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I want to show that the Killing $k$ form of $u(\mathbb{C},3)$ is such that $k(x,y)=0$ $\forall x,y \in u(\mathbb{C},3)$.

I have used the basis {$e_{12}, e_{13}, e_{23}$}, and found that the adjoint matrices are given by $$ad(e_{12})=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, ad(e_{13})=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},ad(e_{23})=\begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

I then found that the matrix of the killing form is just the $0$ matrix, but I was wondering is this enough to say that $k(x,y)=0 \hskip 0.5em \forall x,y \in u(\mathbb{C},3)$?

UsernameInvalid
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    It seems that you are pretty close to showing what you want to show, but what exactly do you mean by "the matrix of the Killing form"? (A shortcut would be though that the Killing form of every nilpotent Lie algebra is identically $0$ -- do you know/can you use that?) – Torsten Schoeneberg Feb 17 '19 at 04:36
  • Thanks for the reply, by the matrix of the killing form I mean the matrix $(k(x,y))$ where $x,y$ are basis elements, so I basically found that for the $x,y$ in the basis $k(x,y)=0$. I don't believe we explicitly learnt about said statement you provided that the killing form of nilpotent Lie algebras are 0, however this seems familiar to Engel's theorem but I am unsure how I could apply that. – UsernameInvalid Feb 17 '19 at 14:15
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    I think you are done then. Of course the Killing form is constantly $0$ iff it is so an all pairs of basis elements iff the matrix defined as you say is the zero matrix. -- As for the second remark, well, strictly upper triangular matrices, as yours, indeed by Engel's theorem are standard examples of nilpotent Lie algebras. – Torsten Schoeneberg Feb 17 '19 at 19:40

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