Diagonalizabilty of ad(adjoint map)?

Question

let $\mathsf{g}$ be a finite dimensional lie algebra and $\xi\in\mathsf{g}$.

Under which conditions the adjoint map $ad_\xi :\mathsf{g}\longrightarrow \mathsf{g}$ is diagonalizable? what about in infinite dimentional case?

Answer 1

The conditions for a linear map $ad(x)$ to be diagonalizable are the same as for any endomorphism of a (finite)-dimensional vector space. In a given Lie algebra, some adjoint operators are diagonalizable, others not. For $\mathfrak{sl}_2(K)$ for example, with standard basis $x,y,h$, the operator $ad(h)$ is diagonalizable, whereas $ad(x)$ and $ad(y)$ are not.

For a semisimple Lie algebra, there is a unique decomposition of a given element $x$ as $x=s+n$ such that $ad(s)$ is diagonalizable, $ad(n)$ is nilpotent, $[s,n]=0$, and $ad(x)=ad(s)+ad(n)$. This decomposition, the Jordan-Chevalley decomposition also holds for some classes of infinite-dimensional Lie algebras, e.g., see here.