# Does every locally compact group $G$ have a nontrivial homomorphism into $\mathbb{R}$?

Does every locally compact group (second countable and Hausdorff) topological group $G$ that is not compact have a nontrivial continuous homomorphism into $\mathbb{R}$?

Obviously for compact groups it is not possible since continuous functions send compact sets to compact sets, and there is only one (trivial) compact subgroup of $\mathbb{R}$.

• Maybe you want to consider homomorphisms into $\mathbb{C}$. – egreg Jul 15 '16 at 20:13
• @egreg Any nontrivial homomorphism into $\mathbb{R}^2$ has a nontrivial vector in its image, hence we may postcompose with projection onto the line containing that vector to get a nontrivial homomorphism into $\mathbb{R}$. – arctic tern Jul 15 '16 at 20:22
• $\mathrm{SL}_2(\mathbb{R})$ is an example (it's only proper nontrivial normal subgroup is $\{\pm I\}$). – arctic tern Jul 15 '16 at 20:35
• @arctictern Sorry, I was thinking to $\mathbb{C}^\times$, the multiplicative group – egreg Jul 15 '16 at 21:24
• $\mathrm{SL}_2(\mathbb{R}$ (or any other connected semisimple Lie group) doesn't have nontrivial homomorphisms into any abelian group, in particular $\mathbb{C}^\times$. – Cronus Aug 19 '16 at 23:42

## 2 Answers

Suppose $G$ is infinite, discrete, and every element has finite order. Then any homomorphism $h : G \to \mathbb R$ would map each element to something of finite order since $h(g)^n = h(g^n) = h(e)=e$ for some positive $n \in \mathbb N$. But only the zero element of $\mathbb R$ has finite order. This implies $h$ is the trivial homomorphism. An example of such a $G$ is the group of all permutations of $\mathbb N$ that fix all but finitely many elements.

• So maybe add the hypothesis $G$ is connected too. – arctic tern Jul 15 '16 at 20:17
• Great example, should have noticed this sooner. – nullUser Jul 15 '16 at 20:48
• Do you know the answer if the group is connected? – Daron Jul 16 '16 at 11:52

For a connected example one can take $G=\mathrm{SL}_2(\mathbb{R})$ (or any connected simple Lie group). If $f:G\to \mathbb{R}$ is a continuous homomorphism then $f(G)$ is a connected simple subgroup of $\mathbb{R}$, hence trivial.

EDIT: I see now someone already mentioned this in the comments.