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The outer automorphism group of a topological group $G$ is constructed by the short exact sequence $$ 1\longrightarrow \operatorname{Inn}(G) \longrightarrow \operatorname{Aut}(G) \longrightarrow \operatorname{Out}(G) \longrightarrow 1. $$ This sequence does not always split, see Non-split Aut(G) $\to$ Out(G)?, for example for the discrete group $G = A_6$.

I am interested in the case where $G$ is a compact, connected Lie group. Does the sequence always split in this case? (If $G$ has a simple Lie algebra $\mathfrak{g}$ then I believe the answer is yes.)

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  • $\begingroup$ In the two paragraphs of your question, $\mathrm{Aut}(G)$ seems to have two meanings: automorphism group as group vs automorphism group as topological group. For nontrivial tori the automorphism group as discrete group is much larger. $\endgroup$
    – YCor
    Commented Dec 6, 2020 at 8:00
  • $\begingroup$ @YCor Thanks for pointing that out, I hadn't thought of it. However, I think my entire question can be understood to be about topological groups, and I have rephrased accordingly. $\endgroup$ Commented Dec 6, 2020 at 15:34

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Yes, $\operatorname{Aut}(G) \to \operatorname{Out}(G)$ always splits. The proof is just as in my answer to your question Classification of (not necessarily connected) compact Lie groups: regard $\operatorname{Aut}(G)$ as an extension of $\operatorname{Inn}(G) = G/\operatorname Z(G)$ by a discrete group $\operatorname{Out}(G)$, and lift $\operatorname{Out}(G)$ to $\operatorname{Aut}(G)$ as the automorphisms that preserve a pinning in the sense of that answer. (These are often called "diagram automorphisms".) Over in that other question we did not get an honest section of the component group inside the Lie group because you did not assume that the identity component was centreless, but since the adjoint group $\operatorname{Inn}(G)$ is centreless, everything is fine here.

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  • $\begingroup$ I think that this question has appeared here before, but I can't find it right now. $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 18:56
  • $\begingroup$ So then the sequence splits whether or not $G$ is compact, correct? $\endgroup$ Commented Dec 5, 2020 at 19:23
  • $\begingroup$ For this argument to work, we need some appropriate notion of a pinning. The usual notion is for a complex group, where, instead of considering real rays in root groups, we just consider individual non-$0$ root vectors; and this works just as well for any real Lie group that is quasisplit, i.e., contains a Borel subgroup (where now we require $X_{\overline\alpha} = \overline{X_\alpha}$). Without a Borel subgroup, or some appropriate substitute (such as a maximal torus in a compact group), I don't know how to construct an appropriate lifting (but I don't know that it doesn't exist). $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 19:27
  • $\begingroup$ @anniemarieheart's question points up some of the difficulties that can arise. $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 19:28
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    $\begingroup$ Maybe it would help to start saying the answer is "yes"? $\endgroup$
    – YCor
    Commented Dec 6, 2020 at 15:38

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