A Note on the Reversibility of Finsler Manifolds

Yin Songting,1,2

 基金资助: 国家自然科学基金.  11471246安徽省自然科学基金.  1608085MA03铜陵学院人才科研启动基金项目.  2015tlxyrc09应用数学福建省高校重点实验室（莆田学院）开放课题.  SX201805

Received: 2017-12-4

 Fund supported: the NSFC.  11471246the AHNSF.  1608085MA03the TLXYRC.  2015tlxyrc09the KLAMFPU.  SX201805

Abstract

For a Finsler manifold with the weighted Ricci curvature bounded from below, we give Cheng type and Mckean type comparison theorems for the first eigenvalue of Finsler Laplacian. When the weighted Ricci curvature is nonnegative, we also obtain Calabi-Yau type volume growth theorem. These generalize and improve some recent literatures. Especially, by using the relationship of the counterparts between a Finsler metric and its reverse metric, we remove some restrictions on the reversibility.

Keywords： Finsler manifold ; Reversibility ; The first eigenvalue ; Comparison theorem ; Volume growth

Yin Songting. A Note on the Reversibility of Finsler Manifolds. Acta Mathematica Scientia[J], 2019, 39(3): 423-430 doi:

1 引言

$(M, F)$为一芬斯勒流形. Rademacher在文献[5]中定义了可反系数如下

$\overleftarrow{\lambda}_F=\lambda_F$,且由文献[2]可知$\lambda^*_F=\lambda_F$.

2 主要定理及其证明

$(M, F, {\rm d}\mu)$$n维芬斯勒流形. B^+_{x_0}(r)$$B^-_{x_0}(r)$分别表示以$x_0$为中心$r$为半径的向前(向后)测地球.对于正数$N$,定义$\overline{N}$

$\begin{eqnarray}\label{1}\overline{N}:=\min\{m\in{\Bbb N}|m\geq N\}.\end{eqnarray}$

若$V_{\overline{N}}(k, r)$为黎曼空间形式中测地球,则令$\varphi$为Laplacian在$\overline{V_{\overline{N}}(k, r)}$中第一Dirichlet特征函数.由于单连通空间形式是两点齐性的,故$\varphi$是径向函数,即$\varphi(\bar{x})=\varphi(d_R(\bar{x}_{0}, \bar{x}))$,其中$\bar{x}_{0}$$\overline{V_{\overline{N}}(k, r)}的中心, d_R为黎曼空间形式中距离函数.进而,我们有[1] \begin{eqnarray}\label{1.1} \left\{\begin{array}{l} \varphi^{\prime\prime}(t)+(\overline{N}-1){\rm{ct}}_{k}(t)\varphi^{\prime}(t) + \lambda_{1}(V_{\overline{N}}(k, r))\varphi(t)=0, \\\varphi(r)=0, \qquad\varphi^{\prime}(t)>0, \quad t\in(0, r). \end{array}\right. \end{eqnarray} 此处 我们断言:若V^+_{\overline{N}}(k, r)为如上所述的芬斯勒空间形式中向前测地球, \varphi(d_F)为芬斯勒Laplacian第一特征函数,其中d_F为芬斯勒流形上距离函数,则方程(2.2)仍然成立.事实上,由Hessian比较定理[7],可得\Delta d_F=(\overline{N}-1){\rm{ct}}_{_{k}}(d_F).从而 下面我们证明\lambda_{1}(V^+_{\overline{N}}(k, r))=\lambda_{1}(V^-_{\overline{N}}(k, r)).注意到若\varphi(d_F)$$\Delta$的一个特征函数且特征值为$\lambda$,则$\varphi(d_{ \overleftarrow{F}})$必为$\overleftarrow{\Delta}$的一个特征函数,且特征值为$\lambda$.反之亦然.此处$\overleftarrow{\Delta}$$d_{ \overleftarrow{F}}为关于反向芬斯勒度量\overleftarrow{F}的Laplacian及距离函数.事实上,若\lambda_{1}(V^+_{\overline{N}}(k, r)) < \lambda_{1}(V^-_{\overline{N}}(k, r)),则对于\lambda_{1}(V^+_{\overline{N}}(k, r)),可以通过解方程(2.2)找到特征函数\varphi(d_F),进一步得到\varphi(d_{ \overleftarrow{F}}).因此, \varphi(d_{ \overleftarrow{F}})$$V^-_{\overline{N}}(k, r)$中的第一特征函数,且第一特征值等于$\lambda_{1}(V^+_{\overline{N}}(k, r))$.这与前面的假定矛盾.类似地,相反的不等式也不成立.故两者必相等.

$\rho(x)=d_{F}(x_0, x)$$(M, F)的向前距离函数, u(x)=\varphi(\rho(x)).因为du=\varphi^{\prime}{\rm d}\rho$$\varphi^{\prime}>0$,所以, $\nabla u=\varphi^{\prime}\nabla\rho$.利用Laplacian比较定理[4],我们有

设$x_{1}, x_{2}\in M$满足$d_{F}(x_{1}, x_{2})=d$.选取$r=\frac{d}{2}$,则$B^+_{x_{1}}(r)\cap B^-_{x_{2}}(r)=\emptyset$.事实上,若存在点$x\in B^+_{x_{1}}(r)\cap B^-_{x_{2}}(r)$,则导致以下矛盾:

$u_{1}$$u_{2}延拓至流形M上,使得在B^+_{x_{1}}(\frac{d}{2})$$B^-_{x_{2}}(\frac{d}{2})$外部分别为零.于是存在两个常数$a_{1}>0$$a_{2} < 0使得 因此 定理2.2证毕. 引理2.1 设(M, F, {\rm d}\mu)$$n$维芬斯勒流形, $\Omega\subset M$$M上具有非空边界及紧致闭包的区域.假定f为芬斯勒Laplacian在\Omega中的第一Dirichlet特征函数, X$$\Omega$上一个向量场,且满足$\inf\limits_{\Omega}{\rm div}(X)>0$.则我们有下面的不等式:

(1)若存在一点$x_0\in\Omega$使得$f(x_0) < 0$,则

(2)若存在一点$x_0\in\Omega$使得$f(x_0)>0$,则

对任意的$\xi\in T^*M, X\in TM$,由基本不等式可得$\xi(X)\leq F^*(\xi)F(X)$.从而,我们有$-\xi(X)\leq F^*(-\xi)F(X)$$\xi(-X)\leq F^*(\xi)F(-X).于是 如所知,若f为芬斯勒Laplacian在\Omega上的一个第一Dirichlet特征函数,令f=f^++f^-,则它的正部f^+与负部f^-均为第一Dirichlet特征函数.因此,我们不妨假定f\leq0$$f\geq0$.

$(M, g)$$n维完备黎曼流形.若其截面曲率K_M\geq-a^2,则由文献[1]或[3]可知, Laplacian第一特征值满足 \begin{eqnarray}\label{2.5} \lambda_{1}(M)\leq\frac{(n-1)^2 a^2}{4}. \end{eqnarray} 根据推论2.1,若芬斯勒流形的旗曲率K\geq-a^2$$S=0$,则(2.5)式在芬斯勒情形下仍然成立.因此,结合定理2.3易得:

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