Ricci Flow on a 3-Manifold with Positive Scalar Curvature I

Ricci ﬂow on a 3-manifold with positive scalar curvature I

By Zhongmin Qian Mathematical Institute, University of Oxford

December 2003

Abstract In this paper we consider Hamilton’s Ricci flow on a 3-manifold having a metric of positive scalar curvature. We establish several a priori estimates for the Ricci flow which we believe are important in understanding possible singularities of the Ricci flow. For Ricci flow with initial metric of positive scalar curvature, we obtain a sharp estimate on the norm of the Ricci curvature in terms of the scalar curvature (which is not trivial even if the initial metric has non-negative Ricci curvature, a fact which is essential in Hamilton’s estimates [8]), some L2- estimates for the gradients of the Ricci curvature, and finally the Harnack type estimates for the Ricci curvature. These results are established through careful (and rather complicated and lengthy) computations, integration by parts and the maximum principles for parabolic equations. (Revised version).

Contents

1 Introduction 2 1.1 Description of main results ...... 3 1.2 Example ...... 5

2 Deformation of metrics 7

3 Hamilton’s Ricci ﬂow 9

4 Control the curvature tensor 14

5 Control the norm of the Ricci tensor 16 5.1 Gradient estimate for scalar curvature ...... 20

1 6 Estimates on the scalar curvature 24

7 L2-estimates for scalar curvature 27 7.1 Some applications ...... 29 7.2 L2-estimates for the Ricci tensor ...... 33

8 Harnack estimate for Ricci ﬂow 35 8.1 Some formulae about Ricci tensors ...... 35 8.2 Harnack inequality for Ricci curvature ...... 37

1 Introduction

In the seminal paper [8], R. S. Hamilton has introduced an evolution equation for metrics on a manifold, the Ricci flow equation, in order to obtain a “better metric” by deforming a Riemannian metric in a way of improving positivity of the Ricci curvature. Hamilton devised his heat flow type equation (originally motivated by Eells-Sampson’s work [6] on the heat flow method for harmonic mappings) by considering the gradient vector field of the total scalar curvature functional on the space M(M) of metrics on a manifold M Z E(g) = Rgdµg ; g ∈ M(M) M where Rg and dµg are the scalar curvature and the volume measure with respect to the metric g, respectively. While Hamilton’s Ricci flow is not the gradient flow of the energy functional E which is ill-posed, rather Hamilton has normalized the gradient flow equation of E to the following heat flow type equation ∂ 2 g = −2R + σ g (1.1) ∂t ij ij n g ij −1 R where σg denotes the average of the scalar curvature Vg M Rdµg. The variation characteristic of the Ricc flow (1.1) has been revealed by G. Perelman in recent works [15], [16]. In these papers Perelman presented powerful and substantial new ideas in order to understand the singularities of solutions to (1.1), for further information, see the recent book [5]. It is clear that the Ricci flow (1.1) preserves the total mass. After change of the space variable scale and re-parametrization of t (1.1) is equivalent to ∂ g = −2R (1.2) ∂t ij ij which however is not volume-preserving. We will refer this equation as the Ricci flow. The parameter t (which has no geometric significance) is suppressed from notations if no confusion may arise.

2 If we express the Ricci tensor Rij in terms of the metric tensor gij, we may see that the Ricci flow equation (1.2) is a highly non-linear system of second-order partial differential equations, which is not strictly parabolic, for which there is no general theory in hand to solve this kind of PDE. Even worse, there are known topological obstructions to the existence of a solution to (1.2). The system (1.2) has a great interest by its own from a point-view of PDE theory, it however has significance in the resolution of the Poincaréconjecture. The Poincaré conjecture claims that a simply-connected 3-manifold is a sphere. If we run the Ricci flow (1.2) on a 3-manifold, and if we could show a limit metric exists with controlled bounds of the curvature tensor, then we could hope the limit metric must have constant Ricci curvature (see (1.1)), therefore the manifold must be a sphere. Thus Ricci flow is a very attractive approach to a possible positive answer to the Poincaréconjecture. This approach was worked out in the classical paper [8] for 3-manifolds with positive Ricci curvature by proving a series of striking a priori estimates for solutions of the Ricci flow.

Theorem 1.1 (Hamilton [8], Main Theorem 1.1, page 255) On any 3-manifold with a metric of positive Ricci curvature, there is an Einstein metric with positive scalar curvature.

1.1 Description of main results Hamilton proved his theorem 1.1 through three a priori estimates for the Ricci ﬂow with initial metric of positive curvature. The essential feature in this particular case is that the scalar curvature dominates the whole curvature tensor. Indeed if the Ricci tensor Rij is positive then we have the following elementary fact 1 R2 ≤ |R |2 ≤ R2 . (1.3) 3 ij

While in general case we still have the lower bound for |Rij|, but there is no way to control |Rij| by its trace R. The first estimate Hamilton proved is the expected one: the Ricci flow improves the positivity of the Ricci tensor, thus if the initial metric has positive Ricci curvature then it remains so as long as the Ricci flow alive. This conclusion is proved by using a maximum principle for solutions to the tensor type parabolic equations (which indeed follows from the use of the classical maximum principle in parabolic theory). One of our results shows that under the Ricci flow on a 3-manifold with positive scalar curvature, the squared norm of the Ricci tensor can be controlled in terms of its scalar curvature and the initial date. Indeed we prove a comparison theorem for 2 2 the quantity |Rij| /R . We then deduce precise bounds on the scalar curvature for the Ricci flow with positive scalar curvature.

3 By removing the positivity assumption of the Ricci curvature, three eigenvalues of the Ricci curvature under the Ricci flow may develop into a state of dispersion: one of eigenvalues may go to −∞ while another to +∞ but still keep the scalar curvature R bounded. Our above result just excludes this case if the initial metric has positive scalar curvature. The second key estimate in [8], also the most striking result in [8], is an estimate which shows (after re-parametrization) the eigenvalues of the Ricci flow at each point approach each other. Hamilton achieved this claim by showing the variance of the three eigenvalues of the Ricci tensor decay like Rκ where κ ∈ (1, 2) depending on the positive lower bound of the Ricci curvature of the initial metric. It is not easy to see the curvature explodes in time for normalized Ricci flow (the volumes of the manifold are scaled to tend to zero), the variance of three eigenvalues of the Ricci curvature 2 2 2 2 are easily seen as |Rij| − R /3, by (1.3) one might guess the variance |Rij| − R /3 possesses the same order of R2. The striking fact is that indeed

2 2 |Rij| − R /3 Rκ

2 2 2 is bounded for some κ < 2! That is to say, |Rij| −R /3 explodes much slower then R , so that, after re-scaling back to the un-normalized Ricci flow, it shows the variance of the eigenvalues of the Ricci tensor goes to zero, hence proves the claim. The positive Ricci curvature assumption is washed down to an elementary fact recorded in Lemma 5.2. This core estimate (see [8], Theorem 10.1, page 283 ) is definitely false if the positivity assumption on the Ricci tensor is removed, and it seems no replacement could be easily recognized without an assumption on the Ricci curvature. Finally in order to show a smooth metric does exist, and has constant Ricci curvature, Hamilton [8] established an important gradient estimate for |∇R| in terms of R and |Rij| ([8], Theorem 11.1, page 287), which in turn implies that, for the un- normalized Ricci flow, |∇R| goes to zero. This estimate was proved by using the 2 2 previous crucial estimate on the variance |Rij| − R /3 and a clever use of the Bianchi identity: |∇Ric|2 ≥ 7|∇R|2/20, instead of the trivial one |∇Ric|2 ≥ |∇R|2/3. The key question we would ask is what kind of gradient estimates for the scalar curvature and for the Ricci curvature can we expect without the essential estimates on the variance of three eigenvalues of the Ricci tensor? This paper gives some partial answers to this question: we establish a weighted integral estimate for |∇R|2 and a Harnack type estimate for the Ricci curvature. We hope these estimates would help us to understand the singularities in the Ricci flow on a 3-manifold with positive scalar curvature, and complete the Hamilton’s program [11] for these 3-manifolds. For recent exciting development, see the papers by G. Perelman and Chow and Knopf’s excellent recent book [5], and also [18] and Ecker’s book for the mean curvature flow.

4 The main simpliﬁcation for 3-manifolds comes from the fact that the full curvature tensor Rijkl may be read out from the Ricci tensor (Rij):

Rijkl = Rikgjl − Rilgjk + Rjlgik − Rjkgil 1 − R (g g − g g ) . (1.4) 2 jl ik jk il In all computations the Bianchi identity will play an essential role. The following are the several forms we will need. The ﬁrst one is

∇mRijkl + ∇kRijlm + ∇lRijmk = 0 . (1.5) Taking trace over indices m and j we obtain

ab g ∇aRibkl = ∇lRik − ∇kRil (1.6) and taking trace again over i and k we thus have

ij ∇kR = 2g ∇jRik . (1.7)

a 1 a That is ∇ Rka = 2 ∇kR. The equation (1.7) may be write as ∇ G(Ric)ak = 0 for all k, where 1 G(V ) = V − tr (V )g ij ij 2 g ij

for a symmetric tensor (Vij).

1.2 Example In order to appreciate the difficulty induced by the topology for the Ricci flow, let us examine a typical example not covered by the theorem of Hamilton [8]. Let M = S2 × S1 endowed with the standard product metric. Choose a coordinate 2 system, and write (hij) to be the standard metric on the sphere S with sectional curvature 1. The standard product metric on M may be written as (h ) 0 (g(0) ) = ij ij 0 1 and the Ricci curvature tensor (h ) 0 (R(0) ) = ij ij 0 0 so that the scalar curvature R(0) = 2, a constant. Consider the Ricci flow with the initial metric (g(0)ij): ∂ 2 (g(t) ) = −2R(t) + σ(t)g(t) . ∂t ij ij 3 ij

5 Let us assume that the solution ﬂow has the following form

a(t)(h ) 0 (g(t) ) = ij . (1.8) ij 0 f(t)

Since (h ) 0 (R(t) ) = ij ij 0 0 so that 2 R(t) = g(t)ijR(t) = . ij a(t) 2 Therefore σ(t) = a(t) . Thus the Ricci ﬂow equation is equivalent to the following system of ordinary diﬀerential equations ∂ 2 a(t) = − ; a(0) = 1 ; ∂t 3 ∂ 4 f(t) f(t) = ; f(0) = 1 ∂t 3 a(t)

which have solutions 2 1 a(t) = 1 − t ; f(t) = . 3 2 2 1 − 3 t

3 3 The condition that a > 0 then yields that t < 2 . Notice that as t ↑ 2 , then σ(t) explodes to +∞, though the volumes still keep as a constant. On the other hand the Ricci tensor for all t has constant eigenvalues, and therefore ∇kRij = 0. Let us write down the solution ﬂow as the following explicit form

2t (1 − 3 )(hij) 0 (g(t)ij) = 2 −2 0 1 − 3 t

2 or in terms of the mean value of the scalar curvature σ(t) = 2/ 1 − 3 t we have the solution metric 2 ! σ(t) (hij) 0 (g(t)ij) = σ(t)2 . 0 4 2 1 The Ricci flow (g(t)ij) on S × S explains the difficulty about Ricci flow on a general 3-manifold: a limit metric (even after a scaling) to the Ricci flow may not exist. However we may also consider the following form of Ricci flow ∂ 2 (g(t) ) = −2R(t) + (ασ(t) + β) g(t) (1.9) ∂t ij ij 3 ij

6 where α and β are two constants to be chosen according to the type of 3-manifolds. 2 We still search for a solution ﬂow with the form (1.8), which leads to again σ(t) = a(t) and ∂ 2 a(t) = −2 + (2α + βa(t)) ; a(0) = 1 ∂t 3 ∂ 2 2 f(t) = α + β f(t); f(0) = 1 . ∂t 3 a(t)

3 2 1 3 Obviously α = 2 is a good choice for the manifold S × S . With this choice α = 2 and any β ≥ 0 we have 2 a(t) = exp βt 3 and 3 3 2 2 f(t) = exp − exp − βt + βt . β β 3 3 Thus the solution ﬂow to (1.9) exists for all time t, the eigenvalues of the Ricci tensor are constant (in space variables), and 2 σ(t) = 2 exp − βt 3

3 goes to zero as t → ∞ if β > 0. In particular if α = 2 and β = 0, then a(t) = 1 ; f(t) = exp (2t).

The solution ﬂow exists, but no limit exist as t → ∞.

2 Deformation of metrics

Let (g(t)ij) be a family of metrics on M satisfying the following equation ∂ g = −2h ∂t ij ij

where hij is a family of symmetric tensors depending on t maybe on (g(t)ij) as well. If no confusion may arise, the parameter t will be suppressed. The inverse of (gij) then evolves according to the equation ∂ gij = 2gibgajh (2.1) ∂t ab

∂ ij ij ij which may be written as ∂t g = 2h (the indices in h(t) are lifted with respect to the metric g(t)ij, this remark applies to other similar notations).

7 j j ∂Γik Although for each t the Christoﬀel symbol Γik is not a tensor, is however ∂t its derivative in t. Recall that j 1 jp ∂gip ∂gkp ∂gik Γik = g + − 2 ∂xk ∂xi ∂xp so that ∂Γj ik = −gjp {(∇ h ) + (∇ h ) − (∇ h )} . (2.2) ∂t k ip i kp p ik

∂Rik From which it follows immediately the variation for the Ricci curvature tensor, ∂t . Indeed under a normal coordinate system

∂R ∂Γj ∂Γj ik = ∂ ik − ∂ jk ∂t j ∂t i ∂t ∂Γj ∂Γj = ∇ ik − ∇ jk j ∂t i ∂t which together (2.2) implies the following

∂ R = ∆h + ∇ ∇ tr (h ) − (∇a∇ h + ∇a∇ h ) . (2.3) ∂t ij ij i j g ab j ia i ja After taking trace ∂ R = 2∆tr (h ) + 2h Rab − 2∇a∇bh . ∂t g ab ab ab We may exchange the order of taking co-variant derivatives in the last term of (2.3) via the following

Lemma 2.1 Let (hij) be a symmetric tensor on a 3-manifold. Then

a a (∇ ∇jhia + ∇ ∇ihja) a a ab = (∇i∇ hja + ∇j∇ hia) − 2R habgij − 2trg(hab)Rij ab −Rhij + Rtrg(hab)gij + 3g (hiaRjb + hjaRib) . (2.4)

Proof. Indeed, by the Ricci identity for symmetric tensors,

ab ab ∇j∇khil = ∇k∇jhil + halg Rkjbi + hiag Rkjbl

and ab ab ∇l∇ihkj = ∇i∇lhkj + hajg Rilbk + hkag Rilbj

8 we thus have

jl g (∇j∇khil + ∇l∇ihkj) jl = g (∇i∇lhkj + ∇k∇jhil) jl ab jl ab jl ab +g hajg (Rilbk + Riblk) + hkag g Rijbl + hiag g Rkjbl jl = g ∇i∇lhkj + ∇k∇jhil jl ab jl ab jl ab −2hajg g Ribkl + hkag g Rijbl + hiag g Rkjbl . (2.5)

However in 3-manifolds, the full Ricci curvature Rijkl may be expressed in terms of the Ricci tensor, equation (1.4), we may easily verify the followings

1 1 h gjlgabR = gjlgabh R − Rtr (h ) g + Rh aj ibkl aj bl 2 g ab ik 2 ik ab ab −g haiRbk − g hkbRia + trg(hab)Rik,

jl ab ab hkag g Rijbl = g hkaRib and jl ab ab hiag g Rkjbl = hiag Rkb , substituting these equations into (2.5) to get the conclusion. Therefore

∂ Lemma 2.2 On 3-manifolds, if ∂t gij = −2hij, then ∂ R = ∆h + ∇ ∇ tr (h ) − {∇ ∇ah + ∇ ∇ah } ∂t ij ij i j g ab i ja j ia ab + 2habR − Rtrg(hab) gij + Rhij ab +2trg(hab)Rij − 3g (hiaRjb + hjaRib) . (2.6)

∂ Corollary 2.3 On 3-manifolds, if ∂t gij = −2hij, then ∂ R = 2∆tr (h ) + 2h Rab − 2 ∇b∇ah . (2.7) ∂t g ab ab ab 3 Hamilton’s Ricci ﬂow

Although most of our results will be stated only for the normalized Ricci ﬂow ∂ g = −2R ∂t ij ij

9 it may be useful to consider a general evolution equation for metrics. Let M(M) denote the space of all metrics on the manifold M of dimension 3, and let 1 h = R − (ασ + β) g (3.1) ij ij 3 ij where α and β are two constant, Rij is the Ricci tensor of gij, and σ is the mean value of the scalar curvature 1 Z σ = Rgdµg Vg M which is a functional on M(M). Therefore (hij) is a symmetric tensor which may be seen as a functional on M(M). It is obvious by deﬁnition

trg(hij) = R − (ασ + β)

The Ricci ﬂow (with parameters α and β) is the following equation on the metrics (gij): ∂ g = −2h ∂t ij ij 1 = −2 R − (ασ + β) g . (3.2) ij 3 ij

The Ricci ﬂow (3.2) may be written in diﬀerent forms:

∂ 1 gij = 2 Rij − (ασ + β) gij (3.3) ∂t 3 and ∂ 1 ∆ − g = −2 R − (ασ + β) g ∂t ij ij 3 ij ij where ∆ denotes the trace Laplacian g ∇i∇j associated with the metric (g(t)ij). In what follows we assume that (g(t)ij) (but t will be suppressed from notations, unless speciﬁed) is the maximum solution to the Ricci ﬂow (3.2). For simplicity we use A(t) to denote ασ(t) + β, and Vg denote the volume of (M, g). Let λ1 ≥ λ2 ≥ λ3 denote the three eigenvalues of the Ricci tensor (Rij). Then

2 2 2 R = λ1 + λ2 + λ3 ; S = λ1 + λ2 + λ3

2 where S = |Rij| . We also deﬁne

3 3 3 4 4 4 T = λ1 + λ2 + λ3, U = λ1 + λ2 + λ3 . Then 1 λ λ + λ λ + λ λ = (R2 − S), 1 2 2 3 1 3 2

10 1 1 1 λ λ λ = R3 − RS + T 1 2 3 6 2 3 and 4 1 1 U = RT − R2S + S2 + R4 . 3 2 6 1 2 The variance of λ1, λ2 and λ3 is S − 3 R which will be denoted by Y . ab ab Let Sij = g RibRaj and Tij = g RibSaj. Then S = trg(Sij) and T = trg(Tij). By equation (2.6) we easily see that

∂ ∆ − R = 6S − 3RR + R2 − 2S g (3.4) ∂t ij ij ij ij which has a form independent of α or β. From here we may compute the evolution for the scalar curvature ∂ 1 ∆ − R = −2 S − AR . (3.5) ∂t 3

We next compute the evolution equations for the tensors (Sij) and (Tij). ∂ ∂ ∂ ∆ − S = gabR ∆ − R − R R gab ∂t ij aj ∂t ib ib aj ∂t ∂ +gabR ∆ − R + 2gabh∇R , ∇R i ib ∂t aj ib aj ab ab ab = 6g RajSib + 6g RibSaj − 2RibRajR 2 −6RS + 2 R2 − 2S R + AS ij ij 3 ij ab +2g h∇Rib, ∇Raji

that is ∂ ∆ − S = 10T − 6RS + 2 R2 − 2S R ∂t ij ij ij ij 2 + AS + 2gab(∇ R )(∇kR ) . (3.6) 3 ij k ib aj Taking trace we obtain ∂ ∆ − S = 2 R3 − 5RS + 4T ∂t 4 + AS + 2|∇ R |2 . (3.7) 3 k ij These are the evolution equations computed in Hamilton [8] (in the case α = β = 0) via his evolution equation for the full curvature tensor.

11 Similarly we have

∂ ∆ − T = 14gabR T − 9RT + 3 R2 − 2S S ∂t ij ib aj ij ij 4 + AT + 2Rab(∇kR )(∇ R ) 3 ij ib k aj k ab +2(∇ R ) {Rib(∇kRaj) + Raj(∇kRib)} . (3.8)

It follows that ∂ ∆ − T = 12U − 9RT + 3 R2 − 2S S + 2AT ∂t ab k ij +6g Ria(∇ R )(∇kRbj)

together with the fact that 4 1 1 U = RT − R2S + S2 + R4 3 2 6 we thus establish the following evolution equation

∂ ∆ − T = 7RT − 9R2S + 2R4 + 2AT ∂t ab k ij +6g Ria(∇ R )(∇kRbj) . (3.9)

In what follows, we only consider 3-manifolds with a metric of positive scalar curvature. Let (g(t)ij) be the maximum solution to the Ricci ﬂow

∂ 1 ∆ − g = −2 R − Ag ∂t ij ij 3 ij

on a 3-manifold M with initial metric g(0)ij of positive constant scalar curvature R(0), unless otherwise speciﬁed. Let µt denote the volume measure associated with the solution metric (g(t)ij) (at time t), and let s dµt det(g(t)ij) Mt = = . dµ0 det(g(0)ij) Then an elementary computation shows that ∂ log M = −tr (h ) ∂t t g ij = −R + A .

12 Therefore d d Z Vg(t) = Mtdµ0 dt dt M Z d = log Mt dµt M dt Z = (−R + A) dµt M = (β + (α − 1) σ) Vg(t) so that the following

Lemma 3.1 The volume function t → Vg(t) is constant if β = 0 and α = 1. For general α and β we have d log V = β + (α − 1) σ(t) dt g(t) where 1 Z σ(t) = Rg(t)dµt . Vg(t) M Since ∂ 1 log M(t) = −tr R − Ag ∂t g ij 3 ij = −R(t, ·) + A(t) and therefore Z t log M(t) = − (R(s, ·) − A(s)) ds . (3.10) 0 Taking Laplacian of logM(t) in the last equation we obtain Z t ∆ log M(t) = − (∆R)(s, ·)ds 0 Z t 1 = R(0) − R + 2 S − AR 0 3 so that ∂ Z t 1 ∆ − log M(t) = R(0) − A(t) + 2 S − AR . (3.11) ∂t 0 3 The scalar curvature R remains positive if the initial metric possesses positive scalar curvature. To see this let us consider the function

2 R t A(s)ds K = e 3 0 R , (3.12)

13 and we can easily see the evolution equation for K is given as ∂ 2 R t A(s)ds ∆ − K = −2e 3 0 S (3.13) ∂t so the claim follows easily from the maximum principle.

4 Control the curvature tensor

Consider the maximum solution (g(t)ij) to the Ricci ﬂow

∂ 1 g = −2 R − Ag (4.1) ∂t ij ij 3 ij

where A(t) = ασ(t) + β as before. Recall that the scalar curvature R satisﬁes the following parabolic equation

∂ 2 ∆ − R = −2S + AR ∂t 3

so that ∂ − 2 R t A(s)ds − 2 R t A(s)ds ∆ − e 3 0 R = −2e 3 0 S , ∂t therefore by the maximum principle, it follows that R remains positive if for the initial metric R(0) is positive. Our ﬁrst result shows that if the initial metric possesses positive scalar curvature, then the full curvature tensor of (g(t)ij) may be controlled by its scalar curvature. Let Vij = Rij − εRgij where ε is a constant. Then by the evolution equations for (Rij), R and the Ricci ﬂow (4.1) we deduce that

∂ ∆ − V = 6gpqV V + (10ε − 3) RV ∂t ij pj iq ij 2 2 + 2 (ε − 1) S + 4ε − 3ε + 1 R gij . (4.2)

The very nice feature of this identity is that it involves α, β and σ only through the symmetric tensor Vij, which takes the same form for all α, β.

Theorem 4.1 Let M be a closed 3-manifold. If ε ≤ 1/3 is a constant such that Rij − εRgij ≥ 0 at t = 0, so does it remain. If ε ≥ 1 and at t = 0, Rij − εRgij ≤ 0, then the inequality remains to hold for all t > 0.

14 Proof. We prove these conclusions by using the maximum principle to the tensor type parabolic equation (4.2) and (Vij) (see [8], Theorem 9.1, page 279). Let us prove the ﬁrst conclusion. Suppose λ ≥ µ ≥ ν are eigenvalues of the Ricci tensor Rij. Then 2 2 2 j S = λ + µ + ν and R = λ + µ + ν. If ξ 6= 0 such that Vijξ = 0 for all i. Then one of the eigenvalues of Vij is zero. Since the eigenvalues of Vij are λ − εR, µ − εR, ν − εR, so we may assume that ν − εR = 0. Hence

λ + µ = (1 − ε)R , S = λ2 + µ2 + ε2R2 .

Since

2 (ε − 1) S + 4ε2 − 3ε + 1 R2 = 2 (ε − 1) λ2 + µ2 + ε2R2 + 4ε2 − 3ε + 1 R2 = 2 (ε − 1) (λ2 + µ2) + 2 (ε − 1) ε2R2 + 4ε2 − 3ε + 1 R2 ≤ (ε − 1) (λ + µ)2 + 2 (ε − 1) ε2R2 + 4ε2 − 3ε + 1 R2 = (ε − 1) (1 − ε)2R2 + 2 (ε − 1) ε2R2 + 4ε2 − 3ε + 1 R2 = (3ε − 1) ε2R2 ≤ 0 ,

where the ﬁrst inequality follows from the Cauchy inequality and the last one follows our assumption ε ≤ 1/3. Now the conclusion follows from equation (4.2) and the maximum principle. To show the second conclusion, we apply the maximum principle to −Vij and use the fact that ε ≥ 1. The only thing then one should notice is the following inequality

2 (ε − 1) S + 4ε2 − 3ε + 1 R2 ≥ 7R2/16 ≥ 0 .

Corollary 4.2 On a closed 3-manifold M and suppose R(0) > 0 (so R(t) > 0 for all t). 2 2 1. If R(0)ij ≥ 0, then R /3 ≤ S ≤ R . If ε > 0 such that

R(0)ij ≥ −εR(0)g(0)ij

then R2/3 ≤ S ≤ (1 + 4ε + 6ε2)R2 . (4.3)

2. If R(0)ij ≥ bR(0)g(0)ij for some constant 0 ≤ b ≤ 1/3, then

R2/3 ≤ S ≤ (1 − 4b + 6b2)R2 .

15 Proof. By Theorem 4.1, Rij ≥ −εRgij as long as the solution exists. Let λ ≥ µ ≥ ν be the eigenvalues of the Ricci tensor Rij. Then λ, µ, ν ≥ −εR. Suppose µ, ν < 0, then

λ2 + µ2 + ν2 ≤ (R + |µ| + |v|)2 + |µ|2 + |v|2 = (R + 2εR)2 + 2ε2R2 = 1 + 4ε + 6ε2 R2 ,

and similarly when λ ≥ µ ≥ 0 and ν < 0 we have

λ2 + µ2 + ν2 ≤ (λ + µ)2 + v2 = (R + |ν|)2 + |v|2 ≤ (R + εR)2 + ε2R2

and while ν ≥ 0 we then clearly have λ2 + µ2 + ν2 ≤ R2. Therefore the conclusion follows. The estimate (4.3) shows that the scalar curvature dominates the full curvature if the initial metric has positive scalar curvature, which however is a rough estimate. We will give a sharp estimate in next section (Theorem 5.6 below).

5 Control the norm of the Ricci tensor

In this section we work with the Ricci ﬂow on a 3-manifold ∂ g = −2R ∂t ij ij with initial metric such that R(0) > 0. Thus R ≥ R(0) > 0 and the variance of three 1 2 eigenvalues of the Ricci tensor (Rij), Y ≡ S − 3 R is non-negative as well. The third variable in our mind is not T whose sign is diﬃculty to determined. Motivated by the fundamental work [8], we consider the polynomial of the eigenvalues of (Rij): 1 5 P = S2 + R4 − R2S + 2RT (5.1) 2 2 as the third independent variable, which is non-negative, as showed in [8]. Let us begin with a geometric explanation about the polynomial P . In terms of the three eigenvalues λ1 ≥ λ2 ≥ λ3 of the Ricci tensor (Rij) ([8], Lemma 10.6, page 285)

2 2 2 P = (λ1 − λ2) λ1 + (λ1 + λ2)(λ2 − λ3) + λ3(λ1 − λ3)(λ2 − λ3) . (5.2)

Hence we establish the following

16 Lemma 5.1 We have 1) P ≥ 0 ; 2) P = 0 if and only if 2a) λ1 = λ2 = λ3; or 2b) one of eigenvalues λi is zero, the other two are equal number.

Together with this lemma, the following lemma explains why the assumption of positive Ricci curvature is special.

Lemma 5.2 Suppose that λ1 + λ2 + λ3 ≥ 0 and

λ1 ≥ λ2 ≥ λ3 ≥ ε (λ1 + λ2 + λ3) (5.3) for some ε ∈ [0, 1/3], then P ≥ ε2S S − R2/3 . (5.4)

Therefore all evolution equations must be written in terms of these variables.

∂ ∆ − R = −2S , (5.5) ∂t ∂ ∆ − S = 2R3 − 10RS + 8T + 2|∇ R |2 . (5.6) ∂t k ij To take care the homogeneity, we may use R2 instead of R

∂ ∆ − R2 = −4RS + 2|∇R|2 , ∂t

1 2 and for S we prefer to use Y = S − 3 R then S: ∂ 26 2 ∆ − Y = 2R3 − RS + 8T + 2|∇ R |2 − |∇R|2 . ∂t 3 k ij 3

Therefore, for any constant κ,

∂ Y 1 ∂ ∂ ∆ − = ∆ − Y + Y ∆ − R−κ ∂t Rκ Rκ ∂t ∂t +2h∇Y, ∇R−κi 13 S 2 = R3 − RS + 4T + κ Y 3 R Rκ 1 κ(κ + 1) |∇R|2 2 + |∇ R |2 − |∇R|2 + Y k ij 3 2 R2 Rκ +2h∇Y, ∇R−κi .

17 Lemma 5.3 We have the following elementary facts 2 1 1 1 h∇Y, ∇R−κi = κ|∇R|2 − κ h∇S, ∇Ri (5.7) 3 Rκ R Rκ and |∇R|2 1 Y h∇Y, ∇R−κi = −κ2Y − κh∇ log R, ∇ i . (5.8) R2 Rκ Rκ Therefore we may write the evolution equation for Y/Rκ into diﬀerent forms. For the one we need, we decompose 2h∇Y, ∇R−κi into 2θh∇Y, ∇R−κi + 2(1 − θ)h∇Y, ∇R−κi |∇R|2 2 Y = −θκ2Y − 2θκh∇ log R, ∇ i R2 Rκ Rκ 2 1 2 + (1 − θ)κ|∇R|2 − κ(1 − θ) h∇S, ∇Ri 3 R Rκ so that ∂ Y 13 S 2 ∆ − = R3 − RS + 4T + κ Y ∂t Rκ 3 R Rκ κ(1 − θ) 2(1 − θ)κ − 1 2 + |∇ R |2 − h∇S, ∇Ri + |∇R|2 k ij R 3 Rκ |∇R|2 Y Y +κ {(1 − 2θ) κ + 1} − 2θκh∇ log R, ∇ i . (5.9) R2 Rκ Rκ To eliminate the term h∇S, ∇Ri we use the following elementary fact Lemma 5.4 For any constant a we have

2 2 2 2 |∇kRij − aRij∇kR| = |∇kRij| + a S|∇R| − ah∇R, ∇Si .

18 Choose θ so that κ (1 − θ) = 1 which will eliminate the non-positive term

2 |∇R|2 (κ (1 − θ) − 1)2 , 3 Rκ that is 1 θ = 1 − . κ With this value of θ,

∂ Y 13 S 2 ∆ − = R3 − RS + 4T + κ Y ∂t Rκ 3 R Rκ 2 Y |∇R|2 +|∇ R − aR ∇ R|2 + (2 − κ)(κ − 1) k ij ij k Rκ R2 Rκ Y −2 (κ − 1) h∇ log R, ∇ i . (5.11) Rκ Finally we replace T by via P , R and S

P S2 4T = 2 − 2 − R3 + 5RS (5.12) R R and thus establish ∂ Y 2 − κ 4 1 Y |∇R|2 ∆ − = P − SY + (2 − κ)(κ − 1) ∂t Rκ 2 R Rκ R2 Rκ Y 2 −2(κ − 1)h∇ log R, ∇ i + |∇ R − aR ∇ R|2 . (5.13) Rκ k ij ij k Rκ

Lemma 5.5 For any constant κ, set Lκ = ∆ + 2(κ − 1)∇ log R.∇. Then

∂ Y 2 − κ 4 Y |∇R|2 Rκ L − = P − SY + (2 − κ)(κ − 1) κ ∂t Rκ 2 R R 2 Rij +2 ∇kRij − ∇kR . (5.14) R

In particular for any κ ∈ [1, 2], the following diﬀerential inequality holds

∂ Y 2 − κ 4 Rκ L − ≥ P − SY . (5.15) κ ∂t Rκ 2 R

19 Indeed this lemma was proved in Hamilton [8] in order to prove his most striking estimate ([8], Theorem 10.1, page 283). We give a detailed computation to exhibit the reason why the diﬀerential inequality (5.15) takes this form, which we need to prove our estimate on S. In particular if κ = 2 then 2 2 ∂ Y 4 Rij R L2 − = P + 2 ∇kRij − ∇kR ∂t R2 R R which is always non-negative, which allows us to derive a sharp estimate on the scalar 2 function |Rij| . Indeed the above equation yields the following diﬀerential inequality ∂ Y R2 L − ≥ 0 2 ∂t R2 so that the maximum principle for parabolic equations yields the following

Theorem 5.6 Under the Ricci ﬂow, and suppose R(0) > 0. Then as long as the Ricci ﬂow exists we always have S S(0) ≤ max . (5.16) R2 M R(0)2 The estimate in this theorem is sharp. Another observation is the following.

Corollary 5.7 We have for any constants κ and η ∂ Y ∂ Y S Rκ+η L − = Rκ L − + 2η Y κ+η ∂t Rκ+η κ ∂t Rκ R Y |∇R|2 + (3 − 2κ − η) η . (5.17) R In particular d ∂ Y S |∇R|2 Rκ L − = 2 − (2κ − 3) Y . (5.18) dκ κ ∂t Rκ R R

5.1 Gradient estimate for scalar curvature Next we want to treat the gradient of the Ricci tensor, begin with the scalar curvature R. In general if a scalar function F satisﬁes the following parabolic equation (under ∂ the Ricci ﬂow ∂t gij = −2Rij) ∂ ∆ − F = B , (5.19) ∂t F

20 then by the Bochner identity and the Ricci ﬂow equation

∂ ∆ − |∇F |2 = 2|∇∇F |2 + 2h∇B , ∇F i . (5.20) ∂t F

It is better to look at the evolution equation for |∇F |2/R which can be found by using the chain rule. It is given indeed as the following

∂ |∇F |2 2 2S 2 ∆ − = h∇B , ∇F i + + |∇R|2 |∇F |2 (5.21) ∂t R R F R2 R3 2 2 + |∇∇F |2 − h∇R, ∇|∇F |2i . R R2 In order to use the hessian term which is non-negative, we observe

2 ab ∇k|∇F | = 2g (∇aF )(∇k∇bF ) so that 2 k l h∇R, ∇|∇F | i = 2(∇ R) ∇ F (∇k∇lF ). It follows that

2 2 |∇∇F | = |∇k∇lF − a(∇kR)(∇lF ) | −a2|∇R|2|∇F |2 + ah∇R, ∇|∇F |2i .

Therefore ∂ |∇F |2 2 2S 2 ∆ − = h∇B , ∇F i + + |∇R|2 |∇F |2 ∂t R R F R2 R3 2 + |∇ ∇ F − a(∇ R)(∇ F ) |2 R k l k l 2a2 2 − |∇R|2|∇F |2 + a h∇R, ∇|∇F |2i R R 2 − h∇R, ∇|∇F |2i . (5.22) R2 In particular, to eliminate the term h∇R, ∇|∇F |2i we choose a = 1/R which then gives us the following

∂ |∇F |2 2 2S ∆ − = h∇B , ∇F i + |∇F |2 ∂t R R F R2

2 −1 2 + ∇k∇lF − R (∇kR)(∇lF ) . (5.23) R

21 In particular, if we apply the formula to the scalar curvature R, we establish

1 ∂ |∇R|2 2 S ∆ − = − h∇S, ∇Ri + |∇R|2 2 ∂t R R R2

1 −1 2 + ∇k∇lR − R (∇kR)(∇lR) . (5.24) R Together with the equations

1 ∂ 2 ∆ − R2 = −2RY − R3 + |∇R|2 2 ∂t 3 and 1 ∂ P S 1 ∆ − Y = 2 − 2 Y + |∇ R |2 − |∇R|2 , 2 ∂t R R k ij 3 for function |∇R|2 H = + ξY + ηR2 R where ξ and η are two constants, we have 1 ∂ 2 1 −1 2 ∆ − H = − h∇S, ∇Ri + ∇k∇lR − R (∇kR)(∇lR) 2 ∂t R R P 2Y 2 + 2ξ − ξS + ηR2 − ηR3 R R 3 1 S + ξ|∇ R |2 − ξ|∇R|2 + η|∇R|2 + |∇R|2 . k ij 3 R2

In this formula, the only term we have to deal is h∇S, ∇Ri, which we handle as the following.

2 2 2 2 2 |b∇kRij − aRij∇kR| = b |∇kRij| + a S|∇R| − abh∇S, ∇Ri

2 2 in which we set ab = R , i.e. a = bR , and thus 2 2 2 − h∇S, ∇Ri = b∇kRij − Rij∇kR R bR 4 S −b2|∇ R |2 − |∇R|2 . k ij b2 R2

22 Therefore 1 ∂ 1 −1 2 ∆ − H = ∇k∇lR − R (∇kR)(∇lR) 2 ∂t R 2 2 + b∇kRij − Rij∇kR bR P 2Y 2 + 2ξ − ξS + ηR2 − ηR3 R R 3 1 + ξ − b2 |∇ R |2 − ξ|∇R|2 + η|∇R|2 k ij 3 4 S + 1 − |∇R|2 . (5.25) b2 R2 We next need to decide the signs of three constants ξ, η and b. It is suggested that 2 2 ξ ≥ b (otherwise we can not control |∇kRij| ). Under this choice, and we are expected with the good choices of these three constants, we will lose nothing from the ﬁrst two terms on the right-hand side of equation (5.25), we thus simply drop these two, and use the inequality ([8], Lemma 11.6, page 288) 7 |∇ R |2 ≥ |∇R|2 k ij 20 so that 1 ∂ P 2Y 2 ∆ − H ≥ 2ξ − ξS + ηR2 − ηR3 2 ∂t R R 3 1 7 + ξ − b2 + η |∇R|2 60 20 4 S + 1 − |∇R|2 . (5.26) b2 R2

Obviously a simple choice for b is b = 2 so that the last term in (5.26) is dropped. Thus 1 ∂ P 2 S η ∆ − H ≥ 2ξ − ηR3 − 2R + ξY 2 ∂t R 3 R2 ξ 1 7 + ξ − + η |∇R|2 . (5.27) 60 5

This inequality is enough to prove Hamilton’s estimate ([8], Theorem 11.1, page 287). Replace Y by |∇R|2 ξY = H − − ηR2 R

23 to obtain 1 ∂ P η 1 η ∆ − H ≥ 2ξ + 2 δ + − ηR3 − 2R δ + H 2 ∂t R ξ 3 ξ 1 η 7 + ξ + 2δ + 2 − + η |∇R|2 (5.28) 60 ξ 5

in which we choose η = 0 and ξ = 84 we have

1 ∂ ∆ − H ≥ −2δRH . (5.29) 2 ∂t

We may use the Kac formula to deduce an estimate for the scalar curvature. s,x Let (Xt, P ) be the diﬀusion process associated with the time dependent elliptic 1 operator 2 ∆. Then we have

∂ Theorem 5.8 Under the Ricci ﬂow ∂t gij = −2Rij with initial metric of positive scalar curvature R(0) > 0. Then

R t 0,x 2δ R(s,Xs)ds H(t, x) ≤ P H(0,Xs)e 0 , (5.30)

|∇R|2 where H = R + 84S. In particular if R(t, ·) ≤ θ(t) on M, then

2δ R t θ(s)ds H(t, x) ≤ e 0 max H(0, ·) . M 6 Estimates on the scalar curvature

Thanks to the resolution of the Yamabe problem (e.g. , for a 3-manifold with a metric of positive scalar curvature, we may run the general Ricci ﬂow with an initial metric (g(0)ij) such that R(0) is a positive constant. Under such an initial metric, and if we run the Ricci ﬂow ∂ 1 g = −2 R − σg ∂t ij ij 3 ij

(i.e. for the case α = 1 and β = 0), then all volume measures µt associated with the Ricci ﬂow (g(t)ij) have the same volume. We may choose a non-negative constant ε such that R(0)ij ≥ −εR(0)g(0)ij, which is possible since R(0) > 0. We will of course choose the least one ε ≥ 0 for a given initial data. Then R > 0 as long as the Ricci ﬂow exists, and by Theorem 4.1

R(t)ij ≥ −εR(t)g(t)ij

24 1 for all t. Furthermore there is a constant δ ≥ 3 such that R(t)2/3 ≤ S(t) ≤ δR(t)2 . (6.1)

For example, any δ ≥ 1 + 4ε + 6ε2 will do. However in the case of the Ricci ﬂow ∂ 2 ∂t gij = −2Rij, the optimal choice for δ is the maximum of S(0)/R(0) . With the last estimate we may deduce several elementary estimates on the scalar curvature R. Deﬁne 3 Z t ρ(t) = exp − A(s)ds , ρ(0) = 1 . 2 0

− 3 1 Lemma 6.1 Let ψ(r) =e 2 r and set

Z t F = ψ(K) exp − ρ(s)ds 0 3 1 Z t = exp − ρ(t) + ρ(s)ds 2 R(t) 0 where K is given in (3.12). Then

ψ00 (K) ∂ ∆ − ∇K − F ≤ 0 . (6.2) ψ0(K) ∂t

Proof. The diﬀerential inequality relies on the evolution (3.13) for K. By the chain ∂ ∂ ψ00 (K) ∆ − ψ(K) = ψ0(K) ∆ − K + ∇K.∇ψ(K) ∂t ∂t ψ0(K) 00 0 2 R t A(s)ds ψ (K) = −2ψ (K)e 3 0 S + ∇K.∇ψ(K) ψ0(K) 00 2 0 2 − 2 R t A(s)ds ψ (K) ≤ − ψ (K)K e 3 0 + ∇K.∇ψ(K) 3 ψ0(K) 00 − 2 R t A(s)ds ψ (K) = −e 3 0 ψ(K) + ∇K.∇ψ(K) ψ0(K)

which implies (6.2). By the maximum principle applying to (6.2) and the fact that R(0) is constant, we have 3 ρ(t) Z t 3 1 + ρ(s)ds ≤ . 2 R(t) 0 2 R(0)

25 ρ(t) Since we always have R(t) > 0 so that Z t 3 1 ρ(s)ds ≤ 0 2 R(0) for all t and 3R(0)ρ(t) R(t) ≥ . R t 3 − 2R(0) 0 ρ(s)ds for all t provides the Ricci ﬂow exists. Similarly, if we use the fact that S ≤ δR2 we then have

− 1 1 Lemma 6.2 Let ψ(r) =e 2δ r and Z t F = ψ(K) exp − ρ(s)ds 0 1 Z t = exp − ρ(t) − ρ(s)du . 2δR 0 Then ψ00 (K) ∂ ∆ − ∇K − F ≥ 0 . (6.3) ψ0(K) ∂t Again the maximum principle then implies that ρ(t) Z t 1 + ρ(s)ds ≥ 2δR 0 2δR(0) and therefore R(0)ρ(t) R(t) ≤ . R t 1 − 2δ 0 ρ(s)ds Theorem 6.3 Under the general Ricci ﬂow with initial metric having positive constant scalar curvature R(0) and 3 Z t ρ(t) = exp − A(s)ds . 2 0 Then Z t 3 1 ρ(s)ds < 0 2 R(0) and 3R(0)ρ(t) R(0)ρ(t) ≤ R(t) ≤ R t R t 3 − 2R(0) 0 ρ(s)ds 1 − 2δ 0 ρ(s)ds as long as the Ricci ﬂow (g(t)ij) exists.

26 7 L2-estimates for scalar curvature

∂ 2,1 If (gij) is a solution ﬂow to an evolution equation: ∂t gij = −2hij, and if f ∈ C (R+ × M), then

d Z Z ∂ Z ∂ f(t, ·)dµt = f(t, ·)dµt + f(t, ·) (log Mt) dµt dt M M ∂t M ∂t Z ∂ Z = f(t, ·)dµt − trg(hij)f(t, ·)dµt . (7.1) M ∂t M R Since M is compact, so that M ∆f(t, ·)dµt = 0 (where ∆ is the Laplace-Beltrami operator associated with g(t)ij, if no confusion may arise we will suppress the parameter t), and thus

d Z Z ∂ Z f(t, ·)dµt = − ∆ − f(t, ·)dµt − trg(hij)f(t, ·)dµt . (7.2) dt M M ∂t M

In particular, under the Ricci ﬂow (3.2), trg(hij) = R − A so that

d Z Z ∂ f(t, ·)dµt = − ∆ − f(t, ·)dµt dt M M ∂t Z Z − Rf(t, ·)dµt + A f(t, ·)dµt . (7.3) M M For the gradient of f we may use the Bochner identity

2 ∆|∇f| = 2Γ2(f) + 2h∇∆f, ∇fi = 2|∇∇f|2 + 2Ric(∇f, ∇f) + 2h∇∆f, ∇fi ,

∂ while by using the fact that ∂t gij = −2hij we have ∂ ∂ ∂f ∂f ∂ |∇f|2 = gij + 2h∇ f , ∇fi ∂t ∂t ∂xi ∂xj ∂t ∂f ∂f ∂ = 2hij + 2h∇ f , ∇fi , ∂xi ∂xj ∂t so that ∂ ∂f ∂f ∆ − |∇f|2 = 2|∇∇f|2 + 2 Rij − hij ∂t ∂xi ∂xj ∂ +2h∇ ∆ − f, ∇fi . (7.4) ∂t

27 Next we consider the solution flow to the Ricci flow (3.2). In this case 1 Rij − hij = Agij 3 where A(t) = ασ(t) + β, and therefore ∂ 2 ∂ ∆ − |∇f|2 = 2|∇∇f|2 + A|∇f|2 + 2h∇ ∆ − f, ∇fi . (7.5) ∂t 3 ∂t Applying (7.2) to |∇f|2 we obtain Z Z d 2 ∂ 2 |∇f| dµt = − ∆ − |∇f| dµt dt M M ∂t Z Z 2 2 − R|∇f| dµt + A |∇f| dµt M M Z Z Z 2 2 1 2 = −2 |∇∇f| dµt − R|∇f| dµt + A |∇f| dµt M M 3 M Z ∂ −2 h∇ ∆ − f, ∇fidµt , M ∂t and integration by parts we establish the following interesting equality Z Z Z d 2 2 2 |∇f| dµt = −2 |∇∇f| dµt − R|∇f| dµt dt M M M Z Z 1 2 ∂ + A |∇f| dµt + 2 (∆f) ∆ − fdµt . (7.6) 3 M M ∂t Theorem 7.1 Under the Ricci flow (3.2) Z Z Z d 2 2 1 2 |∇f| dµt ≤ − R|∇f| dµt + A |∇f| dµt dt M M 3 M 3 Z ∂ 2 + ∆ − f dµt . (7.7) 2 M ∂t 2 1 2 Proof. On 3-manifolds |∇∇f| ≥ 3 (∆f) so that by (7.6) Z Z Z d 2 2 1 2 |∇f| dµt ≤ − R|∇f| dµt + A |∇f| dµt dt M M 3 M Z Z 2 2 ∂ − (∆f) dµt + 2 (∆f) ∆ − fdµt 3 M M ∂t Z Z 2 1 2 ≤ − R|∇f| dµt + A |∇f| dµt M 3 M 3 Z ∂ 2 + ∆ − f dµt . 2 M ∂t

28 Since under the Ricci ﬂow (3.2)

∂ 2 ∆ − R = −2S + AR ∂t 3

we therefore have the following

Corollary 7.2 Under the Ricci ﬂow (3.2) for the scalar curvature R we have the following energy estimate Z Z Z d 2 2 1 2 |∇R| dµt ≤ − R|∇R| dµt + A |∇R| dµt dt M M 3 M Z Z 2 2 2 2 +6 S dµt + A R dµt M 3 M Z −4A RSdµt . (7.8) M

∂ In particular if ∂t gij = −2Rij with initial metric g(0)ij such that R(0) is a positive constant, then Z Z t Z 2 −R(0)(t−s) 2 |∇R| dµt ≤ 6 e S dµs . (7.9) M 0 M

7.1 Some applications Another application of equation (7.6) is to obtain information on spectral gaps. Let us prove the following

∂ Theorem 7.3 Let (g(t)ij) be the solution to the Ricci ﬂow ∂t gij = −2Rij on the 3- manifold M. 1) If λ(t) denotes the ﬁrst non-negative eigenvalue of (M, g(t)ij), then Z Z Z d 2 2 2 2 λ = 2λ + λ Rf dµt − 2 |∇∇f| dµt − R|∇f| dµt (7.10) dt M M M

R 2 where f is an eigenvector: ∆f = −λf such that M f dµt = 1. 1 2) If λ(t) denotes the ﬁrst non-negative eigenvalue of ∆ + 4 R, then d 19 λ ≤ λ2. (7.11) dt 8

29 Proof. Let us make computations under the Ricci ﬂow (3.2). Indeed, consider an eigenvector f with eigenvalue λ (both depending smoothly on t): Z 2 (∆ + V ) f = −λf ; f dµt = 1 M where V is some potential (depending on t as well) to be chosen later. Then, by equation (7.6) Z Z 2 Z Z d 2 ∂f 2 2 |∇f| dµt = (V + λ) dµt − 2 |∇∇f| dµt − R|∇f| dµt dt M M ∂t M M Z Z 2 2 1 2 +2 (V + λ) f dµt + A |∇f| dµt . (7.12) M 3 M On the other hand Z Z d 2 d |∇f| dµt = (−∆f)fdµt dt M dt M Z d 2 = (V + λ) f dµt dt M Z Z 2 2 d ∂f = f (V + λ) dµt + (V + λ) dµt M dt M ∂t Z 2 ∂ + (V + λ) f log Mt dµt M ∂t Z Z 2 2 d ∂f = f (V + λ) dµt + (V + λ) dµt M dt M ∂t Z 2 − (R − A)(V + λ) f dµt , M R 2 combining with equation (7.12), the facts that M f dµt = 1 and Z Z 2 |∇f| dµt = − (∆f) fdµt M M Z 2 = (λ + V )f dµt , M we deduce the following Z Z Z d 2 2 2 λ = (R − A)(V + λ) f dµt − 2 |∇∇f| dµt − R|∇f| dµt dt M M M Z Z Z 2 2 1 2 2 ∂V +2 (V + λ) f dµt + A |∇f| dµt − f dµt M 3 M M ∂t Z Z Z 2 2 2 2 = −2 |∇∇f| dµt − R|∇f| dµt + 2 (V + λ) f dµt M M M Z Z 2 ∂V 2 2 − f dµt + R − A (V + λ) f dµt . M ∂t M 3

30 We then choose V = −ϕ(t)R(t, ·) where ϕ depends on t only. Then ∂V ∂ = −ϕ0R − ϕ R ∂t ∂t so that Z Z Z 2 ∂V 0 2 2 ∂ f dµt = − ϕ Rf dµt − ϕ f Rdµt . M ∂t M M ∂t R 2 ∂ To treat the term M f ∂t Rdµt we use integration by parts again, thus Z Z Z 2 1 2 R|∇f| dµt = h∇(Rf), ∇fidµt − h∇R, ∇f idµt M M 2 M Z Z 2 1 2 = R(V + λ)f dµt + f (∆R)dµt M 2 M Z Z Z 2 1 2 ∂ 1 ∂R 2 = R(V + λ)f dµt + f ∆ − Rdµt + f dµt M 2 M ∂t 2 M ∂t Z Z 2 1 2 ∂R = R(V + λ)f dµt + f dµt M 2 M ∂t Z 2 1 + f −S + AR dµt . M 3 In other words Z Z Z 2 ∂R 2 1 2 f dµt = 2 f S − AR dµt + 2 R|∇f| dµt M ∂t M 3 M Z 2 −2 R(V + λ)f dµt , M therefore Z Z Z 2 ∂V 0 2 2 1 − f dµt = ϕ Rf dµt + 2ϕ f S − AR dµt M ∂t M M 3 Z Z 2 2 −2ϕ R|∇f| dµt + 2ϕ R(ϕR + λ)f dµt . M M Z Z d 2 2 λ = −2 |∇∇f| dµt − (1 + 2ϕ) R|∇f| dµt dt M M Z 2 0 2 + 2ϕS − ϕAR + ϕ R f dµt M 3 Z 2 2 +ϕ R R + 4ϕR − A + 2λ f dµt M 3 Z 2 2 +λ R + 4ϕR − A + 2λ f dµt . M 3

31 ∂ In particular, under the Ricci ﬂow ∂t gij = −2Rij, then A = 0, and if choose ϕ to be a constant as well, then Z Z d 2 2 λ = −2 |∇∇f| dµt − (1 + 2ϕ) R|∇f| dµt dt M M Z Z 2 2 +2ϕ Sf dµt + ϕ R (R + 4ϕR + 2λ) f dµt M M Z 2 +λ (R + 4ϕR + 2λ) f dµt . (7.13) M For example, if we choose ϕ = 0, then Z Z Z d 2 2 2 2 λ = 2λ + λ Rf dµt − 2 |∇∇f| dµt − R|∇f| dµt . (7.14) dt M M M However if we choose ϕ = −1/4 then Z Z d 2 1 2 λ = −2 |∇∇f| dµt − R|∇f| dµt dt M 2 M Z Z 2 1 2 1 2 +2λ − Sf dµt − λ Rf dµt. 2 M 2 M 1 2 While S ≥ 3 R so that Z Z d 2 1 2 λ = −2 |∇∇f| dµt − R|∇f| dµt dt M 2 M Z 2 1 2 2 +2λ − R − 3λR f dµt 6 M 19 ≤ λ2. 8 Let us come back to equation (7.13). By the Bochner identity

2 Γ2(f) = |∇∇f| + Ric(∇f, ∇f) so that Z Z Z 2 Γ2(f)dµt = |∇∇f| dµt + Ric(∇f, ∇f)dµt, M M M while by integration by parts Z Z Γ2(f)dµt = − h∇∆f, ∇fidµt M M Z 2 = (∆f) dµt M Z = (ϕR + λ)2f 2 M

32 hence Z Z Z 2 2 2 − |∇∇f| dµt = − (ϕR + λ) f + Ric(∇f, ∇f)dµt . M M M Inserting this fact into (7.13) we obtain, after simpliﬁcation, Z Z d 2 2 λ = (1 + 2ϕ) λ Rf dµt − (1 + 2ϕ) R|∇f| dµt dt M M Z Z 2 2 2 2 + 2ϕ + ϕ R f dµt + 2ϕ Sf dµt M M Z +2 Ric(∇f, ∇f)dµt . (7.15) M While under the Ricci ﬂow Ric(∇f, ∇f) ≥ −εR|∇f|2 where ε ∈ [0, 1/3] so that Z Z d 2 2 λ ≥ (1 + 2ϕ) λ Rf dµt − (1 + 2ϕ + 2ε) R|∇f| dµt dt M M Z Z 2 2 2 2 + 2ϕ + ϕ R f dµt + 2ϕ Sf dµt . M M In particular if ϕ = 0 then Z Z Z d 2 2 λ = λ Rf dµt − R|∇f| dµt + 2 Ric(∇f, ∇f)dµt . dt M M M

7.2 L2-estimates for the Ricci tensor Our next goal is to improve these L2-estimates to weighted forms. By (7.5) and the evolution equations (3.5, 3.7, 3.9) we may easily establish the following Lemma 7.4 Under the Ricci ﬂow (3.2) 1) For the scalar curvature ∂ ∆ − |∇R|2 = 2|∇∇R|2 − 4h∇S, ∇Ri + 2A|∇R|2 . (7.16) ∂t 2) ∂ 10 ∆ − |∇S|2 = 2|∇∇S|2 + A|∇S|2 − 20R|∇S|2 ∂t 3 +4 3R2 − 5S h∇R, ∇Si 2 +16h∇T, ∇Si + 4h∇|∇kRij| , ∇Si . (7.17)

33 3) ∂ 1 ∆ − |∇T |2 = 2|∇∇T |2 + 14 A + R |∇T |2 − 18R2h∇S, ∇T i ∂t 3 + 14T − 36RS + 16R3 h∇R, ∇T i ab k ij +12h∇ g Ria(∇ R )(∇kRbj) , ∇T i . (7.18) By chain rule, (7.16) and the evolution equation for the scalar curvature we deduce easily the following ∂ S 2γ ∆ − Rγ|∇R|2 = −2γ + + 2 A |∇R|2Rγ ∂t R 3 +γ (γ − 1) Rγ−2|∇R|4 + 2|∇∇R|2Rγ −4h∇S, ∇RiRγ + 2h∇Rγ, ∇|∇R|2i .

Then we apply (7.3) to function Rγ|∇R|2, after simpliﬁcation we obtain Z Z d 2 γ S 2γ 2 γ |∇R| R dµt = 2γ − + 1 A − R |∇R| R dµt dt M M R 3 Z Z γ−2 4 2 γ −γ (γ − 1) R |∇R| dµt − 2 |∇∇R| R dµt M M Z Z γ γ 2 +4 h∇S, ∇RiR dµt − 2 h∇R , ∇|∇R| idµt .(7.19) M M Using integration by parts in the last two integrals we then deduce Z Z Z γ S 2 4 h∇S, ∇RiR dµt = −4 (S∆R) dνt − 4γ |∇R| dνt M M M R and Z Z γ 2 −1 2 −2 h∇R , ∇|∇R| idµt = 2γ R (∆R) |∇R| dνt M M Z −2 4 +2γ(γ − 1) R |∇R| dνt M γ so that, with dνt = R dµt, Z Z d 2 S 2γ 2 |∇R| dνt = − 2γ + + 1 A + R |∇R| dνt dt M M R 3 Z Z −2 4 2 +γ(γ − 1) R |∇R| dνt − 2 |∇∇R| dνt M M Z Z −1 2 +2γ R |∇R| (∆R) dνt − 4 S (∆R) dνt . M M

34 2 1 2 It follows that, since |∇∇R| ≥ 3 (∆R) , Z Z d 2 S 2γ 2 |∇R| dνt ≤ − 2γ + + 1 A + R |∇R| dνt dt M M R 3 Z −2 4 +γ(γ − 1) R |∇R| dνt M Z 2 2 3γ 2 − (∆R) − 2 |∇R| − 3S (∆R) dνt 3 M 2R Z S 2γ 2 ≤ − 8γ + + 1 A + R |∇R| dνt M R 3 Z 4 Z 5 |∇R| 2 −γ 1 − γ 2 dνt + 6 S dνt 2 M R M

R 2 which allows us to establish an estimate for M |∇R| dνt, and here again γ = 2/5 seems to be the best choice.

Theorem 7.5 Under the Ricci ﬂow (3.2) with initial metric of positive scalar curvature, and let γ ∈ [0, 2/5], then Z Z d 2 γ S 2 2 γ |∇R| R dµt ≤ − 8γ + γ + 1 A + R |∇R| R dµt dt M M R 3 Z 2 γ +6 S R dνt . M In particular if A = 0 and R(0) is a positive constant, then

Z Z t Z 2 γ −( 8γ +1)R(0)(t−s) 2 γ |∇R| R dµt ≤ 6 e 3 S R dµs . (7.20) M 0 M 8 Harnack estimate for Ricci ﬂow

We in this section prove Harnack estimates for the Ricci curvature under the Ricci ﬂow.

8.1 Some formulae about Ricci tensors

On a 3-manifold with a family of metrics (g(t)ij) evolving with the equation ∂g ij = −2h ∂t ij

35 and suppose that (Vij) is a symmetric tensor depending smoothly in t, then it is clear that ∂ ∂V tr (V ) = 2hijV + tr ( ab ) . (8.1) ∂t g ij ij g ∂t Since in a local coordinate system

p p ∇bVij = ∂bVij − VpjΓbi − VipΓbj , ∂Γp bi = −gpc {(∇ h ) + (∇ h ) − (∇ h )} ∂t i bc b ic c bi and ∂ ∂V ∂Γp ∂Γp (∇ V ) = ∇ ij − V bi − V bj ∂t b ij b ∂t pj ∂t ip ∂t we have the following

∂ Lemma 8.1 If ∂t gij = −2hij then ∂ ∂V (∇ V ) = ∇ ij + gpqV (∇ h + ∇ h − ∇ h ) ∂t b ij b ∂t qj i pb b pi p ib pq +Viqg (∇jhpb + ∇bhpj − ∇phjb) . (8.2)

∂ Lemma 8.2 If ∂t gij = −2hij then ∂ ∂V ∇ ∇ V = ∇ ∇ ij + V gpq (∇ ∇ h + ∇ ∇ h − ∇ ∇ h ) ∂t a b ij a b ∂t pj a i qb a b qi a q ib pq +Vipg (∇a∇bhqj + ∇a∇jhqb − ∇a∇qhbj) pq +g (∇aVpj)(∇ihqb + ∇bhqi − ∇qhib) pq +g (∇aVip)(∇bhqj + ∇jhqb − ∇qhbj) pq +g (∇pVij)(∇bhqa + ∇ahqb − ∇qhba) pq +g (∇bVpj)(∇ihqa + ∇ahqi − ∇qhia) pq +g (∇bVip)(∇ahqj + ∇jhqa − ∇qhaj) so that ∂ ∂V (∆V ) = ∆ ij + 2h ∇a∇bV − V ∇a∇bh − V ∇a∇bh ∂t ij ∂t ab ij bj ia ib aj pq a a +Vipg (∇ ∇ahqj + ∇ ∇jhqa) pq a a +Vpjg (∇ ∇ihqa + ∇ ∇ahqi) b a + ∇ Vij (2∇ hab − ∇btrg(hkl)) pq a a b +2g (∇ Vpj)(∇ihqa + ∇ahqi) − 2 (∇ Vbj) ∇ hia pq a a b +2g (∇ Vip)(∇ahqj + ∇jhqa) − 2 (∇ Vib) ∇ haj .

36 8.2 Harnack inequality for Ricci curvature In what follows we are working with the Ricci flow ∂ g = −2R (8.3) ∂t ij ij with initial metric (g(0)ij) such that the scalar curvature R(0) is a positive constant. Then R ≥ R(0) and 1 S S(0) ≤ ≤ max 3 R2 M R(0)2 as long as the solution flow to (8.3) exists. 2 Next we compute the evolution equation for |∇kRij| . We consider the Ricci flow (g(t)ij): the maximum solution to the Ricci flow ∂ g = −2R ∂t ij ij 2 In this section we deduce the evolution equation for |∇kRij| . Let us make computation in a normal coordinate system at the point we evaluate the geometric quantities. ∂ ∂ ∆ − |∇ R |2 = 2 ∇kRij ∆ − (∇ R ) ∂t k ij ∂t k ij ka ij jq ip k −2R (∇kRij) ∇aR − 2g R (∇kRij) ∇ Rpq ip jq k k ij −2g R (∇kRij) ∇ Rpq + 2h∇ (∇kRij) , ∇ ∇ R i . By the Ricci identity for symmetric tensor we have

Lemma 8.3 If (Tij) is a symmetric tensor, then on 3-manifolds we have

∇c (∆Tij) − ∆ (∇cTij) a a ab ab = Taj (∇ Rci) + Tia (∇ Rcj) − g Tia (∇jRcb) − g Taj (∇iRcb) a a a + (∇ Tai) (2Rcj − Rgcj) + (∇ Taj) (2Rci − Rgci) − (∇ Tij) Rca ab −2g Rcb {(∇iTaj) + (∇jTai)} + (∇jTci + ∇iTcj) R pq a a +2g (∇ Tpi) Raqgcj − 2 (∇ Tci) Raj pq a a +2g (∇ Tpj) Raqgci − 2 (∇ Tcj) Rai . (8.4) Proof. In fact a direct computation (though a little bit complicated) shows that

∇c∇a∇bTij − ∇a∇b∇cTij p p p = (∇pTij) Racb + (∇aTip) Rbcj + (∇aTpj) Rbci p p p p + (∇bTip) Racj + (∇bTpj) Raci + Tip ∇aRbcj + Tpj (∇aRbci) (8.5)

37 and

∇b∇a∇cTij − ∇c∇a∇bTij p p p = Tpj∇aRcbi + Tip∇aRcbj + (∇pTij) Rcba p p p + (∇aTpj) Rcbi + (∇aTip) Rcbj + (∇bTpj) Rcai p p p + (∇bTip) Rcaj + (∇cTpj) Rabi + (∇cTip) Rabj . Taking trace over indices a and b we thus obtain

∇c (∆Tij) − ∆ (∇cTij) a a p a p = − (∇ Tij) Rca + 2 (∇ Tip) Racj + 2 (∇ Tpj) Raci a a ab ab +Tia (∇ Rcj) + Taj(∇ Rci) − g Tia (∇jRcb) − g Taj (∇iRcb) . (8.6) Now on a 3-manifold we may replace the full curvature tensor by the Ricci curvature via equation (1.4), which then yields (8.4). Applying (8.4) to the Ricci symmetric tensor Rij together with the Bianchi identity a 1 ∇ Raj = 2 ∇jR we obtain r r ∆ (∇kRij) = ∇k (∆Rij) − Rrj (∇ Rki) − Rir (∇ Rkj) rb rb r +g Rir (∇jRkb) + g Rrj (∇iRkb) + (∇ Rij) Rkr 1 1 − (∇ R) R − Rg − (∇ R) R − Rg i kj 2 kj j ki 2 ki rb +2g Rkb {(∇iRrj) + (∇jRri)} − (∇jRki + ∇iRkj) R pq r r −2g (∇ Rpi) Rrqgkj + 2 (∇ Rki) Rrj pq r r −2g (∇ Rpj) Rrqgki + 2 (∇ Rkj) Rri .

On the other hand, applying (8.2) to Rij we have ∂ ∂R (∇ R ) = ∇ ij + grbR {(∇ R ) + (∇ R )} − grbR (∇ R ) ∂t k ij k ∂t rj i bk k bi rj b ik rb rb +g Rir {(∇jRbk) + (∇kRbj)} − g Rir(∇bRjk). Putting the previous two equations, after simpliﬁcation, gives us the following ∂ ∂ ∆ − (∇ R ) = ∇ ∆ − R − R (∇rR ) − R (∇rR ) ∂t k ij k ∂t ij rj ki ir kj 1 1 − (∇ R) R − Rg − (∇ R) R − Rg i kj 2 kj j ki 2 ki rb +2g Rkb {(∇iRrj) + (∇jRri)} − (∇jRki + ∇iRkj) R rs rs r −2gkjR (∇rRsi) − 2gkiR (∇rRsj) + Rkr (∇ Rij) rb +g Rrj {3(∇bRik) − (∇kRbi)} rb +g Rir {3(∇bRjk) − (∇kRbj)} .

38 We are now in a position to show the following ∂ Proposition 8.4 Under the Ricci ﬂow ∂t gij = −2Rij it holds that ∂ ∆ − (∇ R ) = gabR {5(∇ R ) + 2 (∇ R )} ∂t k ij aj k bi b ki ab +g Ria {5(∇kRbj) + 2 (∇bRkj)} ab +g Rka {2 (∇iRbj) + 2 (∇jRbi) + (∇bRij)}

−3Rij(∇kR) + 2Rgij(∇kR) − 2gij(∇kS)

−R {(∇jRki) + (∇iRkj) + 3(∇kRij)} 1 1 − (∇ R) R − Rg − (∇ R) R − Rg i kj 2 kj j ki 2 ki ab ab −2gkjR (∇aRbi) − 2gkiR (∇aRbj) . (8.7) Proof. Diﬀerentiating the evolution equation (3.4) for the Ricci curvature gives that ∂ ∇ ∆ − R = 6gabR (∇ R ) + 6gabR (∇ R ) k ∂t ij aj k ib ib k aj

−3Rij(∇kR) − 3R(∇kRij)

+2Rgij(∇kR) − 2gij(∇kS), ∂ together with the previous equation for ∆ − ∂t (∇kRij) yield the result. Taking trace in equation (8.7) we establish the following Corollary 8.5 Under the Ricci ﬂow (8.3) ∂ ∆ − (∇ R) = −2(∇ S) + gijR (∇ R) . (8.8) ∂t k k kj i The identity (8.8) yields the following ∂ ∆ − |∇R|2 = 2|∇∇R|2 − 4h∇R, ∇Si ∂t obtained via the Bochner identity. 2 We now proceed to deduce the evolution for |∇kRij| . By (8.7) ∂ 2(∇kRij) ∆ − (∇ R ) ∂t k ij 2 k ij = −6R|∇kRij| − 4R(∇ R )(∇iRkj) k 2 k ij −7(∇ R)(∇kS) + 5R|∇R| − 8Rkj(∇ R )(∇iR) ab k ij +g Raj(∇ R ) {20(∇kRbi) + 8 (∇bRki)} ab k ij +g Rka(∇ R ) {8 (∇iRbj) + 2 (∇bRij)} . (8.9) Therefore

39 ∂ Lemma 8.6 Under the Ricci ﬂow ∂t gij = −2Rij we have ∂ ∆ − |∇ R |2 ∂t k ij 2 k ij = −6R|∇kRij| − 4R(∇ R )(∇iRkj) k 2 k ij −7(∇ R)(∇kS) + 5R|∇R| − 8Rkj(∇ R )(∇iR) ab k ij +16g Raj(∇ R ) {(∇kRbi) + (∇bRki)} k ij +2h∇ (∇kRij) , ∇ ∇ R i . (8.10)

2 We next deduce the equation for |∇kRij| /R. By chain rule

∂ |∇ R |2 2S 7 ∆ − k ij = − 6 |∇ R |2 + 5|∇R|2 − h∇S, ∇Ri ∂t R R2 k ij R 8 −4(∇kRij)(∇ R ) − R (∇kRij)(∇ R) i kj R kj i 16 + gabR (∇kRij) {(∇ R ) + (∇ R )} R ai b jk k jb 2 |∇ R |2 + h∇ (∇ R ) , ∇ ∇kRiji − 2h∇ log R, ∇ k ij i .(8.11) R k ij R This identity allows us to deduce a gradient estimate for the Ricci curvature. To 2 simplify the notations set F = |∇kRij| /R, and use a normal coordinate which diago- nalizes the Ricci curvature (Rij), and set Xkij = ∇kRij, then equation (8.11) may be written as the following

∂ 2S 7 ∆ − F = − 6 RF + 5|∇R|2 − h∇S, ∇Ri ∂t R2 R 16 16 8 + λ X2 + λ X X − λ X X − 4X X R i kij R i kij ikj R i iji jaa kij ikj 2 + h∇ (∇ R ) , ∇ ∇kRiji − 2h∇ log R, ∇F i . (8.12) R k ij We then use the following identity, for any constant ξ 6= 0 1 1 −h∇S, ∇Ri = |∇ R − ξR ∇ R|2 − |∇ R |2 − ξS|∇R|2 (8.13) ξ k ij ij k ξ k ij

40 so that, for ξ > 0 we have 7 1 7 1 7 − h∇S, ∇Ri = |∇ R − ξR ∇ R|2 − |∇ R |2 R ξ R k ij ij k ξ R k ij S −7ξ |∇R|2 R 1 7 S ≥ − |∇ R |2 − 7ξ |∇R|2 ξ R k ij R 1 7 ≥ − |∇ R |2 − 7ξδR|∇R|2 . ξ R k ij

5 By choosing 7ξδR = 5, i.e. ξ = 7δR we deduce that ∂ 2S 49 16 16 ∆ − F ≥ − 6 − δ RF + λ X2 + λ X X ∂t R2 5 R i kij R i kij ikj 8 − λ X X − 4X X − 2h∇ log R, ∇F i . R i iji jaa kij ikj Since v u !2 √ uX X |λiXkijXikj| ≤ St XkijXikj i j,k v √ u s s uX X 2 X 2 ≤ St Xkij Xikj i j,k j,k √ 2 ≤ 3S|∇kRij|

41 Therefore, as λi ≥ −εR, ∂ 2S 49 16√ ∆ − F ≥ − 6 − δ RF − 16εRF − 3S|∇ R |2 ∂t R2 5 R k ij 8 √ − S|∇ R |2 − 4|∇ R |2 − 2h∇ log R, ∇F i R k ij k ij 2 √ √ 49 ≥ − 10 − 16 3δ − 8 δ − δ − 16ε RF 3 5 −2h∇ log R, ∇F i .

Theorem 8.7 Under the Ricci ﬂow (8.3) on the 3-manifold with initial metric with positive constant scalar curvature R(0), such that R(0)ij ≥ −εR(0)g(0)ij for some ε ∈ [0, 1/3]. Then ∂ L − F ≥ −C(ε)RF ∂t 2 where L = ∆ + 2∇ log R, F = |∇kRij| /R and 28 √ √ 49 C(ε) = + 16 3δ + 8 δ + δ + 16ε . 3 5 In particular, suppose R(t, ·) ≤ θ(t) then ∂ L − F ≥ −C(ε)θ(t)F ∂t so that 2 2 |∇kRij| C(ε) R t θ(s)ds |∇kR(0)ij| ≤ e 0 . (8.14) R R(0)

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