Valence formula- fundamental domain- why truncation height?

[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?

Many thanks in advance

 

[jvicens]

for any help!
The requirement for a finite truncation height is necessary in order to ensure that the fundamental domain is bounded. This is important because we are integrating around the boundary of the fundamental domain, and if the domain is not bounded, the integral would not converge.

Now, for the comment about poles and zeros lying below some finite height: this is related to the meromorphic nature of the function. A function is considered meromorphic at infinity if it has a Laurent series expansion about infinity that has only finitely many negative powers of the variable. This means that all poles and zeros of the function must lie below some finite height, otherwise the Laurent series expansion would have infinitely many negative powers, which is not allowed for a meromorphic function at infinity.

In terms of the proof of the valence formula, this requirement is important because we are integrating around the boundary of the fundamental domain, and if there were poles or zeros of the function above the truncation height, they would not be accounted for in the integral. This could potentially lead to an incorrect result. Therefore, it is necessary to ensure that all poles and zeros of the function lie below the truncation height in order for the proof to be valid.

I hope this helps to clarify the connection between the truncation height and the requirement for poles and zeros of the function to lie below a finite height. Let me know if you have any further questions.