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I learned from several places that in defining a character of a topological group $G$, we often require it to be continuous, i.e. $\omega:G\to \mathbb{C}^{\times}$ is a continuous group homomorphism. This is particularly the case when $G$ is the $p$-adic field $\mathbb{Q}_p$ or the idele group $\mathbb{A}^{\times}_{\mathbb{Q}}$. Is there any significance to the continuity here? Besides, is here $\mathbb{C}^{\times}$ always equipped with the induced topology from $\mathbb{C}$? Or rather, what is the topology of $\mathbb{C}^{\times}$ here?

I know that in the case of $G=\mathbb{Q}_p$, $\mathbb{C}^{\times}$ is given the discrete topology. Yet I am not very clear about the general case. Will someone be kind enough to say something on this? Thank you very much!

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I cleaned up your text a little bit. It's not a huge deal, but consider placing a space after each period for readability. – Dylan Moreland Sep 6 '11 at 20:15
Strictly, it is impossible to answer your question because it is not complete: why are they required to be continuous to do what? Presumably, you have something in mind, and you want to know why ---in order to do that--- characters are required to be continuous. – Mariano Suárez-Alvarez Sep 6 '11 at 21:50

Discontinuous homomorphisms between uncountable groups

  • usually have no proof of existence without axiom of choice

  • if they exist, can have extremely unfavorable properties, such as not being measurable.

The case of characters of $\mathbb R$ is paradigmatic, with $\log |\omega|$ a solution of $f(x+y)=f(x)+f(y)$.

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Not only can they be non-measurable, they must be non-measurable (at least in all the cases the OP appears to be interested in), say $G$ locally compact second countable. – t.b. Sep 6 '11 at 20:29

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