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Let $L$ be a four-dimensional nilpotent Lie Algebra. Suppose $\text{dim}(Z(L))=1$ where $Z(L)$ is the center of the Lie Algebra. There is a claim that $\text{dim}([L,L])<3$, which I don't understand. It is clear that $\text{dim}([L,L])\leq 3$ since dimension of the center is 1. Maybe this is easy to do but I am missing something I think. Thanks in advance for the clarifications.

Riju
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  • What do you know about the relationship between the commutator sub-algebra and the centre? – Matt Aug 10 '21 at 11:14
  • Are you asking me what I know or are you saying if there is a relation in the question? I didn't understand. If you are asking me I don't know any particular relationship between the center and derived subalgebra. – Riju Aug 10 '21 at 11:25

2 Answers2

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For every nilpotent Lie algebra $L$ we have $$ \dim L/[L,L]\ge 2 $$ So we cannot have $\dim [L,L]=3$ for a nilpotent $4$-dimensional Lie algebra. So this has nothing to do with $\dim Z(L)=1$.

Reference:

codimension of the derived algebra of a nilpotent Lie algebra

Dietrich Burde
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$Z(L)$ is an ideal generated by $x$, $L/Z(L)$ is nilpotent, its center $Z(L/Z(L))$ is not trivial, let $y$ whose image by the quotient map $L\rightarrow L/Z(L)$ is in the center of $L/Z(L)$.

Complete $x,y$ to a basis $x,y,z,t$ of $L$.

$[L,L]$ is generated by the images of $ad_x,ad_y, ad_z,ad_t$.

The image of $ad_x$ is $\{0\}$.

The image of $ad_y$ is contained in $Vect(x)$.

Write $u=[z,t]$, the image of $ad_z$ and $ad_t$ are contained in $Vect(x, u)$, implies that $dim([L,L])\leq 2$.

Tsemo Aristide
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