Let $\phi_{2}$ be the $3$-current defined in Lemma 3.21 such that

(3.243)

$$\Delta\phi_{2}=2\pi\delta_{U}+\mathfrak{B}_{1},$$

with $\mathfrak{B}_{1}=O^{\prime}(1)$. So Lemma 3.22 implies that there is some $3$-current $\mathfrak{R}=O^{\prime}(r^{2})$ such that

(3.244)

$$\Delta\mathfrak{R}=\mathfrak{B}_{1}$$

and hence the $3$-current

(3.245)

$$G_{U}\equiv\phi_{2}+\mathfrak{R}$$

satisfies the equation

(3.246)

$$\Delta G_{U}=2\pi\delta_{U}.$$

Moreover, by Lemma 3.21, $G_{U}$ has the expansion

	$\displaystyle G_{U}=$	$\displaystyle\frac{1}{2r}(1-\frac{H^{\alpha}y_{\alpha}}{2})dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r}y_{\beta}A_{i\alpha\beta}dx_{i}\wedge dy_{\widehat{\alpha}}-\frac{1}{4}A_{ij\alpha\beta}r\cdot dy_{\widehat{\alpha\beta}}\wedge dx_{i}\wedge dx_{j}$
(3.247)			$\displaystyle+\frac{3}{16}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})d(ry_{\alpha})\wedge dx_{i}\wedge dx_{j}+r^{-3}\Pi_{3}^{(4)}+O^{\prime}(r^{2}).$

The proof of Theorem 3.10 is done.

∎