Let V and W be subalgebras of a Lie algebra $\mathcal{L}$. I want to show that $[V,W]$ is not always a subalgebra of $\mathcal{L}$.
Lie product of a two subalgebras
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0Yes I saw that thanks! – 2012-11-27
1 Answers
Consider the Lie algebra $\mathcal L=\{X\in M_4(\mathbb R)\,|\,X^T=-X\}$ with the product $[X,Y]=XY-YX$. A basis of this algebra is given by the matrices $u_{i,j}=e_{i,j}-e_{j,i}$ for $i
, where the matrix $e_{i,j}$ has all its coefficients equal to zero but the $(i,j)$-th coefficient which is one. As an example $u_{1,2}=\left( \begin{array}{cccc} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right)\,.$ Note the formulae, for $i,j,k$ all different, $[u_{i,j},u_{j,k}]=-u_{i,k}$ and for $i,j,k,l$ all different $[u_{i,j},u_{k,l}]=0$ allow us to compute easily with those matrices. Define the subalgebras $V=Span(u_{1,2},u_{1,3},u_{2,3})\quad\text{and}\quad W=Span(u_{2,3},u_{2,4},u_{3,4})\,.$ Then $[V,W]=Span(u_{1,2},u_{1,3},u_{1,4,},u_{2,4},u_{3,4})$ Note that $[u_{1,3},u_{1,2}]=u_{2,3}$ and thus $[V,W]$ is not a Lie algebra, since $u_{2,3}\notin[V,W]$.
Well I am not a big fan of my own example so if anyone had anything more simple or more elegant, I would be interested too.
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1Thanks a lot, I don't think there is any elegant way to prove it... :) – 2012-11-27