The proof consists of two steps.

The first step focuses on the computation for $d^{*}d\phi_{1}$. Starting with the expansion of $\phi_{1}$ in (3.113), we have

(3.120)

$$d\phi_{1}=\frac{-1+H_{\alpha}y_{\alpha}+\widetilde{O}(r^{2})}{2r^{3}}\cdot rdr\wedge\dvol_{N}+\frac{1}{2r}d\Big((1-H^{\alpha}y_{\alpha}+{\widetilde{O}}(r^{2}))\dvol_{N}\Big).$$

To deal with the first term, we use Lemma 3.11 and (3.13) in Lemma 3.2, then

(3.121)		$\displaystyle y_{\alpha}\eta_{\alpha}$	$\displaystyle=$	$\displaystyle y_{\alpha}dy_{\alpha}-y_{\alpha}p_{i\alpha}dx_{i}$
			$\displaystyle=$	$\displaystyle rdr-y_{\alpha}g_{i\alpha}dx_{i}+\widetilde{O}(r^{4})$
			$\displaystyle=$	$\displaystyle rdr+\widetilde{O}(r^{4}),$

which yields

(3.122)

$$rdr\wedge\dvol_{N}=(y_{\alpha}\eta_{\alpha})\wedge\dvol_{N}+\widetilde{O}(r^{4})=\widetilde{O}(r^{4}).$$

So it follows that

(3.123)

$$d\phi_{1}=\frac{1}{2r}d\Big((1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2}))\dvol_{N}\Big)+O^{\prime}(r).$$

It is easy to see that

(3.124)

$$d(\widetilde{O}(r^{2})\dvol_{N})=\widetilde{O}(r^{2}).$$

So we obtain

(3.125)

$$d\phi_{1}=\frac{1}{2r}d\Big((1-H^{\alpha}y_{\alpha})\dvol_{N}\Big)+O^{\prime}(r).$$

Next we will compute the expansion for $d(\dvol_{N})$. By definition,

(3.126)

$$d(\dvol_{N})=d(\eta_{1}\wedge\eta_{2}\wedge\eta_{3})=d\eta_{\alpha}\wedge\eta_{\widehat{\alpha}}.$$

By (3.86),

(3.127)		$\displaystyle d\eta_{\alpha}$	$\displaystyle=$	$\displaystyle d(p_{i\alpha})\wedge dx_{i}$
			$\displaystyle=$	$\displaystyle A_{i\alpha\beta}dy_{\beta}\wedge dx_{i}+A_{ji\alpha\beta}y_{\beta}dx_{j}\wedge dx_{i}+B_{i\alpha\beta\gamma}y_{\beta}dy_{\gamma}\wedge dx_{i}+\widetilde{O}(r^{2}).$

So we have

(3.128)

$$d(\dvol_{N})=A_{i\alpha\beta}dy_{\beta}\wedge dx_{i}\wedge\eta_{\widehat{\alpha}}+y_{\beta}(A_{ji\alpha\beta}dx_{j}\wedge dx_{i}+B_{i\alpha\beta\gamma}dy_{\gamma}\wedge dx_{i})\wedge dy_{\widehat{\alpha}}+\widetilde{O}(r^{2}).$$

Now we need to rearrange the above expansion. Since $A_{i\alpha\beta}$ is skew symmetric in $\alpha$ and $\beta$, we have for $\alpha\in\{1,2,3\}$,

(3.129)

$$A_{i\alpha\alpha}=0,$$

so the leading order in the first term vanishes, hence

(3.130)		$\displaystyle A_{i\alpha\beta}dy_{\beta}\wedge dx_{i}\wedge\eta_{\widehat{\alpha}}$	$\displaystyle=$	$\displaystyle A_{i\alpha\beta}A_{j\mu\gamma}y_{\gamma}dy_{\beta}\wedge dx_{i}\wedge dx_{j}\wedge dy_{\widehat{\alpha\mu}}+\widetilde{O}(r^{2})$
			$\displaystyle=$	$\displaystyle A_{i\alpha\beta}A_{j\beta\gamma}y_{\gamma}dy_{\widehat{\alpha}}\wedge dx_{i}\wedge dx_{j}+\widetilde{O}(r^{2})$
			$\displaystyle=$	$\displaystyle\frac{1}{2}(A_{i\alpha\beta}A_{j\beta\gamma}-A_{i\gamma\beta}A_{j\beta\alpha})y_{\gamma}dy_{\widehat{\alpha}}\wedge dx_{i}\wedge dx_{j}+\widetilde{O}(r^{2}).$

Therefore,

(3.131)

$$d(\dvol_{N})=\Omega_{ij\alpha\beta}\cdot y_{\beta}\cdot dy_{\widehat{\alpha}}\wedge dx_{i}\wedge dx_{j}+B_{i\alpha\beta\alpha}\cdot y_{\beta}\cdot dy_{1}\wedge dy_{2}\wedge dy_{3}\wedge dx_{i}+\widetilde{O}(r^{2}).$$

By (3.92) we have

(3.132)

$$d(H^{\alpha}y_{\alpha})\wedge\dvol_{N}=(H^{\alpha}A_{i\alpha\beta}-\partial_{i}(H^{\beta}))\cdot y_{\beta}\cdot dy_{1}\wedge dy_{2}\wedge dy_{3}\wedge dx_{i}+\widetilde{O}(r^{2}).$$

Now substituting (3.131) and (3.132) into (3.125),

	$\displaystyle d\phi_{1}$	$\displaystyle=\frac{1}{2r}\Omega_{ij\alpha\beta}\cdot y_{\beta}\cdot dy_{\widehat{\alpha}}\wedge dx_{i}\wedge dx_{j}$
(3.133)			$\displaystyle+\frac{1}{2r}\Big(B_{i\alpha\beta\alpha}-(H^{\alpha}A_{i\alpha\beta}-\partial_{i}(H^{\beta})\Big)\cdot y_{\beta}\cdot dy_{1}\wedge dy_{2}\wedge dy_{3}\wedge dx_{i}+O^{\prime}(r).$

Now we need to take $d^{*}$ of this. Notice that the leading order of $d^{*}d\phi_{1}$ can be computed by using the operators in the Euclidean case, so we obtain

(3.134)

$\displaystyle d^{*}d\phi_{1}$

$\displaystyle=$

$\displaystyle\Omega_{ij\alpha\beta}(\frac{1}{2r}dy_{\widehat{\beta\alpha}}-\frac{1}{2r^{3}}y_{\mu}y_{\beta}dy_{\widehat{\mu\alpha}})dx_{i}\wedge dx_{j}+r^{-3}\Pi_{3}^{(2)}+O^{\prime}(1).$

In our next step, we will compute $dd^{*}\phi_{1}$. First,

(3.135)

$$*\phi_{1}=\frac{1}{2r}\dvol_{T}.$$

Notice that $d(\dvol_{T})=0$, so

(3.136)

$$d*\phi_{1}=-\frac{1}{2r^{3}}\cdot rdr\wedge\dvol_{T}.$$

By (3.121), $rdr=y_{\alpha}\eta_{\alpha}+{\widetilde{O}}(r^{4})$, then

(3.137)

$$d*\phi_{1}=-\frac{y_{\alpha}}{2r^{3}}{}\eta_{\alpha}\wedge\dvol_{T}+O^{\prime}(r).$$

Applying Item (2) of Lemma 3.13,

(3.138)

$$*d*\phi_{1}=-\frac{y_{\alpha}}{2r^{3}}{}(1-H^{\beta}y_{\beta}+\widetilde{O}(r^{2}))\eta_{\widehat{\alpha}}+O^{\prime}(r).$$

So it follows that

(3.139)

$\displaystyle d^{*}\phi_{1}$

$\displaystyle=$

$\displaystyle-*d*\phi_{1}=\frac{y_{\alpha}}{2r^{3}}{}(1-H^{\beta}y_{\beta})\eta_{\widehat{\alpha}}{}+\frac{\widetilde{O}(r^{2})}{r^{3}}y_{\alpha}dy_{\widehat{\alpha}}+O^{\prime}(r).$

Taking $d$ and applying Lemma 3.2,

	$\displaystyle dd^{*}\phi_{1}=$	$\displaystyle(1-H^{\beta}y_{\beta})\Big(-\frac{3y_{\alpha}}{2r^{5}}{}rdr\wedge\eta_{\widehat{\alpha}}+\frac{1}{2r^{3}}d(y_{\alpha}\eta_{\widehat{\alpha}})\Big){}-\frac{H^{\beta}y_{\alpha}}{2r^{3}}\eta_{\widehat{\alpha}}\wedge dy_{\beta}+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1)$
	$\displaystyle=$	$\displaystyle(1-H^{\beta}y_{\beta})\Big(-\frac{3y_{\alpha}}{2r^{5}}{}rdr\wedge\eta_{\widehat{\alpha}}+\frac{1}{2r^{3}}d(y_{\alpha}\eta_{\widehat{\alpha}})\Big)-\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}$
(3.140)		$\displaystyle+$	$\displaystyle r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

Now we simplify this expression. By (3.121),

(3.141)

$$-\frac{3y_{\alpha}}{2r^{5}}{}rdr\wedge\eta_{\widehat{\alpha}}=-\frac{3y_{\alpha}y_{\beta}}{2r^{5}}{}\eta_{\beta}\wedge\eta_{\widehat{\alpha}}+O^{\prime}(1)=-\frac{3}{2r^{3}}{}\dvol_{N}+{O}^{\prime}(1).$$

Also

(3.142)		$\displaystyle\frac{1}{2r^{3}}d(y_{\alpha}\eta_{\widehat{\alpha}})$	$\displaystyle=$	$\displaystyle\frac{1}{2r^{3}}dy_{\alpha}\wedge\eta_{\widehat{\alpha}}+\frac{1}{2r^{3}}y_{\alpha}d\eta_{\widehat{\alpha}}$
			$\displaystyle=$	$\displaystyle\frac{1}{2r^{3}}(\eta_{{\alpha}}-p_{i\alpha}dx_{i})\wedge\eta_{\widehat{\alpha}}+\frac{1}{2r^{3}}y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})$
			$\displaystyle=$	$\displaystyle\frac{3}{2r^{3}}\dvol_{N}-\frac{1}{2r^{3}}(A_{i\alpha\beta}y_{\beta}+\frac{1}{2}B_{i\alpha\beta\gamma}y_{\beta}y_{\gamma})dx_{i}\wedge\eta_{\widehat{\alpha}}$
			$\displaystyle+$	$\displaystyle\frac{1}{2r^{3}}y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})+{O^{\prime}}(1).$

So it follows that

	$\displaystyle dd^{*}\phi_{1}=$	$\displaystyle-\frac{1}{2r^{3}}\Big(A_{i\alpha\beta}\cdot y_{\beta}\cdot dx_{i}\wedge\eta_{\widehat{\alpha}}-y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})\Big)$
(3.143)		$\displaystyle-$	$\displaystyle\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

Next, we will show a crucial cancellation for the first term of the above $dd^{*}\phi_{1}$, which gives a further order improvement.

Now we return to the expansion of $dd^{*}\phi_{1}$ given by (3.143). Applying Lemma 3.16, finally we obtain

	$\displaystyle dd^{*}\phi_{1}$	$\displaystyle=-\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r^{3}}A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}\wedge dx_{i}\wedge dx_{j}$
(3.153)			$\displaystyle+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

In the last step of the proof, we will further simplify $d^{*}d\phi_{1}$ and $dd^{*}\phi_{1}$. For this purpose, we need the following lemma.

Applying the above lemma, now (3.134) and (3.153) can be simplified as follows,

(3.157)		$\displaystyle d^{*}d\phi_{1}$	$\displaystyle=-\Omega_{ij\alpha\beta}(\frac{1}{4r}dy_{\widehat{\alpha\beta}}+\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr)\wedge dx_{i}\wedge dx_{j}+r^{-3}\Pi_{3}^{-2}+O^{\prime}(1),$
	$\displaystyle dd^{*}\phi_{1}$	$\displaystyle=-\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}+A_{ij\alpha\beta}(\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr-\frac{1}{4r}dy_{\widehat{\alpha\beta}})\wedge dx_{i}\wedge dx_{j}$
(3.158)			$\displaystyle+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

Therefore,

	$\displaystyle\Delta\phi_{1}=$	$\displaystyle(d^{*}d+dd^{*})\phi_{1}$
	$\displaystyle=$	$\displaystyle-\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}-\Omega_{ij\alpha\beta}(\frac{1}{4r}dy_{\widehat{\alpha\beta}}+\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr)\wedge dx_{i}\wedge dx_{j}$
(3.159)			$\displaystyle+A_{ij\alpha\beta}(\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr-\frac{1}{4r}dy_{\widehat{\alpha\beta}})\wedge dx_{i}\wedge dx_{j}+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

The proof is done.

∎