Timeline for Mixing solids and liquids
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2022 at 10:09 | answer | added | Peter Scholze | timeline score: 12 | |
| Sep 10, 2022 at 21:08 | comment | added | Brian Shin | Ah, that is certainly a good point | |
| Sep 10, 2022 at 19:03 | history | edited | LSpice | CC BY-SA 4.0 |
Names of links
|
| Sep 10, 2022 at 18:39 | history | edited | Wojowu | CC BY-SA 4.0 |
added 232 characters in body
|
| Sep 10, 2022 at 18:37 | comment | added | Wojowu | @BrianShin You are certainly right. However, this restricted product makes it so that spectra of adelic algebras are essentially disjoint unions of those over local fields, while one usually considers groups of adelic points which really are restricted products of sets of points over local fields, and you really want to consider those when e.g. describing Shimura varieties as adelic quotients. It is not clear to be how they would be incorporated in that framework. | |
| Sep 10, 2022 at 17:28 | comment | added | Brian Shin | Perhaps someone should come up with a "restricted (direct) product" for analytic rings which would yield adeles as such an object. | |
| Sep 10, 2022 at 17:23 | comment | added | Brian Shin | The ring of adeles is built on something more like the direct product rather than the tensor product. | |
| Sep 10, 2022 at 17:15 | comment | added | Wojowu | @Z.M Thank you for both of these comments! RE first, yeah, obviously the tensor product in the category of analytic rings is more relevant here than just that of (solid) abelian groups, and your argument shows its triviality. RE second, this is a good point. I imagine tensoring those with each other or with $\mathbb R_{Liq}$ also gives zero by essentially the same argument? | |
| Sep 10, 2022 at 16:00 | comment | added | Z. M | Let me mention that there is a liquid structure on $\mathbb Q_\ell$ described in Remark 5.5 in liquid tensor experiments. | |
| Sep 10, 2022 at 15:05 | comment | added | Z. M | I am not sure whether your argument is valid on the nose, but it could be fixed as follows: consider the map $\mathbb R\to\mathbb R_{\operatorname{Liq}}$ of analytic rings where the first is the condensed ring equipped with the trivial analytic structure. Taking the coproduct with $\mathbb Z_{p,\blacksquare}$ in the category of analytic rings, you get a map of analytic rings, whose source is the zero ring, and therefore the target is also zero. | |
| Sep 10, 2022 at 14:31 | history | edited | YCor | CC BY-SA 4.0 |
remove modern slang, formatting
|
| Sep 10, 2022 at 13:17 | history | asked | Wojowu | CC BY-SA 4.0 |