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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?

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    I see you haven't commented on my answer. If you need more clarifications, just tell ;)2012-04-17

1 Answers 1

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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).