UNIQUENESS OF CONFORMAL RICCI FLOW AND BACKWARD RICCI
FLOW ON HOMOGENEOUS 4-MANIFOLDS
by
THOMAS BELL
A DISSERTATION
Presented to the Department of Mathematics
and the Graduate School of the University of Oregon
in partial fulfillment of the requirements
for the degree of
Doctor of Philosophy
June 2013
DISSERTATION APPROVAL PAGE
Student: Thomas Bell
Title: Uniqueness of Conformal Ricci Flow and Backward Ricci Flow on
Homogeneous 4-Manifolds
This dissertation has been accepted and approved in partial fulfillment of the
requirements for the Doctor of Philosophy degree in the Department of Mathematics
by:
Dr. Peng Lu
Dr. James Isenberg
Dr. Boris Botvinnik
Dr. Huaxin Lin
Dr. Davison Soper
Chairperson
Core Member
Core Member
Core Member
Institutional Representative
and
Kimberly Andrews Espy Vice President for Research and
Innovation/ Dean of the Graduate
School
Original approval signatures are on file with the University of Oregon Graduate
School.
Degree awarded June 2013
ii
c©June 2013
Thomas Bell
iii
DISSERTATION ABSTRACT
Thomas Bell
Doctor of Philosophy
Department of Mathematics
June 2013
Title: Uniqueness of Conformal Ricci Flow and Backward Ricci Flow on
Homogeneous 4-Manifolds
In the first chapter we consider the question of uniqueness of conformal Ricci
flow. We use an energy functional associated with this flow along closed manifolds
with a metric of constant negative scalar curvature. Given initial conditions we use
this functional to demonstrate the uniqueness of the solution to both the metric
and the pressure function along conformal Ricci flow.
In the next chapter we study backward Ricci flow of locally homogeneous
geometries of 4-manifolds which admit compact quotients. We describe the long-
term behavior of each class and show that many of the classes exhibit the same
behavior near the singular time. In most cases, these manifolds converge to a
sub-Riemannian geometry after suitable rescaling.
iv
CURRICULUM VITAE
NAME OF AUTHOR: Thomas Bell
GRADUATE AND UNDERGRADUATE SCHOOLS ATTENDED:
University of Oregon, Eugene, Oregon
Brigham Young University, Provo, Utah
DEGREES AWARDED:
Doctor of Philosophy in Mathematics, 2013, University of Oregon
Master of Science in Mathematics, 2007, Brigham Young University
Bachelor of Science in Mathematics, 2005, Brigham Young University
AREAS OF SPECIAL INTEREST:
Geometric Flows
Gradient Ricci Solitons
Geometric Optimization
PROFESSIONAL EXPERIENCE:
Graduate Teaching Fellow,
University of Oregon, September 2007 - present
Graduate Instructor,
Brigham Young University, September 2005 - April 2007
Assistant to REU,
Brigham Young University, Summer 2005, Summer 2006
Undergraduate Research Assistant,
Brigham Young University, September 2004 - April 2005
v
PUBLICATIONS:
John Bell, Thomas Bell, Hannabeth Franchino, Jennifer Nelson and
Evan Stevans, “Wavelength Dependence of Patman Equilibration
Dynamics in Phosphatidylcholine Bilayers” BBA - Biomembranes.
1828, 2, 877-886, (2013).
Thomas Bell, “Uniqueness of Conformal Ricci Flow using Energy
Methods” arXiv:13015052, (2013).
vi
TABLE OF CONTENTS
Chapter Page
I. UNIQUENESS OF CONFORMAL RICCI FLOW . . . . . . . . . . . . . . . . . . . . . . . 1
The Differences Between g(t) and g˜(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Preliminary Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Bounds on Time Derivatives of h, A and S . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Bounds on q and Its Spatial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Energy Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Evolution of H(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Evolution of A(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Evolution of S(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Proof of Main Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
II. BACKWARD RICCI FLOW OF HOMOGENEOUS
4-GEOMETRIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
The Bianchi Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
A1. Class U1b(1, 1, 1)c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
A2. Class U1b1, 1, 1c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
A3. Class U1bZ, Z¯, 1c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
A4. Class U1b2, 1c, µ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
A5. Class U1b2, 1c, µ = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
A6. Class U1b3c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
A7. Class U3I0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
A8. Class U3I2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
A9. Class U3S1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
A10. Class U3S3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
The Non-Bianchi Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
vii
Chapter Page
B1. H3 × R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
B2. S2 × R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
B3. H2 × R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
B4. S2 × S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
B5. S2 ×H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
B6. H2 ×H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
B7. CP 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
B8. CH2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
B9. S4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
B10. H4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
REFERENCES CITED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
viii
LIST OF TABLES
Table Page
2.1. End Behavior of Backward Ricci Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
ix
CHAPTER I
UNIQUENESS OF CONFORMAL RICCI FLOW
The uniqueness of Ricci Flow on closed manifolds was originally proven by
Hamilton [9]. Later on, Chen and Zhu proved the uniqueness on complete
noncompact manifolds with bounded curvature [5]. The method employed in [5]
utilizes DeTurck Ricci Flow. Recently Kotschwar used energy techniques to give
another proof of the uniqueness on complete manifolds [12]. Kotchwar’s proof does
not rely on DeTurck Ricci Flow. A natural question is whether similar techniques
can be applied to demonstrate the uniqueness of other geometric flows. One of
these flows we have in mind is Conformal Ricci Flow, introduced by Fischer [8].
Ricci Flow preserves many important properties of metrics, but it generally does
not preserve the property of constant scalar curvature. Conformal Ricci Flow is a
modification of Ricci Flow which is intended for this purpose and for this reason
it is restricted to the class of metrics of constant scalar curvature. Conformal
Ricci Flow is, like Ricci Flow, a weakly parabolic flow of the metric on manifolds,
except that Conformal Ricci Flow is coupled with an elliptic equation. Unlike Ricci
Flow, Conformal Ricci Flow is restricted to the class of metrics of constant scalar
curvature.
Let (Mn, g0) be a smooth n-dimensional Riemannian manifold with a metric g0
1
of constant scalar curvature s0. Conformal Ricci Flow on M is defined as follows:
∂g
∂t
= −2Ricg(t) + 2s0
n
g(t)− 2p(t)g(t)
s
(
g(t)
)
= s0
on M × [0, T ]. (I.1)
Here g(t), t ∈ [0, T ], is a family of metrics on M with g(0) = g0, s
(
g(t)
)
is the
scalar curvature of g(t), and p(t), t ∈ [0, T ], is a family of functions on M . In [8]
and [13] we see that (I.1) is equivalent to the following system:
∂g
∂t
= −2Ricg(t) + 2s0
n
g(t)− 2p(t)g(t)(
(n− 1)∆g(t) + s0
)
p(t) = −
〈
Ricg(t) − s0
n
g(t),Ricg(t) − s0
n
g(t)
〉 (I.2)
Throughout this chapter we will use V to denote the following symmetric 2-
tensor:
V (t) = Ricg(t) − s0
n
g(t) + p(t)g(t) (I.3)
In this chapter we use Kotchwar’s energy techniques to give a proof of the
uniqueness of Conformal Ricci Flow for closed manifolds with metrics of constant
negative scalar curvature. It is worth noting similarities to the study of certain
elliptic-hyperbolic systems done by Andersson and Moncrief in [1]. The existence
of solutions to Conformal Ricci Flow has been shown by Fischer /citeFi and by Lu,
Qing and Zheng, [13], the latter paper using DeTurck Conformal Ricci Flow. More
precisely we will prove the following uniqueness theorem of Conformal Ricci Flow:
2
Theorem I.1. Let (Mn, g0) be a closed manifold with constant negative scalar
curvature s0. Suppose
(
g(t), p(t)
)
and
(
g˜(t), p˜(t)
)
are two solutions of (I.1) on
M × [0, T ] with g˜(0) = g(0). Then (g˜(t), p˜(t)) = (g(t), p(t)) for 0 ≤ t ≤ T .
The Differences Between g(t) and g˜(t)
Let g(t) and g˜(t) be as in Theorem I.1. We will treat g as our background metric
and g˜ as our alternative metric. Let ∇, ∇˜ be the Riemannian connections of g and
g˜ respectively. Similarly, let R, R˜ represent the full Riemannian curvature tensors
of g and g˜ respectively.
Let h = g − g˜. Let A = ∇− ∇˜. Explicitly, Aijk = Γijk − Γ˜ijk where Γijk and Γ˜ijk
are the Christoffel symbols of ∇ and ∇˜ respectively. Also let S = R− R˜, q = p− p˜.
In this section we find bounds on h, A, S, q, ∇q and ∇∇q (see Propositions
I.2 and I.4). Throughout this chapter we will use the convention X ∗ Y to denote
any finite sum of tensors of the form X · Y . We use C(X) to denote a finite sum of
tensors of the form X.
Preliminary Calculations
First we calculate some useful expressions for quantities which will arise in the
proofs of Propositions I.2 and I.4. We calculate
gij − g˜ij = gik(g˜j`g˜k`)− g˜j`(gikgk`) = −gikg˜j`hk`,
i.e.
g−1 − g˜−1 = g˜−1 ∗ h.
3
If X is any tensor which is not a function we have
(∇− ∇˜)X = A ∗X.
We check this when X is a (1, 1)-tensor. Calculating in local coordinates we see
(∇i − ∇˜i)Xkj = ∂iXkj − Γ`ijXk` + Γki`X`j − ∂iXkj + Γ˜`ijXk` − Γ˜ki`X`j
= Aki`X
`
j − A`ijXk` = A ∗X.
If f is a function however, then we have the following:
(∇i − ∇˜i)f = (gij − g˜ij)∂jf = −gikg˜j`hk`∂jf = −gikhk`∇˜`f,
or in other words (∇− ∇˜)f = h ∗ ∇˜f.
We now calculate
∇g˜−1 = (∇− ∇˜)g˜−1 = g˜−1 ∗ A.
The following calculation will also be important.
∇ihjk = ∇igjk −∇ig˜jk = −
(∇i − ∇˜i)g˜jk.
Thus we have
∇h = g˜ ∗ A.
4
Now we are able to calculate the following for a function f .
∇(∇− ∇˜)f = ∇(h ∗ ∇˜f)
= ∇h ∗ ∇˜f + h ∗ (∇− ∇˜)∇˜f + h ∗ ∇˜∇˜f
= g˜ ∗ A ∗ ∇˜f + h ∗ A ∗ ∇˜f + h ∗ ∇˜∇˜f.
Now let
Uaijk` = g
ab∇bR˜ijk` − g˜ab∇˜bR˜ijk` (I.4)
= gab(∇b − ∇˜b)R˜ijk` +
(
gab − g˜ab)∇˜bR˜ijk`
= A ∗ R˜ + g˜−1 ∗ h ∗ ∇˜R˜,
and we may calculate
∇a
(
gab∇bR− g˜ab∇˜bR˜
)
= ∇a
(
gab∇bR˜− g˜ab∇˜bR˜
)
+ gab∇a∇b
(
R− R˜)
= divU + ∆S.
We summarize the above calculations in the following Lemma:
5
Lemma I.1. Using the notation defined at the beginning of this section,
g−1 − g˜−1 = g˜−1 ∗ h (I.5)(∇− ∇˜)X = A ∗X (I.6)(∇− ∇˜)f = h ∗ ∇˜f (I.7)
∇g˜−1 = g˜−1 ∗ A (I.8)
∇h = g˜ ∗ A (I.9)
∇(∇− ∇˜)f = g˜ ∗ A ∗ ∇˜f + h ∗ A ∗ ∇˜f + h ∗ ∇˜∇˜f (I.10)
U = A ∗ R˜ + g˜−1 ∗ h ∗ ∇˜R˜ (I.11)
∇a
(
gab∇bR− g˜ab∇˜bR˜
)
= divU + ∆S (I.12)
where U is defined in (I.4).
Bounds on Time Derivatives of h, A and S
In this subsection we derive bounds on the time derivatives of h, A and S. In
particular we will prove the following proposition. Here, as well as throughout this
chapter, C will denote a constant dependent only upon n while N will denote a
constant with further dependencies.
Proposition I.2. Let
(
g(t), p(t)
)
and
(
g˜(t), p˜(t)
)
be two solutions of (I.1) on
M × [0, T ]. Using the notation defined at the beginning of this section, there exist
6
constants Nh, NA and NS such that∣∣∣∣ ∂∂th
∣∣∣∣ ≤ Nh|h|+ C(|S|+ |q|) (I.13)∣∣∣∣ ∂∂tA
∣∣∣∣ ≤ NA(|h|+ |A|)+ C(|∇S|+ |∇q|) (I.14)∣∣∣∣ ∂∂tS −∆S − divU
∣∣∣∣ ≤ NS(|h|+ |A|+ |S|+ |q|)+ C|∇∇q| (I.15)
where U is defined in (I.4).
Proof. We start with the time derivative of h. By (I.1) we have
∂
∂t
hij = −2(Rij − R˜ij) + 2s0
n
(gij − g˜ij)− 2(p gij − p˜ g˜ij)
= −2Skkij + 2
s0
n
hij − 2
[
(p− p˜)gij + p˜(gij − g˜ij)
]
= −2Skkij + 2
s0
n
hij − 2q gij − 2p˜ hij.
Hence
∂
∂t
h = C(S) + C(s0h) + C(q) + p˜ ∗ h
and ∣∣∣∣ ∂∂th
∣∣∣∣ ≤ C((|s0|+ |p˜|)|h|+ |S|+ |q|). (I.16)
This proves (I.13).
Recall the definition of V from (I.3):
V (t) = Ricg(t) − s0
n
g(t) + p(t)g(t). (I.17)
We may define V˜ similarly using our alternate metric g˜. Since V and V˜ are
symmetric 2-tensors, then by [7, p. 108] we may calculate
∂
∂t
Akij = g˜
k`
(∇˜iV˜j` + ∇˜jV˜i` − ∇˜`V˜ij)− gk`(∇iVj` +∇jVi` −∇`Vij). (I.18)
7
We proceed to calculate
g˜k`∇˜iV˜j` − gk`∇iVj`
=g˜k`(∇˜iR˜j`)− gk`(∇iRj`) + g˜k`∇˜i(p˜ g˜j`)− gk`∇i(p gj`)
=
(
g˜k` − gk`)∇˜iR˜j` + gk`(∇˜i −∇i)R˜j` − gk`∇i(Smmj`) + δkj ∇˜ip˜− δkj∇ip
=g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + h ∗ ∇˜p˜+ C(∇q), (I.19)
where we have used (I.7) to get the last equality. Similarly we find
g˜k`∇˜jV˜i` − gk`∇jVi`
=g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + h ∗ ∇˜p˜+ C(∇q). (I.20)
Now we consider
− g˜k`∇˜`V˜ij + gk`∇`Vij
=g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + g˜k`g˜ij∇˜`p˜− gk`gij∇`p
=g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + (g˜k` − gk`)g˜ij∇˜`p˜+ gk`(g˜ij − gij)∇˜`p˜
+ gk`gij(∇˜` −∇`)p˜+ gk`gij∇`(p˜− p)
=g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + g˜−1 ∗ h ∗ g˜ ∗ ∇˜p˜+ h ∗ ∇˜p˜+ C(∇q). (I.21)
Hence by (I.18), (I.19), (I.20) and (I.21),
∂
∂t
A = g˜−1 ∗ h ∗ ∇˜R˜ + A ∗ R˜ + C(∇S) + h ∗ ∇˜p˜+ C(∇q) + g˜−1 ∗ h ∗ g˜ ∗ ∇˜p˜
and∣∣∣∣ ∂∂tA
∣∣∣∣ ≤ C((|g˜−1||∇˜R˜|+ |∇˜p˜|+ |g˜−1||g˜||∇˜p˜|)|h|+ |R˜||A|+ |∇S|+ |∇q|). (I.22)
8
This proves (I.14).
By [7, eqn. (2.67)] we have
∂
∂t
R`ijk = g
`m
(∇i∇kVjm −∇i∇mVjk −∇j∇kVim +∇j∇mVik)
− g`m(RrijkVrm +RqijmVkq)
= g`m
(−∇i∇kRjm +∇i∇mRjk +∇j∇kRim −∇j∇mRik)
+ g`m
(−gjm∇i∇kp+ gjk∇i∇mp+ gim∇j∇kp− gik∇j∇mp)
+ g`m
(
RrijkRrm +R
r
ijmRkr
)− s0
n
g`m
(
Rrijkgrm +R
r
ijmgkr
)
p
+ g`m
(
Rrijkgrm +R
r
ijmgkr
)
p. (I.23)
Following the calculations in [7, p. 119-120] we have
∆R`ijk = g
ab∇a∇bR`ijk = gab
(−∇a∇iR`jbk −∇a∇jR`bik)
= gab
(−∇i∇aR`jbk +RmaijR`mbk +RmaibR`jmk +RmaikR`jbm −R`aimRmjbk
−∇j∇aR`bik +RmajbR`mik +RmajiR`bmk +RmajkR`bim −R`ajmRmbik
)
= g`m
(−∇i∇kRjm +∇i∇mRjk +∇j∇kRim −∇j∇mRik)
+ gmr
(−RirR`jmk −RjrR`mik)
+ gab
(
RmaijR
`
mbk +R
m
aikR
`
jbm −R`aimRmjbk
+RmajiR
`
bmk +R
m
ajkR
`
bim −R`ajmRmbik
)
. (I.24)
9
Combining (I.23) and (I.24) we have
∂
∂t
R`ijk = ∆R
`
ijk + g
mr
(
RirR
`
jmk +RjrR
`
mik
)
+ gab
(−RmaijR`mbk −RmaikR`jbm +R`aimRmjbk
−RmajiR`bmk −RmajkR`bim +R`ajmRmbik
)
+ g`m
(−gjm∇i∇kp+ gjk∇i∇mp+ gim∇j∇kp− gik∇j∇mp)
+ g`m
(
RrijkRrm +R
r
ijmRkr
)− s0
n
g`m
(
Rrijkgrm +R
r
ijmgkr
)
+ g`m
(
Rrijkgrm +R
r
ijmgkr
)
p. (I.25)
Hence the evolution of S is
∂
∂t
S`ijk = ∆R
`
ijk − ∆˜R˜`ijk
+ gmr
(
RirR
`
jmk +RjrR
`
jmk
)− g˜mr(R˜irR˜`jmk + R˜jrR˜`mik)
+ gab
(−RmaijR`mbk −RmaikR`jbm +R`aimRmjbk
−RmajiR`bmk −RmajkR`bim +R`ajmRmbik
)
− g˜ab(−R˜maijR˜`mbk − R˜maikR˜`jbm + R˜`aimR˜mjbk
− R˜majiR˜`bmk − R˜majkR˜`bim + R˜`ajmR˜mbik
)
+ g`m
(−gjm∇i∇kp+ gjk∇i∇mp+ gim∇j∇kp− gik∇j∇mp)
− g˜`m(−g˜jm∇˜i∇˜kp˜+ g˜jk∇˜i∇˜mp˜+ g˜im∇˜j∇˜kp˜− g˜ik∇˜j∇˜mp˜)
+ g`m
(
RrijkRrm +R
r
ijmRkr
)− g˜`m(R˜rijkR˜rm + R˜rijmR˜kr)
− s0
n
g`m
(
Rrijkgrm +R
r
ijmgkr
)
+
s0
n
g˜`m
(
R˜rijkg˜rm + R˜
r
ijmg˜kr
)
+ g`m
(
Rrijkgrm +R
r
ijmgkr
)
p− g˜`m(R˜rijkg˜rm + R˜rijmg˜kr)p˜. (I.26)
10
Looking at the individual components, we see
∆R− ∆˜R˜
=gab∇a∇bR− g˜ab∇˜a∇˜bR˜
=∇a(gab∇bR)−∇a(g˜ab∇˜bR˜) + (∇a − ∇˜a)(g˜ab∇˜bR˜)
=∇a
(
gab∇bR− g˜ab∇˜bR˜
)
+ g˜−1 ∗ A ∗ ∇˜R˜, (I.27)
while
g−1RR− g˜−1R˜R˜
=(g−1 − g˜−1)(R˜R˜) + g−1(RR− R˜R˜)
=g˜−1 ∗ h ∗ R˜ ∗ R˜ + g−1(R− R˜)R˜ + g−1(RR−RR˜)
=g˜−1 ∗ h ∗ R˜ ∗ R˜ + S ∗ R˜ + S ∗R, (I.28)
and
g−1g∇∇p− g˜−1g˜∇˜∇˜p˜
=(g−1 − g˜−1)g˜∇˜∇˜p˜+ g−1(g − g˜)∇˜∇˜p˜+ g−1g(∇∇p− ∇˜∇˜p˜)
=g˜−1 ∗ h ∗ g˜ ∗ ∇˜∇˜p˜+ h ∗ ∇˜∇˜p˜+ g−1g(∇− ∇˜)(∇˜p˜) + g−1g(∇∇p−∇∇˜p˜)
=g˜−1 ∗ h ∗ g˜ ∗ ∇˜∇˜p˜+ h ∗ ∇˜∇˜p˜+ A ∗ ∇˜p˜+ g−1g∇(∇− ∇˜)p˜+ g−1g∇∇(p− p˜)
=g˜−1 ∗ h ∗ g˜ ∗ ∇˜∇˜p˜+ h ∗ ∇˜∇˜p˜+ A ∗ ∇˜p˜+ h ∗ A ∗ ∇˜p˜+ C(∇∇q), (I.29)
11
where in the last equality we used (I.10). We also have
g−1gR− g˜−1g˜R˜
=(g−1 − g˜−1)g˜R˜ + g−1(g − g˜)R˜ + g−1g(R− R˜)
=g˜−1 ∗ h ∗ g˜ ∗ R˜ + h ∗ R˜ + C(S), (I.30)
and lastly
g−1gRp− g˜−1g˜R˜p˜
=(g−1 − g˜−1)g˜R˜p˜+ g−1(g − g˜)R˜p˜+ g−1g(R− R˜)p˜+ g−1gR(p− p˜)
=g˜−1 ∗ h ∗ g˜ ∗ R˜ ∗ p˜+ h ∗ R˜ ∗ p˜+ S ∗ p˜+R ∗ q. (I.31)
Now by (I.26), (I.27), (I.28), (I.29), (I.30) and (I.31) we see
∂
∂t
S = ∇a
(
gab∇bR− g˜ab∇˜bR˜
)
+ g˜−1 ∗ A ∗ ∇˜R˜ + g˜−1 ∗ h ∗ R˜ ∗ R˜
+ S ∗ R˜ + S ∗R + g˜−1 ∗ h ∗ g˜ ∗ ∇˜∇˜p˜+ h ∗ ∇˜∇˜p˜+ A ∗ ∇˜p˜
+ h ∗ A ∗ ∇˜p˜+ C(∇∇q) + g˜−1 ∗ h ∗ g˜ ∗ R˜ + h ∗ R˜ + C(S)
+ g˜−1 ∗ h ∗ g˜ ∗ R˜ ∗ p˜+ h ∗ R˜ ∗ p˜+ S ∗ p˜+R ∗ q.
12
Hence by (I.12) we have
∣∣∣∣ ∂∂tS −∆S − div U
∣∣∣∣
≤C
((
|g˜−1||R˜|2 + |g˜−1||g˜||∇˜∇˜p˜|+ |∇˜∇˜p˜|
+ |g˜−1||g˜||R˜|+ |R˜|+ |g˜−1||g˜||R˜||p˜|+ |R˜||p˜|
)
|h|
+
(
|g˜−1||∇˜R˜|+ |∇˜p˜|+ |h||∇˜p˜|
)
|A|
+
(
|R˜|+ |R|+ 1 + |p˜|
)
|S|+ |R||q|+ |∇∇q|
)
. (I.32)
This proves (I.15).
Remark I.3. Upon closer observation we notice the following dependencies:
Nh = Nh
(
n, s0, |p˜|
)
,
NA = NA
(
n, s0, |g˜|, |g˜−1|, |R˜|, |∇˜R˜|, |∇˜p˜|
)
,
NS = NS
(
n, s0, |g˜|, |g˜−1|, |h|, |R|, |R˜|, |∇˜R˜|, |p˜|, |∇˜p˜|, |∇˜∇˜p˜|
)
.
M is closed, so M × [0, T ] is compact. Thus, given two metrics g and g˜, all of these
quantities will be bounded.
Bounds on q and Its Spatial Derivatives
We turn our attention now to finding bounds on the differences between our
pressure functions p and p˜. We have the following proposition:
Proposition I.4. Let
(
g(t), p(t)
)
and
(
g˜(t), p˜(t)
)
be two solutions of (I.1) on M ×
13
[0, T ]. Then there exist constants Nq and Nˆq such that
∫
M
|q|2dµ ≤ Nq
∫
M
(|h|2 + |A|2 + |S|2)dµ (I.33)∫
M
|∇q|2dµ ≤ Nq
∫
M
(|h|2 + |A|2 + |S|2)dµ (I.34)∫
M
|∇∇q|2dµ ≤ Nˆq
∫
M
(|h|2 + |A|2 + |S|2)dµ (I.35)
Proof. We let f represent any smooth function or tensor. In particular we will let
f be represented by the function q, the difference of the pressure functions. Since
M is compact we have
∫
M
(
(n− 1)∆ + s0
)
(f) · f dµ
=s0
∫
M
|f |2dµ− (n− 1)
∫
M
〈∇f,∇f〉dµ.
Since s0 < 0, taking the absolute value gives∣∣∣∣∫
M
(
(n− 1)∆ + s0)(f) · fdµ
∣∣∣∣ = |s0|∫
M
|f |2dµ+ (n− 1)
∫
M
|∇f |2dµ (I.36)
Now we deal specifically with p, p˜ and q. By (I.2) we have the following
equations for the pressure functions p and p˜:
(
(n− 1)∆ + s0
)
p = −
〈
Ric− s0
n
g,Ric− s0
n
g
〉
(I.37)
(
(n− 1)∆˜ + s0
)
p˜ = −
〈
R˜ic− s0
n
g˜, R˜ic− s0
n
g˜
〉
. (I.38)
14
Now we calculate
∆p− ∆˜p˜ = gab∇a∇bp− g˜ab∇˜a∇˜bp˜
= (g−1 − g˜−1)∇˜∇˜p˜+ g−1(∇− ∇˜)∇˜p˜+ g−1∇(∇− ∇˜)p˜+ ∆(p− p˜)
= g˜−1 ∗ h ∗ ∇˜∇˜p˜+ A ∗ ∇˜p˜+ h ∗ A ∗ ∇˜p˜+ ∆q. (I.39)
We also compute
−
〈
Ric− s0
n
g,Ric− s0
n
g
〉
+
〈
R˜ic− s0
n
g˜, R˜ic− s0
n
g˜
〉
=− (gikgj`RijRk` − g˜ikg˜j`R˜ijR˜k`)+ 2s0
n
(
gijRij − g˜ijR˜ij
)
=− (g−1 − g˜−1)g˜−1R˜R˜− g−1(g−1 − g˜−1)R˜R˜− g−1g−1(R− R˜)R˜
− g−1g−1R(R− R˜) + 2s0
n
(g−1 − g˜−1)R˜ + 2s0
n
g−1(R− R˜)
=g˜−1 ∗ g˜−1 ∗ h ∗ R˜ ∗ R˜ + g˜−1 ∗ h ∗ R˜ ∗ R˜
+ S ∗ R˜ + S ∗R + g˜−1 ∗ h ∗ R˜ + C(S). (I.40)
Combining (I.37), (I.38), (I.39) and (I.40), we see that q satisfies the following
elliptic equation at each time t ∈ [0, T ]:
Lq =
(
(n− 1)∆ + s0
)
(q)
= g˜−1 ∗ h ∗ ∇˜∇˜p˜+ A ∗ ∇˜p˜+ h ∗ A ∗ ∇˜p˜+ g˜−1 ∗ g˜−1 ∗ h ∗ R˜ ∗ R˜
+ g˜−1 ∗ h ∗ R˜ ∗ R˜ + S ∗ R˜ + S ∗R + g˜−1 ∗ h ∗ R˜ + C(S) (I.41)
Hence
|Lq| = ∣∣((n− 1)∆ + s0)(q)∣∣ ≤ N(|h|+ |A|+ |S|). (I.42)
15
To find estimates for q and ∇q, we combine (I.36) and (I.42):
|s0|
∫
M
|q|2dµ+ (n− 1)
∫
M
|∇q|2dµ
=
∣∣∣∣∫
M
(
(n− 1)∆ + s0
)
(q) · q dµ
∣∣∣∣
≤
∫
M
N
(|h|+ |A|+ |S|)|q| dµ
≤|s0|
2
∫
M
|q|2dµ+N
∫
M
(|h|2 + |A|2 + |S|2)dµ.
Thus
|s0|
2
∫
M
|q|2dµ+ (n− 1)
∫
M
∣∣∇q|2dµ ≤ N ∫
M
(|h|2 + |A|2 + |S|2)dµ,
and we proved (I.33) and (I.34).
To find an appropriate bound for |∇∇q| we must turn to Interior Regularity
Theory for Elliptic PDE. From (I.41) we see that Lq = f is an Elliptic Equation.
We then have the following estimate from [15, p. 229].
|q|H2(W ) ≤ K
(|Lq|L2(M) + |q|H1(M)),
where W is any compactly supported open subset of M and K depends only upon
the coefficients of the operator L, the subset W and the manifold M . Since M is a
closed manifold we may in fact choose W = M . Thus we have
|q|H2(M) ≤ K
(|Lq|L2(M) + |q|H1(M)). (I.43)
Upon squaring both sides we observe∫
M
|∇∇q|2dµ ≤ |q|2H2(M) ≤ K2
(∫
M
|Lq|2dµ+ |q|2H1(M)
)
. (I.44)
16
Now (I.33) and (I.34) imply that
|q|2H1(M) ≤ N
∫
M
(|h|2 + |A|2 + |S|2)dµ. (I.45)
Combining (I.42), (I.44) and (I.45) we have
∫
M
|∇∇q|2dµ ≤ N
∫
M
(|h|2 + |A|2 + |S|2)dµ,
and we proved (I.35).
Remark I.5. We observe the following dependencies:
Nq = Nq
(
n, s0, |g˜−1|, |h|, |R|, |R˜|, |∇˜p˜|, |∇˜∇˜p˜|
)
Nˆq = Nˆq
(
n, s0, |g˜−1|, |h|, |R|, |R˜|, |∇˜p˜|, |∇˜∇˜p˜|, K
)
where K is from (I.43).
Energy Estimates
Now we shall approximate the energy
E(t) =
∫
M
(|h|2 + |A|2 + |S|2)dµ. (I.46)
17
We also define the following:
H(t) =
∫
M
|h|2dµ (I.47)
A(t) =
∫
M
|A|2dµ (I.48)
S(t) =
∫
M
|S|2dµ (I.49)
D(t) =
∫
M
|∇S|2dµ (I.50)
Note that E(t) = H(t)+A(t)+S(t). We now estimate the evolution of the energy
functional under Conformal Ricci Flow, E ′(t), by first estimating the evolutions of
H, A and S.
Evolution of H(t)
In [13], Lu, Qing and Zheng give the evolution of the volume element under
Conformal Ricci Flow:
∂
∂t
dµg(t) = −np(t)dµg(t) (I.51)
Hence by (I.13) and (I.47) we have
H′(t) ≤ N
∫
M
|h|2dµ+
∫
M
2
〈
∂h
∂t
, h
〉
dµ
≤ NH(t) +
∫
M
2|h|
∣∣∣∣∂h∂t
∣∣∣∣dµ
≤ NH(t) +N
∫
M
(|S||h|+ |h|2 + |q||h|)dµ.
18
Now we know that N
(|S||h|+ |q||h|) ≤ N(|h|2 + |S|2 + |q|2). Hence
H′(t) ≤ NH(t) +N
∫
M
(|S|2 + |q|2)dµ
≤ NH(t) +N
∫
M
(|S|2 + |h|2 + |A|2)dµ
≤ NH(t) +NS(t) +NA(t) = NE(t). (I.52)
Evolution of A(t)
By (I.14), (I.48) and (I.51) we have
A′(t) ≤ NA(t) +
∫
M
2|A|
∣∣∣∣∂A∂t
∣∣∣∣dµ
≤ NA(t) +
∫
M
(
N |h||A|+N |A|2 + C|∇S||A|+ C|∇q||A|
)
dµ.
Now
N |h||A|+ C|∇S||A|+ C|∇q||A| ≤ N |h|2 +N |A|2 + |∇S|2 + |∇q|2.
hence we have that
A′(t) ≤ NA(t) +
∫
M
(
N |h|2 +N |A|2 + |∇S|2 + |∇q|2
)
dµ
≤ NA(t) +NH(t) +D(t) +N
∫
M
(|h|2 + |A|2 + |S|2)dµ
≤ NA(t) +NH(t) +NS(t) +D(t) = NE(t) +D(t). (I.53)
19
Evolution of S(t)
By (I.15), (I.49) and (I.51) we have
S ′(t) ≤ N
∫
M
|S|2dµ+
∫
M
2
〈
∂S
∂t
, S
〉
dµ
≤ NS(t) +
∫
M
(
2
〈
∆S + div V, S
〉
+N
(|h|+ |A|+ |S|+ |q|)|S|+ C|∇∇q||S|)dµ
≤ NS(t) +
∫
M
(
2
〈
∆S + div V, S
〉
+N
(|h|2 + |A|2 + |S|2 + |q|2 + |∇∇q|2))dµ.
Now by (I.33) and (I.35) we have
S ′(t) ≤ NS(t) +NH(t) +NA(t)
+
∫
M
(
2
〈
∆S + div V, S
〉
+N
(|A|2 + |S|2 + |h|2))dµ
≤ NS(t) +NH(t) +NA(t) +
∫
M
2
〈
∆S + div V, S
〉
dµ.
Upon integrating by parts we get
S ′(t) ≤ NE(t)− 2
∫
M
〈∇S + V,∇S〉dµ
≤ NE(t)− 2
∫
M
|∇S|2dµ+
∫
M
2|V ||∇S|dµ.
Now we know that
2|V ||∇S| ≤ |∇S|2 + |V |2 ≤ |∇S|2 +N(|h|2 + |A|2),
hence
S ′(t) ≤ NE(t) +N
∫
M
(|h|2 + |A|2)dµ− ∫
M
|∇S|2dµ ≤ NE(t)−D(t). (I.54)
20
Proof of Main Theorem
Now we are ready to prove Theorem 1:
Proof. By (I.54), (I.52) and (I.53) we know that
H′(t) ≤ NE(t), A′(t) ≤ NE(t) +D(t), S ′(t) ≤ NE(t)−D(t),
so
E ′(t) ≤ NE(t).
Our initial condition g˜(0) = g(0) tells us that at t = 0 we have |h| = |A| =
|S| = 0. Therefore by the smoothness and integrability of our solutions we know
lim
t→0+
E(t) = 0,
so by Gronwall’s Inequality we know that E ≡ 0 on [0, T ]. Thus for t ∈ [0, T ] we
have that h ≡ 0 and g(t) ≡ g˜(t). Also, E ≡ 0 implies A ≡ 0 and S ≡ 0, so (I.33)
forces q ≡ 0. Thus p(t) ≡ p˜(t). Therefore (g˜(t), p˜(t)) = (g(t), p(t)), t ∈ [0, T ].
21
CHAPTER II
BACKWARD RICCI FLOW OF HOMOGENEOUS 4-GEOMETRIES
Ricci flow on a manifold, (M, g0) is an evolution equation on the metric tensor
given by
∂g
∂t
= −2Ricg(t). (II.1)
This equation was first introduced by Richard Hamilton in [9], where he
demonstrated that one can always expect short time existence as long as (M, g0) is
a smooth manifold. Backward Ricci flow is described by
∂g
∂t
= 2Ricg(t). (II.2)
In general, we cannot expect short time existence of solutions to this equation.
However, in the case of locally homogeneous manifolds, Ricci flow reduces to a
system of Ordinary Differential Equations. As mentioned in [4], this eliminates all
barriers to backward short-time existence of Ricci flow in these manifolds.
In [10], Isenberg and Jackson studied the behavior of solutions to Ricci flow
along locally homogeneous 3-manifolds. Later, in [11], Isenberg Jackson and Lu
studied Ricci flow along locally homogeneous 4-manifolds which admit compact
22
quotients. Subsequently, in [4], Cao and Saloff-Coste used calculations from [6]
and [10] to study backward Ricci flow on the homogeneous 3-manifolds.
In this chapter we use many of the calculations in [11] to examine backward
Ricci flow of compact locally homogeneous geometries on 4-manifolds. The analysis
in this chapter is often very similar to that in [11], and we will use many of the
same calculations, some of which are included for completeness.
The classes of locally homogeneous manifolds are described in [11] and [14]. For
the geometries which are also Lie Groups we will choose a particular basis for the
Lie Algebra, {X1, X2, X3, X4}, that satisfies certain bracket relations. More detail
on these classes can be found in [14]. Letting {φi}4i=1 be the frame of 1-forms dual
to {Xi} we can form a metric g0 = Aijφi ⊗ φj.
Our solutions will represent diagonalized Riemannian metrics. Thus they only
exist as long as all unknowns remain positive and finite. We denote by T0 the
postiive time at which the various solutions to (II.2) fail to exist.
We divide this chapter into two main sections. First is the non-trivial section
where we describe the Bianchi cases. The Lie Group structure has trivial Isotopy
group, so the manifold is in itself a Lie Group. In the next section we give a
23
quick description of each of the 4-dimensional non-Bianchi cases. These are all
metrics of constant sectional or holomorphic bi-sectional curvature or products of
such metrics. The evolution of these metrics is well understood, but we include it
in this thesis for completeness. In the last section we summarize and compare the
behaviors under backward Ricci flow of the various geometries under.
The Bianchi Cases
The reduction of Ricci flow, equation (II.1), on the Bianchi classes of 4-
manifolds, to a system of ODE was done in [11] using the following Ricci curvature
formula for unimodular Lie groups from [2, p. 184]. Recall that the elemenets
X, Y, Z,W come from the Lie Algebra g of our Lie Group G which represents our
manifold.
Ric(W,W ) = −1
2
∑
i
∣∣[W,Yi]∣∣2 − 1
2
∑
i
〈[
W, [W,Yi]
]
, Yi
〉
+
1
2
∑
i<j
〈
[Yi, Yj],W
〉2
.
(II.3)
The only difference in calculating the system of ODE for backwards Ricci flow,
equation (II.2), is that the evolutions of the various metrics are negative of
those found in [11]. In this section I use these systems of ODE without further
explanation.
A formula to calculate the sectional curvature on Lie Groups, with corresponding
24
Lie Algebras, is found in [2, p. 183]:
〈
R(X, Y )X, Y
〉
= −3
4
∣∣[X, Y ]∣∣2 − 1
2
〈[
X, [X, Y ]
]
, Y
〉− 1
2
〈[
Y, [Y,X]
]
, X
〉
+
∣∣U(X, Y )∣∣2 − 〈U(X,X), U(Y, Y )〉, (II.4)
where U is defined by
〈
U(X, Y ), Z
〉
=
1
2
〈
[Z,X], Y
〉
+
1
2
〈
X, [Z, Y ]
〉
for all Z ∈ g. (II.5)
The classification notation I use comes from [14] and [11]. See [14] for more
details on this classification.
A1. Class U1b(1, 1, 1)c
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = 0
[X2, X3] = 0 [X2, X4] = 0 [X3, X4] = 0.
Our Lie Group structure is (M,G) = (R4,R4). The metric g is flat, hence also
Ricci flat and thus remains constant.
A2. Class U1b1, 1, 1c
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
25
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = X1
[X2, X3] = 0 [X2, X4] = kX2 [X3, X4] = −(k + 1)X3.
When k = 0 this corresponds to the geometry (M,G) =
(
˜Sol3 × R, ˜Sol3 × R).
When k = 1 this corresponds to the geometry (M,G) =
(
Sol40, Sol
4
0
)
.
If k 6= 0, 1 and there is some number α > 0 with β = kα, γ = −(k+ 1)α such that
eα, eβ and eγ are roots of λ3 − mλ2 + nλ − 1 = 0, then this corresponds to the
geometry (M,G) =
(
Sol4m,n, Sol
4
m,n
)
.
We diagonalize our initial metric g0 by letting Yi = Upsilon
k
iXk for constants
Υki . Now letting {θi} be the frame of 1-forms dual to {Yi} we may write the metric
as
g0 = λ1θ
2
1 + λ2θ
2
2 + λ3θ
2
3 + λ4θ
2
4.
In [11] we find outlines as to when exactly the Ricci tensor is also diagonal
under these same coordinates. This is exactly when the metric will remain diagonal
under Ricci flow and also backward Ricci flow. The property of a metric to remain
diagonal under this flow is essential to our calculations, and we will only consider
those families which satisfy this property. In each remaining subsection we will
describe a flow which is of the form
g(t) = A(t)θ21 +B(t)θ
2
2 + C(t)θ
2
3 +D(t)θ
2
4,
26
where
A(0) = λ1, B(0) = λ2, C(0) = λ3, and D(0) = λ4.
The systems of ODE governing the evolution of the quantities A, B, C and D
were calculated in [11].
In the class U1[(1, 1, 1)] we may diagonalize the metric by letting Yi = Υ
k
iXk
with
Υ =

1 0 0 0
a1 1 0 0
a2 a3 1 0
a4 a5 a6 1
 .
By Proposition 1 in [11], if k = 1 then the metric g(t) remains diagonal in the
basis Yi if and only if a1 = a2 = a3 = 0, and if k 6= 1 then g(t) remains diagonal in
the basis Yi if and only if a2 = a3 = 0. In either of these cases we find that {Yi}
satisfies the following bracket relations:
[Y1, Y2] = 0 [Y1, Y3] = 0 [Y1, Y4] = Y1
[Y2, Y3] = 0 [Y2, Y4] = kY2 [Y3, Y4] = −(k + 1)Y3.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow (II.2) reduces to the following system of
equations by [11]:
dA
dt
=
dB
dt
=
dC
dt
= 0,
dD
dt
= −4(k2 + k + 1). (II.6)
27
The solution is
A(t) = λ1
B(t) = λ2
C(t) = λ3
D(t) = λ4 − 4
(
k2 + k + 1
)
t.
(II.7)
It is clear that
T0 =
λ4
4
(
k2 + k + 1
)−1
, (II.8)
and that as t→ T0 the volume normalized flow will approach the hyperplane, R3.
The sectional curvatures, also calculated in [11], are as follows:
K(Y1, Y2) = − k
D
K(Y1, Y3) =
k + 1
D
K(Y1, Y4) =
k(k + 1)
D
K(Y2, Y3) = − 1
D
K(Y2, Y4) = −k
2
D
K(Y3, Y4) = −(k + 1)
2
D
.
Thus, as D approaches 0 linearly in t we see that the non-zero curvatures approach
infinity with a singularity of the form (T0 − t)−1.
A3. Class U1bZ, Z¯, 1c
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = kX1 +X2
[X2, X3] = 0 [X2, X4] = −X1 + kX2c [X3, X4] = −2kX3.
28
This corresponds to the geometry (M,G) = (R4, E(2) × R2). Here we diagonalize
the metric by letting Yi = Υ
k
iXk with
Υ =

1 a2 a3 0
0 1 a1 0
0 0 1 0
a4 a5 a6 1
 .
By Proposition 2 in [11], the metric g(t) remains diagonal in the basis Yi if and
only if a1 = a2 = a3 = 0. We find that {Yi} satisfies the following bracket relations:
[Y1, Y2] = 0 [Y1, Y3] = 0 [Y1, Y4] = kY1 + Y2
[Y2, Y3] = 0 [Y2, Y4] = −Y1 + kY2 [Y3, Y4] = −2kY3.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
equations:
dA
dt
=
A2 −B2
BD
dB
dt
=
B2 − A2
AD
dC
dt
= 0
dD
dt
= −(A−B)
2 + 12k2AB
AB
.
(II.9)
Clearly, C(t) = λ3. Also,
d
dt
(
A−B) = (A+B)2
ABD
(A−B),
so the conditions A = B, A > B and A < B are all preserved.
29
First assume λ1 = λ2. Then A = B for all time t, and our solution is
A(t) = λ1
B(t) = λ1
C(t) = λ3
D(t) = λ4 − 12k2t.
(II.10)
Thus
T0 =
λ4
12k2
, (II.11)
and it is clear that as t → T0, the volume normalized flow will approach the
hyperplane, R3.
Now assume λ1 6= λ2. By the symmetry of (II.9), we may assume that λ1 > λ2.
Then A(t) ≥ B(t) for all time t. Now,
d
dt
(
AB
)
= A
B2 − A2
AD
+B
A2 −B2
BD
= 0,
so AB = λ1λ2 is constant. It is also clear that T0 < ∞, because dD
dt
< −12k2
implies T0 <
λ4
12k2
. With AB = λ1λ2, we see that
dA
dt
=
A4 − λ21λ22
λ1λ2AD
,
dD
dt
= −
(
A−B)2
λ1λ2
− 12k2,
hence
dD
dA
= −
(
A−B)2 + 12λ1λ2k2
λ1λ2
· λ1λ2AD
A4 − λ21λ22
,
30
so
1
D
dD
dA
= −
[(
A− λ1λ2
A
)2
+ 12λ1λ2k
2
]
A
A4 − λ21λ22
=
−A4 + (2− 12k2)λ1λ2A2 − λ21λ22
A(A+
√
λ1λ2)(A−
√
λ1λ2)(A2 + λ1λ2)
. (II.12)
Under the substitution A˜ =
√
λ1λ2A, we have the following:
1
D
dD
dA˜
=
−A˜4 + (2− 12k2)A˜2 − 1
A˜(A˜+ 1)(A˜− 1)(A˜2 + 1)
=
1
A˜
− 3k
2
A˜+ 1
− 3k
2
A˜− 1 −
2(1− 3k2)A˜
A˜2 + 1
.
Solving this equation we find
D = Λ
(
A˜
A˜2 + 1
)(
A˜2 + 1
A˜2 − 1
)3k2
= Λ
( √
λ1λ2A
A2 + λ1λ2
)(
A2 + λ1λ2
A2 − λ1λ2
)3k2
, (II.13)
where
Λ = λ4
(
λ1 + λ2√
λ1λ2
)(
λ1 − λ2
λ1 + λ2
)3k2
. (II.14)
We see that D → 0 as t → T0 if and only if A → ∞ as t → T0. But since
AB is constant, we know that B → 0 if and only if A → ∞. Thus we know that
D → 0, B → 0 and A→∞ at the same time, T0 < λ4
12k2
<∞.
As A→∞, we see that
D ≈ Λ
√
λ1λ2
A
=
Λ√
λ1λ2
B. (II.15)
31
Now, as t→ T0, we may approximate
dA
dt
=
A2 −B2
BD
∼ A
4
Λ(λ1λ2)3/2
. (II.16)
Thus
d
dt
(
1
A3
)
→ − 3
Λ(λ1λ2)3/2
as t→ T0. Since A→∞ as t→ T0, we see that
1
A3
=
3
Λ(λ1λ2)3/2
(T0 − t)
(
1 + o(T0 − t)
)
.
Thus
A =
(
3
Λ(λ1λ2)3/2
(T0 − t)
)−1/3 (
1 + o(T0 − t)
)−1/3
=
√
λ1λ2
(
3
Λ
(T0 − t)
)1/3 (
1 + o(T0 − t)
)
Now by (II.15) we have the following behavior as t→ T0:
A ≈
√
λ1λ2
(
3
Λ
(
T0 − t
))−1/3
B ≈
√
λ1λ2
(
3
Λ
(
T0 − t
))1/3
C = λ3
D ≈ Λ
(
3
Λ
(
T0 − t
))1/3
.
(II.17)
where Λ is given by (II.14).
The volume normalized solution will converge to the plane R2.
32
The sectional curvatures are as follows:
K(Y1, Y2) = −
A
B
+ B
A
− 2− 4k2
4D
K(Y1, Y3) =
2k2
D
K(Y1, Y4) =
A
B
− 3B
A
+ 2− 4k2
D
K(Y2, Y3) =
2k2
D
K(Y2, Y4) =
−3A
B
+ B
A
+ 2− 4k2
D
K(Y3, Y4) = −4k
2
D
.
Thus we see that the sectional curvatures perpendicular to Y3 approach infinity
at a rate of (T0 − t)−1. If k = 0 the sectional curvatures parallel to Y3 remain 0,
while if k 6= 0 then these curvatures approach infinity at a rate of (T0 − t)−1/3.
A4. Class U1b2, 1c, µ = 0
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = X2
[X2, X3] = 0 [X2, X4] = 0 [X3, X4] = 0.
This corresponds to the geometry (M,G) = (Nil3 × R, Nil3 × R). Here we
diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 a2 a3 0
0 1 0 0
0 a1 1 0
a4 a5 a6 1
 .
By Proposition 3 in [11], the metric g(t) remains diagonal in the basis Yi for all t.
33
We find that {Yi} satisfies the following bracket relations:
[Y1, Y2] = 0 [Y1, Y3] = 0 [Y1, Y4] = Y2
[Y2, Y3] = 0 [Y2, Y4] = −Y1 + kY2 [Y3, Y4] = −2kY3.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
equations:
dA
dt
= −B
D
dB
dt
=
B2
AD
dC
dt
= 0
dD
dt
= −B
A
.
(II.18)
We calculate
d
dt
(AB) = −BB
D
+ A
B2
AD
= 0
and
d
dt
(
A
D
)
=
1
D2
(
−DB
D
+ A
B
A
)
= 0.
Thus we have
B =
λ1λ2
A
, and D =
λ4
λ1
A, (II.19)
so
A2
dA
dt
= −ABA
D
= −λ
2
1λ2
λ4
, (II.20)
hence
A3 = λ31 −
3λ21λ2
λ4
t. (II.21)
34
Thus by (II.18), (II.19) and (II.21), we have the following solution to (II.18):
A = λ1
(
1− 3λ2
λ1λ4
t
)1/3
B = λ2
(
1− 3λ2
λ1λ4
t
)−1/3
C = λ3
D = λ4
(
1− 3λ2
λ1λ4
t
)1/3
.
(II.22)
We see that
T0 =
λ1λ4
3λ2
, (II.23)
and that the volume normalized solution will converge to the plane R2.
The sectional curvatures are as follows:
K(Y1, Y2) =
B
4AD
K(Y1, Y3) = 0 K(Y1, Y4) = − 3B
4AD
K(Y2, Y3) = 0 K(Y2, Y4) =
B
4AD
K(Y3, Y4) = 0.
Thus all non-zero curvatures will approach infinity near t = T0 at a rate of (T0−t)−1.
A5. Class U1b2, 1c, µ = 1
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = −1
2
X1 +X2
[X2, X3] = 0 [X2, X4] = −1
2
X2 [X3, X4] = X3.
35
This does not correspond to any of the compact homogeneous geometries. Here
we diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 a2 a3 0
0 1 a1 0
0 0 1 0
a4 a5 a6 1
 .
By Proposition 4 in [11], the metric g(t) remains diagonal in the basis Yi if and
only if a1 = a3 = 0. We find that {Yi} satisfies the following bracket relations:
[Y1, Y2] = 0 [Y1, Y3] = 0 [Y1, Y4] = −1
2
Y1 + Y2
[Y2, Y3] = 0 [Y2, Y4] = −1
2
Y2 [Y3, Y4] = Y3.
Let g(t) = A(t)θ21 + B(t)θ
2
2 + C(t)θ
2
3 + D(t)θ
2
4 where A(0) = λ1, B(0) =
λ2, C(0) = λ3 and D(0) = λ4. Then backward Ricci flow reduces to the following
system of equations:
dA
dt
= −B
D
dB
dt
=
B2
AD
dC
dt
= 0
dD
dt
= −3− B
A
.
(II.24)
Since
dD
dt
< −3 we see that there is a maximal time T0 < λ4
3
. Also,
d
dt
(AB) = A
(
B2
AD
)
+B
(
−B
D
)
= 0, (II.25)
so
AB = λ1λ2. (II.26)
36
Thus (II.24) reduces to
dA
dt
= −λ1λ2
AD
B =
λ2
λ1
A
C = λ3
dD
dt
=
−3A2 − λ1λ2
A2
.
(II.27)
Now we calculate
dD
dA
=
3A2 + λ1λ2
A2
· AD
λ1λ2
,
1
D
· dD
dA
=
3
λ1λ2
A+
1
A
.
Solving gives us
D = Λ · Ae
(
3A2
2λ1λ2
)
, (II.28)
where
Λ =
λ4
λ1
e
(
− 3λ1
2λ2
)
. (II.29)
By (II.26) and (II.28) we see that A → 0, B → ∞ and D → 0 all together as
t→ T0.
To describe the behavior near t = T0 we observe that as A approaches 0, (II.24),
(II.26) and (II.28) tell us
A2
dA
dt
→ −λ1λ2
Λ
,
37
hence
A3 =
3λ1λ2
Λ
(T0 − t)
(
1 + o(T0 − t)
)
. (II.30)
Thus, by (II.24), (II.26) and (II.28), we have the following solutions to (II.24):
A ≈
(
3λ1λ2
Λ
(
T0 − t
))1/3
B ≈ λ1λ2
(
3λ1λ2
Λ
(
T0 − t
))−1/3
C = λ3
D ≈ Λ
(
3λ1λ2
Λ
(
T0 − t
))1/3
.
(II.31)
where Λ is given in (II.29). Again we notice that as t approaches T0 the renormalized
flow approaches the plane R2.
The sectional curvatures are as follows:
K(Y1, Y2) =
−1 + B
A
4D
K(Y1, Y3) =
1
2D
K(Y1, Y4) = −
1 + 3B
A
4D
K(Y2, Y3) =
1
2D
K(Y2, Y4) =
−1 + B
A
4D
K(Y3, Y4) = − 1
D
.
Thus curvatures perpendicular to Y3 will have a singularity t = T0 of the form
(T0− t)−1, while those parallel to Y3 will have a singularity of the form (T0− t)−1/3.
A6. Class U1b3c
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
38
Lie bracket is of the form
[X1, X2] = 0 [X1, X3] = 0 [X1, X4] = X2
[X2, X3] = 0 [X2, X4] = X3 [X3, X4] = 0.
This corresponds to the geometry (M,G) = (Nil4, Nil4).
Here we diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 a2 a3 0
0 1 0 0
0 a1 1 0
a4 a5 a6 1
 .
By Proposition 5 in [11], the metric g(t) remains diagonal in the basis Yi if and only
if a1 = a2. We find in this case that {Yi} satisfies the following bracket relations:
[Y1, Y2] = 0 [Y1, Y3] = 0 [Y1, Y4] = Y2
[Y2, Y3] = 0 [Y2, Y4] = Y3 [Y3, Y4] = 0.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
equations:
dA
dt
= −B
D
dB
dt
=
B2 − AC
AD
dC
dt
=
C2
BD
dD
dt
= −B
A
− C
B
.
(II.32)
39
We calculate
d
dt
(
ABC
)
= AB · C
2
BD
+ AC · B
2 − AC
AD
−BC · B
D
= 0
and
d
dt
(
A
CD
)
=
−CD · B
D
+ AC
(
B
A
+ C
B
)− AD · C2
BD
C2D2
= 0,
so we have
ABC = λ1λ2λ3, (II.33)
A
CD
=
λ1
λ3λ4
. (II.34)
Now we define
E =
B
AD
F =
C
BD
. (II.35)
Then
dE
dt
=
AD · B2−AC
AD
+ AB
(
B
A
+ C
B
)
+BD · B
D
(AD)2
=
3B2
(AD)2
= 3E2, (II.36)
and
dF
dt
=
BD · C2
BD
+BC
(
B
A
+ C
B
)− CD · B2−AC
AD
(BD)2
=
3C2
(BD)2
= 3F 2. (II.37)
Solving (II.36) and (II.37) gives us
E(t) =
E(0)
1− 3E(0)t (II.38)
40
and
F (t) =
F (0)
1− 3F (0)t . (II.39)
Now by (II.32), (II.35) and (II.38) we have
1
A
dA
dt
= −E(t) = E(0)
3E(0)t− 1 ,
so
ln(A) =
1
3
ln
∣∣3E(0)t− 1∣∣+ ln(λ1)
and
A = λ1
(
1− 3E(0)t)1/3. (II.40)
Similarly, by (II.32), (II.35) and (II.39), we have
1
C
dC
dt
= F (t) =
F (0)
1− 3F (0)t , (II.41)
so
C = λ3
(
1− 3F (0)t)−1/3. (II.42)
Now using (II.33), (II.34) and (II.35) we have the following solution to (II.32):
A = λ1
(
1− 3λ2
λ3λ4
t
)1/3
B = λ2
(
1− 3λ2
λ3λ4
t
)−1/3(
1− 3λ3
λ2λ4
t
)1/3
C = λ3
(
1− 3λ3
λ2λ4
t
)−1/3
D = λ4
(
1− 3λ2
λ3λ4
t
)1/3(
1− 3λ3
λ2λ4
t
)1/3
.
(II.43)
41
where
T0 = min
{
λ3λ4
3λ2
,
λ2λ4
3λ3
}
. (II.44)
The sectional curvatures are as follows:
K(Y1, Y2) =
B
4AD
K(Y1, Y3) = 0 K(Y1, Y4) =
−3B
4AD
K(Y2, Y3) =
C
4BD
K(Y2, Y4) =
B
A
− 3C
B
4D
K(Y3, Y4) =
C
4BD
.
If λ2 < λ3, then T0 =
λ2λ4
3λ3
, and as t→ T0 we have
A ≈ k1
B ≈ k2(T0 − t)1/3
C = k3(T0 − t)−1/3
D ≈ k4(T0 − t)1/3.
(II.45)
Also, K(Y1, Y2) will approach a positive constant, K(Y1, Y4) will approach a
negative constant, and all curvatures perpendicular to Y1 will have singularities of
the form (T0 − t)−1.
If λ2 > λ3, then T0 =
λ3λ4
3λ2
, and as t→ T0 we have
A = k1(T0 − t)1/3
B ≈ k2(T0 − t)−1/3
C ≈ k3
D ≈ k4(T0 − t)1/3.
(II.46)
42
Here K(Y2, Y3) and K(Y3, Y4) will approach positive constants while each
curvature perpendicular to Y3 will have a singularity of the form (T0 − t)−1.
If λ2 = λ3, then T0 =
λ4
3
, and as t→ T0 we have
A = k1(T0 − t)1/3
B = λ2
C = k3(T0 − t)−1/3
D = k4(T0 − t)2/3.
(II.47)
Here all non-zero curvatures will have a singularity of the form (T0 − t)−1. It is
interesting to note that if 3λ1 = λ2 = λ3, then K(Y2, Y4) = 0 for all t, while in all
other cases it explodes near the singular time.
In all three cases the volume normalized solution approaches the plane R2.
A7. Class U3I0
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = −X3 [X1, X3] = −X2 [X1, X4] = 0
[X2, X3] = X4 [X2, X4] = 0 [X3, X4] = 0.
This corresponds to the geometry (M,G) = (Sol4, Sol4). Here we diagonalize the
43
metric by letting Yi = Υ
k
iXk with
Υ =

1 a4 a5 a6
0 1 a2 a3
0 0 1 a1
0 0 0 1
 .
Let α = a2, β = a1a2 −3 −a4 and γ = a1 − a1a22 + a2a3 + a2a4 − a5. Proposition
6 in [11], the metric g(t) remains diagonal in the basis Yi if and only if one of the
following hold:
(i) α = β = γ = 0
(ii) β = γ = 0 and λ2 = (1− α2)λ3
We analyze these cases separately.
A7(i)
Here we consider the case where α = β = γ = 0. We find that {Yi} satisfies the
following bracket relations:
[Y1, Y2] = −Y3 [Y1, Y3] = −Y2 [Y1, Y4] = 0
[Y2, Y3] = Y4 [Y2, Y4] = 0 [Y3, Y4] = 0.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
44
equations:
dA
dt
= −B
C
− C
B
− 2
dB
dt
= −C
A
− D
C
+
B2
AC
dC
dt
= −B
A
− D
B
+
C2
AB
dD
dt
=
D2
BC
.
(II.48)
By the symmetry of B and C in (II.48), we may assume that λ2 ≥ λ3. We
calculate
d
dt
(
BCD2
)
= 2BCD · D
2
BC
+ CD2
(
B2
AC
− C
A
− D
C
)
+BD2
(
C2
AB
− B
A
− D
B
)
= 0,
so
BCD2 = λ2λ3λ
2
4. (II.49)
Now we observe
dD
dt
=
D2
BC
=
D4
λ2λ3λ24
,
so
1
D4
· dD
dt
=
1
λ2λ3λ24
,
and
D = λ4
(
1− 3λ4
λ2λ3
t
)−1/3
. (II.50)
45
We may also calculate
d
dt
(
AD(B − C)) = AD( B2
AC
− C
A
− D
C
+
B
A
+
D
B
− C
2
AB
)
+ A(B − C) · D
2
BC
−D(B − C)
(
B
C
+
C
B
+ 2
)
= 0,
so
AD(B − C) = λ1λ4(λ2 − λ3). (II.51)
Now by (II.49) and (II.51) we have
(B − C)2
BC
A2 =
(
AD(B − C))2
BCD2
=
λ21(λ2 − λ3)2
λ2λ3
= 4k2. (II.52)
Now
dA
dt
= −(B + C)
2
BC
= −
(
(B − C)2
BC
A2
)
· 1
A2
− 4 = −4
(
k2 + A2
)
A2
, (II.53)
hence (
1− k
2
A2 + k2
)
dA
dt
= −4, (II.54)
so
A− k tan−1
(
A
k
)
= −4t+ λ1 − k tan−1
(
λ1
k
)
, (II.55)
where k is given in (II.52). Now by (II.50) and (II.55) we have
T0 = min
{
λ2λ3
3λ4
,
λ1
4
− λ1(λ2 − λ3)
2
√
λ2λ3
tan−1
(
2
√
λ2λ3
λ2 − λ3
)}
= min
{
T1, T2
}
. (II.56)
46
Using (II.49) and (II.51) we can calculate
B =
λ4
2AD
(
λ1(λ2 − λ3) +
√
λ21(λ2 − λ3)2 + 4A2λ2λ3
)
, (II.57)
and
C =
λ4
2AD
(
λ1(λ3 − λ2) +
√
λ21(λ2 − λ3)2 + 4A2λ2λ3
)
. (II.58)
Now we calculate
lim
A→0
(
A− k tan−1(A
k
)
A3
)
=
1
3k2
.
Thus by (II.55), near A = 0 we have
A ≈ (12k2(T2 − t))1/3, (II.59)
where k is given in (II.52).
The sectional curvatures are as follows:
K(Y1, Y2) =
B
C
− 3C
B
− 2
4A
K(Y1, Y3) =
C
B
− 3B
C
− 2
4A
K(Y1, Y4) = 0 K(Y2, Y3) =
B2 + C2 + 2BC − 3AD
4ABC
K(Y2, Y4) =
D
4BC
K(Y3, Y4) =
D
4BC
.
If T0 = T1 < T2, then we have the following behavior as t→ T0:
A ≈ k1
B ≈ k2(T0 − t)1/3
C ≈ k3(T0 − t)1/3
D = k4(T0 − t)−1/3.
(II.60)
47
In this case the curvatures parallel to Y1 will approach constants while those
perpendicular to Y1 will have a singularity of the form (T0 − t)−1.
If T0 = T2 < T1 and λ2 = λ3, then as t→ T0,
A ≈ k1(T0 − t)1/3
B = C ≈ k2
D ≈ k4.
(II.61)
Here the curvatures perpendicular to Y4 will approach infinity at a rate of
(T0 − t)−1/3 while those curvatures parallel to Y4 will approach constants.
If T0 = T2 < T1 and λ2 > λ3, then as t→ T0,
A ≈ k1(T0 − t)1/3
B ≈ k2(T0 − t)−1/3
C ≈ k3(T0 − t)1/3
D ≈ k4.
(II.62)
In this case the curvatures perpendicular to Y4 will approach infinity at a rate
of (T0 − t)−1 while those curvatures parallel to Y4 will approach constants.
48
If T0 = T1 = T2 and λ2 = λ3, then as t→ T0, AD → constant, so
A ≈ k1(T0 − t)1/3
B = C ≈ k2(T0 − t)1/3
D = k4(T0 − t)−1/3.
(II.63)
Here, K(Y1, Y2) and K(Y1, Y3) will have singularities at t = T0 of the form
(T0 − t)−1/3. All other non-zero sectional curvatures will have singularities of the
form (T0 − t)−1.
If T0 = T1 = T2 and λ2 > λ3, then as t→ T0, AD → constant, so
A ≈ k1(T0 − t)1/3
B ≈ k2
C ≈ k3(T0 − t)2/3
D = k4(T0 − t)−1/3.
(II.64)
Here, all non-zero curvatures will have singularities of the form (T0 − t)−1.
In the special case where both T0 = T2 < T1 and λ1 = λ2, the normalized
solution will approach the hyperplane R3. In all other cases the normalized solution
will approach R2.
49
A7(ii)
Here we consider the case where β = γ = 0 and λ2 = (1− α2)λ3. We find that
{Yi} satisfies the following bracket relations:
[Y1, Y2] = −αY2 [Y1, Y3] = αY3 − Y2 [Y1, Y4] = 0
[Y2, Y3] = Y4 [Y2, Y4] = 0 [Y3, Y4] = 0.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
equations:
dA
dt
= −B
2 + 2(1 + α2)BC + (1− α2)2C2
BC
dB
dt
=
−AD +B2 − (1− α2)2C2
AC
dC
dt
=
−AD −B2 + (1− α2)2C2
AB
dD
dt
= − D
2
BC
λ2 = (1− α2)λ3.
(II.65)
We observe
d
dt
(
B − (1− α2)C) = 0,
so B = (1− α2)C is preserved under (II.65), which then reduces to
dA
dt
= −4
dB
dt
= −(1− α2)D
B
B = (1− α2)C
dD
dt
= (1− α2)D
2
B2
.
(II.66)
50
Now
d
dt
(
BD
)
= 0,
so
BD = λ2λ4. (II.67)
Thus
dB
dt
= −(1− α2)λ2λ4
B2
,
so
1
3
B3 =
1
3
λ32 − (1− α2)λ2λ4t,
and we have the following solution to (II.65):
A = λ1 − 4t
B =
(
λ32 − 3(1− α2)λ2λ4t
)1/3
C =
1
1− α2
(
λ32 − 3(1− α2)λ2λ4t
)1/3
D = λ2λ4
(
λ32 − 3(1− α2)λ2λ4t
)−1/3
,
(II.68)
where
T0 = min
{
λ1
4
,
λ22
3(1− α2)λ4
}
= min
{
T1, T2
}
. (II.69)
The sectional curvatures are as follows:
K(Y1, Y2) = − 1
A
K(Y1, Y3) = − 1
A
K(Y1, Y4) = 0
K(Y2, Y3) =
4BC − 3AD
4ABC
K(Y2, Y4) =
D
4BC
K(Y3, Y4) =
D
4BC
.
51
If T0 = T1 < T2, then near t = T0 we have
A = 4(T0 − t)
B ≈ k2
C ≈ k3
D ≈ k4.
(II.70)
In this case all sectional curvatures perpendicular to Y4 will have a singularity
at T0 of the form (T0− t)−1 while those parallel to Y4 will approach constants. The
normalized solution will converge to R3.
If T0 = T2 < T1, then near t = T0 we have
A ≈ k1
B ≈ k2(T0 − t)1/3
C ≈ k3(T0 − t)1/3
D ≈ k4(T0 − t)−1/3.
(II.71)
Here all sectional curvatures perpendicular to Y1 will have a singularity at T0
of the form (T0 − t)−1 while those parallel to Y1 will approach constants. The
normalized solution will converge to R2.
52
If T0 = T1 = T2, then near t = T0 we have
A = 4(T0 − t)
B ≈ k2(T0 − t)1/3
C ≈ k3(T0 − t)1/3
D ≈ k4(T0 − t)−1/3.
(II.72)
Here all non-zero curvatures will have a singularity of the form (T0 − t)−1, and
the normalized solution will converge to a product metric M2 × R.
A8. Class U3I2
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = X3 [X1, X3] = −X2 [X1, X4] = 0
[X2, X3] = −X4 [X2, X4] = 0 [X3, X4] = 0.
This does not correspond to any of the compact homogeneous geometries.
Here we diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 a4 a5 a6
0 1 a2 a3
0 0 1 a1
0 0 0 1
 .
By Proposition 7 in [11], the metric g(t) remains diagonal in the basis Yi if and
only if a2 = 0, a1 = a5 and a3 = a4. We find in this case that {Yi} satisfies the
53
following bracket relations:
[Y1, Y2] = Y3 [Y1, Y3] = −Y2 [Y1, Y4] = 0
[Y2, Y3] = −Y4 [Y2, Y4] = 0 [Y3, Y4] = 0.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
equations:
dA
dt
= −(B − C)
2
BC
dB
dt
=
B2 − C2 − AD
AC
dC
dt
=
C2 −B2 − AD
AB
dD
dt
=
D2
BC
.
(II.73)
By the symmetry of B and C in (II.73) we may assume that λ2 ≥ λ3. Note also
that the equations for B,C and D are identical to those in (II.48), so by (II.49)
and (II.50) we have
BCD2 = λ2λ3λ
2
4 (II.74)
and
D = λ4
(
1− 3λ4
λ2λ3
t
)−1/3
. (II.75)
Similar calculations as those used to compute (II.51) show that
AD(B + C) = λ1λ4(λ2 + λ3). (II.76)
54
Using (II.73) and (II.76), we can solve for A:
dA
dt
=
−4(k2 + A2)
A2
,
and
A2
k2 − A2 ·
dA
dt
= −4,
where
λ21(λ2 + λ3)
2
λ2λ3
= 4k2. (II.77)
Now we integrate to find
k tanh−1
(
A
k
)
− A = −4t+ k tanh−1
(
λ1
k
)
− λ1, (II.78)
where k is given in (II.77). We then calculate
lim
A→0
k tanh−1
(
A
k
)− A
A3
=
1
3k2
, (II.79)
hence near t = T0 we have
A ≈ (12k2(T0 − t))1/3. (II.80)
Using (II.74) and (II.76) we find
B =
λ4
2AD
(
λ1(λ2 + λ3) +
√
λ21(λ2 + λ3)
2 − 4A2λ2λ3
)
, (II.81)
C =
λ4
2AD
(
λ1(λ2 + λ3)−
√
λ21(λ2 + λ3)
2 − 4A2λ2λ3
)
, (II.82)
and
T0 = min
{
λ2λ3
3λ4
,
1
4
(
k tanh−1
(
λ1
k
)
− λ1
)}
. (II.83)
55
The sectional curvatures are as follows:
K(Y1, Y2) =
B
C
− 3C
B
+ 2
4A
K(Y1, Y3) =
C
B
− 3B
C
+ 2
4A
K(Y1, Y4) = 0 K(Y2, Y3) =
B2 + C2 − 2BC − 3AD
4ABC
K(Y2, Y4) =
D
4BC
K(Y3, Y4) =
D
4BC
.
If T0 = T1 < T2, then we have the following behavior as t→ T0:
A ≈ k1
B ≈ k2(T0 − t)1/3
C ≈ k3(T0 − t)1/3
D = k4(T0 − t)−1/3.
(II.84)
In this case the curvatures parallel to Y1 will approach constants while those
perpendicular to Y1 will have a singularity of the form (T0 − t)−1.
If T0 = T2 < T1, then as t→ T0,
A ≈ k1(T0 − t)1/3
B ≈ k2(T0 − t)−1/3
C ≈ k3(T0 − t)1/3
D ≈ k4.
(II.85)
In this case the curvatures perpendicular to Y4 will approach infinity at a rate of
(T0 − t)−1 while those curvatures parallel to Y4 will approach constants.
56
If T0 = T1 = T2, then as t→ T0, AD → constant, so
A ≈ k1(T0 − t)1/3
B ≈ k2
C ≈ k3(T0 − t)2/3
D = k4(T0 − t)−1/3.
(II.86)
Here, all non-zero curvatures will have singularities of the form (T0 − t)−1.
A9. Class U3S1
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = −X3 [X1, X3] = −X2 [X1, X4] = 0
[X2, X3] = X1 [X2, X4] = 0 [X3, X4] = 0.
This Lie Algebra structure is a direct sum sl2⊕R, and the Lie Group structure
structure is (M,G) = (SˆL(2,R)× R, SˆL(2,R)× R).
Here we diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 0 0 0
0 1 0 0
0 0 1 0
a1 a2 a3 1
 .
Remark II.1. There are initial metrics in this class which cannot be diagonalized
57
so easy. However, the situations presented in these cases provide us with very
complicated situations which we will not address here. See [11, p. 376-377].
Here we calculate the curvatures of this diagonalized metric in general form
so that we can use it to the specific cases outlined below. First we calculate the
operator U using (II.5):
U(Yi, Yi) = 0 for all i
U(Y1, Y2) =
B − A
2C
Y3 +
a3(B − A)
2D
Y4
U(Y1, Y3) =
A+ C
2B
Y2 +
A2(A+ C)
2D
Y4
U(Y1, Y4) =
a3A
2B
Y2 − a2A
2C
Y3
U(Y2, Y3) = −B + C
2A
Y1 − a1(B + C)
2D
Y4
U(Y2, Y4) = −a3B
2A
Y1 +
a1B
2C
Y3
U(Y3, Y4) = −a2C
2A
Y1 +
a1C
2B
Y2.
58
The sectional curvatures may then be calculated using (II.4):
K(Yi, Yi) = 0 for all i
K(Y1, Y2) =
1
4AB
[
−3C − 2B − 2A+ (A−B)2
(
1
C
+
a23
D
)]
K(Y1, Y3) =
1
4AC
[
−3B − 2C + 2A+ (A+ C)2
(
1
B
+
a22
D
)]
K(Y1, Y4) =
1
4AD
[
−3(a23B + a22C) + 2A(a23 − a22) + A2
(
a23
B
+
a22
C
)]
K(Y2, Y3) =
1
4BC
[
−3A− 2C + 2B + (B + C)2
(
1
A
+
a21
D
)]
K(Y2, Y4) =
1
4BD
[
−3(a23A+ a21C) + 2B(a23 − a21) +B2
(
a23
A
+
a21
C
)]
K(Y3, Y4) =
1
4CD
[
−3(a22A+ a21B)− 2C(a21 + a22) + C2
(
a22
A
+
a21
B
)]
.
(II.87)
Proposition 8 in [11] tells us that when g0 can in fact be diagonalized as above:
(i) If λ1 6= λ2, the metric remains diagonal if and only if a1 = a2 = a3 = 0
(ii) If λ1 = λ2, the metric remains diagonal if and only if a1 = a2 = 0.
A9(i)
Here we consider the case where λ1 6= λ2 and a1 = a2 = a3 = 0. In this case,
Υ = I, and the Lie Bracket relations remain the same. It also happens that this
metric is just the product metric S˜L(2,R)×R, so backwards Ricci flow reduces to
the case of the three dimensional flow on S˜L(2,R). The volume-normalized version
of this flow has been discussed in [4].
59
In this case the non-zero curvatures are given by
K(Y1, Y2) =
1
4ABC
[−3C2 − 2BC − 2AC + A2 − 2AB +B2]
K(Y1, Y3) =
1
4ABC
[−3B2 − 2BC + 2AB + A2 + 2AC + C2]
K(Y2, Y3) =
1
4ABC
[−3A2 − 2AC + 2AB +B2 + 2BC + C2] .
(II.88)
Let Yi = Xi, and let g(t) = A(t)θ
2
1 + B(t)θ
2
2 + C(t)θ
2
3 + D(t)θ
2
4 where A(0) =
λ1, B(0) = λ2, C(0) = λ3 and D(0) = λ4. Then backward Ricci flow reduces to
the following system of equations:
dA
dt
=
A2 − (B + C)2
BC
dB
dt
=
B2 − (A+ C)2
AC
dC
dt
=
C2 − (A−B)2
AB
dD
dt
= 0.
(II.89)
By the symmetry of the system we may assume that λ1 > λ2. We also have
d
dt
(A−B) = 2
ABC
(A−B)(A+B + C),
so A > B is preserved by (II.89).
We set
Q =
{
(λ1, λ2, λ3) ∈ R3
∣∣a ≥ b > 0, c > 0}. (II.90)
The result is that there exists a partition of Q into Q1, Q2, S0, with S0 a
hypersurface in R3 and Q1, Q2 open and connected such that:
60
If (λ1, λ2, λ3) ∈ Q1, then
A(t) ≈ k1(T0 − t)1/3
B(t) ≈ k2(T0 − t)1/3
C(t) ≈ k3(T0 − t)−1/3
D(t) = λ4.
(II.91)
If (λ1, λ2, λ3) ∈ Q2, then
A(t) ≈ k1(T0 − t)−1/3
B(t) ≈ k2(T0 − t)1/3
C(t) ≈ k3(T0 − t)1/3
D(t) = λ4.
(II.92)
If (λ1, λ2, λ3) ∈ S0, then
A(t) ≈ k1
B(t) ≈ 4(T0 − t)
C(t) ≈ k1
D(t) = λ4.
(II.93)
In all three cases all the non-zero curvatures have a singularity at T0 of the
form (T0 − t)−1.
61
We first define
Q1 =
{
(λ1, λ2, λ3) ∈ Q
∣∣C(t1) ≥ A(t1) for some time t1 ≥ 0} (II.94)
Q2 =
{
(λ1, λ2, λ3) ∈ Q
∣∣C(t1) ≤ A(t1)−B(t1) for some time t1 ≥ 0} (II.95)
S0 =
{
(λ1, λ2, λ3) ∈ Q
∣∣A(t)−B(t) ≤ C(t) ≤ A(t) for all time t ≥ 0}. (II.96)
Now, if C(t) ≥ A(t), then
d
dt
(
C − A) = 1
ABC
[
(C − A)((C + A)2 −B2)+ 4ABC] > 4,
so C ≥ A is preserved.
Similarly, if C(t) ≤ A(t)−B(t), then
d
dt
(
A−B−C) = 1
ABC
[
(A−B−C)(A(A+ 2B+ 2C) + (B−C)2)+ 8ABC] > 8,
so C ≤ A−B is preserved.
Thus Q1, Q2 and S0 are mutually exclusive sets whose union is all of Q. The
facts that Q1 and Q2 are open and S0 is a hypersurface in Q are shown in [3].
These results are presented in [4] for the normalized Ricci flow, so we give an
un-normalized version here to better describe the behavior.
We first consider the set Q1, given by (II.94). Then C(t) ≥ A(t) for all t ≥ t1,
so we have
dA
dt
=
A2 − (B + C)2
BC
≤ −B − 2C
C
< −2,
62
hence A is decreasing, and T0 < ∞. Since B < A < C, either lim
t→T0
C(t) = ∞ or
lim
t→T0
B(t) = 0. We show that in fact both of these situations happen at the same
time. Since A and B are decreasing, then for C large enough we have
d
dt
(AC) = 2(A+B − C) < 0. (II.97)
Thus AC is bounded above. Thus if C → ∞ at T0, then it must be the case
that A→ 0 at T0, hence also B → 0 since B < A. Now we observe
d
dt
(
B(C − A)) = 4B > 0, (II.98)
so
(
B(C − A)) is bounded below. Thus if B → 0 at T0, then C − A → ∞, hence
also C → ∞. Thus we see that A → 0, B → 0 and C → ∞ all together at time
t = T0. Now,
d
dt
(AC − AB) = 4A, (II.99)
which is positive and approaches 0 at t = T0 < ∞. Therefore we know that
AC − AB → kA > 0 as t→ T0. Since AB → 0, we see that
AC → kA. (II.100)
Similarly,
d
dt
(BC − AB) = 4B → 0 (II.101)
63
implies that as t→ T0,
BC → kB > 0. (II.102)
Now by (II.89) we see that
dC
dt
∼ C
2
AB
∼ kC4, (II.103)
hence by the same argument used to solve (II.16), we see
C ≈ k3(T0 − t)−1/3. (II.104)
Now combining (II.100), (II.102) and (II.104), we have the solution to (II.89):
A(t) ≈ k1(T0 − t)1/3
B(t) ≈ k2(T0 − t)1/3
C(t) ≈ k3(T0 − t)−1/3
D(t) = λ4.
(II.105)
The normalized solution will approach the plane R2.
Now we consider the set Q2 given by (II.95). Here there is some time t1 such
that for t ≥ t1 ≥ 0, (A − B) < C. We denote A1 = A(t1), B1 = B(t1) and
C1 = C(t1). We see that for t ≥ t1
dC
dt
=
C2 − (A+B)2
AB
< 0,
and
dA
dt
=
A2 − (B − C)2
BC
> 0,
64
so C is decreasing and A is increasing. Thus, as t→ T0, either A→∞, B → 0 or
C → 0. We calculate
d
dt
(
A(B + C)
)
= −4B − 4C < 0, (II.106)
so
(
A(B + C)
)
is bounded above. Thus if A → ∞ as t → T0, then we know
(B + C)→ 0 at t = T0, hence B → 0 and C → 0. Now for t ≥ t1,
d
dt
(
B
C
)
=
2
A
(B + C)(B − C − A) < 0, (II.107)
so
B
C
is bounded above and if C → 0 then B → 0. Lastly,
d
dt
(
B(A−B − C)) = 2B
AC
(A2 + C2 −B2) > 0, (II.108)
so
(
B(A − B − C) is bounded below, and if B → 0 then (A − B − C) → ∞,
hence also A→∞. Thus we see that at t = T0 we have A→∞, B → 0 and C → 0.
Now by (II.108) we see that (AB − BC − B2) is increasing for t ≥ t1.Thus we
see that AB > B1(A1 −B1 − C1). However,
d
dt
(AB) = −2(A+B + C) < 0, (II.109)
so AB is decreasing and bounded below by a positive number. Thus at t→ T0, we
have
AB → kAB (II.110)
for some positive constant kAB. Now
d
dt
(AC) = 2(A−B − C) > 0, (II.111)
65
so (AC) is increasing and bounded below. Thus, since B and C are approaching
0 and A is approaching infinity at T0, then by (II.89) and (II.110) there is some
positive number k such that
dA
dt
=
A2 − (B + C)2
BC
≤ kA4, (II.112)
hence we know
A <
(
A−31 − 3kt
)−1/3
. (II.113)
By (II.111) we see that
d
dt
(AC) < A <
(
A−31 − 3kt
)−1/3
. (II.114)
Upon integrating we discover
AC < A1C1 +
1
2A21
− 1
2k
(
A−31 − 3kt
)−1/3
<∞. (II.115)
Since AC is increasing, then we see that at t→ T0, we have
AC → kAC (II.116)
for some positive constant kAC . Now by (II.89), (II.110) and (II.116),
A ∼ kA4. (II.117)
Solving (II.117) the same way we solved (II.16), and then using (II.110) and
66
(II.116), gives the following end behavior of the solutions to (II.89):
A(t) ≈ k1(T0 − t)−1/3
B(t) ≈ k2(T0 − t)1/3
C(t) ≈ k3(T0 − t)1/3
D(t) = λ4.
(II.118)
The normalized solution will approach the plane R2.
Now we consider the set S0 given by (II.96). Here we know
A(t)−B(t) < C(t) < A(t)
for all time 0 ≤ t < T0. Thus we have
dA
dt
=
A2 − (B + C)2
BC
< 0,
and
dC
dt
=
C2 − (A−B)2
AB
> 0,
so A is decreasing while C is increasing.
However, A(t) > C(t), so both A and C are approaching constants. Now
dB
dt
=
B2 − (A+ C)2
AC
< −2,
so B approaches 0 in finite time T0 < ∞. Since A − B < C < A, then in fact A
and C approach the same constant, k1, at time t = T0. Thus near t = T0 we have
67
the approximation:
dB
dt
≈ −(A+ C)
2
AC
≈ −4. (II.119)
Thus near t = T0 we have the following behavior:
A ≈ k1
B ≈ −4(t− T0)
C ≈ k1
D = λ4.
(II.120)
The normalized solution will approach R3.
A9(ii)
We find in this case that {Yi} satisfies the following bracket relations:
[Y1, Y2] = −Y3 [Y1, Y3] = −Y2 [Y1, Y4] = −a3Y2
[Y2, Y3] = Y1 [Y2, Y4] = a3Y1 [Y3, Y4] = 0.
Let g(t) = A(t)θ21+B(t)θ
2
2+C(t)θ
2
3+D(t)θ
2
4 where A(0) = λ1, B(0) = λ2, C(0) = λ3
and D(0) = λ4. Then backward Ricci flow reduces to the following system of
68
equations:
dA
dt
=
A2 − (B + C)2
BC
+
A2 −B2
BD
a23
dB
dt
=
B2 − (A+ C)2
AC
+
B2 − A2
AD
a23
dC
dt
=
C2 − (A−B)2
AB
dD
dt
= −(A+B)
2
AB
a23
λ1 = λ2.
(II.121)
We calculate
d
dt
(
A−B) = [ 1
ABC
(−C2 + A2 + 2AB +B2)
+
a23
ABD
(
A2 + 2AB +B2
)](
A−B),
so A = B for all t.
Now (II.121) reduces to
dA
dt
= −C
A
− 2
B = A
dC
dt
=
C2
A2
dD
dt
= −4a23.
(II.122)
This equations for A, B and C in (II.122) are just a special case of the
corresponding equations in (II.89). However, in this special case we may derive
more explicit solutions.
69
It is clear that D = λ4 − 4a23t, so T0 ≤
λ4
4a23
. Also, letting f ′ denote
df
dt
we have
A′′ =
−A
(
C2
A2
)
+ C
(−C
A
− 2)
A2
= −2C
2
A3
− 2C
A
= − 2
A
(
(A′ + 2)2 − A′ + 2)
= − 2
A
(
(A′ + 1)(A′ + 2)
)
,
hence
A′
(A′ + 1)(A′ + 2)
A′′ = − 2
A
A′.
Integrating gives
2 ln |A′ + 2| − ln |A′ + 1| = −2 ln |A|+ ln(Λ),
where
Λ =
λ23λ1
λ3 + λ1
. (II.123)
Solving for A′ gives us
A′ =
−Λ− 4A2 ±√Λ2 + 4ΛA2
2A2
. (II.124)
Comparing equations (II.122) and (II.124) at t = 0, we confirm in fact that
A′ = −Λ + 4A
2 +
√
Λ2 + 4ΛA2
2A2
,
hence (
2A2
Λ + 4A2 +
√
Λ2 + 4ΛA2
)
A′ = −1. (II.125)
70
Now we observe
2A2
Λ + 4A2 +
√
Λ2 + 4ΛA2
=
1
2
−
√
Λ
2
√
Λ + 4A2
,
so (
1
2
−
√
Λ
2
√
Λ + 4A2
)
A′ = −1,
and
A
2
−
√
Λ
4
sinh−1
(
2A√
Λ
)
= T2 − t. (II.126)
Now we calculate that
lim
A→0
A
2
−
√
Λ
4
sinh−1
(
2A√
Λ
)
A3
=
1
3eΛ
. (II.127)
Thus using (II.125) and (II.127) we find that near t = T2,
A ≈ (3eΛ(T2 − t))1/3 . (II.128)
By (II.122) we know that C = −A
(
dA
dt
+ 2
)
. Thus near A = 0 we know that
C ≈ Λ (3Λ(T2 − t))−1/3 . (II.129)
We observe that
T0 = min
{
λ4
4a23
,
λ1
2
− λ3
4
√
λ1
λ1 + λ3
sinh−1
(
2
λ3
√
λ1(λ1 + λ3)
)}
(II.130)
= min
{
T1, T2
}
. (II.131)
We calculate the sectional curvatures using (II.87) with a1 = a2 = 0 and A = B:
K(Y1, Y2) = −4A+ 3C
4A2
K(Y1, Y3) =
C
4A2
K(Y1, Y4) = 0
K(Y2, Y3) =
C
4A2
K(Y2, Y4) = 0 K(Y3, Y4) = 0.
71
If T0 = T1 < T2, then near t = T0 we have
A = B ≈ k1
C ≈ k3
D = k4(T0 − t).
(II.132)
Here all sectional curvatures will also approach constants as t → T0, and the
volume normalized solution will approach the hyperplane R3.
If T0 = T2 < T1, then near t = T0 we have
A = B ≈ k1(T0 − t)1/3
C ≈ k3(T0 − t)−1/3
D ≈ k4.
(II.133)
Here all non-zero sectional curvatures will have a singularity of the form
(T0 − t)−1. The volume normalized solution will approach the plane R2.
If T0 = T1 = T2, then near t = T0 we have
A = B ≈ k1(T0 − t)1/3
C ≈ k3(T0 − t)−1/3
D = k4(T0 − t).
(II.134)
Here all non-zero sectional curvatures will have a singularity of the form
(T0 − t)−1, and the volume normalized solution will approach a product metric
72
M2 × R.
A10. Class U3S3
Here we may choose a basis for the Lie Albegra {X1, X2, X3, X4} such that the
Lie bracket is of the form
[X1, X2] = X3 [X1, X3] = −X2 [X1, X4] = 0
[X2, X3] = X1 [X2, X4] = 0 [X3, X4] = 0.
This Lie Algebra structure is a direct sum su(2) ⊕ R, and the Lie Group
structure structure is (M,G) = (S3 × R, SU(2)× R).
Here we diagonalize the metric by letting Yi = Υ
k
iXk with
Υ =

1 0 0 0
0 1 0 0
0 0 1 0
a1 a2 a3 1
 .
Here I calculate the curvatures of this diagonalized metric in general form so
that I can use it to the specific cases outlined below in Proposition 9. First I
73
calculate the operator U using (II.5):
U(Yi, Yi) = 0 for all i
U(Y1, Y2) =
B − A
2C
Y3 +
a3(B − A)
2D
Y4
U(Y1, Y3) =
A− C
2B
Y2 +
A2(A− C)
2D
Y4
U(Y1, Y4) =
a3A
2B
Y2 − a2A
2C
Y3
U(Y2, Y3) =
C −B
2A
Y1 +
a1(C −B)
2D
Y4
U(Y2, Y4) = −a3B
2A
Y1 +
a1B
2C
Y3
U(Y3, Y4) =
a2C
2A
Y1 − a1C
2B
Y2.
The sectional curvatures may then be calculated using (II.4):
K(Yi, Yi) = 0 for all i
K(Y1, Y2) =
1
4AB
[
−3C + 2B + 2A+ (A−B)2
(
1
C
+
a23
D
)]
K(Y1, Y3) =
1
4AC
[
−3B + 2C + 2A+ (A− C)2
(
1
B
+
a22
D
)]
K(Y1, Y4) =
1
4AD
[
−3(a23B + a22C) + 2A(a22 + a23) + A2
(
a23
B
+
a22
C
)]
K(Y2, Y3) =
1
4BC
[
−3A+ 2C + 2B + (B − C)2
(
1
A
+
a21
D
)]
K(Y2, Y4) =
1
4BD
[
−3(a23A+ a21C) + 2B(a23 + a21) +B2
(
a23
A
+
a21
C
)]
K(Y3, Y4) =
1
4CD
[
−3(a22A+ a21B) + 2C(a21 + a22) + C2
(
a22
A
+
a21
B
)]
.
(II.135)
Proposition 9 in [11] breaks this down into three situations:
74
(i) If λ1 = λ2 = λ3, them the metric remains diagonal for all choices of a1, a2
and a3.
(ii) If λi 6= λj = λk for some permutation {i, j, k} of {1, 2, 3}, then the metric
remains diagonal if and only if aj = ak = 0.
(iii) If λ1, λ2 and λ3 are all different, then the metric will remain diagonal if
and only if a1 = a2 = a3 = 0.
In all three of these cases, we have that backward Ricci flow reduces to the
system :
dA
dt
=
A2 − (B − C)2
BC
dB
dt
=
B2 − (A− C)2
AC
dC
dt
=
C2 − (A−B)2
AB
dD
dt
= 0.
(II.136)
Thus this reduces to 3 dimensions, and the normalized flow is analyzed in [4].
By symmetry, we may in fact assume λ1 ≥ λ2 ≥ λ3. Now
d
dt
(A−B) = 1
ABC
(A−B)((A+B)2 − C2), (II.137)
so the conditions A(t) ≥ B(t) and A(t) = B(t) are preserved under backwards Ricci
flow. Similarly, the conditions B(t) ≥ C(t) and B(t) = C(t) are preserved.
75
A10(i)
If λ1 = λ2 = λ3, then A ≡ B ≡ C. Then (II.136) reduces to
dA
dt
=
dB
dt
=
dC
dt
= 1
dD
dt
= 0,
(II.138)
and the solution, which exists for all time, is
A(t) = B(t) = C(t) = λ1 + t,
D(t) = λ4.
(II.139)
Using (II.135) with A = B = C we find the non-zero curvatures are
K(Y1, Y2) = K(Y1, Y3) = K(Y2, Y3) =
1
4A
. (II.140)
Thus all non-zero curvatures approach 0 at a rate of 4t−1.
The solution here is actually just a product of an expanding 3-sphere with a
line. The volume normalized solution converges to the line, R.
A10(ii)
If λ1 = λ2 > λ3, then A(t) = B(t) > C(t) for all 0 < t < T0. Then (II.136)
reduces to
dA
dt
= 2− C
A
B = A
dC
dt
=
C2
A2
dD
dt
= 0.
(II.141)
76
Denoting
df
dt
by f ′ we have
A′′ =
1
A2
(
C
(
2− C
A
)
− C
2
A
)
=
1
A
(
(2− A′)A′ − (2− A′)2)
= − 2
A
(A′ − 1)(A′ − 2),
hence
−A′
(A′ − 1)(A′ − 2)A
′′ =
2
A
A′,
and thus
ln
∣∣∣∣ A′ − 1(A′ − 2)2
∣∣∣∣ = ln(A2) + ln(Λ), (II.142)
where
Λ =
λ1 − λ3
λ1λ23
. (II.143)
Since A′ > 1, (II.142) becomes
A′ − 1
(A′ − 2)2 = ΛA
2.
Solving we have
A′ =
1 + 4ΛA2 −√1 + 4ΛA2
2ΛA2
. (II.144)
Note here that since A′ = 2− C
A
by (II.136), then
C =
−1 +√1 + 4ΛA2
2ΛA
. (II.145)
Continuing on, (II.144) becomes
2ΛA2
1 + 4ΛA2 −√1 + 4ΛA2A
′ = 1.
77
Upon integrating we have
1
2
A+
1
4
√
Λ
sinh−1
(
2
√
ΛA
)
= t+
1
2
+
1
4
√
Λ
sinh−1
(
2
√
Λλ1
)
(II.146)
Now we observe that
lim
A→∞
1
2
A+ 1
4
√
Λ
sinh−1
(
2
√
ΛA
)
A
=
1
2
, (II.147)
so as A approaches infinity, t ≈ 1
2
A.
Now by equations (II.136) and (II.145), we have the following behavior as t→
∞:
A(t) = B(t) ≈ 2t
C(t)→ 1√
Λ
D(t) = λ4.
(II.148)
where Λ is given in (II.143).
Using (II.135) with a1 = a2 = 0 and A = B we find the non-zero curvatures are
given by
K(Y1, Y2) =
4A− 3C
4A2
K(Y1, Y3) =
C
4A2
K(Y2, Y3) =
C
4A2
.
(II.149)
Thus all non-zero curvatures parallel to Y3 approach 0 at a rate of t
−2 while the
remaining nonzero curvatures approach 0 at a rate of t−1. The normalized solution
78
will converge to the plane R2.
If λ1 > λ2 = λ3, then A(t) > B(t) = C(t) for all 0 < t < T0, and (II.136)
reduces to the following:
dA
dt
=
A2
B2
dB
dt
= 2− A
B
C = B
dD
dt
= 0.
(II.150)
Similarly to the case A = B > C we calculate
B′′ =
1
B2
(
A
(
2− A
B
)
− A
2
B
)
=
1
B
(
(2−B′)B′ − (2−B′)2)
= − 2
B
(B′ − 1)(B′ − 2),
hence
−B′
(B′ − 1)(B′ − 2)B
′′ =
2
B
B′,
and thus
ln
∣∣∣∣ B′ − 1(B′ − 2)2
∣∣∣∣ = ln(B2) + k, (II.151)
where
Λ =
λ1 − λ2
λ21λ2
. (II.152)
Since B′ < 1, (II.151) becomes
1−B′
(B′ − 2)2 = ΛB
2.
79
Solving we have
B′ =
4ΛB2 − 1−√1− 4ΛB2
2ΛB2
. (II.153)
Note here that since B′ = 2− A
B
, then
A =
1 +
√
1− 4ΛB2
2ΛB
. (II.154)
Now (II.153) becomes
2ΛB2
4ΛB2 − 1−√1− 4ΛB2B
′ = 1.
Upon integrating we have
1
2
B − 1
4
√
Λ
sin−1
(
2
√
ΛB
)
= t− T0, (II.155)
where
T0 =
λ2
2
− λ1
√
λ2
4
√
λ1 − λ2
sin−1
(
2
√
λ2(λ1 − λ2)
λ1
)
. (II.156)
Now we calculate
lim
B→0
1
2
B − 1
4
√
Λ
sin−1
(
2
√
ΛB
)
B3
= −Λ
3
,
so as B approaches 0, we have t−T0 ≈ −Λ
3
B3, so by equations (II.150) and (II.154),
we have the following behavior as t→ T0:
A(t) ≈ 1
Λ
(
3
Λ
(T0 − t)
)−1/3
B(t) = C(t) ≈
(
3
Λ
(T0 − t)
)1/3
D(t) = λ4.
(II.157)
80
where Λ is given in (II.152), and T0 is given in (II.156).
With a2 = a3 = 0 and B = C, (II.135) tells us the non-zero curvatures are given
by
K(Y1, Y2) =
A
4B2
K(Y1, Y3) =
A
4B2
K(Y2, Y3) =
4B − 3A
4B2
.
(II.158)
Thus all non-zero curvatures are of the form (T0− t)−1. The volume normalized
solution will approach the plane R2.
A10(iii)
In the case λ1 > λ2 > λ3, we have by [4] that the end behavior near t = T0 is the
same as when λ1 > λ2 = λ3. However, we do not have more explicit solutions like
we do in the special case. Since the solutions in [4] are using normalized backward
Ricci flow, I shall present a slightly different argument here.
We know that for all t we have A(t) > B(t) > C(t). From (II.136) we calculate
dC
dt
=
C2 − A2 + 2AB −B2
AB
< 2− A
B
< 1, (II.159)
81
so C(t) < λ3 + t. Now we may calculate
d
dt
(
A−B
C
)
=
2
ABC
(A−B)(A2 +B2 − C2)
>
2A2(A−B)
ABC2
> 2
(
A−B
C
)
· 1
C
> 2
(
A−B
C
)
· 1
t+ λ3
,
so
ln
(
A−B
C
)
> 2 ln(t+ λ3) + ln
(
λ1 − λ2
λ3
)
,
and we have that (
A−B
C
)
>
(
λ1 − λ2
λ3
)
(t+ λ3)
2.
Therefore
A−B
C
increases at least quadratically. Thus either T0 < ∞ or
dC
dt
=
C2 − (A−B)2
AB
< 0 for all t large enough.
Either way, we know by (II.159) that C is bounded, hence C ≤ kC for some
0 < kC <∞. Now
d
dt
(
A
B
)
=
2(A−B)
B2C
(A+B − C)
>
2A(A−B)
kCB2
=
2
kC
(
A
B
)(
A
B
− 1
)
. (II.160)
Thus we see that
A
B
→ ∞ in finite time, so we know T0 < ∞. Since for all t
82
we know A > B > C, then near t = T0 either A → ∞ or C → 0. Now we may
calculate
d
dt
(
B
C
)
=
2
AC2
(B − C)(B + C − A),
which is negative for t close enough to T0. Thus we know by monotone convergence
that
lim
t→T0
B
C
= kBC (II.161)
for some kBC ≥ 1. Thus if C → 0 then B → 0 as well. Now we observe
d
dt
(ABC) = −(A2 +B2 + C2) + 2(AB + AC +BC), (II.162)
which is negative for A large enough and B and C bounded. Thus if A → ∞ as
t→ T0, then ABC is bounded, hence C → 0. Similarly,
d
dt
(AC) = 2(A+ C −B), (II.163)
which is positive for B and C close enough to 0 and A bounded below. Hence if
C → 0 then also A→∞. Thus we know that as t→ T0 we have A→∞, B → 0
and C → 0.
Now by (II.136) we observe
dA
dt
=
A2 − (B − C)2
BC
<
A2
BC
. (II.164)
83
Also, by (II.163) we know that AC is bounded from below. Similarly, AB is
bounded from below, so (II.164) becomes
dA
dt
< kA4,
hence
A(t) <
(
λ−31 − 3kt
)−1/3
. (II.165)
Now we have
d
dt
(AB) = 2(A+B − C) < k˜A < k˜(λ−31 − 3kt)−1/3. (II.166)
Integrating (II.166) gives us
AB < λ1λ2 +
k˜
2λ21
− k˜
2
(
λ−31 − 3kt
)2/3
.
hence
lim
t→T0
(AB) = kAB. (II.167)
for some 0 < kAB <∞. Combining (II.161) and (II.167) we conclude
lim
t→T0
(AC) =
kAB
kBC
. (II.168)
Thus we see that near t = T0,
BC ∼ k
2
AB
kBCA2
,
hence (II.136) tells us
dA
dt
∼ kBC
k2AB
A4. (II.169)
84
Solving as we did for equation (II.16) gives
A ≈ k1(T0 − t)−1/3. (II.170)
Now using (II.167) and (II.168) we have the end behavior of (II.136) near t = T0:
A(t) ≈ k1(T0 − t)−1/3
B(t) ≈ k2(T0 − t)1/3
C(t) ≈ k3(T0 − t)1/3
D(t) = λ4.
(II.171)
From (II.135) with a1 = a2 = a3 = 0 we have that the non-zero curvatures are
K(Y1, Y2) =
1
4ABC
(−3C2 + 2AC + 2BC + A2 − 2AB +B2)
K(Y1, Y3) =
1
4ABC
(−3B2 + 2AB + 2BC + A2 − 2AC + C2)
K(Y2, Y3) =
1
4ABC
(−3A2 + 2AB + 2BC +B2 − 2BC + C2).
(II.172)
Thus all the non-zero curvatures have a singularity of the form (T0 − t)−1, and
the volume-normalized solution approaches R2.
The Non-Bianchi Cases
In this section we examine the compact locally homogeneous geometries whose
isotropy group is not trivial, so the dimension of the Lie Group is higher than the
dimension of the manifold. Again, these cases are well-understood, but I include
them here for completion. The following can be found in [11].
85
B1. H3 × R
Any initial metric can be written as
g0 = R
2gH3 + du
2 (II.173)
for some R > 0. The solution to backward Ricci flow is given by
g(t) = (R2 − 4t)gH3 + du2, −∞ < t < R
2
4
. (II.174)
B2. S2 × R2
Any initial metric can be written as
g0 = R
2gS3 + du
2
1 + du
2
2 (II.175)
for some R > 0. The solution to backward Ricci flow is given by
g(t) = (R2 + 2t)gS2 + du
2
1 + du
2
2, −
R2
2
< t <∞. (II.176)
B3. H2 × R2
Any initial metric can be written as
g0 = R
2gH2 + du
2
1 + du
2
2 (II.177)
86
for some R > 0. The solution to backward Ricci flow is given by
g(t) = (R2 − 2t)gH2 + du21 + du22, −∞ < t <
R2
2
. (II.178)
B4. S2 × S2
Any initial metric can be written as
g0 = R
2
1gS2 +R
2
2gS2 (II.179)
for some R1 > 0, R2 > 0. The solution to backward Ricci flow is given by
g(t) = (R21 + 2t)gS2 + (R
2
2 + 2t)gS2 , −min
{
R21
2
,
R22
2
}
< t <∞. (II.180)
B5. S2 ×H2
Any initial metric can be written as
g0 = R
2
1gS2 +R
2
2gH2 (II.181)
for some R1 > 0, R2 > 0. The solution to backward Ricci flow is given by
g(t) = (R21 + 2t)gS2 + (R
2
2 − 2t)gH2 , −
R21
2
< t <
R22
2
. (II.182)
87
B6. H2 ×H2
Any initial metric can be written as
g0 = R
2
1gH2 +R
2
2gH2 (II.183)
for some R1, R2 > 0, R1 6= R2. The solution to backward Ricci flow is given by
g(t) = (R21 − 2t)gH2 + (R22 − 2t)gH2 , −∞ < t < min
{
R21
2
,
R22
2
}
. (II.184)
B7. CP 2
Any initial metric can be written as
g0 = R
2gFS (II.185)
for some R > 0, where gFS is the Fubini-Study metric on complex projective space,
CP 2, with constant holomorphic bisectional curvature 1. The solution to backward
Ricci flow is given by
g(t) = (R2 + 6t)gFS, −R
2
6
< t <∞. (II.186)
B8. CH2
Any initial metric can be written as
g0 = R
2gCH2 (II.187)
88
for some R > 0, where gCH2 is the Ka¨hler metric on complex hyperbolic space, CH2,
with constant holomorphic bisectional curvature −1. The solution to backward
Ricci flow is given by
g(t) = (R2 − 6t)gCH2 , −∞ < t < R
2
6
. (II.188)
B9. S4
Any initial metric can be written as
g0 = R
2gS4 , (II.189)
for some R > 0. The solution to backward Ricci flow is given by
g(t) = (R2 + 6t)gS4 , −R
2
6
< t <∞. (II.190)
B10. H4
Any initial metric can be written as
g0 = R
2gH4 (II.191)
for some R > 0. The solution to backward Ricci flow is given by
g(t) = (R2 − 4t)gH4 , −∞ < t < R
2
6
. (II.192)
89
Conclusions
Our conclusion is that the end behavior of locally homogeneous manifolds under
backward Ricci flow is fairly consistent. In general however, the Bianchi classes have
very different, and more interesting, behavior than the non-Bianchi classes. Recall
that all of our solutions are Riemannian metrics of the form
g(t) = A(t)θ21 +B(t)θ
2
2 + C(t)θ
2
3 +D(t)θ
2
4
where we call A,B,C and D the metric coefficients. Each class of manifolds studied
in this chapter has one of the following types of behavior near the end:
Expanding-1: All metric coefficients grow linearly for all time. The volume-
normalize metric approaches an Einstein metric with positive Ricci curvature.
Expanding-2: One metric coefficient approaches a constant while the other
three grow linearly for all time. The volume-normalized metric approaches the
span of the three two-forms whose coefficients grow linearly.
Expanding-3: Two metric coefficients approach constants while the other two
grow linearly for all time. The volume-normalized metric approaches the span of
the two two-forms whose coefficients grow linearly.
90
Trivial: Each metric coefficient approaches a constant. Thus, the metric
approaches an Einstein metric with constant 0. Among the classes considered, this
only happens for trivial metrics which are just quotients of Euclidean Space.
Pancake-1: One metric coefficient approaches 0 linearly while the other three
approach constants. The volume-normalized solution will approach the sub-
riemannian geometry R3.
Pancake-2: One metric coefficient approaches 0 at a rate of (T0 − t)1/3, one
approaches a constant, and the last two approach infinity at a rate of (T0 − t)−1/3.
The volume-normalized solution will also approach R3.
Pancake-3: One metric coefficient approaches a constant, two approach 0 on the
order of (T0 − t)1/3, and the final approaches infinity on the order of (T0 − t)−1/3.
Here the volume-normalized solution will approach the plane R2.
Pancake-4: One metric coefficient approaches a constant, one approaches
infinity on the order of (T0 − t)−1/3, and the last two approach 0 on the order
of (T0 − t)1/3 and (T0 − t)2/3 respectively. The volume-normalized solution will
approach the plane R2. Nearby initial conditions exhibit Pancake-3 behavior.
91
Pancake-5: Two metric coefficients approach 0 linearly while the other two
approach constants. The volume-normalized solution will approach the plane R2.
Line: One metric coefficient approaches a constant while the others approach 0
linearly. The volume-normalized solution will approach the line R.
Point: All metric coefficients approach 0 linearly. Any compact quotients will
collapse to a point at T0. Among the classes of manifolds we have considered, this
happens only for the Einstein manifolds CH2 and H4. Thus the volume-normalized
solutions are constant solutions.
Tube-1: One metric coefficient approaches 0 linearly while two others approach
0 at the rate of (T0 − t)1/3. The final metric coefficient is of the form (T0 − t)−1/3.
The volume-normalized solution will approach a product metric g = M2 × R.
Nearby initial conditions will exhibit Pancake-1 and Pancake-3 behavior.
Tube-2: Two metric coefficients approach constants, while the other two have
behaviors (T0 − t)1/3 → 0 and (T0 − t)−1/3 respectively. The volume-normalized
solution will approach a product metric g = M2 × R. Nearby initial conditions
exhibit Pancake-2 and Pancake-3 end behavior.
92
Class Lie Group Structure End Behavior
A1 (R4,R4, {0}) Trivial
A2
(S˜ol
3 × R, S˜ol3 × R, e)
(Sol40, Sol
4
0, e)
(Sol4m,n, Sol
4
m,n, e)
Pancake-1
A3 (R4, E(2)× R2, e) Pancake-1
Pancake-3
A4 (Nil3 × R, Nil3 × R, e) Pancake-3
A5 No Compact Geometries Pancake-3
A6 (Nil4, Nil4, e)
Pancake-3
Pancake-4
A7 (Sol4, Sol4, e)
Pancake-1
Pancake-2
Pancake-3
Pancake-4
Tube-1
Tube-2
A8 No Compact Geometries
Pancake-2
Pancake-3
Tube-2
A9 (SˆL(2,R)× R, SˆL(2,R)× R, e)
Pancake-1
Pancake-3
Tube-1
A10 (S3 × R, SU(2)× R, e)
Expanding-2
Expanding-3
Pancake-3
B1 (H3 × R, H(3)× R, SO(3)× {0}) Line
B2 (S2 × R2, SO(3)× R2, SO(2)× {0}) Expanding-3
B3 (H2 × R2, S0(3)× R2, SO(2)× {0}) Pancake-5
B4 (S2 × S2, SO(3)× SO(3), SO(2)× SO(2)) Expanading-1
B5 (S2 ×H2, S0(3)×H(2), SO(2)× SO(2)) Pancake-5
B6 (H2 ×H2, H(2)×H(2), SO(2)× SO(2)) Pancake-5
B7 (CP 2, SU(3), U(2)) Expanding-1
B8 (CH2, SU(1, 2), U(2)) Point
B9 (S4, SO(5), SO(4)) Expanding-1
B10 (H4, H(4), SO(4)) Point
Table 2.1. End Behavior of Backward Ricci Flow
93
In the non-Bianchi cases, it is clear that the shrinking behavior under backward
Ricci flow of negatively curved spaces and the expanding behavior of positively
curved spaces is exactly opposite of that seen in forward Ricci flow. Similarly, the
behaviors in classes A10 and B1 are reversed.
In the Bianchi cases, we notice that in forward Ricci flow all solutions exist for
all time except for a certain case of A9: SˆL(2,R)×R, where the volume-normalized
solution collapses to a plane and exhibits pancake-like behavior. In contrast, the
backward Ricci flow produces finite-time singularities in all cases except for special
cases of A10: SU(2)× R, where the solution increases linearly for all time.
It is also worth mentioning that Pancake-1 type behavior in backward Ricci
flow occurs as a possibility in exactly the same classes of manifolds which exhibit
linear growth in one or more metric coefficient in forward Ricci flow.
In generic cases, end-behavior of solutions to the differential equations in forward
Ricci flow are of the forms kt±1/3, k or kt. Conversely, end-behavior of solutions to
the differential equations in backward Ricci flow are of the forms k(T0 − t)±1/3, k
or k(T0 − t).
94
REFERENCES CITED
[1] Lars Andersson and Vince Mancrief. Elliptic-hyperbolic systems and the
einstein equations. arXiv:gr-qc/0110111, 10 2008.
[2] Arthur Besse. Einstein manifolds. Springer-Verlag, Berlin, 1987.
[3] Xiaodong Cao, John Guckenheimer, and Laurent Saloff-Coste. The backward
behavior of the ricci and cross curvature flow on sl(2,R). Comm. Anal. Geom.,
6 2009.
[4] Xiaodong Cao and Laurent Saloff-Coste. Backward ricci flow on locally
homogeneous three-manifolds. arXiv:0810.3352v2, 10 2008.
[5] Bing-Long Chen and Xi-Ping Zhu. Uniqueness of the ricci flow on complete
noncompact manifolds. J. Differential Geom., 74:119–154, 2006.
[6] Bennett Chow and Dan Knopf. The ricci flow: an introduction. Mathematical
Surveys and Monographs, 110, 2004.
[7] Bennett Chow, Peng Lu, and Lei Ni. Hamilton’s Ricci Flow. Science Press
and AMS, Beijing, 2006.
[8] Arthur Fischer. An introduction to conformal ricci flow. Classical Quantum
Gravity, 21:S171–S218, 2004.
[9] Richard Hamilton. Three-manifolds with positive ricci curvature. J.
Differential Geom., 17:255–306, 1982.
[10] James Isenberg and Martin Jackson. Ricci flow of locally homogeneous
geometries on closed manifolds. J. Differential Geom., 35:733–741, 1992.
[11] James Isenberg, Martin Jackson, and Peng Lu. Ricci flow on locally
homogeneous closed 4-manifolds. Comm. Anal. Geom., 14(2):345–386, 2006.
[12] Brett Kotschwar. An energy approach to the problem of uniqueness for the
ricci flow. arXiv:1206.3225, 6 2012.
[13] Peng Lu, Jie Qing, and Yu Zheng. A note on conformal ricci flow.
arXiv:1206.3225, 9 2010.
95
[14] Malcolm A.H. MacCallum. On the classification of the real four-dimensional
lie algebra. In A. Harvey, editor, On Einstein’s path, pages 299–317. Springer,
1992.
[15] Jeffrey Rauch. Partial Differential Equations. Springer-Verlag, New York,
1991.
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