How do steady solitons look from far away?

September 15, 2026

These are my notes for my talk at Rutgers University Geometric Analysis seminar, Sept 15, Fall 2026.

Everything in this talk is joint work with my adviser, Natasa Sesum, which can be found here. The main motivating question for this talk is the following.

Question. Does the scalar curvature vanish at infinity on a $\kappa$-noncollapsed $4$D steady soliton with $\operatorname{Rm}>0$?

Although there are several mathematical reasons why this question is interesting, some of which we will discuss in the talk, I would like to recall an anecdote on why I started working on it. My adviser gave me two problems last summer and one of them was this. She warned me that this might be a difficult problem and advised me not to work on it right away. But somehow I felt curious about this one. One reason is how easy it is to state this! Within a month, I was hooked onto this problem and I thought about this for a long time that culminated in the paper. Today I hope to show why steady solitons are interesting objects to study.

To begin with, we will define steady solitons. Let $(M^4,g)$ be a Riemannian manifold. By this we mean at each point $p\in M$, $g_p$ is an inner product on $T_p M$ and $p\mapsto g_p$ is smooth over $M$. Using $g$, one can define the Riemann $\operatorname{Rm}$, Ricci $\operatorname{Ric}$, and scalar curvature $R$. We say $(M^4,g,f)$ is a steady soliton if $f:M\to \mathbb{R}$ is smooth and on $M$,

$$ \operatorname{Ric}+\nabla^2 f=0. $$

The function $f$ is called the potential. Adding a constant to $f$, or scaling the metric, or composing by an isometry, doesn’t change the equation.

These generalize Ricci flat manifolds, in the sense that Ricci flat manifolds are trivial steady solitons. But there is nothing trivial about $4$D Ricci flat manifolds! In particular, the flat $\mathbb{R}^4$ is a steady soliton with the Euclidean metric and this is called the Gaussian steady soliton.

In studying any big class of mathematical objects (such as solutions to a PDE), understanding the simple sub-class of objects paves the way to understand more complex objects within the class. In a sense, the complex objects may be composed of these simple objects!

Steady solitons play an important role in a PDE called the Ricci flow because they serve as one class of prototypical solutions.

Given a manifold $\mathcal{M}$ and metrics $g(t),t\in [a,b]$ on $\mathcal{M}$, we say $(\mathcal{M},g(t))$ is a solution to the Ricci flow if
$$ \frac{\partial}{\partial t}g(t)=-2\operatorname{Ric}_{g(t)}. $$
In other words, you are moving the metrics $g(t)$ by their Ricci curvature. You can obtain Ricci flow solutions given a steady solitons as follows: Let $\Phi_t$ denote the one-parameter family of diffeomorphisms generated by $\nabla f$ with $\Phi_0=\operatorname{id}_M$ and set $g(t):=\Phi_t^* g$. Then, $(M,g(t))_{t\in (-\infty,\infty)}$ is an eternal solution to the Ricci flow, called the canonical Ricci flow induced by $g$.

We say these are eternal because they have no finite time singularities! In general Ricci flow behaves in a very complicated way in $\ge 3$ dimensions. But for steady solitons, we can track precisely how the metric evolves! This is the reason why I call these prototypical.

More generally if there exists $\lambda \in \mathbb{R}$ such that $\operatorname{Ric}+\nabla^2 f=\lambda g$, we call $(M,g,f)$ as solitons, which all serve as prototypical solutions to Ricci flow. These also move by diffeomorphisms and additionally by scaling.

Low dimensional classification

In the study of Ricci flow, dimensions 2 and 3 are often called as the low dimensions. One of the main reasons this is easier is that the Ricci curvature controls the full Riemann curvature in these dimensions. When I say they are easy, I by no means am implying that they are trivial! I only mean that they are easier when compared to dimension 4.

Dimension 2

Lets begin with dimension 2. Hamilton constructed the cigar soliton,

$$ g_{\Sigma}=4\frac{d x^{2}+d y^{2}}{1+x^{2}+y^{2}}=4\frac{d r^{2}+r^{2} d \theta^{2}}{1+r^{2}} $$

It turns out by a simple computation that this is a complete nonflat steady soliton. Note that the metric is rotationally symmetric in the sense that it is invariant under $O(2)$-action on the manifold. We can write this metric in a more suitable form by putting

$$ s=2\int_{-\infty}^{r} \sqrt{\frac{1}{1+r'^{2}}} d r^{\prime}=2\operatorname{arcsinh} r=2\log \left(r+\sqrt{1+r^{2}}\right) $$

and after a computation, get

$$ g_{\Sigma}=d s^{2}+\varphi(s)^2 d \theta^{2}\qquad \varphi:=4\tanh ^{2} (\frac{s}{2}) $$

The potential function for the cigar soliton is

$$ f(s)=-2\log(\cosh(\frac{s}{2}))=-2\log(1+r^2) $$

The variable $s\sim d(o,\cdot)$ here serves as distance to the origin. One interesting observation is as follows: the scalar curvature is given by

$$ R_{\Sigma}=-2 \frac{\varphi^{\prime \prime}(s)}{\varphi(s)}=\frac{4}{\left(e^{s/2}+e^{-s/2}\right)^{2}} $$

which shows that the scalar curvature is exponentially decaying $R_\Sigma \sim e^{-d(o,\cdot)}$. Because $\varphi^2$ converges exponentially fast to $1$, when you move to infinity, the metric converges exponentially fast (in $s$) to $g_{cyl}=ds^2+4d\theta^2$. In particular, we have for any $x_k\to \infty$,

$$ (\mathbb{R}^2,g_\Sigma,x_k)\to ( \mathbb{R}\times S^1,ds^2+4d\theta^2,x_\infty) $$

in the smooth sense. In other words, the manifold looks like a cigar!

This is a good place to define what is noncollapsed-ness in the question statement above. Perelman observed that Ricci flow on compact manifolds have a very important property.

(Perelman) Let $\left(\hat{\mathcal{M}}^n, \hat{g}\right)$ be a complete Riemannian manifold. Given $\rho \in(0, \infty]$ and $\kappa>0$, we say that the metric $\hat{g}$ is $\kappa$-noncollapsed below the scale $\rho$ if for any metric ball $B(x, r)$ with $r<\rho$ satisfying $|R(y)|\leq r^{-2}$ for all $y \in B(x, r)$, we have
$$ \frac{\operatorname{Vol} B(x, r)}{r^n} \geq \kappa $$
By $\kappa$-noncollapsed, we mean $\kappa$-noncollapsed below the scale of $\rho$ for any $\rho>0$.

Roughly being $\kappa$-noncollapsed means that the manifold doesn’t collapse at places where the scalar curvature is well controlled.

Perelman observed that this naturally holds in compact Ricci flows.

(Perelman) Let $\left(\mathcal{M}^n, g(t)\right), t \in[0, T), T<\infty$, be a solution to the Ricci flow on a closed Riemannian manifold. Suppose $T<\infty$. For any $\rho \in(0, \infty)$ there exists $\kappa=\kappa(g(0), T, \rho)>0$ such that $g(t)$ is $\kappa$-noncollapsed below the scale $\rho$ for all $t \in[0, T)$:

This theorem is one of most important theorems proved by Perelman in his first paper in 2002.

Testing this for the cigar soliton, we can conclude that since $g_\Sigma$ looks asymptotically like $S^1 \times \mathbb{R}$, which is flat!, $g_\Sigma$ is collapsed (meaning there doesn’t exist any $\kappa>0$ such that $g_\Sigma$ is $\kappa$-noncollapsed). Very importantly this means that these don’t arise in the Ricci flow near the singularity due to Perelman’s theorem above.

Continuing with our story, one can show the following:

(Hamilton) Any complete two-dimensional gradient steady Ricci soliton is either flat or isometric, up to scaling, to the cigar soliton.

Key idea. One first proves that any such soliton is rotationally symmetric: $g=du^2+\alpha (u)^2g_{S^1}$ and the potential function only depends on $u$. The PDE satisfied by $(g,f)$ turns to an ODE system in $\alpha$. It turns out that $\alpha$ is unique.

Dimension 3

Moving on to dimension 3, R. Bryant constructed a complete nonflat noncompact rotationally symmetric steady soliton, which are now called the Bryant soliton, as follows.

Consider the ansatz $\tilde g=dr^2+w(r)^2g_{S^2}$ and radial $f(r)$ on $\mathbb{R}^3$. We put the following conditions on $(\tilde g,f)$: $(M,\tilde g,f)$ is a complete steady soliton. The steady soliton equation then gives the following nonlinear system in $(w,f)$

$$ f^{\prime \prime} =2 w^{\prime \prime} / w \qquad\qquad w w^{\prime} f^{\prime} =w w^{\prime \prime}+\left(\left(w^{\prime}\right)^2-1\right). $$

This ODE system is invariant under $r\mapsto r+$const and $f\mapsto f+$const. One can simplify this a lot further. Studying this system using ODE techniques (such as linearization), Bryant showed that the solution $(w,f)$ is exists and is unique. Using ODE expansions, we can get the behavior of $w$, $f$, but it is still unknown (and unlikely) if there is an exact form for $w,f$. This method actually works in all dimensions gives us a steady soliton $\operatorname{Bry}^n$ and its properties are summarized below.

(Bryant soliton '05) For each $n \geq 3$ there exists a complete nonflat rotationally symmetric steady soliton $(\mathbb{R}^n,\tilde g)$ called the Bryant soliton, written $\operatorname{Bry}^n$. If $\tilde g=dr^2+w(r)^2g_{S^{n-1}}$, then
$$ w(r)=\mathrm{O}\left(r^{1 / 2}\right), \quad w^{\prime}(r)=\mathrm{O}\left(r^{-1 / 2}\right), \quad-w^{\prime \prime}(r)=\mathrm{O}\left(r^{-3 / 2}\right) $$
the sectional curvatures satisfy
$$ K_{\mathrm{orb}}=\mathrm{O}\left(r^{-1}\right) , \quad K_{\mathrm{rad}}=\mathrm{O}\left(r^{-2}\right) $$
and the scalar curvature satisfies
$$ R\sim \frac{n-1}{2 r} $$
Moreover, the metric has positive curvature operator everywhere.

Notice that the scalar curvature decays linearly, i.e. $R\sim d(o,\cdot)^{-1}$ which is very different from the exponential decay for the cigar soliton. Using the asymptotics for $w$, we see that

$$ r\tilde g=rdr^2+rw(r)^2g_{S^{n-1}}\sim dr'^2+g_{S^{n-1}} $$

as $r\to \infty$ where the rate is polynomial of some order. Making this argument rigorous, one concludes that if $x_k\to \infty$ on $\operatorname{Bry}^n$, then

$$ (M,d(x_k,o)^{-1}g,x_k)\to (\mathbb{R}\times S^{n-1},dz^2+g_{S^{n-1}},x_\infty) $$

smoothly. Since $R(x_k)\sim d(x_k,o)^{-1}$, we have

$$ (M,R(x_k)g,x_k)\to (\mathbb{R}\times S^{n-1},dz^2+g_{S^{n-1}},x_\infty) $$

smoothly. In other words, after rescaling, the Bryant soliton is asymptotically cylindrical. Because $\mathbb{R}\times S^{n-1}$ is $\kappa_n$-noncollapsed for some $\kappa_n$, it follows that $\operatorname{Bry}^n$ is $\kappa_n’$-noncollapsed for some $\kappa_n,\kappa_n’$, the constant is not important.

For a long time, much of the classification theory aimed to see whether these properties that the Bryant soliton satisfies forces a steady soliton to be rotationally symmetric, and hence isometric (up to scaling) to the Bryant soliton. The following theorem of Brendle has had lasting implications in the study of 3D Ricci flow.

(Brendle '13) Any complete nonflat $\kappa$-noncollapsed three-dimensional steady gradient Ricci soliton is isometric, up to scaling, to the Bryant soliton.

Note that there really is no other requirement in $3$D other than just being noncollapsed! You may ask, “anyway, why do we need to classify these steady solitons”. I will come to that question very soon. But before that, when the techniques that go into the above theorem are pushed in higher dimensions, Brendle obtained the following conclusion.

(Brendle '13) Any $\kappa$-noncollapsed asymptotically cylindrical steady soliton with $\sec >0$ is also isometric, up to scaling, to the Bryant soliton.

Other than having $\sec>0$, the theorem assumes asympttoical cylindricity. It was thought by some mathematicians for some time maybe there are no other steady solitons? Or if that’s not true, maybe the scalar curvature on every steady soliton has an exponential decay like the cigar soliton, or linear like the Bryant soliton. This was later named the Dichotomy conjecture, which remains open to this day.

Before we move on to the next part of the story, I would like to share another interesting thing about the above theorem. Suppose we keep the asymptotical cylindricity but weaken the positive sectional curvature. What happens then? There is a very recent result due to Law which improves Brendle’s theorem.

In 2020, Yi Lai constructed the flying wing steady solitons. This was a new class of examples that are not asymptotically cylindrical!

(Lai '20) For every $n\ge 3$, there exists a family of $\mathbb{Z}_2\times O(n-1)$-symmetric but non-rotationally symmetric steady solitons with positive curvature operator. When $n\geq 4$, these are $\kappa$-noncollapsed.

Note that in dimension 3, it cannot be $\kappa$-noncollapsed, by Brendle’s theorem. The main idea in constructing such solitons is to take well behaved expanding solitons, i.e. solitons $g_i$ with $\operatorname{Ric}_{g_i}+\nabla^2 f_i=\lambda_i g_i$ and take $\lambda_i\to 0$. Because of such an indirect construction, we really don’t get our hands on any specific properties that the soliton has. One of the major questions here is: what does it look like at infinity? Moreover, in any dimension, it cannot be asymptotically cylindrical due to Brendle’s theorem.

Yi Lai solved the problem of understanding the geometry of these flying wings completely in 3 dimensions. The following is due to Lai.

(Lai '22) Let $(M^3,g,f)$ be any 3D steady soliton with nonnegative curvature operator that is NOT the Bryant soliton. Then,
  • $M$ is collapsed.
  • The asymptotic limits are the Cigar soliton and $S^1\times \mathbb{R}$.
  • The asymptotic cone of $M$ is a 2D sector. More precisely, if $r_k\to 0$,
    $$ (M,r_kg,o)\to (X,d_X,x) $$
    in the Gromov Hausdorff sense where $(X,d_X)$ is a metric space given by the sector of angle $\alpha\in (0,\pi)$ in the Euclidean plane.
  • There are two edges $\Gamma_1,\Gamma_2$ such that along these edges, the scalar curvature converges to a positive limit depending on $\alpha$.

These results are in complete contrast to Bryant soliton and the cigar soliton! One can compute using the asymptotic cylindricity that the asymptotic cone of $\operatorname{Bry}^n$ is a ray $\mathbb{R}_+:=\lbrace x|x\geq 0\rbrace $. The name flying wing comes because these solitons have a sector as their asymptotic cone. Moreover, there is now a real possibility that given a general steady soliton, the scalar curvature need not actually vanish at infinity! This is why the question I mentioned in the beginning of the talk makes sense: is it even possible that the scalar curvature doesn’t vanish in 4D noncollapsed setting? The precise formulation of the question I started in the beginning of the talk is the following conjecture due to Yi Lai.

Conjecture. (Lai‘23 ) The only non-collapsed steady gradient solitons with non-negative curvature operator are the 4D Bryant soliton, and the family of $\mathbb{Z}_2 \times O(3)$-symmetric solitons constructed by Lai. Moreover, the blow-down of each of the $\mathbb{Z}_2 \times O(3)$ -symmetric soliton is a ray.

If I had to guess why this conjecture could be true, I would say: maybe that the kind of behavior non-Bryant 3D steady solitons are showing is because they are collapsed? And moreover, its dimension 3, which means that the flat limits such as $S^1\times \mathbb{R}$ may arise. A similar phenomenon has been established by Haslhofer in the mean curvature flow where certain wing-like structure can never occur.

If one removes either nonnegative curvature operator or the noncollapsed assumptions, then the class of $4$D steady solitons is wild! It feels like there is really no end to the possible structures that exist. I will very briefly describe the solitons I am aware of here.

Need for classification

As you can see, these questions on the classification are interesting in their own right. But the techniques that go into all the theorems I mentioned heavily use Ricci flow. A natural question is: Why should we study steady solitons in the context of Ricci flow? Why should we classify them?

In a word, singularity analysis!

The central theme in Hamilton-Perelman’s 3D Ricci flow is to understand how closed Ricci flow singularities form. More so, what happens just before the singularity arises in the Ricci flow. I intend to write a longer post on singularity models later and update it here. For now, we move on to the setting of 4D steady solitons.

Setting

From now on, we will consider $(M^4,g,f)$, which are complete noncompact gradient steady Ricci soliton such that

The only known soliton satisfying (A1)-(A4) is the one-parameter family of $\mathbb{Z}_2\times O(3)$-symmetric steady solitons constructed by Lai. The question we started with still remains open in the class of steadies satisfying (A1)-(A4).

Hamilton showed that after rescaling, we can take \begin{equation} R+|\nabla f|^2=1 \qquad\text{on }M. \end{equation} Since $R>0$ we get \begin{equation} 0<R\le 1,\qquad |\nabla f|^2\leq 1 \qquad\text{on }M. \end{equation} So, $-f$ can grow atmost linearly. With a little more work, one can use the fact that $f$ is concave to obtain two sided linear bounds on $-f$: for some $c_0,c_1$, we have for all $x\in M$, \begin{equation} d_g(x,o)\geq f(o)-f(x)\ge c_0\,d_g(o,x)+c_1. \end{equation} In particular, $f(x)\to -\infty$ along any divergent curve, and $f^{-1}([a,b])$ is compact for all $a\leq b<f(o)$. We denote the level sets of $f$ by

$$ \Sigma_s:=f^{-1}(s)\qquad \text{ for }s<f(o), $$

each of which is compact and diffeomorphic to $\mathbb{S}^3$ so that $M$ is diffeomorphic to $\mathbb{R}^4$. Fix $s_0<f(o)$ and let $\Sigma:=\lbrace f=s_0\rbrace$.

(Chan-Ma-Zhang,'23) Assume (A1)-(A4). Consider any sequence $x_i\in M$ with $d(x_i,o)\to \infty$. By passing to a subsequence,
$$ \left(M^4,R(x_i) g,x_i\right) \to \operatorname{Bry}^3\times \mathbb{R}\text{ or }S^2 \times R^2, $$
smoothly.

Under (A1)-(A4), the behavior of scalar curvature remains open. The only known fact about the scalar curvature is due to Deng and Zhu:

$$ \lim_{x\to \infty}R(x)d(x,o)=+\infty $$

It is unknown whether $R(x)\to 0$. Now, let us mention the main results of this talk, which is due to Natasa Sesum and myself. The first theorem shows that away from two curves, the scalar curvature decays to zero atleast linearly.

(Main Theorem - I, N. Sesum-AGH) There exists $C>0$ and two curves $\Gamma_1,\Gamma_2$ starting at $o$ and going off to infinity such that rescaling along these curves gives $\operatorname{Bry}^3\times \mathbb{R}$ with the basepoint at the tip and setting $\Gamma=\Gamma_1\cup \Gamma_2$, we have
$$ R(x)d(x,\Gamma)\leq C\qquad \text{ for all }x\in M. $$
Here , $d(x,\Gamma)$ denotes the distance from $x$ to the set $\Gamma$. Further, if \begin{equation} \tag{*} \lim_{d(x,o)\to \infty}R(x)=0, \end{equation} then the following stronger bound holds:
$$ \lim_{d(x,o)\to \infty}R(x)d(x,\Gamma)=0. $$
We call $\Gamma_1,\Gamma_2$ as the edges of the soliton.
(Main Theorem - II, N. Sesum-AGH) Let $\left(M^4, g, f\right)$ satisfy (A1)-(A4). Then the following are equivalent:
  • $\lim _{d(x, o) \rightarrow \infty} R(x)=0$;
  • $\lim _{d(x, o) \rightarrow \infty} R(x) d(x, \Gamma)=0$;
  • the asymptotic cone of $(M, g)$ is a ray.

This question of whether scalar curvature on the steady soliton is important because of Deruelle’s result, which in the present setting shows that if $\lim _{d(x, o) \rightarrow \infty} R(x)=0$, then $(M,g)$ is dynamically stable.

We observe that the conclusion that the asymptotic cone of $(M, g)$ is a ray implies $\lim _{d(x, o) \rightarrow \infty} R(x)=0$ holds more generally in all dimensions.

Before proceeding to the idea behind the proof, let us mention one more result concerning the behavior of scalar curvature.

Recall that $\Phi_t$ is the flow of $\nabla f$ with $\Phi_0=id_M$ and $g(t):=\Phi_t^* g$ is the canonical Ricci flow. Given $p\in M,p\neq o$, $p\mapsto \Phi_{-t}p$, as $t\geq 0$, is in the direction of $-\nabla f$, on which $f$ decreases to $-\infty$, as $t\to -\infty$. Thus as you move forward in time, the point $p$ goes nearer to $o$.

Define the function $G:\Sigma\to [0,1)$ given by

$$ G(q):=\lim_{t\to -\infty}R(\Phi_tq) $$
(Zero-set of $G$) $\#\lbrace G\neq 0\rbrace\leq 2$.

Let us now briefly explain how these conclusions are proved and what the proof shows about the structure of manifold.

In order to make Cheeger-Gromov convergence, we consider the concept of closeness to cylinder.

Say $x$ is an $\epsilon$-center if $(M,R(x)g,x)$ is $\epsilon$-close in the $C^{[\epsilon^{-1}]}$-norm to $S^2 \times \mathbb{R}^2$ on a ball of radius $\epsilon^{-1}$.

Recall that due to the work of Chan-Ma-Zhang, whenever $d(x_i,o)\to \infty$, by passing to a subsequence, we have

$$ \left(M^4,R(x_i) g,x_i\right) \to \operatorname{Bry}^3\times \mathbb{R}\text{ or }S^2 \times R^2, $$

One consequence of this is that for every $\varepsilon>0$, there exists $N>0$ such that if $f(x)<-N$ and $\frac{\lambda_2}{R}(x)<\theta_\varepsilon$, then $x$ is an $\varepsilon$-center.

Passing to the convergence in level sets, one shows that when $R(x_i)\to 0$,

$$ \left(\Sigma_{f(x_i)},R(x_i) g,x_i\right) \to \operatorname{Bry}^3\text{ or }S^2 \times R. $$

On the other hand, if $R(x_i)\to r_\infty>0$, then

$$ \left(M^4, g,x_i\right) \to \operatorname{Bry}^3\times \mathbb{R}, $$

and if $h$ is a soliton potential for the Bryant, one has

$$ \left(\Sigma_{f(x_i)}, g,x_i\right) \to \mathcal{S}_h:=\lbrace (w,z)\mid Az+h(w)=0\rbrace \subset \operatorname{Bry}^3\times \mathbb{R}. $$

If additionally, $x_i$ lying in a single integral curve $\lbrace \Phi_tq:t\leq 0\rbrace $, we observe that

$$ \nabla R\cdot \nabla f|_{\Phi_t p}=2\operatorname{Ric}(\nabla f,\nabla f)|_{\Phi_t p}=\frac{d}{dt}R(\Phi_tp)\to 0, $$

implying that the limit of $x_i$ satifies $\nabla h=0$.

Let us call a point $x\in M$ is called a tip if $\bar \nabla R(x)=0$ and $\frac{\lambda_2}{R}(x)>\frac{1}{6}$. The above computation shows that if the scalar curvature doesn't vanish at infinity on the $\lbrace \Phi_tq:t\leq 0\rbrace $, then there exists tips very close to these integral curves.

Using all these ideas, we show that for all $s\ll 0$, $\Sigma_s$ consists of two Bryant caps and a neck joining them. The proof uses a technique due to Brendle-Dasksalopolous-Sesum.

One of the main questions that arise here is how the model spaces $\operatorname{Bry}^3\times \mathbb{R}$ and $S^2\times \mathbb{R}^2$ are approximately arranged within the manifold. In order to go towards that, we show the following theorem.

(Description of Bubble sheet regions)
  • (bubble sheet persistence) For each $\varepsilon>0$, there exists $\delta_\varepsilon\in (0,\varepsilon)$, $N_\varepsilon>0$ such that if $x\in M$ is a $(\delta_\varepsilon,2)$-center with $d(x,o)>N_\varepsilon$, then $\Phi_{t}(x)$ is an $(\varepsilon,2)$-center for all $t\leq 0$.
  • (bubble sheet improvement) If $\varepsilon\leq \varepsilon_*$ is small enough, then for any $t_k\to -\infty$, we have
    $$ (M,R(\Phi_{t_k}(q))g,\Phi_{t_k}(q))\to S^2 \times \mathbb{R}^2 $$
  • (behavior of scalar curvature) Moreover, for all $t>0$,
    $$ R(\Phi_{-t}(q))\leq \frac{C}{t}, $$
    and
    $$ \lim_{s\to \infty}s\cdot R(\Phi_{-s}(q))=1. $$

Due to lack of time, we shall only illustrate the key idea in the third assertion. It comes from the following identity:

$$ {\left[\frac{1}{R\left(\Phi_{-t}(q)\right)}+f\left(\Phi_{-t}(q)\right)\right]-\left[\frac{1}{R(q)}+f(q)\right]}=\int_0^t\left[2\left(\frac{|\operatorname{Ric}|^2}{R^2}-\frac{1}{2}\right)+R+\frac{\Delta R}{R^2}\right] d s. $$

Note that in the right hand side, all the terms in the integrand are small when $M$ is close to a bubble sheet after rescaling. This shows that the quantity $R+\frac{1}{f}$ is almost constant along the integral curves under consideration.

(Description of tip and edge regions) There exists two distinct points $x_+,x_-\in \Sigma:=\Sigma_{s_0}$ such that the integral curves $\Phi_{-t}x_\pm$ track the two tips along level sets. Given any $t_k\to -\infty$, we have that $(M,R(\Phi_{-t_k}(x_\pm))g,\Phi_{-t_k}(x_\pm))$ converges to $\operatorname{Bry}^3\times \mathbb{R}$ with basepoint at the tip. Moreover, $G\equiv 0$ on $\Sigma\setminus \lbrace x_+,x_-\rbrace $.

As a result, rescaling around $\Phi_{-t}x_\pm$ gives us $\operatorname{Bry}^3\times \mathbb{R}$ with basepoint at the tip. Since $G$ has at most two zeroes and the level sets have at most two tips, they could not possibly occur at any point other than $x_\pm$, that is, $\lbrace G=0\rbrace \subset \lbrace x_+,x_-\rbrace$. The two tips allow us to define the edge integral curves.

For $i=1,2$, let $\Gamma_i:[0,\infty)\to M$ be continuous curves such that $\Gamma_i(0)=o$, $\Gamma_i((0,\infty))$ is an integral curve of $-\nabla f/|\nabla f|$, $x_+\in \Gamma_1((0,\infty))$, and $x_-\in \Gamma_2((0,\infty))$. We set \[ \Gamma=\Gamma_1([0,\infty))\cup \Gamma_2([0,\infty)), \] and call $\Gamma_1$ and $\Gamma_2$ the edges of the soliton .

The key idea in proving Main Theorem - I is as follows. Fix $p$ and let us consider two integral curves: $\Phi_{-t} p$ and $\Phi_{-t} x_+$.

Suppose that rescaling around $p$ gives us the bubble sheet. Then, the first variation of distance shows that

$$ \partial_s^{+} d_g\left(\Phi_{-s}(p), \Phi_{-s}(x_+)\right) \leq C \max \left(\sqrt{R_g\left(\Phi_{-s}(p)\right)}, \sqrt{R_g\left(\Phi_{-s}(x_+)\right)}\right) $$

Therefore,

$$ d_g\left(\Phi_{-s}(p), \Phi_{-s}(x_+)\right)=O(s) $$

If $(*)$ holds this identity shows us that

$$ d_g\left(\Phi_{-s}(p), \Phi_{-s}(x_+)\right)=o(s) $$

Since $R(\Phi_{-s}p)=O(s^{-1})$, we obtain in general that

$$ R(\Phi_{-s}p)d_g\left(\Phi_{-s}(p), \Phi_{-s}(x_+)\right)=O(1) $$

and when $(*)$ holds, this is $o(1)$. The main technical details is to show that this estimate holds more generally for any sequences of points going off to infinity and not just along integral curves.

Now let us illustrate the key ideas in proving Main Theorem - II. If $\lim _{d(x, o) \rightarrow \infty} R(x)=0$, then we recall that

$$ \frac{d_g\left(\Phi_{-s}(p), \Phi_{-s}(x_+)\right)}{s}=o(1) $$

One then converts this bound to geodesic rays $\gamma_1,\gamma_2$ going off to infinity starting at $o$:

$$ \frac{d_g\left(\gamma_1(s), \gamma_2(s)\right)}{s}=o(1) $$

This shows that the asymptotic cone of $(M, g)$ is a ray.

The opposite consequence is actually more general and holds in any dimensions. In the paper, we consider the following angle function $\alpha:T_o M\to \mathbb{R}$ given by

$$ \alpha (v):=\lim_{r\to \infty}\frac{f(o)-f(\gamma_v(r))}{r} $$

where $\gamma_v$ is a geodesic ray starting at $v$. The advantage to this function is that when the asymptotic cone is a ray, $\alpha (v)$ is constant on those $v$ such that the geodesic ray starting at $o$ with velocity $v$ is minimizing and this forces $G$ is constant on $\Sigma$. The only possible way $G$ is constant on $\Sigma$ is $G\equiv 0$.

It remains open whether $R(x)\to 0$ at infinity and we do conjecture that it does. Perhaps thats a question for another day.

Thank you!