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.

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    $\begingroup$ 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. $\endgroup$
    – ChrisLazda
    Sep 5 '20 at 7:54
  • $\begingroup$ I was indeed answering something completely different. I will give it another try. $\endgroup$ Sep 5 '20 at 11:17
  • $\begingroup$ That's great thanks! $\endgroup$
    – ChrisLazda
    Sep 7 '20 at 10:14

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