On the Completeness of Gradient Ricci Solitons

A gradient Ricci soliton is a triple $(M,g,f)$ satisfying $R_{ij}+\nabla_i\nabla_j f=\lambda g_{ij}$ for some real number $\lambda$. In this paper, we will show that the completeness of the metric $g$ implies that of the vector field $\nabla f$.


Introduction
Definition 1.1. Let (M, g, X) be a smooth Riemannian manifold with X a smooth vector field. We call M a Ricci soliton if Ric + 1 2 L X g = λg for some real number λ. It is called shrinking when λ > 0, steady when λ = 0, and expanding when λ < 0. If (M, g, f ) is a smooth Riemannian manifold where f is a smooth function, such that (M, g, ∇f ) is a Ricci soliton, i.e. R ij + ∇ i ∇ j f = λg ij , we call (M, g, f ) a gradient Ricci soliton and f the soliton function.
On the other hand, there has the following definition (see chapter 2 of [3]). Definition 1.2. Let (M, g(t), X) be a smooth Riemannian manifold with a solution g(t) of the Ricci flow on a time interval (a, b) containing 0, where X is smooth vector field. We call (M, g(t), X) self-similar solution if there exist scalars σ(t) such that g(t) = σ(t)ϕ * t (g 0 ), where the diffeomorphisms ϕ t is generated by X. If the vector field X comes from a gradient of a smooth function f , then we call (M, g(t), f ) a gradient self-similar solution.
It is easy to see that if (M, g(t), f ) is a complete gradient self-similar solution, then (M, g(0), f ) must be a complete gradient Ricci soliton. Conversely, when (M, g, f ) is a complete gradient Ricci soliton and in addition, the vector field ∇f is complete, it is well known (see for example Theorem 4.1 of [2]) that there is a complete gradient self-similar solution (M, g(t), f ), t ∈ (a, b) (with 0 ∈ (a, b)), such that g(0) = g. Here we say that a vector field ∇f is complete if it generates a family of diffeomorphisms ϕ t of M for t ∈ (a, b).
So when the vector field is complete, the definitions of gradient Ricci soliton and gradient self-similar solution are equivalent. In literature, people sometimes confuse the gradient Ricci solitons with the gradient self-similar solutions. Indeed, if the gradient Ricci soliton has bounded curvature, then it is not hard to see that the vector field ∇f is complete. But, in general the soliton does not have bounded curvature.
The purpose of this paper is to show that the completeness of the metric g of a gradient Ricci soliton (M, g, f ) implies that of the vector field ∇f , even though the soliton does not have bounded curvature. Our main result is the following Date: July 10, 2008. Key words and phrases. completeness, gradient Ricci soliton, gradient self-similar solution. Theorem 1.3. Let (M, g, f ) be a gradient Ricci soliton. Suppose the metric g is complete, then we have: Indeed, we will show that the vector field ∇f grows at most linearly and so it is integrable. Hence the above Definition 1.1 and 1.2 are equivalent when the metric is complete.
Acknowledgement I would like to thank my advisor Professor X.P.Zhu for many helpful suggestions and discussions.

Gradient Ricci Solitons
Let (M, g, f ) be a gradient Ricci soliton, i.e., R ij + ∇ i ∇ j f = λg ij . By using the contracted second Bianchi identity we get the equation R + |∇f | 2 − 2λf = const.
Definition 2.1. Let (M, g, f ) be a gradient shrinking or expanding soliton. By rescaling g and changing f by a constant we can assume λ ∈ {− 1 2 , 1 2 } and R + |∇f | 2 − 2λf = 0. We call such a soliton normalized, and f a normalized soliton function.
, for some positive numbers r 0 and K. Then for arbitrary point x, outside B r0 (p), we have Proof. (i) By using the soliton equation and the contracted second Bianchi identity (ii) Let γ : [0, d(x)] → M be a shortest normal geodesic from p to x. We may assume that x and p are not conjugate to each other, otherwise we can understand the differential inequality in the barrier sense. Let {γ(0), e 1 , · · · , e n−1 } be an orthonormal basis of T p M . Extend this basis parallel along γ to form a parallel orthonormal basis {γ(t), e 1 (t), · · · , e n−1 (t)} along γ.
Let X i (t), i = 1, 2, · · · , n − 1, be the Jacobian fields along γ with X i (0) = 0 and X i (d(x)) = e i (d(x)). Then it is well-known that (see for example [4]) Define vector fields Y i , i = 1, 2, · · · , n − 1, along γ as follows Then by using the standard index comparison theorem we have On the other hand, Using the soliton equation, we have . We divide the argument into three steps.
Step 1 We want to prove a curvature estimate in the following assertion. Claim For any gradient Ricci soliton, we have: (i) If the soliton is shrinking or steady, then R ≥ 0; (ii) If the soliton is expanding, then there exist a nonnegative constant C = C(n) such that R ≥ λC.
The proof of (ii) is similar.
Step 2 We next want to show that the gradient field grows at most linearly. Claim For any gradient Ricci soliton, there exist constants a and b depending only on the soliton, such that For any point x on M , we connect p and x by a shortest normal geodesic γ(t), t ∈ [0, d(x)].
Step 3 Since the gradient field ∇f grows at most linearly, it must be integrable. Thus we have proved theorem 1.3 .