2
$\begingroup$

Proof that there exists non-zero elements $x, y$ in a solvable Lie algebra $g$ such that $[y,x]=x$. I have seen an answer from a lecture notes on google, but I can't find it now. Anyway, can someone give a more insight proof? Note that you can't use Lie's theorem for solvable Lie algebra $g$ over complex numbers which says a representation of $g$ contains a common eigenvector because our solvable Lie algebra is a general one(over any field say real numbers).

remark: YOU can use the fact that $g$ is solvable if only if it's derived algebra $[g,g]$ is nolptent. (this is ture for real Lie algebras!)

  • 0
    I don't think that this result is true. An abelian Lie algebra is solvable, and all Lie brackets vanish there, so...2011-12-25
  • 0
    You are wright,consider the non-abelian one.2011-12-26

1 Answers 1