Casimir Operator of $\mathfrak{sl}_n(\mathbb{C})$. If I have the Lie algebra $\mathfrak{g} = \mathfrak{sl}_n(\mathbb{C})$ and the trace form $B(x,y) = \text{tr}(x,y)$ for $x, y \in \mathfrak{g}$ how does one calculate by which scalar the Casimir element $C \in Z(\mathscr{U}(\mathfrak{g}))$ acts on the highest weight module $V(\lambda)$?
I know that one defines the Casimir by $C = \sum_i x_i x_i^*$ where $\{x_i\}$ is a basis for the Lie algebra and $\{x_i^*\}$ a dual basis, which are arbitrary (i.e. C is independent of the choice of bases).
I have calculated by hand some simple examples (n=3 acts by the scalar $\frac{8}{3}$ for example on V(2)). How should one proceed in general?
Thanks.
 A: 
The way to calculate the action of $C$ is to make a smart choice of dual bases and then observe that it suffices to check the action of $C$ on a highest weight vector. (The fact that you use the trace form rather than the Killing form shows up only in an overall scaling of $C$.) Basically, you have to observe that the Cartan subalgebra $\mathfrak h$ is orthogonal to all root spaces with respect to $B$ while for two roots $\alpha$ and $\beta$, the restriction of $B$ to $\mathfrak g_{\alpha}\times\mathfrak g_{\beta}$ is non-zero if and only if $\beta=-\alpha$. First choose a basis $\{H_i\}$ of the space $\mathfrak h$ of diagonal matrices which is orthonormal with respect to $B$. Next, for $i<j$ consider the elementary matrix $E_{ij}$. Then it follows that $\{E_{ij}^t,H_i,E_{ij}\}$ is a basis of $\mathfrak{sl}(n,\mathbb C)$ with dual basis $\{E_{ij},H_i,E_{ij}^t\}$. Otherwise put, you get 
$$
C=\sum_{i=1}^{n-1}H_iH_i+\sum_{i<j}(E_{ij}^tE_{ij}+E_{ij}E_{ij}^t). 
$$ 
Acting with an element in the second sum on any representation $V$, you get for $v\in V$:
\begin{align*}
E_{ij}^t\cdot E_{ij}\cdot v+E_{ij}\cdot E_{ij}^t\cdot v=&2E_{ij}^t\cdot E_{ij}\cdot v+(E_{ij}\cdot E_{ij}^t\cdot v-E_{ij}^t\cdot E_{ij}\cdot v)\\
=&2E_{ij}^t\cdot E_{ij}\cdot v+[E_{ij},E_{ij}^t]\cdot v. 
\end{align*}
Since you know in advance that $C$ has to act by a scalar on any irreducible representation, it suffices to compute $C\cdot v_0$ for a highest weight vector $v_0\in V(\lambda)$. But by Definition $E_{ij}\cdot v_0=0$ and for $H\in\mathfrak h$, we get $H\cdot v_0=\lambda(H)v_0$. Hence you get 
$$
C\cdot v_0=\left(\sum_{i=1}^{n-1}\lambda(H_i)^2+\sum_{i<j}\lambda([E_{ij},E_{ij}^t])\right)v_0.
$$
The scalar can be expressed as $\langle\lambda,\lambda\rangle+2\langle\lambda,\rho\rangle$, where $\rho$ is half of the sum of all positive roots and the inner product is induced by $B$. 
