Weak stability of Ricci expanders with positive curvature operator

Abstract

We prove the $L^{\infty }$ stability of expanding gradient Ricci solitons with positive curvature operator and quadratic curvature decay at infinity.

Appendix: Soliton equations

The next lemma gathers well-known Ricci soliton identities together with the (static) evolution equations satisfied by the curvature tensor.

Recall first that an expanding gradient Ricci soliton is said normalized if $\int _Me^{-f}d\mu _g=(4\pi )^{n/2}$ (whenever it makes sense).

Lemma 6.5

Let $(M^n,g,\nabla ^g f)$ be a normalized expanding gradient Ricci soliton. Then the trace and first order soliton identities are:

$$\begin{aligned}&\Delta _g f = \mathop {\mathrm{R}}\nolimits _g+\frac{n}{2}, \end{aligned}$$

(24)

$$\begin{aligned}&\nabla ^g \mathop {\mathrm{R}}\nolimits _g+ 2\mathop {\mathrm{Ric}}\nolimits (g)(\nabla ^g f)=0, \end{aligned}$$

(25)

$$\begin{aligned}&|\nabla ^g f |^2+\mathop {\mathrm{R}}\nolimits _g=f+\mu (g), \end{aligned}$$

(26)

$$\begin{aligned}&\mathop {\mathrm{div}}\nolimits _g\mathop {\mathrm{Rm}}\nolimits (g)(Y,Z,T)=\mathop {\mathrm{Rm}}\nolimits (g)(Y,Z, \nabla f,T), \end{aligned}$$

(27)

for any vector fields Y, Z, T and where $\mu (g)$ is a constant called the entropy.

The evolution equations for the curvature operator, the Ricci tensor and the scalar curvature are:

$$\begin{aligned}&\Delta _f \mathop {\mathrm{Rm}}\nolimits (g)+\mathop {\mathrm{Rm}}\nolimits (g)+\mathop {\mathrm{Rm}}\nolimits (g)*\mathop {\mathrm{Rm}}\nolimits (g)=0, \end{aligned}$$

(28)

$$\begin{aligned}&\quad \Delta _f\mathop {\mathrm{Ric}}\nolimits (g)+\mathop {\mathrm{Ric}}\nolimits (g)+2\mathop {\mathrm{Rm}}\nolimits (g)*\mathop {\mathrm{Ric}}\nolimits (g)=0, \end{aligned}$$

(29)

$$\begin{aligned}&\quad \Delta _f\mathop {\mathrm{R}}\nolimits _g+\mathop {\mathrm{R}}\nolimits _g+2|\mathop {\mathrm{Ric}}\nolimits (g)|^2=0, \end{aligned}$$

(30)

where, if A and B are two tensors, $A*B$ denotes any linear combination of contractions of the tensorial product of A and B.

Proof

See [Chap.1, [5]] for instance. $\square$

Proposition 6.6

Let $(M^n,g,\nabla ^g f)$ be an expanding gradient Ricci soliton.

• If $(M^n,g,\nabla ^g f)$ is non Einstein, if $v:=f+\mu (g)+n/2$,

  $$\begin{aligned} \Delta _fv=v,\quad v>|\nabla v|^2. \end{aligned}$$

  (31)

• Assume $\mathop {\mathrm{Ric}}\nolimits (g)\ge 0$ and assume $(M^n,g,\nabla ^g f)$ is normalized. Then $M^n$ is diffeomorphic to ${\mathbb {R}}^n$ and

  $$\begin{aligned}&v\ge \frac{n}{2}>0. \end{aligned}$$

  (32)
  $$\begin{aligned}&\quad \frac{1}{4}r_p(x)^2+\min _{M}v\le v(x)\le \left( \frac{1}{2}r_p(x)+\sqrt{\min _{M}v}\right) ^2,\quad \forall x\in M, \end{aligned}$$

  (33)
  $$\begin{aligned}&\quad \mathop {\mathrm{AVR}}\nolimits (g):=\lim _{r\rightarrow +\infty }\frac{\mathop {\mathrm{Vol}}\nolimits B(q,r)}{r^n}>0,\quad \forall q\in M, \end{aligned}$$

  (34)
  $$\begin{aligned}&\quad -C(n,V_0,A_0)\le \min _{M}f\le 0;\quad \mu (g)\ge \max _{M}\mathop {\mathrm{R}}\nolimits _g\ge 0, \end{aligned}$$

  (35)

where $V_0$ is a positive number such that $\mathop {\mathrm{AVR}}\nolimits (g)\ge V_0$, $A_0$ is such that $\sup _{M}\mathop {\mathrm{R}}\nolimits _g\le A_0$ and $p\in M$ is the unique critical point of v.

• Assume $\mathop {\mathrm{Ric}}\nolimits (g)=\textit{O}(r_p^{-2})$ where $r_p$ denotes the distance function to a fixed point $p\in M$. Then the potential function is equivalent to $r_p^2/4$ (up to order 2).

For a proof, see [9] and the references therein.

Lemma 6.7

(Distance distortions) Let $(M^n,g_0(t))_{t\ge 0}$ be a Type III solution of the Ricci flow with nonnegative Ricci curvature, i.e.

$$\begin{aligned} |\mathop {\mathrm{Rm}}\nolimits (g_0(t))|_{g_0(t)}\le \frac{A_0}{1+t},\quad \mathop {\mathrm{Ric}}\nolimits (g_0(t))\ge 0,\quad t\ge 0. \end{aligned}$$

Then,

$$\begin{aligned} \left( \frac{1+s}{1+t}\right) ^{c(A_0)}g_0(s)\le & {} g_0(t)\le g_0(s),\\ d_{g_0(s)}(x,y)-c(A_0)\left( \sqrt{t}-\sqrt{s}\right) \le d_{g_0(t)}(x,y)\le & {} d_{g_0(s)}(x,y),\quad s\le t,\quad x,y\in M, \end{aligned}$$

for a positive constant $c(A_0)$, and any $0\le s\le t$.

In particular,

$$\begin{aligned} -\frac{c(A_0)}{\sqrt{t}}\le \partial _td_{g_0(t)}(x,y)\le 0, \quad t>0,\quad x,y \in M, \end{aligned}$$

for a positive constant $c(A_0)$.

Proof

The upper bound comes from the fact that the Ricci curvature is nonnegative, hence, $g_0(t)\le g_0(s)$ for any $s\le t$ in the sense of symmetric 2-tensors.

The lower bound has been proved by Hamilton and a proof can be found for instance in [lemma 8.33, [6]]. $\square$