If a Lie algebra L decomposes as a direct sum of its derived subalgebra and its center, is L reductive?

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 $L = L'\oplus Z(L)$?

• 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. Jan 30, 2015 at 14:10
• @TobiasKildetoft this is the definition used in Structure and Geometry of Lie Groups, by Hilgert and Neeb, for instance. Jan 30, 2015 at 17:37

We cannot deduce that. Let $$L$$ be a finite dimensional semisimple Lie algebra and $$V$$ be an irreducible representation of $$L$$ where $$1<\dim V<\infty$$. Define $$L \times V$$ a Lie algebra structure by $$([X, u], [Y, v])=([X,Y], Xv-Yu)$$. Then you can check the followings:

1. $$[L \times V, L \times V]=L \times V$$, i.e. $$L$$ is perfect.
2. $$Z(L \times V)=0$$
3. $$V=\{ 0 \} \times V$$ is an ideal of $$L \times V$$. $$\nexists$$ ideal $$I$$ of $$L \times V$$ such that $$L \times V = V \oplus I$$.

This shows that $$L$$ is not semisimple (as it is not reductive). We can also prove that $$L$$ is not semisimple by calculating its radical.

Claim. $$Rad(L \times V) = V$$.
proof. It suffices to show $$Rad(L \times V) \subset V$$. If $$(X, u) \in Rad(L \times V)$$ with $$X \neq 0$$, we can prove inductively that there exists $$0 \neq X_n \in L$$ such that $$(Y, 0) \subset (Rad(L \times V))^{(n)}$$ which contradicts to the solvability, by the following lemma:

Lemma. Let $$I$$ be a nonzero ideal of a semisimple Lie algebra $$L$$. Then $$[I, I] \neq0$$.
proof. By Weyl's theorem, there exists an ideal $$J$$ of $$L$$ such that $$L=I \oplus J$$. If $$[I, I]=0$$, we have $$L=[L,L]=[I,I]+[J,J] \subset [J,J] \subset J$$ so contradiction occurs.

In fact, you can deduce $$L=[L, L] \oplus Z(L)$$ if $$[L, L]$$ is semisimple. It follows that $$L$$ is reductive if $$[L, L]$$ is semisimple. To prove this, view $$L$$ as a $$[L, L]$$-module. By Weyl's theorem, $$L=[L, L]\oplus M$$ for some $$[L, L]$$-submodule $$M$$ of $$L$$. Since $$[L, L] \cap Z(L)=0$$, it suffices to show the following claim:

Claim. $$M \subset Z(L)$$.
proof. Let $$x \in M$$ be given, and suppose there exists $$y \in L$$ such that $$[x, y] \neq 0$$. Since $$[L, L]$$ is semisimple and $$0 \neq [x, y] \in [L, L]$$, there exists $$z \in [L, L]$$ such that $$[z, [x, y]] \neq 0$$. However, Jacobi identity yields $$[z, [x, y]] = [x, [z, y]] + [y, [x, z]]$$. Here $$[x, [z,y]]=0$$ since $$[z,y] \in [L, L]$$ and $$L=[L, L] \oplus M$$. Similarly, $$[y,[x,z]]=[y,0]=0$$. Hence $$[z, [x,y]]=0$$; contradiction.

• It seems I misunderstood the question, thank you for pointing it out. I deleted my (wrong) answer. Jun 29, 2020 at 15:01