aa r X i v : . [ m a t h . DG ] N ov Biharmonic hypersurfaces in hemispheres
Matheus Vieira
Abstract
In this paper we consider the Balmuş-Montaldo-Oniciuc’s conjecturein the case of hemispheres. We prove that a compact non-minimal bi-harmonic hypersurface in a hemisphere of S n +1 must be the small hyper-sphere S n (cid:0) / √ (cid:1) , provided that n − H does not change sign. It is well known that minimal hypersurfaces can be seen as hypersurfaces whosecanonical inclusion is a harmonic map. Thus it is natural to study hypersurfaceswhose canonical inclusion is a biharmonic map, known as biharmonic hypersur-faces (for more information see Section 2). From the point of view of findingnew examples and classification results, the theory of biharmonic hypersurfacesseems to be more interesting when the ambient space has positive curvature.There are many papers studying biharmonic hypersurfaces in the sphere (forexample [1], [2], [4], [5], [6], [8], [12], [13], [14], [15]).Let S n +1 be the unit Euclidean sphere. Balmuş-Montaldo-Oniciuc [1] con-jectured that a non-minimal biharmonic hypersurface in S n +1 must be an openpart of the small hypersphere S n (cid:0) / √ (cid:1) of radius / √ or of a generalizedClifford torus S k (cid:0) / √ (cid:1) × S n − k (cid:0) / √ (cid:1) with k = n/ . Since a generalizedClifford torus cannot lie in a hemisphere, it is natural to ask whether a com-pact non-minimal biharmonic hypersurface in a hemisphere must be the smallhypersphere. We give an affirmative answer to this question when n − H doesnot change sign. Theorem 1.
Let M n be a compact biharmonic hypersurface in a closed hemi-sphere of S n +1 . If n − H does not change sign then either M n is the equatorof the hemisphere or it is the small hypersphere S n (cid:0) / √ (cid:1) . Note that n − H ≥ when | A | ≤ n (by the Cauchy-Schwarz inequality)or H is constant (by Theorem 2). This is related to some previous results (seefor example Section 1.4 in [15]).The case n − H ≤ was proved in a direct way by Balmuş-Oniciuc (Corol-lary 3.3 in [2]). Our proof is completely different. It is based on a formulafor the bilaplacian of the restriction of a function defined on the ambient space(Theorem 4).The author would like to thank Detang Zhou, Cezar Oniciuc and Dorel Fetcufor their support. 1 Preliminaries
First we give the notations and conventions of the paper.Let M be a Riemannian manifold and let ∇ be the Levi-Civita connection.We define the Riemann curvature by Riem ( u, v ) w = ∇ u ∇ v w − ∇ v ∇ u w − ∇ [ u,v ] w. If f is a function on M we define the Hessian of f by ∇∇ f ( u, v ) = h∇ u ∇ f, v i , and the Laplacian of f by ∆ f = tr M ∇∇ f. Let ¯ M n +1 be a Riemannian manifold and let M n be a hypersurface in ¯ M n +1 .We define the second fundamental form by A ( u, v ) = (cid:10) ¯ ∇ u v, N (cid:11) , and the mean curvature by H = tr M A, where N is the normal vector. Note that H is not normalized. We denote thegeometric quantities of ¯ M n +1 with a bar. For example ¯ ∇ is the Levi-Civitaconnection of ¯ M n +1 , ¯ Riem is the Riemann curvature of ¯ M n +1 and ¯ Ric is theRicci curvature of ¯ M n +1 . We denote the geometric quantities of M n without abar. For example ∇ is the Levi-Civita connection of M n and ∆ is the Laplacianof M n .Now we recall the definition of biharmonic maps and hypersurfaces.Let φ : M → N be a map between Riemannian manifolds. We say that φ isharmonic if it is a critical point of the functional E ( φ ) = Z M | dφ | . Critical points of E satisfy τ ( φ ) = 0 , where τ ( φ ) = tr ∇ dφ. We say that φ is biharmonic if it is a critical point of the functional E ( φ ) = Z M | τ ( φ ) | . Critical points of E satisfy τ ( φ ) = 0 , where τ ( φ ) = ∆ τ ( φ ) + tr ¯ Riem ( τ ( φ ) , dφ ) dφ. Now let ¯ M n +1 be a Riemannian manifold and let M n be a hypersurface in ¯ M n +1 .It is well known that M n is minimal if and only if the canonical inclusion is a2armonic map. We say that M n is biharmonic if the canonical inclusion is abiharmonic map. For more information about harmonic and biharmonic mapsand submanifolds see [7], [9], [10].In the rest of this section we obtain a formula for the bilaplacian of therestriction of a function defined on the ambient space.The next result was applied to submanifolds of the sphere for the first timeby Oniciuc (Theorem 3.1 in [14]). Later it was applied to hypersurfaces ingeneral ambient spaces by Ou (Theorem 2.1 in [16]). See also Remark 4.10 in[11]. This is an important result in the theory of biharmonic submanifolds. Theorem 2. ([14], [16]) Let ¯ M n +1 be a Riemannian manifold and let M n bea hypersurface in ¯ M n +1 . Then M n is biharmonic if and only if B N and B T vanish, where B N = ∆ H − H | A | + H ¯ Ric ( N, N ) , and B T = 2 A ( ∇ H ) + H ∇ H − H (cid:0) ¯ Ric ( N ) (cid:1) T . Here we identify (1 , tensors and (0 , tensors, and ( · ) T is the projectionon T M n . In particular A ( ∇ H ) = n X i =1 A ( ∇ H, e i ) e i , and (cid:0) ¯ Ric ( N ) (cid:1) T = n X i =1 ¯ Ric ( N, e i ) e i , where { e i } ni =1 is a local orthonormal frame on M n .The next result is well known. We prove it for the sake of completeness. Lemma 3.
Let ¯ M n +1 be a Riemannian manifold and let M n be a hypersurfacein ¯ M n +1 . If ¯ f is a function on ¯ M n +1 and f = ¯ f | M n then ∇∇ f ( u, v ) = ¯ ∇ ¯ ∇ ¯ f ( u, v ) + (cid:10) ¯ ∇ ¯ f , N (cid:11) A ( u, v ) . In particular ∆ f = tr M ¯ ∇ ¯ ∇ ¯ f + (cid:10) ¯ ∇ ¯ f , N (cid:11) H. Proof.
Let { e i } be a local orthonormal frame on M such that ∇ e i e j = 0 at afixed point of M . At this point we have ∇∇ f ( e i , e j ) = e i e j f = e i e j ¯ f , and ¯ ∇ ¯ ∇ ¯ f ( e i , e j ) = e i e j ¯ f − D ¯ ∇ ¯ f , (cid:0) ¯ ∇ e i e j (cid:1) ⊥ E = e i e j ¯ f − (cid:10) ¯ ∇ ¯ f , N (cid:11) A ( e i , e j ) . The result follows from these equations.3he proof of Theorem 1 is based on next result, which may be of independentinterest.
Theorem 4.
Let ¯ M n +1 be a Riemannian manifold and let M n be a hypersurfacein ¯ M n +1 . If ¯ f is a function on ¯ M n +1 and f = ¯ f | M n then ∆∆ f = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) + Htr M (cid:0) ¯ ∇ N ¯ ∇ ¯ ∇ ¯ f (cid:1) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − H (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) + 2 ¯ ∇ ¯ ∇ ¯ f ( ∇ H, N ) + (cid:10) B N N − B T , ¯ ∇ ¯ f (cid:11) . Proof.
By Lemma 3 we have ∆ f = tr M ¯ ∇ ¯ ∇ ¯ f + (cid:10) ¯ ∇ ¯ f , N (cid:11) H. Taking the Laplacian we obtain ∆∆ f = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) + H ∆ (cid:10) ¯ ∇ ¯ f , N (cid:11) + 2 (cid:10) ∇ (cid:10) ¯ ∇ ¯ f , N (cid:11) , ∇ H (cid:11) + (cid:10) ¯ ∇ ¯ f , N (cid:11) ∆ H. Let { e i } be a local orthonormal frame on M . We have e i (cid:10) ¯ ∇ ¯ f , N (cid:11) = (cid:10) ¯ ∇ e i ¯ ∇ ¯ f , N (cid:11) + (cid:10) ¯ ∇ ¯ f , ¯ ∇ e i N (cid:11) = ¯ ∇ ¯ ∇ ¯ f ( e i , N ) − A ( e i , ∇ f ) . We find that ∆∆ f = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) + H ∆ (cid:10) ¯ ∇ ¯ f , N (cid:11) + 2 ¯ ∇ ¯ ∇ ¯ f ( ∇ H, N ) − A ( ∇ H, ∇ f ) + (cid:10) ¯ ∇ ¯ f , N (cid:11) ∆ H. We can assume that ∇ e i e j = 0 at a fixed point of M . At this point we have ∆ (cid:10) ¯ ∇ ¯ f , N (cid:11) = X i e i e i (cid:10) ¯ ∇ ¯ f , N (cid:11) = X i e i (cid:0) ¯ ∇ ¯ ∇ ¯ f ( e i , N ) (cid:1) − X i e i ( A ( e i , ∇ f ))= X i (cid:0) ¯ ∇ e i ¯ ∇ ¯ ∇ ¯ f (cid:1) ( e i , N ) + X i ¯ ∇ ¯ ∇ ¯ f (cid:16)(cid:0) ¯ ∇ e i e i (cid:1) ⊥ , N (cid:17) + X i ¯ ∇ ¯ ∇ ¯ f (cid:0) e i , ¯ ∇ e i N (cid:1) − X i ( ∇ e i A ) ( e i , ∇ f ) − X i A ( e i , ∇ e i ∇ f )= X i (cid:0) ¯ ∇ e i ¯ ∇ ¯ ∇ ¯ f (cid:1) ( e i , N ) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) − X i ( ∇ e i A ) ( e i , ∇ f ) − h∇∇ f, A i . By Lemma 3 we have ∆ (cid:10) ¯ ∇ ¯ f , N (cid:11) = X i (cid:0) ¯ ∇ e i ¯ ∇ ¯ ∇ ¯ f (cid:1) ( e i , N ) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) − X i ( ∇ e i A ) ( e i , ∇ f ) − (cid:10) ¯ ∇ ¯ f , N (cid:11) | A | .
4y the Ricci identity and the Codazzi equation we have (cid:0) ¯ ∇ e i ¯ ∇ ¯ ∇ ¯ f (cid:1) ( e i , N ) = (cid:0) ¯ ∇ e i ¯ ∇ ¯ ∇ ¯ f (cid:1) ( N, e i )= (cid:0) ¯ ∇ N ¯ ∇ ¯ ∇ ¯ f (cid:1) ( e i , e i ) − (cid:10) ¯ Riem ( e i , N ) e i , ¯ ∇ ¯ f (cid:11) , and ( ∇ e i A ) ( e i , ∇ f ) = ( ∇ e i A ) ( ∇ f, e i )= ( ∇ ∇ f A ) ( e i , e i ) + (cid:10) ¯ Riem ( e i , ∇ f ) e i , N (cid:11) . We find that ∆ (cid:10) ¯ ∇ ¯ f , N (cid:11) = tr M (cid:0) ¯ ∇ N ¯ ∇ ¯ ∇ ¯ f (cid:1) + ¯ Ric (cid:0) ¯ ∇ ¯ f , N (cid:1) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) − h∇ f, ∇ H i + ¯ Ric ( ∇ f, N ) − (cid:10) ¯ ∇ ¯ f , N (cid:11) | A | . We conclude that ∆∆ f = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) + Htr M (cid:0) ¯ ∇ N ¯ ∇ ¯ ∇ ¯ f (cid:1) + H ¯ Ric (cid:0) ¯ ∇ ¯ f , N (cid:1) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − H (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) − H h∇ f, ∇ H i + H ¯ Ric ( ∇ f, N ) − H (cid:10) ¯ ∇ ¯ f , N (cid:11) | A | + 2 ¯ ∇ ¯ ∇ ¯ f ( ∇ H, N ) − A ( ∇ H, ∇ f ) + (cid:10) ¯ ∇ ¯ f , N (cid:11) ∆ H = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) + Htr M (cid:0) ¯ ∇ N ¯ ∇ ¯ ∇ ¯ f (cid:1) + H ¯ ∇ ¯ ∇ ¯ f ( N, N ) − H (cid:10) ¯ ∇ ¯ ∇ ¯ f , A (cid:11) + 2 ¯ ∇ ¯ ∇ ¯ f ( ∇ H, N ) +
D(cid:16) ∆ H − H | A | (cid:17) N, ¯ ∇ ¯ f E − (cid:10) A ( ∇ H ) + H ∇ H, ¯ ∇ ¯ f (cid:11) + H ¯ Ric (cid:0) ¯ ∇ ¯ f , N (cid:1) + H ¯ Ric ( ∇ f, N ) . The result follows since ∆ H − H | A | = B N − H ¯ Ric ( N, N ) , A ( ∇ H ) + H ∇ H = B T + 2 H (cid:0) ¯ Ric ( N ) (cid:1) T , and ¯ Ric (cid:0) ¯ ∇ ¯ f , N (cid:1) = ¯ Ric ( ∇ f, N ) + (cid:10) ¯ ∇ ¯ f , N (cid:11) ¯ Ric ( N, N ) . Proof of Theorem 1.
Proof.
Since M is biharmonic, by Theorem 2 we have B N = 0 and B T = 0 . Let ¯ f = h X, V i | S n +1 , where X is the position vector of R n +2 and V is a fixedvector in R n +2 . We have ¯ ∇ ¯ ∇ ¯ f = − ¯ f ¯ g, ¯ ∇ ¯ ∇ ¯ ∇ ¯ f = − d ¯ f ⊗ ¯ g, where ¯ g = g S n +1 . Let f = h X, V i | M . By Theorem 4 we have ∆∆ f = ∆ (cid:0) tr M ¯ ∇ ¯ ∇ ¯ f (cid:1) − nH (cid:10) ¯ ∇ ¯ f , N (cid:11) + H f. Integrating and using the divergence theorem we obtain − Z M nH (cid:10) ¯ ∇ ¯ f , N (cid:11) + Z M H f. By Lemma 3 we have ∆ f = − nf + (cid:10) ¯ ∇ ¯ f , N (cid:11) H. Multiplying by n , integrating and using the divergence theorem we obtain − Z M n f + Z M nH (cid:10) ¯ ∇ ¯ f , N (cid:11) . By the above equations we have Z M (cid:0) n − H (cid:1) f = 0 . Since M lies in a closed hemisphere of S n +1 we can find V in R n +2 such that f ≥ . Since n − H does not change sign we have (cid:0) n − H (cid:1) f = 0 . First suppose H = n everywhere. In this case by Theorem 2.10 in [1] weconclude that M is the small hypersphere S n (cid:0) / √ (cid:1) . Now suppose H = n at some point of M . In this case we have f = 0 on an open subset of M , thatis, an open subset of M lies in the equator of the hemisphere. Since this subsetis minimal, by Theorem 1.3 in [3] we find that the whole M is minimal. Weconclude that M is the equator of the hemisphere. Let M n be a hypersurface in S n +1 and let x i = h X, E i i | M n , where X is the position vector of R n +2 and { E i } n +2 i =1 is the standard basis of R n +2 .Takahashi [17] proved that M n is minimal if and only if ∆ x i = − nx i for ≤ i ≤ n + 2 . As an application of Theorem 4 we obtain a similar result forbiharmonic hypersurfaces. 6 heorem 5. Let M n be a hypersurface in S n +1 . Then M n is biharmonic ifand only if ∆∆ x i = (cid:0) n + H (cid:1) x i − nH h N, E i i f or ≤ i ≤ n + 2 . Proof.
Let ¯ f = h X, E i i | S n +1 . We have ¯ ∇ ¯ ∇ ¯ f = − ¯ f ¯ g, and ¯ ∇ ¯ ∇ ¯ ∇ ¯ f = − d ¯ f ⊗ ¯ g, where ¯ g = g S n +1 . Let f = h X, E i i | M . By Theorem 4 and Lemma 3 we have ∆∆ f = − n ∆ f − nH (cid:10) ¯ ∇ ¯ f , N (cid:11) + H f + (cid:10) B N N − B T , ¯ ∇ ¯ f (cid:11) , and ∆ f = − nf + (cid:10) ¯ ∇ ¯ f , N (cid:11) H. We find that ∆∆ f = (cid:0) n + H (cid:1) f − nH (cid:10) ¯ ∇ ¯ f , N (cid:11) + (cid:10) B N N − B T , ¯ ∇ ¯ f (cid:11) . Since ¯ ∇ ¯ f = proj T S n +1 E i we have ∆∆ x i = (cid:0) n + H (cid:1) x i − nH h N, E i i + (cid:10) B N N − B T , E i (cid:11) f or ≤ i ≤ n + 2 . The result follows from Theorem 2.This result was obtained in a completely different way by Caddeo-Montaldo-Oniciuc (Proposition 4.1 in [5]). The statement is different but equivalent.As an application of Theorem 5 we obtain a sufficient condition for a bihar-monic hypersurface in S n +1 to be minimal. Corollary 6.
Let M n be a biharmonic hypersurface in S n +1 . If there exists afunction φ on M n such that ∆∆ x i = φx i f or ≤ i ≤ n + 2 , then M n is minimal.Proof. By Theorem 5 we have φ h X, E i i = (cid:0) n + H (cid:1) h X, E i i − nH h N, E i i = f or ≤ i ≤ n + 2 . We find that nHN = (cid:0) n + H − φ (cid:1) X. The result follows since N is tangent to S n +1 and X is normal to S n +1 .7 eferenceseferences