Modular Forms - Valance Formula Proof - Truncation height question- integrating boundary of fundamental domain

Question

In the proof of the valence formula you integrate $f'(t)/f(t)$ around the boundary of the fundamental domain, with some modifications such as :

• omitting the points $i, \omega $
• since equivalence is allowed on the boundary, modifying as needed to ensure such points that are equivalent only occur once in the interior
• truncating the fundamental domain at some finite height, rather than letting it run up to $\infty$, let this finite height be called $T$

My question:

My notes say that all poles and zeros of $f$ must lie below some finite $T$, if this were not the case $f$ would not be meromorphic about $\infty$

I don't understand this comment at all. So the fundamental domain is a sketch over $t$ and not its associated variable $q=e^{2\pi i t}$, and for the definition of meromorphic at $\infty$ (that is meromorphic at $q \to 0 $ ) I have :

The expansion about $q=0$ is:

$ \sum\limits_{n>>\infty} a_n q^n $

i.e. as long as the pole at infinity is not of infinite order, otherwise it is a essential pole (I think is the term).

I'm struggling to see the connection to the requirement of a truncation height? The only thing that is needed is that the pole at $\infty$ is of finite order?

Many thanks in advance

Answer 1

The singularities of a holomorphic function on a domain $U\subseteq \mathbb{C}$ must be isolated, they cannot cluster otherwise the function would not be well defined.

In your case, $f$ is meromorphic on the upper half plane (or on the punctured disk if you prefer) so its singularities there cannot cluster. If you had poles over any height $T$, then you would have a sequence of poles converging to infinity (or to $0$ in the punctured disk). This is impossible because infinity would be a cluster point for the poles, which is a contradiction. This argument works regardless of the behaviour of $f$ at infinity.

A similar reasoning holds for the zeros: if you had zeros over any height $T$, then you would have a sequence of zeros converging to infinity, which is impossible unless $f$ is identically zero (by the identity principle).