Two rigidity theorems for fully nonlinear equations

Differential Geometry and its Applications 37 (2014) 17–32

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Differential Geometry and its Applications www.elsevier.com/locate/difgeo

Two rigidity theorems for fully nonlinear equations ✩ Haizhong Li a,b , Changwei Xiong a,∗ a b

Department of Mathematical Sciences, Tsinghua University, 100084, Beijing, PR China Mathematical Sciences Center, Tsinghua University, 100084, Beijing, PR China

a r t i c l e

i n f o

Article history: Received 14 April 2014 Available online xxxx Communicated by F. Fang MSC: 53C21 53C20

a b s t r a c t This paper is concerned with the fully nonlinear equation σ2 (g) = aσ1 (g) + b. The first result is to obtain the entire solutions of the equation for conformally flat metric on Rn under some additional assumptions, which generalizes the famous result of Chang–Gursky–Yang in [3]. The second one is to classify compact locally conformally flat manifolds with nonnegative Ricci curvature and the metric g satisfying the equation. © 2014 Elsevier B.V. All rights reserved.

Keywords: Liouville Theorem Fully nonlinear equation Locally conformally flat

1. Introduction Let (M n , g) be an n-dimensional compact Riemannian manifold. It is well-known that the Riemannian curvature tensor has the following decomposition: Rijkl = Wijkl +

1 (Aik gjl + Ajl gik − Ail gjk − Ajk gil ), n−2

where W is the Weyl curvature tensor and A is the Schouten tensor defined as Ag = Ric g −

Rg g. 2(n − 1)

(1)

Since W is conformally invariant, the conformal transformation of the Riemannian curvature tensor is consistent with that of the Schouten tensor Ag . In view of that, the Schouten tensor Ag has always been playing a central role in conformal geometry. ✩

The research of the authors was supported by NSFC No. 11271214.

* Corresponding author. E-mail addresses: [email protected] (H. Li), [email protected] (C. Xiong). http://dx.doi.org/10.1016/j.difgeo.2014.08.005 0926-2245/© 2014 Elsevier B.V. All rights reserved.

18

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

Associated with Schouten tensor Ag , J.A. Viaclovsky in [15] introduced the so-called σk (g) curvature, which is defined to be   σk (g) := σk g −1 · Ag =



λi1 · · · λik ,

(2)

i1 <···
where λ1 , · · · , λn are the eigenvalues of g −1 · Ag , i.e., σk (g) is the k-th elementary symmetric polynomial with respect to the eigenvalues of the matrix g −1 · Ag . Note that σ1 (g) = tr g (Ag ) is just the scalar curvature Rg up to a constant multiple, and σ2 (g) has the following expression: σ2 (g) =

2  1  tr g (Ag ) − |Ag |2g . 2

(3)

Furthermore, he found that the metric g˜ satisfying σk (˜ g ) = const is in fact related to a variational problem, which can be stated precisely as follows. Denote by [g] the conformal class of the metric g and by [g]1 the restriction of [g] to the unit volume set. Then if k = n/2 and (M n , g) is locally conformally flat, a metric g˜ ∈ [g]1 is a critical point of the functional  Fk (˜ g) =

σk (˜ g )dvg˜

(4)

M

restricted to [g]1 if and only if σk (˜ g ) = const.

(5)

And if M is not locally conformally flat, the statement is true for k = 1 and k = 2. From the last assertion, we can see that the cases k = 1 and k = 2 are special, and for fixed a ∈ R, the metric g˜ ∈ [g]1 satisfying σ2 (˜ g ) − aσ1 (˜ g ) = const

(6)

will correspond to the critical point of some linear combination of the functionals F1 and F2 restricted to [g]1 . See also [8,9]. This is the motivation of the present paper. The issue involved in the first part of this paper can be originated from the classical theorem of Obata [13]. There he proved that any metric conformal to the standard metric on the n-sphere Sn with constant scalar curvature is isometric to the standard metric. Or equivalently, he got the entire solutions to the equation σ1 (g) = const,

g ∈ [g0 ] on Sn ,

(7)

where g0 is the standard metric on Sn . Via stereographic projection, equations on Sn are equivalent to those on Rn with the assumption that the singularity at infinity is removable. So equations on Rn without any assumption are less restricted and more interesting. In this aspect, by using the method of moving planes, Caffarelli, Gidas and Spruck [1] classified the entire solutions of the equation n+2

−Δu = n(n − 2)u n−2 ,

u > 0 on Rn ,

(8)

to prove that the solutions must be  u(x) =

2λ 2 λ + |x − x0 |2

2  n−2

(9)

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

19

4

for some x0 ∈ Rn and λ > 0; or equivalently the metric u(x) n−2 dx2 is isometric to the standard metric on Sn via stereographic projection. Note that in our notation, Eq. (8) is equivalent to the following equation for k = 1: σk (g) = const,

 g ∈ dx2 on Rn ,

(10)

where dx2 is the standard Euclidean metric. Thus after the celebrated work of Caffarelli et al. [1], it is natural to consider the equation for general k. This kind of results is also referred to as Liouville-type theorems. For general k ≥ 2, J.A. Viaclovsky [16] got the Liouville Theorem but under a strong growth condition at infinity. For the important case k = 2, in [2,3], A. Chang, M. Gursky and P. Yang proved the Liouville Theorem for n = 4, 5 and for n ≥ 6 with a finite volume assumption. At last, A. Li and Y.Y. Li [12] dealt with general case k ≥ 2 without additional assumption. See also the work of Guan and Wang [7] on general locally conformally flat manifold. In this paper, along the approach of Obata [13] and Chang et al. [3], we prove the following: Theorem 1. Let g = v −2 dx2 be a conformal metric on Rn , n ≥ 4, satisfying σ2 (g) = aσ1 (g) + b, Moreover, if n ≥ 6, we assume vol(g) =

Rn

b > 0.

 v −n dx < ∞. If n = 5, we assume a > − 57 b. Then v =

p|x|2 + qi xi + r for constants p, qi , r. In particular, g is obtained by pulling back to Rn the round metric on Sn . When a = 0, Theorem 1 reduces to the result in [3]. And our result is another example showing the power of Obata’s method. In the second part of this paper we study the problem to classify compact locally conformally flat Riemannian manifolds with σ2 (g) − aσ1 (g) = b being constant. The corresponding problem for σ1 (g) i.e. scalar curvature being constant is handled by Tani [14], Wegner [17] and Cheng [4] (see Theorem 7 below for precise statement). And the case σk (g)(k ≥ 2) being constant was considered by Z.J. Hu, H. Li and U. Simon [10]. See also [5,6,11] for relevant work in the hypersurface context. By applying the method of [10], we prove the following theorem. Theorem 2. Let (M n , g) (n ≥ 3) be a compact locally conformally flat manifold with σ2 (g) = aσ1 (g) + b,

(11)

where a and b are constants satisfying a2 + 2b ≥ 0. Moreover, assume the Ricci tensor is semi-positive definite. Then (M n , g) is either a space form or a space S1 × N n−1 with N a space form. When a = 0, the result reduces to that of [10]. The present paper is built up as follows. Some notations and lemmas for proving Theorem 1 are prepared in Section 2. And Section 3 is devoted to the proof of Theorem 1. Relatively independent, Section 4 includes the proof of Theorem 2. 2. Preliminaries Let (M n , g) be an n-dimensional Riemannian manifold. The Schouten tensor Ag is defined as Ag = Ric g −

Rg g, 2(n − 1)

where Ric g is the Ricci tensor and Rg the scalar curvature.

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

20

Recall the σk (g) scalar curvature is given by   σk (g) = σk g −1 · Ag , where σk is the k-th elementary symmetric polynomial on Rn . For locally conformally flat manifolds (M n , g), it is well known that Ag is a Codazzi tensor. That is, it satisfies Aij,k = Aik,j . Using this property, J.A. Viaclovsky [16] proved that the k-th Newtonian tensor Tk is divergence-free. In this paper we are only concerned with the first two Newtonian tensors T1 and T2 , which are defined as follows: T1 = σ1 (g)g − Ag , T2 = σ2 (g)g − σ1 (g)Ag + A2g . Now go back to our equation σ2 (Ag ) = aσ1 (Ag ) + b. Related to our equation is the following tensor: T = T2 −

n−2 aT1 . n−1

Denote by L the trace-free part of T . That is, L = T2 −

n−2 n−2 aT1 − (σ2 − aσ1 )g. n−1 n

It is easy to see that when σ2 −aσ1 = b is a constant, L is divergence-free. Furthermore, L has two important properties as follows. Proposition 3. Denote by E the trace-free part of Ricci tensor. Assume that σ2 (g) − aσ1 (g) + n and σ1 (g) − n−1 a > 0. Then (1) L, E ≤ 0, with equality if and only if E = 0. n (2) |L|2 ≤ −2(n−2) (σ1 (g) − n−1 a)L, E. n Proof. (1) Let A˜ = Ag −

a n−1 g.

And for simplicity, we write σi for σi (g), i = 1, 2. Then ˜ = σ1 − σ1 (A)

n a > 0, n−1

˜ = σ2 − aσ1 + σ2 (A)

na2 > 0. 2(n − 1)

And by direct computation we have ˜ = σ2 (A)g ˜ − σ1 (A) ˜ A˜ + A˜2 T2 (A) = T2 − Therefore,

n−2 (n − 2)a2 aT1 + g. n−1 2(n − 1)

na2 2(n−1)

>0

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

L, E =

n−2 n−2 aT1 − (σ2 − aσ1 )g, E T2 − n−1 n

2 ˜ − (n − 2)a g − n − 2 (σ2 − aσ1 )g, E = T2 (A) 2(n − 1) n

˜ − n − 2 σ2 (A)g, ˜ A˜ , = T2 (A) n

where in the last line we have used that L is trace-free. Using the contracted formula tr(Tk A) = (k + 1)σk+1 and Newton–Maclaurin inequality, we obtain ˜ − n − 2 σ2 (A)σ ˜ 1 (A) ˜ ≤ 0, L, E = 3σ3 (A) n and the equality holds if and only if A˜ = λg, that is E = 0. (2) In terms of the trace-free Ricci tensor, we have 1 σ1 g, n 2 1 A2g = E 2 + σ1 E + 2 σ12 g, n n Ag = E +

which implies T1 = σ1 g − Ag =

n−1 σ1 g − E, n

T2 = σ2 g − σ1 Ag + A2g

  n−2 n−1 2 σ1 E + σ2 − =E − σ g. n n2 1 2

Consequently,  2   n−2 n−2  aT1 − (σ2 − aσ1 )g  |L| = T2 − n−1 n 2     2 tr E 2  n−2 n−2  a− σ1 E − g = E + n−1 n n   2   2 n−2 n−2 1 n−2 n−2 4 3 a− σ1 tr E + a− σ1 tr E 2 − tr E 2 . = tr E + 2 n−1 n n−1 n n 2

Meanwhile, we have  L, E = tr E 3 +

 n−2 n−2 a− σ1 tr E 2 . n−1 n

Therefore,  |L|2 = tr E 4 + 2

  2 2 n−2 n−2 n−2 n−2 1 a− σ1 L, E − a− σ1 tr E 2 − tr E 2 . n−1 n n−1 n n

Now by Lemma 2.3 in [3], there holds tr E 4 ≤

2 n2 − 3n + 3  tr E 2 . n(n − 1)

21

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

22

Hence, we obtain 2 2 n−2 n−2 1 a− σ1 tr E 2 − tr E 2 − n−1 n n 2   n−2 (n − 2)2  n−2 2 2 tr E a− σ1 tr E 2 − n(n − 1) n−1 n  2   σ1 (n − 2)2 a 2 2 tr E tr E − n(n − 1) − n(n − 1) n−1 n   n (n − 2)2 tr E 2 −2σ2 + 2aσ1 − a2 ≤ 0. n(n − 1) n−1 

tr E

4

≤ = =

Then the conclusion follows immediately. 2 Next we need a proposition to determine the range which our σ1 (g) belongs to. 4

Proposition 4. (See [2] for n = 4.) There exists no conformal metric g = u n−2 dx2 on Rn such that its scalar curvature Rg ≤ −c for fixed positive constant c. 4

Proof. Assume that there exists such g. Under the transformation g = u n−2 dx2 , we have n+2 n+2 n−2 Rg u n−2  −u n−2 , 4(n − 1)

−Δu =

where and thereafter the notation “ ( resp.)” means “≤ (≥ resp.) up to a harmless constant multiple”. For ρ > 0, let η be a cut off function supported in the ball B(2ρ) with η ≡ 1 on B(ρ). And assume |∇η|  ρ1 . Then multiplying the above inequality by uη n and integrating it on Rn , there holds  u

2n n−2





η dx 

Rn



Δu · uη dx = −

n

|∇u| η dx −

n

Rn

Rn





≤−

|∇u|2 η n dx +

Rn

nη n−1 u∇u, ∇ηdx

2 n

Rn

  |∇u|2 η n + C1 η n−2 u2 |∇η|2 dx

Rn



η n−2 u2 |∇η|2 dx

= C1 Rn

1  2 ρ

 η n−2 u2 dx.

ρ≤|x|≤2ρ

Next using the Hölder inequality, we get  u Rn

2n n−2

1 η dx  2 ρ





n

n

η u

2n n−2

  n−2 n

1dx

ρ≤|x|≤2ρ







 n2



dx ρ≤|x|≤2ρ

2n

η n u n−2 dx

 n−2 n ,

ρ≤|x|≤2ρ

2n from which it is easy to see that Rn u n−2 dx < ∞. Then letting ρ → ∞ in the above inequality, we get

2n u n−2 dx = 0. So u ≡ 0, which is a contradiction. 2 Rn

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

23

Now with this proposition at hand, we can determine the range of σ1 (g). Note that aσ1 (g) + b = σ2 (g) ≤

n−1 2 σ (g). 2n 1

Solving this inequality, it follows that n a− σ1 (g) ≤ n−1



2 2n n a + b, n−1 n−1



2 2n n a + b. n−1 n−1

or n a+ σ1 (g) ≥ n−1

n−2 Since b > 0 and σ1 (g) = 2(n−1) Rg , we see that Rg satisfies either Rg ≤ −c or Rg ≥ c. But we can exclude the former case by Proposition 4. Therefore we obtain

σ1 (g) =

n−2 Rg ≥ 0, 2(n − 1)

and n a≥ σ1 (g) − n−1



2 n 2n a + b > 0. n−1 n−1

The last ingredient we need is a “tail” estimate, whose proof relies on the following well-known technical lemma. Lemma 5. (See [3].) Let g = v −2 dx2 be a conformal metric on Rn with Rg ≥ 0. Then there exists a constant c such that v(x) ≤ c|x|2 for all |x| sufficiently large. Now we can state our “tail” estimate and prove it. Proposition 6. Let g = v −2 dx2 be a conformal metric on Rn , n ≥ 4.Suppose that σ2 (g) − aσ1 (g) = b > 0.

Moreover, if n ≥ 6, assume Rn v −n dx  1. If n = 5, assume a > − 57 b. Then there holds   Aρ

 n σ1 (g) − a |∇v|2 v 1−n dx  ρ2 , n−1

for ρ 1,

where Aρ = B(2ρ) \ B(ρ). Proof. Denote g0 = dx2 . Then we have ∇2 v 1 |∇v|2 1 1 Ag = − Ag g0 + n−2 v 2 v2 n−2 0 = Therefore,

∇2 v 1 |∇v|2 − g0 . v 2 v2

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

24

 σ2

1 Ag n−2



 = v 4 σ2 g0−1 ·

1 Ag n−2

 =

2       1  v4 1 tr 2 Ag −  Ag  2 n−2 n−2

  2  v4 n(n − 1) −4 −v −2 ∇2 v  + v −2 (Δv)2 − (n − 1)v −3 Δv|∇v|2 + v |∇v|4 . 2 4

=

Multiplying both sides by 2v −n and using the Bochner formula  2 2 ∇ v  = 1 Δ|∇v|2 − ∇v, ∇Δv, 2 we arrive at 1 2 σ2 (Ag )v −n = − v 2−n Δ|∇v|2 + v 2−n ∇v, ∇Δv + v 2−n (Δv)2 2 (n − 2) 2 1 − (n − 1)v 1−n Δv|∇v|2 + n(n − 1)v −n |∇v|4 . 4 Take a new cut-off function η supported on B( 25 ρ) \ B( 12 ρ) with η ≡ 1 on Aρ and |∇k η|  ρ−k , k = 1, 2. Multiply the above equality by η 4 v α and integrate over Rn . Next do the integration by parts to eliminate the derivative terms of v of degree bigger than 2. After some tedious computation, we obtain the following equality. For more details, we refer to [3]. 2 (n − 2)2  +

 σ2 (Ag )v

n − 4 − 3α η dx = 2

Rn

v 1+α−n |∇v|2 Δv · η 4 dx Rn



1 1 n(n − 1) − (2 + α − n)(1 + α − n) 4 2

1 − (2 + α − n) 2



v α−n |∇v|4 η 4 dx Rn



v 1+α−n |∇v|2 ∇v · ∇η 4 dx +

Rn

 −



α−n 4

v 2+α−n ∇2 η 4 (∇v, ∇v)dx Rn

v 2+α−n |∇v|2 Δη 4 dx. Rn

Substituting 1 n |∇v|2 + σ1 (g)v −1 2 v n−2   1 n n |∇v|2 n 1 + σ1 (g) − a v −1 + av −1 = 2 v n−2 n−1 n−2n−1

Δv =

into the first line, we get 2 (n − 2)2

 σ2 (g)v α−n η 4 dx Rn

  na η 4 dx v α−n |∇v|2 σ1 (g) − n−1



= aα,n Rn



+ bα,n Rn

v α−n |∇v|4 η 4 dx + M1 + M2 +

na(n − 4 − 3α) 2(n − 1)(n − 2)

 v α−n |∇v|2 η 4 dx, Rn

(12)

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

25

where n − 4 − 3α , 2(n − 2)

aα,n =

bα,n =

1 M1 = − (2 + α − n) 2



1 (α + 1)(n − 2α − 4), 4

v 1+α−n |∇v|2 ∇v · ∇η 4 dx,

Rn





v 2+α−n ∇2 η 4 (∇v, ∇v)dx −

M2 = Rn

v 2+α−n Δη 4 |∇v|2 dx. Rn

On the other hand, we have n 1 σ1 (g) = vΔv − |∇v|2 , n−2 2 which implies    n 1 α−n 4 1+α−n 4 σ1 (g)v η dx = v η Δvdx − v α−n |∇v|2 η 4 dx n−2 2 Rn

Rn



n =− 1+α− 2

Rn

 v

α−n



|∇v| η dx −

v 1+α−n ∇v · ∇η 4 dx

2 4

Rn

Rn

   n 1 =− 1+α− v α−n |∇v|2 η 4 dx + v 2+α−n · Δη 4 dx. 2 2+α−n Rn

Rn

2a Multiplying the above equality by − n−2 and adding it to (12) gives rise to

2 (n − 2)2



  σ2 (g) − aσ1 (g) v α−n η 4 dx

Rn





v

= aα,n Rn

α−n

|∇v|

2



+ bα,n

v

α−n

Rn

+

2a(1 + α − n−2

 na σ1 (g) − η 4 dx n−1

na(n − 4 − 3α) |∇v| η dx + M1 + M2 + 2(n − 1)(n − 2)

 v α−n |∇v|2 η 4 dx

4 4

n 2)

 v α−n |∇v|2 η 4 dx − Rn

Rn

2a (n − 2)(2 + α − n)

 v 2+α−n · Δη 4 dx,

Rn

or simplified to 2 (n − 2)2





Rn

 σ2 (g) − aσ1 (g) v α−n η 4 dx

  na η 4 dx v α−n |∇v|2 σ1 (g) − n−1



= aα,n Rn

 v α−n |∇v|4 η 4 dx + M1 + M2

+ bα,n Rn



+ cα,n Rn

 v α−n |∇v|2 η 4 dx + dα,n Rn

v 2+α−n · Δη 4 dx,

(13)

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

26

where cα,n =

a(−n2 + nα − 4α + 2n − 4) , 2(n − 1)(n − 2)

dα,n = −

2a . (n − 2)(2 + α − n)

Note that M1 and M2 can be estimated as follows. |M1 | 

1 ρ



1 ρ

|∇v|3 η 3 v 1+α−n  Rn



1 |M2 |  2 ρ

|∇v| η v

2 2 2+α−n

Rn

 |∇v|4 v α−n η 4

 34  

Rn

 14 ,

spt η



1  2 ρ

v 4+α−n

|∇v| v

4 α−n 4

 12  

η

v

Rn

4+α−n

 12 ,

spt η

where and thereafter spt η denotes the support set of η and the Euclidean volume element dx is omitted. Next we have to do the analysis case by case. Case 1 (n ≥ 6). We choose α = 0. Then a0,n > 0 and b0,n > 0. It follows that 

v −n η 4  a0,n

Rn

  σ1 (g) −

Rn



− |c0,n |

  na |∇v|2 v −n η 4 + b0,n |∇v|4 v −n η 4 − |M1 | − |M2 | n−1

η 4 v −n |∇v|2 − |d0,n |

Rn

Under the assumption that



Rn



v 2−n |Δη 4 |.

Rn

v −n  1, we get

Rn

1 |M1 |  ρ 

1 ρ



4 −n 4

|∇v| v

 34  

η

v

Rn

|∇v|4 v −n η 4

 34  

Rn

v −n

 n1

 n−4  4n 1

spt η

|∇v|4 v −n η 4



 14

spt η





4−n

spt η

 34 .

Rn

Similarly, there hold  |M2 |  

η v

v Rn

2−n



|∇v| 

Rn



|∇v| v Rn

4 −n

4 −n 4

1 |Δη |  2 ρ



4

Rn

v spt η

η

−n

,

4 −n 4

|∇v| v

2

 12

 12

η

,

 n−2  n η spt η

n

 n2

 1.

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

27

Then by the Young inequality 1 1 + = 1, p q

ab ≤ εap + Cε bq , it is easy to see that |M1 |, |M2 | and the term conclusion, we arrive at



  σ1 (g) − Rn

Then using Lemma 5 and σ1 (g) −   Aρ

n n−1 a

η 4 v −n |∇v|2 can be absorbed into b0,n

Rn

Rn

|∇v|4 v −n η 4 . In

 na |∇v|2 v −n η 4  1. n−1

> 0, we obtain

    na na 2 1−n 2 |∇v| v |∇v|2 v −n η 4  ρ2 . σ1 (g) − σ1 (g) − ρ n−1 n−1 Rn

Case 2 (n = 5). We choose α = 1. Then Eq. (13) yields 2 9

 Rn



 1 σ2 (g) − aσ1 (g) v −4 η 4 + 3

3 = M1 + M2 − a 4



 Rn

   5 1 v −4 |∇v|2 σ1 (g) − a η 4 + v −4 |∇v|4 η 4 4 2

v −4 |∇v|2 η 4 +

Rn

a 3



Rn

v −2 · Δη 4 .

Rn

First since |M1 | 

1 ρ



1 |M2 |  2 ρ

|∇v|4 v −4 η 4

 34  

Rn

 14

 ε

1

Rn

spt η



4 −4 4

|∇v| v Rn

 12  

 12

η

|∇v|4 v −4 η 4 + Cε ρ,

 ε

1

|∇v|4 v −4 η 4 + Cε ρ,

Rn

spt η

M1 and M2 can be absorbed into the term 12 Rn |∇v|4 v −4 η 4 . Second we have    4 −2   v · Δη  ε v −4 η 4 + Cε ρ. Rn

Rn

Thus this term can be absorbed into 29 Rn (σ2 (g) − aσ1 (g))v −4 η 4 .

Last if a ≥ 0, we can drop out the term − 43 a Rn v −4 |∇v|2 η 4 and go through. If a < 0, the assumption  a > − 57 b implies b > 75 a2 . Then by the previous analysis we get 5 σ1 (g) − a ≥ 4



2

5 a 4

5 + b 2

9 =: (1 + δ) (−a), 4 where δ is a positive number. Next direct computation shows that

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

28

    3 1 δ 5 5 1 σ1 (g) − a + a ≥ σ1 (g) − a . 3 4 4 31+δ 4



Thus − 34 a Rn v −4 |∇v|2 η 4 can be absorbed into 13 Rn v −4 |∇v|2 (σ1 (g) − 54 a)η 4 . In conclusion, we derive    5 −4 2 v |∇v| σ1 (g) − a η 4  ρ  ρ2 . 4 Rn

Case 3 (n = 4). We choose α = 1. Then Eq. (13) yields 1 2

 Rn

  3 σ2 (g) − aσ1 (g) v −3 η 4 + 4 

= M1 + M2 − a

 Rn

   4 v −3 |∇v|2 σ1 (g) − a η 4 + v −3 |∇v|4 η 4 3

v −3 |∇v|2 η 4 + a

Rn

Rn



v −1 · Δη 4 .

Rn

First since |M1 | 

|M2 | 

1 ρ



1 ρ2

|∇v|4 v −3 η 4

 34  

Rn



 14 v

 ε Rn

spt η

|∇v|4 v −3 η 4

Rn

 12  

|∇v|4 v −3 η 4 + Cε ρ2 ,

 12

 ε

v

|∇v|4 v −3 η 4 + Cε ρ2 ,

Rn

spt η

M1 and M2 can be absorbed into the term Rn |∇v|4 v −3 η 4 . Second we have   −1 4 v · |Δη |  ε v −3 η 4 + Cε ρ. Rn

Rn

Thus this term can be absorbed into 12 Rn (σ2 (g) − aσ1 (g))v −3 η 4 . Last recall σ2 (g) − aσ1 (g) = b. Again if a ≥ 0 we go through. If a < 0, by the previous analysis, we get 4 σ1 (g) − a ≥ 3



2

4 a 3

8 + b 3

4 =: (1 + δ) (−a), 3 where δ is a positive number. Then direct computation leads to     3 δ 4 4 3 σ1 (g) − a + a ≥ σ1 (g) − a . 4 3 41+δ 3



Thus −a Rn v −3 |∇v|2 η 4 can be absorbed into 34 Rn v −3 |∇v|2 (σ1 (g) − 43 a)η 4 . In summary, we obtain    4 −3 2 v |∇v| σ1 (g) − a η 4  ρ2 . 3 Rn

So far we have completed the proof of this proposition. 2

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

29

3. Proof of Theorem 1 For ρ > 0 let η be a cut off function supported in the ball B(2ρ) with η ≡ 1 on B(ρ). And assume |∇η|  ρ1 . Since g = v −2 dx2 , we get   n−2   vΔg v −1 g. Eg = −(n − 2)v∇2g v −1 + n Pairing both sides with v −1 η 2 L and integrating the equality on Rn , there holds 

−L, Ev −1 η 2 dvg = (n − 2)

Rn





  η 2 L, ∇2g v −1 dvg ,

Rn

where we have used that L is trace-free. Using the divergence theorem and noting that L is also divergence-free, we derive 

−L, Ev −1 η 2 dvg = −(n − 2)

Rn



     L ∇g v −1 , ∇g η 2 dvg

Rn

 

     |L|∇g v −1 ∇g η 2 dvg

Rn



|L||∇g v||∇g η|v −2 ηdvg .

 Rn

Taking into account Proposition 3 and the fact spt |∇η| ⊂ Aρ , it follows that  −L, Ev

 

−1 2

η dvg 

Rn

Aρ



 12

n σ1 (g) − a n−1

  L, Ev −1 η 2 dvg



1  L, E 2 |∇g v||∇g η|v −2 ηdvg

 12  

σ1 (g) −

Aρ



Aρ

  L, Ev −1 η 2 dvg

=

  12 n a |∇g v|2 |∇g η|2 v −3 dvg n−1

 12  

Aρ

Aρ

  12 n 2 2 1−n σ1 (g) − a |∇v| |∇η| v dx , n−1

where we have used that |∇g v|2 = v 2 |∇v|2 , |∇g η|2 = v 2 |∇η|2 and dvg = v −n dx. Thanks to |∇η|  ρ1 , we arrive at  −L, Ev Rn

−1 2



η dvg 

  L, Ev −1 η 2 dvg

 12

Aρ

   σ1 (g) − × ρ−2 Aρ

Now by Proposition 6, we know that the integral ρ−2 with respect to ρ 1. Then it is easy to see that

Aρ

  12 n a |∇v|2 v 1−n dx . n−1

(14)

n (σ1 (g) − n−1 a)|∇v|2 v 1−n dx is uniformly bounded

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

30



−L, Ev −1 dvg < ∞,

Rn

which implies that 

  L, Ev −1 η 2 dvg → 0,

as ρ → ∞.

Aρ

This together with (14) immediately yields 

−L, Ev −1 dvg = 0.

Rn

As a consequence, we get E = 0. Then the conclusion of Theorem 1 follows from a standard argument [13]. 4. Proof of Theorem 2 First recall the following well known theorem. Theorem 7. (See Tani [14], Wegner [17] and Cheng [4].) Let (M n , g) (n ≥ 3) be a compact locally conformally flat Riemannian manifold with constant scalar curvature. If the Ricci tensor is semi-positive definite, then (M n , g) is either a space form or a space S1 × N n−1 with N a space form. Therefore, to prove Theorem 2, it suffices to show that M has constant scalar curvature. For that purpose, first we introduce some notations. For any local orthonormal basis {e1 , · · · , en }, denote Rijkl the components of the Riemannian curvature tensor. And at a point, we choose {ei } as eigenvectors of Schouten tensor such that Aij = λi δij . Now we can state the lemmas which will be used in the proof. The first lemma is due to Tani [14] (see also [4,10,17]). Lemma 8. (See Tani [14].) Let (M n , g) be a locally conformally flat manifold with Ric ≥ 0, then 1 (λi − λj )2 Rijij ≥ 0. 2 i,j

(15)

Remark 9. From the proof of Lemma 8 (see [10]), we can check that, if the equality in (15) holds, then the eigenvalues {ρi } of Ricci tensor must be either {ρ1 = ρ2 = · · · = ρn = ρ(x) ≥ 0} or {ρ1 = 0, ρ2 = · · · = ρn = ρ(x) > 0}. In addition, we need the following inequality of Kato type. Lemma 10. (See [11] for a = 0.) Assume σ2 (g) = aσ1 (g) + b with a2 + 2b ≥ 0. Then we have  i,j,k

(Aij,k ) ≥ 2

 k

2 Ajj,k

.

j

Proof. First we have  i,j

A2ij = σ12 (g) − 2σ2 (g) = σ12 (g) − 2aσ1 (g) − 2b.

(16)

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

31

Taking covariant derivative on the above equality with respect to ek yields 

  Aij Aij,k = σ1 (g) − a ∇k σ1 (g).

i,j

Then using the Cauchy–Schwarz inequality we obtain  2 2    2    σ1 (g) − a ∇σ1 (g) = Aij Aij,k k

≤

k

=

i,j

  i,j

A2ij

i,j

A2ij





 (Aij,k )2

i,j

(Aij,k )2 .

(17)

i,j,k

At last note that 

A2ij = σ12 (g) − 2aσ1 (g) − 2b

i,j

≤ σ12 (g) − 2aσ1 (g) + a2  2 = σ1 (g) − a . So if σ1 (g) = a at the point, the conclusion follows. Otherwise an approximation argument will let us through. 2 Now we are in a position to prove Theorem 2. Proof of Theorem 2. By direct computation, we can obtain (see the formula (3.8) in [10]) 



Δσ2 (g) = tr T1 ◦ Hess σ1 (g)



2   1 2 2 − (λi − λj ) Rijij − (Aij,k ) + Ajj,k , 2 i,j j i,j,k

(18)

k

where T1 is the first Newton tensor associated to Ag , “◦” denotes the composition of two (1, 1)-tensors, and  so tr(T1 ◦ Hess(σ1 (g))) = i,j (T1 )ij (σ1 (g))ij . Noting that T1 is divergence-free and M is compact without boundary, after integrating the above formula we get   M

2    1 2 2 − dvg = 0. (λi − λj ) Rijij − (Aij,k ) + Ajj,k 2 i,j j i,j,k

k

On the other hand, Lemma 8 and Lemma 10 show that 2   1 2 2 − (λi − λj ) Rijij − (Aij,k ) + Ajj,k ≤ 0. 2 i,j j i,j,k

k

Thus we must have 1 (λi − λj )2 Rijij ≡ 0. 2 i,j

32

H. Li, C. Xiong / Differential Geometry and its Applications 37 (2014) 17–32

Then Remark 9 shows that, the eigenvalues {ρi } of Ricci tensor must be either {ρ1 = ρ2 = · · · = ρn = ρ(x) ≥ 0} or {ρ1 = 0, ρ2 = · · · = ρn = ρ(x) > 0}. Since M is connected and the eigenvalues of Ricci tensor are continuous, only one of the two cases can occur on M . Furthermore, in view of σ2 (g) = aσ1 (g) + b, we know ρ(x) is constant and its value is determined by the equation. Then the scalar curvature is constant. Now as mentioned before, this fact together with Theorem 7 gives rise to the desired result. 2 Acknowledgement The authors would like to thank the referee for some helpful comments which made this paper more readable. References [1] L.A. Caffarelli, B. Gidas, J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Commun. Pure Appl. Math. 42 (1989) 271–297. [2] A. Chang, M. Gursky, P. Yang, An a priori estimate for a fully nonlinear equation on four-manifolds, J. Anal. Math. 87 (2002) 151–186. [3] A. Chang, M. Gursky, P. Yang, Entire solutions of a fully nonlinear equation, in: Lectures on Partial Differential Equations in Honor of Louis Nirenberg’s 75th Birthday, Inter. Press, 2003, pp. 43–60. [4] Q.M. Cheng, Compact locally conformally flat Riemannian manifolds, Bull. Lond. Math. Soc. 33 (2001) 459–465. [5] Q.M. Cheng, Compact hypersurfaces in a unit sphere with infinite fundamental group, Pac. J. Math. 212 (2003) 49–56. [6] Q.M. Cheng, Topology and geometry of complete submanifolds in Euclidean spaces, Banach Cent. Publ. 69 (2005) 67–80. [7] P. Guan, G. Wang, A fully nonlinear conformal flow on locally conformally flat manifolds, J. Reine Angew. Math. 557 (2003) 219–238. [8] B. Guo, H. Li, Some variational problems in conformal geometry, in: Proc. of 5-th International Congress of Chinese Mathematicians, vol. 51, American Mathematical Soc., 2012, pp. 221–231. [9] Z.J. Hu, H. Li, A new variational characterization of n-dimensional space forms, Trans. Am. Math. Soc. 356 (2004) 3005–3023. [10] Z.J. Hu, H. Li, U. Simon, Schouten curvature functions on locally conformally flat Riemannian manifolds, J. Geom. 88 (2008) 75–100. [11] H. Li, Global rigidity theorems of hypersurfaces, Ark. Mat. 35 (1997) 327–351. [12] A. Li, Y.Y. Li, On some conformally invariant fully nonlinear equations, Commun. Pure Appl. Math. 56 (2003) 1414–1464. [13] M. Obata, Certain conditions for a Riemannian manifold to be isometric with a sphere, J. Math. Soc. Jpn. 14 (1962) 333–340. [14] M. Tani, On a conformally flat Riemannian space with positive Ricci curvature, Tohoku Math. J. 19 (1967) 227–231. [15] J.A. Viaclovsky, Conformal geometry, contact geometry and the calculus of variations, Duke Math. J. 101 (2000) 283–316. [16] J.A. Viaclovsky, Conformally invariant Monge–Ampère partial differential equations: global solutions, Trans. Am. Math. Soc. 352 (2000) 4371–4379. [17] B. Wegner, Kennzeichnungen von Räumen konstanter Krümmung unter lokal konform Euklidischen Riemannschen Mannigfaltigkeiten, Geom. Dedic. 2 (1973) 269–281.