Is adjoint map invertible?

Question

I've already studied the group of automorphisms of a simple lie algebra on a finite field, but according to the definition of an adjoint representation of a Lie algebra, can we claim an adjoint map is invertible? Can we say that an adjoint map is a member of automorphism group of a lie algebra?

Answer 1

This is an older question, but let's see if we can add some clarification. There are two adjoint representations, one of the group and one of the algebra. You have said algebra, but the only context in which your answer has a chance of making sense is in the group. Let's clarify both.

Let $G$ be your Lie group with Lie algebra $\mathfrak g$. If $g\in G$ define the conjugation map $C_g: G \to G$ by $C_g(h) = ghg^{-1}$. One can quickly check that this is an (inner) automorphism of the group. If $e$ is the identity element, then $C_g(e) = e$ meaning that if we differentiate $C_g$ at the identity, we will get a map $ d_e C_g: \mathfrak g \to \mathfrak g$. Moreover, since $C_g$ was an isomorphism (of Lie groups) then $d_e C_g$ will also be an isomorphism (of Lie algebras). For the sake of notation, let $\operatorname{Ad}_g = d_e C_g$, and note that in particular, $\operatorname{Ad}_g \in \operatorname{Aut}(\mathfrak g)$. The adjoint representation of the Lie group is thus the map $$ \operatorname{Ad}: G \to \operatorname{Aut}(\mathfrak g), \quad g \mapsto \operatorname{Ad}_g.$$ The fact that each $\operatorname{Ad}_g \in \operatorname{Aut}(\mathfrak g)$ is the only possible way that an "adjoint map can be invertible." To see why there is no reason to suspect that $\operatorname{Ad}$ is invertible, note that the image of this map is $\operatorname{Ad}(G)$, and the best we can ask for is that $\operatorname{Ad}(G) = (\operatorname{Aut}(\mathfrak g))_0$, the connected component of the identity.

In the case of Lie algebra representations (which is the case you mention in your question), one recognizes that $\operatorname{Ad}$ is in fact a Lie group homomorphism. Once again differentiating at the identity, one gets the adjoint representation of the Lie algebra $$ \operatorname{ad}:=d_e\operatorname{Ad}: \mathfrak g \to \operatorname{End}(\mathfrak g)$$ where we have used the fact that the Lie algebra of the automorphism group is the endomorphism algebra. The image of $\operatorname{ad}$ lives in the derivation algebra which is certainly not all of $\operatorname{End}(\mathfrak g)$. Each individual $\operatorname{ad}(X)$ is also not invertible since it typically has a large kernel.