# non nilpotent Lie algebras

Let $$L$$ be a nonperfect Lie algebra (i.e. $$L \ne [L, L]$$) which is not nilpotent. Is there a ideal of $$L$$ such as $$M$$ such that intersection drived subalgebra of $$L$$ and $$Z(M)$$ is nonzero?

## 1 Answer

Yes, there is such an ideal. Take $$L$$ to be the $$2$$-dimensional nonabelian Lie algebra with basis $$e_1,e_2$$ and Lie bracket $$[e_1,e_2]=e_1$$ and the ideal $$M=\langle e_1\rangle$$. Then $$L$$ is not nilpotent, $$L\neq [L,L]$$ and $$M$$ is an ideal with $$Z(M)=[L,L]\neq 0$$.

In general, however, this need not be true. Take $$L=\mathfrak{gl}_2(\Bbb{C})=\mathfrak{sl}_2(\Bbb{C})\oplus \Bbb{C}$$. Then $$L$$ is not perfect and not nilpotent, but $$Z(M)$$ is either zero or $$\Bbb{C}$$ for all ideals $$M$$, and the intersection with $$[L,L]$$ is zero.

• Thank you for your help. – Afsaneh Dec 18 '18 at 11:35
• Let L be a non nilpotent Lie algebra. Is there a non abelain nilpotent subalgebra of L? – Afsaneh Dec 18 '18 at 11:38
• In general, no. Take $L$ to be the $2$-dimensional nonabelian Lie algebra as above. – Dietrich Burde Dec 18 '18 at 12:18