Complete non-compact gradient Ricci solitons with nonnegative Ricci curvature

Abstract

In this paper, we give a rigidity theorem for a complete non-compact expanding (or steady) Ricci soliton with nonnegative Ricci curvature and certain scalar curvature decay condition. As an application, we prove that any complete non-compact expanding (or steady) Kähler-Ricci solitons with positively pinched Ricci curvature should be Ricci flat. The result answers a question proposed by Chow, Lu and Ni in case of Kähler-Ricci solitons.

Appendix

In this appendix, we compute the curvature decay of a family of expanding Käher-Ricci solions on ${\mathbb {C}}^n$ constructed by Cao in [5]. The curvature decay of these solitons implies that the curvature decay condition in Theorem 1.3 is almost optimal (see Proposition 6.1). Cao has also constructed the steady Käher-Ricci solitons in [4]. Similarly, one can show that these steady solitons have an exponential decay in case of $n=1$ and a linear decay in case of $n\ge 2$, pointwisely.

We first recall Cao’s construction. Let $(z_{1},z_{2},\ldots ,z_{n})$ be the standard holomorphic coordinates on ${\mathbb {C}}^{n}$. Assume that $g=(g_{i\bar{j}})$ is a U(n)-invariant metric on ${\mathbb {C}}^{n}$ and the corresponding Kähler potential is given by $u(t)$, where $u(t)$ is a strictly increasing and convex function on $(-\infty ,\infty )$ and $t=\ln |z|^{2}$. By a direct computation,

$$\begin{aligned} g_{i\bar{j}}&=\partial _{i}\partial _{\bar{j}}u(t)=e^{-t}u'(t)\delta _{ij}+e^{-2t}\bar{z}_{i}z_{j}(u''(t)-u'(t)),\\ g^{i\bar{j}}&=\partial _{i}\partial _{\bar{j}}u(t)=e^{t}u'(t)^{-1}\delta _{ij}+z_{i}\bar{z}_{j}(u''(t)-u'(t)), \end{aligned}$$

and

$$\begin{aligned} f(t)\triangleq -\ln \det (g_{i\bar{j}})=nt-(n-1)\ln u'(t)-\ln u''(t). \end{aligned}$$

Then

$$\begin{aligned} R_{i\bar{j}}=\partial _{i}\partial _{\bar{j}}f(t)=e^{-t}f'(t)\delta _{ij}+e^{-2t}\bar{z}_{i}z_{j}(f''(t)-f'(t)). \end{aligned}$$

One may check that $g_{i\bar{j}}$ is an expanding soliton if and only if

$$\begin{aligned} v^{i}\frac{\partial }{\partial z_i}= g^{i\bar{j}}\partial _{\bar{j}}(f+u)\frac{\partial }{\partial z_i}=\left( z_{i}\frac{f'+u'}{u''}\right) \frac{\partial }{\partial z_i} \end{aligned}$$

is a holomorphic vector field, which is equivalent to

$$\begin{aligned} f'+u'=\lambda u'', \end{aligned}$$

(6.1)

for some constant $\lambda$.

Let $\phi =u'$. Then, by (6.1), we get an equation for $\phi$,

$$\begin{aligned} \frac{\phi ''}{\phi '}+\left( \frac{n-1}{\phi }+\lambda \right) \phi '=n+\phi . \end{aligned}$$

(6.2)

Cao solved (6.2) by

$$\begin{aligned} \phi '=\frac{1}{\lambda ^{n+1}\phi ^{n-1}}\left( \lambda ^{n}\phi ^{n}+\sum _{j=0}^{n-1}(-1)^{n-j}\frac{n!}{j!}(1-\lambda )\lambda ^{j}\phi ^{j}+ce^{-\lambda \phi }\right) , \end{aligned}$$

(6.3)

where $c$ is a constant. It was proved that $g_{i\bar{j}}$ is a complete expanding Kähler-Ricci soliton by taking $c=(-1)^{n+1}n!(1-\lambda )$ for all $\lambda >0$. Moreover, these solitons have positive sectional curvature in case of $\lambda >1$ and negative sectional curvature in case of $0<\lambda <1$. For all $\lambda >0$, we prove the following proposition.

Proposition 6.1

The scalar curvature of Cao’s expanding solitons satisfies the following curvature property:

$$\begin{aligned} \frac{\lambda -1}{e^{\epsilon _{2} r^{2}(x)}}&\le R(x)\le \frac{\lambda -1}{e^{\epsilon _{1} r^{2}(x)}},\quad \mathrm{if}\; n=1;\end{aligned}$$

(6.4)

$$\begin{aligned} C_1\frac{\lambda -1}{1+r^{2}(x)}&\le R(x)\le C_2\frac{\lambda -1}{1+r^{2}(x)},\quad \mathrm{if}\;n\ge 2; \end{aligned}$$

(6.5)

and

$$\begin{aligned} C_1\frac{\lambda -1}{1+r^{2}}\le \frac{1}{\mathrm{vol}(B_{r}(o))}\int _{B_{r}(o)}R\mathrm{dv}\le C_2\frac{\lambda -1}{1+r^{2}},~\mathrm{if}~n\ge 1. \end{aligned}$$

(6.6)

Proof

By (6.3), it is easy to see

$$\begin{aligned} t=\int _{0}^{\phi }\frac{\lambda \mathrm{d}s}{s+\text{ higher } \text{ order } \text{ term }}. \end{aligned}$$

This means that $\phi$ is asymptotic to $e^{\frac{t}{\lambda }}$ as $t\rightarrow +\infty$, and $\phi '$ is asymptotic to $\frac{1}{\lambda }e^{\frac{t}{\lambda }}$ as $t\rightarrow +\infty$. Let $o=(0,0,\ldots ,0)$ and $p=(z_{1},0,\ldots ,0)$. Since the metric $g$ is U(n)-invariant, $(sz_{1},0,\ldots ,0)$ $(o\le s\le 1)$ is a geodesic curve which connecting $o$ and $p$. Thus

$$\begin{aligned} r(p)=d(o,p)=\int _{0}^{1}\sqrt{g_{1\bar{1}}}\mathrm{ds}=\frac{1}{2}\int _{-\infty }^{t}\sqrt{\phi '}\mathrm{dt}. \end{aligned}$$

This shows that $r$ is asymptotic to $\sqrt{\lambda }e^{\frac{t}{2\lambda }}$ as $t\rightarrow +\infty$. Hence, we get

$$\begin{aligned} \phi = O(r^{2}),~ \phi '=O(r^{2}). \end{aligned}$$

(6.7)

Also, we have

$$\begin{aligned}&g^{1\bar{1}}(p)=e^{t}(\phi ')^{-1},\quad R_{1\bar{1}}(p)=-e^{-t}\left( \left( n-1\right) \left( \frac{\phi '}{\phi }\right) '+\left( \frac{\phi ''}{\phi '}\right) '\right) ,\\&g^{i\bar{j}}(p)=0,\quad R_{i\bar{j}}(p)=0, ~\quad \forall \quad i \ne j,\\&g^{i\bar{i}}(p)=e^{t}(\phi )^{-1},\quad R_{i\bar{i}}(p)=e^{-t}\left( n-\left( n-1\right) \left( \frac{\phi '}{\phi }\right) -\left( \frac{\phi ''}{\phi '}\right) \right) , ~i\ge 2. \end{aligned}$$

In case $n=1$. By (6.7), we have

$$\begin{aligned} \mathrm{vol}(B_{r}(o))&=\int _{0}^{r}\int _{\partial B_{s}(o)}\det (g_{i\bar{j}})r\mathrm{d\sigma }\mathrm{dr}=2\pi \int _{0}^{r}\frac{\phi '}{r}\mathrm{dr}=O(r^2). \end{aligned}$$

(6.8)

On the other hand, differentiating (6.3), we have

$$\begin{aligned} \frac{\phi ''}{\phi '}=\frac{1}{\lambda }+\frac{\lambda -1}{\lambda }e^{-\lambda \phi }. \end{aligned}$$

Then

$$\begin{aligned} R(p)=g^{1\bar{1}}(p)R_{1\bar{1}}(p)=-(\phi ')^{-1}\left( \frac{\phi ''}{\phi '}\right) '=(\lambda -1)e^{-\lambda \phi }. \end{aligned}$$

It follows

$$\begin{aligned} \int _{B_{r}(o)}R\mathrm{dv}=2\pi \int _{0}^{r}(\lambda -1)e^{-\lambda \phi }\frac{\phi '}{r}\mathrm{dr}. \end{aligned}$$

Thus by (6.7), we get (we may assume $\lambda \ge 1$ for simplicity)

$$\begin{aligned} \frac{1}{C'}(\lambda -1)\le \int _{B_{r}(o)}R\mathrm{dv}\le C' (\lambda -1). \end{aligned}$$

(6.9)

Hence, (6.4) follows from (6.8) and (6.9).

In case $n\ge 2$. Differentiating (6.3) and (6.2), respectively, we have

$$\begin{aligned} \frac{\phi ''}{\phi '}&=\frac{1}{\lambda }+\frac{1}{\lambda ^{n+1}}\left( \sum _{j=0}^{n-2}(-1)^{n-j}(j-n+1)\frac{n!}{j!}(1-\lambda )\lambda ^{j}\phi ^{j-n} \right) +o_{1},\\ \left( \frac{\phi ''}{\phi '}\right) '&=\frac{\phi '}{\lambda ^{n+1}}\left( \sum _{j=0}^{n-2}(-1)^{n-j}(j-n+1)(j-n)\frac{n!}{j!}(1-\lambda )\lambda ^{j}\phi ^{j-n-1}\right) +o_{2}, \end{aligned}$$

where

$$\begin{aligned} o_{1}&= -\frac{c}{\lambda ^{n+1}}\left( \frac{n-1}{\phi ^{n}}+\frac{\lambda }{\phi ^{n-1}}\right) e^{-\lambda \phi },\\ o_{2}&= \frac{c}{\lambda ^{n+1}}\left( \frac{n(n-1)}{\phi ^{n+1}}+\frac{2\lambda (n-1)}{\phi ^{n}}\frac{\lambda }{\phi ^{n-1}}+\frac{\lambda ^{2}}{\phi ^{n-1}}\right) e^{-\lambda \phi }. \end{aligned}$$

On the other hand, by (6.2), we have

$$\begin{aligned} \frac{\phi '}{\phi }&= \frac{1}{\lambda }+\frac{1}{\lambda ^{n+1}}\left( \sum _{j=0}^{n-1}(-1)^{n-j}\frac{n!}{j!}(1-\lambda )\lambda ^{j}\phi ^{j-n}+c\phi ^{-n}e^{-\lambda \phi }\right) ,\\ \left( \frac{\phi '}{\phi }\right) '&= \frac{\phi '}{\lambda ^{n+1}}\left( \sum _{j=0}^{n-2}(-1)^{n-j}(j-n)\frac{n!}{j!}(1-\lambda )\lambda ^{j}\phi ^{j-n-1}\right) +o_{3}, \end{aligned}$$

where

$$\begin{aligned} o_{3}=-\frac{c}{\lambda ^{n+1}}\left( \frac{n}{\phi ^{n+1}}+\frac{\lambda }{\phi ^{n}}\right) e^{-\lambda \phi }. \end{aligned}$$

Thus we get

$$\begin{aligned} \frac{\phi ''}{\phi '}\rightarrow \frac{1}{\lambda }, \left( \frac{\phi ''}{\phi '}\right) '\rightarrow 0,\; \frac{\phi '}{\phi }\rightarrow \frac{1}{\lambda }, \; \left( \frac{\phi '}{\phi }\right) '\rightarrow 0,\quad \mathrm{as}\; t\rightarrow +\infty . \end{aligned}$$

(6.10)

Since

$$\begin{aligned} \phi R(p)=\phi g^{i\bar{j}}R_{i\bar{j}}=-\frac{\phi }{\phi '}\left( \left( \frac{\phi '}{\phi }\right) '+\left( \frac{\phi ''}{\phi '}\right) '\right) +(n-1)\left( n-(n-1)\frac{\phi '}{\phi }-\frac{\phi ''}{\phi '}\right) , \end{aligned}$$

by (6.10), we obtain

$$\begin{aligned} \phi R(p)\rightarrow n(n-1)\frac{\lambda -1}{\lambda },~\mathrm{as}~~t\rightarrow +\infty . \end{aligned}$$

It follows

$$\begin{aligned} C_1\frac{\lambda -1}{1+r^{2}} \le R(p)\le C_2\frac{\lambda -1}{1+r^{2}}. \end{aligned}$$

This proves (6.5). (6.6) follows from (6.4)and (6.5) by using the argument in Lemma 3.1.

$\square$