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I know there is a thread discussing this here but I have a problem in a different step. So let $L\subset \mathfrak{gl}(V)$ be a solvable Lie algebra over an algebraically closed field of characteristic 0. As the thread above discusses, one can follow that $\mathrm{tr}_V(ad(x)ad(y))=0$ for all $x\in [L,L]$ and $y\in L.$ So far so good. But Cartan's theorem asserts that $\mathrm{tr}_V(xy)=0$ for all $x\in [L,L]$ and $y\in L$ which is a different statement. How do I arrive there?

Edit: to clear up confusion, here is the version of Cartan's criterion we proved: enter image description here

Of course the usual version of Cartan's criterion follows (see Deif's comment) , by considering the image of L under the adjoint representation. ad(L) is solvable if $tr_V(ad(x)ad(y))=0$ is zero for $x\in [L,L]$ and $y\in L$ . But then considering the exact sequence $$ 0\rightarrow Z(L)\rightarrow L \rightarrow ad(L)\rightarrow 0$$ one sees that L is solvable.

Callum
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Adronic
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Since your Lie algebra $L$ is solvable, by Lie's theorem we may assume it is upper triangular (in some basis). Then $[L,L]$ is strictly upper triangular and so $xy$ is strictly upper triangular for $x \in [L,L], y \in L$ and the trace of strictly upper triangular (thus nilpotent) endomorphisms is $0$.

Indeed this is exactly the proof in the post you linked to. It is only in terms of $\mathrm{ad}$ there because that is how that question was asked.

Callum
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