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Let $\mathbb{C}^\times$ and $\mathbb{T}$ be the multiplicative topological group whose elements are $\mathbb{C}\setminus\{0\}$ and $\{z\in\mathbb{C}:|z|=1\}$ respectively. For a topological group $G$, let $\operatorname{Hom}(G,\mathbb{C}^\times)$ (respectively $\operatorname{Hom}(G,\mathbb{T})$) be the multiplicative group of continuous groups homomorphisms $G\to\mathbb{C}^\times$ (respectively $G\to\mathbb{T}$).

Are $\operatorname{Hom}(G,\mathbb{C}^\times)$ and $\operatorname{Hom}(G,\mathbb{T})$ isomorphic as groups?

Note that we have $\operatorname{Hom}(G,\mathbb{C}^\times)\cong\operatorname{Hom}(G,\mathbb{T})$ if $\mathbb{C}^\times$, $\mathbb{T}$ and $G$ are merely groups, because $\mathbb{C}^\times$ and $\mathbb{T}$ are isomorphic as groups.

For the case of topological groups, we have $\rho\mapsto\left(\log|\rho|,\dfrac{\rho}{|\rho|}\right):\operatorname{Hom}(G,\mathbb{C}^\times)\to\operatorname{Hom}(G,\mathbb{R})\times\operatorname{Hom}(G,\mathbb{T})$ being an isomorphism, where $\mathbb{R}$ and $\operatorname{Hom}(G,\mathbb{R})$ are additive groups, but it is still possible that $\operatorname{Hom}(G,\mathbb{C}^\times)\cong\operatorname{Hom}(G,\mathbb{T})$ as groups, For example, for $G=\mathbb{R}$, we have $\operatorname{Hom}(\mathbb{R},\mathbb{R})=\{x\mapsto\lambda x:\lambda\in\mathbb{R}\}\cong\mathbb{R}$ (consider the Cauchy equation) and $\operatorname{Hom}(\mathbb{R},\mathbb{T})=\{x\mapsto{\rm e}^{{\rm i}\lambda x}:\lambda\in\mathbb{R}\}\cong\mathbb{R}$, and we know that $\mathbb{R}$ and $\mathbb{R}\times\mathbb{R}$ are isomorphic as additive groups (although not canonically).

Thank you in advance.

Jianing Song
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  • @MarianoSuárez-Álvarez To your first comment: That's why I said "isomorphic as groups", as here the topology is not relevant :) – Jianing Song Dec 15 '23 at 06:49
  • @MarianoSuárez-Álvarez if there is a non-trivial homomorphism to $\mathbb{R}$ then there is also a non-trivial map to $\mathbb{T}$ - the composition with complex $x\mapsto exp(rx)$ for some $r$. – freakish Dec 15 '23 at 15:50

1 Answers1

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I don't know the answer for an arbitrary topological group, but if the group $G$ is a locally compact abelian group, it seems that indeed the groups $\operatorname{Hom}(G,\mathbb{C}^*)$ and $\operatorname{Hom}(G,\mathbb{T})$ are isomorphic as abstract groups.

If $G$ is a locally compact abelian group, then $\operatorname{Hom}(G,\mathbb{T})$ is the Pontryagin dual group and we will denote it by the symbol $\widehat{G}$. Next, $\operatorname{Hom}(G,\mathbb{C}^*)$ and $\widehat{G}\times\operatorname{Hom}(G,\mathbb{R})$ are isomorphic as topological groups (actually, that's what you noted, too). If $G$ is a locally compact abelian group, then $\operatorname{Hom}(G,\mathbb{R})$ is a torsion-free divisible groups.

Let us consider two cases:

  1. Let $\widehat{G}$ contain the nontrivial connected component of $e$. We denote it $C$. So, $\widehat{G}\cong H\times C$ as abstract groups (where $H$ is some subgroup of $G$ not necessarily closed) and if $F=\operatorname{Hom}(G,\mathbb{R})$ then $$ \operatorname{Hom}(G,\mathbb{C}^*)\cong \widehat{G}\times\operatorname{Hom}(G,\mathbb{R})\cong H\times C\times F\cong H\times C\cong\widehat{G} $$ as abstract groups.

  2. If $\widehat{G}$ is a totally disconnected group, then the group $G$ has a compact subgroup $H$ such that $G/H$ is a discrete torsion group. In this case, $\operatorname{Hom}(G,\mathbb{R})=0$ and $\operatorname{Hom}(G,\mathbb{C}^*)=\widehat{G}$.

kabenyuk
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