# Lie algebra semisimple if and only if perfect?

If $$L$$ is a semisimple lie algebra then $$L=[L,L]$$. Is the opposite true?

No. A Lie algebra satisfying that property is called perfect. For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $$L$$ and a nontrivial irreducible representation $$V$$ of $$L$$, 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 $$\text{Rad}(L \times V) = V$$.
(Note (YCor): this is a semidirect product, hence usually denoted $$L\ltimes V$$.)
• Note that the one-dimensional vector space $V$ for which $\rho : L \to \mathfrak{gl}(V)$ is zero is also a simple $L$-module, and in this case $[L \times V, L \times V] = L \subsetneq L \times V$. You need to exclude that case, otherwise it works fine since $[L,V]$ contains a non-zero vector, hence is equal to $V$. Commented Feb 15, 2017 at 6:15
• A remark on this answer, is that every non-semisimple perfect Lie algebra (fin. dim., in char 0) has a quotient of the form you're describing. So, in a sense, these are the minimal counterexamples (well, it's minimal if the representation is faithful, in the sense that in this case $L\ltimes V$ is not semisimple but all its proper quotients are semisimple).