Scalar curvature along the Ricci flow

Abstract

In this note, we prove a well-known conjecture on the Ricci flow under a curvature condition, which is a pinching between the Ricci and Weyl tensors divided by suitably translated scalar curvature, motivated by Cao’s result (Commun Anal Geom 19(5):975–990, 2011).

1 Introduction

Consider the Ricci flow

$$\begin{aligned} \partial _{t}g(t)=-2\textrm{Ric}_{g(t)}, \ \ \ g(0)=g, \ \ \ t\in [0,T), \end{aligned}$$

(1.1)

on a given closed n-dimensional Riemannian manifold (M, g). Here T is the maximal time of (1.1) which is finite or infinite according to Hamilton’s result [8]. In this paper, we assume

$$\begin{aligned} T\in (0,\infty ). \end{aligned}$$

(1.2)

In this case, we have

$$\begin{aligned} \lim _{t\rightarrow T}\max _{M}|\textrm{Rm}_{g(t)}|_{g(t)}=\infty \end{aligned}$$

(1.3)

by Hamilton [8], and

$$\begin{aligned} \lim _{t\rightarrow T}\max |\textrm{Ric}_{g(t)}|_{g(t)}=\infty \end{aligned}$$

(1.4)

by Sesum [10] (for another proof see [9]). For scalar curvature, Cao [3] proved that

$$\begin{aligned} { either} \ \lim _{t\rightarrow T}\max _{M}R_{g(t)}=\infty \ \ \ { or} \ \ \ \lim _{t\rightarrow T}\max _{M}R_{g(t)}<\infty \ \text {and} \ \lim _{t\rightarrow T}\frac{|W_{g(t)}|_{g(t)}}{R_{g(t)}+C}=\infty , \end{aligned}$$

(1.5)

where C is a positive constant such that $\min _{M}R_{g}+C>0$ (hence, $R_{g(t)} +C\ge R\ge \min _{M}R_{g(t)}+C>0$ by the evolution equation of $R_{g(t)}$) and, $W_{g(t)}$ is the Weyl tensor of g(t). A well-know conjecture on scalar curvature is

Conjecture 1.1

Under the condition (1.2), the Ricci flow (1.1) has the following property

$$\begin{aligned} \lim _{t\rightarrow T}\max _{M}R_{g(t)}=\infty . \end{aligned}$$

(1.6)

The above conjecture was proved for Kähler-Ricci flow by Zhang[11] and for type-I maximal solution of Ricci flow by Enders, Müller and Topping [7].

In a very recent paper, Buzano and Di Matteo obtained an important result about Conjecture 1.1 in [2], where they showed that (see Corollary 1.12 in [2]) under an extra condition on injective radius bound of Ricci flow (i.e., $\textrm{inj}(M, g(t))\ge \alpha (\sup _{M\times [0,t]}|\textrm{Ric}_{g(s)}|_{g(s)})^{-1/2}$ for some $\alpha >0$) and $n<8$, Conjecture 1.1 is true. When $n\ge 8$, they also studied the singularities (see Theorem 1.13, [2]).

On the other hand, under the condition of boundedness of scalar curvature and finite T, Bamler [1] proved that there exists an open subset $\Sigma$ of M such that g(t) converges in $C^{\infty }(\Sigma )$ to a Riemannian metric $g_{T}$ on $\Sigma$ as $t\rightarrow T$, and the Hausdorff dimension of $M\setminus \Sigma$, with respect to some pseudo-length metric $d_{T}$ (i.e., the limit of the induced length metric $d_{t}$ of g(t)) on M, is not greater than $n-4$.

In this paper, we give a partial answer of Conjecture 1.1. Given an arbitrary positive number $\epsilon$, we choose a positive constant $C:=C_{\epsilon }$ such that $\min _{M}R_{g}+C\ge \epsilon >0$, and then $R_{g(t)}+C\ge \epsilon$. Define two quantities along the Ricci flow

$$\begin{aligned} f:=\frac{|W|^{2}}{(R+C)^{2}}=\frac{|W_{g(t)}|^{2}_{g(t)}}{ (R_{g(t)}+C)^{2}}, \ \ \ h:=\frac{|\textrm{Ric}|^{2}}{(R+C)^{2}}=\frac{|\textrm{Ric}_{g(t)}|^{2}_{g(t)}}{(R_{g(t)}+C)^{2}}. \end{aligned}$$

(1.7)

Cao [3] proved that, for any $T'\in (0,T)$, the inequality

$$\begin{aligned} h\le C_{1}+\frac{1}{\epsilon }\max _{M\times [0,T']}f^{1/2} \end{aligned}$$

(1.8)

holds on $M\times [0,T']$, where $C_{1}$ is a universal constant depending only on $M, g, \epsilon$, and n. Using (1.8) we can easily deduce (1.5).

According to Proposition 1.1 in [4], we can obtain

$$\begin{aligned} \Box |W|^{2}=-2|\nabla W|^{2} +\frac{8}{n-2}W_{ijk\ell }R^{ik}R^{j\ell }+8(W^{ijk\ell } +W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q} \end{aligned}$$

(1.9)

where $\Box =\Box _{g(t)}:=\partial _{t}-\Delta _{g(t)} =\partial _{t}-\Delta$. There exists a positive constant $C_{n}$, depending only on n, so that $C_{n}> \frac{\epsilon }{4}$ and

$$\begin{aligned} 8(W^{ijk\ell } +W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q}\le C_{n}|W|^{3}. \end{aligned}$$

(1.10)

Motivated by (1.8), we make the following assumption: for any $T'\in [0,T)$, the inequality

$$\begin{aligned} h\ge C_{n,\epsilon }\max _{[0,T']}f^{1/2}-C_{2}, \end{aligned}$$

(1.11)

holds on $M\times [0,T']$, for some $\epsilon$ and universal constants $C_{2}$ and $C_{n,\epsilon }$ with $\frac{1}{\epsilon }> C_{n,\epsilon }>\frac{1}{4}C_{n}$.

Theorem 1.2

Under the condition (1.11), Conjecture (1.1) holds. More precisely, there exist constants $C_{n,\epsilon }$ and $C_{2}$ such that if a Ricci flow (1.1) is singular at a finite time T, and satisfying the pinching condition (1.11), then its scalar curvature must blow up at T.

Going through the following proof of Theorem 1.2, we observe that (1.11) can be replaced by the following condition

$$\begin{aligned} |\textrm{Ric}|^{2}\ge C_{n,\epsilon }(R+C)|W| \end{aligned}$$

(1.12)

along the Ricci flow (1.1), where $C=C_{\epsilon }$ satisfies $\min _{M}R_{g}+C\ge \epsilon >0$.

Remark 1.3

We now suppose $R\ge R_{\min }:=\min _{M} R\in (0,4/C_{n})$, where $C_{n}$ is determined by (1.10). In this case, we can take $\epsilon =R_{\min }$ and $C_{\epsilon } =0$. Then the conclusion (1.5) implies that

$$\begin{aligned} \frac{|W_{g(t)}|_{g(t)}}{R_{g(t)}}\le C\Longrightarrow \lim _{t\rightarrow T}\max _{M} R_{g(t)}=\infty , \end{aligned}$$

(1.13)

while (1.12) becomes

$$\begin{aligned} \frac{|W_{g(t)}|_{g(t)}}{R_{g(t)}} \le C'_{n}\frac{|\textrm{Ric}_{g(t)}|^{2}_{g(t)}}{R^{2}_{g(t)}}\Longrightarrow \lim _{t\rightarrow T} \max _{M}R_{g(t)}=\infty . \end{aligned}$$

(1.14)

Here $C'_{n}$ is a constant satisfying $R_{\min }<C'_{n}<4/C_{n}$. In our situation, it is clear that the condition in (1.13) is stronger than that in (1.14), e.g., $|W_{g(t)}|_{g(t)}/R_{g(t)} \le C$ (for some C satisfying $nC\le C'_{n}$) implies $|W_{g(t)}|_{g(t)}/R_{g(t)} \le C'_{n}|\textrm{Ric}_{g(t)}|^{2}_{g(t)}/R^{2}_{g(t)}$. Choosing normal coordinates we can assume that $\textrm{Ric}_{g(t)}=\textrm{diag}(\lambda _{1}, \ldots ,\lambda _{n})$. From

$$ R^{2}_{g(t)}=\left( \sum _{1\le i\le n}\lambda _{i}\right) ^{2} \le n\sum _{1\le i\le n}\lambda ^{2}_{i}=n|\textrm{Ric}_{g(t)}|^{2}_{g(t)} $$

we can conclude that $|\textrm{Ric}_{g(t)}|^{2}_{g(t)}/R^{2}_{g(t)} \ge 1/n$.

2 Proof of Theorem 1.2

We start from an elementary identity.

Lemma 2.1

For any functions F, G we have

$$\begin{aligned} \Box \left( \frac{F}{G}\right) =\frac{\Box F}{G}-\frac{F\Box G}{G^{2}} +2\frac{\langle \nabla F,\nabla G\rangle }{G^{2}} -2\frac{F}{G^{3}}|\nabla G|^{2}. \end{aligned}$$

(2.1)

Proof

Compute

$$\begin{aligned} \Box \left( \frac{F}{G}\right)= & {} (\partial _{t}-\Delta )\left( \frac{F}{G}\right) \\= & {} \frac{\partial _{t}F\cdot G-F\cdot \partial _{t}G}{G^{2}} -\nabla ^{i}\left( \frac{\nabla ^{i}F\cdot G-F\cdot \nabla _{i}G}{G^{2}} \right) \\= & {} \frac{\partial _{t}F}{G} -\frac{F}{G^{2}}\partial _{t}G-\nabla ^{i} \left( \frac{\nabla _{i}F}{G}-\frac{F}{G^{2}}\nabla _{i}G\right) \\= & {} \frac{\partial _{t}F}{G}-\frac{F}{G^{2}}\Box G -\frac{\Delta F\cdot G-\langle \nabla F,\nabla G\rangle }{G^{2}} +\frac{\nabla ^{i}F\cdot G^{2}-2FG\nabla ^{i}G}{G^{4}}\nabla _{i}G \end{aligned}$$

which yields (2.1). $\square$

Now we choose

$$\begin{aligned} F=|W|^{2}, \ \ \ G:=(R+C)^{2} \end{aligned}$$

(2.2)

where as before $C=C_{\epsilon }$ is a positive constant so that $\min _{M}R+C\ge \epsilon >0$. We then get from (2.1) that

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{2}}\right)= & {} \frac{\Box |W|^{2}}{(R+C)^{2}} -\frac{|W|^{2}}{(R+C)^{4}}\Box (R+C)^{2} +2\frac{\langle \nabla |W|^{2},\nabla (R+C)^{2}\rangle }{(R+C)^{4}}\nonumber \\{} & {} - \ 2\frac{|W|^{2}}{(R+C)^{6}}|\nabla (R+C)^{2}|^{2}. \end{aligned}$$

(2.3)

Thanks to

$$\begin{aligned} \nabla (R+C)^{2}= & {} 2(R+C)\nabla (R+C),\\ \Box (R+C)^{2}= & {} (\partial _{t}-\Delta )(R+C)^{2} \ \ = \ \ 2(R+C)\partial _{t}R-\nabla ^{i}[2(R+C)\nabla _{i}(R+C)]\\= & {} 2(R+C)\partial _{t}R-2|\nabla (R+C)|^{2} -2(R+C)\Delta (R+C), \end{aligned}$$

we arrive at

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{2}}\right)= & {} \frac{\Box |W|^{2}}{(R+C)^{2}} -\frac{|W|^{2}}{(R+C)^{4}} \left[ 2(R+C)\Box R-2|\nabla (R+C)|^{2}\right] \\{} & {} + \ 4\frac{\langle \nabla |W|^{2},(R+C)\nabla (R+C)\rangle }{(R+C)^{4}} -8\frac{|W|^{2}}{(R+C)^{4}}|\nabla (R+C)|^{2} \end{aligned}$$

or

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{2}}\right)= & {} \frac{\Box |W|^{2}}{(R+C)^{2}} -2\frac{|W|^{2}\Box R}{(R+C)^{3}} -6\frac{|W|^{2}}{(R+C)^{2}}|\nabla \ln (R+C)|^{2}\nonumber \\{} & {} + \ \frac{4}{(R+C)^{4}} \langle \nabla |W|^{2},(R+C)\nabla (R+C)\rangle . \end{aligned}$$

(2.4)

On the other hand,

$$\begin{aligned} \nabla \left( \frac{|W|^{2}}{(R+C)^{2}}\right)= & {} \frac{\nabla |W|^{2}\cdot (R+C)^{2}-2|W|^{2}(R+C)\nabla (R+C)}{(R+C)^{4}}\\= & {} \frac{\nabla |W|^{2}}{(R+C)^{2}}-2\frac{|W|^{2}}{(R+C)^{3}}\nabla (R+C)\\= & {} \frac{\nabla |W|^{2}}{(R+C)^{2}}-2\frac{|W|^{2}}{(R+C)^{2}} \nabla \ln (R+C). \end{aligned}$$

If we introduce the tensor $Z_{aijk\ell }:=(R+C)\nabla _{a}W_{ijk\ell } -\nabla _{a}R\cdot W_{ijk\ell }$, then

$$\begin{aligned} |Z|^{2}=(R+C)^{2}|\nabla W|^{2} +|\nabla R|^{2}|W|^{2} -\langle \nabla |W|^{2},(R+C)\nabla (R+C)\rangle . \end{aligned}$$

For any $\gamma \in [0,4]$, we obtain from (2.4) and the evolution equation for the scalar curvature that

$$\begin{aligned} \Box f= & {} \frac{\Box |W|^{2}}{(R+C)^{2}} -4f\frac{|\textrm{Ric}|^{2}}{R+C} -6f|\nabla \ln (R+C)|^{2}+(4-\gamma )\langle \nabla f,\nabla \ln (R+C) \rangle \nonumber \\{} & {} + \ 2(4-\gamma )f|\nabla \ln (R+C)|^{2} +\gamma \frac{|\nabla W|^{2}}{(R+C)^{2}} +\gamma \frac{|\nabla R|^{2}|W|^{2}}{(R+C)^{4}} -\gamma \frac{|Z|^{2}}{(R+C)^{4}}.\nonumber \\ \end{aligned}$$

(2.5)

It then follows from (1.9) and (2.5) that

$$\begin{aligned} \Box f= & {} (\gamma -2)\frac{|\nabla W|^{2}}{(R+C)^{2}} -4f\frac{|\textrm{Ric}|^{2}}{R+C} +(4-\gamma )\langle \nabla f,\nabla \ln (R+C)\rangle \nonumber \\{} & {} + \ (2-\gamma )f|\nabla \ln (R+C)|^{2}-\gamma \frac{|Z|^{2}}{(R+C)^{4}}\nonumber \\{} & {} + \ \frac{8(W^{ijk\ell }+W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q} +\frac{8}{n-2}W_{ijk\ell }R^{ik}R^{j\ell }}{(R+C)^{2}}. \end{aligned}$$

(2.6)

Choosing $\gamma =2$ and using (1.7), (1.10), we have

$$\begin{aligned} \Box f\le & {} -4f h(R+C)-2\frac{|Z|^{2}}{(R+C)^{4}}+2\langle \nabla f, \nabla \ln (R+C)\rangle \nonumber \\{} & {} + \ C_{n}f^{3/2}(R+C)+\frac{8}{n-2}f^{1/2}h(R+C). \end{aligned}$$

(2.7)

Proof of Theorem 1.2

Given $T'\in (0,T)$ and consider the time interval $[0,T']$. Suppose f achieves its maximum at a point $(x_{0}, t_{0})\in M\times [0,T']$. The condition (1.11) now implies

$$\begin{aligned} h\ge C_{n,\epsilon }f^{1/2}_{0}-C_{2}, \ \ \ f_{0}:=f(x_{0},t_{0}), \end{aligned}$$

at this point $(x_{0}, t_{0})$. Plugging it into (2.7) and using (1.8), we find

$$\begin{aligned} 0\le & {} -4f_{0}(C_{n,\epsilon }f^{1/2}_{0}-C_{2})(R+C) +C_{n}f^{3/2}_{0}(R+C)\\{} & {} + \ \frac{8}{n-2}f^{1/2}_{0}\left( C_{1}+\frac{1}{\epsilon }f^{1/2}_{0} \right) (R+C)\\= & {} (C_{n}-4C_{n,\epsilon })f^{3/2}_{0}(R+C) +\left( 4C_{2}+\frac{8}{n-2}\frac{1}{\epsilon } \right) f_{0}(R+C)\\{} & {} + \ \frac{8C_{1}}{n-2}f^{1/2}_{0}(R+C) \end{aligned}$$

at $(x_{0}, t_{0})$. Because $R+C\ge \epsilon >0$ and $C_{n,\epsilon } >\frac{1}{4}C_{n}$, we can conclude that

$$\begin{aligned} (4C_{n,\epsilon }-C_{n})f^{3/2}_{0} \le \left( 4C_{2}+\frac{8}{n-2}\frac{1}{\epsilon }\right) f_{0} +\frac{8C_{1}}{n-2}f^{1/2}_{0} \end{aligned}$$

at $(x_{0}, t_{0})$. Hence $f_{0}\le C(n,\epsilon )$ at $(x_{0}, t_{0})$; explicitly,

$$\begin{aligned} f_{0}\le \max \left\{ 1,\left[ \frac{C_{2}+\frac{2}{n-2} (C_{1}+\frac{1}{\epsilon })}{C_{n, \epsilon }-\frac{1}{4}C_{n}}\right] ^{2}\right\} =:C(n,\epsilon ). \end{aligned}$$

Consequently, we get $f\le C(n,\epsilon )$ in $M\times [0,T']$ and hence in $M\times [0,T)$. According to (1.5), we must have $\lim _{t\rightarrow T}\max _{M}R_{g(t)}=\infty$. $\square$

3 A remark on four-dimensional case

When $n=4$, we can make the constant $C_{n,\epsilon }$ in (1.11) explicitly. Recall first the following property for closed 4-manifold (M, g) in [6]. The Weyl tensor W defines a symmetric operator $\mathcal {W}: \wedge ^{2}M\rightarrow \wedge ^{2}M$, that is,

$$\begin{aligned} (\mathcal {W}\alpha )_{k\ell }:=\frac{1}{2}\alpha ^{ij}W_{ijk\ell }, \end{aligned}$$

and then, by the Hodge star operator, splits into two operators $\mathcal {W}^{\pm }: \wedge ^{2,\pm }M\rightarrow \wedge ^{2,\pm }M$ with $\textrm{tr}\mathcal {W}^{\pm } =0$, which induce tensors $W^{\pm }$. In this notation, we can write $\mathcal {W}=\textrm{diag}(\mathcal {W}^{+}, \mathcal {W}^{-})$.

For any point $x\in M$, we can choose an oriented orthogonal basis $\omega ^{+}, \eta ^{+}, \theta ^{+}$ (resp. $\omega ^{-}, \eta ^{-}$, $\theta ^{-}$) of $\wedge ^{2,+}_{x}M$ (resp. $\lambda ^{2,-}_{x}M$), consisting of eigenvectors of $\mathcal {W}^{\pm }$ that $||\omega ^{\pm }|| =||\eta ^{\pm }||=||\theta ^{\pm }||=\sqrt{2}$ and ($\lambda ^{\pm } \le \mu ^{\pm }\le \nu ^{\pm }$)

$$\begin{aligned} W^{\pm }= & {} \frac{1}{2}\left( \lambda {^\pm }\omega ^{\pm }\otimes \omega ^{\pm } +\mu ^{\pm }\eta ^{\pm }\otimes \eta ^{\pm } +\nu ^{\pm }\theta ^{\pm }\otimes \theta ^{\pm }\right) ,\\ \mathcal {W}^{\pm }= & {} \begin{bmatrix} \lambda ^{\pm } &{} 0 &{} 0 \\ 0 &{} \mu ^{\pm } &{} 0\\ 0 &{} 0 &{} \nu ^{\pm } \end{bmatrix}, \ \ \ 0=\lambda ^{\pm }+\mu ^{\pm }+\nu ^{\pm }. \end{aligned}$$

Here $||T||^{2}:=\frac{1}{p}T_{i_{1}\cdots i_{p}}T^{i_{1} \cdots i_{p}}=\frac{1}{p}|T|^{2}$ for $T\in \wedge ^{p}M$. Moreover, $\omega ^{\pm }, \eta ^{\pm }, \theta ^{\pm }$ form a quaternionic structure on $T_{x}M$:

$$\begin{aligned} g^{pq}\omega ^{\pm }_{ip}\omega ^{\pm }_{qj} =g^{pq}\eta ^{\pm }_{ip}\eta ^{\pm }_{qj} =g^{pq}\theta ^{\pm }_{ip}\theta ^{\pm }_{qj}=-g_{ij} \end{aligned}$$

and

$$\begin{aligned} g^{pq}\omega ^{\pm }_{ip}\eta ^{\pm }_{qj} =\theta ^{\pm }_{ij}, \ \ \ g^{pq}\eta ^{\pm }_{ip}\theta ^{\pm }_{qj} =\omega ^{\pm }_{ij}, \ \ \ g^{pq}\theta ^{\pm }_{ip}\omega ^{\pm }_{qj} =\eta ^{\pm }_{ij}. \end{aligned}$$

Using this decomposition, we can prove (see for example [5])

$$\begin{aligned} W_{ijk\ell }W^{i}{}_{p}{}^{k}{}_{q} W_{j}{}^{p\ell q}=\frac{1}{2}W_{ijk\ell } W^{ij}{}_{pq}W^{k\ell pq}, \ \ \ W_{ipjq}W_{k\ell }{}^{pq} =\frac{1}{2}W_{ijpq}W_{k\ell }{}^{pq}. \end{aligned}$$

(3.1)

Now we can simplify the evolution Eq. (1.9) in dimension $n=4$:

$$\begin{aligned} \Box |W|^{2}=-2|\nabla W|^{2}+4W_{ijk\ell }R^{ik}R^{j\ell } +8(W^{ijk\ell }+W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q}. \end{aligned}$$

(3.2)

We may take normal coordinates. Then

$$\begin{aligned} (W^{ijk\ell }+W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q}= & {} (W_{ijk\ell }+W_{ikj\ell })W_{pijq}W_{pk\ell q}\\= & {} W_{ikj\ell }W_{ipjq}W_{kp\ell q}+W_{ijk\ell }W_{pijq}W_{pk\ell q}\\= & {} \frac{1}{2}W_{ijk\ell }W_{ijpq}W_{k\ell pq}+W_{ijk\ell }W_{pijq}W_{pk\ell q} \end{aligned}$$

by the first identity in (3.1). For $A:=W_{ijk\ell }W_{pijq}W_{pk\ell q}$ we have

$$\begin{aligned} A= & {} -W_{ijk\ell }W_{pk\ell q}(W_{ijpq}+W_{jpiq}) \ \ = \ \ W_{ijk\ell }W_{kp\ell q}W_{ijpq} -W_{ijk\ell }W_{pk\ell q}W_{jpiq}\\= & {} \frac{1}{2}W_{ijk\ell }W_{ijpq}W_{k\ell pq} +W_{ijk\ell }W_{kp\ell q}W_{jpiq}\\= & {} \frac{1}{2}W_{ijk\ell }W_{ijpq}W_{k\ell pq} +\left( \frac{1}{2}W_{pqk\ell }W_{ijk\ell }\right) W_{qipj}\\= & {} \frac{1}{2}W_{ijk\ell }W_{ijpq}W_{k\ell pq} +\frac{1}{2}W_{pqk\ell }\left( \frac{1}{2}W_{qpij}W_{k\ell ij}\right) \\= & {} \frac{1}{2}W_{ijk\ell }W_{ijpq}W_{k\ell pq}-\frac{1}{4} W_{pqk\ell }W_{k\ell ij}W_{ijpq} \ \ = \ \ \frac{1}{4}W_{ijk\ell }W_{k\ell pq}W_{pqij}. \end{aligned}$$

Hence

$$\begin{aligned} (W^{ijk\ell }+W^{ikj\ell })W_{pijq}W^{p}{}_{k\ell }{}^{q} =(W_{ijk\ell }+W_{ikj\ell })W_{pijq}W_{pk\ell q} =\frac{3}{4}W_{ijk\ell }W_{ijpq}W_{k\ell pq} \end{aligned}$$

and

$$\begin{aligned} \Box |W|^{2}=-2|\nabla W|^{2} +6W_{ijk\ell }W^{ijpq}W_{pq}{}^{k\ell } +4W_{ijk\ell }R^{ik}R^{j\ell }, \ \ \ n=4. \end{aligned}$$

(3.3)

It is clear that

$$\begin{aligned} 6W_{ijk\ell }W^{ijpq}W_{pq}{}^{k\ell } =\frac{6}{8}\left[ (\lambda ^{+})^{3} +(\mu ^{+})^{3}+(\nu ^{+})^{3}+(\lambda ^{-})^{3} +(\mu ^{-})^{3}+(\nu ^{-})^{3}\right] \end{aligned}$$

and

$$\begin{aligned} \frac{1}{4}|W|^{2}=(\lambda ^{+})^{2} +(\mu ^{+})^{2}+(\nu ^{+})^{2}+(\lambda ^{-})^{2} +(\mu ^{-})^{2}+(\nu ^{-})^{2}. \end{aligned}$$

Then

$$\begin{aligned} \left| 6W_{ijk\ell }W^{ijpq}W_{pq}{}^{k\ell }\right| \le \frac{6}{8}\times 6\times \frac{1}{4^{3/2}}|W|^{3} =\frac{36}{64}|W|^{3}. \end{aligned}$$

Therefore we can take $C_{4}=1$ in (1.10) and then any constant $C_{4,\epsilon }$ in $(1/4,1/\epsilon )$.

4 A remark on the proof of Theorem 1.2

If we choose $F=|W|^{2}$ and $G=(R+C)^{\alpha }$ with $\alpha >0$ in (2.1), then

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{\alpha }}\right)= & {} \frac{\Box |W|^{2}}{(R+C)^{\alpha }} -\frac{|W|^{2}}{(R+C)^{2\alpha }}\Box (R+C)^{\alpha }\nonumber \\{} & {} + \ 2\frac{\langle \nabla |W|^{2},\nabla (R+C)^{\alpha }\rangle }{(R+C)^{2\alpha }} -2\frac{|W|^{2}}{(R+C)^{3\alpha }}|\nabla (R+C)^{\alpha }|^{2}. \end{aligned}$$

(4.1)

From $\nabla (R+C)^{\alpha }=\alpha (R+C)^{\alpha -1} \nabla R$ and

$$ \Box (R+C)^{\alpha }=\alpha (R+C)^{\alpha -1} \Box R-\alpha (\alpha -1)(R+C)^{\alpha -2} |\nabla R|^{2} $$

we from (4.1) that

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{\alpha }}\right)= & {} \frac{\Box |W|^{2}}{(R+C)^{\alpha }} -\alpha \frac{|W|^{2}\Box R}{(R+C)^{\alpha +1}} -\alpha (\alpha +1)\frac{|W|^{2}}{(R+C)^{\alpha +2}} |\nabla \ln (R+C)|^{2}\nonumber \\{} & {} + \ \frac{2\alpha }{(R+C)^{2\alpha }} \langle \nabla |W|^{2},(R+C)^{\alpha -1}\nabla (R+C)\rangle . \end{aligned}$$

(4.2)

With the same tensor Z as in Sect. 2, The following two identities

$$\begin{aligned} \nabla \left( \frac{|W|^{2}}{(R+C)^{\alpha }} \right)= & {} \frac{\nabla |W|^{2}}{(R+C)^{\alpha }} -\alpha \frac{|W|^{2}}{(R+C)^{\alpha }} \nabla \ln (R+C),\\ |Z|^{2}= & {} (R+C)^{2}|\nabla W|^{2}+|\nabla R|^{2} |W|^{2}-\langle \nabla |W|^{2},(R+C)\nabla (R+C)\rangle \end{aligned}$$

show that

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{\alpha }}\right)= & {} \gamma \left\langle \nabla \left( \frac{|W|^{2}}{(R+C)^{\alpha }} \right) ,\nabla \ln (R+C)\right\rangle +\frac{\Box |W|^{2}}{(R+C)^{\alpha }}\nonumber \\{} & {} -\ \alpha \frac{|W|^{2}\Box R}{(R+C)^{\alpha +1}} +(2\alpha -\gamma )\frac{|\nabla W|^{2}}{(R+C)^{\alpha }} -(2\alpha -\gamma )\frac{|Z|^{2}}{(R+C)^{\alpha +2}}\nonumber \\{} & {} - \ (\alpha -\gamma )(\alpha -1)\frac{|W|^{2}}{(R+C)^{\alpha }} |\nabla \ln (R+C)|^{2}, \end{aligned}$$

(4.3)

where $0\le \gamma \le 2\alpha$. In particular, choosing $\gamma =0$,

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{\alpha }} \right)= & {} \frac{\Box |W|^{2}}{(R+C)^{\alpha }} -\alpha \frac{|W|^{2}\Box R}{(R+C)^{\alpha +1}} +2\alpha \frac{|\nabla W|^{2}}{(R+C)^{\alpha }}\nonumber \\{} & {} - \ 2\alpha \frac{|Z|^{2}}{(R+C)^{\alpha +2}} -\alpha (\alpha -1)\frac{|W|^{2}}{(R+C)^{\alpha }} |\nabla \ln (R+C)|^{2}. \end{aligned}$$

(4.4)

Putting $\alpha =1$ in (4.4) yields

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{R+C}\right) =\frac{\Box |W|^{2}+2|\nabla W|^{2}}{R+C} -2\frac{|W|^{2}|\textrm{Ric}|^{2}}{(R+C)^{2}} -2\frac{|Z|^{2}}{(R+C)^{3}}. \end{aligned}$$

(4.5)

If we choose $\gamma =2\alpha -2$ with $\alpha \ge 1$ in (4.3), we get

$$\begin{aligned} \Box \left( \frac{|W|^{2}}{(R+C)^{\alpha }}\right)= & {} \frac{\Box |W|^{2}+2|\nabla W|^{2}}{(R+C)^{\alpha }} -2\alpha \frac{|W|^{2}|\textrm{Ric}|^{2}}{(R+C)^{\alpha +1}} -2\frac{|Z|^{2}}{(R+C)^{\alpha +2}}\nonumber \\{} & {} + \ 2(\alpha -1)\left\langle \nabla \left( \frac{|W|^{2}}{(R+C)^{\alpha }} \right) ,\nabla \ln (R+C)\right\rangle \nonumber \\{} & {} + \ (\alpha -1)(\alpha -2) \frac{|W|^{2}}{(R+C)^{\alpha }}|\nabla \ln (R+C)|^{2}. \end{aligned}$$

(4.6)