# Why do rigid spaces have “not enough points”?

In Brian Conrad's notes here for the 2007 Arizona winter school, bottom of p18, he says that there is an affinoid rigid-analytic space and a sheaf of abelian groups on it equipped with a non-zero section such that all stalks vanish (at all the "usual" points corresponding to maximal ideals in the affinoid algebra). He uses this to motivate Berkovich spaces etc, and explains why the existence of such a section does not contradict anything (the resulting open cover on which the section vanishes might not be an admissible cover) but does not give an explicit example of such an affinoid/sheaf/section.

What is an explicit example?

## 1 Answer

If you are familiar with Berkovich spaces, you can do the following construction. Let $X$ be an affinoid space of positive dimension and pick a point $x$ in $X$ that is not a rigid point. Consider the inclusion map $i\colon x \to X$. Then the sheaf $F = i_*\mathbb{Z}$ does the job. Since the space $X$ is Hausdorff, the point $x$ is closed and the sheaf $F$ has no section on the open set $X\setminus \{x\}$.

Let me try to rewrite this example purely in terms of rigid geometry in a simple case: $X$ is the closed unit disc over an algebraically closed field $k$ and $x$ is the point at its boundary (the Gauss point). The previous sheaf may then be described as follows: for an affinoid domain $V$ of $X$, we have $F(V)=0$ if every connected component of $V$ is contained in an open unit disc (with any center) and $F(V)=\mathbb{Z}$ otherwise.

• Two more useful examples: 1) A descending sequence of discs with empty intersection gives another such skyscraper sheaf (which also comes from a Berkovich point). 2) There are (non-"overconvergent") examples that don't come from Berkovich points, but do come from points in Huber's adic space. For example, let $F(V)=\mathbb{Z}$ if $V$ contains the open unit disc with finitely many open discs of radius $<1$ removed, and $F(V)=0$ otherwise. One way to think of this second example is: the skyscraper sheaf at the origin of the canonical reduction, $\mathbb{A}^1_{\tilde{k}}$. – Andrew Dudzik May 20 '13 at 23:11
• You are right, thanks for the comment Andrew. And since we know we get every point of the topos from the points of the Huber's space, we will not get any other example for the disc. – Jérôme Poineau May 21 '13 at 5:25
• Many thanks for this Jerome and Andrew (and sorry no accents Jerome). So Tate had the right topos, but in some sense not the right site. Jerome -- are you saying that you can recover the adic space (the "Huber's space") from the topos via some "points" construction? Now I understand how to think about the question, the solution is easy. For example if $r$ is a real not in $\sqrt{|k^\times|}$ then we can try to build a sheaf supported on ${|z|=r}$ by setting $F(U)=\mathbb{Z}$ iff $U$ contains an annulus $\{a<|z|<b\}$ with $a<r<b$ and $a,b\in\sqrt{|k^\times|}$. – user34143 May 22 '13 at 10:12
• @ampit6: Yes, there is a notion of point of a topos: it is a morphism from the punctual topos to the topos (see SGA4, IV.6). And if the topos is the topos associated to a topological space, yes you can recover such a space from the topos (see SGA4, IV.7.1). Moreover, if the topological space is sober (every irreducible subspace has exactly one generic point), then it is unique (see SGA4, IV.4.2). And Huber's construction provides such a sober space associated to the rigid topos (see his paper "Continuous valuations" Math. Z. 212 (1993)). – Jérôme Poineau May 22 '13 at 20:54
• Let me also add a warning: your example is probably the simplest, but it will not work if $\sqrt{|k^\times|}=\mathbb{R}_{>0}$. – Jérôme Poineau May 22 '13 at 20:57