# Unique Asymptotics of Steady Ricci Solitons with Symmetry

Zilu Ma, Hamidreza Mahmoudian, and Nataša Šešum

###### Abstract

In this paper we study 4d gradient steady Ricci solitons, which are weak
$\kappa$-solutions, and admit $\mathrm{O}(3)$-symmetry. Under a weak curvature
decay condition, we find precise geometric asymptotics of such solitons, which
are similar to those for 3d compact $\kappa$-solutions found in [ABDS22]. This
is the first step towards the classification of 4d gradient steady Ricci
solitons and more general ancient Ricci flows.

##  1 Introduction

We call a triple $(M^{4},g,f)$ a complete gradient steady Ricci soliton, if
$(M^{4},g)$ is a complete Riemannian manifold and

$\operatorname{Ric}=\nabla^{2}f.$

By a classical result due to Hamilton, we may normalize the metric so that

$R+|\nabla f|^{2}=1.$

If $(\Phi_{t})_{t\in\mathbb{R}}$ denotes the $1$-parameter group of
diffeomorphisms generated by $-\nabla f,$ then
$(M^{4},g_{t})_{t\in\mathbb{R}}$ is a Ricci flow, called the canonical form
induced by the steady soliton $(M^{4},g,f).$

Steady solitons model Type II singularities under Ricci flow ([Ham93e],
[Bre20]). Recall that Hamilton ([Ham93]) distinguished between type I and type
II finite time singularities at time $T$, depending on whether
$(T-t)\,|{\operatorname{Rm}}|$ stays uniformly bounded, or not. By [EMT11] it
is known that type I singularities are modeled on shrinkers, and in general
they are easier to analyze than type II singularities. In an attempt to better
understand type II singularities, steady Ricci solitons received considerable
attention lately, but no classification of steady solitons in general case has
been obtained in dimensions $n\geq 4$.

Let us recall the notion of $\kappa$-solutions as introduced by Perelman in
[Per02].

######  Definition 1.1.

A $\kappa$-solution is an ancient solution of the Ricci flow, which has
nonnegative curvature operator, positive scalar curvature, bounded curvature
on compact time intervals and is $\kappa$-noncollapsed on all scales.

Note that $\kappa$-solutions arise as blowup limits near cylindrical
singularities. They played an important role in Perelman’s proof of the
geometrization conjecture in [Per03a] and [Per03b]. The complete
classification of these solutions in 3d Ricci flow has been recently obtained
in [Bre20] and [BDS21], which are, up to scaling and finite quotients, the
round $\mathbb{S}^{3}$, the round shrinking $\mathbb{R}\times\mathbb{S}^{2}$,
the 3d Bryant soliton, or the 3d Perelman oval.

In this paper, we shall weaken the notion of $\kappa$-solutions and in turn
study the following weak $\kappa$-solutions.

######  Definition 1.2.

A weak $\kappa$-solution is an ancient solution of the Ricci flow, which has
nonnegative sectional curvature, positive scalar curvature, bounded curvature
on compact time intervals and is $\kappa$-noncollapsed on all scales.

Let us focus now on dimension $n=4$. In [Bre14] it has been shown that a
steady gradient Ricci soliton of dimension $n\geq 4$, which has positive
sectional curvature, and is asymptotically cylindrical is the Bryant soliton.
Known examples of 4d steady Ricci solitons up to now include the
$\mathrm{O}(4)$-symmetric 4d steady soliton constructed by Bryant, the 3d
Bryant soliton times a line, and the 1-parameter family of
$\mathbb{Z}_{2}\times\mathrm{O}(3)$ symmetric steady solitons constructed by
Lai ([Lai20]). Solutions found by Lai are the analogous of oval bowl
translators in the mean curvature flow constructed by Hoffman-Ilmanen-Martin-
White. Note that the 4d Bryant soliton has a round cylinder,
$\mathbb{S}^{3}\times\mathbb{R}$ as its tangent flow, whereas the last two
examples have a bubble sheet $\mathbb{S}^{2}\times\mathbb{R}^{2}$ as their
tangent flows. Inspired by an analogous classification of noncollapsed
translators in $\mathbb{R}^{4}$ ([CHH23]) one conjectures (see also [Has23])
the following.

######  Conjecture 1.3.

Any weak $\kappa$-solution in 4d Ricci flow, that is also a steady Ricci
soliton, is up to scaling and finite quotients, given by one of the following:
4d Bryant soliton, the 3d Bryant soliton times a line, or belongs to the
1-parameter family of $\mathbb{Z}_{2}\times\mathrm{O}(3)$-symmetric steady
solitons constructed by Lai ([Lai20]).

######  Remark 1.4.

By the work in [BCDMZ21], see Theorem 2.1 below, gradient steady Ricci soliton
$(M^{4},g,f)$, that also induces a weak $\kappa$-solution, has a unique
tangent flow at infinity. By [CMZ23, Corollary 5.4], if
$\operatorname{Ric}>0$, the unique tangent flow at infinity is either
$\mathbb{S}^{3}\times\mathbb{R}$ or $\mathbb{S}^{2}\times\mathbb{R}^{2}$. In
the former case, by [Bre14, Theorem 1.2] combined with [CMZ23, Theorem 1.5],
we know $(M^{4},g,f)$ is isometric to the Bryant soliton, c.f. [DZ20a, Theorem
1.5] and [ZZ23, Corollary 0.3]. Hence, we can assume from now on that the
Ricci gradient steady soliton we consider here has
$\mathbb{S}^{2}\times\mathbb{R}^{2}$ as the tangent flow at infinity. We also
remark that by the work in [CMZ21] this tangent flow at infinity (in Bamler’s
sense) is identical to Perelman’s asymptotic shrinker, since
$\mathbb{S}^{2}\times\mathbb{R}^{2}$ is smooth.

######  Remark 1.5.

As Perelman mentioned in [Per02, §11.1] which was proved by Zhang in [Z20],
for an ancient flow with $\operatorname{Rm}\geq 0,$ $\kappa$-noncollapsedness
is equivalent to bounded Nash entropy. See also [MZ21, Theorem 1.13], where
for steady solitons that have positive sectional curvature,
$\kappa$-noncollapsedness is equivalent to bounded Nash entropy. Therefore, in
our setting instead of assuming a soliton being $\kappa$-noncollapsed, we may
also assume that the canonical Ricci flow induced by $(M^{4},g,f)$ is
noncollapsed, i.e.,

$\mu_{\infty}:=\inf_{\tau>0}\mathcal{N}_{o,0}(\tau)>-\infty,$

where $\mathcal{N}_{o,0}(\tau)$ denotes the pointed Nash entropy based at a
space-time point $(o,0)$ at scale $\tau$, see, e.g., [Bam20a].

Assumptions: In what follows we will consider GSRS (gradient steady Ricci
soliton), $(M^{4},g,f)$ that is as well a weak $\kappa$-solution, and also has

  1. (A1)

$\operatorname{Ric}>0$.

  2. (A2)

$(M^{4},g)$ admits an $\mathrm{O}(3)$-symmetry, that is,
$\mathrm{O}(3)\subset\rm Isom(M^{4},g)$.

  3. (A3)

A soliton potential function $f$ has a critical point.

Next, denote by

$\mathcal{M}:=\\{\mbox{GSRS}\,\,(M,g,f)\,\,\mbox{that { induces a
weak}}\,\,\,\kappa\,\mbox{-solution, and satisfies assumptions
{\bf(A1)-(A3)}}\\}.$

By Remark 1.4 we may assume that tangent flow of any solution in $\mathcal{M}$
is $\mathbb{S}^{2}\times\mathbb{R}^{2}$.

Note that (A1) implies a critical point of $f$ must be unique, and we denote
this point by $o$. (This holds since $\nabla^{2}f=\operatorname{Ric}>0$ and
$f$ is strictly convex.) It is also known that the level sets
$\Sigma_{s}:=\\{f=s\\}$ are 3d compact manifolds, which are in fact
diffeomorphic to $\mathbb{S}^{3}$ by (A1). See, for example, [DZ21, Lemma
2.3].

The analysis is somewhat similar to [ABDS22]. We describe our approach here in
order to make the comparison with [ABDS22] easier. The main idea is that when
we compare different level sets $\Sigma_{s}$ for large $s$, the induced metric
changes in a way that closely resembles Ricci flow. Consider the level sets
$\Sigma_{s}$, for $s\geq s_{0}$, where $s_{0}$ is a large number that will be
chosen later. Denote by $\Sigma:=\\{f=s_{0}\\}$. Let $\chi_{s}$ be a family of
diffeomorphisms satisfying

$\partial_{s}\chi_{s}=\frac{\nabla f}{|\nabla f|^{2}},\quad{\chi_{s_{0}}={\rm
id}_{\Sigma}.}$

Clearly, $\Sigma_{s}=\\{f=s\\}=\chi_{s}(\Sigma)$ for $s\geq s_{0}.$ We define

$\bar{g}_{s}:=\chi_{s}^{*}(g|_{\Sigma_{s}}),$ (1.6)

for $s\geq 0$. Then $(\Sigma,\bar{g}_{s})$ is isometric to
$(\Sigma_{s},g|_{\Sigma_{s}})$, for $s\geq s_{0}.$ So,

$\partial_{s}\bar{g}_{s}=\chi_{s}^{*}\left(\frac{1}{|\nabla
f|^{2}}\left(\mathcal{L}_{\nabla
f}g\right)|_{\Sigma_{s}}\right)=\frac{2\chi_{s}^{*}(\operatorname{Ric}|_{\Sigma_{s}})}{\chi_{s}^{*}|\nabla
f|^{2}}.$

Thus, we may write

$\partial_{s}\bar{g}_{s}=2\operatorname{Ric}_{\bar{g}_{s}}+2\bar{\mathcal{E}}(s),$
(1.7)

where

$\chi_{s*}\bar{\mathcal{E}}(s)=\frac{\operatorname{Ric}|_{\Sigma_{s}}}{|\nabla
f|^{2}}-\overline{\operatorname{Ric}}.$ (1.8)

Here and in the following, we denote by
$\overline{\operatorname{Ric}},\bar{R}$ the curvatures of the induced metric
$g|_{\Sigma_{s}}.$

Our goal in this paper is to complete the first step needed in completing
Conjecture 1.3, in the case the tangent flow is the bubble sheet. More
precisely, we describe precise asymptotics of $(M^{4},g,f)$. In other words,
our result shows that under suitable scalar curvature decay, the level sets
$\Sigma_{s}$ have similar matched asymptotics as the ones of Perelman’s 3d
solution ([BDS21]). We actually establish unique asymptotics for the profile
function of level sets, which in the light of our assumption (A2) are
$\mathrm{O}(3)$ symmetric (see Lemma 2.6 below).

######  Theorem 1.9.

Let $(M^{4},g,f)\in\mathcal{M}$. Assume in addition that

$\sup_{\partial B_{r}(o)}R=O(r^{-\eta}),$

for some $\eta\in{(\frac{1}{3},1)}$. Then there exists a reference point
$q\in\Sigma$ so that the following holds. Let $F(z,s)$ denote the radius of
the sphere of symmetry in $(\Sigma,\bar{g}_{s})$, which has signed distance
$z$ from the reference point $q$. Then the profile function $F(z,s)$ has the
following asymptotic expansions.

  1. (i)

Fix a large number $L$. Then, as $\tau\to-\infty$,

$F(z,s)=\sqrt{2s}-\frac{1}{4\sqrt{2}\log
s\sqrt{s}}\,(z^{2}-2s)+o\Big{(}s^{-1/2}(\log s)^{-1}\Big{)},$

for $|z|\leq L\,\sqrt{s}$.

  2. (ii)

Fix a small number $\theta\in(0,\frac{1}{2})$, and a large number $M\geq 20$.
If $s\geq s(\theta,M)$, then we have

$2s-\frac{M^{2}+C(\theta)}{M^{2}-2}\,\frac{z^{2}}{2\log s}\leq F^{2}(z,s)\leq
2s-\frac{M^{2}-C(\theta)}{M^{2}}\,\frac{z^{2}}{2\log s},$

whenever $z\geq M\sqrt{s}$ and $F(z,s)\geq\theta\,\sqrt{2s}$.

  3. (iii)

The reference point $q$ has distance $(2+o(1))\sqrt{s\log s}$ from each tip of
$(\Sigma,\bar{g}_{s})$. The scalar curvature at each tip is given by
$(1+o(1))\,\frac{\log s}{s}$. Finally, if we rescale the solution around one
of the tips, then the rescaled solutions converge to the Bryant soliton as
$s\to\infty$.

Without the scalar curvature decay assumption asymptotics could look quite
differently from what is stated in Theorem 1.9, and we describe those in
Section 6. Nevertheless, we still expect Theorem 1.9 to hold even without the
scalar curvature decay assumption. Based on Brendle’s work in [Bre14], Deng
and Zhu in [DZ20a] proved that 4d Bryant is the only positively curved steady
soliton with linear curvature decay, i.e., $\eta=1$. Our work seems to be the
first to analyze steady solitons with slower curvature decay.

Organization of the paper is as follows. In Section 2 we prove some geometric
facts about our soliton, and also analyze the structure of level sets, which
up to an error, turn out to have similar behavior to 3d Perelman’s solution.
In Section 3 we study more carefully asymptotics of the level sets in the
cylindrical region. We adopt methods from [Bre20] and [ABDS22] to show that we
have dichotomy in asymptotic behaviors of projections of our soliton onto
eigenspaces of linearized Ricci flow operator around the cylinder. In Section
4 we show that under additional curvature decay assumption we have unique
asymptotic behavior of our soliton in the cylindrical region. In Section 5 we
give precise matched asymtotics of our solution in the case we have curvature
decay assumption, and hence neutral mode dominating. In Section 6 we drop the
assumption on scalar curvature decay, and instead impose the assumption on
reflection symmetry. Without the scalar curvature decay assumption we are not
able to rule out the case of positive modes dominating in dichotomy result
stated in Proposition 3.41. In the case of positive modes dominating we give
precise asymptotics of our solution in the cylindrical region, hoping that can
be used to eventually rule out positive mode dominating at all.

Conventions. For constants, we shall follow conventions similar to those in
[Bam20a, 2.1]. Throughout our paper, constants $c,C,$ depend on the steady
soliton that we consider. For example, constants of the form $C(a,b)$ depend
on $a,b$, and the steady soliton in question.

We write “if $\epsilon\leq\bar{\epsilon}(a,b),\cdots$” to mean that “for any
$a,b,$ there is a small constant $\bar{\epsilon}(a,b)>0$ depending on $a,b$,
and the steady soliton in question, such that if
$\epsilon\leq\bar{\epsilon}(a,b),\cdots$”. Similarly, we write “if
$s\geq\underline{s}(a,b),\cdots$”.

##  2 Preliminaries

###  2.1 Coarse estimates

Let $(M^{4},g,f)$ be a steady gradient Ricci soliton inducing a weak
$\kappa$-solution. By rescaling $g$ if necessary, we can assume the
normalization

$R+|\nabla f|^{2}=1.$

Since $M$ is nonnegatively curved, we have

$0\leq\sec\leq R\leq 1,$

Thanks to technical justifications made in [Bam21, Appendix A], the theory
developed by Bamler in [Bam20a] and [Bam20c] applies for complete Ricci flows
with bounded curvature on compact intervals. Thus, we may use the following
result in [BCDMZ21].

######  Theorem 2.1 ([BCDMZ21]).

Let $(M^{4},g,f)$ be a complete nontrivial and noncollapsed steady gradient
Ricci soliton with bounded curvature. Then its tangent flow at inifinity is
unique and is one of the following:

$(\mathbb{S}^{3}/\Gamma)\times\mathbb{R},\quad\mathbb{S}^{2}\times\mathbb{R}^{2},\quad(\mathbb{S}^{2}\times_{\mathbb{Z}_{2}}\mathbb{R})\times\mathbb{R}.$

As discussed in Section 1, we may assume that the steady soliton has
$\mathbb{S}^{2}\times\mathbb{R}^{2}$ as its unique tangent flow at infinity.

We will use a splitting result for $4$-dimesional steady solitons proven in
[CMZ23]. We make the following definition before stating it.

######  Definition 2.2 (Dimension Reduction).

Let $(M^{n},g)$ be a complete Riemannian manifold. We say $(M^{n},g)$
dimension reduces to $(N^{n-1},h)$ along some sequence $x_{i}\to\infty$, if
(after possibly passing to a subsequence)

$(M^{n},R(x_{i})g,x_{i})\to(N^{n-1}\times\mathbb{R},h+{\rm
d}z^{2},(x_{\infty},0)),$

locally smoothly in the sense of Cheeger-Gromov convergence.

######  Theorem 2.3 ([CMZ23, Theorem 1.2]).

Let $(M^{4},g,f)$ be a complete noncollapsed steady soliton with $\sec\geq
0,\operatorname{Ric}>0$. Suppose $M$ has $\mathbb{S}^{2}\times\mathbb{R}^{2}$
as its tangent flow at infinity. Then along any sequence $x_{i}\to\infty$,
$(M,g)$ dimension reduces to either $M^{3}_{\rm Bry}$, the 3d Bryant’s steady
soliton, or $\mathbb{S}^{2}\times\mathbb{R}$, the round cylinder with scalar
curvature $1.$

Now assume $(M,g,f)\in\mathcal{M}$ is a steady soliton satisfying our
assumptions. For $x\in M$, let $r_{\rm sym}(x)$ denote the radius of the
symmetric $\mathbb{S}^{2}$ passing through $x$. Using the Theorem above we can
show the following.

######  Lemma 2.4.

There is a constant $C$ depending on $(M^{4},g,f)\in\mathcal{M}$ such that

$R(x)\leq Cr_{\rm sym}^{-2}(x),$

for any $x\in M.$

###### Proof.

Suppose there exist $x_{i}\to\infty$ such that

$R(x_{i})r_{\rm sym}^{2}(x_{i})\to\infty.$

By Theorem 2.3, $(M,R(x_{i})g,x_{i})$ converges to some limit pointed manifold
$(M_{\infty},g_{\infty},x_{\infty})$, locally smoothly. Here,
$(M_{\infty},g_{\infty})$ is either $M^{3}_{\rm Bry}\times\mathbb{R}$ or
$\mathbb{S}^{2}\times\mathbb{R}^{2}$. Along the convergence,

$r_{\rm sym}^{2}(x_{i})R(x_{i})\to r_{\rm sym}^{2}(x_{\infty})<\infty,$

which is a contradiction. ∎

We will also need the following version of Perelman’s result (c.f. [KL08,
Corollary 42.1-(4)]) proven in the appendix of [CMZ23]. As mentioned in Remark
1.5, the assumption of bounded entropy can be replaced by being
$\kappa$-noncollapsed by [MZ21, Theorem 1.9].

######  Theorem 2.5 ([CMZ23, Corollary A.9]).

Let $(M^{n},g_{t})_{t\leq 0}$ be a complete ancient solution to Ricci flow
with Nash entropy bounded from below by $\mu_{\infty}$. Suppose
$(M^{n},g_{t})_{t\leq 0}$ has bounded curvature on compact time-intervals, and
assume

$\operatorname{Ric}\geq 0\,,\quad|{\operatorname{Rm}}|\leq\Lambda R$

on $M\times(-\infty,0]$ for some constant $\Lambda<\infty$. Then for any
$k\geq 0$, there is a constant $C_{k}=C(n,\mu_{\infty},\Lambda,k)<\infty$ such
that

$|\nabla^{k}{\operatorname{Rm}}|\leq C_{k}R^{\frac{k+2}{2}}$

on $M\times(-\infty,0]$.

###  2.2 Structure of the level sets

The goal of this subsection is to describe some geometric properties of the
level sets of the potential function $f$. We show that each level set is
rotationally symmetric, asymptotically cylindrical and has exactly two tips
resembling the Bryant soliton. Throughout this subsection, we assume
$(M^{4},g,f)\in\mathcal{M}$ unless otherwise stated. For simplicity, we also
assume that the scalar curvature decays uniformly.

In the Lemma that follows we show that under our Assumptions (A1)-(A3), the
level sets $\Sigma_{s}=\\{f=s\\}$ of $f$ are rotationally symmetric.

######  Lemma 2.6.

Let $o$ be a critical point of the potential $f$. For any $\phi\in{\rm
Isom}(g),$

$\phi(o)=o,\quad\phi^{*}f=f.$

As a consequence, for any level set $\Sigma_{s}$ with the induced metric,

${\rm Isom}(g)|_{\Sigma_{s}}\subseteq{\rm Isom}(\Sigma_{s}).$

###### Proof.

By Assumptions (A1) and (A3), the point $o$ is the unique point with $R(o)=1$.
Since we have $\phi^{*}R=R$, we must have $\phi(o)=o$. Note also that
$\phi^{*}f$ is a potential function for $(M,g)$. Hence
$\nabla^{2}(\phi^{*}f-f)=0$ and $\phi^{*}f=f+C$ for some $C$, otherwise
$(M,g)$ locally splits off a line in the direction of $\nabla(\phi^{*}f-f)$
which contradicts (A1). Finally we have

$C=(\phi^{*}f)(o)-f(o)=0.$

∎

We now discuss asymptotic cylindricity of the levels sets, in the following
sense.

######  Definition 2.7 ($(\epsilon,k)$-centers).

Let $(M^{n},g)$ be a Riemannian manifold. We say that a point $x\in M$ is an
$(\epsilon,k)$-center, or that $M$ is $\epsilon$-close to
$\mathbb{S}^{n-k}\times\mathbb{R}^{k}$ at $x$, if there is a smooth embedding
$\Phi:V\to M$ with $B((\bar{x},0),1/\epsilon)\subseteq
V\subseteq\mathbb{S}^{n-k}\times\mathbb{R}^{k}$ and $\Phi(\bar{x},0)=x$, such
that for $\tilde{g}=R(x)g$ we have

$\left\|\Phi^{*}\tilde{g}-\bar{g}\right\|_{C^{[1/\epsilon]}}<\epsilon,$

where $\bar{g}$ denotes the standard round metric on
$\mathbb{S}^{m}\times\mathbb{R}^{k-m}$ with constant scalar curvature $1.$

######  Lemma 2.8.

For any $\epsilon>0$, there exists a $\delta(\epsilon)>0$ such that if $x$ is
a $(\delta,2)$-center with $\delta\leq\delta(\epsilon)$, then $\Sigma_{f(x)}$
is $\epsilon$-close to $\mathbb{S}^{2}\times\mathbb{R}$ at $x$.

###### Proof.

Suppose not. Then there exists $\epsilon>0$, $\delta_{i}\to 0$, $x_{i}\in M$
such that $x_{i}$ is a $(\delta_{i},2)$-center, but $\Sigma_{f(x_{i})}$ is not
$\epsilon$-close to $\mathbb{S}^{2}\times\mathbb{R}$ at $x_{i}$. Let
$r_{i}=R^{-\frac{1}{2}}(x_{i})$ and $g_{i}=r_{i}^{-2}g$. Since $x_{i}$ are
$(\delta_{i},2)$-necks and $\delta_{i}\to 0$, Theorem 2.3 implies

$(M,r_{i}^{-2}g,x_{i})\to\mathbb{S}^{2}\times\mathbb{R}^{2}.$

By [CMZ23, Lemma 4.3] we have $r_{i}\to\infty$ and

$(f-f(x_{i}))/r_{i}\to z,$

where $z$ is the coordinate function of some $\mathbb{R}$-factor. By rotation,
we may assume $z=z_{2}$, where $(z_{1},z_{2})$ are the coordinates of the
$\mathbb{R}^{2}$-factor in $\mathbb{S}^{2}\times\mathbb{R}^{2}$. Thus,
$(\Sigma_{f(x_{i})},r_{i}^{-2}g,x_{i})$ converges to
$\mathbb{S}^{2}\times\mathbb{R}$. This is a contradiction. ∎

Let

$0<\lambda_{1}\leq\cdots\leq\lambda_{4}$

denote the eigenvalues of $\operatorname{Ric}.$

######  Lemma 2.9.

For any $\epsilon>0$, there is a $\theta=\theta(\epsilon)\in(0,1)$ such that
$\lambda_{2}(x)\leq\theta R(x)$ implies $x$ is an $(\epsilon,2)$-center and
$\Sigma_{f(x)}$ is $\epsilon$-close to $\mathbb{S}^{2}\times\mathbb{R}$ at
$x$.

###### Proof.

We argue by contradiction. Suppose there exists an $\epsilon>0$ and
$x_{i}\to\infty$ such that $\lambda_{2}(x_{i})/R(x_{i})\to 0$ but $x_{i}$ are
not $(\epsilon,2)$-centers. By Theorem 2.3, $M$ dimension reduces to $M_{\rm
Bry}^{3}$ along $x_{i}$ (since $x_{i}$ are not $(\epsilon,2)$-centers), which
contradicts to the fact that $\lambda_{2}(x_{i})/R(x_{i})\to 0$. By reducing
$\epsilon$ is necessary, we can use Lemma 2.8 to conclude $\epsilon$-closeness
of $\Sigma_{f(x)}$. ∎

The next few results demonstrate that for large $s$, the level sets
$\Sigma_{s}$ look similar to compact ancient 3d solutions.

######  Proposition 2.10.

Consider an arbitrary sequence $s_{k}\to\infty$. For $k\geq\underline{k}$,
there are two disjoint compact domains
$\Omega^{1}_{k}\,,\,\Omega^{2}_{k}\subset\Sigma_{s_{k}}$ with the following
properties.

  * •

$\Omega^{1}_{k},\Omega^{2}_{k}$ are diffeomorphic to $\mathbb{B}^{3}.$

  * •

$\lambda_{2}(x)<\theta R(x)$ for any $x\in\Sigma_{s_{k}}\setminus\Omega_{k}$
,where $\Omega_{k}=\Omega^{1}_{k}\cup\Omega^{2}_{k}$.

  * •

$\lambda_{2}(x)>\frac{1}{2}\theta R(x),$ for any $x\in\Omega_{k}.$

  * •

$\partial\Omega_{k}$ are two leaves of Hamilton’s CMC foliation in
$\Sigma_{s_{k}}$.

  * •

There is a leaf $\Gamma_{k}$ of the CMC foliation in $\Sigma_{s_{k}},$ such
that $\Omega^{1}_{k}\,,\,\Omega^{2}_{k}$ lie in the two different components
of $\Sigma_{s_{k}}\setminus\Gamma_{k},$ and
$\sup_{\Gamma_{k}}\frac{\lambda_{2}}{R}\to 0.$

  * •

For $i=1,2$ and for any sequence of points $x^{i}_{k}\in\Omega^{i}_{k}$ ,
$\big{(}\Omega^{i}_{k}\,,\,R(x^{i}_{k})g\,,\,x^{i}_{k}\big{)}$ converges to a
bounded domain in $M_{\rm Bry}^{3}$ which contains the tip of $M_{\rm
Bry}^{3}.$

###### Proof.

The proof is almost the same as that of [BDS21, Proposition 2.4] now that we
have [CMZ23, Theorem 1.3]. ∎

For a fixed $s$, we let $\overline{\nabla}$ denote the projection of $\nabla$
onto the tangent space of $\Sigma_{s}$.

######  Definition 2.11.

We say that $p\in\Sigma_{s}$ is a tip of $\Sigma_{s}$, if
$\lambda_{2}(p)>R(p)/6$ and $\overline{\nabla}R(p)=0$.

######  Corollary 2.12.

Consider an arbitrary sequence $s_{k}\to\infty$. Let
$\Omega^{1}_{k},\,\Omega^{2}_{k}\subset\Sigma_{s_{k}}$ be the two disjoint
compact domains given by Proposition 2.10. For $k\geq\underline{k}$, there are
exactly two tips in $\Sigma_{s_{k}}$, one in $\Omega^{1}_{k}$ and the other
one in $\Omega^{2}_{k}$.

###### Proof.

The proof is almost the same as that of [BDS21, Corollary 2.5]. ∎

######  Proposition 2.13.

Consider an arbitrary sequence $s_{k}\to\infty$. Let $p^{1}_{k}\,,\,p^{2}_{k}$
be the two tips in $\Sigma_{s_{k}}$ for $k\geq\underline{k}$. Then for
$i=1,2$, as $k\to\infty$, we have

$(\Sigma_{s_{k}},\,R(p^{i}_{k})g\,,\,p^{i}_{k})\to M_{\rm Bry}^{3}.$

###### Proof.

The proof is almost the same as that of [BDS21, Proposition 2.6]. ∎

######  Proposition 2.14.

Consider an arbitrary sequence $s_{k}\to\infty$. Let $p^{1}_{k}\,,\,p^{2}_{k}$
be the two tips in $\Sigma_{s_{k}}$ for $k\geq\underline{k}.$ Then for
$i=1,2,$ as $k\to\infty,$

$R(p^{i}_{k})d(p^{1}_{k}\,,\,p^{2}_{k})^{2}\to\infty.$

###### Proof.

The proof is almost the same as that of [BDS21, Proposition 2.7]. ∎

######  Proposition 2.15.

Consider an arbitrary sequence $s_{k}\to\infty$ and let
$x_{k}\in\Sigma_{s_{k}}$. Let $p^{1}_{k}\,,\,p^{2}_{k}$ be the two tips in
$\Sigma_{s_{k}}$ for $k\geq\underline{k}$. If
$R(p^{1}_{k})d_{g}(x_{k}\,,\,p^{1}_{k})^{2}\to\infty$ and
$R(p^{2}_{k})d_{g}(x_{k}\,,\,p^{2}_{k})^{2}\to\infty$ then
$\tfrac{\lambda_{2}}{R}(x_{k})\to 0$.

###### Proof.

The proof is almost the same as that of [BDS21, Proposition 2.8]. ∎

By combining Corollary 2.12, Proposition 2.13, Proposition 2.14, and
Proposition 2.15 we can draw the following conclusion.

######  Corollary 2.16.

  1. (i)

If $s\geq\underline{s},\Sigma_{s}$ has exactly two tips, denoted by
$p^{1}_{s},p^{2}_{s}$, and they vary continuously in $s.$ We denote by

$r_{i,s}:=R(p^{i}_{s})^{-\frac{1}{2}}$

the scalar curvature radii at the tips, $i=1,2.$ Then

$\lim_{s\to\infty}\frac{r_{i,s}}{\sqrt{s}}\to 0.$ (2.17)

  2. (ii)

For any $A<\infty,$ if $s\geq\underline{s}(A),$

$B(p^{1}_{s},Ar_{1,s})\cap B(p^{2}_{s},Ar_{2,s})=\emptyset.$

  3. (iii)

For any $A<\infty,\epsilon>0,$ if $s\geq\underline{s}(A,\epsilon),$
$\big{(}B(p^{i}_{s},Ar_{i,s}),r_{i,s}^{-2}g,p^{i}_{s}\big{)}$ is
$\epsilon$-close to the corresponding piece in $M_{\rm
Bry}^{3}\times\mathbb{R},i=1,2.$

  4. (iv)

For any $\epsilon>0,$ there are $\underline{s}(\epsilon),A(\epsilon)<\infty$
with the following property. If $s\geq\underline{s},$ and $x\notin
B(p^{1}_{s},Ar_{1,s})\cup B(p^{2}_{s},Ar_{2,s}),$ then $x$ is an
$\epsilon$-center.

######  Proposition 2.18 ([ABDS22, Proposition 2.8]).

Let $(M^{4},g,f)\in\mathcal{M}$. For any $\epsilon>0$ and $\delta>0$, there
exists $\underline{s}(\epsilon,\delta)$ with the following property: If
$s\geq\underline{s}(\epsilon,\delta)$ and $x\in\Sigma_{s}$ satisfies $r_{\rm
sym}(x)\geq\delta\sqrt{s}$, then $x$ is an $\epsilon$-center.

###### Proof.

The proof is almost the same as that of [ABDS22, Proposition 2.8], and uses
Theorem 2.3 and (2.17). ∎

###  2.3 Brendle’s barrier function

As in [ABDS22], the barrier function constructed by Brendle in [Bre20,
Proposition 2.7] plays a key role in our analysis. In order to control our
error terms, we refine the statement of properties of Brendle’s barrier
slightly and obtain the following.

######  Proposition 2.19.

There exist universal constants $r_{*}\in(0,1)$, $N$ and $\underline{a}\gg 1$,
and $0<\theta\ll 1$ with the following properties: For $a\geq\underline{a}$,
we can find a $C^{1}$ function

$\psi_{a}=\psi_{a}:\left[r_{*}a^{-1},1+\tfrac{1}{100}a^{-2}\right]\to\mathbb{R},$

satisfying

$\psi_{a}\psi_{a}^{\prime\prime}-\tfrac{1}{2}\psi_{a}^{\prime
2}+s^{-2}(1-\psi_{a})(s\psi_{a}^{\prime}+2\psi_{a})-s\psi_{a}^{\prime}\leq\left\\{\begin{array}[]{ll}-\tfrac{1}{4}a^{-4}s^{-5},&\text{
on }[Na^{-1},1+\frac{1}{100}a^{-2}]\\\ -\tfrac{1}{2}a,&\text{ on
}[r_{*}a^{-1},Na^{-1}]\end{array}\right.$ (2.20)

Moreover,

  * •

$\psi_{a}(s)\leq Ca^{-2}$ for
$s\in\left[\frac{1}{10},1+\tfrac{1}{100}a^{-2}\right]$

  * •

$\psi_{a}(s)\leq 2a^{-2}s^{-2}$ for
$s\in\left[Na^{-1},1+\tfrac{1}{100}a^{-2}\right]$

  * •

$\psi_{a}(s)\geq\frac{1}{32}a^{-4}$ for
$s\in\left[r_{*}a^{-1},1+\tfrac{1}{100}a^{-2}\right]$

  * •

$\psi_{a}(s)\geq a^{-2}(s^{-2}-1)+\frac{1}{16}a^{-4}$, for
$s\in\left[1-\theta,1+\tfrac{1}{100}a^{-2}\right]$

  * •

$\psi_{a}(r_{*}a^{-1})\geq\frac{3}{2}$.

We shall use the following parabolic version.

######  Corollary 2.21.

Let $\psi_{a}$ be the function constructed in Proposition 2.19. Then the
function

$\Psi_{a}(r,s):=\Psi_{a}(r,s):=\psi_{a}\left(\frac{r}{\sqrt{2s}}\right)$

satisfies

$-\Psi_{a,s}>\Psi_{a}\Psi_{a,rr}-\tfrac{1}{2}\Psi_{a,r}^{2}+r^{-2}(1-\Psi_{a})(r\Psi_{a,r}+2\Psi_{a})+Q$

for $s\geq\underline{s}$, where

$Q\geq\left\\{\begin{array}[]{ll}\tfrac{1}{8}a^{-4}(r/\sqrt{2s})^{-5}s^{-1},&\text{
for }r/\sqrt{2s}\in[Na^{-1},1+\frac{1}{100}a^{-2}]\\\
\tfrac{1}{4}as^{-1},&\text{ for
}r/\sqrt{2s}\in[r_{*}a^{-1},Na^{-1}]\end{array}\right.$

Additionally, $\Psi_{a}$ satisfies the other properties stated in Proposition
2.19.

###### Proof.

The corollary follows immediately from Proposition 2.19. ∎

##  3 Asymptotic analysis of level sets near the cylinder

In this section we discuss fine asymptotic behaviors of the level sets
$\Sigma_{s}=\\{f=s\\}$ as $s\to\infty$. Let $s_{0}>0$ be a large number to be
determined. We proceed to define a family of metrics on the level set

$\Sigma=\Sigma_{s_{0}}=\\{f=s_{0}\\}$

Consider the family $\chi_{s}$ of diffeomorphisms on our steady soliton $M$
satisfying

$\partial_{s}\chi_{s}=\frac{\nabla f}{|\nabla f|^{2}},\quad\chi_{s_{0}}={\rm
id}.$

This family preserves the level sets in the sense that
$\Sigma_{s}=\chi_{s}(\Sigma)$. Thus we defined in (1.6) a metric

$\bar{g}_{s}:=\chi_{s}^{*}(g|_{\Sigma_{s}})$

on $\Sigma$ so that for $s\geq s$, $(\Sigma,\bar{g}_{s})$ is isometric to
$(\Sigma_{s},g|_{\Sigma_{s}})$.

Recall that if $\Phi_{t}$ is the flow of $-\nabla f$, then
$g_{t}=\Phi_{t}^{*}g$ evolves by Ricci flow. Therefore, a simple computation
shows

$\displaystyle\partial_{s}\bar{g}_{s}=2{\chi_{s}}^{*}\left(\frac{\operatorname{Ric}|_{\Sigma_{s}}}{|\nabla
f|^{2}}\right)=2\operatorname{Ric}_{\bar{g}_{s}}+2\bar{\mathcal{E}}(s),$ (3.1)

where

$\displaystyle\bar{\mathcal{E}}(s)=\chi_{s*}\left(\frac{\operatorname{Ric}|_{\Sigma_{s}}}{|\nabla
f|^{2}}-\operatorname{Ric}_{\bar{g}_{s}}\right).$

We shall focus on the evolution equation (3.1) for $(\Sigma,\bar{g}_{s})$. We
will show that the contribution from the error term $\,\overline{\mathcal{E}}$
is small, and thus (3.1) can be considered as an almost ancient Ricci flow.

By Lemma 2.6, we know that $g|_{\Sigma_{s}}$ and hence $\bar{g}_{s}$ is
$O(3)$-invariant. We can therefore write

$\bar{g}_{s}={\rm d}z^{2}+F^{2}(z,s)\,g_{\mathbb{S}^{2}}.$ (3.2)

Here $g_{\mathbb{S}^{2}}$ denotes the standard round metric on
$\mathbb{S}^{2}$ with radius $1$, and $z=z(s)$ represents the signed distance
on $\Sigma$ from some fixed point $q$ (to be determined), measured with
respect to $\bar{g}_{s}$. We can also decompose the error term
$\,\overline{\mathcal{E}}$ into its radial and orbital parts as

$\,\overline{\mathcal{E}}=\,\overline{\mathcal{E}}^{\rm rad}{\rm
d}z^{2}+\,\overline{\mathcal{E}}^{\rm orb}g_{\mathbb{S}^{2}}.$ (3.3)

In the rest of this section, we consider the evolution equation for the
function $F$, and derive estimates for $\,\overline{\mathcal{E}}^{\rm rad}$
and $\,\overline{\mathcal{E}}^{\rm orb}$. It will then be possible to carry
out an analysis similar to [ABDS22] for an appropriately chosen reference
point $q$.

###  3.1 Choice of Reference Point

As in [ABDS22], we shall first fix our reference point $q$. We define
$\Phi_{t}$ by

$\partial_{t}\Phi_{t}=-\nabla f\left(\Phi_{t}\right),\quad\Phi_{0}={\rm id},$

so that the Ricci flow solution induced by the steady soliton is
$g_{t}=\Phi_{t}^{*}g$. As before, we shall denote the distance between $x$ and
$y$ with respect to the soliton metric $g$ by $d_{g}(x,y).$

######  Proposition 3.4 ([ABDS22, Proposition 3.1]).

There is a $q\in\Sigma,$ such that

$\limsup_{s\to\infty}sR(\Phi_{-s}(q))\leq 100.$

###### Proof.

The proof is similar to that of [ABDS22, Proposition 3.1] and we shall only
sketch a few differences. Suppose on the contrary that no such $q\in\Sigma$
exists. By [CMZ23, Proposition 4.5], for any $\epsilon>0,$ if
$s\geq\underline{s}(\epsilon),$ there is $x_{s}\in\Sigma_{s}$ that is an
$(\epsilon,2)$-center. It is not hard to see that
$y_{s}=\Phi_{\sigma_{s}}(x_{s})\in\Sigma$ for some $\sigma_{s}>0,$ by e.g. the
proof of [CMZ21, Proposition 3.1]. Thus, for the induced Ricci flow, there
exist $(q_{k},s_{k})\in M\times(-\infty,0)$ that are
$(\epsilon_{k},2)$-centers, where $\epsilon_{k}\to
0,q_{k}\in\Sigma,s_{k}\to-\infty.$ The rest of the proof follows verbatim as
in [ABDS22, Proposition 3.1], where Perelman’s compactness theorem and the
existence of asymptotic shrinkers [Per02, Section 11] were applied for 3d
$\kappa$-solutions. In our case, we cannot directly apply Perelman’s original
statements as they require $\operatorname{Rm}\geq 0$ in dimension $4$ and
higher, and we shall use the version given in [MZ21, Theorem 1.13]. ∎

######  Proposition 3.5 ([ABDS22, Proposition 3.2]).

For the induced Ricci flow $(M,g_{t})_{t\in\mathbb{R}},$ for any
$t_{k}\to-\infty$, if we dilate the flow around $(q,t_{k})$ by factor
$(-t_{k})^{-\frac{1}{2}},$ then the rescaled manifolds converge to
$\mathbb{S}^{2}\times\mathbb{R}^{2}$ of radius $\sqrt{2}.$

###### Proof.

The proof follows verbatim as that of [ABDS22, Proposition 3.2]. ∎

######  Proposition 3.6.

There is a $q\in\Sigma,$ such that for any $s_{k}\to\infty,$ after passing to
a subsequence,

$(M,s_{k}^{-1}g,\chi_{s_{k}}(q))\to(\mathbb{S}^{2}\times\mathbb{R}^{2},2g_{\mathbb{S}^{2}}\oplus
g_{\mathbb{R}^{2}},q_{\infty}),$

locally smoothly. In particular, for any $\epsilon>0,A<\infty,$ if
$s\geq\underline{s}(\epsilon,A),$ $|y,\chi_{s}(q)|\leq A\sqrt{s},$ then

$1-\epsilon\leq s\cdot R(y)\leq 1+\epsilon.$

###### Proof.

Let $q\in\Sigma$ be chosen as above. Since $s\mapsto\chi_{s}(q)$ is a
reparametrization of $t\mapsto\Phi_{t}(q),$ for any $s\geq s_{0},$ there is
$t_{s}<0$ such that $\chi_{s}(q)=\Phi_{t_{s}}(q).$ It suffices to show that
$-t_{s}\sim s.$

By the proposition above, $-tR(\Phi_{t}(q))\to 1,$ as $t\to-\infty.$ In
particular, $|\nabla f|^{2}(\Phi_{t}(q))=1-R(\Phi_{t}(q))=1-o(1).$ For any
$\epsilon>0,$ if $t\leq\bar{t}=\bar{t}(\epsilon)$, then $|\nabla
f|^{2}(\Phi_{t}(q))\geq 1-\epsilon.$ Then for large $s,$

$\displaystyle s=f(\Phi_{t_{s}}(q))$
$\displaystyle=f(\Phi_{\bar{t}}(q))-\int_{t_{s}}^{\bar{t}}\partial_{t}f(\Phi_{t}(q))\,{\rm
d}t$ $\displaystyle=\bar{s}+\int_{t_{s}}^{\bar{t}}|\nabla
f|^{2}(\Phi_{t}(q))\,{\rm d}t\geq(1-\epsilon)(\bar{t}-t_{s})+\bar{s},$

where $\bar{s}=f(\Phi_{\bar{t}}(q)).$ On the other hand, since $|\nabla f|\leq
1,$

$s=f(\Phi_{t_{s}}(q))=f(q)+\int_{t_{s}}^{0}|\nabla f|^{2}(\Phi_{t}(q))\,{\rm
d}t\leq s_{0}+|t_{s}|.$

Thus, $-t_{s}\sim s,$ and the conclusions follow. ∎

We may take such $q$ as a reference point in the following.

###  3.2 Analysis of the error terms

In this subsection, we estimate the size of the error terms in (3.3) together
with their derivatives. Let $\\{e_{i}\\}_{i=1}^{4}$ be a local orthonormal
frame in the region $\\{x\in M:s=f(x)\geq s_{0},r_{\rm sym}(x)>0\\}.$ For
simplicity, assume that $e_{3}=\partial_{z},e_{4}=\frac{\nabla f}{|\nabla
f|},$ and $e_{1},e_{2}$ are tangent to the symmetric sphere. We follow the
convention

$R_{ijk\ell}=R(e_{i},e_{j},e_{k},e_{\ell}),\qquad
R_{ij}=\operatorname{Ric}(e_{i},e_{j}),\qquad\cdots$

and denote the quantities related to $\bar{g}_{s}$ by $\bar{R}$,
$\bar{\nabla}$, etc. and those related to $g$ by $R$, $\nabla$, etc. For
convenience, we shall also stop explicitly writing the differmorphism
$\chi_{s}$, and use $\Sigma_{s}$ and $\bar{g}_{s}$ interchangeably to refer to
either an arbitrary level set $\\{f=s\\}$ with the induced metric, or our
fixed level set $\Sigma$ equipped with $\bar{g}_{s}$. We will use the estimate
$F^{m}\partial_{z}^{m+1}F\leq C_{m}$ for $m\geq 0$ in Lemma 3.12. This is
proven in the next section. However, the proof only uses results of Section 2.
We have chosen to present the results in this order for the sake of clarity.

By the standard Gauss formula, we have (see, for example, [DZ20b, (3.11)])

$R_{ij}=\overline{R}_{ij}+R(\mathbf{n},e_{i},e_{j},\mathbf{n})-\frac{1}{|\nabla
f|^{2}}\sum_{k=1}^{3}(R_{ij}R_{kk}-R_{ik}R_{kj}).$ (3.7)

Recall that we have normalized the metric $g$ so that $|\nabla f|^{2}+R=1$.
Hence

$\displaystyle\,\overline{\mathcal{E}}_{ij}$
$\displaystyle=\frac{R_{ij}}{|\nabla
f|^{2}}-\overline{R}_{ij}=\frac{1}{|\nabla
f|^{2}}RR_{ij}+R_{ij}-\overline{R}_{ij}$ $\displaystyle=\frac{1}{|\nabla
f|^{2}}RR_{ij}+R_{4ij4}-\frac{1}{|\nabla
f|^{2}}R_{ij}(R-R_{44})+\frac{1}{|\nabla f|^{2}}\sum_{k=1}^{3}R_{ik}R_{kj}$
$\displaystyle=R_{4ij4}+\frac{1}{|\nabla f|^{2}}R_{ij}R_{44}+\frac{1}{|\nabla
f|^{2}}\sum_{k=1}^{3}R_{ik}R_{kj}.$ (3.8)

We first show that $\,\overline{\mathcal{E}}^{\rm
orb},\,\overline{\mathcal{E}}^{\rm rad}$ have a sign in the case we consider.

######  Lemma 3.9.

Assume $(M^{4},g,f)\in\mathcal{M}$. Then for $s\geq\underline{s}$,

$\,\overline{\mathcal{E}}^{\rm orb}\geq
0\quad,\quad\,\overline{\mathcal{E}}^{\rm rad}\geq 0.$

###### Proof.

By the definition of $\,\mathcal{E}^{\rm orb}$,

$\displaystyle F^{-2}\,\overline{\mathcal{E}}^{\rm orb}$
$\displaystyle=\,\overline{\mathcal{E}}_{11}=R_{1441}+\frac{1}{|\nabla
f|^{2}}R_{11}R_{44}+\frac{1}{|\nabla f|^{2}}\sum_{k=1}^{3}R_{1k}^{2}.$

Similarly,

$\,\overline{\mathcal{E}}^{\rm rad}=R_{3443}+\frac{1}{|\nabla
f|^{2}}R_{33}R_{44}+\frac{1}{|\nabla f|^{2}}\sum_{k=1}^{3}R_{3k}^{2}.$

Non-negative sectional curvature implies $R_{3443}\geq 0$, $R_{1441}\geq 0$,
and by assumption $R_{ii}>0$. So the assertion follows. ∎

We can also estimate the derivatives of $\,\overline{\mathcal{E}}$. Since
$\nabla^{2}f=\operatorname{Ric}>0$, $f$ has a unique critical point. Moreover,
if $\Phi_{t}$ denotes the flow of $-\nabla f,$

$\partial_{t}R(\Phi_{t})=-2\operatorname{Ric}(\nabla f,\nabla f)<0,$

and thus $R<1$ away from the critical point. Hence

$\frac{1}{|\nabla f|^{2}}=\frac{1}{1-R}=1+O(R).$ (3.10)

In order to better understand the error terms we carefully analyze the terms
in (3.7). By [DZ20b, Lemma 3.1], for a steady soliton we have

$|\nabla
f|^{2}R(\mathbf{n},e_{i},e_{j},\mathbf{n})=-\tfrac{1}{2}\nabla^{2}_{ij}R-\sum_{k=1}^{3}R_{ik}R_{kj}+(\Delta\operatorname{Ric})_{ij}+2\sum_{k,l=1}^{3}R_{iklj}R_{kl}=O(R^{2}).$
(3.11)

We then have the following rough estimates for $\,\overline{\mathcal{E}}^{\rm
rad}$ and $\,\overline{\mathcal{E}}^{\rm orb}$.

######  Lemma 3.12.

For $s\geq\underline{s},k\geq 0,$

$\left|\,\overline{\mathcal{E}}^{\rm rad}\right|\leq
CF^{-4},\quad\left|\partial_{z}^{k}\,\overline{\mathcal{E}}^{\rm
orb}\right|\leq C_{k}F^{-2-k}.$

###### Proof.

Since $\tfrac{1}{|\nabla f|^{2}}=1+O(R)$ and $R\leq CF^{-2}$ on our region by
Lemma 2.4, the expression (3.2) for $\,\overline{\mathcal{E}}$ together with
identity (3.11) and Theorem 2.5 imply

$|\nabla_{g}^{k}\,\overline{\mathcal{E}}(s)|_{g}\leq
C_{k}R^{2+\frac{k}{2}}\leq C_{k}F^{-4-k}.$ (3.13)

On the other hand, since
$\bar{g}_{s}=\mathrm{d}z^{2}+F^{2}g_{\mathbb{S}^{2}}$, we have

$\displaystyle\left|\,\overline{\mathcal{E}}^{\rm
rad}\right|^{2}+2\left|F^{-2}\,\overline{\mathcal{E}}^{\rm
orb}\right|^{2}=|\,\overline{\mathcal{E}}|_{\bar{g}_{s}}^{2}\leq CF^{-4},$
$\displaystyle F^{-2}\left|\partial_{z}^{k}\,\overline{\mathcal{E}}^{\rm
orb}\right|\leq C\left|\nabla_{g}^{k}\,\overline{\mathcal{E}}\right|_{g}\leq
C_{k}R_{g}^{2+\frac{k}{2}}\leq C_{k}F^{-4-k}.$

∎

######  Lemma 3.14.

On the region $\\{r_{\rm sym}\geq\sqrt{s}\\},$ as $s\to\infty,$

$R_{1221}=\left(\tfrac{1}{2}+o(1)\right)R,\quad R_{a\beta\gamma b}=o(1)R,\quad
R_{ijk4}=O(R^{\frac{3}{2}}),$

where $a,b\in\\{1,2\\},\beta,\gamma\in\\{3,4\\}$, and $i,j,k\in\\{1,2,3,4\\}$.
Consequently,

$R_{11}=R_{22}=\left(\tfrac{1}{2}+o(1)\right)R,\quad R_{33}=o(1)R.$

Moreover,

$R_{a44a}=\left(\tfrac{1}{4}+o(1)\right)R^{2},\quad
R_{44}=\left(\tfrac{1}{2}+o(1)\right)R^{2}.$

###### Proof.

We will prove the first asymptotics statement, the proofs of the rest are
similar. By Proposition 2.18, for any $\epsilon>0,$ if $s\geq\bar{s},$ and
$r_{\rm sym}(x)\geq\sqrt{s},$ then $x$ is an $(\epsilon,2)$-center. So,
$|R^{-1}R_{1221}-\tfrac{1}{2}|\leq\epsilon$. Furthermore, since
$R_{ij}=f_{ij},$

$|\nabla
f|R_{ijk4}=R_{ijk\ell}\nabla_{\ell}f=-\nabla_{i}\nabla_{j}\nabla_{k}f+\nabla_{j}\nabla_{i}\nabla_{k}f=-\nabla_{i}R_{jk}+\nabla_{j}R_{ik}=O(R^{\frac{3}{2}}),$

where we have used Theorem 2.5. It follows that

$R_{11}=R_{22}=R_{1221}+R_{1331}+R_{1441}=\tfrac{1}{2}R+o(1)R.$

By (3.11), (3.10), since $R_{1221}=(\tfrac{1}{2}+o(1))R,$

$\displaystyle R_{1441}$
$\displaystyle=-R_{11}^{2}+2R_{1221}R_{22}+o(1)R^{2}=(-\tfrac{1}{4}+\tfrac{1}{2}+o(1))R^{2}$
$\displaystyle=\tfrac{1}{4}R^{2}+o(1)R^{2}.$

∎

Finally, we refine our estimate for the error term
$\,\overline{\mathcal{E}}^{\rm orb}.$

######  Lemma 3.15.

$\,\overline{\mathcal{E}}^{\rm
orb}(z,s)=\left(\tfrac{1}{2}+o(1)\right)F^{2}R^{2}>0,$ whenever
$s\geq\underline{s},$ and $F(z,s)\geq\sqrt{s}.$

###### Proof.

We identify $\chi_{s*}\,\overline{\mathcal{E}}$ and $\,\overline{\mathcal{E}}$
when there is no ambiguity. Recall that $R+|\nabla f|^{2}=1,$ and
$1=\bar{g}_{s}(e_{1},e_{1})=F^{2}g_{\mathbb{S}^{2}}(e_{1},e_{1}).$ By (3.7),

$\displaystyle F^{-2}\,\overline{\mathcal{E}}^{\rm orb}$
$\displaystyle=\,\overline{\mathcal{E}}(e_{1},e_{1})=\tfrac{1}{|\nabla
f|^{2}}\operatorname{Ric}(e_{1},e_{1})-\operatorname{Ric}_{\bar{g}_{s}}(e_{1},e_{1})$
$\displaystyle=(1+O(R))R\overline{R}_{11}+(1+O(R))\left(R_{1441}-(1+O(R))(R_{11}(R-R_{44})-R_{11}^{2}-o(1)R^{2})\right)$
$\displaystyle=R_{1441}+R(\overline{R}_{11}-R_{11})+R_{11}^{2}+o(1)R^{2}$
$\displaystyle=\tfrac{1}{2}R^{2}+o(1)R^{2}>0.$

∎

###  3.3 Estimates for The Warping Function and Its Derivatives

In this section we prove apriori estimates for the warping function $F$ and
its derivatives. These will play an important role in deriving asymptotics for
the soliton in the cylindrical region in the next sections. Unless otherwise
stated, we are restricting everything to $\Sigma$ as discussed before.
However, as mentioned before, we will oocasionally use the notation
$\Sigma_{s}$ to refer to $\Sigma$ equipped with $\bar{g}_{s}$.

First of all note that by standard computations for warped products, we have

$\operatorname{Ric}_{\bar{g}_{s}}=-2\frac{F_{zz}}{F}\mathrm{d}z^{2}+\left(-\frac{F_{zz}}{F}+\frac{1-F_{z}^{2}}{F^{2}}\right)F^{2}g_{\mathbb{S}^{2}}$

Using the evolution equation (3.1) for $\bar{g}_{s}$, we compute the evolution
equation for the warping function $F(z,s)$ in the Proposition below.

######  Proposition 3.16.

In commuting variables $z$ and $s$, we have

$\displaystyle-F_{s}$
$\displaystyle=F_{zz}-F^{-1}(1+F_{z}^{2})+2F_{z}\left\\{\frac{F_{z}}{F}(0,s)-\int_{0}^{z}\frac{F_{z}^{2}}{F^{2}}(\zeta,s)\,\mathrm{d}\zeta\right\\}$
$\displaystyle\qquad-F^{-1}\,\overline{\mathcal{E}}^{\rm
orb}+F_{z}\int_{0}^{z}\,\overline{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta.$

###### Proof.

Fix $p\in\Sigma$ and let $B$ be the $O(3)$-orbit passing through $p$. Recall
that $z(p,s)$ is the signed distance form our basepoint $q$ to $B$ measured
with respect to $\bar{g}_{s}$. On one hand,

$\left.\partial_{s}\right\rvert_{p=const.}F=\left.\partial_{s}\right\rvert_{z=const.}F+z_{s}F_{z},$

and by standard distance distortion,

$z_{s}=\partial_{s}z(p,s)=\int_{0}^{z}\tfrac{1}{2}(\partial_{s}\bar{g}_{s})(\partial_{z},\partial_{z})\,ds=-2\int_{0}^{z}\frac{F_{zz}}{F}(\zeta,s)\,\mathrm{d}\zeta+\int_{0}^{z(s)}\,\overline{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta.$

On the other hand,

$\left.\partial_{s}\right\rvert_{p=const.}(F^{2}\,g_{\mathbb{S}^{2}})=2(-F_{zz}F+1-F_{z}^{2}+\,\overline{\mathcal{E}}^{\rm
orb})g_{\mathbb{S}^{2}},$

and thus

$\left.\partial_{s}\right\rvert_{p=const.}F=-F_{zz}+F^{-1}(1-F_{z}^{2})+F^{-1}\,\overline{\mathcal{E}}^{\rm
orb}$

Hence, in commuting variables $z$ and $s$ we have

$\displaystyle-F_{s}$
$\displaystyle=F_{zz}-F^{-1}(1-F_{z}^{2})-2F_{z}\int_{0}^{z}\frac{F_{zz}}{F}(\zeta,s)\,\mathrm{d}\zeta$
$\displaystyle\qquad-F^{-1}\,\overline{\mathcal{E}}^{\rm
orb}+F_{z}\int_{0}^{z}\,\overline{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta$

The conclusion follows by integration by parts. ∎

We prove the following Lemma that will be implicitly used.

######  Lemma 3.17.

For $s\geq\underline{s}$,

$\displaystyle|F_{z}|\leq 1,\qquad F_{zz}<0,\qquad
F^{m}\partial_{z}^{m+1}F\leq C_{m}(1+F|F_{zz}|)^{m}$

for every $m\geq 0$ and some constant $C_{m}$.

###### Proof.

By Gauss formula, $\sec_{\Sigma_{s}}>\sec_{M}\geq 0,$ see [CMZ23, Lemma 2.1]
for example. Recall that the sectional curvature of $\Sigma_{s}$ in the radial
direction is $-F_{zz}/F$, and the curvature in the orbital direction is
$(1-F_{z}^{2})/F^{2}$. So, $F_{zz}<0$ and $|F_{z}|\leq 1.$ By (3.7), (3.11)
and Perelman’s derivative estimates in Theorem 2.5,
$|\nabla_{k}(R-\overline{R})|=O(R^{2+\frac{k}{2}}).$ The last assertion
follows verbatim as the three-dimensional case in [Bre20, Proposition 3.4]. ∎

We need the following rough estimate.

######  Lemma 3.18.

Let $r_{\max}(s):=\sup_{z}F(z,s)$ denote the maximum radius of the symmetric
spheres in $\Sigma_{s}$. After possibly shifting the initial time by $s_{0}$,
we have

$2s\leq r_{\max}^{2}(s+s_{0})\leq(1+o(1))2s,$

as $s\to\infty.$

###### Proof.

Since $F(0,s)=(1+o(1))\sqrt{2s}$ as $s\to\infty$, we may assume

$r_{\max}(s)\geq F(0,s)\geq\sqrt{s}$

Let $z_{0}$ be the point that achieves the maximum radius on $\Sigma_{s}$. At
$(z_{0},s)$,

$F=r_{\max}(s),\quad F_{z}=0,\quad F_{zz}\leq 0.$

So at $(z_{0},s)$, the evolution equation of $F$ and the fact that
$\,\overline{\mathcal{E}}^{\rm orb}\geq 0$ (Lemma 3.15) imply

$\displaystyle-r_{\max}^{\prime}(s)$
$\displaystyle=-F_{s}\leq-F^{-1}-\,\overline{\mathcal{E}}^{\rm orb}F^{-1}\leq-
r_{\max}^{-1},$

which can be integrated to

$r_{\max}^{2}(s)\geq 2s-C.$

for large $s$. On the other hand, since $F(z_{0},s)\geq\sqrt{s}$, we can use
Proposition 2.18. So for any given $\epsilon$, we can assume $z_{0}$ to be an
$(\epsilon,2)$-center if $s\geq\underline{s}(\epsilon)$. This implies

$F_{zz}\geq-\epsilon F^{-1},$

and at $(z_{0},s)$, Lemma 3.12 gives

$-F_{s}\geq-\epsilon F^{-1}-F^{-1}-CF^{-3}\geq-2\epsilon F^{-1}-F^{-1}.$

Thus, for $s\geq\underline{s}$

$r^{2}_{\max}(s)\leq 2(1+2\epsilon)s.$

By considering $\bar{g}_{s_{0}+s}$ instead of $\bar{g}_{s}$ in the definition
(1.6) for sufficiently large $s_{0},$ we may assume $C=0$ while having the
same upper bound. More precisely, we have $r^{2}_{\max}(s+s_{0})\geq 2s$.

∎

We are now ready to show the bound on the first derivative of $F$.

######  Proposition 3.19.

There exist a universal number $\underline{a}$ and a large $D$ (depending only
on $M^{4}$) such that the following holds: If $a\geq\underline{a}$ is chosen
large enough such that

$\frac{r_{\max}(s)}{\sqrt{2s}}\leq 1+\tfrac{1}{100}a^{-2}$

holds for $s\geq Da^{3}$, then

$F_{z}^{2}(z,s)\leq\psi_{a}\left(\frac{F(z,s)}{\sqrt{2s}}\right)$

whenever $s\geq Da^{3}$ and $F(z,s)\geq r_{*}a^{-1}\sqrt{2s}$.

###### Proof.

The proof is similar to that of [ABDS22, Proposition 3.8], except we have to
deal with extra error terms here. we will outline the details below.

Take $\underline{a}$ large enough so that the barrier construction in 2.19
works and $s\geq\underline{a}^{3}$ implies all the previous estimates, and fix
$a\geq\underline{a}$ in the statement. Let $\mathcal{I}$ be the set of all
$s\geq Da^{3}$ such that the conclusion holds. Suppose, on the contrary,
$\mathcal{I}$ is not all of $[Da^{3},\infty)$. By Proposition 2.18 and the
lower bound for $\psi_{a}$ in 2.19, $\mathcal{I}\supseteq[C(a),\infty)$ holds
for some constant $C(a)$ depending on $a$. Let
$\bar{s}=\sup\left([Da^{3},\infty)\setminus\mathcal{I}\right)$. Then there is
a point $(\bar{z},\bar{s})$ with

$F_{z}^{2}=\psi_{a}\left(\frac{F}{\sqrt{2\bar{s}}}\right),\quad
F(\bar{z},\bar{s})\geq r_{*}a^{-1}\sqrt{2\bar{s}}$

We can argue as in the proof of [ABDS22, Proposition 3.8] to see that in a
neighborhood of $(\bar{z},\bar{s})$, we may find a smooth function $u$ such
that

$F_{z}^{2}(z,s)=u(F(z,s),s).$

The function $u$ satisfies the equation

$\displaystyle-u_{s}$
$\displaystyle=uu_{rr}-\tfrac{1}{2}u_{r}^{2}+r^{-2}(1-u)(ru_{r}+2u)$
$\displaystyle\qquad+2u^{2}\tilde{\mathcal{E}}^{\rm
rad}+r^{-1}u_{r}\tilde{\mathcal{E}}^{\rm orb}-2u(\tilde{\mathcal{E}}^{\rm
orb}/r)_{r}$

We postpone deriving this equation to the end of the proof.

We shall carefully estimate the contribution from $\,\overline{\mathcal{E}}$
in the equation for $u$. First, $u\leq 1$ (Lemma 3.17) and
$\,\overline{\mathcal{E}}$ estimates in Lemma 3.12 imply

$u|\tilde{\mathcal{E}}^{\rm rad}|\leq Cr^{-4},\quad|\tilde{\mathcal{E}}^{\rm
orb}|\leq Cr^{-2},\quad\sqrt{u}|\partial_{r}\tilde{\mathcal{E}}^{\rm orb}|\leq
Cr^{-3}.$ (3.20)

Now, (3.20) along with $u\leq 1$, $u_{r}<0$ (Lemma 3.17) and
$\,\mathcal{E}^{\rm orb}\geq 0$ (Lemma 3.9) imply

$2u^{2}\tilde{\mathcal{E}}^{\rm rad}\leq C\sqrt{u}r^{-4},\qquad
r^{-1}u_{r}\tilde{\mathcal{E}}^{\rm orb}<0.$

Using (3.20) again we get

$\displaystyle-2u(\tilde{\mathcal{E}}^{\rm orb}/r)_{r}$
$\displaystyle=2u\tilde{\mathcal{E}}^{\rm
orb}r^{-2}-2u\partial_{r}\tilde{\mathcal{E}}^{\rm orb}/r\leq
Cur^{-4}+C\sqrt{u}\,r^{-4}\leq 2C\sqrt{u}\,r^{-4}.$

Thus, $u$ satisfies

$-u_{s}\leq
uu_{rr}-\tfrac{1}{2}u_{r}^{2}+r^{-2}(1-u)(ru_{r}+2u)+C_{0}\sqrt{u}\,r^{-4},$
(3.21)

for some constant $C_{0}$.

On the other hand, by Corollary 2.21 $\Psi_{a}(r,s):=\psi_{a}(r/\sqrt{2s})$
satisfies

$-\Psi_{a,s}>\Psi_{a}\Psi_{a,rr}-\tfrac{1}{2}\Psi_{a,r}^{2}+r^{-2}(1-\Psi_{a})(r\Psi_{a,r}+2\Psi_{a})+Q,$
(3.22)

where

$Q\geq\left\\{\begin{array}[]{ll}\tfrac{1}{8}a^{-4}s^{-1}(r/\sqrt{2s})^{-5},&\text{
for }r/\sqrt{2s}\in[Na^{-1},1+\frac{1}{100}a^{-2}]\\\
\tfrac{1}{4}as^{-1},&\text{ for
}r/\sqrt{2s}\in[r_{*}a^{-1},Na^{-1}]\end{array}\right.$

It is enough to show

$Q\geq 2C_{0}\sqrt{u}\,r^{-4},$ (3.23)

where $C_{0}$ is the same constant as in (3.21). If (3.23) holds, we can use
parabolic maximum principle as in Proposition 3.8 of [ABDS22] and compare
(3.21) and (3.22) to get a contradiction. This would imply that $\bar{s}$ as
described above does not exist, hence concluding the proof of the Proposition.

Let $\bar{r}=F(\bar{z},\bar{s})$. We first consider the case
$\bar{r}/\sqrt{2\bar{s}}\in[Na^{-1},1+\frac{1}{100}a^{-2}]$. On one hand, by
$s\geq Da^{3}$ and Proposition 2.19,

$\displaystyle\bar{r}^{4}\,Q\geq\bar{r}^{4}\,\big{(}\tfrac{1}{8}\bar{s}^{-1}a^{-4}(\bar{r}/\sqrt{2\bar{s}})^{-5}\big{)}=\tfrac{1}{2}\bar{s}a^{-4}(\bar{r}/\sqrt{2\bar{s}})^{-1}\geq\tfrac{1}{2}\bar{D}a^{-1}(\bar{r}/\sqrt{2\bar{s}})^{-1}$

On the other hand, at $(\bar{r},\bar{s})$ we have

$u(\bar{r},\bar{s})=\psi_{a}(\bar{r}/\sqrt{2\bar{s}})\leq
2a^{-2}(\bar{r}/\sqrt{2\bar{s}})^{-2}$

So we have $Q\geq 2C_{0}\sqrt{u}r^{-4}$ if we choose $D>8C_{0}$. So (3.23)
holds in this case.

We then consider the case where
$\bar{r}/\sqrt{2\bar{s}}\in(r_{*}a^{-1},Na^{-1}].$ We have

$\tfrac{1}{4}a\bar{s}^{-1}\bar{r}^{4}\geq\tfrac{1}{4}a\bar{s}^{-1}(r_{*}a^{-1}\sqrt{2\bar{s}})^{4}=r_{*}^{4}a^{1-4}\bar{s}\geq
r_{*}^{4}D>2C_{0},$

if we choose $D$ large enough. So, $Q\geq 2C_{0}r^{-4}\geq
2C_{0}\sqrt{u}\,r^{-4}$ in this case as well.

Thus, (3.23) holds for both cases. This concludes the proof of the Proposition
as discussed above.

Finally, we compute the equation for $u$. After the $s$-dependant change of
variable $r=F(z,s)$, $\bar{g}_{s}$ can be written as

$\tilde{g}=u(r,s)^{-1}\mathrm{d}r^{2}+r^{2}\,g_{\mathbb{S}^{2}}$

and for some $V=v(r,s)\partial_{r}$

$\partial_{s}\tilde{g}=2\operatorname{Ric}_{\tilde{g}}+2\,\overline{\mathcal{E}}+\mathcal{L}_{V}\tilde{g}$

Similar to [Bre20], we have

$\displaystyle\mathcal{L}_{V}\tilde{g}=\left(-u^{-2}u_{r}v+2u^{-1}v_{r}\right)\mathrm{d}r^{2}+2rv\,g_{\mathbb{S}^{2}}$
$\displaystyle\operatorname{Ric}_{\tilde{g}}=-\frac{u_{r}}{ru}\mathrm{d}r^{2}+\left(1-u-\tfrac{1}{2}ru_{r}\right)g_{\mathbb{S}^{2}}$

and we can write

$\tilde{\mathcal{E}}=:\tilde{\mathcal{E}}^{\rm
rad}\,\mathrm{d}r^{2}+\tilde{\mathcal{E}}^{\rm orb}\,g_{\mathbb{S}^{2}}$

Since

$\partial_{s}\tilde{g}=-u^{-2}u_{s}\,\mathrm{d}r^{2}$

comparing the spherical component gives

$v=\frac{u-1}{r}+\frac{u_{r}}{2}-\frac{\tilde{\mathcal{E}}^{\rm orb}}{r}$

and in turn comparing the radial component gives

$\displaystyle-u^{-2}u_{s}$
$\displaystyle=-\frac{2u_{r}}{ru}+2\tilde{\mathcal{E}}^{\rm
rad}-u^{-2}u_{r}\left(\frac{u-1}{r}+\frac{u_{r}}{2}-\frac{\tilde{\mathcal{E}}^{\rm
orb}}{r}\right)+2u^{-1}\left(\frac{u-1}{r}+\frac{u_{r}}{2}-\frac{\tilde{\mathcal{E}}^{\rm
orb}}{r}\right)_{r}$

After simplification, we get

$\displaystyle-u_{s}$
$\displaystyle=uu_{rr}-\tfrac{1}{2}u_{r}^{2}+r^{-2}(1-u)(ru_{r}+2u)$
$\displaystyle\qquad+2u^{2}\tilde{\mathcal{E}}^{\rm
rad}+r^{-1}u_{r}\tilde{\mathcal{E}}^{\rm orb}-2u(\tilde{\mathcal{E}}^{\rm
orb}/r)_{r}$

∎

Let us now perform parabolic rescaling

$G(\xi,\tau)=e^{\frac{\tau}{2}}F\left(e^{-\frac{\tau}{2}}\xi,e^{-\tau}\right)-\sqrt{2}.$

with the following change of variable:

$z=e^{-\frac{\tau}{2}}\xi,\quad s=e^{-\tau}.$

Note that

$G_{\xi}=F_{z},\quad G_{\xi\xi}=e^{-\frac{\tau}{2}}F_{zz}.$

Accordingly, we define

$\displaystyle\,\mathcal{E}^{\rm rad}(\xi,\tau)$
$\displaystyle:=e^{-\tau}\,\overline{\mathcal{E}}^{\rm
rad}\left(e^{-\frac{\tau}{2}}\xi,e^{-\tau}\right),$
$\displaystyle\,\mathcal{E}^{\rm orb}(\xi,\tau)$
$\displaystyle:=\,\overline{\mathcal{E}}^{\rm
orb}\left(e^{-\frac{\tau}{2}}\xi,e^{-\tau}\right),$

so that

$\,\mathcal{E}^{\rm rad}d\xi^{2}+\,\mathcal{E}^{\rm
orb}g_{\mathbb{S}^{2}}=\,\overline{\mathcal{E}}^{\rm
rad}\mathrm{d}z^{2}+\,\overline{\mathcal{E}}^{\rm
orb}g_{\mathbb{S}^{2}}=\,\overline{\mathcal{E}}\left(e^{-\frac{\tau}{2}}\xi,e^{-\tau}\right).$

By convergence to the cylinder (see also Lemma 3.18) we have

$G(\cdot,\tau)\to 0,$

as $\tau\to-\infty,$ locally smoothly.

We compute the evolution equation for $G$.

######  Proposition 3.24.

The function $G$ satisfies

$\displaystyle G_{\tau}$
$\displaystyle=G_{\xi\xi}-\tfrac{\xi}{2}G_{\xi}+\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}(1+G_{\xi}^{2})+2G_{\xi}\left\\{\frac{G_{\xi}}{\sqrt{2}+G}(0,\tau)-\int_{0}^{\xi}\frac{G_{\xi}^{2}(x,\tau)}{(\sqrt{2}+G)^{2}}\,\mathrm{d}x\right\\}$
$\displaystyle\qquad-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb}+G_{\xi}\int_{0}^{\xi}\,\mathcal{E}^{\rm rad}(x,\tau)\,\mathrm{d}x.$

###### Proof.

Then

$\displaystyle G_{\tau}$
$\displaystyle=\tfrac{1}{2}(\sqrt{2}+G)-\tfrac{1}{2}\xi
F_{z}-e^{-\frac{\tau}{2}}F_{s}$
$\displaystyle=\tfrac{1}{2}(\sqrt{2}+G)-\tfrac{\xi}{2}G_{\xi}+e^{-\frac{\tau}{2}}F_{zz}-e^{-\frac{\tau}{2}}F^{-1}(1+F_{z}^{2})+2e^{-\frac{\tau}{2}}F_{z}\left\\{\frac{F_{z}}{F}(0,s)-\int_{0}^{z}\frac{F_{z}^{2}}{F^{2}}(\zeta,s)\,\mathrm{d}\zeta\right\\}$
$\displaystyle\qquad-e^{-\frac{\tau}{2}}F^{-1}\,\overline{\mathcal{E}}^{\rm
orb}+e^{-\frac{\tau}{2}}F_{z}\int_{0}^{z}\,\overline{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta$
$\displaystyle=G_{\xi\xi}-\tfrac{\xi}{2}G_{\xi}+\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}(1+G_{\xi}^{2})+2G_{\xi}\left\\{\frac{G_{\xi}}{\sqrt{2}+G}(0,\tau)-\int_{0}^{\xi}\frac{G_{\xi}^{2}}{(\sqrt{2}+G)^{2}}(x,\tau)\,\mathrm{d}x\right\\}$
$\displaystyle\qquad-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb}+G_{\xi}\int_{0}^{\xi}\,\mathcal{E}^{\rm rad}(x,\tau)\,\mathrm{d}x.$

∎

######  Definition 3.25.

For each $\tau$, define

$\rho_{\max}(\tau):=e^{\frac{1}{4}\tau}+\sup_{\xi}G(\xi,\tau)=e^{\frac{\tau}{2}}r_{\max}(e^{-\tau})-\sqrt{2}+e^{\frac{1}{4}\tau}$

and

$\rho(\tau):=\sup_{\sigma\leq\tau}\,\rho_{\max}(\sigma),\qquad\delta(\tau):=\rho(\tau)+\sup_{\sigma\leq\tau}|G(0,\tau)|.$

Note that by Lemma 3.18, for $\tau\leq\bar{\tau}$ we have

$e^{\frac{\tau}{2}}r_{\max}(e^{-\tau})\geq
e^{\frac{\tau}{2}}\sqrt{2e^{-\tau}}=\sqrt{2}$

and therefore

$\rho_{\max}(\tau)\geq
e^{\frac{1}{4}\tau}>0,\qquad\delta(\tau)\geq\rho(\tau)\geq
e^{\frac{1}{4}\tau}>0.$

if $-\bar{\tau}$ is large. On the other hand, convergence to a cylinder
implies

$\rho(\tau)=o(1),\,\,\,\,\delta(\tau)=o(1)\qquad\mbox{as}\qquad\tau\to-\infty.$

By the definitions above and Proposition 3.19 we have the following.

######  Proposition 3.26.

There is $-\bar{\tau}$ large enough such that

$G_{\xi}^{2}(\xi,\tau)\leq\psi_{a}\left(1+\frac{G(\xi,\tau)}{\sqrt{2}}\right)\leq
C\rho(\tau),$

whenever $\tau\leq\bar{\tau}$ and $\sqrt{2}+G(\xi,\tau)\geq
r_{*}a^{-1}\sqrt{2}$, where $a=10^{-1}\rho^{-\frac{1}{2}}(\tau)$.

###### Proof.

We perform the change of variable $s=e^{-\tau},z=e^{-\frac{\tau}{2}}\xi$. Fix
$-\bar{\tau}$ large enough so that $\rho(\tau)\geq e^{\frac{1}{4}\tau}$ for
$\tau\leq\bar{\tau}$, and take an arbitrary $\tau_{*}\leq\bar{\tau}$. Taking
$a=10^{-1}\rho^{-\frac{1}{2}}(\tau_{*})$, we can assume $a\geq\underline{a}$
since $\rho(\tau)\to 0$. For $\tau\leq\tau_{*}$

$s=e^{-\tau}\geq e^{-\tau_{*}}\geq De^{-\frac{3}{8}\tau_{*}}\geq
D\rho^{-\frac{3}{2}}(\tau_{*})\geq Da^{3}$

if $-\bar{\tau}$ is large enough, and

$1+\frac{G(\cdot,\tau)}{\sqrt{2}}\leq 1+\rho_{\max}(\tau)\leq
1+\frac{1}{100}a^{-2}.$

The first inequality follows from Proposition 3.19 for $\tau\leq\tau_{*}$ and
$\sqrt{2}+G(\xi,\tau)\geq 10r_{*}\sqrt{2}\rho(\tau_{*})^{\frac{1}{2}}$. By the
properties of the barrier function $\psi_{a}$ (see Proposition 2.19), we have
$\psi_{a}\leq Ca^{-2}=C\rho(\tau_{*})$. We get both estimates by choosing
$\tau=\tau_{*}$, and since $\tau_{*}\leq\bar{\tau}$ was arbitrary, we are
done. ∎

######  Lemma 3.27.

If $\tau\leq\bar{\tau}$, $|\xi|\leq\rho^{-\frac{1}{100}}(\tau)$ and
$a=10^{-1}\rho(\tau)^{\frac{-1}{2}}$, then

$G(\xi,\tau)\geq-1>(r_{*}a^{-1}-1)\sqrt{2}.$

###### Proof.

The second inequality follows from $\rho(\tau)\to 0$. As $\tau\to-\infty$, we
have $G(\xi,\tau)\to 0$ uniformly on compact sets in $\xi$. So we can assume
$G(0,\tau)>\frac{-1}{4}$ near the origin. We argue by contradiction. Assume
there is a point with $|\xi|\leq\rho^{-\frac{1}{100}}(\tau)$ such that
$G(\xi,\tau)<-1$. Then we can find a point $\bar{\xi}$ closest to the origin
satisfying $|\bar{\xi}|\leq\rho^{-\frac{1}{100}}(\tau)$ and
$G(\bar{\xi},\tau)=-1$. Without loss of generality we may assume
$\bar{\xi}>0$. For $\xi\in[0,\bar{\xi})$, we have
$G(\xi,\tau)>-1>(r_{*}a^{-1}-1)\sqrt{2}$, and by Proposition 3.26 we get

$G_{\xi}^{2}(\xi,\tau)\leq C\rho(\tau).$

Since $\delta(\tau)\to 0$ and $\rho(\tau)\to 0$ as $\tau\to-\infty$, it
follows that

$G(\bar{\xi},\tau)=G(0,\tau)+\int_{0}^{\bar{\xi}}G_{\xi}(x,\tau)\,\mathrm{d}x\geq-\delta(\tau)-C\rho^{\frac{1}{2}}(\tau)\rho^{-\frac{1}{100}}(\tau)\geq-1/2,$

which is a contradiction. ∎

######  Lemma 3.28.

For $-\bar{\tau}$ sufficiently large, for $\tau\leq\bar{\tau}$ and
$|\xi|\leq\delta^{-\frac{1}{100}}(\tau)$, we have

$\left|G(\xi,\tau)\right|+|G_{\xi}(\xi,\tau)|\leq
C\delta^{\frac{1}{4}}(\tau).$

###### Proof.

Note first that $|G_{\xi}|^{2}\leq C\delta(\tau)$ follows from Lemma 3.27 and
Proposition 3.26. By the definition of $\delta(\tau)$ we have
$|G(0,\tau)|\leq\delta(\tau)$. So $\left|G(\xi,\tau)\right|\leq
C\delta^{\frac{1}{4}}(\tau)$ follows by integrating the bound on $|G_{\xi}|$.
∎

For simplicity, in the following we write

$I_{\tau}:=\left\\{\xi:|\xi|\leq\delta^{-\frac{1}{100}}(\tau)\right\\},\qquad\mathrm{d}\nu(\xi)=(4\pi)^{-\frac{1}{2}}e^{-\frac{\xi^{2}}{4}}\,d\xi.$

and we assume $-\bar{\tau}$ is large enough so that the previous lemmas hold.
We are now ready to prove the higher order derivative estimates for
$G(\xi,\tau)$. Given Theorem 2.5, the proofs are similar to the three
dimensional case.

######  Lemma 3.29.

For $\tau\leq\bar{\tau}$ and $\xi\in I_{\tau}$, we have

$|G_{\xi\xi}(\xi,\tau)|\leq C\delta^{\frac{1}{8}}(\tau).$

###### Proof.

The proof is similar to [Bre20, Lemma 3.8], [ABDS22, Lemma 3.14]. ∎

######  Lemma 3.30.

For $\tau\leq\bar{\tau}$, and $\xi\in I_{\tau}$,

$|\partial_{\xi}^{m}G(\xi,\tau)|\leq C(m),$

for each $m\geq 0.$

###### Proof.

The proof is similar to [Bre20, Lemma 3.9], [ABDS22, Lemma 3.15]. ∎

######  Lemma 3.31.

For $\tau\leq\bar{\tau},$

$\left|G_{\xi}(0,\tau)\right|^{4}\leq
C\delta^{\frac{1}{100}}(\tau)\int_{I_{\tau}}\left|G(\xi,\tau)\right|^{2}\,\mathrm{d}\nu(\xi),$

and

$\int_{I_{\tau}}\left|G_{\xi}(\xi,\tau)\right|^{4}\,\mathrm{d}\nu(\xi)\leq
C\delta^{\frac{1}{100}}(\tau)\int_{I_{\tau}}\left|G(\xi,\tau)\right|^{2}\,\mathrm{d}\nu(\xi)+C\exp\left(-\tfrac{1}{8}\delta^{-\frac{1}{50}}(\tau)\right).$

###### Proof.

The proof is similar to c.f. [Bre20, Lemma 3.10], [ABDS22, Lemma 3.16]. ∎

######  Lemma 3.32.

For $\tau\leq\bar{\tau}$ and any $k\geq 0$,

$\left|\,\mathcal{E}^{\rm rad}\right|\leq
Ce^{\tau}(\sqrt{2}+G)^{-4},\qquad\left|\partial_{\xi}^{k}\,\mathcal{E}^{\rm
orb}\right|\leq C_{k}e^{\tau}(\sqrt{2}+G)^{-2}.$

###### Proof.

This follows directly by rescaling the error terms and Lemma 3.12. ∎

Putting the above estimates together, we get the following Lemma.

######  Lemma 3.33.

For $\tau\leq\bar{\tau},$

$\displaystyle\int_{I_{\tau}}\Big{|}G_{\tau}-G_{\xi\xi}+\frac{\xi}{2}G_{\xi}-G\Big{|}^{2}$
$\displaystyle(\xi,\tau)\,d\nu(\xi)\leq$ $\displaystyle
C\delta^{\frac{1}{100}}(\tau)\int_{I_{\tau}}\left|G(\xi,\tau)\right|^{2}\,d\nu(\xi)+C\exp\left(-\tfrac{1}{8}\delta^{-\frac{1}{50}}(\tau)\right)+Ce^{2\tau}.$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [Bre20, Lemma 3.11],
[ABDS22, Lemma 3.17]. We only need to estimate the extra error terms. For
$\xi\in I_{\tau}$ Lemma 3.32 implies

$\left|(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm orb}\right|\leq Ce^{\tau},$

and Lemma 3.28 and Lemma 3.32 imply

$\displaystyle\left|G_{\xi}\int_{0}^{\xi}\,\mathcal{E}^{\rm
rad}(x,\tau)\,\mathrm{d}x\right|$
$\displaystyle\leq|G_{\xi}|e^{\tau}\delta^{-\frac{1}{100}}(\tau)\leq
Ce^{\tau}.$

∎

###  3.4 Dichotomy in asymptotic behavior

As in [Bre20, ABDS22], we consider the operator

$\mathcal{L}G:=G_{\xi\xi}-\tfrac{1}{2}\xi G_{\xi}+G$

on the weighted $L^{2}$ Hilbert space $\mathcal{H}$ with the inner product

$\langle u,v\rangle_{\mathcal{H}}:=\int_{\mathbb{R}}uv\,d\nu.$

and the norm $\|G\|_{\mathcal{H}}=\langle
G,G\rangle_{\mathcal{H}}^{\frac{1}{2}}$.

Recall that $\mathcal{L}$ admits a spectral decomposition. The eigenvalues of
$\mathcal{L}$ are $1-\frac{n}{2}$ for $n\geq 1$, and the corresponding
eigenfunctions are $H_{n}(\xi/2)$, where $H_{n}$ is the $n^{\rm th}$ Hermite
polynomial. We let

$h_{0}(\xi):=1,\quad h_{1}(\xi):=\tfrac{1}{\sqrt{2}}\xi,\quad
h_{2}(\xi):=\tfrac{1}{2\sqrt{2}}(\xi^{2}-2),\quad
h_{3}(\xi):=\tfrac{1}{4\sqrt{3}}(\xi^{3}-6\xi),\quad\cdots$

be the unit eigenfunctions of $\mathcal{L}$ corresponding to eigenvalues
$1,\frac{1}{2},0,-\frac{1}{2},\cdots$. We also let
$\mathcal{H}=\mathcal{H}_{+}\oplus\mathcal{H}_{0}\oplus\mathcal{H}_{-}$ to be
the decomposition of $\mathcal{H}$ into subspaces generated by $h_{n}$ similar
to [Bre20], and $P_{+},P_{0},P_{-}$ to be the orthogonal projections unto
those subspaces.

Let $\eta$ be a smooth function on $\mathbb{R}$ such that $\eta=1$ over
$[-\frac{1}{2},\frac{1}{2}]$, $\eta(s)=0$ for $s\in\mathbb{R}\setminus[-1,1]$,
and $s\eta^{\prime}(s)\leq 0$ for all $s\in\mathbb{R}$. Now let

$\phi(\xi,\tau):=\eta\left(\delta^{\frac{1}{100}}(\tau)\xi\right),\quad\widehat{G}(\xi,\tau):=\phi(\xi,\tau)G(\xi,\tau)$

and define

$\displaystyle\gamma(\tau)$
$\displaystyle:=\int_{\mathbb{R}}|\widehat{G}|^{2}(\xi,\tau)\,d\nu(\xi),$
$\displaystyle\gamma^{+}(\tau)$
$\displaystyle:=\int_{\mathbb{R}}|P_{+}\widehat{G}|^{2}(\xi,\tau)\,d\nu(\xi),$
$\displaystyle\gamma^{0}(\tau)$
$\displaystyle:=\int_{\mathbb{R}}|P_{0}\widehat{G}|^{2}(\xi,\tau)\,d\nu(\xi),$
$\displaystyle\gamma^{-}(\tau)$
$\displaystyle:=\int_{\mathbb{R}}|P_{-}\widehat{G}|^{2}(\xi,\tau)\,d\nu(\xi).$

######  Lemma 3.34.

The quantities $\gamma$, $\gamma^{+}$ etc. defined above satisfy

$\displaystyle\gamma^{+}(\tau-1)$ $\displaystyle\leq
e^{-1}\gamma^{+}(\tau)+C\delta^{\frac{1}{200}}(\tau)\sup_{[\tau-1,\tau]}\gamma+C\exp\left(-\frac{1}{64}\delta^{-\frac{1}{50}}(\tau)\right)+Ce^{2\tau}$
$\displaystyle\left|\gamma^{0}(\tau-1)-\gamma^{0}(\tau)\right|$
$\displaystyle\leq\delta^{\frac{1}{200}}(\tau)\sup_{[\tau-1,\tau]}\gamma+C\exp\left(-\frac{1}{64}\delta^{-\frac{1}{50}}(\tau)\right)+Ce^{2\tau}$
$\displaystyle\gamma^{-}(\tau-1)$ $\displaystyle\geq
e\,\gamma^{-}(\tau)-C\delta^{\frac{1}{200}}(\tau)\sup_{[\tau-1,\tau]}\gamma-C\exp\left(-\frac{1}{64}\delta^{-\frac{1}{50}}(\tau)\right)-Ce^{2\tau}$

###### Proof.

the proof is similar to [ABDS22, Lemma 3.18] and uses Lemma3.33. ∎

We now analyze the evolution of the function $\rho_{\max}(\tau)$. Recall that
by Lemma 3.18, for $\tau\leq\bar{\tau}$ we have

$\rho_{\max}(\tau)=e^{\frac{1}{4}\tau}+\sup_{\xi}G(\xi,\tau)\geq
e^{\frac{1}{4}\tau}.$

Let $\xi^{*}_{\tau}$ denote the unique point in space where the function
$G(\cdot,\tau)$ attains its maximum.

######  Lemma 3.35.

For $\tau\leq\bar{\tau}$,

$0<-G_{\xi\xi}(\xi^{*}_{\tau},\tau)\leq C\gamma^{\frac{1}{4}}(\tau).$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Lemma 3.19] once we replace their Proposition 2.8 with our Proposition 2.18. ∎

######  Lemma 3.36.

For $\tau\leq\bar{\tau}$,

$\tfrac{{\rm d}}{{\rm
d}\tau}\rho_{\max}(\tau)\geq\tfrac{3}{4}\rho_{\max}-C\rho_{\max}^{2}-C\gamma^{\frac{1}{4}}(\tau).$

###### Proof.

First, note that at $\xi^{*}_{\tau}$ we have $\sqrt{2}+G\geq C>0$ since the
same is true near the origin. Hence
$\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}\geq G-CG^{2}$ at the maximum. By
the evolution equation of $G$, at $(\xi^{*}_{\tau},\tau)$ we have

$\displaystyle\tfrac{{\rm d}}{{\rm d}\tau}\rho_{\max}(\tau)$
$\displaystyle\geq\tfrac{1}{4}e^{\frac{1}{4}\tau}+G_{\xi\xi}+\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb}$
$\displaystyle\geq\tfrac{1}{4}e^{\frac{1}{4}\tau}+\rho_{\max}-e^{\frac{1}{4}\tau}-C(\rho_{\max}-e^{\frac{1}{4}\tau})^{2}-C\gamma^{\frac{1}{4}}-Ce^{\tau}$
$\displaystyle\geq\tfrac{7}{8}\rho_{\max}-C\rho_{\max}^{2}-C\gamma^{\frac{1}{4}}-\tfrac{1}{2}e^{\frac{9}{10}\tau}$
$\displaystyle\geq\tfrac{6}{8}\rho_{\max}-C\rho_{\max}^{2}-C\gamma^{\frac{1}{4}},$

where we have absorbed all the exponential terms into $e^{\frac{1}{4}\tau}$
and used the fact that $e^{\frac{1}{4}\tau}\leq\rho_{\max}(\tau)$ in the last
line. ∎

######  Lemma 3.37.

For $\tau\leq\bar{\tau}$,

$\rho_{\max}(\tau-1)\leq
e^{-\frac{1}{2}}\rho_{\max}(\tau)+C\sup_{[\tau-1,\tau]}\gamma^{\frac{1}{4}}.$

###### Proof.

By the previous Lemma, for $\tau\leq\bar{\tau}$ we have

$\tfrac{{\rm d}}{{\rm
d}\tau}\rho_{\max}(\tau)\geq\tfrac{1}{2}\rho_{\max}(\tau)-C\gamma^{\frac{1}{4}}(\tau).$

The conclusion follows by integration since $\rho_{\max}\to 0$. ∎

######  Lemma 3.38.

For $\tau\leq\bar{\tau},$

$\displaystyle\gamma^{+}(\tau-1)+\rho_{\max}^{8-\frac{1}{200}}(\tau-1)$
$\displaystyle\leq$ $\displaystyle\
e^{-1}\left(\gamma^{+}(\tau)+\rho_{\max}^{8-\frac{1}{200}}(\tau)\right)+C\delta^{\frac{1}{200}}(\tau)\sup_{[\tau-1,\tau]}\gamma+C\exp\left(-\frac{1}{64}\delta^{-\frac{1}{50}}(\tau)\right)+C_{1}e^{2\tau}.$

###### Proof.

The proof follows from Lemma 3.33 as in [ABDS22]. ∎

Define

$\displaystyle\Gamma(\bar{\tau})$
$\displaystyle:=\sup_{\tau\leq\bar{\tau}}\Big{(}\gamma(\tau)+\rho_{\max}^{8-\frac{1}{200}}(\tau)\Big{)},$
$\displaystyle\Gamma^{+}(\bar{\tau})$
$\displaystyle:=\sup_{\tau\leq\bar{\tau}}\Big{(}\gamma^{+}(\tau)+\rho_{\max}^{8-\frac{1}{200}}(\tau)\Big{)},$
$\displaystyle\Gamma^{0}(\bar{\tau})$
$\displaystyle:=\sup_{\tau\leq\bar{\tau}}\gamma^{0}(\tau),$
$\displaystyle\Gamma^{-}(\bar{\tau})$
$\displaystyle:=\sup_{\tau\leq\bar{\tau}}\gamma^{-}(\tau).$

######  Lemma 3.39.

For $\tau\leq\bar{\tau},$

$\displaystyle\Gamma^{+}(\tau-1)$ $\displaystyle\leq
e^{-1}\Gamma^{+}(\tau)+C\delta^{\frac{1}{200}}(\tau)\Gamma(\tau),$
$\displaystyle\left|\Gamma^{0}(\tau-1)-\Gamma^{0}(\tau)\right|$
$\displaystyle\leq C\delta^{\frac{1}{200}}(\tau)\Gamma(\tau),$
$\displaystyle\Gamma^{-}(\tau-1)$ $\displaystyle\geq
e\,\Gamma^{-}(\tau)-C\delta^{\frac{1}{200}}(\tau)\Gamma(\tau).$

###### Proof.

By the standard interpolation inequalities, $|G(0,\tau)|\leq
C\gamma^{\frac{1}{4}}(\tau)$, and thus

$\sup_{\sigma\leq\tau}|G(0,\sigma)|\leq C\Gamma^{\frac{1}{4}}(\tau).$

On one hand, $e^{\frac{1}{4}\tau}\leq\rho(\tau)\leq\delta(\tau)$, so
$e^{2\tau}\leq\delta^{8}(\tau)$. On the other hand,
$\rho_{\max}^{8-\frac{1}{200}}(\sigma)\leq\Gamma(\tau)$ for $\sigma\leq\tau$
and hence $\rho(\tau)^{8-\frac{1}{200}}\leq\Gamma(\tau)$. So we have

$\delta(\tau)^{8-\frac{1}{200}}\leq
C\rho(\tau)^{8-\frac{1}{200}}+C\sup_{\sigma\leq\tau}\left|G(0,\sigma)\right|^{8-\frac{1}{200}}\leq
C\Gamma(\tau).$

So, for $\tau\leq\bar{\tau},$

$C_{1}e^{2\tau}+\exp\left(-\frac{1}{64}\delta^{-\frac{1}{50}}(\tau)\right)\leq
C\delta^{8}(\tau)\leq C\delta(\tau)^{\frac{1}{200}}\Gamma(\tau).$

The rest of the proof follows verbatim as in the three-dimensional case, c.f.
[ABDS22, Page 16]. ∎

We record the following corollary of the proof of Lemma 3.39 for easier
reference.

######  Corollary 3.40.

In the equation $\partial_{\tau}\hat{G}=\mathcal{L}\hat{G}+E$, the source term
$E$ satisfies $\|E\|\leq C\delta^{\frac{1}{200}}\Gamma$.

###### Proof.

Consider the proof of the previous Lemma and the right hand side expression in
3.33. ∎

Now that we have Lemma 3.39, we get the following Lemma inspired by Merle and
Zaag, similar to the three-dimensional case.

######  Proposition 3.41 ([ABDS22, Proposition 3.23]).

We have either

$\Gamma^{+}(\tau)+\Gamma^{-}(\tau)\leq
o(1)\Gamma^{0}(\tau)\qquad\text{or}\qquad\Gamma^{0}(\tau)+\Gamma^{-}(\tau)\leq
o(1)\Gamma^{+}(\tau)$

as $\tau\to-\infty.$

##  4 Ruling out the case when the positive mode dominates

Throughout this section, in addition to Assumptions (A1)-(A3) we also assume

$\sup_{\partial B_{r}(o)}R=O(r^{-\eta}),$ (4.1)

for some $\eta\in(\frac{2}{3},1)$. We show that in this case, the neutral term
$\Gamma^{0}$ has to dominate as $\tau\to-\infty$. We argue by contradiction.
In view of the dichotomy in Proposition 3.41, we may assume that the positive
term dominates the other two, i.e.,

$\Gamma^{0}(\tau)+\Gamma^{-}(\tau)=o(1)\,\Gamma^{+}(\tau),$

as $\tau\to-\infty.$ By Lemma 3.39,

$\Gamma^{+}(\tau-1)\leq
e^{-1}\Gamma^{+}(\tau)+C\delta^{\frac{1}{200}}(\tau)\Gamma^{+}(\tau).$

Iterating this yields

$\Gamma^{+}(\tau)\leq O\left(e^{\frac{\tau}{2}}\right).$

and by definition of $\Gamma^{+}$, we get

$\sup_{\xi}G(\xi,\tau)\leq\rho_{\max}(\tau)\leq
O\left(e^{\frac{\tau}{16}}\right)$

We try to follow the arguments in [ABDS22, Section 4], but we need to modify
them due to extra error terms in our case.

######  Definition 4.2.

Given $0<\alpha<1$, we say that condition $(\star_{\alpha})$ holds if

$r_{\max}^{2}(s)\leq 2s\left(1+O(s^{-\alpha})\right).$ ($\star_{\alpha}$)

As mentioned above, the condition ($\star_{\alpha}$) holds for
$\alpha=\frac{1}{16}$. We use an iteration argument similar to [ABDS22] to
show that ($\star_{\alpha}$) holds for $\alpha=2/3$. this would imply a
diameter bound for the level set that contradicts (4.1).

######  Proposition 4.3.

Suppose ($\star_{\alpha}$) holds for some $\tfrac{1}{16}<\alpha<1$. Choose
$1\leq\gamma\leq 2$ such that

$1/\gamma\geq\max\left(1-\alpha,\tfrac{3}{2}\alpha\right).$

Then

$F_{z}^{2}(z,s)\leq Cs^{-\gamma\alpha},$

holds whenever $s\geq\underline{s}(\alpha)$ and and $F(z,s)\geq\sqrt{s}$.

###### Proof.

We first prove the following Claim.

Claim: There is a large number $K$ and a small number $\theta$ with the
following property: For any $a\geq K$ and any $1\leq\gamma\leq 2$ satisfying
$\frac{1}{\gamma}\geq 1-\alpha$, $s\geq K^{2}a^{\frac{2}{\gamma\alpha}}$
implies

$1-\theta\leq\frac{r_{\max}(s)}{\sqrt{2s+Ka^{\frac{2}{\gamma\alpha}}}}\leq
1+\tfrac{1}{100}a^{-2},$

###### Proof of the Claim..

Let $\underline{s}$ be large such that the results of previous sections and
$r^{2}_{\max}(s)\geq 2s$ (c.f. Lemma 3.18) hold for $s\geq\underline{s}$.
Choose $K$ large enough so that $a\geq K,\alpha<1,\gamma\leq 2$ and $s\geq
K^{2}a^{\frac{2}{\gamma\alpha}}$ imply $s\geq\underline{s}$.

For $s\geq K^{2}a^{\frac{2}{\gamma\alpha}},$ Lemma 3.18 implies

$\frac{r_{\max}(s)}{\sqrt{2s+Ka^{\frac{2}{\gamma\alpha}}}}\geq\frac{\sqrt{2s-C}}{\sqrt{2s+s/K}}\geq
1-\theta,$

if $K$ is sufficiently large (depending on $C$). Let us now prove the upper
bound.

Since $r^{2}_{\max}(s)-2s\leq O(s^{1-\alpha})$, and $r^{2}_{\max}(s)\geq 2s$,
we have

$r^{2}_{\max}(s)-2s\leq O(r^{2-2\alpha}_{\max}(s)).$

Because $\alpha>c>0$, by choosing $K$ large enough we have

$\frac{r_{\max}^{2}(s)-2s}{r_{\max}^{2}(s)}\leq\frac{K^{\alpha}}{100^{1-\alpha}r_{\max}^{2\alpha}(s)}$

Now we can apply Young’s inequality $x+y\geq x^{\alpha}y^{1-\alpha}$ to obtain

$\frac{Ka^{\frac{2}{\gamma\alpha}}}{r^{2}_{\max}(s)}+\frac{1}{100}a^{-2}\geq\frac{K^{\alpha}}{100^{1-\alpha}r^{2\alpha}_{\max}(s)}a^{\frac{2}{\gamma}-2(1-\alpha)}\geq\frac{r_{\max}^{2}(s)-2s}{r_{\max}^{2}(s)}.$

It follows that if $s\geq K^{2}a^{\frac{2}{\gamma\alpha}}$,

$\displaystyle\frac{2s+Ka^{\frac{2}{\gamma\alpha}}}{r_{\max}^{2}(s)}-1+\frac{1}{100}a^{-2}=\frac{Ka^{\frac{2}{\gamma\alpha}}}{r_{\max}^{2}(s)}+\frac{1}{100}a^{-2}-\frac{r_{\max}^{2}(s)-2s}{r_{\max}^{2}(s)}\geq
0$

Thus,

$\frac{r_{\max}(s)}{\sqrt{2s+Ka^{\frac{2}{\gamma\alpha}}}}\leq\left(1-\frac{1}{100}a^{-2}\right)^{-\frac{1}{2}}\leq
1+\frac{1}{100}a^{-2}.$

∎

The rest of the proof follows similar to [ABDS22, Proposition 4.2]. Since we
are assuming $1/\gamma\geq\frac{3}{2}\alpha$, we can guarantee

$K^{2}a^{\frac{2}{\gamma\alpha}}\geq K^{2}a^{3}\geq Da^{3},$

by taking $K^{2}\geq D,$ where $D$ is given by Proposition 3.19. Hence, we can
apply the barrier argument to conclude

$F_{z}^{2}(z,s)\leq\psi_{a}\left(\frac{F(z,s)}{\sqrt{2s+Ka^{\frac{2}{\gamma\alpha}}}}\right)\leq
Ca^{-2},$

whenever $s\geq K^{2}a^{\frac{2}{\gamma\alpha}}$ and $F(z,s)\geq\sqrt{s}.$ The
conclusion then follows by putting $s=K^{2}a^{\frac{2}{\gamma\alpha}}.$ ∎

Here, a major difference from [ABDS22] is that $\alpha$ cannot be any number
close to $1$. In order for Proposition 4.3 to imply an improvement in
estimates, we need to have $\gamma\geq 1$. The issue is the restriction $s\geq
Da^{3}$ in our barrier, which forces $1\geq 1/\gamma\geq\frac{3}{2}\alpha$.
Thus, we can only consider $\alpha\in(0,2/3]$.

######  Proposition 4.4.

Suppose ($\star_{\alpha}$) holds for some $\alpha\in(0,2/3].$ If
$(q_{0},s_{0})$ satisfies $F\geq\sqrt{2s_{0}}$ and $s_{0}$ is sufficiently
big, we have

$-(F^{2})_{zz}\leq Cs_{0}^{-(1+\frac{\alpha^{2}}{200})\alpha},$

at $(q_{0},s_{0})$.

###### Proof.

Fix $1\leq\gamma\leq\frac{1}{\max(\frac{3}{2}\alpha,1-\alpha)}.$ We denote by
$\widetilde{F}(z,t)$ the radius of the sphere of symmetry that has signed
distance $z$ from the point $q_{0}$. The function $\widetilde{F}(z,t)$
satisfies the evolution equation

$\displaystyle-\widetilde{F}_{s}$
$\displaystyle=\widetilde{F}_{zz}-\widetilde{F}^{-1}(1+\widetilde{F}_{z}^{2})+2\widetilde{F}_{z}\left\\{(\ln\widetilde{F})_{z}(0,s)-\int_{0}^{z}(\ln\widetilde{F})^{2}_{z}(\zeta,s)\mathrm{d}\zeta\right\\}$
$\displaystyle\qquad-\widetilde{F}^{-1}\widetilde{\mathcal{E}}^{\rm
orb}+\widetilde{F}_{z}\int_{0}^{z}\widetilde{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta.$

Similar to Lemma 3.15, $\widetilde{\mathcal{E}}^{\rm orb}>0$, whenever
$\widetilde{F}(z,s)\geq\sqrt{s}.$ In particular, since $\widetilde{F}_{zz}<0,$

$-\widetilde{F}_{s}(0,s)\leq-\widetilde{F}^{-1}(0,s)(1-\widetilde{F}_{z}^{2}(0,s)).$

By Proposition 4.3,

$\tfrac{1}{2}\partial_{s}\widetilde{F}^{2}(0,s)\geq
1-\widetilde{F}_{z}^{2}(0,s)\geq 1-C\,s^{-\gamma\alpha}\geq
1-C\,s^{-\alpha}>\tfrac{1}{2},$

whenever $\widetilde{F}(0,s)\geq\sqrt{s},s\geq\underline{s}.$ Since
$\widetilde{F}(0,s_{0})\geq\sqrt{2s_{0}},$ a standard continuity argument
gives

$\widetilde{F}^{2}(0,s)\geq 2s(1-Cs^{-\alpha}),$

if $s\geq s_{0}\geq\underline{s}.$ In the following, we put
$\epsilon:=\frac{\alpha^{2}}{100}.$ By Proposition 4.3, for $s\geq
s_{0},|z|\leq s^{\frac{1+\epsilon}{2}}$,

$\widetilde{F}^{2}(z,s)\geq\widetilde{F}^{2}(0,s)-Cs^{\frac{1}{2}+\frac{1+\epsilon}{2}-\gamma\alpha}\geq
2s(1-Cs^{-\frac{\alpha}{8}}).$

On the other hand, the condition ($\star_{\alpha}$) gives

$\widetilde{F}^{2}(z,s)\leq 2s(1+Cs^{-\alpha}).$

We next consider the parabolic cylinder

$Q:=\left[-s_{0}^{\frac{1+\epsilon}{2}},s_{0}^{\frac{1+\epsilon}{2}}\right]\times[s_{0},s_{0}+s_{0}^{1+\epsilon}].$

Define

$\widetilde{H}(z,s):=\tfrac{1}{2}\widetilde{F}^{2}(z,s)-s.$

By the assumption, $\widetilde{H}(0,s_{0})\geq 0.$ Moreover, the preceding
arguments imply that we can find a positive constant $L$ such that

$-Ls^{1-\frac{\alpha}{8}}\leq\widetilde{H}(z,s)\leq Ls^{1-\alpha},$

in $Q.$ In particular,

$-L(2s_{0})^{(1+\epsilon)(1-\frac{\alpha}{8})}\leq\widetilde{H}(z,s)\leq
Ls^{(1+\epsilon)(1-\alpha)},$

in $Q.$ $\widetilde{H}$ satisfies

$-\widetilde{H}_{s}-\widetilde{H}_{zz}=-S,$

where

$\displaystyle S(z,s)$
$\displaystyle=2\widetilde{F}^{2}_{z}-2\widetilde{F}\widetilde{F}_{z}\left\\{(\ln\widetilde{F})_{z}(0,s)-\int_{0}^{z}(\ln\widetilde{F})^{2}_{z}(\zeta,s)\mathrm{d}\zeta\right\\}$
$\displaystyle\qquad-\widetilde{\mathcal{E}}^{\rm
orb}+\widetilde{F}\widetilde{F}_{z}\int_{0}^{z}\widetilde{\mathcal{E}}^{\rm
rad}(\zeta,s)\,\mathrm{d}\zeta.$

By Proposition 4.3 and Lemma 3.12 we have

$|S|\leq Cs^{-\gamma\alpha}\leq Cs_{0}^{-\gamma\alpha},$

in $Q.$ The rest of the proof follows similar to the three-dimensional case.
See [ABDS22, Proposition 4.4]. ∎

######  Corollary 4.5.

Suppose that ($\star_{\alpha}$) holds for some $\alpha\in(0,2/3].$ If
$0<\tilde{\alpha}<(1+\frac{\alpha^{2}}{200})\alpha,$ then
${\rm(\star_{\tilde{\alpha}})}$ holds.

###### Proof.

For any $s\geq\underline{s}$, consider the point $z^{*}$ where $F(\cdot,s)$
attains its maximum. At $(z^{*},s),F=r_{\max}(s)\geq\sqrt{2s},F_{z}=0$, and

$-(F^{2})_{zz}\leq Cs^{-(1+\frac{\alpha^{2}}{200})\alpha},$

by the previous Proposition. Using the evolution equation of $F$, at
$(z^{*},s)$ we have

$\tfrac{1}{2}\tfrac{\partial}{\partial
s}F^{2}=-(F^{2})_{zz}+1+\,\overline{\mathcal{E}}^{\rm orb}\leq
1+Cs^{-(1+\frac{\alpha^{2}}{200})\alpha}.$

So ${\rm(\star_{\tilde{\alpha}})}$ holds by integration. ∎

######  Lemma 4.6.

($\star_{\alpha}$) holds for $\alpha=2/3.$

###### Proof.

As we have discussed at the beginning of this section, $\rho_{\max}(\tau)\leq
O(e^{\frac{\tau}{16}}).$ So ($\star_{\alpha}$) holds for
$\alpha=\frac{1}{16}.$ Iterating Corollary 4.5 finitely many times, we get
that ($\star_{\alpha}$) holds for $\alpha=2/3.$ ∎

######  Corollary 4.7.

Suppose $\Gamma^{+}$ dominates. Then

${\rm diam}(\bar{g}_{s})\geq c\,s^{\frac{5}{6}},$

for some constant $c>0$ and large $s$.

###### Proof.

By Lemma 4.6, ($\star_{\alpha}$) holds for $\alpha=2/3,$ and we take
$\gamma=\frac{2}{3\alpha}=1.$ Let $z^{*}$ be the maximum point of $F(\cdot,s)$
and let $z<z^{*}$ be the point such that $F(z,s)=\sqrt{s}$. By Proposition
4.3,

$\displaystyle\sqrt{2s}-\sqrt{s}$ $\displaystyle\leq
F(z^{*},s)-F(z,s)=\int_{z}^{z^{*}}F_{z}(\zeta,s)\,\mathrm{d}\zeta$
$\displaystyle\leq C\,(z^{*}-z)s^{-\frac{\gamma\alpha}{2}}\leq C\,{\rm
diam}(\bar{g}_{s})s^{-\frac{1}{3}}.$

The conclusion follows. ∎

Finally, we show that the positive modes cannot dominate under our curvature
decay assumption (4.1).

######  Proposition 4.8.

Let $R$ denote the scalar curvature of $M^{4}$, and let $o$ be the (unique)
critical point of the potential $f$. Suppose

$\sup_{\partial B_{r}(o)}R=O(r^{-\eta}),$

as $r\to\infty$, for some $\eta\in(\frac{1}{3},1).$ Then $\Gamma^{+}$ cannot
dominate.

###### Proof.

Similar to Perelman’s estimates in [Per02, Section 8], we have

$\partial_{s}d_{\bar{g}_{s}}(x,y)\leq C\sqrt{R_{\max}(s)},$

where $R_{\max}(s):=\max_{\Sigma_{s}}R$. Since $\operatorname{Ric}\geq 0$, for
any $x\in\Sigma$

$\partial_{s}R(\chi_{s}(x))=\langle\nabla R,\tfrac{\nabla f}{|\nabla
f|^{2}}\rangle=-\tfrac{2}{|\nabla f|^{2}}\operatorname{Ric}(\nabla f,\nabla
f)\leq 0,$

and thus $R_{\max}(s)$ is decreasing in $s.$ By [CDM22, Theorem 2.1],
$f(x)\geq c\,d_{g}(o,x)$ for some $c>0$. So for any
$x\in\Sigma_{s},d_{g}(o,x)\leq Cs,$ and thus

$R_{\max}(s)=\max_{\Sigma_{s}}R\leq O(s^{-\eta}),$

for large $s.$ So, by integration, for $s\geq\underline{s}$,

${\rm diam}(\bar{g}_{s})\leq{\rm
diam}(\bar{g}_{\underline{s}})+C\int_{\underline{s}}^{s}\sqrt{R_{\max}(\sigma)}\,\mathrm{d}\sigma\leq
Cs^{1-\frac{\eta}{2}},$

which contradicts Corollary 4.7. Hence, $\Gamma^{+}$ cannot dominate. ∎

##  5 Asymptotics In The Case When Neutral Mode Dominates

In this section, we consider the first possibility in Proposition 3.41. Thus,
throughout this section we assume

$\Gamma^{+}(\tau)+\Gamma^{-}(\tau)\leq o(1)\Gamma^{0}(\tau).$ (5.1)

###  5.1 Asymptotics In The Cylindrical Region

Under assumption (5.1), we obtain precise asymptotics of $G$ in the
cylindrical region following arguments similar to [ABDS22, Section 5].
Following [ABDS22], we define

$\|u\|_{\mathcal{D}}^{2}:=\int(u^{2}+u_{\xi}^{2})\,\mathrm{d}\nu=\frac{1}{\sqrt{4\pi}}\int\left(u^{2}+u_{\xi}^{2}\right)e^{-\frac{|\xi|^{2}}{4}}\,d\xi.$

Note that if $u$ has compact support, then

$\|u\|_{\mathcal{D}}^{2}=\langle u,(2-\mathcal{L})u\rangle_{\mathcal{H}}.$

As before, we consider the spectral decomposition for the operator
$\mathcal{L}$, and we let $\mathcal{H}_{0}$ be the subspace spanned by
$h_{2}(\xi)=\frac{1}{2\sqrt{2}}(\xi^{2}-2)$, the unit eigenfunction
corresponding to eigenvalue $0$. We consider the projection of $\widehat{G}$
to $\mathcal{H}_{0}$, and write

$P_{0}\widehat{G}(\xi,\tau)=\alpha(\tau)h_{2}(\xi),$

where

$\alpha(\tau):=\langle\widehat{G}(\cdot,\tau),h_{2}\rangle_{\mathcal{H}}.$

Let

$A(\tau):=\sup_{\sigma\leq\tau}|\alpha(\sigma)|.$

Clearly, $\frac{1}{C}A^{2}\leq\Gamma^{0}\leq CA^{2}$, for some constant $C$.
Hence, assumption 5.1 implies $\frac{1}{C}A^{2}\leq\Gamma\leq CA^{2}$. By the
proof of Lemma 3.39, for $\tau\leq\bar{\tau}$ we have

$\delta(\tau)\leq\Gamma^{\frac{1}{80}}(\tau)\leq CA^{\frac{1}{40}}(\tau).$

Similar to [ABDS22] we can show

$\displaystyle\|P_{+}\widehat{G}(\cdot,\tau)\|_{\mathcal{H}}\leq o(1)A(\tau)$
$\displaystyle\|P_{-}\widehat{G}(\cdot,\tau)\|_{\mathcal{H}}+\|P_{-}\widehat{G}(\cdot,\tau)\|_{\mathcal{D}}\leq
C\delta^{\frac{1}{400}}(\tau)A(\tau)$

Note that in the proof of the last inequality we have used the estimate

$\int(\partial_{\tau}-\mathcal{L})\widehat{G}\,\mathrm{d}\nu\leq
C\delta^{\frac{1}{100}}(\tau)\Gamma(\tau)\leq
C\delta^{\frac{1}{100}}(\tau)A^{2}(\tau),$

which holds for $\tau\leq\bar{\tau}$ according to 3.40 and the above
estimates.

######  Lemma 5.2.

As $\tau\to-\infty$, we have

$\int_{\mathbb{R}}(1+|\xi|)^{4}\left|\widehat{G}(\xi,\tau)-\alpha(\tau)h_{2}(\xi)\right|^{2}\,\mathrm{d}\nu(\xi)\leq
o(1)A^{2}(\tau),$

and

$\int_{\mathbb{R}}(1+|\xi|)^{4}\left|\partial_{\xi}\left(\widehat{G}(\xi,\tau)-\alpha(\tau)h_{2}(\xi)\right)\right|^{2}\,\mathrm{d}\nu(\xi)\leq
o(1)A^{2}(\tau).$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Lemma 5.4].
∎

######  Lemma 5.3.

As $\tau\to-\infty$, we have $|\widehat{G}_{\xi}(0,\tau)|\leq o(1)A(\tau).$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Lemma 5.5].
∎

######  Proposition 5.4.

Let $E=(\partial_{\tau}-\mathcal{L})G$. Then as $\tau\to-\infty$, we have

$\int_{I_{\tau}}E(\cdot,\tau)h_{2}\,\mathrm{d}\nu=-2\alpha^{2}(\tau)+o(A^{2}(\tau))+O(e^{\tau}),$

where
$I_{\tau}=[-\delta^{-\frac{1}{100}}(\tau),\delta^{-\frac{1}{100}}(\tau)].$

###### Proof.

The proof is similar to [ABDS22, Proposition 5.6], and we only need to control
the extra error terms. By Proposition 3.24,

$\displaystyle E$
$\displaystyle=-G+\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}(1+G_{\xi}^{2})+2G_{\xi}\left\\{\frac{G_{\xi}}{\sqrt{2}+G}(0,\tau)-\int_{0}^{\xi}\frac{G_{\xi}^{2}}{(\sqrt{2}+G)^{2}}(x,\tau)\,\mathrm{d}x\right\\}$
$\displaystyle\qquad-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb}+G_{\xi}\int_{0}^{\xi}\,\mathcal{E}^{\rm rad}(x,\tau)\,\mathrm{d}x$
$\displaystyle=-\tfrac{1}{2\sqrt{2}}G^{2}-\tfrac{1}{\sqrt{2}}G_{\xi}^{2}+E_{1}+E_{2}+E_{3}+E_{4},$

where $E_{1},E_{2},E_{3}$ are defined similar to [ABDS22], and $E_{4}$ is
defined as

$\displaystyle E_{4}=-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb}+G_{\xi}\int_{0}^{\xi}\,\mathcal{E}^{\rm rad}(x,\tau)\,\mathrm{d}x.$

The terms $E_{1},E_{2}$ and $E_{3}$ can be treated the same way as in
[ABDS22]. Regarding our extra error term $E_{4}$, for $\tau\leq\bar{\tau}$ and
$\xi\in I_{\tau}$ we have

$|E_{4}|(\xi,\tau)\leq Ce^{\tau}+|G_{\xi}|e^{\tau}|\xi|.$

Hence, by Lemma 5.2,

$\left|\int_{I_{\tau}}E_{4}h_{2}\,\mathrm{d}\nu\right|\leq
Ce^{\tau}+Ce^{\tau}\int_{I_{\tau}}|G_{\xi}||\xi|(\xi^{2}+2)\,\mathrm{d}\nu\leq
Ce^{\tau}+Ce^{\tau}o(1)A(\tau)\leq Ce^{\tau}.$

The conclusion follows by combining the estimates similar to [ABDS22]. ∎

######  Corollary 5.5.

For $\tau\leq\bar{\tau},$

$\alpha^{\prime}(\tau)=-2\alpha^{2}(\tau)+o(A^{2}(\tau)).$

###### Proof.

Directly by the previous Proposition, we have

$\alpha^{\prime}(\tau)=-2\alpha^{2}(\tau)+o(A^{2}(\tau))+O(e^{\tau}).$

It suffices to show that $e^{\tau}\leq o(1)A^{2}(\tau).$

Since we are assuming the neutral mode dominates, for $\tau\leq\bar{\tau}$ we
have

$\Gamma^{+}(\tau)+\Gamma^{-}(\tau)\leq o(1)\Gamma^{0}(\tau)\leq
o(1)A^{2}(\tau)\Rightarrow\Gamma(\tau)\leq(1+o(1))\Gamma^{0}(\tau).$

For any $\beta\in(0,1),$ pick $\epsilon>0$ satisfying $1-\epsilon=e^{-\beta}$.
By Lemma 3.39, for $\tau\leq\bar{\tau},$

$\Gamma^{0}(\tau)\geq(1-\epsilon)\Gamma^{0}(\tau+1)\geq(1-\epsilon)^{2}\Gamma^{0}(\tau+2)\geq\cdots\geq(1-\epsilon)^{k}\Gamma^{0}(\tau_{*}),$

where $\tau+k=\tau_{*}\in[\bar{\tau}-1,\bar{\tau}].$ Thus, for $\tau\leq
2\bar{\tau},$

$\Gamma^{0}(\tau)\geq
e^{\beta(\tau-\bar{\tau})}\min_{[\bar{\tau}-1,\bar{\tau}]}\Gamma^{0}\geq
c\,e^{\beta\tau},$

for some constant $c>0.$ By the discussions at the beginning of this section,
$\Gamma^{0}(\tau)\leq CA^{2}(\tau),$ and thus,

$A^{2}(\tau)\geq c\,\Gamma^{0}(\tau)\geq c\,e^{\beta\tau}.$

It follows that $e^{\tau}\leq o(1)A^{2}(\tau),$ and

$\alpha^{\prime}(\tau)=-2\alpha^{2}(\tau)+o(A^{2}(\tau)).$

∎

######  Corollary 5.6.

As $\tau\to-\infty,$

$\alpha(\tau)=\frac{1}{(2+o(1))\tau}.$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Corollary
5.8, Corollary 5.9]. ∎

######  Proposition 5.7.

As $\tau\to-\infty,$

$-\tau\,G(\xi,\tau)\to-\frac{1}{4\sqrt{2}}(\xi^{2}-2),$

in $C_{\rm loc}^{\infty}.$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Proposition
5.10]. ∎

######  Corollary 5.8.

As $\tau\to-\infty,$ the domain of the function $G(\cdot,\tau)$ is an interval
of length at most $o(-\tau).$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Corollary
5.11], but we shall include some more details.

Let $(-d_{\tau}^{-},d_{\tau}^{+})$ be the domain of $G(\cdot,\tau)$. Recall
that $G(\cdot,\tau)$ is concave. For any large constant $L\gg 1$, as
$\tau\to-\infty$ we have

$\displaystyle\frac{2-L^{2}+o(1)}{4\sqrt{2}}$
$\displaystyle\geq|\tau|\,G(L,\tau)\geq\left(1-\frac{2L}{d_{\tau}^{+}}\right)|\tau|\,G(0,\tau)+\frac{2L}{d_{\tau}^{+}}|\tau|\,G(d_{\tau}^{+}/2,\tau)$
$\displaystyle\geq\frac{2-o(1)}{4\sqrt{2}}\left(1-\frac{2L}{d_{\tau}^{+}}\right)-\sqrt{2}|\tau|\frac{2L}{d_{\tau}^{+}}.$

Hence

$-L^{2}+2+o(1)\geq 2-o(1)-\frac{2L}{d_{\tau}^{+}}\left(8|\tau|+2+o(1)\right),$

which implies

$d_{\tau}^{+}\leq\frac{2L}{L^{2}-o(1)}\left(8|\tau|+2+o(1)\right).$

Therefore, since $L$ can be arbitrarily large, $d_{\tau}^{+}\leq o(-\tau)$.
Inequality $d^{-}_{\tau}\leq o(-\tau)$ can be similarly shown. ∎

######  Corollary 5.9.

Let $\xi_{\tau}^{*}$ be the unique point in space where $G(\cdot,\tau)$
attains its maximum. $\xi_{\tau}^{*}\to 0$ as $\tau\to-\infty$.

###  5.2 Asymptotics in the intermediate region

We now describe the asymptotics in the intermediate region, defined by
$|z|\geq M\sqrt{s}$ and $F(z,s)\geq\theta\sqrt{2s}$, for some large constant
$M$ and some small constant $\theta$. We closely follow [ABDS22, Section 6].

######  Proposition 5.10.

Fix a small number $\theta\in(0,1/2)$ and a large number $M\geq 10.$ For
$s\geq\underline{s}(\theta,M),$

$F_{z}^{2}(z,s)\leq\frac{M^{2}+C(\theta)}{M^{2}-2}\frac{1}{2\log
s}\left(\frac{2s}{F^{2}(z,s)}-1\right),$

whenever $|z|\geq M\sqrt{s},$ and $F(z,s)\geq\theta\sqrt{2s}.$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Proposition 6.1]. The only place that requires justification is where we apply
the barrier argument in Proposition 3.19, when $\log
s\geq\frac{M^{2}+2}{M^{2}-2}\frac{a^{2}}{2},|z|\geq M\sqrt{s},$ and
$F(z,s)\geq\theta\sqrt{2s}.$ Since $s=e^{\log s}\geq e^{a^{2}/4}\gg a^{3},$ we
can apply Proposition 3.19 with $\mu=0$. ∎

######  Corollary 5.11.

Fix a small constant $\theta\in(0,1/2).$ If $s\geq\underline{s}(\theta),$

$\partial_{s}F^{2}(z,s)\geq 2-\frac{C(\theta)}{\log s},$

whenever $|z|\geq 10\sqrt{s}$ and $F(z,s)\geq\theta\sqrt{2s}.$

###### Proof.

The proof is similar to the three-dimensional case, c.f. [ABDS22, Corollary
6.2]. By Corollary 5.9,

$F_{z}(z,s)>0,\quad\text{for }z\leq-\sqrt{s};\qquad
F_{z}(z,s)<0,\quad\text{for }z\geq\sqrt{s}.$

Recall that by Lemma 3.15, $F_{zz}\leq 0,\,\overline{\mathcal{E}}^{\rm
orb}\geq 0,\,\overline{\mathcal{E}}^{\rm rad}\geq 0,$ for large $s.$ By the
evolution equation for $F$, we have

$\displaystyle-F_{s}$
$\displaystyle=F_{zz}-F^{-1}(1-F_{z}^{2})-2F_{z}\int_{0}^{z}\frac{F_{zz}}{F}(\zeta,s)\,{\rm
d}\zeta$ $\displaystyle\qquad-F^{-1}\,\overline{\mathcal{E}}^{\rm
orb}+F_{z}\int_{0}^{z}\,\overline{\mathcal{E}}^{\rm rad}(\zeta,s)\,{\rm
d}\zeta$ $\displaystyle\leq-F^{-1}(1-F_{z}^{2}),$

for $|z|\geq\sqrt{s}.$ Applying Proposition 5.10 with $M=10,$

$F_{z}^{2}(z,s)\leq C(\theta)/\log s,$

whenever $|z|\geq 10\sqrt{s}$ and $F(z,s)\geq\theta\sqrt{2s}.$ The assertion
follows. ∎

######  Proposition 5.12.

Fix a small constant $\theta\in(0,\frac{1}{2})$ and a large number $M\geq 20.$
If $s\geq\underline{s}(\theta,M),$ then

$F^{2}(z,s)\geq 2s-\frac{M^{2}+C(\theta)}{M^{2}-2}\frac{z^{2}}{2\log s},$

whenever $|z|\geq M\sqrt{s}$ and $F(z,s)\geq\theta\sqrt{2s}.$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Proposition 6.3]. ∎

######  Proposition 5.13.

Fix a small constant $\theta\in(0,\frac{1}{2})$ and a large number $M\geq 20.$
If $s\geq\underline{s}(\theta,M),$ then

$F^{2}(z,s)\leq 2s-\frac{M^{2}-C(\theta)}{M^{2}}\frac{z^{2}}{2\log s},$

whenever $|z|\geq M\sqrt{s}$ and $F(z,s)\geq\theta\sqrt{2s}.$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Proposition 6.4]. ∎

As in [ABDS22], combining Proposition 5.12 and Proposition 5.13 and sending
$M\to\infty$, we have the following.

######  Corollary 5.14.

Fix a small constant $\theta\in(0,\frac{1}{2})$. If
$s\geq\underline{s}(\theta),$ we have

$\left\\{z:F(z,s)\geq\theta\sqrt{2s}\right\\}=[-\bar{z}_{1}(\theta,s),\bar{z}_{2}(\theta,s)],$

and

$\displaystyle\bar{z}_{1}(\theta,s)$
$\displaystyle=(2+o(1))\sqrt{1-\theta^{2}}\sqrt{s\log s},$
$\displaystyle\bar{z}_{2}(\theta,s)$
$\displaystyle=(2+o(1))\sqrt{1-\theta^{2}}\sqrt{s\log s}.$

######  Corollary 5.15.

Fix a small constant $\theta\in(0,\frac{1}{2})$. If
$s\geq\underline{s}(\theta),$ we have

$F^{2}(z,s)=2s-\frac{z^{2}}{2\log s}+o(s),$

for $|z|\leq 2\sqrt{1-\theta^{2}}\sqrt{s\log s}.$

###  5.3 Asymptotics in the Tip Region

Following [ABDS22, §7], we analyze the asymptotics near each tip. Throughout
this section, we assume that the neutral mode dominates. For large $s,$ the
function $z\mapsto F(z,s)$ is defined on the interval $[-d_{{\rm
tip},1}(s),d_{{\rm tip},2}(s)]$, where $d_{{\rm tip},1}(s)$ and $d_{{\rm
tip},2}(s)$ denote the distance of the reference point $q$ from each tip
measured with $\bar{g}_{s}.$ We first derive the asymptotics of $d_{{\rm
tip},i}(s)$.

######  Proposition 5.16.

For $i=1,2,$

$\lim_{s\to\infty}\frac{d_{{\rm tip},i}(s)}{\sqrt{s\log s}}=2.$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Proposition 7.1], because it only relies on Corollary 5.14 and the concavity
of $F.$ ∎

We then analyze the asymptotic behaviors of the scalar curvature at each tip.
For simplicity, we assume that $R$ decays uniformly and we believe that the
conclusions should still hold without assuming the decay. We write

$R_{{\rm tip},i}(s)=R(p_{s}^{i}),$

where recall that $p_{s}^{1},p_{s}^{2}$ denote the two tips of $\Sigma_{s}$
given by Corollary 2.16.

######  Lemma 5.17.

Suppose that $R$ decays uniformly. Then for $i=1,2,$

$\tfrac{\rm d}{{\rm d}s}d_{{\rm tip},i}(s)=(1+o(1))R_{{\rm
tip},i}^{\frac{1}{2}}(s).$

###### Proof.

If $R\to 0$ at infinity, $(\Sigma,R(p^{i}_{s})\bar{g}_{s},p^{i}_{s})$
converges to the three-dimensional Bryant soliton as $s\to\infty.$ The proof
follows verbatim as in the three-dimensional case, c.f. [ABDS22, Lemma 7.3]. ∎

######  Proposition 5.18.

Suppose that $R$ decays uniformly. Then for $i=1,2,$ as $s\to\infty,$

$R_{{\rm tip},i}(s)=(1+o(1))\frac{\log s}{s}.$

###### Proof.

The proof follows verbatim as in the three-dimensional case, c.f. [ABDS22,
Proposition 7.4]. ∎

##  6 Asymptotics When the Positive Modes Dominate

We now consider the second case in Proposition 3.41. More specifically, we
assume

$\Gamma^{0}(\tau)+\Gamma^{-}(\tau)=o(1)\,\Gamma^{+}(\tau),$

as $\tau\to-\infty$.

In addition to $(M^{4},g,f)\in\mathcal{M}$, we assume that $(M,g)$ has
reflexive symmetry, i.e., $F(\cdot,s)$ is an even function.

Under the assumptions above we give precise asymptotics of our solution in the
cylindrical region. The hope is that we would be able to use this asymptotics
to eventually to exclude the case of positive mode dominating at all.

######  Theorem 6.1.

When the positive modes dominate and $(M,g)$ is reflexive symmetric,

$F(z,s)=\frac{\log s}{\sqrt{2}s}+o\left(\frac{\log s}{s}\right),$

uniformly on compact sets of the form $|z|\leq L\sqrt{s}.$

By Lemma 3.39,

$\Gamma^{+}(\tau-1)\leq
e^{-1}\Gamma^{+}(\tau)+C\delta^{\frac{1}{200}}(\tau)\Gamma^{+}(\tau).$

Iterating this yields

$\Gamma^{+}(\tau)\leq O\left(e^{\frac{\tau}{2}}\right).$

Thus, as a rough estimate we get $\Gamma^{+}(\tau)\leq
O\left(e^{\frac{\tau}{2}}\right)$.

We need the following coarse estimate.

######  Lemma 6.2.

For $\tau\leq\bar{\tau},$

$\rho_{\max}(\tau)\geq G(0,\tau)\geq-
Ce^{\frac{\tau}{2}}-C\rho\left(\tfrac{\tau}{2}\right).$

As a consequence,

$\delta(\tau)\leq Ce^{\frac{\tau}{2}}+C\rho\left(\tfrac{\tau}{2}\right).$

###### Proof.

Since $0$ is the reference point, we have that for any $\epsilon>0,$ if
$\tau\leq\bar{\tau}(\epsilon),$ then $|G(0,\tau)|\leq\epsilon.$ By the
evolution equation of $G$ given in Proposition 3.24, at $(0,\tau),$

$\displaystyle G_{\tau}$
$\displaystyle\leq\tfrac{1}{2}(\sqrt{2}+G)-(\sqrt{2}+G)^{-1}+(\sqrt{2}+G)^{-1}G_{\xi}^{2}-(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb},$ $\displaystyle\partial_{\tau}(\sqrt{2}+G)^{2}$
$\displaystyle\leq(\sqrt{2}+G)^{2}-2+C\rho(\tau),$

where we applied Lemma 3.9 and Proposition 3.26. By integration, if $\tau\leq
2\bar{\tau}(\epsilon),$

$\displaystyle e^{-\tau}\left((\sqrt{2}+G)^{2}-2\right)(0,\tau)$
$\displaystyle\geq-10\epsilon
e^{-\frac{\tau}{2}}-C\int_{\tau}^{\frac{\tau}{2}}e^{-\sigma}\rho(\sigma)\,d\sigma$
$\displaystyle\geq-e^{-\frac{\tau}{2}}-C\rho\left(\tfrac{\tau}{2}\right)(e^{-\tau}-e^{-\frac{\tau}{2}}).$

Then

$\displaystyle(\sqrt{2}+G)^{2}(0,\tau)$ $\displaystyle\geq
2-e^{\frac{\tau}{2}}-C\rho\left(\tfrac{\tau}{2}\right),$ $\displaystyle
G(0,\tau)$
$\displaystyle\geq-\sqrt{2}+\sqrt{2-e^{\frac{\tau}{2}}-C\rho\left(\tfrac{\tau}{2}\right)}\geq-
Ce^{\frac{\tau}{2}}-C\rho\left(\tfrac{\tau}{2}\right).$

Thus, the conclusion follows.

∎

It follows by Lemma 6.2 that

$\rho_{\max}(\tau)\leq O\left(e^{\frac{\tau}{16}}\right),\quad\rho(\tau)\leq
O\left(e^{\frac{\tau}{16}}\right),\quad\delta(\tau)\leq\rho(\tau/2)\leq
O\left(e^{\frac{\tau}{32}}\right).$

Let $h_{k}$ be the eigenfunctions of $\mathcal{L}$ corresponding to
eigenvalues $1-\frac{k}{2},k\in\mathbb{N}.$ Write

$\widehat{G}(\xi,\tau)=\sum_{k\geq 0}a_{k}(\tau)h_{k}(\xi),$

where

$a_{k}(\tau)=\langle\widehat{G}(\cdot,\tau),h_{k}\rangle_{\mathcal{H}}.$

Since we assumed reflexive symmetry,

$a_{1}(\tau)=\langle\widehat{G}(\cdot,\tau),h_{1}\rangle_{\mathcal{H}}\equiv
0,\qquad P_{+}\widehat{G}=a_{0}(\tau).$

Finally we conclude the following.

######  Lemma 6.3.

$a_{0}(\tau)=-\tfrac{1}{\sqrt{2}}\tau e^{\tau}+o(\tau e^{\tau}).$

###### Proof.

By Lemma 3.15,

$\,\mathcal{E}^{\rm orb}=(\tfrac{1}{2}+o(1))e^{-\tau}(\sqrt{2}+G)^{2}R^{2}.$

Then it is not hard to see that, by Lemma 3.33,

$\displaystyle a_{0}^{\prime}(\tau)$
$\displaystyle=\langle(\partial_{\tau}-\mathcal{L})\widehat{G}+\mathcal{L}\widehat{G},h_{0}\rangle_{\mathcal{H}}$
$\displaystyle=a_{0}(\tau)-\langle(\sqrt{2}+G)^{-1}\,\mathcal{E}^{\rm
orb},h_{0}\rangle+O(e^{\epsilon\tau})a_{0}(\tau)$
$\displaystyle=a_{0}(\tau)-\left(\tfrac{1}{\sqrt{2}}+o(1)\right)e^{\tau}+O(e^{\epsilon\tau})a_{0}(\tau),$

where $\epsilon>0$ is a small constant. By integration,

$a_{0}(\tau)=-\tfrac{1}{\sqrt{2}}\tau e^{\tau}+o(\tau e^{\tau}).$

∎

Theorem 6.1 follows by the standard interpolation inequalities.

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