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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.)

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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 '11 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 '11 at 12:58
    
Of course, that does it. I'm clearly just having a stupid day today. –  David Loeffler Jun 3 '11 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 '11 at 13:36
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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. –  David Loeffler Jan 4 '12 at 21:49

1 Answer 1

up vote 13 down vote accepted

Indeed, for any congruence subgroup $\Gamma \subset SL_2(\mathbb Z)$, (of any level) the graded ring $M_k(\Gamma)$ is generated in weight at most 6, with relations in weight at most 12. Furthermore, in the case that $M_3(\Gamma) \neq 0,$ i.e., there exists a modular form of odd weight, (see Remark 1.6 of Landesman, Ruhm, and Zhang linked below,) $M_k(\Gamma)$ is generated in weight at most 5, with relations in weight at most 10. In the further case that $M_3(\Gamma) \neq 0$ and the genus of the Riemann surface corresponding to $\Gamma$ is $0$ or $1$, then $M(\Gamma)$ is generated in weight at most 4 with relations in weight at most 8.

A proof follows from combining results of two recent articles, namely Theorem 1.4 (the main result) and Theorem 9.3.1 from Voight and Zureick-Brown http://arxiv.org/abs/1501.04657 and Example 1.7 from an article by myself, Ruhm, and Zhang http://arxiv.org/abs/1507.02643.

To spell out how these results fit together, there are two cases. The case that there exists a nonzero odd weight modular form is covered by Example 1.7 of Landesman, Ruhm, and Zhang. So, it suffices to show that if there is no odd weight modular form, then the ring of modular forms is generated in weight at most 6, with relations in weight at most 12. First, in the case that $g > 0$, this follows immediately from the last sentence of Theorem 1.4 of Voight and Zureick-Brown because $2g -2 \geq 0$ and congruence subgroups can only have elliptic points of orders 2 and 3, and so $3 = \max(3,e)$. Second, if $g = 0$, note that because $\Gamma$ is a congruence subgroup it has some cusp, so $\delta \geq 1$ (where $\delta$ is the number of cusps). Since all the exceptional signatures listed in the table in the statement of Theorem 9.3.1 have $\delta = 0,$ they do not occur for congruence subgroups. Therefore, by the last sentence of Theorem 9.3.1, such congruence subgroups are generated in weight at most $2e = 6$ with relations in weight at most $2 \cdot 2 e = 12$.

One final note: in Voight and Zureick-Brown, it says the ring is generated in degree $e$, but degree $e$ is the same as weight $2e$ by their grading convention.

Edit: The above statements hold over any perfect field (or more generally when the stacky curve associated to $\Gamma$, as discussed in Voight and Zureick-Brown Chapter 5, is tame and separably rooted). However, as pointed out by John Voight in the comments, generation in weight 6 and relations in weight 12 still holds over more general base rings. See Voight and Zureick-Brown Proposition 11.3.1 for more.

Edit: The above answer is now fully explained in Example 1.7 of Landesman, Ruhm, and Zhang, referenced above.

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What are "minimal relations"? I think you just mean "relations", since you say "at most 12" anyway. JV –  John Voight Jul 12 at 17:56
    
Right, that was unclear. I removed the word minimal. I probably should say something like "a set of generators of the ideal of relations" to be perfectly clear, but that should be understood. –  Aaron Landesman Jul 12 at 17:58
    
You may also want to refer to DZB's Proposition 11.3.1, since extends Theorem 9.3.1 to more general base rings. JV –  John Voight Jul 12 at 17:58
    
I haven't read chapter 11 carefully, but it looks like that only applies to the base ring $\mathbb Z[1/6N].$ How do you extend it to arbitrary base rings? Are you saying that when you tensor up, the generator and relation degrees will be preserved? I know this works for fields, but I haven't thought through it for general commutative rings with unit. –  Aaron Landesman Jul 12 at 18:06
    
Well, the result over $\mathbb{Z}[1/6N]$ implies the result over $\mathbb{Q}$ and therefore any field of characteristic $0$, so in particular it answers the original question (for the subring in even weight). More generally, one can apply flat base change; but I don't know how far we want to get into this in these comments. –  John Voight Jul 12 at 22:57

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