The Shapovalov determinant formula

The preliminary factorization gives nonnegative integers $m_{\alpha ,n}(\beta )$ supported where $n\alpha \le \beta$. Its top PBW term gives $\mathrm{deg}{D}_{\beta }={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}{\sum }_{n\ge 1}K(\beta -n\alpha )={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}{\sum }_{n\ge 1}{m}_{\alpha ,n}(\beta )$. Both sums are finite.

Fix a pair $(\alpha ,n)$ for which $m_{\alpha ,n}(\beta )>0$. The hyperplane $H_{\alpha ,n}$ is then a factor hyperplane of $D_{\beta }$, so the generic-radical lemma applies. At a generic point of this hyperplane the kernel in weight $\lambda -\beta$ has dimension $K(\beta -n\alpha )$, and the transverse-pairing lemma says the determinant's order along a transverse line is exactly this dimension. Other affine factors are nonzero at the chosen generic point. Thus $m_{\alpha ,n}(\beta )=K(\beta -n\alpha )$.

For every pair with $m_{\alpha ,n}(\beta )=0$ one has $m_{\alpha ,n}(\beta )\le K(\beta -n\alpha )$ trivially; step 2.1 gives equality for every positive multiplicity. The equality of finite total sums in step 1.1 therefore forces equality at every pair, including all pairs with vanishing preliminary multiplicity. Substituting these exponents into the factorization proves the formula.