Consider a torsion-free congruence subgroup $\Gamma\le SL_2(\mathbb{Z})$ (for example, $\Gamma(N)$ for $N\ge 3$, or $\Gamma_1(N)$ for $N\ge 4$).

By a meromorphic modular form for $\Gamma$ of weight $k$, I mean a holomorphic function $f$ on $\mathcal{H}$ with $f(\gamma z) = (cz+d)^kf(z)$ for $\gamma\in\Gamma$, which is meromorphic at the cusps.

Let $M_k(\Gamma)$ be the $\mathbb{C}$-vector space of meromorphic modular forms for $\Gamma$ of weight $k$. Is the graded ring $\bigoplus_{k\ge 0} M_k(\Gamma)$ generated in degree 1?

I believe this follows from the geometric interpretation of modular forms as sections of line bundles on moduli stacks (at least when the moduli stack is representable), but I just wanted to check that this is also known from an analytic perspective as well.

Also, if the above is true, then can someone explain roughly why this fails if $\Gamma$ is not torsion-free? (ie, are line bundles on stacks, viewed as the Spec of the symmetric algebra of the corresponding invertible module not necessarily generated in degree 1? Perhaps I don't have the right definition of a line bundle on a stack)

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    $\begingroup$ I've asked a couple of questions before about the graded ring of holomorphic modular forms: see math.stackexchange.com/questions/96395 (for $\Gamma$ sufficiently small) and mathoverflow.net/questions/66819 (for general $\Gamma$). This is a slightly different question, of course, but the methods might help for your question too. $\endgroup$ Sep 25, 2016 at 8:21
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    $\begingroup$ I believe that this is true. If $X$ is the modular curve minus the cusps, there is a line bundle $\omega$ on it and the set of meromorphic modular forms of weight $k$ is the same as the set of (algebraic) sections of the line bundle $\omega^{\otimes k}$. However, $X$ is an affine scheme. This means that "taking sections" gives an equivalence between line bundles on $X$ and projective modules of rank 1 over $H^0(X;\cal{O})$, so that the group of forms of weight k is always the k-fold tensor of the group of sections of weight 1 over $H^0(X;{\cal O})$. $\endgroup$ Sep 25, 2016 at 16:45
  • $\begingroup$ This doesn't work in the stack case because that you lose affineness; you also lose the property that "taking global sections" is an equivalence of categories. $\endgroup$ Sep 25, 2016 at 16:47

1 Answer 1


Tyler basically answered the question in the comments, but I might as well fill in an answer. Your definition of "meromorphic modular form" is often called "weakly holomorphic modular form" in the literature, to distinguish such forms from those that may have poles in the interior of the half-plane.

Weakly holomorphic modular forms of weight $k$ are algebraic sections of some line bundle $\Omega^{k/2}$ on the affine quotient $Y = \Gamma \backslash \mathfrak{H}$, and by affineness, the space of sections $\Gamma(Y, \Omega^{k/2}) \cong M_k(\Gamma)$ is given by the $k$th tensor power of $\Gamma(Y, \Omega^{1/2})$ as a $\Gamma(Y, \mathcal{O})$-module. Multiplication then yields a ring isomorphism $\operatorname{Sym}^*_{\Gamma(Y, \mathcal{O})} \Gamma(Y, \Omega^{1/2}) \cong \bigoplus_{k \geq 0} M_k(\Gamma)$.

When $\Gamma$ has torsion, the stack quotient is no longer an affine curve, and the line bundle $\Omega^{1/2}$ is nontrivial. We can see this concretely as follows: if $\Gamma$ has an element $\left( \begin{smallmatrix} a&b \\ c & d \end{smallmatrix} \right)$ of finite order $n > 1$, then it has a fixed point $x \in \mathfrak{H}$, and $(cx + d)^1$ is not equal to 1. Thus, all elements of $M_1(\Gamma)$ vanish at $x$. For example, if $\Gamma$ contains $\left(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$, then any weight 1 form satisfies $$f(\tau) = f\left(\frac{-\tau}{-1}\right)= (-1)^1 f(\tau)=-f(\tau),$$ at all $\tau$, so weight 1 forms for $\Gamma$ necessarily vanish everywhere. Since there are positive weight level 1 forms like $\Delta$ that vanish nowhere, there must exist generators whose weight is greater than 1.

  • $\begingroup$ I'm reading your answer again, and I'm wondering - Are you claiming that when $\Gamma$ is torsion-free, then $\Omega^{1/2}$ is the trivial bundle? Why is that true? $\endgroup$ Oct 24, 2017 at 4:04
  • $\begingroup$ @rtz Sorry, my answer was poorly worded, and I should have emphasized something other than the nontriviality of $\Omega^{1/2}$ in the non-affine case. Unfortunately, I do not know enough in general to either make the triviality claim or say it is false. There are plenty of punctured Riemann surfaces of positive genus with nontrivial $Pic^0$, so I don't know an easy way to back up such a claim. $\endgroup$
    – S. Carnahan
    Oct 24, 2017 at 8:07

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