The first transverse Shapovalov pairing is perfect

Let $B_t$ be the restricted form in PBW coordinates, and let $R_{\beta }=\ker B_0$ in weight $\lambda -\beta$. For $x,y\in R_{\beta }$, the value $B_0^{\mathrm{\prime }}(x,y)$ is independent of their chosen lifts to first order, because $B_0$ vanishes whenever one argument is in $R_{\beta }$. Differentiating contravariance shows that these first derivatives assemble into a contravariant bilinear form on $R$: the extra terms from differentiating the module action are paired by $B_0$ with a radical vector and vanish. The form is symmetric.

Since $R\cong M(\mu )$, any contravariant form on it is determined by its value on the one-dimensional highest-weight line: move every negative-root operator across the form and use the highest-vector annihilation relations. Thus $B_0^{\mathrm{\prime }}{\mathrm{\mid }}_R$ is a scalar multiple of the Shapovalov form of $M(\mu )$. That form is nondegenerate because $M(\mu )$ is simple by the generic-radical lemma. It remains to prove the derivative's value on a nonzero highest vector $u_0\in R_{n\alpha }$ is nonzero.

Assume $B_0^{\mathrm{\prime }}(u_0,u_0)=0$. The matrix $B_0$ in weight $\mu$ is symmetric with kernel $\mathbb{C}{u}_0$, by the generic-radical lemma. The linear functional $-B_0^{\mathrm{\prime }}(u_0,-)$ annihilates that kernel, so it lies in the image of $B_0$. Choose a weight-$\mu$ vector $u_1$ with $B_0(u_1,-)=-B_0^{\mathrm{\prime }}(u_0,-)$ and set $u(t)=u_0+t{u}_1$. Then $B_t(u(t),w)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$ for every vector $w$ in that PBW weight space.

For each simple positive-root vector $e_i$ and any vector $w$ of the adjacent weight, contravariance gives $B_t(e_{i}u(t),w)=B_t(u(t),f_{i}w)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$. The specialized form is invertible at weight $\mu +{\alpha }_i$, since $R\cong M(\mu )$ has no weight above $\mu$. Therefore $e_{i}u(t)=0(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$ for every $i$, and the same holds for all positive-root vectors because the $e_i$ generate ${\mathfrak{n}}^{+}$.

In the deformed module the Cartan weight of $u(t)$ is $\mu +t\delta$. Step 4.1 and the usual Casimir calculation on a highest vector give $Cu(t)=(\mu +t\delta ,\mu +t\delta +2\rho )u(t)(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)$. But the same central element acts throughout $M(\lambda +t\delta )$ by $(\lambda +t\delta ,\lambda +t\delta +2\rho )$. Their constant terms agree because $\lambda \in H_{\alpha ,n}$; their linear coefficients differ by $2(\lambda -\mu ,\delta )=n(\alpha ,\alpha )\mathrm{\ne }0$. Since $u_0\mathrm{\ne }0$, these two congruences contradict each other. Thus $B_0^{\mathrm{\prime }}(u_0,u_0)\mathrm{\ne }0$, and step 2.1 proves perfectness on every radical weight space.

Choose bases adapted to the radical and a complementary subspace. The specialized matrix has a zero radical block and an invertible complementary block. The radical block of $B_t$ is $t{B}_0^{\mathrm{\prime }}{\mathrm{\mid }}_{R_{\beta }}+O(t^2)$, while its mixed blocks are $O(t)$. Since $B_0^{\mathrm{\prime }}{\mathrm{\mid }}_{R_{\beta }}$ and the complementary block are invertible, a block determinant expansion gives ${\mathrm{ord}}_{t=0}\det B_t=\dim R_{\beta }$. The generic-radical lemma identifies this dimension as $K(\beta -n\alpha )$, proving the asserted order.