The uniqueness follows from the fact that the difference of any two Green’s currents for $P$ differ by a harmonic form, which must vanish by the asymptotic condition (3.349).

Now we focus on the proof of the global existence of $G_{P}$ on $Q$. Let $\{\phi_{j}\}_{j=0}^{\infty}$ be a complete orthonormal basis of eigenvectors for the Hodge Laplacian acting on real-valued $2$-forms on $D$, and let $\Lambda_{D}\equiv\{\lambda_{j}\}_{j=0}^{\infty}$ be the corresponding spectrum. Our basic strategy is to first obtain a formal series expression of $G_{P}$ and then prove the convergence of this series.

To begin with, the Dirac $3$-current $\delta_{P}$ of $P\subset Q$ has a formal expansion along the $D$ direction

(3.352)

$$\delta_{P}=\sum_{j=0}^{\infty}f_{j}(z)\phi_{j}\wedge dz,$$

where $f_{j}(z)$ is a $0$-current on $\mathbb{R}$ and given by

(3.353)

$$f_{j}(z)\equiv 2\pi\Big(\int_{H}*_{D}\phi_{j}\Big)\delta_{0}(z),$$

where $\delta_{0}(z)$ is the standard Dirac $0$-current acting on functions on $\mathbb{R}$, supported at the slice $\{z=0\}$. For $\lambda_{j}>0$, by Hodge theory, $f_{j}$ is non-zero only when $\phi_{j}$ is a closed real $(1,1)$-form because $H$ is a closed complex submanifold in $D$. So we only restrict to the subset of such $\lambda_{j}$’s. Furthermore, if $\phi_{j}$ is harmonic, then

(3.354)

$$\int_{H}*_{D}\phi_{j}=\int_{D}\frac{d}{2\pi}\omega_{D}\wedge*_{D}\phi_{j}=\frac{d}{2\pi}\langle\omega_{D},\phi_{j}\rangle_{L^{2}(D)}.$$

It follows that there is exactly one $j$, which we may assume to be $0$, such that $\lambda_{j}=0$ and $f_{j}$ is non-zero. The corresponding eigenform is normalized to be

(3.355)

$$\phi_{0}\equiv\frac{1}{(\int_{D}\omega_{D}^{n}/n!)^{1/2}}\cdot\omega_{D}.$$

Now let $G_{P}$ be the formal series

(3.356)

$$G_{P}=\sum\limits_{j=0}^{\infty}h_{j}\cdot(\phi_{j}\wedge dz),$$

where $h_{j}$ satisfies

(3.357)

$$-h_{j}^{\prime\prime}(z)+\lambda_{j}\cdot h_{j}(z)=f_{j}(z).$$

For each $j\in\mathbb{Z}_{+}$, we can write a formal solution

	$\displaystyle h_{j}(z)$	$\displaystyle=\frac{\pi}{-\sqrt{\lambda_{j}}}\Big(e^{-\sqrt{\lambda_{j}}\cdot z}\int_{-\infty}^{z}e^{\sqrt{\lambda_{j}}\cdot u}f_{j}(u)du+e^{\sqrt{\lambda_{j}}z}\int_{z}^{\infty}e^{-\sqrt{\lambda_{j}}\cdot u}f_{j}(u)du\Big)$
(3.358)			$\displaystyle=\begin{cases}\frac{\pi}{-\sqrt{\lambda_{j}}}\cdot e^{-\sqrt{\lambda_{j}}\cdot z}\cdot\int_{H}*_{D}(\phi_{j}),&z>0,\\ \frac{\pi}{-\sqrt{\lambda_{j}}}\cdot e^{\sqrt{\lambda_{j}}\cdot z}\cdot\int_{H}*_{D}(\phi_{j}),&z\leq 0.\end{cases}$

For $j=0$, a solution is given by a piecewise linear function

(3.359)

$\displaystyle h_{0}(z)=\begin{cases}k_{+}z,&z\geq 0,\\ k_{-}z,&z\leq 0.\end{cases}$

Notice that the formal solution $h_{0}(z)$ is unique up to the addition of a linear function in $z$. Fixing a choice of $h_{0}$ we then obtain a formal solution $G_{P}$.

Next we show that the above formal series $G_{P}$ is well-defined by showing the formal solution indeed converges in the weak sense and has some exponential decaying rate as $|z|$ large, which consists of two steps.

In the first step, we claim that globally the formal expansion

(3.360)

$$G_{P}\equiv\sum\limits_{j=0}^{\infty}h_{j}\cdot(\phi_{j}\wedge dz)$$

in fact gives a well-defined $3$-current on $Q$ and the series converges in the following sense: for any test form $\chi\in\Omega_{0}^{m-3}(Q)$,

(3.361)

$\displaystyle\sum\limits_{j=0}^{N}\Big(h_{j}(z)\phi_{j}\wedge dz,\chi\Big)\longrightarrow(G_{P},\chi)\ \text{as}\ N\to\infty.$

It suffices to show that for any smooth test form $\chi\in\Omega_{0}^{m-3}(Q)$ and for any $N\in\mathbb{N}$,

(3.362)

$$\sum\limits_{j=0}^{N}\Big|\Big(h_{j}(z)\phi_{j}\wedge dz,\chi\Big)\Big|\leq C_{0},$$

where $C_{0}>0$ is independent of $N$. To see this, for each $j$, we write

(3.363)		$\displaystyle\Big(h_{j}(z)\phi_{j}\wedge dz,\chi\Big)$	$\displaystyle=$	$\displaystyle\int_{Q}h_{j}(z)\phi_{j}\wedge dz\wedge\chi$
			$\displaystyle=$	$\displaystyle\int_{\mathbb{R}}\Big(h_{j}(z)\cdot\int_{D}\langle\chi,*_{D}(\phi_{j})\rangle\Big)dz.$

The estimate (3.363) can be accomplished in the following manner. To begin with, we will show that the integral $\int_{D}\langle\chi,*_{D}(\phi_{j})\rangle$ has an uniform bound which is independent of $z$. In fact, notice that $(\Delta)^{k}\phi_{j}=(\lambda_{j})^{k}\phi_{j}$ holds for any $k\in\mathbb{Z}_{+}$, then

(3.364)		$\displaystyle\int_{D}\langle\chi(z),*_{D}(\phi_{j})\rangle\dvol_{g}$	$\displaystyle=$	$\displaystyle\frac{1}{(\lambda_{j})^{k}}\int_{D}\langle\chi(z),*_{D}(\Delta)^{k}(\phi_{j})\rangle\dvol_{g}$
			$\displaystyle=$	$\displaystyle\frac{1}{(\lambda_{j})^{k}}\int_{D}\langle\chi(z),(\Delta)^{k}*_{D}(\phi_{j})\rangle\dvol_{g}$
			$\displaystyle=$	$\displaystyle\frac{1}{(\lambda_{j})^{k}}\int_{D}\langle(\Delta)^{k}\chi(z),*_{D}(\phi_{j})\rangle.$

Lemma 3.32 implies

(3.365)

$$\|\phi_{j}\|_{C^{0}(D)}\leq C\cdot(\lambda_{j})^{\frac{n}{2}},$$

where $C$ depends only on $n$ and the metric $g$. So it follows that

(3.366)

$$\Big|\int_{D}\langle\chi(z),*_{D}(\phi_{j})\rangle\dvol_{g}\Big|\leq C\cdot\|\chi\|_{C^{2k}(Q)}\cdot\frac{1}{(\lambda_{j})^{k-\frac{n}{2}}}.$$

Next, we will estimate the integral $\int_{\mathbb{R}}h_{j}(z)dz$. To this end, for each $j\in\mathbb{Z}_{+}$, let $\phi_{j}$ satsify

(3.367)

$\displaystyle\begin{cases}\Delta\phi_{j}=\lambda_{j}\cdot\phi_{j}\\ \|\phi_{j}\|_{L^{2}(D)}=1.\end{cases}$

By Lemma 3.32, for each $k\in\mathbb{N}$,

(3.368)

$$|\nabla^{k}_{\omega_{D}}\phi_{j}|\leq C(\lambda_{j})^{\frac{k+n}{2}}.$$

For fixed constant $z_{0}>10^{2}$, applying (3.358) and (3.368),

(3.369)		$\displaystyle\Big|\int_{\mathbb{R}}h_{j}(z)dz\Big|$	$\displaystyle\leq$	$\displaystyle\int_{-\infty}^{-z_{0}}|h_{j}(z)|dz+\int_{-z_{0}}^{z_{0}}|h_{j}(z)|dz+\int_{z_{0}}^{+\infty}|h_{j}(z)|dz$
			$\displaystyle\leq$	$\displaystyle C(1+\lambda_{j}^{\frac{n}{2}-1}).$

Combining the above estimates, we have

(3.370)

$$|(h_{j}(z)\phi_{j}\wedge dz,\chi)|=\Big|\int_{\mathbb{R}}\Big(h_{j}(z)\cdot\int_{D}\langle\chi,*_{D}(\phi_{j})\rangle\Big)dz\Big|\leq C\cdot\|\chi\|_{C^{2k}(Q)}\cdot\frac{C(1+\lambda_{j}^{\frac{n}{2}-1})}{(\lambda_{j})^{k-\frac{n}{2}}}.$$

Then applying Weyl’s law, if $k$ is sufficiently large, then the above series converges as stated in (3.362), which completes the proof of the claim.

At our next stage, we will study the exponential decaying behavior of the current $G_{P}$ defined in (3.360). For any $j\in\mathbb{Z}_{+}$ and for any number $z>10^{n^{2}+k^{2}}$, we have

(3.371)

$$|\nabla^{k}_{Q}(h_{j}(z)\cdot\phi_{j})|\leq C(\lambda_{j})^{\frac{2n+k-1}{2}}e^{-\sqrt{\lambda_{j}}z}.$$

Notice that by elementary computations, for each $\delta\in(0,10^{-2})$, there is some $z_{0}>10^{n^{2}+k^{2}}$ such that for all $z\in(z_{0},+\infty)$ and $j\in\mathbb{Z}_{+}$,

(3.372)

$$(\lambda_{j})^{\frac{2n+k-1}{2}}\cdot e^{-\delta\cdot\sqrt{\lambda_{j}}z}\leq(\lambda_{j})^{-6n}.$$

This implies that

(3.373)

$$\sum\limits_{j=0}^{\infty}|\nabla^{k}_{Q}(h_{j}(z)\cdot\phi_{j})|\leq Ce^{-(1-\delta)\sqrt{\lambda_{1}}z}\cdot\sum\limits_{j=0}^{\infty}(\lambda_{j})^{-6n}.$$

By Weyl’s law implies that the above numerical series converges, and hence for each $k\in\mathbb{N}$ $\nabla_{Q}^{k}(G_{P}-h_{0}(z))$ has an exponential decaying rate as $z\rightarrow+\infty$. The argument is identical for $z<0$.

The only remaining part is to show that the series $G_{P}$ defined by (3.360) satisfies the current equation

(3.374)

$$\Delta G_{P}=2\pi\delta_{P}$$

in the distributional sense, i.e., for any $\chi\in\Omega_{0}^{m-3}(Q)$,

(3.375)

$$(G_{P},\Delta\chi)=2\pi\int_{P}\chi.$$

Applying the definition of $f_{j}$, $h_{j}$ and integration by parts, it is straightforward that for each $j\in\mathbb{N}$,

(3.376)

$$(h_{j}(z)\phi_{j}\wedge dz,\Delta\chi)=(f_{j}(z)\phi_{j}\wedge dz,\chi).$$

Since $\chi\in\Omega_{0}^{m-3}(Q)$, the smooth $2$-form $*_{D}(\chi)$ has the following $L^{2}$-expansion on the slice $D\times\{0\}$,

(3.377)

$$*_{D}(\chi(0))=\sum\limits_{j=0}^{\infty}\langle*_{D}(\chi(0)),\phi_{j}\rangle_{D}\cdot\phi_{j}=\sum\limits_{j=0}^{\infty}\langle\chi(0),*_{D}(\phi_{j})\rangle_{D}\cdot\phi_{j}$$

and hence

(3.378)

$$\chi(0)=\sum\limits_{j=0}^{\infty}\Big(\int_{D}\phi_{j}\wedge\chi(0)\Big)*_{D}(\phi_{j}).$$

This implies that

(3.379)

$$\Big(\sum\limits_{j=0}^{\infty}f_{j}\cdot\phi_{j}\wedge dz,\chi\Big)=2\pi\sum\limits_{j=0}^{\infty}\Big(\int_{P}*_{D}(\phi_{j})\Big)\cdot\int_{D}\phi_{j}\wedge\chi(0)=\int_{P}\chi(0)=2\pi\int_{P}\chi.$$

Therefore,

(3.380)

$$(G_{P},\Delta\chi)=\Big(\sum\limits_{j=0}^{\infty}h_{j}\cdot\phi_{j}\wedge dz,\Delta\chi\Big)=\Big(\sum\limits_{j=0}^{\infty}f_{j}\cdot\phi_{j}\wedge dz,\chi\Big)=2\pi\int_{P}\chi=2\pi\delta_{P}(\chi),$$

which completes the proof. ∎

First, we prove Item (1). Since $Q$ is a Riemannian product, we have for $z\neq 0$,

(3.382)

$$\frac{d^{2}}{dz^{2}}\psi(z)=\Delta_{D}\psi(z)=dd^{*}\psi(z)$$

is exact, which implies that the cohomology class $[\psi(z)]\in H^{2}(D;\mathbb{R})$ is locally constant for $z\in\mathbb{R}\setminus\{0\}$. On the other hand, by the exponential decay property in (3.349) we see that

(3.383)

$$\lim_{z\rightarrow\pm\infty}[\psi(z)]=\lim_{z\rightarrow\infty}k_{\pm}[\omega_{D}].$$

For Item (2), denote

(3.384)

$$\tilde{\psi}(z)\equiv\psi(-z)+(k_{-}+k_{+})z\omega_{D}.$$

Then $\tilde{\psi}\wedge dz$ is also a Green current for $P$ and it is also asymptotic to $k_{\pm}z\cdot\omega_{D}$ as $z\rightarrow\pm\infty$. Therefore by uniqueness, $\tilde{\psi}(z)=\psi(z)$. Taking the $z$-derivative at $z=0$ we get the conclusion. ∎