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 <i_k} \omega^{i_1}\wedge\cdots \wedge \omega^{i_k}\otimes \omega^{i_1}\wedge\cdots \wedge \omega^{i_k}}.

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 <i_p\\j_1<\cdots <j_q}} \omega^{i_1}\wedge \cdots \wedge \omega^{i_p}\otimes \omega^{j_1}\wedge \cdots \wedge \omega^{j_q}} 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<j, k<l}(R_{ijkl})^2=\frac 14\sum_{i,j,k,l}(R_{ijkl})^2=\frac 14 |R|^2_{\mathcal T}} 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

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