Let $g$ be a Lie algebra. How to compute Casimir elements of $g \otimes g$? I am asking this question because in the book a guide to quantum groups, page 80, there is an equation $r_{12} + r_{21}=t$, where $t$ is the Casimir of $g \otimes g$. We have $ g \otimes g = h \otimes h \oplus n_- \otimes n_+ \oplus n_+ \otimes n_-$. Let $t_0$ be the $h \otimes h$ component of $t$. On page 83, example 3.1.5, it is said that in the case of $g = sl_3$, $$ t_0 = \frac{2}{3} H_{\alpha} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\alpha} + \frac{1}{3} H_{\alpha} \otimes H_{\beta} + \frac{2}{3} H_{\beta} \otimes H_{\beta}.$$ My question is: how to compute $t$ and $t_0$? Any help will be greatly appreciated!
Let $g$ be a semisimple Lie algebra with the root system $\Delta$. We fix a Cartan subalgebra $\mathfrak{h}$ of $g$. For each $\alpha \in \Delta$, we define $E_{\alpha} \in g$ by the formula $[h, E_{\alpha}] = \alpha(h) E_{\alpha}$ for all $h \in \mathfrak{h}$. Then for $\alpha, \beta \in \Delta$, \begin{align} & [h, [E_{\alpha}, E_{\beta}]] = -[ E_{\alpha}, [E_{\beta}, h] ] - [E_{\beta}, [h, E_{\alpha}]] = - [E_{\alpha}, -\beta(h) E_{\beta}] - [E_{\beta}, \alpha(h)E_{\alpha}] \\ & = (\alpha(h) + \beta(h))[E_{\alpha}, E_{\beta}] = (\alpha + \beta)(h)[E_{\alpha}, E_{\beta}]. \end{align} It follows that if $\alpha +\beta \in \Delta$, then $[E_{\alpha}, E_{\beta}] = E_{\alpha+\beta}$ and $0$ if $\alpha + \beta \not\in \Delta$.
By definition, a Casimir element in $g \otimes g \cong g \otimes g^*$ is an element $t = \sum_{i} x_i \otimes x_i^*$, where $x_i$, $i \in I$, is a basis of $g$. Now we choose $\mathfrak{h} = \text{Span}\{H_{\alpha_i}: i \in I\}$, where $\alpha_i$, $i \in I$, is the set of simple roots. A basis of $g$ is $\{H_{\alpha_i}, i \in I, E_{\alpha}, \alpha \in \Delta\}$. We have the Killing from on $g$ which is given by $$ (x, y) = \text{Tr}(\text{ad}_x \text{ad}_y), \quad x, y \in g. $$ We have \begin{align} & (E_{\alpha}, E_{\beta}) = \text{Tr}(\text{ad}_{E_{\alpha}} \text{ad}_{E_{\beta}}). \\ & (\text{ad}_{E_{\alpha}}\text{ad}_{E_{\beta}})(E_{\gamma}) \\ & = [E_{\alpha}, [E_{\beta}, E_{\gamma}]] \\ & = E_{\alpha + \beta + \gamma}. \end{align} We have $E_{\alpha+\beta+\gamma} = E_{\gamma}$ if and only if $\alpha+\beta=0$. Therefore \begin{align} & (E_{\alpha}, E_{\beta}) = \text{Tr}(\text{ad}_{E_{\alpha}} \text{ad}_{E_{\beta}}) \neq 0 \end{align} if and only if $\alpha = - \beta$. Therefore the Casimir is \begin{align} t=\sum_{\alpha \in \Delta} E_{-\alpha} \otimes E_{\alpha} + t_0, \end{align} where $t_0 \in \mathfrak{h} \otimes \mathfrak{h}$ is the Cartan part of $t$.
Now we compute $t_0$ in the case of $sl_3$. We have \begin{align} & ad(H_i)ad(H_j)(H_k)=0,\\ & ad(H_1)ad(H_2)(E_1) = -2E_1, \\ & ad(H_1)ad(H_2)(E_2) = -2E_2, \\ & ad(H_1)ad(H_2)([E_1, E_2]) = [E_1, E_2], \\ & ad(H_1)ad(H_2)(F_1) = -2F_1, \\ & ad(H_1)ad(H_2)(F_2) = -2F_2, \\ & ad(H_1)ad(H_2)([F_1, F_2]) = [F_1, F_2]. \end{align} Therefore $$ (H_1, H_2) = -6. $$ \begin{align} & ad(H_i)ad(H_j)(H_k)=0,\\ & ad(H_1)ad(H_1)(E_1) = 4E_1, \\ & ad(H_1)ad(H_1)(E_2) = E_2, \\ & ad(H_1)ad(H_1)([E_1, E_2]) = [E_1, E_2], \\ & ad(H_1)ad(H_1)(F_1) = 4 F_1, \\ & ad(H_1)ad(H_1)(F_2) = F_2, \\ & ad(H_1)ad(H_1)([F_1, F_2]) = [F_1, F_2]. \end{align} $$ (H_1, H_1) = 12. $$ Similarly, $$ (H_2, H_1) = -6, (H_2, H_2) = 12. $$ Therefore we can take $$ t_0 = \frac{1}{3} ( H_1 \otimes H_1 + H_2 \otimes H_2 ) + \frac{1}{6}(H_1 \otimes H_2 + H_2 \otimes H_1) $$ if we normalize the Killing from such that $(x, y) = 2$ for $x, y \in g$. But it seems there is a problem of the signs in the above computations: $(H_2, H_1) = -6, (H_2, H_2) = 12$.