What is an example of compact topological group $ G $ which is abstractly simple ($ G $ has no proper nontrivial normal subgroups) but is not a Lie group?


2 Answers 2


kabenyuk's answer is correct but we don't have to appeal to the Gleason-Yamabe theorem. By the Peter-Weyl theorem a compact (Hausdorff) group $G$ has the property that its finite-dimensional unitary representations separate points; in particular if $G$ is nontrivial it admits a continuous homomorphism $\rho : G \to U(n)$ with nontrivial image. So if $G$ is simple (or even just topologically simple) it is a closed subgroup of $U(n)$ for some $n$ and so must be a Lie group by the closed subgroup theorem.

(Abstractly, pushing this argument a little further shows that compact (Hausdorff) groups are pro-(compact Lie).)


Let $G$ be a compact simple group. Since the connected component $G_0$ of the group $G$ is normal, either $G_0=1$ or $G_0=G$. If $G_0=1$, then $G$ is a totally disconnected group and hence has any small open normal subgroups. So $G$ is finite.

If $G_0=G$, then $G$ is connected and by the Gleason-Yamabe theorem there exists a normal subgroup $H$ such that the factor $G/H$ is a Lie group. Then $G$ is a Lie group.

Thus a compact simple group is either finite or a connected Lie group.

  • 1
    $\begingroup$ Does this argument still work if $ G $ is quasisimple ( i.e. $ G $ is perfect and $ G/Z(G) $ is simple)? $\endgroup$ Sep 23, 2022 at 15:59
  • $\begingroup$ could there be some weird counterexample that is a perfect extension of a compact connected simple Lie group by a central solenoid/protorus? $\endgroup$ Sep 23, 2022 at 16:25
  • 1
    $\begingroup$ Your question from the first comment is very interesting. I will think about it. The counterexample from the second comment I think I saw somewhere. I will try to look for it. I may be wrong, though. $\endgroup$
    – kabenyuk
    Sep 23, 2022 at 17:13
  • 2
    $\begingroup$ It's not due to Gleason-Yamabe for $G$ compact, but to Peter-Weyl 25 years earlier (Edit: Qiaochu Yuan noticed this too). $\endgroup$
    – YCor
    Sep 23, 2022 at 19:27

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