Timeline for Berkovich space including both archimedean and non-archimedean worlds
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2023 at 8:15 | comment | added | Jérôme Poineau | @Nico Probably not the right place indeed. Anyway, I do not see any problems carrying this construction with adic spaces. However, you would then miss the Archimedean places, which is a shame. I think the most interesting things happen at the interplay between Archimedean (Schottky uniformization) and non-Archimedean (Mumford uniformization). | |
| Sep 14, 2023 at 5:34 | comment | added | Nico | @Jérôme Poineau maybe this is the wrong place to ask, but can you carry out the "universal Tate curve" construction with adic spaces? | |
| Mar 27, 2020 at 18:04 | comment | added | Jérôme Poineau | Changed that, thanks! | |
| Mar 27, 2020 at 18:04 | history | edited | Jérôme Poineau | CC BY-SA 4.0 |
seminars -> seminorms
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| Mar 27, 2020 at 14:27 | comment | added | LSpice | One of your 'seminorms' became 'seminars'. | |
| Aug 15, 2018 at 12:28 | vote | accept | Hang | ||
| Aug 15, 2018 at 7:51 | comment | added | Jérôme Poineau | I added a few words about $\mathbf{G}_m$ in the answer. Hope it helps. | |
| Aug 15, 2018 at 7:50 | history | edited | Jérôme Poineau | CC BY-SA 4.0 |
Added some details.
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| Aug 15, 2018 at 7:41 | comment | added | Jérôme Poineau | Something else you can do it consider $\mathbb{C}$ endowed with the norm that is the maximum of the trivial absolute value and the usual absolute value. This way, you get usual complex spaces that "degenerate" onto a non-archimedean fiber (over $\mathbb{C}$ with trivial valuation). This has been used by Berkovich to show that (under some conditions) the weight 0 part of the cohomology of a family of complex spaces can be computed by the singular cohomology of a suitable Berkovich space. | |
| Aug 15, 2018 at 7:35 | comment | added | Jérôme Poineau | In general, Berkovich spaces are defined over complete normed rings. If $A$ is such a ring, $\mathcal{M}(A)$ is the set of mult. seminorms bounded by the given norm. So you can define Berkovich spaces over $\mathbb{C}[[t]]$ or $\mathbb{C}((t))$ endowed with some $t$-adic norm. (For $\mathbb{Z}$, one can endow it with the usual abs. value and then all mult. seminorms are bounded by it.) But beware that you will not find usual complex spaces in this case since the usual absolute on $\mathbb{C}$ is not bounded by the trivial absolute value on $\mathbb{C}$ (induced by the $t$-adic absolute value). | |
| Aug 14, 2018 at 14:47 | comment | added | Hang | This is really a great answer. I particularly love the last example. I have a few questions: (1) what if we replace $\mathbb Z$ by something else, say $\mathbb C[[t]]$ or $\mathbb C((t))$? I guess we may have similar things, right? For example, what should the possible $\mathcal H(x)$ in this case? (2) By $G_m$ do you simply mean $\mathbb Z^\times$, or $k^\times$ for some $k$? | |
| Aug 14, 2018 at 7:56 | history | answered | Jérôme Poineau | CC BY-SA 4.0 |