Timeline for modular forms, invertible sheaves, and quotients
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2014 at 6:31 | answer | added | Nadim Rustom | timeline score: 1 | |
| Aug 11, 2014 at 21:35 | vote | accept | Nadim Rustom | ||
| Aug 10, 2014 at 7:57 | answer | added | David Loeffler | timeline score: 5 | |
| Aug 9, 2014 at 15:57 | history | edited | Nadim Rustom | CC BY-SA 3.0 |
clarification of assumptions
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| Aug 9, 2014 at 14:39 | comment | added | Nadim Rustom | I am interested in cases where there are no irregular cusps, only possibly elliptic curves. I am also interested in positive characteristic, but for now sticking to characteristic 0. I didn't understand your last comment, did you mean that $\pi_{\ast}^G \mathcal{L}$ will be invertible? Doesn't that contradict the situtation for modular forms? (since we said these sheaves are not necessarily invertible). | |
| Aug 9, 2014 at 14:35 | comment | added | Nadim Rustom | Dear eric, thank you for your reply. I should have quoted precisely the statement from Diamond and Im, which says "$\mathcal{G}_k$ is invertible unless $-1 \in \Gamma$ and $k$ is odd". So I was well aware of the case you mentioned. However, even when $k$ is even, I think the statement is still false, because of the argument I mentioned? | |
| Aug 9, 2014 at 12:23 | comment | added | eric | Finally, for your group theory question, if you're interested only in characteristic zero, then remember that a finite group acts semisimply in any situation where its order is invertible on the base, so everything is easy here. Are you interested in modular curves over finite fields or not? | |
| Aug 9, 2014 at 12:21 | comment | added | eric | Completely off-topic -- you need to be careful at cusps too. I am guessing that by $X(\Gamma)$ you mean the compatified curve, but then you have to be careful about defining $G_k$ at the cusps. For example if $k=1$ then there are problems with the "middle" cusp on $\Gamma_1(4)$ even though $\Gamma_1(4)$ doesn't contain any elements of finite order other than the identity. The problem is the irregular cusp, where $G_1$ doesn't behave too well. But this is a side issue. | |
| Aug 9, 2014 at 12:19 | comment | added | eric | If $k$ is odd then the invertible sheaf $G_k$ will exist for sufficiently small $\Gamma$, for example $\Gamma(N)$ with $N>=3$ (no elliptic elements, no $-1$). But if you start trying to descend it to level 1, for example, you will definitely get the zero sheaf (in characteristic away from 2) because $-1$ acts as $-1$ which has no non-zero fixed points. So there's a trivial counterexample to the assertion that $G_k$ is invertible. | |
| Aug 9, 2014 at 10:36 | history | asked | Nadim Rustom | CC BY-SA 3.0 |