=1}^n U_i$ to describe the Banach topology on $\mathscr{O}_X(U)$, a basic open set has the form $\{F\in\mathscr{O}_X(U):F\vert_{U_i}\in A_i,1\leq i\leq n\}$ for some open subsets $A_i\subseteq\mathscr{O}_X(U_i)$. But since the $U_i$ are among the $W$ in the projective limit, this is identified with a basic open subset of the projective limit, namely $\{(F_W)_W:F_{U_i}\in A_i,1\leq i\leq n\}$. The map $\mathscr{O}_X(U)\to\varprojlim_W\mathscr{O}_X(W)$ is therefore open, hence a topological isomorphism, so the Banach topology on $\mathscr{O}_X(U)$ coincides with its intrinsic topology as a projective limit. In particular, the answer to my question is ``yes."


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