Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm reading Berkovich's book on analytic spaces. The notion of relative interior confuses me. Is there anyway to see how it "looks like"? For instance, if $r <1$, what is the relative interior of \begin{equation} M ( \mathbb{C}_p \{r^{-1} T \} ) \to M ( \mathbb{C}_p \{ T \} ) \end{equation} and how one should view it on the pictorial description of $M ( \mathbb{C}_p \{ T \} )$, i.e. the famous picture which looks like an infinite tree.

share|improve this question
add comment

1 Answer

Let me begin with the absolute interior. Let me fix first a complete non-archimedean field $k$ (take $k=\mathbb{C}_p$ if you like). For a closed disk $D^+(0;r_1,\dots,r_n)$ (center 0, polyradius $(r_1,\dots,r_n)$), I will call ``naive interior'' the open disk $D^-(0;r_1,\dots,r_n)$.

Now, let $Y=\mathcal{M}(B)$ be a $k$-affinoid space. By definition, you may realize it as a Zariski-closed subset of some closed disk $D^+$. The first idea one could have is to define the interior of $Y$ as the set of points that belong to the naive interior of $D^+$. Actually, this depends on the chosen presentation. So you are led to define the absolute interior Int$(Y/k)$ of $Y$ as the set of points that belong to the naive interior of a closed disk $D^+$ for some presentation of $Y$ as as Zariski-closed subset of $D^+$.

Let us consider a simple example where $k=\mathbb{C}_p$ and $B =\mathbb{C}_p\{T\}$. Topologically, this disk is a tree. The root (i.e. the Gauss point), I will denote $\eta_1$. If you write $Y=D^+(0;1)$, you see that $D^-(0;1)$ belong to the interior of $Y$. This is an open branch out of the point $\eta_1$. Changing coordinate gives you another presentation and another open disk $D^-(a;1)$ (i.e. another open branch) in the interior. This way, you show that $Y\setminus\{\eta_1\}$ belongs to the interior of $Y$. It is actually equal to it.

As regards the relative interior, you just have to replace $k$ by a $k$-affinoid algebra $A$, $B$ by an $A$-affinoid algebra and disks by relative disks over $A$. In your situation, by the same kind of arguments, you will find that the relative interior is the complement of the point $\eta_r$. Actually, Berkovich shows more generally that if $Y \to X$ is the inclusion of an affinoid domain, then Int$(Y/X)$ coincides with the topological interior of $Y$in $X$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.