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Given a topological group $G$, I need to prove that

$\hat{G_1} \times \hat{G_2} \cong \widehat {G_1 \times G_2}$

where $\hat{G} = \{ \chi | \ \chi : G \rightarrow S^1 \}$ $ \ \ , $ ($\chi $ is a continuous group homomorphism and $S^1$ is the multiplicative group of all complex numbers of absolute value one). $\hat{G}$ itself is a topological group with the compact open topology.

Any help with this is appreciated!

Dark_Knight
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  • Well, they aren't technically equal, they're isomorphic. What do you think the isomorphism is? (Also, are you treating the character group itself as a topological group, using the compact-open topology?) – anon Nov 24 '16 at 02:24
  • Yes, $\hat{G}$ with the compact open topology is a topological group. I have made an edit. – Dark_Knight Nov 24 '16 at 03:05

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