# Killing form

## Killing form

The goal of this video is to explain why the weight lattice of $SU(3)$ is drawn as the triangular lattice. The reason is that the weight lattice has an intrinsic notion of geometry (dot product, lengths, angles) associated to it, coming from a bilinear form called the Killing form

Definition:

Let $\mathfrak{g}$ be a Lie algebra. The Killing form of $\mathfrak{g}$ is the map $K\colon\mathfrak{g}\times\mathfrak{g}\to\mathbf{C}$ defined by $K(X,Y)=\mathrm{Tr}(\mathrm{ad}_{X}\mathrm{ad}_{Y}).$

Remember that $\mathrm{ad}_{X}\colon\mathfrak{g}\to\mathfrak{g}$ is the map $\mathrm{ad}_{X}(Z)=[X,Z].$ Therefore $\mathrm{ad}_{X}\mathrm{ad}_{Y}\colon\mathfrak{g}\to\mathfrak{g}$ is the map $Z\mapsto[X,[Y,Z]]$ . This is a linear map, so if we pick a basis of $\mathfrak{g}$ we can think of it as a matrix. The Killing form of $X$ with $Y$ is the trace of this matrix.

Remark:

The choice of basis doesn't affect the answer: if we change basis, this conjugates the matrix of the linear map by a change of basis matrix, and trace is invariant under conjugation.

Lemma:

The Killing form is a symmetric bilinear form. This means that $K(X,Y)=K(Y,X)$ and that $K(a_{1}X_{1}+a_{2}X_{2},b_{1}Y_{1}+b_{2}Y_{2})=\sum_{i=1}^{2}\sum_{j=1}^{2}a_{% i}b_{j}K(X_{i},Y_{j})$ .

Proof:

It's symmetric because $\mathrm{Tr}(MN)=\mathrm{Tr}(NM)$ for any two matrices $M$ and $N$ .

It's bilinear because (a) $\mathrm{ad}_{X}$ is linear in $X$ , (b) $\mathrm{ad}_{Y}$ is linear in $Y$ , and (c) the trace is linear on matrices.

Remark:

A more famous symmetric bilinear form is the dot product on Euclidean space. So I want to think of $K$ as like a dot product, and use it to define lengths of and angles between vectors in the same way (using the same formulas) that we do for dot product. That's what I mean when I say that certain roots make a 120 degree angle with one another, for example. For this to be true, we will need the Killing form to have certain positivity properties when restricted to subspaces of $\mathfrak{g}$ . We will discuss this later. For now, let's work out an example.

## Example

Example:

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

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.

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 $\mathrm{ad}_{H_{13}}(E_{12})=E_{12},\ \mathrm{ad}_{H_{13}}(E_{21})=-E_{21}$ $\mathrm{ad}_{H_{13}}(E_{13})=2E_{13},\ \mathrm{ad}_{H_{13}}(E_{31})=-2E_{31}$ $\mathrm{ad}_{H_{13}}(E_{23})=E_{23},\ \mathrm{ad}_{H_{13}}(E_{32})=-E_{32}.$

As a matrix, therefore, $\mathrm{ad}_{H_{13}}$ is the diagonal matrix: $\mathrm{ad}_{H_{13}}=\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&1&&&&&\\ &&&-1&&&&\\ &&&&2&&&\\ &&&&&-2&&\\ &&&&&&1&\\ &&&&&&&-1\end{pmatrix}.$

Similarly, we find that $\mathrm{ad}_{H_{23}}=\begin{pmatrix}0&&&&&&&\\ &0&&&&&&\\ &&-1&&&&&\\ &&&1&&&&\\ &&&&1&&&\\ &&&&&-1&&\\ &&&&&&2&\\ &&&&&&&-2\end{pmatrix}.$

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

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

## Pre-class exercise

Exercise:

Consider the basis of $\mathfrak{su}(2)$ given by the Pauli matrices $\sigma_{1}=\begin{pmatrix}i&0\\ 0&-i\end{pmatrix},\ \sigma_{2}=\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\ \sigma_{3}=\begin{pmatrix}0&i\\ i&0\end{pmatrix}$ Find all possible Killing dot products $K(\sigma_{i},\sigma_{j})$ .