= Adjoint semisimplicity in the symplectic Lie algebra
{title2=$\operatorname{ad}a\text{ diagonalizable}\ \Longleftrightarrow\ a\text{ diagonalizable on }V$}
Let $V$ be a finite-dimensional <symplectic vector space> over an <algebraically closed field> of <characteristic zero>. For $a\in\mathfrak{sp}(V)$, write its <Jordan–Chevalley decomposition> as $a=S+N$. Applying the <symplectic adjoint> to this decomposition and using $a^*=-a$, uniqueness gives $S^*=-S$ and $N^*=-N$. Thus both parts belong to the <symplectic Lie algebra>. The operators $\operatorname{ad}S$ and $\operatorname{ad}N$ commute. The former is <diagonalizable> on $\operatorname{End}(V)$ in a <matrix unit> <basis> for an <eigenbasis> of $S$, and the latter is a <nilpotent linear map>, as follows from
$$
(\operatorname{ad}N)^m(T)=\sum_{j=0}^m(-1)^j\binom mjN^{m-j}TN^j.
$$
The restrictions to the invariant <symplectic Lie algebra> retain these properties. Hence these restrictions form the <Jordan–Chevalley decomposition> of $\operatorname{ad}a$.
If $\operatorname{ad}a$ is <diagonalizable>, $\operatorname{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega(v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega(T_vw,u)+\omega(w,T_vu)$ cancel by alternation. For $v\ne0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A <linear map> preserving every line is scalar: compare its action on two <linearly independent> vectors and on their sum. Since $N$ is a <nilpotent linear map>, that scalar is zero. Thus $a=S$ is <diagonalizable>. Conversely, if $a$ is <diagonalizable> on $V$, its <operator commutator> action is <diagonalizable> on $\operatorname{End}(V)$ and therefore on its invariant <symplectic Lie algebra>. This proves both directions, including $V=0$.
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