Timeline for Generators of the graded ring of modular forms
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 14, 2017 at 14:48 | answer | added | François Brunault | timeline score: 4 | |
| Jul 12, 2015 at 20:35 | vote | accept | David Loeffler | ||
| Jul 12, 2015 at 17:12 | answer | added | Aaron Landesman | timeline score: 18 | |
| Feb 28, 2012 at 15:16 | comment | added | David Loeffler | @Dror: Using QLiu's answer to a question of mine on math.SE I can prove weight 6 suffices whenever $\Gamma$ has no elliptic points. | |
| Feb 28, 2012 at 10:36 | comment | added | Dror Speiser | Hey David, have you had any luck with this? | |
| Jan 4, 2012 at 21:54 | comment | added | David Loeffler | (PS: In fact there are no subgroups other than $SL_2(\mathbb{Z})$ for which every cusp has width 1.) | |
| Jan 4, 2012 at 21:49 | comment | added | David Loeffler | I looked back at this just now, because I needed to use it for something, and I realised that it doesn't work. Multiplication by $\Delta$ is not an isomorphism between $M_k(\Gamma, \mathbb{C})$ and $S_{k+12}(\Gamma, \mathbb{C})$ in general. This only works if the pullback of $\Delta$ to $X(\Gamma)$ has a simple zero at every cusp, i.e. if every cusp of $\Gamma$ has width 1, which is a pretty rare occurrence. Otherwise the spaces $M_k(\Gamma)$ and $S_{k+12}(\Gamma)$ don't even have the same dimension. | |
| Jun 3, 2011 at 16:17 | comment | added | A. Pacetti | You can even take an Eisenstein series of weight 4 with value 1 at one cusp and 0 at the others. If you raise them to the third power, you get the missing part in the previous map. In weight 6 you can do the same for the Eisenstein part... | |
| Jun 3, 2011 at 13:36 | comment | added | Kevin Buzzard | I guess the Eisenstein series argument goes like this: in weight 4 you already know that there's an Eisenstein series which takes any given values at the cusps, so there's one which takes the value 1 at each cusp, and that can be used to get from one weight to another via an easy argument now. This part of the argument works for weight at most 6 I guess. David: I don't know offhand why weight at most 6 seems to work in the cuspidal case for $\Gamma_0(N)$. 6 is of course the magic number when $N=1$... | |
| Jun 3, 2011 at 12:59 | comment | added | David Loeffler | Of course, that does it. I'm clearly just having a stupid day today. | |
| Jun 3, 2011 at 12:58 | comment | added | A. Pacetti | Actually multiplication by $\Delta$ is an isomorphism between $M_k(\Gamma,\mathbb C)$ and $S_{k+12}(\Gamma, \mathbb C)$. Then you are led to prove that the Eisenstein series of weight up to $12$ span the whole space of Eisenstein series. | |
| Jun 3, 2011 at 12:53 | comment | added | Qiaochu Yuan | Isn't this because multiplication by $\Delta$ is an isomorphism $M_k(\Gamma, \mathbb{C}) \to M_{k+12}(\Gamma, \mathbb{C})$? | |
| Jun 3, 2011 at 12:48 | history | asked | David Loeffler | CC BY-SA 3.0 |