At the first stage, we will analyze the regularity of $\omega$. By definition,

(4.68)

$$T^{\frac{n-2}{n}}\omega=T\pi^{*}\omega_{D}+\pi^{*}\psi+dz\wedge\Theta.$$

To start with, let us compute the lifting $\pi^{*}\psi$. By (3.264),

(4.69)

$$\pi^{*}\psi=\pi^{*}\tilde{\omega}_{0}+\frac{1}{2r}(yd\bar{y}+\bar{y}dy)\wedge\Gamma+rd\Gamma+\pi^{*}(O^{\prime}(r)dy+O^{\prime}(r)d\bar{y})+\pi^{*}O^{\prime}(r^{2}),$$

where

(4.70)

$$\tilde{\omega}_{0}=\frac{\sqrt{-1}}{4r}dy\wedge d\bar{y}$$

is the standard form in the model setting (2.45). We also notice that

(4.71)		$\displaystyle\pi^{*}(O^{\prime}(r)dy+O^{\prime}(r)d\bar{y})$	$\displaystyle=sO^{\prime}(s^{2}),$
(4.72)		$\displaystyle\pi^{*}O^{\prime}(r^{2})$	$\displaystyle=O^{\prime}(s^{4}).$

Now by definition

(4.73)

$$\Theta=\Theta_{0}+\frac{z}{r}\Gamma+\frac{k_{-}+k_{+}}{k_{-}-k_{+}}\Gamma+\theta+\theta_{f}=\Theta_{0}+\frac{z}{r}\Gamma+O^{\prime}(s^{2}).$$

Moreover, according to the discussions in Section 2, we have

(4.74)

$$\pi^{*}\tilde{\omega}_{0}+dz\wedge\Theta_{0}=\omega_{\mathbb{C}^{2}},$$

where $\omega_{\mathbb{C}^{2}}=\frac{\sqrt{-1}}{2}(du_{1}\wedge d\bar{u}_{1}+du_{2}\wedge d\bar{u}_{2})$ is the standard Kähler form of $\mathbb{C}^{2}$. Therefore,

(4.75)

$\displaystyle\pi^{*}\psi+dz\wedge\Theta=$

$\displaystyle\omega_{\mathbb{C}^{2}}+rd\Gamma+\frac{1}{2r}(yd\bar{y}+\bar{y}dy)\wedge\Gamma+dz\wedge(\frac{z}{r}\Gamma)+O^{\prime}(s^{3}).$

Using the relation $r^{2}=|y|^{2}+z^{2}$ and the simple computation

(4.76)

$$d(r\Gamma)=rd\Gamma+dr\wedge\Gamma=rd\Gamma+\frac{1}{2r}(yd\bar{y}+\bar{y}dy)\wedge\Gamma+dz\wedge(\frac{z}{r}\Gamma),$$

we have

	$\displaystyle\pi^{*}\psi+dz\wedge\Theta=$	$\displaystyle\omega_{\mathbb{C}^{2}}+d(r\Gamma)+O^{\prime}(s^{3})$
(4.77)		$\displaystyle=$	$\displaystyle\omega_{\mathbb{C}^{2}}+O^{\prime}(s^{3}),$

where we use the fact that $r=\frac{1}{2}s^{2}$ and hence $r\Gamma=s^{2}\Gamma$ is smooth on $\mathbb{L}$. Then it follows that

(4.78)

$$T^{\frac{n-2}{n}}\omega=T\pi^{*}\omega_{D}+\omega_{\mathbb{C}^{2}}+O^{\prime}(s^{3}).$$

Hence we see the $(1,1)$-form $\omega$ locally extends to a $C^{2,\alpha}$-form across the subset $\{u_{1}=u_{2}=0\}$.

Now we analyze the regularity of the holomorphic volume form $\Omega$ which is given by

(4.79)

$$\Omega=\sqrt{-1}(hdz+\sqrt{-1}\Theta)\wedge\pi^{*}\Omega_{D}.$$

By Lemma 3.30, locally we have

(4.80)

$$\pi^{*}\Omega_{D}=F(u_{1}du_{2}+u_{2}du_{1}+2\sqrt{-1}u_{1}u_{2}\Gamma)\wedge\pi^{*}\Omega_{H}+\widetilde{O}(s^{2})(u_{1}du_{2}+u_{2}du_{1})+\widetilde{O}(s^{3}).$$

Also

(4.81)

$$hdz+\sqrt{-1}\Theta=q(z)dz+\frac{1}{|u_{1}|^{2}+|u_{2}|^{2}}(-\bar{u}_{2}du_{2}+\bar{u}_{1}du_{1}+\sqrt{-1}(|u_{1}|^{2}-|u_{2}|^{2})\Gamma)+O^{\prime}(s^{2}).$$

Therefore,

(4.82)

$$\Omega=Fdu_{1}\wedge du_{2}\wedge\Omega_{H}+\sqrt{-1}F(u_{2}du_{1}-u_{1}du_{2})\wedge\Gamma\wedge\Omega_{H}+\widetilde{O}(s^{2})+sO^{\prime}(s^{2}).$$

This implies that $\Omega$ also extends to a $C^{2,\alpha}$ form across $\{u_{1}=u_{2}=0\}$. This is equivalent to saying that the almost complex structure $J$ determined by $\Omega$ extends to a $C^{2,\alpha}$ almost complex structure on $\mathcal{M}$. ∎