Let $ G $ be a compact topological group which is quasisimple in the sense that
$$
[G,G]=G
$$
and
$$
G/Z(G) 
$$
is simple as an abstract group. Must $ G $ be a Lie group?


This is a follow-up question to
<https://math.stackexchange.com/questions/4537401/compact-simple-group-which-is-not-a-lie-group>


By Peter-Weyl theorem there must be a nontrivial continuous homomorphism from $ G/Z(G) $ to some $ U\_n $. Since $ G/Z(G) $ is abstractly simple then this is an embedding and so we realize $ G/Z(G) $ as a compact and thus closed subspace of some $ U\_n $. Thus $ G/Z(G) $ is a closed subgroup of $ U\_n $ so it is a Lie group. So the question is equivalent to: let the compact topological group $ G $ be a perfect central extension of a compact simple Lie group. Must $ G $ be a Lie group?



Is every compact quasisimple group a Lie group?
The group $G/Z(G)$ being simple, is a Lie group by Peter-Weyl. Hence it is either finite or connected. If it is finite, $G$ has center of finite index, hence has a finite derived subgroup. Since $G$ is perfect, this means that $G$ is finite.


Now suppose that $G/Z(G)$ is connected. Let $H$ be any Lie quotient of $G$ that is a cover of $G/Z(G)$. Then the image of $Z(G)$ in $H$ is central, and since the quotient has trivial center, we deduce that this image is precisely $Z(H)$. In particular, we see that $H=Z(H)H^0$. Since $G$ is perfect, so is $H$, and we deduce that $H=H^0$. Thus $H$ is a connected cover of $G/Z(G)$. Since $G/Z(G)$ has a finite fundamental group, say of order $d$, we deduce that $|Z(H)|\le d$.


This concludes: indeed if by contradiction $|Z(G)|>d$, then there exists a Lie quotient $H$ of $G$ in which the image of $Z(G)$ has cardinal $>d$ (by Peter-Weyl) and the above yields a contradiction.


