# Is the ring of meromorphic modular forms on a fine modular curve generated in degree 1?

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)

• 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. – David Loeffler Sep 25 '16 at 8:21
• 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 $H^0(X;{\cal O})$. – Tyler Lawson Sep 25 '16 at 16:45
• 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. – Tyler Lawson Sep 25 '16 at 16:47

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.
• 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? – stupid_question_bot Oct 24 '17 at 4:04
• @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. – S. Carnahan Oct 24 '17 at 8:07