4
$\begingroup$

I recently saw the following claim: Let $\mathbf{C}$ denote the field of complex numbers together with its usual topology. If $\epsilon:\mathbf{C}^\times\to \mathbf{C}^\times$ is a continuous character, then there are complex numbers $a,b$ such that $a-b\in\mathbf{Z}$ and $\epsilon(z) = z^a (\overline{z})^b$ for all $z\in \mathbf{C}^\times$.

Does anyone know a proof? As I heard it, this was first observed by Langlands.

  • 0
    Here's a [related post](http://math.stackexchange.com/q/52319/5363). See also the links therein.2012-05-31

1 Answers 1

0

Tate's thesis has a proof in it.