Question

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.

Answer 1

$SU(n)$ for $n>2$ has complex fundamental representations. Complex conjugation is an automorphism which exchanges the fundamental representation with its complex conjugate, hence it cannot be an inner automorphism.

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Upon further reflection (no pun intended), I think that this is all: basically for simply connected simple Lie groups, the outer automorphisms come from the automorphisms of the Dynkin diagram and for $SU(n)$, $n>2$, the only automorphism is reflection along the midpoint of the diagram. This sends the module with highest weight $(1,0,...,0)$ to $(0,...,0,1)$, hence the fundamental representation to its complex conjugate.

Answer 2

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(y), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective [edit: no it isn't, as pointed out by Theo, but it is locally] and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism [edit: this only works for $g$ sufficiently close to the unit; in general write $y$ as a product of the exponentials and apply this to each factor]). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Answer 3

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?