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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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    In general, if $f$ is a function, $f^3=f\circ f\circ f$. In particular, you have $\mbox{ad } e(y)=[e[e[e,y]]]$.2012-04-03
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    I see you haven't commented on my answer. If you need more clarifications, just tell ;)2012-04-17

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