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Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $ad_y$ is not nilpotent.

Then pick a maximal subalgebra $M$ containing $L_{0,y} := \{x \in L |(ad_y)^n(x) = 0 \space \text{for some} \space n\in \mathbb{N}\}$

Then $M$ is an ideal by assumption. And dim$(L/M) = 1$.

But $y \in L_{0,y} \leq M$ and so $[y,x] \in M \quad \forall x\in L$.

So $ad_y$ acts like $0$ on $L/M$.

$\bf{Question}$ : How can I then deduce that $ad_y$ has eigenvalue $0$ on $L/L_{0,y}$?

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Great question, I'm trying to think of an answer, but I think I only know what to do in the case of a semisimple Lie algebra $L$. In that case, I think you can show that you modded out all nilpotent elements under the adjoint operator and are thus left with only semisimple elements under the adjoint operator. Then it will follow if you can prove that $\text{ad}_{y}(x)$ has eigenvalue 0 on all semisimple elements $x \in L$. –  Samuel Reid May 26 '13 at 5:25
    
What is $y$? It appears all of a sudden without having been introduced. –  Mariano Suárez-Alvarez May 27 '13 at 3:48
    
Hi, sorry about that. $y$ is just an element in $L$. Essentially I am wanting to show that every adjoint map of $L$ is nilpotent. And so by Engel, we could conclude $L$ is nilpotent. I.e I am using a 'proof by contradiction' argument. –  user58514 May 27 '13 at 5:08
    
If $K$ is an ideal, $L/K$ is nilpotent and $ad_{x}|_{K}$ is zero for all $x \in L$, then the Lie algebra $L$ is nilpotent. This is an exercise of Humphreys Lie algebras Page 14. Does it help? –  Vishal May 28 '13 at 3:45
    
Hi, Vishal. If I can show that your statement is true, then of course I would be done. How have you proved this? –  user58514 May 28 '13 at 4:03

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