My imperfect understanding is that, by the work of various authors (Gleason, Yamabe, Montgomery, Zippin ...), the following result is known:


Let $G$ be a connected, locally compact, Hausdorff group, and let $U$ be an open neighbourhood of the identity in $G$. Then $G$ has a compact normal subgroup K satisfying $K\subseteq U$, such that the quotient $L=G/K$ is a (connected) Lie group.


I'm aware this isn't the sharpest version, which relates to what I am about to ask.


Suppose we are given a compact group $K$ and a connected Lie group $L$. Has there been any work to investigate what possible extensions $K\to G \to L$ can arise, with $G$ connected and all homomorphisms being continuous with closed range? What if we require $K$ to be "small" in $G$ in the sense of the Lie approximation theorem above?


I'd be happy with suggestions of places to look in the literature.


A related question, which might have an easy answer (I am not very experienced in the structure theory of locally compact groups):


What is a good example to have in mind, of a connected l.c.h. group $G$ that does not admit any semidirect product decomposition $G\cong K \rtimes L$ where $K$ is compact and $L$ is Lie?



Structure of extensions arising in Lie approximation of connected groups
For $G$ a topological group denote $G^\circ$ its unit component.


Say that a topological group $K$ is compact-semisimple if it is compact, connected and has a dense commutator subgroup. It actually follows that $K$ is a perfect group (abstractly), and that $K$ is quotient of a (possibly infinite) product of simple compact Lie groups by a central subgroup. Moreover, for every compact connected group $K$, the commutator subgroup $[K,K]$ is closed and compact-semisimple, and moreover $K=[K,K]Z(K)$, $Z(K)$ being its center. In addition, if $K$ is compact-semisimple, then $\mathrm{Aut}(K)^\circ$ consists of inner automorphisms.


Lemma: let $K$ be a compact group with $K^\circ$ abelian. Then $\mathrm{Aut}(K)$ is totally disconnected. (The topology being the compact-open, hence, in the compact case, the topology of uniform convergence.)


Proof: if $K$ is profinite this is clear. If $K$ is abelian, then its Pontryagin dual being discrete, this is clear again. Hence, in general, $\mathrm{Aut}(K)^\circ$ acts trivially on both $K^\circ$ and $K/K^\circ$, hence each $\beta\in\mathrm{Aut}(K)^\circ$ acts as $g\mapsto gu\_\beta(g)$, with $u\_\beta$ some homomorphism $K\to K^\circ$ trivial on $K^\circ$; $\beta\mapsto u\_\beta$ is continuous. If $L$ is the Hausdorff abelianization of $K/K^\circ$, then $\mathrm{Hom}(K/K^\circ,K^\circ)=\mathrm{Hom}(L,K^\circ)$. Again by Pontryagin duality, $\mathrm{Hom}(L,K^\circ)$ is totally disconnected. Hence by connectedness $\beta\mapsto u\_\beta$ is the constant $0$. So $\mathrm{Aut}(K)^\circ=\{\mathrm{id}\}$. $\Box$


For $G$ connected locally compact group, let $W$ be a compact normal subgroup (there's a maximal one). Write $S\_W=S\_W(G)=[W^\circ,W^\circ]$, and let $H\_W$ be its centralizer, and $W'=W\cap H\_W$. Hence we can apply the lemma to $W/S\_W$, and thus by connectedness of $G$, we deduce that $W/S\_W$ is central in $G$. Since $S\_{W'}(H)=1$), we have $W'=W\cap H\_W$ central in $H\_W$.


Also, since $S\_W$ is compact-semisimple, by connectedness of $G$, the conjugation action is by inner automorphisms, which implies that $G=S\_WH\_W$. The intersection $S\_WH\_W$ is a central profinite subgroup of $G$.



> 
> Hence a recipe to produce all connected locally compact groups is: consider a compact-semisimple group (as above, produced as a product modulo some central subgroup), consider a connected locally compact group that is central-by-Lie, and mod out by a diagonal central profinite subgroup intersecting trivially each of the two direct factors.
> 
> 
> 


Essentially, this reduces everything to an understanding of central extensions by compact kernels.


