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

Question

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 $\mathbb{R}$.

@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 $\mathbb{R}$. — anon, Jul 15 '16 at 20:22
$\mathrm{SL}_2(\mathbb{R})$ is an example (it's only proper nontrivial normal subgroup is ${\pm I}$). — anon, 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

Daron · Accepted Answer · 2016-07-15T20:09:39.507

6

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.

edited Jul 15 '16 at 20:09

answered Jul 15 '16 at 20:03

Daron

10,300

So maybe add the hypothesis $G$ is connected too. – anon 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

score 3 · Answer 2 · answered Aug 19 '16 at 01:33

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.

score 0 · Answer 3 · answered Jul 31 '19 at 22:01

Suppose that $G$ is a locally compact group with Property FH. Then, every continuous homomorphism from $G$ to $\mathbb{R}^n$ or $\mathbb{Z}^n$ is trivial.

This follows from a corollary in the book "Kazhdan's Property T" by Bekka, de la Harpe and Valette.

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

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