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?

  • 2
    $\begingroup$ 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 $(\hat{\mathbb Z}\times \mathbb R)/\mathbb Z$. $\endgroup$ Aug 14, 2019 at 19:13
  • 1
    $\begingroup$ 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
    Aug 14, 2019 at 20:26
  • 1
    $\begingroup$ And in turn, $K/K^0$ being profinite has a totally disconnected automorphism group and hence is central in $G/K^0$. $\endgroup$
    – YCor
    Aug 14, 2019 at 20:26

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.