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 :(adsbygoogle = window.adsbygoogle || []).push({});

- 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

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

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