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Choose a algebraically closed field $k$ with $\text{char}(k)= 0$ and a finite dimensional solvable Lie algebra $L$ over $k.$

As an corollary from Lie's theorem I have proved that there is a flag $(L_i)$ of $L$ such that each $L_i$ is an ideal of $L$. Consider $u\in [L,L]$. I have read a remark that claims $$\text{ad}(u)(L_i)\subset L_{i-1}$$ and $$\text{Tr}_L(\text{ad}(x)\text{ad}(y))=0~\forall x\in [L,L],y\in L.$$

Since this was marked as a remark the proof of this two statements shouldn't be such hard but I don't get along well.

  • I am very confused by the first part of this. Lie's theorem does not hold in non-zero characteristic. But the claim you make follows immediately from the definition of a solvable Lie algebra anyway. – Tobias Kildetoft Apr 28 '17 at 09:13
  • I am very sorry, I just made a typo. The characteristic is asumed to be $0$. How do the two parts follow now? –  Apr 28 '17 at 09:24

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By Lie theorem, $L$ has a basis such that $ad(x)$ is an upper triangular matrix relative to this basis. Hence $$ ad ([x,y]) = ad(x)ady-ad(y)ad(x) $$ is a strictly upper triangular matrix. Then $ad[xy]ad(y)$ is a strictly upper triangular matrix for all $x,y$, hence we have $$tr(ad[xy]ady) = 0$$ for all $x,y$. This shows the second claim. The first one also follows directly from Lie's Theorem.

Dietrich Burde
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  • Got it! Thanky a lot. Do you want to write a line for the first part? I thought Lie theorem gives us $ad(u)L_i\subset L_i$ instead of $ad(u)L_i\subset L_{i-1}$. –  Apr 28 '17 at 16:09
  • Yes, but again we have that $ad(u)$ is strictly upper-triangular, hence $ad(u)(L_i)\subseteq L_{i-1}$. – Dietrich Burde Apr 28 '17 at 16:27