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Let $L$ be the Lie algebra $\mathfrak{sl}_{2}$ (char ${F}$ $\neq$ 2).

Take as standard basis for $L$ the three matrices:

$x=\begin{pmatrix} 0&1\\0&0\end{pmatrix}, y= \begin{pmatrix} 0&0\\1&0\end{pmatrix}$ and $h=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$.

When using the adjoint representation, the three matrices become:

$ad_{x}=\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}, ad_{y}=\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix}$ and $ad_{h}=\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix}$.

Therefore $\kappa$ (the Killing form) has matrix

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

Can someone explain how the adjoint and further, the Killing form, are calculated here?

I see now how to calculate the adjoint and the Killing form.

$ad_{x}(x)$ = $[xx -xx] = 0$

$ad_{x}(h) = [xh - hx]$ = $\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 1&0\\0&-1\end{pmatrix}$ - $\begin{pmatrix} 1&0\\0&-1\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}$ = $\begin{pmatrix} 0&-2\\0&0\end{pmatrix} = -2x$

$ad_{x}(y) = [xy - yx]$ = $\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 0&0\\1&0\end{pmatrix}$ - $\begin{pmatrix} 0&0\\-1&0\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}$ = $\begin{pmatrix} 1&0\\0&-1\end{pmatrix} = h$

Then, these become the three columns of $ad_{x}$. Do the same for $y$ and $h$.

For the Killing form,

$\kappa(x,x) = tr(ad_{x}ad_{x})$
$\kappa(x,h) = tr(ad_{x}ad_{h})$
$\kappa(x,y) = tr(ad_{x}ad_{y})$

$tr(ad_{x}ad_{x}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0$.

$tr(ad_{x}ad_{h}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0$.

$tr(ad_{x}ad_{y}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix})= tr(\begin{pmatrix} 2&&\\&2&\\&&0\end{pmatrix}) = 4$.

Thus, these three entries are the entries for the first row.

Repeat for $h$ and $y$.

Dazzler95
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  • They are computed as you said, just follow the definitions. So we have ${\rm ad}(v)(w)=[v,w]$, which defines the adjoint matrices. Do you know the definition of the Killing form? – Dietrich Burde Nov 06 '17 at 19:42
  • I know the definition of the Killing form - $\kappa(x,y)$ = Tr$([ad_{x},ad_{y}]ad_{h})$, and according to my calculations, I get 8, and the trace is therefore the same, but the matrix is different. – Dazzler95 Nov 06 '17 at 20:05
  • Sorry, this is the definition of the Killing form I used: $\kappa([x,y],h) = Tr([ad_{x},ad_{y}]ad_{h})$ – Dazzler95 Nov 06 '17 at 20:22
  • This is not the matrix of the Killing form w.r.t. the basis $(x,y,h)$. – Dietrich Burde Nov 06 '17 at 22:46
  • I can now work out the first part - how the three 2x2 matrices correspond to the 3x3 matrices. Can you please explain how the matrix of the Killing form is calculated? – Dazzler95 Nov 07 '17 at 10:36
  • As I said already, use the correct definition. If $(e_1,e_2,e_3)$ is a vector space basis, then the $(i,j)$ entry of this matrix is $\kappa(e_i,e_j)=tr(ad(e_i)ad(e_j))$.. – Dietrich Burde Nov 07 '17 at 12:07
  • Thank you very much for your last comment, it was extremely helpful. I have posted above how I think one goes about obtaining the adjoint representation and the Killing form. – Dazzler95 Nov 08 '17 at 13:28

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