3.2. Green’s currents for Riemannian submanifolds [04ZJ]

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3.2. Green’s currents for Riemannian submanifolds

First we recall and introduce the basic terminology. Let $(Q,g)$ be an oriented Riemannian manifold of dimension $m$ . Denote by $\Omega^{l}_{0}(Q)$ the space of differential $l$ -forms with compact supports in $Q$ .

Definition 3.5 ( $k$ -current).

A $k$ -current on $Q$ is a linear functional $T:\Omega^{m-k}_{0}(Q)\to\mathbb{R}$ which is continuous in the sense of distributions, i.e. suppose $\chi_{j}\in\Omega_{0}^{m-k}(Q)$ is a sequence of differential forms with all derivatives uniformly converging to $0$ as $j\to\infty$ , then $\lim\limits_{j\to\infty}(T,\chi_{j})=0$ .

The notion of currents unifies the notion of differential forms and submanifolds. In particular, a locally integrable $k$ -form $\beta$ can be naturally viewed as a $k$ -current via the pairing

(3.36)

(\beta,\chi)\equiv\int_{Q}\beta\wedge\chi,\ \ \ \ \chi\in\Omega^{m-k}_{0}(Q),

and an oriented submanifold $P$ of co-dimension $k$ also defines a $k$ -current via

(3.37)

(\delta_{P},\chi)\equiv\int_{P}\chi,\ \ \ \ \chi\in\Omega^{m-k}_{0}(Q).

The usual exterior differential $d$ and the Hodge star operator $*$ on differential forms then naturally extend to currents. Given a $k$ -current $T$ and $\chi\in\Omega_{0}^{m-k}(Q)$ , then we define

(3.38)		$\displaystyle(dT,\chi)\equiv(-1)^{k+1}(T,d\chi),$
(3.39)		$\displaystyle(T,\chi)\equiv(-1)^{k(m-k)}(T,\chi).$

Let $d^{*}$ be the codifferential operator and denote by $\Delta\equiv dd^{*}+d^{*}d$ the Hodge Laplacian, then it follows that for every $k$ -current $T$ and $\chi\in\Omega_{0}^{m-k}(Q)$ ,

(3.40)		$\displaystyle(d^{}T,\chi)=(-1)^{k}(T,d^{}\chi),$
(3.41)		$\displaystyle(\Delta T,\chi)=(T,\Delta\chi).$

A $k$ -current $T$ is called harmonic if $\Delta T=0$ . It follows from the standard elliptic regularity theory that a harmonic $k$ -current can be represented by a smooth harmonic $k$ -form.

Now let $P\subset Q$ be a (not necessarily closed) embedded oriented submanifold. Although the following discussion applies to more general setting, for our purpose in the following we will only consider the case when $P$ is of co-dimension $3$ in $Q$ . The importance of the co-dimension $3$ case in our setting is related to the fact that there is a Hopf fibration $\mathbb{R}^{4}\rightarrow\mathbb{R}^{3}$ which is a singular $S^{1}$ fibration with a smooth total space and co-dimension discriminant locus on the base. The co-dimension 3 condition also appears in other geometric settings, for example, Hitchin’s theory of Gerbes [Hit01].

Definition 3.6 (Green’s current).

Suppose $P$ is of co-dimension 3 in $Q$ . A Green’s current $G_{P}$ for $P$ in $Q$ is a locally integrable $3$ -form which solves the following current equation on $Q$

(3.42)

\Delta G_{P}=2\pi\cdot\delta_{P}.

Example 3.7.

The above normalization constant is chosen such that in the case $Q\equiv\mathbb{R}^{3}$ and $P\equiv 0^{3}\in\mathbb{R}^{3}$ , then

(3.43)

G_{P}=\frac{1}{2|y|}dy_{1}\wedge dy_{2}\wedge dy_{3}

solves the current equation $\Delta_{0}G_{P}=2\pi\cdot\delta_{P}$ for the standard Hodge Laplacian $\Delta_{0}$ on $\mathbb{R}^{3}$ .

In particular $G_{P}$ is harmonic outside $P$ hence is smooth. Notice a Green’s current $G_{P}$ for $P$ is not unique, but it is unique up to the addition of a harmonic $3$ -form, so the singular behavior near $P$ does not depend on the particular choice of $G_{P}$ . Also it is clear that if $Q^{\prime}\subset Q$ is an open submanifold, then the restriction of $G_{P}$ to $Q^{\prime}$ is a Green’s current for $P^{\prime}=P\cap Q^{\prime}$ in $Q^{\prime}$ , so that we can study the regularity problem locally. Our goal in this subsection is to understand the local existence and regularity of $G_{P}$ via approximation by the standard model, which is the product space $\mathbb{R}^{3}\times\mathbb{R}^{n-3}$ .

To begin with, we have the following simple regularity result for $d(G_{P})$ .

Proposition 3.8.

Given a Green’s current $G_{P}$ , its differential $d(G_{P})$ extends to a smooth $4$ -form across $P$ .

Proof.

This is a local result so we can work with the geodesic ball $B_{r}(p)$ for any $p\in P$ such that $\overline{B_{r}(p)}\subset\subset Q$ and $\overline{B_{r}(p)}\cap P\subset\subset P$ . We will show that the $4$ -current $dG_{P}$ is a harmonic in $B_{r}(p)$ in the distributional sense. In fact, for any test form $\chi\in\Omega^{m-4}_{0}(B_{r}(p))$ we have

(3.44)

(\Delta(d(G_{P})),\chi)=(d\Delta G_{P},\chi)=(\Delta G_{P},d\chi)=(2\pi\delta_{P},d\chi)=2\pi\int_{P}d\chi=2\pi\int_{\partial B_{r}(p)\cap P}\chi=0.

Therefore, $d(G_{P})$ is a harmonic $4$ -current in $B_{r}(p)$ and hence it is smooth in $B_{r}(p)$ . ∎

In the rest of this section, we will frequently use the following notation.

Notation 3.9.

Given $k\in\mathbb{Z}_{+}$ , a capital Greek letter with index $k$ , $\Pi_{p}^{(k)}$ , always denotes a general local $p$ -form with $0\leq p\leq 3$ , which is of the form

(3.45)

\Pi_{p}^{(k)}\equiv P_{i,j_{1},\ldots,j_{p-1}}^{(k)}(x,y)(dy_{j_{1}}\wedge\ldots\wedge dy_{j_{p-1}})\wedge dx_{i}+Q_{j_{1},\ldots,j_{p}}^{(k)}(x,y)dy_{j_{1}}\wedge\ldots\wedge dy_{j_{p}},

where $P_{i,j_{1},\ldots,j_{p-1}}^{(k)}(x,y)$ and $Q_{j_{1},\ldots,j_{p}}^{(k)}(x,y)$ are homogeneous polynomial functions in $y$ of degree $k$ whoses coefficient functions are smooth on $U\subset P$ .

Remark 3.9.1.

Notice this expression depends on the choice of local coordinates, but under a change of coordinates, a $p$ -form $\Pi_{p}^{(k)}$ will still have such an expression, modulo a term which is of order $\widetilde{O}(r^{k+1})$ .

Now we are ready to state the first main theorem of this section, which gives a local existence for Green’s current, and its leading singular behavior.

Theorem 3.10.

If $P\subset Q$ is an embedded submanifold of co-dimension $3$ , then for any $p\in P$ , there is a neighborhood $\mathcal{U}$ of $p$ in $Q$ , and a Green’s current $G_{U}$ for $U=\mathcal{U}\cap P$ in $\mathcal{U}$ satisfying

(3.46)

\displaystyle\Delta G_{U}=2\pi\cdot\delta_{U}\ \ \ \ \text{in}\ \ \ \ \ \mathcal{U},

and the local expansion

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

where $\overrightarrow{H}=H^{\alpha}e_{\alpha}$ is the mean curvature of $P\subset Q$ and the $3$ -form $\Pi_{3}^{(4)}$ is defined in (3.45).

Here and in the following we use the notation that for $\alpha\in\{1,2,3\}$ , $\widehat{\alpha}=(\alpha+1,\alpha+2)$ (with the convention $3+1=1$ ), $dy_{\alpha\beta}\equiv dy_{\alpha}\wedge dy_{\beta}$ , and that

(3.48)

\displaystyle dy_{\widehat{\alpha\beta}}\equiv\begin{cases}dy_{\alpha+2},&\beta=\alpha+1\\ -dy_{\alpha+1},&\beta=\alpha+2\\ 0,&\beta=\alpha.\end{cases}

Remark 3.10.1.

By the above discussion, if $G_{P}$ is any Green’s current for $P$ in $Q$ , then locally near $p\in P$ , $G_{P}$ will also have an expansion of the form (3.10).

Remark 3.10.2.

At a given point $p\in U$ , we can always choose a special frame $\{e_{\alpha}\}$ such that $A_{i\alpha\beta}=0$ at $p$ . On the other hand, since the singular behavior of $G_{U}$ does not depend on the choice of coordinates, one sees that in general we need the second term in the expansion.

Remark 3.10.3.

By Proposition 3.8, $dG_{U}$ is smooth. This is compatible with the above expansion. For example, from the expansion we see the leading term involving forms of type $dx_{i}\wedge dx_{j}\wedge dy_{\beta}\wedge dy_{\gamma}$ is given by

(3.49)

(\frac{1}{2r}y_{\beta}A_{ij\alpha\beta}dy_{\hat{\alpha}}-\frac{1}{4}A_{ij\alpha\beta}dr\wedge dy_{\widehat{\alpha\beta}})\wedge dx_{i}\wedge dx_{j}.

Elementary calculation shows this vanishes.

Remark 3.10.4.

In the proof we shall not keep track of the explicit form of $\Pi_{3}^{(4)}$ because it is not needed in our applications. However, it is possible to obtain the precise expression with more work. Given the above expansion, there are also some constraint for $\Pi_{3}^{(4)}$ following from the fact that $d(G_{U})$ is smooth by Proposition 3.8.

Before starting the proof of Theorem 3.10, we need some preparations. For the convenience of our calculations, we introduce three $1$ -forms

(3.50)

\eta_{\alpha}\equiv dy_{\alpha}+p_{i\alpha}dx_{i}

such that

(3.51)

\langle\eta_{\alpha},dx_{j}\rangle=0

for all $j$ and $\alpha$ at all points of $\mathcal{U}$ . Then the linear span of the $\eta_{\alpha}$ ’s is orthogonal to the linear span of the $dx_{i}$ ’s.

The lemma below is s crucial in the proof of Theorem 3.10.

Lemma 3.11.

For any $1\leq k\leq m-3$ and $1\leq\alpha\leq 3$ ,

(3.52)

p_{k\alpha}=g_{k\alpha}+\widetilde{O}(r^{3}).

Proof.

We write the full matrix expression of the metric $g$ as

(3.53)

g=\left[{\begin{array}[]{cc}g_{\alpha\beta}&0\\ 0&g_{ij}\\ \end{array}}\right]+\left[{\begin{array}[]{cc}0&S\\ S^{t}&0\\ \end{array}}\right],

where $S=(g_{\alpha i})=\widetilde{O}(r)$ . We denote by $(h_{\alpha\beta})$ and $(h_{ij})$ the inverse matrix of $(g_{\alpha\beta})$ and $(g_{ij})$ respectively. Then by elementary consideration

(3.74)	$\displaystyle g^{-1}$	$\displaystyle=$	$\displaystyle\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]-\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]\left[{\begin{array}[]{cc}0&S\\ S^{t}&0\\ \end{array}}\right]\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]$
		$\displaystyle+$	$\displaystyle\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]\left[{\begin{array}[]{cc}0&S\\ S^{t}&0\\ \end{array}}\right]\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]\left[{\begin{array}[]{cc}0&S\\ S^{t}&0\\ \end{array}}\right]\left[{\begin{array}[]{cc}h_{\alpha\beta}&0\\ 0&h_{ij}\\ \end{array}}\right]$
		$\displaystyle+$	$\displaystyle\widetilde{O}(r^{3}).$

Notice the third term does not have off-diagonal contributions, so the inverse matrix $g^{-1}=(g^{IJ})$ satisfies

(3.75)	$\displaystyle g^{i\alpha}$	$\displaystyle=-h_{ij}g_{j\beta}h_{\beta\alpha}+\widetilde{O}(r^{3})$
(3.76)	$\displaystyle g^{ij}$	$\displaystyle=h_{ij}+\widetilde{O}(r^{2}),$
(3.77)	$\displaystyle g^{\alpha\beta}$	$\displaystyle=h_{\alpha\beta}+\widetilde{O}(r^{2})=\delta_{\alpha\beta}+\widetilde{O}(r^{2}),$

where we used Lemma 3.4. The definition of $\eta_{\alpha}$ requires $\langle\eta_{\alpha},dx_{j}\rangle=0$ , which implies that

(3.78)

p_{i\alpha}g^{ij}+g^{j\alpha}=0.

Let $(\hat{g}_{ij})$ be the inverse of the matrix $(g^{ij})$ with $1\leq i,j\leq m-3$ such that $g^{ij}\hat{g}_{jk}=\delta_{ik}$ . Multiplying by $\hat{g}_{jk}$ , we have

(3.79)

p_{i\alpha}g^{ij}\hat{g}_{jk}+g^{j\alpha}\hat{g}_{jk}=0,

and hence

(3.80)

p_{k\alpha}=-g^{j\alpha}\hat{g}_{jk}.

We claim that for any $1\leq i,j\leq m-3$ ,

(3.81)

\hat{g}_{ij}-g_{ij}=\widetilde{O}(r^{2}).

In fact, since

(3.82)

g^{ij}g_{jk}+g^{i\alpha}g_{\alpha k}=\delta_{ik}.

Multipling by the inverse of the submatrix $(g^{ij})$ ,

(3.83)

\hat{g}_{li}(g^{ij}g_{jk}+g^{i\alpha}g_{\alpha k})=\hat{g}_{lk}.

So this implies that

(3.84)

\hat{g}_{lk}-g_{lk}=\hat{g}_{li}g^{i\alpha}g_{\alpha k}=\widetilde{O}(r^{2}).

Therefore, combining (3.75),(3.77), (3.80) and (3.84), we obtain

(3.85)	$\displaystyle p_{k\alpha}$	$\displaystyle=$	$\displaystyle-(g_{kj}+\widetilde{O}(r^{2}))g^{j\alpha}$
		$\displaystyle=$	$\displaystyle-g_{kj}g^{j\alpha}+\widetilde{O}(r^{3})$
		$\displaystyle=$	$\displaystyle h_{\alpha\beta}g_{k\beta}+\widetilde{O}(r^{3})$
		$\displaystyle=$	$\displaystyle g_{k\alpha}+\widetilde{O}(r^{3}).$

∎

The following symmetry property of $p_{i\alpha}$ will be frequently used in our later calculations. By Lemma 3.4 and Lemma 3.11, we may write

(3.86)

p_{i\alpha}=A_{i\alpha\beta}y_{\beta}+\frac{1}{2}B_{i\alpha\beta\gamma}y_{\beta}y_{\gamma}+\widetilde{O}(r^{3}).

Here the connection term $A_{i\alpha\beta}\equiv\langle\nabla_{\partial_{x_{i}}}\partial_{y_{\beta}},y_{\alpha}\rangle|_{(x,0)}$ is skew-symmetric in $\alpha$ , $\beta$ , and the curvature term $B_{i\alpha\beta\gamma}\equiv-\frac{4}{3}R_{i\beta\gamma\alpha}$ is symmetric in $\beta$ , $\gamma$ .

Lemma 3.12.

For every $1\leq\alpha,\beta\leq 3$ ,

(3.87)

\langle\eta_{\alpha},\eta_{\beta}\rangle=\delta_{\alpha\beta}+\widetilde{O}(r^{2}).

Proof.

By definition

(3.88)

\langle\eta_{\alpha},\eta_{\beta}\rangle=\langle dy_{\alpha}+p_{i\alpha}dx_{i},dy_{\beta}+p_{j\beta}dx_{j}\rangle.

By (3.77) we get

(3.89)

\langle dy_{\alpha},dy_{\beta}\rangle=\delta_{\alpha\beta}+\widetilde{O}(r^{2}).

Also we have $p_{i\alpha}=\widetilde{O}(r)$ and $\langle dx_{i},dy_{\beta}\rangle=\widetilde{O}(r)$ for all $i$ and $\alpha$ . The conclusion then follows. ∎

Using the above differential forms $\eta_{\alpha}$ ’s, we can decompose the volume form $\dvol_{g}$ in the horizontal and vertical directions, which will substantially simplify the computations regarding the Hodge Laplacian. The volume form of $g$ is given by

(3.90)

\dvol_{g}=\sqrt{\det(g)}\cdot dy_{1}\wedge dy_{2}\wedge dy_{3}\wedge dx_{1}\wedge\cdots\wedge dx_{m-3},

where we have used the orientation fixed above. We define the normal and tangential volume forms by

(3.91)

\displaystyle\begin{cases}{\dvol}_{N}\equiv\eta_{1}\wedge\eta_{2}\wedge\eta_{3}\\ {\dvol}_{T}\equiv\sqrt{\det(g_{ij}^{P})}\cdot dx_{1}\wedge\cdots\wedge dx_{m-3}.\end{cases}

By the expansion formula (3.86), the normal volume form $\dvol_{N}$ has the following expansion,

	$\displaystyle{\dvol}_{N}=dy_{1}\wedge dy_{2}\wedge dy_{3}+(A_{i\alpha\beta}y_{\beta}+\frac{1}{2}B_{i\alpha\beta\gamma}y_{\beta}y_{\gamma})dx_{i}\wedge dy_{\widehat{\alpha}}$
(3.92)		$\displaystyle+A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}y_{\beta}y_{\gamma}dy_{\alpha}\wedge dx_{i}\wedge dx_{j}+\widetilde{O}(r^{3}).$

In addition, by the definition of $\eta_{\alpha}$ ’s, it holds that for each $1\leq\alpha\leq 3$ ,

(3.93)

\eta_{\alpha}\wedge\dvol_{T}=dy_{\alpha}\wedge\dvol_{T},

and hence

(3.94)

\dvol_{N}\wedge\dvol_{T}=dy_{1}\wedge dy_{2}\wedge dy_{3}\wedge\dvol_{T}=\frac{\sqrt{\det(g_{ij}^{P})}}{\sqrt{\det(g)}}\cdot\dvol_{g}.

Lemma 3.13.

Denote by $*:\Omega^{k}(Q)\to\Omega^{m-k}(Q)$ the Hodge $*$ operator, then we have the following:

(1)

(3.95) $*\dvol_{T}=(-1)^{m+1}\dvol_{N}\cdot(1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2})),$

(2)

For any $\alpha\in\{1,2,3\}$ ,

(3.96)

*(\eta_{\alpha}\wedge\dvol_{T})=\lambda_{1}\eta_{\widehat{\alpha}}+\lambda_{2}\eta_{\widehat{\alpha+1}}+\lambda_{3}\eta_{\widehat{\alpha+2}},

where $\lambda_{1}=1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2})$ , $\lambda_{2}=\widetilde{O}(r^{2})$ , $\lambda_{3}=\widetilde{O}(r^{2})$ .

Proof.

First, we prove Item (1). By (3.51) we have

(3.97)

(-1)^{m+1}*\dvol_{T}=\lambda\cdot\dvol_{N}

for a function $\lambda>0$ . The function $\lambda$ is given by

(3.98)

\lambda=\frac{|\dvol_{T}|^{2}\sqrt{\det(g)}}{\sqrt{\det(g_{ij}^{P})}}=\sqrt{\det(g)}\det(g^{ij})\sqrt{\det(g_{ij}^{P})}.

Now we compute the expansion of $\lambda$ . Applying the expansions of $g_{ij}$ , $g_{\alpha\beta}$ and $g_{i\alpha}$ in Lemma 3.4, one can directly obtain the following,

(3.99)	$\displaystyle\det(g)$	$\displaystyle=\det(g_{\alpha\beta})\cdot\det(g_{ij})+\widetilde{O}(r^{2}),$
(3.100)	$\displaystyle\det(g_{\alpha\beta})$	$\displaystyle=1+\widetilde{O}(r^{2}),$
(3.101)	$\displaystyle\det(g_{ij})$	$\displaystyle=\det(g_{ij}^{P})\cdot(1+2H^{\alpha}y_{\alpha})+\widetilde{O}(r^{2}).$

Plugging (3.100) and (3.101) into (3.99),

(3.102)

\det(g)=\det(g_{ij}^{P})\cdot(1+2H^{\alpha}y_{\alpha})+\widetilde{O}(r^{2}).

Let $(h_{ij})$ be the inverse of the matrix $(g_{ij})$ . Since $g^{ij}=h_{ij}+\widetilde{O}(r^{2})$ by (3.76), so it follows that

(3.103)

\det(g^{ij})=\det(h_{ij})+\widetilde{O}(r^{2})=(\det(g_{ij}))^{-1}+\widetilde{O}(r^{2}).

Plugging (3.101) into the above,

(3.104)

\det(g^{ij})=\det(g_{ij}^{P})^{-1}\cdot(1-2H^{\alpha}y_{\alpha})+\widetilde{O}(r^{2}).

Therefore, substituting (3.102) and (3.104) into (3.98),

(3.105)

\lambda=1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2}),

which completes the proof of Item (1).

Now we prove Item (2). For each $\alpha\in\{1,2,3\}$ , we can write

(3.106)

*(\eta_{\alpha}\wedge\dvol_{T})=\lambda_{1}\cdot\eta_{\widehat{\alpha}}+\lambda_{2}\cdot\eta_{\widehat{\alpha+1}}+\lambda_{3}\cdot\eta_{\widehat{\alpha+2}}.

Taking point-wise wedge product with $\eta_{\alpha}\wedge\dvol_{T}$ , and noticing $\eta_{\widehat{\alpha+1}}\wedge\eta_{\alpha}$ , $\eta_{\widehat{\alpha+2}}\wedge\eta_{\alpha}$ are both zero, then we obtain

(3.107)

\lambda_{1}\cdot\eta_{\widehat{\alpha}}\wedge\eta_{\alpha}\wedge\dvol_{T}=(\eta_{\alpha}\wedge\dvol_{T})\wedge*(\eta_{\alpha}\wedge\dvol_{T}).

(3.108)

\displaystyle\lambda_{1}\cdot\dvol_{N}\wedge\dvol_{T}=|\eta_{\alpha}\wedge\dvol_{T}|^{2}\dvol_{g}

Therefore, by (3.51) and (3.87) we get

(3.109)

\lambda_{1}=\frac{|\eta_{\alpha}\wedge\dvol_{T}|^{2}\sqrt{\det(g)}}{\sqrt{\det(g_{ij}^{P})}}=1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2}),

Similarly taking wedge product with $\eta_{\alpha+1}\wedge\dvol_{T}$ and $\eta_{\alpha+2}\wedge\dvol_{T}$ respectively, and again by (3.87) we obtain that

(3.110)

\lambda_{2}=\widetilde{O}(r^{2}),\lambda_{3}=\widetilde{O}(r^{2}).

These imply that

(3.111)

*(\eta_{\alpha}\wedge\dvol_{T})=\lambda_{1}\cdot\eta_{\widehat{\alpha}}+\lambda_{2}\cdot\eta_{\widehat{\alpha+1}}+\lambda_{3}\cdot\eta_{\widehat{\alpha+2}},

where $\lambda_{1}=1+H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2})$ , $\lambda_{2}=\widetilde{O}(r^{2})$ and $\lambda_{3}=\widetilde{O}(r^{2})$ . ∎

Now we proceed to prove Theorem 3.10. This will be done in several steps.

Step 1. We start by defining a 3-form

(3.112)

\phi_{1}\equiv(-1)^{m+1}\frac{1}{2r}*\dvol_{T}.

By Item (1) of Lemma 3.13, immediately we have

(3.113)

\displaystyle\phi_{1}=\frac{\dvol_{N}}{2r}\cdot(1-H^{\alpha}y_{\alpha}+\widetilde{O}(r^{2})).

Then applying the expansion of $\dvol_{N}$ in (3.92), $\phi_{1}$ has a further expansion,

	$\displaystyle\phi_{1}$	$\displaystyle=\frac{1-H^{\alpha}y_{\alpha}}{2r}dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r}A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\widehat{\alpha}}$
		$\displaystyle+\frac{1}{2r}A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}y_{\beta}y_{\gamma}dy_{\alpha}\wedge dx_{i}\wedge dx_{j}+r^{-1}\Pi_{3}^{(2)}+O^{\prime}(r^{2})$
		$\displaystyle=\frac{1-H^{\alpha}y_{\alpha}}{2r}dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r}A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\hat{\alpha}}$
(3.114)			$\displaystyle+\frac{1}{4}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}\cdot dr\wedge dx_{i}\wedge dx_{j}+r^{-1}\Pi_{3}^{(2)}+O^{\prime}(r^{2}),$

where $\Pi_{3}^{(2)}$ is the $3$ -form introduced in Notation 3.9 and the last step can be achieved by applying the following lemma:

Lemma 3.14 (Rearrangement Lemma).

(3.115)

A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}y_{\beta}y_{\gamma}dy_{\alpha}\wedge dx_{i}\wedge dx_{j}=\frac{1}{2}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}\cdot rdr\wedge dx_{i}\wedge dx_{j}.

Proof.

First by writing out the terms and re-arranging the subscripts and using the skew symmetry of $A_{i\alpha\beta}$ we get

(3.116)		$\displaystyle A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}y_{\beta}y_{\gamma}dy_{\alpha}$
	$\displaystyle=$	$\displaystyle\Big(A_{i,\alpha+1,\alpha}A_{j,\alpha+2,\alpha}y_{\alpha}^{2}+A_{i,\alpha+1,\alpha}A_{j,\alpha+2,\alpha+1}y_{\alpha}y_{\alpha+1}$
		$\displaystyle+A_{i,\alpha+1,\alpha+2}A_{j,\alpha+2,\alpha}y_{\alpha}y_{\alpha+2}+A_{i,\alpha+1,\alpha+2}A_{j,\alpha+2,\alpha+1}y_{\alpha+1}y_{\alpha+2}\Big)dy_{\alpha}$
	$\displaystyle=$	$\displaystyle A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}y_{\alpha}^{2}dy_{\alpha}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1}y_{\alpha}y_{\alpha+2}dy_{\alpha+2}$
		$\displaystyle-A_{i,\alpha,\alpha+2}A_{j,\alpha,\alpha+1}y_{\alpha}y_{\alpha+1}dy_{\alpha+1}-A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+1}y_{\alpha}y_{\alpha+1}dy_{\alpha+2}$

Now we can skew-symmetrize with respect to $i$ and $j$

(3.117)

A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}y_{\beta}y_{\gamma}dy_{\alpha}\wedge dx_{i}\wedge dx_{j}=\frac{1}{2}(A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}-A_{j,\alpha+1,\beta}A_{i,\alpha+2,\gamma})y_{\beta}y_{\gamma}dy_{\alpha}\wedge dx_{i}\wedge dx_{j}

Correspondingly by skew-symmetrizing each term of (3.116) with respect to $i$ and $j$ , we get

(3.118)		$\displaystyle\frac{1}{2}(A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}-A_{j,\alpha+1,\beta}A_{i,\alpha+2,\gamma})y_{\beta}y_{\gamma}dy_{\alpha}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}(y_{\alpha}dy_{\alpha}+y_{\alpha+1}dy_{\alpha+1}+y_{\alpha+2}dy_{\alpha+2})$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}\cdot rdr$

∎

Step 2. In this step will explicitly compute the singular (unbounded) terms of $\Delta\phi_{1}$ . Mainly, we will prove the following proposition.

Proposition 3.15.

Let $\phi_{1}$ be the $3$ -form defined in (3.112), then $\Delta\phi_{1}$ has the following expansion,

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

Proof.

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

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.

Lemma 3.16 (Cancellation Lemma).

		$\displaystyle A_{i\alpha\beta}\cdot y_{\beta}\cdot dx_{i}\wedge\eta_{\widehat{\alpha}}-y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})$
(3.144)		$\displaystyle=$	$\displaystyle-A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}\wedge dx_{i}\wedge dx_{j}+\Pi_{3}^{(2)}+{\widetilde{O}}(r^{3}).$

Proof.

Directly applying the definition of $\eta_{\alpha}$ , then we have

		$\displaystyle A_{i\alpha\beta}\cdot y_{\beta}\cdot dx_{i}\wedge\eta_{\widehat{\alpha}}$
(3.145)		$\displaystyle=$	$\displaystyle A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\widehat{\alpha}}+A_{i\alpha\beta}y_{\beta}y_{\gamma}(A_{j,\alpha+1,\gamma}dy_{\alpha+2}-A_{j,\alpha+2,\gamma}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}+\widetilde{O}(r^{3}).$

By (3.127), we get

		$\displaystyle y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})$
	$\displaystyle=$	$\displaystyle y_{\alpha}(A_{i,\alpha+1,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+1})$
	$\displaystyle+$	$\displaystyle y_{\alpha}y_{\gamma}(A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}-A_{i,\alpha+2,\beta}A_{j,\alpha+1,\gamma})dy_{\beta}\wedge dx_{i}\wedge dx_{j}$
	$\displaystyle+$	$\displaystyle y_{\alpha}y_{\beta}(A_{ij,\alpha+1,\beta}dy_{\alpha+2}-A_{ij,\alpha+2,\beta}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}$
(3.146)		$\displaystyle+$	$\displaystyle\Pi_{3}^{(2)}+\widetilde{O}(r^{3}).$

Rearranging the subscripts of the first groups of terms in (3.146),

(3.147)		$\displaystyle y_{\alpha}(A_{i,\alpha+1,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+1})$
(3.148)	$\displaystyle=$	$\displaystyle y_{\alpha}(A_{i,\alpha+1,\alpha}dy_{\alpha}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\alpha}dy_{\alpha}\wedge dx_{i}\wedge dy_{\alpha+1})$
(3.149)	$\displaystyle=$	$\displaystyle y_{\alpha+2}A_{i,\alpha,\alpha+2}dy_{\alpha+2}\wedge dx_{i}\wedge dy_{\alpha+1}-y_{\alpha+1}A_{i,\alpha,\alpha+1}dy_{\alpha+1}\wedge dx_{i}\wedge dy_{\alpha+2}$
(3.150)	$\displaystyle=$	$\displaystyle A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\widehat{\alpha}},$

which matches the first term of (3.145). As in the proof of Lemma 3.14, one can see that the second groups of terms in (3.145) and (3.146) are both equal to

(3.151)

(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}\cdot rdr\wedge dx_{i}\wedge dx_{j}.

Next, the third group of terms in (3.146) can be rewritten as follows,

(3.152)		$\displaystyle y_{\alpha}y_{\beta}(A_{ij,\alpha+1,\beta}dy_{\alpha+2}-A_{ij,\alpha+2,\beta}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}$
	$\displaystyle=$	$\displaystyle A_{ij\alpha\beta}y_{\beta}(y_{\alpha+2}dy_{\alpha+1}-y_{\alpha+1}dy_{\alpha+2})\wedge dx_{i}\wedge dx_{j}$
	$\displaystyle=$	$\displaystyle A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}\wedge dx_{i}\wedge dx_{j}.$

The conclusion just follows.

∎

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.

Lemma 3.17.

(3.154)		$\displaystyle\Omega_{ij\alpha\beta}(\frac{1}{2r}dy_{\widehat{\beta\alpha}}-\frac{1}{2r^{3}}y_{\mu}y_{\beta}dy_{\widehat{\mu\alpha}})$	$\displaystyle=-\Omega_{ij\alpha\beta}(\frac{1}{4r}dy_{\widehat{\alpha\beta}}+\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr),$
(3.155)		$\displaystyle\frac{1}{2r^{3}}A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}$	$\displaystyle=A_{ij\alpha\beta}(\frac{1}{4r^{2}}y_{\widehat{\alpha\beta}}dr-\frac{1}{4r}dy_{\widehat{\alpha\beta}}).$

Proof.

We only prove (3.154) because the other equality follows from the same computations. Using the fact that $\Omega_{ij\alpha\beta}=-\Omega_{ij\beta\alpha}$ , we can write out the left hand side as

		$\displaystyle\Omega_{ij\alpha\beta}(\frac{1}{2r}dy_{\widehat{\beta\alpha}}-\frac{1}{2r^{3}}y_{\mu}y_{\beta}dy_{\widehat{\mu\alpha}})$
	$\displaystyle=$	$\displaystyle\Omega_{ij\alpha,\alpha+1}\Big(-\frac{1}{r}dy_{\alpha+2}-\frac{1}{2r^{3}}(y_{\alpha+2}y_{\alpha+1}dy_{y_{\alpha+1}}-y_{\alpha+1}^{2}dy_{\alpha+2})+\frac{1}{2r^{3}}(y_{\alpha}^{2}dy_{\alpha+2}-y_{\alpha}y_{\alpha+2}dy_{\alpha})\Big)$
	$\displaystyle=$	$\displaystyle\Omega_{ij\alpha,\alpha+1}\Big(-\frac{1}{2r}dy_{\alpha+2}-\frac{1}{2r^{2}}y_{\alpha+2}dr\Big)$
(3.156)		$\displaystyle=$	$\displaystyle-\frac{1}{2}\Omega_{ij\alpha,\beta}\Big(\frac{1}{2r}dy_{\widehat{\alpha\beta}}+\frac{1}{2r^{2}}y_{\widehat{\alpha\beta}}dr\Big).$

∎

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.

∎

Step 3. In this step we modify $\phi_{1}$ to kill the unbounded terms on the right hand side of (3.119). We first we recall some elementary computations involving the standard Euclidean Hodge Laplacian.

Lemma 3.18.

Let $\Delta_{0}$ be the standard Hodge Laplacian on the Euclidean space $\mathbb{R}^{3}$ , then the following holds:

(1)

Let $\{y_{1},y_{2},y_{3}\}$ be the Cartesian coordinates of $\mathbb{R}^{3}$ , then

(3.160)

\displaystyle\begin{cases}\Delta_{0}r=-\frac{2}{r},\\ \Delta_{0}(\frac{y_{\alpha}y_{\beta}}{r})=\frac{4y_{\alpha}y_{\beta}}{r^{3}},&\alpha\neq\beta,\\ \Delta_{0}((\frac{y_{\alpha}^{2}}{r}-r))=\frac{4y_{\alpha}^{2}}{r^{3}},\\ \Delta_{0}(\frac{y_{\alpha}}{r})=\frac{2y_{\alpha}}{r^{3}}.\end{cases}

(2)

Denote by $\mathcal{P}_{4}$ the space of all homogeneous degree 4 polynomials on $\mathbb{R}^{3}$ , then the operator

(3.161) $\square:\mathcal{P}_{4}\rightarrow\mathcal{P}_{4};f\mapsto r^{5}\Delta_{0}(r^{-3}f)$

is an isomorphism.

Proof.

The first item is a direct calculation. An convenient way to see this is to use the following two facts

(1)

A homogeneous polynomial degree $k$ polynomial restricts to an eigenfunction of the Hodge-Laplacian $\Delta_{S^{2}}$ on the unit sphere, with eigenvalue $k(k+1)$ .
(2)

Given an eigenfunction $h$ of $\Delta_{S^{2}}$ on the unit sphere with eigenvalue $k$ , for any $l$ , we can extend $h$ to a homogeneous function $h_{l}$ on $\mathbb{R}^{3}\setminus\{0\}$ of degree $l$ , and

(3.162) $\Delta_{0}h_{l}=r^{-2}(k-l(l+1))h_{l}$

For the second item it is possible to write down an explicit inverse to $\Delta_{0}$ . Here we provide a quick abstract proof. First we notice $\square:\mathcal{P}_{4}\to\mathcal{P}_{4}$ is a well-defined linear map. This follows from the standard computations

	$\displaystyle r^{5}\Delta_{0}(r^{-3}f)$	$\displaystyle=$	$\displaystyle r^{5}\Delta_{0}(r^{-3})\cdot f-2r^{5}\nabla(r^{-3})\cdot\nabla f+r^{2}\Delta_{0}f$
		$\displaystyle=$	$\displaystyle-6f-3\nabla(r^{2})\cdot\nabla f+r^{2}\Delta_{0}f.$

Since each term in the above formula is a polynomial in $\mathcal{P}_{4}$ , so $\square f\in\mathcal{P}_{4}$ .

Now to prove $\square$ is an isomorphism it suffices to prove it has a trivial kernel in $\mathcal{P}_{4}$ . Let $u\equiv r^{-3}f$ , then $u=O(r)$ for both $r\to 0$ and $r\to\infty$ . If $\Delta_{0}(u)=0$ , then $u$ is harmonic on $\mathbb{R}^{3}\setminus\{0\}$ . The removable singularity theorem implies that $u$ extends smoothly on $\mathbb{R}^{3}$ . Since $u=O(r)$ as $r\to\infty$ , applying the standard derivative estimate for harmonic functions, we conclude $\nabla^{2}u\equiv 0$ . Therefore, $u$ must be a linear function. Noticing $f\in\mathcal{P}_{4}$ , we conclude $f\equiv 0$ . The proof is done.

∎

Next, we want to find a bounded correction $3$ -form $\mathfrak{B}_{0}=O^{\prime}(1)$ such that $\Delta\phi_{1}$ is corrected to a bounded term on $\mathcal{U}\setminus P$ , i.e.,

(3.163)

\Delta(\phi_{1}+\mathfrak{B}_{0})=O^{\prime}(1)\ \text{on}\ \mathcal{U}\setminus P.

Now the main part is to eliminate the unbounded terms in $\Delta\phi_{1}$ which relies on the following explicit calculations for $\Delta\mathfrak{B}_{0}$ . In fact, the leading terms of $\Delta\mathfrak{B}_{0}$ are exactly given by the Euclidean Laplacian $\Delta_{0}$ acting on the normal components such that the explicit computations in Lemma 3.18 can be effectively used in our context. Precisely, we have the following lemma.

Lemma 3.19.

Let $\Delta_{0}$ be the Hodge Laplacian on $\mathbb{R}^{3}$ , then the following holds:

(1)

Denote by $\nu_{y}$ one of the following differential forms $dy_{\alpha}$ , $dy_{\alpha}\wedge dy_{\beta}$ or $dy_{1}\wedge dy_{2}\wedge dy_{3}$ . Similarly, let $\tau_{x}$ be a tangential $p$ -form given by $\tau_{x}\equiv dx_{1}\wedge\ldots\wedge dx_{\alpha_{p}}$ with $0\leq p\leq m-3$ . Let

(3.164) $\omega\equiv f(x)h(y)\nu_{y}\wedge\tau_{x},$

where $f(x)$ is a smooth function defined on $U$ and $h(y)=O^{\prime}(|y|^{k})$ for some $k\in\mathbb{Z}_{+}$ , then

(3.165) $\Delta\omega-f(x)\cdot\Delta_{0}(h(y))\cdot\nu_{y}\wedge\tau_{x}=O^{\prime}(|y|^{k-1}).$
(2)

Let $f$ be a smooth function defined on $U\subset P$ and let

(3.166) $\omega\equiv f(x)\cdot\frac{y_{\alpha}}{r}dy_{1}\wedge dy_{2}\wedge dy_{3},$

then

(3.167) $\displaystyle\Delta\omega=2r^{-2}\omega+r^{-5}\Gamma_{3}^{(4)}+O^{\prime}(1),$

where the definition of the $3$ -form $\Gamma_{3}^{(4)}$ is in (3.45) of Notation 3.9.

Proof.

First, we prove Item (1). By definition, $\Delta=dd^{*}+d^{*}d$ . We only prove the case $\nu_{y}=dy_{\alpha}$ for $1\leq\alpha\leq 3$ and $1\leq p\leq m-3$ . The proof of the remaining cases is identical.

First, we compute $d^{*}d\omega$ .

(3.168)

\displaystyle d\omega

\displaystyle=f(x)\cdot\frac{\partial h(y)}{\partial y_{\beta}}dy_{\beta}\wedge dy_{\alpha}\wedge\tau_{x}+\frac{\partial f(x)}{\partial x_{j}}\cdot h(y)\cdot dx_{j}\wedge\nu_{y}\wedge\tau_{x},

which implies that

	$\displaystyle*d\omega$	$\displaystyle=(-1)^{p}f(x)\cdot\frac{\partial h(y)}{\partial y_{\beta}}dy_{\widehat{\beta\alpha}}\wedge*_{T}(\tau_{x})+O^{\prime}(\|y\|^{k})$
(3.169)			$\displaystyle=(-1)^{p}f(x)\Big(\frac{\partial h(y)}{\partial y_{\alpha-1}}dy_{\alpha+1}-\frac{\partial h(y)}{\partial y_{\alpha+1}}dy_{\alpha-1}\Big)\wedge*_{T}(\tau_{x})+O^{\prime}(\|y\|^{k}).$

Differentiating the above equality,

	$\displaystyle d*d\omega$	$\displaystyle=(-1)^{p}f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha-1}}dy_{\alpha}\wedge dy_{\alpha+1}-\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha+1}}dy_{\alpha}\wedge dy_{\alpha-1}$
(3.170)			$\displaystyle+\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}^{2}}+\frac{\partial^{2}h}{\partial y_{\alpha+1}^{2}}\Big)dy_{\alpha-1}\wedge dy_{\alpha+1}\Big)\wedge*_{T}(\tau_{x})+O^{\prime}(\|y\|^{k-1}).$

Then it follows that

	$\displaystyle d^{*}d\omega$	$\displaystyle=(-1)^{mp+m+1}dd\omega$
		$\displaystyle=f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha-1}}dy_{\alpha-1}+\frac{\partial^{2}h}{\partial y_{\alpha}\partial y_{\alpha+1}}dy_{\alpha+1}-\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}^{2}}+\frac{\partial^{2}h}{\partial y_{\alpha+1}^{2}}\Big)dy_{\alpha}\Big)\wedge\tau_{x}$
(3.171)			$\displaystyle+O^{\prime}(\|y\|^{k-1}).$

On the other hand,

(3.172)

\displaystyle*\omega=f(x)h(y)dy_{\widehat{\alpha}}\wedge*_{T}(\tau_{x}),

which implies

(3.173)

d*\omega=f(x)\frac{\partial h}{\partial y_{\alpha}}\dvol_{N}\wedge*_{T}(\tau_{x})+O^{\prime}(|y|^{k}).

So it follows that

(3.174)

d^{*}\omega=(-1)^{mp+1}*d*\omega=-f\cdot\frac{\partial h}{\partial y_{\alpha}}\cdot\tau_{x}+O^{\prime}(|y|^{k}),

and hence

(3.175)

dd^{*}\omega=-f\cdot\Big(\frac{\partial^{2}h}{\partial y_{\alpha-1}\partial y_{\alpha}}dy_{\alpha-1}+\frac{\partial^{2}h}{\partial y_{\alpha}^{2}}dy_{\alpha}+\frac{\partial^{2}h}{\partial y_{\alpha+1}\partial y_{\alpha}}dy_{\alpha+1}\Big)\cdot dy_{\alpha}\wedge\tau_{x}+O^{\prime}(|y|^{k-1}).

Therefore, combining (3.171) and (3.175),

(3.176)

\Delta\omega=(d^{*}d+dd^{*})\omega=-f\cdot(\Delta_{0}h(y))dy_{\alpha}\wedge\tau_{x}+O^{\prime}(|y|^{k-1}),

where $\Delta_{0}(h(y))=-\frac{\partial^{2}h}{\partial y_{1}^{2}}-\frac{\partial^{2}h}{\partial y_{2}^{2}}-\frac{\partial^{2}h}{\partial y_{3}^{2}}$ . The proof of (1) is done.

Now we prove Item (2). Let $\omega=f\cdot\frac{y_{\alpha}}{r}dy_{1}\wedge dy_{2}\wedge dy_{3}$ and the first step is to compute the term $d^{*}d\omega$ . By Lemma 3.2, $dr=\frac{y_{\gamma}dy_{\gamma}}{r}$ , then

(3.177)

d(\frac{y_{\alpha}}{r})\wedge dy_{1}\wedge dy_{2}\wedge dy_{3}=0.

This implies that

(3.178)

d\omega=\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}}{r}dx_{i}\wedge dy_{1}\wedge dy_{2}\wedge dy_{3},

and hence

(3.179)

*d\omega=-\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}}{r}*_{T}(dx_{i})+O^{\prime}(r).

Differentiating the above equality and applying Lemma 3.2 again,

(3.180)

\displaystyle d*d\omega=-\frac{\partial f}{\partial x_{i}}\Big(\frac{dy_{\alpha}}{r}-\frac{y_{\alpha}y_{\beta}\cdot dy_{\beta}}{r^{3}}\Big)*(dx_{i})+O^{\prime}(1).

It follows that

(3.181)

\displaystyle*d*d\omega=(-1)^{m+1}\cdot\frac{\partial f}{\partial x_{i}}\cdot\frac{dy_{\widehat{\alpha}}\wedge dx_{i}}{r}+(-1)^{m}\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dy_{\widehat{\beta}}\wedge dx_{i}+O^{\prime}(1)

Therefore,

	$\displaystyle d^{*}d\omega$	$\displaystyle=(-1)^{m+1}dd\omega$
(3.182)			$\displaystyle=\frac{\partial f}{\partial x_{i}}\cdot\frac{dy_{\widehat{\alpha}}\wedge dx_{i}}{r}-\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dy_{\widehat{\beta}}\wedge dx_{i}+O^{\prime}(1).$

Now we compute $dd^{*}\omega$ . By Lemma 3.13 and the expansion of $\dvol_{N}$ in (3.92),

	$\displaystyle*(dy_{1}\wedge dy_{2}\wedge dy_{3})$	$\displaystyle=*(\dvol_{N}+A_{i\gamma\beta}y_{\gamma}dy_{\widehat{\gamma}}\wedge dx_{i})+\widetilde{O}(r^{2})$
(3.183)			$\displaystyle=(1+H^{\beta}y_{\beta})\dvol_{T}+A_{i\gamma\beta}y_{\gamma}dy_{\gamma}\wedge*_{T}(dx_{i})+\widetilde{O}(r^{2}),$

so we have

	$\displaystyle*\omega$	$\displaystyle=f\cdot\frac{y_{\alpha}}{r}\cdot\dvol_{T}+f\cdot H^{\beta}\cdot\frac{y_{\alpha}y_{\beta}}{r}\cdot\dvol_{T}+f\cdot A_{i\gamma\beta}\frac{y_{\alpha}y_{\gamma}}{r}dy_{\gamma}\wedge*_{T}(dx_{i})+O^{\prime}(r^{2})$
(3.184)			$\displaystyle\equiv\FT_{1}+\FT_{2}+\FT_{3}+O^{\prime}(r^{2}).$

By collecting the leading terms, it is easy to compute the leading term in the above equality,

(3.185)		$\displaystyle d*d(\FT_{1})$	$\displaystyle=\frac{\partial f}{\partial x_{i}}\Big(\frac{dx_{i}\wedge dy_{\widehat{\alpha}}}{r}-\frac{y_{\alpha}y_{\beta}dx_{i}\wedge dy_{\widehat{\beta}}}{r^{3}}\Big)-2\omega+O^{\prime}(1)$
(3.186)		$\displaystyle d*d(\FT_{2})$	$\displaystyle=r^{-5}\Pi_{3}^{(4)},\ d*d(\FT_{3})=r^{-5}\Pi_{3}^{(4)}.$

Therefore,

(3.187)

dd^{*}\omega=-\frac{\partial f}{\partial x_{i}}\cdot\frac{dx_{i}\wedge dy_{\widehat{\alpha}}}{r}+\frac{\partial f}{\partial x_{i}}\cdot\frac{y_{\alpha}y_{\beta}}{r^{3}}\cdot dx_{i}\wedge dy_{\widehat{\beta}}+2\omega+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).

By (3.182) and (3.187) we obtain the expansion

(3.188)

\displaystyle\Delta\omega=(d^{*}d+dd^{*})\omega=2\omega+r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).

So the proof is done.

∎

Now we finish Step 2 by proving the following

Proposition 3.20.

There is some $3$ -form $\Lambda_{3}^{(4)}$ (given in Notation 3.9) such that if we choose

	$\displaystyle\mathfrak{B}_{0}$	$\displaystyle\equiv\frac{H^{\alpha}y_{\alpha}}{4r}dy_{1}\wedge dy_{2}\wedge dy_{3}$
(3.189)			$\displaystyle+\Big(\Omega_{ij\alpha\beta}(\frac{1}{16}y_{\widehat{\alpha\beta}}dr-\frac{3}{16}rdy_{\widehat{\alpha\beta}})-\frac{1}{16}A_{ij\alpha\beta}(y_{\widehat{\alpha\beta}}dr+rdy_{\widehat{\alpha\beta}})\Big)\wedge dx_{i}\wedge dx_{j}+r^{-3}\Lambda_{3}^{(4)},$

then the corrected $3$ -form of $\phi_{1}$ ,

(3.190)

\displaystyle\phi_{2}\equiv

\displaystyle\phi_{1}+\mathfrak{B}_{0}

satisfies

(3.191)

\Delta\phi_{2}=O^{\prime}(1)\ \text{on}\ \mathcal{U}\setminus P,

and has the expansion

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

Proof.

Let $\mathfrak{b}_{0}\equiv\frac{H^{\alpha}y_{\alpha}}{4r}dy_{1}\wedge dy_{2}\wedge dy_{3}$ , then Item (2) of Lemma 3.19 tells us that

(3.193)

\Delta\mathfrak{b}_{0}=\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}+r^{-5}\Gamma_{3}^{(4)}+O^{\prime}(1).

Let $\Pi^{(4)}$ be the $3$ -form in the expansion of $\Delta\phi_{1}$ given by (3.119) in Proposition 3.15.

Next, Lemma 3.18 and Lemma 3.19 tell us that there are $3$ -forms $\widehat{\Gamma}_{3}^{(4)}$ and $\widehat{\Pi}_{3}^{(4)}$ which are also of the form as in (3.45) such that

(3.194)		$\displaystyle\Delta(r^{-3}\widehat{\Gamma}_{3}^{(4)})=-r^{-5}\Gamma_{3}^{(4)}+O^{\prime}(1),$
(3.195)		$\displaystyle\Delta(r^{-3}\widehat{\Pi}_{3}^{(4)})=-r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

Now let

(3.196)

\mathfrak{b}_{1}\equiv r^{-3}\widehat{\Gamma}_{3}^{(4)}+r^{-3}\widehat{\Pi}_{3}^{(4)},

then the correction term $\mathfrak{b}_{0}+\mathfrak{b}_{1}$ is chosen as the above such that $\Delta(\mathfrak{b}_{0}+\mathfrak{b}_{1})$ in fact eliminates the $O^{\prime}(r^{-2})$ -term and implicit $O^{\prime}(r^{-1})$ -terms in the expansion of $\Delta\phi_{1}$ (see Proposition 3.15).

In the following, we will make a further correction such that those explicit $O^{\prime}(r^{-1})$ -terms will be cancelled out as well. In fact, we define

(3.197)

\mathfrak{b}_{2}\equiv\Big(\Omega_{ij\alpha\beta}(\frac{1}{16}y_{\widehat{\alpha\beta}}dr-\frac{3}{16}rdy_{\widehat{\alpha\beta}})-\frac{1}{16}A_{ij\alpha\beta}(y_{\widehat{\alpha\beta}}dr+rdy_{\widehat{\alpha\beta}})\Big)\wedge dx_{i}\wedge dx_{j},

applying Lemma 3.18 and Lemma 3.19 again, then

(3.198)

\Delta\mathfrak{b}_{2}=\Omega_{ij\alpha\beta}\Big(\frac{1}{4r}dy_{\widehat{\beta\alpha}}+\frac{y_{\widehat{\alpha\beta}}}{4r^{2}}dr\Big)\wedge dx_{i}\wedge dx_{j}-A_{ij\alpha\beta}\Big(\frac{y_{\widehat{\alpha\beta}}}{4r^{2}}dr-\frac{1}{4r}dy_{\widehat{\alpha\beta}}\Big)\wedge dx_{i}\wedge dx_{j},

and hence

(3.199)

\Delta(\phi_{1}+\mathfrak{b}_{0}+\mathfrak{b}_{1}+\mathfrak{b}_{2})=O^{\prime}(1).

Therefore, it suffices to choose the correction term

(3.200)

\mathfrak{B}_{0}\equiv\mathfrak{b}_{0}+\mathfrak{b}_{1}+\mathfrak{b}_{2},

which gives $\Delta(\phi_{1}+\mathfrak{B}_{0})=O^{\prime}(1)$ .

Notice that, $\mathfrak{b}_{2}$ has a further cancellation,

	$\displaystyle\mathfrak{b}_{2}=$	$\displaystyle\Big(\Omega_{ij\alpha\beta}(\frac{1}{16}y_{\widehat{\alpha\beta}}dr-\frac{3}{16}rdy_{\widehat{\alpha\beta}})-\frac{1}{16}A_{ij\alpha\beta}(y_{\widehat{\alpha\beta}}dr+rdy_{\widehat{\alpha\beta}})\Big)\wedge dx_{i}\wedge dx_{j},$
	$\displaystyle=$	$\displaystyle-\frac{1}{4}A_{ij\alpha\beta}rdy_{\widehat{\alpha\beta}}\wedge dx_{i}\wedge dx_{j}-\frac{1}{16}(A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}-A_{i,\alpha,\alpha+2}A_{j,\alpha,\alpha+1})y_{\alpha}dr\wedge dx_{i}\wedge dx_{j}$
(3.201)			$\displaystyle+\frac{3}{16}(A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}-A_{i,\alpha,\alpha+2}A_{j,\alpha,\alpha+1})rdy_{\alpha}\wedge dx_{i}\wedge dx_{j}.$

Therefore,

	$\displaystyle\phi_{2}=$	$\displaystyle\phi_{1}+\mathfrak{B}_{0}$
	$\displaystyle=$	$\displaystyle\frac{1}{2r}(1-\frac{H^{\alpha}y_{\alpha}}{2})dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r}y_{\beta}A_{i\alpha\beta}dx_{i}\wedge dy_{\widehat{\alpha}}-\frac{1}{4}A_{ij\alpha\beta}r\cdot dy_{\widehat{\alpha\beta}}\wedge dx_{i}\wedge dx_{j}$
(3.202)			$\displaystyle+\frac{3}{16}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})d(ry_{\alpha})\wedge dx_{i}\wedge dx_{j}+r^{-3}\Pi_{3}^{(4)}+O^{\prime}(r^{2}),$

and

(3.203)

\Delta\phi_{2}=O^{\prime}(1).

∎

Step 4. In this step we compute $\Delta\phi_{2}$ as a current on $\mathcal{U}$ .

Lemma 3.21.

In $\mathcal{U}$ , we have

(3.204)

\Delta\phi_{2}=2\pi\delta_{U}+O^{\prime}(1).

Proof.

Suppose we are given a compactly supported test form $\chi\in\Omega_{0}^{m-3}(\mathcal{U})$ , then we apply integration by parts once and we have

(3.205)

(\phi_{2},\Delta\chi)=\int_{\mathcal{U}}\phi_{2}\wedge(dd^{*}+d^{*}d)\chi=\int_{\mathcal{U}}d\phi_{2}\wedge d^{*}\chi-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi.

Here there is no boundary term because $\phi_{2}=O(r^{-1})$ . Notice that (3.133) and (3.190) implies $d\phi_{2}=O^{\prime}(1)$ , so

(3.206)

\int_{\mathcal{U}}d\phi_{2}\wedge d^{*}\chi=\int_{\mathcal{U}}d^{*}d\phi_{2}\wedge\chi.

On the other hand, by (3.139),

(3.207)

d^{*}\phi_{2}=-\frac{y_{\alpha}}{2r^{3}}{}dy_{\hat{\alpha}}+\zeta,

where $\zeta$ is a $2$ -form satisfying $\zeta=O^{\prime}(r^{-1})$ . Denote by $S_{\epsilon}^{2}$ the normal geodesic sphere bundle $\{r=\epsilon\}$ , then we get that

(3.208)		$\displaystyle-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi$	$\displaystyle=$	$\displaystyle\int_{\mathcal{U}}dd^{}\phi_{2}\wedge\chi+\int_{S_{\epsilon}^{2}}d^{}\phi_{2}\wedge\chi$
(3.208)			$\displaystyle=$	$\displaystyle\int_{\mathcal{U}}dd^{*}\phi_{2}\wedge\chi+\lim_{\epsilon\rightarrow 0}\frac{1}{2\epsilon^{3}}{}\int_{S_{\epsilon}}(y_{\alpha}dy_{\hat{\alpha}}+\epsilon^{2}\zeta)\wedge\chi.$

By direct calculation of the last term on the right hand side we obtain

(3.209)

-\int_{\mathcal{U}}d^{*}\phi_{2}\wedge d\chi=\int_{\mathcal{U}}dd^{*}\phi_{2}\wedge\chi+2\pi{}\int_{P}\chi.

This concludes the proof. ∎

Step 5. Now we solve the Laplace equation with right hand side in $O^{\prime}(1).$

Lemma 3.22.

Given a local 3-form $v$ defined on a neighborhood $\mathcal{U}$ of $p$ in $Q$ with $v=O^{\prime}(1)$ , then there is some smaller neighborhood $\mathcal{V}\subset\subset\mathcal{U}$ such that there exists a local solution $T$ to the equation

(3.210)

\Delta T=v,

with $T=O^{\prime}(r^{2})$ . Here $r$ is the distance to the submanifold $P$ .

Proof.

We just need to establish the following:

(1)

(General derivatives estimate) For each $k\in\mathbb{N}$ and $\epsilon>0$ , it holds that

(3.211) $|\nabla^{k+2}T|=O(r^{-(k+\epsilon)}).$
(2)

(Mixed derivatives estimate) For each $k\in\mathbb{N}$ , $\ell\in\mathbb{N}$ and $\epsilon>0$ , it holds that

(3.212) $|(\nabla^{t})^{k}\nabla^{\ell+2}T|=O(r^{-(\ell+\epsilon)}),$

where $\nabla^{t}$ denotes the tangential derivative.

The above estimates will be proved by induction.

First, we will prove the following order estimate for $\nabla^{2}T$ in a smaller neighborhood $\mathcal{U}^{\prime}\subset\subset\mathcal{U}$

(3.213)

|\nabla^{2}T|=O(r^{-\epsilon}).

This can be viewed as the base step for carrying out the inductive argument.

To begin with, by definition, for any $p>0$ , we have $|v|_{L^{p}(\mathcal{U})}\leq C_{p}$ . Applying the standard elliptic $W^{2,p}$ -estimate, for each $p>0$ , there is some constant $C_{p}>0$ such that in a smaller neighborhood $\mathcal{U}_{1}\subset\subset\mathcal{U}$ such that

(3.214)

\|\nabla^{2}T\|_{L^{p}(\mathcal{U}_{1})}\leq C_{p}.

Then Sobolev embedding theorem tells us that

(3.215)

T\in W^{2,p}(\mathcal{U}_{1})\cap C^{1,\alpha}(\mathcal{U}_{1})

for any $p>1$ and $0<\alpha<1$ .

To prove (3.213), we need to differentiate the equation, which schematically yields that

(3.216)

\Delta\nabla^{2}T+Q_{1}*\nabla^{3}T+Q_{2}*\nabla^{2}T+Q_{3}*\nabla T=w,

where $|w|=O^{\prime}(r^{-2})$ and $Q_{i}$ ’s are smooth terms arising from differentiating the coefficients of $\Delta$ . The above equation can be viewed as an elliptic system in terms of the Hessian of $T$ . Let $\Phi\equiv\nabla^{2}T$ , noticing $\nabla T\in C^{\alpha}(\mathcal{U}_{1})$ , so the terms involving $\nabla T$ can be absorbed to the right hand side of the equation. Then $\Phi$ can be treated as vector valued functions, once we fix a local frame. So it follows that

(3.217)

\Delta\Phi+Q_{1}*\nabla\Phi+Q_{2}*\Phi=\eta,

where $|\eta|=O^{\prime}(r^{-2})$ . Since $\eta$ has unbounded $L^{p}$ -norm for large $p$ , the standard $W^{2,p}$ -estimate for $\Phi$ does not directly apply.

For improving the regularity of $\nabla^{2}T$ , we will rescale the metric $g$ . For each $x$ in an even smaller neighborhood $\mathcal{U}_{2}$ with $r(x)=r_{x}$ , we rescale the metric $g$ in $B_{r_{x}}(x)$ by letting

(3.218)

\tilde{g}=(r_{x})^{-2}g,

then the following equation holds in the rescaled geodesic ball $B_{1}^{\tilde{g}}(x)$ ,

(3.219)

\widetilde{\Delta}\widetilde{\Phi}+r_{x}\widetilde{Q}_{1}*\widetilde{\nabla}\widetilde{\Phi}+(r_{x})^{2}\widetilde{Q}_{2}*\widetilde{\Phi}=\tilde{\eta},\quad|\tilde{\eta}|=O^{\prime}(1),

where $\widetilde{\Phi}(y)=\Phi(r_{x}\cdot y)$ for each $y\in B_{1}^{\tilde{g}}(x)$ . In the above equation, all the coefficients are uniformly bounded independent of $x$ . Since we have shown $|\Phi|_{L^{p}}\leq C_{p}$ in (3.214), so simple rescaling gives rise to the following estimate for any $p>0$ ,

(3.220)

|\widetilde{\Phi}|_{L^{p}(B_{1}^{\tilde{g}}(x))}\leq C_{p}\cdot(r_{x})^{-\frac{n}{p}}.

Now applying the $W^{2,p}$ -estimate for $\widetilde{\Phi}$ , then for each $p>0$

(3.221)

|\widetilde{\Phi}|_{W^{2,p}(B_{1/2}^{\tilde{g}}(x))}\leq C_{p}\cdot(r_{x})^{-\frac{n}{p}}

with $C_{p}>0$ independent of $x$ . By the Sobolev embedding

(3.222)

|\widetilde{\Phi}|_{C^{1,\alpha}(B_{1/4}^{\tilde{g}}(x))}\leq C_{\alpha,p}\cdot(r_{x})^{-\frac{n}{p}}

with $C_{\alpha,p}>0$ independent of $x$ . Scale back to the original metric, for any $p>0$ , there is some $C_{\alpha,p}>0$ independent of the base point $x\in\mathcal{U}_{2}$ such that

(3.223)

|\Phi|_{L^{\infty}(B_{r_{x}/4}(x))}+|r_{x}\nabla\Phi|_{L^{\infty}(B_{r_{x}/4}(x))}\leq C_{\alpha,p}\cdot(r_{x})^{-\frac{n}{p}}.

This completes the proof of (3.213).

Now we will finish the proof of Item (1) by using the induction. Based on (3.223), the key induction step is to prove the following: Given any $\ell\in\mathbb{Z}_{+}$ , if for each $\epsilon>0$ and $0\leq k\leq\ell-1$ ,

(3.224)

|\nabla^{k}\Phi|=O(r^{-(k+\epsilon)}),\quad|\widetilde{\nabla}^{k}\widetilde{\Phi}|_{W^{2,p}(B_{1/2}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}\cdot(r_{x})^{-\epsilon},

then for each $\epsilon>0$ , we have

(3.225)

|\nabla^{\ell}\Phi|=O(r^{-(\ell+\epsilon)}),\quad|\widetilde{\nabla}^{\ell}\widetilde{\Phi}|_{W^{2,p}(B_{1/4}^{\tilde{g}}(x))}\leq C_{\ell,p,\epsilon}\cdot(r_{x})^{-\epsilon}.

Indeed, then differentiating (3.216) by $\nabla^{k}$ ,

(3.226)

\Delta(\nabla^{\ell}\Phi)+\sum\limits_{j=1}^{\ell+1}Q_{j}*\nabla^{j}\Phi=w_{\ell},

where $|w_{\ell}|=O^{\prime}(r^{-(\ell+2)})$ . As before, we rescale the metric $g$ by taking $\tilde{g}=(r_{x})^{-2}g$ , then

(3.227)

\widetilde{\Delta}(\widetilde{\nabla}^{\ell}\widetilde{\Phi})+\sum\limits_{j=1}^{\ell+1}(r_{x})^{\ell-j+2}\cdot Q_{j}*\nabla^{j}\widetilde{\Phi}=\tilde{w}_{\ell},

where $|\tilde{w}_{\ell}|=O^{\prime}(1)$ . Let $k=\ell-1$ , applying the induction hypothesis (3.224) and Sobolev embedding, we have

(3.228)

|\widetilde{\nabla}^{\ell}\widetilde{\Phi}|_{L^{\infty}(B_{1/2}^{\tilde{g}}(x))}=O((r_{x})^{-\epsilon}).

The above enables us to apply the $W^{2,p}$ -elliptic estimate, so we obtain the following estimate for each $\epsilon>0$ ,

(3.229)

|\widetilde{\nabla}^{\ell}\widetilde{\Phi}|_{W^{2,p}(B_{1/4}^{\tilde{g}}(x))}\leq C_{\ell,p,\epsilon}\cdot(r_{x})^{-\epsilon},

where $C_{\ell,p,\epsilon}>0$ is independent of the base point $x$ . Applying the Sobolev embedding $W^{2,p}\subset C^{1,\alpha}$ and scaling back to the original metric $g$ ,

(3.230)

|\nabla^{\ell}\Phi|_{L^{\infty}(B_{r_{x}/8}(x))}\leq C_{\ell,\epsilon}\cdot(r_{x})^{-(\ell+\epsilon)}

for each $\epsilon>0$ . So we complete the proof of Item (1).

Now we are ready to finish the proof of Item (2). We only focus on the case $\ell=0$ and the case for $\ell>0$ can be directly achieved by applying the above rescaling arguments. To this end, we need the following claim for the tangential derivatives estimate.

Claim. Let $|\widetilde{\Phi}|_{W^{2,p}(B_{1/2}^{\tilde{g}}(x))}\leq C_{p,\epsilon}\cdot(r_{x})^{-\epsilon}$ for any $\epsilon>0$ and for any $x\in\mathcal{U}$ . Assume that $\widetilde{\Phi}$ solves the elliptic equation

(3.231)

\Delta\widetilde{\Phi}+Q_{1}*\nabla\widetilde{\Phi}+Q_{2}*\widetilde{\Phi}=\zeta,

where $Q_{i}$ ’s are smooth coefficients, $|\zeta|\in O^{\prime}(1)$ . Then for any $x\in\mathcal{U}$ , the estimate

(3.232)

|(\nabla^{t})^{k}\widetilde{\Phi}|_{W^{2,p}(B_{1/8}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}\cdot(r_{x})^{-\epsilon},

holds for all $k\geq 1$ , $p>1$ and $\epsilon>0$ .

Taking the first tangential derivative $\nabla^{t}$ for $\Delta\widetilde{\Phi}$ ,

(3.233)

(\nabla^{t})(\Delta\widetilde{\Phi})=\Delta\nabla^{t}\widetilde{\Phi}+Q_{1}*\nabla^{2}\widetilde{\Phi}+Q_{2}*\nabla\widetilde{\Phi},

where $Q_{i}$ ’s are smooth functions. Hence differentiating (3.231) once by the tangential derivative $\nabla^{t}$ , we have

(3.234)

\Delta(\nabla^{t}\widetilde{\Phi})+Q_{1}*\nabla^{2}\widetilde{\Phi}+Q_{2}*\nabla\widetilde{\Phi}=w_{1},

where $Q_{i}$ ’s are smooth functions, $w_{1}\equiv\nabla^{t}\zeta$ and $|w_{1}|\in O^{\prime}(1)$ . Since we have already assumed $|\nabla^{2}\widetilde{\Phi}|_{L^{p}(B_{1/2}^{\tilde{g}}(x))}\leq C_{p,\epsilon}\cdot(r_{x})^{-\epsilon}$ for all $\epsilon>0$ , applying the standard $W^{2,p}$ -estimate, then for any $p>1$ and $\epsilon>0$ ,

(3.235)

|\nabla^{t}\widetilde{\Phi}|_{W^{2,p}(B_{1/3}^{\tilde{g}}(x))}\leq C_{p,\epsilon}\cdot(r_{x})^{-\epsilon}.

Now we prove the higher order mixed derivatives estimate by induction. Repeat taking the tangential derivatives and let $\Psi^{(k)}\equiv(\nabla^{t})^{k}\widetilde{\Phi}$ for all $k>1$ . Assume that $|\Psi^{(j)}|_{W^{2,p}(B_{1/4}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}(r_{x})^{-\epsilon}$ holds for all $j\leq k-1$ and $p>1$ , then

(3.236)

\Delta(\Psi^{(k)})+\sum_{\begin{subarray}{c}\mu=1,2\\ 1\leq\mu+\nu\leq k\end{subarray}}Q_{\mu\nu}*\nabla^{\mu}(\Psi^{(\nu)})=w_{k},

where $w_{k}\equiv(\nabla^{t})^{k}\zeta$ and $|w_{k}|=O^{\prime}(1)$ . Applying the induction hypothesis, it follows that for any $k\geq 1$ ,

(3.237)

|\Delta(\Psi^{(k)})|_{L^{p}(B_{1/4}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}\cdot(r_{x})^{-\epsilon}

Therefore, for any $k\geq 1$ , $p>1$ and $\epsilon>0$ , there is some constant $C_{k,p,\epsilon}>0$ such that

(3.238)

|(\nabla^{t})^{k}\widetilde{\Phi}|_{W^{2,p}(B_{1/8}^{\tilde{g}}(x))}=|\Psi^{(k)}|_{W^{2,p}(B_{1/8}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}\cdot(r_{x})^{-\epsilon}.

This completes the proof of the claim.

Now we are in a position to finish the proof of the lemma by completing the induction arguments for Item (2). As before, for any $x$ , under the rescaled metric $\tilde{g}=(r_{x})^{-2}g$ , we start with the equation for $\widetilde{\Phi}$ in the rescaled geodesic ball $B_{1}^{\tilde{g}}(x)$ ,

(3.239)

\widetilde{\Delta}\widetilde{\Phi}+r_{x}\widetilde{Q}_{1}*\widetilde{\nabla}\widetilde{\Phi}+(r_{x})^{2}\widetilde{Q}_{2}*\widetilde{\Phi}=\tilde{\eta},\ 0<r_{x}<1,

where $\widetilde{Q}_{i}$ ’s are smooth functions and $|\tilde{\eta}|=O^{\prime}(1)$ . The above claim tells us that for any $k\geq 1$ , $p>1$ and $\epsilon>0$ ,

(3.240)

|(\nabla^{t})^{k}\widetilde{\Phi}|_{W^{2,p}(B_{1/2}^{\tilde{g}}(x))}\leq C_{k,p,\epsilon}\cdot(r_{x})^{-\epsilon},

where $C_{k,p,\epsilon}>0$ is independent of $x$ . Applying the Sobolev embedding, then for any $0<\alpha<1$ ,

(3.241)

|(\nabla^{t})^{k}\widetilde{\Phi}|_{C^{1,\alpha}(B_{1/4}^{\tilde{g}}(x))}\leq C_{k,\alpha,\epsilon}\cdot(r_{x})^{-\epsilon}.

In particular, $|(\nabla^{t})^{k}\widetilde{\Phi}|=O(r^{-\epsilon})$ for all $\epsilon>0$ . Notice that the above estimate is independent of the choice of $x$ . Rescaling back to the original metric, then for each $\epsilon>0$ ,

(3.242)

|(\nabla^{t})^{k}\nabla^{2}T|=O(r^{-\epsilon}).

The proof of the lemma is done.

∎

With all the above preparations, now we are ready to finish the proof of Theorem 3.10.

Proof of Theorem 3.10.

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.

∎

For our purpose later, we also need the following lemma.

Lemma 3.23.

Let $\Delta$ denote the Hodge Laplacian, then

(3.248)

\Delta(r^{2})=-6+\widetilde{O}(r^{2}).

Proof.

This follows from similar, and simpler arguments as above. First,

(3.249)

dr^{2}=2rdr=2y_{\alpha}\eta_{\alpha}+\widetilde{O}(r^{4}),

so it follows that

(3.250)

*dr^{2}=2y_{\alpha}\eta_{\widehat{\alpha}}\wedge\dvol_{T}+\widetilde{O}(r^{3})

and

(3.251)

d*dr^{2}=6\dvol_{N}\wedge\dvol_{T}+\widetilde{O}(r^{2})=6\dvol_{g}+\widetilde{O}(r^{2}).

Hence

(3.252)

\Delta(r^{2})=d^{*}dr^{2}=-*d*dr^{2}=-6+\widetilde{O}(r^{2}).

∎

3.2. Green’s currents for Riemannian submanifolds [04ZJ]

3.2. Green’s currents for Riemannian submanifolds

Definition 3.5 (kk-current).

Definition 3.6 (Green’s current).

Example 3.7.

Proposition 3.8.

Proof.

Notation 3.9.

Remark 3.9.1.

Theorem 3.10.

Remark 3.10.1.

Remark 3.10.2.

Remark 3.10.3.

Remark 3.10.4.

Lemma 3.11.

Proof.

Lemma 3.12.

Proof.

Lemma 3.13.

Proof.

Lemma 3.14 (Rearrangement Lemma).

Proof.

Proposition 3.15.

Proof.

Lemma 3.16 (Cancellation Lemma).

Proof.

Lemma 3.17.

Proof.

Lemma 3.18.

Proof.

Lemma 3.19.

Proof.

Proposition 3.20.

Proof.

Lemma 3.21.

Proof.

Lemma 3.22.

Proof.

Proof of Theorem 3.10.

Lemma 3.23.

Proof.

Definition 3.5 ( $k$ -current).