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Denote by $\mathrm{Hom}$ continuous group homomorphisms. Fix the quotient homomorphism $\mathbb{R}\to S^1$.

Can one characterize those topological (e.g., locally compact, and in particular discrete) groups $G$ such that the induced map $\mathrm{Hom}(G,\mathbb{R})\to\mathrm{Hom}(G,S^1)$ is bijective?

Clearly this holds for $G$ if and only if it holds for the abelianization $G/\overline{[G,G]}$, so we can assume that $G$ is abelian and Hausdorff.

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  • $\begingroup$ For any homomorphism from G to R, there are uncountably many scaled copies of it. This is not true for homomorphisms from G to S^1, is it? $\endgroup$
    – Anthony Quas
    Commented Mar 8, 2014 at 0:14
  • $\begingroup$ @Anthony: check "Pontryagin duality". For instance for $G=\mathbf{R}/\mathbf{Z}$, the group $Hom(G,\mathbf{R}/\mathbf{Z})$ is infinite and countable (indeed cyclic). $\endgroup$
    – YCor
    Commented Mar 9, 2014 at 10:58

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If $G$ is a simply-connected Lie group, then any Lie group morphism $G\rightarrow S^1$ factors through the universal cover $\mathbb{R}\rightarrow S^1$ to give a Lie group morphism $G\rightarrow\mathbb{R}$ (by requiring the identity in $G$ to be sent to $0$). This gives a bijection between the Lie group $Hom$ sets. Is this the sort of answer you were seeking?

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  • $\begingroup$ Well, there may be a lot more homomorphisms than the smooth ones, so this does not answer the question even in the setting of Lie groups... $\endgroup$
    – Igor Rivin
    Commented Mar 7, 2014 at 22:06
  • $\begingroup$ Yes, that's certainly the case. For that reason, I indicated that the bijection existed only between Lie group Hom sets. $\endgroup$
    – Peter Crooks
    Commented Mar 7, 2014 at 23:18
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I guess you mean continuous homomorphisms. Note that the condition only depends on the abelianization $G/\overline{[G,G]}$; let $H$ be its Pontryagin dual. By Pontryagin duality it's equivalent to determine for which LCA groups $H$ every element lies in a 1-parameter subgroup. These groups $H$ have been characterized by Dixmier (*): those of the form $\mathbf{R}^k\times D$ where $D$ is a discrete abelian group such that $\mathrm{Ext}^1_\mathbf{Z}(D,\mathbf{Z})=0$. This fully characterize groups with the required property.

(*) J. Dixmier. Quelques propriétés des groupes abéliens localement compacts. Bull. Sci. Math. (2) 81 (1957) 38-48.

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  • $\begingroup$ Why do you assume continuous? $\endgroup$
    – Igor Rivin
    Commented Mar 8, 2014 at 1:27
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    $\begingroup$ I think it is reasonable to assume continuity holds, at the very least. How often does one consider the groups $\mathbb{R}$ and $S^1$ outside of a topological context? $\endgroup$
    – Peter Crooks
    Commented Mar 8, 2014 at 3:46
  • $\begingroup$ $\mathbf{R}$ as a discrete group is just $\mathbf{Q}^{(c)}$ (free $\mathbf{Q}$-module of continuum rank) and $\mathbf{R}/\mathbf{Z}$ is $\mathbf{Q}^{(c)}\times\mathbf{Q}/\mathbf{Z}$. $\endgroup$
    – YCor
    Commented Mar 9, 2014 at 10:55
  • $\begingroup$ @IgorRivin insofar as we consider arbitrary topological groups, continuity is not a restriction but a generalization. If you want to consider arbitrary homomorphisms from a given topological group $G$ to the reals, then just change the topology of $G$. (On the other hand it would be a restriction if we were sticking, say, to separable topological groups, since this is not stable under the operation of changing to the discrete topology.) $\endgroup$
    – YCor
    Commented Dec 4, 2016 at 1:47

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