$ S_4 $ subgroups and $ \operatorname{SO}_3(\mathbb{R}) $

Question

$\DeclareMathOperator\SO{SO}$I posted this on MSE 10 days ago and it got 3 upvotes but no answers or comments, so I'm cross-posting to MO.

Background: The group of rotations $ \SO_3(\mathbb{R}) $ has a 24 element subgroup of integer points $ \SO_3(\mathbb{Z}) $ which is isomorphic to the symmetric group $ S_4 $.

Original Question: (answered almost immediately by LSpice with a great subgroup of $ \SO_5(\mathbb{R}) $)

Let $ G $ be a connected Lie group. For every injective group homomorphism $ \phi:\SO_3(\mathbb{Z}) \to G $ does there exist a unique extension $ \tilde{\phi} : \SO_3(\mathbb{R}) \to G $?

New Question:

Let $ G $ be a connected Lie group. Let $ \phi:\SO_3(\mathbb{Z}) \to G $ be a group homomorphism whose image is a maximal closed subgroup of $ G $. Does there exist a unique extension $ \tilde{\phi} : \SO_3(\mathbb{R}) \to G $?

Note that the only quotients of $ S_4 $ are $ S_4 $, cyclic 2 group trivial group and the six element dihedral group $ D_6 $. Of these only $ S_4 $ itself is a maximal closed subgroup of $ \SO_3(\mathbb{R}) $.

For $ G= \SO_3(\mathbb{R}) $ the answer is yes to both questions. Here is a proof:

First we show existence of an extension. Define $ \Gamma:=\SO_3(\mathbb{Z}) $ and $ \Gamma':=\phi(\SO_3(\mathbb{Z})) $. Isomorphic finite subgroups of $ \SO_3(\mathbb{R}) $ are conjugate. So there always exists some $ g \in G $ such that $ g \Gamma g^{-1}=\Gamma' $. So the map conjugation by $ g $ agrees with $ \phi $ up to some automorphism. From there we can conjugate by some $ x \in \Gamma $ to get an exact match (recall every automorphism of the symmetric group $ S_4 $ is inner). So conjugation by $ xg $ is an automorphism of $ G $ extending $ \phi $.

Now we show uniqueness of extensions. Any two extensions $ \phi_1 $ and $ \phi_2 $ of $ \phi $ are automorphisms of $ \SO_3(\mathbb{R}) $. Every automorphism of $ G $ is inner, so the extensions $ \phi_1$, $\phi_2 $ correspond to conjugation by $ g_1$, $g_2 $. Thus conjugation by $ g_2^{-1}g_1 $ fixes $ \Gamma $ pointwise. In other words $ g_2^{-1}g_1 $ centralizes $ \Gamma $. So in particular $ g_2^{-1}g_1 \in N_G(\Gamma) $. But $ \Gamma $ is self-normalizing $ N_G(\Gamma)=\Gamma $. So really $ g_2^{-1}g_1 \in \Gamma $. However $ \Gamma $ also has trivial center $ Z(\Gamma)=1 $. So we must have $ g_1=g_2 $ and thus uniqueness is proven.

EDIT: As Lspice points out the answer to the first question is no there does not always exist an extension and for the second question Lspice notes that if such an extension exists then $ G $ must be $ SO_3(\mathbb{R}) $.

First we show uniqueness. If $ \phi $ is trivial there is nothing to prove. If $ \phi $ is trivial and there exists an extension $ \tilde{\phi} $ then $ \tilde{\phi} $ must be injective since $ \SO_3(\mathbb{R}) $ is simple so since $ \Gamma':=\phi(\SO_3(\mathbb{Z})) $ is maximal closed and contained in $ \tilde{\phi}(\SO_3(\mathbb{R}))\cong \SO_3(\mathbb{R})$ we can conclude that $ G=\tilde{\phi}(\SO_3(\mathbb{R})) $.

Now we show existence. Suppose that $ \Gamma' $ is a maximal closed subgroup. Then $ G $ must be compact since $ \Gamma' $ is contained in a maximal compact. Also $ \Gamma' $ must contain the whole center otherwise it is not maximal closed (and $ G $ cannot be abelian because then it contains no maximal closed subgroups). So $ G $ must be semisimple since $ \Gamma' $ is finite. In fact I claim $ G $ must be simple. Otherwise we can pass to the universal cover and say $ G_1\times G_2 $ are two different semi simple factors with the lift of $ \Gamma' $ intersecting $ G_1 $ non trivially. Then the direct product of $ G_2 $ with the projection onto $ G_1 $ of lift of $ \Gamma' $ is a closed subgroup. So descending back to $ G $ we have that $ \Gamma' $ is not maximal closed, a contradiction. This is a general proof that a Lie group with a finite maximal closed subgroup must be simple and compact. For here we specialize to the case of $ S_4 $ subgroup. The only quotients are $ 1, C_2,D_6, S_4 $. It is fairly clear that of these only $ S_4 $ can be maximal closed in a Lie group since only an embedded $ S_4 $ can act irreducibly with respect to the adjoint representation of a compact simple Lie group. Moreover, the only compact simple Lie group for which this is true. Thus any connected Lie group containing a quotient of $ S_4 $ as a maximal closed subgroup must just be $ SO_3(\mathbb{R}) $.

Answer 1

Consider the group of matrices of the form $w \oplus \det(w)^{-1}$, with $w$ a $4\times4$ permutation matrix. This is an $\operatorname S_4$ inside $\operatorname{SL}_5(\mathbb R) \subseteq \operatorname{SL}_5(\mathbb C)$, and $\mathbb C^5$ decomposes as a sum of the trivial, the sign, and the reflection representation of $\operatorname S_4$. In particular, the space of fixed vectors for $\operatorname S_4$ in $\mathbb C^5$ is $1$-dimensional.

If this $\operatorname S_4$ were contained in an $\operatorname{SO}_3(\mathbb R)$ in $\operatorname{SL}_5(\mathbb R)$, or even in $\operatorname{SL}_5(\mathbb C)$, then the resulting (complex) $5$-dimensional representation of $\operatorname{SO}_3(\mathbb R)$ would be $0^5$, $2 + 0^2$, or $4$.

Since $\sum_{g \in \operatorname{SO}_3(\mathbb Z)} \chi_4(g) = 1(5) + 6(1) + 3(1) + 8(-1) + 6(-1)$ equals $0$, there are no fixed vectors for $\operatorname{SO}_3(\mathbb Z)$ in $4$. Clearly, there is at least a $2$-dimensional space of fixed vectors for $\operatorname{SO}_3(\mathbb Z)$ in $2 + 0^2$ and in $0^5$.