# A Valence formula- fundamental domain- why truncation height?

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1. May 7, 2017

### binbagsss

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?

2. May 12, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. May 12, 2017

ta bot bruv'