## A note on double forms and an almost Schur theorem

In this note I am going to introduce the notion of double forms on a Riemannian manifold and use it to define the so called $(p,q)$-curvatures; finally I will prove an almost-Schur type theorem involving $(p,q)$-curvatures. This extends one of the results in a previous post.

1. Preliminaries for double forms

1.1. Algebra structures

Let ${(M^n,g)}$ be a smooth Riemannian manifold. Let ${\Omega^*(M)=\bigoplus_{p\ge0}\Omega^p(M)}$ be the ring of differential forms on ${M}$. Considering the tensor product over the space of smooth functions, we define ${\mathcal D^{p,q}= \Omega^p(M)\otimes \Omega^q(M)}$. The ring of double forms is then defined as ${\mathcal D=\bigotimes _{p,q\ge 0}\mathcal D^{p,q}= \Omega^*(M)\otimes \Omega^*(M)}$.

We define a multiplication on ${\mathcal D}$ as follows. For ${\alpha= \theta_1\otimes \theta_2\in \mathcal D^{p,q}}$ and ${\beta= \theta_3\otimes \theta_4\in \mathcal D^{r,s}}$, we define

$\displaystyle \alpha\cdot \beta= (\theta_1\wedge \theta_3)\otimes (\theta_2\wedge \theta_4)\in \mathcal D^{p+r,q+s}$

and extend it linearly.

Naturally, ${\omega\in \mathcal D^{p,q}}$ can be regarded as a multilinear form which is skew symmetric in the first ${p}$ arguments and also in the last ${q}$ arguments. Under this identification, we have

$\displaystyle \begin{array}{rl} \alpha\cdot \beta(x_1\wedge \cdots \wedge x_{p+r}&, y_1\wedge\cdots \wedge y_{q+s})\\ = \frac 1{p!r!q!s!} \displaystyle\sum _{\sigma\in S_{p+r}, \tau\in S_{q+s}}& \epsilon(\sigma)\epsilon(\tau) \alpha(x_{\sigma(1)}\wedge \cdots\wedge x_{\sigma(p)}, y_{\tau(1)}\wedge \cdots \wedge y_{\tau(q)})\\ & \beta(x_{\sigma(p+1)}\wedge \cdots \wedge x_{\sigma(p+r)}, y_{\tau(q+1)}\wedge \cdots \wedge y_{\tau(q+s)}). \end{array}$

Here ${\epsilon(\sigma)}$ is the sign of the permutation ${\sigma}$ and ${S_k}$ is the permutation group of ${k}$ elements. Whenever possible we will omit the dot and write ${\overbrace{\omega\cdots \omega}^{k }}$ as ${\omega^k}$.

Naturally we can regard the Riemannian metric ${g\in \mathcal D^{1,1}}$ and regard the Riemannian curvature tensor ${R\in \mathcal D^{2,2}}$. For example, if ${\lbrace \omega^i\rbrace_{i=1}^n}$ is an orthonormal coframe for ${M}$, then ${\displaystyle g^k = k!\sum _{i_1<\cdots .

The contraction operator ${c}$ is defined as follows. If ${\omega\in \mathcal D^{p,q}}$, then ${c\omega=0}$ if ${p=0}$ or ${q=0}$, otherwise ${c\omega\in \mathcal D^ {p-1,q-1}}$ is defined by

$\displaystyle c\omega(x_1\wedge\cdots \wedge x_{p-1}, y_1\wedge\cdots \wedge y_{q-1})= \sum_{i=1}^n \omega(e_i \wedge x_1\wedge \cdots \wedge x_{p-1},e_i\wedge y_1\wedge \cdots\wedge y_{q-1})$

where ${\lbrace e_i\rbrace_{i=1}^n}$ is a local orthonormal frame on ${(M^n,g)}$. We then define ${c^r\omega}$ inductively.

We can give an inner product on ${\mathcal D}$ by declaring ${\mathcal D^{p,q}\perp \mathcal D^{r,s}}$ if ${(p,q)\ne (r,s)}$ and elements of the form ${\displaystyle\sum_{\substack{i_1<\cdots to be an orthonormal basis for ${\mathcal D^{p,q}}$.

 Remark 1 It should be noted that the inner product on ${\mathcal D^{p,q}}$ is generally different from the inner product on the space of ${(p+q)}$-tensors by a multiplicative constant. For example, using local orthonormal frame, ${\displaystyle |R|^2_{\mathcal D^{2,2}}= \sum_{i where ${|\cdot|_{\mathcal T}}$ is the norm in the tensor algebra ${\mathcal T(M)}$. However, these two inner products are the same on ${\mathcal D^{p,q}}$ for ${0\le p,q\le 1}$. The only place where we will use ${\langle\cdot, \cdot\rangle_\mathcal T}$ is the proof of Theorem 6, and since in that proof the two inner products coincide, there will be no confusion.

Finally, the Hodge star operator ${*: \mathcal D^{p,q}\rightarrow \mathcal D^{n-p,n-q}}$ is defined by

$\displaystyle *(\theta_1\otimes \theta_2)= *\theta_1\otimes *\theta_2$

(the $*$ on the RHS is just the ordinary Hodge star operator) and extending it linearly.

The following basic but important identity will be used repeatedly ([Labbi1] Theorem 3.1):

$\displaystyle \langle g \alpha, \beta\rangle=\langle \alpha, c\beta\rangle\textrm{ for }\alpha,\beta\in \mathcal D. \ \ \ \ \ (1)$

 Lemma 1 Suppose ${\alpha\in \mathcal D^{1,1}}$, then we have $\displaystyle \langle g,\alpha\rangle= \frac{(n-k)!}{k! (n-1)!}\langle g^k, g^{k-1}\alpha\rangle \textrm{ for all }k\ge 1 \ \ \ \ \ (2)$ and $\displaystyle c^k(g^l)= \frac{l!(n+k-l)!}{(n-l)!}\frac{g^{l-k}}{(l-k)!}\textrm{ for }k\le l \ \ \ \ \ (3)$ where ${g^0=1}$ by definition.

Proof: Equation (3) is a direct consequence of [Labbi1] Lemma 2.1 by substituting ${\omega=1}$ there. By the same lemma again we have

$\displaystyle c^k(g^{k-1}\alpha)= g^{k-1}c^k\alpha + k! \prod_{i=1}^{k-2} (n-1-i)c(\alpha).$

Equation (2) is clearly true for ${k=1}$, otherwise for ${k>1}$, by applying (1) and the above equation we have

$\displaystyle \begin{array}{rl} \langle g^k , g^{k-1}\alpha\rangle= \langle g^0 , c^k (g^{k-1}\alpha)\rangle &= c^{k}g^{k-1}\alpha\\ &= k! (n-1)(n-2)\cdots (n-k+1)c(\alpha)\\ &= \frac{k! (n-1)!}{(n-k)!}\langle g,\alpha\rangle. \end{array}$

$\Box$

 Lemma 2 If ${\alpha\in \mathcal D^{1,1}}$ is trace-free, i.e. ${c(\alpha)=0}$, then $\displaystyle |g^k \alpha | ^2 = \frac{k! (n-2)!}{(n-k-2)!}|\alpha|^2.$

Proof: By [Labbi1] Lemma 2.1, we have ${ c(g^k \alpha)= k(n-1-k) g^{k-1}\alpha}$. Thus

$\displaystyle \begin{array}{rl} |g^{k} \alpha|^2 = \langle g^{k-1}\alpha, cg^k \alpha\rangle &= k(n-1-k) |g^{k-1}\alpha|^2=\cdots \\ &= k! (n-2)(n-3)\cdots (n-k-1)|\alpha|^2\\ &= \frac{k! (n-2)!}{(n-k-2)!}|\alpha|^2. \end{array}$

$\Box$

1.2. Curvature structures

Let ${\mathcal C^p}$ denotes the symmetric elements of ${\mathcal D^{p,p}}$. For ${\omega\in \mathcal C^p}$, we define its sectional curvature ${K_\omega}$ as follows. Let ${P}$ be a ${p}$-dimensional plane in ${T_pM}$, then we define

$\displaystyle K_\omega(P)= \omega(e_1\wedge \cdots \wedge e_p, e_1\wedge \cdots \wedge e_p)$

where ${\lbrace e_i\rbrace _{i=1}^n}$ is an orthonormal basis of ${P}$.

 Definition 3 For ${1\le q\le \frac n2}$ and ${0\le p\le n-2q}$, the ${(p,q)}$-curvature ${s_{p,q}}$ is defined as the sectional curvature of the following ${(p,q)}$-curvature tensor: $\displaystyle R_{(p,q)}=\frac 1 {(n-2q-p)!} * (g^{n-2q-p}R^q).$ Equivalently, ${s_{(p,q)}(P)}$ is the sectional curvature of ${\frac {g^{n-2q-p}R^q}{(n-2q-p)!}}$ on the orthogonal complement of ${P}$.

 Remark 2 For ${q=1}$, ${s_{(p,1)}}$ is called the ${p}$-curvature. In particular, ${s_{(0,1)}}$ is half of the scalar curvature and ${s_{(n-2,1)}(P)}$ is the sectional curvature of ${P^\perp}$. Up to a constant, ${s_{(n-2q,q)}}$ is the Killing-Lipschitz curvature of ${P^\perp }$, which are also called the ${2p}$-sectional curvatures as defined by Thorpe [Thorpe]. For ${p=0}$, we have ([Labbi1] Theorem 4.1) $\displaystyle s_{(0,q)}= * \frac {1}{(n-2q)!} g ^{n-2q}R^q=\frac 1 {(2q)!} c^{2q}R^q$ which are scalar functions generalizing the scalar curvature. They are called the ${2q}$-scalar curvature functions, which we will denote by ${h_{2q}}$, i.e. $\displaystyle h_{2q} = \frac 1 {(2q)!} c^{2q}R^q. \ \ \ \ \ (4)$ We define the ${2q}$-Einstein tensor ${T_{2q}}$ by $\displaystyle T_{2q}= *\frac 1 {(n-2q-1)!} g^{n-2q-1}R^q.$ For ${p=1}$, ${s_{(1,q)}}$ are then the curvatures of the generalized Einstein tensors (See also here). By [Labbi1] Theorem 4.1, we have $\displaystyle T_{2q}= *\frac 1 {(n-2q-1)!} g^{n-2q-1}R^q= h_{2q}g -\frac 1{(2q-1)!}c^{2q-1}R^q. \ \ \ \ \ (5)$

The following result can be found, for example, in [Labbi2] p.179.

 Lemma 4 We have $\displaystyle \mathrm{div} (T_{2q})=0\mathrm{ \quad and \quad }\mathrm{tr}(T_{2q})=(n-2q)h_{2q}.$
 Proposition 5 Let ${\stackrel\circ T_{2q}=T_{2q}-\frac{c(T_{2q})}n g}$ be the traceless part of the generalized Einstein tensor ${T_{2q}}$. Then we have $\displaystyle \begin{array}{rl} |\stackrel\circ T_{2q}|^2 & \le {{n-2}\choose {2q-1}}\left|R^q - \frac{(n-2q)! h_{2q}}{n!}g^{2q}\right|^2\\ \end{array}$

Proof:

By Lemma 4 and (5), it is easy to compute

$\displaystyle \stackrel\circ T_{2q}= \frac {c^{2q-1}R^q}{ (2q-1)!} - \frac{2q}n h_{2q}g. \ \ \ \ \ (6)$

By iterating (1), we have for any ${\alpha\in \mathcal D}$

$\displaystyle \langle \alpha, c^{2q-1}R^q\rangle=\langle g^{k-1}\alpha,c^{2q-k}R^q\rangle \textrm{ for all }1\le k \le 2q.$

In particular,

$\displaystyle \langle \stackrel\circ T_{2q}, c^{2q-1}R^q\rangle=\langle g^{2q-1}\stackrel\circ T_{2q},R^q\rangle . \ \ \ \ \ (7)$

Combining this with (6) and Lemma 1, we have

$\displaystyle \begin{array}{rl} |\stackrel\circ T_{2q}|^2 &= \langle \frac {c^{2q-1}R^q}{(2q-1)!}- \frac {2q}nh_{2q}g, \stackrel\circ T_{2q}\rangle\\ &= \langle \frac {R^q}{(2q-1)!} - \frac{(n-2q)! h_{2q}}{n! (2q-1)!}g^{2q}, g^{2q-1}\stackrel \circ T_{2q}\rangle\\ &\le \left|\frac {R^q}{(2q-1)!} - \frac{(n-2q)! h_{2q}}{n! (2q-1)!}g^{2q}\right|\left| g^{2q-1}\stackrel \circ T_{2q}\right|\\ &= \left|\frac {R^q}{(2q-1)!} - \frac{(n-2q)! h_{2q}}{n! (2q-1)!}g^{2q}\right|\left| g^{2q-1}\stackrel \circ T_{2q}\right|\\ &= \left(\frac{(2q-1)!(n-2)!}{ (n-2q-1)!}\right)^\frac 12\left|\frac {R^q}{(2q-1)!} - \frac{(n-2q)! h_{2q}}{n! (2q-1)!}g^{2q}\right||\stackrel \circ T_{2q}| \end{array}$

We conclude that

$\displaystyle \begin{array}{rl} |\stackrel\circ T_{2q}|^2 & \le \frac{(n-2)!}{ (n-2q-1)!(2q-1)!}\left|R^q - \frac{(n-2q)! h_{2q}}{n!}g^{2q}\right|^2\\ & = {{n-2}\choose {2q-1}}\left|R^q - \frac{(n-2q)! h_{2q}}{n!}g^{2q}\right|^2. \end{array}$

$\Box$

2. Main result

 Theorem 6 Let ${(M^n,g)}$ (${n\ge 3}$) be a closed oriented Riemannian manifold with ${Ric\ge -(n-1)K}$, ${K\ge0}$ and ${1\le q<\frac n2}$, then $\displaystyle \begin{array}{rcl} \int_M |h_{2q}-\overline h_{2q}|^2 &\stackrel{\mathrm{(i)}}\le &\frac{ n(n-1)}{(n-2q)^2} (1+\frac{nK}\lambda)\int_M |\stackrel \circ {T_{2q}}|^2\\ &\stackrel{\mathrm{(ii)}}\le &\frac {2q}{n-2q}{n\choose {2q}} (1+\frac{nK}\lambda)\int_M |R^q - \frac{(n-2q)! h_{2q}}{n!}g^{2q}|^2. \end{array} \ \ \ \ \ (8)$ Here ${\overline h_{2q}}$ is the average of ${h_{2q}}$.

Proof: We follow the ideas in [Lellis-Topping] and [Cheng-Zhou]. We assume ${h_{2q}-\overline h_{2q}}$ is not vanishing everywhere, otherwise there is nothing to prove. Let ${f}$ be the solution to

$\displaystyle \begin{cases} \Delta f = h_{2q}- \overline h_{2q}\\ \int_M f =0. \end{cases} \ \ \ \ \ (9)$

The solution exists because ${\int_M h_{2q}- \overline h_{2q}=0}$. As ${\stackrel \circ {T_{2q}} = T_{2q}- \frac{(n-2q)}n h_{2q} g}$ and ${\textrm{div}(T_{2q})=0}$ by Lemma 4, we have

$\displaystyle \mathrm{div}(\stackrel\circ {T_{2q}})=-\frac {n-2q}n\nabla h_{2q}.$

Then (with Remark 1 in mind)

$\displaystyle \begin{array}{rcl} \int_M |\Delta f|^2 = \int_M (h_{2q}- \overline h_{2q}) \Delta f &=& -\int_M \langle \nabla h_{2q},\nabla f\rangle\\ &=&\frac n{n-2q} \int_M \langle \mathrm{div}(\stackrel \circ {T_{2q}}),\nabla f\rangle\\ &=& \frac n{n-2q}\int_M \langle -\stackrel \circ {T_{2q}} , \nabla ^2f\rangle\\ &=& \frac n{n-2q}\int_M \langle -\stackrel \circ {T_{2q}} , \nabla ^2f- \frac {\Delta f }ng\rangle\\ &\le&\frac n{n-2q} \|\stackrel \circ {T_{2q}} \|_{L^2}\|\nabla ^2 f- \frac{\Delta f}n g\|_{L^2}. \end{array} \ \ \ \ \ (10)$

We have

$\displaystyle \begin{array}{rcl} \|\nabla ^2 f -\frac {\Delta f}n g\|_{L^2}^2 &=& \int_M |\nabla ^2 f|^2 +\frac 1n \int_M |\Delta f|^2 -\frac 2n \int_M |\Delta f|^2\\ &=& \int_M |\nabla ^2 f|^2 -\frac 1n \int_M |\Delta f|^2. \end{array} \ \ \ \ \ (11)$

By Bochner formula, we have

$\displaystyle \int_M |\nabla ^2f|^2 = \int_ M |\Delta f|^2 -\int_M Ric(\nabla f, \nabla f) \le \int_M |\Delta f|^2 +(n-1)K\int_M |\nabla f|^2. \ \ \ \ \ (12)$

Consider

$\displaystyle \begin{array}{rl} \int_M | \nabla f|^2 =-\int_M f \Delta f &\le (\int_M |f|^2)^\frac 12\left(\int_M |\Delta f|^2\right)^\frac12\\ &\le \left(\frac {\int_M | \nabla f|^2}{\lambda}\right)^\frac 12\left(\int_M |\Delta f|^2\right)^\frac12. \end{array}$

Thus

$\displaystyle \int_M |\nabla f|^2 \le \frac 1{\lambda} \int_M |\Delta f|^2. \ \ \ \ \ (13)$

Here we have used the fact that the first eigenvalue ${\lambda=\min \{ \frac {\int_M | \nabla \phi|^2}{\int_M \phi^2}: \int_M \phi=0, \phi\ne0\}}$. So (11) becomes

$\displaystyle \|\nabla ^2 f -\frac {\Delta f}n g\|_{L^2}^2 \le (\frac{n-1}n) (1+\frac{nK}\lambda)\int_M |\Delta f|^2 . \ \ \ \ \ (14)$

Substitute this into (10), we obtain inequality (i) of (8):

$\displaystyle \int_M |h_{2q}-\overline h_{2q}|^2 \le \frac{ n(n-1)}{(n-2q)^2} (1+\frac{nK}\lambda)\int_M |\stackrel \circ {T_{2q}}|^2.$

The inequality (ii) of (8) can be obtained by applying Proposition 5.

As ${T_{2q}- \frac{n-2q}n \overline h_{2q} g= \stackrel\circ {T_{2q}}+\frac{n-2q}n (h_{2q}-\overline h_{2q})g}$, we have

$\displaystyle |T_{2q}- \frac{n-2q}n \overline h_{2q} g |^2= |\stackrel \circ {T_{2q}}|^2 +\frac {(n-2q)^2}n|h_{2q}-\overline h_{2q}|^2.$

Therefore (8) can be rephrased as

$\displaystyle \int_M |T_{2q} - \frac{n-2q}n \overline h_{2q} g|^2 \le n (1+(n-1)K\lambda^{-1}) \int_M |T_{2q} - \frac{n-2q}n h_{2q} g|^2.$

$\Box$