On an inequality of De Lellis and Topping

This is a sequel to my previous post.

It is a classical result of Schur that if the Ricci curvature {Ric=\frac Rn g} on a Riemannian manifold {(M^n,g)}, then the scalar curvature {R} must be constant for {n\ge 3}. This is a simple consequence of the twice contracted Bianchi identity {\mathrm{div}(Ric-\frac R2g)=0}. It is interesting to see how {R} differs from a constant if {Ric-\frac Rng} is close to zero. In this direction Lellis and Topping proved in [Lellis-Topping] that

Theorem 1 If {(M^n,g)} is a closed Riemannian manifold {(n\ge 3)} with nonnegative Ricci curvature, then

\displaystyle \int_M (R-\overline R)^2 \le \frac{4n(n-1)}{(n-2)^2}\int_M |Ric - \frac Rng|^2.

Here {\overline R} is the average of {R} on {M}.  

In this note, we will show an analogous result for the higher {r}-th mean curvature for closed submanifolds in space forms (Theorem 4). In particular we will show that our result implies Theorem 1 and some other results in [Cheng-Zhou] and [Cheng1]. We will also establish another version of this type of result which involves the Riemannian curvature tensor (Theorem 11), which gives a quantitative version of another result of Schur: if {(M^n,g)} {(n\ge 3)} has sectional curvature which depends on its base point only, then its curvature is constant.

1. Submanifolds in space forms

Let {\Sigma^n} be an immersed submanifold in a space form {(N^m,h)}, {n<m}. The second fundamental form of {\Sigma } in {N } is defined by {A(X,Y)=-({\overline \nabla _XY})^\perp} and is normal-valued. Here {\overline \nabla } is the connection on {N}. We denote {A(e_i, e_j)} by {A_{ij}}, where {\lbrace e_i\rbrace_{i=1}^n} is a local orthonormal frame on {\Sigma}.

We define the {r}-th mean curvature as follows. If {r} is even,

\displaystyle  H _r = \frac 1{r!}\sum_{\substack{i_1,\cdots, i_r\\ j_1, \cdots, j_r}} \delta _{j_1\cdots j_r}^{i_1\cdots i_r}h(A_{i_1j_1},A_{i_2j_2})\cdots h(A_{i_{r-1}j_{r-1}},A_{i_rj_r}).

If {r} is odd, the {r}-th mean curvature is a normal vector field defined by

\displaystyle  H _r =\frac 1{r!}\sum_{\substack{i_1,\cdots, i_r\\ j_1, \cdots, j_r}} \delta _{j_1\cdots j_r}^{i_1\cdots i_r}h(A_{i_1j_1},A_{i_2j_2})\cdots h(A_{i_{r-2}j_{r-2}},A_{i_{r-1}j_{r-1}})A_{i_rj_r}.

Here {\delta_{i_1 \cdots i_r}^{j_1\cdots j_r}} is the generalized Kronecker delta symbol. In the codimension one case, i.e. {\Sigma} is a hypersurface, by taking the inner product with a unit normal if necessary, we can assume {H_r} is scalar valued. In this case the value of {H_r} is given by

\displaystyle  H_r =\sum_{i_1<\cdots< i_r}k_{i_1}\cdots k_{i_r} \ \ \ \ \ (1)

where {k_i} are the principal curvatures. This definition of {H_r} will be used whenever {\Sigma } is a hypersurface.

Following [Grosjean] and [Reilly], we define the (generalized) {r}-th Newton transformation {T^r} of {A} (as a {(1,1)} tensor, possibly vector-valued) as follows.
If {r} is even,

\displaystyle  {(T^r)}_j^{\,i}= \frac 1 {r!} \sum_{\substack{i,i_1,\cdots, i_r\\j, j_1, \cdots, j_r}} \delta^{i i_1 \ldots i_r}_{j j_1 \ldots j_r} h(A_{i_1j_1},A_{i_2j_2})\cdots h(A_{i_{r-1}j_{r-1}},A_{i_rj_r}).

If {r} is odd,

\displaystyle {(T^r)}_j^{\,i}= \frac 1 {r!} \sum_{\substack{i,i_1,\cdots, i_r\\j, j_1, \cdots, j_r}} \delta^{i i_1 \ldots i_r}_{j j_1 \ldots j_r} h(A_{i_1j_1},A_{i_2j_2})\cdots h(A_{i_{r-2}j_{r-2}},A_{i_{r-1}j_{r-1}})A_{i_rj_r}.

Again in the codimension one case, we can assume {T^r} is an ordinary {(1,1)} tensor and if {\lbrace e_i\rbrace_{i=1}^n} are the eigenvectors of {A}, then

\displaystyle T^r(e_i)=\frac 1{{n\choose r}}\sum_{\substack{i_1<\cdots <i_r\\ i\ne i_l}}k_{i_1}\cdots k_{i_r}e_i.

This definition of {T^r} will be used whenever {\Sigma} is a hypersurface.

Lemma 2 If {\Sigma^n} is an immersed submanifold in a space form {N^m}, let {\mathrm{div}} and {\mathrm{tr}} denotes the divergence and the trace on {\Sigma} respectively, then

\displaystyle \mathrm{div} (T^r)=0 \textrm{\quad and \quad}\mathrm{tr}(T^{r})= (n-r)H_r.

 

Proof: The first assertion follows from the proof of [Grosjean] Lemma 2.1. But since it has only been shown in the case where {r} is even in that paper, let us assume the odd case, and for the sake of demonstration let {r=3}. Using local orthonormal frame, the first assertion follows from

\displaystyle  \begin{array}{rcl}  &&r! \nabla _i{ (T^{r})}_{i}^j\\ &=& \delta_{i \, i_1 i_1 i_3}^{j \,j_1 j_2 j_3} (h(\nabla_iA_{i_1j_1 }, A_{i_2j_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, \nabla_iA_{i_2j_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, A_{i_2j_2 }) \nabla_i A_{i_3j_3 })\\ &=& \delta_{i \, i_1 i_1 i_3}^{j \,j_1 j_2 j_3} (h(\nabla_{i_1}A_{ij_1 }, A_{i_2j_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, \nabla_{i_1}A_{ij_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, A_{i_2j_2 }) \nabla_{i_3} A_{ij_3 })\\ &=& -\delta_{i \, i_1 i_1 i_3}^{j \,j_1 j_2 j_3} (h(\nabla_iA_{i_1j_1 }, A_{i_2j_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, \nabla_iA_{i_2j_2 }) A_{i_3j_3 }+h(A_{i_1j_1 }, A_{i_2j_2 }) \nabla_i A_{i_3j_3 })\\ &=&-r! \nabla _i T^{r}_{ij} \end{array}

where we have used the Codazzi equation {\nabla _i A_{jk}=\nabla _j A_{ik}} (as {N} has constant curvature). The second assertion is straightforward. \Box

By Lemma 2 we immediately have the following Schur-type theorem which is perhaps well-known to experts (we will write {V s} instead of {V\otimes s} for a vector valued function {V} and a {(1,1) } tensor {s}):

Theorem 3 Let {\Sigma^n} be a closed (compact without boundary) immersed submanifold in a space form {N^m} and {r\in\lbrace1, \cdots, n-1\rbrace}. If the traceless part {\stackrel{\circ}{T^{r}}} of {T^{r}} vanishes, i.e. {T^r= \frac {n-r}nH_r I}, then {H_r} is parallel. (In particular it is constant in the case where it can be defined as a scalar. )  

Proof:

By Lemma 2, as {\stackrel \circ {T^r} = T^r- \frac{(n-r)}n H_r I}, we have {0=\mathrm{div}(\stackrel\circ {T^r})=-\frac {n-r}n\nabla H_r }. i.e. {H_r} is parallel. \Box

Theorem 4 Let {\Sigma^n} ({n\ge 2}) be a closed immersed orientable submanifold in a space form {N^m}, {m>n}. Let {r\in \lbrace 1,\cdots, n-1\rbrace}. Assume that either

  1. {r} is even, or
  2. {r} is odd and {N=\mathbb R^m}, or
  3. {\Sigma} is of codimension one, i.e. a hypersurface.

Let {\overline H_r=\frac 1{ \mathrm{Area}(\Sigma)}\int_\Sigma H_r} be the average of {H_r} (which is a vector when {r} is odd and is defined by (1) in the codimension one case) and {\stackrel{\circ}{T^{r}}= T^{r}- \frac{(n-r)}n H_r I} be the traceless part of {T^{r}}. Let {\lambda} be the first eigenvalue for the Laplacian on {\Sigma} and suppose the Ricci curvature of {\Sigma } is bounded from below by {-(n-1)K}, {K\ge 0}, then for {r=1,\cdots, n-1}, we have

\displaystyle  \int_\Sigma |H_r-\overline H_r|^2 \le \frac{ n(n-1)}{(n-r)^2} (1+\frac{nK}\lambda)\int_\Sigma |\stackrel \circ {T^{r}}|^2, \ \ \ \ \ (2)

or equivalently,

\displaystyle  \int_\Sigma |T^{r} - \frac{n-r}n \overline H_r I|^2 \le n (1+\frac{(n-1)K}\lambda) \int_\Sigma |T^{r} - \frac{n-r}n H_r I|^2.

 

Proof: We follow the ideas in [Lellis-Topping] and [Cheng-Zhou]. We do the the case where {r} is odd (and thus {N=\mathbb R^m}) first. We can assume {H_r-\overline H_r} is not vanishing everywhere, otherwise there is nothing to prove. Let {F=(F^1,\cdots, F^m)} be the solution to

\displaystyle  \begin{cases} \Delta F = H_r- \overline H_r\\ \int_\Sigma F =0. \end{cases} \ \ \ \ \ (3)

The solution exists because {\int_\Sigma H- \overline H_r=0}. As {\stackrel \circ {T^r} = T^r- \frac{(n-r)}n H_r I} and {\textrm{div}(T^r)=0} by Lemma 2, (as a vector-valued {1}-form) we have

\displaystyle  \mathrm{div}(\stackrel\circ {T^r})=-\frac {n-r}n\nabla H_r.

Let us denote the dot product in {\mathbb R^m} by {\cdot} and the intrinsic inner product on {\Sigma } by {\langle \cdot, \cdot\rangle}. Then

\displaystyle  \begin{array}{rcl} \int_\Sigma |\Delta F|^2 = \int_\Sigma (H_r- \overline H_r)\cdot\Delta F &=& -\int_\Sigma \langle \nabla H_r,\nabla F\rangle\\ &=&\frac n{n-r} \int_\Sigma \langle \mathrm{div}(\stackrel \circ {T^r}),\nabla F\rangle\\ &=& \frac n{n-r}\int_\Sigma \langle -\stackrel \circ {T^r} , \nabla ^2F\rangle\\ &= &\frac n{n-r}\int_\Sigma \langle -\stackrel \circ {T^r} , \nabla ^2F- \frac {\Delta F }nI\rangle\\ &\le&\frac n{n-r} \|\stackrel \circ {T^r} \|_{L^2}\|\nabla ^2 F- \frac{\Delta F}n I\|_{L^2}. \end{array} \ \ \ \ \ (4)

We have

\displaystyle  \begin{array}{rcl} \|\nabla ^2 F -\frac {\Delta F}n I\|_{L^2}^2 &= &\int_\Sigma |\nabla ^2 F|^2 +\frac 1n \int_\Sigma |\Delta F|^2 -\frac 2n \int_\Sigma |\Delta F|^2\\ &= &\int_\Sigma |\nabla ^2 F|^2 -\frac 1n \int_\Sigma |\Delta F|^2. \end{array} \ \ \ \ \ (5)

By Bochner formula, we have

\displaystyle  \int_\Sigma |\nabla ^2F|^2 = \int_ M |\Delta F|^2 -\int_\Sigma Ric(\nabla F, \nabla F) \le \int_\Sigma |\Delta F|^2 +(n-1)K\int_\Sigma |\nabla F|^2.

Here by {Ric(\nabla F,\nabla F)} we mean the quantity {\displaystyle \sum_{\substack{1\le i,j\le n\\ 1\le l\le m}} R_{ij}\nabla _iF^l \nabla _j F^l}. Consider

\displaystyle  \begin{array}{rcl}  \int_\Sigma | \nabla F|^2 =-\int_\Sigma F \cdot\Delta F &\le (\int_\Sigma |F|^2)^\frac 12\left(\int_\Sigma |\Delta F|^2\right)^\frac12\\ &\le \left(\frac {\int_\Sigma | \nabla F|^2}{\lambda}\right)^\frac 12\left(\int_\Sigma |\Delta F|^2\right)^\frac12. \end{array}

Thus

\displaystyle  \int_\Sigma |\nabla F|^2 \le \frac 1{\lambda} \int_\Sigma |\Delta F|^2. \ \ \ \ \ (6)

Here we have used the fact that the first eigenvalue {\lambda=\min \{ \frac {\int_\Sigma | \nabla \phi|^2}{\int_\Sigma \phi^2}: \int_\Sigma \phi=0, \phi\ne0\}}. So (5) becomes

\displaystyle  \begin{array}{rcl} \|\nabla ^2 F -\frac {\Delta F}n I\|_{L^2}^2 &=\frac{n-1}n \int_\Sigma |\Delta F|^2 - \int_\Sigma Ric(\nabla F,\nabla F)\\ &\le (\frac{n-1}n) (1+\frac{nK}\lambda)\int_\Sigma |\Delta F|^2 . \end{array} \ \ \ \ \ (7)

Substitute this into (4), we obtain (2):

\displaystyle  \int_\Sigma |H_r-\overline H_r|^2 \le \frac{ n(n-1)}{(n-r)^2} (1+\frac{nK}\lambda)\int_\Sigma |\stackrel \circ {T^{r}}|^2.

As {T^{r}- \frac{n-r}n \overline H_r I= \stackrel\circ {T^{r}}+\frac{n-r}n (H_r-\overline H_r)I}, we have

\displaystyle |T^{r}- \frac{n-r}n \overline H_r I |^2= |\stackrel \circ {T^{r}}|^2 +\frac {(n-r)^2}n|H-\overline H_r|^2.

Therefore (2) can be rephrased as

\displaystyle  \int_\Sigma |T^{r} - \frac{n-r}n \overline H_r I|^2 \le n (1+\frac{(n-1)K}\lambda) \int_\Sigma |T^{r} - \frac{n-r}n H_r I|^2.

For the remaining cases where {r} is even or {\Sigma} is a hypersurface, as {H_r} is scalar valued, just replace {F} in (3) by a scalar valued function {f} and run through the same proof, we can get the result. \Box

Remark 1 When {r=1} and {\Sigma} is a hypersurface, as {T^1= H_1 I-A}, it is easy to see that {\stackrel \circ {T^{1}}= -\stackrel \circ A} where {\stackrel \circ A} is the traceless part of the second fundamental form. This generalizes [Perez] Theorem 3.1 (see also [Cheng-Zhou]). In the codimension one case, this recovers [Cheng1] Theorem 1.10.  

Remark 2 For a embedded hypersurface in Euclidean space, having nonnegative Ricci curvature is equivalent to {A\ge 0} (i.e. convex), see [Perez] p. 48. So when {K=0}, the curvature assumptions in Theorem 4 can be replaced by {\Sigma} being convex when it is an embedded hypersurface.  

2. Relations with Lovelock curvatures

In this section, we investigate the relation between Theorem 4 and an analogous result of Ge-Wang-Xia [Ge-Wang-Xia] when {r} is even. Following [Ge-Wang-Xia], we define the Lovelock curvatures (or the so called {2k}-dimensional Euler density in Physics) of a Riemannian manifold {(M^n,g)}, {k<\frac n2}, by

\displaystyle  R^{(k)}= \frac 1{2^k} \delta_{ i_1 \ldots i_{2k}}^{j_1 \ldots j_{2k}}{R_{j_1 j_2}}^{i_1i_2}\cdots {R_{j_{2k-1} j_{2k}}}^{i_{2k-1} i_{2k}}. \ \ \ \ \ (8)

It can be easily seen that {R^{(1)}} is the sclar curvature. We can also define the generalized Einstein {2}-tensor by defining

\displaystyle  {E^{(k)}}_{i}^j=-\frac 1{2^{k+1}} \delta_{ ii_1 \ldots i_{2k}}^{jj_1 \ldots j_{2k}}{R_{j_1 j_2}}^{i_1i_2}\cdots {R_{j_{2k-1} j_{2k}}}^{i_{2k-1} i_{2k}}.

We have the following analogue of Lemma 2 for {E^{(k)}}:

Lemma 5 We have

\displaystyle  \mathrm{tr}(E^{(k)})= -\frac{n-2k}2R^{(k)}\textrm{\quad and \quad }\nabla ^i {E^{(k)}}_i^j=0.

 

Proof: The first assertion is a straightforward calculation. For the second assertion, for the sake of demonstration let {k=1}. Then this follows from

\displaystyle  \begin{array}{rcl}  -2^{k+1}\nabla ^i {E^{(k)}}_i^j &=&\delta_{i i_1 i_2}^{j j_1j_2}\nabla^i{R_{j_1 j_2}}^{i_1i_2}\\ &=&-\delta_{i i_1 i_2}^{j j_1j_2}(\nabla^{i_1}{R_{j_1 j_2}}^{i_2i}+\nabla^{i_2}{R_{j_1 j_2}}^{ii_1})\\ &=&-\delta_{i_2 i i_1 }^{j j_1j_2}\nabla^i{R_{j_1 j_2}}^{i_1i_2}-\delta_{i_1i_2i }^{j j_1j_2}\nabla^i{R_{j_1 j_2}}^{i_1i_2}\\ &=&-2\delta_{i i_1 i_2}^{j j_1j_2}\nabla^i{R_{j_1 j_2}}^{i_1i_2}. \end{array}

Here we have used the Bianchi identity in the second line. \Box

We see that {E^{(k)}} is divergence free and indeed {E^{(1)}} is the Einstein tensor. By Lemma 5, it is clear that we can prove the analogue of Theorem 4 in this setting. Indeed, by using the same method, Ge, Wang and Xia [Ge-Wang-Xia] proved that (they have assumed {Ric\ge 0}, but their result can be easily extended to the version below):

Theorem 6 Let {(M,g)} be a closed Riemannian manifold with {Ric\ge -(n-1)K}, {K\ge0} and {1\le k<\frac n2}, then

\displaystyle  \int_M |R^{(k)}-\overline R^{(k)}|^2 \le \frac{4n(n-1)}{(n-2k)^2}((1+\frac{nK}\lambda)\int_M |\stackrel{\circ}{E^{(k)}}|^2.

Here {\overline R^{(k)}} is the average of {R^{(k)}}.  

We will show that Theorem 4 is equivalent to Theorem 6 in the case where {r} is even.

Proposition 7 For an immersed submanifold {\Sigma\subset \mathbb R^{n+1}}, for {k=1, \cdots, \lfloor \frac n2\rfloor}, we have

\displaystyle  E^{(k)}=-\frac { k!}2T^{2k}.

 

Proof: We use a local orthnormal frame for computations. We also denote the dot product in {\mathbb R^m} by {\cdot}. By Gauss equation, we have

\displaystyle {R_{ij}}^{kl}= A_{ik} \cdot A_{jl}-A_{il} \cdot A_{jk}.

Consider

\displaystyle  \begin{array}{rcl}  & &\delta_{i i_1 \ldots i_{2k}}^{jj_1 \ldots j_{2k}}{R_{j_1 j_2}}^{i_1i_2}\cdots {R_{j_{2k-1} j_{2k}}}^{i_{2k-1} i_{2k}}\\ & =&\delta_{i i_1 \ldots i_{2k}}^{jj_1 \ldots j_{2k}}(A_{i_1j_1} \cdot A_{i_2j_2}-A_{i_2j_1} \cdot A_{i_1j_2} )\cdots (A_{i_{2k-1}j_{2k-1}}\cdot A_{i_{2k}j_{2k}}-A_{i_{2k}j_{2k-1}}\cdot A_{i_{2k-1}j_{2k}})\\ & =&2^k\delta_{i i_1 \ldots i_{2k}}^{jj_1 \ldots j_{2k}}(A_{i_1j_1}\cdot A_{i_2j_2})\cdots (A_{i_{2k-1}j_{2k-1}}\cdot A_{i_{2k}j_{2k}})\\ & =&2^kk!{(T^{2k})}_i^{\,j}. \end{array}

This implies the result. \Box

Proposition 8 Theorem 4 is equivalent to Theorem 6 when {r=2k} and {N=\mathbb R^m}.  

Proof: To see that Theorem 4 is equivalent to Theorem 6 when {r} is even, we just observe that by Proposition 7, clearly Theorem 6 implies our result in the even case and when {N} is Euclidean. On the other hand, since any manifold can be isometrically embedded into some {\mathbb R^m} for {m} large enough, we see that our result also implies Theorem 6. \Box

3. Equality case of Theorem 4

It is easy to see from the proof that if the Ricci curvature assumption in Theorem 4 is strengthened to {Ric>-(n-1)Kg}, then {H_r=\overline H_r} (as {F} is constant). On the other hand, it is more subtle if we omit this assumption and so far we have only got some partial results in this case. The equality case for {r=1, K=0} when {\Sigma} is an embedded hypersurface has been considered in [Cheng-Zhou], in which they prove that {\Sigma} is a distance sphere. It seems that their proof cannot be modified in our case because in their proof it is essential that {\Sigma} contains a point whose Ricci curvature is positive, which is not true for submanifold in higher codimension in general.
The case for {r=2}.
Suppose {\Sigma^n} is immersed in a space form {N^m} of curvature {c}, using the same computations as in the proof of Proposition 7, we can get that

\displaystyle  E^{(1)}= -T^2- {{n-2}\choose 2 }cI. \ \ \ \ \ (9)

On the other hand, it is not hard to deduce that {E^{1}= Ric-\frac R2 g} is the Einstein tensor (one way of seeing that without computing is to observe that {E^{(1)}\in \mathcal T^2} contains only linear term involving the curvature and is divergence free, thus, up to constant, it must be the Einstein tensor). Thus

\displaystyle -\stackrel{\circ }{T^2}=\stackrel\circ{Ric}=Ric -\frac Rn g.

Note also that {R^{(1)}= R} and is equal to {H_2} up to a constant which only depends on {c} and {n}. Thus our result in this case is reduced to Theorem 1.2 of [Cheng2] or the {k=1} case of Theorem 6. By the rigidity case of [Cheng2] Theorem 1.2, we deduce that the equality holds in Theorem 4 in the {r=2} case if and only if {\Sigma} is Einstein. Therefore {R} is constant and so is {H_2} by Gauss equation. In particular, if {\Sigma} is an embedded surface in {\mathbb R^3}, then by [Ros] it is a sphere.

4. Another form of almost-Schur type theorem

In this section, we derive another form of Schur-type theorem which gives a quantitative version of the following classical result of Schur: if {(M^n,g)} {(n\ge 3)} has sectional curvature which depends on its base point only, then its curvature is constant.

We first set up some notations. Let {\mathcal T^r(M)} denote the space of covariant {r}-tensor on {M} (e.g. {g\in \mathcal T^2(M)}). We define a product

\displaystyle \odot: \mathcal T^2(M)\times \mathcal T^2(M)\rightarrow \mathcal T^4(M)

by

\displaystyle (\alpha\odot \beta)(X,Y,Z,W)= \alpha(X,Z)\beta(Y,W)-\alpha(X,W)\beta(Y,Z).

For our purpose, our product is different from the Kulkarni-Nomizu product (ours has only half of the terms in their product). We define {B= g\odot g}. It is easy to see that {B} is the Riemann curvature tensor of a space form with curvature {1} (we use the convention that {R_{ijij}} is the sectional curvature). In local coordinates, it is given by

\displaystyle  B_{ijkl}=g_{ik}g_{jl}-g_{il}g_{jk}.

Lemma 9 Let {h} be a symmetric {2}-tensor, then

\displaystyle 2\langle Ric , h\rangle=\langle Rm, g\odot h\rangle \textrm{\quad and \quad} 2\langle \frac Rng, g\odot h\rangle=\langle \frac R{n(n-1)}g\odot g, h\rangle.

In particular,

\displaystyle  \langle Ric-\frac Rn g, h\rangle= \frac 12\langle Rm-\frac R{n(n-1)g}g\odot g, g\odot h\rangle.

 

Proof: The first assertion follows from

\displaystyle \langle Rm, g\odot h\rangle= R_{ijkl} (g_{ik}h_{jl}-g_{il}h_{jk})=2R_{ij}h_{ij}= 2\langle Ric, h\rangle.

Define the contraction map {C: \mathcal T^4(M)\rightarrow \mathcal T^2(M)} by {C(A)_{ij}=g^{kl}A_{kilj}}. Then {C(B)= (n-1)g}. By similar calculations, { \langle B, g\odot h\rangle= 2(n-1)g_{ij}h_{ij}= 2\langle C(B),h\rangle}. Thus

\displaystyle \langle \frac R{n(n-1)}B, h\rangle= 2\langle \frac Rng, g\odot h\rangle.

\Box

Lemma 10 Let {h} be a symmetric {2}-tensor, then

\displaystyle  |g\odot h|^2 =2(n-1)|h|^2.

 

Proof: Let {e_i} be an orthonormal basis such that {h_{ij}= \lambda_j \delta_{ij}} (no summation). Then if it easy to see that

\displaystyle  (g\odot h)_{ijkl}=g_{ik}h_{jl}-g_{il}h_{jk}= \begin{cases} \lambda_j\textrm{\quad if }(i,j)=(k,l)\textrm{ and }i\ne j\\ -\lambda_j\textrm{\quad if }(i,j)=(l,k)\textrm{ and }i\ne j\\ 0\textrm{\quad otherwise.} \end{cases}

Thus { |g\odot h|^2 = \sum (g\cdot h)_{ijkl} ^2 = 2\sum_{i\ne j}\lambda_j^2=2(n-1)\sum_j \lambda_j^2= 2(n-1)|h|^2}. \Box

Theorem 11 Suppose {(M^n,g)} ({n\ge 3}) is a closed Riemannian manifold such that its Ricci curvature is bounded from below by {-K}, {K\ge 0}, then we have

\displaystyle  \begin{array}{rcl} \int _M (R-\overline R) ^2&\stackrel{\mathrm{(i)}}\le &\frac {4n(n-1) } {(n-2)^2}(1+\frac{nK}\lambda) \int_M |Ric-\frac Rn g|^2 \\ &\stackrel{\mathrm{(ii)}}\le &\frac {2n(n-1)^2 } {(n-2)^2} (1+\frac{nK}\lambda)\int_M |Rm-\frac {R }{n(n-1)} g\odot g|^2. \end{array} \ \ \ \ \ (10)

where {\overline R} is the average of its scalar curvature {R}. The equality sign in (i) holds if and only if {(M,g)} is Einstein. The equality sign in (ii) holds if and only if {(M,g)} has constant curvature {\frac{\overline R}{n(n-1)}}.  

Proof: From [Cheng2], we have

\displaystyle  \int _M (R-\overline R) ^2 \le \frac {4n(n-1) } {(n-2)^2}(1+\frac{nK}\lambda) \int_M |Ric-\frac Rn g|^2. \ \ \ \ \ (11)

By Lemmas 9, 10, as {\stackrel\circ {Ric}=Ric-\frac Rn g},

\displaystyle  \begin{array}{rcl}  |\stackrel\circ {Ric}|^2 &=&\frac 12 \langle Rm-\frac R{n(n-1)}g\odot g, g\odot \stackrel\circ{Ric}\rangle\\ &\le& \frac 12 |Rm-\frac R{n(n-1)}g\odot g||g\odot \stackrel\circ{Ric}|\\ &= &\frac {\sqrt{2(n-1)}}2 |Rm-\frac R{n(n-1)}g\odot g ||\stackrel\circ{Ric}|. \end{array}

Thus

\displaystyle  |\stackrel\circ {Ric}|^2\le \frac{n-1}2 |Rm-\frac R{n(n-1)}g\odot g|^2.

Plugging this into (11), we can get the result.

If the inequality in (i) becomes an equality, then by [Cheng2], {(M,g)} is Einstein. If the equality in (ii) holds, then from the proof we have

\displaystyle  Rm - \frac {R}{n(n-1)}g\odot g= \alpha (g\odot (Ric -\frac Rn g))

for some constant {\alpha}. By taking the trace we deduce that {\alpha=\frac 1{n-1}} and

\displaystyle  Rm= \frac 1{n-1}g\odot Ric.

Let {\lbrace e_i\rbrace} be an orthonormal basis such that {Ric(e_i,e_j)=(n-1)\lambda_i\delta _{ij}}. (In this proof repeated indices will not be summed over. ) Then in this frame we have

\displaystyle  R_{ijkl}= \lambda_j \delta_{il}\delta_{jk}- \lambda_j \delta_{ik}\delta_{jl} .

From this we see that {R_{ijij}=\lambda_j} ({i\ne j}) is independent of {i}, and thus the sectional curvature only depends on the base point. We thus conclude our result using either the classical Schur’s theorem or using (10). \Box

Corollary 12 With the same assumptions as in Theorem 11, we have

\displaystyle  \int_M |Rm-\frac{\overline R}{n(n-1)}g\odot g|^2 \le \frac{n^2+4(n-1)Kn\lambda^{-1}}{(n-2)^2}\int_M |Rm-\frac{R}{n(n-1)}g\odot g|^2.

 

Proof: It is easy to see that

\displaystyle  \langle Rm, g\odot g\rangle=2R \textrm{\quad and \quad }|g\odot g|^2= 2n(n-1)

which implies

\displaystyle  \langle Rm- \frac {R}{n(n-1)}g\odot g, g\odot g\rangle=0.

As

\displaystyle Rm- \frac {\overline R}{n(n-1)}g\odot g=(Rm- \frac { R}{n(n-1)}g\odot g)+\frac {1}{n(n-1)}(R-\overline R)g\odot g,

we have

\displaystyle  |Rm- \frac {\overline R}{n(n-1)}g\odot g|^2=|Rm- \frac { R}{n(n-1)}g\odot g|^2+\frac 2{n(n-1)}(R-\overline R)^2.

Combining this with (10), we can get the result. \Box

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