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

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  • $\begingroup$ 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? $\endgroup$ – gen Jun 12 at 19:00
  • $\begingroup$ @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. $\endgroup$ – Dietrich Burde Jun 13 at 8:35
  • $\begingroup$ thank you (few more chars) $\endgroup$ – gen Jun 13 at 9:40

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