Surjectivity of the map $\operatorname{Ad}:\mathfrak{g}\to \operatorname{Der}(\mathfrak{g})$ if $\mathfrak{g}$ is semisimple

Question

Let $\mathfrak{g}$ be a semisimple Lie algebra over $K$ with $\operatorname{char}(K)=0$ and let $\operatorname{Der}(\mathfrak{g})$ denote the space of all derivations on $\mathfrak{g}$. I showed that the map $\operatorname{Ad}:\mathfrak{g}\to \operatorname{end}(\mathfrak{g})$ is injective and that $\operatorname{Ad}(X)$ is a derivation for all $X\in \mathfrak{g}$. Next, I want to show surjectivity of the map $Ad$. Now I think it suffices to prove that the dimension of $\operatorname{Der}(\mathfrak{g})$ is the dimension of $\mathfrak{g}$. If $\{e_1,..,e_n\}$ is a basis for $\mathfrak{g}$, consider $\{\delta_1,...,\delta_n\}\subseteq$ Der$(\mathfrak{g})$ with $\delta_i(e_j)=[e_i,e_j]$. I want to show that this is a basis for Der$(\mathfrak{g})$. Then each $\delta_i$ is a derivation, I have also shown that they are linearly independant, so Der($\mathfrak{g}$) has dimension $\geq n$. What argument could I use that Der($\mathfrak{g}$) cannot have dimension strictly greater than $n$?

Answer 1

This is a particular case of the Whitehead lemma for the ajoint representation. You can see a proof in Jacobson.

https://en.wikipedia.org/wiki/Whitehead%27s_lemma_(Lie_algebras)