I am a bit confused about the connection between ad-diagonal/nilpotent and diagonal/nilpotent property of Lie algebra elements. Suppose $L$ is a complex Lie algebra and $x\in L$. If $L$ is semisimple complex Lie algebra, say $\mathfrak{sl}(2,\mathbb{C})$, then Jordan-Chevalley decomposition says we can always write this as $x=d+n$ where $d$ is diagonal (because $L$ is complex and hence the field is algebraically closed) and $n$ is nilpotent. It can then be shown that this also implies that $\text{ad}(x)=\text{ad}(d)+\text{ad}(n)$.
Should I understand this as saying that $\text{ad}(x)$ is diagonal iff $x$ is diagonal (same for nilpotent case)? If not, what are the conditions that relate $x$ and its adjoint representation $\text{ad}(x)$?
At least it looks to me that for arbitrary complex Lie algebra this is not true: take the identity matrix $I\in L=\mathfrak{gl}(2,\mathbb{C})$, and clearly $\text{ad}(I)\in \text{gl}(L)$ is a nilpotent matrix even though $I$ is not nilpotent matrix.
Best Answer
The Jordan-Chevalley decompositions holds for semisimple Lie algebras. The Lie algebra $\mathfrak{gl}_n(\Bbb C)$ is not semisimple, since it has a non-trivial abelian ideal, its center.
Furthermore, $d$ need not be diagonal. It is only diagonalizable. And indeed, if $x$ is a non-nilpotent endomorphism in a linear Lie algebra, $ad(x)$ can be nilpotent. For $x=I$ in $\mathfrak{gl}_n(\Bbb C)$, we have $ad(I)=0$, the zero endomorphism, because $[I,B]=IB-BI=0$ for all $B\in \mathfrak{gl}_n(\Bbb C)$. An element $x\in L$ is called ad-nilpotent, if $ad(x)$ is nilpotent.
"Should I understand this as saying that $ad(x)$ is nilpotent iff $x$ is nilpotent?" No, this is not true, as you have seen yourself.