The $\mu_{1},\mu_{2}$ derivatives are clear. For the $\bar{\eta}$ derivative, we can apply the component form of the distributional equation (2.4) away from $\mathfrak{D}$, to see

$$\begin{split}\frac{\partial\beta_{1}}{\partial\bar{\eta}}&=2\lim\int\frac{\partial^{2}v^{11}}{\partial\eta\partial\bar{\eta}}d\mu_{1}+\frac{\partial^{2}v^{12}}{\partial\eta\partial\bar{\eta}}d\mu_{2}\\ &=-\frac{1}{2}\lim\int\frac{\partial^{2}w}{\partial\mu_{1}\partial\mu_{1}}d\mu_{1}+\frac{\partial^{2}w}{\partial\mu_{1}\partial\mu_{2}}d\mu_{2}=-\frac{1}{2}\frac{\partial w}{\partial\mu_{1}},\end{split}$$

where in the last equality we compare the asymptotic values at infinity to show there is no constant term depending on $\eta$. Likewise with $\beta_{2},\beta_{0}$. ∎

By Lemma 2.7 the sum $\beta_{1}+\beta_{2}+\beta_{0}$ is independent of $\mu_{1},\mu_{2}$. Given $\eta\neq 0$, we shall evaluate this sum at the limit point $(\mu_{1}\to-\infty,\mu_{2}\to-\infty,\mu_{1}-\mu_{2}\to+\infty)$. Then $\beta_{0}$ has no contribution, while $\beta_{1}$ contributes

$$2\lim_{\mu_{1}-\mu_{2}\to+\infty}\int_{\mu_{1}=+\infty,\text{fix $\mu_{1}-\mu_{2}$}}^{\mu_{1}=-\infty}\frac{\partial\alpha_{1}}{\partial{\eta}}d\mu_{1},$$

and $\beta_{2}$ contributes

$$2\lim_{\mu_{1}\to-\infty}\int_{\mu_{2}=+\infty,\text{fix $\mu_{1}$}}^{\mu_{2}=-\infty}\frac{\partial\alpha_{2}}{\partial{\eta}}d\mu_{2}$$

plus

$$-2\lim_{\mu_{1}\to-\infty}\int_{\mu_{1}-\mu_{2}=-\infty,\text{fix $\mu_{1}$}}^{\mu_{1}-\mu_{2}=+\infty}\frac{\partial\alpha_{3}}{\partial{\eta}}d(\mu_{1}-\mu_{2}).$$

Observe

$$\lim_{\mu_{1}\to-\infty,\text{fix $\mu_{1}-\mu_{2}$}}\alpha_{3}(\mu_{1},\mu_{2},\eta)=\frac{1}{2\sqrt{(\mu_{1}-\mu_{2})^{2}+(a_{11}+2a_{12}+a_{22})|\eta|^{2}}},$$

$$\lim_{\mu_{1}\to-\infty,\text{fix $\mu_{1}-\mu_{2}$}}\frac{\partial\alpha_{3}}{\partial\eta}=\frac{-(a_{11}+2a_{12}+a_{22})\bar{\eta}}{4((\mu_{1}-\mu_{2})^{2}+(a_{11}+2a_{12}+a_{22})|\eta|^{2})^{3/2}}$$

Using Lebesgue dominated convergence theorem, the third integral contribution is equal to

$$\frac{(a_{11}+2a_{12}+a_{22})\bar{\eta}}{2}\int_{-\infty}^{+\infty}\frac{1}{(x^{2}+(a_{11}+2a_{12}+a_{22})|\eta|^{2})^{3/2}}dx=\frac{1}{\eta}.$$

The other two contributions are zero by similar arguments. ∎

We focus on the neighbourhood of $\mathfrak{D}_{1}$. Modulo smooth terms

$$\alpha_{1}\sim\frac{1}{2\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}},\quad\frac{\partial\alpha_{1}}{\partial\eta}\sim-\frac{a_{22}\bar{\eta}}{4({\mu_{1}^{2}+a_{22}|\eta|^{2}})^{3/2}},$$

so by the integral definitions, along $\mathfrak{D}_{1}$ the function $\beta_{2}$ is non-singular, and

$$\beta_{1}\sim\frac{-1}{2\eta}(\frac{\mu_{1}}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}-1),\quad\beta_{0}\sim\frac{1}{\eta}-\beta_{1}=\frac{1}{2\eta}(\frac{\mu_{1}}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}+1).$$

Now

$$\begin{split}&d\log|z_{1}|=(a_{11}+v^{11})d\mu_{1}+(a_{12}+v^{12})d\mu_{2}+\text{Re}(\beta_{1}d\eta)\\ &\sim\frac{1}{2\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}d\mu_{1}+\frac{1}{2}(1-\frac{\mu_{1}}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}})d\log|\eta|+a_{11}d\mu_{1}+a_{12}d\mu_{2}\\ &=\frac{1}{2}d\log|\eta|+\frac{1}{2}d\sinh^{-1}(\frac{\mu_{1}}{\sqrt{a_{22}}|\eta|})+d(a_{11}\mu_{1}+a_{12}\mu_{2}),\end{split}$$

hence up to multiplying by a smooth function

$$|z_{1}|\sim\text{const}\cdot(\frac{\mu_{1}}{\sqrt{a_{22}}}+\sqrt{\frac{\mu_{1}^{2}+a_{22}|\eta|^{2}}{{a_{22}}}})^{1/2}e^{a_{11}\mu_{1}+a_{12}\mu_{2}}\to 0$$

as the point moves to $\mathfrak{D}_{1}$. Similarly

$$|z_{0}|\sim\text{const}\cdot(-\frac{\mu_{1}}{\sqrt{a_{22}}}+\sqrt{\frac{\mu_{1}^{2}+a_{22}|\eta|^{2}}{{a_{22}}}})^{1/2}e^{-a_{11}\mu_{1}-a_{12}\mu_{2}-a_{21}\mu_{1}-a_{22}\mu_{2}}\to 0.$$

The function $\log z_{2}$ encounters no singularity along $\mathfrak{D}_{1}$. These calculations guarantee the continuous extension of the holomorphic functions $z_{0},z_{1},z_{2}$ over $\mathfrak{D}_{1}$. Since the complex structure is compatible with the smooth topology by Section 2.3, these holomorphic functions in fact extend smoothly along $\mathfrak{D}_{1}$. We remark that what happens in these calculations is essentially identical to the Taub-NUT metric near the origin. ∎

To see the main ideas, let us focus on $\eta=0$ and let $\mu_{1},\mu_{2}\to 0$. By construction $d\log|z_{1}|=V^{1j}_{(1)}d\mu_{j}+\text{Re}(\beta_{1}d\eta).$ Restricted to $\eta=0$,

$$d\log|z_{1}|=(a_{11}+v^{11})d\mu_{1}+(a_{11}+v^{12})d\mu_{2}=d(a_{11}\mu_{1}+a_{12}\mu_{2})+\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2}),$$

hence

$$|z_{1}|=\text{const}\cdot e^{a_{11}\mu_{1}+a_{12}\mu_{2}}\exp\left(\int^{(\mu_{1},\mu_{2})}\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})\right).$$

We need to show $|z_{1}|\to 0$ as $\mu_{1},\mu_{2}\to 0$, namely $\int\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})\to-\infty.$ Now because $\alpha_{1},\alpha_{3}$ are positive, this integral viewed as a function of $\mu_{1}$ and $\mu_{1}-\mu_{2}$ is an increasing functions of both variables, so it suffices to show this integral decreases to $-\infty$ as $(\mu_{1},\mu_{2})\to 0$ along the ray $\mathfrak{D}_{2}$:

$$\begin{split}\int\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})=&\log|\mu_{1}|\{\frac{1}{2}+\frac{1}{2\pi}\arctan(\frac{a_{12}}{\sqrt{A}})+\frac{1}{2\pi}\arctan(\frac{-a_{11}-a_{12}}{\sqrt{A}})\}\\ =&\log|\mu_{1}|\{\frac{1}{4}+\frac{1}{2\pi}\arctan(\frac{a_{12}+a_{22}}{\sqrt{A}})\}\to-\infty,\end{split}$$

where the first equality uses that the arctan functions are constant on the ray $\mathfrak{D}_{2}$, and the second equality is an elementary trignometric identity.

In the more general case of $\eta\neq 0$ the $\arctan$ factor would no longer be exactly constant, but one can still make $|z_{1}|$ arbitrarily small for sufficiently small $|\eta|,\mu_{1},\mu_{2}$. The cases of $z_{0}$ and $z_{2}$ are completely analogous. ∎

To identify the $T^{2}$-action we examine the Hamiltonian vector field action. Recall that $\{\frac{\partial}{\partial\theta_{i}}\}_{i=1,2}$ is dual to the connection $\{\vartheta_{i}\}_{i=1,2}$. We compute

$$\mathcal{L}_{\frac{\partial}{\partial\theta_{1}}}z_{i}=dz_{i}(\frac{\partial}{\partial\theta_{1}})=z_{i}\zeta_{i}^{\prime}(\frac{\partial}{\partial\theta_{1}}),$$

in particular

$$\mathcal{L}_{\frac{\partial}{\partial\theta_{1}}}z_{1}=z_{1}\sqrt{-1}\vartheta_{1}(\frac{\partial}{\partial\theta_{1}})=\sqrt{-1}z_{1},\quad\mathcal{L}_{\frac{\partial}{\partial\theta_{1}}}z_{0}=-\sqrt{-1}z_{0},\quad\mathcal{L}_{\frac{\partial}{\partial\theta_{1}}}z_{2}=0,$$

from which the first circle action is clear. Likewise with the second circle action.

The holomorphic (3,0)-form $\Omega$ is uniquely determined by the condition that $\Omega(\frac{\partial}{\partial\theta_{1}},\frac{\partial}{\partial\theta_{2}},\cdot)=d\eta$. But

$$-dz_{0}\wedge dz_{1}\wedge dz_{2}(\frac{\partial}{\partial\theta_{1}},\frac{\partial}{\partial\theta_{2}},\cdot)=z_{0}z_{1}dz_{2}+z_{1}z_{2}dz_{0}+z_{0}z_{2}dz_{1}=d\eta$$

where in the last step we used the functional equation $z_{0}z_{1}z_{2}=\eta$. This shows $\Omega=-dz_{0}\wedge dz_{1}\wedge dz_{2}$. In particular the map $M\setminus\{0\}\to\mathbb{C}^{3}\setminus\{0\}$ is locally invertible.

To show the map is a homeomorphism, we notice that it is compatible with the fibration structure $M\to\mathbb{C}_{\eta}$ and $\mathbb{C}^{3}\to\mathbb{C}_{\eta}$, so it suffices to show the fibres are identified, which follows from looking at the complexification of the $T^{2}$-action into a $(\mathbb{C}^{*})^{2}$-action.

Combining the above proves the biholomorphism claim. ∎