2
$\begingroup$

Let $L$ be a non-abelian nilpotent Lie algebra and $Z(L)$ be its center. Is it possible for $Z(L)$ to be a maximal ideal in $L$?


An attempt: Since $L$ is not abelian then there exists $x\in L, x\notin Z(L)$, now $\langle x\rangle_F + Z(L)$ is an ideal of $L$, by maximality of $Z(L)$ this ideal is $L$. Thus $dim\ L/Z(L)=1$ and the nilpotency class of $L$ is $2$ i.e $[L,[L,L]]=0$ if this implies that $Z(L)=[L,L]$ then we are done. since for a nilpotent Lie algebra $L$ we must have $dim\ L/[L,L]>1$.

$\endgroup$
  • $\begingroup$ what about L abelian? $\endgroup$ – Tim kinsella Feb 16 '16 at 8:00
  • $\begingroup$ sorry , i forgot to exclude abelian case! $\endgroup$ – Ronald Feb 16 '16 at 8:00
  • $\begingroup$ By centralizer you mean center? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '16 at 8:22
  • $\begingroup$ Yes .. I changed it $\endgroup$ – Ronald Feb 16 '16 at 8:24
  • $\begingroup$ Why is what you say is an ideal an ideal? $\endgroup$ – Mariano Suárez-Álvarez Feb 16 '16 at 8:38
1
$\begingroup$

Suppose such an $L$. Then $[L,L]+Z(L)$ is an ideal (why is $[L,L]+Z$ not all of $L$? because $[[L,L],L]$ must be strictly smaller than $[L,L]$, since $L$ is nilpotent.. rt?). Since $Z$ is maximal, $[L,L]\subset Z$, but then $Z$ must have codimension one if it is to be maximal. But then $L$ is abelian. contradiction.

$\endgroup$
  • $\begingroup$ I have written my attempt under the question. Nice answer $\endgroup$ – Ronald Feb 16 '16 at 8:31
  • $\begingroup$ Sorry, could you please explain why if $[L,L]\subset Z$ then $L$ is abelian? $\endgroup$ – Ronald Feb 16 '16 at 8:41
  • 1
    $\begingroup$ Suppose $[L,L]\subset Z$. Every linear subspace containing $[L,L]$ is an ideal, so if $Z$ is going to be maximal, it cannot be properly contained in a proper subspace of $L$. So $dim(Z) = dim(L)-1$. So $L= Z\oplus \mathbb{R}x$ where $x\notin Z$. Now $[\mathbb{R}x, Z] = [\mathbb{R}x, \mathbb{R}x]= [Z,Z]=0$, so $[L,L]=0$. $\endgroup$ – Tim kinsella Feb 16 '16 at 8:45

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.