This essentially follows from the fact that $P$ is located on the slice $\{z=0\}$ and $H$ is a complex submanifold of $D$. Indeed, we can decompose

(3.266)

$$G_{P}=\psi_{1}\wedge dz+\mathcal{R}_{1},$$

where $\mathcal{R}_{1}$ does not involve $dz$. Given any compactly supported test form $\chi\in\Omega_{0}^{2n-4}(Q)$, we can write

(3.267)

$$\chi=\beta\wedge dz+\gamma,$$

where $\gamma$ does not involve $dz$. Immediately,

(3.268)

$$(\mathcal{R}_{1},\Delta\gamma)=\int_{Q}\mathcal{R}_{1}\wedge\Delta\gamma=0$$

and

(3.269)

$$(\psi_{1}\wedge dz,\Delta(\beta\wedge dz))=\int_{Q}\psi_{1}\wedge dz\wedge\Delta(\beta\wedge dz)=0.$$

So it follows that

(3.270)

$$(\mathcal{R}_{1},\Delta\chi)=(\mathcal{R}_{1},\Delta(\beta\wedge dz))=(G_{P},\Delta(\beta\wedge dz))=2\pi\int_{P}(\beta\wedge dz)=0.$$

This implies that $\Delta\mathcal{R}_{1}=0$ in the distributional sense. By the standard elliptic regularity, we have $\mathcal{R}_{1}\in C^{\infty}$.

Now write

(3.271)

$$\psi_{1}=\psi+\psi_{2},$$

where $\psi$ is $J_{D}$-invariant, i.e. of type $(1,1)$ in $D$, and $\psi_{2}$ is anti-$J_{D}$-invariant. Since $H$ is a complex submanifold of $D$, the Dirac current $\delta_{P}$ is $J_{D}$-invariant, hence $J_{D}(G_{P})$ is also a Green’s current for $P$, so we see that $\psi_{2}=\frac{1}{2}(G_{P}-J(G_{P}))$ is smooth. Then we have

(3.272)

$$G_{P}=\psi\wedge dz+\mathcal{R},$$

where $\mathcal{R}=\mathcal{R}_{1}+\psi_{2}\wedge dz$ is smooth. Similarly since the $\delta_{P}$ is invariant under $z\mapsto-z$, the difference $\psi(z)-\psi(-z)$ is smooth.

To see the expansion of $\psi$, we notice that $H$ is a Kähler, in particular minimal, submanifold of $D$. So the mean curvature of $H$ in $D$ vanishes. Also notice $\partial_{z}$ is parallel on $Q$ so $A_{i\alpha\beta}=0$ if either $\alpha=3$ or $\beta=3$. This then implies that

(3.273)

$$\psi=\frac{\sqrt{-1}}{4r}dy\wedge d\bar{y}+\frac{1}{2r}(yd\bar{y}+\bar{y}dy)\wedge\Gamma+r\cdot\mathbb{A}+r^{-3}\Pi_{2}^{(4)}+\widetilde{O}(r^{2}),$$

where

(3.274)		$\displaystyle\Gamma$	$\displaystyle=-\frac{1}{2}A_{i12}dx_{i},$
(3.275)		$\displaystyle\mathbb{A}$	$\displaystyle=-\frac{1}{4}A_{ij\alpha\beta}dx_{i}\wedge dx_{j}.$

In particular, $\mathbb{A}=d\Gamma$. Re-writing

(3.276)

$$A_{i12}=\langle\nabla_{\partial_{x_{i}}}\partial_{y_{2}},\partial_{y_{1}}\rangle$$

in terms of the complex coordinates $y,\bar{y}$ and bearing in mind (3.261) we obtain the desired formula for $\Gamma$. ∎