Matrix of Killing form a Lie algebra

Question

Let $L$ be the Lie algebra with basis $B = \{u,v,w\}$, with $[u,v] = w, [v,w] = u, [w,u] = v$.

Question : Find the matrix of the Killing form $\kappa$ of $L$ with respect to $B$.

I have come across this question in a past exam paper for my Lie algebras course. It is not worth many marks so I am guessing this is an easy question, however I can't seem to convince myself of an answer.

My guess is that the matrix of $\kappa$ is given by

\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}

Can anyone confirm this is correct, or tell me the correct way to do it?

Edit: After thinking about this again, I am pretty sure my guess is incorrect. Any guidance is appreciated.

Answer 1

Following the definition/notation here, we need to compute the matrices for $\DeclareMathOperator{\ad}{ad}$ $\DeclareMathOperator{\tr}{trace}$ $\ad(u),\ad(v),\ad(w)$ with respect to $B$. In particular, we have $$ \ad(u) = \pmatrix{ 0&0&0\\ 0&0&-1\\ 0&1&0}\\ \ad(v) = \pmatrix{ 0&0&1\\ 0&0&0\\ -1&0&0}\\ \ad(w) = \pmatrix{ 0&-1&0\\ 1&0&0\\ 0&0&0} $$ The entries of the Killing form can then be found as $\tr(\ad(e_i)\ad(e_j))$. In particular, we have $$ L = \pmatrix{ -2&0&0\\ 0&-2&0\\ 0&0&-2 } $$ for this computation, it helps to note that $\tr(A^TB)$ is the "dot-product" of the matrices $A$ and $B$. Note that the Killing form should always be symmetric.