On reduced L² cohomology of hypersurfaces in spheres with finite total curvature

Abstract

In this paper, we prove that the dimension of the second space of reduced L² cohomology of M is finite if is a complete noncompact hypersurface in a sphere 𝕊ⁿ⁺¹and has finite total curvature (n≥3).

total curvature; reduced L² cohomology; hypersurface in sphere; L² harmonic 2-form

Introduction

For a complete manifold $M^n$, the $p$-th space of reduced $L^2$-cohomology is defined, for $0\le p\le n$ in Carron (²⁰⁰⁷). It is interesting and importantto discuss the finiteness of the dimension of these spaces.Carron (¹⁹⁹⁹) proved that if $M^n$ $(n\ge 3)$ is a complete noncompact submanifold of ${ℝ}^{n+p}$with finite total curvature and finite mean curvature (i. e., the $L^n$-norm of the mean curvature vector is finite),then each $p$-th space of reduced $L^2$-cohomology on $M$ has finite dimension, for $0\le p\le n$.The reduced $L^2$ cohomology is related with the $L^2$ harmonic forms (^{Carron 2007}). In fact, several mathematicians studied the space of $L^2$harmonic $p$-forms for $p=1,2$.If $M^n$ $(n\ge 3)$ is a complete minimal hypersurface in ${ℝ}^{n+1}$ with finite index, Li and Wang (²⁰⁰²) proved that the dimension of the space of the $L^2$ harmonic $1$-forms $M$ is finite and $M$ has finitely many ends.More generally, Zhu (²⁰¹³) showed that:suppose that $N^{n+1}$ $(n\ge 3)$ is a complete simply connected manifold with non-positive sectional curvature and$M^n$ is a complete minimal hypersurface in $N$ with finite index. If the bi-Ricci curvature satisfies

$$b-\overline{Ric}(X,Y)+\frac{1}{n}{|A|}^2\ge 0,$$

for all orthonormal tangent vectors $X,Y$ in $T_{p}N$ for $p\in M$, then the dimension of the space of the $L^2$ harmonic $1$-forms $M$ is finite. Furthermore, following the idea of Cheng and Zhou (²⁰⁰⁹); Zhu (²⁰¹³) gave a result on finitely many ends of complete manifolds with a weighted Poincaréinequality by use of the space of $L^2$ harmonic functions.Cavalcante et al. (²⁰¹⁴) discussed a complete noncompact submanifold $M^n$ $(n\ge 3)$ isometrically immersed in a Hadamard manifold$N^{n+p}$ with sectional curvature satisfying $-k^2\le K_N\le 0$ for some constant $k$ and showed that if the total curvature is finiteand the first eigenvalue of the Laplacian operator of $M$ is bounded from below by a suitable constant, then the dimension of the space of the $L^2$ harmonic $1$-forms on $M$ is finite.Fu and Xu (²⁰¹⁰) studied a complete submanifold $M^n$ in a sphere ${𝕊}^{n+p}$ with finite total curvature and bounded mean curvature and proved that the dimension of the space of the $L^2$ harmonic $1$-forms on $M$ is finite. Zhu and Sw. (²⁰¹⁴) proved Fu-Xu’s result without the restriction on the mean curvature vector and therefore obtained that the first space of reduced $L^2$-cohomology on $M$ has finite dimension.Zhu (²⁰¹¹) studied the existence of the symplectic structure and $L^2$ harmonic $2$-forms on complete noncompact manifolds by use of a special version of Bochner formula.

Motivated by above results, we discuss a complete noncompact hypersurface $M^n$ in a sphere ${𝕊}^{n+1}$ with finite total curvature in this paper.We obtain the following finiteness results on the space of all $L^2$ harmonic $2$-forms and the second space of reduced $L^2$cohomology:

Theorem 1. Let $M^n$ $\mathrm{(}n\mathrm{\ge }\mathrm{3}\mathrm{)}$ be an $n$-dimensional complete noncompact oriented manifold isometrically immersed in an $\mathrm{(}n\mathrm{+}\mathrm{1}\mathrm{)}$-dimensional sphere${𝕊}^{n\mathrm{+}\mathrm{1}}$. If the total curvature is finite, then the space of all $L^{\mathrm{2}}$ harmonic $\mathrm{2}$-forms has finite dimension.

Corollary 2. Let $M^n$ $\mathrm{(}n\mathrm{\ge }\mathrm{3}\mathrm{)}$ be an $n$-dimensional complete noncompact oriented manifold isometrically immersed in${𝕊}^{n\mathrm{+}\mathrm{1}}$. If the total curvature is finite, then the dimension of the second space of reduced $L^{\mathrm{2}}$cohomology of $M$ is finite.

Remark 3.Under the same condition of Corollary 2, we conjecture thatthe p-th space of reduced $L^{\mathrm{2}}$ cohomology of $M$ has finite dimension for $\mathrm{3}\mathrm{\le }p\mathrm{\le }n\mathrm{-}\mathrm{3}$.

PRELIMINARIES

In this section, we recall some relevant definitions and results. Suppose that $M^n$ is an $n$-dimensional complete Riemannian manifold.The Hodge operator$*:{\wedge }^p(M)\to {\wedge }^{n-p}(M)$ is defined by

$$*e^{i_1}\wedge \mathrm{\cdots }\wedge e^{i_p}=\mathrm{sgn}\sigma (i_1,i_2,\mathrm{\cdots },i_n)e^{i_{p+1}}\wedge \mathrm{\cdots }\wedge e^{i_n},$$

where $\sigma (i_1,i_2,\mathrm{\cdots },i_n)$ denotes a permutation ofthe set $(i_1,i_2,\mathrm{\cdots },i_n)$ and $\mathrm{sgn}\sigma$ is the sign of$\sigma$. The operator $d^{*}:{\wedge }^p(M)\to {\wedge }^{p-1}(M)$is given by

$$d^{*}\omega ={(-1)}^{(nk+k+1)}*d*\omega .$$

The Laplacian operator is defined by

$$\mathrm{△}\omega =-d{d}^{*}\omega -d^{*}d\omega .$$

A $p$-form $\omega$ is called $L^2$ harmonic if $\mathrm{△}\omega =0$and

We denote by $H^p(L^2(M))$the space of all $L^2$ harmonic $p$-forms on $M$.Let

$$Z_2^p(M)=\{\alpha \in L^2({\wedge }^p(T^{*}M)):d\alpha =0\}$$

and

$$D^p(d)=\{\alpha \in L^2({\wedge }^p(T^{*}M)):d\alpha \in L^2({\wedge }^{p+1}(T^{*}M))\}.$$

We define the $p$-th space of reduced $L^2$ cohomology by

$$H_2^p(M)=\frac{Z_2^p(M)}{\overline{D^{p-1}(d)}}.$$

Suppose that $x:M^n\to {𝕊}^{n+1}$ is an isometric immersion of an $n$-dimensional manifold $M$ in an $(n+1)$-dimensionalsphere.Let $A$ denote the second fundamental form and $H$ the mean curvature of the immersion $x$.Let

$$\mathrm{\Phi }(X,Y)=A(X,Y)-H⟨X,Y⟩,$$

for all vector fields $X$ and $Y$, where $⟨,⟩$ is the induced metric of $M$.We say the immersion $x$ has finite total curvature if

We state several results which will be used to prove Theorem 1.

Proposition 4. (^{Carron 2007})Let $\mathrm{(}M\mathrm{,}g\mathrm{)}$ is a complete Riemannian manifold, then the space of $L^{\mathrm{2}}$ harmonic $p$-forms $H^p\mathit{}\mathrm{(}{L}^{\mathrm{2}}\mathit{}\mathrm{(}M\mathrm{)}\mathrm{)}$ is isomorphic tothe $p$-th space of reduced $L^{\mathrm{2}}$ cohomology $H_{\mathrm{2}}^p\mathit{}\mathrm{(}M\mathrm{)}$.

Lemma 5. (^{Li 1993}) If $\mathrm{(}{M}^n\mathrm{,}g\mathrm{)}$ is a Riemannianmanifold and $\omega \mathrm{=}{a}_I\mathit{}{\omega }_I\mathrm{\in }{\mathrm{\wedge }}^p\mathrm{(}M\mathrm{)}$, then

$\mathrm{△}{|\omega |}^2=2⟨\mathrm{△}\omega ,\omega ⟩+2{|\nabla \omega |}^2+2⟨E(\omega ),\omega ⟩,$

where $E\mathit{}\mathrm{(}\omega \mathrm{)}\mathrm{=}{R}_{k_{\beta }\mathit{}{i}_{\beta }\mathit{}{j}_{\alpha }\mathit{}{i}_{\alpha }}\mathit{}{a}_{i_{\mathrm{1}}\mathit{}\mathrm{\cdots }\mathit{}{k}_{\beta }\mathit{}\mathrm{\cdots }\mathit{}{i}_p}\mathit{}{e}^{i_p}\mathrm{\wedge }\mathrm{\dots }\mathrm{\wedge }{e}^{j_{\alpha }}\mathrm{\wedge }\mathrm{\dots }\mathrm{\wedge }{e}^{i_{\mathrm{1}}}\mathit{.}$

Proposition 6. (^{Hoffman and Spruck 1974}, ^{Zhu and Fang 2014})Let $M^n$ be a complete noncompact oriented manifold isometrically immersed in a sphere${𝕊}^{n\mathrm{+}\mathrm{1}}$. Then

${({\displaystyle {\int }_M}{|f|}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le C_0({\displaystyle {\int }_M}{|\nabla f|}^2+n^2{\displaystyle {\int }_M}(H^2+1)f^2)$

for each $f\mathrm{\in }{C}_{\mathrm{0}}^{\mathrm{1}}\mathit{}\mathrm{(}M\mathrm{)}$, where $C_{\mathrm{0}}$ depends only on $n$ and $H$ is the mean curvature of $M$ in ${𝕊}^{n\mathrm{+}\mathrm{1}}$.

AN INEQUALITY FOR $L^2$ HARMONIC $2$-FORMS

In this section, we show an inequality for $L^2$ harmonic $2$-forms on hypersurfaces in a sphere ${𝕊}^{n+1}$,which plays an important role in the proof of main results.

Proposition 7. Let $M^n$ $\mathrm{(}n\mathrm{\ge }\mathrm{3}\mathrm{)}$ be an $n$-dimensional complete noncompact hypersurface isometrically immersed in an $\mathrm{(}n\mathrm{+}\mathrm{1}\mathrm{)}$-dimensional sphere ${𝕊}^{n\mathrm{+}\mathrm{1}}$. If $\omega \mathrm{\in }{H}^{\mathrm{2}}\mathit{}\mathrm{(}{L}^{\mathrm{2}}\mathit{}\mathrm{(}M\mathrm{)}\mathrm{)}$, then

$$h\mathrm{△}h\ge {|\nabla h|}^2+2h^2-{|\mathrm{\Phi }|}^2{h}^2+\frac{3}{2}{H}^2{h}^2,$$

for $n\mathrm{=}\mathrm{3}$ and

$$h\mathrm{△}h\ge \frac{1}{n-2}{|\nabla h|}^2+2(n-2)h^2-\frac{n-2}{2}{|\mathrm{\Phi }|}^2{h}^2+n{H}^2{h}^2,$$

for $n\mathrm{\ge }\mathrm{4}$, where $h\mathrm{=}\mathrm{|}\omega \mathrm{|}$.

Proof. Suppose that $\omega \in H^2(L^2(M))$. Then we have

$\mathrm{△}{|\omega |}^2=2{|\nabla |\omega ||}^2+2|\omega |\mathrm{△}|\omega |.$ (1)

By Lemma 5, we get that:

$\mathrm{△}{|\omega |}^2$

$=2⟨\mathrm{△}\omega ,\omega ⟩+2{|\nabla \omega |}^2+2⟨E(\omega ),\omega ⟩$

$=2{|\nabla \omega |}^2+2⟨E(\omega ),\omega ⟩.$ (2)

Combining (1) with (AN INEQUALITY FOR $L^2$ HARMONIC $2$-FORMS), we obtain that

$|\omega |\mathrm{△}|\omega |={|\nabla \omega |}^2-{|\nabla |\omega ||}^2+⟨E(\omega ),\omega ⟩.$ (3)

There exists the Kato inequality for $L^2$ harmonic $2$-forms as follows (^{Cibotaru and Zhu 2012}, ^{Wang 2002}):

${\displaystyle \frac{n-1}{n-2}}{|\nabla |\omega ||}^2\le {|\nabla \omega |}^2.$ (4)

By (3) and (4), we get that

$|\omega |\mathrm{△}|\omega |\ge {\displaystyle \frac{1}{n-2}}{|\nabla |\omega ||}^2+⟨E(\omega ),\omega ⟩.$ (5)

Now, we give the estimate of the term $⟨E(\omega ),\omega ⟩$. Let ${\omega }_1=b_{i_1{i}_2}{e}^{i_2}\wedge e^{i_1}\in {\wedge }^2(M)$ and${\omega }_2=c_{i_1{i}_2}{e}^{i_2}\wedge e^{i_1}\in {\wedge }^2(M)$, where$b_{i_1{i}_2}=-b_{i_2{i}_1}$ and $c_{i_1{i}_2}=-c_{i_2{i}_1}$. By Lemma 5, we obtain that

$E({\omega }_1)$

$=R_{k_1{i}_1{j}_1{i}_1}{b}_{k_1{i}_2}{e}^{i_2}\wedge e^{j_1}+R_{k_2{i}_2{j}_2{i}_2}{b}_{i_1{k}_2}{e}^{j_2}\wedge e^{i_1}$

$+R_{k_2{i}_2{j}_1{i}_1}{b}_{i_1{k}_2}{e}^{i_2}\wedge e^{j_1}+R_{k_1{i}_1{j}_2{i}_2}{b}_{k_1{i}_2}{e}^{j_2}\wedge e^{i_1}$

$=Ri{c}_{k_1{j}_1}{b}_{k_1{i}_2}{e}^{i_2}\wedge e^{j_1}+Ri{c}_{k_2{j}_2}{b}_{i_1{k}_2}{e}^{j_2}\wedge e^{i_1}$

$+R_{k_2{i}_2{j}_1{i}_1}{b}_{i_1{k}_2}{e}^{i_2}\wedge e^{j_1}+R_{k_1{i}_1{j}_2{i}_2}{b}_{k_1{i}_2}{e}^{j_2}\wedge e^{i_1}.$

So, we get that

$⟨E({\omega }_1),{\omega }_2⟩=$

$Ri{c}_{k_1{j}_1}{b}_{k_1{i}_2}{c}_{j_1{i}_2}+Ri{c}_{k_2{j}_2}{b}_{i_1{k}_2}{c}_{i_1{j}_2}$

$+R_{k_2{i}_2{j}_1{i}_1}{b}_{i_1{k}_2}{c}_{j_1{i}_2}+R_{k_1{i}_1{j}_2{i}_2}{b}_{k_1{i}_2}{c}_{i_1{j}_2},$

which implies that

$⟨E(\omega ),\omega ⟩=$

$Ri{c}_{k_1{j}_1}{a}_{k_1{i}_2}{a}_{j_1{i}_2}+Ri{c}_{k_2{j}_2}{a}_{i_1{k}_2}{a}_{i_1{j}_2}$

$+R_{k_2{i}_2{j}_1{i}_1}{a}_{i_1{k}_2}{a}_{j_1{i}_2}+R_{k_1{i}_1{j}_2{i}_2}{a}_{k_1{i}_2}{a}_{i_1{j}_2}.$ (6)

By Gauss equation, we have that

$R_{ijkl}=({\delta }_{ik}{\delta }_{jl}-{\delta }_{il}{\delta }_{jk})+h_{ik}{h}_{jl}-h_{il}{h}_{jk}.$

A direct computation shows that

$Ri{c}_{k_1{j}_1}=(n-1){\delta }_{k_1{j}_1}+nH{h}_{k_1{j}_1}-h_{k_{1}i}{h}_{i{j}_1};$ (7)

$Ri{c}_{k_2{j}_2}=(n-1){\delta }_{k_2{j}_2}+nH{h}_{k_2{j}_2}-h_{k_{2}i}{h}_{i{j}_2};$ (8)

$R_{k_2{i}_2{j}_1{i}_1}=({\delta }_{k_2{j}_1}{\delta }_{i_2{i}_1}-{\delta }_{k_2{i}_1}{\delta }_{i_2{j}_1})+h_{k_2{j}_1}{h}_{i_2{i}_1}-h_{k_2{i}_1}{h}_{i_2{j}_1}$ (9)

and

$R_{k_1{i}_1{j}_2{i}_2}=({\delta }_{k_1{j}_2}{\delta }_{i_1{i}_2}-{\delta }_{k_1{i}_2}{\delta }_{i_1{j}_2})+h_{k_1{j}_2}{h}_{i_1{i}_2}-h_{k_1{i}_2}{h}_{i_1{j}_2}.$ (10)

Since the curvature operator $E$ is linear andzero order, and hence tensorial, it is sufficient to compute $⟨E(\omega ),\omega ⟩$ at a point $p$.We can choose an orthonormal frame $\{e_i\}$ such that$h_{ij}={\lambda }_i{\delta }_{ij}$ at $p$. Obviously,

$$nH={\lambda }_1+\mathrm{\cdots }+{\lambda }_n.$$

By (6)-(10), we have

$⟨E(\omega ),\omega ⟩=$

$(n-1){\displaystyle \sum {(a_{j_1{i}_2})}^2}+{\displaystyle \sum nH{\lambda }_{k_1}{(a_{k_1{i}_2})}^2}-{\displaystyle \sum {\lambda }_{k_1}^2{(a_{k_1{i}_2})}^2}$

$+$

$(n-1){\displaystyle \sum {(a_{i_1{j}_2})}^2}+{\displaystyle \sum nH{\lambda }_{k_2}{(a_{i_1{k}_2})}^2}-{\displaystyle \sum {\lambda }_{k_2}^2{(a_{i_1{k}_2})}^2}$

$+$

${\displaystyle \sum a_{i_1{j}_1}{a}_{j_1{i}_1}}-{\displaystyle \sum {\lambda }_{k_2{\lambda }_{i_2}}{(a_{k_2{i}_2})}^2}$

$+$

${\displaystyle \sum a_{j_2{i}_2}{a}_{i_2{j}_2}}-{\displaystyle \sum {\lambda }_{j_2{\lambda }_{i_2}}{(a_{j_2{i}_2})}^2}$

$=$

$2{\displaystyle \sum _{i\ne j}}\left((n-2)+({\lambda }_1+\mathrm{\cdots }+{\lambda }_n){\lambda }_i-{\lambda }_i^2-{\lambda }_i{\lambda }_j\right){(a_{ij})}^2.$

Note that

$${|A|}^2={|\mathrm{\Phi }|}^2+n{H}^2.$$

For $n=3$, we have that

$⟨$

$E(\omega ),\omega ⟩=2{\displaystyle \sum }{}_{i\ne j}(1+({\lambda }_1+{\lambda }_2+{\lambda }_3){\lambda }_i-{\lambda }_i^2-{\lambda }_i{\lambda }_j)(a_{ij}){}^2$

$={\displaystyle \sum _{i\ne j}}\left(2+({\lambda }_1+{\lambda }_2+{\lambda }_3)({\lambda }_i+{\lambda }_j)-({\lambda }_i^2+{\lambda }_j^2)-2{\lambda }_i{\lambda }_j\right){(a_{ij})}^2$

$={\displaystyle \sum _{i\ne j}}\left(2+{\displaystyle \frac{1}{2}}{(3H)}^2-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1,k\ne i,j}^3}{\lambda }_k^2-{\displaystyle \frac{1}{2}}{({\lambda }_i+{\lambda }_j)}^2\right){(a_{ij})}^2$

$\ge {\displaystyle \sum _{i\ne j}}\left(2+{\displaystyle \frac{1}{2}}{(3H)}^2-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{k=1,k\ne i,j}^3}{\lambda }_k^2-({\lambda }_i^2+{\lambda }_j^2)\right){(a_{ij})}^2$

$\ge {\displaystyle \sum _{i\ne j}}\left(2+{\displaystyle \frac{9}{2}}{H}^2-{|A|}^2\right){(a_{ij})}^2$

$=(2+{\displaystyle \frac{3}{2}}{H}^2-{|\mathrm{\Phi }|}^2){|\omega |}^2.$

For $n\ge 4$, we obtain that

$⟨$

$E(\omega ),\omega ⟩=2{\displaystyle \sum }{}_{i\ne j}((n-2)+({\lambda }_1+\mathrm{\cdots }+{\lambda }_n){\lambda }_i-{\lambda }_i^2-{\lambda }_i{\lambda }_j)(a_{ij}){}^2$

$={\displaystyle \sum _{i\ne j}}\left(2(n-2)+({\lambda }_1+\mathrm{\cdots }+{\lambda }_n)({\lambda }_i+{\lambda }_j)-({\lambda }_i^2+{\lambda }_j^2)-2{\lambda }_i{\lambda }_j\right){(a_{ij})}^2$

$={\displaystyle \sum _{i\ne j}}\left(2(n-2)+({\lambda }_1+\mathrm{\cdots }+\widehat{{\lambda }_i}+\mathrm{\cdots }+\widehat{{\lambda }_j}+\mathrm{\cdots }+{\lambda }_n)({\lambda }_i+{\lambda }_j)\right){(a_{ij})}^2$

$={\displaystyle \sum _{i\ne j}}\left(2(n-2)+{\displaystyle \frac{1}{2}}{(nH)}^2-{\displaystyle \frac{1}{2}}{({\displaystyle \sum _{k=1,k\ne i,j}^n}{\lambda }_k)}^2-{\displaystyle \frac{1}{2}}{({\lambda }_i+{\lambda }_j)}^2\right){(a_{ij})}^2$

$\ge {\displaystyle \sum _{i\ne j}}\left(2(n-2)+{\displaystyle \frac{1}{2}}{(nH)}^2-{\displaystyle \frac{n-2}{2}}({\displaystyle \sum _{k=1,k\ne i,j}^n}{\lambda }_k^2)-({\lambda }_i^2+{\lambda }_j^2)\right){(a_{ij})}^2$

$\ge {\displaystyle \sum _{i\ne j}}\left(2(n-2)+{\displaystyle \frac{1}{2}}{(nH)}^2-{\displaystyle \frac{n-2}{2}}{|A|}^2\right){(a_{ij})}^2$

$=\left(2(n-2)+{\displaystyle \frac{1}{2}}{(nH)}^2-{\displaystyle \frac{n-2}{2}}{|A|}^2\right){|\omega |}^2$

$=\left(2(n-2)+n{H}^2-{\displaystyle \frac{n-2}{2}}{|\mathrm{\Phi }|}^2\right){|\omega |}^2.$

By (5), we have that:

$h\mathrm{△}h\ge {|\nabla h|}^2+2h^2-{|\mathrm{\Phi }|}^2{h}^2+{\displaystyle \frac{3}{2}}{H}^2{h}^2,$

for $n=3$ and

$h\mathrm{△}h\ge {\displaystyle \frac{1}{n-2}}{|\nabla h|}^2+2(n-2)h^2-{\displaystyle \frac{n-2}{2}}{|\mathrm{\Phi }|}^2{h}^2+n{H}^2{h}^2,$

for $n\ge 4$.∎

Remark 8. If $\omega$ is $\mathrm{1}$-form , then the term $E\mathit{}\mathrm{(}\omega \mathrm{,}\omega \mathrm{)}$ is equal to $R\mathit{}i\mathit{}c\mathit{}\mathrm{(}\omega \mathrm{,}\omega \mathrm{)}$.The corresponding estimate for this term was given by Leung (¹⁹⁹²).

PROOF OF MAIN RESULTS

In this section, we prove Theorem 1 and Corollary 2.

If $\eta$ is a compactly supported piecewise smooth function on $M$, then

$div({\eta }^{2}h\nabla h)$

$={\eta }^{2}h\mathrm{△}h+⟨\nabla ({\eta }^{2}h),\nabla h⟩$

$={\eta }^{2}h\mathrm{△}h+{\eta }^2{|\nabla h|}^2+2\eta h⟨\nabla \eta ,\nabla h⟩.$

Integrating by parts on $M$, we obtain that

${\displaystyle {\int }_M}{\eta }^{2}h\mathrm{△}h+{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+2{\displaystyle {\int }_M}\eta h⟨\nabla \eta ,\nabla h⟩=0.$ (11)

Case I: $n\mathrm{=}\mathrm{3}$. By Proposition 7 and (11), we obtain that

$-$

$2{\displaystyle {\int }_M}\eta h⟨\nabla \eta ,\nabla h⟩-2{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2-2{\displaystyle {\int }_M}{\eta }^2{h}^2$

$+{\displaystyle {\int }_M}{|\mathrm{\Phi }|}^2{\eta }^2{h}^2-{\displaystyle \frac{3}{2}}{\displaystyle {\int }_M}{H}^2{h}^2{\eta }^2\ge 0.$ (12)

Note that

$-2{\displaystyle {\int }_M}\eta h⟨\nabla \eta ,\nabla h⟩\le a_1{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{\displaystyle \frac{1}{a_1}}{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (13)

for any positive real number $a_1$.Now we give an estimate of the term ${\int }_M{|\mathrm{\Phi }|}^2{\eta }^2{h}^2$ as follows:set ${\varphi }_1(\eta )={\left({\int }_{Supp\eta }{|\mathrm{\Phi }|}^3\right)}^{\frac{1}{3}}$. Then there exists

${\displaystyle {\int }_M}{|\mathrm{\Phi }|}^2{\eta }^2{h}^2\le {\left({\displaystyle {\int }_{Supp\eta }}{\left({|\mathrm{\Phi }|}^2\right)}^{\frac{3}{2}}\right)}^{\frac{2}{3}}\cdot {\left({\displaystyle {\int }_M}{\left({\eta }^2{h}^2\right)}^3\right)}^{\frac{1}{3}}$

$={\varphi }_1{(\eta )}^2\cdot {\left({\displaystyle {\int }_M}{\left(\eta h\right)}^6\right)}^{\frac{1}{3}}$

$\le C_0{\varphi }_1{(\eta )}^2\cdot \left({\displaystyle {\int }_M}{|\nabla (\eta h)|}^2+9{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2\right)$

$\le C_0{\varphi }_1{(\eta )}^2\cdot \left((1+{\displaystyle \frac{1}{b_1}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+(1+b_1){\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+9{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2\right),$ (14)

for any positive real number $b_1$, where the second inequality holds because of Proposition 6.By (12)-(14), we obtain that

${𝒜}_1{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{ℬ}_1{\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2+{𝒞}_1{\displaystyle {\int }_M}{\eta }^2{h}^2\le {𝒟}_1{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (15)

where

${𝒜}_1:$

$=(2-C_0{\varphi }_1{(\eta )}^2)-(a_1+b_1{C}_0{\varphi }_1{(\eta )}^2),$

${ℬ}_1:$

$={\displaystyle \frac{3}{2}}-9C_0{\varphi }_1{(\eta )}^2,$

${𝒞}_1:$

$=2-9C_0{\varphi }_1{(\eta )}^2$

and

${𝒟}_1:={\displaystyle \frac{1}{a_1}}+C_0{\varphi }_1{(\eta )}^2(1+{\displaystyle \frac{1}{b_1}}).$

Since the total curvature ${\parallel \mathrm{\Phi }\parallel }_{L^3(M)}$ is finite, we can choose a fixed$r_0$ such that

Set

${\widetilde{𝒜}}_1:$

$=(2-C_0{\delta }_1^2)-(a_1+b_1{C}_0{\delta }_1^2),$

${\widetilde{ℬ}}_1:$

$={\displaystyle \frac{3}{2}}-9C_0{\delta }_1^2,$

${\widetilde{𝒞}}_1:$

$=2-9C_0{\delta }_1^2$

and

${\widetilde{𝒟}}_1:={\displaystyle \frac{1}{a_1}}+C_0{\delta }_1^2(1+{\displaystyle \frac{1}{b_1}}).$

Thus,

${\widetilde{𝒜}}_1{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{\widetilde{ℬ}}_1{\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2+{\widetilde{𝒞}}_1{\displaystyle {\int }_M}{\eta }^2{h}^2\le {\widetilde{𝒟}}_1{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (16)

for any $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$.By Proposition 6, we have

${\displaystyle \frac{1}{C_0}}{\left({\displaystyle {\int }_M}{\left(\eta h\right)}^6\right)}^{\frac{1}{3}}\le {\displaystyle {\int }_M}{|\nabla (\eta h)|}^2+9{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2$

$\le (1+{\displaystyle \frac{1}{c_1}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+(1+c_1){\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+9{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2,$ (17)

for any positive real number $c_1$.By (16) and (17), we have

${\displaystyle \frac{1}{C_0}}{\left({\displaystyle {\int }_M}{\left(\eta h\right)}^6\right)}^{\frac{1}{3}}$

$\le (1+{\displaystyle \frac{1}{c_1}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+(1+c_1){\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+9{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2$

$\le (1+{\displaystyle \frac{1}{c_1}}+(1+c_1){\displaystyle \frac{{\widetilde{𝒟}}_1}{{\widetilde{𝒜}}_1}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+(9-(1+c_1){\displaystyle \frac{{\widetilde{ℬ}}_1}{{\widetilde{𝒜}}_1}}){\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2$

$+(9-(1+c_1){\displaystyle \frac{{\widetilde{𝒞}}_1}{{\widetilde{𝒜}}_1}}){\displaystyle {\int }_M}{\eta }^2{h}^2.$ (18)

Choose a sufficient large $c_1$ such that

and

Then (18) implies that

${({\displaystyle {\int }_M}{(\eta h)}^6)}^{\frac{1}{3}}\le \tilde{A}{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (19)

for any $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$. where $\tilde{A}$ is a positive constant.

Case II: $n\mathrm{\ge }\mathrm{4}$. By Proposition 7 and (11), we obtain that

$-$

$2{\displaystyle {\int }_M}\eta h⟨\nabla \eta ,\nabla h⟩-{\displaystyle \frac{n-1}{n-2}}{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2-2(n-2){\displaystyle {\int }_M}{\eta }^2{h}^2$

$+{\displaystyle \frac{n-2}{2}}{\displaystyle {\int }_M}{|\mathrm{\Phi }|}^2{\eta }^2{h}^2-n{\displaystyle {\int }_M}{H}^2{h}^2{\eta }^2\ge 0.$ (20)

Note that

$-2{\displaystyle {\int }_M}\eta h⟨\nabla \eta ,\nabla h⟩\le a_2{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{\displaystyle \frac{1}{a_2}}{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (21)

for any positive real number $a_2$. Weset ${\varphi }_2(\eta )={\left({\int }_{Supp\eta }{|\mathrm{\Phi }|}^n\right)}^{\frac{1}{n}}$ and obtain that

${\displaystyle {\int }_M}{|\mathrm{\Phi }|}^2{\eta }^2{h}^2\le {\left({\displaystyle {\int }_{Supp\eta }}{\left({|\mathrm{\Phi }|}^2\right)}^{\frac{n}{2}}\right)}^{\frac{2}{n}}\cdot {\left({\displaystyle {\int }_M}{\left({\eta }^2{h}^2\right)}^{\frac{n}{n-2}}\right)}^{\frac{n-2}{n}}$

$={\varphi }_2{(\eta )}^2\cdot {\left({\displaystyle {\int }_M}{\left(\eta h\right)}^{\frac{2n}{n-2}}\right)}^{\frac{n-2}{n}}$

$\le C_0{\varphi }_2{(\eta )}^2\cdot \left({\displaystyle {\int }_M}{|\nabla (\eta h)|}^2+n^2{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2\right)$

$\le C_0{\varphi }_2{(\eta )}^2\cdot \left({\displaystyle {\int }_M}(1+{\displaystyle \frac{1}{b_2}})h^2{|\nabla \eta |}^2+(1+b_2){\eta }^2{|\nabla h|}^2+n^2{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2\right),$ (22)

for any positive real number $b_2$, where the second inequality holds because of Proposition 6.By (20)-(22), there exists

${𝒜}_2{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{ℬ}_2{\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2+{𝒞}_2{\displaystyle {\int }_M}{\eta }^2{h}^2\le {𝒟}_2{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (23)

where

${𝒜}_2:$

$=\left({\displaystyle \frac{n-1}{n-2}}-{\displaystyle \frac{n-2}{2}}{C}_0{\varphi }_2{(\eta )}^2\right)-\left(a_2+{\displaystyle \frac{n-2}{2}}{b}_2{C}_0{\varphi }_2{(\eta )}^2\right),$

${ℬ}_2:$

$=n-{\displaystyle \frac{n^2(n-2)}{2}}{C}_0{\varphi }_2{(\eta )}^2,$

${𝒞}_2:$

$=2(n-2)-{\displaystyle \frac{n^2(n-2)}{2}}{C}_0{\varphi }_2{(\eta )}^2$

and

${𝒟}_2:={\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{n-2}{2}}(1+{\displaystyle \frac{1}{b_2}})C_0{\varphi }_2{(\eta )}^2.$

Since the total curvature ${\parallel \mathrm{\Phi }\parallel }_{L^n(M)}$ is finite, we can choose a fixed$r_0$ such that

${\widetilde{𝒜}}_2:$

$=\left({\displaystyle \frac{n-1}{n-2}}-{\displaystyle \frac{n-2}{2}}{C}_0{\delta }_2^2\right)-\left(a_2+{\displaystyle \frac{n-2}{2}}{b}_2{C}_0{\delta }_2^2\right),$

${\widetilde{ℬ}}_2:$

$=n-{\displaystyle \frac{n^2(n-2)}{2}}{C}_0{\delta }_2^2,$

${\widetilde{𝒞}}_2:$

$=2(n-2)-{\displaystyle \frac{n^2(n-2)}{2}}{C}_0{\delta }_2^2$

and

${\widetilde{𝒟}}_2:={\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{n-2}{2}}(1+{\displaystyle \frac{1}{b_2}})C_0{\delta }_2^2.$

Obviously, ${\widetilde{𝒜}}_2$, ${\widetilde{ℬ}}_2$, ${\widetilde{𝒞}}_2$ and ${\widetilde{𝒟}}_2$ are positive.Thus,

${\widetilde{𝒜}}_2{\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+{\widetilde{ℬ}}_2{\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2+{\widetilde{𝒞}}_2{\displaystyle {\int }_M}{\eta }^2{h}^2\le {\widetilde{𝒟}}_2{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (24)

for any $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$.Combining with Proposition 6, we get that

${\displaystyle \frac{1}{C_0}}$

${({\displaystyle {\int }_M}{|\eta h|}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le {\displaystyle {\int }_M}{|\nabla (\eta h)|}^2+n^2{\displaystyle {\int }_M}(H^2+1){(\eta h)}^2$

$\le (1+c_2){\displaystyle {\int }_M}{\eta }^2{|\nabla h|}^2+(1+{\displaystyle \frac{1}{c_2}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+n^2{\displaystyle {\int }_M}(H^2+1){\eta }^2{h}^2,$ (25)

for any positive real number $c_2$.By (24) and (25), we have

${\displaystyle \frac{1}{C_0}}{({\displaystyle {\int }_M}{|\eta h|}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}$

$\le (1+{\displaystyle \frac{1}{c_2}}+(1+c_2){\displaystyle \frac{{\widetilde{𝒟}}_2}{{\widetilde{𝒜}}_2}}){\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2+(n^2-(1+c_2){\displaystyle \frac{{\widetilde{ℬ}}_2}{{\widetilde{𝒜}}_2}}){\displaystyle {\int }_M}{H}^2{\eta }^2{h}^2$

$+(n^2-(1+c_2){\displaystyle \frac{{\widetilde{𝒞}}_2}{{\widetilde{𝒜}}_2}}){\displaystyle {\int }_M}{\eta }^2{h}^2.$ (26)

We choose a sufficient large $c_2$ such that

and

Then (26) implies that

${({\displaystyle {\int }_M}{(\eta h)}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le \tilde{A}{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (27)

for any $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$, where $\tilde{A}$ is a positive constant depending only on $n$.

By Case I and Case II, we have that

${({\displaystyle {\int }_M}{(\eta h)}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le \tilde{A}{\displaystyle {\int }_M}{h}^2{|\nabla \eta |}^2,$ (28)

for any $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$, where $\tilde{A}$ is a positive constant depending only on $n$ $(n\ge 3)$.

Next, the proof follows standard techniques (after inequality (33) in Cavalcante et al. (²⁰¹⁴ and uses a Moser iteration argument (lemma 11 in Li (¹⁹⁸⁰)).We include a concise proof here for the sake of completeness.Choose $r>r_0+1$ and $\eta \in C_0^{\mathrm{\infty }}(M-B_{r_0})$ such that

for some positive constant $\tilde{c}$. Then (28) becomes that

${({\displaystyle {\int }_{B_r-B_{r_0+1}}}{h}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le \tilde{A}{\displaystyle {\int }_{B_{r_0+1}-B_{r_0}}}{h}^2+{\displaystyle \frac{\tilde{A}}{r^2}}{\displaystyle {\int }_{B_{2r}-B_r}}{h}^2.$

Letting $r\to \mathrm{\infty }$ and noting that $h\in L^2(M)$, we obtain that

${({\displaystyle {\int }_{M-B_{r_0+1}}}{h}^{\frac{2n}{n-2}})}^{\frac{n-2}{n}}\le \tilde{A}{\displaystyle {\int }_{B_{r_0+1}-B_{r_0}}}{h}^2.$ (29)

By Hölder inequality

$${\int }_{B_{r_0+2}-B_{r_0+1}}{h}^2\le {\left({\int }_{B_{r_0+2}-B_{r_0+1}}{h}^{\frac{2n}{n-2}}\right)}^{\frac{n-2}{n}}\cdot {\left({\int }_{B_{r_0+2}-B_{r_0+1}}{1}^{\frac{n}{2}}\right)}^{\frac{2}{n}},$$

we get that

${\displaystyle {\int }_{B_{r_0+2}}}{h}^2\le (1+\tilde{A}Vol{(B_{r_0+2})}^{\frac{2}{n}}){\displaystyle {\int }_{B_{r_0+1}}}{h}^2.$ (30)

Set

$$\mathrm{\Psi }=\{\begin{array}{cc}|2-{|\mathrm{\Phi }|}^2+\frac{3}{2}{H}^2|,forn=3,\hfill & \\ |2(n-2)-\frac{n-2}{2}{|\mathrm{\Phi }|}^2+n{H}^2|,forn\ge 4.\hfill & \end{array}$$

Fix $x\in M$ and take $\tau \in C_0^1(B_1(x))$. Proposition 7 implies that

$$h\mathrm{△}h\ge \alpha {|\nabla h|}^2-\mathrm{\Psi }{h}^2,$$

where

$$\alpha =\{\begin{array}{cc}\frac{1}{2},forn=3,\hfill & \\ \frac{1}{n-2},forn\ge 4.\hfill & \end{array}$$

Then, for $p>2$, there exists

$${\int }_M{\tau }^2{h}^{p-1}\mathrm{△}h\ge \alpha {\int }_M{\tau }^2{h}^{p-2}{|\nabla h|}^2-{\int }_M{\tau }^2\mathrm{\Psi }{h}^p.$$

That is,

$-2{\displaystyle {\int }_{B_1(x)}}\tau h^{p-1}⟨\nabla \tau ,\nabla h⟩$

$\ge (\alpha +(p-1)){\displaystyle {\int }_{B_1(x)}}{\tau }^2{h}^{p-2}{|\nabla h|}^2$

$-{\displaystyle {\int }_{B_1(x)}}{\tau }^2\mathrm{\Psi }{h}^p.$ (31)

Note that

$-2\tau h^{p-1}⟨\nabla \tau ,\nabla h⟩$

$=-2⟨h^{\frac{p}{2}}\nabla \tau ,\tau h^{\frac{p}{2}-1}\nabla h⟩$

$\le {\displaystyle \frac{1}{\alpha }}{h}^p{|\nabla \tau |}^2+\alpha {\tau }^2{h}^{p-2}{|\nabla h|}^2.$

Combining with (31), we obtain that

$(p-1){\displaystyle {\int }_{B_1(x)}}{\tau }^2{h}^{p-2}{|\nabla h|}^2\le {\displaystyle {\int }_{B_1(x)}}\mathrm{\Psi }{\tau }^2{h}^p+{\displaystyle \frac{1}{\alpha }}{\displaystyle {\int }_{B_1(x)}}{|\nabla \tau |}^2{h}^p.$ (32)

Combining Cauchy-Schwarz inequality with (32), we obtain that

${\displaystyle {\int }_{B_1(x)}}{|\nabla (\tau h^{\frac{p}{2}})|}^2\le {\displaystyle {\int }_{B_1(x)}}𝒜\mathrm{\Psi }{\tau }^2{h}^p+ℬ{|\nabla \tau |}^2{h}^p,$ (33)

where $𝒜=\frac{1}{p-1}(\frac{p^2}{4}+\frac{p}{2})$ and $ℬ=(1+\frac{p}{2})+\frac{1}{\alpha (p-1)}(\frac{p^2}{4}+\frac{p}{2}).$Choose $f=\tau h^{\frac{p}{2}}$ in Proposition 6. Combining with (33), we obtain that

${({\displaystyle {\int }_{B_1(x)}}{(\tau h^{\frac{p}{2}})}^{\frac{2n}{n-2}})}^{\frac{n-2}{2}}\le p𝒞{\displaystyle {\int }_{B_1(x)}}({\tau }^2+{|\nabla \tau |}^2)h^p,$ (34)

where $𝒞$ depends on $n$ and ${sup}_{B_1(x)}\mathrm{\Psi }$.Set $p_k=\frac{2n^k}{{(n-2)}^k}$ and ${\rho }_k=\frac{1}{2}+\frac{1}{2^{k+1}}$ for $k=0,1,2,\mathrm{\cdots }$.Take a function ${\tau }_k\in C_0^{\mathrm{\infty }}(B_{{\rho }_k(x)})$ satisfying:

$$\{\begin{array}{cc}0\le {\tau }_k\le 1,\hfill & \\ {\tau }_k=1on{B}_{{\rho }_{k+1}}(x),\hfill & \\ |\nabla {\tau }_k|\le 2^{k+3}.\hfill & \end{array}$$

Choosing $p=p_k$ and $\tau ={\tau }_k$ in (34), we obtain that

${({\displaystyle {\int }_{B_{{\rho }_{k+1}}(x)}}{h}^{p_{k+1}})}^{\frac{1}{p_{k+1}}}\le {(𝒞p_k{4}^{k+4})}^{\frac{1}{p_k}}{({\displaystyle {\int }_{B_{{\rho }_k}(x)}}{h}^{p_k})}^{\frac{1}{p_k}}.$ (35)

By recurrence, we have

${\parallel h\parallel }_{L^{p_{k+1}}(B_{\frac{1}{2}}(x))}\le {\displaystyle \prod _{i=0}^k}{p}_i^{\frac{1}{p_i}}{4}^{\frac{i}{p_i}}{(𝒞4^4)}^{\frac{1}{p_i}}{\parallel h\parallel }_{L^2(B_1(x))}\le 𝒟{\parallel h\parallel }_{L^2(B_1(x))},$ (36)

where $𝒟$ is a positive constant depending only on $n$, $Vol(B_{r_0+2})$ and ${sup}_{B_{r_0+2}}\mathrm{\Psi }$. Letting $k\to \mathrm{\infty }$,we get

${\parallel h\parallel }_{L^{\mathrm{\infty }}(B_{\frac{1}{2}}(x))}\le 𝒟{\parallel h\parallel }_{L^2(B_1(x))}.$ (37)

Now, choose $y\in {\overline{B}}_{r_0+1}$ such that ${sup}_{B_{r_0+1}}{h}^2=h{(y)}^2$. Note that $B_1(y)\subset B_{r_0+2}$.(37) implies that

$\underset{B_{r_0+1}}{sup}{h}^2\le 𝒟{\parallel h\parallel }_{L^2(B_1(y))}^2\le 𝒟{\parallel h\parallel }_{L^2(B_{r_0+2})}^2.$ (38)

By (30), we have

$\underset{B_{r_0+1}}{sup}{h}^2\le ℱ{\parallel h\parallel }_{L^2(B_{r_0+1})}^2,$ (39)

where $ℱ$ depends only on $n$, $Vol(B_{r_0+2})$ and ${sup}_{B_{r_0+2}}\mathrm{\Psi }$.In order to show the finiteness of the dimension of $H^2(L^2(M))$, it suffices to provethat the dimension of any finite dimensional subspaces of $H^2(L^2(M))$ is bounded above by a fixed constant. Combining (39)with Lemma 11 in Li (¹⁹⁸⁰), we show that .By Proposition 4, we obtain that the dimension of the second space of reduced $L^2$cohomology of $M$ is finite.

Remark 9.For the case of $n\mathrm{=}\mathrm{3}$, Theorem 1 can also be obtained by a different method. In fact, Yau (¹⁹⁷⁶) proved that if $\omega \mathrm{\in }{H}^{\mathrm{2}}\mathit{}\mathrm{(}{L}^{\mathrm{2}}\mathit{}\mathrm{(}M\mathrm{)}\mathrm{)}$, then$\omega$ is closed and coclosed. By use of the Hodge-$\mathrm{*}$ operator, we obtain the dimensions of $H^{\mathrm{2}}\mathit{}\mathrm{(}{L}^{\mathrm{2}}\mathit{}\mathrm{(}M\mathrm{)}\mathrm{)}$ and $H^{\mathrm{1}}\mathit{}\mathrm{(}{L}^{\mathrm{2}}\mathit{}\mathrm{(}M\mathrm{)}\mathrm{)}$ are equal.By Theorem 1.1 in Zhu and Fang (²⁰¹⁴), we obtain the desired result.

ACKNOWLEDGMENTS

The author would like to thank professor Detang Zhou for useful suggestions.The work was partially supported by NSFC Grants 11471145, 11371309 and Qing Lan Project.

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Publication Dates

• Publication in this collection
  Dec 2016

History

• Received
  6 Feb 2015
• Accepted
  17 June 2016