I would like to know how to compute the tensor product of the matrices below and how to deal with duality of vector spaces.
The vector space I concern is the Lie algebra $\mathscr{sl_2}$ with basis $\{H, E, F\}$ and whose bracket is given by $[H, E]=2E, [H, F]=-2F, [E,F]=H$. With respect to this basis, the Killing form $B:\mathscr{sl_2}\otimes \mathscr{sl_2} \to \mathbb{C}$ is given by the matrix
$$\begin{bmatrix} 0& 0 &4\\ 0 & 8& 0 \\ 4&0&0 \end{bmatrix}$$
What I am confused is the following computations.
We regard $B$ as an element in $(\mathscr{sl_2})^*\otimes (\mathscr{sl_2})^*$. (How?)
Then the dual element of $B$ is given by $$\begin{bmatrix} E & H& F \end{bmatrix}\otimes \begin{bmatrix}0& 0 &4\\ 0 & 8& 0 \\ 4&0&0 \end{bmatrix}^{-1} \begin{bmatrix} E\\H\\F \end{bmatrix}=\frac{1}{4}E\otimes F+\frac{1}{8}H\otimes H+ \frac{1}{4} F\otimes E.$$ I would like to know why the element on the left hand side of the equation is the dual of $B$. Also I would like to know how to show this equality.
The point is that the dual basis to $(E,F,H)$ with respect to the Killing form is just given by $(F/4,E/4,H/8)$. This follows directly from the definition of a dual basis $(b_i^*)$ to a basis $(b_i)$ with respect to a non-degenerate symmetric bilinear form. You can see this from the matrix of the Killing form. Hence the Casimir operator $C=C_B=\sum_i b_i\otimes b_i^*$ is given by $$ C=\frac{E\otimes F}{4}+\frac{F\otimes E}{4}+\frac{H\otimes H}{8}. $$