Consider the left-regular representation $\lambda : G \to B(L^2(G))$, $\lambda\_g f(h) = f(g^{-1}h)$, for a locally compact group.


It is well-known that this is a unitary faithful and strongly-continuous representation, but is it also a homeomorphism onto its image $\lambda(G)$ (equipped with the strong-operator topology?


It would suffice to show that for any net $g\_\alpha$, convergence in the strong operator topology of $\lambda\_{g\_\alpha}$ to the identity $I\_{L^2(G)}$ implies convergence of $g\_\alpha$ to the neutral element $e$ in $G$.



Is the left-regular representation of a locally compact group a homeomorphism onto its image?
Yes. It's more generally true for every faithful $C^0$ unitary representation $\pi$ of $G$. (Recall that a unitary representation $\pi$ is $C^0$ if for all $v,w$ in the Hilbert space, one has $\langle \pi(g)v,w\rangle\to 0$ when $g$ leaves compact subsets of $G$.)


Indeed if this is not a homeomorphism onto its image, there exists an ultrafilter $\eta$ on $G$, not converging to $1$, such that $\lim\_{g\to\eta}\lambda\_g=\mathrm{id}$ (for the strong topology, i.e., $\lim\_{g\to\eta}\pi(g)v=v$ for every $v$ in the Hilbert space.


If $\eta$ is unbounded (i.e. no compact subset of $G$ is in $\eta$), then we get a contradiction, since the $C^0$ property implies $\lim\_{g\to\eta}\langle\pi(g)v,v\rangle=0$ for every $v$ in the Hilbert space.


If $\eta$ is bounded, then $\eta$ has a limit $g\_0$ (which by assumption is not $1\_G$), and we deduce $\lim\_{g\to\eta}\pi(g)=\pi(g\_0)$. So $\pi(g\_0)=\mathrm{id}$, contradicting the faithfulness assumption.


