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When $\operatorname{char}F=0$, show that each classical algebra $L=A_l,B_l,C_l$ or $D_l$ is equal to $[LL]$.

Since $L$ is a Lie algebra, we know that if $x,y\in L$, then $[x,y]=xy-yx\in L$. So $[LL]\in L$. We have to show that $L\in [LL]$, i.e. every element $z$ of $L$ is equal to $xy-yx$ for some $x,y\in L$. How might we choose $x,y\in L$ so that $xy-yx=z$?

PJ Miller
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    That's not what you required to show. You are required to show that every element $z$ is a linear combination of commutators. Therefore it is enough to show this for all $z$ belonging to a basis of $L$. For classical algebras the first thing you should do is to calculate $$[e_{ij},e_{k\ell}],$$ where $e_{ij}$ stands for the $n\times n$ matrix with a single $1$ at column $j$, row $i$, and zeros elsewhere. That is only a starting point though. – Jyrki Lahtonen Aug 11 '13 at 06:46

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It should be helpful to know that $L$ is a simple Lie algebra, and that $[LL]$ is a Lie ideal in $L$. This means that there are only two possibilities; either $[LL]=L$ or $[LL]=0$. Why can we exclude the latter?

This answers your question, although it doesn't show how to compute $x$ and $y$ given $z$ as you've asked. Such a constructive proof may be much more difficult.

Jared
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  • Thanks, Jared. As to why we can exclude the latter, $[LL]=0$ would mean $xy=yx$ for all $x,y\in L$, so it suffices to provide a pair of matrices in $L$ that don't commute. – PJ Miller Aug 11 '13 at 06:46
  • Hmm. Judging from the OP's question history, this exercise may have appeared before the simplicity of the classical algebras has been established. Can't tell for sure, and my copy of Humphreys is in my office :-/ – Jyrki Lahtonen Aug 11 '13 at 06:48
  • @JyrkiLahtonen Yes, it does appear before the simplicity of the classical algebras. It would be great if you have another method. Otherwise, it's okay, as I'm just learning by myself. I can sort of "skip ahead" :) – PJ Miller Aug 11 '13 at 06:49
  • @PJMiller: You're right. In general, a Lie algebra satisfying $[LL]=0$ is called abelian, and the classical lie algebras are not abelian. – Jared Aug 11 '13 at 06:54
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    @PJMiller: That other method involves calculations with those simple matrices. It is a useful exercise as it gives a preview of the phenomenon: commutators of vectors from root spaces land in the root space corresponding to the sum of the roots. The previous sentence may mean nothing to you at this point, but just do it! The case $A_\ell$ is easy, and follows from the rule $$[e_{ij},e_{k\ell}]=\delta_{jk}e_{i\ell}-\delta_{\ell i}e_{kj}.$$ The other cases require a bit of case-by-case analysis. – Jyrki Lahtonen Aug 13 '13 at 08:23
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    @JyrkiLahtonen For $A_l$: To get the matrix $e_{ab}$ ($a\ne b$) we use $[e_{ac}, e_{cb}]$. To get the matrix $e_{ii}-e_{i+1,i+1}$ we use $[e_{i,i+1},e_{i+1,i}]$. – PJ Miller Aug 13 '13 at 12:23
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    Well done, @PJMiller! I don't remember how you hanlde the other classical algebras, shouldn't be too difficult. – Jyrki Lahtonen Aug 13 '13 at 13:17