If $\zeta$ is a function of characters what does it mean for it to be regular?

58 Views Asked by At

This is from lemma 2.4.1 of Tate's thesis.

Lemma 2.4.1: A $\zeta$-function is regular in the "domain" of all quasi-characters of exponent greater than $0$.

proof: We must show that for each $c$ of exponent $>0$ the integral $\int f(\alpha)c(\alpha)|\alpha|^s\mathrm{d}\alpha$ represents a regular function of $s$ near $0$.

Does regular mean that it has a derivative? In the proof, that's what Tate seems to be arguing. Moreover, if this is indeed the case, why is it obvious that the above integral has a derivative for large $s$ but a priori not for $s$ near zero?

1

There are 1 best solutions below

2
On

"regular" here means holomorphic. Each connected component of the space of quasi-characters is just a copy of the complex plane; if I understand your notation correctly, then $c(\alpha)$ is some (unitary) character in this component, and then all the other quasi-chars. in the component have the form $c(\alpha)|\alpha|^s$.

The $\zeta$-function can thus be viewed as a function of $s$, and Tate is determining where it does (or doesn't) have poles.