A Lie algebra $L$ is said to be reductive if for any ideal $\mathfrak{a}$ of $L$, there is an ideal $\mathfrak{b}$ of $L$ such that $L=\mathfrak{a}\oplus\mathfrak{b}$.

It is known that a reductive Lie algebra decomposes as $L = L'\oplus Z(L)$, where $L'$ is the derived subalgebra of $L$ and $Z(L)$ is the center of $L$. Is the converse true?

The answer is "yes" if $L'$ is semisimple; can we deduce that $L'$ is semisimple just from the decomposition $L = L'\oplus Z(L)$?

  • $\begingroup$ Could you provide a reference for defining reductive like this? The definition I am used to is being the direct sum of the radical and the center. $\endgroup$ – Tobias Kildetoft Jan 30 '15 at 14:10
  • $\begingroup$ @TobiasKildetoft this is the definition used in Structure and Geometry of Lie Groups, by Hilgert and Neeb, for instance. $\endgroup$ – essay Jan 30 '15 at 17:37

Suppose $L^\prime$ is not semisimple; then it has a solvable ideal $R$ by definition. Then $R+Z(L)$ is also a solvable ideal in $L$, contradicting the fact the biggest solvable ideal of $L$ is its centre (this is an equivalent formulation for reductivity; it is proved in Proposition 5.7.3 in Structure and Geometry of Lie Groups by Hilgert and Neeb, which I saw you referenced).

It's been a while since I studied Lie algebras, so it's not unlikely I made some stupid mistake... You should check me.

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