Timeline for Lie algebra admitting some hyperbolic automorphism is nilpotent

6 events

when toggle format	what		by	license	comment
Mar 13, 2012 at 2:01	comment	added	Pengfei		I see! Just realized that the identity is valid for all $u$ and $v$. So we can iterate.
Mar 13, 2012 at 0:49	vote	accept	Pengfei
Mar 12, 2012 at 13:45	comment	added	Vladimir Dotsenko		$\gamma$ should be $\mu$, or vice versa - your choice of notation clashed with my notational habits, sorry!
Mar 12, 2012 at 13:39	comment	added	Vladimir Dotsenko		You did all the job already! From $(\phi-\lambda\gamma)[u,v]=[(\phi-\lambda)u,v]+\lambda[u,(\phi-\mu)v]$, you can deduce by induction $(\phi-\lambda\gamma)^L[u,v]=\sum_{i=0}^L\lambda^{L-i}[(\phi-\lambda)^iu,(\phi-\mu)^{L-i}v]$, so taking $L=N+M$ would show that if $u$ and $v$ are generalised eigenvectors, then their bracket is too, with eigenvalue being the product of eigenvalues.
Mar 12, 2012 at 13:24	comment	added	Pengfei		Thank you! I got some feeling of the propositions. But I can only prove the trivial case $N=M=1$ since $(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma v)]$ I do not how to prove '$\mathrm{ad}(x)^K(y)$ is a generalised eigenvector with generalised eigenvalue $\alpha^K\beta$' for general cases.
Mar 12, 2012 at 8:44	history	answered	Vladimir Dotsenko	CC BY-SA 3.0