# Compact simple group which is not a Lie group

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?

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.

• Does this argument still work if $G$ is quasisimple ( i.e. $G$ is perfect and $G/Z(G)$ is simple)? Sep 23, 2022 at 15:59
• could there be some weird counterexample that is a perfect extension of a compact connected simple Lie group by a central solenoid/protorus? Sep 23, 2022 at 16:25
• 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. Sep 23, 2022 at 17:13
• It's not due to Gleason-Yamabe for $G$ compact, but to Peter-Weyl 25 years earlier (Edit: Qiaochu Yuan noticed this too).
– YCor
Sep 23, 2022 at 19:27