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On compact Riemannian manifolds with convex boundary and Ricci curvature bounded from below

We propose a new approach to the study of compact Riemannian manifolds with nonnegative Ricci curvature and strictly convex boundary or positive Ricci curvature and convex boundary. Several conjectures are formulated. Some partial results that support these conjectures are established.


Introduction
For a compact Riemannian manifold (M n , g) with nonempty boundary Σ = ∂M , it is interesting to study connections between the intrinsic geometry g| Σ and the extrinsic geometry (the 2nd fundamental form), under a lower bound for scalar curvature or Ricci curvature. We refer to [ST1,ST2,WY1,MW] and references therein for recent works in this direction. Some of these works are motivated by problems in general relativity, in particular about understanding various definitions of quasi-local mass. The following fundamental result was proved by Shi and Tam [ST1].
Theorem 1. Let (M n , g) be a compact Riemannian manifold with scalar curvature R ≥ 0 and with a connected boundary Σ. Suppose • M is spin, • the mean curvature H of Σ is positive, • there exists an isometric embedding ι : Σ → R n as a strictly convex hypersurface. Then where H 0 is the mean curvature of ι : Σ → R n . Moreover, if equality holds, then M is isometric to the Euclidean domain enclosed by ι : Σ → R n .
The right hand side of (1.1) is determined by the intrinsic geometry of Σ. Therefore by the inequality the extrinsic geometry of Σ is constrained by its intrinsic geometry. But the assumption that there is an isometric embedding of Σ into R n as a strictly convex hypersurface imposes severe restriction on the kind of intrinsic geometry of Σ for which the theorem is applicable. In the more recent work [MW] Miao and I proved a slightly different inequality under the stronger condition Ric ≥ 0, but without any restriction on the intrinsic geometry of the boundary.
Theorem 2. Let (M n , g) be a compact Riemannian manifold with Ric ≥ 0 and with a connected boundary Σ that has positive mean curvature H. Let ι : Σ → R m be an isometric embedding. Then where − → H 0 is the mean curvature vector of ι : Σ → R m . Moreover, if equality holds, then ι (Σ) is contained in an n-dimensional plane of R m and M is isometric to the Euclidean domain enclosed by ι (Σ) in that n-dimensional plane.
Notice that an isometric embedding ι : Σ → R m always exists by Nash's famous theorem.
When the scalar curvature has a negative lower bound, results similar to Theorem 1 were proved by [WY1] and [ST2]. The counterexample to the Min-Oo conjecture by Brendle, Marques and Neves [BMN] shows that no such result holds when the scalar curvature has a positive lower bound. Results similar to Theorem 2 are also established when the Ricci curvature has a positive or negative lower bound. But in these two cases the inequalities obtained are not sharp. Some rigidity results under stronger assumptions on the boundary were proved in [MW].
In all of these studies the result is basically an estimate on an integral involving the mean curvature. It is natural to ask if one can bound the area of the boundary, the volume of the interior and other more direct geometric or analytic quantities. It is easy to see that for such results to hold a lower bound for the mean curvature is not enough. For example, for any closed Σ n−2 , h with nonnegative Ricci curvature, M := B 2 × Σ with the product metric dx 2 + h has nonnegative Ricci curvature and mean curvature H ≥ 1 while the area of ∂M can be arbitrarily large. Therefore we will in this paper mostly consider compact Riemannian manifolds (M n , g) with nonnegative Ricci curvature and with a connected boundary Σ whose 2nd fundamental form has a positive lower bound. Motivated by a uniqueness theorem in [BVV], we study positive harmonic functions on M that satisfy a semilinear Neumann condition on the boundary. We formulate a conjecture which has important geometric implications. We will prove some partial results that support this conjecture. Another case we consider is when M has positive Ricci curvature, by scaling we can always assume Ric ≥ n − 1 and the boundary Σ is convex in the sense that its 2nd fundamental form is nonnegative. There is similarly a natural conjecture on the area of the boundary.
The paper is organized as follows. In section 2 we discuss some natural PDEs on a compact manifold with boundary. We formulate a uniqueness conjecture on a semilinear Neumann problem in the nonnegative Ricci case and discuss its geometric implications. In Section 3 we prove some topological results. In Section 4 we present some partial results and several other conjectures.
Aknowledgement. The work of the author is partially supported by a Simons Foundation Collaboration Grant for Mathematicians.
This theorem was proved by Bidaut-Veron and Veron [BVV] when ∂M = ∅ and by Ilias when ∂M = ∅ using the same method. It has some important corollaries. We recall the Yamabe problem on a compact Riemannian manifold (M n , g) with boundary. The conformal Laplacian is defined to be L g = −c n ∆ g + R g , with c n = 4(n−1) n−2 . If g = φ 4/(n−2) g, then Under the conformal deformation the mean curvature of the boundary transforms according to the following formula 2 (n − 1) n − 2 ∂φ ∂ν + Hφ = Hφ n/(n−2) .
We consider the following functional This functional is conformally invariant: E g (u) = E g (uφ). If u is positive then where g = u 4/(n−2) g We define The sign of λ is conformally invariant. The Yamabe invariant is defined to be Escobar [E] and Cherrier [C] proved that Y (M, g) ≤ Y S n + = n (n − 1) (|S n | /2) 2/n and moreover the infimum is achieved if it is strict. Let (M n , g) be a compact Riemannian manifold with convex boundary and Ric ≥ (n − 1). Then for 2 < q ≤ (n + 2) / (n − 2) we have • Q (M, g) ≥ n (n − 1) V 2/n . Moreover, equality holds iff g is Einstein with totally geodesic boundary.
Theorem 4. Let (M n , g) be a compact Riemannian manifold with convex boundary and Ric ≥ (n − 1). Then Q (M, g) ≥ n (n − 1) V 2/n . Moreover, equality holds iff g is Einstein with totally geodesic boundary.
Corollary 1. Let (M n , g) be a compact Riemannian manifold with convex boundary and Ric ≥ (n − 1). Then V ≤ |S n | /2. Moreover, equality holds iff M is isometric to S n + .
Given a compact Riemannian problem (M n , g) with nonempty boundary, the type II Yamabe problem studied by Escobar is whether one can find a conformal metric g = φ 4/(n−2) g with zero scalar curvature and constant mean curvature on the boundary. This leads to the following equation Assuming λ (M, g) > 0 Escobar introduced the following minimization Motivated by Theorem we propose to study positive solutions of the following equation ∆u = 0 on M, ∂u ∂ν + λu = u q on ∂M, where λ > 0 and 1 < q < n/ (n − 2), and make the following conjecture.
Conjecture 1. Let (M n , g) be a compact Riemannian manifold with Ric ≥ 0 and Π ≥ 1 on ∂M . If 0 < λ ≤ 1/ (q − 1), then any positive solution u of the above equation must be constant unless q = n/ (n − 2), M is isometric to B n ⊂ R n and u corresponds to At the moment this conjecture is completely open. But in dimension 2 an analogous problem was studied by the author [W2] in which the following result was proved.
Theorem 5. Let (Σ, g) be a compact surface with Gaussian curvature K ≥ 0 and on the boundary the geodesic curvature κ ≥ 1. Consider the following equation where λ is a positive constant. If λ < 1 then u is constant; if λ = 1 and u is not constant, then Σ is isometric to the unit disc B 2 and u is given by for some a ∈ B 2 .
In summary Conjecture 1 implies the following conjecture.
Conjecture 2. Let (M n , g) be a compact Riemannian manifold with Ric ≥ 0 and Π ≥ 1 on ∂M . Then |∂M | ≤ S n−1 . The result should be well known. A proof using minimal surfaces for the 2nd part was given by Fraser and Li [FL]. We explain the standard argument with harmonic forms. By the Hodge theory for compact Riemannian manifolds with boundary
Remark 1. In the second part if we only assume Π ≥ 0, then the same argument proves that a harmonic form α ∈ H 1 A (M ) must be parallel. As α (ν) = 0 on the boundary Σ we can write α = X, · on Σ, where X is a vector field on Σ. As α is parallel it is easy to see that X is a parallel vector field on Σ. Therefore we conclude that either H 1 (M ) = 0 or there exists a nonzero parallel vector field on Σ.
In dimension 3 we have the following consequence.
Corollary 2. Let M 3 , g be a compact Riemannian 3-manifold with boundary Σ. Suppose Ric ≥ 0 and the boundary is strictly convex. Then the boundary Σ is topologically a sphere.
In fact the same argument combined with Remark 1 yields Proposition 2. Let M 3 , g be a compact Riemannian 3-manifold with boundary Σ. Suppose Ric ≥ 0 and the boundary is convex. Then Σ is either a topological sphere or a flat torus.
Therefore the boundary cannot be a Riemann surface of higher genus. The above two results in dimension 3 may be deduced from the work of Meeks-Simon-Yau [MSY], but the argument here is much more elementary.
The same argument works for the following situation.  The proof is more complicated. We first recall the following result which can be proved by classic methods.
Proposition 5. Let (M n , g) be a compact Riemannian manifold with Ric ≥ − (n − 1). Let ρ be the distance function to the boundary. Suppose the mean curvature of the boundary satisfies H ≥ (n − 1). Then in the support sense ∆ρ ≤ − (n − 1) .
By the Bochner formula we have Let f = |α| (n−2)/(n−1) . Direct calculation yields Suppose that f is not identically zero. By the maximum principle u must achieve its positive maximum somewhere on the boundary and furthermore at this point we must have ∂u ∂ν ≥ 0.
On the other hand on the boundary, as H ≥ n − 1 This is strictly negative when n ≥ 4 and hence a contradiction. Therefore α is identically zero. When n = 3 and if α is not identically zero, then by the Hopf lemma u must be a positive constant. By scaling α we can assume that u ≡ 1 or φ = f . Therefore −∆φ = φ. By elliptic regularity φ is smooth. From the proof of (3.1) it follows that ρ is smooth everywhere and |∇ρ| ≡ 1. But this is impossible as ρ is not smooth at a cut point, e.g. at a point where it achieves its maximum. Therefore we must have α = 0 too when n = 3.
The same argument can be used to prove the following: Let (M n , g) be a compact Riemannian manifold with boundary Σ. Suppose Ric ≥ − (n − 1). If the second fundamental form of Σ satisfies Π > (n − 1) / 2 (n − 2), then H 1 (M ) = 0. The only difference is that at the end we have for α ∈ H 1 Is the constant sharp? It seems reasonable to expect H 1 (M ) = 0 if Π > 1.

On the size of the boundary
In this section we prove some estimates on the size of the boundary, in particular we show that Conjecture 2 is true in dimension 3. First we recall a result in Xia [X].
Proposition 6. Let (M, g) be a compact Riemannian manifold with boundary Σ. Suppose Ric(g) ≥ 0 and Π ≥ 1. Then λ 1 (Σ) ≥ n − 1 and the equality holds iff (M, g) is isometric to the unit ball in Euclidean space R n .
The proof is based on Reilly's formula [Re]. For completeness and comparison later, we present the proof. Let u be the solution of the following equation where f is a first eigenfunction on Σ, i.e. − △ Σ f = λ 1 f . Let χ = ∂u ∂ν with ν being the outer unit normal. By Reilly's formula Therefore λ 1 ≥ n − 1.
If λ 1 = n− 1, then we must have D 2 u = 0, χ = f and Π = g| Σ . As a consequence we have D 2 f = −f g on ∂Σ. By the well known Obata theorem ∂M is isometric to the standard sphere S n−1 . Let f 1 , · · · , f n be a standard basis of the first eigenspace on ∂M ∼ = S n−1 and u 1 , · · · , u n the corresponding harmonic extensions on M . We know that ∇u 1, · · · , ∇u n are parallel vector fields on M . It is then easy to see that U = (u 1 , · · · , u n ) isometrically embeds M into R n with the image the unit ball. This is basically the same argument used by Choi and Wang [CW] to prove that the 1st eigenvalue of an embedded minimal hypersurface Σ n−1 ⊂ S n is at least (n − 1) /2. But a conjecture of Yau the 1st eigenvalue should equal to n − 1. By the same argument we have the following estimate for a general compact Riemannian manifold with Ric ≥ n and with a convex boundary.
Remark 2. In view of Yau's conjecture, we also conjecture that in this case the best lower bound is n − 1.
Proof. Let u be the solution of the following equation We will also need the following result due to Ros [Ros], which was also proved by Reilly's formula.
Theorem 6. (Ros) Let (M, g) be a compact Riemannian manifold with boundary. If Ric ≥ 0 and the mean curvature H of ∂M is positive, then The equality holds iff M is isometric to an Euclidean ball.
We can now prove the following result in dimension 3.
Theorem 7. Let (M 3 , g) be a compact Riemannian manifold with boundary Σ. Suppose Ric(g) ≥ 0 and Π ≥ g| Σ . Then Moreover if equality holds in either case, M is isometric to the unit ball B 3 ⊂ R 3 .
Example 1. Let S n−2 , h be compact Riemannian manifold with nonnegative Ricci curvature. Then B 2 ×S has nonnegative Ricci curvature and the boundary has mean curvature H = 1. This show that the conjecture is not true if the condition on 2nd fundamental form is weakened to a condition on the mean curvature.
In the case of positive Ricci curvature we make the following Moreover if equality holds then (M n , g) is isometric to the hemisphere S n In [HW] the following rigidity result was established.
Theorem 8. Let (M n , g) (n ≥ 2) be a compact Riemannian manifold with nonempty boundary Σ = ∂M . Suppose • Ric≥ (n − 1) g, • (Σ, g| Σ ) is isometric to the standard sphere S n−1 ⊂ R n , • Σ is convex in M in the sense that its second fundamental form is nonnegative. Then (M, g) is isometric to the hemisphere S n + . Therefore the conjecture, if true, is a far-reaching generalization of the above rigidity result. When n = 2 the above theorem can be reformulated as follows.
Theorem 9. Let (M 2 , g) be compact surface with boundary and the Gaussian curvature K ≥ 1. Suppose the geodesic curvature k of the boundary γ satisfies k ≥ 0. Then L(γ) ≤ 2π. Moreover equality holds iff (M, g) is isometric to S 2 + . It implies a classic result of Toponogov [T]: Let (M 2 , g) be a closed surface with Gaussian curvature K ≥ 1. Then any simple closed geodesic in M has length at most 2π. Moreover if there is one with length 2π, then M is isometric to the standard sphere S 2 . We refer to [HW] for more details. In view of this connection, Conjecture 3 can be viewed as a generalization of Toponogov's theorem in higher dimensions. We note that Marques and Neves [MN] offered a generalization of Toponogov's theorem in dimension 3 in terms of the scalar curvature.
As an evidence for Conjecture 3 we show that it is true under the stronger condition that sectional curvatures are at least one.
Moreover if |Σ| = S n−1 , then Σ is isometric to S n−1 . By Theorem 8 M is isometric to the hemisphere S n + . Similarly we have the following parallel result when sectional curvature is nonnegative.
Moreover if equality holds then (M n , g) is isometric to the hemisphere B n = {x ∈ R n : |x| ≤ 1}.
Using Proposition 7 one can easily prove the following by the same method used to prove Theorem 7.