0
$\begingroup$

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. enter image description here

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Not all cusps are equivalent to $i \infty$ since $\Gamma$ is just a normal subgroup of finite index of the modular group $PSL(2,Z)$. $\endgroup$ Mar 29 at 17:17
  • $\begingroup$ @Neilhawking Since $\Gamma$ is assumed to be normal in $\mathrm{PSL}_2(\mathbf{Z})$, the group $\mathrm{PSL}_2(\mathbf{Z})$ acts by conjugation on $\Gamma$ and thus acts all the cusps are equivalent by automorphisms of $R$, which is exactly what I said. The computation is at any rate local and the same at each cusp. $\endgroup$
    – user902362
    Mar 30 at 3:01
  • $\begingroup$ Do you have any idea regarding the second question that is" A differential has a residue at a cusp $\alpha$ if the corresponding form does not vanish at $\alpha$?". $\endgroup$ Apr 26 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.