Finite maximal closed subgroups of Lie groups

Question

$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\SO{SO}$Let $\G$ be a Lie group.

I am interested in finite maximal closed subgroups of $ G $.

I'm guessing that $ \G $ has a finite maximal closed subgroup if and only if $ \G $ is simple and compact. Does anyone have other examples of finite maximal closed subgroups?

Answer 1

WARNING: This answer has lots of true things in it but is also riddled with false claims, I keep meaning to fix it but don't get around to it. $\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}$There exist Lie groups $ \G $ which are not compact or simple but have finite maximal closed subgroups. For example the five element cyclic subgroup $ C_5 $ of $ \mathbb{R}^2 \rtimes C_5 $ described at my answer to Finite maximal closed subgroups of Lie groups.

These sorts of examples are interesting in their own right but are not really what I was thinking about when I asked the question. So from now on I will confine myself to the case that $ \G $ is connected.

In other words I will consider the statement "A connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ if and only if $ \G $ is compact and simple."

The first implication is true.

Claim 1: Let $ G $ be a connected Lie group. Suppose that $ G $ has a finite subgroup which is maximal among the closed subgroups of $ G $. Then $ G $ must be compact and simple.

Proof:

Call such a finite maximal closed subgroup $ \Gamma $. $ \Gamma $ is finite thus compact. So $ \Gamma $ must be contained in some maximal compact subgroup $ K $ of $ G $. Since $ \Gamma $ is maximal then we have that either $ K=\Gamma $ or $ K=G $. Suppose for the sake of contradiction that $ K=\Gamma $. But according to

[https://mathoverflow.net/questions/140622/are-maximal-compact-subgroups-of-connected-groups-connected]

the maximal compact subgroup $ K $ of a connected group $ G $ is always itself connected. So if $ \Gamma=K $ then $ K $ is connected and finite thus trivial. But the trivial subgroup can never be a maximal closed subgroup in a connected Lie group $ G $. The trivial subgroup of a connected Lie group will always be contained in a closed 1-parameter subgroup, which is either a circle or line. So by maximality of $ \Gamma $ that implies the whole group $ G $ is either a circle or a line. But the trivial subgroup is not a maximal closed subgroup of $ S^1 $ or $ \mathbb{R} $. Thus we have a contradiction. So we can conclude that $ K \neq \Gamma $ and thus $$ K=G $$ that is, $ G $ is compact. Now we will show that $ G $ must be simple. Suppose for the sake of contradiction that $ G $ is not simple. Then $ G $ must have some normal subgroup $ N $ such that $$ 0 < dim(N) < dim(G) $$ . So there is a surjective map $$ \pi: G \to G/N $$ Then $ \pi(\Gamma) $ is a maximal closed subgroup of $ G/N $ ( $ \pi(\Gamma) \neq G/N $ since $ dim(\pi(\Gamma))=0< dim(G/N))$ ). Note that $$ \pi^{-1}[\pi(\Gamma)] $$ is a closed subgroup of $ G $ containing $ \Gamma $. But $ \pi^{-1}[\pi(\Gamma)] $ contains $ N $ so $$ 0 <dim(N) < dim(\pi^{-1}[\pi(\Gamma)]) $$ thus $ \Gamma \neq \pi^{-1}[\pi(\Gamma)] $ so by maximality $$ \pi^{-1}[\pi(\Gamma)]=G $$ But that implies that the image of $ \pi $ is contained in $ \pi (\Gamma) $. However that is impossible because $ \pi $ is surjective and $ \pi(\Gamma) $ is zero dimensional while $$ Im(\pi)=G/N $$ is positive dimensional. Thus if a connected Lie group $ G $ has a finite maximal closed subgroup then we can conclude that $ G $ is both compact and simple.

This significantly narrows down the the possibilities for $ G $ to the well known classification of compact simple Lie groups.

However the reverse implication does not hold: $ \operatorname{SU}_{15} $ is an example of a compact connected simple Lie group with no finite maximal closed subgroups.

To see why this is the case it is important to note that

Claim 2: For a compact connected simple Lie group $ \G $, $ G $ is a finite maximal closed subgroup of $ \G $ if and only if $ G $ is Ad-irreducible and $ G $ is a maximal finite subgroup of $ \G $.

The implication Ad-irred+ max finite implies finite and max closed follows from Corollary 3.5 of Sawicki and Karnas - Universality of single qudit gates. The reverse implication, that finite and max closed implies Ad-irred+ max finite follows from the following argument (no it doesn't the reverse implication is false see the EDIT): Let $ \Gamma $ be finite and maximal closed. SSOC that $ \Gamma $ is Ad-reducible. Then $ \Gamma $ acting by conjugation preserves some proper nonzero Lie subalgebra $ \mathfrak{h} $ (EDIT: THIS IS FALSE, $ \mathfrak{h} $ is a sub vector space but not necessarily a subalgebra, so the entire reverse implication falls apart, in particular I believe that the $ SL(2,8) $ subgroup of the exceptional Lie group $ G_2 $ is an example of a finite maximal closed subgroup which is not Ad-irreducible) . So $ \Gamma $ acting by conjugation preserves some connected nontrivial proper subgroup $ H $ of $ G $ with Lie algebra $ \mathfrak{h} $. That implies that $ \Gamma $ normalizes $ H $. That is, $ \Gamma $ is contained in $ N(H) $. But $ \Gamma $ is finite while $ N(H) $ contains the positive dimensional subgroup $ H $. Thus the closed subgroup $ N(H) $ properly contains $ \Gamma $. But $ \Gamma $ is maximal by assumption so that implies $ N(H)=G $. However $ 0 < dim(H) < dim(G) $ and $ H $ is normal, so that contradicts the fact that $ G $ is simple.

Since a finite subgroup of $ \SU_n $ is Ad-irreducible if and only if it is a unitary 2-design we have

Claim 3: $ G $ is a finite maximal closed subgroup of $ \SU_n $ if and only if $ G $ is a maximal unitary 2-group in $ \SU_n $.

By inspecting Theorem 3 of Bannai, Navarro, Rizo, and Pham Huu Tiep - Unitary $t$-groups one immediately determines that $ \SU_{15} $ has no finite maximal closed subgroups.

Some of the main examples of finite maximal closed subgroups of $ \SU_n $ include the normalizer in $ \SU(p^n) $ of an extra-special group $ p^{2n+1} $. Here $ p $ is an odd prime. There is also a similar construction $ p=2 $. Then there are infinite families of examples relating to the Weil module for $ \operatorname{PSp}_{2n}(3) $ and another family related to $ \operatorname{PSU}_n(2) $. Plus many exceptional cases.