If Lie algebra is equal to its commutant, is it semisimple?

Question

I am searching for a prove or a counterexample for this statement:

If finite-dimensional complex Lie algebra is equal to its commutant, then it is semisimple.

I suppose it is not true, because otherwise I would be able to find this beautiful result somewhere.

Answer 1

No, it is a strictly weaker property. If Lie algebra is equal to its commutant, it is called perfect. See here for a counterexample.

Answer 2

Nonsemisimple perfect Lie algebras arise when contracting, or degenerating semisimple Lie algebras. A well-known example in physics is that the semisimple Lie algebra $\mathfrak{so}(4,1)$ (the de Sitter Lie algebra) contracts to the Poincaré Lie algebra, which is perfect, but not semisimple. Of course, this Lie algebra is just $\mathbb{R}\rtimes \mathfrak{so}(3,1)$.