TITLE: What is the outer automorphism group of SU(n)?
QUESTION [18 upvotes]: All the automorphisms of $SU(2)$ seem to be inner, which would mean that $\mathrm{Out}$ $SU(2)$ is trivial. Is that correct? Is this true in general $SU(n)$? I can't quite see -- any thoughts would be helpful.

REPLY [3 votes]: I found the previous answers somewhat unsatisfying. If one takes a complex semisimple Lie group (e.g. $G=GL(n,\mathbb{C})$), then $Out(G)$ is huge, since it is an algebraic group over $\mathbb{Q}$, and therefore acted on by $Aut(\mathbb{C}/\mathbb{Q})$, which is huge. In fact, I don't know what the full automorphism group is. The same applies to $\mathbb{H}^{\ast}$, the non-zero quaternions, when viewed as the Cayley-Dickson construction applied to $\mathbb{C}$. However, taking the unit quaternions $\mathbb{H}^1\cong SU(2)$ requires complex conjugation, which must preserve $\mathbb{R}$, and therefore gets rid of all of $Aut(\mathbb{C}/\mathbb{Q})$ except for complex conjugation. However, this argument doesn't exclude other outer automorphisms. So I think one can reduce your question to (modulo the previous answers): when does the real form of a Lie group have only continuous (and therefore analytic) automorphisms?