The quadratic Casimir eigenvalue on a highest-weight module is $(\lambda ,\lambda +2\rho )$

The supplied root-space decomposition [F1] makes every ${\mathrm{ad}}_h$, $h\in \mathfrak{h}$, diagonalizable, and its zero weight space is $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. The latter equality makes $\mathfrak{h}$ abelian and forces every toral enlargement of $\mathfrak{h}$ to equal $\mathfrak{h}$; thus $\mathfrak{h}$ is maximal toral. Apply [F2]: every root space is one-dimensional, $B{\mathrm{\mid }}_{\mathfrak{h}}$ is nondegenerate, and opposite root spaces are perfectly paired. Invariance makes distinct nonopposite weight spaces orthogonal. Choose a basis $h_j$ of $\mathfrak{h}$ with Killing-dual basis $h^j$, and for each positive root choose $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$ and $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ with $B(e_{\alpha },f_{\alpha })=1$. These are full dual bases, so $C={\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}(e_{\alpha }{f}_{\alpha }+f_{\alpha }{e}_{\alpha })$. All choices are finite.

Since $e_{\alpha }v=0$ for every positive root, one has $f_{\alpha }{e}_{\alpha }v=0$ and $e_{\alpha }{f}_{\alpha }v=[e_{\alpha },f_{\alpha }]v$. By [F2] this bracket belongs to $\mathfrak{h}$. Let $H_{\alpha }\in \mathfrak{h}$ be the unique vector satisfying $B(H_{\alpha },h)=\alpha (h)$ for every $h\in \mathfrak{h}$; it exists by nondegeneracy of $B{\mathrm{\mid }}_{\mathfrak{h}}$. Killing invariance gives $B([e_{\alpha },f_{\alpha }],h)=B(e_{\alpha },[f_{\alpha },h])=\alpha (h)B(e_{\alpha },f_{\alpha })=\alpha (h)$. Thus $[e_{\alpha },f_{\alpha }]=H_{\alpha }$, and step 1.1 gives $Cv=({\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}{H}_{\alpha })v$.

The Cartan part acts on $v$ by $(\lambda ,\lambda )$, and the root contribution acts by ${\sum }_{\alpha >0}\lambda (H_{\alpha })=2(\lambda ,\rho )$. Therefore $Cv=(\lambda ,\lambda +2\rho )v$. Because $C$ is central by The quadratic Casimir element is central and $M=U(\mathfrak{g})v$, this same scalar acts on every element of $M$.