One can give the cross product as a Lie bracket on $\mathbb{R}^3$ and the matrix commutator to $\mathbb{R}^{n^2}$ ($n \ge 2$). They both give a perfect Lie algebra structure.

However, every Lie algebra of dimension $1$ is abelian, and for $2$-dim we can write any nontrivial Lie bracket as $[x, y]=x$ (if $\{x, y\}$ is a basis) but this Lie algebra is not perfect.

So I'm wondering which vector spaces have a perfect Lie algebra structure. Does every $\mathbb{R}^n$ ($n \ge 3$) have a Lie bracket which makes it into a perfect Lie algebra?

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    $\begingroup$ I don't think $M_n(\mathbb{R})$ is perfect, all commutators have trace zero? $\endgroup$ – Najib Idrissi Nov 24 '16 at 13:28

First, every semisimple Lie algebra is perfect. So in each dimension $n\ge 3$, where we have a semisimple Lie algebra of dimension $n$, we have a Lie bracket which makes $\mathbb{R}^n$ into a perfect Lie algebra. These dimensions start with $n=3,6,8,9,\ldots$. On the other hand, there are also perfect Lie algebras, which are not semisimple. For example take a semisimple Lie algebra $L$ of dimension $n$ and an irreducible representation $V$ of $L$, of dimension $m \ge 2$ and define a bracket on $L \times V$ by $$ [(X,v),(Y,u)] := ([X,Y],Xu-Yv). $$ This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \times V) = V$. The dimension is $n+m$.

However, not all dimensions can occur. For example, there is no perfect Lie algebra in dimension $4$.

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    $\begingroup$ To complete the conclusion, over any field of characteristic zero, the dimensions of perfect Lie algebras are 0,3,5,6,7\dots. In other words, the only dimensions with no perfect Lie algebra are 1,2,4. (Indeed 0 is achieved by the zero algebra, 3 by $\mathfrak{sl}_2$, and $3+k$ with $k\ge 2$ by the semidirect product of $\mathbf{sl}_2$ by a $k$-dimensional representation.) $\endgroup$ – YCor Nov 24 '16 at 17:49

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