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I am new to rigid analytic spaces (over non-archimedean fields) and I am confused about the notions of closed and open immersions. My question is are these two notions are "complement" of each other like the case of schemes? I know that this is the case at least when we have a closed immersion $Z \hookrightarrow X$ then we have an open complement defined set-theoretically by $U = X \setminus Z$ and we can endow $U$ with a rigid structure so that the inclusion $U \hookrightarrow X$ is an open immersion (= an admissible open). For instance, the standard example that I keep in mind for this case is the puncture disk inside the Tate unit ball (which is the complement of the origin), which is a union of annuli with increasing radii. However, I am doubtful about the reverse direction: given an admissible open $U \subset X$, can we endow $Z = X \setminus U$ with a rigid structure so that the inclusion $Z \hookrightarrow X$ is a closed immersion of rigid spaces?

Update: In case $Z$ cannot be given a structure such that $Z \hookrightarrow X$ is a closed immersion. Is $Z$ still a rigid space? For example, this is a case that I am particularly interested in: let $R = k[[t]]$, $K = k((t))$ with $k$ a field, let $X$ be a $R$-variety, then there is an open immersion $(\hat{X})_{\eta} \hookrightarrow (X_K)^{\text{an}}$ from the Raynaud's generic fiber of the $t$-adic completion $\hat{X}$ into the analytification of $X_K$? Is $(X_K)^{\text{an}} \setminus (\hat{X})_{\eta}$ a rigid space? If so, can we write down a concrete admissible covering?

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  • $\begingroup$ No, you cannot, and this is indicative of an important reality: rigid geometry acts quite differently from algebraic geometry. Let me interpret your question in turns of adic geometry, as this is more clarifying in my opinion. If $X=\mathbb{D}^1_{\mathbb{C}_p}$ is the closed unit disk over $\mathbb{C}_p$. There is a classification of the points of $X$ into five types. This is recorded quite clearly in Example 2.20 of Scholze's paper Perfectoid Spaces. I will assume you will look there for the meaning of the following terms. Let $x$ be a point of Type 5 in $X$. Then, $x$ is closed. $\endgroup$ Commented Apr 24, 2024 at 18:54
  • $\begingroup$ That said, there is no adic space $Z$ and morphism $f\colon Z\to X$ whose image is $\{x\}$. The reason is that $Z$ is necessarily an analytic adic space (as it would admit a map to $X$ which is analytic) and then we use the following fact which greatly separates algebraic and analytic geometry: any map of analytic adic spaces is generalizing (e.g., see Lemma 1.1.10 of Huber's book on etale cohomology). In particular, what this tells us is that if $f(z)=x$ then $f(G(z))=G(x)$, where $G(-)$ denotes the set of generalizations of the point in the relevant space. $\endgroup$ Commented Apr 24, 2024 at 18:54
  • $\begingroup$ But, we run into obvious trouble. Namely, $G(x)$ contains a unique point of Type 2, but this cannot possible be in $f(G(z))\subseteq f(Z)=\{x\}$. Of course, this is not an issue for some closed subsets (e.g., Zariski closed ones), but this issue is also why Huber defines the notion of a 'pseudo-adic space'. This essentially is an adic space $X$ and a (topologically nice) subset $S$ such that the 'geometry of $(X,S)$ approximates the geometry of $S$'. $\endgroup$ Commented Apr 24, 2024 at 18:57
  • $\begingroup$ If $Z\subseteq X$ is generalizing, then you can at least endow it with some geometric structure: that of a small v-sheaf (a la Scholze). Namely, you can consider $X\times_{|X|}|Z|$, where for a space $T$ we take $|T|(S):=\mathrm{Hom}_\mathrm{cont.}(|S|,|T|)$ for a perfectoid space $S$ over $\mathbb{F}_p$. $\endgroup$ Commented Apr 24, 2024 at 19:16
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    $\begingroup$ It's late here, and so I will double-check this tomorrow, and post it as an answer. $\endgroup$ Commented Apr 24, 2024 at 19:48

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No, it's not true that if $X$ is a rigid $K$-space and $U\subseteq X$ is open, that $Z:=X-U$ can be given the structure of an adic space (let alone a rigid $K$-space) such that the tautological map $i\colon Z\to X$ of topological spaces upgrades to a morphism of adic spaces.

The issue is that morphisms of analytic adic spaces (note any adic space over $X$ is analytic as $X$ is) are generalizing. Namely, suppose that $f\colon Y\to X$ is a morphism of analytic adic spaces. Then, [Huber, Lemma 1.1.10] shows that for any $y$ in $Y$ with $x=f(y)$ the equality $f(G_Y(y))=G_X(x)$ holds. Here $G_S(s)$ denotes the set of generalizations of the point $s$ in the space $S$.

So, if $i\colon Z\to X$ could be upgraded to a map of adic spaces, then $Z$ would be need to be generalizing: if $z\in Z$ then we'd have that $Z\supseteq i(G_Z(z))=G_X(z)$. Equivalently, we would have to have that $U\subseteq X$ is a so-called overconvergent (aka partially proper, aka Berkovich) open subset. This just means that $U$ is closed under specializations. Note that this is quite restrictive. For instance, if $U\to X$ is quasi-compact then $U$ is actually clopen (e.g., apply [Fujiwara--Kato, Chapter 0, Corollary 2.2.27]).

So, let's consider a concrete example conforming to your stated interests. Let $\mathcal{X}=\mathbb{A}^1_{k[[t]]}$, and let $\widehat{\mathcal{X}}$ be its $t$-adic completion. Then, as per usual (e.g., see [Huber, §1.9]) we have an open embedding $\widehat{\mathcal{X}}_\eta\hookrightarrow \mathcal{X}_\eta^\mathrm{an}$. In this case this concretely corresponds to the open embedding $\mathbb{D}^1_{k((t))}\hookrightarrow\mathbb{A}^{1,\mathrm{an}}_{k((t))}$. The complement of this open embedding does not have the structure of an adic space as $\mathbb{D}^1_{k((t))}$ is not an overconvergent open in $\mathbb{A}^{1,\mathrm{an}}_{k((t))}$. One no-thinking way to see this based off what I said above: this inclusion is quasi-compact but the image is not closed. More concretely, this open subset is not closed under specialization since its image is missing the 'outward facing Type 5 point' around the Gauss point (I am using terminology from [Scholze, Example 2.20]).

So, what to do? There are two options:

(1) one can try to utilize Huber's theory of pseudo-adic spaces (see [Huber]) which essentially tries to develop the geometry of (certain) subsets of an adic space even when they do not themselves possess the structure of an adic space,

(2) when $Z\subseteq X$ is generalizing, there is always a diamond $Z^\lozenge$ (in the sense of [Scholze2]) and a map $Z^\lozenge\to X^\lozenge$ recovers the map $Z\to X$ on topological spaces. Namely, one takes $Z^\lozenge:=X\times_{|X|}|Z|$, with notation as in [AGLR, §2.1].

EDIT: Given my impression of your interests, it may be worth noting that the (etale topos of the) pseudo-adic space of $Z$ does form the closed complement of the open subtopos $U_\mathrm{et}\subseteq X_\mathrm{et}$ (e.g., see [Huber, Lemma 2.3.11]). I would guess this is enough for your purposes.

References:

[Fujiwara--Kato] Fujiwara, K. and Kato, F., 2018. Foundations of rigid geometry I.

[Huber] Huber, R., 2013. Étale cohomology of rigid analytic varieties and adic spaces (Vol. 30). Springer.

[Scholze] Scholze, P., 2012. Perfectoid spaces. Publications mathématiques de l'IHÉS, 116(1), pp.245-313.

[Scholze2] Scholze, P., 2017. Étale cohomology of diamonds. arXiv preprint arXiv:1709.07343.

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    $\begingroup$ Hi Alex, thanks for the answer. Btw, we met in Hanoi few months ago, I was the guy saying that your blog is useful for me in the first day. I still wait for the second part of your blog post on rigid geometry. $\endgroup$ Commented Apr 28, 2024 at 11:51
  • $\begingroup$ @AlexeyDo ah of course! It’s a little hard to find time these days. But I do hope to do it at some point. :) $\endgroup$ Commented Apr 28, 2024 at 12:03
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    $\begingroup$ @AlexeyDo Everything changes but only superficially. Like, it's analogous to saying that for a finite type K-algebra R, that two subsets $S,T\subseteq \mathrm{Spec}(R)$ can have the property that $S\cap \mathrm{MaxSpec}(R)=T\cap\mathrm{MaxSpec}(R)$ while also $S\ne T$. Like in this MaxSpec world you are missing points, and so set-theoretic equality can be misleading (this is an issue of constructability of $S,T$ , but this is also what's happening secretly in this rigid world). In particular, the complement in the Berkovich world misses a higher rank point in the complement $\endgroup$ Commented Jun 6, 2024 at 2:29
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    $\begingroup$ in the adic world. So, you can work in this Berkovich world and if you have something like an open/closed decomposition of topoi, things are fine, but you have to work a priori harder because this missing point contains information. But, in the end, there is no actual contradiction because this open/closed decomposition for the pseudo-adic space $Z$ should roughly capture the same information as the open/closed decomposition of topoi in this Berkovich world. $\endgroup$ Commented Jun 6, 2024 at 2:31
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    $\begingroup$ So in short: they are different points of view, but some care must be taken in the change of perspectives, similar to 'classical varieties' vs. 'schemes' change of view. $\endgroup$ Commented Jun 6, 2024 at 2:31

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