4
$\begingroup$

There are at least two ways of defining the inner automorphisms of a real Lie algebra $\mathfrak{g}$. One is the algebraic definition: an inner automorphism is $\exp (\text{ad} X)$, where $X$ is an nilpotent element of $\mathfrak{g}$. The other is the analytic definition: the automorphism group $\text{Aut}\ \mathfrak{g}$ of $\mathfrak{g}$ being a Lie group, has the Lie algebra $\text{Der}\ \mathfrak{g}$ which consists of derivations of $\mathfrak{g}$; then an inner automorphism is an element of the connected Lie subgroup of $\text{Aut}\ \mathfrak{g}$ associated to the Lie subalgebra $\text{ad} \ \mathfrak{g}\subset \text{Der}\ \mathfrak{g}$.

My question is, do these two definitions match for $\mathfrak{g}$ semisimple. Of course, any algebraic inner automorphism is an analytic one. For $\mathfrak{g}$ complex, I believe the question is answered positively and one may use the machinery of root systems to prove it, though I am not sure about the details, which I hope someone could elucidate. Also what about when $\mathfrak{g}$ is real?

$\endgroup$
2
$\begingroup$

For split real semisimple Lie algebras, yes. Otherwise, not necessarily.

Namely, if I'm not mistaken, in the notation of Bourbaki's Groupes et algèbres de Lie, your first ("algebraic") definition is denoted as $Aut_{e}(\mathfrak{g})$ (cf. ch. VII §3 no. 1 definition 1 and ch. VIII §5 no. 2). The second ("analytic") definition is denoted as $Int(\mathfrak{g})$ (cf. ch. III §6 no. 2 definition 2). And they prove that for a semisimple split Lie algebra over $\Bbb R$ or $\Bbb C$, we have $Aut_e(\mathfrak{g}) = Int(\mathfrak{g})$, in ch. VIII, §5 no. 5 proposition 11(iv).

$\mathfrak{g}$ being split here means that there exists a splitting Cartan subalgebra (i.e. one that consists of ad-diagonalisable instead of just ad-semisimple elements). Over $\Bbb C$, that is of course an empty condition, but over $\Bbb R$, it severely restricts the scope of the theorem, as it excludes the plethora of non-split real semisimple Lie algebras which exist.

For those, the theorem is certainly not true in general. As an extreme case, take any compact real Lie algebra, e.g. $\mathfrak{su}_n$. Those contain no nonzero nilpotent elements at all, so that $Aut_e(\mathfrak{su}_n)$ is trivial. However, if I understand this MO post correctly, we have e.g. $Aut(\mathfrak{su}_2) = Int(\mathfrak{su}_2) \simeq SU(2)/\pm I \simeq SO(3)$.


Finally, let me note that I find ch. VIII §5 of that Bourbaki volume one of the best treatises on automorphisms of simple Lie algebras, and to add to the existing ambiguity, it's actually interesting to look at a third possible group they call $Aut_0(\mathfrak g)$, which are the automorphisms which become elementary after scalar extension to an algebraic closure. This is related to the real Lie algebra $\mathfrak{sl}_2(\Bbb R)$ having a "non-inner" automorphism called $\varphi$ in this answer, even though one usually thinks of $\mathfrak{sl}_2$ as not having outer automorphisms. (This one, in this case, is in $Aut(\mathfrak g)= Aut_0(\mathfrak g)$ but not in $Aut_e(\mathfrak g)$.)

$\endgroup$
  • $\begingroup$ Salute to Bourbaki! $\endgroup$ – shrinklemma Apr 9 at 20:27

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.