3

This question originates from the comments made by @Torsten Schoeneberg in my previous question.

It is well-known that there is a one-to-one correspondence between the outer automorphisms of a complex semisimple Lie algebra and the diagram automorphisms of its Dynkin diagram.

My question: Is there an analogous one-to-one correspondence between the outer automorphisms of a real semisimple Lie algebra and the diagram automorphisms of its Satake-Tits diagram, or the complexification of its Satake-Tits diagram?

Thank you!

Borromean
  • 641
  • 1
    Well we first have to decide what the outer automorphism group is. Surely we want it to be $Aut(\mathfrak g)/Inn(\mathfrak g)$, then we have to say what counts as INNer automorphisms, and then we can choose between what Bourbaki calls $Aut_e(\mathfrak g)$ or what they call $Aut_0(\mathfrak g)$, cf. https://math.stackexchange.com/a/3180478/96384 and links therein. If we opt for $Aut_0$, I have a hunch (but not checked) that the final answer might be as easy as the diagram automorphisms of the underlying Dynkin diagram. If we opt for just $Aut_e$, the quotient in general will be much bigger. – Torsten Schoeneberg Jun 02 '22 at 20:08
  • @TorstenSchoeneberg Thank you very much! I originally thought that the definition of inner automorphisms of a Lie algebra is the one given in Wikipedia https://en.wikipedia.org/wiki/Inner_automorphism#Lie_algebra_case. After reading the Refs you cited, I realize that there are at least 3 definitions: $Int(\mathfrak{g})$, $Aut_e(\mathfrak{g})$, and $Aut_0(\mathfrak{g})$. Which one is more close to the definition in Wikipedia? – Borromean Jun 04 '22 at 07:32

0 Answers0