3
$\begingroup$

Let $G$ be a Lie group, $\operatorname{Aut}(G)$ the group of diffeomorphisms of $G$ that are also homomorphisms. Denote by $\operatorname{Inn}(G) \unlhd \operatorname{Aut}(G)$ the group of automorphisms which correspond to conjugation by some $h \in G$. Define $$\operatorname{Out}(G) := \operatorname{Aut}(G) / \operatorname{Inn}(G).$$ I would like to show that $$\operatorname{Out}(\operatorname{SL}_3(\mathbb{R}))\cong \mathbb{Z}/2 \mathbb{Z}.$$ I have already seen computations for different Lie groups, namely $\operatorname{SL}_2(\mathbb{C})$ and $\operatorname{SL}_2(\mathbb{R})$. The common idea seems to be to use the fact that if $\Phi: G \to G$ is an automorphism, then its differential at the identity $d_e \Phi: \mathfrak{g} \to \mathfrak{g}$ is an automorphism as well and then to use the structure on the corresponding Lie algebra.

I hardly know anything about the Lie algebra $\mathfrak{sl}_3(\mathbb{R})$. What would be a good approach to show the claim above? Can it be done in a similar way or is there an easier approach?

$\endgroup$
  • $\begingroup$ The non-trivial outer automorphism corresponds to a graph automorphism of the Dynkin diagram of the root system. At the level of matrices the automorphism is simply $g\mapsto (g^{-1})^T$. Taking the inverse and taking the transpose are both antiautomorphisms, so... $\endgroup$ – Jyrki Lahtonen Jan 8 '16 at 12:06
  • $\begingroup$ @JyrkiLahtonen: I haven't studied Dynkin diagrams or root systems yet. Can you maybe recommend a reference for a novice? At this point, I don't really understand anything about your comment. :-( $\endgroup$ – user302234 Jan 8 '16 at 12:08
3
$\begingroup$

For the Lie algebra $\mathbb{sl}_n(\mathbb{C})$ the inner automorphism group is isomorphic to $PGL_n(\mathbb{C})$, and the outer automorphism group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ for $n\ge 3$. This is proved in Jacobson's book on Lie algebras, chapter $IX$. Indeed, the outer automorphism different from the identity is simply given by $X\mapsto -X^T$. The result is proved for an arbitrary algebraically closed field of characteristic zero.

$\endgroup$
  • $\begingroup$ Thank you for your answer. Do you possibly know a reference publicly available or one that I can access via SpringerLink? $\endgroup$ – user302234 Jan 7 '16 at 14:38
  • $\begingroup$ Jacobson's book is available here; otherwise check the references given here. $\endgroup$ – Dietrich Burde Jan 7 '16 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.