Singular Yamabe problem for scalar flat metrics on the sphere

Abstract

Let $\Omega$ be a domain on the unit n-sphere ${\mathbb {S}}^n$ and $\overset{{\,}_\circ }{g}$ the standard metric of ${\mathbb {S}}^n$, $n\ge 3$. We show that there exists a conformal metric g with vanishing scalar curvature $R(g)=0$ such that $(\Omega , g)$ is complete if and only if the Bessel capacity ${\mathcal {C}}_{\alpha , q}({\mathbb {S}}^n\setminus \Omega )=0$, where $\alpha =1+\frac{2}{n}$ and $q=\frac{n}{2}$. Our analysis utilizes some well known properties of capacity and Wolff potentials, as well as a version of the Hopf–Rinow theorem for the divergent curves.

1 Introduction

Let $\Omega$ be an open subset of the unit sphere ${\mathbb {S}}^n, n\ge 3$, and $\overset{{\,}_\circ }{g}$ the standard metric of ${\mathbb {S}}^n$ induced by the embedding ${\mathbb {S}}^n\hookrightarrow {\mathbb {R}}^{n+1}$. Let $K\subset {\mathbb {S}}^n$ be a closed subset. We want to characterize the open sets $\Omega ={\mathbb {S}}^n{\setminus } K$ with the following property: there exists a metric g, conformal to $\overset{{\,}_\circ }{g}$ such that $(\Omega , g)$ is complete and g has vanishing scalar curvature. This question was studied by Schoen and Yau [17, 18]. If we are given a compact Riemannian manifold (M, g) then the problem of existence of a conformal deformation of g into a metric ${\bar{g}}$ with constant scalar curvature is known as the Yamabe problem [21]. Note that for compact manifolds there no need to specify that conformal metric to be complete. However, the completeness of the conformal metric is required in the singular Yamabe problem. Yamabe’s original approach was to formulate the existence of ${\bar{g}}$ in terms of a variational problem. Later contributions of Trudinger [20], Aubin [2] and Schoen [16] helped to complete Yamabe’s original approach.

One of the first results towards solving the singular Yamabe problem says that if there exists a complete g conformal to $\overset{{\,}_\circ }{g}$ such that the scalar curvature $R(g)\ge 0$ then the Hausdorff dimension of $\partial \Omega$ must be at most $\frac{n}{2}$, [18]. To obtain existence of g, one can impose structural assumptions on $K:={\mathbb {S}}^n\setminus \Omega$. In particular, if $K={\mathbb {S}}^n\setminus K$ is a finite union of Lipschitz submanifolds of dimension $k\le (n-2)/2$ then there is a g, solving the singular Yamabe problem in the case $R(g)=0$, see [4, 10, 13]. See also [3] for the periodic setting with equator as singular set in $n\ge 5$ sphere. Some discussion on this and related open problems is contained in [14]. For a survey of related recent results see [6, 11] and the references therein.

Given a domain $\Omega \subset {{\mathbb {S}}}^n$. The scalar curvature of R(g) of a metric g after a conformal change $g=u^{\frac{4}{n-2}} \overset{{\,}_\circ }{g}$ is

$$\begin{aligned} R(g)=u^{-\frac{n+2}{n-2}}\left( -\frac{4(n-1)}{n-2}\Delta u +R( \overset{{\,}_\circ }{g})u\right) , \end{aligned}$$

(1.1)

where $\Delta u$ is the Laplace–Beltrami operator on the Riemannian manifold $({\mathbb {S}}^n, \overset{{\,}_\circ }{g})$. Suppose g is scalar flat, that is, $R(g)=0$, then from (1.1) and the observation that $R( \overset{{\,}_\circ }{g})=n(n-1)$ we arrive at the following problem:

$$\begin{aligned} \frac{4(n-1)}{n-2}\Delta u -n(n-1)u =0&\ \textrm{in}\ {}&\Omega ,\nonumber \\ u>0&\ \textrm{in}\ {}&\Omega ,\nonumber \\ u^{4/(n-2)} \overset{{\,}_\circ }{g}\ \mathrm{is \ complete\ metric}&\ \textrm{in}\ {}&\Omega . \end{aligned}$$

(1.2)

For the negative scalar case $R(g)=-1$ Labutin showed that a sufficient and necessary condition for the existence of g is

$$\begin{aligned} \int _{0}^{\frac{1}{2}}\left( \frac{{\mathcal {C}}(B(x, r)\cap K)}{{\mathcal {C}}(B(x, r)}\right) ^{\frac{2}{n-2}} dr=+\infty \end{aligned}$$

for any $x\in K$, where ${\mathcal {C}}$ is the Bessel capacity for the Sobolev space $W^{2, \frac{n+2}{4}}({\mathbb {R}}^n)$, see [11]. His result can be seen as a way of measuring the thinness of K, which comes from the classical potential theory, namely Wiener’s test. Recall that Wiener’s theorem states that the Dirichlet problem

$$\begin{aligned} \left\{ \begin{array}{rcl} \Delta w =0&{}\textrm{in}&{}D, \\ w=f&{}\textrm{on}&{}\partial D, \end{array} \right. \end{aligned}$$

in a bounded domain $D\subset \textbf{R}^n$, $n\ge 3$, is solvable for all boundary data $f\in C(\partial D)$ if and only if

$$\begin{aligned} \int _0^\delta \frac{cap_2 (B(x, r)\setminus D)}{cap_2 (B(x, r))} \, \frac{dr}{r} =+\infty \quad \mathrm{for \quad any} \quad x\in \partial D. \end{aligned}$$

(1.3)

Here the upper limit $\delta >0$ is some fixed constant (e.g. one can take $\delta =1$), and $cap_2$ is the classical (electrostatic) capacity.

For $R(g)=0$, our goal is to find a correct quantity measuring the thinness of K, similar to (1.3), such that it gives a complete characterization of K for which the existence of a solution to (1.2) follows. It turns out that Wolff’s potential ${\mathcal {W}}^\mu _{\alpha , q}$ is the correct quantity, as the inequalities (4.6) and (4.7) suggest, in this sense with $\alpha =1+\frac{2}{n}, q=\frac{n}{2}$, and hence the corresponding Bessel integral capacity ${\mathcal {C}}_{1+\frac{2}{n}, \frac{n}{2}}$ is not the same as the standard Sobolev capacity, defined via the $L^q$ norm of the gradient [11] (2.9). These capacities, however, agree if $\alpha$ is an integer, see [8] page 161. In [11] Labutin explicitly constructs a divergent curve (which are used to formulate a version of the Hopf-Rinow theorem for noncompact manifolds) to show that if $R(g)=-1$ and K is not thin at some point then g cannot be complete. This is the most technical part of [11]. However, for $R(g)=0$ the construction of a divergent curve simplifies if we use the Schoen-Yau estimate of the Hausdorff dimension of $\partial \Omega$.

The aim of this work is to give a complete characterization of open set $\Omega$ without any assumption on the structure of $K={\mathbb {S}}^n\setminus \Omega$. In what follows $\textrm{cap}(\cdot ): ={\mathcal {C}}_{1+\frac{2}{n}, \frac{n}{2}}(\cdot )$ stands for Bessel’s capacity (see Sect. 2 for precise definitions). Our main result is the following theorem:

Theorem 1.1

Let $\Omega \subset {\mathbb {S}}^n, n\ge 3$, be an open set and $K={{\mathbb {S}}}^n {\setminus } \Omega$. Then the following properties are equivalent:

• (i) In $\Omega$ there exists a scalar flat complete metric conformal to $\overset{{\,}_\circ }{g}$.

• (ii) $\textrm{cap}(K)=0$.

The paper is organized as follows: Sect. 2 contains some background material: first we use the steregraphic projection to reformulate the problem on ${\mathbb {R}}^n$. Then we introduce the Wolff potentials and give a characterization of polar sets, i.e. sets of vanishing capacity, see Proposition 2.1. We close the section by stating a non-standard version of the Hopf-Rinow theorem (for noncompact manifolds) formulated in terms of divergent curves, see Theorem 2.3. This allows to link the finiteness of Wolff’s potential with the completeness of the metric.

In Sect. 3 we show that the existence of a solution to scalar flat singular Yamabe problem implies that the capacity of K is zero. A part of the argument is based on the representation of positive harmonic functions in terms of Martin kernels [1, 9]. The crucial step in the proof is the estimate (3.3). In the construction of the divergence curve we also use the Schoen-Yau estimate of the Hausdorff dimension of $\partial \Omega$.

Finally, in Sect. 4 we prove the implication $(ii)\Rightarrow (i)$ in Theorem 1.1.

2 Background

This section contains some background results from conformal geometry and potential theory, see [8, 19].

2.1 Stereographic projection and reduction to ${\mathbb {R}}^n$

Let (M, g) be a Riemannian manifold of dimension $n\ge 3$. Let R(g) be the scalar curvature of the metric g and $\Delta _g$ the Laplace–Beltrami operator. The operator

$$\begin{aligned} {\mathcal {L}}_g=-4\frac{n-1}{n-2}\Delta _g u+R(g) \end{aligned}$$

(2.1)

is called conformal Laplacian.

It is well known that under conformal change of metric $\widehat{g}=\phi ^\frac{4}{n-2}g, \phi \in C^\infty (M), \phi >0$, we have

$$\begin{aligned} R({\widehat{g}})= & {} \phi ^{-\frac{n+2}{n-2}}{\mathcal {L}}_g\phi , \end{aligned}$$

(2.2)

$$\begin{aligned} {\mathcal {L}}_{\widehat{g}} v= & {} \phi ^{-\frac{n+2}{n-2}}{\mathcal {L}}_g(\phi v). \end{aligned}$$

(2.3)

Suppose $\widetilde{M}$ is another manifold with metric $\widetilde{g}$, and let $f:M\rightarrow \widetilde{M}$ be a diffeomorphism changing the metric conformally. If $f^*\widetilde{g}=\widetilde{g}\circ f$ is the pull-back then

$$\begin{aligned} f^*\widetilde{g}=\phi ^{\frac{4}{n-2}}g. \end{aligned}$$

Consequently, from (2.2) and (2.3) it follows that

$$\begin{aligned} f^*(R_{\widetilde{g}})= & {} \phi ^{-\frac{n+2}{n-2}}{\mathcal {L}}_g\phi , \nonumber \\ f^*({\mathcal {L}}_{\widetilde{g}}v)= & {} \phi ^{-\frac{n+2}{n-2}}{\mathcal {L}}_g(\phi f^*v). \end{aligned}$$

(2.4)

Using the stereographic projection $\sigma :{\mathbb {S}}^n{\setminus }\{N\}\rightarrow {\mathbb {R}}^n$, where N is the north pole, we can rewrite the transformation equations on ${\mathbb {R}}^n$. Indeed, $\sigma$ is a diffeomorphism between $({\mathbb {S}}^n\setminus \{N\}, \overset{{\,}_\circ }{g})$ and $({\mathbb {R}}^n, g_E)$, because

$$\begin{aligned} (\sigma ^{-1})^* \overset{{\,}_\circ }{g}= & {} \left( \frac{2}{1+|x|^2}\right) ^2 g_E\\= & {} U^{\frac{4}{n-2}} g_E, \end{aligned}$$

where

$$\begin{aligned} U(x):=\left( \frac{2}{1+|x|^2}\right) ^{\frac{n-2}{2}}\quad x\in {\mathbb {R}}^n. \end{aligned}$$

Since we consider the scalar flat case, i.e. $R({\widetilde{g}})=0$, then (2.4) yields

$$\begin{aligned} {{\mathcal {L}}}_{ \overset{{\,}_\circ }{g}} v=0, \quad v>0\ \ \hbox {in} \ \ \Omega . \end{aligned}$$

Introduce the function

$$\begin{aligned} u(x)= & {} U(x)(\sigma ^{-1})^*v(x)\nonumber \\ {}= & {} U(x)v(\sigma ^{-1}x), \quad x\in {\mathbb {R}}^n. \end{aligned}$$

Then from (2.3) we obtain

$$\begin{aligned} {{\mathcal {L}}}_{ g_E} v=0, \quad v>0\ \ \hbox {in}\ \ \sigma (\Omega )\subset {\mathbb {R}}^n. \end{aligned}$$

Since $R({ g_E})=0$ then we get

$$\begin{aligned} \Delta _{ g_E}v=0, \quad v>0\ \ \hbox {in}\ \ \sigma (\Omega )\subset {\mathbb {R}}^n. \end{aligned}$$

2.2 Characterization of capacity

For $\alpha >0, 1<q\le \frac{n}{\alpha }$ we define the Bessel capacity of $E\subset {\mathbb {R}}^n$ as follows

$$\begin{aligned} {\mathcal {C}}_{\alpha , q}(E)=\inf \left\{ \int _{{\mathbb {R}}^n}\psi ^q;\ \psi \ge 0, G_\alpha *\psi \ge 1\ \text {for all}\ x\in E\right\} , \end{aligned}$$

where $G_\alpha$ is the Bessel kernel, best defined as the inverse Fourier transform of $(1+|\xi |^2)^{-\frac{\alpha }{2}}$, see [8]. For given Radon measure $\mu$ the Wolff potentials are defined as follows

$$\begin{aligned} {\mathcal {W}}_{\alpha , q}^\mu (x)=\int _0^1\left( \frac{\mu (B(x, \delta ))}{\delta ^{n-\alpha q}}\right) ^{p-1}\frac{d\delta }{\delta }, \end{aligned}$$

(2.5)

where $p+q=pq$.

An important fact is that ${\mathcal {W}}^\mu _{\alpha , q}$ bounds the nonlinear potential ${\mathcal {V}}_{\alpha , q}^\mu (x):=G_\alpha *(G_\alpha *\mu )^{p-1}$ from below (see [8], page 164), i.e. there is a constant $A>0$ such that

$$\begin{aligned} {\mathcal {W}}_{\alpha , q}^\mu (x)\le A {\mathcal {V}}_{\alpha , q}^\mu (x). \end{aligned}$$

(2.6)

In what follows we take $\alpha =1+\frac{2}{n}, q=\frac{n}{2}$ and denote the resulted capacity by $\textrm{cap}(\cdot )={\mathcal {C}}_{1+\frac{2}{n}, \frac{n}{2}}(\cdot )$. Then for this choice of parameters the Wolff potential, which we denote by ${\mathcal {W}}^{\mu }$ for short, takes the following form

$$\begin{aligned} {\mathcal {W}}^\mu _{1+\frac{2}{n}, \frac{n}{2}}(x):= {\mathcal {W}}^\mu (x)= \int _0^1 \left( \frac{\mu \left( B(x,r)\right) }{r^{\frac{n-2}{2}}} \right) ^{\frac{2}{n-2}}\frac{dr}{r}. \end{aligned}$$

The corresponding capacity on ${\mathbb {S}}^n$ is defined accordingly. By abuse of notation, we continue to use the same notation $\textrm{cap}(\cdot )$. The following characterization of vanishing capacity compacts is hard to find in the literature, so we give a proof for the reader’s convenience.

Proposition 2.1

Let $K\subset {\mathbb {R}}^n$ be a compact set then $\textrm{cap}(K)=0$ if and only if there exists a Radon measure $\mu$, $\Vert \mu \Vert =1$, such that ${ \mathrm supp}\mu \subset K$ and

$$\begin{aligned} {\mathcal {W}}^\mu ( x)=+\infty \quad \mathrm{for \quad all \quad } x\in K. \end{aligned}$$

(2.7)

Proof

Suppose there is a measure $\mu$ such that ${ \mathrm supp}\mu \subset K, \Vert \mu \Vert =1$ and ${\mathcal {W}}(\mu , x)=+\infty$ for all $x\in K$. From (2.6) it follows that $V_{\alpha , q}^\mu (x)=\infty$ for all $x\in K$. Note that $f=(G_\alpha *\mu )^{p-1}\in L^q({\mathbb {R}}^n)$, see [8] page 163. By Theorem 1.4 [15] $\textrm{cap}(E)=0$ if and only if there is a nonnegative $f\in L^q$ such that $G_\alpha *f=\infty$ for all $x\in E$. Since by definition $G_\alpha *f={\mathcal {V}}_{\alpha , q}^\mu$, and thanks to (2.6), it follows that $\textrm{cap}(E)=0$.

Now we prove the converse statement. Suppose $\textrm{cap}(E)=0$. By Proposition 4 [8] we have

$$\begin{aligned} 0=\inf _\mu \left\{ \int {\mathcal {W}}^\mu d\mu : \ {\mathcal {W}}^\mu \ge 1 \ \text {on}\ E\right\} . \end{aligned}$$

Let $\mu _j$ be a minimizing sequence such that $\int {\mathcal {W}}^{\mu _j}d\mu _j\rightarrow 0$ and $\Vert \mu _j\Vert >0$. Such sequence exists because ${\mathcal {W}}^\mu \ge 1$ on E, see (2.5). This inequality also implies that the restricted measures $\Vert \mu _j|_E\Vert \le \int {\mathcal {W}}^{\mu _j}d\mu _j\rightarrow 0$ have vanishing mass in the limit. Introduce the unit mass measures ${\widehat{\mu }}_j=\mu _j|_{E}/\mu _j(E)$. Then from the scaling property of $W^\mu$ we get that

$$\begin{aligned} W^{{\widehat{\mu }}_j}_{\alpha , q}\ge \frac{1}{(\mu _j(E))^{p-1}}\rightarrow \infty \quad \text{ on }\ E. \end{aligned}$$

Then the existence follows from a customary compactness argument for ${\widehat{\mu }}_j$ and the semicontinuity of the potential ${\mathcal {W}}^{\mu }$. $\square$

2.3 Divergent curves and Hopf–Rinow theorem

In order to characterize the completeness of the metric by Wolff’s potential we state a version of the Hopf-Rinow theorem formulated in terms of divergent curves.

Definition 2.2

A divergent curve in a Riemannian manifold M is a differentiable mapping $c: [0,T)\rightarrow M$ such that for any compact set $K\subset M$ there exists $t_0 \in (0, T)$ with $c(t) \not \in K$ for all $t > t_0$.

In other words, if c is a divergent curve then it "escapes" every compact set in M. Define the length of a divergent curve by

$$\begin{aligned} L_g(c)=\lim _{t\rightarrow T}\int _0^t|c'(\tau )|d\tau . \end{aligned}$$

(2.8)

Then we have the following version of the Hopf-Rinow theorem.

Theorem 2.3

Let M be a noncompact Riemannian manifold. Then M is complete if and only if the length of any divergent curve is unbounded.

Proof

Suppose M is complete then the classical Hopf-Rinow theorem holds. Consequently, the closed ball $\overline{B(0, N)}$ is compact for every $N=1, 2, \dots$. Let $c:[0, T)\rightarrow M$ be a divergent curve, then for every N there is $t_N\in (0, T)$ such that $c(t_N)\not \in \overline{B(0, N})$. Therefore, from (2.8) we get that

$$\begin{aligned} L_g(c|_{[0, t_N)})\ge N\rightarrow \infty , \end{aligned}$$

and, hence c is a divergent curve.

Now suppose that every divergent curve has infinite length. We want to show that then this implies that M is complete. If M is not complete then there is a geodesic $c: [0, T)\rightarrow M$ such that c cannot be extended further than T. Observe that c has constant speed since

$$\begin{aligned} c'g(c', c')=2g(\nabla _{c'}c', c')=0, \end{aligned}$$

where $\nabla$ is the Riemannian connection on M. Therefore $L_g(c)\le \theta T$ for some constant $\theta$. To finish the proof it is enough to show that c is a divergent curve, hence c cannot have finite length. Thus, suppose that c is contained in some compact K. Let us take $t_k\in (0, T)$ so that $p_k:=c(t_k)\in K$ and $\lim _{k\rightarrow \infty }t_k=T$. We can extract a subsequence so that $c(t_{k_m})\rightarrow p$ for some $p\in K\subset M$. Let W be a totally normal neighborhood of p, that is, W is a normal neighborhood of all of its points. The existence of W is well known, see [5], page 72. Consequently, if $\epsilon >0$ is small so that $c(T-\epsilon )\in W$, then the geodesic joining $c(T-\epsilon )$ and p can be continued further than p, which is a contradiction. Thus c escapes any compact and hence it is a divergent curve. $\square$

3 Proof of $(i)\Rightarrow (ii)$: existence of u implies $cap(K)=0$

We can use the stereographic projection and, thanks to the conformal homothety on ${\mathbb {R}}^n$, without loss of generality assume that the north pole $N\in \Omega$ such that $\sigma (K)\subset B(0, 1/2)$, and there is $u:{\mathbb {R}}^n\setminus \sigma (K)\rightarrow {\mathbb {R}}$ such that $g=u^{\frac{4}{n-2}} \overset{{\,}_\circ }{g}$ is complete, where

$$\begin{aligned} \Delta u =0, \quad \text {in}\ \ \sigma (\Omega )=\{u>0\}. \end{aligned}$$

Recall that for a point $x=(x_1, \dots , x_d)$ on the unit sphere in ${\mathbb {R}}^d$ its stereographic projection is defined as $y=\frac{(x_1, \dots , x_{d-1})}{1-x_d}\in {\mathbb {R}}^{d-1}$. We claim that there exists a Radon measure $\mu$, with ${ \mathrm supp}\mu \subset K$, such that the following representation of u is true

$$\begin{aligned} u(x)=\int _{{\mathbb {R}}^n} k(x,y)d\mu (y)\quad \forall x\in B(0,3)\setminus K, \end{aligned}$$

(3.1)

where k(x, y) is the Martin kernel (see [1] Theorem 8.4.1 or Chapter 12 [9] p 251) and k is locally integrable in ${\mathbb {R}}^n\times {\mathbb {R}}^n$. Moreover, there exists a universal constant $C_M$ such that

$$\begin{aligned} 0\le k(x,y)\le \frac{C_M}{|x-y|^{n-2}}, \end{aligned}$$

(3.2)

see [9], Chapter 12.

Suppose $\textrm{cap}(K)>0.$ By Proposition 2.1 the Wolff potential of $\mu$ must be finite at some point $x_0\in K$. Without loss of generality we assume that $x_0=0$ and $\mu$ is a probability measure such that

$$\begin{aligned} {\mathcal {W}}^\mu (0)<+\infty , \quad 0\in K. \end{aligned}$$

We first establish a useful estimate.

Lemma 3.1

Let $u>0$ be as above and ${ \mathrm supp}\mu \subset B(0, 1/2)$. Then there is a constant $C>0$ such that

$$\begin{aligned} \int _{B(0, 1)} \frac{\left( u(x)\right) ^{\frac{2}{n-2}}}{|x|^{n-1}} \, dx \le C {\mathcal {W}}^\mu ( 0). \end{aligned}$$

(3.3)

Proof

Denote $D_m=B(0, \rho _{m}){\setminus } B(0, \rho _{m+1}), \rho _m=2^{-m}$. We have

$$\begin{aligned} \int _{B(0,2)} \frac{u(x)^{\frac{2}{n-2}}}{|x|^{n-1}} \,dx= & {} \sum _{m=0}^\infty \int _{D_m}\frac{u(x)^{\frac{2}{n-2}}}{|x|^{n-1}}\,dx\nonumber \\{} & {} + \int _{B(0,2)\setminus B(0,1)}\frac{u(x)^{\frac{2}{n-2}}}{|x|^{n-1}}\,dx\nonumber \\= & {} I_1+I_2. \end{aligned}$$

(3.4)

Since $u=k*\mu$ then from (3.2) we see that

$$\begin{aligned} u\le C \quad \textrm{in}\quad B(0, 2)\setminus B(0,1). \end{aligned}$$

Hence

$$\begin{aligned} I_2 \le C. \end{aligned}$$

(3.5)

As for $I_1$ in (3.4), we have

$$\begin{aligned}{} & {} \int _{D_m}\dfrac{1}{\left| x\right| ^{n-1}}\left( \int _{B(0, {2})}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}\right) ^{\frac{2}{n-2}}dx\\{} & {} =\int _{D_m}\dfrac{1}{\left| x\right| ^{n-1}} \left[ \int \limits _{B\left( 0,\rho _{m+2}\right) }\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}} + \int \limits _{B\left( 0,\rho _{m-2}\right) \setminus B(0, \rho _{m+2})}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}} \right. \\{} & {} \left. + \int \limits _{B(0, 1)\setminus B\left( 0,\rho _{m-2}\right) }\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}} \right] ^{\frac{2}{n-2}}dx. \end{aligned}$$

For $x\in D_{m}$ we have

$$\begin{aligned} \int _{B\left( 0,\rho _{m+2}\right) }\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}\le \dfrac{1}{\rho ^{n-2}_{m+2}}\mu \left( B\left( 0,\rho _{m+1}\right) \right) , \end{aligned}$$

and

$$\begin{aligned} \int _{B(0, 1)\backslash B\left( 0,\rho _{m-2}\right) }\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}=\sum ^{m-3}_{k=0}\int _{D_k}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}. \end{aligned}$$

Noting that

$$\begin{aligned} \dfrac{1}{2^{k+1}}-\dfrac{1}{2^{m}}=\dfrac{1}{2^{k+1}}\left( 1-\dfrac{1}{2^{m-k-1}}\right) \ge \dfrac{3}{4}\dfrac{1}{2^{k+1}}, \end{aligned}$$

we get

$$\begin{aligned} \sum ^{m-3}_{k=0}\int _{D_k}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}\le \left( \dfrac{8}{3}\right) ^{n-2}\sum ^{m-3}_{k=0}\dfrac{\mu \left( B\left( 0,\rho _{k}\right) \right) }{\rho ^{n-2}_{k}}. \end{aligned}$$

Combining, we obtain the estimate

$$\begin{aligned} I_1\le & {} C \left( n\right) \sum ^{\infty }_{m=0} \Bigg \{ \int \limits _{D_m}\dfrac{1}{\left| x\right| ^{n-1}}\left( \sum ^{m-3}_{k=0}\dfrac{\mu \left( B\left( 0,\rho _{k}\right) \right) }{\rho ^{n-2}_{k}}\right) ^{\frac{2}{n-2}} + \left( \frac{\mu \left( B\left( 0,\rho _{m+1}\right) \right) }{\rho ^{n-2}_{m+2}} \right) ^{\frac{2}{n-2}}\\ {}{} & {} + \left( \int \limits _{B\left( 0,\rho _{m-2}\right) \setminus B(0, \rho _{m+2})}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}} \right) ^{\frac{2}{n-2}} \Bigg \}\\ {}\le & {} C \left( n\right) \sum ^{\infty }_{m=0} \left\{ \rho _m\left( \sum ^{m}_{k=0}\dfrac{\mu \left( B\left( 0,\rho _{k}\right) \right) }{\rho ^{n-2}_{k}}\right) ^{\frac{2}{n-2}}\right. \\{} & {} \left. +\frac{1}{\rho _m^{n-1}} \int _{D_m}\left( \int \limits _{B\left( 0,\rho _{m-2}\right) \setminus B(0, \rho _{m+2})}\dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}\right) ^{\frac{2}{n-2}}dx \right\} \\= & {} C( I_3+I_4). \end{aligned}$$

For $n=3$ we take a sequence of smooth functions $f_i$ weakly converging to $\mu$ in $\widetilde{D_m}: =B\left( 0,\rho _{m-2}\right) {\setminus } B(0, \rho _{m+2})$ (see Lemma 0.2 [12]) then applying lemma 7.12 from [7] to

$$\begin{aligned} V_sf_i\left( x\right) =\int _{\widetilde{D}_m}\left| x-y\right| ^{n\left( s-1\right) }f_i(y)dy \end{aligned}$$

with $q=2,p=1, \delta =1-\dfrac{1}{q}=\dfrac{1}{2}$, and $s=\delta +\dfrac{1}{6}=\frac{2}{3}$, we get

$$\begin{aligned} \int _{\widetilde{D}_m}\left| V_{2/3}f_{i}\right| ^{2}\le C\left( \mathrm{{ Vol}}_{ g_E} (\widetilde{D}_{m})\right) ^{\frac{1}{3}}\left( \int _{\widetilde{D}_m}f_{i}\right) ^{2}. \end{aligned}$$

After letting $i\rightarrow \infty$ this yields

$$\begin{aligned} I_4\le \sum ^{\infty }_{m=0}\dfrac{1}{\rho _{m}}\left( \mu \left( B\left( 0,\rho _{m}\right) \right) \right) ^{2}\le C\int ^{1}_{0}\dfrac{\left( \mu \left( B\left( 0,t\right) \right) \right) ^{2}}{t^{2}}dt. \end{aligned}$$

Moreover, denoting $m(t)=\mu (B(0, t))$ and using integration by parts together with Cauchy-Schwarz inequality we get

$$\begin{aligned} \int ^{1}_{0}\left( \int ^{1}_{t}\dfrac{m\left( \tau \right) }{\tau ^{2}}d\tau \right) ^{2}dt= & {} 2\int ^{1}_{0}\dfrac{m\left( t\right) }{t}\left( \int ^{1}_{t}\dfrac{m\left( \tau \right) }{\tau ^{2}}d\tau \right) dt\\\le & {} 2\left[ \int ^{1}_{0}\left( \dfrac{m\left( t\right) }{t}\right) ^2dt\int _0^1\left( \int ^{1}_{t}\dfrac{m\left( \tau \right) }{\tau ^{2}}d\tau \right) ^2 dt\right] ^{\frac{1}{2}}, \end{aligned}$$

implying that

$$\begin{aligned} \int ^{1}_{0}\left( \int ^{1}_{t}\dfrac{m\left( \tau \right) }{\tau ^{2}}d\tau \right) ^{2}dt \le 4\int ^{1}_{0}\left( \dfrac{m\left( t\right) }{t}\right) ^2dt. \end{aligned}$$

Hence

$$\begin{aligned} I_3\le C\int ^{1}_{0}\left( \int ^{1}_{t}\dfrac{m\left( \tau \right) }{\tau ^{2}}d\tau \right) ^{2}dt \le 4C\int ^{1}_{0}\left( \dfrac{m\left( t\right) }{t}\right) ^2dt=4C{\mathcal {W}}^\mu (0). \end{aligned}$$

If $n=4$ we have

$$\begin{aligned} \int _{\widetilde{D_m}}d\mu \left( y\right) \int _{D_m}\dfrac{dx}{\left| x-y\right| ^{2}}\le \mu \left( B\left( 0,\rho _{m}\right) \right) \rho ^{2}_{m}, \end{aligned}$$

and then from Fubini’s theorem we get, as above, the bound

$$\begin{aligned} I_4\le \sum ^{\infty }_{m=0}\dfrac{1}{\rho _{m}}\mu \left( B\left( 0,\rho _{m}\right) \right) \le C\int ^{1}_{0}\dfrac{\mu \left( B\left( 0,t\right) \right) }{t^{2}}dt.\ \end{aligned}$$

The estimate for $I_3$ follows from integration by parts.

Finally, let us consider the case $n\ge 5$. We have

$$\begin{aligned} \int _{D_m}\left( \int _{\widetilde{D}_m} \dfrac{d\mu \left( y\right) }{\left| x-y\right| ^{n-2}}\right) ^{\frac{2}{n-2}}\le & {} \left( \int _{D_m}\int _{\widetilde{D}_m} \dfrac{d\mu }{\left| x-y\right| ^{n-2}}\right) ^ \frac{2}{n-2}\left( \mathrm{{ Vol}}_{ g_E}( D_{m})\right) ^{1-\frac{2}{n-2}}\\\le & {} C\left( \rho ^{2}_{m}\mu (B(0, \rho _m)) \right) ^{\frac{2}{n-2}}\rho ^{n\left( 1-\frac{2}{n-2}\right) }_{m}\\= & {} C\left( \mu (B(0, \rho _m)) \right) ^{\frac{2}{n-2}}\rho ^{n-2}_{m}. \end{aligned}$$

Thus

$$\begin{aligned} I_4\le \sum ^{\infty }_{m=0}\dfrac{1}{\rho _{m}}\left( \mu \left( B\left( 0,\rho _{m}\right) \right) \right) ^{\frac{2}{n-2}}\le C\int ^{1}_{0}\dfrac{\left( \mu \left( B\left( 0,t\right) \right) \right) ^{\frac{2}{n-2}}}{t^{2}}dt. \end{aligned}$$

As for $I_3$, one can easily see that

$$\begin{aligned} \left( \sum ^{m}_{k=0}\dfrac{\mu \left( B\left( 0,\rho _{k}\right) \right) }{\rho ^{n-2}_{k}}\right) ^{\frac{2}{n-2}} \le C\sum ^{m}_{k=0}\left( \dfrac{\mu \left( B\left( 0,\rho _{k}\right) \right) }{\rho ^{n-2}_{k}}\right) ^{\frac{2}{n-2}}, \end{aligned}$$

and consequently after integration by parts we get

$$\begin{aligned} I_3\le C \int ^{1}_{0} \int ^{1}_{t}\left( \dfrac{m\left( \tau \right) }{\tau ^{n-2}} \right) ^{\frac{2}{n-2}}\frac{d\tau }{\tau }{dt} \le C \int ^{1}_{0} \left( \dfrac{m\left( t \right) }{t ^{n-2}} \right) ^{\frac{2}{n-2}}{dt}. \end{aligned}$$

The proof of lemma is complete. $\square$

Proof of (i)$\Rightarrow$(ii) in Theorem 1.1.

We claim that there exists a smooth curve $c:[0,1]\rightarrow B(0, 2){\setminus } K$ such that

$$\begin{aligned}{} & {} L_g(c)=\int _0^1(u(c(t)))^{\frac{2}{n-2}}\left| {c'}(t)\right| dt<+\infty , \nonumber \\ {}{} & {} \mathrm{and\ }\gamma (t)\rightarrow 0\ \textrm{as}\ t\rightarrow 1. \end{aligned}$$

(3.6)

Observe that (3.6) is impossible if $u^{4/(n-2)} g_E$ is complete, thanks to Theorem 2.3. Hence to finish the proof we have to establish (3.6). So it’s left to show the existence of such curve. Take $\omega \in \partial B$ and let $\ell (\omega )$ be the interval

$$\begin{aligned} \ell (\omega )= \left\{ x\in \textbf{R}^n:\quad x=s\omega , \quad 0<s\le 1 \right\} , \end{aligned}$$

where $\pi : x\rightarrow \frac{x}{|x|}$ is the projection on $\partial B$. Put

$$\begin{aligned} \Xi = \partial B \setminus \pi (K \setminus \{0\}), \end{aligned}$$

and observe that

$$\begin{aligned} \pi ^{-1}( \Xi ) \subset B\setminus K. \end{aligned}$$

We claim that ${\mathcal {H}}^{n-1}(\Xi )>0$, otherwise this means that ${\mathcal {H}}^{n-1}(\partial \Omega )>0$ But this will be in contradiction with the Schoen-Yau estimate of the Hausdorff dimension of $\partial \Omega$ which can be at most $\frac{n}{2}$, see [18], Theorem 2.7.

Switching to polar coordinates $(r,\omega )$, $r>0$, $\omega \in \partial B$, we get from (3.3) that

$$\begin{aligned} +\infty> & {} \int _{\pi ^{-1}( \Xi ) } u(x)^{\frac{2}{n-2}}\,\frac{1}{|x|^{n-1}}\, dx\\= & {} \int _{\Xi }\int _0^1 u(x(r,\omega ))^{\frac{2}{n-2}}\,\frac{1}{r^{n-1}}\,r^{n-1}\, dr\, d{\mathcal {H}}^{n-1}(\omega )\\= & {} \int _{\Xi }\left( \int _{\ell (\omega )}u^{\frac{2}{n-2}}\, ds\right) \, d{\mathcal {H}}^{n-1}(\omega ). \end{aligned}$$

Consequently

$$\begin{aligned} \int _{\ell (\omega _0)}u^{\frac{2}{n-2}}\, ds <+\infty \quad \mathrm{for\quad some\quad } \omega _0 \in \Xi . \end{aligned}$$

By our definitions

$$\begin{aligned} \ell (\omega _0)\cap K =\emptyset , \end{aligned}$$

and we conclude that (3.6) holds for the curve $\gamma =\ell (\omega _0)$. This finishes the proof.

4 Proof of $(ii)\Rightarrow (i)$: $\textrm{cap}(K)=0$ implies existence of metric

Proof of (ii)$\Rightarrow$(i) in Theorem 1.1.

In what follows we assume, without loss of generality, that the north pole $N\in \Omega$. Since $\sigma (K)\subset {\mathbb {R}}^n$ is the image of K under stereographic projection then it is compact such that

$$\begin{aligned} \textrm{cap}(\sigma (K))=0. \end{aligned}$$

From Proposition 2.1 it follows that there is a probability measure $\mu$, such that ${ \mathrm supp}\mu \subset \sigma (K)$ and

$$\begin{aligned} {\mathcal {W}}^\mu ( x)=+\infty \quad \mathrm{for \quad all \quad }x\in \sigma (K). \end{aligned}$$

(4.1)

The convolution

$$\begin{aligned} u(x) = \int \frac{d\mu (y)}{|x-y|^{n-2}} \end{aligned}$$

solves

$$\begin{aligned} \Delta u =0, \quad u>0\quad \textrm{in}\quad {\mathbb {R}}^n\setminus \sigma (K). \end{aligned}$$

To finish the proof we have to show that

$$\begin{aligned} \sigma ^* \left( u^{\frac{4}{n-2}} g_E\right) \ \mathrm{is \ a \ complete\ metric}\ \textrm{in}\ \Omega . \end{aligned}$$

(4.2)

Since u is harmonic in $\{u>0\}$, and $u(\sigma (N))>0$ then $\sigma ^*(u^{4/(n-2)}) g_E$ gives a metric on ${\mathbb {S}}^n$ which is smooth at N. To check (4.2), we use a version of the Hopf-Rinow theorem formulated in terms of divergent curves, see Theorem 2.3. Let us take a divergent curve c in $\Omega$, and denote $\widetilde{c}: [0,+\infty )\rightarrow {\mathbb {R}}^n{\setminus } \sigma (K)$ its stereographic projection. Clearly c is divergent curve in ${\mathbb {R}}^n\setminus \sigma (K)$. Since by assumption $N\in \Omega$ then $\widetilde{c}$ is contained in some ball in ${\mathbb {R}}^n$. Recall the arc length formula

$$\begin{aligned} L_{g}(c)=\int _0^\infty \sqrt{g(c'(t), c'(t))}dt= \int _0^\infty u(\widetilde{c}(t))^{\frac{2}{n-2}}\,\left| {\widetilde{c}\,{}'}(t)\right| \,dt, \end{aligned}$$

(4.3)

where $g=u^{\frac{4}{n-2}} g_E.$

By assumption c (and hence $\widetilde{c}$) is a divergent curve, therefore there exists $x_0\in K$ such that

$$\begin{aligned} { \mathrm dist}_{ \overset{{\,}_\circ }{g}}(x_0, c(t_k)) \rightarrow 0 \quad \mathrm{for\quad a \quad sequence} \quad \{t_k\}, \quad k\rightarrow +\infty . \end{aligned}$$

(4.4)

For $m\in {\mathbb {N}}$, we let $\gamma _m=D_m\cap \widetilde{c}$, where

$$\begin{aligned} D_m=\left\{ x\in {{\mathbb {R}}}^n:\frac{1}{2^m}<|x-\widetilde{x}_0|<\frac{1}{2^{m-1}}\right\} . \end{aligned}$$

If $m\ge m_0$, for sufficiently large $m_0$, it follows from the smoothness of c that $\gamma _m$ is at most a countable union of open smooth curves. Moreover, from (4.4) we see that $\gamma _m\ne \emptyset$, and

$$\begin{aligned} L_{ g_E}(\gamma _m)\ge \frac{1}{2^{m-1}}-\frac{1}{2^{m}}=\frac{1}{2^{m}} \quad \mathrm{for \quad all } \quad m \ge m_0. \end{aligned}$$

For $y\in B(\widetilde{x}_0, 2^{-(k+2)})$ and $x\in D_k$ we have $|x-y|\le \frac{1}{2^k}+\frac{1}{2^{k+2}}=\frac{5}{2^{k+2}}$. Therefore

$$\begin{aligned} u(x)\ge & {} \int _{B(\widetilde{x}_0, \rho _{k+2})}\frac{d\mu (y)}{|x-y|^{n-2}} \nonumber \\\ge & {} \frac{1}{5^{n-2}}\frac{\mu \left( B(\widetilde{x}_0, \rho _{k+2})\right) }{\rho _{k+2}^{n-2}} \quad \mathrm{for\ all}\ x\in D_k, \end{aligned}$$

(4.5)

where we set $\rho _i=2^{-i}.$ Let $I_k\subset (0,+\infty )$ denote the open set such that

$$\begin{aligned} \widetilde{c}: I_k\rightarrow \sigma (\Omega )\cap D_k. \end{aligned}$$

Then we derive that

$$\begin{aligned} L_g(c)= & {} \int _0^\infty u(\widetilde{c}(t))^{\frac{2}{n-2}}\,\left| {\widetilde{c}\,{}'}(t)\right| \,dt\nonumber \\ {}= & {} \sum _{k=0}^\infty \int _{I_k} u(\widetilde{c}(t))^{\frac{2}{n-2}}\,\left| {\widetilde{c}\,{}'}(t)\right| \,dt\nonumber \\ {}\ge & {} \sum _{k=0}^\infty \left( \inf _{D_k} u \right) ^{\frac{2}{n-2}}L_{ g_E}(\gamma _k)\nonumber \\ {}\ge & {} \frac{2}{5^{n-2}}\sum _{k=0}^\infty \left( \frac{\mu \left( B(\widetilde{x}_0, \rho _{k+2})\right) }{\rho _{k+2}^{n-2}}\right) ^{\frac{2}{n-2}} \rho _k, \end{aligned}$$

(4.6)

where the last line follows from (4.5). Recalling the definition of ${\mathcal {W}}^\mu$ we see that

$$\begin{aligned} {\mathcal {W}}^\mu (x)= & {} \int _0^1 \left( \frac{\mu \left( B(x,r)\right) }{r^{n-2}}\right) ^{\frac{2}{n-2}}dr =\sum _{k=0}^\infty \int _{2^{-(k+1)}}^{2^{-k}} \left( \frac{\mu \left( B(x,r)\right) }{r^{n-2}} \right) ^{\frac{2}{n-2}}dr\nonumber \\ {}\le & {} 4\sum _{k=0}^\infty \left( \frac{\mu \left( B(x, \rho _k)\right) }{\rho _k^{n-2}} \right) ^{\frac{2}{n-2}}\rho _k. \end{aligned}$$

(4.7)

Comparing the inequalities for $L_g(c)$ and ${\mathcal {W}}^\mu$ and recalling (4.1)

$$\begin{aligned} {\mathcal {W}}^\mu (\widetilde{x}_0)=+\infty , \end{aligned}$$

we obtain that ${{\widetilde{c}}}$ (and hence c) has infinite length, and thus Theorem 2.3 implies that (4.2) is true. $\square$