Let $G$ be a Lie group.

I am interested in maximal closed subgroups $ G $ which happen to be finite.

The octohedral group $ S_4 $ and the icosahedral group $ A_5 $ are both maximal closed subgroups of $\SO_3(\mathbb{R})$. Lifting to the cover $\SU_2 $ we have that the binary octohedral and binary icosahedral groups maximal closed subgroups of $\SU(2).$

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

The octahedral group coincides with the first Clifford group $ C_1 $. This generalizes to the Clifford group $ C_n $ being a a maximal closed subgroup of the projective special unitary group $\PSU(2^n) $.

Lifting to the universal cover $\SU(2^n)$ gives an maximal closed subgroup which is often invoked when people in quantum computing say that "adding any unitary operation to the Clifford group gives a universal gate set on $n$ qubits".

Here I am using the convention that elements of $ C_n $ are automorphisms of the Pauli group $ P_n $. Equivalently, $ C_n $ is the image of the normalizer $ N_{\mathrm{U}(2^n)}(P_n) $ under the adjoint representation of $ U(2^n) $.

Take the qudit (qubit is case $d=2$) analogue with qudit Pauli group and qudit Clifford group and repeat and the result is (I think) a finite maximal closed subgroup of $\SU(d^n) $ for any $ d $ prime.

Hopefully, for any $N$ we can take prime factorization of $ N= d_1 \dots d_k $ and then look at a pseudo Pauli group given as tensor product of single qudit Pauli groups for each $d_i$. Then we can take the normalizer in $U_n$ and get a pseudo Clifford group and thus eventually produce an analogous finite maximal closed subgroup of $\SU(N) $ for any $N$.

  1. Are Clifford groups maximal closed subgroups of special unitary groups? Please help with any proofs or references you have.

  2. What are some other examples of maximal closed subgroups that also happen to be finite?

  3. Show me a counter example I would love to see a compact simple group with a proof that it has no maximal closed finite subgroups... Or even better I would love an example of at least one maximal closed finite subgroup in every compact simple group.

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    $\begingroup$ 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). $\endgroup$
    – YCor
    Oct 11 at 7:58
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    $\begingroup$ @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 $\endgroup$ Oct 11 at 13:51
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Oct 11 at 13:55
  • $\begingroup$ 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) $\endgroup$ Oct 11 at 14:02
  • $\begingroup$ @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? $\endgroup$ Oct 11 at 14:07

Ok so here goes an attempt at an answer to my own question. Everything here is either just a guess without proof or is based off of other peoples thoughts. I tried to give proper attribution. If any conclusion here is wrong its my fault. Most of the good ideas in this answer are not my own. I'm still hoping someone more knowledgeable will give a better answer that I can accept instead of this one.

Question 1: When does $ G $ admit a subgroup which is both maximal closed and finite?

Answer 1: Three cases: either (1) $ G $ is compact and simple (see argument from @YCor given in comments which I didn't really understand but hopefully he will explain more) or (2) $ G $ is finite or (3) $ G $ has a finite maximal compact subgroup $ K $ (which is the same size as the component group $ G/G^\circ $) and moreover $ K $ acts on $ G^\circ $ in such a weird way that adding any element of $ G^\circ $ to $ K $ basically generates the whole group. (again see comment from @YCor about $ C_5 \ltimes \mathbb{R}^2 $)

~At this point I will specialize to the case that $ G $ is the real points of a linear algebraic group. I think that rules out the weird case (3) above, also I will ignore the case of $ G $ finite because that is just a finite group theory question~

Question 2: Examples of maximal closed finite subgroups?

Answer 2: Recall that now $ G $ is the group of real points of linear algebraic group and moreover $ G $ is simple and compact. In a simple group, maximal closed= maximal discrete + Zariski dense (see comment by @YCor on Is the group of integer points of a simple real linear algebraic group a maximal closed subgroup? ). Since $ G $ is compact then for subgroups discrete=finite. So it is enough to find Zariski dense maximal finite subgroups of $ G $. To find Zariski dense subgroups just find lots of integral forms for $ G $ and look at the group of integer points $ G_\mathbb{Z} $ for all your different integral forms. If you choose reasonable integral forms of $ G $ then you should get nice big discrete (finite) subgroups that are Zariski dense. If $ G_\mathbb{Z} $ is Zariski dense but not maximal among the finite groups then just pass to a larger group ( perhaps take the normalizer of $ G_\mathbb{Z} $?) and you should get a maximal finite subgroup. For example the normalizer of the tetrahedral group $ \cong A_4 $ in $ SO_3(\mathbb{R}) $ is the octohedral group $ \cong S_4 $, which is maximal closed.

Question 3: Counterexample of some compact simple group with no maximal closed finite subgroups?

Answer: No counterexample exists. $ G $ is the real points of a simple linear algebraic group. Taking some reasonable integer form for $ G $ (like the standard integer form for $ SO_n $) then the discrete (finite) group $ G_\mathbb{Z} $ will be Zariski dense. Then passing to a larger group yields a maximal discrete (finite) group which is still Zariski dense and thus is maximal closed.

TL;DR The group $ G $ of real points of a linear algebraic group has a discrete maximal closed subgroup if and only if it is simple. Maximal closed= maximal discrete+ Zariski dense for simple groups. So to find discrete maximal closed subgroups you go hunting for reasonable integer forms of $ G $ whose group of integer points is Zariski dense in $ G $. If the group of integer points $ G_\mathbb{Z} $ is maximal discrete then it is maximal closed and you are done. Pass to a larger group which is maximal discrete and still Zariski dense and thus maximal closed and you are done.


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