# Finite maximal closed subgroups of Lie groups

Cross-posted from MSE

https://math.stackexchange.com/questions/4272017/finite-maximal-closed-subgroups-of-lie-groups

$$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\SO{SO}$$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?

• I'd rather guess a compact Lie group has a finite "almost dense" subgroup iff it's semisimple. It's not hard to check it's a necessary condition. But producing such subgroups can require a case-by-case study (esp. in the non-connected case) and there might be exceptions. For a connected Lie group, it can be shown that being compact is a necessary condition. But not for a virtually connected Lie group: if $C_5$ is cyclic of order 5 acting by rotation on the plane, then $C_5$ is "almost dense" in $C_5\ltimes\mathbf{R}^2$. Also among discrete groups (these are Lie) this happens (Tarski monsters).
– YCor
Oct 11, 2021 at 7:58
• @Ycor I don't think semi simple is sufficient for a compact group to have an almost dense subgroup. Here is my logic. Let G be a compact simple group. Then $G \times G$ is semi simple. However for any finite $\Gamma$ in $G \times G$ then the projection onto the second factor say $\pi_2(\Gamma)$ is finite so $G \times \pi_2(\Gamma)$ is a closed subgroup properly containing $\Gamma$ but is not all of $G$. So $G \times G$ has no almost dense finite subgroups Oct 11, 2021 at 13:51
• Oh, indeed, you're right: if $G$ is a compact Lie group with a finite almost dense subgroup, then the action of $G/G_0$ in the set of simple factors of $G$ is transitive. In particular, if $G$ is connected, then it has to be simple.
– YCor
Oct 11, 2021 at 13:55
• In fact I think in general by passing to a simply connected cover this proves that simplicity is a necessary condition for having an almost dense finite subgroup (assuming connectedness) Oct 11, 2021 at 14:02
• @Ycor also can you explain your example more. I don't really get it. Why did you pick $C_5$ not $C_3$? And you are just thinking of this as a subgroup of isometries of the plane? Oh wait is it because C_2 and C_4 act on the square lattice and C_3 acts on triangular/ hexagonal lattice? Oct 11, 2021 at 14:07

$$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}}$$Question: When does $$\G$$ admit a finite maximal closed subgroup?

Answer : Must be one of the following two cases

1. $$\G$$ is compact and simple
2. $$\G$$ is not compact in which case $$\G$$ cannot be connected and moreover the component group $$\G/\G^\circ$$ does not preserve any nontrivial proper closed subgroup (see comment from YCor about $$C_5 \ltimes \mathbb{R}^2$$).

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: If a connected Lie group $$\G$$ has a finite maximal closed subgroup $$G$$ then $$\G$$ must be compact and simple.

Proof: Let $$\G$$ be a connected Lie group and $$G$$ a finite maximal closed subgroup. Since $$G$$ is finite then $$G$$ is a compact subgroup of $$\G$$ so must be contained in a maximal compact subgroup, call it $$\K$$. But $$G$$ is a maximal closed subgroup thus we must have that $$\K=\G$$ (note that $$\K$$ cannot equal $$G$$ since $$\K$$ is connected (the maximal compact of a connected group is always connected)). So $$\G$$ must be compact. If $$\G$$ is not simple then there exists some morphism $$\pi: \G \to \G_i$$ with positive dimensional kernel (here $$\G_i$$ is basically one of the semisimple factors of $$\G$$). Then $$\pi^{-1}(\pi(G))$$ is a closed positive dimensional subgroup containing $$G$$, contradicting the fact that $$G$$ is a finite maximal closed subgroup. Thus if a connected Lie group $$\G$$ has a finite maximal closed subgroup then we can conclude that $$\G$$ is simple.

However the reverse implication does not hold: $$SU_{15}$$ 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 $$\G$$.

this follows from Corollary 3.5 of Sawicki and Karnas - Universality of single qudit gates.

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 $$p=2$$. These are known as (complex) Clifford groups. Then there are infinite families of examples relating to the Weil module for $$\operatorname{PSp}_{2n}(3)$$ and another family related to $$U_n(2)$$. Plus many exceptional cases.

A similar normalizer construction to the above gives finite maximal closed subgroups of all the $$\operatorname{SO}(2^n)$$ as normalizers of an extra-special group $$2^{2n+1}$$. This is known as the real Clifford group. For details about real and complex Clifford groups see Nebe, Rains, and Sloane - Self-Dual Codes and Invariant Theory.

• Why, if compact, must $G$ be simple or finite? Wouldn't arbitrary finite product of groups of this type also work? Jun 16, 2022 at 20:57
• @Wojowu no see my comment above "I don't think semi simple is sufficient for a compact group to have an almost dense subgroup. Here is my logic..." For example a product like $A_5 \times A_5$ in $SO_3 \times SO_3$ won't be a maximal closed subgroup since it is contained in $A_5 \times SO_3$ (here $A_5$ is the icosahedral subgroup of $SO_3$). Jun 16, 2022 at 21:19
• Ah apologies, I misread the claim, I thought it asked for maximal (closed and finite). In hindsight that makes little sense to state like that since finite groups are always closed... Jun 17, 2022 at 7:39
• @Wojowu No problem! Thanks for your interest. If you want to know some examples of these finite maximal closed groups it turns out that for $SU_n$ they are exactly the finite subgroups of $SU_n$ listed here arxiv.org/abs/1810.02507 Jun 17, 2022 at 11:07
• The integers form a zero dimensional Lie group containing a maximal closed subgroup which is finite, the origin, but the integers are noncompact and they have an infinite group of components. Jun 21, 2022 at 9:46