Bull. Sci. math. 130 (2006) 330–336www.elsevier.com/locate/bulsci

Some properties of non-compact completeRiemannian manifolds

Li Ma

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

Received 3 December 2005; accepted 19 December 2005

Available online 3 February 2006

Abstract

In this paper, we study the volume growth property of a non-compact complete Riemannian manifold M .We improve the volume growth theorem of Calabi (1975) and Yau (1976), Cheeger, Gromov and Taylor(1982). Then we use our new result to study gradient Ricci solitons. We also show that on M , for anyq ∈ (0,∞), every non-negative Lq subharmonic function is constant under a natural decay condition on theRicci curvature.© 2006 Elsevier SAS. All rights reserved.

MSC: 53Cxx

Keywords: Volume growth; Gradient estimate; Ricci soliton

1. Introduction

The motivation for this paper comes from the interest in the understanding the Ricci soli-tons [8]. However, at this moment, almost all works in this direction are about Gradient RicciSolitons. See [2,4,8,11]. Generally speaking, a non-compact Ricci soliton may not be a gradientRicci soliton. So it may be interesting to consider problems related to Ricci solitons.

In this paper, we consider the volume growth properties of the non-compact complete Rie-mannian manifold (M,g) under a natural Ricci curvature condition. We can improve the volumegrowth theorem of Calabi [3] and Yau [14]. Then we use our new result to study gradient Riccisoliton. We also show that on M , for any q ∈ (0,∞), every non-negative Lq subharmonic func-tion is constant under a natural decay condition on the Ricci curvature.

E-mail address: lma@math.tsinghua.edu.cn (L. Ma).

0007-4497/$ – see front matter © 2006 Elsevier SAS. All rights reserved.doi:10.1016/j.bulsci.2005.12.002

L. Ma / Bull. Sci. math. 130 (2006) 330–336 331

Result about Gradient Solitons is stated in section four. In section two, we generalize theresult of Calabi [3] and Yau [14] on infinite volume property for Riemannian manifolds with non-negative Ricci curvature. Calabi and Yau’s result was generalized by Cheeger–Gromov–Taylor(see Theorem 4.9 in [6]) to Riemannian manifolds with lower bound like

Rc � − νn

ρ2(x),

for ρ(x) � 1 and some restricted dimensional constant νn, where Rc is the Ricci curvature ofthe metric g, and ρ(x) is the distance function from some fixed point p. We can remove thisrestriction to the dimensional constant νn. Our result is

Theorem 1. Let (M,g) be a complete non-compact Riemannian manifold. Assume that its Riccicurvature has the lower bound

Rc(g) � −Cρ−2(x),

for ρ(x) � 1, where C is a constant. Then for any x ∈ M with ρ(x) � 1 and r > 1, there is aconstant C(n,VolB1(x)) such that it holds

VolBr(x) � C(n,VolB1(x)

)r.

At the first glance, one may think that Theorem 1 is false by looking at the following example:Let M = R2 with metric

g = dr2 + 1

r2dθ2, r � 1,

which is smoothly extended to r � 1. In this case, we have

K = −2r−2, dA = r−1 dr dθ, r � 1.

Hence, V (Br(0)) ∼ ln r , r → +∞. Actually, this corresponds to p = 0 and

(R − 1)Vol(BR+1(p) − BR−1(p)

)� 2(n + 1 + C)VolBR+1(p)

in Section 2 below.In section three, we give some remarks on Yau’s gradient estimate and vanishing properties for

subharmonic functions. We may use the letter C to denote various constants in different places.

2. Proof of Theorem 1

We let D and R(., .). be the Levi-Civita derivative and Riemannian curvature of the metricg respectively. Let γ (t) be the minimizing geodesic from p to a point x, parametrized by thearc-length parameter such that γ (0) = p and γ (r) = x. Assume that the point x is inside the cutlocus of p ∈ M . Let Y0 be a vector in TxM with g(Y0,

∂∂t

) = 0. Then we can get an Jacobi field Y

by extending Y0 along γ (see [1] or [5]). Let I r0 (., .) be the index form along γ . Then the Hessian

of ρ at x

H(ρ)(Y0, Y0) = YYρ − DY Yρ

can be written asr∫ (|DtY |2 − g

(R(Y,Dt )Dt , Y

))dt

0

332 L. Ma / Bull. Sci. math. 130 (2006) 330–336

which is the index form I r0 (Y,Y ). We now extend Y0 along γ and get a parallel vector field X.

Then by the minimizing property of the index form I r0 we have that

I r0 (Y,Y ) � I r

0

(t

rX,

t

rX

),

and the right side of the above inequality is

1

r− 1

r2

r∫0

t2g(R(X,Dt)Dt ,X

)dt.

To compute the Laplacian �ρ of the distance function ρ, we choose vector fields{∂

∂t,X1, . . . ,Xn−1

}

as an orthonormal basis of Tγ (t)M and parallel along γ . Then we have

�ρ =n−1∑i=1

H(ρ)(Xi,Xi),

which is bounded above by

n − 1

r− 1

r2

r∫0

t2 Rc(Dt ,Dt )dt.

Using the assumption that

Rc � C(ρ(x)

),

where C(ρ(x)) = −Cρ−2(x) for ρ � 1, we get that

�ρ � n − 1

r− 1

r2

r∫0

t2C(t)dt.

Since

− 1

r2

r∫0

t2C(t)dt � C

r,

we have

�ρ � n − 1 + C

ρ.

Now it is standard to verify (see also p. 7 in [12] or [7]) that in the distributional sense, it holdson M

�ρ � n − 1 + C

ρ.

Then it holds in the distributional sense that

�ρ2 = 2ρ�ρ + 2 � 2(n + 1 + C).

L. Ma / Bull. Sci. math. 130 (2006) 330–336 333

That is, for any non-negative function φ ∈ C∞0 (M),∫

M

ρ2�φ � 2(n + 1 + C)

∫M

φ. (1)

Given any R > 0. Let x0 ∈ ∂BR(x) ∩ γ , where γ is the minimizing geodesic passing through p

to a point x with d(x,p) � 1. So the computation above is also true for the distance functiond(x, x0), and we may assume that ρ = d(x, x0) and x0 = p.

We now follow the argument of Schoen and Yau (see [12]). Let r = R. By approximation, wecan let φ in (1) be a Lipschitz function with compact support. Choose φ(x) = ξ(ρ(x)), whereξ(ρ) = 1 for ρ � R − 1, = 0 for ρ � R + 1, and ξ ′(ρ) = − 1

2 for R − 1 � ρ � R + 1. By directcomputation we have∫

M

ρ2�φ = −2∫

BR+1(p)

ξ ′ρ

since |Dρ| = 1. Then we get∫M

ρ2�φ =∫

BR+1(p)−BR−1(p)

ρ.

From this we clearly have,∫M

ρ2�φ � (R − 1)Vol(BR+1(p) − BR−1(p)

).

Note that∫M

φ � VolBR+1(p).

So we have

(R − 1)Vol(BR+1(p) − BR−1(p)

)� 2(n + 1 + C)VolBR+1(p).

Note that

B1(x) ⊂ BR+1(p) − BR−1(p).

Then by (1) we have

(R − 1)VolB1(x) � 2(n + 1 + C)VolBR+1(p).

Since

BR+1(p) ⊂ B2(R+1)(x),

we obtain that

(R − 1)VolB1(x) � 2(n + 1 + C)VolB2(R+1)(x).

This implies Theorem 1.

334 L. Ma / Bull. Sci. math. 130 (2006) 330–336

3. Remarks on harmonic functions

In [13], Yau proved that

Theorem 2. Let (M,g) be an n (� 2) dimensional complete Riemannian manifold with its Riccicurvature Rc(g) � −(n − 1)K , where K � 0 is a constant. Assume u is a positive harmonicfunction on M . Let BR be a geodesic ball in M . Then it holds on BR/2 that

|D logu| � Cn

(1 + R

√K

R

),

where Cn is a constant depending only on n.

We now use this gradient estimate to study the positive solution of the equation on the mani-fold M :

�u = −μ2u, (2)

where μ � 0 is a constant.

Proposition 3. Let (M,g) be an n (� 2) dimensional complete Riemannian manifold withRc(g) � −(n − 1)K , where K � 0 is a constant. Assume u is a positive solution of (2) on M .Let BR be a geodesic ball in M . Then it holds on BR/2 that

|D logu| � Cn

(1 + R

√K

R

),

where Cn is a constant depending only on n.

Proof. Let

N = M ×Rhave the product metric. Then we still have the lower bound for the Ricci curvature of N :

Rc(N) � −(n − 1)K

provided Rc(M) := Rc(g) � −(n − 1)K . We write DN as the covariant derivative on N . Let

w(x, t) = eμtu(x).

Then

�Nw = 0.

By Yau’s estimate we have that

|D logu| = |D logw| � |DN logw| � Cn

(1 + R

√K

R

). � (3)

The important matter for us is that the constant Cn is independent of the constant μ. This isan important thing for us to study the Ricci soliton.

We also need to study the positive solution of the following equation on the manifold M :

�u = μ2u � 0, (4)

L. Ma / Bull. Sci. math. 130 (2006) 330–336 335

where μ � 0 is a constant. Note that non-negative solutions to (3) are non-negative subharmonicfunctions. So we now use our Theorem 1 to study Lq non-negative subharmonic functions oncomplete Riemannian manifolds. As in the proof of Theorem 2.5 in [9], we have

Proposition 4. Suppose (M,g) is a complete Riemannian manifold of dimension n. Assume thatthere is a constant C > 0 such that the Ricci curvature has the bound

Rc(g) � −Cρ−2(x)

for large ρ � 1. Then for any q ∈ (0,∞), M has no Lq non-negative subharmonic functionexcept the constants.

4. Ricci soliton with Ricci curvature quadratic decay

In this section, we assume that the non-compact complete Riemannian manifold (M,g) is agradient Ricci soliton, that is, there is a smooth function f such that

Rc(g) = D2f,

on M . The classification for Ricci solitons is important to the research for Ricci flow, see [2,4,8,10,11]. Generally speaking, a non-compact Ricci soliton may not be a gradient Ricci soliton.

For a gradient Ricci soliton, as showed by R. Hamilton [8], we have a constant c such that

|Df |2 + s = c,

where s is the scalar curvature of the metric g. We assume that f is not a constant, so M maynot be Ricci flat.

We now come to the question: Whether the constant c is bounded under a nice curvaturecondition?

From the definition of the gradient Ricci soliton, it is clear that

s = �f.

Set

u = ef ,

which is a positive function on M . Then we have [11] that

�u = cu.

If c is negative, then we can write it as c = −μ2, and then we can use the gradient estimate inProposition 3. Note that

D logu = Df.

Then in this case, we have the bound

|Df | � Cn

(1 + R

√K

R

),

for every R > 0, provided Rc(M) � −(n − 1)K on the ball BR and

−n(n − 1)K � c � 0.

If c = μ2 for some constant μ � 0, we can use our Proposition 4.In conclusion we have

336 L. Ma / Bull. Sci. math. 130 (2006) 330–336

Theorem 5. Assume (M,g) is a complete non-compact gradient Ricci soliton such that

Rc = D2f.

Suppose M is not Ricci flat. Assume the Ricci curvature has the bound

Rc � C(ρ(x)

) := −Cρ−2(x)

for ρ � 1. Then either (1) we have c = 0, and u := ef is a positive harmonic function on M; or(2) c > 0 and for any q ∈ (0,∞), we have∫

M

uq = ∞.

Proof. In fact we have two cases when (i) c � 0 or (ii) c > 0. In case (i), we clearly have that

s < |Df |2 + s = c � 0.

The lower bound for s follows easily from the assumption that

s(x) � nC(ρ(x)

)for ρ(x) large. By this we have s(x) → 0 as ρ(x) → ∞, and c = 0. In case (ii), u is a positivesubharmonic function, and we use Proposition 4 to get∫

M

uq = ∞. �

Acknowledgement

This work is supported in part by the Key 973 project of ministry of Science and Technologyof China. The author is grateful to Prof. Th. Aubin for helpful comments.

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