# When is the Killing form null?

When is the Killing form $\kappa$ of a Lie algebra $\mathfrak g$ null, i.e. $\kappa(\cdot,\cdot)=0$? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't seem to find any nontrivial examples.

• You can go a bit better than trivial bracket: for any nilpotent Lie algebra the Killing form vanishes. – Willie Wong Feb 21 '13 at 16:27
• I see, thank you. And is also the converse true, in general? That is, does $\kappa$ null imply $\mathfrak g$ nilpotent? – Linda Serafim Feb 21 '13 at 16:41
• In this Wikipedia page it is mentioned that the converse is false. An example is given but I haven't checked it myself. – Willie Wong Feb 21 '13 at 16:51
• @WillieWong: I have checked all $3$-dimensional solvable Lie algebras -see below. – Dietrich Burde Dec 18 '13 at 17:58
• @DietrichBurde: I saw, very nice. (Where did you think you got that upvote from? :-) ) – Willie Wong Dec 19 '13 at 8:35

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.
• when you say "by Lie's theorem we may assume that all matrices of ad(L) are upper-triangular" do you mean that there is a basis of L such that for all $x\in L$ the matrix of $ad(x)$ with respect to this matrix is upper triangular? Is this true generally for all finite dim. lie algebras? What statement of Lie's theorem do you use? – gen Jun 12 at 19:00
• @gen I use the following corollary to Lie's theorem: For all representations $\phi$ of a solvable Lie algebra $L$ on a finite dimensional vector space $V$ over an algebraically closed field $K$ of characteristic zero there exists a basis for $V$ for which the matrices of $\phi(x)$ for all $x\in L$ are upper triangular. You will find this in books on Lie algebras. – Dietrich Burde Jun 13 at 8:35