Timeline for Are some congruence subgroups better than others?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2023 at 6:21 | vote | accept | Coherent Sheaf | ||
| Sep 25, 2023 at 17:06 | comment | added | Coherent Sheaf | @JoeSilverman. That definitely makes a lot of sense. Thank you. | |
| Sep 25, 2023 at 16:25 | comment | added | Joe Silverman | @CoherentSheaf I don't know a lot about Shimura varieties, much less perfectoid spaces. But it seems that even in such cases, you are looking at a group-like object with either a marked point of order $N$ or a marked cyclic subgroup of order $N$, and these are both natural, concrete, and among the simplest sort of level structures that one might consider. Of course, "group of order $N$" may need to be interpreted as "group scheme of order $N$" or something more complicated, but still with that general idea. | |
| Sep 25, 2023 at 16:19 | comment | added | Joe Silverman | @LSpice abx already clarified (thanks), but I also find it enlightening to consider the map $X_1(N)\to X_0(N)$ between the associated modular curves, where the map (at the level of moduli problems) is $$ (E,P) \longrightarrow (E,\text{subgroup generated by $P$}). $$ | |
| Sep 25, 2023 at 16:08 | comment | added | LSpice | @abx, re, I see, it's not classifying "elliptic curve that has a cyclic subgroup of order $N$" as I originally read it, but "pairs of an elliptic curve and a cyclic subgroup of order $N$" (and analogously for points). Thanks! | |
| Sep 25, 2023 at 16:00 | comment | added | Coherent Sheaf | @Joe Silverman. Just to clarify, are you saying that pedagogical/understanding purposes could be further reason to study these special congruence subgroups? But are there any mathematical benefits of considering these? Because I have seen similar things happen in notes on Shimura varieties, where one sometime assume $G$ has an integral model $\mathcal{G}$, and then we consider the kernel of $\mathcal{G}(\mathbb{Z})\rightarrow \mathcal{G}(\mathbb{Z}/n\mathbb{Z})$ (reference: Cariani, Perfectoid Shimura varieties). Also, thank you very much for your answer. I am a big fan! | |
| Sep 25, 2023 at 15:54 | comment | added | abx | @LSpice: point of order $N$ = cyclic subgroup of order $N$ + generator. | |
| Sep 25, 2023 at 15:49 | comment | added | LSpice | What is the difference between a cyclic subgroup of order $N$ and a point of order $N$? | |
| Sep 25, 2023 at 15:49 | history | edited | LSpice | CC BY-SA 4.0 |
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| Sep 25, 2023 at 15:37 | history | answered | Joe Silverman | CC BY-SA 4.0 |