# finite number of vertices of the polyhedron of variation of an invertible function on a Berkovich curve

The paper of Ducros "Cohomologie non-ramifiée sur une courbe p-adique lisse" mentions a theorem (1.21) about the existence of a "polyhedron of variation" of an invertible function on a Berkovich curve. Here is the statement:

Let $Y$ be an analytic curve over a non-Archimedean non-trivially valued field $k$, and let $f$ be an invertible analytic function on $Y$. Then there exists a polyhedron $P$ in $Y^{an}$ such that $|f|$ is constant on each connected component of the complement $Y \setminus P$.

(here, "polyhedron" is 1-dimensional and is understood as a realization of a graph as a topological space)

My question is: is it true that $P$ has finitely many vertices, i.e. points that do not have a neighbourhood homeomorphic to an interval $(a,b) \subset \mathbb{R}$? If this is not to be expected in general, is it true if one assumes $Y$ to be a subspace of a smooth projective curve $X$ over $k$ and $f$ to be the restriction of a rational function on $X$ to $Y$?

• it is worth making a remark that it follows from the proof in Ducros' paper that the number of vertices is finite if $Y$ is compact, so my question is mainly about non-compact case – Dima Sustretov Nov 24 '17 at 17:05
• "compact" in the comment above refers to the analytification $Y^{an}$ of $Y$ – Dima Sustretov Nov 24 '17 at 17:29

## 1 Answer

It is not true in general that $P$ is finite. To see this, take an open disk $D$ and a non-zero function $f$ on it with infinitely many zeroes. In this case, your polyhedron of variation is infinite: it is a tree where all those zeroes are leaves. If you insist on having an invertible function, just remove the zero locus of $f$ from $D$.

With the additional assumptions you make, the result is true. Let us assume that $Y$ is the analytification of a smooth projective curve and $f$ is a non-constant rational function on it. Remark that we may extend the scalars so as to assume that $k$ is algebraically closed. (The image of a finite polyhedron will still be a finite polyhedron.) The function $f$ has only finitely many zeroes and poles. Each such point $x$ has a neighborhood $D_x$ in $Y$ that is isomorphic to an open disk (because $Y$ is smooth and $k$ algebraically closed). Removing all the $D_x$'s from $Y$, you find a compact curve to which Ducros's result applies. It remains to show that the polyhedron of variation in each $D_x$ is finite. Since $D_x$ is a disk, this is quite clear. (In fact, if you assume that $D_x$ contains no zero or pole other than $x$ itself, then this polyhedron is just going to be a segment joining $x$ to the boundary of $D_x$.)

Alternatively, you could also prove the result directly using the semistable reduction theorem in the form of the existence of a skeleton of $Y$: there exists a finite polyhedron $P_Y$ in $Y$ whose complement is a disjoint union of open disks (if $k$ is algebraically closed). And then you are back to the case of an open disk again. Note also that $|f|$ is constant on a disk where $f$ has neither zeroes nor poles.