Action of the Casimir on highest weight modules for Kac-Moody algebra Let $g$ be a Kac-Moody algebra with a symmetrizable Cartan matrix, and let $\{u_j\}$ and $\{u^j\}$ be bases of $g$ dual with respect to a nondegenerate invariant bilinear form $(\cdot|\cdot)$ on $g$, and consistent with the triangular decomposition of $g$. Let $L(\Lambda)$ be an integrable representation of $g$ with highest weight $\Lambda$, and let $v_\Lambda$ be its highest weight vector. Denote the Casimir $\Omega=\sum_j u_j\otimes u^j$.

Question. Why $\Omega(v_\Lambda\otimes v_\Lambda)=(\Lambda|\Lambda)v_\Lambda\otimes v_\Lambda$? Could someone
give some explanation or some references?

 A: You should be a bit careful, as this isn't precisely the action of the Casimir on $v \otimes v$, but instead follows from it.
For each positive root $\alpha$, let $e_\alpha^{(1)}, \dots, e_\alpha^{(n_\alpha)}$ be a basis of the root space $\mathfrak{g}_\alpha$, and let $\{f_\alpha^{(i)}\}$ be the corresponding dual space for $\mathfrak{g}_{-\alpha}$. Define $\Omega_\alpha:= \sum_{i=1}^{n_\alpha} f_\alpha^i e_\alpha^i$. Also, let $\{h_i, h^i\}$ be a dual basis for $\mathfrak{h}$, and set $\Omega_0:= \sum_i h_ih^i$. Then the Casimir element is given by
$$ \Omega= 2\nu^{-1}(\rho)+\Omega_0 +2\sum_{\alpha \in \Delta^+} \Omega_\alpha$$
where $\rho$ is a Weyl vector defined by $\rho(\alpha_i^\vee)=1$ for all simple coroots $\alpha_i^\vee$, and $\nu: \mathfrak{h} \to \mathfrak{h}^\ast$ is the isomorphism determined by the nondegenerate bilinear form.
A key fact about the Casimir operator is that, for any $v \in L(\mu)$ for any dominant weight $\mu$, we have that $\Omega(v)=(\mu | \mu+2\rho)v$. Now, by just expanding out over the tensor product using the above definition of $\Omega$, we get for any $v \in L(\Lambda)$
$$
\Omega(v \otimes v) = (\Omega(v)) \otimes v + v \otimes (\Omega(v)) +2 \sum_{\alpha \in \Delta \sqcup \{0\}} \sum_{i=1}^{n_\alpha} e_\alpha^i(v) \otimes f_\alpha^i(v).
$$
Now, if $v \in G(v_\Lambda)$ is a vector in the $G$-orbit of $v_\Lambda$ (one such example is precisely $v_\Lambda$), we have $v \otimes v \in L(2\Lambda)$. Then the left-hand side is precisely $(2\Lambda | 2\Lambda +2\rho) (v\otimes v)$. The right-hand side is similarly
$$
2(\Lambda | \Lambda+2\rho)(v \otimes v) + 2 (\text{the term you are interested in}).$$
Solving for the term you want gives precisely $(\Lambda | \Lambda)(v\otimes v)$.
