# Bach flow on homogeneous products

Dylan Helliwell  Department of Mathematics, Seattle University.
helliwed(at)seattleu.edu

###### Abstract.

Qualitative behavior of Bach flow is established on compact four-dimensional
locally homogeneous product manifolds. This is achieved by lifting to the
homogeneous universal cover and, in most cases, capitalizing on the resultant
group structure. The resulting system of ordinary differential equations is
carefully analyzed on a case-by-case basis, with explicit solutions found in
some cases. Limiting behavior of the metric and the curvature are determined
in all cases. The behavior on quotients of $\mathbb{R}\times\mathbb{S}^{3}$
proves to be the most challenging and interesting.

###### Key words and phrases:

High-order geometric flows, Bach flow, locally homogeneous manifold, three-
dimensional Lie group

###### 2010 Mathematics Subject Classification:

53C44, (53C30, 34C40)

##  1\. Introduction

In four dimensions, Bach flow is a solution to

$\left\\{\begin{aligned} \partial_{t}g&=B+\frac{1}{12}\Delta Sg\\\
g(0)&=h.\end{aligned}\right.$

where the Bach tensor $B$ is the gradient of the Weyl energy functional. This
serves as a concrete motivating example of a higher-order intrinsic curvature
flow. Such flows, including flow by the ambient obstruction tensor, flow by
the gradient of the total curvature energy functional, have been of interest
recently. See for example[BH11], [BH15], and with related work found in
[Str08], [Bou10], [Lop18], and [Ho18].

Our goal is to understand Bach flow on $(M,g)$ where $M=\mathbb{S}^{1}\times
N$, $(N,\tilde{g})$ is a closed locally homogeneous three-dimensional
Riemannian manifold, and $g=g_{\mathbb{S}^{1}}+\tilde{g}$ is the product
metric. By lifting to the universal cover $\widehat{M}$ of $M$, this analysis
reduces to analysis of Bach flow on one of nine simply connected homogeneous
spaces.

The specific details for each of the nine cases can be found in Sections 5 and
6. As a summary, we find:

  * •

if $\widehat{N}=\mathbb{R}^{3}$ or $\mathbb{H}^{3}$, Bach flow is static;

  * •

if $\widehat{N}=Nil$, $\widehat{SL}(2,\mathbb{R})$,
$\mathbb{R}\times\mathbb{S}^{2}$, or $\mathbb{R}\times\mathbb{H}^{2}$, Bach
flow collapses to a flat surface;

  * •

if $\widehat{N}=Solv$, Bach flow collapses to a curve;

  * •

if $\widehat{N}=E(2)$, Bach flow converges to a flat four-dimensional
manifold;

  * •

if $\widehat{N}=\mathbb{S}^{3}$, Bach flow can collapse to a flat three-
dimensional manifold, collapse to a flat surface, or converge to a curved
four-dimensional manifold, depending on the initial conditions.

In this paper, $\mathbb{S}^{n}$ is the $n$-dimensional sphere,
$\mathbb{H}^{n}$ is $n$-dimensional hyperbolic space, $Nil$ is the Heisenberg
group consisting of $3\times 3$ upper triangular matrices with 1’s on the
diagonal, $Solv$ is the Poincaré group for 2-D Minkowski space
$\mathbb{R}^{2}\rtimes O(1,1)$, $E(2)$ is the group of Euclidean
transformations of the plane, and $\widehat{SL}(2,\mathbb{R})$ is the
universal cover of $SL(2,\mathbb{R})$.

The method here is similar to that of [IJ92] and [IJL06], where the
qualitative behavior of volume-normalized Ricci flow on locally homogeneous
three- and four-dimensional manifolds was determined. See [GP10] for an
alternative approach to analyzing Ricci flow on homogeneous three-dimensional
manifolds.

Here, for most spaces, the analysis is a bit more challenging than for Ricci
flow, since the polynomials in the systems determined by Ricci flow are third
order while those for Bach flow are seventh order. These higher order
expressions are more difficult to analyze for the purposes of qualitative
analysis.

Determining the behavior of Bach flow on model spaces has so far been limited
to flow on locally homogeneous $2\times 2$ products by [DK12] and
$\mathbb{S}^{1}\times Solv$ by [Ho18]. Additionally, in [Str08], flow by the
gradient of the total curvature energy functional was analyzed on two specific
four-dimensional homogeneous spaces: $\mathbb{S}^{2}\times\mathbb{H}^{2}$ and
$\mathbb{R}\times\mathbb{S}^{3}$. Bach flow is related to this flow, and
comparing and contrasting the qualitative behavior of these flows helps to
understand this relationship. On $\mathbb{S}^{2}\times\mathbb{H}^{2}$ the
equations determined by the two flows are essentially the same. On
$\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, only round metrics on
$\mathbb{S}^{3}$ were considered in [Str08], and on compact quotients, the
resulting product metric was found to collapse to a three-dimensional space,
with the $\mathbb{S}^{1}$ slice shrinking. Here, we find that Bach flow is
static in this case.

The general approach to understanding Bach flow on the spaces of interest is
similar to that found in [IJ92]. The universal cover $\widehat{N}$ is either a
Lie group or it is not. In the case where $\widehat{N}$ is a Lie group, the
set of homogeneous metrics can be identified with the set of left-invariant
metrics on $\widehat{N}$, which in turn are identified with the set of inner
products on the tangent space at the identity. Curvature can then be expressed
in terms of the structure constants and the inner product. The Lie groups of
interest have the property that a basis can be found where the inner product
is diagonal and the structure constants can be written in a convenient form.
As was true for the Ricci tensor in [IJ92], we find here that the Bach tensor
in this setting is diagonal and so Bach flow preserves this structure. The
resulting system is analyzed, with explicit solutions found in some cases, and
limiting behavior is determined. If $\widehat{N}$ is not a Lie group, the
analysis proves to be somewhat simpler, owing to the fact that there are fewer
homogeneous metrics on these spaces. The resulting systems can all be solved
explicitly.

This paper is organized as follows: In Section 2, the Bach tensor and Bach
flow are discussed. Additionally, formulas for the Bach tensor on products are
provided. In Section 3, details surrounding the locally homogeneous spaces and
Lie groups of interest are provided, including curvature formulas in terms of
structure constants. In Section 4, useful results about ordinary differential
equations are provided. Section 5 is devoted to the derivation and analysis of
Bach flow on locally homogeneous $1\times 3$ products and in Section 6, Bach
flow is analyzed on locally homogeneous $2\times 2$ products. Finally, in
Section 7, the qualitative results for Bach flow are compared and contrasted
with those of Ricci flow.

##  2\. The Bach tensor and Bach flow

On a four-dimensional Riemannian manifold $(M,g)$, the Bach tensor $B$ is
given by

$B_{jk}=g^{lq}P_{jk;lq}-g^{lq}P_{jl;kq}+P^{il}W_{ijkl}$

where $P$ is the Schouten tensor, which, for an $n$-dimensional manifold is
defined as

$P=\frac{1}{n-2}\left(Ric-\frac{S}{2(n-1)}g\right)$

and $W$ is the Weyl tensor. Throughout this paper, curvature and index
conventions follow those found in [Lee97]. The Bach tensor is a symmetric,
trace free, divergence free tensor that is fourth order in the metric and is
conformally invariant: if $\bar{g}=\rho^{2}g$, then $\bar{B}=\rho^{-2}B$. It
can be realized as $-\frac{1}{4}\mathrm{grad}(\mathcal{W})$ where
$\mathcal{W}$ is the Weyl energy functional:

$\mathcal{W}=\int_{M}|W|^{2}d\mu$

with $|W|^{2}=g^{ip}g^{jq}g^{kr}g^{ls}W_{ijkl}W_{pqrs}$.

In [BH11] and [BH15] short time existence and uniqueness are established for
solutions to the geometric flow

$\left\\{\begin{aligned} \partial_{t}g&=B+\frac{1}{12}\Delta Sg\\\
g(0)&=h.\end{aligned}\right.$

Here, and throughout, $\Delta=g^{ij}\nabla_{i}\nabla_{j}$. The positive
multiple of $\Delta Sg$ is included to ensure that the resulting flow is well
posed.

In this paper, this flow is analyzed on locally homogeneous product manifolds.
The local homogeneity ensures that the scalar curvature is constant so the
flow reduces to

(1) $\left\\{\begin{aligned} \partial_{t}g&=B\\\ g(0)&=h.\end{aligned}\right.$

One useful consequence of the fact that the Bach tensor is trace free is that
the volume form is constant along the flow:

$\partial_{t}d\mu_{g}=0.$

If fixed coordinates are chosen, this is equivalent to saying $\det g(t)=\det
h$ is constant in time.

###  2.1. Bach tensor on products

In general, the Bach flow equations lead to a complicated nonlinear system.
Making use of the product structure significantly simplifies the resulting
equations.

Let $(N^{(1)},\tilde{g}^{(1)})$ and $(N^{(2)},\tilde{g}^{(2)})$ be Riemannian
manifolds. Let $M=N^{(1)}\times N^{(2)}$. The product metric $g$ on $M$ is

(2) $\displaystyle g=g^{(1)}+g^{(2)}$

where $g^{(i)}=\pi_{i}^{*}(\tilde{g}^{(i)})$ are the pullbacks of the
component metrics by the natural projections. Greek indices ($\alpha$,
$\beta$, $\gamma$, etc.) will be used for $N^{(1)}$, and lower case roman
indices ($i$, $j$, $k$, etc.) will be used for $N^{(2)}$. Abusing notation
slightly, the tildes used above will be dropped. To clarify when dealing with
an object on $N^{(1)}$ or $N^{(2)}$ (as opposed to $M$) a parenthetical
superscript will be used to indicate the component.

For a general product,

$\displaystyle Ric_{\alpha\beta}$ $\displaystyle=Ric^{(1)}_{\alpha\beta}$
$\displaystyle Ric_{jk}$ $\displaystyle=Ric^{(2)}_{jk}$ $\displaystyle
Ric_{\alpha k}$ $\displaystyle=0$

and

$S=S^{(1)}+S^{(2)}.$

In particular, for $1\times 3$ products $Ric_{00}=0$ and $S=S^{(2)}$.

The Bach tensor is somewhat more complicated. While the Bach tensor splits in
the sense that the components with mixed indices are zero, the components
corresponding to one factor depend on the curvature from the other factor.

The $1\times 3$ and $2\times 2$ cases are as follows: If $\dim(N^{(1)})=1$ and
$\dim(N^{(2)})=3$ then

(3)
$B_{00}=\left(-\frac{1}{12}(\Delta^{(2)}S^{(2)})-\frac{1}{4}\left[(|Ric|^{(2)})^{2}-\frac{1}{3}(S^{(2)})^{2}\right]\right)g_{00},$
(4)
$\begin{split}B_{jk}&=\frac{1}{2}\Delta^{(2)}Ric^{(2)}_{jk}-\frac{1}{12}\Delta^{(2)}S^{(2)}g_{jk}-\frac{1}{6}S^{(2)}_{;jk}\\\
&\qquad-2\mathrm{tr}^{(2)}(Ric^{(2)}\otimes
Ric^{(2)})_{jk}+\frac{7}{6}S^{(2)}Ric^{(2)}_{jk}\\\
&\qquad+\frac{3}{4}(|Ric|^{(2)})^{2}g_{jk}-\frac{5}{12}(S^{(2)})^{2}g_{jk},\end{split}$

and

$B_{0k}=0.$

Here, $\mathrm{tr}(Ric\otimes Ric)_{jk}=g^{il}Ric_{ij}Ric_{lk}$. If
$\dim(N^{(1)})=\dim(N^{(2)})=2$ then

(5)
$B_{\alpha\beta}=-\frac{1}{6}S^{(1)}_{;\alpha\beta}+\frac{1}{6}\left[\Delta^{(1)}S^{(1)}-\frac{1}{2}\Delta^{(2)}S^{(2)}+\frac{1}{4}\left((S^{(1)})^{2}-(S^{(2)})^{2}\right)\right]g_{\alpha\beta}$

and similarly

(6)
$B_{jk}=-\frac{1}{6}S^{(2)}_{;jk}+\frac{1}{6}\left[\Delta^{(2)}S^{(2)}-\frac{1}{2}\Delta^{(1)}S^{(1)}+\frac{1}{4}\left((S^{(2)})^{2}-(S^{(1)})^{2}\right)\right]g_{jk}$

and

$B_{\alpha k}=0.$

Using the formulation for the Bach tensor in the $1\times 3$ setting, we have
the following:

######  Proposition 2.1.

Let $\dim(N^{(1)})=1$ and $\dim(N^{(2)})=3$, and suppose $S=S^{(2)}$ is
constant. Then $B=0$ if and only if $g^{(2)}$ is Einstein.

###### Proof.

If $g^{(2)}$ is Einstein then $Ric^{(2)}=\frac{S^{(2)}}{3}g^{(2)}$ and as a
result, equations (3) and (4) both simplify to zero.

In the other direction, if $g^{(2)}$ is not Einstein, then

$(|Ric|^{(2)})^{2}>\frac{(S^{(2)})^{2}}{3}.$

Since $\Delta^{(2)}S^{(2)}=0$, then in particular, $B_{00}<0$. ∎

An immediate consequence of this result and its proof is the following:

######  Proposition 2.2.

Let $M$ be a $1\times 3$ product with product metric $g$ solving equation (1).
Suppose that for all time, the scalar curvature is constant on $M$. Then
$g_{00}$ is static if and only if $B=0$, in which case, all components of the
metric or static. Otherwise, $g_{00}$ is strictly decreasing.

##  3\. Locally homogeneous spaces and Lie groups

A Riemannian manifold $(M,g)$ is locally homogeneous if for all points $p$,
$q$ in $M$, there exist neighborhoods $U$ and $V$ about $p$ and $q$
respectively, and an isometry

$\varphi:U\rightarrow V$

with $\varphi(p)=q$. If, for all pairs of points, the isometry can be chosen
to be global, so that $\varphi:M\rightarrow M$, then $(M,g)$ is homogeneous.
If $M$ is closed and locally homogeneous, then its universal cover is
homogeneous. A straightforward, but useful, result is that if a manifold is
locally homogeneous, its scalar curvature is constant. There are nine three-
dimensional simply connected homogeneous manifolds with compact quotients, six
of which are Lie groups. The Lie groups support a larger class of homogeneous
metrics and require a more sophisticated analysis than the three non-Lie
groups. See [IJ92] for more details surrounding these definitions and results.

###  3.1. Structure constants and curvature

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $\\{e_{i}\\}$
be a left-invariant basis for $\mathfrak{g}$. The bracket can be expressed in
terms of structure constants ${C_{ij}}^{k}$

$[e_{i},e_{j}]={C_{ij}}^{k}e_{k}.$

Given a left-invariant metric $g$, and working with a left-invariant frame,
covariant derivatives, and then curvature can be expressed in terms of
structure constants. The Ricci and scalar curvatures are

(7)
$Ric_{jk}=-\frac{1}{2}({{C^{l}}_{j}}^{p}+{{C^{p}}_{j}}^{l})C_{lkp}+\frac{1}{4}{C^{lp}}_{j}C_{lpk}+\frac{1}{2}{C^{lp}}_{l}(C_{pjk}+C_{pkj})$

and

(8)
$S=-\frac{1}{4}C^{lkp}C_{lkp}-\frac{1}{2}C^{pkl}C_{lkp}-{C^{lp}}_{l}{C^{k}}_{pk}.$

Additionally, with a bit more calculation, the Laplacian of a left-invariant
tensor can be expressed in terms of structure constants. For this paper, the
Laplacian of a left-invariant symmetric $\binom{2}{0}$-tensor $T$ is needed
and we have the following:

(9) $\begin{split}(\Delta
T)_{ij}&=\frac{1}{2}T_{pq}\Bigl{(}{{C^{k}}_{i}}^{p}{C_{kj}}^{q}+{C^{kp}}_{i}{{C_{k}}^{q}}_{j}+{C_{i}}^{pk}{{C_{j}}^{q}}_{k}\\\
&\qquad\qquad\qquad-{{C^{k}}_{i}}^{p}{{C_{k}}^{q}}_{j}-{{C^{k}}_{j}}^{p}{{C_{k}}^{q}}_{i}\\\
&\qquad\qquad\qquad-{{C^{k}}_{i}}^{p}{{C_{j}}^{q}}_{k}-{{C^{k}}_{j}}^{p}{{C_{i}}^{q}}_{k}\\\
&\qquad\qquad\qquad+{C^{kp}}_{i}{{C_{j}}^{q}}_{k}+{C^{kp}}_{j}{{C_{i}}^{q}}_{k}\Bigr{)}\\\
&\qquad+\frac{1}{4}T_{qj}\Bigl{(}({C^{kp}}_{i}-{{C^{k}}_{i}}^{p}+{C_{i}}^{pk})({{C_{k}}^{q}}_{p}-{C_{kp}}^{q}+{{C_{p}}^{q}}_{k})\\\
&\qquad\qquad\qquad\qquad+2{C^{kp}}_{k}({{C_{p}}^{q}}_{i}-{C_{pi}}^{q})\Bigr{)}\\\
&\qquad+\frac{1}{4}T_{qi}\Bigl{(}({C^{kp}}_{j}-{{C^{k}}_{j}}^{p}+{C_{j}}^{pk})({{C_{k}}^{q}}_{p}-{C_{kp}}^{q}+{{C_{p}}^{q}}_{k})\\\
&\qquad\qquad\qquad\qquad+2{C^{kp}}_{k}({{C_{p}}^{q}}_{j}-{C_{pj}}^{q})\Bigr{)}.\end{split}$

###  3.2. Three-dimensional Lie groups

As seen in [RS75], the six three-dimensional simply connected Lie groups with
compact quotients are all unimodular and all have the property that for each
group, there is a basis for the Lie algebra such that the structure constants
can be represented by

${C_{ij}}^{k}=\varepsilon_{ijs}E^{ks}$

where $\varepsilon_{ijk}$ is the Levi-Civita symbol which captures the parity
of the permutation generating “$ijk$” with $\varepsilon_{123}=1$, and where
$E$ is a $3\times 3$ matrix specific to the group. See Figure 1.

Type | Group | $E$
---|---|---
I | $\mathbb{R}^{3}$ | 0
II | $Nil$ | $\mathrm{diag}(1,0,0)$
VI0 | $Solv$ | $\mathrm{diag}(-1,1,0)$
VII0 | $E(2)$ | $\displaystyle\mathrm{diag}(-1,-1,0)$
VIII | $\widehat{SL}(2,\mathbb{R})$ | $\displaystyle\mathrm{diag}(-1,1,1)$
IX | $\mathbb{S}^{3}$ | $id$
Figure 1. The six three-dimensional simply connected Lie groups with compact
quotients. The $3\times 3$ matrix $E$ encapsulates the structure constants.

To simplify the later analysis, the Bach flow equations will be determined in
a basis where the structure constants have the form indicated here and where
the initial metric is diagonal. As shown in [Mil76], such an initial set-up is
always possible:

######  Theorem 3.1.

Given a three-dimensional Lie algebra with structure constants of the form

(10) ${C_{ij}}^{k}=\varepsilon_{ijl}E^{lk},$

and an inner product $g$, there is a basis where

  * •

$g$ is diagonal (the basis is orthogonal),

  * •

the structure constants can still be written in the form (10),

  * •

the matrix $E$ is unchanged.

The proof of this theorem follows from the Principal Axis Theorem and the fact
that structure constants can be rescaled by rescaling the basis. We call the
basis guaranteed by Theorem 3.1 a diagonalizing basis.

In light of the structure afforded by Theorem 3.1, we note the following
general facts about the Ricci tensor and Bach tensor:

######  Proposition 3.2.

Let $N$ be a three-dimensional Lie group with structure constants of the form
of equation (10), and left-invariant metric $g$. Then, in a basis where $E$
and $g$ are diagonal, the Ricci tensor is diagonal and, on
$\mathbb{S}^{1}\times N$, the Bach tensor is diagonal.

For the Ricci tensor, this was established in [IJ92] for the specific matrices
in Figure 1. This proof shows that the property is a consequence of the
diagonal structure of $E$ and $g$, and not specific to particular matrices.
The proof of the general result follows from careful accounting of the indices
in each term found in the formulas for the Ricci and Bach tensors, using the
fact that in three dimensions, the indices are restricted to just three
values.

###### Proof.

First observe that if $E$ is diagonal then equation (10) shows that
${C_{ij}}^{k}$ can only be nonzero if $i$, $j$, and $k$ are all different.
Moreover, if $g$ is diagonal, then the same must be true for any raising or
lowering of any of the indices. Hence any structure constant with a repeated
index must be zero, and in any double sum involving a pair of structure
constants, the two free indices must be equal in order for the result to be
nonzero. Based on these observations, every term in equation (7) must be zero
unless $j=k$ so $Ric$ must be diagonal.

The analysis for the Bach tensor is similar. Looking at equation (4), note
first that the second and third terms are zero since scalar curvature is
constant, and the fifth, sixth, and seventh terms are diagonal since
$Ric^{(2)}$ and $g$ are diagonal. So the only terms to check are the first and
the fourth. For the fourth term, we have

$\mathrm{tr}(Ric\otimes Ric)_{jk}=g^{il}Ric_{ij}Ric_{lk}.$

Since $g$ is diagonal, the terms in this sum are only nonzero when $i=l$, and
then, since $Ric$ is diagonal, we can only have a nonzero term when $j=k$.

Finally for the first term, we use Equation 9 with $T=Ric^{(2)}$ in order to
analyze $(\Delta^{(2)}Ric^{(2)})_{ij}$. Equation (9) has three large terms in
it. For the first term, since $T$ is diagonal, the only way any of the sums of
products can be nonzero is if $p=q$, but then each product becomes a double
sum and so must be zero unless $i=j$. For the second and third terms, one
piece is zero because of a structure constant with a repeated index. For the
rest, the double sums again require the third pair of indices to match in
order to produce something nonzero, and since $T$ is diagonal, the only
nonzero terms appear when $i=j$. ∎

To help with the analysis of curvature along the flow, we have the following:

######  Lemma 3.3.

Let $\\{e_{1},e_{2},e_{3}\\}$ be an orthogonal basis for the tangent space of
a point in a 3-dimensional manifold. Then at that point, the sectional
curvatures are given by

$K(e_{i},e_{j})=\frac{Ric_{ii}}{g_{ii}}+\frac{Ric_{jj}}{g_{jj}}-\frac{S}{2}.$

###### Proof.

On a 3-dimensional manifold, Riemann curvature can be expressed completely in
terms of Ricci and scalar curvature as

$R=Ric\circ g-\frac{S}{4}g\circ g$

where $A\circ B$ is the Kulkarni-Nomizu product. Using an orthogonal basis,
this reduces to

$R_{ijji}=Ric_{ii}g_{jj}+Ric_{jj}g_{ii}-\frac{S}{2}g_{ii}g_{jj}.$

Sectional curvature is given by

$K(v,w)=\frac{R(v,w,w,v)}{|v|^{2}|w|^{2}-\langle v,w\rangle^{2}}$

so

$\displaystyle K(e_{i},e_{j})$
$\displaystyle=\frac{R(e_{i},e_{j},e_{j},e_{i})}{|e_{i}|^{2}|e_{j}|^{2}-\langle
e_{i},e_{j}\rangle^{2}}$
$\displaystyle=\frac{Ric_{ii}g_{jj}+Ric_{jj}g_{ii}-\frac{S}{2}g_{ii}g_{jj}}{g_{ii}g_{jj}}$
$\displaystyle=\frac{Ric_{ii}}{g_{ii}}+\frac{Ric_{jj}}{g_{jj}}-\frac{S}{2}$

as desired. ∎

Lemma 3.3 is all that is needed in this paper since the four-dimensional
manifolds considered are $1\times 3$ products, so the formula above can be
used for the three-dimensional slice, and the sectional curvatures involving
the one-dimensional slice are zero.

##  4\. Ordinary Differential Equations

The ordinary differential equations to which Bach flow reduces on homogeneous
products are analyzed using standard techniques which are recalled here.
First, we appeal to existence and uniqueness of solutions regularly and
without mention. In some instances, the equations of interest are separable
and explicit solutions may be found. When such explicit solutions cannot be
found, the Escape Lemma, which states that if a maximal flow does not exist
for all time then it cannot lie in a compact set, may be used to help
determine the qualitative behavior of solutions. See [Lee03] for details
surrounding these results. In addition to these methods, we make use of a
couple more specialized results which follow.

The following lemma provides a technique for determining the traces of
solutions to a system of two equations involving homogeneous functions.

######  Lemma 4.1.

Let $(x(t),y(t))$ solve the following system

$\frac{dx}{dt}=p(x,y),\ \ \ \ \ \frac{dy}{dt}=q(x,y)$

where $p$ and $q$ are both homogeneous of degree $k$. Suppose $x\neq 0$,
$p(x,y)\neq 0$, and $\frac{q(x,y)}{p(x,y)}\neq\frac{y}{x}$. Then $(x(t),y(t))$
will trace out a subset of the curve

$x=\eta e^{\Psi\left(\frac{y}{x}\right)}$

where

$\Psi(v)=\int\frac{1}{\frac{q(1,v)}{p(1,v)}-v}dv$

and $\eta$ is a constant depending on the initial conditions.

This result, while perhaps a bit obscure, is not difficult to prove. Briefly,
express $\frac{dy}{dx}$ in terms of $p$ and $q$ and then compute the
derivative of $v=\frac{y}{x}$ with respect to $x$, substitute and use
separation of variables.

In general, an integral curve for a vector field can be bounded but fail to
converge to a limit. The next lemma shows that if a coordinate of a bounded
solution does converge, then that component of the vector field must go to
zero. The proof is left to the reader.

######  Lemma 4.2.

Let $x(t)=(x^{1}(t),\ldots,x^{n}(t))$ be a bounded solution to

$\frac{d}{dt}x=V(x)$

where $V$ is a continuous vector field on a domain $D$. Suppose

$\lim_{t\rightarrow\infty}x^{i}(t)=L$

and let $\\{x_{k}\\}=\\{x(t_{k})\\}$, $t_{k}\rightarrow\infty$ be a sequence
of points on the curve that converges to $x_{\infty}\in D$. Then
$V^{i}(x_{\infty})=0$.

##  5\. Bach Flow on locally homogeneous $1\times 3$ products

In this section, the main results of this paper are proved for $1\times 3$
products that are not also $2\times 2$ products. For each universal cover,
explicit formulas for the Bach tensor are found and the evolution of the
metric under Bach flow is determined. In some cases, explicit solutions are
found. When explicit solutions are not found, qualitative behavior is
determined. Limiting behaviors of both the metric and its curvature are also
found.

The general method is as follows: Given an initial metric, a diagonalizing
basis is found so that the metric is diagonal

$h=\mathrm{diag}(h_{00},h_{11},h_{22},h_{33}).$

Its Ricci and scalar curvatures are calculated using equations (7) and (8),
and then using equations (3), (4), and (9) the Bach tensors are calculated. As
indicated by Proposition 3.2, the Bach tensor is also diagonal, so the fact
that the metric is diagonal is preserved along the flow. The solution will be
denoted

$g=\mathrm{diag}(g_{00},g_{11},g_{22},g_{33}).$

One quantity that makes a regular appearance is

$\beta=\frac{1}{6(\det g)^{2}}=\frac{1}{6(\det h)^{2}}.$

This quantity depends on the initial metric, but is constant along the flow.
As a consequence, once an initial metric is chosen, $\beta$ can be treated as
a constant for the whole system.

It turns out that in general the nonzero components of the Bach tensor have
the form

$B_{ii}=\alpha_{i}\,\beta\,p_{i}(g_{11},g_{22},g_{33})\,(g_{00})^{2}g_{ii}$

where $\alpha_{i}$ is a constant and $p_{i}$ is a homogeneous fourth degree
polynomial.

One general fact is that $g_{00}$ is decreasing, as indicated by Proposition
2.2. This fact will not be explicitly included in the specific theorems for
each space. Another general fact is that the flow is defined (at least) on the
interval $[0,\infty)$. In cases where the flow remains bounded, this follows
from the Escape Lemma. In cases where the flow does not remain bounded, this
is discovered after the analysis of each flow is completed and follows from
the work in [Lop18], which shows that the maximal time is finite only if there
is curvature blow-up, and the fact that in all of our cases, curvature remains
bounded.

###  5.1. $\mathbb{R}^{3}$

For this manifold the matrix $E$ used to determine the structure constants in
Theorem 3.1 is the zero matrix, so regardless of the initial metric, the
structure constants are all zero. Hence the Ricci tensor, scalar curvature are
zero, and on $\mathbb{R}\times\mathbb{R}^{3}$ the Bach tensor is zero and so
the metric is static under Bach flow.

###  5.2. $Nil$

For this manifold the matrix used to determine the structure constants in
Theorem 3.1 is

$E=\mathrm{diag}(1,0,0).$

For any metric $g$, using a diagonalizing basis, the Ricci tensor is diagonal
with

$\displaystyle Ric_{11}$ $\displaystyle=\frac{(g_{11})^{2}}{2g_{22}g_{33}}$
$\displaystyle Ric_{22}$ $\displaystyle=-\frac{g_{11}}{2g_{33}}$
$\displaystyle Ric_{33}$ $\displaystyle=-\frac{g_{11}}{2g_{22}}$

and scalar curvature is

$S=-\frac{g_{00}(g_{11})^{2}}{2\det g}.$

The Bach tensor on $\mathbb{R}\times Nil$ is diagonal with

$\displaystyle B_{00}$ $\displaystyle=-\beta\,(g_{00})^{3}(g_{11})^{4}$
$\displaystyle B_{11}$ $\displaystyle=-5\beta\,(g_{00})^{2}(g_{11})^{5}$
$\displaystyle B_{22}$ $\displaystyle=3\beta\,(g_{00})^{2}(g_{11})^{4}g_{22}$
$\displaystyle B_{33}$ $\displaystyle=3\beta\,(g_{00})^{2}(g_{11})^{4}g_{33}.$

With the Bach tensor in hand, we have the following theorem:

######  Theorem 5.1.

On $\widehat{M}=\mathbb{R}\times Nil$ the solutions to equation (1) in a
diagonalizing basis for $h$ are given by

$\displaystyle g_{00}(t)$ $\displaystyle=\left(\gamma
t+(h_{00})^{-22}\right)^{-\frac{1}{22}}$ $\displaystyle g_{11}(t)$
$\displaystyle=\alpha\left(\gamma t+(h_{00})^{-22}\right)^{-\frac{5}{22}}$
$\displaystyle g_{22}(t)$ $\displaystyle=h_{22}(h_{00})^{3}\left(\gamma
t+(h_{00})^{-22}\right)^{\frac{3}{22}}$ $\displaystyle g_{33}(t)$
$\displaystyle=h_{33}(h_{00})^{3}\left(\gamma
t+(h_{00})^{-22}\right)^{\frac{3}{22}}$

where

$\alpha=\frac{h_{11}}{(h_{00})^{5}}$

and

$\gamma=22\alpha^{4}\beta=\frac{11}{3(\det
h)^{2}}\left(\frac{h_{11}}{(h_{00})^{5}}\right)^{4}.$

###### Proof.

Note that the first and second equations are coupled and the third and fourth
equations depend on the first and second solutions, but are otherwise
uncoupled. Because everything is multiplicative, we can solve explicitly for
$g_{00}$, and $g_{11}$, and then $g_{22}$ and $g_{33}$.

We do this by making a guess

$g_{11}=\alpha(g_{00})^{k}$

and then comparing the resulting differential equations to get

$g_{00}(t)=\left(\gamma t+(h_{00})^{-22}\right)^{-\frac{1}{22}}$

and

$g_{11}(t)=\alpha\left(\gamma t+(h_{00})^{-22}\right)^{-\frac{5}{22}}$

where

$\gamma=22\alpha^{4}\beta.$

Then we can solve for $g_{22}$ and $g_{33}$:

$g_{22}(t)=h_{22}(h_{00})^{3}\left(\gamma
t+(h_{00})^{-22}\right)^{\frac{3}{22}}$

and

$g_{33}(t)=h_{33}(h_{00})^{3}\left(\gamma
t+(h_{00})^{-22}\right)^{\frac{3}{22}}.$

∎

With these solutions in hand, we find two dimensions collapse in the limit as
$t\rightarrow\infty$. The “$g_{00}$” direction collapses more slowly than the
first dimension in $Nil$. Meanwhile, the other two dimensions grow at the same
rate, preserving their aspect ratio. These solutions are immortal, but not
ancient.

All components of the Ricci tensor converge to zero in the limit, and using
Lemma 3.3 we have the following:

######  Theorem 5.2.

Let $M$ be a compact quotient of $\mathbb{R}\times Nil$ and let $p\in M$. Let
$g$ solve equation (1) where $h$ is locally homogeneous. Then $(M,g,p)$
collapses to a flat surface in the pointed Gromov-Hausdorff topology.

  * $\Box$

read through. Done up to lemma 5.23

  * $\Box$

remove commented parts

###  5.3. $Solv$

For this manifold the matrix used to determine the structure constants in
Theorem 3.1 is

$E=\mathrm{diag}(-1,1,0).$

For any metric $g$, using a diagonalizing basis, the Ricci tensor is diagonal
with

$\displaystyle Ric_{11}$
$\displaystyle=\frac{(g_{11})^{2}-(g_{22})^{2}}{2g_{22}g_{33}}$ $\displaystyle
Ric_{22}$ $\displaystyle=\frac{(g_{22})^{2}-(g_{11})^{2}}{2g_{11}g_{33}}$
$\displaystyle Ric_{33}$
$\displaystyle=-\frac{(g_{11}+g_{22})^{2}}{2g_{11}g_{22}}$

and scalar curvature is

$S=-\frac{(g_{11}+g_{22})^{2}}{2g_{11}g_{22}g_{33}}.$

The Bach tensor is diagonal with

$\displaystyle B_{00}$ $\displaystyle=-\beta\,p(g_{11},g_{22})\,(g_{00})^{3}$
$\displaystyle B_{11}$
$\displaystyle=-\beta\,q(g_{11},g_{22})\,(g_{00})^{2}g_{11}$ $\displaystyle
B_{22}$ $\displaystyle=-\beta\,q(g_{22},g_{11})\,(g_{00})^{2}g_{22}$
$\displaystyle B_{33}$
$\displaystyle=3\beta\,p(g_{11},g_{22})\,(g_{00})^{2}g_{33}$

where

$\displaystyle p(x,y)$ $\displaystyle=x^{4}+x^{3}y+xy^{3}+y^{4}$
$\displaystyle q(x,y)$ $\displaystyle=5x^{4}+3x^{3}y-xy^{3}-3y^{4}.$

With the Bach tensor in hand, we have the following theorem:

######  Theorem 5.3.

On $\widehat{M}=\mathbb{R}\times Solv$ Every solution to equation (1) in a
diagonalizing basis has the following properties:

  * •

$g_{00},g_{11},g_{22}\rightarrow 0$;

  * •

$g_{33}\rightarrow\infty$ monotonically;

  * •

$\frac{g_{11}}{g_{22}}\rightarrow 1$.

If $h_{11}=h_{22}$, then

$\displaystyle g_{00}$ $\displaystyle=\mu^{\frac{1}{2}}\Bigl{(}24\mu\beta
t+(h_{11})^{-6}\Bigr{)}^{-\frac{1}{6}}$ $\displaystyle g_{11}$
$\displaystyle=g_{22}=\Bigl{(}24\mu\beta
t+(h_{11})^{-6}\Bigr{)}^{-\frac{1}{6}}$ $\displaystyle g_{33}$
$\displaystyle=(h_{11})^{3}h_{33}\Bigl{(}24\mu\beta
t+(h_{11})^{-6}\Bigr{)}^{\frac{1}{2}}.$

Otherwise, if (without loss of generality) $h_{11}<h_{22}$, then

  * •

$g_{11}<g_{22}$ for the entire flow;

  * •

$g_{22}$ is decreasing;

  * •

$\frac{g_{11}}{g_{22}}$ is increasing;

  * •

$g_{11}$ and $g_{22}$ are related by

$(g_{11}g_{22})^{25}=\eta\,(g_{22}-g_{11})^{4}\left(2(g_{22})^{2}+g_{11}g_{22}+2(g_{11})^{2}\right)^{3}$

where

$\eta=\frac{(h_{11}h_{22})^{25}}{(h_{22}-h_{11})^{4}\left(2(h_{22})^{2}+h_{11}h_{22}+2(h_{11})^{2}\right)^{3}}.$

It turns out that the two polynomials $p$ and $q$ also make an appearance in
the next section so we establish some facts about them here.

######  Lemma 5.4.

The polynomial $p(x,y)$ is symmetric, homogeneous of degree 4, positive when
$x$ or $y$ is nonzero, and can be factored as

$p(x,y)=(x+y)^{2}(x^{2}-xy+y^{2}).$

The polynomial $q(x,y)$ is homogeneous of degree 4 and can be factored as

$q(x,y)=(x+y)(5x^{3}-2x^{2}y+2xy^{2}-3y^{3}).$

The cubic factor has exactly one real factor $(\alpha x-y)$ where $\alpha$ is
about 1.225. If $x>y$, $q(x,y)>0$.

With these facts about $p$ and $q$ established, we proceed with the proof of
Theorem 5.3.

###### Proof.

(of Theorem 5.3) Since $B_{00}$ and $B_{33}$ are so similar, we can compute:

$\frac{\frac{d}{dt}g_{33}}{\frac{d}{dt}g_{00}}=\frac{B_{33}}{B_{00}}=\frac{-3g_{33}}{g_{00}}$

which implies

(11) $g_{33}=\gamma(g_{00})^{-3}.$

where $\gamma=(h_{00})^{3}h_{33}$. Since, by Proposition 2.2, $g_{00}$ is
decreasing, this shows $g_{33}$ must be increasing. Since $\det g$ is
constant, using equation (11) we have

(12) $(g_{00})^{2}=\mu g_{11}g_{22}$

where

$\mu=\frac{\gamma}{\det h}.$

Incorporating these identities into the formulas for $B_{11}$ and $B_{22}$ we
have

(13) $\frac{d}{dt}g_{11}=-\mu\beta\,q(g_{11},g_{22})\,(g_{11})^{2}g_{22}$

and

$\frac{d}{dt}g_{22}=-\mu\beta\,q(g_{22},g_{11})\,(g_{22})^{2}g_{11}.$

Because of the symmetry in these equations, we may, without loss of
generality, restrict our attention to the region defined by $0\leq g_{11}\leq
g_{22}$.

If $h_{11}=h_{22}$ then, focusing on $g_{11}$ we get

$\frac{d}{dt}g_{11}=-4\mu\beta g_{11}^{7}$

which is separable. Solving, we get

$g_{11}=g_{22}=\Bigl{(}24\mu\beta t+(h_{11})^{-6}\Bigr{)}^{-\frac{1}{6}}.$

Next, we solve for $g_{00}$ and $g_{33}$ using equations (11) and (12). We
have

$g_{00}=\mu^{\frac{1}{2}}\Bigl{(}24\mu\beta
t+(h_{11})^{-6}\Bigr{)}^{-\frac{1}{6}}$

and

$g_{33}=(h_{11})^{3}h_{33}\Bigl{(}24\mu\beta
t+(h_{11})^{-6}\Bigr{)}^{\frac{1}{2}}.$

In this special case, we see that under Bach flow, any compact quotient of
$\mathbb{R}\times Solv$ collapses to a curve in the limit.

If $h_{11}<h_{22}$ then $g_{11}<g_{22}$ by existence and uniqueness, since we
have a solution that preserves the equality $g_{11}=g_{22}$. With this
inequality preserved, from the properties of $q$, looking at equation (13) we
find that $g_{22}$ is decreasing. Also

$\displaystyle\frac{d}{dt}\left(\frac{g_{11}}{g_{22}}\right)$
$\displaystyle=\frac{\left(\frac{d}{dt}g_{11}\right)g_{22}-g_{11}\left(\frac{d}{dt}g_{22}\right)}{(g_{22})^{2}}$
$\displaystyle=4\mu\beta\,(g_{11}+g_{22})(g_{22}-g_{11})\bigl{(}2(g_{11})^{2}+g_{11}g_{22}+2(g_{22})^{2}\bigr{)}\,(g_{11})^{2}$
$\displaystyle>0$

so we find that $\frac{g_{11}}{g_{22}}$ is increasing. This fraction is
bounded above by 1, so it must converge. We will see below that it converges
to 1.

The fact that this fraction is increasing also implies that $g_{11}$ cannot
converge to zero unless $g_{22}$ does as well. This, combined with Lemma 4.2
implies that $g_{22}$ and hence $g_{11}$ must converge to 0 since the only
points in the domain of interest where $\frac{d}{dt}g_{22}$ is zero are along
the $g_{22}$ axis.

Next, letting $v=\frac{g_{11}}{g_{22}}$ we have

$\displaystyle\frac{dg_{11}}{dg_{22}}$
$\displaystyle=\frac{q(v,1)v^{2}}{q(1,v)v}$
$\displaystyle=v\frac{(5v^{3}-2v^{2}+2v-3)}{(5-2v+2v^{2}-3v^{3})}.$

Therefore by Lemma 4.1, the solution curves for our original differential
equation satisfy the equation

$g_{22}=\frac{\tilde{\eta}\left(1-\frac{g_{11}}{g_{22}}\right)^{\frac{1}{10}}\left(2+\frac{g_{11}}{g_{22}}+2\left(\frac{g_{11}}{g_{22}}\right)^{2}\right)^{\frac{3}{40}}}{\left(\frac{g_{11}}{g_{22}}\right)^{\frac{5}{8}}}$

where $\tilde{\eta}$ is a constant determined by the initial conditions.
Multiplying both sides by $\left(\frac{g_{11}}{g_{22}}\right)^{\frac{5}{8}}$,
we have

(14)
$\left(\frac{g_{11}}{g_{22}}\right)^{\frac{5}{8}}g_{22}=\tilde{\eta}\left(1-\frac{g_{11}}{g_{22}}\right)^{\frac{1}{10}}\left(2+\frac{g_{11}}{g_{22}}+2\left(\frac{g_{11}}{g_{22}}\right)^{2}\right)^{\frac{3}{40}}.$

This can be rewritten as

$(g_{11}g_{22})^{25}=\eta(g_{22}-g_{11})^{4}\left(2(g_{22})^{2}+g_{11}g_{22}+2(g_{11})^{2}\right)^{3}$

where $\eta=\tilde{\eta}^{40}$. This is true in particular at $t=0$, so

$\eta=\frac{(h_{11}h_{22})^{25}}{(h_{22}-h_{11})^{4}\left(2(h_{22})^{2}+h_{11}h_{22}+2(h_{11})^{2}\right)^{3}}.$

Taking the limit as $t\rightarrow\infty$, the left side of equation (14) must
be zero, and therefore so must the right. Since the second factor is positive,
it follows that
$\lim_{t\rightarrow\infty}\left(1-\frac{g_{11}}{g_{22}}\right)^{\frac{1}{10}}=0$
and so $\lim_{t\rightarrow\infty}\frac{g_{11}}{g_{22}}=1$.

From the analysis above, we know that in general, $g_{11}$ and $g_{22}$ go to
zero as $t$ goes to infinity. From equations (12) and (11), we then know that
$g_{00}$ also goes to zero and $g_{33}$ grows to infinity. Therefore, under
Bach flow, any compact quotient of $\mathbb{R}\times Solv$ collapses to a
curve in the limit. These facts, combined with the fact that
$\frac{g_{11}}{g_{22}}$ goes to 1 imply that a general solution approaches the
specific solution found above in the limit. ∎

We now have the following:

######  Theorem 5.5.

Let $M$ be a compact quotient of $\mathbb{R}\times Solv$ and let $p\in M$. Let
$g$ solve equation (1) where $h$ is locally homogeneous. Then $(M,g,p)$
collapses to a line in the pointed Gromov-Hausdorff topology.

###### Proof.

From the previous theorem, we know that three dimensions collapse, while one
expands. Moreover, working in a diagonalizing basis, we find that $Ric_{11}$
and $Ric_{22}$ converge to zero while $Ric_{33}$ converges to $-2$. Therefore,
by Lemma 3.3, all the sectional curvatures go to zero in the limit. ∎

###  5.4. $E(2)$

For this manifold the matrix used to determine the structure constants in
Theorem 3.1 is

$E=\mathrm{diag}(-1,-1,0).$

For any metric $g$, using a diagonalizing basis, the Ricci tensor is diagonal
with

$\displaystyle Ric_{11}$
$\displaystyle=\frac{(g_{11})^{2}-(g_{22})^{2}}{2g_{22}g_{33}}$ $\displaystyle
Ric_{22}$ $\displaystyle=\frac{(g_{22})^{2}-(g_{11})^{2}}{2g_{11}g_{33}}$
$\displaystyle Ric_{33}$
$\displaystyle=-\frac{(g_{11}-g_{22})^{2}}{2g_{11}g_{22}}$

and scalar curvature is

$S=-\frac{(g_{11}-g_{22})^{2}}{2g_{11}g_{22}g_{33}}$

The Bach tensor is diagonal with

$\displaystyle B_{00}$ $\displaystyle=-\beta\,p(-g_{11},g_{22})\,(g_{00})^{3}$
$\displaystyle B_{11}$
$\displaystyle=-\beta\,q(-g_{11},g_{22})\,(g_{00})^{2}g_{11}$ $\displaystyle
B_{22}$ $\displaystyle=-\beta\,q(g_{22},-g_{11})\,(g_{00})^{2}g_{22}$
$\displaystyle B_{33}$
$\displaystyle=3\beta\,p(-g_{11},g_{22})\,(g_{00})^{2}g_{33},$

where $p$ and $q$ were defined is Section 5.3.

######  Theorem 5.6.

On $\widehat{M}=\mathbb{R}\times E(2)$ Every solution to equation (1) in a
diagonalizing basis has the following properties:

  * •

$g_{11}$ and $g_{22}$ are related by

$(g_{11}g_{22})^{25}=\eta(g_{22}+g_{11})^{4}\left(2(g_{22})^{2}-g_{11}g_{22}+2(g_{11})^{2}\right)^{3}$

where

$\eta=\frac{(h_{11}h_{22})^{25}}{(h_{22}+h_{11})^{4}\left(2(h_{22})^{2}-h_{11}h_{22}+2(h_{11})^{2}\right)^{3}};$

  * •

the flow exists for all time and as $t\rightarrow\infty$,

$\displaystyle g_{11},g_{22}$
$\displaystyle\rightarrow(432\eta)^{\frac{1}{40}}$ $\displaystyle g_{00}$
$\displaystyle\rightarrow\left(\frac{(h_{00})^{3}h_{33}}{\det
h}\right)^{\frac{1}{2}}(432\eta)^{\frac{1}{40}}$ $\displaystyle g_{33}$
$\displaystyle\rightarrow\left(\frac{(\det
h)^{3}}{(h_{00})^{3}h_{33}}\right)^{\frac{1}{2}}(432\eta)^{-\frac{3}{40}};$

  * •

$g_{33}$ is increasing.

If $h_{11}=h_{22}$, then the solution is static. Otherwise, if (without loss
of generality) $h_{11}<h_{22}$, then

  * •

$g_{11}<g_{22}$ for the entire flow,

  * •

$g_{11}$ is increasing,

  * •

$g_{22}$ is decreasing.

###### Proof.

Since the only difference between this system and that of $\mathbb{R}\times
Solv$ is the minus sign on one of the variables in $p$ and $q$, most of the
initial analysis of the previous section carries over and we have

(15) $g_{33}=\gamma(g_{00})^{-3}.$

with $\gamma=(h_{00})^{3}h_{33}$. Because of this inverse relationship, since
$g_{00}$ is decreasing, we find that $g_{33}$ must be increasing. We also have

(16) $(g_{00})^{2}=\mu g_{11}g_{22}$

where

$\mu=\frac{\gamma}{\det h}.$

Incorporating these identities into the formulas for $B_{11}$ and $B_{22}$ we
have

$\displaystyle\frac{d}{dt}g_{11}$
$\displaystyle=-\mu\beta\,q(-g_{11},g_{22})\,(g_{11})^{2}g_{22}$
$\displaystyle\frac{d}{dt}g_{22}$
$\displaystyle=-\mu\beta\,q(g_{22},-g_{11})\,(g_{22})^{2}g_{11}.$

Because of the symmetry in these equations, we may, without loss of
generality, restrict our attention to the region defined by $0\leq g_{11}\leq
g_{22}$. If $h_{11}=h_{22}$, then $B=0$ and we have a set of stationary
solutions corresponding to the flat metrics on $E(2)$. If $h_{11}<h_{22}$ then
$g_{11}<g_{22}$ for all time and from the poperties of $q$, we find that
$g_{11}$ is increasing and $g_{22}$ is decreasing. Therefore, both must
converge and by Lemma 4.2 this can only happen at a point where
$g_{11}=g_{22}$.

As with $Solv$, we can say a bit more about the curves traced out by the
solutions using Lemma 4.1. Except for two minus signs, the analysis here is
almost identical to that for $Solv$ and we find

(17)
$(g_{11}g_{22})^{25}=\eta(g_{22}+g_{11})^{4}\left(2(g_{22})^{2}-g_{11}g_{22}+2(g_{11})^{2}\right)^{3}$

where

$\eta=\frac{(h_{11}h_{22})^{25}}{(h_{22}+h_{11})^{4}\left(2(h_{22})^{2}-h_{11}h_{22}+2(h_{11})^{2}\right)^{3}}.$

Let $g_{ii}(\infty)$ be the limit of $g_{ii}$ as $t\rightarrow\infty$. Then we
know that $g_{11}(\infty)=g_{22}(\infty)$ and using equation (17) we find

$\bigl{(}g_{11}(\infty)\bigr{)}^{50}=\eta\,\bigl{(}2g_{11}(\infty)\bigr{)}^{4}\bigl{(}3(g_{11}(\infty))^{2}\bigr{)}^{3}$

and so

$g_{11}(\infty)=g_{22}(\infty)=(432\eta)^{\frac{1}{40}}.$

Then by equation (16)

$g_{00}(\infty)=\mu^{\frac{1}{2}}(432\eta)^{\frac{1}{40}}=\left(\frac{(h_{00})^{3}h_{33}}{\det
h}\right)^{\frac{1}{2}}(432\eta)^{\frac{1}{40}}$

and by equation (15)

$g_{33}(\infty)=\gamma\bigl{[}\mu^{\frac{1}{2}}(432\eta)^{\frac{1}{40}}\bigr{]}^{-3}=\left(\frac{(\det
h)^{3}}{(h_{00})^{3}h_{33}}\right)^{\frac{1}{2}}(432\eta)^{-\frac{3}{40}}.$

∎

Finally, we have the following

######  Theorem 5.7.

Let $M$ be a compact quotient of $\mathbb{R}\times E(2)$. Let $g$ solve
equation (1) where $h$ is locally homogeneous. Then $(M,g)$ converges to a
flat four-dimensional manifold in the Gromov-Hausdorff topology.

###### Proof.

None of the components of the metric converge to zero, so there is no
collapse. Since $g_{11}-g_{22}\rightarrow 0$ as $t\rightarrow\infty$, looking
at the Ricci curvature, we find that the manifold becomes Ricci-flat in the
limit. By Lemma 3.3, so do the sectional curvatures. ∎

###  5.5. $\widehat{SL}(2,\mathbb{R})$

For this manifold the matrix used to determine the structure constants in
Theorem 3.1 is

$E=\mathrm{diag}(-1,1,1).$

For any metric $g$, using a diagonalizing basis, the Ricci tensor is diagonal
with

$\displaystyle Ric_{11}$
$\displaystyle=\frac{(g_{11})^{2}-(g_{22}-g_{33})^{2}}{2g_{22}g_{33}}$
$\displaystyle Ric_{22}$
$\displaystyle=\frac{(g_{22})^{2}-(g_{11}+g_{33})^{2}}{2g_{11}g_{33}}$
$\displaystyle Ric_{33}$
$\displaystyle=\frac{(g_{33})^{2}-(g_{11}+g_{22})^{2}}{2g_{11}g_{22}}$

and scalar curvature is

$S=-\frac{(g_{11})^{2}+(g_{22})^{2}+(g_{33})^{2}+2(g_{11}g_{22}+g_{11}g_{33}-g_{22}g_{33})}{2g_{11}g_{22}g_{33}}.\\\
$

The Bach tensor is diagonal with

$\displaystyle B_{00}$
$\displaystyle=-\beta\,p(-g_{11},g_{22},g_{33})\,(g_{00})^{3}$ $\displaystyle
B_{11}$ $\displaystyle=-\beta\,q(-g_{11},g_{22},g_{33})\,(g_{00})^{2}g_{11}$
$\displaystyle B_{22}$
$\displaystyle=-\beta\,q(g_{22},-g_{11},g_{33})\,(g_{00})^{2}g_{22}$
$\displaystyle B_{33}$
$\displaystyle=-\beta\,q(g_{33},-g_{11},g_{22})\,(g_{00})^{2}g_{33}$

where

$\displaystyle p(x,y,z)$ $\displaystyle=x^{4}-x^{3}(y+z)+x^{2}yz$
$\displaystyle\quad+x(-y^{3}+y^{2}z+yz^{2}-z^{3})$
$\displaystyle\quad+y^{4}-y^{3}z-yz^{3}+z^{4}$

and

$\displaystyle q(x,y,z)$ $\displaystyle=5x^{4}-3x^{3}(y+z)+x^{2}yz$
$\displaystyle\quad+x(y^{3}-y^{2}z-yz^{2}+z^{3})$
$\displaystyle\quad-3y^{4}+3y^{3}z+3yz^{3}-3z^{4}.$

The sign choices made in the formulas for the Bach tensor here come from the
fact that $p$ and $q$ are also used in the next section for $\mathbb{S}^{3}$,
where no minus signs are needed in the expressions for the Bach tensor.

With the Bach tensor in hand, we have the following theorem:

######  Theorem 5.8.

On $\widehat{M}=\mathbb{R}\times\widehat{SL}(2,\mathbb{R})$, every solution to
equation (1) in a diagonalizing basis has the following properties:

  * •

$g_{00},g_{11}\rightarrow 0$;

  * •

$g_{22},g_{33}\rightarrow\infty$;

  * •

$g_{33}-g_{22}\rightarrow 0$.

The two polynomials $p$ and $q$ also make an appearance in the next section so
we establish some facts about them here. For $p$, we have the following lemma,
the proof of which is left to the reader.

######  Lemma 5.9.

The polynomial $p$ has the following properties:

  * •

It is invariant under permutation of the variables;

  * •

$p(-x,-y,-z)=p(x,y,z)$;

  * •

it is always nonnegative;

  * •

it is equal to zero if and only if $x=y=z$ or two variables are equal and the
third is zero.

Note that $q$ is symmetric in the last two variables. Because of this, and the
fact that the flow equations for $g_{22}$ and $g_{33}$ are essentially the
same, we say that without loss of generality, $g_{22}\leq g_{33}$.

The qualitative behavior of the flow is determined through a number of
estimates which arise from monotonicity of various quantities. To keep things
clear, these monotonicity results are presented in the following lemmas.

######  Lemma 5.10.

Suppose $h_{22}<h_{33}$. Then $\frac{g_{33}}{g_{22}}$ decreases along the
flow.

###### Proof.

Writing out the quotient rule and plugging in the differential equations for
$g_{22}$ and $g_{33}$, we have

$\frac{d}{dt}\frac{g_{33}}{g_{22}}=-\beta\,\Bigl{(}q(g_{33},-g_{11},g_{22})-q(g_{22},-g_{11},g_{33})\Bigr{)}\,(g_{00})^{2}\frac{g_{33}}{g_{22}}.$

Writing out and simplifying the factor involving the $q$’s, we get

$\displaystyle q(g_{33},-g_{11},g_{22})$
$\displaystyle-q(g_{22},-g_{11},g_{33})$
$\displaystyle=2(g_{33}-g_{22})[4(g_{33})^{3}+2(g_{33})^{2}g_{22}+2g_{33}(g_{22})^{2}+4(g_{22})^{3}$
$\displaystyle\qquad\qquad\qquad\quad+g_{11}(3(g_{33})^{2}+2g_{33}g_{22}+3(g_{22})^{2})+(g_{11})^{3}]$

and this is positive since $g_{22}<g_{33}.$ ∎

######  Lemma 5.11.

Suppose $h_{22}\leq h_{33}$. Then $g_{00}g_{22}$ increases along the flow.

###### Proof.

Writing out the product rule and plugging in the differential equations for
$g_{00}$ and $g_{22}$, we have

$\frac{d}{dt}(g_{00}g_{22})=-\beta\,\Bigl{(}p(-g_{11},g_{22},g_{33})+q(g_{22},-g_{11},g_{33})\Bigr{)}\,(g_{00})^{3}g_{22}.$

Writing out and simplifying the factor involving the $p$ and $q$, we get

$\displaystyle p(-g_{11},g_{22},g_{33})$
$\displaystyle+q(g_{22},-g_{11},g_{33})$
$\displaystyle=-2\Bigl{[}(g_{33}-g_{22})[(g_{33})^{3}+(g_{33})^{2}g_{22}+g_{33}(g_{22})^{2}+3(g_{22})^{3}$
$\displaystyle\qquad\qquad\qquad\qquad+g_{11}((g_{33})^{2}+g_{33}g_{22}+2(g_{22})^{2})]$
$\displaystyle\qquad\qquad\quad+(g_{11})^{3}(g_{11}+g_{33})\Bigr{]}.$

This is negative when $g_{22}\leq g_{33}$ so $\frac{d}{dt}(g_{00}g_{22})$ is
positive. ∎

Since $g_{00}$ decreases along the flow, an immediate consequence of this
lemma is that if $h_{22}\leq h_{33}$ then $g_{22}$ by itself increases along
the flow. In fact, a consequence of the proof is that $\frac{d}{dt}g_{22}$ is
positive.

######  Lemma 5.12.

Suppose $h_{22}\leq h_{33}$. Then ($g_{11})^{\frac{3}{5}}g_{22}$ increases
along the flow.

###### Proof.

Writing out the product rule and plugging in the differential equations for
$g_{11}$ and $g_{22}$, we have

$\frac{d}{dt}\left[(g_{11})^{\frac{3}{5}}g_{22}\right]=-\frac{\beta}{5}\,\Bigl{(}3q(-g_{11},g_{22},g_{33})+5q(g_{22},-g_{11},g_{33})\Bigr{)}\,(g_{00})^{2}(g_{11})^{\frac{3}{5}}g_{22}.$

Writing out and simplifying the factor involving $q$ we get

$\displaystyle 3q(-g_{11},g_{22},g_{33})$
$\displaystyle+5q(g_{22},-g_{11},g_{33})$
$\displaystyle=-2\Bigl{[}(g_{33}-g_{22})[12(g_{33})^{3}+5(g_{33})^{2}g_{22}+5g_{33}(g_{22})^{2}+8(g_{22})^{3}$
$\displaystyle\qquad\qquad\qquad\qquad+g_{11}(9(g_{33})^{2}+5g_{33}g_{22}+6(g_{22})^{2})]$
$\displaystyle\qquad\qquad\quad+g_{11}^{2}[g_{33}g_{22}+g_{11}(3g_{33}-2g_{22})]\Bigr{]}.$

This is always negative since $g_{22}\leq g_{33},$ so
$\frac{d}{dt}\left[(g_{11})^{\frac{3}{5}}g_{22}\right]$ is positive. ∎

######  Lemma 5.13.

Suppose $g_{22}$ and $g_{33}$ diverge and $g_{11}$ converges to zero. Then the
quantity $\frac{g_{33}-g_{22}}{g_{11}}$ converges to zero along the flow.

###### Proof.

First note that if $g_{22}=g_{33}$ then the result is true immediately. If,
without loss of generality, $g_{22}<g_{33}$, we proceed in two steps. The
first step is similar to the lemmas above. We have

$\frac{d}{dt}(g_{33}-g_{22})=-\beta\,s(g_{11},g_{22},g_{33})\,(g_{00})^{2}(g_{33}-g_{22})$

where

$\displaystyle s(x,y,z)$ $\displaystyle=-3x^{4}-x^{3}(y+z)-x^{2}yz$
$\displaystyle\quad+x(3y^{3}+5y^{2}z+5yz^{2}+3z^{3})$
$\displaystyle\quad+5y^{4}+5y^{3}z+4y^{2}z^{2}+5yz^{3}+5z^{4}.$

With this we have

$\frac{d}{dt}\frac{g_{33}-g_{22}}{(g_{11})^{2}}=-\beta\,\Bigl{[}s(g_{11},g_{22},g_{33})-2q(-g_{11},g_{22},g_{33})\Bigr{]}\,(g_{00})^{2}\,\frac{g_{33}-g_{22}}{(g_{11})^{2}}.$

Looking at the polynomial in brackets, we have

$\displaystyle s(g_{11},g_{22},g_{33})$
$\displaystyle-2q(-g_{11},g_{22},g_{33})$
$\displaystyle=g_{33}[(g_{33})^{2}+(g_{22})^{2}](g_{33}-g_{22})$
$\displaystyle\qquad+7g_{33}[(g_{33})^{3}-(g_{11})^{3}]+7g_{22}[(g_{22})^{3}-(g_{11})^{3}]$
$\displaystyle\qquad+3g_{22}g_{33}[g_{22}g_{33}-(g_{11})^{2}]$
$\displaystyle\qquad+g_{11}[5(g_{22})^{3}+3(g_{22})^{2}g_{33}+g_{22}(g_{33})^{2}+5(g_{33})^{3}]$
$\displaystyle\qquad+3(g_{33})^{4}+4(g_{22})^{4}-13(g_{11})^{4}.$

Since $g_{22}$ and $g_{33}$ go to infinity, and $g_{11}$ goes to zero, this
must eventually become and stay positive and so the fraction
$\frac{g_{33}-g_{22}}{(g_{11})^{2}}$ must eventually decrease. Since
$\frac{1}{g_{11}}$ diverges, this impiles that $\frac{g_{33}-g_{22}}{g_{11}}$
must converge to zero. ∎

We are now in a position to prove Theorem 5.8. The proof requires considering
a few different possibilities and ruling out any option other than what is
described in the theorem.

###### Proof.

(of Theorem 5.8) Without loss of generality, we may restrict our attention to
flows that satisfy $g_{22}\leq g_{33}$. Now, first suppose $g_{00}\rightarrow
n>0$, with the goal of ruling this possibility out. Consider two
possibilities. Suppose first that $g_{22}$ remains bounded above. By Lemma
5.10 $g_{33}$ must remain bounded as well. From this, since $\det g$ is
constant, we know that $g_{11}$ remains bounded above and also below by some
positive number.

By Lemma 5.11, $g_{22}$ is increasing, since $g_{00}$ is decreasing, so since
$g_{22}$ is bounded, it must converge. By Lemma 4.2, there must be a point
where $\frac{d}{dt}g_{22}=0$. This contradicts the fact, from the proof of
Lemma 5.11, that $\frac{d}{dt}g_{22}$ is positive in the given domain.

This implies that $g_{22}$ goes to infinity, and so must $g_{33}$. Since we
are still working with the possibility that $g_{00}$ does not go to zero, we
may conclude that $g_{11}$ converges to zero, again since $\det g$ is
constant.

Now consider the product $g_{00}(g_{11})^{\frac{6}{5}}g_{22}g_{33}$. Note that
this is equal to $(\det g)(g_{11})^{\frac{1}{5}}$ which must go to zero since
$\det g$ is constant and $g_{11}$ goes to zero. On the other hand

$\displaystyle g_{00}(g_{11})^{\frac{6}{5}}g_{22}g_{33}$
$\displaystyle=g_{00}[(g_{11})^{\frac{3}{5}}g_{22}][(g_{11})^{\frac{3}{5}}g_{33}]$
$\displaystyle\geq g_{00}[(g_{11})^{\frac{3}{5}}g_{22}]^{2}.$

By Lemma 5.12, the squared factor is increasing. But this implies that
$g_{00}$ must go to zero, a contradiction.

So we may conclude that $g_{00}$ converges to zero. Knowing this, since
$g_{00}g_{22}$ is increasing by Lemma 5.11, $g_{22}$ and hence $g_{33}$ must
both diverge to $\infty$. But then, again by Lemma 5.11, $g_{00}g_{22}g_{33}$
diverges and so $g_{11}$ must go to zero since $g_{00}g_{11}g_{22}g_{33}=\det
g$ is constant.

Finally, we have now established the hypotheses for Lemma 5.13 so we may
conclude that $g_{33}-g_{22}\rightarrow 0$. ∎

With the limiting behavior of the metric established, the next step is to
determine the curvature.

######  Proposition 5.14.

On $\widehat{M}=\mathbb{R}\times\widehat{SL}(2,\mathbb{R})$ for every solution
to equation (1) in a diagonalizing basis, $\frac{Ric_{11}}{g_{11}}$ converges
to 0, and $Ric_{22}$ and $Ric_{33}$ both converge to $-1$. The scalar
curvature converges to 0 as well.

###### Proof.

We have

$\frac{Ric_{11}}{g_{11}}=\frac{(g_{11})^{2}-(g_{22}-g_{33})^{2}}{2g_{11}g_{22}g_{33}}.$

By Theorem 5.8, the numerator goes to zero and, since the determinant is
constant, the denominator goes to infinity.

For $Ric_{22}$ we rewrite:

$\displaystyle Ric_{22}$
$\displaystyle=\frac{(g_{22})^{2}-(g_{11}+g_{33})^{2}}{2g_{11}g_{33}}$
$\displaystyle=\frac{g_{22}-g_{33}}{g_{11}}\frac{1}{2}\left(\frac{g_{22}}{g_{33}}+1\right)-\frac{g_{11}}{2g_{33}}-1.$

By Lemma 5.13, the first factor in the first term goes to zero and the rest of
the term is bounded. The middle term also goes to zero. The computation for
$Ric_{33}$ is similar.

For scalar curvature, we rewrite to get:

$S=-\frac{g_{11}}{2g_{22}g_{33}}-\frac{g_{22}-g_{33}}{g_{11}}\frac{g_{22}-g_{33}}{g_{22}g_{33}}-\frac{1}{g_{33}}-\frac{1}{g_{22}}$

and, by Theorem 5.8 and Lemma 5.13, all these terms go to zero. ∎

Finally, we have the following:

######  Theorem 5.15.

Let $M$ be a compact quotient of $\mathbb{R}\times\widehat{SL}(2,\mathbb{R})$
and let $p\in M$. Let $g$ solve equation (1) where $h$ is locally homogeneous.
Then $(M,g,p)$ converges to a flat surface in the pointed Gromov-Hausdorff
topology.

###### Proof.

From the previous proposition, and by Lemma 3.3, the sectional curvatures all
go to zero along the flow. ∎

###  5.6. $\mathbb{S}^{3}$

For this manifold the matrix used to determine the structure constants in
Theorem 3.1 is

$E=id.$

For any metric $g$, using a diagonalizing basis, the Ricci tensor is diagonal
with

$\displaystyle Ric_{11}$
$\displaystyle=\frac{(g_{11})^{2}-(g_{22}-g_{33})^{2}}{2g_{22}g_{33}}$
$\displaystyle Ric_{22}$
$\displaystyle=\frac{(g_{22})^{2}-(g_{11}-g_{33})^{2}}{2g_{11}g_{33}}$
$\displaystyle Ric_{33}$
$\displaystyle=\frac{(g_{33})^{2}-(g_{11}-g_{22})^{2}}{2g_{11}g_{22}}$

and scalar curvature is

$S=-\frac{(g_{11})^{2}+(g_{22})^{2}+(g_{33})^{2}-2(g_{11}g_{22}+g_{11}g_{33}+g_{22}g_{33})}{2g_{11}g_{22}g_{33}}.$

The Bach tensor is diagonal with

$\displaystyle B_{00}$
$\displaystyle=-\beta\,p(g_{11},g_{22},g_{33})\,(g_{00})^{3}$ $\displaystyle
B_{11}$ $\displaystyle=-\beta\,q(g_{11},g_{22},g_{33})\,(g_{00})^{2}g_{11}$
$\displaystyle B_{22}$
$\displaystyle=-\beta\,q(g_{22},g_{33},g_{11})\,(g_{00})^{2}g_{22}$
$\displaystyle B_{33}$
$\displaystyle=-\beta\,q(g_{33},g_{11},g_{22})\,(g_{00})^{2}g_{33}$

where $p$ and $q$ were defined in Section 5.5.

On this space, there are a variety of possibilities for Bach flow, depending
on the initial conditions. To accommodate this richer structure, we break the
results into a number of theorems. Because of the symmetry in the equations,
we may suppose, without loss of generality, that $h_{11}\leq h_{22}\leq
h_{33}$. We first analyze the cases where at least two of the initial
conditions are equal. These results will begin to illustrate the complexity of
the situation and begin to provide some context for the remaining cases.

######  Theorem 5.16.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1) in
a diagonalizing basis with $h_{11}=h_{22}=h_{33}$. Then $g$ is static.

Note that in this case, the three-dimensional slice is a round sphere.

######  Theorem 5.17.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1) in
a diagonalizing basis with $h_{11}=h_{22}<h_{33}$ or
$h_{11}<h_{22}=h_{33}<4h_{11}$. Then

  * •

$g_{00}\rightarrow\left(\frac{\kappa}{3}\right)^{\frac{3}{8}}(\det
h)^{-\frac{1}{2}}$;

  * •

$g_{11}$, $g_{22}$,
$g_{33}\rightarrow\left(\frac{3}{\kappa}\right)^{\frac{1}{8}}(\det
h)^{\frac{1}{2}}$;

  * •

the components of $g_{00}$ and $g_{22}$ are related by

$(4\det h-g_{00}(g_{22})^{3})(g_{00})^{3}g_{22}=\kappa$

where

$\kappa=(4\det h-h_{00}(h_{22})^{3})(h_{00})^{3}h_{22};$

  * •

if $h_{11}=h_{22}$, then $g_{11}$ and $g_{22}$ are increasing, and $g_{33}$ is
decreasing;

  * •

if $h_{22}=h_{33}$, then $g_{11}$ is increasing.

######  Theorem 5.18.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1) in
a diagonalizing basis with $4h_{11}=h_{22}=h_{33}$. Then

$\displaystyle g_{00}(t)$ $\displaystyle=4(\det
h)\left(\frac{1}{2^{6}}t+(h_{33})^{2}\right)^{-\frac{3}{2}}$ $\displaystyle
4g_{11}(t)=g_{22}(t)$
$\displaystyle=g_{33}(t)=\left(\frac{1}{2^{6}}t+(h_{33})^{2}\right)^{\frac{1}{2}}.$

Note that in this case, $g_{00}\rightarrow 0$ and the three dimensional slice
is self-similar as it expands.

######  Theorem 5.19.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1) in
a diagonalizing basis with $4h_{11}<h_{22}=h_{33}$. Then

  * •

$g_{00},\,g_{11}\rightarrow 0$;

  * •

$g_{22},\,g_{33}\rightarrow\infty$;

  * •

the components of $g_{00}$ and $g_{22}$ are related by

$(g_{00}(g_{22})^{3}-4\det h)(g_{00})^{3}g_{22}=\kappa$

where

$\kappa=(h_{00}(h_{22})^{3}-4\det h)(h_{00})^{3}h_{22};$

  * •

$g_{22}$ and $g_{33}$ are increasing.

Before proving these theorems, we introduce some new structure to help with
the analysis. To capitalize on the fact that $\det g$ is constant along the
flow, and to exploit the symmetry among the equations for $g_{11}$, $g_{22}$,
and $g_{33}$, we introduce three new variables:

$\displaystyle a$ $\displaystyle=(g_{00})^{\frac{1}{3}}g_{11},$ $\displaystyle
b$ $\displaystyle=(g_{00})^{\frac{1}{3}}g_{22},$ $\displaystyle c$
$\displaystyle=(g_{00})^{\frac{1}{3}}g_{33}$

so that $abc=\det g$, and we rewrite our system using these. We have

(18) $\left\\{\begin{aligned}
\frac{d}{dt}g_{00}&=-\beta\,p(a,b,c)\,(g_{00})^{\frac{5}{3}}\\\
\frac{d}{dt}a&=-\frac{2\beta}{3}\,r(a,b,c)(g_{00})^{\frac{2}{3}}a\\\
\frac{d}{dt}b&=-\frac{2\beta}{3}\,r(b,a,c)\,(g_{00})^{\frac{2}{3}}b\\\
\frac{d}{dt}c&=-\frac{2\beta}{3}\,r(c,a,b)\,(g_{00})^{\frac{2}{3}}c\end{aligned}\right.$

where

$\displaystyle r(x,y,z)$ $\displaystyle=8x^{4}-5x^{3}(y+z)+2x^{2}yz$
$\displaystyle\qquad+x(y^{3}-y^{2}z-yz^{2}+z^{3})$
$\displaystyle\qquad-4y^{4}+4y^{3}z+4yz^{3}-4z^{4}.$

For this new system, the solution curves lie in the surface $\\{abc=\det
h\\}$. Moreover, because of the symmetry in the equations, we may restrict our
attention to solutions that satisfy $a\leq b\leq c$. From the determinant
constraint, we know $a=\frac{\det h}{bc}$ so this inequality becomes
$\sqrt{\frac{\det h}{c}}\leq b\leq c$. Thus, the flow is analyzed on the
domain

$D=\left\\{(g_{00},b,c):g_{00}\geq 0,\sqrt{\frac{\det h}{c}}\leq b\leq
c\right\\}.$

In the following, while $a$ can be eliminated, we find that it is useful to
use in the analysis. As such, $a$ should always be thought of as a function of
$b$ and $c$. Let $P_{0}$ be the point in $D$ where $g_{00}=0$ and $a=b=c$, let
$L_{0}$ be the ray where $g_{00}\geq 0$ and $a=b=c$, let $\partial D_{0}$ be
the set of points in $D$ where $g_{00}=0$, let $\partial D_{a=b}$ be the set
of points in $D$ where $a=b$, let $\partial D_{b=c}$ be the set of points in
$D$ where $b=c$. Note that $\partial D=\partial D_{0}\cup\partial
D_{a=b}\cup\partial D_{b=c}$. Let $P_{1}$ be the point in $D$ where $g_{00}=0$
and $4a=b=c$, and let $L_{1}$ be the ray where $g_{00}\geq 0$ and $4a=b=c$.
See Figures 2 and 3.

$b$$c$$P_{0}$$P_{1}$$\partial D_{0}\cap\partial D_{a=b}$$M_{S}$$M_{U}=\partial
D_{0}\cap\partial D_{b=c}$ Figure 2. Essential features of $\partial D_{0}$
$b$$c$$g_{00}$$P_{0}$$P_{1}$$L_{0}$$L_{1}$$\partial D_{0}$$\partial
D_{b=c}$$\partial D_{a=b}$$D_{S}$ Figure 3. Essential features of $D$. The
regions $D_{L_{0}}$ and $D_{\infty}$, not labelled, are determined by the
surfaces shown, with $D_{L_{0}}$ between $\partial D_{a=b}$ and $D_{S}$.

With this notation in place, before proving the theorems above, we note that
$\partial D_{0}$ corresponds to degenerate metrics and points in $\partial
D_{0}$ are not really achievable from the perspective of the original system.
However, once an initial metric $h$ is chosen, determining $\beta$, system
(18) is well defined on $\partial D_{0}$, and it is useful to explore the
behavior here because it informs the behavior on the interior. All solutions
starting here are static and as a consequence, it is conceivable that
nondegenerate solutions converge to these points. We will find that with the
exception of solutions converging to $P_{1}$, this is not the case. In the
following proofs, unless otherwise indicated, we restrict our attention to
initial conditions with $g_{00}>0$.

We now have the following:

###### Proof.

(of Theorem 5.16) This case corresponds to $L_{0}$. Here,
$r(a,b,c)=p(a,b,c)=0$, and we have static solutions. ∎

###### Proof.

(of Theorem 5.17) First, we consider the case where $h_{11}=h_{22}$, which
corresponds to $\partial D_{a=b}$. Here, $c=\frac{\det h}{b^{2}}$ and, with
this, the system reduces to two variables:

$\displaystyle\frac{d}{dt}g_{00}$
$\displaystyle=-\beta\,p(b,b,c)(g_{00})^{\frac{5}{3}}$
$\displaystyle\frac{d}{dt}b$
$\displaystyle=-\frac{2\beta}{3}\,r(b,b,c)(g_{00})^{\frac{2}{3}}b.$

Both $p$ and $r$ simplify substantially:

$\displaystyle p(b,b,c)$ $\displaystyle=c^{2}(b-c)^{2}$ $\displaystyle
r(b,b,c)$ $\displaystyle=-c^{2}(b-c)(b-4c)$

and from this, we can see that $b$ is increasing, since $b<c$. Hence $a$ is
also increasing. This implies that $g_{11}$ and $g_{22}$ are also increasing.
Furthermore, since $abc$ is constant, $c$ is decreasing. By Lemma 4.2 the flow
must converge to a point in either $L_{0}$ or $\partial D_{0}\cap\partial
D_{a=b}$. This second possibility will be ruled out below.

Next, since $r(b,b,c)$ is not zero, we have

$\frac{dg_{00}}{db}=\frac{\frac{d}{dt}g_{00}}{\frac{d}{dt}b}=-\frac{3}{2}\frac{b-c}{b-4c}\frac{g_{00}}{b}.$

This is separable and we get

(19) $g_{00}=\Bigl{[}\kappa\,\bigl{[}(4\det
h-b^{3})b\bigr{]}^{-1}\Bigr{]}^{\frac{3}{8}}$

where

$\displaystyle\kappa$ $\displaystyle=(h_{00})^{\frac{8}{3}}(4\det
h-(b(0))^{3})b(0)$ $\displaystyle=(4\det
h-h_{00}(h_{22})^{3})(h_{00})^{3}h_{22}.$

Substituting for $b$ in equation (19) and rearranging gives us the desired
relationship between $g_{00}$ and $g_{22}$.

The relationship given by equation (19) shows us two things. First since

$g_{33}=(g_{00})^{-\frac{1}{3}}c=\det h(g_{00})^{-\frac{1}{3}}b^{-2}$

we can substitute and then differentiate with respect to $b$ to find that
$g_{33}$ is decreasing.

Second, in the limit, we find that $g_{00}$ stays positive, so these solutions
stay nondegenerate. In the limit, $a$, $b$, and $c$ converge to $(\det
h)^{\frac{1}{3}}$ so

$\displaystyle g_{00}$ $\displaystyle\rightarrow 3^{-\frac{3}{8}}\kappa\,(\det
h)^{-\frac{1}{2}}$ $\displaystyle g_{11},g_{22},g_{33}$
$\displaystyle\rightarrow 3^{\frac{1}{8}}\kappa^{-\frac{1}{3}}(\det
h)^{\frac{1}{2}}.$

Next, we consider the case where $h_{22}=h_{33}$, which, accounting for the
allowable values for $h_{11}$, corresponds to those points in $\partial
D_{b=c}$ that lie between $L_{0}$ and $L_{1}$. Here, $a=\frac{\det h}{b^{2}}$
and, as above, the system reduces to two variables:

$\displaystyle\frac{d}{dt}g_{00}$
$\displaystyle=-\beta\,p(a,b,b)(g_{00})^{\frac{5}{3}}$
$\displaystyle\frac{d}{dt}b$
$\displaystyle=-\frac{2\beta}{3}\,r(b,a,b)(g_{00})^{\frac{2}{3}}b.$

Again, both $p$ and $r$ simplify substantially:

$\displaystyle p(a,b,b)$ $\displaystyle=a^{2}(b-a)^{2}$ $\displaystyle
r(b,a,b)$ $\displaystyle=-a^{2}(b-a)(b-4a).$

Algebraically, this system is identical to the previous case, so the analysis
is quite similar, and the resulting relationship between $g_{00}$ and $b$ is
determined by the same equation (19). This implies the same relationship for
$g_{00}$ and $g_{22}$. Important differences arises when analyzing the
qualitative behavior however. First, note that here, $\frac{d}{dt}b<0$ so $b$
and $c$ are decreasing. This implies that $a$ is increasing, and so $g_{11}$
must be increasing as well. We cannot conclude that $g_{22}$ and $g_{33}$ are
decreasing however, and it turns out that if the initial conditions are close
enough to $L_{1}$ then in fact $g_{22}$ and $g_{33}$ will increase for a while
before eventually decreasing. The transition occurs when
$g_{22}=g_{33}=3g_{11}$, which is found by analyzing the equation for
$\frac{d}{dt}g_{22}$ directly. Finally, while the qualitative behavior differs
somewhat from the previous case, the limiting behavior is the same. ∎

###### Proof.

(of Theorem 5.18) This case corresponds to $L_{1}$. Here,
$r(a,b,b)=r(b,a,b)=0$ which implies that
$\frac{d}{dt}a=\frac{d}{dt}b=\frac{d}{dt}c=0$, and $g_{22}=g_{33}=4g_{11}$ for
all time. Then, using the fact that $\det
g=g_{00}g_{11}g_{22}g_{33}=\frac{1}{4}g_{00}(g_{33})^{3}$ and focusing on the
equation for $g_{33}$, we have

$\frac{d}{dt}g_{33}=-\beta\,q\\!\left(g_{33},\frac{1}{4}g_{33},g_{33}\right)(g_{00})^{2}g_{33}=\frac{1}{2^{7}}(g_{33})^{-1}.$

This is separable and we have

$g_{33}=\left(\frac{1}{2^{6}}t+h_{33}^{2}\right)^{\frac{1}{2}}.$

Once this is known, the other three components are also known. We have

$g_{22}=g_{33},\ g_{11}=\frac{1}{4}g_{33}$

and

$g_{00}=4(\det h)\left(\frac{1}{2^{6}}t+h_{33}^{2}\right)^{-\frac{3}{2}}.$

∎

###### Proof.

(of Theorem 5.19) This case corresponds to those points in $\partial D_{b=c}$
that do not lie between $L_{0}$ and $L_{1}$. Algebraically, the system is the
same as for the second case in Theorem 5.17. In this case, since $b>4a$, $b$
and $c$ are increasing, and $a$ is decreasing, so $g_{22}$ and $g_{33}$ must
be increasing as well.

Since $b$ is increasing, if it were bounded, it would have to converge to a
point where $r(b,a,b)=0$ or where $g_{00}=0$, by Lemma 4.2. We will find below
that because of the algebraic relationship between $g_{00}$ and $b$, $g_{00}$
is positive as long as $b<\infty$ so the only possibility is $r(b,a,b)=0$.
Since there are no points where this occurs other than $b=a$ and $b=4a$, we
find that $b$ and $c$, and hence $g_{22}$ and $g_{33}$ must diverge in the
limit. The fact that $g_{11}\rightarrow 0$ follows from Lemma 5.24 which
appears later and is used for solutions starting at other points in $D$ as
well.

For the algebraic relationship between $g_{00}$ and $g_{22}$, the system is
the same as for the second case in Theorem 5.17, and the analysis is
essentially the same. Again, the fact that $b>4a$ alters the formula for the
trace of the solution so that instead of equation (19), we have

$g_{00}=\Bigl{[}\kappa\,\bigl{[}(b^{3}-4\det
h)b\bigr{]}^{-1}\Bigr{]}^{\frac{3}{8}}$

where

$\displaystyle\kappa$ $\displaystyle=(h_{00})^{\frac{8}{3}}((b(0))^{3}-4\det
h)b(0)$ $\displaystyle=(h_{00}(h_{22})^{3}-4\det h)(h_{00})^{3}h_{22}.$

With this small change made, substituting for $b$ and rearranging produces the
result. ∎

Our next goal is to determine the qualitative behavior of solutions with
initial conditions that do not lie on the boundary. In light of the results
above, we introduce a bit more notation and structure before stating the
theorems. First, observe that there are no equilibria aside from those found
on the boundary above. To see this, note that to have
$\frac{d}{dt}b=\frac{d}{dt}c=0$, we must have $r(b,a,c)=r(c,a,b)=0$ and so in
particular,

$r(c,a,b)-r(b,a,c)=0.$

Writing the left side out explicitly, we have

$r(c,a,b)-r(b,a,c)=3(c-b)(4c^{3}+2c^{2}b+2cb^{2}+4b^{3}-3c^{2}a-2abc-3b^{2}a-a^{3}).$

Under the condition that $a\leq b\leq c$, we find that the large factor on the
right is always positive so the only way we can have an equilibrium point is
if $b=c$.

As mentioned earlier, the fact that all the points in $\partial D_{0}$ are
equilibria is problematic. To resolve this, we adjust the system again.
Specifically, we rescale the system by multiplying the right hand sides by the
nonzero factor $\beta^{-1}(g_{00})^{-\frac{2}{3}}$ to produce the new system

(20) $\left\\{\begin{aligned} \frac{d}{dt}g_{00}&=-p(a,b,c)\,g_{00},\\\
\frac{d}{dt}b&=-\frac{2}{3}\,r(b,a,c)\,b,\\\
\frac{d}{dt}c&=-\frac{2}{3}\,r(c,a,b)\,c.\end{aligned}\right.$

The solutions to this system will just be reparameterizations of solutions to
system (18). Moreover, this system extends to a (mostly) nonzero system on
$\partial D_{0}$ and, since $\frac{d}{dt}g_{00}$ is still zero here, solutions
on this part of the boundary stay in this part of the boundary.

Restricting attention to $\partial D_{0}$, note that, consistent with the
observations above, there are two equilibrium points $P_{0}$ and $P_{1}$.
Disregarding the equation for $g_{00}$, the linearization at $P_{0}$ is

$\frac{d}{dt}\begin{pmatrix}b\\\ c\end{pmatrix}=6(\det
h)^{\frac{4}{3}}\begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix}\begin{pmatrix}b\\\
c\end{pmatrix}$

and we have a stable equilibrium. The linearization at $P_{1}$ is

$\frac{d}{dt}\begin{pmatrix}b\\\ c\end{pmatrix}=2^{-\frac{7}{3}}3(\det
h)^{\frac{4}{3}}\begin{pmatrix}-106&107\\\
107&-106\end{pmatrix}\begin{pmatrix}b\\\ c\end{pmatrix}$

resulting in a saddle.

The unstable manifold $M_{U}$ for the saddle is the line $\\{b=c\\}$. The
stable manifold $M_{S}$ is a curve that approaches $P_{1}$ perpendicularly to
$M_{U}$. See Figure 2.

Analyzing the ratio $\frac{c}{b}$, we have

$\frac{d}{dt}\frac{c}{b}=-\frac{2}{3}\,\bigl{[}r(c,a,b)-r(b,a,c)\bigr{]}\,\frac{c}{b}.$

As shown above, $r(c,a,b)-r(b,a,c)$ is always positive when $a<b<c$ so the
fraction $\frac{c}{b}$ decreases as $t$ increases. From this, we find that the
solutions approach the boundary $b=c$.

The set $\partial D_{0}\backslash M_{S}$ comprises two components. By Lemma
4.2, and the fact that $\frac{c}{b}$ is decreasing, solutions starting in the
component that includes $P_{0}$ converge to $P_{0}$ as $t\rightarrow\infty$
while solutions starting in the other component converge to $a=0,\
b=c=\infty$.

Motivated by these observations, let $D_{S}$ be the set of points in $D$ where
$g_{00}\geq 0$ and $(0,b,c)\in M_{S}$, and note that $D\backslash D_{S}$
comprises two components. Let $D_{L_{0}}$ be the component that includes
$L_{0}$, and let $D_{\infty}$ be the component that avoids $L_{0}$. See Figure
3.

We now have

######  Theorem 5.20.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1).
In a diagonalizing basis, suppose $h$ corresponds to a point in $D_{S}$. Then

  * •

$g_{00}\rightarrow 0$;

  * •

$g_{11},\,g_{22},\,g_{33}\rightarrow\infty$;

  * •

$\frac{g_{22}}{g_{11}}\rightarrow 4$, and $\frac{g_{22}}{g_{33}}\rightarrow
1$.

######  Theorem 5.21.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1).
In a diagonalizing basis, suppose $h$ corresponds to a point in $D_{\infty}$.
Then

  * •

$g_{00},\,g_{11}\rightarrow 0$;

  * •

$g_{22},\,g_{33}\rightarrow\infty$;

  * •

$g_{33}-g_{22}\rightarrow 0$.

######  Theorem 5.22.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ solve equation (1).
In a diagonalizing basis, suppose $h$ corresponds to a point in $D_{L_{0}}$.
Then

  * •

$g_{00}$ does not converge to zero;

  * •

$g_{11}$, $g_{22}$, and $g_{33}$ converge to the same value.

We prove these by analyzing the behavior of system (20) and we note the
following general structure for its solutions. Let $(0,b(t),c(t))$ be a
solution in $\partial D_{0}$ and consider the solution with initial condition
$(h_{00},b(0),c(0))$. Since the equations for $b$ and $c$ do not depend on
$g_{00}$, $b(t)$ and $c(t)$ still solve this system. Then

$\frac{d}{dt}g_{00}=-p(a(t),b(t),c(t))\,g_{00}$

which is separable and we have

$g_{00}=h_{00}\,e^{P(t)}$

where

(21) $P(t)=\int_{0}^{t}-p(a(\tau),b(\tau),c(\tau))\,d\tau.$

Note that, since $P$ does not depend on $g_{00}$, the ratio of two solutions
with initial conditions that differ only in $h_{00}$ will be constant.

We can now prove Theorems 5.20, 5.21, and 5.22. While Theorem 5.20 is
straightforward, it turns out that Theorems 5.21 and 5.22 are fairly subtle.

###### Proof.

(of Theorem 5.20) Since $h$ corresponds to a point in $D_{S}$, we know that
$(0,b(0),c(0))\in M_{S}$. Hence $4a(t)$, $b(t)$, and $c(t)$ all converge to
the same value. The theorem will then be proved once it is established that
$g_{00}\rightarrow 0$. Since the solution is bounded for $t\geq 0$, the
interval on which it is defined includes $[0,\infty)$. Moreover, since the
solution is converging to $L_{1}$, $p(a(t),b(t),c(t))$ is bounded below by a
positive constant, so $P(t)\rightarrow-\infty$ as $t\rightarrow\infty$.
Therefore $g_{00}\rightarrow 0$, as desired. ∎

For Theorem 5.21, we first establish a couple lemmas.

######  Lemma 5.23.

Suppose $a\leq b\leq c$ and $b$ and $c$ diverge (so $a$ converges to zero).
Then for all $m\in\mathbb{R}$, $\frac{c-b}{a^{m}}\rightarrow 0$.

###### Proof.

Observe first that if $\frac{c-b}{a^{m^{\prime}}}$ is bounded for a particular
exponent $m^{\prime}$, then the result is true for all $m<m^{\prime}$.
Therefore, it is enough to show that $\frac{c-b}{a^{m}}$ is bounded for all
$m\geq 1$.

The equation solved by $c-b$ is

$\frac{d}{dt}(c-b)=-\frac{2}{3}u(a,b,c)(c-b)$

where

$\displaystyle u(x,y,z)$ $\displaystyle=-4x^{4}+x^{3}(y+z)-x^{2}yz$
$\displaystyle\qquad-x(5y^{3}+7y^{2}z+7yz^{2}+5z^{3})$
$\displaystyle\qquad+8y^{4}+7y^{3}z+6y^{2}z^{2}+7yz^{3}+8z^{4},$

and so

$\frac{d}{dt}\frac{c-b}{a^{m}}=-\frac{2}{3}[u(a,b,c)-m\,r(a,b,c)]\frac{c-b}{a^{m}}.$

Writing out $u(a,b,c)-m\,r(a,b,c)$, we find

$\displaystyle u(a,b,c)-m\,r(a,b,c)$ $\displaystyle=3b^{3}+6b^{2}c^{2}+3c^{4}$
$\displaystyle\qquad+b^{2}(5b+7c)(b-a)+c^{2}(7b+5c)(c-a)$
$\displaystyle\qquad+m(c-b)^{2}[3b^{2}+3bc+4c^{2}+(b-a)(b+c)]$
$\displaystyle\qquad+(1+5m)a^{3}(b+c)$
$\displaystyle\qquad-a(1+2m)(abc+4a^{3}).$

As $a\rightarrow 0$ and $b$ and $c$ diverge to $\infty$, the first and second
lines are positive and diverge, and the third and fourth lines are positive.
Only the last line is negative, but it converges to zero (since $abc=\det h$
is constant). From this, we find that $\frac{c-b}{a^{m}}$ is eventually
decreasing, and hence bounded above. ∎

######  Lemma 5.24.

Suppose $a\leq b\leq c$ and $b$ and $c$ diverge (so $a$ converges to zero).
Then $g_{11}\rightarrow 0$.

###### Proof.

We already know that $b=(g_{00})^{\frac{1}{3}}g_{22}$ diverges. In fact it can
be shown that $(g_{00})^{k}g_{22}$ eventually increases for all $k$ in the
interval $[\frac{1}{3},1)$. Here we work specifically with $k=\frac{2}{3}$. We
have

$\frac{d}{dt}\left[(g_{00})^{\frac{2}{3}}g_{22}\right]=-\frac{\beta}{3}\,s(g_{11},g_{22},g_{33})\,(g_{00})^{\frac{8}{3}}g_{22}$

where

$\displaystyle s(x,y,z)$ $\displaystyle=2p(x,y,z)+3q(y,x,z)$
$\displaystyle=-[(z-y)(7z^{3}+6yz^{2}+6y^{2}z+17y^{3}-x(7z^{2}+6yz+11y^{2}))$
$\displaystyle\qquad+x^{2}(yz-xy-7xz+7x^{2})].$

Analyzing, $s(g_{11},g_{22},g_{33})$, the large factor inside the first term
is positive if $g_{11}\leq g_{22}\leq g_{33}$. For the second term, note that
$\frac{g_{22}}{g_{11}}=\frac{b}{a}$ which diverges for the solutions under
consideration. This implies that eventually, $g_{22}$ becomes, and stays,
larger than $8g_{11}$. This, combined with the fact that $g_{22}\leq g_{33}$
implies

$g_{22}g_{33}-g_{11}g_{22}-7g_{11}g_{33}\geq 0.$

From this, we find that $s(g_{11},g_{22},g_{33})$ eventually becomes, and
stays, negative and so $(g_{00})^{\frac{2}{3}}g_{22}$ eventually increases.

Using this fact, rewrite $\det g$ as follows:

$\det g=g_{00}g_{11}g_{22}g_{33}=g_{11}[(g_{00})^{\frac{2}{3}}g_{22}]c.$

Since $\det g$ is constant, $(g_{00})^{\frac{2}{3}}g_{22}$ is increasing, and
$c\rightarrow\infty$, it must be the case that $g_{11}\rightarrow 0$. ∎

We are now ready to prove Theorem 5.21

###### Proof.

(of Theorem 5.21) Since $h$ corresponds to a point in $D_{\infty}$, we know
that $b(t)$ and $c(t)$ diverge, and so $a(t)\rightarrow 0$. Since $g_{00}$ is
decreasing, it must be the case that $g_{22}$ and $g_{33}$ diverge. By Lemma
5.24, $g_{11}\rightarrow 0$, so the theorem will then be proved once it is
established that $g_{00}\rightarrow 0$ as well.

For metrics in the given domain, $r(b,a,c)<0$ so for any flow in this setting,
$b$ is strictly increasing. Using this, we make a substitution to rewrite
equation (21) to get

$-P(T)=\int_{0}^{T}p(a(t),b(t),c(t))\,dt=\int_{b(0)}^{b(T)}\frac{p(a,b,c)}{-\frac{2}{3}r(b,a,c)b}db$

where we recognize that $a$ and $c$ are now functions of $b$. We now estimate
$p$ and $r$ along the flow. For $p$, we have the following:

$\displaystyle p(a,b,c)$
$\displaystyle=a^{2}(b-a)(c-a)+(b-c)^{2}[a(b+c)+b^{2}+bc+c^{2}]$
$\displaystyle\geq a^{2}(b-a)(c-a).$

For $r$, we have

$\displaystyle-r(b,a,c)$ $\displaystyle=a^{2}(b-4a)(c-a)$
$\displaystyle\qquad+(c-b)[-a(5b^{2}+3bc+4c^{2})+8b^{3}+3b^{2}c+3bc^{2}+4c^{3}]$
$\displaystyle=a^{2}(b-4a)(c-a)+\frac{(c-b)}{a^{5}}a^{5}\bar{r}(a,b,c)$
$\displaystyle=a^{2}[(b-4a)(c-a)+\frac{(c-b)}{a^{5}}a^{3}\bar{r}(a,b,c)]$

where $\bar{r}$ is a cubic polynomial. Since $abc$ is constant, $ab$ and $ac$
go to zero along the flow, and this implies that
$a^{3}\bar{r}(a,b,c)\rightarrow 0$. By Lemma 5.23,
$\frac{(c-b)}{a^{5}}\rightarrow 0$ as well. Hence,
$\frac{(c-b)}{a^{5}}a^{3}\bar{r}(a,b,c)$ is bounded by a positive constant $K$
along the flow (for $t\geq 0$) and so

$\displaystyle-r(b,a,c)$ $\displaystyle\leq a^{2}([b-4a)(c-a)+K]$
$\displaystyle\leq La^{2}(b-4a)(c-a)$ $\displaystyle\leq La^{2}b(c-a)$

where the second inequality follows from the fact that $(b-4a)(c-a)$ is
bounded below by a positive constant along the flow.

Combining the estimate for $p$ and for $r$, we have

$\displaystyle\int_{b(0)}^{\infty}\frac{p(a,b,c)}{-\frac{2}{3}r(b,a,c)b}db$
$\displaystyle\geq\frac{3}{2}\int_{b(0)}^{\infty}\frac{a^{2}(b-a)(c-a)}{La^{2}b(c-a)b}db$
$\displaystyle=\frac{3L}{2}\left(\int_{b(0)}^{\infty}\frac{1}{b}db-\int_{b(0)}^{\infty}\frac{a}{b^{2}}db\right).$

The first integral diverges while the second integral stays finite so
$P(t)\rightarrow-\infty$ along the flow, and $g_{00}\rightarrow 0$. ∎

Before proving Theorem 5.22, we establish some estimates for $p$ and $r$ near
$P_{0}$. In the following, keep in mind that since $a=\frac{\det h}{bc}$, its
value changes when comparing the functions in question at different points.

######  Lemma 5.25.

There is a neighborhood $U$ of $P_{0}$ such that for all points $(0,b,c)$ in
$U\cap\partial D_{0}$,

$r(c,a,b)\geq r(c,a,c)\geq 0.$

###### Proof.

As a first step, we show that $r(c,a,b)>0$ for points in the given domain near
$a=b=c$. In fact, to help with the argument later, we show that for each $c$
the function is minimized at $b=c$. First, note that when
$b=a=\sqrt{\frac{\det h}{c}}$,

$r(c,a,a)=2c^{2}(4c-a)(c-a)$

and when $b=c$, so that $a=\frac{\det h}{c^{2}}$,

$r(c,a,c)=a^{2}(c-a)(4a-c)$

These are both positive as long as $a<c<4a$. Next we compute the derivative
with respect to $b$. Since $a=\frac{\deg h}{bc}$, we have
$\partial_{b}a=-\frac{\det h}{c}b^{-2}=-ab^{-1}$ so

$\displaystyle\partial_{b}r(c,a,b)$
$\displaystyle=-5c^{3}-abc-3a^{3}b^{-1}c-16b^{3}-8a^{3}$
$\displaystyle\qquad+5ab^{-1}c^{3}+3b^{2}c+8ab^{2}+16a^{4}b^{-1}+2a^{2}c$

and the second derivative

$\displaystyle\partial_{b}^{2}r(c,a,b)$
$\displaystyle=12a^{3}b^{-2}c-48b^{2}+24a^{3}b^{-1}-10ab^{-2}c^{3}$
$\displaystyle\qquad+6bc+8ab-80a^{4}b^{-2}-2a^{2}b^{-1}c.$

This is negative at $P_{0}$, so must be negative in a neighborhood of this
point. This implies that as $b$ varies, $r(c,a,b)$ is minimized at one of the
endpoints above.

To determine which endpoint is the minimum, comparing the two expressions
algebraically proves difficult. To more easily compare, let $c=v(\det
h)^{\frac{1}{3}}$ (and note that $v=1$ corresponds to the point $P_{0}$). Then
define

$\displaystyle f_{a}(v)$ $\displaystyle=r(c,a,a)$ $\displaystyle=(\det
h)^{\frac{4}{3}}(8v^{4}-10v^{\frac{5}{2}}+2v).$

Then

$(f_{a})^{\prime}(v)=(\det h)^{\frac{4}{3}}(32v^{3}-25v^{\frac{3}{2}}+2)$

and so $(f_{a})^{\prime}(1)=9(\det h)^{\frac{4}{3}}$. Also

$(f_{a})^{\prime\prime}(v)=(\det
h)^{\frac{4}{3}}\left(96v^{2}-\frac{75}{2}v^{\frac{1}{2}}\right)$

and so $(f_{a})^{\prime\prime}(1)=\frac{117}{2}(\det h)^{\frac{4}{3}}$.

On the other hand define

$\displaystyle f_{c}(v)$ $\displaystyle=r(c,c,a)$ $\displaystyle=(\det
h)^{\frac{4}{3}}(-4v^{-8}+5v^{-5}-v^{-2}).$

Then

$(f_{c})^{\prime}(v)=(\det h)^{\frac{4}{3}}(32v^{-9}-25v^{-5}+2v^{-3})$

and so $(f_{c})^{\prime}(1)=9(\det h)^{\frac{4}{3}}$, which matches
$(f_{a})^{\prime}(1)$. Also

$(f_{c})^{\prime\prime}(v)=(\det
h)^{\frac{4}{3}}(-288v^{-10}+125v^{-6}-6v^{-4})$

so $(f_{c})^{\prime\prime}(1)=-139(\det h)^{\frac{4}{3}}$. This shows that
$f_{a}$ and $f_{c}$ agree to first order, but that near $P_{0}$, $f_{a}$
eventually grows faster and we can conclude that for each $c$ close to
$P_{0}$, $r(c,a,b)$ is minimized when $b=c$. ∎

######  Lemma 5.26.

There is a neighborhood $U$ of $P_{0}$ such that for all points $(0,b,c)$ in
$U\cap\partial D_{0}$,

$0\leq p(a,b,c)\leq p(a,c,c).$

###### Proof.

We have

$p(a,a,c)=c^{2}(c-a)^{2}$

and

$p(a,c,c)=a^{2}(c-a)^{2}.$

For points between these two, we compute the partial derivative with respect
to $b$. As in the previous lemma, $a=\frac{\deg h}{bc}$ so
$\partial_{b}a=-ab^{-1}$ and we have

$\displaystyle\partial_{b}p(a,b,c)$
$\displaystyle=-4a^{4}b^{-1}+2a^{3}+3a^{3}b^{-1}c-a^{2}c$
$\displaystyle\qquad-2ab^{2}+abc+ab^{-1}c^{3}+4b^{3}-3b^{2}c-c^{3}.$

The second derivative is

$\displaystyle\partial_{b}^{2}p(a,b,c)$
$\displaystyle=20a^{4}b^{-2}-6a^{3}b^{-1}-12a^{3}b^{-2}c+2a^{2}b^{-1}c$
$\displaystyle\qquad-2ab-2ab^{-2}c^{3}+12b^{2}-6bc.$

At $P_{0}$, this is positive so $p$ is concave up near $P_{0}$ and we may
conclude that it is maximized at one endpoint. To determine which endpoint is
larger, let $c=v(\det h)^{\frac{1}{3}}$ and define

$f_{a}(v)=\bigl{(}p(a,a,c)\bigr{)}^{\frac{1}{2}}=c(c-a)=(\det
h)^{\frac{2}{3}}(v^{2}-v^{\frac{1}{2}})$

and

$f_{c}(v)=\bigl{(}p(a,c,c)\bigr{)}^{\frac{1}{2}}=a(c-a)=(\det
h)^{\frac{2}{3}}(v^{-1}-v^{-4}).$

Then $f_{a}(1)=f_{c}(1)$. Computing derivatives, we have

$(f_{a})^{\prime}(v)=(\det h)^{\frac{2}{3}}(2v-\frac{1}{2}v^{-\frac{1}{2}})$

and

$(f_{c})^{\prime}(v)=(\det h)^{\frac{2}{3}}(-v^{-2}+4v^{-5}).$

Hence $(f_{a})^{\prime}(1)=\frac{3}{2}$ while $(f_{c})^{\prime}(1)=3$ and we
may conclude that near $P_{0}$, $f_{c}$ grows faster, so $p$ is maximized when
$b=c$. ∎

We are now ready to prove Theorem 5.22.

###### Proof.

(of Theorem 5.22) Since $h$ corresponds to a point in $D_{L0}$, we know that
$a(t)$, $b(t)$, and $c(t)$ converge to the same value. The theorem will then
be proved once it is established that $g_{00}$ does not go to zero. Note that,
since the solution is bounded, the interval on which it is defined includes
$[0,\infty)$.

By Lemma 5.25, $r(c,b,a)$ is positive near $a=b=c$, so $\frac{d}{dt}c$ is
negative and $c$ is strictly decreasing. From this, we can reparameterize the
integral above and then use Lemmas 5.25 and 5.26 to get

$\displaystyle-P(T)$ $\displaystyle=\int_{0}^{T}p(a(t),b(t),c(t))\,dt$
$\displaystyle=\int_{c(0)}^{c(T)}\frac{p(a,b,c)}{-\frac{2}{3}r(c,a,b)c}dc$
$\displaystyle=\frac{3}{2}\int_{c(T)}^{c(0)}\frac{p(a,b,c)}{r(c,a,b)c}dc$
$\displaystyle\leq\frac{3}{2}\int_{c(T)}^{c(0)}\frac{p(a,c,c)}{r(c,a,c)c}dc$
$\displaystyle=\frac{3}{2}\int_{c(T)}^{c(0)}\frac{a^{2}(c-a)^{2}}{a^{2}(c-a)(4a-c)c}dc$
$\displaystyle=\frac{3}{2}\int_{c(T)}^{c(0)}\frac{(c-a)}{(4a-c)c}dc.$

The integrand on the last line is bounded and the interval of integration
stays bounded, so the integral stays finite as $T\rightarrow\infty$. (We are
allowed to cancel the factor $a^{2}(c-a)$ because the $a$ being used is the
same for the numerator and the denominator, since the estimates are both taken
on the same side of the boundary.) ∎

With the limiting behavior of the metric established, the next step is to
determine curvature.

######  Proposition 5.27.

On $\widehat{M}=\mathbb{R}\times\mathbb{S}^{3}$, let $g$ be a solution to
equation (1) with initial metric $h$. Then in a diagonalizing basis,

  * •

if $h$ corresponds to a point in $D_{S}$, then $Ric_{11}$ converges to
$\frac{1}{32}$, $Ric_{22}$ and $Ric_{33}$ both converge to $\frac{7}{8}$, and
$S$ converges to 0;

  * •

if $h$ corresponds to a point in $D_{\infty}$, then $\frac{Ric_{11}}{g_{11}}$
converges to zero, $Ric_{22}$ and $Ric_{33}$ both converge to 1, and $S$
converges to 0;

  * •

if $h$ corresponds to a point in $D_{L_{0}}$, then $Ric_{11}$, $Ric_{22}$, and
$Ric_{33}$ all converge to $\frac{1}{2}$, and $S$ converges to a positive
value.

###### Proof.

Since the Ricci tensor is invariant under uniform rescaling of the metric, we
can use the components of the metric directly, or we can use $a$, $b$, and $c$
to determine the Ricci curvature. We have three cases.

  * •

If $h$ corresponds to a point in $D_{S}$, then we can simply plug in the fact
that, in the limit, $4a=b=c$ to get the values indicated for the Ricci tensor.
For scalar curvature, we have

$\displaystyle S$
$\displaystyle=-\frac{(g_{11})^{2}+(g_{22})^{2}+(g_{33})^{2}-2(g_{11}g_{22}+g_{11}g_{33}+g_{22}g_{33})}{2g_{11}g_{22}g_{33}}$
$\displaystyle=-(g_{00})^{\frac{1}{3}}\frac{a^{2}+b^{2}+c^{2}-2(ab+ac+bc)}{2\det
h}$ $\displaystyle\rightarrow 0$

since $g_{00}\rightarrow 0$.

  * •

If $h$ corresponds to a point in $D_{\infty}$, we have

$\displaystyle Ric_{33}$ $\displaystyle=\frac{c^{2}-(b-a)^{2}}{2ab}$
$\displaystyle=\frac{(c-b)(c+b)}{2ab}+1-\frac{a}{2b}.$

The first term goes to zero since $\frac{c-b}{a}\rightarrow 0$ by Lemma 5.23,
and $\frac{c+b}{b}$ stays bounded since $\frac{c}{b}\rightarrow 1$. The last
term also goes to zero so $Ric_{33}\rightarrow 1$. The analysis for $Ric_{22}$
is similar.

For $\frac{Ric_{11}}{g_{11}}$ we have

$\frac{Ric_{11}}{g_{11}}=\frac{a^{2}-(c-b)^{2}}{2g_{11}bc}=\frac{g^{\frac{1}{3}}[a^{2}-(c-b)^{2}]}{2abc}.$

Here, the numerator goes to zero while the denominator stays constant.

For scalar curvature, we have

$\displaystyle S$
$\displaystyle=-\frac{(g_{11})^{2}+(g_{22})^{2}+(g_{33})^{2}-2(g_{11}g_{22}+g_{11}g_{33}+g_{22}g_{33})}{2g_{11}g_{22}g_{33}}$
$\displaystyle=-(g_{00})^{\frac{1}{3}}\frac{a^{2}+b^{2}+c^{2}-2(ab+ac+bc)}{2\det
g}$ $\displaystyle=-(g_{00})^{\frac{1}{3}}\frac{(c-b)^{2}+a^{2}-2a(c+b)}{2\det
g}$ $\displaystyle=\frac{-(g_{00})^{\frac{1}{3}}}{2\det
g}[(c-b)^{2}+a^{2}]+(g_{00})^{\frac{1}{3}}\left(\frac{1}{b}+\frac{1}{c}\right).$

In this form, we can see that both terms go to zero.

  * •

If $h$ corresponds to a point in $D_{L_{0}}$, then in the limit, $a=b=c$, and
we get the desired values for the Ricci tensor. For scalar curvature,

$\displaystyle S$
$\displaystyle=-\frac{(g_{11})^{2}+(g_{22})^{2}+(g_{33})^{2}-2(g_{11}g_{22}+g_{11}g_{33}+g_{22}g_{33})}{2g_{11}g_{22}g_{33}}$
$\displaystyle=-(g_{00})^{\frac{1}{3}}\frac{a^{2}+b^{2}+c^{2}-2(ab+ac+bc)}{2\det
h}$
$\displaystyle\rightarrow\frac{3}{2}\frac{a^{2}(g_{00})^{\frac{1}{3}}}{\det
h}$ $\displaystyle=\frac{3}{2g_{11}}$

which is positive.

∎

Finally, we have the following:

######  Theorem 5.28.

Let $M$ be a quotient of $\mathbb{R}\times\mathbb{S}^{3}$ and let $p\in M$.
Let $g$ solve equation (1) where $h$ is locally homogeneous. Then

  * •

if $h$ corresponds to a point in $D_{S}$, then $(M,g,p)$ collapses to a flat
three-dimensional manifold in the pointed Gromov-Hausdorff topology;

  * •

if $h$ corresponds to a point in $D_{\infty}$, then $(M,g,p)$ collapses to a
flat surface in the pointed Gromov-Hausdorff topology;

  * •

if $h$ corresponds to a point in $D_{L_{0}}$, then $(M,g)$ converges to a
quotient of the product of a circle and the round sphere in the Gromov-
Hausdorff topology.

###### Proof.

These results follow from the previous proposition and Lemma 3.3. ∎

###  5.7. $\mathbb{H}^{3}$

This space is not a Lie group so the techniques used above do not apply. In
fact, the analysis here is much simpler. There is a one parameter family of
homogeneous metrics for $\mathbb{H}^{3}$ and they are all constant scalar
multiples of the standard hyperbolic metric and hence Einstein. Therefore, by
Proposition 2.1, $\mathbb{R}\times\mathbb{H}^{3}$ is static under Bach flow.

###  5.8. $\mathbb{R}\times\mathbb{S}^{2}$ and
$\mathbb{R}\times\mathbb{H}^{2}$

While these spaces can be thought of as three-dimensional factors for various
$1\times 3$ products, they are more naturally viewed in terms of $2\times 2$
products which are discussed in the next section.

##  6\. Bach flow on locally homogeneous $2\times 2$ products

Bach flow on products of homogeneous surfaces was explored in [DK12]. We
include the analysis here for completeness because of the fact that three
families of $1\times 3$ products can also be viewed as $2\times 2$ manifolds,
namely quotients of $\mathbb{R}\times\widehat{N}$ where $\widehat{N}$ is
$\mathbb{R}\times\mathbb{S}^{2}$, $\mathbb{R}\times\mathbb{H}^{2}$, or
$\mathbb{R}^{3}$. Working as before on the universal cover, it seems at first
that there are essentially six different cases to consider:
$\mathbb{R}^{2}\times\mathbb{R}^{2}$, $\mathbb{R}^{2}\times\mathbb{S}^{2}$,
$\mathbb{R}^{2}\times\mathbb{H}^{2}$, $\mathbb{S}^{2}\times\mathbb{S}^{2}$,
$\mathbb{S}^{2}\times\mathbb{H}^{2}$, $\mathbb{H}^{2}\times\mathbb{H}^{2}$.
However, it turns out that the Bach tensor does a poor job of distinguishing
the spherical and hyperbolic slices and the analysis reduces to three cases,
one of which is trivial (since the Bach tensor vanishes). To see this, because
of the constancy of the scalar curvatures, equations (5) and (6) reduce to

$B_{\alpha\beta}=\frac{1}{24}\left((S^{(1)})^{2}-(S^{(2)})^{2}\right)g_{\alpha\beta}$

and

$B_{jk}=\frac{1}{24}\left((S^{(2)})^{2}-(S^{(1)})^{2}\right)g_{jk}.$

Since the scalar curvatures of the slices are squared, there is no way to
distinguish between a positively curved space and a negatively curved space.

For any of these spaces we can write a homogeneous product metric as

$g=f_{1}\,g^{(1)}+f_{2}\,g^{(2)}$

where $g^{(i)}$ is the standard metric for the $i$th slice and $f_{i}>0$.

The constancy of the volume form along the flow implies that

$\gamma=f_{1}(t)f_{2}(t)=f_{1}(0)f_{2}(0)$

is a constant determined by the initial metric.

###  6.1. $\mathbb{R}^{2}\times\mathbb{R}^{2}$,
$\mathbb{R}^{2}\times\mathbb{S}^{2}$, and $\mathbb{R}^{2}\times\mathbb{H}^{2}$

These spaces can also be thought of as $1\times 3$ products and the analysis
here completes the $1\times 3$ product cases. The space
$\mathbb{R}^{2}\times\mathbb{R}^{2}$ is flat, so the metric is static under
Bach flow. For the remaining two spaces spaces, we have the following:

######  Theorem 6.1.

Let $M$ be a compact quotient of $\mathbb{R}^{2}\times\mathbb{S}^{2}$ or
$\mathbb{R}^{2}\times\mathbb{H}^{2}$ and let $p\in M$. Let $g$ solve equation
(1) where $h$ is locally homogeneous. Then $(M,g,p)$ converges to a flat
surface in the pointed Gromov-Hausdorff topology.

###### Proof.

Using the structure introduced above, the curvature of the first slice is zero
and that of the second slice is

$S^{(2)}=\pm 2f_{2}^{-1},$

(positive for $\mathbb{S}^{2}$ and negative for $\mathbb{H}^{2}$). Bach flow
then reduces to

$\frac{d}{dt}f_{1}=-\frac{1}{6}f_{2}^{-2}f_{1},\ \ \ \ \
\frac{d}{dt}f_{2}=\frac{1}{6}f_{2}^{-1}.$

The second equation is separable. Once its solution is found, it can be
plugged into the first and we can solve the resulting separable equation for
$f_{1}$. We end up with

$\displaystyle f_{1}(t)$
$\displaystyle=\gamma\left(\frac{1}{3}t+f_{2}(0)^{2}\right)^{-\frac{1}{2}}$
$\displaystyle f_{2}(t)$
$\displaystyle=\left(\frac{1}{3}t+f_{2}(0)^{2}\right)^{\frac{1}{2}}.$

We find that the solutions are immortal but not ancient. The flat slice
shrinks while the curved slice blows up and its scalar curvature goes to zero.
∎

###  6.2. $\mathbb{S}^{2}\times\mathbb{S}^{2}$,
$\mathbb{S}^{2}\times\mathbb{H}^{2}$, and $\mathbb{H}^{2}\times\mathbb{H}^{2}$

These spaces are not $1\times 3$ products. They are included here for
completeness.

######  Theorem 6.2.

Let $M$ be a compact quotient of $\mathbb{R}^{2}\times\mathbb{S}^{2}$ or
$\mathbb{R}^{2}\times\mathbb{H}^{2}$ and let $p\in M$. Let $g$ solve equation
(1) where $h$ is locally homogeneous. Then $(M,g,p)$ converges to a flat
surface in the pointed Gromov-Hausdorff topology.

######  Theorem 6.3.

Let $M$ be a compact quotient of $\mathbb{S}^{2}\times\mathbb{S}^{2}$,
$\mathbb{S}^{2}\times\mathbb{H}^{2}$, or $\mathbb{H}^{2}\times\mathbb{H}^{2}$.
Let $g$ solve equation (1) where $h$ is locally homogeneous. Then $(M,g)$
converges to a Bach-flat four-dimensional manifold in the Gromov-Hausdorff
topology. The difference in the magnitude of the curvature of each slice
converges to zero, and the scalar curvature converges to

  * •

$4\gamma^{-\frac{1}{2}}$ on $\mathbb{S}^{2}\times\mathbb{S}^{2}$,

  * •

$0$ on $\mathbb{S}^{2}\times\mathbb{H}^{2}$,

  * •

$-4\gamma^{-\frac{1}{2}}$ on $\mathbb{H}^{2}\times\mathbb{H}^{2}$.

###### Proof.

For these spaces the scalar curvatures are

$S^{(i)}=\pm 2f_{i}^{-1},$

positive for spheres and negative for hyperbolic spaces, and the Bach tensor
can be written

$\displaystyle B_{\alpha\beta}$
$\displaystyle=\frac{1}{6}(f_{1}^{-2}-f_{2}^{-2})\,f_{1}\,(g^{(1)})_{\alpha\beta}$
$\displaystyle B_{jk}$
$\displaystyle=\frac{1}{6}(f_{2}^{-2}-f_{1}^{-2})\,f_{2}\,(g^{(2)})_{jk}.$

Note that if $f_{1}(0)=f_{2}(0)$, then the solution is constant. Otherwise,
suppose $f_{1}(0)<f_{2}(0)$. The constant $\gamma$ allows us to reduce the
system (1) to a single equation.

$\frac{d}{dt}f_{1}=\frac{1}{6}(f_{1}^{-1}-\gamma^{-2}f_{1}^{3}).$

This is separable and we can solve to get

$\displaystyle f_{1}$
$\displaystyle=\sqrt{\gamma\tanh\left(\frac{t}{3\gamma}+\mu\right)}$
$\displaystyle f_{2}$
$\displaystyle=\sqrt{\gamma\coth\left(\frac{t}{3\gamma}+\mu\right)}$

where

$\mu=\frac{1}{2}\ln\left(\frac{f_{1}(0)+f_{2}(0)}{f_{2}(0)-f_{1}(0)}\right).$

If $f_{1}(0)>f_{2}(0)$ the solutions are swapped. From this we see that,
again, solutions are immortal, but not ancient. Here, the absolute values of
the curvatures of the slices converge to the same value, and the manifold
converges to a Bach-flat four-dimensional manifold. The limiting scalar
curvatures can be calculated directly. ∎

##  7\. Comparison with Ricci flow

We finish with a comparison of the qualitative behavior between Ricci flow, as
determined in [IJ92] and [IJL06], and Bach flow. There are a number of ways
this might be done. First, there is the choice of whether to include the one-
dimensional component for Ricci flow. Also, there is the choice of whether or
not to use volume-normalized Ricci flow or unmodified Ricci flow. For a
product metric, the Ricci tensor splits and on a one-dimensional manifold, the
Ricci tensor is zero. This implies that unmodified Ricci flow on on
$\mathbb{S}^{1}\times N$ leaves the one-dimensional component fixed and the
behavior on the three-dimensional slice is the same as for Ricci flow on just
$N$. For volume-normalized Ricci flow, the behavior on $\mathbb{S}^{1}\times
N$ will be somewhat different from that of volume-normalized Ricci flow on
just $N$. First, there is a dimensional constant in the modifying term, and
second, volume normalized flow on $\mathbb{S}^{1}\times N$ does not preserve
the volume of $N$. In the end, the differences in all these flows are somewhat
cosmetic. Rescaling space and time in appropriate ways allows the solution to
one of these problems to be modified so as to solve another. For a bit more
detail, see the discussions in [IJL06] including the analysis for those cases
that relate to the results in [IJ92].

At first, it might seem most natural to compare Bach flow on
$\mathbb{S}^{1}\times N$ to volume-normalized Ricci flow on
$\mathbb{S}^{1}\times N$ since both flows are acting on the same space, and
both flows preserve volume. However for volume-normalized Ricci flow the
behavior of the one-dimensional slice depends on the scalar curvature of $N$,
while by Proposition 2.2, under Bach flow the one-dimensional slice never
expands. On the other hand, for volume-normalized Ricci flow on $N$ the static
solutions are Einstein. Similarly, the static solutions
$g_{\mathbb{S}^{1}}\\!+\tilde{g}$ for Bach flow are those for which
$\tilde{g}$ is Einstein on $N$ by Proposition 2.1. As such, the qualitative
behavior can more easily be compared. We find that on most spaces, the
qualitative behavior is the same, but there are two notable differences. If
$N=\mathbb{S}^{3}$, volume-normalized Ricci flow always converges to the round
sphere, while for Bach flow, the eventual qualitative behavior depends on the
initial metric. If $N=\mathbb{S}^{1}\times\mathbb{S}^{2}$, volume-normalized
Ricci flow experiences curvature blow-up in finite time, while Bach flow
collapses to a flat surface as $t\rightarrow\infty$.

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