Let $X$ be a smooth affinoid rigid space over a nonarchimedean field $K$. Does $X$ admit an etale map to a $K$-affinoid ball? The answer is almost certainly "no" in general, but I don't know any counterexample.
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asked Sep 16, 2020 at 19:51
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1$\begingroup$ is this true for affine curves? $\endgroup$– user164751Commented Sep 16, 2020 at 20:55
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4$\begingroup$ If this is true, then the cotangent bundle of $X$ is trivial, in fact trivialized by exact differentials. Now it can’t possibly be true that every smooth affinoid satisfies this... can it? $\endgroup$– Piotr AchingerCommented Sep 16, 2020 at 21:40
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3$\begingroup$ My guess would be to take the “sphere” $x^2+y^2+z^2=1$ in the 3-dim polydisc, and maybe odd residue characteristic. It it is fibered over $\mathbb{P}^1$ with fibers being discs and some Chern classes will give you an obstruction. $\endgroup$– Piotr AchingerCommented Sep 16, 2020 at 21:44
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$\begingroup$ @Piotr: Very nice idea! $\endgroup$– David HansenCommented Sep 16, 2020 at 21:47
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3$\begingroup$ In algebraic geometry, there's already an obstruction at the level of the canonical bundle. Let $X \subset Y = \mathbf{P}^2$ be the (smooth affine) complement of a smooth degree $d$ curve $C$. Then $\mathrm{Pic}(X) = \mathrm{Pic}(Y)/\langle \mathcal{O}(C) \rangle$, which is a cyclic subgroup of order $d$ generated by $\mathcal{O}(1)$. Now $K_X = (K_{\mathbf{P}^2})|_X$ is the class of $\mathcal{O}(-3)$, so if $d$ does not divide $3$ then $K_X \neq \mathcal{O}_X$. Perhaps a smooth affinoid whose good reduction is such an $X$ gives a counterexample? $\endgroup$– AnonymousCommented Sep 17, 2020 at 18:02
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