#  Eigenvalues under the backward Ricci flow on locally homogeneous closed
3-manifolds

Songbo Hou  Department of Applied Mathematics, College of Science, China
Agricultural University, Beijing, 100083, P.R. China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the evolving behaviors of the first eigenvalue of
Laplace-Beltrami operator under the normalized backward Ricci flow, construct
various quantities which are monotonic under the backward Ricci flow and get
upper and lower bounds. We prove that in cases where the backward Ricci flow
converges to a sub-Riemannian geometry after a proper rescaling, the
eigenvalue evolves toward zero.

###### Key words and phrases:

Homogeneous 3-manifold; Backward Ricci flow; Eigenvalue, Estimate

###### 2010 Mathematics Subject Classification:

53C44.

##  1\. Introduction

The Ricci flow on a closed manifold is a flow of Riemannian metric $g(t)$
evolved by the equation

$\frac{\partial g}{\partial t}=-2\textup{Rc},\,\,g(0)=g_{0},$

where Rc is the Ricci curvature tensor. The customary normalization is setting
$\widetilde{g}(\widetilde{t})=\psi(t)g(t)$,
$\widetilde{t}=\int_{0}^{t}\psi(s)ds$ with
$\frac{1}{\psi}\frac{\partial\psi}{\partial t}=\frac{2r}{3}$, where $r$ is the
average of the scalar curvature $R$. Setting
$\widetilde{\psi}(\widetilde{t})=\psi(t)$, then we have
$\frac{1}{\widetilde{\psi}}\frac{\partial\widetilde{\psi}}{\partial\widetilde{t}}=\frac{2\widetilde{r}}{3}$
and

$\frac{\partial\widetilde{g}}{\partial\widetilde{t}}=-2\widetilde{\textup{Rc}}+\left(\frac{\partial\ln\widetilde{\psi}}{\partial\widetilde{t}}\right)\widetilde{g}=-2\widetilde{\textup{Rc}}+\frac{2\widetilde{r}}{3}\widetilde{g},\,\,\widetilde{g}(0)=g_{0}.$

We often write the following flow

(1.1) $\frac{\partial g}{\partial
t}=-2\textup{Rc}+\frac{2r}{3}g,\,\,g(0)=g_{0},$

which is called the normalized Ricci flow keeping the volume constant. In
dimension 3, Hamilton [7] proves that the solution to the Ricci flow converges
to the constant curvature metric on a 3-sphere if the Ricci curvature of
initial metric is positive.

The eigenvalues of geometric operators under the Ricci flow are important to
understand geometry and topology of manifolds. In [18], Perelman proves the
the first eigenvalue of the Laplace-Beltrami operator with potential R,
i.e.,$-\Delta+4R$ is nondecreasing under the Ricci flow, where $R$ denotes the
scalar curvature of metric $g$. He also applies this to show that there are no
nontrivial steady or expanding breathers on closed manifolds. Later, Cao [1]
showes that the eigenvalues of $-\Delta+\frac{R}{2}$ are nondecreasing under
the Ricci flow on manifolds with nonnegative curvature operator. Using the
same techniques, Li [15] extends Cao’s result to manifolds without nonnegative
curvature operator. Similar results hold for the first eigenvalue of
$-\Delta+aR(a\geq\frac{1}{4})$ along the Ricci flow [2, 15]. In fact, the
eigenvalues $\lambda$ are no longer differentiable in time $t$. If we denote
$u$ the eigenfunction of the eigenvalue then $\lambda(u,t)=\lambda(t)$. By the
eigenvalue perturbation theory, there is a $C^{1}$-family of smooth
eigenvalues and eigenfunctions [11]. One can assume that the first eigenvalue
$\lambda(t)$ and the corresponding eigenfunction $u(x,t)$ are smooth along the
Ricci flow.

Cao, the author and Ling [4] derive a monotonicity formula for the first
eigenvalue of $-\Delta+aR(0<a\leq\frac{1}{2})$ on a closed surface under the
Ricci flow and obtain various monotonicity formulae and estimates along the
normalized Ricci flow. Our results indicate that although it is difficult to
get better estimates for the eigenvalue under the Ricci flow, one can get
interesting results if the problem can be dealt with by ODE techniques.

For the Laplace-Beltrami operator, Ma [16] prove the first eigenvalue on
domains with Dirichlet boundary condition is nondecreasing along the Ricci
flow. Ling [14] gets a sharp bound of the first eigenvalue under the
normalized Ricci flow, and proves that an appropriate multiple is monotonic.
The author [8] considers the eigenvalue of Laplace-Beltrami operator under the
normalized Ricci flow on locally homogeneous closed 3-manifolds, constructs
various monotonic quantities and gets estimates for upper and lower bounds.

For p-Laplace operator, Wu, Wang and Zheng [20] proves the first p-eigenvalue
is strictly increasing along the Ricci flow under some curvature assumption,
and construct various monotonic quantities on closed Riemannian surface. Wu
[19] proves the first eigenvalue monotoncity of the p-Laplace operator along
the Ricci flow on closed Riemannian manifolds under some different curvature
assumptions.

A Riemannian manifold $(M,g)$ is called to be locally homogeneous if for every
two points $x,y\in M$, there are neighborhoods $U$ of $x$ and $V$ of $y$, and
an isometry $\phi$ from $(U,g|_{U})$ to $(V,g|_{V})$ with $\phi(x)=y$.
Furthermore, $(M,g)$ is called to be homogeneous if the isometry group is
transitive i.e., $U=V=M$ for all $x$ and $y$. A result of Singer [17] says
that the universal cover of a locally homogeneous manifold is homogeneous. We
may study the Ricci flow of homogeneous models instead of locally homogeneous
manifolds since Ricci flow commutes with the cover map.

The locally homogeneous 3-manifolds contains nine classes which can be divided
into two sets. The first set consists of classes $H(3)$,
$H(2)\times\mathbb{R}$ and $\text{SO(3)}\times\mathbb{R}$, where $H(n)$
denotes the group of isometries of hyperbolic n-space. The second set includes
$\mathbb{R}$, SU(2), $\text{SL}(2,\mathbb{R})$, Heisenberg, $E(1,1)$ (the
group of isometries of the plane with flat Lorentz metric) and $E(2)$ (the
group of isometries of the Euclidian plane), and these are called Bianchi
classes in [9].

In Bianich classes, given a initial metric $g_{0}$, there is a Milnor frame
$(f_{1},f_{2},f_{3})$ such that the metric and Ricci tensor are diagonalized.
Since this property is preserved by the Ricci flow, we often write

$g=Af^{1}\otimes f^{1}+Bf^{2}\otimes f^{2}+Cf^{3}\otimes f^{3},$

where $(f^{1},f^{2},f^{3})$ is a dual frame of Milnor frame. Then the Ricci
flow reduces to an ODE system involving A, B and C. Since the homogeneous
3-manifolds are models of geometrization conjecture, Isenberg and Jackson [9]
study the Ricci flow on such manifolds and describe their characteristic
behaviors by analyzing the corresponding system. There are three types of
behaviors depending on the geometry type.

Later Knopf and McLeod [12] study quasi-convergence equivalence of model
geometries under the Ricci flow.

If we assume $g(t)$, $t\in(-T_{b},T_{f})$, is the maximal solution of the
forward Ricci flow (1.1), it is interesting to consider the behaviors of
$g(t)$ as $t$ goes to $-T_{b}$. For convenience, we reverse time and consider
the following backward Ricci flow equation

(1.2) $\frac{\partial g}{\partial
t}=2\textup{Rc}-\frac{2r}{3}g,\,\,g(0)=g_{0}.$

The behaviors of backward Ricci flow are described in [5] and [3], and the
interesting phenomena is that the backward Ricci flow converges to a sub-
Riemannian geometry after a proper rescaling. We use the simple example to
explain the term of sub-Riemannian geometry. In section three, the evolving
metric is

$g=Af^{1}\otimes f^{1}+Bf^{2}\otimes f^{2}+Cf^{3}\otimes f^{3}.$

As $t$ goes to $-3/(16R_{0})$, we get

$A(t)\rightarrow+\infty,\,\,\,B(t)\rightarrow 0,\,\,\,C(t)\rightarrow 0.$

Then the rescaled metric $\bar{g}(t)=(C_{0}/C(t))g(t)$ converges to

$\infty f^{1}\otimes f^{1}+B_{0}f^{2}\otimes f^{2}+C_{0}f^{3}\otimes f^{3}.$

Look at the dual tensor defined on the co-tangent bundle

$Q=A^{-1}f_{1}\otimes f_{1}+B^{-1}f_{2}\otimes f_{2}+C^{-1}f_{3}\otimes
f_{3}.$

Then the tensor $Q$ tends to

$Q_{*}=B_{0}^{-1}f_{2}\otimes f_{2}+C_{0}^{-1}f_{3}\otimes f_{3}.$

It turns out that $[f_{2},f_{3}]=2f_{1}$. Then the tensor $Q_{*}$ induces a
natural distance function $d_{*}$ defined on $M$. We take the infimum of the
length of all curves staying tangent to the linear span of $f_{2},\,f_{3}$,
which are called horizontal curves, to computed the distance. The associated
geometry is called a sub-Riemannian geometry.

In this paper, we study the first eigenvalue of the Laplace-Beltrami operator
under the backward Ricci flow on locally homogeneous 3-manifolds in Bianchi
classes.

######  Theorem 1.1.

Let $(M,g(t)),t\in[0,T_{+})$ be a solution to the backward Ricci flow in
Bianchi classes, where $T_{+}\in[0,+\infty]$ is the maximal existence time.
Assume that $\lambda(t)$ is the first eigenvalue of $-\Delta$ with respect to
$g(t)$. Then in cases where $g(t)$ converges to a sub-Riemannian geometry
after a proper re-scaling, $\lambda(t)$ goes to zero as $t$ approaches
$T_{+}$.

We construct various monotonic quantities and get upper and lower bounds for
the eigenvalue. The behaviors of the eigenvalue are not very diverse. In many
cases, the eigenvalue is decreasing after a time and goes to zero. The next of
this paper is arranged as follows. In section two, we derive an evolution
equation of the eigenvalue which is important to estimates. From section three
to section seven, we analyze the behaviors of the eigenvalue and get estimates
case by case.

##  2\. Evolution equation of the eigenvalue

In this section, we get the following theorem.

######  Theorem 2.1.

Let $(M,g(t)),t\in[0,T_{+})$ be a solution to the backward Ricci flow on a
locally homogeneous 3-manifold. Assume that $\lambda(t)$ is the first
eigenvalue of $-\Delta$ and $u(x,t)>0$ satisfies

(2.1) $-\Delta u=\lambda u,$

with $\int u^{2}(x,t)d\mu=1.$ Then along the the backward Ricci flow, we
obtain

(2.2)
$\displaystyle\frac{d}{dt}\lambda=\frac{2}{3}R\lambda-\int(2R_{ij}\nabla_{i}u\nabla_{j}u)d\mu.$

The proof of this theorem is similar to Lemma 3.1 in [4].

###### Proof.

By a direct calculation as in [1], we have

$\displaystyle\frac{d}{dt}\lambda=\int\left(2uR_{ij}\nabla_{i}\nabla_{j}u-\frac{2r}{3}u\Delta
u\right)d\mu$

where

$r=\frac{\int_{M}Rd\mu}{\int_{M}d\mu}$

is the average of the scalar curvature, which equals $R$ on locally
homogeneous manifolds. Integrating by parts and using contracted Bianchi
identity, we have

$\displaystyle\int
2uR_{ij}\nabla_{i}\nabla_{j}ud\mu=-\int\left[(2u\nabla_{i}R_{ij})\nabla_{j}u\right]d\mu-\int(2R_{ij}\nabla_{i}u\nabla_{j}u)d\mu,$

and

$\displaystyle-\int(2\nabla_{i}R_{ij})u\nabla_{j}ud\mu$ $\displaystyle=-\int
u(\nabla_{j}R)\nabla_{j}ud\mu=\int Ru\Delta ud\mu+\int R|\nabla u|^{2}d\mu$
$\displaystyle=-\lambda R+\lambda R=0.$

We arrive at

$\displaystyle\frac{d}{dt}\lambda=\frac{2}{3}R\lambda-\int(2R_{ij}\nabla_{i}u\nabla_{j}u)d\mu.$

∎

##  3\. Heisenberg

In this class, given a metric $g_{0}$, there is fixed Milnor frame such that

$\left[f_{2},f_{3}\right]=2f_{1},\,\,\,\left[f_{3},f_{1}\right]=0,\,\,\,\left[f_{1},f_{2}\right]=0.$

Under the normalization $A_{0}B_{0}C_{0}=4$, the curvature components for
metrics are (see examples on page 171 in [6])

(3.1) $\left\\{\begin{aligned} &R_{11}=\frac{1}{2}A^{3},\\\
&R_{22}=-\frac{1}{2}A^{2}B\\\ &R_{33}=-\frac{1}{2}A^{2}C,\\\
&R=-\frac{1}{2}A^{2}.\end{aligned}\right.$

The backward Ricci flow equations are then

(3.2) $\left\\{\begin{aligned} &\frac{d}{dt}A=\frac{4}{3}A^{3},\\\
&\frac{d}{dt}B=-\frac{2}{3}A^{2}B,\\\
&\frac{d}{dt}C=-\frac{2}{3}A^{2}C,\end{aligned}\right.$

and the solution is

(3.3) $\left\\{\begin{aligned} &A=A_{0}(1+\frac{16}{3}R_{0}t)^{-1/2},\\\
&B=B_{0}(1+\frac{16}{3}R_{0}t)^{1/4},\\\
&C=C_{0}(1+\frac{16}{3}R_{0}t)^{1/4}.\end{aligned}\right.$

where $R_{0}=-\frac{1}{2}A_{0}^{2}$. The metric $\bar{g}(t)=(C_{0}/C(t))g(t)$
converges to a sub-Riemannian geometry.

######  Theorem 3.1.

Let $\lambda(t)$ be the first eigenvalue of $-\Delta$. Assume that $B_{0}\geq
C_{0}$. Then $\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is
nondecreasing along the backward Ricci flow, and
$\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.
Moreover, we have

$e^{\frac{3A_{0}}{4}[1-(1+\frac{16R_{0}t}{3})^{-1/2}]}(1+\frac{16R_{0}t}{3})^{1/8}\lambda(0)\leq\lambda(t)\leq\lambda(0)e^{\frac{3B_{0}}{2}[1-(1+\frac{16R_{0}t}{3})^{1/4}]}(1+\frac{16R_{0}t}{3})^{1/8}.$

for $t\in[0,-3/(16R_{0})).$ As $t$ goes to $-3/(16R_{0})$, $\lambda(t)$ goes
to zero.

###### Proof.

Assume that $B_{0}\geq C_{0}$. By (2.2) and (3.1), have

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{22}\lambda.$

Then $\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.

Integrating from $0$ to $t$ yields

$e^{\frac{3A_{0}}{4}[1-(1+\frac{16R_{0}t}{3})^{-1/2}]}(1+\frac{16R_{0}t}{3})^{1/8}\lambda(0)\leq\lambda(t)\leq\lambda(0)e^{\frac{3B_{0}}{2}[1-(1+\frac{16R_{0}t}{3})^{1/4}]}(1+\frac{16R_{0}t}{3})^{1/8}.$

∎

##  4\. $\text{SU}(2)$

Given a metric $g_{0}$, we choose a Milnor frame such that

$\left[f_{2},f_{3}\right]=2f_{1},\,\,\,\left[f_{3},f_{1}\right]=2f_{2},\,\,\,\left[f_{1},f_{2}\right]=2f_{3}.$

Under the normalization $A_{0}B_{0}C_{0}=4$, then the nonzero components of
Ricci tensor are (see examples on page 171 in [6])

(4.1) $\left\\{\begin{aligned} &R_{11}=\frac{1}{2}A[A^{2}-(B-C)^{2}],\\\
&R_{22}=\frac{1}{2}B[B^{2}-(A-C)^{2}],\\\
&R_{33}=\frac{1}{2}C[C^{2}-(A-B)^{2}],\end{aligned}\right.$

and the scalar curvature is

(4.2)
$R=\frac{1}{2}[A^{2}-(B-C)^{2}]+\frac{1}{2}[B^{2}-(A-C)^{2}]+\frac{1}{2}[C^{2}-(A-B)^{2}].$

The backward Ricci flow equations are

(4.3) $\left\\{\begin{aligned}
&\frac{dA}{dt}=-\frac{2}{3}A\left[-A(2A-B-C)+(B-C)^{2}\right],\\\
&\frac{dB}{dt}=-\frac{2}{3}B\left[-B(2B-A-C)+(A-C)^{2}\right],\\\
&\frac{dC}{dt}=-\frac{2}{3}C\left[-C(2C-A-B)+(A-B)^{2}\right].\end{aligned}\right.$

Assume that $A_{0}\geq B_{0}\geq C_{0}$. Cao [5] proves the following theorem

######  Theorem 4.1.

(1) If $A_{0}=B_{0}=C_{0}$, then $T_{+}=\infty$ and $g(t)=g_{0}$.

(2) If $A_{0}=B_{0}>C_{0}$, then $T_{+}=\infty$, $A=B>C$ and, as $t$ goes to
infinity,

$A\sim\frac{8}{3}t,\,C\sim\frac{9}{16}t^{-2}$.

(3) If $A_{0}>B_{0}\geq C_{0}$, then $T_{+}<\infty$, $A>B\geq C$ and there are
constants

$\eta_{1},\,\eta_{2}\in(0,\infty)$ such that

$A\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,B\sim\eta_{1}(T_{+}-t)^{1/4},\,C\sim\eta_{2}(T_{+}-t)^{1/4}.$

In case (3), $\bar{g}(t)=(B_{0}/B(t))g(t)$ converges to a sub-Riemannian
geometry.

In the following, $\tau$, $c_{1}$ and $c_{2}$ denote constants which may vary
from line to line and from section to section.

######  Theorem 4.2.

Let $\lambda(t)$ be the first eigenvalue of $-\Delta$. Then we get

(1) If $A_{0}=B_{0}=C_{0}$, then $\lambda(t)=\lambda(0)$.

(2) If $A_{0}=B_{0}>C_{0}$, there is a time $\tau$ such that
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing. We
also get the further estimate

$\lambda(\tau)e^{8(\tau-t)}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{t}{\tau}\right)^{c_{1}}.$

(3)If $A_{0}>B_{0}\geq C_{0}$, then there is a time $\tau$ such that
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing. We
get the following estimate

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

###### Proof.

(1) If $A_{0}=B_{0}=C_{0}$, then $\lambda(t)=\lambda(0)$.

(2) If $A_{0}=B_{0}>C_{0}$, then by (4.1) and (2) in Theorem 4.1 we have

$R_{11}=R_{22}>0,\,\,R_{33}>0$

after a time $\tau$. It also follows that

$\displaystyle R_{22}-R_{33}$
$\displaystyle=\frac{1}{2}[B^{3}-B(A-C)^{2}-C^{3}+C(A-B)^{2}]$
$\displaystyle=\frac{1}{2}(B-C)(B^{2}++2BC+C^{2}-A^{2})$ $\displaystyle>0$

for $t\geq\tau$. Thus we derive

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{33}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.

Moreover, by (2) in Theorem 4.1 we estimate

$\displaystyle\frac{2}{3}R-2R_{33}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2AB+2BC+2CA-A^{2}-B^{2}-C^{2})-[C^{3}-C(A-B)^{2}]$
$\displaystyle=$ $\displaystyle\frac{1}{3}(4AC-C^{2})-C^{3}$
$\displaystyle\leq$ $\displaystyle c_{1}t^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2AB+2BC+2CA-A^{2}-B^{2}-C^{2})-[A^{3}-A(B-C)^{2}]$
$\displaystyle=$ $\displaystyle\frac{1}{3}(4AC-C^{2})-A^{3}+A(B-C)^{2}$
$\displaystyle\geq$ $\displaystyle-8$

for $t\geq\tau$.

Thus we arrive at

$-8\leq\frac{1}{\lambda}\frac{d}{dt}\lambda\leq c_{1}t^{-1}.$

Integrating from $\tau$ to $t$ gives

$\lambda(\tau)e^{8(\tau-t)}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{t}{\tau}\right)^{c_{1}}.$

(3) If $A_{0}>B_{0}\geq C_{0}$, then by (3) in Theorem 4.1 we have

$R_{11}>0,\,\,R_{22}<0,\,\,R_{33}<0,$

after a time $\tau$. It is easy to see that

$\displaystyle R_{22}-R_{33}$
$\displaystyle=\frac{1}{2}[B^{3}-B(A-C)^{2}-C^{3}+C(A-B)^{2}]$
$\displaystyle=\frac{1}{2}(B-C)(B^{2}+2BC+C^{2}-A^{2})$ $\displaystyle\leq 0$

if $t\geq\tau$.

Thus we have

$R_{11}>R_{33}\geq R_{22}$

and

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{22}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.

Furthermore, we conclude from the behavior of metric that

$\displaystyle\frac{2}{3}R-2R_{22}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2AB+2BC+2CA-A^{2}-B^{2}-C^{2})-[B^{3}-B(A-C)^{2}]$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2AB+2BC+2CA-A^{2}-B^{2}-C^{2})-[A^{3}-A(B-C)^{2}]$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-3/2}$

for $t\geq\tau$.

Thus we obtain

$-c_{2}(T_{+}-t)^{-3/2}\leq\frac{1}{\lambda}\frac{d}{dt}\lambda\leq-
c_{1}(T_{+}-t)^{-1}.$

Integrating from $\tau$ to $t$ gives

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$. ∎

##  5\. E(1,1)

Given a metric $g_{0}$, we choose a fixed Milnor frame such that

$\left[f_{2},f_{3}\right]=2f_{1},\,\,\,\left[f_{3},f_{1}\right]=0\,\,\,\left[f_{1},f_{2}\right]=-2f_{3}.$

Under the normalization $A_{0}B_{0}C_{0}=4$, the nonzero curvature components
of the metric are

(5.1) $\left\\{\begin{aligned} &R_{11}=\frac{1}{2}A(A^{2}-C^{2}),\\\
&R_{22}=-\frac{1}{2}B(A+C)^{2},\\\ &R_{33}=\frac{1}{2}C(C^{2}-A^{2}),\\\
&R=-\frac{1}{2}(A+C)^{2}.\end{aligned}\right.$

The backward Ricci flow equations are

(5.2) $\left\\{\begin{aligned}
&\frac{dA}{dt}=\frac{2}{3}A(2A^{2}+AC-C^{2}),\\\
&\frac{dB}{dt}=-\frac{2}{3}B(A+C)^{2},\\\
&\frac{dC}{dt}=\frac{2}{3}C(2C^{2}+AC-A^{2}).\end{aligned}\right.$

Assume that $A_{0}\geq C_{0}$. Cao [5] proved the following theorem.

######  Theorem 5.1.

(1) If $A_{0}=C_{0}$, then $T_{+}=\frac{3}{32}B_{0}$ and

$A(t)=C(t)=\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\;\;B(t)=\frac{32}{3}(T_{+}-t),\;\;t\in[0,T_{+}).$

(2) If $A_{0}>C_{0}$, then $T_{+}<\infty$, and there exist constants
$\eta_{1},\eta_{2}\in(0,\infty)$ such that

$A\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,B(t)\sim\eta_{1}(T_{+}-t)^{1/4},\,C(t)\sim\eta_{2}(T_{+}-t)^{1/4},$

as $t$ goes to $T_{+}$.

In case (2), $\bar{g}(t)=(B_{0}/B(t))g(t)$ converges to a sub-Riemannian
geometry.

######  Theorem 5.2.

Let $\lambda(t)$ be the first eigenvalue of $-\Delta$. Then we get

(1) If $A_{0}=C_{0}$, then $\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R)dt}$ is
nondecreasing along the backward Ricci flow, and
$\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.
Moreover, we get

$\lambda(0)\left(\frac{T_{+}-t}{T_{+}}\right)^{1/2}\leq\lambda(t)\leq\lambda(0)\left(\frac{T_{+}-t}{T_{+}}\right)^{1/2}e^{16t}.$

(2) If $A_{0}>C_{0}\geq B_{0}$, then there is time $\tau$ such that
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.
Moreover, we have

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

Remark. In case (2) of the above theorem, if $B_{0}>C_{0}$, we can get the
similar estimate.

###### Proof.

(1) If $A_{0}=C_{0}$, then by (5.1) we have

$R_{11}=0,\,\,R_{22}<0,\,\,R_{33}=0.$

Thus we have

$R_{11}=R_{33}>R_{22}$

and

$\frac{2}{3}R\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{22}\lambda.$

Then $\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R)dt}$ is nondecreasing along the
backward Ricci flow, and $\lambda(t)e^{\int_{0}^{t}(-\frac{2}{3}R+2R_{22})dt}$
is nonincreasing.

Next, we do further estimates

$\displaystyle\frac{2}{3}R-2R_{22}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}A^{2}-\frac{1}{3}C^{2}-\frac{2}{3}AC+BA^{2}+BC^{2}+2ABC$
$\displaystyle=$ $\displaystyle-\frac{1}{2}(T_{+}-t)^{-1}+16$

and

$\displaystyle\frac{2}{3}R=-\frac{1}{2}(T_{+}-t)^{-1}.$

Thus we arrive at

$-\frac{1}{2}(T_{+}-t)^{-1}\leq\frac{1}{\lambda}\frac{d}{dt}\lambda\leq-\frac{1}{2}(T_{+}-t)^{-1}+16.$

Integration from $0$ to $t$ gives

$\lambda(0)\left(\frac{T_{+}-t}{T_{+}}\right)^{1/2}\leq\lambda(t)\leq\lambda(0)\left(\frac{T_{+}-t}{T_{+}}\right)^{1/2}e^{16t}.$

(2) If $A_{0}>C_{0}$, then by (5.1) and (2) in Theorem 5.1 we have

$R_{11}>0,\,\,R_{22}<0,\,\,R_{33}<0$

after a time $\tau$.

Assume that $C_{0}\geq B_{0}$. By (5.2) we get

$\displaystyle\frac{d}{dt}\ln\frac{C}{B}$ $\displaystyle=2(AC+C^{2}),$

which implies $C(t)>B(t)$ for all $t>0$. It is easy to see that

$\displaystyle R_{22}-R_{33}$
$\displaystyle=\frac{1}{2}(CA^{2}-BA^{2}-C^{3}-BC^{2}-2ABC)$ $\displaystyle>0$

after a time $\tau$.

So we arrive at

$R_{11}>R_{22}>R_{33}$

and

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{33}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.

Moreover, we obtain

$\displaystyle\frac{2}{3}R-2R_{33}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}A^{2}-\frac{1}{3}C^{2}-\frac{2}{3}AC-C^{3}+CA^{2}$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}A^{2}-\frac{1}{3}C^{2}-\frac{2}{3}AC-A^{3}+AC^{2}$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-3/2}$

with $t\geq\tau$.

Thus we get

$-c_{2}(T_{+}-t)^{-3/2}\leq\frac{1}{\lambda}\frac{d}{dt}\lambda\leq-
c_{1}(T_{+}-t)^{-1}.$

Integration from $\tau$ to $t$ yields

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$. ∎

##  6\. $\text{E}(2)$

Given a metric $g_{0}$, we choose a Milnor frame such that

$\left[f_{2},f_{3}\right]=2f_{1},\,\,\,\left[f_{3},f_{1}\right]=2f_{2},\,\,\,\left[f_{1},f_{2}\right]=0.$

Under the normalization $A_{0}B_{0}C_{0}=4$, the nonzero curvature components
are

(6.1) $\left\\{\begin{aligned} &R_{11}=\frac{1}{2}A(A^{2}-B^{2}),\\\
&R_{22}=\frac{1}{2}B(B^{2}-A^{2}),\\\ &R_{33}=-\frac{1}{2}C(A-B)^{2},\\\
&R=-\frac{1}{2}(A-B)^{2}.\end{aligned}\right.$

Then the backward Ricci flow equations are

(6.2) $\left\\{\begin{aligned} &\frac{dA}{dt}=\frac{2}{3}A(2A+B)(A-B),\\\
&\frac{dB}{dt}=-\frac{2}{3}B(2B+A)(A-B),\\\
&\frac{dC}{dt}=-\frac{2}{3}C(A-B)^{2}.\end{aligned}\right.$

Assume that $A_{0}\geq B_{0}$, Cao [5] proved the following theorem.

######  Theorem 6.1.

(1) If $A_{0}=B_{0}$, then $T_{+}=\infty$, and $g(t)=g_{0}$ for
$t\in[0,+\infty)$.

(2) If $A_{0}>B_{0}$, then $T_{+}<\infty$, there exist two positive constants
$\eta_{1},\,\eta_{2}$ such that

$A\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,B(t)\sim\eta_{1}(T_{+}-t)^{1/4},\,C(t)\sim\eta_{2}(T_{+}-t)^{1/4},$

as $t$ goes to $T_{+}$.

In case (2), $\bar{g}(t)=(B_{0}/B(t))g(t)$ converges to a sub-Riemannian
geometry.

We will prove the following theorem.

######  Theorem 6.2.

Let $\lambda(t)$ be the first eigenvalue of $-\Delta$. Then we get

(1) If $A_{0}=B_{0}$, then $g(t)=g_{0}$, and $\lambda(t)$ is a constant.

(2) If $A_{0}>B_{0}$ and $C_{0}\geq B_{0}$, then there is a time $\tau$ such
that $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.
Moreover, we have

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

Remark. In case (2) of the above theorem, if $C_{0}<B_{0}$, we can get the
similar estimate.

###### Proof.

(1) If $A_{0}=B_{0}$, then $g(t)=g_{0}$, and $\lambda(t)$ is independent of
$t$.

(2) If $A_{0}>B_{0}$, then by (6.2) we have

$\frac{d}{dt}(A-B)=\frac{4}{3}(A-B)(A^{2}+AB+B^{2}).$

So $A-B$ is increasing and $A(t)>B(t)$. This and (6.1) yields

$R_{11}>0,\,\,R_{22}<0,\,\,R_{33}<0.$

Assume that $C_{0}\geq B_{0}$. From the equations for $B$, $C$ in (6.2), we
get

$\displaystyle\frac{d}{dt}\ln\frac{C}{B}$
$\displaystyle=\frac{2}{3}[(2B+A)(A-B)-(A-B)^{2}]$ $\displaystyle=2B(A-B)$

and conclude that $\frac{C}{B}$ is increasing and $C(t)>B(t)$ for $t>0$.

By (6.1) and (2) in Theorem 6.1, we get

$\displaystyle R_{22}-R_{33}$ $\displaystyle=\frac{1}{2}(A-B)(AC-BC-AB-B^{2})$
$\displaystyle=\frac{1}{2}(A-B)[A(C-B)-BC-B^{2}]$ $\displaystyle>0$

if $t\geq\tau$. So we obtain $R_{11}>R_{22}>R_{33}$ and

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{33}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.

Further computations show that

$\displaystyle\frac{2}{3}R-2R_{33}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}(A-B)^{2}+C(A-B)^{2}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}A^{2}-\frac{1}{3}B^{2}+\frac{2}{3}AB++CA^{2}+CB^{2}-2ABC$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle-\frac{1}{3}(A-B)^{2}-A(A^{2}-B^{2})$ $\displaystyle=$
$\displaystyle-\frac{1}{3}A^{2}-\frac{1}{3}B^{2}+\frac{2}{3}AB-A^{3}+AB^{2}$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-3/2}$

after a time $\tau$. Thus we arrive at

$-c_{2}(T_{+}-t)^{-3/2}\leq\frac{1}{\lambda}\frac{d}{dt}\lambda\leq-
c_{1}(T_{+}-t)^{-1}.$

Integration from $\tau$ to $t$ gives

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

∎

##  7\. $\text{SL}(2,\mathbb{R})$

Given a metric $g_{0}$, we choose a Milnor frame such that

$\left[f_{2},f_{3}\right]=-2f_{1},\,\,\,\left[f_{3},f_{1}\right]=2f_{2},\,\,\,\left[f_{1},f_{2}\right]=2f_{3}.$

Under the normalization $A_{0}B_{0}C_{0}=4$, the nonzero curvature components
are

(7.1) $\left\\{\begin{aligned} &R_{11}=\frac{1}{2}A[A^{2}-(B-C)^{2}],\\\
&R_{22}=\frac{1}{2}B[B^{2}-(A+C)^{2}],\\\
&R_{33}=\frac{1}{2}C[C^{2}-(A+B)^{2}],\\\
&R=\frac{1}{2}[A^{2}-(B-C)^{2}]+\frac{1}{2}[B^{2}-(A+C)^{2}]+\frac{1}{2}[C^{2}-(A+B)^{2}].\end{aligned}\right.$

Then the backward Ricci flow equations are

(7.2) $\left\\{\begin{aligned}
&\frac{dA}{dt}=-\frac{2}{3}[-A^{2}(2A+B+C)+A(B-C)^{2}],\\\
&\frac{dB}{dt}=-\frac{2}{3}[-B^{2}(2B+A-C)+B(A+C)^{2}],\\\
&\frac{dC}{dt}=-\frac{2}{3}[-C^{2}(2C+A-B)+C(A+B)^{2}]\end{aligned}\right.$

Under the assumption $B_{0}\geq C_{0}$, Cao [5, 3] proved the following
theorem.

######  Theorem 7.1.

The maximal existence time $T_{+}$ is finite. Moreover,

(1) If there exists a time $t_{0}$ such that $A(t_{0})\geq B(t_{0})$, then

$A\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,B(t)\sim\eta_{1}(T_{+}-t)^{1/4},\,C(t)\sim\eta_{2}(T_{+}-t)^{1/4}$

with positive constants $\eta_{i}$, $i=1,2.$

(2) If there exists a time $t_{0}$ such that $A(t_{0})\leq B(t_{0})-C(t_{0})$,
then

$A\sim\eta_{1}(T_{+}-t)^{1/4},\,B(t)\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,C(t)\sim\eta_{2}(T_{+}-t)^{1/4}$

with positive constants $\eta_{i}$, $i=1,2.$

(3) If $A<B<A+C$ for all time $t\in[0,T_{+})$, we arrive at

$A\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,B(t)\sim\frac{\sqrt{6}}{4}(T_{+}-t)^{-1/2},\,C(t)\sim\frac{32}{3}(T_{+}-t).$

In all cases, the metric $g(t)$ converges to a sub-Riemannian geometry after a
proper rescaling.

We have the following theorem.

######  Theorem 7.2.

Let $\lambda(t)$ be the first eigenvalue of $-\Delta$. Then we get

(1) If there is a time $t_{0}$ such that $A(t_{0})\geq B(t_{0})$, then there
exist a time $\tau$ such that
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.
Moreover, we have

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

(2) If there exists a time $t_{0}$ such that $A(t_{0})\leq B(t_{0})-C(t_{0})$,
and a $t_{1}$ such that $A(t_{1})>C(t_{1})$, then there is a time $\tau$ such
that $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nonincreasing.
Moreover, we have

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}},$

and $\lambda(t)$ approaches $0$ as $t$ goes to $T_{+}$.

(3) If $A<B<A+C$ for all time $t\in[0,T_{+})$, then there is time $\tau$ such
that $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.
Moreover, we have

$\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{2}}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

Obviously, $\lambda(t)$ approaches $0$, as $t$ goes to $T_{+}$.

Remark. If $A(t)\leq C(t)$ for all $t$ in case (2) of the above theorem, we
can get the similar estimate.

###### Proof.

(1) If there is a time $t_{0}$ such that $A(t_{0})>B_{0}$, then by (7.1) and
(1) in Theorem 7.1, we have

$R_{11}>0,\,\,R_{22}<0,\,\,R_{33}<0$

after a time $\tau$.

Next we compare $R_{22}$ with $R_{33}$:

$\displaystyle R_{22}-R_{33}$
$\displaystyle=\frac{1}{2}B[B^{2}-(A+C)^{2}]-\frac{1}{2}C[C^{2}-(A+B)^{2}]$
$\displaystyle=\frac{1}{2}(B-C)[(B+C)^{2}-A^{2}].$

The evolution equation of $B-C$ is

$\frac{d}{dt}(B-C)=\frac{2}{3}[2(B^{3}-C^{3})+A(B^{2}-C^{2})-A^{2}(B-C)],$

from which it follows that $B\geq C$ for all $t$.

Thus we obtain

$R_{11}>R_{33}\geq R_{22}$

and

$\frac{2}{3}R\lambda-2R_{11}\lambda\leq\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{22}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nonincreasing.

Moreover, we get

$\displaystyle\frac{2}{3}R-2R_{22}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-B[B^{2}-(A+C)^{2}]$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-A[A^{2}-(B-C)^{2}]$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-3/2}$

after a time $\tau$.

Integrating from $\tau$ to $t$ gives

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

As $t$ goes to $T_{+}$, $\lambda(t)$ approaches $0$.

(2) If there exists a time $t_{0}$ such that $A(t_{0})\leq B(t_{0})-C(t_{0})$
, the second case in Theorem 7.1 implies

$R_{11}<0,\,\,R_{22}>0,\,\,R_{33}<0$

after a time $\tau$.

It follows from equations for both $A$ and $C$ that

$\frac{d}{dt}\ln(A/C)=2(A+B-C)(A+C),$

which implies that $\frac{d}{dt}\ln(A/C)$ is increasing since $B\geq C$ is
preserved.

Assume that there is time $t_{1}$ such that $A(t_{1})>C(t_{1})$. Then
$A(t)>C(t)$ after $t_{1}$.

The behaviors of $A$, $B$ and $C$ yield that

$\displaystyle R_{11}-R_{33}$
$\displaystyle=\frac{1}{2}A(A^{2}-(B-C)^{2}]-\frac{1}{2}C[C^{2}-(A+B)^{2}]$
$\displaystyle=\frac{1}{2}(C-A)B^{2}+\frac{1}{2}(A^{3}-C^{3}+CA^{2}-AC^{2}+4ABC)$
$\displaystyle<0$

after a time $\tau$.

Thus we arrive at

$R_{22}>R_{33}>R_{11}$

and

$\frac{2}{3}R\lambda-2R_{22}\lambda<\frac{d}{dt}\lambda<\frac{2}{3}R\lambda-2R_{11}\lambda$

with $t\geq\tau$.

Then $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{22})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nonincreasing.

Now we estimate

$\displaystyle\frac{2}{3}R-2R_{22}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-B[B^{2}-(A+C)^{2}]$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-3/2}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-A[A^{2}-(B-C)^{2}]$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

if $t\geq\tau$.

The above estimates imply

$\lambda(\tau)e^{2c_{2}\left[(T_{+}-\tau)^{-1/2}-(T_{+}-t)^{-1/2}\right]}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}},$

and $\lambda(t)$ approaches $0$, as $t$ goes to $T_{+}$.

(3) If $A<B<A+C$ for all time $t\in[0,T_{+})$, it follows from (7.1) and
Theorem 7.1 that

$R_{11}>0,\,\,R_{22}<0,\,\,R_{33}<0$

after a time $\tau$.

Since $A<B<A+C$ , we obtain

$\displaystyle R_{22}-R_{33}$
$\displaystyle=\frac{1}{2}(B-C)[(B+C)^{2}-A^{2}]\geq 0.$

Thus we have

$R_{11}>R_{22}\geq R_{33}$

and

$\frac{2}{3}R\lambda-2R_{11}\lambda<\frac{d}{dt}\lambda\leq\frac{2}{3}R\lambda-2R_{33}\lambda$

with $t\geq\tau$.

Thus $\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{11})dt}$ is nondecreasing
along the backward Ricci flow, and
$\lambda(t)e^{\int_{\tau}^{t}(-\frac{2}{3}R+2R_{33})dt}$ is nonincreasing.

Moreover, we compute

$\displaystyle\frac{2}{3}R-2R_{33}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-C[C^{2}-(A+B)^{2}]$
$\displaystyle\leq$ $\displaystyle-c_{1}(T_{+}-t)^{-1}$

and

$\displaystyle\frac{2}{3}R-2R_{11}$ $\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})-A[A^{2}-(B-C)^{2}]$
$\displaystyle=$
$\displaystyle\frac{1}{3}(2BC-2AB-2AC-A^{2}-B^{2}-C^{2})+A[A+(B-C)][B-A-C]$
$\displaystyle\geq$ $\displaystyle-c_{2}(T_{+}-t)^{-1}$

after a time $\tau$ since in this case the fact $\lim_{T_{+}}C=0$ yields
$\lim_{T_{+}}(B-A)=0$.

After integrating from $\tau$ to $T_{+}$, we get

$\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{2}}\leq\lambda(t)\leq\lambda(\tau)\left(\frac{T_{+}-t}{T_{+}-\tau}\right)^{c_{1}}.$

Obviously, $\lambda(t)$ approaches $0$, as $t$ goes to $T_{+}$. ∎

Acknowledgement

The author was supported by NSFC (11001268) and Chinese University Scientific
Fund (2014QJ002). The author would like to thank Professor Xiaodong Cao and
Professor Laurent Saloff-Coste for their suggestions and interests in this
work.

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