Timeline for Why are Berkovich spaces locally connected?
Current License: CC BY-SA 4.0
7 events
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| Sep 28, 2022 at 19:54 | comment | added | Jérôme Poineau | @TimCampion I don't think $\mathcal{M}(A)$ should be locally connected in general, just as $\operatorname{Spec}(A)$ is not. Yes, $\mathcal{M}(\mathbb{Z})$ is connected and locally connected, which may be seen directly. Actually, all Berkovich spaces over $\mathbb{Z}$ are locally (path-)connected, see Théorème 7.2.17 in arxiv.org/abs/2010.08858. | |
| Sep 28, 2022 at 17:10 | vote | accept | Tim Campion | ||
| Sep 28, 2022 at 12:40 | comment | added | Tim Campion | Note to self: this answer shows that in the affinoid case, the connected components of $\mathcal M(A)$ are open in $\mathcal M(A)$. This doesn't compete the proof of local connectedness -- one needs to show that each point has a neighborhood basis of connected open sets. But that's exactly what Bekovich's 2.2.8 does. | |
| Sep 28, 2022 at 12:38 | comment | added | Tim Campion | Thanks! I was indeed under the impression that $\mathcal M(A)$ should be locally connected for any discrete ring or Banach ring $A$, but I see now that it's important here to restrict to the affinoid case. So for example, the general theorem does not explain the fact (also true, I think!) that $\mathcal M(\mathbb Z)$ is connected and locally connected. | |
| Sep 27, 2022 at 20:47 | comment | added | Jérôme Poineau | @Z.M: I am rather familiar with the construction but still not sure what the question is. Is $\mathcal{M}(A)$ locally connected for an arbitrary Banach ring $A$? Too general to be true I guess. | |
| Sep 27, 2022 at 20:22 | comment | added | Z. M | I think that the Berkovich spectrum of a ring is simply the usual one, namely, the set of seminorms on a ring with weak topology. For example, $\mathcal M(\mathbb Z)$ looks like a star, with all archimedean and non-archimedean seminorms. | |
| Sep 27, 2022 at 20:10 | history | answered | Jérôme Poineau | CC BY-SA 4.0 |