Back to the Example

Let’s go back to our explicit example of $L=\mathfrak{sl}(2,\mathbb{F})$ and look at its Killing form. We first recall our usual basis:

$\displaystyle\begin{aligned}x&=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\\y&=\begin{pmatrix}0&0\\1&0\end{pmatrix}\\h&=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\end{aligned}$

which lets us write out matrices for the adjoint action:

$\displaystyle\begin{aligned}\mathrm{ad}(x)&=\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\\\mathrm{ad}(y)&=\begin{pmatrix}0&0&0\\ 0&0&2\\-1&0&0\end{pmatrix}\\\mathrm{ad}(h)&=\begin{pmatrix}2&0&0\\ 0&-2&0\\ 0&0&0\end{pmatrix}\end{aligned}$

and from here it’s easy to calculate the Killing form. For example:

$\displaystyle\begin{aligned}\kappa(x,y)&=\mathrm{Tr}\left(\mathrm{ad}(x)\mathrm{ad}(x)\right)\\&=\mathrm{Tr}\left(\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\begin{pmatrix}0&0&0\\ 0&0&2\\-1&0&0\end{pmatrix}\right)\\&=\mathrm{Tr}\left(\begin{pmatrix}2&0&0\\ 0&0&0\\ 0&0&2\end{pmatrix}\right)\\&=4\end{aligned}$

We can similarly calculate all the other values of the Killing form on basis elements.

$\displaystyle\begin{aligned}\kappa(x,x)&=0\\\kappa(x,y)=\kappa(y,x)&=4\\\kappa(x,h)=\kappa(h,x)&=0\\\kappa(y,y)&=0\\\kappa(y,h)=\kappa(h,y)&=0\\\kappa(h,h)&=8\end{aligned}$

So we can write down the matrix of $\kappa$ :

$\displaystyle\begin{pmatrix}0&4&0\\4&0&0\\ 0&0&8\end{pmatrix}$

And we can test this for degeneracy by taking its determinant to find $-128$ . Since this is nonzero, we conclude that $\kappa$ is nondegenerate, which we know means that $\mathfrak{sl}(2,\mathbb{F})$ is semisimple — at least in fields where $1+1\neq0$ .

September 7, 2012 - Posted by John Armstrong | Algebra, Lie Algebras

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