Rigidity and Non-rigidity Results on the Sphere

Rigidity and Non-rigidity Results on the Sphere Fengbo Hang Xiaodong Wang Department of Mathematics Michigan State University Oct., 00 1 Introduction It is a simple consequence of the maximum principle that a superharmonic function u on R n (i. e. u 0) which is 1 near infinity is identically 1 on R n (throughout this paper, n 3). Geometrically this means that one can not conformally deform the Euclidean metric in a bounded region without decreasing the scalar curvature somewhere. In fact there is a much stronger result: one can not have any compact deformation of the Euclidean metric without decreasing the scalar curvature somewhere, i. e., if g is a metric on R n which has nonnegative scalar curvature and is the Euclidean metric near infinity, then g is the Euclidean metric on R n. This is a simple version of the positive mass theorem ([9, 1]). Another implication of the positive mass theorem is the following rigidity theorem for the unit ball in R n. Theorem 1.1 Let (M, g) be an n-dimensional compact Riemannian manifold with boundary and the scalar curvature R 0. The boundary is isometric to the standard sphere S n 1 and has mean curvature n 1. Then (M, g) is isometric to the unit ball in R n. (If n > 7, we also assume M is spin.) The proof uses a generalized version of the positive mass theorem, see Shi and Tam [11] and Miao [6]. On the other hand, there are nontrivial metrics on R n which agree with the Euclidean metric near infinity and have nonpositive scalar curvature by the work of Lohkamp [5]. fhang@math.msu.edu. Supported by NSF Grant DMS-00950 xwang@math.msu.edu 1

One can establish parallel results for the hyperbolic space H n by analogous methods. It is natural to wonder about the other space form S n. The following conjecture was posed by Min-Oo in 1995. Conjecture 1 Let (M, g) be an n-dimensional compact Riemannian manifold with boundary and the scalar curvature R n(n 1). The boundary is isometric to the standard sphere S n 1 and is totally geodesic. Then (M, g) is isometric to the hemisphere S n +. This is an intriguing conjecture. It seems extremely difficult. The formulation given here is probably over-ambitious, but any progress under some extra assumptions would be interesting. In an unpublished manuscript [7] Min-Oo attempted to prove the conjecture by a Witten type argument under the assumption that M is spin. Unfortunately his attempt has been unsuccessful. To the authors knowledge the conjecture is even open under the stronger assumption Ric n 1. Inspired by this conjecture, we study some special cases and related questions. We first prove that on the standard sphere (S n, g 0 ) we can even conformally deform g 0 without decreasing the scalar curvature and with the deformation supported in any given open geodesic ball of radius > π/. In other words, the corresponding rigidity for a geodesic ball of radius > π/ fails even among conformal deformations. Without restricting to conformal deformation we also construct a rotationally symmetric g on S n such that its sectional curvature 1 and strict somewhere and near the north pole and south pole g = g 0. These results are interesting in view of the work of Corvino [3]. We then verify the rigidity for the hemisphere among conformal deformations. In fact in this situation we have some stronger results. In the last section we establish the rigidity in the Einstein case. Acknowledgment: The second author wants to thank Professors Lars Andersson and Mingliang Cai for their interest in this work and for useful conversations on the subject. He also wants to thank Professor Min-Oo for sending his manuscript. Nonrigidity when the boundary is non-convex We first introduce some notations. Let (S n, g S n) be the unit sphere in the Euclidean space R n+1 with the induced metric. We denote the north pole by N and the south pole by S. For r (0, π) let B(N, r) be the open geodesic ball of radius r with center N. Its boundary is umbilic with principal

curvatures all equal to cos r sin r. Therefore the boundary is non-convex if and only if r > π/. The closed upper hemisphere is denoted by S+. n Theorem.1 For any r ( π, π) there is a smooth metric g = eφ g S n S n with the following properties on R g n(n 1), Supp (φ) B(N, r). φ 0. Remark. Since φ 0, the inequality R g n(n 1) must be strict somewhere inside B(N, r). To put the above theorem in a context, we mention the following theorem due to Corvino [3] which has shed new light on the positive mass theorem. Theorem. (Corvino) Let Ω be a compactly contained smooth domain in a Riemannian manifold (M, g 0 ). Suppose the linearization L g0 of the scalar curvature map R : C (Ω) C (Ω) has an injective formal L - adjoint L g 0 on Ω. Then ɛ > 0 such that for any smooth function f which equals R(g 0 ) in a neighborhood of Ω and f R(g 0 ) C 1 < ɛ, there is a smooth metric g on M with R(g) = f and g g 0 outside Ω. The main point is that if g 0 is non-static (i. e. Ker L g 0 = 0) then there are compact deformations of g 0 with the scalar curvature going either direction. This is in contrast with R n, which is static, where one can not have compact deformations without decreasing the scalar curvature somewhere. The sphere (S n, g S n) is also static. In fact L g S n f = f g S n + D f (n 1)f g S n and its kernel is spanned by the n + 1 coordinate functions x 1,..., x n+1 (also the first eigenspace). Theorem.1 shows that one still can deform g S n without decreasing the scalar curvature on any geodesic ball of radius r > π/. To prove Theorem.1 we need a technical lemma. Lemma.1 Assume f 1, f C ([ 1, 1]) and f 1 (0) = f (0), f 1 (0) < f (0). Let g (x) = f1 (x), 1 x 0; f (x), 0 x 1. 3

Then for any ε > 0 small, there exists a g ε C ([ 1, 1], R) such that g ε (x) = g (x) for x ε; g (x) g ε (x) g (x) + ε for x ε; g (x) g ε (x) for x 0; g (x) < g ε (x) for some x 0. Proof. Without loss of generality, we may assume f 1 0. Then f (0) = 0, a = f (0) > 0. Let f 3 (x) = f (x) f (0) x, then f 3 (0) = f 3 (0) = 0. We may find some M > 0 such that f 3 (x) M for x 1. Let k (x) = 0, x 0; ax, 0 x; r (x) = 0, 1 x 0; f 3 (x), 0 x 1. Denote ρ (x) = c 0 e 1 1 x, x 1; 0, x > 1. Here c 0 is a positive constant such that ρ (x) dx = 1. Fix δ > 0 small, then we let k δ (x) = (ρ δ k) (x), here ρ δ (x) = δ 1 ρ (x/δ). It is clear that k δ (x) = k (x) for x δ, k (x) < k δ (x) k (x)+aδ for x < δ and k δ (x) = aρ δ (x). Fix a smooth function η on R such that 0 η 1, η (x) = 0 for x 0 and η (x) = 1 for x 1. For 0 < τ < δ/, we let r τ (x) = η ( ) x τ f3 (x). Then for 0 x τ, we have r τ (x) = 1 ( x ) τ η τ f 3 (x) + ( x ) ( x ) τ η f 3 (x) + η f τ τ 3 (x) cm. Here c is an absolute constant. On the other hand, for 0 x τ, r τ (x) r (x) M τ. Hence for 0 < x τ, we have k δ (x) + r τ (x) a ( ) 1 δ ρ cm M > g (x)

if δ is small enough. For x > τ, we have Moreover, for 0 x τ, we have for some σ > 0, hence k δ (x) + r τ (x) g (x). k δ (x) > k (x) + σ g (x) + aδ + M τ = k (x) + r (x) + aδ + M τ k δ (x) + r τ (x) k (x) + r (x) + σ M τ g (x) when τ is small enough. For other x, we clearly have g (x) + aδ = k (x) + r (x) + aδ k δ (x) + r τ (x) k (x) + r (x) = g (x). The lemma follows by taking g ε (x) = k δ (x) + r τ (x). We now present the proof of Theorem.1. The stereographic projection from the south pole is given by π S (y) = y 1 + y n+1 for y = ( y, y n+1) S n. On R n, we have standard coordinates x 1,, x n, polar coordinates r, θ and cylindrical coordinates t, θ, where r = e t. We have ( ) π 1 S gs n = (1 + x ) n dx i dx i ( = dr dr + r ) (1 + r ) g S n 1 i=1 = (cosh t) (dt dt + g S n 1). Let g be the metric we are looking for, then ( ) π 1 S g = u n n i=1 The scalar curvature of g is given by (n 1) R = n [ = v n+ n n+ u n u (n 1) n dx i dx i = v n (dt dt + gs n 1). ] (v tt + S n 1v) + (n 1) (n ) v. 5

For λ > 0, let d λ x = λx be the dilation, then d λ ( ) π 1 S gs n = λ (1 + λ x ) n dx i dx i. i=1 Denote u λ (x) = ( 1 λ n n + λ x ), then u λ = We need to solve the following n (n ) n+ n uλ. u C (R n ), u > 0, u u 1, u (x) = u 1 (x) for x > a; (1) u n(n ) u n+ n. Claim 1 For any a > 1, (1) has at least one solution. Remark. It is interesting to note here that for a 1, (1) has no solution. This is implied by the Theorem 3.1 below. Proof. The rough idea is the following, let ua (x), x a u (x) = min u a (x), u 1 (x)} = ; u 1 (x), x a. Then clearly u n(n ) n in weak sense. One may get a smooth u by suitable smoothing procedure. More precisely, we may do the following, let f (t) = (cosh t) n, then u n+ f n (n ) = f n+ (n ) n f. For δ > 0 small, let g (t) = f (t + δ), then f ( δ) = g ( δ), f ( δ) > g ( δ). Let g (t), t δ; h (t) = f (t), t δ. 6

By Lemma.1 for ε > 0 tiny, we may find a smooth function h ε such that Hence for t δ, h ε (t) = h (t) for t + δ ε; h (t) ε h ε (t) h (t) for t + δ ε; h ε (t) h (t) for t δ. h ε (t) h (t) = n (n ) h ε (t) n+ n (n ) h (t) n+ (n ) n (n ) n h (t) h ε (t), observing that h (t) is very close to 1 when ) t + δ ε. Now g = πs (h ε (t) n (dt dt + gs n 1) is the needed metric. If we do not restrict ourselves to conformal deformations, we can even construct a deformation without decreasing the sectional curvatures. Claim For any 0 < a < b < π, there exists a function f C ([0, b], R) such that sin x, 0 x a, f (x) = sin (b x), b a x b, Proof. Denote f (x) = f (b x) for 0 < x < b, f f > 0 on (0, b), f > f somewhere, 1 f + f on [0, b], 1 > f + f somewhere. ρ (x) = ce 1 1 x, x < 1, 0, x 1, here c is a positive constant such that R ρ (x) dx = 1. For δ > 0, ρ δ (x) = 1 δ ρ ( ) x δ. For 0 < δ < π/, denote then c δ = δ δ δ δ ρ δ (x) cos xdx (0, 1), ρ δ (y) sin (x y) dy = c δ sin x. 7

Let For 0 < δ < b a, let g (x) = sin x, x b, sin (b x), b x. f (x) = 1 δ ρ δ (y) g (x y) dy, c δ δ then f satisfies all the requirements. Theorem.3 For any a ( 0, π ), there exists a smooth metric g on S n such that g = g S n on B (S, a) B (N, a) and the sectional curvature of g is at least 1 and larger than 1 somewhere. Proof. Fix a number b ( a, π ). Let f be as in the Claim. Consider the metric g = dr dr + f (r) g S n 1. Let e 1,, e n 1 be a local orthonormal frame on S n 1, then the curvature operator of g is given by Q ( r e i ) = f Q (e i e j ) = 1 f f r e i, f e i e j, for 1 i, j n 1. By Claim, we see the sectional curvature of g is at least 1. Next we will construct a smooth function φ : [0, π] [0, b] such that φ (r) = r, 0 r a, r + b π, π a r π, φ (r) > 0 and φ (r) = φ (π r). Indeed, let α (x) = ce 1 x(1 x), 0 < x < 1, 0, x 0 or x 1, here c is chosen such that 1 0 α (x) dx = 1. Let β (x) = x 0 α (t) dt. Fix a λ > 0 such that a + λ (π a) < b. For 0 < ε < π a, let δ = 8

min ε, π a ε} and λ + (1 λ) β g ε (x) = λ + (1 λ) β ( π ε x ( δ x π ε δ ), 0 x π ), π, x π. Then for some ε we have π 0 g ε (x) dx = b. We may put φ (x) = x 0 g ε (x) dx and φ satisfies all the requirements. Let r be the distance function on S n to N, then we may put g = dφ (r) dφ (r) + f (φ (r)) g S n 1. It satisfies all the requirements in the claim. 3 Conformal deformation on the hemisphere The assumption r > π/ in Theorem.1 is optimal as it turns out that it is impossible to localize the deformation in the hemisphere. be a C metric on S n + satisfying the assump- Theorem 3.1 Let g = e φ g S n tions R g n(n 1), the boundary is totally geodesic and is isometric to the standard S n 1. Then g is isometric to g S n. Remark. This verifies Conjecture 1 among conformal deformations. Proof. By the assumption g S n 1 = e φ S n 1 g S n 1 is isometric to g S n 1. By the Obata theorem, there exist λ 1 and ζ S n 1 such that g S n 1 = ψλ,ζ g S n 1, where ψ λ,ζ is the conformal transformation of S n which is dilation by λ when we identify S n with R n by the stereographic projection from ζ. Replacing g by (ψ 1 λ,ζ ) g, we can assume g S n 1 = g S n 1, i. e. φ S n 1 0. We are to prove φ 0 on S+. n As in the proof the Theorem.1, we work on R n via the stereographic projection from the south pole. We write ( ) π 1 S g = u n n i=1 dx i dx i = v n (dt dt + gs n 1). 9

Then u C (B 1 ) is positive and satisfies u n(n ) u n+ n in B 1 ; u = 1 on B 1 u r = n on B 1. () (The Neumann boundary condition is the geometric assumption that the boundary is totally geodesic.) Claim 3 The only solution to () is u 1 (x) = ( We work with v in cylindrical coordinates. v tt S n 1v + (n ) v ) n 1+ x n (n ) v n+ n. Let f (t) = 1 v (t, θ) ds (θ), nω n S n 1 here ω n is the volume of the unit ball in R n. By Holder s inequality, we have Therefore we have Denote 1 nω n S n 1 v (t, θ) n+ n ds (θ) f(t) n+ n. (3) f C ([0, ), R), f > 0 f (t) n(n ) f (t) n+ n (n ) f (t), f (0) = 1, f (0) = 0. e (t) = f n (n ) (t) f (t) n+ (n ) n + f (t) 0. Since f (0) < 0, we see f (t) < 0 for t > 0 small. Assume b > 0 such that f (t) < 0 on (0, b), then for 0 t b, we have f (t) = t 0 f (s) e (s) ds + (n ). ) (f (t) f (t) n n. In particular, f (b) > 0. This implies that f (t) < 0 for any t. 10

Assume e is not identically zero, then for some b > 0, e is not identically zero on (0, b), then for any t > b, we have f (t) c > 0. This implies f (t) c and hence lim t f (t) =, a contradiction. Hence e (t) 0. This shows f (t) = (cosh t) n for t 0. Moreover the inequality (3) must be an equality. This implies that v(t, θ) = f(t) = (cosh t) n. Hence u = ( ) n 1+ x. With a little improvement of our argument we can remove the assumption that the boundary is totally geodesic. be a C metric on S n + satisfying the assump- Theorem 3. Let g = e φ g S n tions R g n(n 1), the boundary is isometric to the standard S n 1. Then g is isometric to g S n. By the same argument, we can reduce the problem to a partial differential inequality on B 1 R n. In fact we establish the following stronger result Claim Assume u C ( B 1, R ), u > 0 and Then u = u 1. u n(n ) n in B1 ; u B1 1. u n+ (The proof of Theorem 3. only requires the special case u B1 = 1.) To prove this claim, we take an approach different from our previous method. First, we observe that if we solve v such that v = n(n ) u n+ n in B1 ; v B1 = 1. 11

Then 1 v u and v n(n ) v n+ n. If we can prove v = u1, then n (n ) u n+ n = v = u1 = n (n ) n+ n u1. Hence u = u 1. Therefore from now on, we may assume u B1 = 1. We consider the following PDE v = n(n ) v n+ n in B1 ; v B1 = 1. () We claim the only positive solution is v = u 1. Indeed, it follows from the moving plane method of Gidas, Ni and Nirenberg [] that v (x) = f ( x ) for some f, moreover f + n 1 r f + n(n ) f n+ n = 0; f (0) = a > 0, f (0) = 0. It is clear g (r) = a ) (1 + a n n r is a solution to the problem. On the other hand, since f satisfies f (r) = a n (n ) r 0 dt t 0 ( s t ) n 1 f (s) n+ n ds. It follows from contraction mapping theorem that for some ε > 0, f = g on [0, ε]. Hence f = g. Since f (1) = 1, we see a = n. Hence f (r) = n (1+r ) n and v = u 1. Since 1 is a subsolution for () and u is a supersolution with u 1, by the standard method of iteration we may find a solution v for () and 1 v u. Since the only positive solution is u 1, we see u 1 = v u. If u 1 u, then since (u u 1 ) 0 and (u u 1 ) B1 = 0, we see u > u 1 in B 1. Moreover, it follows from Hopf maximum principle that for some c 1 > 0, u (x) u 1 (x) c 1 (1 x ). This implies that for some c > 0, u (x) [ n u 1 (x) 1 + c (1 x )]. 1

On the other hand, for λ > 1, we have ( ) uλ (x) n u 1 (x) = 1 + (λ 1) ( 1 λ x ) 1 + λ x 1 + (λ 1) ( 1 x ) 1 + λ x 1 + c (1 x ) ( ) u (x) n u 1 (x) if λ 1 is small enough. Hence for some λ > 1, u λ u. Since u λ is also a subsolution, we may find a solution v for () such that u λ v u. It follows from previous discussion that v = u 1. Hence u λ u 1, this contradicts with the fact λ > 1. The Einstein case In this section we prove the following uniqueness theorem: Theorem.1 Let (M, g) be a smooth n-dimensional compact Einstein manifold with boundary Σ. If Σ is totally geodesic and is isometric to S n 1 with the standard metric, then (M, g) is isometric to the hemisphere S n + with the standard metric. This verifies Conjecture 1 in the special case that g is Einstein. Given local coordinates ξ 1,..., ξ n 1 on the boundary, we can introduce local coordinates on a collar neighborhood of Σ in M as follows. For ξ Σ let γ ξ (t) = γ(t, ξ) be the normal geodesic starting at ξ with initial velocity ν(ξ), the unit inner normal vector at ξ. Then t, ξ 1,..., ξ n 1 form local coordinates on a collar neighborhood of Σ in M. Let h ij = γ, γ. By ξ i ξ j the Gauss lemma the metric g takes the form g = dt + h ij (t, ξ)dξ i dξ j, where Latin indices i, j,... run from 1 to n 1. Greek indices α, β,... will be used to run from 0 to n 1. We denote the curvature tensors of M and Σ by R and K, respectively. Since Σ is totally geodesic, by the Gauss equation we have R ikjl = K ikjl. 13

Then the Ricci tensor is given by R ij = R i0j0 + K ij. Taking trace we get the scalar curvature R = R 00 + (n 1)(n ) = R/n + (n 1)(n ), hence R = n(n 1). Thus Ric (g) = (n 1)g. The second fundamental form of the t-hypersurface is given by A ij = D γ t ξ i, γ ξ j = 1 h ij t. We also need to know the second derivative of h ij in t. 1 h ij t = D γ t ξ i, γ ξ i + D t γ ξ i, D γ t ξ j = R( γ t, γ ξ i, γ t, γ ξ j ) + 1 h hkl ik t h jl t, where in the last step we use the fact that γ is a Jacobi field along the ξ i geodesic γ(t, ξ). As g is Einstein, the above equation can be written as 1 h ij t = (n 1)h ij + 1 hkl h ik t h jl t + h kl R ikjl. (5) Claim 5 Infinitesimally h ij (t, ξ) equals cos (t)h ij (0, ξ). Remark. It is clear that g S n = dt + cos (t)g S n 1 We prove by induction that h ij (t, ξ) = cos (t)h ij (0, ξ) + O(t m ), as t 0 for any integer m > 0. The case m = 1 is trivial. Suppose it is true for m. We assume without loss of generality that h ij (0, ξ) = δ ij. We first have R ikjl = cos (t)(δ ij δ kl δ il δ kj ) + O(t m 1 ) (this is true because R ijkl only involves differentiating the metric in t once). By (5) we get 1 h ij t = (n 1) cos (t)δ ij + 1 δkl sin (t) cos (t) δ ikδ jl + cos (t)δ kl (δ ij δ kl δ il δ kj ) + O(t m 1 ) = (n 1) cos (t)δ ij + sin (t)δ ij + (n ) cos (t)δ ij + O(t m 1 ) = cos(t)δ ij + O(t m 1 ). 1

This implies that h ij (t, ξ) = cos (t)h ij (0, ξ) + O(t m+1 ). Consider S n + with the standard metric g 0. It is easy to see that g 0 = dt + cos (t)h, where t is the distance to the boundary S n 1 and h is the standard metric on S n 1. We form a closed manifold M by joining M and S n + along their boundary. In view of Claim 5 we get a smooth Riemannian manifold with a totally geodesic hypersurface Σ which is isometric to S n 1. The metric, also denoted by g, is of course Einstein. By [], g is real analytic in harmonic coordinates. We define Ω to be the set of points where g has constant curvature 1 in a neighborhood. This is an open set by definition. If it is not the whole manifold, we take a point p on its boundary and choose local harmonic coordinates x 1,..., x n on a connected neighborhood U. The analytic functions R ikjl g ij g kl + g il g jk vanish on an open subset of U for U Ω, hence vanish identically on U. Then p Ω, a contradiction. Therefore g has constant sectional curvature 1 everywhere. It is then easy to see that (M, g) is isometric to S n and (M, g) is isometric to S n +. References [1] A. Besse, Einstein Manifolds, Springer-Verlag, New York (1987). [] D. DeTurck and J. Kazdan, Some Regularity Theorems in Riemannian Geometry, Ann. Sci. cole Norm. Sup. () 1 (1981) 9-60. [3] J. Corvino, Scalar curvature deformation and a gluing construction for the Einstein constraint equations, Comm. Math. Phys. 1 (000), no. 1, 137 189. [] B. Gidas, W.M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Physics 68(1979) 03-3. [5] J. Lohkamp, Scalar curvature and hammocks, Math. Ann. 313 (1999), no. 3, 385 07. [6] P. Miao, Positive mass theorem on manifolds admitting corners along a hypersurface, Adv. Theor. Math. Phys. 6 (00), no. 6, 1163 118 (003). [7] M. Min-Oo, Scalar curvature rigidity of the hemisphere, unpublished, 1995. 15

[8] R. Schoen, Variational theory for the total scalar curvature functional for Riemannian metrics and related topics. Topics in calculus of variations (Montecatini Terme, 1987), 10 15, Lecture Notes in Math., 1365, Springer, Berlin, 1989. [9] R. Schoen and S.-T. Yau, On the proof of the positive mass conjecture in general relativity, Comm. Math. Phys. 65 (1979) 5 76. [10] R. Schoen and S.-T. Yau, Lectures on Differential Geometry, International Press, Cambridge, MA 199. [11] Y. Shi and L. Tam, Positive mass theorem and the boundary behaviors of compact manifolds with nonnegative scalar curvature, J. Differential Geom. 6 (00), no. 1, 79 15. [1] E. Witten, A new proof of the positive energy theorem, Comm. Math. Phys. 80 (1981) 381 0. 16