Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 3 '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

1 Answer 1

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.