Let $g$ be a lie algebra let $D: g-> g$ and $D$ a derivation on g then $(ad D)^k$ is a commutator its trace is zero. I am wondering how to prove it?I think we can use $ad(DX)=[D,ad(X)]$ this implies $ad(D(ad(DX)))=[D,ad(ad(DX))]$. Afterthat what to do?
2026-04-09 15:59:33.1775750373
Let $g$ be a lie algebra let $D: g-> g$ and $D$ a derivation on g then $(ad D)^k$ is a commutator its trace is zero.
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Just continue like this. Starting with $ad(D(x))=[D,ad(x)]$ we obtain $ad(D)^k(x)=[D, ad(ad(\ldots (ad(D(x))\ldots )]$. This is a commutator, and the trace of every commutator is zero. A formal proof is an induction over $k$.