Differential and Modular Form on a Compact Riemann Surface

Question

Let $\Gamma$ be a normal subgroup of finite index of the modular group $PSL(2,\mathbb{Z})$.

Let $\mathbb{H}$ be the upper half-plane.

Let $C$ be the set of all cusps of $\Gamma$.

Let $R = (\mathbb{H}\cup C)/\Gamma$ be the associated compact Riemann surface.

It is well-known that there is a correspondence between the modular forms of dimension $-2$ for $\Gamma$ and differentials on $R$. (See, Elliptic Modular Functions, An Introduction, B. Schoenberg, Chapter V, Pages 125 and 126).

The real questions are, how to prove that:

1. Differentials on $R$ having no poles outside the cups and degree at most $-1$ at the cusps are represented by entire modular forms of dimension $-2$?
2. A differential has a residue at a cusp $\alpha$ if the corresponding form does not vanish at $\alpha$?

One can take a look at Asymptotic Winding of the Geodesic Flow on Modular Surfaces and Continuous Fractions. Y. Guivarc'h and Y. Le Jan. Page 26.

Answer 1

On the open part of the modular curve, the differential form associated to $f(\tau)$ is just $\omega = f(\tau) d \tau$. To understand what happens at a cusp, you need to know the complex structure of the curve at that cusp. Since all cusps are equivalent by automorphisms to $i \infty$, let's just think of that cusp.

Suppose the stabilizer of that cusp is generated by the matrix $\displaystyle{\left(\begin{matrix} 1 & N \\ 0 & 1 \end{matrix} \right)}$. Then a uniformizer of the at the cusp is given by $t = q^{1/N}$ with $q = e^{2 \pi i \tau}$. This vanishes at the cusp, it is well-defined in a neighbourhood of the cusp because it is invariant under the stabilizer, and it vanishes to order $1$.

Now one has to expand $f(\tau) d \tau$ in terms of the local parameter $q$. We have (using $q = e^{2 \pi i \tau} = t^N$):

$$d q =2 \pi i e^{2 \pi i \tau} d \tau = 2 \pi i q d \tau,$$ $$d q = N t^{N-1} dt = N q t^{-1} dt,$$ and so

$$d \tau = \frac{d q}{2 \pi i q} = \frac{N dt}{2 \pi i t}.$$

Hence if you write $$f(\tau) = \sum a_n q^{n/N} = \sum a_n t^n,$$ then $$f(\tau) = \frac{N}{2 \pi i} \sum \frac{a_n}{t^{n+1}} dt.$$

By definition, to be a modular form, you want $a_n = 0$ for $n < 0$, and additionally $a_0 = 0$ (at all cusps) to be a cusp form. But you see that the residue at the cusp is $a_0$ and to be a modular form you don't want any higher order poles.