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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!)

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  • $\begingroup$ I don't think that this result is true. An abelian Lie algebra is solvable, and all Lie brackets vanish there, so... $\endgroup$ – Jyrki Lahtonen Dec 25 '11 at 11:36
  • $\begingroup$ You are wright,consider the non-abelian one. $\endgroup$ – hnzt Dec 26 '11 at 1:57
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$\newcommand\ad{\operatorname{ad}}\newcommand\g{\mathfrak g}$Notice that this cannot happen in a nilpotent Lie algebra, so you need to exclude all of them, not just the abelian ones. So let's show that

a non-nilpotent Lie algebra $\g$ contains elements $x$, $y$ such that $[x,y]=y$.

Now, if $\g$ is not nilpotent, Engel's theorem tells us that there is an $x\in\g$ such that the map $\ad(x):y\in\g\mapsto[x,y]\in\g$ is not nilpotent. In particular, there exists an $x\in\g$ such that $0$ is not the only eigenvalue of $\ad(x)$. Fix that $x$, and let $\lambda\in\mathbb C$ be a non-zero eigenvalue of $\ad(x)$ and let $y\in g$ be an eigenvector of $\ad(x)$ for the eigenvalue $\lambda$. The pair of elements $x'=\lambda^{-1}x$, $y$ are then such that $[x',y]=y$.

I did not need to assume the algebra is solvable, by the way.

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  • $\begingroup$ That should work over any field, right? $\endgroup$ – Ehsan M. Kermani Dec 26 '11 at 2:41
  • $\begingroup$ I assumed the existence of eigenvalues and eigenvectors, so the field should be algebraically closed. The characteristic is irrelevant, though, for the same is true for Engel's theorem/ $\endgroup$ – Mariano Suárez-Álvarez Dec 26 '11 at 2:42
  • $\begingroup$ Yes, that's right. $\endgroup$ – Ehsan M. Kermani Dec 26 '11 at 2:43
  • $\begingroup$ That's the way we do it in proving Lie's theorem, but it is not true for real Lie algebras. $\endgroup$ – hnzt Dec 26 '11 at 6:44

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