CONFORMALLY FLAT 3-MANIFOLDS

WITH CONSTANT SCALAR CURVATURE

Qing-Ming Cheng*, Susumu Ishikawa & Katsuhiro Shiohama**

Abstract. We classify complete conformally flat three dimensional Riemannian manifolds with constant scalar curvature and constant squared norm of Ricci curvature tensor by applying the Generalized Maximum Principle due to H. Omori.

1. Introduction

It is interesting to investigate the structure of complete and conformally flat Riemannian manifolds with constant scalar curvature. The class of such manifolds is very wide. In fact, from Theorem due to Yamabe, Trudinger, Aubin and Schoen in [1] any compact Riemannian manifold can be deformed into a Riemannian manifold with constant scalar curvature by a conformal transformation. So the metric structure on conformally flat Riemannian manifold M^{n} will be specified under certain restrictions on the behavior of Ricci curvature tensor of M^{n} for the classification of them. Throughout this article, let M^{n} be a connected complete and conformally flat Riemannian manifold of constant scalar curvature r without boundary. The Riemannian product M^{n−1}(c) × N^{1} are well known as such examples, where M^{n−1}(c) is (n − 1)- dimensional manifold of constant sectional curvature c and N^{1} = S^{1} or R. The scalar curvature r of M^{n−1}(c) × N^{1} is positive (negative, respectively) according as c > 0 (c < 0, respectively).

In [3] (cf. [2]), S. T. Goldberg essentially proved that every complete conformally flat Riemannian manifold M^{n} with positive constant scalar curvature r is a space form if the Ricci curvature tensor of M^{n} satisfies the inequality

sup ∑_{i,j} R_{ij}^{2} < \frac{r^{2}}{n − 1},

where ∑_{i,j} R_{ij}^{2} is the squared norm of the Ricci curvature tensor of M^{n}. We should remark that the condition of r being positive is essential in the proof of Goldberg’s Theorem.

The purpose of this article is to study the metric structure of M^{n} with (not necessarily positive) constant scalar curvature such that the squared norm of the Ricci curvature tensor is constant. The basic tool used here is the Generalized Maximum Principle due to H. Omori:

Typeset by AMS-TEX

Key words and phrases. space form, conformally flat manifold, constant scalar curvature and Ricci curvature tensor.

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

**Research partially Supported by Grant-in-Aid for Co-operative Research No. 06302007 from the Ministry of Education, Science and Culture

1991 Mathematics Subject Classification 53C20

Generalized Maximum Principle (H. Omori [5]). Let X be an n dimensional complete Riemannian manifold whose sectional curvature is bounded from below. If f is a C^{2}-function bounded from above on X, then there exists a sequence {p_{m}} of points in X such that

lim_{m→∞} f(p_{m}) = sup f, lim_{m→∞} |grad f|(p_{m}) = 0,

lim_{m→∞} sup ∇_{l}∇_{l}f(p_{m}) ≤ 0 for l = 1, 2, ⋯, n.

Here grad f is the gradient vector field for function f and ∇ is the operator of covariant differentiation on X.

Our Main Theorems in this article are stated as follows:

Main Theorem 1. Let M^{3} be a 3-dimesional complete conformally flat Riemannian manifold with constant scalar curvature and constant squared norm of the Ricci curvature tensor. Then we have

(1) If the scalar curvature r is nonnegative, M^{3} is either isometric to a space form or else the Riemannian product M^{2}(c) × N^{1} (c ≥ 0).

(2) If the scalar curvature r is negative, either M^{3} is isometric to a space form or else the squared norm of the Ricci curvature tensor of M^{3} lies in (r^{2}/3, r^{2}/2].

Remark 1. It is obvious that the 3-dimesional hyperbolic space form H^{3}(c) is conformally flat and it satisfies S = r^{2}/3, where S is the squared norm of the Ricci curvature tensor of H^{3}(c). We also know the conformally flat space H^{2}(c)×R satisfies S = r^{2}/2. But we do not know whether there exist complete conformally flat Riemannian manifolds with negative constant scalar curvature r and constant squared norm S of the Ricci curvature tensor which satisfies r^{2}/3 < S < r^{2}/2. From Theorem 3 in the section 3, we propose the following conjecture.

Conjecture. Let M^{3} be a 3-dimesional complete conformally flat Riemannian manifold with negative constant scalar curvature and constant squared norm of the Ricci curvature tensor. Then, M^{3} is isometric to a space form or else the Riemannian product H^{2}(c) × N^{1} (c < 0).

Main Theorem 2. There is no 3-dimensional compact conformally flat Riemannian manifold M^{3} such that the following conditions (1) and (2) are satisfied:

(1) The scalar curvature of M^{3} is negative constant.

(2) The squared norm of the Ricci curvature tensor of M^{3} is constant and the eigen-values of Ricci curvature tensor are all distinct at every point of M^{3}.

Acknowledgement. Authors would like to express their gratitude to the referee for his valuable suggestion.

2. Preliminaries

In this section we shall prepare some local formulas for conformally flat Riemannian manifolds with constant scalar curvature. Let M^{n} be a conformally flat Riemannian manifold and e_{1}, ⋯, e_{n} an orthonormal basis of T_{p}M^{n}, where T_{p}M^{n} is the tangent space at a point p to M^{n}. We denote the components of curvature tensor of M^{n} by R_{ijkl}. Since M^{n} is conformally flat, we have

(2.1)  R_{ijkl} = \frac{1}{n − 2}(R_{ik}δ_{jl} − R_{il}δ_{jk} + δ_{ik}R_{jl} − δ_{il}R_{jk})

− \frac{r}{(n − 1)(n − 2)}(δ_{ik}δ_{jl} − δ_{il}δ_{jk}),

where R_{ij}’s are the components of Ricci curvature tensor and r the scalar curvature of M^{n}.

Next we assume that r is constant. Since M is conformally flat, we have

(2.2)  ∇_{k}R_{ij} = ∇_{j}R_{ik},

where ∇ is the operator of covariant differentiation on M^{n}. At each point p, we can choose an orthonormal basis e_{1}, ⋯, e_{n} such that R_{ij} = λ_{i}δ_{ij}. We put S = ∑_{i,j} R_{ij}^{2} = ∑_{i} λ_{i}^{2} and B = ∑_{i}(λ_{i} − r/n)^{2}. Then B = S − r^{2}/n. By letting R_{ij,k} := ∇_{k}R_{ij}, R_{ij,kl} := ∇_{l}∇_{k}R_{ij}, etc., we have the Ricci formulas:

(2.3-i)  R_{ij,kl} − R_{ij,lk} = ∑_{t} R_{tj}R_{tikl} + ∑_{t} R_{it}R_{tjkl},
(2.3-ii)  R_{ij,klm} − R_{ij,kml} = ∑_{t} R_{tj,k}R_{tilm} + ∑_{t} R_{it,k}R_{tjlm}

+ ∑_{t} R_{ij,t}R_{tklm}.

Because of B = S − r^{2}/n, the Ricci formula (2.3-i) and (2.1) yield

\frac{1}{2}ΔB = ∑_{i,j,k} R_{ij,k}^{2} + ∑_{i,j} R_{ij}ΔR_{ij}

= ∑_{i,j,k} R_{ij,k}^{2} + ∑_{i,j,k,m} R_{ij}R_{mk}R_{mijk} + ∑_{i,j,k,m} R_{ij}R_{im}R_{mkjk}

= ∑_{i,j,k} R_{ij,k}^{2} + \frac{1}{n − 2}(n ∑_{i} λ_{i}^{3} − \frac{2n − 1}{n − 1}rS + \frac{r^{3}}{n − 1})

and

(2.4)  \frac{1}{2}Δ ∑_{i,j,k} R_{ij,k}^{2} = ∑_{i,j,k,l} R_{ij,kl}^{2} + ∑_{i,j,k} R_{ij,k}ΔR_{ij,k}

where Δ is the Laplace operator on M^{n}. By directly calculating ΔR_{ij} and ΔR_{ij,k} and using Ricci formulas (2.3-i), (2.3-ii), we obtain the equations

(2.5)  ΔR_{ij} = ∑_{l} R_{ij,ll} = ∑_{l,t} R_{tl}R_{tijl} + ∑_{t} R_{it}R_{tj}

and

(2.6)  ΔR_{ij,k} = ∑_{l} R_{ij,kll} = ∑_{l} ∇_{l}R_{ij,kl}

= ∑_{l} R_{ij,lkl} + ∑_{l} ∇_{l}(∑_{t} R_{tj}R_{tikl})+ ∑_{l} ∇_{l}(∑_{t} R_{ti}R_{tjkl})

= ∑_{l} R_{ij,llk} + ∑_{l,t} R_{tj,l}R_{tikl} + ∑_{l,t} R_{it,l}R_{tjkl}

+ ∑_{l,t} R_{ij,t}R_{tlkl} + ∑_{l,t} R_{tj,l}R_{tikl} + ∑_{l,t} R_{it,l}R_{tjkl}

+ ∑_{l,t} R_{tj}R_{tikl,l} + ∑_{l,t} R_{it}R_{tjkl,l}.

Hence, by using (2.5) and (2.6), we have

(2.7) ∑_{i,j,k} R_{ij,k}ΔR_{ij,k} = 3 ∑_{i,j,k} λ_{i}R_{ij,k}^{2} + 5 ∑_{i,j,k,l,t} R_{ij,k}R_{tj,l}R_{tikl}

+ 2 ∑_{i,j,k,l,t} R_{ti}R_{ij,k}R_{tjkl,l} + ∑_{i,j,k,l,t} R_{tl}R_{ij,k}R_{tijl,k},

where R_{ijkl,t} = ∇_{t}R_{ijkl}. Since

(2.8)  R_{ijkl,m} = \frac{1}{n − 2}(R_{ik,m}δ_{jl} − R_{il,m}δ_{jk} + δ_{ik}R_{jl,m} − δ_{il}R_{jk,m})

in the case of r =constant, we also obtain the relations

(2.9)  ∑_{i,j,k,l,t} R_{ti}R_{ij,k}R_{tjkl,l} = 0,
(2.10)  ∑_{i,j,k,l,t} R_{tl}R_{ij,k}R_{tijl,k} = \frac{2}{n − 2} ∑_{i,j,k} λ_{i}R_{ij,k}^{2} − \frac{r}{n − 2} ∑_{i,j,k} R_{ij,k}^{2}

and

∑_{i,j,k,l,t} R_{ij,k}R_{tj,l}R_{tikl} = \frac{2}{n − 2} ∑_{i,j,k} λ_{i}R_{ij,k}^{2} − \frac{r}{(n − 1)(n − 2)} ∑_{i,j,k} R_{ij,k}^{2}.

Hence, we have finally

\frac{1}{2}Δ ∑_{i,j,k} R_{ij,k}^{2} = ∑_{i,j,k,l} R_{ij,kl}^{2} + \frac{3n + 2}{n − 2} ∑_{i,j,k} λ_{i}R_{ij,k}^{2} − \frac{(n + 2)r}{(n − 1)(n − 2)} ∑_{i,j,k} R_{ij,k}^{2}.

Summing up the above computations, we have proved the following Proposition 1:

Proposition 1. On conformally flat Riemannian manifold M^{n} with constant scalar curvature r, we have the following formulae (2.11) and (2.12):

(2.11)  \frac{1}{2}ΔB = ∑_{i,j,k} R_{ij,k}^{2} + \frac{1}{n − 2}(n ∑_{i} λ_{i}^{3} − \frac{2n − 1}{n − 1}rS + \frac{r^{3}}{n − 1})

and

(2.12)  \frac{1}{2}Δ ∑_{i,j,k} R_{ij,k}^{2} = ∑_{i,j,k,l} R_{ij,kl}^{2} + \frac{3n + 2}{n − 2} ∑_{i,j,k} λ_{i}R_{ij,k}^{2}

− \frac{(n + 2)r}{(n − 1)(n − 2)} ∑_{i,j,k} R_{ij,k}^{2}.

Remark 2. Using the formula (2.11) and the Generalized Maximum Principle of Omori, we can conclude the result due to Goldberg [3] denoted in section 1 and the following result of Tani [6]: Any compact conformally flat Riemannian manifold with constant scalar curvature r and positive Ricci curvature is of constant curvature.

Now, we consider the case of n = 3 and S=constant. Then (2.11) and (2.12) become

(2.13)  ∑_{i,j,k} R_{ij,k}^{2} = −3 ∑_{i} λ_{i}^{3} + \frac{5}{2}rS − \frac{1}{2}r^{3}

and

(2.14)  Δ ∑_{i,j,k} R_{ij,k}^{2} = −3Δ ∑_{i} λ_{i}^{3}

respectively. Setting μ_{i} = λ_{i} − r/3, we have

(2.15)  ∑_{i} μ_{i} = 0, B = ∑_{i} μ_{i}^{2} = S − \frac{1}{3}r^{2},

(2.16)  ∑_{i} λ_{i}^{3} = ∑_{i} μ_{i}^{3} + rB + \frac{1}{9}r^{3},

(2.17)  ∑_{i} λ_{i}^{4} = \frac{1}{2}B^{2} + \frac{4}{3}r ∑_{i} μ_{i}^{3} + \frac{2}{3}r^{2}B + \frac{1}{27}r^{4}

and (2.13) can be rewritten as

(2.18)  ∑_{i,j,k} R_{ij,k}^{2} = −3 ∑_{i} μ_{i}^{3} − \frac{1}{2}rB.

Since

∑_{i} λ_{i}^{3} = ∑_{i,j,k} R_{ik}R_{kj}R_{ji},

we obtain from (2.16) and (2.5)

(2.19)  \frac{1}{3}Δ ∑_{i} μ_{i}^{3} = 2 ∑_{i,j,k} λ_{i}R_{ij,k}^{2} + ∑_{i,k} λ_{i}^{2}R_{ii,kk}

= 2 ∑_{i,j,k} μ_{i}R_{ij,k}^{2} + \frac{2}{3}r ∑_{i,j,k} R_{ij,k}^{2} + \frac{5}{2}r ∑_{i} μ_{i}^{3}

+ \frac{1}{2}B^{2} + \frac{1}{3}r^{2}B

= \frac{1}{2}(B^{2} + r ∑_{i} μ_{i}^{3} + 4 ∑_{i,j,k} μ_{i}R_{ij,k}^{2}).

Therefore we have from (2.18) and (2.19)

(2.20)  Δ ∑_{i,j,k} R_{ij,k}^{2} = −\frac{9}{2}(B^{2} + r ∑_{i} μ_{i}^{3} + 4 ∑_{i,j,k} μ_{i}R_{ij,k}^{2}).

Moreover we have from (2.12), (2.19), (2.20)

(2.21)  ∑_{i,j,k,l} R_{ij,kl}^{2} = −11 ∑_{i,j,k} λ_{i}R_{ij,k}^{2} + \frac{5}{2}r ∑_{i,j,k} R_{ij,k}^{2} + \frac{1}{2}Δ ∑_{i,j,k} R_{ij,k}^{2}

= −20 ∑_{i,j,k} μ_{i}R_{ij,k}^{2} − \frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{2} − \frac{9}{4}B(B − \frac{1}{6}r^{2}).

Summing up the above computations, we have proved the following Proposition 2:

Proposition 2. If the squared norm of Ricci curvature tensor of the conformally flat Riemannian manifold M^{3} with constant scalar curvature r is constant, then we have the following three formulae:

∑_{i,j,k} R_{ij,k}^{2} = −3 ∑_{i} μ_{i}^{3} − \frac{1}{2}rB,

Δ ∑_{i} μ_{i}^{3} = \frac{3}{2}(B^{2} + r ∑_{i} μ_{i}^{3} + 4 ∑_{i,j,k} μ_{i}R_{ij,k}^{2}),

∑_{i,j,k,l} R_{ij,kl}^{2} = −20 ∑_{i,j,k} μ_{i}R_{ij,k}^{2} − \frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{2} − \frac{9}{4}B(B − \frac{1}{6}r^{2}).

3. Theorems and their proofs

First of all, we consider the case of r ≥ 0.

Theorem 1. In addition to the assumptions in Main Theorem 1, if r ≥ 0, then M^{3} is either isometric to a space form or the Riemannian product M^{2} × N^{1}, where M^{2} and N^{1} are of constant curvature with dimension 2 and 1, respectively.

Proof. Since M^{3} is conformally flat and the squared norm of Ricci curvature tensor is constant, we know from (2.1) that the sectional curvature of M^{3} is bounded. Applying the Generalized Maximum Principle to the function B_{3} := ∑_{i} μ_{i}^{3}, there exists a sequence {p_{m}} of points in M^{3} such that

(3.1)  lim_{m→∞} B_{3}(p_{m}) = sup B_{3}, lim_{m→∞} |grad B_{3}|(p_{m}) = 0,
(3.2)  lim_{m→∞} sup ∇_{l}∇_{l}B_{3}(p_{m}) ≤ 0 for l = 1, 2, 3.

From (2.18) and (2.21) it follows that R_{ij,k} and R_{ij,kl} are bounded. Thus, by taking a subsequence if necessary, we see that the sequences {λ_{i}(p_{m})}, {μ_{i}(p_{m})}, {B_{3}(p_{m})}, {R_{ij,k}(p_{m})} and {R_{ij,kl}(p_{m})} have their limits λ_{i}^{∗}, μ_{i}^{∗}, B_{3}^{∗}, R_{ij,k}^{∗} and R_{ij,kl}^{∗}, respectively.

First we note the following lemma:

Lemma 1.

−\frac{B^{\frac{3}{2}}}{\sqrt{6}} ≦ B_{3} ≦ \frac{B^{\frac{3}{2}}}{\sqrt{6}}

and the equality holds if and only if at least two of μ_{i} are equal.

Proof. Because of ∑_{i} μ_{i} = 0 and ∑_{i} μ_{i}^{2} = B, by making use of Lagrange multiplier method, it follows that −B^{\frac{3}{2}} /\sqrt{6} ≦ B_{3} ≦ B^{\frac{3}{2}} /\sqrt{6} and the equality holds if and only if at least two of μ_{i} are equal. This finishes the proof.

Now, if S ≤ r^{2}/2, then we know from the result in [2] due to Wu and first-named author that M^{3} is either isometric to a space form or Riemannian product M^{2} × N^{1}. Thus, Theorem 1 is true for the case of S ≤ r^{2}/2.

Next we shall prove that S > r^{2}/2 is impossible. In fact, by (2.15) and the assumptions for r and S, we have ∑_{i} R_{ii,k} = 0 and ∑_{i} μ_{i}R_{ii,k} = 0 on M^{3}. In particular, we get

(3.3)  ∑_{i} R_{ii,k}^{∗} = 0 for all k

and

(3.4)  ∑_{i} μ_{i}^{∗}R_{ii,k}^{∗} = 0 for all k.

∑_{i} μ_{i}^{∗}^{2}R_{ii,k}^{∗} = 0 is obtained from (3.1). So we have, together with (3.3) and (3.4), the following system of linear equations:

(3.5)  \{ \begin{matrix}∑_{i} R_{ii,k}^{∗} = 0 \\ ∑_{i} μ_{i}^{∗}R_{ii,k}^{∗} = 0 \\ ∑_{i} μ_{i}^{∗}^{2}R_{ii,k}^{∗} = 0 \end{matrix}\}

for k = 1, 2 and 3.

(i) If μ_{1}^{∗}, μ_{2}^{∗} and μ_{3}^{∗} are all distinct, then R_{ii,k}^{∗} = 0 for i, k = 1, 2, 3. This fact and (2.15) mean

(3.6)  ∑_{i,j,k} μ_{i}^{∗}R_{ij,k}^{∗}^{2} = \frac{1}{3} ∑_{i≠j≠k≠i}(μ_{i}^{∗} + μ_{j}^{∗} + μ_{k}^{∗})R_{ij,k}^{∗}^{2}

+ ∑_{i≠j}(2μ_{i}^{∗} + μ_{j}^{∗})R_{ii,j}^{∗}^{2} + ∑_{i} μ_{i}^{∗}R_{ii,i}^{∗}^{2}

= 0.

Therefore (2.21) and (3.6) imply

(3.7)  ∑_{i,j,k,l} R_{ij,kl}^{∗}^{2} = −\frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{∗}^{2} − \frac{9}{4}B(B − \frac{1}{6}r^{2}).

This is a contradiction because B = S − r^{2}/3 > r^{2}/6 and r ≥ 0.

(ii) If μ_{1}^{∗}, μ_{2}^{∗} and μ_{3}^{∗} are not distinct with each other, then we have μ_{1}^{∗} = μ_{2}^{∗} = μ_{3}^{∗} or 2μ_{1}^{∗} = 2μ_{2}^{∗} = −μ_{3}^{∗}.

(ii-1) μ_{1}^{∗} = μ_{2}^{∗} = μ_{3}^{∗} case. In this case, (2.15) implies μ_{1}^{∗} = μ_{2}^{∗} = μ_{3}^{∗} = 0 and hence we have S = r^{2}/3.

(ii-2) 2μ_{1}^{∗} = 2μ_{2}^{∗} = −μ_{3}^{∗} case. In this case, we have from (2.15)

B_{3}^{∗} = ±\frac{B^{\frac{3}{2}}}{\sqrt{6}} .

(a) If

B_{3}^{∗} = −\frac{B^{\frac{3}{2}}}{\sqrt{6}} ,

then, Lemma 1 implies B_{3} =constant. Hence the λ_{i}’s are constant. Thus, we have ∇_{l}R_{ij} = 0. So we see from (2.18)

(3.8)  0 = −3B_{3}^{∗} − \frac{1}{2}rB = B(\frac{3\sqrt{B}}{\sqrt{6}} − \frac{1}{2}r).

Thus, B = 0 or 3\sqrt{B}/\sqrt{6} = r/2, that is, we have S = r^{2}/3 or S = r^{2}/2.

(b) If

B_{3}^{∗} = \frac{B^{\frac{3}{2}}}{\sqrt{6}} ,

then we see from (2.18)

0 ≥ \frac{3}{\sqrt{6}}B^{\frac{3}{2}} + \frac{1}{2}rB = 3B(\begin{matrix}√ & \overline{B} \\ √ & \overline{6} \end{matrix} + \frac{1}{6}r)≥ 0.

This means B = 0, that is, we have S = r^{2}/3. Thus, the proof of Theorem 1 is completed. □

Next we consider the case of r < 0.

Theorem 2. In addition to the assumptions in Main Theorem 1, if r < 0, then the squared norm S of the Ricci curvature tensor satisfies the following inequality:

(3.9)  \frac{r^{2}}{3} ≤ S ≤ \frac{r^{2}}{2} .

Proof. By rewriting (2.19) in term of B_{3}, we have

(3.10)  ΔB_{3} = \frac{3}{2}(B^{2} + rB_{3} + 4 ∑_{i,j,k} μ_{i}R_{ij,k}^{2}).

Hence, by using (3.2), we have

(3.11)  B^{2} + rB_{3}^{∗} + 4 ∑_{i,j,k} μ_{i}^{∗}R_{ij,k}^{∗}^{2} ≤ 0.

(i) If μ_{i}^{∗}’s are distinct with each other, then by making use of the similar method to the proof of Theorem 1, we have

(3.12)  ∑_{i,j,k} μ_{i}^{∗}R_{ij,k}^{∗}^{2} = 0

and hence the inequality

(3.13)  B^{2} + rB_{3}^{∗} ≤ 0.

On the other hand, by using (2.18) we know

(3.14)  r^{2}B + 6rB_{3}^{∗} ≥ 0.

Thus, (3.13) and (3.14) imply the inequality

−\frac{1}{2}r^{2}B ≤ 3rB_{3}^{∗} ≤ −3B^{2}.

Therefore, the proposed relation

\frac{r^{2}}{3} ≤ S ≤ \frac{r^{2}}{2}

in Theorem 2 is valid in this case.

(ii) If μ_{i}^{∗}’s are not distinct with each other, then by the same reasons as in Theorem 1 we have S = r^{2}/3 or

(3.15)  B_{3}^{∗} = ±\frac{B^{\frac{3}{2}}}{\sqrt{6}} .

(a) If

B_{3}^{∗} = −\frac{B^{\frac{3}{2}}}{\sqrt{6}} ,

then ∇_{l}R_{ij} = 0 by the same reason as in the case of (ii-2) of the proof of Theorem 1. Therefore we have from (2.18)

(3.16)  0 = −3B_{3}^{∗} − \frac{1}{2}rB = B(\frac{3\sqrt{B}}{\sqrt{6}} − \frac{r}{2}).

Because of r < 0, we see 3\sqrt{B}/\sqrt{6} − r/2 > 0. Hence, B = 0, that is, S = r^{2}/3 is valid. It should be remarked that in this case r < 0, we can only get S = r^{2}/3 and we can not conclude S = r^{2}/2 from the above equality (3.16).

(b) If

B_{3}^{∗} = \frac{B^{\frac{3}{2}}}{\sqrt{6}} ,

then we see from (2.18)

3\frac{B^{\frac{3}{2}}}{\sqrt{6}} + \frac{1}{2}rB = 3B(\frac{1}{\sqrt{6}}\sqrt{B} + \frac{r}{6})≤ 0.

This means B ≤ r^{2}/6. Hence we conclude S ≤ r^{2}/2. This completes the proof of Theorem 2. □

Proof of Main Theorem 1. From Theorem 1, we know that the assertion (1) in Main Theorem 1 is true. When r < 0, we have the inequality r^{2}/3 ≤ S ≤ r^{2}/2 from Theorem 2. If S = r^{2}/3, then we know that

∑_{ij}(R_{ij} − \frac{1}{3}rδ_{ij})^{2}= 0.

Hence the conformally flat Riemannian manifold M^{3} is Einstein space. So we can conclude that M^{3} is space form by which the assertion (2) in Main Theorem 1 is also true. This completes the proof of Main Theorem 1. □

In order to prove Main Theorem 2, we will prepare the following Theorem 3.

Theorem 3. In addition to the assumptions in Main Theorem 1, if r < 0, then we have

sup B_{3} = \frac{B^{\frac{3}{2}}}{\sqrt{6}} .

Proof. We use the same notations above. suppose B_{3}^{∗} := sup B_{3} ≠ B^{\frac{3}{2}} /\sqrt{6}, then we shall get a contradiction. To do it, we first prepare the following lemma:

Lemma 2. If B_{3}^{∗} ≠ B^{\frac{3}{2}} /\sqrt{6}, then we have (1) μ_{1}^{∗}, μ_{2}^{∗}, μ_{3}^{∗} are all distinct.

(2) 0 < B < \frac{r^{2}}{6} .

(3) ∑_{i,j,k,l} R_{ij,kl}^{∗}^{2} = −\frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{∗}^{2} − \frac{9}{4}B(B − \frac{r^{2}}{6} ).

(4) B_{3}^{∗} ≥ −\frac{B^{2}}{r} > 0.

Proof. As a consequence of Lemma 1 noted in the proof of Theorem 1, the statement (1) is obtained.

In view of Theorem 2 and its proof, the statement (1) yields the inequality

\frac{r^{2}}{3} < S < \frac{r^{2}}{2}

that is, the relation (2) holds:

(3.17)  0 < B < \frac{r^{2}}{6} .

The relation (3) is proved as follows: From the statement (1), we have, by using (3.5),

(3.18)  R_{ii,k}^{∗} = 0 for i, k = 1, 2, 3.

Hence (2.18) means

(3.19)  ∑_{i,j,k} R_{ij,k}^{∗}^{2} = 6R_{12,3}^{∗}^{2} = −3 ∑_{i} μ_{i}^{∗}^{3} − \frac{r}{2}B.

By differentiating (2.18) and using (3.1), we get

(3.20)  ∑_{i,j,k} R_{ij,k}^{∗}R_{ij,kl}^{∗} = −\frac{3}{2} lim_{m→∞} ∇_{l}B_{3}(p_{m}) = 0 for l = 1, 2, 3.

According to (3.18) and (3.20), we infer

(3.21)  R_{ij,kl}^{∗} = 0 for i ≠ j ≠ k ≠ i.

From the Ricci formula (2.3-i) and (2.1), we have

(3.22)  R_{ij,kk}^{∗} − R_{kk,ij}^{∗} = ∑_{t} R_{it}^{∗}R_{tkjk}^{∗} + ∑_{t} R_{tk}^{∗}R_{tijk}^{∗} = 0 for i ≠ j ≠ k ≠ i.

Thus, (3.21) and (3.22) yield

(3.23)  R_{ii,jk}^{∗} = 0 for i ≠ j ≠ k ≠ i.

Because of

(3.24)  0 = \frac{1}{2}∇_{l}∇_{k}S = ∑_{i,j} R_{ij}R_{ij,kl} + ∑_{i,j} R_{ij,k}R_{ij,l}

and

0 = ∇_{l}∇_{k}r = ∑_{i} R_{ii,kl},

we have

(3.25)  ∑_{i} R_{ii,kl}^{∗} = 0

and

(3.26)  ∑_{i} λ_{i}^{∗}R_{ii,kl}^{∗} + ∑_{i,j} R_{ij,k}^{∗}R_{ij,l}^{∗} = 0.

In view of Ricci formula (2.3-i) and (2.1), the relations (3.23), (3.25) and (3.26) imply

(3.27)  R_{jj,jk}^{∗} = R_{kk,jk}^{∗} = 0 for j ≠ k.

Thus, (2.21), (3.6), (3.21), (3.23) and (3.27) yield the relation (3) of this Lemma 2:

(3.28)  ∑_{i,j,k,l} R_{ij,kl}^{∗}^{2} = 3 ∑_{i,l} R_{ii,ll}^{∗}^{2} − 2 ∑_{i} R_{ii,ii}^{∗}^{2}

= −\frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{∗}^{2} − \frac{9}{4}B(B − \frac{1}{6}r^{2}).

Finally, the inequality (4) is proved as follows: By using (2.19), we have

0 ≥ \frac{1}{3} lim_{m→∞} sup Δ ∑_{i} μ_{i}^{3}(p_{m}) = lim_{m→∞} sup(−Δ ∑_{i,j,k} R_{ij,k}^{2}(p_{m}))

= \frac{9}{2}(B^{2} + r ∑_{i} B_{3}^{∗} + 4 ∑_{i,j,k} μ_{i}^{∗}R_{ij,k}^{∗}^{2}).

Since μ_{i}^{∗}’s are all distinct, (3.18) is also true. Hence,

∑_{i,j,k} μ_{i}^{∗}R_{ij,k}^{∗}^{2} = 0.

Thus the inequality

(3.29)  B_{3}^{∗} ≥ −\frac{B^{2}}{r} > 0

holds because r < 0. Thus the proof of Lemma 2 is finished.

Now we continue to prove the Theorem 3. By taking k = l in (3.25) and (3.26), we have

(3.30)  ∑_{i} R_{ii,ll}^{∗} = 0 for l = 1, 2, 3

and

(3.31)  ∑_{i} μ_{i}^{∗}R_{ii,ll}^{∗} = − ∑_{i,j} R_{ij,l}^{∗}^{2} = −2R_{12,3}^{∗}^{2}

= \frac{1}{6}rB + B_{3}^{∗} for l = 1, 2, 3,

respectively. On the other hand, the Ricci formula (2.3-i) and (2.1) yield

(3.32)  R_{ij,ij}^{∗} − R_{ji,ji}^{∗} = (μ_{i}^{∗} − μ_{j}^{∗})(μ_{i}^{∗} + μ_{j}^{∗} + \frac{1}{6}r) for l = 1, 2, 3.

In view of (3.32), we can solve the rank 5 linear system (3.30) and (3.31) of six equations and six unknowns R_{ii,ll}^{∗}. In fact, the solutions are

(3.33)  R_{ii,ll}^{∗} = μ_{i}^{∗}^{2} + \frac{1}{6}rμ_{i}^{∗} − \frac{1}{3}B + yg_{i}g_{l}.

Here

(3.34)  g_{i} = μ_{i}^{∗}^{2} − \frac{B_{3}^{∗}}{B} μ_{i}^{∗} − \frac{1}{3}B

and y is a quantity which need not to be determined. Hence we get

(3.35)  ∑_{i,l} R_{ii,ll}^{∗}^{2} = ∑_{i,l}(μ_{i}^{∗}^{2} + \frac{1}{6}rμ_{i}^{∗} − \frac{B}{3} + yg_{i}g_{l})^{2}

= \frac{B}{2} (B + \frac{1}{6}r^{2})+x^{2} + rB_{3}^{∗}

and

(3.36)  ∑_{i} R_{ii,ii}^{∗}^{2} = ∑_{i}(μ_{i}^{∗}^{2} + \frac{1}{6}rμ_{i}^{∗} − \frac{1}{3}B + yg_{i}^{2})^{2}

= \frac{1}{2}B^{2} + \frac{1}{36}r^{2}B + \frac{B^{2}}{3} + \frac{1}{2}x^{2} + \frac{1}{3}rB_{3}^{∗} − \frac{2}{3}B^{2}

+ 2y ∑_{i} μ_{i}^{∗}^{2}g_{i}^{2} + \frac{1}{3}yr ∑_{i} μ_{i}^{∗}g_{i}^{2} − \frac{2}{3}Bx

= \frac{1}{2}x^{2} − \frac{1}{3}x(B + \frac{1}{B} rB_{3}^{∗})+\frac{1}{3}rB_{3}^{∗} + \frac{1}{6}B^{2} + \frac{1}{36}r^{2}B,

where

x = y(\frac{B^{2}}{6} − \frac{B_{3}^{∗}^{2}}{B} ).

Here we used

∑_{i} μ_{i}^{∗}^{2}g_{i}^{2} = \frac{B}{6} (\frac{B^{2}}{6} − \frac{B_{3}^{∗}^{2}}{B} )

and

∑_{i} μ_{i}^{∗}g_{i}^{2} = \frac{B_{3}^{∗}}{B} (\frac{B_{3}^{∗}^{2}}{B} − \frac{B^{2}}{6} ).

By using (3) in Lemma 2, (3.35) and (3.36), we have

(3.37)  ∑_{i,j,k,l} R_{ij,kl}^{∗}^{2} = 2x^{2} + \frac{2}{3}x(B + r \frac{B_{3}^{∗}}{B} )+\frac{3}{2}B(B + \frac{1}{6}r^{2})

− \frac{B^{2}}{3} − r^{2} \frac{B}{18} + 3rB_{3}^{∗} − \frac{2}{3}rB_{3}^{∗}

= 2x^{2} + \frac{2}{3}x(B + r \frac{B_{3}^{∗}}{B} )+\frac{7}{6}B(B + \frac{1}{6}r^{2})+\frac{7}{3}rB_{3}^{∗}

− \frac{5}{12}r ∑_{i,j,k} R_{ij,k}^{∗}^{2} − \frac{9}{4}B(B − \frac{1}{6}r^{2})

= \frac{5}{4}rB_{3}^{∗} − \frac{9}{4}B(B − \frac{1}{6}r^{2})+\frac{5}{24}r^{2}B.

Therefore, we obtain the quadratic equation of x

(3.38)  x^{2} + \frac{1}{3}x(B + r \frac{B_{3}^{∗}}{B} )+\frac{13}{24}rB_{3}^{∗} + \frac{41}{24}B^{2} − \frac{7}{36}r^{2}B = 0

of which the solutions are

(3.39)  x = −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} )±\frac{1}{6}\sqrt{r^{2}(\frac{B_{3}^{∗}}{B} )^{2}−\frac{35}{2}rB_{3}^{∗} + 7r^{2}B − \frac{121}{2}B^{2}}.

Since μ_{i}^{∗}’s are all distinct as noted in Lemma 2, we can conclude, by using the inequality (4) in Lemma 2

(3.40)  μ_{1}^{∗} < μ_{2}^{∗} < 0 and μ_{3}^{∗} > 0

and

(3.41)  \frac{1}{6}B < μ_{1}^{∗}^{2} < \frac{1}{2}B + \frac{\sqrt{2}}{3} \frac{B^{\frac{3}{2}}}{r} , 0 < μ_{2}^{∗}^{2} < \frac{1}{6}B, \frac{1}{2}B < μ_{3}^{∗}^{2} < \frac{2}{3}B.

Here we assumed without loss of generality μ_{1}^{∗} < μ_{2}^{∗} < μ_{3}^{∗} and used the formulas B_{3}^{∗} = 3μ_{j}^{∗}(μ_{j}^{∗}^{2} − B/2) for j =1, 2 and 3.

On the other hand, by making use of the similar computation as in (2.19), we have from (3.2) the inequality

0 ≥ lim_{m→∞} \frac{1}{3} sup ∇_{l}∇_{l}B_{3}(p_{m})

= ∑_{i} λ_{i}^{∗}^{2}R_{ii,ll}^{∗} + 2 ∑_{i,j} λ_{i}^{∗}R_{ij,l}^{∗}^{2}.

From (3.18) and (3.19), we obtain

2 ∑_{i,j} μ_{i}^{∗}R_{ij,l}^{∗}^{2} = ∑_{i,j}(μ_{i}^{∗} + μ_{j}^{∗})R_{ij,l}^{∗}^{2} = −2μ_{l}^{∗}R_{12,3}^{∗}

= (\frac{1}{6}rB + B_{3}^{∗})μ_{l}^{∗}.

Hence, by using (3.30), (3.33) and this relation, we get the inequality

0 ≥ ∑_{i} λ_{i}^{∗}^{2}R_{ii,ll}^{∗} + 2 ∑_{i,j} μ_{i}^{∗}R_{ij,l}^{∗}^{2} + \frac{2}{3}r ∑_{i,j} R_{ij,l}^{∗}^{2} (3.42)

= ∑_{i}(μ_{i}^{∗}^{2} + \frac{2}{3}rμ_{i}^{∗})(μ_{i}^{∗}^{2} + \frac{1}{6}rμ_{i}^{∗} − \frac{1}{3}B + yg_{i}g_{l})

+ (\frac{1}{6}rB + B_{3}^{∗})μ_{l}^{∗} + \frac{2}{3}r(−\frac{1}{6}rB − B_{3}^{∗})

= (\frac{1}{6}rB + B_{3}^{∗})μ_{l}^{∗} + xg_{l} + \frac{1}{6}B^{2} + \frac{1}{6}rB_{3}^{∗}

= (−\frac{3}{B} x(μ_{l}^{∗}^{2} − \frac{2}{3}B) + 3μ_{l}^{∗}^{2} + \frac{1}{2}rμ_{l}^{∗} − B)(μ_{l}^{∗}^{2} − \frac{1}{6}B).

Next, we estimate x by using (4) in Lemma 2 and so on.

Lemma 3. If B_{3}^{∗} ≠ B^{\frac{3}{2}} /\sqrt{6}, then

x ≤ −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} ).

Proof. If x > −(B + r \frac{B_{3}^{∗}}{B})/6, then, by taking l = 1 in (3.42) and using (3.17), (3.29) and (3.41), we have

0 ≥ −\frac{3}{B} x(μ_{1}^{∗}^{2} − \frac{2}{3}B)+3μ_{1}^{∗}^{2} + \frac{1}{2}rμ_{1}^{∗} − B3

≥ \frac{3}{2B} (μ_{1}^{∗}^{2} − \frac{2}{3}B)(B + r \frac{B_{3}^{∗}}{B} )+3μ_{1}^{∗}^{2} + \frac{1}{2}rμ_{1}^{∗} − B

≥ 3μ_{1}^{∗}^{2} + \frac{1}{2}rμ_{1}^{∗} − B

> −\frac{1}{2}(B + \frac{1}{\sqrt{6}}r\sqrt{B})> 0.

This is a contradiction. Thus the proof of lemma 3 is finished.

Now, Lemma 3 means that

(3.43)  x = −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B}) − \frac{1}{6}\sqrt{r^{2}(\frac{B_{3}^{∗}}{B})^{2} − \frac{35}{2}rB_{3}^{∗} + 7Br^{2} − \frac{121}{2}B^{2}}.

Finally we estimate B as follows.

Lemma 4. If B_{3}^{∗} ≠ B^{3/2}/\sqrt{6}, then

B < \frac{16}{(47)^{2} · 9}r^{2}.

Proof. By taking l = 2 in (3.42) and using the inequality μ_{2}^{∗}^{2} − B/6 < 0 in (3.41), we have

(3.44)  0 ≤ −\frac{3}{B} x(μ_{2}^{∗}^{2} − \frac{2}{3}B)+3μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − B.

First of all, we assert B < r^{2}/7. In fact, since B and r are constant by the assupmtion and B < r^{2}/6 as noted in Lemma 2, there exists a constant a > 0 such that (1+a)B < r^{2}/6. If a ≥ 1/6, then B < r^{2}/7. In this case our assertion holds. If a < 1/6, then we have

(3.45)  x = −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} )−\frac{1}{6}\sqrt{((1 − 6a)B + r \frac{B_{3}^{∗}}{B} )^{2}+G},

where G is defined by

G = −(1 − 6a)^{2}B^{2} − 2(1 − 6a)rB_{3}^{∗} − \frac{35}{2}rB_{3}^{∗} + 7Br^{2} − \frac{121}{2}B^{2}.

According to rB_{3}^{∗} ≤ −B^{2} in Lemma 2 and (1 + a)B < r^{2}/6 we have

G ≥ −(1 − 6a)^{2}B^{2} + 2(1 − 6a)B^{2} + \frac{35}{2}B^{2} + 42(1 + a)B^{2} − \frac{121}{2}B^{2}
= B^{2}(42a − 36a^{2}) > 0.

Therefore, we obtain

x ≤ −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} )+\frac{1}{6}((1 − 6a)B + r \frac{B_{3}^{∗}}{B} )= −aB.

Using this inequalty, μ_{2}^{∗}^{2} − B/6 < 0 in (3.41) and μ_{2}^{∗} < 0, (3.44) implies

0 ≤ 3a(μ_{2}^{∗}^{2} − \frac{2}{3}B)+3μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − B

= 3(a + 1)μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − (2a + 1)B

< \frac{1}{2}(a + 1)B − \frac{1}{2\sqrt{6}}\sqrt{B} − (2a + 1)B.

Hence, we have

(1 + 6a)B < \frac{r^{2}}{6} .

If 6a ≥ 1/6, then our assertion is true. If 6a < 1/6, then by applying the above procedures for a_{1} = 6a, we get

(1 + 6^{2}a)B < \frac{r^{2}}{6} .

If we repeat the above procedure k times so that 6^{k}a ≥ 1/6, then we have

(1 + 6^{k}a)B < \frac{r^{2}}{6} .

Thus we obtain B < r^{2}/7, that is, our assertion holds.

Next, by taking account of B < r^{2}/7, we will show B < r^{2}/24 and B < (4r)^{2}/(3·47)^{2} finally. In fact, from (3.43), rB_{3}^{∗} ≤ −B^{2} of (4) in Lemma 2 and B < r^{2}/7, we have

(3.46)  x < −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} )−\frac{1}{6}(B − r \frac{B_{3}^{∗}}{B} )= −\frac{B}{3} .

Hence, by substituting the above inequality (3.46) into (3.44), from 0 < μ_{2}^{∗}^{2} < B/6 and μ_{2}^{∗} < 0, we get

(3.47)  0 ≤ (μ_{2}^{∗}^{2} − \frac{2}{3}B)+3μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − B = 4μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − \frac{5}{3}B

< \frac{2}{3}B − \frac{1}{2\sqrt{6}}r\sqrt{B} − \frac{5}{3}B = −\sqrt{B}(\sqrt{B} + \frac{r}{2\sqrt{6}}).

Thus we have

B < \frac{r^{2}}{24}.

By applying the above method to B < r^{2}/24, which introduced (3.46) and (3.47), we get

B < \frac{16}{(47)^{2} · 9}r^{2}.

Thus the proof of Lemma 4 is completed.

In view of Lemma 4 and the inequality rB_{3}^{∗} ≤ −B^{2} of (4) in Lemma 2, we obtain

(−\frac{35}{2}r − \frac{4r^{2}}{3\sqrt{B}} )B_{3}^{∗} + \frac{19}{3}r^{2}B − \frac{121}{2}B^{2} + \frac{1}{3}r^{2} \frac{B_{3}^{∗}^{2}}{B^{2}}

= \frac{1}{3}(\frac{rB_{3}^{∗}}{B} − \frac{105B}{4} − 2r\sqrt{B})^{2}−\frac{1}{3}(\frac{105B}{4} + 2r\sqrt{B})^{2}+\frac{19}{3}r^{2}B − \frac{121}{2}B^{2} > 0.

Therefore, we have

(3.48)  x < −\frac{1}{6}(B + r \frac{B_{3}^{∗}}{B} )−\frac{1}{6}\sqrt{\frac{2}{3}(r\sqrt{B} + r \frac{B_{3}^{∗}}{B} )^{2}}

= −\frac{1}{6}B + \frac{1}{3\sqrt{6}}r\sqrt{B} − \frac{1}{6}(1 − \frac{2}{\sqrt{6}})r \frac{B_{3}^{∗}}{B} .

Hence, by substituting the above inequality (3.48) into (3.44) and using the inequality rB_{3}^{∗} ≤ −B^{2} of (4) in Lemma 2, we get finally

0 ≤ −\frac{3}{B} (μ_{2}^{∗}^{2} − \frac{2}{3}B)(−\frac{1}{6}B + \frac{1}{3\sqrt{6}}r\sqrt{B} − \frac{1}{6}(1 − \frac{2}{\sqrt{6}})r \frac{B_{3}^{∗}}{B} )+3μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − B

≤ \frac{3}{B} (μ_{2}^{∗}^{2} − \frac{2}{3}B)(\frac{1}{6}B − \frac{1}{3\sqrt{6}}r\sqrt{B} − \frac{1}{6}(1 − \frac{2}{\sqrt{6}})B)+3μ_{2}^{∗}^{2} + \frac{1}{2}rμ_{2}^{∗} − B

< −\frac{1}{2\sqrt{6}}B + \frac{1}{2}B − \frac{1}{2\sqrt{6}}rB − B < 0.

But this is impossible. Thus the proof of Theorem 3 is completed. □

Proof of Main Theorem 2. From Lemma 1 in the proof of Theorem 1, it follows that B_{3}^{∗} = B^{3/2}/\sqrt{6} if and only if two of μ_{i}^{∗}’s are equal with each other. Therefore, from Theorem 3 and the assumptions (1) and (2) in Main Theorem 2 we complete the proof of Main Theorem 2. □

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Department of Mathematics, Faculty of Science, Josai University, Saitama, Sakado 350-02, Japan

Department of Mathematics, Saga University, Saga 840, Japan

Graduate school of Mathematics, Kyushu University, Fukuoka 812, Japan