Killing form

Example:

4.05 Let $\mathfrak{g}=\mathfrak{sl}(3,\mathbf{C})$ and let $H_{13}=\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&-1\end{pmatrix}$ and $H_{23}=\begin{pmatrix}0&0&0\\ 0&1&0\\ 0&0&-1\end{pmatrix}$ . Together, these span the subspace

$\mathfrak{h}_{\mathbf{R}}=\left\{H_{\theta}:=\theta_{1}H_{13}+\theta_{2}H_{23}% =\begin{pmatrix}\theta_{1}&0&0\\ 0&\theta_{2}&0\\ 0&0&-\theta_{1}-\theta_{2}\end{pmatrix}\ :\ \theta_{1},\theta_{2}\in\mathbf{R}\right\}$ of real diagonal matrices in

$\mathfrak{sl}(3,\mathbf{C})$ . I want to compute all possible Killing "dot products" between these two. We will see that

$K(H_{13},H_{13})=K(H_{23},H_{23})=12$ and

$K(H_{13},H_{23})=6.$

5.30 If we think of this as a dot product, then this would be telling us

$|H_{13}|=|H_{23}|=\sqrt{12}$ and the cosine of the angle between them would be

$60$ degrees.

6.15 Let's proceed with the calculation. What is $\mathrm{ad}_{H_{13}}\colon\mathfrak{sl}(3,\mathbf{C})\to\mathfrak{sl}(3,% \mathbf{C})$ ? Let's calculate the matrix of this linear map with respect to the basis:

$H_{13},H_{23},E_{12},E_{21},E_{13},E_{31},E_{23},E_{32}.$ We have

$\mathrm{ad}_{H_{13}}(H_{13})=[H_{13},H_{13}]=0$

$\mathrm{ad}_{H_{13}}(H_{23})=[H_{13},H_{23}]=0$ because the matrices

$H_{ij}$ are diagonal and commute with one another. We also have

$\mathrm{ad}_{H_{\theta}}(E_{ij})=(\theta_{i}-\theta_{j})E_{ij}$ and

$\theta=(1,0,-1)$ for

$H_{13}$ , so

$\mathrm{ad}_{H_{13}}(E_{12})=E_{12},\ \mathrm{ad}_{H_{13}}(E_{21})=-E_{21}$

$\mathrm{ad}_{H_{13}}(E_{13})=2E_{13},\ \mathrm{ad}_{H_{13}}(E_{31})=-2E_{31}$

$\mathrm{ad}_{H_{13}}(E_{23})=E_{23},\ \mathrm{ad}_{H_{13}}(E_{32})=-E_{32}.$

9.00 As a matrix, therefore, $\mathrm{ad}_{H_{13}}$ is the diagonal matrix:

$\mathrm{ad}_{H_{13}}=\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&1&&&&&\\ &&&-1&&&&\\ &&&&2&&&\\ &&&&&-2&&\\ &&&&&&1&\\ &&&&&&&-1\end{pmatrix}.$

9.40 Similarly, we find that

$\mathrm{ad}_{H_{23}}=\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&-1&&&&&\\ &&&1&&&&\\ &&&&1&&&\\ &&&&&-1&&\\ &&&&&&2&\\ &&&&&&&-2\end{pmatrix}.$

10.35 Therefore $K(H_{13},H_{13})$ is the trace of

$\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&1&&&&&\\ &&&-1&&&&\\ &&&&2&&&\\ &&&&&-2&&\\ &&&&&&1&\\ &&&&&&&-1\end{pmatrix}^{2}=\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&1&&&&&\\ &&&1&&&&\\ &&&&4&&&\\ &&&&&4&&\\ &&&&&&1&\\ &&&&&&&1\end{pmatrix}$ which is

$12$ . Similarly for

$K(H_{23},H_{23})$ .

12.00 Finally, we have

$K(H_{13},H_{23})=\mathrm{Tr}\left(\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&1&&&&&\\ &&&-1&&&&\\ &&&&2&&&\\ &&&&&-2&&\\ &&&&&&1&\\ &&&&&&&-1\end{pmatrix}\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&-1&&&&&\\ &&&1&&&&\\ &&&&1&&&\\ &&&&&-1&&\\ &&&&&&2&\\ &&&&&&&-2\end{pmatrix}\right)=-1-1+2+2+2+2=6.$

Killing form

Example

Pre-class exercise