0

How do you compute the group of inner automorphisms of a given Lie algebra?

It seems as a dumb question to me, but I wasn't able to find the answer anywhere. I know that the group of inner automorphisms of Lie algebra $L$ is generated by elements $\exp({\rm ad}_x)$ for $x\in L$.

Is there some condition for the Lie algebra assuring that every inner automorphism is of the form $\exp({\rm ad}_x)$ (i.e. this set is closed w.r.t. multiplication)? Is there some condition for the Lie algebra assuring that every inner automorphism is of the form $\prod\exp(t_i{\rm ad}_{e_i})$, where $(e_i)$ is the basis (because this leads to much simpler results than $\exp(\sum t_i{\rm ad}_{e_i})$)?

glS
  • 6,818
Daniel
  • 448
  • The group ${\rm Int}(L)$ of inner automorphisms of $L$ is generated by elements of the form $\exp(ad(x))$ with $x\in L$ such that $ad(x)$ is nilpotent, and not by all elements $\exp(ad(x))$. – Dietrich Burde Mar 10 '17 at 12:16
  • @DietrichBurde , why nilpotent only? For any $x\in L$, we have $\exp({\rm ad}x)={\rm Ad}{{\rm e}^x}$, so it is an inner automorphism. – Daniel Mar 10 '17 at 12:40
  • Why nilpotent only? Because this is the definition. If $ad(x)$ is nilpotent, the exponential series $\exp(ad(x))$ is finite. – Dietrich Burde Mar 10 '17 at 12:44
  • @DietrichBurde, well, that is maybe one of possible definitions. For example, here they define inner automorphism for general elements, not only the nilponent. Another possible definition of inner automorphism was mentioned by fulges, which is obviously equivalent to my original definition (as mentioned in the linked book). But maybe it holds that your definition is also equivalent, i.e. the nilpotent elements generate the non-nilpotent ones. I don't know and I would like to learn more about those relationships. – Daniel Mar 10 '17 at 12:58
  • Well, the definitions cannot be equivalent (check e.g. the simplest case $[e_1,e_2]=e_1$) and I am indeed interested in the definition that I have presented. – Daniel Mar 10 '17 at 13:17

1 Answers1

1

I am not sure about the technicalities, nor if this is an answer to your question, but the general idea should be the following (at least over $\mathbb{C}$).

For a Lie algebra $\mathfrak{g}$ there is a unique connected, simply connected, Lie group $\tilde{G}$ such that $Lie(\tilde{G}) = T_{id} \tilde{G} = \mathfrak{g}$ (the tangent space at the identity element).

The action by conjugation of $\tilde{G}$ on itself induces an action on $\mathfrak{g}$ and by definition an automorphism of $\mathfrak{g}$ is inner if and only if it is realized by this action.

The kernel of this action is the center of $\tilde{G}$, therefore the group of inner automorphisms is $ad(\tilde{G}) := \tilde{G}/Z(\tilde{G})$ (the adjoint form of $\tilde{G}$).

fulges
  • 915