Timeline for Topology of ring of global sections of finite union of affinoid opens in a rigid analytic space
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2015 at 18:38 | vote | accept | Keenan Kidwell | ||
| Jun 30, 2015 at 18:33 | answer | added | Keenan Kidwell | timeline score: 2 | |
| Jun 29, 2015 at 0:09 | comment | added | Keenan Kidwell | Okay, I expected the basic version was adequate. I think I understand. Thank you. | |
| Jun 29, 2015 at 0:07 | comment | added | grghxy | We define the topology on $O(U)$ to be that transported from the equalizer, and directly prove that this is independent of all choices. One has no need for any kind of open mapping theorem beyond the setting of Banach spaces. | |
| Jun 28, 2015 at 23:37 | comment | added | Keenan Kidwell | Dear @grghxy, Perhaps I'm missing the point, and this is overkill, but the most general version of the open mapping theorem I know of has the source an LF-space (locally convex inductive limit of a sequence of Fréchet spaces), but I don't see why $\mathscr{O}(U)$ has this structure without further hypotheses. I'm assuming you're suggesting applying the open mapping theorem to the continuous bijection $\mathscr{O}(U)\to E$, where $E$ is the equalizer equipped with its Banach structure and deducing that this map is open. | |
| Jun 28, 2015 at 19:30 | history | edited | Keenan Kidwell | CC BY-SA 3.0 |
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| Jun 28, 2015 at 19:28 | comment | added | Keenan Kidwell | Dear @grghxy, Thanks for the response and for pointing out my oversight. I was thinking of affinoid spaces. | |
| Jun 28, 2015 at 18:54 | comment | added | grghxy | Why is $U$ admissible (if using rigid-analytic spaces, not adic spaces)? Assume $X$ is quasi-separated, so (exercise!) $U$ is admissible with $\{U_i\}$ an admissible covering, so $U$ is qcqs. Hence, in such cases $O(U)$ is the equalizer of the natural map $\prod O(U_i) \rightarrow \prod O(V_{ijh})$ where $\{V_{ijh}\}$ is a finite (admissible) affinoid open cover of the quasi-compact $U_i \cap U_j$. You can check via the Banach open mapping theorem that the natural Banach space structure on the equalizer (a closed subspace of $\prod O(U_i)$) is the intrinsic topology you are considering. | |
| Jun 28, 2015 at 17:33 | history | edited | Keenan Kidwell | CC BY-SA 3.0 |
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| Jun 28, 2015 at 17:23 | history | asked | Keenan Kidwell | CC BY-SA 3.0 |