# Topological and algebraic covering spaces in Berkovich geometry

Let $$k$$ be a complete, non-archimedean field, and $$X$$ a Berkovich space over $$k$$ (as nice as you like, for arguments sake let's say strictly $$k$$-analytic, good, and geometrically connected). As discussed in this article of de Jong, covering spaces of $$X$$ come in two slightly different flavours. One the one hand you can take finite etale covers $$Y\rightarrow X$$ as you would for schemes, on the other hand you can take a covering space $$Y\rightarrow |X|$$ of the underlying topological space of $$X$$, and, roughly speaking, use the Berkovich space structure of $$X$$ to put one on $$Y$$. Following de Jong, let us call the first of these 'algebraic' and the second 'topological'. A general covering space is then some kind of mixture of the two.

If $$k$$ is not separably closed, then one way of producing algebraic covering spaces is via finite separable extensions of $$k$$: if $$L/k$$ is such an extension then $$X_L \rightarrow X$$ is a finite etale map of Berkovich spaces, where $$X_L$$ denotes the base change of $$X$$ to $$L$$. My question is then the following:

Question: Is it possible that $$X_L \rightarrow X$$ is a topological covering space, for some non-trivial extension $$L/k$$?

It's not to hard to see this can't be the case if $$X$$ has a $$k$$-rational point (since the fibre of $$X_L\rightarrow X$$ over this point will have cardinality 1), but I'm particularly interested in the case when we might have $$X(k)=\emptyset$$. Concretely, I'm interested in the case when $$X$$ is (the analyitification of) a smooth projective conic over $$k$$, without a rational point, and $$L/k$$ is a quadratic extension over which $$X$$ does admit a rational point.

In your particular case, $$X_L$$ has a point, so it is isomorphic to $$P^{1,\mathrm{an}}_L$$, hence simply connected. If your covering $$X_L \to X$$ were a covering, it would then be a universal covering. But we know that Berkovich curves retract by deformation onto graphs, so the topological fundamental group of $$X$$ is a free group. In particular, the universal covering of $$X$$ is either $$X$$ itself or of infinite degree, and we get a contradiction.
• Hi Jerome, I was actually about to email you this question, so I'm glad you've popped up here! I think I perhaps didn't explain it very well - the question was more about whether $X_L$ can ever be a topological covering space of $X$ - i.e. a covering space map on the underlying topological spaces. It boils down to the question of whether or not every point of $X$ (of any Type) has precisely $[L:k]$ preimages in $X_L$. It feels like this is unlikely - for conics, I feel as though it should be possible to cook up some Type II point with only one preimage, but I didn't manage to do so. Sep 5 '20 at 7:54