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.
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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.
gen
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Dietrich Burde
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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 '19 at 19:00
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@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 '19 at 08:35
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:-)) – Willie Wong Dec 19 '13 at 08:35