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Aug 31, 2019 at 7:35 comment added Jérôme Poineau Yes, opens in rigid geometry need not be open in Berkovich theory, so you should really be clear on which setting you use.
Aug 30, 2019 at 21:50 comment added Hang Thank you very much for the clarification! Now I think I understand. The issue is we should be more careful for Berkovich topology, but this is ok for rigid geometry.
Aug 30, 2019 at 21:03 comment added Jérôme Poineau OK. I had the Berkovich picture in mind where affinoid domains are closed but not open in general. Anyway, $\mathrm{val}^{-1}(\Delta') \subset \mathrm{val}^{-1}(\Delta')$ is an embedding of an affinoid domain, so it would indeed be an open immersion in rigid geometry.
Aug 30, 2019 at 16:21 comment added Hang @JérômePoineau Let's say we are in the category of the classical rigid analytic space and let's even consider the special case $\Delta'=\{0\}$. It seems strange to me why $\mathrm{val}^{-1}(\Delta')$ is not open in $\mathrm{val}^{-1}(\Delta)$, because both are open in the whole $\mathbb G_m^n$, if we consider subspace topology (or maybe not).
Aug 30, 2019 at 15:41 comment added Jérôme Poineau You should make clear which space and which topology you use: the Berkovich analytification of $\mathbb{G}_m^n$? Then, I do not see how $\textrm{val}^{-1}(\Delta')$ could be ever be open except when it is the whole thing. Do you have any example?
Aug 30, 2019 at 14:11 comment added Hang @JérômePoineau Thanks! So, let put it in a simpler way: if $\Delta'\subset \Delta$ is not a closed face, then is $\mathrm{val}^{-1}(\Delta')$ an open subspace of $\mathrm{val}^{-1}(\Delta)$?
Aug 30, 2019 at 6:31 comment added Jérôme Poineau You got the statement wrong: the open immersion is not the map between the Weierstrass domains, but the induced map between their reductions.
Aug 30, 2019 at 3:53 history asked Hang CC BY-SA 4.0