0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.