# Improved classification of compact Lie groups

This question is a follow-up to Classification of (not necessarily connected) compact Lie groups. In the answer to that question, @LSpice proved that any compact, not necessarily connected Lie group $$G$$ takes the form $$G = \frac{G_0 \rtimes R}{P}$$ where $$G_0$$ is the identity component of $$G$$, $$R$$ is a finite group, and $$P$$ is a finite, common subgroup of $$G_0$$ and $$R$$ that is central within $$G_0$$ (but need not be central within $$R$$).

Nonetheless, there are many possibilities for the semi-direct product. To narrow the list, it would be convenient to separate out those elements of $$R$$ that act by non-trivial outer automorphisms on $$G_0$$ and modify the rest so that they commute with $$G_0$$.

UPDATE: my original hypothesis (below) is false. A weaker, possibly correct version is:

Hypothesis: $$R$$ and $$P$$ can be chosen above such that every element of $$R$$ either (1) acts by a non-trivial outer automorphism on $$G_0$$ or (2) acts trivially on $$G_0$$.

UPDATE 2: @LSpice proved this in the updated answer to Classification of (not necessarily connected) compact Lie groups. A concise rephrasing of the proof is given in my answer below.

By comparison, this is false:

Hypothesis: Any compact Lie group $$G$$ can be written in the form $$G = \frac{(G_0 \times H) \rtimes R}{P}$$ where $$H, R, P$$ are finite groups and non-trivial elements of $$R$$ act by non-trivial outer automorphisms on $$G_0$$.

Counterexample: consider $$G = U(1) \rtimes \mathbb{Z}_4$$, where the generator $$r$$ of $$\mathbb{Z}_4$$ acts by the charge conjugation'' outer automorphism $$r^{-1} e^{i \theta} r = e^{-i \theta}$$ on $$U(1)$$. In any finite extension $$G'$$ of this group, elements of $$\pi_0(G)$$ that act by charge conjugation will never square to the identity in $$G'$$, so $$G'$$ never takes the required $$(G\times H) \rtimes \mathbb{Z}_2$$ form with $$\mathbb{Z}_2$$ acting on $$U(1)$$ by charge conjugation.

• I have deleted my wrong answer after you produced a counterexample. Commented Dec 5, 2020 at 19:53
• Do you agree that my answer to your other question also answers the revised version of this question? Commented Dec 6, 2020 at 0:21
• @LSpice Yes. I'm also writing a short alternative answer to this question to explain the proof method I had in mind (very similar to your proof but not identical). Commented Dec 6, 2020 at 6:06

@LSpice has already proven my revised conjecture in the updated answer to Classification of (not necessarily connected) compact Lie groups, but let me give another, closely related proof.

Since $$1\to \mathrm{Inn}(G_0) \to \mathrm{Aut}(G_0) \to \mathrm{Out}(G_0) \to 1$$ always splits, see Does Aut(G) → Out(G) always split for a compact, connected Lie group G?, we can choose a subgroup $$R_0 \subseteq \mathrm{Aut}(G_0)$$ for which the restriction of $$\mathrm{Aut}(G_0) \to \mathrm{Out}(G_0)$$ is an isomorphism. The inverse image of $$R_0$$ under the map $$f:G \to \mathrm{Aut}(G_0)$$ induced by conjugation is a subgroup $$K \subseteq G$$ whose intersection with $$G_0$$ is $$Z(G_0)$$.

Multiplying any $$g\in G$$ by arbitrary $$h \in G_0$$ multiplies the associated $$f(g) \in \mathrm{Aut}(G_0)$$ by an arbitrary inner automorphism $$f(h) \in \mathrm{Inn}(G_0)$$, without changing $$g$$'s connected component. Thus, $$K$$ meets every connected component of $$G$$.

Using the result of In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component?, $$K$$ has a finite subgroup $$R$$ that meets every component of $$K$$, hence it meets every component of $$G$$ as well, and intersects $$G_0$$ within $$Z(G_0)$$. By design, the elements of $$R$$ either act by non-trivial outer automorphisms on $$G_0$$ or they act trivially on $$G_0$$. This proves my (revised) conjecture.

COMMENT ADDED: An interesting, yet false, generalization is stated and disproven below.

It is well known that any compact, connected Lie group $$G_0$$ takes the form $$G_0 = \frac{T^k \times G_1 \times \ldots \times G_\ell}{P}$$ where $$T^k$$ denotes a $$k$$-torus, $$G_1, \ldots, G_\ell$$ are compact, simply connected, simple Lie groups, and $$P$$ is central. One might think that the quotients in the expressions for $$G$$ and $$G_0$$ could be combined, so that any compact Lie group $$G$$ would take the form: $$G = \frac{(T^k \times G_1 \times \ldots \times G_\ell) \rtimes R}{P}$$ where as before each element of $$R$$ acts by a non-trivial outer or acts trivially on $$T^k \times G_1 \times \ldots \times G_\ell$$. However, this is false.

Counterexample: Consider $$G=(\mathrm{SO}(2k) \rtimes \mathbb{Z}_4) / \mathbb{Z}_2$$, where the generator $$r \in \mathbb{Z}_4$$ acts by parity on $$\mathrm{SO}(2k)$$ and $$r^2 = -1 \in SO(2k)$$. Now let $$G’=(\mathrm{Spin}(2k) \rtimes R)/P$$ be a cover of $$G$$ whose connected component is $$G_0'=\mathrm{Spin}(2k)$$. There is some element $$r'$$ of $$R$$ that projects to $$r$$, hence $$r’$$ acts on $$\mathrm{Spin}(2k)$$ by parity. If $$k$$ is odd, then $$Z(G_0') = \mathbb{Z}_4$$, and $$(r’)^2$$ must be one of the two elements of order 4 in $$Z(G_0')$$ to project to $$(r)^2 = -1$$. However, parity exchanges these two elements, so we find $$(r’)^{-1} (r’)^2 r’ \ne (r’)^2$$, which is a contradiction. The case of even $$k$$ is very similar.

• If you use the subgroup $R_0$ from the answer to mathoverflow.net/questions/378218, then your $K$ is $\ker(p)$ from the answer to mathoverflow.net/questions/150949. Commented Dec 6, 2020 at 12:31
• @LSpice Great, so this approach really is very closely related to your original proof, as I suspected. Commented Dec 6, 2020 at 15:37
• Since @LSpice is not going to duplicate the updated proof from mathoverflow.net/questions/378218 here, I've accepted my own answer to indicate that it is correct (and essentially summarizes LSpice's proof in alternate form.) Commented Dec 7, 2020 at 16:07