1 Introduction

Let \((M^n,g)\) be an n-dimensional Riemannian manifold with \(n\ge 3\). It is well-known that for \(n\ge 4\), the metric g is conformally flat if and only if its Weyl curvature tensor is zero. If \(n=3\), then it is conformally flat if and only if the Cotton tensor is zero. In the last years, the classifications of conformally flat manifolds under some geometrical or topological assumptions have been paid much attention. For example, in [17], Tani proved that any closed conformally flat manifold with positive Ricci curvature and constant scalar curvature is covered isometrically by \({\mathbb {S}}^n\) with the round metric. For complete conformally flat manifolds with non-negative Ricci curvature, Carron and Herzlich [2] gave the following classifications: they are either flat, or locally isometric to \({\mathbb {R}}\times {\mathbb {S}}^{n-1}\) with the product metric; or are globally conformally equivalent to \({\mathbb {R}}^n\) or to a spherical space form. For closed conformally flat manifolds satisfying some integral pinching conditions, see [5, 7, 8, 16, 18]. On the other hand, for some classifications with point-wise pinching condition on the Ricci curvature, see [4, 14, 19] and references therein.

Throughout this paper, all the calculations are carried out under the normal coordinates. Denote by R, \(R_{ij}\) the scalar curvature and the Ricci curvature respectively. We let \(\mathring{R}_{ij}=R_{ij}-\frac{R}{n}g_{ij}\) be the trace-less Ricci curvature. With the help of the properties of the Codazzi tensor, Catino [3] studied closed conformally flat manifolds with positive constant scalar curvature (in this case, the Ricci curvature is a Codazzi tensor) and satisfying an optimal integral pinching condition. He proved the following

Theorem A

Let \((M^n,g)\) be a closed conformally flat Riemannian manifold with positive constant scalar curvature. Then

$$\begin{aligned} \int \limits _{M}\Big (R -\sqrt{n(n-1)}|\mathring{R}_{ij}|\Big )|\mathring{R}_{ij}|^{\frac{n-2}{n}}\le 0, \end{aligned}$$
(1.1)

and equality occurs if and only if \((M^n,g)\) is covered isometrically by either \({\mathbb {S}}^n\) with the round metric, \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with the product metric, or \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with a rotationally symmetric Derdziński metric.

Generalizing the above results of Catino, Huang and Ma [11] studied manifolds with harmonic curvature tensor and positive scalar curvature. They proved

Theorem B

Let \((M^n,g)\) be a closed Riemannian manifold with harmonic curvature tensor and positive scalar curvature. Then

$$\begin{aligned} \int \limits _{M}\Big (R-\sqrt{n(n-1)}|\mathring{R}_{ij}|\Big )|\mathring{R}_{ij}|^{\frac{n-2}{n}}\le \sqrt{\frac{(n-1)(n-2)}{2}}\int \limits _{M} |W||\mathring{R}_{ij}|^{\frac{n-2}{n}}, \end{aligned}$$
(1.2)

and equality occurs if and only if \((M^n,g)\) is either Einstein or isometrically covered by one of:

  1. (1)

    \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with a product metric;

  2. (2)

    \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with a rotationally symmetric Derdziński metric.

The Cotton tensor is defined by

$$\begin{aligned} \begin{aligned} C_{ijk}=&R_{kj,i}-R_{ki,j}-\frac{1}{2(n-1)}(R_{,i}g_{jk}-R_{,j}g_{ik}), \end{aligned} \end{aligned}$$
(1.3)

where the indices after the comma denotes the covariant derivatives, which is related to the Weyl curvature tensor by

$$\begin{aligned} -\frac{n-3}{n-2}C_{ijk}=W_{ijkl,l}. \end{aligned}$$
(1.4)

Thus, it is easy to see that for conformally flat manifolds with constant scalar curvature and \(n\ge 4\), the Ricci curvature must be a Codazzi tensor, and hence the curvature tensor is harmonic (since \(R_{ij,k}-R_{ik,j}=R_{jkil,l}\), the curvature tensor being harmonic is equivalent to the Ricci curvature being a Codazzi tensor with constant scalar curvature). That is to say, conditions on harmonic curvature are weaker than those on conformal flatness. On the other hand, the key to prove Theorem A and Theorem B is the fact that the Ricci curvature becomes a Codazzi tensor under assumptions.

In this note, we will continue to generalize the above results by removing the conditions on constant scalar curvature and that the Ricci curvature is a Codazzi tensor. Our main results are stated as follows:

Theorem 1.1

Let \((M^n,g)\) be a closed Riemannian manifold with positive scalar curvature, where \(n\ge 3\). Then

$$\begin{aligned} \begin{aligned} \int \limits _{M}\Bigg [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -&\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Bigg ]|\mathring{R}_{ij}|^2\\ \le&\frac{(n-2)^2}{4n}\int \limits _{M}|\nabla R|^2+\frac{n-1}{2}\int \limits _{M}|C_{ijk}|^2, \end{aligned} \end{aligned}$$
(1.5)

and equality occurs if and only if \((M^n,g)\) is either Einstein or isometrically covered by \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with the product metric.

The following modified Schouten tensor with a parameter \(\tau \) was introduced by Gursky and Viaclovsky [6] (see also Li and Li [13]):

$$\begin{aligned} A^{\tau }_{ij}=R_{ij}-\frac{\tau R}{2(n-1)}g_{ij}, \end{aligned}$$
(1.6)

where \(\tau \in {\mathbb {R}}\) is a constant. When \(\tau =1\), the tensor \(A^{1}_{ij}\) is exactly the Schouten tensor. We denote by \(\sigma _2(A^{\tau })\) the 2nd-elementary symmetric function of the eigenvalues of the so-called modified Schouten tensor. Then, for manifolds with non-negative constant \(\sigma _2(A^{\tau })\), we have the following

Theorem 1.2

Let \((M^n,g)\) be a closed Riemannian manifold with positive scalar curvature, where \(n\ge 3\). If the function \(\sigma _2(A^{\tau })\) is a non-negative constant, where \(\tau <1\) or \(\tau >3-\frac{4}{n}\), then

$$\begin{aligned} \begin{aligned} \int \limits _{M}\Bigg [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -&\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Bigg ]|\mathring{R}_{ij}|^2\le \frac{n-1}{2}\int \limits _{M}|C_{ijk}|^2, \end{aligned} \end{aligned}$$
(1.7)

and equality occurs if and only if \((M^n,g)\) is either Einstein or isometrically covered by \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with the product metric.

Remark 1.1

If we add the conditions \(W_{ijkl}=0\) and that the scalar curvature is constant in Theorem 1.1, then (1.5) becomes

$$\begin{aligned} \int \limits _{M}(R-\sqrt{n(n-1)}|\mathring{R}_{ij}|)|\mathring{R}_{ij}|^2\le 0. \end{aligned}$$
(1.8)

On the other hand, if we add the condition that the curvature tensor is harmonic in Theorem 1.1, then (1.5) becomes

$$\begin{aligned} \begin{aligned} \int \limits _{M}\Big [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Big ]|\mathring{R}_{ij}|^2 \le 0. \end{aligned} \end{aligned}$$
(1.9)

Comparing (1.8) and (1.9) with (1.1) and (1.2), respectively, our Theorem 1.1 gives a new estimate.

Remark 1.2

Since \(R=\sqrt{n(n-1)}|\mathring{R}_{ij}|\) on \({\mathbb {S}}^1\times {\mathbb {S}}^{n-1}\) with the product metric, Theorem 1.1 can also be interpreted as an interpolation curvature estimate.

Remark 1.3

In [1], Bo and Sheng gave some rigidity characterization for conformally flat manifolds with constant quotient curvature of \(\sigma _k(A^{\tau })\). Our Theorem 1.2 gives an integral pinching condition with the \(\sigma _2(A^{\tau })\) assumption, which can be seen as a complement.

2 Proof of the results

2.1 Proof of Theorem 1.1

Recall that the Riemannian curvature tensor and Weyl curvature tensor are related by

$$\begin{aligned} \begin{aligned} W_{ijkl}&=R_{ijkl}-\frac{1}{n-2}(R_{ik}g_{jl}-R_{il}g_{jk} +R_{jl}g_{ik}-R_{jk}g_{il})\\&\quad +\frac{R}{(n-1)(n-2)}(g_{ik}g_{jl}-g_{il}g_{jk})\\&=R_{ijkl}-\frac{1}{n-2}(\mathring{R}_{ik}g_{jl}-\mathring{R}_{il}g_{jk} +\mathring{R}_{jl}g_{ik}-\mathring{R}_{jk}g_{il})\\&\quad -\frac{R}{n(n-1)}(g_{ik}g_{jl}-g_{il}g_{jk}). \end{aligned} \end{aligned}$$
(2.1)

Using formula (2.1), it is easy to check that

$$\begin{aligned} \begin{aligned} \mathring{R}_{kl}R_{ikjl}=&\mathring{R}_{kl}W_{ikjl}+\frac{1}{n-2}(|\mathring{R}_{ij}|^2g_{ij} -2\mathring{R}_{ik}\mathring{R}_{jk})-\frac{1}{n(n-1)}R \mathring{R}_{ij}. \end{aligned}\end{aligned}$$
(2.2)

It follows that

$$\begin{aligned} \begin{aligned} \mathring{R}_{kj,ik}&=\mathring{R}_{kj,ki}+\mathring{R}_{lj}R_{lkik}+\mathring{R}_{kl}R_{ljik}\\&=\frac{n-2}{2n}R_{,ij}+\mathring{R}_{ik}\mathring{R}_{jk}+\frac{1}{n}R\mathring{R}_{ij}-\Big [\mathring{R}_{kl}W_{ikjl} \\&\quad +\frac{1}{n-2}(|\mathring{R}_{ij}|^2g_{ij}-2\mathring{R}_{ik}\mathring{R}_{jk}) -\frac{1}{n(n-1)}R \mathring{R}_{ij}\Big ]\\&=\frac{n-2}{2n}R_{,ij}+\frac{n}{n-2}\mathring{R}_{ik}\mathring{R}_{jk}+\frac{1}{n-1}R\mathring{R}_{ij}\\&\quad -\mathring{R}_{kl}W_{ikjl}-\frac{1}{n-2}|\mathring{R}_{ij}|^2g_{ij}. \end{aligned}\end{aligned}$$
(2.3)

On the other hand, formula (1.3) is equivalent to

$$\begin{aligned} \begin{aligned} C_{ijk}=&\mathring{R}_{kj,i}-\mathring{R}_{ki,j}+\frac{n-2}{2n(n-1)}(R_{,i}g_{jk}-R_{,j}g_{ik}). \end{aligned} \end{aligned}$$
(2.4)

Therefore, from (2.3) and (2.4), we have

$$\begin{aligned} \begin{aligned} C_{kij,k}&=\Delta \mathring{R}_{ij}-\mathring{R}_{kj,ik}+\frac{n-2}{2n(n-1)}(g_{ij}\Delta R-R_{,ij})\\&=\Delta \mathring{R}_{ij}-\Big (\frac{n-2}{2n}R_{,ij}+\frac{n}{n-2}\mathring{R}_{ik}\mathring{R}_{jk} +\frac{1}{n-1}R\mathring{R}_{ij}-\mathring{R}_{kl}W_{ikjl}\\&\quad -\frac{1}{n-2}|\mathring{R}_{ij}|^2g_{ij}\Big )+\frac{n-2}{2n(n-1)}(g_{ij}\Delta R-R_{,ij}), \end{aligned}\end{aligned}$$
(2.5)

which shows

$$\begin{aligned} \begin{aligned} \Delta \mathring{R}_{ij}=&\frac{n-2}{2n}R_{,ij}+\frac{n}{n-2}\mathring{R}_{ik}\mathring{R}_{jk} +\frac{1}{n-1}R\mathring{R}_{ij}-\mathring{R}_{kl}W_{ikjl}\\&-\frac{1}{n-2}|\mathring{R}_{ij}|^2g_{ij}-\frac{n-2}{2n(n-1)}(g_{ij}\Delta R-R_{,ij})+C_{kij,k}, \end{aligned}\end{aligned}$$
(2.6)

and hence

$$\begin{aligned} \begin{aligned} \frac{1}{2}\Delta |\mathring{R}_{ij}|^2=&|\nabla \mathring{R}_{ij}|^2+\mathring{R}_{ij}\Delta \mathring{R}_{ij}\\ =&|\nabla \mathring{R}_{ij}|^2+\frac{n}{n-2}\mathring{R}_{ij}\mathring{R}_{ik}\mathring{R}_{jk} -W_{ikjl}\mathring{R}_{ij}\mathring{R}_{kl}+\frac{1}{n-1}R|\mathring{R}_{ij}|^2\\&+\frac{n-2}{2(n-1)}R_{,ij}\mathring{R}_{ij}+C_{kij,k}\mathring{R}_{ij}. \end{aligned}\end{aligned}$$
(2.7)

We recall the following inequalies (cf. [12, Lemma 3.4]):

$$\begin{aligned} |W_{ikjl}\mathring{R}_{ij}\mathring{R}_{kl}|\le \sqrt{\frac{n-2}{2(n-1)}}|W||\mathring{R}_{ij}|^2, \end{aligned}$$
(2.8)

and

$$\begin{aligned} \mathring{R}_{ij}\mathring{R}_{ik}\mathring{R}_{jk}\ge -\frac{n-2}{\sqrt{n(n-1)}}|\mathring{R}_{ij}|^3, \end{aligned}$$
(2.9)

with equality in (2.9) at some point \(p\in M\) if and only if \(\mathring{R}_{ij}\) can be diagonalized at p and the eigenvalue multiplicity of \(\mathring{R}_{ij}\) is at least \(n-1\) (see also [9, 10], or [15]). Thus, (2.7) becomes

$$\begin{aligned} \begin{aligned} \frac{1}{2}\Delta |\mathring{R}_{ij}|^2\ge&|\nabla \mathring{R}_{ij}|^2-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}|^3 -\sqrt{\frac{n-2}{2(n-1)}}|W||\mathring{R}_{ij}|^2+\frac{1}{n-1}R|\mathring{R}_{ij}|^2\\&+\frac{n-2}{2(n-1)}R_{,ij}\mathring{R}_{ij}+C_{kij,k}\mathring{R}_{ij}. \end{aligned}\end{aligned}$$
(2.10)

Integrating both sides of (2.10) gives

$$\begin{aligned} 0\ge & {} \int \limits _M|\nabla \mathring{R}_{ij}|^2+\int \limits _M\Big [-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}|^3 -\sqrt{\frac{n-2}{2(n-1)}}|W||\mathring{R}_{ij}|^2+\frac{1}{n-1}R|\mathring{R}_{ij}|^2\Big ]\nonumber \\&+\frac{n-2}{2(n-1)}\int \limits _MR_{,ij}\mathring{R}_{ij}+\int \limits _MC_{kij,k}\mathring{R}_{ij}. \end{aligned}$$
(2.11)

Using the definition of the Cotton tensor given by (2.4) and the fact that the Cotton tensor is trace-less in any two indices, we obtain

$$\begin{aligned} \int \limits _MC_{kij,k}\mathring{\mathrm{R}}_{ij}=-\int \limits _MC_{kij}\mathring{\mathrm{R}}_{ij,k}=-\frac{1}{2}\int \limits _M|C_{ijk}|^2. \end{aligned}$$

On the other hand,

$$\begin{aligned} \int \limits _MR_{,ij}\mathring{R}_{ij}=-\int \limits _MR_{,i}\mathring{R}_{ij,j}=-\frac{n-2}{2n}\int \limits _M|\nabla R|^2, \end{aligned}$$

where we used the second Bianchi identity \(\mathring{R}_{ij,j}=\frac{n-2}{2n}R_{,i}\). Hence, (2.11) becomes

$$\begin{aligned} 0\ge & {} \int \limits _M|\nabla \mathring{R}_{ij}|^2+\int \limits _M\Big [-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}| -\sqrt{\frac{n-2}{2(n-1)}}|W|+\frac{1}{n-1}R\Big ]|\mathring{R}_{ij}|^2\nonumber \\&-\frac{(n-2)^2}{4n(n-1)}\int \limits _M|\nabla R|^2-\frac{1}{2}\int \limits _M|C_{ijk}|^2\nonumber \\\ge & {} \int \limits _M\Big [-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}| -\sqrt{\frac{n-2}{2(n-1)}}|W|+\frac{1}{n-1}R\Big ]|\mathring{R}_{ij}|^2\nonumber \\&-\frac{(n-2)^2}{4n(n-1)}\int \limits _M|\nabla R|^2-\frac{1}{2}\int \limits _M|C_{ijk}|^2, \end{aligned}$$
(2.12)

which yields the desired estimate (1.5).

Now, we consider the case of equality in (1.5), that is

$$\begin{aligned} \begin{aligned} \int \limits _{M}\Bigg [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -&\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Bigg ]|\mathring{R}_{ij}|^2\\ =&\frac{(n-2)^2}{4n}\int \limits _{M}|\nabla R|^2+\frac{n-1}{2}\int \limits _{M}|C_{ijk}|^2. \end{aligned} \end{aligned}$$
(2.13)

In this case, the inequalities (2.8), (2.9), and (2.12) become equalities. In particular, the second equality in (2.12) implies

$$\begin{aligned} \begin{aligned} \nabla \mathring{R}_{ij}=0, \end{aligned}\end{aligned}$$
(2.14)

which shows that \(\mathring{R}_{ij,j}=0\) and hence the scalar curvature is constant. Furthermore, the Ricci curvature is parallel and hence the metric g has harmonic curvature. Thus, (2.13) becomes

$$\begin{aligned} \int \limits _{M}\Bigg [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Bigg ]|\mathring{R}_{ij}|^2=0. \end{aligned}$$
(2.15)

As stated in the lines following (2.9), \(\mathring{R}_{ij}\) has, at each point p, an eigenvalue of multiplicity \(n-1\) or n. Writing \(\mathring{R}_{ij}=av_iv_j+bg_{ij}\) at p, with some scalars a, b and a vector v, we see that the left-hand side of (2.8) is zero at every point p. As (2.8) is an equality, the metric g must be conformally flat or Einstein. Our claim about the equality case now follows from Theorem B of Huang and Ma (or see the proof of Catino’s Theorem A) since the Ricci curvature is parallel.

This completes the proof of Theorem 1.1.

2.2 Proof of Theorem 1.2

By virtue of the definition of \(\sigma _2(A^{\tau })\), we have

$$\begin{aligned} \begin{aligned} \sigma _2(A^{\tau })=&\frac{1}{2}[(\mathrm{tr}A^{\tau })^2-|A^{\tau }_{ij}|^2]\\ =&\frac{1}{2}\Big [\frac{4(n-1)(1-\tau )+n\tau ^2}{4(n-1)}R^2-|R_{ij}|^2\Big ]\\ =&\frac{1}{2}\Big [\frac{[2(n-1)-n\tau ]^2}{4n(n-1)}R^2-|\mathring{R}_{ij}|^2\Big ]. \end{aligned}\end{aligned}$$
(2.16)

Since \(\sigma _2(A^{\tau })\) is a non-negative constant, we have (see [1, Proposition 2.11],

$$\begin{aligned} \begin{aligned} |\nabla A^{\tau }_{ij}|^2\ge |\nabla (\mathrm{tr}A^{\tau })|^2. \end{aligned}\end{aligned}$$
(2.17)

By a direct calculation, we have

$$\begin{aligned} \begin{aligned} |\nabla A^{\tau }_{ij}|^2=&|\nabla R_{ij}|^2+\frac{n\tau ^2-4(n-1)\tau }{4(n-1)^2}|\nabla R|^2\\ =&|\nabla \mathring{R}_{ij}|^2+\frac{[2(n-1)-n\tau ]^2}{4n(n-1)^2}|\nabla R|^2, \end{aligned}\end{aligned}$$
(2.18)

and

$$\begin{aligned} \begin{aligned} |\nabla (\mathrm{tr}A^{\tau })|^2=\frac{[2(n-1)-n\tau ]^2}{4(n-1)^2}|\nabla R|^2, \end{aligned}\end{aligned}$$
(2.19)

then (2.17) is equivalent to

$$\begin{aligned} \begin{aligned} |\nabla \mathring{R}_{ij}|^2\ge \frac{[2(n-1)-n\tau ]^2}{4n(n-1)}|\nabla R|^2. \end{aligned}\end{aligned}$$
(2.20)

Inserting

$$\begin{aligned} |\nabla R|^2\le \frac{4n(n-1)}{[2(n-1)-n\tau ]^2}|\nabla \mathring{R}_{ij}|^2 \end{aligned}$$

with \(\tau \ne 2-\frac{2}{n}\) into (2.12) yields

$$\begin{aligned} \begin{aligned} 0\ge&\Bigg (1-\frac{(n-2)^2}{[2(n-1)-n\tau ]^2}\Bigg )\int \limits _M|\nabla \mathring{R}_{ij}|^2+\int \limits _M\Big [-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}|\\&-\sqrt{\frac{n-2}{2(n-1)}}|W|+\frac{1}{n-1}R\Big ]|\mathring{R}_{ij}|^2-\frac{1}{2}\int \limits _M|C_{ijk}|^2. \end{aligned}\end{aligned}$$
(2.21)

It is easy to check that

$$\begin{aligned} \frac{(n-2)^2}{[2(n-1)-n\tau ]^2}<1 \end{aligned}$$

is equivalent to \(\tau <1\) or \(\tau >3-\frac{4}{n}\). In this case, we have from (2.21),

$$\begin{aligned} \begin{aligned} 0\ge&\int \limits _M\Big [-\sqrt{\frac{n}{n-1}}|\mathring{R}_{ij}| -\sqrt{\frac{n-2}{2(n-1)}}|W|+\frac{1}{n-1}R\Big ]|\mathring{R}_{ij}|^2-\frac{1}{2}\int \limits _M|C_{ijk}|^2, \end{aligned}\end{aligned}$$
(2.22)

and the desired estimate (1.7) is attained.

If the equality in (1.7) holds, then

$$\begin{aligned} \begin{aligned} \int \limits _{M}\Bigg [R-\sqrt{n(n-1)}|\mathring{R}_{ij}| -&\sqrt{\frac{(n-1)(n-2)}{2}}|W|\Bigg ]|\mathring{R}_{ij}|^2=\frac{n-1}{2}\int \limits _{M}|C_{ijk}|^2. \end{aligned} \end{aligned}$$
(2.23)

In this case, the inequalities (2.8), (2.9), and (2.22) become equalities, which also shows that

$$\begin{aligned} \begin{aligned} \nabla \mathring{R}_{ij}=0, \end{aligned}\end{aligned}$$
(2.24)

and the Ricci curvature is parallel. Since the rest of proof is the same as that of Theorem 1.1, we omit it here.

Therefore, we complete the proof of Theorem 1.2.