Expanding Ricci solitons asymptotic to cones

Abstract

We show that an expanding gradient Ricci soliton which is asymptotic to a cone at infinity in a certain sense must be rotationally symmetric.

Appendix A: Expanding Bryant solitons

In this appendix, we describe the rotationally symmetric expanding solitons (with positive sectional curvature) constructed by Bryant in the unpublished note [5]. In particular, we check below that they satisfy the conditions of Definition 1.1. We remark that Bryant’s family extends past the Gaussian (flat) soliton to continue into a family of negatively curved rotationally symmetric expanding gradient solitons, which we do not discuss here (see the discussion in [5, Corollary 3]).

It is standard (see, e.g. [19, Section 2.3]) that for a warped product metric of the form $g = dt^{2} + a(t)^{2} g_{S^{n-1}}$,

$$\begin{aligned} {{\mathrm{Ric}}}= -(n-1)\frac{a''(t)}{a(t)}dt^{2} + ((n-2)-a(t)a''(t)-(n-2)a'(t)^{2})g_{S^{n-1}}. \end{aligned}$$

In fact, we recall for later use that the metric $g$ has sectional curvature in the radial direction given by $-\frac{a''(t)}{a(t)}$ and for planes tangent to the orbits of rotation given by $\frac{1-(a'(t))^{2} }{a(t)^{2} }$. Furthermore, for a function $f(t)$, the Hessian of $f$ with respect to the metric $g$ is given by

$$\begin{aligned} D^{2} f = f'' (t)dt^{2} + a(t)a' (t)f'(t) g_{S^{n-1}}. \end{aligned}$$

Thus, we see that the soliton equations $2D^{2}f = g + 2{{\mathrm{Ric}}}$ are equivalent to the following family of ODEs

$$\begin{aligned} 2 f''(t)&= 1 - 2(n-1)\frac{a''(t)}{a(t)}\\ 2 a(t) a'(t) f'(t)&= a(t)^{2}+ 2 ((n-2)-a(t)a''(t) - (n-2)a'(t)^{2}). \end{aligned}$$

Supposing that there is a fixed point of the rotation, i.e. $t_{0} \in \mathbb R$ so that $a(t_{0}) =0$ (by translating, we may assume that $t_{0} = 0$), the second soliton equation clearly implies that $a'(0) = \pm 1$. By reversing the $t$-variables if necessary, we thus may assume that $a'(t) > 0$ on $[0,T)$ for some $T>0$. In this region, it is convenient to change radial coordinates, from $t$ to $a = a(t)$. The metric in these coordinates may now be written

$$\begin{aligned} g = \frac{da^{2}}{\omega (a^{2})} + a^{2}g_{S^{n-1}} \end{aligned}$$

where $\omega (a^{2})$ is defined implicitly by

$$\begin{aligned} a'(t) = \sqrt{\omega (a(t)^{2})}. \end{aligned}$$

The Ricci tensor in these coordinates is given by

$$\begin{aligned} {{\mathrm{Ric}}}= -(n-1) \omega '(a^{2}) \frac{da^{2} }{ \omega (a^{2})} +((n-2) - \omega '(a^{2})a^{2} - (n-2) \omega (a^{2})) g_{S^{n-1}}, \end{aligned}$$

and the Hessian of $f(a^{2})$ by

$$\begin{aligned} D^{2}f&= \left( 4 f''(a^{2}) a^{2}\omega (a^{2})+ 2 f'(a^{2})\omega (a^{2}) + 2 f'(a^{2})a^{2} \omega '(a^{2}) \right) \frac{da^{2}}{ \omega (a^{2})} \\&+ 2f'(a^{2}) \omega (a^{2}) a^{2} g_{S^{n-1}}. \end{aligned}$$

In particular, the expanding soliton equations imply that the following system of ODEs must hold

$$\begin{aligned} \begin{aligned} 1 - 2(n-1) \omega '(s)&= 8 f''(s) s \omega (s)+ 4 f'(s)\omega (s) + 4 f'(s)s \omega '(s)\\ 4 f'(s) s \omega (s)&= s + 2 ((n-2) - \omega '(s) s - (n-2) \omega (s)), \end{aligned} \end{aligned}$$

(7)

where we have set $s=a^{2}$. Differentiating the second equation in $s$, we may eliminate the dependence on $f$ in the first equation, obtaining

$$\begin{aligned} 4 s^{2}\omega (s) \omega ''(s) = 2(n-2)\omega (s)(\omega (s)-1)+s\omega '(s)(2s\omega '(s)-s-2(n-2)). \end{aligned}$$

(8)

Lemma A.1

[5, Lemma 1] For $\omega (s)$ a positive solution of (8), defined for $s \in [0,M) \subset (0,\infty )$. Then, either $\omega \equiv 1$ or $\omega$ has at most one critical point in $(0,M)$ which is nondegenerate if it exists. Furthermore, if $\omega '(s_{0}) \ge 0$ and $\omega (s_{0}) >1$ for $s_{0}\in (0,M)$, then $\omega '(s)>$ for $s \in (s_{0},M)$. Similarly, if $\omega '(s_{0}) \le 0$ and $\omega (s_{0}) <1$ then $\omega '(s) < 0$ on $(s_{0},M)$.

To prove this, one may observe that (8) shows that if $s_{0}$ is a critical point of $\omega$, then

$$\begin{aligned} \omega ''(s_{0}) = \frac{(n-2)(\omega (s_{0}) -1)}{2s_{0}^{2}}. \end{aligned}$$

This shows that at a critical point of $h$, $\omega ''(s_{0})>0$ is equivalent to $\omega (s_{0})>1$ and $\omega ''(s_{0})<0$ is equivalent to $\omega (s_{0})<1$ ($\omega (s_{0})=1$ implies that $\omega \equiv 1$ by ODE uniqueness), so one may consider various cases to check the asserted properties.

Proposition A.2

[5, Proposition 4] If $\omega (s)$ is a solution of (8) with $\omega (0) = 1$ and $\omega '(0) < 0$ and $\omega (s)$ is defined on a maximally extended interval $[0,M) \subset [0,\infty )$, then necessarily $M = \infty$.

Proof

We first claim that if $M < \infty$, then $\lim _{s\nearrow M} \omega (s) = 0$. To see this, note that (8) implies

$$\begin{aligned} \omega ''(s) \ge -\frac{n-2}{2s^{2}} + \frac{1}{2} (\omega '(s))^{2} - \frac{1}{4} \omega '(s) \left( 1+\frac{2(n-2)}{s}\right) . \end{aligned}$$

From this, it is clear that there is some $C>0$ so that if $\omega '(s) \le -C$ for some $s \ge 1$, then $\omega ''(s) > 0$. This implies that $\omega '(s)$ must be uniformly bounded from below on $[0,M)$. Because we have assumed that $M<\infty$, it must be that $\lim _{s\nearrow M} \omega (s) = 0$, otherwise we could extend the solution $\omega (s)$ past $M$.

Now, (8) also implies that for $s >0$, then

$$\begin{aligned} \omega ''(s) > - \frac{n-2}{2s^{2}} - \frac{1}{4} \frac{\omega '(s)}{\omega (s)}. \end{aligned}$$

Integrating from $s_{0}$ to $s < M$, this implies that

$$\begin{aligned} \omega '(s) -\omega '(s_{0}) > -\frac{n-2}{2}\left( \frac{1}{s_{0}} - \frac{1}{s}\right) + \frac{1}{4} \log \left( \frac{\omega (s_{0})}{\omega (s)}\right) . \end{aligned}$$

Letting $s\nearrow M$, the left hand side must tend to infinity, because $\lim _{s\nearrow M}\omega (s) = 0$, but the right hand side is bounded above, a contradiction. $\square$

Lemma A.3

For $\omega (s)$ a solution of (8) with $\omega (0) =1$ and $\omega '(0) < 0$, we have that $\omega '(s),\omega ''(s) = o(1)$.

Proof

Rewriting (8) as

$$\begin{aligned} \omega ''(s) = \frac{(n-2)(\omega (s)-1)}{2s^{2}} - \frac{\omega '(s)(s+2(n-2))}{4s \omega (s)} + \frac{(\omega '(s))^{2}}{2 \omega (s)}, \end{aligned}$$

we see that for a fixed $\delta >0$, there is $s_{0}=s_{0}(\delta )$ large enough so that if $\omega '(s) < -\delta$ for $s \ge s_{0}$, then $\omega ''(s) > 0$. On the other hand, we must be able to find $s_{1} > s_{0}$ so that $\omega '(s_{1}) > - \delta$ (otherwise $\omega (s)$ could not converge). As such, for $s \ge s_{1}$, $\omega '(s) \ge -\delta$ (we have just shown that $-\delta$ is a barrier for $\omega '(s)$). This clearly shows that $\omega '(s) = o(1)$. Using this in (8) gives $\omega ''(s) = o(1)$.

Corollary A.4

[5, Corollary 2] A solution of (8) with $\omega (0) =1$ and $\omega '(0) < 0$ exists for all $s \ge 0$ and is monotonically decreasing with a positive lower bound.

proof

As in the proof of Proposition A.2,

$$\begin{aligned} \omega '(s) -\omega '(s_{0}) > -\frac{n-2}{2}\left( \frac{1}{s_{0}} - \frac{1}{s}\right) + \frac{1}{4} \log \left( \frac{\omega (s_{0})}{\omega (s)}\right) . \end{aligned}$$

By the previous lemma, we have that $\omega '(s) - \omega '(s_{0}) \le -\omega '(s_{0}) = o(1)$. Thus, it cannot happen that $\omega (s) \rightarrow 0$ as $s\rightarrow \infty$. $\square$

Now, we show that $\omega (s)$ and $f(s)$ agree with their formal asymptotic expansions up to second order; this will allow us to study the rate at which the Bryant solitons approach a cone.

Proposition A.5

For a solution of (8) with $\omega (0) =1$ and $\omega '(0) < 0$, by Corollary A.4, there is some $\alpha \in [0,1)$ so that $\lim _{s\rightarrow \infty } \omega (s) = 1-\alpha$. With this choice of $\alpha$, we have the asymptotic expansion of $\omega (s)$,

$$\begin{aligned} \omega (s) = 1-\alpha + \frac{2(n-2)\alpha (1-\alpha )}{s} + \varphi (s), \end{aligned}$$

where $\varphi (s)$ satisfies $\varphi (s) = O(s^{-2})$, $\varphi '(s) ,\varphi ''(s) = O(s^{-3})$. Furthermore, we have that $f(s)$ satisfies the expansion (up to addition of a constant)

$$\begin{aligned} f(s) = \frac{s}{4(1-\alpha )} + \psi (s), \end{aligned}$$

where $\psi (s) = O(s^{-1})$, $\psi '(s) = O(s^{-2})$, and $\psi ''(s) = O(s^{-3})$.

Proof

By (8), we have that

$$\begin{aligned} \omega ''(s) + \frac{1}{4} \omega '(s) > - \frac{C}{s^{2}}. \end{aligned}$$

We may use an integrating factor to rewrite this as

$$\begin{aligned} \frac{d}{ds}\left( e^{s/4} \omega '(s) \right) \ge - \frac{C}{s^{2}} e^{s/4}. \end{aligned}$$

Integrating from $1$ to $s$ thus yields

$$\begin{aligned} e^{s/4}\omega '(s) - \omega '(1) \ge - C \int _{1}^{s} \frac{e^{x/4}}{x^{2}} dx. \end{aligned}$$

Now, because

$$\begin{aligned} \int _{1}^{s} \frac{e^{(x-s)/4}}{x^{2}}dx&= \int _{1}^{s/2} \frac{e^{(x-s)/4}}{x^{2}}dx + \int _{s/2}^{s} \frac{e^{(x-s)/4}}{\tau ^{2}}dx\\&\le \int _{1}^{s/2} e^{(x-s)/4} dx + \frac{4}{s^{2}} \int _{s/2}^{s} e^{(x-s)/4} dx\\&\le 4 \left( e^{-s/8}-e^{(1-s)/4}\right) + \frac{16}{s^{2}} \left( 1 - e^{-s/8}\right) = O(s^{-2}), \end{aligned}$$

we have that $\omega '(s) = O(s^{-2})$. This implies $\omega (s)-1+\alpha = O(s^{-1})$ and from (8) it is not hard to see that also $\omega ''(s) = O(s^{-2})$.

We now define a function $\varphi (s)$ by

$$\begin{aligned} \omega (s) = 1-\alpha + \frac{2(n-2)\alpha (1-\alpha )}{s}+ \varphi (s). \end{aligned}$$

Here, the choice of second order term comes from formally expanding $\omega (s)$ in a power series in $s^{-k}$ and solving for the $s^{-1}$ term (the power series does not converge, cf. [5, Remark 11], so we are simply using the truncated expansion to cancel the highest order term in the ODE). By the above asymptotics of $\omega (s)$, we see that $\varphi (s) = O(s^{-1})$ and $\varphi '(s),\varphi ''(s) = O(s^{-2})$. Using this, one may show (as above, except the ODE for $\varphi (s)$ decays one order faster in $s$, as we have just explained)

$$\begin{aligned} \varphi ''(s) + \frac{1}{4} \varphi '(s) \ge - \frac{C}{s^{3}}, \end{aligned}$$

and then the same argument implies that $\varphi (s) = O(s^{-2})$ and $\varphi '(s),\varphi ''(s) = O(s^{-3})$, as desired.

Now, by the bottom line of (7), we see that

$$\begin{aligned} 4 \left[ f(s) - \frac{s}{4(\alpha -1)}\right] ' \omega (s)&= \frac{1-\alpha -\omega (s)}{1-\alpha } + 2 \left( \frac{n-2}{s}(1-\omega (s)) - \omega ' (s)\right) \\&= - \frac{\varphi (s)}{1-\alpha } - \frac{4(n-2)^{2}\alpha (1-\alpha )}{s^{2}} \\&- \frac{2(n-2)\varphi (s)}{s} + \frac{2(n-2)\alpha (1-\alpha )}{s^{2}} - \varphi '(s)\\&=: 4 \psi '(s) \omega (s). \end{aligned}$$

Here, we may choose $\psi (s)$ so that $\psi (s)\rightarrow 0$ as $s \rightarrow \infty$. In particular, we easily see that $\psi (s) = O(s^{-1})$, $\psi '(s) = O(s^{-2})$, $\psi ''(s) = O(s^{-3})$, as desired. $\square$

It is clear from the proof that it is possible to show that $\omega (s)$ and $f(s)$ agree with their formal power series at infinity up to any finite number of terms. However, as remarked above, the power series does not converge.

Proposition A.6

Each of the solutions $\omega (s)$ of (8) with $\omega (0)=0$ and $\omega '(0) < 0$ define a rotationally symmetric soliton with positive sectional curvature which is asymptotically conical as a soliton, in the sense of Definition 1.1.

Proof

We have just shown that any solution of (8) with $\omega (0)=0$ and $\omega '(0) < 0$ exists for all time and is monotonically decreasing with a positive lower bound. Fix $\alpha \in [0,1)$ so that $\lim _{s\rightarrow \infty }\omega (s) = 1-\alpha$. As the radial sectional curvature is $-\omega '(a^{2})$ and the sectional curvatures tangent to the orbits of rotations are $\frac{1-\omega (a^{2})}{a^{2}}$, these solutions have positive sectional curvature. That the soliton is asymptotically conical as a soliton follows readily from the asymptotics of $f(s)$ and $\omega (s)$ in the previous proposition. $\square$