Find smallest $m$ for which $L^{(m)} = 0$ for 3 dimensional Lie algebra $L$

Question

Let $L$ be a three dimensional Lie algebra. $L'$ be the derived algebra with $\dim(L') = 2$. Define the following - $$ L^{(1)} := L' , \space L^{(n)} := [ L^{(n-1)}, L^{(n-1)}] $$

How do I find the smallest $m \in \mathbb{N}$ such that $ L^{(m)} = 0$ ?

Answer 1

Let $y,z$ be a basis of $L'$. It suffices to show that $[y,z]=0$ (why?). Since $[y,z]\in L'$, $[y,z]=\alpha y+\beta z$. Now write the matrix of $\text{ad}_y$ w.r.t. to a basis $\left\{x,y,z\right\}$ of $L$. You should find that $\text{Tr}(\text{ad}_y)=\beta$. Since $y\in L'$, $\text{Tr}(\text{ad}_y)=0$ (why?). Do the same argument for $\text{ad}_z$ and conclude that $\alpha=0$. What does this tell you about the minimal $m$?

Answer 2

We can reduce the question as follows to the nilpotent case: if $L$ is an $n$-dimensional solvable Lie of solvability class $m$, then $[L,L]$ is a nilpotent Lie algebra of dimension at most $n-1$ and solvability class $m-1$. If $n=3$, then nilpotent Lie algebras of dimension at most $n-1=2$ have solvability class $1$, because they are in fact abelian. Hence $m\le1+1=2$ for solvable Lie algebras of dimension $3$.

Answer 3

I don't know which assumption you have in mind on the underlying field, but in the following I make no assumption, and use no particular nontrivial theorem.

The 2-dimensional non-abelian Lie algebra is complete, in the sense that every derivation is inner, and has trivial center (these are straightforward verification). It follows that whenever it occurs as an ideal in a Lie algebra, it is a direct factor. In particular, since in addition it's not a perfect Lie algebra, it cannot be the derived subalgebra of any Lie algebra.

Thus, in your case the derived Lie algebra has to be abelian and hence the derived length is 2.