4.6. Complex geometric perspective [045Q]

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4.6. Complex geometric perspective

The goal of this Section is to identify the holomorphic structure of the Kähler ansatz $(J^{(1)},\Omega^{(1)})$ . Recall the (1,0)-form $\zeta=V_{(1)}d\mu+\sqrt{-1}\vartheta$ and formula (1.14) for its differential. The main idea is to produce holomorphic differentials

(4.22)

\begin{cases}\zeta_{3}=\zeta+\beta_{13}d\eta_{1}+\beta_{23}d\eta_{2},\\ \zeta_{4}=-\zeta+\beta_{14}d\eta_{1}+\beta_{24}d\eta_{2},\end{cases}

by solving for the unknown functions $\beta_{13},\beta_{23},\beta_{14},\beta_{24}$ . The requirement for $d\zeta_{3}=d\zeta_{4}=0$ translates into an overdetermined and underdetermined system of equations

(4.23)

\begin{cases}\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}=-\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu},\quad&\frac{\partial\beta_{p3}}{\partial\mu}=2\frac{\partial v}{\partial\eta_{p}},\\ \frac{\partial\beta_{p4}}{\partial\bar{\eta}_{q}}=\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu},\quad&\frac{\partial\beta_{p4}}{\partial\mu}=-2\frac{\partial v}{\partial\eta_{p}},\\ \frac{\partial\beta_{p3}}{\partial\eta_{q}}=\frac{\partial\beta_{q3}}{\partial\eta_{p}},\quad&\frac{\partial\beta_{p4}}{\partial\eta_{q}}=\frac{\partial\beta_{q4}}{\partial\eta_{p}},\quad p,q=1,2.\end{cases}

The overdetermined nature is closely related to the integrability of the complex structure. The underdetermined nature is related to the fact that we can add certain holomorphic functions of $\eta_{1},\eta_{2}$ to $\beta_{p3},\beta_{p4}$ and solve the same equations; to eliminate this ambiguity one has to impose more growth conditions. Our strategy for solving this system is a direct construction using integral representations, and the main technical difficulty is to extract finite expressions out of divergent integrals.

We use the shorthand notation $|\eta|_{a}=(a_{p\bar{q}}\eta_{p}\bar{\eta}_{q})^{1/2}$ and $|y|_{a}=(a_{p\bar{q}}y_{p}y_{q})^{1/2}$ . We introduce two auxiliary functions

\gamma_{\pm}(\eta_{1},\eta_{2},\mu)=\frac{1}{3}\frac{A^{1/2}\mu}{|(\eta_{1},\eta_{2},\mu)|_{a}^{3}|\eta|_{a}^{2}}+\frac{2}{3}\frac{1}{|\eta|_{a}^{4}}(\frac{A^{1/2}\mu}{|(\eta_{1},\eta_{2},\mu)|_{a}}\mp 1),

and define for $p=1,2$ the series

(4.24)

\begin{cases}\gamma_{p3}(\eta_{1},\eta_{2},\mu)=\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{3a_{p\bar{q}}(\bar{\eta}_{q}+n_{q})}{8\pi^{2}A^{1/2}}\gamma_{+}(\eta_{1}+n_{1},\eta_{2}+n_{2},\mu),\\ \gamma_{p4}(\eta_{1},\eta_{2},\mu)=-\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{3a_{p\bar{q}}(\bar{\eta}_{q}+n_{q})}{8\pi^{2}A^{1/2}}\gamma_{-}(\eta_{1}+n_{1},\eta_{2}+n_{2},\mu).\end{cases}

These series converge absolutely for $(\eta_{1},\eta_{2})\notin\mathbb{Z}^{2}$ , and are 1-periodic in $\eta_{1},\eta_{2}$ variables. When $\eta_{1}=\eta_{2}=0$ , the series are designed so that $\gamma_{+}$ and $\gamma_{p3}$ extend smoothly over $\{\mu>0\}$ , while $\gamma_{-}$ and $\gamma_{p4}$ extend smoothly over $\{\mu<0\}$ . We will later use $\gamma_{p3}$ and $\gamma_{p4}$ as integrands to construct $\beta_{p3}$ and $\beta_{p4}$ .

Lemma 4.20.

(Differential identities)

\frac{\partial\gamma_{p3}}{\partial\mu}=-\frac{\partial\gamma_{p4}}{\partial\mu}=2\frac{\partial\gamma}{\partial\eta_{p}},\quad p=1,2,

and morever

\frac{\partial\gamma_{p3}}{\partial\eta_{q}}=\frac{\partial\gamma_{q3}}{\partial\eta_{p}},\quad\frac{\partial\gamma_{p4}}{\partial\eta_{q}}=\frac{\partial\gamma_{q4}}{\partial\eta_{p}},\quad p,q=1,2.

Proof.

We differentiate the series definition (4.9) of $\gamma$ to get

\frac{\partial\gamma}{\partial\eta_{p}}=\frac{3}{16\pi^{2}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{a_{p\bar{q}}(\bar{\eta}_{q}+n_{q})}{|(\eta_{1}+n_{1},\eta_{2}+n_{2},\mu)|_{a}^{5}}.

Using the elementary formula for indefinite integrals

\int\frac{1}{(s^{2}+t^{2})^{5/2}}ds=\frac{1}{3}\frac{s}{(s^{2}+t^{2})^{3/2}t^{2}}+\frac{2}{3}\frac{1}{t^{4}}(\frac{s}{\sqrt{s^{2}+t^{2}}}\mp 1),

we see

\int\frac{1}{|(\eta_{1},\eta_{2},\mu)|_{a}^{5}}d\mu=A^{-1/2}\gamma_{\pm},

or equivalently

\frac{\partial}{\partial\eta_{p}}\left(-\frac{1}{8\pi^{2}}\frac{1}{|(\eta_{1},\eta_{2},\mu)|_{a}^{3}}\right)=\frac{\partial}{\partial\mu}\left(\frac{3a_{p\bar{q}}\bar{\eta}_{q}}{16\pi^{2}A^{1/2}}\gamma_{\pm}\right).

Thus after summation

\frac{\partial\gamma}{\partial\eta_{p}}=\frac{\partial}{\partial\eta_{p}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\left(-\frac{1}{8\pi^{2}}\frac{1}{|(\eta_{1}+n_{1},\eta_{2}+n_{2},\mu)|_{a}^{3}}\right)=\frac{1}{2}\frac{\partial\gamma_{p3}}{\partial\mu}=-\frac{1}{2}\frac{\partial\gamma_{p4}}{\partial\mu}.

The ‘morever’ statement follows from summing over the elementary differential relations

(a_{p\bar{r}}\bar{\eta}_{r})\frac{\partial\gamma_{\pm}}{\partial\eta_{q}}(\eta_{1},\eta_{2},\mu)=(a_{q\bar{r}}\bar{\eta}_{r})\frac{\partial\gamma_{\pm}}{\partial\eta_{p}}(\eta_{1},\eta_{2},\mu),\quad p,q=1,2.

∎

By the periodicity of $\gamma_{p3},\gamma_{p4}$ , we may assume $|x_{1}|,|x_{2}|\leq\frac{1}{2}$ . In order to integrate $\gamma_{p3}$ and $\gamma_{p4}$ we need to bound these functions. It is convenient to introduce some closely related integrals:

(4.25)

\begin{cases}I_{01}(y_{1},y_{2},\mu)=\int_{\mathbb{R}^{2}}\frac{1}{|(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{3}a_{p\bar{q}}(s_{p}+\sqrt{-1}y_{p})(s_{q}-\sqrt{-1}y_{q})}ds_{1}ds_{2},\\ I_{02}(y_{1},y_{2},\mu)=\int_{\mathbb{R}^{2}}\frac{1}{|a_{p\bar{q}}(s_{p}+\sqrt{-1}y_{p})(s_{q}-\sqrt{-1}y_{q})|^{2}|(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)|_{a}}ds_{1}ds_{2},\\ I_{03}(y_{1},y_{2},\mu)=\int_{\mathbb{R}^{2}}\frac{1}{|a_{p\bar{q}}(s_{p}+\sqrt{-1}y_{p})(s_{q}-\sqrt{-1}y_{q})|^{2}}ds_{1}ds_{2},\end{cases}

and we can express

\begin{split}I_{p3}&=\frac{-3a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}A^{1/2}}\int_{\mathbb{R}^{2}}\gamma_{+}(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)ds_{1}ds_{2}\\ &=\frac{-a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}}(\mu I_{01}+2\mu I_{02}-2A^{-1/2}I_{03}),\end{split}

and

\begin{split}I_{p4}&=\frac{3a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}A^{1/2}}\int_{\mathbb{R}^{2}}\gamma_{-}(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)ds_{1}ds_{2}\\ &=\frac{a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}}(\mu I_{01}+2\mu I_{02}+2A^{-1/2}I_{03}).\end{split}

Lemma 4.21.

These integrals admit the simplified formulae:

\begin{cases}I_{01}=\frac{\pi}{\sqrt{\mathbb{A}}}\int_{\frac{A}{\mathbb{A}}|y|_{a}^{2}}^{\infty}\frac{1}{s(s+A\mu^{2})^{3/2}}ds,\\ I_{02}=-\frac{1}{2}I_{01}+\frac{\pi\sqrt{\mathbb{A}}}{A\varrho|y|_{a}^{2}}\\ I_{03}=\frac{\pi\sqrt{\mathbb{A}}}{A|y|_{a}^{2}}.\end{cases}

Consequently

\begin{cases}I_{p3}=\frac{a_{p\bar{q}}\sqrt{-1}y_{q}}{4\pi A^{1/2}\sqrt{\mathbb{A}}}\frac{1}{\varrho(\varrho+A^{1/2}\mu)},\\ I_{p4}=\frac{a_{p\bar{q}}\sqrt{-1}y_{q}}{4\pi A^{1/2}\sqrt{\mathbb{A}}}\frac{1}{\varrho(\varrho-A^{1/2}\mu)}.\end{cases}

Proof.

To evaluate these integrals, we introduce a radial variable

s=a_{p\bar{q}}(s_{p}+\sqrt{-1}y_{p})(s-\sqrt{-1}y_{q}),

and then elementary calculations in polar coordinates give

\begin{split}I_{01}=\frac{\pi}{\sqrt{\mathbb{A}}}\int_{\frac{A}{\mathbb{A}}|y|_{a}^{2}}^{\infty}\frac{1}{s(s+A\mu^{2})^{3/2}}ds,\end{split}

and similarly

\begin{split}I_{02}=\frac{\pi}{\sqrt{\mathbb{A}}}\int_{\frac{A}{\mathbb{A}}|y|_{a}^{2}}^{\infty}\frac{1}{s^{2}(s+A\mu^{2})^{1/2}}ds=-\frac{1}{2}I_{01}+\frac{\pi\sqrt{\mathbb{A}}}{A\varrho|y|_{a}^{2}},\end{split}

together with the formula for $I_{03}$ . The formulae for $I_{p3}$ and $I_{p4}$ follow from taking linear combinations. ∎

Lemma 4.22.

(Estimating integrands I) For $|y_{1}|+|y_{2}|\gtrsim 1$ and $|x_{1}|,|x_{2}|\leq\frac{1}{2}$ , we have the estimate

|\gamma_{p3}-\gamma_{p4}|\leq\frac{CA^{-1/4}|\mu|}{|y|_{a}\varrho}.

Morever there are improved estimates for $\gamma_{p3},\gamma_{p4}$ depending on the sign of $\mu$ :

\begin{cases}|\gamma_{p3}|\leq\frac{CA^{-3/4}|y|_{a}}{\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\quad&\mu\geq 0,\\ |\gamma_{p4}|\leq\frac{CA^{-3/4}|y|_{a}}{\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\quad&\mu\leq 0.\end{cases}

Proof.

Consider first the special case where $x_{1}=x_{2}=0,\eta_{p}=\sqrt{-1}y_{p}$ . By pairing $(n_{1},n_{2})$ with $(-n_{1},-n_{2})$ in the summation, we obtain

\begin{cases}\gamma_{p3}(\eta_{1},\eta_{2},\mu)=\frac{-3a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}A^{1/2}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\gamma_{+}(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu),\\ \gamma_{p4}(\eta_{1},\eta_{2},\mu)=\frac{3a_{p\bar{q}}\sqrt{-1}y_{q}}{8\pi^{2}A^{1/2}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\gamma_{-}(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu).\end{cases}

By the Cauchy integral test,

\begin{split}&\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{1}{|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{3}a_{p\bar{q}}(n_{p}+\sqrt{-1}y_{p})(n_{q}-\sqrt{-1}y_{q})}\\ \leq&CI_{01}\leq CA^{-1/2}\frac{1}{\varrho^{3}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})).\end{split}

Similarly,

\begin{split}&|\sum_{n_{1},n_{2}}\frac{1}{(a_{p\bar{q}}(n_{p}+\sqrt{-1}y_{p})(n_{q}-\sqrt{-1}y_{q}))^{2}|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}}|\\ \leq&CI_{02}\leq CA^{-1/2}\frac{1}{|y|_{a}^{2}\varrho}.\end{split}

Combining these two estimates,

|\gamma_{p3}-\gamma_{p4}|(\eta_{1},\eta_{2},\mu)\leq CA^{-1/2}|a_{pq}y_{q}||\mu|\frac{1}{|y|_{a}^{2}\varrho}\leq\frac{CA^{-1/4}|\mu|}{|y|_{a}\varrho}.

Morever, when $\mu\geq 0$ ,

\begin{split}&\frac{1}{a_{p\bar{q}}(n_{p}+\sqrt{-1}y_{p})(n_{q}-\sqrt{-1}y_{q})}|\frac{A^{1/2}\mu}{|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}}-1|\\ \leq&C\frac{1}{|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{2}},\end{split}

whence

\begin{split}&|\sum_{n_{1},n_{2}}\frac{1}{|a_{p\bar{q}}(n_{p}+\sqrt{-1}y_{p})(n_{q}-\sqrt{-1}y_{q})|^{2}}\left(\frac{A^{1/2}\mu}{|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}}-1\right)|\\ \leq&C\sum_{n_{1},n_{2}}\frac{1}{a_{p\bar{q}}(n_{p}+\sqrt{-1}y_{p})(n_{q}-\sqrt{-1}y_{q})|(n_{1}+\sqrt{-1}y_{1},n_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{2}}\\ \leq&C\int_{\mathbb{R}^{2}}\frac{1}{a_{p\bar{q}}(s_{p}+\sqrt{-1}y_{p})(s_{q}-\sqrt{-1}y_{q})|(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{2}}ds_{1}ds_{2}\\ \leq&CA^{-1/2}\frac{1}{\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})).\end{split}

This leads to

|\gamma_{p3}(\eta_{1},\eta_{2},\mu)|\leq\frac{CA^{-1}|a_{p\bar{q}}y_{q}|}{\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\quad\mu\geq 0.

Similarly

|\gamma_{p4}(\eta_{1},\eta_{2},\mu)|\leq\frac{CA^{-1}|a_{p\bar{q}}y_{q}|}{\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\quad\mu\leq 0.

For general $|x_{1}|,|x_{2}|\leq\frac{1}{2}$ , the difference $\gamma_{p3}(\eta_{1},\eta_{2},\mu)-\gamma_{p3}(\sqrt{-1}y_{1},\sqrt{-1}y_{2},\mu)$ , respectively $\gamma_{p4}(\eta_{1},\eta_{2},\mu)-\gamma_{p4}(\sqrt{-1}y_{1},\sqrt{-1}y_{2},\mu)$ , can be estimated by termwise comparing the two series using the methods above. The result is

\begin{cases}|\gamma_{p3}(\eta_{1},\eta_{2},\mu)-\gamma_{p3}(\sqrt{-1}y_{1},\sqrt{-1}y_{2},\mu)|\leq\frac{C(|x_{1}|+|x_{2}|)}{A^{1/2}\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\mu\geq 0,\\ |\gamma_{p4}(\eta_{1},\eta_{2},\mu)-\gamma_{p4}(\sqrt{-1}y_{1},\sqrt{-1}y_{2},\mu)|\leq\frac{C(|x_{1}|+|x_{2}|)}{A^{1/2}\varrho^{2}}\max(1,\log(\frac{A\mu^{2}}{|y|_{a}^{2}})),\mu\leq 0,\end{cases}

and

\begin{split}&|(\gamma_{p3}-\gamma_{p4})(\eta_{1},\eta_{2},\mu)-(\gamma_{p3}-\gamma_{p4})(\sqrt{-1}y_{1},\sqrt{-1}y_{2},\mu)|\leq\frac{C|\mu|(|x_{1}|+|x_{2}|)}{|y|_{a}^{2}\varrho},\end{split}

so the claims in the Lemma reduces to the special case $x_{1}=x_{2}=0$ above. ∎

Next we examine

(4.26)

\begin{split}\gamma_{p3}+\gamma_{p4}=&-\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{a_{p\bar{q}}(\bar{\eta}_{q}+n_{q})}{2\pi^{2}A^{1/2}|a_{i\bar{j}}(\eta_{i}+n_{i})(\bar{\eta}_{j}+n_{j})|^{2}}\\ =&\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{\partial}{\partial\eta_{p}}\frac{1}{2\pi^{2}A^{1/2}a_{i\bar{j}}(\eta_{i}+n_{i})(\bar{\eta}_{j}+n_{j})}.\end{split}

Lemma 4.23.

(Estimating integrands II) For $|y_{1}|+|y_{2}|\gtrsim 1$ and $|x_{1}|,|x_{2}|\leq\frac{1}{2}$ , we have the estimate

|\gamma_{p3}+\gamma_{p4}-\frac{\sqrt{-1}a_{p\bar{q}}y_{q}\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y|_{a}^{2}}|\leq\frac{C}{A^{1/2}|y|_{a}^{2}}.

Morever,

|\gamma_{p3}-I_{p3}|+|\gamma_{p4}-I_{p4}|\leq\frac{C|\mu|}{|y|_{a}^{2}\varrho}.

Proof.

Using the same strategy as in Lemma 4.22, we reduce to the special case $x_{1}=x_{2}=0,\eta_{p}=\sqrt{-1}y_{p}$ . Pairing $(n_{1},n_{2})$ with $(-n_{1},-n_{2})$ in the series (4.26),

(\gamma_{p3}+\gamma_{p4})(\eta_{1},\eta_{2},\mu)=\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{\sqrt{-1}a_{p\bar{q}}y_{q}}{2\pi^{2}A^{1/2}|a_{i\bar{j}}(n_{i}+\sqrt{-1}y_{i})(n_{j}-\sqrt{-1}y_{j})|^{2}}.

We compare this series expression of $\gamma_{p3}+\gamma_{p4}$ to the closely related integral (cf. Lemma 4.21)

\begin{split}\frac{\sqrt{-1}a_{p\bar{q}}y_{q}}{2\pi^{2}A^{1/2}}I_{03}=\frac{\sqrt{-1}a_{p\bar{q}}y_{q}\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y|_{a}^{2}}.\end{split}

The deviation between the series and the integral is bounded by

C|a_{p\bar{q}}y_{q}|A^{-1/4}\int_{\mathbb{R}^{2}}\frac{1}{|a_{i\bar{j}}(s_{i}+\sqrt{-1}y_{i})(s_{j}-\sqrt{-1}y_{j})|^{5/2}}ds_{1}ds_{2}\leq\frac{C}{A^{1/2}|y|_{a}^{2}},

using the same type of Cauchy integral test argument as Lemma 4.1.

The ‘morever’ statement is a minor variant of the proof of Lemma 4.22, where in the application of the Cauchy integral test we use the mean value inequality to estimate the difference between the series and the integral, similar to the argument in Lemma 4.1. ∎

We would like to use Lemma 4.22, 4.23 to construct functions $\beta_{p3},\beta_{p4}$ as integrals:

(4.27)

\begin{cases}\beta_{p3}(\eta_{1},\eta_{2},\mu)=-2\pi A^{1/2}\int_{S}\gamma_{p3}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}),\\ \beta_{p4}(\eta_{1},\eta_{2},\mu)=-2\pi A^{1/2}\int_{S}\gamma_{p4}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).\end{cases}

where we recall $d\mathcal{A}$ is the area form on $S$ . The problem is that these integrals diverge at the three ends of $S$ , and we need to extract some convergent limit to make sense of $\beta_{p3},\beta_{p4}$ , in a fashion rather similar to (4.11).

The ends of $S$ are up to exponentially small errors approximately $\mathfrak{D}_{i}\times S^{1}$ for $i=1,2,3$ . By Lemma 4.22, the expression $\beta_{p3}-\beta_{p4}$ makes sense as an ordinary integral with integrand $\gamma_{p3}-\gamma_{p4}$ thanks to the convergence of $\int^{\infty}\frac{1}{y^{2}}dy$ . It suffices to makes sense of $\beta_{p3}+\beta_{p4}$ . We consider the integral over large bounded regions with a cutoff scale $\Lambda$ ,

\int_{S\cap\{|y^{\prime}|_{a}<\Lambda\}}(\gamma_{p3}+\gamma_{p4})(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).

Lemma 4.23 tells us the exact nature of divergence. At the end $\mathfrak{D}_{1}\times S^{1}$ ,

\gamma_{p3}+\gamma_{p4}\sim-\frac{\sqrt{-1}a_{p\bar{q}}y_{q}^{\prime}\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y^{\prime}|_{a}^{2}}\sim-\frac{\sqrt{-1}a_{p\bar{2}}\sqrt{\mathbb{A}}}{2\pi A^{3/2}a_{2\bar{2}}y_{2}^{\prime}},\quad d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})\sim a_{2\bar{2}}dx_{2}^{\prime}\wedge dy_{2}^{\prime},

so the divergence behaviour of the integral is $\sim-\frac{\sqrt{-1}a_{p\bar{2}}\sqrt{\mathbb{A}}}{2\pi A^{3/2}}\log\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}$ at $\mathfrak{D}_{1}\times S^{1}$ . Similarly, the divergence behaviour is $\sim-\frac{\sqrt{-1}a_{p\bar{1}}\sqrt{\mathbb{A}}}{2\pi A^{3/2}}\log\frac{\Lambda}{\sqrt{a_{1\bar{1}}}}$ at $\mathfrak{D}_{2}\times S^{1}$ , and is $\sim\frac{\sqrt{-1}(a_{p\bar{1}}+a_{p\bar{2}})\sqrt{\mathbb{A}}}{2\pi A^{3/2}}\log\frac{\Lambda}{\sqrt{a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}}}}$ at $\mathfrak{D}_{3}\times S^{1}$ . The remarkable fact is that the divergent parts cancel out so that

\lim_{\Lambda\to\infty}\int_{S\cap\{|y^{\prime}|_{a}<\Lambda\}}(\gamma_{p3}+\gamma_{p4})(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})

converges; geometrically this cancellation comes from some balancing condition on the 3 directional vectors along $\mathfrak{D}_{1},\mathfrak{D}_{2},\mathfrak{D}_{3}$ . The upshot is that $\beta_{p3}$ and $\beta_{p4}$ make sense as improper integrals. The domain of definition for $\beta_{p3}$ is $(\mathbb{C}^{*})^{2}\times\mathbb{R}_{\mu}\setminus\{f_{S}=0,\mu\leq 0\}$ , and for $\beta_{p4}$ it is $(\mathbb{C}^{*})^{2}\times\mathbb{R}_{\mu}\setminus\{f_{S}=0,\mu\geq 0\}$ .

Lemma 4.24.

(Asymptotes as $\mu\to\pm\infty$ ) For any fixed $\eta_{1},\eta_{2}$ ,

\begin{cases}\lim_{\mu\to+\infty}\beta_{p3}(\eta_{1},\eta_{2},\mu)=0,\\ \lim_{\mu\to-\infty}\beta_{p4}(\eta_{1},\eta_{2},\mu)=0.\end{cases}

Morever

\begin{cases}\lim_{\mu\to+\infty}\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{p}}(\eta_{1},\eta_{2},\mu)=0,\\ \lim_{\mu\to-\infty}\frac{\partial\beta_{p4}}{\partial\bar{\eta}_{p}}(\eta_{1},\eta_{2},\mu)=0.\end{cases}

Proof.

We focus on the $\beta_{p3}$ case, and consider $A^{1/4}\mu\gg|\eta_{1}|+|\eta_{2}|+1$ . Using Lemma 4.22, the contribution to $\int\gamma_{p3}d\mathcal{A}$ from the region $\{|\eta_{1}-\eta_{1}^{\prime}|+|\eta_{2}-\eta_{2}^{\prime}|\leq(A^{1/4}\mu)^{1-\epsilon}\}\cap S$ is negligible, where $0<\epsilon\ll 1$ is any small given number. Outside this region $S$ is asymptotic to $\mathfrak{D}_{i}\times S^{1}$ along the three ends up to exponentially small error, and furthermore Lemma 4.23 allows us to replace $\gamma_{p3}$ by $I_{p3}$ without affecting the $\mu\to+\infty$ limit.

We are now left to consider the improper integral

\int_{\cup(\mathfrak{D}_{i}\times S^{1})}I_{p3}(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).

Using the formula of $I_{p3}$ in Lemma 4.21, we can simplify further by setting $y_{1}=y_{2}=0$ without affecting the $\mu\to+\infty$ limit. Along the $\mathfrak{D}_{1}\times S^{1}$ end,

\int_{\mathfrak{D}_{1}\times S^{1}\cap\{y_{2}^{\prime}<\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}\}}I_{p3}(-y_{1}^{\prime},-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=\int_{0}^{\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}}I_{p3}(0,-y_{2}^{\prime},\mu)a_{2\bar{2}}dy_{2}^{\prime},

which we compute as

\begin{split}&\frac{-a_{p\bar{2}}\sqrt{-1}}{4\pi A^{1/2}\sqrt{\mathbb{A}}}\int_{0}^{\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}}a_{2\bar{2}}y_{2}^{\prime}dy_{2}^{\prime}\frac{1}{|(0,-y_{2}^{\prime},\mu)|_{a}^{\prime}(|(0,-y_{2}^{\prime},\mu)|_{a}^{\prime}+A^{1/2}\mu)}\\ =&\frac{-a_{p\bar{2}}\sqrt{-1}\sqrt{\mathbb{A}}}{8\pi A^{3/2}}\int_{A\mu^{2}}^{\frac{A}{\mathbb{A}}\Lambda^{2}+A\mu^{2}}\frac{ds}{s^{1/2}(s^{1/2}+A^{1/2}\mu)}\\ =&\frac{-a_{p\bar{2}}\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right).\end{split}

Similarly, the integrals from $\mathfrak{D}_{2}\times S^{1}$ and $\mathfrak{D}_{3}\times S^{1}$ are respectively

\frac{-a_{p\bar{1}}\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right)

and

\frac{(a_{p\bar{1}}+a_{p\bar{2}})\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right).

Summing over the three contributions and take the limit $\Lambda\to+\infty$ ,

\int_{\cup(\mathfrak{D}_{i}\times S^{1})}I_{p3}(-y_{1}^{\prime},-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=0.

This proves $\lim_{\mu\to+\infty}\beta_{p3}(\eta_{1},\eta_{2},\mu)=0$ . Likewise with the $\beta_{p4}$ case.

The ‘morever’ statement follows from a simpler argument. The key is that higher derivatives of the integrand have faster decay at large distance, so that the divergence issues do not arise. ∎

Lemma 4.25.

The explicit formula for $\beta_{p3}+\beta_{p4}$ is

\beta_{p3}+\beta_{p4}=\frac{-2\pi ie^{2\pi i\eta_{p}}}{1-e^{2\pi i\eta_{1}}-e^{2\pi i\eta_{2}}}+K_{p}(a)=\frac{-2\pi ie^{2\pi i\eta_{p}}}{f_{S}}+K_{p}(a),\quad p=1,2.

where the constant $K_{p}(a)$ is

\begin{split}K_{p}(a)=&\frac{\sqrt{-1}(a_{p\bar{2}}\text{Re}(a_{1\bar{2}})-a_{p\bar{1}}a_{2\bar{2}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{2\bar{2}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}))\\ +&\frac{\sqrt{-1}(a_{p\bar{1}}\text{Re}(a_{1\bar{2}})-a_{p\bar{2}}a_{1\bar{1}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{1\bar{1}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}})).\end{split}

Proof.

The basic strategy is a Liouville theorem argument: we will construct a function with the same distributional $\Delta_{a}$ -Laplacian as $\beta_{p3}+\beta_{p4}$ , and then argue they must be equal.

We start with the Poincaré-Lelong formula

S=\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|1-e^{2\pi i\eta_{1}}-e^{2\pi i\eta_{1}}|^{2}=\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|f_{S}|^{2},

from which we obtain the equality of measures

\begin{split}\int_{S}d\mathcal{A}=&\int_{S}\frac{\sqrt{-1}}{2}a_{p\bar{q}}d\eta_{p}\wedge d\bar{\eta}_{q}\\ =&\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|f_{S}|^{2}\wedge\frac{\sqrt{-1}}{2}a_{p\bar{q}}d\eta_{p}\wedge d\bar{\eta}_{q}\\ =&\frac{1}{4\pi}(\Delta_{a}\log|f_{S}|^{2})d\text{Vol}_{a}.\end{split}

The periodic Newtonian potential on $(\mathbb{C}^{*})^{2}$ with the $g_{a}$ -metric is

\gamma_{5}(\eta_{1},\eta_{2},\mu)=-\frac{1}{4\pi^{2}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\{\frac{1}{a_{p\bar{q}}(\eta_{p}+n_{p})(\bar{\eta}_{q}+n_{q})}-\frac{1}{a_{p\bar{q}}n_{p}n_{q}}\}.

Thus for any large cutoff scale $\Lambda$ , the Green’s representation

\int_{S\cap\{|y^{\prime}|_{a}<\Lambda\}}\gamma_{5}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})

has the same distributional $\Delta_{a}$ -Laplacian as that of $\frac{1}{4\pi}\log|f_{S}|^{2}$ in the large compact region. Taking the $\eta_{p}$ derivative and taking the $\Lambda\to\infty$ limit shows that the $\Delta_{a}$ -Laplacian of $\frac{-ie^{2\pi i\eta_{p}}}{2f_{S}}$ agrees with that of the improper integral

\int_{S}\frac{\partial}{\partial\eta_{p}}\gamma_{5}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}),

which by formula (4.26) is the same as the improper integral

-\frac{1}{2}A^{1/2}\int_{S}(\gamma_{p3}+\gamma_{p4})(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=\frac{1}{4\pi}(\beta_{p3}+\beta_{p4}).

The upshot is that $\beta_{p3}+\beta_{p4}$ differs from $\frac{-2\pi ie^{2\pi i\eta_{p}}}{f_{S}}$ by a globally smooth $\Delta_{a}$ -harmonic function on $(\mathbb{C}^{*})^{2}$ . It is also easy to show using techniques in this Section that this difference can have at most log growth in $y_{1},y_{2}$ variables. Thus it has to be a constant.

The rest of this proof is to pin down precisely this constant, by considering the limit $(\eta_{1},\eta_{2})=(\sqrt{-1}b,\sqrt{-1}b)$ for $b\to+\infty$ . This uses techniques similar to the proof of Lemma 4.24. Without affecting the limit, we can replace $S$ with $\cup\mathfrak{D}_{i}\times S^{1}$ and replace $(\gamma_{p3}+\gamma_{p4})$ with

\frac{\sqrt{-1}a_{p\bar{q}}(b-y_{q}^{\prime})\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y-y^{\prime}|_{a}^{2}}.

This leads to an asymptotic expression for $b\gg 1$ ,

(\beta_{p3}+\beta_{p4})(\sqrt{-1}b,\sqrt{-1}b)\sim\int_{\cup\mathfrak{D}_{i}\times S^{1}}\frac{\sqrt{-1}a_{p\bar{q}}(-b+y_{q}^{\prime})\sqrt{\mathbb{A}}}{A|y-y^{\prime}|_{a}^{2}}d\mathcal{A},

where the RHS is understood as an improper integral. To evaluate this integral we fix $b$ and calculate the $\Lambda\to+\infty$ asymptotic expression of the integral over the large bounded domain $(\cup\mathfrak{D}_{i}\times S^{1})\cap\{|y^{\prime}|_{a}<\Lambda\}.$ The contribution from the end $\mathfrak{D}_{1}\times S^{1}$ is

\begin{split}&\frac{\sqrt{-1}a_{p\bar{2}}\sqrt{\mathbb{A}}}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})\\ +&\frac{\sqrt{-1}(a_{p\bar{2}}\text{Re}a_{1\bar{2}}-a_{p\bar{1}}a_{2\bar{2}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{2\bar{2}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}))+o(1).\end{split}

The contribution from $\mathfrak{D}_{2}\times S^{1}$ is

\begin{split}&\frac{\sqrt{-1}a_{p\bar{1}}\sqrt{\mathbb{A}}}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})\\ +&\frac{\sqrt{-1}(a_{p\bar{1}}\text{Re}a_{1\bar{2}}-a_{p\bar{2}}a_{1\bar{1}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{1\bar{1}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}))+o(1).\end{split}

The contribution from $\mathfrak{D}_{3}\times S^{1}$ is

-\frac{\sqrt{-1}\sqrt{\mathbb{A}}(a_{p\bar{1}}+a_{p\bar{2}})}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})+o(1).

Summing up, the log terms cancel out, so the improper integral

\int_{\cup\mathfrak{D}_{i}\times S^{1}}\frac{\sqrt{-1}a_{p\bar{q}}(-b+y_{q}^{\prime})\sqrt{\mathbb{A}}}{A|y-y^{\prime}|_{a}^{2}}d\mathcal{A},

is equal to the constant $K_{p}(a)$ defined in the statement of the Lemma. This shows limiting value

\lim_{b\to+\infty}(\beta_{p3}+\beta_{p4})(\sqrt{-1}b,\sqrt{-1}b)=K_{p}(a).

Comparing this with

\lim_{b\to+\infty}\frac{-2\pi ie^{2\pi i\eta_{p}}}{f_{S}}(\sqrt{-1}b,\sqrt{-1}b)=0

determines the constant. ∎

Remark 4.6.

The trigonometric factors in $K_{p}(a)$ have elementary geometric interpretations. The Euclidean metric $g_{a}^{\prime}$ induces an inner product on $\mathbb{R}^{2}_{y_{1},y_{2}}$ . Then the angles between the asymptotic directions of $S$ are

\begin{cases}\angle(\mathfrak{D}_{1},\mathfrak{D}_{3})=\frac{\pi}{2}+\arctan(\frac{a_{2\bar{2}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}),\\ \angle(\mathfrak{D}_{2},\mathfrak{D}_{3})=\frac{\pi}{2}+\arctan(\frac{a_{1\bar{1}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}).\end{cases}

Remark 4.7.

We have chosen a special ray $(\eta_{1},\eta_{2})=(\sqrt{-1}b,\sqrt{-1}b)$ to calculate the asymptotic value of $\beta_{p3}+\beta_{p4}$ . More generally $\mathfrak{D}_{1},\mathfrak{D}_{2},\mathfrak{D}_{3}$ divide the plane $\mathbb{R}^{2}_{y_{1},y_{2}}$ into three sectors, and the asymptotic value of function

\beta_{p3}+\beta_{p4}=\frac{-2\pi ie^{2\pi i\eta_{p}}}{1-e^{2\pi i\eta_{1}}-e^{2\pi i\eta_{2}}}+K_{p}(a)

along the ray $\{(y_{1},y_{2})=s\vec{e}\text{ for }s>0\}$ specified by a directional vector $\vec{e}$ depends on which sector $\vec{e}$ belongs to, and can have a jumping discontinuity as we cross $\mathfrak{D}_{i}$ . This is known as Stokes phenomenon in complex analysis.

Proposition 4.26.

The functions $\beta_{p3}$ and $\beta_{p4}$ solve the overdetermined system (4.23). Equivalently, the $(1,0)$ -forms $\zeta_{3},\zeta_{4}$ defined by (4.22) are holomorphic differentials.

Proof.

Starting from the definition of the function $v$ in terms of $\gamma_{i}$ (cf. (4.11)), we can differentiate with respect to $\eta_{p}$ to get

\frac{\partial v}{\partial\eta_{p}}=-2\pi A^{1/2}\int_{S}\frac{\partial\gamma}{\partial\eta_{p}}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).

Using the differential relations in Lemma 4.20,

\begin{split}2\frac{\partial v}{\partial\eta_{p}}&=-2\pi A^{1/2}\int_{S}\frac{\partial\gamma_{p3}}{\partial\mu}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})\\ &=-2\pi A^{1/2}\frac{\partial}{\partial\mu}\int_{S}\gamma_{p3}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})\\ &=\frac{\partial\beta_{p3}}{\partial\mu},\end{split}

and similarly $2\frac{\partial v}{\partial\eta_{p}}=-\frac{\partial\beta_{p4}}{\partial\mu}.$

Next we study $\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}$ in the complement of $\{f_{S}=0,\mu\leq 0\}$ . We have

\frac{\partial^{2}\beta_{p3}}{\partial\bar{\eta}_{q}\partial\mu}=2\frac{\partial^{2}v}{\partial\eta_{p}\partial\bar{\eta}_{q}}=-\frac{1}{2}\frac{\partial^{2}w^{p\bar{q}}}{\partial\mu\partial\mu},

where the second equality uses the distributional equation (4.6). But by Lemma 4.24, for fixed $\eta_{1},\eta_{2}$ ,

\lim_{\mu\to+\infty}\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}(\eta_{1},\eta_{2},\mu)=0,

and the asymptotes we obtained in Section 4.2, 4.3 easily imply

\lim_{\mu\to+\infty}\frac{\partial w^{p\bar{q}}}{\partial\mu}(\eta_{1},\eta_{2},\mu)=0.

Thus we can integrate from $\mu=+\infty$ to obtain

\frac{\partial\beta_{p3}}{\partial\bar{\eta}_{q}}=-\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu}.

A completely parallel argument shows

\frac{\partial\beta_{p4}}{\partial\bar{\eta}_{q}}=\frac{1}{2}\frac{\partial w^{p\bar{q}}}{\partial\mu}.

Finally by integrating the second part of Lemma 4.20 we see

\frac{\partial\beta_{p3}}{\partial\eta_{q}}=\frac{\partial\beta_{q3}}{\partial\eta_{p}},\quad\frac{\partial\beta_{p4}}{\partial\eta_{q}}=\frac{\partial\beta_{q4}}{\partial\eta_{p}},\quad p,q=1,2.

∎

To compute the periods of the integrals $\int\zeta_{3}$ and $\int\zeta_{4}$ , we recall from the topological description (cf. review Section 1.1.4, 1.1.5) that there are 3 generating $S^{1}$ -cycles in $H_{1}(T^{3})$ , one of which is the $S^{1}$ -fibre, and the other two come from lifting $T^{2}\subset(\mathbb{C}^{*})^{2}\times\mathbb{R}_{\mu}$ to the total space, which involve monodromy issues.

Lemma 4.27.

For appropriate choices of constants $0\leq\theta_{31}^{\infty},\theta_{32}^{\infty}\leq 2\pi$ , the $T^{3}$ -periods of the holomorphic differentials

(4.28)

\begin{cases}d\log z_{3}=\zeta_{3}-\sqrt{-1}(\theta_{13}^{\infty}d\eta_{1}+\theta_{23}^{\infty}d\eta_{2})\\ d\log z_{4}=\zeta_{4}+\sqrt{-1}(\theta_{13}^{\infty}d\eta_{1}+\theta_{23}^{\infty}d\eta_{2})-(K_{1}(a)d\eta_{1}-K_{2}(a)d\eta_{2})\end{cases}

take values in $2\pi\sqrt{-1}\mathbb{Z}$ ; here $K_{p}(a)$ are the constants defined in Lemma 4.25. In particular, the holomorphic functions $z_{3}$ and $z_{4}$ are defined without multivalue issues. For a suitable choice of multiplicative normalisation on $z_{3},z_{4}$ we have the functional equation

(4.29)

z_{3}z_{4}=f_{S}=1-z_{1}-z_{2}=1-e^{2\pi i\eta_{1}}-e^{2\pi i\eta_{2}}.

Proof.

This Lemma is parallel to Lemma 3.10, so we will only highlight the key issues. The constants $\theta_{13}^{\infty}$ and $\theta_{23}^{\infty}$ are the asymptotic holonomy as $\mu\to+\infty$ of the $S^{1}$ -connection $\vartheta$ , along the $S^{1}$ -cycles in the base corresponding to the $x_{1}$ and $x_{2}$ variables respectively. These are introduced in order to cancel the twist of $\vartheta$ by a flat connection.

The functional equation follows from

\begin{split}\log z_{3}+\log z_{4}=&(\beta_{13}+\beta_{14}-K_{1}(a))d\eta_{1}+(\beta_{23}+\beta_{24}-K_{2}(a))d\eta_{2}\\ =&-2\pi i\frac{e^{2\pi i\eta_{1}}d\eta_{1}+e^{2\pi i\eta_{2}}d\eta_{2}}{f_{S}}\\ =&d\log f_{S}.\end{split}

which crucially uses Lemma 4.25. ∎

We have thus defined a holomorphic map away from the singular locus of the $S^{1}$ -fibration on the negative vertex $M^{-}$ :

M^{-}\setminus S\to\{z_{3}z_{4}=1-z_{1}-z_{2}\}\subset\mathbb{C}^{*}_{z_{1}}\times\mathbb{C}^{*}_{z_{2}}\times\mathbb{C}^{2}_{z_{3},z_{4}}.

Here the functional equation allows us to extend the map holomorphically across $\{\mu\neq 0,f_{S}=0\}$ . However the complex structure on $M^{-}$ is not a priori defined along $S\cap M^{-}$ .

Lemma 4.28.

The holomorphic functions $z_{3},z_{4}$ on $M^{-}$ extend continuously over the singular locus $S$ where they attain the value zero. Morever $z_{3},z_{4}$ are $C^{2,\alpha}$ -regular with respect to $g^{(1)}$ -metric.

Proof.

By construction $\log|z_{3}|$ is a function of $\eta_{1},\eta_{2},\mu$ with differential

d\log|z_{3}|=V_{(1)}d\mu+\text{Re}(\beta_{13}d\eta_{1}+\beta_{23}d\eta_{2})+\text{Im}(\theta^{\infty}_{13}d\eta_{1}+\theta_{23}^{\infty}d\eta_{2}).

In particular the positivity of $V_{(1)}$ in $M^{-}$ means $\log|z_{3}|$ is increasing in $\mu$ . Around a given point $P\in S\cap M^{-}$ , we first show continuity of $z_{3}$ at $P$ . Observe

\log|z_{3}|(\eta_{1},\eta_{2},\mu)=\log|z_{3}|(\eta_{1},\eta_{2},A^{-1/4})+\int_{A^{-1/4}}^{\mu}V_{(1)}d\mu.

Here $\log|z_{3}|(\eta_{1},\eta_{2},\mu=A^{-1/4})$ is locally $L^{\infty}$ by smoothness of $\log z_{3}$ in $\{\mu>0\}$ . Applying Proposition 4.16 and neglecting all locally bounded terms, as $(\eta_{1},\eta_{2},\mu)\to(\eta_{1}(P),\eta_{2}(P),0)$ ,

\log|z_{3}|\sim\int_{A^{-1/4}}^{\mu}\frac{A^{1/2}}{2R}d\mu\sim\frac{1}{2}\log(\frac{A^{1/2}\mu+R}{A^{1/4}})\to-\infty,

or equivalently $|z_{3}|\to 0$ as required. The case of $z_{4}$ is completely analogous.

Since $(g^{(1)},\Omega^{(1)})$ is $C^{1,\alpha}$ -regular by Proposition 4.19, holomorphicity implies that $z_{3},z_{4}$ are $C^{2,\alpha}$ -regular in the local chart of Section 4.4. ∎

Proposition 4.29.

The map $M^{-}\to\{z_{3}z_{4}=1-z_{1}-z_{2}\}$ is a holomorphic open embedding. The $S^{1}$ -action is identified as

e^{i\theta}\cdot(z_{1},z_{2},z_{3},z_{4})=(z_{1},z_{2},e^{i\theta}z_{3},e^{-i\theta_{1}}z_{4}),

and the holomorphic volume form is $\Omega^{(1)}=-\frac{\sqrt{-1}}{4\pi^{2}z_{1}z_{2}}dz_{2}\wedge dz_{3}\wedge dz_{4}$ .

Proof.

The $S^{1}$ -action can be identified as in Proposition 2.11. The holomorphic volume form is characterised by $\iota_{\frac{\partial}{\partial\theta}}\Omega^{(1)}=d\eta_{1}\wedge d\eta_{2}$ , which is compared to

\begin{split}\iota_{\frac{\partial}{\partial\theta}}(\sqrt{-1}dz_{2}\wedge dz_{3}\wedge dz_{4})=&-d(z_{3}z_{4})\wedge dz_{2}=-d(1-z_{1}-z_{2})\wedge dz_{2}=dz_{1}\wedge dz_{2}\\ =&(2\pi\sqrt{-1}z_{1}d\eta_{1})\wedge(2\pi\sqrt{-1}z_{2}d\eta_{2})=-4\pi^{2}z_{1}z_{2}d\eta_{1}\wedge d\eta_{2},\end{split}

to yield $\Omega^{(1)}=-\frac{\sqrt{-1}}{4\pi^{2}z_{1}z_{2}}dz_{2}\wedge dz_{3}\wedge dz_{4}$ .

This holomorphic volume form formula in particular shows the map $M^{-}\to\{z_{3}z_{4}=1-z_{1}-z_{2}\}$ is a local biholomorphism wherever the complex structure is defined. We finally need to show this map is injective. Since both $M^{-}$ and $\{z_{3}z_{4}=1-z_{1}-z_{2}\}$ fibre over $\mathbb{C}^{*}_{z_{1}}\times\mathbb{C}^{*}_{z_{2}}$ in a compatible way, it suffices to compare the $\mathbb{C}^{*}$ -fibres. The map between the fibres is equivariant with respect to the $S^{1}$ -action, so to conclude injectivity we only need to recall from the proof of Lemma 4.28 that $\log|z_{3}|$ is a monotone function of $\mu$ . ∎