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Let $\Gamma$ be a finite-index subgroup of $\operatorname{SL}_2(\mathbb{Z})$ . I've seen it stated (in a comment in the code of a computer program) that the graded ring $$ M(\Gamma, \mathbb{C}) = \bigoplus_{k \ge 0} M_k(\Gamma, \mathbb{C}),$$ where $M_k(\Gamma, \mathbb{C})$ is the space of modular forms of weight $k$ and level $\Gamma$, is always generated as a $\mathbb{C}$-algebra by forms of weight $\le 12$.

Why is this true? Moreover, can one improve on the bound of 12? (For the subgroups $\Gamma_0(N)$, weight $\le 6$ always seems to be sufficient.)

flag	asked Jun 3 2011 at 12:48 David Loeffler 8,603●15●40

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Isn't this because multiplication by $\Delta$ is an isomorphism $M_k(\Gamma, \mathbb{C}) \to M_{k+12}(\Gamma, \mathbb{C})$? – Qiaochu Yuan Jun 3 2011 at 12:53

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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. – A. Pacetti Jun 3 2011 at 12:58

Of course, that does it. I'm clearly just having a stupid day today. – David Loeffler Jun 3 2011 at 12:59

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 $N=1$... – Kevin Buzzard Jun 3 2011 at 13:36

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... – A. Pacetti Jun 3 2011 at 16:17

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Generators of the graded ring of modular forms

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