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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.

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    This was known earlier (e.g., to Hecke). Write ${\mathbf C}^\times$ as ${\mathbf R}{>0} \times S^1$, corr. to $z = |z|(z/|z|)$: what are continuous homs from ${\mathbf R}{>0}$ and $S^1$ to ${\mathbf C}^\times$? The first kind are $t \mapsto t^{w}$ and the second kind are $u \mapsto u^n$ with $n \in {\mathbf Z}$. So the formula on $z$ is $z \mapsto |z|^w(z/|z|)^n = z^n|z|^{w-n} = z^n(|z|^2)^{(w-n)/2} = z^n(z\overline{z})^{(w-n)/2} = z^{(w+n)/2}\overline{z}^{(w-n)/2}$. Let $a=(w+n)/2, b=(w-n)/2$. The last eqn. is purely formal: $z^a$ is ill-defined for general $z$ if $a \not\in {\mathbf Z}$. – KCd May 31 '12 at 13:23
  • Note that $z^a\bar z^b$ isn't even well-defined for $a,b$ not integers. But you can use the factorization of $C^\times$ that @KCd provides to make sense of this expression for all $z$ when $a-b\in\mathbb Z$. – Thomas Andrews May 31 '12 at 13:39
  • Here's a related post. See also the links therein. – t.b. May 31 '12 at 13:50

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Tate's thesis has a proof in it.

Marc Palm
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