HYPERSURFACES IN A UNIT SPHERE S^{n+1}(1)

WITH CONSTANT SCALAR CURVATURE*

Qing-Ming Cheng

Abstract. In this paper, we consider n-dimensional hypersurfaces with constant scalar curvature of a unit sphere S^{n+1}(1). We characterize the hyersurface S^{k}(c_{1}) × S^{n−k}(c_{2}) in a unit sphere S^{n+1}(1) and show that there exist many compact hypersurfaces with constant scalar curvature in a unit sphere S^{n+1}(1) which are not congruent to each other in it. In particular, we prove that let M is an n-dimensional (n > 3) complete locally conformally flat hypersurface with constant scalar curvature n(n − 1)r in a unit sphere S^{n+1}(1), then r > 1 − \frac{2}{n} and (1) when r ≠ \frac{n−2}{n−1} ,

if S ≥ (n − 1) \frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} , then M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c), where S is the squared norm of the second fundamental form of M; (2) there are no complete hypersurfaces in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures one of which is simple such that r = \frac{n−2}{n−1} and

S > (n − 1) \frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n.

1. Introduction

Let M be an n-dimensional hypersurface in a unit sphere S^{n+1}(1) of dimension n + 1. In [5], S.Y. Cheng and Yau proved that let M be an n-dimensional compact hypersurface with constant scalar curvature n(n − 1)r, if r ≥ 1 and the sectional curvature of M is non-negative, then M is isometric to either a totally umbilical hypersurface or a Riemannian product S^{k}(c_{1})×S^{n−k}(c_{2}) 1 ≤ k ≤ n−1, where S^{k}(c) denotes the sphere of radius c. By making use of the similar method which was used by Nakagawa and the author in [4] and a differential operator introduced by S.Y. Cheng and Yau, Li [9] proved that let M be an n-dimensional compact hypersurface with constant scalar curvature n(n−1)r, if r ≥ 1 and S ≤ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to either a totally umbilical hypersurface or the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) with c^{2} = \frac{n−2}{nr} ≤ \frac{n−2}{n}, where S is the squared norm of the second fundamental form of M. We should notice that the condition

Key words and phrases. hypersurface with constant scalar curvature, principal curvature and locally conformally flat hypersurface.

*Research partially supported by a Grant-in-Aid for Scientific Research from the Ministry of Education, Science and Culture and by a Grant-in-Aid for Scientific Research from Josai University.

1991 Mathematics Subject Classification 53C20, 53C42

Typeset by AMS-TEX

r ≥ 1 plays an essential role in their proofs of Theorems. On the other hand, for any 0 < c < 1, by considering the standard immersions S^{n−1}(c) ⊂ R^{n}, S^{1}(\sqrt{1 − c^{2}}) ⊂ R^{2} and taking the Riemannian product immersion S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) ↪ R^{2} × R^{n}, we obtain a compact hypersurface S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) in S^{n+1}(1) with constant scalar curvature n(n − 1)r, where r > 1 − \frac{2}{n}. Hence, not all Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) appear in these results of S. Y. Cheng and Yau [5] and Li [9]. Since the Riemannian product S^{1}(\sqrt{1 − c^{2}})×S^{n−1}(c) has only two distinct principal curvatures and its scalar curvature n(n − 1)r is constant and satisfies r > 1 − \frac{2}{n}. Hence, we would like to ask the following:

Problem 1. Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r in S^{n+1}(1). If M is of only two distinct principal curvatures one of which is simple, then, does r > 1 − \frac{2}{n} hold?

In section 3, we give an affirmative answer for this Problem 1 (see. Theorem 3.1). And if r ≠ \frac{n−2}{n−1}, we prove that S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) is the only complete hypersurfaces in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures one of which is simple such that S ≥ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}.

Furthermore, when r = \frac{n−2}{n−1}, we infer that S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n and we prove that there exist no complete hypersurfaces in S^{n+1}(1) with constant scalar curvature n(n−1)r and with two distinct principal curvatures one of which is simple such that S > (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n. On the other hand, Clifford torus

S^{1}(\sqrt{\frac{1}{n}})×S^{n−1}(\sqrt{\frac{n−1}{n}}) is a compact minimal hypersurface in S^{n+1}(1) with r = \frac{n−2}{n−1} and with S = n. Hence, we can consider the following:

Problem 2. Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r (r = \frac{n−2}{n−1}) in S^{n+1}(1). If M is of only two distinct principal

curvatures one of which is simple, then is M isometric to the Clifford torus S^{1}(\sqrt{\frac{1}{n}})×

S^{n−1}(\sqrt{\frac{n−1}{n}})?

This Problem 2 is still open.

From our Theorem 3.1, Theorem 3.2 in the Section 3 and the assertion above, it is interesting to generalize these results due to S.Y. Cheng and Yau [5] and Li [9] to the case r > 1 − \frac{2}{n}. That is, it is interesting to prove the following:

Problem 3. Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r in S^{n+1}(1). If r > 1 − \frac{2}{n} and S ≤ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to either a totally umbilical hypersurface or the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

When r = \frac{n−2}{n−1}, we answer this Problem 3 affirmatively (see. Theorem 3.3). For the general case, we can not answer it yet.

On the other hand, every compact locally conformally flat hypersurface in a unit sphere S^{n+1}(1) (n > 3) is conformally equivalent to the standard sphere or a classical Schottky manifold (cf. Suyama [12] and Pinkall [11]). We know that the standard sphere and Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) are compact locally conformally flat hypersurfaces in a unit sphere S^{n+1}(1) with constant scalar curvature. It is natural to ask whether there exist compact locally conformally flat hypersurfaces in a unit sphere S^{n+1}(1) with constant scalar curvature which are not congruent to the standard sphere and Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). In this paper, we consider n-dimensional hypersurfaces with constant scalar curvature of a unit sphere S^{n+1}(1). We show that there exist many compact hypersurfaces with constant scalar curvature in a unit sphere S^{n+1}(1) which are not congruent to each other in it and characterize the hyersurface S^{k}(c_{1}) × S^{n−k}(c_{2}) in a unit sphere S^{n+1}(1). In particular, we prove that if M is an n-dimensional (n > 3) complete locally conformally flat hypersurface with constant scalar curvature n(n − 1)r in a unit sphere S^{n+1}(1), then r > 1 − \frac{2}{n} and when r ≠ \frac{n−2}{n−1}, if S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c), where S is the squared norm of the second fundamental form of M.

2. Hypersurfaces with two distinct principal curvatures and with constant scalar curvature in a unit sphere S^{n+1}(1)

Let M be an n-dimensional hypersurface in a unit sphere S^{n+1}(1) with constant scalar curvature n(n − 1)r. We take a local orthonormal frame field {e_{1}, …, e_{n+1}} in S^{n+1}(1), restricted to M, so that e_{1}, …, e_{n} are tangent to M. Let ω_{1}, ⋯, ω_{n+1} denote the dual coframe fields in S^{n+1}(1). We shall make use of the following convention on the ranges of indices:

1 ≤ A, B, C, ⋯, ≤ n + 1; 1 ≤ i, j, k, ⋯, ≤ n.

Then the structure equations of S^{n+1}(1) are given by

dω_{A} = ∑_{B=1}^{n+1} ω_{AB} ∧ ω_{B}, ω_{AB} + ω_{BA} = 0,

dω_{AB} = ∑_{C=1}^{n+1} ω_{AC} ∧ ω_{CB} + Ω_{AB}, Ω_{AB} = −\frac{1}{2} ∑_{C,D=1}^{n+1} R_{ABCD}ω_{C} ∧ ω_{D},

R_{ABCD} = (δ_{AC}δ_{BD} − δ_{AD}δ_{BC}),

where Ω_{AB} (resp. R_{ABCD}) denotes the curvature form (resp. the components of the curvature tensor) of S^{n+1}(1). Then, in M,

ω_{n+1} = 0.

It follows from Cartan’s Lemma that

(2.1)  ω_{n+1i} = ∑_{j} h_{ij}ω_{j}, h_{ij} = h_{ji}.

The second fundamental form α and the mean curvature of M are defined by

α = ∑_{i,j} h_{ij}ω_{i}ω_{j}e_{n+1} and nH = ∑_{i} h_{ii},

respectively. M is said to be totally umbilical if the h_{ij} can be expressed as h_{ij} = Hδ_{ij}. The structure equations of M are given by

dω_{i} = ∑_{k=1}^{n} ω_{ik} ∧ ω_{k}, ω_{ij} + ω_{ji} = 0,

dω_{ij} = ∑_{k=1}^{n} ω_{ik} ∧ ω_{kj} + Ω_{ij}, Ω_{ij} = −\frac{1}{2} ∑_{k,l=1}^{n} R_{ijkl}ω_{k} ∧ ω_{l},

(2.2)  R_{ijkl} = (δ_{ik}δ_{jl} − δ_{il}δ_{jk}) + (h_{ik}h_{jl} − h_{il}h_{jk}),

where Ω_{ij} (resp. R_{ijkl}) denotes the Riemannian curvature form (resp. the components of the Riemannian curvature tensor) of M. From the above equation, we have

(2.3)  n(n − 1)r = n(n − 1) + n^{2}H^{2} − S

where n(n − 1)r is the scalar curvature of M and S = ∑_{ij=1}^{n} h_{ij}^{2} is the squared norm of the second fundamental form of M.

In [10], Otsuki proved the following:

Theorem ( Otsuki [10]). Let M be a hypersurface in a unit sphere S^{n+1}(1) such that the multiplicities of the principal curvatures are constant. Then the distribution of the space of the principal vectors correspending to each principal curvature is completely integrable. In particular, if the multiplicity of a principal curvature is greater than 1, then this principal curvature is constant on each integral submanifold of the corresponding to distribution of the space of the principal vectors.

From the above theorem due to Otsuki, we have

Corollary 1. Let M be an n-dimensional compact hypersurface in a unit sphere S^{n+1}(1) with constant scalar curvature and with two distinct principal curvatures. If the multiplicities of these two distinct principal curvatures are greater than 1, then M is isometric to Riemannian product S^{k}(c_{1}) × S^{n−k}(c_{2}), 1 < k < n − 1.

Proof. Since M has only two distinct principal curvatures and the scalar curvature is constant, from the Theorem due to Otsuki, we obtain that these two distinct principal curvatures are constant on M. Hence, Corollary 1 is true from a result due to E. Cartan [1].

From now on, we assume that M is an n-dimensional complete hypersurface with constant scalar curvature and with two distinct principal curvatures in S^{n+1}(1). If multiplicities of these two distinct principal curvatures are all greater than 1, from Corollary 1 we know that M is isometric to Riemannian product S^{k}(c_{1}) × S^{n−k}(c_{2}), 1 < k < n − 1. Hence, we shall assume that one of these two distinct principal curvatures is simple, that is, we assume

λ_{1} = λ_{2} = ⋯ = λ_{n−1} = λ and λ_{n} = μ,

where λ_{i} for i = 1, 2, ⋯, n are the principal curvatures of M. Since the scalar curvature n(n − 1)r is constant, from Gauss equation (2.3), we obtain

(2.4)  n(n − 1)(r − 1) = (n − 1)(n − 2)λ^{2} + 2(n − 1)λμ.

If λ = 0 at some point, then r = 1 at this point. Since r is constant, we know r = 1 on M. Since these principal curvatures λ and μ are continuous on M, from (2.4) and r = 1 we obtain λ = 0 on M. Hence, the sectional curvature of M is not less than 1 from the Gauss equation (2.2). According to the result due to S.Y. Cheng and Yau in [5], we know that M is a totally umbilical hypersurface. Hence, we can assume λ ≠ 0 on M. Thus, (2.4) yields

(2.5)  μ = \frac{n(r − 1)}{2λ} − \frac{n − 2}{2}λ.

From the formula

λ − μ = n\frac{λ^{2} − (r − 1)}{2λ} ≠ 0,

we know λ^{2} − (r − 1) ≠ 0. If λ^{2} − (r − 1) < 0, from the above formula we know

λ^{2} − μλ = \frac{n}{2}\{λ^{2} − (r − 1)\} < 0,

and r > 1. Hence,

μλ > λ^{2}.

According to the Gauss equation (2.2), we know that the sectional curvature of M is greater than 1 + λ^{2} ≥ 1 and r > 1. Thus, M is a totally umbilical hypersurface. This is impossible because we assumed that M is of two distinct principal curvatures. Hence, λ^{2} − (r − 1) > 0. Put w = \{λ^{2} − (r − 1)\}^{− \frac{1}{n}}. By making use of the similar methods in [10], we can prove the following:

Theorem 2.1. If M is an n-dimensional hypersurface (n > 2) in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures and the space of principal vectors corresponding to one of them is of one dimension, then M is a locus of the moving n − 1-dimensional submanifold M_{1}^{n−1}(s) along which the principal curvature λ of multiplicity n − 1 is constant and which is locally isometric to an n − 1-dimensional sphere S^{n−1}(c(s)) = E^{n}(s) ∩ S^{n+1}(1) of constant

curvature \{^{\frac{d\{log |λ^{2}−(r−1)|^{\frac{1}{n}} \}}{ds}}\}^{2} + λ^{2} + 1 and w = \{λ^{2} − (r − 1)\}^{− \frac{1}{n}} satisfies the ordinary differential equation of order 2

(2.6)  \frac{d^{2}w}{ds^{2}} − w(\frac{n − 2}{2} \frac{1}{w^{n}} − r) = 0.

Where E^{n}(s) is an n-dimensional linear subspace in the Euclidean space R^{n+2} which are parallel to a fixed E^{n}.

By integrating the equation (2.6), we have

(2.7)  (\frac{dw}{ds})^{2} = C − rw^{2} − \frac{1}{w^{n−2}} ,

where C is the integral constant. The same assertion as in [10] infers the following:

Corollary 2. In S^{n+1}(1), there exist many hypersurfaces with constant scalar curvature which are not congruent to each other in it.

Remark 1. By making use of the same construction as in Otsuki [10], we can prove that there exist many compact hypersurfaces in S^{n+1}(1) with constant scalar curvature of the type in Theorem 2.1 which are not congruent with each other in it.

3. A characterization of Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c)

In this section, we shall characterize the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). At first, we give an affirmative answer for this Problem 1 which is stated in the Introduction and if r ≠ \frac{n−2}{n−1}, we prove that if S ≥ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) is the only complete hypersurfaces in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures one of which is simple. Furthermore, when r = \frac{n−2}{n−1}, we infer that S ≥ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n and we prove that there exist no complete hypersurfaces in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures one of which is simple such that S > (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n.

Theorem 3.1. Let M be an n-dimensional complete hypersurface in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures one of which is simple. Then r > 1 − \frac{2}{n} and, when r ≠ \frac{n−2}{n−1}, if S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

Corollary 3. Let M be an n-dimensional complete hypersurface in S^{n+1}(1) with constant scalar curvature n(n − 1)r and with two distinct principal curvatures. If r ≠ \frac{n−2}{n−1} and S ≥ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to the Riemannian product S^{k}(c_{1}) × S^{n−k}(c_{2}) 1 ≤ k ≤ n − 1.

Proof. According to Corollary 1 and Theorem 3.1, we know that Corollary 3 is true obviously.

Every n-dimensional (n > 3) compact locally conformally flat hypersurface in a unit sphere S^{n+1}(1) (n > 3) is conformally equivalent to the standard sphere or a classical Schottky manifold (cf. Suyama [12] and Pinkall [11]). We know that the standard sphere and Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) are compact locally conformally flat hypersurfaces in a unit sphere S^{n+1}(1) with constant scalar curvature. In the Section 2, we showed that there exist many locally conformally flat hypersurfaces in a unit sphere S^{n+1}(1) with constant scalar curvature which are not congruent to the standard sphere and Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). The following Corollary 4 give a characterization of the class of hypersurfaces.

Corollary 4. Let M be an n-dimensional (n > 3) complete locally conformally flat hypersurface with constant scalar curvature n(n − 1)r in S^{n+1}(1). Then r > 1 − \frac{2}{n} and, when r ≠ \frac{n−2}{n−1}, if S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

Proof. Since M is a locally conformally flat hypersurface in S^{n+1}(1) and n > 3, from a result in [7], we know that M is of at most two distinct principal curvatures and one of them is of at least n − 1 multiplicities. If at some point p, these principal curvatures are equal with each other, then S = nH^{2} at this point p. From the Gauss equation, we have S = nH^{2} = n(r − 1) ≥ 0. Since S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2},

we have n(r − 1) ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}. This is impossible. Hence M is of two distinct principal curvatures and one of them is of n − 1 multiplicities. From the Theorem 3.1, we know that M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). That is, Corollary 4 is proved.

Theorem 3.2.

(1) Let M be a complete hypersurface with two distinct principal curvatures one of which is simple in S^{n+1}(1). If r = \frac{n−2}{n−1}, then S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n and if S = n on M, then M is isometric to the Clifford torus

S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}});

(2) There are no complete hypersurfaces with two distinct principal curvatures one of which is simple in S^{n+1}(1) such that r = \frac{n−2}{n−1} and S > (n−1)\frac{n(r−1)+2}{n−2} +\frac{n−2}{n(r−1)+2} = n.

Remark 2. We know that Clifford torus S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}) is a complete mini-

mal hypersurface in S^{n+1}(1) such that r = \frac{n−2}{n−1} and S = (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n. But we do not know whether a complete hypersurface in S^{n+1}(1) with two distinct principal curvatures one of which is simple and with r = \frac{n−2}{n−1} is only the Clifford

torus S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}).

Corollary 5. There are no complete locally conformally flat n-dimensional (n > 3) hypersurfaces in S^{n+1}(1) such that r = \frac{n−2}{n−1} and S > (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}.

7

Proof. By making use of Theorem 3.2 and the same argument in Corollary 4, we know that Corollary 5 is true.

Remark 3. It is well-known that a locally conformally flat hypersurface in S^{n+1}(1) (n > 3) is of at most two distinct principal curvatures. But when n = 3, this assertion is not true. Hence, we would like to ask whether these assertions in Corollary 4 and Corollary 5 hold when n = 3. But we think that under the condition that scalar curvature of M is constant, it should be true that a locally conformally flat hypersurface in S^{4}(1) is of at most two distinct principal curvatures. Hence, we conjecture that these assertions in Corollary 4 and Corollary 5 hold.

In order to prove the Theorem 3.1, we shall prepare several Lemmas.

Lemma 1. If M is of two distinct principal curvatures and λ is the principal curvature of n − 1 multiplicities, then S = (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if w^{−n} = λ^{2} − r + 1 = \frac{2r}{n−2}.

Proof. From the definition of S and (2.5), we have

S = (n − 1)λ^{2} + μ^{2}

= (n − 1)λ^{2} + (\frac{n(r − 1)}{2λ} − \frac{n − 2}{2}λ)^{2}

= (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}.

Hence, λ is constant. Since w = \{λ^{2} − (r − 1)\}^{− \frac{1}{n}} is also constant, from (2.6) we know that \frac{n−2}{2} \frac{1}{w^{n}} − r = 0. Therefore, w^{−n} = λ^{2} − (r − 1) = \frac{2r}{n−2}. Lemma 1 is proved.

Lemma 2. If M is of two distinct principal curvatures and λ is the principal curvature of n − 1 multiplicities, we obtain the following:

(1) When r − 1 ≥ 0, S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if w^{−n} = λ^{2} − r + 1 ≥ \frac{2r}{n−2}.

(2) When 1 − \frac{2}{n} < r < 1 and r ≠ \frac{n−2}{n−1}, S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if

and only if one of the following conditions holds (a). w^{−n} = λ^{2} − r + 1 ≥ \frac{2r}{n−2} and λ^{2} > 1 − r (b). w^{−n} = λ^{2} − r + 1 ≤ \frac{2r}{n−2} and λ^{2} < 1 − r.

Proof. At first, we consider a function

f(t) = \frac{n^{2}}{4} t + \frac{n^{2}(r − 1)^{2}}{4t} − \frac{n(n − 2)}{2}(r − 1)

for t > 0. Since

\frac{df(t)}{dt} = \frac{n^{2}}{4} − \frac{n^{2}(r − 1)^{2}}{4t^{2}} ,

we know that t^{2} ≥ (r − 1)^{2} holds if and only if f(t) is an increasing function and t^{2} ≤ (r − 1)^{2} holds if and only if f(t) is a decreasing function. And f(t) obtains its minimum at t^{2} = (r − 1)^{2}.

From the arguments in the Section 2, we know λ^{2} −(r −1) > 0 and we can assume λ > 0. Since

(3.1)  S = (n − 1)λ^{2} + μ^{2}

= (n − 1)λ^{2} + (\frac{n(r − 1)}{2λ} − \frac{n − 2}{2}λ)^{2}

= \frac{n^{2}}{4} λ^{2} + \frac{n^{2}(r − 1)^{2}}{4λ^{2}} − \frac{n(n − 2)}{2}(r − 1),

we conclude that S is an increasing function of λ^{2} if λ^{2} > |r−1| and S is a decreasing function of λ^{2} if 0 < λ^{2} < |r − 1|.

(a). If r − 1 ≥ 0, we know that S is an increasing function of λ^{2} because of λ^{2} > r − 1 = |r − 1|. Hence, from the Lemma 1 and the property that S is an increasing function of λ^{2}, we obtain that S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if w^{−n} = λ^{2} − r + 1 ≥ \frac{2r}{n−2}.

(b). In the case that 1 − \frac{2}{n} < r < 1 and r ≠ \frac{n−2}{n−1}. According to the arguments in Section 2, we know that λ^{2} is a continuous function of s. If there exist two points s_{1} and s_{2} such that λ^{2}(s_{1}) > 1 − r and λ^{2}(s_{2}) < 1 − r, then there must exist a point s_{0} such that λ^{2}(s_{0}) = 1 − r. we substitute λ^{2}(s_{0}) = 1 − r into (3.1), we obtain

S = n(n − 1)(1 − r) < (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}

because of r ≠ \frac{n−2}{n−1}. Hence, in this case,

S ≥ (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}

holds if and only if only one of λ^{2} > 1 − r and λ^{2} < 1 − r holds. If λ^{2} > 1 − r, S is an increasing function of λ^{2}. Hence, S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if w^{−n} = λ^{2} − r + 1 ≥ \frac{2r}{n−2}. If λ^{2} < 1 − r, S is a decreasing function of λ^{2}. Hence, S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if w^{−n} = λ^{2} − r + 1 ≤ \frac{2r}{n−2}. We finished the proof of Lemma 2.

Lemma 3. Let M be of two distinct principal curvatures and let λ denote the principal curvature of n − 1 multiplicities. If r ≤ 1 − \frac{2}{n}, then \frac{dw(s)}{ds} is a monotone increasing function of s.

Proof. According to the argument in the Section 2, we know that w^{−n} = λ^{2} − r + 1 satisfies the equation (2.6), that is,

(3.2)  \frac{d^{2}w}{ds^{2}} − w(\frac{n − 2}{2} \frac{1}{w^{n}} − r) = 0.

Hence,

\frac{d^{2}w}{ds^{2}} = w(\frac{n − 2}{2} \frac{1}{w^{n}} − r)

= w\{\frac{n − 2}{2}(λ^{2} − r + 1) − r\}

= w\{\frac{n − 2}{2}λ^{2} + \frac{(n − 2) − nr}{2} \} > 0

because of r ≤ 1 − \frac{2}{n} and λ^{2} > 0. Hence, \frac{dw(s)}{ds} is a monotone increasing function of s. The proof of Lemma 3 is finished.

Proposition. Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r and with two distinct principal curvatures in S^{n+1}(1). Let λ denote the principal curvature of n − 1 multiplicities. Then w(s) is constant if \frac{dw(s)}{ds} is a monotone function of s.

Proof. From the assumption and by making use of the similar computations as in the section 3 of [10], we can obtain

ω_{jn} = \frac{d\{log |λ^{2} − (r − 1)|^{\frac{1}{n}} \}}{ds}ω_{j}.

Hence, ∇_{e_{n}}e_{n} = ∑_{k=1}^{n} ω_{ni}(e_{n})e_{i} = 0. From the definition of geodesic, we know that any integral curve of principal vector field corresponding to μ is a geodesic. If w is not constant, we shall obtain a contradiction. In fact, since w^{−n} = λ^{2} −r+1 satisfies the equation (2.6), that is

\frac{d^{2}w}{ds^{2}} − w(\frac{n − 2}{2} \frac{1}{w^{n}} − r) = 0,

we know that w(s) is a function defined in (−∞, +∞) because M is complete and the integral curve of principal vector field corresponding to μ is a geodesic. Integrating the equation (2.6), we have

(3.3)  (\frac{dw}{ds})^{2} = C − rw^{2} − \frac{1}{w^{n−2}} ,

where C is an integral constant. Hence, we must have

(3.4)  C ≥ min_{w>0}(rw^{2} + \frac{1}{w^{n−2}} ) = \frac{n}{2}(\frac{2r}{n − 2})^{\frac{n−2}{n}} ,

since g(w) = rw^{2} + \frac{1}{w^{n−2}} is an increasing function if 0 < w^{−n} < \frac{2r}{n−2} and g(w) = rw^{2} + \frac{1}{w^{n−2}} is a decreasing function if w^{−n} > \frac{2r}{n−2} and g(w) = rw^{2} + \frac{1}{w^{n−2}} obtains its minimum \frac{n}{2}(\frac{2r}{n−2})^{\frac{n−2}{n}} at w^{−n} = \frac{2r}{n−2}. And C = \frac{n}{2}(\frac{2r}{n−2})^{\frac{n−2}{n}} if and only if w is constant. Thus we infer that, for any non-constant solution w of (2.6),

(3.5)  C > \frac{n}{2}(\frac{2r}{n − 2})^{\frac{n−2}{n}} .

Since g(w) = rw^{2} + \frac{1}{w^{n−2}} is an increasing function if 0 < w^{−n} < \frac{2r}{n−2} and g(w) = rw^{2} + \frac{1}{w^{n−2}} is a decreasing function if w^{−n} > \frac{2r}{n−2} and C > min g(w), we obtain that the following equation has two distinct positive roots w_{1}, w_{2}:

rw^{2} + \frac{1}{w^{n−2}} − C = 0.

And

(3.6)  w_{1} < (\frac{2r}{n − 2})^{− \frac{1}{n}} < w_{2}.

Since

(\frac{dw}{ds})^{2} = C − rw^{2} − \frac{1}{w^{n−2}} ,

and \frac{dw}{ds} is monotone, we know that \frac{dw}{ds} has at most one zero point in (−∞, +∞).

(a). If \frac{dw}{ds} has no zero point in (−∞, +∞), then w(s) is a monotone function of s in (−∞, +∞).

(b). If \frac{dw}{ds} has only one zero point in (−∞, +∞), denoting it by s_{0}, then w(s) is a monotone function of s both in (−∞, s_{0}] and [s_{0}, +∞).

In any case, w(s) is a monotone function of s both in (−∞, s_{0}] and in [s_{0}, +∞). Since

(\frac{dw}{ds})^{2} = C − rw^{2} − \frac{1}{w^{n−2}} ≥ 0,

we know that w(s) is bounded, that is, w_{1} ≤ w(s) ≤ w_{2}. Without loss of generality, we can assume that \frac{dw(s)}{ds} is a monotone increasing function. Then for some s_{0}, from the mean value theorem, we have that for any sϵ[s_{0}, +∞), there exists a tϵ[s_{0}, s] such that

\frac{dw(t)}{ds} = \frac{w(s) − w(s_{0})}{s − s_{0}}.

Hence,

lim_{s→∞} \frac{dw(s)}{ds} = 0

holds because \frac{dw(s)}{ds} is a monotone increasing function and w(s) is bounded. Thus, we infer \frac{dw(s)}{ds} < 0 for any sϵ(−∞, +∞) and sup \frac{dw(s)}{ds} = 0. By making use of the same argument, we consider \frac{dw(s)}{ds} on sϵ(−∞, s_{0}] and we can conclude inf \frac{dw(s)}{ds} = 0. This is impossible because \frac{dw(s)}{ds} < 0 for any sϵ(−∞, +∞) and \frac{dw(s)}{ds} is a monotone increasing function. Hence, we infer that w(s) is constant. Thus, the proof of Proposition is finished.

Proof of Theorem 3.1. From Lemma 3 and Proposition, we have r > 1 − \frac{2}{n}. In fact, if r ≤ 1 − \frac{2}{n} holds, then w as well as λ must be constant. From the result of E. Cartan [1], we obtain that M is isometric to S^{1}(\sqrt{1 − c^{2}})×S^{n−1}(c) for some constant 0 < c < 1, which infers r > 1 − \frac{2}{n}. This is a contradition. Therefore, r > 1 − \frac{2}{n} holds. If r ≠ \frac{n−2}{n−1} and S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} are satisfied, from Lemma 2,

we infer that \frac{d^{2}w}{ds^{2}} ≥ 0 is valid in the case r ≥ 1 and one of \frac{d^{2}w}{ds^{2}} ≥ 0 and \frac{d^{2}w}{ds^{2}} ≤ 0 is valid in the case 1 − \frac{2}{n} < r < 1. Hence, either \frac{dw}{ds} is an increasing function of s or \frac{dw}{ds} is a decreasing function of s. Proposition yields that w as well as λ must be constant. From the result of E. Cartan [1], we conclude that Theorem 3.1 is true.

Proof of Theorem 3.2.

(1). Since r = \frac{n−2}{n−1}, we have

S = (n − 1)λ^{2} + μ^{2}

= (n − 1)λ^{2} + (\frac{n(r − 1)}{2λ} − \frac{n − 2}{2}λ)^{2}

= \frac{n^{2}}{4} λ^{2} + \frac{n^{2}(r − 1)^{2}}{4λ^{2}} − \frac{n(n − 2)}{2}(r − 1)

= \frac{n^{2}}{4} (λ − \frac{1}{(n − 1)λ})^{2} + n

≥ n.

And S = n at some point if and only if λ − \frac{1}{(n−1)λ} = 0, that is, λ^{2} = \frac{1}{n−1} at this point. From (2.3), we know that M is minimal at this point. Hence, if S = n on M holds, then M is a minimal hypersurface. From the result due to Chern, do Carmo and Kobayashi [6] (cf. [3]), we know that M is isometric to the Clifford torus

S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}).

(2). Since r = \frac{n−2}{n−1}, by making use of the same proof as one in Theorem 3.1, we know that S > (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds if and only if one of the following conditions holds

(a). w^{−n} = λ^{2} − r + 1 > \frac{2r}{n−2} and λ^{2} > 1 − r

(b). w^{−n} = λ^{2} − r + 1 < \frac{2r}{n−2} and λ^{2} < 1 − r.

and S = (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n holds if and only if λ^{2} = 1 − r holds.

Since S > (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds and λ^{2} is a continuous function of s, we conclude that λ^{2} ≠ 1 − r. Hence, we infer that either \frac{dw}{ds} is an increasing function of s or \frac{dw}{ds} is a decreasing function of s if w is not constant. Hence, we infer that w(s) is constant from Proposition. From (2.6) we have λ^{2} = \frac{1}{n−1}. Hence, S = (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} = n. This is impossible. The proof of Theorem 3.2 is completed.

From Theorem 3.2 (1), we have the following:

Theorem 3.3. Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r (r = \frac{n−2}{n−1}) in S^{n+1}(1). If S ≤ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2},

then M is isometric to the Clifford torus S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}).

Acknowledgement. The author would like to thank professor K. Shiohama and professor S. Ishikawa for their valuable suggestions. The author would also like to express his gratitude to the referee for valuable comments.

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Qing-Ming Cheng Department of Mathematics, Faculty of Science and Engineering, Saga University, Saga 840-8502, Japan e-mail: cheng@ms.saga-u.ac.jp