Adjoint endomorphism on $\mathfrak{sl}(2,k)$

Question

Let $\{e,h,f\}$ be the standard basis of the Lie algebra $\mathfrak{sl}(2,k)$. Prove that $(\mbox{ad }e)^3=0$

http://en.wikipedia.org/wiki/Special_linear_Lie_algebra

First I computed $(\mbox{ad }e)(y)$. Let $y=ah+be+cf$, then $(\mbox{ad }e)(y)=[e,y]=ch-ae$. However, I don't really know what $(\mbox{ad }e)^3=0$ means in notation as we only defined $\mbox{ad}$ is not to the power of three.

So anyone got any ideas on what it means and how you actually prove it?

In general, if $f$ is a function, $f^3=f\circ f\circ f$. In particular, you have $\mbox{ad } e(y)=[e[e[e,y]]]$. — M Turgeon, Apr 03 '12 at 17:06
I see you haven't commented on my answer. If you need more clarifications, just tell ;) — M Turgeon, Apr 17 '12 at 15:48

score 0 · Answer 1 · answered Apr 03 '12 at 17:15

Recall the following relations : $[e,f]=h$, $[h,f]=-2f$, and $[h,e]=2e$. Given an element $y=ah+be+df$ of your Lie algebra $\mathfrak{sl}(2,k)$, you first have $$[e,y]=[e,ah+be+df]=a[e,h]+b[e,e]+c[e,f]=-2ae+ch.$$ Continuing on, we have $$[e,[e,y]]=-2a[e,e]+c[e,h]=-2ce.$$ Applying $\mbox{ad }e$ once more gives $$[e,[e,[e,y]]]=[e,-2ce]=0.$$ Therefore, we conclude that $(\mbox{ad }e)^3=0$ (since $y$ was an arbitrary element).

Adjoint endomorphism on $\mathfrak{sl}(2,k)$

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