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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?

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    $\begingroup$ The typical example is that a group with fundamental group $\mathbb Z$ has a central extension by $\hat{\mathbb Z}$. eg the inverse limit of all finite covers of $S^1$ is an extension of $S^1$ by $\hat{\mathbb Z}$. Another description of it is $(\hat{\mathbb Z}\times \mathbb R)/\mathbb Z$. $\endgroup$
    – Ben Wieland
    Commented Aug 14, 2019 at 19:13
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    $\begingroup$ In general if $K$ is a compact group, $S_K=[K^0,K^0]$ is a projective limit of compact semisimple group, essentially a product thereof, and has the property that $\mathrm{Aut}(S_K)^0$ consists of inner automorphisms. It follows that if $G$ is a connected locally compact group and $K$ a compact normal subgroup, then $G=HS_K$, where $H$ is the centralizer of $S_K$. Hence (modulo a central kernel) this essentially reduces to understanding the case when $K^0$ is abelian. In turn, if $K$ is a compact abelian group then its automorphism group is connected, so $K^0$, if abelian, is central. $\endgroup$
    – YCor
    Commented Aug 14, 2019 at 20:26
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    $\begingroup$ And in turn, $K/K^0$ being profinite has a totally disconnected automorphism group and hence is central in $G/K^0$. $\endgroup$
    – YCor
    Commented Aug 14, 2019 at 20:26

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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.

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