COMPACT HYPERSURFACES IN A UNIT SPHERE

WITH INFINITE FUNDAMENTAL GROUP*

Qing-Ming Cheng

Abstract. It is our purpose to study curvature structures of compact hypersurfaces in the unit sphere S^{n+1}(1). 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 infinite fundamental group, which satisfy 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}. In particular, we obtained that the Riemannian product S^{1}(\sqrt{1 − c^{2}})×S^{n−1}(c) is the only compact hypersurfaces with infinite fundamental group in S^{n+1}(1) if the sectional curvatures are non-negative.

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 the investigation on curvature structures of compact hypersurfaces is important and interesting. In 1977, S.Y. Cheng and Yau [4] studied compact hypersurfaces with constant scalar curvature in the unit sphere S^{n+1}(1). They proved that let M be an n-dimensional compact hypersurface with constant scalar curvature n(n − 1)r, if r ≥ 1 and the sectional curvatures of M are nonnegative, then M is isometric to the totally umbilical hypersurface S^{n}(c) or the Riemannian product S^{k}(c_{1}) × S^{n−k}(c_{2}) 1 ≤ k ≤ n − 1, where S^{k}(c) denote the sphere of radius c. In order to prove this theorem, they introduced a differential operator □ defined by

□f = ∑_{i,j=1}^{n}(nHδ_{ij} − h_{ij})∇_{i}∇_{j}f,

for any C^{2}-function f on M. Where h_{ij} and H are components of the second fundamental form and the mean curvature of M, respectively. We should notice the following:

(1) the differential operator □ is self-adjoint.

Typeset by AMS-TEX

Key words and phrases. compact hypersurface, scalar curvature, principal curvatures, Ricci curvature and fundamental group.

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

2001 Mathematics Subject Classification 53C42

(2) the differential operator □ is degenerate elliptic if r ≥ 1. Therefore, in order to prove their theorem, they must make use of the properties that the differential operator □ is self-adjoint and degenerate elliptic. And in order to obtain the estimate ∑_{i,j,k=1}^{n} h_{ijk}^{2} ≥ n^{2}|gradH|^{2}, which is very important in the proof of their theorem, the condition of r ≥ 1 and the assumption of constant scalar curvature is essential. Where h_{ijk}’s are components of the covariant differentiation of the second fundamental form. Hence, the condition r ≥ 1 and the assumption of constant scalar curvature play an essential role in the proof of their theorem. Further, by making use of the similar method which was used by Nakagawa and the author in [3] and the differential operator introduced by S.Y. Cheng and Yau, Li [5] 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}, where S is the squared norm of the second fundamental form of M. These properties that the differential operator □ is self-adjoint and degenerate elliptic are indispensable in the proof of his theorem again. And the estimate ∑_{i,j,k=1}^{n} h_{ijk}^{2} ≥ n^{2}|gradH|^{2} is also essential in the proof of his theorem.

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 = \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 these results of S. Y. Cheng and Yau [4] and Li [5]. Moreover, Cheng [2] proved

Theorem C. (Cheng[2]). 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}.

From the assertion above, it is natural and interesting to generalize these results due to S.Y. Cheng and Yau [4] and Li [5] to the case r > 1 − \frac{2}{n}. That is, it is interesting to prove the following:

Problem 1 (cf. Cheng [2]). 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 M is isometric to the totally umbilical hypersurface or the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c).

It is our purpose to try to solve this problem above. Since the problem seems to be a very hard problem, we shall try to solve it under a topological condition. It is known that S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) has infinite fundamental group. Hence, we shall consider compact hypersurfaces with infinite fundamental group in the unit sphere S^{n+1}(1). The following theorems will be proved.

Theorem 1. 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}.

Theorem 2. 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).

Remark. In our theorems 1 and 2, we do not assume that the scalar curvature is constant. And in our theorem 2, we do not assume any condition on scalar curvature.

2. Proofs of theorems

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)  R_{ij} = (n − 1)δ_{ij} + nHh_{ij} − ∑_{k=1}^{n} h_{ik}h_{kj},

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

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

Proof of Theorem 1. Since r ≥ \frac{n−2}{n−1} and S ≤ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, we infer

(2.5)  n + 2(n − 1)(r − 1) − \frac{n − 2}{n}S ≥ 0.

In fact,

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

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

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

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

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

We consider function f(t) = \frac{(n−2)^{2}}{n} + \frac{n−1}{n}t − \frac{(n−2)^{2}}{nt}. By a direct computation, we have f(t) ≥ 0 if t ≥ \frac{n−2}{n−1}. Since r ≥ \frac{n−2}{n−1}, we have n(r − 1) + 2 ≥ \frac{n−2}{n−1}. Thus, we infer

n + 2(n − 1)(r − 1) − \frac{n − 2}{n}S ≥ 0.

Therefore, when r ≥ \frac{n−2}{n−1}, we know that

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

is equivalent to

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

Indeed, we can prove that

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)\}.

Since (2.5) holds, we obtain that (2.7) is true.

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

(2.8)  \{S − n(r − 1)\} = \frac{n}{n − 1}(S − nH^{2}).

Hence, from (2.7) and (2.8), we obtain

(2.9)  n + 2nH^{2} − S ≥ \frac{n − 2}{\sqrt{n(n − 1)}}\sqrt{n^{2}H^{2}(S − nH^{2})}.

For any point p and any unit vector u⃗ ∈ T_{p}M, we choose a local orthonormal frame field {e_{1}, ⋯, e_{n}} such that e_{n} = u⃗, we have, from Gauss equation (2.3),

(2.10)  Ric(u⃗) = (n − 1) + nHh_{nn} − ∑_{i=1}^{n} h_{in}^{2}.

Since

(nH − h_{nn})^{2} = (∑_{i=1}^{n−1} h_{ii})^{2} ≤ (n − 1) ∑_{i=1}^{n−1} h_{ii}^{2},

we have

n^{2}H^{2} − (n − 1) ∑_{i=1}^{n} h_{ii}^{2} + nh_{nn}^{2} − 2nHh_{nn} ≤ 0.

From ∑_{i}(h_{ii} − H) = 0 and ∑_{i=1}^{n}(h_{ii} − H)^{2} = ∑_{i=1}^{n} h_{ii}^{2} − nH^{2}, we have, for any i,

(h_{ii} − H)^{2} ≤ \frac{n − 1}{n}(∑_{i=1}^{n} h_{ii}^{2} − nH^{2}).

0 ≥ n(h_{nn}^{2} − nHh_{nn}) + (n − 2)nH(h_{nn} − H)

+ 2(n − 1)nH^{2} − (n − 1) ∑_{i=1}^{n} h_{ii}^{2}

≥ n(h_{nn}^{2} − nHh_{nn}) − (n − 2)n|H|\sqrt{\frac{n − 1}{n}(∑_{i=1}^{n} h_{ii}^{2} − nH^{2})}

+ 2(n − 1)nH^{2} − (n − 1) ∑_{i=1}^{n} h_{ii}^{2},

namely,

(2.11) (h_{nn}^{2} − nHh_{nn})

≤ (n − 2)|H|\sqrt{\frac{n − 1}{n}(∑_{i=1}^{n} h_{ii}^{2} − nH^{2})} − 2(n − 1)H^{2} + \frac{n − 1}{n} ∑_{i=1}^{n} h_{ii}^{2}.

From (2.10) and (2.11), we have

Ric(u⃗) ≥ (n − 1) − (n − 2)|H|\sqrt{\frac{n − 1}{n}(∑_{i=1}^{n} h_{ii}^{2} − nH^{2})} + 2(n − 1)H^{2} − \frac{n − 1}{n} ∑_{i,j=1}^{n} h_{ij}^{2}

because of \frac{n−1}{n} > \frac{1}{2}. Thus, we obtain, from the above inequality and S = ∑_{i,j=1}^{n} h_{ij}^{2},

(2.12)  Ric(u⃗) ≥ \frac{n − 1}{n}\{n + 2nH^{2} − S − \frac{n − 2}{\sqrt{n(n − 1)}}\sqrt{n^{2}H^{2}(S − nH^{2})}\}.

From (2.9), we have Ric(u⃗) ≥ 0. In particular, from the assertions above, we know that if S < (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds, then Ric(u⃗) > 0. Thus, if there exists

point p in M such that S < (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, then, at the point p, the Ricci curvature is positive. From the following Lemma due to Aubin [1], we know that there exists a metric on M such that the Ricci curvature is positive on M. According to Myers theorem, we know that the fundamental group is finite. This is impossible because M has infinite fundamental group.

Lemma. (cf. Aubin [1, p. 344)]. If the Ricci curvature of a compact Riemannian manifold is non-negative and positive at somewhere, then the manifold carries a metric with positive Ricci curvature.

Thus, we must have S = (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}. And at each point, there exists a unit vector u⃗ such that Ric(u⃗) = 0. From the assertions above, we infer that these inequalities above are equalities. That is, we must have h_{ij} = 0 if i ≠ j and h_{11} = ⋯ = h_{n−1n−1},

(h_{nn} − H)^{2} = \frac{n − 1}{n}(∑_{i=1}^{n} h_{ii}^{2} − nH^{2}) = \frac{n − 1}{n}(S − nH^{2})

and

(h_{11} − H)^{2} = ⋯ = (h_{n−1n−1} − H)^{2} = \frac{1}{n(n − 1)}(S − nH^{2}).

Hence, we conclude that M has only two distinct principal curvatures one of which is simple. Let {e_{1}, ⋯, e_{n}} be a local orthonormal frame field such that h_{ij} = λ_{i}δ_{ij}, where λ_{i}’s are principal curvatures on M. Without loss of generality, we can assume μ = λ_{n}, λ = λ_{1} = ⋯ = λ_{n−1}. From Gauss equation (2.2) and the definition of the Ricci curvature, we have 1 + μλ = 0 because of 1 + λ_{i}λ_{j} = 1 + λ^{2} > 0, for any i, j = 1, ⋯, n − 1. From (2.4), we have

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

Hence λ^{2} = \frac{n(r−1)+2}{n−2} and μ^{2} = \frac{n−2}{n(r−1)+2}.

We consider the integral submanifold of the corresponding distribution of the space of principal vectors corresponding to the principal curvature λ. Since the multiplicity of the principal curvature λ is greater than 1, we know that the principal curvature λ is constant on this integral submanifold (cf. Otsuki [6]). From λ^{2} = \frac{n(r−1)+2}{n−2} and μ^{2} = \frac{n−2}{n(r−1)+2}, we know that the scalar curvature n(n − 1)r and the principal curvature μ are constant. Thus, we obtain that M is isoparametric. Therefore, M is isometric to the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c) because S = (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds. This completes the proof of Theorem 1.

Proof of Theorem 2. Since the sectional curvatures are non-negative, we have that the Ricci curvature is non-negative. From the arguments in the proof of theorem 1, we infer that at each point, there exists a unit vector u⃗ such that Ric(u⃗) = 0.

Let {e_{1}, ⋯, e_{n}} be a local orthonormal frame field such that h_{ij} = λ_{i}δ_{ij}, where λ_{i}’s are principal curvatures on M. Then, from Gauss equation (2.2), we have 1 + λ_{i}λ_{j} ≥ 0 for i ≠ j. Further, there exists an i such that ∑_{j≠i}(1 + λ_{i}λ_{j}) = 0 from the definition of Ricci curvature. Hence, we must have 1 + λ_{i}λ_{j} = 0 for j ≠ i.

Therefore, M has only two distinct principal curvatures one of which is simple. Let μ = λ_{i} and λ = λ_{j} for j ≠ i. From (2.4), we have

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

Since 1 + μλ = 0 and (2.13) hold, we have λ^{2} = \frac{n(r−1)+2}{n−2} and μ^{2} = \frac{n−2}{n(r−1)+2}. Hence, we have

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

By making use of the same assertion as in the proof of Theorem 1, we infer that M is isometric to the Riemannian product S^{1}(\sqrt{1 − c^{2}}) × S^{n−1}(c). This completes the proof of theorem 2.

References

1. Aubin, T., Some nonlinear problems in Riemannian geometry, Springer-Verlag, Berlin, New York., 1998.

2. Cheng, Q.M., Hypersurfaces in a unit sphere S^{n+1}(1) with constant scalar curvature, J. London Math. Soc. 64 (2001), 755-768.

3. Cheng, Q.M. and Nakagawa, H., Totally umbilical hypersurfaces, Hiroshima Math. J. 20 (1990), 1-10.

4. Cheng, S.Y. and Yau, S.T., Hypersurfaces with constant scalar curvature, Math. Ann. 225 (1977), 195-204.

5. Li, H., Hypersurfaces with constant scalar curvature in space forms, Math. Ann. 305 (1996), 665-672.

6. Otsuki, T., Minimal hypersurfaces in a Riemannian manifold of constant curvature, Amer. J. Math. 92 (1970), 145-173.

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