COMPACT HYPERSURFACES IN A UNIT SPHERE

QING-MING CHENG^{1}, SHICHANG SHU^{2} AND YOUNG JIN SUH^{3}

Abstract. It is our purpose to study curvature structures of compact hypersurfaces in the unit sphere S^{n+1}(1) with two distinct principal curvatures. First of all, we proved that the Riemannian product S^{1}(\sqrt{1 − c^{2}}) ×S^{n−1}(c) is the only compact hypersurfaces in S^{n+1}(1) with two distinct principal curvatures one of which is simple and satisfying r > 1 − \frac{2}{n}, r ≠ \frac{n−2}{n−1} and S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, where n(n − 1)r is the scalar curvature of hypersurfaces and c^{2} = \frac{n−2}{nr}. This generalized the result in Cheng [4], where the scalar curvature is constant is assumed. Secondly, we proved that the Riemannian product S^{1}(\sqrt{1 − c^{2}}) ×S^{n−1}(c) is the only compact hypersurfaces with nonzero mean curvature in S^{n+1}(1) with two distinct principal curvatures one of which is simple and satisfying r > 1 − \frac{2}{n} and S ≤ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}. This gives a partial anwser for the problem proposed in [4].

1. Introduction

Let M be an n-dimensional hypersurface in a unit sphere S^{n+1}(1) of dimension n+1. It is well known that a compact minimal hypersurface with S = n in S^{n+1}(1) is isometric to a Clifford torus S^{1}(\sqrt{\frac{1}{n}})×S^{n−1}(\sqrt{\frac{n−1}{n}}), where S is the squared norm of the second fundamental form of the hypersurface (cf. [6], [7] and [10]). Furthermore, Otsuki [12] investigated the converse problem. He proved that Riemannian prod-

uct S^{k}(\sqrt{\frac{k}{n}}) × S^{n−k}(\sqrt{\frac{n−k}{n}}) are the only compact minimal hypersurfaces with two distinct principal curvatures whose multiplicities are greater than 1. For compact minimal hypersurfaces with two distinct principal curvatures one of which is simple, Otsuki constructed many such examples which are not congruent to each other. But they have the same topological types as S^{n−1} × S^{1} when they are compact. The first author in [3] investigated compact minimal hypersurfaces with two distinct principal curvatures one of which is simple in detail and proved that Cilifford torus S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}) is the only compact minimal hypersurface in S^{n+1}(1) with n ≤ S ≤ C(n), where C(n) is a constant only depending on n. As a generalization

2001 Mathematics Subject Classification: 53C42, 53C20. Key words and phrases: hypersurface, scalar curvature, second fundamental form and sphere. 1 The author’s research was partially supported by a Grant-in-Aid for Scientific Research from

the Japan Society for the Promotion of Science.

2 The author’s research was partially supported by the Natural Science Foundation of China and NSF of Shaanxi.

3 The author’s research was partially supported by R14-2002-003-01001-0 from the Korea Science and Engineering Foundation.

2 QING-MING CHENG^{1}, SHICHANG SHU^{2} AND YOUNG JIN SUH^{3}

of the result of Cheng [3], Hasanis and Vlachos [9] droped out the above bounded of S, that is, they proved that Clifford toru S^{1}(\sqrt{\frac{1}{n}}) × S^{n−1}(\sqrt{\frac{n−1}{n}}) is the only compact minimal hypersurfaces in S^{n+1}(1) with two distinct principal curvatures one of which is simple if S ≥ n.

Recently, the first author in [4] studied n-dimensional complete hypersurfaces in S^{n+1}(1) with constant scalar curvature and proved

Theorem 1.1. (Cheng [4]) Let M be an n-dimensional complete hypersurface with constant scalar curvature n(n − 1)r in S^{n+1}(1). If M has only two distinct principal curvatures one of which is simple, then, r > 1− \frac{2}{n} holds and M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) 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}, where c^{2} = \frac{n−2}{nr}.

Further, he constructed infinitely many isometrically distinct compact hypersurfaces in S^{n+1}(1) with constant scalar curvature which are not congruent to each other.

On the other hand, it is natural to ask whether the condition of constant scalar curvature in Theorem 1.1 of Cheng [4] is necessary? In this paper, first of all, we shall answer the problem and prove the following:

Theorem 1.2. Let M be an n-dimensional compact hypersurface in S^{n+1}(1) with two distinct principal curvatures one of which is simple. Then, M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) if r > 1 − \frac{2}{n}, r ≠ \frac{n−2}{n−1} and S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, where c^{2} = \frac{n−2}{nr} and n(n − 1)r denotes the scalar curvature.

Remark 1.3. In our theorem, we do not know whether the condition r ≠ \frac{n−2}{n−1} is necessary, which was also assumed in the theorem 1.1. From the examples constructed by Cheng [4], we know that the condition S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} is essential.

It is well known that the 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) satisfying r = \frac{n−2}{n−1}.

On the other hand, Li [11] 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 the 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}.

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 = \frac{n−2}{nc^{2}} > 1 − \frac{2}{n}. Hence, some of Riemannian products S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) do not appear in the results of Li [11].

From the above theorem 1.1 of Cheng [4], it is natural and interesting to generalize these results due to Li [11] to the case r > 1 − \frac{2}{n}. That is, it is interesting to prove the following:

Problem 1.4. (cf. Cheng [4]) Let M be an n-dimensional compact 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, is M isometric to the totally umbilical hypersurface or the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

In order to solve this problem above, Cheng [5] proved.

Theorem 1.5. Let M be an n-dimensional compact hypersurface with infinite fundamental group in S^{n+1}(1). 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^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c), where n(n − 1)r is the scalar curvature of M and c^{2} = \frac{n−2}{nr}.

In the following theorem, we will study the problem 1.4 and prove

Theorem 1.6. Let M be an n-dimensional compact hypersurface in S^{n+1}(1) with nonzero mean curvature and with two distinct principal curvatures one of which is simple. Then, M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) if r > 1 − \frac{2}{n} and S ≤ (n−1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, where c^{2} = \frac{n−2}{nr} and n(n−1)r denotes the scalar curvature.

2. Prelimenary

Throughout this paper, all manifolds are assumed to be smooth and connected without boundary. Let φ : M → S^{n+1}(1) be an isometric immersion of n-dimensional Riemannian manifold M into a unit sphere S^{n+1}(1). We choose a local orthonormal frame {e⃗_{1}, ⋯, e⃗_{n}, e⃗_{n+1}} and the dual coframe {ω_{1}, ⋯, ω_{n}, ω_{n+1}} in such a way that {e⃗_{1}, ⋯, e⃗_{n}} is a local orthonormal frame on M. Hence, we have

ω_{n+1} = 0

on M. From Cartan’s Lemma, we have

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

The mean curvature H and the second fundamental form α⃗ of M are defined, respectively, by

H = \frac{1}{n} ∑_{i=1}^{n} h_{ii}, α⃗ = h_{ij}ω_{i} ⊗ ω_{j}e⃗_{n+1}.

When the mean curvature H of M is identically zero, we recall that M is by definition a minimal hypersurface.

From the structure equations of M, Guass equation and Codazzi equation are given by

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

h_{ijk} = h_{ikj}. (2.3)

From (2.2), we have

n(n − 1)r = n(n − 1) + n^{2}H^{2} − S, (2.4) where n(n − 1)r and S denote the scalar curvature and the squared norm of the second fundamental form of M, respectively. We choose a local field of orthonormal frames e⃗_{1}, ⋯, e⃗_{n} on M such that, at the point that we consider,

h_{ij} = \{ \begin{matrix}λ_{i} & if i = j, \\ 0 & if i ≠ j, \end{matrix} (2.5)

where λ_{i}’s are the principal curvatures of M.

Next, we consider parallel hypersurfaces φ_{θ} : M → S^{n+1}(1) of M in S^{n+1}(1) given by map

φ_{θ} = cos θφ + sin θN, (2.6)

where N is the Gauss map of φ : M → S^{n+1}(1). It is well known that when cot θ is not a principal curvature of φ, φ_{θ} is an isometric immersion if we endow M with the pullback metric g_{θ} via φ_{θ}. We take cot θ ≠ λ_{i} for any i = 1, 2, ⋯, n. Thus, we know that the shape operator A_{θ} of the immersion φ_{θ} with respect to the unit normal vector N_{θ} = cos θN − sin θφ is given by

A_{θ} = {cot θA_{N} + I}{cot θI − A_{N}}^{−1}, (2.7)

where A_{N} is the shape operator with respect to the unit normal vector N and I is the identity mapping. Hence, the principal curvatures of φ_{θ} : M → S^{n+1}(1) are given by

λ_{i}(θ) = \frac{cot θλ_{i} + 1}{cot θ − λ_{i}}. (2.8)

3. Proofs of Theorem 1.2 and Theorem 1.6

Before we prove our theorem, we shall review some known results which will be used in the proof of our theorems.

In [5], the first author studied compact hypersurfaces with infinite fundamental group in S^{n+1}(1) and proved

Theorem 3.1. Let M be an n-dimensional compact hypersurface with infinite fundamental group in S^{n+1}(1). If the sectional curvatures are non-negative, then M is isometric to the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

The following theorem due to do Carmo, Dajczer and Mercuri [8] will be used

Theorem 3.2. Let M be an n(n ≥ 3)-dimensional compact orentiable hypersurface in S^{n+1}(1) with two distinct principal curvatures one of which is simple. Then, M is homeomorphic to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

Remark 3.3. We should remark that in the statement of theorem due to do Carmo, Dajczer and Mercuri, they considered that M is an n(n ≥ 4)-dimensional conformal flat hypersurface in S^{n+1}(1). But in their proof, they only used the condition that the hypersurface has two distinct principal curvatures one of which is simple. Hence, their result is also true for n = 3.

Proof of Theorem 1.2. Since M has two distinct principal curvatures, we denote them by λ and μ. Then, we have

(n − 1)λ + μ = nH, S = (n − 1)λ^{2} + μ^{2}. (3.1)

From (3.1) and Gauss equation (2.4), we have

λμ = (n − 1)(r − 1) − (n − 2)H^{2} ± (n − 2)H\sqrt{H^{2} − (r − 1)}.

Since S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} and r ≠ \frac{n−2}{n−1}, we know that H ≠ 0. In fact, if H = 0 holds at some point, we have, at the point, S = −n(n − 1)(r − 1). Hence,

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

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

Since the scalar curvature is continuous, we have r > \frac{n−2}{n−1} on M or r < \frac{n−2}{n−1} on M. If r > \frac{n−2}{n−1} holds, then n(r − 1) + 2 > \frac{n−2}{n−1}. If r < \frac{n−2}{n−1} holds, then n(r − 1) + 2 < \frac{n−2}{n−1}. We consider a function f(t) defined by

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

We can prove that f(t) is an increasing function of t when t > \frac{n−2}{n−1}. Hence, we infer a contradiction because f(\frac{n−2}{n−1}) = 0. If t < \frac{n−2}{n−1}, then f(t) is a decreasing function. This is also impossible. Hence, we have H ≠ 0.

Thus, we can assume H > 0 on M. Therefore, M is orentiable. Since the principal curvatures λ and μ are continuous, we have

λμ = (n − 1)(r − 1) − (n − 2)H^{2} + (n − 2)H\sqrt{H^{2} − (r − 1)}

or

λμ = (n − 1)(r − 1) − (n − 2)H^{2} − (n − 2)H\sqrt{H^{2} − (r − 1)}

on M. From Gauss equation (2.4), we have

λμ = \frac{2(n − 1)}{n}(r − 1) − \frac{(n − 2)}{n^{2}}S + \frac{(n − 2)}{n^{2}} \sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)}

or

λμ = \frac{2(n − 1)}{n}(r − 1) − \frac{(n − 2)}{n^{2}}S − \frac{(n − 2)}{n^{2}} \sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)}

on M.

(1). First of all, we consider the case where

λμ = \frac{2(n − 1)}{n}(r − 1) − \frac{(n − 2)}{n^{2}}S + \frac{(n − 2)}{n^{2}} \sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)}

holds on M. In this case, we can prove λμ + 1 ≥ 0 on M. Indeed, if λμ + 1 < 0 at some point, then

\frac{(n − 2)}{n^{2}} \sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)} < −1 − \frac{2(n − 1)}{n}(r − 1) + \frac{(n − 2)}{n^{2}}S.

From S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} and a direct computation, we have a contradiction. Hence, we have λμ + 1 ≥ 0 on M. From Gauss equation (2.2), we infer that the sectional curvatures of M are non-negative since M has two distinct principal curvatures and one of which is simple. From theorem 3.2, we have that M is homeomorphic to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). Thus, the fundamental group of M is infinite. From the above theorem 3.1 of Cheng, we know that φ(M) is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

(2) Secondly, we consider the case where

(3.2)  λμ = \frac{2(n − 1)}{n}(r − 1) − \frac{(n − 2)}{n^{2}}S

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

holds on M. From S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, we have λ ≠ 0 and μ ≠ 0. In fact, if λ = 0, we have n^{2}H^{2} = S. From the Guass equation (2.4), we have r = 1. Hence, from the formula (3.2) of λμ, we have a contradiction. If μ = 0, we have n^{2}H^{2} = (n−1)S. From the Guass equation (2.4), we have (n−2)S = n(n−1)(r−1).

It is impossible because of S ≥ (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2} and r > 1 − \frac{2}{n}.

Next we shall prove that λμ + 1 ≤ 0. If \frac{2(n−1)}{n}(r − 1) − \frac{(n−2)}{n^{2}}S + 1 ≤ 0, then, from (3.2), it is obvious that λμ + 1 ≤ 0 holds. If \frac{2(n−1)}{n}(r − 1) − \frac{(n−2)}{n^{2}}S + 1 > 0, then, we can prove that λμ + 1 ≤ 0 holds also. In fact, From S ≥ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, we have

(n − 2)(n(r − 1) + 2)S ≥ (n − 1)n^{2}(r − 1)^{2} + 4n(n − 1)(r − 1) + n^{2},

that is,

(n − 2){4n(n − 1)(r − 1) + 2n^{2} + (n − 2)^{2}n(r − 1)}S
≥ {2n(n − 1)(r − 1) + n^{2}}^{2} + (n − 2)^{2}n^{2}(n − 1)(r − 1)^{2}.

Hence,

\{n + 2(n − 1)(r − 1) − \frac{n − 2}{n}S\}^{2} ≤ \frac{(n − 2)^{2}}{n^{2}} \{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}.

Since \frac{2(n−1)}{n}(r − 1) − \frac{(n−2)}{n^{2}}S + 1 > 0 holds, we have

2n(n − 1)(r − 1) + n^{2} − (n − 2)S ≤ (n − 2)\sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)}.

From

λμ = \frac{2(n − 1)}{n}(r − 1) − \frac{(n − 2)}{n^{2}}S − \frac{(n − 2)}{n^{2}} \sqrt{S + n(n − 1)(r − 1)}\sqrt{S − n(r − 1)},

we infer λμ + 1 ≤ 0. Hence, we have \frac{1}{λμ} + 1 ≥ 0.

Since λμ ≠ 0, we know that φ\frac{π}{2} is an isometric immersion from M into S^{n+1}(1). Therefore, from (2.8), we know that the sectional curvatures K\frac{π}{2} of φ\frac{π}{2} with respect to the plane spanned by e⃗_{j} and e⃗_{n} for j = 1, ⋯, n−1, satisfy K\frac{π}{2} (e⃗_{j}, e⃗_{n}) = \frac{1}{λμ} +1 ≥ 0 and K\frac{π}{2} (e⃗_{i}, e⃗_{j}) = 1+ \frac{1}{λ^{2}} > 1 for i, j = 1, 2, ⋯, n−1. It is obvious that the immersion φ\frac{π}{2} has only two distinct principal curvatures and one of them is simple. From the theorem 3.2, we infer that M is homeomorphic to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). Thus, the fundamental group of M is infinite. From the theorem 3.1, we know that φ\frac{π}{2} (M) is

isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). Hence, the principal curvatures \frac{1}{λ} and \frac{1}{μ} of φ\frac{π}{2}

are constant. Thus, we have that λ and μ are constant and λμ + 1 = 0. Hence, the sectional curvatures of M are non-negative. From the theorem 3.1, we infer that φ(M) is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). Proof of Theorem 1.6. Since M has two distinct principal curvatures one of which is simple, from the theorem 3.2 we know that M is homeomorphic to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). Hence, the fundamental group of M is infinite. Since

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

holds if and only if

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

From Gauss equation n(n − 1)r = n(n − 1) + n^{2}H^{2} − S, we conclude

(S − S_{−})(S − S_{+}) ≥ 0,

where

S_{±} = n + \frac{n^{3}}{2(n − 1)}H^{2} ± \frac{n − 2}{2(n − 1)}\sqrt{n^{4}H^{4} + 4n^{2}(n − 1)H^{2}}.

Since the mean curvatue H ≠ 0 holds on M, we have S_{+} > S_{−} on M. Since S is a continuous function on M, we infer that only one of S ≥ S_{+} on M and S ≤ S_{−} on M is satisfied.

If the S ≤ S_{−} on M holds, from the result of [1] and the fundamental group is infinite, we know that M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

If the S ≥ S_{+} on M holds, from the result of [2], we know that M is isometric to S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

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

Shichang Shu Department of Mathematics, Weinan Teachers’ College, Weinan 714000, Shaanxi, P. R. China xysxssc@yahoo.com.cn Present address: Department of Mathematics, Faculty of Science and Engineering, Saga University, Saga 840-8502, Japan

Young Jin Suh Department of Mathematics, Kyungpook National University, Taegu 702-701, South Korea yjsuh@wmail.knu.ac.kr