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 $\exp(\sum t_i{\rm ad}_{e_i})$)?
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 $\tilde{G}$).