# Rigidity of the round cylinders in Ricci shrinkers

Yu Li and Bing Wang

###### Abstract

In this paper, we prove that the round cylinders are rigid in the space of
Ricci shrinkers. Namely, any Ricci shrinker that is sufficiently close to
$S^{n-1}\times\mathbb{R}$ in the pointed-Gromov-Hausdorff topology must itself
be isometric to $S^{n-1}\times\mathbb{R}$.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Estimates on the neck region
  4. 4 Estimates on the cap region
  5. 5 Proof of the main theorem
  6. 6 Further discussion
  7. A Steady soliton conifold
  8. B Improvement theorems of symmetry

##  1 Introduction

A Ricci shrinker is a triple $(M^{n},g,f)$ of smooth manifold $M^{n}$,
Riemannian metric $g$ and a smooth function $f$ satisfying

$\displaystyle Rc+\text{Hess}\,f=\frac{1}{2}g,$

where the potential function $f$ is normalized so that

$\displaystyle R+|\nabla f|^{2}$ $\displaystyle=f.$ (1.1)

The Ricci shrinkers play essential roles in studying the singularities of the
Ricci flow. For example, it was proved by Enders-Müller-Topping [26] that any
proper blowup sequence from a type-I Ricci flow converges smoothly to a
nontrivial Ricci shrinker.

Due to its importance, the classification of Ricci shrinkers has attracted
extensive attentions. In dimension $2$ or $3$, we know that
$\mathbb{R}^{2},S^{2},\mathbb{R}^{3},S^{3},S^{2}\times\mathbb{R}$, and their
quotients make the complete list of all Ricci shrinkers (e.g.,
[32][54][57][13]). In higher dimensions, there exist many nontrivial, non-
product Ricci shrinkers (e.g., [37][11][27]), and the classification of Ricci
shrinkers is only achieved when extra assumptions are assumed. Such
assumptions include non-negativity of curvatures (e.g., [52][46][55][45][48]),
restriction of the Weyl curvatures (e.g., [59][50][12][23]), restriction of
asymptotic behavior at infinity (e.g., [40][41]), Kähler conditions (e.g.,
[56][22][20]), and others. In general, much less is known if no extra
assumptions are assumed.

We can consider all Ricci shrinkers as one moduli space $\mathcal{M}$ equipped
with pointed-Gromov-Hausdorff topology. On each Ricci shrinker, $f$ always
achieves its minimum value (cf. [14]) at some point $p$, which can be assigned
as a base point. A natural question is: which Ricci shrinker is rigid in
$\mathcal{M}$, in the sense that there is no nearby (in the pointed Gromov-
Hausdorff distance) Ricci shrinker other than itself? For instance, it follows
from [35] and the weak-compactness theory developed in [44] that the spherical
space form $S^{n}/\Gamma$ is rigid since any Ricci shrinker is a self-similar
Ricci flow solution. Other rigid examples include $\mathbb{CP}^{2n}$ proved by
Kröncke [42] and $S^{2}\times S^{2}$ proved by Sun-Zhu [66] very recently.
Their proofs depend on a delicate local structure theory of $\mathcal{M}$ (cf.
[62][36]), which is not available for non-compact Ricci shrinkers. This makes
the rigidity problem for non-compact Ricci shrinkers much more involved. Up to
now, the only non-compact Ricci shrinkers known to be rigid are the Gaussian
solitons $(\mathbb{R}^{n},g_{E})$, whose rigidity is proved through an
entropy-gap argument (cf. [69][70][48]).

Inspired by the fundamental work of Colding-Ilmanen-Minicozzi [21] on mean
curvature flow, and our earlier research on 4-dimensional Ricci shrinkers
[47], we are interested in figuring out whether the generalized cylinders
$S^{k}\times\mathbb{R}^{n-k},2\leq k\leq n-1$ are rigid in the moduli space of
Ricci shrinkers. In this article, we confirm the rigidity of the round
cylinders $S^{n-1}\times\mathbb{R}$, i.e., the cases $k=n-1$.

######  Theorem 1.1 (Main Theorem).

There exists a small constant $\hat{\epsilon}=\hat{\epsilon}(n)>0$ satisfying
the following property.

Suppose $(M^{n},p,g,f)$ is a Ricci shrinker such that

$\displaystyle
d_{PGH}\left\\{(M^{n},p,g),(S^{n-1}\times\mathbb{R},p_{c},g_{c})\right\\}<\hat{\epsilon},$
(1.2)

then $(M,g)$ is isometric to $(S^{n-1}\times\mathbb{R},g_{c})$. Here $d_{PGH}$
means the pointed-Gromov-Hausdorff distance, $p$ is a minimum point of $f$ and
$p_{c}$ is a fixed point of $S^{n-1}\times\mathbb{R}$.

The proof of Theorem 1.1 relies heavily on the symmetry improvement technique
of S. Brendle et al. (cf. [3][5][6][7][9]), the classification result of B.
Kotschwar [38] and the weak-compactness theory developed by H. Li, S. Huang
and the authors ([44] [34] [48]).

###### Outline of the proof of Theorem 1.1.

The proof consists of four steps.

Step 1. The base point $p$ is in an $\epsilon$-neck.

Condition (1.2) means that the base point $p$ has a neighborhood very close to
the standard cylinder in the Gromov-Hausdorff topology, which is very rough.
However, by the weak-compactness theory developed in [44], we can improve the
Gromov-Hausdorff topology to the $C^{\infty}$-Cheeger-Gromov topology (cf.
Proposition 7.4 of [44]). Therefore, by choosing $\hat{\epsilon}$ sufficiently
small, we are able to show (cf. Proposition 2.6) that $p$ is actually in the
center of an evolving normalized $\epsilon$-neck (cf. Definition 2.5). We may
understand that the point $p$ has $\epsilon$-symmetry with respect to the
model space $S^{n-1}\times\mathbb{R}$.

Step 2. Each point $x\in M$ is in an $\epsilon$-neck or a region
$\epsilon$-close to the Bryant soliton, after proper rescaling.

This step is the technical core of this paper. We characterize the
$\epsilon$-symmetry in terms of curvature and potential function estimates and
show that these estimates are almost preserved along the flow line of $\nabla
f$. Roughly speaking, if the level set of $f$ is uniformly away from being
critical (i.e., $\nabla f=0$), then each point locates in an $\epsilon$-neck.
Moreover, the critical points are isolated, and the region is very close to
the Bryant soliton near the critical point of $f$. Therefore, each point “far
away” from the critical set is in an $\epsilon$-neck, and each point “near”
critical set is in a region $\epsilon$-close to the Bryant soliton. These
descriptions can be made precise, which is the key new ingredient of this
argument.

Step 3. $(M,g)$ is rotationally symmetric.

This step utilizes the celebrated symmetry improvement technique developed by
S. Brendle et al. (cf. Appendix B for details). Suppose $\epsilon$ is
sufficiently small and every point on $(M,g)$ has $\epsilon$-symmetry (with
respect to either the cylinder or Bryant soliton), then every point has
$\frac{\epsilon}{2}$-symmetry. Consequently, this process will run forever and
yield that every point $x\in M$ has $2^{-k}\epsilon$-symmetry for each
positive integer $k$. Therefore, $(M,g)$ is rotationally symmetric.

Step 4. $(M,g)$ is isometric to $S^{n-1}\times\mathbb{R}$.

Since $(M,g)$ is rotationally symmetric already, the classification result of
Kotschwar [38] applies. We know that $(M,g)$ is isometric to $S^{n}$,
$S^{n-1}\times\mathbb{R}$ or $\mathbb{R}^{n}$. By condition (1.2), we conclude
that $(M,g)$ must be isometric to $S^{n-1}\times\mathbb{R}$. ∎

From the discussion above, it is clear that the most difficult part of the
proof is to analyze precisely the propagation of the $\epsilon$-symmetry along
the flow line of $\nabla f$. Due to this propagation, if there is a neck
region far away from the base point and close to the round cylinder, then the
behavior of the far-end can be described precisely. Since
$f\sim\frac{d^{2}(p,\cdot)}{4}$, the value of $f$ indicates the distance to
the base point. For each pair $t\leq s$, we define $\Sigma(t,s)$ to be a
connected component of the level set

$\displaystyle\\{x\in M\,\mid t\leq f(x)\leq s\\}.$ (1.3)

In the particular case $t=s$, we define

$\displaystyle\Sigma(t)\coloneqq\Sigma(t,t)=\\{x\in M\,\mid f(x)=t\\}.$ (1.4)

If a point locates in a cylinder-like neighborhood, we say it is the center of
$\epsilon$-neck. Alternatively, if a point locates in a neighborhood close to
a steady soliton with mild singularities, we say it is the center of
$\epsilon$-steady soliton conifold. Note that the conifold here is the one
introduced by Chen-Wang (cf. Definition 1.2 of [24]) to denote the Riemannian
space with mild singularities. For the exact definitions of the aforementioned
concepts, see Definition 2.5, Definition 4.3 and Definition A.1. With these
terminologies, we can precisely describe the propagation of the
$\epsilon$-symmetry in the following theorem.

######  Theorem 1.2 (Propagation of almost symmetry).

For any positive constants $n,A,B,\epsilon$ and $\delta_{0}\in(0,1)$, there
exist positive constants $\sigma=\sigma(n,\delta_{0},A,B)$,
$L=L(n,\delta_{0},A,B)$ and $\eta=\eta(n,\delta_{0},A,B,\epsilon)$ satisfying
the following property.

Let $(M^{n},g,f)$ be a Ricci shrinker with

$\displaystyle\begin{cases}|\nabla^{i}Rm|\leq
BR^{\frac{i}{2}+1},\quad\forall\,0\leq i\leq
4\quad&\text{on}\quad\Sigma(t_{0},s_{0});\\\
R\leq\epsilon_{1}f\quad&\text{on}\quad\Sigma(t_{0},s_{0});\\\
|\overset{\circ}{Rm_{\Sigma}}|\leq\epsilon_{1}^{\prime}R\quad&\text{on}\quad\Sigma(t_{0});\\\
\left|R-\frac{n-1}{2}\right|\leq\epsilon^{\prime}_{1}\quad&\text{on}\quad\Sigma(t_{0});\\\
(1-\delta_{0})s_{0}\geq t_{0}\geq n\epsilon_{1}^{-1};\\\
\boldsymbol{\mu}(g)\geq-A;\\\ \Sigma(t_{0})\text{ is diffeomorphic to
}S^{n-1}.\end{cases}$ (1.5)

If $\epsilon_{1}\leq\sigma$ and $\epsilon_{1}^{\prime}\leq\eta$, then one of
the following statements holds.

  * (a).

There exists an end $E$ with $\partial E=\Sigma(t_{0})$ such that any point in
$\Sigma(\eta^{-1}t_{0},\infty)\subset E$ is the center of an evolving
normalized $\epsilon$-neck. Moreover, $E$ is asymptotic to the round cylinder
with rate $O(r^{-\tau(n)})$.

  * (b).

There exists an end $E$ with $\partial E=\Sigma(t_{0})$ such that any point in
$\Sigma(\eta^{-1}t_{0},\infty)\subset E$ is the center of an evolving
$\epsilon$-neck. Moreover, $E$ is asymptotic to a regular cone with cross
section diffeomorphic to $S^{n-1}$.

  * (c).

There exists a compact set $E$ with $\partial E=\Sigma(t_{0})$, a number
$s\geq s_{0}$ and a point $q\in\Sigma(s)$ such that $R(q)=\sigma s$. Moreover,
any point in $\Sigma(\eta^{-1}t_{0},s)\subset E$ is the center of an evolving
$\epsilon$-neck and any point in the cap $D:=E\backslash\Sigma(t_{0},s)$ is
the center of an $\epsilon$-steady soliton conifold. Furthermore,

$\displaystyle\emph{diam}_{g}D\leq\frac{L}{\sqrt{s}},\quad\sup_{D}|f-s|\leq
L\quad\text{and}\quad L^{-1}s\leq\inf_{D}R\leq\sup_{D}R\leq s+L.$ (1.6)

Note that (1.5) on $\Sigma(t_{0},s_{0})$ is satisfied if the region is very
close to the cylinder, where the constant $\epsilon_{1}^{\prime}$ can be
chosen as zero. Since the function $f$ contains no critical point on the
region $\Sigma(t_{0},s_{0})$, one can follow Munteanu-Wang [53] to define the
tangential curvature operator $Rm_{\Sigma}$ restricted on each level set of
$f$ and the term $\overset{\circ}{Rm_{\Sigma}}$ is the “traceless” part of
$Rm_{\Sigma}$ (cf. (2.6), (2.7)). If $R/f$ is small, the value of
$|\overset{\circ}{Rm_{\Sigma}}|$ measures how close $Rm_{\Sigma}$ resembles
the curvature operator of the standard $S^{n-1}$. Theorem 1.2 implies that if
a Ricci shrinker has a neck region sufficiently close to the cylinder in some
weak sense, one can move forward along the direction of $\nabla f$ to
determine the future geometric property. If we regard a Ricci shrinker as a
Ricci flow solution, the control of the geometric behavior at spatial infinity
amounts to that near the singular time $t=1$. Recall that the traditional
pseudolocality on Ricci shrinkers (cf. [48, Corollary $10.6$]) means that the
almost Euclidean property is preserved by the flow until time $t=1$. Theorem
1.2 claims instead that the $\epsilon$-symmetric property is almost preserved
by the flow until time $t=1$. Therefore, Theorem 1.2 can be understood as a
generalized form of pseudolocality theorem on Ricci shrinkers. See Figure 1
for rough intuition. We emphasize here that the global bound of the curvature
is not needed. With some extra assumptions like the boundedness of the
curvature, Munteanu-Wang [53, Theorem $1.6$] has already obtained case (a) in
Theorem 1.2.

Figure 1: Three types of end

We briefly discuss the proof of Theorem 1.2. We first show on the neck region
satisfying (1.5), the estimate
$|\overset{\circ}{Rm_{\Sigma}}|\leq\epsilon_{0}R$ holds on
$\Sigma(t_{0},s_{0})$, where $\epsilon_{0}$ is a constant depending only on
$\epsilon_{1}$ and $\epsilon^{\prime}_{1}$. Since $f\geq t_{0}$ is large and
$R/f\leq\epsilon_{1}$ is small, all normal directional part of $Rm$ is small
and insignificant. Moreover, we can show that the scalar curvature $R$ is
almost constant along any level set $\Sigma$ of $f$ contained in the neck
region. From the elliptic equation of $R$, one obtains an ODE by examining the
behavior of $R$ along the gradient flow of $f$.

One can extend the neck region from $\Sigma(t_{0},s_{0})$ to $\Sigma(t_{0},s)$
with $s$ to be maximal so that (1.5) still holds. Then we show that $s$ is
finite only if there exists a point $q\in\Sigma(s)$ with
$R(q)=\epsilon_{1}f(q)$. Here, we need to use the pseudolocality for Ricci
shrinkers proved in [48] to show that any point $x\in\Sigma(\eta^{-1}t_{0},s)$
is the center of an evolving $\epsilon$-neck. Depending on the behavior of $R$
along the gradient flow of $f$, only three cases can happen, as described in
Theorem 1.2. If $R$ stays close to $\frac{n-1}{2}$, then $s=\infty$ and we
obtain the case (a). If at some point in the neck $R$ is smaller than
$\frac{n-1}{2}$ by some detectable small number, then $R$ will be
quadratically decaying along the flow. In this case, $s=\infty$ and we obtain
the case (b). If at some point in the neck $R$ is bigger than $\frac{n-1}{2}$
by some detectable small number, then $R$ will be increasing along the flow
until it reaches the boundary $\Sigma(s)$. If this happens, $s<\infty$ and we
obtain case (c). In this case, it is important to show that there exists a
compact cap $D$ with $\partial D=\Sigma(s)$. The reason is that the behavior
of the Ricci shrinker near the boundary $\Sigma(s)$ is modeled on a Ricci
steady soliton conifold (cf. Definition A.1) at a proper scale. This can be
proved through a compactness argument (cf. [44][34]). Since each model steady
soliton conifold has only one end (cf. Theorem A.22), the cap region $D$ is
compact. Further analysis implies the estimates in (1.6).

Now we continue to discuss Step 2 in the outline proof of Theorem 1.1 in more
detail. First, one notices that any Ricci shrinker close to the cylinder
satisfies the conditions of Theorem 1.2 for two neck regions. By extending
those two neck regions, there are three possibilities.

  * (1).

The Ricci shrinker has two ends so that each point is the center of an
evolving $\epsilon$-neck.

  * (2).

The Ricci shrinker has exactly one end so that outside a compact cap $D$, any
point is the center of an evolving $\epsilon$-neck.

  * (3).

The Ricci shrinker is compact and it consists of two caps $D_{1}$ and $D_{2}$
such that outside those two caps, any point is the center of an evolving
$\epsilon$-neck.

In order to realize Step 2 in the proof of Theorem 1.1, the last key
ingredient needed is the observation that the cap is modeled on the Bryant
soliton in cases (2), (3) above. This fact follows from the classification
result (cf. Theorem 4.7) of any asymptotically cylindrical Ricci steady
soliton conifold with the curvature condition PIC2, which is a generalization
of the celebrated work of S. Brendle [3]. On the other hand, if one considers
the Ricci flow associated with a Ricci shrinker close to the cylinder, we show
that locally the Ricci flow almost preserves the nonnegativity of some
curvature conditions (cf. Theorem 5.4) defined by S. Brendle. By analyzing the
geometry along the gradient flow of $f$, one can show that the blowup limit
near the cap region must have the PIC2 condition. Consequently, the blowup
limit must be a steady soliton conifold with PIC2 condition, which can be
classified and can only be the Bryant soliton. Therefore, the cap regions are
modeled on Bryant solitons. In conclusion, for a Ricci shrinker sufficiently
close to a cylinder, each point lies in a region either $\epsilon$-close to
the cylinder or the Bryant soliton. Roughly speaking, this means that the
shrinker itself is $\epsilon$-symmetric (cf. Definition B.1, Definition B.3 or
Definition B.4 for precise definitions). Therefore, we have collected
sufficient technical preparation to apply the symmetry improvement argument of
S. Brendle et al.

This paper is organized as follows. In Section 2, we discuss some basic
properties for Ricci shrinkers and the associated Ricci flows. In Section 3,
we obtain all the necessary technical estimates on the neck region. Section 4
focuses on the cap region and shows that it can be modeled on a Ricci steady
soliton conifold. As a conclusion of the estimates in Section 3 and Section 4,
we prove Theorem 1.2 at the end of Section 4. In Section 5, we prove Theorem
1.1. In the last section, we discuss possible generalizations of our theorems.

Acknowledgements:

Yu Li is supported by YSBR-001 and a research fund from USTC (University of
Science and Technology of China). Bing Wang is supported by NSFC-11971452,
NSFC-12026251, YSBR-001 and a research fund from USTC.

##  2 Preliminaries

For any Ricci shrinker $(M^{n},g,f)$, the scalar curvature $R\geq 0$ by [17,
Corollary $2.5$]. Moreover, it follows from the strong maximum principle that
$R>0$ unless $(M,g)$ is the Gaussian soliton.

We recall the following fundamental estimate of the potential function $f$.

######  Lemma 2.1 ([14] [33]).

Let $(M^{n},g,f)$ be a Ricci shrinker. Then there exists a point $p\in M$
where $f$ attains its infimum and $f$ satisfies the quadratic growth estimate

$\displaystyle\frac{1}{4}\left(d(x,p)-5n\right)^{2}_{+}\leq
f(x)\leq\frac{1}{4}\left(d(x,p)+\sqrt{2n}\right)^{2}$

for all $x\in M$, where $a_{+}:=\max\\{0,a\\}$.

For any Ricci shrinker $(M^{n},g,f)$ with the normalization (1.1), the entropy
is defined as

$\displaystyle\boldsymbol{\mu}=\boldsymbol{\mu}(g)\coloneqq\log\int\frac{e^{-f}}{(4\pi)^{n/2}}\,dV.$

We remark that $e^{\boldsymbol{\mu}}$ is comparable to the volume of the unit
ball $B(p,1)$, see [44, Lemma 2.5].

Next we recall the following identities and elliptic equations on Ricci
shrinkers (see [59] [53]):

$\displaystyle R_{ik}f_{k}=$
$\displaystyle\frac{1}{2}\nabla_{i}R=\nabla_{k}R_{ki},$ (2.1) $\displaystyle
R_{ijkl}f_{l}=$
$\displaystyle\nabla_{j}R_{ik}-\nabla_{i}R_{jk}=\nabla_{l}R_{ijkl},$ (2.2)
$\displaystyle\Delta_{f}R=$ $\displaystyle R-2|Rc|^{2},$ (2.3)
$\displaystyle\Delta_{f}R_{ij}=$ $\displaystyle R_{ij}-2R_{ikjl}R_{kl}.$ (2.4)

Here, the weighted Laplacian $\Delta_{f}\coloneqq\Delta-\langle\nabla
f,\nabla\cdot\rangle$.

We will only consider the case that $\Sigma(t)=\Sigma(t,s)\cap\\{x\in M\,\mid
f(x)=t\\}$. Notice that $\Sigma(t,\infty)$ can be either compact or non-
compact and it represents a cap or an end of the Ricci shrinker, respectively.

For the rest of the paper, we will mainly focus on $\Sigma(t)$ when $t$ is a
noncritical value of $f$. Also, we omit $t$ in $\Sigma(t)$, if there is no
confusion.

On $\Sigma$, we denote the unit normal vector $|\nabla f|^{-1}\nabla f$ by
$e_{n}$ and assume $\\{e_{1},e_{2},\cdots,e_{n}\\}$ is a local orthonormal
frame such that $\\{e_{1},\cdots,e_{n-1}\\}$ are tangent to $\Sigma$. By using
the same convention as in [53], we use subscript $n$ to denote $e_{n}$ and
$a,b,c,d,\cdots$ to denote all tangential directions.

We have the following estimates of the curvature along the normal direction.
The proof follows from direct calculations based on (2.1)-(2.4).

######  Proposition 2.2.

For any Ricci shrinker $(M^{n},g,f)$ such that $R\leq\frac{3}{4}f$ on a level
set $\Sigma$ of $f$, the following estimates hold on $\Sigma$.

$\displaystyle|R_{an}|\leq$ $\displaystyle Cf^{-\frac{1}{2}}|\nabla R|,$
$\displaystyle|R_{nn}|\leq$ $\displaystyle
Cf^{-1}\left(R+|Rc|^{2}+|\nabla^{2}R|\right),$ $\displaystyle|R_{abcn}|\leq$
$\displaystyle Cf^{-\frac{1}{2}}|\nabla Rc|,$ $\displaystyle|R_{ancn}|\leq$
$\displaystyle Cf^{-1}\left(|Rc|+|\nabla^{2}Rc|+|Rc|\,|Rm|\right),$

where $C=C(n)>0$.

###### Proof.

From (1.1) and our assumption on $R$, we have $|\nabla
f|\geq\frac{1}{2}\sqrt{f}$. We compute on $\Sigma$,

$\displaystyle|R_{an}|=\frac{|Rc(e_{a},\nabla f)|}{|\nabla f|}\leq
Cf^{-\frac{1}{2}}|\langle\nabla R,e_{a}\rangle|\leq Cf^{-\frac{1}{2}}|\nabla
R|.$

from (2.1). Moreover, it follows from (2.1) and (2.3) that

$\displaystyle|R_{nn}|=$ $\displaystyle|\nabla f|^{-2}|Rc(\nabla f,\nabla f)|$
$\displaystyle\leq$ $\displaystyle Cf^{-1}|\langle\nabla R,\nabla f\rangle|$
$\displaystyle=$ $\displaystyle Cf^{-1}\left|\Delta R-R+2|Rc|^{2}\right|$
$\displaystyle\leq$ $\displaystyle
Cf^{-1}\left(R+|Rc|^{2}+|\nabla^{2}R|\right).$

Similarly, we compute from (2.2),

$\displaystyle|R_{abcn}|\leq|\nabla f|^{-1}|R_{abcl}f_{l}|\leq
Cf^{-\frac{1}{2}}|\nabla Rc|.$

For the last inequality, we have

$\displaystyle|R_{ancn}|\leq$ $\displaystyle|\nabla
f|^{-2}|R_{akcl}f_{k}f_{l}|$ $\displaystyle\leq$ $\displaystyle
Cf^{-1}|(\nabla_{k}R_{ac}-\nabla_{a}R_{kc})f_{k}|$ $\displaystyle=$
$\displaystyle
Cf^{-1}|\nabla_{k}R_{ac}f_{k}-\nabla_{a}(R_{kc}f_{k})+R_{kc}f_{ka}|$
$\displaystyle=$ $\displaystyle Cf^{-1}|\Delta
R_{ac}-R_{ac}+2R_{akcl}R_{kl}-\nabla^{2}_{ac}R/2+R_{kc}(g_{ka}/2-R_{ka})|$
$\displaystyle=$ $\displaystyle
Cf^{-1}|2R_{akcl}R_{kl}-R_{kc}R_{ka}-R_{ac}/2+\Delta
R_{ac}-\nabla^{2}_{ac}R/2|$ $\displaystyle\leq$ $\displaystyle
Cf^{-1}\left(|Rc|+|\nabla^{2}Rc|+|Rc|\,|Rm|\right).$

In sum, the proof is complete. ∎

As in [53], we define the following tensors on $\Sigma$:

$\displaystyle\overset{\circ}{Rc_{\Sigma}}=\overset{\circ}{R}_{ab}:=R_{ab}-\frac{1}{n-1}R\,g_{ab},$
(2.5) $\displaystyle
U=U_{abcd}:=\frac{1}{\left(n-1\right)\left(n-2\right)}R\left(g_{ac}g_{bd}-g_{ad}g_{bc}\right),$
(2.6)
$\displaystyle\overset{\circ}{Rm_{\Sigma}}=\overset{\circ}{R}_{abcd}:=R_{abcd}-U_{abcd},$
(2.7) $\displaystyle
V=V_{abcd}:=\frac{1}{n-3}\left(\overset{\circ}{R}_{ac}g_{bd}+\overset{\circ}{R}_{bd}g_{ac}-\overset{\circ}{R}_{ad}g_{bc}-\overset{\circ}{R}_{bc}g_{ad}\right),$
(2.8) $\displaystyle W=W_{abcd}:=R_{abcd}-U_{abcd}-V_{abcd}.$ (2.9)

Notice that all tensors
$\overset{\circ}{Rc_{\Sigma}},U,\overset{\circ}{Rm_{\Sigma}},V$ and $W$ can be
extended to tensors on $\Sigma(s,t)$ by requiring they vanish on the normal
direction $e_{n}$. By this convention, all tensors can be written in a
coordinate-free form. For instance, we have

$\displaystyle\overset{\circ}{Rc_{\Sigma}}=Rc-\frac{R}{n-1}g-T$ (2.10)

where

$\displaystyle T\coloneqq\frac{1}{2|\nabla f|^{2}}(dR\otimes df+df\otimes
dR)-\left(\frac{R}{(n-1)|\nabla f|^{2}}+\frac{Rc(\nabla f,\nabla f)}{|\nabla
f|^{4}}\right)df\otimes df.$ (2.11)

Next, we have the following lower bound estimate of the scalar curvature,
which essentially follows from [19, Theorem $1$].

######  Proposition 2.3.

For any Ricci shrinker $(M^{n},g,f)$, suppose $\bar{t}\geq 2n$ and

$\displaystyle\min_{\Sigma(\bar{t})}\frac{Rf}{1+nf^{-1}}=\alpha>0.$

Then for any $x\in\Sigma(\bar{t},\infty)$,

$\displaystyle R(x)f(x)\geq\frac{R(x)f(x)}{1+nf^{-1}(x)}\geq\alpha.$

###### Proof.

We define $\phi=R-\alpha f^{-1}-\alpha nf^{-2}$. Then it follows from direct
computations, see [19, Equation (6)] for details, that on
$\Sigma(\bar{t},\infty)$

$\displaystyle\Delta_{f}\phi\leq\phi-\alpha
nf^{-3}\left(\frac{f}{2}-n\right)-\alpha f^{-4}(2f+6n)|\nabla f|^{2}\leq\phi$

since $f\geq\bar{t}\geq 2n$. By our assumption, $\phi\geq 0$ on the boundary
$\Sigma(\bar{t})$. If $\Sigma(\bar{t},\infty)$ is compact, then we conlude
that $\phi\geq 0$ on $\Sigma(\bar{t},\infty)$ from the maximum principle. If
$\Sigma(\bar{t},\infty)$ is non-compact, the maximum principle also applies
since $\liminf_{x\to\infty}\phi(x)\geq 0$ from its definition.

In sum, we have proved that on $\Sigma(\bar{t},\infty)$,

$\displaystyle\frac{Rf}{1+nf^{-1}}\geq\alpha$

and the proof is complete. ∎

### Ricci flow associated with a Ricci shrinker

Recall that any Ricci shrinker $(M^{n},g,f)$ can be regarded as a self-similar
solution of the Ricci flow. Let ${\psi^{t}}:M\to M$ be a family of
diffeomorphisms generated by $X(t)=\dfrac{1}{1-t}\nabla f$, and
$\psi^{0}=\text{id}$. In other words, we have

$\displaystyle\frac{\partial}{\partial t}{\psi^{t}}(x)=\frac{1}{1-t}\nabla
f\left({\psi^{t}}(x)\right).$ (2.12)

It is well known that the rescaled pull-back metric
$g(t)\coloneqq(1-t)(\psi^{t})^{*}g$ satisfies the Ricci flow equation

$\displaystyle\partial_{t}g=-2Rc(g(t))$

for any $-\infty<t<1$. In particular, $g(0)=g$. For any Ricci shrinker, the
associated Ricci flow is implicitly understood.

######  Example 2.4 (Round cylinder).

The round cylinder $(S^{n-1}\times\mathbb{R},g_{c})$ is a Ricci shrinker where
the scalar curvature is identically $\dfrac{n-1}{2}$ and the potential
function $f_{c}=\dfrac{z^{2}}{4}+\dfrac{n-1}{2}$, where $z$ is the coordinate
in the $\mathbb{R}$-factor. The associated Ricci flow is
$\\{(S^{n-1}\times\mathbb{R},g_{c}(t)),\,-\infty<t<1\\}$, where

$\displaystyle g_{c}(t)=2(n-2)(1-t)g_{S^{n-1}}+dz\otimes dz,\quad
g_{c}(0)=g_{c}.$

Next, we have the following definition, which measures how close a parabolic
neighborhood of a point in a Ricci flow is to the round cylinder, see also
[58, Section 11.8].

######  Definition 2.5 (Center of an evolving $\epsilon$-neck).

Let $(M^{n},g(t))$ be a Ricci flow solution and let $(\bar{x},\bar{t})$ be a
point in space-time with $R(\bar{x},\bar{t})=\frac{n-1}{2}r^{-2}$. We say that
$(\bar{x},\bar{t})$ is the center of an evolving $\epsilon$-neck
$S^{n-1}\times\mathbb{R}$ if, after rescaling the metric by the factor
$r^{-2}$, the parabolic neighborhood
$B_{g(\bar{t})}(\bar{x},\epsilon^{-1}r)\times[\bar{t}-\epsilon^{-1}r^{2},\bar{t}]$
is $\epsilon$-close in $C^{[\epsilon^{-1}]}$-topology to
$\left(S^{n-1}\times\mathbb{R},g_{c}(t)\right)$. The evolving neck is called
an evolving normalized $\epsilon$-neck, if we further require that
$|r-1|<\epsilon$. For any Ricci shrinker $(M^{n},g,f)$, we say $x\in M$ is the
center of an evolving (normalized) $\epsilon$-neck if $(x,0)$ is the center of
an evolving (normalized) $\epsilon$-neck.

From the weak-compactness theory developed in [44], we prove the following
result.

######  Proposition 2.6.

For any $n$ and $\epsilon>0$, there exists a constant
$\eta_{1}=\eta_{1}(\epsilon,n)>0$ satisfying the following property.

Suppose $(M^{n},p,g,f)$ is a Ricci shrinker such that

$\displaystyle
d_{PGH}\left\\{(M^{n},p,g),\left(S^{n-1}\times\mathbb{R},p_{c},g_{c}\right)\right\\}<\eta_{1},$

then $p$ is the center of an evolving normalized $\epsilon$-neck.

###### Proof.

Suppose otherwise, there exists a number $\bar{\epsilon}$ and a sequence of
Ricci shrinkers $(M^{n}_{i},p_{i},g_{i},f_{i})$ such that

$\displaystyle\lim_{i\to
0}d_{PGH}\left\\{(M_{i}^{n},p_{i},g_{i}),\left(S^{n-1}\times\mathbb{R},p_{c},g_{c}\right)\right\\}=0$

and $p_{i}$ is not the center of an $\bar{\epsilon}$-neck. By Proposition 5.8
of [44], the entropy $\boldsymbol{\mu}(g_{i})$ is uniformly bounded from
below. Using Theorem 1.1 of [44], the above convergence can be improved to be
in the smooth topology

$\displaystyle(M_{i},p_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}\left(S^{n-1}\times\mathbb{R},p_{c},g_{c}\right).$

From the definition of the Ricci flow associated with a Ricci shrinker, it is
clear that the corresponding sequence of Ricci flows
$(M_{i},p_{i},g_{i}(t))_{t\leq 0}$ converges smoothly to
$(S^{n-1}\times\mathbb{R},g_{c}(t))_{t\leq 0}$. Therefore, we obtain a
contradiction. ∎

##  3 Estimates on the neck region

Throughout this section, we consider a Ricci shrinker $(M^{n},p,g,f)$ such
that the following assumptions as in Theorem 1.2 are satisfied.

$\displaystyle\begin{cases}|\nabla^{i}Rm|\leq
BR^{\frac{i}{2}+1},\quad\forall\,0\leq i\leq
4\quad&\text{on}\quad\Sigma(t_{0},s);\\\
R\leq\epsilon_{1}f\quad&\text{on}\quad\Sigma(t_{0},s);\\\
|\overset{\circ}{Rm_{\Sigma}}|\leq\epsilon_{1}^{\prime}R\quad&\text{on}\quad\Sigma(t_{0});\\\
\left|R-\frac{n-1}{2}\right|\leq\epsilon^{\prime}_{1}\quad&\text{on}\quad\Sigma(t_{0});\\\
(1-\delta_{0})s\geq t_{0}\geq n\epsilon_{1}^{-1};\\\
\boldsymbol{\mu}(g)\geq-A;\\\ \Sigma(t_{0})\text{ is diffeomorphic to
}S^{n-1}.\end{cases}$ ($\dagger$)

Moreover, we set

$\displaystyle\epsilon_{0}:=\max\\{\epsilon_{1}^{\prime},\epsilon_{1}^{\frac{1}{2}}\\}.$
(3.1)

Here, we assume $0<\delta_{0}<1$ and $A,B\geq 1$. With fixed $n,\delta_{0},A$
and $B$, we regard $\epsilon_{1}$ and $\epsilon_{1}^{\prime}$ as two small
positive parameters. In addition, the constant $s$ can be finite or $\infty$.
Notice that for the round cylinder $S^{n-1}\times\mathbb{R}$, ($\dagger$ ‣ 3)
is satisfied for any $\epsilon_{1}$ and $\epsilon_{1}^{\prime}$ provided that
$t_{0}$ is sufficiently large. In the following analysis, we set

$\displaystyle E\coloneqq\Sigma(t_{0},s)$ (3.2)

and use the notation $\phi_{1}=O(\phi_{2})$ if $|\phi_{1}|\leq C|\phi_{2}|$
for some constant $C=C(n,A,B)>0$.

One important consequence of $\boldsymbol{\mu}(g)\geq-A$ is the following no-
local-collapsing result from [48].

######  Theorem 3.1.

_([48, Theorem 22])_ For any Ricci shrinker $(M^{n},g,f)$ with
$\boldsymbol{\mu}(g)\geq-A$, there exists a $\kappa=\kappa(n,A)>0$ such that
for any $B(q,r)\subset M$ with $R\leq r^{-2}$, we have

$\displaystyle|B(q,r)|\geq\kappa r^{n}.$

We next prove

######  Lemma 3.2.

On $E$, we have

$\displaystyle Rf$ $\displaystyle\geq t_{0},$ (3.3)
$\displaystyle|R_{an}|,\,|R_{abcn}|$
$\displaystyle=O(f^{-\frac{1}{2}}R^{\frac{3}{2}}),$ (3.4)
$\displaystyle|R_{nn}|,\,|R_{ancn}|$ $\displaystyle=O(f^{-1}R(1+R)).$ (3.5)

###### Proof.

The inequality (3.3) follows immediately from Proposition 2.3, since by
($\dagger$ ‣ 3),

$\displaystyle\inf_{\Sigma(t_{0})}\frac{Rf}{1+nf^{-1}}\geq\frac{(\frac{n-1}{2}-\epsilon_{1}^{\prime})t_{0}}{1+nt^{-1}_{0}}\geq
t_{0}.$

The equations (3.4) and (3.5) follow from Proposition 2.2 and ($\dagger$ ‣ 3).
∎

The next lemma follows from Lemma 3.2 and direct calculations from the
definition (2.5).

######  Lemma 3.3.

On $E$, we have

$\displaystyle|Rc|^{2}=|\overset{\circ}{Rc_{\Sigma}}|^{2}+\frac{R^{2}}{n-1}+O(f^{-1}R^{2}(1+R)),$
$\displaystyle\langle U,V\rangle\,,\langle V,W\rangle,\,\langle
W,U\rangle=O(f^{-1}R^{2}(1+R)),$
$\displaystyle|U|^{2}=\frac{2}{(n-1)(n-2)}R^{2},$
$\displaystyle|V|^{2}=\frac{4}{n-3}|\overset{\circ}{Rc_{\Sigma}}|^{2}+O(f^{-1}R^{2}(1+R)),$
$\displaystyle|Rm|^{2}=|\overset{\circ}{Rm_{\Sigma}}|^{2}+\frac{2}{(n-1)(n-2)}R^{2}+O(f^{-1}R^{2}(1+R)),$
$\displaystyle|Rm|^{2}=|U|^{2}+|V|^{2}+|W|^{2}+O(f^{-1}R^{2}(1+R)).$

######  Proposition 3.4.

Under the assumption ($\dagger$ ‣ 3), there exists
$\sigma_{1}=\sigma_{1}(n,B)>0$ such that

$\displaystyle|\overset{\circ}{Rm_{\Sigma}}|=O(\epsilon_{0}R)$ (3.6)

on $E$ if $\epsilon_{0}\leq\sigma_{1}$.

###### Proof.

In the proof, all constants $c_{i},i=1,2,\cdots$ are positive and depend only
on $n$ and $B$. We define $s_{0}\in[t_{0},s]$ to be the largest number such
that

$\displaystyle\frac{|\overset{\circ}{Rm_{\Sigma}}|^{2}}{R^{2}}\leq\tau\epsilon^{2}_{0}$

on $\Sigma(t_{0},s_{0})$, where $\tau>1$ is a large constant to be determined
later. By the assumption ($\dagger$ ‣ 3), we have

$\displaystyle|\overset{\circ}{Rm_{\Sigma}}|\leq\epsilon_{1}^{\prime}R$

on $\Sigma(t_{0})$ and hence $s_{0}>t_{0}$ as
$\epsilon_{0}\geq\epsilon_{1}^{\prime}$. If $s_{0}=s$, then the conclusion
follows. Otherwise, we assume $s_{0}<s$.

Following [53], we define

$\displaystyle G\coloneqq$
$\displaystyle\frac{|Rm|^{2}}{R^{2}}-\frac{2}{(n-1)(n-2)}=\frac{|\overset{\circ}{Rm_{\Sigma}}|^{2}}{R^{2}}+O(f^{-1}(1+R))$
$\displaystyle=$
$\displaystyle\frac{|\overset{\circ}{Rm_{\Sigma}}|^{2}}{R^{2}}+O(t_{0}^{-1}+\epsilon_{1})=\frac{|\overset{\circ}{Rm_{\Sigma}}|^{2}}{R^{2}}+O(\epsilon_{0}^{2}).$
(3.7)

where the equalities follow from Lemma 3.3 and ($\dagger$ ‣ 3). Therefore,
there exists $s_{1}\in(t_{0},s_{0}]$ such that

$\displaystyle\max_{\Sigma(t_{0},s_{0})}G=\max_{\Sigma(s_{1})}G\in[\frac{\tau}{2}\epsilon_{0}^{2},2\tau\epsilon_{0}^{2}].$
(3.8)

It follows from a direct calculation, see [53, (4.10),(4.11)] for details,
that

$\displaystyle\Delta_{f}G\geq-2\langle\nabla G,\nabla\log R\rangle+4R^{-3}P,$
(3.9)

where

$\displaystyle
P:=-2RR_{ijkl}R_{piqk}R_{pjql}-\frac{1}{2}RR_{ijkl}R_{ijpq}R_{pqkl}+|Rm|^{2}|Rc|^{2}.$

By similar calculations as [53, (4.16)] and our assumptions, we obtain

$\displaystyle P\geq$
$\displaystyle\frac{1}{n-1}R^{2}|W|^{2}+\frac{4}{(n-1)(n-2)(n-3)}R^{2}|\overset{\circ}{Rc_{\Sigma}}|^{2}-\frac{5}{2}R|W|^{3}$
$\displaystyle-\frac{8}{(n-3)^{2}}R|\overset{\circ}{Rc_{\Sigma}}|^{3}-\frac{6}{n-3}R|W||\overset{\circ}{Rc_{\Sigma}}|^{2}-c_{1}f^{-1}R^{4}(1+R).$
(3.10)

From Lemma 3.3, we obtain on $\Sigma(t_{0},s_{0})$ that

$\displaystyle|W|^{2}+\frac{4}{n-3}|\overset{\circ}{Rc_{\Sigma}}|^{2}=|\overset{\circ}{Rm_{\Sigma}}|^{2}+O(f^{-1}R^{2}(1+R))\leq\tau\epsilon_{0}^{2}R^{2}+O((t_{0}^{-1}+\epsilon_{1})R^{2})\leq
c_{2}\epsilon_{0}^{2}R^{2}.$ (3.11)

Now, it follows from (3.10) and (3.11) that

$\displaystyle
P\geq\left(\frac{1}{n-1}-c_{3}\epsilon_{0}\right)R^{2}|W|^{2}+\left(\frac{4}{(n-1)(n-2)(n-3)}-c_{3}\epsilon_{0}\right)R^{2}|\overset{\circ}{Rc_{\Sigma}}|^{2}-c_{3}f^{-1}R^{4}(1+R).$
(3.12)

Therefore, if $\epsilon_{0}$ is sufficiently small, we obtain

$\displaystyle P\geq
c_{4}R^{2}|\overset{\circ}{Rm_{\Sigma}}|^{2}-c_{5}f^{-1}R^{4}(1+R).$

It follows from (3.9) and (3.12) that

$\displaystyle\Delta_{f}G\geq-2\langle\nabla G,\nabla\log
R\rangle+4c_{4}RG-4c_{5}f^{-1}R(1+R).$ (3.13)

If we denote the Laplacian and inner product on the level set $\Sigma$ by
$\Delta_{\Sigma}$ and $\langle\,,\,\rangle_{\Sigma}$ respectively, then

$\displaystyle\Delta G=\Delta_{\Sigma}G+G_{nn}+HG_{n},$

where $H:=\text{Tr}_{\Sigma}\frac{\nabla^{2}f}{|\nabla f|}$ is the mean
curvature of the level set $\Sigma$. From the fact that $|\nabla^{i}Rm|\leq
BR^{\frac{i}{2}+1}$ for $0\leq i\leq 4$, the term $G_{nn}$ can be estimated as
(see [53, (4.26)] for details)

$\displaystyle G_{nn}\leq c_{6}f^{-1}(1+R)|\langle\nabla G,\nabla
f\rangle|+O(f^{-1}R(1+R)).$ (3.14)

Since

$\displaystyle H=\frac{\Delta f-f_{nn}}{|\nabla
f|}=\frac{\frac{n-1}{2}-R+R_{nn}}{|\nabla
f|}=\frac{\frac{n-1}{2}-R+O(f^{-1}R(1+R))}{|\nabla f|},$

we have

$\displaystyle HG_{n}=O(f^{-1}(1+R)|\langle\nabla G,\nabla f\rangle|).$ (3.15)

Moreover,

$\displaystyle\langle\nabla G,\nabla\log R\rangle=$
$\displaystyle\langle\nabla G,\nabla\log R\rangle_{\Sigma}+\frac{1}{|\nabla
f|^{2}}\langle\nabla G,\nabla f\rangle\langle\nabla\log R,\nabla f\rangle$
$\displaystyle=$ $\displaystyle\langle\nabla G,\nabla\log
R\rangle_{\Sigma}+O(f^{-1}(1+R))|\langle\nabla G,\nabla f\rangle|$ (3.16)

since $\langle\nabla\log R,\nabla f\rangle=2R^{-1}Rc(\nabla f,\nabla
f)=O(R^{-1}fR_{nn})=O(1+R)$.

Combining (3.13), (3.14), (3.15) and (3.16), we derive on
$\Sigma(t_{0},s_{0})$ that

$\displaystyle\Delta_{\Sigma}G\geq\langle\nabla G,\nabla f\rangle-
c_{7}f^{-1}(1+R)|\langle\nabla G,\nabla f\rangle|-2\langle\nabla G,\nabla\log
R\rangle_{\Sigma}+c_{8}RG-c_{7}f^{-1}R(1+R).$ (3.17)

By our assumption, there exists a point $z\in\Sigma(s_{1})$ such that

$\displaystyle
G(z)=\max_{\Sigma(s_{1})}G=\max_{\Sigma(t_{0},s_{0})}G\in[\frac{\tau}{2}\epsilon_{0}^{2},2\tau\epsilon_{0}^{2}].$

From (3.17) and the maximum principle, we conclude at $z$ that

$\displaystyle 0\geq\langle\nabla G,\nabla f\rangle-
c_{7}f^{-1}(1+R)|\langle\nabla G,\nabla f\rangle|+c_{8}RG-c_{7}f^{-1}R(1+R).$

Since $\langle\nabla G,\nabla f\rangle\geq 0$ at $z$, we have

$\displaystyle G\leq c_{9}f^{-1}(1+R)$

at $z$. Therefore, it follows from (3.8) that

$\displaystyle\frac{\tau}{2}\epsilon_{0}^{2}\leq G(z)\leq c_{9}f^{-1}(1+R)\leq
c_{10}\epsilon_{0}^{2},$

where for the last inequality we have used $t_{0}\geq n\epsilon_{1}^{-1}$ and
hence $t_{0}^{-1}=O(\epsilon_{0}^{2})$. However, it contradicts our assumption
($\dagger$ ‣ 3) if $\tau$ is sufficiently large. ∎

######  Remark 3.5.

From the proof of Proposition 3.4, we have shown that on $E$

$\displaystyle|\overset{\circ}{Rm_{\Sigma}}|\leq C\epsilon_{0}R$

for some constant $C$ depending only on $n$ and $B$.

Next, we prove

######  Lemma 3.6.

With the same assumptions above, we have on $E$,

$\displaystyle\begin{dcases}&|\overset{\circ}{Rc_{\Sigma}}|=O(\epsilon_{0}R),\\\
&|\nabla\overset{\circ}{Rc_{\Sigma}}|=O(\epsilon_{0}^{\frac{1}{2}}R^{\frac{3}{2}}),\\\
&|\nabla^{2}\overset{\circ}{Rc_{\Sigma}}|=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).\end{dcases}$
(3.18)

###### Proof.

In the proof, all constants $C_{i},i=1,2,\cdots$ are positive and depend only
on $n,A$ and $B$.

From the definition of $\overset{\circ}{Rc_{\Sigma}}$, we obtain

$\displaystyle\overset{\circ}{R}_{ac}=\sum_{k=1}^{n-1}\overset{\circ}{R}_{akck}+R_{ancn}.$

Therefore, it follows from (3.5) and (3.6) that

$\displaystyle|\overset{\circ}{Rc_{\Sigma}}|=O(\epsilon_{0}R)+O(f^{-1}R(1+R))=O(\epsilon_{0}R).$
(3.19)

In addition, we have

$\displaystyle\begin{dcases}&|\nabla\overset{\circ}{Rc_{\Sigma}}|=O(R^{\frac{3}{2}}),\\\
&|\nabla^{2}\overset{\circ}{Rc_{\Sigma}}|=O(R^{2}),\\\
&|\nabla^{3}\overset{\circ}{Rc_{\Sigma}}|=O(R^{\frac{5}{2}}).\end{dcases}$
(3.20)

Indeed, it follows from (2.10) and ($\dagger$ ‣ 3) that

$\displaystyle|\nabla\overset{\circ}{Rc_{\Sigma}}|=$ $\displaystyle O(|\nabla
Rc|)+O(f^{-1}|\nabla
R|(1+R))+O(f^{-\frac{1}{2}}|\nabla^{2}R|)+O(f^{-\frac{1}{2}}|Rc|(1+R))$
$\displaystyle=$ $\displaystyle
O(R^{\frac{3}{2}})+O(f^{-1}R^{\frac{3}{2}}(1+R))+O(f^{-\frac{1}{2}}R^{2})+O(f^{-\frac{1}{2}}R(1+R))=O(R^{\frac{3}{2}}).$

Here, we have used (3.3). The other two equalities in (3.20) can be proved
similarly by using $|\nabla^{i}Rm|\leq BR^{\frac{i}{2}+1}$ for $0\leq i\leq
4$.

For any $q\in E$, we set $R(q)=r^{-2}$ and $r_{1}=C_{1}^{-1}r$ for a large
constant $C_{1}>1$ determined later.

Case 1: $B(q,2r_{1})\cap\Sigma(s)=\emptyset$.

From the fact that $|\nabla R|^{2}=O(R^{3})$, we may assume that $C_{1}$ is
large enough so that

$\displaystyle C_{1}^{-1}r^{-2}\leq R\leq C_{1}r^{-2}$

on $B(q,2r_{1})$. It follows from Theorem 3.1 and the Bishop-Gromov volume
comparison that

$\displaystyle C_{2}^{-1}\leq\frac{|B(q,r_{1})|}{r_{1}^{n}}\leq C_{2}$

for some $C_{2}>1$. By the standard interpolation inequality (see [30, Theorem
$7.28$]) that for any $\epsilon>0$

$\displaystyle\left(\int_{B(q,r_{1})}|\nabla\overset{\circ}{Rc_{\Sigma}}|^{2}r_{1}^{6-n}\,dV\right)^{\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\epsilon\left(\int_{B(q,r_{1})}|\overset{\circ}{Rc_{\Sigma}}|^{2}r_{1}^{4-n}\,dV+\int_{B(q,r_{1})}|\nabla\overset{\circ}{Rc_{\Sigma}}|^{2}r_{1}^{6-n}\,dV+\int_{B(q,r_{1})}|\nabla^{2}\overset{\circ}{Rc_{\Sigma}}|^{2}r_{1}^{8-n}\,dV\right)^{\frac{1}{2}}$
$\displaystyle+C_{3}\epsilon^{-1}\left(\int_{B(q,r_{1})}|\overset{\circ}{Rc_{\Sigma}}|^{2}r_{1}^{4-n}\,dV\right)^{\frac{1}{2}}\leq
C_{4}\epsilon+C_{4}\epsilon^{-1}\epsilon_{0},$

where we have used (3.19) and (3.20). Plugging
$\epsilon=\epsilon_{0}^{\frac{1}{2}}$ into the above inequality, we obtain

$\displaystyle\int_{B(q,r_{1})}|\nabla\overset{\circ}{Rc_{\Sigma}}|^{2}\,dV\leq
2C_{4}\epsilon_{0}r_{1}^{n-6}.$ (3.21)

Since $|\nabla^{2}\overset{\circ}{Rc_{\Sigma}}|\leq C_{5}r_{1}^{-4}$ by
(3.20), it follows from mean value formula and (3.21) that

$\displaystyle|\nabla\overset{\circ}{Rc_{\Sigma}}|\leq
C_{6}\epsilon_{0}^{\frac{1}{2}}r_{1}^{-3}.$

In other words, we have proved that
$|\nabla\overset{\circ}{Rc_{\Sigma}}|=O(\epsilon_{0}^{\frac{1}{2}}R^{\frac{3}{2}})$.
The estimate of $|\nabla^{2}\overset{\circ}{Rc_{\Sigma}}|$ can be proved
similarly.

Case 2: $B(q,2r_{1})\cap\Sigma(s)\neq\emptyset$.

In this case, there exists $q^{\prime}\in\Sigma(s)$ which can be connected by
a shortest geodesic in $E$ from $q$ to $q^{\prime}$ such that its length is
smaller than $2r_{1}$. From $|\nabla R|^{2}=O(R^{3})$, $R(q^{\prime})$ is
almost equal to $r^{-2}$, if $C_{1}$ is sufficiently large.

Next, we denote the induced metric on $\Sigma(s)$ by $g_{\Sigma}$ and define
$\delta_{1}>0$ to be the largest number such that $R\leq 2r^{-2}$ on
$\Omega:=B_{g_{\Sigma}}(q^{\prime},\delta_{1}r)$. In addition, we define
$\rho=2\sqrt{f}$ and $\phi_{t}$ to be the family of diffeomorphisms generated
by $\nabla\rho/|\nabla\rho|^{2}$.

Now we define a map

$\displaystyle\Phi:\Omega\times(-\delta_{2}r,0]\longrightarrow E$

by $\Phi(z,t)=\phi_{t}(z)$. We claim that there exists $C_{7}>1$ such that
$\delta_{2}\geq C_{7}^{-1}$ and $\Phi^{*}g$ is close to
$\bar{g}:=g_{\Sigma}+dt^{2}$ in $C^{2}$ sense. More precisely,

$\displaystyle|\Phi^{*}g-\bar{g}|_{C^{k}(\bar{g})}=O(\epsilon^{\frac{1}{2}}_{1}r^{-k})$
(3.22)

for $k=0,1,2$. We only prove the case $k=0$, and the other cases are similar.
We first compute for any $t\in(-\delta_{2}r,0]$,

$\displaystyle\Phi^{*}g(\partial_{t},\partial_{t})=\frac{1}{|\nabla\rho|^{2}}=\frac{f}{|\nabla
f|^{2}}=1+\frac{R}{f-R}=1+O(\epsilon_{1}).$ (3.23)

Next, we set $g(t)$ to be $\Phi^{*}g$ restricted on $\Omega\times\\{t\\}$.
From our definition of $\phi_{t}$, we compute

$\displaystyle\partial_{t}g(t)=\frac{2\sqrt{f}}{|\nabla
f|^{2}}\nabla^{2}f=\frac{2\sqrt{f}}{|\nabla
f|^{2}}(g(t)-2Rc)=O(f^{-\frac{1}{2}}R).$

By integration, we have for any $t\in(-\delta_{2}r,0]$,

$\displaystyle|g(t)-g_{\Sigma}|=O(\delta_{2}rf^{-\frac{1}{2}}R)=O(\epsilon^{\frac{1}{2}}_{1}),$
(3.24)

where for the last equality we have used $|\nabla R|^{2}=O(R^{3})$ and (3.23).
Combining (3.23) and (3.24), (3.22) is proved.

From (3.22), it follows that $g$ is almost a product metric on a neighborhood
of $\Sigma(s)$. By $|\nabla R|^{2}=O(R^{3})$ again, there exists $C_{8}>1$
such that $\delta_{1}\geq C_{8}^{-1}$. Moreover, we can guarantee that
$B(q,2r_{1})\cap E\subset\Phi\left(\Omega\times(-\delta_{2}r,0]\right)$.

Therefore, the rest of the proof can be done similarly to Case 1 by
considering a product neighborhood of $q$ with size $r_{1}$ instead. ∎

Next, we prove that the scalar curvature is almost constant on any level set
contained in $E$. We first prove

######  Lemma 3.7.

With the same assumptions above, on any level set $\Sigma$ of $f$ contained in
$E$, we have

$\displaystyle|\nabla_{\Sigma}R|=O(\epsilon_{0}^{\frac{1}{2}}R^{\frac{3}{2}}).$

###### Proof.

Recall the decomposition of $Rc$ from (2.10). Direct calculation implies

$\displaystyle\frac{1}{2}\nabla_{a}R=$
$\displaystyle\sum_{b=1}^{n-1}\nabla_{b}R_{ab}+\nabla_{n}R_{an}$
$\displaystyle=$
$\displaystyle\sum_{b=1}^{n-1}\nabla_{b}\left(\overset{\circ}{R}_{ab}+\frac{R}{n-1}g_{ab}+T_{ab}\right)+\nabla_{n}R_{an}$
$\displaystyle=$
$\displaystyle\sum_{b=1}^{n-1}\left(\nabla_{b}\overset{\circ}{R}_{ab}+\nabla_{b}T_{ab}\right)+\frac{\nabla_{a}R}{n-1}+\nabla_{n}R_{an}.$

Therefore, we obtain

$\displaystyle\frac{n-3}{2(n-1)}\nabla_{a}R=\sum_{b=1}^{n-1}\left(\nabla_{b}\overset{\circ}{R}_{ab}+\nabla_{b}T_{ab}\right)+\nabla_{n}R_{an}.$
(3.25)

It is clear from the definition that

$\displaystyle\nabla_{b}T_{ab}=\frac{f_{ab}\nabla_{b}R}{|\nabla
f|^{2}}=\frac{(g_{ab}/2-R_{ab})\nabla_{b}R}{|\nabla f|^{2}}.$

Since
$R_{ab}=\overset{\circ}{R}_{ab}+\frac{R}{n-1}g_{ab}=\frac{R}{n-1}g_{ab}+O(\epsilon_{0}R)$,
we have

$\displaystyle\sum_{b=1}^{n-1}\nabla_{b}T_{ab}=$
$\displaystyle\frac{1}{|\nabla
f|^{2}}\left(\frac{\nabla_{a}R}{2}-\frac{R\nabla_{a}R}{n-1}+O(\epsilon_{0}R^{\frac{5}{2}})\right)$
$\displaystyle=$ $\displaystyle
O(f^{-1}R^{\frac{3}{2}}(1+R))+O(\epsilon_{0}f^{-1}R^{\frac{5}{2}})=O(\epsilon_{0}R^{\frac{3}{2}}).$
(3.26)

Moreover, we compute

$\displaystyle\nabla_{n}R_{an}=$ $\displaystyle\frac{\langle\nabla
R_{an},\nabla f\rangle}{|\nabla f|}=\frac{\Delta
R_{an}-R_{an}+2R_{aknl}R_{kl}}{|\nabla f|}$ $\displaystyle=$ $\displaystyle
O(f^{-\frac{1}{2}}R^{2})+O(f^{-1}R^{\frac{3}{2}})=O(\epsilon_{0}R^{\frac{3}{2}}).$
(3.27)

Combining (3.25),(3.26), (3.27) and Lemma 3.6, we immediately obtain

$\displaystyle|\nabla_{a}R|=O(\epsilon_{0}^{\frac{1}{2}}R^{\frac{3}{2}}).$

∎

######  Lemma 3.8.

With the same assumptions above, for any level set $\Sigma$ contained in $E$
and $x,y\in\Sigma$, we have

$\displaystyle\left|\frac{R(x)}{R(y)}-1\right|=O(\epsilon_{0}^{\frac{1}{2}}).$

Moreover, the diameter of $\Sigma$ is $O(R^{-\frac{1}{2}}(x))$.

###### Proof.

For the second fundamental form $h$ of $\Sigma$, we have

$\displaystyle h_{ab}=\frac{f_{ab}}{|\nabla
f|}=\frac{\frac{g_{ab}}{2}-R_{ab}}{|\nabla f|}=O(f^{-\frac{1}{2}}(1+R))$

and hence the mean curvature

$\displaystyle H=\sum h_{aa}=O(f^{-\frac{1}{2}}(1+R)).$

We denote the intrinsic Ricci curvature and scalar curvature of $\Sigma$ by
$\tilde{R}_{ab}$ and $\tilde{R}$, respectively. It follows from the Gauss
curvature equation that

$\displaystyle\tilde{R}_{ab}=$ $\displaystyle
R_{ab}-R_{anbn}+Hh_{ab}-h_{ac}h_{cb}$ $\displaystyle=$ $\displaystyle
R_{ab}+O(f^{-1}R(1+R))+O(f^{-1}(1+R)^{2})=R_{ab}+O(\epsilon_{0}^{2}R)$ (3.28)

where we have used (3.3) and (3.5).

Similarly,

$\displaystyle\tilde{R}=$ $\displaystyle R-R_{nn}+H^{2}-|h|^{2}$
$\displaystyle=$ $\displaystyle R+O(\epsilon_{0}^{2}R).$ (3.29)

Combining (3.28), (3.29) and Lemma 3.6, we have

$\displaystyle\tilde{R}_{ab}=$
$\displaystyle\frac{R}{n-1}g_{ab}+\overset{\circ}{R_{ab}}+O(\epsilon_{0}^{2}R)$
$\displaystyle=$
$\displaystyle\frac{R}{n-1}g_{ab}+O(\epsilon_{0}R)=\frac{\tilde{R}}{n-1}g_{ab}+O(\epsilon_{0}\tilde{R}).$
(3.30)

Now we fix a point $x\in\Sigma$ and set $R(x)=r^{-2}$. By Lemma 3.7, we have
$|\nabla_{\Sigma}R^{-\frac{1}{2}}|\leq C_{1}\epsilon_{0}^{\frac{1}{2}}$ and
hence

$\displaystyle|R^{-\frac{1}{2}}(y)-r|\leq C_{1}L\epsilon_{0}^{\frac{1}{2}},$
(3.31)

provided that $d_{\Sigma}(x,y)\leq L$. From (3.29), (3.30) and (3.31), the
proof of the Myers theorem indicates that

$\displaystyle\text{diam}\,\Sigma=O(r)$

and hence by (3.31),

$\displaystyle R(y)=\left(1+O(\epsilon_{0}^{\frac{1}{2}})\right)r^{-2}.$

In sum, the proof is complete. ∎

For later applications, we prove

######  Lemma 3.9.

With the same assumptions above, we have on $E$,

$\displaystyle|\Delta R|=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).$

###### Proof.

It follows from definition that

$\displaystyle\Delta R=\sum_{a=1}^{n-1}\nabla^{2}_{aa}R+\nabla^{2}_{nn}R.$
(3.32)

We shall estimate the different terms of the right-hand side of (3.32)
separately.

Firstly, it is clear that

$\displaystyle\nabla^{2}_{nn}R=$ $\displaystyle\frac{\nabla^{2}R(\nabla
f,\nabla f)}{|\nabla f|^{2}}$ $\displaystyle=$ $\displaystyle\frac{1}{|\nabla
f|}\nabla_{n}\langle\nabla R,\nabla f\rangle-\frac{1}{|\nabla
f|^{2}}\nabla^{2}f(\nabla R,\nabla f)$ $\displaystyle=$
$\displaystyle\frac{1}{|\nabla f|}\nabla_{n}\left(\Delta
R-R+2|Rc|^{2}\right)-\frac{1}{|\nabla f|^{2}}\left(\frac{\langle\nabla
R,\nabla f\rangle}{2}-Rc(\nabla R,\nabla f)\right)$ $\displaystyle=$
$\displaystyle
O(f^{-\frac{1}{2}}R^{\frac{3}{2}}(1+R))+O(f^{-1}R(1+R))=O(\epsilon_{0}R^{2}).$
(3.33)

Secondly, we move on to estimate $\nabla^{2}_{aa}R$. It follows from (3.25)
that

$\displaystyle\frac{n-3}{2(n-1)}\nabla_{ca}^{2}R=\sum_{b=1}^{n-1}\left(\nabla_{c}\nabla_{b}\overset{\circ}{R}_{ab}-\nabla_{c}\nabla_{b}T_{ab}\right)+\nabla_{c}\nabla_{n}R_{an}.$
(3.34)

In light of (3.20), we have

$\displaystyle|\nabla_{c}\nabla_{b}\overset{\circ}{R}_{ab}|=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).$
(3.35)

Recall that $T$ is defined in (2.11). Direct calculation implies that

$\displaystyle|\nabla_{c}\nabla_{b}T_{ab}|=$ $\displaystyle O(|\nabla
f|^{-2}|\nabla R||\nabla^{3}f|)+O(|\nabla
f|^{-2}|\nabla^{2}R||\nabla^{2}f|)+O(|\nabla f|^{-2}|Rc||\nabla^{2}f|^{2})$
$\displaystyle=$ $\displaystyle O(f^{-1}|\nabla R||\nabla
Rc|)+O(f^{-1}|\nabla^{2}R|(1+R))+O(f^{-1}|Rc|(1+R)^{2})$ $\displaystyle=$
$\displaystyle
O(f^{-1}R^{3})+O(f^{-1}R^{2}(1+R))+O(f^{-1}R(1+R)^{2})=O(\epsilon_{0}^{2}R^{2}).$
(3.36)

Here, we have used the fact that some components of $\nabla^{2}T$ along the
normal direction vanish. Furthermore, we compute

$\displaystyle\nabla_{c}\nabla_{n}R_{an}=$
$\displaystyle\nabla_{c}\left(\frac{\langle\nabla R_{an},\nabla
f\rangle}{|\nabla f|}\right)=\nabla_{c}\left(\frac{\Delta
R_{an}-R_{an}+2R_{aknl}R_{kl}}{|\nabla f|}\right)$ $\displaystyle=$
$\displaystyle O(|\nabla f|^{-1}(|\nabla^{3}Rc|+|\nabla Rc|+|\nabla
Rm||Rm|))+O(|\nabla f|^{-2}|\nabla^{2}f|(R+R^{2}))$ $\displaystyle=$
$\displaystyle
O(f^{-\frac{1}{2}}R^{\frac{3}{2}}(1+R))+O(f^{-1}R(1+R)^{2})=O(\epsilon_{0}R^{2}).$
(3.37)

Combining (3.34), (3.35), (3.36) and (3.37), we obtain

$\displaystyle\nabla^{2}_{ca}R=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).$ (3.38)

Finally, it is clear that (3.32) follows from the combination of (3.33) and
(3.38). The proof of the lemma is complete. ∎

For later applications, we prove

######  Lemma 3.10.

With the same assumptions above, we have on $E$,

$\displaystyle|\Delta R_{ab}|=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).$

###### Proof.

By definition,

$\displaystyle\Delta
R_{ab}=\Delta\left(\overset{\circ}{R}_{ab}+\frac{R}{n-1}g_{ab}-T_{ab}.\right)$
(3.39)

It is clear from (3.18) and Lemma 3.9 that

$\displaystyle\Delta\left(\overset{\circ}{R}_{ab}+\frac{R}{n-1}g_{ab}\right)=O(\epsilon_{0}^{\frac{1}{3}}R^{2}).$

On the other hand, one can compute as (3.36) that $|\Delta
T_{ab}|=O(\epsilon_{0}^{2}R^{2})$. Now the conclusion follows from (3.39). ∎

Now we consider a family of diffeomorphisms $\varphi_{t}$ defined by

$\displaystyle\begin{dcases}&\frac{d\varphi_{t}}{dt}=\frac{\nabla f}{|\nabla
f|^{2}},\\\
&\varphi_{t_{0}}=\text{id}\quad\text{on}\quad\Sigma(t_{0}).\end{dcases}$
(3.40)

By abuse of notation, we set $R=R(t)=R(\varphi_{t}(x))$ for
$x\in\Sigma(t_{0})$. Direct calculation yields that

$\displaystyle R_{t}=\frac{\langle\nabla R,\nabla f\rangle}{|\nabla
f|^{2}}=\frac{\Delta R-R+2|Rc|^{2}}{t-R}.$ (3.41)

On the one hand, we compute on $E$,

$\displaystyle|Rc|^{2}=$
$\displaystyle\sum_{a,b=1}^{n-1}R_{ab}^{2}+2\sum_{a=1}^{n-1}R^{2}_{an}+R^{2}_{nn}$
$\displaystyle=$
$\displaystyle\sum_{a,b=1}^{n-1}\left(\overset{\circ}{R}_{ab}+\frac{R}{n-1}g_{ab}\right)^{2}+2\sum_{a=1}^{n-1}R^{2}_{an}+R^{2}_{nn}$
$\displaystyle=$
$\displaystyle\frac{R^{2}}{n-1}+O(\epsilon_{0}R^{2})+O(f^{-1}R^{3})+O(f^{-2}R^{2}(1+R)^{2})$
$\displaystyle=$ $\displaystyle\frac{R^{2}}{n-1}+O(\epsilon_{0}R^{2}),$ (3.42)

where we have used (3.4) and (3.5).

On the other hand, $\Delta R=O(\epsilon_{0}^{\frac{1}{3}}R^{2})$ from Lemma
3.9. Combining this with (3.42), we obtain from (3.41) that

$\displaystyle tR_{t}=\left(\frac{2}{n-1}+X\right)R^{2}-R$ (3.43)

where $X$ is a function such that

$\displaystyle
X=\frac{O(\epsilon_{0}^{\frac{1}{3}})t-1+\frac{2}{n-1}R}{t-R}=O(\epsilon_{0}^{\frac{1}{3}}).$
(3.44)

Now we define

$\displaystyle\epsilon_{2}\coloneqq\epsilon_{0}^{\frac{1}{4}}$ (3.45)

and analyze different cases depending on whether $R$ stays in the interval
$[\frac{n-1}{2}-\epsilon_{2},\frac{n-1}{2}+\epsilon_{2}]$. Notice that by the
assumption ($\dagger$ ‣ 3), on $\Sigma(t_{0})$, $R$ stays in this interval.

### Cylindrical end

Without loss of generality, we assume that $s$ is the largest number such that
$E=\Sigma(t_{0},s)$ satisfies ($\dagger$ ‣ 3). Throughout this subsection, we
assume on $E$

$\displaystyle\left|R-\frac{n-1}{2}\right|<\epsilon_{2}.$ (3.46)

######  Proposition 3.11.

Under the assumption (3.46), there exists
$\sigma_{2}=\sigma_{2}(n,A,B,\delta_{0})>0$ such that if
$\epsilon_{0}\leq\sigma_{2}$, then $s=\infty$.

###### Proof.

We assume $s<\infty$ and derive a contradiction. By the definition of $s$,
there exists a point $q\in\Sigma(s)$ such that at $q$ either

$\displaystyle
R=\epsilon_{1}f\quad\text{or}\quad|\nabla^{k}Rm|=BR^{\frac{k}{2}+1}$

for some $0\leq k\leq 4$. We claim that the first case cannot happen. Indeed,
by (3.46) that

$\displaystyle\frac{n-1}{2}+\epsilon_{2}\geq
R(q)=\epsilon_{1}f(q)\geq\epsilon_{1}t_{0}\geq n$

which is impossible.

To exclude the second case, we assume there exists a sequence of Ricci
shrinkers $(M^{n}_{i},g_{i},f_{i})$ with the end $E_{i}=\Sigma(t_{0,i},s_{i})$
satisfying ($\dagger$ ‣ 3) and $\epsilon_{0,i}\to 0$. Moreover, there exists a
$q_{i}\in\Sigma(s_{i})$ at which

$\displaystyle|\nabla^{k}Rm_{i}|=BR_{i}^{\frac{k}{2}+1}$ (3.47)

for a fixed $0\leq k\leq 4$.

Now we fix a small number $\delta<\delta_{0}/2$ to be determined later and
define $x_{i}$ to be a point such that $\psi_{i}^{\delta}(x_{i})=q_{i}$, where
$\psi^{t}$ is a family of diffeomorphisms defined in (2.12). We claim that
such $x_{i}$ must exist and lie in $E_{i}$. Indeed, from (2.12) we have for
any $t\geq 0$ that

$\displaystyle\frac{df_{i}(\psi_{i}^{t}(x_{i}))}{dt}=\frac{|\nabla
f_{i}|^{2}(\psi^{t}_{i}(x_{i}))}{1-t}=\frac{f_{i}(\psi^{t}_{i}(x_{i}))-R_{i}(\psi^{t}_{i}(x_{i}))}{1-t}\leq\frac{f_{i}(\psi^{t}_{i}(x_{i}))}{1-t}.$

Therefore,

$\displaystyle t_{i}:=f_{i}(x_{i})\geq s_{i}(1-\delta).$ (3.48)

In particular, $x_{i}\in E_{i}$ since $(1-\delta_{0})s_{i}\geq t_{0,i}$ by our
assumption ($\dagger$ ‣ 3). On the other hand, by (3.46) we have for any
$0\leq t\leq\delta$,

$\displaystyle\frac{df_{i}(\psi_{i}^{t}(x_{i}))}{dt}=\frac{f_{i}(\psi^{t}_{i}(x_{i}))-R_{i}(\psi^{t}_{i}(x_{i}))}{1-t}\geq\frac{f_{i}(\psi^{t}_{i}(x_{i}))-n}{1-t}.$

By solving the above ODE, we obtain

$\displaystyle t_{i}\leq(s_{i}-n)(1-\delta)+n=s_{i}(1-\delta)+n\delta.$ (3.49)

Claim.

$\displaystyle(M_{i},x_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(S^{n-1}\times\mathbb{R},p_{c},g_{c}).$

_Proof of the Claim:_ From (3.48) and (3.49), for any $L>1$ we have
$B_{g_{i}}(x_{i},L)\subset E_{i}$, if $i$ is sufficiently large. Indeed, since
$2\sqrt{f_{i}}$ is $1$-Lipschitz, we have for any $y\in B_{g_{i}}(x_{i},L)$,

$\displaystyle|2\sqrt{f_{i}(y)}-2\sqrt{t_{i}}|=|2\sqrt{f_{i}(y)}-2\sqrt{f_{i}(x_{i})}|\leq
L.$ (3.50)

Since $\lim_{i\to\infty}t_{i}=+\infty$ by (3.48), we obtain

$\displaystyle\lim_{i\to\infty}\frac{f_{i}(y)}{t_{i}}=1$ (3.51)

uniformly for $y\in B_{g_{i}}(x_{i},L)$. By our assumptions ($\dagger$ ‣ 3)
and (3.46), the Riemannian curvature $|Rm_{i}|$ is uniformly bounded on
$E_{i}$. Moreover, one can obtain the higher order estimates of $Rm_{i}$ on
$B_{g_{i}}(x_{i},L)$. Indeed, we take $\delta_{1}:=\delta_{0}/10$ and for any
$(y,t)\in B_{g_{i}}(x_{i},L)\times[-\delta_{1},0]$, we compute

$\displaystyle|Rm_{i}|(y,t)=\frac{1}{1-t}|Rm_{i}|(\psi_{i}^{t}(y),0).$ (3.52)

Similar to (3.48), if $i$ is sufficiently large, we obtain

$\displaystyle f_{i}(y)\geq
f_{i}(\psi_{i}^{t}(y))\geq\frac{f_{i}(y)}{1+\delta_{1}}\geq t_{0,i}.$

In other words, $\psi_{i}^{t}(y)\in E_{i}$. Therefore, it is clear from (3.52)
that $|Rm_{i}|$ is uniformly bounded on
$B_{g_{i}}(x_{i},L)\times[-\delta_{1},0]$. Now the higher order estimates of
$|Rm_{i}|$ follow from Shi’s local estimates [63].

From the above estimates of the curvature and Theorem 3.1, we conclude that

$\displaystyle(M^{n}_{i},x_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M^{n}_{\infty},x_{\infty},g_{\infty}),$

where the limit is a complete smooth Riemannian manifold with bounded
curvature. On the other hand, if we define
$\tilde{f}_{i}:=t_{i}^{-\frac{1}{2}}(f_{i}-t_{i})$, then

$\displaystyle|\nabla_{g_{i}}\tilde{f}_{i}|^{2}_{g_{i}}=\frac{|\nabla_{g_{i}}f_{i}|_{g_{i}}^{2}}{t_{i}}=\frac{f_{i}-R_{i}}{t_{i}}.$
(3.53)

In addition, we obtain

$\displaystyle\text{Hess}_{g_{i}}\tilde{f}_{i}=t_{i}^{-\frac{1}{2}}\text{Hess}_{g_{i}}f_{i}=\frac{1}{2}t_{i}^{-\frac{1}{2}}g_{i}-t_{i}^{-\frac{1}{2}}Rc_{i}.$
(3.54)

From the definition of $\tilde{f}_{i}$ and (3.50), we compute for any $y\in
B_{g_{i}}(x_{i},L)$ that

$\displaystyle\tilde{f}_{i}(y)=t_{i}^{-\frac{1}{2}}(f_{i}(y)-t_{i})=t_{i}^{-\frac{1}{2}}(\sqrt{f_{i}(y)}-\sqrt{t_{i}})(\sqrt{f_{i}(y)}+\sqrt{t_{i}})\leq\frac{3}{2}L.$
(3.55)

Therefore, it follows from (3.53), (3.54), (3.55) and the standard regularity
theorem for elliptic equations that $\tilde{f}_{i}$ converges smoothly to a
smooth function $f_{\infty}$ on $M_{\infty}$. Moreover, we conclude from
(3.54) that

$\displaystyle\text{Hess}_{g_{\infty}}f_{\infty}=0.$ (3.56)

In addition, $f_{\infty}$ is not a constant since by (3.46) and (3.53),

$\displaystyle|\nabla_{g_{\infty}}f_{\infty}|^{2}=1.$

From (3.56), we conclude that $(M_{\infty},g_{\infty})$ is isometric to
$(\Sigma_{\infty}\times\mathbb{R},g^{\prime}_{\infty}\times g_{E})$, where
$\Sigma_{\infty}$ is the level set of $f_{\infty}$. Since the second
fundamental form of $\Sigma_{\infty}$ vanishes, it follows from (3.6) that the
sectional curvature of $g^{\prime}_{\infty}$ on $\Sigma_{\infty}$ is
identically $\frac{1}{2(n-2)}$. Therefore, it is clear that
$(M_{\infty},g_{\infty})$ is isometric to $(S^{n-1}\times\mathbb{R},g_{c})$
and Claim is proved.

Now we fix the parameter $\delta$ small enough such that the pseudolocality
Theorem [48, Theorem 24] can be applied. Combining with [17, Theorem $3.1$],
we conclude that for any $L>0$,

$\displaystyle|Rm_{i}|\leq C(n)$

on $B_{g_{i}}(x_{i},L)\times[0,\delta]$, if $i$ is sufficiently large.
Therefore, we conclude that

$\displaystyle(M_{i},x_{i},g_{i}(t))_{t\in[0,\delta]}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M_{\infty},x_{\infty},g_{\infty}(t))_{t\in[0,\delta]}$

such that $(M_{\infty},g_{\infty}(t))$ has unifomrly bounded curvature. From
the uniqueness of the Ricci flow with bounded curvature [25] and the Claim, we
have

$\displaystyle(M_{\infty},x_{\infty},g_{\infty}(t))_{t\in[0,\delta]}=(S^{n-1}\times\mathbb{R},p_{c},g_{c}(t))_{t\in[0,\delta]}.$

Therefore, for $0\leq k\leq 4$,

$\displaystyle\lim_{i\to\infty}\frac{|\nabla^{k}Rm_{i}|}{R^{\frac{k}{2}+1}}(q_{i},0)=\lim_{i\to\infty}\frac{|\nabla^{k}Rm_{i}|}{R^{\frac{k}{2}+1}}(x_{i},\delta)=\frac{|\nabla^{k}Rm_{g_{c}}|}{R_{g_{c}}^{\frac{k}{2}+1}}=\begin{cases}\sqrt{\frac{2}{(n-1)(n-2)}}\quad&\text{if}\quad
k=0,\\\ 0\quad&\text{if}\quad k>0.\end{cases}$

However, this contradicts (3.47) since $B\geq 1$. In sum, we obtain a
contradiction, and the proof is complete. ∎

From Proposition 3.11, $E$ is an end of the Ricci shrinker. Next, we show that
any point in $E$ far away from $\Sigma(t_{0})$ is the center of an evolving
normalized $\epsilon$-neck.

######  Proposition 3.12.

With the above assumptions ($\dagger$ ‣ 3) and (3.46), for any $\epsilon>0$,
there exists a constant $\eta_{3}=\eta_{3}(\epsilon,n,\delta_{0},A,B)>0$ such
that if $\epsilon_{0}\leq\eta_{3}$, then any point
$x\in\Sigma(\eta_{3}^{-1}t_{0},\infty)\subset E$ is the center of an evolving
normalized $\epsilon$-neck.

###### Proof.

We prove it by contradiction. Suppose for fixed $\epsilon$ the conclusion does
not hold. Then there exists a sequence of Ricci shrinkers
$(M^{n}_{i},p_{i},g_{i},f_{i})$ satisfying all the assumptions with
$\epsilon_{0,i}\to 0$ and $x_{i}\in\Sigma(i^{2}t_{0,i},\infty)$ which is not
the center of an evolving normalized $\epsilon$-neck. By the identical proof
as that of Proposition 3.11, we conclude that

$\displaystyle(M_{i},x_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(S^{n-1}\times\mathbb{R},p_{c},g_{c}).$

For a fixed number $L>0$, we consider any spacetime point $(y,\bar{t})\in
B_{g_{i}}(x_{i},L)\times[-L,0]$. From

$\displaystyle
0\leq\frac{df_{i}(\psi_{i}^{t}(y))}{dt}=\frac{f_{i}(\psi^{t}_{i}(y))-R_{i}(\psi^{t}_{i}(y))}{1-t}\leq\frac{f_{i}(\psi^{t}_{i}(y))}{1-t},$

we have

$\displaystyle f_{i}(y)\geq
f_{i}(\psi_{i}^{\bar{t}}(y))\geq\frac{f_{i}(y)}{1-\bar{t}}\geq\frac{f_{i}(y)}{1+L}.$
(3.57)

On the other hand, if we set $f_{i}(x_{i})=t_{i}$, then similar to (3.51),

$\displaystyle\lim_{i\to\infty}\frac{f_{i}(y)}{t_{i}}=1$ (3.58)

for $y\in B_{g_{i}}(x_{i},L)$ uniformly. In particular, (3.57) and (3.58)
imply $\psi_{i}^{\bar{t}}(y)\in E_{i}$ since $t_{i}\geq i^{2}t_{0,i}$. Now we
compute

$\displaystyle|Rm_{i}|(y,\bar{t})=\frac{|Rm_{i}|(\psi_{i}^{\bar{t}}(y),0)}{1-\bar{t}}$

which is uniformly bounded for $(y,\bar{t})\in
B_{g_{i}}(x_{i},L)\times[-L,0]$. Combining this fact with Theorem 3.1, we
conclude that

$\displaystyle(M_{i},x_{i},g_{i}(t))_{t\leq 0}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M^{n}_{\infty},x_{\infty},g_{\infty}(t))_{t\leq 0}$

such that $(M^{n}_{\infty},g_{\infty}(t))_{t\leq 0}$ has uniformly bounded
curvature. By the same proof as in Proposition 3.11,
$(M^{n}_{\infty},g_{\infty}(0))=(S^{n-1}\times\mathbb{R},g_{c})$. Therefore,
it follows from the backward uniqueness of the Ricci flow [39, Theorem $1.1$]
that $g_{\infty}(t)=g_{c}(t)$ for any $t\leq 0$. However, this implies that
$x_{i}$ is the center of an evolving normalized $\epsilon$-neck for large $i$,
which is a contradiction. ∎

Next, we show that $E$ is asymptotic to the cylinder at some fixed rate
depending only on $n$.

From Proposition 3.12, there exists a large number $t^{\prime}>t_{0}$ such
that on $E^{\prime}:=\Sigma(t^{\prime},\infty)$ we have

$\displaystyle|\nabla^{i}Rm|\leq C(n)R^{\frac{i}{2}+1}$

for any $0\leq i\leq 4$ and Theorem 3.1 holds for some $\kappa$ depending only
on $n$. Therefore, it follows from (3.17) that on $E^{\prime}$ we have

$\displaystyle\Delta_{\Sigma}G\geq\langle\nabla G,\nabla f\rangle-
c_{2}f^{-1}(1+R)|\langle\nabla G,\nabla f\rangle|-2\langle\nabla G,\nabla\log
R\rangle_{\Sigma}+c_{1}RG-c_{2}f^{-1}R(1+R),$

where $G=\frac{|Rm|^{2}}{R^{2}}-\frac{2}{(n-1)(n-2)}$ and $c_{1},c_{2}$ depend
only on $n$. If we set $k(t)=\sup_{\Sigma(t)}G$, then for any $t\geq
t^{\prime}$, there exists a maximum point $z\in\Sigma(t)$ such that at $z$,

$\displaystyle 0\geq\langle\nabla G,\nabla f\rangle-c_{4}t^{-1}|\langle\nabla
G,\nabla f\rangle|+c_{3}Rk-c_{4}Rt^{-1},$ (3.59)

for $c_{3}$ and $c_{4}$ depending on $n$, where we have used the fact that $R$
is uniformly bounded on $E$.

If $k(t)\geq 2c_{4}c_{3}^{-1}t^{-1}$, then by (3.59) we have

$\displaystyle 0\geq\langle\nabla G,\nabla f\rangle-c_{3}t^{-1}|\langle\nabla
G,\nabla f\rangle|+\frac{c_{4}}{2}Rk.$ (3.60)

It is clear from (3.60) that $\langle\nabla G,\nabla f\rangle\leq 0$ at $z$.
Therefore, we obtain

$\displaystyle k^{\prime}(t)\leq\frac{\langle\nabla G,\nabla f\rangle}{|\nabla
f|^{2}}(z)\leq-\frac{c_{4}Rk(t)}{2|\nabla
f|^{2}(1-c_{3}t^{-1})}(z)\leq-\tau_{1}k(t)t^{-1}$ (3.61)

for $\tau_{1}=\tau_{1}(n)\in(0,1)$, where the last inequality follows from
(3.46). If we consider $\max\\{k(t),2c_{4}c_{3}^{-1}t^{-1}\\}$ and (3.61), it
is easy to see that for any $t\geq t^{\prime}$,

$\displaystyle k(t)=O(t^{-\tau_{1}}).$ (3.62)

Combining (3.62) with (3.7), we conclude that

$\displaystyle\frac{|\overset{\circ}{Rm_{\Sigma}}|}{R}=O(t^{-\tau_{2}}),$
(3.63)

on $\Sigma(t)$, where $\tau_{2}=\tau_{1}/2$.

Next, we prove

######  Proposition 3.13.

With the above assumptions, for any sequence $x_{i}\in E$ with
$x_{i}\to\infty$, we have

$\displaystyle(M,x_{i},g)\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}\left(S^{n-1}\times\mathbb{R},p_{c},g_{c}\right).$

###### Proof.

For any sequence $x_{i}\to\infty$ with $t_{i}=f(x_{i})$, we conclude that

$\displaystyle(M_{i},x_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M_{\infty},x_{\infty},g_{\infty}),$

where the limit is a complete smooth Riemannian manifold. Here, the smooth
convergence follows from the same proof as that of the Claim in Proposition
3.11. Moreover, by considering $f_{i}=t_{i}^{-\frac{1}{2}}(f_{i}-t_{i})$, we
can show as in Proposition 3.11 that the limit $(M_{\infty},g_{\infty})$
splits isometrically as
$(\Sigma_{\infty}\times\mathbb{R},g^{\prime}_{\infty}\times g_{E})$. From
(3.63), it is clear that $(\Sigma_{\infty},g^{\prime}_{\infty})$ is isometric
to $(S^{n-1},g_{S})$ up to scaling, where $g_{S}$ is the standard metric on
$S^{n-1}$ such that the scalar curvature is $\frac{n-1}{2}$. In particular, if
the limit $(M_{\infty},g_{\infty})$ is a Ricci shrinker, then
$(\Sigma_{\infty},g^{\prime}_{\infty})$ is exactly isometric to
$(S^{n-1},g_{S})$.

Therefore, we conclude from [54, Proposition 5.2] that for any
$x_{i}\to\infty$

$\displaystyle(M_{i},x_{i},g_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(S^{n-1}\times\mathbb{R},p_{c},g_{c})$

and the proof is complete. ∎

If we denote the induced metric on $\Sigma(t)$ by $\tilde{g}(t)$, then it
follows from (3.63), Proposition 3.13 and the same proof as in [53, Theorem
$4.1$, Page 919-922] that

$\displaystyle\tilde{g}(t)=g_{S}+O(t^{-\tau}),$ (3.64)

where $\tau=\tau(n)>0$. Similar estimates also hold for the higher derivatives
of $\tilde{g}(t)$. Now we define a map

$\displaystyle\Psi:S^{n-1}\times[t_{0},\infty)\longrightarrow E$

by $\Psi(z,t)=\varphi_{t}(z)$. Here $\varphi_{t}$ is defined in (3.40) and we
have identified $S^{n-1}$ with $\Sigma(t_{0})$ by a diffeomorphism.

Therefore, it follows from (3.64) that on $S^{n-1}\times\\{t\\}$,

$\displaystyle|\Psi^{*}g-g_{c}|=O(t^{-\tau}).$

If we consider the estimates of the second fundamental form, then the
estimates for all higher derivatives can also be derived. More precisely, we
have on $S^{n-1}\times\\{t\\}$ that

$\displaystyle|\nabla^{k}_{g_{c}}(\Psi^{*}g)|=O(t^{-\tau_{k}})$

for some $\tau_{k}=\tau_{k}(n)>0$. In other words, the end $E$ is smoothly
asymptotic to $S^{n-1}\times\mathbb{R}$.

######  Remark 3.14.

In _[53, Theorem $1.6$]_, Munteanu and Wang have obtained a similar result
with an extra assumption that the Ricci shrinker considered has uniformly
bounded curvature. This assumption plays an important role in _[53, Theorem
$3.1$]_ for getting the higher-order estimates $|\nabla^{k}Rm|\leq
B_{k}R^{\frac{k}{2}+1}$ for any $k\geq 0$. Notice that we drop the uniformly
bounded curvature condition by applying the pseudolocality theorem (cf. _[48,
Corollary 10.6], [58, Theorem 10.1], [67, Theorem 1.2 ]_).

### Conical end

In this subsection, we assume that (3.46) is violated in one direction and
hence there exists a $t_{1}\in(t_{0},\infty)$ such that

$\displaystyle\left|R-\frac{n-1}{2}\right|<\epsilon_{2}$ (3.65)

on $\Sigma(t_{0},t_{1})$ and

$\displaystyle\min_{\Sigma(t_{1})}R=\frac{n-1}{2}-\epsilon_{2}.$ (3.66)

We define

$\displaystyle E^{1}\coloneqq\Sigma(t_{1},s)\subset E.$ (3.67)

######  Proposition 3.15.

With the assumptions above, there exist positive constants $\hat{c}_{1}$ and
$\hat{c}_{2}$ depending only on $n,A$ and $B$ such that if $\epsilon_{0}$ is
sufficiently small, then

$\displaystyle\frac{1}{\frac{2}{n-1}+\hat{c}_{2}\epsilon_{2}t_{1}^{-1}t}\leq
R(x)\leq\frac{1}{\frac{2}{n-1}+\hat{c}_{1}\epsilon_{2}t_{1}^{-1}t}$ (3.68)

for any $x\in E^{1}$, where $t=f(x)$.

###### Proof.

In the proof, all constants $c_{i},i=1,2,\cdots$ depend only on $n,A$ and $B$.

By our assumption, there exists $x\in\Sigma(t_{1})$ such that

$\displaystyle R(x)=\frac{n-1}{2}-\epsilon_{2}.$

From Lemma 3.8, we may assume on $\Sigma(t_{1})$,

$\displaystyle\frac{n-1}{2}-\frac{3}{2}\epsilon_{2}\leq
R\leq\frac{n-1}{2}-\frac{1}{2}\epsilon_{2}.$ (3.69)

In addition, it follows from (3.43) and (3.44) that

$\displaystyle\left(\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R\leq
tR_{t}\leq\left(\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R$
(3.70)

for some constant $c_{1}>0$. By considering the ODE with the given initial
values in (3.69), we obtain

$\displaystyle\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}+c_{3}\epsilon_{2}t_{1}^{-1}t}\leq
R(t)\leq\frac{1}{\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}+c_{2}\epsilon_{2}t_{1}^{-1}t}$
(3.71)

for any $t\geq t_{1}$. Here, the constants $c_{2}$ and $c_{3}$ are determined
by

$\displaystyle\frac{1}{\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}+c_{2}\epsilon_{2}}=\frac{n-1}{2}-\frac{1}{2}\epsilon_{2}$
(3.72)

and

$\displaystyle\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}+c_{3}\epsilon_{2}}=\frac{n-1}{2}-\frac{3}{2}\epsilon_{2}.$
(3.73)

It is clear from (3.71), (3.72), (3.73) and the definition of $\epsilon_{2}$
that (3.68) holds if $\epsilon_{0}$ is sufficiently small. ∎

Similar to Proposition (3.11), we have the following result.

######  Proposition 3.16.

Under the assumptions ($\dagger$ ‣ 3), (3.65) and (3.66), there exists
$\sigma_{3}=\sigma_{3}(n,\sigma_{0},A,B)>0$ such that if
$\epsilon_{0}\leq\sigma_{3}$, then $s=\infty$.

###### Proof.

We prove it by contradiction. Similar to Proposition (3.11), we assume there
exists a sequence of Ricci shrinkers $(M_{i},g_{i},f_{i})$ with the end
$E_{i}=\Sigma(t_{0,i},s_{i})$ satisfying ($\dagger$ ‣ 3), (3.65) and (3.66)
with $\epsilon_{0,i}\to 0$. Moreover, there exists a point
$q_{i}\in\Sigma(s_{i})$ at which

$\displaystyle|\nabla^{k}Rm_{i}|=BR_{g_{i}}^{\frac{k}{2}+1}$

for a fixed $0\leq k\leq 4$.

Claim 1.

$\displaystyle\liminf_{i\to\infty}\epsilon_{2,i}\frac{s_{i}}{t_{1,i}}>0.$
(3.74)

_Proof of Claim 1:_ Otherwise, it follows from the assumption (3.65) and
(3.68) that

$\displaystyle\limsup_{i\to\infty}\max_{\Sigma(t_{0,i},s_{i})}\left|R_{i}-\frac{n-1}{2}\right|=0.$

Therefore, the same proof of Proposition 3.11 yields a contradiction.

Now we set $r^{-2}_{i}=\frac{2}{n-1}R_{i}(q_{i})$, then it follows from (3.68)
that

$\displaystyle r_{i}=\sqrt{1+\frac{n-1}{2}\sigma_{i}s_{i}}\geq 1$

where

$\displaystyle\hat{c}_{1}\epsilon_{2,i}t_{1,i}^{-1}\leq\sigma_{i}\leq\hat{c}_{2}\epsilon_{2,i}t_{1,i}^{-1}.$

Next we fix a constant $\delta=\delta(n)\in(0,1/20)$ determined later and
prove the following result.

Claim 2. There exists a constant $c_{1}=c_{1}(n)>0$ such that for any
$t\in[-2\delta r_{i}^{2},0]$,

$\displaystyle\frac{s_{i}}{1+|t|}\leq
f_{i}(\psi_{i}^{t}(q_{i}))\leq\frac{s_{i}-c_{1}}{1+|t|}+c_{1}$ (3.75)

_Proof of Claim 2:_ For any $t\in[-2\delta r_{i}^{2},0]$ we have

$\displaystyle\frac{df_{i}(\psi_{i}^{t}(q_{i}))}{dt}=\frac{|\nabla_{g_{i}}f_{i}|_{g_{i}}^{2}}{1-t}(\psi_{i}^{t}(q_{i}))=\frac{f_{i}-R_{i}}{1-t}(\psi_{i}^{t}(q_{i}))$

Since $0\leq R_{i}(\psi_{i}^{t}(q_{i}))\leq c_{1}$, it is clear that

$\displaystyle\frac{f_{i}(\psi_{i}^{t}(q_{i}))-c_{1}}{1-t}\leq\frac{df_{i}(\psi_{i}^{t}(q_{i}))}{dt}\leq\frac{f_{i}(\psi_{i}^{t}(q_{i}))}{1-t}.$

By solving the above ODE, we immediately obtain (3.75).

Now we set $q_{i}^{t}:=\psi_{i}^{t}(q_{i})$ and
$s_{i}^{t}:=f_{i}(\psi_{i}^{t}(q_{i}))$. Moreover, for any fixed number $L>1$
and $t\in[-2\delta r_{i}^{2},0]$, we consider $y\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)$ and obtain

$\displaystyle|2\sqrt{f_{i}(y)}-2\sqrt{s^{t}_{i}}|=|2\sqrt{f_{i}(y)}-2\sqrt{f_{i}(q^{t}_{i})}|\leq\frac{Lr_{i}}{\sqrt{1+|t|}}=L\sqrt{\frac{1+\frac{n-1}{2}\sigma_{i}s_{i}}{1+|t|}}.$

From (3.75) and the fact that $\lim_{i\to\infty}\sigma_{i}=0$, we obtain

$\displaystyle\lim_{i\to\infty}\frac{f_{i}(y)}{s^{t}_{i}}=1$ (3.76)

uniformly for $(y,t)\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)\times[-2\delta
r_{i}^{2},0]$.

In particular, for $(y,t)\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)\times[-2\delta
r_{i}^{2},-L^{-1}r_{i}^{2}]$, we obtain

$\displaystyle s_{i}^{t}\geq\frac{s_{i}}{1+|t|}\geq\frac{s_{i}}{1+2\delta
r_{i}^{2}}\geq t_{1,i},$ (3.77)

where we have used (3.74) for the last inequality. Moreover, since $r_{i}\geq
1$, we have

$\displaystyle
s_{i}^{t}\leq\frac{s_{i}-c_{1}}{1+|t|}+c_{1}\leq\frac{s_{i}-c_{1}}{1+L^{-1}r_{i}^{2}}+c_{1}\leq\frac{s_{i}-c_{1}}{1+L^{-1}}+c_{1}.$

Therefore, it follows from (3.76) that

$\displaystyle t_{1,i}\leq f_{i}(y)\leq s_{i}$

for sufficiently large $i$, which implies that $y\in E^{1}_{i}$.

Claim 3. If we set $\tilde{g}_{i}(t)=r_{i}^{-2}g_{i}(r_{i}^{2}t)$, then

$\displaystyle(M_{i},q_{i},\tilde{g}_{i}(t))_{t\in[-\delta,0)}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M^{n}_{\infty},q_{\infty},g_{\infty}(t))_{t\in[-\delta,0)}$

for a complete smooth Ricci flow
$(M^{n}_{\infty},g_{\infty}(t))_{t\in[-\delta,0)}$ with uniformly bounded
curvature.

_Proof of Claim 3:_ For any $L>1$ and $(y,t)\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)\times[-2\delta
r_{i}^{2},-L^{-1}r_{i}^{2}]$ it follows from (3.68) and (3.76) that

$\displaystyle
R_{i}(y)\leq\frac{1}{\frac{2}{n-1}+\hat{c}_{1}\frac{\epsilon_{2,i}}{t_{1,i}}f_{i}(y)}\leq\frac{2}{\frac{2}{n-1}+\hat{c}_{1}\frac{\epsilon_{2,i}}{t_{1,i}}s_{i}^{t}}.$

Therefore, we obtain

$\displaystyle\frac{1}{1+|t|}R_{i}(y)\leq\frac{s_{i}^{t}}{s_{i}}R_{i}(y)\leq\frac{2}{s_{i}}\frac{1}{\frac{2}{n-1}(s^{t}_{i})^{-1}+\hat{c}_{1}\frac{\epsilon_{2,i}}{t_{1,i}}}\leq\frac{K}{\frac{2}{n-1}+\hat{c}_{2}\frac{\epsilon_{2,i}}{t_{1,i}}s_{i}}\leq
KR_{i}(q_{i})=\frac{n-1}{2}Kr_{i}^{-2}.$ (3.78)

Here, $K:=2(1+\hat{c}_{2}/\hat{c}_{1})$ and the third inequality follows from
(3.74).

From (3.78), it is clear that $R_{\tilde{g}_{i}}(y,t)$ is uniformly bounded
for any $t\in[-2\delta,-L^{-1}]$ and $y\in B_{\tilde{g}_{i}(t)}(q_{i},L)$.
Since $|Rm_{\tilde{g}_{i}}|$ is controlled by $R_{\tilde{g}_{i}}$ by our
assumption ($\dagger$ ‣ 3), we immediately conclude from the Shi’s local
estimates and Theorem 3.1 that

$\displaystyle(M_{i},q_{i},\tilde{g}_{i}(t))_{t\in[-\delta,0)}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M^{n}_{\infty},q_{\infty},g_{\infty}(t))_{t\in[-\delta,0)}$

for a complete smooth Ricci flow
$(M^{n}_{\infty},g_{\infty}(t))_{t\in[-\delta,0)}$ with uniformly bounded
curvature.

We have

Claim 4. There exists a constant $\lambda\geq K^{-1}$ such that

$\displaystyle(M_{\infty},g_{\infty}(-\delta))=(S^{n-1}\times\mathbb{R},\lambda
g_{c}),$

where $K$ is the same constant in (3.78).

_Proof of Claim 4:_ We set

$\displaystyle\tilde{f}_{i}:=r_{i}^{-1}t_{i}^{-\frac{1}{2}}(F_{i}(-\delta
r_{i}^{2})-t_{i}),$

where $F_{i}(t):=(1-t)f_{i}(t)$ and $t_{i}:=F_{i}(-\delta r_{i}^{2})(q_{i})$.
From a direct calculation,

$\displaystyle|\nabla_{\tilde{g}_{i}(-\delta)}\tilde{f}_{i}|^{2}_{\tilde{g}_{i}(-\delta)}=\frac{|\nabla_{g_{i}(-\delta
r_{i}^{2})}F_{i}(-\delta r_{i}^{2})|_{g_{i}(-\delta
r_{i}^{2})}^{2}}{t_{i}}=\frac{F_{i}(-\delta r_{i}^{2})-(1+\delta
r_{i}^{2})^{2}R_{g_{i}(-\delta r_{i}^{2})}}{t_{i}},$ (3.79)

where we have used the identity $(1-t)^{2}R_{i}+|\nabla F_{i}|^{2}=F_{i}$.
From (3.78), we obtain $(1+\delta r_{i}^{2})R_{g_{i}(-\delta r_{i}^{2})}$ is
uniformly bounded on $B_{\tilde{g}_{i}(-\delta)}(q_{i},L)$. Moreover,

$\displaystyle t_{i}=(1+\delta r_{i}^{2})f_{i}(q^{-\delta
r_{i}^{2}}_{i})=(1+\delta r_{i}^{2})s_{i}^{-\delta r_{i}^{2}},$ (3.80)

Therefore, it follows from (3.77) that

$\displaystyle\lim_{i\to\infty}\frac{(1+\delta r_{i}^{2})^{2}R_{g_{i}(-\delta
r_{i}^{2})}}{t_{i}}=0$ (3.81)

uniformly on $B_{\tilde{g}_{i}(-\delta)}(q_{i},L)$. On the other hand, it
follows from (3.79) that $F_{i}(-\delta r_{i}^{2})$ is $r_{i}/2$-Lipschitz
with respect to $\tilde{g}_{i}(-\delta)$. Therefore, for any $y\in
B_{\tilde{g}_{i}(-\delta)}(q_{i},L)$,

$\displaystyle\tilde{f}_{i}(y)=$ $\displaystyle
r_{i}^{-1}t_{i}^{-\frac{1}{2}}(F_{i}(y,-\delta r_{i}^{2})-t_{i})$
$\displaystyle=$ $\displaystyle
r_{i}^{-1}t_{i}^{-\frac{1}{2}}(\sqrt{F_{i}(y,-\delta
r_{i}^{2})}-\sqrt{t_{i}})(\sqrt{F_{i}(y,-\delta r_{i}^{2})}+\sqrt{t_{i}})$
$\displaystyle\leq$ $\displaystyle
r_{i}^{-1}t_{i}^{-\frac{1}{2}}\frac{L}{2}r_{i}3\sqrt{t_{i}}\leq\frac{3}{2}L,$
(3.82)

where we have used the fact that $\lim_{i\to\infty}t_{i}^{-1}r_{i}^{2}=0$ from
(3.80). Then it is clear that

$\displaystyle\lim_{i\to\infty}\frac{F_{i}(-\delta
r_{i}^{2})}{t_{i}}=\lim_{i\to\infty}\frac{\tilde{f}_{i}r_{i}t_{i}^{\frac{1}{2}}+t_{i}}{t_{i}}=1$

uniformly on $B_{\tilde{g}_{i}(-\delta)}(q_{i},L)$. In addition, we compute

$\displaystyle\text{Hess}_{\tilde{g}_{i}(-\delta)}\tilde{f}_{i}$
$\displaystyle=$ $\displaystyle
r_{i}^{-1}t_{i}^{-\frac{1}{2}}\text{Hess}_{g_{i}(-\delta r_{i}^{2})}F_{i}$
$\displaystyle=$
$\displaystyle\frac{1}{2}r^{-1}_{i}t_{i}^{-\frac{1}{2}}g_{i}(-\delta
r_{i}^{2})-(1+\delta r_{i}^{2})r_{i}^{-1}t_{i}^{-\frac{1}{2}}Rc_{g_{i}(-\delta
r_{i}^{2})}$ $\displaystyle=$
$\displaystyle\frac{1}{2}r_{i}t_{i}^{-\frac{1}{2}}\tilde{g}_{i}(-\delta)-(1+\delta
r_{i}^{2})r_{i}^{-1}t_{i}^{-\frac{1}{2}}Rc_{\tilde{g}_{i}(-\delta)}.$ (3.83)

Therefore, it follows from (3.82), (3.83) and the standard regularity theorem
for elliptic equations that $\tilde{f}_{i}$ converges smoothly to a smooth
function $f_{\infty}$ on $M_{\infty}$ and

$\displaystyle\text{Hess}_{g_{\infty}}f_{\infty}=0,$ (3.84)

since $\displaystyle\lim_{i\to\infty}t_{i}^{-1}r_{i}^{2}=0$. Furthermore, it
follows from the combination of (3.79), (3.80) and (3.81) that

$\displaystyle|\nabla_{g_{\infty}}f_{\infty}|^{2}=1.$ (3.85)

Therefore, $f_{\infty}$ is not a constant. From (3.84) and (3.85), we conclude
that $(M_{\infty},g_{\infty})$ is isometric to
$(\Sigma_{\infty}\times\mathbb{R},g^{\prime}_{\infty}\times g_{E})$, where
$\Sigma_{\infty}$ is the level set of $f_{\infty}$. It follows immediately
from Lemma 3.8 that $(M_{\infty},g_{\infty})=(S^{n-1}\times\mathbb{R},\lambda
g_{c})$ for some constant $\lambda>0$. Notice that $(M_{\infty},g_{\infty})$
cannot be flat $\mathbb{R}^{n}$ since by (3.68) and (3.75) that
$R_{i}(q_{i}^{-\delta})/R_{i}(q_{i})\geq C>0$ uniformly. From the estimate
(3.78), it is clear that $\lambda\geq K^{-1}$.

Now we fix the parameter $\delta=\delta(n)>0$ small enough such that the
pseudolocality Theorem [48, Theorem 24] can be applied. Combining with [17,
Theorem $3.1$], we conclude that for any $L>0$,

$\displaystyle|Rm_{\tilde{g}_{i}}|\leq C(n)$

on $B_{\tilde{g}_{i}}(q_{i},L)\times[-\delta,0]$, if $i$ is sufficiently
large. Therefore, we conclude that

$\displaystyle(M_{i},q_{i},\tilde{g}_{i}(t))_{t\in[-\delta,0]}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(S^{n-1}\times\mathbb{R},p_{c},g_{c}(t))_{t\in[-\delta,0]}$

since $R_{\tilde{g}_{i}}(q_{i},0)=\frac{n-1}{2}$. Therefore, we can derive the
contradiction as in the proof of Proposition 3.11. ∎

Proposition 3.16 implies that $E$ is an end of the Ricci shrinker. Moreover,
it follows from (3.68) and ($\dagger$ ‣ 3) that $|Rm|$ is quadratically
decaying. It is well-known (e.g., [40]) that the tangent cone at infinity
concerning this end is a metric cone over a smooth cross-section.

Next, we show that any point in $E$ far away from $\Sigma(t_{0})$ is the
center of an evolving $\epsilon$-neck. We only sketch the proof as it is
similar to the proofs of Proposition 3.12 and Proposition 3.16.

######  Proposition 3.17.

Under the assumptions ($\dagger$ ‣ 3), (3.65) and (3.66), for any
$\epsilon>0$, there exists a constant
$\eta_{4}=\eta_{4}(\epsilon,n,\delta_{0},A,B)>0$ such that if
$\epsilon_{0}\leq\eta_{4}$, then any point
$x\in\Sigma(\eta_{4}^{-1}t_{0},\infty)\subset E^{1}$ is the center of an
evolving $\epsilon$-neck.

###### Proof.

We prove it by contradiction. Suppose for fixed $\epsilon,n$ and $A$ the
conclusion does not hold. Then there exists a sequence of Ricci shrinkers
$(M^{n}_{i},p_{i},g_{i},f_{i})$ satisfying all the assumptions and
$x_{i}\in\Sigma(i^{2}t_{0,i},\infty)$ which is not the center of an evolving
$\epsilon$-neck. Now we set $t_{i}=f_{i}(x_{i})$. We claim that

$\displaystyle\liminf_{i\to\infty}\epsilon_{2,i}\frac{t_{i}}{t_{1,i}}>0.$
(3.86)

Otherwise, it follows from the assumption (3.65) and (3.68) that

$\displaystyle\limsup_{i\to\infty}\max_{\Sigma(t_{0,i},t_{i})}\left|R_{i}-\frac{n-1}{2}\right|=0.$

Therefore, the same proof of Proposition (3.12) shows that $x_{i}$ is the
center of an evolving normalized $\epsilon$-neck for sufficiently large $i$.

By the same proof as in that of Proposition 3.16 we have

$\displaystyle(M_{i},x_{i},\tilde{g}_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}\left(S^{n-1}\times\mathbb{R},p_{c},g_{c}\right)$ (3.87)

where $\tilde{g}_{i}=r_{i}^{-2}g_{i}$ and
$R_{i}(x_{i})=\frac{n-1}{2}r_{i}^{-2}$. For a fixed number $L>1$ and $y\in
B_{g_{i}}(x_{i},Lr_{i})$, one can prove as (3.76) that

$\displaystyle\lim_{i\to\infty}\frac{f_{i}(y)}{t_{i}}=1$

uniformly for $y$, where $t_{i}=f_{i}(x_{i})$. If we set
$\psi_{i}^{t}(y)=y^{t}$, then we can prove as (3.75) that

$\displaystyle\frac{9}{10}\frac{t_{i}}{1+|t|}\leq
f_{i}(y^{t})\leq\frac{11}{10}\frac{t_{i}}{1+|t|}$ (3.88)

for any $(y,t)\in B_{g_{i}}(x_{i},Lr_{i}^{2})\times[-Lr_{i}^{2},0]$. Since

$\displaystyle r_{i}=\sqrt{1+\frac{n-1}{2}\sigma_{i}t_{i}}$

for some

$\displaystyle\hat{c}_{1}\epsilon_{2,i}t_{1,i}^{-1}\leq\sigma_{i}\leq\hat{c}_{2}\epsilon_{2,i}t_{1,i}^{-1},$

it is clear from (3.86) and (3.88) that

$\displaystyle
f_{i}(y^{t})\geq\frac{9}{10}\frac{t_{i}}{1+Lr_{i}^{2}}=\frac{9}{10}\frac{t_{i}}{1+L(1+\frac{n-2}{2}\sigma_{i}t_{i})}\geq
t_{1,i}.$

In particular, $y^{t}\in E^{1}_{i}$. Therefore, we can apply (3.68) and prove
exactly as (3.78) that

$\displaystyle\frac{1}{1+|t|}R_{i}(y^{t})\leq\frac{f_{i}(y^{t})}{f_{i}(y)}R_{i}(y^{t})\leq\frac{2}{t_{i}}\frac{1}{\frac{2}{n-1}u_{i}^{-1}+\hat{c}_{1}\frac{\epsilon_{2,i}}{t_{1,i}}}\leq\frac{K}{\frac{2}{n-1}+\hat{c}_{2}\frac{\epsilon_{2,i}}{t_{1,i}}t_{i}}\leq
Kr_{i}^{-2}$

for $(y,t)\in B_{g_{i}}(x_{i},Lr_{i}^{2})\times[-Lr_{i}^{2},0]$, where
$u_{i}=f_{i}(y^{t})$. Therefore, we conclude that for
$\tilde{g}_{i}(t):=r_{i}^{-2}g_{i}(r_{i}^{2}t)$, $R_{\tilde{g}_{i}}(y,t)$ is
uniformly bounded on $B_{\tilde{g}_{i}(0)}(y,L)\times[-L,0]$ since

$\displaystyle R_{\tilde{g}_{i}}(y,t)=\frac{r_{i}^{2}}{1+|t|}R_{i}(y^{t})\leq
K.$

Then it is clear as before that

$\displaystyle(M_{i},x_{i},\tilde{g}_{i}(t))_{t\leq 0}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(M^{n}_{\infty},x_{\infty},g_{\infty}(t))_{t\leq 0}$

for a complete smooth Ricci flow solution $(M_{\infty},g_{\infty}(t))_{t\leq
0}$ with uniformly bounded curvature. From (3.87) and the backward uniqueness
of the Ricci flow [39, Theorem $1.1$], we conclude that

$\displaystyle(M_{\infty},g_{\infty}(t))_{t\leq
0}=(S^{n-1}\times\mathbb{R},g_{c}(t))_{t\leq 0}.$

However, this implies that $x_{i}$ is the center of an evolving
$\epsilon$-neck for large $i$, which is a contradiction. ∎

### Cap region

In this subsection, we assume that (3.46) is violated in the other direction
and hence there exists a $t_{2}\in(t_{0},\infty)$ such that on
$\Sigma(t_{0},t_{2})$

$\displaystyle\left|R-\frac{n-1}{2}\right|<\epsilon_{2}$ (3.89)

and

$\displaystyle\max_{\Sigma(t_{2})}R=\frac{n-1}{2}+\epsilon_{2}.$ (3.90)

Now we set $E^{2}=\Sigma(t_{2},s)\subset E$ and prove the following result.

######  Proposition 3.18.

With the assumptions above, there exists a positive constants
$\hat{c}_{3},\hat{c}_{4}$ depending only on $n,A$ and $B$ such that

$\displaystyle\frac{1}{\frac{2}{n-1}-\hat{c}_{3}\epsilon_{2}t_{2}^{-1}t}\leq
R(x)\leq\frac{1}{\frac{2}{n-1}-\hat{c}_{4}\epsilon_{2}t_{2}^{-1}t}$ (3.91)

for any $x\in E^{2}$, where $t=f(x)$. Moreover, $R(\psi^{t}(x))$ is increasing
for $t\geq 0$ as long as $\psi^{t}(x)$ stays in $E^{2}$.

###### Proof.

By our assumption, there exists $z\in\Sigma(t_{2})$ such that

$\displaystyle R(z)=\frac{n-1}{2}+\epsilon_{2}.$

From Lemma 3.8, we may assume on $\Sigma(t_{2})$ that

$\displaystyle\frac{n-1}{2}+\frac{1}{2}\epsilon_{2}\leq
R\leq\frac{n-1}{2}+\frac{3}{2}\epsilon_{2}.$ (3.92)

Recall from (3.70) we have

$\displaystyle\left(\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R\leq
tR_{t}\leq\left(\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R$
(3.93)

for some constant $c_{1}=c_{1}(n)>0$. By considering the ODE with the given
initial values in (3.92), we obtain

$\displaystyle\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}-c_{2}\epsilon_{2}t_{2}^{-1}t}\leq
R(t)\leq\frac{1}{\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}-c_{3}\epsilon_{2}t_{2}^{-1}t}$
(3.94)

for any $t\geq t_{2}$. Here, the constants $c_{2}$ and $c_{3}$ are determined
by

$\displaystyle\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}-c_{2}\epsilon_{2}}=\frac{n-1}{2}+\frac{1}{2}\epsilon_{2}.$
(3.95)

and

$\displaystyle\frac{1}{\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}-c_{3}\epsilon_{2}}=\frac{n-1}{2}+\frac{3}{2}\epsilon_{2}.$
(3.96)

It is clear from (3.94), (3.95), (3.96) and the definition of $\epsilon_{2}$
that (3.91) holds if $\epsilon_{0}$ is sufficiently small. Now the last
statement follows from (3.91) and (3.93) since on $E^{2}$,

$\displaystyle\left(\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R>0.$

∎

Parallel to Proposition (3.11) and Proposition 3.16, we have the following
result.

######  Proposition 3.19.

Under the assumptions ($\dagger$ ‣ 3), (3.89) and (3.90), there exists
$\sigma_{4}=\sigma_{4}(n,\delta_{0},A,B)>0$ such that if
$\epsilon_{0}\leq\sigma_{4}$, then $s<\infty$ and there exists a point
$q\in\Sigma(s)$ such that

$\displaystyle R(q)=\epsilon_{1}f(q),$ (3.97)

if $\epsilon_{0}$ is sufficiently small.

###### Proof.

By (3.91), the scalar curvature blowup before the time
$t=\frac{2t_{2}}{(n-1)\epsilon_{2}\hat{c}_{4}}$. Therefore, we must have
$s<\infty$. We move on to show (3.97).

From our definition of $E$ in ($\dagger$ ‣ 3), there exists a point
$q\in\Sigma(s)$ such that at $q$ either

$\displaystyle
R=\epsilon_{1}f\quad\text{or}\quad|\nabla^{k}Rm|=BR^{\frac{k}{2}+1}$

for some $0\leq k\leq 4$. Notice that the second possibility can be excluded
exactly as Proposition 3.16, if $\epsilon_{0}$ is sufficiently small and we
sketch the proof below.

Assume there exists a sequence of Ricci shrinkers $(M_{i},g_{i},f_{i})$ with
the end $E_{i}=\Sigma(t_{0,i},s_{i})$ satisfying ($\dagger$ ‣ 3), (3.89) and
(3.90) with $\epsilon_{0,i}\to 0$. Moreover, there exists a
$q_{i}\in\Sigma(s_{i})$ at which

$\displaystyle|\nabla^{k}Rm_{i}|=BR_{i}^{\frac{k}{2}+1}$

for a fixed $0\leq k\leq 4$.

Claim 1.

$\displaystyle\liminf_{i\to\infty}\epsilon_{2,i}\frac{s_{i}}{t_{2,i}}>0.$
(3.98)

_Proof of Claim 1:_ Otherwise, it follows from the assumption (3.89) and
(3.91) that

$\displaystyle\limsup_{i\to\infty}\max_{\Sigma(t_{0,i},s_{i})}\left|R_{i}-\frac{n-1}{2}\right|=0.$

Therefore, the same proof of Proposition (3.11) yields a contradiction.

Now we set $r^{-2}_{i}=\frac{2}{n-1}R_{i}(q_{i})$. Then it is clear from
(3.91) that $r_{i}\leq 1$. Moreover, we define
$q_{i}^{t}=\psi_{i}^{t}(q_{i})$, $s_{i}^{t}=f_{i}(q_{i}^{t})$ and fix a
constant $\delta=\delta(n)\in(0,1/20)$ determined later.

Claim 2. For any $t\in[-2\delta r_{i}^{2},0]$, we have

$\displaystyle\frac{s_{i}}{1+|t|}\leq
s_{i}^{t}\leq\frac{s_{i}}{(1+|t|)^{1-\epsilon_{1,i}}}.$ (3.99)

_Proof of Claim 2:_ For any $t\in[-2\delta r_{i}^{2},0]$ we have

$\displaystyle\frac{df_{i}(q_{i}^{t})}{dt}=\frac{|\nabla_{g_{i}}f_{i}|_{g_{i}}^{2}}{1-t}(q_{i}^{t})=\frac{f_{i}-R_{i}}{1-t}(q_{i}^{t}).$

Since $0\leq R_{i}(q_{i}^{t})\leq\epsilon_{1,i}f_{i}(q_{i}^{t})$ by our
assumptions, it is clear that

$\displaystyle(1-\epsilon_{1,i})\frac{f_{i}(q_{i}^{t})}{1-t}\leq\frac{df_{i}(q_{i}^{t})}{dt}\leq\frac{f_{i}(q_{i}^{t})}{1-t}.$

By solving the above ODE, we immediately obtain (3.99).

Claim 3. For any fixed number $L>1$, $t\in[-2\delta
r_{i}^{2},-L^{-1}r_{i}^{2}]$ and $y\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)$, we have

$\displaystyle\frac{s_{i}}{1+2\delta}\leq f_{i}(y)\leq s_{i}$ (3.100)

for sufficiently large $i$.

_Proof of Claim 3:_ We compute

$\displaystyle|2\sqrt{f_{i}(y)}-2\sqrt{s^{t}_{i}}|=|2\sqrt{f_{i}(y)}-2\sqrt{f_{i}(q^{t}_{i})}|\leq\frac{Lr_{i}}{\sqrt{1+|t|}}.$

Therefore, it follows from (3.99) that

$\displaystyle 2\sqrt{f_{i}(y)}\leq
2\sqrt{s^{t}_{i}}+\frac{Lr_{i}}{\sqrt{1+|t|}}\leq\frac{2\sqrt{s_{i}}}{(1+|t|)^{\frac{1-\epsilon_{1,i}}{2}}}+\frac{Lr_{i}}{\sqrt{1+|t|}}.$
(3.101)

Since $L^{-1}r_{i}^{2}\leq|t|\leq\delta r_{i}^{2}\leq\delta$, there exists a
universal constant $c_{1}>0$ such that

$\displaystyle
2\sqrt{s_{i}}-\frac{2\sqrt{s_{i}}}{(1+|t|)^{\frac{1-\epsilon_{1,i}}{2}}}\geq
c_{1}|t|\sqrt{s_{i}}\geq
c_{1}L^{-1}r_{i}^{2}\sqrt{s_{i}}\geq\frac{Lr_{i}}{\sqrt{1+|t|}}.$ (3.102)

Here, the last inequality follows from the fact that
$\limsup_{i\to\infty}s_{i}^{-1}r^{-2}_{i}=0$ since
$R_{i}\leq\epsilon_{1,i}f_{i}$ on $E_{i}$. Therefore, the second inequality in
Claim 3 follows from (3.101) and (3.102). Similarly, we compute

$\displaystyle 2\sqrt{f_{i}(y)}\geq
2\sqrt{s^{t}_{i}}-\frac{Lr_{i}}{\sqrt{1+|t|}}\geq\frac{2\sqrt{s_{i}}-L}{\sqrt{1+|t|}}\geq\frac{2\sqrt{s_{i}}}{\sqrt{1+2\delta}}$

since $\lim_{i\to\infty}s_{i}=\infty$ and $|t|\leq\delta$. Therefore, the
first inequality in Claim 3 follows.

Combining (3.98) and (3.100), we conclude that $y\in E^{2}_{i}$.

Claim 4. If we set $\tilde{g}_{i}(t)=r_{i}^{-2}g_{i}(r_{i}^{2}t)$, then

$\displaystyle(M_{i},q_{i},\tilde{g}_{i}(t))_{t\in[-\delta,0]}\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}(S^{n-1}\times\mathbb{R},p_{c},g_{c}(t))_{t\in[-\delta,0]}$

for a complete smooth Ricci flow
$(M^{n}_{\infty},g_{\infty}(t))_{t\in[-\delta,0)}$ with uniformly bounded
curvature.

_Proof of Claim 4:_ For any $L>1$ and $(y,t)\in
B_{g_{i}}\left(q^{t}_{i},L(1+|t|)^{-\frac{1}{2}}r_{i}\right)\times[-2\delta
r_{i}^{2},-L^{-1}r_{i}^{2}]$, it follows from Proposition 3.18 that $R_{i}(y)$
is bounded above by $R_{i}(\phi^{u}(y))$, where $u=s-f(y)$. Since the scalar
curvature is almost constant on $\Sigma(s)$ by Lemma 3.8, we have

$\displaystyle\frac{1}{1+|t|}R_{i}(y)\leq(n-1)r_{i}^{-2}.$

The rest of the proof follows exactly like that of Proposition 3.16, and we
can derive a contradiction as before.

In sum, the proof is complete. ∎

######  Corollary 3.20.

Under the assumptions of Proposition 3.19, there exists
$\hat{c}_{5}=\hat{c}_{5}(n,A,B)>0$ such that

$\displaystyle s\geq\frac{\hat{c}_{5}}{\epsilon_{2}}t_{2}\quad\text{and}\quad
R(q)\geq\frac{\hat{c}_{5}}{\epsilon_{2}}.$ (3.103)

###### Proof.

By our assumption ($\dagger$ ‣ 3), $R(q)=\epsilon_{1}f(q)\geq n$. Therefore,
it follows from (3.91) that

$\displaystyle s\geq\frac{n+1}{n(n-1)\hat{c}_{4}}\frac{t_{2}}{\epsilon_{2}}.$

Consequently,

$\displaystyle
R(q)=\epsilon_{1}s\geq\frac{n+1}{n(n-1)\hat{c}_{4}}\frac{\epsilon_{1}t_{2}}{\epsilon_{2}}\geq\frac{n+1}{(n-1)\hat{c}_{4}}\frac{1}{\epsilon_{2}}.$

∎

Next, we show that any point in $E$ far away from $\Sigma(t_{0})$ is the
center of an evolving $\epsilon$-neck. We omit the proof as it is similar to
the proofs of Proposition 3.17 and Proposition 3.19.

######  Proposition 3.21.

Under the assumptions ($\dagger$ ‣ 3), (3.65) and (3.66), for any
$\epsilon>0$, there exists $\eta_{5}=\eta_{5}(\epsilon,n,\delta_{0},A,B)>0$
such that if $\epsilon_{0}\leq\eta_{5}$, then any point
$x\in\Sigma(\eta_{5}^{-1}t_{0},s)\subset E^{2}$ is the center of an evolving
$\epsilon$-neck.

For later applications, we show that any point in $E^{2}$ has a positive
curvature operator.

######  Proposition 3.22.

Under the assumptions ($\dagger$ ‣ 3), (3.65) and (3.66), there exists
$\sigma_{5}=\sigma_{5}(n,A,B)>0$ such that if $\epsilon_{0}\leq\sigma_{5}$,
then

$\displaystyle Rm\geq\delta\epsilon_{2}\frac{R^{2}}{f}$

for $\delta=\delta(n)>0$ on $E^{2}$.

###### Proof.

From (3.6), we have

$\displaystyle
R_{abcd}=\frac{R}{(n-1)(n-2)}(g_{ac}g_{bd}-g_{ad}g_{bc})+O(\epsilon_{0}R).$

Moreover, it follows from (2.2) that

$\displaystyle R_{abcn}=\frac{R_{abcl}f_{l}}{|\nabla
f|}=\frac{\nabla_{b}R_{ac}-\nabla_{a}R_{bc}}{|\nabla
f|}=\frac{\nabla_{b}\overset{\circ}{R}_{ac}-\nabla_{a}\overset{\circ}{R}_{bc}+\frac{\nabla_{b}Rg_{ac}}{n-1}-\frac{\nabla_{a}Rg_{bc}}{n-1}}{|\nabla
f|}.$

Therefore, it is clear from (3.18) and Lemma 3.7 that

$\displaystyle
R_{abcn}=O(\epsilon_{0}^{\frac{1}{2}}f^{-\frac{1}{2}}R^{\frac{3}{2}}).$

In addition, from the computation in the proof of Proposition 2.2, we have

$\displaystyle R_{ancn}=\frac{2R_{akcl}R_{kl}-R_{kc}R_{ka}-R_{ac}/2+\Delta
R_{ac}-\nabla^{2}_{ac}R/2}{|\nabla f|^{2}}.$ (3.104)

From direct calculations,

$\displaystyle 2R_{akcl}R_{kl}=$ $\displaystyle
2R_{abcd}R_{bd}+2R_{abcn}R_{bn}+2R_{ancn}R_{nn}$ $\displaystyle=$
$\displaystyle 2R_{abcd}R_{bd}+O(f^{-1}R^{3})+O(f^{-2}R^{4})$ $\displaystyle=$
$\displaystyle\frac{2R}{(n-1)(n-2)}(g_{ac}g_{bd}-g_{ad}g_{bc})R_{bd}+O(\epsilon_{0}R^{2})+O(\epsilon_{1}R^{2})$
$\displaystyle=$
$\displaystyle\frac{2R^{2}g_{ac}}{(n-1)(n-2)}-\frac{2RR_{ac}}{(n-1)(n-2)}-\frac{2RR_{nn}}{(n-1)(n-2)}+O(\epsilon_{0}R^{2})$
$\displaystyle=$
$\displaystyle\frac{2R^{2}g_{ac}}{(n-1)^{2}}+O(\epsilon_{0}R^{2}),$ (3.105)

where we have used (3.4) and (3.5). In addition,

$\displaystyle
R_{kc}R_{ka}=R_{bc}R_{ab}+R_{cn}R_{an}=\frac{R^{2}}{(n-1)^{2}}+O(\epsilon_{0}R^{2}).$
(3.106)

Combining (3.105), (3.106), (3.38) and Lemma 3.10, it follows from (3.104)
that

$\displaystyle R_{ancn}=\frac{1}{|\nabla
f|^{2}}\left(\frac{R^{2}}{(n-1)^{2}}-\frac{R}{2(n-1)}\right)g_{ac}+O(\epsilon_{0}^{\frac{1}{3}}f^{-1}R^{2}).$
(3.107)

Since $R\geq\frac{n-1}{2}+\frac{\epsilon_{2}}{2}$ on $E^{2}$, we obtain from
(3.107) that

$\displaystyle R_{anan}\geq c_{1}\epsilon_{2}\frac{R^{2}}{f}$

for some constant $c_{1}=c_{1}(n)>0$. For any two-form $w$, we decompose it as

$\displaystyle w=\tilde{w}+x_{a}e^{a}\wedge e^{n},$

where $\tilde{w}$ is the two-form on $\Sigma$. If we set
$|x|=\sqrt{\sum_{a=1}^{n-1}x_{a}^{2}}$, then we compute

$\displaystyle Rm(w,w)=$ $\displaystyle
Rm(\tilde{w},\tilde{w})+2x_{a}Rm(\tilde{w},e^{a}\wedge
e^{n})+x_{a}x_{b}R_{anbn}$ $\displaystyle\geq$
$\displaystyle\frac{R}{(n-1)(n-2)}|\tilde{w}|^{2}-c_{2}\epsilon_{0}R|\tilde{w}|^{2}-c_{2}|x||\tilde{w}|\epsilon_{0}^{\frac{1}{2}}f^{-\frac{1}{2}}R^{\frac{3}{2}}+x_{a}x_{b}R_{anbn}$
$\displaystyle\geq$
$\displaystyle\frac{R}{(n-1)(n-2)}|\tilde{w}|^{2}-c_{2}\epsilon_{0}R|\tilde{w}|^{2}-c_{2}|x||\tilde{w}|\epsilon_{0}^{\frac{1}{2}}f^{-\frac{1}{2}}R^{\frac{3}{2}}+c_{1}|x|^{2}\epsilon_{2}f^{-1}R^{2}-c_{3}|x|^{2}\epsilon_{0}^{\frac{1}{3}}f^{-1}R^{2}$

for some constants $c_{2}$ and $c_{3}$ depending only on $n,A$ and $B$. Since
$\epsilon_{2}=\epsilon_{0}^{\frac{1}{4}}$, it follows from the mean value
inequality that

$\displaystyle
Rm(w,w)\geq\frac{c_{1}}{2}\epsilon_{2}f^{-1}R^{2}(|x|^{2}+|\tilde{w}|^{2})=\frac{c_{1}}{2}\epsilon_{2}f^{-1}R^{2}|w|^{2},$

if $\epsilon_{0}$ is sufficiently small. ∎

##  4 Estimates on the cap region

Throughout this section, we assume ($\dagger$ ‣ 3), (3.89), (3.90) and
$\epsilon_{0}$ is small enough such that Proposition 3.19 holds. We will fix
$\epsilon_{1}$ and choose $\epsilon_{1}^{\prime}$ to be a small parameter
which may depend on $\epsilon_{1}$. In particular, we assume
$\epsilon_{1}^{\prime}\leq\epsilon_{1}^{\frac{1}{2}}$ and hence
$\epsilon_{0}=\epsilon_{1}^{\frac{1}{2}}$ which is also fixed.

Define

$\displaystyle D\coloneqq\textrm{the connected component of
$M\backslash\text{Int}\,\Sigma(t_{0},s)$ containing $\Sigma(s)$.}$ (4.1)

From Proposition 3.18 there exists a point $q\in\Sigma(s)$ such that

$\displaystyle R(q)=\epsilon_{1}f(q)=\epsilon_{1}s.$

First we prove

######  Proposition 4.1.

Under the above assumptions, there exist positive constants $m_{1},m_{2}$
depending only on $n$ such that

$\displaystyle\frac{0.9}{\epsilon_{1}^{-1}+m_{2}L}\leq\frac{R(y)}{s}\leq\frac{1.1}{\epsilon_{1}^{-1}+m_{1}L}$
(4.2)

for any $L\in[0,s/2]$ and $y\in\Sigma(s-L)$.

###### Proof.

We first prove $\Sigma(s/2,s)\subset E^{2}$. Otherwise, there exist
$s_{1}\in[s/2,s]$ and a point $x\in\Sigma(s_{1})$ such that
$\Sigma(s_{1},s)\subset E^{2}$ and

$\displaystyle R(x)=\frac{n-1}{2}+\epsilon_{2}.$ (4.3)

From (3.70) we have on $E$ that

$\displaystyle
tR_{t}\geq\left(\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R.$

Consequently, solving the ODE yields that

$\displaystyle
R(q^{t})\geq\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}-\delta_{1}t}$
(4.4)

for any $t\in[s_{1},s]$, where $q^{t}=\phi^{t-s}(q)\in\Sigma(t)$ and
$\delta_{1}$ is determined by

$\displaystyle\epsilon_{1}s=R(q)=\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}-\delta_{1}s}.$
(4.5)

Since $R(q)\gg 1$ by (3.103), it follows from (4.4) and (4.5) that

$\displaystyle R(q^{s_{1}})\geq\frac{n}{2}.$ (4.6)

However, (4.6) contradicts (4.3) by Lemma 3.8.

Now it is clear that (4.4) holds for any $t\in[s/2,s]$. Therefore, it follows
from (4.4) and (4.5) that for $L\in[0,s/2]$,

$\displaystyle
R(q^{s-L})\geq\frac{1}{\frac{2}{n-1}-c_{1}\epsilon_{0}^{\frac{1}{3}}-\delta_{1}(s-L)}=\frac{1}{\frac{1}{\epsilon_{1}s}+\delta_{1}s-\delta_{1}(s-L)}=\frac{1}{\frac{1}{\epsilon_{1}s}+\delta_{1}L}$

and hence

$\displaystyle\frac{R(q^{s-L})}{R(q)}\geq\frac{1}{1+\delta_{1}s\epsilon_{1}L}.$
(4.7)

From (4.5) and the fact that $R(q)\gg 1$, we know that
$\delta_{1}s\approx\frac{2}{n-1}$. Combining (4.7) and Lemma 3.8, we conclude
that

$\displaystyle\min_{\Sigma(s-L)}R\geq\frac{0.9R(q)}{1+m_{2}\epsilon_{1}L}$

for some constant $m_{2}=m_{2}(n)\approx\frac{2}{n-1}$. Therefore,

$\displaystyle\min_{\Sigma(s-L)}R\geq\frac{0.9s}{\epsilon_{1}^{-1}+m_{2}L}.$

Now the other inequality in (4.2) can be proved similarly by considering

$\displaystyle
tR_{t}\leq\left(\frac{2}{n-1}+c_{1}\epsilon_{0}^{\frac{1}{3}}\right)R^{2}-R,$

which follows from (3.70). ∎

Next, we prove

######  Proposition 4.2.

Under the above assumptions, for any small $\epsilon_{1}>0$, there exist
$\tau_{1}=\tau_{1}(\epsilon_{1},n,\delta_{0},A,B)>0$ and
$L_{1}=L_{1}(\epsilon_{1},n,\delta_{0},A,B)>0$ such that if
$\epsilon_{1}^{\prime}\leq\tau_{1}$, then

$\displaystyle\emph{diam}_{g}D\leq\frac{L_{1}}{\sqrt{s}}\quad\text{and}\quad\sup_{D}|f-s|\leq
L_{1}.$

In particular, $D$ is compact and $\partial D=\Sigma(s)$.

###### Proof.

We prove it by contradiction. Suppose for fixed $\epsilon_{1}$ the conclusion
does not hold. Then there exists a sequence of Ricci shrinkers
$(M^{n}_{i},g_{i},f_{i})$ satisfying all the assumptions with
$\epsilon^{\prime}_{1,i}\to 0$ and

$\displaystyle\textrm{diam}_{g_{i}}D_{i}\geq\frac{i^{2}}{\sqrt{s_{i}}},\quad\textrm{or}\quad\sup_{D_{i}}|f_{i}-s_{i}|\geq
i^{2}.$ (4.8)

From Proposition 3.18, there exists a point $q_{i}\in\Sigma(s_{i})$ satisfying

$\displaystyle R_{i}(q_{i})=\epsilon_{1}f_{i}(q_{i})=\epsilon_{1}s_{i}.$

Now we define $\tilde{f}_{i}:=f_{i}-s_{i}$ and $\tilde{g}_{i}:=s_{i}g_{i}$.
From direct calculations, we have

$\displaystyle
R_{\tilde{g}_{i}}+|\nabla_{\tilde{g}_{i}}\tilde{f}_{i}|^{2}_{\tilde{g}_{i}}=\frac{R_{i}+|\nabla_{g_{i}}f_{i}|^{2}_{g_{i}}}{s_{i}}=\frac{f_{i}}{s_{i}}=1+\frac{\tilde{f}_{i}}{s_{i}}$
(4.9)

and

$\displaystyle\text{Hess}_{\tilde{g}_{i}}\tilde{f}_{i}+Rc_{\tilde{g}_{i}}=\text{Hess}_{g_{i}}f_{i}+Rc_{i}=\frac{g_{i}}{2}=\frac{\tilde{g}_{i}}{2s_{i}}.$
(4.10)

For any $L>1$ and $y\in B_{\tilde{g}_{i}}(q_{i},L)$, it follows from (4.9)
that

$\displaystyle|\tilde{f}_{i}(y)|\leq 2L\quad\text{and}\quad
R_{\tilde{g}_{i}}(y)\leq 2$ (4.11)

if $i$ is sufficiently large. Therefore, it follows from (4.11) and Theorem
3.1 that

$\displaystyle|B_{\tilde{g}_{i}}(q_{i},1)|_{\mu_{i}}\geq v_{0}$ (4.12)

for some $v_{0}=v_{0}(n,A)>0$, where
$\mu_{i}:=e^{-\tilde{f}_{i}}dV_{\tilde{g}_{i}}$ is the weighted measure.
Combining (4.9)-(4.12), it follows from the main result [44, Theorem $1.1$]
and its variant [44, Theorem $10.2$] that by taking a subsequence if
necessary,

$\displaystyle(M_{i},q_{i},\tilde{g}_{i},\tilde{f}_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{pointed-Gromov-
Hausdorff}\left(X_{\infty},q_{\infty},d_{\infty},f_{\infty}\right),$ (4.13)

where $(X_{\infty},d_{\infty})$ is a length space, $f_{\infty}$ is a Lipschitz
function on $(X_{\infty},d_{\infty})$. The space $X_{\infty}$ has a natural
regular-singular decomposition $X_{\infty}=\mathcal{R}\cup\mathcal{S}$
satisfying

  * (a).

The singular part $\mathcal{S}$ is a closed set of Minkowski codimension at
least $4$. Namely, we have

$\displaystyle\dim_{\mathcal{M}}\mathcal{S}\leq n-4.$ (4.14)

  * (b).

The regular part $\mathcal{R}$ is an open manifold with smooth metric
$g_{\infty}$ satisfying Ricci steady soliton equation

$\displaystyle Rc_{\infty}+\text{Hess}_{g_{\infty}}f_{\infty}=0.$ (4.15)

  * (c).

The convergence (4.13) can be improved to

$\displaystyle(M_{i},q_{i},\tilde{g}_{i},\tilde{f}_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{pointed-\hat{C}^{\infty}-Cheeger-
Gromov}\left(X_{\infty},q_{\infty},g_{\infty},f_{\infty}\right).$ (4.16)

  * (d).

On the regular part $\mathcal{R}$,

$\displaystyle R_{\infty}+|\nabla f_{\infty}|^{2}=1.$ (4.17)

Note that (4.16) implies that away from $\mathcal{S}$, the convergence is
smooth. This follows from the bootstrapping argument based on (4.9) and
(4.10), see [44]. Moreover, (4.15) and (4.17) follow from (4.10) and (4.9),
respectively, by taking the limit.

In addition, it follows from [34, Theorem $4.10$] that the regular part
$\mathcal{R}$ is geodesically convex. Therefore, the limit
$(X_{\infty},q_{\infty},d_{\infty},f_{\infty})$ is a Ricci steady soliton
conifold in the sense of Definition A.1.

By our assumptions, any point $y\in B_{\tilde{g}_{i}}(q_{i},L)\cap E^{2}_{i}$
satisfies ($\dagger$ ‣ 3) and (4.2). Furthermore, $y$ is an evolving
$\epsilon$-neck by Proposition 3.21. Therefore, the corresponding limit set
$E_{\infty}$ of $E^{2}_{i}$ under the convergence (4.16) is an end of
$(X_{\infty},d_{\infty})$. On the other hand, it follows from Theorem A.22
that $(X_{\infty},d_{\infty})$ has only one end. In other words,
$D_{\infty}:=X_{\infty}\backslash E_{\infty}$ is a compact set, which fact in
turn yields that $\displaystyle\text{diam}_{d_{\infty}}D_{\infty}<\infty$.
Since $f_{\infty}$ is $1$-Lipschitz and $f_{\infty}(q_{\infty})=0$, it is
clear that $\displaystyle\sup_{D_{\infty}}|f_{\infty}|<\infty$. Therefore, for
some $L_{1}>0$ and sufficiently large $i$, we have

$\displaystyle\text{diam}_{\tilde{g}_{i}}D_{i}\leq
L_{1},\quad\sup_{D_{i}}|f_{i}-s_{i}|\leq L_{1},$

which contradicts (4.8). The proof is established by this contradiction. ∎

We have the following definition, which measures how close the neighborhood of
a point in a Ricci shrinker is to a Ricci steady soliton conifold.

######  Definition 4.3 (Center of an $\epsilon$-steady soliton conifold).

Let $(M^{n},g,f)$ be a Ricci shrinker and let $\bar{x}\in M$ such that
$f(\bar{x})=s$ . We say that $\bar{x}$ is the center of an $\epsilon$-steady
soliton conifold if, after rescaling the metric by the factor $s$, the
neighborhood $B_{g}(\bar{x},\epsilon^{-1}s^{-\frac{1}{2}})$ is
$\epsilon$-close in the Gromov-Hausdorff sense to a nontrivial
$(X,q,d,f)\in\mathfrak{S}(n)$.

From the proof of Proposition 4.2, we immediately have

######  Corollary 4.4.

Under the above assumptions, for any $\epsilon,\epsilon_{1}>0$, there exists
$\tau_{2}=\tau_{2}(\epsilon,\epsilon_{1},n,\delta_{0},A,B)>0$ such that if
$\epsilon_{1}^{\prime}\leq\tau_{2}$, then any point in $D$ is the center of an
$\epsilon$-steady soliton conifold.

Next, we estimate the scalar curvature on the cap region. Before that, we
recall the following Sobolev inequality proved in [48].

######  Theorem 4.5.

_([48, Theorem 1])_ For any Ricci shrinker $(M^{n},g,f)$ with
$\boldsymbol{\mu}(g)\geq-A$, there exists a $C=C(n,A)>0$ such that for any
compactly supported locally Lipschitz function $u$, we have

$\displaystyle\left(\int u^{\frac{n}{n-2}}\,dV\right)^{\frac{n-2}{n}}\leq
C\int 4|\nabla u|^{2}+Ru^{2}\,dV.$

Notice that the Sobolev inequality is scaling-invariant and also holds at
different time slices of the Ricci flow associated with a Ricci shrinker. Now
we have

######  Proposition 4.6.

Under the above assumptions, for any small $\epsilon_{1}>0$, there exist
positive numbers $\tau_{3}=\tau_{3}(\epsilon_{1},n,\delta_{0},A,B)$ and
$L_{2}=L_{2}(\epsilon_{1},n,\delta_{0},A,B)$ such that if
$\epsilon_{1}^{\prime}\leq\tau_{3}$, then

$\displaystyle L_{2}^{-1}s\leq R\leq s+L_{1}$ (4.18)

on $D$, where $L_{1}$ is the same constant in Proposition 4.2.

###### Proof.

Without loss of generality, we assume $\tau_{3}\leq\tau_{1}$ and the
conclusion of Proposition 4.2 holds. In particular, we have

$\displaystyle R(y)\leq f(y)\leq s+L_{1},\quad\forall\;y\in D.$ (4.19)

For the other direction, we assume the inequality does not hold. Then there
exists a sequence of Ricci shrinkers $(M^{n}_{i},q_{i},g_{i},f_{i})$
satisfying all the assumptions with $\epsilon^{\prime}_{1,i}\to 0$ such that

$\displaystyle R_{i}(x_{i})<i^{-2}s_{i}$ (4.20)

for some $x_{i}\in D_{i}$. For $t\leq 0$, we define

$\displaystyle\tilde{g}_{i}(t)\coloneqq
s_{i}g_{i}(s_{i}^{-1}t),\quad\tilde{F}_{i}(t)\coloneqq
F_{i}(s_{i}^{-1}t)-s_{i},\quad\tilde{\square}_{i}\coloneqq\partial_{t}-\Delta_{\tilde{g}_{i}(t)},$

where $F_{i}(t)=(1-t)f_{i}(t)$. Direct calculation shows that

$\displaystyle|\nabla_{\tilde{g}_{i}(t)}\tilde{F}_{i}(t)|^{2}_{\tilde{g}_{i}(t)}=$
$\displaystyle
s_{i}^{-1}|\nabla_{g_{i}(s_{i}^{-1}t)}F_{i}(s_{i}^{-1}t)|^{2}_{g_{i}(s_{i}^{-1}t)}=\frac{F_{i}(s_{i}^{-1}t)-(1-s_{i}^{-1}t)^{2}R_{g_{i}(s_{i}^{-1}t)}}{s_{i}}.$

Thus we obtain

$\displaystyle|\nabla_{\tilde{g}_{i}(t)}\tilde{F}_{i}(t)|^{2}_{\tilde{g}_{i}(t)}+(1-s_{i}^{-1}t)^{2}R_{\tilde{g}_{i}(t)}=1+\frac{\tilde{F}_{i}(t)}{s_{i}}.$

Moreover, we compute

$\displaystyle\partial_{t}\tilde{F}_{i}(t)=s_{i}^{-1}\partial_{t}F_{i}(s_{i}^{-1}t)=-s_{i}^{-1}(1-s_{i}^{-1}t)R_{g_{i}(s_{i}^{-1}t)}=-(1-s_{i}^{-1}t)R_{\tilde{g}_{i}(t)}$
(4.21)

and

$\displaystyle\Delta_{\tilde{g}_{i}(t)}\tilde{F}_{i}(t)=s_{i}^{-1}\Delta_{g_{i}(s_{i}^{-1}t)}F_{i}(s_{i}^{-1}t)=s_{i}^{-1}\left(\frac{n}{2}-(1-s_{i}^{-1}t)R_{g_{i}(s_{i}^{-1}t)}\right)=\frac{n}{2s_{i}}-(1-s_{i}^{-1}t)R_{\tilde{g}_{i}(t)}.$
(4.22)

Combining (4.21) and (4.22), we obtain

$\displaystyle\tilde{\square}_{i}\tilde{F}_{i}(t)=(\partial_{t}-\Delta_{\tilde{g}_{i}(t)})\tilde{F}_{i}(t)=-\frac{n}{2s_{i}}.$
(4.23)

Now we define the following parabolic balls

$\displaystyle P^{1}_{i}$
$\displaystyle\coloneqq\\{(y,t)\,\mid|\tilde{F}_{i}(y,t)|\leq
2L_{1},\,-2L_{1}\leq t\leq 0\\},$ $\displaystyle P^{2}_{i}$
$\displaystyle\coloneqq\\{(y,t)\,\mid|\tilde{F}_{i}(y,t)|\leq
4L_{1},\,-4L_{1}\leq t\leq 0\\}.$

In the following proof, all positive constants $C_{i}$ in the following proof
depend only on $n,\delta_{0},A,B$ and $\epsilon_{1}$ and the corresponding
inequalities hold for sufficiently large $i$. It follows from (4.2) and (4.19)
that

$\displaystyle 0\leq R_{\tilde{g}_{i}(t)}\leq C_{1}$ (4.24)

on $P^{2}_{i}$, since by the definition we know

$\displaystyle
R_{\tilde{g}_{i}(t)}(z)=s_{i}^{-1}R_{g_{i}(s_{i}^{-1}t)}(z)=s_{i}^{-1}(1-s_{i}^{-1}t)^{-1}R_{g_{i}(0)}(\psi_{i}^{s_{i}^{-1}t}(z))$

for any point $z$. Since $\tilde{F}_{i}(q_{i},0)=0$, it follow from (4.21) and
(4.24) that

$\displaystyle 0\leq\tilde{F}_{i}(q_{i},t)\leq C_{2}$

for $t\in[-4L_{1},0]$.

Claim 1. Suppose $u$ is a nonnegative function such that

$\displaystyle\tilde{\square}_{i}u\leq 0,\quad\textrm{on}\;P^{2}_{i}.$

Then we have

$\displaystyle\max_{P^{1}_{i}}u^{2}\leq
C_{3}\iint_{P^{2}_{i}}u^{2}\,dV_{\tilde{g}_{i}(t)}dt$ (4.25)

for some $C_{3}>0$.

_Proof of Claim 1:_ The proof follows verbatim as [44, Lemma $9.7$] by using
the Moser iteration. Notice that the Sobolev inequality and the control of
$R_{\tilde{g}_{i}(t)}$ on $P^{2}_{i}$ are guaranteed by Theorem 4.5 and (4.24)
respectively. Moreover, the cutoff functions $\eta_{k}$ in the proof of [44,
Lemma $9.7$] can be defined similarly by using $\tilde{F}_{i}$ and $\eta_{k}$
can be estimated similarly by using (4.23). We omit the details.

Next we define for any $\tau>0$,

$\displaystyle P_{i}(\tau)=P^{2}_{i}\cap\\{(z,t)\,\mid
R_{\tilde{g}_{i}(t)}(z)<\tau\\}.$

Claim 2. There exists a constant $C_{4}>0$ such that

$\displaystyle|P_{i}(2i^{-2})|\geq C_{4}.$ (4.26)

Here the volume is with respect to $dV_{\tilde{g}_{i}(t)}dt$.

_Proof of Claim 2:_ It follows from (4.20) that $(x_{i},0)\in P^{1}_{i}$. Then
the proof follows exactly as that of [44, Lemma $9.8$] by using (4.25).

It is clear from (4.26) that for any $i$ sufficiently large, there exists
$t_{i}\in[-2L_{1},0]$ such that

$\displaystyle|\Sigma_{i}|_{dV_{\tilde{g}_{i}(t_{i})}}\geq C_{5},$

where $\Sigma_{i}=M_{i}\times\\{t_{i}\\}\cap P_{i}(2i^{-2})$ and
$C_{5}=\frac{C_{4}}{2L_{1}}$. Now, we define

$\displaystyle\Omega_{i}:=\psi_{i}^{s_{i}^{-1}t_{i}}\left(\text{pr}(\Sigma_{i})\right)$

where pr is the projection onto $M_{i}$. Then we compute

$\displaystyle|\Omega_{i}|_{dV_{\tilde{g}_{i}(0)}}\geq\frac{1}{2}(1-s_{i}^{-1}t_{i})^{\frac{n}{2}}|\Omega_{i}|_{dV_{\tilde{g}_{i}(0)}}=\frac{1}{2}|\Sigma_{i}|_{dV_{\tilde{g}_{i}(t_{i})}}\geq
C_{6},$ (4.27)

where $C_{6}=C_{5}/2$. On the other hand, for any $(z,t_{i})\in\Sigma_{i}$, we
have

$\displaystyle\tilde{F}_{i}(z,t_{i})=F_{i}(z,t)-s_{i}=(1-s_{i}^{-1}t_{i})f_{i}(z^{\prime},0)-s_{i}$

where $z^{\prime}=\psi_{i}^{s_{i}^{-1}t_{i}}(z)$. Therefore, it follows from
the definition of $P^{2}_{i}$ and (4.18) that

$\displaystyle|\tilde{f}_{i}(z^{\prime})|=|f_{i}(z^{\prime},0)-s_{i}|\leq
4L_{1}+1.$ (4.28)

As in the proof of Proposition 4.2, by taking a subsequence if necessary, we
have

$\displaystyle(M_{i},q_{i},\tilde{g}_{i},\tilde{f}_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{pointed-\hat{C}^{\infty}-Cheeger-
Gromov}\left(M_{\infty},q_{\infty},g_{\infty},f_{\infty}\right),$

where $(M_{\infty},q_{\infty},g_{\infty},f_{\infty})$ is a steady soliton
conifold. Notice that $\lim_{z\to\infty}f_{\infty}(z)=-\infty$ by our
assumptions. Combining (4.27) and (4.28), we conclude that there exists a
point $z\in\mathcal{R}$ such that

$\displaystyle R_{g_{\infty}}(z)=0.$

By applying the strong maximum principle for
$\Delta_{f_{\infty}}R_{\infty}=-2|Rc_{\infty}|^{2}\leq 0$, we conclude that
$R_{\infty}\equiv 0$ on $\mathcal{R}$. However, this contradicts the fact that
$R_{\infty}(q_{\infty})=\epsilon_{1}$.

In sum, the proof is complete. ∎

As observed above, the cap region is modeled on a steady soliton conifold if
$\epsilon_{1}^{\prime}$ is sufficiently small. In general, it is difficult to
classify all steady soliton conifolds even though there are no singularities.
However, under some special conditions, we can classify all such steady
soliton conifolds. For the definition of PIC2, see [2, Section $7.5$].

######  Theorem 4.7.

Let $(X,p,d,f)\in\mathfrak{S}(n)$ be non-compact such that $(\mathcal{R},g)$
has _PIC2_. Suppose there exists a compact set $C$ satisfying

  * (a).

The singular part $\mathcal{S}\subset C$.

  * (b).

For any $x\in X\backslash C$, the scalar curvature satisfies
$\dfrac{c_{1}}{d(p,x)}\leq R(x)\leq\dfrac{c_{2}}{d(p,x)}$, where $c_{1},c_{2}$
are two positive constants.

  * (c).

For any sequence $q_{i}\to\infty$, $q_{i}$ is the center of an evolving
$\epsilon_{i}$-neck with $\epsilon_{i}\to 0$.

Then $(X,d)$ is isometric to the Bryant soliton up to scaling.

###### Proof.

Since $(R,g)$ has PIC2 on $\mathcal{R}$, it implies that $sec\geq 0$ on
$\mathcal{R}$. Combined with the identity $R+|\nabla f|^{2}=1$ as $(X,d)$ is
nontrivial, we conclude that $|Rm|$ is uniformly bounded on $\mathcal{R}$. For
any $x\in\mathcal{R}$, we define $\lambda=\lambda(x)$ to be the largest
constant such that $S:=Rm-\lambda I$ has weakly PIC2. By our assumption,
$\lambda>0$ on $\mathcal{R}$.

Claim 1. $\lambda$ satisfies the following inequality on $\mathcal{R}$.

$\displaystyle\Delta_{f}\lambda\leq 0$

_Proof of the Claim 1:_ By our definition, $S$ is contained in the boundary of
PIC2 cone. For fixed $x\in\mathcal{R}$, there exists a four-frame
$\\{e_{1},e_{2},e_{3},e_{4}\\}\in T_{x}M$ and constants $u,v\in[0,1]$ such
that

$\displaystyle
S_{1313}+u^{2}S_{1414}+v^{2}S_{2323}+u^{2}v^{2}S_{2424}-2uvS_{1234}=0.$

Therefore, it follows from [2, Proposition $7.21$] that

$\displaystyle
Q(S)_{1313}+u^{2}Q(S)_{1414}+v^{2}Q(S)_{2323}+u^{2}v^{2}Q(S)_{2424}-2uvQ(S)_{1234}\geq
0,$ (4.29)

where $Q$ is a quadratic term of the curvature tensor, see [31] for its
precise definition.

In addition, we have

$\displaystyle Q(S)=Q(Rm)-2\lambda
Rc\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}\text{id}+2(n-1)\lambda^{2}I,$
(4.38)

where
$\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}$
is the Kulkarni-Nomizu product. Therefore, it follows from (4.29) and (4.38)
that

$\displaystyle
Q(Rm)_{1313}+u^{2}Q(Rm)_{1414}+v^{2}Q(Rm)_{2323}+u^{2}v^{2}Q(Rm)_{2424}-2uvQ(Rm)_{1234}$
$\displaystyle\geq$ $\displaystyle
2\lambda\left(R_{11}+R_{33}+u^{2}(R_{11}+R_{44})+v^{2}(R_{22}+R_{33})+u^{2}v^{2}(R_{22}+R_{44})\right)$
$\displaystyle-2(n-1)\lambda^{2}(1+u^{2}+v^{2}+u^{2}v^{2})\geq 0,$

where for the last inequality, we have used $R_{ii}\geq(n-1)\lambda$ since $S$
has weakly PIC2. Therefore, we conclude that

$\displaystyle\Delta_{f}\lambda\leq$
$\displaystyle\frac{\Delta_{f}(Rm)_{1313}+u^{2}\Delta_{f}(Rm)_{1414}+v^{2}\Delta_{f}(Rm)_{2323}+u^{2}v^{2}\Delta_{f}(Rm)_{2424}-2uv\Delta_{f}(Rm)_{1234}}{(1+u^{2})(1+v^{2})}$
$\displaystyle=$
$\displaystyle-\frac{Q(Rm)_{1313}+u^{2}Q(Rm)_{1414}+v^{2}Q(Rm)_{2323}+u^{2}v^{2}Q(Rm)_{2424}-2uvQ(Rm)_{1234}}{(1+u^{2})(1+v^{2})}\leq
0$

at $x$, where we have used the equation $\Delta_{f}Rm=-Q(Rm)$.

Hence, we have on $\mathcal{R}$,

$\displaystyle\Delta_{f}\lambda\leq 0.$

Since $\lambda$ is positive on $\mathcal{R}$, it follows from Proposition A.10
that

$\displaystyle\inf_{K\cap\mathcal{R}}\lambda>0$ (4.39)

for any compact set $K$. Since the regular part $\mathcal{R}$ is geodesically
convex and $(X,d)$ is the completion of $(\mathcal{R},g)$, it follows from
[61] that $(X,d)$ is an Alexandrov space with $sec\geq 0$.

Claim 2. $f$ is a concave function on $(X,d)$ with a unique critical point.

_Proof of the Claim 2:_ Since $\text{Hess}\,f=-Rc<0$, it is clear that $f$ is
concave on $(\mathcal{R},g)$ in the sense that for any geodesic $\gamma(t)$
contained in $\mathcal{R}$, $f(\gamma(t))$ is concave. Now we fix a geodesic
$\gamma(t)_{t\in[0,L]}$ in $X$ such that $\gamma(0)=x$ and $\gamma(L)=y$. For
any small number $\epsilon>0$, we set $x^{\prime}=\gamma(\epsilon)$ and
$y^{\prime}=\gamma(L-\epsilon)$. Since $\mathcal{R}$ is dense, we can take two
sequences $x_{n}\to x^{\prime}$ and $y_{n}\to y^{\prime}$ such that $x_{n}$
and $y_{n}$ are contained in $\mathcal{R}$. From the geodesical convexity of
$\mathcal{R}$, there exists a geodesic $\gamma_{n}\subset\mathcal{R}$
connecting $x_{n}$ and $y_{n}$. Since the geodesic in an Alexandrov space is
not branching, $\gamma_{n}$ converges to
$\gamma(t)_{t\in[\epsilon,L-\epsilon]}$. By taking the limit, we immediately
conclude that $f(\gamma(t))$ is concave for $t\in[\epsilon,L-\epsilon]$. Since
$\epsilon>0$ is arbitrary, $f(\gamma(t))$ is concave for any $t\in[0,L]$.
Therefore, $f$ is concave in the sense of [60, Definition $1.1$].

We fix a point $x\in X$. Starting from $x$, we denote the gradient flow of $f$
by $\alpha(t)$ for $t\geq 0$, see [60, Section 2]. If
$\alpha(t)_{t\in[0,\infty)}$ is contained in a compact set $K$ of $X$, then
there exists a sequence $t_{i}\to\infty$ such that $|\nabla
f|(\alpha(t_{i}))\to 0$, since $f$ is bounded on $K$. By taking a subsequence,
we may assume $\alpha(t_{i})$ converges to a point $y\in K$. By the lower
semicontinuity of $|\nabla f|$ [60, Corollary $1.3.5$], $y$ is a critical
point of $f$. If $\alpha(t)_{t\in[0,\infty)}$ is not contained in any compact
set, there exists a sequence $s_{i}$ such that $\alpha(s_{i})\to\infty$. By
our assumption (b), $R(\alpha(s_{i}))\to 0$. However, this contradicts the
fact that $R$ is increasing along $\alpha(t)$ as long as $\alpha(t)$ is
outside $C$ since $\langle\nabla R,\nabla f\rangle=2Rc(\nabla f,\nabla f)\geq
0$. In any case, $f$ has a critical point. Without loss of generality, we set
$p$ to be a critical point of $f$.

It follows from (4.39) that for any constant $c$, there exists a constant
$\delta>0$ such that $(X_{c},d)$ is an Alexandrov space with $sec\geq\delta$,
where $X_{c}=\\{x\in C\mid f(x)\geq c\\}$. By this reason, $f$ is a strictly
concave function on $X$. Therefore, it is a standard fact that $p$ is the
unique critical point which is also the maximum point.

Claim 3. $X\backslash\\{p\\}\subset\mathcal{R}$.

_Proof of the Claim 3:_ Otherwise, we set $q\in\mathcal{S}$ and $q\neq p$ such
that $f(q)=t$ is minimal. Notice that by our assumption (b), the choice of $t$
is possible. By our assumption, $|\nabla f|(q)>0$ since $p$ is the only
critical point of $f$. From the lower semicontinuity of $|\nabla f|$ [60,
Corollary $1.3.5$], we conclude that there exists a small $\delta>0$ such that

$\displaystyle|\nabla f|\geq\delta$ (4.40)

on $B(q,\delta)$. Now we take any positive sequence $r_{i}\to 0$ and define
$d_{i}=r_{i}^{-1}d$, $g_{i}=r_{i}^{-2}g$ and $f_{i}=r_{i}^{-1}(f-f(q))$. From
the definition, it is clear that on $\mathcal{R}$

$\displaystyle\text{Hess}_{g_{i}}f_{i}=-r_{i}^{-1}Rc_{i}.$ (4.41)

Now we have

$\displaystyle(X,q,d_{i},f_{i})\ext@arrow
0099{\arrowfill@--\rightarrow}{}{pointed-Gromov-
Hausdorff}\left(K_{q},0,d_{\infty},f_{\infty}\right),$

where $(K_{q},d_{\infty})$ is the tangent cone at $q$ and $f_{\infty}$ is a
locally Lipschitz function. In addition, it follows from (4.41) and the same
arguments in Claim 1 that $f_{\infty}$ is both concave and convex on $K_{q}$.

For any $s<t$, it is clear from our choice of $t$ that $f^{-1}(s)\cap
B(q,\delta)\subset\mathcal{R}$ on which the $|Rm|$ is uniformly bounded.
Moreover, for any $x\in f^{-1}(s)\cap B(q,\delta)$, it follows from (4.40)
that the second fundamental form at $x$, which is
$\frac{\text{Hess}\,f}{|\nabla f|}(x)$, is uniformly bounded. Therefore, we
conclude that

$\displaystyle\left(f_{i}^{-1}((-\infty,0)),q,g_{i},f_{i}\right)\ext@arrow
0099{\arrowfill@--\rightarrow}{}{C^{\infty}-Cheeger-
Gromov}\left(f_{\infty}^{-1}((-\infty,0)),0,d_{\infty},f_{\infty}\right).$

Here, for the smooth convergence, we have used $\Delta_{f}Rm=-Q(Rm)$ to obtain
the higher order estimates of $Rm$ on $f^{-1}(s)\cap B(q,\delta)$. Moreover,
we have

$\displaystyle(f_{\infty}^{-1}((-\infty,0)),d_{\infty})=(\mathbb{R}^{n}\cap\\{x_{n}<0\\},g_{E}).$

In particular, it follows from the smooth convergence of $f_{i}$ to
$f_{\infty}$ on $f_{\infty}^{-1}((-\infty,0))$ and (4.40) that $f_{\infty}$ is
not a constant. Since $f_{\infty}$ is both convex and concave, it is clear
(see [60, Lemma $2.1.4$]) that $(K_{q},d_{\infty})$ is isometric to
$(Y\times\mathbb{R},d^{\prime}\times g_{E})$, where $Y$ is a metric cone
isometric to any level set of $f_{\infty}$. Therefore, it is immediate from
(4.41) that $Y$ is smooth and hence isometric to $\mathbb{R}^{n-1}$. From
this, we conclude that $(K_{q},d_{\infty})$ is isometric to
$(\mathbb{R}^{n},g_{E})$. However, this implies that $q\in\mathcal{R}$, which
contradicts our assumption.

Claim 4. $p\in\mathcal{R}$.

_Proof of the Claim 4:_ From Claim 2, any point in $X\backslash\\{p\\}$ is
regular with uniformly bounded $|Rm|$. Therefore, for any sequence $r_{i}\to
0$, the convergence

$\displaystyle(X,p,r_{i}^{-1}d)\ext@arrow
0099{\arrowfill@--\rightarrow}{}{pointed-Gromov-
Hausdorff}\left(K_{p},0,d_{\infty}\right)$

is smooth away from the vertex. Moreover,

$\displaystyle(K_{p},d)=(\mathbb{R}^{n}/\Gamma,g_{E})$ (4.42)

where $\Gamma\leq O(n)$ is a finite subgroup acting freely on $S^{n-1}$. If we
set the nontrivial level set of $f$ by $\Sigma$, then near $p$ we have two
foliations by $\Sigma$ and $S^{n-1}/\Gamma$ respectively. Then it is easy to
see that $\Sigma$ and $S^{n-1}/\Gamma$ are $h$-cobordant. Indeed, by our
assumption, there exist a region in $M$ homeomorphic to $\Sigma\times[0,4]$
and an embedding of $N\times[1,3]$ into the interior of $\Sigma\times[1,4]$
such that $N\times\\{k\\}$ separates $\Sigma\times\\{k\\}$ and
$\Sigma\times\\{k+1\\}$ for $1\leq k\leq 3$. By an easy topological argument,
the compact manifold bounded by $N\times\\{1\\}$ and $\Sigma\times\\{3\\}$ is
a cobordism such that the boundary inclusions are homotopy equivalences. In
particular, $\Sigma$ and $N$ are homotopic and hence $\Gamma$ must be trivial.
From (4.42), we have

$\displaystyle(K_{p},d)=(\mathbb{R}^{n},g_{E})$

and hence $p\in\mathcal{R}$.

In sum, we have proved that $\mathcal{S}=\emptyset$ and hence $(X,p,d,f)$ is a
smooth steady soliton. The rest of the proof follows from the classification
theorem of Brendle [3, Theorem $1.2$]. ∎

Combining the results in Section 3 and Section 4, we are able to prove Theorem
1.2 now. Let us first recall and fix notations.

For any constants $n,\delta_{0},A,B$ and $\epsilon$, we define the following
constants:

  * •

Let
$\bar{\sigma}\coloneqq\min\\{\sigma_{1}(n,B),\sigma_{2}(n,\delta_{0},A,B),\sigma_{3}(n,\delta_{0},A,B),\sigma_{4}(n,\delta_{0},A,B),\sigma_{5}(n,A,B)\\}$,
where $\sigma_{1}$, $\sigma_{2}$, $\sigma_{3}$, $\sigma_{4}$ and $\sigma_{5}$
are functions in Proposition 3.4, Proposition 3.11, Proposition 3.16,
Proposition 3.19 and Proposition 3.22, respectively.

  * •

Let
$\bar{\eta}\coloneqq\min\\{\eta_{3}(\epsilon,n,\delta_{0},A,B),\eta_{4}(\epsilon,n,\delta_{0},A,B),\eta_{5}(\epsilon,n,\delta_{0},A,B)\\}$,
where $\eta_{3}$, $\eta_{4}$ and $\eta_{5}$ are functions in Proposition 3.12,
Proposition 3.17 and Proposition 3.21, respectively.

  * •

Let $\sigma\coloneqq\min\\{\bar{\sigma}^{2},\bar{\eta}^{2}\\}$.

  * •

Let
$\eta\coloneqq\min\\{\sigma^{\frac{1}{2}},\bar{\eta},\tau_{1}(\sigma,n,\delta_{0},A,B),\tau_{2}(\sigma,n,\delta_{0},A,B),\tau_{3}(\sigma,n,\delta_{0},A,B)\\}$,
where $\tau_{1}$, $\tau_{2}$ and $\tau_{3}$ are functions in Proposition 4.2,
Corollary 4.4 and Proposition 4.6, respectively.

  * •

Let
$L\coloneqq\max\\{L_{1}(\sigma,n,\delta_{0},A,B),L_{2}(\sigma,n,\delta_{0},A,B)\\}$,
where $L_{1}$ and $L_{2}$ are functions in Proposition 4.2 and Proposition
4.6, respectively.

We close this section with the proof of Theorem 1.2.

###### Proof of Theorem 1.2.

From our assumptions, the condition (1.5) in particular holds for
$\epsilon_{1}=\sigma$ and $\epsilon_{1}^{\prime}=\eta$. With our choice of
$\sigma,\eta$ and $L$, it is clear from the results of Section $3$ and Section
$4$ that the conclusion holds. ∎

##  5 Proof of the main theorem

The purpose of this section is to prove Theorem 1.1. As we discussed in the
introduction, the key ingredient is to show each point in the Ricci shrinker
is either in a cylinder-like neighborhood or in a Bryant-soliton-like
neighborhood, then we apply the symmetry improvement argument.

We recall the following definition of the curvature cone from [8]. Here
$\mathscr{C}_{B}(\mathbb{R}^{n})$ denotes the vector space of algebraic
curvature tensor.

######  Definition 5.1.

A cone $\hat{C}$ in $\mathscr{C}_{B}(\mathbb{R}^{n})$ has the property $(*)$
if it satisfies the following conditions:

  * •

$\hat{C}$ is closed, convex, $O(n)$-invariant, and of full-dimension.

  * •

$\hat{C}$ is transversally invariant under Hamilton’s ODE:
$\dfrac{dRm}{dt}=Q(Rm)$.

  * •

Every algebraic curvature tensor $Rm\in\hat{C}\backslash\\{0\\}$ has positive
scalar curvature.

  * •

The identity $I$ lies in the interior of $\hat{C}$.

For later applications, we mainly consider the following example from [4,
Definition $2.2$].

######  Example 5.2.

For any parameter $\sigma\in(0,2)$ and $\theta\geq 0$, we define the cone
$C_{\sigma,\theta}$ as the set

$\displaystyle\left\\{Rm=S+H\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}\emph{id}\mid
S\in\emph{PIC2}\,,Rc_{0}(S)=0,\,\emph{tr}(H)\emph{id}-(n-2\sigma)H\geq
0,\,\emph{tr}(H)-\theta\emph{scal}(S)\geq 0\right\\},$

where $H$ is a symmetric bilinear form,
$\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}$
is the Kulkarni-Nomizu product, $Rc_{0}$ is the traceless Ricci, and _PIC2_
denotes the cone consists of curvature tensor satisfying the weakly PIC2
condition. It follows from _[4, Theorem $2.5$]_ and the definition that
$C_{\sigma,\theta}$ has the property $(*)$ if $\sigma\in(1,2)$ and
$\theta\in(0,\bar{\theta}(n))$ for a constant $\bar{\theta}$ depending only on
$n$.

We denote $C_{1+\theta,\theta}$ by $\hat{C}(\theta)$ for
$\theta\in(0,\min\\{1/2,\bar{\theta}\\})$. It is clear from the definition
that

$\displaystyle Rc\geq-C(n)\theta R$ (5.1)

for any $Rm\in\hat{C}(\theta)$. Furthermore, it is easy to see that the
curvature tensor of $S^{n-1}\times\mathbb{R}$ lies in $\hat{C}(\theta)$.
Indeed, we can write the curvature tensor as
$Rm=H\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}\emph{id}$,
where $H_{ij}=\delta_{ij}$ for $i,j\geq 2$ and $H_{1i}=-\delta_{1i}$ and hence
it is easy to check from the definition.

If a cone $\hat{C}$ has the property $(*)$, we have the following result (see,
e.g., [55, Lemma $4.2$]).

######  Lemma 5.3.

Let $\hat{C}$ in $\mathscr{C}_{B}(\mathbb{R}^{n})$ be a cone with property
$(*)$. Then there exists two positive constants $\tau$ and $K$ depending only
on the cone such that

$\displaystyle Q(Rm)-\tau R^{2}I\in T_{Rm}\hat{C}\quad\text{and}\quad|Rm|\leq
KR$

for any $Rm\in\hat{C}$.

We prove the following theorem, which implies that locally the Ricci flow
associated with a Ricci shrinker almost preserves the cone $\hat{C}$. Recall
that on the Ricci flow associated with a Ricci shrinker, $F(x,t):=(1-t)f(x,t)$
for any $t<1$. Moreover, the constant $\tau$ and $K$ are from Lemma 5.3 which
depends only on the cone $\hat{C}$.

######  Theorem 5.4.

Let $\hat{C}$ in $\mathscr{C}_{B}(\mathbb{R}^{n})$ be a cone with $(*)$. There
exist constants $\delta=\delta(n,\tau,K)>0$ and $C=C(n)>0$ satisfying the
following property. For any Ricci flow $(M^{n},g(t))_{t<1}$ associated with a
Ricci shrinker, we define $\lambda(x,t)$ be the minimal number such that
$Rm(x,t)+\lambda(x,t)I\in\hat{C}$, for any $(x,t)\in M\times(-\infty,1)$. For
any $r>0,\epsilon>0$, if $\epsilon+r^{-2}\leq\delta R$ on $P_{2}:=\\{(x,t)\mid
F(x,t)\leq 2r,\,0\leq t<1\\}$ and $\lambda\leq\epsilon$ on $\\{(x,0)\mid
F(x,0)\leq 2r\\}$, then

$\displaystyle\lambda\leq\epsilon+Cr^{-2}$

on $P_{1}:=\\{(x,t)\,\mid F(x,t)\leq r,0\leq t<1\\}$.

###### Proof.

We fix a smooth function $\eta$ on $[0,\infty)$ such that $0\leq\eta\leq 1$,
$\eta=1$ on $[0,1]$ and $\eta=0$ on $[2,\infty)$. Furthermore,
$|\eta^{\prime}|^{2}\leq C\eta^{\frac{3}{2}}$ and $|\eta^{\prime\prime}|\leq
C\eta^{\frac{1}{2}}$ for a universal constant $C$. If we set
$\phi=\phi_{r}=\eta\left(\frac{F}{r}\right)$, it follows from the identities
of $F$ (see [48, Section $3$] for details) that

$\displaystyle\quad|\nabla\phi|^{2}\leq
C\phi^{\frac{3}{2}}r^{-1}\quad\text{and}\quad|\square\phi|\leq
C\phi^{\frac{1}{2}}r^{-1}$ (5.2)

for some constant $C$ depending only on $n$.

From our definition, the curvature operator

$\displaystyle S\coloneqq Rm+\lambda I$ (5.3)

lies on the boundary of $\hat{C}$. Therefore, it follows from Lemma 5.3 that

$\displaystyle Q(S)-\tau\text{scal}^{2}(S)I\in T_{S}\hat{C},$ (5.4)

where $\text{scal}(S)$ is the scalar curvature of $S$. Direct calculation
yields that

$\displaystyle Q(S)=Q(Rm)+2\lambda
Rc\mathchoice{\mathbin{\ooalign{$\displaystyle\bigcirc$\cr$\displaystyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\textstyle\bigcirc$\cr$\textstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptstyle\bigcirc$\cr$\scriptstyle\wedge$\cr}}{}}{\mathbin{\ooalign{$\scriptscriptstyle\bigcirc$\cr$\scriptscriptstyle\wedge$\cr}}{}}\text{id}+2(n-1)\lambda^{2}I,$
(5.13) $\displaystyle\text{scal}(S)=R+n(n-1)\lambda.$ (5.14)