Consider $\gamma_{4}$ at a given point in the region (4.14). Its integral formula (4.11) can be split into two parts, corresponding to far away sources $|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime}\gtrsim A^{1/4}$ and nearby sources $|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime}\lesssim A^{1/4}$.

For far away sources, we use Lemma 4.1 to write the integrand $\gamma$ as a dominant term $-\frac{1}{4\pi\sqrt{\mathbb{A}}|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime}}$ plus a remainder term estimated by $\frac{C}{|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime 3}}$. The dominant term does not contribute to $\gamma_{4}-\bar{\gamma}_{4}$ because it is constant in the $x_{1},x_{2}$ direction. The remainder term contribution to $\gamma_{4}$ is bounded by

$$CA^{1/2}\text{Im}\int_{S\cap\{\text{dist}\gtrsim A^{1/4}\}}\frac{1}{|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime 3}}\sqrt{-1}d\eta_{2}^{\prime}\wedge d\bar{\eta}_{1}^{\prime}\leq CA^{-1/4}.$$

The contribution from nearby sources only arises if our given point of interest is too close to $S$ along one of $\mathfrak{D}_{1}$, $\mathfrak{D}_{2}$ or $\mathfrak{D}_{3}$ directions; we focus on $\mathfrak{D}_{1}$. Lemma 4.2 allows us to write $\gamma$ as $-\frac{1}{8\pi^{2}|(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)|_{a}^{3}}$ plus a well controlled remainder term. By the exponential decay property of the measure $\text{Im}\sqrt{-1}d\eta_{2}^{\prime}\wedge d\bar{\eta}_{1}^{\prime}$,

$$\begin{split}&CA^{1/2}e^{-2\pi y_{2}}\int_{S\cap\{\text{dist}\lesssim A^{1/4}\}}\frac{\sqrt{-1}d\eta_{1}^{\prime}\wedge d\bar{\eta}_{1}^{\prime}}{|(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)|_{a}^{3}}\leq Ce^{-2\pi y_{2}}R^{-1}\leq CA^{-1/4}.\end{split}$$

Combining the above shows $|\gamma_{4}-\bar{\gamma}_{4}|\leq CA^{-1/4}$.

All these arguments carry through to $\gamma_{1},\gamma_{2},\gamma_{3}$ except the exponential decay of the measure. This is compensated by staying sufficiently far from $\mathfrak{D}_{i}$. ∎

This proof is parallel to Proposition 3.5, so will be sketchy. We focus on $\gamma_{4}-\bar{\gamma}_{4}$ as the same arguments work for $\gamma_{i}-\bar{\gamma}_{i}$.

We perform Fourier decomposition in the periodic variables $x_{1},x_{2}$,

$$\gamma_{4}-\bar{\gamma}_{4}=\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}\setminus\{0\}}h_{n_{1},n_{2}}(y_{1},y_{2},\mu)e^{2\pi in_{1}x_{1}+2\pi in_{2}x_{2}}.$$

The zeroth Fourier mode vanishes by construction. Parseval identity combined with Lemma 4.12 shows

$$\sum_{n_{1},n_{2}}|h_{n_{1},n_{2}}|^{2}=\int_{0}^{1}\int_{0}^{1}|\gamma_{4}-\bar{\gamma}_{4}|^{2}dx_{1}dx_{2}\leq CA^{-1/2}.$$

Over the region (4.14), according to Proposition 4.4 and (4.16)

$$\Delta_{a}(\gamma_{4}-\bar{\gamma}_{4})=0,$$

which translates into the Helmholtz type equations

$$\Delta_{a}^{\prime}h_{n_{1},n_{2}}-4\pi\sqrt{-1}\text{Im}(a^{1\bar{2}})(n_{1}\frac{\partial h_{n_{1},n_{2}}}{\partial y_{2}}-n_{2}\frac{\partial h_{n_{1},n_{2}}}{\partial y_{1}})-4\pi^{2}(a^{p\bar{q}}n_{p}n_{q})h_{n_{1},n_{2}}=0.$$

After the variable substitution

$$\tilde{h}_{n_{1},n_{2}}=h_{n_{1},n_{2}}\exp\left(\frac{2\pi i\text{Im}(a_{1\bar{2}})}{\mathbb{A}}(a_{1\bar{1}}n_{2}y_{1}-\text{Re}(a_{1\bar{2}})n_{1}y_{1}+\text{Re}(a_{1\bar{2}})n_{2}y_{2}-a_{2\bar{2}}n_{1}y_{2})\right),$$

these equations become the Helmholtz equations on $\mathbb{R}^{2}_{y_{1},y_{2}}\times\mathbb{R}_{\mu}$,

$$\Delta_{a}^{\prime}\tilde{h}_{n_{1},n_{2}}=k_{n_{1},n_{2}}^{2}\tilde{h}_{n_{1},n_{2}},$$

where we denote $k_{n_{1},n_{2}}=2\pi(a^{p\bar{q}}n_{p}n_{q})^{1/2}(\frac{A}{\mathbb{A}})^{1/2}$.

The rest of the argument is substantially similar to Proposition 3.5. Observe that $|\text{dist}_{g_{a}^{\prime}}(\cdot,\text{Im}(S))-R|\leq CA^{1/4}$. An upper barrier supersolution to the Helmholtz equation is directly constructed as

$$\begin{split}h_{n_{1},n_{2}}^{\prime}=A^{-1/4}&\{\int_{\mathfrak{D}_{1}}e^{-k_{n_{1},n_{2}}|(y_{1},y_{2}-s,\mu)|_{a}^{\prime}}ds+\int_{\mathfrak{D}_{2}}e^{-k_{n_{1},n_{2}}|(y_{1}-s,y_{2},\mu)|_{a}^{\prime}}ds\\ &+\int_{\mathfrak{D}_{3}}e^{-k_{n_{1},n_{2}}|(y_{1}-s,y_{2}-s,\mu)|_{a}^{\prime}}ds\},\end{split}$$

where $ds=dy_{2},dy_{1},-dy_{1}$ on $\mathfrak{D}_{1},\mathfrak{D}_{2},\mathfrak{D}_{3}$ respectively. Comparing the real and imaginary parts of $\tilde{h}_{n_{1},n_{2}}$ with $C(|n_{1}|+|n_{2}|)h_{n_{1},n_{2}}^{\prime}$ yields the result. ∎

As a preliminary remark, although $\omega^{(1)}$ is a Kähler form only in a bounded region, it makes sense as a closed 2-form over the entire $(\mathbb{C}^{*})^{2}\times\mathbb{R}_{\mu}$. The integral $\int_{T^{2}}\omega^{(1)}$ is a cohomological invariant, which can be evaluated asymptotically on a $T^{2}$-cycle as $\eta_{1},\eta_{2}$ stay bounded and $\mu\to+\infty$. By choosing the $T^{2}$-cycle on which $\mu,y_{1},y_{2}$ are constants,

$$\begin{split}\int_{T^{2}}\omega^{(1)}=&\frac{\sqrt{-1}}{2}\int_{T^{2}}W^{p\bar{q}}_{(1)}d\eta_{p}\wedge d\bar{\eta}_{q}=\frac{\sqrt{-1}}{2}\int_{T^{2}}(W^{1\bar{2}}_{(1)}-W^{2\bar{1}}_{(1)})dx_{1}\wedge dx_{2}\\ =&\int_{T^{2}}\text{Im}(W^{2\bar{1}}_{(1)})dx_{1}\wedge dx_{2}=\text{Im}(a_{2\bar{1}})+\int_{T^{2}}\gamma_{4}dx_{1}\wedge dx_{2}.\end{split}$$

But by Lemma 4.10 and Proposition 4.13, the quantity $\gamma_{4}\to 0$ as $\mu\to+\infty$, so the only contribution is $\int_{T^{2}}\omega^{(1)}=\text{Im}(a_{2\bar{1}}).$ ∎