Conformal scalar curvature rigidity on Riemannian manifolds

TL;DR

This paper investigates conformal scalar curvature rigidity on Riemannian manifolds, proving that under certain boundary conditions, the conformal metric must coincide with the original metric on specific domains, extending previous results.

Contribution

It extends existing scalar curvature rigidity results to more general Riemannian manifolds with boundary conditions, showing conformal metrics are uniquely determined on certain domains.

Findings

Conformal scalar curvature rigidity holds on some smooth domains.

The original metric is uniquely determined under boundary conditions.

Extension of previous rigidity results to broader Riemannian settings.

Abstract

Let $(M, \bar g)$ be an $n$-dimensional complete Riemannian manifold. In this paper, we considers the following conformal scalar curvature rigidity problem: Given a compact smooth domain $\Omega$ with $\partial \Omega$, can one find a conformal metric $g$ whose scalar curvature $R[g]\ge R[\bar g]$ on $\Omega$ and the mean curvature $H[g] \ge H[ \bar g]$ on $\partial \Omega$ with $\bar g = g$ on $\partial \Omega$? We prove that $\bar g = g$ on some smooth domains in a general Riemannian manifold, which is an extension of the previous results given by Qing and Yuan, and Hang and Wang.

Equations

$$\begin{array}{rl}g& =e^{2u}\bar{g},R[g]=e^{-2u}\left(R[\bar{g}]-2\Delta u\right)\text{ and}\\ H[g]& =e^{-u}[H[\bar{g}]+{\partial }_{\nu }u]\text{ when}n=2,\end{array}$$

$$\begin{split}g&=u^{\frac{4}{n-2}}\bar{g},R[g]=u^{-\frac{n+2}{n-2}}\left(R[\bar{g}]u-\frac{4(n-1)}{n-2}\Delta u\right)\text{ and}\\ H[g]&=u^{-\frac{n}{n-2}}\Big{[}H[\bar{g}]+\frac{2}{n-2}\partial_{\nu}u\Big{]}\text{ when}\,n\geq 3.\end{split}$$

$$\begin{array}{r}R[\bar{g}]e^{2u}\le \left(R[\bar{g}]-2\Delta u\right)\text{ when}n=2\text{ and}\\ R[\bar{g}]u^{\frac{n+2}{n-2}}\le \left(R[\bar{g}]u-\frac{4(n-1)}{n-2}\Delta u\right)\text{ when}n\ge 3.\end{array}$$

$$\begin{array}{r}{\partial }_{\nu }u\ge 0.\end{array}$$

$$-\Delta f-\frac{R[\bar{g}]}{n-1}f=0\text{ and }f>0\text{ in Ω+}$$

$$\nabla f\ne 0\text{ at}\partial {\Omega }^{+}=\{x\in M^n|f(x)=0\},$$

$$Q(M,g)\equiv \underset{0\ne u\in C_0^{\infty }(M)}{\inf }\frac{{\int }_M|\nabla u{|}^2+\frac{(n-2)}{4(n-1)}{R}_g{u}^{2}d{V}_g}{{\left({\int }_M|u{|}^{2n/(n-2)}d{V}_g\right)}^{(n-2)/n}}$$

$$Q(\Omega ,g)\equiv \underset{0\ne u\in C_0^{\infty }(\Omega )}{\inf }\frac{{\int }_M|\nabla u{|}^2+\frac{(n-2)}{4(n-1)}{R}_g{u}^{2}d{V}_g}{{\left({\int }_M|u{|}^{2n/(n-2)}d{V}_g\right)}^{(n-2)/n}}.$$

$$A(x)=\frac{\frac{n-2}{4(n-1)}R[\bar{g}]u(x)\left(u(x{)}^{\frac{4}{n-2}}-1\right)}{u(x)-1}.$$

$$-\Delta v-A(x)v\ge 0$$

$${\int }_{{\Omega }_1}|\nabla v{|}^2-A(x)v^{2}d{V}_{\bar{g}}\le 0.$$

$$\begin{split}Q(\Omega,\bar{g}){\left(\int_{\Omega}|v|^{\frac{2n}{n-2}}dV_{\bar{g}}\right)^{\frac{n-2}{n}}}&\leq\int_{\Omega_{1}}|\nabla v|^{2}+{{(n-2)}\over{4(n-1)}}R_{\bar{g}}v^{2}dV_{\bar{g}}\\ &\leq\int_{\Omega_{1}}\Big{(}A(x)+{{(n-2)}\over{4(n-1)}}R_{\bar{g}}\Big{)}v^{2}dV_{\bar{g}}\\ &\leq\Big{(}{\frac{1}{n-1}+\frac{(n-2)}{4(n-1)}}\Big{)}\int_{\Omega_{1}}R^{+}_{\bar{g}}v^{2}dV_{\bar{g}}\\ &\leq{\frac{(n+2)}{4(n-1)}}\int_{\Omega_{1}}|R^{+}|_{\bar{g}}v^{2}dV_{\bar{g}}\\ &\leq{\frac{(n+2)}{4(n-1)}}\Big{[}\int_{\Omega_{1}}|R^{+}|_{\bar{g}}^{\frac{n}{2}}dV_{\bar{g}}\Big{]}^{\frac{2}{n}}\Big{[}\int_{\Omega_{1}}|v|^{\frac{2n}{n-2}}dV_{\bar{g}}\Big{]}^{\frac{n-2}{n}},\end{split}$$

$$R(\bar{g},D)=\underset{\{p\in D\}}{\sup }R[\bar{g}](p)$$

$${\lambda }_1(D)=\underset{u\in H}{\inf }\frac{{\int }_D|\nabla u{|}^{2}d{V}_{\bar{g}}}{{\int }_D{u}^{2}d{V}_{\bar{g}}}$$

$$\begin{array}{rl}0\ge & \Delta v+A(x)v\\ =& △(u_1\beta )+A(x)u_1\beta \\ =& u_1△\beta +\Delta u_1\beta +\nabla u_1\cdot \nabla \beta +A(x)u_1\beta \\ =& u_1△\beta -\lambda u_1\beta +\nabla u_1\cdot \nabla \beta +A(x)u_1\beta .\end{array}$$

$$\begin{array}{rl}0\ge & △\beta -\lambda \beta +\frac{\nabla u_1}{u_1}\cdot \nabla \beta +A(x)\beta \\ 0\ge & △\beta +(A(x)-\lambda )\beta +\frac{\nabla u_1}{u_1}\cdot \nabla \beta .\end{array}$$

Full text

1. Introduction

Let $(M,\bar{g})$ be an $n$-dimensional complete Riemannian manifold and $\Omega$ be a smooth domain in $(M,g)$ with smooth boundary $\partial\Omega$. Denote $R[\bar{g}]$ by the scalar curvature of $\bar{g}$ and $H[\bar{g}]$ the mean curvature of $\partial\Omega$. In this paper, we considers the following problem: Given a compact smooth domain $\Omega$ with $\partial\Omega$, can one find a conformal metric $g$ whose scalar curvature $R[g]\geq R[\bar{g}]$ on $\Omega$ and the mean curvature $H[g]\geq H[\bar{g}]$ on $\partial\Omega$ with $\bar{g}=g$ on $\partial\Omega$?

For the conformal metric $g$ of the given metric $(M,\bar{g})$, scalar curvature $R[g]$ and mean curvature $H[g]$ on the boundary of the domain change in the following ways:

[TABLE]

[TABLE]

Therefore the given condition $R[g]\geq R[\bar{g}]$ on $\Omega$ is equivalent to

[TABLE]

The condition $H[g]\geq H[\bar{g}]$ with $\bar{g}=g$ on $\partial\Omega$ is equivalent to

[TABLE]

This problem is a conformal version of Min-Oo’s conjecture. Uniqueness and non-uniqueness of conformal metric with prescribed scalar curvature on $\Omega$ with the mean curvature condition on $\partial\Omega$ was studied by Escobar. He proved that on the annulus $A_{a,b}=\{x\in R^{n}|0<a<|x|<b\}$ with the Euclidean metric $\delta_{ij}$ admits a conformal metrics of the form $u^{4/(n-2)}(|x|)\delta_{ij}$ that there exist several metrics with the same constant scalar curvature and the same constant mean curvature on the boundary if $\frac{b}{a}$ is big enough [3]. Therefore we can not expect to get a conformal scalar curvature rigidity to our question in general. However, if $\frac{b}{a}$ is small and $\bar{g}=g$ on $|x|=a$ and $|x|=b$, our result implies that $\bar{g}=g$ on $A_{a,b}$. Hang and Wang obtained the conformal deformations rigidity of metrics on the hemisphere $D=S^{+}$ in the standard sphere $(S^{n},\bar{g})$ where $\bar{g}=g_{0}$. They proved that (3) and $\bar{g}=g$ on $\partial D$ imply $\bar{g}=g$ on $D$ [5]. This result was recently generalized to the domains in the vacuum static spaces by Qing and Yuan.

Theorem 1**.**

[7, Theorem 5.1]** Let $(M^{n},\bar{g},f)$ be a complete $n$-dimensional static space with $R_{\bar{g}}>0$ ($n\geq 2$). Assume $\Omega^{+}\equiv\{p\in M|f(p)>0\}$ is a pre-compact subset in $M$. Then, if a metric $g\in[\bar{g}]$ on $M$ satisfies that

• •

$R[g]\geq R[\bar{g}]$* in $\Omega^{+}$,*

• •

$g$* and $\bar{g}$ induced the same metric on $\partial\Omega^{+}$, and*

• •

$H[g]=H[\bar{g}]$* on $\partial\Omega^{+}$,*

then $g=\bar{g}$.

For the proof of Theorem 1, they used the existence of the lapse functions $f$ with the following properties:

[TABLE]

and

[TABLE]

where $\Omega^{+}$ is the maximal subset where the conformal rigidity holds. The existence of lapse function comes from the vacuum static space $(M,g)$ and (6) holds for $f$ (see: [4, Theorem 1]). In this paper, we extend the previous conformal scalar curvature rigidity results to the domains in a general Riemannian manifold with the conformal invariant.

2. Conformal Rigidity of Scalar curvature

Let $(M,g)$ be a complete Riemannian manifold of dimension $n\geq 3$ with scalar curvature $R[g]$. The Sobolev constant $Q(M,g)$ of $(M,g)$ and $Q(\Omega,g)$ of a smooth domain $\Omega\subset(M,g)$ are defined by

[TABLE]

and

[TABLE]

Note that $Q(M,g)$ and $Q(\Omega,g)$ are conformal invariant and $Q(\Omega,g)\geq Q(M,g)$. There are domains in a complete Riemannian manifolds with positive Sobolev constant. For example, any simply connected domain in a complete locally conformally flat manifold has positive Sobolev constant [9]. It is known that for any smooth domain $\Omega\subset(M,g)$, $Q(\Omega,g)\leq Q(S^{n},g_{0})$ where $(S^{n},g_{0})$ is the standard sphere (see [1]). Using $Q(\Omega,g)$ and $Q(M,g)$, we obtain conformal rigidity phenomena of scalar curvature. Let $R^{+}(x)=\sup(0,R[\bar{g}](x))$.

Theorem 2**.**

Let $(M^{n},\bar{g})$ be a complete Riemannian $n$-manifold with scalar curvature $R[\bar{g}]$ ($n\geq 3$) and $\Omega$ be a smooth domain in $(M^{n},\bar{g})$ with positive $Q(\Omega,\bar{g})>0$. Assume that {\frac{(n+2)}{4(n-1)}}\Big{[}\int_{\Omega}|R^{+}|_{\bar{g}}^{\frac{n}{2}}dV_{\bar{g}}\Big{]}^{\frac{2}{n}}<Q(\Omega,\bar{g}). Then, if a conformal metric $g\in[\bar{g}]$ on $M$ satisfies that

• •

$R[g]\geq R[\bar{g}]$* in $\Omega$,*

• •

$g$* and $\bar{g}$ induced the same metric on $\partial\Omega$, and*

• •

$H[g]\geq H[\bar{g}]$* on $\partial\Omega$,*

then $g=\bar{g}$.

Proof.

Since $R[\bar{g}]\leq R[g]$ we have (3). Take $v=u-1$, $\Omega_{1}=\{x\in\Omega|u(x)<1\}$ and $\Omega_{2}=\{x\in\Omega|u(x)\geq 1\}$. We shall show that $\Omega_{1}=\phi$. If then, $u>1$ on $\Omega$ by the maximum principle. Since the mean curvature at the boundary is increasing, $\partial_{\nu}u\geq 0$ on $\partial\Omega$. However, this contradicts to the strong maximum principle since $u>1$ on $\Omega$ and $u=1$ on the $\partial\Omega$ (see [7]). To show that $\Omega_{1}=\phi$, we let

[TABLE]

The given conditions imply that

[TABLE]

on $\Omega$, $v=0$ on $\partial\Omega$ and $\partial_{\nu}v=0$ on $\partial\Omega$. Note that $A(x)\leq\frac{1}{n-1}R^{+}[\bar{g}]$ on $\Omega_{1}$. Multiplying $v$ on (8) on $\Omega_{1}$,

[TABLE]

We may consider $v$ as a function defined on $\Omega$ by extending the domain. By using the Sobolev constant $Q(\Omega,g)$ of $\Omega$,

[TABLE]

where (9) is used. Therefore {\frac{(n+2)}{4(n-1)}}\Big{[}\int_{\Omega_{1}}|R^{+}|_{\bar{g}}^{\frac{n}{2}}dV_{\bar{g}}\Big{]}^{\frac{2}{n}}<Q(\Omega,g) implies $v\equiv 0$ on ${\Omega_{1}}$. ∎

Remark 3*.*

Let $\Omega$ be a simply connected domain in a locally conformally flat manifold $(M,\bar{g})$, then $Q(\Omega,g)=Q(S^{n},g_{0})$. If {\frac{(n+2)}{4(n-1)}}\Big{[}\int_{\Omega}|R^{+}|_{\bar{g}}^{\frac{n}{2}}dV_{\bar{g}}\Big{]}^{\frac{2}{n}}<Q(S^{n},g_{0}), then Theorem 2 holds for $\Omega$. If $(M,g)$ is locally conformally flat with constant positive scalar curvature, then the rigidity holds for $\Omega$ with sufficiently small $|\Omega|$.

Remark 4*.*

When $(M^{n},\bar{g})$ is a compact Einstein manifold with positive scalar curvature, $Q(M^{n},\bar{g})={{(n-2)}\over{4(n-1)}}R[\bar{g}][Vol(M,g)]^{\frac{2}{n}}$ [6, page 48]. Since $Q(M,\bar{g})\leq Q(\Omega,\bar{g})$, Theorem 2 holds for any smooth domain $\Omega$ with $|\Omega|<[\frac{n-2}{n+2}]^{\frac{n}{2}}[Vol(M,g)]$.

For $n\geq 2$, let $(M^{n},\bar{g})$ be a Riemannian space with positive scalar curvature $R[\bar{g}]>0$. Next we prove that for each given point $p$ there exist domain $D\ni p$ in a general manifold, on which conformal scalar curvature rigidity holds by applying the techniques of [7]. For a domain $D\subset(M,\bar{g})$, we let

[TABLE]

and $\lambda_{1}(D)$ be the 1-st nonzero eigenvalue of domain $D$ with Dirichlet condition with respect to the metric $\bar{g}$, i.e.,

[TABLE]

where $u\in H={H}_{0}^{1,2}({D})$. It is known that for a given point $p\in(M,g)$, we can find $D\ni p$ with sufficiently large $\lambda_{1}(D)$.

Theorem 5**.**

Let $(M^{n},\bar{g})$ be a complete $n$-dimensional Riemannian space with $R_{\bar{g}}>0$ ($n\geq 2$). Assume $D$ is a smooth pre-compact subset in $M$ with $\lambda_{1}(D)>\frac{R(\bar{g},D)}{n-1}$. Then, if a metric $g\in[\bar{g}]$ on $M$ satisfies that

• •

$R[g]\geq R[\bar{g}]$* in $D$,*

• •

$g$* and $\bar{g}$ induced the same metric on $\partial D$, and*

• •

$H[g]=H[\bar{g}]$* on $\partial D$,*

then $g=\bar{g}$.

Proof.

Let $g=u^{\frac{4}{n-2}}\bar{g}$. To prove the rigidity on the domains in a general Riemannian manifold, we construct a positive smooth function on a suitable domain $D$ with the properties similar to (5, 6). For this, we take any smooth domain $D\subset\subset M$ with $\frac{R(\bar{g},D)}{n-1}<\lambda_{1}(D)$ and the eigenfunction $u_{1}$ with $\Delta u_{1}+\lambda u_{1}=0$ on $D$. Take $v=u-1$. Since $u_{1}$ is positive on $D$, we can express $v=u_{1}\beta$ with some function $\beta$ on $D$. From (8),

[TABLE]

Since $u_{1}>0$ on $D$,

[TABLE]

Using L’hospital’s rule, $\beta=0$ on $\partial D$. If there exists a minimum point $p\in D$ with $\beta(p)<0$ and $\nabla\beta(p)=0$, then $\triangle\beta(p)<0$ since $A(x)\leq\lambda_{1}(D)$ if $v\leq 0$. This contradicts to the Maximum principle (see [2]). Therefore $v\geq 0$ on $D$. Then by the Maximum principle again, it does not satisfy the boundary condition $\partial_{\nu}u=0$ on $\partial D$ if $v$ is not identically zero on $D$. ∎

Note that on a standard hemisphere $S^{+}$, $R[\bar{g}]=n(n-1)$ $\lambda_{1}(S^{+})=n$ (see [8]), which provides maximal domain. Any smaller domain than the hemisphere $D\subset\subset S^{+}$ satisfies $\lambda_{1}(D)>n$, on which Theorem 5 holds.