The Shapovalov determinant formula

In PBW bases, commute each positive-root factor past negative-root factors before evaluating on $v_{\lambda }$. The diagonal terms are the only terms of maximal total degree. Counting, for each occurrence of a root $\alpha$, the PBW monomials in which that occurrence can be removed gives the following leading term.

In particular, $D_{\beta }$ is nonzero and has the total degree displayed on the right.

Suppose $D_{\beta }(\lambda )=0$. The radical is then nonzero in weight $\lambda -\beta$. By The Shapovalov radical is the maximal submodule and Every nonzero Verma submodule contains a singular vector, it contains a singular vector of some weight $\lambda -\gamma$, with $\gamma \in Q^{+}\setminus \{0\}$. The universal property gives a nonzero map $M(\lambda -\gamma )\to M(\lambda )$. The Casimir has the same scalar on its source and image, giving the following identity.

Consequently every irreducible factor of $D_{\beta }$ is an affine linear form with normal direction $\gamma$. Comparing its leading direction with the product in step 1.1 shows that $\gamma =n\alpha$ for a positive root $\alpha$ and an integer $n\ge 1$. Substitution in step 3.1 then gives $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=n$. Thus, for some integers $m_{n\alpha }(\beta )\ge 0$, the following factorization holds.

This is the claimed preliminary factorization.

Fix $\alpha ,n$ and choose $\lambda$ generically on the hyperplane $H_{\alpha ,n}$. The generic-hyperplane result [L1] gives an injective map $M(\lambda -n\alpha )\to M(\lambda )$ whose image is exactly the Shapovalov radical. Under PBW, its part in weight $\lambda -\beta$ has dimension $K(\beta -n\alpha )$, including dimension $0$ when $\beta -n\alpha \mathrm{\notin }{Q}^{+}$.

Choose $\delta \in {\mathfrak{h}}^{\ast }$ with $⟨\delta ,{\alpha }^{\vee }⟩=1$ and restrict the form in weight $\lambda -\beta$ along $\lambda +t\delta$. Its kernel at $t=0$ is the space in step 6.1. By [L1], the first derivative of the form is nondegenerate on that kernel. Equivalently, a vector pairing to order $t^2$ would force the two relevant Casimir scalars to agree modulo $t^2$, although their difference is the following nonzero linear term.

This difference is nonzero modulo $t^2$, a contradiction. The elementary determinant lemma obtained by choosing bases adapted to the kernel now says that the transverse order of $D_{\beta }$ along $H_{\alpha ,n}$ is the kernel dimension. Therefore $m_{n\alpha }(\beta )=K(\beta -n\alpha )$.

Substitution in step 5.1 proves the formula up to the nonzero basis scalar. Finally $\beta -n\alpha \in Q^{+}$ bounds $n$ by the height of $\beta$, so only finitely many displayed exponents are nonzero.