I apologize in advance if this is well-known, but I can't seem to find the answer in the literature. Let me be precise about my question. I am looking for concrete examples of locally compact Hausdorff groups $G$ such that that there exists $a, b \in G$ with $a \ne b$, but for any continuous representation $\pi : G \to \operatorname{GL}\_n(k)$ with $k$ a valuated field and $n$ a positive integer, we have $\pi(a) = \pi(b)$.


I am also interested in a weaker form: the case where $k$ is the field $\mathbb{C}$ of complex numbers.


If it turns out that there are no such groups, please indicate a reference or sketch a proof.


Also a relevant statement which my gut tells me should be true, but I don't know enough Lie theory yet to give a complete proof: for every real Lie group $G$, there are enough finite dimensional, continuous complex representations of $G$ to distinguish points of $G$. I was thinking maybe this can be proved by exploiting the close relationship between representations of a Lie group and representations of its Lie algebra. It would be nice if someone could give a sketch if this can indeed be achieved or describe why this "hand-waving" actually does not work.


Edit: Sorry for the poor choice of terminology. To truly get the "weaker form" as YCor pointed out, I should replace "valuated field" by "a field $k$ with an absolute value" $|\cdot|: k \to \mathbb{R}\_{\ge 0}$, such that (1) $|ab| = |a| |b|$; (2) $|a| = 0$ if and only if $a=0$; (3) $|1 + a| \le 2$ for all $a$ with $|a| \le 1$ (or equivalently by a well-known argument, the triangle inequality). To exclude the discrete topology induced by this absolute value, we exclude the trivial absolute value that $|a| = 1$ for all $a \in k^\times$.



Examples of locally compact groups that do not admit enough finite dimensional representations
There is an example which satisfies something much stronger: there exist nontrivial groups $G$ such that *any* homomorphism (not even necessarily continuous) $\pi:G\to GL\_n(k)$ for *any* field $k$ (and, in fact, any commutative integral domain) is trivial, so in particular for *any* $a,b\in G$ we have $\pi(a)=\pi(b)$. Since any group can be given a locally compact Hausdorff topology, namely the discrete topology, these will in particular answer your question.


As for examples of such groups $G$, we can take any finitely generated which has no finite quotients, e.g. the [Higman group](https://en.wikipedia.org/wiki/Higman_group) mentioned in the comment by Terry Tao.


This result is apparently due to Mal'cev, but I don't have a reference at hand so here is a sketch of an argument. It is enough to show that if you have a finitely generated subgroup $\Gamma$ of a group $GL\_n(R)$ for some integral domain $R$ (e.g. the image of $G$ under a representation), then $\Gamma$ is [residually finite](https://en.wikipedia.org/wiki/Residually_finite_group) (hence trivial, if $\Gamma$ is a quotient of $G$ which has no finite quotients). The main idea of the proof is to replace $R$ by some ring which is finitely generated over $\mathbb Z$, meaning it is a quotient of a polynomial algebra over $\mathbb Z$. Then $R$ is a [Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring), so in particular the intersection of all maximal ideals of $R$ is trivial, and moreover for all maximal ideals $m$ of $R$ we have $R/m$ finite. Now if $\gamma\in\Gamma$ is any nontrivial element, then there is a maximal ideal $m$ not containing all coefficients of $\gamma$, so $\gamma$ has nontrivial image in the finite group $GL\_n(R/m)$.


