Regarding nilpotent 3-dim Lie-algebra

Question

A nilpotent three-dimensional Lie algebra is either abelian ("commutative") or isomorphic to $n_3$.

• We say that a lie-algeba $L$ is nilpotent if there exists $N$ such that $C^N(L) = 0$ where

$L=C^1(L) > [L,L]=C^2 (L) > [L,[L,L]]=C^3 (L)>...$

• $n_3={A \in gl_3 : a_{ij}=0 , i \geq j}$.

  $n_3$= span {$E_{12} , E_{23}, E_{13}$}, Here $E_{12}$ denotes the $3\times 3$ matrix with $1$ if $(i,j)=(1,2)$ and $0$ otherwise (same for the two other matrices).

• Every abelian algebra is nilpotent.

• There is a theorem : $L$ is a solvable lie algebra if and only if there is a sequence of ideals $L > I_ 0 > I_ 1>...>I_ n = 0$ such that $I_k / I_{k+1}$ is an abelian quotient. And knowing that every nilpotent Lie- algebra is solvable.

Can this theorem be used to show the above claim?

Answer 1

There is a direct proof by using the dimension of $[L,L]$. So let $m=\dim([L,L])$. If $m=0$, then $L$ is abelian. If $m=3$, then $L=[L,L]$ is perfect and hence not solvable. So then $L$ also is not nilpotent. If $m=2$, then we also can show that $L$ is not nilpotent. Indeed, then $L$ has zero center and hence cannot be nilpotent.

So we are left with $m=1$. Suppose first that $[L,L]\subset Z(L)$ and let $[L,L]=Fz$, where $F$ is the field. We can extend $z$ to a basis $(x,y,z)$ of $L$. Because of $z\in Z(L)$ we have $[x,z]=[y,z]=0$. We may assume that $[x,y]=z$. Then $L$ is isomorphic to the Heisenberg Lie algebra $\mathfrak{n}_3(F)$.

If $[L,L]=Fy$ is not contained in the center of $L$, then there exists an $x'\in L$ with $[x',y]\neq 0$. Since $\dim [L,L]=1$, we have $[x',y]=\alpha y$, and therefore $[x,y]=y$ with $x=\alpha^{-1}x'$. The subalgebra $\mathfrak{a}=Fx+Fy$ hence is isomorphic to $\mathfrak{r}_2(F)$, the unique non-abelian Lie algebra over $F$ of dimension $2$. It is not nilpotent. This implies that $L$ is not nilpotent.