The goal is to show $A_{0}=1$ and $A_{1}=0$ in the expansion (3.317). We work in the above special coordinates centered at $p\in H$.

The first step is to show that the $O^{\prime}(1)$-term in the expansion of $\psi$ given by Proposition 3.24 in fact vanishes along $N_{0}(p)$. To this end, notice that $\partial_{y}=\sigma=\partial_{w_{1}}$ at $p\in H$ and hence by Lemma 3.27,

(3.324)

$$w_{1}=y+a_{2}y^{2}+\widetilde{O}(|y|^{3}).$$

Since the only non-trivial Christofell symsbols at $p$ are $\Gamma_{ij}^{1}$ and $\Gamma_{\bar{i}\bar{j}}^{\bar{1}}$ for $i,j\geq 2$, it easily follows that

(3.325)

$$d|\sigma|(p)=0.$$

Combining (3.325) and Lemma 3.25,

(3.326)

$$\Gamma(p)=\frac{1}{2}(d_{H}^{c}\log|\sigma|)(p)=0,$$

for each $p\in H$. Therefore, along the fiber $N_{0}(p)$ of the normal bundle $N_{0}(p)$, the expansion of $\psi$ in Proposition 3.24 becomes

(3.327)

$$\psi=\frac{\sqrt{-1}}{4|y|}dy\wedge d\bar{y}+O(|y|).$$

Next, we will compute the coefficients $A_{0}(p)$ and $A_{1}(p)$ in (3.317). As in the proof of Lemma 3.27, we obtain that

(3.328)

$$\frac{\partial w_{j}}{\partial y}(p)=\frac{\partial w_{j}}{\partial\bar{y}}=0,\ \ j\geq 2,$$

and

(3.329)

$$\frac{\partial^{2}w_{j}}{\partial y^{2}}(p)=\frac{\partial^{2}w_{j}}{\partial y\partial\bar{y}}(p)=\frac{\partial^{2}w_{j}}{\partial\bar{y}^{2}}(p)=0,\ \ j\geq 1.$$

This particularly implies that $a_{2}(p)=0$ and along the fiber $N_{0}(p)$,

(3.330)

$$w_{j}=O(|y|^{3}),\ \ j\geq 2.$$

By Lemma 3.29, $\partial_{w_{1}}g_{i\bar{j}}(p)=0$ for all $1\leq i,j\leq n-1$, then the expansion of $\omega_{D}$ along the fiber $N_{0}(p)$ is at least quadratic in the $w_{1}$-direction, i.e.

(3.331)		$\displaystyle\omega_{D}$	$\displaystyle=$	$\displaystyle\frac{\sqrt{-1}}{2}\Big(dw_{1}\wedge d\bar{w}_{1}+\sum_{j=2}^{n-1}dw_{j}\wedge d\bar{w}_{j}\Big)+O\Big(\sqrt{\sum_{j=2}^{n-1}|w_{j}|^{2}}\Big)+O(|w_{1}|^{2})$
			$\displaystyle=$	$\displaystyle\frac{\sqrt{-1}}{2}(dw_{1}\wedge d\bar{w}_{1}+\sum_{j=2}^{n-1}dw_{j}\wedge d\bar{w}_{j})+O(|y|^{2}).$

By (3.324) and (3.330), along the fiber $N_{0}(p)$, we have

(3.332)

$$dw_{1}=dy+O(|y|^{2}).$$

and

(3.333)

$$dw_{j}=dw_{j}^{\prime}+O(|y|^{2}),\ \ j\geq 2.$$

So we get

(3.334)

$$\omega_{D}=\frac{\sqrt{-1}}{2}(dy\wedge d\bar{y}+\sum_{j=2}^{n-1}dw_{j}^{\prime}\wedge d\bar{w}_{j}^{\prime})+O(|y|^{2})$$

Since by definition,

(3.335)

$$\Big(\Tr_{\omega_{D}}\psi\Big)\cdot\frac{\omega_{D}^{n-1}}{(n-1)!}=\psi\wedge\frac{\omega_{D}^{n-2}}{(n-2)!}.$$

by elementary manipulations we get that $A_{0}(p)=1$ and $A_{1}(p)=0$. ∎