Let $G$ be a locally compact group, and $X$ a Hilbert space. A unitary representation $\varphi:G\to B(X)$ is said to be norm-continuous, if it is continuous with respect to the norm in $B(X)$: $$ x_i\to x\quad\Longrightarrow\quad ||\varphi(x_i)-\varphi(x)||\to 0, $$ My question is the following:
Is it true that each norm-continuous unitary representaion $\varphi:G\to B(X)$ can be factored through a Lie quotient group?
I.e. is there a Lie quotient group $G/H$ such that $$ \varphi=\varphi_H\circ\pi_H, $$ where $\pi_H:G\to G/H$ is the quotient map, and $\varphi_H:G/H\to B(X)$ a new norm-continuous unitary representaion?
An equivalent formulation:
Let $\varphi:G\to B(X)$ be a norm-continuous unitary representaion, and let $$ \operatorname{Ker}\varphi=\{g\in G:\ \varphi(g)=1\} $$ be its kernel. Is the quotient group $G/\operatorname{Ker}\varphi$ always a Lie group?
I thought that this follows from the results on the 5-th Hilbert problem, but when I tried to write the proof I understood that I can prove this only in the case when $G$ is commutative or compact.
I think there must be a trick that I don't know, or, on the contrary, there are some counterexamples. Could you, please, help me?
P.S. I asked this also at MathOverFlow.
P.P.S. Perhaps, the following reformulation will simplify the question:
Can the group $U(X)$ of unitary operators on a Hilbert space $X$ (not necessarily separable) with the norm topology contain a closed subgroup $G$ which is locally compact, but not locally Euclidean (with respect to the topology induced from $U(X)$)?
I am sorry, I have just understood that this is simple, the answer is "yes", see my explanations at MO.