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In my Lie Algebra notes I don't understand the following (I cite):

"To compute this Killing form [of $\mathfrak{sl} (2, \mathbb{C})$], we look at a special basis of $\mathfrak{sl} (2, \mathbb{C})$ [...]. Put $H= \begin{pmatrix} 1 & 0\\ 0 &-1 \end{pmatrix}$, $G= \begin{pmatrix} 0 & 1\\ 0 &0 \end{pmatrix}$ and $F= \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$. [...] one immediately computes that $[H,E] = 2E$, $[H,F] = -F$ and $[E,F]=H$. This easily implies that in the basis $\{E,H,F\}$ we get the matrix representations

$$\text{ad}(H) = \begin{pmatrix} 2 & 0 & 0\\ 0 & 0 & 0 \\ 0&0&-2 \end{pmatrix} \qquad \text{ad}(E) = \begin{pmatrix} 0 & -2 & 0\\ 0 & 0 & 1 \\ 0&0&0 \end{pmatrix} \qquad \text{ad}(F) =\begin{pmatrix} 0 & 0 & 0\\ -1 & 0 & 0 \\ 0&2&0 \end{pmatrix}$$"

Now I don't understand where these $3 \times 3$-matrices come from. As far as I know the adjoint representation maps from $\mathfrak{sl}$ to $\mathfrak{gl}(V)$ for some vector space $V$. Where do extra dimensions get involved?

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These matrices $H= \begin{pmatrix} 1 & 0\\ 0 &-1 \end{pmatrix}$, $E= \begin{pmatrix} 0 & 1\\ 0 &0 \end{pmatrix}$ and $F= \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$ are a base of the algebra, you can consider them as 3 indipendent vectors which forms a 3-dimensional vectorial space.

Instead $ad(H)$, $ad(E)$, $ad(F)$ are linear transformations of the above 3 dimensional vectorial space into the same space. So what you're actually looking for is to find a matrix representation for these linear transformations in the basis elements given above.

Let's say you want to find the matrix representation for $ad(H)$ which is the linear transformation given by $[H,.]$. Then you have to calculate $[H,E]$, $[H,F]$ and $[H,H]$ that are: $$[H,E] = 2E, [H,F]=-2F, [H,H]=0.$$

Considering the order of the basis E,F,H, you have the following matrix $$\text{ad}(H) = \begin{pmatrix} 2 & 0 & 0\\ 0 & -2 & 0 \\ 0&0&0 \end{pmatrix} $$ because $[H,E] = 2E+0F+0H$ and so the first row is $(2,0,0)$, the second row is given by $[H,F] = 0E-2F+0H$ and the third row is given by $[H,H]=0E+0F+0H$.

I think now you should compute the other two matrices by your own to understand correctly the mechanism.

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  • $\begingroup$ Thanks, that's exactly the insight I was looking for! $\endgroup$ – Jakob Elias Jan 15 '17 at 0:50

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