When is the Killing form null?

Solution 1:

Suppose that $L$ is a finite-dimensional solvable Lie algebra over the complex numbers. By Lie's theorem we may assume that all matrices of $ad(L)$ are upper-triangular. If $L$, and hence $ad(L)$ is actually nilpotent, then they are even strictly upper-triangular, so that $\kappa(x,y)=tr (ad (x)ad(y))=0$ for all $x,y\in L$. Hence nilpotent Lie algebras have vanishing Killing form.

Conversely, let us consider the family $\mathfrak{r}_3(\lambda)$ of $3$-dimensional solvable, non-nilpotent Lie algebras, given by the brackets $[e_1,e_2]=e_2$ and $[e_1,e_3]=\lambda e_3$, with $\lambda\in \mathbb{C}$. Then the Killing form satisfies $\kappa(e_i,e_j)=0$, except for $\kappa(e_1,e_1)=1+\lambda^2$. Now just take $\lambda=i$, with $i^2=-1$, and we have a solvable, non-nilpotent Lie algebra with vanishing Killing form.